ESSENTIALLY COMMUTING HANKEL AND TOEPLITZ OPERATORS

ESSENTIALLY COMMUTING HANKEL AND TOEPLITZ OPERATORS KUNYU GUO AND DECHAO ZHENG Abstract. We characterize when a Hankel operator a Toeplitz operator have a compact commutator. Let dσ(w) be the normalized Lebesgue measure on the unit circle D. The Hardy space H 2 is the subspace of L 2 ( D, dσ), denoted by L 2, which is spanned by the space of analytic polynomials. So there is an orthogonal projection P from L 2 onto the Hardy space H 2, the so-called Hardy projection. Let f be in L. The Toeplitz operator T f the Hankel operator H f with symbol f are defined by T f h = P (fh), H f h = P (Ufh), for h in H 2. Here U is the unitary operator on L 2 defined by Uh(w) = w h(w). Clearly, H f = H f, where f (w) = f( w). U is a unitary operator which maps H 2 onto [H 2 ] has the following useful property: UP = (1 P )U. These two classes of operators, Hankel operators Toeplitz operators have played an especially prominent role in function theory on the unit circle there are many fascinating problems about those two classes of operators [7], [16], [17], [18] [19]. It is natural to ask about the relationships between these two classes of operators. In this paper we shall concentrate mainly on the following problem: When is the commutator [H g, T f ] = H g T f T f H g of the Hankel operator H g Toeplitz operator T f compact? This problem is motivated by Martinez-Avendaño s recent paper [15] solving the problem of when a Hankel operator commutes with a Toeplitz operator. Martinez- Avendaño showed that H g commutes with T f if only if either g H or there exists a constant λ such that f + λg is in H, both f + f f f are constants. Here f(z) denotes the function f( z). An equvialent statement is : H g T f commute if only if one of the following three conditions is satisfied: (M1). g is in H. (M2). f f are in H. (M3). There exists a nonzero constant λ such that f + λg f + f f f are in H. 0 The first author is partially supported by NNSFC(10171019), Shuguang project in Shanghai, Young teacher Fund of higher school of National Educational Ministry of China. The second author was partially supported by the National Science Foundation. 1

2 KUNYU GUO AND DECHAO ZHENG Note that f is in H whenever f is in H. Clearly, (M2) means that f is constant; (M3) implies that f + f f f are constant since f + f = f + f, f f = f f. One may conjecture that H g T f have a compact commutator if only if Martinez-Avendaño s conditions hold on the boundary of the unit disk in some sense. In Theorem 2 we confirm this conjecture with (M2) replaced by the following condition: (M2 ). f f are in H, (f f)g is in H. To state our main results we will also need results about Douglas algebras. Let H be the subalgebra of L consisting of bounded analytic functions on the unit disk D. A Douglas algebra is, by definition, a closed subalgebra of L that contains H. Let H [f] denote the Douglas algebra generated by the function f in L, H [f, g, h] the Douglas algebra generated by the functions f, g h in L. Theorem 1. The commutator [H g, T f ] = H g T f T f H g of the Hankel operator H g Toeplitz operator T f is compact if only if (0.1) H [g] H [f, f, (f f)g] λ >0 H [f + λg, f + f, f f] H + C. Here H + C denotes the minimal Douglas algebra, i.e., the sum of H the algebra C( D) of continuous functions on the unit circle. This theorem completely solves the problem we mentioned before. In Section 3, we show that (0.1) can be restated as a local condition involving support sets (see Section 3 for the definition). Theorem 2. The commutator [H g, T f ] = H g T f T f H g of the Hankel operator H g Toeplitz operator T f is compact if only if for each support set S, one of the following holds: (1). g S is in H S. (2). f S, f S [(f f)g] S are in H S. (3). There exists a nonzero constant λ S such that [f + λ S g] S, [f + f] S [f f] S are in H S. Theorems 1 2 are applications of the main result in [12], which characterizes when those compact perturbations of Toeplitz operators on the Hardy space that can be written as a finite sum of finite products of Toeplitz operators. Examples exist [2] of some f g such that K = H g T f T f H g is not in the Toeplitz algebra, the C -algebra generated by the bounded Toeplitz operators; see Section 2. Clearly, such a K is not a finite sum of finite products of Toeplitz operators. But we will show that K K is a finite sum of finite products of Toeplitz operators. Our work is inspired by the following beautiful theorem of Axler, Chang Sarason [1] Volberg [21], stated below, on the compactness of the semicommutator T fg T f T g of two Toeplitz operators. if Axler-Chang-Sarason-Volberg Theorem. T fg T f T g is compact if only H [ f] H [g] H + C. One of our motivations is the solution of the compactness of the commutator T f T g T g T f of two Toeplitz operators T f T g in [9]:

