PRODUCTS OF BLOCK TOEPLITZ OPERATORS

CAIXING GU AND DECHAO ZHENG

Abstract. In this paper we characterize when the product of two block Toeplitz operators is a compact perturbation of a block Toeplitz operator on the Hardy space of the open unit disk. Necessary and sufficient conditions are given for the commutator of two block Toeplitz operators to be compact.

1. Introduction

Let D be the open unit disk in the complex plane and ∂D the unit circle. Let dσ(w) be the normalized Lebesgue measure on the unit circle. We denote by L^{2}(C^{n}) (L^{2} for n = 1) the space of C^{n}-valued Lebesgue square integrable functions on the unit circle. The Hardy space H^{2}(C^{n}) (H^{2} for n = 1) is the closed linear span of C^{n}-valued analytic polynomials. We observe that L^{2}(C^{n}) = L^{2} ⊗ C^{n} and H^{2}(C^{n}) = H^{2} ⊗ C^{n}, where ⊗ denotes the Hilbert space tensor product. Let M_{n×n} be the set of n × n complex matrices. L_{n×n}^{∞} denotes the space of M_{n×n}-valued essentially bounded Lebesgue measurable functions on the unit circle and H_{n×n}^{∞} denotes the space of M_{n×n}-valued essentially bounded analytic functions in the disk.

Let P be the projection of L^{2}(C^{n}) onto H^{2}(C^{n}). For F ∈ L_{n×n}^{∞}, the block Toeplitz operator T_{F} : H^{2}(C^{n}) → H^{2}(C^{n}) with symbol F is defined by the rule T_{F} h = P (F h). The Hankel operator H_{F} : H^{2}(C^{n}) → L^{2}(C^{n})⊖H^{2}(C^{n}) with symbol F is defined by H_{F} h = (I −P)(F h). The block Toeplitz operator T_{F} has the following matrix representation:

[ \begin{matrix}A_{0} & A_{−1} & A_{−2} & ⋯ \\ A_{1} & A_{0} & A_{−1} & ⋯ \\ A_{2} & A_{1} & A_{0} & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ \end{matrix} ]

where A_{i} belongs to M_{n×n}. The word ”block” refers to the fact that in the above matrix representation the entries are not scalars but linear transformations on C^{n}. In this paper the word ”block” will often be omitted. For more details on the block Toeplitz operators and Hankel operators, see [7], [9] and [3].

1991 Mathematics Subject Classification. Primary 47B35.

Key words and phrases. Block Toeplitz operators, Hankel operators.

Gu’s research is supported in part by NSF Grant DMS-9022140 during residence at MSRI. Zheng is supported in part by a NSF grant.

If we set H^{2}(C^{n}) = H^{2} ⊕...⊕H^{2}, then we see that the Toeplitz operator T_{F} has the form

T_{F} = [ \begin{matrix}T_{f_{11}} & T_{f_{12}} & ⋯ & T_{f_{1n}} \\ T_{f_{21}} & T_{f_{22}} & ⋯ & T_{f_{2n}} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ T_{f_{n1}} & T_{f_{n2}} & ⋯ & T_{f_{nn}} \end{matrix} ]

and the Hankel operator H_{F} has the form

H_{F} = [ \begin{matrix}H_{f_{11}} & H_{f_{12}} & ⋯ & H_{f_{1n}} \\ H_{f_{21}} & H_{f_{22}} & ⋯ & H_{f_{2n}} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ H_{f_{n1}} & H_{f_{n2}} & ⋯ & H_{f_{nn}} \end{matrix} ] ,

where

F = [ \begin{matrix}f_{11} & f_{12} & ⋯ & f_{1n} \\ f_{21} & f_{22} & ⋯ & f_{2n} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ f_{n1} & f_{n2} & ⋯ & f_{nn} \end{matrix} ] .

Thus as in the scalar case, the block Toeplitz operators and Hankel operators are connected by the following important relation:

T_{F G} −T_{F} T_{G} = H_{F^{∗}}^{∗}H_{G}.

The map ξ: F → T_{F}, which is called the Toeplitz quantization, carries L_{n×n}^{∞} into the C^{∗}−algebra of bounded operators on H^{2}(C^{n}). It is a contractive *-linear mapping [7]. However it is not multiplicative in general. Indeed Brown and Halmos [4] showed that for scalar functions f and g, T_{f}T_{g} = T_{fg} if and only either f^{∗} or g is in H^{∞}. It is not difficult to see that in the matrix case T_{F} T_{G} = T_{F G} still holds if either F^{∗} or G is in H_{n×n}^{∞}. But the converse is not valid in the matrix case. We will characterize F and G such that T_{F} T_{G} = T_{F G}; see Theorem 6 below for details. On the other hand, Douglas [7] 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}(C^{n}), onto L_{n×n}^{∞}. So modulo the commutator ideal of the Toeplitz algebra, ξ is multiplicative.

The main question to be considered in this paper is when the product of two Toeplitz operators is a compact perturbation of a Toeplitz operator. This problem is connected with the spectral theory of Toeplitz operators; see [7], [9] and [3]. It follows from a theorem of Douglas [7] that T_{F} T_{G} can be a compact perturbation of a Toeplitz operator only when it is a compact perturbation of T_{F G}. Thus it suffices to study when the semicommutator T_{F G} −T_{F} T_{G} is compact. When F = f and G = g are scalar functions, the problem was solved by Axler, Chang, and Sarason [2] and Volberg [14]. Their beautiful result is that T_{fg} −T_{f}T_{g} is compact if and only if H^{∞}[f̄] ⋂ H^{∞}[g] ⊂ H^{∞} +C(∂D); here H^{∞}[g] denotes the closed subalgebra of L^{∞} generated by H^{∞} and g.

Recently, Zheng [15] proved that T_{fg} −T_{f}T_{g} is compact if and only if

lim_{|z|→1} ∥Hf̄k_{z}∥_{2}∥H_{g}k_{z}∥_{2} = 0;

here k_{z} denotes the normalized reproducing kernel in H^{2} for point evaluation at z. If we write f = f_{+} +f_{−} for each f ∈ L^{∞} where f_{+} and \overline{f_{−}} are in H^{2}, then the above condition is equivalent to

lim_{|z|→1} |\overline{f_{+}} −\overline{f_{+}}(z)|^{2}(z)|g_{−} −g_{−}(z)|^{2}(z) = 0,

where h(z) denotes the harmonic extension of h at z ∈ D for h ∈ L^{1}, via the Poisson integral

h(z) = ∫_{∂D} h(w)\frac{(1−|z|^{2})}{|1−w\overline{z}|^{2}}dσ(w).

For the block Toeplitz operators, we will show that T_{F G} − T_{F} T_{G} is compact if and only if

lim_{|z|→1} ∥[|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)]^{1/2}∥ = 0,

where we write F = (f_{ij})_{n×n} as F = F_{+} +F_{−} with F_{+} = ((f_{ij})_{+})_{n×n} and F_{−} = ((f_{ij})_{−})_{n×n}, and

H(z) = (h_{ij}(z))_{n×n}

if H = (h_{ij})_{n×n}. For a matrix A, we define |A|^{2} = AA^{∗}. Several other equivalent conditions, in particular a condition in the spirit of the result of Axler, Chang, and Sarason

[2] and Volberg [14], will be given.

In [11] Gorkin and Zheng characterized when the commutator T_{f}T_{g} −T_{g}T_{f} is compact for scalar functions f and g. In this paper, by considering block Toeplitz operators, we will give an unified approach for the study of compactness of boths semi-commutators and commutators. Namely, by a theorem of Douglas [7], we have that the commutator T_{F} T_{G} −T_{G}T_{F} of block Toeplitz operators T_{F} and T_{G} is compact if and only if F G = GF and the semi-commutator T_{BC} −T_{B}T_{C} is compact, where

B = [ \begin{matrix}F & −G \\ 0 & 0 \end{matrix} ] , C = [ \begin{matrix}G & 0 \\ F & 0 \end{matrix} ];

see Theorem 7 below for details. Thus we will show the commutator T_{F} T_{G} − T_{G}T_{F} is compact if and only if F G = GF and

lim_{|z|→1} ∥ [ \begin{matrix}|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z) & −(((F_{+})^{∗} −(F_{+})^{∗}(z))(G_{+} −G_{+}(z)))(z) \\ −(((G_{+})^{∗} −(G_{+})^{∗}(z))(F_{+} −F_{+}(z)))(z) & |(G_{+})^{∗} −(G_{+})^{∗}(z)|^{2}(z) \end{matrix} ]^{1/2} ×

[ \begin{matrix}|G_{−} −G_{−}(z)|^{2}(z) & ((G_{−} −G_{−}(z))((F_{−})^{∗} −(F_{−})^{∗}(z)))(z) \\ ((F_{−} −F_{−}(z))((G_{−})^{∗} −(G_{−})^{∗}(z)))(z) & |F_{−} −F_{−}(z)|^{2}(z) \end{matrix} ]^{1/2} ∥ = 0.

2. A necessary condition for compactness

In this section we will obtain a necessary condition for the compactness of the semicommutator T_{F G} − T_{F} T_{G}. This will also motivate a necessary and sufficient condition for T_{F G} −T_{F} T_{G} to be zero. The question when T_{F G} −T_{F} T_{G} is zero will be discussed in the next two sections.

First we introduce an antiunitary operator V on L^{2} by defining (V h)(w) = \overline{wh(w}). The operator enjoys many nice properties such as V^{−1}(I − P)V = P and V = V^{−1}. These properties leads easily to the relation V^{−1}H_{f}V = H_{f}^{∗}.

Let x and y be two vectors in L^{2}. x⊗y is the operator of rank one defined by

(x⊗y)(f) =< f, y > x.

Observe that the norm of the operator x⊗y is ∥x∥_{2}∥y∥_{2}.

For z in D, let k_{z} be the normalized reproducing kernel \frac{(1−|z|)^{1/2}}{(1−\overline{z}w)} for point evaluation at z, and ϕ_{z} the Möbius map on the unit disk,

ϕ_{z}(w) = \frac{z −w}{1−\overline{z}w}.

ϕ_{z} can also be viewed as a function on the unit circle. Let Φ_{z} denote the function diag\{ϕ_{z}, ⋯, ϕ_{z}\} ∈ H_{n×n}^{∞}. The product T_{ϕ_{z}}T\overline{ϕ_{z}} is the orthogonal projection onto H^{2} ⊖ \{k_{z}\}. Thus 1 − T_{ϕ_{z}}T\overline{ϕ_{z}} is the operator k_{z} ⊗ k_{z} of rank one. This leads to the following lemma.

Lemma 1. Let F = (f_{ij})_{n×n} and G = (g_{ij})_{n×n} be in L_{n×n}^{∞}. Then the operator H_{F}^{∗} H_{G} − T_{Φ_{z}}^{∗}H_{F}^{∗} H_{G}T_{Φ_{z}} is anti-unitary equivalent to

(∑_{j=1}^{n} H_{f_{ji}}k_{z} ⊗H_{g_{jk}}k_{z})_{n×n}.

Proof. Let F = (f_{ij})_{n×n} and G = (g_{ij})_{n×n}. Then it is easy to check that

H_{F}^{∗} H_{G} = (∑_{j=1}^{n} H_{f_{ji}}^{∗}H_{g_{jk}})_{n×n}

and

T_{Φ_{z}}^{∗}H_{F}^{∗} H_{G}T_{Φ_{z}} = (∑_{j=1}^{n} T_{ϕ_{z}}^{∗}H_{f_{ji}}^{∗}H_{g_{jk}}T_{ϕ_{z}})_{n×n}.

