Products of Toeplitz Operators on the 2-Analytic Bergman Space

<jats:p>Let <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M1">
                        <mi>f</mi>
                     </math>
                  </jats:inline-formula> and <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M2">
                        <mi>g</mi>
                     </math>
                  </jats:inline-formula> be bounded functions, and let <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M3">
                        <msub>
                           <mrow>
                              <mi>T</mi>
                           </mrow>
                           <mrow>
                              <mi>f</mi>
                           </mrow>
                        </msub>
                     </math>
                  </jats:inline-formula> and <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M4">
                        <msub>
                           <mrow>
                              <mi>T</mi>
                           </mrow>
                           <mrow>
                              <mi>g</mi>
                           </mrow>
                        </msub>
                     </math>
                  </jats:inline-formula> be Toeplitz operators on <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M5">
                        <msubsup>
                           <mrow>
                              <mi>A</mi>
                           </mrow>
                           <mrow>
                              <mn>2</mn>
                           </mrow>
                           <mrow>
                              <mn>2</mn>
                           </mrow>
                        </msubsup>
                        <mfenced open="(" close=")">
                           <mrow>
                              <mi mathvariant="double-struck">D</mi>
                           </mrow>
                        </mfenced>
                     </math>
                  </jats:inline-formula>. We show that if the product <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M6">
                        <msub>
                           <mrow>
                              <mi>T</mi>
                           </mrow>
                           <mrow>
                              <mi>f</mi>
                           </mrow>
                        </msub>
                        <msub>
                           <mrow>
                              <mi>T</mi>
                           </mrow>
                           <mrow>
                              <mi>g</mi>
                           </mrow>
                        </msub>
                     </math>
                  </jats:inline-formula> equals zero and one of <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M7">
                        <mi>f</mi>
                     </math>
                  </jats:inline-formula> and <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M8">
                        <mi>g</mi>
                     </math>
                  </jats:inline-formula> is a radial function satisfying a Mellin transform condition, then the other function must be zero.</jats:p>


Introduction
Let D be the open unit disk in ℂ equipped with the normalized Lebesgue area measure dAðzÞ = ð1/πÞdxdy, and let L 2 = L 2 ðD, dAÞ denote the Lebesgue space on D. For n ∈ ℤ + , let A 2 n denote the n-analytic Bergman space, that is, the subspaces of L 2 consisting of n-differentiable functions such that ∂ n z f = 0, where As we know, A 2 n is a Hilbert subspace with the inner product where f , g ∈ A 2 n . The planar Beurling transform is the singular integral operator given by It is well known that the Beurling transform is a unitary operator acting on L 2 ðℂ, dAÞ (see [1], p. 364). For D ⊂ ℂ, the compression of the Beurling transform to L 2 is a bounded linear operator acting on L 2 defined by The n-analytic Bergman projection P n is defined to be the orthogonal projection of L 2 onto A 2 n . The singular integral operator S D is related to P n , and it is known (see [2]) that For a function u ∈ L ∞ , the Toeplitz operator T u with symbol u on A 2 n is defined by n-analytic functions play an important role in mathematical, and the space A 2 n has been intensively studied. More details about the structure of these spaces can be found in paper [3][4][5] and Balk's book [6].
Zero-product problem is a very important question in the operator theory. For Toeplitz operators, we have the general zero-product problem. Namely, if f and g are bounded functions such that T f T g = 0, then must one of the functions be zero? Ahern and Cučković (see [7]) obtained an affirmative answer for Toeplitz operators on A 2 1 when one of the functions is radial. Le (see [8,9]) generalized this result to more than two functions. Cučković and Le (see [10]) gave a positive answer when both functions are harmonic. While the general zero-product problem (even on A 2 1 ) is still far from being solved, it is known that Toeplitz operators with radial symbols are diagonal with respect to the standard orthonormal basis of A 2 1 . However, this is not the case on A 2 n when n ≥ 2. Then, Cučković and Le (see [10]) raised the following open question: Question 1. Let f and g be bounded functions, one of which is radial. If T f T g = 0 on A 2 n (or more generally, T f T g has finite rank), must one of these functions be zero?
In this paper, we give a partial answer to this question on the 2-analytic Bergman space A 2 2 . We show that if g is a radial function satisfying a Mellin transform condition, then T f T g = 0 if and only if f is a zero function.

