Abstract

We derive a necessary condition for exponent functions 𝑝,𝛽 such that the variable exponent Hardy inequality 𝑥𝛽(𝑥)1𝑥0𝑓(𝑡)𝑑𝑡𝐿𝑝(.)(0,𝑙)𝐶𝑥𝛽(𝑥)𝑓𝐿𝑝(.)(0,𝑙) holds.

1. Introduction

A sufficient condition on measurable functions 𝑝(0,𝑙)[1,), 𝛽(0,𝑙)(,) for which the variable exponent Hardy inequality|𝑥|𝛽()1𝐻𝑓𝐿𝑝()(0,𝑙)𝐶|𝑥|𝛽()𝑓𝐿𝑝()(0,𝑙),𝐻𝑓(𝑥)=𝑥0𝑓(𝑡)𝑑𝑡(1.1) holds for all 𝑓0 have been known (see [13]). 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,𝛽Λ, where Λ means a class of measurable functions 𝑔(0,𝑙)(,) such that 𝑔(0),𝐶>0||||1𝑔(𝑥)𝑔(0)ln1|𝑥|𝐶,0<𝑥<2.(1.2)

The purpose of this paper is to prove that a weaker continuity condition on 𝑝() 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:||||1𝑔(2𝑥)𝑔(𝑥)ln𝑥1𝐶,0<𝑥<2.(1.3)

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)<11/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 [311].

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 spaces 𝐿𝑝(), 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)=𝑙0|𝑓(𝑥)|𝑝(𝑥)𝑑𝑥-modular. A norm in 𝐿𝑝()(0,𝑙) is given in the form𝑓𝐿𝑝()(0,𝑙)=inf𝜆>0𝐼𝑝𝑓𝜆1.(2.1) There exists a relation between modular and norm, which is expressed by the following inequalities:𝑓𝑝+𝐿𝑝()(0,𝑙)𝐼𝑝(𝑓)𝑓𝑝𝐿𝑝()(0,𝑙),1𝑓𝑝(),(2.2)𝑓𝑝𝐿𝑝()(0,𝑙)𝐼𝑝(𝑓)𝑓𝑝+𝐿𝑝()(0,𝑙),1𝑓𝑝().(2.3) 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. Let 𝛽 a function 𝑝(0,𝑙)[1,) be increasing on (0,𝜀) and such that 𝑝(0)=lim𝑥0𝑝(𝑥) exists, 𝛽(0)<11/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. Let 𝑝, let 𝛽(0,𝑙)[,) be a function decreasing on (0,𝜀) such that 𝛽(0)=lim𝑥0𝛽(𝑥) exists, and let the conditions 𝛽(0)<11/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. Let 𝛽, and 𝛿𝑛=4𝑛,𝑛. There exist a sequence of functions {𝑓𝑛} and a function 𝑝(0,𝑙)[1,) satisfying the conditions 𝛽<11/𝑝(0), 𝑝>1 such that lim𝑛||𝑝𝛿𝑛||1𝑝(0)ln𝛿𝑛=,(2.4) and inequality (1.1) is violated.

Theorem 2.4. Let 𝑝>1, 𝛿𝑛=4𝑛,𝑛. Then there exist a sequence of functions {𝑓𝑛} and a function 𝛽(0,𝑙)(,) satisfying the conditions 𝛽(0)<11/𝑝, lim𝑛||𝛽𝛿𝑛||1𝛽(0)ln𝛿𝑛=,(2.5) such that inequality (1.1) is violated.

3. Proofs of Main Results

Proof of Theorem 2.1. Denote 𝐼𝑝()(𝑓)=𝑙0|𝑓(𝑡)|𝑝(𝑡)𝑑𝑡. By (2.2) note that the condition 𝐼𝑝(𝑓)1 is equivalent to 𝑓𝐿𝑝(.)(0,𝑙)1.
Put 𝛿𝑘=𝜀4𝑘,𝑘, and 𝑓𝑘(𝑥)=𝑥1/𝑝(𝑥)𝛽𝜒(𝛿𝑘,2𝛿𝑘)(𝑥),𝑥(0,𝑙). Then for sufficiently large 𝑘, 𝐼𝑝()𝑥𝛽(𝑥)𝑓𝑘=(𝑥)2𝛿𝑘𝛿𝑘𝑡𝛽𝑡1/𝑝(𝑡)𝛽𝑝(𝑡)=𝑑𝑡2𝛿𝑘𝛿𝑘𝑡1𝑑𝑡=ln2.(3.1) Also 𝐼𝑝()𝑥𝛽(𝑥)1𝐻𝑓𝑘(𝑥)4𝛿𝑘3𝛿𝑘𝑥(𝛽1)2𝛿𝑘𝛿𝑘𝑡(1/𝑝(𝑡))𝛽𝑑𝑡𝑝(𝑥)𝑑𝑥𝐶4𝛿𝑘3𝛿𝑘𝛿(11/𝑝(2𝛿𝑘𝑘)𝛽)𝑥(𝛽1)𝑝(2𝛿𝑘)𝑑𝑥𝐶𝛿1𝑝(3𝛿𝑘)/𝑝(2𝛿𝑘)𝑘=𝐶𝑒(1/𝑝+)[𝑝(3𝛿𝑘)𝑝(2𝛿k)]ln(1/2𝛿𝑘).(3.2) Applying inequality (1.1), we have ||𝑝3𝛿𝑘𝑝2𝛿𝑘||1ln2𝛿𝑘𝐶,𝑘,(3.3) which by using of monotony of 𝑝 and its boundedness implies (1.3).
This completes the proof of Theorem 2.1.