So the difference H_{F}^{∗} H_{G} −T_{Φ_{z}}^{∗}H_{F}^{∗} H_{G}T_{Φ_{z}} is

(∑_{j=1}^{n}[H_{f_{ji}}^{∗}H_{g_{jk}} −T_{ϕ_{z}}^{∗}H_{f_{ji}}^{∗}H_{g_{jk}}T_{ϕ_{z}}])_{n×n}.

PRODUCTS OF BLOCK TOEPLITZ OPERATORS 5

Hence it is sufficient to check that

∑_{j=1}^{n}[H_{f_{ji}}^{∗}H_{g_{jk}} −T_{ϕ_{z}}^{∗}H_{f_{ji}}^{∗}H_{g_{jk}}T_{ϕ_{z}}]

is unitary equivalent to

∑_{j=1}^{n} H_{f_{ji}}k_{z} ⊗H_{g_{jk}}k_{z}.

Applying V and V^{−1} to the difference

∑_{j=1}^{n}[H_{f_{ji}}^{∗}H_{g_{jk}} −T_{ϕ_{z}}^{∗}H_{f_{ji}}^{∗}H_{g_{jk}}T_{ϕ_{z}}]

we have

V^{−1} ∑_{j=1}^{n}[H_{f_{ji}}^{∗}H_{g_{jk}} −T_{ϕ_{z}}^{∗}H_{f_{ji}}^{∗}H_{g_{jk}}T_{ϕ_{z}}]V

= V^{−1} ∑_{j=1}^{n}[H_{f_{ji}}^{∗}H_{g_{jk}} −H_{f_{ji}ϕ_{z}}^{∗}H_{g_{jk}ϕ_{z}}]V.

Because V^{−1}H_{f}V = H_{f}^{∗}, the above equality simplifies to

= V^{−1} ∑_{j=1}^{n}[H_{f_{ji}}^{∗}H_{g_{jk}} −H_{f_{ji}ϕ_{z}}^{∗}H_{g_{jk}ϕ_{z}}]V

= ∑_{j=1}^{n}[H_{f_{ji}}H_{g_{jk}}^{∗} −H_{f_{ji}ϕ_{z}}H_{g_{jk}ϕ_{z}}^{∗}]

= ∑_{j=1}^{n} H_{f_{ji}}(1−T_{ϕ_{z}}T_{ϕ_{z}}^{∗})H_{g_{jk}}^{∗}.

Since 1−T_{ϕ_{z}}T_{ϕ_{z}}^{∗} = k_{z} ⊗k_{z}, the right hand side of the above equation is

∑_{j=1}^{n} H_{f_{ji}}k_{z} ⊗H_{g_{jk}}k_{z}

for all z in D. This completes the proof of the lemma.

Let trace be the trace on the trace class of operators on H^{2} and tr denote the trace on the n by n matrices. Lemma 2. Let F and G be in L_{n×n}^{∞}. Let T = H_{F}^{∗} H_{G} −T_{Φ_{z}}^{∗}H_{F}^{∗} H_{G}T_{Φ_{z}}. Then

trace{T^{∗}T } = tr[|F_{−} −F_{−}(z)|^{2}(z)|G_{−} −G_{−}(z)|^{2}(z)].

Proof. Let F = (f_{ij})_{n×n} and G = (g_{ij})_{n×n}. By Lemma 1, H_{F}^{∗} H_{G} −T_{Φ_{z}}^{∗}H_{F}^{∗} H_{G}T_{Φ_{z}} is antiunitary equivalent to

T_{1} = (∑_{j=1}^{n} H_{f_{ji}}k_{z} ⊗H_{g_{jk}}k_{z})_{n×n}.

So we need to computer the trace of the operator T_{1}^{∗}T_{1}. It is easy to see that

T_{1}^{∗}T_{1} = (∑_{j_{1}}(∑_{j} H_{f_{jj_{1}}}k_{z} ⊗H_{g_{jk}}k_{z})^{∗}(∑_{μ} H_{f_{μj_{1}}}k_{z} ⊗H_{g_{μl}}k_{z}))

= (∑_{j_{1}} ∑_{j} ∑_{μ} < H_{f_{μj_{1}}}k_{z}, H_{f_{jj_{1}}}k_{z} > H_{g_{jk}}k_{z} ⊗H_{g_{μl}}k_{z}),

where the second equality above follows from the fact that

(x_{1} ⊗y_{1})^{∗}(x_{2} ⊗y_{2}) =< x_{2}, x_{1} > y_{1} ⊗y_{2}.

Thus

trace{T_{1}^{∗}T_{1}} = ∑_{l} ∑_{j_{1}} ∑_{j} ∑_{μ} < H_{f_{μj_{1}}}k_{z}, H_{f_{jj_{1}}}k_{z} >< H_{g_{jl}}k_{z}, H_{g_{μl}}k_{z} > .

If we write f_{ij} = (f_{ij})_{+} + (f_{ij})_{−} and g_{ij} = (g_{ij})_{+} + (g_{ij})_{−} for (f_{ij})_{+}, (g_{ij})_{+} ∈ H^{2} and (f_{ij})_{−}, (g_{ij})_{−} ∈ \overline{zH^{2}}, then H_{f_{ij}}k_{z} = ((f_{ij})_{−} −(f_{ij})_{−}(z))k_{z} and H_{g_{ij}}k_{z} = ((g_{ij})_{−} −(g_{ij})_{−}(z))k_{z}. Therefore, by a change of the order of summations, we have

trace{T_{1}^{∗}T_{1}} = ∑_{l} ∑_{j_{1}} ∑_{j} ∑_{μ} < [(f_{μj_{1}})_{−} −(f_{μj_{1}})_{−}(z)]k_{z}, [(f_{jj_{1}})_{−} −(f_{jj_{1}})_{−}(z)]k_{z} >

× < [(g_{jl})_{−} −(g_{jl})_{−}(z)]k_{z}, [(g_{μl})_{−} −(g_{μl})_{−}(z)]k_{z} >

= ∑_{μ} ∑_{j}(∑_{j_{1}} < [(f_{μj_{1}})_{−} −(f_{μj_{1}})_{−}(z)]k_{z}, [(f_{jj_{1}})_{−} −(f_{jj_{1}})_{−}(z)]k_{z} >)×

(∑_{l} < [(g_{jl})_{−} −(g_{jl})_{−}(z)]k_{z}, [(g_{μl})_{−} −(g_{μl})_{−}(z)]k_{z} >).

Note that

(∑_{j_{1}} < [(f_{μj_{1}})_{−} −(f_{μj_{1}})_{−}(z)]k_{z}, [(f_{jj_{1}})_{−} −(f_{jj_{1}})_{−}(z)]k_{z} >)_{n×n} = |F_{−} −F_{−}(z)|^{2}(z),

and similarly

(∑_{l} < [(g_{jl})_{−} −(g_{jl})_{−}(z)]k_{z}, [(g_{μl})_{−} −(g_{μl})_{−}(z)]k_{z} >)_{n×n} = |G_{−} −G_{−}(z)|^{2}(z).

Hence

trace{T_{1}^{∗}T_{1}} = ∑_{u} ∑_{j}(|F_{−} −F_{−}(z)|^{2}(z))_{μj}(|G_{−} −G_{−}(z)|^{2}(z))_{jμ}

= ∑_{u}[|F_{−} −F_{−}(z)|^{2}(z))(|G_{−} −G_{−}(z)|^{2}(z))]_{μμ}

= tr[|F_{−} −F_{−}(z)|^{2}(z)|G_{−} −G_{−}(z)|^{2}(z)].

Here we note that for a matrix A, |A|^{2} = AA^{∗} and (A)_{ij} denotes the (i, j)−th entry of A. This completes the proof of this lemma.

Theorem 3. Let F and G be in L_{n×n}^{∞}. If H_{F}^{∗} H_{G} is compact, then

lim_{|z|→1} ∥{[|F_{−} −F_{−}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)]^{1/2}}∥ = 0.

Proof. Set H^{2}(C^{n}) = H^{2} ⊕ ... ⊕ H^{2}. Let (A_{ij}) := H_{F}^{∗} H_{G} be the operator-valued n × n matrix representation of H_{F}^{∗} H_{G} with respect to the above decomposition of H^{2}(C^{n}). Since H_{F}^{∗} H_{G} compact, each entry A_{ij} of H_{F}^{∗} H_{G} is compact on H^{2}, by Lemma 2 [15], we have

lim_{|z|→1} ∥A_{ij} −T_{ϕ_{z}}^{∗}A_{ij}T_{ϕ_{z}}∥ = 0.

Hence

lim_{|z|→1} ∥H_{F}^{∗} H_{G} −T_{Φ_{z}}^{∗}H_{F}^{∗} H_{G}T_{Φ_{z}}∥ = 0.

By Lemma 1, H_{F}^{∗} H_{G} −T_{Φ_{z}}^{∗}H_{F}^{∗} H_{G}T_{Φ_{z}} is a finite rank operator. Therefore, we have

lim_{|z|→1} trace{(H_{F}^{∗} H_{G} −T_{Φ_{z}}^{∗}H_{F}^{∗} H_{G}T_{Φ_{z}})^{∗}(H_{F}^{∗} H_{G} −T_{Φ_{z}}^{∗}H_{F}^{∗} H_{G}T_{Φ_{z}})} = 0

because the norm of a finite rank positive operator is equivalent to its trace. By Lemma 2, we obtain

lim_{|z|→1} tr[|F_{−} −F_{−}(z)|^{2}(z)|G_{−} −G_{−}(z)|^{2}(z)] = 0.

On the other hand,

tr[|F_{−} −F_{−}(z)|^{2}(z)|G_{−} −G_{−}(z)|^{2}(z)]
= tr{[|F_{−} −F_{−}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)][|F_{−} −F_{−}(z)|^{2}(z)]^{1/2}}
= tr{[|F_{−} −F_{−}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)]^{1/2}×
{[|F_{−} −F_{−}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)]^{1/2}}^{∗}}.

As is well-known, for all n×n matrices A,

trA^{∗}A ≥ C∥A∥^{2}

for some constant C > 0. Hence we conclude

lim_{|z|→1} ∥{[|F_{−} −F_{−}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)]^{1/2}}∥ = 0.

This completes the proof of the theorem.

3. Finite sum of the product of Hankel operators

In this section we will discuss when the finite sum of the products of Hankel operators with scalar symbols is zero. This is needed in the next section for characterizing F and G in L_{n×n}^{∞} such that T_{F G} = T_{F} T_{G}.

Let M_{n×n} be the set of n×n matrices. Let A = [a_{ij}] ∈ M_{n×n}, define

∥A∥_{∞} = sup_{1≤i,j≤n}|a_{ij}|

and (M_{n×n})_{1} denotes the closed unit ball of M_{n×n} in the above norm. Let P_{n} be the set of n×n permutation matrices.

Proposition 4. Let f_{k} = (f_{k1}, ⋯, f_{kn})^{T} for k = 1, ⋯, m and g = (g_{1}, ⋯, g_{n})^{T} ∈ L^{2}(C^{n}). Let

S_{kn}:= f_{k1} ⊗g_{1} +⋯+f_{kn} ⊗g_{n}, k = 1, ⋯, m.

Then S_{kn} = 0 for all k = 1, ⋯, m if and only if there are a matrix A ∈ (M_{n×n})_{1} and a permutation matrix R such that

(R−A)f_{k} = 0, k = 1, ⋯, m and A^{∗}g = 0.