Some Preliminary Results
We adopt the following boundary conditions for the binomial coefficients: An orthogonal basis in the space A 2 n is given by (see [3,11]) where k = 1, 2, ⋯and j = 1, 2, ⋯, n. The orthogonal basis can also be written as where k = 1, 2, ⋯and j = 1, 2, ⋯, n. For n = 2, we have the following lemma.
For each z ∈ D, since the point evaluation at z is a bounded linear functional on A 2 n , there exists a unique reproducing kernel function Kðz, ωÞ ∈ A 2 n such that The Mellin transformĝ of a function g ∈ L 1 ð½0, 1, rdrÞ is defined byĝ It is easy to see thatĝ is well defined and analytic on the right half-plane fz : Re z ≥ 2g. Cučković and Rao (see [12]) first used the Mellin transform to study Toeplitz operators on the classical Bergman space.

Lemma 3. Let g be a bounded radial function. Then, for each
where Proof. For each p = 1, 2, ⋯, since g is a bounded radial function, thus It is well known that radial Toeplitz operators acting on A 2 1 are diagonal, and radial Toeplitz operators acting on A 2 n can be represented as matrix sequences (see [13]). In the following, we give the exact expression of radial Toeplitz operators on A 2 2 .

Lemma 4.
Let g be a bounded radial function. Then, for each p ∈ ℤ + , Proof. Since g is a bounded radial function, for each p ∈ ℤ + , using Lemma 3, we get Applying Lemma 4, we conclude that radial Toeplitz operators on A 2 2 are not diagonal. The following corollary is an immediate consequence of Lemma 4.

Corollary 5.
Let g be a bounded radial function. Then, for each p, q ∈ ℤ + ,

Products of Two Toeplitz Operators
A bounded function f is said to be quasihomogeneous of where gðrÞ is a radial function (see [14]). For any function f ∈ L 2 ðD, dAÞ, it has the polar decomposition, i.e., where f k ðrÞ are radial functions in L 2 ð ½0, 1, rdrÞ (see [12]). A direct calculation gives the following lemma.
Lemma 6. Let f be a bounded function. Then, for each p, q ∈ ℤ + ,

Journal of Function Spaces
Proof. For all p, q ∈ ℤ + , it is easy to verify that Similarly, the rest of the lemma can be proved.
When considering the product of two Toeplitz operators, we often use the Mellin convolution. If f , g ∈ L 1 ð ½0, 1 , rdrÞ, then their Mellin convolution is given by The Mellin convolution theorem (see [15]) states that and if f and g are bounded, then so is f * g.
It is well known that the Mellin transform is uniquely determined by its value on an arithmetic sequence of integers. The following results (see [15], p. 102, [16]) will be needed later.
Theorem 7. Suppose f is a bounded analytic function on fz : Re z > 0g which vanishes at the pairwise distinct points z 1 , Then, f vanishes identically on fz : Re z > 0g.
For p ∈ ℤ + , we obtain the numbersĝðpÞ can also be called the moment Mellin sequence of g. Let AðpÞ is closed related to the moment Mellin sequence of g, and we have the following lemma. To prove (iv), in fact, for a fixed p ∈ ℤ + , It follows that jAðp + 1Þj = 0 if and only if Usingĝð2p + 2Þ = d r 2 gð2pÞ,ĝð2p + 4Þ = d r 4 gð2pÞ, and Mellin convolution (24), we get the above equality is equivalent to Proof. If the function g = 0, then r 2 g * r 2 g = r 4 g * g = 0, for each p ∈ ℤ + , This proves the sufficient condition.
Theorem 11. Let f be a bounded function and g be a bounded radial function. Then, T f T g = 0 on A 2 2 if and only if for each p, q ∈ ℤ + , AðpÞBðp − 1, qÞ = 0.
Proof. Using the fact that f is a bounded function, we have If T f T g = 0, then for each p, q ∈ ℤ + , from which we conclude that Since p is arbitrary, it follows that Analogously, for each p, q ∈ ℤ + , it is easily verified that thus, we get The above equations are equivalent to This completes the proof of the theorem.
For p = 1, 2, ⋯, firstly if a p = c p = 0, thenĝð2pÞ = 0, using Remark 8, we get g = 0. Now, if b p = 0, then 2p + 2 ð Þĝ 2p + 2 ð Þ− 2pĝ 2p ð Þ = 0: ð48Þ Letting ζ = 2p, we have ζĝ ζ ð Þ = ζ + 2 ð Þĝ ζ + 2 ð Þ: ð49Þ It is easy to see that the function ζĝðζÞ is a periodic function with a period 2. Using the same argument as the one at the end of Section 2 in [12], we conclude that ζĝðζÞ must be a constant function. Hence,ĝ where C is a constant and it is clear that g is also a constant. Finally, if d p = 0, then that is,