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General Toeplitz operators on weighted Bloch-type spaces in the unit ball of C n

Abstract

In this paper, we consider the weighted Bloch-type spaces B ω α , β ( B n ) with α>0 and β0 in the unit ball of C n . We present some basic properties of the spaces B ω α , β ( B n ), then we consider the Toeplitz operator T μ α , β ; ω acting between B ω α , β ( B n ) spaces, where μ is a positive Borel measure in the unit ball B n . Moreover, we characterize complex measures μ for which the Toeplitz operator T μ α , β ; ω is bounded or compact on B ω α , β ( B n ).

MSC: 47B35, 32A18.

1 Introduction

We start here with some terminology, notations and definitions of various classes of analytic functions defined on the unit ball of C n .

Let B n be the unit ball of the n-dimensional complex Euclidean space C n . The boundary of B n is denoted by S n and is called the unit sphere in C n . Occasionally, we will also need the closed unit ball B ¯ n . We denote the class of all holomorphic functions on the unit ball B n by H( B n ). The ball centered at z C n with radius r is denoted by B(z,r). For α>1, let d ν α (z)= c α ( 1 | z | 2 ) α dν, where is the normalized Lebesgue volume measure on B n and c α = Γ ( n + α + 1 ) n ! Γ ( α + 1 ) (where Γ denotes the gamma function) so that ν α ( B n )1. The surface measure on S n is denoted by . Once again, we normalize σ so that σ α ( S n )1. For any z=( z 1 , z 2 ,, z n ),w=( w 1 , w 2 ,, w n ) C n , the inner product is defined by

z,w= k = 1 n z k w ¯ k .

For every point a B n , the Möbius transformation φ a : B n B n is defined by

φ a (z)= a P a ( z ) S a Q a ( z ) 1 z , a ,z B n ,

where S a = 1 | a | 2 , P a (z)= a z , a | a | 2 , P 0 =0 and Q a =I P a . The map φ a has the following properties that φ a (0)=a, φ a (a)=0, φ a = φ a 1 and

1 φ a ( z ) , φ a ( w ) = ( 1 | a | 2 ) ( 1 z , w ) ( 1 z , a ) ( 1 a , w ) ,

where z and w are arbitrary points in B n . In particular,

1 | φ a ( z ) | 2 = ( 1 | a | 2 ) ( 1 | z | 2 ) | 1 z , a | 2 .

For fH( B n ), the holomorphic gradient of f at z is defined by

f(z)= ( f z 1 ( z ) , f z 2 ( z ) , , f z n ( z ) )

and the radial derivative of f at z is defined by

Rf(z)=f, z ¯ = j = 1 n z j f ( z ) z j .

Similarly, the Möbius invariant complex gradient of f at z is defined by

˜ f(z)=(f φ z )(0).

For α>0, a function fH( B n ) is said to belong to the α-Bloch spaces B α ( B n ) if (see [1])

b α (f)( B n )= sup z B n | f ( z ) | ( 1 | z | 2 ) α <.

The little Bloch space B 0 α ( B n ) consists of all f B α ( B n ) such that

lim | z | 1 | f ( z ) | ( 1 | z | 2 ) α =0.

With the norm f B α =|f(0)|+ b α (f)( B n ), we know that B α ( B n ) becomes a Banach space and B 0 α ( B n ) is its closed subspace (see [1]). For α=1, the spaces B 1 ( B n ) and B 0 1 ( B n ) become the Bloch and the little Bloch space, respectively (see, for example, [25]). Zhu in [5] says that the norm f B ( B n ) is equivalent to

| f ( 0 ) | + sup z B n | R f ( z ) | ( 1 | z | 2 ) .

For α>1 and 0<p<, the weighted Bergman space A α p ( B n ) consists of holomorphic functions f L p ( B n ,d ν α ) such that

f A α p ( B n ) p := B n | f ( z ) | p d ν α (z)<,

that is, A α p ( B n )= L p ( B n ,d ν α )H( B n ). When the weight α=0, we simply write A p ( B n ) for A 0 p ( B n ). In the special case when p=2, A α 2 ( B n ) is a Hilbert space. It is well known that for α>1, the Bergman kernel of A α 2 ( B n ) is given by

K α (z,w)= 1 ( 1 z , w ) n + α + 1 ,z,w B n .

For α>1, a complex measure μ is such that

| B n ( 1 | w | 2 ) α d μ ( w ) | = | B n d μ α ( w ) | <.

The general Bergman projection P α is the orthogonal projection of the measure μ from L 2 ( B n ,d ν α ) into A α 2 ( B n ) defined by

P α (μ)(z)= c α B n ( 1 | w | 2 ) α ( 1 z , w ) n + α + 1 dμ(w)= c α B n d μ α ( w ) ( 1 z , w ) n + α + 1 .

The general Bergman projection of the function f is

P α f(z)= c α B n f ( w ) ( 1 | w | 2 ) α ( 1 z , w ) n + α + 1 dν(w)= c α B n f ( w ) d ν α ( w ) ( 1 z , w ) n + α + 1 .

Let ω:(0,1](0,) be a right-continuous and nondecreasing function. For a complex measure μ,α>1, β0, and f L 1 ( B n ,d ν α + β ), define weighted general Toeplitz operator as follows:

T μ α , β ; ω f ( z ) = c α + β ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) B n ( 1 | w | 2 ) α + β f ( w ) ( 1 z , w ) n + α + β + 1 d μ ( w ) = c α + β ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) B n f ( w ) d μ α + β ( w ) ( 1 z , w ) n + α + β + 1 .

Thus P α + β ; ω (μ)(z)= T μ α , β ; ω (1)(z), where 1 stands for a constant function.

