# A Local Estimate for the Maximal Function in Lebesgue Spaces with EXP-Type Exponents

Authors: Alberto Fiorenza
Journal: Journal of Function Spaces
 (2015)

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## Abstract

It is proven that if1≤p(·)<∞ in a bounded domain Ω⊂Rn and if p(·)∈EXPa(Ω) for some a>0, then given f∈Lp(·)(Ω), the Hardy-Littlewood maximal function of f, Mf, is such that p(·)log(Mf)∈EXPa/(a+1)(Ω). Because a/(a+1)<1, the thesis is slightly weaker than (Mf)λp(·)∈L1(Ω) for some λ>0. The assumption that p(·)∈EXPa(Ω) for some a>0 is proven to be optimal in the framework of the Orlicz spaces to obtain p(·)log(Mf) in the same class of spaces.


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## Body

## 1. Introduction

LetΩ⊂Rn be a bounded domain. By p(·), we denote a finite exponent function on Ω, that is, a Lebesgue measurable function p(·):Ω→[1,∞[. The space Lp(·)(Ω) is defined as the set of Lebesgue measurable functions f on Ω such that (1)∫Ωμfxpxdx<∞,for some μ>0.The Hardy-Littlewood maximal operator of a functionf∈Lloc1(Rn) is defined by (2)Mfx=supQ∋x⁡⨏Q⁡fydy,x∈Rn,where the supremum is taken over all cubes Q⊂Rn that contain x and whose sides are parallel to the coordinate axes (the symbol ⨏Q denotes 1/|Q|, and for a Lebesgue measurable E⊂Rn, the symbol |E| denotes its Lebesgue measure). If f∈L1(Ω) (and, in particular, also if f∈Lp(·)(Ω); see [1, Proposition 2.41 page 36]), then by Mf, we mean the maximal operator computed on the extension of f by zero outside Ω; in this case, it is known (see, e.g., [2, 8.15 page 43] or [3]) that(3)Mfq∈L1E∀0<q<1,∀E⊂Rn,E<∞,(4)∫EMfxqdx≤cqE1-q∫Ωfxdxq.The main classical result regarding the maximal operator is that it is bounded in every Lebesgue space with (constant) exponent greater that 1 (see, e.g., [2, 4–7] and references therein for the main results regarding the maximal operator). The extension of this result in the framework of variable Lebesgue spaces theory has been intensively studied (see, e.g., [1, Chapter 3] or [8, Chapter 4], [9, Chapter 1] for a collection of results regarding this topic; see also the survey [10]).Assume thatp(·) is such that p+=esssupΩ⁡p(·)<∞. It is well-known that M is not necessarily bounded in Lp(·)(Ω) even if, in addition, p-=essinfΩ⁡p(·)>1 and p(·) is continuous (see [11]; the same example is also presented in [1, Example 4.43 page 160]). A well-known sufficient condition for boundedness of the maximal operator for exponents p(·) such that p->1 is expressed in terms of the so-called log-Hölder continuity, which may be required to the function 1/p(·) (see [12] and references therein for details). However, sufficient conditions that do not require continuity do exist (see, e.g., [13, 14]). In [15], another class of exponents that ensures the boundedness of the maximal operator for bounded domains has been presented.However, we can assert that(5)p+<∞,f∈Lp·Ω⟹∃λ>0:Mfλp·∈L1Ω.In fact, by (3), for 0<q<1, we have (6)f∈Lp·Ω⟹f∈L1Ω⟹Mfq=Mfp+q/p+∈L1Ω⟹Mfp·q/p+∈L1Ω.Note that from (5), one gets that, for 0<β<1, if p+<∞,(7)f∈Lp·Ω⟹∃λ>0:exp⁡λp·log⁡Mfβ∈L1Ω.For unbounded (and possibly discontinuous) exponents, (7) is not generically true (see the example presented below). In this note, we prove that, for some 0<β<1, (7) holds true under an assumption of exponential integrability of the exponent. We observe that some type of high integrability of the exponent is needed; in fact, the result is not true if it is only assumed that p(·) is contained in a Lebesgue space with any finite, constant exponent. For instance, if p(x)=x-b+1 on (0,1), b>0, then f(x)=(1/x)1/(x-b+1), x∈(0,1), is such that f∈Lp(·)(0,1) but for every λ>0, it is (8)1<∫01fxdxλ<∞.Therefore, for any 0<β<1, (9)∫01exp⁡λpxlog⁡Mfxβdx≥∫01exp⁡λx-b+1log⁡∫01fβdx≥cb,β∫01λx-b+1log⁡∫01fβ1/bβdx=cb,β∫01λ1/bx-b+11/blog1/b⁡∫01fdx=∞.For other results regarding the maximal operator that are specific to unbounded exponents, see [16].


