ON THE HALF-SPACE - Korea Universityelie.korea.ac.kr/~choebr/papers/schattenherz(halfspace).pdfBergman kernels. In Section 3 we investigate weighted Lp-behavior of averaging functions

POSITIVE SCHATTEN(-HERZ) CLASS TOEPLITZ OPERATORSON THE HALF-SPACE

BOO RIM CHOE, HYUNGWOON KOO, AND YOUNG JOO LEE

ABSTRACT. On the harmonic Bergman space of the half-space, we give characterizationsfor an arbitrary positive Toeplitz operator to be a Schatten (or Schatten-Herz) class operatorin terms of averaging functions and Berezin transforms. Examples are provided to showthat various results are sharp.

1. INTRODUCTION

For a positive integern ≥ 2, let H = Rn−1 ×R+ denote the upper half-space inRn

whereR+ is the set of all positive real numbers. We will often write pointsz ∈ H asz = (z′, zn) wherez′ ∈ Rn−1 andzn ∈ R+.

The harmonic Bergman spaceb2 is the space of all complex-valued harmonic functionsf onH such that ∫

H

|f |2 dV <∞

whereV denotes the Lebesgue measure onH. The spaceb2 is a closed subspace ofL2(V )and hence a Hilbert space. We letR be the Hilbert space orthogonal projection fromL2(V )onto b2. Each point evaluation is easily verified to be a bounded linear functional onb2.Hence, for eachz ∈ H, there exists a unique functionRz (known to be real) inb2 whichhas the following reproducing property:

(1.1) f(z) =∫H

fRz dV, z ∈ H

for all f ∈ b2. LetR(z, w) = Rz(w). The kernelR(z, w) is the well-known harmonicBergman kernel whose explicit formula is given by

(1.2) R(z, w) =4nσn

n(zn + wn)2 − |z − w|2

|z − w|n+2, z, w ∈ H

whereσn is the volume of the unit ball ofRn andw = (w′,−wn). See [1] for moreinformation and related facts. The reproducing formula (1.1) yields the following integralrepresentation of the projectionR:

(1.3) Rψ(z) =∫H

ψ(w)R(z, w) dw, z ∈ H

for functionsψ ∈ L2(V ). Here and elsewhere, we use the notationdw = dV (w) forsimplicity.

2000Mathematics Subject Classification.Primary 47B35, Secondary 31B05.Key words and phrases.Toeplitz operator, Harmonic Bergman space, Schatten class, Schatten-Herz class.This research was supported by KOSEF(R01-2003-000-10243-0).

1

2 B. R. CHOE, H. KOO, AND Y. J. LEE

From the explicit formula (1.2) ofR(z, w), it is easily seen that

|R(z, w)| ≤ C

|z − w|n, z, w ∈ H(1.4)

for some constantC = C(n). So, we can easily see thatR extends to the class of positiveBorel measuresµ (we simply writeµ ≥ 0) satisfying(1 + |z|n)−1 ∈ L1(µ). Namely, forsuchµ, the integral

Rµ(z) =∫H

R(z, w) dµ(w), z ∈ H

defines a function harmonic onH.LetM+ be the set of allµ ≥ 0 such that(1 + |z|β)−1 ∈ L1(µ) for someβ > 0. For

µ ∈M+, theToeplitz operatorwith symbolµ is defined by

Tµf = R(fdµ)

for f ∈ b2. Let M be the set of all “formal” finite linear combinations of measures inM+. We extend the notion ofTµ for µ ∈ M by linearity. Note thatTµ may not even bedefined on all ofb2, but it is always densely defined by the fact ([13]) that, for eachβ > n,harmonic functionsf onH such thatf(z) = O

[(1 + |z|β)−1

]form a dense subset ofb2.

For positive Toeplitz operators on harmonic Bergman spaces, basic operator theoreticproperties such as boundedness, compactness and the membership in the Schatten classesSp (see Section 4),1 ≤ p < ∞, have been studied on various settings as in [4], [5], [6],[11] and [12]. Also, the notion of the so-called Schatten-Herz classesSp,q (see Section 5),1 ≤ p, q ≤ ∞, has been recently introduced and studied in [2] and [7]. In this paper weextend the characterizations ([4], [7]) for Schatten(-Herz) class positive Toeplitz operatorsto the full range of indicesp andq.

To state our results we briefly introduce some notation. Givenµ ≥ 0, µr denotesthe averaging function over pseudohyperbolic balls of radiir and µ denotes the Berezintransform (see Section 3). Motivated by the factR(z, z) = cnz

−nn , we letλ denote the

measure defined bydλ(z) = z−n

n dz.

We also letKpq(λ) denote the so-called Herz spaces (see Section 5).

The following characterizations were proved in [4] and [7], respectively.

Theorem A ([4]). Let 1 ≤ p < ∞, 0 < r < 1 and µ ∈ M+. Then the followingstatements are equivalent.

(a) Tµ ∈ Sp.(b) µr ∈ Lp(λ).(c) µ ∈ Lp(λ).

Theorem B ([7]). Let1 ≤ p ≤ ∞, 0 < r < 1 andµ ∈M+. Also, assume1 ≤ q ≤ ∞ orq = 0. Then the following statements are equivalent.

(a) Tµ ∈ Sp,q.(b) µr ∈ Kp

q(λ).(c) µ ∈ Kp

q(λ).

The holomorphic version of the equivalence of Conditions (a) and (b) in Theorem Awas originally proved by Luecking[9] for the full range0 < p < ∞ on the disk. LaterZhu[14] extended Luecking’s result to bounded symmetric domains and added Condition(c), but with the restricted range1 ≤ p < ∞. Quite recently, Zhu[15] has studied similarcharacterizations for the range0 < p < 1 on the ball. His results reveal an interesting

SCHATTEN CLASS TOEPLITZ OPERATORS 3

aspect that, while the full range0 < p < 1 is good for the averaging-function character-ization (as one may naturally expect in view of results in [9]), this isnot the case for theBerezin-transform characterization. That is, thep-index range turns out to have a sharplower bound, depending on the complex dimensions of underlying spaces, for the Berezin-transform characterization.

Motivated by this recent work by Zhu, we here extend Theorem A to the range0 <p < 1. Then, using that result, we extend Theorem B to the full range of indicesp andq. It turns out that the lower-bound phenomena for the Berezin transform characterizationcontinue to hold for our setting of the harmonic Bergman space on the half-space as in thenext theorems, which are the main results of this paper.

Theorem 1.1. Let0 < p < 1, 0 < r < 1 andµ ∈M+. Then the following two statementsare equivalent.

(a) Tµ ∈ Sp.(b) µr ∈ Lp(λ).

Moreover, ifn−1n < p < 1, then the above statements are also equivalent to

(c) µ ∈ Lp(λ).

Theorem 1.2. Let0 < p ≤ ∞, 0 ≤ q ≤ ∞, 0 < r < 1 andµ ∈ M+. Then the followingtwo statements are equivalent.


q(λ).

Moreover, ifn−1n < p ≤ ∞, then the above statements are also equivalent to

(c) µ ∈ Kpq(λ).

The cut-off pointn−1n is sharp in both theorems. While the main scheme of our proof

of Theorem 1.1 is adapted from [15], the unboundedness of our domain causes variousdifficulties and substantial amount of extra work.

In Section 2 we collect some basic results on the pseudohyperbolic balls and harmonicBergman kernels. In Section 3 we investigate weightedLp-behavior of averaging functionsand Berezin transforms. In Section 4 we prove Theorem 1.1. In Section 5 we prove The-orem 1.2. In Section 6 we provide examples indicating that the parameter ranges requiredfor all the results of the previous sections are best possible.

Constants.In the rest of the paper the same letterC will denote various positive con-stants, unless otherwise specified, which may change at each occurrence. The constantCmay often depend on the dimensionn and some other parameters likeδ, p, k, or r, etc, butit will be always independent of particular functions, measures, points or sequences underconsideration. We will often abbreviate inessential constants involved in inequalities bywriting X . Y or Y & X for positive quantitiesX andY if the ratioX/Y has a positiveupper bound. Also, we writeX ≈ Y if X . Y andX & Y .

