Research ArticleHigher Order Sobolev-Type Spaces on the Real Line

Bogdan Bojarski1 Juha Kinnunen2 and Thomas Zuumlrcher34

1 Institute of Mathematics Polish Academy of Sciences 00-956 Warsaw Poland2Department of Mathematics School of Science and Technology Aalto University PO Box 11100 00076 Aalto Finland3Department of Mathematics and Statistics University of Jyvaskyla PO Box 35 (MaD) 40014 Jyvaskyla Finland4Mathematics Institute University of Warwick Coventry CV4 7AL UK

Correspondence should be addressed to Juha Kinnunen juhakkinnunenaaltofi

Received 16 July 2014 Accepted 6 October 2014 Published 25 November 2014

Academic Editor Dachun Yang

Copyright copy 2014 Bogdan Bojarski et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

This paper gives a characterization of Sobolev functions on the real line by means of pointwise inequalities involving finitedifferences This is also shown to apply to more general Orlicz-Sobolev Lorentz-Sobolev and Lorentz-Karamata-Sobolev spaces

1 Introduction

The general opinion in the literature on Sobolev spaces119882

119898119901

(R119899

) and their generalizations is that the one-dimensional case 119899 = 1 may not be interesting In the firstorder case 119898 = 1 the theory is essentially contained in thefundamental theorem of calculus

119891 (119909) minus 119891 (119910) = int

119910

119909

1198911015840

(119905) 119889119905 119909 119910 isin R (1)

and the higher order case 119898 gt 1 reduces recursively to thefirst order case according to the following definition

Definition 1 (Sobolev spaces) Assume thatΩ sub R is an openinterval with finite Lebesgue measure in R and let 119883 be avector space of functions One says that 119891 isin 119882

119898

119883if there

exists 119865 isin 119883 that agrees with 119891 almost everywhere and hasweak derivatives of order119898 in119883 and the derivatives of order119903 0 le 119903 le 119898 minus 1 are absolutely continuous

The main goal in this paper is to point out that there areinteresting phenomena related to the Sobolev spaces alreadyin the one-dimensional case More precisely we give a directand relatively elementary proof of the following result

Theorem2 Suppose thatΩ sub R is an open interval with finitemeasure A real valued (continuous) function 119891 Ω rarr R is inthe Sobolev space 119882

119898119901

(Ω) = 119882119898

119871119901(Ω) (119898 ge 1 119901 gt 1) if and

only if for each 119909 and 119910 isin R the 119898th difference of the function119891 at the point 119909 with step ℎ = (119910 minus 119909)119898

Δ119898

119891 (119909 119910) = Δ119898

ℎ119891 (119909) =

119898

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + 119895ℎ) (2)

satisfies the inequality

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

[119886119891(119909) + 119886

119891(119910)] (3)

for some function 119886119891isin 119871

119901

(Ω)

Note that this result is false for domains with infinitemeasure as nonzero constant functions satisfy (3) with 119886

119891= 0

but are not in 119871119901

(Ω) (we leave it to the reader to formulate anappropriate version of Theorem 2 for homogeneous Sobolevspaces which avoids this deficiency) We will also considerthis result in the more general framework of Banach functionspaces However the material is presented in such a mannerthat it is accessible to readers who are only interested inthe classical case Inequality (3) in Theorem 2 is one of theexamples of pointwise inequalities characterising Sobolevfunctions proved and propagated by Bojarski and his stu-dents in a series of papers over the last almost 30 years see forexample [1ndash6]The case119898 = 1 lies at the initial concepts andcornerstones of analysis on general measure metric spaces[7]

Hindawi Publishing CorporationJournal of Function SpacesVolume 2014 Article ID 261565 13 pageshttpdxdoiorg1011552014261565

2 Journal of Function Spaces

Let us give an overview of the paper The proof ofTheorem 2 is presented in a series of results We first givea detailed proof of the fact that Sobolev functions satisfythe pointwise inequality (3) The idea is to consider theTaylor expansion of the function Using the linearity of thefinite differences we only need to compute and bound thefinite differences of polynomials (119909 minus 119910)

119898minus1 and of termslike (119909 minus 119910)

119898minus1

+ The so-called Hermite-Genocchi formula is

an integral representation of the finite differences (actuallydivided differences) which makes the computation rathereasy It turns out that the functions 119886

119891are multiples of the

Hardy-Littlewood maximal function of the correspondingderivatives Reference [8] studies the possibility to find 119886

119891

whose definition does not involve derivatives at allThe otherdirection is easily shown to hold for smooth functions andthe general result follows from this Thereafter we recall thetheorems from literature necessary to be able to deduce thatour main results also apply to Orlicz Lorentz and Lorentz-Karamata spaces

2 Sobolev Functions SatisfyPointwise Inequality

We refer the reader to Chapter 7 in [9] for the definition andproperties of absolutely continuous functions We write 119891 isin

119871119898

1[119886 119887] to express that the function119891 [119886 119887] rarr R is (119898minus1)-

times differentiable and that the derivative of order 119898 minus 1 isabsolutely continuous The aim of this section is to show thatmembership in a Sobolev space implies (3) where the finitedifference is replaced by divided differences

Definition 3 (divided difference) Let 1199101 119910

119903+1be points in

a domain Ω sub R One assumes that if 119910119894has multiplicity 119899

then119891 Ω rarr R is (119899minus1)-times differentiable in 119910119894 First one

considers the case where 1199101le 119910

2le sdot sdot sdot le 119910

119903+1are ordered

One defines for 119903 ge 1

[1199101] 119891 = 119891 (119910

1)

[1199101 119910

119903+1] 119891

=

[1199102 119910

119903+1] 119891 minus [119910

1 119910

119903] 119891

119910119903+1

minus 1199101

if 1199101

= 119910119903+1

119891(119903)

(1199101)

119903otherwise

(4)

We extend the definition to unordered tuples in the onlypossible way that makes the divided difference independentof the ordering

Remark 4 In the recursion formula we omitted 1199101in the

nominator of the minuend and 119910119903+1

in the nominator of thesubtrahend and then divided by 119910

119903+1minus 119910

119903 We claim that we

obtain the same result if we replace 1199101by 119910

119896and 119910

119903+1by 119910

119898as

long as 119910119896

= 119910119898 This can be seen by induction We look here

only at one of the two more difficult cases when one of theomitted points is either the largest or the smallest (and there

are at least three points) The omitted points are denoted by ahat We also omit the function Consider[119910

1] minus [119910

119898]

119910119898

minus 1199101

= ([119910

1 119910

119898] minus [119910

1 119910

119903+1]

119910119903+1

minus 119910119898

minus[119910

1 119910

119898] minus [119910

119898 119910

119903+1]

119910119903+1

minus 1199101

)

times (119910119898

minus 1199101)minus1

= ((119910119903+1

minus 1199101) [119910

1 119910

119898] minus (119910

119903+1minus 119910

1) [119910

1 119910

119903+1]

minus (119910119903+1

minus 119910119898) [119910

1 119910

119898] + (119910

119903+1minus 119910

119898) [119910

119898 119910

119903+1])

times ((119910119903+1

minus 119910119898) (119910

119898minus 119910

1) (119910

119903+1minus 119910

1))

minus1

(5)

We collect the first and third term and expand the second one

[1199101] minus [119910

119898]

119910119898

minus 1199101

= ((119910119898

minus 1199101) [119910

1 119910

119898] minus (119910

119898minus 119910

1) [119910

1 119910

119903+1]

minus (119910119903+1

minus 119910119898) [119910

1 119910

119903+1] + (119910

119903+1minus 119910

119898) [119910

119898 119910

119903+1])

times ((119910119903+1

minus 119910119898) (119910

119898minus 119910

1) (119910

119903+1minus 119910

1))

minus1

(6)

Collecting the first two terms and the second two cancellingand applying the induction hypothesis give

[1199101] minus [119910

119898]

119910119898

minus 1199101

=[119910

1]

119910119903+1

minus 1199101

minus[119910

119903+1]

119910119903+1

minus 1199101

= [1199101 119910

119903+1] (7)

The following formula is a useful tool to compute divideddifferences

Lemma 5 (Hermite-Genocchi formula) Let 1199101 119910

119903+1be

points in a domain Ω sub R One supposes that 119891 Ω rarr R

is such that 119891(119899minus1) is absolutely continuous if not all 119910119894coincide

and that additionally 119891(119899)

(1199101) exists if all points are the same

Then[119910

0 119910

119899] 119891

= int

1

0

1198891199051int

1199051

0

1198891199052

sdot sdot sdot int

119905119899minus1

0

119891(119899)

(1199100+ (119910

1minus 119910

0) 119905

1

+ sdot sdot sdot + (119910119899minus 119910

119899minus1) 119905

119899) 119889119905

119899

(8)

Proof (see page 122 in [10] as well) If 119899 = 1 we have that

int

1

0

1198911015840

(1199100+ (119910

1minus 119910

0) 119905

1) 119889119905

1

=

119891 (1199101) minus 119891 (119910

0)

1199101minus 119910

0

1199100

= 1199101

1198911015840

(1199100) 119910

0= 119910

1

= [119910

0 119910

1] 119891 = [119910

1 119910

0] 119891 119910

0= 119910

1

[1199100 119910

0] 119891 119910

0= 119910

1

(9)

Journal of Function Spaces 3

Assume now that the claim is true for some 119899 We assumehere that 119910

119899+1= 119910

119899and leave the case when 119910

119899and 119910

119899+1

coincide to the reader (computation (14) might be helpful)Then evaluating the right most integral

int

1

0

1198891199051int

1199051

0

1198891199052

sdot sdot sdot int

119905119899

0

119891(119899+1)

(1199100+

119899

sum

119896=1

(119910119896minus 119910

119896minus1) 119905

119896

+ (119910119899+1

minus 119910119899) 119905

119899+1)119889119905

119899+1

= int

1

0

1198891199051int

1199051

0

1198891199052

sdot sdot sdot int

119905119899minus1

0

(119891(119899)

(1199100+

119899minus1

sum

119896=1

(119910119896minus 119910

119896minus1) 119905

119896

+ (119910119899+1

minus 119910119899minus1

) 119905119899)

minus119891(119899)

(1199100+

119899

sum

119896=1

(119910119896minus 119910

119896minus1) 119905

119896))119889119905

119899+1

sdot1

119910119899+1

minus 119910119899

=1

119910119899+1

minus 119910119899

([1199100 119910

119899minus1 119910

119899+1] 119891 minus [119910

0 119910

119899] 119891)

= [1199100 119910

119899+1] 119891

(10)

We use the Hermite-Genocchi formula to compute finitedifferences of polynomials

Lemma 6 Let 119886 be a point in R and 119898 isin N0 We set

119901119898119886

(119910) = (119910 minus 119886)119898

(11)

Then

[1199100 119910

119899] 119901

119898119886=

0 0 le 119898 lt 119899

1 119898 = 119899(12)

Proof As 119901119898119886

is smooth we can use the Hermite-Genocchiformula Lemma 5 If 0 le 119898 lt 119899 then 119901

(119899)

119898119886= 0 hence the

Hermite-Genocchi formula gives the claim If 119898 = 119899 then119901(119898)

119898119886= 119898 we compute the integral in the Hermite-Genocchi

formula via induction over119898We let first for fixed 119905

0ge 0

⟨1199100 119910

119898⟩1199050

= int

1199050

0

1198891199051int

1199051

0

1198891199052sdot sdot sdot int

119905119898minus1

0

1 119889119905119898 (13)

We claim that

⟨1199100 119910

119898⟩1199050

=119905119898

0

119898 119898 isin N (14)

For 119898 = 1 this can easily be seen Assume that the claimholds for some119898 Then

⟨1199100 119910

119898+1⟩1199050

= int

1199050

0

1198891199051⟨119910

0 119910

119898⟩1199051

= int

1199050

0

119905119898

1

119898119889119905

1=

119905119898+1

0

(119898 + 1)

(15)

The Hermite-Genocchi formula tells us [1199100 119910

119898]119901119898119886

=

⟨1199100 119910

119898⟩1sdot 119898 giving the proof

The following notation is on page 15 in [11]

Definition 7 Set

(119909 minus 119910)0

+=

1 119909 ge 119910

0 119909 lt 119910(16)

and for119898 gt 1

(119909 minus 119910)119898minus1

+=

(119909 minus 119910)119898minus1

119909 ge 119910

0 119909 lt 119910(17)

The definition below is a variant of Definition 412 in [11]However as our starting point is different than Schumakerrsquosthere is a slight regularity issue when 119909 does not differ fromat least two 119910

119894 However we are only interested in cases where

119876119898

119894lives under an integral andwe agree here to set119876119898

119894(119909) = 0

in points where we do not have enough regularity

Definition 8 (B-spline) Let sdot sdot sdot le 119910minus1

le 1199100le 119910

1le 119910

2le sdot sdot sdot

be a sequence of real numbers Given integers 119894 and 119898 gt 0one defines (note that 119909 is fixed so that the finite difference iswith respect to 119910)

119876119898

119894(119909) =

(minus1)119898

[119910119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+ if 119910

119894lt 119910

119894+119898

0 otherwise(18)

We call 119876119898

119894the 119898th order B-spline associated with the knots

119910119894 119910

119894+119898

This is Theorem 22 in [11] Schumaker does not give theproof but writes that it is similar to the one ofTheorem 21 inhis book

Theorem 9 (Dual Taylor Expansion) Let 119891 isin 119871119898

1[119886 119887] Then

for all 119886 le 119910 le 119887

119891 (119910) =

119898minus1

sum

119895=0

(minus1)119895

119863119895

119891 (119887) (119887 minus 119910)119895

119895

+ int

119887

119886

(119909 minus 119910)119898minus1

+(minus1)

119898

119863119898

119891 (119909)

(119898 minus 1)119889119909

(19)

4 Journal of Function Spaces

Proof The proof uses induction and integration by partsNote that by replacing 119886 by 119910 in the integral we may deletethe + in (119909 minus 119910)

+ The case119898 = 1 reduces to the fundamental

theorem of calculus for absolutely continuous functions seefor example Theorem 7115 in [9] Let us assume that thestatement holds for some119898 and fix 119891 isin 119871

119898+1

1[119886 119887] Then

int

119887

119910

(119909 minus 119910)119898

(minus1)119898+1

119863119898+1

119891 (119909)

119898119889119909

=(119887 minus 119910)

119898

(minus1)119898+1

119863119898

119891 (119887)

119898

minus int

119887

119910

(119909 minus 119910)119898minus1

(minus1)119898+1

119863119898

119891 (119909)

(119898 minus 1)119889119909

= minus(119887 minus 119910)

119898

(minus1)119898

119863119898

119891 (119887)

119898

minus

119898minus1

sum

119895=0

(minus1)119895

119863119895

119891 (119887) (119887 minus 119910)119895

119895+ 119891 (119910)

(20)

We now cite a part ofTheorem 423 in [11] Note that119898 isa positive integer and 119863

119895

+means that we take the right hand

derivative of order 119895 We only need and prove the equality for119895 = 0 We also assume that not all points coincide

Theorem 10 (Peano representation) Fix 0 le 119895 le 119898minus119899 where119899 is the maximum multiplicity of 119910

119894 119910

119894+119898 Then

[119910119894 119910

119894+119898] 119891 = int

119910119894+119898

119910119894

(minus1)119895

119863119895

+119876

119898

119894(119909)119863

119898minus119895

119891 (119909)

(119898 minus 1)119889119909 (21)

for all 119891 isin 119871119898minus119895

1[min

119894119910

119894max

119894119910

119894]

Proof of the Peano Representation for 119895 = 0 We take thedual Taylor expansion from Theorem 9 Then we apply[119910

119894 119910

119894+119898] on both sides and remember that we computed

the divided differences of polynomials in Lemma 6

Our goal is to use the Peano representation to obtainan upper bound for the divided differences We start withestablishing an upper bound of the factor 119863

+119876

119898

119894in the

integrand

Lemma 11 Suppose that 119910119894le sdot sdot sdot le 119910

119894+119898and 119910

119894lt 119910

119894+119898 Then

there is a constant 119862 such that

1003816100381610038161003816119863+119876

119898

119894(119909)

1003816100381610038161003816 =10038161003816100381610038161003816[119910

119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+

10038161003816100381610038161003816le

119862

119910119894+119898

minus 119910119894

(22)

Proof We want to estimate

[119910119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+

=[119910

119894+1 119910

119894+119898] (119909 minus 119910)

119898minus1

+minus [119910

119894 119910

119894+119898minus1] (119909 minus 119910)

119898minus1

+

119910119894+119898

minus 119910119894

(23)

Assume 119896 lt 119898 minus 1 We claim that

119889119896

119889119910119896

(119909 minus 119910)119898minus1

+= (minus1)

119896

(119898 minus 1) sdot sdot sdot (119898 minus 119896) (119909 minus 119910)119898minus1minus119896

+

(24)

This claim is easily seen to be true for 119910 = 119909 That it is truealso for 119910 = 119909 follows by induction We also note that

119889119898minus2

119889119910119898minus2(119909 minus 119910)

119898minus1

+(25)

is absolutely continuous on bounded intervals and that

(minus1)(119898minus1)

(119898 minus 1) (119909 minus 119910)0

+

(26)

is a weak derivative that equals the derivative whenever119910 = 119909 We first assume that not all 119910

119895 119910

119895+119898minus1coincide

An application of the Hermite-Genocchi formula Lemma 5gives

[119910119895 119910

119895+119898minus1] (119909 minus 119910)

119898minus1

+

= int

1

0

1198891199051int

1199051

0

1198891199052

sdot sdot sdot int

119905119898minus2

0

(minus1)(119898minus1)

(119898 minus 1)

sdot (119909 minus (119910119895+

119898minus1

sum

119896=1

(119910119895+119896

minus 119910119895+119896minus1

) 119905119896))

0

+

119889119905119898minus1

(27)

We use now the fact that the integrand is bounded from aboveby (119898 minus 1) and the computation (14) to obtain

10038161003816100381610038161003816[119910

119895 119910

119895+119898minus1] (119909 minus 119910)

119898minus1

+

10038161003816100381610038161003816le 1 (28)

Suppose now that 119910119895

= sdot sdot sdot = 119910119895+119898minus1

= 119909 Then|[119909

119895 119909

119895+119898minus1](119909 minus 119910)

119898minus1

+| is bounded from above by 1 if

119909 gt 119910119894and vanishes if 119910

119894lt 119909 In the end we find a constant

119862 such that10038161003816100381610038161003816119863

+[119910

119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+

10038161003816100381610038161003816le

119862

119910119894+119898

minus 119910119894

(29)

In the next result we use the notation

119872119892(119909) = sup119911

minusint

119911

119909

10038161003816100381610038161198921003816100381610038161003816 (30)

for the famous Hardy-Littlewood maximal function of 119892

introduced in the seminal paper [12] of Hardy-Littlewood inActa Math (1930)

Theorem 12 Suppose 1199100

le sdot sdot sdot le 119910119898 119910

0lt 119910

119898and 119891 is in

119871119898

1[119910

0 119910

119898] for some 119898 isin N Then there is some constant 119862

such that1003816100381610038161003816[1199100 119910119898] 119891

1003816100381610038161003816 le 119862 (119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (1199100) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910119898))

(31)

Journal of Function Spaces 5

Proof We use the Peano representation Theorem 10 with119894 = 119895 = 0 Then using the bound of |119863

+119876

119898

119894(119909)| found in

Lemma 11

1003816100381610038161003816[1199100 119910119898] 1198911003816100381610038161003816 le

1

(119898 minus 1)int

119910119898

1199100

1003816100381610038161003816119863+119876

119898

119894(119909)

1003816100381610038161003816

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

le119862

119910119898

minus 1199100

int

119910119898

1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

(32)

For 119910 isin [1199100 119910

119898] we have

1003816100381610038161003816[1199100 119910119898] 1198911003816100381610038161003816

le119862

119910119898

minus 1199100

(int

119910

1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909 + int

119910119898

119910

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909)

(33)

and further1003816100381610038161003816[1199100 119910119898] 119891

1003816100381610038161003816

le 119862(minusint

119910

1199100

119910 minus 1199100

119910119898

minus 1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

+minusint

119910119898

119910

119910119898

minus 119910

119910119898

minus 1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909)

(34)

The first factors in both integrals are bounded from above by1 We see that the integrals are bounded from above by thecorresponding maximal operators This gives the claim

The finite differences defined in (2) are up to a factordivided differences for equispaced nodes

Lemma 13 Let 119909 119910 be two points in a domain Ω sub R Let119898 isin N and 119909

119896= 119909

0+ ℎ119896 where ℎ = (119910 minus 119909)119898 Then each

119891 Ω rarr R satisfies

[1199090 119909

119898] 119891 =

1

ℎ119898119898Δ119898

119891 (119909 119910) (119909 = 1199090 119909

119898= 119910)

(35)

Proof We use induction For119898 = 1 we have

[1199090 119909

1] 119891 =

119891 (1199091) minus 119891 (119909

0)

1199091minus 119909

0

(36)

As ℎ = 1199091minus 119909

0 119909 = 119909

0 119910 = 119909

1 and 119898 = 1 we obtain the

claim Suppose that the claim is true for some119898 isin N Then

Δ119898+1

ℎ119891 (119909)

=

119898+1

sum

119895=0

(minus1)119898+1minus119895

(119898 + 1

119895)119891 (119909 + 119895ℎ)

=

119898+1

sum

119895=0

(minus1)119898+1minus119895

((119898

119895 minus 1) + (

119898

119895))119891 (119909 + 119895ℎ)

=

119898+1

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + (119895 + 1) ℎ)

minus

119898+1

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + 119895ℎ)

=

119898

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909

1+ 119895ℎ)

minus

119898

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + 119895ℎ)

= ℎ119898

119898 ([1199091 119909

119898+1] 119891 minus [119909

0 119909

119898] 119891)

= ℎ119898

119898 (ℎ (119898 + 1)) [1199090 119909

119898+1] 119891

(37)

We can now formulate Theorem 12 in terms of finitedifferences

Corollary 14 (Corollary of Theorem 12) Let Ω sub R be adomain Suppose 119891 isin 119871

119898

1(Ω) for some 119898 isin N then there is

a constant 119862 = 119862(119898) such that

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816

le 1198621003816100381610038161003816119909 minus 119910

1003816100381610038161003816

119898

(119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

119909 119910 isin Ω

(38)

In particular we have the following special case Suppose 119883

and 119884 are normed spaces of functions such that the maximaloperator 119872 119883 rarr 119884 is bounded If 119891 isin 119882

119898

119883(Ω) for some

119898 ge 1 then

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (39)

for some function 119886119891isin 119884 depending on 119891

Proof We first use Lemma 13 and thenTheorem 12 with stepℎ = (119910 minus 119909)119898 There is some constant 119862 = 119862(119898) such that

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816

= ℎ119898

1198981003816100381610038161003816[1199090

119909119898] 119891

1003816100381610038161003816

le 119862ℎ119898

(119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

le1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

119898

(119862119872(1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119862119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

(40)

Proof of Theorem 2 (Sobolev functions satisfy pointwiseinequality) We combine Corollary 14 with the boundednessof the maximal operator which is by now classical and can befound for example in Theorem 3310 in [13]

6 Journal of Function Spaces

3 Pointwise Inequality Implies Membershipin Sobolev Space

In this section our aim is to show that the pointwiseinequality (3) implies the existence of the highest orderderivative in the correct space We will take care of theintermediate derivatives in the next section In this sectionthe properties of the Lebesgue spaces play an important roleOur point of view is however to extract the properties ofthe Lebesgue spaces and base our proofs solely on theseproperties We hope that the disadvantage of being slightlymore abstract is later on compensatedwhenwe generalize theresults We start with some auxiliary results

Lemma 15 Let Ω sub R be a domain and 120593 isin C119899

0(Ω) that

is 120593 has compact support in Ω and is 119899-times continuouslydifferentiable Then

1003816100381610038161003816100381610038161003816100381610038161003816

[1199090 119909

119899] 120593 minus

120593(119899)

(119910)

119899

1003816100381610038161003816100381610038161003816100381610038161003816

997888rarr 0 (41)

uniformly inΩ as 1199090 119909

119899rarr 119910

Proof By Taylorrsquos formula for 119909 isin conv1199090 119909

119899 there is

120585119909isin conv119909

0 119909

119899 such that

120593 (119909) = 120593 (119910) + 1205931015840

(119910) (119909 minus 119910) + sdot sdot sdot + 120593(119899minus1)

(119910)(119909 minus 119910)

119899minus1

(119899 minus 1)

+1

119899120593(119899)

(120585119909) (119909 minus 119910)

119899

(42)

We apply [1199090 119909

119899] on both sides (with variable 119909) and use

the computation of the divided differences of polynomialscarried out in Lemma 6 to obtain

[1199090 119909

119899] 120593 (119909) =

1

119899120593(119899)

(120585119909) (43)

As 120593(119899) is uniformly continuous on compact intervals the

claim follows

Lemma 16 Assume Ω sub R is open and 119870 sub Ω is compactThen there is a neighborhood of119870 that is a subset of Ω

Proof For each 119909 isin 119870 we find a ball 119861(119909 119903119909) sub Ω Suppose

the statement of the lemma is falseThen for every 119899 isin N thereis 119911

119899isin Ω 119870 with dist(119911

119899 119870) lt 1119899 Thus for each 119899 isin N

there is 119910119899

isin 119870 with 119889(119911119899 119910

119899) lt 1119899 Since 119870 is compact

there is a subsequence (119910119899119896)119896of (119910

119899)119899that converges to some

119910 isin 119870 Now 119910 lies in one of the balls 119861(119909 119903119909) sub Ω For 119896 large

enough we have that 119910119899119896and 119911

119899119896are in 119861(119909 119903

119909) sub Ω This is

a contradiction

Lemma 17 SupposeΩ sub R is open and119870 sub Ω compactThenthere exists 120576 gt 0 such that

(Ω minus ℎ) cap 119870 = Ω cap 119870 (44)

for |ℎ| lt 120576

Proof We first show that any member of the left hand side iscontained in the right hand side this is true for any ℎ isin R If119911 is a member of the left hand side then it is in 119870 and since119870 is a subset of Ω it is also a member of the right hand sideTo prove the other inclusion note that by Lemma 16 there is120576 gt 0 such that the 120576-neighborhood 119870

120576of 119870 lies in Ω Now

suppose that 119911 is in the right hand side and |ℎ| lt 120576 Then for119909 = 119911+ℎ we have119909 isin 119870

120576sub Ω implying that 119911 = 119909minusℎ isin Ωminusℎ

That 119911 isin 119870 is clear This gives the proof

Next we state basically Definition 311 in [14] Readersfollowing [13] might want to have a look at Definitions 111and 113 in their book To gain familiarity with the conceptof Banach function norm the reader may want to verify thatLebesgue spaces are examples of Banach function spaces Aswe are here only interested in Lebesgue measure we adaptthe notion to this case The set M+ denotes the cone ofmeasurable functions whose values lie in [0infin]

Definition 18 (Banach function norm Banach functionspace) Assume that 119877 sub R A Banach function norm on 119877

is a map 120588 from M+

(119877) to [0infin] such that for all 119891 119892 and119891119899(119899 isin N) in M+

(119877) all scalars 120582 ge 0 and all measurablesubsets 119864 of 119877 the following are true where 120583 is the Lebesguemeasure

(P1) 120588(119891) = 0 if and only if 119891 = 0 120583-almost everywhere120588(120582119891) = 120582120588(119891) 120588(119891 + 119892) le 120588(119891) + 120588(119892)

(P2) if 0 le 119892 le 119891 120583-almost everywhere then 120588(119892) le 120588(119891)(P3) if 0 le 119891

119899uarr 119891 120583-almost everywhere we have 120588(119891

119899) uarr

120588(119891)(P4) if 120583(119864) lt infin then 120588(120594

119864) lt infin

(P5) if120583(119864) lt infin then there is a constant119862 = 119862(119864 120588) lt infin

such that int119864

119891119889119909 le 119862120588(119891)

Given such a function norm 120588 the set 119883 = 119883(120588) ofall extended measurable functions 119891 (identifying functionsequal 120583-almost everywhere) such that 120588(|119891|) lt infin is called aBanach function space (BFS for short) and we define

10038171003817100381710038171198911003817100381710038171003817119883

= 120588 (1003816100381610038161003816119891

1003816100381610038161003816) 119891 isin 119883 (45)

We leave it to the reader to verify that Lebesgue spacessatisfy the following properties (1198831015840 can be chosen as the dualspace)

Properties 19 Given a Banach function space 119883 we requirethe existence of a Banach function space119883

1015840 such that

(B) 119883 contains each essentially bounded function whosesupport has finite measure

(HI) Holderrsquos inequality int |119891119892|119889119909 le 119891119883119892

1198831015840

(C) 1198831015840 is complete

(D) 119892119883

= supintΩ

|119892|ℎ ℎ isin M+

ℎ1198831015840 le 1

(AC) absolutely continuous norm if (120594119864119899)119899is a sequence of

characteristic functions converging pointwise almosteverywhere to 0 then 119891120594

1198641198991198831015840 converges to zero for

all 119891 isin 1198831015840

Journal of Function Spaces 7

(A) each function 119891 isin 1198831015840 can be approximated by simple

functions (in1198831015840-norm)

(NC) if 119891 isin 1198831015840 and ℎ

119899are step functions with 0 le ℎ

119899uarr |119891|

then1003817100381710038171003817119891 minus ℎ

119899sgn (119891)

10038171003817100381710038171198831015840997888rarr 0 as 119899 997888rarr infin (46)

Lemma 20 Suppose Ω sub R is open and 119883 is a Banachfunction space fulfilling the items listed in Properties 19 Assumethat 119891 Ω rarr R is in 119883 We let 120582

119891 119883

1015840

cap C119903+1

0(Ω) rarr R be

defined as

120582119891(120593) = int

Ω

119891120593(119903)

119889119909 (47)

If1003816100381610038161003816Δ

119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)]

for almost every 119909 119910 isin Ω

(48)

for some 119886119891

isin 119883 then 120582119891has a continuous extension to 119883

1015840

with norm bounded from above by 2119903119903

119886119891119883

Proof Our goal is to show1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816le 2119903

11990310038171003817100381710038171003817119886119891

10038171003817100381710038171003817119883

100381710038171003817100381712059310038171003817100381710038171198831015840

(49)

because 120582119891extends then by the theorem of Hahn-Banach to

a bounded functional on1198831015840 with norm bounded from above

by 2119903119903

119886119891119883 Without loss of generality we may assume that

119886119891is nonnegative If 119891 were be smooth we could just use

integration by parts and use as bound the limiting case of thepointwise inequality (48) As 119891 is not necessarily smooth wehave to operate on the level of finite differences

Let us choose 120593 isin C0(Ω) and denote its support by 119870

Lemma 17 ensures the existence of some positive 120576 gt 0 suchthat (Ωminus119895ℎ)cap119870 = Ωcap119870 for 0 le 119895 le 119903 as long as |ℎ| lt 120576119903Weassume in the following that ℎ has been chosen accordingly

The following integrals exist by Holderrsquos inequalitybecause 120593 and thus Δ

119903

minusℎ120593(119909) are bounded with compact

support and 119891 is integrable Consider

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

119891 (119909) (minusℎ)minus119903

Δ119903

minusℎ120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(50)

We compute the following integral

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

= intΩ

119891 (119909)

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119909) 120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωminus119895ℎ

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

(Ωminus119895ℎ)cap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

(51)

We have chosen 120576 gt 0 such that (Ω minus 119895ℎ) cap 119870 = Ω cap 119870 Wecontinue

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωcap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

(52)

With the aid of (52) we can now rewrite (50) as

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

(minusℎ)minus119903

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(53)

Inequality (48) leads to

1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816

le 119903119903

intΩ

(119886119891(119909) + 119886

119891(119909 + 119903ℎ))

1003816100381610038161003816120593 (119909)1003816100381610038161003816 119889119909

+ intΩ

1003816100381610038161003816119891 (119909)1003816100381610038161003816

10038161003816100381610038161003816120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)

10038161003816100381610038161003816119889119909

(54)

Using Holderrsquos inequality (HI) for the first summand weare left to show that the second summand tends to 0 asℎ approaches 0 We note first that since the second factorhas support in a set of finite measures we can replace theintegration domain by a set of finite measuresThen we applya (1198711

119871infin

)-Holder inequality and the uniform convergence ofthe divided difference to 120593

(119903)

(119909)119903 established in Lemmas 13and 15

After having verified that 120582119891has a continuous extension

to 1198831015840 we want to show that this extension has a representa-

tion as integral

8 Journal of Function Spaces

Lemma 21 Assume Ω sub R is open Suppose the propertieslisted in Properties 19 are met Let 120582

119891be defined as in (47) and

bounded on 1198831015840 Then there exists 119892 isin 119883 with 119892

119883le 120582

119891

such that

120582119891(120593) = (minus1)

119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (55)

for all 120593 isin 1198831015840 especially for all 120593 isin Cinfin

0(Ω)

Proof The proof is heavily influenced by the proof of Theo-rem 141 in [13]We first want to find a signedmeasure ] suchthat

120582119891(120593) = int

Ω

120593119889] (56)

for each 120593 isin 1198831015840 Then we want to apply the Lebesgue-

Radon-Nikodym theorem to replace 119889] by (minus1)119903

119892(119909)119889119909 forsome integrable function 119892 for which we then show that it iscontained in119883

We start by assuming that the measure of Ω is finite Foreach 120583-measurable set 119860 we set

] (119860) = 120582119891(120594

119860) (57)

We start by proving the 120590-additivity of ] (for the 120583-measurable sets) In the first step we just show the additivityBy induction it is enough to consider only two sets

Let 1198601and 119860

2be two disjoint 120583-measurable sets Then

] (1198601cup 119860

2) = 120582

119891(120594

1198601cup1198602) = 120582

119891(120594

1198601+ 120594

1198602)

= 120582119891(120594

1198601) + 120582

119891(120594

1198602) = ] (119860

1) + ] (119860

2)

(58)

We move to the verification of the 120590-additivity Let119860

1 119860

2 be countablymany pairwise disjoint120583-measurable

sets We first want to show thatsum119894120594119860119894

converges in1198831015840

We let 119864119896= cup

119896

119894=1119860

119894and 119864 = cup

infin

119894=1119860

119894 Then since (119864

119896)119896is

increasing

120583 (119864) = 120583(

infin

⋃

119896=1

119864119896) = lim

119896rarrinfin

120583 (119864119896) (59)

As 119864 has finite measure given 120576 gt 0 there exists 1198960isin N such

that

(1 minus 120576) 120583 (119864) lt 120583 (1198641198960) (60)

By the disjointness of the sets 119860119894 we see that

120583 (119864) = 120583 (1198641198960) + 120583(

infin

⋃

119894=1198960+1

119860119894) (61)

It follows that

120583(

infin

⋃

119894=1198960+1

119860119894) = 120583 (119864) minus 120583 (119864

1198960) lt 120583 (119864) minus (1 minus 120576) 120583 (119864)

= 120576120583 (119864)

(62)

We have for119898 lt 119899

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=1

120594119860119894

minus

119898

sum

119894=1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

=

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=119898+1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

le10038171003817100381710038171003817120594cupinfin119894=119898+1

119860119894

100381710038171003817100381710038171198831015840 (63)

By the considerations before and the fact that sdot 1198831015840 is

absolutely continuous as stipulated in (AC) we see thatsum

119899

119894=1120594119860119894

is a Cauchy sequence Now as 1198831015840 is complete by

(C) the Cauchy sequence converges in 1198831015840 We have by the

continuity of 120582119891

] (cup119894119860

119894) = 120582

119891(120594

cup119894119860119894) = 120582

119891(

infin

sum

119894=1

120594119860119894)

= lim119899rarrinfin

120582119891(

119899

sum

119894=1

120594119860119894) = lim

119899rarrinfin

119899

sum

119894=1

120582119891(120594

119860119894)

= lim119899rarrinfin

119899

sum

119894=1

] (119860119894) =

infin

sum

119894=1

] (119860119894)

(64)

Thus ] is 120590-additiveIn order to apply the theorem of Lebesgue-Radon-

Nikodym see for example Section 69 in [15] and conclud-ing the existence of 119892 isin 119871

1

(Ω) such that

] (119860) = int119860

119892119889119909 119860 120583-measurable (65)

we need to show that ] is absolutely continuous with respectto 120583 To do so assume that 120583(119873) = 0 Then ](119873) = 120582

119891(120594

119873)

By the absolute continuity of the normof1198831015840 required in (AC)we see that 120594

1198731198831015840 = 0 and therefore ](119873) = 0 Replacing 119892

by minus119892 if necessary we obtain

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909 (66)

Now we assume that Ω does not necessarily have finitemeasureWewriteΩ = cup

119899Ω

119899as union of open sets with finite

measure such that Ω119899sub Ω

119899+1 By the considerations above

we find for each 119899 isin N a 119892119899such that

120582119891(120594

119860) = (minus1)

119903

intΩ

119892119899(119909) 120594

119860(119909) 119889119909 (67)

for each measurable set 119860 sub Ω119899 It is easily seen that 119892

119899

and 119892119899+1

agree almost everywhere on Ω119899 This gives rise to

a function 119892 satisfying

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909

119860 sub

119873

⋃

119899=1

Ω119899for some 119873

(68)

By linearity we have for each simple function 119904

120582119891(119904) = (minus1)

119903

intΩ

119892 (119909) 119904 (119909) 119889119909 (69)

To show that 119892 isin 119883 we want to use property (D)

Journal of Function Spaces 9

Let ℎ be a nonnegative simple function with supportin some Ω

119899 Then ℎ sgn(119892) is also a simple function with

support inΩ119899 It is a finite linear combination of characteristic

functions of sets in Ω119899 Hence

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = intℎ sdot sgn (119892) 119892 119889119909 = 120582119891(ℎ sgn (119892)) (70)

Using the boundedness of 120582119891 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (71)

If ℎ is an arbitrary function in1198831015840 thenwemay construct a

sequence (ℎ119899)infin

119899=1of simple functions each ℎ

119899having support

in Ω119899 such that 0 le ℎ

119899uarr |ℎ| 120583-almost everywhere By the

monotone convergence theorem we have by (71) that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = lim119899rarrinfin

int1003816100381610038161003816ℎ119899119892

1003816100381610038161003816 119889119909 le lim sup119899rarrinfin

10038171003817100381710038171003817120582119891

10038171003817100381710038171003817

1003817100381710038171003817ℎ11989910038171003817100381710038171198831015840

(72)

By (P3) in Definition 18 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (73)

and Property (D) gives that 119892 isin 119883 In the last step we want toupgrade the integral representation (69) to hold for arbitraryfunctions in 120593 isin 119883

1015840 As before we approximate 120593 by simplefunctions ℎ

119899with support contained in Ω

119899and 0 le ℎ

119899uarr |120593|

in 120583-almost everywhere point Define 119904119899

= ℎ119899sgn(120593) By

(NC) 119904119899converges to 120593 in119883

1015840 and as both 120582119891and ldquoits integral

representationrdquo are continuous and agree on the set of stepfunctions they agree on119883

1015840

Having shown the existence of the weak derivativesof highest order we are left to show the existence of theintermediate derivatives which we do in Proposition 22 inthe next section To keep the flow going we give here theproof that the pointwise inequality implies membership inthe corresponding Sobolev space under the assumption thatwe already know that Proposition 22 holds

Proof of Theorem 2 (pointwise inequality implies membershipin Sobolev space) We combine Lemmas 20 and 21 and applyProposition 22

4 Equivalent Definitions of Sobolev Spaces

The aim of this section is to prove the following equivalence

Proposition 22 Let Ω sub R be a domain with finite Lebesguemeasure inR Suppose119883 is a Banach function space satisfying(B) and (D) Assume that 119898 isin N Then the following setscoincide

(a) 119860 = 119891 isin 119883 there exists 119865 isin 119883 where 119865 agrees almosteverywherewith119891 and hasweak derivatives up to order119898 in119883 where the derivatives of order 119903 0 le 119903 le 119898minus1are absolutely continuous

(b) 119861 = 119891 isin 119883 119891(119896)

isin 119883 119896 isin 0 119898 where 119891(119896) arethe weak derivatives of 119891

(c) 119862 = 119891 isin 119883 119891(119898)

isin 119883 where 119891(119898) is the weak

derivative of 119891 of order 119898

We see that 119860 sub 119861 sub 119862 We now show that 119862 sub 119860Let us give an outline of the proof Given 119892 Ω rarr R

(we will choose 119892 = 119891(119898)) by integrating we construct in

Lemma 23 an absolutely continuous function 119866119898together

with its weak derivatives 1198660 119866

119898minus1 which are absolutely

continuous and 119866(119898)

119898= 119892 However 119866

119898and 119891 might

not agree almost everywhere We verify however in theproof of Theorem 26 (with the help of Theorem 24) thatwe find a polynomial 119875 such that 119866

119898+ 119875 agrees with 119865

almost everywhere and shares with 119866119898the other required

properties We conclude the proof of Theorem 26 by settlingthe membership questions with the aid of Proposition 25

Lemma 23 Assume Ω sub R is a domain We further assumethat 119892 Ω rarr R is in 119871

1

loc(Ω) One fixes 119909lowast

isin Ω and defines

1198660(119909

0) = 119892 (119909

0)

119866119896(119909

119896) = int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119896 isin N

(74)

where the integrals are Lebesgue integrals and we use thenotation

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

=

int[119909lowast119909119896]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909lowast

le 119909119896

minusint[119909119896 119909lowast]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909119896lt 119909

lowast

(75)

Then

(a) 119866119896is locally absolutely continuous for all 119896 isin N

especially locally integrable(b) if 120593 isin Cinfin

0(Ω) and 0 le 119903 le 119896 one has

intΩ

119866119896120593(119903)

= (minus1)119903

intΩ

119866119896minus119903

120593 (76)

Proof Let us prove (a) by induction Assume 119896 = 1 Note that1198661(119909

lowast

) = 0 Thus we have for 119909lowast

le 1199091

1198661(119909

1) minus 119866

1(119909

lowast

) = 1198661(119909

1) = int

1199091

119909lowast

119892 (1199090) 119889119909

0 (77)

As 119892 isin 1198711

([119909lowast

1199091]) we see that 119866

1is absolutely continuous

The case 1199091lt 119909

lowast is treated similarlyAssume now that the claim is true for some 119896 Now

119866119896+1

(119909119896+1

) is the integral of a continuous function and thuslocally absolutely continuous

Let us have a look at (b) Note that the statement is true for119896 = 0 thus we may assume that 119896 ge 1 We make an inductionover 119903 The case 119903 = 0 is clear Let us choose an interval[119886 119887] sub Ω containing a neighborhood of the support of 120593 As119866119896and 120593

(119903) are both absolutely continuous in [119886 119887] we have

10 Journal of Function Spaces

by the integration by part formula for absolutely continuousfunctions (see for exampleTheorem7147 in [9]) and the factthat 120593(119903)

(119886) = 120593(119903)

(119887) = 0

intΩ

119866119896120593(119903+1)

= int[119886119887]

119866119896120593(119903+1)

= (minus1) int[119886119887]

1198661015840

119896120593(119903)

(78)

Noting that 1198661015840

119896= 119866

119896minus1almost everywhere we see that

intΩ

119866119896120593(119903+1)

= (minus1) int[119886119887]

119866119896minus1

120593(119903)

= (minus1) intΩ

119866119896minus1

120593(119903)

(79)

Using the induction hypothesis the claim follows

We cite Theorem 253 in [10] without giving its proof

Theorem 24 Let 119860 = [119886 119887] R or R+and 119903 ge 0 If the

function 119891 isin 1198711

loc(119860) satisfies

int119860

119891120593(119903)

119889119909 = 0 (80)

for all 120593 isin Cinfin

0(119860) then there is a polynomial 119875 of degree lt 119903

such that 119891 = 119875 almost everywhere on 119860

To show that 119866119896is in 119883 we need to verify that (if there

is danger of confusion with respect to which variable we takethe norm we use notations as119883(119909

119896))

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

lt infin (81)

under the assumption that 119866119896minus1

is in 119883 Actually bounding119866119896from above by its absolute value this means that we have

first to take the1198711-norm and then the norm in119883We howeverwould like to switch the norms

We adapt Proposition 21 in [16] to our setting

Proposition 25 Let 120588 be a Banach function norm satisfyingproperty (D) If 119891 ΩtimesΩ rarr R is a measurable function andone has 119891

119909and 119891

119910 functions onΩ such that 119891119909(119910) = 119891(119909 119910) =

119891119910

(119909) then

120588 (1003817100381710038171003817119891119909

10038171003817100381710038171) le

1003817100381710038171003817120588 (119891119910

)10038171003817100381710038171

(82)

The cases 119883 isin 1198711

119871infin

are easy to prove If 119883 = 119871119901 for

some 1 lt 119901 lt infin then the result can be proven by usingFubinirsquos theorem together with property (D) For the generalresults we refer to Scheprsquos proof

The next result builds uponTheorem 254 in [10]

Theorem 26 Let Ω sub R be a domain and let 119903 ge 1 Suppose119883 is a Banach function space satisfying (B) and (D) If 119891 islocally in 119883 and has a generalized 119903th derivative 119892 locally in119883 then there is 119865 agreeing almost everywhere with 119891 havingabsolutely continuous intermediate weak derivatives and theweak derivative of order 119903 agrees almost everywhere with 119892Furthermore the derivatives of 119865 are locally in 119883 If Ω hasfinite measure one obtains the result with ldquolocallyrdquo replaced byldquogloballyrdquo

We thank Pilar Silvestre for pointing out how to prove themembership of the intermediate derivatives in119883

Proof Choose 119909lowast

isin Ω and let 119866119896be defined as in Lemma 23

Assume 120593 isin Cinfin

0(Ω) Then using the fact that 119892 is a weak

derivative of 119891 of order 119903 and (b) in Lemma 23

intΩ

(119891 minus 119866119903) 120593

(119903)

119889119909 = intΩ

119891120593(119903)

119889119909 minus intΩ

119866119903120593(119903)

119889119909

= (minus1)119903

intΩ

119892120593119889119909 minus (minus1)119903

intΩ

119892120593119889119909 = 0

(83)

UsingTheorem 24 we see that119891minus119866119903= 119875 almost everywhere

where119875 is a polynomial of degreelt 119903Thus119891 = 119875+119866119903almost

everywhere and for 120593 isin Cinfin

0(Ω) and 0 le 119898 le 119903

intΩ

119891120593(119898)

119889119909 = intΩ

(119875 + 119866119903) 120593

(119898)

= (minus1)119898

intΩ

(119866119903minus119898

+ 119875(119898)

) 120593 119889119909

(84)

Thus119866119903minus119898

+119875(119898) is the119898th generalized derivative of119891 As it is

locally absolutely continuous (if119898 lt 119903) it lies locally in119883 Inthe case 119898 = 119903 the same conclusion holds by the assumptionon 119892

Moreover there exists 119865 for example 119865 = 119875 + 119866119903 that

agrees with 119891 almost everywhere such that its generalizedderivatives up to order 119903 minus 1 are locally absolutely continuousand the 119903th generalized derivative equals 119892 almost every-where

As 119892 is locally in 119883 and the other derivatives are locallyabsolutely continuous they are locally in 119871

infin and thus locallyin119883 by (B)

Suppose now that Ω has additionally finite measureAssume that 119891 and 119892 are in119883 globally We want to show thatthe intermediate derivatives are in 119883 as well

Note in our setting that by (P5) in Definition 18membership in 119883 implies membership in 119871

1 We proceedby induction Suppose that 119866

119896minus1isin 119883 We will apply

Proposition 25 We set denoting by conv119909lowast

119909119896 the closed

interval with endpoints 119909lowast and 119909119896

119891 (119909119896minus1

119909119896) = 119866

119896minus1(119909

119896minus1) 120594conv119909lowast 119909119896 (119909119896minus1

) (85)

