INTEGRAL OPERATORS ON NONHOMOGENEOUS SPACE

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 onR^d with nondoubling measures has been developed very rapidly; here, by a doubling measure, we mean a Radon measure μon R^d 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 atxof radiusr. 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:

μis a Radon measure onR^dwithμB(x,r)≤c0rⁿ for somec0>0, 0< n≤d.

(1.1) A growth measure is a Radon measureμsatisfying (1.1). We define the fractional inte- gral operatorI_αassociated with the growth measureμas

I_αf(x) :=

R^d

f(y)

|x−y|^nαdμ(y), 0< α <1. (1.2)

Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 92470, Pages1–16 DOI 10.1155/JIA/2006/92470

Let 1/q=1/ p−(1−α) with 1< p < q <∞.L^p(μ)-L^q(μ) boundedness ofIαin 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 definingIαf(x) does not converge absolutely forμ- a.e., iff ∈L^1/(1−α)(μ). Garc´ıa-Cuerva and Gatto considered some mod- ified operator and showed its boundedness fromL^1/(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× ··· ×

xd−r,xd+r, x=

x1,...,xd

∈R^d, 0< r≤ ∞. (1.3) Given a cubeQ=Q(x,r),κ >0, we denoteκQ:=Q(x,κr) and(Q)=2r. We defineᏽ(μ) by

ᏽ(μ) :=

Q⊂R^d:Qis 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 byᏹq^p(k,μ) a set of L^q_loc(μ) functions f for which the quasinorm

f :ᏹq^p(k,μ) := sup

Q∈ᏽ(μ)

μ(kQ)^{1/ p−1/q}

Q

f(y)^qdμ(y) 1/q

<∞. (1.5) Note that this definition does not involve the growth condition (1.1). So in this paper, we assumeμis 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 existsC_d,k¹,k²,qso that, for everyμ-measurable function f,

f :ᏹq^p

k2,μ ≤ f :ᏹq^p

k1,μ ≤Cd,k1,k2,q f :ᏹq^p

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 <∞andk >1. Then

f :ᏹq^p1(k,μ) ≤ f :ᏹq^p2(k,μ) ≤ f :ᏹ^pp(k,μ) = f :L^p(μ) . (1.7)

(2) Letμ(R^d)<∞and 0< q≤p1≤p2<∞. Then

f :ᏹq^p1(k,μ) ≤μR^d1/ p1−1/ p2 f :ᏹq^p2(k,μ) . (1.8)

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

As for (1.8), thanks to the finiteness ofμwriting out the left-hand side in full, we have f :ᏹq^p1(k,μ) = sup

Q∈ᏽ(μ)μ(kQ)^{1/ p1−1/q}

Q

f(y)^qdμ(y) 1/q

≤ sup

Q∈ᏽ(μ)

μR^d1/ p1−1/ p2

μ(k Q)^{1/ p2−1/q}

Q

f(y)^qdμ(y) 1/q

=μR^d1/ p¹−1/ p² f :ᏹq^p2(k,μ) .

(1.9)

Lemma 1.3is therefore proved.

KeepingProposition 1.2in mind, for simplicity, we denote

ᏹq^p(μ) :=ᏹq^p(2,μ), ·:ᏹq^p(μ) := ·:ᏹq^p(2,μ) . (1.10) In [10, Theorem 3.3], we showed thatI_αis bounded fromᏹq^p(μ) toᏹ^st(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 ofIαas “p→1/(1−α)” or “s→ ∞,” wherepandssatisfy (1.11).

Problem 1.4. Let 0< α <1 and assume that μis a finite growth measure. Find a nice function spaceXto whichIαsendsᏹ^1/(1q ^−α)(μ) continuously, where 1< q≤1/(1−α).

Although the Morrey spaces are the function spaces coming with two parameters, we arrangeᏹq^p(μ) toᏹ_βp^p (μ) withβ∈(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_βp^p (μ)}_p∈(0,∞). We also consider the generalized version ofProblem 1.4.

Problem 1.5. Letμbe finite and 0< p0< p < r <∞, 0< β≤1, 1/s=1/ p−1/r. Suppose that we are given an operatorTfrom_p>p0ᏹ_βp^p (μ) to_s>0ᏹ^s_βs(μ). Assume, restrictingT toᏹ_βp^p (μ), we have a precise estimate

T f :ᏹ^s_βs(μ) ≤c(s) f :ᏹ_βp^p (μ) , (1.12) where 1/s=1/ p−1/rwithp,r,s >0. Then what can we say about the boundedness ofT on the limit function spaceᏹ^r_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.Iαappearing 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 asIαto 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 theL^pspaces.