Toeplitz operators have been studied extensively on the Bergman spaces by many authors. For references, see [6] and [7]. Boundedness and compactness of the general Toeplitz operators T μ α on the α-Bloch B α (D) spaces have been investigated in [8] on the unit disk D for 0<α<. Also, in [9], the authors extended the general Toeplitz operator T μ α to B α ( B n ) with 1α<2. Recently, in [10], the general Toeplitz operators T μ α on the analytic Besov B p (D) spaces with 1p< have been investigated. Under a prerequisite condition, the authors characterized a complex measure μ on the unit disk for which T μ α is bounded or compact on the Besov space B p (D). For more studies on the Toeplitz operator, we refer to [1117].

In this paper, we consider the weighted Bloch-type spaces B ω α , β ( B n ) with α>0 and β0 in the unit ball of C n . We prove a certain integral representation theorem that is used to determine the degree of growth of the functions in the space B ω α , β ( B n ). It is also proved that the space B ω α , β ( B n ) is a Banach space for each weight α>0, β0, and the Banach dual of the Bergman space A 1 ( B n ) is B ω α , β ( B n ) for each α1, β0. Further, we extend the Toeplitz operator T μ α , β ; ω to B ω α , β ( B n ) in the unit ball of C n and completely characterize the positive Borel measure μ such that T μ α , β ; ω is bounded or compact in B ω α , β ( B n ) spaces with α+β1.

Throughout the paper, we say that the expressions A and B are equivalent, and write AB whenever there exist positive constants C 1 and C 2 such that C 1 AB C 2 A. As usual, the letter C denotes a positive constant, possibly different on each occurrence. Hereafter, ω stands for a right-continuous and nondecreasing function.

Theorem 1.1 (see [[5], Theorem 1.12])

Suppose b is real and s>1. Then the integrals

I b (z)= S n d σ ( ζ ) | 1 z , ζ | n + b ,z B n

and

J b , s (z)= B n ( 1 | w | 2 ) s d ν ( w ) | 1 z , w | n + 1 + s + b ,z B n ,

have the following asymptotic properties.

(1) If b<0, then I b (z) and J b , s (z) are both bounded in B n .

(2) If b=0, then

I b (z) I b , s (z)log 1 1 | z | 2 as |z| 1 1 .

(3) If b>0, then

I b (z) J b , s (z) ( 1 | z | 2 ) b as |z| 1 1 .

Lemma 1.1 (see [[5], Lemma 3.3])

Suppose γ is a real constant and g L 1 ( B n ,dν). If

u(z)= ( 1 | φ a ( z ) | 2 ) β B n g ( w ) d ν ( w ) ( 1 z , w ) γ ,z B n ,

then

| ˜ u ( z ) | 2 |γ| ( 1 | z | 2 ) 1 2 B n g ( w ) d ν ( w ) | 1 z , w | γ + 1 2 ,z B n .

Let β(,) be the Bergman metric on B n . Denote the Bergman metric ball at w ( j ) by B( w ( j ) ,r)={z B n :β( w ( j ) ,z)<r}, where w ( j ) B n and r>0.

Lemma 1.2 (see [[5], Theorem 2.23])

For fixed r>0, there is a sequence { w ( j ) } B n such that:

j = 1 B( w ( j ) ,r)= B n ;

there is a positive integer N such that each z B n is contained in at most N of the sets B( w ( j ) ,2r).

The following characterization of Carleson measures can be found in [6], or in [5].

A positive Borel measure μ on the unit ball B n is said to be a Carleson measure for the Bergman space A α p ( B n ) if

B n | f ( z ) | p d ν α (z)C f A α p ( B n ) p ,f A α p ( B n ).

It is well known that a positive Borel measure μ is a Carleson measure if and only if there is a positive constant C such that

sup w ( j ) B n μ ( B ( w ( j ) , r ) ) ν ( B ( w ( j ) , r ) ) <,

where { w ( j ) } is the sequence in Lemma 1.2. If μ satisfies that

lim j μ ( B ( w ( j ) , r ) ) ν ( B ( w ( j ) , r ) ) =0,

then μ is called a vanishing Carleson measure.

For a given reasonable function ω:(0,1](0,), the weighted Bloch space B ω of several complex variables is defined as the set of all analytic functions f on B n satisfying

( 1 | z | ) α | f ( z ) | Cω ( 1 | z | ) ,z B n , where α(0,),

for some fixed C= C f >0. In the special case where ω1, B ω reduces to the classical Bloch space in C n . This class of functions extends and generalizes the well known Bloch space. Now, we define the space B α , β ; ω ( B n ) in the unit ball B n . For α>0 and β0, a function fH( B n ) is said to belong to the (α,β;ω)-Bloch space B α , β ; ω ( B n ) if

b α , β ; ω (f)( B n )= sup a , z B n ( 1 | z | 2 ) α + β ( 1 | φ a ( z ) | 2 ) β ω ( 1 | z | ) | f ( z ) | <.

The little (α,β;ω)-Bloch space B α , β ; ω , 0 ( B n ) is a subspace of B α , β ; ω ( B n ) consisting of all f B α , β ; ω ( B n ) such that

lim | a | 1 lim | z | 1 ( 1 | z | 2 ) α + β ( 1 | φ a ( z ) | 2 ) β ω ( 1 | z | ) | f ( z ) | =0.

If β=0, ω(1|z|)=1, then we get the α-Bloch space B α ( B n ) and the little α-Bloch space B 0 α ( B n ). If ω(1|z|)=1, α=1 and β=0, then we get the classical Bloch space B( B n ) and B 0 ( B n ). These classes extend the weighted Bloch spaces defined in [18] to the setting of several complex variables.

