The distribution function inequality for a finite sum of finite products of Toeplitz operators

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1 Journal of Functional Analysis 218 (2005) The distribution function inequality for a finite sum of finite products of Toeplitz operators Kunyu Guo a, Dechao Zheng b, a Department of Mathematics, Fudan University, Shanghai, , The People s Republic of China b Department of Mathematics, Verbilt University, Nashville, TN 37240, USA Received 4 March 2002; received in revised form 22 March 2004; accepted 22 June 2004 Available online 13 October 2004 Abstract A generalized area function associated with a finite sum of finite products of Toeplitz operators is introduced. A distribution function inequality is established for the generalized area function. By using the distribution function inequality, we characterize when a finite sum of finite products of Toeplitz operators on the Hardy space is a compact perturbation of a Toeplitz operator Elsevier Inc. All rights reserved. 1. Introduction Let D be the open unit disk in the complex plane D the unit circle. dσ(w) denotes the normalized Lebesgue measure on the unit circle. Let L 2 denote the Lebesgue square integrable functions on the unit circle. For 1 p <, f(z) an analytic function on D, wesayf H p if sup f(re iθ ) p dσ(e iθ ) f p p <. r D H denotes the set of bounded analytic functions on the unit disk. Corresponding author. addresses: kyguo@fudan.edu.cn (K. Guo), zheng@math.verbilt.edu (D. Zheng) /$ - see front matter 2004 Elsevier Inc. All rights reserved. doi: /j.jfa

2 2 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Let P be the Hardy projection of L 2 onto H 2.ForA L, the Toeplitz operator T A : H 2 H 2 with symbol A is defined by T A h P (Ah). The Hankel operator H A : H 2 L 2 H 2 with symbol A is defined by H A h (I P )(Ah). For more details on Toeplitz operators, see [4,7,8,20,21]. The map ξ: A T A, which is called the Toeplitz quantization, carries L into the C -algebra of bounded operators on H 2. It is a contractive *-linear mapping [8]. However it is not multiplicative in general. On the other h, Douglas [8] showed that ξ is actually a cross-section for a *-homomorphism from the Toeplitz algebra, the C -algebra generated by all bounded Toeplitz operators on H 2, onto L. So modulo the commutator ideal of the Toeplitz algebra, ξ is multiplicative. Studying the Toeplitz algebra has shed light on the theory of Toeplitz operators [7,8,20]. In this paper we will study the (not closed) algebra of finite sums of finite products of Toeplitz operators, which is dense in the Toeplitz algebra. The main question to be considered in this paper is when a finite sum of finite products of Toeplitz operators is a compact perturbation of a Toeplitz operator. This problem is connected with the spectral theory of Toeplitz operators; see [4,7,8,20]. A theorem of Douglas [8] implies that L Il T A can be a compact perturbation of a Toeplitz operator only when it is a compact perturbation of T L Il A In this paper we will introduce a generalized area function associated with a finite sum of finite products of Toeplitz operators establish a distribution function inequality for the area function. By means of the key distribution function inequality we will prove that a finite sum T of finite products of Toeplitz operators is a compact perturbation of a Toeplitz operator if only if Here z denotes the Möbius map,. z 1 T T z TT z 0. (1) z (w) z w 1 zw. The above result is a variant of Theorem 4 in [14]. However, some crucial details are omitted from the proof in [14], especially, details in the proof of a key distribution function inequality. One of our motivations is the result of Axler the second author [2] that if an operator S on the Bergman space equals a finite sum of finite products of Toeplitz

3 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) operators, then S is compact if only if the Berezin transform of S vanishes on the boundary of the unit disk. One may expect that the Berezin transform gives the analogous characterization for a finite sum of finite products of Toeplitz operators to be compact on the Hardy space. However, we will use examples from [12] to show that even if an operator T on the Hardy space equals a finite sum of finite products of Toeplitz operators, the vanishing of the Berezin transform of T does not have to imply that T is compact. Another motivation is the solution of the problem of characterizing when the product of two Toeplitz operator on the Hardy space H 2 is a compact perturbation of a Toeplitz operator, by Axler et al. [1] Volberg [22]. Their beautiful result is that T f T g is a compact perturbation of a Toeplitz operator if only if H [ f ] H [g] H + C( D); here H [g] denotes the closed subalgebra of L generated by H g. Recently, the second author [23] showed that T f T g is a compact perturbation of a Toeplitz operator if only if z 1 H f k z 2 H g k z 2 0; here k z denotes the normalized reproducing kernel in H 2 for point evaluation at z. This is equivalent to [T f T g T fg ] T z 1 [T f T g T fg ]T z 0. z The semicommutator T f T g T fg can be written as a product of two bounded Hankel operators. To study a finite sum of finite products of Toeplitz operators we will decompose the finite sum as a finite sum of products of two (unbounded) Hankel operators in Section 3. Clearly, a much more involved cancellation may happen in the sum of products of two Hankel operators. We need to take care of the cancellation by introducing a generalized area integral associated with the sum in Section 4. Even in some special cases [13,15] some generalized area integral functions were introduced. Gorkin the second author [13] have shown that the commutator [T f,t g ]( T f T g T g T f ) of two Toeplitz operators is compact on H 2 if only if [T f,t g ] T z 1 [T f,t g ]T z 0. z Condition (1) not only unifies the results on the compactness of commutators or semi-commutators of Toeplitz operators, but is also useful in understing the Toeplitz algebra. In Section 7 we will give applications of our main result to the following two questions: Question 1. For an inner function b, characterize the operators X on H 2 such that T b XT b X is compact. Question 2. For an inner function b, characterize the operators X on H 2 such that the commutator [T b,x] is compact.

4 4 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 These questions are closely related to inspired by the following Douglas problems: Douglas problem 1. If X is an operator on H 2 such that T b XT b X is compact for every inner function b, then is X T ψ + K for some ψ in L compact operator K? [7] Douglas problem 2. If the commutator [T b,x] is compact for each b in H +C, then is X T ψ + K for some ψ in H + C compact operator K? [9]. Douglas showed [9] that the solution of the first problem will give the solution of the second problem. Douglas [9] solved the first problem in the case that X is in the Toeplitz algebra. Although the Douglas problem 1 remains open, Davidson [6] has solved the second problem. Clearly, the above questions localize the Douglas problems in some sense. Another application of our main result is the solution of the problem of when a Hankel operator essentially commutes with a Toeplitz operator [16]. 2. Examples maximal ideal space In this section we will recall examples from [12] to show that the Berezin transform does not characterize the compactness of a finite sum of finite products of Toeplitz operators on the Hardy space. Let T be a bounded operator on H 2. The Berezin transform of T is defined by ˆT(z) Tk z,k z for z in D. Perhaps the most important tool in the study of the Toeplitz algebra, the norm-closed algebra of operators generated by the Toeplitz operators, is the existence of a homomorphism, the so-called symbol mapping σ, from the Toeplitz algebra to L such that σ(t f ) f for every f L. The key point here is that σ is multiplicative. The symbol mapping was discovered exploited by Douglas [7]. Barría Halmos [3] showed the symbol mapping σ is well defined for asymptotic Toeplitz operators. Recently Englis [10] showed that the nontangential it of the Berezin transform of T equals the symbol of T, for T in the Toeplitz algebra. To present the examples in [12], we need to introduce the maximal ideal space of H. Let M(H ) be the set of the multiplicative linear functionals on H.IfB is a Douglas algebra, i.e., a subalgebra of L that contains H, then M(B) can be identified with the set of nonzero linear functionals in M(H ) whose representing measures (on M(L )) are multiplicative on B. We identify a function f in B with its Gelf transform on M(B). In particular, M(H +C) M(H ) D, a function f H may be thought of as a continuous function on M(H ).