2. PRELIMINARIES

In this section we introduce notation and collect several basic lemmas which will beused in later sections. The hyperbolic distance between two pointsz, w ∈ H is defined by

β(z, w) =12

log1 + ρ(z, w)1− ρ(z, w)

(2.1)


where

ρ(z, w) =|z − w||z − w|

.

Thisβ is indeed a distance function onH. Furthermore,ρ is also a distance function onHand called the pseudohyperbolic distance. We shall work withρ throughout the paper. Notethat the pseudohyperbolic distance is horizontal translation invariant and dilation invariant.In particular, we have

ρ(z, w) = ρ(φa(z), φa(w)

), z, w ∈ H

whereφa, a ∈ H, denotes the mapping defined by

φa(z) =(z′ − a′

an,zn

an

)for z = (z′, zn) ∈ H.

For z ∈ H andr ∈ (0, 1), let Er(z) denote the pseudohyperbolic ball centered atzwith radiusr. A straightforward calculation yields

(2.2) Er(z) = B

((z′,

1 + r2

1− r2zn

),

2r1− r2

zn

).

Here,B(z, δ) denotes the euclidean ball centered atz with radiusδ. This leads to the nexttwo lemmas taken from Lemmas 3.2 and 3.3 of [3].

Lemma 2.1. The inequality

1− ρ(z, w)1 + ρ(z, w)

≤ zn

wn≤ 1 + ρ(z, w)

1− ρ(z, w)

holds forz, w ∈ H.

Lemma 2.2. The inequality

1− ρ(z, w)1 + ρ(z, w)

≤ |z − a||w − a|

≤ 1 + ρ(z, w)1− ρ(z, w)

holds forz, w, a ∈ H.

Note that we have by (2.2)

|Er(z)| = σn

(2r

1− r2zn

)n

(2.3)

for anyz ∈ H andr ∈ (0, 1). Here and elsewhere,|E| = V (E) for Borel setsE ⊂ H.Consequently, we obtain by Lemma 2.1 an estimate on size of pseudohyperbolic balls:Givenδ, t ∈ (0, 1), there is a constantC = C(δ, t) > 0 such that

(2.4) C−1 ≤ |Er1(z)||Er2(w)|

≤ C, w ∈ Er3(z)

wheneverr1, r2, r3 < δ andt < r1/r2 < t−1.For each integerk ≥ 0, define

Rk(z, w) =(−2)k

k!wk

nDkwnR(z, w)

for z, w ∈ H whereDwn denotes the differentiation with respect to the last componentof w. Note thatR0(z, w) = R(z, w). Also, note that the kernelRk(z, w) is harmonicin z-variable. We remark in passing that the kernelRk(z, w) also has the reproducing


property asR(z, w) does ([13]). A straightforward calculation yieldsR(φa(z), φa(w)

)=

annR(z, w) and thus

Rk

(φa(z), φa(w)

)= an

nRk(z, w)(2.5)

for a, z, w ∈ H. By induction one can check thatRk(z, w) can be written in the form

Rk(z, w) = wkn

k+2∑m=0

ckm(zn + wn)m

|z − w|n+k+m

for some coefficientsckm. It is clear from this identity that we have

(2.6) |Rk(z, w)| ≤ Cwk

n

|z − w|n+k, z, w ∈ H

for some constantC = C(n, k).Note that the kernelR(z, w) has zeros unlike the (holomorphic) Bergman kernel on

the ball ofCn. Thus, it is less clear that the kernelRk(z, w) has the following positivityproperty. In what follows we let

z0 = (0′, 1).

Lemma 2.3. Given an integerk ≥ 0, there exist somerk = rk(n) ∈ (0, 1) and a constantC = C(n, k) such that

C−1 ≤ Rk(z, a)ann ≤ C

whenevera ∈ H, 0 < r ≤ rk andz ∈ Er(a).

Proof. By the invariance property (2.5) we may assumea = z0. We need to show thatRk(z, z0) is positive forz nearz0. By continuity it is sufficient to showRk(z0, z0) > 0.Now, we computeRk(z0, z0). By (1.2) we haveR(z0, w) = 4

nσng(|w′|, 1 + wn) where

g(s, t) = [(n − 1)t2 − s2](s2 + t2)−(n+2)/2. Thus, usingg(0, t) = (n − 1)t−n andreal-analyticity, we have

Rk(z0, z0) =4nσn

(−2)k

k!

[dk

dtkg(0, t)

]t=2

=4

n2nσn

(n+ k − 1)!(n− 1)!k!

> 0,

as required. The proof is complete.

The next two integral identities are easily verified via elementary repeated integrations.

Lemma 2.4. Givenα > −1 ands > 0, there exists a constantC = C(n, s, α) such that∫H

wαn

|z − w|n+α+sdw = Cz−s

n

for z ∈ H.

Lemma 2.5. Givenα > −1, there exists a constantC = C(n, α) such that∫S

dw

|z − w|n+α= C

∫ t

s

dwn

(zn + wn)α+1, z ∈ H

for any horizontal stripeS = z ∈ H : 0 ≤ s < zn < t ≤ ∞.


3. AVERAGING FUNCTIONS ANDBEREZIN TRANSFORM

In this section we collect the weightedLp-behavior of averaging functions (as well asits discretized version) and Berezin transform.

Givenµ ≥ 0 andr ∈ (0, 1), theaveragingfunctionµr and theBerezin transformµ aredefined by

µr(z) =µ[Er(z)]|Er(z)|

and

µ(z) = znn

∫H

|Rz|2 dµ

for z ∈ H. While it is customary to putR(z, z)−1 in place ofznn in the definition of

the Berezin transform, we adopted the above definition for simplicity. For measurablefunctionsf , we definefr andf similarly, whenever they are well defined.

We introduce more notation. Givenα real, we letLpα = Lp(Vα) whereVα denotes the

weighted measure defined bydVα(z) = zαn dz. Forα = 0, we letLp = Lp

0 = Lp(V ).Noteλ = V−n. Givenr ∈ (0, 1), it is easily seen from (2.3) and Lemma 2.1 that

Vα[Er(z)] ≈ zn+αn(3.1)

for z ∈ H. Also, given a sequencea = am in H, we let`p,α(a) denote the one-sidedp-summable sequence space weighted byaα

mn. Forα = 0, we let`p = `p,0(a).We now recall the notion of lattices. Letam be a sequence inH andr ∈ (0, 1). We

say thatam is r-separatedif the ballsEr(am) are pairwise disjoint or simply say thatam is separatedif it is r-separated for somer. Also, we say thatam is anr-lattice ifit is r

2 -separated andH = ∪mEr(am). Note that any ‘maximal’r2 -separated sequence isanr-lattice.

We begin with a lemma.

Lemma 3.1. Let k ≥ 0 be an integer,µ ≥ 0 and 0 < r < rk whererk is the numberprovided by Lemma 2.3. Then there exists a constantC = C(n, k, r) > 0 such that

µr(a) ≤ Cann

∫H

|Rk(z, a)|2 dµ(z), a ∈ H.

In particular, µr ≤ Cµ for 0 < r < r0.

Proof. Let a ∈ H be given. By Lemma 2.3 and (2.3), we have∫H

|Rk(z, a)|2 dµ(z) ≥∫

Er(a)

|Rk(z, a)|2 dµ(z) & a−2nn µ[Er(a)] ≈ a−n

n µr(a).

The constants suppressed above depend only onn, k andr. This completes the proof of thefirst part of the lemma. The second part is clear (withk = 0). The proof is complete.

As a preliminary step, we first prove that theLpα-behavior of averaging functions of

positive measures is independent of radii.

Proposition 3.2. Let 0 < p ≤ ∞, r, δ ∈ (0, 1) andα be real. Assumeµ ≥ 0. Thenµr ∈ Lp

α if and only ifµδ ∈ Lpα.

