# On the Logarithmic Regularity Conditions for the Variable Exponent Hardy Type Inequality

Authors: Aziz Harman; Mustafa Özgür Keleş
Journal: International Journal of Analysis
 (2014)

---

## Abstract

We discuss a logarithmic
regularity condition in a neighborhood of the origin and infinity on the
exponent functionsq
x
≥
p
x and β
x for the variable exponent
Hardy inequality x
β
·
-
1
∫
0
x
f
t
d
t
L
p
·
0
,
l
≤
C
x
β
·
f
L
p
·
0
,
l to hold.


---

## Body

## 1. Introduction

There are several papers devoted to the variable exponent Hardy inequality:(1)
∥
v
1
/
q
(
·
)
∫
0
x
f
(
t
)
d
t
∥
L
q
(
·
)
(
0
,
∞
)
≤
∥
ω
1
/
p
(
·
)
f
(
·
)
∥
L
p
(
·
)
(
0
,
∞
)
.Letv
,
w
:
ℝ
→
[
0
,
∞
) be positive measurable functions and p
-
=
inf
⁡
{
p
(
x
)
:
x
∈
ℝ
}
>
1, p
+
=
sup
⁡
{
p
(
x
)
:
x
∈
ℝ
}
<
∞; then inequality (1) and (2) hold under certain conditions on the weight functions v, ω and the exponents p
(
·
), q
(
·
). 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 p
(
x
)
≤
q
(
x
), in [2] for cases q
(
0
)
≥
p
(
0
) and q
(
∞
)
≥
p
(
∞
), in [3] for cases q
(
0
)
<
p
(
0
) and q
(
∞
)
<
p
(
∞
), and in [4] for mixed cases q
(
0
)
≥
p
(
0
) and q
(
∞
)
<
p
(
∞
)  (q
(
0
)
<
p
(
0
) and q
(
∞
)
≥
p
(
∞
)). Some special cases of (1) are studied in [5–10] too.So, the inequality(2)
∥
x
β
(
·
)
-
1
H
f
(
·
)
∥
L
p
(
·
)
(
0
,
∞
)
≤
C
∥
x
β
(
·
)
f
(
·
)
∥
L
p
(
·
)
(
0
,
∞
)
is a particular case of (1) when p
(
x
)
=
q
(
x
), v
(
x
)
=
x
β
(
x
)
-
1 and w
(
x
)
=
x
β
(
x
), where H
f
(
x
)
=
∫
0
x
‍
f
(
t
)
d
t is the Hardy operator. For the constant exponents p, β this inequality holds if β
<
1
-
(
1
/
p
), p
>
1 (see, e.g., [11]). Necessary and sufficient conditions on the β
(
·
), p
(
·
) for inequality (2) to hold are β
(
0
)
<
1
-
(
1
/
p
(
0
)
) and β
(
∞
)
<
1
-
(
1
/
p
(
∞
)
) if the exponents p, q are continuous near the origin and infinity such that the conditions p
,
β
∈
Λ
0
∩
Λ
∞ are satisfied (see, e.g., [5–7, 9]). In [8] (see also [10]) it was proved that the condition p
,
β
∈
Λ
1 is necessary for inequality (2) to hold if one of these exponents is a constant. Also, it was proved in [10] that the condition Λ
1 is sufficient for inequality (2) to hold on bounded interval (
0
,
l
) if a constant A
1 in the condition Λ
1 for the p satisfies A
1
<
p
(
0
)
(
p
(
0
)
-
1
) and the β is zero. The condition Λ
1 is weaker then Λ
0. Also the function p
(
x
)
=
C
/
ln
⁡
(
1
/
x
) satisfies the condition Λ
1 but does not satisfy Λ
