Commuting Quasihomogeneous Toeplitz Operators

Complex Anal. Oper. Theory (2013) 7:1267–1285 DOI 10.1007/s11785-012-0223-0

Complex Analysis and Operator Theory

Commuting Quasihomogeneous Toeplitz Operators on the Harmonic Bergman Space Xing-Tang Dong · Ze-Hua Zhou

Received: 29 May 2011 / Accepted: 8 February 2012 / Published online: 22 February 2012 © Springer Basel AG 2012

Abstract In this paper, we obtain a symmetry number for the commutator of quasihomogeneous Toeplitz operators on the harmonic Bergman space. Then we use it to characterize the commuting Toeplitz operators with quasihomogeneous symbols. Also, we show that a Toeplitz operator with an analytic or co-analytic monomial symbol commutes with another Toeplitz operator only in the trivial case. Keywords Toeplitz operators · Harmonic Bergman space · Quasihomogeneous symbols Mathematics Subject Classification (2000)

Primary 47B35

1 Introduction Let d A denote the Lebesgue area measure on the unit disk D, normalized so that the measure of D equals 1. L 2 (D, d A) is the Hilbert space of Lebesgue square integrable functions on D with the inner product

Communicated by H. Turgay Kaptanoglu. The work of the first author was supported by the TianYuan Special Funds of the National Natural Science Foundation of China (Grant No. 11126164). The work of the second author was supported by the National Natural Science Foundation of China (Grant No. 10971153). X.-T. Dong · Z.-H. Zhou (B) Department of Mathematics, Tianjin University, Tianjin 300072, People’s Republic of China e-mail: [email protected] X.-T. Dong e-mail: [email protected]

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X.-T. Dong, Z.-H. Zhou

f, g =

f (z)g(z)d A(z). D

The harmonic Bergman space L 2h is the closed subspace of L 2 (D, d A) consisting of the harmonic functions on D. We will write Q for the orthogonal projection from L 2 (D, d A) onto L 2h . Each point evaluation is easily verified to be a bounded linear functional on L 2h . Hence, for each z ∈ D, there exists a unique function Rz (called the harmonic Bergman kernel) in L 2h that has the following reproducing property: f (z) = f, Rz for every f ∈ L 2h . Given z ∈ D, let K z (w) = L a2

1 (1−wz)2

be the well-known reproducing kernel for

the analytic Bergman space consisting of all L 2 -analytic functions on D. The well-known Bergman projection P is then the integral operator P f (z) = f (w)K z (w)d A(w) D

for f ∈ L 2 (D, d A). Since L 2h = L a2 + L a2 , it is easily checked that Rz = K z + K z − 1. Thus, Q can be represented by Q f = P f + P f − P f (0). For u ∈ L 1 (D, d A), the Toeplitz operator Tu with symbol u is the operator on L 2h defined by Tu f = Q(u f )

(1.1)

for f ∈ L 2h . This operator is always densely defined on the polynomials and not bounded in general. We are interested in the case where this densely defined operator is bounded in the L 2h norm. Inspired by [11], we give the following definitions. Definition 1.1 Let F ∈ L 1 (D, d A). (a) We say that F is a T-function if the equation (1.1), with u = F, defines a bounded operator on L 2h . (b) If F is a T-function, we write TF for the continuous extension of the operator defined by equation (1.1). We say that TF is a Toeplitz operator if and only if TF is defined in this way. (c) If there is an r ∈ (0, 1) such that F is (essentially) bounded on the annulus {z : r < |z| < 1}, then we say that F is “nearly bounded”. Generally, the T-functions form a proper subset of L 1 (D, d A) which contains all bounded and “nearly bounded” functions. A function f is said to be quasihomogeneous of degree k ∈ Z if and only if