5 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) Examples. Let b be any interpolating Blaschke product with zeros {z n }. Choose a sequence of positive integers l n such that Let l n (1 z n )<. n1 b 1 n1 z n z n ( ) zn z ln 1 z n z denote the corresponding Blaschke product. It was proved in [12] that for each m M(H + C), This is equivalent to m(b 1 b) m(b 1 )m( b). [ b 1 b(z) b 1 (z)b(z)] 0, z 1 where b 1 b(z) is the harmonic extension of b 1 b at z given by b 1 b(z) b 1 (w) b(w) k z (w) 2 dσ(w). D Let T T b1 b T b 1 T b. Clearly, T is a finite sum of finite products of Toeplitz operators. An easy calculation gives that the Berezin transform of T is Thus ˆT(z) [T b1 b T b 1 T b ]k z,k z b 1 b(z) b 1 (z)b(z). z 1 ˆT(z) 0. Since b 1 b b 1 b is in H,wehave T b 1 b T T b 1 b (T b 1 b T b 1 T b ) I T bt b is an infinite dimensional projection, hence T is not compact.

6 6 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Hoffman [17,18] has shown that for each m M(H +C), m has a unique extension to L, which is given by m(f ) fdμ m S m for f L. Here S m is the (closed) support of the representing measure dμ m. A subset S of M(L ) is called a support set if it is the (closed) support of the representing measure for a functional in M(H + C). Let H 2 (m) be the closure of H in L 2 (dμ m ). Let H0 2(m){f H 2 (m): S fdμ m 0}. Hoffman [17, p. 289] proved that L 2 (dμ m ) H 2 (m) H0 2(m). An inner function in H 2 (m) is a function q H (m) with q 1 a.e. on S m.an outer function in H 2 (m) is a function o such that H o is dense in H 2 (m). But Theorem 22 [19] says that every function f in H 2 (m) with f (m) 0 has the factorization qo for an inner function q an outer function o. The following lemma will be needed in Section 7. Lemma 1. If m M(H +C) b is an inner function in H not equal to a constant on the support set S m, then 1 b is an outer function in H 2 (m). Proof. We assume that b does not identically equal 1 on the support set S m. Let E {x S m : b(x) 1}, a subset of S m of positive measure. For 0 <r<1, the function (1 rb) 1 is in H, (1 rb) 1 (1 b) χ E pointwise boundedly on S m as r 1. Hence χ E is in the H 2 (m)-closure of (1 b)h, also in H (m). Since μ m is multiplicative on H (m), wehave ( ) 2 μ m (E) 2 χ E dμ m χ 2 E dμ m μ m (E), giving μ m (E) 1 (since μ m (E) 0). Hence the constant function 1 is in the H 2 (m)- closure of (1 b)h, showing that b is outer in H 2 (m). We thank D. Sarason for his suggesting the above proof. 3. Decomposition Although our main concern is with bounded Toeplitz operators Hankel operators, we will need to make use of densely defined unbounded Toeplitz operators Hankel operators. Given two operators S 1 S 2 densely defined on H 2, we say that S 1 S 2 if S 1 p S 2 p, for each p in the set P of analytic polynomials.

7 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) As in [14], in this section we will show that a finite sum of finite products of Toeplitz operators can be written as a finite sum of products of two Toeplitz operators. The key here is a simple useful idea used in [14] T A1 T A2 T A3 T A1 [(A 2 ) + +c 1 ]T A3 + T A1 T [(A2 ) c 1 ]A 3, for three bounded functions A 1, A 2 A 3, a constant c 1. Here A + P (A) A (I P )(A). For four bounded functions A 1, A 2, A 3 A 4 ; three constants c 1, c 2 c 3,wehave T A1 T A2 T A3 T A4 [T A1 [(A 2 ) + +c 1 ]T A3 + T A1 T [(A2 ) c 1 ]A 3 ]T A4 T A1 [(A 2 ) + +c 1 ][(A 3 ) + +c 2 ]T A4 + T A1 [(A 2 ) + +c 1 ]T [(A3 ) c 2 ]A 4 + T A1 {[[(A 2 ) c 1 ]A 3 ] + +c 3 }T A4 + T A1 T {[[(A2 ) c 1 ]A 3 ] c 3 }A 4. Clearly, for an integer m 2, by induction, we see that a product of m Toeplitz operators with bounded symbols can be written in a sum of 2 m 2 terms that are products of two Toeplitz operators with (perhaps unbounded) symbols, the decomposition is not unique. In order to deal with a finite sum of products of two Toeplitz operators with unbounded symbols we need to introduce systematic decompositions of the finite products. To do so, let λ {λ(l, k)} be a sequence of complex numbers. For a sequence of functions A 1,A 2,...,A n in L, we inductively define for k 2 i 1. λa 0 1 A 1, λb1 0 A 2 λa i 2k 1 λa i 1 k [( λ Bk i 1 ) + + λ(i 1,k)], λ B2k 1 i A i+2 [( λ Bk i 1 ) λ(i 1,k)]A i+2, λa i 2k λ A i 1 k, λb i 2k Lemma 2. Let λ {λ(l, k)} be a sequence of complex numbers. If A 1,A 2,...,A m are of functions in L, then λ A i j λb i j defined above are in >p>1 L p. Moreover, T A1 T A2 T Am 2 m 2 T λ A m 2 T j λ Bj m 2 A 1 A 2 A m 2 m 2 λa m 2 j λbj m 2.

8 8 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Proof. We use induction to prove the theorem. When n 2, from our definition we have T A1 T A2 T λ A 0 1 T λb 0 1 A 1 A 2 λ A 0 1λB 0 1. For n m, we assume that T A1 T A2 T Am 2 m 2 T λ A m 2 T j λ Bj m 2 (2) A 1 A 2 A m 2 m 2 λa m 2 j λbj m 2. (3) Now 2 m 1 λa m 1 j λb m 1 2 m 2 j 2 m 2 { [ λ A m 1 2j 1λB m 1 λa m 2 j 2j 1 + λa m 1 2j λb2j m 1 ] [( λ Bj m 2 ) + + λ(m 2,k)]A m+1 + λ A m 2 j } [( λ Bj m 2 ) λ(m 2,k)] λ A m+1 2 m 2 λa m 2 j [( λ Bj m 2 ) + + λ(m 2,k)]+ λ A m 2 j [( λ B m 2 j 2 m 2 λa m 2 j ) λ(m 2,k)] λ A m+1 [( λ Bj m 2 ) + + λ(m 2,k)

9 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) ( λ B m 2 j 2 m 2 ) λ(m 2,k)] λ A m+1 λa m 2 j λb m 2 j A 1 A 2 A m A m+1. λ A m+1 The last equality follows from (3). Note that both ( λ Bk m 2 ) + + λ(m 2,k) [( λ Bk m 2 ) λ(m 2,k)] are in H 2. Thus T λ A m 2 j So by (2) we obtain Hence we conclude T [(λ B m 2 j ) + +λ(m 2,k)] T λa m 2 j [( λ Bj m 2 ) + +λ(m 2,k)], T [(λ Bj m 2 ) λ(m 2,k)] T A m+1 T [(λ Bj m 2 ) λ(m 2,k)]A m+1. 2 m 2 T A1 T A2 T Am T Am+1 2 m 2 T λ A m 2 j T λ Bj m 2 T Am+1 [T λ A m 2 T j [(λ Bj m 2 ) + +λ(m 2,k)] T A m+1 + T λ A m 2 T j [(λ Bj m 2 ) λ(m 2,k)] T A m+1 ] 2 m 2 T A1 T A2 T Am T Am+1 [T λ A m 2 j [( λ Bj m 2 ) + +λ(m 2,k)] T A m+1 + T λ A m 2 T j [(λ Bj m 2 ) λ(m 2,k)]A m+1 ]. 2 m 1 T λ A m 1 j T λ Bj m 1. Note that >p>1 L p is an algebra, i.e., both fg f + g are in >p>1 L p if f g are in >p>1 L p. In addition, P + P are bounded on L p for 1 <p<,

