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Research Article Lipschitz Spaces and Fractional Integral Operators Associated with Nonhomogeneous Metric Measure Spaces Jiang Zhou and Dinghuai Wang Department of Mathematics, Xinjiang University, Urumqi 830046, China Correspondence should be addressed to Jiang Zhou; [email protected] Received 1 December 2013; Revised 3 April 2014; Accepted 5 April 2014; Published 17 April 2014 Academic Editor: S. A. Mohiuddine Copyright © 2014 J. Zhou and D. Wang. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. e fractional operator on nonhomogeneous metric measure spaces is introduced, which is a bounded operator from () into the space ,∞ (). Moreover, the Lipschitz spaces on nonhomogeneous metric measure spaces are also introduced, which contain the classical Lipschitz spaces. e authors establish some equivalent characterizations for the Lipschitz spaces, and some results of the boundedness of fractional operator in Lipschitz spaces are also presented. 1. Introduction As we know, the theory on spaces of homogeneous type is needed to assume that measure of metric spaces (X, , ) satisfies the doubling measure condition, which means that there exists a constant , such that, for every ball (, ) of center and radius , ((, 2)) ≤ ((, )). In recent years, many classical theories have been proved still valid without the assumption of doubling measure condition; see [1–12]. Recall that a Radon measure on is said to only satisfy the polynomial growth condition, if there exists a positive constant such that, for all ∈ R and >0, ((, )) ≤ , where is some fixed number in (0, ] and (, ) = { ∈ R : | − | < }. e analysis associated with such nondoubling measures is proved to play a striking role in solving the long-standing open Painlev´ e’s problem by Tolsa [13]. Obviously, the nondoubling measure with the polynomial growth condition may not satisfy the well- known doubling condition, which is a key assumption in harmonic analysis on spaces of homogeneous type. In 2010, Hyt¨ onen [14] introduced a new class of metric measure spaces satisfying both the so-called geometrically doubling and the upper doubling conditions (see the definition below), which are called nonhomogeneous spaces. Recently, many classical results have been proved still valid if the underlying spaces are replaced by the nonhomogeneous spaces of Hyt¨ onen (see [4–6, 9–12]). Let (X, , ) be a nonhomogeneous metric measure space in the sense of Hyt¨ onen [14]. In this paper, we establish the definition of fractional operator on nonhomogeneous metric measure spaces, which contains the classical fractional integral operator introduced by Garc´ ıa-Cuerva and Gatto [7], and similar to the definition introduced by Fu et al. [11], then we get the ( (), ,∞ ())-boundedness for frac- tional integral operator on nonhomogeneous metric measure spaces. In Section 3, we also establish the definition of Lip- schitz spaces on nonhomogeneous metric measure spaces, which contains the classical Lipschitz spaces. We establish some equivalent characterizations for the Lipschitz spaces. In Section 4, we present some results of the boundedness of fractional operator in Lipschitz spaces. To state the main results of this paper, we first recall some necessary notions and remarks. Definition 1 (see [15]). A metric space (X, , ) is said to be geometrically doubling if there exists some 0 ∈ N such that, for any ball (, ) ⊂ X, there exist a finite ball covering {( , /2)} of (, ) such that the cardinality of this cov- ering is at most 0 . Definition 2 (see [14]). A metric measure space (X, , ) is said to be upper doubling if is a Borel measure on X and there exist a dominating function : X × (0, ∞) → (0, ∞) Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 174010, 8 pages http://dx.doi.org/10.1155/2014/174010

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Page 1: Research Article Lipschitz Spaces and Fractional Integral ...downloads.hindawi.com/journals/aaa/2014/174010.pdf · Research Article Lipschitz Spaces and Fractional Integral Operators

Research ArticleLipschitz Spaces and Fractional Integral Operators Associatedwith Nonhomogeneous Metric Measure Spaces

Jiang Zhou and Dinghuai Wang

Department of Mathematics Xinjiang University Urumqi 830046 China

Correspondence should be addressed to Jiang Zhou zhoujiangshuxue126com

Received 1 December 2013 Revised 3 April 2014 Accepted 5 April 2014 Published 17 April 2014

Academic Editor S A Mohiuddine

Copyright copy 2014 J Zhou and D WangThis is an open access article distributed under theCreative CommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The fractional operator on nonhomogeneous metric measure spaces is introduced which is a bounded operator from 119871119901 (120583) intothe space 119871119902infin (120583) Moreover the Lipschitz spaces on nonhomogeneous metric measure spaces are also introduced which containthe classical Lipschitz spaces The authors establish some equivalent characterizations for the Lipschitz spaces and some results ofthe boundedness of fractional operator in Lipschitz spaces are also presented

1 Introduction

As we know the theory on spaces of homogeneous type isneeded to assume that measure 120583 of metric spaces (X 119889 120583)satisfies the doubling measure condition which means thatthere exists a constant 119862 such that for every ball 119861(119909 119903) ofcenter 119909 and radius 119903 120583(119861(119909 2119903)) le 119862120583(119861(119909 119903)) In recentyears many classical theories have been proved still validwithout the assumption of doubling measure condition see[1ndash12] Recall that a Radon measure 120583 on 119877119889 is said to onlysatisfy the polynomial growth condition if there exists apositive constant 119888 such that for all 119909 isin R119889 and 119903 gt 0120583(119861(119909 119903)) le 119888119903

119899 where 119899 is some fixed number in (0 119889] and119861(119909 119903) = 119910 isin R119889

|119909 minus 119910| lt 119903 The analysis associatedwith such nondoublingmeasures120583 is proved to play a strikingrole in solving the long-standing open Painleversquos problemby Tolsa [13] Obviously the nondoubling measure 120583 withthe polynomial growth condition may not satisfy the well-known doubling condition which is a key assumption inharmonic analysis on spaces of homogeneous type In 2010Hytonen [14] introduced a new class ofmetricmeasure spacessatisfying both the so-called geometrically doubling and theupper doubling conditions (see the definition below) whichare called nonhomogeneous spaces Recently many classicalresults have been proved still valid if the underlying spacesare replaced by the nonhomogeneous spaces of Hytonen (see[4ndash6 9ndash12])

Let (X 119889 120583) be a nonhomogeneousmetricmeasure spacein the sense of Hytonen [14] In this paper we establishthe definition of fractional operator on nonhomogeneousmetricmeasure spaces which contains the classical fractionalintegral operator introduced byGarcıa-Cuerva andGatto [7]and similar to the definition introduced by Fu et al [11]then we get the (119871119901(120583) 119871119902infin(120583))-boundedness for frac-tional integral operator on nonhomogeneousmetric measurespaces In Section 3 we also establish the definition of Lip-schitz spaces on nonhomogeneous metric measure spaceswhich contains the classical Lipschitz spaces We establishsome equivalent characterizations for the Lipschitz spacesIn Section 4 we present some results of the boundedness offractional operator in Lipschitz spaces

To state the main results of this paper we first recall somenecessary notions and remarks

Definition 1 (see [15]) A metric space (X 119889 120583) is said to begeometrically doubling if there exists some1198730 isin N such thatfor any ball 119861(119909 119903) sub X there exist a finite ball covering119861(119909119894 1199032)119894 of 119861(119909 119903) such that the cardinality of this cov-ering is at most1198730

Definition 2 (see [14]) A metric measure space (X 119889 120583) issaid to be upper doubling if 120583 is a Borel measure on X andthere exist a dominating function 120582 X times (0infin) rarr (0infin)

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 174010 8 pageshttpdxdoiorg1011552014174010

2 Abstract and Applied Analysis

and a positive constant 119888120582 such that for each 119909 isin X 119903 rarr120582(119909 119903) is nondecreasing and

120583 (119861 (119909 119903)) le 120582 (119909 119903) le 119888120582120582 (119909119903

2) forall119909 isin X 119903 gt 0

(1)

A metric measure space (X 119889 120583) is called a nonhomo-geneous metric measure space if (X 119889 120583) is geometricallydoubling and (X 119889 120583) is upper doubling

Remark 3 (i) Obviously a space of homogeneous type isa special case of upper doubling spaces where we take thedominating function120582(119909 119903) = 120583(119861(119909 119903)) On the other handthe Euclidean space R119889 with any Radon measure 120583 as in (1)is also an upper doubling space by taking the dominatingfunction 120582(119909 119903) = 119862119903119896

(ii) Let (X 119889 120583) be upper doubling with 120582 being thedominating function onX times (0infin) as in Definition 2 It wasproved in [6] that there exists another dominating function such that le 120582 and for all 119909 119910 isin X with 119889(119909 119910) le 119903

(119909 119903) le 119862 (119910 119903) (2)

Thus in this paper we always suppose that 120582 satisfies (2)

Definition 4 (see [14]) Let 120572 120573120572 isin (1infin) A ball 119861 sub 119883 iscalled (120572 120573)-doubling if 120583(120572119861) le 120573120572120583(119861)

As stated in lemma of [4] there exist plenty of doublingballs with small radii and with large radii In the rest of thepaper unless 120572 and 120573120572 are specified otherwise by an (120572 120573120572)-doubling ball wemean a (6 1205736)-doublingwith a fixed number1205736 gt max1198623log26

120582 6

119899 where 119899 = log

21198730 is viewed as a geo-

metric dimension of the spaces

Definition 5 (see [11]) Let 120598 isin (0infin) A dominating function120582 is satisfying the 120598-weak reverse doubling condition if forall 119903 isin (0 2 diam(X)) and 119886 isin (1 2 diam(X)119903) there existsa number 119862(119886) isin [1infin) depending only on 119886 and X suchthat for all 119909 isin X

120582 (119909 119886119903) ge 119862 (119886) 120582 (119909 119903) (3)

and moreoverinfin

sum

119896=1

1

[119862(119886119896)]120598 lt infin (4)

Remark 6 (i) It is easy to see that if 1205981 lt 1205982 and 120582 satisfies the1205981-weak reverse doubling condition then 120582 also satisfies the1205982-weak reverse doubling condition

(ii) Assume that diam(X) = infin For any fixed 119909 isin X weknow that

lim119903rarr0120582 (119909 119903) = 0 lim

119903rarrinfin120582 (119909 119903) = infin (5)

(ii) It is easy to see that the 120598-weak reverse doubling con-dition ismuchweaker than the assumption introduced by BuiandDuong in [4 Subsection 73] there exists119898 isin (0infin) suchthat for all 119909 isin X and 119886 119903 isin (0infin) 120582(119909 119886119903) = 119886119898120582(119909 119903)

Definition 7 (see [14]) For any two balls 119861 sub 119878 define

119870119861119878 = 1 + int2119878119861

1

120582 (119888119861 119889 (119909 119888119861))119889120583 (119909) (6)

where 119888119861 is the center of the ball 119861

Remark 8 The following discrete version 119861119878 of 119870119861119878defined in Definition 7 was first introduced by Bui andDuong [4] in nonhomogeneous metric measure spaceswhich is more close to the quantity119870119876119877 introduced by Tolsa[1] in the setting of nondoubling measures For any two balls119861 sub 119878 let 119861119878 be defined by

119861119878 = 1 +

119873119861119878

sum

119896=1

120583 (6119896119861)

120582 (119888119861 6119903119896119903119861) (7)

where 119903119861 and 119903119878 respectively denote the radii of the balls119861 and 119878 and 119873119861119878 denotes the smallest integer satisfying6119873119861119878119903119861 ge 119903119878 Obviously 119870119861119878 le 119862119861119878 As was pointed by Bui

and Duong [4] in general it is not true that119870119861119878 sim 119862119861119878

Definition 9 (see [14]) Let 120588 isin (1infin) A function119891 isin 1198711loc(120583)is said to be in the space RBMO(120583) if there exist a positiveconstant 119862 and for any ball 119861 sub X a number 119891119861 such that

1

120583 (120588119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816 119889120583 (119909) le 119862 (8)

for any two balls 119861 and 1198611 such that 119861 sub 1198611

10038161003816100381610038161003816119891119861 minus 119891119861

1

10038161003816100381610038161003816le 119862119870119861119861

1

(9)

The infimum of the positive constants 119862 satisfying abovetwo inequalities is defined to be the RBMO(120583) norm of119891 anddenoted by 119891RBMO(120583)

From [14 Lemma 46] it follows that the space RBMO(120583)is independent of 120588 isin (1infin)

In this paper we consider a variant of the fractionalintegrals from [7 Definition 41]

Definition 10 Let 0 lt 120572 lt 119899 and 0 lt 120575 le 1 A function 119870120572 isin1198711

loc(X timesX (119909 119910) 119909 = 119910) is said to be a fractional kernelof order 120572 and regularity 120575 if it satisfies the following twoconditions

(i) for all 119909 119910 isin X with 119909 = 119910

1003816100381610038161003816119870120572 (119909 119910)1003816100381610038161003816 le 119862

1

[120582 (119909 119889 (119909 119910))]1minus120572119899 (10)

(ii) for all 119909 119909 119910 isin X with 120582(119909 119889(119909 119910)) ge 2120582(119909 119889(119909 119909))

1003816100381610038161003816119870120572 (119909 119910) minus 119870120572 (119909 119910)1003816100381610038161003816 +1003816100381610038161003816119870120572 (119910 119909) minus 119870120572 (119910 119909)

1003816100381610038161003816

le 119862[120582 (119909 119889 (119909 119909))]

120575119899

[120582 (119909 119889 (119909 119910))]1minus120572119899+120575119899

(11)

Abstract and Applied Analysis 3

A linear operator 119868120572 is called fractional integral operatorwith 119870120572 satisfying (10) and (11)

119868120572119891 (119909) = intX

119870120572 (119909 119910) 119891 (119910) 119889120583 (119910) (12)

Remark 11 By taking 120582(119909 119889(119909 119910)) = 119889(119909 119910)119899 it is easy tosee that Definition 10 in this paper contains Definition 41introduced by Garcıa-Cuerva and Gatto in [7] and Defini-tion 10 is similar to Definition 19 introduced by Fu et al in[11]

Definition 12 Let 119870120572 be a fractional kernel of order 120572 andregularity 120575 119891 isin 119871119901(120583) and 0 lt 120572 minus 119899119901 lt 120575 We define

119868120572119891 (119909) = intX

119870120572 (119909 119910) minus 119870120572 (1199090 119910) 119891 (119910) 119889120583 (119910)

(13)

where 1199090 is some fixed point ofXWe observe that the integral in (13) converges both

locally and atinfin as a consequence of (10) (11) and Holderrsquosinequality Of course the function just defined depends on theelection of 1199090 but the difference between any two functionsobtained in (13) for different elections of 1199090 is just a constant

From now on we will assume that 120583(X) = infinThe resultsbelow are also true when 120583(X) lt infin

Now we state the first main theorem of this paper

Theorem 13 Let 1 le 119901 lt 119899120572 and 1119902 = 1119901 minus 120572119899If 120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min120572119899 (1119901 minus 120572119899)1199011015840) then

120583 (119909 isin X 1003816100381610038161003816119868120572119891 (119909)

1003816100381610038161003816 gt ]) le (11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

])

119902

(14)

that is 119868120572 is a bounded operator from 119871119901(120583) into the space119871119902infin(120583)

Next let us introduce Lipschitz spaces on nonhomoge-neous metric measure spaces

Definition 14 Given that 120573 isin (0 1] we say that the function119891 X rarr C satisfies a Lipschitz condition of order 120573provided that

1003816100381610038161003816119891 (119909) minus 119891 (119910)1003816100381610038161003816 le 119862[120582 (119909 119889 (119909 119910))]

120573119899 for every 119909 119910 isin X(15)

and the smallest constant in inequality (15) will be denoted by119891Lip(120573) It is easy to see that the linear space with the norm sdot Lip(120573) is a Banach space and we will call it Lip(120573)

Remark 15 Lipschitz condition can also be defined by

1003816100381610038161003816119891 (119909) minus 119891 (119910)1003816100381610038161003816 le 119862[120582 (119910 119889 (119909 119910))]

120573119899 for every 119909 119910 isin X(15)

1015840

by (2) it is easy to see that (15) and (15)1015840 are equivalent

The secondmain result of this paper is the following someequivalent characterizations for the Lipschitz spaces

Theorem 16 For a function 119891 isin 1198711119897119900119888(120583) the conditions (A)

(B) and (C) are equivalent as follows(A) There exist some constant 1198621 and a collection of num-

bers of119891119861 one for each119861 such that these two propertieshold for any all 119861 with radius 1199031

120583 (2119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816 119889120583 (119909) le 1198621120582(119909 119903)

120573119899 (16)

and for any ball 119880 such that 119861 isin 119880 and radius (119880) le2119903

1003816100381610038161003816119891119861 minus 1198911198801003816100381610038161003816 le 1198621120582(119909 119903)

120573119899 (17)

(B) There is a constant 1198622 such that1003816100381610038161003816119891 (119909) minus 119891 (119910)

1003816100381610038161003816 le 1198622120582(119909 119889(119909 119910))120573119899 (18)

for 120583-almost every 119909 and 119910 in the support of 120583(C) For any given 119901 1 le 119901 le infin there is a constant 119862(119901)

such that for every ball 119861 of radius 119903 one has

(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 119898119861 (119891)1003816100381610038161003816

119901119889120583 (119909))

1119901

le 119862 (119901) 120582(119909 119903)120573119899

(19)

where 119898119861 = (1120583(119861)) int119861119891(119910)119889120583(119910) and also for any

ball 119880 such that 119861 sub 119880 and radius (119880) le 21199031003816100381610038161003816119898119861 (119891) minus 119898119880 (119891)

1003816100381610038161003816 le 119862 (119901) 120582(119909 119903)120573119899 (20)

In addition the quantities inf 1198621 inf 1198622 and inf 119862(119901)witha fixed 119901 are equivalent

Now we state the third main result of this paper

Theorem 17 Let119870120572 be a fractional kernel 119899120572 lt 119901 le infin and120572 minus 119899119901 lt 120575 If 120582 satisfy the 120598-weak reverse doubling conditionwith 120598 isin (0min(1 minus ((120572 minus 120575)119899))1199011015840 (120572119899 minus 1119901)1199011015840) then 119868120572maps 119871119901(120583) boundedly into Lip(120572 minus 119899119901)

Theorem 18 Let 119870120572 be a fractional kernel and 120572 + 120573 lt 120575if 120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min(120572+120573)119899 (120575minus120572minus120573)119899) then 119868120572 maps Lip(120573) bound-edly into Lip(120572 + 120573) if and only if 119868120572(1) = 0

Finally we present a result which can be viewed eitheras an extension of the case 119901 = infin of the Theorem 17 or asextension of the case 120573 = 0 of Theorem 18

Theorem 19 Let 119870120572 be a fractional kernel and 0 lt 120572 lt 120575If 120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min120572119899 (120575 minus 120572)119899) then 119868120572 maps RBMO(120583) boundedlyinto Lip(120572) if and only if 119868120572(1) = 0

Finally we make some conventions on notationThroughout the whole paper 119862 stands for a positive con-stant which is independent of the main parameters but itmay vary from line to line

4 Abstract and Applied Analysis

2 Proof of Theorem 13

Proof of Theorem 13 We are going to adapt to our context ofthe proof given by Garcıa-Cuerva and Gatto [7] Consider

1003816100381610038161003816119868120572119891 (119909)1003816100381610038161003816 le int

X

1003816100381610038161003816119891 (119910)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910)

le int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910)

+ intX119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910)

= 1198681 + 1198682

(21)

By Holderrsquos inequality if 119901 gt 1 then 1 minus (1 minus 120572119899)1199011015840 lt 0

100381610038161003816100381611986821003816100381610038161003816 le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (intX119861(119909119903)

1

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

119889120583 (119910))

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (

infin

sum

119895=1

int21198951198612119895minus1119861

1

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

119889120583 (119910))

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (

infin

sum

119895=1

120583 (119861 (119909 2119895119903))

