THE SEMI-COMMUTATOR OF TOEPLITZ OPERATORS ON THE BIDISC

CAIXING GU AND DECHAO ZHENG

Abstract. In this paper we characterize when the semi-commutator T_{f}T_{g} − T_{fg} of two Toeplitz operators T_{f} and T_{g} on the Hardy space of the bidisc is zero. We also show that there is no nonzero finite rank semi-commutator on the bidisc. Furthermore explicit examples of compact semi-commutators with symbols continuous on the bitorus T^{2} are given.

1. Introduction

Let D be the open unit disk in C. Its boundary is the unit circle T. The bidisc D^{2} and the torus T^{2} are the subsets of C^{2} which are Cartesian products of two copies D and T, respectively. Let dσ(z) be the normalized Haar measure on T^{2}. The Hardy space H^{2}(D^{2}) is the closure of the polynomials in L^{2}(T^{2}, dσ) (or L^{2}(T^{2})). Let P be the orthogonal projection from L^{2}(T^{2}) onto H^{2}(D^{2}). The Toeplitz operator with symbol f in L^{∞}(T^{2}) is defined by T_{f}(h) = P (fh), for all h ∈ H^{2}(D^{2}) and the Hankel operator with symbol f is defined by H_{f}(h) = (I − P)(fh), for all h ∈ H^{2}(D^{2}).

Let f and g be two bounded functions on T^{2}. In this paper we study the the semicommutator T_{f}T_{g} − T_{fg} of two Toeplitz operators T_{f} and T_{g} on the bidisc. As in the case of the unit disk, the semi-commutator is connected to the Hankel operators by the following relation.

T_{f}T_{g} − T_{fg} = −H∗ ; f̄H_{g}. (1)

To motivate the problems to be considered in the paper, we shall recall some classical results for the semi-commutators of Toeplitz operators on the Hardy space H^{2}(D) of the unit disk. Let f_{1} and g_{1} be bounded functions on the unit circle T. For two Toeplitz operators T_{f_{1}} and T_{g_{1}} on the Hardy space H^{2}(D) of the unit disk, Brown-Halmos [2] shows that the semi-commutator T_{f_{1}}T_{g_{1}} − T_{f_{1}g_{1}} is zero if and only if either f̄_{1} or g_{1} is analytic. In other words, H∗ ; f̄_{1}H_{g_{1}} is zero if and only if either Hf̄_{1} or H_{g_{1}} is zero.

In this paper, we will characterize when the semi-commutator T_{f}T_{g} − T_{fg} of two Toeplitz operators T_{f} and T_{g} on the bidisc is zero. In particular, we note that unlike on the unit disk one can have both Hf̄ and H_{g} are not zero, but their product H∗ ; f̄H_{g} is zero. Furthermore we will see that there is no finite rank semi-commutator on the bidisc. But this is false on the unit disk. Indeed it was shown in [1] that for Toeplitz operators T_{f_{1}} and T_{g_{1}} on H^{2}(D), T_{f_{1}}T_{g_{1}} − T_{f_{1}g_{1}} is of finite rank if and only if either f̄_{1} or g_{1} is an analytic function plus a rational function.

1991 Mathematics Subject Classification. Primary 47B35.

Key words and phrases. Block Toeplitz operators, Hankel operators, semi-commutators, bidisc.

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.

The main question to be considered here is when the semi-commutator T_{f}T_{g} −T_{fg} on the bidisc is compact. This problem is connected with the spectral theory of Toeplitz operators on the bidisc and various applications, see [7], [8] and reference therein.

For Toeplitz operators T_{f_{1}} and T_{g_{1}} on the unit disk, this hard problem was solved by the combined efforts of Axler, Chang, and Sarason [1] and Volberg [12]. Their beautiful result is that that T_{f_{1}g_{1}} − T_{f_{1}}T_{g_{1}} is compact if and only if H^{∞}[f̄_{1}] ⋂ H^{∞}[g_{1}] ⊂ H^{∞}(D) + C(T); here H^{∞}[g_{1}] denotes the closed subalgebra of L^{∞}(T) generated by H^{∞}(D) and g_{1} and C(T) the continuous functions on T.

The function theory on the bidisc is quite different from and much less understood than the function theory on the unit disk [9], [3] and [5]. The proof of the above result on the unit disk relies on some deep results and techniques from function theory on the unit disk which are not available from function theory on the bidisc.

In this paper we content ourselves with some partial results for the compactness of T_{f}T_{g} − T_{fg}. By carefully analyzing the action of T_{f}T_{g} − T_{fg} on the reproducing kernel of H^{2}(D^{2}) and exploiting the harmonicity of certain functions, we will get a necessary condition for the compactness of T_{f}T_{g} − T_{fg}. This shows that for a large class of functions f and g, T_{f}T_{g} − T_{fg} is compact if and only if it is zero. For example, if f is a trigonometric polynomial on T^{2} and g is an arbitrary bounded function on T^{2}, then for such f and g, there is no compact semi-commutator T_{f}T_{g} − T_{fg}. Also as a corollary of this condition we see that there are no compact Hankel operators on bidisc, which was proved by M. Cotlar and C. Sadosky [4] using a completely different method. It is natural to guess that for f and g in L^{∞}(T^{2}), T_{f}T_{g} − T_{fg} is compact if and only if T_{f}T_{g} − T_{fg} is zero. But it is false.

For the class of bounded functions f and g on T^{2} of the form f(z_{1}, z_{2}) = f_{1}(z_{1})f_{2}(z_{2}) and g(z_{1}, z_{2}) = g_{1}(z_{1})g_{2}(z_{2}), we will show that T_{f}T_{g} − T_{fg} is compact if and only if the following two conditions hold.

(1) f_{1}(z_{1})g_{1}(z_{1}) = 0 and f_{2}(z_{2})g_{2}(z_{2}) = 0 on T, and

(2) lim_{z→T^{2}} ∥H\overline{f}k_{z}∥_{2}∥H_{g}k_{z}∥_{2} = 0, where k_{z} denotes the normalized reproducing kernel of H^{2}(D^{2}). Here the Littlewood-Paley theory on the bidisc and a certain distribution function inequality in Zheng [13] play a key role. We remark that only one condition similar to Condition (2) above is needed for the compactness of the semi-commutator T_{f_{1}}T_{g_{1}} − T_{f_{1}g_{1}} on the Hardy space H^{2}(D) of the unit disk. Indeed it was shown in [13] that T_{f_{1}}T_{g_{1}} − T_{f_{1}g_{1}} is compact on H^{2}(D) if and only if lim_{z_{1}→T} ∥H\overline{f_{1}}k_{z_{1}}∥_{2}∥H_{g_{1}}k_{z_{1}}∥_{2} = 0, where k_{z_{1}} denotes the normalized reproducing kernel of H^{2}(D).

Now we outline the plan of the paper. In Section 2 we study when the semicommutator of two Toeplitz operators is zero or finite rank. In Section 3 we derive a necessary condition for the compactness of T_{f}T_{g} − T_{fg}. In Section 4, for f and g of the form f(z_{1}, z_{2}) = f_{1}(z_{1})f_{2}(z_{2}) and g(z_{1}, z_{2}) = g_{1}(z_{1})g_{2}(z_{2}), we characterize when T_{f}T_{g} − T_{fg} is compact. This result shows that the necessary condition for the compactness of T_{f}T_{g} − T_{fg} obtained in Section 3 is not sufficient in general. This result also provides us with explicit examples of compact semi-commutators T_{f}T_{g} − T_{fg} with symbols f and g continuous on the bitorus T^{2}.

