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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo


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Volume 16, Issue 1

Issues

Volume 13 (2015)

Regularity of one-sided multilinear fractional maximal functions

Feng Liu
  • Corresponding author
  • College of Mathematics and Systems Sciences, Shandong University of Science and Technology, Qingdao 266590, China
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  • De Gruyter OnlineGoogle Scholar
/ Lei Xu
Published Online: 2018-12-31 | DOI: https://doi.org/10.1515/math-2018-0129

Abstract

In this paper we introduce and investigate the regularity properties of one-sided multilinear fractional maximal operators, both in continuous case and in discrete case. In the continuous setting, we prove that the one-sided multilinear fractional maximal operatorsMβ+andMβmap W1,p1 (ℝ)×· · ·×W1,pm (ℝ) into W1,q(ℝ) with 1 < p1, … , pm < ∞, 1 ≤ q < ∞ and 1/q=i=1m1/piβ, boundedly and continuously. In the discrete setting, we show that the discrete one-sided multilinear fractional maximal operators are bounded and continuous from 1(ℤ)×· · ·×1(ℤ) to BV(ℤ). Here BV(ℤ) denotes the set of functions of bounded variation defined on ℤ. Our main results represent significant and natural extensions of what was known previously.

Keywords: One-sided multilinear fractional maximal operators; Sobolev spaces; discrete maximal operators; bounded variation; continuity

MSC 2010: 42B25; 46E35

1 Introduction and the main results

Over the last several years a considerable amount of attention has been given to investigate the behavior of differentiability of maximal function. A good start was due to Kinnunen [1] who showed that the usual centered Hardy-Littlewood maximal function Mis bounded on W1,p(ℝd) for all 1 < p ≤ ∞, where W1,p(ℝd) is the first order Sobolev space, which consists of functions f ϵ Lp(ℝd), whose first weak partial derivatives Dif , i = 1, 2, … , d, belong to Lp(ℝd). We endow W1,p(ℝd) with the norm

f1,p=fLp(Rd)+fLp(Rd),

where ▽f = (D1f , D2f , … , D df ) is the weak gradient of f . Later on, Kinnunen’s result was extended to a local version in [2], to a fractional version in [3], to a multilinear version in [4, 5] and to a one-sided version in [6]. Meanwhile, the continuity of M : W1,pW1,p for 1 < p < ∞ was proved by Luiro in [7] and in [8] for its local version. Since Kinnunen’s result does not hold for p = 1, an important question was posed by Hajłasz and Onninen in [9]: Is the operator f ➝ |ΔMf| bounded from W1,1(ℝd) to L1(ℝd)? Progress on the above problem has been restricted to dimension d = 1. In 2002, Tanaka [10] showed that if fW1,1(ℝ), then the uncentered Hardy-Littlewood maximal function M̃f is weakly differentiable and

(M~f)L1(R)2fL1(R).(1.1)

This result was later sharpened by Aldaz and Pérez Lázaro [11] who proved that if f is of bounded variation on ℝ, then M̃f is absolutely continuous and its total variation satisfies

Var(M~f)Var(f).(1.2)

The above result implies directly (1.1) with constant C = 1 (also see [12] for a simple proof). In remarkable work [13], Kurka obtained that (1.1) and (1.2) hold for M(with constant C = 240, 004). Recently, Carneiro and Madrid [14] extended (1.1) and (1.2) to a fractional setting. Very recently, Liu and Wu [15] extended the partial result of [14] to a multilinear setting. For other interesting works related to this theory, we refer the reader to [16, 17, 18, 19, 20, 21, 22, 23, 24, 25], among others.

In this paper we focus on the regularity properties of the one-sided multilinear fractional maximal operators. More precisely, let m be a positive integer. For 0 ≤ β < m, we define the one-sided multilinear fractional maximal operators by Mβ+ and Mβ

Mβ+(f)(x)=sups>01smβi=1mxx+s|fi(y)|dyandMβ(f)(x)=supr>01rmβi=1mxrx|fi(y)|dy,

where f=(f1,,fm)with eachfiLloc1(R)When β = 0, the operator Mβ+(resp.,Mβreduces to the one-sided multilinear Hardy-Littlewood maximal operator M+ (resp., M). When m = 1, the operator Mβ+(resp.,Mβ)reduces to the one-sided fractional maximal operator Mβ+(resp.,Mβ).Especially, the one-sided Hardy-Littlewood maximal operator M+ (resp., M) corresponds to the operator Mβ+(resp.,Mβ)in this case β = 0.

As we all known, the reasons to study one-sided operators involve not only the generalization of the theory of the two-sided operators but also the close connection between the one-sided operators and two-sided operators. The one-sided Hardy-Littlewood maximal operator M+ can be seen as the special case of the ergodic maximal operator. Furthermore, there is a close connection between the one-sided fractional maximal functions and the well-known Riemann-Liourille fractional integral that can be viewed as the one-sided version of Riesz potential and the Weyl fractional integral (see [26]). It was known that both Mβ+and Mβare of type (p, q) for 1 < p < 1, 0 ≤ β < 1/p and q = p/(1 − ). Moreover, bothMβ+andMβare of weak type (1, q) for 0 ≤ β < 1 and q = 1/(1 − β). Observing that the following inequalities are valid:

Mβ+(f)(x)i=1mMβi+fi(x),xR,(1.3)

where = (f1, … , fm) and β=i=1mβiwithβi0(i=1,2,,m)By (1.3), the Lp bounds for Mβ+and Hölder’s inequality, one has

Mβ+(f)Lq(R)C(β,p1,,pm)i=1mfiLpi(R)(1.4)

for 1/q=i=1m1/piβ,provided that (i) β = 0, i1 mq ≤ ∞ and 1 < p1, … , pm ≤ ∞; (ii) 0 < β < m, 1 ≤ q < ∞ and 1 < p1, … , pm < ∞. The same result holds forMβ.