[120582 (119909 2119895minus1119903)](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (

infin

sum

119895=1

120582 (119909 2119895119903)

[120582 (119909 2119895minus1119903)](1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=1

[120582 (119909 2119895minus1119903)]

1minus(1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=1

1

[119862 (2119895)](1minus120572119899)1199011015840minus1

)

11199011015840

times [120582 (119909 119903)]120572119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)[

120582 (119909 119903)]120572119899minus1119901

(22)

which holds even for 119901 = 1 We can and assume that119891

119871119901(120583)= 1 By (5) we can choose 119903 isin (0infin) such that

119862[120582(119909 119903)]120572119899minus1119901

= ]2 Then

119909 isin X 1003816100381610038161003816119868120572119891 (119909)

1003816100381610038161003816 gt ] sub 119909 isin X 100381610038161003816100381611986811003816100381610038161003816 gt

]2

cup 119909 isin X 100381610038161003816100381611986821003816100381610038161003816 gt

]2

(23)

By the relation between 119903 and ] 120583(119909 isin X |1198682| gt ]2) = 0We use Holderrsquos inequality once more to obtain

100381610038161003816100381611986811003816100381610038161003816 le (int

119861(119909119903)

119889120583 (119910)

[120582 (119909 119889 (119909 119910))]1minus120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

120583 (119861 (119909 2minus119895119903))

[120582 (119909 2minus119895minus1119903)]1minus120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

120582 (119909 2119895119903)

[120582 (119909 2minus119895minus1119903)]1minus120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

[120582 (119909 2minus119895minus1119903)]

120572119899

)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

1

[119862 (2119895)]120572119899[120582 (119909 119903)]

120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862[120582 (119909 119903)]1205721199011015840119899

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

(24)

Abstract and Applied Analysis 5

Then by applying Tchebichevrsquos inequality we have

120583 (119909 isin X 1003816100381610038161003816119868120572119891 (119909)

1003816100381610038161003816 gt ])

le 120583 (119909 isin X 100381610038161003816100381611986811003816100381610038161003816 gt

]2)

le 119862[120582(119909 119903)]1205721199011199011015840119899]minus119901

times intX

int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910) 119889120583 (119909)

= 119862[120582 (119909 119903)]1205721199011199011015840119899]minus119901

times intX

int119861(119910119903)

119889120583 (119909)

[120582 (119909 119889 (119909 119910))]1minus120572119899

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901119889120583 (119910)

le 119862[120582(119909 119903)]1205721199011199011015840119899]minus119901[120582(119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817119871119901(120583)

= 119862 [120582 (119909 119903)] = 119862]minus119902(25)

This completes the proof of Theorem 13

Corollary 20 Let 1 lt 119901 lt 119899120572 and 1119902 = 1119901 minus 120572119899 If120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min120572119899 (1119901 minus 120572119899)1199011015840) then

10038171003817100381710038171198681205721198911003817100381710038171003817119871119902(120583)

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(26)

Proof It suffices to apply Marcinkiewiczrsquos interpolation the-orem with indices slightly bigger and slightly smaller than119901

3 Proof of Theorem 16

Before we give the proof of Theorem 16 we first introduce atechnical lemma from [8 Lemma 32]

Lemma 21 Let 119891 isin 1198711119897119900119888(120583) If 1205732 gt 2119889 then for almost every

119909 with respect to 120583 there exists a sequence of (2 1205732)-doublingballs 119861119895 = 119861(119909 119903119895) with 119903119895 rarr 0 such that

lim119895rarrinfin

1

120583 (119861119895)

int119861119895

119891 (119910) 119889120583 (119910) = 119891 (119909) (27)

Proof of Theorem 16 (A) rArr (B) Consider 119909 as in the lemmaand let 119861119895 = 119861(119909 119903119895) 119895 ge 1 a sequence of (2 1205732)-doublingballs with 119903119895 rarr 0 Consider

100381610038161003816100381610038161003816119898119861119895

(119891) minus 119891119861119895

100381610038161003816100381610038161003816le

1

120583 (119861119895)

int119861119895

100381610038161003816100381610038161003816119891 (119910) minus 119891119861

119895

100381610038161003816100381610038161003816119889120583 (119910)

le

120583 (2119861119895)

120583 (119861119895)

1

120583 (2119861119895)

int119861119895

100381610038161003816100381610038161003816119891 (119910) minus 119891119861

119895

100381610038161003816100381610038161003816119889120583 (119910)

le 1205731198621[120582 (119909 119903119895)]120573119899

(28)

and by (5) Lemma 21 we obtain that

lim119895rarrinfin

119891119861119895

= 119891 (119909) (29)

Let 119909 and 119910 be two points as in the lemma take 119861 = 119861(119909 119903)any ball with 119903 le 119889(119909 119910) and let 119880 = 119861(119909 2119889(119909 119910)) Nowdefine 119861119896 = 119861(119909 2

119896119903) for 0 le 119896 le where is the first inte-

ger such that 2119903 ge 119889(119909 119910) Then

1003816100381610038161003816119891119861 minus 1198911198801003816100381610038161003816 le

minus1

sum

119896=0

10038161003816100381610038161003816119891119861119896

minus 119891119861119896+1

10038161003816100381610038161003816+10038161003816100381610038161003816119891119861

minus 119891119880

10038161003816100381610038161003816

le 1198621

sum

119896=0

[120582 (119909 2119896119903)]

120573119899

le 1198621

sum

119896=0

119888120573119896119899

120582[120582 (119909 119903)]

120573119899

le 11986210158401198621120582(119909 119889 (119909 119910))

120573119899

(30)

where 1198621015840 is independent of 119909 and 119889(119909 119910)A similar argument can be made for the point 119910 with any

ball 1198611015840 = 119861(119910 119904) such that 119904 le 119889(119909 119910) and119881 = 119861(119910 3119889(119909 119910))Therefore1003816100381610038161003816119891119861 minus 1198911198611015840

1003816100381610038161003816 le1003816100381610038161003816119891119861 minus 119891119880

1003816100381610038161003816 +1003816100381610038161003816119891119880 minus 119891119881

1003816100381610038161003816 +1003816100381610038161003816119891119881 minus 1198911198611015840

1003816100381610038161003816

le 119862101584010158401198621120582(119909 119889 (119909 119910))

120573119899

(31)

Take two sequences of (2 1205732)-doubling balls 119861119895 = 119861(119909 119903119895)and 1198611015840 = 119861(119910 119904119895) with 119903119895 rarr 0 and 119904119895 rarr 0 We have

1003816100381610038161003816119891 (119909) minus 119891 (119910)1003816100381610038161003816 = lim

119895rarr0

100381610038161003816100381610038161003816119891119861119895

minus 1198911198611015840119895

100381610038161003816100381610038161003816le 119862

101584010158401198621120582(119909 119889 (119909 119910))

120573119899

(32)

(B) rArr (C) For 1199090 isin 119861 = 119861(119909 119903) by the properties offunction 120582 and Holderrsquos inequality we obtain

(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 119898119861 (119891)1003816100381610038161003816

119901119889120583 (119909))

1119901

le (1

120583 (119861)int119861

10038161003816100381610038161003816100381610038161003816

1

120583 (119861)int119861

(119891 (119909) minus 119891 (119910)) 119889120583 (119910)

10038161003816100381610038161003816100381610038161003816

119901

119889120583 (119909))

1119901

le (1

120583 (119861)int119861

[1198622120582 (119909 119903)]120573119901119899119889120583 (119909))

1119901

le (1

120583 (119861)int119861

[1198622119862120582120582 (1199090 119903)]120573119901119899119889120583 (119909))

1119901

le 1198622119862120582[120582 (1199090 119903)]120573119899le 1198622119862

2

120582[120582 (119909 119903)]120573119899

(33)

By the similar argument for any ball 119880 such that 119861 sub 119880 andradius 119880 le 2119903

1003816100381610038161003816119898119861 (119891) minus 119898119880 (119891)1003816100381610038161003816 le 1198622119862

2

120582[120582 (119909 2119903)]120573119899

le 119888120573119899

1205821198622119862

2

120582[120582 (119909 119903)]120573119899

(34)

6 Abstract and Applied Analysis

(C) rArr (A) Define first 119891119861 = 119898119861(119891) Then (17) is exactly(20) To prove (16) we write

1

120583 (2119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816 119889120583 (119909)

le1

120583 (2119861)(int

119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816

119901119889120583 (119909))

1119901

120583(119861)11199011015840

le120583 (119861)

120583 (2119861)(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816

119901119889120583 (119909))

1119901

le 119862 (119901) 120582(119909 119903)120573119899

(35)

This concludes the proof of the theorem

Remark 22 Theorem 16 is also true if the number 2 in con-dition (A) is replaced by any fixed 120588 gt 1 In that case theproof uses (120588 120573120588)-doubling balls that is balls satisfying120583(120588119861) le 120573120588120583(119861)

4 Proofs of Theorems 17ndash19

Proof of Theorem 17 Without loss of generality we assumethat 119901 lt infin Consider 119909 = 119910 and let 119861 be the ball with center119909 and radius 119903 = 119889(119909 119910) Then we have

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le int

2119861

1003816100381610038161003816119870120572 (119909 119911)1003816100381610038161003816

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

+ int

2119861

1003816100381610038161003816119870120572 (119910 119911)1003816100381610038161003816

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

+ int

X2119861

1003816100381610038161003816119870120572 (119909 119911) minus 119870120572 (119910 119911)1003816100381610038161003816

times1003816100381610038161003816119891 (119911)

1003816100381610038161003816 119889120583 (119911)

= 1198691 + 1198692 + 1198693

(36)

For the first term by 119899120572 lt 119901 then 1 minus (1 minus 120572119899)1199011015840 gt 0

1198691 le int2119861

1003816100381610038161003816119891 (119911)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119911)

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

120583 (119861 (119909 2minus119895+1119903))

[120582 (119909 119889 (119909 2minus119895119903))](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

120582 (119909 2minus119895+1119903)

[120582 (119909 119889 (119909 2minus119895119903))](1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

[120582 (119909 119889 (119909 2minus119895119903))]

1minus(1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

1

[119862 (2119895)](120572119899minus1119901)1199011015840

)

11199011015840

times [120582 (119909 119903)]120572119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)[

120582 (119909 119903)]120572119899minus1119901

(37)

The second term is estimated in a similar way after noting that2119861 sub 119861(119910 3119903)

Next by using (2) Holderrsquos inequality and 120572 minus 119899119901 lt 120575we get

1198693 le 119862intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817119871119901(120583)

times (intX2119861

1

[120582 (119909 119889 (119909 119911))](1minus(120572minus120575)119899)1199011015840

119889120583 (119911))

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120575119899

times (intX2119861

1

[120582 (119909 119889 (119909 119911))](1minus(120572minus120575)119899)1199011015840

119889120583 (119911))

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572minus120575)119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120572119899minus1119901

(38)

Putting together the three estimates

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120572119899minus1119901

(39)

then 119868120572 maps 119871119901(120583) boundedly into Lip(120572 minus 119899119901)

Proof of Theorem 18 If 119868120572 isin Lip(120573) then by the continuity ofthe operator 119868120572 implies that 119868120572(1) must be constant that is119868120572(1)(119909) = 119868120572(1)(1199090) = 0

On the other hand we can observe that

119868120572 (1) = 0 lArrrArr 119868120572 (1) (119909) minus 119868120572 (1) (119910) = 0 (40)

this implies that

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) 119889120583 (119911) = 0 (41)

Abstract and Applied Analysis 7

Thus we can write

119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

= intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= int2119861(119909119903)

119870120572 (119909 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ int2119861(119909119903)

minus119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ intX2119861(119909119903)

119870120572 (119909 119911) minus 119870120572 (119910 119911)

times (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= 1198721 +1198722 +1198723

(42)

where 119903 = 119889(119909 119910) For1198721

10038161003816100381610038161198721

1003816100381610038161003816 le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(43)

Similarly we know that 2119861(119909 119903) sub 3119861(119910 119903) using Holderrsquosinequality (1119905 + 11199051015840 = 1) and letting 119905 gt 119899120572 then we have

10038161003816100381610038161198722

1003816100381610038161003816

le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862(int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119905119899119889120583 (119911))

1119905

times (int2119861(119909119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862(

infin

sum

119895=0

int2minus119895+1119861(119909119903)2minus119895119861(119909119903)

[120582 (119909

119889 (119909 2minus119895+1119903))]

120573119905119899

119889120583(119911))

1119905

times (int3119861(119910119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862[120582 (119909 119889 (119909 119910))]120572119899[120582 (119910 119889 (119909 119910))]

120573119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(44)

In order to estimate1198723 we use (2) to obtain

10038161003816100381610038161198723

1003816100381610038161003816

le 119862intX2119861(119909119903)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119911))]

120573119899

[120582 (119909 119889 (119909 119911))]1minus(120572minus120575)119899

119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times intX2119861(119909119903)

1

[120582 (119909 119889 (119909 119911))]1minus(120572+120573minus120575)119899

119889120583 (119911)

le 119862[120582(119909 119889(119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572+120573minus120575)119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(45)

Combining the estimates for11987211198722 and1198723 we obtain

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 119862[120582 (119909 119889 (119909 119910))]

(120572+120573)119899 (46)

this finishes the proof

Proof of Theorem 19 119868120572(119891) isin Lip(120572) implies that 119868120572(1) = 0and equivalently that

intX

(119870120572 (119909 119910) minus 119870120572 (1199090 119910)) 119889120583 (119910) = 0 (47)

for all 119909Take two points 119909 = 119910 and let119861 = 119861(119909 119903)with 119903 = 119889(119909 119910)

Then

10038161003816100381610038161003816119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

10038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

le

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119909 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

X2119861

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

= 1198731 + 1198732 + 1198733

(48)

8 Abstract and Applied Analysis

For the first term by Holderrsquos inequality with some 119901 gt 119899120572

1198731 le int2119861

1

[120582 (119909 119889 (119909 119911))]1minus120572119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le (int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119911))](1minus120572119899)1199011015840

)

11199011015840

times (int2119861

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816

119901119889120583 (119911))

1119901

le 119862[120582 (119909 119903)]120572119899minus1119901

120583(120588119861)11199011003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

le 119862[120582 (119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(49)

Using 2119861 sub 119861(119910 3119903) the second term can be dealt with inthe same way as the 1198731 then 1198732 le 119862[120582(119909 119903)]

120572119899119891RBMO(120583)

It is easy to see that |1198912119896+1119861 minus 1198912119861| le 119891RBMO(120583)11987021198612119896+1119861 le

119896119891RBMO(120583) then

1198733 le intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times

infin

sum

119896=1

(int2119896+11198612119896119861

1003816100381610038161003816119891 (119911) minus 1198912119896+11198611003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

119889120583 (119911)

+

10038161003816100381610038161198912119896+1119861 minus 11989121198611003816100381610038161003816

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

120583 (2119896+1119861))

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=1

120583 (1205882119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

+

infin

sum

119896=1

119896

120583 (2119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

)

le 119862[120582(119909 119889(119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=0

(119896 + 1) [120582 (119909 119889 (119909 2119896119903))]

120572119899minus120575119899

)

le 119862[120582 (119909 119889 (119909 119910))]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(50)

Thus the proof of Theorem 19 is completed

Conflict of Interests

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

Acknowledgments

Jiang Zhou is supported by the National Science Foundationof China (Grants nos 11261055 and 11161044) and by theNational Natural Science Foundation of Xinjiang (Grantsnos 2011211A005 and BS120104)

References

[1] X Tolsa ldquoBMO1198671 and Calderon-Zygmund operators for nondoubling measuresrdquoMathematische Annalen vol 319 no 1 pp89ndash149 2001

[2] X Tolsa ldquoLittlewood-Paley theory and the 119879(1) theorem withnon-doublingmeasuresrdquoAdvances inMathematics vol 164 no1 pp 57ndash116 2001

[3] X Tolsa ldquoThe space 1198671 for nondoubling measures in terms ofa grand maximal operatorrdquo Transactions of the American Math-ematical Society vol 355 no 1 pp 315ndash348 2003

[4] T A Bui and X T Duong ldquoHardy spaces regularized BMOspaces and the boundedness of Calderon-Zygmund operatorson non-homogeneous spacesrdquo Journal of Geometric Analysisvol 23 no 2 pp 895ndash932 2013

[5] T Hytonen S Liu D Yang and D Yang ldquoBoundedness of Cal-deron-Zygmund operators on non-homogeneous metric mea-sure spacesrdquo Canadian Journal of Mathematics vol 64 no 4pp 892ndash923 2012

[6] THytonenD Yang andD Yang ldquoTheHardy space1198671 onnon-homogeneous metric spacesrdquo Mathematical Proceedings of theCambridge Philosophical Society vol 153 no 1 pp 9ndash31 2012

[7] J Garcıa-Cuerva and A E Gatto ldquoBoundedness properties offractional integral operators associated to non-doubling mea-suresrdquo Studia Mathematica vol 162 no 3 pp 245ndash261 2004

[8] J Garcıa-Cuerva and A E Gatto ldquoLipschitz spaces and Cal-deron-Zygmund operators associated to non-doubling mea-suresrdquo Publicacions Matematiques vol 49 no 2 pp 285ndash2962005

[9] Y Cao and J Zhou ldquoMorrey spaces for non-homogeneousmetricmeasure spacesrdquoAbstract andApplied Analysis vol 2013Article ID 196459 8 pages 2013

[10] X Fu D Yang and W Yuan ldquoBoundedness of multilinearcommutators of Calderon-Zygmund operators onOrlicz spacesover non-homogeneous spacesrdquo Taiwanese Journal of Mathe-matics vol 16 no 6 pp 2203ndash2238 2012

[11] X Fu D Yang and W Yuan ldquoGeneralized fractional integralsand their commutators over non-homogeneous spacesrdquo Tai-wanese Journal of Mathematics vol 18 no 2 pp 509ndash557 2014

[12] H Lin andD Yang ldquoEquivalent boundedness ofMarcinkiewiczintegrals on non-homogeneous metric measure spacesrdquo ScienceChina Mathematics vol 57 no 1 pp 123ndash144 2014

[13] X Tolsa ldquoPainleversquos problem and the semiadditivity of analyticcapacityrdquo Acta Mathematica vol 190 no 1 pp 105ndash149 2003

[14] T Hytonen ldquoA framework for non-homogeneous analysis onmetric spaces and the RBMO space of Tolsardquo PublicacionsMatematiques vol 54 no 2 pp 485ndash504 2010

[15] R R Coifman and G Weiss Analyse Harmonique Non-Com-mutative sur Certains Espaces Homogenes vol 242 of LectureNotes in Mathematics Springer Berlin Germany 1971

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Stochastic AnalysisInternational Journal of

Page 2: Research Article Lipschitz Spaces and Fractional Integral ...downloads.hindawi.com/journals/aaa/2014/174010.pdf · Research Article Lipschitz Spaces and Fractional Integral Operators

2 Abstract and Applied Analysis

and a positive constant 119888120582 such that for each 119909 isin X 119903 rarr120582(119909 119903) is nondecreasing and

120583 (119861 (119909 119903)) le 120582 (119909 119903) le 119888120582120582 (119909119903

2) forall119909 isin X 119903 gt 0

(1)

A metric measure space (X 119889 120583) is called a nonhomo-geneous metric measure space if (X 119889 120583) is geometricallydoubling and (X 119889 120583) is upper doubling

Remark 3 (i) Obviously a space of homogeneous type isa special case of upper doubling spaces where we take thedominating function120582(119909 119903) = 120583(119861(119909 119903)) On the other handthe Euclidean space R119889 with any Radon measure 120583 as in (1)is also an upper doubling space by taking the dominatingfunction 120582(119909 119903) = 119862119903119896

