Holder functions and Itsproperties

Ashis Kumar PatiUnder the supervision of

Dr. SaugataBandyopadhyay

Department Of Mathematicsand Statistics

IISER Kolkataashisinmath@gmail.com

Abstract

The Holder continuous functions are useful to study the clas-sical theory of partial differential equation. First , We defineHolder continuous functions and discuss basic properties ofHolder continuous functions. Then we proved compact em-bedding for Ck, functions. Our main aim is to discuss theextension of Ck, functions from open Rn to Rn. In thisregard we prove Extension theorem using Dminated conver-gence theorem and Smoothness of Domain and some otherconcepts.

1

Acknowledgement

At the outset I would like to extend my humble gratitude to my guideDr. Saugata Bandyopadhaya for being such an amazing humanbeingand teacher. He has been a friend, teacher and an exemplary guideat the same time. He has been infinitely patient with me, wheneverI showed signs of ignorance, kindly reemphasizing the concepts againand again. He showed the way and guided me along. For this I will bealways grateful to him. It is great experience learning from him.

Also I sincerely thank all my teachers Himadri sir, Satyaki sir, andall of them, Mrinmay Bhaiya, Amit Bhaiya and all my seniors De-banjan, Bibhash, Abhraneel, Guddu, Ciny and all my friends for theirsupport and encouragement. And I cant move forward without mak-ing a commendation on IISER Kolkata, for giving me this opertunityin promoting the values and ethics of science among young minds andproviding sholarship.

Last but not the least I thank my parents my brother and sister forbeing there always with me in my depression time.

2

3

Certificate

This is to certify that the thesis entitled Holder Space and Its proper-ties , being submitted to the Department of Mathematics and Statis-tics of Indian Institute of Science Education and Research Kolkata inpartial fulfillment of the requirements for the award of Integrated MS-PhD degree, embodies the research work carried out by Ashis KumarPati under my supervision at IISER Kolkata. The work presented inthis thesis is original and has not been submitted so far, in part or full,for any other degree or diploma of any other university/institute.

Date: Signature of the Supervisor

4

5

Dedication

This may be a very small work But what ever I have done its your mercy, it isdedicated to you Krishna.

6

Contents

1 Preliminaries 81.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.2.1 Continuous and Holder continuous . . . . . . . . . . . . . . . 121.2.2 Lipschitz and Holder continuous . . . . . . . . . . . . . . . . . 121.2.3 Uniform and Holder continuous . . . . . . . . . . . . . . . . . 121.2.4 Bounded variation and Holder continuous . . . . . . . . . . . 131.2.5 One Important example . . . . . . . . . . . . . . . . . . . . . 151.2.6 Some Special Caes . . . . . . . . . . . . . . . . . . . . . . . . 16

1.3 Some Elementary Properties Related Norm . . . . . . . . . . . . . . . 171.3.1 Equivalence Of Norm And Some Properties . . . . . . . . . . 19

2 Embeddings and Extension 212.1 Compact Embeddings . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2 Continuous Extension . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.2.1 Tietze Extension and McShane Lemma . . . . . . . . . . . . . 242.2.2 Two Main Theorems Of Extension . . . . . . . . . . . . . . . 26

2.3 Semicontinuity Result . . . . . . . . . . . . . . . . . . . . . . . . . . 41

7

Chapter 1

Preliminaries

1.1 Definitions

First of all , let us summarize the basic notations and statements which will be usedlater in this context. Well include all the definitions , which we are going to use.

Definition 1.1.1. Let(X,) be a metric space. A subset Y is compact if and onlyiff for each open cover of Y has a finite subcover.

Definition 1.1.2. Let (X,) be a metric space. A subset A X is called sequentiallycompact if every sequence {xn}nNhas a converging subsequence {xnk} whose limitx lies in A.

Definition 1.1.3. A is called relatively compact if A is compact.

Remark 1.1.4. Compactness and sequentially compactness are equivalent in a metricspace.

Definition 1.1.5. Let (X, ||.||X) is a normed space. A subset A X is calledbounded if there exists a real constant C such that||x||X C x ADefinition 1.1.6. Let A Rn. A function f : A Rm is called a Lipshitz functionif there exists a nonnegative C such that

|f(x) f(y)| C|x y| x, y A. (1.1)We denote the smallest such C for which the above condition is satisfied by

Lip(f) = sup{|f(x)f(y)||xy| , x, y A, x 6= y

}Definition 1.1.7. for x= (x1,...,xn) Rn , we will write |x| = max {|xi|}

8

Now well introduce Cr spaces

Definition 1.1.8. f : X R/C then well say f C(X) if f is continuous and andsup|f(x)|

Definition 1.1.17. Assume that H.Then, for a function f : K R we definethe Holder seminorm of f as

|f | := supx,yK,x6=y

|f(x) f(y)|(|x y|) (1.4)

We denote by C0,(K) the space of all real valued functions f on K such that|f |

Definition 1.1.22. The set Cr,0 () denotes the set of functions in Cr,() with

compact support in

Well setcr = cr,0

[f ]c0,0 = 0

||f ||Cr,0 = ||f ||CrDefinition 1.1.23. Let(X, ||.||X) and (Y , ||.||Y ) be two normed spaces. We say thatthe space X is contuously embedded into Y and denote this by X Y if it holdsthat X Y and there exists such that

||x||Y b||x||X x X (1.6)Definition 1.1.24. we say X is compactly embedded into Y and denote this byX Y if X Y and every bounded set in (X,||.||X) is relatively compact in (Y ,||.||Y )

Let us recall some important theorem which we are going to use.