Then1003817100381710038171003817119866119896

(119909119896)1003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119891(119909119896minus1

119909119896)119889119909

119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)10038171003817100381710038171198711(119909119896minus1)

10038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)

le10038171003817100381710038171003817

1003816100381610038161003816119866119896minus1(119909

119896minus1)1003816100381610038161003816

1003817100381710038171003817120594Ω1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)lt infin

(86)

Journal of Function Spaces 11

5 Banach Function Space Setting

Here our main focus is a more advanced readership thatlikely knows already Banach function spaces Therefore weadopt a much briefer style pointing merely to the literaturewhere the corresponding definitions and results can be foundWe start with giving generalized versions of the necessity andsufficiency of Theorem 2

We list the results and prove them simultaneously bysplitting the proofs into parts according to the functionspaces

Theorem 27 Assume Ω sub R is a domain 119883 is a normedspace and 119898 isin N If the maximal function 119872 119883 rarr 119883 isbounded then there is a constant 119862 = 119862(119883119898) such that foreach continuous 119891 isin 119882

119898

119883there is 119886

119891isin 119883with 119886

119891 le 119862119891

119882119898119883

and

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (87)

In particular such 119886119891exists if

(i) 1 lt 119901 le infin and 119883 is the Lebesgue space 119871119901

(Ω)

(ii) 1 lt 119901 lt infin 1 le 119902 le infin and 119883 is the Lorentz space119871119901119902

(Ω)

(iii) 1 lt 119901 lt infin 1 le 119902 le infin 119887 is slowly varying and 119883 isthe Lorentz-Karamata space 119871

119901119902119887(Ω)

(iv) 120601 is an 119873-function with inf119904gt0

(119904120593(119904)120601(119904)) gt 1 where120593 is the density function of 120601 and 119883 is the Orlicz space119871lowast

120601(Ω)

(v) Ω = R and 119883 is rearrangement-invariant with upperindex 120572

119883lt 1

Theorem 28 Let Ω sub R be open Suppose 119883 is a Banachfunction space such that (AC) holds Assume further that thefunction 119891 Ω rarr R is in 119883 If for some 119903 isin N

0 the following

inequality holds

1003816100381610038161003816Δ119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)] (88)

for some 119886119891isin 119883 then there exists119892 isin 119883with 119892

119883le 2119903

119903

119886119891119883

such that

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909 = (minus1)119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (89)

for all 120593 isin Cinfin

0(Ω)

In particular the result holds if119883 is

(i) the Lebesgue space 119871119901

(Ω) for some 1 lt 119901 lt infin

(ii) the Lorentz space 119871119901119902

(Ω) where 1 lt 119901 lt infin and 1 le

119902 lt infin

(iii) the Lorentz-Karamata space 119871119901119902119887

(Ω) where 119887 isslowly varying 1 lt 119901 lt infin and 1 lt 119902 le infin

(iv) the Orlicz space 119871lowast

120601(Ω) for some 119873-function 120601 satisfy-

ing

inf119904gt0

119904120593 (119904)

120601 (119904)gt 1 (90)

where 120593 is the density function of 120601

To help with orientation let us give an overview ofthe remaining material in this section With respect toTheorem 27 note that we carried out the proof of Corollary 14in a high level of abstraction Hence we are left to verifythat the spaces under question are normed spaces (we willshow that they are even Banach function spaces) and theboundedness of the maximal operator

Concerning the proof Theorem 28 we proved the cor-responding part of Theorem 2 under the assumptions thatthe spaces under question are Banach function spaces andfurther satisfy the requirements listed in Properties 19 Theexistence of 1198831015840 is classical (Proposition 29) as is the fact that(AC) implies (A) and (NC) (Proposition 30) Hence we onlyneed to point to the literature where (AC) was verified

Proposition 29 Given a Banach function space there is aspace 119883

1015840 such that properties (B) (HI) (C) and (D) aresatisfied

Proof We refer the reader to Sections 11 and 12 in [13] or toSection 31 in [14]

The definition of absolutely continuous norm can befound as Definition 3111 in [14] and Definition 131 in [13]

Proposition 30 A Banach function space that has absolutelycontinuous norm satisfies (A) and (NC)

Proof The first statement follows fromTheorem 1313 in [13]and the second fromProposition 136 in the same source

51 General Statements

Proof of Theorem 27 (general statement) We basically havealready proven the general statement in Corollary 14

Proof of Theorem 28 (general statement) We gave the proofof the Lebesgue space case in such generality that thegeneral statement follows from the proof of Theorem 2 andPropositions 29 and 30

52 Lorentz-Karamata Spaces Lorentz-Karamata spaces aredetailed in [14] mainly Section 343

Proof of Theorem 27 (ii) and (iii) The part that is left toprove is the boundedness of the maximal operator which isguaranteed by Remark 3517 in [14]

Proof of Theorem 28 (ii) and (iii) By Theorem 3441 in [14]we see that 119883 is a rearrangement-invariant Banach space Acombination of Lemma 3439 in [14] together with Lemma3443 in [14] gives the absolute continuity (AC) of the normof1198831015840

12 Journal of Function Spaces

53 Orlicz Spaces The setting is the one of [17]

Lemma 31 If 120601 is an 119873-function satisfying the Δ2-condition

then the Orlicz space 119871lowast

120601has absolutely continuous norm

Proof Let 119891 isin 119871lowast

120601(Ω) We verify first that

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 lt infin (91)

Let 120576 gt 0 By the Δ2-condition there exists a constant 120581 =

120581(119891(120601)

+ 120576) such that

120601 ((1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576) 119904) le 120581120601 (119904) (92)

for all 119904 ge 0 Thus

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 = intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576(1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576))119889119909

le 120581intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576)119889119909 le 120581

(93)

Suppose that 119864 is a measurable set Then

10038171003817100381710038171198911205941198641003817100381710038171003817(120601)

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891120594119864

1003816100381610038161003816) 119889119909 le 1

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 1

(94)

Let us fix 120582 gt 0 Again using the Δ2-condition we have

120601(120582minus1

119904) le 120581(120582minus1

)120601(119904) Hence

intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 (95)

Example 133 in [13] tells us that 1198711 has absolutely continuousnorm Thus if 120594

119864119899converges 120583-almost everywhere to 0 then

intΩ

120601(|119891|)120594119864119899

converges to zero as well Thus we have for 119899

large enough that

120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119899119889119909 le 1 (96)

implying that 119891120594119864119899(120601)

le 120582 Since 120582 gt 0 was arbitrary theabsolute continuity of 119891 follows

Proof of Theorem 27 (iv) We conclude from Gallardorsquos resultTheorem 21 in [17] that it is sufficient that the complemen-tary 119873-function satisfies the Δ

2-condition in [0infin) Again

resorting to Gallardo this time to Proposition 14 in [17] weobtain the conclusion

Proof of Theorem 28 (iv) That 119883 is a Banach functionspace follows from Theorem 3416 in [14] To prove that119883

1015840 has absolutely continuous norm we note that inequal-ity (90) together with Proposition 14 in [17] ensuresthat the complementary 119873-function of 120601 satisfies the Δ

2-

condition Lemma 31 verifies that 1198831015840 has absolutely contin-uous norm

54 Rearrangement-Invariant Spaces The necessary defini-tions can be found in [13]

Proof of Theorem 27 (v) We are only left to show theboundedness of the maximal operator However this is thecontent of a result by Lorentz and Shimogaki for examplestated as Theorem 3517 in [13]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank Pilar Silvestre for hersuggestions of how to prove the existence of intermediatederivatives The research is supported by the Academy ofFinland The third author thanks the Department of Math-ematics and Systems Analysis at Aalto University for itshospitality

References

[1] B Bojarski ldquoRemarks on some geometric properties of Sobolevmappingsrdquo in Functional Analysis amp Related Topics (Sapporo1990) pp 65ndash76 World Scientific River Edge NJ USA 1991

[2] B Bojarski PHajlasz andP Strzelecki ldquoImproved119862119896120582 approxi-

mation of higher order Sobolev functions in normand capacityrdquoIndiana University Mathematics Journal vol 51 no 3 pp 507ndash540 2002

[3] B Bojarski ldquoPointwise characterization of Sobolev classesrdquoProceedings of the Steklov Institute of Mathematics vol 255 no1 pp 65ndash81 2006

[4] B Bojarski ldquoWhitneyrsquos jets for Sobolev functionsrdquo UkrainianMathematical Journal vol 59 no 3 pp 345ndash358 2007

[5] B Bojarski ldquoTaylor expansion and Sobolev spacesrdquo Bulletin ofthe Georgian National Academy of Sciences New Series vol 5no 2 pp 5ndash10 2011

[6] B Bojarski ldquoSobolev spaces and Lagrange interpolationrdquoGeorgian Academy of Sciences Proceedings of A RazmadzeMathematical Institute vol 158 pp 1ndash12 2012

[7] P Hajłasz ldquoSobolev spaces on an arbitrary metric spacerdquoPotential Analysis vol 5 no 4 pp 403ndash415 1996

[8] B Bojarski ldquoSobolev spaces and averaging Irdquo httparxivorgabs13085171

[9] R Kannan and C K Krueger Advanced Analysis on the RealLine Universitext Springer New York NY USA 1996

[10] R A DeVore and G G Lorentz Constructive Approximationvol 303 of Fundamental Principles of Mathematical SciencesSpringer Berlin Germany 1993

[11] L L Schumaker Spline Functions Basic Theory CambridgeMathematical Library CambridgeUniversity Press CambridgeUK 3rd edition 2007

[12] G H Hardy and J E Littlewood ldquoA maximal theorem withfunction-theoretic applicationsrdquo Acta Mathematica vol 54 no1 pp 81ndash116 1930

[13] C Bennett andR Sharpley Interpolation ofOperators vol 129 ofPure and Applied Mathematics Academic Press Boston MassUSA 1988

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Stochastic AnalysisInternational Journal of

2 Journal of Function Spaces

Let us give an overview of the paper The proof ofTheorem 2 is presented in a series of results We first givea detailed proof of the fact that Sobolev functions satisfythe pointwise inequality (3) The idea is to consider theTaylor expansion of the function Using the linearity of thefinite differences we only need to compute and bound thefinite differences of polynomials (119909 minus 119910)

119898minus1 and of termslike (119909 minus 119910)

119898minus1

+ The so-called Hermite-Genocchi formula is

an integral representation of the finite differences (actuallydivided differences) which makes the computation rathereasy It turns out that the functions 119886

119891are multiples of the

Hardy-Littlewood maximal function of the correspondingderivatives Reference [8] studies the possibility to find 119886

119891

whose definition does not involve derivatives at allThe otherdirection is easily shown to hold for smooth functions andthe general result follows from this Thereafter we recall thetheorems from literature necessary to be able to deduce thatour main results also apply to Orlicz Lorentz and Lorentz-Karamata spaces

2 Sobolev Functions SatisfyPointwise Inequality

We refer the reader to Chapter 7 in [9] for the definition andproperties of absolutely continuous functions We write 119891 isin

119871119898

1[119886 119887] to express that the function119891 [119886 119887] rarr R is (119898minus1)-

times differentiable and that the derivative of order 119898 minus 1 isabsolutely continuous The aim of this section is to show thatmembership in a Sobolev space implies (3) where the finitedifference is replaced by divided differences

Definition 3 (divided difference) Let 1199101 119910

119903+1be points in

a domain Ω sub R One assumes that if 119910119894has multiplicity 119899

then119891 Ω rarr R is (119899minus1)-times differentiable in 119910119894 First one

considers the case where 1199101le 119910

2le sdot sdot sdot le 119910

119903+1are ordered

One defines for 119903 ge 1

[1199101] 119891 = 119891 (119910

1)

[1199101 119910

119903+1] 119891

=

[1199102 119910

119903+1] 119891 minus [119910

1 119910

119903] 119891

119910119903+1

minus 1199101

if 1199101

= 119910119903+1

119891(119903)

(1199101)

119903otherwise

(4)

We extend the definition to unordered tuples in the onlypossible way that makes the divided difference independentof the ordering

Remark 4 In the recursion formula we omitted 1199101in the

nominator of the minuend and 119910119903+1

in the nominator of thesubtrahend and then divided by 119910

119903+1minus 119910

119903 We claim that we

obtain the same result if we replace 1199101by 119910

119896and 119910

119903+1by 119910

119898as

long as 119910119896

= 119910119898 This can be seen by induction We look here

only at one of the two more difficult cases when one of theomitted points is either the largest or the smallest (and there

are at least three points) The omitted points are denoted by ahat We also omit the function Consider[119910

1] minus [119910

119898]

119910119898

minus 1199101

= ([119910

1 119910

119898] minus [119910

1 119910

119903+1]

119910119903+1

minus 119910119898

minus[119910

1 119910

119898] minus [119910

119898 119910

119903+1]

119910119903+1

minus 1199101

)

times (119910119898

minus 1199101)minus1

= ((119910119903+1

minus 1199101) [119910

1 119910

119898] minus (119910

119903+1minus 119910

1) [119910

1 119910

119903+1]

minus (119910119903+1

minus 119910119898) [119910

1 119910

119898] + (119910

119903+1minus 119910

119898) [119910

119898 119910

119903+1])

times ((119910119903+1

minus 119910119898) (119910

119898minus 119910

1) (119910

119903+1minus 119910

1))

minus1

(5)

We collect the first and third term and expand the second one

[1199101] minus [119910

119898]

119910119898

minus 1199101

= ((119910119898

minus 1199101) [119910

1 119910

119898] minus (119910

119898minus 119910

1) [119910

1 119910

119903+1]

minus (119910119903+1

minus 119910119898) [119910

1 119910

119903+1] + (119910

119903+1minus 119910

119898) [119910

119898 119910

119903+1])

times ((119910119903+1

minus 119910119898) (119910

119898minus 119910

1) (119910

119903+1minus 119910

1))

minus1

(6)

Collecting the first two terms and the second two cancellingand applying the induction hypothesis give

[1199101] minus [119910

119898]

119910119898

minus 1199101

=[119910

1]

119910119903+1

minus 1199101

minus[119910

119903+1]

119910119903+1

minus 1199101

= [1199101 119910

119903+1] (7)

The following formula is a useful tool to compute divideddifferences

Lemma 5 (Hermite-Genocchi formula) Let 1199101 119910

119903+1be

points in a domain Ω sub R One supposes that 119891 Ω rarr R

is such that 119891(119899minus1) is absolutely continuous if not all 119910119894coincide

and that additionally 119891(119899)

(1199101) exists if all points are the same

Then[119910

0 119910

119899] 119891

= int

1

0

1198891199051int

1199051

0

1198891199052

sdot sdot sdot int

119905119899minus1

0

119891(119899)

(1199100+ (119910

1minus 119910

0) 119905

1

+ sdot sdot sdot + (119910119899minus 119910

119899minus1) 119905

119899) 119889119905

119899

(8)

Proof (see page 122 in [10] as well) If 119899 = 1 we have that

int

1

0

1198911015840

(1199100+ (119910

1minus 119910

0) 119905

1) 119889119905

1

=

119891 (1199101) minus 119891 (119910

0)

1199101minus 119910

0

1199100

= 1199101

1198911015840

(1199100) 119910

0= 119910

1

= [119910

0 119910

1] 119891 = [119910

1 119910

0] 119891 119910

0= 119910

1

[1199100 119910

0] 119891 119910

0= 119910

1

(9)

Journal of Function Spaces 3

Assume now that the claim is true for some 119899 We assumehere that 119910

119899+1= 119910

119899and leave the case when 119910

119899and 119910

119899+1

coincide to the reader (computation (14) might be helpful)Then evaluating the right most integral

int

1

0

1198891199051int

1199051

0

1198891199052

sdot sdot sdot int

119905119899

0

119891(119899+1)

(1199100+

119899

sum

119896=1

(119910119896minus 119910

119896minus1) 119905

119896

+ (119910119899+1

minus 119910119899) 119905

119899+1)119889119905

119899+1

= int

1

0

1198891199051int

1199051

0

1198891199052

sdot sdot sdot int

119905119899minus1

0

(119891(119899)

(1199100+

119899minus1

sum

119896=1

(119910119896minus 119910

119896minus1) 119905

119896

+ (119910119899+1

minus 119910119899minus1

) 119905119899)

minus119891(119899)

(1199100+

119899

sum

119896=1

(119910119896minus 119910

119896minus1) 119905

119896))119889119905

119899+1

sdot1

119910119899+1

minus 119910119899

=1

119910119899+1

minus 119910119899

([1199100 119910

119899minus1 119910

119899+1] 119891 minus [119910

0 119910

119899] 119891)

= [1199100 119910

119899+1] 119891

(10)

We use the Hermite-Genocchi formula to compute finitedifferences of polynomials

Lemma 6 Let 119886 be a point in R and 119898 isin N0 We set

119901119898119886

(119910) = (119910 minus 119886)119898

(11)

Then

[1199100 119910

119899] 119901

119898119886=

0 0 le 119898 lt 119899

1 119898 = 119899(12)

Proof As 119901119898119886

is smooth we can use the Hermite-Genocchiformula Lemma 5 If 0 le 119898 lt 119899 then 119901

(119899)

119898119886= 0 hence the

Hermite-Genocchi formula gives the claim If 119898 = 119899 then119901(119898)

119898119886= 119898 we compute the integral in the Hermite-Genocchi

formula via induction over119898We let first for fixed 119905

0ge 0

⟨1199100 119910

119898⟩1199050

= int

1199050

0

1198891199051int

1199051

0

1198891199052sdot sdot sdot int

119905119898minus1

0

1 119889119905119898 (13)

We claim that

⟨1199100 119910

119898⟩1199050

=119905119898

0

119898 119898 isin N (14)

For 119898 = 1 this can easily be seen Assume that the claimholds for some119898 Then

⟨1199100 119910

119898+1⟩1199050

= int

1199050

0

1198891199051⟨119910

0 119910

119898⟩1199051

= int

1199050

0

119905119898

1

119898119889119905

1=

119905119898+1

0

(119898 + 1)

(15)

The Hermite-Genocchi formula tells us [1199100 119910

119898]119901119898119886

=

⟨1199100 119910

119898⟩1sdot 119898 giving the proof

The following notation is on page 15 in [11]

Definition 7 Set

(119909 minus 119910)0

+=

1 119909 ge 119910

0 119909 lt 119910(16)

and for119898 gt 1

(119909 minus 119910)119898minus1

+=

(119909 minus 119910)119898minus1

119909 ge 119910

0 119909 lt 119910(17)

The definition below is a variant of Definition 412 in [11]However as our starting point is different than Schumakerrsquosthere is a slight regularity issue when 119909 does not differ fromat least two 119910

119894 However we are only interested in cases where

119876119898

119894lives under an integral andwe agree here to set119876119898

119894(119909) = 0

in points where we do not have enough regularity

Definition 8 (B-spline) Let sdot sdot sdot le 119910minus1

le 1199100le 119910

1le 119910

2le sdot sdot sdot

be a sequence of real numbers Given integers 119894 and 119898 gt 0one defines (note that 119909 is fixed so that the finite difference iswith respect to 119910)

119876119898

119894(119909) =

(minus1)119898

[119910119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+ if 119910

119894lt 119910

119894+119898

0 otherwise(18)

We call 119876119898

119894the 119898th order B-spline associated with the knots

119910119894 119910

119894+119898

This is Theorem 22 in [11] Schumaker does not give theproof but writes that it is similar to the one ofTheorem 21 inhis book

Theorem 9 (Dual Taylor Expansion) Let 119891 isin 119871119898

1[119886 119887] Then

for all 119886 le 119910 le 119887

119891 (119910) =

119898minus1

sum

119895=0

(minus1)119895

119863119895

119891 (119887) (119887 minus 119910)119895

119895

+ int

119887

119886

(119909 minus 119910)119898minus1

+(minus1)

119898

119863119898

119891 (119909)

(119898 minus 1)119889119909

(19)

4 Journal of Function Spaces

Proof The proof uses induction and integration by partsNote that by replacing 119886 by 119910 in the integral we may deletethe + in (119909 minus 119910)

+ The case119898 = 1 reduces to the fundamental

theorem of calculus for absolutely continuous functions seefor example Theorem 7115 in [9] Let us assume that thestatement holds for some119898 and fix 119891 isin 119871

119898+1

1[119886 119887] Then

int

119887

119910

(119909 minus 119910)119898

(minus1)119898+1

119863119898+1

119891 (119909)

119898119889119909

=(119887 minus 119910)

119898

(minus1)119898+1

119863119898

119891 (119887)

119898

minus int

119887

119910

(119909 minus 119910)119898minus1

(minus1)119898+1

119863119898

119891 (119909)

(119898 minus 1)119889119909

= minus(119887 minus 119910)

119898

(minus1)119898

119863119898

119891 (119887)

119898

minus

119898minus1

sum

119895=0

(minus1)119895

119863119895

119891 (119887) (119887 minus 119910)119895

119895+ 119891 (119910)

(20)

We now cite a part ofTheorem 423 in [11] Note that119898 isa positive integer and 119863

119895

+means that we take the right hand

derivative of order 119895 We only need and prove the equality for119895 = 0 We also assume that not all points coincide

Theorem 10 (Peano representation) Fix 0 le 119895 le 119898minus119899 where119899 is the maximum multiplicity of 119910

119894 119910

119894+119898 Then

[119910119894 119910

119894+119898] 119891 = int

119910119894+119898

119910119894

(minus1)119895

119863119895

+119876

119898

119894(119909)119863

119898minus119895

119891 (119909)

(119898 minus 1)119889119909 (21)

for all 119891 isin 119871119898minus119895

1[min

119894119910

119894max

119894119910

119894]

Proof of the Peano Representation for 119895 = 0 We take thedual Taylor expansion from Theorem 9 Then we apply[119910

119894 119910

119894+119898] on both sides and remember that we computed

the divided differences of polynomials in Lemma 6

Our goal is to use the Peano representation to obtainan upper bound for the divided differences We start withestablishing an upper bound of the factor 119863

+119876

119898

119894in the

integrand

Lemma 11 Suppose that 119910119894le sdot sdot sdot le 119910

119894+119898and 119910

119894lt 119910

119894+119898 Then

there is a constant 119862 such that

1003816100381610038161003816119863+119876

119898

119894(119909)

1003816100381610038161003816 =10038161003816100381610038161003816[119910

119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+

10038161003816100381610038161003816le

119862

119910119894+119898

minus 119910119894

(22)

Proof We want to estimate

[119910119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+

=[119910

119894+1 119910

119894+119898] (119909 minus 119910)

119898minus1

+minus [119910

119894 119910

119894+119898minus1] (119909 minus 119910)

119898minus1

+

119910119894+119898

minus 119910119894

(23)

Assume 119896 lt 119898 minus 1 We claim that

119889119896

119889119910119896

(119909 minus 119910)119898minus1

+= (minus1)

119896

(119898 minus 1) sdot sdot sdot (119898 minus 119896) (119909 minus 119910)119898minus1minus119896

+

(24)

This claim is easily seen to be true for 119910 = 119909 That it is truealso for 119910 = 119909 follows by induction We also note that

119889119898minus2

119889119910119898minus2(119909 minus 119910)

119898minus1

+(25)

is absolutely continuous on bounded intervals and that

(minus1)(119898minus1)

(119898 minus 1) (119909 minus 119910)0

+

(26)

is a weak derivative that equals the derivative whenever119910 = 119909 We first assume that not all 119910

119895 119910

119895+119898minus1coincide

An application of the Hermite-Genocchi formula Lemma 5gives

[119910119895 119910

119895+119898minus1] (119909 minus 119910)

119898minus1

+

= int

1

0

1198891199051int

1199051

0

1198891199052

sdot sdot sdot int

119905119898minus2

0

(minus1)(119898minus1)

(119898 minus 1)

sdot (119909 minus (119910119895+

119898minus1

sum

119896=1

(119910119895+119896

minus 119910119895+119896minus1

) 119905119896))

0

+

119889119905119898minus1

(27)

We use now the fact that the integrand is bounded from aboveby (119898 minus 1) and the computation (14) to obtain

10038161003816100381610038161003816[119910

119895 119910

119895+119898minus1] (119909 minus 119910)

119898minus1

+

10038161003816100381610038161003816le 1 (28)

Suppose now that 119910119895

= sdot sdot sdot = 119910119895+119898minus1

= 119909 Then|[119909

119895 119909

119895+119898minus1](119909 minus 119910)

119898minus1

+| is bounded from above by 1 if

119909 gt 119910119894and vanishes if 119910

119894lt 119909 In the end we find a constant

119862 such that10038161003816100381610038161003816119863

+[119910

119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+

10038161003816100381610038161003816le

119862

119910119894+119898

minus 119910119894

(29)

In the next result we use the notation

119872119892(119909) = sup119911

minusint

119911

119909

10038161003816100381610038161198921003816100381610038161003816 (30)

for the famous Hardy-Littlewood maximal function of 119892

introduced in the seminal paper [12] of Hardy-Littlewood inActa Math (1930)

Theorem 12 Suppose 1199100

le sdot sdot sdot le 119910119898 119910

0lt 119910

119898and 119891 is in

119871119898

1[119910

0 119910

119898] for some 119898 isin N Then there is some constant 119862

such that1003816100381610038161003816[1199100 119910119898] 119891

1003816100381610038161003816 le 119862 (119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (1199100) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910119898))

(31)

Journal of Function Spaces 5

Proof We use the Peano representation Theorem 10 with119894 = 119895 = 0 Then using the bound of |119863

+119876

119898

119894(119909)| found in

Lemma 11

1003816100381610038161003816[1199100 119910119898] 1198911003816100381610038161003816 le

1

(119898 minus 1)int

119910119898

1199100

1003816100381610038161003816119863+119876

119898

119894(119909)

1003816100381610038161003816

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

le119862

119910119898

minus 1199100

int

119910119898

1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

(32)

For 119910 isin [1199100 119910

119898] we have

1003816100381610038161003816[1199100 119910119898] 1198911003816100381610038161003816

le119862

119910119898

minus 1199100

(int

119910

1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909 + int

119910119898

119910

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909)

(33)

and further1003816100381610038161003816[1199100 119910119898] 119891

1003816100381610038161003816

le 119862(minusint

119910

1199100

119910 minus 1199100

119910119898

minus 1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

+minusint

119910119898

119910

119910119898

minus 119910

119910119898

minus 1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909)

(34)

The first factors in both integrals are bounded from above by1 We see that the integrals are bounded from above by thecorresponding maximal operators This gives the claim

The finite differences defined in (2) are up to a factordivided differences for equispaced nodes

Lemma 13 Let 119909 119910 be two points in a domain Ω sub R Let119898 isin N and 119909

119896= 119909

0+ ℎ119896 where ℎ = (119910 minus 119909)119898 Then each

119891 Ω rarr R satisfies

[1199090 119909

119898] 119891 =

1

ℎ119898119898Δ119898

119891 (119909 119910) (119909 = 1199090 119909

119898= 119910)

(35)

Proof We use induction For119898 = 1 we have

[1199090 119909

1] 119891 =

119891 (1199091) minus 119891 (119909

0)

1199091minus 119909

0

(36)

As ℎ = 1199091minus 119909

0 119909 = 119909

0 119910 = 119909

1 and 119898 = 1 we obtain the

claim Suppose that the claim is true for some119898 isin N Then

Δ119898+1

ℎ119891 (119909)

=

119898+1

sum

119895=0

(minus1)119898+1minus119895

(119898 + 1

119895)119891 (119909 + 119895ℎ)

=

119898+1

sum

119895=0

(minus1)119898+1minus119895

((119898

119895 minus 1) + (

119898

119895))119891 (119909 + 119895ℎ)

=

119898+1

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + (119895 + 1) ℎ)

minus

119898+1

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + 119895ℎ)

=

119898

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909

1+ 119895ℎ)

minus

119898

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + 119895ℎ)

= ℎ119898

119898 ([1199091 119909

119898+1] 119891 minus [119909

0 119909

119898] 119891)

= ℎ119898

119898 (ℎ (119898 + 1)) [1199090 119909

119898+1] 119891

(37)

We can now formulate Theorem 12 in terms of finitedifferences

Corollary 14 (Corollary of Theorem 12) Let Ω sub R be adomain Suppose 119891 isin 119871

119898

1(Ω) for some 119898 isin N then there is

a constant 119862 = 119862(119898) such that

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816

le 1198621003816100381610038161003816119909 minus 119910

1003816100381610038161003816

119898

(119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

119909 119910 isin Ω

(38)

In particular we have the following special case Suppose 119883

and 119884 are normed spaces of functions such that the maximaloperator 119872 119883 rarr 119884 is bounded If 119891 isin 119882

119898

119883(Ω) for some

119898 ge 1 then

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (39)

for some function 119886119891isin 119884 depending on 119891

Proof We first use Lemma 13 and thenTheorem 12 with stepℎ = (119910 minus 119909)119898 There is some constant 119862 = 119862(119898) such that

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816

= ℎ119898

1198981003816100381610038161003816[1199090

119909119898] 119891

1003816100381610038161003816

le 119862ℎ119898

(119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

le1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

119898

(119862119872(1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119862119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

(40)

Proof of Theorem 2 (Sobolev functions satisfy pointwiseinequality) We combine Corollary 14 with the boundednessof the maximal operator which is by now classical and can befound for example in Theorem 3310 in [13]

6 Journal of Function Spaces

3 Pointwise Inequality Implies Membershipin Sobolev Space

In this section our aim is to show that the pointwiseinequality (3) implies the existence of the highest orderderivative in the correct space We will take care of theintermediate derivatives in the next section In this sectionthe properties of the Lebesgue spaces play an important roleOur point of view is however to extract the properties ofthe Lebesgue spaces and base our proofs solely on theseproperties We hope that the disadvantage of being slightlymore abstract is later on compensatedwhenwe generalize theresults We start with some auxiliary results

Lemma 15 Let Ω sub R be a domain and 120593 isin C119899

0(Ω) that

is 120593 has compact support in Ω and is 119899-times continuouslydifferentiable Then

1003816100381610038161003816100381610038161003816100381610038161003816

[1199090 119909

119899] 120593 minus

120593(119899)

(119910)

119899

1003816100381610038161003816100381610038161003816100381610038161003816

997888rarr 0 (41)

uniformly inΩ as 1199090 119909

119899rarr 119910

Proof By Taylorrsquos formula for 119909 isin conv1199090 119909

119899 there is

120585119909isin conv119909

0 119909

119899 such that

120593 (119909) = 120593 (119910) + 1205931015840

(119910) (119909 minus 119910) + sdot sdot sdot + 120593(119899minus1)

(119910)(119909 minus 119910)

119899minus1

(119899 minus 1)

+1

119899120593(119899)

(120585119909) (119909 minus 119910)

119899

(42)

We apply [1199090 119909

119899] on both sides (with variable 119909) and use

the computation of the divided differences of polynomialscarried out in Lemma 6 to obtain

[1199090 119909

119899] 120593 (119909) =

1

119899120593(119899)

(120585119909) (43)

As 120593(119899) is uniformly continuous on compact intervals the

claim follows

Lemma 16 Assume Ω sub R is open and 119870 sub Ω is compactThen there is a neighborhood of119870 that is a subset of Ω

Proof For each 119909 isin 119870 we find a ball 119861(119909 119903119909) sub Ω Suppose

the statement of the lemma is falseThen for every 119899 isin N thereis 119911

119899isin Ω 119870 with dist(119911

119899 119870) lt 1119899 Thus for each 119899 isin N

there is 119910119899

isin 119870 with 119889(119911119899 119910

119899) lt 1119899 Since 119870 is compact

there is a subsequence (119910119899119896)119896of (119910

119899)119899that converges to some

119910 isin 119870 Now 119910 lies in one of the balls 119861(119909 119903119909) sub Ω For 119896 large

enough we have that 119910119899119896and 119911

119899119896are in 119861(119909 119903

119909) sub Ω This is

a contradiction

Lemma 17 SupposeΩ sub R is open and119870 sub Ω compactThenthere exists 120576 gt 0 such that

(Ω minus ℎ) cap 119870 = Ω cap 119870 (44)

for |ℎ| lt 120576

Proof We first show that any member of the left hand side iscontained in the right hand side this is true for any ℎ isin R If119911 is a member of the left hand side then it is in 119870 and since119870 is a subset of Ω it is also a member of the right hand sideTo prove the other inclusion note that by Lemma 16 there is120576 gt 0 such that the 120576-neighborhood 119870

120576of 119870 lies in Ω Now

suppose that 119911 is in the right hand side and |ℎ| lt 120576 Then for119909 = 119911+ℎ we have119909 isin 119870

120576sub Ω implying that 119911 = 119909minusℎ isin Ωminusℎ

That 119911 isin 119870 is clear This gives the proof

Next we state basically Definition 311 in [14] Readersfollowing [13] might want to have a look at Definitions 111and 113 in their book To gain familiarity with the conceptof Banach function norm the reader may want to verify thatLebesgue spaces are examples of Banach function spaces Aswe are here only interested in Lebesgue measure we adaptthe notion to this case The set M+ denotes the cone ofmeasurable functions whose values lie in [0infin]

Definition 18 (Banach function norm Banach functionspace) Assume that 119877 sub R A Banach function norm on 119877

is a map 120588 from M+

(119877) to [0infin] such that for all 119891 119892 and119891119899(119899 isin N) in M+

(119877) all scalars 120582 ge 0 and all measurablesubsets 119864 of 119877 the following are true where 120583 is the Lebesguemeasure

(P1) 120588(119891) = 0 if and only if 119891 = 0 120583-almost everywhere120588(120582119891) = 120582120588(119891) 120588(119891 + 119892) le 120588(119891) + 120588(119892)

(P2) if 0 le 119892 le 119891 120583-almost everywhere then 120588(119892) le 120588(119891)(P3) if 0 le 119891

119899uarr 119891 120583-almost everywhere we have 120588(119891

119899) uarr

120588(119891)(P4) if 120583(119864) lt infin then 120588(120594

119864) lt infin

(P5) if120583(119864) lt infin then there is a constant119862 = 119862(119864 120588) lt infin

such that int119864

119891119889119909 le 119862120588(119891)

Given such a function norm 120588 the set 119883 = 119883(120588) ofall extended measurable functions 119891 (identifying functionsequal 120583-almost everywhere) such that 120588(|119891|) lt infin is called aBanach function space (BFS for short) and we define

10038171003817100381710038171198911003817100381710038171003817119883

= 120588 (1003816100381610038161003816119891

1003816100381610038161003816) 119891 isin 119883 (45)

We leave it to the reader to verify that Lebesgue spacessatisfy the following properties (1198831015840 can be chosen as the dualspace)

Properties 19 Given a Banach function space 119883 we requirethe existence of a Banach function space119883

1015840 such that

(B) 119883 contains each essentially bounded function whosesupport has finite measure

(HI) Holderrsquos inequality int |119891119892|119889119909 le 119891119883119892

1198831015840

(C) 1198831015840 is complete

(D) 119892119883

= supintΩ

|119892|ℎ ℎ isin M+

ℎ1198831015840 le 1

(AC) absolutely continuous norm if (120594119864119899)119899is a sequence of

characteristic functions converging pointwise almosteverywhere to 0 then 119891120594

1198641198991198831015840 converges to zero for

all 119891 isin 1198831015840

Journal of Function Spaces 7

(A) each function 119891 isin 1198831015840 can be approximated by simple

functions (in1198831015840-norm)

(NC) if 119891 isin 1198831015840 and ℎ

119899are step functions with 0 le ℎ

119899uarr |119891|

then1003817100381710038171003817119891 minus ℎ

119899sgn (119891)

10038171003817100381710038171198831015840997888rarr 0 as 119899 997888rarr infin (46)

Lemma 20 Suppose Ω sub R is open and 119883 is a Banachfunction space fulfilling the items listed in Properties 19 Assumethat 119891 Ω rarr R is in 119883 We let 120582

119891 119883

1015840

cap C119903+1

0(Ω) rarr R be

defined as

120582119891(120593) = int

Ω

119891120593(119903)

119889119909 (47)

If1003816100381610038161003816Δ

119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)]

for almost every 119909 119910 isin Ω

(48)

for some 119886119891

isin 119883 then 120582119891has a continuous extension to 119883

1015840

with norm bounded from above by 2119903119903

119886119891119883

Proof Our goal is to show1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816le 2119903

11990310038171003817100381710038171003817119886119891

10038171003817100381710038171003817119883

100381710038171003817100381712059310038171003817100381710038171198831015840

(49)

because 120582119891extends then by the theorem of Hahn-Banach to

a bounded functional on1198831015840 with norm bounded from above

by 2119903119903

119886119891119883 Without loss of generality we may assume that

119886119891is nonnegative If 119891 were be smooth we could just use

integration by parts and use as bound the limiting case of thepointwise inequality (48) As 119891 is not necessarily smooth wehave to operate on the level of finite differences

Let us choose 120593 isin C0(Ω) and denote its support by 119870

Lemma 17 ensures the existence of some positive 120576 gt 0 suchthat (Ωminus119895ℎ)cap119870 = Ωcap119870 for 0 le 119895 le 119903 as long as |ℎ| lt 120576119903Weassume in the following that ℎ has been chosen accordingly

The following integrals exist by Holderrsquos inequalitybecause 120593 and thus Δ

119903

minusℎ120593(119909) are bounded with compact

support and 119891 is integrable Consider

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

119891 (119909) (minusℎ)minus119903

Δ119903

minusℎ120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(50)

We compute the following integral

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

= intΩ

119891 (119909)

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119909) 120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωminus119895ℎ

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

(Ωminus119895ℎ)cap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

(51)

We have chosen 120576 gt 0 such that (Ω minus 119895ℎ) cap 119870 = Ω cap 119870 Wecontinue

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωcap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

(52)

With the aid of (52) we can now rewrite (50) as

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

(minusℎ)minus119903

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(53)

Inequality (48) leads to

1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816

le 119903119903

intΩ

(119886119891(119909) + 119886

119891(119909 + 119903ℎ))

1003816100381610038161003816120593 (119909)1003816100381610038161003816 119889119909

+ intΩ

1003816100381610038161003816119891 (119909)1003816100381610038161003816

10038161003816100381610038161003816120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)

10038161003816100381610038161003816119889119909

(54)

Using Holderrsquos inequality (HI) for the first summand weare left to show that the second summand tends to 0 asℎ approaches 0 We note first that since the second factorhas support in a set of finite measures we can replace theintegration domain by a set of finite measuresThen we applya (1198711

119871infin

)-Holder inequality and the uniform convergence ofthe divided difference to 120593

(119903)

(119909)119903 established in Lemmas 13and 15

After having verified that 120582119891has a continuous extension

to 1198831015840 we want to show that this extension has a representa-

tion as integral

8 Journal of Function Spaces

Lemma 21 Assume Ω sub R is open Suppose the propertieslisted in Properties 19 are met Let 120582

119891be defined as in (47) and

bounded on 1198831015840 Then there exists 119892 isin 119883 with 119892

119883le 120582

119891

such that

120582119891(120593) = (minus1)

119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (55)

for all 120593 isin 1198831015840 especially for all 120593 isin Cinfin

0(Ω)

Proof The proof is heavily influenced by the proof of Theo-rem 141 in [13]We first want to find a signedmeasure ] suchthat

120582119891(120593) = int

Ω

120593119889] (56)

for each 120593 isin 1198831015840 Then we want to apply the Lebesgue-

Radon-Nikodym theorem to replace 119889] by (minus1)119903

119892(119909)119889119909 forsome integrable function 119892 for which we then show that it iscontained in119883

We start by assuming that the measure of Ω is finite Foreach 120583-measurable set 119860 we set

] (119860) = 120582119891(120594

119860) (57)

We start by proving the 120590-additivity of ] (for the 120583-measurable sets) In the first step we just show the additivityBy induction it is enough to consider only two sets

Let 1198601and 119860

2be two disjoint 120583-measurable sets Then

] (1198601cup 119860

2) = 120582

119891(120594

1198601cup1198602) = 120582

119891(120594

1198601+ 120594

1198602)

= 120582119891(120594

1198601) + 120582

119891(120594

1198602) = ] (119860

1) + ] (119860

2)

(58)

We move to the verification of the 120590-additivity Let119860

1 119860

2 be countablymany pairwise disjoint120583-measurable

sets We first want to show thatsum119894120594119860119894

converges in1198831015840

We let 119864119896= cup

119896

119894=1119860

119894and 119864 = cup

infin

119894=1119860

119894 Then since (119864

119896)119896is

increasing

120583 (119864) = 120583(

infin

⋃

119896=1

119864119896) = lim

119896rarrinfin

120583 (119864119896) (59)

As 119864 has finite measure given 120576 gt 0 there exists 1198960isin N such

that

(1 minus 120576) 120583 (119864) lt 120583 (1198641198960) (60)

By the disjointness of the sets 119860119894 we see that

120583 (119864) = 120583 (1198641198960) + 120583(

infin

⋃

119894=1198960+1

119860119894) (61)

It follows that

120583(

infin

⋃

119894=1198960+1

119860119894) = 120583 (119864) minus 120583 (119864

1198960) lt 120583 (119864) minus (1 minus 120576) 120583 (119864)

= 120576120583 (119864)

(62)

We have for119898 lt 119899

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=1

120594119860119894

minus

119898

sum

119894=1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

=

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=119898+1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

le10038171003817100381710038171003817120594cupinfin119894=119898+1

119860119894

100381710038171003817100381710038171198831015840 (63)

By the considerations before and the fact that sdot 1198831015840 is

absolutely continuous as stipulated in (AC) we see thatsum

119899

119894=1120594119860119894

is a Cauchy sequence Now as 1198831015840 is complete by

(C) the Cauchy sequence converges in 1198831015840 We have by the

continuity of 120582119891

] (cup119894119860

119894) = 120582

119891(120594

cup119894119860119894) = 120582

119891(

infin

sum

119894=1

120594119860119894)

= lim119899rarrinfin

120582119891(

119899

sum

119894=1

120594119860119894) = lim

119899rarrinfin

119899

sum

119894=1

120582119891(120594

119860119894)

= lim119899rarrinfin

119899

sum

119894=1

] (119860119894) =

infin

sum

119894=1

] (119860119894)

(64)

Thus ] is 120590-additiveIn order to apply the theorem of Lebesgue-Radon-

Nikodym see for example Section 69 in [15] and conclud-ing the existence of 119892 isin 119871

1

(Ω) such that

] (119860) = int119860

119892119889119909 119860 120583-measurable (65)

we need to show that ] is absolutely continuous with respectto 120583 To do so assume that 120583(119873) = 0 Then ](119873) = 120582

119891(120594

119873)

By the absolute continuity of the normof1198831015840 required in (AC)we see that 120594

1198731198831015840 = 0 and therefore ](119873) = 0 Replacing 119892

by minus119892 if necessary we obtain

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909 (66)

Now we assume that Ω does not necessarily have finitemeasureWewriteΩ = cup

119899Ω

119899as union of open sets with finite

measure such that Ω119899sub Ω

119899+1 By the considerations above

we find for each 119899 isin N a 119892119899such that

120582119891(120594

119860) = (minus1)

119903

intΩ

119892119899(119909) 120594

119860(119909) 119889119909 (67)

for each measurable set 119860 sub Ω119899 It is easily seen that 119892

119899

and 119892119899+1

agree almost everywhere on Ω119899 This gives rise to

a function 119892 satisfying

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909

119860 sub

119873

⋃

119899=1

Ω119899for some 119873

(68)

By linearity we have for each simple function 119904

120582119891(119904) = (minus1)

119903

intΩ

119892 (119909) 119904 (119909) 119889119909 (69)

To show that 119892 isin 119883 we want to use property (D)

Journal of Function Spaces 9

Let ℎ be a nonnegative simple function with supportin some Ω

119899 Then ℎ sgn(119892) is also a simple function with

support inΩ119899 It is a finite linear combination of characteristic

functions of sets in Ω119899 Hence

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = intℎ sdot sgn (119892) 119892 119889119909 = 120582119891(ℎ sgn (119892)) (70)

Using the boundedness of 120582119891 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (71)

If ℎ is an arbitrary function in1198831015840 thenwemay construct a

sequence (ℎ119899)infin

119899=1of simple functions each ℎ

119899having support

in Ω119899 such that 0 le ℎ

119899uarr |ℎ| 120583-almost everywhere By the

monotone convergence theorem we have by (71) that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = lim119899rarrinfin

int1003816100381610038161003816ℎ119899119892

1003816100381610038161003816 119889119909 le lim sup119899rarrinfin

10038171003817100381710038171003817120582119891

10038171003817100381710038171003817

1003817100381710038171003817ℎ11989910038171003817100381710038171198831015840

(72)

By (P3) in Definition 18 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (73)

and Property (D) gives that 119892 isin 119883 In the last step we want toupgrade the integral representation (69) to hold for arbitraryfunctions in 120593 isin 119883

1015840 As before we approximate 120593 by simplefunctions ℎ

119899with support contained in Ω

119899and 0 le ℎ

119899uarr |120593|

in 120583-almost everywhere point Define 119904119899

= ℎ119899sgn(120593) By

(NC) 119904119899converges to 120593 in119883

1015840 and as both 120582119891and ldquoits integral

representationrdquo are continuous and agree on the set of stepfunctions they agree on119883

1015840

Having shown the existence of the weak derivativesof highest order we are left to show the existence of theintermediate derivatives which we do in Proposition 22 inthe next section To keep the flow going we give here theproof that the pointwise inequality implies membership inthe corresponding Sobolev space under the assumption thatwe already know that Proposition 22 holds

Proof of Theorem 2 (pointwise inequality implies membershipin Sobolev space) We combine Lemmas 20 and 21 and applyProposition 22

4 Equivalent Definitions of Sobolev Spaces

The aim of this section is to prove the following equivalence

Proposition 22 Let Ω sub R be a domain with finite Lebesguemeasure inR Suppose119883 is a Banach function space satisfying(B) and (D) Assume that 119898 isin N Then the following setscoincide

(a) 119860 = 119891 isin 119883 there exists 119865 isin 119883 where 119865 agrees almosteverywherewith119891 and hasweak derivatives up to order119898 in119883 where the derivatives of order 119903 0 le 119903 le 119898minus1are absolutely continuous

(b) 119861 = 119891 isin 119883 119891(119896)

isin 119883 119896 isin 0 119898 where 119891(119896) arethe weak derivatives of 119891

(c) 119862 = 119891 isin 119883 119891(119898)

isin 119883 where 119891(119898) is the weak

derivative of 119891 of order 119898

We see that 119860 sub 119861 sub 119862 We now show that 119862 sub 119860Let us give an outline of the proof Given 119892 Ω rarr R

(we will choose 119892 = 119891(119898)) by integrating we construct in

Lemma 23 an absolutely continuous function 119866119898together

with its weak derivatives 1198660 119866

119898minus1 which are absolutely

continuous and 119866(119898)

119898= 119892 However 119866

119898and 119891 might

not agree almost everywhere We verify however in theproof of Theorem 26 (with the help of Theorem 24) thatwe find a polynomial 119875 such that 119866

119898+ 119875 agrees with 119865

almost everywhere and shares with 119866119898the other required

properties We conclude the proof of Theorem 26 by settlingthe membership questions with the aid of Proposition 25

Lemma 23 Assume Ω sub R is a domain We further assumethat 119892 Ω rarr R is in 119871

1

loc(Ω) One fixes 119909lowast

isin Ω and defines

1198660(119909

0) = 119892 (119909

0)

119866119896(119909

119896) = int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119896 isin N

(74)

where the integrals are Lebesgue integrals and we use thenotation

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

=

int[119909lowast119909119896]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909lowast

le 119909119896

minusint[119909119896 119909lowast]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909119896lt 119909

lowast

(75)

Then

(a) 119866119896is locally absolutely continuous for all 119896 isin N

especially locally integrable(b) if 120593 isin Cinfin

0(Ω) and 0 le 119903 le 119896 one has

intΩ

119866119896120593(119903)