(ii) Let (X 119889 120583) be upper doubling with 120582 being thedominating function onX times (0infin) as in Definition 2 It wasproved in [6] that there exists another dominating function such that le 120582 and for all 119909 119910 isin X with 119889(119909 119910) le 119903

(119909 119903) le 119862 (119910 119903) (2)

Thus in this paper we always suppose that 120582 satisfies (2)

Definition 4 (see [14]) Let 120572 120573120572 isin (1infin) A ball 119861 sub 119883 iscalled (120572 120573)-doubling if 120583(120572119861) le 120573120572120583(119861)

As stated in lemma of [4] there exist plenty of doublingballs with small radii and with large radii In the rest of thepaper unless 120572 and 120573120572 are specified otherwise by an (120572 120573120572)-doubling ball wemean a (6 1205736)-doublingwith a fixed number1205736 gt max1198623log26

120582 6

119899 where 119899 = log

21198730 is viewed as a geo-

metric dimension of the spaces

Definition 5 (see [11]) Let 120598 isin (0infin) A dominating function120582 is satisfying the 120598-weak reverse doubling condition if forall 119903 isin (0 2 diam(X)) and 119886 isin (1 2 diam(X)119903) there existsa number 119862(119886) isin [1infin) depending only on 119886 and X suchthat for all 119909 isin X

120582 (119909 119886119903) ge 119862 (119886) 120582 (119909 119903) (3)

and moreoverinfin

sum

119896=1

1

[119862(119886119896)]120598 lt infin (4)

Remark 6 (i) It is easy to see that if 1205981 lt 1205982 and 120582 satisfies the1205981-weak reverse doubling condition then 120582 also satisfies the1205982-weak reverse doubling condition

(ii) Assume that diam(X) = infin For any fixed 119909 isin X weknow that

lim119903rarr0120582 (119909 119903) = 0 lim

119903rarrinfin120582 (119909 119903) = infin (5)

(ii) It is easy to see that the 120598-weak reverse doubling con-dition ismuchweaker than the assumption introduced by BuiandDuong in [4 Subsection 73] there exists119898 isin (0infin) suchthat for all 119909 isin X and 119886 119903 isin (0infin) 120582(119909 119886119903) = 119886119898120582(119909 119903)

Definition 7 (see [14]) For any two balls 119861 sub 119878 define

119870119861119878 = 1 + int2119878119861

1

120582 (119888119861 119889 (119909 119888119861))119889120583 (119909) (6)

where 119888119861 is the center of the ball 119861

Remark 8 The following discrete version 119861119878 of 119870119861119878defined in Definition 7 was first introduced by Bui andDuong [4] in nonhomogeneous metric measure spaceswhich is more close to the quantity119870119876119877 introduced by Tolsa[1] in the setting of nondoubling measures For any two balls119861 sub 119878 let 119861119878 be defined by

119861119878 = 1 +

119873119861119878

sum

119896=1

120583 (6119896119861)

120582 (119888119861 6119903119896119903119861) (7)

where 119903119861 and 119903119878 respectively denote the radii of the balls119861 and 119878 and 119873119861119878 denotes the smallest integer satisfying6119873119861119878119903119861 ge 119903119878 Obviously 119870119861119878 le 119862119861119878 As was pointed by Bui

and Duong [4] in general it is not true that119870119861119878 sim 119862119861119878

Definition 9 (see [14]) Let 120588 isin (1infin) A function119891 isin 1198711loc(120583)is said to be in the space RBMO(120583) if there exist a positiveconstant 119862 and for any ball 119861 sub X a number 119891119861 such that

1

120583 (120588119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816 119889120583 (119909) le 119862 (8)

for any two balls 119861 and 1198611 such that 119861 sub 1198611

10038161003816100381610038161003816119891119861 minus 119891119861

1

10038161003816100381610038161003816le 119862119870119861119861

1

(9)

The infimum of the positive constants 119862 satisfying abovetwo inequalities is defined to be the RBMO(120583) norm of119891 anddenoted by 119891RBMO(120583)

From [14 Lemma 46] it follows that the space RBMO(120583)is independent of 120588 isin (1infin)

In this paper we consider a variant of the fractionalintegrals from [7 Definition 41]

Definition 10 Let 0 lt 120572 lt 119899 and 0 lt 120575 le 1 A function 119870120572 isin1198711

loc(X timesX (119909 119910) 119909 = 119910) is said to be a fractional kernelof order 120572 and regularity 120575 if it satisfies the following twoconditions

(i) for all 119909 119910 isin X with 119909 = 119910

1003816100381610038161003816119870120572 (119909 119910)1003816100381610038161003816 le 119862

1

[120582 (119909 119889 (119909 119910))]1minus120572119899 (10)

(ii) for all 119909 119909 119910 isin X with 120582(119909 119889(119909 119910)) ge 2120582(119909 119889(119909 119909))

1003816100381610038161003816119870120572 (119909 119910) minus 119870120572 (119909 119910)1003816100381610038161003816 +1003816100381610038161003816119870120572 (119910 119909) minus 119870120572 (119910 119909)

1003816100381610038161003816

le 119862[120582 (119909 119889 (119909 119909))]

120575119899

[120582 (119909 119889 (119909 119910))]1minus120572119899+120575119899

(11)

Abstract and Applied Analysis 3

A linear operator 119868120572 is called fractional integral operatorwith 119870120572 satisfying (10) and (11)

119868120572119891 (119909) = intX

119870120572 (119909 119910) 119891 (119910) 119889120583 (119910) (12)

Remark 11 By taking 120582(119909 119889(119909 119910)) = 119889(119909 119910)119899 it is easy tosee that Definition 10 in this paper contains Definition 41introduced by Garcıa-Cuerva and Gatto in [7] and Defini-tion 10 is similar to Definition 19 introduced by Fu et al in[11]

Definition 12 Let 119870120572 be a fractional kernel of order 120572 andregularity 120575 119891 isin 119871119901(120583) and 0 lt 120572 minus 119899119901 lt 120575 We define

119868120572119891 (119909) = intX

119870120572 (119909 119910) minus 119870120572 (1199090 119910) 119891 (119910) 119889120583 (119910)

(13)

where 1199090 is some fixed point ofXWe observe that the integral in (13) converges both

locally and atinfin as a consequence of (10) (11) and Holderrsquosinequality Of course the function just defined depends on theelection of 1199090 but the difference between any two functionsobtained in (13) for different elections of 1199090 is just a constant

From now on we will assume that 120583(X) = infinThe resultsbelow are also true when 120583(X) lt infin

Now we state the first main theorem of this paper

Theorem 13 Let 1 le 119901 lt 119899120572 and 1119902 = 1119901 minus 120572119899If 120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min120572119899 (1119901 minus 120572119899)1199011015840) then

120583 (119909 isin X 1003816100381610038161003816119868120572119891 (119909)

1003816100381610038161003816 gt ]) le (11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

])

119902

(14)

that is 119868120572 is a bounded operator from 119871119901(120583) into the space119871119902infin(120583)

Next let us introduce Lipschitz spaces on nonhomoge-neous metric measure spaces

Definition 14 Given that 120573 isin (0 1] we say that the function119891 X rarr C satisfies a Lipschitz condition of order 120573provided that

1003816100381610038161003816119891 (119909) minus 119891 (119910)1003816100381610038161003816 le 119862[120582 (119909 119889 (119909 119910))]

120573119899 for every 119909 119910 isin X(15)

and the smallest constant in inequality (15) will be denoted by119891Lip(120573) It is easy to see that the linear space with the norm sdot Lip(120573) is a Banach space and we will call it Lip(120573)

Remark 15 Lipschitz condition can also be defined by

1003816100381610038161003816119891 (119909) minus 119891 (119910)1003816100381610038161003816 le 119862[120582 (119910 119889 (119909 119910))]

120573119899 for every 119909 119910 isin X(15)

1015840

by (2) it is easy to see that (15) and (15)1015840 are equivalent

The secondmain result of this paper is the following someequivalent characterizations for the Lipschitz spaces

Theorem 16 For a function 119891 isin 1198711119897119900119888(120583) the conditions (A)

(B) and (C) are equivalent as follows(A) There exist some constant 1198621 and a collection of num-

bers of119891119861 one for each119861 such that these two propertieshold for any all 119861 with radius 1199031

120583 (2119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816 119889120583 (119909) le 1198621120582(119909 119903)

120573119899 (16)

and for any ball 119880 such that 119861 isin 119880 and radius (119880) le2119903

1003816100381610038161003816119891119861 minus 1198911198801003816100381610038161003816 le 1198621120582(119909 119903)

120573119899 (17)

(B) There is a constant 1198622 such that1003816100381610038161003816119891 (119909) minus 119891 (119910)

1003816100381610038161003816 le 1198622120582(119909 119889(119909 119910))120573119899 (18)

for 120583-almost every 119909 and 119910 in the support of 120583(C) For any given 119901 1 le 119901 le infin there is a constant 119862(119901)

such that for every ball 119861 of radius 119903 one has

(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 119898119861 (119891)1003816100381610038161003816

119901119889120583 (119909))

1119901

le 119862 (119901) 120582(119909 119903)120573119899

(19)

where 119898119861 = (1120583(119861)) int119861119891(119910)119889120583(119910) and also for any

ball 119880 such that 119861 sub 119880 and radius (119880) le 21199031003816100381610038161003816119898119861 (119891) minus 119898119880 (119891)

1003816100381610038161003816 le 119862 (119901) 120582(119909 119903)120573119899 (20)

In addition the quantities inf 1198621 inf 1198622 and inf 119862(119901)witha fixed 119901 are equivalent

Now we state the third main result of this paper

Theorem 17 Let119870120572 be a fractional kernel 119899120572 lt 119901 le infin and120572 minus 119899119901 lt 120575 If 120582 satisfy the 120598-weak reverse doubling conditionwith 120598 isin (0min(1 minus ((120572 minus 120575)119899))1199011015840 (120572119899 minus 1119901)1199011015840) then 119868120572maps 119871119901(120583) boundedly into Lip(120572 minus 119899119901)

Theorem 18 Let 119870120572 be a fractional kernel and 120572 + 120573 lt 120575if 120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min(120572+120573)119899 (120575minus120572minus120573)119899) then 119868120572 maps Lip(120573) bound-edly into Lip(120572 + 120573) if and only if 119868120572(1) = 0

Finally we present a result which can be viewed eitheras an extension of the case 119901 = infin of the Theorem 17 or asextension of the case 120573 = 0 of Theorem 18

Theorem 19 Let 119870120572 be a fractional kernel and 0 lt 120572 lt 120575If 120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min120572119899 (120575 minus 120572)119899) then 119868120572 maps RBMO(120583) boundedlyinto Lip(120572) if and only if 119868120572(1) = 0

Finally we make some conventions on notationThroughout the whole paper 119862 stands for a positive con-stant which is independent of the main parameters but itmay vary from line to line

4 Abstract and Applied Analysis

2 Proof of Theorem 13

Proof of Theorem 13 We are going to adapt to our context ofthe proof given by Garcıa-Cuerva and Gatto [7] Consider

1003816100381610038161003816119868120572119891 (119909)1003816100381610038161003816 le int

X

1003816100381610038161003816119891 (119910)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910)

le int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910)

+ intX119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910)

= 1198681 + 1198682

(21)

By Holderrsquos inequality if 119901 gt 1 then 1 minus (1 minus 120572119899)1199011015840 lt 0

100381610038161003816100381611986821003816100381610038161003816 le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (intX119861(119909119903)

1

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

119889120583 (119910))

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (

infin

sum

119895=1

int21198951198612119895minus1119861

1

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

119889120583 (119910))

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (

infin

sum

119895=1

120583 (119861 (119909 2119895119903))

[120582 (119909 2119895minus1119903)](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (

infin

sum

119895=1

120582 (119909 2119895119903)

[120582 (119909 2119895minus1119903)](1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=1

[120582 (119909 2119895minus1119903)]

1minus(1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=1

1

[119862 (2119895)](1minus120572119899)1199011015840minus1

)

11199011015840

times [120582 (119909 119903)]120572119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)[

120582 (119909 119903)]120572119899minus1119901

(22)

which holds even for 119901 = 1 We can and assume that119891

119871119901(120583)= 1 By (5) we can choose 119903 isin (0infin) such that

119862[120582(119909 119903)]120572119899minus1119901

= ]2 Then

119909 isin X 1003816100381610038161003816119868120572119891 (119909)

1003816100381610038161003816 gt ] sub 119909 isin X 100381610038161003816100381611986811003816100381610038161003816 gt

]2

cup 119909 isin X 100381610038161003816100381611986821003816100381610038161003816 gt

]2

(23)

By the relation between 119903 and ] 120583(119909 isin X |1198682| gt ]2) = 0We use Holderrsquos inequality once more to obtain

100381610038161003816100381611986811003816100381610038161003816 le (int

119861(119909119903)

119889120583 (119910)

[120582 (119909 119889 (119909 119910))]1minus120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

120583 (119861 (119909 2minus119895119903))

[120582 (119909 2minus119895minus1119903)]1minus120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

120582 (119909 2119895119903)

[120582 (119909 2minus119895minus1119903)]1minus120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

[120582 (119909 2minus119895minus1119903)]

120572119899

)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

1

[119862 (2119895)]120572119899[120582 (119909 119903)]

120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862[120582 (119909 119903)]1205721199011015840119899

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

(24)

Abstract and Applied Analysis 5

Then by applying Tchebichevrsquos inequality we have

120583 (119909 isin X 1003816100381610038161003816119868120572119891 (119909)

1003816100381610038161003816 gt ])

le 120583 (119909 isin X 100381610038161003816100381611986811003816100381610038161003816 gt

]2)

le 119862[120582(119909 119903)]1205721199011199011015840119899]minus119901

times intX

int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910) 119889120583 (119909)

= 119862[120582 (119909 119903)]1205721199011199011015840119899]minus119901

times intX

int119861(119910119903)

119889120583 (119909)

[120582 (119909 119889 (119909 119910))]1minus120572119899

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901119889120583 (119910)

le 119862[120582(119909 119903)]1205721199011199011015840119899]minus119901[120582(119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817119871119901(120583)

= 119862 [120582 (119909 119903)] = 119862]minus119902(25)

This completes the proof of Theorem 13

Corollary 20 Let 1 lt 119901 lt 119899120572 and 1119902 = 1119901 minus 120572119899 If120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min120572119899 (1119901 minus 120572119899)1199011015840) then

10038171003817100381710038171198681205721198911003817100381710038171003817119871119902(120583)

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(26)

Proof It suffices to apply Marcinkiewiczrsquos interpolation the-orem with indices slightly bigger and slightly smaller than119901

3 Proof of Theorem 16

Before we give the proof of Theorem 16 we first introduce atechnical lemma from [8 Lemma 32]

Lemma 21 Let 119891 isin 1198711119897119900119888(120583) If 1205732 gt 2119889 then for almost every

119909 with respect to 120583 there exists a sequence of (2 1205732)-doublingballs 119861119895 = 119861(119909 119903119895) with 119903119895 rarr 0 such that

lim119895rarrinfin

1

120583 (119861119895)

int119861119895

119891 (119910) 119889120583 (119910) = 119891 (119909) (27)

Proof of Theorem 16 (A) rArr (B) Consider 119909 as in the lemmaand let 119861119895 = 119861(119909 119903119895) 119895 ge 1 a sequence of (2 1205732)-doublingballs with 119903119895 rarr 0 Consider

100381610038161003816100381610038161003816119898119861119895

(119891) minus 119891119861119895

100381610038161003816100381610038161003816le

1

120583 (119861119895)

int119861119895

100381610038161003816100381610038161003816119891 (119910) minus 119891119861

119895

100381610038161003816100381610038161003816119889120583 (119910)

le

120583 (2119861119895)

120583 (119861119895)

1

120583 (2119861119895)

int119861119895

100381610038161003816100381610038161003816119891 (119910) minus 119891119861

119895

100381610038161003816100381610038161003816119889120583 (119910)

le 1205731198621[120582 (119909 119903119895)]120573119899

(28)

and by (5) Lemma 21 we obtain that

lim119895rarrinfin

119891119861119895

= 119891 (119909) (29)

Let 119909 and 119910 be two points as in the lemma take 119861 = 119861(119909 119903)any ball with 119903 le 119889(119909 119910) and let 119880 = 119861(119909 2119889(119909 119910)) Nowdefine 119861119896 = 119861(119909 2

119896119903) for 0 le 119896 le where is the first inte-

ger such that 2119903 ge 119889(119909 119910) Then

1003816100381610038161003816119891119861 minus 1198911198801003816100381610038161003816 le

minus1

sum

119896=0

10038161003816100381610038161003816119891119861119896

minus 119891119861119896+1

10038161003816100381610038161003816+10038161003816100381610038161003816119891119861

minus 119891119880

10038161003816100381610038161003816

le 1198621

sum

119896=0

[120582 (119909 2119896119903)]

120573119899

le 1198621

sum

119896=0

119888120573119896119899

120582[120582 (119909 119903)]

120573119899

le 11986210158401198621120582(119909 119889 (119909 119910))

120573119899

(30)

where 1198621015840 is independent of 119909 and 119889(119909 119910)A similar argument can be made for the point 119910 with any

ball 1198611015840 = 119861(119910 119904) such that 119904 le 119889(119909 119910) and119881 = 119861(119910 3119889(119909 119910))Therefore1003816100381610038161003816119891119861 minus 1198911198611015840

1003816100381610038161003816 le1003816100381610038161003816119891119861 minus 119891119880

1003816100381610038161003816 +1003816100381610038161003816119891119880 minus 119891119881

1003816100381610038161003816 +1003816100381610038161003816119891119881 minus 1198911198611015840

1003816100381610038161003816

le 119862101584010158401198621120582(119909 119889 (119909 119910))

120573119899

(31)

Take two sequences of (2 1205732)-doubling balls 119861119895 = 119861(119909 119903119895)and 1198611015840 = 119861(119910 119904119895) with 119903119895 rarr 0 and 119904119895 rarr 0 We have

1003816100381610038161003816119891 (119909) minus 119891 (119910)1003816100381610038161003816 = lim

119895rarr0

100381610038161003816100381610038161003816119891119861119895

minus 1198911198611015840119895

100381610038161003816100381610038161003816le 119862

101584010158401198621120582(119909 119889 (119909 119910))

120573119899

(32)

(B) rArr (C) For 1199090 isin 119861 = 119861(119909 119903) by the properties offunction 120582 and Holderrsquos inequality we obtain

(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 119898119861 (119891)1003816100381610038161003816

119901119889120583 (119909))

1119901

le (1

120583 (119861)int119861

10038161003816100381610038161003816100381610038161003816

1

120583 (119861)int119861

(119891 (119909) minus 119891 (119910)) 119889120583 (119910)

10038161003816100381610038161003816100381610038161003816

119901

119889120583 (119909))

1119901

le (1

120583 (119861)int119861

[1198622120582 (119909 119903)]120573119901119899119889120583 (119909))

1119901

le (1

120583 (119861)int119861

[1198622119862120582120582 (1199090 119903)]120573119901119899119889120583 (119909))

1119901

le 1198622119862120582[120582 (1199090 119903)]120573119899le 1198622119862

2

120582[120582 (119909 119903)]120573119899

(33)

By the similar argument for any ball 119880 such that 119861 sub 119880 andradius 119880 le 2119903

1003816100381610038161003816119898119861 (119891) minus 119898119880 (119891)1003816100381610038161003816 le 1198622119862

2

120582[120582 (119909 2119903)]120573119899

le 119888120573119899

1205821198622119862

2

120582[120582 (119909 119903)]120573119899

(34)

6 Abstract and Applied Analysis

(C) rArr (A) Define first 119891119861 = 119898119861(119891) Then (17) is exactly(20) To prove (16) we write

1

120583 (2119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816 119889120583 (119909)

le1

120583 (2119861)(int

119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816

119901119889120583 (119909))