Theorem 1.1.25. Arzella ascoli theorem:-Let X be a compact metric space, Let C(X,K) be given with the sup norm metric.

(K is either R or C). Then a set B C(X) is compact iff B is bounded closed andequicontinuous.

Theorem 1.1.26. Let J R be a bounded interval and let (fn) be a sequence offunctions on J to R. Suppose that there exists x0 J such that (fn(x0)) converges,and that the sequence f

n of derivatives exists on J and converges uniformly on J to

a function g.Then the sequence (fn) converges uniformly on J to a function f that has a

derivative at everypoint of J and f= g

Proof. can be found in Introduction to Real Analysis by Bartle and Sherbert.

11

1.2 Examples

1.2.1 Continuous and Holder continuous

Every Holder continuous functions are continuous but all continuous functions arenot Holder for an example

f : R R where f(x) = x2

1.2.2 Lipschitz and Holder continuous

Every Lipschitz continuous functions are Holder continuous functions of Holder con-stant 1.

But there are Holder continuous functions that are not Lipschitz function.For an examplef : (0, 1) R where f(x) = x.See that f is 1

2Holder continuous.

|xy|2 |x y| |xy| |x y| 12

But it is not Lipschitz, if so then for x 6= y|xy||x y| =

1

|x+y| C |x+y| 1

c

But this is not possible for any constant C, when x, y 0.

1.2.3 Uniform and Holder continuous

Every Holder continuous functions are unniform continuous.But there are functions uniform continuous but not Holder continuous.f : [0, 1

2] R

f(x) =

{1

log(x)if 0 < x 1

2

0 if x = 0

As log(x) is continuous and log(x) 6= 0 in 0 < x 12

so 1log(x)

is cts on [0,1/2] and

since [0,1/2] is compact so it is uniformly continuous on [0,1/2].But well show that f is not Holder continuous for any > 0 at x=0.If it were there would exist positive c and such that

12

|0 1log(x)

| c|x| , x (0, 12

)

c|x||log(x)| 1 x (0, 12

]

which cant happen aslimx0+

c|x||logx| = 0

1.2.4 Bounded variation and Holder continuous

Its easy to say that Every Bounded variation function is Holder continuous but theconverse is not true.

for example lets take the functionf : R R

f(x) =

{xsin 1

xif x 6= 0

0 if x = 0

this function is 12

Holder continuos ,but this is not bounded variation. For Holdercontinuity choose two points x1 < x2 in [0,1].

Let x [x1, x2) be the greatest number for which

f(x1) = f(x).

Its easy to see that (x)1 (x2)1 + 2pi

This implies that|f(x2) f(x1)| = |f(x2) f(x,,)|

x2x,,

|f (x)|dx

Applying Holders inequality x2x,,

|f (x)|dx

|x2 x,,|12

( x2x| sin(1/x) cos(1/x) 1

x|2dx

) 12

|x2 x1| 12( x2

x|1 + 1/x|2dx

) 12

13

|x2 x1| 12(x2 x + 2 log(x2/x1)

1

x2+

1

x

) 12

|x2 x1| 12 (1 + 2 log(1 + 2pix2) + 2pi)12

As x2 1

|f(x2) f(x1)| |x2 x1| 12 (1 + 2 log(1 + 2pi) + 2pi)12

which implies it is 12Holder continuous where holder constant

c = (1 + 2 log(1 + 2pi) + 2pi)12

Now for Bounded variation,take partition P of [0,1] by points

ak =2

(2k + 1)pik 0

WhereP = {ak}nk=1 {0, 1}

f(x) =

{ak if k is even

ak if k is oddtherefore

Nk=1

|f(ak) f(ak 1)|

=Nk=1

(ak + ak1)

>N1k=1

ak >N1k=1

(1

pi)(

1

k + 1/2)

1

pi

N1k=1

1

k + 1

That means we can take N large enough to make variations as large as we want.

14

1.2.5 One Important example

Let 0 < , 1 is given. Then x is -Holder continious for , not for > .Proof : Let . Consider h [0, 1].Then h h.

