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Volume 2010, Article ID 837951, 14 pagesdoi:10.1155/2010/837951 Research Article On Boundedness of Weighted Hardy Operator in Aziz Harman1 and Farman Imran Mamedov1, 2 1 Education Facult

Volume 2010, Article ID 837951, 14 pages

doi:10.1155/2010/837951

Research Article

On Boundedness of Weighted Hardy Operator in

Aziz Harman1 and Farman Imran Mamedov1, 2

1 Education Faculty, Dicle University, 21280 Diyarbakir, Turkey

2 Institute of Mathematics and Mechanics of National Academy of Science, Azerbaijan

Correspondence should be addressed to Farman Imran Mamedov,m.farman@dicle.edu.tr

Received 22 September 2010; Accepted 26 November 2010

Academic Editor: P J Y Wong

Copyrightq 2010 A Harman and F I Mamedov This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

We give a new proof for power-type weighted Hardy inequality in the norms of generalized

Lebesgue spaces L p·Rn Assuming the logarithmic conditions of regularity in a neighborhood

of zero and at infinity for the exponents p x ≤ qx, βx, necessary and sufficient conditions are proved for the boundedness of the Hardy operator Hfx  |y|≤|x| f ydy from L p·

|x| β·Rn into

L q·

|x| β ·−n/p·−n/q·RN Also a separate statement on the exactness of logarithmic conditions at zero

and at infinity is given This shows that logarithmic regularity conditions for the functions β, p at

the origin and infinity are essentially one

1 Introduction

The object of this investigation is the Hardy-type weighted inequality



|x| β ·−n/p·−n/q· Hf

L q· Rn≤ C|x| β·f

L p· Rn, Hf x 



|y|≤|x| f

in the norms of generalized Lebesgue spaces L p·Rn This subject was investigated in the papers1 7 For the one-dimensional Hardy operator in 1, the necessary and sufficient

condition was obtained for the exponents β, p, q We give a new proof for this result in

more general settings for the multidimensional Hardy operator Also we prove that the logarithmic regularity conditions are essential one for such kind of inequalities to hold In that proposal, we improve a result sort of8 since, there is an estimation by the maximal function|x| −n Hf x ≤ CMfx.

At the beginning, a one-dimensional Hardy inequality was considered assuming the the local log condition at the finite interval 0, l Subsequently, the logarithmic condition

was assumed in an arbitrarily small neighborhood of zero, where an additional restriction

p x ≥ p0 was imposed on the exponent In 3,9 it was shown that it is sufficient to assume the logarithmic condition only at the zero point In10 the case of an entire semiaxis was

considered without using the condition px ≥ p0 However, a more rigid condition β <

1− 1/p− was introduced for a range of exponents The exact condition was found in 1 They proved this result by using of interpolation approaches In this paper, we use other approaches, analogous to those in10, based on the property of triangles for px-norms and

binary decomposition near the origin and infinity We consider the multidimensional case,

and the condition βx  const is not obligatory, while the necessary and sufficient condition

is obtained by a set of exponents p, q, β without imposing any preliminary restrictions on their

valuesTheorems3.1and3.2 InTheorem 3.3, it has been proved that logarithmic conditions

at zero and at infinity are exact for the Hardy inequality to be valid in the case q  p.

Problems of the boundedness of classical integral operators such as maximal and singular operators, the Riesz potential, and others in Lebesgue spaces with variable exponent,

as well as the investigation of problems of regularity of nonlinear equations with nonstandard growth condition have become of late the arena of an intensive attack of many authorssee

11–18

2 Lebesgue Spaces with a Variable Exponent

As to the basic properties of spaces L p·, we refer to19 Throughout this paper, it is assumed

that px is a measurable function in Ω, where Ω ∈ R nis an open domain, taking its values from the interval1, ∞ with p  supx∈Rn p < ∞ The space of functions L p·Ω is introduced

as the class of measurable functions f x in Ω, which have a finite I p f : Ω|fx| p x

dx-modular A norm in L p·Ω is given in the form

f

L p· Ω inf



λ > 0 : I p



f λ

For p−> 1, p< ∞ the space L p·Ω is a reflexive Banach space

Denote byΛ a class of measurable functions f : R n → R satisfying the following conditions:

