SCHATTEN p CLASS HANKEL OPERATORS ON THE SEGAL BARGMANN SPACE H 2 (C n, dµ) FOR 0 < p < 1

J. OPERATOR THEORY 66:1(2011), 145 160 Copyright by THETA, 2011 SCHATTEN p CLASS HANKEL OPERATORS ON THE SEGAL BARGMANN SPACE H 2 (C n, dµ) FOR 0 < p < 1 J. ISRALOWITZ This paper is dedicated to the memory of Laura Jane Wisewell, whose kindness and generosity will always be remembered. May you finally rest in peace forever. Communicated by Nikolai K. Nikolski ABSTRACT. We consider Hankel operators on the Segal Bargmann space H 2 (C n, dµ). We obtain necessary and sufficient conditions for the simultaneous membership of H f and H f in the Schatten class S p for 0 < p < 1. In particular, we show that the necessary and sufficient conditions obtained by J. Xia and D. Zheng for the case 1 p < extends to the case 0 < p < 1. KEYWORDS: Schatten class, Hankel operators, Segal Bargmann space. MSC (2000): 47B32, 32A36. 1. INTRODUCTION Let dµ be the normalized Gaussian measure on C n centered at 0, so that dµ(z) = π n e z 2 dv(z). Recall that the Segal Bargmann space H 2 (C n, dµ) is defined as { f L 2 (C n, dµ) : f is analytic on C n }. It is well known that {(k 1! k n!) 1/2 z k 1 1 zk n n : k 1 0,..., k n 0} forms an orthonormal basis for H 2 (C n, dµ) and that the orthogonal projection P : L 2 (C n, dµ) H 2 (C n, dµ) is an integral operator on L 2 (C n, dµ) with kernel e z,w. Here and in what follows, we write z, w = z 1 w 1 + + z n w n. For each ν C n, let τ ν : C n C n be the translation τ ν (w) = w + ν w C n,

146 J. ISRALOWITZ and define T (C n ) = { f L 2 (C n, dµ) : f τ ν L 2 (C n, dµ) for every ν 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,ν belongs to L 2 (C n, dµ) for every ν C n. This means that if f T (C n ), then the set {h H 2 (C n, dµ) : f h L 2 (C n, dµ)} is a dense, linear subspace of H 2 (C n, dµ). Recall that the Hankel operator H f : L 2 (C n, dµ) L 2 (C n, dµ) with symbol f is defined by the formula H f = (I P)M f P. Thus, if f T (C n ), then H f has at least a dense domain in L 2 (C n, dµ). Given a ϕ L 2 (C n, dµ), let SD(ϕ) denote its standard deviation with respect to the probability measure dµ, which is defined by { ϕ SD(ϕ) = ϕdµ 2 } 1/2 { dµ = ϕ 2 dµ ϕdµ 2 } 1/2. When f T (C n ), it was shown in [1] that H f and H f are simultaneously bounded if and only if ξ SD( f τ ξ ) is a bounded function on C n, and H f and H f are simultaneously compact if and only if lim SD( f τ ξ ) = 0 (it should be noted ξ that the later was proved in [2] for bounded measurable symbols, where in this setting, it was also proved that H f is compact if and only if H f is compact). Therefore, for f T (C n ), it is reasonable to think that the simultaneous Schatten p class membership of H f and H f for 1 p < would be characterized by an L p condition involving the standard deviation. In fact, it was shown in [6] that for f T (C n ) and 1 p <, H f and H f are simultaneously members of S p if and only if C n {SD( f τ ξ )} p dv(ξ) <. With this in mind, the following is the main result of this paper. THEOREM 1.1. Let 0 < p < 1 and f T (C n ). Let H f and H f be the corresponding Hankel operators from L 2 (C n, dµ) to L 2 (C n, dµ). Then we have the simultaneous membership of H f and H f in S p if and only if C n {SD( f τ ξ )} p dv(ξ) <. In particular, we will show that the quantity C n {SD( f τ ξ )} p dv(ξ) is comparable to H f Sp + H f Sp with a constant that is independent of f. We close this section with a sketch of the proof. The sufficiency direction is proved by an argument that is identical to the proof for the case 1 p < 2, and