1119901

120583(119861)11199011015840

le120583 (119861)

120583 (2119861)(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816

119901119889120583 (119909))

1119901

le 119862 (119901) 120582(119909 119903)120573119899

(35)

This concludes the proof of the theorem

Remark 22 Theorem 16 is also true if the number 2 in con-dition (A) is replaced by any fixed 120588 gt 1 In that case theproof uses (120588 120573120588)-doubling balls that is balls satisfying120583(120588119861) le 120573120588120583(119861)

4 Proofs of Theorems 17ndash19

Proof of Theorem 17 Without loss of generality we assumethat 119901 lt infin Consider 119909 = 119910 and let 119861 be the ball with center119909 and radius 119903 = 119889(119909 119910) Then we have

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le int

2119861

1003816100381610038161003816119870120572 (119909 119911)1003816100381610038161003816

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

+ int

2119861

1003816100381610038161003816119870120572 (119910 119911)1003816100381610038161003816

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

+ int

X2119861

1003816100381610038161003816119870120572 (119909 119911) minus 119870120572 (119910 119911)1003816100381610038161003816

times1003816100381610038161003816119891 (119911)

1003816100381610038161003816 119889120583 (119911)

= 1198691 + 1198692 + 1198693

(36)

For the first term by 119899120572 lt 119901 then 1 minus (1 minus 120572119899)1199011015840 gt 0

1198691 le int2119861

1003816100381610038161003816119891 (119911)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119911)

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

120583 (119861 (119909 2minus119895+1119903))

[120582 (119909 119889 (119909 2minus119895119903))](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

120582 (119909 2minus119895+1119903)

[120582 (119909 119889 (119909 2minus119895119903))](1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

[120582 (119909 119889 (119909 2minus119895119903))]

1minus(1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

1

[119862 (2119895)](120572119899minus1119901)1199011015840

)

11199011015840

times [120582 (119909 119903)]120572119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)[

120582 (119909 119903)]120572119899minus1119901

(37)

The second term is estimated in a similar way after noting that2119861 sub 119861(119910 3119903)

Next by using (2) Holderrsquos inequality and 120572 minus 119899119901 lt 120575we get

1198693 le 119862intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817119871119901(120583)

times (intX2119861

1

[120582 (119909 119889 (119909 119911))](1minus(120572minus120575)119899)1199011015840

119889120583 (119911))

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120575119899

times (intX2119861

1

[120582 (119909 119889 (119909 119911))](1minus(120572minus120575)119899)1199011015840

119889120583 (119911))

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572minus120575)119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120572119899minus1119901

(38)

Putting together the three estimates

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120572119899minus1119901

(39)

then 119868120572 maps 119871119901(120583) boundedly into Lip(120572 minus 119899119901)

Proof of Theorem 18 If 119868120572 isin Lip(120573) then by the continuity ofthe operator 119868120572 implies that 119868120572(1) must be constant that is119868120572(1)(119909) = 119868120572(1)(1199090) = 0

On the other hand we can observe that

119868120572 (1) = 0 lArrrArr 119868120572 (1) (119909) minus 119868120572 (1) (119910) = 0 (40)

this implies that

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) 119889120583 (119911) = 0 (41)

Abstract and Applied Analysis 7

Thus we can write

119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

= intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= int2119861(119909119903)

119870120572 (119909 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ int2119861(119909119903)

minus119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ intX2119861(119909119903)

119870120572 (119909 119911) minus 119870120572 (119910 119911)

times (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= 1198721 +1198722 +1198723

(42)

where 119903 = 119889(119909 119910) For1198721

10038161003816100381610038161198721

1003816100381610038161003816 le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(43)

Similarly we know that 2119861(119909 119903) sub 3119861(119910 119903) using Holderrsquosinequality (1119905 + 11199051015840 = 1) and letting 119905 gt 119899120572 then we have

10038161003816100381610038161198722

1003816100381610038161003816

le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862(int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119905119899119889120583 (119911))

1119905

times (int2119861(119909119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862(

infin

sum

119895=0

int2minus119895+1119861(119909119903)2minus119895119861(119909119903)

[120582 (119909

119889 (119909 2minus119895+1119903))]

120573119905119899

119889120583(119911))

1119905

times (int3119861(119910119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862[120582 (119909 119889 (119909 119910))]120572119899[120582 (119910 119889 (119909 119910))]

120573119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(44)

In order to estimate1198723 we use (2) to obtain

10038161003816100381610038161198723

1003816100381610038161003816

le 119862intX2119861(119909119903)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119911))]

120573119899

[120582 (119909 119889 (119909 119911))]1minus(120572minus120575)119899

119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times intX2119861(119909119903)

1

[120582 (119909 119889 (119909 119911))]1minus(120572+120573minus120575)119899

119889120583 (119911)

le 119862[120582(119909 119889(119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572+120573minus120575)119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(45)

Combining the estimates for11987211198722 and1198723 we obtain

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 119862[120582 (119909 119889 (119909 119910))]

(120572+120573)119899 (46)

this finishes the proof

Proof of Theorem 19 119868120572(119891) isin Lip(120572) implies that 119868120572(1) = 0and equivalently that

intX

(119870120572 (119909 119910) minus 119870120572 (1199090 119910)) 119889120583 (119910) = 0 (47)

for all 119909Take two points 119909 = 119910 and let119861 = 119861(119909 119903)with 119903 = 119889(119909 119910)

Then

10038161003816100381610038161003816119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

10038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

le

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119909 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

X2119861

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

= 1198731 + 1198732 + 1198733

(48)

8 Abstract and Applied Analysis

For the first term by Holderrsquos inequality with some 119901 gt 119899120572

1198731 le int2119861

1

[120582 (119909 119889 (119909 119911))]1minus120572119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le (int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119911))](1minus120572119899)1199011015840

)

11199011015840

times (int2119861

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816

119901119889120583 (119911))

1119901

le 119862[120582 (119909 119903)]120572119899minus1119901

120583(120588119861)11199011003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

le 119862[120582 (119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(49)

Using 2119861 sub 119861(119910 3119903) the second term can be dealt with inthe same way as the 1198731 then 1198732 le 119862[120582(119909 119903)]

120572119899119891RBMO(120583)

It is easy to see that |1198912119896+1119861 minus 1198912119861| le 119891RBMO(120583)11987021198612119896+1119861 le

119896119891RBMO(120583) then

1198733 le intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times

infin

sum

119896=1

(int2119896+11198612119896119861

1003816100381610038161003816119891 (119911) minus 1198912119896+11198611003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

119889120583 (119911)

+

10038161003816100381610038161198912119896+1119861 minus 11989121198611003816100381610038161003816

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

120583 (2119896+1119861))

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=1

120583 (1205882119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

+

infin

sum

119896=1

119896

120583 (2119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

)

le 119862[120582(119909 119889(119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=0

(119896 + 1) [120582 (119909 119889 (119909 2119896119903))]

120572119899minus120575119899

)

le 119862[120582 (119909 119889 (119909 119910))]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(50)

Thus the proof of Theorem 19 is completed

Conflict of Interests

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

Acknowledgments

Jiang Zhou is supported by the National Science Foundationof China (Grants nos 11261055 and 11161044) and by theNational Natural Science Foundation of Xinjiang (Grantsnos 2011211A005 and BS120104)

References

[1] X Tolsa ldquoBMO1198671 and Calderon-Zygmund operators for nondoubling measuresrdquoMathematische Annalen vol 319 no 1 pp89ndash149 2001

[2] X Tolsa ldquoLittlewood-Paley theory and the 119879(1) theorem withnon-doublingmeasuresrdquoAdvances inMathematics vol 164 no1 pp 57ndash116 2001

[3] X Tolsa ldquoThe space 1198671 for nondoubling measures in terms ofa grand maximal operatorrdquo Transactions of the American Math-ematical Society vol 355 no 1 pp 315ndash348 2003

[4] T A Bui and X T Duong ldquoHardy spaces regularized BMOspaces and the boundedness of Calderon-Zygmund operatorson non-homogeneous spacesrdquo Journal of Geometric Analysisvol 23 no 2 pp 895ndash932 2013

[5] T Hytonen S Liu D Yang and D Yang ldquoBoundedness of Cal-deron-Zygmund operators on non-homogeneous metric mea-sure spacesrdquo Canadian Journal of Mathematics vol 64 no 4pp 892ndash923 2012

[6] THytonenD Yang andD Yang ldquoTheHardy space1198671 onnon-homogeneous metric spacesrdquo Mathematical Proceedings of theCambridge Philosophical Society vol 153 no 1 pp 9ndash31 2012

[7] J Garcıa-Cuerva and A E Gatto ldquoBoundedness properties offractional integral operators associated to non-doubling mea-suresrdquo Studia Mathematica vol 162 no 3 pp 245ndash261 2004

[8] J Garcıa-Cuerva and A E Gatto ldquoLipschitz spaces and Cal-deron-Zygmund operators associated to non-doubling mea-suresrdquo Publicacions Matematiques vol 49 no 2 pp 285ndash2962005

[9] Y Cao and J Zhou ldquoMorrey spaces for non-homogeneousmetricmeasure spacesrdquoAbstract andApplied Analysis vol 2013Article ID 196459 8 pages 2013

[10] X Fu D Yang and W Yuan ldquoBoundedness of multilinearcommutators of Calderon-Zygmund operators onOrlicz spacesover non-homogeneous spacesrdquo Taiwanese Journal of Mathe-matics vol 16 no 6 pp 2203ndash2238 2012

[11] X Fu D Yang and W Yuan ldquoGeneralized fractional integralsand their commutators over non-homogeneous spacesrdquo Tai-wanese Journal of Mathematics vol 18 no 2 pp 509ndash557 2014

[12] H Lin andD Yang ldquoEquivalent boundedness ofMarcinkiewiczintegrals on non-homogeneous metric measure spacesrdquo ScienceChina Mathematics vol 57 no 1 pp 123ndash144 2014

[13] X Tolsa ldquoPainleversquos problem and the semiadditivity of analyticcapacityrdquo Acta Mathematica vol 190 no 1 pp 105ndash149 2003

[14] T Hytonen ldquoA framework for non-homogeneous analysis onmetric spaces and the RBMO space of Tolsardquo PublicacionsMatematiques vol 54 no 2 pp 485ndash504 2010

[15] R R Coifman and G Weiss Analyse Harmonique Non-Com-mutative sur Certains Espaces Homogenes vol 242 of LectureNotes in Mathematics Springer Berlin Germany 1971

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: Research Article Lipschitz Spaces and Fractional Integral ...downloads.hindawi.com/journals/aaa/2014/174010.pdf · Research Article Lipschitz Spaces and Fractional Integral Operators

Abstract and Applied Analysis 3

A linear operator 119868120572 is called fractional integral operatorwith 119870120572 satisfying (10) and (11)

119868120572119891 (119909) = intX

119870120572 (119909 119910) 119891 (119910) 119889120583 (119910) (12)

Remark 11 By taking 120582(119909 119889(119909 119910)) = 119889(119909 119910)119899 it is easy tosee that Definition 10 in this paper contains Definition 41introduced by Garcıa-Cuerva and Gatto in [7] and Defini-tion 10 is similar to Definition 19 introduced by Fu et al in[11]

Definition 12 Let 119870120572 be a fractional kernel of order 120572 andregularity 120575 119891 isin 119871119901(120583) and 0 lt 120572 minus 119899119901 lt 120575 We define

119868120572119891 (119909) = intX

119870120572 (119909 119910) minus 119870120572 (1199090 119910) 119891 (119910) 119889120583 (119910)

(13)

where 1199090 is some fixed point ofXWe observe that the integral in (13) converges both

locally and atinfin as a consequence of (10) (11) and Holderrsquosinequality Of course the function just defined depends on theelection of 1199090 but the difference between any two functionsobtained in (13) for different elections of 1199090 is just a constant

From now on we will assume that 120583(X) = infinThe resultsbelow are also true when 120583(X) lt infin

Now we state the first main theorem of this paper

Theorem 13 Let 1 le 119901 lt 119899120572 and 1119902 = 1119901 minus 120572119899If 120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min120572119899 (1119901 minus 120572119899)1199011015840) then

120583 (119909 isin X 1003816100381610038161003816119868120572119891 (119909)

1003816100381610038161003816 gt ]) le (11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

])

119902

(14)

that is 119868120572 is a bounded operator from 119871119901(120583) into the space119871119902infin(120583)

Next let us introduce Lipschitz spaces on nonhomoge-neous metric measure spaces

Definition 14 Given that 120573 isin (0 1] we say that the function119891 X rarr C satisfies a Lipschitz condition of order 120573provided that

1003816100381610038161003816119891 (119909) minus 119891 (119910)1003816100381610038161003816 le 119862[120582 (119909 119889 (119909 119910))]

120573119899 for every 119909 119910 isin X(15)

and the smallest constant in inequality (15) will be denoted by119891Lip(120573) It is easy to see that the linear space with the norm sdot Lip(120573) is a Banach space and we will call it Lip(120573)

Remark 15 Lipschitz condition can also be defined by

1003816100381610038161003816119891 (119909) minus 119891 (119910)1003816100381610038161003816 le 119862[120582 (119910 119889 (119909 119910))]

120573119899 for every 119909 119910 isin X(15)

1015840

by (2) it is easy to see that (15) and (15)1015840 are equivalent

The secondmain result of this paper is the following someequivalent characterizations for the Lipschitz spaces

Theorem 16 For a function 119891 isin 1198711119897119900119888(120583) the conditions (A)

(B) and (C) are equivalent as follows(A) There exist some constant 1198621 and a collection of num-

bers of119891119861 one for each119861 such that these two propertieshold for any all 119861 with radius 1199031

120583 (2119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816 119889120583 (119909) le 1198621120582(119909 119903)

120573119899 (16)

and for any ball 119880 such that 119861 isin 119880 and radius (119880) le2119903

1003816100381610038161003816119891119861 minus 1198911198801003816100381610038161003816 le 1198621120582(119909 119903)

120573119899 (17)

(B) There is a constant 1198622 such that1003816100381610038161003816119891 (119909) minus 119891 (119910)

1003816100381610038161003816 le 1198622120582(119909 119889(119909 119910))120573119899 (18)

for 120583-almost every 119909 and 119910 in the support of 120583(C) For any given 119901 1 le 119901 le infin there is a constant 119862(119901)

such that for every ball 119861 of radius 119903 one has

(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 119898119861 (119891)1003816100381610038161003816

119901119889120583 (119909))

1119901

le 119862 (119901) 120582(119909 119903)120573119899

(19)

where 119898119861 = (1120583(119861)) int119861119891(119910)119889120583(119910) and also for any

ball 119880 such that 119861 sub 119880 and radius (119880) le 21199031003816100381610038161003816119898119861 (119891) minus 119898119880 (119891)

1003816100381610038161003816 le 119862 (119901) 120582(119909 119903)120573119899 (20)

In addition the quantities inf 1198621 inf 1198622 and inf 119862(119901)witha fixed 119901 are equivalent

Now we state the third main result of this paper

Theorem 17 Let119870120572 be a fractional kernel 119899120572 lt 119901 le infin and120572 minus 119899119901 lt 120575 If 120582 satisfy the 120598-weak reverse doubling conditionwith 120598 isin (0min(1 minus ((120572 minus 120575)119899))1199011015840 (120572119899 minus 1119901)1199011015840) then 119868120572maps 119871119901(120583) boundedly into Lip(120572 minus 119899119901)

Theorem 18 Let 119870120572 be a fractional kernel and 120572 + 120573 lt 120575if 120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min(120572+120573)119899 (120575minus120572minus120573)119899) then 119868120572 maps Lip(120573) bound-edly into Lip(120572 + 120573) if and only if 119868120572(1) = 0

Finally we present a result which can be viewed eitheras an extension of the case 119901 = infin of the Theorem 17 or asextension of the case 120573 = 0 of Theorem 18

Theorem 19 Let 119870120572 be a fractional kernel and 0 lt 120572 lt 120575If 120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min120572119899 (120575 minus 120572)119899) then 119868120572 maps RBMO(120583) boundedlyinto Lip(120572) if and only if 119868120572(1) = 0

Finally we make some conventions on notationThroughout the whole paper 119862 stands for a positive con-stant which is independent of the main parameters but itmay vary from line to line

4 Abstract and Applied Analysis

2 Proof of Theorem 13

Proof of Theorem 13 We are going to adapt to our context ofthe proof given by Garcıa-Cuerva and Gatto [7] Consider

1003816100381610038161003816119868120572119891 (119909)1003816100381610038161003816 le int

X

1003816100381610038161003816119891 (119910)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910)

le int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910)

+ intX119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910)

= 1198681 + 1198682

(21)

By Holderrsquos inequality if 119901 gt 1 then 1 minus (1 minus 120572119899)1199011015840 lt 0

100381610038161003816100381611986821003816100381610038161003816 le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (intX119861(119909119903)

1

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

119889120583 (119910))

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (

infin

sum

119895=1

int21198951198612119895minus1119861

1

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

119889120583 (119910))

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (

infin

sum

119895=1

120583 (119861 (119909 2119895119903))

[120582 (119909 2119895minus1119903)](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (

infin

sum

119895=1

120582 (119909 2119895119903)

[120582 (119909 2119895minus1119903)](1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=1

[120582 (119909 2119895minus1119903)]

1minus(1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=1

1

[119862 (2119895)](1minus120572119899)1199011015840minus1

)

11199011015840

times [120582 (119909 119903)]120572119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)[

120582 (119909 119903)]120572119899minus1119901

(22)

which holds even for 119901 = 1 We can and assume that119891

119871119901(120583)= 1 By (5) we can choose 119903 isin (0infin) such that

119862[120582(119909 119903)]120572119899minus1119901

= ]2 Then

119909 isin X 1003816100381610038161003816119868120572119891 (119909)

1003816100381610038161003816 gt ] sub 119909 isin X 100381610038161003816100381611986811003816100381610038161003816 gt

]2

cup 119909 isin X 100381610038161003816100381611986821003816100381610038161003816 gt

]2

(23)

By the relation between 119903 and ] 120583(119909 isin X |1198682| gt ]2) = 0We use Holderrsquos inequality once more to obtain

100381610038161003816100381611986811003816100381610038161003816 le (int

119861(119909119903)

119889120583 (119910)

[120582 (119909 119889 (119909 119910))]1minus120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

120583 (119861 (119909 2minus119895119903))

[120582 (119909 2minus119895minus1119903)]1minus120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

120582 (119909 2119895119903)

[120582 (119909 2minus119895minus1119903)]1minus120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

[120582 (119909 2minus119895minus1119903)]

120572119899

)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

1

[119862 (2119895)]120572119899[120582 (119909 119903)]

120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862[120582 (119909 119903)]1205721199011015840119899

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

(24)

Abstract and Applied Analysis 5

Then by applying Tchebichevrsquos inequality we have

120583 (119909 isin X 1003816100381610038161003816119868120572119891 (119909)

1003816100381610038161003816 gt ])

le 120583 (119909 isin X 100381610038161003816100381611986811003816100381610038161003816 gt

]2)

le 119862[120582(119909 119903)]1205721199011199011015840119899]minus119901

times intX

int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910) 119889120583 (119909)

= 119862[120582 (119909 119903)]1205721199011199011015840119899]minus119901

times intX

int119861(119910119903)

119889120583 (119909)

[120582 (119909 119889 (119909 119910))]1minus120572119899

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901119889120583 (119910)

le 119862[120582(119909 119903)]1205721199011199011015840119899]minus119901[120582(119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817119871119901(120583)

= 119862 [120582 (119909 119903)] = 119862]minus119902(25)

This completes the proof of Theorem 13

Corollary 20 Let 1 lt 119901 lt 119899120572 and 1119902 = 1119901 minus 120572119899 If120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min120572119899 (1119901 minus 120572119899)1199011015840) then