∃m ∈



0,1

2

, ∃f0 ∈ R, sup

x ∈B0,m

f x − f0 ln 1

|x| < ∞, 2.2

∃M > 1, ∃f∞ ∈ R, sup

x ∈R n \B0,M

f x − f∞ ln|x| < ∞. 2.3

For the exponential functions βx, px, and qx, we further assume β, p, q ∈ Λ.

We will many times use the following statement in the proof of main results

Lemma 2.1 Let s ∈ Λ be a measurable function such that −∞ < s−, s < ∞ Then the condition

2.2 for the function sx is equivalent to the estimate

C−13 |x| s0≤ |x| s x ≤ C3|x| s0 2.4

when |x| ≤ m and the condition 2.3 for sx is equivalent to the estimate

C4−1|x| s∞ ≤ |x| s x ≤ C4|x| s∞ 2.5

when |x| ≥ M Where the constants C3, C4> 1 depend on s 0, s∞, s−, s, s 0, s∞, m, M, C1,

C2.

To proveLemma 2.1, for example2.4, it suffices to rewrite the inequality 2.4 in the form

and pass to logarithmic in this inequalitysee also, 1,7,17

For 1 < p < ∞, p denotes the conjugate number of p, p  p/p − 1 It is further assumed that p ∞ for p  1, and p 1 for p  ∞, 1/∞  0, 1/0  ∞ We denote by C, C1, C2

various positive constants whose values may vary at each appearance Bx, r denotes a ball with center at x and radius r > 0 We write u ∼ v if there exist positive constants C3, C4such

that C3u x ≤ vx ≤ C4u x By χ E , we denote the characteristic function of the set E.

3 The Main Results

The main results of the paper are contained in the next statements The theorem below gives

a solution of the two-weighted problem for the multidimensional Hardy operator in the case

of power-type weights

Theorem 3.1 Let qx ≥ px and βx be measurable functions taken from the class Λ Let the

following conditions be fulfilled:

0 < p−≤ px, q x ≤ q < ∞, −∞ < β−≤ βx ≤ β< ∞. 3.1

p 0 > 1, p∞ > 1, β 0 < n



1− 1

, β ∞ < n



1− 1

We have the following analogous result for the conjugate Hardy operator Hf x 



|y|≥|x| f ydy.

Theorem 3.2 Let qx ≥ px and βx be measurable functions taken from the class Λ Let the

operator Hf is fulfilled if and only if

p 0 > 1, p∞ > 1, β 0 > n



1− 1

, β ∞ > n



1− 1

In the next theorem, we prove that the logarithmic conditions near zero and at infinity are essentially one

Theorem 3.3 If condition 2.2 or 2.3 does not hold, then there exists an example of functions p, β,

and a sequence f below index k violating the inequality



|x| β ·−n Hf

L p· Rn ≤ C|x| β·f

4 Proofs of the Main Results

Sufficiency Let fx ≥ 0 be a measurable function such that



|x| β·f

We will prove that



|x| β ·−n/p·−n/q· Hf

L q· Rn≤ C5. 4.2

Assume that 0 < δ < m is a sufficiently small number such that n/px > n/p0 − ε for all x ∈ B0, δ, where ε  n/p0−β0/2 Let, furthermore, M < N < ∞ be a sufficiently large number such that n/px > n/p∞−δ1for all x∈ Rn \B0, N, where δ1 n/p∞−

β ∞/2.

By Minkowski inequality, for px-norms, we have



|x| β ·−n/p·−n/q· Hf

L q· Rn≤|x| β ·−n/p·−n/q· Hf

L q·B0,δ

|x| β ·−n/p·−n/q· Hf

L q·B0,N\B0,δ





|x|β ·−n/p

·−n/q·

{t:|t|<N} f tdt





L q· Rn \B0,N





|x|β ·−n/p

·−n/q·

{t:N<|t|<|x|} f tdt





L q· RN \B0,N

: i1 i2 i3 i4.