Commuting Quasihomogeneous Toeplitz Operators

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f (r eiθ ) = eikθ ϕ(r ), where ϕ is a radial function (see [15]). In this case the associated Toeplitz operator T f is also called quasihomogeneous Toeplitz operator of degree k. In 1964, Brown and Halmos [2] proved that T f Tg = Tg T f on the classical Hardy space H 2 if and only if: (I) both f and g are analytic, or (II) both f and g are analytic, ˇ cković [1] showed that a or (III) one is a linear function of the other. Then Axler and Cuˇ Brown–Halmos type result holds for Toeplitz operators with harmonic symbols on L a2 . ˇ cković and Rao used the Mellin transform to characterize all Toeplitz operIn [6] Cuˇ ators on L a2 which commute with Tei pθ r m for (m, p) ∈ N × N. Later in [12], Louhichi and Zakariasy gave a partial characterization of commuting Toeplitz operators on L a2 with quasihomogeneous symbols. Recently, we studied some algebraic properties of quasihomogeneous Toeplitz operators on the analytic Bergman space of the unit ball in [8] and [16]. The theory of Toeplitz operators on L 2h is quite different from that on L a2 . Choe and Lee [3] showed that two analytic Toeplitz operators on L 2h commute only when their symbols and the constant function 1 are linearly dependent. Then [4] and [7] showed that an analytic Toeplitz operator and a co-analytic Toeplitz operator on L 2h can commute only when one of their symbols is constant. In [9] we considered when the product of two quasihomogeneous Toeplitz operators on L 2h is a Toeplitz operator, and in this paper we shall study the commuting problem for quasihomogeneous Toeplitz operators on L 2h . Some new methods and complex calculating technics are used in this paper. 2 Preliminaries One of the most useful tools in the following calculations will be the Mellin transform. The Mellin transform ϕˆ of a function ϕ ∈ L 1 ([0, 1], r dr ) is defined by the equation: 1 ϕ(s)s z−1 ds.

ϕ(z) ˆ = 0

It is clear that ϕˆ is well defined on the right half-plane {z : Rez ≥ 2} and analytic on {z : Rez > 2}. If f and g are in L 1 ([0, 1], r dr ), then their Mellin convolution is defied by 1 ( f ∗ M g)(r ) =

f

r t

g(t)

dt , 0 ≤ r < 1. t

r

The Mellin convolution theorem states that f ∗ M g(s) = f (s) g (s),

(2.1)

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and that, if f and g are in L 1 ([0, 1], r dr ) then so is f ∗ M g. It is important and helpful to know that the Mellin transform is uniquely determined by its value on an arithmetic sequence of integers. In fact we have the following classical theorem (see [14, p. 102]). Theorem 2.1 Suppose that f is a bounded analytic function on {z : Rez > 0} which vanishes at the pairwise distinct points z 1 , z 2 , . . . , where (i) inf{|z n |} >10 and (ii) n≥1 Re( z n ) = ∞. Then f vanishes identically on {z : Rez > 0}. In [9] we proved the following results which we shall use often in this paper. Lemma 2.2 Let k ∈ Z and let ϕ be a radial T-function. Then for each n ∈ N, Teikθ ϕ (z n ) = Teikθ ϕ (z ) = n

2(n + k + 1) ϕ (2n + k + 2)z n+k 2(−n − k + 1) ϕ (−k + 2)z −n−k

if n ≥ −k if n < −k;

2(n − k + 1) ϕ (2n − k + 2)z n−k 2(k − n + 1) ϕ (k + 2)z k−n

if n ≥ k if n < k.

Remark 2.3 Let k1 , k2 ∈ Z. Then Lemma 2.2 implies that the image of r |k2 | eik2 θ by a quasihomogeneous Toeplitz operator of degree k1 is λk1 ,k2 r |k1 +k2 | ei(k1 +k2 )θ for some constant λk1 ,k2 . 3 Commuting Quasihomogeneous Toeplitz Operators In this section, we will characterize commuting Toeplitz operators on L 2h with quasihomogeneous symbols. First, we give the following definition. Definition 3.1 Let l ∈ Z. Then T on L 2h if and only if

l 2

is said to be the symmetry number for an operator

T r |k| eikθ = 0 ⇐⇒ T r |−k+l| ei(−k+l)θ = 0 for any k ∈ Z and l ilθ T r | 2 |e 2 ≡ 0 when l is even. It is well known that