10 10 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 map L into BMO. The John-Nirenberg theorem tells us that BMO is contained in the intersection >p>1 L p. These imply that λ A i j λbj i are products of functions in >p>1 L p. So they are also in >p>1 L p. This completes the proof. The above lemma gives the following proposition. The decompositions of A i are different from those in [14]. Proposition 3. Let λ {λ(l, k)} be a sequence of complex numbers. T A1 T A2 T Am T A1 A 2 A m 2 m 2 H λa m 2 j H λ Bj m 2. Proof. By Lemma 2, wehave T A1 T A2 T Am 2 m 2 T λ A m 2 T j λ Bj m 2 Because A 1 A 2 A m 2 m 2 λa m 2 j λbj m 2. T A T B T AB H Ā H B, we get 2 m 2 [ T A1 T A2 T Am T A1 A 2 A m This completes the proof. 2 m 2 T λ A m 2 j H λa m 2 j T λ B m 2 j H λ Bj m 2. T λ A m 2 j λbj m 2 Although the representation of a finite product of Toeplitz operators as a sum of products of two Toeplitz operators is not unique, it has the advantage of letting us to ]

11 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) choose λ(j, k). In order to establish our distribution function inequality we need to choose those constants λ(j, k) appropriately at each point z D. The following lemma tells us that we can do so. Let A 1,...,A m be in L. Given a point z D, inductively define a sequence {λ(l, k)} of complex numbers λ(i 1,k) ( λ Bk i 1 ) (z). From the definition of ( λ B i 1 k ) (z), it depends on only λ(j, k) for j<i 1. Lemma 4. Let A 1,...,A m be in L. Suppose that sup A i M i for some constant M. For a fixed z in D, let λ(i 1,k) ( λ Bk i 1 ) (z). Then for 1 <p< there are constants M pi, such that max j max{ λ A i 2 j z p, λ B i 2 j z p } M pi. Moreover M pi depends on M p, but does not depend on z. Proof. We will prove this lemma by induction. When i 2, we have λa 0 1 A 1, λ B 0 1 A 2. For each 1 <p<, λ A 0 1 z p A 1 z p A 1 M λ B 0 1 z p A 2 z p A 2 M. When i n, for each 1 <p<, assume max j max{ λ A n 2 j z p, λ B n 2 j z p } M pn.

12 12 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Let N p be the positive constant such that for f L p. When i n + 1, Clearly, λa2k 1 n 1 z λak n 2 max k λb n 1 2k P + f p N p f p, P f p N p f p z [( λ Bk n 2 ) + z + λ(n 2,k)], 2k 1 z A n+1 z, λb n 1 z [( λ Bk n 2 ) z λ(n 2,k)]A n+1 z, λa n 1 2k z λ Ak n 2 z. max{ λ A n 1 2k Note that for each function f L 2, z p, λ B n 1 2k 1 z p} max{m pn,m}. f + z (f z ) + f (z), f z (f z ) + f (z). Thus ( λ Bk n 2 ) + z ( λ Bk n 2 z ) + ( λ Bk n 2 ) (z), By our choice, we have So ( λ Bk n 2 ) z ( λ Bk n 2 z ) + ( λ Bk n 2 ) (z). λ(n 2,k) ( λ Bk n 2 ) (z). ( λ Bk n 2 ) + z + λ(n 2,k) ( λ Bk n 2 z ) +, ( λ Bk n 2 ) z λ(n 2,k) ( λ Bk n 2 z ).

13 Hence we conclude ( λ B n 2 k K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) λ A n 1 2k 1 z p λ A n 2 k λ B n 1 2k λ A n 2 k λ A n 2 k N p M 2 (2p)n, z [( λ Bk n 2 ) + z + λ(n 2,k)] p z ( λ B n 2 k z 2p ( λ B n 2 k z ) + p z ) + 2p z p [( λ Bk n 2 ) z λ(n 2,k)]A n+1 z p z ) A n+1 z p ( λ B n 2 k z ) p A n+1 z N p M pn M. The last inequality follows because the Hardy projection is bounded on L p for 1<p<. Letting M p(n+1) max{n p M 2 (2p)n,N pm pn M,M pn,m}, we complete the proof. Summarily, Proposition 3 suggests the first part of the following theorem Lemma 4 gives the second part of the following theorem. Theorem 5. Let M be a positive constant. Suppose that T is a finite sum of finite products of Toeplitz operators, i.e., for A in L with max l,j A M, T L I l T A. (1) For any sequence λ {λ(l, j)} of complex numbers, then T T L Il A L H λa I l 2 HλB I l 2. (2) For each z D we can find a sequence λ z {λ(l, j)(z)} of complex numbers so that for 1<p< max l,j max{ λz A I l 2 z p, λz B I l 2 z p } M p, for some constant M p depending only on M p.

14 14 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) A generalized area integral function For a point w of D, let Γ(w) denote the angle with vertex w opening π/2 which is bisected by the radius to w. The set of points z in Γ(w) satisfying z w < ε will be denoted by Γ ε (w). Forh in L 1 ( D), define the truncated Lusin area integral of h to be [ 1/2 A ε (h)(w) grad h(z) da(z)] 2, Γ ε (w) where (grad h)(z) denotes the gradient of the harmonic extension h at z x + iy: ( ) h h grad h(z) (z), x y (z), da(z) denotes the Lebesgue measure on the unit disk D h(z) denotes the harmonic extension of h at z D, via the Poisson integral h(z) h(w) (1 z 2 ) D 1 wz 2 dσ(w). Observe that if h is holomorphic, A ε (h)(w) equals the area of the image of Γ ε (w) under the mapping z h(z), with points counted according to their multiplicity. Suppose that T is a finite sum of finite products of Toeplitz operators, i.e., for some functions A in L, T L I l T A. By Theorem 5, for any sequence λ of complex numbers, we have the representation T T L Il A L H [ λ A I l 2 ] HλB I l 2.

15 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) Let λ 0 be the sequence {λ(l, j)} with λ(l, j) 0 for l, j. Let u v be in the class P of analytic polynomials on the unit disk. Define a generalized area integral by L 2 I l 2 T B ε (u, v)(w) ((grad Γ ε (w) Hλ 0 B I l 2 u)(z)) ((grad v)(z)) Hλ 0 A I l 2 da(z). Here ((grad Hλ 0 B I l 2 complex vectors ((grad Hλ 0 B I l 2 u)(z)) ((grad v)(z)) denotes the inner product of the two Hλ 0 A I l 2 v)(z)). u)(z)) ((grad Hλ 0 A I l 2 The main result in this section is that T B ε (u, v)(w) does not depend on λ 0. That is, for any sequence λ of complex numbers, Note that both HλB I l 2 T B ε (u, v)(w) L 2 2 z z Γ ε (w) L 2 I l 2 ((grad HλB I l 2 u)(z)) ((grad HλA I l 2 v)(z)) da(z). u v are in H HλA 2. Thus I l 2 ((grad HλB I l 2 L 2 I l 2 u)(z)) ((grad v)(z)) HλA I l 2 ((HλB I l 2 u)(z))((hλa I l 2 v)(z)). (4) So T B ε (u, v)(w)2 Γ ε (w) ((Hλ 0 A I l 2 2 L 2 I l 2 (( ) ) z z Hλ 0 B I l 2 u (z) v)(z)) da(z). (5) We need to introduce some notation. For x y two vectors in L 2. x y is the operator of rank one defined by (x y)(f ) f, y x.

