# On Necessary Condition for the Variable Exponent Hardy Inequality

Authors: Aziz Harman
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

We derive a necessary condition for exponent functionsp,β such that the variable exponent Hardy inequality ∥xβ(x)-1∫0xf(t)dt∥Lp(.)(0,l)≤C∥xβ(x)f∥Lp(.)(0,l) holds.


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## Body

## 1. Introduction

A sufficient condition on measurable functionsp:(0,l)→[1,∞), β:(0,l)→(-∞,∞) for which the variable exponent Hardy inequality(1.1)‖|x|β(⋅)-1Hf‖Lp(⋅)(0,l)≤C‖|x|β(⋅)f‖Lp(⋅)(0,l),Hf(x)=∫0xf(t)dt
holds for all f≥0 have been known (see [1–3]). According to mentioned works, if β(0)<1-(1/p(0)), p-:=inf{p(x):x∈(0,l)}>1, then a sufficient condition is p,β∈Λ, where Λ means a class of measurable functions g:(0,l)→(-∞,∞) such that ∃g(0),C>0(1.2)|g(x)-g(0)|ln⁡1|x|≤C,0<x<12.The purpose of this paper is to prove that a weaker continuity condition onp(·) and β(·) is necessary for the norm inequality to hold provided that p(·) and β(·) are monotone (see Theorems 2.1 and 2.2 for a precise statement), which is the following condition:(1.3)|g(2x)-g(x)|ln⁡1x≤C,0<x<12.Note that condition (1.3) is strictly weaker than (1.2). For example, it is satisfied by p(x)=C/|log(x)|1/2. This condition is new and somewhat surprising. Since in the corresponding theorem for the maximal operator, it is known that p(·) need not be continuous, and the problem of determining which exponent conditions are necessary and/or sufficient is an open one.If the powers are not monotone, it follows from the results of the paper [2] that condition (1.2) is close to be sharp. Also in [2], the necessity of conditions p->1 and β(0)<1-1/p(0) was proved. Recently, there have been quite a number of papers discussing the Hardy inequality in norms of the variable exponent Lebesgue spaces [3–11].For problems of boundedness of classical integral operators in variable exponent Lebesgue spaces and regularity results for nonlinear equations with nonstandard growth condition, see monograph [12] and references therein.


## 2. Main Results

As to the basic properties of spacesLp(·), we refer to [13]. Throughout this paper it is assumed that p(x) is a measurable function in (0,l) taking its values from the interval [1,∞) with p+:=sup{p(x):x∈(0,l)}<∞. The space of functions Lp(·)(0,l) is introduced as the class of measurable functions f(x) in (0,l), which have a finite Ip(f):=∫0l|f(x)|p(x)dx-modular. A norm in Lp(·)(0,l) is given in the form(2.1)‖f‖Lp(⋅)(0,l)=inf⁡{λ>0:Ip(fλ)≤1}.
There exists a relation between modular and norm, which is expressed by the following inequalities:(2.2)‖f‖Lp(⋅)(0,l)p+≤Ip(f)≤‖f‖Lp(⋅)(0,l)p-,1≥‖f‖p(⋅),(2.3)‖f‖Lp(⋅)(0,l)p-≤Ip(f)≤‖f‖Lp(⋅)(0,l)p+,1≤‖f‖p(⋅).
Such estimates allow us to perform our estimates in terms of a modular. In the following two theorems, we show that if functions p,β are monotone, then condition (1.3) for them is necessary for inequality (1.1) to hold.Theorem 2.1.
Letβ∈ℝ a function p:(0,l)→[1,∞) be increasing on (0,ɛ) and such that p(0)=limx→0p(x) exists, β(0)<1-1/p(0), p->1. Then for inequality (1.1) to hold, it is necessary that for the function p(·) condition (1.3) is satisfied.Theorem 2.2.
Letp∈ℝ, let β:(0,l)→[-∞,∞) be a function decreasing on (0,ɛ) such that β(0)=limx→0β(x) exists, and let the conditions β(0)<1-1/p(0), p->1 be satisfied. Then for inequality (1.1) to hold, it is necessary that for the function β(·) condition (1.3) is satisfied.The following two theorems show that the logarithmic regularity conditions (1.2) for the functions p,β are essential for inequality (1.1) to hold.Theorem 2.3.
Letβ∈ℝ, and δn=4-n,n∈ℕ. There exist a sequence of functions {fn} and a function p:(0,l)→[1,∞) satisfying the conditions β<1-1/p(0), p->1 such that
(2.4)lim⁡n→∞|p(δn)-p(0)|ln⁡1δn=∞,
and inequality (1.1) is violated.Theorem 2.4.
Letp>1, δn=4-n,n∈ℕ. Then there exist a sequence of functions {fn} and a function β:(0,l)→(-∞,∞) satisfying the conditions β(0)<1-1/p,
(2.5)lim⁡n→∞|β(δn)-β(0)|ln⁡1δn=∞,
such that inequality (1.1) is violated.


