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      ON THE GRAPHS OF PRODUCTS OF CONTINUOUS FUNCTIONS AND FRACTAL DIMENSIONS?

      2023-04-25 01:41:36劉佳石賽賽張遠(yuǎn)
      關(guān)鍵詞:劉佳

      (劉佳) (石賽賽) (張遠(yuǎn)))

      Institute of Statistics and Applied Mathematics, Anhui University of Finance and Economics,Bengbu 233030, China

      E-mails: liujia860319@163.com; saisai shi@126.com; 120210066@aufe.edu.cn

      Abstract In this paper,we consider the graph of the product of continuous functions in terms of Hausdorffand packing dimensions.More precisely,we show that,given a real number 1 ≤β ≤2,any real-valued continuous function in C([0,1])can be decomposed into a product of two real-valued continuous functions,each having a graph of Hausdorffdimension β.In addition,a product decomposition result for the packing dimension is obtained.This work answers affirmatively two questions raised by Verma and Priyadarshi [14].

      Key words Hausdorffdimension;packing dimension;graph of function;product of functions

      1 Introduction

      LetXbe a compact metric space,we denote byC(X) the collection of all real-valued continuous functions onX.ThenC(X) is a Banach space when endowed with the maximum norm.The graph of a functionf ∈C(X) is defined as the set

      For simplicity,we write thatGf=Gf(X)as long as this causes no confusion.Throughout this paper,the Hausdorff,packing,lower box and upper box dimensions of a setAare denoted by dimHA,dimP A,,respectively(see[6]for definitions and related properties).

      The aim of this paper is to consider the product decomposition of continuous functions in terms of the Hausdorffdimension,as well as the packing dimension.Problems of decomposition of continuous functions in terms of fractal dimensions were first concerned by Mauldin and Williams [13],who proved that any continuous function can be written as a sum of two continuous functions with their graphs having a Hausdorffdimension one.Wingren [15]provided a constructive proof of this result and moreover showed that any continuous function can be decomposed into a sum of two continuous functions such that both of them have graphs with a lower box dimension one.After that,some related problems have been studied in [16,17].Generally,for any functionf ∈C([0,1]) andβ ∈[1,2],are there functionsg,h ∈C([0,1]) such that

      In 2013,Bayart and Heurteaux [4]proved that Equation (1.1) holds forβ=2.Later,Liu and Wu solved the problem for the remaining case 1<β<2.Versions of Result (1.1)for other fractal dimensions were also studied in [1,7–11,17].Recently,motivated by above decomposition result,Verma and Priyadarshi began to study fractal dimensional results for graph of the product of two continuous functions.They obtained some results related to the product decomposition of continuous functions in terms of the upper box dimension.More precisely,they proved the following result,which is an analogue of Corollary 1.6 in [11]:

      Theorem 1.1([14]) Letβ ∈[1,2]andf ∈C([0,1]) withf(x)≠ 0,?x ∈[0,1].Then there exist functionsg,h ∈C([0,1]) such that

      However,some analogous decomposition results regarding the Hausdorffdimension [12],the packing dimension [11]and the lower box dimension [10]are still open.Thus Verma and Priyadarshi posed some open problems: see Questions 4.1,4.2,4.3 in [14].

      In this paper,we give affirmative solutions to Question 4.1 and Question 4.2 from[14],and prove the following two results:

      Theorem 1.2Givenβ ∈[1,2]andf ∈C([0,1]),there are functionsg,h ∈C([0,1]) such that

      In addition,we obtain the counterpart decomposition theorem for the packing dimension,by a suitable modification to the proof of Theorem 1.2.

      Theorem 1.3Givenβ ∈[1,2]andf ∈C([0,1]),if dimP Gf ≤β,then there are functionsg,h ∈C([0,1]) such that

      The following proposition shows that not any functionf ∈C([0,1]) can be decomposed into a product of two functions whose graphs have a packing dimension

      Proposition 1.4Assume thatf ∈C([0,1]) satisfies the condition that dimP Gf(U)=dimP Gffor any nonempty open setU ?[0,1].Then,for any functionsg,h ∈C([0,1]) such thatf=g·h,we have that

      2 Preliminaries

      In this section,we cite some useful lemmas which will be used in the next section.Throughout the paper,for a setK ?R and a numberβ>0,we writeHβ(K),Pβ(K) as

      Lemma 2.1([11]) LetX ?Rdbe a compact set,f ∈C(X) and letαbe a real number.Iffor any open setV ?RdwithV ∩X≠?,then we have that

      Lemma 2.2([13])H1([0,1]) is co-meager inC([0,1]).