## 2. The Local Estimate

In the following, three prerequisites are necessary. The first is the well-known Fefferman-Stein inequality ([17]; see also [18])(10)∫RnMfxqgxdx≤cq∫RnfxqMgxdx,f,g∈Lloc⁡1Rn,which is true for q>1 (here q is constant). The second is a well-known extrapolation characterisation (see, e.g., [19–23]) of EXPa(Ω), a>0, the Orlicz space of the functions, which can be characterised in one of the two following equivalent ways:(11)limsupk→∞⁡1k1/a⨏Ω⁡hxkdx1/k<∞⟺∃μ>0:⨏Ω⁡exp⁡μhxadx<∞.We remark that the growth of integrability of the functions in EXPa(Ω), 0<a<1, is somehow connected with the A∞ property for Young functions; see [24]. Finally, the third tool is an elementary inequality. For k≥1, α≥1, and a positive constant (12)γk,α=kαk-1+1exp⁡-kα,(as usual, subscripts indicate the dependence of the constant on the variables involved, which may change from line to line), it is(13)logk⁡ξ≤kαk+γk,αξ-exp⁡kα∀ξ≥exp⁡kα.This inequality is a direct consequence of the concavity of the kth power of the logarithm in (exp⁡(k-1),∞).We are now ready for the proof of the following theorem.Theorem 1.
LetΩ be a bounded domain in Rn, and let p(·)∈EXPa(Ω) for some a>0 be such that 1≤p(·)<∞. If f∈Lp(·)(Ω), f≢0, then p·log⁡(Mf)∈EXPa/(a+1)(Ω).Proof.
Let us first assume thatp->1. Considering that p(·)∈EXPa(Ω), for every k≥log⁡fL1(Ω) we have (14)1Ω∫x∈Ω:Mfx≤exp⁡kpxklog⁡Mfxkdx1/k≤1Ω·∫x∈Ω:Mfx≤exp⁡kpxklogk⁡exp⁡kdx1/k=1Ω∫x∈Ω:Mfx≤exp⁡kpxkkkdx1/k≤k⨏Ω⁡pxkdx1/k≤kcp·k1/a=cp·k1+1/a.On the other hand, for any q>1 and r>1 such that qr<p-, the following inequalities hold (we are going to apply (13) with the choices α=1 and ξ=Mf(x)q; note that because we work on the set in which Mf(x)>exp⁡(k), it is also true that Mf(x)q>exp⁡(k)):(15)1Ω∫x∈Ω:Mfx>exp⁡kpxklog⁡Mfxkdx1/k=1q1Ω∫x∈Ω:Mfx>exp⁡kpxk·logk⁡Mfxqdx1/k≤1q1Ω·∫x∈Ω:Mfx>exp⁡kpxk·kk+γk,1Mfxq-exp⁡kdx1/k=1q1Ω·∫x∈Ω:Mfx>exp⁡kpxkkkdx+1Ω·∫x∈Ω:Mfx>exp⁡kpxk·γk,1Mfxq-exp⁡kdx1/k≤1q1Ω·∫x∈Ω:Mfx>exp⁡kpxkkkdx1/k+1q1Ω·∫x∈Ω:Mfx>exp⁡kpxk·γk,1Mfxq-exp⁡kdx1/k,and because γk,11/k=k/e, by again using the fact that p(·)∈EXPa(Ω) and by applying (10),(16)≤kq⨏Ω⁡pxkdx1/k+keq1Ω·∫x∈Ω:Mfx>exp⁡kpxkMfxqdx1/k≤kq·cp·k1/a+keq1Ω∫RnMfxqpxk·χx∈Ω:Mfx>exp⁡kxdx1/k≤cp·qk1+1/a+keqcqΩ∫Rnfxq·Mpxkχx∈Ω:Mfx>exp⁡kxdx1/k≤cp·q·k1+1/a+keqcqΩ∫ΩfxqMpxkdx1/k.By Hölder’s inequality and the boundedness of the maximal operator in Lr/(r-1)(Ω),(17)≤cp·qk1+1/a+keqcq1/k⨏Ω⁡fxqrdx1/kr·⨏Ω⁡Mpxkr/r-1dxr-1/kr≤cp·qk1+1/a+cq1/kkeq⨏Ω⁡fxqrdx1/qr·q/k·cr⨏Ω⁡pxkr/r-1dxr-1/kr≤cp·qk1+1/a+cq1/kcrkeq⨏Ω⁡fxp-dx1/p-·q/k·cp·krr-11/a≤cp·qk1+1/a+cacq1/kcr·cp·q⨏Ω⁡fxp-dx1/p-·q/kk1+1/a≤Ck1+1/a,where C>0 is a constant that depends on all parameters and f∈Lp(·)(Ω) but is independent of k≥1.