2. Zero semi-commutators

Let Z denote the set of all integers, Z_{+} the set of all nonnegative integers and Z_{−} the set of all negative integers. As in [10] we consider multiple Fourier series on the bitorus T^{2}. The multiple Fourier series on the bitorus T^{2} can be viewed as the Fourier transformation on L^{1}(T^{2}). For f in L^{1}(T^{2}), the Fourier transformation of f on Z × Z is given by

f_{m} = (\frac{1}{2π})^{2} ∫_{0}^{2π} ∫_{0}^{2π} f(e^{iθ_{1}}, e^{iθ_{1}})e^{i(m,θ)}dθ_{1}dθ_{2}

where m = (m_{1}, m_{2}) ∈ Z × Z, θ = (θ_{1}, θ_{2}) and (m, θ) = m_{1}θ_{1} + m_{2}θ_{2}. By Theorem 1.7 in [SW], the Fourier transformation is injective, i.e. If f ∈ L^{1}(T^{2}) and f_{m} = 0 for all m ∈ Z × Z, then f ≡ 0. We recall also that by using multiple Fourier series,

L^{2}(T^{2}) = {f : f = ∑_{m∈Z×Z} f_{m}e^{i(m,x)}, ∑_{m∈Z×Z} |f_{m}|^{2} < ∞}

H^{2}(D^{2}) = {h : h = ∑_{m∈Z_{+}×Z_{+}} h_{m}e^{i(m,x)}, ∑_{m∈Z_{+}×Z_{+}} |h_{m}|^{2} < ∞}

and

P f = ∑_{m∈Z_{+}×Z_{+}} f_{m}e^{i(m,x)} for f = ∑_{m∈Z×Z} f_{m}e^{i(m,x)} ∈ L^{2}(T^{2}).

Theorem 1. Let f and g be two bounded functions on the torus T^{2}. The following are equivalent.

(1) T_{f}T_{g} − T_{fg} is a finite rank operator.

(2) T_{f}T_{g} − T_{fg} is zero.

(3) For each i (i = 1, 2) either f̄ or g is analytic in z_{i}.

Proof. We first show that (3) implies (2). Assume that for each i (i = 1, 2) either f̄ or g is analytic in z_{i}. Without loss of generality, assume that f̄ is analytic in z_{1} for and g is analytic in z_{2}. Then it is easy to see that

(I − P)(gh_{1}) = ∑_{m=(m_{1},m_{2})∈Z_{−}×Z_{+}} a_{m}z^{m}

for all h_{1} ∈ H^{2}(D^{2}) and

(I − P)(f̄h_{2}) = ∑_{m=(m_{1},m_{2})∈Z_{+}×Z_{−}} b_{m}z^{m}

for all h_{2} ∈ H^{2}(D^{2}). Therefore

(H_{g}h_{1}, Hf̄h_{2}) = 0

for all h_{1}, h_{2} ∈ H^{2}(D^{2}). That is, H∗ ; f̄H_{g} = 0. So T_{f}T_{g} − T_{fg} = −H∗ ; f̄H_{g} = 0

It is obvious that (2) implies (1). Now we prove that (1) implies (3). That is, assume that T_{f}T_{g} − T_{fg} is a finite rank operator, we will show that for each i (i = 1, 2) either f̄ or g is analytic in z_{i}. Without loss of generality, we assume that i = 1. We write f and g as

f̄ = ∑_{i=−∞}^{∞} f_{i}(z_{2})z_{1}^{i} = ∑_{i=−∞}^{∞} f_{i}z_{1}^{i}

g = ∑_{i=−∞}^{∞} g_{i}(z_{2})z_{1}^{i} = ∑_{i=−∞}^{∞} g_{i}z_{1}^{i}

Let α, β, k, l ∈ Z_{+}. Then by a straightforward computation we have

(H_{f}^{∗}H_{g}z_{1}^{k}z_{2}^{α}, z_{1}^{l}z_{2}^{β})
= (H_{g}z_{2}^{α}z_{1}^{k}, H_{f}z_{2}^{β}z_{1}^{l})
= ∑_{i≤−(k+1)}(g_{i}(z_{2})z_{2}^{α}, f_{−(i+k−l)}(z_{2})z_{2}^{β}) + ∑_{i≥−k}((I_{2} − P_{2})g_{i}(z_{2})z_{2}^{α}, f_{−(i+k−l)}(z_{2})z_{2}^{β})

where I_{2} is the identity on L^{2}(T) and P_{2} is the projection from L^{2}(T) onto H^{2}(D). Therefore

(H_{g}z_{1}^{k}z_{2}^{α}, H_{f}z_{1}^{l}z_{2}^{β}) − (H_{g}z_{1}^{k+1}z_{2}^{α}, H_{f}z_{1}^{l+1}z_{2}^{β}))
= (g_{−(k+1)}(z_{2})(z_{2})^{α}, f_{−(l+1)}(z_{2})(z_{2})^{β}) − ((I_{2} − P_{2})g_{−(k+1)}(z_{2})(z_{2})^{α}, f_{−(l+1)}(z_{2})(z_{2})^{β})
= (P_{2}g_{−(k+1)}(z_{2})(z_{2})^{α}, f_{−(l+1)}(z_{2})(z_{2})^{β}).

Let S_{1} denote the multiplication by z_{1} on H^{2}(D^{2}), i.e., S_{1}h = z_{1}h for h ∈ H^{2}(D^{2}). The above relation implies that

(S_{1}^{∗l}H_{f}^{∗}H_{g}S_{1}^{k} − S_{1}^{∗(l+1)}H_{f}^{∗}H_{g}S_{1}^{k+1})h_{2}(z_{2}) = T\overline{f_{−(l+1)}}T_{g_{−(k+1)}}h_{2}(z_{2})

for all h_{2} ∈ H^{2}(D). Therefore, if H_{f}^{∗}H_{g} is a finite rank operator on H^{2}(D^{2}), then T\overline{f_{−(l+1)}}T_{g_{−(k+1)}} is a finite rank operator on H^{2}(D). By a result in [2], we have that either f_{−(l+1)} or g_{−(k+1)} = 0. Hence either f_{−(l+1)} = 0 for all k ≥ 0 or g_{−(k+1)} = 0 for all l ≥ 0. That is either f̄ or g is analytic in z_{1}. This finishes the proof of the theorem.

3. A necessary condition for compactness

In this section we will give a necessary condition for the compactness of the semicommutator T_{f}T_{g} − T_{fg}.

Before going to the main result of this section, we need some notations and definitions. We use z to denote the vector (z_{1}, z_{2}) in C^{2} of two dimensional complex plane. Since for any z in D^{2}, the pointwise evaluation of functions in H^{2}(D^{2}) at z is a bounded functional, there is a function K_{z} in H^{2}(D^{2}) such that

f(z) = (f, K_{z})

for all f in H^{2}(D^{2}). K_{z}(w) is called the reproducing kernel for H^{2}(D^{2}) and sometimes we use K(z,w) to denote K_{z}(w).

Let K_{z_{1}} denote the reproducing kernel of the Hardy space H^{2}(D) of the unit disk at the point z_{1} in D, and k_{z_{1}} the normalized reproducing kernel of the Hardy space H^{2}(D) at the point z_{1} ∈ D. Namely,

K_{z_{1}} = \frac{1}{(1 − \overline{z_{1}}w_{1})}, k_{z_{1}} = \frac{(1 − |z_{1}|^{2})^{1/2}}{(1 − \overline{z_{1}}w_{1})} .