Proof. We first prove that S_{kn} = 0, k = 1, ⋯, m if and only if there are a matrix A_{0} ∈ (M_{n×n})_{1} and a permutation σ such that

(I −A_{0})f_{kσ} = 0, k = 1, ⋯, m and A_{0}^{∗}g_{σ} = 0,

where f_{kσ} = (f_{kσ(1)}, ⋯, f_{kσ(n)})^{T}. For any A ∈ (M_{n×n})_{1}, any permutation σ, set

x_{k}:= (x_{k1}, ⋯, x_{kn})^{T} = (I −A)f_{kσ}, k = 1, ⋯, m and y:= (y_{1}, ⋯, y_{n})^{T} = A^{∗}g_{σ},

then we have

S_{kn} = ∑ f_{kσ_{i}} ⊗g_{σ_{i}} = x_{k1} ⊗g_{σ(1)} +⋯+x_{kn} ⊗g_{σ(n)} +f_{kσ(1)} ⊗y_{1} +⋯+f_{kσ(n)} ⊗y_{n}.

The sufficiency follows from the above relation. To prove the necessity we use induction. It is clear that for n = 1, the result is true with A = 1 or A = 0. Now assume the result is true for n−1. Without loss of generality, assume that

max_{1≤i≤n}∥g_{i}∥_{2} = ∥g_{j}∥_{2} > 0

for some j. Note that if S_{kn} = 0, k = 1, ⋯, m, then

S_{kn}g_{j} = ∑_{i=1}^{n} < g_{j}, g_{i} > f_{i} = 0, k = 1, ⋯, m.

That is

f_{kj} +∑_{i≠j} a_{j}f_{ki} = 0,

where a_{i} =< g_{i}, g_{j} > / < g_{i}, g_{j} >, |a_{i}| ≤ 1 for i ≠ j. Now we rewrite S_{kn} as

S_{kn} = (f_{kj} +∑_{i≠j} a_{j}f_{ki})⊗g_{j} +∑_{i≠j} f_{ki} ⊗(g_{i} −\overline{a_{j}}g_{j}).

¿From above analysis we have that

∑_{i≠j} f_{ki} ⊗(g_{i} −\overline{a_{j}}g_{j}) = 0, k = 1, ⋯, m.

By induction there exist A_{1} ∈ (M_{n−1×n−1})_{1} and a permutation ω of {1, ⋯, j − 1, j + 1, ⋯, n} such that

(I −A_{1})(f_{kω(1)}, ⋯, f_{kω(j−1)}, f_{kω(j+1)}, ⋯, f_{kω(n)})^{T} = 0, k = 1, ⋯, m,
A_{1}^{∗}(g_{ω(1)} −\overline{a_{ω(1)}}g_{ω(j)}, ⋯, g_{ω(j−1)} −\overline{a_{ω(j−1)}}g_{ω(j)},
g_{ω(j+1)} −\overline{a_{ω(j+1)}}g_{ω(j)}, ⋯, g_{ω(n)} −\overline{a_{ω(n)}}g_{ω(j)})^{T} = 0.

Let

A_{0} = [ \begin{matrix}0 & a \\ 0 & A_{1} \end{matrix} ] , where a = [−a_{ω(1)}, ⋯, −a_{ω(j−1)}, −a_{ω(j+1)}, ⋯, −a_{ω(n)}].

Take σ to be such that σ(1) = j, σ(i) = ω(i + 1) for 2 ≤ i ≤ j − 1 and σ(i) = ω(i) for j +1 ≤ i ≤ n. It is easy to check that such A_{0} and σ are what we need. Now let R be the permuation matrix such that f_{kσ} = Rf_{k} for all k = 1, ⋯, m and A = R^{∗}A_{0}, then it is easy to check that for such A and R

(R−A)f_{k} = 0, k = 1, ⋯, m and A^{∗}g = 0

if and only if

(I −A_{0})f_{kσ} = 0, k = 1, ⋯, m and A_{0}^{∗}g_{σ} = 0.

The proof is complete.

Next we discuss when the finite sum of the products of Hankel operators is zero.

Theorem 5. Let f_{k} = (f_{k1}, ⋯, f_{kn})^{T} , k = 1, ⋯, m and g = (g_{1}, ⋯, g_{n})^{T} for f_{ij} and g_{i} in L^{∞}. Let

T_{kn}:= ∑_{i=1}^{n}H_{f_{ki}}^{∗}H_{g_{i}}, k = 1, ⋯, m.

Then T_{kn} = 0 for all k = 1, ⋯, m if and only if there are a matrix A ∈ (M_{n×n})_{1} and a permutation matrix R such that

(R−A)f_{k} ∈ H_{n×1}^{∞}, k = 1, ⋯, m and A^{∗}g ∈ H_{n×1}^{∞}. (1)

Proof. To prove the necessity, we recall the following identity proved in Lemma 1.

V^{−1}[∑_{i=1}^{n}(H_{f_{ki}}^{∗}H_{g_{i}} −T_{ϕ_{z}}^{∗}H_{f_{ki}}H_{g_{i}}T_{ϕ_{z}})]V = ∑_{i=1}^{n}H_{f_{ki}}k_{z} ⊗H_{g_{i}}k_{z}, k = 1, ⋯, m. (2)

Therefore if ∑_{i=1}^{n}H_{f_{ki}}^{∗}H_{g_{i}} = 0, then

∑_{i=1}^{n}H_{f_{ki}}k_{z} ⊗H_{g_{i}}k_{z} = 0.

In particular for z = 0 (i.e., k_{0} = 1), by Proposition 4, the above equation implies that there exist a matrix A ∈ (M_{n×n})_{1} and a permutation matrix R such that

(R−A)[H_{f_{k1}}1, ⋯, H_{f_{kn}}1]^{T} = 0, k = 1, ⋯, m and A^{∗}[H_{g_{1}}1, ⋯, H_{g_{n}}1]^{T} = 0.

That is (1) holds. To prove the sufficiency, as in the proof of Proposition 4, we note that for any A ∈ (M_{n×n})_{1}, any permutation matrix R, if we set

x_{k} = (x_{k1}, ⋯, x_{kn})^{T} = (R−A)f_{kσ} and y = (y_{1}, ⋯, y_{n})^{T} = A^{∗}g_{σ},

then we have

T_{kn} = H_{x_{k1}}^{∗}H_{g_{1}} +⋯+H_{x_{kn}}^{∗}H_{g_{n}} +H_{f_{k1}}^{∗}H_{y_{1}} +⋯+H_{f_{kn}}^{∗}H_{y_{n}}, k = 1, ⋯, m. (3)

The above formula and the fact that H_{b} is zero when b ∈ H^{∞} prove the sufficiency part of our theorem.

4. Zero semi-commutator or commutator

Brown and Halmos [4] showed that the semi-commutator T_{ϕψ} −T_{ϕ}T_{ψ} is zero exactly when either ϕ or ψ is analytic for the scalar functions ϕ and ψ. Halmos [12] also characterized when the commutator T_{ϕ}T_{ψ} −T_{ψ}T_{ϕ} is zero. In this section, we will characterize when the semi-commutator T_{F G} − T_{F} T_{G} or the commutator T_{F} T_{G} − T_{G}T_{F} is zero for block Toeplitz operators with matrix symbols F and G.

Let E_{j} be the n×n matrix unit with (j, j)-th entry equal to one and all other entries equal to zero. Note that for a m×n matrix B, BE_{j} is basically the j-th column of B.

Theorem 6. Let F, G ∈ L_{n×n}^{∞}. The following are equivalent.

(1) The semi-commutator T_{F} T_{G} −T_{F G}(= −H_{F^{∗}}^{∗}H_{G}) is zero.

(2) There exist matrices A_{j} ∈ (M_{n×n})_{1} and R_{j} ∈ P_{n}, j = 1, ⋯, n such that

(R_{j} −A_{j})F^{∗} ∈ H_{n×n}^{∞} and A_{j}^{∗}GE_{j} ∈ H_{n×n}^{∞}, j = 1, ⋯, n.

(3)

[|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)]^{1/2} = 0

for all z ∈ D. (4)

[|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)]^{1/2} = 0

for some z ∈ D.

Proof. (1)⟺(2). Let F = (f_{ij}) and G = (g_{ij}). Since T_{F} T_{G} − T_{F G} = H_{F^{∗}}^{∗}H_{G}, T_{F} T_{G} − T_{F G} = 0 if and only if for each j = 1, ⋯, n

∑_{i=1}^{n}H∗ ; f̄_{ik}H_{g_{ij}} = 0, k = 1, ⋯, n.

By Theorem 5, this is equivalent to that for each j = 1, ⋯, n, there exist matrix A_{j} ∈ (M_{n×n})_{1} and R_{j} ∈ P_{n} such that

(R_{j} −A_{j})F^{∗} ∈ H_{n×n}^{∞} and A_{j}^{∗}GE_{j} ∈ H_{n×n}^{∞}.

(1)⟹ (3). Since T_{F} T_{G} −T_{F G} = H_{F^{∗}}^{∗}H_{G}, by Lemma 2, for all z ∈ D, we have

trace{[H_{F^{∗}}^{∗}H_{G} −T_{Φ_{z}}^{∗}[H_{F^{∗}}^{∗}H_{G}]T_{Φ_{z}}]^{∗}[H_{F^{∗}}^{∗}H_{G} −T_{Φ_{z}}^{∗}[H_{F^{∗}}^{∗}H_{G}]T_{Φ_{z}}]}
= tr[(|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z))(|G_{−} −G_{−}(z)|^{2}(z))]
= tr[[|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)]^{1/2} ×
[|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)]^{1/2})^{∗}].

Hence

[|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)]^{1/2} = 0.

(3)⟹(4). This is obvious.

(4)⟹(1). For a given z ∈ D, define a unitary operator U_{z} on H^{2} by U_{z}h = h∘ϕ_{z}k_{z}. Let U_{z} = diag{U_{z}, ⋯, U_{z}}. Then it is easy to check that

U_{z}^{∗}T_{F} U_{z} = T_{F ∘ϕ_{z}}.

Therefore

U_{z}^{∗}[H_{F^{∗}}^{∗}H_{G} −T_{Φ_{z}}^{∗}H_{F^{∗}}^{∗}H_{G}T_{Φ_{z}}]U_{z}
= [H_{F^{∗}∘ϕ_{z}}^{∗}H_{G∘ϕ_{z}} −T_{Φ_{0}}^{∗}H_{F^{∗}∘ϕ_{z}}^{∗}H_{G∘ϕ_{z}}T_{Φ_{0}}].

So it is sufficient to prove that H_{F^{∗}}^{∗}H_{G} = 0 if we assume that

[|(F_{+})^{∗} −(F_{+})^{∗}(0)|^{2}(0)]^{1/2}[|G_{−} −G_{−}(0)|^{2}(0)]^{1/2} = 0.

By Lemma 2, we have

trace{[H_{F^{∗}}^{∗}H_{G} −T_{Φ_{0}}^{∗}H_{F^{∗}}^{∗}H_{G}T_{Φ_{0}}]^{∗}[H_{F^{∗}}^{∗}H_{G} −T_{Φ_{0}}^{∗}H_{F^{∗}}^{∗}H_{G}T_{Φ_{0}}]}
= tr[[|(F_{+})^{∗} −(F_{+})^{∗}(0)|^{2}(0)]^{1/2}[|G_{−} −G_{−}(0)|^{2}(0)]^{1/2} ×
([|(F_{+})^{∗} −(F_{+})^{∗}(0)|^{2}(0)]^{1/2}[|G_{−} −G_{−}(0)|^{2}(0)]^{1/2})^{∗}].