The logarithmic (α,β;ω)-Bloch space LB ω α , β ( B n ) is the space of holomorphic functions f such that

sup a , z B n ( 1 | z | 2 ) α + β ( 1 | φ a ( z ) | 2 ) β ω ( 1 | z | ) ( ln 2 1 | z | 2 ) | f ( z ) | <.

Correspondingly, the little logarithmic (α,β;ω)-Bloch space LB ω ; 0 α , β ( B n ) is a subspace of LB ω α , β ( B n ) consisting of all functions f such that

lim | a | 1 lim | z | 1 ( 1 | z | 2 ) α + β ( 1 | φ a ( z ) | 2 ) β ω ( 1 | z | ) ( ln 2 1 | z | 2 ) | f ( z ) | =0.

If ω(1|z|)=1 and β=0, then we get the logarithmic α-Bloch space LB α ( B n ) and the little logarithmic α-Bloch space LB 0 α ( B n ). If ω(1|z|)=1, α=1 and β=0, then we get the logarithmic Bloch space LB( B n ) and LB 0 ( B n ) (see [19]).

2 Holomorphic (α,β;ω)-Bloch space in the unit ball

In this section, we study the general (α,β;ω)-Bloch space B ω α , β ( B n ) in the unit ball of C n by giving some characterizations of (α,β;ω)-Bloch space, then we present several auxiliary results, which play important roles in the proofs of our main results.

Lemma 2.1 Let α,β(0,) and f B ω α , β ( B n ). Suppose that

0 1 ω ( 1 t | z | ) | z | d t ( 1 t 2 | z | 2 ) α + β <.
(1)

Then

| f ( z ) | | f ( 0 ) | + f B ω α , β ( B n ) .

Proof Let z B n , 0t<1 and f B ω α , β ( B n ). By the definition of B ω α , β ( B n ) and |z|> 1 2 , we have that

| f ( z ) f ( z 2 ) | = | 1 2 1 f ( t z ) , z d t | | 1 2 1 R f ( t z ) d t t | b α , β ; ω ( f ) 0 1 ( 1 | φ a ( t z ) | 2 ) β ω ( 1 t | z | ) ( 1 t 2 | z | 2 ) α + β | z | d t b α , β ; ω ( f ) 0 1 ( 1 | a | 2 ) β ω ( 1 t | z | ) | 1 t z , a | 2 β ( 1 t 2 | z | 2 ) α | z | d t .

Since (1|a|)|1tz,a| and (1t|z|)|1tz,a|, a,z B n , we get

| f ( z ) f ( z 2 ) | b α , β ; ω ( f ) 0 1 ( 1 | a | 2 ) β ω ( 1 | z | ) ( 1 | a | ) β ( 1 t | z | ) β ( 1 t 2 | z | 2 ) α | z | d t 4 β b α , β ( f ) 0 1 ω ( 1 t | z | ) | z | d t ( 1 t 2 | z | 2 ) α + β ,

from which the result follows. □

Theorem 2.1 For each 0<α,β<, γ>1 and fH( B n ). Then the following conditions are equivalent:

(i) f B ω α , β ( B n );

(ii) The function ( 1 | z | 2 ) α + β ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) |Rf(z)| is bounded in B n ;

(iii) There exists a function g L ( B n ) such that

f(z)= ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) B n g ( w ) d ν γ ( w ) ( 1 z , w ) n + α + β + γ ,z B n .

Proof By the Cauchy-Schwarz inequality in C n , we have

| R f ( z ) | |z| | f ( z ) | | f ( z ) | .

This proves that (i) (ii).

If (ii) holds, then the function

g(z)= c α + β + γ c γ ( 1 | z | 2 ) α + β ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) ( f ( z ) + R f ( z ) n + α + β + γ )

is bounded in B n . For z B n consider the holomorphic function

F ( z ) = B n g ( w ) ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) d ν γ ( w ) ( 1 z , w ) n + α + β + γ = B n ω ( 1 | w | ) ( 1 z , w ) n + α + β + γ ( f ( w ) + R f ( w ) n + α + β + γ ) d ν α + β + γ ( w ) .

As in the proof of Theorem 7.1 in [5], we have F=f.

This shows that (ii) implies (iii). That (iii) implies (i) follows from differentiating under the integral sign and then applying Theorem 1.12 in [5]. □

Theorem 2.2 For each α>0, β0, α+β>0 and s=α+β1. If s>1, then the Banach dual of A 1 ( B n ) can be identified with B ω α , β ( B n ) (with equivalent norms) under the following integral pairing:

f , g s = B n f(z) g ( z ) ¯ d ν s (z),f A 1 ( B n ),g B ω α , β ( B n ).
(2)

Proof It is easy to see that 1(α+β)+s>1. If g B ω α , β ( B n ), then by Theorem 2.1, there exists a function h L ( B n ) such that

g(z)= 1 ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) B n h ( w ) d ν 1 ( α + β ) + s ( w ) ( 1 z , w ) n + 1 + s ,z,w B n ,

and h C g B ω α , β ( B n ) , where C is a positive constant independent of g. By Fubini’s theorem,

f , g s = B n f ( z ) h ( z ) ¯ ( 1 | z | 2 ) d ν 1 ( α + β ) + s ( z ) = c 1 ( α + β ) + s B n f ( z ) h ( z ) ¯ d ν ( z ) .

Applying Lemma 2.15 in [5] for all f A 1 ( B n ), we have

B n | f ( z ) | dν(z) f A 1 ( B n ) .

Combining this, we see that

| f , g s | h f A 1 ( B n ) C g B ω α , β ( B n ) f A 1 ( B n ) .

Conversely, if F is a bounded linear functional on A 1 ( B n ) and f A 1 ( B n ), then

f r (z)= 1 ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) B n f r ( w ) d ν s ( w ) ( 1 z , w ) n + 1 + s for 0<r<1.