2. Orlicz-Morrey spacesᏹ^Φ_β(μ)

In this section, we introduce function spacesᏹ_β^Φ(μ) to formulate our main results. E.

Nakai definedᏹ^Φ_β(μ) for Lebesgue measureμ=dx. We denote by|E|the volume of a measurable setE. LetΦ: [0,∞)→[0,∞) be a Young function, that is,Φis convex with Φ(0)=0 and lim_x→∞Φ(x)= ∞.

Forβ∈(0, 1], E. Nakai has defined the Orlicz-Morrey spaces: the spaceᏹ^Φ_β(dx) con- sists of all measurable functions f for which the norm

f :ᏹ_β^Φ(dx) :=inf

λ >0 : sup

Q∈ᏽ(dx)

|Q|^β−1

QΦf(y) λ

dy≤1

<∞. (2.1)

For details, we refer to [6].

Motivated by this definition and that of ᏹq^p(μ) with 0< q≤p <∞, we define the Orlicz-Morrey spacesᏹ^Φ_β(μ) as follows.

Definition 2.1. Letβ∈(0, 1],k >1, andΦbe a Young function. Then define f :ᏹ^Φ_β(k,μ) :=inf

λ >0 : sup

Q∈ᏽ(μ)

μ(kQ)^β−1

QΦ f(y) λ

dμ(y)≤1

. (2.2)

We define the function spaceᏹ^Φ_β(k,μ) as a set ofμ-measurable functionsf for which the norm is finite.

The function spaceᏹ^Φ_β(k,μ) is independent ofk >1. More precisely, we have the fol- lowing.

Proposition 2.2. Letk1> k2>1. Then there exists constantC_d,k1,k2such that

f :ᏹ^Φ_βk1,μ ≤ f :ᏹ_β^Φk2,μ ≤C_d,k1,k2 f :ᏹ^Φ_βk1,μ . (2.3)

Here,Cd,k1,k2>0 is independent of f.

Proof. By the monotonicity off :ᏹ^Φ_β(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 :μk2Q^β−1

QΦf(x) λ

dμ(x)≤1

(2.4)

byλ0:= f :ᏹ^Φ_β(k1,μ)uniformly overQ.

BisectQinto 2^d cubes and labelQ1,Q2,...,QLto those inᏽ(μ), then the distance be- tween the boundary ofk2Qand the center ofQjis

k2

2 ⁻ 1 4

(Q)=k1

4(Q). (2.5)

Consequently, we havek1Qj⊂k2Qfor j=1, 2,...,L. This inclusion gives us that μk2Q^β−1

QΦf(x) λ0

dμ(x)≤^L

j=1

μk1Q_j^β−1

Qj

Φf(x) λ0

dμ(x)≤2^d. (2.6)

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

Q∈ᏽ(μ)

μk2Q^β−1

QΦf(x) 2^dλ0

dμ(x)≤1. (2.7)

Thus we have obtained

f :ᏹ_β^Φk2,μ ≤2^dλ0=2^d f :ᏹ^Φ_βk1,μ . (2.8) Hence we have established that we can takeCd,2k2−1,k2=2^d. Keeping this proposition in mind, we setᏹ_β^Φ(μ) :=ᏹ^Φ_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 thatμis finite and

0< p0< p < r <∞, 0< β≤1, 1 s ⁼

1 p⁻

1

r. (3.1)

LetTbe an operator fromᏹ_βp^p (μ) toᏹ^s_βs(μ) with a precise estimate

T f :ᏹ_βs^s(μ) ≤cs^ρ f :ᏹ_βp^p (μ) , ρ >0. (3.2) Then we can say that the limit result of

T:ᏹ_βp^p (μ)−→ᏹ_βs^s(μ), p0< p < r, 1 s ⁼

1 p⁻

1

r, (3.3)

asp→r,s→ ∞, is

T:ᏹ^r_βr(μ)−→ᏹ_β^Φ(μ), (3.4)

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

Theorem 3.1. Supposeμ(R^d)<∞. Let 0< p0< r, 0< ρ≤1, and 0< β≤1. Suppose that the sublinear operatorTsatisfies

T f :ᏹ^s_βs(μ) ≤C0s^ρ f :ᏹ_βp^p (μ) ∀f ∈ᏹ_βp^p (μ) (3.5)

for eachp0≤p < rwith 1/s=1/ p−1/r. Here,C0>0 is a constant independent ofpands.