√

n + 1z n

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∞ √ n=0

n + 1z n

∞ n=1

is an orthonormal basis for the harmonic Bergman space. Therefore, the above definition shows that an operator T on L 2h having a symmetry number must be zero if T taking on “nearly half” of the orthonormal basis to be equal to zero. The following theorem will show the symmetry number for the commutator of quasihomogeneous Toeplitz operators on L 2h . Theorem 3.2 Let k1 , k2 ∈ Z and let ϕ1 , ϕ2 be two radial T-functions. (a) If k1 + k2 ≥ 0, then we can get the following properties. (I) For any n ∈ N, Teik1 θ ϕ1 Teik2 θ ϕ2 (z n ) = Teik2 θ ϕ2 Teik1 θ ϕ1 (z n ) ⇐⇒ Teik1 θ ϕ1 Teik2 θ ϕ2 z n+k1 +k2 = Teik2 θ ϕ2 Teik1 θ ϕ1 z n+k1 +k2 . (II) For any l ∈

1, . . . ,

k1 +k2

2l

function, Teik1 θ ϕ1 Teik2 θ ϕ2 z

, here [ · ] denotes the greatest integer = Teik2 θ ϕ2 Teik1 θ ϕ1 (z l )

⇐⇒ Teik1 θ ϕ1 Teik2 θ ϕ2 z k1 +k2 −l = Teik2 θ ϕ2 Teik1 θ ϕ1 z k1 +k2 −l . Moreover, if k1 + k2 is even, then k1 +k2 k1 +k2 ≡ Teik2 θ ϕ2 Teik1 θ ϕ1 z 2 . Teik1 θ ϕ1 Teik2 θ ϕ2 z 2 (b) If k1 + k2 < 0, then we can get the following properties.

(I) For any n ∈ N, Teik1 θ ϕ1 Teik2 θ ϕ2 z n−k1 −k2 = Teik2 θ ϕ2 Teik1 θ ϕ1 z n−k1 −k2

⇐⇒ Teik1 θ ϕ1 Teik2 θ ϕ2 z n = Teik2 θ ϕ2 Teik1 θ ϕ1 z n .

(II) For any l ∈ 1,. . ., −k12−k2 ,Teik1 θ ϕ1 Teik2 θ ϕ2 z l = Teik2 θ ϕ2 Teik1 θ ϕ1 z l ⇐⇒ Teik1 θ ϕ1 Teik2 θ ϕ2 z −k1 −k2 −l = Teik2 θ ϕ2 Teik1 θ ϕ1 z −k1 −k2 −l . Moreover, if −k1 − k2 is even, then −k1 −k2 −k1 −k2 ≡ Teik2 θ ϕ2 Teik1 θ ϕ1 z 2 . Teik1 θ ϕ1 Teik2 θ ϕ2 z 2 Proof First suppose that k1 + k2 ≥ 0. Without loss of generality, we can also assume that k1 ≥ 0. For any n ∈ N, by Lemma 2.2 we can get

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Teik1 θ ϕ1 Teik2 θ ϕ2 z n ⎧ ⎨ 2(n + k1 + k2 + 1)ϕ1 (2n + k1 + 2k2 + 2) ×2(n + k2 + 1)ϕ2 (2n + k2 + 2)z n+k1 +k2 = ⎩ 2(n + k1 +k2 +1)ϕ1 (k1 +2)2(−n− k2 + 1)ϕ2 (−k2 + 2)z n+k1 +k2

Teik2 θ ϕ2 Teik1 θ ϕ1 z n

if n ≥−k2 if n 0, for otherwise we could take the adjoints. It follows from Theorem 3.2 that Tei pθ ϕ1 Tei pθ ϕ2 = Tei pθ ϕ2 Tei pθ ϕ1 if and only if ϕ1 (2n + 3 p + 2)ϕ2 (2n + p + 2) = ϕ1 (2n + p + 2)ϕ2 (2n + 3 p + 2)

(3.4)

for any n ∈ N and ϕ1 (3 p − 2l + 2)ϕ2 ( p + 2) = ϕ1 ( p + 2)ϕ2 (3 p − 2l + 2) for any l ∈ {1, . . . , p}. According to Theorem 2.1, (3.4) holds if and only if ϕ1 (z + 2 p)ϕ2 (z) = ϕ1 (z)ϕ2 (z + 2 p),

z ∈ {z : Rez > 2}.

