Open Access

Hyponormal Toeplitz operators with polynomial symbols on weighted Bergman spaces

Journal of Inequalities and Applications20142014:335

https://doi.org/10.1186/1029-242X-2014-335

Received: 16 May 2014

Accepted: 26 August 2014

Published: 2 September 2014

Abstract

In this note we consider the hyponormality of Toeplitz operators T φ on weighted Bergman space A α 2 ( D ) with symbol in the class of functions f + g ¯ with polynomials f and g.

MSC:47B20, 47B35.

Keywords

Toeplitz operators hyponormal weighted Bergman space polynomial symbols

1 Introduction

Let D be the open unit disk in the complex plane. For 1 < α < , the weighted Bergman space A α 2 ( D ) of the unit disk D is the space of analytic functions in L 2 ( D , d A α ) , where
d A α ( z ) = ( α + 1 ) ( 1 | z | 2 ) α d A ( z ) .
The space L 2 ( D , d A α ) is a Hilbert space with the inner product
f , g α = D f ( z ) g ( z ) ¯ d A α ( z ) ( f , g L 2 ( D , d A α ) ) .
If α = 0 then A 0 2 ( D ) is the Bergman space A 2 ( D ) . For any nonnegative integer n, let
e n ( z ) = Γ ( n + α + 2 ) Γ ( n + 1 ) Γ ( α + 2 ) z n ( z D ) ,
where Γ ( s ) stands for the usual Gamma function. It is easy to check that { e n } is an orthonormal basis for A α 2 ( D ) [1]. For φ L ( D ) , the Toeplitz operator T φ , and the Hankel operator H φ on A α 2 ( D ) are defined by
T φ f : = P α ( φ f ) and H φ f : = ( I P α ) ( φ f ) ( f A α 2 ( D ) ) ,
where P α denotes the orthogonal projection that maps from L 2 ( D , d A α ) onto A α 2 ( D ) . The reproducing kernel in A α 2 ( D ) is given by
K z ( α ) ( ω ) = 1 ( 1 z ω ¯ ) 2 + α ,
for z , ω D . We thus have
( T φ f ) ( z ) = D φ ( ω ) f ( ω ) ( 1 z ω ¯ ) 2 + α d A α ( ω ) ,

for f A α 2 ( D ) and ω D .

A bounded linear operator A on a Hilbert space is said to be hyponormal if its selfcommutator [ A , A ] : = A A A A is positive (semidefinite). The hyponormality of Toeplitz operators on the Hardy space H 2 ( T ) of the unit circle T = D has been studied by Cowen [2], Curto, Hwang and Lee [35] and others [6]. Recently, in [7] and [8], the hyponormality of T φ on the weighted Bergman space A α 2 ( D ) was studied. In [2], Cowen characterized the hyponormality of Toeplitz operator T φ on H 2 ( T ) by properties of the symbol φ L ( T ) . Here we shall employ an equivalent variant of Cowen’s theorem that was first proposed by Nakazi and Takahashi [9].

Cowen’s theorem ([2, 9])

For φ L ( T ) , write
E ( φ ) : = { k H : k 1  and  φ k φ ¯ H ( T ) } .

Then T φ is hyponormal if and only if E ( φ ) is nonempty.

The solution is based on a dilation theorem of Sarason [10]. For the weighted Bergman space, no dilation theorem (similar to Sarason’s theorem) is available. In [11], the first named author characterized the hyponormality of T φ on A α 2 ( D ) in terms of the coefficients of the trigonometric polynomial φ under certain assumptions as regards the coefficients of φ on the weighted Bergman space when α 0 and in [12], extended for all 1 < α < .

Theorem A ([12])

Let φ ( z ) = g ( z ) ¯ + f ( z ) , where f ( z ) = a 1 z + a 2 z 2 g ( z ) = a 1 z + a 2 z 2 . If a 1 a 2 ¯ = a 1 a 2 ¯ and 1 < α < , then
T φ  on  A α 2 ( D )  is hyponormal  { 1 α + 3 ( | a 2 | 2 | a 2 | 2 ) 1 2 ( | a 1 | 2 | a 1 | 2 ) if  | a 2 | | a 2 | , 4 ( | a 2 | 2 | a 2 | 2 ) | a 1 | 2 | a 1 | 2 if  | a 2 | | a 2 | .
In this note we consider the hyponormality of Toeplitz operators T φ on A α 2 ( D ) with symbol in the class of functions f + g ¯ with polynomials f and g. Since the hyponormality of operators is translation invariant we may assume that f ( 0 ) = g ( 0 ) = 0 . The following relations can easily be proved:
T φ + ψ = T φ + T ψ ( φ , ψ L ) ;
(1.1)
T φ = T φ ¯ ( φ L ) ;
(1.2)
T φ ¯ T ψ = T φ ¯ ψ if  φ  or  ψ  is analytic .
(1.3)

The purpose of this paper is to prove Theorem A for the Toeplitz operators on A α 2 ( D ) when f and g of degree N.