0.In this note, we will focus on the results of sufficiency and necessity of regularity conditionsΛ
0, Λ
∞, and Λ
1 below for inequality (2) to hold.The space of functionsL
p
(
·
)
(
0
,
∞
) is introduced as the class of measurable functions f
(
x
) in (
0
,
∞
), which have a finite I
p
(
f
)
:
=
∫
0
∞
‍
|
f
(
x
)
|
p
(
x
)
d
x modular. A norm in L
p
(
·
)
(
0
,
∞
) is given in the form
(3)
∥
f
∥
L
p
(
·
)
(
0
,
∞
)
=
inf
⁡
{
λ
>
0
,
I
p
(
f
λ
)
≤
1
}
.
As to the basic properties of spaces L
p
(
·
), 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,s
(
x
)
:
(
0
,
∞
)
→
ℝ is in the class Λ
0 if
(4)
Λ
0
:
=
limsup
⁡
x
→
0
|
s
(
x
)
-
s
(
0
)
|
ln
⁡
1
|
x
|
<
∞
,
is in the class Λ
∞ if
(5)
Λ
∞
:
=
limsup
⁡
x
→
∞
|
s
(
x
)
-
s
(
∞
)
|
ln
⁡
|
x
|
<
∞
,
and is in the class Λ
1 if
(6)
Λ
1
:
=
limsup
⁡
x
→
0
|
s
(
x
)
-
s
(
x
2
)
|
ln
⁡
1
|
x
|
<
∞
.Theorem 1 (see [8]).
Supposeβ
∈
ℝ and p
:
(
0
,
l
)
→
[
1
,
∞
) is an increasing function on (
0
,
ϵ
) such that p
(
x
) is continuous at x
=
0 and β
<
1
-
(
1
/
p
(
0
)
), p
-
>
1; then for the inequality
(7)
∥
x
β
(
x
)
-
1
H
f
∥
L
p
(
·
)
(
0
,
l
)
≤
C
∥
x
β
(
x
)
f
∥
L
p
(
·
)
(
0
,
l
)
to hold it is necessary that p
(
·
)
∈
Λ
1.Theorem 2 (see [8]).
Supposep
∈
ℝ and β
:
(
0
,
l
)
→
ℝ is a decreasing function on (
0
,
ϵ
) such that β
(
x
) is continuous at x
=
0 and β
(
0
)
<
1
-
(
1
/
p
), p
-
>
1; then for the inequality
(8)
∥
x
β
(
·
)
-
1
H
f
∥
L
p
(
·
)
(
0
,
l
)
≤
C
∥
x
β
(
·
)
f
∥
L
p
(
·
)
(
0
,
l
)
to hold it is necessary that β
(
·
)
∈
Λ
1.Theorem 3.
Supposep
,
β
:
(
0
,
∞
)
→
ℝ is measurable functions such that 1
<
β
-, β
+
<
∞, 1
<
p
-, p
+
<
∞; then for the inequality
(9)
∥
x
β
(
·
)
-
1
H
f
(
·
)
∥
L
p
(
·
)
(
0
,
∞
)
≤
C
∥
x
β
(
·
)
f
(
·
)
∥
L
p
(
·
)
(
0
,
∞
)
to hold it is sufficient that β
,
p
∈
Λ
0
∩
Λ
∞, whenever β
(
0
)
<
1
-
(
1
/
p
(
0
)
) and β
(
∞
)
<
1
-
(
1
/
p
(
∞
)
).Theorem 4.
Letβ
∈
ℝ, p
:
(
0
,
∞
)
→
[
1
,
∞
) and let p
-
>
1. There exists a sequence f
n and a function p, satisfying the conditions β
<
1
-
(
1
/
p
(
0
)
), β
<
1
-
(
1
/
p
(
∞
)
) and
(10)
lim
⁡
n
→
∞
|
p
(
δ
n
)
-
p
(
0
)
|
ln
⁡
1
δ
n
=
∞
:
δ
n
=
4
-
n
or
(11)
lim
⁡
n
→
∞
|
β
(
μ
n
)
-
β
∞
|
ln
⁡
μ
n
=
∞
:
μ
n
=
4
n
violating inequality (2).Theorem 5.
Letp
∈
ℝ, β
:
(
0
,
∞
)
→
ℝ and let p
>
1. Then there exists a sequence f
n and a function β satisfying the conditions β
0
<
1
-
(
1
/
p
), β
∞
<
1
-
(
1
/
p
), and
(12)
lim
⁡
n
→
∞
|
β
(
δ
n
)
-
β
0
|
ln
⁡
1
δ
n
=
∞
:
δ
n
=
4
-
n
or
(13)
lim
⁡
n
→
∞
|
β
(
μ
n
)
-
β
∞
|
ln
⁡
μ
n
=
∞
:
μ
n
=
4
n
violating inequality (2).


