Products of Toeplitz operators on the Fock space

Let $f$ and $g$ be functions, not identically zero, in the Fock space $F^2$ of $C_n$. We show that the product $T_fT_{\bar g}$ of Toeplitz operators on $F^2$ is bounded if and only if $f(z)=e^{q(z)}$ and $g(z)=ce^{-q(z)}$, where $c$ is a nonzero constant and $q$ is a linear polynomial.


INTRODUCTION
Let C n be the complex n-space. For points z = (z 1 , · · · , z n ) and w = (w 1 , · · · , w n ) in C n we write Let dv be ordinary volume measure on C n . For any positive parameter α we consider the Gaussian measure dλ α (z) = α π n e −α|z| 2 dv(z).
The Fock space F 2 α is the closed subspace of entire functions in L 2 (C n , dλ α ). The orthogonal projection P : L 2 (C n , dλ α ) → F 2 α is given by where K(z, w) = e αz·w is the reproducing kernel of F 2 α . We say that f satisfies Condition (G) if the function z → f (z)e αz·w belongs to L 1 (C n , dλ α ) for every w ∈ C n . Equivalently, f satisfies Condition (G) if every translate of f , z → f (z + w), belongs to L 1 (C n , dλ α ). If f ∈ F 2 α , then there exists a constant C > 0 such that |f (z)| ≤ Ce This clearly implies that f satisfies Condition (G). If f satisfies Condition (G), we can define a linear operator T f on F 2 α by T f g = P (f g), where is any finite linear combination of kernel functions. It is easy to see that the set of all finite linear combinations of kernel functions is dense in F 2 α . Here P (f g) is to be interpreted as the following integral: Therefore, for g in a dense subset of F 2 α , T f g is a well-defined entire function (not necessarily in F 2 α though). The purpose of this article is to study the Toeplitz product T f T g , where f and g are functions in F 2 α . Such a product is well defined on the set of finite linear combinations of kernel functions. Our main concern is the following: what conditions on f and g will ensure that the Toeplitz product T f T g extends to a bounded (or compact) operator on F 2 α ? This problem was first raised by Sarason in [5] in the context of Hardy and Bergman spaces. It was partially solved for Toeplitz operators on the Hardy space of the unit circle in [9], on the Bergman space of the unit disk in [6], on the Bergman space of the polydisk in [7], and on the Bergman space of the unit ball in [4,8]. In all these cases, the necessary and/or sufficient condition for T f T g to be bounded is where ε is any positive number and f denotes the Berezin transform of f . Note that in the Hardy space case, the Berezin transform is nothing but the classical Poisson transform.
Here we obtain a much more explicit characterization for T f T g to be bounded on the Fock space.
Main Theorem. Let f and g be functions in F 2 α , not identically zero. Then T f T g is bounded on F 2 α if and only if f = e q and g = ce −q , where c is a nonzero complex constant and q is a complex linear polynomial.
Furthermore, our proof reveals that when T f T g is bounded, it must be a constant times a unitary operator. Consequently, T f T g is never compact unless it is the zero operator.
As another by-product of our analysis, we will construct a class of unbounded, densely defined, operators on the Fock space whose Berezin transform is bounded. It has been known that such operators exist, but our examples are very simple products of Toeplitz operators.
We wish to thank Boorim Choe and Hyungwoon Koo for useful conversations.