HANKEL AND TOEPLITZ OPERATORS 3 Theorem 3. The commutator [T f, T g ] of two Toeplitz operators is compact if only if H [f, g] H [f, g] a + b >0 H [af + bg, af + bg] H + C. Another motivation is the characterization of the compactness of a finite sum of products of two Hankel operators in [11]. 1. Elementary identities In this section we will obtain some identities needed in the proof of Theorem 2. Hankel operators Toeplitz operators are closely related. First we introduce some notation. For each z in the unit disk D, let k z denote the normalized reproducing kernel at z: k z (w) = (1 z 2 ) 1/2 1 zw, φ z be the Möbius transform: φ z (w) = z w 1 zw. Define a unitary operator U z on L 2 by U z f(w) = f(φ z (w))k z (w), for f L 2. Since φ z φ z (w) = w k z φ z k z = 1, we have U z = U z = U 1 z. For each f L 2, we use f + to denote P (f) f to denote (1 P )(f). The operator U z has the following useful properties: Lemma 4. For each z D, (1) U z commutes with P, (2) U z U = UU z. Proof. First we show that U z commutes with P. Let f be in L 2. Thus U z P (f) = f + (φ z )k z, P U z (f) = P (f + (φ z )k z + f (φ z )k z ) = f + (φ z )k z. The last equality follows because f + (φ z )k z is in H 2 is perpendicular to H 2. So we obtain φ z (w)k z (w) = wk z ( w) U z P (f) = P U z (f) for each f L 2. Hence U z commutes with P. Next we turn to the proof of the statement (2). For each f in L 2, an easy calculation gives U z Uf = U z ( w f) = φ z k z f(φz ) = wk z ( w)f(φ z ) = wk z ( w)f(φ z ( w)), UU z f = U(f(φ z )k z ) = wf(φ z ( w))k z ( w).

4 KUNYU GUO AND DECHAO ZHENG This implies U z Uf = UU z f. So we conclude that U z U = UU z, to complete the proof of the lemma. Let x y be two vectors in L 2. Define x y to be the following operator of rank one: for f L 2, (x y)(f) = f, y x. Lemma 5. For fixed z D, H φz = k z k z. Proof. Let {w n } 0 be the basis for H 2. For n 0, This gives for n > 1, Hence we have P U( ww n ) = P ( w n ). H w w n = P U( ww n ) = 0 H w 1 = P U( w) = 1. H w = 1 1. By Lemma 4, we have that U z commutes with P giving U z U = UU z U z H w U z = H φz, since U 2 z = I. On the other h, an easy calculation leads to U z [1 1]U z = [U z 1] [U z 1] = k z k z. This completes the proof. To get the relationship between these two classes of operators, we consider the multiplication operator M φ on L 2 for φ L, defined by M φ h = φh for h L 2. If M φ is expressed as an operator matrix with respect to the decomposition L 2 = H 2 [H 2 ], the result is of the form ( ) Tφ H (1.1) M φ = φu. UH φ UT φu If f g are in L, then M fg = M f M g, therefore (multiply matrices compare upper or lower left corners) (1.2) T fg = T f T g + H f H g (1.3) H fg = T f H g + H f T g. The second equality implies that if f is in H, then (1.4) T f H g = H g T f, for g L. These identities can be found in [3] [17]. They will play an important role be used often in this paper

HANKEL AND TOEPLITZ OPERATORS 5 Lemma 6. Suppose that f g are in L. For each z D, T φz H g T f T φz = H g T f [H g T f k z ] k z + [H g k z ] [T φz H f k z ]. Proof. Since φ z is in H for each z D, (1.4) gives So we obtain T φz H g = H g T φz. T φz H g T f T φz = H g T φz T f T φz = H g T f T φz T φz H g H φz H f T φz. The last equality follows from the consequence of (1.2): T φz T f = T fφz H φz H f = T f T φz H φz H f, since φ z is in H. By (1.2) again, we obtain Lemma 5 implies Therefore we conclude T φz T φz = 1 H φz H φz. H φz = k z k z, H φz = H φ z = k z k z. T φz H g T f T φz = H g T f [H g T f k z ] k z + [H g k z ] [T φz H f k z ]. Lemma 7. Suppose that f g are in L. For each z D, T φz T f H g T φz = T f H g [T f H g k z ] k z [H f k z ] [T φz H g k z ]. Proof. Let z be in D. (1.2) gives T φz T f H g T φz = T f T φz H g T φz + H f H φz H g T φz = T f H g T φz T φz + H f H φz H g T φz. The last equality comes from (1.4). As in the proof of Lemma 6, by Lemma 5 we obtain T φz T f H g T φz = T f H g [T f H g k z ] k z [H f k z ] [T φz H g k z ]. This gives the desired result. The next lemma will be used in the proof of Theorem 2. Lemma 8. Suppose that f g are in L. Let K = H g T f T f H g. Then (1) For each z D, KT φz = T φz K [H g k z ] [H f k z ] [H f k z ] [H g k z ]. (2) Let λ 0 be a constant. For each z D, λkt φz = T φz λk + [H f λg k z ] [H f k z ] [H f k z ] [H f+λgk z ].