= (minus1)119903

intΩ

119866119896minus119903

120593 (76)

Proof Let us prove (a) by induction Assume 119896 = 1 Note that1198661(119909

lowast

) = 0 Thus we have for 119909lowast

le 1199091

1198661(119909

1) minus 119866

1(119909

lowast

) = 1198661(119909

1) = int

1199091

119909lowast

119892 (1199090) 119889119909

0 (77)

As 119892 isin 1198711

([119909lowast

1199091]) we see that 119866

1is absolutely continuous

The case 1199091lt 119909

lowast is treated similarlyAssume now that the claim is true for some 119896 Now

119866119896+1

(119909119896+1

) is the integral of a continuous function and thuslocally absolutely continuous

Let us have a look at (b) Note that the statement is true for119896 = 0 thus we may assume that 119896 ge 1 We make an inductionover 119903 The case 119903 = 0 is clear Let us choose an interval[119886 119887] sub Ω containing a neighborhood of the support of 120593 As119866119896and 120593

(119903) are both absolutely continuous in [119886 119887] we have

10 Journal of Function Spaces

by the integration by part formula for absolutely continuousfunctions (see for exampleTheorem7147 in [9]) and the factthat 120593(119903)

(119886) = 120593(119903)

(119887) = 0

intΩ

119866119896120593(119903+1)

= int[119886119887]

119866119896120593(119903+1)

= (minus1) int[119886119887]

1198661015840

119896120593(119903)

(78)

Noting that 1198661015840

119896= 119866

119896minus1almost everywhere we see that

intΩ

119866119896120593(119903+1)

= (minus1) int[119886119887]

119866119896minus1

120593(119903)

= (minus1) intΩ

119866119896minus1

120593(119903)

(79)

Using the induction hypothesis the claim follows

We cite Theorem 253 in [10] without giving its proof

Theorem 24 Let 119860 = [119886 119887] R or R+and 119903 ge 0 If the

function 119891 isin 1198711

loc(119860) satisfies

int119860

119891120593(119903)

119889119909 = 0 (80)

for all 120593 isin Cinfin

0(119860) then there is a polynomial 119875 of degree lt 119903

such that 119891 = 119875 almost everywhere on 119860

To show that 119866119896is in 119883 we need to verify that (if there

is danger of confusion with respect to which variable we takethe norm we use notations as119883(119909

119896))

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

lt infin (81)

under the assumption that 119866119896minus1

is in 119883 Actually bounding119866119896from above by its absolute value this means that we have

first to take the1198711-norm and then the norm in119883We howeverwould like to switch the norms

We adapt Proposition 21 in [16] to our setting

Proposition 25 Let 120588 be a Banach function norm satisfyingproperty (D) If 119891 ΩtimesΩ rarr R is a measurable function andone has 119891

119909and 119891

119910 functions onΩ such that 119891119909(119910) = 119891(119909 119910) =

119891119910

(119909) then

120588 (1003817100381710038171003817119891119909

10038171003817100381710038171) le

1003817100381710038171003817120588 (119891119910

)10038171003817100381710038171

(82)

The cases 119883 isin 1198711

119871infin

are easy to prove If 119883 = 119871119901 for

some 1 lt 119901 lt infin then the result can be proven by usingFubinirsquos theorem together with property (D) For the generalresults we refer to Scheprsquos proof

The next result builds uponTheorem 254 in [10]

Theorem 26 Let Ω sub R be a domain and let 119903 ge 1 Suppose119883 is a Banach function space satisfying (B) and (D) If 119891 islocally in 119883 and has a generalized 119903th derivative 119892 locally in119883 then there is 119865 agreeing almost everywhere with 119891 havingabsolutely continuous intermediate weak derivatives and theweak derivative of order 119903 agrees almost everywhere with 119892Furthermore the derivatives of 119865 are locally in 119883 If Ω hasfinite measure one obtains the result with ldquolocallyrdquo replaced byldquogloballyrdquo

We thank Pilar Silvestre for pointing out how to prove themembership of the intermediate derivatives in119883

Proof Choose 119909lowast

isin Ω and let 119866119896be defined as in Lemma 23

Assume 120593 isin Cinfin

0(Ω) Then using the fact that 119892 is a weak

derivative of 119891 of order 119903 and (b) in Lemma 23

intΩ

(119891 minus 119866119903) 120593

(119903)

119889119909 = intΩ

119891120593(119903)

119889119909 minus intΩ

119866119903120593(119903)

119889119909

= (minus1)119903

intΩ

119892120593119889119909 minus (minus1)119903

intΩ

119892120593119889119909 = 0

(83)

UsingTheorem 24 we see that119891minus119866119903= 119875 almost everywhere

where119875 is a polynomial of degreelt 119903Thus119891 = 119875+119866119903almost

everywhere and for 120593 isin Cinfin

0(Ω) and 0 le 119898 le 119903

intΩ

119891120593(119898)

119889119909 = intΩ

(119875 + 119866119903) 120593

(119898)

= (minus1)119898

intΩ

(119866119903minus119898

+ 119875(119898)

) 120593 119889119909

(84)

Thus119866119903minus119898

+119875(119898) is the119898th generalized derivative of119891 As it is

locally absolutely continuous (if119898 lt 119903) it lies locally in119883 Inthe case 119898 = 119903 the same conclusion holds by the assumptionon 119892

Moreover there exists 119865 for example 119865 = 119875 + 119866119903 that

agrees with 119891 almost everywhere such that its generalizedderivatives up to order 119903 minus 1 are locally absolutely continuousand the 119903th generalized derivative equals 119892 almost every-where

As 119892 is locally in 119883 and the other derivatives are locallyabsolutely continuous they are locally in 119871

infin and thus locallyin119883 by (B)

Suppose now that Ω has additionally finite measureAssume that 119891 and 119892 are in119883 globally We want to show thatthe intermediate derivatives are in 119883 as well

Note in our setting that by (P5) in Definition 18membership in 119883 implies membership in 119871

1 We proceedby induction Suppose that 119866

119896minus1isin 119883 We will apply

Proposition 25 We set denoting by conv119909lowast

119909119896 the closed

interval with endpoints 119909lowast and 119909119896

119891 (119909119896minus1

119909119896) = 119866

119896minus1(119909

119896minus1) 120594conv119909lowast 119909119896 (119909119896minus1

) (85)

Then1003817100381710038171003817119866119896

(119909119896)1003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119891(119909119896minus1

119909119896)119889119909

119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)10038171003817100381710038171198711(119909119896minus1)

10038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)

le10038171003817100381710038171003817

1003816100381610038161003816119866119896minus1(119909

119896minus1)1003816100381610038161003816

1003817100381710038171003817120594Ω1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)lt infin

(86)

Journal of Function Spaces 11

5 Banach Function Space Setting

Here our main focus is a more advanced readership thatlikely knows already Banach function spaces Therefore weadopt a much briefer style pointing merely to the literaturewhere the corresponding definitions and results can be foundWe start with giving generalized versions of the necessity andsufficiency of Theorem 2

We list the results and prove them simultaneously bysplitting the proofs into parts according to the functionspaces

Theorem 27 Assume Ω sub R is a domain 119883 is a normedspace and 119898 isin N If the maximal function 119872 119883 rarr 119883 isbounded then there is a constant 119862 = 119862(119883119898) such that foreach continuous 119891 isin 119882

119898

119883there is 119886

119891isin 119883with 119886

119891 le 119862119891

119882119898119883

and

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (87)

In particular such 119886119891exists if

(i) 1 lt 119901 le infin and 119883 is the Lebesgue space 119871119901

(Ω)

(ii) 1 lt 119901 lt infin 1 le 119902 le infin and 119883 is the Lorentz space119871119901119902

(Ω)

(iii) 1 lt 119901 lt infin 1 le 119902 le infin 119887 is slowly varying and 119883 isthe Lorentz-Karamata space 119871

119901119902119887(Ω)

(iv) 120601 is an 119873-function with inf119904gt0

(119904120593(119904)120601(119904)) gt 1 where120593 is the density function of 120601 and 119883 is the Orlicz space119871lowast

120601(Ω)

(v) Ω = R and 119883 is rearrangement-invariant with upperindex 120572

119883lt 1

Theorem 28 Let Ω sub R be open Suppose 119883 is a Banachfunction space such that (AC) holds Assume further that thefunction 119891 Ω rarr R is in 119883 If for some 119903 isin N

0 the following

inequality holds

1003816100381610038161003816Δ119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)] (88)

for some 119886119891isin 119883 then there exists119892 isin 119883with 119892

119883le 2119903

119903

119886119891119883

such that

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909 = (minus1)119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (89)

for all 120593 isin Cinfin

0(Ω)

In particular the result holds if119883 is

(i) the Lebesgue space 119871119901

(Ω) for some 1 lt 119901 lt infin

(ii) the Lorentz space 119871119901119902

(Ω) where 1 lt 119901 lt infin and 1 le

119902 lt infin

(iii) the Lorentz-Karamata space 119871119901119902119887

(Ω) where 119887 isslowly varying 1 lt 119901 lt infin and 1 lt 119902 le infin

(iv) the Orlicz space 119871lowast

120601(Ω) for some 119873-function 120601 satisfy-

ing

inf119904gt0

119904120593 (119904)

120601 (119904)gt 1 (90)

where 120593 is the density function of 120601

To help with orientation let us give an overview ofthe remaining material in this section With respect toTheorem 27 note that we carried out the proof of Corollary 14in a high level of abstraction Hence we are left to verifythat the spaces under question are normed spaces (we willshow that they are even Banach function spaces) and theboundedness of the maximal operator

Concerning the proof Theorem 28 we proved the cor-responding part of Theorem 2 under the assumptions thatthe spaces under question are Banach function spaces andfurther satisfy the requirements listed in Properties 19 Theexistence of 1198831015840 is classical (Proposition 29) as is the fact that(AC) implies (A) and (NC) (Proposition 30) Hence we onlyneed to point to the literature where (AC) was verified

Proposition 29 Given a Banach function space there is aspace 119883

1015840 such that properties (B) (HI) (C) and (D) aresatisfied

Proof We refer the reader to Sections 11 and 12 in [13] or toSection 31 in [14]

The definition of absolutely continuous norm can befound as Definition 3111 in [14] and Definition 131 in [13]

Proposition 30 A Banach function space that has absolutelycontinuous norm satisfies (A) and (NC)

Proof The first statement follows fromTheorem 1313 in [13]and the second fromProposition 136 in the same source

51 General Statements

Proof of Theorem 27 (general statement) We basically havealready proven the general statement in Corollary 14

Proof of Theorem 28 (general statement) We gave the proofof the Lebesgue space case in such generality that thegeneral statement follows from the proof of Theorem 2 andPropositions 29 and 30

52 Lorentz-Karamata Spaces Lorentz-Karamata spaces aredetailed in [14] mainly Section 343

Proof of Theorem 27 (ii) and (iii) The part that is left toprove is the boundedness of the maximal operator which isguaranteed by Remark 3517 in [14]

Proof of Theorem 28 (ii) and (iii) By Theorem 3441 in [14]we see that 119883 is a rearrangement-invariant Banach space Acombination of Lemma 3439 in [14] together with Lemma3443 in [14] gives the absolute continuity (AC) of the normof1198831015840

12 Journal of Function Spaces

53 Orlicz Spaces The setting is the one of [17]

Lemma 31 If 120601 is an 119873-function satisfying the Δ2-condition

then the Orlicz space 119871lowast

120601has absolutely continuous norm

Proof Let 119891 isin 119871lowast

120601(Ω) We verify first that

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 lt infin (91)

Let 120576 gt 0 By the Δ2-condition there exists a constant 120581 =

120581(119891(120601)

+ 120576) such that

120601 ((1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576) 119904) le 120581120601 (119904) (92)

for all 119904 ge 0 Thus

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 = intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576(1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576))119889119909

le 120581intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576)119889119909 le 120581

(93)

Suppose that 119864 is a measurable set Then

10038171003817100381710038171198911205941198641003817100381710038171003817(120601)

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891120594119864

1003816100381610038161003816) 119889119909 le 1

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 1

(94)

Let us fix 120582 gt 0 Again using the Δ2-condition we have

120601(120582minus1

119904) le 120581(120582minus1

)120601(119904) Hence

intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 (95)

Example 133 in [13] tells us that 1198711 has absolutely continuousnorm Thus if 120594

119864119899converges 120583-almost everywhere to 0 then

intΩ

120601(|119891|)120594119864119899

converges to zero as well Thus we have for 119899

large enough that

120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119899119889119909 le 1 (96)

implying that 119891120594119864119899(120601)

le 120582 Since 120582 gt 0 was arbitrary theabsolute continuity of 119891 follows

Proof of Theorem 27 (iv) We conclude from Gallardorsquos resultTheorem 21 in [17] that it is sufficient that the complemen-tary 119873-function satisfies the Δ

2-condition in [0infin) Again

resorting to Gallardo this time to Proposition 14 in [17] weobtain the conclusion

Proof of Theorem 28 (iv) That 119883 is a Banach functionspace follows from Theorem 3416 in [14] To prove that119883

1015840 has absolutely continuous norm we note that inequal-ity (90) together with Proposition 14 in [17] ensuresthat the complementary 119873-function of 120601 satisfies the Δ

2-

condition Lemma 31 verifies that 1198831015840 has absolutely contin-uous norm

54 Rearrangement-Invariant Spaces The necessary defini-tions can be found in [13]

Proof of Theorem 27 (v) We are only left to show theboundedness of the maximal operator However this is thecontent of a result by Lorentz and Shimogaki for examplestated as Theorem 3517 in [13]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank Pilar Silvestre for hersuggestions of how to prove the existence of intermediatederivatives The research is supported by the Academy ofFinland The third author thanks the Department of Math-ematics and Systems Analysis at Aalto University for itshospitality

References

[1] B Bojarski ldquoRemarks on some geometric properties of Sobolevmappingsrdquo in Functional Analysis amp Related Topics (Sapporo1990) pp 65ndash76 World Scientific River Edge NJ USA 1991

[2] B Bojarski PHajlasz andP Strzelecki ldquoImproved119862119896120582 approxi-

mation of higher order Sobolev functions in normand capacityrdquoIndiana University Mathematics Journal vol 51 no 3 pp 507ndash540 2002

[3] B Bojarski ldquoPointwise characterization of Sobolev classesrdquoProceedings of the Steklov Institute of Mathematics vol 255 no1 pp 65ndash81 2006

[4] B Bojarski ldquoWhitneyrsquos jets for Sobolev functionsrdquo UkrainianMathematical Journal vol 59 no 3 pp 345ndash358 2007

[5] B Bojarski ldquoTaylor expansion and Sobolev spacesrdquo Bulletin ofthe Georgian National Academy of Sciences New Series vol 5no 2 pp 5ndash10 2011

[6] B Bojarski ldquoSobolev spaces and Lagrange interpolationrdquoGeorgian Academy of Sciences Proceedings of A RazmadzeMathematical Institute vol 158 pp 1ndash12 2012

[7] P Hajłasz ldquoSobolev spaces on an arbitrary metric spacerdquoPotential Analysis vol 5 no 4 pp 403ndash415 1996

[8] B Bojarski ldquoSobolev spaces and averaging Irdquo httparxivorgabs13085171

[9] R Kannan and C K Krueger Advanced Analysis on the RealLine Universitext Springer New York NY USA 1996

[10] R A DeVore and G G Lorentz Constructive Approximationvol 303 of Fundamental Principles of Mathematical SciencesSpringer Berlin Germany 1993

[11] L L Schumaker Spline Functions Basic Theory CambridgeMathematical Library CambridgeUniversity Press CambridgeUK 3rd edition 2007

[12] G H Hardy and J E Littlewood ldquoA maximal theorem withfunction-theoretic applicationsrdquo Acta Mathematica vol 54 no1 pp 81ndash116 1930

[13] C Bennett andR Sharpley Interpolation ofOperators vol 129 ofPure and Applied Mathematics Academic Press Boston MassUSA 1988

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

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Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 3

Assume now that the claim is true for some 119899 We assumehere that 119910

119899+1= 119910

119899and leave the case when 119910

119899and 119910

119899+1

coincide to the reader (computation (14) might be helpful)Then evaluating the right most integral

int

1

0

1198891199051int

1199051

0

1198891199052

sdot sdot sdot int

119905119899

0

119891(119899+1)

(1199100+

119899

sum

119896=1

(119910119896minus 119910

119896minus1) 119905

119896

+ (119910119899+1

minus 119910119899) 119905

119899+1)119889119905

119899+1

= int

1

0

1198891199051int

1199051

0

1198891199052

sdot sdot sdot int

119905119899minus1

0

(119891(119899)

(1199100+

119899minus1

sum

119896=1

(119910119896minus 119910

119896minus1) 119905

119896

+ (119910119899+1

minus 119910119899minus1

) 119905119899)

minus119891(119899)

(1199100+

119899

sum

119896=1

(119910119896minus 119910

119896minus1) 119905

119896))119889119905

119899+1

sdot1

119910119899+1

minus 119910119899

=1

119910119899+1

minus 119910119899

([1199100 119910

119899minus1 119910

119899+1] 119891 minus [119910

0 119910

119899] 119891)

= [1199100 119910

119899+1] 119891

(10)

We use the Hermite-Genocchi formula to compute finitedifferences of polynomials

Lemma 6 Let 119886 be a point in R and 119898 isin N0 We set

119901119898119886

(119910) = (119910 minus 119886)119898

(11)

Then

[1199100 119910

119899] 119901

119898119886=

0 0 le 119898 lt 119899

1 119898 = 119899(12)

Proof As 119901119898119886

is smooth we can use the Hermite-Genocchiformula Lemma 5 If 0 le 119898 lt 119899 then 119901

(119899)

119898119886= 0 hence the

Hermite-Genocchi formula gives the claim If 119898 = 119899 then119901(119898)

119898119886= 119898 we compute the integral in the Hermite-Genocchi

formula via induction over119898We let first for fixed 119905

0ge 0

⟨1199100 119910

119898⟩1199050

= int

1199050

0

1198891199051int

1199051

0

1198891199052sdot sdot sdot int

119905119898minus1

0

1 119889119905119898 (13)

We claim that

⟨1199100 119910

119898⟩1199050

=119905119898

0

119898 119898 isin N (14)

For 119898 = 1 this can easily be seen Assume that the claimholds for some119898 Then

⟨1199100 119910

119898+1⟩1199050

= int

1199050

0

1198891199051⟨119910

0 119910

119898⟩1199051

= int

1199050

0

119905119898

1

119898119889119905

1=

119905119898+1

0

(119898 + 1)

(15)

The Hermite-Genocchi formula tells us [1199100 119910

119898]119901119898119886

=

⟨1199100 119910

119898⟩1sdot 119898 giving the proof

The following notation is on page 15 in [11]

Definition 7 Set

(119909 minus 119910)0

+=

1 119909 ge 119910

0 119909 lt 119910(16)

and for119898 gt 1

(119909 minus 119910)119898minus1

+=

(119909 minus 119910)119898minus1

119909 ge 119910

0 119909 lt 119910(17)

The definition below is a variant of Definition 412 in [11]However as our starting point is different than Schumakerrsquosthere is a slight regularity issue when 119909 does not differ fromat least two 119910

119894 However we are only interested in cases where

119876119898

119894lives under an integral andwe agree here to set119876119898

119894(119909) = 0

in points where we do not have enough regularity

Definition 8 (B-spline) Let sdot sdot sdot le 119910minus1

le 1199100le 119910

1le 119910

2le sdot sdot sdot

be a sequence of real numbers Given integers 119894 and 119898 gt 0one defines (note that 119909 is fixed so that the finite difference iswith respect to 119910)

119876119898

119894(119909) =

(minus1)119898

[119910119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+ if 119910

119894lt 119910

119894+119898

0 otherwise(18)

We call 119876119898

119894the 119898th order B-spline associated with the knots

119910119894 119910

119894+119898

This is Theorem 22 in [11] Schumaker does not give theproof but writes that it is similar to the one ofTheorem 21 inhis book

Theorem 9 (Dual Taylor Expansion) Let 119891 isin 119871119898

1[119886 119887] Then

for all 119886 le 119910 le 119887

119891 (119910) =

119898minus1

sum

119895=0

(minus1)119895

119863119895

119891 (119887) (119887 minus 119910)119895

119895

+ int

119887

119886

(119909 minus 119910)119898minus1

+(minus1)

119898

119863119898

119891 (119909)

(119898 minus 1)119889119909

(19)

4 Journal of Function Spaces

Proof The proof uses induction and integration by partsNote that by replacing 119886 by 119910 in the integral we may deletethe + in (119909 minus 119910)

+ The case119898 = 1 reduces to the fundamental

theorem of calculus for absolutely continuous functions seefor example Theorem 7115 in [9] Let us assume that thestatement holds for some119898 and fix 119891 isin 119871

119898+1

1[119886 119887] Then

int

119887

119910

(119909 minus 119910)119898

(minus1)119898+1

119863119898+1

119891 (119909)

119898119889119909

=(119887 minus 119910)

119898

(minus1)119898+1

119863119898

119891 (119887)

119898

minus int

119887

119910

(119909 minus 119910)119898minus1

(minus1)119898+1

119863119898

119891 (119909)

(119898 minus 1)119889119909

= minus(119887 minus 119910)

119898

(minus1)119898

119863119898

119891 (119887)

119898

minus

119898minus1

sum

119895=0

(minus1)119895

119863119895

119891 (119887) (119887 minus 119910)119895

119895+ 119891 (119910)

(20)

We now cite a part ofTheorem 423 in [11] Note that119898 isa positive integer and 119863

119895

+means that we take the right hand

derivative of order 119895 We only need and prove the equality for119895 = 0 We also assume that not all points coincide

Theorem 10 (Peano representation) Fix 0 le 119895 le 119898minus119899 where119899 is the maximum multiplicity of 119910

119894 119910

119894+119898 Then

[119910119894 119910

119894+119898] 119891 = int

119910119894+119898

119910119894

(minus1)119895

119863119895

+119876

119898

119894(119909)119863

119898minus119895

119891 (119909)

(119898 minus 1)119889119909 (21)

for all 119891 isin 119871119898minus119895

1[min

119894119910

119894max

119894119910

119894]

Proof of the Peano Representation for 119895 = 0 We take thedual Taylor expansion from Theorem 9 Then we apply[119910

119894 119910

119894+119898] on both sides and remember that we computed

the divided differences of polynomials in Lemma 6

Our goal is to use the Peano representation to obtainan upper bound for the divided differences We start withestablishing an upper bound of the factor 119863

+119876

119898

119894in the

integrand

Lemma 11 Suppose that 119910119894le sdot sdot sdot le 119910

119894+119898and 119910

119894lt 119910

119894+119898 Then

there is a constant 119862 such that

1003816100381610038161003816119863+119876

119898

119894(119909)

1003816100381610038161003816 =10038161003816100381610038161003816[119910

119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+

10038161003816100381610038161003816le

119862

119910119894+119898

minus 119910119894

(22)

Proof We want to estimate

[119910119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+

=[119910

119894+1 119910

119894+119898] (119909 minus 119910)

119898minus1

+minus [119910

119894 119910

119894+119898minus1] (119909 minus 119910)

119898minus1

+

119910119894+119898

minus 119910119894

(23)

Assume 119896 lt 119898 minus 1 We claim that

119889119896

119889119910119896

(119909 minus 119910)119898minus1

+= (minus1)

119896

(119898 minus 1) sdot sdot sdot (119898 minus 119896) (119909 minus 119910)119898minus1minus119896

+

(24)

This claim is easily seen to be true for 119910 = 119909 That it is truealso for 119910 = 119909 follows by induction We also note that

119889119898minus2

119889119910119898minus2(119909 minus 119910)

119898minus1

+(25)

is absolutely continuous on bounded intervals and that

(minus1)(119898minus1)

(119898 minus 1) (119909 minus 119910)0

+

(26)

is a weak derivative that equals the derivative whenever119910 = 119909 We first assume that not all 119910

119895 119910

119895+119898minus1coincide

An application of the Hermite-Genocchi formula Lemma 5gives

[119910119895 119910

119895+119898minus1] (119909 minus 119910)

119898minus1

+

= int

1

0

1198891199051int

1199051

0

1198891199052

sdot sdot sdot int

119905119898minus2

0

(minus1)(119898minus1)

(119898 minus 1)

sdot (119909 minus (119910119895+

119898minus1

sum

119896=1

(119910119895+119896

minus 119910119895+119896minus1

) 119905119896))

0

+

119889119905119898minus1

(27)

We use now the fact that the integrand is bounded from aboveby (119898 minus 1) and the computation (14) to obtain

10038161003816100381610038161003816[119910

119895 119910

119895+119898minus1] (119909 minus 119910)

119898minus1

+

10038161003816100381610038161003816le 1 (28)

Suppose now that 119910119895

= sdot sdot sdot = 119910119895+119898minus1

= 119909 Then|[119909

119895 119909

119895+119898minus1](119909 minus 119910)

119898minus1

+| is bounded from above by 1 if

119909 gt 119910119894and vanishes if 119910

119894lt 119909 In the end we find a constant

119862 such that10038161003816100381610038161003816119863

+[119910

119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+

10038161003816100381610038161003816le

119862

119910119894+119898

minus 119910119894

(29)

In the next result we use the notation

119872119892(119909) = sup119911

minusint

119911

119909

10038161003816100381610038161198921003816100381610038161003816 (30)

for the famous Hardy-Littlewood maximal function of 119892

introduced in the seminal paper [12] of Hardy-Littlewood inActa Math (1930)

Theorem 12 Suppose 1199100

le sdot sdot sdot le 119910119898 119910

0lt 119910

119898and 119891 is in

119871119898

1[119910

0 119910

119898] for some 119898 isin N Then there is some constant 119862

such that1003816100381610038161003816[1199100 119910119898] 119891

1003816100381610038161003816 le 119862 (119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (1199100) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910119898))

(31)

Journal of Function Spaces 5

Proof We use the Peano representation Theorem 10 with119894 = 119895 = 0 Then using the bound of |119863

+119876

119898

119894(119909)| found in

Lemma 11

1003816100381610038161003816[1199100 119910119898] 1198911003816100381610038161003816 le

1

(119898 minus 1)int

119910119898

1199100

1003816100381610038161003816119863+119876

119898

119894(119909)

1003816100381610038161003816

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

le119862

119910119898

minus 1199100

int

119910119898

1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

(32)

For 119910 isin [1199100 119910

119898] we have

1003816100381610038161003816[1199100 119910119898] 1198911003816100381610038161003816

le119862

119910119898

minus 1199100

(int

119910

1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909 + int

119910119898

119910

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909)

(33)

and further1003816100381610038161003816[1199100 119910119898] 119891

1003816100381610038161003816

le 119862(minusint

119910

1199100

119910 minus 1199100

119910119898

minus 1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

+minusint

119910119898

119910

119910119898

minus 119910

119910119898

minus 1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909)

(34)

The first factors in both integrals are bounded from above by1 We see that the integrals are bounded from above by thecorresponding maximal operators This gives the claim

The finite differences defined in (2) are up to a factordivided differences for equispaced nodes

Lemma 13 Let 119909 119910 be two points in a domain Ω sub R Let119898 isin N and 119909

119896= 119909

0+ ℎ119896 where ℎ = (119910 minus 119909)119898 Then each

119891 Ω rarr R satisfies

[1199090 119909

119898] 119891 =

1

ℎ119898119898Δ119898

119891 (119909 119910) (119909 = 1199090 119909

119898= 119910)

(35)

Proof We use induction For119898 = 1 we have

[1199090 119909

1] 119891 =

119891 (1199091) minus 119891 (119909

0)

1199091minus 119909

0

(36)

As ℎ = 1199091minus 119909

0 119909 = 119909

0 119910 = 119909

1 and 119898 = 1 we obtain the

claim Suppose that the claim is true for some119898 isin N Then

Δ119898+1

ℎ119891 (119909)

=

119898+1

sum

119895=0

(minus1)119898+1minus119895

(119898 + 1

119895)119891 (119909 + 119895ℎ)

=

119898+1

sum

119895=0

(minus1)119898+1minus119895

((119898

119895 minus 1) + (

119898

119895))119891 (119909 + 119895ℎ)

=

119898+1

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + (119895 + 1) ℎ)

minus

119898+1

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + 119895ℎ)

=

119898

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909

1+ 119895ℎ)

minus

119898

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + 119895ℎ)

= ℎ119898

119898 ([1199091 119909

119898+1] 119891 minus [119909

0 119909

119898] 119891)

= ℎ119898

119898 (ℎ (119898 + 1)) [1199090 119909

119898+1] 119891

(37)

We can now formulate Theorem 12 in terms of finitedifferences

Corollary 14 (Corollary of Theorem 12) Let Ω sub R be adomain Suppose 119891 isin 119871

119898

1(Ω) for some 119898 isin N then there is

a constant 119862 = 119862(119898) such that

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816

le 1198621003816100381610038161003816119909 minus 119910

1003816100381610038161003816

119898

(119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

119909 119910 isin Ω

(38)

In particular we have the following special case Suppose 119883

and 119884 are normed spaces of functions such that the maximaloperator 119872 119883 rarr 119884 is bounded If 119891 isin 119882

119898

119883(Ω) for some

119898 ge 1 then

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (39)

for some function 119886119891isin 119884 depending on 119891

Proof We first use Lemma 13 and thenTheorem 12 with stepℎ = (119910 minus 119909)119898 There is some constant 119862 = 119862(119898) such that

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816

= ℎ119898

1198981003816100381610038161003816[1199090

119909119898] 119891

1003816100381610038161003816

le 119862ℎ119898

(119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

le1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

119898

(119862119872(1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119862119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

(40)

Proof of Theorem 2 (Sobolev functions satisfy pointwiseinequality) We combine Corollary 14 with the boundednessof the maximal operator which is by now classical and can befound for example in Theorem 3310 in [13]

6 Journal of Function Spaces

3 Pointwise Inequality Implies Membershipin Sobolev Space

In this section our aim is to show that the pointwiseinequality (3) implies the existence of the highest orderderivative in the correct space We will take care of theintermediate derivatives in the next section In this sectionthe properties of the Lebesgue spaces play an important roleOur point of view is however to extract the properties ofthe Lebesgue spaces and base our proofs solely on theseproperties We hope that the disadvantage of being slightlymore abstract is later on compensatedwhenwe generalize theresults We start with some auxiliary results

Lemma 15 Let Ω sub R be a domain and 120593 isin C119899

0(Ω) that

is 120593 has compact support in Ω and is 119899-times continuouslydifferentiable Then

1003816100381610038161003816100381610038161003816100381610038161003816

[1199090 119909

119899] 120593 minus

120593(119899)

(119910)

119899

1003816100381610038161003816100381610038161003816100381610038161003816

997888rarr 0 (41)

uniformly inΩ as 1199090 119909

119899rarr 119910

Proof By Taylorrsquos formula for 119909 isin conv1199090 119909

119899 there is

120585119909isin conv119909

0 119909

119899 such that

120593 (119909) = 120593 (119910) + 1205931015840

(119910) (119909 minus 119910) + sdot sdot sdot + 120593(119899minus1)

(119910)(119909 minus 119910)

119899minus1

(119899 minus 1)

+1

119899120593(119899)

(120585119909) (119909 minus 119910)

119899

(42)

We apply [1199090 119909

119899] on both sides (with variable 119909) and use

the computation of the divided differences of polynomialscarried out in Lemma 6 to obtain

[1199090 119909

119899] 120593 (119909) =

1

119899120593(119899)

(120585119909) (43)

As 120593(119899) is uniformly continuous on compact intervals the

claim follows

Lemma 16 Assume Ω sub R is open and 119870 sub Ω is compactThen there is a neighborhood of119870 that is a subset of Ω

Proof For each 119909 isin 119870 we find a ball 119861(119909 119903119909) sub Ω Suppose

the statement of the lemma is falseThen for every 119899 isin N thereis 119911

119899isin Ω 119870 with dist(119911

119899 119870) lt 1119899 Thus for each 119899 isin N

there is 119910119899

isin 119870 with 119889(119911119899 119910

119899) lt 1119899 Since 119870 is compact

there is a subsequence (119910119899119896)119896of (119910

119899)119899that converges to some

119910 isin 119870 Now 119910 lies in one of the balls 119861(119909 119903119909) sub Ω For 119896 large

enough we have that 119910119899119896and 119911

119899119896are in 119861(119909 119903

119909) sub Ω This is

a contradiction

Lemma 17 SupposeΩ sub R is open and119870 sub Ω compactThenthere exists 120576 gt 0 such that

(Ω minus ℎ) cap 119870 = Ω cap 119870 (44)

for |ℎ| lt 120576

Proof We first show that any member of the left hand side iscontained in the right hand side this is true for any ℎ isin R If119911 is a member of the left hand side then it is in 119870 and since119870 is a subset of Ω it is also a member of the right hand sideTo prove the other inclusion note that by Lemma 16 there is120576 gt 0 such that the 120576-neighborhood 119870

120576of 119870 lies in Ω Now

suppose that 119911 is in the right hand side and |ℎ| lt 120576 Then for119909 = 119911+ℎ we have119909 isin 119870

120576sub Ω implying that 119911 = 119909minusℎ isin Ωminusℎ

That 119911 isin 119870 is clear This gives the proof

Next we state basically Definition 311 in [14] Readersfollowing [13] might want to have a look at Definitions 111and 113 in their book To gain familiarity with the conceptof Banach function norm the reader may want to verify thatLebesgue spaces are examples of Banach function spaces Aswe are here only interested in Lebesgue measure we adaptthe notion to this case The set M+ denotes the cone ofmeasurable functions whose values lie in [0infin]

Definition 18 (Banach function norm Banach functionspace) Assume that 119877 sub R A Banach function norm on 119877

is a map 120588 from M+

(119877) to [0infin] such that for all 119891 119892 and119891119899(119899 isin N) in M+

(119877) all scalars 120582 ge 0 and all measurablesubsets 119864 of 119877 the following are true where 120583 is the Lebesguemeasure

(P1) 120588(119891) = 0 if and only if 119891 = 0 120583-almost everywhere120588(120582119891) = 120582120588(119891) 120588(119891 + 119892) le 120588(119891) + 120588(119892)

(P2) if 0 le 119892 le 119891 120583-almost everywhere then 120588(119892) le 120588(119891)(P3) if 0 le 119891

119899uarr 119891 120583-almost everywhere we have 120588(119891

119899) uarr

120588(119891)(P4) if 120583(119864) lt infin then 120588(120594

119864) lt infin

(P5) if120583(119864) lt infin then there is a constant119862 = 119862(119864 120588) lt infin

such that int119864

119891119889119909 le 119862120588(119891)

Given such a function norm 120588 the set 119883 = 119883(120588) ofall extended measurable functions 119891 (identifying functionsequal 120583-almost everywhere) such that 120588(|119891|) lt infin is called aBanach function space (BFS for short) and we define

10038171003817100381710038171198911003817100381710038171003817119883

= 120588 (1003816100381610038161003816119891

1003816100381610038161003816) 119891 isin 119883 (45)

We leave it to the reader to verify that Lebesgue spacessatisfy the following properties (1198831015840 can be chosen as the dualspace)

Properties 19 Given a Banach function space 119883 we requirethe existence of a Banach function space119883

1015840 such that

(B) 119883 contains each essentially bounded function whosesupport has finite measure

(HI) Holderrsquos inequality int |119891119892|119889119909 le 119891119883119892

1198831015840

(C) 1198831015840 is complete

(D) 119892119883

= supintΩ

|119892|ℎ ℎ isin M+

ℎ1198831015840 le 1

(AC) absolutely continuous norm if (120594119864119899)119899is a sequence of

characteristic functions converging pointwise almosteverywhere to 0 then 119891120594

1198641198991198831015840 converges to zero for

all 119891 isin 1198831015840

Journal of Function Spaces 7

(A) each function 119891 isin 1198831015840 can be approximated by simple

functions (in1198831015840-norm)

(NC) if 119891 isin 1198831015840 and ℎ

119899are step functions with 0 le ℎ

119899uarr |119891|

then1003817100381710038171003817119891 minus ℎ

119899sgn (119891)

10038171003817100381710038171198831015840997888rarr 0 as 119899 997888rarr infin (46)

Lemma 20 Suppose Ω sub R is open and 119883 is a Banachfunction space fulfilling the items listed in Properties 19 Assumethat 119891 Ω rarr R is in 119883 We let 120582

119891 119883

1015840

cap C119903+1

0(Ω) rarr R be

defined as

120582119891(120593) = int

Ω

119891120593(119903)

119889119909 (47)

If1003816100381610038161003816Δ

119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)]

for almost every 119909 119910 isin Ω

(48)

for some 119886119891

isin 119883 then 120582119891has a continuous extension to 119883

1015840

with norm bounded from above by 2119903119903

119886119891119883

Proof Our goal is to show1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816le 2119903

11990310038171003817100381710038171003817119886119891

10038171003817100381710038171003817119883

100381710038171003817100381712059310038171003817100381710038171198831015840

(49)

because 120582119891extends then by the theorem of Hahn-Banach to

a bounded functional on1198831015840 with norm bounded from above

by 2119903119903

119886119891119883 Without loss of generality we may assume that

119886119891is nonnegative If 119891 were be smooth we could just use

integration by parts and use as bound the limiting case of thepointwise inequality (48) As 119891 is not necessarily smooth wehave to operate on the level of finite differences

Let us choose 120593 isin C0(Ω) and denote its support by 119870

Lemma 17 ensures the existence of some positive 120576 gt 0 suchthat (Ωminus119895ℎ)cap119870 = Ωcap119870 for 0 le 119895 le 119903 as long as |ℎ| lt 120576119903Weassume in the following that ℎ has been chosen accordingly

The following integrals exist by Holderrsquos inequalitybecause 120593 and thus Δ

119903

minusℎ120593(119909) are bounded with compact

support and 119891 is integrable Consider

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

119891 (119909) (minusℎ)minus119903

Δ119903

minusℎ120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(50)

We compute the following integral

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

= intΩ

119891 (119909)

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119909) 120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωminus119895ℎ

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

(Ωminus119895ℎ)cap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

(51)

We have chosen 120576 gt 0 such that (Ω minus 119895ℎ) cap 119870 = Ω cap 119870 Wecontinue

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωcap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

(52)

With the aid of (52) we can now rewrite (50) as

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

(minusℎ)minus119903

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(53)

Inequality (48) leads to

1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816

le 119903119903

intΩ

(119886119891(119909) + 119886

119891(119909 + 119903ℎ))

1003816100381610038161003816120593 (119909)1003816100381610038161003816 119889119909

+ intΩ

1003816100381610038161003816119891 (119909)1003816100381610038161003816

10038161003816100381610038161003816120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)

10038161003816100381610038161003816119889119909

(54)

Using Holderrsquos inequality (HI) for the first summand weare left to show that the second summand tends to 0 asℎ approaches 0 We note first that since the second factorhas support in a set of finite measures we can replace theintegration domain by a set of finite measuresThen we applya (1198711

119871infin

)-Holder inequality and the uniform convergence ofthe divided difference to 120593

(119903)

(119909)119903 established in Lemmas 13and 15

After having verified that 120582119891has a continuous extension

to 1198831015840 we want to show that this extension has a representa-

tion as integral

8 Journal of Function Spaces

Lemma 21 Assume Ω sub R is open Suppose the propertieslisted in Properties 19 are met Let 120582

119891be defined as in (47) and

bounded on 1198831015840 Then there exists 119892 isin 119883 with 119892

119883le 120582

119891

such that

120582119891(120593) = (minus1)

119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (55)

for all 120593 isin 1198831015840 especially for all 120593 isin Cinfin

0(Ω)

Proof The proof is heavily influenced by the proof of Theo-rem 141 in [13]We first want to find a signedmeasure ] suchthat

120582119891(120593) = int

Ω

120593119889] (56)

for each 120593 isin 1198831015840 Then we want to apply the Lebesgue-

Radon-Nikodym theorem to replace 119889] by (minus1)119903

119892(119909)119889119909 forsome integrable function 119892 for which we then show that it iscontained in119883

We start by assuming that the measure of Ω is finite Foreach 120583-measurable set 119860 we set

] (119860) = 120582119891(120594

119860) (57)

We start by proving the 120590-additivity of ] (for the 120583-measurable sets) In the first step we just show the additivityBy induction it is enough to consider only two sets

Let 1198601and 119860

2be two disjoint 120583-measurable sets Then

] (1198601cup 119860

2) = 120582

119891(120594

1198601cup1198602) = 120582

119891(120594

1198601+ 120594

1198602)

= 120582119891(120594

1198601) + 120582

119891(120594

1198602) = ] (119860

1) + ] (119860

2)

(58)

We move to the verification of the 120590-additivity Let119860

1 119860

2 be countablymany pairwise disjoint120583-measurable

sets We first want to show thatsum119894120594119860119894

converges in1198831015840

We let 119864119896= cup

119896

119894=1119860

119894and 119864 = cup

infin

119894=1119860

119894 Then since (119864

119896)119896is

increasing

120583 (119864) = 120583(

infin

⋃

119896=1

119864119896) = lim

119896rarrinfin

120583 (119864119896) (59)

As 119864 has finite measure given 120576 gt 0 there exists 1198960isin N such

that

(1 minus 120576) 120583 (119864) lt 120583 (1198641198960) (60)

By the disjointness of the sets 119860119894 we see that

120583 (119864) = 120583 (1198641198960) + 120583(

infin

⋃

119894=1198960+1

119860119894) (61)

It follows that

120583(

infin

⋃

119894=1198960+1

119860119894) = 120583 (119864) minus 120583 (119864

1198960) lt 120583 (119864) minus (1 minus 120576) 120583 (119864)

= 120576120583 (119864)

(62)

We have for119898 lt 119899

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=1

120594119860119894

minus

119898

sum

119894=1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

=

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=119898+1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

le10038171003817100381710038171003817120594cupinfin119894=119898+1

119860119894

100381710038171003817100381710038171198831015840 (63)

By the considerations before and the fact that sdot 1198831015840 is

absolutely continuous as stipulated in (AC) we see thatsum

119899

119894=1120594119860119894

is a Cauchy sequence Now as 1198831015840 is complete by

(C) the Cauchy sequence converges in 1198831015840 We have by the

continuity of 120582119891

] (cup119894119860

119894) = 120582

119891(120594

cup119894119860119894) = 120582

119891(

infin

sum

119894=1

120594119860119894)

= lim119899rarrinfin

120582119891(

119899

sum

119894=1

120594119860119894) = lim

119899rarrinfin

119899

sum

119894=1

120582119891(120594

119860119894)

= lim119899rarrinfin

119899

sum

119894=1

] (119860119894) =

infin

sum

119894=1

] (119860119894)

(64)

Thus ] is 120590-additiveIn order to apply the theorem of Lebesgue-Radon-

Nikodym see for example Section 69 in [15] and conclud-ing the existence of 119892 isin 119871

1

(Ω) such that

] (119860) = int119860

119892119889119909 119860 120583-measurable (65)

we need to show that ] is absolutely continuous with respectto 120583 To do so assume that 120583(119873) = 0 Then ](119873) = 120582

119891(120594

119873)

By the absolute continuity of the normof1198831015840 required in (AC)we see that 120594

1198731198831015840 = 0 and therefore ](119873) = 0 Replacing 119892

by minus119892 if necessary we obtain

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909 (66)

Now we assume that Ω does not necessarily have finitemeasureWewriteΩ = cup

119899Ω

119899as union of open sets with finite

measure such that Ω119899sub Ω

119899+1 By the considerations above

we find for each 119899 isin N a 119892119899such that

120582119891(120594

119860) = (minus1)

119903

intΩ

119892119899(119909) 120594

119860(119909) 119889119909 (67)

for each measurable set 119860 sub Ω119899 It is easily seen that 119892

119899

and 119892119899+1

agree almost everywhere on Ω119899 This gives rise to

a function 119892 satisfying

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909

119860 sub

119873

⋃

119899=1

Ω119899for some 119873

(68)

By linearity we have for each simple function 119904

120582119891(119904) = (minus1)

119903

intΩ

119892 (119909) 119904 (119909) 119889119909 (69)

To show that 119892 isin 119883 we want to use property (D)

Journal of Function Spaces 9

Let ℎ be a nonnegative simple function with supportin some Ω

119899 Then ℎ sgn(119892) is also a simple function with

support inΩ119899 It is a finite linear combination of characteristic

functions of sets in Ω119899 Hence

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = intℎ sdot sgn (119892) 119892 119889119909 = 120582119891(ℎ sgn (119892)) (70)

Using the boundedness of 120582119891 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (71)

If ℎ is an arbitrary function in1198831015840 thenwemay construct a

sequence (ℎ119899)infin

119899=1of simple functions each ℎ

119899having support

in Ω119899 such that 0 le ℎ

119899uarr |ℎ| 120583-almost everywhere By the

monotone convergence theorem we have by (71) that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = lim119899rarrinfin

int1003816100381610038161003816ℎ119899119892

1003816100381610038161003816 119889119909 le lim sup119899rarrinfin

10038171003817100381710038171003817120582119891

10038171003817100381710038171003817

1003817100381710038171003817ℎ11989910038171003817100381710038171198831015840

(72)

By (P3) in Definition 18 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (73)

and Property (D) gives that 119892 isin 119883 In the last step we want toupgrade the integral representation (69) to hold for arbitraryfunctions in 120593 isin 119883

1015840 As before we approximate 120593 by simplefunctions ℎ

119899with support contained in Ω

119899and 0 le ℎ

119899uarr |120593|

in 120583-almost everywhere point Define 119904119899

= ℎ119899sgn(120593) By

(NC) 119904119899converges to 120593 in119883

1015840 and as both 120582119891and ldquoits integral

representationrdquo are continuous and agree on the set of stepfunctions they agree on119883

1015840

Having shown the existence of the weak derivativesof highest order we are left to show the existence of theintermediate derivatives which we do in Proposition 22 inthe next section To keep the flow going we give here theproof that the pointwise inequality implies membership inthe corresponding Sobolev space under the assumption thatwe already know that Proposition 22 holds

Proof of Theorem 2 (pointwise inequality implies membershipin Sobolev space) We combine Lemmas 20 and 21 and applyProposition 22

4 Equivalent Definitions of Sobolev Spaces

The aim of this section is to prove the following equivalence

Proposition 22 Let Ω sub R be a domain with finite Lebesguemeasure inR Suppose119883 is a Banach function space satisfying(B) and (D) Assume that 119898 isin N Then the following setscoincide

(a) 119860 = 119891 isin 119883 there exists 119865 isin 119883 where 119865 agrees almosteverywherewith119891 and hasweak derivatives up to order119898 in119883 where the derivatives of order 119903 0 le 119903 le 119898minus1are absolutely continuous

(b) 119861 = 119891 isin 119883 119891(119896)

isin 119883 119896 isin 0 119898 where 119891(119896) arethe weak derivatives of 119891

(c) 119862 = 119891 isin 119883 119891(119898)

isin 119883 where 119891(119898) is the weak

derivative of 119891 of order 119898

We see that 119860 sub 119861 sub 119862 We now show that 119862 sub 119860Let us give an outline of the proof Given 119892 Ω rarr R

(we will choose 119892 = 119891(119898)) by integrating we construct in

Lemma 23 an absolutely continuous function 119866119898together

with its weak derivatives 1198660 119866

119898minus1 which are absolutely

continuous and 119866(119898)

119898= 119892 However 119866

119898and 119891 might

not agree almost everywhere We verify however in theproof of Theorem 26 (with the help of Theorem 24) thatwe find a polynomial 119875 such that 119866

119898+ 119875 agrees with 119865

almost everywhere and shares with 119866119898the other required

properties We conclude the proof of Theorem 26 by settlingthe membership questions with the aid of Proposition 25

Lemma 23 Assume Ω sub R is a domain We further assumethat 119892 Ω rarr R is in 119871