It is easy to check that the reproducing kernel K_{z} of the Hardy space H^{2}(D^{2}) of the bidisc is the product K_{z_{1}}(w_{1})K_{z_{2}}(w_{2}) of the reproducing kernels of H^{2}(D). So the normalized reproducing kernel k_{z} of H^{2}(D^{2}) is also the product k_{z_{1}}(w_{1})k_{z_{2}}(w_{2}) of the normalized reproducing kernels of H^{2}(D). We observe that k_{z} weakly converges to zero in H^{2}(D^{2}) as z tends to the boundary of D^{2}. For a function f in the space L^{2}(T^{2}), i.e.,

f(w) = ∑_{m∈Z×Z} f_{m}w^{m}, w ∈ T^{2},

where f_{m} is a sequence of numbers such that

∑_{m∈Z×Z} |f_{m}|^{2} < ∞,

the bi-harmonic (in short, harmonic) extension f(z) of f is defined via the Possion integral

f(z) := ∫_{T} ∫_{T} f(w)\frac{(1 − |z_{1}|^{2})}{|1 − \overline{z_{1}}w_{1}|^{2}} \frac{(1 − |z_{2}|^{2})}{|1 − \overline{z_{2}}w_{2}|^{2}} dσ(w_{1})dσ(w_{2})

= (f(w)k_{z}(w), k_{z}(w)).

We write the power series expansion of the harmonic extension f(z) of f as follows:

f(z) = ∑_{m∈Z×Z} f_{m}z^{m}

:= f_{++}(z) + f_{+−}(z) + f_{−+}(z) + f_{−−}(z), (2) where

f_{++}(z) := ∑_{m∈Z_{+}×Z_{+}} f_{m}z^{m}, f_{+−}(z) := ∑_{m∈Z_{+}×Z_{−}} f_{m}z^{m}

f_{−+}(z) := ∑_{m∈Z_{−}×Z_{+}} f_{m}z^{m}, f_{−−}(z) := ∑_{m∈Z_{−}×Z_{−}} f_{m}z^{m}.

Also let

f_{+}(z) = f_{++}(z) + f_{+−}(z), f_{−}(z) = f_{−+}(z) + f_{−−}(z), (3)

where for example, m = (m_{1}, m_{2}) ∈ Z_{−} × Z_{+} means that m_{1} ∈ Z_{+} and m_{2} ∈ Z_{−}, z^{m} the product z̄_{1}^{|m_{1}|}z_{2}^{m_{2}}.

Theorem 2. Let f and g be two bounded functions on the bitorus T^{2}. Then T_{f}T_{g} − T_{fg} is compact implies that the following two statements hold.

(1) For all z_{1} ∈ D and all z_{2} ∈ T,

\frac{∂f}{∂z_{1}} (z_{1}, z_{2})\frac{∂g}{∂z̄_{1}} (z_{1}, z_{2}) = 0.

(1) For all z_{2} ∈ D and all z_{1} ∈ T,

\frac{∂f}{∂z_{2}} (z_{1}, z_{2})\frac{∂g}{∂z̄_{2}} (z_{1}, z_{2}) = 0.

Proof. Without loss of generality, we prove statement (1). The proof of statement (2) is similar. We write f and g as (see above for the notations)

f(z) := f_{++}(z) + f_{+−}(z) + f_{−+}(z) + f_{−−}(z)
:= f_{+}(z) + f_{−}(z)

and similarly

g(z) := g_{++}(z) + g_{+−}(z) + g_{−+}(z) + g_{−−}(z)
:= g_{+}(z) + g_{−}(z).

We compute the action of a Toeplitz operator T_{g} for g as above on the normalized reproducing kernel k_{z}(w) as follows:

T_{g}k_{z}(w_{1}, w_{2})
(4)  = g_{++}(w_{1}, w_{2})k_{z} + g_{+−}(w_{1}, z_{2})k_{z} + g_{−+}(z_{1}, w_{2})k_{z} + g_{−−}(z_{1}, z_{2})k_{z}

for z = (z_{1}, z_{2}) in D^{2}, where for example

f_{+−}(w_{1}, z_{2}) = ∑_{(m_{1},m_{2})∈Z_{+}×Z_{−}} g_{m}w_{1}^{m_{1}}z̄_{2}^{|m_{2}|}

for all w_{1} ∈ T and z_{2} ∈ D. We remark that in fact the above formula (4) holds for any g ∈ L^{2}(T^{2}).

For convenience, we consider Tf̄T_{g} − T_{f̄g} instead of T_{f}T_{g} − T_{fg}. We write that

((Tf̄T_{g} − T_{f̄g})k_{z}, k_{z})
= − ∫_{T^{2}} \overline{f(w)}g(w)|k_{z}(w)|^{2}dσ(w) + (T_{g}k_{z}, T_{f}k_{z})
(5)  := h_{1}(z) + h(z)

for any z in D^{2}, where

h_{1}(z) = − ∫_{T^{2}} \overline{f(w)}g(w)|k_{z}(w)|^{2}dσ(w),
h(z) = (Tf̄T_{g}k_{z}, k_{z}).

Furthermore by (5), we write h(z) as

h(z) = (Tf̄T_{g}k_{z}, k_{z}) = (Tf̄T_{g}k_{z}, k_{z})
:= h_{11}(z) + h_{12}(z) + h_{21}(z) + h_{22}(z)

where

h_{11}(z) := (T_{g_{+}}k_{z}(w), T_{f_{+}}k_{z}(w)), h_{12}(z) := (T_{g_{+}}k_{z}(w), T_{f_{−}}k_{z}(w)), (6)

h_{21}(z) := (T_{g_{−}}k_{z}(w), T_{f_{+}}k_{z}(w)), h_{22}(z) := (T_{g_{−}}k_{z}(w), T_{f_{−}}k_{z}(w)). (7)

Let m = (m_{1}, m_{2}) ∈ Z × Z be fixed. Let θ = (θ_{1}, θ_{2}) and

U_{θ} = diag{e^{iθ_{1}}, e^{iθ_{2}}}.

Set

H_{1}(z) := ∫_{T^{2}} h_{1}(U_{θ}z)e^{i(m,θ)}dθ

H_{ij}(z) := ∫_{T^{2}} h_{ij}(U_{θ}z)e^{i(m,θ)}dθ, i, j = 1, 2

for z ∈ D^{2}, where h_{1}(U_{θ}z) and h_{ij}(U_{θ}z) are obtained by replacing z with U_{θ}z in the definition of h_{1} and h_{ij} respectively. For example, by (4), we have

H_{12}(z) := ∫_{T^{2}}(g_{++}(w_{1}, w_{2})k_{U_{θ}z} + g_{+−}(w_{1}, e^{iθ_{2}}z_{2})k_{U_{θ}z},

f_{−+}(e^{iθ_{1}}z_{1}, w_{2})k_{U_{θ}z} + f_{−−}(e^{iθ_{1}}z_{1}, e^{iθ_{2}}z_{2})k_{U_{θ}z})e^{i(m,θ)}dθ. (8)

For a fixed z_{1} ∈ D and u_{2} ∈ T, let z_{2α} converge to the point u_{2} on T, then z_{α} = (z_{1}, z_{2α}) converges to the boundary point (z_{1}, u_{2}) of the bidisc D^{2}. We claim that H_{1}(z), H_{11}(z), H_{12}(z), H_{21}(z) and H_{22}(z) are continuous on \overline{D^{2}}; and furthermore if we denote the limits of H_{1}(z), H_{11}(z), H_{12}(z), H_{21}(z) and H_{22}(z) as z_{α} converges to the boundary point (z_{1}, u_{2}) by H_{1}(z_{1}, u_{2}), H_{11}(z_{1}, u_{2}), H_{12}(z_{1}, u_{2}), H_{21}(z_{1}, u_{2}) and H_{22}(z_{1}, u_{2}), respectively. Then H_{1}(z_{1}, u_{2}), H_{11}(z_{1}, u_{2}), H_{12}(z_{1}, u_{2}) and H_{21}(z_{1}, u_{2}) are harmonic in z_{1} for any u_{2} ∈ T. However H_{22}(z_{1}, u_{2}) is not necessary harmonic in z_{1}. We postpone the proof of the claim and first continue with the proof of the theorem.