      Lemma 2.3([1,3]) LetKbe an uncountable compact metric space and letd ∈N.Then,for a prevalent continuous functionf ∈C(K,Rd),we have that

      where dim expresses any one of the Hausdorffor packing dimensions andC(K,Rd) denotes the set of all continuous mapsf:K →Rd.

      Lemma 2.4([2]) LetX,Ybe two complete metric spaces and letR:X →Ybe a continuous,open and surjective mapping.Then,1.ifA ?Xis of second category(co-meager),R(A)?Yis of second category(co-meager);2.ifB ?Yis of second category (co-meager),R-1(B)?Xis of second category (comeager).

      Lemma 2.5([5]) LetG1,G2be abelian Polish groups and let Φ:G1→G2be a continuous onto homomorphism.IfS ?G2is prevalent,then so is Φ-1(S)?G1.

      Note that Verma and Priyadarshi showed the following two lemmas for the case in whichX=[0,1]:

      Lemma 2.6LetXbe a nonempty compact metric space and letf ∈C(X) be a continuous function such thatf(x)≠0 for allx ∈X.Then

      Then it is easy to see thatTis bi-Lipschitz.Combining this and the bi-Lipschitz invariance property of these dimensions,the lemma follows.

      Lemma 2.7LetXbe a nonempty compact metric space and letf ∈C(X) be a continuous function satisfying the condition thatf(x)≠0 for allx ∈X.Then

      where dim is any one of dimH,dimP,

      ProofConsider the mapping Φ:Gf →Gf2defined by

      Clearly Φ is bi-Lipschitz.

      Lemma 2.8([14]) Letf ∈C([0,1]).Then

      Lemma 2.9LetXbe a nonempty compact metric space and letf ∈C(X)be a Lipschitz function such thatf(x)≠0 for allx ∈X.Then,for anyg ∈C(X),we have that

      where dim is any one of dimH,dimP,

      ProofIn fact,the mapping Φ:Gg →Gf·gdefined by

      is bi-Lipschitz.

      3 Proofs of the Main Theorems

      In this section we prove our main results.

      Proposition 3.1Letf ∈C([0,1]) be a continuous function such thatf(x)≠ 0 for allx ∈[0,1].Then there exist functionsg,h ∈H1([0,1]) such that

      ProofConsider the mappingTf:C([0,1])→C([0,1]),defined by

      Then,clearly,Tfis a continuous,open and surjective mapping.Thus,by Lemmas 2.2,2.4,we have thatTf(H1([0,1]))=f·H1([0,1]) is co-meager.This gives that the setH1([0,1])∩f·H1([0,1]) is also co-meager and thus dense inC([0,1]).Without loss of generality,we assume thatf(x)>0,?x ∈[0,1](otherwise,we may consider-f).Let I(x)=1,?x ∈[0,1],so there exists a functiong ∈H1([0,1])∩f·H1([0,1])∩B(I(x),).Here,B(I(x),) denotes the ball inC([0,1]) of radiuscentered at the function I(x).Thus there exists a real valued functionh1∈H1([0,1]) such thatg=f·h1.Note that,sincef(x)>0 andg(x)>,thush1(x)>0 for allx ∈[0,1].This gives thatf=g·h,whereh=.Due to Lemma 2.6,we have thath ∈H1([0,1]).

      Proposition 3.2LetX ?[0,1]be an uncountable compact set with dimHX=s(dimP X=s).Then,for any functionf ∈C(X) withf(x)≠ 0 for allx ∈X,there are functionsg,h ∈Hs+1(X)(g,h ∈Ps+1(X))such that

      ProofBecause of the packing dimensional case can be proved in the same way.Thus we only need to prove this proposition for the Hausdorffdimensional case.Define mapping Γf:C(X)→C(X) by

      It is easy to show that Γfis a continuous onto homomorphism,so combining Lemma 2.3 and Lemma 2.5,we see that is prevalent inC(X).Thusf·Hs+1(X)∩Hs+1(X) is also prevalent and of course is dense inC([0,1]).Similarly to the discussion in the proof of Proposition 3.1,there are functionsg,h2∈Hs+1(X) withg(x)>0,?x ∈Xsuch thatf·h2=g(note thath2(x)≠ 0 for anyx ∈X).Writing,we have thatf=g·h.Applying Lemma 2.6 again,we have thath ∈Hs+1(X).

