On the commutativity of a certain class of Toeplitz operators

In this paper we prove that if the polar decomposition of a symbol $f$ is truncated above, i.e., $f(re^{i\theta} )=\sum_{k=-\infty}^Ne^{ik\theta} f_k (r)$ where the $f_k$'s are radial functions, and if the associated Toeplitz operator $T_f$ commutes with $T_{z^2+\bar{z}^2}$, then $T_f=Q(T_{z^2+\bar{z}^2})$ where $Q$ is a polynomial of degree at most $1$. This gives a partial answer to an open problem by S. Axler, Z. Cuckovic and N. V. Rao [2, p. 1953].


Introduction
Let D be the unit disk of the complex plane C, and dA = rdr dθ π , where (r, θ) are polar coordinates, be the normalized Lebesgue measure, so that the area of D is one. We define the analytic Bergman space, denoted L 2 a (D), to be the set of all analytic functions on D that are square integrable with respect to the measure dA. It is well know that L 2 a (D) is a closed subspace of the Hilbert space L 2 (D, dA) and has the set { √ n + 1z n | n ≥ 0} as an orthonormal basis (see [4]). Thus, L 2 a (D) is itself a Hilbert space with the usual inner product of L 2 (D, dA). Moreover the orthogonal projection, denoted P , from L 2 (D, dA) onto L 2 a (D), often called the Bergman projection, is well defined. Let f be a bounded function on D. We define on L 2 a (D) the Toeplitz operator T f with symbol f by T f (u) = P (f u), for any u ∈ L 2 a (D). A natural question to ask is under which conditions is the product (in a sense of composition) of two Toeplitz operators commutative? In other words, when is T f T g = T g T f for given two Toeplitz operators T f and T g ? It is easy to see from the definition of Toeplitz operators that if the symbol f is analytic and bounded on D, then T f is simply the multiplication operator by f , i.e., T f (u) = f u for all u ∈ L 2 a (D). Thus, any two analytic Toeplitz operators (i.e., Toeplitz operators with analytic symbols) commute with each other. Again from the definition of Toeplitz operators, we have that the adjoint of T f is Tf wheref is the complex conjugate of f . It follows that if f is antianalytic (i.e.,f is analytic), then T * is the multiplication operator byf . Hence, if two symbols f , and g are antianalytic, then their associated Toeplitz operators commute since their adjoints commute. This situation in which the symbols are both analytic (resp. antianalytic) is known to us as the trivial situation. One might ask what if the symbols were harmonic but not necessarily analytic or antianalytic. The answer to this question was given by S. Axler andZ.Cucković in [1]. They proved the following: So basically if both symbols are harmonic, then the product is commutative only in the trivial case. In fact, the sufficient condition (a) (resp. (b)) says that the operators T f and T g (resp. their adjoints Tf and Tḡ) are multiplication operators and so they commute. For the sufficient condition (c), since Toeplitz operators are linear with respect to their symbol, we can write T f = αT g + βI where I = T 1 is the identity operator on L 2 a (D), and hence, since T g commutes with itself and with the identity, T g commutes with T f . The next natural step was to relax the hypothesis of the previous theorem in order to obtain results for a larger class of symbols. In [2], S. Axler,Z.Cucković, and N. V. Rao proved that analytic Toeplitz operators commute only with other such operators. Their result can be stated as follows: For Theorem 2, the authors do not ask the function f to be harmonic but only bounded. However this was not without cost. In fact the hypothesis on the symbol g is stronger than the one in Theorem 1 since here g has to be analytic. Finally, the authors conclude [2] by asking the following open problem: "Suppose g is a bounded harmonic function in D that is neither analytic nor antianalytic. If f is a bounded function in D such that T f and T g commute, must f be of the form αg + β for some constants α, β?" The first partial answers to this problem can be found in [6] and [7].

Quasihomogeneous Toeplitz operators
Definition 1. A symbol f is said to be quasihomogeneous of order p, and the associated Toeplitz operator T f is also called a quasihomogeneous Toeplitz operator of order p, if f (re iθ ) = e ipθ φ(r), where φ is an arbitrary radial function.
The motivation behind considering such a family of symbols is that any function f in L 2 (D, dA) has the following polar decomposition (Fourier series) where R = L 2 ([0, 1], rdr). In other words f (re iθ ) = k∈Z e ikθ f k (r), where the f k 's are radial functions in R. So the study of quasihomogeneous Toeplitz operators will allow us to obtain interesting results about Toeplitz operators with more general symbols.
Another interesting property of a quasihomogeneous operator is that it acts on the elements z n of the orthogonal basis of L 2 a (D) as a shift operator with weight. In fact, if k ∈ Z + (the case where k is a negative integer can be done in the exact same way) and f k is a bounded radial function, then for any n ≥ 0 we have The integral 1 0 f k (r)r 2n+k+1 dr that appears in the weight is known as the Mellin transform.