The investigation on the regularity of one-sided maximal operator began with Tanaka [10] in 2002 when he observed that if f ∈ W1,1(ℝ), then the distributional derivatives of M+f and Mf are integrable functions, and

(M+f)L1(R)fL1(R)and(Mf)L1(R)fL1(R).

By a combination of arguments in [10, 12], both M +f and M f are absolutely continuous on ℝ. Recently, Liu and Mao [6] proved that both M+ and M are bounded and continuous on W1,p(ℝ) for 1 < p < ∞. Very recently, Liu [27] extended the main results of [6] to the fractional case. More precisely, Liu proved the following result.

Theorem A

([27]). Let 1 < p < ∞, 0 ≤ β < 1/p and q = p/(1 − ). Then both Mβ+andMβmap W1,p(ℝ) into W1,q(ℝ) boundedly and continuously. Moreover, if f ϵ W1,p(ℝ), then

|(Mβ+f)(x)|Mβ+f(x)and|(Mβf)(x)|Mβf(x)

for almost every x ∈ ℝ.

In this paper we shall extended Theorem A to the multilinear case. We now formulate our main results as follows.

Theorem 1.1

Let 1<p1,,pm<,0β<i=1m1/pi,1/q=i=1m1/piβand1q<.ThenMβ+maps W1,p1 (ℝ) ×· ··× W1,pm (ℝ) into W1,q(ℝ) boundedly and continuously. Especially, if f ➝= (f1, … , fm) with each fi ∈ W1,pi (ℝ), then the weak derivative (Mβ+(f))exists almost everywhere. More precisely,

|(Mβ+(f))(x)|j=1mMβ+(fj)(x)

for almost every x ϵ, where fj=(f1,,fj1,fj,fj+1,,fm).Moreover,

Mβ+(f)1,qC(β,p1,,pm)i=1mfi1,pi.

The same results hold forMβ.

Theorem 1.2

Let f=(f1,,fm)with each fiLpi(R)for1<p1,,pm<and1β<i=1m1/pi.

  • (i)

    Then the weak derivative (Mβ+(f))exists almost everywhere. Precisely,

|(Mβ+(f))(x)|C(m,β)Mβ1+(f)(x)

for almost every x ϵ ℝ.

  • (ii)

    Let 1/q=i=1m1/piβ+1.Then

(Mβ+(f))Lq(R)C(m,β,p1,,pm)i=1mfiLpi(R).

The same results hold for Mβ.

Remark 1.1

Theorem 1.1 extends Theorems 1.1-1.2 in [6], which correspond to the case m = 1 and β = 0. Theorem 1.1 also extends Theorem A, which corresponds to the case m = 1.

On the other hand, the investigation of the regularity properties of discrete maximal operators has also attracted the attention of many authors (see [6, 14, 16, 27, 28, 29, 30, 31, 32, 33] for example). Let us recall some notation and relevant results. For 1 ≤ p < ∞ and a discrete function f : ℤ ➝ ℝ, we define the p-norm and the -norm of f by ∥f∥ℓp(ℤ) = (ΣnϵZ |f(n)|p)1/p and f(Z)=supnZ|f(n)|.We also define the first derivative of f by f '(n) = f(n + 1) − f(n) for any n ϵ ℤ. For f : ℤ ➝ ℝ, we define the total variation of f by

Var(f)=f1(Z).

We denote by BV(ℤ) the set of all functions f : ℤ ➝ ℝ satisfying Var(f) < ∞.

In 2011, Bober et al. [28] first studied the regularity properties of discrete Hardy-Littlewood maximal operators and proved that

Var(M~f)Var(f)(1.5)

and

Var(Mf)(2+146315)f1(Z).(1.6)

Here M (resp.,M~)denotes the discrete centered (resp., uncentered) Hardy-Littlewood maximal operator, which are defined by

Mf(n)=suprN12r+1k=rr|f(n+k)|andM~f(n)=supr,sN1r+s+1k=rs|f(n+k)|,

where ℕ = {0, 1, 2, 3, … , }. We note that inequality (1.5) is sharp. It was known that inequality Var(Mf)≤ 294, 912, 004Var(f ) was established by Temur in [32]. Inequality (1.6) is not optimal, and it was asked in [28] whether the sharp constant for (1.6) is in fact C = 2, which was addressed by Madrid in [31]. Recently, Carneiro and Madrid [14] extended (1.5) to the fractional setting. They also pointed out that the discrete fractional maximal operators MβandM~βare bounded and continuous from 1(ℤ) to BV(ℤ) (also see [29, 34]). Here M β and M~βare the discrete centered and uncentered fractional maximal operators, which are defined by

Mβf(n)=suprN1(2r+1)1βk=rr|f(n+k)|andM~βf(n)=supr,sN1(r+s+1)1βk=rs|f(n+k)|.

Our second aim of this paper is to consider the discrete one-sided multilinear fractional maximal operators

Mβ+(f)(n)=supsN1(s+1)mβi=1mk=0s|fi(n+k)|,Mβ(f)(n)=suprN1(r+1)mβi=1mk=r0|fi(n+k)|,

where 0β<mandf=(f1,,fm)with each fiLloc1(Z).When β = 0, the operators Mβ+andMβreduce to the discrete one-sided multilinear Hardy-Littlewood maximal operators M+ and M, respectively. When m = 1, the operators Mβ+andMβreduce to the discrete one-sided fractional maximal operators Mβ+andMβrespectively. Particularly, the discrete one-sided Hardy-Littlewood maximal operators M+ and M correspond to the special case of Mβ+andMβwhen β = 0, respectively. Recently, Liu and Mao [6] proved that both +−M+ and M are bounded and continuous from 1(ℤ) to BV(ℤ). Moreover, if f ∈ BV(ℤ), then

max{Var(M+f),Var(Mf)}Var(f).(1.7)

We notice that the constant C = 1 in inequality (1.7) is sharp. Very recently, Liu [27] pointed out that Mβ+and Mβare not bounded from BV(ℤ) to BV(ℤ) when 0 < β < 1. However, Liu established the following result.