16 16 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Observe that the norm of the operator x y equals x 2 y 2. We thank D. Sarason for suggesting the following lemma that gives a way to estimate the norm of the operators with finite rank. Let trace be the trace on the trace class of operators on a Hilbert space. Lemma 6. Let x 1,...,x N,y 1,...,y N be vectors in a Hilbert space, let S N i1 x i y i. Then there is an N N unitary matrix U such that S N i1 x i ỹ i (6) N tracess x i 2 ỹ i 2, (7) i1 where x 1.. U x N x 1.. x N, ỹ 1.. ỹ N U y 1.. y N. For the proof of the above lemma, a computation shows (6) holds for any N N unitary matrix U. To get (7) one just takes U to diagonalize the Grammian matrix of the vectors y 1,...,y N. The details are left to the reader. Note that if f 1,...,f N are in L p, U is an N N unitary matrix, then h 1. h N U f 1. f N, h j p N max f i p j for j 1,...,N. Let x i H fi k z y i H gi k z. Applying the above lemma, we obtain the following lemma.

17 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) Lemma 7. Let S N i1 H fi k z H gi k z. Then there is a unitary N N matrix U z (a ij (z)) N N such that where ( f i ) T U z (f i ) T tracess N i1 H f i k z 2 2 H g i k z 2 2, ( g i ) T U z (g i ) T. Moreover, S N i1 H f i k z H gi k z, if for some p (1, ), there is a positive constant M p such that then max i max{ f i z p, g i z p } M p, max i Define an antiunitary operator V on L 2 by max{ f i z p, g i z p } NM p. (V h)(w) wh(w). The operator enjoys many nice properties such as V 1 (I P)V P V V 1. These properties easily leads to the relation V 1 H f V H f. To show that T B ε (u, v)(w) does not depend on λ 0, we need the following lemma. Lemma 8. Let ψ be polynomials in z. Suppose that f g are in p>1 L p. Then (1 z 2 )H g (z)h f ψ(z) z 2 [VH f k z VH g k z ], ψ. Proof. For each z D, f f(z) is a bounded linear functional on [H 2 ], { w n } is an orthonormal basis for [H 2 ]. Thus the reproducing kernel at z is given by w n z n z wk z ( w) zv K z. n1

18 18 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 So H g (z) z H g,vk z H f ψ(z) z H f ψ,vk z. This gives H g (z)h f ψ(z) z 2 H g,vk z H f ψ,vk z z 2,Hg Vk z ψ,hf Vk z z 2 [Hf Vk z] [Hg Vk z], ψ z 2 [VH f k z VH g k z ], ψ, to complete the proof. The proof of Lemma 1 [23] leads to the following lemma. Lemma 9. Suppose that f g are in p>1 L p. Then the operator H f H g T z H f H gt z equals [VH f k z ] [VH g k z ] Theorem 10. For any sequence λ {λ(l, j)} of complex numbers, L 2 I l 2 (( ) ) T B ε (u, v)(w) grad HλB Γ ε (w) I l 2 u (z) (( ) grad v (z)) da(z). HλA I l 2 Proof. Let T L I l T A.

19 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) By Theorem 5, for any sequence λ {λ(l, j)} of complex numbers, we have the following representation: Note that for each η D, Thus for each η D, T T L Il A L H [ λ A I l 2 ] HλB I l 2 T T L Il η A T η T L Il A.. T T η TT η L By Lemma 9, weget L H [ λ A I l 2 ] HλB I l 2 T η H [ λ A I l 2 ] HλB I l 2 T η L L H [ λ A I l 2 ] HλB I l 2 H [ λ A I l 2 ] HλB I l 2 T η. (8) T η L [VH k [λ A I l η] [VH 2 ] λb I l 2 k η ]. (9) By Lemma 8, wehave (1 η 2 ) L L η 2 ( HλB I l 2 u ) (V HλB I l 2 v)(η) (η)(hλa I l 2 k η ) (V HλA I l 2 k η ) u, v. (10) Combining (9) with (10) gives (1 η 2 ) L (HλB I l 2 v)(η) u)(η)(hλa I l 2

20 20 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 L η 2 L H [ λ A I l 2 ] HλB I l 2 T η T η H [ λ A I l 2 ] η 2 [T T η TT η ]u, v. HλB I l 2 u, v The last equality follows from (8). Clearly, the last term does not involve λ. Hence we conclude that 2 η η L (HλB I l 2 u)(η)(hλa I l 2 v)(η) does not depend on the choice of λ. That is, 2 η η L 2 η η (HλB I l 2 L 2 I l 2 u)(η)(hλa I l 2 (Hλ 0 B I l 2 v)(η) u)(η)(hλ 0 A I l 2 v)(η). Hence (5) gives that T B ε (u, v)(w) 2 2 L 2 I l 2 Γ ε (w) η η 2 2 L 2 I l 2 Γ ε (w) η η L 2 I l 2 Γ ε (w) (Hλ 0 B I l 2 (HλB I l 2 ((grad HλB I l 2 u)(η)(hλ 0 A I l 2 u)(η)(hλa I l 2 u)(η)) ((grad HλA I l 2 v)(η) da(η) v)(η) da(η) v)(η)) da(η). The last equality follows from (4). This completes the proof.

21 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) A distribution function inequality In this section we will establish a distribution function inequality for the generalized area integral introduced in Section 4. The distribution function inequality involves the Lusin area integral the Hardy Littlewood maximal function. The idea to use distribution function inequalities in the theory of Toeplitz operators Hankel operators first appeared in [1]. Chang [5] also used a distribution function inequality to study the commutator of the Szegö projection multiplication operators. Write I for the length of an arc I. The Hardy Littlewood maximal function of h is Mh(e iθ ) sup e iθ I 1 h(e iθ ) dσ(e iθ ) I I for h integrable on the unit circle D. The Hardy Littlewood maximal theorem ([11, Theorem 4.3]) states that for 1 <p, Mh p N p h p for h L p where N p is a constant depending only on p. Forr>1, let Λ r h(e iθ ) [M h r (e iθ )] 1/r. Then Λ r h p N 1 r p h p, r for p>r. For z D, we let I z denote the closed subarc of D with center z z length δ(z) 1 z. The Lebesgue measure of a subset E of D will be denoted by E. Recall the area integral function A ε (h)(w) for a function h in L 1 : [ 1/2 A ε (h)(w) grad h(z) da(z)] 2, Γ ε (w) where h(z) denotes the harmonic extension of h at z D. The following distribution function inequality was established in [23] The distribution function inequality Let f g be in L 2, ψ in the Hardy space H 2. Fix s > 2. Then there are numbers p, r (1, 2) with 1 s + 1 r p 1, such that for z > 1/2 a>0

22 22 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 sufficiently large, { w I z : A 2δ(z) (H f )(w)a 2δ(z) (H g ψ)(w) <a 2 f z f (z) s g z g (z) s } inf Λ r ( )(w) inf Λ r (ψ)(w) C a I z. w I z w I z Moreover, the constant C a 1 C a p C is a constant depending only on s. For each f in L 2, write f f + + f.givenz in D, an easy calculation gives H f k z [f f (z)]k z. Thus by a change of variable, we have H f k z 2 [f f (z)]k z 2 f z f (z) 2. If f is in p>1 L p, by the Cauchy Schwarz inequality, for s>2, we have to get f s s f(w) s dσ(w) D [ f(w) 2 dσ(w) D ] 1/2 [ 1/2 f(w) dσ(w)] 2s 2, D f z f (z) s f z f (z) 1/s 2 f z f (z) (s 1)/s 2s 2. The above distribution function inequality implies the following form, which will be needed later on. LetfgbeinL 2, ψ in the Hardy space H 2. Suppose that for some s>2 there is a constant M 2s 2 such that sup max{ f z f (z) 2s 2, g z g (z) 2s 2 } M 2s 2. z D