Then there exists a constantδ >0 such that sup

Q

Q

exp

δ T f(x) f :ᏹ^r_βr(μ)

^1/ρ

−1

dμ(x) μ(2Q)^1−β

≤1 ∀f ∈ᏹ^r_βr(μ) (3.6)

or equivalently

T f :ᏹ_β^Φ(μ) ≤δ^−1/ρ f :ᏹ^r_βr(μ) ∀f ∈ᏹ^r_βr(μ) (3.7)

forΦ(t)=exp(t^1/ρ)−1.

More can be said about this theorem: the case whenβ=1 corresponds to the Zygmund- type extrapolation theorem (see [15]). SetL^Φ(μ)=ᏹ^Φ1(μ).

Corollary 3.2. Keep to the same assumption asTheorem 3.1onμ,ρ,p0,r, andT. Suppose T f :L^s(μ) ≤C0s^ρ f :L^p(μ) ∀f ∈L^p(μ) (3.8)

fors,pwith 1/s=1/ p−1/r. Here,C0>0 is a constant independent ofpands. Then there exists some constantδ >0 such that

R^d

exp

δ T f(x) f :L^r(μ)

^1/ρ

−1

dμ(x)≤1 ∀f ∈L^r(μ) (3.9)

or equivalently

T f :L^Φ(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 thatΩis a bounded open set inR^d. ApplyingT=Iαwithμ=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 ofTheorem 3.1. By subadditivity, it can be assumed thatf :ᏹ^r_βr(μ) =1. From (3.5) andLemma 1.3, we haveT f :ᏹ_βs^s(μ) ≤cs^ρf :ᏹ_βp^p (μ) ≤cs^ρ.

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

Q

expδT 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 :ᏹ^k/ρβ_k/ρ (μ) ^k/ρ

=^L

k=1

δ^k k!

T f :ᏹ^k/ρβ_k/ρ (μ) ^k/ρ+ ∞ k=L+1

δ^k k!

T f :ᏹ_k/ρ^k/ρβ(μ) ^k/ρ,

(3.11)

whereLis the largest integer not exceedingβρp0. If we invokeLemma 1.3, we see L

k=1

δ^k k!

T f :ᏹ^k/ρβ_k/ρ (μ) ^k/ρ≤c L k=1

δ^k k!

T f :ᏹ^L/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 :ᏹ^k/ρβ_k/ρ (μ) ^k/ρ≤ ∞ k=L+1

(cδ)^kk^k

k! . (3.13)

We put (3.12) and (3.13) together,

Q

expδT f(x)^1/ρ−1 dμ(x) μ(2Q)^{1−β ≤}

∞ k=1

(cδ)^kk^k

k! . (3.14)

lim_k→∞(k^k/k!)^1/k=eimplies that the functionψ(δ) :=_∞

k=1((C0δ)^kk^k/k!) is a contin- uous function in the neighborhood of 0 in [0, 1) withψ(0)=0. Consequently, ifδis small enough, then

Q

expδT f(x)^1/ρ−1 dμ(x)

μ(2Q)^{1−β ≤}ψ(δ)≤1 (3.15) for all f ∈ᏹ^r_βr(μ) withf :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, ifμis 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=Iα. The estimates involve the modified uncentered maximal operator given by

M_κf(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 μx∈R^d:M_κf(x)> λ≤Cd,κ

λ

R^d

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 ofMκ:

Mκ

L^p(μ)_→L^p(μ)≤Cd,κp

p−1. (4.3)

Finally, examining the proof in [10, Theorem 2.3] gives us the estimate of the operator norm onᏹq^p(μ):

Mκ

ᏹq^p(μ)→ᏹq^p(μ)≤Cd,κq

q−1. (4.4)

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 onμis 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. Fork∈Z, takeᏽ^(k)⊂ ᏽ(μ) that satisfies the following.

(1) For allQ∈ᏽ^(k), 2^k< μ(κ²Q)≤2^k+1. (2) sup_x∈Rd

Q∈ᏽ^(k)χκQ(x)≤Nκ<∞, whereNκdepends only onκandd.

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

By the way of{ᏽ^(k)}_k∈Z, for f ∈L¹_loc(μ), define the operatorJ_α,κas Jα,κf(x) :=

R^d

k∈Z

Q∈ᏽ^(k)

χκQ(x)χκQ(y)

2^kα f(y)dμ(y). (4.5)

If

jα,κ(x,y) :=

k∈Z

Q∈ᏽ^(k)

χ_κQ(x)χ_κQ(y)

2^kα , (4.6)

then one can writeJα,κf(x)=

R^djα,κ(x,y)f(y)dμ(y) in terms of the integral kernel.