We will first show that if Tf̄T_{g} − T_{f̄g} is compact, then H_{22}(z) is harmonic in z_{1}. Replacing z by U_{θ}z in (5) yields

((Tf̄T_{g} − T_{f̄g})k_{U_{θ}z}, k_{U_{θ}z}) = h_{1}(U_{θ}z) + h(U_{θ}z).

Multiplying the above equation by e^{i(m,θ)} and then integrating with respect to θ give that

∫_{T^{2}}((Tf̄T_{g} − T_{f̄g})k_{U_{θ}z}, k_{U_{θ}z})e^{i(m,θ)}dθ
= H_{1}(z) + H_{11}(z) + H_{12}(z) + H_{21}(z) + H_{22}(z).

Now note that k_{z} weakly converges to zero as z goes to the boundary of D^{2}. Therefore if Tf̄T_{g} − T_{f̄g} is compact, then

lim_{(z)_{α}→(z_{1},u_{2})} ∫_{T^{2}}((Tf̄T_{g} − T_{f̄g})k_{U_{θ}z_{α}}, k_{U_{θ}z_{α}})e^{i(m,θ)}dθ = 0.

Hence by (5), we conclude that

lim_{z_{α}→(z_{1},u_{2})}(H_{1}(z) + H_{11}(z) + H_{12}(z) + H_{21}(z) + H_{22}(z)) = 0

for any z_{1} ∈ D and u_{2} ∈ T. That is

−H_{22}(z_{1}, u_{2}) = H_{1}(z_{1}, u_{2}) + H_{11}(z_{1}, u_{2}) + H_{12}(z_{1}, u_{2}) + H_{21}(z_{1}, u_{2}).

Since by our claim H_{1}(z_{1}, u_{2}), H_{11}(z_{1}, u_{2}), H_{12}(z_{1}, u_{2}) and H_{21}(z_{1}, u_{2}) are harmonic in z_{1} for any u_{2} ∈ T, so H_{22}(z_{1}, u_{2}) is harmonic in z_{1} for any u_{2} ∈ T. Thus Δ_{z_{1}}H_{22}(z_{1}, u_{2}) = 0.

On the other hand, if we write

f_{−}(z_{1}, z_{2}) = ∑_{k=1}^{∞} \frac{1}{k!} \frac{∂^{k}f_{−}}{∂z̄_{1}^{k}} (0, z_{2})z̄_{1}^{k}

and

g_{−}(z_{1}, z_{2}) = ∑_{k=0}^{∞} \frac{1}{k!} \frac{∂^{k}g_{−}}{∂z̄_{1}^{k}} (0, z_{2})z̄_{1}^{k}.

Let m = (0, m_{2}). We note that

H_{22}(0, u_{2}) =

∑_{k=1}^{∞} \frac{1}{(k!)^{2}} ∫_{T} \overline{\frac{∂^{k}f_{−}}{∂z̄_{1}^{k}} (0, u_{2}e^{iθ_{2}})}\frac{∂^{k}g_{−}}{∂z̄_{1}^{k}} (0, u_{2}e^{iθ_{2}})e^{im_{2}θ_{2}}dθ_{2}|z_{1}|^{2k}.

Since H_{22}(z_{1}, u_{2}) is harmonic in z_{1} for any u_{2} ∈ T. Thus Δ_{z_{1}}H_{22}(z_{1}, u_{2}) = 0. In particular, Δ_{z_{1}}H_{22}(z_{1}, u_{2}) = 0 for z_{1} = 0. Hence

∫_{T} \overline{\frac{∂f_{−}}{∂z̄_{1}} (0, u_{2}e^{iθ_{2}})}\frac{∂g_{−}}{∂z̄_{1}} (0, u_{2}e^{iθ_{2}})e^{im_{2}θ_{2}}dθ_{2} = 0

for all m_{2} ∈ Z and u_{2} ∈ T. We observe that both \frac{∂f_{−}}{∂z̄_{1}} (0, u_{2}) and \frac{∂g_{−}}{∂z̄_{1}} (0, u_{2}) are

in L^{2}(T). So \overline{\frac{∂f_{−}}{∂z̄_{1}} (0, u_{2})}\frac{∂g_{−}}{∂z̄_{1}} (0, u_{2}) is in L^{1}(T) and the Fourier transformation of

\overline{\frac{∂f_{−}}{∂z̄_{1}} (0, u_{2})}\frac{∂g_{−}}{∂z̄_{1}} (0, u_{2}) on Z is zero. The injection of the Fourier transformation im-

plies that

\overline{\frac{∂f_{−}}{∂z̄_{1}} (0, u_{2})}\frac{∂g_{−}}{∂z̄_{1}} (0, u_{2}) = 0

for all u_{2} ∈ T. But this is same as

\overline{\frac{∂f}{∂z̄_{1}} (0, u_{2})}\frac{∂g}{∂z̄_{1}} (0, u_{2}) = 0

for all u_{2} ∈ T.

Next by using the Möbius transform of the bidisc, we will show that statement (1) in the theorem holds. Let ϕ_{z}(w) denote the Möbius transform

ϕ_{z}(w) = (ϕ_{z_{1}}(w_{1}), ϕ_{z_{2}}(w_{2}))

in the bidisc D^{2} for each point z = (z_{1}, z_{2}) ∈ D^{2}. For a fixed point z ∈ D^{2}, we define a unitary operator U_{z} on L^{2}(D^{2}) by

U_{z}h(w) = h ∘ ϕ_{z}(w)k_{z}(w)

for all h ∈ H^{2}(D^{2}). It is easy to see that U_{z}^{∗}T_{f}U_{z} = T_{f∘ϕ_{z}}. Thus

T_{f∘ϕ_{z}}T_{g_{∘}ϕ_{z}} − T_{(f∘ϕ_{z})(g∘ϕ_{z})} = U_{z}^{∗}(T_{f}T_{g} − T_{fg})U_{z}.

Therefore if T_{f}T_{g} − T_{fg} is compact, then T_{f∘ϕ_{z}}T_{g_{∘}ϕ_{z}} − T_{(f∘ϕ_{z})(g∘ϕ_{z})} is compact.

Replacing f and g respectively by f ∘ ϕ_{(z_{1},0,··· ,0)} and g ∘ ϕ_{(z_{1},0,··· ,0)} in above analysis, we get that

\overline{\frac{∂(f ∘ ϕ_{(z_{1},0)})}{∂z̄_{1}}(0, u_{2})}\frac{∂(g ∘ ϕ_{(z_{1},0)})}{∂z̄_{1}}(0, u_{2}) = 0

for all z_{1} ∈ D and u_{2} ∈ T. But

\overline{\frac{∂(f ∘ ϕ_{(z_{1},0,··· ,0)})}{∂z̄_{1}}(0, u_{2})} = (|z_{1}|^{2} − 1)\overline{\frac{∂f}{∂z̄_{1}} (z_{1}, u_{2})}.

Therefore

\overline{\frac{∂f}{∂z̄_{1}} (z_{1}, u_{2})}\frac{∂g}{∂z̄_{1}} (z_{1}, u_{2}) = 0

for u_{2} ∈ T and z_{1} ∈ D. This completes the proof of the theorem except we still need to prove our claim.

We first discuss H_{1}(z). Let

A(z) := ∫_{T^{2}} \overline{f(U_{θ}z)}g(U_{θ}z)e^{i(m,θ)}dθ (9)

for z ∈ D^{2}. Since both f and g are in L^{∞}(T^{2}), we write the harmonic extension of f and g in D^{2} as

f = ∑_{l∈Z×Z} f_{l}z^{l} and g = ∑_{j∈Z×Z} g_{j}z^{j}

where f_{l} and g_{j} are two sequences of numbers satisfying

∑_{l∈Z×Z} |f_{l}|^{2} < ∞ and ∑_{j∈Z×Z} |g_{j}|^{2} < ∞.