For 1 ≤ p ≤ ∞, this proposition is already proved in [4, Corollary 3.3] and the proofbelow provides another proof. In the course of the proof we will use the following change-of-variable formula for eacha ∈ H:∫

H

f(φ−1

z (a))zαn dz = a−α−1

n

∫H

f(w)wαn dw(3.2)


for functionsf ∈ L1α. In order to see this, note that the Jacobian of the mappingz 7→

φ−1z (a) = (z′ + zna

′, znan) is simplyan.

Proof. By symmetry it suffices to establish the estimate

(3.3) ‖µr‖Lpα

. ‖µδ‖Lpα.

Since the caser ≤ δ is easily treated by (2.4), we may assumeδ < r. Choose a finite seta1, a2, . . . , aN inEr(z0) which is maximal subject to the conditionEδ/2(ai)

⋂Eδ/2(aj)

= ∅ for i 6= j. We then haveEr(z0) ⊂ ∪Nj=1Eδ(aj) by maximality and hence

Er(z) = φ−1z Er(z0) ⊂

N⋃j=1

φ−1z Eδ(aj) =

N⋃j=1

Eδ

(φ−1

z (aj))

for everyz ∈ H. Sinceφ−1z (aj) ∈ Er(z), we note by (2.4)

∣∣Eδ

(φ−1

z (aj))∣∣ ≈ |Er(z)| for

all j andz ∈ H. It follows that

µr(z) ≤1

|Er(z)|

N∑j=1

µ[Eδ

(φ−1

z (aj))]

=N∑

j=1

∣∣Eδ

(φ−1

z (aj))∣∣

|Er(z)|µδ

(φ−1

z (aj))

≈N∑

j=1

µδ

(φ−1

z (aj))

for everyz ∈ H. This implies (3.3) forp = ∞. Forp <∞, the above yields

µr(z)p . max1, Np−1N∑

j=1

µδ

(φ−1

z (aj))p

for all z ∈ H. Now, integrating both sides of the above against the measurezαndz and then

using (3.2), we conclude (3.3). The proof is complete.

For the discretized version of Proposition 3.2, we need a couple of known facts. Thefollowing is taken from [7, Theorem 3.3].

Lemma 3.3. Let1 ≤ p ≤ ∞ andα be real. Then the Berezin transform is bounded onLpα

if and only if−n < (α+ 1)/p < 1.

The following is a special case of [4, Lemma 3.8].

Lemma 3.4. Givenr ∈ (0, 1), there exists a constantC = C(r) > 0 such thatµ ≤ C (µr)for µ ≥ 0.

Theorem 3.5. Let 0 < p ≤ ∞, r, δ ∈ (0, 1) andα be real. Letµ ≥ 0 anda = am bean r-lattice. Then the following two statements are equivalent.

(a) µδ ∈ Lpα.

(b) µr(am) ∈ `p,n+α(a).Moreover, if

max

1 + α, 1 +α

n,−α+ 1

n

< p ≤ ∞,(3.4)

then the above statements are also equivalent to


(c) µ ∈ Lpα.

It is clear from (3.1) that (b) is a discretized version of (a). Thus it is not too surprisingto see the equivalence of (a) and (b). However, it seems quite interesting to see that therestricted range (3.4) is best possible for (c); see Section 6.

Proof. (a) =⇒ (b): Assume (a). Pick a positive numberε < min r2 , 1−r. By Proposition

3.2, in order to prove (b), it is sufficient to establish the estimate

‖µr(am)‖`p,n+α(a) . ‖µr+ε‖Lpα.(3.5)

Sinceε < r/2 andam is anr-lattice, the ballsEε(am) are pairwise disjoint. Also, givenz ∈ Eε(am), we haveEr(am) ⊂ Er+ε(z) and thus

µr+ε(z) ≥µ[Er(am)]|Er+ε(z)|

=|Er(am)||Er+ε(z)|

µr(am) ≈ µr(am)

by (2.4). This implies (3.5) forp = ∞. Forp <∞, it follows from (3.1) that∫H

µr+ε(z)pzαn dz &

∑m

∫Eε(am)

µr+ε(z)pzαn dz &

∑m

µr(am)pan+αmn

and thus (3.5) holds.(b) =⇒ (a): Assume (b). In order to prove (a), we may assumeδ < 1−r by Proposition

3.2. Givenz ∈ H, letN(z) = m : Er(am) ∩ Eδ(z) 6= ∅. Sinceam is anr-lattice,we haveEδ(z) ⊂ ∪m∈N(z)Er(am) and thusµ[Eδ(z)] ≤

∑m∈N(z) µ[Er(am)]. It follows

from this and (2.4) that

µδ(z) .∑

m∈N(z)

|Er(am)||Eδ(z)|

µr(am) ≈∑

m∈N(z)

µr(am).(3.6)

Meanwhile, form ∈ N(z), since the ballsEε(am) are pairwise disjoint and are all con-tained inEr+δ+ε(z) for a fixed positive numberε < min r

2 , 1− r− δ, we see from (2.4)that

N = N(r, δ, ε) := supz∈H

]N(z) <∞

where]N(z) denotes the number of elements in the setN(z). Thus, we have‖µδ‖L∞(λ) .‖µr(am)‖L∞(λ) by (3.6). Also, forp <∞, we have by (3.6)

µδ(z)p . max1, Np−1∑

m∈N(z)

µr(am)p

for all z ∈ H. Now, integrating both sides of the above against the measurezαn dz and then

applying Fubini’s theorem, we obtain∫H

µδ(z)pzαn dz .

∫H

∑m∈N(z)

µr(am)pzαn dz =

∑m

µr(am)pVα[Q(m)]

whereQ(m) = z ∈ H : Er(am) ∩ Eδ(z) 6= ∅. Note thatQ(m) ⊂ Er+δ(am) and thusVα[Q(m)] ≤ Vα[Er+δ(am)] ≈ an+α

mn by (3.1). Combining this with the above estimate,we conclude that (a) holds.

(c) =⇒ (a): This follows from Lemma 3.1 and Proposition 3.2.


Now, we assume (3.4) and prove that either (a) or (b) implies (c). Ifp ≥ 1, one may useLemmas 3.3 and 3.4 to see that (a) implies (c). Ifp < 1, we haveα < 0 by (3.4) and thus(3.4) reduces to

max

1 +α

n,−α+ 1

n

< p < 1.(3.7)

We now assume this and prove below that (b) implies (c).

(b) =⇒ (c): Assumeµr(am) ∈ `p,n+α(a). By (2.6) and Lemma 2.2, we note

|R(z, w)| . 1|am − z|n

, w ∈ Er(am)

for all m andz ∈ H. It follows from this that

µ(z) = znn

∫H

|R(z, w)|2 dµ(w)

≤ znn

∑m

∫Er(am)

|R(z, w)|2 dµ(w)

.∑m

µ[Er(am)]znn

|am − z|2n

≈∑m

anmnµr(am)

znn

|am − z|2n

and thus (recallp < 1)

µ(z)p .∑m

anpmnµr(am)p znp

n

|am − z|2np

for all z ∈ H. Now, sincenp + α > −1 andnp − n − α > 0 by (3.7), we obtain byLemma 2.4∫

H

µ(z)pzαn dz .

∑m

µr(am)panpmn

∫H

znp+αn

|am − z|2npdz ≈

∑m

µr(am)pan+αmn

and thus (c) holds. The proof is complete.

As a consequence of Theorem 3.5, we have the following.

Corollary 3.6. Let 0 < p ≤ ∞, r, δ ∈ (0, 1) and α be real. Assumeµ ≥ 0. Thenµr(am) ∈ `p,α(a) for all r-latticesa = am if and only if µδ(am) ∈ `p,α(a) forsomeδ-latticea = am.

4. SCHATTEN CLASSTOEPLITZ OPERATORS

In this section we prove Theorem 1.1. We first briefly review the notion of Schattenclass operators. For a positive compact operatorT on a separable Hilbert spaceX, thereexist an orthonormal setem in X and a sequenceλm that decreases to 0 such that

Tx =∑m

λm〈x, em〉em


for all x ∈ X where〈 , 〉 denotes the inner product onX. For0 < p < ∞, we say that apositive operatorT belongs to the Schattenp-classSp(X) if

‖T‖p :=

∑m

λpm

1/p

<∞.