So

H_{F^{∗}}^{∗}H_{G} −T_{Φ_{0}}^{∗}H_{F^{∗}}^{∗}H_{G}T_{Φ_{0}} = 0.

Thus it follows from a theorem [6] that there is a matrix valued function M in L_{n×n}^{∞} such that H_{F^{∗}}^{∗}H_{G} = T_{M}. But by the Douglas theorem [7] we have that M = 0. Hence H_{F^{∗}}^{∗}H_{G} = 0.

Next we study the commutator T_{F} T_{G} −T_{G}T_{F} by reducing it to the semi-commutator case. To see this, first note that

T_{F} T_{G} −T_{G}T_{F} = T_{F} T_{G} −T_{F G} +T_{GF} −T_{G}T_{F} +T_{(F G−GF)}
= −(H_{F^{∗}}^{∗}H_{G} −H_{G^{∗}}^{∗}H_{F})+T_{(F G−GF)}.

Let

B = [ \begin{matrix}F & −G \\ 0 & 0 \end{matrix} ] , C = [ \begin{matrix}G & 0 \\ F & 0 \end{matrix} ] .

A simple calculation gives that

H_{B^{∗}}^{∗}H_{C} = [ \begin{matrix}H_{F^{∗}}^{∗}H_{G} −H_{G^{∗}}^{∗}H_{F} & 0 \\ 0 & 0 \end{matrix} ] .

Therefore, by the Douglas theorem [7], T_{F} T_{G} −T_{G}T_{F} = 0 if and only if H_{B^{∗}}^{∗}H_{C} = 0 and F G = GF. A straightforward computation shows that

|(B_{+})^{∗} −(B_{+})^{∗}(z)|^{2}(z) (4)

= [ \begin{matrix}|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z) & −(((F_{+})^{∗} −(F_{+})^{∗}(z))(G_{+} −G_{+}(z)))(z) \\ −(((G_{+})^{∗} −(G_{+})^{∗}(z))(F_{+} −F_{+}(z)))(z) & |(G_{+})^{∗} −(G_{+})^{∗}(z)|^{2}(z) \end{matrix} ]

|C_{−} −C_{−}(z)|^{2}(z) (5)

= [ \begin{matrix}|G_{−} −G_{−}(z)|^{2}(z) & ((G_{−} −G_{−}(z))((F_{−})^{∗} −(F_{−})^{∗}(z)))(z) \\ ((F_{−} −F_{−}(z))((G_{−})^{∗} −(G_{−})^{∗}(z)))(z) & |F_{−} −F_{−}(z)|^{2}(z) \end{matrix} ]

This leads to the following result.

Theorem 7. Let F, G ∈ L_{n×n}^{∞}. The following are equivalent.

(1) The commutator T_{F} T_{G} −T_{G}T_{F} is zero.

(2) GF = F G and there exist matrices A_{j} ∈ (M_{2n×2n})_{1} and R_{j} ∈ P_{2n}, j = 1, ⋯, n such that

(R_{j} −A_{j}) [ \begin{matrix}F^{∗} \\ −G^{∗} \end{matrix} ] ∈ H_{2n×n}^{∞} and A_{j}^{∗} [ \begin{matrix}G \\ F \end{matrix} ] E_{j} ∈ H_{2n×n}^{∞}, j = 1, ⋯, n.

(3) GF = F G. And

[ \begin{matrix}|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z) & −(((F_{+})^{∗} −(F_{+})^{∗}(z))(G_{+} −G_{+}(z)))(z) \\ −(((G_{+})^{∗} −(G_{+})^{∗}(z))(F_{+} −F_{+}(z)))(z) & |(G_{+})^{∗} −(G_{+})^{∗}(z)|^{2}(z) \end{matrix} ]^{1/2} ×

[ \begin{matrix}|G_{−} −G_{−}(z)|^{2}(z) & ((G_{−} −G_{−}(z))((F_{−})^{∗} −(F_{−})^{∗}(z)))(z) \\ ((F_{−} −F_{−}(z))((G_{−})^{∗} −(G_{−})^{∗}(z)))(z) & |F_{−} −F_{−}(z)|^{2}(z) \end{matrix} ]^{1/2} = 0

for some z ∈ D.

We remark that as in Theorem 6, the matrix in the statement (3) of Theorem 7 is zero for some z ∈ D if and only if it is zero for all z ∈ D.

An operator A is said to be normal if AA^{∗} − A^{∗}A = 0. We observe that by taking G = F^{∗} in Theorem 7 and noting that T_{F}^{∗} = T_{F^{∗}}, we have the following characterization of normal block Toeplitz operators.

Corollary 8. Let F ∈ L_{n×n}^{∞}. The following are equivalent.

(1) T_{F} is normal.

(2) F^{∗}F = F F^{∗}. And there exist matrices A_{j} ∈ (M_{2n×2n})_{1} and R_{j} ∈ P_{2n}, j = 1, ⋯, n such that

(R_{j} −A_{j}) [ \begin{matrix}F^{∗} \\ −F \end{matrix} ] ∈ H_{2n×n}^{∞} and A_{j}^{∗} [ \begin{matrix}F^{∗} \\ F \end{matrix} ] E_{j} ∈ H_{2n×n}^{∞}, j = 1, ⋯, n.

(3) F^{∗}F = F F^{∗}. And

[ \begin{matrix}|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z) & −((F_{−} −F_{−}(z))(F_{+} −F_{+}(z))(z) \\ −(((F_{+})^{∗} −(F_{+})^{∗}(z))((F_{−})^{∗} −(F_{−})^{∗}(z))(z) & |F_{−} −F_{−}(z)|^{2}(z) \end{matrix} ]^{1/2} ×

[ \begin{matrix}|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z) & ((F_{−} −F_{−}(z))(F_{+} −F_{+}(z))(z) \\ (((F_{+})^{∗} −(F_{+})^{∗}(z))((F_{−})^{∗} −(F_{−})^{∗}(z))(z) & |F_{−} −F_{−}(z)|^{2}(z) \end{matrix} ]^{1/2} = 0

for some z ∈ D.

Proof. Let |(B_{+})^{∗} − (B_{+})^{∗}(z)|^{2}(z) and |C_{−} − C_{−}(z)|^{2} be defined by (4) and (5) with G = F^{∗}. The corollary follows from Theorem 7 by noting that

(G_{+})^{∗} −(G_{+})^{∗}(z) = F_{−} −F_{−}(z), G_{−} −G_{−}(z) = (F_{+})^{∗} −(F_{+})^{∗}(z).

5. A distribution function inequality

Recall that a necessary condition for the compactness of the semi-commutator T_{F G} − T_{F} T_{G} is obtained in Theorem 3. Namely, the compactness of T_{F G} −T_{F} T_{G} implies that

lim_{|z|→1} ∥[|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)]^{1/2}∥ = 0.

To prove that the above condition is also a sufficient condition for the compactness of T_{F G} − T_{F} T_{G}, we need a certain distribution function inequality. The distribution function inequality involves the Lusin area integral and the Hardy-Littlewood maximal function. The idea to use distribution function inequalities in the theory of Toeplitz operators and Hankel operators first appeared in [2]. In this section we will get such a distribution function inequality.

For w a point of ∂D, we let Γ_{w} denote the angle with vertex w and 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_{w,ϵ}. For h in L^{1}(∂D), we define the Lusin area integral of h to be

A_{ϵ}(h)(w) = [∫_{Γ_{w,ϵ}} |gradh(z)|^{2}dA(z)]^{1/2}

where h(z) means the harmonic extension of h on D and dA(z) denotes the Lebesgue measure on the unit disk D. The Hardy-Littlewood maximal function of the function h will be denoted Mh, and for r > 1, we let Λ_{r}h = [M|h|^{r}]^{1/r}. For z ∈ D, we let I_{z} denote the closed subarc of ∂D with center \frac{z}{|z|} and measure δ(z) = 1 − |z|. The Lebesgue measure of the subset E of ∂D will be denoted by |E|.

Let f_{i} and g_{i} (i = 1, ⋯, n) be in L^{∞}(∂D) and u and v be in H^{2}(∂D). We define a generalized area integral to be

B_{ϵ}(u, v)(w) = ∫_{Γ_{w,ϵ}} |∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}v)}|dA(z).

For l > 2 and z ∈ D, define

Ξ_{l}(z) = inf{∑_{i=1}^{n}(∥x_{i} ∘ϕ_{z} −P (x_{i} ∘ϕ_{z})∥_{l} +∥y_{i} ∘ϕ_{z} −P (g_{i} ∘ϕ_{z})∥_{l}) : A ∈ (M_{n×n})_{1}, R ∈ P_{n}},

where x = (x_{1}, ⋯, x_{n})^{T} = (R−A)f and y = (y_{1}, ⋯, y_{n})^{T} = A^{∗}g,

Γ_{l}(z) = ∑_{i=1}^{n}(∥f_{i} ∘ϕ_{z} −P (f_{i} ∘ϕ_{z})∥_{l} +∥g_{i} ∘ϕ_{z} −P (g_{i} ∘ϕ_{z})∥_{l}).

We have the following distribution function inequality.

Theorem 9. Let f_{i} and g_{i} be in L^{∞}, i = 1, ⋯, n, and u and v in the Hardy space H^{2}. Let z be a point in D such that |z| > 1/2. Then for any l > 2, for a > 0 sufficiently large and δ(z) = 1−|z|

|{w ∈ I_{z}: B_{2δ(z)}(u, v)(w) < aΞ_{l}(z)Γ_{l}(z) inf_{w∈I_{z}} Λ_{r}(u)(w) inf_{w∈I_{z}} Λ_{r}(v)(w)}| ≥ C_{a}|I_{z}|

where C_{a} depends only on l and a, lim_{a→∞}C_{a} = 1, and 1/l +1/r = 2/p for some 1 < p < 2 and 1 < r < 2.

Proof. Let f_{i} and g_{i} be in L^{∞} and u and v in H^{2}. By our definition

B_{ϵ}(u, v)(w) = ∫_{Γ_{w,ϵ}} |∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}v)}|dA(w).

For a fixed z ∈ D, let χ_{2I_{z}} denotes the characteristic function of the subset 2I_{z} of ∂D and write H_{f_{i}}u = u_{i1} +u_{i2}, H_{g_{i}}v = v_{i1} +v_{i2}, where

u_{i1} = (I −P)[(f_{i} −P (f ∘ϕ_{z}))∘ϕ_{z})χ_{2I_{z}}u],
u_{i2} = (I −P)[(f_{i} −P (f_{i} ∘ϕ_{z}))∘ϕ_{z})(1−χ_{2I_{z}})u],

v_{i1} = (I −P)[(g_{i} −P (g_{i} ∘ϕ_{z}))∘ϕ_{z})χ_{2I_{z}}v], v_{i2} = (I −P)[(g_{i} −P (g_{i} ∘ϕ_{z}))∘ϕ_{z})(1−χ_{2I_{z}})v].

Thus we have

B_{ϵ}(u, v)(w) ≤ ∑_{i=1}^{n}(A_{ϵ}(u_{i1})+A_{ϵ}(u_{i2}))(A_{ϵ}(v_{i1})+A_{ϵ}(v_{i2})) (6)

¿From now on we use ϵ to denote 2δ(z). Let i (1 ≤ i ≤ n) be fixed. We first recall the following estimates from [15] for the terms A_{ϵ}(u_{i1}) and A_{ϵ}(u_{i2}) (similarly A_{ϵ}(v_{i1}) and A_{ϵ}(v_{i2})).

