COMPACT PERTURBATIONS OF HANKEL OPERATORS

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1 COMPACT PERTURBATIONS OF HANKEL OPERATORS XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG Abstract. We give some necessary sufficient conditions of when the product of two Hankel operators is a compact perturbation of a Hankel operator on the Hardy space. For f in L, the Hankel operator with symbol f is the operator H f on the Hardy space H 2 of the unit circle, defined by H f h = P (Ufh), for h in H 2. Here P is the orthogonal projection from L 2 onto H 2 U is a unitary operator on L 2 defined by Uh(w) = wh( w). There are many fascinating problems about the Hankel operator [30], [32]. In this paper we will concentrate on the problem: Main Problem. For what symbols f, g is the product H f H g of two Hankel operators a compact perturbation of a Hankel operator? There are many motivations for us to study the above problem. On one h, the problem involves another important class of operators, Toeplitz operators. The Toeplitz operator induced by the function f in L is the operator T f on H 2 defined by T f h = P (fh). Hankel operators Toeplitz operators are closedly related by the important fact that the product H f H g of two Hankel operators equals the semicommutator T fg T f T g of two Toeplitz operators. Here f(w) = f( w). The main problem is more general than inspired by the problem about semicommutator: For what symbols f, g is the product T f T g of two Toeplitz operators a compact perturbation of a Toeplitz operator? If T f T g is a compact perturbation of the Toeplitz operator T h, the Douglas symbol mapping [12] gives that h must equal fg. Thus the semicommutator T fg T f T g is compact. The above problem is equivalent to the problem of when the semicommutator is compact, which arose in the Fredholm theory of Toeplitz operators [12], [28], [34]. Fortunately, the semicommutator problem was solved by the combined efforts by Axler, Chang, Sarason Volberg [2], [37]. They proved the following beautiful result: T fg T f T g is compact if only if for each support set S, either f S or g S is in H S. On the other h, another motivation is the problem of when the product of two Hankel operators equals another Hankel operator. It was shown in [31], [39] that the product of two Hankel operators is rarely a Hankel 2000 Mathematics Subject Classification. Primary 47B35, 47B47. 1

2 2 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG operator, namely, it is if only if both operators are scalar multiples of H φλ for some Blaschke factor φ λ = z λ a number λ in the unit disk D. 1 λz The product is then also a scalar multiple of H φλ. From the result mentioned above the Axler, Chang, Sarason Volberg theorem one may guess that the product of two Hankel operators is a compact perturbation of a Hankel operator if only if the product is compact. Unfortunately, in Section 3, we will show that there are products of two Hankel operators which are compact perturbations of a noncompact Hankel operator. So the main problem turns out to be quite subtle. Another motivation is the problem when a Hankel operator is in the Toeplitz algebra, the C -algebra generated by bounded Toeplitz operators [5], [6]. The fact that the square of every Hankel operators lies in the Toeplitz C -algebra suggests that perhaps the Hankel operators themselves belong. This is the case for positive Hankel operators since they are the unique roots of their squares. So the Hankel operator associated with the Hilbert matrix is in the Toeplitz algebra [6]. But it is not so in general. Axler [31] first observed that it is necessary H φ essentially commutes with the unilateral shift, i.e., H φ T z T z H φ is compact if H φ is in the Toeplitz algebra. Barría Halmos [6] asked a natural question whether a Hankel operator is in the Toeplitz algebra if it essentially commutes with the unilateral shift. X. Chen F. Chen [10] first proved that there are Hankel operators, which essentially commute with the unilateral shift but are not in the Toeplitz algebra. Later such concrete examples of Hankel operators are constructed in [5] [11]. In Section 4 we will present a concrete example a short proof of the fact. In Section 3, we will obtain examples that noncompact Hankel operators are even compact perturbations of a product of two Hankel operators by thin Blaschke products. These examples are inspired by the Volberg solution on Nikolskii s conjectures on bases consisting of rational fractions [37]. For a complex number z = x + iy, let us denote by R(z) I(z), respectively, the real part x the imaginary part y of the complex number z. The following theorem is our main result. Theorem 1. Suppose that B 1 B 2 are two inner functions. H B1 H B2 is a compact perturbation of a Hankel operator if only if for each support set S m, one of the following holds: (1) Either B 1 Sm or B 2 Sm is constant. (2) m is a thin part in the fibre M 1 (H ) with the following properties: (2a) m is in the closure of a sequence {z n } in D satisfying 1 z n 1 z n 2 < M for n. Here M is a positive constant; (2b) B 1 Sm = cb 2 Sm B 2 φ m (λ) = ξλ for some unimodular constants c ξ;

3 COMPACT PERTURBATIONS OF HANKEL OPERATORS 3 (2c) If m is in the closure of some sequence {w n } D, then ρ(r(w α ), w α ) 0 whenever the subnet {w α } converges to m. (3) m is a thin part in the fibre M 1 (H ) with the following properties: (3a) m is in the closure of a sequence {z n } in D satisfying 1 + z n 1 z n 2 < M for n. Here M is a positive constant; (3b) B 1 Sm = cb 2 Sm B 2 φ m (λ) = ξλ for some unimodular constants c ξ; (3c) If m is in the closure of some sequence {w n } D, then ρ(r(w α ), w α ) 0 whenever the subnet {w α } converges to m. Fibres M 1 (H ) M 1 (H ) play the privilege roles in the above theorem since 1 1 are the fixed points of the reflection map z z the map is used in the definition of the Hankel operator. Some notation in the above theorem will be introduced in Section 1. The proof of Theorem 1 is long so it is divided into two parts, in Sections 5 6. We will discuss Theorem 1 in Section 4. Many ideas in [2], [18], [23], [22], [37] [40] are useful for us to study the main problem. Two important properties of thin Blaschke sequences will play an important role in this paper: (1) Sundberg Wolff ([36]) proved that a sequence is thin interpolating if only if it is an interpolating sequence for QA = H V MO, where V MO is the space of functions on the unit circle with vanishing mean oscillation; (2) Volberg [37] proved that {z n } is a thin interpolating sequence if only if {k zn } is a U + K basis where 1 zn k zn is the normalized reproducing kernel 2 1 z nz. 1. Some Notation Some notation is needed. The unit disk will be denoted by D the unit circle by D. We shall regard functions in L 2 as extended harmonically into D by means of Poisson s formula: g(z) = D g(e iθ ) 1 z 2 1 ze iθ 2 dσ(θ), for z D. Thus the Poisson integral gives that for each z D, g g(z) is bounded linear functional on H 2. So there is a function K z (w) in H 2 such that g(z) = g, K z.