More generally, given a compact operatorT onX, we say thatT ∈ Sp(X) if the positiveoperator|T | = (T ∗T )1/2 belongs toSp(X) and we define‖T‖p = ‖ |T | ‖p. Of course,we will takeX = b2 in our applications below and, in that case, we putSp = Sp(b2).

First, we recall a couple of basic facts aboutSp(X), which we need later. The followinglemma is taken from [9, Lemma 5] where further references are given.

Lemma 4.1. Let0 < p ≤ 2 andT be a compact operator onX. Then

‖T‖pp ≤

∑i

∑j

|〈Tei, ej〉|p

for any orthonormal basisek ofX.

The proof of the following lemma is implicit in the proof of [14, Theorem 1.4.7].

Lemma 4.2. Let0 < p < 1 andT be a positive compact operator onX. Then

‖T‖pp ≤ sup

∑m

〈Tem, em〉p

where sup is taken over all orthonormal basesem ofX.

We also need three known results involving separated sequences. The next two lemmasare taken from [4, Lemmas 3.4 and 4.2].

Lemma 4.3. For α > 0 andr ∈ (0, 1) withαr < 1, there is a constantN = N(α, r) withthe following property: Ifam is anr-separated sequence, then more thanN of the ballsEαr(am) contain no point in common.

Lemma 4.4. Given0 < r < δ < 1, there exists a positive integerN = N(δ, r) withthe following property: Anyr-separated sequence can be decomposed intoN δ-separatedsubsequences.

Let am be a sequence inH andk ≥ 0 be an integer. Forξ = ξm ∈ `2, letQk(ξ)denote the series defined by

(4.1) Qk(ξ)(z) =∑m

ξman2mnRk(z, am), z ∈ H.

The following lemma is taken from [3, Proposition 4.5 and Theorem 4.6].

Lemma 4.5. Letk ≥ 0 be an integer. Then the following statements hold.

(a) If am is a separated sequence, thenQk : `2 → b2 is bounded.(b) There existsδk > 0 with the following property: Ifam is aδ-lattice withδ < δk,

thenQk : `2 → b2 is onto.

We also need a lemma which is useful in estimating certain quantities related to theoperatorsQk.


Lemma 4.6. Lets, t be real andr, δ ∈ (0, 1). Then

btn|z − b|s

≈∫

Er(b)

wtn

|a− w|sdλ(w)

wheneverb ∈ H andz ∈ Eδ(a). The constants suppressed above depend only onn, s, t, randδ.

Proof. Let a, b ∈ H. By Lemma 2.2, we note

btn|z − b|s

≈ btn|a− b|s

for z ∈ Eδ(a). On the other hand, we have by Lemmas 2.1 and 2.2

btn|a− b|s

≈ wtn

|a− w|s

for w ∈ Er(b). Thus, we conclude the lemma by (3.1). One can check that all the con-stants suppressed above depend on parameters, but not on particular points. The proof iscomplete.

We are now ready to prove the following version of Theorem 1.1.

Theorem 4.7. Let 0 < p < 1, µ ∈ M+ and assume thatam is anr-lattice. Then thefollowing three statements are equivalent.

(a) Tµ ∈ Sp.(b) µr(am) ∈ `p.(c) µr ∈ Lp(λ).

Moreover, ifn−1n < p < 1, then the above statements are also equivalent to

(d) µ ∈ Lp(λ).

By Theorem 3.5 (withα = −n) we only need to prove that (a) and (b) are equivalent.

Proof of(a) =⇒ (b). Assume (a). By Corollary 3.6, it suffice to considerr sufficientlysmall. Fix a large integerk > 0 such that

k >n− 1p

− n

2

and letrk be the constant provided by Lemma 2.3. To begin with, fixr < minrk, 12 and

let 2r < δ < 1. Later the constantδ will chosen to be sufficiently close to 1.By Lemma 4.4, we can decompose the latticeam into finitely many, sayN , δ

2 -separated subsequences. Letbm be one of suchδ2 -separated subsequences and definea measureν by

dν =∑m

χEr(bm) dµ.

Here and elsewhere,χE denotes the characteristic function ofE ⊂ H. Note that the ballsEr(bm) are pairwise disjoint, because2r < δ. Also, note thatν ∈ M+ and ||Tν ||p ≤||Tµ||p, because0 ≤ ν ≤ µ.

Now, we introduce some auxiliary operators. Fix an orthonormal basisem for b2

and letJ : b2 → `2 be the unitary operator defined byJf = 〈f, em〉 for f ∈ b2. PutA = QkJ whereQk is the operator associated with the sequencebm as in (4.1). By

Lemma 4.5 (a),A : b2 → b2 is bounded. Puthm = Aem = bn2mnRk(·, bm) for simplicity.


PutT = A∗TνA. NoteT ≥ 0. SinceA is bounded andTν ∈ Sp, we haveT ∈ Sp with‖T‖p ≤ ‖A‖2‖Tν‖p ≤ ‖A‖2‖Tµ‖p (see, for example, [10]).

Note that

〈Tej , em〉 = 〈Tνhj , hm〉 =∫H

hjhm dν(4.2)

for eachm andj. The change of integrations, which is implicit in the last equality of theabove, is justified by [4, Lemma 4.5]. Sinceν = µ on the ballsEr(bm), we see fromLemma 3.1 that

(4.3)∑m

〈Tem, em〉p ≥ C1

∑m

µr(bm)p

for some constantC1 = C1(n, p, k, r) > 0. We claim that there exists a constantCδ =Cδ(n, p, k, r) > 0 such that∑

m6=j

|〈Tej , em〉|p ≤ Cδ

∑m

µr(bm)p(4.4)

andCδ → 0 asδ → 1.Now, we prove (4.4). Letm andj be given. Since

〈Tej , em〉 =∑

i

∫Er(bi)

hjhm dµ,

we have

|〈Tej , em〉|p ≤∑

i

∫Er(bi)

|hjhm| dµ

p

.(4.5)

Note that we have by (2.6) and Lemma 4.6

|hj(z)|p .

b

n2 +kjn

|z − bj |n+k

p

≈ Iij , z ∈ Er(bi)

for all i, j where

Iij =∫

Er(bj)

w

n2 +kn

|bi − w|n+k

p

dλ(w).

It follows that∫Er(bi)

|hjhm| dµ

p

. µ[Er(bi)]pIijIim ≈ bnpin µr(bi)pIijIim

for all i, j andm. Thus, by (4.5), we have

|〈Tej , em〉|p .∑

i

bnpin µr(bi)pIijIim.

Now, summing up both sides of the above over allm andj with m 6= j and then changingthe order of summations, we obtain∑

m6=j

|〈Tej , em〉|p .∑

i

bpnin µr(bi)pIi(4.6)


whereIi =∑

m6=j IijIim. Since the ballsEr(bm) are pairwise disjoint, we have

Ii =∑m6=j

∫Er(bj)

∫Er(bm)

z

n2 +kn

|bi − z|n+k

wn2 +kn

|bi − w|n+k

p

dλ(z) dλ(w)

=∫∫

Ωr,δ

z

n2 +kn

|z − bi|n+k

wn2 +kn

|w − bi|n+k

p

dλ(z) dλ(w)

whereΩr,δ =

⋃j 6=m

Er(bj)× Er(bm).

Note that, for(z, w) ∈ Ωr,δ, we have

β(z, w) ≥ 12

log1 + δ

1− δ− log

1 + r

1− r;(4.7)

recall thatβ denotes the hyperbolic distance given by (2.1). Thus, denoting byGr,δ the setof all (z, w) satisfying (4.7), we obtain

Ii .∫∫

Gr,δ

z

n2 +kn

|z − bi|n+k

wn2 +kn

|w − bi|n+k

p

dλ(z) dλ(w).

Note that the regionGr,δ is invariant under transformationsφa. Also, note that the measuredλ is invariant under transformationsφa. Thus, making change of variablesx = φbi(z)andy = φbi

(w), we have

Ii . Cδb−pnin

where

Cδ = Cδ(n, p, k, r) =∫∫

Gr,δ

x

n2 +kn

|x− z0|n+k

yn2 +kn

|y − z0|n+k

p

dλ(x) dλ(y).