(3.5)


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Then by Lemma 6 of [10], the above equation is equivalent to ϕ1 (z) = C ϕ2 (z) for some constant C. Moreover, if ϕ1 (z) = C ϕ2 (z) then (3.5) holds, and hence condition (a) holds. Similarly, Theorem 3.2 shows that Tei pθ ϕ1 Te−i pθ ϕ2 = Te−i pθ ϕ2 Tei pθ ϕ1 if and only if (n − p+1)ϕ1 (2n − p+2)ϕ2 (2n − p+2) = (n + p+1)ϕ1 (2n + p+2)ϕ2 (2n + p+2) (3.6)

for any n ∈ N such that n ≥ p and (−n+ p+1)ϕ1 ( p+2)ϕ2 ( p+2) = (n+ p+1)ϕ1 (2n+ p+2)ϕ2 (2n+ p+2) (3.7) for any n ∈ {0, 1, . . . , p − 1}. Suppose that Tei pθ ϕ1 commutes with Te−i pθ ϕ2 . Denote H (z) = 2(z + 1)2(z − p + 1)ϕ1 (2z − p + 2)ϕ2 (2z − p + 2), then H is analytic and bounded on {z : Rez > p} since ϕ1 and ϕ2 are two T-functions on D. Moreover, (3.6) implies that H (n) = H (n + p), ∀ n ∈ N, n ≥ p. Then by Theorem 2.1, H (z) = H (z + p),

z ∈ {z : Rez > p},

which implies that H (z) is a constant C. Therefore ϕ1 (2z − p + 2)ϕ2 (2z − p + 2) =

C , 2(z + 1)2(z − p + 1)

(3.8)

Moreover, from (3.7) and (3.8) we can deduce −n + p + 1 n+ p+1 = , p+1 (n + p + 1)(n + 1) and hence n(n − p) = 0 for any n ∈ {0, 1, . . . , p − 1}, which implies that p = 1. On the other hand, a direct calculation gives 1 −1 (2z + 1). = r (2z + 1)r 2(z + 1)2z

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Hence, it follows from (2.1) and (3.8) that ϕ1 ∗ M ϕ2 = C(r ∗ M r −1 ). Conversely, if p = 1 and ϕ1 ∗ M ϕ2 = C(r ∗ M r −1 ), then by a direct calculation, one can easily get (3.6) and (3.7), and hence condition (b) holds. If p = 0, then by Theorem 3.2, one can easily get that Tei pθ ϕ1 and Tei pθ ϕ2 are

commute for any radial T-functions ϕ1 and ϕ2 . This completes the proof. Now we would like to study the commutativity of two quasihomogeneous Toeplitz operators Teik1 θ r m and Teik2 θ ϕ(r ) . First, we give some lemmas. Lemma 3.5 Let k2 ∈ Z such that k2 = 0 and let ϕ1 , ϕ2 be two radial T-functions. If Tϕ1 Teik2 θ ϕ2 = Teik2 θ ϕ2 Tϕ1 , then ϕ2 = 0 or ϕ1 is a constant. Proof Without loss of generality, we can assume that k2 > 0. Then the equality Tϕ1 Teik2 θ ϕ2 (z n ) = Teik2 θ ϕ2 Tϕ1 (z n ) for each n ∈ N together with (3.1) gives (n + k2 + 1)ϕ1 (2n + 2k2 +2)ϕ2 (2n + k2 + 2) = (n + 1)ϕ1 (2n+2)ϕ2 (2n + k2 + 2). Using the same reasoning as in the proof of Proposition 6 of [12], then it follows that ϕ2 = 0 or ϕ1 is a constant. This completes the proof.

Lemma 3.6 Let k2 ∈ Z and let m ∈ N such that m = 0. For a nonzero radial T-function ϕ on D, assume Tz m Teik2 θ ϕ = Teik2 θ ϕ Tz m , then we can get the following properties. (a) If k2 > 0, then eik2 θ ϕ = C z m for some constant C; (b) If k2 = 0, then eik2 θ ϕ = C for some constant C. (c) If k2 < 0, then m = 1 and eik2 θ ϕ = C z −1 for some constant C. Proof Suppose Tz m and Teik2 θ ϕ commute, then by Lemma 2.2, for any n ∈ N such that n ≥ −k2 , Tz m Teik2 θ ϕ (z n ) = Teik2 θ ϕ Tz m (z n ) gives 2(n + k2 + 1) ϕ (2n + k2 + 2) = 2(n + m + k2 + 1) ϕ (2n + 2m + k2 + 2), which implies that (z + k2 ) ϕ (z) = C, z ∈ {z : Rez > max{2, 2 − k2 }} for some constant C, and hence ϕ = Cr k2 .


