Standard Deviation and Schatten Class Hankel Operators on the Segal-Bargmann Space

Transcription

1 1381 Indiana University Mathematics Journal c, Vol. 53, No. 5 (2004) tandard Deviation and chatten Class Hankel Operators on the egal-bargmann pace JINGBO XIA & DECHAO ZHENG ABTRACT. We consider Hankel operators on the egal-bargmann space H 2 (C n,dµ). Our main result is a necessary and sufficient condition for the simultaneous membership of H f and H f in the chatten class C p,1 p<. We will explain that, since this condition is valid in the case 1 p 2 as well as in the case 2 p<, this result reflects the structural difference between the egal-bargmann space and other reproducing-kernel spaces such as the Bergman space L 2 a(b n,dv). 1. INTRODUCTION Let dµ be the Gaussian measure on C n centered at 0 and normalized so that µ(c n ) = 1. In terms of the standard volume measure dv on C n R 2n we can write dµ as dµ(z) = π n e z 2 dv (z). Recall that the egal-bargmann space (also called Fock space) H 2 (C n,dµ) is defined to be the subspace {f L 2 (C n,dµ) f is analytic on C n } of L 2 (C n,dµ). Throughout the paper, we let P : L 2 (C n,dµ) H 2 (C n,dµ) be the orthogonal projection. It is well known that H 2 (C n,dµ) is the L 2 (C n,dµ)-closure of the polynomials in the variables z 1,...,z n. Therefore {(k 1! k n!) 1/2 z k 1 1 zk n n k 1 0,...,k n 0}

2 1382 JINGBO XIA & DECHAO ZHENG is an orthonormal basis in H 2 (C n,dµ). Thus P has e z,w as its kernel function, i.e., (Pϕ)(z) = e z,w ϕ(w) dµ(w), ϕ L 2 (C n,dµ). Here and in what follows, we write z, w =z 1 w 1 + +z n w n for z = (z 1,...,z n )and w = (w 1,...,w n )in C n. The function K z (w) = e w,z is called the reproducing kernel for the egal-bargmann space H 2 (C n,dµ),which simply reflects the fact that (Pϕ)(z) = ϕ, K z, ϕ L 2 (C n,dµ).lateronwe will also need the normalized reproducing kernel k z = K z / K z. For each ζ C n,letτ ζ :C n C n be the translation (1.1) τ ζ (w) = w + ζ, w C n. Define T (C n ) ={f L 2 (C n,dµ) f τ ζ L 2 (C n,dµ)for every ζ C n }. It is easy to see that a measurable function f on C n belongs to T (C n ) if and only if the function w f(w)e w,ζ belongs to L 2 (C n,dµ) for every ζ C n. This in particular means that, if f T(C n ), then the set {h H 2 (C n,dµ) fh L 2 (C n,dµ)}is a dense, linear subspace of H 2 (C n,dµ). Recall that the Hankel operator H f : H 2 (C n,dµ) L 2 (C n,dµ) H 2 (C n,dµ) with symbol function f is defined by the formula H f = (1 P)M f P. Thus if f T(C n ), then H f has at least a dense domain in H 2 (C n,dµ). When considering Hankel operators on the egal-bargmann space H 2 (C n,dµ),onealways needs to impose some growth condition on the symbol function. This is due to the fact that, unlike other reproducing-kernel Hilbert spaces, there are no bounded functions in H 2 (C n,dµ)other than constants. There is another kind of Hankel operators, i.e., the so-called small Hankel operators h f = PMf P, where P is the projection from L 2 (C n,dµ)to H2 (C n,dµ)= { f f H 2 (C n,dµ)}. uch operators were studied extensively in, e.g. [6, 12] and are not the concern of the present paper. In other words, in this paper we will deal with H f only. The compactness of Hankel operators with bounded symbols on the egal- Bargmann H 2 (C n,dµ)was characterized by C. Berger and L. Coburn in [2] and by K. troethoff in [7] (also see [3]). Normally one would expect that a characterization of the membership of such operators in the chatten p-classes would soon