6 KUNYU GUO AND DECHAO ZHENG Proof. Since φ z is in H φ z is in H for each z D, by (1.2) (1.4), (1.5) T f T φz = T φz T f + H φz H f, (1.6) T f T φz = T φz T f H f H φz, (1.7) T φz H f = H f T φz. Thus we have KT φz = H g T f T φz T f H g T φz = H g T φz T f + H g H φz H f T f T φz H g = T φz H g T f + H g H φz H f T φz T f H g + H f H φz H g = T φz K + H g H φz H f + H f H φz H g = T φz K [H g k z ] [H f k z ] [H f k z ] [H g k z ]. The second equality comes from (1.5) (1.7). The third equality follows from (1.6) (1.7). The last equality follows from Lemma 5. This completes the proof of (1). To prove (2), for the given constant λ 0, by (1.3), write to obtain H ff = T f H f + H f T f = λ[h g T f T f H g ] + T f H f+λg + H f λg T f, λk = H ff T f H f+λg H f λg T f. Similarly, use of (1.5), (1.6) (1.7) gives λkt φz = T φz λk + [H f λg k z ] [H f k z ] [H f k z ] [H f+λgk z ]. This completes the proof. 2. Compact operators We begin with a necessary condition for a bounded operator to be compact on H 2. The proof of the following lemma is analogous to the proof of Lemma 6.1 in [20]. Lemma 9. If K : H 2 H 2 is a compact operator, then K T KT z 1 φz φz = 0. Proof. Since operators of finite rank are dense in the set of compact operators, given ɛ > 0 there exist f 1,, f n g 1,, g n in H 2 so that n K f i g i < ɛ. i=1 Thus this lemma follows once we prove it for operators of rank one. If f L 2 z 1, then for every w on D we have z φ z (w) = (1 z 2 )w/(1 zw) 0 z φ z (w) = (1 z 2 ) w/(1 z w), so by the Lebesgue Dominated Convergence Theorem, zf φ z f 2 0 zf φ z f 2 0, as z 1. It follows that ζf φ z f 2 0 ζf φ z f 2 0, if z D tends to ζ.

HANKEL AND TOEPLITZ OPERATORS 7 If f H 2, we apply the Hardy projection P to obtain ζf T φz f 2 = ζf P (φ z f) 2 0, ζf T φz f 2 = ζf P ( φ z f) 2 0, as z in D tends to ζ. If f, g H 2, then writing f g T φz (f g)t φz = (ζf) (ζg) (T φz f) (T φz g) we see that (ζf T φz f) (ζg) + (T φz f) (ζg T φz g) ζf T φz f 2 g 2 + f 2 ζg T φz g 2, f g T φz (f g)t φz 0 as z 1. This completes the proof of the lemma. By making use of the above lemma, we obtain a necessary condition for H g T f T f H g to be compact. Lemma 10. Suppose that f g are in L. If H g T f T f H g is compact, then [H g k z ] [H z 1 f k z ] + [H f k z ] [Hg k z ] = 0. Proof. Suppose that H g T f T f H g is compact. Letting K = H g T f T f H g, by Lemma 9, we obtain K T KT z 1 φz φz = 0. On the other h, Lemmas 6 7 give T φz KT φz = [H g T f [H g T f k z ] k z + [H g k z ] [T φz H f k z ]] [T f H g [T f H g k z ] k z [H f k z ] [T φz H g k z ]] = K [Kk z ] k z + [H g k z ] [T φz H f k z ] + [H f k z ] [T φz H g k z ]. Noting that k z converges weakly to zero as z 1, we have giving Since we conclude that Kk z 0 z 1 [H gk z ] [T φz H f k z ] + [H f k z ] [T φz H g k z ] = 0. [H g k z ] [H f k z ] + [H f k z ] [H g k z ] = [[H g k z ] [T φz H f k z ] + [H f k z ] [T φz H g k z ]]T φz [[H g k z ] [T φz H f k z ] + [H f k z ] [T φz H g k z ]] T φz, [H g k z ] [H z 1 f k z ] + [H f k z ] [Hg k z ] = 0. The next lemma gives a close relationship between H f H f. Lemma 11. Suppose that f is in L. For each z D, H f k z 2 = H f k z 2. The above lemma is the special case of the following lemma with g = k z. We thank the referee for suggesting the following lemma.