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Therefore, if k2 > 0, then it follows that eik2 θ ϕ = C z k2 . Then by Theorem 5 of [3], Tz m Tz k2 = Tz k2 Tz m yields that k2 = m. If k2 = 0, then it follows that eik2 θ ϕ = C. Now, assume k2 < 0. Since ϕ is a T-function, ϕ = Cr k2 ∈ L 1 (D, d A) which implies that k2 ≥ −1, and hence k2 = −1. Note that m ∈ N and m = 0, so m +k2 ≥ 0. Then by (3.2), Tz m Teik2 θ r k2 (1) = Teik2 θ r k2 Tz m (1) gives that −k2 + 1 1 = , 2m + 2 2m + 2k2 + 2 which implies that m = −k2 = 1. This completes the proof.

Lemma 3.7 Let ϕ1 and ϕ2 be two radial T-functions and let k1 , k2 ∈ Z such that k1 > 0 and |k2 | > 0. If ϕ2 (|k2 | + 2) = 0 and Teik1 θ ϕ1 Teik2 θ ϕ2 (z n ) = Teik2 θ ϕ2 Teik1 θ ϕ1 (z n ) for any integer n ≥ max{−k2 , 0}, then either ϕ1 = 0 or ϕ2 = 0. Proof For any integer n ≥ max{−k2 , 0}, by Lemma 2.2 we get (n + k2 + 1)ϕ1 (2n + k1 + 2k2 + 2)ϕ2 (2n + k2 + 2) = (n + k1 + 1)ϕ1 (2n + k1 + 2)ϕ2 (2n + 2k1 + k2 + 2).

(3.9)

Denote a0 = max{−k2 , 0}, then it follows from ϕ2 (|k2 | + 2) = 0 that ϕ2 (2a0 + k2 + 2) = 0, and hence (3.9) implies ϕ1 (2a0 + k1 + 2)ϕ2 (2a0 + 2k1 + k2 + 2) = 0. If ϕ1 (2a0 + k1 + 2) = 0, we will let a1 = a0 + |k2 |, otherwise, let a1 = a0 + k1 , then a direct calculation from (3.9) shows that ϕ1 (2a1 + k1 + 2)ϕ2 (2a1 + 2k1 + k2 + 2) = 0.

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So we can find a sequence {am }m∈N , which is defined by am+1 = am + |k2 | or am + k1 , such that ϕ1 (2am + k1 + 2)ϕ2 (2am + 2k1 + k2 + 2) = 0. It is clear that 1 = ∞. am

m∈N

Then by Theorem 2.1, one can easily see that ϕ1 = 0 or ϕ2 = 0. This completes the proof.

Theorem 3.8 Let k1 , k2 ∈ Z such that |k1 | ≤ |k2 | and let m be a real number greater than or equal to −1. Then for a nonzero radial T-function ϕ on D, Teik1 θ r m Teik2 θ ϕ = Teik2 θ ϕ Teik1 θ r m , if and only if one of the following conditions is holds: (1) (2) (3) (4)

k1 = m = 0. k1 = k2 = 0. k1 = k2 and ϕ = Cr m −1 − k1 k2 = −1 and ϕ = C m+1 2 r

m−1 2 r

.