3 chatten Class Hankel Operators 1383 follow. A search of the literature reveals, however, that the latter characterization is still lacking. The purpose of this paper is to fill this void. Recall that, for any 1 p<, the chatten class C p consists of operators T with T p <, where the p-norm is defined by the formula T p ={tr( T p )} 1/p ={tr((t T) p/2 )} 1/p. In this paper we will completely determine the membership H f C p and H f C p in terms of the symbol function f. Given a ϕ L 2 (C n,dµ),letd(ϕ) denote its standard deviation with respect to the probability measure dµ. Recall that the formula for standard deviation is { 2 } 1/2 D(ϕ) = ϕ ϕdµ dµ { = ϕ 2 1/2 dµ ϕdµ 2} ={ ϕ 2 ϕ, 1 2 } 1/2. When f L (C n,dµ),h f is compact if and only if H f is compact [2]. It follows from [7, Theorem 12] that the condition lim ζ D(f τ ζ ) = 0 is equivalent to the compactness of H f and H f. The following is the main result of the paper. Theorem 1.1. Let 1 p<. Let f T(C n )and let H f and H f be the corresponding Hankel operators from H 2 (C n,dµ) to {H 2 (C n,dµ)} defined above. Then we have the simultaneous membership H f C p and H f C p if and only if (1.2) {D(f τ ζ )} p dv (ζ) <, where, as we recall, dv denotes the standard volume measure on C n R 2n. ince [M f,p] = H f H f and [M f,p]p = H f, an alternate statement of Theorem 1.1 is that the membership [M f,p] C p is equivalent to (1.2). Theorem 1.1 is motivated by K. Zhu s characterization [11] of chatten class Hankel operators on the Bergman space L 2 a (B n,dv) of the unit ball of C n. A detailed comparison of our result with Zhu s is appropriate at this point. However, since such a comparison cannot avoid technical definitions associated with the Bergman space, definitions which have nothing to do with the rest of the paper, we will defer it until ection 6 at the end of the paper. We close this section with a brief sketch of the ideas involved in the proof of Theorem 1.1. It is fairly easy to show (Lemma 3.1 below) that, when 1 p 2, the simultaneous membership H f C p and H f C p implies (1.3) b Z 2n { Q p/2 f(z+b) f(w +b) 2 dv (w) dv (z)} <, Q

4 1384 JINGBO XIA & DECHAO ZHENG where Q is the standard cubeinC n R 2n. The key here is the realization that (1.2) can be deduced from (1.3). And this deduction is the main content of ection 3. Thus, in the case 1 p 2, (1.2) is necessary for H f C p and H f C p. It is also quite easy to show (Lemma 4.1) that(1.2) implies (1.3). As it turns out, in the case 2 p<, we can deduce [M f,p] C p from (1.3)by an explicit decomposition v Z 2n Y v of the commutator and by direct estimates of Y v p, which are given in ection 4. Thus, when 2 p<,(1.2)issufficient for H f C p and H f C p. The opposite directions in the cases 1 p 2and 2 p<, which will be taken care of in ection 5, are both easy. Indeed the material contained in ection 5 is simply a specialization to the egal-bargmann space of what is well known for reproducing-kernel Hilbert spaces [10]. Acknowledgement. The authors wish to thank Lewis A. Coburn for the many illuminating conversations related this work. Note added January 31, From the conversations mentioned above arose the following problem: For f L (C n,dµ) and 1 p <, does the membership H f C p imply H f C p?this problem was motivated by a theorem of Berger and Coburn [2] which asserts that, for f L (C n,dµ),h f is compact if and only if H f is compact. We were able to solve the problem for the special case where p = 2 and n = 1. When this paper was first completed in eptember 2000, it contained Theorem 1.3: If f is a bounded measurable function on the complex plane C, thenh f C 2 if and only if H f C 2.Theorem 1.3 has since been extended to f L (C n,dµ)for all complex dimensions n 1 by W. Bauer (Hilbert-chmidt Hankel operators on the egal-bargmann space, preprint, April, 2002). In eptember 2002, the authors received the preprint Hilbert-chmidt Hankel operators on the Fock space by K. troethoff, which also contained the same generalization of Theorem 1.3 to all complex dimensions n 1. In view of these developments, and of the fact that our method only covered the case n = 1and was rather ad hoc, the authors have decided not to include Theorem 1.3 in the present version of the paper. But we would like to mention that, for all values p 2andn 1, the problem stated above remains open and appears to be rather challenging. 2. PRELIMINARIE We begin with the lattice Z 2n in R 2n C n. A subset s ={p 0,...,p k } of Z 2n with k 1 is said to be a discrete segment in Z 2n if there exist a j {1,...,2n} and a z Z 2n such that p i = z + ie j, 0 i k, where e j is the vector in Z 2n whose j-th component is 1 and whose other components are 0. Furthermore, in this setting we will say that p 0 and p k are the end points of s. Also, for this s, we define its length s to be k, i.e., the distance