(x+ h) xh

(x+ h) x

h.h

Case-1 h < x(x+ h) x

h (x+ h)

xh.h1

Now applying Mean value theorem h0 (0, h)

=1

h1(x+ h0)

1

= (h

x+ h0)1 < 1

As

h < x < x+ h0 ( hx+ h0

) < 1 ( hx+ h0

)1 < < 1

Case-2 h x, 2h x+ h(x+ h) x

h

(2h) hh

2 1So, we see that for all 2 cases our Holder norm is bounded.

Now let us consider >

sup|f(x) f(y)||x y|

sup0

1.2.6 Some Special Caes

Theorem 1.2.1. Let U Rn is Open. If u : U C and [u] 1then u is constant on each connected component of U .

Proof. Let x U h Rn then

|u(x+ th) u(x)t

| [u]t

t

and[u]t

t 0 t 0, > 1

Which shows hu(x) = 0 for all x U .let x0 U Consider the set

S = {x U : u(x) = u(x0)}Note that S is closed in U. So,its enough to show S is open. So then this will

show that u is constant.Now let x U consider the open ball B around x such that x B U .For all y B a smooth path(straight line)joing x and y.If y U is in the same connected component as x, there exists a smooth curve : [0, 1] U such that (0) = x and (1) = y. So by fundamental theorem of

calculus and the chain rule

u(y) u(x) = 1

0

d

dtu((t))dt =

10

0dt = 0

u(x) = u(y) on the connected component

Remark 1.2.2. So because of this we dont consider the case when > 1.

16

1.3 Some Elementary Properties Related Norm

Note that [f ]C0, is a seminorm. since,

[f + g]C0, = supx,yUx 6=y

|(f + g)(x) (f + g)(y)||x y|

= supx,yU

(|f(x) f(y)||x y| +

|g(x) g(y)||x y| )

supx 6=y

|f(x) f(y)||x y| + supx 6=y

|g(x) g(y)||x y|

= [f ]c0, + [g]C0,

[f ]C0,

= supx,yU

|f(x) f(y)||x y|

= sup(x 6=y)U

|f(x) f(y)||x y|

We know that ||.||C0 is norm.So Holder Norm which is defined as

||f ||Ck, =ka=0

||af ||C0 + [kf ]C0, 0 N N such that

17

||un um|| n,m > N ||un um||Ck + [k(un um)]Co, <

||un um||Ck < |a|k||aun um||C0

||aun aum||C0 < |a| Kwhich shows un is cauchy in C

k() and {aun}nN is cauchy in C() for all |a| kWe know (C(), sup) is a banach spaceUsing this well show Ck is complete.step-2To show Ck() is completeAs un is cauchy in C

K() we want to show that there exists u Ck and un uin Ck

Now note that (C(), sup) is a banach space so there exists u C() such thatun u in C().

claim- au C() when |a| = k and ||un u||Ck 0As we have shown that {aun}nN is a cauchy in C() and C() is complete

ga C() such thatlimn||aun ga||C0 = 0 for|a| k (1.8)

Given aun ga uniformly in C()and un u uniformly in C() and au = ga By the uniqueness of the limit

aun au uniformly. So ga C() implies au C()Note that here we are using Theorem(1.1.26) and induction steps.step-3 we want to show this u is the limit of un in C

k,

That means we want to show u Ck, and ||un u||CK, 0claim u Ck,We know u Ck so we want to show au Co, for |a| = k that means for

[au]c0,()

lim|aun(x) aun(y)|

|x y| lim[aun]C0,

lim |aun(x) aun(y)||x y|

lim[aun]C0,

=|au(x) au(y)|

|x y| limsup[aun(x)]C0,

More over

||f ||Cs, ||f ||Cs,? ||f ||Cs,and therefore the ||.||Cs,? and the ||.||Cs, Norms are equivalent.In particularly if 0 < < < 1, We can say that

Cr+1 Cr,1 Cr, Cr, Cr

And the imbeddings are continuous.

20

Chapter 2

Embeddings and Extension

2.1 Compact Embeddings

Theorem 2.1.1. Let Rn be a bounded open lipschitz set. Let s r 0 beintegers and 0 , 1 with

r + < s+

then the embeddingCs, Cr,

is compact.

Proof. So we want to prove is that Every bounded set in Cs, is relatively compactin Cr, Let {fv}vN Cs,() with||fv||Cs, C for all v.We have to show that we can extract a convergent subsequence in Cr,() Well

prove this in two steps.step-1 Well assume r=s and thus < step 1.1 r=s=0Now fv is a bounded sequence in C

0,. which implies

||fv||C0, < C||fv||+ [fv]C0, < C

So this is bounded in C0

Now [fv]C0, < Cfor x 6= y and |x y| < |fv(x) fv(y)| < C

21

which can be made smaller than and this shows inf fv is equicontinuous.So now by applying Arzella-Ascoli theorem we can find a subsequence , still

denoted by fv (for notational convinience) , which converges to f in C0()