Proof If one of the conditions (1)–(3) holds, then it is clear that Teik1 θ r m commutes with Teik2 θ ϕ . Assume condition (4) holds. A direct calculation shows that

m r ∗M

m + 1 −1 m − 1 r − r 2 2

=

1 2

1 −r r

= r ∗ M r −1 ,

then Proposition 3.4 implies that Teik1 θ r m Teik2 θ ϕ = Teik2 θ ϕ Teik1 θ r m . Conversely, assume that Teik1 θ r m commutes with Teik2 θ ϕ . To prove this theorem we need to discuss several cases. Case 1. Suppose k1 = 0. Then it follows from Lemma 3.5 that m = 0 or k2 = 0. Thus conditions (1) or (2) holds. Case 2. Suppose k1 k2 > 0. Without loss of generality, we can also assume that k1 > 0, for otherwise we could take the adjoints. Then it follows from |k1 | ≤ |k2 | that 0 < k1 ≤ k2 . Thus by (3.3), Teik1 θ r m Teik2 θ ϕ (z k1 ) = Teik2 θ ϕ Teik1 θ r m (z k1 ) gives 1 k2 − k1 + 1 ϕ (k2 + 2) = ϕ (k2 + 2). m + 2k2 − k1 + 2 m + k1 + 2

(3.10)


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Noting that ϕ is nonzero, by Lemma 3.7 we obtain ϕ (k2 + 2) = 0. Then it follows from (3.10) that m = −k1 or k1 = k2 . If m = −k1 , then by (3.1), Teik1 θ r m Teik2 θ ϕ (z n ) = Teik2 θ ϕ Teik1 θ r m (z n ) shows that ϕ (2n + 2k1 + k2 + 2) 2(n + 1) ϕ (2n + k2 + 2) = 2(n + k1 + 1) for any n ∈ N, which implies ϕ = Cr −k2 . Using the fact that m ≥ −1, k1 > 0 and ϕ is a T-function, we can get k1 = k2 = −m = 1. Therefore, in this case we can always get k1 = k2 , then Proposition 3.4 shows that ϕ = Cr m , and hence condition (3) holds. Case 3. Suppose k1 k2 < 0. Similarly, we can also assume that k1 > 0, Then it follows that k2 < 0 and −k1 − k2 ≥ 0. Thus by Lemma 2.2, Teik1 θ r m Teik2 θ ϕ (z −k1 −k2 ) = Teik2 θ ϕ Teik1 θ r m (z −k1 −k2 ) gives −k2 + 1 k1 + 1 ϕ (−k2 + 2) = ϕ (−k2 + 2). m + k1 + 2 m − k1 − 2k2 + 2

(3.11)

Since ϕ is nonzero, Lemma 3.7 shows that ϕ (−k2 + 2) = 0. Therefore, it follows from (3.11) that m = k1 or k1 = −k2 . If m = k1 , then by Lemma 3.6 we can also get that k1 = −k2 . Then Proposition 3.4 shows that k1 = −k2 = 1 and r m ∗ M ϕ = C(r ∗ M r −1 ).

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An easy computation yields ϕ(r ) = C

m + 1 −1 m − 1 r − r . 2 2

Thus we conclude that condition (4) holds. This completes the proof.

Remark 3.9 One naturally want to know the corresponding results when |k1 | > |k2 |. In fact, if we further assume m ∈ N and assume ϕ is bounded, then using the results of [6], we can give a description for the commutant of Teik1 θ r m and Teik2 θ ϕ for any k1 , k2 ∈ Z. Denote n 0 = max {−k1 , −k2 , −k1 − k2 , 0}, then the equality

Teik1 θ r m Teik2 θ ϕ z n = Teik2 θ ϕ Teik1 θ r m z n for each n ∈ Z such that n ≥ n 0 together with Lemma 2.2 gives ϕ (2n + 2k1 + k2 + 2) = ϕ (2n + k2 + 2)

(2n + 2k2 + 2)(2n + k1 + m + 2) , (2n + 2k1 + 2)(2n + k1 + 2k2 + m + 2)

which is same to Equation (2.4) of [6], then it follows that

ϕ (z) = C

z+k2 2k1

z+2k1 −k2 2k1

z+m+k1 −k2 2k1

z+m+k1 +k2 2k1

(3.12)

for some constant C. Therefore, by Theorem 3.2 we get that Teik1 θ r m Teik2 θ ϕ = Teik2 θ ϕ Teik1 θ r m if and only if (3.12) holds and Teik1 θ r m Teik2 θ ϕ r |l| eilθ = Teik2 θ ϕ Teik1 θ r m r |l| eilθ for any l ∈ Z such that

−k1 −k2 2

(3.13)

< l < n0.