One can check that all the constants suppressed so far depend only onn, p, r andk. Now,sincep(n

2 + k)− n > −1 by our choice ofk, we see from Lemma 2.4 that∫H

x

n2 +kn

|x− z0|n+k

p

dλ(x) <∞.

Therefore, by (4.7) and the Lebesgue dominate convergence theorem, we haveCδ → 0 asδ → 1. Finally, fixing δ sufficiently close1, we see from (4.6) that (4.4) holds.

We introduce some additional auxiliary operators. Associated withT is the diagonaloperatorD : b2 → b2 whose diagonal components are inherited fromT . More precisely,D is defined by

Df =∑m

〈Tem, em〉〈f, em〉em

for f ∈ b2. Note thatD is compact by [14, Proposition 1.3.10] and positive. PutE =T −D. Note that

〈Dem, em〉 = 〈Tem, em〉 and 〈Eej , em〉 = 〈Tej , em〉(4.8)

for anym andj with m 6= j.Assume for a moment thatµ is supported on a compact set. Then a standard volume

argument as in the proof of Theorem 3.5 shows that only finitely many of the ballsEr(am)can intersect the support ofµ. Sinceµ ∈ M+, it is easily checked thatµ with compact


support is a finite measure. Thus it follows from (4.4), (4.8) and Lemma 4.1 thatE ∈ Sp

with

‖E‖pp ≤

∑m6=j

|〈Eej , em〉|p ≤ Cδ

∑m

µr(bm)p <∞.

SinceT,E ∈ Sp, we also haveD ∈ Sp with

‖D‖pp ≥ C1

∑m

µr(bm)p

by (4.3) and (4.8). Thus the triangle inequality‖D‖pp − ‖E‖p

p ≤ ‖T‖pp for 0 < p < 1

yields

(C1 − Cδ)∑m

µr(bm)p ≤ ‖T‖pp ≤ ‖A‖2p‖Tµ‖p

p.

Now, choosingδ sufficiently close to 1 so thatCδ < C1 and then applying the above toeach one of theN subsequencesbm, we conclude∑

m

µr(am)p ≤ N

C1 − Cδ‖A‖2p‖Tµ‖p

p.

This estimate holds for allµ ∈ M+ with compact support and thus for arbitraryµ ∈ M+

by an approximation argument. The proof is complete.

Proof of(b) =⇒ (a). Assume (b). The proof is quite similar to that of (4.4). Fix a largeintegerk > 0 such that

k >n− 12p

− n

2.

By Corollary 3.6 again, we may assumer < minδk, 12 whereδk is the constant as in

Lemma 4.5(b). Letem be an arbitrary orthonormal basis forb2 and letA = QkJ asin the proof of (a)=⇒ (b). But this time the operatorQk is associated with the sequenceam. Puthm = Aem = a

n2mnRk(·, am) as before. LetT = A∗TµA. Note thatA is

bounded and surjective onb2 by Lemma 4.5(b). Thus, there exists a bounded right inverseof A, sayB, so thatTµ = B∗TB. It follows thatTµ ∈ Sp if and only if T ∈ Sp. We nowproveT ∈ Sp in the rest of the proof.

Sinceµr(am) → 0 asm → ∞, we see from (3.6) thatµδ(z) → 0 as |z| → ∞ orzn → 0 for δ sufficiently small. Thus,Tµ is compact by [4, Theorem 5.5] and so isT .NoteT ≥ 0. So, in order to proveT ∈ Sp, it is sufficient to show that∑

m

〈Tem, em〉p ≤ C∑

j

µr(aj)p(4.9)

for some constantC = C(n, p, r) by Lemma 4.2. To prove (4.9), we first note (as in (4.2))

〈Tem, em〉p =∫

H

|hm|2 dµp

≤∑

j

∫Er(aj)

|hm|2 dµ

p

for eachm, because0 < p < 1 andEr(aj) is a covering ofH. Letm andj be given.By (2.6) and Lemma 4.6, we have

|hm(z)|2p .

an+2k

mn

|z − am|2n+2k

p

≈∫

Er(am)

wn+2k

n

|aj − w|2n+2k

p

dλ(w)


for all z ∈ Er(aj). It follows that

〈Tem, em〉p .∑

j

µ[Er(aj)]p∫

Er(am)

wn+2k

n

|aj − w|2n+2k

p

dλ(w)

for all m. Sinceµ[Er(aj)] ≈ anjnµr(aj), summing up both sides of the above over allm

and then changing the order of summations, we obtain∑m

〈Tem, em〉p .∑

j

anpjn µr(aj)pIj(4.10)

where

Ij =∑m

∫Er(am)

wn+2k

n

|aj − w|2n+2k

p

dλ(w).

Now, sincep(n+2k)−n > −1 by our choice ofk, we obtain by Lemma 4.3 (withα = 1)and Lemma 2.4

Ij . N

∫H

zn+2kn

|aj − z|2n+2k

p

dλ(z) ≈ a−npjn

for all j whereN = N(1, r) is the number provided by Lemma 4.3. Hence, combiningthis estimate with (4.10), we conclude (4.9). The proof is complete.

5. SCHATTEN-HERZ CLASSTOEPLITZ OPERATORS

In this section we prove Theorem 1.2. Before introducing Schatten-Herz classes, werecall the Herz spaces and their discrete versions. We decomposeH into a family ofhorizontal stripesAk : k ∈ Z where

Ak = z ∈ H : 2k−1 < zn ≤ 2k

andZ is the set of all integers. For eachk, we letχk = χAk. Recall thatχE denotes the

characteristic function ofE ⊂ H. Also, givenµ ∈M, we letµχk stand for the restrictionof µ to Ak. For 0 < p, q ≤ ∞ andα real, the Herz spaceKp,α

q consists of functionsϕ ∈ Lp

loc(V ) for which

‖ϕ‖Kp,αq

:=∥∥∥

2kα‖ϕχk‖Lp

∥∥∥κq<∞

whereκq denotes the two-sidedq summable sequence space. Also, we letKp,α0 be the

space of all functionsϕ ∈ Kp,α∞ such that

2kα‖ϕχk‖Lp

∈ κ0 whereκ0 denotes the

subspace ofκ∞ consisting of all sequences vanishing at±∞. Note thatKp,αq ⊂ Kp,α

0 forall q <∞. For more information on the Herz spaces, see [8] and references therein.

Let 0 < p <∞ andα be real. Then, sincezn ≈ 2k for z ∈ Ak andk ∈ Z, we have

2kαp‖ϕχk‖pLp =

∫Ak

2kαp|ϕ(z)|p dz ≈∫

Ak

|ϕ(z)|pzαpn dz

and thus

‖ϕ‖Kp,αq

≈∥∥∥‖ϕχk‖Lp

αp

∥∥∥κq

(5.1)

for 0 < q ≤ ∞. In particular, we have

‖ϕ‖K

p,−np

q

≈∥∥∥‖ϕχk‖Lp(λ)

∥∥∥κq


and this estimate is valid even forp = ∞, because−n/p = 0 if p = ∞. For this reason,we put

Kpq(λ) = Kp,−n

pq and Kp

0(λ) = Kp,−np

0

for the full range0 < p, q ≤ ∞. Note thatKpp(λ) ≈ Lp(λ) for 0 < p ≤ ∞. That is, these

two spaces are the same as sets and have equivalent norms.Next, we introduce a discrete version of Herz spaces. Leta = am be an arbitrary

lattice. Given a complex sequenceξ = ξm andk ∈ Z, let ξχk denote the sequencedefined by(ξχk)m = ξmχk(am). Now, given0 < p, q ≤ ∞ andα real, we letκp,α

q (a) bethe mixed-norm space of all complex sequencesξ such that

‖ξ‖kp,αq (a) :=

∥∥∥2kα‖ξχk‖`p∥∥∥

κq<∞.