4.3

The estimate near zero i1.

By Minkowski inequality, we have the inequalities

i1≤



|x|β ·−n/p

·−n/q·∞

k0

  {t:2 −k−1 |x|<|t|<2 −k |x|}f tdt







L q·B0,δ

≤∞

k0





|x|β ·−n/p

·−n/q·

{t:2 −k−1 |x|<|t|<2 −k |x|} f tdt





L q·B0,δ

.

4.4

Denote B x,k  {y ∈ R n : 2−k−1 |x| < |y| < 2 −k |x|} and p−

x,k  minpx, inf y ∈B x,k p y.

By2.2 andLemma 2.1, for x ∈ B0, δ, t ∈ B x,k, we have|x| β x ∼ 2kβ0t β t To prove this equivalence, we use that|t| ∼ |x|2 −k , |x| β x ∼ |x| β0and|t| β t ∼ |t| β0 Therefore, and due to

Holder’s inequality, for x ∈ B0, δ, we get

|x| β x−n/px−n/qx

B x,k

f tdt

≤ C62kβ0|x| −n/px−n/qx

B x,k

|t| β t f tdt

≤ C62kβ0|x| −n/p0−n/qx

B x,k



|t| β t f tp x,k−

dt

1/p−

x,k

2−k |x|n/ p−x,k 

.

4.5

a If p−

x,k /  px, then by 2.2 andLemma 2.1,



2−k |x|n/ p x,k−  

∼ t n/pt ∼ t n/p0∼ 2−kn/p 0|x| n/p0∼ 2−kn/p 0|x| n/px 4.6

Demonstrate details in proof of4.6 For t ∈ B x,k and x ∈ B0, δ, we have 2 −k−1 |x| <

|t| ≤ 2 −k |x| Then



2−k |x|n/ p x,k− 

∼ |t| n/ p−

x,k 

By hypothesisa, p−

x,k attains in the interval B x,k , because there exists a point y ∈ B x,kwhere

p−x,k ∼ py Obviously, the point y depends on x, k Then |t| n/ p−

x,k ∼ |t| n/py By virtue of

2−k−1 |x| < |y| ≤ 2 −k−1 |x|, we have |t|/2 < |y| ≤ 2|t| Hence, |t| n/py ∼ |y| n/py, byLemma 2.1,

|y| n/py ∼ |y| n/p0∼ |t| n/p0

b If p−

x,k  px, then by choice of δ,



2−k |x|n/ p−x,k 

∼ 2−kn/px |x| n/px≤ 2−kn/p0εk |x| n/px; x ∈ B0, δ. 4.8

Applying estimate4.8 to both hypotheses a and b, by choosing of ε and δ, the right-hand

part of4.5 is less than

C7|x| −n/qx2−kε

B x,k



|t| β t f tp−x,k

dt

1/p−

x,k

Simultaneously,



B x,k



|t| β t f tp−x,k

dt

≤



B x,k ∩{t∈R n:|t|β t f t≥1}



|t| β t f tp t dt



B x,k

dt≤ 1  2−kn δ n  C8.