It is easy to verify (using the homogeneous expansion of the kernel function) that

F( f r )= B n f r (w) F z [ 1 ( 1 z , w ) n + 1 + s ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) ] d ν s (w).

Define a function g on B n by

g ( w ) ¯ = F z [ 1 ( 1 z , w ) n + 1 + s ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) ] .

Then

F( f r )= B n f r (w) g ( w ) ¯ d ν s (w)= f , g s .

It remains for us to show that g B ω α , β .

We interchange differentiation and the application of F, which can be justified by using the homogeneous expansion of the kernel. The result is

R g ( w ) ¯ =(n+1+s) F z [ 1 ( 1 z , w ) n + 1 + s ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) ] .

Since F is bounded on A 1 ( B n ), we have

| R g ( w ) | C F ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) B n d ν ( w ) | 1 z , w | n + 2 + s .

An application of Theorem 1.1 for s+1=α+β then shows that

| R g ( w ) | C F ( 1 | z | 2 ) α + β ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) .

This shows that g B ω α , β ( B n ) and completes the proof of the theorem. □

Lemma 2.2 If n>1, α+β> 1 2 , then f B ω α , β ( B n ) if and only if the function

( 1 | z | 2 ) α + β 1 ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) | ˜ f ( z ) |

is bounded in B n .

Proof Recall from Lemma 2.14 in [5] that

( 1 | z | 2 ) | f ( z ) | | ˜ f ( z ) | ,z B n .

So, the boundedness of

( 1 | z | 2 ) α + β 1 ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) | ˜ f ( z ) |

implies that of

( 1 | z | 2 ) α + β ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) | f ( z ) | .

On the other hand, if f B ω α , β ( B n ), then by Theorem 2.1,

f(z)= ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) B n g ( w ) d ν ( w ) ( 1 z , w ) n + α + β ,z B n ,

where g is a function in L ( B n ). Now we let f(z)=h(z)u(z), where h(z)= ( 1 | φ w ( z ) | 2 ) β ω(1|z|) and

u(z)= B n g ( w ) d ν ( w ) ( 1 z , w ) n + α + β .

An application of Lemma 1.1 gives

| ˜ u ( z ) | |n+α+β| 2 ( 1 | z | 2 ) 1 2 B n g ( w ) d ν ( w ) | 1 z , w | n + α + β + 1 2 ,z B n .

Since g(z) is bounded, by Theorem 1.1 we have

B n g ( w ) d ν ( w ) | 1 z , w | n + α + β + 1 2 ( 1 | z | 2 ) 1 2 ( α + β ) .

So,

| ˜ u ( z ) | C ( 1 | z | 2 ) 1 ( α + β ) .

It is easy to check that ˜ h(z)=(h φ z )(0)=0.

Using the product rule, we have

| ˜ f ( z ) | | ˜ h ( z ) | | u ( z ) | + | h ( z ) | | ˜ u ( z ) | | ˜ h ( z ) | | u ( z ) |

and we have

| ˜ f ( z ) | | n + α + β | 2 ( 1 | z | 2 ) 1 2 ( 1 | φ a ( z ) | 2 ) β B n g ( w ) d ν ( w ) | 1 z , w | n + α + β + 1 2 for all  z B n .

Hence,

( 1 | z | 2 ) α + β 1 ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) | ˜ f ( z ) |

is bounded in B n . This completes the proof. □

Lemma 2.3 Let 0<α+β2. Let λ be any real number satisfying the following properties:

0λα+β if 0<α+β<1;

0<λ<1 if α+β=1;

α+β1λ1 if 1<α+β2.

Then, for all z,w B n a holomorphic function f B ω α , β ( B n ) if and only if

sup z w ( 1 | z | 2 ) λ ( 1 | w | ) α + β λ ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) | f ( z ) f ( w ) | | z w | <.
(3)

Proof Let f B ω α , β ( B n ). By a similar proof to the one for Theorem 3.1 in [20], we have

| f ( z ) f ( w ) | = n |zw| 0 1 | f ( t z ( 1 t ) w ) | dt

for any z,w B n with zw. We know that

f B ω α , β ( B n ) sup w , z B n ( 1 | z | 2 ) α + β ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) | f ( z ) | .

Thus, there is a constant C>0 such that

| f ( z ) f ( w ) | | z w | C f B ω α , β ( B n ) 0 1 ( 1 | φ a ( t z ( 1 t ) a ) | 2 ) β ω ( 1 | t z ( 1 t ) w | ) ( 1 | t z ( 1 t ) w | 2 ) α + β dt.

Since (1|z|)|1w,z| and

1 | t z + ( 1 t ) w | 2 1 | t z + ( 1 t ) w | 1|w|+ ( | w | | z | ) t,

we get

( 1 | φ w ( t z ( 1 t ) w ) | 2 ) β = ( 1 | w | 2 ) β ( 1 | t z ( 1 t ) w | 2 ) β | 1 t z ( 1 t ) w , w | 2 β ( 1 | w | 2 ) β ( 1 | t z ( 1 t ) w | 2 ) β ( 1 | w | ) 2 β ( 1 + | w | ) β ( 1 | t z ( 1 t ) w | 2 ) β ( 1 | w | ) β ( 1 | t z ( 1 t ) w | 2 ) β ( 1 | w | ) β .

Thus

| f ( z ) f ( w ) | | z w | C f B ω α , β ( B n ) 0 1 1 ( 1 | w | + ( | w | | z | ) t ) α ( 1 | w | ) β dt.
(4)

If |z|=|a|, then

| f ( z ) f ( w ) | | z w | C f B ω α , β ( B n ) 0 1 1 ( 1 | w | ) α + β dt C f B ω α , β ( B n ) ( 1 | z | 2 ) λ ( 1 | w | 2 ) α + β λ .
(5)

Now suppose |z||w| as in [21], there is a constant C>0 such that this integral in (4) is dominated by

C ( 1 | z | 2 ) λ ( 1 | w | 2 ) α + β λ .

