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Norm of an integral operator on some analytic function spaces on the unit disk

Abstract

If f is an analytic function in the unit disc , a class of integral operators is defined as follows:

I f (h)(z)= 0 z f(w) h (w)dw,hH(D),zD.

The norm of I f on some analytic function spaces is computed in this paper.

MSC:47B38, 32A35.

1 Introduction

Let D={z:|z|<1} be the unit disk of a complex plane . Denote by H(D) the class of functions analytic in . Let denote the normalized Lebesgue area measure in and g(a,z) the Green function with logarithmic singularity at a, i.e., g(a,z)=log| φ a (z)|, where φ a (z)=(az)/(1 a ¯ z) is the Möbius transformation of .

For 0<p<, the Q p is the space of all functions fH(D), for which

f Q p 2 = | f ( 0 ) | 2 + sup a D D | f ( z ) | 2 ( 1 | φ a ( z ) | 2 ) p dσ(z)<.
(1.1)

We know that Q 1 =BMOA, the space of all analytic functions of bounded mean oscillation [1, 2]. For all p>1, the space Q p is the same and equal to the Bloch space , consisting of analytic functions f in such that

f B = | f ( 0 ) | + sup z D | f ( z ) | ( 1 | z | 2 ) <.
(1.2)

See [3, 4] for the theory of Bloch functions.

For α>0, the α-Bloch space, denoted by B α , is the space of all functions f in , for which

f B α = | f ( 0 ) | + sup z D | f ( z ) | ( 1 | z | 2 ) α <.
(1.3)

Obviously, B α 1 B B α 2 for 0< α 1 <1< α 2 <.

For any fH(D), the next two integral operators on H(D) are induced as follows:

I f (h)(z)= 0 z h (w)f(w)dwand J f (h)(z)= 0 z h(w) f (w)dw(zD).

Let M f denote the multiplication operator, that is, M f (h)=fh.

Let fH(D). Then

( I f + J f )h=fhf(0)h(0)= M f (h)f(0)h(0).

If f is a constant, then all results about I f , J f or M f are trivial. In general, f is assumed to be non-constant. Both integral operators have been studied by many authors. See [521] and the references therein.

Norm of composition operator, weighted composition operator and some integral operators have been studied extensively by many authors, see [2234] for example. Recently, Liu and Xiong discussed the norm of integral operators I f and J f on the Bloch space, Dirichlet space, BMOA space and so on in [35].

In this paper, we study the norm of integral operator I f . The norm of I f on several analytic function spaces is computed.

2 Main results

In this section, we state and prove our main results. In order to formulate our main results, we need an auxiliary result which is incorporated in the following lemma.

Lemma 2.1 Let 0<p<1. For any z 0 D, the function

g z 0 (z)= z 0 z 1 z ¯ 0 z z 0
(2.1)

is analytic in and g z 0 Q p =1/ ( p + 1 ) 1 / 2 .

Proof By (1.1) and [[1], Proposition 1, p.109], we have

g z 0 Q p 2 = sup a D D | g z 0 ( z ) | 2 ( 1 | φ a ( z ) | 2 ) p d σ ( z ) = sup b D D ( 1 | φ b ( z ) | 2 ) p d σ ( z ) ,

where b= φ z 0 (a). Taking w= φ b (z), we have

g z 0 Q p 2 = sup b D ( 1 | b | 2 ) 2 D ( 1 | w | 2 ) p | 1 b ¯ w | 4 dσ(w).

Since

1 ( 1 b ¯ w ) 2 = n = 0 Γ ( n + 2 ) n ! Γ ( 2 ) b ¯ n w n = n = 0 Γ ( n + 2 ) n ! b ¯ n w n ,

we have

D ( 1 | w | 2 ) p | 1 b ¯ w | 4 d σ ( w ) = n = 0 + Γ ( n + 2 ) 2 ( n ! ) 2 | b | 2 n D ( 1 | w | 2 ) p | w | 2 n d σ ( w ) = n = 0 + Γ ( n + 2 ) 2 ( n ! ) 2 | b | 2 n 0 1 ( 1 r ) p r n d r = n = 0 + Γ ( n + 2 ) 2 ( n ! ) 2 Γ ( p + 1 ) Γ ( n + 1 ) Γ ( n + p + 2 ) | b | 2 n = n = 0 + Γ ( p + 1 ) Γ ( n + 2 ) 2 n ! Γ ( n + p + 2 ) | b | 2 n .