Definition 2. We define the Mellin transform of a function φ in L 1 ([0, 1], rdr), denoted φ, to be It is well known that the Mellin transform is related to the Laplace transform via the change of variable r = e −u . Moreover, for φ ∈ L 1 ([0, 1], rdr), φ is bounded in the right-half plane {z ∈ C|ℜz ≥ 2} and analytic in {z ∈ C|ℜz > 2}.
Using the Mellin transform, we can rewrite Equation (1) as follows Therefore, we can summarize the above calculation in the following lemma which we shall be using often.
Lemma 1. Let k ∈ Z and n ∈ N be two integers, and let φ be a bounded radial function in D. If k ≥ 0, then The Mellin transform is going to play a major role in our arguments for the proofs. In fact a function is well determined by its Mellin transform on any arithmetic sequence. We have the following important lemma that can be found in [5, Remark 2.p 1466] Lemma 2. If φ ∈ L 1 ([0, 1], rdr) is such that φ(a n ) = 0, where (a n ) n is a sequence of integers satisfying the condition n 1 an = ∞, then φ(z) = 0 on {z ∈ C|ℜz > 2}, and therefore φ is the zero function.
In other words, the lemma is saying that the Mellin transform is injective, and so two functions whose Mellin transforms coincide on an arithmetic sequence will be equal to each other.
Another classical lemma which we shall use often can be stated as follows: , H is p-periodic, then H must be constant.
When dealing with the product of quasihomogeneous Toeplitz operators, we are often confronted with the Mellin convolution of the radial functions in their quasihomogeneous symbols. We define the Mellin convolution of two radial functions φ and ψ in It is well known that the Mellin transform converts the Mellin convolution into a product of Mellin transforms. In fact and so if φ and ψ are in L 1 (D, dA), then so is φ * M ψ. We are now ready to present our main result.

Commutant of T z 2 +z 2
In this section we shall extend the work started in [6] and [7]. We consider the Toeplitz T z 2 +z 2 (the symbol z 2 +z 2 is harmonic but neither analytic nor antianalytic). It is known to us that such operator raised to any power n ≥ 2 is not a Toeplitz operator. We shall prove that if the symbol f has truncated polar decomposition i.e., f (re iθ ) = N k=−∞ e ikθ f k (r) where N is a positive integer, and if T f commutes with T z 2 +z 2 , then T f is polynomial of degree at most one in T z 2 +z 2 . This result goes in the direction of the open problem we mentioned previously. We would like to emphasize the fact that though we are using the same tools and techniques as in [6], new ideas and tricks were needed to overcome numerous obstacles we faced in the proof of the main result.
In our presentation of the main theorem, we shall proceed as follows: First we prove that if f (re iθ ) = N k=−∞ e ikθ f k (r) is such that T f commutes with T z 2 +z 2 , then N has to be an even number. Second, we shall demonstrate that this same N cannot exceed 4. Finally, we shall exhibit all the radial functions f k for k ≤ 4, and shall show that f k (r) = 0 for k = {−2, 0, 2}, f k (r) = cr 2 for k = {2, −2}, and f 0 (r) = c 0 where c, c 0 are constants. Hence, by reconstructing the symbol f , we shall obtain that and therefore T f = cT z 2 +z 2 + c 0 I.
In the above equation, the term with the highest degree is z n+N +2 . It comes on the left hand side from the product T e iN θ fN T z 2 (z n ) only, and on the right hand side from the product T z 2 T e iN θ fN (z n ) only. Thus, by equality, we must have Since z 2 is analytic, e iN θ f N must be analytic too. Which is possible if and only if Redoing the same argument for the term in z of degree N + n − 2, we obtain which, using Lemma 1, is equivalent to for all n ≥ 2. Thus, Lemma 2 implies for ℜz ≥ 4. Now, we introduce the function (4) f * (r) = −4c N r 2 1 − r 2N 1 − r 4 . By direct calculation and simple algebraic operations, we can see that We denote by F (z) = (z + 2N − 6) r N −2 f N −4 (z) and G(z) = f * (z). Then Equation (3) can be rewritten as and so Lemma 3 implies Since 1 z+2N −6 = r 2N −6 (z), the above equation becomes (5) r N −2 f N −4 (z) = c N −4 r 2N −6 (z) + r 2N −6 (z) f * (z).
Since by Equation (2), r 2N −6 (z) f * (z) = (r 2N −6 * M f * )(z), we have Let us denote by Next, we need to determine the conditions on N under which 1 r N −2 I N , as a function of r, is in L 1 ([0, 1], rdr). Otherwise c N must be zero, in which case f N (r) = 0 and we are done. Since But since N is an odd positive integer, we have either N = 3 or N = 5. We recall that the previous inequality was obtained after the assumption N ≥ 3, and so we shall look at the case where N = 1 separately.