Theorem B

([27]). Let 0 ≤ β < 1. ThenMβ+is bounded and continuous from ℓ1(ℤ) to BV(ℤ).Moreover, if f ϵ ℓ1(ℤ), then

Var(Mβ+f)2f1(Z),

and the constant C = 2 is the best possible. The same results hold for Mβ.

In this paper we shall extended Theorem B to the following.

Theorem 1.3

Let 0 ≤ β < m. Then Mβ+is bounded and continuous from ℓ1(ℤ) ×· ··× 1(ℤ) to BV(ℤ). Moreover, if f=(f1,,fm)with each fi ϵ ℓ1(ℤ), then

Var(Mβ+(f))2mi=1mfi1(Z).

The same results hold for Mβ.

Remark 1.2

When m = 1, Theorem 1.3 implies Theorem B.

The rest of this paper is organized as follows. Section 2 contains some notation and preliminary lemmas, which can be used to prove the continuity part in Theorem 1.1. Motivated by the ideas in [5, 7], we give the proofs of Theorems 1.1-1.2 in Section 3. Finally, we prove Theorem 1.3 in Section 4. It should be pointed out that the proof of the boundedness part in Theorem 1.3 is based on the method of [31]. The proof of the continuity part in Theorem 1.3 relies on the previous boundedness result and a useful application of the Brezis-Lieb lemma in [35]. Throughout this paper, the letter C, sometimes with additional parameters, will stand for positive constants, not necessarily the same one at each occurrence but independent of the essential variables.

2 Preliminary notation and lemmas

In this section we shall introduce some notation and lemmas, which play key roles in the proof of the continuity part in Theorem 1.1. Let A ⊂ ℝ and r ϵ ℝ. We define

d(r,A):=infaA|ra|andA(λ):={xR:d(x,A)λ}forλ0.

Denote ∥f ∥p,A by the Lp-norm of A for all measurable sets A ⊂ ℝ. Let f=(f1,,fm)with each fi ϵ Lpi (ℝ) for 1 < pi < ∞ and 1 ≤ < ∞ with 1/q=i=1m1/piβ.In what follows, we only consider the operatorMβ+and the other case is analogous. Fix x ∈ ℝ, we define the set Rβ+(f)(x)by

Rβ+(f)(x):={s0:Mβ+(f)(x)=lim supk1skmβi=1mxx+sk|fi(y)|dyforsomesk>0,sks}.

We also define the function ux,f,β+:[0,)Rby

ux,f,β+(0)={i=1m|fi(x)|,ifβ=0;0,if0<β<m,ux,f,β+(s)=1smβi=1mxx+s|fi(y)|dyfors(0,).

We notice that the followings are valid.

  1. ux,f,β+is continuous on (0,∞) for all x ∈ ℝ and at r = 0 for almost everywhere x ∈ ℝ ;

  2. limsux,f,β+(s)=0sinceux,f,β+(s)i=1mfiLpi(R)s1/q

  3. The set Rβ+(f)(x)is nonempty and closed for any x ∈ ℝ;

  4. Almost every point is a Lebesgue point.

From the above observations we have

Mβ+(f)(x)=ux,f,β+(s)if0<sRβ+(f)(x),xR,

Mβ+(f)(x)=ux,f,β+(0)foralmosteveryxRsuchthat0Rβ+(f)(x).

Lemma 2.1

Let 1 < p1, … , pm < ∞ and 1 ≤ q < ∞ with 1/q=i=1m1/piβ.Letfj=(f1,j,,fm,j)and f̄ = (f1, … , fm) such that fi,jfi in Lpi (ℝ) when j ➝ ∞. Then, for all ℝ > 0 and λ > 0, it holds that

limj|{x(R,R):Rβ+(fj)(x)Rβ+(f)(x)(λ)}|=0.(2.1)

Proof. Without loss of generality, we may assume that all fi,j ≥ 0 and fi ≥ 0. By the similar argument as in the proof of Lemma 2.2 in [7], we can conclude that the set {xR:Rβ+(fj)(x)Rβ+(f)(x)(λ)}is measurable for any j ϵ ℤ. Let λ > 0 and R > 0. We first claim that for almost every x ∈ (−R, R), there exists γ(x) ϵ ℕ \ {0} such that

ux,f,β+(s)<Mβ+(f)(x)1γ(x)whend(s,Rβ+(f)(x))>λ.(2.2)

Otherwise, for almost every x ϵ (−R, R), there exists a bounded sequence of radii {sk}k=1such that

limkux,f,β+(sk)=Mβ+(f)(x)andd(sk,Rβ+(f)(x))>λ.

We can choose a subsequence {rk}k=1of{sk}k=1such that rks as k ➝ ∞. Then we have sRβ+(f)(x)and d(s,Rβ+(f)(x))λ,which is a contradiction. Thus (2.2) holds. Given ϵ ∈ (0, 1), (2.2) yields that there exists γ = γ(R, λ, ϵ) ∈ ℕ \ {0} and a measurable set E with |E| < ϵ such that

(R,R){xR:ux,f,β+(s)<Mβ+(f)(x)γ1ifd(s,Rβ+(f)(x))>λ}E.

Notice that

Mβ+(f)(x)ux,f,β+(s)|Mβ+(fj)(x)Mβ+(f)(x)|+|ux,fj,β+(s)ux,f,β+(s)|+Mβ+(fj)(x)ux,fj,β+(s).

It yields that

{xR:ux,f,β+(s)<Mβ+(f)(x)γ1ifd(s,Rβ+(f)(x))>λ}A1,jA2,jA3,j,

where

A1,j:={xR:|Mβ+(fj)(x)Mβ+(f)(x)|(4γ)1},A2,j:={xR:|ux,fj,β+(s)ux,f,β+(s)|(2γ)1forsomessuchthatd(s,Rβ+(f)(x))>λ},A3,j:={xR:ux,fj,β+(s)<Mβ+(fj)(x)(4γ)1ifd(s,Rβ+(f)(x))>λ}.

