Báo cáo hóa học: "LIMITING CASE OF THE BOUNDEDNESS OF FRACTIONAL INTEGRAL OPERATORS ON NONHOMOGENEOUS SPACE" pptx

INTEGRAL OPERATORS ON NONHOMOGENEOUS SPACEYOSHIHIRO SAWANO, TAKUYA SOBUKAWA, AND HITOSHI TANAKA Received 13 April 2006; Accepted 12 June 2006 We show the boundedness of fractional integr

INTEGRAL OPERATORS ON NONHOMOGENEOUS SPACE

YOSHIHIRO SAWANO, TAKUYA SOBUKAWA, AND HITOSHI TANAKA

Received 13 April 2006; Accepted 12 June 2006

We show the boundedness of fractional integral operators by means of extrapolation We also show that our result is sharp

Copyright © 2006 Yoshihiro Sawano et al 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

1 Introduction

Recently, harmonic analysis onRd with nondoubling measures has been developed very rapidly; here, by a doubling measure, we mean a Radon measure μ onRd satisfying

μ(B(x,2r)) ≤ c0μ(B(x,r)), x ∈supp(μ), r > 0 In what follows, B(x,r) is the closed ball

centered atx of radius r In this paper, we deal with measures which do not necessarily

satisfy the doubling condition

We can list [7,8,11] as important works in this field Tolsa proved subadditivity and bi-Lipschitz invariance of the analytic capacity [12,13] Many function spaces and many linear operators for such measures stem from their works For example, Tolsa has defined the Hardy spaceH1(μ) [11] Han and Yang have defined the Triebel-Lizorkin spaces [3]

In the present paper, we mainly deal with the fractional integral operators We occa-sionally postulate the growth condition onμ:

μ is a Radon measure onRdwithμ

B(x,r)

≤ c0r n for somec0> 0, 0 < n ≤ d.

(1.1)

A growth measure is a Radon measureμ satisfying (1.1) We define the fractional inte-gral operatorI αassociated with the growth measureμ as

I α f (x) : =



Rd

f (y)

| x − y | nα dμ(y), 0< α < 1. (1.2)

Hindawi Publishing Corporation

Journal of Inequalities and Applications

Volume 2006, Article ID 92470, Pages 1 16

DOI 10.1155/JIA/2006/92470

Let 1/q =1/ p −(1− α) with 1 < p < q < ∞. L p(μ)-L q(μ) boundedness of I αin a more general form was proved by Kokilashvili [4] On general nonhomogeneous spaces, that

is, on metric measure spaces, it was also proved in [5] (see [1]) In [2], the limit case

p =1/(1 − α) was considered In general, the integral defining I α f (x) does not converge

absolutely forμ- a.e., if f ∈ L1/(1 − α)(μ) Garc´ıa-Cuerva and Gatto considered some

mod-ified operator and showed its boundedness fromL1/(1 − α)(μ) to some BMO-like space

de-fined in [11]

This paper deals mainly with the Morrey spaces By a cube, we mean a set of the form

Q(x,r) : =x1− r,x1+r

× ··· ×x d − r,x d+r

, x =x1, ,x d

∈ R d, 0< r ≤ ∞

(1.3)

Given a cubeQ = Q(x,r), κ > 0, we denote κQ : = Q(x,κr) and (Q) =2r We define ᏽ(μ)

by

ᏽ(μ) :=Q ⊂ R d:Q is a cube with 0 < μ(Q) < ∞. (1.4) Now we are in the position of describing the Morrey spaces for nondoubling measures

Definition 1.1 (see [10, Section 1]) Let 0< q ≤ p < ∞, k > 1 Denote by ᏹ q p(k,μ) a set of

L qloc(μ) functions f for which the quasinorm

f : ᏹ p(k,μ) := sup

Q ∈ ᏽ(μ) μ(kQ)1/ p −1/q

Q

f (y) q

dμ(y)

1/q

< ∞ (1.5)

Note that this definition does not involve the growth condition (1.1) So in this paper,

we assumeμ is just a Radon measure unless otherwise stated.