For l > 2, there are a positive constant C and r ∈ (1, 2) such that

[∫_{I_{z}} A_{ϵ}(u_{i1})^{p}dσ(w)]^{1/p} ≤ C|I_{z}|^{1/p}∥f_{i} ∘ϕ_{z} −P (f_{i} ∘ϕ_{z})∥_{l} inf_{w∈I_{z}} Λ_{r}u(w) (7)

for some p > 1 so that 1/l +1/r = 1/p.

For l > 2, on I_{z},

A_{ϵ}(u_{i2}) ≤ C∥f_{i} ∘ϕ_{z} −P (f_{i} ∘ϕ_{z})∥_{l} inf_{w∈I_{z}} Λ_{l′}u(w), (8)

for some C > 0 and 1/l +1/l′ = 1.

For completeness we give a proof here. We first write u_{ij} = (I −P)u_{j} ( j = 1, 2) where

u_{1} = (f_{i−} −f_{i−}(z))χ_{2I_{z}}u, u_{2} = (f_{i−} −f_{i−}(z))(1−χ_{2I_{z}}u)

and f_{i−} = (I −P)f_{i}.

We first prove (7). Note that for l > 2, we can always find l′ > 2 and p > 1 so that l = l′p and r = p\frac{l′}{l′−1} < 2. By the theorem of Marcinkiewicz and Zygmund, the truncated Lusin area integral A_{ϵ}f(w) is L^{p}-bounded for 1 < p < ∞. So for l > 2, we have

∫_{I_{z}}[A_{ϵ}(I −P)(u_{1})(w)]^{p}dσ(w) ≤ C ∫_{∂D} |u_{1}|^{p}dσ(w)

= C ∫_{2I_{z}} |f_{i−} −f_{i−}(z)|^{p}|u(w)|^{p}dσ(w)

≤ |2I_{z}|[\frac{1}{|2I_{z}|} ∫_{2I_{z}} |f_{i−}(w)−f_{i−}(z)|^{pl′}dσ(w)]^{1/l′}[\frac{1}{|2I_{z}|} ∫_{2I_{z}} |u|^{r}dσ(w)]^{p/r}.

Let P (z, w) denote the Poisson kernel for the point z. Since

[\frac{1}{|2I_{z}|} ∫_{2I_{z}} |u|^{r}dσ(w)]^{1/r} ≤ Λ_{r}u(w)

for each w ∈ 2I_{z}, and an elementary estimate shows that for w ∈ 2I_{z}, P (z, w) > \frac{C}{|2I_{z}|}, it follows that

[∫_{I_{z}} A_{ϵ}((I −P)(u_{1})(w))^{p}dσ(w)]^{1/p} ≤ C|I_{z}|^{1/p}[|f_{i−} −f_{i−}(z)|^{l}(z)]^{1/l} inf_{w∈I_{z}} Λ_{r}u(w).

Thus (7) follows from the following identity.

[|f_{i−} −f_{i−}(z)|^{l}(z)]^{1/l} = ∥f_{i} ∘ϕ_{z} −P (f_{i} ∘ϕ_{z})∥_{l} (9)

Now we prove (8). For u_{2}, we shall use a pointwise estimate of the norm of the gradient of (I −P)u_{2}. It is easy to see that

(I −P)(u_{2})(w) = \frac{1}{2π} ∫ \frac{\overline{w}ξu_{2}(ξ)}{1−\overline{w}ξ}dσ(ξ).

Thus

|grad(I −P)u_{2}(w)| ≤ C ∫ \frac{|u_{2}(ξ)|}{|1−\overline{w}ξ|^{2}} dσ(ξ)

≤ C ∫_{∂D/2I_{z}} \frac{|[f_{i−}(ξ)−f_{i−}(z)]u(ξ)|}{|1−\overline{w}ξ|^{2}}dσ(ξ)

On the other hand, there is a constant C > 0 so that

|\frac{1−\overline{z}ξ}{1−\overline{w}ξ} | ≥ C

for all ξ in ∂D/2I_{z} and w in I_{z}. Thus we obtain

|grad(I −P)u_{2}(w)| ≤ C ∫_{∂D/2I_{z}} \frac{|[f_{i−}(ξ)−f_{i−}(z)]u(ξ)|}{|1−\overline{z}ξ|^{2}}dσ(ξ).

Applying the Hölder inequality yields

|grad(I −P)u_{2}(w)| ≤ \frac{C}{1−|z|^{2}} [|f_{i−} −f_{i−}(z)|^{l}(z)]^{1/l}[(|u|^{l′})(z)]^{1/l′}.

Because the nontangential maximal function is bounded by a constant times the Hardy-Littlewood maximal function, and because z belongs to Γ_{w,ϵ}, the last factor on the right is no larger than CΛ_{l′}u(w), and again the desired inequality is established by noting (9).

Now we can estimate the products A_{ϵ}(u_{ij})A_{ϵ}(v_{ij}), j = 1, 2 by using (7) and (8). By Hölder inequality, we have

[∫_{I_{z}}[A_{ϵ}(u_{ij})A_{ϵ}(v_{ij})]^{p/2}dσ(w)] ≤ [∫_{I_{z}}[A_{ϵ}(u_{ij})]^{p}dσ(w)]^{1/2}[∫_{I_{z}}[A_{ϵ}(v_{ij})]^{p}dσ(w)]^{1/2}.

By using estimate (7) for integral of A_{ϵ}(u_{i1}) or A_{ϵ}(v_{i1}) and estimate (8) for integral of A_{ϵ}(u_{i2}) or A_{ϵ}(v_{i2}), we get that for l > 2, there is r with 1 < r < 2 such that

[∫_{I_{z}}[A_{ϵ}(u_{ij})A_{ϵ}(v_{ij})]^{p/2}dσ(w)]^{2/p} ≤ C|I_{z}|^{2/p}∥f_{i} ∘ϕ_{z} −P (f_{i} ∘ϕ_{z})∥_{l}∥g_{i} ∘ϕ_{z} −P (g_{i} ∘ϕ_{z}∥_{l}×

(10)  inf_{w∈I_{z}} Λ_{r}u(w) inf_{w∈I_{z}} Λ_{r}v(w)

for some p > 1 so that 1/l +1/r = 2/p, and a constant C depends on only l.

Now we are going to finish the proof by summing up the estimates as above. ¿From (6) we have

[∫_{I_{z}} B(u, v)^{p/2}dσ(w)]^{2/p} ≤ C(∫_{I_{z}} ∑_{i=1}^{n} ∑_{j=1}^{2}{[A_{ϵ}(u_{ij})(w)A_{ϵ}(v_{ij})(w)]^{p/2})^{2/p}.

By estimate (10), we have

∫_{I_{z}} B(u, v)^{p/2}dσ(w) ≤ C|I_{z}|×

(11) (∑_{i=1}^{n}∥(f_{i} ∘ϕ_{z} −P (f_{i} ∘ϕ_{z})∥_{l}^{p/2}∥g_{i} ∘ϕ_{z} −P (g_{i} ∘ϕ_{z})∥_{l}^{p/2})( inf_{w∈I_{z}} Λ_{s}u(w) inf_{w∈I_{z}} Λ_{s}v(w))^{p/2}.

Next for any A ∈ (M_{n×n})_{1} and any permutation matrix R ∈ P_{n}, let

x = (x_{1}, ⋯, x_{n})^{T} = (R−A)f and y = (y_{1}, ⋯, y_{n})^{T} = A^{∗}g,

where f = (f_{1}, ⋯, f_{n})^{T} and g = (g_{1}, ⋯, g_{n})^{T}. We apply the above inequality (11) to the vector-valued functions x and y. We note that the B_{ϵ}(u, v) corresponding to f and g is the same as the B_{ϵ}(u, v) corresponding to x and y; more precisely,

B_{ϵ}(u, v)(w) = ∫_{Γ_{w,ϵ}} |∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}v)}|dA(w)

= ∫_{Γ_{w,ϵ}} |∑_{i=1}^{n}(grad(H_{x_{i}}u)\overline{grad(H_{g_{i}}v)}+grad(H_{f_{i}}u)\overline{grad(H_{y_{i}}v)})|dA(w).

By formula (11), we have

∫I_{z}B(u, v)^{p/2}dσ(w) ≤ C|I_{z}|( inf_{w∈I_{z}} Λ_{s}u(w) inf_{w∈I_{z}} Λ_{s}v(w)]^{p/2})×

(∑_{i=1}^{n}(∥x_{i} ∘ϕ_{z} −P (x_{i} ∘ϕ_{z})∥_{l}^{p/2}∥g_{i} ∘ϕ_{z} −P (g_{i} ∘ϕ_{z})∥_{l}^{p/2} +

∥f_{i} ∘ϕ_{z} −P (f_{i} ∘ϕ_{z})∥_{l}^{p/2}∥y_{i} ∘ϕ_{z} −P (y_{i} ∘ϕ_{z})∥_{l}^{p/2})).

Therefore

[∫_{I_{z}} B_{ϵ}(u, v)^{p/2}dσ(w)]^{2/p} ≤ C|I_{z}|^{2/p}Ξ_{l}(z)Γ_{l}(z) inf_{w∈I_{z}} Λ_{s}u(w) inf_{w∈I_{z}} Λ_{s}v(w).

Next for a fixed z in D and a > 0, let E(a) be the set of points in I_{z} where

B_{ϵ}(u, v) ≤ aΞ_{l}(z)Γ_{l}(z) inf_{w∈I_{z}} Λ_{s}h(w) inf_{w∈I_{z}} Λ_{s}v(w).

Then

|I_{z}/E(a)|^{2/p}aΞ_{l}(z)Γ_{l}(z) inf_{w∈I_{z}} Λ_{s}h(w) inf_{w∈I_{z}} Λ_{s}v(w) ≤ [∫_{I_{z}} B_{ϵ}(u, v)dσ(w)]^{2/p}

≤ C|I_{z}|^{2/p}Ξ_{l}(z)Γ_{l}(z) inf_{w∈I_{z}} Λ_{s}h(w) inf_{w∈I_{z}} Λ_{s}v(w).

So

|I_{z}/E(a)| ≤ Ca^{−p/2}|I_{z}|.

Therefore for a sufficient large a > 0, we have

|E(a)| ≥ (1−Ca^{−p/2})|I_{z}|.

Let C_{a} = 1−Ca^{−p/2}. This completes the proof of the theorem.

6. Compact finite sum of products

Before proceeding to our main results in this section, we need to introduce some notations involving the maximal ideal space of an algebra. Let M be the maximal ideal space of H^{∞}, which is defined to be the set of multiplicative linear maps from H^{∞} onto the field of complex numbers. Each multiplicative linear functional ϕ ∈ M has norm 1 (as an element of the dual of H^{∞}). If we think of M has a subset of the dual space H^{∞} with weak-star topology then M becomes a compact Hausdorff space. For z ∈ D the evaluation functional f → f(z) is a multiplicative functional. So we can think of D as a subset of M. The Carleson corona theorem tells us that D is dense in M.

By using the Gelfand transform, we can think of H^{∞} as a subset of C(M), the continuous, complex-valued functions on the maximal ideal space of H^{∞}. Explicitly, for f ∈ H^{∞}, we extend f from D to M by defining

f(τ) = τ(f)

for every τ ∈ M. Note that this definition is consistent with our earlier identification of D with a subset of M.