Combining with (5), we get that whenever zw,

| f ( z ) f ( w ) | | z w | C f B ω α , β ( B n ) ( 1 | z | 2 ) λ ( 1 | w | 2 ) α + β λ .

This proves the necessity. The proof of the sufficiency condition is much akin to the corresponding one in [21], so the proof is omitted. □

Proposition 2.1 Suppose f B ω α , β ( B n ) and 1α+β2. Let λ be any real number satisfying:

0<λ<1 if α+β=1;

α+β1λ1 if 1<α+β2.

Then

sup z , w B n ( 1 | z | 2 ) 2 α + 2 β λ 1 ( 1 | w | 2 ) α + β λ | f ( z ) f ( w ) | ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) | 1 w , z | 2 ( α + β ) ( 2 λ + 1 ) | z P z ( w ) S z Q z ( w ) | C f B ω α , β ( B n ) .
(6)

Proof Let z=0 in (3), then we have

( 1 | w | 2 ) α + β λ | f ( 0 ) f ( w ) | | w | C b α , β ; ω (f)( B n ),w B n {0}.

Now, replacing f by f φ w , we get

( 1 | u | 2 ) α + β λ | f φ w ( 0 ) f φ w ( u ) | | u | C b α , β ; ω (f φ w )( B n ),u B n {0}.
(7)

Since

| ˜ ( f φ w ) ( z ) | = | ˜ f ( φ w ( z ) ) | ,

by Lemma 2.2, we obtain that

b α , β ; ω ( f φ w ) ( B n ) sup z , w B n ( 1 | w | 2 ) α + β 1 ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) | ˜ ( f φ w ) ( z ) | = sup z , w B n ( 1 | w | 2 ) α + β 1 ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) | ˜ f ( φ w ( z ) ) | = sup z , w B n ( 1 | w | 2 ) α + β 1 ( 1 | φ w ( z ) | 2 ) α + β 1 ( 1 | φ w ( z ) | 2 ) α + β 1 ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) | ˜ f ( φ w ( z ) ) | .

Then

b α , β ; ω ( f φ z ) ( B n ) C f B ω α , β ( B n ) ( 1 | w | 2 ) α + β 1 ( 1 | φ z ( w ) | 2 ) α + β 1 ω ( 1 | w | ) C f B ω α , β ( B n ) ( 1 | z | 2 ) α + β 1 .

Letting u= φ w (z) and wz in (7), we obtain

| f ( z ) f ( w ) | | φ w ( z ) | ω ( 1 | z | ) ( 1 | φ w ( z ) | 2 ) α + β λ C f B ω α , β ( B n ) ( 1 | z | 2 ) α + β 1 .

Since

1 | φ z ( w ) | 2 = ( 1 | z | 2 ) ( 1 | w | 2 ) | 1 w , z | 2

and

φ z (w)= z P z ( w ) S z Q z ( w ) 1 w , z .

Consequently,

( 1 | z | 2 ) 2 α + 2 β λ 1 ( 1 | w | 2 ) α + β λ | f ( z ) f ( w ) | ω ( 1 | z | ) ( 1 | φ z ( w ) | 2 ) α + β λ | 1 w , z | 2 ( α + β ) ( 2 λ + 1 ) | z P z ( w ) S z Q z ( w ) | C f B ω α , β ( B n ) .

 □

3 Boundedness of general Toeplitz operators

In this section, we study the boundedness of general Toeplitz operators acting on the weighted Bloch-type spaces B ω α , β ( B n ) in the unit ball of C n .

Theorem 3.1 Let μ be a positive Borel measure on B n . Then we have

(1) if α+β=1, then T μ α , β ; ω is bounded on B ω α , β ( B n ) if and only if P α + β 1 ; ω (μ) LB ω ( B n ) and μ is a Carleson measure;

(2) if α=β=1, then T μ α , β ; ω is bounded on B ω α , β ( B n ) if and only if P α + β 1 ; ω (μ) B ω ( B n ) LB ω 2 ( B n ) and μ is a Carleson measure;

(3) if α>1, β>1, then T μ α , β ; ω is bounded on B ω α , β ( B n ) if and only if P α + β 1 ; ω (μ) B ω α , β ( B n ) and μ is a Carleson measure.

Proof Since the Banach dual of A 1 ( B n ) is B ω α , β ( B n ) under the pairing (2), to prove the boundedness of T μ α , β ; ω , it suffices to show that

| f , T μ α , β ; ω ( g ) α | C f A 1 ( B n ) g B ω α , β ( B n )

for all f A 1 ( B n ) and g B ω α , β ( B n ), where C is a positive constant that does not depend on f or g.

For s=α+β1, by Fubini’s theorem, we get

f , T μ α , β ; ω g s = B n f ( z ) T μ α , β ; ω g ( z ) ¯ d ν s ( z ) = c α + β 1 B n f ( z ) g ( z ) ¯ ( 1 | z | 2 ) α + β 1 d μ ( z ) .

Using the operator P α + β ; ω , we have

f , T μ α , β ; ω g s = c α + β 1 B n ( I z , w ; ω P α + β ; ω ) ( f g ¯ ) ( z ) ( 1 | z | 2 ) α + β 1 d μ ( z ) + c α + β 1 B n P α + β ; ω ( f g ¯ ) ( z ) ( 1 | z | 2 ) α + β 1 d μ ( z ) = I 1 + I 2 ,

where I z , w ; ω = I ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) , and I is the identity operator. Also,

( I z , w ; ω P α + β ; ω ) ( f g ¯ ) ( z ) = f ( z ) g ( z ) ¯ ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) c α + β ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) B n f ( w ) g ( w ) ¯ ( 1 | w | 2 ) α + β ( 1 z , w ) n + α + β + 1 d ν ( w ) = c α + β ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) × B n ( g ( z ) ¯ g ( w ) ¯ ) f ( w ) ( 1 | w | 2 ) α + β ( 1 z , w ) n + α + β + 1 ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) d ν ( z ) .