SCHATTEN CLASS HANKEL OPERATORS 147 so we refer the reader to [6] for the details (more precisely, the proof follows from standard reproducing kernel Hilbert space techniques that are specialized to the Segal Bargmann space.) For the other direction, let Z 2n be treated as a lattice in C n. We will first prove that C n {SD( f τ ξ )} p dv(ξ) C b 1 N Z2n { f (z + b) f (w + b) 2 dv(w)dv(z) } p/2 where is the cube [ 1 N, 2 N )2n in R 2n and C depends only on n, N, and p. The proof is very similar to the proof of Lemma 3.4 in [6], though we include it for the sake of the reader. We will next show that for large enough N and each b 1 N Z2n, f (z + b) f (w + b) 2 dv(w)dv(z) C ( f (z) f (w))e z,w e w 2 /2 e iim b,w dv(w) 2 e z 2 dv(z) +b +b where C only depends on n, N, and p. To prove this, we will make crucial use of the fact that N is chosen to be sufficiently large, and it is for this reason alone that we work with the lattice 1 N Z2n rather than Z 2n. Next, it is easy to see that H f and H f are both in S p for all 0 < p < if and only if the first order commutator [M f, P] is in S p. With this in mind, we will fix some N large enough and estimate W Sp directly, where W = A[M f, P]B and where A and B are some bounded operators that will depend on our fixed N. This will be done by the standard general method employed to handle estimating the Schatten p quasinorm of special classes of integral operators for 0 < p < 1 (for example, see [4] and [5]). For each M N, we will first appropriately decompose 1 N Z2n as the disjoint union of lattices {Λ M j } j {1,...,M} 2n, and analogously decompose W = W j. We will break up each W j = D j + E j where D j is a diago- j {1,...,M} 2n nal operator and E j is an off-diagonal operator, so that W j p S p D j p S p E j Sp. Finally, we will show that our choices of A and B, and the results from above, give us that D j p j {1,...,M} 2n S p is bounded below by C {SD( f τ ξ )} p dv(ξ), and that C n E j p j {1,...,M} 2n S p is bounded above by C M {SD( f τ ξ )} p dv(ξ), where C > 0 C n is a constant that only depends on n, N and p, and C M is a constant depending on n, N, p, and M with lim C M = 0. Thus, with this fixed N, we can complete M the proof by setting M large enough.

