Abstract

We discuss a logarithmic regularity condition in a neighborhood of the origin and infinity on the exponent functions and for the variable exponent Hardy inequality to hold.

1. Introduction

There are several papers devoted to the variable exponent Hardy inequality:

Let be positive measurable functions and , ; then inequality (1) and (2) hold under certain conditions on the weight functions , and the exponents , . Two types of conditions arise here: a balance condition on the weights and a regularity condition on the exponents (see below). Necessary and sufficient conditions for the validity of general inequality (1) were found in [1] for the case of , in [2] for cases and , in [3] for cases and , and in [4] for mixed cases and   ( and ). Some special cases of (1) are studied in [5–10] too.

So, the inequality is a particular case of (1) when , and , where is the Hardy operator. For the constant exponents , this inequality holds if , (see, e.g., [11]). Necessary and sufficient conditions on the , for inequality (2) to hold are and if the exponents , are continuous near the origin and infinity such that the conditions are satisfied (see, e.g., [5–7, 9]). In [8] (see also [10]) it was proved that the condition is necessary for inequality (2) to hold if one of these exponents is a constant. Also, it was proved in [10] that the condition is sufficient for inequality (2) to hold on bounded interval if a constant in the condition for the satisfies and the is zero. The condition is weaker then . Also the function satisfies the condition but does not satisfy .

In this note, we will focus on the results of sufficiency and necessity of regularity conditions , , and below for inequality (2) to hold.

The space of functions is introduced as the class of measurable functions in , which have a finite modular. A norm in is given in the form As to the basic properties of spaces , we refer to [12].

2. Main Results

We will state some sufficiency and necessity assertions concerning inequality (2). Along the way, it will be given a proof for two elementary estimates that we had used.

Let us introduce the following classes of measurable functions. We say, is in the class if is in the class if and is in the class if

Theorem 1 (see [8]). Suppose and is an increasing function on such that is continuous at and , ; then for the inequality to hold it is necessary that .

Theorem 2 (see [8]). Suppose and is a decreasing function on such that is continuous at and , ; then for the inequality to hold it is necessary that .

Theorem 3. Suppose is measurable functions such that , , , ; then for the inequality to hold it is sufficient that , whenever and .

Theorem 4. Let , and let . There exists a sequence and a function , satisfying the conditions , and or violating inequality (2).

Theorem 5. Let , and let . Then there exists a sequence and a function satisfying the conditions , , and or violating inequality (2).

3. Proof of Main Results

For the proof of Theorems 1 and 2 we refer to [8]. Other proofs of these theorems are given in [10]. The proof of Theorem 3 also is given in [8]. Here we derive an alternative proof of that theorem using the general results of [2, 4].

In the proof of main results we use the following elementary Lemma.

Lemma 6. Suppose is a measurable function such that and , ; then it holds the estimate for and the estimate for .

Proof of Theorem 3. To prove Theorem 3 we apply the results of [2, 4], where it was proved that the following conditions are necessary and sufficient for inequality (1) to hold if , , and the regularity conditions are satisfied: where , .
In Theorem 3, we have accepted that and , . It is easy to show that the conditions imply . Therefore, it follows from Lemma 6 that for by some . Also the conditions imply . Therefore, it follows from Lemma 6 that for by some . Integrating these inequalities over the intervals and , respectively, we get
To complete the proof of Theorem 3 it suffices to apply estimates (20) to verify conditions (16) and (17). Now, Theorem 3 follows from the upper refereed results of the works [2, 4].

Proof of Lemma 6. Let and let ; then we have , where is a certain number from the interval . If and then by condition we have Therefore, for we have the estimation where the positive constant depends on , , , . Same inequality holds for the function . Indeed, for and we have . If and by condition we have By using these inequalities and by the representation we have estimate (14).
To show estimate (15) note that for and we have , where the is a certain number. If and then by the condition we have Combining the estimates for the functions , for by the presentation we get estimate (15). To complete the proof of Lemma 6, note that the condition is equivalent to and the is equivalent to respectively.

Proof of Theorem 4. Let us assume that ; . Fix . We define the step function Here is a sequence of positive numbers that satisfies the condition as . Then as and condition (10) is fulfilled for the function . We have as . The last relation shows violating of inequality (2) for sufficiently large .
We define for . Fix . We also define the step function where as . We have as and condition (11) holds for the function . Furthermore, as , which contradicts (2) for sufficiently large .

Proof of Theorem 5. Let us assume that ; . Fix . We define the step function as where . Then, as . The last relation contradicts the validity of inequality (2).
We define ; . Fix , where the function is defined as where . Then, as which contradicts inequality (2).