5 chatten Class Hankel Operators 1385 between the end points p 0 and p k. In other words, for a discrete segment s, we have s = card(s) 1. Let v = (v 1,...,v 2n ) Z 2n and suppose that v hasatleastonenonzero component. Then we can enumerate the integers {j v j 0, 1 j 2n} as j 1,..., j m in ascending order. That is, j 1 < <j m in the event m>1. We set z 0 (v) = (0,...,0), the zero vector in Z 2n. We then inductively define z t (v) = z t 1 (v) + v jt e jt for every t {1,...,m}. Note that z m (v) = v. Let s t (v) be the discrete segment in Z 2n which has z t 1 (v) and z t (v) as its end points, t {1,...,m}. The union of the discrete segments s 1 (v),..., s m (v) will be denoted by Γ (v). We will call Γ (v) the discrete path in Z 2n from 0 to v. Furthermore, we define the length Γ (v) of Γ (v) to be s 1 (v) + + s m (v). In other words, the length of the discrete path Γ (v) is simply the sum of the lengths of the discrete segmentswhich make up Γ (v). This defines the discrete path Γ (v) and its length Γ (v) in the case v 0. In the case v = 0, we define the discrete path form 0 to 0 to be the singleton set Γ (0) ={0}. Naturally, the length Γ (0) of Γ (0) is defined to be 0. We will consider Z 2n as a lattice in C n in the obvious sense. That is, we will identify (k 1,l 1,...,k n,l n ) with (k 1 + il 1,...,k n +il n ) for any k 1, l 1,...,k n,l n Z. We conclude this section with a few more conventions which will be used in the rest of the paper. The symbol will denote the unit cube in C n.thatis, (2.1) ={(x 1 + iy 1,...,x n +iy n ) x 1,y 1,...,x n,y n [0,1)}. The symbol Q is also reserved to denote a particular cube in C n : (2.2) Q ={(x 1 + iy 1,...,x n +iy n ) x 1,y 1,...,x n,y n [ 1,2)}. For any f L 2 local(c n,dv),wewillwrite (2.3) J(f) = f(z) f(w) 2 dv (z) dv (w). Q Q As we will see in the next two sections, this notation saves a lot of writing. For example, with this notation, condition (1.3) is now simply b Z 2n {J(f τ b )} p/2 <. Finally, if E is a Borel set with 0 <V(E)<, we will denote the mean value of f on E by f E.Thatis, (2.4) f E = 1 fdv. V(E) E

6 1386 JINGBO XIA & DECHAO ZHENG In particular, f = fdvsince V() = 1. We emphasize that the measure which appears in (2.3)and(2.4)isdV,notdµ. 3. NECEITY IN THE CAE 1 p 2 The purpose of this section is to establish that the simultaneous membership H f C p and H f C p implies (1.2) in the case 1 p 2. Lemma 3.1. uppose that 1 p 2. Iff T(C n )is such that H f C p and H f C p,then v Z 2n{J(f τ v)} p/2 <, whereτ v is the translation defined by (1.1). Proof. Given f as above, for each v Z 2n we define the operator (K v ψ)(z) = χ Q+v (z) ( f(w) f (z))e z,w ψ(w) dµ(w), ψ L 2 (C n,dµ). Q+v Let us first consider the case 1 <p 2. In this case we set q = p/(p 1). Itis obvious that K v C 2 for every v. inceq 2, we have (3.1) Kv 2 q Kv 2 2 = f(w) f(z) 2 e z,w 2 dµ(w) dµ(z) Q+v Q+v = π 2n f(w) f(z) 2 e w z 2 dv (w) dv (z) Q+v Q+v = π 2n f(w +v) f(z+v) 2 e w z 2 dv (w) dv (z) Q Q J(f τ v ). For each v Z 2n,defineα v ={J(f τ v )} (p 2)/2 in the case J(f τ v )>0and define α v = 0inthecaseJ(f τ v ) = 0. Given any finite subset F Z 2n,we define K F = v F α v K v. We claim that (3.2) K F q 3 2n{ (J(f τ v )) p/2} 1/q. v F To prove (3.2), let us introduce the subset Λ ={(λ 1,...,λ 2n ) λ j {0,1,2}, 1 j 2n} of the additive group Z 2n. It is clear that λ Λ{(3Z) 2n + λ} =Z 2n and that {(3Z) 2n + λ} {(3Z) 2n +λ }= for λ λ in Λ. Thus we can write (3.3) K F = λ Λ K F,λ, where K F,λ = v F {(3Z) 2n +λ} α v K v.

7 chatten Class Hankel Operators 1387 Note that each K v maps the subspace L 2 (Q+v,dµ) to itself. Because the length of each side of Q is 3, for any fixed λ Λ, the families of subspaces {L 2 (Q+v,dµ) v (3Z) 2n + λ} are pairwise orthogonal. Thus, recalling (3.1), for any λ Λ, we obtain (3.4) { K F,λ q = v F {(3Z) 2n +λ} αv q Kv q q } 1/q { αv(j(f q τ v )) q/2} 1/q { = (J(f τ v )) p/2} 1/q, v F where the second = follows from the definition of α v and from the fact that (q(p 2)/2) + (q/2) = p/2. Therefore (3.2) follows from (3.3), (3.4) andthe fact that card(λ) = 3 2n. Now [M f,p] = [M f,p]p +[M f,p](1 P) = H f (H f ) C p under our assumption. ince [M f,p] has (f (z) f (w))e z,w as its kernel function, for any v Z 2n, tr([m f,p]k v )= f(w) f(z) 2 e z,w 2 dµ(w) dµ(z) Q+v Q+v = π 2n f(w +v) f(z+v) 2 e (w+v) (z+v) 2 dv (w) dv (z) Q Q π 2n e 18n J(f τ v ), where the follows from the fact that w z 3 2nfor any z, w Q. Hence v F [Mf,P] p KF q tr([m f,p]k F ) π 20n v F α v J(f τ v ) = π 20n v F(J(f τ v )) p/2. Combining this with (3.2), we now have (J(f τ v )) p/2 π 22pn [Mf,P] p p. v F ince this holds for any finite subset F Z 2n, we conclude that, in the case 1 <p 2, we have v Z 2n(J(f τ v)) p/2 <. Let us now consider the case p = 1. In this case [M f,p] C 1. For any 1 <t 2, since [M f,p] t [M f,p] 1, the above provides (J(f τ v )) t/2 π 22tn [Mf,P] t 1. v F Taking the limit t 1, we obtain the finiteness of v Z 2n(J(f τ v)) 1/2 under the assumption H f C 1 and H f C 1.