By the Hahn-Banach theorem each τ ∈ M extends to a linear functional τ′ on L^{∞}. In fact, there is a unique representing measure dμ supported on M(L^{∞}), the maximal ideal space of L^{∞}, such that for each g ∈ L^{∞}, τ′(g) = ∫_{supp(τ′)} gdμ. A subset of M(L^{∞}) will be called a support set, denoted by suppτ, if it is the (closed) support set of the representing measure for the extension of a functional τ in M(H^{∞} +C).

For f ∈ L^{∞}, we let H^{∞}[f] denote the closed subalgebra of L^{∞} generated by H^{∞} and the function f. If f = (f_{1}, ⋯, f_{n})^{T}, we still use H^{∞}[f] to denote the closed subalgebra of L^{∞} generated by H^{∞} and functions f_{1}, ⋯, f_{n}. Recall that

Ξ_{2}(z) = inf{∑_{i=1}^{n}(∥x_{i} ∘ϕ_{z} −P (x_{i} ∘ϕ_{z})∥_{2} +∥y_{i} ∘ϕ_{z} −P (g_{i} ∘ϕ_{z})∥_{2}) : A ∈ (M_{n×n})_{1} R ∈ P_{n}},

where

x = (x_{1}, ⋯, x_{n})^{T} = (R−A)f and y = (y_{1}, ⋯, y_{n})^{T} = A^{∗}g.

Theorem 10. Let f = (f_{1}, ⋯, f_{n})^{T} and g = (g_{1}, ⋯, g_{n})^{T} for f_{i} and g_{i} in L^{∞}. The following are equivalent.

(1) H_{f_{1}}^{∗}H_{g_{1}} +⋯+H_{f_{n}}^{∗}H_{g_{n}} is compact.

(2) lim_{z→∂D} ∥∑_{i=1}^{n}(H_{f_{i}}k_{z})⊗(H_{g_{i}}k_{z})∥ = 0.

(3) lim_{z→∂D} Ξ_{2}(z) = 0.

(4) For each m ∈ M(H^{∞} + C), there exist a matrix A ∈ (M_{n×n})_{1} and a permutation matrix R ∈ P_{n} such that (R−A)f|suppm ∈ H^{∞}|suppm and A^{∗}g|suppm ∈ H^{∞}|suppm (5) The following relation holds.

∩_{{A∈(M_{n×n})_{1}, R∈P_{n}}}H^{∞}[(R−A)f, A^{∗}g] ⊂ H^{∞} +C.

Proof. Without loss of generality we may assume that ∥f_{i}∥_{∞} < 1/2 and ∥g_{i}∥_{∞} < 1/2 for all i = 1, ⋯, n.

(1) ⟹ (2). Assume that ∑_{i=1}^{n}H_{f_{i}}^{∗}H_{g_{i}} is compact. By Lemma 2 [15] we obtain

lim_{|z|→1} ∥∑_{i=1}^{n}H_{f_{i}}^{∗}H_{g_{i}} −T_{ϕ_{z}}^{∗}(∑_{i=1}^{n}H_{f_{i}}^{∗}H_{g_{i}})T_{ϕ_{z}}∥ = 0. (12)

But by the proof of Lemma 1,

V^{−1}(∑_{i=1}^{n}H_{f_{i}}^{∗}H_{g_{i}} −T_{ϕ_{z}}^{∗}(∑_{i=1}^{n}H_{f_{i}}^{∗}H_{g_{i}})T_{ϕ_{z}})V = ∑_{i=1}^{n}H_{f_{i}}k_{z} ⊗H_{g_{i}}k_{z},

where recall that V is the antiunitary operator on L^{2}(∂𝔻) defined by V h(e^{iθ}) = e^{−iθ}\overline{h(e^{iθ}}). Thus

lim_{|z|→1} ∥∑_{i=1}^{n}H_{f_{i}}k_{z} ⊗H_{g}k_{z}∥ = 0. (13)

That is (2) holds.

(2) ⟹ (3). Assume now that (2) holds. Suppose that (3) does not hold. That is, there are δ > 0 and a net {z} ⊂ D accumulating a point in ∂D such that

Ξ_{2}(z) ≥ δ.

We will get a contradiction. We may assume that the net {z} converges to some nontrivial point m ∈ M(H^{∞} +C).

Let H_{m}^{∞} denote the algebra H^{∞}|_{suppm} on suppm, and L_{m}^{∞} denote the algebra L^{∞}|_{suppm}. Then L_{m}^{∞}/H_{m}^{∞} is a vector space. For a function ψ in L^{∞}, let [ψ]_{m} denote the element in L_{m}^{∞}/H_{m}^{∞} which contains ψ. For f = (f_{1}, ⋯, f_{n})^{T}, let [f]_{m} = ([f_{1}]_{m}, ⋯, [f_{n}]_{m})^{T} and f ∈ H_{m}^{∞} means that f_{i} ∈ H_{m}^{∞} for all i = 1, ⋯, n. Let g = (g_{1}, ⋯, g_{n})^{T}. For convenience we also introduce the following notations.

∥H_{f}k_{z}∥_{2} := ∑_{i=1}^{n} ∥H_{f_{i}}k_{z}∥_{2},

H_{f}k_{z} ⊗H_{g}k_{z}:= ∑_{i=1}^{n} H_{f_{i}}k_{z} ⊗H_{g_{i}}k_{z}.

Suppose that the dimension of the space spanned by [f_{1}]_{m}, ..., [f_{n}]_{m} is N ≤ n. We may assume that {[f_{1}]_{m}, ..., [f_{N}]_{m}} is a basis such that ([f_{1}]_{m}, ..., [f_{n}]_{m})^{T} = B([f_{1}]_{m}, ..., [f_{N}]_{m})^{T} with B = (b_{ij}) and |b_{ij}| ≤ 1, see the proof of Proposition 4 for details. Let A be the matrix (B, 0)_{n×n}. Then f −Af is in H_{m}^{∞} on the support set suppm. By Lemma 3 [15],

lim_{z→m} ∥H_{f−Af}k_{z}∥_{2} = 0.

On the other hand,

H_{f}k_{z} ⊗H_{g}k_{z} = H_{f−Af}k_{z} ⊗H_{g}k_{z} +H_{f}k_{z} ⊗H_{A^{∗}g}k_{z}
= H_{f−Af}k_{z} ⊗H_{g}k_{z} +H_{f_{(N)}}k_{z} ⊗H_{B^{∗}g}k_{z},

where f_{(N)} = (f_{1}, ..., f_{N}). As z goes to m, the first term in the right hand side of the above equation goes to zero. Hence

lim_{z→m} ∥H_{f_{(N)}}k_{z} ⊗H_{B^{∗}g}k_{z}∥ = 0.

We are going to show that

lim_{z→m} ∥H_{B^{∗}g}k_{z}∥_{2} = 0.

Suppose that this is not true. We may assume that

lim_{z→m} ∥H_{g_{1}}k_{z}∥_{2} > 0.

Let a_{i}(z) =< H_{g_{1}}k_{z}, H_{g_{i}}k_{z} >. Note that |a_{i}(z)| ≤ \sqrt{∥g_{1}∥_{∞}}\sqrt{∥g_{i}∥_{∞}} ≤ 1. We may assume that a_{i}(z) converges to a_{i} as z goes to m. By our assumption a_{1} ≠ 0. But

lim_{z→m} ∥(H_{f_{(N)}}k_{z} ⊗H_{B^{∗}g}k_{z})H_{g_{1}}k_{z}∥_{2} = 0

implies that

lim_{z→m} ∥H_{∑_{i=1}^{N} a_{i}f_{i}}k_{z}∥_{2} = 0.

By Lemma 3 [15], ∑_{i=1}^{N} a_{i}f_{i} is in H^{∞} on suppm. This contradicts the fact that {[f_{1}]_{m}, ..., [f_{N}]_{m}} is a basis. Therefore

lim_{z→m} ∥H_{B^{∗}g}k_{z}∥_{2} = 0.

Hence

lim_{z→m} ∥H_{A^{∗}g}k_{z}∥_{2} = 0.

But

∥H_{(I−A)f}k_{z}∥+∥H_{A^{∗}g}k_{z}∥ ≥ Ξ_{2}(z).

Hence

lim_{z→m}(∥H_{(I−A)f}k_{z}∥+∥H_{A^{∗}g}k_{z}∥) ≥ δ.

This is a contradiction.

(3) ⟹ (4). We are going to show that for each m ∈ M(H^{∞} +C), there exist matrices A_{m} ∈ (M_{n×n})_{1} and R_{m} ∈ P_{n} such that

[(R_{m} −A_{m})f]_{m} = 0, [A_{m}^{∗}g]_{m} = 0.

Let z be a net in D converging to m. By condition (3), there are matrices A_{z} ∈ (M_{n×n})_{1} and R_{z} ∈ P_{n} such that

lim_{z→m}(∥H_{(R_{z}−A_{z})f}k_{z}∥_{2} +∥H_{A_{z}^{∗}g}k_{z}∥_{2}) = 0.

Since (M_{n×n})_{1} is compact and the permutation group P_{n} is also compact, we may assume that A_{z} converges to A_{m} and R_{z} converges to R_{m}. Hence

lim_{z→m}(∥H_{(R_{m}−A_{m})f}k_{z}∥_{2} +∥H_{A_{m}^{∗}g}k_{z}∥_{2}) = 0.

By Lemma 3 [15], we have

[(R_{m} −A_{m})f]_{m} = 0, [A_{m}^{∗}g]_{m} = 0.

(4) ⟹ (5). By the Chang-Marshall theorem [8], we need only to show that

M(H^{∞} +C) ⊂ M(∩_{{A∈(M_{n×n})_{1}, R∈P_{n}}}H^{∞}[(R−A)f, A^{∗}g]).

Condition (4) states exactly that

M(H^{∞} +C) ⊂ ∪_{{A∈(M_{n×n})_{1}, R∈P_{n}}}M(H^{∞}[(R−A)f, A^{∗}g]).

By the Sarason theorem [11],

M(∩_{{A∈(M_{n×n})_{1}, R∈P_{n}}}H^{∞}[(R−A)f, A^{∗}g])

= Closure of ∪_{{A∈(M_{n×n})_{1}, R∈P_{n}}} M(H^{∞}[(R−A)f, A^{∗}g]). (14)

Hence

M(H^{∞} +C) ⊂ M(∩_{{A∈(M_{n×n})_{1}, R∈P_{n}}}H^{∞}[(R−A)f, A^{∗}g]).

(5) ⟹ (3). Suppose that (3) does not hold. There are δ > 0 and a net z in D converging to some m ∈ M(H^{∞} +C) such that

Ξ_{2}(z) ≥ δ.

By condition (5) and Sarason’s Theorem [11] as in (14), there are a net m_{α} ∈ M(H^{∞}) and matrices A_{α}, R_{α} such that m_{α} converges to m and

[(R_{α} −A_{α})f]_{m_{α}} = 0, [A_{α}^{∗}g]_{m_{α}} = 0.

We may assume that A_{α} converges to some A_{m} and R_{α} converges to some R_{m}. We claim that

[(R_{m} −A_{m})f]_{m} = 0, [A_{m}^{∗}g]_{m} = 0.