10038171003817100381710038171198681205721198911003817100381710038171003817119871119902(120583)

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(26)

Proof It suffices to apply Marcinkiewiczrsquos interpolation the-orem with indices slightly bigger and slightly smaller than119901

3 Proof of Theorem 16

Before we give the proof of Theorem 16 we first introduce atechnical lemma from [8 Lemma 32]

Lemma 21 Let 119891 isin 1198711119897119900119888(120583) If 1205732 gt 2119889 then for almost every

119909 with respect to 120583 there exists a sequence of (2 1205732)-doublingballs 119861119895 = 119861(119909 119903119895) with 119903119895 rarr 0 such that

lim119895rarrinfin

1

120583 (119861119895)

int119861119895

119891 (119910) 119889120583 (119910) = 119891 (119909) (27)

Proof of Theorem 16 (A) rArr (B) Consider 119909 as in the lemmaand let 119861119895 = 119861(119909 119903119895) 119895 ge 1 a sequence of (2 1205732)-doublingballs with 119903119895 rarr 0 Consider

100381610038161003816100381610038161003816119898119861119895

(119891) minus 119891119861119895

100381610038161003816100381610038161003816le

1

120583 (119861119895)

int119861119895

100381610038161003816100381610038161003816119891 (119910) minus 119891119861

119895

100381610038161003816100381610038161003816119889120583 (119910)

le

120583 (2119861119895)

120583 (119861119895)

1

120583 (2119861119895)

int119861119895

100381610038161003816100381610038161003816119891 (119910) minus 119891119861

119895

100381610038161003816100381610038161003816119889120583 (119910)

le 1205731198621[120582 (119909 119903119895)]120573119899

(28)

and by (5) Lemma 21 we obtain that

lim119895rarrinfin

119891119861119895

= 119891 (119909) (29)

Let 119909 and 119910 be two points as in the lemma take 119861 = 119861(119909 119903)any ball with 119903 le 119889(119909 119910) and let 119880 = 119861(119909 2119889(119909 119910)) Nowdefine 119861119896 = 119861(119909 2

119896119903) for 0 le 119896 le where is the first inte-

ger such that 2119903 ge 119889(119909 119910) Then

1003816100381610038161003816119891119861 minus 1198911198801003816100381610038161003816 le

minus1

sum

119896=0

10038161003816100381610038161003816119891119861119896

minus 119891119861119896+1

10038161003816100381610038161003816+10038161003816100381610038161003816119891119861

minus 119891119880

10038161003816100381610038161003816

le 1198621

sum

119896=0

[120582 (119909 2119896119903)]

120573119899

le 1198621

sum

119896=0

119888120573119896119899

120582[120582 (119909 119903)]

120573119899

le 11986210158401198621120582(119909 119889 (119909 119910))

120573119899

(30)

where 1198621015840 is independent of 119909 and 119889(119909 119910)A similar argument can be made for the point 119910 with any

ball 1198611015840 = 119861(119910 119904) such that 119904 le 119889(119909 119910) and119881 = 119861(119910 3119889(119909 119910))Therefore1003816100381610038161003816119891119861 minus 1198911198611015840

1003816100381610038161003816 le1003816100381610038161003816119891119861 minus 119891119880

1003816100381610038161003816 +1003816100381610038161003816119891119880 minus 119891119881

1003816100381610038161003816 +1003816100381610038161003816119891119881 minus 1198911198611015840

1003816100381610038161003816

le 119862101584010158401198621120582(119909 119889 (119909 119910))

120573119899

(31)

Take two sequences of (2 1205732)-doubling balls 119861119895 = 119861(119909 119903119895)and 1198611015840 = 119861(119910 119904119895) with 119903119895 rarr 0 and 119904119895 rarr 0 We have

1003816100381610038161003816119891 (119909) minus 119891 (119910)1003816100381610038161003816 = lim

119895rarr0

100381610038161003816100381610038161003816119891119861119895

minus 1198911198611015840119895

100381610038161003816100381610038161003816le 119862

101584010158401198621120582(119909 119889 (119909 119910))

120573119899

(32)

(B) rArr (C) For 1199090 isin 119861 = 119861(119909 119903) by the properties offunction 120582 and Holderrsquos inequality we obtain

(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 119898119861 (119891)1003816100381610038161003816

119901119889120583 (119909))

1119901

le (1

120583 (119861)int119861

10038161003816100381610038161003816100381610038161003816

1

120583 (119861)int119861

(119891 (119909) minus 119891 (119910)) 119889120583 (119910)

10038161003816100381610038161003816100381610038161003816

119901

119889120583 (119909))

1119901

le (1

120583 (119861)int119861

[1198622120582 (119909 119903)]120573119901119899119889120583 (119909))

1119901

le (1

120583 (119861)int119861

[1198622119862120582120582 (1199090 119903)]120573119901119899119889120583 (119909))

1119901

le 1198622119862120582[120582 (1199090 119903)]120573119899le 1198622119862

2

120582[120582 (119909 119903)]120573119899

(33)

By the similar argument for any ball 119880 such that 119861 sub 119880 andradius 119880 le 2119903

1003816100381610038161003816119898119861 (119891) minus 119898119880 (119891)1003816100381610038161003816 le 1198622119862

2

120582[120582 (119909 2119903)]120573119899

le 119888120573119899

1205821198622119862

2

120582[120582 (119909 119903)]120573119899

(34)

6 Abstract and Applied Analysis

(C) rArr (A) Define first 119891119861 = 119898119861(119891) Then (17) is exactly(20) To prove (16) we write

1

120583 (2119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816 119889120583 (119909)

le1

120583 (2119861)(int

119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816

119901119889120583 (119909))

1119901

120583(119861)11199011015840

le120583 (119861)

120583 (2119861)(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816

119901119889120583 (119909))

1119901

le 119862 (119901) 120582(119909 119903)120573119899

(35)

This concludes the proof of the theorem

Remark 22 Theorem 16 is also true if the number 2 in con-dition (A) is replaced by any fixed 120588 gt 1 In that case theproof uses (120588 120573120588)-doubling balls that is balls satisfying120583(120588119861) le 120573120588120583(119861)

4 Proofs of Theorems 17ndash19

Proof of Theorem 17 Without loss of generality we assumethat 119901 lt infin Consider 119909 = 119910 and let 119861 be the ball with center119909 and radius 119903 = 119889(119909 119910) Then we have

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le int

2119861

1003816100381610038161003816119870120572 (119909 119911)1003816100381610038161003816

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

+ int

2119861

1003816100381610038161003816119870120572 (119910 119911)1003816100381610038161003816

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

+ int

X2119861

1003816100381610038161003816119870120572 (119909 119911) minus 119870120572 (119910 119911)1003816100381610038161003816

times1003816100381610038161003816119891 (119911)

1003816100381610038161003816 119889120583 (119911)

= 1198691 + 1198692 + 1198693

(36)

For the first term by 119899120572 lt 119901 then 1 minus (1 minus 120572119899)1199011015840 gt 0

1198691 le int2119861

1003816100381610038161003816119891 (119911)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119911)

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

120583 (119861 (119909 2minus119895+1119903))

[120582 (119909 119889 (119909 2minus119895119903))](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

120582 (119909 2minus119895+1119903)

[120582 (119909 119889 (119909 2minus119895119903))](1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

[120582 (119909 119889 (119909 2minus119895119903))]

1minus(1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

1

[119862 (2119895)](120572119899minus1119901)1199011015840

)

11199011015840

times [120582 (119909 119903)]120572119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)[

120582 (119909 119903)]120572119899minus1119901

(37)

The second term is estimated in a similar way after noting that2119861 sub 119861(119910 3119903)

Next by using (2) Holderrsquos inequality and 120572 minus 119899119901 lt 120575we get

1198693 le 119862intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817119871119901(120583)

times (intX2119861

1

[120582 (119909 119889 (119909 119911))](1minus(120572minus120575)119899)1199011015840

119889120583 (119911))

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120575119899

times (intX2119861

1

[120582 (119909 119889 (119909 119911))](1minus(120572minus120575)119899)1199011015840

119889120583 (119911))

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572minus120575)119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120572119899minus1119901

(38)

Putting together the three estimates

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120572119899minus1119901

(39)

then 119868120572 maps 119871119901(120583) boundedly into Lip(120572 minus 119899119901)

Proof of Theorem 18 If 119868120572 isin Lip(120573) then by the continuity ofthe operator 119868120572 implies that 119868120572(1) must be constant that is119868120572(1)(119909) = 119868120572(1)(1199090) = 0

On the other hand we can observe that

119868120572 (1) = 0 lArrrArr 119868120572 (1) (119909) minus 119868120572 (1) (119910) = 0 (40)

this implies that

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) 119889120583 (119911) = 0 (41)

Abstract and Applied Analysis 7

Thus we can write

119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

= intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= int2119861(119909119903)

119870120572 (119909 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ int2119861(119909119903)

minus119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ intX2119861(119909119903)

119870120572 (119909 119911) minus 119870120572 (119910 119911)

times (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= 1198721 +1198722 +1198723

(42)

where 119903 = 119889(119909 119910) For1198721

10038161003816100381610038161198721

1003816100381610038161003816 le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(43)

Similarly we know that 2119861(119909 119903) sub 3119861(119910 119903) using Holderrsquosinequality (1119905 + 11199051015840 = 1) and letting 119905 gt 119899120572 then we have

10038161003816100381610038161198722

1003816100381610038161003816

le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862(int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119905119899119889120583 (119911))

1119905

times (int2119861(119909119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862(

infin

sum

119895=0

int2minus119895+1119861(119909119903)2minus119895119861(119909119903)

[120582 (119909

119889 (119909 2minus119895+1119903))]

120573119905119899

119889120583(119911))

1119905

times (int3119861(119910119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862[120582 (119909 119889 (119909 119910))]120572119899[120582 (119910 119889 (119909 119910))]

120573119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(44)

In order to estimate1198723 we use (2) to obtain

10038161003816100381610038161198723

1003816100381610038161003816

le 119862intX2119861(119909119903)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119911))]

120573119899

[120582 (119909 119889 (119909 119911))]1minus(120572minus120575)119899

119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times intX2119861(119909119903)

1

[120582 (119909 119889 (119909 119911))]1minus(120572+120573minus120575)119899

119889120583 (119911)

le 119862[120582(119909 119889(119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572+120573minus120575)119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(45)

Combining the estimates for11987211198722 and1198723 we obtain

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 119862[120582 (119909 119889 (119909 119910))]

(120572+120573)119899 (46)

this finishes the proof

Proof of Theorem 19 119868120572(119891) isin Lip(120572) implies that 119868120572(1) = 0and equivalently that

intX

(119870120572 (119909 119910) minus 119870120572 (1199090 119910)) 119889120583 (119910) = 0 (47)

for all 119909Take two points 119909 = 119910 and let119861 = 119861(119909 119903)with 119903 = 119889(119909 119910)

Then

10038161003816100381610038161003816119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

10038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

le

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119909 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

X2119861

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

= 1198731 + 1198732 + 1198733

(48)

8 Abstract and Applied Analysis

For the first term by Holderrsquos inequality with some 119901 gt 119899120572

1198731 le int2119861

1

[120582 (119909 119889 (119909 119911))]1minus120572119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le (int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119911))](1minus120572119899)1199011015840

)

11199011015840

times (int2119861

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816

119901119889120583 (119911))

1119901

le 119862[120582 (119909 119903)]120572119899minus1119901

120583(120588119861)11199011003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

le 119862[120582 (119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(49)

Using 2119861 sub 119861(119910 3119903) the second term can be dealt with inthe same way as the 1198731 then 1198732 le 119862[120582(119909 119903)]

120572119899119891RBMO(120583)

It is easy to see that |1198912119896+1119861 minus 1198912119861| le 119891RBMO(120583)11987021198612119896+1119861 le

119896119891RBMO(120583) then

1198733 le intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times

infin

sum

119896=1

(int2119896+11198612119896119861

1003816100381610038161003816119891 (119911) minus 1198912119896+11198611003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

119889120583 (119911)

+

10038161003816100381610038161198912119896+1119861 minus 11989121198611003816100381610038161003816

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

120583 (2119896+1119861))

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=1

120583 (1205882119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

+

infin

sum

119896=1

119896

120583 (2119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

)

le 119862[120582(119909 119889(119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=0

(119896 + 1) [120582 (119909 119889 (119909 2119896119903))]

120572119899minus120575119899

)

le 119862[120582 (119909 119889 (119909 119910))]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(50)

Thus the proof of Theorem 19 is completed

Conflict of Interests

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

Acknowledgments

Jiang Zhou is supported by the National Science Foundationof China (Grants nos 11261055 and 11161044) and by theNational Natural Science Foundation of Xinjiang (Grantsnos 2011211A005 and BS120104)

References

[1] X Tolsa ldquoBMO1198671 and Calderon-Zygmund operators for nondoubling measuresrdquoMathematische Annalen vol 319 no 1 pp89ndash149 2001

[2] X Tolsa ldquoLittlewood-Paley theory and the 119879(1) theorem withnon-doublingmeasuresrdquoAdvances inMathematics vol 164 no1 pp 57ndash116 2001

[3] X Tolsa ldquoThe space 1198671 for nondoubling measures in terms ofa grand maximal operatorrdquo Transactions of the American Math-ematical Society vol 355 no 1 pp 315ndash348 2003

[4] T A Bui and X T Duong ldquoHardy spaces regularized BMOspaces and the boundedness of Calderon-Zygmund operatorson non-homogeneous spacesrdquo Journal of Geometric Analysisvol 23 no 2 pp 895ndash932 2013

[5] T Hytonen S Liu D Yang and D Yang ldquoBoundedness of Cal-deron-Zygmund operators on non-homogeneous metric mea-sure spacesrdquo Canadian Journal of Mathematics vol 64 no 4pp 892ndash923 2012

[6] THytonenD Yang andD Yang ldquoTheHardy space1198671 onnon-homogeneous metric spacesrdquo Mathematical Proceedings of theCambridge Philosophical Society vol 153 no 1 pp 9ndash31 2012

[7] J Garcıa-Cuerva and A E Gatto ldquoBoundedness properties offractional integral operators associated to non-doubling mea-suresrdquo Studia Mathematica vol 162 no 3 pp 245ndash261 2004

[8] J Garcıa-Cuerva and A E Gatto ldquoLipschitz spaces and Cal-deron-Zygmund operators associated to non-doubling mea-suresrdquo Publicacions Matematiques vol 49 no 2 pp 285ndash2962005

[9] Y Cao and J Zhou ldquoMorrey spaces for non-homogeneousmetricmeasure spacesrdquoAbstract andApplied Analysis vol 2013Article ID 196459 8 pages 2013

[10] X Fu D Yang and W Yuan ldquoBoundedness of multilinearcommutators of Calderon-Zygmund operators onOrlicz spacesover non-homogeneous spacesrdquo Taiwanese Journal of Mathe-matics vol 16 no 6 pp 2203ndash2238 2012

[11] X Fu D Yang and W Yuan ldquoGeneralized fractional integralsand their commutators over non-homogeneous spacesrdquo Tai-wanese Journal of Mathematics vol 18 no 2 pp 509ndash557 2014

[12] H Lin andD Yang ldquoEquivalent boundedness ofMarcinkiewiczintegrals on non-homogeneous metric measure spacesrdquo ScienceChina Mathematics vol 57 no 1 pp 123ndash144 2014

[13] X Tolsa ldquoPainleversquos problem and the semiadditivity of analyticcapacityrdquo Acta Mathematica vol 190 no 1 pp 105ndash149 2003

[14] T Hytonen ldquoA framework for non-homogeneous analysis onmetric spaces and the RBMO space of Tolsardquo PublicacionsMatematiques vol 54 no 2 pp 485ndash504 2010

[15] R R Coifman and G Weiss Analyse Harmonique Non-Com-mutative sur Certains Espaces Homogenes vol 242 of LectureNotes in Mathematics Springer Berlin Germany 1971

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 4: Research Article Lipschitz Spaces and Fractional Integral ...downloads.hindawi.com/journals/aaa/2014/174010.pdf · Research Article Lipschitz Spaces and Fractional Integral Operators

4 Abstract and Applied Analysis

2 Proof of Theorem 13

Proof of Theorem 13 We are going to adapt to our context ofthe proof given by Garcıa-Cuerva and Gatto [7] Consider

1003816100381610038161003816119868120572119891 (119909)1003816100381610038161003816 le int

X

1003816100381610038161003816119891 (119910)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910)

le int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910)

+ intX119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910)

= 1198681 + 1198682

(21)

By Holderrsquos inequality if 119901 gt 1 then 1 minus (1 minus 120572119899)1199011015840 lt 0

100381610038161003816100381611986821003816100381610038161003816 le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (intX119861(119909119903)

1

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

119889120583 (119910))

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (

infin

sum

119895=1

int21198951198612119895minus1119861

1

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

119889120583 (119910))

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (

infin

sum

119895=1

120583 (119861 (119909 2119895119903))

[120582 (119909 2119895minus1119903)](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

times (

infin

sum

119895=1

120582 (119909 2119895119903)

[120582 (119909 2119895minus1119903)](1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=1

[120582 (119909 2119895minus1119903)]

1minus(1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=1

1

[119862 (2119895)](1minus120572119899)1199011015840minus1

)

11199011015840

times [120582 (119909 119903)]120572119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)[

120582 (119909 119903)]120572119899minus1119901

(22)

which holds even for 119901 = 1 We can and assume that119891

119871119901(120583)= 1 By (5) we can choose 119903 isin (0infin) such that

119862[120582(119909 119903)]120572119899minus1119901

= ]2 Then

119909 isin X 1003816100381610038161003816119868120572119891 (119909)

1003816100381610038161003816 gt ] sub 119909 isin X 100381610038161003816100381611986811003816100381610038161003816 gt

]2

cup 119909 isin X 100381610038161003816100381611986821003816100381610038161003816 gt

]2

(23)

By the relation between 119903 and ] 120583(119909 isin X |1198682| gt ]2) = 0We use Holderrsquos inequality once more to obtain

100381610038161003816100381611986811003816100381610038161003816 le (int

119861(119909119903)

119889120583 (119910)

[120582 (119909 119889 (119909 119910))]1minus120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

120583 (119861 (119909 2minus119895119903))

[120582 (119909 2minus119895minus1119903)]1minus120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

120582 (119909 2119895119903)

[120582 (119909 2minus119895minus1119903)]1minus120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

[120582 (119909 2minus119895minus1119903)]

120572119899

)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862(

infin

sum

119895=0

1

[119862 (2119895)]120572119899[120582 (119909 119903)]

120572119899)

11199011015840

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

le 119862[120582 (119909 119903)]1205721199011015840119899

times (int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119910))

1119901

(24)

Abstract and Applied Analysis 5

Then by applying Tchebichevrsquos inequality we have

120583 (119909 isin X 1003816100381610038161003816119868120572119891 (119909)

1003816100381610038161003816 gt ])

le 120583 (119909 isin X 100381610038161003816100381611986811003816100381610038161003816 gt

]2)

le 119862[120582(119909 119903)]1205721199011199011015840119899]minus119901

times intX

int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910) 119889120583 (119909)

= 119862[120582 (119909 119903)]1205721199011199011015840119899]minus119901

times intX

int119861(119910119903)

119889120583 (119909)

[120582 (119909 119889 (119909 119910))]1minus120572119899

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901119889120583 (119910)

le 119862[120582(119909 119903)]1205721199011199011015840119899]minus119901[120582(119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817119871119901(120583)

= 119862 [120582 (119909 119903)] = 119862]minus119902(25)

This completes the proof of Theorem 13

Corollary 20 Let 1 lt 119901 lt 119899120572 and 1119902 = 1119901 minus 120572119899 If120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min120572119899 (1119901 minus 120572119899)1199011015840) then

10038171003817100381710038171198681205721198911003817100381710038171003817119871119902(120583)