Therefore, if |k1 | > |k2 |, then we can use the above conditions to describe the commutant of Teik1 θ r m and Teik2 θ ϕ . However, it raises a natural question: Does there exist a nontrivial ϕ satisfying (3.12) and (3.13). Here, we give an example for this problem.


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Example Let m ∈ N and let ϕ be a bounded radial function on D. Then Remark 3.9 shows that Tei2θ r m Teiθ ϕ = Teiθ ϕ Tei2θ r m if and only if ϕ (z) =

z+1

z+m+1 4 4 C z+3 z+m+3 4 4

(3.14)

2 ϕ (5) = ϕ (3).

(3.15)

and

However, in the following we will show that (3.14) and (3.15) imply C = 0 for any m ∈ N, and hence Tei2θ r m commutes with Teiθ ϕ if and only if ϕ = 0. If C = 0, then it follows from (3.14) and (3.15) that

2+m

4 m4

√ m m+4 . =√ 2π (m + 2)

(3.16)

Noting that m ∈ N, we consider four cases. In each case, we will show that there are no such m satisfying (3.16). Case 1. Assume m = 4n for any n ∈ N. Then a direct calculation shows that (3.16) holds if and only if √ √ 2n+1 2 n + 1n! . π= 1 · · · (2n − 1)(2n + 1) However, one can easily see that the above equation is impossible. Case 2. Assume m = 4n + 1 for any n ∈ N. Using the fact that

√ 1 3 = 2π, 4 4

we can get that (3.16) holds if and only if 2

3 1 3 · · · (4n − 1)(4n + 3) 2π 2 = . √ 4 1 · · · (4n − 3)(4n + 1) 4n + 5

It is know that 41 is transcendental, and up to 4 digits, the numerical values of it is 3

2 √2π 41 = 3.6256 · · ·. Denote an = 3···(4n−1)(4n+3) 1···(4n−3)(4n+1) 4n+5 , then it is clear that sequence {an }n∈N is monotonically increasing. Therefore, for any n ∈ N,

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X.-T. Dong, Z.-H. Zhou 3

6π 2 an ≥ a0 = √ > 14 > 2 5

1 , 4

which is a contradiction. Case 3. Assume m = 4n + 2 for any n ∈ N. Then (3.16) holds if and only if 1=

2n+1 (n + 1)!

√ 1 · · · (2n − 1)(2n + 1) 2n + 3

Obviously, the above equation is impossible. Case 4. Assume m = 4n + 3 for any n ∈ N. Then (3.16) holds if and only if √ 1 3 · · · (4n + 3)(4n + 7) 4 π = . √ 4 1 · · · (4n + 1)(4n + 5) 4n + 7 2

√

4 π √ Denote bn = 3···(4n+3)(4n+7) 1···(4n+1)(4n+5) 4n+7 , then it is clear that sequence {bn }n∈N is monotonically decreasing. Therefore, for any n ∈ N,

√ 1 12 7π bn ≤ b0 = < 12 < 2 , 5 4 which is a contradiction. The following corollary will characterize commutativity of Toeplitz operators with bounded monomial symbols. Corollary 3.10 Let l1 , l2 > 0 and let k1 , k2 ∈ Z such that k1 k2 = 0. Then Teik1 θ r l1 Teik2 θ r l2 = Teik2 θ r l2 Teik1 θ r l1 if and only if k1 = k2 and l1 = l2 . 4 Commuting Toeplitz Operators with Arbitrary Bounded Symbols ˇ cković and Rao [6] showed that any function f in L 2 (D, d A) has the following Cuˇ polar decomposition f (r eiθ ) =

eikθ f k (r ),

k∈Z

where f k are radial functions in L 2 ([0, 1], r dr ). Moreover, if f ∈ L ∞ (D, d A) then the functions f k (r ) are bounded on D. This polar decomposition plays a vital role in the research of commuting Toeplitz operators with arbitrary bounded symbols. Similarly as Lemma 10 of [12], we first give the following lemma.