4 4 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG K z (w) is called the reproducing kernel of H 2 at z. We use k z to denote the 1 z 2 normalized reproducing kernel at z. In fact, k z (w) = 1 zw. By H we denote the usual Hardy space on D of boundary functions for bounded holomorphic functions in D. The space of continuous functions on D will be denoted by C. The algebra QC is defined by QC = (H + C) (H + C). By QA we denote QC H. Let B be a commutative Banach algebra. The Gelf space (space of nonzero multiplicative linear functionals) of the algebra B will be denoted by M(B). If we think of M(H ) as a subset of the dual of H with the weak-star topology, then M(H ) becomes a compact Hausdorff space. Explicitly, a net {φ α } in M(H ) converges to φ in M(H ) if only if φ α (f) φ(f) for every f H. If z is a point in the unit disk D, then the point evaluation at z is a multiplicative linear functional on H, so we can think of z as an element of M(H ). Carleson s Corona Theorem [9] implies that the unit disk D is a dense subset of M(H ). The maximal ideal space M(H ) of H is unraveled by interpolating sequences their Blaschke products. An interpolating sequence is a sequence {z n } in D such that for every bounded sequence {c n } of complex numbers, there is a function f H such that f(z n ) = c n for every positive integer n. Carleson [8] proved that a sequence {z n } in D is interpolating if only if inf z n z m n 1 z m z n > 0. m=1 m n For a sequence {z n } n in D with n=1 (1 z n ) <, there corresponds a Blaschke product b(z) = n=1 z n z z n z n 1 z n z, z D. Blaschke products play an important role in the study of H. A sequence {z n } n an associated Blaschke product are called thin if lim z n z k = 1. n 1 z k z n k n If b is a thin Blaschke product with zeros {z n } n, then b(z j ) 1 for every sequence {z j } j in D satisfying ρ(z j, {z n } n ) 1 as j. Two important properties of thin Blaschke sequences will play an important role in the paper: (1) Sundberg Wolff ([36]) proved that a sequence {z n } is thin interpolating if only if it is an interpolating sequence for QA, i.e, for each

5 COMPACT PERTURBATIONS OF HANKEL OPERATORS 5 sequence {w n } l, there is a function h QA such that h(z n ) = w n. (2) Volberg [37] proved that {z n } is a thin interpolating sequence if only if {k zn } is a U + K basis, i.e., {k zn } is near an orthogonal basis in the following sense: k zn = (V + K)e n, where {e n } is the stard orthogonal basis of l 2, V is a unitary operator K is a compact operator. A Douglas algebra is, by definition, a closed subalgebra of L which contains H. 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 each m in the maximal ideal space M(H + C), we use S m to denote the support set for m. The fiber of M(H ) above the point λ of D is the set {x M(H ) : x(z) = λ} will be denoted by M λ (H ). The pseudohyperbolic distance between two points m 1 m 2 in M(H ) is given by ρ(m 1, m 2 ) = sup{ f(m 2 ) : f H, f 1, f(m 1 ) = 0}. The Gleason part of a point m 1 M(H ), denoted by P (m 1 ) is given by P (m 1 ) = {m : ρ(m, m 1 ) < 1}. It is well known that each Gleason part of M(H ) is either one point or an analytic disc. When the Gleason part of m consists of one point, m is said to be a trivial point. Otherwise m is a nontrivial point. A continuous mapping F : D M(H ) is analytic if f F is analytic on D whenever f H. An analytic disk in M(H ) is the image F (D) where F is a one-to-one analytic map from D to M(H ). A theorem from the general theory of logmodular algebras implies that each Gleason part of M(H ) is either a one-point part or an analytic disk [25]. For each nontrivial point m, Hoffman [26] constructed a canonical map φ m of the disk D onto the part P (m). This map is defined by taking a net {z α } D such that z α m, defining (φ m (z))(f) = lim f( z α z α 1 z α z ) for z D f H. The above limit exists is independent of the net {z α }, provided that z α m. Hoffman [26] showed the following remarkable properties of φ m analytic disks: (H0) Let b be an interpolating Blaschke product with the zero sequence {z n } in D. Then m is in Z H +C(b) if only if m lies in the closure {z n }.

6 6 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG Here Z H +C(b) denotes the zero set {m M(H + C) : b(m) = 0} of b in M(H + C). (H1) For m in M(H ) \ D, P (m) is an analytic disk if only if there is an interpolating sequence whose closure contains m. (H2) If z α m, then for each bounded harmonic function h on D, h φ zα (z) h φ m (z) uniformly on each compact subset of D where φ zα (z) = zα z 1 z. αz Recall some notation facts about abstract H p theory on a support set. Let m be in M(H + C) let dµ m denote the unique representing measure for m with support S. That is, (i) m(fg) = S fgdµ m = S fdµ m S gdµ m for all f, g H. (ii) If h is an a.e. nonnegative function in L 1 (dµ m ) such that S fhdµ m = S fdµ m for all f H, then h = 1 a.e. dµ m. Define H p (m) be the closure of H in L p (dµ m ). Let Hm = {f H : m(f) = 0} H0 2(m) = {f H2 (m) : S fdµ m = 0}. Hoffman ([25], page 289) proved that (H3) H + Hm is dense in L 2 (dµ m ). (H4) L 2 (dµ m ) = H 2 (m) H0 2(m). 2. Hankel operators which are products of two Hankel operators In this section we present a proof of the result of when the product of two Hankel operators equals a Hankel operator [31], [39]. The relationship between Hankel operators Toeplitz operators is not just formal but, in fact, rather intimate. To get the relationship, we consider the multiplication operator M φ on L 2 for φ L, defined by M φ h = φh for h L 2. By the property that U is a unitary operator which maps H 2 onto [H 2 ] UP = (1 P )U, 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) 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) (2) T fg = T f T g + H f H g (3) H fg = T f H g + H f T g. The second equality gives that if f is in H, then (4) T f H g = H g T f,

7 COMPACT PERTURBATIONS OF HANKEL OPERATORS 7 for g L. The above Hankel Toeplitz relations have been known before [6],[7], [13] [31]. Let x y be two functions in L 2. x y is the operator of rank one defined by (x y)(f) = f, y x, for f L 2. Now we are ready to present a proof of the following theorem [31], [39]. Theorem 2. For three functions f, g, h in L, H f H g = H h if only if H f = c f H φλ, H g = c g H φλ, H h = c h H φλ for some constants c f, c g c h a point λ D. Proof. For a fixed λ in D, the long division for the rational function φ λ (z) = λ z 1 λz zk λ gives φ λ (z) = λ z 1 λz = 1 λ + (λ 1 λ)k λ zk λ = 1 λ + 1 λk λ. It is easy to verify that H φλ = c λ H zkλ. So proving this theorem is equivalent to proving that H f H g = H h if only if H f = c f H zkλ, H g = c g H zkλ, H h = c h H zkλ for some constants c f, c g c h a fixed point λ D. To do this, simply compute to verify that H zkλ = K λ K λ. Take product of both sides of the above equality to obtain H zkλ H zkλ = 1/[1 λ 2 ] H zkλ. Now we will show that it is only the above case if the product of two Hankel operators is a Hankel operator. Let f, g be co-analytic such that f(0) = g(0) = 0, H f H g = H h. Noting that the commutator I T z T z of the unilateral shift equals the rank one operator 1 1, we have H f (1 1)H g = z f zḡ = H h H f T z T z H g = H h(1 z 2 ). Thus the Hankel operator H h(1 z 2 ) is of rank one, so ker H h(1 z 2 ) is an invariant subspace with codimension 1. The Beurling theorem [12] gives that for some λ D, ker H h(1 z 2 ) = {K λ }, to obtain ḡ = c 1 zk λ.