148 J. ISRALOWITZ 2. PRELIMINARIES For each N N, let 1 N Z2n denote the set {( k 1 N,..., k 2n N ) R2n : k i Z}. A subset S = {p 0,..., p k } of 1 N Z2n with k 1 is said to be a discrete segment in 1 N Z2n if there exists j {1,..., 2n} and z Z 2n such that p i = z + i N e j, 0 i k where e j is the standard j th basis vector of R 2n. In this setting, we say that p 0 and p k are the endpoints of S. Also, we define the length of S to be S = k. Let ν = ( ν 1 N,..., ν 2n N ) and ν = ( ν 1 N,..., ν 2n N ) be elements of N 1 Z2n where ν = ν. We can enumerate the integers {j : ν j = ν j, 1 j 2n} as j 1,..., j m in ascending order, so that j 1 < < j m when m > 1. Set z 0 (ν, ν ) = ν, and inductively define z t (ν, ν ) = z t 1 (ν, ν ) + ν j ν t jt N e j t for t {1,..., m}. Note that z m (ν, ν ) = ν. Let S t (ν, ν ) be the discrete segment in N 1 Z2n which has z t 1 (ν, ν ) and z t (ν, ν ) as its endpoints. The union of the discrete segments S 1 (ν, ν ),..., S m (ν, ν ) will be denoted by Γ(ν, ν ). We call Γ(ν, ν ) the discrete path in N 1 Z2n from ν to ν. Furthermore, we define the length Γ(ν, ν ) of Γ(ν, ν ) to be S 1 (ν, ν ) + + S m (ν, ν ). That is, the length of Γ(ν, ν ) is just the sum of the lengths of the discrete segments which make up Γ(ν, ν ). If ν = 0, we let Γ(ν) denote Γ(ν, ν ). In the case ν = ν, we define the discrete path from ν to ν to be the singleton set Γ(ν, ν) = {ν}. Let S N denote the cube S N = [0, N 1 )2n and let be the cube [ N 1, N 2 )2n. For any f L 2 loc (Cn, dv), write J N ( f ) = f (z) f (w) 2 dv(z)dv(w). If E is a Borel set with 0 < V(E) <, we will denote the mean value of f on E by f E. That is, f E = 1 f dv. V(E) E Universal constants will be denoted by C 1, C 2,... and will represent different values in the proofs of different results. To keep better track of the dependence of the various constants encountered, we will use subscripts to denote what a particular constant depends on (though we implicitly assume that all universal constants may depend on n and p). Finally, we conclude this section by reviewing some necessary facts about Schatten class ideals, all of which can be found in [3]. Recall that for any 0 < p <, the Schatten p class S p B(H) consists of operators T satisfying the condition T Sp <, where Sp is defined by T Sp = {tr( T p )} 1/p = {tr((t T )} 1/p.

SCHATTEN CLASS HANKEL OPERATORS 149 When p 1, Sp defines a norm. However, when 0 < p < 1, Sp only defines a quasinorm, which means that we have the following: LEMMA 2.1. If 0 < p < 1, then for any Schatten p class operators T and S, we have that T + S p S p T p S p + S p S p. For all 0 < p <, it is well known that S p is a two sided ideal of the ring of bounded operators B(H). More precisely, if A, B B(H) and T S p, then ATB S p with ATB Sp A Op T Sp B Op. If 0 < p 2, then for any T S p and any orthonormal basis { f n } of H (where H is a separable Hilbert space), we have that T p S p n=1 k=1 T f n, f k p. 3. MAIN RESULT : NECESSITY FOR 0 < p < 1. We will now follow the outline discussed in the introduction. As stated before, the details for sufficiency can be found in [6]. The results and proofs of the next three lemmas are very similar to Lemmas 3.2 3.4 in [6], though we include proofs for the sake of the reader. (3.1) LEMMA 3.1. For any f L 2 loc (Cn, dv) and ν 1 N Z2n, we have S N f τ ν f SN 2 dv (N 2n + 2 N4n ) 3 2n Γ(ν) J N ( f τ a ). a Γ(ν) Proof. The case ν = 0 is trivial. If ν = 0, enumerate the points in Γ(ν) as a 0, a 1,..., a l with l = γ(ν) in such a way that a 0 = 0, a l = ν, and By the triangle inequality, (3.2) {S N + a j 1 } {S N + a j } + a j 1, 1 j l. ( f τ aj ) SN ( f τ aj 1 ) SN ( f τ aj ) SN ( f τ aj 1 ) QN + ( f τ aj 1 ) QN ( f τ aj 1 ) SN for any 1 j l. Since V(S N + a j ) = 1 and S N 2n N + a j + a j 1, we have ( f τ aj ) SN ( f τ aj 1 ) QN 2 = N 4n { f f QN +a j 1 }dv 2 S N +a j