8 1388 JINGBO XIA & DECHAO ZHENG Lemma 3.2. For any f L 2 local(c n,dv)and v Z 2n,wehave (3.5) f τ v f 2 dv (1 + 2 Γ (v) ) J(f τ a ), a Γ (v) where Γ (v) is the discrete path in Z 2n from 0 to v and Γ (v) is the length of Γ (v) as defined in ection 2. Proof. The case v = 0 is trivial. uppose that v 0. Then by ection 2 we can enumerate the points in Γ (v) as a 0, a 1,...,a l in such a way that l = Γ(v), a 0 = 0, a l = v,and {+a j 1 } {+a j } Q+a j 1, 1 j l. Obviously, (3.6) (f τ aj ) (f τ aj 1 ) (f τ aj ) (f τ aj 1 ) Q + (f τ aj 1 ) Q (f τ aj 1 ) for any 1 j l. incev( +a j ) = 1and+a j Q+a j 1,wehave (f τ aj ) (f τ aj 1 ) Q 2 2 = {f f Q+aj 1 } dv +a j f f Q+aj 1 2 dv +a j f f Q+aj 1 2 dv Q+a j 1 = f τ aj 1 (f τ aj 1 ) Q 2 dv Q J(f τ aj 1 ) =. 2V(Q) imilarly, (f τ aj 1 ) Q (f τ aj 1 ) 2 J(f τ aj 1 )/2V(Q). Thus, by (3.6), (3.7) (f τ aj ) (f τ aj 1 ) 2 J(f τ aj 1 ), 1 j l. Now (3.8) f τ v f 2 dv 2 { f τ v (f τ v ) 2 + (f τ v ) f 2 } dv.

9 chatten Class Hankel Operators 1389 We have 2 f τ v (f τ v ) 2 dv = f(w +v) f(z+v) 2 dv (w) dv (z) J(f τ v ) = J(f τ al ) and, by (3.7), (3.9) (f τ v ) f 2 = (f τ al ) (f τ a0 ) 2 { l l l j=1 } 2 (f τ aj ) (f τ aj 1 ) l (f τ aj ) (f τ aj 1 ) 2 j=1 l J(f τ aj 1 ). j=1 Thus, upon noting that l = Γ(v),(3.5) follows from (3.8)and(3.9). Lemma 3.3. There exists a constant 0 <C 3.3 < such that 2 (3.10) sup f τ z f τ z dµ dµ C 3.3 e v 2 /3 J(f τ a ) z v Z 2n a Γ (v) for any f T(C n ),whereγ(v) is the discrete path in Z 2n from 0 to v as in Lemma 3.2. Proof. For any z, wehave (3.11) 2 f τ z f τ z dµ dµ f τ z f 2 dµ = 1 v Z 2n π n f(w) f 2 e w z 2 dv (w) +v = 1 v Z 2n π n (f τ v )(w) f 2 e (w z)+v 2 dv (w) f τ v f 2 dv, v Z 2n d(v)