1

loc(Ω) One fixes 119909lowast

isin Ω and defines

1198660(119909

0) = 119892 (119909

0)

119866119896(119909

119896) = int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119896 isin N

(74)

where the integrals are Lebesgue integrals and we use thenotation

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

=

int[119909lowast119909119896]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909lowast

le 119909119896

minusint[119909119896 119909lowast]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909119896lt 119909

lowast

(75)

Then

(a) 119866119896is locally absolutely continuous for all 119896 isin N

especially locally integrable(b) if 120593 isin Cinfin

0(Ω) and 0 le 119903 le 119896 one has

intΩ

119866119896120593(119903)

= (minus1)119903

intΩ

119866119896minus119903

120593 (76)

Proof Let us prove (a) by induction Assume 119896 = 1 Note that1198661(119909

lowast

) = 0 Thus we have for 119909lowast

le 1199091

1198661(119909

1) minus 119866

1(119909

lowast

) = 1198661(119909

1) = int

1199091

119909lowast

119892 (1199090) 119889119909

0 (77)

As 119892 isin 1198711

([119909lowast

1199091]) we see that 119866

1is absolutely continuous

The case 1199091lt 119909

lowast is treated similarlyAssume now that the claim is true for some 119896 Now

119866119896+1

(119909119896+1

) is the integral of a continuous function and thuslocally absolutely continuous

Let us have a look at (b) Note that the statement is true for119896 = 0 thus we may assume that 119896 ge 1 We make an inductionover 119903 The case 119903 = 0 is clear Let us choose an interval[119886 119887] sub Ω containing a neighborhood of the support of 120593 As119866119896and 120593

(119903) are both absolutely continuous in [119886 119887] we have

10 Journal of Function Spaces

by the integration by part formula for absolutely continuousfunctions (see for exampleTheorem7147 in [9]) and the factthat 120593(119903)

(119886) = 120593(119903)

(119887) = 0

intΩ

119866119896120593(119903+1)

= int[119886119887]

119866119896120593(119903+1)

= (minus1) int[119886119887]

1198661015840

119896120593(119903)

(78)

Noting that 1198661015840

119896= 119866

119896minus1almost everywhere we see that

intΩ

119866119896120593(119903+1)

= (minus1) int[119886119887]

119866119896minus1

120593(119903)

= (minus1) intΩ

119866119896minus1

120593(119903)

(79)

Using the induction hypothesis the claim follows

We cite Theorem 253 in [10] without giving its proof

Theorem 24 Let 119860 = [119886 119887] R or R+and 119903 ge 0 If the

function 119891 isin 1198711

loc(119860) satisfies

int119860

119891120593(119903)

119889119909 = 0 (80)

for all 120593 isin Cinfin

0(119860) then there is a polynomial 119875 of degree lt 119903

such that 119891 = 119875 almost everywhere on 119860

To show that 119866119896is in 119883 we need to verify that (if there

is danger of confusion with respect to which variable we takethe norm we use notations as119883(119909

119896))

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

lt infin (81)

under the assumption that 119866119896minus1

is in 119883 Actually bounding119866119896from above by its absolute value this means that we have

first to take the1198711-norm and then the norm in119883We howeverwould like to switch the norms

We adapt Proposition 21 in [16] to our setting

Proposition 25 Let 120588 be a Banach function norm satisfyingproperty (D) If 119891 ΩtimesΩ rarr R is a measurable function andone has 119891

119909and 119891

119910 functions onΩ such that 119891119909(119910) = 119891(119909 119910) =

119891119910

(119909) then

120588 (1003817100381710038171003817119891119909

10038171003817100381710038171) le

1003817100381710038171003817120588 (119891119910

)10038171003817100381710038171

(82)

The cases 119883 isin 1198711

119871infin

are easy to prove If 119883 = 119871119901 for

some 1 lt 119901 lt infin then the result can be proven by usingFubinirsquos theorem together with property (D) For the generalresults we refer to Scheprsquos proof

The next result builds uponTheorem 254 in [10]

Theorem 26 Let Ω sub R be a domain and let 119903 ge 1 Suppose119883 is a Banach function space satisfying (B) and (D) If 119891 islocally in 119883 and has a generalized 119903th derivative 119892 locally in119883 then there is 119865 agreeing almost everywhere with 119891 havingabsolutely continuous intermediate weak derivatives and theweak derivative of order 119903 agrees almost everywhere with 119892Furthermore the derivatives of 119865 are locally in 119883 If Ω hasfinite measure one obtains the result with ldquolocallyrdquo replaced byldquogloballyrdquo

We thank Pilar Silvestre for pointing out how to prove themembership of the intermediate derivatives in119883

Proof Choose 119909lowast

isin Ω and let 119866119896be defined as in Lemma 23

Assume 120593 isin Cinfin

0(Ω) Then using the fact that 119892 is a weak

derivative of 119891 of order 119903 and (b) in Lemma 23

intΩ

(119891 minus 119866119903) 120593

(119903)

119889119909 = intΩ

119891120593(119903)

119889119909 minus intΩ

119866119903120593(119903)

119889119909

= (minus1)119903

intΩ

119892120593119889119909 minus (minus1)119903

intΩ

119892120593119889119909 = 0

(83)

UsingTheorem 24 we see that119891minus119866119903= 119875 almost everywhere

where119875 is a polynomial of degreelt 119903Thus119891 = 119875+119866119903almost

everywhere and for 120593 isin Cinfin

0(Ω) and 0 le 119898 le 119903

intΩ

119891120593(119898)

119889119909 = intΩ

(119875 + 119866119903) 120593

(119898)

= (minus1)119898

intΩ

(119866119903minus119898

+ 119875(119898)

) 120593 119889119909

(84)

Thus119866119903minus119898

+119875(119898) is the119898th generalized derivative of119891 As it is

locally absolutely continuous (if119898 lt 119903) it lies locally in119883 Inthe case 119898 = 119903 the same conclusion holds by the assumptionon 119892

Moreover there exists 119865 for example 119865 = 119875 + 119866119903 that

agrees with 119891 almost everywhere such that its generalizedderivatives up to order 119903 minus 1 are locally absolutely continuousand the 119903th generalized derivative equals 119892 almost every-where

As 119892 is locally in 119883 and the other derivatives are locallyabsolutely continuous they are locally in 119871

infin and thus locallyin119883 by (B)

Suppose now that Ω has additionally finite measureAssume that 119891 and 119892 are in119883 globally We want to show thatthe intermediate derivatives are in 119883 as well

Note in our setting that by (P5) in Definition 18membership in 119883 implies membership in 119871

1 We proceedby induction Suppose that 119866

119896minus1isin 119883 We will apply

Proposition 25 We set denoting by conv119909lowast

119909119896 the closed

interval with endpoints 119909lowast and 119909119896

119891 (119909119896minus1

119909119896) = 119866

119896minus1(119909

119896minus1) 120594conv119909lowast 119909119896 (119909119896minus1

) (85)

Then1003817100381710038171003817119866119896

(119909119896)1003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119891(119909119896minus1

119909119896)119889119909

119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)10038171003817100381710038171198711(119909119896minus1)

10038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)

le10038171003817100381710038171003817

1003816100381610038161003816119866119896minus1(119909

119896minus1)1003816100381610038161003816

1003817100381710038171003817120594Ω1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)lt infin

(86)

Journal of Function Spaces 11

5 Banach Function Space Setting

Here our main focus is a more advanced readership thatlikely knows already Banach function spaces Therefore weadopt a much briefer style pointing merely to the literaturewhere the corresponding definitions and results can be foundWe start with giving generalized versions of the necessity andsufficiency of Theorem 2

We list the results and prove them simultaneously bysplitting the proofs into parts according to the functionspaces

Theorem 27 Assume Ω sub R is a domain 119883 is a normedspace and 119898 isin N If the maximal function 119872 119883 rarr 119883 isbounded then there is a constant 119862 = 119862(119883119898) such that foreach continuous 119891 isin 119882

119898

119883there is 119886

119891isin 119883with 119886

119891 le 119862119891

119882119898119883

and

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (87)

In particular such 119886119891exists if

(i) 1 lt 119901 le infin and 119883 is the Lebesgue space 119871119901

(Ω)

(ii) 1 lt 119901 lt infin 1 le 119902 le infin and 119883 is the Lorentz space119871119901119902

(Ω)

(iii) 1 lt 119901 lt infin 1 le 119902 le infin 119887 is slowly varying and 119883 isthe Lorentz-Karamata space 119871

119901119902119887(Ω)

(iv) 120601 is an 119873-function with inf119904gt0

(119904120593(119904)120601(119904)) gt 1 where120593 is the density function of 120601 and 119883 is the Orlicz space119871lowast

120601(Ω)

(v) Ω = R and 119883 is rearrangement-invariant with upperindex 120572

119883lt 1

Theorem 28 Let Ω sub R be open Suppose 119883 is a Banachfunction space such that (AC) holds Assume further that thefunction 119891 Ω rarr R is in 119883 If for some 119903 isin N

0 the following

inequality holds

1003816100381610038161003816Δ119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)] (88)

for some 119886119891isin 119883 then there exists119892 isin 119883with 119892

119883le 2119903

119903

119886119891119883

such that

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909 = (minus1)119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (89)

for all 120593 isin Cinfin

0(Ω)

In particular the result holds if119883 is

(i) the Lebesgue space 119871119901

(Ω) for some 1 lt 119901 lt infin

(ii) the Lorentz space 119871119901119902

(Ω) where 1 lt 119901 lt infin and 1 le

119902 lt infin

(iii) the Lorentz-Karamata space 119871119901119902119887

(Ω) where 119887 isslowly varying 1 lt 119901 lt infin and 1 lt 119902 le infin

(iv) the Orlicz space 119871lowast

120601(Ω) for some 119873-function 120601 satisfy-

ing

inf119904gt0

119904120593 (119904)

120601 (119904)gt 1 (90)

where 120593 is the density function of 120601

To help with orientation let us give an overview ofthe remaining material in this section With respect toTheorem 27 note that we carried out the proof of Corollary 14in a high level of abstraction Hence we are left to verifythat the spaces under question are normed spaces (we willshow that they are even Banach function spaces) and theboundedness of the maximal operator

Concerning the proof Theorem 28 we proved the cor-responding part of Theorem 2 under the assumptions thatthe spaces under question are Banach function spaces andfurther satisfy the requirements listed in Properties 19 Theexistence of 1198831015840 is classical (Proposition 29) as is the fact that(AC) implies (A) and (NC) (Proposition 30) Hence we onlyneed to point to the literature where (AC) was verified

Proposition 29 Given a Banach function space there is aspace 119883

1015840 such that properties (B) (HI) (C) and (D) aresatisfied

Proof We refer the reader to Sections 11 and 12 in [13] or toSection 31 in [14]

The definition of absolutely continuous norm can befound as Definition 3111 in [14] and Definition 131 in [13]

Proposition 30 A Banach function space that has absolutelycontinuous norm satisfies (A) and (NC)

Proof The first statement follows fromTheorem 1313 in [13]and the second fromProposition 136 in the same source

51 General Statements

Proof of Theorem 27 (general statement) We basically havealready proven the general statement in Corollary 14

Proof of Theorem 28 (general statement) We gave the proofof the Lebesgue space case in such generality that thegeneral statement follows from the proof of Theorem 2 andPropositions 29 and 30

52 Lorentz-Karamata Spaces Lorentz-Karamata spaces aredetailed in [14] mainly Section 343

Proof of Theorem 27 (ii) and (iii) The part that is left toprove is the boundedness of the maximal operator which isguaranteed by Remark 3517 in [14]

Proof of Theorem 28 (ii) and (iii) By Theorem 3441 in [14]we see that 119883 is a rearrangement-invariant Banach space Acombination of Lemma 3439 in [14] together with Lemma3443 in [14] gives the absolute continuity (AC) of the normof1198831015840

12 Journal of Function Spaces

53 Orlicz Spaces The setting is the one of [17]

Lemma 31 If 120601 is an 119873-function satisfying the Δ2-condition

then the Orlicz space 119871lowast

120601has absolutely continuous norm

Proof Let 119891 isin 119871lowast

120601(Ω) We verify first that

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 lt infin (91)

Let 120576 gt 0 By the Δ2-condition there exists a constant 120581 =

120581(119891(120601)

+ 120576) such that

120601 ((1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576) 119904) le 120581120601 (119904) (92)

for all 119904 ge 0 Thus

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 = intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576(1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576))119889119909

le 120581intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576)119889119909 le 120581

(93)

Suppose that 119864 is a measurable set Then

10038171003817100381710038171198911205941198641003817100381710038171003817(120601)

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891120594119864

1003816100381610038161003816) 119889119909 le 1

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 1

(94)

Let us fix 120582 gt 0 Again using the Δ2-condition we have

120601(120582minus1

119904) le 120581(120582minus1

)120601(119904) Hence

intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 (95)

Example 133 in [13] tells us that 1198711 has absolutely continuousnorm Thus if 120594

119864119899converges 120583-almost everywhere to 0 then

intΩ

120601(|119891|)120594119864119899

converges to zero as well Thus we have for 119899

large enough that

120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119899119889119909 le 1 (96)

implying that 119891120594119864119899(120601)

le 120582 Since 120582 gt 0 was arbitrary theabsolute continuity of 119891 follows

Proof of Theorem 27 (iv) We conclude from Gallardorsquos resultTheorem 21 in [17] that it is sufficient that the complemen-tary 119873-function satisfies the Δ

2-condition in [0infin) Again

resorting to Gallardo this time to Proposition 14 in [17] weobtain the conclusion

Proof of Theorem 28 (iv) That 119883 is a Banach functionspace follows from Theorem 3416 in [14] To prove that119883

1015840 has absolutely continuous norm we note that inequal-ity (90) together with Proposition 14 in [17] ensuresthat the complementary 119873-function of 120601 satisfies the Δ

2-

condition Lemma 31 verifies that 1198831015840 has absolutely contin-uous norm

54 Rearrangement-Invariant Spaces The necessary defini-tions can be found in [13]

Proof of Theorem 27 (v) We are only left to show theboundedness of the maximal operator However this is thecontent of a result by Lorentz and Shimogaki for examplestated as Theorem 3517 in [13]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank Pilar Silvestre for hersuggestions of how to prove the existence of intermediatederivatives The research is supported by the Academy ofFinland The third author thanks the Department of Math-ematics and Systems Analysis at Aalto University for itshospitality

References

[1] B Bojarski ldquoRemarks on some geometric properties of Sobolevmappingsrdquo in Functional Analysis amp Related Topics (Sapporo1990) pp 65ndash76 World Scientific River Edge NJ USA 1991

[2] B Bojarski PHajlasz andP Strzelecki ldquoImproved119862119896120582 approxi-

mation of higher order Sobolev functions in normand capacityrdquoIndiana University Mathematics Journal vol 51 no 3 pp 507ndash540 2002

[3] B Bojarski ldquoPointwise characterization of Sobolev classesrdquoProceedings of the Steklov Institute of Mathematics vol 255 no1 pp 65ndash81 2006

[4] B Bojarski ldquoWhitneyrsquos jets for Sobolev functionsrdquo UkrainianMathematical Journal vol 59 no 3 pp 345ndash358 2007

[5] B Bojarski ldquoTaylor expansion and Sobolev spacesrdquo Bulletin ofthe Georgian National Academy of Sciences New Series vol 5no 2 pp 5ndash10 2011

[6] B Bojarski ldquoSobolev spaces and Lagrange interpolationrdquoGeorgian Academy of Sciences Proceedings of A RazmadzeMathematical Institute vol 158 pp 1ndash12 2012

[7] P Hajłasz ldquoSobolev spaces on an arbitrary metric spacerdquoPotential Analysis vol 5 no 4 pp 403ndash415 1996

[8] B Bojarski ldquoSobolev spaces and averaging Irdquo httparxivorgabs13085171

[9] R Kannan and C K Krueger Advanced Analysis on the RealLine Universitext Springer New York NY USA 1996

[10] R A DeVore and G G Lorentz Constructive Approximationvol 303 of Fundamental Principles of Mathematical SciencesSpringer Berlin Germany 1993

[11] L L Schumaker Spline Functions Basic Theory CambridgeMathematical Library CambridgeUniversity Press CambridgeUK 3rd edition 2007

[12] G H Hardy and J E Littlewood ldquoA maximal theorem withfunction-theoretic applicationsrdquo Acta Mathematica vol 54 no1 pp 81ndash116 1930

[13] C Bennett andR Sharpley Interpolation ofOperators vol 129 ofPure and Applied Mathematics Academic Press Boston MassUSA 1988

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Journal of Function Spaces

Proof The proof uses induction and integration by partsNote that by replacing 119886 by 119910 in the integral we may deletethe + in (119909 minus 119910)

+ The case119898 = 1 reduces to the fundamental

theorem of calculus for absolutely continuous functions seefor example Theorem 7115 in [9] Let us assume that thestatement holds for some119898 and fix 119891 isin 119871

119898+1

1[119886 119887] Then

int

119887

119910

(119909 minus 119910)119898

(minus1)119898+1

119863119898+1

119891 (119909)

119898119889119909

=(119887 minus 119910)

119898

(minus1)119898+1

119863119898

119891 (119887)

119898

minus int

119887

119910

(119909 minus 119910)119898minus1

(minus1)119898+1

119863119898

119891 (119909)

(119898 minus 1)119889119909

= minus(119887 minus 119910)

119898

(minus1)119898

119863119898

119891 (119887)

119898

minus

119898minus1

sum

119895=0

(minus1)119895

119863119895

119891 (119887) (119887 minus 119910)119895

119895+ 119891 (119910)

(20)

We now cite a part ofTheorem 423 in [11] Note that119898 isa positive integer and 119863

119895

+means that we take the right hand

derivative of order 119895 We only need and prove the equality for119895 = 0 We also assume that not all points coincide

Theorem 10 (Peano representation) Fix 0 le 119895 le 119898minus119899 where119899 is the maximum multiplicity of 119910

119894 119910

119894+119898 Then

[119910119894 119910

119894+119898] 119891 = int

119910119894+119898

119910119894

(minus1)119895

119863119895

+119876

119898

119894(119909)119863

119898minus119895

119891 (119909)

(119898 minus 1)119889119909 (21)

for all 119891 isin 119871119898minus119895

1[min

119894119910

119894max

119894119910

119894]

Proof of the Peano Representation for 119895 = 0 We take thedual Taylor expansion from Theorem 9 Then we apply[119910

119894 119910

119894+119898] on both sides and remember that we computed

the divided differences of polynomials in Lemma 6

Our goal is to use the Peano representation to obtainan upper bound for the divided differences We start withestablishing an upper bound of the factor 119863

+119876

119898

119894in the

integrand

Lemma 11 Suppose that 119910119894le sdot sdot sdot le 119910

119894+119898and 119910

119894lt 119910

119894+119898 Then

there is a constant 119862 such that

1003816100381610038161003816119863+119876

119898

119894(119909)

1003816100381610038161003816 =10038161003816100381610038161003816[119910

119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+

10038161003816100381610038161003816le

119862

119910119894+119898

minus 119910119894

(22)

Proof We want to estimate

[119910119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+

=[119910

119894+1 119910

119894+119898] (119909 minus 119910)

119898minus1

+minus [119910

119894 119910

119894+119898minus1] (119909 minus 119910)

119898minus1

+

119910119894+119898

minus 119910119894

(23)

Assume 119896 lt 119898 minus 1 We claim that

119889119896

119889119910119896

(119909 minus 119910)119898minus1

+= (minus1)

119896

(119898 minus 1) sdot sdot sdot (119898 minus 119896) (119909 minus 119910)119898minus1minus119896

+

(24)

This claim is easily seen to be true for 119910 = 119909 That it is truealso for 119910 = 119909 follows by induction We also note that

119889119898minus2

119889119910119898minus2(119909 minus 119910)

119898minus1

+(25)

is absolutely continuous on bounded intervals and that

(minus1)(119898minus1)

(119898 minus 1) (119909 minus 119910)0

+

(26)

is a weak derivative that equals the derivative whenever119910 = 119909 We first assume that not all 119910

119895 119910

119895+119898minus1coincide

An application of the Hermite-Genocchi formula Lemma 5gives

[119910119895 119910

119895+119898minus1] (119909 minus 119910)

119898minus1

+

= int

1

0

1198891199051int

1199051

0

1198891199052

sdot sdot sdot int

119905119898minus2

0

(minus1)(119898minus1)

(119898 minus 1)

sdot (119909 minus (119910119895+

119898minus1

sum

119896=1

(119910119895+119896

minus 119910119895+119896minus1

) 119905119896))

0

+

119889119905119898minus1

(27)

We use now the fact that the integrand is bounded from aboveby (119898 minus 1) and the computation (14) to obtain

10038161003816100381610038161003816[119910

119895 119910

119895+119898minus1] (119909 minus 119910)

119898minus1

+

10038161003816100381610038161003816le 1 (28)

Suppose now that 119910119895

= sdot sdot sdot = 119910119895+119898minus1

= 119909 Then|[119909

119895 119909

119895+119898minus1](119909 minus 119910)

119898minus1

+| is bounded from above by 1 if

119909 gt 119910119894and vanishes if 119910

119894lt 119909 In the end we find a constant

119862 such that10038161003816100381610038161003816119863

+[119910

119894 119910

119894+119898] (119909 minus 119910)

119898minus1

+

10038161003816100381610038161003816le

119862

119910119894+119898

minus 119910119894

(29)

In the next result we use the notation

119872119892(119909) = sup119911

minusint

119911

119909

10038161003816100381610038161198921003816100381610038161003816 (30)

for the famous Hardy-Littlewood maximal function of 119892

introduced in the seminal paper [12] of Hardy-Littlewood inActa Math (1930)

Theorem 12 Suppose 1199100

le sdot sdot sdot le 119910119898 119910

0lt 119910

119898and 119891 is in

119871119898

1[119910

0 119910

119898] for some 119898 isin N Then there is some constant 119862

such that1003816100381610038161003816[1199100 119910119898] 119891

1003816100381610038161003816 le 119862 (119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (1199100) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910119898))

(31)

Journal of Function Spaces 5

Proof We use the Peano representation Theorem 10 with119894 = 119895 = 0 Then using the bound of |119863

+119876

119898

119894(119909)| found in

Lemma 11

1003816100381610038161003816[1199100 119910119898] 1198911003816100381610038161003816 le

1

(119898 minus 1)int

119910119898

1199100

1003816100381610038161003816119863+119876

119898

119894(119909)

1003816100381610038161003816

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

le119862

119910119898

minus 1199100

int

119910119898

1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

(32)

For 119910 isin [1199100 119910

119898] we have

1003816100381610038161003816[1199100 119910119898] 1198911003816100381610038161003816

le119862

119910119898

minus 1199100

(int

119910

1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909 + int

119910119898

119910

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909)

(33)

and further1003816100381610038161003816[1199100 119910119898] 119891

1003816100381610038161003816

le 119862(minusint

119910

1199100

119910 minus 1199100

119910119898

minus 1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

+minusint

119910119898

119910

119910119898

minus 119910

119910119898

minus 1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909)

(34)

The first factors in both integrals are bounded from above by1 We see that the integrals are bounded from above by thecorresponding maximal operators This gives the claim

The finite differences defined in (2) are up to a factordivided differences for equispaced nodes

Lemma 13 Let 119909 119910 be two points in a domain Ω sub R Let119898 isin N and 119909

119896= 119909

0+ ℎ119896 where ℎ = (119910 minus 119909)119898 Then each

119891 Ω rarr R satisfies

[1199090 119909

119898] 119891 =

1

ℎ119898119898Δ119898

119891 (119909 119910) (119909 = 1199090 119909

119898= 119910)

(35)

Proof We use induction For119898 = 1 we have

[1199090 119909

1] 119891 =

119891 (1199091) minus 119891 (119909

0)

1199091minus 119909

0

(36)

As ℎ = 1199091minus 119909

0 119909 = 119909

0 119910 = 119909

1 and 119898 = 1 we obtain the

claim Suppose that the claim is true for some119898 isin N Then

Δ119898+1

ℎ119891 (119909)

=

119898+1

sum

119895=0

(minus1)119898+1minus119895

(119898 + 1

119895)119891 (119909 + 119895ℎ)

=

119898+1

sum

119895=0

(minus1)119898+1minus119895

((119898

119895 minus 1) + (

119898

119895))119891 (119909 + 119895ℎ)

=

119898+1

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + (119895 + 1) ℎ)

minus

119898+1

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + 119895ℎ)

=

119898

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909

1+ 119895ℎ)

minus

119898

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + 119895ℎ)

= ℎ119898

119898 ([1199091 119909

119898+1] 119891 minus [119909

0 119909

119898] 119891)

= ℎ119898

119898 (ℎ (119898 + 1)) [1199090 119909

119898+1] 119891

(37)

We can now formulate Theorem 12 in terms of finitedifferences

Corollary 14 (Corollary of Theorem 12) Let Ω sub R be adomain Suppose 119891 isin 119871

119898

1(Ω) for some 119898 isin N then there is

a constant 119862 = 119862(119898) such that

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816

le 1198621003816100381610038161003816119909 minus 119910

1003816100381610038161003816

119898

(119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

119909 119910 isin Ω

(38)

In particular we have the following special case Suppose 119883

and 119884 are normed spaces of functions such that the maximaloperator 119872 119883 rarr 119884 is bounded If 119891 isin 119882

119898

119883(Ω) for some

119898 ge 1 then

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (39)

for some function 119886119891isin 119884 depending on 119891

Proof We first use Lemma 13 and thenTheorem 12 with stepℎ = (119910 minus 119909)119898 There is some constant 119862 = 119862(119898) such that

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816

= ℎ119898

1198981003816100381610038161003816[1199090

119909119898] 119891

1003816100381610038161003816

le 119862ℎ119898

(119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

le1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

119898

(119862119872(1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119862119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

(40)

Proof of Theorem 2 (Sobolev functions satisfy pointwiseinequality) We combine Corollary 14 with the boundednessof the maximal operator which is by now classical and can befound for example in Theorem 3310 in [13]

6 Journal of Function Spaces

3 Pointwise Inequality Implies Membershipin Sobolev Space

In this section our aim is to show that the pointwiseinequality (3) implies the existence of the highest orderderivative in the correct space We will take care of theintermediate derivatives in the next section In this sectionthe properties of the Lebesgue spaces play an important roleOur point of view is however to extract the properties ofthe Lebesgue spaces and base our proofs solely on theseproperties We hope that the disadvantage of being slightlymore abstract is later on compensatedwhenwe generalize theresults We start with some auxiliary results

Lemma 15 Let Ω sub R be a domain and 120593 isin C119899

0(Ω) that

is 120593 has compact support in Ω and is 119899-times continuouslydifferentiable Then

1003816100381610038161003816100381610038161003816100381610038161003816

[1199090 119909

119899] 120593 minus

120593(119899)

(119910)

119899

1003816100381610038161003816100381610038161003816100381610038161003816

997888rarr 0 (41)

uniformly inΩ as 1199090 119909

119899rarr 119910

Proof By Taylorrsquos formula for 119909 isin conv1199090 119909

119899 there is

120585119909isin conv119909

0 119909

119899 such that

120593 (119909) = 120593 (119910) + 1205931015840

(119910) (119909 minus 119910) + sdot sdot sdot + 120593(119899minus1)

(119910)(119909 minus 119910)

119899minus1

(119899 minus 1)

+1

119899120593(119899)

(120585119909) (119909 minus 119910)

119899

(42)

We apply [1199090 119909

119899] on both sides (with variable 119909) and use

the computation of the divided differences of polynomialscarried out in Lemma 6 to obtain

[1199090 119909

119899] 120593 (119909) =

1

119899120593(119899)

(120585119909) (43)

As 120593(119899) is uniformly continuous on compact intervals the

claim follows

Lemma 16 Assume Ω sub R is open and 119870 sub Ω is compactThen there is a neighborhood of119870 that is a subset of Ω

Proof For each 119909 isin 119870 we find a ball 119861(119909 119903119909) sub Ω Suppose

the statement of the lemma is falseThen for every 119899 isin N thereis 119911

119899isin Ω 119870 with dist(119911

119899 119870) lt 1119899 Thus for each 119899 isin N

there is 119910119899

isin 119870 with 119889(119911119899 119910

119899) lt 1119899 Since 119870 is compact

there is a subsequence (119910119899119896)119896of (119910

119899)119899that converges to some

119910 isin 119870 Now 119910 lies in one of the balls 119861(119909 119903119909) sub Ω For 119896 large

enough we have that 119910119899119896and 119911

119899119896are in 119861(119909 119903

119909) sub Ω This is

a contradiction

Lemma 17 SupposeΩ sub R is open and119870 sub Ω compactThenthere exists 120576 gt 0 such that

(Ω minus ℎ) cap 119870 = Ω cap 119870 (44)

for |ℎ| lt 120576

Proof We first show that any member of the left hand side iscontained in the right hand side this is true for any ℎ isin R If119911 is a member of the left hand side then it is in 119870 and since119870 is a subset of Ω it is also a member of the right hand sideTo prove the other inclusion note that by Lemma 16 there is120576 gt 0 such that the 120576-neighborhood 119870

120576of 119870 lies in Ω Now

suppose that 119911 is in the right hand side and |ℎ| lt 120576 Then for119909 = 119911+ℎ we have119909 isin 119870

120576sub Ω implying that 119911 = 119909minusℎ isin Ωminusℎ

That 119911 isin 119870 is clear This gives the proof

Next we state basically Definition 311 in [14] Readersfollowing [13] might want to have a look at Definitions 111and 113 in their book To gain familiarity with the conceptof Banach function norm the reader may want to verify thatLebesgue spaces are examples of Banach function spaces Aswe are here only interested in Lebesgue measure we adaptthe notion to this case The set M+ denotes the cone ofmeasurable functions whose values lie in [0infin]

Definition 18 (Banach function norm Banach functionspace) Assume that 119877 sub R A Banach function norm on 119877

is a map 120588 from M+

(119877) to [0infin] such that for all 119891 119892 and119891119899(119899 isin N) in M+

(119877) all scalars 120582 ge 0 and all measurablesubsets 119864 of 119877 the following are true where 120583 is the Lebesguemeasure

(P1) 120588(119891) = 0 if and only if 119891 = 0 120583-almost everywhere120588(120582119891) = 120582120588(119891) 120588(119891 + 119892) le 120588(119891) + 120588(119892)

(P2) if 0 le 119892 le 119891 120583-almost everywhere then 120588(119892) le 120588(119891)(P3) if 0 le 119891

119899uarr 119891 120583-almost everywhere we have 120588(119891

119899) uarr

120588(119891)(P4) if 120583(119864) lt infin then 120588(120594

119864) lt infin

(P5) if120583(119864) lt infin then there is a constant119862 = 119862(119864 120588) lt infin

such that int119864

119891119889119909 le 119862120588(119891)

Given such a function norm 120588 the set 119883 = 119883(120588) ofall extended measurable functions 119891 (identifying functionsequal 120583-almost everywhere) such that 120588(|119891|) lt infin is called aBanach function space (BFS for short) and we define

10038171003817100381710038171198911003817100381710038171003817119883

= 120588 (1003816100381610038161003816119891

1003816100381610038161003816) 119891 isin 119883 (45)

We leave it to the reader to verify that Lebesgue spacessatisfy the following properties (1198831015840 can be chosen as the dualspace)

Properties 19 Given a Banach function space 119883 we requirethe existence of a Banach function space119883

1015840 such that

(B) 119883 contains each essentially bounded function whosesupport has finite measure

(HI) Holderrsquos inequality int |119891119892|119889119909 le 119891119883119892

1198831015840

(C) 1198831015840 is complete

(D) 119892119883

= supintΩ

|119892|ℎ ℎ isin M+

ℎ1198831015840 le 1

(AC) absolutely continuous norm if (120594119864119899)119899is a sequence of

characteristic functions converging pointwise almosteverywhere to 0 then 119891120594

1198641198991198831015840 converges to zero for

all 119891 isin 1198831015840

Journal of Function Spaces 7

(A) each function 119891 isin 1198831015840 can be approximated by simple

functions (in1198831015840-norm)

(NC) if 119891 isin 1198831015840 and ℎ

119899are step functions with 0 le ℎ

119899uarr |119891|

then1003817100381710038171003817119891 minus ℎ

119899sgn (119891)

10038171003817100381710038171198831015840997888rarr 0 as 119899 997888rarr infin (46)

Lemma 20 Suppose Ω sub R is open and 119883 is a Banachfunction space fulfilling the items listed in Properties 19 Assumethat 119891 Ω rarr R is in 119883 We let 120582

119891 119883

1015840

cap C119903+1

0(Ω) rarr R be

defined as

120582119891(120593) = int

Ω

119891120593(119903)

119889119909 (47)

If1003816100381610038161003816Δ

119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)]

for almost every 119909 119910 isin Ω

(48)

for some 119886119891

isin 119883 then 120582119891has a continuous extension to 119883

1015840

with norm bounded from above by 2119903119903

119886119891119883

Proof Our goal is to show1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816le 2119903

11990310038171003817100381710038171003817119886119891

10038171003817100381710038171003817119883

100381710038171003817100381712059310038171003817100381710038171198831015840

(49)

because 120582119891extends then by the theorem of Hahn-Banach to

a bounded functional on1198831015840 with norm bounded from above

by 2119903119903

119886119891119883 Without loss of generality we may assume that

119886119891is nonnegative If 119891 were be smooth we could just use

integration by parts and use as bound the limiting case of thepointwise inequality (48) As 119891 is not necessarily smooth wehave to operate on the level of finite differences

Let us choose 120593 isin C0(Ω) and denote its support by 119870

Lemma 17 ensures the existence of some positive 120576 gt 0 suchthat (Ωminus119895ℎ)cap119870 = Ωcap119870 for 0 le 119895 le 119903 as long as |ℎ| lt 120576119903Weassume in the following that ℎ has been chosen accordingly

The following integrals exist by Holderrsquos inequalitybecause 120593 and thus Δ

119903

minusℎ120593(119909) are bounded with compact

support and 119891 is integrable Consider

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

119891 (119909) (minusℎ)minus119903

Δ119903

minusℎ120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(50)

We compute the following integral

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

= intΩ

119891 (119909)

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119909) 120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωminus119895ℎ

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

(Ωminus119895ℎ)cap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

(51)

We have chosen 120576 gt 0 such that (Ω minus 119895ℎ) cap 119870 = Ω cap 119870 Wecontinue

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωcap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

(52)

With the aid of (52) we can now rewrite (50) as

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

(minusℎ)minus119903

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(53)

Inequality (48) leads to

1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816

le 119903119903

intΩ

(119886119891(119909) + 119886

119891(119909 + 119903ℎ))

1003816100381610038161003816120593 (119909)1003816100381610038161003816 119889119909

+ intΩ

1003816100381610038161003816119891 (119909)1003816100381610038161003816

10038161003816100381610038161003816120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)

10038161003816100381610038161003816119889119909

(54)

Using Holderrsquos inequality (HI) for the first summand weare left to show that the second summand tends to 0 asℎ approaches 0 We note first that since the second factorhas support in a set of finite measures we can replace theintegration domain by a set of finite measuresThen we applya (1198711

119871infin

)-Holder inequality and the uniform convergence ofthe divided difference to 120593

(119903)

(119909)119903 established in Lemmas 13and 15

After having verified that 120582119891has a continuous extension

to 1198831015840 we want to show that this extension has a representa-

tion as integral

8 Journal of Function Spaces

Lemma 21 Assume Ω sub R is open Suppose the propertieslisted in Properties 19 are met Let 120582

119891be defined as in (47) and

bounded on 1198831015840 Then there exists 119892 isin 119883 with 119892

119883le 120582

119891

such that

120582119891(120593) = (minus1)

119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (55)

for all 120593 isin 1198831015840 especially for all 120593 isin Cinfin

0(Ω)

Proof The proof is heavily influenced by the proof of Theo-rem 141 in [13]We first want to find a signedmeasure ] suchthat

120582119891(120593) = int

Ω

120593119889] (56)

for each 120593 isin 1198831015840 Then we want to apply the Lebesgue-

Radon-Nikodym theorem to replace 119889] by (minus1)119903

119892(119909)119889119909 forsome integrable function 119892 for which we then show that it iscontained in119883

We start by assuming that the measure of Ω is finite Foreach 120583-measurable set 119860 we set

] (119860) = 120582119891(120594

119860) (57)

We start by proving the 120590-additivity of ] (for the 120583-measurable sets) In the first step we just show the additivityBy induction it is enough to consider only two sets

Let 1198601and 119860

2be two disjoint 120583-measurable sets Then

] (1198601cup 119860

2) = 120582

119891(120594

1198601cup1198602) = 120582

119891(120594

1198601+ 120594

1198602)

= 120582119891(120594

1198601) + 120582

119891(120594

1198602) = ] (119860

1) + ] (119860

2)

(58)

We move to the verification of the 120590-additivity Let119860

1 119860

2 be countablymany pairwise disjoint120583-measurable

sets We first want to show thatsum119894120594119860119894

converges in1198831015840

We let 119864119896= cup

119896

119894=1119860

119894and 119864 = cup

infin

119894=1119860

119894 Then since (119864

119896)119896is

increasing

120583 (119864) = 120583(

infin

⋃

119896=1

119864119896) = lim

119896rarrinfin

120583 (119864119896) (59)

As 119864 has finite measure given 120576 gt 0 there exists 1198960isin N such

that

(1 minus 120576) 120583 (119864) lt 120583 (1198641198960) (60)

By the disjointness of the sets 119860119894 we see that

120583 (119864) = 120583 (1198641198960) + 120583(

infin

⋃

119894=1198960+1

119860119894) (61)

It follows that

120583(

infin

⋃

119894=1198960+1

119860119894) = 120583 (119864) minus 120583 (119864

1198960) lt 120583 (119864) minus (1 minus 120576) 120583 (119864)

= 120576120583 (119864)

(62)

We have for119898 lt 119899

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=1

120594119860119894

minus

119898

sum

119894=1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

=

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=119898+1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

le10038171003817100381710038171003817120594cupinfin119894=119898+1

119860119894

100381710038171003817100381710038171198831015840 (63)

By the considerations before and the fact that sdot 1198831015840 is

absolutely continuous as stipulated in (AC) we see thatsum

119899

119894=1120594119860119894

is a Cauchy sequence Now as 1198831015840 is complete by

(C) the Cauchy sequence converges in 1198831015840 We have by the

continuity of 120582119891

] (cup119894119860

119894) = 120582

119891(120594

cup119894119860119894) = 120582

119891(

infin

sum

119894=1

120594119860119894)

= lim119899rarrinfin

120582119891(

119899

sum

119894=1

120594119860119894) = lim

119899rarrinfin

119899

sum

119894=1

120582119891(120594

119860119894)

= lim119899rarrinfin

119899

sum

119894=1

] (119860119894) =

infin

sum

119894=1

] (119860119894)

(64)

Thus ] is 120590-additiveIn order to apply the theorem of Lebesgue-Radon-

Nikodym see for example Section 69 in [15] and conclud-ing the existence of 119892 isin 119871

1

(Ω) such that

] (119860) = int119860

119892119889119909 119860 120583-measurable (65)

we need to show that ] is absolutely continuous with respectto 120583 To do so assume that 120583(119873) = 0 Then ](119873) = 120582

119891(120594

119873)

By the absolute continuity of the normof1198831015840 required in (AC)we see that 120594

1198731198831015840 = 0 and therefore ](119873) = 0 Replacing 119892

by minus119892 if necessary we obtain

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909 (66)

Now we assume that Ω does not necessarily have finitemeasureWewriteΩ = cup

119899Ω

119899as union of open sets with finite

measure such that Ω119899sub Ω

119899+1 By the considerations above

we find for each 119899 isin N a 119892119899such that

120582119891(120594

119860) = (minus1)

119903

intΩ

119892119899(119909) 120594

119860(119909) 119889119909 (67)

for each measurable set 119860 sub Ω119899 It is easily seen that 119892

119899

and 119892119899+1

agree almost everywhere on Ω119899 This gives rise to

a function 119892 satisfying

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909

119860 sub

119873

⋃

119899=1

Ω119899for some 119873

(68)

By linearity we have for each simple function 119904

120582119891(119904) = (minus1)

119903

intΩ

119892 (119909) 119904 (119909) 119889119909 (69)

To show that 119892 isin 119883 we want to use property (D)

Journal of Function Spaces 9

Let ℎ be a nonnegative simple function with supportin some Ω

119899 Then ℎ sgn(119892) is also a simple function with

support inΩ119899 It is a finite linear combination of characteristic

functions of sets in Ω119899 Hence

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = intℎ sdot sgn (119892) 119892 119889119909 = 120582119891(ℎ sgn (119892)) (70)

Using the boundedness of 120582119891 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (71)

If ℎ is an arbitrary function in1198831015840 thenwemay construct a

sequence (ℎ119899)infin

119899=1of simple functions each ℎ

119899having support

in Ω119899 such that 0 le ℎ

119899uarr |ℎ| 120583-almost everywhere By the

monotone convergence theorem we have by (71) that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = lim119899rarrinfin

int1003816100381610038161003816ℎ119899119892

1003816100381610038161003816 119889119909 le lim sup119899rarrinfin

10038171003817100381710038171003817120582119891

10038171003817100381710038171003817

1003817100381710038171003817ℎ11989910038171003817100381710038171198831015840

(72)

By (P3) in Definition 18 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (73)

and Property (D) gives that 119892 isin 119883 In the last step we want toupgrade the integral representation (69) to hold for arbitraryfunctions in 120593 isin 119883

1015840 As before we approximate 120593 by simplefunctions ℎ

119899with support contained in Ω

119899and 0 le ℎ

119899uarr |120593|

in 120583-almost everywhere point Define 119904119899

= ℎ119899sgn(120593) By

(NC) 119904119899converges to 120593 in119883

1015840 and as both 120582119891and ldquoits integral

representationrdquo are continuous and agree on the set of stepfunctions they agree on119883

1015840

Having shown the existence of the weak derivativesof highest order we are left to show the existence of theintermediate derivatives which we do in Proposition 22 inthe next section To keep the flow going we give here theproof that the pointwise inequality implies membership inthe corresponding Sobolev space under the assumption thatwe already know that Proposition 22 holds

Proof of Theorem 2 (pointwise inequality implies membershipin Sobolev space) We combine Lemmas 20 and 21 and applyProposition 22

4 Equivalent Definitions of Sobolev Spaces

The aim of this section is to prove the following equivalence

Proposition 22 Let Ω sub R be a domain with finite Lebesguemeasure inR Suppose119883 is a Banach function space satisfying(B) and (D) Assume that 119898 isin N Then the following setscoincide

(a) 119860 = 119891 isin 119883 there exists 119865 isin 119883 where 119865 agrees almosteverywherewith119891 and hasweak derivatives up to order119898 in119883 where the derivatives of order 119903 0 le 119903 le 119898minus1are absolutely continuous

(b) 119861 = 119891 isin 119883 119891(119896)

isin 119883 119896 isin 0 119898 where 119891(119896) arethe weak derivatives of 119891

(c) 119862 = 119891 isin 119883 119891(119898)

isin 119883 where 119891(119898) is the weak

derivative of 119891 of order 119898

We see that 119860 sub 119861 sub 119862 We now show that 119862 sub 119860Let us give an outline of the proof Given 119892 Ω rarr R

(we will choose 119892 = 119891(119898)) by integrating we construct in

Lemma 23 an absolutely continuous function 119866119898together

with its weak derivatives 1198660 119866

119898minus1 which are absolutely

continuous and 119866(119898)

119898= 119892 However 119866

119898and 119891 might

not agree almost everywhere We verify however in theproof of Theorem 26 (with the help of Theorem 24) thatwe find a polynomial 119875 such that 119866

119898+ 119875 agrees with 119865

almost everywhere and shares with 119866119898the other required

properties We conclude the proof of Theorem 26 by settlingthe membership questions with the aid of Proposition 25

Lemma 23 Assume Ω sub R is a domain We further assumethat 119892 Ω rarr R is in 119871

1

loc(Ω) One fixes 119909lowast

isin Ω and defines

1198660(119909

0) = 119892 (119909

0)

119866119896(119909

119896) = int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119896 isin N

(74)

where the integrals are Lebesgue integrals and we use thenotation

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

=

int[119909lowast119909119896]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909lowast

le 119909119896

minusint[119909119896 119909lowast]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909119896lt 119909

lowast

(75)

Then

(a) 119866119896is locally absolutely continuous for all 119896 isin N

especially locally integrable(b) if 120593 isin Cinfin

0(Ω) and 0 le 119903 le 119896 one has

intΩ

119866119896120593(119903)

= (minus1)119903

intΩ

119866119896minus119903

120593 (76)

Proof Let us prove (a) by induction Assume 119896 = 1 Note that1198661(119909

lowast

) = 0 Thus we have for 119909lowast

le 1199091

1198661(119909

1) minus 119866

1(119909

lowast

) = 1198661(119909

1) = int

1199091

119909lowast

119892 (1199090) 119889119909

0 (77)

As 119892 isin 1198711

([119909lowast

1199091]) we see that 119866

1is absolutely continuous

The case 1199091lt 119909

lowast is treated similarlyAssume now that the claim is true for some 119896 Now

119866119896+1

(119909119896+1

) is the integral of a continuous function and thuslocally absolutely continuous

Let us have a look at (b) Note that the statement is true for119896 = 0 thus we may assume that 119896 ge 1 We make an inductionover 119903 The case 119903 = 0 is clear Let us choose an interval[119886 119887] sub Ω containing a neighborhood of the support of 120593 As119866119896and 120593

(119903) are both absolutely continuous in [119886 119887] we have

10 Journal of Function Spaces

by the integration by part formula for absolutely continuousfunctions (see for exampleTheorem7147 in [9]) and the factthat 120593(119903)

(119886) = 120593(119903)

(119887) = 0

intΩ

119866119896120593(119903+1)

= int[119886119887]

119866119896120593(119903+1)

= (minus1) int[119886119887]

1198661015840

119896120593(119903)

(78)

Noting that 1198661015840

119896= 119866

119896minus1almost everywhere we see that

intΩ

119866119896120593(119903+1)

= (minus1) int[119886119887]

119866119896minus1

120593(119903)

= (minus1) intΩ

119866119896minus1

120593(119903)

(79)

Using the induction hypothesis the claim follows

We cite Theorem 253 in [10] without giving its proof

Theorem 24 Let 119860 = [119886 119887] R or R+and 119903 ge 0 If the

function 119891 isin 1198711

loc(119860) satisfies

int119860

119891120593(119903)

119889119909 = 0 (80)

for all 120593 isin Cinfin

0(119860) then there is a polynomial 119875 of degree lt 119903

such that 119891 = 119875 almost everywhere on 119860

To show that 119866119896is in 119883 we need to verify that (if there

is danger of confusion with respect to which variable we takethe norm we use notations as119883(119909

119896))

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

lt infin (81)

under the assumption that 119866119896minus1

is in 119883 Actually bounding119866119896from above by its absolute value this means that we have

first to take the1198711-norm and then the norm in119883We howeverwould like to switch the norms

We adapt Proposition 21 in [16] to our setting

Proposition 25 Let 120588 be a Banach function norm satisfyingproperty (D) If 119891 ΩtimesΩ rarr R is a measurable function andone has 119891

119909and 119891

119910 functions onΩ such that 119891119909(119910) = 119891(119909 119910) =

119891119910

(119909) then

120588 (1003817100381710038171003817119891119909

10038171003817100381710038171) le

1003817100381710038171003817120588 (119891119910

)10038171003817100381710038171

(82)

The cases 119883 isin 1198711

119871infin

are easy to prove If 119883 = 119871119901 for

some 1 lt 119901 lt infin then the result can be proven by usingFubinirsquos theorem together with property (D) For the generalresults we refer to Scheprsquos proof

The next result builds uponTheorem 254 in [10]