A simple computation shows

Γ ( p + 1 ) Γ ( n + 2 ) 2 n ! Γ ( n + p + 2 ) = ( n + 1 ) ! ( n + 1 ) ( p + 1 ) ( p + 2 ) ( p + n + 1 ) .

Also, it is easy to see

1 p + 1 n + 1 p + n + 1 ( n + 1 ) ! ( n + 1 ) ( p + 1 ) ( p + 2 ) ( p + n + 1 ) n + 1 p + 1 .

Thus,

g z 0 Q p 2 sup b D ( 1 | b | 2 ) 2 p + 1 n = 0 + (n+1) | b | 2 n = sup b D ( 1 | b | 2 ) 2 p + 1 1 ( 1 | b | 2 ) 2 = 1 p + 1 ,

and

g z 0 Q p 2 sup b D ( 1 | b | 2 ) 2 p + 1 n = 0 + | b | 2 n = sup b D ( 1 | b | 2 ) 2 p + 1 1 1 | b | 2 = 1 p + 1 .

Then the proof is complete. □

First, we consider the norm of I f on Q p , 0<p<1.

Theorem 2.2 Let 0<p<1. If fH(D), then I f is bounded on Q p if and only if f H . Moreover,

I f = f H .

Proof For any h Q p with h Q p =1, it is trivial that I f f H . To prove the converse, define c= sup z D |f(z)|. Given any ϵ>0, there exists z 1 D such that |f( z 1 )|>cϵ. Let h(z)= g z 1 (z)/ g z 1 Q p , where

g z 1 (z)= z 1 z 1 z ¯ 1 z z 1 .

It is easy to see that

h Q p =1, | h ( z 1 ) | ( 1 | z 1 | 2 ) =1/ g z 1 Q p .

Henceforth,

I f 2 I f h Q p 2 = sup a D D | h ( z ) f ( z ) | 2 ( 1 | φ a ( z ) | 2 ) p d σ ( z ) = sup a D D | h ( φ a ( w ) ) f ( φ a ( w ) ) φ a ( w ) | 2 ( 1 | w | 2 ) p d σ ( w ) .

Taking w= re i θ and by the subharmonicity of | h ( φ a ( w ) ) f ( φ a ( w ) ) φ a ( w ) | 2 , we obtain

I f 2 sup a D D | h ( z ) f ( z ) | 2 ( 1 | φ a ( z ) | 2 ) p d σ ( z ) = sup a D 0 1 1 π 0 2 π | h ( φ a ( re i θ ) ) f ( φ a ( re i θ ) ) φ a ( re i θ ) | 2 ( 1 r 2 ) p r d r d θ sup a D | h ( a ) f ( a ) | 2 ( 1 | a | 2 ) 2 2 0 1 ( 1 r 2 ) p r d r = 1 p + 1 sup a D | h ( a ) f ( a ) | 2 ( 1 | a | 2 ) 2 1 p + 1 | h ( z 1 ) f ( z 1 ) | 2 ( 1 | z 1 | 2 ) 2 1 p + 1 | f ( z 1 ) | 2 g z 1 Q p 2 .
(2.2)

By Lemma 2.1 we have

I f | f ( z 1 ) | >cϵ.

Since ϵ is arbitrary, we have I f sup z D |f(z)| and the proof is complete. □

Next, we consider the norm of I f from Q p (0<p<1) to .

Theorem 2.3 Let 0<p<1. If fH(D), then I f is bounded from Q p space to space if and only if f H . Moreover, we have

I f = ( p + 1 ) 1 / 2 f H .