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Lemma 4.1 Let ei pθ φ be a bounded quasihomogeneous function and let f (r eiθ ) =

eikθ f k (r ) ∈ L ∞ (D, d A).

k∈Z

Then T f Tei pθ φ = Tei pθ φ T f ⇐⇒ Teikθ fk Tei pθ φ = Tei pθ φ Teikθ fk , ∀ k ∈ Z. Proof Since for any k1 ∈ Z T f Tei pθ φ (r |k1 | eik1 θ ) =

k∈Z

Teikθ fk Tei pθ φ r |k1 | eik1 θ

and Tei pθ φ T f (r |k1 | eik1 θ ) =

k∈Z

Tei pθ φ Teikθ fk r |k1 | eik1 θ ,

then for each couple (k1 , k2 ) ∈ Z2 , by Remark 2.3 we can get T f Tei pθ φ r |k1 | eik1 θ , r |k2 | eik2 θ = Tei(k2 −k1 − p)θ fk −k − p Tei pθ φ r |k1 | eik1 θ , r |k2 | eik2 θ 2

1

and Tei pθ φ T f r |k1 | eik1 θ , r |k2 | eik2 θ = Tei pθ φ Tei(k2 −k1 − p)θ fk

2 −k1

On the other hand, Teikθ fk Tei pθ φ r |k1 | eik1 θ , r |k2 | eik2 θ 0 = Tei(k2 −k1 − p)θ fk −k − p Tei pθ φ r |k1 | eik1 θ , r |k2 | eik2 θ 2

1

|k1 | ik1 θ |k2 | ik2 θ . r e , r e −p

if k = k2 − k1 − p if k = k2 − k1 − p.

and Tei pθ φ Teikθ fk r |k1 | eik1 θ , r |k2 | eik2 θ 0 = Tei pθ φ Tei(k2 −k1 − p)θ fk −k − p r |k1 | eik1 θ , r |k2 | eik2 θ

2

1

if k = k2 − k1 − p if k = k2 − k1 − p.

Therefore T f Tei pθ φ r |k1 | eik1 θ , r |k2 | eik2 θ = Tei pθ φ T f r |k1 | eik1 θ , r |k2 | eik2 θ ⇐⇒ Teikθ fk Tei pθ φ r |k1 | eik1 θ , r |k2 | eik2 θ = Tei pθ φ Teikθ fk r |k1 | eik1 θ , r |k2 | eik2 θ

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for any k ∈ Z, and hence the desired result is obvious.

ˇ cković [5] showed that if Tz n and Tψ commute on L a2 , then ψ is an analytic. On Cuˇ 2 L h , Ohno [13] showed that for an analytic symbol f, T f commutes with Tz if and only if f is a polynomial of degree at most 1. The following result will give a partial result on commuting Toeplitz operators whose symbols are arbitrary bounded functions and (co-)analytic monomials. Theorem 4.2 Let k ∈ Z such that k = 0 and let f be a bounded function on D. Then T f Tr |k| eikθ = Tr |k| eikθ T f if and only if a nontrivial linear combination of f and r |k| eikθ is constant on D. Proof Assume T f and Tr |k| eikθ commute. Without loss of generality, wecan also assume that k > 0, for otherwise we could take the adjoints. Let f (r eiθ ) = l∈Z eilθ fl (r ) then Lemma 4.1 implies Teilθ fl and Tr k eikθ commute for any l ∈ Z. Then by Lemma 3.6 one can easily get that fl = Cr k for l = k and fl = 0 for any nonzero integer l such that l = k, and hence a nontrivial linear combination of f and r |k| eikθ is constant on D. The converse implication is clear. This completes the proof.

It was shown in [6], that a Toeplitz operator on L a2 with radial symbol commutes only with other such operators. The following theorem will show this result is also true on L 2h . Theorem 4.3 Let ϕ be a nontrivial bounded radial function and let f be a bounded function on D. Then T f Tϕ = Tϕ T f if and only if f is radial. Proof Assume T f and Tϕ commute. Let f (r eiθ ) = k∈Z eikθ f k (r ), then Lemma 4.1 implies Teikθ fk and Tϕ commute for any k ∈ Z. Then it follows from Lemma 3.5 that f k = 0 for any k = 0, and hence f is radial. Conversely, assume f is radial, then Proposition 3.4 shows that T f and Tϕ commute. This completes the proof.

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