10 1390 JINGBO XIA & DECHAO ZHENG where d(v) = exp{ inf w,ζ (w ζ) + v 2 }.ince (w ζ) + v 2 v 2 + w ζ 2 2 w ζ v ( v 2 /2) w ζ 2, there is a B = B(n) > 0such that d(v) Be v 2 /2. Obviously, v dominates the length of every segment in Γ (v). This means that Γ (v) 2n v. Therefore (1 + 2 Γ (v) )d(v) B(1 + 4n v )e v 2 /2 C 3.3 e v 2 /3, where C 3.3 depends only on n. Applying (3.5) and this inequality in (3.11), (3.10) follows. The main result of the section is the following. Lemma 3.4. uppose that 1 p 2. Iff T(C n )is such that then {D(f τ ζ )} p dv (ζ) <. b Z 2n {J(f τ b )} p/2 <, Proof. ince { + u u Z 2n }=C n and V( + u) = 1, it suffices to prove (3.12) sup u Z 2n ζ +u {D(f τ ζ )} p = u Z 2n sup z {D(f τ u τ z )} p < under the assumptions of the lemma. ince p/2 1, ( t ν ) p/2 t p/2 ν if every t ν is positive. Applying this elementary inequality to the sum in (3.10), we obtain sup z {D(f τ u τ z )} p C p/2 3.3 v Z 2n a Γ(v) e p v 2 /6 {J(f τ u τ a )} p/2, where Γ (v) is the discrete path from 0 to v. inceτ u τ a =τ u+a,wehave u Z 2n sup z C p/2 3.3 = C p/2 3.3 = C p/2 3.3 {D(f τ u τ z )} p u Z 2n v Z 2n a Γ(v) e p v 2/6 v Z 2n a Γ(v) e p v 2 /6 {J(f τ u+a )} p/2 {J(f τ u+a )} p/2 u Z 2n e p v 2/6 card(γ (v)) {J(f τ b )} p/2. v Z 2n b Z 2n ince card(γ (v)) = 1 + Γ(v) 1 +2n v,(3.12) follows.

11 chatten Class Hankel Operators UFFICIENCY IN THE CAE 2 p< In this section we prove that (1.2) implies H f C p and H f C p in the case 2 p<. Lemma 4.1. Let 1 p< and let f T(C n ).If {D(f τ ζ )} p dv (ζ) <, then v Z 2n{J(f τ v)} p/2 <. Proof. We first show that there is a c>0 such that (4.1) {D(f τ ζ )} 2 cj(f τ v ) for any f T(C n ), v Z 2n,ζ +v. For this we recall the well-known fact that ϕ ϕdν 2 dν = inf α C ϕ α 2 dν if dν is a probability measure and ϕ L 2 (dν). By the definition (2.4), we have J(f τ v ) = Q ( = 2 f(z+v) f(w +v) 2 dv (z) dv (w) Q V(Q) f 2 2) dv fdv Q+v =2(V (Q)) 2 1 V(Q) 2(V (Q)) 2 1 = 2V(Q) Q+v ζ Q+v Q+v f f Q+v 2 dv 2 V(Q) f f τ ζ dµ dv Q+v 2 f τ ζ f τ ζ dµ dv. If ζ + v, then Q + v ζ Q ={x y x Q, y }. Hence J(f τ v ) 2V(Q) Q 2 f τ ζ f τ ζ dµ dv, whenever ζ + v. Obviously there is a δ = δ(n) > 0 such that e w 2 δ for every w Q. ince dµ(w) = π n e w 2 dv (w), weseethatc =δ/2π n V(Q) will do for (4.1). But from (4.1) it follows immediately that {D(f τ ζ )} p dv (ζ) = v Z 2n c p/2 {D(f τ ζ )} p dv (ζ) +v {J(f τ v )} p/2. v Z 2n

12 1392 JINGBO XIA & DECHAO ZHENG Lemma 4.2. For each 2 p<, there exists a constant 0 <C 4.2 < such that ( ) [M f,p] p C 4.2 {J(f τ b )} p/2 1/p for all f L 2 local(c n,dv). b Z 2n Proof. If f L 2 local(c n,dv), than M χe [M f,p]m χe C 2 whenever E is a bounded Borel set in C n. Therefore it suffices to produce a constant 0 <C 4.2 <, which depends only on n and p, such that ( ) (4.2) M χe [M f,p]m χe p C 4.2 {J(f τ b )} p/2 1/p b Z 2n for any bounded Borel set E C n. Given such an E, there is a finite subset F = F(E) Z 2n such that { + v v F} E.ince XYX p X Y p X and since M χe M χ {+v v F} = M χe,itsuffices to estimate the p-norm of Y = Y v = u,u F M χ+u [M f,p]m χ+u = v Z 2n Y v, u Z 2n χ F F (u, u + v)m χ+u [M f,p]m χ+u+v. where Note that, for each given v, the families {L 2 ( + u + v,dµ) u Z 2n } of subspaces are pairwise orthogonal. ince T p T 2 when p 2, for each v Z 2n,wehave (4.3) Yv p p = χ F F (u, u + v) u Z 2n Mχ+u [M f,p]m χ+u+v p 2. u Z 2n Mχ+u [M f,p]m χ+u+v p p ince [M f,p]has (f (z) f (w))e z,w as its kernel function, we have (4.4) Mχ+u [M f,p]m χ+u+v 2 2 = (f (z) f (w))e z,w 2 dµ(w) dµ(z) +u +u+v = π 2n f(z) f(w) 2 e w z 2 dv (w) dv (z) +u +u+v d(v) (f τ u )(z) (f τ u )(w) 2 dv (w) dv (z), +v where d(v) = exp( inf w,z (w z) + v 2 ) Be v 2 /2 as in the proof of Lemma 3.3.