Theorem 26 Let Ω sub R be a domain and let 119903 ge 1 Suppose119883 is a Banach function space satisfying (B) and (D) If 119891 islocally in 119883 and has a generalized 119903th derivative 119892 locally in119883 then there is 119865 agreeing almost everywhere with 119891 havingabsolutely continuous intermediate weak derivatives and theweak derivative of order 119903 agrees almost everywhere with 119892Furthermore the derivatives of 119865 are locally in 119883 If Ω hasfinite measure one obtains the result with ldquolocallyrdquo replaced byldquogloballyrdquo

We thank Pilar Silvestre for pointing out how to prove themembership of the intermediate derivatives in119883

Proof Choose 119909lowast

isin Ω and let 119866119896be defined as in Lemma 23

Assume 120593 isin Cinfin

0(Ω) Then using the fact that 119892 is a weak

derivative of 119891 of order 119903 and (b) in Lemma 23

intΩ

(119891 minus 119866119903) 120593

(119903)

119889119909 = intΩ

119891120593(119903)

119889119909 minus intΩ

119866119903120593(119903)

119889119909

= (minus1)119903

intΩ

119892120593119889119909 minus (minus1)119903

intΩ

119892120593119889119909 = 0

(83)

UsingTheorem 24 we see that119891minus119866119903= 119875 almost everywhere

where119875 is a polynomial of degreelt 119903Thus119891 = 119875+119866119903almost

everywhere and for 120593 isin Cinfin

0(Ω) and 0 le 119898 le 119903

intΩ

119891120593(119898)

119889119909 = intΩ

(119875 + 119866119903) 120593

(119898)

= (minus1)119898

intΩ

(119866119903minus119898

+ 119875(119898)

) 120593 119889119909

(84)

Thus119866119903minus119898

+119875(119898) is the119898th generalized derivative of119891 As it is

locally absolutely continuous (if119898 lt 119903) it lies locally in119883 Inthe case 119898 = 119903 the same conclusion holds by the assumptionon 119892

Moreover there exists 119865 for example 119865 = 119875 + 119866119903 that

agrees with 119891 almost everywhere such that its generalizedderivatives up to order 119903 minus 1 are locally absolutely continuousand the 119903th generalized derivative equals 119892 almost every-where

As 119892 is locally in 119883 and the other derivatives are locallyabsolutely continuous they are locally in 119871

infin and thus locallyin119883 by (B)

Suppose now that Ω has additionally finite measureAssume that 119891 and 119892 are in119883 globally We want to show thatthe intermediate derivatives are in 119883 as well

Note in our setting that by (P5) in Definition 18membership in 119883 implies membership in 119871

1 We proceedby induction Suppose that 119866

119896minus1isin 119883 We will apply

Proposition 25 We set denoting by conv119909lowast

119909119896 the closed

interval with endpoints 119909lowast and 119909119896

119891 (119909119896minus1

119909119896) = 119866

119896minus1(119909

119896minus1) 120594conv119909lowast 119909119896 (119909119896minus1

) (85)

Then1003817100381710038171003817119866119896

(119909119896)1003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119891(119909119896minus1

119909119896)119889119909

119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)10038171003817100381710038171198711(119909119896minus1)

10038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)

le10038171003817100381710038171003817

1003816100381610038161003816119866119896minus1(119909

119896minus1)1003816100381610038161003816

1003817100381710038171003817120594Ω1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)lt infin

(86)

Journal of Function Spaces 11

5 Banach Function Space Setting

Here our main focus is a more advanced readership thatlikely knows already Banach function spaces Therefore weadopt a much briefer style pointing merely to the literaturewhere the corresponding definitions and results can be foundWe start with giving generalized versions of the necessity andsufficiency of Theorem 2

We list the results and prove them simultaneously bysplitting the proofs into parts according to the functionspaces

Theorem 27 Assume Ω sub R is a domain 119883 is a normedspace and 119898 isin N If the maximal function 119872 119883 rarr 119883 isbounded then there is a constant 119862 = 119862(119883119898) such that foreach continuous 119891 isin 119882

119898

119883there is 119886

119891isin 119883with 119886

119891 le 119862119891

119882119898119883

and

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (87)

In particular such 119886119891exists if

(i) 1 lt 119901 le infin and 119883 is the Lebesgue space 119871119901

(Ω)

(ii) 1 lt 119901 lt infin 1 le 119902 le infin and 119883 is the Lorentz space119871119901119902

(Ω)

(iii) 1 lt 119901 lt infin 1 le 119902 le infin 119887 is slowly varying and 119883 isthe Lorentz-Karamata space 119871

119901119902119887(Ω)

(iv) 120601 is an 119873-function with inf119904gt0

(119904120593(119904)120601(119904)) gt 1 where120593 is the density function of 120601 and 119883 is the Orlicz space119871lowast

120601(Ω)

(v) Ω = R and 119883 is rearrangement-invariant with upperindex 120572

119883lt 1

Theorem 28 Let Ω sub R be open Suppose 119883 is a Banachfunction space such that (AC) holds Assume further that thefunction 119891 Ω rarr R is in 119883 If for some 119903 isin N

0 the following

inequality holds

1003816100381610038161003816Δ119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)] (88)

for some 119886119891isin 119883 then there exists119892 isin 119883with 119892

119883le 2119903

119903

119886119891119883

such that

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909 = (minus1)119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (89)

for all 120593 isin Cinfin

0(Ω)

In particular the result holds if119883 is

(i) the Lebesgue space 119871119901

(Ω) for some 1 lt 119901 lt infin

(ii) the Lorentz space 119871119901119902

(Ω) where 1 lt 119901 lt infin and 1 le

119902 lt infin

(iii) the Lorentz-Karamata space 119871119901119902119887

(Ω) where 119887 isslowly varying 1 lt 119901 lt infin and 1 lt 119902 le infin

(iv) the Orlicz space 119871lowast

120601(Ω) for some 119873-function 120601 satisfy-

ing

inf119904gt0

119904120593 (119904)

120601 (119904)gt 1 (90)

where 120593 is the density function of 120601

To help with orientation let us give an overview ofthe remaining material in this section With respect toTheorem 27 note that we carried out the proof of Corollary 14in a high level of abstraction Hence we are left to verifythat the spaces under question are normed spaces (we willshow that they are even Banach function spaces) and theboundedness of the maximal operator

Concerning the proof Theorem 28 we proved the cor-responding part of Theorem 2 under the assumptions thatthe spaces under question are Banach function spaces andfurther satisfy the requirements listed in Properties 19 Theexistence of 1198831015840 is classical (Proposition 29) as is the fact that(AC) implies (A) and (NC) (Proposition 30) Hence we onlyneed to point to the literature where (AC) was verified

Proposition 29 Given a Banach function space there is aspace 119883

1015840 such that properties (B) (HI) (C) and (D) aresatisfied

Proof We refer the reader to Sections 11 and 12 in [13] or toSection 31 in [14]

The definition of absolutely continuous norm can befound as Definition 3111 in [14] and Definition 131 in [13]

Proposition 30 A Banach function space that has absolutelycontinuous norm satisfies (A) and (NC)

Proof The first statement follows fromTheorem 1313 in [13]and the second fromProposition 136 in the same source

51 General Statements

Proof of Theorem 27 (general statement) We basically havealready proven the general statement in Corollary 14

Proof of Theorem 28 (general statement) We gave the proofof the Lebesgue space case in such generality that thegeneral statement follows from the proof of Theorem 2 andPropositions 29 and 30

52 Lorentz-Karamata Spaces Lorentz-Karamata spaces aredetailed in [14] mainly Section 343

Proof of Theorem 27 (ii) and (iii) The part that is left toprove is the boundedness of the maximal operator which isguaranteed by Remark 3517 in [14]

Proof of Theorem 28 (ii) and (iii) By Theorem 3441 in [14]we see that 119883 is a rearrangement-invariant Banach space Acombination of Lemma 3439 in [14] together with Lemma3443 in [14] gives the absolute continuity (AC) of the normof1198831015840

12 Journal of Function Spaces

53 Orlicz Spaces The setting is the one of [17]

Lemma 31 If 120601 is an 119873-function satisfying the Δ2-condition

then the Orlicz space 119871lowast

120601has absolutely continuous norm

Proof Let 119891 isin 119871lowast

120601(Ω) We verify first that

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 lt infin (91)

Let 120576 gt 0 By the Δ2-condition there exists a constant 120581 =

120581(119891(120601)

+ 120576) such that

120601 ((1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576) 119904) le 120581120601 (119904) (92)

for all 119904 ge 0 Thus

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 = intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576(1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576))119889119909

le 120581intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576)119889119909 le 120581

(93)

Suppose that 119864 is a measurable set Then

10038171003817100381710038171198911205941198641003817100381710038171003817(120601)

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891120594119864

1003816100381610038161003816) 119889119909 le 1

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 1

(94)

Let us fix 120582 gt 0 Again using the Δ2-condition we have

120601(120582minus1

119904) le 120581(120582minus1

)120601(119904) Hence

intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 (95)

Example 133 in [13] tells us that 1198711 has absolutely continuousnorm Thus if 120594

119864119899converges 120583-almost everywhere to 0 then

intΩ

120601(|119891|)120594119864119899

converges to zero as well Thus we have for 119899

large enough that

120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119899119889119909 le 1 (96)

implying that 119891120594119864119899(120601)

le 120582 Since 120582 gt 0 was arbitrary theabsolute continuity of 119891 follows

Proof of Theorem 27 (iv) We conclude from Gallardorsquos resultTheorem 21 in [17] that it is sufficient that the complemen-tary 119873-function satisfies the Δ

2-condition in [0infin) Again

resorting to Gallardo this time to Proposition 14 in [17] weobtain the conclusion

Proof of Theorem 28 (iv) That 119883 is a Banach functionspace follows from Theorem 3416 in [14] To prove that119883

1015840 has absolutely continuous norm we note that inequal-ity (90) together with Proposition 14 in [17] ensuresthat the complementary 119873-function of 120601 satisfies the Δ

2-

condition Lemma 31 verifies that 1198831015840 has absolutely contin-uous norm

54 Rearrangement-Invariant Spaces The necessary defini-tions can be found in [13]

Proof of Theorem 27 (v) We are only left to show theboundedness of the maximal operator However this is thecontent of a result by Lorentz and Shimogaki for examplestated as Theorem 3517 in [13]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank Pilar Silvestre for hersuggestions of how to prove the existence of intermediatederivatives The research is supported by the Academy ofFinland The third author thanks the Department of Math-ematics and Systems Analysis at Aalto University for itshospitality

References

[1] B Bojarski ldquoRemarks on some geometric properties of Sobolevmappingsrdquo in Functional Analysis amp Related Topics (Sapporo1990) pp 65ndash76 World Scientific River Edge NJ USA 1991

[2] B Bojarski PHajlasz andP Strzelecki ldquoImproved119862119896120582 approxi-

mation of higher order Sobolev functions in normand capacityrdquoIndiana University Mathematics Journal vol 51 no 3 pp 507ndash540 2002

[3] B Bojarski ldquoPointwise characterization of Sobolev classesrdquoProceedings of the Steklov Institute of Mathematics vol 255 no1 pp 65ndash81 2006

[4] B Bojarski ldquoWhitneyrsquos jets for Sobolev functionsrdquo UkrainianMathematical Journal vol 59 no 3 pp 345ndash358 2007

[5] B Bojarski ldquoTaylor expansion and Sobolev spacesrdquo Bulletin ofthe Georgian National Academy of Sciences New Series vol 5no 2 pp 5ndash10 2011

[6] B Bojarski ldquoSobolev spaces and Lagrange interpolationrdquoGeorgian Academy of Sciences Proceedings of A RazmadzeMathematical Institute vol 158 pp 1ndash12 2012

[7] P Hajłasz ldquoSobolev spaces on an arbitrary metric spacerdquoPotential Analysis vol 5 no 4 pp 403ndash415 1996

[8] B Bojarski ldquoSobolev spaces and averaging Irdquo httparxivorgabs13085171

[9] R Kannan and C K Krueger Advanced Analysis on the RealLine Universitext Springer New York NY USA 1996

[10] R A DeVore and G G Lorentz Constructive Approximationvol 303 of Fundamental Principles of Mathematical SciencesSpringer Berlin Germany 1993

[11] L L Schumaker Spline Functions Basic Theory CambridgeMathematical Library CambridgeUniversity Press CambridgeUK 3rd edition 2007

[12] G H Hardy and J E Littlewood ldquoA maximal theorem withfunction-theoretic applicationsrdquo Acta Mathematica vol 54 no1 pp 81ndash116 1930

[13] C Bennett andR Sharpley Interpolation ofOperators vol 129 ofPure and Applied Mathematics Academic Press Boston MassUSA 1988

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

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Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 5

Proof We use the Peano representation Theorem 10 with119894 = 119895 = 0 Then using the bound of |119863

+119876

119898

119894(119909)| found in

Lemma 11

1003816100381610038161003816[1199100 119910119898] 1198911003816100381610038161003816 le

1

(119898 minus 1)int

119910119898

1199100

1003816100381610038161003816119863+119876

119898

119894(119909)

1003816100381610038161003816

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

le119862

119910119898

minus 1199100

int

119910119898

1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

(32)

For 119910 isin [1199100 119910

119898] we have

1003816100381610038161003816[1199100 119910119898] 1198911003816100381610038161003816

le119862

119910119898

minus 1199100

(int

119910

1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909 + int

119910119898

119910

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909)

(33)

and further1003816100381610038161003816[1199100 119910119898] 119891

1003816100381610038161003816

le 119862(minusint

119910

1199100

119910 minus 1199100

119910119898

minus 1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909

+minusint

119910119898

119910

119910119898

minus 119910

119910119898

minus 1199100

1003816100381610038161003816119863119898

1198911003816100381610038161003816 (119909) 119889119909)

(34)

The first factors in both integrals are bounded from above by1 We see that the integrals are bounded from above by thecorresponding maximal operators This gives the claim

The finite differences defined in (2) are up to a factordivided differences for equispaced nodes

Lemma 13 Let 119909 119910 be two points in a domain Ω sub R Let119898 isin N and 119909

119896= 119909

0+ ℎ119896 where ℎ = (119910 minus 119909)119898 Then each

119891 Ω rarr R satisfies

[1199090 119909

119898] 119891 =

1

ℎ119898119898Δ119898

119891 (119909 119910) (119909 = 1199090 119909

119898= 119910)

(35)

Proof We use induction For119898 = 1 we have

[1199090 119909

1] 119891 =

119891 (1199091) minus 119891 (119909

0)

1199091minus 119909

0

(36)

As ℎ = 1199091minus 119909

0 119909 = 119909

0 119910 = 119909

1 and 119898 = 1 we obtain the

claim Suppose that the claim is true for some119898 isin N Then

Δ119898+1

ℎ119891 (119909)

=

119898+1

sum

119895=0

(minus1)119898+1minus119895

(119898 + 1

119895)119891 (119909 + 119895ℎ)

=

119898+1

sum

119895=0

(minus1)119898+1minus119895

((119898

119895 minus 1) + (

119898

119895))119891 (119909 + 119895ℎ)

=

119898+1

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + (119895 + 1) ℎ)

minus

119898+1

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + 119895ℎ)

=

119898

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909

1+ 119895ℎ)

minus

119898

sum

119895=0

(minus1)119898minus119895

(119898

119895)119891 (119909 + 119895ℎ)

= ℎ119898

119898 ([1199091 119909

119898+1] 119891 minus [119909

0 119909

119898] 119891)

= ℎ119898

119898 (ℎ (119898 + 1)) [1199090 119909

119898+1] 119891

(37)

We can now formulate Theorem 12 in terms of finitedifferences

Corollary 14 (Corollary of Theorem 12) Let Ω sub R be adomain Suppose 119891 isin 119871

119898

1(Ω) for some 119898 isin N then there is

a constant 119862 = 119862(119898) such that

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816

le 1198621003816100381610038161003816119909 minus 119910

1003816100381610038161003816

119898

(119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

119909 119910 isin Ω

(38)

In particular we have the following special case Suppose 119883

and 119884 are normed spaces of functions such that the maximaloperator 119872 119883 rarr 119884 is bounded If 119891 isin 119882

119898

119883(Ω) for some

119898 ge 1 then

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (39)

for some function 119886119891isin 119884 depending on 119891

Proof We first use Lemma 13 and thenTheorem 12 with stepℎ = (119910 minus 119909)119898 There is some constant 119862 = 119862(119898) such that

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816

= ℎ119898

1198981003816100381610038161003816[1199090

119909119898] 119891

1003816100381610038161003816

le 119862ℎ119898

(119872 (1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

le1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

119898

(119862119872(1003816100381610038161003816119863

119898

1198911003816100381610038161003816) (119909) + 119862119872(

1003816100381610038161003816119863119898

1198911003816100381610038161003816) (119910))

(40)

Proof of Theorem 2 (Sobolev functions satisfy pointwiseinequality) We combine Corollary 14 with the boundednessof the maximal operator which is by now classical and can befound for example in Theorem 3310 in [13]

6 Journal of Function Spaces

3 Pointwise Inequality Implies Membershipin Sobolev Space

In this section our aim is to show that the pointwiseinequality (3) implies the existence of the highest orderderivative in the correct space We will take care of theintermediate derivatives in the next section In this sectionthe properties of the Lebesgue spaces play an important roleOur point of view is however to extract the properties ofthe Lebesgue spaces and base our proofs solely on theseproperties We hope that the disadvantage of being slightlymore abstract is later on compensatedwhenwe generalize theresults We start with some auxiliary results

Lemma 15 Let Ω sub R be a domain and 120593 isin C119899

0(Ω) that

is 120593 has compact support in Ω and is 119899-times continuouslydifferentiable Then

1003816100381610038161003816100381610038161003816100381610038161003816

[1199090 119909

119899] 120593 minus

120593(119899)

(119910)

119899

1003816100381610038161003816100381610038161003816100381610038161003816

997888rarr 0 (41)

uniformly inΩ as 1199090 119909

119899rarr 119910

Proof By Taylorrsquos formula for 119909 isin conv1199090 119909

119899 there is

120585119909isin conv119909

0 119909

119899 such that

120593 (119909) = 120593 (119910) + 1205931015840

(119910) (119909 minus 119910) + sdot sdot sdot + 120593(119899minus1)

(119910)(119909 minus 119910)

119899minus1

(119899 minus 1)

+1

119899120593(119899)

(120585119909) (119909 minus 119910)

119899

(42)

We apply [1199090 119909

119899] on both sides (with variable 119909) and use

the computation of the divided differences of polynomialscarried out in Lemma 6 to obtain

[1199090 119909

119899] 120593 (119909) =

1

119899120593(119899)

(120585119909) (43)

As 120593(119899) is uniformly continuous on compact intervals the

claim follows

Lemma 16 Assume Ω sub R is open and 119870 sub Ω is compactThen there is a neighborhood of119870 that is a subset of Ω

Proof For each 119909 isin 119870 we find a ball 119861(119909 119903119909) sub Ω Suppose

the statement of the lemma is falseThen for every 119899 isin N thereis 119911

119899isin Ω 119870 with dist(119911

119899 119870) lt 1119899 Thus for each 119899 isin N

there is 119910119899

isin 119870 with 119889(119911119899 119910

119899) lt 1119899 Since 119870 is compact

there is a subsequence (119910119899119896)119896of (119910

119899)119899that converges to some

119910 isin 119870 Now 119910 lies in one of the balls 119861(119909 119903119909) sub Ω For 119896 large

enough we have that 119910119899119896and 119911

119899119896are in 119861(119909 119903

119909) sub Ω This is

a contradiction

Lemma 17 SupposeΩ sub R is open and119870 sub Ω compactThenthere exists 120576 gt 0 such that

(Ω minus ℎ) cap 119870 = Ω cap 119870 (44)

for |ℎ| lt 120576

Proof We first show that any member of the left hand side iscontained in the right hand side this is true for any ℎ isin R If119911 is a member of the left hand side then it is in 119870 and since119870 is a subset of Ω it is also a member of the right hand sideTo prove the other inclusion note that by Lemma 16 there is120576 gt 0 such that the 120576-neighborhood 119870

120576of 119870 lies in Ω Now

suppose that 119911 is in the right hand side and |ℎ| lt 120576 Then for119909 = 119911+ℎ we have119909 isin 119870

120576sub Ω implying that 119911 = 119909minusℎ isin Ωminusℎ

That 119911 isin 119870 is clear This gives the proof

Next we state basically Definition 311 in [14] Readersfollowing [13] might want to have a look at Definitions 111and 113 in their book To gain familiarity with the conceptof Banach function norm the reader may want to verify thatLebesgue spaces are examples of Banach function spaces Aswe are here only interested in Lebesgue measure we adaptthe notion to this case The set M+ denotes the cone ofmeasurable functions whose values lie in [0infin]

Definition 18 (Banach function norm Banach functionspace) Assume that 119877 sub R A Banach function norm on 119877

is a map 120588 from M+

(119877) to [0infin] such that for all 119891 119892 and119891119899(119899 isin N) in M+

(119877) all scalars 120582 ge 0 and all measurablesubsets 119864 of 119877 the following are true where 120583 is the Lebesguemeasure

(P1) 120588(119891) = 0 if and only if 119891 = 0 120583-almost everywhere120588(120582119891) = 120582120588(119891) 120588(119891 + 119892) le 120588(119891) + 120588(119892)

(P2) if 0 le 119892 le 119891 120583-almost everywhere then 120588(119892) le 120588(119891)(P3) if 0 le 119891

119899uarr 119891 120583-almost everywhere we have 120588(119891

119899) uarr

120588(119891)(P4) if 120583(119864) lt infin then 120588(120594

119864) lt infin

(P5) if120583(119864) lt infin then there is a constant119862 = 119862(119864 120588) lt infin

such that int119864

119891119889119909 le 119862120588(119891)

Given such a function norm 120588 the set 119883 = 119883(120588) ofall extended measurable functions 119891 (identifying functionsequal 120583-almost everywhere) such that 120588(|119891|) lt infin is called aBanach function space (BFS for short) and we define

10038171003817100381710038171198911003817100381710038171003817119883

= 120588 (1003816100381610038161003816119891

1003816100381610038161003816) 119891 isin 119883 (45)

We leave it to the reader to verify that Lebesgue spacessatisfy the following properties (1198831015840 can be chosen as the dualspace)

Properties 19 Given a Banach function space 119883 we requirethe existence of a Banach function space119883

1015840 such that

(B) 119883 contains each essentially bounded function whosesupport has finite measure

(HI) Holderrsquos inequality int |119891119892|119889119909 le 119891119883119892

1198831015840

(C) 1198831015840 is complete

(D) 119892119883

= supintΩ

|119892|ℎ ℎ isin M+

ℎ1198831015840 le 1

(AC) absolutely continuous norm if (120594119864119899)119899is a sequence of

characteristic functions converging pointwise almosteverywhere to 0 then 119891120594

1198641198991198831015840 converges to zero for

all 119891 isin 1198831015840

Journal of Function Spaces 7

(A) each function 119891 isin 1198831015840 can be approximated by simple

functions (in1198831015840-norm)

(NC) if 119891 isin 1198831015840 and ℎ

119899are step functions with 0 le ℎ

119899uarr |119891|

then1003817100381710038171003817119891 minus ℎ

119899sgn (119891)

10038171003817100381710038171198831015840997888rarr 0 as 119899 997888rarr infin (46)

Lemma 20 Suppose Ω sub R is open and 119883 is a Banachfunction space fulfilling the items listed in Properties 19 Assumethat 119891 Ω rarr R is in 119883 We let 120582

119891 119883

1015840

cap C119903+1

0(Ω) rarr R be

defined as

120582119891(120593) = int

Ω

119891120593(119903)

119889119909 (47)

If1003816100381610038161003816Δ

119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)]

for almost every 119909 119910 isin Ω

(48)

for some 119886119891

isin 119883 then 120582119891has a continuous extension to 119883

1015840

with norm bounded from above by 2119903119903

119886119891119883

Proof Our goal is to show1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816le 2119903

11990310038171003817100381710038171003817119886119891

10038171003817100381710038171003817119883

100381710038171003817100381712059310038171003817100381710038171198831015840

(49)

because 120582119891extends then by the theorem of Hahn-Banach to

a bounded functional on1198831015840 with norm bounded from above

by 2119903119903

119886119891119883 Without loss of generality we may assume that

119886119891is nonnegative If 119891 were be smooth we could just use

integration by parts and use as bound the limiting case of thepointwise inequality (48) As 119891 is not necessarily smooth wehave to operate on the level of finite differences

Let us choose 120593 isin C0(Ω) and denote its support by 119870

Lemma 17 ensures the existence of some positive 120576 gt 0 suchthat (Ωminus119895ℎ)cap119870 = Ωcap119870 for 0 le 119895 le 119903 as long as |ℎ| lt 120576119903Weassume in the following that ℎ has been chosen accordingly

The following integrals exist by Holderrsquos inequalitybecause 120593 and thus Δ

119903

minusℎ120593(119909) are bounded with compact

support and 119891 is integrable Consider

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

119891 (119909) (minusℎ)minus119903

Δ119903

minusℎ120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(50)

We compute the following integral

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

= intΩ

119891 (119909)

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119909) 120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωminus119895ℎ

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

(Ωminus119895ℎ)cap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

(51)

We have chosen 120576 gt 0 such that (Ω minus 119895ℎ) cap 119870 = Ω cap 119870 Wecontinue

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωcap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

(52)

With the aid of (52) we can now rewrite (50) as

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

(minusℎ)minus119903

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(53)

Inequality (48) leads to

1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816

le 119903119903

intΩ

(119886119891(119909) + 119886

119891(119909 + 119903ℎ))

1003816100381610038161003816120593 (119909)1003816100381610038161003816 119889119909

+ intΩ

1003816100381610038161003816119891 (119909)1003816100381610038161003816

10038161003816100381610038161003816120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)

10038161003816100381610038161003816119889119909

(54)

Using Holderrsquos inequality (HI) for the first summand weare left to show that the second summand tends to 0 asℎ approaches 0 We note first that since the second factorhas support in a set of finite measures we can replace theintegration domain by a set of finite measuresThen we applya (1198711

119871infin

)-Holder inequality and the uniform convergence ofthe divided difference to 120593

(119903)

(119909)119903 established in Lemmas 13and 15

After having verified that 120582119891has a continuous extension

to 1198831015840 we want to show that this extension has a representa-

tion as integral

8 Journal of Function Spaces

Lemma 21 Assume Ω sub R is open Suppose the propertieslisted in Properties 19 are met Let 120582

119891be defined as in (47) and

bounded on 1198831015840 Then there exists 119892 isin 119883 with 119892

119883le 120582

119891

such that

120582119891(120593) = (minus1)

119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (55)

for all 120593 isin 1198831015840 especially for all 120593 isin Cinfin

0(Ω)

Proof The proof is heavily influenced by the proof of Theo-rem 141 in [13]We first want to find a signedmeasure ] suchthat

120582119891(120593) = int

Ω

120593119889] (56)

for each 120593 isin 1198831015840 Then we want to apply the Lebesgue-

Radon-Nikodym theorem to replace 119889] by (minus1)119903

119892(119909)119889119909 forsome integrable function 119892 for which we then show that it iscontained in119883

We start by assuming that the measure of Ω is finite Foreach 120583-measurable set 119860 we set

] (119860) = 120582119891(120594

119860) (57)

We start by proving the 120590-additivity of ] (for the 120583-measurable sets) In the first step we just show the additivityBy induction it is enough to consider only two sets

Let 1198601and 119860

2be two disjoint 120583-measurable sets Then

] (1198601cup 119860

2) = 120582

119891(120594

1198601cup1198602) = 120582

119891(120594

1198601+ 120594

1198602)

= 120582119891(120594

1198601) + 120582

119891(120594

1198602) = ] (119860

1) + ] (119860

2)

(58)

We move to the verification of the 120590-additivity Let119860

1 119860

2 be countablymany pairwise disjoint120583-measurable

sets We first want to show thatsum119894120594119860119894

converges in1198831015840

We let 119864119896= cup

119896

119894=1119860

119894and 119864 = cup

infin

119894=1119860

119894 Then since (119864

119896)119896is

increasing

120583 (119864) = 120583(

infin

⋃

119896=1

119864119896) = lim

119896rarrinfin

120583 (119864119896) (59)

As 119864 has finite measure given 120576 gt 0 there exists 1198960isin N such

that

(1 minus 120576) 120583 (119864) lt 120583 (1198641198960) (60)

By the disjointness of the sets 119860119894 we see that

120583 (119864) = 120583 (1198641198960) + 120583(

infin

⋃

119894=1198960+1

119860119894) (61)

It follows that

120583(

infin

⋃

119894=1198960+1

119860119894) = 120583 (119864) minus 120583 (119864

1198960) lt 120583 (119864) minus (1 minus 120576) 120583 (119864)

= 120576120583 (119864)

(62)

We have for119898 lt 119899

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=1

120594119860119894

minus

119898

sum

119894=1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

=

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=119898+1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

le10038171003817100381710038171003817120594cupinfin119894=119898+1

119860119894

100381710038171003817100381710038171198831015840 (63)

By the considerations before and the fact that sdot 1198831015840 is

absolutely continuous as stipulated in (AC) we see thatsum

119899

119894=1120594119860119894

is a Cauchy sequence Now as 1198831015840 is complete by

(C) the Cauchy sequence converges in 1198831015840 We have by the

continuity of 120582119891

] (cup119894119860

119894) = 120582

119891(120594

cup119894119860119894) = 120582

119891(

infin

sum

119894=1

120594119860119894)

= lim119899rarrinfin

120582119891(

119899

sum

119894=1

120594119860119894) = lim

119899rarrinfin

119899

sum

119894=1

120582119891(120594

119860119894)

= lim119899rarrinfin

119899

sum

119894=1

] (119860119894) =

infin

sum

119894=1

] (119860119894)

(64)

Thus ] is 120590-additiveIn order to apply the theorem of Lebesgue-Radon-

Nikodym see for example Section 69 in [15] and conclud-ing the existence of 119892 isin 119871

1

(Ω) such that

] (119860) = int119860

119892119889119909 119860 120583-measurable (65)

we need to show that ] is absolutely continuous with respectto 120583 To do so assume that 120583(119873) = 0 Then ](119873) = 120582

119891(120594

119873)

By the absolute continuity of the normof1198831015840 required in (AC)we see that 120594

1198731198831015840 = 0 and therefore ](119873) = 0 Replacing 119892

by minus119892 if necessary we obtain

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909 (66)

Now we assume that Ω does not necessarily have finitemeasureWewriteΩ = cup

119899Ω

119899as union of open sets with finite

measure such that Ω119899sub Ω

119899+1 By the considerations above

we find for each 119899 isin N a 119892119899such that

120582119891(120594

119860) = (minus1)

119903

intΩ

119892119899(119909) 120594

119860(119909) 119889119909 (67)

for each measurable set 119860 sub Ω119899 It is easily seen that 119892

119899

and 119892119899+1

agree almost everywhere on Ω119899 This gives rise to

a function 119892 satisfying

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909

119860 sub

119873

⋃

119899=1

Ω119899for some 119873

(68)

By linearity we have for each simple function 119904

120582119891(119904) = (minus1)

119903

intΩ

119892 (119909) 119904 (119909) 119889119909 (69)

To show that 119892 isin 119883 we want to use property (D)

Journal of Function Spaces 9

Let ℎ be a nonnegative simple function with supportin some Ω

119899 Then ℎ sgn(119892) is also a simple function with

support inΩ119899 It is a finite linear combination of characteristic

functions of sets in Ω119899 Hence

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = intℎ sdot sgn (119892) 119892 119889119909 = 120582119891(ℎ sgn (119892)) (70)

Using the boundedness of 120582119891 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (71)

If ℎ is an arbitrary function in1198831015840 thenwemay construct a

sequence (ℎ119899)infin

119899=1of simple functions each ℎ

119899having support

in Ω119899 such that 0 le ℎ

119899uarr |ℎ| 120583-almost everywhere By the

monotone convergence theorem we have by (71) that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = lim119899rarrinfin

int1003816100381610038161003816ℎ119899119892

1003816100381610038161003816 119889119909 le lim sup119899rarrinfin

10038171003817100381710038171003817120582119891

10038171003817100381710038171003817

1003817100381710038171003817ℎ11989910038171003817100381710038171198831015840

(72)

By (P3) in Definition 18 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (73)

and Property (D) gives that 119892 isin 119883 In the last step we want toupgrade the integral representation (69) to hold for arbitraryfunctions in 120593 isin 119883

1015840 As before we approximate 120593 by simplefunctions ℎ

119899with support contained in Ω

119899and 0 le ℎ

119899uarr |120593|

in 120583-almost everywhere point Define 119904119899

= ℎ119899sgn(120593) By

(NC) 119904119899converges to 120593 in119883

1015840 and as both 120582119891and ldquoits integral

representationrdquo are continuous and agree on the set of stepfunctions they agree on119883

1015840

Having shown the existence of the weak derivativesof highest order we are left to show the existence of theintermediate derivatives which we do in Proposition 22 inthe next section To keep the flow going we give here theproof that the pointwise inequality implies membership inthe corresponding Sobolev space under the assumption thatwe already know that Proposition 22 holds

Proof of Theorem 2 (pointwise inequality implies membershipin Sobolev space) We combine Lemmas 20 and 21 and applyProposition 22

4 Equivalent Definitions of Sobolev Spaces

The aim of this section is to prove the following equivalence

Proposition 22 Let Ω sub R be a domain with finite Lebesguemeasure inR Suppose119883 is a Banach function space satisfying(B) and (D) Assume that 119898 isin N Then the following setscoincide

(a) 119860 = 119891 isin 119883 there exists 119865 isin 119883 where 119865 agrees almosteverywherewith119891 and hasweak derivatives up to order119898 in119883 where the derivatives of order 119903 0 le 119903 le 119898minus1are absolutely continuous

(b) 119861 = 119891 isin 119883 119891(119896)

isin 119883 119896 isin 0 119898 where 119891(119896) arethe weak derivatives of 119891

(c) 119862 = 119891 isin 119883 119891(119898)

isin 119883 where 119891(119898) is the weak

derivative of 119891 of order 119898

We see that 119860 sub 119861 sub 119862 We now show that 119862 sub 119860Let us give an outline of the proof Given 119892 Ω rarr R

(we will choose 119892 = 119891(119898)) by integrating we construct in

Lemma 23 an absolutely continuous function 119866119898together

with its weak derivatives 1198660 119866

119898minus1 which are absolutely

continuous and 119866(119898)

119898= 119892 However 119866

119898and 119891 might

not agree almost everywhere We verify however in theproof of Theorem 26 (with the help of Theorem 24) thatwe find a polynomial 119875 such that 119866

119898+ 119875 agrees with 119865

almost everywhere and shares with 119866119898the other required

properties We conclude the proof of Theorem 26 by settlingthe membership questions with the aid of Proposition 25

Lemma 23 Assume Ω sub R is a domain We further assumethat 119892 Ω rarr R is in 119871

1

loc(Ω) One fixes 119909lowast

isin Ω and defines

1198660(119909

0) = 119892 (119909

0)

119866119896(119909

119896) = int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119896 isin N

(74)

where the integrals are Lebesgue integrals and we use thenotation

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

=

int[119909lowast119909119896]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909lowast

le 119909119896

minusint[119909119896 119909lowast]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909119896lt 119909

lowast

(75)

Then

(a) 119866119896is locally absolutely continuous for all 119896 isin N

especially locally integrable(b) if 120593 isin Cinfin

0(Ω) and 0 le 119903 le 119896 one has

intΩ

119866119896120593(119903)

= (minus1)119903

intΩ

119866119896minus119903

120593 (76)

Proof Let us prove (a) by induction Assume 119896 = 1 Note that1198661(119909

lowast

) = 0 Thus we have for 119909lowast

le 1199091

1198661(119909

1) minus 119866

1(119909

lowast

) = 1198661(119909

1) = int

1199091

119909lowast

119892 (1199090) 119889119909

0 (77)

As 119892 isin 1198711

([119909lowast

1199091]) we see that 119866

1is absolutely continuous

The case 1199091lt 119909

lowast is treated similarlyAssume now that the claim is true for some 119896 Now

119866119896+1

(119909119896+1

) is the integral of a continuous function and thuslocally absolutely continuous

Let us have a look at (b) Note that the statement is true for119896 = 0 thus we may assume that 119896 ge 1 We make an inductionover 119903 The case 119903 = 0 is clear Let us choose an interval[119886 119887] sub Ω containing a neighborhood of the support of 120593 As119866119896and 120593

(119903) are both absolutely continuous in [119886 119887] we have

10 Journal of Function Spaces

by the integration by part formula for absolutely continuousfunctions (see for exampleTheorem7147 in [9]) and the factthat 120593(119903)

(119886) = 120593(119903)

(119887) = 0

intΩ

119866119896120593(119903+1)

= int[119886119887]

119866119896120593(119903+1)

= (minus1) int[119886119887]

1198661015840

119896120593(119903)

(78)

Noting that 1198661015840

119896= 119866

119896minus1almost everywhere we see that

intΩ

119866119896120593(119903+1)

= (minus1) int[119886119887]

119866119896minus1

120593(119903)

= (minus1) intΩ

119866119896minus1

120593(119903)

(79)

Using the induction hypothesis the claim follows

We cite Theorem 253 in [10] without giving its proof

Theorem 24 Let 119860 = [119886 119887] R or R+and 119903 ge 0 If the

function 119891 isin 1198711

loc(119860) satisfies

int119860

119891120593(119903)

119889119909 = 0 (80)

for all 120593 isin Cinfin

0(119860) then there is a polynomial 119875 of degree lt 119903

such that 119891 = 119875 almost everywhere on 119860

To show that 119866119896is in 119883 we need to verify that (if there

is danger of confusion with respect to which variable we takethe norm we use notations as119883(119909

119896))

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

lt infin (81)

under the assumption that 119866119896minus1

is in 119883 Actually bounding119866119896from above by its absolute value this means that we have

first to take the1198711-norm and then the norm in119883We howeverwould like to switch the norms

We adapt Proposition 21 in [16] to our setting

Proposition 25 Let 120588 be a Banach function norm satisfyingproperty (D) If 119891 ΩtimesΩ rarr R is a measurable function andone has 119891

119909and 119891

119910 functions onΩ such that 119891119909(119910) = 119891(119909 119910) =

119891119910

(119909) then

120588 (1003817100381710038171003817119891119909

10038171003817100381710038171) le

1003817100381710038171003817120588 (119891119910

)10038171003817100381710038171

(82)

The cases 119883 isin 1198711

119871infin

are easy to prove If 119883 = 119871119901 for

some 1 lt 119901 lt infin then the result can be proven by usingFubinirsquos theorem together with property (D) For the generalresults we refer to Scheprsquos proof

The next result builds uponTheorem 254 in [10]

Theorem 26 Let Ω sub R be a domain and let 119903 ge 1 Suppose119883 is a Banach function space satisfying (B) and (D) If 119891 islocally in 119883 and has a generalized 119903th derivative 119892 locally in119883 then there is 119865 agreeing almost everywhere with 119891 havingabsolutely continuous intermediate weak derivatives and theweak derivative of order 119903 agrees almost everywhere with 119892Furthermore the derivatives of 119865 are locally in 119883 If Ω hasfinite measure one obtains the result with ldquolocallyrdquo replaced byldquogloballyrdquo

We thank Pilar Silvestre for pointing out how to prove themembership of the intermediate derivatives in119883

Proof Choose 119909lowast

isin Ω and let 119866119896be defined as in Lemma 23

Assume 120593 isin Cinfin

0(Ω) Then using the fact that 119892 is a weak

derivative of 119891 of order 119903 and (b) in Lemma 23

intΩ

(119891 minus 119866119903) 120593

(119903)

119889119909 = intΩ

119891120593(119903)

119889119909 minus intΩ

119866119903120593(119903)

119889119909

= (minus1)119903

intΩ

119892120593119889119909 minus (minus1)119903

intΩ

119892120593119889119909 = 0

(83)

UsingTheorem 24 we see that119891minus119866119903= 119875 almost everywhere

where119875 is a polynomial of degreelt 119903Thus119891 = 119875+119866119903almost

everywhere and for 120593 isin Cinfin

0(Ω) and 0 le 119898 le 119903

intΩ

119891120593(119898)

119889119909 = intΩ

(119875 + 119866119903) 120593

(119898)

= (minus1)119898

intΩ

(119866119903minus119898

+ 119875(119898)

) 120593 119889119909

(84)

Thus119866119903minus119898

+119875(119898) is the119898th generalized derivative of119891 As it is

locally absolutely continuous (if119898 lt 119903) it lies locally in119883 Inthe case 119898 = 119903 the same conclusion holds by the assumptionon 119892

Moreover there exists 119865 for example 119865 = 119875 + 119866119903 that

agrees with 119891 almost everywhere such that its generalizedderivatives up to order 119903 minus 1 are locally absolutely continuousand the 119903th generalized derivative equals 119892 almost every-where

As 119892 is locally in 119883 and the other derivatives are locallyabsolutely continuous they are locally in 119871

infin and thus locallyin119883 by (B)

Suppose now that Ω has additionally finite measureAssume that 119891 and 119892 are in119883 globally We want to show thatthe intermediate derivatives are in 119883 as well

Note in our setting that by (P5) in Definition 18membership in 119883 implies membership in 119871

1 We proceedby induction Suppose that 119866

119896minus1isin 119883 We will apply

Proposition 25 We set denoting by conv119909lowast

119909119896 the closed

interval with endpoints 119909lowast and 119909119896

119891 (119909119896minus1

119909119896) = 119866

119896minus1(119909

119896minus1) 120594conv119909lowast 119909119896 (119909119896minus1

) (85)

Then1003817100381710038171003817119866119896

(119909119896)1003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119891(119909119896minus1

119909119896)119889119909

119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)10038171003817100381710038171198711(119909119896minus1)

10038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)

le10038171003817100381710038171003817

1003816100381610038161003816119866119896minus1(119909

119896minus1)1003816100381610038161003816

1003817100381710038171003817120594Ω1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)lt infin

(86)

Journal of Function Spaces 11

5 Banach Function Space Setting

Here our main focus is a more advanced readership thatlikely knows already Banach function spaces Therefore weadopt a much briefer style pointing merely to the literaturewhere the corresponding definitions and results can be foundWe start with giving generalized versions of the necessity andsufficiency of Theorem 2

We list the results and prove them simultaneously bysplitting the proofs into parts according to the functionspaces

Theorem 27 Assume Ω sub R is a domain 119883 is a normedspace and 119898 isin N If the maximal function 119872 119883 rarr 119883 isbounded then there is a constant 119862 = 119862(119883119898) such that foreach continuous 119891 isin 119882

119898

119883there is 119886

119891isin 119883with 119886

119891 le 119862119891

119882119898119883

and

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (87)

In particular such 119886119891exists if

(i) 1 lt 119901 le infin and 119883 is the Lebesgue space 119871119901

(Ω)

(ii) 1 lt 119901 lt infin 1 le 119902 le infin and 119883 is the Lorentz space119871119901119902

(Ω)

(iii) 1 lt 119901 lt infin 1 le 119902 le infin 119887 is slowly varying and 119883 isthe Lorentz-Karamata space 119871

119901119902119887(Ω)

(iv) 120601 is an 119873-function with inf119904gt0

(119904120593(119904)120601(119904)) gt 1 where120593 is the density function of 120601 and 119883 is the Orlicz space119871lowast

120601(Ω)

(v) Ω = R and 119883 is rearrangement-invariant with upperindex 120572

119883lt 1

Theorem 28 Let Ω sub R be open Suppose 119883 is a Banachfunction space such that (AC) holds Assume further that thefunction 119891 Ω rarr R is in 119883 If for some 119903 isin N

0 the following

inequality holds

1003816100381610038161003816Δ119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)] (88)

for some 119886119891isin 119883 then there exists119892 isin 119883with 119892

119883le 2119903

119903

119886119891119883

such that

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909 = (minus1)119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (89)

for all 120593 isin Cinfin

0(Ω)

In particular the result holds if119883 is

(i) the Lebesgue space 119871119901

(Ω) for some 1 lt 119901 lt infin

(ii) the Lorentz space 119871119901119902

(Ω) where 1 lt 119901 lt infin and 1 le

119902 lt infin

(iii) the Lorentz-Karamata space 119871119901119902119887

(Ω) where 119887 isslowly varying 1 lt 119901 lt infin and 1 lt 119902 le infin

(iv) the Orlicz space 119871lowast

120601(Ω) for some 119873-function 120601 satisfy-

ing

inf119904gt0

119904120593 (119904)

120601 (119904)gt 1 (90)

where 120593 is the density function of 120601

To help with orientation let us give an overview ofthe remaining material in this section With respect toTheorem 27 note that we carried out the proof of Corollary 14in a high level of abstraction Hence we are left to verifythat the spaces under question are normed spaces (we willshow that they are even Banach function spaces) and theboundedness of the maximal operator

Concerning the proof Theorem 28 we proved the cor-responding part of Theorem 2 under the assumptions thatthe spaces under question are Banach function spaces andfurther satisfy the requirements listed in Properties 19 Theexistence of 1198831015840 is classical (Proposition 29) as is the fact that(AC) implies (A) and (NC) (Proposition 30) Hence we onlyneed to point to the literature where (AC) was verified

Proposition 29 Given a Banach function space there is aspace 119883

1015840 such that properties (B) (HI) (C) and (D) aresatisfied

Proof We refer the reader to Sections 11 and 12 in [13] or toSection 31 in [14]

The definition of absolutely continuous norm can befound as Definition 3111 in [14] and Definition 131 in [13]

Proposition 30 A Banach function space that has absolutelycontinuous norm satisfies (A) and (NC)

Proof The first statement follows fromTheorem 1313 in [13]and the second fromProposition 136 in the same source

51 General Statements

Proof of Theorem 27 (general statement) We basically havealready proven the general statement in Corollary 14

Proof of Theorem 28 (general statement) We gave the proofof the Lebesgue space case in such generality that thegeneral statement follows from the proof of Theorem 2 andPropositions 29 and 30

52 Lorentz-Karamata Spaces Lorentz-Karamata spaces aredetailed in [14] mainly Section 343

Proof of Theorem 27 (ii) and (iii) The part that is left toprove is the boundedness of the maximal operator which isguaranteed by Remark 3517 in [14]

Proof of Theorem 28 (ii) and (iii) By Theorem 3441 in [14]we see that 119883 is a rearrangement-invariant Banach space Acombination of Lemma 3439 in [14] together with Lemma3443 in [14] gives the absolute continuity (AC) of the normof1198831015840

12 Journal of Function Spaces

53 Orlicz Spaces The setting is the one of [17]

Lemma 31 If 120601 is an 119873-function satisfying the Δ2-condition

then the Orlicz space 119871lowast

120601has absolutely continuous norm

Proof Let 119891 isin 119871lowast

120601(Ω) We verify first that

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 lt infin (91)

Let 120576 gt 0 By the Δ2-condition there exists a constant 120581 =

120581(119891(120601)

+ 120576) such that

120601 ((1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576) 119904) le 120581120601 (119904) (92)

for all 119904 ge 0 Thus

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 = intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576(1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576))119889119909

le 120581intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576)119889119909 le 120581

(93)

Suppose that 119864 is a measurable set Then

10038171003817100381710038171198911205941198641003817100381710038171003817(120601)

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891120594119864

1003816100381610038161003816) 119889119909 le 1

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 1

(94)

Let us fix 120582 gt 0 Again using the Δ2-condition we have

120601(120582minus1

119904) le 120581(120582minus1

)120601(119904) Hence

intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 (95)

Example 133 in [13] tells us that 1198711 has absolutely continuousnorm Thus if 120594

