β Hausdorff Operator on Lipschitz Space in the Unit Polydisk

http://dx.doi.org/10.4236/apm.2014.45022

How to cite this paper: Hu, R. and Zhang, C.F. (2014) β-Hausdorff Operator on Lipschitz Space in the Unit Polydisk.

Ad-vances in Pure Mathematics, 4, 171-175. http://dx.doi.org/10.4236/apm.2014.45022

β

-Hausdorff Operator on Lipschitz Space

in the Unit Polydisk

*

Rong Hu, Chaofeng Zhang#

School of Mathematics and Finance-Economics, Sichuan University of Arts and Science, Dazhou, China

Email: #[email protected]

Received 27 February 2014; revised 27 March 2014; accepted 2 April 2014

Copyright © 2014 by authors and Scientific Research Publishing Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

Abstract

In this paper, we define β-Hausdorff operator on the unit polydisk and study the boundedness of the operator on Lipschitz space. Firstly, we translate the problem of coefficient into integral of weighted composition operator, then give the sufficient conditions of boundedness, and also ob-tain an upper bound for the operator norm on Lipschitz space.

Keywords

Unit Polydisk, Lipschitz Space, β-Hausdorff Operator, Weighted Composition Operator

1. Introduction

Let ∆ be the forward difference operator defined on sequences

{ }

µn n ₀

∞

= by ∆µn=µn−µn+1. Let operator

k Fβ

be

(

1

)

0

, , , , 1.

k k

n n n n

F_βµ = ∆ F_β−µ F_βµ =µ_β k∈ β∈β ≥

Define the β-Hausdorff matrix Hµβ_n as the lower triangular matrix

( )

cn k,

β

with entries

, , .

n k

n k k

n

c F k n

k β

β− µ

 

=_{ } ≤

 

For β =1, it is the Hausdorff matrix H

( )

µn , see [1].

When µn is the moment sequence of a measure i.e.

( )

1

0 d ,

n

n t t n

µ =

_∫

µ ∈, the matrix arising from a Borel measure µ is denoted by H_µβ, a simple calculation then gives

(

)

( )

1

, ₀ 1 d , .

n k k

n k

n

c t t t k n

k

β   β − _µ

=_{ } − ≤

 

∫

Let Un be the unit polydisk in the complex vector space _n

, H U

( )

n be the space of all holomorphic functions on n

U , and µ =i,i 1,,n be the Borel measures on

( )

0,1 , μ=

(

µ µ1, 2,,µn

)

, d

( )

1d

( )

n j j j

t t

µ = Π = µ .

In [2], the Lipschitz space Lip_α

( )

Un

(

0≤ <α 1

)

is defined on n U by

( )

n

{

( )

n and <

}

Lipα U = f f∈H U f α ∞

where

( )

( )

(

)

1 2 1

0 sup 1

n

n

i i

z U i

f

f f z z

z

α α

−

= ∈

∂

= + −

∂

∑

. It is easy to prove that Lip_α

( )

Un is a Banach space under the norm ._α.

Let f ∈H U

( )

n , suppose

( )

1

1 1 1

, , , , 0

n n n n

m m m m m m

f z a z z

≥

=

∑

_

 

, and H _i

(

cn k,

( )

i

)

,i 1, 2, ,n

β β

µ = µ =  be the β-

Hausdorff matrices arising by Borel measures µi. The β-Hausdorff operator

(

, 1

)

β _{β∈ β ≥}

μ 

 is defined as

follows:

( )( )

( )

1

1 1

, , , 1

, , 00 , 1, , 1

n

n j j

n j j

n

m m

i i m i j n

m m i m j n j

f z a c z z

β β _µ

≥ ≤ ≤ = =

=

∑

∑

_∏

μ 

  

 . For β =1, we obtain the classical

Hausdorff operator _μ, see [3].