8 8 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG Taking adjoint of H h(1 z 2 ) gives to obtain that for some µ D, Use to get zḡ z f = H h (1 z 2 ), f = c 2 zk µ. H zkλ = K λ K λ, H f H g = c 1c 2 1 λµ K µ K λ. Noting that H f H g = H h, we have getting T z [K µ K λ ] = [K µ K λ ]T z, µk µ = λk µ. Hence λ = µ, to complete the proof. Remark. An analogous result to the above theorem was obtained in [21] for small Hankel operators on the Bergman space. 3. Noncompact Hankel operators in the Toeplitz algebra Clearly, compact Hankel operators are in the Toeplitz algebra [12]. In this section we will construct a Hankel operator which is in the Toeplitz algebra but not compact. In fact, we will construct concrete examples that the Hankel operator is a compact perturbation of the product of two Hankel operators. In other words, we obtain examples that the product of two Hankel operators is a compact perturbation of a Hankel operator. To do this, let {x n } be a thin interpolating sequence on the x-axis such that 0 < δ n. Here δ n = m n n=1 x m x n 1 x m x n. Let B be the Blaschke product associated with the sequence {x n }. Because those numbers x n are real numbers, we see that B = B. By the interpolating theorem [36], there is a function h in QA such that m n xm xn 1 x h(x n ) = mx n, δ n for all n.

9 COMPACT PERTURBATIONS OF HANKEL OPERATORS 9 Theorem 3. Suppose that B is the thin Blaschke product defined above. Then H B is in the Toeplitz algebra. Proof. First we show that H BH B H BT h is compact. To estimate [H BH B H BT h]k xn 2, we have H BT hk xn = H h(xn) B k x n = h(x n )U[ Bk xn ] to obtain Thus H BH Bk xn = k xn, [H BH B H BT h]k xn = k xn h(x n )U[ Bk xn ]. [H BH B H BT h]k xn 2 2 = k xn h(x n )U[ Bk xn ] 2 2 = 2 2h(x n )R( k xn, U[ Bk xn ] ) = 2(1 h(x n )R(1 x 2 n)b (x n )) = 2(1 δ n ). The last equality follows from that m n xm xn 1 x h(x n ) = mx n, δ n (1 x 2 n)b (x n ) = m n x m x n 1 x m x n, for all n. By Theorem 3 in [37], {k xn } is a U + K 2 -Riesz basis, that is, there are a unitary operator V a Hilbert-Schmidt operator K such that k xn = (V + K)e n where {e n } is the stard orthogonal basis of l 2. Thus {k zn } is a basis for the kernel of T B. So for each f in kernel of T B, there is a sequence {a n } in l 2 such that f = a n k xn n=1 f 2 [ a n 2 ] 1/2. n=1 Let P n be the projection from the kernel of T B onto the subspace spanned by {k xi } n i=1. Clearly P n is a compact operator on the kernel of T B. Now we

10 10 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG have that for each f KerT B [H BH B H BT h][i P n ]f 2 to obtain i=n+1 [ i=n+1 C f 2 [ a i [H BH B H BT h]k xi 2 a i 2 ] 1/2 [ i=n+1 i=n+1 (1 δ n )] 1/2, [H BH B H BT h][i P n ] C[ (1 δ n )] 1/2 0. i=n+1 [H BH B H BT h]k xi 2 2] 1/2 This shows that H BH B H BT h is compact on the kernel KerT B of T B. In order to prove that H BH B H BT h is compact, we need only to show that [H BH B H BT h] is compact on BH 2. To do so, letting f n be a weak convergence sequence in BH 2, we write f n = Bg n. Thus g n is also a weak convergence sequence in H 2. An easy calculation gives that [H BH B H BT h]f n = H BT hbg n = H BT B T hg n H BH BH hg n = H BH BH hg n 0, to obtain that H BH B H BT h is compact. The last limit comes from that H h is compact on H 2. The second equality follows from (2). Second we show that H BH B H B h is compact on the Hardy space. Since h is in QA, the Hankel operator H h is compact. Using (3), we have H B h = H BT h + T B H h, getting that H BH B H B h is compact. Finally, we show that H B H BH Bh is compact. Noting that h(x n ) 2 = 1 for each n, we have that h(m) 2 1 = 0 for each m in the zero set Z H +C(B) = {m M(H + C) : B(m) = 0}, to obtain that for such m, ( h 2 1) Sm = 0. By the fact that for each m Z H +C(B), each m P (m), m m have the same support set, we have B( h 2 1) S m = B( h 2 1) Sm = 0. For each m M(H + C)/[ m ZH +C (B)P (m)], the Hedenmalm result [24] gives that B S m is constant. Since h is in QA, ( h 2 1) Sm is also

11 COMPACT PERTURBATIONS OF HANKEL OPERATORS 11 constant. Thus B( h 2 1) S m is constant. So we have proved that for each support set S, B( h 2 1) S is constant, getting that B( h 2 1) is in QC. Hence H B( h 2 1) is compact. On the other h, by (4), we have to conclude H B h 2 = H B ht h, H B = H B( h 2 1) + H B h 2 = H B( h 2 1) + H B h T h = H B( h 2 1) + [H B h H BH B]T h + H BH BT h = H BH Bh + K for some compact operator K. This implies that H B is in the Toeplitz algebra since H BH Bh is a semicommutator of two Toeplitz operators the ideal of compact operators is contained in the Toeplitz algebra to complete the proof. Remark. From the last part of the above proof, we see that the product H BH Bh of two Hankel operators is the compact perturbation of the Hankel operator H B. 4. Discussion on Theorem 1 In this section we first give a proof that there is a Hankel operator not in the Toeplitz algebra even if it essentially commutes with the unilateral shift, which was first shown in [10] constructed in [5] [11]. Recall that the Toeplitz algebra is the C -algebra generated by bounded Toeplitz operators. It is well known [12] that the ideal K of compact operators on the Hardy space H 2 is contained in the Toeplitz algebra. First we state some facts, which are known before, e.g., [10]. Fact 1. For f L g QC, T f T g T g T f is compact. Hartman s theorem gives that both H g H g are compact. By (2), we have T f T g T g T f = H g H f H f H g, to obtain that the commutator T f T g T g T f is compact. Fact 2. For each g QC, H f T g T g H f is compact if H f is in the Toeplitz algebra. Since the Toeplitz algebra is the C -algebra generated by bounded Toeplitz operators, we see that if T is in the Toeplitz algebra, then T T g T g T is compact. This leads to that if the Hankel operator H f is in the Toeplitz algebra, then H f T g T g H f is compact.