119864119899converges 120583-almost everywhere to 0 then

intΩ

120601(|119891|)120594119864119899

converges to zero as well Thus we have for 119899

large enough that

120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119899119889119909 le 1 (96)

implying that 119891120594119864119899(120601)

le 120582 Since 120582 gt 0 was arbitrary theabsolute continuity of 119891 follows

Proof of Theorem 27 (iv) We conclude from Gallardorsquos resultTheorem 21 in [17] that it is sufficient that the complemen-tary 119873-function satisfies the Δ

2-condition in [0infin) Again

resorting to Gallardo this time to Proposition 14 in [17] weobtain the conclusion

Proof of Theorem 28 (iv) That 119883 is a Banach functionspace follows from Theorem 3416 in [14] To prove that119883

1015840 has absolutely continuous norm we note that inequal-ity (90) together with Proposition 14 in [17] ensuresthat the complementary 119873-function of 120601 satisfies the Δ

2-

condition Lemma 31 verifies that 1198831015840 has absolutely contin-uous norm

54 Rearrangement-Invariant Spaces The necessary defini-tions can be found in [13]

Proof of Theorem 27 (v) We are only left to show theboundedness of the maximal operator However this is thecontent of a result by Lorentz and Shimogaki for examplestated as Theorem 3517 in [13]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank Pilar Silvestre for hersuggestions of how to prove the existence of intermediatederivatives The research is supported by the Academy ofFinland The third author thanks the Department of Math-ematics and Systems Analysis at Aalto University for itshospitality

References

[1] B Bojarski ldquoRemarks on some geometric properties of Sobolevmappingsrdquo in Functional Analysis amp Related Topics (Sapporo1990) pp 65ndash76 World Scientific River Edge NJ USA 1991

[2] B Bojarski PHajlasz andP Strzelecki ldquoImproved119862119896120582 approxi-

mation of higher order Sobolev functions in normand capacityrdquoIndiana University Mathematics Journal vol 51 no 3 pp 507ndash540 2002

[3] B Bojarski ldquoPointwise characterization of Sobolev classesrdquoProceedings of the Steklov Institute of Mathematics vol 255 no1 pp 65ndash81 2006

[4] B Bojarski ldquoWhitneyrsquos jets for Sobolev functionsrdquo UkrainianMathematical Journal vol 59 no 3 pp 345ndash358 2007

[5] B Bojarski ldquoTaylor expansion and Sobolev spacesrdquo Bulletin ofthe Georgian National Academy of Sciences New Series vol 5no 2 pp 5ndash10 2011

[6] B Bojarski ldquoSobolev spaces and Lagrange interpolationrdquoGeorgian Academy of Sciences Proceedings of A RazmadzeMathematical Institute vol 158 pp 1ndash12 2012

[7] P Hajłasz ldquoSobolev spaces on an arbitrary metric spacerdquoPotential Analysis vol 5 no 4 pp 403ndash415 1996

[8] B Bojarski ldquoSobolev spaces and averaging Irdquo httparxivorgabs13085171

[9] R Kannan and C K Krueger Advanced Analysis on the RealLine Universitext Springer New York NY USA 1996

[10] R A DeVore and G G Lorentz Constructive Approximationvol 303 of Fundamental Principles of Mathematical SciencesSpringer Berlin Germany 1993

[11] L L Schumaker Spline Functions Basic Theory CambridgeMathematical Library CambridgeUniversity Press CambridgeUK 3rd edition 2007

[12] G H Hardy and J E Littlewood ldquoA maximal theorem withfunction-theoretic applicationsrdquo Acta Mathematica vol 54 no1 pp 81ndash116 1930

[13] C Bennett andR Sharpley Interpolation ofOperators vol 129 ofPure and Applied Mathematics Academic Press Boston MassUSA 1988

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Journal of Function Spaces

3 Pointwise Inequality Implies Membershipin Sobolev Space

In this section our aim is to show that the pointwiseinequality (3) implies the existence of the highest orderderivative in the correct space We will take care of theintermediate derivatives in the next section In this sectionthe properties of the Lebesgue spaces play an important roleOur point of view is however to extract the properties ofthe Lebesgue spaces and base our proofs solely on theseproperties We hope that the disadvantage of being slightlymore abstract is later on compensatedwhenwe generalize theresults We start with some auxiliary results

Lemma 15 Let Ω sub R be a domain and 120593 isin C119899

0(Ω) that

is 120593 has compact support in Ω and is 119899-times continuouslydifferentiable Then

1003816100381610038161003816100381610038161003816100381610038161003816

[1199090 119909

119899] 120593 minus

120593(119899)

(119910)

119899

1003816100381610038161003816100381610038161003816100381610038161003816

997888rarr 0 (41)

uniformly inΩ as 1199090 119909

119899rarr 119910

Proof By Taylorrsquos formula for 119909 isin conv1199090 119909

119899 there is

120585119909isin conv119909

0 119909

119899 such that

120593 (119909) = 120593 (119910) + 1205931015840

(119910) (119909 minus 119910) + sdot sdot sdot + 120593(119899minus1)

(119910)(119909 minus 119910)

119899minus1

(119899 minus 1)

+1

119899120593(119899)

(120585119909) (119909 minus 119910)

119899

(42)

We apply [1199090 119909

119899] on both sides (with variable 119909) and use

the computation of the divided differences of polynomialscarried out in Lemma 6 to obtain

[1199090 119909

119899] 120593 (119909) =

1

119899120593(119899)

(120585119909) (43)

As 120593(119899) is uniformly continuous on compact intervals the

claim follows

Lemma 16 Assume Ω sub R is open and 119870 sub Ω is compactThen there is a neighborhood of119870 that is a subset of Ω

Proof For each 119909 isin 119870 we find a ball 119861(119909 119903119909) sub Ω Suppose

the statement of the lemma is falseThen for every 119899 isin N thereis 119911

119899isin Ω 119870 with dist(119911

119899 119870) lt 1119899 Thus for each 119899 isin N

there is 119910119899

isin 119870 with 119889(119911119899 119910

119899) lt 1119899 Since 119870 is compact

there is a subsequence (119910119899119896)119896of (119910

119899)119899that converges to some

119910 isin 119870 Now 119910 lies in one of the balls 119861(119909 119903119909) sub Ω For 119896 large

enough we have that 119910119899119896and 119911

119899119896are in 119861(119909 119903

119909) sub Ω This is

a contradiction

Lemma 17 SupposeΩ sub R is open and119870 sub Ω compactThenthere exists 120576 gt 0 such that

(Ω minus ℎ) cap 119870 = Ω cap 119870 (44)

for |ℎ| lt 120576

Proof We first show that any member of the left hand side iscontained in the right hand side this is true for any ℎ isin R If119911 is a member of the left hand side then it is in 119870 and since119870 is a subset of Ω it is also a member of the right hand sideTo prove the other inclusion note that by Lemma 16 there is120576 gt 0 such that the 120576-neighborhood 119870

120576of 119870 lies in Ω Now

suppose that 119911 is in the right hand side and |ℎ| lt 120576 Then for119909 = 119911+ℎ we have119909 isin 119870

120576sub Ω implying that 119911 = 119909minusℎ isin Ωminusℎ

That 119911 isin 119870 is clear This gives the proof

Next we state basically Definition 311 in [14] Readersfollowing [13] might want to have a look at Definitions 111and 113 in their book To gain familiarity with the conceptof Banach function norm the reader may want to verify thatLebesgue spaces are examples of Banach function spaces Aswe are here only interested in Lebesgue measure we adaptthe notion to this case The set M+ denotes the cone ofmeasurable functions whose values lie in [0infin]

Definition 18 (Banach function norm Banach functionspace) Assume that 119877 sub R A Banach function norm on 119877

is a map 120588 from M+

(119877) to [0infin] such that for all 119891 119892 and119891119899(119899 isin N) in M+

(119877) all scalars 120582 ge 0 and all measurablesubsets 119864 of 119877 the following are true where 120583 is the Lebesguemeasure

(P1) 120588(119891) = 0 if and only if 119891 = 0 120583-almost everywhere120588(120582119891) = 120582120588(119891) 120588(119891 + 119892) le 120588(119891) + 120588(119892)

(P2) if 0 le 119892 le 119891 120583-almost everywhere then 120588(119892) le 120588(119891)(P3) if 0 le 119891

119899uarr 119891 120583-almost everywhere we have 120588(119891

119899) uarr

120588(119891)(P4) if 120583(119864) lt infin then 120588(120594

119864) lt infin

(P5) if120583(119864) lt infin then there is a constant119862 = 119862(119864 120588) lt infin

such that int119864

119891119889119909 le 119862120588(119891)

Given such a function norm 120588 the set 119883 = 119883(120588) ofall extended measurable functions 119891 (identifying functionsequal 120583-almost everywhere) such that 120588(|119891|) lt infin is called aBanach function space (BFS for short) and we define

10038171003817100381710038171198911003817100381710038171003817119883

= 120588 (1003816100381610038161003816119891

1003816100381610038161003816) 119891 isin 119883 (45)

We leave it to the reader to verify that Lebesgue spacessatisfy the following properties (1198831015840 can be chosen as the dualspace)

Properties 19 Given a Banach function space 119883 we requirethe existence of a Banach function space119883

1015840 such that

(B) 119883 contains each essentially bounded function whosesupport has finite measure

(HI) Holderrsquos inequality int |119891119892|119889119909 le 119891119883119892

1198831015840

(C) 1198831015840 is complete

(D) 119892119883

= supintΩ

|119892|ℎ ℎ isin M+

ℎ1198831015840 le 1

(AC) absolutely continuous norm if (120594119864119899)119899is a sequence of

characteristic functions converging pointwise almosteverywhere to 0 then 119891120594

1198641198991198831015840 converges to zero for

all 119891 isin 1198831015840

Journal of Function Spaces 7

(A) each function 119891 isin 1198831015840 can be approximated by simple

functions (in1198831015840-norm)

(NC) if 119891 isin 1198831015840 and ℎ

119899are step functions with 0 le ℎ

119899uarr |119891|

then1003817100381710038171003817119891 minus ℎ

119899sgn (119891)

10038171003817100381710038171198831015840997888rarr 0 as 119899 997888rarr infin (46)

Lemma 20 Suppose Ω sub R is open and 119883 is a Banachfunction space fulfilling the items listed in Properties 19 Assumethat 119891 Ω rarr R is in 119883 We let 120582

119891 119883

1015840

cap C119903+1

0(Ω) rarr R be

defined as

120582119891(120593) = int

Ω

119891120593(119903)

119889119909 (47)

If1003816100381610038161003816Δ

119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)]

for almost every 119909 119910 isin Ω

(48)

for some 119886119891

isin 119883 then 120582119891has a continuous extension to 119883

1015840

with norm bounded from above by 2119903119903

119886119891119883

Proof Our goal is to show1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816le 2119903

11990310038171003817100381710038171003817119886119891

10038171003817100381710038171003817119883

100381710038171003817100381712059310038171003817100381710038171198831015840

(49)

because 120582119891extends then by the theorem of Hahn-Banach to

a bounded functional on1198831015840 with norm bounded from above

by 2119903119903

119886119891119883 Without loss of generality we may assume that

119886119891is nonnegative If 119891 were be smooth we could just use

integration by parts and use as bound the limiting case of thepointwise inequality (48) As 119891 is not necessarily smooth wehave to operate on the level of finite differences

Let us choose 120593 isin C0(Ω) and denote its support by 119870

Lemma 17 ensures the existence of some positive 120576 gt 0 suchthat (Ωminus119895ℎ)cap119870 = Ωcap119870 for 0 le 119895 le 119903 as long as |ℎ| lt 120576119903Weassume in the following that ℎ has been chosen accordingly

The following integrals exist by Holderrsquos inequalitybecause 120593 and thus Δ

119903

minusℎ120593(119909) are bounded with compact

support and 119891 is integrable Consider

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

119891 (119909) (minusℎ)minus119903

Δ119903

minusℎ120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(50)

We compute the following integral

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

= intΩ

119891 (119909)

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119909) 120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωminus119895ℎ

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

(Ωminus119895ℎ)cap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

(51)

We have chosen 120576 gt 0 such that (Ω minus 119895ℎ) cap 119870 = Ω cap 119870 Wecontinue

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωcap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

(52)

With the aid of (52) we can now rewrite (50) as

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

(minusℎ)minus119903

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(53)

Inequality (48) leads to

1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816

le 119903119903

intΩ

(119886119891(119909) + 119886

119891(119909 + 119903ℎ))

1003816100381610038161003816120593 (119909)1003816100381610038161003816 119889119909

+ intΩ

1003816100381610038161003816119891 (119909)1003816100381610038161003816

10038161003816100381610038161003816120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)

10038161003816100381610038161003816119889119909

(54)

Using Holderrsquos inequality (HI) for the first summand weare left to show that the second summand tends to 0 asℎ approaches 0 We note first that since the second factorhas support in a set of finite measures we can replace theintegration domain by a set of finite measuresThen we applya (1198711

119871infin

)-Holder inequality and the uniform convergence ofthe divided difference to 120593

(119903)

(119909)119903 established in Lemmas 13and 15

After having verified that 120582119891has a continuous extension

to 1198831015840 we want to show that this extension has a representa-

tion as integral

8 Journal of Function Spaces

Lemma 21 Assume Ω sub R is open Suppose the propertieslisted in Properties 19 are met Let 120582

119891be defined as in (47) and

bounded on 1198831015840 Then there exists 119892 isin 119883 with 119892

119883le 120582

119891

such that

120582119891(120593) = (minus1)

119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (55)

for all 120593 isin 1198831015840 especially for all 120593 isin Cinfin

0(Ω)

Proof The proof is heavily influenced by the proof of Theo-rem 141 in [13]We first want to find a signedmeasure ] suchthat

120582119891(120593) = int

Ω

120593119889] (56)

for each 120593 isin 1198831015840 Then we want to apply the Lebesgue-

Radon-Nikodym theorem to replace 119889] by (minus1)119903

119892(119909)119889119909 forsome integrable function 119892 for which we then show that it iscontained in119883

We start by assuming that the measure of Ω is finite Foreach 120583-measurable set 119860 we set

] (119860) = 120582119891(120594

119860) (57)

We start by proving the 120590-additivity of ] (for the 120583-measurable sets) In the first step we just show the additivityBy induction it is enough to consider only two sets

Let 1198601and 119860

2be two disjoint 120583-measurable sets Then

] (1198601cup 119860

2) = 120582

119891(120594

1198601cup1198602) = 120582

119891(120594

1198601+ 120594

1198602)

= 120582119891(120594

1198601) + 120582

119891(120594

1198602) = ] (119860

1) + ] (119860

2)

(58)

We move to the verification of the 120590-additivity Let119860

1 119860

2 be countablymany pairwise disjoint120583-measurable

sets We first want to show thatsum119894120594119860119894

converges in1198831015840

We let 119864119896= cup

119896

119894=1119860

119894and 119864 = cup

infin

119894=1119860

119894 Then since (119864

119896)119896is

increasing

120583 (119864) = 120583(

infin

⋃

119896=1

119864119896) = lim

119896rarrinfin

120583 (119864119896) (59)

As 119864 has finite measure given 120576 gt 0 there exists 1198960isin N such

that

(1 minus 120576) 120583 (119864) lt 120583 (1198641198960) (60)

By the disjointness of the sets 119860119894 we see that

120583 (119864) = 120583 (1198641198960) + 120583(

infin

⋃

119894=1198960+1

119860119894) (61)

It follows that

120583(

infin

⋃

119894=1198960+1

119860119894) = 120583 (119864) minus 120583 (119864

1198960) lt 120583 (119864) minus (1 minus 120576) 120583 (119864)

= 120576120583 (119864)

(62)

We have for119898 lt 119899

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=1

120594119860119894

minus

119898

sum

119894=1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

=

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=119898+1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

le10038171003817100381710038171003817120594cupinfin119894=119898+1

119860119894

100381710038171003817100381710038171198831015840 (63)

By the considerations before and the fact that sdot 1198831015840 is

absolutely continuous as stipulated in (AC) we see thatsum

119899

119894=1120594119860119894

is a Cauchy sequence Now as 1198831015840 is complete by

(C) the Cauchy sequence converges in 1198831015840 We have by the

continuity of 120582119891

] (cup119894119860

119894) = 120582

119891(120594

cup119894119860119894) = 120582

119891(

infin

sum

119894=1

120594119860119894)

= lim119899rarrinfin

120582119891(

119899

sum

119894=1

120594119860119894) = lim

119899rarrinfin

119899

sum

119894=1

120582119891(120594

119860119894)

= lim119899rarrinfin

119899

sum

119894=1

] (119860119894) =

infin

sum

119894=1

] (119860119894)

(64)

Thus ] is 120590-additiveIn order to apply the theorem of Lebesgue-Radon-

Nikodym see for example Section 69 in [15] and conclud-ing the existence of 119892 isin 119871

1

(Ω) such that

] (119860) = int119860

119892119889119909 119860 120583-measurable (65)

we need to show that ] is absolutely continuous with respectto 120583 To do so assume that 120583(119873) = 0 Then ](119873) = 120582

119891(120594

119873)

By the absolute continuity of the normof1198831015840 required in (AC)we see that 120594

1198731198831015840 = 0 and therefore ](119873) = 0 Replacing 119892

by minus119892 if necessary we obtain

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909 (66)

Now we assume that Ω does not necessarily have finitemeasureWewriteΩ = cup

119899Ω

119899as union of open sets with finite

measure such that Ω119899sub Ω

119899+1 By the considerations above

we find for each 119899 isin N a 119892119899such that

120582119891(120594

119860) = (minus1)

119903

intΩ

119892119899(119909) 120594

119860(119909) 119889119909 (67)

for each measurable set 119860 sub Ω119899 It is easily seen that 119892

119899

and 119892119899+1

agree almost everywhere on Ω119899 This gives rise to

a function 119892 satisfying

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909

119860 sub

119873

⋃

119899=1

Ω119899for some 119873

(68)

By linearity we have for each simple function 119904

120582119891(119904) = (minus1)

119903

intΩ

119892 (119909) 119904 (119909) 119889119909 (69)

To show that 119892 isin 119883 we want to use property (D)

Journal of Function Spaces 9

Let ℎ be a nonnegative simple function with supportin some Ω

119899 Then ℎ sgn(119892) is also a simple function with

support inΩ119899 It is a finite linear combination of characteristic

functions of sets in Ω119899 Hence

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = intℎ sdot sgn (119892) 119892 119889119909 = 120582119891(ℎ sgn (119892)) (70)

Using the boundedness of 120582119891 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (71)

If ℎ is an arbitrary function in1198831015840 thenwemay construct a

sequence (ℎ119899)infin

119899=1of simple functions each ℎ

119899having support

in Ω119899 such that 0 le ℎ

119899uarr |ℎ| 120583-almost everywhere By the

monotone convergence theorem we have by (71) that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = lim119899rarrinfin

int1003816100381610038161003816ℎ119899119892

1003816100381610038161003816 119889119909 le lim sup119899rarrinfin

10038171003817100381710038171003817120582119891

10038171003817100381710038171003817

1003817100381710038171003817ℎ11989910038171003817100381710038171198831015840

(72)

By (P3) in Definition 18 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (73)

and Property (D) gives that 119892 isin 119883 In the last step we want toupgrade the integral representation (69) to hold for arbitraryfunctions in 120593 isin 119883

1015840 As before we approximate 120593 by simplefunctions ℎ

119899with support contained in Ω

119899and 0 le ℎ

119899uarr |120593|

in 120583-almost everywhere point Define 119904119899

= ℎ119899sgn(120593) By

(NC) 119904119899converges to 120593 in119883

1015840 and as both 120582119891and ldquoits integral

representationrdquo are continuous and agree on the set of stepfunctions they agree on119883

1015840

Having shown the existence of the weak derivativesof highest order we are left to show the existence of theintermediate derivatives which we do in Proposition 22 inthe next section To keep the flow going we give here theproof that the pointwise inequality implies membership inthe corresponding Sobolev space under the assumption thatwe already know that Proposition 22 holds

Proof of Theorem 2 (pointwise inequality implies membershipin Sobolev space) We combine Lemmas 20 and 21 and applyProposition 22

4 Equivalent Definitions of Sobolev Spaces

The aim of this section is to prove the following equivalence

Proposition 22 Let Ω sub R be a domain with finite Lebesguemeasure inR Suppose119883 is a Banach function space satisfying(B) and (D) Assume that 119898 isin N Then the following setscoincide

(a) 119860 = 119891 isin 119883 there exists 119865 isin 119883 where 119865 agrees almosteverywherewith119891 and hasweak derivatives up to order119898 in119883 where the derivatives of order 119903 0 le 119903 le 119898minus1are absolutely continuous

(b) 119861 = 119891 isin 119883 119891(119896)

isin 119883 119896 isin 0 119898 where 119891(119896) arethe weak derivatives of 119891

(c) 119862 = 119891 isin 119883 119891(119898)

isin 119883 where 119891(119898) is the weak

derivative of 119891 of order 119898

We see that 119860 sub 119861 sub 119862 We now show that 119862 sub 119860Let us give an outline of the proof Given 119892 Ω rarr R

(we will choose 119892 = 119891(119898)) by integrating we construct in

Lemma 23 an absolutely continuous function 119866119898together

with its weak derivatives 1198660 119866

119898minus1 which are absolutely

continuous and 119866(119898)

119898= 119892 However 119866

119898and 119891 might

not agree almost everywhere We verify however in theproof of Theorem 26 (with the help of Theorem 24) thatwe find a polynomial 119875 such that 119866

119898+ 119875 agrees with 119865

almost everywhere and shares with 119866119898the other required

properties We conclude the proof of Theorem 26 by settlingthe membership questions with the aid of Proposition 25

Lemma 23 Assume Ω sub R is a domain We further assumethat 119892 Ω rarr R is in 119871

1

loc(Ω) One fixes 119909lowast

isin Ω and defines

1198660(119909

0) = 119892 (119909

0)

119866119896(119909

119896) = int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119896 isin N

(74)

where the integrals are Lebesgue integrals and we use thenotation

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

=

int[119909lowast119909119896]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909lowast

le 119909119896

minusint[119909119896 119909lowast]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909119896lt 119909

lowast

(75)

Then

(a) 119866119896is locally absolutely continuous for all 119896 isin N

especially locally integrable(b) if 120593 isin Cinfin

0(Ω) and 0 le 119903 le 119896 one has

intΩ

119866119896120593(119903)

= (minus1)119903

intΩ

119866119896minus119903

120593 (76)

Proof Let us prove (a) by induction Assume 119896 = 1 Note that1198661(119909

lowast

) = 0 Thus we have for 119909lowast

le 1199091

1198661(119909

1) minus 119866

1(119909

lowast

) = 1198661(119909

1) = int

1199091

119909lowast

119892 (1199090) 119889119909

0 (77)

As 119892 isin 1198711

([119909lowast

1199091]) we see that 119866

1is absolutely continuous

The case 1199091lt 119909

lowast is treated similarlyAssume now that the claim is true for some 119896 Now

119866119896+1

(119909119896+1

) is the integral of a continuous function and thuslocally absolutely continuous

Let us have a look at (b) Note that the statement is true for119896 = 0 thus we may assume that 119896 ge 1 We make an inductionover 119903 The case 119903 = 0 is clear Let us choose an interval[119886 119887] sub Ω containing a neighborhood of the support of 120593 As119866119896and 120593

(119903) are both absolutely continuous in [119886 119887] we have

10 Journal of Function Spaces

by the integration by part formula for absolutely continuousfunctions (see for exampleTheorem7147 in [9]) and the factthat 120593(119903)

(119886) = 120593(119903)

(119887) = 0

intΩ

119866119896120593(119903+1)

= int[119886119887]

119866119896120593(119903+1)

= (minus1) int[119886119887]

1198661015840

119896120593(119903)

(78)

Noting that 1198661015840

119896= 119866

119896minus1almost everywhere we see that

intΩ

119866119896120593(119903+1)

= (minus1) int[119886119887]

119866119896minus1

120593(119903)

= (minus1) intΩ

119866119896minus1

120593(119903)

(79)

Using the induction hypothesis the claim follows

We cite Theorem 253 in [10] without giving its proof

Theorem 24 Let 119860 = [119886 119887] R or R+and 119903 ge 0 If the

function 119891 isin 1198711

loc(119860) satisfies

int119860

119891120593(119903)

119889119909 = 0 (80)

for all 120593 isin Cinfin

0(119860) then there is a polynomial 119875 of degree lt 119903

such that 119891 = 119875 almost everywhere on 119860

To show that 119866119896is in 119883 we need to verify that (if there

is danger of confusion with respect to which variable we takethe norm we use notations as119883(119909

119896))

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

lt infin (81)

under the assumption that 119866119896minus1

is in 119883 Actually bounding119866119896from above by its absolute value this means that we have

first to take the1198711-norm and then the norm in119883We howeverwould like to switch the norms

We adapt Proposition 21 in [16] to our setting

Proposition 25 Let 120588 be a Banach function norm satisfyingproperty (D) If 119891 ΩtimesΩ rarr R is a measurable function andone has 119891

119909and 119891

119910 functions onΩ such that 119891119909(119910) = 119891(119909 119910) =

119891119910

(119909) then

120588 (1003817100381710038171003817119891119909

10038171003817100381710038171) le

1003817100381710038171003817120588 (119891119910

)10038171003817100381710038171

(82)

The cases 119883 isin 1198711

119871infin

are easy to prove If 119883 = 119871119901 for

some 1 lt 119901 lt infin then the result can be proven by usingFubinirsquos theorem together with property (D) For the generalresults we refer to Scheprsquos proof

The next result builds uponTheorem 254 in [10]

Theorem 26 Let Ω sub R be a domain and let 119903 ge 1 Suppose119883 is a Banach function space satisfying (B) and (D) If 119891 islocally in 119883 and has a generalized 119903th derivative 119892 locally in119883 then there is 119865 agreeing almost everywhere with 119891 havingabsolutely continuous intermediate weak derivatives and theweak derivative of order 119903 agrees almost everywhere with 119892Furthermore the derivatives of 119865 are locally in 119883 If Ω hasfinite measure one obtains the result with ldquolocallyrdquo replaced byldquogloballyrdquo

We thank Pilar Silvestre for pointing out how to prove themembership of the intermediate derivatives in119883

Proof Choose 119909lowast

isin Ω and let 119866119896be defined as in Lemma 23

Assume 120593 isin Cinfin

0(Ω) Then using the fact that 119892 is a weak

derivative of 119891 of order 119903 and (b) in Lemma 23

intΩ

(119891 minus 119866119903) 120593

(119903)

119889119909 = intΩ

119891120593(119903)

119889119909 minus intΩ

119866119903120593(119903)

119889119909

= (minus1)119903

intΩ

119892120593119889119909 minus (minus1)119903

intΩ

119892120593119889119909 = 0

(83)

UsingTheorem 24 we see that119891minus119866119903= 119875 almost everywhere

where119875 is a polynomial of degreelt 119903Thus119891 = 119875+119866119903almost

everywhere and for 120593 isin Cinfin

0(Ω) and 0 le 119898 le 119903

intΩ

119891120593(119898)

119889119909 = intΩ

(119875 + 119866119903) 120593

(119898)

= (minus1)119898

intΩ

(119866119903minus119898

+ 119875(119898)

) 120593 119889119909

(84)

Thus119866119903minus119898

+119875(119898) is the119898th generalized derivative of119891 As it is

locally absolutely continuous (if119898 lt 119903) it lies locally in119883 Inthe case 119898 = 119903 the same conclusion holds by the assumptionon 119892

Moreover there exists 119865 for example 119865 = 119875 + 119866119903 that

agrees with 119891 almost everywhere such that its generalizedderivatives up to order 119903 minus 1 are locally absolutely continuousand the 119903th generalized derivative equals 119892 almost every-where

As 119892 is locally in 119883 and the other derivatives are locallyabsolutely continuous they are locally in 119871

infin and thus locallyin119883 by (B)

Suppose now that Ω has additionally finite measureAssume that 119891 and 119892 are in119883 globally We want to show thatthe intermediate derivatives are in 119883 as well

Note in our setting that by (P5) in Definition 18membership in 119883 implies membership in 119871

1 We proceedby induction Suppose that 119866

119896minus1isin 119883 We will apply

Proposition 25 We set denoting by conv119909lowast

119909119896 the closed

interval with endpoints 119909lowast and 119909119896

119891 (119909119896minus1

119909119896) = 119866

119896minus1(119909

119896minus1) 120594conv119909lowast 119909119896 (119909119896minus1

) (85)

Then1003817100381710038171003817119866119896

(119909119896)1003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119891(119909119896minus1

119909119896)119889119909

119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)10038171003817100381710038171198711(119909119896minus1)

10038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)

le10038171003817100381710038171003817

1003816100381610038161003816119866119896minus1(119909

119896minus1)1003816100381610038161003816

1003817100381710038171003817120594Ω1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)lt infin

(86)

Journal of Function Spaces 11

5 Banach Function Space Setting

Here our main focus is a more advanced readership thatlikely knows already Banach function spaces Therefore weadopt a much briefer style pointing merely to the literaturewhere the corresponding definitions and results can be foundWe start with giving generalized versions of the necessity andsufficiency of Theorem 2

We list the results and prove them simultaneously bysplitting the proofs into parts according to the functionspaces

Theorem 27 Assume Ω sub R is a domain 119883 is a normedspace and 119898 isin N If the maximal function 119872 119883 rarr 119883 isbounded then there is a constant 119862 = 119862(119883119898) such that foreach continuous 119891 isin 119882

119898

119883there is 119886

119891isin 119883with 119886

119891 le 119862119891

119882119898119883

and

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (87)

In particular such 119886119891exists if

(i) 1 lt 119901 le infin and 119883 is the Lebesgue space 119871119901

(Ω)

(ii) 1 lt 119901 lt infin 1 le 119902 le infin and 119883 is the Lorentz space119871119901119902

(Ω)

(iii) 1 lt 119901 lt infin 1 le 119902 le infin 119887 is slowly varying and 119883 isthe Lorentz-Karamata space 119871

119901119902119887(Ω)

(iv) 120601 is an 119873-function with inf119904gt0

(119904120593(119904)120601(119904)) gt 1 where120593 is the density function of 120601 and 119883 is the Orlicz space119871lowast

120601(Ω)

(v) Ω = R and 119883 is rearrangement-invariant with upperindex 120572

119883lt 1

Theorem 28 Let Ω sub R be open Suppose 119883 is a Banachfunction space such that (AC) holds Assume further that thefunction 119891 Ω rarr R is in 119883 If for some 119903 isin N

0 the following

inequality holds

1003816100381610038161003816Δ119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)] (88)

for some 119886119891isin 119883 then there exists119892 isin 119883with 119892

119883le 2119903

119903

119886119891119883

such that

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909 = (minus1)119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (89)

for all 120593 isin Cinfin

0(Ω)

In particular the result holds if119883 is

(i) the Lebesgue space 119871119901

(Ω) for some 1 lt 119901 lt infin

(ii) the Lorentz space 119871119901119902

(Ω) where 1 lt 119901 lt infin and 1 le

119902 lt infin

(iii) the Lorentz-Karamata space 119871119901119902119887

(Ω) where 119887 isslowly varying 1 lt 119901 lt infin and 1 lt 119902 le infin

(iv) the Orlicz space 119871lowast

120601(Ω) for some 119873-function 120601 satisfy-

ing

inf119904gt0

119904120593 (119904)

120601 (119904)gt 1 (90)

where 120593 is the density function of 120601

To help with orientation let us give an overview ofthe remaining material in this section With respect toTheorem 27 note that we carried out the proof of Corollary 14in a high level of abstraction Hence we are left to verifythat the spaces under question are normed spaces (we willshow that they are even Banach function spaces) and theboundedness of the maximal operator

Concerning the proof Theorem 28 we proved the cor-responding part of Theorem 2 under the assumptions thatthe spaces under question are Banach function spaces andfurther satisfy the requirements listed in Properties 19 Theexistence of 1198831015840 is classical (Proposition 29) as is the fact that(AC) implies (A) and (NC) (Proposition 30) Hence we onlyneed to point to the literature where (AC) was verified

Proposition 29 Given a Banach function space there is aspace 119883

1015840 such that properties (B) (HI) (C) and (D) aresatisfied

Proof We refer the reader to Sections 11 and 12 in [13] or toSection 31 in [14]

The definition of absolutely continuous norm can befound as Definition 3111 in [14] and Definition 131 in [13]

Proposition 30 A Banach function space that has absolutelycontinuous norm satisfies (A) and (NC)

Proof The first statement follows fromTheorem 1313 in [13]and the second fromProposition 136 in the same source

51 General Statements

Proof of Theorem 27 (general statement) We basically havealready proven the general statement in Corollary 14

Proof of Theorem 28 (general statement) We gave the proofof the Lebesgue space case in such generality that thegeneral statement follows from the proof of Theorem 2 andPropositions 29 and 30

52 Lorentz-Karamata Spaces Lorentz-Karamata spaces aredetailed in [14] mainly Section 343

Proof of Theorem 27 (ii) and (iii) The part that is left toprove is the boundedness of the maximal operator which isguaranteed by Remark 3517 in [14]

Proof of Theorem 28 (ii) and (iii) By Theorem 3441 in [14]we see that 119883 is a rearrangement-invariant Banach space Acombination of Lemma 3439 in [14] together with Lemma3443 in [14] gives the absolute continuity (AC) of the normof1198831015840

12 Journal of Function Spaces

53 Orlicz Spaces The setting is the one of [17]

Lemma 31 If 120601 is an 119873-function satisfying the Δ2-condition

then the Orlicz space 119871lowast

120601has absolutely continuous norm

Proof Let 119891 isin 119871lowast

120601(Ω) We verify first that

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 lt infin (91)

Let 120576 gt 0 By the Δ2-condition there exists a constant 120581 =

120581(119891(120601)

+ 120576) such that

120601 ((1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576) 119904) le 120581120601 (119904) (92)

for all 119904 ge 0 Thus

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 = intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576(1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576))119889119909

le 120581intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576)119889119909 le 120581

(93)

Suppose that 119864 is a measurable set Then

10038171003817100381710038171198911205941198641003817100381710038171003817(120601)

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891120594119864

1003816100381610038161003816) 119889119909 le 1

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 1

(94)

Let us fix 120582 gt 0 Again using the Δ2-condition we have

120601(120582minus1

119904) le 120581(120582minus1

)120601(119904) Hence

intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 (95)

Example 133 in [13] tells us that 1198711 has absolutely continuousnorm Thus if 120594

119864119899converges 120583-almost everywhere to 0 then

intΩ

120601(|119891|)120594119864119899

converges to zero as well Thus we have for 119899

large enough that

120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119899119889119909 le 1 (96)

implying that 119891120594119864119899(120601)

le 120582 Since 120582 gt 0 was arbitrary theabsolute continuity of 119891 follows

Proof of Theorem 27 (iv) We conclude from Gallardorsquos resultTheorem 21 in [17] that it is sufficient that the complemen-tary 119873-function satisfies the Δ

2-condition in [0infin) Again

resorting to Gallardo this time to Proposition 14 in [17] weobtain the conclusion

Proof of Theorem 28 (iv) That 119883 is a Banach functionspace follows from Theorem 3416 in [14] To prove that119883

1015840 has absolutely continuous norm we note that inequal-ity (90) together with Proposition 14 in [17] ensuresthat the complementary 119873-function of 120601 satisfies the Δ

2-

condition Lemma 31 verifies that 1198831015840 has absolutely contin-uous norm

54 Rearrangement-Invariant Spaces The necessary defini-tions can be found in [13]

Proof of Theorem 27 (v) We are only left to show theboundedness of the maximal operator However this is thecontent of a result by Lorentz and Shimogaki for examplestated as Theorem 3517 in [13]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank Pilar Silvestre for hersuggestions of how to prove the existence of intermediatederivatives The research is supported by the Academy ofFinland The third author thanks the Department of Math-ematics and Systems Analysis at Aalto University for itshospitality

References

[1] B Bojarski ldquoRemarks on some geometric properties of Sobolevmappingsrdquo in Functional Analysis amp Related Topics (Sapporo1990) pp 65ndash76 World Scientific River Edge NJ USA 1991

[2] B Bojarski PHajlasz andP Strzelecki ldquoImproved119862119896120582 approxi-

mation of higher order Sobolev functions in normand capacityrdquoIndiana University Mathematics Journal vol 51 no 3 pp 507ndash540 2002

[3] B Bojarski ldquoPointwise characterization of Sobolev classesrdquoProceedings of the Steklov Institute of Mathematics vol 255 no1 pp 65ndash81 2006

[4] B Bojarski ldquoWhitneyrsquos jets for Sobolev functionsrdquo UkrainianMathematical Journal vol 59 no 3 pp 345ndash358 2007

[5] B Bojarski ldquoTaylor expansion and Sobolev spacesrdquo Bulletin ofthe Georgian National Academy of Sciences New Series vol 5no 2 pp 5ndash10 2011

[6] B Bojarski ldquoSobolev spaces and Lagrange interpolationrdquoGeorgian Academy of Sciences Proceedings of A RazmadzeMathematical Institute vol 158 pp 1ndash12 2012

[7] P Hajłasz ldquoSobolev spaces on an arbitrary metric spacerdquoPotential Analysis vol 5 no 4 pp 403ndash415 1996

[8] B Bojarski ldquoSobolev spaces and averaging Irdquo httparxivorgabs13085171

[9] R Kannan and C K Krueger Advanced Analysis on the RealLine Universitext Springer New York NY USA 1996

[10] R A DeVore and G G Lorentz Constructive Approximationvol 303 of Fundamental Principles of Mathematical SciencesSpringer Berlin Germany 1993

[11] L L Schumaker Spline Functions Basic Theory CambridgeMathematical Library CambridgeUniversity Press CambridgeUK 3rd edition 2007

[12] G H Hardy and J E Littlewood ldquoA maximal theorem withfunction-theoretic applicationsrdquo Acta Mathematica vol 54 no1 pp 81ndash116 1930

[13] C Bennett andR Sharpley Interpolation ofOperators vol 129 ofPure and Applied Mathematics Academic Press Boston MassUSA 1988

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

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Volume 2014

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 7

(A) each function 119891 isin 1198831015840 can be approximated by simple

functions (in1198831015840-norm)

(NC) if 119891 isin 1198831015840 and ℎ

119899are step functions with 0 le ℎ

119899uarr |119891|

then1003817100381710038171003817119891 minus ℎ

119899sgn (119891)

10038171003817100381710038171198831015840997888rarr 0 as 119899 997888rarr infin (46)

Lemma 20 Suppose Ω sub R is open and 119883 is a Banachfunction space fulfilling the items listed in Properties 19 Assumethat 119891 Ω rarr R is in 119883 We let 120582

119891 119883

1015840

cap C119903+1

0(Ω) rarr R be

defined as

120582119891(120593) = int

Ω

119891120593(119903)

119889119909 (47)

If1003816100381610038161003816Δ

119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)]

for almost every 119909 119910 isin Ω

(48)

for some 119886119891

isin 119883 then 120582119891has a continuous extension to 119883

1015840

with norm bounded from above by 2119903119903

119886119891119883

Proof Our goal is to show1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816le 2119903

11990310038171003817100381710038171003817119886119891

10038171003817100381710038171003817119883

100381710038171003817100381712059310038171003817100381710038171198831015840

(49)

because 120582119891extends then by the theorem of Hahn-Banach to

a bounded functional on1198831015840 with norm bounded from above

by 2119903119903

119886119891119883 Without loss of generality we may assume that

119886119891is nonnegative If 119891 were be smooth we could just use

integration by parts and use as bound the limiting case of thepointwise inequality (48) As 119891 is not necessarily smooth wehave to operate on the level of finite differences

Let us choose 120593 isin C0(Ω) and denote its support by 119870

Lemma 17 ensures the existence of some positive 120576 gt 0 suchthat (Ωminus119895ℎ)cap119870 = Ωcap119870 for 0 le 119895 le 119903 as long as |ℎ| lt 120576119903Weassume in the following that ℎ has been chosen accordingly

The following integrals exist by Holderrsquos inequalitybecause 120593 and thus Δ

119903

minusℎ120593(119909) are bounded with compact

support and 119891 is integrable Consider

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

119891 (119909) (minusℎ)minus119903

Δ119903

minusℎ120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(50)

We compute the following integral

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

= intΩ

119891 (119909)

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119909) 120593 (119909 minus 119895ℎ) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωminus119895ℎ

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

(Ωminus119895ℎ)cap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

(51)

We have chosen 120576 gt 0 such that (Ω minus 119895ℎ) cap 119870 = Ω cap 119870 Wecontinue

intΩ

119891 (119909) Δ119903

minusℎ120593 (119909) 119889119909

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ωcap119870

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

=

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)int

Ω

119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

119903

sum

119895=0

(minus1)119903minus119895

(119903

119895)119891 (119910 + 119895ℎ) 120593 (119910) 119889119910

= intΩ

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

(52)

With the aid of (52) we can now rewrite (50) as

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

= intΩ

(minusℎ)minus119903

Δ119903

ℎ119891 (119909) 120593 (119909) 119889119909

+ intΩ

119891 (119909) (120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)) 119889119909

(53)

Inequality (48) leads to

1003816100381610038161003816100381610038161003816intΩ

119891 (119909) 120593(119903)

(119909) 119889119909

1003816100381610038161003816100381610038161003816

le 119903119903

intΩ

(119886119891(119909) + 119886

119891(119909 + 119903ℎ))

1003816100381610038161003816120593 (119909)1003816100381610038161003816 119889119909

+ intΩ

1003816100381610038161003816119891 (119909)1003816100381610038161003816

10038161003816100381610038161003816120593(119903)

(119909) minus (minusℎ)minus119903

Δ119903

minusℎ120593 (119909)

10038161003816100381610038161003816119889119909

(54)

Using Holderrsquos inequality (HI) for the first summand weare left to show that the second summand tends to 0 asℎ approaches 0 We note first that since the second factorhas support in a set of finite measures we can replace theintegration domain by a set of finite measuresThen we applya (1198711

119871infin

)-Holder inequality and the uniform convergence ofthe divided difference to 120593

(119903)

(119909)119903 established in Lemmas 13and 15

After having verified that 120582119891has a continuous extension

to 1198831015840 we want to show that this extension has a representa-

tion as integral

8 Journal of Function Spaces

Lemma 21 Assume Ω sub R is open Suppose the propertieslisted in Properties 19 are met Let 120582

119891be defined as in (47) and

bounded on 1198831015840 Then there exists 119892 isin 119883 with 119892

119883le 120582

119891

such that

120582119891(120593) = (minus1)

119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (55)

for all 120593 isin 1198831015840 especially for all 120593 isin Cinfin

0(Ω)

Proof The proof is heavily influenced by the proof of Theo-rem 141 in [13]We first want to find a signedmeasure ] suchthat

120582119891(120593) = int

Ω

120593119889] (56)

for each 120593 isin 1198831015840 Then we want to apply the Lebesgue-

Radon-Nikodym theorem to replace 119889] by (minus1)119903

119892(119909)119889119909 forsome integrable function 119892 for which we then show that it iscontained in119883

We start by assuming that the measure of Ω is finite Foreach 120583-measurable set 119860 we set

] (119860) = 120582119891(120594

119860) (57)

We start by proving the 120590-additivity of ] (for the 120583-measurable sets) In the first step we just show the additivityBy induction it is enough to consider only two sets

Let 1198601and 119860

2be two disjoint 120583-measurable sets Then

] (1198601cup 119860

2) = 120582

119891(120594

1198601cup1198602) = 120582

119891(120594

1198601+ 120594

1198602)

= 120582119891(120594

1198601) + 120582

119891(120594

1198602) = ] (119860

1) + ] (119860

2)

(58)

We move to the verification of the 120590-additivity Let119860

1 119860

2 be countablymany pairwise disjoint120583-measurable

sets We first want to show thatsum119894120594119860119894

converges in1198831015840

We let 119864119896= cup

119896

119894=1119860

119894and 119864 = cup

infin

119894=1119860

119894 Then since (119864

119896)119896is

increasing

120583 (119864) = 120583(

infin

⋃

119896=1

119864119896) = lim

119896rarrinfin

120583 (119864119896) (59)

As 119864 has finite measure given 120576 gt 0 there exists 1198960isin N such

that

(1 minus 120576) 120583 (119864) lt 120583 (1198641198960) (60)

By the disjointness of the sets 119860119894 we see that

120583 (119864) = 120583 (1198641198960) + 120583(

infin

⋃

119894=1198960+1

119860119894) (61)

It follows that

120583(

infin

⋃

119894=1198960+1

119860119894) = 120583 (119864) minus 120583 (119864

1198960) lt 120583 (119864) minus (1 minus 120576) 120583 (119864)

= 120576120583 (119864)

(62)

We have for119898 lt 119899

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=1

120594119860119894

minus

119898

sum

119894=1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

=

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=119898+1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

le10038171003817100381710038171003817120594cupinfin119894=119898+1

119860119894

100381710038171003817100381710038171198831015840 (63)

By the considerations before and the fact that sdot 1198831015840 is

absolutely continuous as stipulated in (AC) we see thatsum

119899

119894=1120594119860119894

is a Cauchy sequence Now as 1198831015840 is complete by

(C) the Cauchy sequence converges in 1198831015840 We have by the

continuity of 120582119891

] (cup119894119860

119894) = 120582

119891(120594

cup119894119860119894) = 120582

119891(

infin

sum

119894=1

120594119860119894)

= lim119899rarrinfin

120582119891(

119899

sum

119894=1

120594119860119894) = lim

119899rarrinfin

119899

sum

119894=1

120582119891(120594

119860119894)

= lim119899rarrinfin

119899

sum

119894=1

] (119860119894) =

infin

sum

119894=1

] (119860119894)

(64)

Thus ] is 120590-additiveIn order to apply the theorem of Lebesgue-Radon-

Nikodym see for example Section 69 in [15] and conclud-ing the existence of 119892 isin 119871

1

(Ω) such that

] (119860) = int119860

119892119889119909 119860 120583-measurable (65)

we need to show that ] is absolutely continuous with respectto 120583 To do so assume that 120583(119873) = 0 Then ](119873) = 120582

119891(120594

119873)

By the absolute continuity of the normof1198831015840 required in (AC)we see that 120594

1198731198831015840 = 0 and therefore ](119873) = 0 Replacing 119892

by minus119892 if necessary we obtain

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909 (66)

Now we assume that Ω does not necessarily have finitemeasureWewriteΩ = cup

119899Ω

119899as union of open sets with finite

measure such that Ω119899sub Ω

119899+1 By the considerations above

we find for each 119899 isin N a 119892119899such that

120582119891(120594

119860) = (minus1)

119903

intΩ

119892119899(119909) 120594

119860(119909) 119889119909 (67)

for each measurable set 119860 sub Ω119899 It is easily seen that 119892

119899

and 119892119899+1

agree almost everywhere on Ω119899 This gives rise to

a function 119892 satisfying

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909

119860 sub

119873

⋃

119899=1

Ω119899for some 119873

(68)

By linearity we have for each simple function 119904

120582119891(119904) = (minus1)

119903

intΩ

119892 (119909) 119904 (119909) 119889119909 (69)

To show that 119892 isin 119883 we want to use property (D)

Journal of Function Spaces 9

Let ℎ be a nonnegative simple function with supportin some Ω

119899 Then ℎ sgn(119892) is also a simple function with

support inΩ119899 It is a finite linear combination of characteristic

functions of sets in Ω119899 Hence

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = intℎ sdot sgn (119892) 119892 119889119909 = 120582119891(ℎ sgn (119892)) (70)

Using the boundedness of 120582119891 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (71)

If ℎ is an arbitrary function in1198831015840 thenwemay construct a

sequence (ℎ119899)infin

119899=1of simple functions each ℎ

119899having support

in Ω119899 such that 0 le ℎ

119899uarr |ℎ| 120583-almost everywhere By the

monotone convergence theorem we have by (71) that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = lim119899rarrinfin

int1003816100381610038161003816ℎ119899119892

1003816100381610038161003816 119889119909 le lim sup119899rarrinfin

10038171003817100381710038171003817120582119891

10038171003817100381710038171003817

1003817100381710038171003817ℎ11989910038171003817100381710038171198831015840

(72)