4.9

By4.5 and 4.9, we have

I q;B 0,δ |x| β ·−n/p·−n/q·

B x,k

f tdt



≤ C92−kεq−



B 0,δ |x| −n

B x,k



|t| β t f tp x,k−

dt

q x/p−

x,k

dx

≤ C9C8q/p−−12−kεq−



B 0,δ B x,k



|t| β t f tp t 1

dt



|x| −n dx

4.10

which, due to Fubini’s theorem, yields

≤ C9C q8/p−−12−kεq−



{t:|t|<2 −k δ}



f t|t| β tp t

B0,2 k1|t|\B0,2 k |t||x|

−n dx



dt

 C102−kεq−ln 2



{t:|t|<2 −k δ}



f t|t| β tp t

 1

4.11

Therefore,





|x|β ·−n/p

·−n/q·

B x,k

f tdt





L q·B0,δ

≤ C122−kεq−/q. 4.12

By4.12 and 4.4, we get

∞

k0

2−kεq−/q C13< ∞. 4.13

Put f N t  ftχ |t|>N Analogously to the case of4.4, we have

k0





|x|β ·−n/p

·−n/q·

{t:2 −k−1 |x|<|t|<2 −k |x|} f N tdt





L q· Rn \B0,N

By|t| ∼ |x|2 −k, condition2.3 andLemma 2.1for x∈ Rn \ B0, N, t ∈ B x,k, we have

|x| β x ∼ |x| β∞∼ 2kβ∞t β∞∼ 2kβ∞t β t 4.15 Therefore, by virtue of Holder’s inequality,

|x| β x−n/px−n/qx

B x,k

f N tdt

≤ C142kβ∞|x| −n/px−n/qx



B x,k

|t| β t f N tdt

≤ C142kβ∞|x| −n/px−n/qx

B x,k



|t| β t f N tp−x,k

dt

1/p−

x,k

2−k |x|n/ p−x,k 

.

4.16

i If p−

x,k /  px and t ∈ B x,k, by2.3 andLemma 2.1, we have



2−k |x|n/ p−x,k

∼ t n/pt ∼ t n/p∞ ∼ 2−kn/p ∞|x| n/p∞∼ 2−kn/p ∞|x| n/px 4.17

ii If p−

x,k  px, then by choice of δ1,



2−k |x|n/ p−x,k 

∼ 2−kn/px |x| n/px≤ 2−kn/p∞δ1k |x| n/px 4.18

In both hypothesesi and ii by choosing of δ1, we have

|x| β x−n/px−n/qx

B x,k

f N tdt ≤ C15|x| −n/qx2−kδ1

B x,k



|t| β t f N tp−x,k

dt

1/p−

x,k

. 4.19

On the other hand,



B x,k



|t| β t f tp x,k−



B x,k ∩{t∈R n:|t|β t f t≥Gt}

|t| β t f t

p−

x,k

G t p−x,k dt



B x,k

G t p−dt,

4.20

where Gt  1/1  t2 Hence,

≤



B x,k



f N t|t| β tp t

G t p−x,k −pt



B x,k

By2.3, for t ∈ B x,k, we have

G t p−x,k −pt≤1 t2p t−p−

x,k

Then4.21 implies



B x,k



|t| β t f N tp−x,k

Therefore,

I q;Rn \B0,N |x| β x−n/px−n/qx

B x,k

f N tdt



≤ C q17/p−2−kδ1q−



Rn \B0,N |x| −n

B x,k



|t| β t f N tp t dt



dx,

4.24

by Fubini’s theorem,

≤ C q/p−−1

17 2−kδ1q−ln 2



{t:|t|>2 −k N}



f N t|t| β tp t

dt ≤ C182−kδ1q−. 4.25

From4.25 and expansion 4.14, we get

∞

k0

We have

i2



|x|β ·−n/p

·−n/q·

{t∈R n:|t|<|x|}f tdt





L q·B0,N\B0,δ

≤

B 0,N f tdt

|x| β ·−n/p·−n/q·

L q·B0,N\B0,δ

≤ C20



B 0,N f tdt,

4.27

from which, by virtue of Holder’s inequality, for px-norms, we obtain the estimate



B 0,N f tdt ≤|t| β·f t

L p·B0,N



|t| −β·

L p·B0,N 4.27

Using t −βtpt ∼ t −β0p 0 by Lemma 2.1for t ∈ B0, N and taking the condition β0 <

n/p0 into account, we find

I p;B0,N



|t| −β·



B 0,N |t| −βtpt dt ≤ C21



B 0,N |t| −β0p0dt  C22. 4.28 From4.27 and 4.28, it follows that

Furthermore, we have

i3≤

B 0,N f tdt

|x| β x−n/px−n/qx

L q· Rn \B0,δ

The boundedness of the first term follows by4.27 Due to 2.3 andLemma 2.1, for x ∈