Hence,

(R,R)A1,jA2,jA3,jE.(2.3)

Let Ā be the set of all points x such that x is a Lebesgue point of all fj. Note that |R \ Ā| = 0 and A3,j Ā {xR:Rβ+(fj)(x)Rβ+(f)(x)(λ)}.This together with (2.3) implies

{x(R,R):Rβ+(fj)(x)Rβ+(f)(x)(λ)}A1,jA2,jE(RA¯).

It follows that

|{x(R,R):Rβ+(fj)(x)Rβ+(f)(x)(λ)}||A1,j|+|A2,j|+ϵ.(2.4)

We can write

|Mβ+(fj)(x)Mβ+(f)(x)|sups>01smβ|i=1mxx+sfi,j(y)dyi=1mxx+sfi(y)dy|l=1msups>01smβμ=1l1xx+sfμ(y)dyν=l+1mxx+sfν,j(y)dyxx+s|fl,j(y)fl(y)|dyl=1mMβ+(fjl)(x)(2.5)

for any x ϵ ℝ, where fjl=(f1,,fl1,fl,jfl,fl+1,j,,fm,j).From (2.5) we have

|A1,j||{xR:l=1mMβ+(fjl)(x)(4γ)1}|l=1m|{xR:Mβ+(fjl)(x)(4mγ)1}|(4mγ)ql=1mMβ+(fjl)Lq(R)q.(2.6)

Since fi,jfi in Lpi (ℝ) as j ➝ ∞, then there exists N0 = N0(ϵ, γ ) ∈ ℕ \ {0} such that

fi,jfiLpi(R)<ϵγandfi,jLpi(R)fiLpi(R)+1,jN0.(2.7)

(2.7) together with (2.6) and (1.4) yields that

|A1,j|C(m,q,β,p1,,pm,f)ϵ,jN0.(2.8)

On the other hand, one can easily check that

|ux,fj,β+(s)ux,f,β+(s)|l=1mMβ+(fjl)(x),s>0.

This together with the argument similar to those used in deriving (2.8) implies

|A2,j|C(m,q,β,p1,,pm,f)ϵ,jN0.(2.9)

It follows from (2.4), (2.8) and (2.9) that

|{x(R,R):Rβ+(fj)(x)Rβ+(f)(x)(λ)}|C(m,q,β,p1,,pm,f)ϵ,jN0,

which gives (2.1) and completes the proof of Lemma 2.1.

We now define the Hausdorff distance between two sets A and B by

π(A,B):=inf{δ>0:AB(δ)andBA(δ)}.

The following result can be obtained by Lemma 2.1 and a similar argument as in the proof of Corollary 2.3 in [7], we omit the details.

Lemma 2.2

Let f=(f1,,fm)with each fi ∈ Lpi (ℝ) for 1 < p1, … , pm < ∞. Let 1 ≤ q < ∞ and 1/q = i=1m1/piβ.Then, for all ℝ > 0 and λ > 0, we have

limh0|{x(R,R):π(Rβ+(f)(x),Rβ+(f)(x+h))>λ}|=0.

The following result presents some formulas for the derivatives of the one-sided multilinear fractional maximal functions, which play the key roles in the proof of the continuity part in Theorem 1.1.

Lemma 2.3

Let f̄ = (f1, … , fm) with each fi ϵ W1,pi (ℝ) for 1 < pi < ∞. Let 1 ≤ q < ∞ and 1/q=i=1m1/piβ.

Then, for almost every x ∈, we have

(Mβ+(f))(x)=l=1m1smβ1jmjlxx+s|fj(y)|dyxx+s|fl|(y)dyforall0<sRβ+(f)(x);(2.10)

(Mβ+(f))(x)=l=1m|fl|(x)1jmjl|fj(x)|,ifβ=0and0Rβ+(f)(x),0,if0<β<mand0Rβ+(f)(x).(2.11)

Proof. We may assume that all fi ≥ 0 since |f| ϵ W1,p(ℝ) if f ∈ W1,p(ℝ) with 1 < p < ∞. By the boundedness part in Theorem 1.1 we see that Mβ+(f)W1,q(R).Invoking Lemma 2.2, we can choose a sequence {sk}k=1sk>0such that limk➝∞ sk = 0 and limkπ(Rβ+(f)(x),Rβ+(f)(x+sk))=0for almost every x ϵ (−R, R). For 1 ≤ im and h ϵ ℝ, we set

fhi(x)=fτ(h)i(x)fi(x)handfτ(h)i(x)=fi(x+h).

It was known that

fτ(sk)ifiLpi(R)0ask,fski(fi)Lpi(R)0ask,

M+(fτ(sk)ifi)Lpi(R)0ask,M+(fskifi)Lpi(R)0ask,(Mβ+(f))sk(Mβ+(f))Lq(R)0ask.

Here (Mβ+(f))sk(x)=1sk(Mβ+(f)(x+sk)Mβ+(f)(x)).Furthermore, there exists a subsequence {hk}k=1of{sk}k=1and a measurable set A1 (−R, R) with |(−R, R)\A1| = 0 such that

  1. fτ(hk)i(x)fi(x),fhki(x)fi(x),M+(fτ(hk)ifi)(x)0,M+(fhkifi)(x)0and(Mβ+(f))hk(x)(Mβ+(f))(x)when k ➝ ∞ for any x ϵ A1 and 1 ≤ im;

  2. limkπ(Rβ+(f)(x),Rβ+(f)(x+hk))=0for any x ∈ A1.

Let

A2:=k=1{xR:Mβ+(f)(x+hk)ux+hk,f,β+(0)},A3:=k=1{xR:Mβ+(f)(x+hk)=ux+hk,f,β+(0)if0Rβ+(f)(x+hk)},A4:={xR:Mβ+(f)(x)=ux,f,β+(0)if0Rβ+(f)(x)}.