8 KUNYU GUO AND DECHAO ZHENG Lemma 12. Let H f be a bounded Hankel operator let g H 2. Then H f g = (H f g) thus H f g = H f g. Proof. Notice that for all g L 2, (Ug) = Ug P g = (P g). Therefore, H f g = H f g = P U(f g ) = P (Ufg) = (P Ufg) = (H f g). Since g = g for all g L 2, we have H f g = (H f g) = H f g, to complete the proof. To prove the sufficiency part of Theorem 2 we need the following result [12] which characterizes when an operator is the compact perturbation of a Toeplitz operator if the operator is a finite sum of finite products of Toeplitz operators. Theorem 13. A finite sum T of finite products of Toeplitz operators is a compact perturbation of a Toeplitz operator if only if (2.1) z 1 T T φ z T T φz = 0. Theorem 13 is a variant of Theorem 4 in the paper [10] of C. Gu. However, some crucial details are omitted from the proof in [10], especially details in the proof of a key distribution function inequality. An alternative proof of Theorem 13 can be found in the authors paper [12]. Theorem 13 can not be applied to H g T f T f H g directly since H g T f T f H g may not be a finite sum of finite products of Toeplitz operators. The following example shows that there are f g in L such that H g T f T f H g is not a finite sum of finite products of Toeplitz operators. Example: Let {a n } be a Blaschke sequence in the unit disk such that a n = 1, n 1 a n 1 a n 2n. Let b be the Blaschke product associated with the sequence. Let f be the function constructed in [2] with the following properties (A) f is in QC(= [H + C] [H + C]). (B) f = f. (C) f(a n ) 1. Let g = b, K = H g T f T f H g. It was shown that K is not compact in [2]. By making use of Theorem 2, we will show that K is not in the Toeplitz algebra. Suppose that K is in the Toeplitz algebra. We will derive a contradiction. By (1.3), we see K = H (f f)g + H f T g T g H f. It is well known that both H f H f are compact. Letting we have that O is compact O = H f T g T g H f, K = H (f f)g + O. By a lemma in [2], which states that if a bounded operator K on H 2 is in the Toeplitz algebra, then KT f T f K is compact for every function f QC, we have that H (f f)g T f T f H (f f)g is compact. Let m be in the closure of {a n } in the

HANKEL AND TOEPLITZ OPERATORS 9 maximal ideal space M(H ) of H. Let S be the support set of m. Noting that g S = b S is not in H S, [(f f)g] S = 2 b S, we have that for any nonzero constant λ [f + λ(f f)g] S = (1 + 2λ b) S is not in H S. By Theorem 2, we see that only Condition (2) in Theorem 2 may hold. That is, (f f) 2 g S H S. But (B) (C) imply that f f = 2f f S = 1. This leads to 4g S H S, which is a contradiction. The following lemma shows that K K is a finite sum of finite products of Toeplitz operators. Lemma 14. Suppose that f g are in L. Let K = H g T f T f H g. Then K K is a finite sum of finite products of Toeplitz operators. Proof. Letting K = H g T f T f H g, by (1.3) we write K as K = H fg + H g T f + H f T g. Taking adjoint both sides of the above equality gives The last equality follows from where f (w) = f(w). This leads to K = H ( fg) + T g H f + T f H g H f = H f, K K = H ( fg) H fg H ( fg) [H g T f + H f T g ] (2.2) [T g H f + T f H g ]H fg + [T g H f + T f H g ][H g T f + H f T g ]. The first term in the right h side of (2.2) is a semicommutator of two Toeplitz operators since for two functions φ ψ in L, by (1.3) H φ H ψ = T φψ T φ T ψ ; both the second the third terms are products of a Toeplitz operator a semicommutator of two Toeplitz operators; the fourth term is the product of two Toeplitz operators a semicommutator of two Toeplitz operators. Therefore (2.2) gives that K K is a finite sum of finite products of Toeplitz operators. This completes the proof of the lemma. We thank the referee for pointing out that any product of Hankel Toeplitz operators that has an even number of Hankel operators is a finite sum of finite products of Toeplitz operators. A symbol mapping was defined on the Toeplitz algebra in [7]. It was extended to a -homomorphism on the Hankel algebra in [3]. One of the important properties of the symbol mapping is that the symbols of both compact operators Hankel operators are zero ([7], [3]). Note K is in the Hankel algebra equals H g T f T f H g. Clearly, the symbol of K is zero. By Theorem 13 we see that K is compact if only if z 1 K K Tφ z K KT φz = 0.