Hausdorff matrix and Hausdorff operator have studied on various space of holomorphic functions, see, e.g., [3]-[9]. In [3], the author obtained that the Hausdorff operator μ is bounded on Hardy space Hp

(

1≤ p<∞

)

,

and in [4] we showed that this conclusion cannot be extended to the Bloch space directly. Then we try to study on the Lipschitz space, found that when the measure is common Lebesgue measure dt, the Hausdorff operator

t

 is unbounded on Lipschitz space Lip_α

( )

Un , see the remark. In this paper, we study the operator which is got by amending the Hausdorff operator and called it β-Hausdorff operator. The results of this paper can be deemed as a continuation of the results in [3] on Lipschitz space.

2. Main Results

The main results in this paper is the following:

Theorem 1 Let µi

(

i=1,,n

)

be finite Borel measures on (0,1) and H i

(

cn k,

( )

i

)

,i 1, ,n

β β

µ = µ =  be cor-

responding β-Hausdorff matrices, _μβ be β-Hausdorff operator. For β≥2, _μβ is bounded on Lip_α

( )

Un if

, < ) ( d ) (1

1 1

1 1) 2( 1

= , 1 =

[0,1]

∏

∑

− − − ∞

∫

t

t t

t _{k l}

n

l k j n

j

n β α µ

In this case, the operator norm satisfies

( )

(

)

[ ]0,1

₍

_{2( 1)}

₎

1

( )

, 1

1

1 1

0,1 d .

1 n

n n n

k l j j

k l

C t

t _{t t}

β

α β

µ ₋ µ

− =

=

≤ +

−

∑

∏

∫

μ



for some constant C.

In order to prove the main results, we need some auxiliary result.

Lemma 1 [2] Let f∈Lipα

( )

Un ,then

( )

( )

(

)

1 2 1

0 in 1 i , n

f z f f z z U

α α

− =

≤ +

∑

− ∈ .

For each t_j∈

( )

0,1 , we note the functions j t

β

φ given by

( ) ( )

1 1

j

j j t j

j j

t z z

t z

β β

φ =

− + ,

Lemma 2 Let µ =i,i 1,,n be finite Borel measures on

( )

0,1 and H i

(

cn k,

( )

i

)

β β

µ = µ be corresponding β-Hausdorff matrices. Suppose

( )

1

( )

1 1

, , 1 , , 0

. n

n n

m

m n

m m n

m m

f z a z z Lip_α U

≥

=

∑

_ ∈

Then,

(a) The power series _μβ

( )

f in (2) represents a holomorphic functions on Un; (b) _μβ

( )

f can be written in terms of weighted composition operators as follows:

( )( )

_{[ ]}

(

)

(

)

(

1

( )

1

( )

)

( )

0,1 1

1

, ,

1 1

n n t t_n n

j j

j

f z f z z t

t z

β _φβ _φβ _µ

= =

+ −

∫ ∏

μ 

 . For each z∈Un.

Proof (a) Let

( )

1, , 1, , ,

0 , 1, , 1

n n j j

j j

n

m m i i m i j

i m j n j

A a cβ µ

≤ ≤ = =

=

∑

_∏

 



. Since Lip_α

( )

Un the sequence of Taylor coeffi- cients of f is bounded by a constant M <∞, then

( )

(

)

( )

( )

1 1

1

, , , , , ₀

0 , 1, , 1 1 0 1

1 j j 0,1 .

j

n n j j

j j j

m j

n n n

m i i

j

m m i i m i j j j j j j

i m j n j j i j j

m

A a c M t t t M

i

β

β _µ − _{µ µ}

≤ ≤ = = = = =

 

≤ ≤ _{ } − ≤

 

∑

∏

∏

∫

∑

∏

 



Hence the coefficients of the series (2) are bounded and consequently _μβ

( )

f is defined and analytic on n

U .