A straightforward computation gives that

A(z) = ∑_{j∈Z×Z} \overline{f_{j+m}}g_{j}z^{−(j+m)}z^{j} (10)

for z ∈ D^{2}, where recall that by our convention z_{1}^{−1} is z̄_{1}, for example. Since

∑_{j∈Z×Z} |\overline{f_{j+m}}g_{j}| ≤ ( ∑_{j∈Z×Z} |f_{j+m}|^{2})^{1/2}( ∑_{l∈Z×Z} |g_{j}|^{2})^{1/2} < ∞,

A(z) is continuous on the closure \overline{D}^{2} of D^{2}; and indeed A(z) is defined by the series expansion in (10) for all z ∈ \overline{D}^{2}. But note that here A(z) for z ∈ D^{2} is not necessary the harmonic extension of A(w) for w ∈ T^{2}. By an abuse of notation, we have that (9) holds for all z ∈ \overline{D}^{2}. It follows from the definition of H_{1}(z) that for all z ∈ D^{2},

H_{1}(z) = − ∫_{T^{2}} ∫_{T^{2}} \overline{f(w)}g(w)|k_{U_{θ}z}(w)|^{2}dσ(w)e^{i(m,θ)}dθ
= − ∫_{T^{2}} ∫_{T^{2}} \overline{f(U_{θ}w)}g(U_{θ}w)e^{i(m,θ)}dθ|k_{z}(w)|^{2}dσ(w)
= − ∫_{T^{2}} A(w)|k_{z}(w)|^{2}dA(w)

where the second equality is obtained by changing the order of integration and a change of variables from w to U_{θ}w. That is, H_{1}(z) in D^{2} is the harmonic extension of the continuous function A(w) on T^{2}. Therefore H_{1}(z) is continuous on \overline{D}^{2}. Let ϕ_{z_{2}}(λ_{2}) be the Möbius map \frac{z_{2} − λ_{2}}{1 − \overline{z_{2}}λ_{2}} . Changing variable w_{2} = ϕ_{z_{2}}(λ_{2}) in the above integral, we

have

H_{1}(z) = − ∫_{T} ∫_{T} A(w_{1}, ϕ_{z_{2}}(λ_{2}))|k_{z_{1}}(w_{1})|^{2}dA(w_{1})dA(λ_{2}).

It follows from Lebesque dominated convergence theorem and the continuity of A(z) on \overline{D^{2}} that

H_{1}(z, u_{2}) = lim_{(z)_{α}→(z_{1},u_{2})} H_{22}(z) = − ∫_{T} A(w_{1}, u_{2})|k_{z_{1}}(w_{1})|^{2}dA(w_{1}).

That is, H_{1}(z, u_{2}) as a function of z_{1} on D is the harmonic extension of A(w_{1}, u_{2}) as a function of w_{1} on T. Therefore H_{1}(z, u_{2}) is harmonic in z_{1} ∈ D for any u_{2} ∈ T.

Next we discuss H_{12}(z). Without loss of generality, we consider a term B(z) of H_{12} as in (8).

B(z) = ∫_{T^{2}} ∫_{T^{2}} \overline{f_{−+}(e^{iθ_{1}}z_{1}, w_{2})}g_{++}(w)|k_{U_{θ}z}(w)|^{2}dA(w)e^{i(m,θ)}dθ
= ∫_{T^{2}} ∫_{T^{2}} \overline{f_{−+}(U_{θ}(z_{1}, w_{2}))}g_{++}(U_{θ}w)e^{i(m,θ)}dθ|k_{z}(w)|^{2}dA(w)

where the second equality is obtained by changing the order of integration and a change of variables from w to U_{θ}w. By integrating with respect to w_{1} and noting that f_{−+}(U_{θ}(z_{1}, w_{2})) is independent of w_{1}, we get that

B(z) = ∫_{T} ∫_{T^{2}} \overline{f_{−+}(U_{θ}(z_{1}, w_{2}))}g_{++}(U_{θ}(z_{1}, w_{2}))e^{i(m,θ)}dθ|k_{z_{2}}(w)|^{2}dA(w_{2}).

Now we define

C(z) = ∫_{T^{2}} \overline{f_{−+}(z_{1}, w_{2})}g_{++}(w)e^{i(m,θ)}dθ. (11)

By the proof of the continuity of A(z), we see that C(z) is continuous on \overline{D^{2}}. By an abuse of notation we see that (11) holds for all z ∈ \overline{D^{2}}. Indeed we have

C(z) = ∑_{j∈Z×Z} \overline{a_{j+m}}b_{j}z^{−(j+m)}z^{j}, (12)

for all z ∈ \overline{D^{2}}, if we write f_{−+}(z) and g_{++}(z) as

f_{−+}(z) = ∑_{l∈Z×Z} a_{l}z^{l} and g_{−−}(z) = ∑_{j∈Z×Z} b_{j}z^{j}

where, since f_{−+}(z) and g_{−−}(z) are in L^{2}(T^{2}), a_{l} and b_{j} are two sequences of numbers satisfying

∑_{l∈Z×Z} |a_{l}|^{2} < ∞, and ∑_{j∈Z×Z} |b_{j}|^{2} < ∞.

Next note that

B(z) = ∫_{T} C(z_{1}, w_{2})|k_{z_{2}}(w)|^{2}dA(w_{2}).

Hence B(z) is continuous on \overline{D^{2}}. Let ϕ_{z_{2}}(λ_{2}) be the Möbius map \frac{z_{2} − λ_{2}}{1 − \overline{z_{2}}λ_{2}} . We have

B(z) = ∫_{T} C(z_{1}, ϕ_{z_{2}}(λ_{2}))dA(λ_{2}).

It follows from the continuity of C(z) that

B(z_{1}, u_{2}) = lim_{(z)_{α}→(z_{1},u_{2})} B(z) = ∫_{T} C(z_{1}, u_{2})dA(λ_{2}) = C(z_{1}, u_{2}).

To prove that C(z_{1}, u_{2}) is harmonic in z_{1} for any u_{2} ∈ T, we look at the series expansion of C(z) more carefully. We first note that a_{l} = 0 for l = (l_{1}, l_{2}) ∈ Z × Z such that l_{1} ≥ 0 and b_{j} = 0 except for j = (j_{1}, j_{2}) ∈ Z_{+} ×Z_{+}. Therefore in fact C(z_{1}, u_{2}) = 0 for m = (m_{1}, m_{2}) ∈ Z_{+} × Z, and

C(z_{1}, u_{2}) = z_{1}^{−m_{1}}u_{2}^{m_{2}}( ∑_{0≤j_{1}<−m_{1}} ∑_{j_{2}∈Z_{+}} \overline{a_{j+m}}b_{j})

for m = (m_{1}, m_{2}) ∈ Z_{−} × Z. This finishes the proof of the claim concerning H_{21}(z).

The proof of the claim concerning H_{11}(z) and H_{21}(z) is similar. So does the proof of the continuity of H_{22}(z). Nevertheless one does not necessarily have that H_{22}(z_{1}, u_{2}) is harmonic in z_{1}. So the claim is established. This completes the proof of the theorem.

We remark that in general the conditions (1) and (2) in Theorem 2 are not sufficient for T_{f}T_{g} − T_{fg} to be compact. See Theorem 3 in next section for examples.

Next we use Theorem 2 to derive a result in [4] about compact Hankel operators on the bidisc; see [4] and reference therein for more related results.

Corollary 1. [4] Let f be a bounded functions on the bitorus T^{2}. The Hankel operators H_{f} is compact if and only if it is zero.