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(26)

Proof It suffices to apply Marcinkiewiczrsquos interpolation the-orem with indices slightly bigger and slightly smaller than119901

3 Proof of Theorem 16

Before we give the proof of Theorem 16 we first introduce atechnical lemma from [8 Lemma 32]

Lemma 21 Let 119891 isin 1198711119897119900119888(120583) If 1205732 gt 2119889 then for almost every

119909 with respect to 120583 there exists a sequence of (2 1205732)-doublingballs 119861119895 = 119861(119909 119903119895) with 119903119895 rarr 0 such that

lim119895rarrinfin

1

120583 (119861119895)

int119861119895

119891 (119910) 119889120583 (119910) = 119891 (119909) (27)

Proof of Theorem 16 (A) rArr (B) Consider 119909 as in the lemmaand let 119861119895 = 119861(119909 119903119895) 119895 ge 1 a sequence of (2 1205732)-doublingballs with 119903119895 rarr 0 Consider

100381610038161003816100381610038161003816119898119861119895

(119891) minus 119891119861119895

100381610038161003816100381610038161003816le

1

120583 (119861119895)

int119861119895

100381610038161003816100381610038161003816119891 (119910) minus 119891119861

119895

100381610038161003816100381610038161003816119889120583 (119910)

le

120583 (2119861119895)

120583 (119861119895)

1

120583 (2119861119895)

int119861119895

100381610038161003816100381610038161003816119891 (119910) minus 119891119861

119895

100381610038161003816100381610038161003816119889120583 (119910)

le 1205731198621[120582 (119909 119903119895)]120573119899

(28)

and by (5) Lemma 21 we obtain that

lim119895rarrinfin

119891119861119895

= 119891 (119909) (29)

Let 119909 and 119910 be two points as in the lemma take 119861 = 119861(119909 119903)any ball with 119903 le 119889(119909 119910) and let 119880 = 119861(119909 2119889(119909 119910)) Nowdefine 119861119896 = 119861(119909 2

119896119903) for 0 le 119896 le where is the first inte-

ger such that 2119903 ge 119889(119909 119910) Then

1003816100381610038161003816119891119861 minus 1198911198801003816100381610038161003816 le

minus1

sum

119896=0

10038161003816100381610038161003816119891119861119896

minus 119891119861119896+1

10038161003816100381610038161003816+10038161003816100381610038161003816119891119861

minus 119891119880

10038161003816100381610038161003816

le 1198621

sum

119896=0

[120582 (119909 2119896119903)]

120573119899

le 1198621

sum

119896=0

119888120573119896119899

120582[120582 (119909 119903)]

120573119899

le 11986210158401198621120582(119909 119889 (119909 119910))

120573119899

(30)

where 1198621015840 is independent of 119909 and 119889(119909 119910)A similar argument can be made for the point 119910 with any

ball 1198611015840 = 119861(119910 119904) such that 119904 le 119889(119909 119910) and119881 = 119861(119910 3119889(119909 119910))Therefore1003816100381610038161003816119891119861 minus 1198911198611015840

1003816100381610038161003816 le1003816100381610038161003816119891119861 minus 119891119880

1003816100381610038161003816 +1003816100381610038161003816119891119880 minus 119891119881

1003816100381610038161003816 +1003816100381610038161003816119891119881 minus 1198911198611015840

1003816100381610038161003816

le 119862101584010158401198621120582(119909 119889 (119909 119910))

120573119899

(31)

Take two sequences of (2 1205732)-doubling balls 119861119895 = 119861(119909 119903119895)and 1198611015840 = 119861(119910 119904119895) with 119903119895 rarr 0 and 119904119895 rarr 0 We have

1003816100381610038161003816119891 (119909) minus 119891 (119910)1003816100381610038161003816 = lim

119895rarr0

100381610038161003816100381610038161003816119891119861119895

minus 1198911198611015840119895

100381610038161003816100381610038161003816le 119862

101584010158401198621120582(119909 119889 (119909 119910))

120573119899

(32)

(B) rArr (C) For 1199090 isin 119861 = 119861(119909 119903) by the properties offunction 120582 and Holderrsquos inequality we obtain

(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 119898119861 (119891)1003816100381610038161003816

119901119889120583 (119909))

1119901

le (1

120583 (119861)int119861

10038161003816100381610038161003816100381610038161003816

1

120583 (119861)int119861

(119891 (119909) minus 119891 (119910)) 119889120583 (119910)

10038161003816100381610038161003816100381610038161003816

119901

119889120583 (119909))

1119901

le (1

120583 (119861)int119861

[1198622120582 (119909 119903)]120573119901119899119889120583 (119909))

1119901

le (1

120583 (119861)int119861

[1198622119862120582120582 (1199090 119903)]120573119901119899119889120583 (119909))

1119901

le 1198622119862120582[120582 (1199090 119903)]120573119899le 1198622119862

2

120582[120582 (119909 119903)]120573119899

(33)

By the similar argument for any ball 119880 such that 119861 sub 119880 andradius 119880 le 2119903

1003816100381610038161003816119898119861 (119891) minus 119898119880 (119891)1003816100381610038161003816 le 1198622119862

2

120582[120582 (119909 2119903)]120573119899

le 119888120573119899

1205821198622119862

2

120582[120582 (119909 119903)]120573119899

(34)

6 Abstract and Applied Analysis

(C) rArr (A) Define first 119891119861 = 119898119861(119891) Then (17) is exactly(20) To prove (16) we write

1

120583 (2119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816 119889120583 (119909)

le1

120583 (2119861)(int

119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816

119901119889120583 (119909))

1119901

120583(119861)11199011015840

le120583 (119861)

120583 (2119861)(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816

119901119889120583 (119909))

1119901

le 119862 (119901) 120582(119909 119903)120573119899

(35)

This concludes the proof of the theorem

Remark 22 Theorem 16 is also true if the number 2 in con-dition (A) is replaced by any fixed 120588 gt 1 In that case theproof uses (120588 120573120588)-doubling balls that is balls satisfying120583(120588119861) le 120573120588120583(119861)

4 Proofs of Theorems 17ndash19

Proof of Theorem 17 Without loss of generality we assumethat 119901 lt infin Consider 119909 = 119910 and let 119861 be the ball with center119909 and radius 119903 = 119889(119909 119910) Then we have

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le int

2119861

1003816100381610038161003816119870120572 (119909 119911)1003816100381610038161003816

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

+ int

2119861

1003816100381610038161003816119870120572 (119910 119911)1003816100381610038161003816

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

+ int

X2119861

1003816100381610038161003816119870120572 (119909 119911) minus 119870120572 (119910 119911)1003816100381610038161003816

times1003816100381610038161003816119891 (119911)

1003816100381610038161003816 119889120583 (119911)

= 1198691 + 1198692 + 1198693

(36)

For the first term by 119899120572 lt 119901 then 1 minus (1 minus 120572119899)1199011015840 gt 0

1198691 le int2119861

1003816100381610038161003816119891 (119911)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119911)

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

120583 (119861 (119909 2minus119895+1119903))

[120582 (119909 119889 (119909 2minus119895119903))](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

120582 (119909 2minus119895+1119903)

[120582 (119909 119889 (119909 2minus119895119903))](1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

[120582 (119909 119889 (119909 2minus119895119903))]

1minus(1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

1

[119862 (2119895)](120572119899minus1119901)1199011015840

)

11199011015840

times [120582 (119909 119903)]120572119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)[

120582 (119909 119903)]120572119899minus1119901

(37)

The second term is estimated in a similar way after noting that2119861 sub 119861(119910 3119903)

Next by using (2) Holderrsquos inequality and 120572 minus 119899119901 lt 120575we get

1198693 le 119862intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817119871119901(120583)

times (intX2119861

1

[120582 (119909 119889 (119909 119911))](1minus(120572minus120575)119899)1199011015840

119889120583 (119911))

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120575119899

times (intX2119861

1

[120582 (119909 119889 (119909 119911))](1minus(120572minus120575)119899)1199011015840

119889120583 (119911))

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572minus120575)119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120572119899minus1119901

(38)

Putting together the three estimates

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120572119899minus1119901

(39)

then 119868120572 maps 119871119901(120583) boundedly into Lip(120572 minus 119899119901)

Proof of Theorem 18 If 119868120572 isin Lip(120573) then by the continuity ofthe operator 119868120572 implies that 119868120572(1) must be constant that is119868120572(1)(119909) = 119868120572(1)(1199090) = 0

On the other hand we can observe that

119868120572 (1) = 0 lArrrArr 119868120572 (1) (119909) minus 119868120572 (1) (119910) = 0 (40)

this implies that

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) 119889120583 (119911) = 0 (41)

Abstract and Applied Analysis 7

Thus we can write

119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

= intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= int2119861(119909119903)

119870120572 (119909 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ int2119861(119909119903)

minus119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ intX2119861(119909119903)

119870120572 (119909 119911) minus 119870120572 (119910 119911)

times (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= 1198721 +1198722 +1198723

(42)

where 119903 = 119889(119909 119910) For1198721

10038161003816100381610038161198721

1003816100381610038161003816 le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(43)

Similarly we know that 2119861(119909 119903) sub 3119861(119910 119903) using Holderrsquosinequality (1119905 + 11199051015840 = 1) and letting 119905 gt 119899120572 then we have

10038161003816100381610038161198722

1003816100381610038161003816

le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862(int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119905119899119889120583 (119911))

1119905

times (int2119861(119909119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862(

infin

sum

119895=0

int2minus119895+1119861(119909119903)2minus119895119861(119909119903)

[120582 (119909

119889 (119909 2minus119895+1119903))]

120573119905119899

119889120583(119911))

1119905

times (int3119861(119910119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862[120582 (119909 119889 (119909 119910))]120572119899[120582 (119910 119889 (119909 119910))]

120573119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(44)

In order to estimate1198723 we use (2) to obtain

10038161003816100381610038161198723

1003816100381610038161003816

le 119862intX2119861(119909119903)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119911))]

120573119899

[120582 (119909 119889 (119909 119911))]1minus(120572minus120575)119899

119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times intX2119861(119909119903)

1

[120582 (119909 119889 (119909 119911))]1minus(120572+120573minus120575)119899

119889120583 (119911)

le 119862[120582(119909 119889(119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572+120573minus120575)119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(45)

Combining the estimates for11987211198722 and1198723 we obtain

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 119862[120582 (119909 119889 (119909 119910))]

(120572+120573)119899 (46)

this finishes the proof

Proof of Theorem 19 119868120572(119891) isin Lip(120572) implies that 119868120572(1) = 0and equivalently that

intX

(119870120572 (119909 119910) minus 119870120572 (1199090 119910)) 119889120583 (119910) = 0 (47)

for all 119909Take two points 119909 = 119910 and let119861 = 119861(119909 119903)with 119903 = 119889(119909 119910)

Then

10038161003816100381610038161003816119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

10038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

le

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119909 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

X2119861

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

= 1198731 + 1198732 + 1198733

(48)

8 Abstract and Applied Analysis

For the first term by Holderrsquos inequality with some 119901 gt 119899120572

1198731 le int2119861

1

[120582 (119909 119889 (119909 119911))]1minus120572119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le (int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119911))](1minus120572119899)1199011015840

)

11199011015840

times (int2119861

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816

119901119889120583 (119911))

1119901

le 119862[120582 (119909 119903)]120572119899minus1119901

120583(120588119861)11199011003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

le 119862[120582 (119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(49)

Using 2119861 sub 119861(119910 3119903) the second term can be dealt with inthe same way as the 1198731 then 1198732 le 119862[120582(119909 119903)]

120572119899119891RBMO(120583)

It is easy to see that |1198912119896+1119861 minus 1198912119861| le 119891RBMO(120583)11987021198612119896+1119861 le

119896119891RBMO(120583) then

1198733 le intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times

infin

sum

119896=1

(int2119896+11198612119896119861

1003816100381610038161003816119891 (119911) minus 1198912119896+11198611003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

119889120583 (119911)

+

10038161003816100381610038161198912119896+1119861 minus 11989121198611003816100381610038161003816

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

120583 (2119896+1119861))

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=1

120583 (1205882119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

+

infin

sum

119896=1

119896

120583 (2119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

)

le 119862[120582(119909 119889(119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=0

(119896 + 1) [120582 (119909 119889 (119909 2119896119903))]

120572119899minus120575119899

)

le 119862[120582 (119909 119889 (119909 119910))]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(50)

Thus the proof of Theorem 19 is completed

Conflict of Interests

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

Acknowledgments

Jiang Zhou is supported by the National Science Foundationof China (Grants nos 11261055 and 11161044) and by theNational Natural Science Foundation of Xinjiang (Grantsnos 2011211A005 and BS120104)

References

[1] X Tolsa ldquoBMO1198671 and Calderon-Zygmund operators for nondoubling measuresrdquoMathematische Annalen vol 319 no 1 pp89ndash149 2001

[2] X Tolsa ldquoLittlewood-Paley theory and the 119879(1) theorem withnon-doublingmeasuresrdquoAdvances inMathematics vol 164 no1 pp 57ndash116 2001

[3] X Tolsa ldquoThe space 1198671 for nondoubling measures in terms ofa grand maximal operatorrdquo Transactions of the American Math-ematical Society vol 355 no 1 pp 315ndash348 2003

[4] T A Bui and X T Duong ldquoHardy spaces regularized BMOspaces and the boundedness of Calderon-Zygmund operatorson non-homogeneous spacesrdquo Journal of Geometric Analysisvol 23 no 2 pp 895ndash932 2013

[5] T Hytonen S Liu D Yang and D Yang ldquoBoundedness of Cal-deron-Zygmund operators on non-homogeneous metric mea-sure spacesrdquo Canadian Journal of Mathematics vol 64 no 4pp 892ndash923 2012

[6] THytonenD Yang andD Yang ldquoTheHardy space1198671 onnon-homogeneous metric spacesrdquo Mathematical Proceedings of theCambridge Philosophical Society vol 153 no 1 pp 9ndash31 2012

[7] J Garcıa-Cuerva and A E Gatto ldquoBoundedness properties offractional integral operators associated to non-doubling mea-suresrdquo Studia Mathematica vol 162 no 3 pp 245ndash261 2004

[8] J Garcıa-Cuerva and A E Gatto ldquoLipschitz spaces and Cal-deron-Zygmund operators associated to non-doubling mea-suresrdquo Publicacions Matematiques vol 49 no 2 pp 285ndash2962005

[9] Y Cao and J Zhou ldquoMorrey spaces for non-homogeneousmetricmeasure spacesrdquoAbstract andApplied Analysis vol 2013Article ID 196459 8 pages 2013

[10] X Fu D Yang and W Yuan ldquoBoundedness of multilinearcommutators of Calderon-Zygmund operators onOrlicz spacesover non-homogeneous spacesrdquo Taiwanese Journal of Mathe-matics vol 16 no 6 pp 2203ndash2238 2012

[11] X Fu D Yang and W Yuan ldquoGeneralized fractional integralsand their commutators over non-homogeneous spacesrdquo Tai-wanese Journal of Mathematics vol 18 no 2 pp 509ndash557 2014

[12] H Lin andD Yang ldquoEquivalent boundedness ofMarcinkiewiczintegrals on non-homogeneous metric measure spacesrdquo ScienceChina Mathematics vol 57 no 1 pp 123ndash144 2014

[13] X Tolsa ldquoPainleversquos problem and the semiadditivity of analyticcapacityrdquo Acta Mathematica vol 190 no 1 pp 105ndash149 2003

[14] T Hytonen ldquoA framework for non-homogeneous analysis onmetric spaces and the RBMO space of Tolsardquo PublicacionsMatematiques vol 54 no 2 pp 485ndash504 2010

[15] R R Coifman and G Weiss Analyse Harmonique Non-Com-mutative sur Certains Espaces Homogenes vol 242 of LectureNotes in Mathematics Springer Berlin Germany 1971

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 5: Research Article Lipschitz Spaces and Fractional Integral ...downloads.hindawi.com/journals/aaa/2014/174010.pdf · Research Article Lipschitz Spaces and Fractional Integral Operators

Abstract and Applied Analysis 5

Then by applying Tchebichevrsquos inequality we have

120583 (119909 isin X 1003816100381610038161003816119868120572119891 (119909)

1003816100381610038161003816 gt ])

le 120583 (119909 isin X 100381610038161003816100381611986811003816100381610038161003816 gt

]2)

le 119862[120582(119909 119903)]1205721199011199011015840119899]minus119901

times intX

int119861(119909119903)

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901

[120582 (119909 119889 (119909 119910))]1minus120572119899

119889120583 (119910) 119889120583 (119909)

= 119862[120582 (119909 119903)]1205721199011199011015840119899]minus119901

times intX

int119861(119910119903)

119889120583 (119909)

[120582 (119909 119889 (119909 119910))]1minus120572119899

1003816100381610038161003816119891 (119910)1003816100381610038161003816

119901119889120583 (119910)

le 119862[120582(119909 119903)]1205721199011199011015840119899]minus119901[120582(119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817119871119901(120583)

= 119862 [120582 (119909 119903)] = 119862]minus119902(25)

This completes the proof of Theorem 13

Corollary 20 Let 1 lt 119901 lt 119899120572 and 1119902 = 1119901 minus 120572119899 If120582 satisfy the 120598-weak reverse doubling condition with 120598 isin(0min120572119899 (1119901 minus 120572119899)1199011015840) then

10038171003817100381710038171198681205721198911003817100381710038171003817119871119902(120583)

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(26)

Proof It suffices to apply Marcinkiewiczrsquos interpolation the-orem with indices slightly bigger and slightly smaller than119901

3 Proof of Theorem 16

Before we give the proof of Theorem 16 we first introduce atechnical lemma from [8 Lemma 32]

Lemma 21 Let 119891 isin 1198711119897119900119888(120583) If 1205732 gt 2119889 then for almost every

119909 with respect to 120583 there exists a sequence of (2 1205732)-doublingballs 119861119895 = 119861(119909 119903119895) with 119903119895 rarr 0 such that

lim119895rarrinfin

1

120583 (119861119895)

int119861119895

119891 (119910) 119889120583 (119910) = 119891 (119909) (27)

Proof of Theorem 16 (A) rArr (B) Consider 119909 as in the lemmaand let 119861119895 = 119861(119909 119903119895) 119895 ge 1 a sequence of (2 1205732)-doublingballs with 119903119895 rarr 0 Consider

100381610038161003816100381610038161003816119898119861119895

(119891) minus 119891119861119895

100381610038161003816100381610038161003816le

1

120583 (119861119895)

int119861119895

100381610038161003816100381610038161003816119891 (119910) minus 119891119861

119895

100381610038161003816100381610038161003816119889120583 (119910)

le

120583 (2119861119895)

120583 (119861119895)

1

120583 (2119861119895)

int119861119895

100381610038161003816100381610038161003816119891 (119910) minus 119891119861

119895

100381610038161003816100381610038161003816119889120583 (119910)

le 1205731198621[120582 (119909 119903119895)]120573119899

(28)

and by (5) Lemma 21 we obtain that

lim119895rarrinfin

119891119861119895

= 119891 (119909) (29)

Let 119909 and 119910 be two points as in the lemma take 119861 = 119861(119909 119903)any ball with 119903 le 119889(119909 119910) and let 119880 = 119861(119909 2119889(119909 119910)) Nowdefine 119861119896 = 119861(119909 2

119896119903) for 0 le 119896 le where is the first inte-

ger such that 2119903 ge 119889(119909 119910) Then

1003816100381610038161003816119891119861 minus 1198911198801003816100381610038161003816 le

minus1

sum

119896=0

10038161003816100381610038161003816119891119861119896

minus 119891119861119896+1

10038161003816100381610038161003816+10038161003816100381610038161003816119891119861

minus 119891119880

10038161003816100381610038161003816

le 1198621

sum

119896=0

[120582 (119909 2119896119903)]

