# Norm of a Volterra Integral Operator on Some Analytic Function Spaces

Authors: Hao Li; Songxiao Li
Journal: ISRN Mathematical Analysis
 (2013)

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## Abstract

Letf be an analytic function in the unit disc 𝔻. The Volterra integral operator If is defined as follows: If(h)(z)=∫0zf(w)h'(w)dw,h∈H(𝔻),z∈𝔻. In this paper, we compute the norm of If on some analytic function spaces.


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## Body

## 1. Introduction

Let𝔻={z:|z|<1} be the unit disk of complex plane ℂ and H(𝔻) the class of functions analytic in 𝔻. Denote by dσ the normalized Lebesgue area measure in 𝔻 and g(a,z) the Green function with logarithmic singularity at a; that is, g(a,z)=-log|φa(z)|, where φa(z)=(a-z)/(1-a-z) is the Möbius transformation of 𝔻.Let0<p<∞. The Qp is the space of all functions f∈H(𝔻) such that
(1)∥f∥Qp2=|f(0)|2+supa∈𝔻∫𝔻‍|f′(z)|2(1-|φa(z)|2)pdσ(z)<∞.
From [1, 2], we see that Q1= BMOA, the space of all analytic functions of bounded mean oscillation. When p>1, the space Qp is the same and equal to the Bloch space 𝔅, which consists of all f∈H(𝔻) for which
(2)∥f∥𝔅=|f(0)|+supz∈𝔻|f′(z)|(1-|z|2)<∞.
See [3, 4] for the theory of Bloch functions.Forα>0, the α-Bloch space, denoted by 𝔅α, is the space of all f∈H(𝔻) such that
(3)∥f∥𝔅α=|f(0)|+supz∈𝔻|f′(z)|(1-|z|2)α<∞.
It is clear that 𝔅α1⫋𝔅⫋𝔅α2 for 0<α1<1<α2<∞.Let1⩽q⩽∞ and let 0⩽α⩽1. The mean Lipschitz space Λ(q,α) consists of those functions f∈H(𝔻) for which
(4)∥f∥Λ(q,α)=|f(0)|+sup0⩽r<1(1-r2)1-α×(12π∫02π‍|f′(reiφ)|qdφ)1/q<∞.
It is obvious that Λ(∞,0) is just the Bloch space 𝔅, which is contained in Λ(q,0) for every 1<q<∞. Note that Λ(q,1/q) increases with q∈(1,∞). We refer to [5] for more information of mean Lipschitz spaces.For0⩽s<∞, we say that an f∈H(𝔻) belongs to the growth space Hs∞ if
(5)∥f∥Hs∞=sup0⩽r<1|f(reiθ)|(1-r2)s<∞.
It is easy to see that H0∞=H∞.For-1<α<∞, an f∈H(𝔻) is said to belong to the 𝒟α space if
(6)∥f∥𝒟α2=|f(0)|2+∫𝔻‍|f′(z)|2(1-|z|2)αdσ(z)<∞.For1<p<∞, the Besov space ℬp is defined to be the space of all analytic functions f in 𝔻 such that
(7)∥f∥ℬpp=|f(0)|p+∫𝔻‍|f′(z)|p(1-|z|2)p-2dσ(z)<∞.Letf∈H(𝔻). The Volterra integral operators If and Jf are defined as follows:
(8)If(h)(z)=∫0z‍h′(w)f(w)dw,Jf(h)(z)=∫0z‍h(w)f′(w)dw,(z∈𝔻).
It is easy to see that
(9)(If+Jf)h+f(0)h(0)=Mf(h),
where Mf denotes the multiplication operator; that is, Mf(h)=fh. If f is a constant, then all results about If, Jf, or Mf are trivial. In this paper, we assume that f is a nonconstant. Both operators have been studied by many authors. See [6–23] and the references therein.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, 24–32], for example. Recently, Liu and Xiong studied the norm of integral operators If and Jf on the Bloch space, Dirichlet space, BMOA space, and so on in [13]. In this paper, we study the norm of integral operator If on some function spaces in the unit disk.