23 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) Then there are numbers p, r (1, 2) with 1 s + 1 r a>0 sufficiently large, { w I z : A 2δ(z) (H f )(w)a 2δ(z) (H g ψ)(w) p 1, such that for z > 1/2 <a 2 M 2s 2 s 2s 2 [ H f k z 2 H g k z 2 ] 1/s inf Λ r ( )(w) inf Λ r (ψ)(w)} w I z w I z C a I z. (11) Moreover, the constant C a 1 C a p C is a constant depending only on s. The following distribution function inequality is the main result in this section is the key to the proof of Theorem 12. Theorem 11. Let M be a positive constant. Suppose that T is a finite sum of finite products of Toeplitz operators, i.e., for A in L with max l,j A M, T L I l T A. LetuvbeinP. Let z be a point in D such that z > 1/2. Then for any s>2, for a>0 sufficiently large δ(z) 1 z, { w I z : T B 2δ(z) (u, v)(w) <a 2 T T z TT z 1/s M 2(s 1)/s 2s 2 [ inf Λ r (v)(w) w I z ]} C a I z, [ ] inf Λ r (u)(w) w I z where C a depends only on s a, a C a 1, 1 s + 1 r p 1 for some p rin(1, 2) M 2s 2 is the constant in Theorem 5 depending only on M s. Proof. Assume that T L I l T A. Let L(T ) denote the integer L I l. Fix s>2. Choose two numbers 1 <p<r<2 such that 1 s + 1 r 1 p. Fix a point z D. Let S z T T z TT z.

24 24 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Since for some positive constant M, max A M, l,j for such z, by Theorem 5, we can choose λ {λ(l, j)(z)} so that T T L Il A L H λa I l 2 HλB I l 2, max l,j max{ λ A I l 2 z 2s 2, λ B I l 2 z 2s 2 } M 2s 2, where M 2s 2 is the constant in Theorem 5, depending only on 2s 2 M. Let E be the subset of I z such that [ ][ ] T B 2δ(z) (u, v)(w) a 2 M 2(s 1)/s 2s 2 S z 1/s inf Λ r (u)(w) inf [Λ r (v)(w). w I z w I z To complete the proof, we need only to prove that E C a I z (12) for some positive constant C a depending only on a, s L(T ) satisfying Because a C a 1. (13) we have S z L T T L Il z A T z T L Il A, H λa I l 2 HλB I l 2 T z L H λa I l 2 HλB I l 2 T z.

25 Lemma 9 gives K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) S z L [VHλA I l 2 k z ] [VHλB I l 2 k z ]. By Lemma 7, there are functions { λ f i } J i1 in the space spanned by { λa I l 2 } L,2I l 2, { λ g i } J i1 in the space spanned by { λb I l 2 } L,2I l 2, such that J S z [VH λ f i k z ] [VH λ g i k z ], i1 J trace(s z Sz ) H λ f i k z 2 2 H λg i k z 2 2. (14) i1 Lemma 7 also gives that J L 2 I l 2. Thus J 2 L(T ). (15) By Lemma 8, we obtain (1 η 2 ) L L η 2 (HλB I l 2 (V HλB I l 2 v)(η) u)(η)(hλa I l 2 k η ) (V HλA I l 2 [ J ] η 2 H λ f i k η ) (V H λ g i k η ) u, v i1(v k η ) u, v η 2 S η u, v (1 η 2 ) J [H λ g i u(η)][h λ f i v(η)]. (16) i1

26 26 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Thus Theorem 10 gives T B 2δ(z) (u, v)(w) Γ 2δ(z) (w) L 2 I l 2 (grad HλB I l 2 u)(η) (grad(hλa I l 2 2 Γ 2δ(z) (w) (HλA I l 2 2 Γ 2δ(z) (w) Γ 2δ(z) (w) v)(η) da(η) 2 L 2 I l 2 η η v)(η) da(η) (HλB I l 2 u)(η) (by (4)) 2 [ J (H η η λ g i u)(η)(h λ f i v)(η)] da(η) i1 (by (16)) J (grad(h λ g i u)(η)) (grad(h λ f i v)(η)) da(η). i1 The last equality also follows from (4). Let E i be the subset of I z such that A 2δ(z) (H λ f i v)(w)a 2δ(z) (H λ g i u)(w) a 2 (J M 2s 2) (2s 2)/s J 1+(2s 2)/s [ H λ f i k z 2 H λ g i k z 2 ] 1/s [ ][ ] inf Λ r (u)(w) inf Λ r (v)(w). w I z w I z Note that Lemma 7 gives, for s>2, max i max{ λ f i z 2s 2, λ g i z 2s 2 } JM 2s 2. The distribution function inequality (11) gives E i (1 a p J p 2 + p(s 1) s C ) I z. (17)

27 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) The Cauchy Schwarz inequality gives T B 2δ(z) (u, v)(w) Γ 2δ(z) (w) i1 J ((grad H λ g i u)(η)) ((grad H λ f i v)(η)) da(η) i1 J ((grad H λ g i u)(η)) ((grad H λ f i v)(η)) da(η) Γ 2δ(z) (w) [ J (grad H λ g i u)(η) 2 da(η) Γ 2δ(z) (w) i1 [ (grad H λ f i v)(η) 2 da(η) Γ 2δ(z) (w) ] 1/2 ] 1/2 J A 2δ(z) (H λ f i v)(w)a 2δ(z) (H λ g i u)(w). i1 Thus for w in the intersection J i1 E i,wehave T B 2δ(z) (u, v)(w) J A 2δ(z) (H λ f i v)(w)a 2δ(z) (H λ g i u)(w) i1 J i1 a 2 M (2s 2)/s 2s 2 [ H J λ f i k z 2 2 H λg i k z 2 2 ]1/(2s) [ ][ ] inf Λ r (u)(w) inf Λ r (v)(w) w I z w I z a2 M (2s 2)/s 2s 2 J { J } [ H λ f i k z 2 2 H λg i k z 2 2 ]1/(2s) i1 [ inf Λ r (u)(w) w I z a2 M (2s 2)/s 2s 2 J 1/(2s) ][ ] inf Λ r (v)(w) w I z { J } 1/(2s) [ H λ f i k z 2 2 H λg i k z 2 2 ] i1 [ inf Λ r (u)(w) w I z ][ ] inf Λ r (v)(w) w I z (by the Hölder inequality)

28 28 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 a2 M (2s 2)/s 2s 2 J 1/(2s) [trace(s z S z )]1/(2s) [ ][ ] inf Λ r (u)(w) inf Λ r (v)(w) w I z w I z (by (14)) [ ][ ] a 2 M (2s 2)/s 2s 2 S z 1/s inf Λ r (u)(w) inf Λ r (v)(w). w I z w I z (18) The last inequality follows from that S z Sz So (18) gives Since E 1 E 2 I z, By induction, we get Thus (17) gives trace(s z S z ) J S zs z J S z 2. J i1 E i E. is a finite rank operator of rank at most J E 1 E 2 E 1 + E 2 E 1 E 2 E 1 + E 2 I z, [ J ] J i1 E i E i (J 1) I z. i1 So By (15) wehave J i1 E i (1 a p J 1+ p 2 + p(s 1) s C ) I z. E (1 a p J 1+ p 2 + p(s 1) s C ) I z. E (1 a p 2 L(T )(1+ p 2 + p(s 1) s ) C ) I z.