2 Main result

In this section we establish a necessary and sufficient condition for the hyponormality of the Toeplitz operator T φ on the weighted Bergman space under a certain additional assumption concerning the symbol φ. The assumption is related on the symmetry, so it is reasonable in view point of the Hardy space [13]. We expect that this approach would provide some clue for the future study of the symmetry case.

Lemma 1 ([11]) For any s, t nonnegative integers,
P α ( z ¯ t z s ) = { Γ ( s + 1 ) Γ ( s t + α + 2 ) Γ ( s + α + 2 ) Γ ( s t + 1 ) z s t if  s t , 0 if  s < t .
For 0 i N 1 , write
k i ( z ) : = n = 0 c N n + i z N n + i .

The following two lemmas will be used for proving the main result of this section.

Lemma 2 For 0 m N , we have
( i ) z ¯ m k i ( z ) α 2 = n = 0 Γ ( N n + i + m + 1 ) Γ ( α + 2 ) Γ ( N n + i + m + α + 2 ) | c N n + i | 2 , ( ii ) P α ( z ¯ m k i ( z ) ) α 2 = { n = 0 Γ ( N n + i + 1 ) 2 Γ ( N n + i m + α + 2 ) Γ ( α + 2 ) Γ ( N n + i + α + 2 ) 2 Γ ( N n + i m + 1 ) | c N n + i | 2 if  m i , n = 1 Γ ( N n + i + 1 ) 2 Γ ( N n + i m + α + 2 ) Γ ( α + 2 ) Γ ( N n + i + α + 2 ) 2 Γ ( N n + i m + 1 ) | c N n + i | 2 if  m > i .
Proof Let 0 m N . Then we have
z ¯ m k i ( z ) α 2 = n = 0 c N n + i z N n + i + m α 2 = n = 0 | c N n + i | 2 z N n + i + m α 2 = n = 0 Γ ( N n + i + m + 1 ) Γ ( α + 2 ) Γ ( N n + i + m + α + 2 ) | c N n + i | 2 .
This proves (i). For (ii), if m i then by Lemma 1 we have
P α ( z ¯ m k i ( z ) ) α 2 = n = 0 Γ ( N n + i + 1 ) Γ ( N n + i m + α + 2 ) Γ ( N n + i + α + 2 ) Γ ( N n + i m + 1 ) c N n + i z N n + i m α 2 = n = 0 Γ ( N n + i + 1 ) 2 Γ ( N n + i m + α + 2 ) Γ ( α + 2 ) Γ ( N n + i + α + 2 ) 2 Γ ( N n + i m + 1 ) | c N n + i | 2 .

If instead m > i , a similar argument gives the result. □

Lemma 3 ([14])

Let f ( z ) = a N 1 z N 1 + a N z N and g ( z ) = a ( N 1 ) z N 1 + a N z N . If a N 1 a N ¯ = a ( N 1 ) a N ¯ , then for i j , we have
H f ¯ k i ( z ) , H f ¯ k j ( z ) α = H g ¯ k i ( z ) , H g ¯ k j ( z ) α .

Our main result now follows.