(b) By the Schwarz lemma we have t_j

( )

zj zj

β

φ ≤ for each t_j∈

( )

0,1 . Hence applying (3) we have

( )

( )

(

1

)

(

)

( )

(

)

1 2

1 1

0 1, 1, , , 1, , 1

, , , , 0 1 ,

sup sup

n

i i i

n

t t n n i

t i n z i n i

f z z f f f z

α

β β

α ζ

φ φ ζ ζ −

< < = ≤ = =

≤ ≤ +

∑

−

   

On the other hand,

(

)

(

)

₁

(

)

1

1 1

, 1

1 1

n n

j j j

j j

z

t z

= =

≤ −

+ −

∏

∏

Hence

( )

_{[ ]}

(

)

(

)

(

1

( )

1

( )

)

( )

0,1

1

1

, ,

1 1

n n t t_n n

j j

j

G z f z z t

t z

β _φβ _φβ _µ

= =

+ −

∫

∏



is finite and analytic on n U .

Now we proof _μβ

( )

f =Gβ

( )

z , in order to avoid tedious calculations, we may assume that n=2, For a fixed z=

(

z z1, 2

)

∈U2, we have

( )

_{[ ]}

(

)

(

)

(

( )

( )

)

( )

[ ]

( ) ( )

(

)

(

)

(

(

)

)

( )

(

)

(

)

(

(

( )

)

)

2 _{1 2}

2

1 2

2 0,1

1

1 2 1 1 2 2

, 1 2 1 1

0,1 , 0

1 1 2 2

1 _{1 1 1} 1 _{2 2 2}

, 1 2 0 1 0 1

, 0

1 1 2 2

1

,

1 1

1 1 1 1

.

1 1 1 1

t t

j j

j

k l k l

k l k l

k l

k l

k l

k l k l

k l

G z f z z t

t z

t t t t

a z z

t z t z

t t t t

a z z

t z t z

β β β

β β

β β

φ φ µ

µ µ

µ µ

=

+ +

≥

+ +

≥ =

+ −

=

+ − + −

=

+ − + −

∫

∏

∑

∫

∑

∫

∫

It easy to see that

( )

(

)

(

)

(

)

( )

( )

1 1

, 1

0 0 1 ,

1 1

k

n k

j j j k n k n k

j j j j j n k j j

k

n k n k

j j

t t n

t t z t c z

k

t z

β

β β

µ

µ µ

∞ ₋ ∞

− −

+

= =

 

= _{ } − =

 

+ −

∑

∑

∫

∫

Hence,

( )

, 1 2 ,

( )

1 1 ,

( )

2 2 , ,

( )

1 ,

( )

2 1 2

, 0 , 0 0 0

.

n m

k l n k m l n m

k l n k m l k l n k m l

k l n k m l m n k l

Gβ z a z z cβ µ z cβ µ z a cβ µ cβ µ z z

∞ ∞

− −

≥ = = ≥ = =

 

= = _{ }

 

Denote Tt

β

as follows

( )( )

(

)

(

)

(

1

( )

1

( )

)

1

1

, , ,

1 1

n

t n t t n

j j

j

T f z f z z

t z

β _φβ _φβ

= =

+ −

∏



where

( )(

1

)

i

t zi i n

β

φ ≤ ≤ is defined in (4).

Now we obtain estimates for the norms of the weighted composition operator Tt

β

. Lemma 3 Suppose β≥2, then Tt

β

is bounded on Lip_α

( )

Un . Further more, there is a constant C> 0 such that

( )

(

)

1 2 1 , 1 1 1 1 1 n n t k l j j k l T C

t _{t t}

β

α β− − =

= ≤

−

∑

∏

. For each t_i∈

( )

0,1 ,i=1,,n.

Proof Let Tt

( )( ) ( )

f z t z f t

( )

z

β _{= φ}β

, in which

( )

(

)

(

)

1 1 1 1 t n j j j z t z = = + −

∏

, t

( )

z

(

t1

( )

z1 , , tn

( )

zn

)

β β β

φ = φ φ ,

and the function

( )(

1

)

i

t zi i n

β

φ ≤ ≤ is defined in (4).