## 3. Proof of Main Results

For the proof of Theorems1 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.
Supposes
:
ℝ
→
(
0
,
∞
) is a measurable function such that s
∈
Λ
0
∩
Λ
∞ and 0
<
s
-, s
+
<
∞; then it holds the estimate
(14)
C
1
-
1
x
s
(
0
)
≤
x
s
(
x
)
≤
C
1
x
s
(
0
)
,
for 0
<
x
<
1 and the estimate
(15)
C
2
-
1
x
s
(
∞
)
≤
x
s
(
∞
)
≤
C
2
x
s
(
∞
)
for x
≥
2.Proof of Theorem3.
To prove Theorem3 we apply the results of [2, 4], where it was proved that the following conditions
(16)
sup
⁡
x
∈
(
0,1
)
V
(
x
)
1
/
q
(
0
)
W
(
x
)
(
(
p
(
0
)
-
1
)
/
p
(
0
)
)
<
∞
,
sup
⁡
x
∈
(
2
,
∞
)
V
(
x
)
1
/
q
(
∞
)
W
(
x
)
(
(
p
(
∞
)
-
1
)
/
p
(
∞
)
)
<
∞
are necessary and sufficient for inequality (1) to hold if q
(
0
)
≥
p
(
0
), q
(
∞
)
≥
p
(
∞
), and the regularity conditions are satisfied:
(17)
limsup
⁡
x
→
0
[
p
(
x
)
-
p
(
0
)
]
ln
⁡
1
W
(
x
)
<
∞
;
limsup
⁡
x
→
∞
[
p
(
x
)
-
p
(
∞
)
]
ln
⁡
1
V
(
x
)
<
∞
,
where V
(
x
)
=
∫
x
∞
‍
v
(
t
)
d
t, W
(
x
)
=
∫
0
x
‍
ω
-
1
/
(
p
(
t
)
-
1
)
d
t.
In Theorem3, we have accepted that q
(
x
)
=
p
(
x
) and V
(
x
)
:
=
V
1
(
x
)
=
∫
x
∞
‍
t
(
β
(
t
)
-
1
)
p
(
t
)
d
t, W
(
x
)
:
=
W
1
(
x
)
=
∫
0
x
‍
t
-
(
β
(
t
)
p
(
t
)
)
/
(
p
(
t
)
-
1
)
d
t. It is easy to show that the conditions p
,
β
∈
Λ
∞ imply (
β
-
1
)
p
∈
Λ
∞. Therefore, it follows from Lemma 6 that
(18)
C
3
-
1
x
(
β
(
∞
)
-
1
)
p
(
∞
)
≤
x
(
β
(
x
)
-
1
)
p
(
x
)
≤
C
3
x
(
β
(
∞
)
-
1
)
p
(
∞
)
for x
≥
2 by some C
3
>
0. Also the conditions p
,
β
∈
Λ
0 imply -
(
β
p
/
(
p
-
1
)
)
∈
λ
0. Therefore, it follows from Lemma 6 that
(19)
C
4
-
1
x
-
(
β
(
0
)
p
(
0
)
/
(
p
(
0
)
-
1
)
)
≤
x
-
(
β
(
x
)
p
(
x
)
/
(
p
(
x
)
-
1
)
)
≤
C
4
x
-
(
β
(
0
)
p
(
0
)
/
(
p
(
0
)
-
1
)
)
for 0
<
x
<
1 by some C
4
>
0. Integrating these inequalities over the intervals (
x
,
∞
) and (