We will show this fv converges to f in C0, and its sufficient to show that [ffv]C0,

converges to 0.for = 0 this is C0 convergence, so well consider > 0.Let > 0 and x 6= y withStep-1.1.1 |x y|

|(f fv)(x) (f fv)(y)||x y|

= lim|(f fv)(x) (f fv)(y)|

|x y| sup

[f fv]C0, |x y|

2Cstep-1.1.2 |x y|

|(f fv)(x) (f fv)(y)||x y|

2||f fv||C0For any given > 0. We can take > 0 small enough to make 2C and

using the convergence in C0()We can take m N such that2||f fv||C0 for every v m.So we have |f fv|C0, for every v m.step-1.2 r = s 1We know that ||f ||Cs, ||f ||Cs, C||f ||Cs,therefore

||fv||Cs, K ||fv||Cs, Kst=0

|| 5t fv||C0, K

22

so || 5t fv||C0, bounded for 0 t s. So now by step 1.1 have gt C0,(;Rnt)such that

5fv gt in C0,as v for0 t s5tfv gt in C0()that means we have uniform convergence in all derivatives so we find g0 such that

gt = 5tg0 and fv g0 in Cs, = Cr,Step-2 Consider the case r < sstep-2.1 r< s and > 0Cs,() Cs,0() this is a compact embdding by the first step.Cs,0() Cr,() is a continuos embedding.The composition of continuous and compact embedding is compact.Cs,() Cr,() is compact.step-2.2, r < s, = 02.2.1 s=r+1Cr,1() Cr,() is a compact embedding by step 1.cr+1,0() Cr,1() is a continuous embedding.So again composition of these two embeddings is compact.Cr+1,0() Cr,() is compact.2.2.2 s > r = 1s > r + 1 and = 0, and thus s 1 r + .cs1,1() Cs1,0() is compact by step 1.Cs,0() Cs1,1()andCs1,0() Cr,() are continuos embdding.so Cs,0() Cr,() embedding is compact.

23

2.2 Continuous Extension

Here Our main intention is to know when can we extend a Cr,() function to aCr,(Rn) first for

r = = 0 we have the Tietze extension theorem.r = 0 and 0 < 1 we have mc shane lemma.

2.2.1 Tietze Extension and McShane Lemma

Theorem 2.2.1 (Tietze extension theorem). If X is a normal topological space andf : A R is continuos map from closed set A of X into the real numbers. then aconcontinuos map F : X R with F(a)=f(a) a A.Theorem 2.2.2 (Mc Shane Lemma). Let D Rn be any set, 0 < 1 andf : D R with = [f ]C0,(D)

f(x) f(y) |x y| y Df(x) f(y) + |x y| y D

so that means f(x) f+(x) by taking infimum over y.And If well choose y = x in the definition of f+ that gives us f(x).so

f+(x) f(x).

thus f+ is indeed an extension of f.step-2: [f+]C0, = Let x, z Rn. Assume without loss of generality, that f+(z) f+(x).For > 0 by the property of infimum(f+ is infimum) there exists yz D such

thatf+(z) + f(yz) + |x y|

+ f(yz) + |x yz| f+(z) f(yz) + |x yz|Using this we can see that

|f+(x) f+(z)|= f+(x) f+(z) f(yz) + |x yz| + f(yz) |z yz| + |x z|

Note that here are we are writing |x||y| |xz| That we can write becausecheck that the function f(t) = (1 t) + t 1 is incresing and at note f(0)=1

Now letting 0 we have [f+]C0, indeed [f+]C0, = As f+|D = fStep-3 Let we have some other extension g of f such that [g]C0, = We therefore have for x Rn and for evry y D.Note f(y) = g(y)[g]C0, = so

|x y| g(x) g(y) = g(x) f(y) |x y|this gives

f(y) |x y| g(x) f(y) + |x y|supyD{f(y) |x y|} g(x) inf

yD{f(y) + |x y|}

f(x) g(x) f+(x)

25

Now well prove the part two.Here our D is bounded and f is a C0, function.so we can find k > 0 such that ||f ||C0(D) k.Since D is bounded, we can find a R > 0 such that D BRand dist(D; BR) = (k )

1

Now well define a continuous extension of f

f1(x) =

{f(x) if x D0 if x (BR)c

.Its easy to see that f1 C0,(D (BR)c)and lets show that [f1]C0,(D (BR)c) = Indeed for x, y D or if x, y (BR)c it is obvious.So lets prove the iinequality for x D and y (BR)cThe crucial point here is that the distance between them.

|x y| dist(D; BR)

|x y| dist(D; BR) = k

We therefore have that

|f1(x) f1(y)| = |f1(x)| = |f(x)| k = k |x y|

that shows [f1]C0,(D (BR)c) = last part of part 2We want to show g is a cotinous extension of f, and will satisfy the required

prperties.Use part 1 to extend the function f1 to Rn.This we can take g=(f1)+ and now we can see that g|D = f and supp(g) is

compact because its subset ofBR.and [g]C0,(Rn) = . g is continuous and we know g has compact support, so

g C0. This with [g]C0,(Rn) = implies that g C0,this concludes our proof.