So, we have

‖ξ‖qkp,α

q (a)=

∑k

2kαp

∑am∈Ak

|ξm|p q

p

, 0 < p, q <∞.

Also, we sayξ ∈ κp,α0 (a) if ‖2kαξχk‖`p ∈ κ0. Finally, we letκp,0

q (a) = κpq(a).

We now introduce the so-called Schatten-Herz class of Toeplitz operators. LetS∞denote the class of all bounded linear operators onb2 and‖ ‖∞ denote the operator norm.Given0 < p, q ≤ ∞, theSchatten-HerzclassSp,q is the class of all Toeplitz operatorsTµ

such thatTµχk∈ Sp for eachk and the sequence‖Tµχk

‖p belongs toκq. The norm ofTµ ∈ Sp,q is defined by

‖Tµ‖p,q =∥∥∥‖Tµχk

‖p∥∥∥

κq.

Also, we sayTµ ∈ Sp,0 if Tµ ∈ Sp,∞ and‖Tµχk‖p ∈ κ0. Note thatSp,q ⊂ Sp,0 for all

q <∞.One key step toward Theorem 1.2 is to establish the Herz space version of Theorem 3.5.

The following observation plays a key role in proving thatKp,αq -behavior orκp,α

q -behaviorof averaging functions are independent of radii.

Theorem 5.1. Let0 < p, q ≤ ∞, r ∈ (0, 1) andα be real. Then the following statementshold forµ, τ ≥ 0.

(a)2kα‖µrχk‖Lp(τ)

∈ κq if and only if

2kα‖(µχk)r‖Lp(τ)

∈ κq.

(b)2kα‖µrχk‖Lp(τ)

∈ κ0 if and only if

2kα‖(µχk)r‖Lp(τ)

∈ κ0.

Before proceeding to the proof, we note the following covering property which is asimple consequence of (2.2): Ifr ∈ (0, 1) andN = N(r) is a positive integer such that

2N−1 ≤ 1 + r

1− r< 2N ,(5.2)

then

Er(z) ⊂k+N⋃

j=k−N

Aj(5.3)

for z ∈ Ak andk ∈ Z.


Proof. Let µ, τ ≥ 0 be given. We prove the proposition only forq < ∞; the caseq = ∞is implicit in the proof below. ChooseN = N(r) as in (5.2) and putγp = γp,N =max1, (2N + 1)p−1. Note that we have by (5.3)

µrχk ≤k+N∑

j=k−N

(µχj)r

for all k ∈ Z. Thus, we have

‖µrχk‖Lp(τ) ≤ γ1pp

k+N∑j=k−N

‖(µχj)r‖Lp(τ)

for all k. Since2kα ≤ 2N |α|2jα for k −N ≤ j ≤ k +N , it follows that

2kα‖µrχk‖Lp(τ) .k+N∑

j=k−N

2jα‖(µχj)r‖Lp(τ)

for all k. This implies one direction of (b). Also, it follows that

2kαq‖µrχk‖qLp(τ) . γq

k+N∑j=k−N

2jαq‖(µχj)r‖qLp(τ)

for all k. Thus, summing up both sides of the above over allk, we have∞∑

k=−∞

2kαq‖µrχk‖qLp(τ) . (2N + 1)

∞∑j=−∞

2jαq‖(µχj)r‖qLp(τ),

which gives one direction of (a).We now prove the other directions. Note thatEr(z) can intersectAk only whenz ∈

∪k+Nj=k−NAj by (5.3). Thus we have

(µχk)r =k+N∑

j=k−N

(µχk)rχj ≤k+N∑

j=k−N

µrχj .

Thus, a similar argument yields the other directions of (a) and (b). The proof is complete.

As an immediate consequence of Proposition 3.2 and Theorem 5.1 (withτ = V ), wehave the following Herz space version of Proposition 3.2.

Corollary 5.2. Let0 < p ≤ ∞, 0 ≤ q ≤ ∞, δ, r ∈ (0, 1) andα be real. Letµ ≥ 0. Thenµr ∈ Kp,α

q if and only ifµδ ∈ Kp,αq .

Also, applying Theorem 5.1 with discrete measuresτ =∑

m δam whereδx denotes thepoint mass atx ∈ H, we have the following.

Corollary 5.3. Let 0 < p, q ≤ ∞, r ∈ (0, 1) andα be real. Letµ ≥ 0 and assume

that a = am is an r-lattice. Putξk = 2kα‖(µχk)r(am)m‖`p for k ∈ Z. Then thefollowing statements hold.

(a) µr(am) ∈ κp,αq (a) if and only ifξk ∈ κq.

(b) µr(am) ∈ κp,α0 (a) if and only ifξk ∈ κ0.


We need one more fact from [7, Theorem 3.6]. In fact the following is proved in [7,Theorem 3.6] only for1 ≤ q ≤ ∞ or q = 0, but the same (actually simpler) proof worksfor 0 < q < 1.

Lemma 5.4. Let1 ≤ p ≤ ∞ andα be real such that−n < α+ 1p < 1. Then the Berezin

transform is bounded onKp,αq for 0 ≤ q ≤ ∞.

We now prove the Herz space version of Theorem 3.5.

Theorem 5.5. Let 0 < p ≤ ∞, 0 ≤ q ≤ ∞, r ∈ (0, 1) andα be real. Letµ ≥ 0 andassume thata = am is anr-lattice. Then the following two statements are equivalent.

(a) µr ∈ Kp,αq .

(b) µr(am) ∈ κp,α+ np

q (a).

Moreover, if

−n− α <1p< min

1− α, 1− α

n

,(5.4)

then the above statements are equivalent to

(c) µ ∈ Kp,αq .

As in Theorem 3.5, the restricted range in (5.4) can’t be improved; see Section 6.

Proof. (a) ⇐⇒ (b): Let ξk = 2k( np +α)‖(µχk)r(am)‖`p for eachk. Note that if

(µχk)r(am) > 0, thenAk ∩ Er(am) 6= ∅ and thusamn ≈ 2k. Thus, forp < ∞, wehave

ξk = ‖(µχk)r(am)anp +αmn ‖`p = ‖(µχk)r(am)‖`p,n+αp(a) ≈ ‖(µχk)r‖Lp

αp(5.5)

for all k by Theorem 3.5; the last norm equivalence is implicit in the proof of Theorem 3.5.It follows that

µr ∈ Kp,αq ⇐⇒ ‖µrχk‖Lp

αp ∈ κq (5.1)

⇐⇒ ‖(µχk)r‖Lpαp ∈ κq (Theorem 5.1)

⇐⇒ ξk ∈ κq (5.5)

⇐⇒ µr(am) ∈ κp, np +α

q (a) (Corollary 5.3),

which completes the proof forp <∞. The proof forp = ∞ is similar.

(c) =⇒ (a): This follows from Lemma 3.1 and Proposition 3.2.

We now assume (5.4) and prove that (a) or (b) implies (c). For1 ≤ p ≤ ∞, one mayuse Lemmas 5.4 and 3.4 to see that (a) implies (c). So, we may further assumep < 1 inthe proof below.

(b) =⇒ (c): Assume (b). First, consider the case0 < q <∞. By the proof of Theorem3.5, we have

µ(z)p .∑m

anpmnµr(am)p znp

n

|am − z|2np(5.6)


for z ∈ H. Note that, foram ∈ Ak andz ∈ Aj , we have

anpmn

∫Aj

znpn

|am − z|2npzαpn dz ≈ 2knp2jp(n+α)

∫ 2j

2j−1

dt

(amn + t)2np−n+1

≈ 2knp2j(np+αp+1)

(2k + 2j)2np−n+1.

Thus, integrating both sides of (5.6) overAj against the measuredz, we obtain∫Aj

µ(z)pzαpn dz .

∑k

ξpk ·

2k(np−n−αp)2j(np+αp+1)

(2k + 2j)2np−n+1

=∑

k

ξpk ·

2(np−n−αp)(k−j)

(2k−j + 1)2np−n+1

whereξpk = 2kp( n

p +α) ∑am∈Ak

µr(am)p. Note that


(2k−j + 1)2np−n+1≤ 2(np−n−αp)(k−j) =

12(np−n−αp)|k−j| for k < j

and


(2k−j + 1)2np−n+1≈ 1

2(np+αp+1)(k−j)=

12(np+αp+1)|k−j| for k ≥ j.