Combining the estimates obtained above, we obtain(18)⨏Ω⁡pxklog⁡Mfxkdx1/k≤Ck1+1/a.That is, using (11) (with a replaced by a/(a+1)), we obtain the assertion.
Now, letp(·)∈EXPa(Ω) for some a>0, and assume p-=1. If p+<∞, we already proved (see (7)) that the assertion is trivially true. If p+=∞, set(19)A=x∈Ω:1≤px≤2a.e.,B=x∈Ω:px>2a.e.,and observe that A is nonempty (because p-=1) and B is nonempty (because p+=∞). By (3), (Mf)2/3∈L1(A). Therefore, (Mf)p(·)/3∈L1(A), from which (20)exp⁡p·3log⁡Mfa/a+1∈L1A.On the other hand, since p(·)∈EXPa(B) and essinfB⁡p(·)≥2, there exists μ>0 such that (21)exp⁡μp·log⁡Mfa/a+1∈L1B.In conclusion, since Ω=A∪B, exp⁡((λp(·)|log(Mf)|)a/(a+1))∈L1(Ω) for λ=min⁡{1/3,μ}.The heart of Theorem1 can be stated as follows: if p(·) belongs to some EXPa(Ω), then for any f∈Lp(·)(Ω) also p·log⁡(Mf) does. From this perspective, our result is optimal: if one assumes that p(·) belongs to any Orlicz space that contains all of the EXPa(Ω)’s, it is possible to construct an exponent that is not in any EXPa(Ω) and a function f∈Lp(·)(0,1) such that p·log⁡(Mf) is also not in any EXPa(Ω). This statement is the essence of the following result (for the definition of Young functions and the classical embedding theorem for Orlicz spaces; see, e.g., [25]).Theorem 2.
LetΦ:[0,∞[→[0,∞[ be a Young function such that for all a∈]0,1](22)Φt≤exp⁡ta∀t>ta.There exists p(·)∈LΦ(0,1), p(·)∉EXPa(0,1)∀a∈]0,1], p->1, and there exists f∈Lp(·)(0,1) such that(23)p·log⁡Mf∉EXPa0,1∀a∈0,1.Proof.
SetA(t)=log⁡(Φexp⁡t) for t large. For every β∈]0,1], set Aβ(t)=exp⁡(βt). Then, (22) reads as (24)Alog⁡t-Aalog⁡t≤0,or, equivalently,(25)A-1t≥Aa-1t=1alog⁡t,for large t. Set (26)px=Φ-1Φ2x=exp⁡A-1log⁡Φ2xx∈0,1.We observe that, trivially, p(·)∈LΦ(0,1) and p-=2>1.
We now prove thatp(·)∉EXPa(0,1)∀a∈]0,1]. Let a∈]0,1] and k>0. Fix β∈]0,a[ (for instance, β=a/2 can be assumed in the following). For sufficiently small x, using (25) with a replaced by β, we see that(27)exp⁡kpxa=exp⁡kexp⁡A-1log⁡Φ2xa≥exp⁡kexp⁡1βlog⁡log⁡Φ2xa=exp⁡klog⁡Φ2x1/βa≥exp⁡21/alog⁡Φ2x1/aa=Φ22x.Therefore,(28)exp⁡kp·a∉L10,1∀a∈0,1,∀k>0.Let us now set (29)fx=1x1/pxx∈0,1,so that f∈Lp(·)(0,1). It remains to prove (23). Fix λ>0, set k=λlog⁡(∫01‍|f(x)|dx), and note that k>0. Thus, (30)∫01exp⁡λpxlog⁡Mfxadx≥∫01exp⁡kpxadx=∞,where the last equality results from (28).



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