12 12 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG Fact 3. For f L g QC, the function f(g g) is in H + C if only if H f T g T g H f is compact. To do this, use (3) to obtain (5) H f T g T g H f = H f(g g) + H g T f T f H g. The Hartman theorem gives that the second third terms on the right h side of the above equality are compact, so H f T g T g H f is compact if only if the Hankel operator H f(g g) is compact. By the Hartman theorem again, we have that H f T g T g H f is compact if only if the function f(g g) is in H + C. Fact 4. For f L, put Then H f T z T z H f M 1 (H + C). This follows from Fact 3. A(f) = {x M(H + C); f S x / H S x }. is compact if only if A(f) M 1 (H + C) Fact 5. Let b be a Blaschke product with zeros {z n } n in D such that z n 1 as n. Then H b T z T z H b is compact if only if cluster points of {z n } n in D are 1 or -1. Since A(b) M 1 (H + C), this follws from Fact 4. Fact 6. There exists a function g in QC such that g g does not vanish on M 1 (H + C). Lemma 4. Let {z n } n be thin. Suppose that Iz n > 0, z n 1, ρ(z n, z n ) 1. Then {z n, z n } n is thin. Proof. Write z n = x n + iy n for real numbers x n y n. Then y n > 0. Using ρ(z n, z m ) = z n z m (x n x m ) = 2 + (y n y m ) 2 1 z n z m (1 x n x m y n y m ) 2 + (x n y m y n x m ) 2, ρ(z n, z m ) = z n z m 1 z n z m = simply compute to verify (x n x m ) 2 + (y n + y m ) 2 (1 x n x m + y n y m ) 2 + (x n y m + y n x m ) 2, ρ(z n, z m ) 2 ρ(z n, z m ) 2 = 4y n y m (1 x 2 n y 2 n)(1 x 2 m y 2 m) [(1 x n x m y n y m ) 2 + (x n y m y n x m ) 2 ][(1 x n x m + y n y m ) 2 + (x n y m + y n x m ) 2 ]. Thus ρ(z n, z m ) ρ(z n, z m ) for n m,

13 k:k n COMPACT PERTURBATIONS OF HANKEL OPERATORS 13 so ( z k z )( n z k z ) n = 1 z k z n 1 z k z n k n k=1 ( )( ) ρ(z n, z k ) ρ(z n, z k ) ρ(z n, z n ) k:k n as n. Similarly, ( )( ρ(z n, z k ) k:k n Hence {z n, z n } n is thin. k:k n 1 ( k:k n ) ρ(z n, z k ) ρ(z n, z n ) 1. ρ(z n, z k )) 2ρ(zn, z n ) Example. Let {z n } n be a sequence given in Lemma 4. By the Sundberg- Wolff interpolation theorem, there is a function g in QA such that g(z n ) = 1 g(z n ) = 0 for every n. Let m be a cluster point of {z n } n. Then m M 1 (H + C) there exists a subnet {z nα } α in {z n } n such that z nα m as α. We have g(m) = 1 g(m) = lim α g(z n α ) = lim α g(z n α ) = 0. Hence (g g)(m) = 1. The above example facts suggest the following result. Theorem 5. There is an interpolating Blaschke product B such that H B not only essentially commutes with T z, but is also not in the Toeplitz algebra. Proof. By Fact 6, there exists g QC such that g g does not vanish on M 1 (H + C). Then there exists an interpolating sequence {z n } n in D δ > 0 such that z n 1 as n (g g)(z n ) δ for every n. Let B be the Blaschke product with zeros {z n } n. By Fact 5, H B T z T z H B is compact. Let m be a cluster point of {z n } n. Then (g g)(m) 0. Since B Sm / H Sm (g g) Sm is nonzero constant, B(g g) / H Sm. Hence B(g g) / H + C. By Fact 3, H B T g T g H B is not compact. Thus by Fact 2, H B is not in the Toeplitz algebra. Now we discuss the reduction of our main result. By making use of results in [38], Guillory Sarason [19] proved that for each inner function u, there are a Blaschke product B an invertible function u 1 in QC such that u = Bu 1. Noting that u 1 Sm is a unimodular constant on each support set S m T u1 Tū1 I, Tū1 T u1 I, Hū H BTū1 are compact, we see that it suffices to prove Theorem 1 in the special case that both B 1 B 2 are Blaschke products. So we assume that B 1 B 2 are Blaschke products in Sections 5 6.

14 14 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG By the Axler, Chang, Sarason Volberg Theorem, condition (1) in Theorem 1 is just the necessary condition for H B1 H B2 to be compact. Axler [31] first observed that it is necessary H φ essentially commutes with the unilateral shift, i.e., H φ T z T z H φ is compact if H φ is in the Toeplitz algebra. But this commutator is compact only when H φ(1 z 2 ) is compact. By Hartman s theorem [34], this occurs only when (z 2 1)φ is in H + C this need not hold in general. This observation implies the following two lemmas. Lemma 6. Suppose H f H g H h is compact. Then H (1 z 2 )h is compact. Proof. By the relationship between the Hankel operators Toeplitz operators H f H g = T fg T f T g, we see that the operator T z essentially commutes with H f H g because every Toeplitz operators essentially commute with T z. This implies that T z H h H h T z is compact. Using the identity [T z H h H h T z ]T z = H h (T z T z )T z = H (1 z 2 )h, we obtain the desired result. Lemma 7. Suppose m is not in M 1 (H ) or M 1 (H ). If H f H g H h is compact, then h Sm is in H Sm either g Sm is in H Sm or f Sm is in H Sm Proof. For each point m in neither M 1 (H ) nor M 1 (H ), we see that (1 z 2 ) Sm is a nonzero constant. Suppose that H f H g H h is compact. By Lemma 6, H (1 z 2 )h is compact. Thus [H f H g H h ]T (1 z 2 ) is compact, so H f H g(1 z 2 ) is compact. By the Axler-Chang-Sarason-Volberg theorem ([2], [37]), the compactness of H f H g(1 z 2 ) implies that either f Sm or g(1 z 2 ) Sm is in H Sm. Hence either f Sm or g Sm is in H Sm. The compactness of H (1 z 2 )h implies that (1 z 2 )h Sm is in H Sm, to conclude that h Sm is in H Sm. The examples in [5] are based on the following lemma. Lemma 8. [5] Let {a n } be a Blaschke sequence in the unit disk such that lim a n = 1, ( lim a n = 1) n n 1 a n 1 a n 2n ( 1 + a n 1 a n 2n ). There is a function f such that (A) f is in QC. (B) f = f. (C) f(a n ) 1. The following lemma gives a necessary condition for H f H g H h to be compact.

15 COMPACT PERTURBATIONS OF HANKEL OPERATORS 15 Lemma 9. If H f H g H h is compact, then for each support set S F in QC, [F F ]h S is in H S either [F F ]g S or ([F F ]f) S is in H S. Proof. Let S be a support set F in QC. The Hartman theorem gives that both H F H F are compact. By (3), we have T F H f + H F T f = H F f = T f H F + H f T F, to obtain T F H f H f T F is compact. Similarly T F H g H g T F is also compact. Thus T F H f H g H f H g T F is compact. By the compactness of H f H g H h, we have that T F H h H h T F is compact. From (3), we see that both T F H h H F h H h T F H F h are compact, getting that H (F F )h is compact. So the Hartman theorem gives that [F F ]h S is in H S. On the other h, the compactness of [H f H g H h ]T F F gives that H f H (F F )g is compact. By the Axler, Chang, Sarason Volberg theorem [2], [37], we have that either (F F )g S or f S is in H S, to obtain that either (F F )g S or ((F F )f) S is in H S. This completes the proof. The above two lemmas suggest Conditions (2a) (3a) in Theorem 1. On the other h, for each thin Blaschke product B each m in the zero set Z H +C(B), Hedenmalm [24] showed that B φ m (λ) = ξλ for a unimodular constant ξ. Those examples in Section 3 suggest Conditions (2b) (3b). 5. Necessary part In this section we will prove the necessary part of Theorem 1. By the definition of the Hankel operator, clearly, H f = H f, where f (w) = f( w). The following lemma follows from a simple computation will be used later. For a function f in L 2, let f + = P f f = (I P )f. Then H f = H f. Lemma 10. Suppose that f is in L. For each z D, H f k z 2 = H f k z 2.