      Theorem 3.3Letβ ∈[1,2]andf ∈C([0,1])withf(x)≠0,?x ∈[0,1].Then there exist functionsg,h ∈Hβ([0,1]) such that

      ProofIt suffices to prove this result forβ ∈(1,2).Without loss of generality,we may assume thatf(x)>0,?x ∈[0,1].The proof is divided into the following two cases:

      Case 1dimHGf([0,1])>β.

      Proposition 3.1 yields that we can find two functions,f1,f2∈C([0,1]),such that

      Now we show thatg,hare the desired functions.Obviouslyg,hare all continuous andf=g·h.Thus,we only need to show that the Hausdorffdimensions of their graphs are all equal toβ.In fact,onK,according to whetherf2>0 or not.Combining this and Lemma 2.7,we get that

      OutsideK,is locally Lipschitz,and so,by Lemma 2.9,

      Therefore,dimHGh([0,1])=β.Similarly,dimHGg([0,1])=β.

      Case 2dimHGf([0,1])≤β.

      Take a nonempty compact subsetMof [0,1]with dimHM=β-1.Then,by Proposition 3.2,there are positive valued functionsg1,g2∈C(M) such that

      As in the previous case,we extendg1,g2linearly to the whole space [0,1]and denote the extensional functions byrespectively,such that(x)>0,(x)>0 for anyx ∈[0,1].Set that

      Clearly,g,hare all continuous.OnM,h=g2andg=g1,so

      OutsideM,sinceare all locally Lipschitz,applying Lemma 2.7 and Lemma 2.9,we have that

      Therefore,dimHGh([0,1])=dimHGg([0,1])=β.

      The proof is complete.

      Proposition 3.4Letβ ∈[1,2]andf ∈C([0,1]) withf(x)≠0,?x ∈[0,1].If dimP Gf ≤β,then there exist functionsg,h ∈Pβ([0,1]) such that

      ProofThe proof of this proposition is quite similar to that used in Case 2 of Theorem 3.3,and so it is omitted.

      Lemma 3.5Let 00,there exists a Lipschitz functions ∈C([0,1]) withs(x)≠0 for allx ∈[a,b]satisfying that

      where Oscf(K) denotes the oscillation of functionfonK,i.e.,

      ProofLetLbe the family of all Lipschitz functions on[a,b].The well-known Weierstrass approximation theorem yields thatLis dense inC([a,b]).Furthermore,g·L={g·f|f ∈L}is also dense inC([a,b]).Then,for any 0

      where I denotes the constant function I(x)=1,?x ∈[a,b].Multiply a linear functions2tog·s1such that (g·s1·s2)(a)=(g·s1·s2)(b)=1.In fact,

      Set thats=s1·s2.Then we have that

      It is directly appeared thatsis a Lipschitz function and that (g·s)(a)=(g·s)(b)=1.Now we will show that the functionssatisfies the oscillation condition of the lemma by taking an appropriate?.

      If Oscf([a,b])=0,we only need to take that,wherec=|f|.Then,by Equation (3.1),it is easy to show that

      Theorem 3.6Letβ ∈[1,2],δ>0 andf ∈C([a,b]) be such thatf(a)=f(b)=0 andf(x)≠0,?x ∈(a,b).Then there are functionsg,h ∈Hβ([a,b]) such that

      ProofFirst,we take two sequences of real numbers,{an}n≥0and{bn}n≥0,in interval(a,b) with the following properties:

      (I) sequence{an} is strictly decreasing with

      (II) sequence{bn} is strictly increasing with

      (III)a0=b0.

      For eachn ≥0,we consider the decompositions of the restricted functionsf|[an+1,an]∈C([an+1,an]) andf|[bn,bn+1]∈C([bn,bn+1]).Applying Theorem 3.3,there exist functions

      Let{δn}n≥0be a sequence of positive real numbers such that

      Second,we define the functionsgandhby

      Note that,for anyn ≥0,g(an)=g(bn)=1,h(an)=f(an) andh(bn)=f(bn).