( )

(

)

_{( )}

₍

₎

( )

₍

_{( )}

₎

₍

₎

_{( )}

₍

_{( )}

₎

( )

₍

₎

( )

₍

_{( )}

₎

₍

₎

_{( )}

( )

(

)

( )

1 2 1 1 1 2 2

1 , 1

1 2

1 1

2 ₂

2

1 1 , 1

, 1 1

1

1 1

1

2 1 1

1 1 1 2 1 l l l l j n t k k k n n t l t

t k t t k

k k k l l k

n n n _k

t l t

t l k t

k k l k l k

t l n n k l j k T f z z z z z f

f z z z z z

z z

z z

z

f z z z f

z z _z

t t β α β α α β β α β α α β

α _β α

ϕ φ φ ω φ φ φ − = − − = = − − − = = = = = ∂ − ∂ ∂ ∂ ∂ ≤ − + − ∂ ∂ ∂  ₋  ∂ ∂   _{ } ≤ _ − _ − + _{ } ∂   ∂ _ − _ ≤

∑

∑

∑

∑

∑

∑

∑

∏

(

( )

)

1

2 1 2

1 1 1 . j n n l k k j

t t f

t α β β α − − − = =   − +     

∏

∑



and Tt

( )( )

f 0 f

( )

0 f

β

α

= ≤ . Hence we obtain that

( )

_{( )}

(

)

1 2 1 , 1 1 1 1 . 1 n n t k l j j k l

T f C f

t _{t t}

β

α α

α ≤

∏

₌

∑

_{= −} β− −

Now we proof the main results.

The Proof of Theorem 1 For each z∈Un,f ∈Lip_α

( )

Un , by (5) we can obtain

( )

(

)

_{( )}

₍

₎

[ ]

( )( ) ( )

(

)

[ ]

(

( )

)

( )

(

)

( )

[ ]

( )

( )

1 1 2 2 0,1 1 1 1 2 0,1 0,1 1 1 1 1 . n n n n n

k t k

k k k k

n

t k t

k k

f

z z T f z t z

z z

T f z z t T f t

z

β

α _β α

α β β α µ µ µ − − = = − = ∂ _∂ − = − ∂ ∂ ∂ ≤ − ≤ ∂

∑

∑

∫

∑

∫

∫

μ 

Then by (1) and (6),

( )

( )( )

(

( )

)

( )

(

)

( )

(

)

[ ]

₍

_{( )}

₎

( )

1 2 1 1

0,1 _{2 1}

, 1 1

0 sup 1

1 1

0,1 d .

1 n n n k k

z U k

n n n k l j j k l f

f f z z

z

f C f t

t _{t t}

β

α

β β

α

α

α α _β

µ µ − = ∈ − − = = ∂ = + − ∂ ≤ + −

∑

∑

∏

∫

μ μ μ   

from which the result follows.

Remark When the Borel measure µ

( )

t is the common Lebesgue measure dt, the Hausdorff operator arising from measure dt is denoted as t. t is bounded on Hardy space

( )

(

1 <

)

p n

However, it is unbounded on Lipschitz space. For example, fix k

(

1≤ ≤k n

)

, and let f z

( )

=ln 1

(

−z_k

)

, it is easy to see that f ∈Lip U

( )

n , then

( )( )

2

(

)

(

)

1

1 1

1 ln 1

ln 2

n

k j

j

k j

j k

f z z z

z = z

≠

 

 

= −  − _− _ −

 

∏

 

t



From this it follows that

( )

(

( )

)

( )

(

)

( )

(

)

(

)

(

)

(

)

2

1 1

2 2

2 0,1

1 ln 1

ln

1 1 .

sup sup

1 2

n

k

k

z r z U k

f r r

f z z C r

z r r r

α α

α

− −

= ∈ ∈

∂ − −

≥ − ≥ + − = ∞

∂ −

t t

 

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