0
,
x
), respectively, we get
(20)
C
5
-
1
x
(
β
(
∞
)
-
1
)
p
(
∞
)
+
1
≤
V
1
(
x
)
≤
C
5
x
(
β
(
∞
)
-
1
)
p
(
∞
)
+
1
,
x
>
2
,
C
6
-
1
x
1
-
(
β
(
0
)
p
(
0
)
/
(
p
(
0
)
-
1
)
)
≤
W
1
(
x
)
≤
C
6
x
1
-
(
β
(
0
)
p
(
0
)
/
(
p
(
0
)
-
1
)
)
,
0
<
x
<
1
.
To complete the proof of Theorem3 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 Lemma6.
Let0
<
x
≤
δ and let s
(
x
)
≥
s
(
0
); then we have x
s
(
x
)
-
s
(
0
)
≤
1
+
δ
s
+
-
s
(
0
), where δ is a certain number from the interval (
0,1
). If 0
<
x
≤
δ and s
(
x
)
<
s
(
0
) then by condition s
∈
Λ
0 we have
(21)
x
s
(
x
)
-
s
(
0
)
=
(
δ
x
)
s
(
0
)
-
s
(
x
)
δ
s
(
x
)
-
s
(
0
)
≤
(
δ
x
)
c
1
/
ln
⁡
(
1
/
x
)
δ
s
(
x
)
-
s
(
0
)
=
e
c
1
δ
s
(
x
)
-
s
(
0
)
+
c
1
/
ln
⁡
(
1
/
x
)
≤
e
c
1
(
1
+
δ
c
1
/
ln
⁡
(
1
/
δ
)
)
(
1
+
(
1
δ
)
s
(
0
)
-
s
-
)
.
Therefore, for 0
<
x
≤
δ we have the estimation
(22)
x
s
(
x
)
-
s
(
0
)
≤
C
,
where the positive constant C depends on s
-, s
+, s
(
0
), δ. Same inequality holds for the function x
s
(
0
)
-
s
(
x
). Indeed, for 0
<
x
≤
δ and s
(
x
)
≥
s
(
0
) we have x
s
(
0
)
-
s
(
x
)
≤
1
+
δ
s
(
0
)
-
s
-. If 0
<
x
≤
δ and s
(
x
)
≥
s
(
0
) by condition s
∈
Λ
0 we have
(23)
x
s
(
0
)
-
s
(
x
)
=
(
δ
x
)
s
(
x
)
-
s
(
0
)
δ
s
(
0
)
-
s
(
x
)
≤
(
δ
x
)
c
1
/
ln
⁡
(
1
/
x
)
δ
s
(
0
)
-
s
(
x
)
=
e
c
1
δ
s
(
0
)
-
s
(
x
)
+
c
1
ln
⁡
(
1
/
x
)
≤
e
c
1
(
1
+
δ
c
1
ln
⁡
(
1
/
x
)
)
(
1
+
(
1
δ
)
s
+
-
s
(
0
)
)
.
By using these inequalities and by the representation
(24)
x
s
(
x
)
=
x
s
(
0
)
x
s
(
x
)
-
s
(
0
)
we have estimate (14).
To show estimate (15) note that for x
≥
M and s
(
x
)
≤
s
(
∞
) we have x
s
(
x
)
-
s
(
∞
)
≤
1
+
(
1
/
M
)
s
(
∞
)
-
s
-, where the M
≥
2 is a certain number. If x
≥
M and s
(
x
)
≥
s
(
∞
) then by the condition s
∈
Λ
∞ we have
(25)
x
s
(
x
)
-
s
(
∞
)
=