120573119899

le 1198621

sum

119896=0

119888120573119896119899

120582[120582 (119909 119903)]

120573119899

le 11986210158401198621120582(119909 119889 (119909 119910))

120573119899

(30)

where 1198621015840 is independent of 119909 and 119889(119909 119910)A similar argument can be made for the point 119910 with any

ball 1198611015840 = 119861(119910 119904) such that 119904 le 119889(119909 119910) and119881 = 119861(119910 3119889(119909 119910))Therefore1003816100381610038161003816119891119861 minus 1198911198611015840

1003816100381610038161003816 le1003816100381610038161003816119891119861 minus 119891119880

1003816100381610038161003816 +1003816100381610038161003816119891119880 minus 119891119881

1003816100381610038161003816 +1003816100381610038161003816119891119881 minus 1198911198611015840

1003816100381610038161003816

le 119862101584010158401198621120582(119909 119889 (119909 119910))

120573119899

(31)

Take two sequences of (2 1205732)-doubling balls 119861119895 = 119861(119909 119903119895)and 1198611015840 = 119861(119910 119904119895) with 119903119895 rarr 0 and 119904119895 rarr 0 We have

1003816100381610038161003816119891 (119909) minus 119891 (119910)1003816100381610038161003816 = lim

119895rarr0

100381610038161003816100381610038161003816119891119861119895

minus 1198911198611015840119895

100381610038161003816100381610038161003816le 119862

101584010158401198621120582(119909 119889 (119909 119910))

120573119899

(32)

(B) rArr (C) For 1199090 isin 119861 = 119861(119909 119903) by the properties offunction 120582 and Holderrsquos inequality we obtain

(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 119898119861 (119891)1003816100381610038161003816

119901119889120583 (119909))

1119901

le (1

120583 (119861)int119861

10038161003816100381610038161003816100381610038161003816

1

120583 (119861)int119861

(119891 (119909) minus 119891 (119910)) 119889120583 (119910)

10038161003816100381610038161003816100381610038161003816

119901

119889120583 (119909))

1119901

le (1

120583 (119861)int119861

[1198622120582 (119909 119903)]120573119901119899119889120583 (119909))

1119901

le (1

120583 (119861)int119861

[1198622119862120582120582 (1199090 119903)]120573119901119899119889120583 (119909))

1119901

le 1198622119862120582[120582 (1199090 119903)]120573119899le 1198622119862

2

120582[120582 (119909 119903)]120573119899

(33)

By the similar argument for any ball 119880 such that 119861 sub 119880 andradius 119880 le 2119903

1003816100381610038161003816119898119861 (119891) minus 119898119880 (119891)1003816100381610038161003816 le 1198622119862

2

120582[120582 (119909 2119903)]120573119899

le 119888120573119899

1205821198622119862

2

120582[120582 (119909 119903)]120573119899

(34)

6 Abstract and Applied Analysis

(C) rArr (A) Define first 119891119861 = 119898119861(119891) Then (17) is exactly(20) To prove (16) we write

1

120583 (2119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816 119889120583 (119909)

le1

120583 (2119861)(int

119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816

119901119889120583 (119909))

1119901

120583(119861)11199011015840

le120583 (119861)

120583 (2119861)(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816

119901119889120583 (119909))

1119901

le 119862 (119901) 120582(119909 119903)120573119899

(35)

This concludes the proof of the theorem

Remark 22 Theorem 16 is also true if the number 2 in con-dition (A) is replaced by any fixed 120588 gt 1 In that case theproof uses (120588 120573120588)-doubling balls that is balls satisfying120583(120588119861) le 120573120588120583(119861)

4 Proofs of Theorems 17ndash19

Proof of Theorem 17 Without loss of generality we assumethat 119901 lt infin Consider 119909 = 119910 and let 119861 be the ball with center119909 and radius 119903 = 119889(119909 119910) Then we have

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le int

2119861

1003816100381610038161003816119870120572 (119909 119911)1003816100381610038161003816

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

+ int

2119861

1003816100381610038161003816119870120572 (119910 119911)1003816100381610038161003816

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

+ int

X2119861

1003816100381610038161003816119870120572 (119909 119911) minus 119870120572 (119910 119911)1003816100381610038161003816

times1003816100381610038161003816119891 (119911)

1003816100381610038161003816 119889120583 (119911)

= 1198691 + 1198692 + 1198693

(36)

For the first term by 119899120572 lt 119901 then 1 minus (1 minus 120572119899)1199011015840 gt 0

1198691 le int2119861

1003816100381610038161003816119891 (119911)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119911)

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

120583 (119861 (119909 2minus119895+1119903))

[120582 (119909 119889 (119909 2minus119895119903))](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

120582 (119909 2minus119895+1119903)

[120582 (119909 119889 (119909 2minus119895119903))](1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

[120582 (119909 119889 (119909 2minus119895119903))]

1minus(1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

1

[119862 (2119895)](120572119899minus1119901)1199011015840

)

11199011015840

times [120582 (119909 119903)]120572119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)[

120582 (119909 119903)]120572119899minus1119901

(37)

The second term is estimated in a similar way after noting that2119861 sub 119861(119910 3119903)

Next by using (2) Holderrsquos inequality and 120572 minus 119899119901 lt 120575we get

1198693 le 119862intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817119871119901(120583)

times (intX2119861

1

[120582 (119909 119889 (119909 119911))](1minus(120572minus120575)119899)1199011015840

119889120583 (119911))

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120575119899

times (intX2119861

1

[120582 (119909 119889 (119909 119911))](1minus(120572minus120575)119899)1199011015840

119889120583 (119911))

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572minus120575)119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120572119899minus1119901

(38)

Putting together the three estimates

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120572119899minus1119901

(39)

then 119868120572 maps 119871119901(120583) boundedly into Lip(120572 minus 119899119901)

Proof of Theorem 18 If 119868120572 isin Lip(120573) then by the continuity ofthe operator 119868120572 implies that 119868120572(1) must be constant that is119868120572(1)(119909) = 119868120572(1)(1199090) = 0

On the other hand we can observe that

119868120572 (1) = 0 lArrrArr 119868120572 (1) (119909) minus 119868120572 (1) (119910) = 0 (40)

this implies that

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) 119889120583 (119911) = 0 (41)

Abstract and Applied Analysis 7

Thus we can write

119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

= intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= int2119861(119909119903)

119870120572 (119909 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ int2119861(119909119903)

minus119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ intX2119861(119909119903)

119870120572 (119909 119911) minus 119870120572 (119910 119911)

times (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= 1198721 +1198722 +1198723

(42)

where 119903 = 119889(119909 119910) For1198721

10038161003816100381610038161198721

1003816100381610038161003816 le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(43)

Similarly we know that 2119861(119909 119903) sub 3119861(119910 119903) using Holderrsquosinequality (1119905 + 11199051015840 = 1) and letting 119905 gt 119899120572 then we have

10038161003816100381610038161198722

1003816100381610038161003816

le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862(int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119905119899119889120583 (119911))

1119905

times (int2119861(119909119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862(

infin

sum

119895=0

int2minus119895+1119861(119909119903)2minus119895119861(119909119903)

[120582 (119909

119889 (119909 2minus119895+1119903))]

120573119905119899

119889120583(119911))

1119905

times (int3119861(119910119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862[120582 (119909 119889 (119909 119910))]120572119899[120582 (119910 119889 (119909 119910))]

120573119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(44)

In order to estimate1198723 we use (2) to obtain

10038161003816100381610038161198723

1003816100381610038161003816

le 119862intX2119861(119909119903)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119911))]

120573119899

[120582 (119909 119889 (119909 119911))]1minus(120572minus120575)119899

119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times intX2119861(119909119903)

1

[120582 (119909 119889 (119909 119911))]1minus(120572+120573minus120575)119899

119889120583 (119911)

le 119862[120582(119909 119889(119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572+120573minus120575)119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(45)

Combining the estimates for11987211198722 and1198723 we obtain

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 119862[120582 (119909 119889 (119909 119910))]

(120572+120573)119899 (46)

this finishes the proof

Proof of Theorem 19 119868120572(119891) isin Lip(120572) implies that 119868120572(1) = 0and equivalently that

intX

(119870120572 (119909 119910) minus 119870120572 (1199090 119910)) 119889120583 (119910) = 0 (47)

for all 119909Take two points 119909 = 119910 and let119861 = 119861(119909 119903)with 119903 = 119889(119909 119910)

Then

10038161003816100381610038161003816119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

10038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

le

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119909 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

X2119861

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

= 1198731 + 1198732 + 1198733

(48)

8 Abstract and Applied Analysis

For the first term by Holderrsquos inequality with some 119901 gt 119899120572

1198731 le int2119861

1

[120582 (119909 119889 (119909 119911))]1minus120572119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le (int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119911))](1minus120572119899)1199011015840

)

11199011015840

times (int2119861

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816

119901119889120583 (119911))

1119901

le 119862[120582 (119909 119903)]120572119899minus1119901

120583(120588119861)11199011003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

le 119862[120582 (119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(49)

Using 2119861 sub 119861(119910 3119903) the second term can be dealt with inthe same way as the 1198731 then 1198732 le 119862[120582(119909 119903)]

120572119899119891RBMO(120583)

It is easy to see that |1198912119896+1119861 minus 1198912119861| le 119891RBMO(120583)11987021198612119896+1119861 le

119896119891RBMO(120583) then

1198733 le intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times

infin

sum

119896=1

(int2119896+11198612119896119861

1003816100381610038161003816119891 (119911) minus 1198912119896+11198611003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

119889120583 (119911)

+

10038161003816100381610038161198912119896+1119861 minus 11989121198611003816100381610038161003816

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

120583 (2119896+1119861))

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=1

120583 (1205882119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

+

infin

sum

119896=1

119896

120583 (2119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

)

le 119862[120582(119909 119889(119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=0

(119896 + 1) [120582 (119909 119889 (119909 2119896119903))]

120572119899minus120575119899

)

le 119862[120582 (119909 119889 (119909 119910))]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(50)

Thus the proof of Theorem 19 is completed

Conflict of Interests

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

Acknowledgments

Jiang Zhou is supported by the National Science Foundationof China (Grants nos 11261055 and 11161044) and by theNational Natural Science Foundation of Xinjiang (Grantsnos 2011211A005 and BS120104)

References

[1] X Tolsa ldquoBMO1198671 and Calderon-Zygmund operators for nondoubling measuresrdquoMathematische Annalen vol 319 no 1 pp89ndash149 2001

[2] X Tolsa ldquoLittlewood-Paley theory and the 119879(1) theorem withnon-doublingmeasuresrdquoAdvances inMathematics vol 164 no1 pp 57ndash116 2001

[3] X Tolsa ldquoThe space 1198671 for nondoubling measures in terms ofa grand maximal operatorrdquo Transactions of the American Math-ematical Society vol 355 no 1 pp 315ndash348 2003

[4] T A Bui and X T Duong ldquoHardy spaces regularized BMOspaces and the boundedness of Calderon-Zygmund operatorson non-homogeneous spacesrdquo Journal of Geometric Analysisvol 23 no 2 pp 895ndash932 2013

[5] T Hytonen S Liu D Yang and D Yang ldquoBoundedness of Cal-deron-Zygmund operators on non-homogeneous metric mea-sure spacesrdquo Canadian Journal of Mathematics vol 64 no 4pp 892ndash923 2012

[6] THytonenD Yang andD Yang ldquoTheHardy space1198671 onnon-homogeneous metric spacesrdquo Mathematical Proceedings of theCambridge Philosophical Society vol 153 no 1 pp 9ndash31 2012

[7] J Garcıa-Cuerva and A E Gatto ldquoBoundedness properties offractional integral operators associated to non-doubling mea-suresrdquo Studia Mathematica vol 162 no 3 pp 245ndash261 2004

[8] J Garcıa-Cuerva and A E Gatto ldquoLipschitz spaces and Cal-deron-Zygmund operators associated to non-doubling mea-suresrdquo Publicacions Matematiques vol 49 no 2 pp 285ndash2962005

[9] Y Cao and J Zhou ldquoMorrey spaces for non-homogeneousmetricmeasure spacesrdquoAbstract andApplied Analysis vol 2013Article ID 196459 8 pages 2013

[10] X Fu D Yang and W Yuan ldquoBoundedness of multilinearcommutators of Calderon-Zygmund operators onOrlicz spacesover non-homogeneous spacesrdquo Taiwanese Journal of Mathe-matics vol 16 no 6 pp 2203ndash2238 2012

[11] X Fu D Yang and W Yuan ldquoGeneralized fractional integralsand their commutators over non-homogeneous spacesrdquo Tai-wanese Journal of Mathematics vol 18 no 2 pp 509ndash557 2014

[12] H Lin andD Yang ldquoEquivalent boundedness ofMarcinkiewiczintegrals on non-homogeneous metric measure spacesrdquo ScienceChina Mathematics vol 57 no 1 pp 123ndash144 2014

[13] X Tolsa ldquoPainleversquos problem and the semiadditivity of analyticcapacityrdquo Acta Mathematica vol 190 no 1 pp 105ndash149 2003

[14] T Hytonen ldquoA framework for non-homogeneous analysis onmetric spaces and the RBMO space of Tolsardquo PublicacionsMatematiques vol 54 no 2 pp 485ndash504 2010

[15] R R Coifman and G Weiss Analyse Harmonique Non-Com-mutative sur Certains Espaces Homogenes vol 242 of LectureNotes in Mathematics Springer Berlin Germany 1971

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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

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 6: Research Article Lipschitz Spaces and Fractional Integral ...downloads.hindawi.com/journals/aaa/2014/174010.pdf · Research Article Lipschitz Spaces and Fractional Integral Operators

6 Abstract and Applied Analysis

(C) rArr (A) Define first 119891119861 = 119898119861(119891) Then (17) is exactly(20) To prove (16) we write

1

120583 (2119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816 119889120583 (119909)

le1

120583 (2119861)(int

119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816

119901119889120583 (119909))

1119901

120583(119861)11199011015840

le120583 (119861)

120583 (2119861)(1

120583 (119861)int119861

1003816100381610038161003816119891 (119909) minus 1198911198611003816100381610038161003816

119901119889120583 (119909))

1119901

le 119862 (119901) 120582(119909 119903)120573119899

(35)

This concludes the proof of the theorem

Remark 22 Theorem 16 is also true if the number 2 in con-dition (A) is replaced by any fixed 120588 gt 1 In that case theproof uses (120588 120573120588)-doubling balls that is balls satisfying120583(120588119861) le 120573120588120583(119861)

4 Proofs of Theorems 17ndash19

Proof of Theorem 17 Without loss of generality we assumethat 119901 lt infin Consider 119909 = 119910 and let 119861 be the ball with center119909 and radius 119903 = 119889(119909 119910) Then we have

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le int

2119861

1003816100381610038161003816119870120572 (119909 119911)1003816100381610038161003816

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

+ int

2119861

1003816100381610038161003816119870120572 (119910 119911)1003816100381610038161003816

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

+ int

X2119861

1003816100381610038161003816119870120572 (119909 119911) minus 119870120572 (119910 119911)1003816100381610038161003816

times1003816100381610038161003816119891 (119911)

1003816100381610038161003816 119889120583 (119911)

= 1198691 + 1198692 + 1198693

(36)

For the first term by 119899120572 lt 119901 then 1 minus (1 minus 120572119899)1199011015840 gt 0

1198691 le int2119861

1003816100381610038161003816119891 (119911)1003816100381610038161003816

[120582 (119909 119889 (119909 119910))]1minus120572119899119889120583 (119911)

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119910))](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

120583 (119861 (119909 2minus119895+1119903))

[120582 (119909 119889 (119909 2minus119895119903))](1minus120572119899)1199011015840

)

11199011015840

le10038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

120582 (119909 2minus119895+1119903)

[120582 (119909 119889 (119909 2minus119895119903))](1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

[120582 (119909 119889 (119909 2minus119895119903))]

1minus(1minus120572119899)1199011015840

)

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

(

infin

sum

119895=0

1

[119862 (2119895)](120572119899minus1119901)1199011015840

)

11199011015840

times [120582 (119909 119903)]120572119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)[

120582 (119909 119903)]120572119899minus1119901

(37)

The second term is estimated in a similar way after noting that2119861 sub 119861(119910 3119903)

Next by using (2) Holderrsquos inequality and 120572 minus 119899119901 lt 120575we get

1198693 le 119862intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911)1003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817119871119901(120583)

times (intX2119861

1

[120582 (119909 119889 (119909 119911))](1minus(120572minus120575)119899)1199011015840

119889120583 (119911))

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120575119899

times (intX2119861

1

[120582 (119909 119889 (119909 119911))](1minus(120572minus120575)119899)1199011015840

119889120583 (119911))

11199011015840

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572minus120575)119899minus1119901

le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120572119899minus1119901

(38)

Putting together the three estimates

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 11986210038171003817100381710038171198911003817100381710038171003817119871119901(120583)

[120582 (119909 119889 (119909 119910))]120572119899minus1119901

(39)

then 119868120572 maps 119871119901(120583) boundedly into Lip(120572 minus 119899119901)

Proof of Theorem 18 If 119868120572 isin Lip(120573) then by the continuity ofthe operator 119868120572 implies that 119868120572(1) must be constant that is119868120572(1)(119909) = 119868120572(1)(1199090) = 0

On the other hand we can observe that

119868120572 (1) = 0 lArrrArr 119868120572 (1) (119909) minus 119868120572 (1) (119910) = 0 (40)

this implies that

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) 119889120583 (119911) = 0 (41)

Abstract and Applied Analysis 7

Thus we can write

119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

= intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= int2119861(119909119903)

119870120572 (119909 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ int2119861(119909119903)

minus119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ intX2119861(119909119903)

119870120572 (119909 119911) minus 119870120572 (119910 119911)

times (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= 1198721 +1198722 +1198723

(42)

where 119903 = 119889(119909 119910) For1198721

10038161003816100381610038161198721

1003816100381610038161003816 le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(43)

Similarly we know that 2119861(119909 119903) sub 3119861(119910 119903) using Holderrsquosinequality (1119905 + 11199051015840 = 1) and letting 119905 gt 119899120572 then we have

10038161003816100381610038161198722

1003816100381610038161003816

le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862(int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119905119899119889120583 (119911))

1119905

times (int2119861(119909119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862(

infin

sum

119895=0

int2minus119895+1119861(119909119903)2minus119895119861(119909119903)

[120582 (119909

119889 (119909 2minus119895+1119903))]

120573119905119899

119889120583(119911))

1119905

times (int3119861(119910119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862[120582 (119909 119889 (119909 119910))]120572119899[120582 (119910 119889 (119909 119910))]

120573119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(44)

In order to estimate1198723 we use (2) to obtain

10038161003816100381610038161198723

1003816100381610038161003816

le 119862intX2119861(119909119903)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119911))]

120573119899

[120582 (119909 119889 (119909 119911))]1minus(120572minus120575)119899

119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times intX2119861(119909119903)

1

[120582 (119909 119889 (119909 119911))]1minus(120572+120573minus120575)119899

119889120583 (119911)

le 119862[120582(119909 119889(119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572+120573minus120575)119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(45)

Combining the estimates for11987211198722 and1198723 we obtain

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 119862[120582 (119909 119889 (119909 119910))]

(120572+120573)119899 (46)

this finishes the proof

Proof of Theorem 19 119868120572(119891) isin Lip(120572) implies that 119868120572(1) = 0and equivalently that

intX

(119870120572 (119909 119910) minus 119870120572 (1199090 119910)) 119889120583 (119910) = 0 (47)

for all 119909Take two points 119909 = 119910 and let119861 = 119861(119909 119903)with 119903 = 119889(119909 119910)

Then

10038161003816100381610038161003816119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

10038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

le

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119909 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