Theorem 4 Let φ ( z ) = g ( z ) ¯ + f ( z ) , where
f ( z ) = a N 1 z N 1 + a N z N and g ( z ) = a ( N 1 ) z N 1 + a N z N .
If a N 1 a N ¯ = a ( N 1 ) a N ¯ and | a N | | a N | , then T φ on A α 2 ( D ) is hyponormal if and only if
1 N + α + 1 ( | a N | 2 | a N | 2 ) 1 N ( | a ( N 1 ) | 2 | a N 1 | 2 ) .
Proof For 0 i < N , put
K i : = { k i ( z ) A α 2 ( D ) : k i ( z ) = n = 0 c N n + i z N n + i } .
Then a straightforward calculation shows that T φ is hyponormal if and only if
( H f ¯ H f ¯ H g ¯ H g ¯ ) i = 0 N 1 k i ( z ) , i = 0 N 1 k i ( z ) α 0 for all  k i K i ( i = 0 , 1 , , N 1 ) .
(2.1)
Also we have
H f ¯ H f ¯ i = 0 N 1 k i ( z ) , i = 0 N 1 k i ( z ) α = i = 0 N 1 H f ¯ k i ( z ) , H f ¯ k i ( z ) α + i j , i , j 0 N 1 H f ¯ k i ( z ) , H f ¯ k k ( z ) α
(2.2)
and
H g ¯ H g ¯ i = 0 N 1 k i ( z ) , i = 0 N 1 k i ( z ) α = i = 0 N 1 H g ¯ k i ( z ) , H g ¯ k i ( z ) α + i j , i , j 0 N 1 H g ¯ k i ( z ) , H g ¯ k k ( z ) α .
(2.3)
Substituting (2.2) and (2.3) into (2.1), it follows from Lemma 3 that
T φ : hyponormal i = 0 N 1 ( H f ¯ H f ¯ H g ¯ H g ¯ ) k i ( z ) , k i ( z ) α 0 i = 0 N 1 ( f ¯ k i α 2 g ¯ k i α 2 + P α ( g ¯ k i ) α 2 P α ( f ¯ k i ) α 2 ) 0 .
Therefore it follows from Lemma 2 that T φ is hyponormal if and only if
( | a N 1 | 2 | a ( N 1 ) | 2 ) [ i = 0 N 2 { Γ ( i + N ) Γ ( α + 2 ) Γ ( i + N + α + 1 ) | c i | 2 + n = 1 ( Γ ( N n + i + N ) Γ ( α + 2 ) Γ ( N n + i + N + α + 1 ) Γ ( N n + i + 1 ) 2 Γ ( N n + i N + α + 3 ) Γ ( α + 2 ) Γ ( N n + i + α + 2 ) 2 Γ ( N n + i N + 2 ) ) | c N n + i | 2 } + n = 0 ( Γ ( N n + 2 N 1 ) Γ ( α + 2 ) Γ ( N n + 2 N + α ) Γ ( N n + N ) 2 Γ ( N n + α + 2 ) Γ ( α + 2 ) Γ ( N n + N + α + 1 ) 2 Γ ( N n + 1 ) ) | c N n + N 1 | 2 ] + ( | a N | 2 | a N | 2 ) [ i = 0 N 1 { Γ ( N + i + 1 ) Γ ( α + 2 ) Γ ( i + n + α + 2 ) | c i | 2 + n = 1 ( Γ ( N n + i + N + 1 ) Γ ( α + 2 ) Γ ( N n + i + N + α + 2 ) Γ ( N n + i + 1 ) 2 Γ ( N n + i N + α + 2 ) Γ ( α + 2 ) Γ ( N n + i + α + 2 ) 2 Γ ( N n + i N + 1 ) ) | c N n + i | 2 } ] 0 ,
or equivalently
( | a N 1 | 2 | a ( N 1 ) | 2 ) { n = 0 N 2 Γ ( N + n ) Γ ( α + 2 ) Γ ( N + n + α + 1 ) | c n | 2 + n = N 1 ( Γ ( N + n ) Γ ( α + 2 ) Γ ( N + n + α + 1 ) Γ ( n + 1 ) 2 Γ ( n N + α + 3 ) Γ ( α + 2 ) Γ ( n + α + 2 ) 2 Γ ( n N + 2 ) ) | c n | 2 } + ( | a N | 2 | a N | 2 ) { n = 0 N 1 Γ ( n + N + 1 ) Γ ( α + 2 ) Γ ( N + n + α + 2 ) | c n | 2 + n = N ( Γ ( N + n + 1 ) Γ ( α + 2 ) Γ ( N + n + α + 2 ) Γ ( n + 1 ) 2 Γ ( n N + α + 2 ) Γ ( α + 2 ) Γ ( n + α + 2 ) 2 Γ ( n N + 1 ) ) | c n | 2 } 0 .
(2.4)
Define ζ α by
ζ α ( n ) : = Γ ( N + n ) Γ ( α + 2 ) Γ ( N + n + α + 1 ) Γ ( n + 1 ) 2 Γ ( n N + α + 3 ) Γ ( α + 2 ) Γ ( n + α + 2 ) 2 Γ ( n N + 2 ) Γ ( N + n + 1 ) Γ ( α + 2 ) Γ ( N + n + α + 2 ) Γ ( n + 1 ) 2 Γ ( n N + α + 2 ) Γ ( α + 2 ) Γ ( n + α + 2 ) 2 Γ ( n N + 1 ) ( n 1 ) .
Then a direct calculation gives
ζ α ( n ) < Γ ( N + n ) Γ ( α + 2 ) Γ ( N + n + α + 1 ) Γ ( N + n + 1 ) Γ ( α + 2 ) Γ ( N + n + α + 2 ) .
Observe that
N + α + 1 N N + n + α + 1 N + n N + N i + α + 1 N + N i ζ α ( N i ) for all  N i N  and  n = 1 , 2 , , N 1 ;
(2.5)
and
N + α + 1 N Γ ( 2 N 1 ) Γ ( α + 2 ) Γ ( 2 N + α ) Γ ( N ) 2 Γ ( α + 2 ) 2 Γ ( N + α + 1 ) 2 Γ ( 2 N ) Γ ( α + 2 ) Γ ( 2 N + α + 1 ) .
Therefore (2.4) and (2.5) show that T φ is hyponormal if and only if
1 N + α + 1 ( | a N | 2 | a N | 2 ) 1 N ( | a ( N 1 ) | 2 | a N 1 | 2 ) .