(
x
M
)
s
(
x
)
-
s
(
∞
)
M
s
(
x
)
-
s
(
∞
)
≤
(
x
M
)
C
2
/
ln
⁡
x
M
s
(
x
)
-
s
(
∞
)
=
e
C
2
M
s
(
x
)
-
s
(
∞
)
-
C
2
/
ln
⁡
x
≤
e
C
2
(
1
+
(
1
M
)
C
2
/
ln
⁡
M
)
(
1
+
M
s
(
∞
)
-
s
+
)
.
Combining the estimates for the functions x
s
(
x
)
-
s
(
∞
), x
s
(
∞
)
-
s
(
x
) for x
≥
M by the presentation
(26)
x
s
(
x
)
=
x
s
(
∞
)
x
s
(
x
)
-
s
(
∞
)
we get estimate (15). To complete the proof of Lemma 6, note that the condition s
∈
Λ
0 is equivalent to
(27)
-
C
≤
|
s
(
x
)
-
s
(
0
)
|
ln
⁡
1
x
≤
C
and the s
∈
Λ
∞ is equivalent to
(28)
-
C
≤
|
s
(
x
)
-
s
(
∞
)
|
ln
⁡
x
≤
C
,
respectively.Proof of Theorem4.
Let us assume thatf
k
(
x
)
=
x
-
(
1
/
β
(
x
)
)
-
β
χ
(
δ
k
,
2
δ
k
)
(
x
); x
∈
(
0
,
∞
). Fix k
∈
ℕ. We define the step function
(29)
p
(
x
)
=
{
p
0
+
a
n
if
x
∈
(
2
δ
n
,
4
δ
n
)
p
0
if
x
∈
(
δ
n
,
2
δ
n
)
,
n
∈
ℕ
.
Here α
n is a sequence of positive numbers that satisfies the condition n
α
n
→
∞ as n
→
∞. Then a
n
·
ln
⁡
⁡
(
1
/
δ
n
)
→
∞ as n
→
∞ and condition (10) is fulfilled for the function p
(
x
). We have
(30)
I
p
(
·
)
(
x
β
(
x
)
·
f
k
(
x
)
)
=
∫
δ
n
2
δ
n
(
t
β
·
t
-
(
1
/
p
0
)
-
β
)
p
0
d
t
=
ln
⁡
2
<
∞
,
I
p
(
·
)
(
x
β
(
x
)
-
1
·
H
f
k
(
x
)
)
≥
∫
2
δ
n
4
δ
n
[
∫
δ
n
2
δ
n
t
-
(
1
/
p
(
t
)
)
-
β
d
t
]
p
0
+
α
n
·
x
(
β
0
-
1
)
·
(
p
0
+
α
n
)
d
x
≥
∫
2
δ
n
4
δ
n
δ
n
(
1
-
(
1
/
p
0
)
-
β
)
(
p
0
+
α
n
)
·
x
(
β
-
1
)
·
(
p
0
+
α
n
)
d
x
≥
C
·
δ
n
-
α
n
/
p
0
=
C
·
e
α
n
/
p
0
ln
⁡
(
1
/
δ
n
)
⟶
∞
as n
→
∞. The last relation shows violating of inequality (2) for sufficiently large n.
We definef
k
(
x
)
=
x
-
(
1
/
p
(
x
)
)
-
β
χ
(
μ
k
,
2
μ
k
)
(
x
) for x
∈
(
0
,
∞
). Fix k
∈
ℕ. We also define the step function
(31)
p
(
x
)
=
{
p
∞
-
a
n
if
x
∈
(
2
μ
n
,
4
μ
n
)
p
∞
if
x
∈
(
μ
n
,
2
μ
n
)
,
n
∈
ℕ
,
where n
α
n
→
∞ as n
→
∞. We have α
n
ln
⁡
μ
n