Therefore, combining these estimates, we have∫Aj

µ(z)pzαpn dz .

∑k

ξpk

2γ|k−j|(5.7)

for all j whereγ = minnp− n− αp, np+ αp+ 1. Noteγ > 0.Now, for p < q ≤ ∞, we have by (5.7) and Young’s inequality

‖µ‖pKp,α

q. ‖ξp

k‖κqp

= ‖ξk‖pκq .

On the other hand, for0 < q ≤ p, we have again by (5.7)

‖µ‖pKp,α

q.

∑j

∑k

ξpk

2γ|k−j|

qp

.∑

k

ξqk

∑j

12γq|k−j|/p

≈ ‖ξk‖qκq

as desired. The caseq = 0 also easily follows from (5.7). The proof is complete.

The following version of Theorem 1.2 is now a simple consequence of what we’veproved so far.

Theorem 5.6. Let 0 < p ≤ ∞, 0 ≤ q ≤ ∞ andµ ∈ M+. Assume thata = am is anr-lattice. Then the following three statements are equivalent.


q(λ).(c) µr(am) ∈ κp

q .

Moreover, ifn−1n < p ≤ ∞, then the above statements are also equivalent to

(d) µ ∈ Kpq(λ).

Proof. The proofs of Theorems 4.7 and 5.1 show that all the associated norms are equiva-lent. We have (a)⇐⇒ (b) by Theorem 4.7 and Corollary 5.3. Hence the theorem followsfrom Theorem 5.5 (withα = −n

p ).


6. EXAMPLES

In this section we show that the parameter ranges (3.4) and (5.4) are sharp and providesome related examples. Throughout the section we consider arbitrary0 < p ≤ ∞ andαreal, unless otherwise specified.

We split the parameterα into two cases depending on its sign.

6.1. The caseα ≤ 0. In this case the parameter ranges for which we need examples areas follows:

p ≤ max

1 +α

n,−α+ 1

n

for (3.4)

and

1p≤ −n− α or

1p≥ 1− α

nfor (5.4).

The general property for Berezin transforms of positive measures stated in the nextproposition covers all the cases except for the critical cases withq = ∞ in (c).

Proposition 6.1. Letµ ≥ 0 andµ(H) > 0. Then the following statements hold:

(a) If p ≤ max1 + αn ,−

α+1n , thenµ /∈ Lp

α.(b) If 1

p < −n− α or 1− αn < 1

p , thenµ /∈ Kp,α∞ .

(c) If 1p = −n− α or 1

p = 1− αn , thenµ /∈ Kp,α

0 .

Proof. We introduce some temporary notation. Givenx ∈ Rn, we let

Kx = w ∈ H : wn − xn ≥ 2|w′ − x′|.

Note thatKx is the intersection ofH and a cone with vertex atx. The following two factsare easily verified:

Kx ⊂ Kz, x ∈ Rn, z ∈ Kx(6.1)

and

R(z, w) ≥ C1

|z − w|n≥ C2

(zn + wn)n, z ∈ H, w ∈ Kz(6.2)

for some positive constantsC1 andC2 depending only onn.Sinceµ(H) > 0 by assumption, there is somea ∈ H andN > 0 such thatµ(Ka,N ) >

0 whereKa,N = w ∈ Ka : wn ≤ N. Givenz ∈ Ka, we have by (6.1) and (6.2)

µ(z) & znn

∫Ka,N

dµ(w)(zn + wn)2n

≥ znn

(N + zn)2nµ(Ka,N ).(6.3)


Accordingly, givenp ≤ max1 + αn ,−

α+1n , we have∫

H

µ(z)pzαn dz &

∫Ka

zpn+αn

(N + zn)2pndz

=∫ ∞

0

∫2|z′−a′|≤an+zn

zpn+αn

(N + zn)2pndz′ dzn

≈∫ ∞

0

zpn+αn (an + zn)n−1

(N + zn)2pndzn

≈∫ ∞

0

zpn+αn

(1 + zn)2pn−n+1dzn

= ∞.

The last equality holds, becausepn + α ≤ −1 or pn − n − α ≤ 0. This completes theproof of (a).

In order to prove (b) and (c), note that

|Ak ∩Ka| ≈ (an + 2k)n−12k ≈ (1 + 2k)n−12k

for all k. It follows from (6.3) that

2kα‖µχk‖Lp &2k(n+α+ 1

p )

(1 + 2k)2n− (n−1)p

≈

2k(α−n+ n

p ) for k ≥ 0,2k(n+α+ 1

p ) for k < 0.

This implies (b) and (c). The proof is complete.

For the missing critical cases in Proposition 6.1(c) withq = ∞, we provide explicitexamples below. First, we provide an example that covers the critical case1

p = 1− αn . In

the following we letK = w ∈ H : |w′| ≤ wn.

Example 6.2. Letf(w) = w−nn χK(w). If 1

p = 1− αn , thenf, fr ∈ Kp,α

∞ for all r ∈ (0, 1)

but f = ∞ on some open set.

Proof. Since1p = 1− α

n and|Ak ∩K| ≈ 2kn, we have

2kα‖fχk‖Lp ≈ 2kα2−kn|Ak ∩K|1p ≈ 1

for all k and thusf ∈ Kp,α∞ . Givenr ∈ (0, 1), note that

fr(z) =λ[K ∩ Er(z)]|Er(z)|

≤ λ[Er(z)]|Er(z)|

=|Er(z0)||Er(z)|

= z−nn

for z ∈ H. Also, it is not hard to see that there exists a constantc = c(r) > 0 sufficientlylarge such thatK ∩ Er(z) = ∅ if |z′| > czn. So, we havefr(z) ≤ cnf(z′, czn) ∈ Kp,α

∞ .We now show thatf = ∞ on Ω := z ∈ H : 1 ≤ |z′| ≤ zn

4 . Note that ifz ∈ Ω,w ∈ K andwn ≤ 1, then2|z′ − w′| ≤ 2(|z′|+ 1) ≤ zn and thus

|R(z, w)| ≈ 1|z − w|n

≈ 1(wn + zn)n

.

It follows that

f(z) & znn

∑k≤0

∫Ak∩K

dw

wnn(wn + zn)2n

&1znn

∑k≤0

1 = ∞

for z ∈ Ω. This completes the proof.


Next, we provide an example that covers the critical case1p = −n−α. In the following

we letΓ = w ∈ Rn : |w′| ≤ 1 and wn ≤ 1.

Example 6.3. Letf(w) = wnnχΓ(w). If 1

p = −n−α, thenf, fr ∈ Kp,α∞ for all r ∈ (0, 1)

but f /∈ Kp,α∞ .

Proof. Before proceeding to the proof, we first note the following representation of theBerezin transform:

g(z) = znn

∫H

g(z′ + (zn + wn)w′, wn

) ∣∣R(z0, (w′, 0)

)∣∣2(zn + wn)n+1

dw(6.4)

for measurable functionsg ≥ 0. This is easily verified by an elementary change of vari-ables. In conjunction with (6.4), we note

|z′ + (zn + wn)w′| ≤ 1(6.5)

for z, w ∈ 12Γ.

We now turn to the proof. Since1p = −n−α, it is straightforward to see2kα‖fχk‖Lp ≈2k(α+n+ 1

p ) = 1 for all k < 0. Thus,f ∈ Kp,α∞ . Givenr ∈ (0, 1), note that

fr(z) =Vn[Γ ∩ Er(z)]

|Er(z)|≤ Vn[Er(z)]

|Er(z)|≈ zn

n

for z ∈ H. Also, it is not hard to see that there exists a constantc = c(r) > 0 sufficientlylarge such thatΓ ∩ Er(z) = ∅ if z /∈ cΓ. So, we havefr(z) . cnf(c−1z) ∈ Kp,α

∞ .Forz ∈ 1

2Γ, we have by (6.4) and (6.5)

f(z) ≥ znn

∫ 12

0

wnn dwn

(zn + wn)n+1

∫|w′|≤ 1

2

∣∣R(z0, (w′, 0)

)∣∣2 dw′& zn

n

∫ 12

zn

1wn

dwn

= znn(| log zn| − log 2).