By Proposition 2.1, we have

| I 1 | = c α + β 1 | B n ( I z , w ; ω P α + β ; ω ) ( f g ¯ ) ( z ) ( 1 | z | 2 ) α + β 1 d μ ( z ) | = c α + β 1 c α + β | B n B n ( g ( z ) ¯ g ( w ) ¯ ) f ( w ) ( 1 | w | 2 ) α + β ( 1 | z | 2 ) α + β 1 ( 1 z , w ) n + α + β + 1 ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) d ν ( w ) d μ ( z ) | = c α + β 1 c α + β | B n f ( w ) ( 1 | w | 2 ) α + β × B n ( g ( z ) ¯ g ( w ) ¯ ) ( 1 | z | 2 ) α + β 1 ( 1 z , w ) n + α + β + 1 ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) d μ ( z ) d ν ( w ) | c α + β 1 c α + β B n | f ( w ) | ( 1 | w | 2 ) λ × B n ( 1 | z | 2 ) 2 ( α + β ) λ 1 ( 1 | w | 2 ) α + β λ | f ( z ) f ( w ) | | 1 w , z | 2 ( α + β ) ( 2 λ + 1 ) | z P z ( w ) S z Q z ( w ) | × | z P z ( w ) S z Q z ( w ) | ( 1 | z | 2 ) λ ( α + β ) | 1 z , w | n ( α + β ) + 2 λ + 2 ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) d μ ( z ) d ν ( w ) C B n g B ω α , β ( B n ) | f ( w ) | ( 1 | w | 2 ) λ B n ( 1 | z | 2 ) λ ( α + β ) | 1 z , w | n ( α + β ) + 2 λ + 1 d μ ( z ) d ν ( w ) .

Since μ is a Carleson measure, taking λ(α+β)>1, then as in [9] or in [[22], Proposition 1.4.10], for fixed r>0, we get

B n ( 1 | z | 2 ) λ ( α + β ) | 1 z , w | n ( α + β ) + 2 λ + 1 d μ ( z ) j = 1 μ ( B ( z ( j ) , r ) ) ν ( B ( z ( j ) , r ) ) B ( z ( j ) , r ) ( 1 | z | 2 ) λ ( α + β ) | 1 z , w | n ( α + β ) + 2 λ + 1 d ν ( z ) C .

Therefore,

| I 1 |C f A 1 ( B n ) g B ω α , β ( B n ) .

Next considering I 2 , we have

| I 2 | = c α + β 1 | B n P α + β ; ω ( f g ¯ ) ( z ) ( 1 | z | 2 ) α + β 1 d μ ( z ) | = c α + β 1 c α + β | B n B n f ( w ) g ( w ) ¯ ( 1 | w | 2 ) α + β ( 1 | z | 2 ) α + β 1 ( 1 z , w ) n + α + β + 1 ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) d ν ( w ) d μ ( z ) | c α + β B n | f ( w ) | ( 1 | w | 2 ) α + β | g ( w ) | × ( c α + β 1 ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) B n ( 1 | z | 2 ) α + β 1 d μ ( z ) | 1 z , w | n + α + β + 1 ) d ν ( w ) C B n f A 1 ( B n ) ( 1 | w | 2 ) α + β | g ( w ) | Q μ α , β ; ω ( w ) d ν ( w ) ,

where

Q μ α , β ; ω (w)= c α + β 1 ( 1 | φ z ( w ) | 2 ) β ω ( 1 | w | ) B n ( 1 | z | 2 ) α + β 1 d μ ( z ) | 1 z , w | n + α + β + 1 .

As in [9], by simple calculation, we have

Q μ α , β ; ω (w)= P α + β 1 ; ω (μ)(w)+ 1 n + α + β R P α + β 1 ; ω (μ)(w).
(8)

It is easy to see that

(1) if α+β=1 and P α + α 1 ; ω (μ) LB ω α , β ( B n ), then

( 1 | w | 2 ) Q μ α , β ; ω (w) ( ln 2 1 | w | 2 ) L ( B n );

(2) if α=β=1, P α + α 1 ; ω (μ) B ω ( B n )LB ω 2 ( B n ), then

( 1 | w | 2 ) Q μ α , β ; ω (w) L ( B n )

and

( 1 | w | 2 ) 2 Q μ α , β ; ω (w) ( ln 2 1 | w | 2 ) L ( B n );

(3) if α>1, β>1, and P α + α 1 ; ω (μ)B ω α , β ( B n ), then

( 1 | w | 2 ) α + β Q μ α , β ; ω (w) L ( B n ).

This implies that | I 2 |C f A 1 ( B n ) g B ω α , β ( B n ) . Hence, T μ α , β ; ω is a bounded operator on B ω α , β ( B n ) with α>0, β0.

Conversely, suppose that T μ α , β ; ω is a bounded operator on B ω α , β ( B n ). Take

f w (z)= ( 1 | w | 2 ) t ( 1 z , w ) n + t + 1 for t>0.

It is clear that f w A 1 ( B n ) C. On the other hand, take

g w (z)= ( 1 | w | 2 ) n + 2 + t ( α + β ) ( 1 z , w ) n + t + 1 ; φ w (z)1andω ( 1 | z | ) 1for t>0.