What is important aboutJα,κis that it is linear, it can be defined for any Radon measure μand, ifμsatisfies the growth condition, it plays a role of the majorant operator ofIα. We give a more simpler fractional maximal operator which substitutes forJ_α,κ.

Definition 4.2. Letα∈(0, 1) andκ >1. Forx,y∈R^d∈supp(μ), set K_α,κ(x,y)= sup

x,y∈Q∈ᏽ(μ)

μ(κQ)^−α. (4.7)

It will be understood thatK_α,κ(x,y)=0 unlessx,y∈supp(μ). For a positiveμ-measurable function f, set

I_α,κf(x)=

R^dK_α,κ(x,y)f(y)dμ(y). (4.8) Suppose thatμsatisfies the growth condition (1.1). Then the comparison of the kernel reveals us thatI_αf(x)≤cI_α,κf(x)μ- a.e. for all positiveμ-measurable functions f.

I_α,κandJ_α,κare comparable in the following sense.

Lemma 4.3. Letα∈(0, 1) andκ >1. There exists constantC >0 so that, for every positive μ-measurable function f,

I_α,κ²f(x)≤J_α,κf(x)≤CI_α,κf(x). (4.9) Proof. It suffices to compare the kernel.

First, we will deal with the left inequality. Suppose thatQ∈ᏽ(μ) containsx, yand satisfies

2^k0< μκ²Q≤2^k0+1, k0∈Z. (4.10) Then byDefinition 4.1, we can findQ^∗∈ᏽ^(k0)such thatQ⊂κQ^∗. SinceκQ^∗contains bothxandy, we obtain

μκ²Q^−α≤2^−k0α=χ_κQ^∗(x)χ_κQ^∗(y)

2^{k0α ≤}jα,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 thatκQcontainsx,y. Then by definition,

μκ²Q^−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)andκQcontains x,y, thenk≥k1. From the definition of j_α,κ, it follows that

jα,κ(x,y)=

k≥k1

Q∈ᏽ^(k)

χ_κQ(x)χ_κQ(y) 2^{kα ≤}cNκ

k≥k1

1

2^{k α =}c2^−k1α≤cK_α,κ(x,y). (4.14)

As a result, the right inequality is proved.

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_αf(x)J_α,κf(x)∼I_α,κf(x), (4.15)

andμ- a.e.x∈R^d, where the implicit constants inand∼depend only onα,κ, andc0in (1.1).

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

Theorem 4.5. Letκ >1, 0< α <1, andp0>1. Assume that p,s >1 satisfy p0≤p, 1

s ⁼ 1

p⁻(1−α). (4.16)

Then there exists a constantC >0 depending only onαandp0so that, for every f ∈L^p(μ), Jα,κf :L^s(μ) ≤Cs^α f :L^p(μ) , (4.17) I_α,κf :L^s(μ) ≤Cs^α f :L^p(μ) . (4.18)

Ifμadditionally satisfies the growth condition (1.1), then

I_αf :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. LetR >0 be fixed. We will splitI_α,κf(x). For fixedx∈supp(μ), let us set

Ᏸj:=

y∈R^d\ {x}: 2^j−1R < inf

x,y∈Q∈ᏽ(μ)μ(κQ)≤2^jR, j∈Z. (4.20) We decomposeI_α,κf(x) by using the partition{Ᏸj}^∞_j=−∞∪ {x}of supp(μ). For the time being, we assume thatμcharges{x}. By definition, we have

I_α,κf(x)= 0 j=−∞

Ᏸj

K_α,κ(x,y)f(y)dμ(y) +

∞ j=1Ᏸj

K_α,κ(x,y)f(y)dμ(y) +μ({x})^1−αf(x).

(4.21) Suppose thatᏰjis nonempty. By the Besicovitch covering lemma, we can findN∈N, independent ofx,j, andR, and a collection of cubesQ1^j,Q2^j,...,Q_N^j which containxsuch thatᏰj⊂√

κQ1^j∪√

κQ2^j∪ ··· ∪√

κQ_N^j andμ(κQ_l^j)≤2^j+1Rfor all 1≤l≤Nand j∈Z. From this covering and the definition ofᏰj, we obtainμ(Ᏸj)≤c2^jR. 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 2^jαR^α

√κQl^j

f(y)dμ(y)≤cR^1−αM^√κf(x).