2.2.2 Two Main Theorems Of Extension

Before going to the proof of these two theorems, well state 3 lemmas.

26

Lemma 2.2.3. There exists C0([1,)) such that for every N N There existsAN > 0 so that

|()| ANN

for every [1,) and for every k 1, 1

()d = 1 andquad

1

k()d = 0

Lemma 2.2.4. Let C0,1(Rn1) and

={x = (x

, xn) Rn1 R : xn > (x)

}and = ()c d(x)=d(x;)=inf

{|x y| : y } Then for any x = (x , xn) (1 + []C0,1).dx (x) xn

More over for every x,y with x 6= y There exists z such that(x, z] (y, z]

|x z|+ |z y| (2 + 4[]C0,1)The same also holds true when is also replaced by .

Proof: Let x = (x, xn) . Note that is closed and bounded so sompact, so

the distance function will attain the minimum on the compact set . So z such that

d(x; ) = |z x| = max1in

{|zi xi|}

As z so z = (y , (y)) for some y Rn1.

d(x; ) = max1in

{|y x|; |(y) xn|

}We then have

(x) xn = |(x) xn|

|(x) (y)|+ |(y) xn| []C0,1|x y |+ |(y) xn|

(1 + []C0,1) max1in

{|y x |; |(y) xn|

}27

= (1 + []C0,1)d(x; )

so we proved our first claim.Now step 2:- let x,y without loss of generality, lets assume that xn ynLet z = (y

, xn 2[]C0,1|x y|), well show this z has the claimed properties.

Claim:|x z|+ |z y| (2 + 4[]C0,1)

Now consider|x z|+ |z Y |

= maxi|xi zi|+ max

i|zi yi|

= max{|x z|, |xn zn|

}+ |yn xn 2[]C0,1|x y| |

max{|x z |, 2[]C0,1|x y |

}+ |yn xn|+ 2[]C0,1|x y |

|x y|+ 2[]C0,1|x y|+ |yn xn|+ 2[]C0,1|x y| 4[]C0,1|x y|+ 2 max

{|x y |, |yn xn|

}= 4[]C0,1|x y |+ 2|x y| (4[]C0,1 + 2)|x y|

So with this we proved the lemma. Now well look to the distance function. Weknow distance function is continuous but not differentiable. so well define regularizeddistance function that is just some modification of our distance function. and itllbe smooth. Before stating the lemma we have one lemma that will imply our lemmadirectly.

lemma: For every closed set F Rn, there exists a constant C and a function4(.;F ) C(F c) such that for every x, y F c

1

Cd(x;F ) 4(x;F ) Cd(x;F )

|Or(4(x;F ))| C(d(x;F ))1r|Or(4(x;F )) Or(4(y;F ))| C|x y| max{d(x;F )1r; d(y;F )1r}

proof Reference [7]Stein E.M.

28

Lemma 2.2.5. C0,1(Rn1), r 0 be an integer and 0 1. Let ={x = (x

, xn) Rn1 R : xn > (x)

}Define = c. Then there exists d? =

d?(x; ) C(; [0,]) and a constant C = C(r, n, []C0,1) such that for everyx = (x

, xn),y = (y

, yn)

d?(x) 2((x) xn)1

Cd(x) d?(x) Cd(x)|Ord?(x)| Cd(x)1r

|Ord?(x) Ord?(y)| C|x y| max{d(x)1r; d(y)1r}Proof: Well take

d?(x) = 2C1(1 + []C0,1)

Its easy to see that all these properties are satisfied by d? because all these propertiesare satisfied by 4 and d? is constant multiple of 4

Now we are ready to go for the proof of our two main extension theorems.

Theorem 2.2.6. Let C0,1(Rn1) and

={x = (x

, xn) Rn1 R : xn > (x)

}Then there exists a continuous linear extension operator

E : Cr,() Cr,(Rn)for any integer r 0 and 0 1. In particular there exists a constant

C = C(r,) such that for every f Cr,()

||E(f)||Cr,(Rn) C||f ||Cr,()proof We define the desired extension

E(f)(x, xn) =

{f(x

, xn) if (x

, xn)

1()f(x())d if (x

, xn) 6

Where is the function defined as in lemma 5.3 And d?(x) as defined in lemma5.5.

Andx() = (x

, xn + d

?(x))

29

To make the Expression for E(f) well defined We want to show that x() ,that means for (x

, xn) c well do some transfer to the nth coordinate so that itll

go to Now for 1, x c

xn + d?(x)

xn + d?(x)= xn + 2((x

) xn)

= (x) + ((x

) xn)

As x c so ((x) xn) > 0So now we have xn+d

?(x) > (x) which shows that x() So the expression

in side the integral is defined.Now f Cr,() so this implies f is bounded in and We have |()| AN

Nso

E(f) is finite for N > 1. so E(f)well defined.Now well show E(f) Cr,(Rn) and

||E(f)||Cr,(Rn) ||f ||Cr,().