Key properties that we are going to use can be summarized as follows

Proposition 1.2 (see [10, Proposition 1.1]) Let 0 < q ≤ p < ∞ , k1> k2> 1 Then there exists C d,k1 ,k2 ,q so that, for every μ-measurable function f ,

f : ᏹ p

q

k2,μ ≤ f : ᏹ q p

k1,μ ≤ C d,k1 ,k2 ,q f : ᏹ p

q

k2,μ . (1.6) The proof is omitted: interested readers may consult [10] However, we deal with sim-ilar assertion whose proof is wholly included in this present paper

Lemma 1.3 (see [10, Section 1]) (1) Let 0 < q1≤ q2≤ p < ∞ and k > 1 Then

f : ᏹ p

q1(k,μ) ≤ f : ᏹ q p2(k,μ) ≤ f : ᏹ p p(k,μ) = f : L p(μ) . (1.7)

(2) Let μ(R d)< ∞ and 0 < q ≤ p1≤ p2< ∞ Then

f : ᏹ p1

q (k,μ) ≤ μ

Rd 1/ p1−1/ p2 f : ᏹ p2

Proof Equation (1.7) is straightforward by using the H¨older inequality.

As for (1.8), thanks to the finiteness ofμ writing out the left-hand side in full, we have

f : ᏹ p1

q (k,μ) = sup

Q ∈ ᏽ(μ) μ(kQ)1/ p1−1/q

Q

f (y) q

dμ(y)

1/q

≤ sup

Q ∈ ᏽ(μ) μ

Rd 1/ p1−1/ p2

μ(k Q)1/ p2−1/q

Q

f (y) q

dμ(y)

1/q

= μ

Rd 1/ p1−1/ p2 f : ᏹ p2

q (k,μ) .

(1.9)

KeepingProposition 1.2in mind, for simplicity, we denote

ᏹp

q(μ) : =ᏹp

q(2,μ), ·:ᏹp

q(μ) := ·:ᏹp

q(2,μ) . (1.10)

In [10, Theorem 3.3], we showed thatI αis bounded fromᏹp

q(μ) to ᏹ s t(μ), if q

p = t

s,

1

s =1

p −(1− α), 1< q ≤ p < ∞, 1 < t ≤ s < ∞, 0 < α < 1. (1.11) Having described the main function spaces, we present our problem In the present paper, from the viewpoint different from [2], we will consider the limit case of the bound-edness ofI αas “p →1/(1 − α)” or “s → ∞,” where p and s satisfy (1.11)

Problem 1.4 Let 0 < α < 1 and assume that μ is a finite growth measure Find a nice

function spaceX to which I αsendsᏹ1/(1 − α)

q (μ) continuously, where 1 < q ≤1/(1 − α).

Although the Morrey spaces are the function spaces coming with two parameters, we arrangeᏹp

q(μ) to ᏹ βp p (μ) with β ∈(0, 1] fixed and regard them as a family of function spaces parameterized only by p We turn our attention to the family of spaces {ᏹ p

βp(μ) } p ∈(0,∞) We also consider the generalized version ofProblem 1.4

Problem 1.5 Let μ be finite and 0 < p0< p < r < ∞, 0 < β ≤1, 1/s =1/ p −1/r Suppose

that we are given an operatorT from p>p0ᏹp

βp(μ) to s>0ᏹs

βs(μ) Assume, restricting T

toᏹp

βp(μ), we have a precise estimate

T f : ᏹ s

βs(μ) ≤ c(s) f : ᏹ p

where 1/s =1/ p −1/r with p,r,s > 0 Then what can we say about the boundedness of T

on the limit function spaceᏹr

βr(μ)?

Here we describe the organization of this paper.Section 2is devoted to the definition

of the function spaces to answer Problems1.4and 1.5 InSection 3, we give a general machinery for Problems1.4and1.5.I αappearing here will be an example of the theorem

inSection 3 BesidesI α, we take up two types of other fractional integral operators The task inSection 4is to determinec(s) in (1.12) precisely We skillfully use two types of fractional integral operators as well asI αto see the size ofc(s) InSection 5, we exhibit an

example showing the sharpness of the estimate ofc(s) obtained inSection 4 The example will reveal us the difference between the Morrey spaces and the Lpspaces

2 Orlicz-Morrey spacesᏹΦ

β(μ)

In this section, we introduce function spacesᏹΦ

β(μ) to formulate our main results E.