It is obvious that |(−R, R)\Aj| = 0 for j = 2, 3, 4. Let x ϵ A1 ∩ A2 ∩ A3 ∩ A4 be a Lebesgue point of all fiandfi.Fix sRβ+(f)(x), there exists radii rkRβ+(f)(x+hk)such that limk➝∞ rk = s. We consider the following two cases:

Case A (s > 0). Without loss of generality we may assume that all rk > 0. Then

(Mβ+(f))(x)=limk1hk(Mβ+(f)(x+hk)Mβ+(f)(x))limk1hk1rkmβ(i=1mx+hkx+hk+rkfi(y)dyi=1mxx+rkfi(y)dy)=l=1mlimk1rkmβμ=1l1xx+rkfμ(y)dyν=l+1mxx+rkfτ(hk)ν(y)dyxx+rkfhkl(y)dy.(2.12)

Since fτ(hk)νχ(x,x+rk)fνχ(x,x+s)and fhklχ(x,x+rk)flχ(x,x+s)in L1(ℝ) as k ➝ ∞. Then (2.12) yields that

(Mβ+(f))(x)l=1m1smβ1jmjlxx+sfj(y)dyxx+sfl(y)dy.(2.13)

On the other hand,

(Mβ+(f))(x)=limk1hk(Mβ+(f)(x+hk)Mβ+(f)(x))limk1hk1smβ(i=1mx+hkx+hk+sfi(y)dyi=1mxx+sfi(y)dy)=l=1mlimk1smβμ=1l1xx+sfμ(y)dyν=l+1mxx+sfτ(hk)ν(y)dyxx+sfhkl(y)dy=l=1m1smβ1jmjlxx+sfj(y)dyxx+sfl(y)dy.(2.14)

Combining (2.14) with (2.13) yields that (2.10) holds for almost every x ∈ (−R, R).

Case B (s = 0). We shall discuss the following two cases:

  • (i)

    When 0 < β < m. SinceMβ+(f)(x)=0,then all fi(y 0 for almost every y ϵ (x,∞). ThusMβ+(f)(y)0for yx. It follows that

(Mβ+(f))(x)=limk1hk(Mβ+(f)(x+hk)Mβ+(f)(x))=0.

This yields that (2.11) holds for almost every x ∈ (−R, R) in this case 0 < β < m.

  • (ii)

    When β = 0. We notice that

limk1hk(i=1mfi(x+hk)i=1mfi(x))=l=1mlimkfhkl(x)μ=1l1fμ(x)ν=l+1mfν(x+hk)=l=1mfl(x)1jmjlfj(x).(2.15)

It follows that

(Mβ+(f))(x)=limk1hk(Mβ+(f)(x+hk)Mβ+(f)(x))limk1hk(i=1mfi(x+hk)i=1mfi(x))=l=1mfl(x)1jmjlfj(x).(2.16)

Below we estimate the upper bound of (Mβ+(f))(x).If there exists k0 ϵ ℕ\ {0} such that sk > 0 for any kk0, then, by the argument similar to those used in deriving (2.12),

(Mβ+(f))(x)l=1mlimk(μ=1l11rkxx+rkfμ(y)dy)(ν=l+1m1rkxx+rkfτ(hk)ν(y)dy)×(1rkxx+rkfhkl(y)dy).(2.17)

Since

|limk1rkxx+rkfτ(hk)ν(y)dyfν(x)|limk1rkxx+rk|fτ(hk)ν(y)fν(y)|dylimkM+(fτ(hk)νfν)(x)=0.

It follows that

limk1rkxx+rkfτ(hk)ν(y)dy=fν(x).(2.18)

Similarly,

limk1rkxx+rkfhkl(y)dy=fl(x).(2.19)

It follows from (2.17)-(2.19) that

(Mβ+(f))(x)l=1mfl(x)1jmjlfj(x).(2.20)

If sk = 0 for infinitely many k, then, by (2.15) we have

(Mβ+(f))(x)=limk1hk(Mβ+(f)(x+hk)Mβ+(f)(x))=limk1hk(i=1mfi(x+hk)i=1mfi(x))=l=1mfl(x)1jmjlfj(x).

This together with (2.16) and (2.20) yields that (2.11) holds for almost every x ∈ (−R, R) in the case β = 0. Since R was arbitrary, this proves Lemma 2.3.

3 Proofs of Theorems 1.1-1.2

In this section we shall prove Theorems 1.1-1.2. Let us begin with the proof of Theorem 1.1.

Proof of Theorem 1.1. We only prove Theorem 1.1 for Mβ+and the other case is analogous. Let {sk}k⪰1 be an enumeration of positive rational numbers. Then we can write

Mβ+(f)(x)=supk11skmβi=1mxx+sk|fi(y)|dy.

Define the family of operators {Tk}k≥1 by

Tk(f)(x)=max1ik1simβj=1mxx+si|fj(y)|dy.

Fix x, h ϵ ℝ, one has

|Tk(f)(x+h)Tk(f)(x)|max1ik1simβ|j=1mx+hx+h+si|fj(y)|dyj=1mxx+si|fj(y)|dy|l=1mmax1ik1simβμ=1l1xx+si|fμ(y)|dyν=l+1mxx+si|fτ(h)ν(y)|dyxx+si|fτ(h)l(y)fl(y)|dy.

It follows that

(Tk(f))(x)l=1mMβ+(fl)(x)(3.1)

for almost every x ∈ ℝ, where fl=(f1,,fl1,fl,fl+1,,fm).Here we used the fact that ||fj|'(x)| = |f'(x)| for almost every x ϵ ℝ. By (3.1), (1.4) and Minkowski’s inequality, we obtain

Tk(f)1,qTk(f)Lq(R)+(Tk(f))Lq(R)Mβ+(f)Lq(R)+l=1mMβ+(fl)Lq(R)C(β,p1,,pm)(i=1mfiLpi(R)+l=1mflLpl(R)1jmjlfjLpj(R))C(m,β,p1,,pm)i=1mfi1,pi.