      Finally,we check that the functionsgandhsatisfy the conditions of the theorem.We claim thatgis continuous on [a,b].Clearly g is continuous on (a,b) from the construction ofgand the continuity ofWe only need to show thatgis continuous at the end pointsx ∈{a,b}.It suffices to show the casex=a,since the casex=bcan be proven in the same way.For any sequence{xk} withthere exists a sequencenktending to infinity such thatxk ∈(ank+1,ank].According to the equations (3.4),(3.5) and (3.6),and the continuity of functionf,we have thatThe continuity ofhcan be proven in a similar way.We conclude that functionsg,hare all continuous.

      Combining equation (3.3) and the definitions ofgandh,it is directly appeared thatf=g·h,g(a)=g(b)=1,h(a)=h(b)=0.By Lemma 2.9,equation (3.2) and the stability of the Hausdorffdimension,we have thatf,g ∈Hβ([0,1]).From the definitions of functionshandgand Equations (3.4),(3.5),(3.6) and (3.7),it is not difficult to see that,for any?>0,there existx0,x1,y0,y1∈(0,1)(without loss of generality,we may assume thaty0∈(an0+1,an0]andy1∈(an1+1,an1]for somen0,n1≥0) such that

      Therefore,we get that Oscg([a,b])≤3Oscf([a,b])+δand that Osch([a,b])≤3Oscf([a,b])+δ.

      Proposition 3.7Letβ ∈[1,2],δ>0 andf ∈C([a,b])be such thatf(a)=f(b)=0 andf(x)≠0,?x ∈(a,b).If dimP Gf ≤β,then there are functionsg,h ∈Pβ([a,b]) such that

      ProofThe proof of the proposition follows from an argument similar to that used in Theorem 3.6 (combined with Proposition 3.4).Thus the proof is here omitted.

      Proof of the Theorem 1.2Let Kerfbe the set of all zero points of functionf ∈C([0,1]),i.e.,let

      Then Kerfis a compact subset of [0,1].If Kerf=[0,1].Let 0

      If Kerf≠[0,1].Write

      It follows from Kerf≠[0,1]that at least one of(ai,bi)is nonempty.Without loss of generality,we may assume that{0,1} ?Kerf(otherwise,we can make a linear extension to a large interval such that functionfvanishes at the endpoints).Take a sequence{δn}n≥1of positive real numbers withδn →0 asn →∞.For each fixed nonempty interval (an,bn),we consider the product decomposition of the functionf|[an,bn].It follows from Theorem 3.6 that there exist functionshn,gn ∈Hβ([an,bn]) such that

      Then we defineh,gby

      We claim thatg,hare continuous.Obviously,gandhare both continuous on every interval(an,bn),thus we only need to show that,for anyx0∈Kerf,andHere we only prove the case wherethe proofs of the remaining cases are omitted.we can now divide things into the following two cases:

      Case 1There is an increasing sequence{yn}?Kerfsuch that

      Letkn=min{i ≥1 : (ai,bi)?[yn,x0])} (we use the convention that min?=∞).Thenkn →∞asn →∞.By the construction,the oscillation ofgon any interval [yn,x0]is not greater than 6Oscf([yn,x0])+2 supj≥kn δj,so we obtain that

      Caes 2(x0-ζ,x0)∩Kerf=?for someζ>0.

      In this case,we have that (x0-ζ,x0)?(ai,bi) for someiand thusx0=bi.Thenby the definition ofg|[ai,bi]=gi.

      We then conclude thatgandhare continuous on [0,1].

      Sinceg|Kerf ≡1 andh|Kerf ≡0 andhn,gn ∈Hβ([an,bn]),so by the countable stability of the Hausdorffdimension,we have thatg,h ∈Hβ([0,1]).

      The proof is now complete.

      Proof of the Theorem 1.3With the help of Propositions 3.4 and Proposition 3.7,an argument similar to that of the proof of Theorem 1.2 is obtained,so detailed proofs will be omitted here.

      Proof of the Proposition 1.4Without loss of generality,we suppose that dimP Gh

      for any open subintervalO ?[0,1],withU ∩O≠?.Applying Lemma 2.1 again,we have that dimP Gg(U)≥dimP Gf.The proof is complete.

      Conflict of InterestThe authors declare no conflict of interest.

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