Nakai definedᏹΦ

β(μ) for Lebesgue measure μ = dx We denote by | E |the volume of a measurable setE Let Φ : [0, ∞) →[0,∞) be a Young function, that is,Φ is convex with Φ(0)=0 and limx →∞ Φ(x) = ∞.

Forβ ∈(0, 1], E Nakai has defined the Orlicz-Morrey spaces: the spaceᏹΦ

β(dx)

con-sists of all measurable functions f for which the norm

f : ᏹΦ

β(dx) :=infλ > 0 : sup

Q ∈ ᏽ(dx) | Q | β −1



QΦ f (y)

λ



dy ≤1



< ∞ (2.1)

For details, we refer to [6]

Motivated by this definition and that of ᏹp

q(μ) with 0 < q ≤ p < ∞, we define the

Orlicz-Morrey spacesᏹΦ

β(μ) as follows.

Definition 2.1 Let β ∈(0, 1],k > 1, and Φ be a Young function Then define

f : ᏹΦ

β(k,μ) :=infλ > 0 : sup

Q ∈ ᏽ(μ) μ(kQ) β −1



QΦ f (y)

λ



dμ(y) ≤1



. (2.2)

We define the function spaceᏹΦ

β(k,μ) as a set of μ-measurable functions f for which the

norm is finite

The function spaceᏹΦ

β(k,μ) is independent of k > 1 More precisely, we have the

fol-lowing

Proposition 2.2 Let k1> k2> 1 Then there exists constant C d,k1 ,k2such that

f : ᏹΦ

β



k1,μ ≤ f : ᏹ βΦ

k2,μ ≤ C d,k1 ,k2 f : ᏹΦ

β



k1,μ . (2.3)

Here, C d,k1 ,k2> 0 is independent of f

Proof By the monotonicity of  f : ᏹΦβ(k,μ) with respect tok, the left inequality is

ob-vious What is essential in (2.3) is the right inequality The monotonicity allows us to assume thatk1=2k2−1 We takeQ ∈ ᏽ(μ) arbitrarily We have to majorize

inf



λ > 0 : μ

k2Qβ −1 

QΦ f (x)

λ



dμ(x) ≤1



(2.4)

byλ0:= f : ᏹΦβ(k1,μ) uniformly overQ.

BisectQ into 2 d cubes and labelQ1,Q2, ,Q Lto those inᏽ(μ), then the distance

be-tween the boundary ofk2Q and the center of Q jis

k2

2 −1

4

(Q) = k1

Consequently, we havek1Q j ⊂ k2Q for j =1, 2, ,L This inclusion gives us that

μ

k2Qβ −1 

QΦ f (x)

λ0



dμ(x) ≤L

j =1

μ

k1Q jβ −1 

Q j

Φ f (x)

λ0



dμ(x) ≤2d (2.6)

Note thatΦ(tx) ≤ tΦ(x) for 0 ≤ t ≤1 by convexity As a result, we obtain

sup

Q ∈ ᏽ(μ) μ

k2Qβ −1 

QΦ f (x)

2d λ0

Thus we have obtained

f : ᏹΦ

β



k2,μ ≤2d λ0=2d f : ᏹΦ

β



Hence we have established that we can takeC d,2k2−1,k2=2d 

Keeping this proposition in mind, we setᏹΦ

β(μ) : =ᏹΦ

β(2,μ) The same argument as

Proposition 2.2works forProposition 1.2

3 Extrapolation theorem on the Morrey spaces

In this section, we will prove the key lemma dealing with an extrapolation theorem on the Morrey spaces Assume thatμ is finite and

0< p0< p < r < ∞, 0< β ≤1, 1

s = 1

p −1

LetT be an operator from ᏹ βp p (μ) to ᏹ s βs(μ) with a precise estimate

T f : ᏹ s

βs(μ) ≤ cs ρ f : ᏹ p

βp(μ) , ρ > 0. (3.2) Then we can say that the limit result of

T : ᏹ βp p (μ) −→ᏹs

βs(μ), p0< p < r, 1

s = 1

p −1

asp → r, s → ∞, is

T : ᏹ r βr(μ) −→ᏹΦ

whereΦ(x) =exp(x1/ρ)−1 More precisely, our main extrapolation theorem is the fol-lowing