Proof. Note that −H_{f}^{∗}H_{f} = Tf̄T_{f} − T_{f̄f}. Thus by Theorem 2, H_{f} is compact implies that for all z_{1} ∈ D and all z_{2} ∈ T,

|\frac{∂f}{∂z̄_{1}} (z_{1}, z_{2})|^{2} = 0

and for all z_{2} ∈ D and all z_{1} ∈ T,

|\frac{∂f}{∂z̄_{2}} (z_{1}, z_{2})|^{2} = 0.

Therefore f is analytic in D^{2}. So H_{f} is zero. The proof is complete.

The following result is an immediate consequence of Theorem 2. See also [11] for related results on the Bergman space of the bidisc.

Corollary 2. Let f and g be two bounded functions on T^{2}. If one of the functions f, f̄, g, ḡ is analytic, then T_{f}T_{g} − T_{g}T_{f} is zero if and only if T_{f}T_{g} − T_{g}T_{f} is compact.

Proof. Without loss of generality, we assume that f is analytic. It is easy to see that in this case T_{f}T_{g} − T_{g}T_{f} = T_{f}T_{g} − T_{fg}.

By Theorem 2, if T_{f}T_{g} − T_{fg} is compact, then for all z_{1} ∈ D and z_{2} ∈ T,

\frac{∂f}{∂z_{1}} (z_{1}, z_{2})\frac{∂g}{∂z̄_{1}} (z_{1}, z_{2}) = 0,

and for all z_{2} ∈ D and all z_{1} ∈ T,

\frac{∂f}{∂z_{2}} (z_{1}, z_{2})\frac{∂g}{∂z̄_{2}} (z_{1}, z_{2}) = 0.

Since f is analytic on the unit disk, we conclude that for each i (i = 1, 2) either f̄ or g is analytic in z_{i}. By Theorem 1, this implies that T_{f}T_{g} − T_{fg} = 0. This completes the proof of the corollary.

We remark that the above corollary is not valid without the assumption that one of the functions f, f̄, g, ḡ is analytic. An example will be shown in next section.

4. Compact semi-commutators

In this section we will characterize when T_{f}T_{g} −T_{fg} is compact for Toeplitz operators T_{f} and T_{g} with symbols of the form f(z_{1}, z_{2}) = f_{1}(z_{1})f_{2}(z_{2}) and g(z_{1}, z_{2}) = g_{1}(z_{1})g_{2}(z_{2}). This will show that the condition in Theorem 2 is in general not a sufficient condition for T_{f}T_{g} − T_{fg} to be compact. Furthermore it provides us with nontrivial compact semi-commutators.

Theorem 3. Let f_{1}, f_{2}, g_{1} and g_{2} be nonzero bounded functions on the unit circle T. Let f and g be bounded functions on T^{2} of the form f(z_{1}, z_{2}) = f_{1}(z_{1})f_{2}(z_{2}) and g(z_{1}, z_{2}) = g_{1}(z_{1})g_{2}(z_{2}). Then T_{f}T_{g} − T_{fg} is a nonzero compact operator on the Hardy space of the bidisc if and only if the following two conditions hold.

(1) f_{1}(z_{1})g_{1}(z_{1}) = 0 and f_{2}(z_{2})g_{2}(z_{2}) = 0 on T, and

(2) lim_{z→T^{2}} ∥H\overline{f}k_{z}∥_{2}∥H_{g}k_{z}∥_{2} = 0.

Proof. We first prove the necessity part of the theorem. Assume that T_{f}T_{g} − T_{fg} is compact. By Theorem 2, we have that for all z_{1} ∈ D and all z_{2} ∈ T,

\frac{∂f_{1}(z_{1})}{∂z_{1}} \frac{∂g_{1}(z_{1})}{∂z̄_{1}}f_{2}(z_{2})g_{2}(z_{2}) = 0 (13)

and for all z_{2} ∈ D and all z_{1} ∈ T,

\frac{∂f_{2}(z_{2})}{∂z_{2}} \frac{∂g_{2}(z_{2})}{∂z̄_{2}}f_{1}(z_{1})g_{1}(z_{1}) = 0. (14)

We claim that this implies that Condition (1) holds. If Condition (1) does not hold. Without loss of generality, assume that f_{1}g_{1} is not zero. Then equation (14) gives that

\frac{∂f_{2}}{∂z_{2}} \frac{∂g_{2}}{∂z̄_{2}} = 0

for all z_{2} ∈ D. That is either f̄_{2} or g_{2} is analytic in z_{2}. Say, g_{2} is analytic in z_{2}. Thus g_{2}f_{2} is not zero since otherwise either f_{2} or g_{2} will be zero. Now equation (13) implies that

\frac{∂f_{1}}{∂z_{1}} \frac{∂g_{1}}{∂z̄_{1}} = 0.

That is either f̄_{1} or g_{1} is analytic in z_{1}. Say, f̄_{1} is analytic in z_{1}. Hence f̄ is analytic in z_{1} and g is analytic in z_{2}. By Theorem 1, we get that T_{f}T_{g} − T_{fg} is zero. This is a contradiction. Therefore Condition (1) holds.

Next we show that Condition (2) holds. We first need some notations. Let P_{1} be the projection defined by

P_{1}(h) = ∑_{m=(m_{1},m_{2})∈Z_{+}×Z} h_{m}z^{m} if h = ∑_{m=(m_{1},m_{2})∈Z×Z} h_{i}(z_{2})z^{m} ∈ L^{2}(T^{2}),

and similarly P_{2} the projection defined by

P_{2}(h) = ∑_{m=(m_{1},m_{2})∈Z×Z_{+}} h_{m}z^{m} if h = ∑_{m=(m_{1},m_{2})∈Z×Z} h_{i}(z_{2})z^{m} ∈ L^{2}(T^{2}),

It is easy to see that P = P_{1}P_{2} = P_{2}P_{1}. Let h_{n}(z_{1}) ∈ H^{2}(D) such that h_{n} converges weakly to zero in H^{2}(D). For any h(z_{2}) in H^{2}(D). Let H_{n}(z_{1}, z_{2}) be h(z_{2})h_{n}(z_{1}). Note that H_{n} converges weakly to zero in H^{2}(D^{2}) as well. Note also by Condition (1),

(T_{f}T_{g} − T_{fg})H_{n}(z_{1}, z_{2}) = T_{f}T_{g}H_{n}(z_{1}, z_{2})

= P_{2}(f_{2}P_{2}g_{2}h)(z_{2})T_{f_{1}}T_{g_{1}}h_{n}(z_{1}) = −P_{2}(f_{2}P_{2}g_{2}h)(z_{2})H∗ ; \overline{f_{1}}H_{g_{1}}h_{n}(z_{1})

and P_{2}(f_{2})P_{2}(g_{2}h)(z_{2}) is not zero for some h ∈ H^{2}(D). Thus if T_{f}T_{g} − T_{fg} is compact, then ∥H∗ ; \overline{f_{1}}H_{g_{1}}h_{n}∥ converges to zero. Therefore H∗ ; \overline{f_{1}}H_{g_{1}} is compact on the Hardy space of the unit disk. The compactness of H∗ ; \overline{f_{2}}H_{g_{2}} can be established similarly. It follows from Theorem 2 in [13] that

lim_{z_{1}→T} ∥H\overline{f_{1}}k_{z_{1}}∥_{2}∥H_{g_{1}}k_{z_{1}}∥_{2} = 0,

and

lim_{z_{2}→T} ∥H\overline{f_{2}}k_{z_{2}}∥_{2}∥H_{g_{2}}k_{z_{2}}∥_{2} = 0.