Therefore, {Tk(f)}is a bounded sequence in W1, q(ℝ) which converges to Mβ+(f)pointwise. The weak compactness of Sobolev spaces implies that Mβ+(f)W1,q(R),Tk(f)converges to Mβ+(f)weakly in Lq(ℝ) and (Tk(f))converges to (Mβ+(f))weakly in Lq(ℝ). This together with (3.1) yields that

|(Mβ+(f))(x)|l=1mMβ+(fl)(x)(3.2)

for almost every x ∈ ℝ. It follows from (3.2) and (1.4) that

Mβ+(f)1,q=Mβ+(f)Lq(R)+(Mβ+(f))Lq(R)C(m,β,p1,,pm)i=1mfi1,pi.

This completes the boundedness part of Theorem 1.1.

We now prove the continuity forMβ+by employing the idea in [20]. Let β, m, p1, … , pm, q be given as in

Theorem 1.1. Let = (f1, … , fm) with each fi ϵ W1,pi (ℝ) and j = (f1,j , … , fm,j) such that fi,jfi in W1,pi (ℝ) when j ➝ ∞. We get from (2.5) that

|Mβ+(fj)(x)Mβ+(f)(x)|l=1mMβ+(fjl)(x)(3.3)

for any x ∈ ℝ, where fjlis given as in (2.5). (3.3) together with (1.4) implies that

Mβ+(fj)Mβ+(f)Lq(R)l=1mMβ+(fjl)Lq(R)C(m,β,p1,,pm)l=1mfl,jflLpl(R)μ=1l1fμLpμ(R)ν=l+1mfν,jLpν(R).

It follows that

Mβ+(fj)Mβ+(f)Lq(R)0whenj.

Hence, to prove the continuity forMβ+,it suffices to show that

(Mβ+(fj))(Mβ+(f))Lq(R)0whenj.(3.4)

Below we prove (3.4). We may assume that all fi,j ≥ 0 and fi ≥ 0. For 1 ≤ lm, we set fl=(f1,,fl1,fl,fl+1,,fm).Fix ϵ ∈ (0, 1). We can choose R > 0 such that l=1mMβ+(fl)q,B1<ϵwith B1 = (−1, −R)(R,∞). The absolute continuity implies that there exists η > 0 such thatl=1mMβ+(fl)q,B<ϵfor any measurable subset B of (−R, R) with |B| < η. As already observed, for almost every x ∈ ℝ, the function ux,fl,β+is uniformly continuous on [0,∞). Therefore, for almost every x ∈ ℝ, the function l=1mux,fl,β+is uniformly continuous on [0,∞). We can find δ(x) > 0 such that

|l=1mux,fl,β+(s1)l=1mux,fl,β+(s2)|<R1/qϵif|s1s2|δ(x).

We can write (−R, R) as

(R,R)=(k=1{x(R,R):δ(x)>1k})N,

where |N| = 0. We can choose δ > 0 such that

|{x(R,R):|l=1mux,fl,β+(s1)l=1mux,fl,β+(s2)|R1/qϵforsomes1,s2with|s1s2|δ}|=:|B2|<η2.

By Lemma 2.1, there exists N1 ∈ ℕ \ {0} such that

|{x(R,R):Rβ+(fj)(x)Rβ+(f)(x)(δ)}|=:|Bj|<η2jN1.

Fix jN1. Let fl,j=(f1,j,,fl1,j,fl,j,fl+1,j,,fm,j)and sRβ+(fj)(x).We consider the following two cases:

  • (i)

    s > 0. We can write

|ux,fl,j,β+(s)ux,fl,β+(s)|=1smβ|1μmμlxx+sfμ,j(y)dyxx+sfl,j(y)dy1μmμlxx+sfμ(y)dyxx+sfl(y)dy|μ=1l1Mβ+(Fμ,j)(x)+ν=l+1mMβ+(Gν,j)(x)+Mβ+(Hl,j)(x)=:Gl,j(x),(3.5)

where

Fμ,j=(f1,,fμ1,fμ,jfμ,fμ+1,j,,fl1,j,fl,j,fl+1,j,,fm,j),Gν,j=(f1,,fl1,fl,fl+1,,fν1,fν,jfν,fν+1,j,,fm,j),Hl,j=(f1,,fl1,fl,jfl,fl+1,j,,fm,j).

  • (ii)

    s = 0. If 0 < β < m, |ux,fl,j,β(s)ux,fl,β(s)|=0.If β = 0, then we have

|ux,fl,j,β+(s)ux,fl,β+(s)|μ=1l1(l1=1μ1fl1(x))(fμ,j(x)fμ(x))(l2=μ+1l1fl2,j(x))|fl,j(x)|(l3=l+1mfl3,j(x))+ν=l+1m(l1=1l1fl1(x))|fl(x)|(l2=l+1ν1fl2(x))|fν,j(x)fν(x)|(l3=ν+1mfl3,j(x))+(l1=1l1fl1(x))|fl,j(x)fl(x)|(l2=l+1mfl2,j(x)).

This together with (3.5) and the Lebesgue differentiation theorem leads to

|ux,fl,j,β+(s)ux,fl,β+(s)|Gl,j(x)(3.6)

for almost every x ∈ ℝ and sRβ+(fj)(x).By (3.6) and Lemma 2.3, we obtain

|(Mβ+(fj))(x)(Mβ+(f))(x)|=|l=1mux,fl,j,β(s1)l=1mux,fl,β(s2)||l=1mux,fl,j,β(s1)l=1mux,fl,β(s1)|+|l=1mux,fl,β(s1)l=1mux,fl,β(s2)|l=1mGl,j(x)+|l=1mux,fl,β(s1)l=1mux,fl,β(s2)|(3.7)

for almost every x ∈ ℝ and any s1Rβ+(fj)(x)ands2Rβ+(f)(x).On can easily check that

limjGl,jLq(R)=0,1lm.