Case N = 1. If we set N = 1, then and we have Now, it is easy to see that f −3 is in L 1 ([0, 1], rdr) if and only if c −3 = 0 and c 1 = 0. Hence f 1 (r) = c 1 r = 0. Case N = 3. If we let N = 3, then the terms in z n+1 comes from the following equality: In particular, for n = 1 we have where f * is obtained from (4) with N = 3, and we have Therefore, (6) implies which is possible if and only if c 3 = 0, and hence f 3 (r) = 0. Case N = 5. If we set N = 5, then the terms in z of degree n+3 comes from the following equation: In particular, for n = 0 and using Lemma 1, we have where f * is obtained from (4)  Now, substituting f 1 (7) and f 1 (3) in Equation (7), we obtain 1 − 12 ln 2 6 which is true if and only if c 5 = 0, and hence f 5 (r) = 0. This completes the proof.
where N is a positive even integer, is such that T f commutes with T z 2 +z 2 , then N ≤ 4.
Proof. If T f commutes with T z 2 +z 2 , then In the above equation, the term with the highest degree is z n+N +2 . On the left-hand side , this term comes from the product T e iN θ fN T z 2 (z n ) only, and on the right-hand side it is obtained from the product T z 2 T e iN θ fN (z n ) only. Thus, by equality, we must have Since z 2 is analytic, e iN θ f N is analytic too. This is possible if and only if f N = c N r N i.e., e iN θ f N = c N z N . Redoing the same argument for the term in z of degree N + n − 2, we obtain which, using Lemma 1 , becomes for n ≥ 2. Thus, Lemma 2 implies for ℜz ≥ 2. Now, if we let F and G to be the the previous equation can be written as Hence, by Lemma 3 we have the following, At this point we shall assume N ≥ 6 and we shall prove that in this case f N −4 will not be in L 1 ([0, 1], rdr) unless c N = 0 and so f N (r) = 0. Therefore N shall be strictly less than 6 and because N is even we shall conclude that N ≤ 4. If N ≥ 6, Hence, Otherwise the constant c N must be zero. In particular, for i = 0, we must have −N + 4 ≥ −1, i.e., N ≤ 5. Therefore N cannot be greater than or equal to 6, otherwise c N = 0. Since N is even and N < 6, we deduce that N ≤ 4, i.e., N = 4 or N = 2 We are now ready to state our main result.
Proof. From the previous propositions we know that N is even and N ≤ 4. We shall prove that f k (r) = 0 for all k = {−2, 0, 2}, f 0 (r) = c 0 , and f 2 (r) = c 2 r 2 = f −2 (r) for some constants c 0 and c 2 .
Since T f commutes with T z 2 +z 2 , we have T e ikθ f k (z n ), ∀n ≥ 0.
In the equation above, the term in z with the highest degree is z n+6 , and it is coming from the product of T e i4θ f4 T z 2 (z n ) on the left hand side, and from T z 2 T e i4θ f4 (z n ) on the right hand side. Thus, by equality, we must have Since z 2 is analytic, e i4θ f 4 must be analytic as well by Theorem 2, which is possible if and only if f 4 (r) = c 4 r 4 , i.e., T e 4iθ f4 = c 4 T z 4 . Next, we shall prove that f 0 = c 0 and f 4 (r) = 0. In (8), the terms in z n+2 come from the following equality which, using Lemma 1 and Lemma 2, is equivalent to If we let F (z) = 2(z +1) f 0 (2z +2) and G(z) = c 4 z − 1 z + 1 + z + 1 z + 3 , then the previous equation can be written as Hence, Lemma 3 implies Since r m (z) = 1 z+m and r m ln r(z) = − 1 (z+m) 2 for any integer m, the above equality becomes (10) f 0 (2z + 2) = (c 0 + c 4 ) 1(2z + 2) + c 4 r 4 (2z + 2) + 4 ln r(2z + 2) . Now, if we take n = 0 in Equation (9) and apply Lemma 1, we obtain 6 f 0 (6) = 6c 4 10 + 2 f 0 (2). Since But, it is easy to see that the above equality is possible if and only if c 4 = 0 and therefore f 4 (r) = 0, while Lemma 2 and Equation (10) imply that f 0 (r) = c 0 . Next, we shall prove that f −4 (r) = 0, and consequently f −4+4k (r) = 0, ∀k ≤ −1. In (8), the terms z n−2 come from the product of T f0 with Tz2 and the product of T z 2 with T e −4iθ f−4 . Thus, we must have T f0 Tz2(z n ) + T e −4iθ f−4 T z 2 (z n ) = Tz2T f0 (z n ) + T z 2 T e −4iθ f−4 (z n ), ∀n ≥ 0.