150 J. ISRALOWITZ Similarly, Thus, by (3.2), N 2n f f QN +a j 1 2 dv +a j 1 = N 2n f τ aj 1 ( f τ aj 1 ) QN 2 dv = ( f τ aj 1 ) QN ( f τ aj 1 ) SN 2 1 2 1 2 N 4n 3 2n J N( f τ aj 1 ). N 4n 3 2n J N( f τ aj 1 ). (3.3) Now, (3.4) ( f τ aj ) SN ( f τ aj 1 ) SN N4n 3 2n J N( f τ aj 1 ) 1 j l. f τ ν f SN 2 dv 2 S N S N { f τ ν ( f τ ν ) SN 2 + ( f τ ν ) SN f SN 2 }dv. However, 2 f τ ν ( f τ ν ) SN 2 1 dv = V(S N ) S N and by (3.3), S N S N f (w + ν) f (z + ν) 2 dv(w)dv(z) 1 V(S N ) J N( f τ ν ) = N 2n J N ( f τ al ) ( f τ ν ) SN f SN 2 = ( f τ al ) SN ( f τ a0 ) SN 2 (3.5) l { l } 2 ( f τ aj ) SN ( f τ aj 1 ) SN j=1 l j=1 ( f τ aj ) SN ( f τ aj 1 ) SN 2 l N4n 3 2n But l = Γ(ν), so that (3.1) follows from (3.4) and (3.5). (3.6) LEMMA 3.2. For f T (C n ), there exists C N > 0 such that sup z S N C n f τ z f τ z dµ 2 dµ C N C n ν N 1 Z2n a Γ(ν) l J N ( f τ aj 1 ). j=1 e ν 2 /3 J N ( f τ a ).

SCHATTEN CLASS HANKEL OPERATORS 151 C n Proof. For any z S N, we have f τ z f τ z dµ 2 dµ C n (3.7) C n f τ z f SN 2 dµ 1 = π n ν N 1 Z2n 1 = π n ν N 1 Z2n S N +ν f (w) f SN 2 e w z 2 dv(w) ( f τ ν )(w) f SN 2 e (w z)+ν 2 dv(w) S N d(ν) π n f τ ν f SN 2 dv, ν N 1 Z2n { where d(ν) = exp inf (w ξ) + ν 2}. Since (w ξ) + ν 2 ν 2 + w,ξ S N w ξ 2 2 w ξ ν ν 2 2 w ξ 2, there exists C 1 N > 0 such that d(ν) C 1 N e ν 2 /2. Obviously, N ν dominates the length of every discrete segment in Γ(ν), so that Γ(ν) 2nN ν. Therefore, we have that (N 2n + 2 N4n ) 3 2n Γ(ν) d(ν) C 1 N (N 2n + 4n N4n+1 ) 3 2n ν e ν 2 /2 C 2 N e ν 2 /3 and so (3.6) follows from the above inequality and plugging (3.1) into (3.7). LEMMA 3.3. For 0 < p 2 and f T (C n ), there exists C N > 0 such that {SD( f τ ξ )} p dv(ξ) C N {J N ( f τ b )} p/2. C n b N 1 Z2n Proof. Since {S N + u} = C n and V(S N + u) = 1, it is enough to u N 1 N 2n Z2n show that sup u N 1 Z2n z S N +u {SD( f τ z )} p = S N u 1 N Z2n sup z S N {SD( f τ z τ u )} p C N b 1 N Z2n {J N ( f τ b )} p/2. Since 0 < p 2, Hölder s inequality applied to (3.6) gives that sup z S N {SD( f τ u τ z )} p C 1 N ν N 1 Z2n a Γ(ν) e p ν 2 /6 {J N ( f τ u τ a )} p/2.