13 chatten Class Hankel Operators 1393 Because V() = 1, for any g L 2 local(c n,dv), { } g(z) g(w) 2 dv (w) dv (z) +v 2( g(z) g 2 + g (g τ v )(w) 2 ) dv (w) dv (z) = 2 g g 2 dv + 2 g τ v g 2 dv (also see (2.4)). Clearly, g g 2 dv g g Q 2 dv 1 2 J(g) by (2.3). Thus, applying Lemma 3.2 to g τ v g 2 dv, we obtain g(z) g(w) 2 dv (w) dv (z) J(g) + 4 Γ(v) J(g τ a ) +v (1 + 4 Γ(v) ) a Γ (v) a Γ (v) J(g τ a ), where Γ (v) is the discrete path in Z 2n from 0 to v. Letting g = f τ u in the above and recalling (4.4), we now have Mχ+u [M f,p]m χ+u+v 2 2 Be v 2 /2 (1 + 4 Γ (v) ) a Γ (v) J(f τ u τ a ). ince p/2 1, we can apply Hölder s inequality in the above to obtain Mχ+u [M f,p]m χ+u+v p 2 h(v) {J(f τ u τ a )} p/2, a Γ (v) where h(v) = (Be v 2 /2 ) p/2 (1 + 8n v ) (p/2)+{(p 2)/2}. (Recall that Γ (v) 2n v.) Thus, combining this with (4.3), we find that Consequently Yv p p u Z 2n h(v) a Γ (v) {J(f τ u+a )} p/2 = h(v) {J(f τ u+a )} p/2 a Γ (v) u Z 2n = h(v) card(γ (v)) {J(f τ b )} p/2 b Z 2n h(v)(1 + 2n v ) b Z 2n {J(f τ b )} p/2. Y p ( Y v p {h(v)(1 + 2n v )} 1/p {J(f τ b )} p/2 v Z 2n v Z 2n b Z 2n ) 1/p.

14 1394 JINGBO XIA & DECHAO ZHENG By the definition of h(v), the constant C 4.2 = v Z 2n{h(v)(1 + 2n v )}1/p is obviously finite. With this C 4.2,(4.2) holds for any bounded Borel set E C n. 5. THE EAY DIRECTION The substantive work for the proof of Theorem 1.1 has already been done in the previous sections. The rest of the proof follows well-established routines for reproducing-kernel Hilbert spaces, which we carry out below in the interest of completeness. We start with an adaptation of a well-known result. ee, e.g., [10, Proposition 7.3.5]. Lemma 5.1. For any g L 2 (C n,dµ),wehave (1 P)g 2 {D(g)} 2 (1 P)g 2 + (1 P)ḡ 2. Proof. ince {D(g)} 2 = g 2 g,1 2 and g,1 2 Pg 2,wehave {D(g)} 2 g 2 Pg 2 = (1 P)g 2. To prove the other inequality, let us write {D(g)} 2 = (1 P)g 2 +{ Pg 2 g,1 2 }.Itsuffices to show that (5.1) Pg 2 g,1 2 (1 P)ḡ 2. Let P (respectively P0 ) denote the orthogonal projection from L 2 (C n,dµ) onto H 2 (C n,dµ) ={ ϕ ϕ H 2 (C n,dµ)} (respectively C, the subspace of constant functions in L 2 (C n,dµ)). It is obvious that Pḡ is the complex conjugate of Pg. That is, Pg 2 = Pḡ 2. Also, g,1 2 = ḡ,1 2 = P 0 ḡ 2. Therefore Pg 2 g,1 2 = ( P P0 )ḡ 2.Thus(5.1) follows from the obvious fact that P P 0 1 P. Recall that K z (w) = e w,z is the reproducing kernel for H 2 (C n,dµ), i.e., (Pϕ)(z) = ϕ, K z, ϕ L 2 (C n,dµ). Let k z denote the normalized reproducing kernel K z / K z.ince K z =e z 2 /2, we have k z (w) = e w,z e z 2 /2. ince K z (z) dµ(z) = π n dv (z), Lemma 5.2 below is well-known. In fact its proof is simply an adaptation of the proof of, e.g., [10, Proposition 6.3.2] to the egal-bargmann space in the obvious way. Lemma 5.2. If A is a trace class operator or a positive operator on H 2 (C n,dµ), then tr(a) = 1 π n Ak z,k z dv (z).