16 16 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG Proof. For each f in L, f = f + + f. Then f = f + + f. Simply compute to verify that for each z D, H f k z 2 = H f k z 2 The last equality follows from that = UH f k z 2 = (I P )f k z 2 = (f f ( z))k z 2 = (f f (z))k z 2 ; H f k z 2 = H f k z 2 = UH f k z 2 = (I P )f k z 2 = (f f (z))k z 2. P (f k z ) = f (z)k z. Combining the above two equalities gives to complete the proof. H f k z = H f k z, Lemma 11. Suppose that H f H g H h is compact. Let S be a support set. If either f S or g S is in H S, then h S is in H S. Proof. Suppose that S is the support set for a point m M(H + C). If either f S or g S is in H S, by Lemma 2.5 [18], we have that either lim H gk z 2 = 0, z m or lim H f k z 2 = 0. z m From the proof of Lemma 10, we see that An easy calculation gives H g k z 2 = (g g (z))k z 2, H f k z 2 = (f f (z))k z 2. H f H g k z 2 = P ((f f (z))((g g (z))k z 2. Since both f g are in BMO, we obtain that lim H f H g k z 2 = 0. z m On the other h, by the compactness of H f H g H h, we have lim [H f H g H h ]k z 2 = 0, z m

17 COMPACT PERTURBATIONS OF HANKEL OPERATORS 17 getting Thus h S is in H S. lim H hk z 2 = 0. z m Lemma 12. Suppose that m is a point in M(H + C) S is the support set for m. If H f H g H h is compact neither f S nor g S is in H S, then there is a point m in the Gleason part P (m) such that [F m(f )] f S, [F m(f )]g S [F m(f )]h S are in H S for each F H. Moreover, the mapping m m is constant on P (y) for each nontrivial point y. Proof. Let m be in M(H + C) S the support set for m. For each F in H, we have T F H h = H F h = H h T F. By the compactness of H f H g H h, we see that T F H f H g H f H g T F = H F f H g H f H F g is also compact. Thus the main result in [23] implies lim T φ z [H F f H g H f H F g ]T φz [H F f H g H f H F g ] = 0. z 1 An easy calculation gives to obtain Thus T φ z [H F f H g ]T φz = H F f T φ z T φ z H g = H F f H g + H F f [T φ z T φ z 1]H g = H F f H g H F f [k z k z ]H g = H F f H g [H F f k z ] [H g k z ], T φ z [H F f H g H f H F g ]T φz [H F f H g H f H F g ] = [H f k z ] [H F g k z] [H F f k z ] [H g k z ]. (6) lim z 1 [H f k z ] [H F gk z ] [H F f k z ] [H g k z ] = 0. Since neither f S nor g S is in H S, by Lemma 2.5 in [18], Letting lim z m H f k z 2 > 0 lim z m H g k z 2 > 0. λ z (F ) = < H F f k z, Hg k z > Hg k z 2, 2

18 18 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG we have λ z (F ) F H f k z 2 H g k z 2, to obtain that λ z (F ) m(f ) for some finite number m(f ) (7) m(f ) C F for some positive constant C. First we show that m is in M(H + C). Apply the operator [H f k z ] [H F g k z] [H F f k z ] [H g k z ] to the function H g k z, solve for H F f k z then use (6) to obtain lim H F f k z m(f )H f k z 2 = 0. z m Subsituting the above limit in (6) gives lim z m H F gk z m(f )H g k z 2 = 0. By Lemma 10, the first limit gives that (F m(f )) f S is also in H S, the second limit gives that (F m(f ))g S is in H S. Noting that H f H g H h is compact, we have [H f H g H h ]T F m(f ) = H f H g(f m(f )) H h(f m(f )) is also compact, getting that lim z m H h(f m(f )) k z = 0. So h(f m(f )) S is in H S. Second we show that m is a bounded linear multiplicative functional on H. Noting that for each F, G in H, (F m(f ))(G m(g)) = F G m(f )(G m(g)) m(g)(f m(f )) m(f ) m(g) we have On the other h, we also have (F G m(f ) m(g))g S H S. (F G m(f G))g S H S, to obtain that m(f G) m(f ) m(g) = 0. Similarly we see that m is linear on H. By (7), we obtain that m is in M(H ). Third we show that m is in the Gleason part P (m). If this is false, then ρ(m, m) = 1. Thus there are a sequence {b k } of functions in the unit ball of H such that b k (m) = 0 b k ( m) 1. Since the unit ball of H is weakly compact, we assume that b k weakly converges to b in H. Clearly, b(m) = 0 b 1. On the other h, [f(b k m(b k ))] Sm is in H Sm. Thus for each H H0 2(m), f(b k m(b k ))Hdµ m = 0, S m

19 COMPACT PERTURBATIONS OF HANKEL OPERATORS 19 S m f(b 1)Hdµ m is a cluster point of { f(b k m(b k ))Hdµ m }, S m so we have S m f(b 1)Hdµ m = 0. to get that f(b 1) Sm is in H Sm. From the proof of Lemma 1 [23] we see that (b 1) is an outer function in H 2 (m), getting that f Sm is in H Sm. This is a contradiction. Finally, we show that for each nontrivial point y, the mapping m m is constant on P (y). If this is false, there are two distinct points m 1 m 2 in P (y) such that g[f m i (F )] Smi is in H Smi for i = 1, 2 each F in H. Since m 1, m 2, y are in the same Gleason part, they have the support set. Thus we have that g[f m i (F )] Sy is in H Sy for i = 1, 2. Noting m 1 does not equal m 2, we see that for some function b in H, getting that m 1 (b) m 2 (b), g[ m 1 (b) m 2 (b)] Sy = g[(b m 2 (b)) (b m 1 (b))] Sy is in H Sy, so g Sy is in H Sy. This is a contradiction, to complete the proof. Lemma 13. If H f H g H h is compact, then for each trivial point m, either f Sm or g Sm is in H Sm. Proof. Assuming that neither f Sm nor g Sm is in H Sm, we will derive a contradiction. First we show that for each nontrivial point y with S y S m, either f Sy or g Sy is in H Sy. Suppose that y is a nontrivial point with S y S m. Thus for some interpolating Blaschke product b y, b y (ỹ) = 0, but m(b y ) 0. Here ỹ is a point in P (y) as in Lemma 12. Noting that m = m, by Lemma 12, we have f [(b y m(b y ))] Sm H Sm. Now we consider two cases. In the first case that m(b y ) = 1, by a lemma [23], we have that f Sm is in H Sm. In the second case that m(b y ) < 1, letting λ = m(b y ) using the function φ λ (z) = z λ, we have 1 λz to obtain f (φ λ (b y )) Sm H Sm, f (φ λ (b y )) Sy H Sy.