→
∞ as n
→
∞ and condition (11) holds for the function p
(
x
). Furthermore,
(32)
I
p
(
·
)
(
x
β
·
f
k
(
x
)
)
=
∫
μ
n
2
μ
n
(
x
β
·
x
-
(
1
/
p
∞
)
-
β
)
p
∞
d
x
=
ln
⁡
2
<
∞
,
I
p
(
·
)
(
x
β
-
1
·
H
f
k
(
x
)
)
≥
∫
2
μ
n
4
μ
n
[
∫
μ
n
2
μ
n
t
-
(
1
/
p
(
t
)
)
-
β
d
t
]
(
p
∞
-
α
n
)
·
x
(
β
-
1
)
(
p
∞
-
α
n
)
d
x
≥
C
μ
n
α
n
/
p
∞
=
C
e
(
α
n
/
p
∞
)
ln
⁡
μ
n
⟶
∞
as n
→
∞, which contradicts (2) for sufficiently large n.Proof of Theorem5.
Let us assume thatf
k
(
x
)
=
x
-
(
1
/
p
)
-
β
(
x
)
χ
(
δ
k
,
2
δ
k
)
(
x
); x
∈
(
0
,
∞
). Fix k
∈
ℕ. We define the step function β as
(33)
β
(
x
)
=
{
β
0
+
a
n
if
x
∈
(
δ
n
,
2
δ
n
)
β
0
if
x
∈
(
2
δ
n
,
4
δ
n
)
,
n
∈
ℕ
,
where a
n
·
ln
⁡
(
1
/
δ
n
)
→
∞. Then,
(34)
I
p
(
·
)
(
x
β
(
x
)
·
f
k
(
x
)
)
=
∫
δ
n
2
δ
n
(
x
β
0
+
α
n
·
x
-
(
1
/
p
)
-
β
0
-
α
n
)
p
d
x
=
ln
⁡
2
<
∞
,
I
p
(
·
)
(
x
β
(
x
)
-
1
·
H
(
f
k
(
x
)
)
)
≥
∫
2
δ
n
4
δ
n
[
∫
δ
n
2
δ
n
t
-
(
1
/
p
)
-
β
0
-
α
n
d
t
]
p
·
x
(
β
0
-
1
)
p
d
x
=
C
·
δ
n
-
p
α
n
=
C
·
e
p
α
n
ln
⁡
(
1
/
δ
n
)
⟶
∞
as n
→
∞. The last relation contradicts the validity of inequality (2).
We definef
k
(
x
)
=
x
-
(
1
/
x
)
-
β
(
x
)
χ
(
μ
k
,
2
μ
k
)
(
x
); x
∈
(
0
,
∞
). Fix k
∈
ℕ, where the function β is defined as
(35)
β
(
x
)
=
{
β
∞
-
a
n
if
x
∈
(
μ
n
,
2
μ
n
)
β
∞
if
x
∈
(
2
μ
n
,
4
μ
n
)
,
n
∈
ℕ
,
where lim
⁡
n
→
∞
a
n
·
ln
⁡
μ
n
=
∞. Then,
(36)
I
p
(
·
)
(
x
β
(
x
)
·
f
k
(
x
)
)
=
∫
μ
n
2
μ
n
(
x
β
∞
-
α
n
·
x
-
(
1
/
p
)
-
β
∞
+
α
n
)
p
d
x
=
ln
⁡
2
<
∞
,
I
p
(
·
)
(
x
β
(
x
)
-
1
·
H
(
f
k
(
x
)
)
)
≥
∫
2
μ
n
4
μ
n
[
∫
μ
n
2
μ
n
t
-
(
1
/
p
)
-
β
∞
+
α
n
d
t
]
p
·
x
(
β
∞
-
1
)
p
d
x
≥
∫
2
μ
n
3
μ
n
μ
n
(
1
-
(
1
/
p
)
-
β
∞
+
α
n
)
p
·
x
(
β
∞
-
1
)
p
d
x
≥
C
·
μ
n
p
α
n
=
C
·
e
p
α
n
ln
⁡
μ
⟶
∞
as n
→
∞ which contradicts inequality (2).



---

*Source: 606012-2014-04-23.xml*