## 2. 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.Lemma 1 (see [5, page 144]).
Iff∈Hp  (0<p⩽∞), then |f(z)|⩽(1-|z|2)-1/p∥f∥p, |z|<1, and the inequality is sharp for each fixed z.Lemma 2.
Let-1<α<∞ and 0<p<∞. For any f∈H(𝔻), the following one has:
(10)|f(a)|p⩽(α+1)∫𝔻‍|f(z)|p(1-|a|2)2+α|1-a-z|4+2α(1-|z|2)αdσ(z),
where a is any point in 𝔻.Proof.
For anyf∈H(𝔻), taking z=reiθ and the subharmonicity of |f(z)|p, we get
(11)|f(0)|p⩽12π∫02π‍|f(reiθ)|pdθ,
and so
(12)|f(0)|p⩽(α+1)1π∫01‍∫02π‍|f(reiθ)|phhhhhhhhhhhhhhh×(1-r2)αdθrdr=(α+1)∫𝔻‍|f(z)|p(1-|z|2)αdσ(z).
For any a∈𝔻, let φa(z)=(a-z)/(1-a-z). Replacing f by f∘φa(z) and applying the change of variable formula give the following:
(13)|f(a)|p⩽(α+1)∫𝔻‍|f∘φa(z)|p(1-|z|2)αdσ(z)=(α+1)∫𝔻‍|f(z)|p(1-|φa(z)|2)α(1-|a|2)2|1-a-z|4dσ(z)=(α+1)∫𝔻‍|f(z)|p(1-|a|2)2+α(1-|z|2)α|1-a-z|2α+4dσ(z).
The proof is complete.Theorem 3.
Letf∈H(𝔻). The integral operator If is bounded on Λ(1,1) if and only if f∈H∞. Moreover, one has
(14)∥If∥=∥f∥H∞.Proof.
Iff∈H∞, by (4), we have
(15)∥Ifh∥Λ(1,1)=sup0⩽r<1(12π∫02π‍|f(reiφ)h′(reiφ)|dφ)⩽∥f∥H∞sup0⩽r<1(12π∫02π‍|h′(reiφ)|dφ)⩽∥f∥H∞∥h∥Λ(1,1).
Thus ∥If∥⩽∥f∥H∞.
On the other hand, denotec=supz∈𝔻|f(z)|. Given any ϵ>0, there exists z1∈𝔻 such that |f(z1)|>c-ϵ. Let
(16)h(z)=z1-z1-z-1z-z1.
Then we have ∥h∥Λ(1,1)=1. In fact, taking z1=r1eiφ1 and z=reiφ and using Poisson integral, we get
(17)∥h∥Λ(1,1)=sup0⩽r<1(12π∫02π‍|h′(reiφ)|dφ)=sup0⩽r<1(12π∫02π‍1-r12|1-rr1ei(φ-φ1)|2dφ)=sup0⩽r<11-r121-r2r12=1.
Taking z1=r1eiφ1, we obtain
(18)∞>∥If∥⩾∥Ifh∥Λ(1,1)=sup0⩽r<1(12π∫02π‍|f(reiφ)h′(reiφ)|dφ)=sup0⩽r<1(1-r12|1-r1rei(φ-φ1)|2dφ12π∫02π‍|f(reiφ)|hhhhhhhhhhhhhh×1-r12|1-r1rei(φ-φ1)|2dφ).
So fh′∈H1 and f∈H1. Thus Theorem 2.6 in [5] yields
(19)∥If∥⩾(12π∫02π‍|f(eiφ)|1-r12|1-r1ei(φ-φ1)|2dφ).
By Theorem 2.12 in [5], we have
(20)∥If∥⩾|f(z1)|>c-ϵ,
so the arbitrariness of ϵ gives ∥If∥⩾∥f∥H∞ and the proof is complete.Lemma 3 in [13] gives the norm of If on Dirichlet space. Here, we consider the norm of If on α-Dirichlet space 𝒟α.Theorem 4.
Letf∈H(𝔻) and -1<α<∞. Then If is bounded on 𝒟α if and only if f∈H∞. Moreover, one has
(21)∥If∥=∥f∥H∞.Proof.
First, we assume thatf∈H∞. Let h∈𝒟α. Then (6) gives
(22)∥Ifh∥𝒟α2=∫𝔻‍|f(z)h′(z)|2(1-|z|2)αdσ(z)⩽∥f∥H∞2∫𝔻‍|h′(z)|2(1-|z|2)αdσ(z)⩽∥f∥H∞2∥h∥𝒟α2,
and so we have ∥If∥⩽∥f∥H∞.
Now we need only to show the reverse inequality. Denotec=supz∈𝔻|f(z)|. Given any ϵ>0, there exists z1∈𝔻 such that |f(z1)|>c-ϵ. Let
(23)h1(z)=∫Γ(z)‍(1-|z1|2)1+(α/2)(1-z-1ζ)2+αdζ,