29 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) Letting C a (1 a p 2 L(T )(1+ p 2 + p(s 1) s ) C ), we obtain (12) (13) to complete the proof. Remark. The above proof shows that the constant C a depends also on the length L(T ) of T. We thank the referee for pointing out the fact. Also the constant M in Theorem 11 may be chosen as max l,j A that is finite. So Theorem 11 holds only for a finite sum T of finite products of Toeplitz operators. Certainly, we would like that Theorem 11 holds for T in the Toeplitz algebra. But it remains open. 6. Finite sums of finite products of Toeplitz operators In this section, using the key distribution function inequality in the previous section, we will prove the main result in this paper about a finite sum of finite products of Toeplitz operators. Theorem 12. A finite sum T of finite products of Toeplitz operators is a compact perturbation of a Toeplitz operator if only if z 1 T T z TT z 0. (19) Proof. Suppose T T A +K where K is a compact operator on H 2 A is a function in L. Note that T A T z T A T z. An easy calculation gives T T z TT z K T z KT z. By Lemma 2 [23], K T z 1 KT z 0. z Thus z 1 T T z TT z 0.

30 30 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Conversely, suppose that T is a finite sum of finite products of Toeplitz operators T T z 1 TT z 0. (20) z We need to prove that T is a compact perturbation of a Toeplitz operators. We may assume that where A are in L satisfy T L I l T A, A M, if M max l,j A. Let λ 0 be the sequence {λ(l, j)} of complex numbers satisfying λ(l, j) 0, for l,j. By Theorem 5, we have the following representation: T T L Il A L H λ 0 A I l 2 Hλ 0 B I l 2. Now let u v be two functions in P. In order to estimate the distance of the operator T T L Il A to the set of compact operators we consider the inner product, [T T L Il A ]u, v L 2 I l 2 H λ 0 A I l 2 Hλ 0 B I l 2 u, v L Hλ 0 B I l 2 u, v. Hλ 0 A I l 2 Since Hλ 0 B I l 2 u is orthogonal to H 2, we see that Hλ 0 B I l 2 u(0) 0.

31 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) By the Littlewood Paley formula ([11, Lemma 3.1]), we have [T T L Il A ]u, v L For each 1/2 <R<1, denote Thus (21) gives D ((grad Hλ 0 B I l 2 u)(z)) ((grad v)(z)) log 1 Hλ 0 A I l 2 z da(z) L D ((grad Hλ 0 B I l 2 u)(z)) ((grad v)(z)) log 1 da(z). (21) Hλ 0 A I l 2 z W R z >R L ((grad Hλ 0 B I l 2 )u(z)) ((grad v)(z)) log 1 Hλ 0 A I l 2 z da(z) Z R z R L ((grad Hλ 0 B I l 2 u)(z)) ((grad v)(z)) log 1 Hλ 0 A I l 2 z da(z). [T T L Il A ]u, v W R + Z R. (22) First we show that there is a compact operator K R such that Z R K R u, v. (23)

32 32 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Note that L 2 2 z z ((grad Hλ 0 B I l 2 L 2 I l 2 u)(z)) ((grad v)(z)) Hλ 0 A I l 2 (Hλ 0 B I l 2 From the proof of Theorem 10, we know that Thus So Let Z R L L (Hλ 0 B I l 2 u)(z)(hλ 0 A I l 2 v)(z) u)(z)(hλ 0 A I l 2 z 2 (1 z 2 ) [T T z TT z ]u, v. ((grad Hλ 0 B I l 2 v)(z). u)(z)) ((grad v)(z)) Hλ 0 A I l 2 [ 2 2 z 2 ] z z (1 z 2 ) [T T TT z ]u, v. z 2 2 { z R} K R [ z 2 ] z z (1 z 2 ) [T T TT z ]u, v z [ z 2 ] (1 z 2 ) [T T TT z ] z 2 2 { z R} z z [ 2 2 { z R} z z z 2 (1 z 2 ) [T T z TT z ] log 1 z da(z) log 1 z da(z)u, v ] log 1 z da(z). Because T is a finite sum of finite products of Toeplitz operators the integral is taken over the compact subset { z R} of the unit disk D, K R is an integral operator with kernel in L 2 (D D,dAdA). Thus it is a compact operator on H 2. This gives (23)..

33 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) For any τ > 0, recall that Γ τ (w) is the cone at w truncated at height τ the generalized area integral is given by T B τ (u, v)(w) Γ τ (w) L 2 I l 2 ((grad Hλ 0 B I l 2 u)(z)) ((grad Hλ 0 A I l 2 v)(z)) da(z). Note that T B τ (u, v)(w) is increasing with τ. We define the Stopping time τ(w) by τ(w)sup { τ > 0 : T B τ (u, v)(w) } M 2(s 1)/s 2s 2 a 2 sup T T TT z 1/s [Λ r (u)(w)][λ r (v)(w)]. z >R z Here M 2s 2 is the constant in Theorem 11 a is sufficiently large so that C a 1 2, for the constant C a in Theorem 11. Forz D, let δ(z) 1 z. The distribution function inequality (Theorem 11) gives that for each z D, {w I z : τ(w) 2δ(z)} C a I z. Let E z {w I z : τ(w) 2 I z }. Let χ w (z) be the characteristic function of the truncated cone Γ τ(w) (w). Now,forw E z, write z te iθ note that τ(w) 3 2 (1 z ). We have t e i θ w t e i θ e i θ + e i θ w (1 z ) + (1 z ) τ(w). 2 Therefore, for w E z, we have that z Γ τ(w) (w) that χ w (z) 1onE z. So, χ w (z)dσ(w) E z C a I z C a (1 z ). (24) D

34 34 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Fubini s theorem gives C a z >R L 2 I l 2 ((grad Hλ 0 B I l 2 L 2 I l 2 χ w (z) z >R D ((grad v)(z)) Hλ 0 A I l 2 dσ(w)da(z) D Γ τ(w) (w) ((grad Hλ 0 A I l 2 u)(z)) ((grad Hλ 0 A I l 2 ((grad Hλ 0 B I l 2 u)(z)) (by (24)) L 2 I l 2 ((grad Hλ 0 B I l 2 u)(z)) v)(z)) da(z)dσ(w) v)(z)) (1 z )da(z) D T B τ(w) (u, v)(w)dσ(w) M 2(s 1)/s 2s 2 a 2 sup T T TT z 1/s [Λ r (u)(w)][λ r (v)(w)]dσ(w) D z >R z M 2(s 1)/s 2s 2 a 2 sup T T TT z 1/s [ Λ r (u) 2 ][ Λ r (v) 2 ] z >R z N 2 r 2r M 2(s 1)/s 2s 2 a 2 sup T T TT z 1/s u 2 v 2. z >R z The last inequality follows from that Λ r u 2 N 1 r 2r u 2 since 2 r > 1. Note that log 1 1 z z

35 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) for 1/2 z < 1. Thus we obtain so (22) (23) give W R M 2(s 1)/s 2s 1 Ca 1 N 2 r 2r a 2 sup T T TT z 1/s u 2 v 2, z >R z T T L Il A because P is dense in H 2. Therefore (20) implies K R M 2(s 1)/s 2s 1 Ca 1 N 2 r 2r a 2 sup T T TT z 1/s, z >R z T R 1 T L Il A K R 0. We conclude that T T L Il A is compact. This completes the proof. 7. Two applications In this section we will completely answer Questions 1 2 if X is a finite sum of finite products of Toeplitz operators. First let the operator S A with symbol A L 2 be densely defined on [H 2 ],by S A h P (Ah). For two functions F G, an easy calculation gives H Ḡ F T GH F + H Ḡ S F, (25) S A H G H AG (26) if A is in H 2. For a function f on the unit disk D m M(H + C), wesay if for every net {z α } D converging to m, f(z) 0 z m f(z α) 0. z α m