This completes the proof. □

Remark 5 Let φ ( z ) = g ( z ) ¯ + f ( z ) , where
f ( z ) = a N 1 z N 1 + a N z N and g ( z ) = a ( N 1 ) z N 1 + a N z N .
If a N 1 a N ¯ = a ( N 1 ) a N ¯ , | a N | | a N | , and T φ on A α 2 ( D ) is hyponormal. Then
| a N | 2 | a N | 2 { 2 N + α 2 N 1 Γ ( N ) 2 Γ ( 2 N + α + 1 ) Γ ( α + 2 ) Γ ( 2 N ) Γ ( N + α + 1 ) 2 } ( | a N 1 | 2 | a ( N 1 ) | 2 ) .
Proof If we let c j = 1 for 0 j N 1 and the other c j ’s be 0 into (2.4), then we have
( | a N 1 | 2 | a ( N 1 ) | 2 ) { n = 0 N 2 Γ ( N + n ) Γ ( α + 2 ) Γ ( N + n + α + 1 ) + ( Γ ( 2 N 1 ) Γ ( α + 2 ) Γ ( 2 N + α ) Γ ( N ) 2 Γ ( α + 2 ) 2 Γ ( N + α + 1 ) 2 Γ ( 1 ) ) } + ( | a N | 2 | a N | 2 ) n = 0 N 1 Γ ( n + N + 1 ) Γ ( α + 2 ) Γ ( N + n + α + 2 ) 0 .
(2.6)
Define ξ α by
ξ α ( n ) : = Γ ( N + n ) Γ ( α + 2 ) Γ ( N + n + α + 1 ) Γ ( N + n + 1 ) Γ ( α + 2 ) Γ ( N + n + α + 2 ) ( 0 n N 1 ) .
Then ξ α ( n ) is a strictly decreasing function and
N + n + α + 1 N + n 2 N + α 2 N 1 2 N + α 2 N 1 Γ ( N ) 2 Γ ( 2 N + α + 1 ) Γ ( α + 2 ) Γ ( 2 N ) Γ ( N + α + 1 ) 2 for all  n = 0 , 1 , , N 1 .
(2.7)
Therefore (2.6) and (2.7) give that if T φ is hyponormal then
{ 2 N + α 2 N 1 Γ ( N ) 2 Γ ( 2 N + α + 1 ) Γ ( α + 2 ) Γ ( 2 N ) Γ ( N + α + 1 ) 2 } ( | a N 1 | 2 | a ( N 1 ) | 2 ) | a N | 2 | a N | 2 .

This completes the proof. □

Example 6 Let φ ( z ) = 2 z ¯ 2 + 3 2 z ¯ + 7 2 z + 6 7 z 2 and α = 0 . Then by Theorem A, T φ is not hyponormal. But φ satisfies the inequality in Remark 5, hence the inverse of Remark 5 is not satisfied.

Remark 7 Let φ ( z ) = n = m N a n z n , where a m and a N are nonzero. Suppose T φ on H 2 ( T ) is hyponormal. It is well known [15] that
N m rank [ T φ , T φ ] N .
However, the result cannot be extended to the case of A 2 ( D ) ; for example, if φ ( z ) = a 1 z ¯ + a 1 z then a straightforward calculation shows that the selfcommutator of Toeplitz operator T φ on A 2 ( D ) is given by
[ T φ , T φ ] = ( | a 1 | 2 | a 1 | 2 ) [ α 1 0 0 0 α 2 0 0 0 α 3 ] ,

where α n = 1 n ( n + 1 ) . Thus rank [ T φ , T φ ] = and the trace of the selfcommutator tr [ T φ , T φ ] = 1 .

Declarations

Acknowledgements

This work was supported by National Research Foundation of Korea Grant funded by the Korean Government (2011-0022577). The authors are grateful to the referee for several helpful suggestions.

Authors’ Affiliations

(1)
Department of Mathematics, Sungkyunkwan University
(2)
Department of Mathematics, Shingyeong University

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© Hwang et al.; licensee Springer. 2014

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