10 KUNYU GUO AND DECHAO ZHENG 3. Proof of main results To prove Theorems 1 2 we need some notation. The Gelf space (space of nonzero multiplicative linear functionals) of the Douglas algebra B will be denoted by M(B). If B is a Douglas algebra, 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 the function f 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 + C). A subset 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). For a function F on the unit disk D m M(H + C), we say F (z) = 0 if for every net {z α } D converging to m, F (z α) = 0. z α m The following lemma in [9] (Lemma 2.5) will be used several times later. Lemma 15. Let f be in L m M(H + C), let S be the support set for m. Then f S H S if only if H f k z 2 = 0. Clearly, Theorem 1 follows from Theorem 2 the following lemma. Lemma 16. Let f, g L. Then (3.1) H [g] H [f, f, (f f)g] λ >0 H [f + λg, f + f, f f] H + C if only if for each support set S one of the following holds: (1). g S is in H S. (2). f S, f S [(f f)g] S are in H S. (3). There exists nonzero constant λ S, such that [f + λ S g] S is in H both [f + f] S [f f] S are in H S. Proof. Without loss of generality we may assume that f < 1/4 g < 1/4. Let A denote the Douglas algebra H [g] H [f, f, (f f)g] λ >0 H [f + λg, f + f, f f]. By the Sarason Theorem (Lemma 1.3 in [9]), we get that M(A) equals M(H [g]) M(H [f, f, (f f)g]) λ >0 M(H [f + λg, f + f, f f]). Suppose that (3.1) holds. Then A H + C, so M(H + C) M(A). Let m M(H + C). Then m is an element of M(H [g]) M(H [f, f, (f f)g]) λ >0 M(H [f + λg, f + f, f f]). If m is in either of the first two sets, Lemma 1.5 in [9] gives that either Condition (1) or Condition (2) holds. Thus, we may assume that m λ >0 M(H [f + λg, f + f, f f]). Note λ >0 H [f + λg, f + f, f f] =

HANKEL AND TOEPLITZ OPERATORS 11 1 λ >0 H [f + λg, f + f, f f] 1 λ >0 H [λf + g, f + f, f f], since (f + λg) = λ( f λ + g). Thus λ >0 M(H [f + λg, f + f, f f]) = 1 λ >0 M(H [f + λg, f + f, f f]) 1 λ >0 M(H [λf + g, f + f, f f]). So m must be either in or in 1 λ >0 M(H [f + λg, f + f, f f]) 1 λ >0 M(H [λf + g, f + f, f f]). Now we only consider the case that m is in 1 λ >0 M(H [f + λg, f + f, f f]). If m is in the second set, use the same argument that we will use below. We shall show that m M(H [f + λg, f + f, f f]) for some λ with λ 1. By Lemma 15 Lemma 1.5 in [9], it suffices to show that for some λ with λ 1, H f+λgk z 2 = 0, H f+ f k z 2 = 0, H f f k z 2 = 0. We only prove the first it; the second third its follow by the same argument. Hence there exist constants λ α points m α M(H [f + λ α g, f + f, f f]) such that m α m. We may assume that λ α λ, for some complex number λ. Clearly, λ 1. Note that since m α M(H [f + λ α g, f + f, f f]), Since α H f+λαgk z 2 = 0. dist L (f + λg, H ) f + λg < 1/2, as a consequence of the Adamian-Arov-Krein Theorem [8], [16], there exists a unimodular function u λ in f + λg + H. Lemma 2 [22] gives (3.2) H f+λg k z 2 (1 u λ (z) 2 ) 1/2 3 H f+λg k z 2, where u λ (z) denotes the value of the harmonic extension of u λ at z. Note H f+λg k z 2 H f+λαgk z 2 + λ λ α. Thus we have (3.2) gives sup H f+λg k z 2 sup H f+λαgk z 2 + λ λ α = λ λ α. α α sup α (1 u λ (z) 2 ) 1/2 3 sup α H f+λg k z 2 3 λ λ α.