36 36 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Let T be the Toeplitz algebra, generated by Toeplitz operators with symbols in L. Theorem 4 in [7] implies that there exists a symbol map from T to L, for an operator in T, its symbol is zero if only if the operator is in the commutator ideal of T. The following theorem answers Question 1 for a finite sum of finite products of Toeplitz operators. Theorem 13. Suppose that X is a finite sum of finite products of Toeplitz operators on H 2 b is an inner function. Then T b XT b X is compact if only if for each m M(H + C) with b(m) < 1, z m X T z XT z 0. Theorem 13 implies the following theorem, which gives the answer to Question 2 for a finite sum of finite products of Toeplitz operators. Theorem 14. Suppose that X is a finite sum of finite products of Toeplitz operators on H 2 b is an inner function. Then T b X XT b is compact if only if there are F L an operator X 1 in the commutator ideal of T such that X T F + X 1 for each m M(H + C) with b(m) < 1, z m X 1 T z X 1 T z 0, z m H F k z 2 0. Proof. Assume that X L I l T A. Let M max A. l,j

37 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) Then M<. Theorem 5 implies that for each z D, there is a sequence λ z of complex numbers such that X T L Il A L H λz AI l 2 Hλz BI l 2, (27) max l,j max{ λz A I l 2 z 4, λz B I l 2 z 4 } M 4. for some positive constant M 4. Let F L Il A X 1 X T F. By Theorem 4 in [8], the symbol of X 1 is zero. Suppose that T b X XT b is compact. We need to show that for each m M(H +C) with b(m) < 1, X 1 T z m X 1 T z 0, (28) z Since T b T b I T b T F T b T F, we obtain that z m H F k z 2 0. (29) T b XT b X T b [XT b T b X] is compact hence T b X 1T b X 1 T b [X T F ]T b [X T F ]T b XT b X is also compact. By Theorem 13, for each m M(H + C) with b(m) < 1, X 1 T z m X 1 T z 0. (30) z We obtain (28). To prove (29), first we show that Condition (30) implies that T b X 1 X 1 T b is compact. This result will be also used at the end of this proof.

38 38 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Let Z T b X 1 X 1 T b. Since T b X 1 X 1 T b is a finite sum of finite products of Toeplitz operators, to prove that Z is compact, by Theorem 12 we need only to show that z 1 Z T z ZT z 0. By the Corona theorem, this is equivalent to the requirement that for each m M(H + C), z m Z T z ZT z 0. (31) Since Z T z ZT z X 1 T z X 1 T z + T b [X 1 T z X 1 T z ]T b X 1 T z X 1 T z + T b X 1 T z X 1 T z T b 2 X 1 T z X 1 T z, for each m M(H + C) satisfying b(m) < 1, by (30), we have z m Z T z ZT z 0. So we need only to prove (31) for m M(H + C) satisfying b(m) 1. In this case, b is constant on the support set of m. Thus z m b b(z) 4 k z 2 dσ 0. Making a change of variable gives z m b z b(z) 4 0. By (27), we have X 1 L H λz AI l 2 Hλz BI l 2. (32)

39 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) Let G be either λz A I l 2 or λz B I l 2. Then, T b b(z) VH Gk z 2 U z T b z b(z)vh G z 1 2 T b z b(z)vh G z 1 2 P [(b z b(z))v H G z 1] 2 (b z b(z))v H G z 1] 2 b z b(z) 4 VH G z 1 4 b z b(z) 4 (1 + N 4 ) G z 4 (1 + N 4 )M 4 b z b(z) 4. Here N 4 is the norm of the Hardy projection P on L 4, U z is a unitary operator defined on L 2 by Similarly, we also have Those give For each z D, (32) gives Thus U z h h z k z. T b b(z) VH G k z 2 (1 + N 4 )M 4 b z b(z) 4. z m max{ T b b(z) VH Gk z 2, T b b(z) VH G k z 2 }0. (33) L Z L [ [ T b H λz AI l 2 Hλz BI l 2 T b b(z) H λz AI l 2 Hλz BI l 2 H λz AI l 2 H λz AI l 2 ] Hλz BI l 2 T b ] Hλz BI l 2 T b b(z). Z T z ZT z L { [H λz AI l 2 [T b b(z) H λz AI l 2 Hλz BI l 2 Hλz BI l 2 T b b(z) T z H λz AI l 2 T z T b b(z) H λz AI l 2 Hλz BI l 2 T b b(z) T z ] Hλz BI l 2 T z ] }.

40 40 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 To prove (31) it suffices to show that for each l,j, T b b(z)h z m λz AI l 2 H z m λz AI l 2 Hλz BI l 2 Hλz BI l 2 T z T b b(z) H λz AI l 2 T b b(z) T z H λz AI l 2 Since T b b(z) T z T z T b b(z), by Lemma 9, wehave Hλz BI l 2 T z 0, (34) Hλz BI l 2 T b b(z) T z 0. (35) H λz AI l 2 { Hλz BI l 2 [VH λz AI l 2 [VH λz AI l 2 T b b(z) T z H λz AI l 2 k z ] [VHλz BI l 2 k z ] Hλz BI l 2 T b b(z) T z } T b b(z) k z ] [Tb b(z) VH λz BI l 2 k z ]. Thus (33) implies (35). Using two well-known identities (see, e.g., (1.2) Lemma 5in[16]), T z T b b(z) T b b(z) T z H b b(z) H z H z Vk z k z, by Lemma 9 again, we have T b b(z) H λz AI l 2 Hλz BI l 2 [T b b(z) VHλz AI l 2 [T b b(z) VHλz AI l 2 [(H λz AI l 2 T z T b b(z) H λz AI l 2 k z ] [VHλz BI l 2 Hλz BI l 2 T z k z ]+H b b(z) H z H λz AI l 2 k z ] [VHλz BI l 2 k z ]+[VH b b(z) k z ] Hλz BI l 2 T z ) k z ]. Thus (33) implies (34). Therefore, we conclude z m Z T z ZT z 0. Hλz BI l 2 T z

41 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) Hence by Theorem 12, Z T b X 1 X 1 T b is compact. Noting that T b T F T F T b T b [X X 1 ] [X X 1 ]T b T b X XT b Z we have that T b T F T F T b is compact. Since T b T F T F T b T b T F T Fb H b H F, by the main result in [23] we obtain Because b z b(z) 2 F z F (z) 2 0. z 1 z m b z b(z) 2 z m (1 b(z) 2 ) 1/2 (1 b(m) 2 ) 1/2 > 0, for each m M(H + C) with b(m) < 1, the above it gives Thus we get F z m z F (z) 2 0. H F k z 2 F z m z m z F (z) 2 0, we get (29), as desired. Conversely, suppose that there are F L an operator X 1 in the commutator ideal of T such that X T F + X 1 for each m M(H + C) with b(m) < 1, z m X 1 T z X 1 T z 0, (36) We need to show that T b X XT b is compact. Since H F k z 2 0. (37) z m T b X XT b T b X 1 X 1 T b + T b T F T F T b,

42 42 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 it suffices to show that both T b X 1 X 1 T b T b T F T F T b are compact. In the proof of (29) we have shown that Condition (36)((30)) implies that Z T b X 1 X 1 T b is compact. Also for each m M(H + C) satisfying b(m) 1, z m H b k z 2 z m (1 b(z) 2 ) 1/2 0, hence (37) gives that for each m M(H + C), z m H b k z 2 H F k z 2 0. Thus by the main result in [23] again, T b T F T F T b is compact. This completes the proof. To prove Theorem 13 we need the following lemmas: Lemma 15. Let {g j } be functions in L 2. Suppose that for a fixed z D, {VH gj k z } N are linearly independent. Let A z ( [VH gi k z ], [VH gj k z ] ) N N, B z ( [VH gi k z ], [VH gj bk z ] ) N N. If c is an eigenvalue of the matrix A 1 z B z, then c 1. Proof. Letting (x 1,...,x N ) T have be the eigenvector for the eigenvalue c of A 1 z B z,we x 1 ca z. B z x N x 1. x N. Taking inner product of (x 1,...,x N ) T we obtain with both sides of the above vector equations c VH N x j g j k z 2 VS b H N x j g j k z,vh N x j g j k z T b VH N x j g j k z,vh N x j g j k z.