152 J. ISRALOWITZ Since τ u τ a = τ u+a, we have u 1 N Z2n sup z S N {SD( f τ u τ z )} p C 1 N u N 1 Z2n ν N 1 Z2n a Γ(ν) = C 1 N ν 1 N Z2n e p ν 2 /6 a Γ(ν) e p ν 2 /6 {J N ( f τ u+a )} p/2 u 1 N Z2n {J N ( f τ u+a )} p/2 = C 1 N e p ν 2 /6 card(γ(ν)) {J N ( f τ b )} p/2. ν N 1 Z2n b N 1 Z2n Since card(γ(ν)) = 1 + Γ(ν) 1 + 2nN ν, it is clear that Lemma 3.3 holds. LEMMA 3.4. There exists N N and C N > 0 such that for any f L 2 loc (Cn ) and ν N 1 Z2n, we have ( f (z) f (w))e z,w e w 2 /2 e iim ν,w dv(w) 2 e z 2 dv(z) C N J N ( f τ ν ). +ν +ν Proof. Since e z 2 /2 e z,w e w 2 /2 = e z 2 /2 e z,w /2 e z,w /2 e w 2 /2 = e z 2 /2 e z,w /2 2 e w 2 /2 e iim z,w = e z w 2 /2 e iim z,w, we have that ( f (z) f (w))e z,w e w 2 /2 e iim ν,w dv(w) 2 e z 2 dv(z) +ν +ν = = +ν +ν ( f (z) f (w))e z w 2 /2 e iim z,w e iim ν,w dv(w) 2 dv(z) ( f τ ν (z) f τ ν (w))e z w 2/2 e iim z,w dv(w) 2 dv(z). Pick some δ > 0 to be determined, and pick N large enough so that (3.8) e z w 2 /2 e iim z,w = 1 + γ z,w where γ z,w < δ for any (z, w). This implies that if z, then ( ) 2 ( f τ ν (z) f τ ν (w))dv(w) ( f τ ν (z) f τ ν (w))γ z,w dv(w) ( f τ ν (z) f τ ν (w))dv(w) 2 2 ( f τ ν (z) f τ ν (w))dv(w) ( f τ ν (z) f τ ν (w))γ z,w dv(w)

( f τ ν (z) f τ ν (w))dv(w) 2 ( 2δ SCHATTEN CLASS HANKEL OPERATORS 153 Therefore, (3.8) and the triangle inequality implies that (3.9) However, 1 (V( )) 2 (3.10) ) 2. f τ ν (z) f τ ν (w) dv(w) ( f τ ν (z) f τ ν (w))e z w 2/2 e iim z,w dv(w) 2 dv(z) [ ( f τ ν (z) f τ ν (w))dv(w) 2 ( ) 2 ] 2δ f τ ν (z) f τ ν (w) dv(w) dv(z). ( f τ ν (z) f τ ν (w))dv(w) 2 dv(z) = f τ ν ( f τ ν ) QN 2 dv 1 = 2V( ) f τ ν (z) f τ ν (w) 2 dv(w)dv(z) so that (3.11) ( f τ ν (z) f τ ν (w))dv(w) 2 dv(z) = 1 2 V() whereas the Cauchy Schwarz inequality gives us ( ) 2dV(z) f τ ν (z) f τ ν (w) dv(w) f τ ν (z) f τ ν (w) 2 dv(w)dv(z) (3.12) V( ) f τ ν (z) f τ ν (w) 2 dv(z)dv(w). Finally, plugging (3.11) and (3.10) into (3.9) gives us that ( f τ ν (z) f τ ν (w))e z w 2/2 e iim z,w dv(w) 2 dv(z)