Theorem 3.1 Suppose μ(R d)< ∞ Let 0 < p0< r, 0 < ρ ≤ 1, and 0 < β ≤ 1 Suppose that the sublinear operator T satisfies

T f : ᏹ s

βs(μ) ≤ C0s ρ f : ᏹ p

βp(μ) ∀ f ∈ᏹp

for each p0≤ p < r with 1/s =1/ p −1/r Here, C0> 0 is a constant independent of p and s Then there exists a constant δ > 0 such that

sup

Q



Q



exp δ

f : ᏹ T f (x) r

βr(μ) 1/ρ−1

 dμ(x)

μ(2Q)1− β



≤1 ∀ f ∈ᏹr

βr(μ) (3.6)

or equivalently

T f : ᏹΦ

β(μ) ≤ δ −1/ρ f : ᏹ r

βr(μ) ∀ f ∈ᏹr

for Φ(t) =exp(t1/ρ)− 1.

More can be said about this theorem: the case whenβ =1 corresponds to the Zygmund-type extrapolation theorem (see [15]) SetLΦ(μ) =ᏹΦ

1(μ).

Corollary 3.2 Keep to the same assumption as Theorem 3.1 on μ, ρ, p0, r, and T Suppose

T f : L s(μ) ≤ C0s ρ f : L p(μ) ∀ f ∈ L p(μ) (3.8)

for s, p with 1/s =1/ p −1/r Here, C0> 0 is a constant independent of p and s Then there exists some constant δ > 0 such that



Rd

 exp δ

f : L T f (x) r(μ) 1/ρ−1



dμ(x) ≤1 ∀ f ∈ L r(μ) (3.9)

or equivalently

T f : LΦ(μ) ≤ δ −1/ρ f : L r(μ) ∀ f ∈ L r(μ). (3.10) Before we come to the proof, a remark may be in order

Remark 3.3 Suppose thatΩ is a bounded open set inRd ApplyingT = I αwithμ = dx |Ω,

Lebesgue measure onΩ, we obtain a result corresponding to the one in [14]

The proof ofTheorem 3.1is after the one of Zygmund’s extrapolation theorem in [15]

Proof of Theorem 3.1 By subadditivity, it can be assumed that  f : ᏹ r

βr(μ)  =1 From (3.5) andLemma 1.3, we have T f : ᏹ βs s(μ)  ≤ cs ρ  f : ᏹ βp p (μ)  ≤ cs ρ

LetQ ∈ ᏽ(μ) Then by Taylor’s expansion,



Q



exp

δ T f (x) 1/ρ

−1 dμ(x)

μ(2Q)1− β

=

∞



k =1

δ k

k!



Q

T f (x) k/ρ dμ(x)

μ(2Q)1− β ≤

∞



k =1

δ k

k!

T f : ᏹk/ρβ k/ρ (μ) k/ρ

=L

k =1

δ k

k!

T f : ᏹk/ρβ

k/ρ (μ) k/ρ

+

∞



k = L+1

δ k

k!

T f : ᏹk/ρβ

k/ρ (μ) k/ρ

,

(3.11)

whereL is the largest integer not exceeding βρp0 If we invokeLemma 1.3, we see

L



k =1

δ k

k!

T f : ᏹk/ρβ k/ρ (μ) k/ρ

≤ c

L



k =1

δ k

k!

T f : ᏹL/ρβ L/ρ (μ) k/ρ

≤ c

L



k =1

δ k (3.12)

By (3.5), we have

∞



k = L+1

δ k

k!

T f : ᏹk/ρβ

k/ρ (μ) k/ρ

≤

∞



k = L+1

(cδ) k k k

We put (3.12) and (3.13) together,



Q

 exp

δ T f (x) 1/ρ

−1 dμ(x)

μ(2Q)1− β ≤

∞



k =1

(cδ) k k k

limk →∞(k k /k!)1/k = e implies that the function ψ(δ) : =∞ k =1((C0δ) k k k /k!) is a

contin-uous function in the neighborhood of 0 in [0, 1) withψ(0) =0 Consequently, ifδ is small

enough, then



Q

 exp

δ T f (x) 1/ρ

−1 dμ(x)

μ(2Q)1− β ≤ ψ(δ) ≤1 (3.15) for all f ∈ᏹr

βr(μ) with  f : ᏹ r

βr(μ)  =1.Theorem 3.1is therefore proved 

Remark 3.4 To obtainTheorem 3.1, the growth condition is unnecessary Thus, the proof

is still available, ifμ is just a finite Radon measure.