Then, we have g w B ω α , β ( B n ) C. Therefore

| f , T μ α g s | = c α + β 1 ( 1 | w | 2 ) n + 2 + 2 t ( α + β ) B n ( 1 | z | 2 ) α + β 1 d μ ( z ) | 1 z , w | 2 ( n + t + 1 ) C T μ α , β ; ω f w A 1 ( B n ) g w B ω α , β ( B n ) C .

Thus,

( 1 | w | 2 ) n + 2 + 2 t ( α + β ) B ( w , r ) ( 1 | z | 2 ) α + β 1 d μ ( z ) | 1 z , w | 2 ( n + t + 1 ) C

for every w B n . This implies that

sup w B n μ ( B ( w , r ) ) ν ( B ( w , r ) ) <.

Hence μ is a Carleson measure on B n .

From the proof of the sufficient condition, we find that there exists a constant C such that

| I 2 | = c α + β | B n f ( w ) ( 1 | w | 2 ) α + β g ( w ) Q μ α , β ; ω ( w ) ¯ d ν ( w ) | C f A 1 ( B n ) g B ω α , β ( B n ) .

This implies that

| g ( w ) Q μ α , β ; ω ( w ) | ( 1 | w | 2 ) α + β C g B ω α , β ( B n ) .

If α+β=1, we have

| g ( w ) Q μ α , β ; ω ( w ) | ( 1 | w | 2 ) C g B ω α , β ( B n ) .

Take g w (z)=ln 2 1 z , w ; φ w (z)1 and ω(1|z|)1. It is clear that g w LB ω ( B n ) C. Taking z=w, then

| Q μ α , β ; ω ( w ) | ( 1 | w | 2 ) ( ln 2 1 | w | 2 ) C.

From (8) we have P α + β 1 (μ) LB ω ( B n ). Let α=β=1, we have

| g ( w ) Q μ α , β ; ω ( w ) | ( 1 | w | 2 ) 2 C g B ω α , β ( B n ) .

Take g w (z)= 1 1 z , w +ln 2 1 z , w ; φ w (z)1 and ω(1|z|)1. It is clear that

g w B ω ( B n ) LB ω 2 ( B n ) C.

Taking z=w, then

| g ( w ) Q μ α , β ; ω ( w ) | ( 1 | w | 2 ) 2 = ( 1 1 | w | 2 + ln 2 1 | w | 2 ) | Q μ α , β ( w ) | ( 1 | w | 2 ) 2 | Q μ α , β ; ω ( w ) | ( 1 | w | 2 ) + | Q μ α , β ; ω ( w ) | ( 1 | w | 2 ) 2 ( ln 2 1 | w | 2 ) C .

By (8), then P α + β 1 ; ω (μ) LB ω 2 ( B n ) and P α + β 1 ; ω (μ) B ω ( B n ).

When α,β>1, taking g w (z)= ( 1 z , w ) 1 ( α + β ) , we have g w B ω α , β ( B n ) C.

From Lemma 2.1, we get

| Q μ α , β ; ω ( w ) | ( 1 | w | 2 ) α + β Cfor w B n .

By (8) it is obvious that P α + β 1 ; ω (μ) B ω α , β ( B n ).

This completes the proof of Theorem 3.1. □

4 Compactness of general Toeplitz operators

In this section, we study the compactness of Toeplitz operators on the weighted Bloch-type spaces B ω α , β ( B n ) in the unit ball of C n . We need the following lemma.

Lemma 4.1 Let 0<α<, 0β< and T μ α , β ; ω be a bounded linear operator from B ω α , β ( B n ) into B ω α , β ( B n ). When 0<α<1, 0β<1 and α+β<1, then T μ α , β ; ω is compact if and only if

lim j T μ α , β ; ω f j B ω α , β ( B n ) =0,

whenever ( f j ) is a bounded sequence in B ω α , β ( B n ) that converges to 0 uniformly on B ¯ n .

Proof This lemma can be proved by Montel’s theorem and Lemma 2.1. □

Theorem 4.1 Let μ be a positive Borel measure on B n . We have the following:

(1) if α+β=1, then T μ α , β ; ω is compact on B ω α , β ( B n ) if and only if P α + β 1 ; ω (μ) LB ω ; 0 ( B n ) and μ is a vanishing Carleson measure;

(2) if α=β=1, then T μ α , β ; ω is compact on B ω α , β ( B n ) if and only if P α + β 1 ; ω (μ) B ω ; 0 ( B n ) LB ω ; 0 2 ( B n ) and μ is a vanishing Carleson measure;

(3) if α>1, β>1, then T μ α , β ; ω is compact on B ω α , β ( B n ) if and only if P α + β 1 (μ) B ω ; 0 α , β ( B n ) and μ is a vanishing Carleson measure.

Proof For α+β1, let ( g j ) be a sequence in B ω α , β ( B n ) satisfying g j B ω α , β ( B n ) 1 and g j converges to 0 uniformly as j on B ¯ n . Suppose f A 1 ( B n ). By duality, we have that T μ α , β ; ω is compact on B ω α , β ( B n ) if and only if

lim j sup f A 1 ( B n ) 1 | f , T μ α , β ; ω ( g j ) | =0.

Similarly, as in the proof of Theorem 3.1 for s=α+β1, we have

f , T μ α , β ; ω g j s = c α + β 1 B n f ( z ) g j ( z ) ¯ ( 1 | z | 2 ) α + β 1 d μ ( z ) = c α + β 1 B n [ ( I z , w ; ω P α + β ; ω ) ( f g ¯ j ) ] ( z ) ( 1 | z | 2 ) α + β 1 d μ ( z ) + c α + β 1 B n P α + β ; ω ( f g ¯ j ) ( z ) ( 1 | z | 2 ) α + β 1 d μ ( z ) = J 1 + J 2 .