43 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) The Cauchy Schwarz inequality gives c VH N x j g j k z 2 T b VH N x j g j k z 2. Thus c 1 because T b 1 VH N x j g j k z 2 0. Lemma 16. Suppose that A is a N N matrix with eigenvalues c i 1 for some positive constant M 4, sup f j z p M 4. z D,j If for m M(H + C) with b(m) < 1, VH f1 k z VH f1 bk z. A. 0, VH fn k z VH fn bk z 2 z m then VH f1 k z. 0. VH fn k z 2 z m Proof. By the Jordan theory there is a unitary matrix U such that Let c ε 21 c U AU ε 31 ε 32 c ε N1 ε N2.. ε NN 1 c N VH f 1 k z. VHf N k z VH f1 k z U.. VH fn k z

44 44 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 We get U VH f1 k z VH f1 bk z. U AUU. VH fn k z VH fn bk z c VHf 1 k z ε 21 c ε 31 ε 32 c VHf N k z ε N1 ε..... N2 εnn 1 c N VHf 1 b k z.. VHf N b k z. (38) The first equality in the above vector equation gives Making a change of variable yields VH z m f 1 (1 c 1 b) k z 2 0. (1 P)[ f 1 z m z (1 c 1 b z )] 2 0. Since c 1 1 b is not constant on the support set of m, by Lemma 1, (1 c 1 b) is an outer function on the support set of m. For any ε > 0, there is a function p H such that p(1 c 1 b) 1 2 dμ m < ε. S m For such ε, there is also a neighborhood W of m such that for z W D, p(1 c 1 b) 1 2 dμ m p(1 c 1 b) 1 2 k z 2 dσ < ε. S m S m Making a change of variable we obtain p z (1 c 1 b z ) 1 2 dσ < 2ε. For t 4 3, the Hölder inequality gives (1 P)( f 1 z [p z (1 c 1 b z ) 1]) t

45 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) C t ( f 1 z [p z (1 c 1 b z ) 1]) t C t f 1 z (2t)/(2 t) p z (1 c 1 b z ) 1 2 C t f 1 z 4 p z (1 c 1 b z ) 1 2 C t M 4 ε 1/2. Thus (1 P) f 1 z t C t M 4 ε 1/2 + (1 P)[ f 1 z (p z )(1 c 1 b z )] t The last inequality follows from So Hence we get This implies because C t M 4 ε 1/2 + p (1 P)( f 1 z (1 c t b z )) 2. (1 P)[ f 1 z (p z )(1 c 1 b z )] Hf 1 z (p z ) (1 c 1b z ) S p z Hf 1 (1 c 1 b z ). (by (26)) (1 P) f 1 z m z t C t M 4 ε 1/2. (1 P) f 1 z m z t 0. VH z m f 1 k z 2 (1 P) f 1 z m z 2 0 (1 P) f 1 z 2 (1 P) f 1 z 1/2 M 4 (1 P) f 1 z 1/2 t. t (1 P) f 1 z 4) The second equality in (38) yields VH z m f 2 (1 c 2 b) k z + ε 21 VHf 1 k z 2 0.

46 46 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Hence Repeating the above argument gives By induction we conclude that for all j. Therefore VH z m f 2 (1 c 2 b) k z 2 0. z m VH f 2 k z 2 0. z m VH f j k z 2 0, VH f1 k z. z m U VH fn k z 2 z m Now we are ready to prove Theorem 13. Proof of Theorem 13. Assume that Let X L I l T A. M max A. l,j VHf 1 k z.. VHf N k z 2 0. Theorem 5 implies that for each z D, there is a sequence λ z of complex numbers such that X T L Il A is a finite sum of products of two Hankel operators: max k N H λz f H k λz g, k k1 max{ λz f k 4, λz g k 4 } M 4.

47 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) Let Y T b XT b X. Then Y is also a finite sum of finite products of Toeplitz operators Y N N Hb λz f k H bλz g k H λz f H k λz g. k k1 k1 Suppose that for each m M(H + C) with b(m) < 1, z m X T z XT z 0. In order to prove that Y is compact, by Theorem 12 we need only to show z 1 Y T z YT z 0. This is equivalent to requirement that for each m M(H + C), Because z m Y T z YT z 0. (39) Y T z YT z T b [X T z YT z ]T b [X T z XT z ] T b X T z YT z T b + X T z XT z 2 X T z XT z, for m satisfying b(m) < 1, we get z m Y T z YT z 0. For m satisfying that b(m) 1, b is constant on the support set of m. Thus b z m z b(m) 4 4 z m b(w) b(m) 4 k z (w) 2 dσ(w) 0. Let f be either λz f k or λz g k. Then sup f z 4 M 4. z D

48 48 K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) 1 53 Thus we have H fb k z b(m)h f k z 2 H f T b b(m) k z 2 (1 P)[f z (b z b(m))] 2 f z (b z b(m)) 2 f z 4 (b z b(m)) 4 M 4 (b z b(m)) 4. This implies By Lemma 9, wehave z m H fbk z b(m)h f k z 2 0. (40) ( K ) Y T YT z {[VH z λz f k bk z ] [VH λz g k bk z ] [VH λz f k k z ] [VH λz g k k z ]}. Thus (40) gives k1 k1 N N [VH z m λz f k bk z ] [VH λz g k bk z ] [VH λz f k b(m)k z ] [VH λz g k b(m)k z ] 0. On the other h, we have k1 N [VH λz f k b(m)k z ] [VH λz g k b(m)k z ] k1 b(m) 2 N k1 [VH λz f k k z ] [VH λz g k k z ] N [VH λz f k k z ] [VH λz g k k z ]. k1 This leads to N N [VH z m λ f k bk z ] [VH λz g k bk z ] [VH λz f k k z ] [VH λz g k k z ] 0. k1 k1

49 Therefore we obtain K. Guo, D. Zheng / Journal of Functional Analysis 218 (2005) Y T z m YT z z N N [VH z m λz f k bk z ] [VH λz g k bk z ] [VH λz f k k z ] [VH λz g k k z ] 0. k1 This completes the proof of (39). Conversely suppose that Y is compact. By Theorem 12, we have k1 z 1 Y T z YT z 0. Thus N N [VH λz f k bk z ] [VH λz g k bk z ] [VH λz f k k z ] [VH λz g k k z ] 0. z 1 k1 k1 Note that X T z XT z N [VH λz f k k z ] [VH λz g k k z ]. It suffices to show that for each m M(H + C) with b(m) < 1, z 1 k1 N [VH λz f k k z ] [VH λz g k k z ] 0. k1 Let S z N [VH λz f j k z ] [VH λz g j k z ]. By Lemma 7, we may assume that {VH λz g j k z } N are orthogonal Since N trace(s z Sz ) VH λz f j k z 2 2 VH λz g k j z 2 2. S z 2 trace(s z S z ) N S z 2,