4 Precise estimate of the fractional integrals

Our task in this section is to see the size ofc(s) in (1.12) withT = I α The estimates involve the modified uncentered maximal operator given by

M κ f (x) : = sup

x ∈ Q ∈ ᏽ(μ)

1

μ(κQ)



Q

f (y) dμ(y), κ > 1. (4.1)

We make a quick view of the size of the constant First, we see that

μ

x ∈ R d:M κ f (x) > λ

≤ C d,κ

λ



Rd

f (x) dμ(x) (4.2)

by Besicovitch’s covering lemma Then thanks to Marcinkiewicz’s interpolation theorem,

we obtain a precise estimate of the operator norm ofM κ:

M κ

L p(μ) → L p(μ) ≤ C d,κ p

Finally, examining the proof in [10, Theorem 2.3] gives us the estimate of the operator norm onᏹp

q(μ):

M κ

ᏹp

(μ) →ᏹp

(μ) ≤ C d,κ q

We will make use of (4.3) and (4.4) in this section

4.1 Fractional integral operatorsJ α,κandI α,κ  For the definition ofI α, the growth con-dition onμ is indispensable However, in [9], the theory of fractional integral operators without the growth condition was developed The construction of the fractional integral operators without the growth condition involves a covering lemma In this present paper,

we intend to define another substitute We take advantage of the simple definition of the new fractional integral operator

Definition 4.1 (see [9, Definitions 13, 14]) Letα ∈(0, 1) andκ > 1 For k ∈ Z, takeᏽ(k) ⊂ ᏽ(μ) that satisfies the following.

(1) For allQ ∈ᏽ(k), 2k < μ(κ2Q) ≤2k+1

(2) supx ∈R d



Q ∈ᏽ (k) χ κQ(x) ≤ N κ < ∞, where N κdepends only onκ and d.

(3) For any cube with 2k −1< μ(κ2Q )≤2k, findQ ∈ᏽ(k)such thatQ ⊂ κQ.

By the way of{ᏽ(k) } k ∈Z, for f ∈ L1

loc(μ), define the operator J α,κas

J α,κ f (x) : =



Rd



k ∈Z



Q ∈ᏽ (k)

χ κQ(x)χ κQ(y)

If

j α,κ(x, y) : =

k ∈Z



Q ∈ᏽ (k)

χ κQ(x)χ κQ(y)

then one can writeJ α,κ f (x) =Rd j α,κ(x, y) f (y)dμ(y) in terms of the integral kernel.

What is important aboutJ α,κis that it is linear, it can be defined for any Radon measure

μ and, if μ satisfies the growth condition, it plays a role of the majorant operator of I α We give a more simpler fractional maximal operator which substitutes forJ α,κ

Definition 4.2 Let α ∈(0, 1) andκ > 1 For x, y ∈ R d ∈supp(μ), set

K α,κ  (x, y) = sup

It will be understood thatK α,κ  (x, y) =0 unlessx, y ∈supp(μ) For a positive μ-measurable

function f , set

I α,κ  f (x) =



Rd K α,κ  (x, y) f (y)dμ(y). (4.8) Suppose thatμ satisfies the growth condition (1.1) Then the comparison of the kernel reveals us thatI α f (x) ≤ cI α,κ  f (x) μ- a.e for all positive μ-measurable functions f

I α,κ  andJ α,κare comparable in the following sense

Lemma 4.3 Let α ∈ (0, 1) and κ > 1 There exists constant C > 0 so that, for every positive μ-measurable function f ,

I α,κ  2f (x) ≤ J α,κ f (x) ≤ CI α,κ  f (x). (4.9)

Proof It suffices to compare the kernel

First, we will deal with the left inequality Suppose thatQ ∈ ᏽ(μ) contains x, y and

satisfies

2k0< μ

κ2Q

Then byDefinition 4.1, we can findQ ∗ ∈ᏽ(k0 )such thatQ ⊂ κQ ∗ SinceκQ ∗contains bothx and y, we obtain