where Γ(z) is any path in 𝔻 from 0 to z. By Theorem 13.11 in [33, page 274], we know h1 is an analytic function in 𝔻 and h1′(z)=(1-|z1|2)1+(α/2)/(1-z-1z)2+α. Also it is easy to check that ∥h1∥𝒟α2=1/(α+1). Indeed, by using the method of the proof of Lemma 4.2.2 in [4], we have
(24)∥h1∥𝒟α2=∫𝔻‍(1-|z1|2)2+α|1-z-1z|4+2α(1-|z|2)αdσ(z)=(1-|z1|2)2+α∫𝔻‍(1-|z|2)α|1-z-1z|4+2αdσ(z)=(1-|z1|2)2+αΓ(α+1)Γ2(α+2)∑n=0∞‍Γ2(n+2+α)n!Γ(n+2+α)|z1|2n=(1-|z1|2)2+αα+1∑n=0∞‍Γ(n+2+α)n!Γ(α+2)|z1|2n=(1-|z1|2)2+αα+11(1-|z1|2)2+α=1α+1.
Let h(z)=h1(z)/∥h1∥𝒟α, and so ∥h∥𝒟α=1. Thus by Lemma 2 we have
(25)∥If∥2≥∥Ifh(z)∥𝒟α2=∫𝔻‍|f(z)h′(z)|2(1-|z|2)αdσ(z)=(α+1)∫𝔻‍|f(z)|2(1-|z1|2)2+α|1-z-1z|4+2α(1-|z|2)αdσ(z)⩾|f(z1)|2>(c-ϵ)2.
Since ϵ is arbitrary, we get
(26)∥If∥⩾∥f∥H∞,
which implies the desired result.Theorem 5.
Letf∈H(𝔻) and let 1<p<∞. The integral operator If is bounded on ℬp if and only if f∈H∞. Moreover, one has
(27)∥If∥=∥f∥H∞.Proof.
Iff∈H∞, then by (7), we have
(28)∥Ifh∥ℬpp=∫𝔻‍|f(z)h′(z)|p(1-|z|2)p-2dσ(z)⩽∥f∥H∞p∫𝔻‍|h′(z)|p(1-|z|2)p-2dσ(z)⩽∥f∥H∞p∥h∥ℬpp,
and so ∥If∥⩽∥f∥H∞.
Now we need only to show the reverse inequality. Denotec=supz∈𝔻|f(z)|. Given any ϵ>0, there exists z1∈𝔻 such that |f(z1)|>c-ϵ. Let
(29)h1(z)=z1-z1-z1z-z1,z∈𝔻.
We see that ∥h1∥ℬpp=1/(p-1). Indeed,
(30)∥h1∥ℬpp=∫𝔻‍(1-|z1|2)p|1-z-1z|2p(1-|z|2)p-2dσ(z)=(1-|z1|2)p∫𝔻‍(1-|z|2)p-2|1-z-1z|2pdσ(z)=(1-|z1|2)pΓ(p-1)Γ2(p)∑n=0∞‍Γ(n+p)2n!Γ(n+p)|z1|2n=(1-|z1|2)pp-1∑n=0∞‍Γ(n+p)n!Γ(p)|z1|2n=(1-|z1|2)pp-11(1-|z1|2)p=1p-1.
Let h(z)=h1(z)/∥h1∥ℬp. Then ∥h∥ℬp=1. Thus by Lemma 2, we have
(31)∥If∥p≥∥Ifh(z)∥ℬpp=∫𝔻‍|f(z)h′(z)|p(1-|z|2)p-2dσ(z)=(p-1)∫𝔻‍|f(z)|p(1-|z1|2)p|1-z-1z|2phhhhhhhhhh×(1-|z|2)p-2dσ(z)⩾|f(z1)|p>(c-ϵ)p.
Since ϵ is arbitrary, we get ∥If∥⩾∥f∥H∞. The proof is complete.Theorem 6.
Letf∈H(𝔻) and let 0<α⩽β<∞. The integral operator If is bounded from 𝔅α to 𝔅β if and only if f∈Hβ-α∞. Moreover, one has
(32)∥If∥=supz∈𝔻|f(z)|(1-|z|2)β-α.Proof.
Iff∈Hβ-α∞, then by (3), we have
(33)∥Ifh∥𝔅β=supz∈𝔻|f(z)h′(z)|(1-|z|2)β⩽∥f∥Hβ-α∞supz∈𝔻|h′(z)|(1-|z|2)α⩽∥f∥Hβ-α∞∥h∥𝔅α.
Hence ∥If∥⩽∥f∥Hβ-α∞.
For the converse, denotec=supz∈𝔻|f(z)|(1-|z|2)β-α. Given any ϵ>0, there exists z1∈𝔻 such that |f(z1)|(1-|z1|2)β-α>c-ϵ. Set
(34)h(z)=∫Γ(z)‍(1-|z1|2)α(1-z-1ζ)2αdζ,
where Γ(z) is any path in 𝔻 from 0 to z. By Theorem 13.11 in [33, page 274], we know that h is an analytic function in 𝔻 and h′(z)=(1-|z1|2)α/(1-z-1z)2α, and it is easy to check that ∥h∥𝔅α=1. Thus