For fixed 0<ε<1, since μ is a vanishing Carleson measure, there exists 0<η<1 such that

( 1 | z | 2 ) λ B n η B n ( 1 | w | 2 ) λ ( α + β ) | 1 z , w | n + 2 λ + 1 ( α + β ) dμ(w)<ε,

where η B n ={z C n ,|z|<η} and λ(α+β)>1. For a positive constant 0<δ<1, as in the proof of Theorem 3.1, by Proposition 2.1, we obtain

| J 1 | = c α + β 1 c α + β | B n B n ( g j ( z ) ¯ g j ( w ) ¯ ) f ( w ) ( 1 | w | 2 ) α + β ( 1 | z | 2 ) α + β 1 ( 1 z , w ) n + α + 1 ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) d ν ( w ) d μ ( z ) | = c α + β 1 c α + β | B n f ( w ) ( 1 | w | 2 ) α + β × B n ( g j ( z ) ¯ g j ( w ) ¯ ) ( 1 | z | 2 ) α + β 1 ( 1 z , w ) n + α + 1 ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) d μ ( z ) d ν ( w ) | c α + β 1 c α + β B n δ B n | f ( w ) | ( 1 | w | 2 ) α + β × B n | g j ( z ) g j ( w ) | ( 1 | z | 2 ) α + β 1 | 1 z , w | n + α + 1 ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) d μ ( z ) d ν ( w ) + c α + β 1 c α + β δ B n | f ( w ) | ( 1 | w | 2 ) α + β × B n | g j ( z ) g j ( w ) | ( 1 | z | 2 ) α + β 1 | 1 z , w | n + α + 1 ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) d μ ( z ) d ν ( w ) = L 1 + L 2 .

Since g j 0 as j on compact subsets of B n , we can choose j big enough so that

| f ( w ) | ( 1 | w | 2 ) α + β <ε.

Therefore,

L 2 ε C δ B n B n | g j ( z ) g j ( w ) | ( 1 | z | 2 ) α + β 1 | 1 z , w | n + α + 1 ( 1 | φ w ( z ) | 2 ) β ω ( 1 | z | ) d μ ( z ) d ν ( w ) ε C g j B ω α , β ( B n ) .

Now, taking δ such that 1 [ ε ( 1 η ) n + 1 + λ ] 1 λ δ<1, then

L 1 C g j B ω α , β ( B n ) B n δ B n | f ( w ) | ( 1 | w | 2 ) λ B n ( 1 | z | 2 ) λ ( α + β ) | 1 z , w | n + 2 λ + 1 ( α + β ) d μ ( z ) d ν ( w ) C B n δ B n | f ( w ) | ( 1 | w | 2 ) λ B n η B n ( 1 | z | 2 ) λ ( α + β ) | 1 z , w | n + 2 λ + 1 ( α + β ) d μ ( z ) d ν ( w ) + C B n δ B n | f ( w ) | ( 1 | w | 2 ) λ η B n ( 1 | z | 2 ) λ ( α + β ) | 1 z , w | n + 2 λ + 1 ( α + β ) d μ ( z ) d ν ( w ) C ε B n δ B n | f ( w ) | d ν ( w ) + C B n δ B n | f ( w ) | ( 1 δ ) λ ( 1 η ) n + 1 + λ d ν ( w ) C ε f A 1 ( B n ) .

Hence | J 1 |<Cε, where C does not depend on f(z), and so

lim j sup f A 1 ( B n ) 1 | J 1 |=0.

Thus, T μ α , β ; ω is compact on B ω α , β ( B n ) if and only if

lim j sup f A 1 ( B n ) 1 | J 2 |=0.

Again, as in the proof of Theorem 3.1, we have

| J 2 |C B n f A 1 ( B n ) ( 1 | w | 2 ) α + β | g ( w ) | Q μ α , β ; ω (w)dν(w).

From (8) it is easy to see that

(1) if α+β=1 and P α + α 1 ; ω (μ) LB ω , 0 α , β ( B n ), then

lim | w | 1 ( 1 | w | 2 ) Q μ α , β ; ω (w) ( ln 2 1 | w | 2 ) =0;

(2) if α=β=1, P α + α 1 ; ω (μ) B ω ; 0 ( B n ) LB ω ; 0 2 ( B n ), then

lim | w | 1 ( 1 | w | 2 ) Q μ α , β ; ω (w)=0

and

lim | w | 1 ( 1 | w | 2 ) 2 Q μ α , β ; ω (w) ( ln 2 1 | w | 2 ) =0;

(3) if α>1, β>1, and P α + α 1 ; ω (μ) B ω ; 0 α , β ( B n ), then

lim | w | 1 ( 1 | w | 2 ) α + β Q μ α , β ; ω (w)=0.

Combined with g j 0 as j on compact subsets of B n , we have

lim j sup f A 1 ( B n ) 1 | J 2 |=0.

Therefore,

lim j T μ α , β ; ω g j B ω α , β ( B n ) =0,

which implies that T μ α , β ; ω is a compact operator.

Next assume that T μ α , β ; ω is a compact operator on B ω α , β ( B n ). Again, as in the proof of Theorem 3.1, we take

f w (z)= ( 1 | w | 2 ) t ( 1 z , w ) n + t + 1 for t>0.

We know that f w A 1 ( B n ) C. On the other hand, take

g w (z)= ( 1 | w | 2 ) n + 2 + t ( α + β ) ( 1 z , w ) n + t + 1 ; φ w (z)1andω ( 1 | z | ) 1for t>0.

Then g w B ω α , β ( B n ) C and g w 0 uniformly on compact subsets of B n , as |w|1,

| f , T μ α , β ; ω g s | = c α + β 1 </