Norm of a Volterra Integral Operator on Some Analytic Function Spaces

Let be an analytic function in the unit disc . The Volterra integral operator is defined as follows: In this paper, we compute the norm of on some analytic function spaces.


Introduction
Let 0 < < ∞. The is the space of all functions ∈ (D) such that From [1,2], we see that 1 = BMOA, the space of all analytic functions of bounded mean oscillation. When > 1, the space is the same and equal to the Bloch space B, which consists of all ∈ (D) for which See [3,4] for the theory of Bloch functions.
For 0 ⩽ < ∞, we say that an ∈ (D) belongs to the growth space ∞ if It is easy to see that ∞ 0 = ∞ .
Norms of some special operators, such as composition operator, weighted composition operator, and some integral operators, have been studied by many authors. The interested readers can refer [13,[24][25][26][27][28][29][30][31][32], for example. Recently, Liu and Xiong studied the norm of integral operators and on the Bloch space, Dirichlet space, BMOA space, and so on in [13]. In this paper, we study the norm of integral operator on some function spaces in the unit disk.

Main Results
In this section, we state and prove our main results. In order to formulate our main results, we need some auxiliary results which are incorporated in the following lemmas.
, and the inequality is sharp for each fixed . Lemma 2. Let −1 < < ∞ and 0 < < ∞. For any ∈ (D), the following one has: where is any point in D.
Proof. For any ∈ (D), taking = and the subharmonicity of | ( )| , we get and so For any ∈ D, let ( ) = ( − )/(1 − ). Replacing by ∘ ( ) and applying the change of variable formula give the following: The proof is complete.
Lemma 3 in [13] gives the norm of on Dirichlet space. Here, we consider the norm of on -Dirichlet space D .
Now we need only to show the reverse inequality. Denote = sup ∈D | ( )|. Given any > 0, there exists 1 ∈ D such that | ( 1 )| > − . Let where Γ( ) is any path in D from 0 to . By Theorem 13.11 in [33, page 274], we know ℎ 1 is an analytic function in D and Also it is easy to check that ‖ ℎ 1 ‖ 2 D = 1/( + 1). Indeed, by using the method of the proof of Lemma 4.2.2 in [4], we have Let ℎ( ) = ℎ 1 ( )/‖ℎ 1 ‖ D , and so ‖ℎ‖ D = 1. Thus by Lemma 2 we have Since is arbitrary, we get which implies the desired result.
Theorem 6. Let ∈ (D) and let 0 < ⩽ < ∞. The integral operator is bounded from B to B if and only if ∈ ∞ − . Moreover, one has Proof. If ∈ ∞ − , then by (3), we have where Γ( ) is any path in D from 0 to . By Theorem 13.11 in [33, page 274], we know that ℎ is an analytic function in D and ℎ ( ) = (1 − | 1 | 2 ) /(1 − 1 ) 2 , and it is easy to check that ‖ℎ‖ B = 1. Thus Since is arbitrary, we obtain the desired result. The proof is complete.
Finally, we consider the norm of from Λ(∞, 1) to some Banach spaces.