20 20 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG We claim that either f Sy or g Sy is in H Sy. If this is false, by Lemma 12 we have that f [φ λ (b y ) ỹ(φ λ (b y ))] Sy H Sy, to obtain Thus either f Sy is in H Sy or But the above equation gives [fỹ(φ λ (b y ))] Sy H Sy. 0 = ỹ(φ λ (b y )) = φ λ (ỹ(b y )). λ = ỹ(b y ) = 0 which contradicts to λ 0. So either f Sy or g Sy is in H Sy, for each nontrivial point y with S y S m. Finally, we derive a contradiction. By Corollary 3.2 [17], there is a net {y α } of nontrivial points with S yα S m such that y α m. Since either f Syα or g Syα is in H Syα, we may assume that f Syα is in H Syα for each α. For each H H, 0 = f (H H(y α ))dµ yα = f Hdµ yα H(y α )f (y α ) f Hdµ m H(m)f (m) = f (H H(m))dµ m. Thus f Sm is in H Sm. This is a contradiction to complete the proof. The following lemma is a consequence of the extension of Beurling s invariant subspace theorem (Theorem 20 on page 137 [27]). Lemma 14. If m is a nontrivial point, then there is an inner function Z in H (m) with Z(m) = 0 H 2 0 (m) = ZH 2 (m). Proof. Since m is a nontrivial point, there is a point m in P (m) distinct from m. Thus we can find a function f H such that m(f) = 0, but m(f) 0. Note that H0 2(m) is a closed subspace of H2 (m) = H 2 ( m) which is invariant under multiplication by H. Since f is in H0 2 (m) m(f) = fdµ m 0, S m The function 1 is not orthogonal to H0 2(m) in L2 (dµ m ). invariant subspace theorem, we deduce that H 2 0 (m) = ZH 2 (dµ m ) = ZH 2 (dµ m ) By Beurling s for some inner function Z in H 2 (dµ m ) = H 2 (dµ m ), to complete the proof.

21 COMPACT PERTURBATIONS OF HANKEL OPERATORS 21 Lemma 15. Suppose that m is a nontrivial Gleason part B is a Blaschke product. If [ B c 0 Z] Sm is in H Sm, for some nonzero constant c 0 inner function Z in H0 2 (m) satisfying Z φ m (λ) = ηλ for some unimodular constant η, then B Sm = c 0 Z Sm. Proof. Let m be a nontrivial Gleason part. Then the support set S m is also nontrivial. Since every real-valued function in H Sm is constant [ B c 0 Z] Sm is in H Sm, we have on S m for some constant c 1 to obtain B c 0 Z = c 1 (8) B φ m (λ) = c 0 Z φ m (λ) + c 1 = c 0 ηλ + c 1, for λ D. On the other h, since B = 1 on S m, we have that c 0 Z + c 1 = 1 Z = 1 on S m. Noting that Z is not constant on S m S m is nontrivial, we see that the intersection of two circles c 0 λ + c 1 = 1 λ = 1 contains at least two points, to obtain that the open unit disk λ < 1 contains an open arc of the circle c 0 λ + c 1 = 1 or c 1 = 0. If the open unit disk λ < 1 contains an open arc of the circle c 0 λ+c 1 = 1, by (8) we have that B φ m (λ) = 1 for some λ D. But B φ m (λ) is analytic on the unit disk B φ m (λ) 1. Thus B φ m (λ) is constant. This contradicts that c 0 is not zero. If c 1 = 0, then B = c 0 Z on S m. The proof is completed. Lemma 16. Suppose that B is a Blaschke product associated with {z n } in D. If m is a nontrivial point so that B φ m (λ) = ηλ for some unimodular constant η, then m is in the closure of {z n }. Proof. Suppose that m is not in the closure of {z n }. For δ > 0, set K δ (B) = n=1{z : ρ(z, z n ) > δ}. According to the Hoffman theorem [26], factor B = B 1 B 2 on K δ (B) with B 1 (m) = B 2 (m) = 0, to obtain that But We conclude that ηλ = B 1 φ m (λ)b 2 φ m (z). B 1 (m) = B 2 (m) = 0. B 1 φ m (λ)b 2 φ m (λ) = λ 2 h(λ) for some analytic function h on D getting for λ D, which is a contradiction. η = λh(λ),

22 22 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG Lemma 17. Suppose 1 z is sufficiently small. For z D, if 1 z 1 z 2 C 1, for some positive constant C 1, then there are positive constants C 2 > 0 0 < C 3 < 1 such that θ 1 r 2 C 2, ρ(x, z) < C 3. where z = re iθ, z = x + iy. Proof. To write z = re iθ z = x+iy in polar coordinate the Cartesian coordinate, respectively, we have that x = r cos θ y = r sin θ. Simply compute to verify that 1 z (1 r cos θ) 1 z 2 = 2 + (r sin θ) 2 1 r 2 1 sin θ/2 (9) = + [2r (1 + r) 2 1 r 2 ] 2, ρ(x, z) = x z 1 xz = iy 1 x 2 ixy = r sin θ r 2 (2 r 2 ) sin 2 θ + 1 r sin θ r 2 (2 r 2 ) sin 2 θ + (1 r 2 ) 2 (10) = r sin θ 1 r 2 (2 r 2 r sin θ )[ ] 1 r Let C 1 be the positive constant such that 1 z 1 z 2 C 1. As 1 z 2 is small, (9) gives that θ is small. constant C 2, depending only on C 1, such that. Thus there is a positive r sin θ 1 r 2 C 2. By (10), we see that for some positive constant C 3, depending only on C 2, to complete the proof. ρ(x, z) < C 3 < 1,

23 COMPACT PERTURBATIONS OF HANKEL OPERATORS 23 The following lemma suggests conditions (2c) (3c) in Theorem 1. Lemma 18. Suppose that H f H g H h is compact m is in the closure of the sequence {z n } with the following property: ρ(r(z n ), z n ) < c for some positive constant c < 1. If for the support set S m, there are constants c f, c g, c h an inner function Z in H (m) with Z( m) = 0, such that [f c f Z] S m, [g c g Z] S m, [h c h Z] S m are in H S m, then one of the following holds (1) c h = 0 either c f = 0 or c g = 0. (2) Z φ m (λ) = ξλ for λ D, where ξ is a unimodular constant ξ, whenever z α m. ρ(r(z α ), z α ) 0 Proof. Because Z is defined only on the support set S m, we can use functions in H to approximate Z. To simplify the proof, we may assume that Z is in H. Suppose that m is in the closure of {z n } ρ(r(z n ), z n ) < c < 1. Choose m in the closure of {R(z n )} so that m = φ m (z 0 ) for some z 0 in cd. Then m m are in the same Gleason part so S m = S m. Since H f H g H h is compact, lim [H f H g H h ]k z 2 = 0. z D Noting that for the support set S m, there are constants c f, c g, c h an inner function Z in H such that [f c f Z] S m, [g c g Z] S m, [h c h Z] S m are in H S m, by Lemma 2.5 [18], we have that for each y P ( m), getting that lim H z y (g c g Z) k z 2 = 0, lim H z y (f c f Z) k z 2 = 0, lim H z y (h c h Z) k z 2 = 0, (11) lim z y [ c f c g H ZH Z c h H Z]k z 2 2 = 0. For each z D, evaluate the Hankel operator H Z on the the normalized reproducing kernel k z to verify that H ZH Zk z = (1 Z(z)Z)k z H Zk z = [Z Z(z)] w k z. Since Z is an inner function in H, we have (1 Z(z)Z)k z 2 2 = 1 Z(z) 2,