By (P3) in Definition 18 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (73)

and Property (D) gives that 119892 isin 119883 In the last step we want toupgrade the integral representation (69) to hold for arbitraryfunctions in 120593 isin 119883

1015840 As before we approximate 120593 by simplefunctions ℎ

119899with support contained in Ω

119899and 0 le ℎ

119899uarr |120593|

in 120583-almost everywhere point Define 119904119899

= ℎ119899sgn(120593) By

(NC) 119904119899converges to 120593 in119883

1015840 and as both 120582119891and ldquoits integral

representationrdquo are continuous and agree on the set of stepfunctions they agree on119883

1015840

Having shown the existence of the weak derivativesof highest order we are left to show the existence of theintermediate derivatives which we do in Proposition 22 inthe next section To keep the flow going we give here theproof that the pointwise inequality implies membership inthe corresponding Sobolev space under the assumption thatwe already know that Proposition 22 holds

Proof of Theorem 2 (pointwise inequality implies membershipin Sobolev space) We combine Lemmas 20 and 21 and applyProposition 22

4 Equivalent Definitions of Sobolev Spaces

The aim of this section is to prove the following equivalence

Proposition 22 Let Ω sub R be a domain with finite Lebesguemeasure inR Suppose119883 is a Banach function space satisfying(B) and (D) Assume that 119898 isin N Then the following setscoincide

(a) 119860 = 119891 isin 119883 there exists 119865 isin 119883 where 119865 agrees almosteverywherewith119891 and hasweak derivatives up to order119898 in119883 where the derivatives of order 119903 0 le 119903 le 119898minus1are absolutely continuous

(b) 119861 = 119891 isin 119883 119891(119896)

isin 119883 119896 isin 0 119898 where 119891(119896) arethe weak derivatives of 119891

(c) 119862 = 119891 isin 119883 119891(119898)

isin 119883 where 119891(119898) is the weak

derivative of 119891 of order 119898

We see that 119860 sub 119861 sub 119862 We now show that 119862 sub 119860Let us give an outline of the proof Given 119892 Ω rarr R

(we will choose 119892 = 119891(119898)) by integrating we construct in

Lemma 23 an absolutely continuous function 119866119898together

with its weak derivatives 1198660 119866

119898minus1 which are absolutely

continuous and 119866(119898)

119898= 119892 However 119866

119898and 119891 might

not agree almost everywhere We verify however in theproof of Theorem 26 (with the help of Theorem 24) thatwe find a polynomial 119875 such that 119866

119898+ 119875 agrees with 119865

almost everywhere and shares with 119866119898the other required

properties We conclude the proof of Theorem 26 by settlingthe membership questions with the aid of Proposition 25

Lemma 23 Assume Ω sub R is a domain We further assumethat 119892 Ω rarr R is in 119871

1

loc(Ω) One fixes 119909lowast

isin Ω and defines

1198660(119909

0) = 119892 (119909

0)

119866119896(119909

119896) = int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119896 isin N

(74)

where the integrals are Lebesgue integrals and we use thenotation

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

=

int[119909lowast119909119896]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909lowast

le 119909119896

minusint[119909119896 119909lowast]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909119896lt 119909

lowast

(75)

Then

(a) 119866119896is locally absolutely continuous for all 119896 isin N

especially locally integrable(b) if 120593 isin Cinfin

0(Ω) and 0 le 119903 le 119896 one has

intΩ

119866119896120593(119903)

= (minus1)119903

intΩ

119866119896minus119903

120593 (76)

Proof Let us prove (a) by induction Assume 119896 = 1 Note that1198661(119909

lowast

) = 0 Thus we have for 119909lowast

le 1199091

1198661(119909

1) minus 119866

1(119909

lowast

) = 1198661(119909

1) = int

1199091

119909lowast

119892 (1199090) 119889119909

0 (77)

As 119892 isin 1198711

([119909lowast

1199091]) we see that 119866

1is absolutely continuous

The case 1199091lt 119909

lowast is treated similarlyAssume now that the claim is true for some 119896 Now

119866119896+1

(119909119896+1

) is the integral of a continuous function and thuslocally absolutely continuous

Let us have a look at (b) Note that the statement is true for119896 = 0 thus we may assume that 119896 ge 1 We make an inductionover 119903 The case 119903 = 0 is clear Let us choose an interval[119886 119887] sub Ω containing a neighborhood of the support of 120593 As119866119896and 120593

(119903) are both absolutely continuous in [119886 119887] we have

10 Journal of Function Spaces

by the integration by part formula for absolutely continuousfunctions (see for exampleTheorem7147 in [9]) and the factthat 120593(119903)

(119886) = 120593(119903)

(119887) = 0

intΩ

119866119896120593(119903+1)

= int[119886119887]

119866119896120593(119903+1)

= (minus1) int[119886119887]

1198661015840

119896120593(119903)

(78)

Noting that 1198661015840

119896= 119866

119896minus1almost everywhere we see that

intΩ

119866119896120593(119903+1)

= (minus1) int[119886119887]

119866119896minus1

120593(119903)

= (minus1) intΩ

119866119896minus1

120593(119903)

(79)

Using the induction hypothesis the claim follows

We cite Theorem 253 in [10] without giving its proof

Theorem 24 Let 119860 = [119886 119887] R or R+and 119903 ge 0 If the

function 119891 isin 1198711

loc(119860) satisfies

int119860

119891120593(119903)

119889119909 = 0 (80)

for all 120593 isin Cinfin

0(119860) then there is a polynomial 119875 of degree lt 119903

such that 119891 = 119875 almost everywhere on 119860

To show that 119866119896is in 119883 we need to verify that (if there

is danger of confusion with respect to which variable we takethe norm we use notations as119883(119909

119896))

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

lt infin (81)

under the assumption that 119866119896minus1

is in 119883 Actually bounding119866119896from above by its absolute value this means that we have

first to take the1198711-norm and then the norm in119883We howeverwould like to switch the norms

We adapt Proposition 21 in [16] to our setting

Proposition 25 Let 120588 be a Banach function norm satisfyingproperty (D) If 119891 ΩtimesΩ rarr R is a measurable function andone has 119891

119909and 119891

119910 functions onΩ such that 119891119909(119910) = 119891(119909 119910) =

119891119910

(119909) then

120588 (1003817100381710038171003817119891119909

10038171003817100381710038171) le

1003817100381710038171003817120588 (119891119910

)10038171003817100381710038171

(82)

The cases 119883 isin 1198711

119871infin

are easy to prove If 119883 = 119871119901 for

some 1 lt 119901 lt infin then the result can be proven by usingFubinirsquos theorem together with property (D) For the generalresults we refer to Scheprsquos proof

The next result builds uponTheorem 254 in [10]

Theorem 26 Let Ω sub R be a domain and let 119903 ge 1 Suppose119883 is a Banach function space satisfying (B) and (D) If 119891 islocally in 119883 and has a generalized 119903th derivative 119892 locally in119883 then there is 119865 agreeing almost everywhere with 119891 havingabsolutely continuous intermediate weak derivatives and theweak derivative of order 119903 agrees almost everywhere with 119892Furthermore the derivatives of 119865 are locally in 119883 If Ω hasfinite measure one obtains the result with ldquolocallyrdquo replaced byldquogloballyrdquo

We thank Pilar Silvestre for pointing out how to prove themembership of the intermediate derivatives in119883

Proof Choose 119909lowast

isin Ω and let 119866119896be defined as in Lemma 23

Assume 120593 isin Cinfin

0(Ω) Then using the fact that 119892 is a weak

derivative of 119891 of order 119903 and (b) in Lemma 23

intΩ

(119891 minus 119866119903) 120593

(119903)

119889119909 = intΩ

119891120593(119903)

119889119909 minus intΩ

119866119903120593(119903)

119889119909

= (minus1)119903

intΩ

119892120593119889119909 minus (minus1)119903

intΩ

119892120593119889119909 = 0

(83)

UsingTheorem 24 we see that119891minus119866119903= 119875 almost everywhere

where119875 is a polynomial of degreelt 119903Thus119891 = 119875+119866119903almost

everywhere and for 120593 isin Cinfin

0(Ω) and 0 le 119898 le 119903

intΩ

119891120593(119898)

119889119909 = intΩ

(119875 + 119866119903) 120593

(119898)

= (minus1)119898

intΩ

(119866119903minus119898

+ 119875(119898)

) 120593 119889119909

(84)

Thus119866119903minus119898

+119875(119898) is the119898th generalized derivative of119891 As it is

locally absolutely continuous (if119898 lt 119903) it lies locally in119883 Inthe case 119898 = 119903 the same conclusion holds by the assumptionon 119892

Moreover there exists 119865 for example 119865 = 119875 + 119866119903 that

agrees with 119891 almost everywhere such that its generalizedderivatives up to order 119903 minus 1 are locally absolutely continuousand the 119903th generalized derivative equals 119892 almost every-where

As 119892 is locally in 119883 and the other derivatives are locallyabsolutely continuous they are locally in 119871

infin and thus locallyin119883 by (B)

Suppose now that Ω has additionally finite measureAssume that 119891 and 119892 are in119883 globally We want to show thatthe intermediate derivatives are in 119883 as well

Note in our setting that by (P5) in Definition 18membership in 119883 implies membership in 119871

1 We proceedby induction Suppose that 119866

119896minus1isin 119883 We will apply

Proposition 25 We set denoting by conv119909lowast

119909119896 the closed

interval with endpoints 119909lowast and 119909119896

119891 (119909119896minus1

119909119896) = 119866

119896minus1(119909

119896minus1) 120594conv119909lowast 119909119896 (119909119896minus1

) (85)

Then1003817100381710038171003817119866119896

(119909119896)1003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119891(119909119896minus1

119909119896)119889119909

119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)10038171003817100381710038171198711(119909119896minus1)

10038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)

le10038171003817100381710038171003817

1003816100381610038161003816119866119896minus1(119909

119896minus1)1003816100381610038161003816

1003817100381710038171003817120594Ω1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)lt infin

(86)

Journal of Function Spaces 11

5 Banach Function Space Setting

Here our main focus is a more advanced readership thatlikely knows already Banach function spaces Therefore weadopt a much briefer style pointing merely to the literaturewhere the corresponding definitions and results can be foundWe start with giving generalized versions of the necessity andsufficiency of Theorem 2

We list the results and prove them simultaneously bysplitting the proofs into parts according to the functionspaces

Theorem 27 Assume Ω sub R is a domain 119883 is a normedspace and 119898 isin N If the maximal function 119872 119883 rarr 119883 isbounded then there is a constant 119862 = 119862(119883119898) such that foreach continuous 119891 isin 119882

119898

119883there is 119886

119891isin 119883with 119886

119891 le 119862119891

119882119898119883

and

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (87)

In particular such 119886119891exists if

(i) 1 lt 119901 le infin and 119883 is the Lebesgue space 119871119901

(Ω)

(ii) 1 lt 119901 lt infin 1 le 119902 le infin and 119883 is the Lorentz space119871119901119902

(Ω)

(iii) 1 lt 119901 lt infin 1 le 119902 le infin 119887 is slowly varying and 119883 isthe Lorentz-Karamata space 119871

119901119902119887(Ω)

(iv) 120601 is an 119873-function with inf119904gt0

(119904120593(119904)120601(119904)) gt 1 where120593 is the density function of 120601 and 119883 is the Orlicz space119871lowast

120601(Ω)

(v) Ω = R and 119883 is rearrangement-invariant with upperindex 120572

119883lt 1

Theorem 28 Let Ω sub R be open Suppose 119883 is a Banachfunction space such that (AC) holds Assume further that thefunction 119891 Ω rarr R is in 119883 If for some 119903 isin N

0 the following

inequality holds

1003816100381610038161003816Δ119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)] (88)

for some 119886119891isin 119883 then there exists119892 isin 119883with 119892

119883le 2119903

119903

119886119891119883

such that

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909 = (minus1)119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (89)

for all 120593 isin Cinfin

0(Ω)

In particular the result holds if119883 is

(i) the Lebesgue space 119871119901

(Ω) for some 1 lt 119901 lt infin

(ii) the Lorentz space 119871119901119902

(Ω) where 1 lt 119901 lt infin and 1 le

119902 lt infin

(iii) the Lorentz-Karamata space 119871119901119902119887

(Ω) where 119887 isslowly varying 1 lt 119901 lt infin and 1 lt 119902 le infin

(iv) the Orlicz space 119871lowast

120601(Ω) for some 119873-function 120601 satisfy-

ing

inf119904gt0

119904120593 (119904)

120601 (119904)gt 1 (90)

where 120593 is the density function of 120601

To help with orientation let us give an overview ofthe remaining material in this section With respect toTheorem 27 note that we carried out the proof of Corollary 14in a high level of abstraction Hence we are left to verifythat the spaces under question are normed spaces (we willshow that they are even Banach function spaces) and theboundedness of the maximal operator

Concerning the proof Theorem 28 we proved the cor-responding part of Theorem 2 under the assumptions thatthe spaces under question are Banach function spaces andfurther satisfy the requirements listed in Properties 19 Theexistence of 1198831015840 is classical (Proposition 29) as is the fact that(AC) implies (A) and (NC) (Proposition 30) Hence we onlyneed to point to the literature where (AC) was verified

Proposition 29 Given a Banach function space there is aspace 119883

1015840 such that properties (B) (HI) (C) and (D) aresatisfied

Proof We refer the reader to Sections 11 and 12 in [13] or toSection 31 in [14]

The definition of absolutely continuous norm can befound as Definition 3111 in [14] and Definition 131 in [13]

Proposition 30 A Banach function space that has absolutelycontinuous norm satisfies (A) and (NC)

Proof The first statement follows fromTheorem 1313 in [13]and the second fromProposition 136 in the same source

51 General Statements

Proof of Theorem 27 (general statement) We basically havealready proven the general statement in Corollary 14

Proof of Theorem 28 (general statement) We gave the proofof the Lebesgue space case in such generality that thegeneral statement follows from the proof of Theorem 2 andPropositions 29 and 30

52 Lorentz-Karamata Spaces Lorentz-Karamata spaces aredetailed in [14] mainly Section 343

Proof of Theorem 27 (ii) and (iii) The part that is left toprove is the boundedness of the maximal operator which isguaranteed by Remark 3517 in [14]

Proof of Theorem 28 (ii) and (iii) By Theorem 3441 in [14]we see that 119883 is a rearrangement-invariant Banach space Acombination of Lemma 3439 in [14] together with Lemma3443 in [14] gives the absolute continuity (AC) of the normof1198831015840

12 Journal of Function Spaces

53 Orlicz Spaces The setting is the one of [17]

Lemma 31 If 120601 is an 119873-function satisfying the Δ2-condition

then the Orlicz space 119871lowast

120601has absolutely continuous norm

Proof Let 119891 isin 119871lowast

120601(Ω) We verify first that

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 lt infin (91)

Let 120576 gt 0 By the Δ2-condition there exists a constant 120581 =

120581(119891(120601)

+ 120576) such that

120601 ((1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576) 119904) le 120581120601 (119904) (92)

for all 119904 ge 0 Thus

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 = intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576(1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576))119889119909

le 120581intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576)119889119909 le 120581

(93)

Suppose that 119864 is a measurable set Then

10038171003817100381710038171198911205941198641003817100381710038171003817(120601)

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891120594119864

1003816100381610038161003816) 119889119909 le 1

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 1

(94)

Let us fix 120582 gt 0 Again using the Δ2-condition we have

120601(120582minus1

119904) le 120581(120582minus1

)120601(119904) Hence

intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 (95)

Example 133 in [13] tells us that 1198711 has absolutely continuousnorm Thus if 120594

119864119899converges 120583-almost everywhere to 0 then

intΩ

120601(|119891|)120594119864119899

converges to zero as well Thus we have for 119899

large enough that

120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119899119889119909 le 1 (96)

implying that 119891120594119864119899(120601)

le 120582 Since 120582 gt 0 was arbitrary theabsolute continuity of 119891 follows

Proof of Theorem 27 (iv) We conclude from Gallardorsquos resultTheorem 21 in [17] that it is sufficient that the complemen-tary 119873-function satisfies the Δ

2-condition in [0infin) Again

resorting to Gallardo this time to Proposition 14 in [17] weobtain the conclusion

Proof of Theorem 28 (iv) That 119883 is a Banach functionspace follows from Theorem 3416 in [14] To prove that119883

1015840 has absolutely continuous norm we note that inequal-ity (90) together with Proposition 14 in [17] ensuresthat the complementary 119873-function of 120601 satisfies the Δ

2-

condition Lemma 31 verifies that 1198831015840 has absolutely contin-uous norm

54 Rearrangement-Invariant Spaces The necessary defini-tions can be found in [13]

Proof of Theorem 27 (v) We are only left to show theboundedness of the maximal operator However this is thecontent of a result by Lorentz and Shimogaki for examplestated as Theorem 3517 in [13]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank Pilar Silvestre for hersuggestions of how to prove the existence of intermediatederivatives The research is supported by the Academy ofFinland The third author thanks the Department of Math-ematics and Systems Analysis at Aalto University for itshospitality

References

[1] B Bojarski ldquoRemarks on some geometric properties of Sobolevmappingsrdquo in Functional Analysis amp Related Topics (Sapporo1990) pp 65ndash76 World Scientific River Edge NJ USA 1991

[2] B Bojarski PHajlasz andP Strzelecki ldquoImproved119862119896120582 approxi-

mation of higher order Sobolev functions in normand capacityrdquoIndiana University Mathematics Journal vol 51 no 3 pp 507ndash540 2002

[3] B Bojarski ldquoPointwise characterization of Sobolev classesrdquoProceedings of the Steklov Institute of Mathematics vol 255 no1 pp 65ndash81 2006

[4] B Bojarski ldquoWhitneyrsquos jets for Sobolev functionsrdquo UkrainianMathematical Journal vol 59 no 3 pp 345ndash358 2007

[5] B Bojarski ldquoTaylor expansion and Sobolev spacesrdquo Bulletin ofthe Georgian National Academy of Sciences New Series vol 5no 2 pp 5ndash10 2011

[6] B Bojarski ldquoSobolev spaces and Lagrange interpolationrdquoGeorgian Academy of Sciences Proceedings of A RazmadzeMathematical Institute vol 158 pp 1ndash12 2012

[7] P Hajłasz ldquoSobolev spaces on an arbitrary metric spacerdquoPotential Analysis vol 5 no 4 pp 403ndash415 1996

[8] B Bojarski ldquoSobolev spaces and averaging Irdquo httparxivorgabs13085171

[9] R Kannan and C K Krueger Advanced Analysis on the RealLine Universitext Springer New York NY USA 1996

[10] R A DeVore and G G Lorentz Constructive Approximationvol 303 of Fundamental Principles of Mathematical SciencesSpringer Berlin Germany 1993

[11] L L Schumaker Spline Functions Basic Theory CambridgeMathematical Library CambridgeUniversity Press CambridgeUK 3rd edition 2007

[12] G H Hardy and J E Littlewood ldquoA maximal theorem withfunction-theoretic applicationsrdquo Acta Mathematica vol 54 no1 pp 81ndash116 1930

[13] C Bennett andR Sharpley Interpolation ofOperators vol 129 ofPure and Applied Mathematics Academic Press Boston MassUSA 1988

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Journal of Function Spaces

Lemma 21 Assume Ω sub R is open Suppose the propertieslisted in Properties 19 are met Let 120582

119891be defined as in (47) and

bounded on 1198831015840 Then there exists 119892 isin 119883 with 119892

119883le 120582

119891

such that

120582119891(120593) = (minus1)

119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (55)

for all 120593 isin 1198831015840 especially for all 120593 isin Cinfin

0(Ω)

Proof The proof is heavily influenced by the proof of Theo-rem 141 in [13]We first want to find a signedmeasure ] suchthat

120582119891(120593) = int

Ω

120593119889] (56)

for each 120593 isin 1198831015840 Then we want to apply the Lebesgue-

Radon-Nikodym theorem to replace 119889] by (minus1)119903

119892(119909)119889119909 forsome integrable function 119892 for which we then show that it iscontained in119883

We start by assuming that the measure of Ω is finite Foreach 120583-measurable set 119860 we set

] (119860) = 120582119891(120594

119860) (57)

We start by proving the 120590-additivity of ] (for the 120583-measurable sets) In the first step we just show the additivityBy induction it is enough to consider only two sets

Let 1198601and 119860

2be two disjoint 120583-measurable sets Then

] (1198601cup 119860

2) = 120582

119891(120594

1198601cup1198602) = 120582

119891(120594

1198601+ 120594

1198602)

= 120582119891(120594

1198601) + 120582

119891(120594

1198602) = ] (119860

1) + ] (119860

2)

(58)

We move to the verification of the 120590-additivity Let119860

1 119860

2 be countablymany pairwise disjoint120583-measurable

sets We first want to show thatsum119894120594119860119894

converges in1198831015840

We let 119864119896= cup

119896

119894=1119860

119894and 119864 = cup

infin

119894=1119860

119894 Then since (119864

119896)119896is

increasing

120583 (119864) = 120583(

infin

⋃

119896=1

119864119896) = lim

119896rarrinfin

120583 (119864119896) (59)

As 119864 has finite measure given 120576 gt 0 there exists 1198960isin N such

that

(1 minus 120576) 120583 (119864) lt 120583 (1198641198960) (60)

By the disjointness of the sets 119860119894 we see that

120583 (119864) = 120583 (1198641198960) + 120583(

infin

⋃

119894=1198960+1

119860119894) (61)

It follows that

120583(

infin

⋃

119894=1198960+1

119860119894) = 120583 (119864) minus 120583 (119864

1198960) lt 120583 (119864) minus (1 minus 120576) 120583 (119864)

= 120576120583 (119864)

(62)

We have for119898 lt 119899

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=1

120594119860119894

minus

119898

sum

119894=1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

=

1003817100381710038171003817100381710038171003817100381710038171003817

119899

sum

119894=119898+1

120594119860119894

10038171003817100381710038171003817100381710038171003817100381710038171198831015840

le10038171003817100381710038171003817120594cupinfin119894=119898+1

119860119894

100381710038171003817100381710038171198831015840 (63)

By the considerations before and the fact that sdot 1198831015840 is

absolutely continuous as stipulated in (AC) we see thatsum

119899

119894=1120594119860119894

is a Cauchy sequence Now as 1198831015840 is complete by

(C) the Cauchy sequence converges in 1198831015840 We have by the

continuity of 120582119891

] (cup119894119860

119894) = 120582

119891(120594

cup119894119860119894) = 120582

119891(

infin

sum

119894=1

120594119860119894)

= lim119899rarrinfin

120582119891(

119899

sum

119894=1

120594119860119894) = lim

119899rarrinfin

119899

sum

119894=1

120582119891(120594

119860119894)

= lim119899rarrinfin

119899

sum

119894=1

] (119860119894) =

infin

sum

119894=1

] (119860119894)

(64)

Thus ] is 120590-additiveIn order to apply the theorem of Lebesgue-Radon-

Nikodym see for example Section 69 in [15] and conclud-ing the existence of 119892 isin 119871

1

(Ω) such that

] (119860) = int119860

119892119889119909 119860 120583-measurable (65)

we need to show that ] is absolutely continuous with respectto 120583 To do so assume that 120583(119873) = 0 Then ](119873) = 120582

119891(120594

119873)

By the absolute continuity of the normof1198831015840 required in (AC)we see that 120594

1198731198831015840 = 0 and therefore ](119873) = 0 Replacing 119892

by minus119892 if necessary we obtain

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909 (66)

Now we assume that Ω does not necessarily have finitemeasureWewriteΩ = cup

119899Ω

119899as union of open sets with finite

measure such that Ω119899sub Ω

119899+1 By the considerations above

we find for each 119899 isin N a 119892119899such that

120582119891(120594

119860) = (minus1)

119903

intΩ

119892119899(119909) 120594

119860(119909) 119889119909 (67)

for each measurable set 119860 sub Ω119899 It is easily seen that 119892

119899

and 119892119899+1

agree almost everywhere on Ω119899 This gives rise to

a function 119892 satisfying

120582119891(120594

119860) = (minus1)

119903

intΩ

119892 (119909) 120594119860(119909) 119889119909

119860 sub

119873

⋃

119899=1

Ω119899for some 119873

(68)

By linearity we have for each simple function 119904

120582119891(119904) = (minus1)

119903

intΩ

119892 (119909) 119904 (119909) 119889119909 (69)

To show that 119892 isin 119883 we want to use property (D)

Journal of Function Spaces 9

Let ℎ be a nonnegative simple function with supportin some Ω

119899 Then ℎ sgn(119892) is also a simple function with

support inΩ119899 It is a finite linear combination of characteristic

functions of sets in Ω119899 Hence

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = intℎ sdot sgn (119892) 119892 119889119909 = 120582119891(ℎ sgn (119892)) (70)

Using the boundedness of 120582119891 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (71)

If ℎ is an arbitrary function in1198831015840 thenwemay construct a

sequence (ℎ119899)infin

119899=1of simple functions each ℎ

119899having support

in Ω119899 such that 0 le ℎ

119899uarr |ℎ| 120583-almost everywhere By the

monotone convergence theorem we have by (71) that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = lim119899rarrinfin

int1003816100381610038161003816ℎ119899119892

1003816100381610038161003816 119889119909 le lim sup119899rarrinfin

10038171003817100381710038171003817120582119891

10038171003817100381710038171003817

1003817100381710038171003817ℎ11989910038171003817100381710038171198831015840

(72)

By (P3) in Definition 18 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (73)

and Property (D) gives that 119892 isin 119883 In the last step we want toupgrade the integral representation (69) to hold for arbitraryfunctions in 120593 isin 119883

1015840 As before we approximate 120593 by simplefunctions ℎ

119899with support contained in Ω

119899and 0 le ℎ

119899uarr |120593|

in 120583-almost everywhere point Define 119904119899

= ℎ119899sgn(120593) By

(NC) 119904119899converges to 120593 in119883

1015840 and as both 120582119891and ldquoits integral

representationrdquo are continuous and agree on the set of stepfunctions they agree on119883

1015840

Having shown the existence of the weak derivativesof highest order we are left to show the existence of theintermediate derivatives which we do in Proposition 22 inthe next section To keep the flow going we give here theproof that the pointwise inequality implies membership inthe corresponding Sobolev space under the assumption thatwe already know that Proposition 22 holds

Proof of Theorem 2 (pointwise inequality implies membershipin Sobolev space) We combine Lemmas 20 and 21 and applyProposition 22

4 Equivalent Definitions of Sobolev Spaces

The aim of this section is to prove the following equivalence

Proposition 22 Let Ω sub R be a domain with finite Lebesguemeasure inR Suppose119883 is a Banach function space satisfying(B) and (D) Assume that 119898 isin N Then the following setscoincide

(a) 119860 = 119891 isin 119883 there exists 119865 isin 119883 where 119865 agrees almosteverywherewith119891 and hasweak derivatives up to order119898 in119883 where the derivatives of order 119903 0 le 119903 le 119898minus1are absolutely continuous

(b) 119861 = 119891 isin 119883 119891(119896)

isin 119883 119896 isin 0 119898 where 119891(119896) arethe weak derivatives of 119891

(c) 119862 = 119891 isin 119883 119891(119898)

isin 119883 where 119891(119898) is the weak

derivative of 119891 of order 119898

We see that 119860 sub 119861 sub 119862 We now show that 119862 sub 119860Let us give an outline of the proof Given 119892 Ω rarr R

(we will choose 119892 = 119891(119898)) by integrating we construct in

Lemma 23 an absolutely continuous function 119866119898together

with its weak derivatives 1198660 119866

119898minus1 which are absolutely

continuous and 119866(119898)

119898= 119892 However 119866

119898and 119891 might

not agree almost everywhere We verify however in theproof of Theorem 26 (with the help of Theorem 24) thatwe find a polynomial 119875 such that 119866

119898+ 119875 agrees with 119865

almost everywhere and shares with 119866119898the other required

properties We conclude the proof of Theorem 26 by settlingthe membership questions with the aid of Proposition 25

Lemma 23 Assume Ω sub R is a domain We further assumethat 119892 Ω rarr R is in 119871

1

loc(Ω) One fixes 119909lowast

isin Ω and defines

1198660(119909

0) = 119892 (119909

0)

119866119896(119909

119896) = int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119896 isin N

(74)

where the integrals are Lebesgue integrals and we use thenotation

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

=

int[119909lowast119909119896]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909lowast

le 119909119896

minusint[119909119896 119909lowast]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909119896lt 119909

lowast

(75)

Then

(a) 119866119896is locally absolutely continuous for all 119896 isin N

especially locally integrable(b) if 120593 isin Cinfin

0(Ω) and 0 le 119903 le 119896 one has

intΩ

119866119896120593(119903)

= (minus1)119903

intΩ

119866119896minus119903

120593 (76)

Proof Let us prove (a) by induction Assume 119896 = 1 Note that1198661(119909

lowast

) = 0 Thus we have for 119909lowast

le 1199091

1198661(119909

1) minus 119866

1(119909

lowast

) = 1198661(119909

1) = int

1199091

119909lowast

119892 (1199090) 119889119909

0 (77)

As 119892 isin 1198711

([119909lowast

1199091]) we see that 119866

1is absolutely continuous

The case 1199091lt 119909

lowast is treated similarlyAssume now that the claim is true for some 119896 Now

119866119896+1

(119909119896+1

) is the integral of a continuous function and thuslocally absolutely continuous

Let us have a look at (b) Note that the statement is true for119896 = 0 thus we may assume that 119896 ge 1 We make an inductionover 119903 The case 119903 = 0 is clear Let us choose an interval[119886 119887] sub Ω containing a neighborhood of the support of 120593 As119866119896and 120593

(119903) are both absolutely continuous in [119886 119887] we have

10 Journal of Function Spaces

by the integration by part formula for absolutely continuousfunctions (see for exampleTheorem7147 in [9]) and the factthat 120593(119903)

(119886) = 120593(119903)

(119887) = 0

intΩ

119866119896120593(119903+1)

= int[119886119887]

119866119896120593(119903+1)

= (minus1) int[119886119887]

1198661015840

119896120593(119903)

(78)

Noting that 1198661015840

119896= 119866

119896minus1almost everywhere we see that

intΩ

119866119896120593(119903+1)

= (minus1) int[119886119887]

119866119896minus1

120593(119903)

= (minus1) intΩ

119866119896minus1

120593(119903)

(79)

Using the induction hypothesis the claim follows

We cite Theorem 253 in [10] without giving its proof

Theorem 24 Let 119860 = [119886 119887] R or R+and 119903 ge 0 If the

function 119891 isin 1198711

loc(119860) satisfies

int119860

119891120593(119903)

119889119909 = 0 (80)

for all 120593 isin Cinfin

0(119860) then there is a polynomial 119875 of degree lt 119903

such that 119891 = 119875 almost everywhere on 119860

To show that 119866119896is in 119883 we need to verify that (if there

is danger of confusion with respect to which variable we takethe norm we use notations as119883(119909

119896))

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

lt infin (81)

under the assumption that 119866119896minus1

is in 119883 Actually bounding119866119896from above by its absolute value this means that we have

first to take the1198711-norm and then the norm in119883We howeverwould like to switch the norms

We adapt Proposition 21 in [16] to our setting

Proposition 25 Let 120588 be a Banach function norm satisfyingproperty (D) If 119891 ΩtimesΩ rarr R is a measurable function andone has 119891

119909and 119891

119910 functions onΩ such that 119891119909(119910) = 119891(119909 119910) =

119891119910

(119909) then

120588 (1003817100381710038171003817119891119909

10038171003817100381710038171) le

1003817100381710038171003817120588 (119891119910

)10038171003817100381710038171

(82)

The cases 119883 isin 1198711

119871infin

are easy to prove If 119883 = 119871119901 for

some 1 lt 119901 lt infin then the result can be proven by usingFubinirsquos theorem together with property (D) For the generalresults we refer to Scheprsquos proof

The next result builds uponTheorem 254 in [10]

Theorem 26 Let Ω sub R be a domain and let 119903 ge 1 Suppose119883 is a Banach function space satisfying (B) and (D) If 119891 islocally in 119883 and has a generalized 119903th derivative 119892 locally in119883 then there is 119865 agreeing almost everywhere with 119891 havingabsolutely continuous intermediate weak derivatives and theweak derivative of order 119903 agrees almost everywhere with 119892Furthermore the derivatives of 119865 are locally in 119883 If Ω hasfinite measure one obtains the result with ldquolocallyrdquo replaced byldquogloballyrdquo

We thank Pilar Silvestre for pointing out how to prove themembership of the intermediate derivatives in119883

Proof Choose 119909lowast

isin Ω and let 119866119896be defined as in Lemma 23

Assume 120593 isin Cinfin

0(Ω) Then using the fact that 119892 is a weak

derivative of 119891 of order 119903 and (b) in Lemma 23

intΩ

(119891 minus 119866119903) 120593

(119903)

119889119909 = intΩ

119891120593(119903)

119889119909 minus intΩ

119866119903120593(119903)

119889119909

= (minus1)119903

intΩ

119892120593119889119909 minus (minus1)119903

intΩ

119892120593119889119909 = 0

(83)

UsingTheorem 24 we see that119891minus119866119903= 119875 almost everywhere

where119875 is a polynomial of degreelt 119903Thus119891 = 119875+119866119903almost

everywhere and for 120593 isin Cinfin

0(Ω) and 0 le 119898 le 119903

intΩ

119891120593(119898)

119889119909 = intΩ

(119875 + 119866119903) 120593

(119898)

= (minus1)119898

intΩ

(119866119903minus119898

+ 119875(119898)

) 120593 119889119909

(84)

Thus119866119903minus119898

+119875(119898) is the119898th generalized derivative of119891 As it is

locally absolutely continuous (if119898 lt 119903) it lies locally in119883 Inthe case 119898 = 119903 the same conclusion holds by the assumptionon 119892

Moreover there exists 119865 for example 119865 = 119875 + 119866119903 that

agrees with 119891 almost everywhere such that its generalizedderivatives up to order 119903 minus 1 are locally absolutely continuousand the 119903th generalized derivative equals 119892 almost every-where

As 119892 is locally in 119883 and the other derivatives are locallyabsolutely continuous they are locally in 119871

infin and thus locallyin119883 by (B)

Suppose now that Ω has additionally finite measureAssume that 119891 and 119892 are in119883 globally We want to show thatthe intermediate derivatives are in 119883 as well

Note in our setting that by (P5) in Definition 18membership in 119883 implies membership in 119871

1 We proceedby induction Suppose that 119866

119896minus1isin 119883 We will apply

Proposition 25 We set denoting by conv119909lowast

119909119896 the closed

interval with endpoints 119909lowast and 119909119896

119891 (119909119896minus1

119909119896) = 119866

119896minus1(119909

119896minus1) 120594conv119909lowast 119909119896 (119909119896minus1

) (85)

Then1003817100381710038171003817119866119896

(119909119896)1003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119891(119909119896minus1

119909119896)119889119909

119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)10038171003817100381710038171198711(119909119896minus1)

10038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)

le10038171003817100381710038171003817

1003816100381610038161003816119866119896minus1(119909

119896minus1)1003816100381610038161003816

1003817100381710038171003817120594Ω1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)lt infin

(86)

Journal of Function Spaces 11

5 Banach Function Space Setting

Here our main focus is a more advanced readership thatlikely knows already Banach function spaces Therefore weadopt a much briefer style pointing merely to the literaturewhere the corresponding definitions and results can be foundWe start with giving generalized versions of the necessity andsufficiency of Theorem 2

We list the results and prove them simultaneously bysplitting the proofs into parts according to the functionspaces

Theorem 27 Assume Ω sub R is a domain 119883 is a normedspace and 119898 isin N If the maximal function 119872 119883 rarr 119883 isbounded then there is a constant 119862 = 119862(119883119898) such that foreach continuous 119891 isin 119882

119898

119883there is 119886

119891isin 119883with 119886

119891 le 119862119891

119882119898119883

and

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (87)

In particular such 119886119891exists if

(i) 1 lt 119901 le infin and 119883 is the Lebesgue space 119871119901

(Ω)

(ii) 1 lt 119901 lt infin 1 le 119902 le infin and 119883 is the Lorentz space119871119901119902

(Ω)

(iii) 1 lt 119901 lt infin 1 le 119902 le infin 119887 is slowly varying and 119883 isthe Lorentz-Karamata space 119871

119901119902119887(Ω)

(iv) 120601 is an 119873-function with inf119904gt0

(119904120593(119904)120601(119904)) gt 1 where120593 is the density function of 120601 and 119883 is the Orlicz space119871lowast

120601(Ω)

(v) Ω = R and 119883 is rearrangement-invariant with upperindex 120572

119883lt 1

Theorem 28 Let Ω sub R be open Suppose 119883 is a Banachfunction space such that (AC) holds Assume further that thefunction 119891 Ω rarr R is in 119883 If for some 119903 isin N

0 the following

inequality holds

1003816100381610038161003816Δ119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)] (88)

for some 119886119891isin 119883 then there exists119892 isin 119883with 119892

119883le 2119903

119903

119886119891119883

such that

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909 = (minus1)119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (89)

for all 120593 isin Cinfin

0(Ω)

In particular the result holds if119883 is

(i) the Lebesgue space 119871119901

(Ω) for some 1 lt 119901 lt infin

(ii) the Lorentz space 119871119901119902

(Ω) where 1 lt 119901 lt infin and 1 le

119902 lt infin

(iii) the Lorentz-Karamata space 119871119901119902119887

(Ω) where 119887 isslowly varying 1 lt 119901 lt infin and 1 lt 119902 le infin

(iv) the Orlicz space 119871lowast

120601(Ω) for some 119873-function 120601 satisfy-

ing

inf119904gt0

119904120593 (119904)

120601 (119904)gt 1 (90)

where 120593 is the density function of 120601

To help with orientation let us give an overview ofthe remaining material in this section With respect toTheorem 27 note that we carried out the proof of Corollary 14in a high level of abstraction Hence we are left to verifythat the spaces under question are normed spaces (we willshow that they are even Banach function spaces) and theboundedness of the maximal operator

Concerning the proof Theorem 28 we proved the cor-responding part of Theorem 2 under the assumptions thatthe spaces under question are Banach function spaces andfurther satisfy the requirements listed in Properties 19 Theexistence of 1198831015840 is classical (Proposition 29) as is the fact that(AC) implies (A) and (NC) (Proposition 30) Hence we onlyneed to point to the literature where (AC) was verified

Proposition 29 Given a Banach function space there is aspace 119883

1015840 such that properties (B) (HI) (C) and (D) aresatisfied

Proof We refer the reader to Sections 11 and 12 in [13] or toSection 31 in [14]

The definition of absolutely continuous norm can befound as Definition 3111 in [14] and Definition 131 in [13]

Proposition 30 A Banach function space that has absolutelycontinuous norm satisfies (A) and (NC)

Proof The first statement follows fromTheorem 1313 in [13]and the second fromProposition 136 in the same source

51 General Statements

Proof of Theorem 27 (general statement) We basically havealready proven the general statement in Corollary 14

Proof of Theorem 28 (general statement) We gave the proofof the Lebesgue space case in such generality that thegeneral statement follows from the proof of Theorem 2 andPropositions 29 and 30

52 Lorentz-Karamata Spaces Lorentz-Karamata spaces aredetailed in [14] mainly Section 343

Proof of Theorem 27 (ii) and (iii) The part that is left toprove is the boundedness of the maximal operator which isguaranteed by Remark 3517 in [14]

Proof of Theorem 28 (ii) and (iii) By Theorem 3441 in [14]we see that 119883 is a rearrangement-invariant Banach space Acombination of Lemma 3439 in [14] together with Lemma3443 in [14] gives the absolute continuity (AC) of the normof1198831015840

12 Journal of Function Spaces

53 Orlicz Spaces The setting is the one of [17]

Lemma 31 If 120601 is an 119873-function satisfying the Δ2-condition

then the Orlicz space 119871lowast

120601has absolutely continuous norm

Proof Let 119891 isin 119871lowast

120601(Ω) We verify first that

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 lt infin (91)

Let 120576 gt 0 By the Δ2-condition there exists a constant 120581 =

120581(119891(120601)

+ 120576) such that

120601 ((1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576) 119904) le 120581120601 (119904) (92)

for all 119904 ge 0 Thus

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 = intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576(1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576))119889119909

le 120581intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576)119889119909 le 120581

(93)

Suppose that 119864 is a measurable set Then

10038171003817100381710038171198911205941198641003817100381710038171003817(120601)

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891120594119864

1003816100381610038161003816) 119889119909 le 1

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 1

(94)

Let us fix 120582 gt 0 Again using the Δ2-condition we have

120601(120582minus1

119904) le 120581(120582minus1

)120601(119904) Hence

intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 (95)

Example 133 in [13] tells us that 1198711 has absolutely continuousnorm Thus if 120594

119864119899converges 120583-almost everywhere to 0 then

intΩ

120601(|119891|)120594119864119899

converges to zero as well Thus we have for 119899

large enough that

120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119899119889119909 le 1 (96)

implying that 119891120594119864119899(120601)

le 120582 Since 120582 gt 0 was arbitrary theabsolute continuity of 119891 follows

Proof of Theorem 27 (iv) We conclude from Gallardorsquos resultTheorem 21 in [17] that it is sufficient that the complemen-tary 119873-function satisfies the Δ

2-condition in [0infin) Again

resorting to Gallardo this time to Proposition 14 in [17] weobtain the conclusion

Proof of Theorem 28 (iv) That 119883 is a Banach functionspace follows from Theorem 3416 in [14] To prove that119883

1015840 has absolutely continuous norm we note that inequal-ity (90) together with Proposition 14 in [17] ensuresthat the complementary 119873-function of 120601 satisfies the Δ

2-

condition Lemma 31 verifies that 1198831015840 has absolutely contin-uous norm

54 Rearrangement-Invariant Spaces The necessary defini-tions can be found in [13]

Proof of Theorem 27 (v) We are only left to show theboundedness of the maximal operator However this is thecontent of a result by Lorentz and Shimogaki for examplestated as Theorem 3517 in [13]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank Pilar Silvestre for hersuggestions of how to prove the existence of intermediatederivatives The research is supported by the Academy ofFinland The third author thanks the Department of Math-ematics and Systems Analysis at Aalto University for itshospitality

References

[1] B Bojarski ldquoRemarks on some geometric properties of Sobolevmappingsrdquo in Functional Analysis amp Related Topics (Sapporo1990) pp 65ndash76 World Scientific River Edge NJ USA 1991

[2] B Bojarski PHajlasz andP Strzelecki ldquoImproved119862119896120582 approxi-

mation of higher order Sobolev functions in normand capacityrdquoIndiana University Mathematics Journal vol 51 no 3 pp 507ndash540 2002

[3] B Bojarski ldquoPointwise characterization of Sobolev classesrdquoProceedings of the Steklov Institute of Mathematics vol 255 no1 pp 65ndash81 2006

[4] B Bojarski ldquoWhitneyrsquos jets for Sobolev functionsrdquo UkrainianMathematical Journal vol 59 no 3 pp 345ndash358 2007

[5] B Bojarski ldquoTaylor expansion and Sobolev spacesrdquo Bulletin ofthe Georgian National Academy of Sciences New Series vol 5no 2 pp 5ndash10 2011

[6] B Bojarski ldquoSobolev spaces and Lagrange interpolationrdquoGeorgian Academy of Sciences Proceedings of A RazmadzeMathematical Institute vol 158 pp 1ndash12 2012

[7] P Hajłasz ldquoSobolev spaces on an arbitrary metric spacerdquoPotential Analysis vol 5 no 4 pp 403ndash415 1996

[8] B Bojarski ldquoSobolev spaces and averaging Irdquo httparxivorgabs13085171

[9] R Kannan and C K Krueger Advanced Analysis on the RealLine Universitext Springer New York NY USA 1996

[10] R A DeVore and G G Lorentz Constructive Approximationvol 303 of Fundamental Principles of Mathematical SciencesSpringer Berlin Germany 1993

[11] L L Schumaker Spline Functions Basic Theory CambridgeMathematical Library CambridgeUniversity Press CambridgeUK 3rd edition 2007

[12] G H Hardy and J E Littlewood ldquoA maximal theorem withfunction-theoretic applicationsrdquo Acta Mathematica vol 54 no1 pp 81ndash116 1930

[13] C Bennett andR Sharpley Interpolation ofOperators vol 129 ofPure and Applied Mathematics Academic Press Boston MassUSA 1988

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 9

Let ℎ be a nonnegative simple function with supportin some Ω

119899 Then ℎ sgn(119892) is also a simple function with

support inΩ119899 It is a finite linear combination of characteristic

functions of sets in Ω119899 Hence

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = intℎ sdot sgn (119892) 119892 119889119909 = 120582119891(ℎ sgn (119892)) (70)

Using the boundedness of 120582119891 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (71)

If ℎ is an arbitrary function in1198831015840 thenwemay construct a

sequence (ℎ119899)infin

119899=1of simple functions each ℎ

119899having support

in Ω119899 such that 0 le ℎ

119899uarr |ℎ| 120583-almost everywhere By the

monotone convergence theorem we have by (71) that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 = lim119899rarrinfin

int1003816100381610038161003816ℎ119899119892

1003816100381610038161003816 119889119909 le lim sup119899rarrinfin

10038171003817100381710038171003817120582119891

10038171003817100381710038171003817

1003817100381710038171003817ℎ11989910038171003817100381710038171198831015840

(72)

By (P3) in Definition 18 we see that

int1003816100381610038161003816ℎ119892

1003816100381610038161003816 119889119909 le10038171003817100381710038171003817120582119891

10038171003817100381710038171003817ℎ

1198831015840 (73)

and Property (D) gives that 119892 isin 119883 In the last step we want toupgrade the integral representation (69) to hold for arbitraryfunctions in 120593 isin 119883

1015840 As before we approximate 120593 by simplefunctions ℎ

119899with support contained in Ω

119899and 0 le ℎ

119899uarr |120593|

in 120583-almost everywhere point Define 119904119899

= ℎ119899sgn(120593) By

(NC) 119904119899converges to 120593 in119883

1015840 and as both 120582119891and ldquoits integral

representationrdquo are continuous and agree on the set of stepfunctions they agree on119883

1015840

Having shown the existence of the weak derivativesof highest order we are left to show the existence of theintermediate derivatives which we do in Proposition 22 inthe next section To keep the flow going we give here theproof that the pointwise inequality implies membership inthe corresponding Sobolev space under the assumption thatwe already know that Proposition 22 holds

Proof of Theorem 2 (pointwise inequality implies membershipin Sobolev space) We combine Lemmas 20 and 21 and applyProposition 22

4 Equivalent Definitions of Sobolev Spaces

The aim of this section is to prove the following equivalence

Proposition 22 Let Ω sub R be a domain with finite Lebesguemeasure inR Suppose119883 is a Banach function space satisfying(B) and (D) Assume that 119898 isin N Then the following setscoincide

(a) 119860 = 119891 isin 119883 there exists 119865 isin 119883 where 119865 agrees almosteverywherewith119891 and hasweak derivatives up to order119898 in119883 where the derivatives of order 119903 0 le 119903 le 119898minus1are absolutely continuous

(b) 119861 = 119891 isin 119883 119891(119896)

isin 119883 119896 isin 0 119898 where 119891(119896) arethe weak derivatives of 119891

(c) 119862 = 119891 isin 119883 119891(119898)

isin 119883 where 119891(119898) is the weak

derivative of 119891 of order 119898

We see that 119860 sub 119861 sub 119862 We now show that 119862 sub 119860Let us give an outline of the proof Given 119892 Ω rarr R