(4.22)

The estimate of the second term will be accomplished by the H¨older inequality,

∞ j=1Ᏸj

K_α,κ(x,y)f(y)dμ(y)

≤

∞ j=1Ᏸj

K_α,κ(x,y)^pdμ(y)

1/ p f :L^p(μ)

= _∞

j=1

Ᏸj

K_α,κ(x,y)^pdμ(y) 1/ p

f :L^p(μ)

≤c _∞

j=1

(2^jR)^1−αp 1/ p

f :L^p(μ) ≤c

α− 1 p

₋1/ p

R^{1/ p−α} f :L^p(μ) ,

(4.23)

where we use an inequality 1/(2^a−1)≤1/(log 2·a), a >0. Taking into account these estimates, we obtain

0 j=−∞

Ᏸj

K_α,κ(x,y)f(y)dμ(y) +

∞ j=1Ᏸj

K_α,κ(x,y)f(y)dμ(y)

≤Cα,κ

R^1−αM^√κf(x) +R^{−(α−1/ p)}

α− 1 p

−1/ p f :L^p(μ)

.

(4.24)

We have to deal withμ({x})^1−αf(x). Ifμ({x})≤R, thenμ({x})^1−αf(x)≤R^1−αM^√_κf(x).

Conversely, ifμ({x})≥R, thenμ({x})^1−αf(x)≤R^{−(α−1/ p)}f :L^p(μ). As a result, we can incorporateμ({x})^1−αf(x) to the above formula. The result is

I_α,κf(x)≤Cα,κ

R^1−αM^√κf(x) +R^{−(α−1/ p)}

α− 1 p

−1/ p f :L^p(μ)

(4.25) for allR∈(0,∞). Taking

R=

(α−1/ p)^{−1/ p} f :L^p(μ) M^√_κf(x)

p

, (4.26)

we have

I_α,κf(x)≤C_α,κ

α− 1

p

₋(1₋α)(p−1)

M^√_κf(x)^{p(α−1/ p)} f :L^p(μ) ^{1−p(α−1/ p)}. (4.27) Recall that 1/s=α−1/ p by assumption. Thus the above estimate can be restated as

I_α,κf(x)≤C_α,κs^{(1−α)(p−1)}M^√_κf(x)^p/s f :L^p(μ) ^1−p/s. (4.28)

Insertingp(1−α)−1= −p/s, we sees^{(1−α)(p−1)}=s^α−p/s≤s^α. As a consequence, we have I_α,κf :L^s(μ) ≤C_α,κ,p0s^α f :L^p(μ) . (4.29)

This is the desired estimate.

Consequently, if we useTheorem 3.1, then we obtain the following.

Theorem 4.6. Assume thatμis a finite Radon measure. LetT be eitherJα,κ or I_α,κ with 0< α <1 andκ >1. Then there existsC >0 so that, for everyf ∈L^1/(1−α)(μ),

T f :L^Φ(μ) ≤C f :L^1/(1−α)(μ) , (4.30)

whereΦ(x)=exp(x^1/α)−1. Ifμsatisfies the growth condition (1.1), then (4.30) is still avail- able forT=I_α.

4.3. Morrey estimates. Now we will prove the Morrey estimates associated with frac- tional integral operators.

Theorem 4.7. Let 0< α <1, 0< β≤1,κ >1, andp0>1/β. Assume thatpandssatisfy p0≤p <∞, 1< s <∞, 1

s ⁼ 1

p⁻(1−α). (4.31)

Then there exists a constant C >0 depending only onα,βand p0 so that, for every f ∈ ᏹ_βp^p (μ),

Jα,κf :ᏹ_βs^s(μ) ≤Cs f :ᏹ_βp^p (μ) , (4.32) I_α,κ f :ᏹ_βs^s(μ) ≤Cs f :ᏹ_βp^p (μ) . (4.33)

Ifμadditionally satisfies the growth condition (1.1), then

Iαf :ᏹ_βs^s(μ) ≤Cs f :ᏹ_βp^p (μ) . (4.34)

Proof. It is enough to prove (4.33) for a positiveμ-measurable function f. We have only to make a minor change of the proof ofTheorem 4.5. So we indicate the necessary change.

Under the notation in the proof ofTheorem 4.5, we change the estimate of

_∞ j=1Ᏸj

K_α,κ(x,y)f(y)dμ(y). (4.35)