Proof If f H , then (1.2) gives

I f h B = sup z D | f ( z ) h ( z ) | ( 1 | z | 2 ) f H sup z D | h ( z ) | ( 1 | z | 2 ) .

From a part of the proof of estimate (2.2) for f1, we see that

sup z D | h ( z ) | ( 1 | z | 2 ) ( p + 1 ) 1 / 2 h Q p ,

and so

I f h B f H ( p + 1 ) 1 / 2 h Q p .

This leads to

I f ( p + 1 ) 1 / 2 f H .

On the other hand, define c= sup z D |f(z)|. Given any ϵ>0, there exists z 1 D such that |f( z 1 )|>cϵ. Let h(z)= g z 1 (z)/ g z 1 Q p , where

g z 1 (z)= z 1 z 1 z ¯ 1 z z 1 .

This together with Lemma 2.1 gives the following:

I f I f h B = sup z D | f ( z ) h ( z ) | ( 1 | z | 2 ) | f ( z 1 ) h ( z 1 ) | ( 1 | z 1 | 2 ) = | f ( z 1 ) | / g z 1 Q p > ( p + 1 ) 1 / 2 ( c ϵ ) .

Since ϵ is arbitrary, we have

I f ( p + 1 ) 1 / 2 sup z D | f ( z ) | = ( p + 1 ) 1 / 2 f H .

The proof is complete. □

Finally, we consider the norm of the integral operator I f on B α , 0<α<1.

Theorem 2.4 Let 0<α<1 and fH(D). Then the integral operator I f is bounded on B α if and only if f H . Moreover,

I f = f H .

Proof For any h B α with h B α =1, by (1.3) we have

I f h B α = sup z D ( 1 | z | 2 ) α | f ( z ) | | h ( z ) | h B α f H .

This implies I f f H .

Now we need to show the reverse inequality. Define c= sup z D |f(z)|. Given any ϵ>0, there exists z 1 D such that |f( z 1 )|>cϵ. Put

h(z)= Γ ( z ) ( 1 | z 1 | 2 ) α ( 1 z ¯ 1 ζ ) 2 α dζ,
(2.3)

where Γ(z) is any path in from 0 to z, and a single-valued analytic branch is specified. By Theorem 13.11 in [[36], p.274], we know h is an analytic function in and h (z)= ( 1 | z 1 | 2 ) α / ( 1 z ¯ 1 z ) 2 α . Also, it is easy to check h B α =1. In fact,

h B α = sup z D | h ( z ) | ( 1 | z | 2 ) α = sup z D ( 1 | z 1 | 2 ) α | 1 z ¯ 1 z | 2 α ( 1 | z | 2 ) α sup z D ( 1 | z 1 | 2 ) α ( 1 | z | 2 ) α ( 1 | z 1 | | z | ) 2 α 1 .
(2.4)

On the other hand, we have

h B α = sup z D | h ( z ) | ( 1 | z | 2 ) α = sup z D ( 1 | z 1 | 2 ) α | 1 z ¯ 1 z | 2 α ( 1 | z | 2 ) α ( 1 | z 1 | 2 ) α ( 1 | z 1 | 2 ) 2 α ( 1 | z 1 | 2 ) α = 1 .
(2.5)

Hence, the assertion follows by (2.4) and (2.5). Thus

I f I f h B α = sup z D | f ( z ) h ( z ) | ( 1 | z | 2 ) α | f ( z 1 ) h ( z 1 ) | ( 1 | z 1 | 2 ) α | f ( z 1 ) | ( 1 | z 1 | 2 ) α ( 1 | z 1 | 2 ) 2 α ( 1 | z 1 | 2 ) α = | f ( z 1 ) | > c ϵ .

Since the ϵ is arbitrary, the proof is complete. □

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Acknowledgements

The first author is supported by the National Natural Science Foundation of China (No. 11126284). The second author is supported by the project of Department of Education of Guangdong Province (No. 2012KJCX0096).

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Li, H., Li, S. Norm of an integral operator on some analytic function spaces on the unit disk. J Inequal Appl 2013, 342 (2013). https://doi.org/10.1186/1029-242X-2013-342

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