So

lim_{z_{1} orz_{2}→T} ∥H\overline{f_{1}}k_{z_{1}}∥_{2}∥H_{g_{1}}k_{z_{1}}∥_{2}∥H\overline{f_{2}}k_{z_{2}}|∥_{2}∥H_{g_{2}}k_{z_{2}}∥_{2} = 0.

THE SEMI-COMMUTATOR OF TOEPLITZ OPERATORS ON THE BIDISC 15

On the other hand,

∥H\overline{f}k_{z}∥_{2}∥H_{g}k_{z}∥_{2} = ∥H\overline{f_{1}}k_{z_{1}}∥_{2}∥H_{g_{1}}k_{z_{1}}∥_{2}∥H\overline{f_{2}}k_{z_{2}}∥_{2}∥H_{g_{2}}k_{z_{2}}∥_{2}.

Hence

lim_{z→T^{2}} ∥H\overline{f}k_{z}∥_{2}∥H_{g}k_{z}∥_{2} = 0

as it is equivalent to

lim_{z_{1}→T} ∥H\overline{f_{1}}k_{z_{1}}∥_{2}∥H_{g_{1}}k_{z_{1}}∥_{2} = 0,

and

lim_{z_{2}→T} ∥H\overline{f_{2}}k_{z_{2}}∥_{2}∥H_{g_{2}}k_{z_{2}}∥_{2} = 0.

We now turn to the proof of the sufficient part of the theorem. By Condition (1), we have T_{f}T_{g} − T_{fg} = T_{f}T_{g}. Let ϕ and ψ be in H^{2}(D^{2}). We observe that

T_{f}T_{g}ϕ = P_{1}P_{2}(f_{1}f_{2}P_{1}P_{2}(g_{1}g_{2}ϕ))

= P_{1}f_{1}P_{1}g_{1}P_{2}(f_{2}P_{2}(g_{2}ϕ))
= P_{1}f_{1}(I − P_{1})g_{1}P_{2}(f_{2}(I − P_{2})(g_{2}ϕ))
= P_{1}f_{1}P_{2}(f_{2}(I − P_{1})g_{1}(I − P_{2})(g_{2}ϕ)),

where the third equality follows from Condition (1). Therefore

< T_{f}T_{g}ϕ, ψ >=< P_{1}f_{1}P_{2}(f_{2}(I − P_{1})g_{1}(I − P_{2})g_{2}ϕ), ψ >

=< (I − P_{1})g_{1}(I − P_{2})g_{2}ϕ, \overline{f_{1}}\overline{f_{2}}ψ >
=< (I − P_{1})g_{1}(I − P_{2})g_{2}ϕ, (I − P_{1})\overline{f_{1}}(I − P_{2})\overline{f_{2}}ψ > .

It is easy to see that (I − P_{1})g_{1}(I − P_{2})g_{2}ϕ is co-analytic in z_{1} and z_{2}. So

▽_{1}(I − P_{1})g_{1}(I − P_{2})g_{2}ϕ = \frac{∂}{∂\overline{z_{1}}} (I − P_{1})g_{1}(I − P_{2})g_{2}ϕ.

Since

∫_{∂D}(I − P_{1})g_{1}(I − P_{2})g_{2}ϕ(z_{1}, z_{2})dσ(z_{1}) = 0

and

∫_{∂D} \frac{∂}{∂\overline{z_{1}}} [(I − P_{1})g_{1}(I − P_{2})g_{2}ϕ(z_{1}, z_{2})]dσ(z_{2}) = 0,

16 CAIXING GU AND DECHAO ZHENG

using the Littlewood-Paley formula, we have

< (I − P_{1})g_{1}(I − P_{2})g_{2}ϕ, (I − P_{1})\overline{f_{1}}(I − P_{2})\overline{f_{2}}ψ >

= ∫_{D} ∫_{D} < ▽_{1} ▽_{2} (I − P_{1})g_{1}(I − P_{2})g_{2}ϕ, ▽_{1} ▽_{2} (I − P_{1})\overline{f_{1}}(I − P_{2})\overline{f_{2}}ψ > ×

log \frac{1}{|z_{1}|} log \frac{1}{|z_{2}|}dA(z_{1})dA(z_{2})

= ∫_{D} ∫_{D} < ▽_{1}(I − P_{1})g_{1} ▽_{2} (I − P_{2})g_{2}ϕ, ▽_{1}(I − P_{1})\overline{f_{1}} ▽_{2} (I − P_{2})\overline{f_{2}}ψ > ×

log \frac{1}{|z_{1}|} log \frac{1}{|z_{2}|}dA(z_{1})dA(z_{2})

= I_{1} + I_{2} + I_{3} where

I_{1} = ∫_{rD} ∫_{rD} < ▽_{1} ▽_{2} (I − P_{1})g_{1}(I − P_{2})g_{2}ϕ, ▽_{1} ▽_{2} (I − P_{1})\overline{f_{1}}(I − P_{2})\overline{f_{2}}ψ >

× log \frac{1}{|z_{1}|} log \frac{1}{|z_{2}|}dA(z_{1})dA(z_{2}),

I_{2} = ∫_{D/rD} ∫_{D} < ▽_{1} ▽_{2} (I − P_{1})g_{1}(I − P_{2})g_{2}ϕ, ▽_{1} ▽_{2} (I − P_{1})\overline{f_{1}}(I − P_{2}) \overline{f_{2}}ψ >

× log \frac{1}{|z_{1}|} log \frac{1}{|z_{2}|}dA(z_{1})dA(z_{2}),

I_{3} = ∫_{rD} ∫_{D/rD} < ▽_{1} ▽_{2} (I − P_{1})g_{1}(I − P_{2})g_{2}ϕ, ▽_{1} ▽_{2} (I − P_{1})\overline{f_{1}}(I − P_{2})\overline{f_{2}}ψ >

× log \frac{1}{|z_{1}|} log \frac{1}{|z_{2}|}dA(z_{1})dA(z_{2}).

It is easy to see that there is a compact operator K_{r} such that

I_{1} =< K_{r}ϕ, ψ > .

To deal with I_{2} and I_{3}, we use the distribution function inequality established in [13]. As in the proof of Theorem 7 in [13], we obtain that

| ∫_{D/rD} < ▽_{1}(I − P_{1})g_{1} ▽_{2} (I − P_{2})g_{2}ϕ, ▽_{1}(I − P_{1})\overline{f_{1}} ▽_{2} (I − P_{2})\overline{f_{2}}ψ > log \frac{1}{|z_{1}|}dA(z_{1})|

≤ C sup_{|z_{1}|>r} ∥H\overline{f_{1}}k_{z_{1}}∥∥H_{g_{1}}k_{z_{1}}∥∥ ▽_{2} (I − P_{2})g_{2}ϕ∥_{z_{1}}∥ ▽_{2} (I − P_{2})\overline{f_{2}}ψ∥_{z_{1}},

where for h ∈ L^{2}(T^{2}) and z_{2} fixed, ∥h∥_{z_{1}} denotes the L^{2} norm of h(z_{1}, z_{2}) as a function of z_{1}. Similarly we also have

| ∫_{D/rD} < ▽_{1} (I − P_{1})g_{1} ▽_{2} (I − P_{2})g_{2}ϕ, ▽_{1}(I − P_{1})\overline{f_{1}} ▽_{2} (I − P_{2})\overline{f_{2}}ψ > log \frac{1}{|z_{2}|} dA(z_{2})|

≤ C sup_{|z_{2}|>r} ∥ ∥H\overline{f_{2}}k_{z_{2}}∥∥H_{g_{2}}k_{z_{2}}∥∥ ▽_{1} (I − P_{1})g_{1}ϕ∥_{z_{2}}∥ ▽_{1} (I − P_{1})\overline{f_{1}}ψ∥_{z_{2}}.