24 24 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG [Z Z(z)] w k z 2 2 = 1 Z(z) 2. By the fact that 1 Z( m) 2 = 1, we have c f c g = c h, to obtain that there is a unimodular constant η such that c f c g η = c h. If c h = 0, then either c f or c g must be zero. In this case, Condition (1) holds. If c h 0, use (11) to obtain Thus so we have lim [ηh z y ZH Z H Z]k z 2 2 = 0. lim η(1 Z(z)Z)k z [Z Z(z)] w k z 2 z y 2 = 0, lim z y [ ηk z + Z(z) w k z 2 2 η Z(z)Zk z Z w k z 2 2] = 0. Easy calculations give lim [ ηk z + Z(z) w k z 2 z y 2 (1 + Z(z) 2 )] = 0, lim [ η Z(z)Zk z Z w k z 2 z y 2 [1 + Z(z) 2 2R{ η Z(z)Zk z, Z w k z }] = 0, to obtain lim R{ η Z(z)Zk z, Z w k z } = 0. z y Now we consider two cases. In the first case that z is a real number, we have η(1 Z(z)Z)k z [Z Z(z)] w k z 2 2 = 2[(1 Z(z) 2 ) (R{η(Z ) (z)(1 z 2 ) + R{ η Z(z)Zk z, Z w k z }). Let z = φ R(zk )(λ) for the fixed real number λ in the unit disk D. Then z = z z φ m (λ), so By (Z φ z ) (0) (Z φ φm(λ)) (0), Z(z) = Z(φ xk (λ)) Z(φ m (λ)). [(1 Z(z) 2 ) + R{η[(Z φ z ) (0)] = [(1 Z(z) 2 ) R{η[(Z ) (z)(1 z 2 )], we have to obtain 1 Z(φ m (λ)) 2 = R{η(Z φ φm(λ)) (0)}, 1 F (λ) 2 = R{F (λ)}(1 λ 2 ),

25 COMPACT PERTURBATIONS OF HANKEL OPERATORS 25 for each real number λ with λ < c. Here F = ηz φ m. In the other words, (12) Since F is analytic function 1 1 λ 2 = R{ F (λ) 1 F (λ) 2 }. F (z) 1 for z D, the Schwarz Lemma (Lemma 1.2 [15]) states that F (z) 1 F (z) z 2, the above equality holds at some z D if only if F (z) is a Möbious transformation. By (12), we see that F (λ) 1 F (λ) 2 = 1 1 λ 2 for real numbers λ with λ < c, to conclude that F is a Möbius transformation. That is, Z φ m (λ) = ξφ z1 (λ), for some unimodular constant ξ a point z 1 D. Since m = φ m (z 0 ) Z( m) = 0 we have 0 = Z φ m (z 0 ) = ξφ z1 (z 0 ), to obtain that z 1 = z 0. Now we show that z 0 is a real number. If this is false, for complex numbers z with z z, we have η(1 Z(z)Z)k z [Z Z(z)] w k z 2 2 = 2[(1 Z(z) 2 (1 z 2 ) (R{η z z (Z( z) Z(z))} + R{ η Z(z)Zk z, Z w k z })]. Let z = φ Rzn (z 0 ) in the above equality take the limit as Rz n m to obtain 1 Z φ m (z 0 ) 2 = R{ 1 z 0 2 [Z φ m ( z 0 ) Z φ m (z 0 )]}. z 0 z 0 Thus 1 = R{ 1 z 0 2 φ z0 ( z 0 )}, z 0 z 0 so 1 = R{ 1 z z 0 2 } to force that z 0 is real. Next we show that z 0 = 0. To do this, let x n = R(z n ). Noting that z n = φ xn (λ n ) z n m = φ m (z 0 ), we have λ n = x n z n 1 x n z n = iy n 1 x 2 n ix n y n

26 26 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG getting Since y n 1 x 2 n = x n( yn ) 2 i yn 1 x 2 n 1 x 2 n 1 + x 2 n( yn ) 1 x 2 z 0, 2 n I(λ n ) = y n 1 x 2 n 1 + x 2 n( yn ) 1 x n < M for some constant M x n 1, we have 1 x 2 0, n to conclude that λ n 0 so z 0 = 0. This gives that y n Z φ m (λ) = ξλ. The above proof also works for any net w α m 1 with sup ρ(r(w α ), w α ) < 1. α If w α m 1, by Lemmas 8 9, we may assume that sup ρ(r(w α ), w α ) < 1, α since f Sm1 is not in H Sm1. Now we are ready to prove the main result in this section, which is the necessary part of Theorem 1. Theorem 19. Suppose that B 1 B 2 are Blaschke products. If H B1 H B2 H h is compact for some h in L, then for each support set S(= S m ), one of the following holds: (1) h S is in H S either B 1 S or B 2 S is in H S. (2) m is a thin part in the fibre M 1 (H ) with the following properties: (2a) m is in the closure of a sequence {z n } in D satisfying 1 z n 1 z n 2 < M for n. Here M is a positive contradiction; (2b) B 1 Sm = cb 2 Sm, [h c Z] Sm is in H Sm, B 2 φ m (λ) = ξλ for some unimodular constants c ξ; (2c) If m is in the closure of some sequence {w n } D, then ρ(r(w α ), w α ) 0 whenever the subnet {w α } converges to m. (3) m is a thin part in the fibre M 1 (H ) with the following properties: (3a) m is in the closure of a sequence {z n } in D satisfying 1 + z n 1 z n 2 < M for n. Here M is a positive contradiction; (3b) B 1 Sm = cb 2 Sm, [h c Z] Sm is in H Sm, B 2 φ m (λ) = ξλ for some unimodular constants c ξ;

27 COMPACT PERTURBATIONS OF HANKEL OPERATORS 27 (3c) If m is in the closure of some sequence {w n } D, then ρ(r(w α ), w α ) 0 whenever the subnet {w α } converges to m. Proof. First we introduce some notation to simplify this proof. Use G 1 to denote the set {m M(H +C) : 1 B 1 (m) 2 = 0 or 1 B 2 (m) 2 = 0} G 2 to denote the set {m M(H +C) : 1 B 1 (m) 2 > 0 1 B 2 (m) 2 > 0}. By a lemma in [18], G 1 = {m M(H + C), B 1 Sm or B 2 Sm is constant}. Suppose that m is a point in M(H + C). We consider two cases. In the first case that m is a trivial point, by Lemma 13, thus m is in G 1. So condition (1) holds for the support set S m, G 2 does not contain any trivial points. In the second case that m is a nontrivial point such that condition (1) does not hold, we show that condition (2) or (3) holds. Clearly, m must be in G 2. By Lemma 7, m is in either M 1 (H ) or M 1 (H ). We consider only the case that m is in M 1 (H ). In the case that m is in M 1 (H ), the argument below also works. Assume that for some positive constant γ such that 1 B 1 (m) 2 > γ 1 B 2 (m) 2 > γ. Let N(m) denote the set {m 1 M(H ) : 1 B 1 (m 1 ) 2 > γ/2 1 B 2 (m 1 ) 2 > γ/2}. Thus N(m) is an open neighborhood of m the Carleson corona theorem [8] gives that the intersection of N(m) the unit disk D is dense in N(m). Let {z n } be the intersection of the zeros 1 z of B 1 N(m) D. Set D k = {z D : (1 z 2 ) 2k }, D k denotes the closure of D k in the maximal ideal space of H. We claim that the intersection of the closure of {z n } in M(H ) k D k is empty. If this is not true, let m 2 be a point in the intersection. Then there is a sequence {w k } with w k D k such that {w k } captures m. By Lemma 8, there is a function u in QC such that u = ũ u(w k ) 1. By the compactness of H (u ũ) B1, we see that (u ũ) B 1 Sm2 is in H m2, getting that B 1 Sm2 is in H Sm2. since u(m 2 ) = 1. This implies that 1 B 1 (m) 2 = 0. On the other h, m 2 is in the closure of {z n } in M(H ) such that (1 B 1 (z n ) 2 ) γ 2, to obtain that 1 B 1 (m 2 ) 2 = lim z n m 2 (1 B 1 (z n ) 2 ) γ 2, which is a contradiction. From now on we assume that {z n } are contained in some D/D k0. That is, 1 z n 1 z n 2 2k 0, for every n.