154 J. ISRALOWITZ ( 1 ) 2 2δ V( ) = 32n N 2n ( 1 2 2δ ) f τ ν (z) f τ ν (w) 2 dv(w)dv(z) f τ ν (z) f τ ν (w) 2 dv(w)dv(z) since V( ) = 32n. Therefore, picking 0 < δ < 1 N 2n 4 completes the proof of Lemma 3.4. and N corresponding to δ 4. PROOF OF THE MAIN RESULT We can now prove our main result. THEOREM 4.1. Let 0 < p < 1 and f T (C n ). If H f and H f S p, then there exists a constant C > 0 independent of f such that C n {SD( f τ ξ )} p dv(ξ) < C( H f p S p + H f p S p ). Proof. Fix N N such that Lemma 3.4 holds. For M N to be determined later and each j = (j 1,..., j 2n ) {1,..., M} 2n, set Λ M j = { ν N 1 Z2n : ν = ( 1 N ν 1,..., 1 N ν 2n) with each ν l j l mod M }. Since [M f, P] = [M f, P]P + [M f, P](I P) = H f (H f ), we have that [M f, P] S p with [M f, P] p S p H f p S p + H f p S p. Let {e ν } ν Λ M j be an orthonormal basis for L 2 (C n, dµ). Let h ν (w) = e w 2 /2 e iim ν,w χ QN +ν(w) and ξ ν (z) = χ +ν(z)([m f, P]hν(z)). χ QN +ν[m f, P]hν Set W j = A j [M f, P]B j where A j e ν = ξ ν and B j e ν = h ν, so that (4.1) W j p 2n 3np S p M N np [M f, P] p 2n 3np S p M N np ( H f p S p + H f p S p ). j {1,...,M} 2n Fix R N and let Z = {ν = (ν 1,..., ν 2n ) 1 N Z2n where each ν i R} so that for any ν, ν Z, we have Γ(ν, ν ) Z. Let Z j = Λ M j Z and let P Zj denote the orthogonal projection onto span{e ν : ν Z j }, so that clearly P Zj W j P Zj f = f, e ν W j e ν, e ν e ν. Let D j be defined by D j f = ν, f, e ν W j e ν, e ν e ν and set E j = P Zj W j P Zj D j so that W j p S p P Zj W j P Zj p S p D j p S p E j p S p.

SCHATTEN CLASS HANKEL OPERATORS 155 Thus, since D j is diagonal, we have that D j p S p = (4.2) A j [M f, P]B j e ν, e ν p = χ QN +ν[m f, P]hν p ( ( f (z) f (w))e z,w e w 2 /2 e iim ν,w dv(w) 2 e z 2 dv(z) = C 1 N +ν +ν {J N ( f τ ν )} p/2, where the last inequality follows from Lemma 3.4. We now get a upper bound for E j p S p. Since 0 < p < 1, we have that (4.3) E j p S p ν Λ M j = = ν Λ M j E j e ν, e ν p = [M f, P]h ν, χ QN +ν [M f, P]h ν ( χ QN +ν [M f, P]h ν χ QN +ν [M f, P]h ν p +ν +ν E j e ν, e ν p p ( f (z) f (w))e z,w e w 2 /2 e iim ν,w dv(w) 2 p/2. e dv(z)) z 2 But by the Cauchy Schwarz inequality, we have that ( +ν +ν 3np N np = 3np N np ( f (z) f (w))e z,w e w 2 /2 e iim ν,w dv(w) 2 e z 2 dv(z) ( +ν +ν f (z) f (w) 2 e z 2 e z,w 2 e w 2 dv(w)dv(z) ( f (z) f (w) 2 e z w 2 dv(w)dv(z) +ν +ν

156 J. ISRALOWITZ (4.4) (np/4)((m 3)/N)2 3np e N np ( +ν +ν C 2 N e (np/4)((m 3)/N)2 ( f (z) f (w) 2 e z w 2 /2 dv(w)dv(z) e p ν ν 2 /5 f τ ν (z) f τ ν (w) 2 dv(w)dv(z). Now, if z and w are both in, then enumerating the points in Γ(ν, ν) Z as {a 0,..., a l } in such a way that a 0 = ν and a l = ν, we have that (4.5) However, {S N + a j 1 } {S N + a j } + a j 1, 1 j l, f τ ν (z) f τ ν (w) f τ ν (z) ( f τ ν ) QN + f τ ν (z) ( f τ ν ) QN + ( f τ ν ) QN f τ ν (w) l j=1 + l j=1 ( f τ aj 1 ) QN ( f τ aj ) QN. ( f τ aj 1 ) QN ( f τ aj ) QN + ( f τ ν ) QN f τ ν (w) {2nN ν ν + 2} 1/2 ( f τ ν (z) ( f τ ν ) QN 2 + ( f τ ν ) QN f τ ν(w) 2 + l j=1 ( f τ aj 1 ) QN ( f τ aj ) QN 2 ) 1/2 which means that ( f τ ν (z) f τ ν (w) 2 dv(w)dv(z) (4.6) 3np N np ( (2nN ν ν + 2) + 32np ( N 2np (2nN ν ν + 2) + 3np N np ( (2nN ν ν + 2) f τ ν ( f τ ν ) QN 2 dv l j=1 ( f τ aj 1 ) QN ( f τ aj ) QN 2 f τ ν ( f τ ν ) QN 2 dv.