12 KUNYU GUO AND DECHAO ZHENG Since u λ (m) is continuous on M(H ) [13], we have (1 u λ (m α ) 2 ) 1/2 3 λ λ α. Taking the it on both sides of the above inequality gives (1 u λ (m) ) 1/2 = sup(1 u λ (m α ) 2 ) 1/2 sup 3 λ λ α = 0. m α m m α m We obtain (1 u λ (m) ) 1/2 = 0. On the other h, (3.2) gives sup H f+λg k z 2 sup(1 u λ (z) ) 1/2 = (1 u λ (m) ) 1/2 = 0. This gives the desired result. Conversely, let S be the support set for an element m M(H +C) suppose that one of Conditions (1), (2) (3) holds for m. Then by Lemma 1.5 in [9], either m M(H [g]) or m M(H [f, f, (f f)g]), or there exists a nonzero constant λ, such that m M(H [f + λg, f + f, f f]). Thus m is in M(H [g] H [f, f, (f f)g] λ >0 H [f + λg, f + f, f f]). Therefore, M(H + C) M(A). By the Chang-Marshall Theorem ([6], [14]) A H + C. The proof of Lemma 16 is completed Let BMO be the space of functions with bounded mean oscillation on the unit circle. If f is in BMO analytic or co-analytic on D, the norm f BMO is equivalent to f(0) + sup f φ z f(z) p z D for p 1. It is well known that the Hardy projection P maps L into BMO ([8] [18]). Lemma 17. Suppose that f g are in L. If then H gk z 2 = 0, H gt f k z 2 = 0. Proof. Write g = g + + g where g + = P (g) g = (1 P )(g). Since U z commutes with the Hardy projection P we get H g k z = H g k z = H g U z 1 = U z H g φ z 1. Thus we have H g k z 2 = U z H g φ z 1 2 = H g φ z 1 2. The last equality follows because U z is a unitary operator on L 2. An easy calculation gives H g φ z 1 = U(g φ z g (z)). Therefore z 1 g φ z g (z) 2 = 0.

HANKEL AND TOEPLITZ OPERATORS 13 Similarly we can also get H g T f k z 2 = U z H g φ z T f φz 1 2 = H g φ z T f φz 1 2 = H g φ z (f + φ z + f (z)) 2 = (1 P )(g φ z g (z))(f + φ z + f (z)) 2. The first equality holds because U z commutes with P the second equality holds because U z is a unitary operator on L 2. The third equality follows from the decomposition of f: The Hölder inequality gives To prove that we need only to show that we have f = f + + f. H g T f k z 2 g φ z g (z) 4 f + φ z + f (z) 4. H gt f k z 2 = 0 f + φ z + f (z) 4 C f, g φ z g (z) 4 = 0. f + φ z + f (z) = f + φ z f + (z) + f(z), f + φ z + f (z) 4 f + φ z f + (z) 4 + f C 1 P (f) BMO + f C f, for some positive constants C C 1. The last inequality follows because P is bounded from L to BMO. The Hölder inequality gives g φ z g (z) 4 g φ z g (z) 1/4 2 g φ z g (z) 3/4 6 C g φ z g (z) 1/4 2 g 1/4 BMO C g φ z g (z) 1/4 2 g 1/4. The last inequality also follows because P is bounded from L to BMO. This gives g φ z g (z) 4 = 0, to complete the proof of the lemma. Combining Lemmas 11 17 we have the following lemma needed in the proof of Theorem 2. Lemma 18. Suppose that f g are in L. If then H g k z 2 = 0, H g T f k z 2 = 0.

14 KUNYU GUO AND DECHAO ZHENG Now we are ready to prove Theorem 2. Proof of Theorem 2 First we prove the necessity part of Theorem 2. Suppose that H g T f T f H g is compact. Without loss of generality we may assume that f < 1/2 g < 1/2. By Lemma 10 we get (3.3) z 1 [H gk z ] [H f k z ] + [H f k z ] [H g k z ] = 0. Let m be in M(H + C), let S be the support set of m. By Carleson s Corona Theorem [5], there is a net z converging to m. Suppose that H g k z 2 = 0. By Lemma 15 we have that g S is in H S. So Condition (1) holds. Suppose that there is a constant c such that H g k z 2 c > 0. Let λ z = H f k z, H g k z / H g k z 2. Then λ z 1 c, so we may assume that λ z c m for some constant c m. Applying the operator [[H g k z ] [Hf k z] + [H f k z ] [Hg k z ]] to H g k z multiplying by 1 H gk z we get 2 2 Thus Lemma 11 implies H f k z + λ z [Hg k z ] 2 = 0. H f+c mgk z 2 = 0. (3.4) H f+cmgk z 2 = 0. Now we consider two cases. Case 1. c m = 0. In this case, we have H f k z 2 = 0. (3.3) gives H f k z 2 = 0. Thus by Lemma 15, we obtain f S H S, f S H S. On the other h, by making use of (1.3) twice we have K = H (f f)g T g H f + H f T g. Since K is compact k z converges to 0 weakly as z m, we have Lemma 17 gives [H (f f)g T gh f + H f T g ]k z 2 = 0. [ T gh f + H f T g ]k z 2 = 0.