(35)∥If∥⩾∥Ifh∥𝔅β=supz∈𝔻|f(z)h′(z)|(1-|z|2)β⩾|f(z1)h′(z1)|(1-|z1|2)β⩾|f(z1)|(1-|z1|2)α(1-|z1|2)2α(1-|z1|2)β=|f(z1)|(1-|z1|2)β-α>c-ϵ.
Since ϵ is arbitrary, we obtain the desired result. The proof is complete.Theorem 7.
Letf∈H(𝔻). The integral operator If is bounded from Λ(1,1) to 𝔅 if and only if f∈H∞. Moreover, one has
(36)∥If∥=∥f∥H∞.Proof.
Iff∈H∞, then by Lemma 1, we have
(37)∥Ifh∥𝔅=supz∈𝔻|f(z)h′(z)|(1-|z|2)⩽∥h∥Λ(1,1)supz∈𝔻(1-|z|2)-1(1-|z|2)|f(z)|=∥h∥Λ(1,1)supz∈𝔻|f(z)|,
and hence ∥If∥⩽∥f∥H∞.
For the converse, denotec=supz∈𝔻|f(z)|. Given any ϵ>0, there exists z1∈𝔻 such that |f(z1)|>c-ϵ. Let
(38)h(z)=z1-z1-z1z-z1,z∈𝔻.
Then by the proof of Theorem 3, we see that ∥h∥Λ(1,1)=1. In the meantime, we know that |h′(z1)|(1-|z1|2)=1, which gives
(39)∥If∥⩾∥Ifh∥𝔅=supz∈𝔻|f(z)h′(z)|(1-|z|2)⩾|f(z1)h′(z1)|(1-|z1|2)=|f(z1)|>c-ϵ.
Since ϵ is arbitrary, we get the desired result. The proof is complete.Finally, we consider the norm ofIf from Λ(∞,1) to some Banach spaces.Theorem 8.
Iff∈H(𝔻), then the following assertions hold.(1)
Let0<p<∞. The integral operator If is bounded from Λ(∞,1) space to Qp space if and only if f satisfies
(40)supa∈𝔻∫𝔻‍|f(z)|2(1-|φa(z)|2)pdσ(z)<∞.
Moreover, one has
(41)∥If∥=supa∈𝔻∫𝔻‍|f(z)|2(1-|φa(z)|2)pdσ(z).(2)
Let0⩽α⩽1 and let 0⩽q⩽∞. The integral operator If is bounded from Λ(∞,1) space to Λ(q,α) space if and only if f satisfies
(42)sup0⩽r<1(1-r2)1-α(12π∫02π‍|f(reiφ)|qdφ)1/q<∞.
Moreover, one has
(43)∥If∥=sup0⩽r<1(1-r2)1-α(12π∫02π‍|f(reiφ)|qdφ)1/q.(3)
Let0<α<∞. The integral operator If is bounded from Λ(∞,1) space to 𝔅α space if and only if f satisfies supz∈𝔻|f(z)|(1-|z|2)α<∞. Moreover, one has
(44)∥If∥=supz∈𝔻|f(z)|(1-|z|2)α.(4)
Let-1<α<∞. The integral operator If is bounded from Λ(∞,1) space to 𝒟α space if and only if f satisfies ∫𝔻‍|f(z)|2(1-|z|2)αdσ(z)<∞. Moreover, one has
(45)∥If∥=∫𝔻‍|f(z)|2(1-|z|2)αdσ(z).Proof.
The assertion (1) will be proved only here, and the conclusions of (2), (3), and (4) follow by using the similar arguments to that used in proving (1), and so the proofs are omitted.
Ifh∈Λ(∞,1), then by (1), we have
(46)∥Ifh∥Qp2=supa∈𝔻∫𝔻‍|f(z)h′(z)|2(1-|φa(z)|2)pdσ(z)⩽∥h∥Λ(∞,1)2supa∈𝔻∫𝔻‍|f(z)|2(1-|φa(z)|2)pdσ(z),
and so
(47)∥If∥⩽supa∈𝔻∫𝔻‍|f(z)|2(1-|φa(z)|2)pdσ(z).
For the converse, leth(z)=z. It is easy to see that ∥h∥Λ(∞,1)=1. Thus
(48)∥If∥⩾∥Ifh∥Qp2=supa∈𝔻∫𝔻‍|f(z)h′(z)|2(1-|φa(z)|2)pdσ(z)=supa∈𝔻∫𝔻‍|f(z)|2(1-|φa(z)|2)pdσ(z).
The desired result follows by (47) and (48). The proof is complete.



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