It follows that

2kα‖fχk‖Lp ≥ 2kα‖fχ(Ak∩Γ)‖Lp & 2k(α+n+ 1p )| log 2k| = |k| log 2

for k < −1 and thusf /∈ Kp,α∞ . The proof is complete.

6.2. The caseα > 0. Since Proposition 6.1, Examples 6.3 and 6.2 are valid for allα, theparameter ranges withα > 0 for which we need examples are as follows:

1 +α

n< p ≤ 1 + α for (3.4)

and

1− α ≤ 1p< 1− α

nfor (5.4).

Note that we havep > 1 in both cases of the above. Forp ≥ 1, it is not hard to verifythat averaging operators are all bounded onLp

α for anyα. Thus, whenp > 1 in addition inboth examples below, we havefr ∈ Lp

α (orKp,αq ), r ∈ (0, 1), wheneverf ∈ Lp

α (orKp,αq ).


The source of our examples is the family of functionsFε, ε > 0, defined by

Fε(w) =χΓ(w)

wn(1 + | logwn|)ε.

whereΓ is the set introduced in Example 6.3. Since|Ak ∩ Γ| ≈ 2k for k ≤ 0, it is easilyseen that

2kα‖Fεχk‖Lp ≈ 1

(1 + |k|)ε2|k|(α−1+ 1p ), k ≤ 0(6.6)

for eachε.Among the statements of the next example, (a) and (c) imply that the range in (3.4) is

best possible forα > 0. Meanwhile, (b) and (c) imply that the range in (5.4) is also bestpossible forα > 0 except for the critical case where1p = 1− α < 1 and0 < q ≤ 1.

Example 6.4. The functionF1 has the following properties.

(a) F1 ∈ Lpα if and only if either0 < p < α+ 1 or 1 < p = 1 + α.

(b) F1 ∈ Kp,αq if and only if

either (i)1p> 1− α, 0 ≤ q ≤ ∞

or (ii)1p

= 1− α, 1 < q ≤ ∞ or q = 0.

(c) F1 = ∞ on 12Γ.

Proof. Note thatw−sn (1 + | logwn|)−t is integrable near0 if and only if eithers < 1 or

s = 1 < t. Thus (a) holds. Also, (b) holds by (6.6).Following the proof of Example 6.3, we have by (6.4) forz ∈ 1

2Γ

F1(z) & znn

∫ 12

0

dwn

(wn + zn)n+1wn(1 + | logwn|)& z−1

n

∫ zn

0

dwn

wn| logwn|= ∞

and thus (c) holds. The proof is complete.

The missing case where1p = 1−α < 1 and0 < q ≤ 1 is covered by the next example.

Example 6.5. Suppose1p = 1− α and0 < q ≤ 1. If 1q < ε ≤ 1

q + 1, thenFε ∈ Kp,αq but

Fε 6∈ Kp,αq .

Proof. We haveFε ∈ Kp,αq by (6.6), whenqε > 1. Also, following the proof of Example

6.4, we have

Fε(z) & z−1n

∫ zn

0

dwn

wn| logwn|ε=

1(ε− 1)zn| log zn|ε−1

for z ∈ 12Γ. This estimate yields

2kα‖Fεχk‖Lp ≥ 2kα‖Fεχ(Ak∩ 12Γ)‖Lp & |k|−ε+1

for k < 0. ThusFε /∈ Kp,αq , whenq(ε− 1) ≤ 1. The proof is complete.


6.3. Remarks. 1. If 1 ≤ p and1 + α ≤ p, then the proof of [7, Lemma 3.3] shows theexistence of a functionf ≥ 0 in Lp

α such thatf = ∞ on some open set. Example 6.4(a)provides such a function explicitly.

2. In view of Theorem 3.5 and Proposition 6.1, one may ask whether the restrictionp ≥ 1 in Lemma 3.3 can be improved top > n−1

n . The answer isno. (Example):Assume0 < p < 1. LetEr(a) be a small ball such thatE2r(a) is contained in the coneK = Kz0 (defined in the proof of Proposition 6.1). Choose a functionf ≥ 0 supportedonEr(a) such thatf ∈ Lp but f /∈ L1. Suchf always exists, becausep < 1. Sincefhas compact support, we havef ∈ Lp

α but f /∈ L1α for anyα. Also, for all z such that

φz(Er(a)) ⊂ E2r(a) ⊂ K, we have by (2.5) and (6.2)

znn f(z) =

∫Er(a)

f(w)|R(z0, φz(w)

)|2 dw &

∫Er(a)

f(w)(1 + wn/zn)2n

dw = ∞

so thatf = ∞ on some open set.3. Note that the example in the second remark above also shows that the restriction

p ≥ 1 in Lemma 5.4 cannot be improved, either. Thus examples in this section show thatthe hypothesis1 ≤ p ≤ ∞, −n < α + 1

p < 1 andq arbitrary is the precise range ofparameters for which the Berezin transform is bounded onKp,α

q .4. In Proposition 6.1, one may wonder whether the conclusions in the critical cases (c)

can be improved as in (b). The answer isno. In fact we haveχE ∈ Kp,α∞ , when either

1p = 1− α

n ≥ −n− α or 1p = −n− α ≤ 1− α

n , for any compact setE ⊂ H. (Sketch ofProof): We may assumeE = E1/2(a) for somea ∈ H. By (1.4) and Lemma 2.2

χE(z) = znn

∫E1/2(a)

|R(z, w)|2 dw .znn

|z − a|2n|E1/2(a)| ≈

znn

|z − a|2n

for all z ∈ H. This estimate and Lemma 2.5 yield

2kα‖χEχk‖Lp .

2k(α−n+ n

p ) for k ≥ 0,2k(n+α+ 1

p ) for k < 0.

Now one can seeχE ∈ Kp,α∞ in either case.

5. Forp ≥ 1 and−n(1 + 1p ) < α < 1− 1

p , we have by [7, Proposition 3.5]

|f(z)| ≤ C(p, α)z−n

p−αn ‖f‖Kp,α

∞ , f ∈ Kp,α∞ .

Moreover, for the end pointα = −n(1 + 1p ) or α = 1− 1

p , the inequality

|f(z)| ≤ C(p, α)z−n

p−αn ‖f‖Kp,α

1, f ∈ Kp,α

1

is implicit in the proof of [7, Proposition 3.5]. Thus, whenp ≥ 1 in Examples 6.3 and6.5, one cannot expect (unlessp = ∞ andα = −n) examples of functions whose Berezintransforms diverge as in Examples 6.2 and 6.4.

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Nagoya Math. J., 151(1998), 51–89[4] B. R. Choe, H. Koo and H. Yi,Positive Toeplitz operators between the harmonic Bergman spaces,Potential

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[11] J. Miao,Reproducing kernels for harmonic Bergman spaces of the unit ball, Monatsh. Math. 125(1998),25–35.

[12] J. Miao,Toeplitz operators on harmonic Bergman space, Integr. Equ. Oper. Theory 27(1997), 426–438.[13] W. Ramey and H. Yi,Harmonic Bergman functions on half-spaces, Trans. Amer. Math. Soc. 348(1996),

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DEPARTMENT OFMATHEMATICS, KOREA UNIVERSITY, SEOUL 136-713, KOREAE-mail address: [email protected]

DEPARTMENT OFMATHEMATICS, KOREA UNIVERSITY, SEOUL 136-713, KOREAE-mail address: [email protected]

DEPARTMENT OFMATHEMATICS, CHONNAM NATIONAL UNIVERSITY, GWANGJU 500-757, KOREAE-mail address: [email protected]

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ON THE HALF-SPACE - Korea Universityelie.korea.ac.kr/~choebr/papers/schattenherz(halfspace).pdfBergman kernels. In Section 3 we investigate weighted Lp-behavior of averaging functions