SCHATTEN CLASS HANKEL OPERATORS 157 we get (4.7) Now plug the first term of (4.4) into (4.3) and noting that {2nN ν ν + 2} p/2 e p ν ν 2 /5 C 5 N e p ν ν 2 /6, C 6 N e (np/4)((m 3)/N)2 ( (2nN ν ν +2) C 7 N e (np/4)((m 3)/N)2 ( e p ν ν 2 /5 f τ ν ( f τ ν ) QN 2 dv e p ν ν 2 /6 f τ ν ( f τ ν ) QN 2 dv = C 8 N e (np/4)((m 3)/N)2 ( = C 9 N e (np/4)((m 3)/N)2 {J N ( f τ ν )} p/2 f τ ν ( f τ ν ) QN 2 dv and by symmetry, we get the exact same estimate by plugging the third term of (4.4) into (4.3). Now we plug in the second term of (4.4) into (4.3). Since 0 < p 1, we only need to estimate the quantity (4.8) e (np/4)((m 3)/N)2 l (ν,ν) e p ν ν 2 /6 ( f τ aj 1 ) QN ( f τ aj ) QN p j=1 where for each ν and ν in the above sum, Γ(ν, ν) = {a 0,..., a l(ν,ν) }. As in the computation from equation (3.3), we have ( f τ aj 1 ) QN ( f τ aj ) QN p = f QN +a j 1 f QN +a j p f QN +a j 1 f SN +a j p + f SN +a j f QN +a j p C 10 N [(J N( f τ aj 1) + (J N ( f τ aj ) ]. Now, if ν = (y 1,..., y 2n ) and ν = (z 1,..., z 2n ) with ν = ν, then by definition, Γ(ν, ν) = 2n l=1 {(y 1,..., y l 1, u, z l+1,..., z 2n ) Z : min{y l, z l } u max{y l, z l }}.

158 J. ISRALOWITZ Therefore, combining the three summations in (4.5) into one single sum, this sum is taken over the set {u, ν, ν : u Z, (ν, ν ) Γ u } where with and Γ u 2n 2n l =1 l=1 Γ l + u Γ l u, Γ l u = {ν, ν Z : u = (u 1,..., u 2n ), ν = (u 1,..., u l 1, u, x l+1,..., x 2n ), ν = (y 1,..., y l 1, u, u l+1,..., u 2n ) where u u l u } Γ l + u = {ν, ν Z : u = (u 1,..., u 2n ), ν = (u 1,..., u l 1, u, x l+1,..., x 2n ), ν = (y 1,..., y l 1, u, u l+1,..., u 2n ) where u u l u }. Thus, after switching the order of summation, (4.5) is smaller than (4.9) C 11 N e (np/4)((m 3)/N)2 {J N ( f τ u )} p/2 e p ν ν 2/6. u Z (ν,ν ) Γ u If we denote ν = (ν 1,..., ν 2n ) and ν = (ν 1,..., ν 2n ), and let π i : C n R be the canonical projection onto the i th factor, then ( ) ( ) e p ν ν 2/6 = e p ν 1 ν 1 2 /6 e p ν 2n ν 2n 2 /6. (ν,ν ) Γ u (ν 1,ν 1 ) π 1 π 1 (Γ u ) For some l {1,..., 2n}, we now get an estimate on ( ) ( e p ν 1 ν 1 2 /6 (4.10) (ν 1,ν 1 ) π 1 π 1 (Γ l + u ) (ν 2n,ν 2n ) π 2n π 2n (Γ u ) (ν 2n,ν 2n ) π 2n π 2n (Γ l + u ) e p ν 2n ν 2n 2 /6 ) for fixed u. The terms that involve some Γ l u s are treated similarly. Assume l is neither 1 nor 2n, since the case where l = 1 and l = 2n is very similar. If j {1,..., l 1}, then ν j ν j = u j ν j, and if j {l + 1,..., 2n}, then ν j ν j = u j ν j. Moreover, since ν l u l ν l, (4.7) is smaller than ( ) ( ) ( ) e p u1 ν 1 2 /6 e p ν l ν l 2 /6... e p ν2n u2n 2/6. ν 1 N 1 Z ν 2n N 1 Z ν l u l ν l Thus, as we are holding u fixed, each of the factors corresponding to j = l in the product above converge to a positive number that depends only on N and p. For the factor that corresponds to j = l, we have e p ν l ν l 2 /6 = ν l u l ν l u l ν l = C 12 N ν l Z k=(u l ν l ) e pk2 /6N 2 e p u l ν l 2 /12N 2. e pk2 /6N 2 ν l Z k= u l ν l