## 3. Proofs of Main Results

Proof of Theorem2.1.
DenoteIp(·)(f)=∫0l|f(t)|p(t)dt. By (2.2) note that the condition Ip(f)≤1 is equivalent to ∥f∥Lp(.)(0,l)≤1.
Putδk=ɛ4-k,k∈ℕ, and fk(x)=x-1/p(x)-βχ(δk,2δk)(x),x∈(0,l). Then for sufficiently large k,
(3.1)Ip(⋅)(xβ(x)fk(x))=∫δk2δk(tβt-1/p(t)-β)p(t)dt=∫δk2δkt-1dt=ln⁡2.
Also
(3.2)Ip(⋅)(xβ(x)-1H(fk(x)))≥∫3δk4δk(x(β-1)∫δk2δkt-(1/p(t))-βdt)p(x)dx≥C∫3δk4δkδk(1-1/p(2δk)-β)x(β-1)p(2δk)dx≥Cδk1-p(3δk)/p(2δk)=Ce(1/p+)[p(3δk)-p(2δk)]ln⁡⁡(1/2δk).
Applying inequality (1.1), we have
(3.3)|p(3δk)-p(2δk)|ln⁡12δk≤C,k∈N,
which by using of monotony of p and its boundedness implies (1.3).
This completes the proof of Theorem2.1.Proof of Theorem2.2.
Putδk=ɛ4-k,k∈ℕ and fk(x)=x-1/p-β(x)χ(δk,2δk)(x),x∈(0,l). Then
(3.4)Ip(⋅)(xβ(x)fk(x))=∫δk2δk(tβ(t)t-1/p-β(t))p(t)dt=∫δk2δkt-1dt=ln⁡2.
Also
(3.5)Ip(⋅)(xβ(x)-1H(fk(x)))≥∫3δk4δk(xβ(x)-1∫δk2δkt-1/p-β(t)dt)p(x)dx≥Cδk[β(3δk)-β(2δk)]p≥Cep[β(3δk)-β(2δk)]ln⁡⁡(1/δk).
Applying inequality (1.1), we have
(3.6)|β(2δk)-β(3δk)|ln⁡1δk≤C,k∈N
which by using monotony of β implies (1.3).
This completes the proof of Theorem2.2.Proof of Theorem2.3.
Let us assume thatfk(x)=x-1/p(x)-βχ(δk,2δk)(x), x∈(0,l). Fix k∈ℕ. We define the step function
(3.7)p(x)={p0+αnifx∈(2δn,4δn),p0ifx∈(δn,2δn),n∈N.
Here {αn}is a sequence of positive numbers that satisfies the condition
(3.8)nαn⟶∞asn⟶∞.
Then αnln(1/δn)→∞ as n→∞; that is, condition (1.2) is not satisfied for the function p(x). Also note that this function p(·) is not monotone. We have
(3.9)Ip(⋅)(xβfk(x))=∫δk2δk(tβt-1/p(t)-β)p(t)dt=∫δk2δkt-1dt=ln⁡2,(3.10)Ip(⋅)(xβ-1H(fk(x)))≥∫2δk4δk(∫δk2δkt-1/p(t)-βdt)(p0+αk)x(β-1)(p0+αk)dx≥C∫2δk4δkδk(1-1/p0-β)(p0+αk)x(β-1)(p0+αk)dx≥Cδk-αk/p0=Ceαk/p0ln⁡⁡(1/δk)⟶∞ask⟶∞.
The last relation shows violating of inequality (1.1) for sufficiently large k.Proof of Theorem2.4.
Let us assume thatfk(x)=x-1/p-β(x)χ(δn,2δn)(x), x∈(0,l), n∈ℕ. Fix n∈ℕ. We define the step function β as
(3.11)β(x)={β0+αnifx∈(δn,2δn),β0ifx∈(2δn,4δn),n∈N,
where αnln(1/δn)→∞; that is, condition (1.2) is not satisfied for the function β. Note that this function β is not monotone.
We have(3.12)Ip(⋅)(xβ(x)-1fn(x))≥Cδn-pαn⟶∞asn⟶∞,(3.13)Ip(⋅)(xβ(x)fn(x))≤C0ln⁡2.
The last relation contradicts to inequality (1.1) for sufficiently large k.



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