PROOF OF MAIN RESULT
For any point a ∈ C n we consider the operator U a : is the normalized reproducing kernel of F 2 α at a. It follows from a change of variables that each U a is a unitary operator on F 2 α . We begin with the very special case of Toeplitz operators induced by kernel functions.
We have In particular, T f T g is bounded on F 2 α . Proof. To avoid triviality we assume that a is nonzero. The Toeplitz operator T f is just multiplication by f , as a densely defined unbounded linear operator. So we focus on the operator T g .
Given any function h ∈ F 2 α , we have Therefore, the Toeplitz operator T g is an operator of translation, and . This proves the desired result.
An immediate consequence of Lemma 1 is that if f = C 1 e q and g = C 2 e −q , where C 1 and C 2 are complex constants and q is a complex linear polynomial, then there exists a complex constant c and a unitary operator U such that T f T g = cU.
To deal with more general symbol functions, we need the following characterization of nonvanishing functions in F 2 α .
Proof. In the case when the dimension n = 1, the Weierstrass factorization of functions in the Fock space F 2 α takes the form f (z) = P (z)e q(z) , where P is the canonical Weierstrass product associated to the zero sequence of f , and q(z) = az 2 + bz + c is a quadratic polynomial with |a| < α 2 . In particular, if f is zero-free, then f = e q for some quadratic polynomial. See [10].
When n > 1, we no longer have such a nice factorization. But the absence of zeros makes a special version of the factorization above still valid. More specifically, if f is any function in F 2 α = F 2 α (C n ) and f is nonvanishing, then the function z 1 → f (z 1 , · · · , z n ) is in F 2 α (C), so by the factorization theorem stated in the previous paragraph, where a, b, and c are holomorphic functions of z 2 , · · · , z n . Repeat this for every independent variable, we conclude that f = e q for some polynomial of degree 2n or less.
Recall that every function f ∈ F 2 α satisfies the pointwise estimate If q is a polynomial of degree N and N > 2, then for any fixed ζ = (ζ 1 , · · · , ζ n ) on the unit sphere of C n with each ζ k = 0, and for z = rζ, where r > 0, we have q(z) ∼ r N as r → ∞, which shows that the estimate |f (z)| ≤ Ce α 2 |z| 2 is impossible to hold. This shows that the degree of q is less than or equal to 2.
We can now prove the main result, which we restate as follows. Proof. If condition (a) holds, then the Toeplitz product T f T g is 0. If condition (b) holds, the boundedness of T f T g follows from Lemma 1.
Next assume that T = T f T g is bounded on F 2 α . Then the Berezin transform T is a bounded function on C n , where It follows from the integral representation of T g and the reproducing property of the kernel function e αz·w that T g k z = g(z) k z . Therefore, Write the inner product above as an integral and apply the reproducing property of the kernel function e αz·w one more time. We obtain T (z) = f (z)g(z). It follows that |f (z)g(z)| ≤ T for all z ∈ C n . But f g is entire, so by Liouville theorem, there is a constant c such that f g = c.
If c = 0, then at least one of f and g must be identically zero, so condition (a) holds.
If c = 0, then both f and g are nonvanishing. By Lemma 2, there exists a complex polynomial q, with deg(q) ≤ 2, such that f = e q and g = ce −q . It remains for us to show that deg(q) ≤ 1.
Let us assume deg(q) = 2, in the hope of reaching a contradition, and write q = q 2 + q 1 , where q 1 is linear and q 2 is a homogeneous polynomial of degree 2. By the boundedness of T = T f T g on F 2 α , the function T (z, w) = T f T g k z , k w , z ∈ C n , w ∈ C n , is bounded on C n × C n . We proceed to show that this is impossible unless q 2 = 0. Again, by the integral representation for Toeplitz operators and the reproducing property of the kernel function e αz·w , it is easy to obtain that for all (z, w) ∈ C n × C n . Using the explict form of f and g, we can write Since q 1 is linear, it is easy to see that there is a point a ∈ C n such that for all z and w.
For the second-degree homogeneous polynomial q 2 we can find a complex matrix A = A n×n , symmetric in the real sense, such that q 2 (z) = Az, z , where , is the real inner product. Fix two points u and v in C n such that Re Au, v = 0. This is possible as long as A = 0. Now let z = ru and w = ru + v, where r is any real number. We have It follows that there exists a positive constant M = M(u, v) such that Since Re Au, v = 0, this shows that T (z, w) cannot be a bounded function on C n × C n . This contradition shows that A = 0 and the polynomial q must be linear.
As a consequence of the analysis above, we obtain an interesting class of unbounded operators on F 2 α whose Berezin transforms are bounded.
Corollary 4. Suppose f (z) = e q and g = e −q , where q is any second-degree homogeneous polynomial whose coefficients are small enough so that f and g belong to F 2 α . Then the Toeplitz product T f T g is unbounded on F 2 α , but its Berezin transform is bounded.
Proof. By Theorem 3, the operator T f T g is unbounded. On the other hand, by the proof of Theorem 3, the Berezin transform of T = T f T g is given by It follows that |T (z)| = |f (z)g(z)| = 1 for all z ∈ C n .
Another consequence of our earlier analysis is the following.
Corollary 5. If f and g are functions in F 2 α , then the following conditions are equivalent: (c) f = 0 or g = 0.
Proof. Combining Lemma 1 and Theorem 3, we see that whenever T f T g is compact on F 2 α , we must have f = 0 or g = 0. This clearly gives the desired result.

FURTHER REMARKS
For any 0 < p ≤ ∞ let F p α denote the Fock space consisting of entire functions f such that the function f (z)e − α 2 |z| 2 belongs to L p (C n , dv). When 0 < p < ∞, the norm in F p α is defined by It is easy to check that the normalized reproducing kernel k a (z) = e αz·a− α 2 |a| 2 is a unit vector in each F p α , where 0 < p ≤ ∞. Also, it can be shown that the set of functions of the form [10]. Therefore, if 0 < p < ∞ and f satisfies Condition (G), we can consider the action of the Toeplitz operator T f (defined using the integral representation in Section 1) on F p α . Also, if f ∈ F p α , then it satisfies the pointwise estimate |f (z)| ≤ Ce α 2 |z| 2 , which implies that f satisfies Condition (G). When 1 < p < ∞ and 1/p + 1/q = 1, the dual space of F p α can be identified with F q α under the integral pairing When 0 < p ≤ 1, the dual space of F p α can be identified with F ∞ α under the same integral pairing above. See [3,10].
Thus for functions f and g in F p α , if the Toeplitz product T = T f T g is bounded on F p α , we can still consider the function T (z, w) = T f T g k z , k w α on C n × C n . Exactly the same arguments from Section 2 will yield the following result.
Theorem 6. Suppose 0 < p < ∞. If f and g are functions in F p α , not identically zero, then the Toeplitz product T f T g is bounded on F p α if and only if f = e q and g = ce −q , where c is a nonzero complex contant and q is a complex linear polynomial.
It would be nice to extend our results here to more general Fock-type spaces. In particular, generalization to the Fock-Sobolev spaces studied in [1,2] should be possible.
It would also be interesting to take a second look at the original Hardy space setting. More specifically, if f and g are functions in the Hardy space H 2 (of the unit disk, for example), the the boundedness of the Toeplitz product T f T g on H 2 implies that the product function f g is in H ∞ . Is it possible to derive more detailed information about f and g, say in terms of inner and outer functions? A more explicit condition on f and g (as opposed to the condition |f | 2+ε |g| 2+ε ∈ L ∞ ) would certainly be more desirable.
We hope that this paper will generate some further interest in this subject.