It follows that there exists N2 ∈ ℕ \ {0} such l=1mGl,jLq(R)<ϵfor any jN2.

If x B1 ∪ B2 ∪ Bj, we can choose s1Rβ+(fj)(x)ands2Rβ+(f)(x)such that |s1s2|≤ δ and

|l=1mux,fl,β(s1)l=1mux,fl,β(s2)|<|R|1/qϵ.

On the other hand, we have that for any l=1,2,,m,s1Rβ+(fj)(x)ands2Rβ+(f)(x),

|l=1mux,fl,β(s1)l=1mux,fl,β(s2)|2l=1mMβ+(fl)(x).

Note that |B2Bj| < η for any jN1. Thus we get from (3.7) that

(Mβ+(fj))(Mβ+(f))Lq(R)l=1mGl,jLq(R)+|R|1/qϵq,(R,R)+2l=1mMβ+(fl)q,B1B2BjCϵ,

for any j ≥ max{N1, N2}, which leads to

limj(Mβ+(fj))(Mβ+(f))Lq(R)=0.

This yields (3.4) and completes the proof of Theorem 1.1.

Proof of Theorem 1.2. The proof is similar to the proof of Theorem 2.3 in [5]. We omit the details.

4 Proof of Theorem 1.3

We only prove Theorem 1.3 for Mβ+and the other case is analogous.

  1. Step 1: proof of the boundedness forMβ+.We shall adopt the method in [31] to prove the boundedness forMβ+.

Let f=(f1,,fm)with each fi ∈ ℓ1(ℤ). Without loss of generality, we may assume fi ≥ 0. For convenience, let Г(x) = (x + 1)βm − (x + 2)βm for any x ≥ 0. One can easily check that Г(x) is decreasing on [0,∞) and Σn∈N Г(n) = 1. Since all fi ∈ ℓ1(ℤ), then, for any n ∈ ℤ, there exists sn ∈ ℕ such that Mβ+(f)(n)=Asn(f)(n),where

As(f)(n)=(s+1)βmi=1mk=0sfi(n+k)

for any s ∈ ℕ and n ∈ ℤ. Let

X+={nZ:Mβ+(f)(n+1)>Mβ+(f)(n)}andX={nZ:Mβ+(f)(n)Mβ+(f)(n+1)}.

Then we can write

Var(Mβ+(f))=nX+(Mβ+(f)(n+1)Mβ+(f)(n))+nX(Mβ+(f)(n)Mβ+(f)(n+1))nX+(Asn+1(f)(n+1)Asn+1+1(f)(n))+nX(Asn(f)(n)Asn+1(f)(n+1)).(4.1)

Fix n ∈ ℤ, by direct computations we obtain

Asn+1(f)(n+1)Asn+1+1(f)(n)=(sn+1+1)βmi=1mk=0sn+1fi(n+1+k)(sn+1+2)βmi=1mk=0sn+1+1fi(n+k)l=1m((sn+1+1)βmk=0sn+1fl(n+1+k)(sn+1+2)βmk=0sn+1+1fl(n+k))×μ=1l1k=0sn+1+1fμ(n+k)ν=lmk=0sn+1fν(n+1+k).(4.2)

Since

(sn+1+1)βmk=0sn+1fl(n+1+k)(sn+1+2)βmk=0sn+1+1fl(n+k)(sn+1+1)βmkZfl(k)χ[n+1,n+sn+1+1](k)(sn+1+2)βmkZfl(k)χ[n,n+sn+1+1](k)kZfl(k)Γ(sn+1)χ[n+1,n+sn+1+1](k)kZfl(k)Γ(kn1)χ(n,)(k).(4.3)

Combining (4.3) with (4.2) yields that

Asn+1(f)(n+1)Asn+1+1(f)(n)l=1m1jmjlfj1(Z)(kZfl(k)Γ(kn1)χ(n,)(k)).(4.4)

On the other hand, one finds

Asn(f)(n)Asn+1(f)(n+1)=(sn+1)βmi=1mk=0snfi(n+k)(sn+2)βmi=1mk=0sn+1fi(n+1+k)=l=1m((sn+1)βmk=0snfl(n+k)(sn+2)βmk=0sn+1fl(n+1+k))×μ=1l1k=0sn+1fμ(n+1+k)ν=l+1mk=0snfν(n+k).

It follows that

Asn(f)(n)Asn+1(f)(n+1)l=1m((sn+1)βmkZfl(k)χ[n,n+sn](k)(sn+2)βmkZfl(k)χ[n+1,n+sn+2](k))×1jmjlfj1(Z)l=1m1jmjlfj1(Z)(kZfl(k)Γ(sn)χ[n+1,n+sn+1](k)+fl(n))l=1m1jmjlfj1(Z)(kZfl(k)Γ(kn1)χ(n,)(k)+fl(n)).(4.5)

(4.1) and (4.4)-(4.5) imply that

Var(Mβ+(f))l=1m1jmjlfj1(Z)(kZfl(k)(nX+n<kΓ(kn1)+nXn<kΓ(kn1))+nXfl(n))l=1m1jmjlfj1(Z)(kZfl(k)n<kΓ(kn1)+fl1(Z))2m1jmfj1(Z).