HANKEL AND TOEPLITZ OPERATORS 15 Thus H (f f)g k z 2 = 0. By Lemma 15, we have [(f f)g] S H S. to get Condition (2). Case 2. c m 0. In this case, by Lemma 15 (3.4), we obtain Now write Because we get Note that (f + c m g) S H S. [H g k z ] [H f k z ] + [H f k z ] [H g k z ] = [H g k z ] [H f+c mgk z ] + [H f cmg k z] [H g k z ]. [H gk z ] [Hf k z ] + [H f k z ] [Hg k z ] = 0, [H f+c mgk z ] 2 = 0, [H f cmg k z] [Hg k z ] = 0. [H g k z ] 2 = H g k z 2, [H f cmg k z] [H g k z ] = [H f cmg k z] 2 [H g k z ] 2. We have [H f cmg k z] 2 = 0. Combining (3.4) with the above it gives since Therefore by Lemma 15, H f+f k z 2 = 0 H f+f k z = H f cmg k z + H f+cmgk z. (f + f) S H S. To prove that (f f) S H S, by (1.3), write H ff = T f H f + H f T f (3.5) = c m [H g T f T f H g ] + T f H f+cmg + H f cmg T f By Lemma 17, we obtain H f cmg T f k z 2 = 0. Apply the bounded operator T f to H f+cmg T f k z, to get T f H f+cmgk z 2 = 0. Since [H g T f T f H g ] is compact k z weakly converges to zero as z m, we have [H gt f T f H g ]k z 2 = 0.

16 KUNYU GUO AND DECHAO ZHENG Therefore (3.5) implies Lemma 15 gives H f f k z 2 = 0. (f f) S H S. So Condition (3) holds. This completes the proof of the necessity part. Next we shall prove the sufficiency part of Theorem 2. Suppose that f g satisfy one of Conditions (1)-(3) in Theorem 2. Let K = H g T f T f H g T = K K. By Lemma 14, T is a finite sum of finite products of Toeplitz operators with zero symbol. By Theorem 13, we need only to show z 1 T T φ z T T φz = 0. By the Carleson Corona Theorem, the above condition is equivalent to the condition that for each m M(H + C), (3.6) T T φ z T T φz = 0. Let m be in M(H + C), let S be the support set of m. By Carleson s Corona Theorem, there is a net z converging to m. Suppose that Condition (1) holds, i.e., g S H S. Lemma 15 gives that By Lemma 11, we have Let the above its give This gives H gk z 2 = 0. H g k z 2 = 0. F z = [H g k z ] [H f k z ] [H f k z ] [H g k z ], F z = 0. (3.7) [K T φz ]F z + Fz [T φz K] + Fz F z = 0, since K <, sup F z <. z D By Lemma 8 we also have to get KT φz = T φz K + F z, Tφ z T T φz = [KT φz ] [KT φz ] = K T φz T φz K + [K T φz ]F z + Fz [T φz K] + Fz F z = K K + [K k z ] [K k z ] + [K T φz ]F z + F z [T φz K] + F z F z. The last equality comes from (3.8) T φz T φz = 1 k z k z. Lemma 18 gives Therefore (3.7) implies (3.6). K k z 2 = 0.

HANKEL AND TOEPLITZ OPERATORS 17 Suppose that Condition (2) holds. By Lemma 15, we have H f k z 2 = 0, H f k z 2 = 0, (3.9) H (f f)g k z 2 = 0. Let F z = [H g k z ] [H f k z ] [H f k z ] [H g k z ]; the above its give This gives F z = 0. (3.10) [K T φz ]F z + F z [T φz K] + F z F z = 0. By Lemma 8 we have KT φz = T φz K + F z, to get T φ z T T φz = [KT φz ] [KT φz ] = K T φz T φz K + [K T φz ]F z + F z [T φz K] + F z F z = K K [K k z ] [K k z ] + [K T φz ]F z + F z [T φz K] + F z F z. The last equality comes from (3.8). By making use of (1.3) twice, we have K = H (f f)g T g H f + H f T g. Lemma 18 (3.9) give K k z 2 = 0. Therefore (3.10) implies (3.6). Suppose that Condition (3) holds. Then for some constant c m 0, (3.11) H f cmgk z 2 = 0, (3.12) H f+ f k z 2 = 0, (3.13) H f f k z 2 = 0. These give (3.14) H f+cmg k z 2 = 0, since f + c m g = f + f (f c m g). By (3.5) Lemma 18, we have (3.15) K k z = 0. Lemma 8 gives (3.16) c m KT φz = T φz c m K [H f cmg k z] [H f k z ] + [H f k z ] [H f+c mgk z ]. Letting G z = [H f cmg k z] [H f k z ] [H f k z ] [H f+c mgk z ],