SCHATTEN CLASS HANKEL OPERATORS 159 Since each e p u l ν l 2 /12N 2 converges to a number that only depends on ν l Z p and N, we have that (4.6) is smaller than which therefore gives us that (4.11) 2n 3np M C 13 N e (np/4)((m 3)/N)2 {J N ( f τ u )} p/2 u Z E j p S p C 14 N e (np/4)((m 3)/N)2 {J N ( f τ u )} p/2. u Z Going back to (4.1) and combining (4.2) with (4.8), we have that N np ( H f p S p + H f p S p ) [M f, P] p S p j {1,...,M} 2n W j p S p ( D j p S p E j p S p ) j {1,...,M} 2n (C 3 N C14 N M2n e (np/4)((m 3)/N)2 ) j {1,...,M} 2n u Z P Z W j P Z p S p {J N ( f τ u )} p/2. Pick M > 0 large enough so that C 3 N C14 N M2n e (np4)((m 3)/N)2 > 0. Thus, since the above estimate holds for any R N (and recalling that Z = {ν = (ν 1,..., ν 2n ) N 1 Z2n with each ν i R}), Lemma 3.3 gives us ( H f p S p + H f p S p ) C 15 N For some constants C 15 N C n {SD( f τ ξ )} p dv(ξ). > 0, which proves the theorem. Acknowledgements. The author wishes to thank L. Coburn for the interesting discussions regarding this work, and J. Xia for his extremely valuable comments. REFERENCES [1] W. BAUER, Mean oscillation and Hankel operators on the Segal Bargmann space, Integral Eqnuations Operator Theory 52(2005), 1 15. [2] C.A. BERGER, L.A. COBURN, Toeplitz operators on the Segal Bargmann space, Trans. Amer. Math. Soc. 301(1987), 813 829. [3] I.C. GOHBERG, M.G. KREIN, Introduction to the Theory of Linear Nonselfadjoint Operators, Transl. Math. Monographs, vol. 18, Amer. Math. Soc., Providence, R.I. 1969. [4] L. PENG, Paracommutators of Schatten von Neumann class S p, 0 < p < 1, Math. Scand. 61(1987), 68 92. [5] D. TIMOTIN, C p estimates for certain kernels: the case 0 < p < 1, J. Funct. Anal. 72(1987), 368 380.

160 J. ISRALOWITZ [6] J. XIA, D. ZHENG, Standard deviation and Schatten class Hankel operators on the Segal Bargmann space, Indiana Univ. Math J. 53(2004), 1381 1399. J. ISRALOWITZ, MATHEMATICS DEPARTMENT, UNIVERSITY AT BUFFALO, BUF- FALO, 14260, U.S.A. E-mail address: jbi2@buffalo.edu Received October 22, 2008.