So

|I_{2}| ≤ C sup_{|z_{1}|>r} ∥ H\overline{f_{1}}k_{z_{1}}∥∥H_{g_{1}}k_{z_{1}}∥ ×

∫_{D} ∥ ▽_{2} (I − P_{2})g_{2}ϕ∥_{z_{1}}∥ ▽_{2} (I − P_{2})\overline{f_{2}}ψ∥_{z_{1}} log \frac{1}{|z_{2}|} dA(z_{2}).

By the Holder inequality,

|I_{2}| ≤ C sup_{|z_{1}|>r} ∥H\overline{f_{1}}k_{z_{1}}∥∥H_{g_{1}}k_{z_{1}}∥ ×

[∫_{D} ∫_{∂D} | ▽_{2} (I − P_{2})g_{2}ϕ(z_{1}, z_{2})|^{2} log \frac{1}{|z_{2}|}dσ(z_{1})dA(z_{2})]^{1/2} ×

[∫_{D} ∫_{∂D} | ▽_{2} (I − P_{2})\overline{f_{2}}ψ(z_{1}, z_{2})|^{2} log \frac{1}{|z_{2}|}dσ(z_{1})dA(z_{2})]^{1/2}.

By the Littlewood-Paley theorem and the fact that g_{2} and f_{2} are bounded,

|I_{2}| ≤ C sup_{|z_{1}|>r} ∥H\overline{f_{1}}k_{z_{1}}∥∥H_{g_{1}}k_{z_{1}}∥∥ϕ∥∥ψ∥.

Similarly we can also get the estimate for I_{3},

|I_{3}| ≤ C sup_{|z_{2}|>r} ∥H\overline{f_{2}}k_{z_{2}}∥∥H_{g_{2}}k_{z_{2}}∥∥ϕ∥∥ψ∥.

In the proof of the necessary part of the theorem we have shown that the second condition is equivalent to

lim_{z_{1}→T} ∥H\overline{f_{1}}k_{z_{1}}∥_{2}∥H_{g_{1}}k_{z_{1}}∥_{2} = 0,

and

lim_{z_{2}→T} ∥H_{\overline{f_{2}}k_{z_{2}}}∥_{2}∥H_{g_{2}}k_{z_{2}}∥_{2} = 0.

Hence we conclude that

lim_{r→1} ∥T_{f}T_{g} − T_{fg} − K_{r}∥ = 0.

Therefore T_{f}T_{g} − T_{fg} is compact. This completes the proof of the theorem.

We now give an example to show that the assumption in Corollary 2 is necessary. Namely, there exist bounded functions f and g such that T_{f}T_{g} − T_{g}T_{f} is compact, but T_{f}T_{g} − T_{g}T_{f} is not zero.

Example. Let f_{1}(z_{1}) and g_{1}(z_{1}) be nonzero continuous functions on the unit circle satisfying the conditions that f_{1}(z_{1})g_{1}(z_{1}) = 0 and the set {z_{1} : f_{1}(z_{1}) = 0} ∩ {z_{1} : g_{1}(z_{1}) = 0} as a subset of T has positive Lebesgue measure. Let f_{2}(z_{2}) and g_{2}(z_{2}) be nonzero continuous functions on the unit circle such that f_{2}(z_{2})g_{2}(z_{2}) = 0.

First we show that T_{f}T_{g} − T_{g}T_{f} is not zero. Note that

(T_{f}T_{g} − T_{g}T_{f})h_{1}(z_{1}) = P_{2}(f_{2})P_{2}(g_{2})(T_{f_{1}}T_{g_{1}} − T_{g_{1}}T_{f_{1}})h_{1}(z_{1})

for all h_{1} ∈ H^{2}(D). It is easy to see that the assumption on f_{2} and g_{2} implies that P_{2}(f_{2})P_{2}(g_{2}) is not zero. Therefore it suffices to show that T_{f_{1}}T_{g_{1}} − T_{g_{1}}T_{f_{1}} is not zero. Let S_{1} denote the multiplication by z_{1} on H^{2}(D). We observe that I − S_{1}S_{1}^{∗} is the rank one operator e_{0} ⊗ e_{0}, where e_{0} = 1 ∈ H^{2}(D). Note that

T_{f_{1}}T_{g_{1}} − T_{g_{1}}T_{f_{1}} − S_{1}^{∗}(T_{f_{1}}T_{g_{1}} − T_{g_{1}}T_{f_{1}})S_{1}
= T_{f_{1}}T_{g_{1}} − S_{1}^{∗}T_{f_{1}}T_{g_{1}}S_{1} − (T_{g_{1}}T_{f_{1}} − S_{1}^{∗}T_{g_{1}}T_{f_{1}}S_{1})
= S_{1}^{∗}T_{f_{1}}S_{1}S_{1}^{∗}T_{g_{1}}S_{1} − S_{1}^{∗}T_{f_{1}}T_{g_{1}}S_{1} − (S_{1}^{∗}T_{g_{1}}S_{1}S_{1}^{∗}T_{f_{1}}S_{1} − S_{1}^{∗}T_{g_{1}}T_{f_{1}}S_{1})
= −S_{1}^{∗}T_{f_{1}}(I − S_{1}S_{1}^{∗})T_{g_{1}}S_{1} + S_{1}^{∗}T_{g_{1}}(I − S_{1}S_{1}^{∗})T_{f_{1}}S_{1}
= −S_{1}^{∗}T_{f_{1}}e_{0} ⊗ S_{1}^{∗}T_{g_{1}}^{∗}e_{0} + S_{1}^{∗}T_{g_{1}}e_{0} ⊗ S_{1}^{∗}T_{f_{1}}^{∗},

where the second equality follows from the fact that for Toeplitz operator T_{f_{1}}, S_{1}^{∗}T_{f_{1}}S_{1} = T_{f_{1}}. Thus if T_{f_{1}}T_{g_{1}} − T_{g_{1}}T_{f_{1}} = 0, then there exists a contant c such that

S_{1}^{∗}T_{f_{1}}e_{0} = cS_{1}^{∗}T_{g_{1}}e_{0}.

That is, there exists a constant d such that P_{1}(f_{1} − cg_{1}) = d. Hence if we denote the function \overline{f_{1}} − \overline{cg_{1}} by h, then h is analytic. But by our assumption, h(z_{1}) = 0 on a subset of T with positive Lebesgue measure. Therefore h_{1} = 0 on T, i.e., f_{1} = cg_{1}. This contradicts to the conditions that both f_{1} and g_{1} are nonzero and f_{1}g_{1} = 0 on T. So T_{f_{1}}T_{g_{1}} − T_{g_{1}}T_{f_{1}} is not zero.

Next we show that T_{f}T_{g} − T_{g}T_{f} is compact. Since f_{1}(z_{1}), g_{1}(z_{1}), f_{2}(z_{2}) and g_{2}(z_{2}) are continuous, so H\overline{f_{1}}H_{g_{1}} and H\overline{f_{2}}H_{g_{2}} are compact. Hence

lim_{z_{1}→T} ∥H\overline{f_{1}}k_{z_{1}}∥_{2}∥H_{g_{1}}k_{z_{1}}∥_{2} = 0,

and

lim_{z_{2}→T} ∥H_{\overline{f_{2}}k_{z_{2}}}∥_{2}∥H_{g_{2}}k_{z_{2}}∥_{2} = 0,

or equivalently Condition (2) in Theorem 3 holds. Therefore T_{f}T_{g} − T_{fg} and T_{g}T_{f} − T_{gf} are compact. Note that

T_{f}T_{g} − T_{g}T_{f} = (T_{f}T_{g} − T_{fg}) − (T_{g}T_{f} − T_{gf}).

Hence T_{f}T_{g} − T_{g}T_{f} is compact.

References

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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