Well prove this result for r = 0, 1.. and we can check on the same for generalcase.

Just to remember We have taken = c

= = ={

(x, (x

)) : x

Rn1}

For g : Rn Rn[g]Cr,0 (); [g]C0,0 ()

D [g]c0,(Rn) 2D

Step-2 Now lets proof the theorem.Case-2.1: r=0First Well show E(f) C0(Rn) and ||E(f)||C0(Rn) C||f ||C0()

E(f)| = f C0()So Its enough to prove for Claim E(f) C0()Let y x in we want to show E(f)(y) E(f)(x) in Note that y x y() x() so

30

f(y()) f(x())We want to show that

limyx,y

E(f)(y) = E(f)(x)

Note that

|()f(x())| |()||f(x)()| C|ANN|

And we know C ANN

is integrable(L1)So using Dominated convergence theorem

limyx

E(f)(y)

= limyx

1

()f(y())

=

1

() limyx

f(y())

=

1

()f(x())

= E(f)(x)

So this shows E(f) C0(Rn)Claim-2

||E(f)||C0(Rn) C||f ||C0()

||E(f)||C0() = ||f ||C0()for

||E(f)||C0()=

1

()f(x())d

= ||f ||C0()

1

()d

= C||f ||C0()||E(f)||C0() C||f ||C0()

31

So we have proved for = 0.Step-2Well prove E(f) C0,(Rn) [E(f)]C0,(Rn) C[f ]C0,()E(f)| = fSo[E(f)]C0,() C[f ]C0,()So it remain to show that

[E(f)]C0,() [f ]C0,()Let x, y

|d?(x) d?(y)| C1|x y|max

{d(x)1, d(y)1

}Lets take = 1

|d?(x) d?(y)| C1|x y|Now

|E(f)(x) E(f)(y)|

1

|()f(x()) ()f(y())|d

1

|()||f(x()) f(y())|

1

|()|[f ]C0,()|x() y()|d

=

1

|()|[f ]C0,()|xn yn + (d?(x , xn) d?(y , yn))|d

1

()[f ]C0,()(|xn yn| + C1 |x y|)d

1

()[f ]C0,()|x y|(1 + C1 )d

[f ]C0, |x y|

1

()(1 + C1 )

C3[f ]C0,()|x y|

32

That means[E(f(x))]C0,() [f ]C0,()

So we proved it for r = 0Step:3(Case r=1)step 3.1We first prove that E(f) C1(Rn) and

||E(f)||C1(Rn) C||f ||C1()Since f C1() and E(f) C1() C1() and by the properties of (), we getfor every x = (x

, xn)

First applying chain rule

xif(x()) = fxi(x()) + fxn(x())

xi(xn + d

?(x))

= fxi(x()) + fxn(x())d?xi

(x)

Hence

E(f)xi(x) =

1

fxi(x())()d+

1

fxn(x())()d?xi

(x)

Where We recall thatx() = (x

, xn + d

?(x))

From the lemma relating () and by the properties of d?(x) that |d?xi | C in similarly as in step 2, applying Dominated convergence theorem, that for everyx ,

limyx,y

E(f)xi(y) = fxi(x)

Using that |()| ANN| We have for every x ,|E(f)xi(x)| C||Of ||C0()

Clearly, E(f) is differentiable in , thus E(f) is differentiable in Rn.Step 3.2We now show that for every 1 i n, E(f)xi C0,(Rn) and

[E(f)xi ]C0,(Rn) C[Of ]C0,()As before its enough to prove that

[E(f)xi ]C0,() C[Of ]C0,()

33

.Let x, y and assume , without loss of generality, that d(x) d(y) so we

have

[E(f)xi(x) E(f)xi(y)]

|

1

()[fxi(x())fxi(y())]d|+|

1

()[fxn(x())d?xi

(x)fxn(y())d?xi(y)]d|

And thus , as in Step 2.2 the first term is readily estimated by

C1[fxi ]C0,(bar)|x y|

The second term is estimated as follows. Since () satisfies those Integrbleconditions so

|

1

()[fxn(x())d?xi

(x) fxn(y())d?xi(y)]d|

|

1

()fxn(x()[d?xi

(x)d?xi(y)]d)|+|

1

()d?xi(y)[fxn(x())fxn(y())]|

= |

1

()[fxn(x()) fxn(y())]d[d?(xi) d?(yi)]d|

+|

1

()d?xi(y)[fxn(x()) fxn(y())]|

and hence the claim, by applying the properties of () and d?(x)

|

1

()[fxn(x())d?xi

(x) fxn(y())d?xi(y)]d|

C1

1

|()|(d?(x))[Of ]C0,d(x)|x y|

+C2

1

(1 + )|psi()|[Of ]C0, |x y|

C3[Of ]C0, |x y|This proves the case r = 1. Similarly we can do for r = 2, 3....

