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Hyponormal Toeplitz operators with polynomial symbols on weighted Bergman spaces

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.

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 dA(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 (zD),

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 (φf)and H φ f:=(I P α )(φf) ( 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,ωD. 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 ωD.

A bounded linear operator A on a Hilbert space is said to be hyponormal if its selfcommutator [ A ,A]:= A AA 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 α0 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 0iN1, 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 0mN, 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 0mN. 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 mi 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 ij, 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 andg(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 0i<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 0for all  k i K i (i=0,1,,N1).
(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 ) (n1).

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 andg(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 0jN1 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 ) (0nN1).

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

Nmrank [ 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.

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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.

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Correspondence to Jongrak Lee.

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Hwang, I.S., Lee, J. & Park, S.W. Hyponormal Toeplitz operators with polynomial symbols on weighted Bergman spaces. J Inequal Appl 2014, 335 (2014). https://doi.org/10.1186/1029-242X-2014-335

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Keywords

  • Toeplitz operators
  • hyponormal
  • weighted Bergman space
  • polynomial symbols