18 KUNYU GUO AND DECHAO ZHENG we have c m KT φz = T φz c m K + G z G z = 0. The last it follows from (3.11) (3.14) gives (3.17) [T φz c m K] G z + G zt φz c m K + G zg z = 0. (3.16) gives [c m KT φz ] [c m KT φz ] = [T φz c m K + G z ] [T φz c m K + G z ] = [T φz c m K] [T φz c m K] + [T φz c m K] G z + G zt φz c m K + G zg z = c m 2 K T φz T φz K + [T φz c m K] G z + G zt φz c m K + G zg z = c m 2 K K c m 2 [K k z ] [K k z ] + [T φz c m K] G z + G zt φz c m K + G zg z. The last equality comes from (3.8). (3.15) implies that the second term on the right h side of the above equality converges to zero (3.17) implies that the third, fourth fifth terms converge to zero. Thus we conclude c m 2 T c m 2 Tφ z T T φz = 0. Since c m 0, the above it gives (3.6). This completes the proof of Theorem 2. ACKNOWLEDGMENTS The authors would like to thank D. Sarason the referee for their useful suggestions. References [1] S. Axler, S.-Y. A. Chang, D. Sarason, Product of Toeplitz operators, Integral Equations Operator Theory 1 (1978), 285-309. [2] J. Barría, On Hankel operators not in the Toeplitz algebra, Proc. Amer. Math. Soc. 124 (1996), 1507-1511. [3] J. Barría P. Halmos, Asymptotic Toeplitz operators, Trans. Amer. Math. Soc. 273 (1982), 621-630. [4] A. Böttcher B. Silbermann, Analysis of Toeplitz operators, Springer-Verlag, 1990. [5] L. Carleson, Interpolations by bounded analytic functions the corona problem, Ann. of Math. 76 (1962), 547-559. [6] S.-Y. A. Chang, A characterization of Douglas subalgebras, Acta Math. 137 (1976), 81-89. [7] R. G. Douglas, Banach algebra techniques in the operator theory, Academic Press, New York London, 1972. [8] J. B. Garnett, Bounded Analytic Functions, Academic Press, New York, 1981. [9] P. Gorkin D. Zheng, Essentially commuting Toeplitz operators, Pacific J. Math., 190 (1999), 87-109. [10] C. Gu, Products of several Toeplitz operators, J. Functional analysis 171(2000), 483-527. [11] C. Gu D. Zheng, Products of block Toeplitz operators, Pacific J. Math. 185 (1998), 115-148. [12] K. Guo D. Zheng, The distribution function inequality block Toeplitz operators, preprint. [13] K. Hoffman, Bounded analytic functions Gleason parts, Ann. of Math. 86 (1967), 74-111. [14] D. E. Marshall, Subalgebras of L containing H, Acta Math. 137(1976), 91-98. [15] R. Martnez-Avendao, When do Toeplitz Hankel operators commute? Integral Equations Operator Theory 37 (2000), 341-349. [16] N. K. Nikolskii, Treatise on the shift operator, Speinger-Verlag, NY etc. 1985. [17] S. Power, Hankel operators on Hilbert space, Pittman Publishing, Boston, 1982.

HANKEL AND TOEPLITZ OPERATORS 19 [18] D. Sarason, Function theory on the unit circle, Virginia Polytechnic Institute State University, Blacksburg, VA, 1979. [19] D. Sarason, Holomorphic spaces: a brief selective survey, in Holomorphic spaces (Berkeley, CA, 1995), 1-34, Math. Sci. Res. Inst. Publ. 33, Cambridge Univ. Press, Cambridge, 1998. [20] K. Stroethoff D. Zheng, Products of Hankel Toeplitz operator on the Bergman space, J. Funct. Anal. 169 (1999), 289-313. [21] A. Volberg, Two remarks concerning the theorem of S. Axler, S.-Y. A. Chang, D. Sarason, J. Operator Theory 8 (1982), 209-218. [22] R. Younis D. Zheng, A distance formula Bourgain algebras, Math. Proc. Cambridge Philos. Soc. 120 (1996), 631-641. [23] D. Zheng, The distribution function inequality products of Toeplitz operators Hankel operators, J. Functional Analysis 138 (1996), 477-501. Department of Mathematics, Fudan University, Shanghai, 200433, The People s Republic of China. E-mail address: kyguo@fudan.edu.cn Department of Mathematics, Verbilt University, Nashville, Tennessee 37240, USA E-mail address: zheng@math.verbilt.edu