μ

κ2Q− α

≤2− k0α = χ κQ ∗(x)χ κQ ∗(y)

2k0α ≤ j α,κ(x, y). (4.11) Consequently, the left inequality is established

We turn to the right inequality Assume that

2− α(k1 +1)≤ K α,κ  (x, y) < 2 − αk1, k1∈ Z (4.12) LetQ ∈ᏽ(k) Suppose thatκQ contains x, y Then by definition,

μ

κ2Q−α

≤ K α,κ  (x, y) < 2 − αk1 (4.13) and henceμ(κ2Q) > 2 k1 SinceQ ∈ᏽ(k), we havek ≥ k1 Thus ifQ ∈ᏽ(k)andκQ contains

x, y, then k ≥ k1 From the definition of j α,κ, it follows that

j α,κ(x, y) = 

k ≥ k1



Q ∈ᏽ (k)

χ κQ(x)χ κQ(y)

2kα ≤ cN κ



k ≥ k1

1

2k α = c2 − k1α ≤ cK α,κ  (x, y). (4.14)

We summarize the relations between three operators

Corollary 4.4 If μ satisfies the growth condition ( 1.1 ), then, for every positive μ-measurable function f ,

I α f (x)J α,κ f (x) ∼ I α,κ  f (x), (4.15)

and μ- a.e x ∈ R d , where the implicit constants inand ∼ depend only on α, κ, and c0in ( 1.1 ).

4.2.L p-estimates Here we will prove theL p-estimates associated with fractional integral operators

Theorem 4.5 Let κ > 1, 0 < α < 1, and p0> 1 Assume that p,s > 1 satisfy

p0≤ p, 1

s =1

Then there exists a constant C > 0 depending only on α and p0so that, for every f ∈ L p(μ),

J α,κ f : L s(μ) ≤ Cs α f : L p(μ) , (4.17)

I 

α,κ f : L s(μ) ≤ Cs α f : L p(μ) . (4.18)

If μ additionally satisfies the growth condition ( 1.1 ), then

I

α f : L s(μ) ≤ Cs α f : L p(μ) . (4.19)

Proof We have only to prove (4.18) The rest is immediate once we prove it We may assume that f is positive Let R > 0 be fixed We will split I α,κ  f (x) For fixed x ∈supp(μ),

let us set

Ᏸj:=y ∈ R d \ { x }: 2j −1R < inf

x,y ∈ Q ∈ ᏽ(μ) μ(κQ) ≤2j R

, j ∈ Z (4.20)

We decomposeI α,κ  f (x) by using the partition {Ᏸ j } ∞

j =−∞ ∪ { x }of supp(μ) For the time

being, we assume thatμ charges { x } By definition, we have

I α,κ  f (x) =

0



j =−∞



Ᏸj

K α,κ  (x, y) f (y)dμ(y) +



∞

j =1 Ᏸj

K α,κ  (x, y) f (y)dμ(y) + μ( { x })1− α f (x).

(4.21) Suppose thatᏰjis nonempty By the Besicovitch covering lemma, we can findN ∈ N,

independent ofx, j, and R, and a collection of cubes Q1j,Q2j, ,Q N j which containx such

thatᏰj ⊂ √ κQ1j ∪ √ κQ2j ∪ ··· ∪ √ κQ N j andμ(κQ l j)≤2j+1 R for all 1 ≤ l ≤ N and j ∈ Z.

From this covering and the definition ofᏰj, we obtainμ(Ᏸ j)≤ c2 j R With these

ob-servations, it follows that

0



j =−∞



Ᏸj

K α,κ  (x, y) f (y)dμ(y) ≤ c

0



j =−∞

N



l =1

1

2jα R α



√

κQ j f (y)dμ(y) ≤ cR1− α M √ κ f (x).

(4.22)

Lebesgue measure on? ?, we obtain a result corresponding to the one in [14]

The proof ofTheorem 3.1is after the one of Zygmund’s extrapolation theorem in [15]

Proof of Theorem 3.1 ... αappearing here will be an example of the theorem

inSection BesidesI α, we take up two types of other fractional integral operators The task inSection 4is to determinec(s)