Proof of Theorem 2.2. Put 𝛿𝑘=𝜀4𝑘,𝑘 and 𝑓𝑘(𝑥)=𝑥1/𝑝𝛽(𝑥)𝜒(𝛿𝑘,2𝛿𝑘)(𝑥),𝑥(0,𝑙). Then 𝐼𝑝()𝑥𝛽(𝑥)𝑓𝑘=(𝑥)2𝛿𝑘𝛿𝑘𝑡𝛽(𝑡)𝑡1/𝑝𝛽(𝑡)𝑝(𝑡)=𝑑𝑡2𝛿𝑘𝛿𝑘𝑡1𝑑𝑡=ln2.(3.4) Also 𝐼𝑝()𝑥𝛽(𝑥)1𝐻𝑓𝑘(𝑥)4𝛿𝑘3𝛿𝑘𝑥𝛽(𝑥)12𝛿𝑘𝛿𝑘𝑡1/𝑝𝛽(𝑡)𝑑𝑡𝑝(𝑥)𝑑𝑥𝐶𝛿[𝛽(3𝛿𝑘)𝛽(2𝛿𝑘𝑘)]𝑝𝐶𝑒𝑝[𝛽(3𝛿𝑘)𝛽(2𝛿𝑘)]ln(1/𝛿𝑘).(3.5) Applying inequality (1.1), we have ||𝛽2𝛿𝑘𝛽3𝛿𝑘||1ln𝛿𝑘𝐶,𝑘(3.6) which by using monotony of 𝛽 implies (1.3).
This completes the proof of Theorem 2.2.

Proof of Theorem 2.3. Let us assume that 𝑓𝑘(𝑥)=𝑥1/𝑝(𝑥)𝛽𝜒(𝛿𝑘,2𝛿𝑘)(𝑥), 𝑥(0,𝑙). Fix 𝑘. We define the step function 𝑝𝑝(𝑥)=0+𝛼𝑛if𝑥2𝛿𝑛,4𝛿𝑛,𝑝0𝛿if𝑥𝑛,2𝛿𝑛,𝑛.(3.7) Here {𝛼𝑛}is a sequence of positive numbers that satisfies the condition 𝑛𝛼𝑛as𝑛.(3.8) 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 𝐼𝑝()𝑥𝛽𝑓𝑘=(𝑥)2𝛿𝑘𝛿𝑘𝑡𝛽𝑡1/𝑝(𝑡)𝛽𝑝(𝑡)=𝑑𝑡2𝛿𝑘𝛿𝑘𝑡1𝐼𝑑𝑡=ln2,(3.9)𝑝()𝑥𝛽1𝐻𝑓𝑘(𝑥)4𝛿𝑘2𝛿𝑘2𝛿𝑘𝛿𝑘𝑡1/𝑝(𝑡)𝛽𝑑𝑡(𝑝0+𝛼𝑘)𝑥(𝛽1)(𝑝0+𝛼𝑘)𝑑𝑥𝐶4𝛿𝑘2𝛿𝑘𝛿(11/𝑝0𝛽)(𝑝0+𝛼𝑘)𝑘𝑥(𝛽1)(𝑝0+𝛼𝑘)𝑑𝑥𝐶𝛿𝛼𝑘/𝑝0𝑘=𝐶𝑒𝛼𝑘/𝑝0ln(1/𝛿𝑘)as𝑘.(3.10) The last relation shows violating of inequality (1.1) for sufficiently large 𝑘.

Proof of Theorem 2.4. Let us assume that 𝑓𝑘(𝑥)=𝑥1/𝑝𝛽(𝑥)𝜒(𝛿𝑛,2𝛿𝑛)(𝑥), 𝑥(0,𝑙), 𝑛. Fix 𝑛. We define the step function 𝛽 as 𝛽𝛽(𝑥)=0+𝛼𝑛𝛿if𝑥𝑛,2𝛿𝑛,𝛽0if𝑥2𝛿𝑛,4𝛿𝑛,𝑛,(3.11) 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 𝐼𝑝()𝑥𝛽(𝑥)1𝑓𝑛(𝑥)𝐶𝛿𝑝𝛼𝑛𝑛𝐼as𝑛,(3.12)𝑝()𝑥𝛽(𝑥)𝑓𝑛(𝑥)𝐶0ln2.(3.13) The last relation contradicts to inequality (1.1) for sufficiently large 𝑘.

Acknowledgment

The author thanks Professor Mamedov F.I. for his valuable discussion of the subject. The author also expresses his acknowledgement to the referee for drawing attention to the log example and comments of stylistic character.