Step 2: proof of the continuity for Mβ+.Letf=(f1,,fm)with each fj ∈ ℓ1(ℤ) and fj=(f1,j,,fm,j)such that fi,jfi in 1(ℤ) as j ➝ 1. By the boundedness part in Theorem 1.3, we know that (Mβ+(f))1(Z).Without loss of generality we may assume that all fi,j ≥ 0 and fi ≥ 0 since |fj| − |f ||≤| fj |. We want to show that

limj(Mβ+(fj))(Mβ+(f))1(Z)=0.(4.6)

Given ϵ ∈ (0, 1), there exists N1=N1(ϵ,f)>0such that

fi,jfi1(Z)<ϵ(4.7)

and

fi,j1(Z)fi,jfi1(Z)+fi1(Z)<fi1(Z)+1(4.8)

for any jN1 and all 1 ≤ im. We get from (4.7)-(4.8) that

|Mβ+(fj)(n)Mβ+(f)(n)|supsN(s+1)βm|i=1mk=0sfi,j(n+k)i=1mk=0sfi(n+k)|supsN(s+1)βml=1mk=0s|fi,j(n+k)fi(n+k)|μ=1l1k=0sfμ(n+k)ν=l+1mk=0sfν,j(n+k)l=1mfl,jfl1(Z)μ=1l1fμ1(Z)ν=l+1mfν,j1(Z)c(f)ϵ

for any n ∈ ℤ and jN1, which implies that Mβ+(fj)Mβ+(f)pointwise as j ➝ ∞ and

limj(Mβ+(fj))(n)=(Mβ+(f))(n)(4.9)

for all n ∈ ℤ. By the fact that (Mβ+(f))1(Z)and the classical Brezis-Lieb lemma in [35], to prove (4.6), it suffices to show that

limj(Mβ+(fj))1(Z)=(Mβ+(f))1(Z).(4.10)

By (4.9) and Fatou’s lemma, one finds

(Mβ+(f))1(Z)lim infj(Mβ+(fj))1(Z).

Thus, to prove (4.10), it suffices to show that

lim supj(Mβ+(fj))1(Z)(Mβ+(f))1(Z).(4.11)

We now prove (4.11). Since each fi Є ℓ1(ℤ), then there exists a sufficiently large positive integer R1 = R1(ϵ, f⃗ ) such that

sup1im|n|R1fi(n)<ϵ.(4.12)

Note that

sup1im|n|R1fi(n)<ϵ.

It follows that there exists an integer R2 = R2(ϵ) > 0 such that Mβ+(f)(n)<ϵfor all |n| ≥ R 2. Moreover, there exists an integer R3 > 0 such that sβm < ∈ if sR3 since β < m. Let R = max{R1, R2, R3}. (4.9) yields that there exists an integer N 2 = N (ϵ, R) > 0 such that

|(Mβ+(fj))(n)(Mβ+(f))(n)|ϵ4R+2(4.13)

for any jN2 and |n| ⪯ 2R. From (4.13) we have

(Mβ+(fj))1(Z)|n|2R|(Mβ+(fj))(n)(Mβ+(f))(n)|+(Mβ+(f))1(Z)+|n|2R|(Mβ+(fj))(n)|(Mβ+(f))1(Z)+ϵ+|n|2R|(Mβ+(fj))(n)|(4.14)

for any jN2. Fix jN2 and set

Xj+={|n|2R:Mβ+(fj)(n+1)>Mβ+(fj)(n)},Xj={|n|2R:Mβ+(fj)(n)Mβ+(fj)(n+1)}.

Since all fi,j1(ℤ), then, for any n ∈ ℤ, there exists rn ∈ ℕ such that Mβ+(fj)(n)=Arn(fj)(n).Then we have

|n|2R|(Mβ+(fj)(n)|=nXj+(Mβ+(fj)(n+1)Mβ+(fj)(n))+nXj(Mβ+(fj)(n)Mβ+(fj)(n+1))nXj+(Arn+1(fj)(n+1)Arn+1+1(fj)(n))+nXj(Arn(fj)(n)Arn+1(fj)(n+1)).(4.15)

By the arguments similar to those used in deriving (4.4) and (4.5), one has

Arn+1(fj)(n+1)Arn+1+1(fj)(n)l=1m1μmμlfμ,j1(Z)(kZfl,j(k)Γ(kn1)χ(n,)(k)),(4.16)

Arn(fj)(n)Arn+1(fj)(n+1)l=1m1μmμlfμ,j1(Z)(kZfl,j(k)Γ(kn1)χ(n,)(k)+fl,j(n)).(4.17)

It follows from (4.13)-(4.15) that

|n|2R|(Mβ+(fj)(n)|l=1m1μmμlfμ,j1(Z)(nXj+kZfl,j(k)Γ(kn1)χ(n,)(k))+l=1m1μmμlfμ,j1(Z)(nXjkZfl,j(k)Γ(kn1)χ(n,)(k)+nXjfl,j(n))l=1m1μmμlfμ,j1(Z)(|n|2RkZfl,j(k)Γ(kn1)χ(n,)(k)+|n|2Rfl,j(n)).(4.18)

By (4.7)-(4.8) and (4.12), we obtain

|n|2RkZfl,j(k)Γ(kn1)χ(n,)(k)kZfl,j(k)|n|2Rn<kΓ(kn1)|k|Rfl,j(k)|n|2Rn<kΓ(kn1)+|k|<Rfl,j(k)n2RΓ(kn1)|k|Rfl,j(k)+|k|<Rfl,j(k)n=2RΓ(nR)fl,jfl1(Z)+flχ|n|2R1(Z)+Rβmfl,j1(Z)C(fl)ϵ(4.19)

for any jN1. It follows from (4.8), (4.12) and (4.18)-(4.19) that

|n|2R|(Mβ+(fj))(n)|C(f)ϵ(4.20)

for any jN1. Combining (4.20) with (4.14) yields that

(Mβ+(fj))1(Z)(Mβ+(f))1(Z)+Cϵ

for any j ≥ max {N1, N 2}. This proves (4.11) and finishes the proof of Theorem 1.3.

Acknowledgement

This work was supported partly by the National Natural Science Foundation of China (Grant No. 11701333) and the Support Program for Outstanding Young Scientific and Technological Top-notch Talents of College of Mathematics and Systems Science (Grant No. Sxy2016K01).

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About the article

Received: 2018-04-26

Accepted: 2018-11-26

Published Online: 2018-12-31


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 1556–1572, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0129.

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© 2018 Lui and Xu, published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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