28 28 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG Write z n = x n + iy n where x n y n are real numbers. If we write z n = r n e iθn, then x n = r n cos θ n y n = r n sin θ n. By Lemma 17, we have θ n 1 rn 2 C 2, ρ(x n, z n ) = x n z n < C 3 < 1 1 z n x n for some positive constants C 2, C 3 to obtain that the closure of {x n } in the maximal ideal space of H intersects with the Gleason part P (m). Let Φ be the Blaschke product associated with {z n }. Next we show that m is in the zero set of Φ Φ is locally thin at each point in Z H +C(Φ). Let m 3 be either m or a point in Z H +C(Φ) S the support set for m 3. Clearly, m 3 is in G 2. That means that neither B 1 S nor B 2 S is in H S. By Lemma 12, we have that for some point m 3 in the Gleason part P (m 3 ), for each F H, [F m 3 (F )] B1 S, (F m 3 (F )) B 2 S (F m 3 (F ))h S are in H S. Noting that m 3 is in P (m 3 ), we have that S m3 = S m3 = S, getting that by Lemma 14, for some inner function Z in H ( m 3 ), Z( m 3 ) = 0 H 2 0 ( m 3 ) = ZH 2 ( m 3 ). Choose a sequence {f n } H such that f n Z H 2 ( m 3 ) 0, to obtain that f n ( m 3 ) Z( m 3 ) = 0. Thus B 1 [f n f n ( m 3 )] B 1 Z H 2 ( m 3 ) 0, so B 1 Z S m3 is in H S m3, because B 1 [f n f n ( m 3 )] S m3 is in H S m3. This implies that for some constant c 1 function L 1 H ( m 3 ), B 1 Z = c 1 + ZL 1 on S m3. Therefore we have that [ B 1 c 1 Z] S m3 is in H S m3. Similarly we have that for some constants c 2 c h, [ B 2 c 2 Z] S m3 [h c h Z] S m3 are in H S m3. Since H ( m 3 ) S m3 does not contain any nonconstant real valued functions, we have [B 1 c 1 Z] S m3 is constant. Thus we assume Z is in H. Lemma 18 gives Z φ m3 (λ) = ηλ, Lemma 15 gives that B 1 S m3 = c 1 Z S m3. So we have to obtain that m 3 = m 3 B 1 φ m3 (λ) = c 1 Z φ m3 (λ) = c 1 ηλ, B 1 φ m3 (λ) = c 1 ηλ, because B 1 (m 3 ) = 0, Z(m 3 ) = 0 Z has only one zero in P (m 3 ). By Lemma 16, m 3 is in the closure of {z n }. Factor B 1 = ΦΨ Φ φ m3 (λ) = λφ 1 (λ),

29 COMPACT PERTURBATIONS OF HANKEL OPERATORS 29 for some Blaschke product Ψ function Φ 1 in H with Thus we have Φ 1 (λ) 1. c 1 ηλ = B 1 φ m3 (λ) = Φ φ m3 (λ)ψ φ m3 (λ) = λφ 1 (λ)ψ φ m3 (λ), getting that Φ 1 (λ) is a constant. Hence Φ φ m3 (λ) = cλ for some unimodular constant c. By Theorem 3.2 [16], Φ is locally thin at m 3. By Lemma 18, we see that ρ(r(z α ), z α ) 0 whenever z α m, sup α ρ(r(z α ), z α ) < 1, to obtain that Φ is a thin Blaschke product m is a thin part. Using the same procedure as above, we obtain that B 2 Sm = cb 1 Sm = c 2 Z Sm [h c h B 1 ] Sm is in H Sm for some unimodular constant c 2, to complete the proof. 6. Sufficient part In this section, we will present the proof of the sufficient part of Theorem 1. Suppose that {x n } is a thin Blaschke sequence in D. As in Section 3, define δ n = x m x n. 1 x m x n m n By the Sundberg-Wolff interpolation theorem [36], there is a function σ in QA such that m n xm xn 1 x σ(x n ) = mx n. δ n For each integer k > 0, we write τ k (t) for the kth Rademacher function defined on [0, 1] by τ k (t) = sign sin 2 k πt. Clearly, {τ k } is orthonormal in L 2 [0, 1] ([29], [14]). The following theorem is inspired by Lemma 7 [37]. Theorem 20. Suppose that {x n } is a thin sequence on the real axis B is a thin Blaschke product associated with {x n }. Let B (n) be the Blaschke product associated with the subsequence {x k } k n. If for each factorization B (n) = B 1 B 2, H B1 H B2 < ɛ, then for each φ = k=n c kk xk [BH 2 ], [H BH B H BT σ ]φ 2 2 8ɛ 1 ɛ φ c k 2 [H BH B H BT σ ]k xk 2 2. k=n

30 30 XIAOMAN CHEN, KUNYU GUO, KEIJI IZUCHI, AND DECHAO ZHENG Proof. Suppose that φ = k=n c kk xk [BH 2 ], for some sequence {c k } in l 2. For each t [0, 1], define L(t)φ = c k τ k (t)[h BH B H BT σ ]k xk = k=n c k τ k (t)[k xk B w k xk ], k=n where {τ k (t)} are Rademacher functions. The last equality follows from that as we did in Section 3, [H BH B H BT σ ]k xk = k xk B w k xk. For each fixed t in [0, 1], let σ + = {k n : τ k (t) = 1} σ = {k n : τ k (t) = 1}. Let B + be the Blaschke product associated with {x k } k σ+ B the Blaschke product associated with {x k } k σ. Thus B + B is the Blaschke product associated with {x k } k n, so Define Then H B+ H B < ɛ. X + = k σ + c k k xk, X = k σ c k k xk, Y + = k σ + c k B z σ(x k ) k xk, Y = k σ c k B z σ(x k ) k xk. L(t)φ = X + + Y (X + Y + ) [H BH B H BT σ ]φ = X + Y + X Y +. Let P + be the projection onto the space spanned by {k xk } k σ+ P the projection onto the space spanned by {k xk } k σ. Since for k l B z k xk, k xl = 0, B z k xk, k xk = (1 x 2 k )B (x k ), we have that X + Y X Y +. An easy calculation gives [H BH B H BT σ ]φ 2 2 = X + Y X Y R{ X + Y, X Y + } = X Y X Y R{ X +, X X +, Y + Y, X + Y, Y + },