As in Lemma 3 [15], let u_{i} (i = 1, ⋯, n) be the unimodular functions such that u_{m} = (u_{1}, ..., u_{n})^{T} in (R_{m} − A_{m})f + H_{n}^{∞} and u_{i} (i = 1, ⋯, n) be the unimodular functions such that v_{m} = (v_{1}, ..., v_{n})^{T} in A_{m}^{∗}g +H_{n}^{∞}. Then by Lemma 3 [15]

∑_{i=1}^{n}[(1−|u_{i}(z)|^{2})+(1−|v_{i}(z)|^{2})]

≤ C[∥A_{α} −A_{m}∥_{∞} +∥R_{α} −R_{m}∥_{∞}]+C[∥H_{(R_{α}−A_{α})f}k_{z}∥_{2} +∥H_{A_{α}^{∗}g}k_{z}∥_{2}]

for all z ∈ D. Hence

∑_{i=1}^{n}[(1−|u_{i}(m_{α})|^{2})+(1−|v_{i}(m_{α})|^{2})] ≤ C[∥A_{α} −A_{m}∥_{∞} +∥R_{α} −R_{m}∥_{∞}].

Since these functions u_{i} and v_{i} are continuous on M(H^{∞}), we have

∑_{i=1}^{n}[(1−|u_{i}(m)|^{2})+(1−|v_{i}(m)|^{2})] = 0.

Therefore

[(R_{m} −A_{m})f]_{m} = 0, [A_{m}^{∗}g]_{m} = 0.

This proves our claim. But again by Lemma 3 [15], this implies that

lim_{z→m}(∥H_{(R_{m}−A_{m})f}k_{z}∥_{2} +∥H_{A_{m}^{∗}g}k_{z}∥_{2}) = 0.

This contradicts to the assumption that

∥H_{(R_{m}−A_{m})f}k_{z}∥_{2} +∥H_{A_{m}^{∗}g}k_{z}∥_{2} ≥ Ξ_{2}(z) > δ,

(3) ⟹ (1). Now we assume that

lim_{z→∂D} Ξ_{2}(z) = 0.

We use the distribution inequality obtained in Section 5 to show that ∑_{i=1}^{n}H_{f_{i}}^{∗}H_{g_{i}} is compact. Since the quantity Ξ_{r}(z) for some r > 2 appears in the distribution inequality, we first need to show that in fact for some r such that 3 > r > 2,

lim_{z→∂D} Ξ_{r}(z) = 0 (15)

Recall that

Ξ_{l}(z) = inf{∑_{i=1}^{n}(∥x_{i} ∘ϕ_{z} −P (x_{i} ∘ϕ_{z})∥_{l} +∥y_{i} ∘ϕ_{z} −P (g_{i} ∘ϕ_{z})∥_{l}) : A ∈ (M_{n×n})_{1} R ∈ P_{n}},

where

x = (x_{1}, ⋯, x_{n})^{T} = (R−A)f and y = (y_{1}, ⋯, y_{n})^{T} = A^{∗}g.

First note that since A ∈ (M_{n×n})_{1}, R ∈ P_{n} and by our assumption ∥f_{i}∥_{∞} < 1/2 and ∥g_{i}∥_{∞} < 1/2, we have ∥x_{i}∥_{∞} ≤ (1+n)/2 and ∥y_{i}∥_{∞} ≤ n/2. Thus

∑_{i=1}^{n}(∥x_{i} ∘ϕ_{z} −P (x_{i} ∘ϕ_{z})∥_{r} +∥y_{i} ∘ϕ_{z} −P (y_{i} ∘ϕ_{z})∥_{r})

≤ C_{r}(∑_{i=1}^{n}(∥x_{i} ∘ϕ_{z} −P (x_{i} ∘ϕ_{z})∥_{2} +∥y_{i} ∘ϕ_{z} −P (y_{i} ∘ϕ_{z})∥_{2})^{2/r})).

for some constant C_{r} dependent only on r and n. Therefore

Ξ_{r}(z) ≤ C_{r}(Ξ_{2}(z))^{2/r}.

Since lim_{z→∂D} Ξ_{2}(z) = 0, we obtain

lim_{z→∂D} Ξ_{r}(z) = 0.

This completes the proof of (15).

Now let u and v be two functions in H^{2}. Since H_{f_{i}}u is orthogonal to H^{2}, we see that (H_{f_{i}}u)(0) = 0. Thus by the Littlewood-Paley formula [8], we have

< u, (∑_{i=1}^{n}H_{f_{i}}^{∗}H_{g_{i}})v >= ∑_{i=1}^{n} < H_{f_{i}}u, H_{g_{i}}v >=

\frac{1}{π} ∫_{𝔻}(∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}v)}) log \frac{1}{|z|} dA(z) = I_{R} +II_{R},

where for 1/2 < R < 1,

I_{R} = ∫_{|z|>R}(∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}v)}) log \frac{1}{|z|} dA(z)

and

II_{R} = ∫_{|z|≤R}(∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}}) log \frac{1}{|z|} dA(z).

One easily checks that there is a compact operator K_{R} such that

II_{R} =< u, K_{R}v > .

Thus, if we show that I_{R} → 0 as R → 1, then ||T_{f}T_{g} −T_{g}T_{f} −K_{R}|| → 0, and we are done. The rest of the proof will be devoted to showing that I_{R} → 0 as R → 1.

Choose z ∈ 𝔻 and fix a constant a ≥ 1 for which the Distribution Inequality holds; that is

|t ∈ I_{z}: {B_{ϵ}(u, v)(t) ≤ aΞ_{r}(z)Γ_{r}(z)Λ_{s}u(t)Λ_{s}v(t)}| ≥ K_{a}|I_{z}|.

For t ∈ ∂𝔻, let

ρ(t) = max{ϵ: B_{ϵ}(u, v)(t) ≤ aΞ_{r}(z)Γ_{r}(z)Λ_{s}u(t)Λ_{s}v(t)}.

Let χ_{t} denote the characteristic function of Γ_{t,ρ(t)}. Then

∫_{∂𝔻} B_{ρ(t)}(u, v)(t) dt ≤ aΞ_{r}(z)Γ_{r}(z) ∫_{∂𝔻} Λ_{s}u(t)Λ_{s}v(t) dt
≤ aΞ_{r}(z)Γ_{r}(z)||Λ_{s}u||_{2}||Λ_{s}v||_{2}.

Since \frac{2}{s} > 1, so by [8]

||Λ_{s}u||_{2} = ||M(|u|^{s})^{1/s}||_{2} = [M(|u|^{s})||_{2/s}]^{1/s} ≤ A_{s}(|| |u|^{s}||_{2/s}||)^{1/s}

So

||Λ_{s}u||_{2} ≤ A_{s}||u||_{2}.

Similarly,

||Λ_{s}v||_{2} ≤ A′_{s}||v||_{2}.

Thus

∫_{∂𝔻} B_{ρ(t)}(u, v)(t)dt ≤ a^{∗}Ξ_{r}(z)Γ_{r}(z)||u||_{2} ||v||_{2}. (16)

On the other hand,

∫_{∂𝔻} B_{ϵ}(u, v)(t) = ∫_{∂𝔻} ∫_{Γ_{t,ρ(t)}} |∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}v)}|dA(z) dt

So

∫_{∂𝔻} B_{ϵ}(u, v)(t) ≥ ∫_{∂𝔻} ∫_{|z|>R} χ_{ω}(z)|∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}v)}|dA(z) dt.

Now the Distribution Function Inequality tells us that ρ(t) ≥ (1 − |z|^{2}) on a subset E_{z} of I_{z} satisfying

|E_{z}| ≥ K_{a}|I_{z}|.

Now, for t ∈ E_{z}, we have t ∈ I_{z}. Thus if we write z = re^{iθ} and note that ρ(t) ≥ \frac{3}{2}(1−|z|) we have

|r e^{i θ} −e^{i t}| ≤ |r e^{i θ} −e^{i θ}|+|e^{i θ} −e^{i t}| ≤ (1−|z|)+ \frac{(1−|z|)}{2}) ≤ ρ(t)

Therefore, for t ∈ E_{z}, we have that z ∈ Γ_{t,ρ(t)} and that χ_{t}(z) = 1 on E_{z}. So,

∫_{∂𝔻} ∫_{|z|>R} χ_{t}(z)|∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}v)}|dA(z) dt

≥ ∫_{|z|>R}[∫_{∂𝔻} χ_{t}(z) dt]||∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}v)}|dA(z)

Since χ_{t}(z) = 1 on E_{z}, we have

∫_{∂𝔻} B_{ϵ}(u, v)(t)dt ≥ ∫_{|z|>R} |E_{z}||∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}v)}|dA(z)

But, |E_{z}| ≥ K_{a}(1−|z|^{2}), so

∫_{∂𝔻} B_{ϵ}(u, v)(t) ≥ K_{a} ∫_{|z|>R} |∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}v)}|(1−|z|^{2})dA(z).

Since

I_{R} = ∫_{|z|>R} |∑_{i=1}^{n}grad(H_{f_{i}}u)\overline{grad(H_{g_{i}}v)}| log \frac{1}{|z|} dA(z),

we have,

∫_{∂𝔻} B_{ϵ}(u, v)(t) ≥ K_{a}|I_{R}|.

Combining this together with (16), we see that |I_{R}| ≤ CΞ_{r}(z)Γ_{r}(z)||u||_{2} ||v||_{2}. But by (15) ,

lim_{z→∂D} Ξ_{r}(z) = 0.

and Γ_{r}(z) is bounded. Hence we have lim_{R→1} |I_{R}| = 0. This finishes the proof.

7. Compact semi-commutator or commutator

In this section by combining the results in Sections 2 and 6, we will show several necessary and sufficient conditions for the semi-commutator or the commutator of the block Toeplitz operators with matrix symbols to be compact. We also give a characterization of essentially normal block Toeplitz operators.

Theorem 11. Let F and G be in L_{n×n}^{∞}. Let F^{∗} = (f_{1}, ⋯, f_{n}) and G = (g_{1}, ⋯, g_{n}). The following are equivalent.

(1) T_{F G} −T_{F} T_{G}(= H_{F^{∗}}^{∗}H_{G}) is compact.

(2) ∪_{i,j} ∩_{{A∈(M_{n×n})_{1}, R∈P_{n}}} H^{∞}[(R−A)f_{i}, A^{∗}g_{j}] ⊂ H^{∞} +C. (3)

lim_{|z|→1} ∥[|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z)]^{1/2}[|G_{−} −G_{−}(z)|^{2}(z)]^{1/2}∥ = 0.

Proof. (1)⟺(2). Let F = (f_{ij})_{n×n} and G = (g_{ij})_{n×n}. Note that T_{F G} −T_{F} T_{G} is compact if and only if each entry of

(∑_{k} H∗ ; \overline{f_{ik}}H_{g_{kj}})_{ij}

is compact. By Theorem 10, this is equivalent to

∪_{i,j} ∩_{{A∈(M_{n×n})_{1}, R∈P_{n}}} H^{∞}[(R−A)f_{i}, A^{∗}g_{j}] ⊂ H^{∞} +C.

(1)⟺ (3). (1)⇒(3) is proved in Theorem 3. We are going to prove that (3)⇒(1). By Lemma 2,

tr(|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z))(|G_{−} −G_{−}(z)|^{2}(z)) =
trace[H_{F^{∗}}^{∗}H_{G} −T_{Φ_{z}}^{∗}H_{F^{∗}}^{∗}H_{G}T_{Φ_{z}}]^{∗}[H_{F^{∗}}^{∗}H_{G} −T_{Φ_{z}}^{∗}H_{F^{∗}}^{∗}H_{G}T_{Φ_{z}}].

Thus by Lemma 1, (3) implies that

lim_{|z|→1} ∥ ∑_{k} H\overline{f_{ik}}k_{z} ⊗H_{g_{kj}}k_{z}∥ = 0

for all i, j. By Theorem 10, we have that ∑_{k} H∗ ; \overline{f_{ik}}H_{g_{kj}} are compact for all i, j. Hence H_{F^{∗}}^{∗}H_{G} is compact. Therefore T_{F G} −T_{F} T_{G} is compact. The proof is complete.