34

Theorem 2.2.7. Let Rn be bounded open lipschitz set. then there exists acontinuous linear extension operator

E : Cr,() Cr,0 (Rn)for any integer r > 0 and every 1 more precisely there exists a constant C > 0such that for every f Cr,

E(f)| = fand supp[E(f)] is compact.

||E(f)||Cr,(RN ) C||f ||Cr,().

Proof. well find a cover for the boundary is lipschitz and bounded. so is bounded and we know that is closed so it

is compact. There are finite number of xi , i > 0, 1 i N , such that

Ni=1Bi(xi)Now is lipschitz so there exists i C0,1(Rn1) , 1 i N .Let us define

i ={x = (x

, xn) Rn1 R : xni(xi)

}Now upto a rotation

Bi(xi) = i Bi(xi)Choose 0 < < min1iNi such that

Ni=1Bi(xi)(Note: this is possible because all the balls are open so they are intersecting)Now lets choose

C = max1iN(i

) > 1 i > istep 2Lets define the auxilary functions from which well get our extension function.For each xi and Bi(xi) let i C(Rn; [0, 1]) be such that

35

i = 1 in Bi 2 (xi)

supp(i) Bi 2 (xi)let0, , + Cinfty(Rn; [0, 1])

0 = 1 in and supp(0) +B 2

+ = 1 in +B 2

and supp(+) +B 2

= 1 in ( +B 2)c and supp()

Now let

+ = 0(+

= + )

and

= 0(

= + )

supp(0) {x Rn;+ + 1}because

x supp(0)x +B

2

x ( +B 2)c or x +B

2

= 1 Or + = 1

so + + 1So we can say that the functions + and Cinfty0 (Rn)Note that + + = 0Now supp(+) +B

2

Because + = 0(+

++)

supp(0) +B 2supp(+) +B

supp(0+) +B 2

But Ni=1Bi(xi) Sosupp(+) +B

2

36

And noting that2i 1 in ni=1Bi 2

Because i = 1 for at least one i.So we can say

+Ni=1

2i

C0 (Rn)

Simplification of the extension Instead of proving this for Cr, well prove this forCr,? . So first well show Its enough to show that Existence of

E : Cr,? () Cr,0 (Rn)And

||E(f)||Cr,(Rn) ||f ||Cr,? (2.1)for every f Cr,? ()

Thus suppose that 36 holds true. So its enough to show that for f Cr,() f Cr,? and ||f ||Cr,? () C||f ||Cr,

(To be written)ConclusionStep-1 Let f Cr,? () and x RnThe desired extension given by

E(f(x) = +(x)

{Ni=1 i(x)Ei(fi)(x)N

i=1 2i (x)

}+ (x)f(x)

Where eachEi : C

r,? (i) Cr,(Rn)

is the extension operator of our last theorem. And we have

||Ei(f)||Cr,? (Rn) C||f ||Cr,(i)and

fi : i Rdefined as

fi(x) =

{if if x i Bi(xi)o if x i\ Bi(xi)

Now well show that E(f) is well defined and has all the desired properties.Step:A We have g,h Cr,? gh Cr, because

||gh||Cr, ||gh||Cr,? ||g||Cr,? ||h||Cr,?Step:B Lets first show that fi Cr,? and for an appropriate C

37

||fi||Cr,? (i) C||f ||Cr,? ()Substeps:B1 We have taken f Cr,? () , i C(Rn)

soif = fi Cr,? (i Bi(xi))

and||fi||r,? ( i Bi(xi)) C||f ||Cr,? (iZX)

Because||fi||Cr,? (iBi (xi))

= ||if ||Cr,? (iBi (xi)) ||i||Cr,? (iBi (xi))||f ||Cr,? (iBi (xi))

But we know (i Bi(xi)) = ( Bi(xi)) so||i||Cr,? (iBi (xi))||f ||Cr,? (iBi (xi))

= ||i||Cr,? (iBi (xi))||f ||Cr,? (Bi (xi)) C||f ||Cr,?

Suubstep:B2 We have supp(i) Bi 4 (x1) and write the function code So

supp(fi) i Bi 4 (xi)

This with fi Cr,? (i Bi(xi)). But see thati Bi 4 (xi) i Bi(xi)

So fi Cr,? (i) this implies fi Cr(i)||fi||Cr(i) C||f ||cr()

Substep:B3 Finally well prove the desired inequallity In the step 2. Its enough to

show that for every integer 0 m r and every x, y i|Omfi(x) Omfi(y)| C||f ||Cr,? |x y|

because this will imply[Omfi]C0, C||f ||Cr,? m

38

rm=0

[Omfi]c0, Cr

m=0

||f ||Cr,?