Rn \ B0, N, we have

|x| βx−n/pxqx−n ∼ |x| β∞−n/p∞qx−n 4.31 Applying condition4.31, we get

I q;Rn /B 0,N



|x| β ·−n/p·−n/q·

≤ C24



Rn \B0,N |x| −n−2δ1dx  C25. 4.32

Then

test function

f τ t  t −n/pt−βt χ B 0,δ/τ\B0,δ/2τ t. 4.34

We come to a contradiction

I p

|t| β·f τ





B 0,δ/τ\B0,δ/2τ |x| −n dx  C0ln 2 < ∞,

I q |t| β ·−n/p·−n/q·

B 0,t f τ



dy



≥



B 0,1\B0,δ/τ |t| βt−n/pt−n/qtqt

B 0,δ/τ\B0,δ/2τ

y −n/p0−β0

dy

q t

dt

≥



δ

2τ

n/p0−β0q−

B 0,1\B0,δ/τ |t| β0−n/p0qt−n dt−→ ∞

4.35

as τ → ∞

If 0 < p0 ≤ 1, then by virtue of inequalities 4.35 and 3.2 we obtain

I q |t| β t−n/pt−n/qt

B 0,t f τ



dy



Also,

I p

|t| β t f τ t C0ln 2, 4.37

and we come to a contradiction

If β∞ ≥ n/p∞, then, using condition 2.3 andLemma 2.1assuming 0 < τ < 1, we

again obtain

I p



|t| β t f τ t C0ln 2,

I q |t| β t−n/pt−n/qt

B 0,t f τ tdy



≥



Rn \B0,δ/τ |t| βt−n/ptqt−n

B 0,δ/τ\B0,δ/2τ

y −n/p∞−β∞ dy



dt

≥



δ

2τ

n/p∞−β∞q

Rn \B0,δ/τ |t| β∞−n/p∞qt−n dt−→ ∞

4.38

as τ → ∞ If β∞  n/p∞, then from 4.38 we have

I q |t| β t−n/pt−n/qt

B 0,t f τ tdy



From4.38 and 3.2, we derive, as above, the necessity of the condition p∞ > 1.

This completes the proof ofTheorem 3.1

The proof ofTheorem 3.2easily follows fromTheorem 3.1by using the equivalence of inequalities



|x| β x−n/px−n/qx Hf x

L q· Rn≤ C|x| β x f x

L p· Rn,



|z| n −βz−2n/qz Hf x



L q· Rn≤ C|z| −βz−2n/pz f z

L p· Rn,

4.40

where px, qx, and βx stand for the functions px/|x|2, qx/|x|2, and βx/|x|2, respectively The equivalence readily follows from the equality

g

L p· Rn|z| −2n/pz g

for any function g :Rn → R, where gz  gz/|z|2, which easily can be proved by changing

of variable x  z/|z|2in the definition of px-norm.

5 Exactness of the Logarithmic Conditions

β x  β0 Define the function p : 0, ∞ → 1, ∞ as

p x 

⎧

⎨

⎩

p0, x ∈ B0, 2δ k  \ B0, δ k ,

p k , x ∈ B0, 4δ k  \ B0, 2δ k , k ∈ N 5.1

where p0 > 1, p k  p0 α k , β0 ∈ R, and {α k} is an arbitrary sequence of positive numbers satisfying the condition

Applying estimate4.8 to both hypotheses a and b, by choosing of ε and δ, the right-hand

part...

from which, by virtue of Holder’s inequality, for px-norms, we obtain the estimate



B... 4.37

and we come to a contradiction

If β∞ ≥ n/p∞, then, using condition 2.3 andLemma 2.1assuming < τ < 1, we

again obtain

I