15 chatten Class Hankel Operators 1395 It is straightforward to verify that, for any given z C n, the formula (U z ϕ)(w) = (ϕ τ z )(w)k z (w) = ϕ(w z)k z (w) defines a unitary operator on L 2 (C n,dµ). It is also easy to verify that [U z,p]=0 for every z C n. In particular this means (5.2) (1 P)(f τ z ) = U z (1 P)(f τ z ) = (1 P)U z (f τ z ) = (1 P)fk z = H f k z for all f T(C n )and z C n. Proof of Theorem 1.1. Let us first consider the case 1 p 2. By Lemmas 3.1 and 3.4, the simultaneous membership H f C p and H f C p implies (1.2). To prove the converse in this case, let us assume that f is a function in T (C n ) for which (1.2) holds. To prove that H f C p, we first need to establish the existence of (H f H f ) p/2 as a positive operator. By Lemma 4.1, (1.2) implies that v Z 2n {J(f τ v )} p/2 <. ince p/2 1, this in turn implies v Z 2n J(f τ v)<. Thus, by Lemma 4.2, [M f,p] C 2. This in particular tells us that H f is bounded and, therefore, the positive operator (H f H f ) p/2 is well defined. Using the condition p/2 1once more, we have (H f H f ) p/2 k z,k z H f H fk z,k z p/2.but H f H fk z,k z p/2 = H f k z p = U z (1 P)(f τ z ) p {D(f τ z )} p,whereweused(5.2) and Lemma 5.1 for the last two steps. Therefore, by Lemma 5.2, Hf p p = tr((h f H f ) p/2 ) = 1 1 π n π n {D(f τ z )} p dv (z) < (H f H f ) p/2 k z,k z dv (z) whenever (1.2) holds. ince D( f τ z ) = D(f τ z ), we also have H f C p if (1.2) holds. This completes the proof in the case 1 p 2. Let us now assume 2 p <. Then it follows from Lemmas 4.1 and 4.2 that H f C p and H f C p whenever (1.2) holds. Conversely, let us suppose that H f C p and H f C p. ince p/2 is now at least 1, we have (H f H f ) p/2 k z,k z H f H fk z,k z p/2 = H f k z p and the same holds with f

16 1396 JINGBO XIA & DECHAO ZHENG in place of f. Therefore, by Lemma 5.2, ( H f k z p + H fk z p )dv(z) {(H f H f ) p/2 + (H f H f ) p/2 }k z,k z dv (z) = π n( Hf p p + H f p p). ince p/2 1, Hölder s inequality yields ( H f k z 2 + H fk z 2 ) p/2 2 (p 2)/2 ( H f k z p + H fk z p ). Now, by (5.2) and Lemma 5.1, ( H f k z 2 + H fk z 2 ) p/2 = ( (1 P)(f τ z ) 2 + (1 P)( f τ z ) 2 ) p/2 {D(f τ z )} p. Thus the condition H f C p and H f C p implies (1.2). 6. ACOMPARION WITH THE BERGMAN PACE Let B n denote the unit ball in C n. Let dv denote the volume measure on B n normalized in such a way that v(b n ) = 1. Recall that the Bergman space L 2 a(b n,dv) is the collection of analytic functions on B n which belong to L 2 (B n,dv). Let P Berg : L 2 (B n,dv) L 2 a(b n,dv) be the orthogonal projection. Then P Berg is givenbytheformula (P Berg ϕ)(z) = Bn The Bergman space Hankel operator H Berg f ϕ(w) dv(w) (1 z, w ) n+1, ϕ L2 (B n,dv). is defined by the formula H Berg f ψ = (1 P Berg )(f ψ), ψ L 2 a(b n,dv). Recall that, for any ϕ L 1 (B n,dv),itsberezin transform ϕ is defined by the formula ϕ(z) = ϕ(w) (1 z 2 ) n+1 1 z, w 2n+2 dv(w), z B n. Bn For any f L 2 (B n,dv),itsmean oscillation MO(f ) is defined by the formula MO(f )(z) ={ f 2 (z) f(z) 2 } 1/2 { = f(w) f(z) 2 (1 z 2 ) n+1 } 1/2 dv(w). 1 z, w 2n+2 Bn What originally led to Theorem 1.1 was the following result due to Zhu.