(we will choose 119892 = 119891(119898)) by integrating we construct in

Lemma 23 an absolutely continuous function 119866119898together

with its weak derivatives 1198660 119866

119898minus1 which are absolutely

continuous and 119866(119898)

119898= 119892 However 119866

119898and 119891 might

not agree almost everywhere We verify however in theproof of Theorem 26 (with the help of Theorem 24) thatwe find a polynomial 119875 such that 119866

119898+ 119875 agrees with 119865

almost everywhere and shares with 119866119898the other required

properties We conclude the proof of Theorem 26 by settlingthe membership questions with the aid of Proposition 25

Lemma 23 Assume Ω sub R is a domain We further assumethat 119892 Ω rarr R is in 119871

1

loc(Ω) One fixes 119909lowast

isin Ω and defines

1198660(119909

0) = 119892 (119909

0)

119866119896(119909

119896) = int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119896 isin N

(74)

where the integrals are Lebesgue integrals and we use thenotation

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

=

int[119909lowast119909119896]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909lowast

le 119909119896

minusint[119909119896 119909lowast]

119866119896minus1

(119909119896minus1

) 119889119909119896minus1

119909119896lt 119909

lowast

(75)

Then

(a) 119866119896is locally absolutely continuous for all 119896 isin N

especially locally integrable(b) if 120593 isin Cinfin

0(Ω) and 0 le 119903 le 119896 one has

intΩ

119866119896120593(119903)

= (minus1)119903

intΩ

119866119896minus119903

120593 (76)

Proof Let us prove (a) by induction Assume 119896 = 1 Note that1198661(119909

lowast

) = 0 Thus we have for 119909lowast

le 1199091

1198661(119909

1) minus 119866

1(119909

lowast

) = 1198661(119909

1) = int

1199091

119909lowast

119892 (1199090) 119889119909

0 (77)

As 119892 isin 1198711

([119909lowast

1199091]) we see that 119866

1is absolutely continuous

The case 1199091lt 119909

lowast is treated similarlyAssume now that the claim is true for some 119896 Now

119866119896+1

(119909119896+1

) is the integral of a continuous function and thuslocally absolutely continuous

Let us have a look at (b) Note that the statement is true for119896 = 0 thus we may assume that 119896 ge 1 We make an inductionover 119903 The case 119903 = 0 is clear Let us choose an interval[119886 119887] sub Ω containing a neighborhood of the support of 120593 As119866119896and 120593

(119903) are both absolutely continuous in [119886 119887] we have

10 Journal of Function Spaces

by the integration by part formula for absolutely continuousfunctions (see for exampleTheorem7147 in [9]) and the factthat 120593(119903)

(119886) = 120593(119903)

(119887) = 0

intΩ

119866119896120593(119903+1)

= int[119886119887]

119866119896120593(119903+1)

= (minus1) int[119886119887]

1198661015840

119896120593(119903)

(78)

Noting that 1198661015840

119896= 119866

119896minus1almost everywhere we see that

intΩ

119866119896120593(119903+1)

= (minus1) int[119886119887]

119866119896minus1

120593(119903)

= (minus1) intΩ

119866119896minus1

120593(119903)

(79)

Using the induction hypothesis the claim follows

We cite Theorem 253 in [10] without giving its proof

Theorem 24 Let 119860 = [119886 119887] R or R+and 119903 ge 0 If the

function 119891 isin 1198711

loc(119860) satisfies

int119860

119891120593(119903)

119889119909 = 0 (80)

for all 120593 isin Cinfin

0(119860) then there is a polynomial 119875 of degree lt 119903

such that 119891 = 119875 almost everywhere on 119860

To show that 119866119896is in 119883 we need to verify that (if there

is danger of confusion with respect to which variable we takethe norm we use notations as119883(119909

119896))

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

lt infin (81)

under the assumption that 119866119896minus1

is in 119883 Actually bounding119866119896from above by its absolute value this means that we have

first to take the1198711-norm and then the norm in119883We howeverwould like to switch the norms

We adapt Proposition 21 in [16] to our setting

Proposition 25 Let 120588 be a Banach function norm satisfyingproperty (D) If 119891 ΩtimesΩ rarr R is a measurable function andone has 119891

119909and 119891

119910 functions onΩ such that 119891119909(119910) = 119891(119909 119910) =

119891119910

(119909) then

120588 (1003817100381710038171003817119891119909

10038171003817100381710038171) le

1003817100381710038171003817120588 (119891119910

)10038171003817100381710038171

(82)

The cases 119883 isin 1198711

119871infin

are easy to prove If 119883 = 119871119901 for

some 1 lt 119901 lt infin then the result can be proven by usingFubinirsquos theorem together with property (D) For the generalresults we refer to Scheprsquos proof

The next result builds uponTheorem 254 in [10]

Theorem 26 Let Ω sub R be a domain and let 119903 ge 1 Suppose119883 is a Banach function space satisfying (B) and (D) If 119891 islocally in 119883 and has a generalized 119903th derivative 119892 locally in119883 then there is 119865 agreeing almost everywhere with 119891 havingabsolutely continuous intermediate weak derivatives and theweak derivative of order 119903 agrees almost everywhere with 119892Furthermore the derivatives of 119865 are locally in 119883 If Ω hasfinite measure one obtains the result with ldquolocallyrdquo replaced byldquogloballyrdquo

We thank Pilar Silvestre for pointing out how to prove themembership of the intermediate derivatives in119883

Proof Choose 119909lowast

isin Ω and let 119866119896be defined as in Lemma 23

Assume 120593 isin Cinfin

0(Ω) Then using the fact that 119892 is a weak

derivative of 119891 of order 119903 and (b) in Lemma 23

intΩ

(119891 minus 119866119903) 120593

(119903)

119889119909 = intΩ

119891120593(119903)

119889119909 minus intΩ

119866119903120593(119903)

119889119909

= (minus1)119903

intΩ

119892120593119889119909 minus (minus1)119903

intΩ

119892120593119889119909 = 0

(83)

UsingTheorem 24 we see that119891minus119866119903= 119875 almost everywhere

where119875 is a polynomial of degreelt 119903Thus119891 = 119875+119866119903almost

everywhere and for 120593 isin Cinfin

0(Ω) and 0 le 119898 le 119903

intΩ

119891120593(119898)

119889119909 = intΩ

(119875 + 119866119903) 120593

(119898)

= (minus1)119898

intΩ

(119866119903minus119898

+ 119875(119898)

) 120593 119889119909

(84)

Thus119866119903minus119898

+119875(119898) is the119898th generalized derivative of119891 As it is

locally absolutely continuous (if119898 lt 119903) it lies locally in119883 Inthe case 119898 = 119903 the same conclusion holds by the assumptionon 119892

Moreover there exists 119865 for example 119865 = 119875 + 119866119903 that

agrees with 119891 almost everywhere such that its generalizedderivatives up to order 119903 minus 1 are locally absolutely continuousand the 119903th generalized derivative equals 119892 almost every-where

As 119892 is locally in 119883 and the other derivatives are locallyabsolutely continuous they are locally in 119871

infin and thus locallyin119883 by (B)

Suppose now that Ω has additionally finite measureAssume that 119891 and 119892 are in119883 globally We want to show thatthe intermediate derivatives are in 119883 as well

Note in our setting that by (P5) in Definition 18membership in 119883 implies membership in 119871

1 We proceedby induction Suppose that 119866

119896minus1isin 119883 We will apply

Proposition 25 We set denoting by conv119909lowast

119909119896 the closed

interval with endpoints 119909lowast and 119909119896

119891 (119909119896minus1

119909119896) = 119866

119896minus1(119909

119896minus1) 120594conv119909lowast 119909119896 (119909119896minus1

) (85)

Then1003817100381710038171003817119866119896

(119909119896)1003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119891(119909119896minus1

119909119896)119889119909

119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)10038171003817100381710038171198711(119909119896minus1)

10038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)

le10038171003817100381710038171003817

1003816100381610038161003816119866119896minus1(119909

119896minus1)1003816100381610038161003816

1003817100381710038171003817120594Ω1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)lt infin

(86)

Journal of Function Spaces 11

5 Banach Function Space Setting

Here our main focus is a more advanced readership thatlikely knows already Banach function spaces Therefore weadopt a much briefer style pointing merely to the literaturewhere the corresponding definitions and results can be foundWe start with giving generalized versions of the necessity andsufficiency of Theorem 2

We list the results and prove them simultaneously bysplitting the proofs into parts according to the functionspaces

Theorem 27 Assume Ω sub R is a domain 119883 is a normedspace and 119898 isin N If the maximal function 119872 119883 rarr 119883 isbounded then there is a constant 119862 = 119862(119883119898) such that foreach continuous 119891 isin 119882

119898

119883there is 119886

119891isin 119883with 119886

119891 le 119862119891

119882119898119883

and

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (87)

In particular such 119886119891exists if

(i) 1 lt 119901 le infin and 119883 is the Lebesgue space 119871119901

(Ω)

(ii) 1 lt 119901 lt infin 1 le 119902 le infin and 119883 is the Lorentz space119871119901119902

(Ω)

(iii) 1 lt 119901 lt infin 1 le 119902 le infin 119887 is slowly varying and 119883 isthe Lorentz-Karamata space 119871

119901119902119887(Ω)

(iv) 120601 is an 119873-function with inf119904gt0

(119904120593(119904)120601(119904)) gt 1 where120593 is the density function of 120601 and 119883 is the Orlicz space119871lowast

120601(Ω)

(v) Ω = R and 119883 is rearrangement-invariant with upperindex 120572

119883lt 1

Theorem 28 Let Ω sub R be open Suppose 119883 is a Banachfunction space such that (AC) holds Assume further that thefunction 119891 Ω rarr R is in 119883 If for some 119903 isin N

0 the following

inequality holds

1003816100381610038161003816Δ119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)] (88)

for some 119886119891isin 119883 then there exists119892 isin 119883with 119892

119883le 2119903

119903

119886119891119883

such that

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909 = (minus1)119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (89)

for all 120593 isin Cinfin

0(Ω)

In particular the result holds if119883 is

(i) the Lebesgue space 119871119901

(Ω) for some 1 lt 119901 lt infin

(ii) the Lorentz space 119871119901119902

(Ω) where 1 lt 119901 lt infin and 1 le

119902 lt infin

(iii) the Lorentz-Karamata space 119871119901119902119887

(Ω) where 119887 isslowly varying 1 lt 119901 lt infin and 1 lt 119902 le infin

(iv) the Orlicz space 119871lowast

120601(Ω) for some 119873-function 120601 satisfy-

ing

inf119904gt0

119904120593 (119904)

120601 (119904)gt 1 (90)

where 120593 is the density function of 120601

To help with orientation let us give an overview ofthe remaining material in this section With respect toTheorem 27 note that we carried out the proof of Corollary 14in a high level of abstraction Hence we are left to verifythat the spaces under question are normed spaces (we willshow that they are even Banach function spaces) and theboundedness of the maximal operator

Concerning the proof Theorem 28 we proved the cor-responding part of Theorem 2 under the assumptions thatthe spaces under question are Banach function spaces andfurther satisfy the requirements listed in Properties 19 Theexistence of 1198831015840 is classical (Proposition 29) as is the fact that(AC) implies (A) and (NC) (Proposition 30) Hence we onlyneed to point to the literature where (AC) was verified

Proposition 29 Given a Banach function space there is aspace 119883

1015840 such that properties (B) (HI) (C) and (D) aresatisfied

Proof We refer the reader to Sections 11 and 12 in [13] or toSection 31 in [14]

The definition of absolutely continuous norm can befound as Definition 3111 in [14] and Definition 131 in [13]

Proposition 30 A Banach function space that has absolutelycontinuous norm satisfies (A) and (NC)

Proof The first statement follows fromTheorem 1313 in [13]and the second fromProposition 136 in the same source

51 General Statements

Proof of Theorem 27 (general statement) We basically havealready proven the general statement in Corollary 14

Proof of Theorem 28 (general statement) We gave the proofof the Lebesgue space case in such generality that thegeneral statement follows from the proof of Theorem 2 andPropositions 29 and 30

52 Lorentz-Karamata Spaces Lorentz-Karamata spaces aredetailed in [14] mainly Section 343

Proof of Theorem 27 (ii) and (iii) The part that is left toprove is the boundedness of the maximal operator which isguaranteed by Remark 3517 in [14]

Proof of Theorem 28 (ii) and (iii) By Theorem 3441 in [14]we see that 119883 is a rearrangement-invariant Banach space Acombination of Lemma 3439 in [14] together with Lemma3443 in [14] gives the absolute continuity (AC) of the normof1198831015840

12 Journal of Function Spaces

53 Orlicz Spaces The setting is the one of [17]

Lemma 31 If 120601 is an 119873-function satisfying the Δ2-condition

then the Orlicz space 119871lowast

120601has absolutely continuous norm

Proof Let 119891 isin 119871lowast

120601(Ω) We verify first that

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 lt infin (91)

Let 120576 gt 0 By the Δ2-condition there exists a constant 120581 =

120581(119891(120601)

+ 120576) such that

120601 ((1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576) 119904) le 120581120601 (119904) (92)

for all 119904 ge 0 Thus

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 = intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576(1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576))119889119909

le 120581intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576)119889119909 le 120581

(93)

Suppose that 119864 is a measurable set Then

10038171003817100381710038171198911205941198641003817100381710038171003817(120601)

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891120594119864

1003816100381610038161003816) 119889119909 le 1

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 1

(94)

Let us fix 120582 gt 0 Again using the Δ2-condition we have

120601(120582minus1

119904) le 120581(120582minus1

)120601(119904) Hence

intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 (95)

Example 133 in [13] tells us that 1198711 has absolutely continuousnorm Thus if 120594

119864119899converges 120583-almost everywhere to 0 then

intΩ

120601(|119891|)120594119864119899

converges to zero as well Thus we have for 119899

large enough that

120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119899119889119909 le 1 (96)

implying that 119891120594119864119899(120601)

le 120582 Since 120582 gt 0 was arbitrary theabsolute continuity of 119891 follows

Proof of Theorem 27 (iv) We conclude from Gallardorsquos resultTheorem 21 in [17] that it is sufficient that the complemen-tary 119873-function satisfies the Δ

2-condition in [0infin) Again

resorting to Gallardo this time to Proposition 14 in [17] weobtain the conclusion

Proof of Theorem 28 (iv) That 119883 is a Banach functionspace follows from Theorem 3416 in [14] To prove that119883

1015840 has absolutely continuous norm we note that inequal-ity (90) together with Proposition 14 in [17] ensuresthat the complementary 119873-function of 120601 satisfies the Δ

2-

condition Lemma 31 verifies that 1198831015840 has absolutely contin-uous norm

54 Rearrangement-Invariant Spaces The necessary defini-tions can be found in [13]

Proof of Theorem 27 (v) We are only left to show theboundedness of the maximal operator However this is thecontent of a result by Lorentz and Shimogaki for examplestated as Theorem 3517 in [13]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank Pilar Silvestre for hersuggestions of how to prove the existence of intermediatederivatives The research is supported by the Academy ofFinland The third author thanks the Department of Math-ematics and Systems Analysis at Aalto University for itshospitality

References

[1] B Bojarski ldquoRemarks on some geometric properties of Sobolevmappingsrdquo in Functional Analysis amp Related Topics (Sapporo1990) pp 65ndash76 World Scientific River Edge NJ USA 1991

[2] B Bojarski PHajlasz andP Strzelecki ldquoImproved119862119896120582 approxi-

mation of higher order Sobolev functions in normand capacityrdquoIndiana University Mathematics Journal vol 51 no 3 pp 507ndash540 2002

[3] B Bojarski ldquoPointwise characterization of Sobolev classesrdquoProceedings of the Steklov Institute of Mathematics vol 255 no1 pp 65ndash81 2006

[4] B Bojarski ldquoWhitneyrsquos jets for Sobolev functionsrdquo UkrainianMathematical Journal vol 59 no 3 pp 345ndash358 2007

[5] B Bojarski ldquoTaylor expansion and Sobolev spacesrdquo Bulletin ofthe Georgian National Academy of Sciences New Series vol 5no 2 pp 5ndash10 2011

[6] B Bojarski ldquoSobolev spaces and Lagrange interpolationrdquoGeorgian Academy of Sciences Proceedings of A RazmadzeMathematical Institute vol 158 pp 1ndash12 2012

[7] P Hajłasz ldquoSobolev spaces on an arbitrary metric spacerdquoPotential Analysis vol 5 no 4 pp 403ndash415 1996

[8] B Bojarski ldquoSobolev spaces and averaging Irdquo httparxivorgabs13085171

[9] R Kannan and C K Krueger Advanced Analysis on the RealLine Universitext Springer New York NY USA 1996

[10] R A DeVore and G G Lorentz Constructive Approximationvol 303 of Fundamental Principles of Mathematical SciencesSpringer Berlin Germany 1993

[11] L L Schumaker Spline Functions Basic Theory CambridgeMathematical Library CambridgeUniversity Press CambridgeUK 3rd edition 2007

[12] G H Hardy and J E Littlewood ldquoA maximal theorem withfunction-theoretic applicationsrdquo Acta Mathematica vol 54 no1 pp 81ndash116 1930

[13] C Bennett andR Sharpley Interpolation ofOperators vol 129 ofPure and Applied Mathematics Academic Press Boston MassUSA 1988

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Journal of Function Spaces

by the integration by part formula for absolutely continuousfunctions (see for exampleTheorem7147 in [9]) and the factthat 120593(119903)

(119886) = 120593(119903)

(119887) = 0

intΩ

119866119896120593(119903+1)

= int[119886119887]

119866119896120593(119903+1)

= (minus1) int[119886119887]

1198661015840

119896120593(119903)

(78)

Noting that 1198661015840

119896= 119866

119896minus1almost everywhere we see that

intΩ

119866119896120593(119903+1)

= (minus1) int[119886119887]

119866119896minus1

120593(119903)

= (minus1) intΩ

119866119896minus1

120593(119903)

(79)

Using the induction hypothesis the claim follows

We cite Theorem 253 in [10] without giving its proof

Theorem 24 Let 119860 = [119886 119887] R or R+and 119903 ge 0 If the

function 119891 isin 1198711

loc(119860) satisfies

int119860

119891120593(119903)

119889119909 = 0 (80)

for all 120593 isin Cinfin

0(119860) then there is a polynomial 119875 of degree lt 119903

such that 119891 = 119875 almost everywhere on 119860

To show that 119866119896is in 119883 we need to verify that (if there

is danger of confusion with respect to which variable we takethe norm we use notations as119883(119909

119896))

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

lt infin (81)

under the assumption that 119866119896minus1

is in 119883 Actually bounding119866119896from above by its absolute value this means that we have

first to take the1198711-norm and then the norm in119883We howeverwould like to switch the norms

We adapt Proposition 21 in [16] to our setting

Proposition 25 Let 120588 be a Banach function norm satisfyingproperty (D) If 119891 ΩtimesΩ rarr R is a measurable function andone has 119891

119909and 119891

119910 functions onΩ such that 119891119909(119910) = 119891(119909 119910) =

119891119910

(119909) then

120588 (1003817100381710038171003817119891119909

10038171003817100381710038171) le

1003817100381710038171003817120588 (119891119910

)10038171003817100381710038171

(82)

The cases 119883 isin 1198711

119871infin

are easy to prove If 119883 = 119871119901 for

some 1 lt 119901 lt infin then the result can be proven by usingFubinirsquos theorem together with property (D) For the generalresults we refer to Scheprsquos proof

The next result builds uponTheorem 254 in [10]

Theorem 26 Let Ω sub R be a domain and let 119903 ge 1 Suppose119883 is a Banach function space satisfying (B) and (D) If 119891 islocally in 119883 and has a generalized 119903th derivative 119892 locally in119883 then there is 119865 agreeing almost everywhere with 119891 havingabsolutely continuous intermediate weak derivatives and theweak derivative of order 119903 agrees almost everywhere with 119892Furthermore the derivatives of 119865 are locally in 119883 If Ω hasfinite measure one obtains the result with ldquolocallyrdquo replaced byldquogloballyrdquo

We thank Pilar Silvestre for pointing out how to prove themembership of the intermediate derivatives in119883

Proof Choose 119909lowast

isin Ω and let 119866119896be defined as in Lemma 23

Assume 120593 isin Cinfin

0(Ω) Then using the fact that 119892 is a weak

derivative of 119891 of order 119903 and (b) in Lemma 23

intΩ

(119891 minus 119866119903) 120593

(119903)

119889119909 = intΩ

119891120593(119903)

119889119909 minus intΩ

119866119903120593(119903)

119889119909

= (minus1)119903

intΩ

119892120593119889119909 minus (minus1)119903

intΩ

119892120593119889119909 = 0

(83)

UsingTheorem 24 we see that119891minus119866119903= 119875 almost everywhere

where119875 is a polynomial of degreelt 119903Thus119891 = 119875+119866119903almost

everywhere and for 120593 isin Cinfin

0(Ω) and 0 le 119898 le 119903

intΩ

119891120593(119898)

119889119909 = intΩ

(119875 + 119866119903) 120593

(119898)

= (minus1)119898

intΩ

(119866119903minus119898

+ 119875(119898)

) 120593 119889119909

(84)

Thus119866119903minus119898

+119875(119898) is the119898th generalized derivative of119891 As it is

locally absolutely continuous (if119898 lt 119903) it lies locally in119883 Inthe case 119898 = 119903 the same conclusion holds by the assumptionon 119892

Moreover there exists 119865 for example 119865 = 119875 + 119866119903 that

agrees with 119891 almost everywhere such that its generalizedderivatives up to order 119903 minus 1 are locally absolutely continuousand the 119903th generalized derivative equals 119892 almost every-where

As 119892 is locally in 119883 and the other derivatives are locallyabsolutely continuous they are locally in 119871

infin and thus locallyin119883 by (B)

Suppose now that Ω has additionally finite measureAssume that 119891 and 119892 are in119883 globally We want to show thatthe intermediate derivatives are in 119883 as well

Note in our setting that by (P5) in Definition 18membership in 119883 implies membership in 119871

1 We proceedby induction Suppose that 119866

119896minus1isin 119883 We will apply

Proposition 25 We set denoting by conv119909lowast

119909119896 the closed

interval with endpoints 119909lowast and 119909119896

119891 (119909119896minus1

119909119896) = 119866

119896minus1(119909

119896minus1) 120594conv119909lowast 119909119896 (119909119896minus1

) (85)

Then1003817100381710038171003817119866119896

(119909119896)1003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119866119896minus1

(119909119896minus1

)119889119909119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

=

10038171003817100381710038171003817100381710038171003817

int

119909119896

119909lowast

119891(119909119896minus1

119909119896)119889119909

119896minus1

10038171003817100381710038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)10038171003817100381710038171198711(119909119896minus1)

10038171003817100381710038171003817119883(119909119896)

le10038171003817100381710038171003817

1003817100381710038171003817119891 (119909119896minus1

119909119896)1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)

le10038171003817100381710038171003817

1003816100381610038161003816119866119896minus1(119909

119896minus1)1003816100381610038161003816

1003817100381710038171003817120594Ω1003817100381710038171003817119883(119909119896)

100381710038171003817100381710038171198711(119909119896minus1)lt infin

(86)

Journal of Function Spaces 11

5 Banach Function Space Setting

Here our main focus is a more advanced readership thatlikely knows already Banach function spaces Therefore weadopt a much briefer style pointing merely to the literaturewhere the corresponding definitions and results can be foundWe start with giving generalized versions of the necessity andsufficiency of Theorem 2

We list the results and prove them simultaneously bysplitting the proofs into parts according to the functionspaces

Theorem 27 Assume Ω sub R is a domain 119883 is a normedspace and 119898 isin N If the maximal function 119872 119883 rarr 119883 isbounded then there is a constant 119862 = 119862(119883119898) such that foreach continuous 119891 isin 119882

119898

119883there is 119886

119891isin 119883with 119886

119891 le 119862119891

119882119898119883

and

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (87)

In particular such 119886119891exists if

(i) 1 lt 119901 le infin and 119883 is the Lebesgue space 119871119901

(Ω)

(ii) 1 lt 119901 lt infin 1 le 119902 le infin and 119883 is the Lorentz space119871119901119902

(Ω)

(iii) 1 lt 119901 lt infin 1 le 119902 le infin 119887 is slowly varying and 119883 isthe Lorentz-Karamata space 119871

119901119902119887(Ω)

(iv) 120601 is an 119873-function with inf119904gt0

(119904120593(119904)120601(119904)) gt 1 where120593 is the density function of 120601 and 119883 is the Orlicz space119871lowast

120601(Ω)

(v) Ω = R and 119883 is rearrangement-invariant with upperindex 120572

119883lt 1

Theorem 28 Let Ω sub R be open Suppose 119883 is a Banachfunction space such that (AC) holds Assume further that thefunction 119891 Ω rarr R is in 119883 If for some 119903 isin N

0 the following

inequality holds

1003816100381610038161003816Δ119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)] (88)

for some 119886119891isin 119883 then there exists119892 isin 119883with 119892

119883le 2119903

119903

119886119891119883

such that

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909 = (minus1)119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (89)

for all 120593 isin Cinfin

0(Ω)

In particular the result holds if119883 is

(i) the Lebesgue space 119871119901

(Ω) for some 1 lt 119901 lt infin

(ii) the Lorentz space 119871119901119902

(Ω) where 1 lt 119901 lt infin and 1 le

119902 lt infin

(iii) the Lorentz-Karamata space 119871119901119902119887

(Ω) where 119887 isslowly varying 1 lt 119901 lt infin and 1 lt 119902 le infin

(iv) the Orlicz space 119871lowast

120601(Ω) for some 119873-function 120601 satisfy-

ing

inf119904gt0

119904120593 (119904)

120601 (119904)gt 1 (90)

where 120593 is the density function of 120601

To help with orientation let us give an overview ofthe remaining material in this section With respect toTheorem 27 note that we carried out the proof of Corollary 14in a high level of abstraction Hence we are left to verifythat the spaces under question are normed spaces (we willshow that they are even Banach function spaces) and theboundedness of the maximal operator

Concerning the proof Theorem 28 we proved the cor-responding part of Theorem 2 under the assumptions thatthe spaces under question are Banach function spaces andfurther satisfy the requirements listed in Properties 19 Theexistence of 1198831015840 is classical (Proposition 29) as is the fact that(AC) implies (A) and (NC) (Proposition 30) Hence we onlyneed to point to the literature where (AC) was verified

Proposition 29 Given a Banach function space there is aspace 119883

1015840 such that properties (B) (HI) (C) and (D) aresatisfied

Proof We refer the reader to Sections 11 and 12 in [13] or toSection 31 in [14]

The definition of absolutely continuous norm can befound as Definition 3111 in [14] and Definition 131 in [13]

Proposition 30 A Banach function space that has absolutelycontinuous norm satisfies (A) and (NC)

Proof The first statement follows fromTheorem 1313 in [13]and the second fromProposition 136 in the same source

51 General Statements

Proof of Theorem 27 (general statement) We basically havealready proven the general statement in Corollary 14

Proof of Theorem 28 (general statement) We gave the proofof the Lebesgue space case in such generality that thegeneral statement follows from the proof of Theorem 2 andPropositions 29 and 30

52 Lorentz-Karamata Spaces Lorentz-Karamata spaces aredetailed in [14] mainly Section 343

Proof of Theorem 27 (ii) and (iii) The part that is left toprove is the boundedness of the maximal operator which isguaranteed by Remark 3517 in [14]

Proof of Theorem 28 (ii) and (iii) By Theorem 3441 in [14]we see that 119883 is a rearrangement-invariant Banach space Acombination of Lemma 3439 in [14] together with Lemma3443 in [14] gives the absolute continuity (AC) of the normof1198831015840

12 Journal of Function Spaces

53 Orlicz Spaces The setting is the one of [17]

Lemma 31 If 120601 is an 119873-function satisfying the Δ2-condition

then the Orlicz space 119871lowast

120601has absolutely continuous norm

Proof Let 119891 isin 119871lowast

120601(Ω) We verify first that

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 lt infin (91)

Let 120576 gt 0 By the Δ2-condition there exists a constant 120581 =

120581(119891(120601)

+ 120576) such that

120601 ((1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576) 119904) le 120581120601 (119904) (92)

for all 119904 ge 0 Thus

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 = intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576(1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576))119889119909

le 120581intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576)119889119909 le 120581

(93)

Suppose that 119864 is a measurable set Then

10038171003817100381710038171198911205941198641003817100381710038171003817(120601)

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891120594119864

1003816100381610038161003816) 119889119909 le 1

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 1

(94)

Let us fix 120582 gt 0 Again using the Δ2-condition we have

120601(120582minus1

119904) le 120581(120582minus1

)120601(119904) Hence

intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 (95)

Example 133 in [13] tells us that 1198711 has absolutely continuousnorm Thus if 120594

119864119899converges 120583-almost everywhere to 0 then

intΩ

120601(|119891|)120594119864119899

converges to zero as well Thus we have for 119899

large enough that

120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119899119889119909 le 1 (96)

implying that 119891120594119864119899(120601)

le 120582 Since 120582 gt 0 was arbitrary theabsolute continuity of 119891 follows

Proof of Theorem 27 (iv) We conclude from Gallardorsquos resultTheorem 21 in [17] that it is sufficient that the complemen-tary 119873-function satisfies the Δ

2-condition in [0infin) Again

resorting to Gallardo this time to Proposition 14 in [17] weobtain the conclusion

Proof of Theorem 28 (iv) That 119883 is a Banach functionspace follows from Theorem 3416 in [14] To prove that119883

1015840 has absolutely continuous norm we note that inequal-ity (90) together with Proposition 14 in [17] ensuresthat the complementary 119873-function of 120601 satisfies the Δ

2-

condition Lemma 31 verifies that 1198831015840 has absolutely contin-uous norm

54 Rearrangement-Invariant Spaces The necessary defini-tions can be found in [13]

Proof of Theorem 27 (v) We are only left to show theboundedness of the maximal operator However this is thecontent of a result by Lorentz and Shimogaki for examplestated as Theorem 3517 in [13]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank Pilar Silvestre for hersuggestions of how to prove the existence of intermediatederivatives The research is supported by the Academy ofFinland The third author thanks the Department of Math-ematics and Systems Analysis at Aalto University for itshospitality

References

[1] B Bojarski ldquoRemarks on some geometric properties of Sobolevmappingsrdquo in Functional Analysis amp Related Topics (Sapporo1990) pp 65ndash76 World Scientific River Edge NJ USA 1991

[2] B Bojarski PHajlasz andP Strzelecki ldquoImproved119862119896120582 approxi-

mation of higher order Sobolev functions in normand capacityrdquoIndiana University Mathematics Journal vol 51 no 3 pp 507ndash540 2002

[3] B Bojarski ldquoPointwise characterization of Sobolev classesrdquoProceedings of the Steklov Institute of Mathematics vol 255 no1 pp 65ndash81 2006

[4] B Bojarski ldquoWhitneyrsquos jets for Sobolev functionsrdquo UkrainianMathematical Journal vol 59 no 3 pp 345ndash358 2007

[5] B Bojarski ldquoTaylor expansion and Sobolev spacesrdquo Bulletin ofthe Georgian National Academy of Sciences New Series vol 5no 2 pp 5ndash10 2011

[6] B Bojarski ldquoSobolev spaces and Lagrange interpolationrdquoGeorgian Academy of Sciences Proceedings of A RazmadzeMathematical Institute vol 158 pp 1ndash12 2012

[7] P Hajłasz ldquoSobolev spaces on an arbitrary metric spacerdquoPotential Analysis vol 5 no 4 pp 403ndash415 1996

[8] B Bojarski ldquoSobolev spaces and averaging Irdquo httparxivorgabs13085171

[9] R Kannan and C K Krueger Advanced Analysis on the RealLine Universitext Springer New York NY USA 1996

[10] R A DeVore and G G Lorentz Constructive Approximationvol 303 of Fundamental Principles of Mathematical SciencesSpringer Berlin Germany 1993

[11] L L Schumaker Spline Functions Basic Theory CambridgeMathematical Library CambridgeUniversity Press CambridgeUK 3rd edition 2007

[12] G H Hardy and J E Littlewood ldquoA maximal theorem withfunction-theoretic applicationsrdquo Acta Mathematica vol 54 no1 pp 81ndash116 1930

[13] C Bennett andR Sharpley Interpolation ofOperators vol 129 ofPure and Applied Mathematics Academic Press Boston MassUSA 1988

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 11

5 Banach Function Space Setting

Here our main focus is a more advanced readership thatlikely knows already Banach function spaces Therefore weadopt a much briefer style pointing merely to the literaturewhere the corresponding definitions and results can be foundWe start with giving generalized versions of the necessity andsufficiency of Theorem 2

We list the results and prove them simultaneously bysplitting the proofs into parts according to the functionspaces

Theorem 27 Assume Ω sub R is a domain 119883 is a normedspace and 119898 isin N If the maximal function 119872 119883 rarr 119883 isbounded then there is a constant 119862 = 119862(119883119898) such that foreach continuous 119891 isin 119882

119898

119883there is 119886

119891isin 119883with 119886

119891 le 119862119891

119882119898119883

and

1003816100381610038161003816Δ119898

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119898

(119886119891(119909) + 119886

119891(119910)) (87)

In particular such 119886119891exists if

(i) 1 lt 119901 le infin and 119883 is the Lebesgue space 119871119901

(Ω)

(ii) 1 lt 119901 lt infin 1 le 119902 le infin and 119883 is the Lorentz space119871119901119902

(Ω)

(iii) 1 lt 119901 lt infin 1 le 119902 le infin 119887 is slowly varying and 119883 isthe Lorentz-Karamata space 119871

119901119902119887(Ω)

(iv) 120601 is an 119873-function with inf119904gt0

(119904120593(119904)120601(119904)) gt 1 where120593 is the density function of 120601 and 119883 is the Orlicz space119871lowast

120601(Ω)

(v) Ω = R and 119883 is rearrangement-invariant with upperindex 120572

119883lt 1

Theorem 28 Let Ω sub R be open Suppose 119883 is a Banachfunction space such that (AC) holds Assume further that thefunction 119891 Ω rarr R is in 119883 If for some 119903 isin N

0 the following

inequality holds

1003816100381610038161003816Δ119903

119891 (119909 119910)1003816100381610038161003816 le

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

119903

[119886119891(119909) + 119886

119891(119910)] (88)

for some 119886119891isin 119883 then there exists119892 isin 119883with 119892

119883le 2119903

119903

119886119891119883

such that

intΩ

119891 (119909) 120593(119903)

(119909) 119889119909 = (minus1)119903

intΩ

119892 (119909) 120593 (119909) 119889119909 (89)

for all 120593 isin Cinfin

0(Ω)

In particular the result holds if119883 is

(i) the Lebesgue space 119871119901

(Ω) for some 1 lt 119901 lt infin

(ii) the Lorentz space 119871119901119902

(Ω) where 1 lt 119901 lt infin and 1 le

119902 lt infin

(iii) the Lorentz-Karamata space 119871119901119902119887

(Ω) where 119887 isslowly varying 1 lt 119901 lt infin and 1 lt 119902 le infin

(iv) the Orlicz space 119871lowast

120601(Ω) for some 119873-function 120601 satisfy-

ing

inf119904gt0

119904120593 (119904)

120601 (119904)gt 1 (90)

where 120593 is the density function of 120601

To help with orientation let us give an overview ofthe remaining material in this section With respect toTheorem 27 note that we carried out the proof of Corollary 14in a high level of abstraction Hence we are left to verifythat the spaces under question are normed spaces (we willshow that they are even Banach function spaces) and theboundedness of the maximal operator

Concerning the proof Theorem 28 we proved the cor-responding part of Theorem 2 under the assumptions thatthe spaces under question are Banach function spaces andfurther satisfy the requirements listed in Properties 19 Theexistence of 1198831015840 is classical (Proposition 29) as is the fact that(AC) implies (A) and (NC) (Proposition 30) Hence we onlyneed to point to the literature where (AC) was verified

Proposition 29 Given a Banach function space there is aspace 119883

1015840 such that properties (B) (HI) (C) and (D) aresatisfied

Proof We refer the reader to Sections 11 and 12 in [13] or toSection 31 in [14]

The definition of absolutely continuous norm can befound as Definition 3111 in [14] and Definition 131 in [13]

Proposition 30 A Banach function space that has absolutelycontinuous norm satisfies (A) and (NC)

Proof The first statement follows fromTheorem 1313 in [13]and the second fromProposition 136 in the same source

51 General Statements

Proof of Theorem 27 (general statement) We basically havealready proven the general statement in Corollary 14

Proof of Theorem 28 (general statement) We gave the proofof the Lebesgue space case in such generality that thegeneral statement follows from the proof of Theorem 2 andPropositions 29 and 30

52 Lorentz-Karamata Spaces Lorentz-Karamata spaces aredetailed in [14] mainly Section 343

Proof of Theorem 27 (ii) and (iii) The part that is left toprove is the boundedness of the maximal operator which isguaranteed by Remark 3517 in [14]

Proof of Theorem 28 (ii) and (iii) By Theorem 3441 in [14]we see that 119883 is a rearrangement-invariant Banach space Acombination of Lemma 3439 in [14] together with Lemma3443 in [14] gives the absolute continuity (AC) of the normof1198831015840

12 Journal of Function Spaces

53 Orlicz Spaces The setting is the one of [17]

Lemma 31 If 120601 is an 119873-function satisfying the Δ2-condition

then the Orlicz space 119871lowast

120601has absolutely continuous norm

Proof Let 119891 isin 119871lowast

120601(Ω) We verify first that

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 lt infin (91)

Let 120576 gt 0 By the Δ2-condition there exists a constant 120581 =

120581(119891(120601)

+ 120576) such that

120601 ((1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576) 119904) le 120581120601 (119904) (92)

for all 119904 ge 0 Thus

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 = intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576(1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576))119889119909

le 120581intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576)119889119909 le 120581

(93)

Suppose that 119864 is a measurable set Then

10038171003817100381710038171198911205941198641003817100381710038171003817(120601)

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891120594119864

1003816100381610038161003816) 119889119909 le 1

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 1

(94)

Let us fix 120582 gt 0 Again using the Δ2-condition we have

120601(120582minus1

119904) le 120581(120582minus1

)120601(119904) Hence

intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 (95)

Example 133 in [13] tells us that 1198711 has absolutely continuousnorm Thus if 120594

119864119899converges 120583-almost everywhere to 0 then

intΩ

120601(|119891|)120594119864119899

converges to zero as well Thus we have for 119899

large enough that

120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119899119889119909 le 1 (96)

implying that 119891120594119864119899(120601)

le 120582 Since 120582 gt 0 was arbitrary theabsolute continuity of 119891 follows

Proof of Theorem 27 (iv) We conclude from Gallardorsquos resultTheorem 21 in [17] that it is sufficient that the complemen-tary 119873-function satisfies the Δ

2-condition in [0infin) Again

resorting to Gallardo this time to Proposition 14 in [17] weobtain the conclusion

Proof of Theorem 28 (iv) That 119883 is a Banach functionspace follows from Theorem 3416 in [14] To prove that119883

1015840 has absolutely continuous norm we note that inequal-ity (90) together with Proposition 14 in [17] ensuresthat the complementary 119873-function of 120601 satisfies the Δ

2-

condition Lemma 31 verifies that 1198831015840 has absolutely contin-uous norm

54 Rearrangement-Invariant Spaces The necessary defini-tions can be found in [13]

Proof of Theorem 27 (v) We are only left to show theboundedness of the maximal operator However this is thecontent of a result by Lorentz and Shimogaki for examplestated as Theorem 3517 in [13]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank Pilar Silvestre for hersuggestions of how to prove the existence of intermediatederivatives The research is supported by the Academy ofFinland The third author thanks the Department of Math-ematics and Systems Analysis at Aalto University for itshospitality

References

[1] B Bojarski ldquoRemarks on some geometric properties of Sobolevmappingsrdquo in Functional Analysis amp Related Topics (Sapporo1990) pp 65ndash76 World Scientific River Edge NJ USA 1991

[2] B Bojarski PHajlasz andP Strzelecki ldquoImproved119862119896120582 approxi-

mation of higher order Sobolev functions in normand capacityrdquoIndiana University Mathematics Journal vol 51 no 3 pp 507ndash540 2002

[3] B Bojarski ldquoPointwise characterization of Sobolev classesrdquoProceedings of the Steklov Institute of Mathematics vol 255 no1 pp 65ndash81 2006

[4] B Bojarski ldquoWhitneyrsquos jets for Sobolev functionsrdquo UkrainianMathematical Journal vol 59 no 3 pp 345ndash358 2007

[5] B Bojarski ldquoTaylor expansion and Sobolev spacesrdquo Bulletin ofthe Georgian National Academy of Sciences New Series vol 5no 2 pp 5ndash10 2011

[6] B Bojarski ldquoSobolev spaces and Lagrange interpolationrdquoGeorgian Academy of Sciences Proceedings of A RazmadzeMathematical Institute vol 158 pp 1ndash12 2012

[7] P Hajłasz ldquoSobolev spaces on an arbitrary metric spacerdquoPotential Analysis vol 5 no 4 pp 403ndash415 1996

[8] B Bojarski ldquoSobolev spaces and averaging Irdquo httparxivorgabs13085171

[9] R Kannan and C K Krueger Advanced Analysis on the RealLine Universitext Springer New York NY USA 1996

[10] R A DeVore and G G Lorentz Constructive Approximationvol 303 of Fundamental Principles of Mathematical SciencesSpringer Berlin Germany 1993

[11] L L Schumaker Spline Functions Basic Theory CambridgeMathematical Library CambridgeUniversity Press CambridgeUK 3rd edition 2007

[12] G H Hardy and J E Littlewood ldquoA maximal theorem withfunction-theoretic applicationsrdquo Acta Mathematica vol 54 no1 pp 81ndash116 1930

[13] C Bennett andR Sharpley Interpolation ofOperators vol 129 ofPure and Applied Mathematics Academic Press Boston MassUSA 1988

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

12 Journal of Function Spaces

53 Orlicz Spaces The setting is the one of [17]

Lemma 31 If 120601 is an 119873-function satisfying the Δ2-condition

then the Orlicz space 119871lowast

120601has absolutely continuous norm

Proof Let 119891 isin 119871lowast

120601(Ω) We verify first that

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 lt infin (91)

Let 120576 gt 0 By the Δ2-condition there exists a constant 120581 =

120581(119891(120601)

+ 120576) such that

120601 ((1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576) 119904) le 120581120601 (119904) (92)

for all 119904 ge 0 Thus

intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 119889119909 = intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576(1003817100381710038171003817119891

1003817100381710038171003817(120601)+ 120576))119889119909

le 120581intΩ

120601(

10038161003816100381610038161198911003816100381610038161003816

10038171003817100381710038171198911003817100381710038171003817(120601)

+ 120576)119889119909 le 120581

(93)

Suppose that 119864 is a measurable set Then

10038171003817100381710038171198911205941198641003817100381710038171003817(120601)

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891120594119864

1003816100381610038161003816) 119889119909 le 1

= inf 120582 gt 0 intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 1

(94)

Let us fix 120582 gt 0 Again using the Δ2-condition we have

120601(120582minus1

119904) le 120581(120582minus1

)120601(119904) Hence

intΩ

120601 (120582minus1 1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 le 120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119889119909 (95)

Example 133 in [13] tells us that 1198711 has absolutely continuousnorm Thus if 120594

119864119899converges 120583-almost everywhere to 0 then

intΩ

120601(|119891|)120594119864119899

converges to zero as well Thus we have for 119899

large enough that

120581intΩ

120601 (1003816100381610038161003816119891

1003816100381610038161003816) 120594119864119899119889119909 le 1 (96)

implying that 119891120594119864119899(120601)

le 120582 Since 120582 gt 0 was arbitrary theabsolute continuity of 119891 follows

Proof of Theorem 27 (iv) We conclude from Gallardorsquos resultTheorem 21 in [17] that it is sufficient that the complemen-tary 119873-function satisfies the Δ

2-condition in [0infin) Again

resorting to Gallardo this time to Proposition 14 in [17] weobtain the conclusion

Proof of Theorem 28 (iv) That 119883 is a Banach functionspace follows from Theorem 3416 in [14] To prove that119883

1015840 has absolutely continuous norm we note that inequal-ity (90) together with Proposition 14 in [17] ensuresthat the complementary 119873-function of 120601 satisfies the Δ

2-

condition Lemma 31 verifies that 1198831015840 has absolutely contin-uous norm

54 Rearrangement-Invariant Spaces The necessary defini-tions can be found in [13]

Proof of Theorem 27 (v) We are only left to show theboundedness of the maximal operator However this is thecontent of a result by Lorentz and Shimogaki for examplestated as Theorem 3517 in [13]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank Pilar Silvestre for hersuggestions of how to prove the existence of intermediatederivatives The research is supported by the Academy ofFinland The third author thanks the Department of Math-ematics and Systems Analysis at Aalto University for itshospitality

References

[1] B Bojarski ldquoRemarks on some geometric properties of Sobolevmappingsrdquo in Functional Analysis amp Related Topics (Sapporo1990) pp 65ndash76 World Scientific River Edge NJ USA 1991

[2] B Bojarski PHajlasz andP Strzelecki ldquoImproved119862119896120582 approxi-

mation of higher order Sobolev functions in normand capacityrdquoIndiana University Mathematics Journal vol 51 no 3 pp 507ndash540 2002

[3] B Bojarski ldquoPointwise characterization of Sobolev classesrdquoProceedings of the Steklov Institute of Mathematics vol 255 no1 pp 65ndash81 2006

[4] B Bojarski ldquoWhitneyrsquos jets for Sobolev functionsrdquo UkrainianMathematical Journal vol 59 no 3 pp 345ndash358 2007

[5] B Bojarski ldquoTaylor expansion and Sobolev spacesrdquo Bulletin ofthe Georgian National Academy of Sciences New Series vol 5no 2 pp 5ndash10 2011

[6] B Bojarski ldquoSobolev spaces and Lagrange interpolationrdquoGeorgian Academy of Sciences Proceedings of A RazmadzeMathematical Institute vol 158 pp 1ndash12 2012

[7] P Hajłasz ldquoSobolev spaces on an arbitrary metric spacerdquoPotential Analysis vol 5 no 4 pp 403ndash415 1996

[8] B Bojarski ldquoSobolev spaces and averaging Irdquo httparxivorgabs13085171

[9] R Kannan and C K Krueger Advanced Analysis on the RealLine Universitext Springer New York NY USA 1996

[10] R A DeVore and G G Lorentz Constructive Approximationvol 303 of Fundamental Principles of Mathematical SciencesSpringer Berlin Germany 1993

[11] L L Schumaker Spline Functions Basic Theory CambridgeMathematical Library CambridgeUniversity Press CambridgeUK 3rd edition 2007

[12] G H Hardy and J E Littlewood ldquoA maximal theorem withfunction-theoretic applicationsrdquo Acta Mathematica vol 54 no1 pp 81ndash116 1930

[13] C Bennett andR Sharpley Interpolation ofOperators vol 129 ofPure and Applied Mathematics Academic Press Boston MassUSA 1988

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 13

[14] D E Edmunds and W D Evans Hardy Operators FunctionSpaces and Embeddings Springer Monographs in MathematicsSpringer Berlin Germany 2004

[15] W Rudin Real and Complex Analysis McGraw-Hill Series inHigher Mathematics McGraw-Hill Book Co New York NYUSA 2nd edition 1974

[16] A R Schep ldquoMinkowskirsquos integral inequality for functionnormsrdquo in Operator Theory in Function Spaces and BanachLattices vol 75 of Operator Theory Advances and Applicationspp 299ndash308 Birkhauser Basel Switzerland 1995

[17] D Gallardo ldquoOrlicz spaces for which the Hardy-Littlewoodmaximal operator is boundedrdquo Publicacions Matematiques vol32 no 2 pp 261ndash266 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of