X2119861

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

= 1198731 + 1198732 + 1198733

(48)

8 Abstract and Applied Analysis

For the first term by Holderrsquos inequality with some 119901 gt 119899120572

1198731 le int2119861

1

[120582 (119909 119889 (119909 119911))]1minus120572119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le (int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119911))](1minus120572119899)1199011015840

)

11199011015840

times (int2119861

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816

119901119889120583 (119911))

1119901

le 119862[120582 (119909 119903)]120572119899minus1119901

120583(120588119861)11199011003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

le 119862[120582 (119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(49)

Using 2119861 sub 119861(119910 3119903) the second term can be dealt with inthe same way as the 1198731 then 1198732 le 119862[120582(119909 119903)]

120572119899119891RBMO(120583)

It is easy to see that |1198912119896+1119861 minus 1198912119861| le 119891RBMO(120583)11987021198612119896+1119861 le

119896119891RBMO(120583) then

1198733 le intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times

infin

sum

119896=1

(int2119896+11198612119896119861

1003816100381610038161003816119891 (119911) minus 1198912119896+11198611003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

119889120583 (119911)

+

10038161003816100381610038161198912119896+1119861 minus 11989121198611003816100381610038161003816

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

120583 (2119896+1119861))

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=1

120583 (1205882119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

+

infin

sum

119896=1

119896

120583 (2119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

)

le 119862[120582(119909 119889(119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=0

(119896 + 1) [120582 (119909 119889 (119909 2119896119903))]

120572119899minus120575119899

)

le 119862[120582 (119909 119889 (119909 119910))]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(50)

Thus the proof of Theorem 19 is completed

Conflict of Interests

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

Acknowledgments

Jiang Zhou is supported by the National Science Foundationof China (Grants nos 11261055 and 11161044) and by theNational Natural Science Foundation of Xinjiang (Grantsnos 2011211A005 and BS120104)

References

[1] X Tolsa ldquoBMO1198671 and Calderon-Zygmund operators for nondoubling measuresrdquoMathematische Annalen vol 319 no 1 pp89ndash149 2001

[2] X Tolsa ldquoLittlewood-Paley theory and the 119879(1) theorem withnon-doublingmeasuresrdquoAdvances inMathematics vol 164 no1 pp 57ndash116 2001

[3] X Tolsa ldquoThe space 1198671 for nondoubling measures in terms ofa grand maximal operatorrdquo Transactions of the American Math-ematical Society vol 355 no 1 pp 315ndash348 2003

[4] T A Bui and X T Duong ldquoHardy spaces regularized BMOspaces and the boundedness of Calderon-Zygmund operatorson non-homogeneous spacesrdquo Journal of Geometric Analysisvol 23 no 2 pp 895ndash932 2013

[5] T Hytonen S Liu D Yang and D Yang ldquoBoundedness of Cal-deron-Zygmund operators on non-homogeneous metric mea-sure spacesrdquo Canadian Journal of Mathematics vol 64 no 4pp 892ndash923 2012

[6] THytonenD Yang andD Yang ldquoTheHardy space1198671 onnon-homogeneous metric spacesrdquo Mathematical Proceedings of theCambridge Philosophical Society vol 153 no 1 pp 9ndash31 2012

[7] J Garcıa-Cuerva and A E Gatto ldquoBoundedness properties offractional integral operators associated to non-doubling mea-suresrdquo Studia Mathematica vol 162 no 3 pp 245ndash261 2004

[8] J Garcıa-Cuerva and A E Gatto ldquoLipschitz spaces and Cal-deron-Zygmund operators associated to non-doubling mea-suresrdquo Publicacions Matematiques vol 49 no 2 pp 285ndash2962005

[9] Y Cao and J Zhou ldquoMorrey spaces for non-homogeneousmetricmeasure spacesrdquoAbstract andApplied Analysis vol 2013Article ID 196459 8 pages 2013

[10] X Fu D Yang and W Yuan ldquoBoundedness of multilinearcommutators of Calderon-Zygmund operators onOrlicz spacesover non-homogeneous spacesrdquo Taiwanese Journal of Mathe-matics vol 16 no 6 pp 2203ndash2238 2012

[11] X Fu D Yang and W Yuan ldquoGeneralized fractional integralsand their commutators over non-homogeneous spacesrdquo Tai-wanese Journal of Mathematics vol 18 no 2 pp 509ndash557 2014

[12] H Lin andD Yang ldquoEquivalent boundedness ofMarcinkiewiczintegrals on non-homogeneous metric measure spacesrdquo ScienceChina Mathematics vol 57 no 1 pp 123ndash144 2014

[13] X Tolsa ldquoPainleversquos problem and the semiadditivity of analyticcapacityrdquo Acta Mathematica vol 190 no 1 pp 105ndash149 2003

[14] T Hytonen ldquoA framework for non-homogeneous analysis onmetric spaces and the RBMO space of Tolsardquo PublicacionsMatematiques vol 54 no 2 pp 485ndash504 2010

[15] R R Coifman and G Weiss Analyse Harmonique Non-Com-mutative sur Certains Espaces Homogenes vol 242 of LectureNotes in Mathematics Springer Berlin Germany 1971

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 7: Research Article Lipschitz Spaces and Fractional Integral ...downloads.hindawi.com/journals/aaa/2014/174010.pdf · Research Article Lipschitz Spaces and Fractional Integral Operators

Abstract and Applied Analysis 7

Thus we can write

119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

= intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= int2119861(119909119903)

119870120572 (119909 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ int2119861(119909119903)

minus119870120572 (119910 119911) (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

+ intX2119861(119909119903)

119870120572 (119909 119911) minus 119870120572 (119910 119911)

times (119891 (119911) minus 119891 (119909)) 119889120583 (119911)

= 1198721 +1198722 +1198723

(42)

where 119903 = 119889(119909 119910) For1198721

10038161003816100381610038161198721

1003816100381610038161003816 le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119909 119889 (119909 119911))]1minus120572119899119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(43)

Similarly we know that 2119861(119909 119903) sub 3119861(119910 119903) using Holderrsquosinequality (1119905 + 11199051015840 = 1) and letting 119905 gt 119899120572 then we have

10038161003816100381610038161198722

1003816100381610038161003816

le int2119861(119909119903)

1003816100381610038161003816119891 (119911) minus 119891 (119909)1003816100381610038161003816

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119899

[120582 (119910 119889 (119910 119911))]1minus120572119899119889120583 (119911)

le 119862(int2119861(119909119903)

[120582 (119909 119889 (119909 119911))]120573119905119899119889120583 (119911))

1119905

times (int2119861(119909119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862(

infin

sum

119895=0

int2minus119895+1119861(119909119903)2minus119895119861(119909119903)

[120582 (119909

119889 (119909 2minus119895+1119903))]

120573119905119899

119889120583(119911))

1119905

times (int3119861(119910119903)

1

[120582 (119910 119889 (119910 119911))](1minus120572119899)1199051015840

119889120583 (119911))

11199051015840

le 119862[120582 (119909 119889 (119909 119910))]120572119899[120582 (119910 119889 (119909 119910))]

120573119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(44)

In order to estimate1198723 we use (2) to obtain

10038161003816100381610038161198723

1003816100381610038161003816

le 119862intX2119861(119909119903)

[120582 (119909 119889 (119909 119910))]120575119899[120582 (119909 119889 (119909 119911))]

120573119899

[120582 (119909 119889 (119909 119911))]1minus(120572minus120575)119899

119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times intX2119861(119909119903)

1

[120582 (119909 119889 (119909 119911))]1minus(120572+120573minus120575)119899

119889120583 (119911)

le 119862[120582(119909 119889(119909 119910))]120575119899[120582 (119909 119889 (119909 119910))]

(120572+120573minus120575)119899

le 119862[120582 (119909 119889 (119909 119910))](120572+120573)119899

(45)

Combining the estimates for11987211198722 and1198723 we obtain

10038161003816100381610038161003816119868120572119891 (119909) minus 119868120572119891 (119910)

10038161003816100381610038161003816le 119862[120582 (119909 119889 (119909 119910))]

(120572+120573)119899 (46)

this finishes the proof

Proof of Theorem 19 119868120572(119891) isin Lip(120572) implies that 119868120572(1) = 0and equivalently that

intX

(119870120572 (119909 119910) minus 119870120572 (1199090 119910)) 119889120583 (119910) = 0 (47)

for all 119909Take two points 119909 = 119910 and let119861 = 119861(119909 119903)with 119903 = 119889(119909 119910)

Then

10038161003816100381610038161003816119868120572 (119891) (119909) minus 119868120572 (119891) (119910)

10038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

intX

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

le

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119909 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816int

2119861

119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

1003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

X2119861

119870120572 (119909 119911) minus 119870120572 (119910 119911) (119891 (119911) minus 1198912119861) 119889120583 (119911)

10038161003816100381610038161003816100381610038161003816

= 1198731 + 1198732 + 1198733

(48)

8 Abstract and Applied Analysis

For the first term by Holderrsquos inequality with some 119901 gt 119899120572

1198731 le int2119861

1

[120582 (119909 119889 (119909 119911))]1minus120572119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le (int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119911))](1minus120572119899)1199011015840

)

11199011015840

times (int2119861

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816

119901119889120583 (119911))

1119901

le 119862[120582 (119909 119903)]120572119899minus1119901

120583(120588119861)11199011003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

le 119862[120582 (119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(49)

Using 2119861 sub 119861(119910 3119903) the second term can be dealt with inthe same way as the 1198731 then 1198732 le 119862[120582(119909 119903)]

120572119899119891RBMO(120583)

It is easy to see that |1198912119896+1119861 minus 1198912119861| le 119891RBMO(120583)11987021198612119896+1119861 le

119896119891RBMO(120583) then

1198733 le intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times

infin

sum

119896=1

(int2119896+11198612119896119861

1003816100381610038161003816119891 (119911) minus 1198912119896+11198611003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

119889120583 (119911)

+

10038161003816100381610038161198912119896+1119861 minus 11989121198611003816100381610038161003816

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

120583 (2119896+1119861))

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=1

120583 (1205882119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

+

infin

sum

119896=1

119896

120583 (2119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

)

le 119862[120582(119909 119889(119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=0

(119896 + 1) [120582 (119909 119889 (119909 2119896119903))]

120572119899minus120575119899

)

le 119862[120582 (119909 119889 (119909 119910))]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(50)

Thus the proof of Theorem 19 is completed

Conflict of Interests

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

Acknowledgments

Jiang Zhou is supported by the National Science Foundationof China (Grants nos 11261055 and 11161044) and by theNational Natural Science Foundation of Xinjiang (Grantsnos 2011211A005 and BS120104)

References

[1] X Tolsa ldquoBMO1198671 and Calderon-Zygmund operators for nondoubling measuresrdquoMathematische Annalen vol 319 no 1 pp89ndash149 2001

[2] X Tolsa ldquoLittlewood-Paley theory and the 119879(1) theorem withnon-doublingmeasuresrdquoAdvances inMathematics vol 164 no1 pp 57ndash116 2001

[3] X Tolsa ldquoThe space 1198671 for nondoubling measures in terms ofa grand maximal operatorrdquo Transactions of the American Math-ematical Society vol 355 no 1 pp 315ndash348 2003

[4] T A Bui and X T Duong ldquoHardy spaces regularized BMOspaces and the boundedness of Calderon-Zygmund operatorson non-homogeneous spacesrdquo Journal of Geometric Analysisvol 23 no 2 pp 895ndash932 2013

[5] T Hytonen S Liu D Yang and D Yang ldquoBoundedness of Cal-deron-Zygmund operators on non-homogeneous metric mea-sure spacesrdquo Canadian Journal of Mathematics vol 64 no 4pp 892ndash923 2012

[6] THytonenD Yang andD Yang ldquoTheHardy space1198671 onnon-homogeneous metric spacesrdquo Mathematical Proceedings of theCambridge Philosophical Society vol 153 no 1 pp 9ndash31 2012

[7] J Garcıa-Cuerva and A E Gatto ldquoBoundedness properties offractional integral operators associated to non-doubling mea-suresrdquo Studia Mathematica vol 162 no 3 pp 245ndash261 2004

[8] J Garcıa-Cuerva and A E Gatto ldquoLipschitz spaces and Cal-deron-Zygmund operators associated to non-doubling mea-suresrdquo Publicacions Matematiques vol 49 no 2 pp 285ndash2962005

[9] Y Cao and J Zhou ldquoMorrey spaces for non-homogeneousmetricmeasure spacesrdquoAbstract andApplied Analysis vol 2013Article ID 196459 8 pages 2013

[10] X Fu D Yang and W Yuan ldquoBoundedness of multilinearcommutators of Calderon-Zygmund operators onOrlicz spacesover non-homogeneous spacesrdquo Taiwanese Journal of Mathe-matics vol 16 no 6 pp 2203ndash2238 2012

[11] X Fu D Yang and W Yuan ldquoGeneralized fractional integralsand their commutators over non-homogeneous spacesrdquo Tai-wanese Journal of Mathematics vol 18 no 2 pp 509ndash557 2014

[12] H Lin andD Yang ldquoEquivalent boundedness ofMarcinkiewiczintegrals on non-homogeneous metric measure spacesrdquo ScienceChina Mathematics vol 57 no 1 pp 123ndash144 2014

[13] X Tolsa ldquoPainleversquos problem and the semiadditivity of analyticcapacityrdquo Acta Mathematica vol 190 no 1 pp 105ndash149 2003

[14] T Hytonen ldquoA framework for non-homogeneous analysis onmetric spaces and the RBMO space of Tolsardquo PublicacionsMatematiques vol 54 no 2 pp 485ndash504 2010

[15] R R Coifman and G Weiss Analyse Harmonique Non-Com-mutative sur Certains Espaces Homogenes vol 242 of LectureNotes in Mathematics Springer Berlin Germany 1971

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

Page 8: Research Article Lipschitz Spaces and Fractional Integral ...downloads.hindawi.com/journals/aaa/2014/174010.pdf · Research Article Lipschitz Spaces and Fractional Integral Operators

8 Abstract and Applied Analysis

For the first term by Holderrsquos inequality with some 119901 gt 119899120572

1198731 le int2119861

1

[120582 (119909 119889 (119909 119911))]1minus120572119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le (int2119861

119889120583 (119911)

[120582 (119909 119889 (119909 119911))](1minus120572119899)1199011015840

)

11199011015840

times (int2119861

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816

119901119889120583 (119911))

1119901

le 119862[120582 (119909 119903)]120572119899minus1119901

120583(120588119861)11199011003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

le 119862[120582 (119909 119903)]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(49)

Using 2119861 sub 119861(119910 3119903) the second term can be dealt with inthe same way as the 1198731 then 1198732 le 119862[120582(119909 119903)]

120572119899119891RBMO(120583)

It is easy to see that |1198912119896+1119861 minus 1198912119861| le 119891RBMO(120583)11987021198612119896+1119861 le

119896119891RBMO(120583) then

1198733 le intX2119861

[120582 (119909 119889 (119909 119910))]120575119899

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

1003816100381610038161003816119891 (119911) minus 11989121198611003816100381610038161003816 119889120583 (119911)

le 119862[120582 (119909 119889 (119909 119910))]120575119899

times

infin

sum

119896=1

(int2119896+11198612119896119861

1003816100381610038161003816119891 (119911) minus 1198912119896+11198611003816100381610038161003816

[120582 (119909 119889 (119909 119911))]1minus120572119899+120575119899

119889120583 (119911)

+

10038161003816100381610038161198912119896+1119861 minus 11989121198611003816100381610038161003816

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

120583 (2119896+1119861))

le 119862[120582 (119909 119889 (119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=1

120583 (1205882119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

+

infin

sum

119896=1

119896

120583 (2119896+1119861)

[120582 (119909 119889 (119909 2119896119903))]1minus120572119899+120575119899

)

le 119862[120582(119909 119889(119909 119910))]1205751198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

times (

infin

sum

119896=0

(119896 + 1) [120582 (119909 119889 (119909 2119896119903))]

120572119899minus120575119899

)

le 119862[120582 (119909 119889 (119909 119910))]1205721198991003817100381710038171003817119891

1003817100381710038171003817RBMO(120583)

(50)

Thus the proof of Theorem 19 is completed

Conflict of Interests

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

Acknowledgments

Jiang Zhou is supported by the National Science Foundationof China (Grants nos 11261055 and 11161044) and by theNational Natural Science Foundation of Xinjiang (Grantsnos 2011211A005 and BS120104)

References

[1] X Tolsa ldquoBMO1198671 and Calderon-Zygmund operators for nondoubling measuresrdquoMathematische Annalen vol 319 no 1 pp89ndash149 2001

[2] X Tolsa ldquoLittlewood-Paley theory and the 119879(1) theorem withnon-doublingmeasuresrdquoAdvances inMathematics vol 164 no1 pp 57ndash116 2001

[3] X Tolsa ldquoThe space 1198671 for nondoubling measures in terms ofa grand maximal operatorrdquo Transactions of the American Math-ematical Society vol 355 no 1 pp 315ndash348 2003

[4] T A Bui and X T Duong ldquoHardy spaces regularized BMOspaces and the boundedness of Calderon-Zygmund operatorson non-homogeneous spacesrdquo Journal of Geometric Analysisvol 23 no 2 pp 895ndash932 2013

[5] T Hytonen S Liu D Yang and D Yang ldquoBoundedness of Cal-deron-Zygmund operators on non-homogeneous metric mea-sure spacesrdquo Canadian Journal of Mathematics vol 64 no 4pp 892ndash923 2012

[6] THytonenD Yang andD Yang ldquoTheHardy space1198671 onnon-homogeneous metric spacesrdquo Mathematical Proceedings of theCambridge Philosophical Society vol 153 no 1 pp 9ndash31 2012

[7] J Garcıa-Cuerva and A E Gatto ldquoBoundedness properties offractional integral operators associated to non-doubling mea-suresrdquo Studia Mathematica vol 162 no 3 pp 245ndash261 2004

[8] J Garcıa-Cuerva and A E Gatto ldquoLipschitz spaces and Cal-deron-Zygmund operators associated to non-doubling mea-suresrdquo Publicacions Matematiques vol 49 no 2 pp 285ndash2962005

[9] Y Cao and J Zhou ldquoMorrey spaces for non-homogeneousmetricmeasure spacesrdquoAbstract andApplied Analysis vol 2013Article ID 196459 8 pages 2013

[10] X Fu D Yang and W Yuan ldquoBoundedness of multilinearcommutators of Calderon-Zygmund operators onOrlicz spacesover non-homogeneous spacesrdquo Taiwanese Journal of Mathe-matics vol 16 no 6 pp 2203ndash2238 2012

[11] X Fu D Yang and W Yuan ldquoGeneralized fractional integralsand their commutators over non-homogeneous spacesrdquo Tai-wanese Journal of Mathematics vol 18 no 2 pp 509ndash557 2014

[12] H Lin andD Yang ldquoEquivalent boundedness ofMarcinkiewiczintegrals on non-homogeneous metric measure spacesrdquo ScienceChina Mathematics vol 57 no 1 pp 123ndash144 2014

[13] X Tolsa ldquoPainleversquos problem and the semiadditivity of analyticcapacityrdquo Acta Mathematica vol 190 no 1 pp 105ndash149 2003

[14] T Hytonen ldquoA framework for non-homogeneous analysis onmetric spaces and the RBMO space of Tolsardquo PublicacionsMatematiques vol 54 no 2 pp 485ndash504 2010

[15] R R Coifman and G Weiss Analyse Harmonique Non-Com-mutative sur Certains Espaces Homogenes vol 242 of LectureNotes in Mathematics Springer Berlin Germany 1971

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

Page 9: Research Article Lipschitz Spaces and Fractional Integral ...downloads.hindawi.com/journals/aaa/2014/174010.pdf · Research Article Lipschitz Spaces and Fractional Integral Operators

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


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