Next we characterize when the commutator T_{F} T_{G} − T_{G}T_{F} is compact. To do this, recall that

T_{F} T_{G} −T_{G}T_{F} == T_{F} T_{G} −T_{F G} +T_{GF} −T_{G}T_{F} +T_{(F G−GF)}
= −(H_{F^{∗}}^{∗}H_{G} −H_{G^{∗}}^{∗}H_{F})+T_{(F G−GF)}.

Therefore by the Douglas theorem [7], T_{F} T_{G} −T_{G}T_{F} is compact if and only if F G = GF and H_{F^{∗}}^{∗}H_{G} −H_{G^{∗}}^{∗}H_{F} is compact. But if we let

B = [ \begin{matrix}F & −G \\ 0 & 0 \end{matrix} ] , C = [ \begin{matrix}G & 0 \\ F & 0 \end{matrix} ] ,

then

H_{B^{∗}}^{∗}H_{C} = [ \begin{matrix}H_{F^{∗}}^{∗}H_{G} −H_{G^{∗}}^{∗}H_{F} & 0 \\ 0 & 0 \end{matrix} ] .

Therefore, T_{F} T_{G} − T_{G}T_{F} is compact if and only if F G = GF and H_{B^{∗}}^{∗}H_{C} is compact. Note that |(B_{+})^{∗} −(B_{+})^{∗}(z)|^{2}(z) and |C_{−} −C_{−}(z)|^{2}(z) can be computed as in (4) and (5). The following result now follows immediately from Theorem 11.

Theorem 12. Let F and G be in L_{n×n}^{∞}. Let

[ \begin{matrix}F^{∗} \\ −G^{∗} \end{matrix} ] = (f_{1}, ⋯, f_{n}), [ \begin{matrix}G \\ F \end{matrix} ] = (g_{1}, ⋯, g_{n}).

The following are equivalent.

(1) T_{F} T_{G} −T_{G}T_{F} is compact.

(2) F G = GF and ∪_{i,j} ∩_{{A∈(M_{n×n})_{1}, R∈P_{n}}} H^{∞}[(R−A)f_{i}, A^{∗}g_{j}] ⊂ H^{∞} +C.

(3) F G = GF and

lim_{|z|→1} ∥ [ \begin{matrix}|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z) & −(((F_{+})^{∗} −(F_{+})^{∗}(z))(G_{+} −G_{+}(z)))(z) \\ −(((G_{+})^{∗} −(G_{+})^{∗}(z))(F_{+} −F_{+}(z)))(z) & |(G_{+})^{∗} −(G_{+})^{∗}(z)|^{2}(z) \end{matrix} ]^{1/2} ×

[ \begin{matrix}|G_{−} −G_{−}(z)|^{2}(z) & ((G_{−} −G_{−}(z))((F_{−})^{∗} −(F_{−})^{∗}(z)))(z) \\ ((F_{−} −F_{−}(z))((G_{−})^{∗} −(G_{−})^{∗}(z)))(z) & |F_{−} −F_{−}(z)|^{2}(z) \end{matrix} ]^{1/2} ∥ = 0.

An operator A is said to be essentially normal if A^{∗}A−AA^{∗} is compact. By taking G = F^{∗}, we immediately get the following characterization of essentially normal block Toeplitz operators.

Corollary 13. Let F be in L_{n×n}^{∞}. Let

[ \begin{matrix}F^{∗} \\ −F \end{matrix} ] = (f_{1}, ⋯, f_{n}), [ \begin{matrix}F^{∗} \\ F \end{matrix} ] = (g_{1}, ⋯, g_{n}).

The following are equivalent.

(1) T_{F} is essentially normal.

(2) F F^{∗} = F^{∗}F and ∪_{i,j} ∩_{{A∈(M_{n×n})_{1}, R∈P_{n}}} H^{∞}[(R−A)f_{i}, A^{∗}g_{j}] ⊂ H^{∞} +C.

(3) F F^{∗} = F^{∗}F and

lim_{|z|→1} ∥ [ \begin{matrix}|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z) & −((F_{−} −F_{−}(z))(F_{+} −F_{+}(z))(z) \\ −(((F_{+})^{∗} −(F_{+})^{∗}(z))((F_{−})^{∗} −(F_{−})^{∗}(z))(z) & |F_{−} −F_{−}(z)|^{2}(z) \end{matrix} ]^{1/2} ×

[ \begin{matrix}|(F_{+})^{∗} −(F_{+})^{∗}(z)|^{2}(z) & ((F_{−} −F_{−}(z))(F_{+} −F_{+}(z))(z) \\ (((F_{+})^{∗} −(F_{+})^{∗}(z))((F_{−})^{∗} −(F_{−})^{∗}(z))(z) & |F_{−} −F_{−}(z)|^{2}(z) \end{matrix} ]^{1/2} ∥ = 0.

For the scalar symbols there were several other sufficient conditions for the product of two Toeplitz operators to be a compact perturbation of a Toeplitz operator. To state those conditions we need some notations. The fiber M_{λ} of M(L^{∞}) above the point λ is the set {τ ∈ M(L^{∞}) : z(τ) = λ}. We recall that a subset of M(L^{∞}) is called an antisymmetric set if any real-valued function in H^{∞} +C is constant on the set.

One of the following conditions implies the compactness of the semi-commutator T\overline{ϕ}T_{ψ} −T_{\overline{ϕ}ψ} of Toeplitz operators with scalar symbols ϕ and ψ.

(1) Either ϕ or ψ is in C(∂D) [5].

(2) ϕ and ψ are piecewise continuous and have no common discontinuities [10]

(3) Either \overline{ϕ} or ψ is in H^{∞} on each fiber M_{z} for z on the circle [13].

(4) Either \overline{ϕ} is in H^{∞} or ψ is in H^{∞} on each set of maximal antisymmetry of H^{∞} +C [1].

It was shown in [2] that H^{∞}[\overline{ϕ}] ⋂ H^{∞}[ψ] ⊂ H^{∞} +C is equivalent to

(5) Either ϕ or ψ is in H^{∞} on each support set.

Next we will show some sufficient conditions for the compactness of the semi-commutator T_{F} T_{G} −T_{F G} of Toeplitz operators with matrix symbols F and G. Those conditions are analogous to the above conditions of the scalar case. Some of them are well known ([7], [9]).

Corollary 14. Let F and G be in L_{n×n}^{∞}. Then one of the following conditions is a sufficient condition for T_{F} T_{G} −T_{F G} to be compact:

(1) Either F^{∗} or G is in C_{n×n}(∂D).

(2) F^{∗} and G are piecewise continuous and have no common discontinuities.

(3) Either F^{∗} or G is in H_{n×n}^{∞} on each fiber M_{z} for z on the circle.

(4) Either F^{∗} or G is in H_{n×n}^{∞} on each maximal antisymmetric set of H^{∞} +C.

(5) Either F^{∗} or G is in H_{n×n}^{∞} on each support set.

(6) H_{n×n}^{∞}[F^{∗}]∩H_{n×n}^{∞}[G] ⊂ H_{n×n}^{∞} +C_{n×n}(∂D), where H_{n×n}^{∞}[G] denotes the subalgebra of L_{n×n}^{∞} generated by H_{n×n}^{∞} and G.

Proof. Notice that Conditions (1) to (6) in the corollary are ordered by weakness. So it is sufficient to show that Condition (6) is stronger than Condition (2) in Theorem 11. Since H_{n×n}^{∞}[G] denotes the subalgebra of L_{n×n}^{∞} generated by H_{n×n}^{∞} and G, we observe that

H_{n×n}^{∞}[G] = (H^{∞}[g_{11}, ⋯, g_{1n}, ⋯, g_{n1}, ⋯, g_{nn}])_{n×n}.

Hence Condition (6) is equivalent to

H^{∞}[\overline{f_{11}}, ⋯, \overline{f_{1n}}, ⋯, \overline{f_{n1}}, ⋯, \overline{f_{nn}}]∩H^{∞}[g_{11}, ⋯, g_{1n}, ⋯, g_{n1}, ⋯, g_{nn}] ⊂ H^{∞} +C.

Let F^{∗} = (f_{1}, ⋯, f_{n}) and G = (g_{1}, ⋯, g_{n}). It is easy to see that

∪_{i,j} ∩_{{A∈(M_{n×n})_{1}, R∈P_{n}}} H^{∞}[(R−A)f_{i}, A^{∗}g_{j}]
⊂ ∪_{i,j}H^{∞}[f_{i}]∩H^{∞}[g_{j}]

But for all i, j, we have

H^{∞}[f_{i}]∩H^{∞}[g_{j}]
⊂ H^{∞}[\overline{f_{11}}, ⋯, \overline{f_{1n}}, ⋯, \overline{f_{n1}}, ⋯, \overline{f_{nn}}]∩H^{∞}[g_{11}, ⋯, g_{1n}, ⋯, g_{n1}, ⋯, g_{nn}]

Hence

∪_{i,j} ∩_{{A∈(M_{n×n})_{1}, R∈P_{n}}} H^{∞}[(R−A)f_{i}, A^{∗}g_{j}] ⊂ H^{∞} +C.

So by Theorem 11, T_{F} T_{G} −T_{F G} is compact. This completes the proof.

Acknowledgment. We thank S. Treil for his useful suggestions.

References

[1] S. Axler, Doctoral dissertation, University of California, Berkeley, 1975.

[2] S. Axler, S.-Y. A. Chang, D. Sarason, Product of Toeplitz operators, Integral Equations and Operator Theory 1 (1978) 285-309.

[3] A. Böttcher and B. Silbermann, Analysis of Toeplitz operators, Springer-Verlag, 1990.

[4] A. Brown and P. R. Halmos, Algebraic properties of Toeplitz operators, J. Reine Angew. Math. 213 (1963) 89-102.

[5] L.A. Coburn, The C^{∗}-algebra generated by an isometry I, Bull. Amer. Math. Soc. 137 (1967) 211-217.

[6] R. G. Douglas, Banach algebra techniques in the operator theory, Academic Press, New York and London, 1972.

[7] R. G. Douglas, Banach algebra techniques in the theory of Toeplitz operators, Regional Conference Series in Mathematics, American Mathematical Society 15, 1972.

[8] J. B. Garnett, Bounded Analytic Functions, Academic Press, New York, 1981.

[9] I. Gohberg, S. Goldberg and M. A. Kaashoek, Classes of Linear Operators, Vol. 2 Birkhauser Verlag Basel, 1993.

[10] I. C. Gohberg and N. J. Krupnik, The algebra generated by Toeplitz matrices, Functional Anal. Appl. 3 (1969) 119-127.

[11] P. Gorkin and D. Zheng, Essentially commuting Toeplitz operators, preprint.

[12] P. R. Halmos, A Hilbert space problem book, New York, Springer-Verlag, 1982.

[13] D. Sarason, On Products of Toeplitz operators, Acta Scien. Math. 35 (1973) 7-12.

[14] A. Volberg, Two remarks concerning the theorem of S. Axler, S.-Y. A. Chang, and D. Sarason, J. Operator Theory 8 (1982) 209-218.

[15] D. Zheng, The distribution function inequality and products of Toeplitz operators and Hankel operators, to appear in J. Functional Analysis.

Caixing Gu, Mathematics Department, University of California Irvine, Irvine, CA 72717

  E-mail address: cgu@math.uci.edu

Dechao Zheng, Mathematics Department, Michigan State University, East Lansing, MI 48824

  E-mail address: dechao@math.msu.edu