17 chatten Class Hankel Operators 1397 Theorem 6.1 ([11]). uppose that 2 p< and f is in L 2 (B n,dv).then we have the simultaneous membership H Berg f C p and H f Berg C p if and only if (6.1) {MO(f )(z)} p dv(z) (1 z 2 <. ) n+1 Bn Note that (1 z 2 ) (n+1) dv(z) is the Möbius invariant measure on B n.on C n, what corresponds to the Möbius group is the group {τ ζ ζ C n }. Therefore (1.2) is just the egal-bargmann space version of (6.1) and, conversely, (6.1) isthe Bergman space version of (1.2). Thus, given Theorem 6.1, it does not take too much imagination to conceive Theorem 1.1. But, given that Theorem 6.1 was published in 1991, it is somewhat surprising that Theorem 1.1 has not appeared in the literature until the present paper. There are, however, some not-too-subtle differences between Theorem 6.1 and Theorem 1.1, which might help explain the chronological gap between the two theorems. First of all, a quick glance of our paper and [11] reveals that the techniques used in the two papers are completely different. The techniques we use in this paper take full advantage of the fact that, as a reproducing-kernel space, the structure of H 2 (C n,dµ) is completely flat. This is why the proof of Lemma 4.2 is much shorter and much more straightforward than its counter part on the Bergman space. But the most obvious difference between Theorem 6.1 and Theorem 1.1 is the range of validity with respect to p: whereas Theorem 1.1 holds true for all 1 p<, in its generality Theorem 6.1 was proved only for 2 p<.only recently was Theorem 6.1 extended to the case 2n/(n + 1) <p<2[8,9]. It is easy to see that, for any n 1, if 1 p 2n/(n + 1), then the condition (6.1) is sufficient but not necessary for the simultaneous membership H Berg f C p, both for trivial reasons. Indeed, we have the elementary estimate H Berg f { MO(f )(z) } 1/2 dv 2n+2 f f(z) 2 (1 z 2 ) n+1 Bn (1+ z 2 ) { (1 z 2 ) (n+1)/2 2 n 1 inf α C Thus there are only two possible ways for (6.1) to hold: Either p(n + 1) 2 (n + 1) > 1, which translates to p>2n/(n + 1), or, failing that, we must have inf f α 2 dv = 0, α C Bn Bn f α 2 dv} 1/2. C p and

18 1398 JINGBO XIA & DECHAO ZHENG which forces f to be a constant on B n. But there are non-constant functions f on B n for which both H Berg f and H f Berg belong to the trace class. For example, if f is bounded and vanishes outside {z B n : z <r}for some 0 <r <1, then H Berg f C 1 and H Berg C f 1. (This is because for any 0 <r <1, there is an F C (C n C n ) such that (1 z, w ) n 1 χ [0,r ) ( z ) = F(z,w)χ [0,r ) ( z ) when z, w B n.) Therefore, when 1 p 2n/(n+1),(6.1)issufficient but not necessary for H Berg f C p and H f Berg C p. This comparison between Theorem 1.1 and Theorem 6.1 seems to suggest that the fundamental difference between the domains C n and B n somehow shows up at the level of operator theory. REFERENCE [1] J. ARAZY,.D.FIHER,andJ.PEETRE, Hankel operators on weighted Bergman spaces,amer.j. Math. 110 (1988), MR 90a:47067 [2] C. A. BERGER and L. A. COBURN, Toeplitz operators on the egal-bargmann space,trans.amer. Math. oc. 301 (1987), MR 88c:47044 [3] C. A. BERGER, L.A.COBURN, andk.h.zhu, Toeplitz operators and function theory in n-dimensions, Pseudodifferential Operators (Oberwolfach, 1986), Lecture Notes in Math., vol. 1256, pringer, Berlin, 1987, pp MR 89a:47039 [4] I. C. GOHBERG and M. G. KREĬ N,Introduction to the theory of linear nonselfadjoint operators, Translated from the Russian by A. Feinstein. Translations of Mathematical Monographs, Vol. 18, American Mathematical ociety, Providence, R.I., MR 39 #7447 [5] JAN JANA, Toeplitz and Hankel operators on Bargmann spaces, Glasgow Math. J. 30 (1988), , MR 90c:47044 [6] VANTE JANON, JAAK PEETRE, andrichard ROCHBERG, Hankel forms and the Fock space, Rev.Mat.Iberoamericana3(1987), MR 91a:47029 [7] KAREL TROETHOFF, Hankel and Toeplitz operators on the Fock space, Michigan Math. J. 39 (1992), MR 93d:47058 [8] JINGBO XIA, Hankel operators in the Bergman space and chatten p-classes: the case 1 <p<2, Proc. Amer. Math. oc. 129 (2001), (electronic), MR 2002f:47064 [9], On the chatten class membership of Hankel operators on the unit ball, Illinois J. Math. 46 (2002), MR 2004b:47058 [10] KE HE ZHU, Operator theory in function spaces, Monographs and Textbooks in Pure and Applied Mathematics, vol. 139, Marcel Dekker Inc., New York, 1990, IBN MR 92c:47031 [11], chatten class Hankel operators on the Bergman space of the unit ball, Amer. J. Math. 113 (1991), MR 91m:47036 [12], Hankel operators on the Bergman space of bounded symmetric domains, Trans. Amer. Math. oc. 324 (1991), MR 92f:47023

19 chatten Class Hankel Operators 1399 JINGBO XIA: Department of Mathematics tate University of New York at Buffalo Buffalo, NY 14260, U..A.. DECHAO ZHENG: Department of Mathematics Vanderbilt University Nashville, TN 37240, U..A.. KEY WORD AND PHRAE: Hankel operator; chatten class MATHEMATIC UBJECT CLAIFICATION: 47B10, 47B32, 47B35. Received: February 20th, 2003; revised: April 21st, 2003.