||fi||Cr,? (Cr + 1)||f ||Cr,?substep :B4 well prove the claim in B. Its enough to prove that for x iBi 4 (xi)and y i (Bi(xi))c. Note that |y x| 4 For any z i Bi 4 (xi)

||x z|| 2(i 4

) 2(i 1

4) 2C

Note that we have taken C = maxii

Now

|Omfi(x) Omfi(y)| |Omfi(x) Omfi(z)|+ |Omfi(z) Omfi(y)|

c||f ||Cr,? |x z| c||f ||Cr,? ()(2C)

8cC||f ||Cr,? (

4)

8cC||f ||Cr,? |x y||Omfi(x) Omfi(y)| 8cC||f ||Cr,? |x y|

Which proves the claim in B.

step:C Now well check the properties of E. Here we want to show E is welldefined. Indeed its because supp() . So (x)f(x) will be defined on

+(x)Ni=1

2i (x) C0 (Rn) , i C(Rn) and Eifi(x) Cr,(Rn) So

E(f)(x) =

(+(x)Ni=1

2i (x)

)(Ni=1

i(x)Ei(f)(x)) + (x)f(x)

is defined.step:D Claim E(f) | = f For every i, 1 i N

iEi(fi)(x)| = ifi = 2i f+ + = 0 = 1

So E(f)(x)| = +(x)ni=1

2i (x)

(n

i=1 2i f(x)) + (x)f(x)

= f(x)(+(x) + (x)) = f(x)

39

Step:E SuppE(f) +B 2

By our construction of + and see that Supp(+) +B

2and supp() So we deduce that

supp(E(f)) +B 2

Step:F Since C0 (Rn) with supp() We get that f C0 (Rn) and||f ||Cr,(Rn) C||f ||Cr,?

Step:G

||Ei(fi)||Cr,? C||f ||Cr,?Ei(fi) is an extension of fi So

||Ei(fi)||Cr,(Rn) ||fi||Cr,(i) ||Ei(fi)||Cr,? (Rn) ||fi||Cr,? (Rn)

And in B we proved that

||fi||Cr,? (i) ||f ||Cr,? ()SO by the previous lines we conclude that

||Ei(fi)||Cr,? (Rn) ||f ||Cr,?

40

2.3 Semicontinuity Result

Theorem 2.3.1. Lower Semicontinuous resultLet r 0 be an integer and 0 < 1. Let b Rn be a bounded open lischitz

set. Let R > 0 andCR =

{f Cr,() : ||f ||Cr, R

}Let fv CR be a sequence such that fv f in C0() as v then f CR and ||f ||Cr, limv inf ||f ||Cr,proof Before going to the proof lets recall that f is called lower semi continuous

at x if for every sequence xt in X with xt x thenlimtinff(xt) f(x)

And the setLy = {x X : f(x) y}

is closed iff f is lower semicontinuous.Here we can see that X = Cr, and f(x) = ||.||Cr,Set

L = limv

inf ||fv||Cr,By the property of lim infimum choose a subsequence such that

L = limi||fvi||Cr,

we know that Cr, Cr is a compact embedding.so fvi is convergent in C

r, means it is bounded so fvij a subsequence that isconvergent in Cr.

fvij f in Cr() as j means

rfvij rf in C0

|rfvij rf |= |rf(x)rfvij(x)rf(y) +rfvij(y)rfvij(y) +rfvij(x)| |rf(x)rfvij(x)|+ |rf(y)rfvij(y)|+ |rfvij(y)rfvij(x)|

2||rf rfvij ||+ [rfvij ]C0,|x y|

41

taking liminfj both sides, our left side is independent of j, and ||rf rfvij || 0 so

|rf(x)rf(y)| liminfj[rfvij ]Co, |x y|

[rf ]C0, liminfj[rfvij ]Co,As fv f in C0() as v so

||f ||Cr, limv

inf ||f ||Cr,

42

Bibliography

[1] Gilbarg D. and Trudinger N.S., Elliptic partial differential equations of secondorder, Springer-Verlag, Berlin, 1977.

[2] Adams R.A., Sobolev spaces, Academic Press, New York, 1975; second editionwith Fournier J.J.F., 2003.

[3] Coddington E.A. and Levinson N., Theory of ordinary differential equations,McGraw-Hill Book Company Inc., New York, 1955.

[4] Dacorogna B., Introduction to the calculus of variations, second edition, Im-perial College Press, London, 2009.

[5] Evans L.C., Partial differential equations, Graduate Studies in Mathematics,19, American Mathematical Society, Providence, RI, 1998.

[6] Hormander L., Linear partial differential operators, Springer-Verlag, Berlin,1963.

[7] Stein E.M., Singular integrals and differentiability properties of functions,Princeton University Press, Princeton, NJ, 1970.

[8] Rudin W., Real and complex analysis, Mc Graw-Hill, New York, 1966

Further readingsGyula Csat,Bernard Dacorogna,Olivier Kneuss The Pullback Equation for Differ-

ential Forms 2012

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