# Extended Cesàro Operators from Logarithmic-Type Spaces to Bloch-Type Spaces

Authors: Dinggui Gu
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

The boundedness and compactness of the extended Cesàro operator
from logarithmic-type spaces to Bloch-type spaces on the unit ball are
completely characterized in this paper.


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## Body

## 1. Introduction

Let𝔹n be the unit ball of ℂn,∂𝔹n the unit sphere of ℂn, H(𝔹n) the space of all holomorphic functions in 𝔹n, and H∞ the space of all bounded holomorphic functions on 𝔹n. For f∈H(𝔹n), let
(1.1)ℜf(z)=∑j=1n‍zj∂f∂zj(z)
denote the radial derivative of f∈H(𝔹n).A positive continuous functionμ on the interval [0,1) is called normal if there is δ∈[0,1) and s and t, 0<s<t such that
(1.2)μ(r)(1-r)sisdecreasingon[δ,1),limr→1μ(r)(1-r)s=0,μ(r)(1-r)tisincreasingon[δ,1),limr→1μ(r)(1-r)t=∞.
If we say that μ:𝔹n→[0,∞) is normal we will also assume that μ(z)=μ(|z|),z∈𝔹n.Letμ:𝔹n→[0,∞) be normal. The Bloch-type space ℬμ=ℬμ(𝔹n) is the space of all functions f∈H(𝔹n) such that
(1.3)bμ(f)=supz∈𝔹nμ(z)|ℜf(z)|<∞.ℬμ is Banach space with the norm ∥f∥ℬμ=|f(0)|+bμ(f). The little Bloch-type space ℬμ,0=ℬμ,0(𝔹n) consists of all f∈H(𝔹n) such that
(1.4)lim|z|→1μ(z)|ℜf(z)|=0.
It is easy to see that ℬμ,0 is a closed subspace of ℬμ. When μ(r)=(1-r2)α, α∈(0,∞), we obtain so-called α-Bloch spaces and little α-Bloch spaces, respectively, which for α=1 are reduced to classical Bloch spaces (see, e.g., [1–4] and the references therein). When μ(r)=(1-r2)ln(e/(1-r2)), we obtain the logarithmic Bloch space ℒℬ=ℒℬ(𝔹n) and the little logarithmic Bloch space ℒℬ0=ℒℬ0(𝔹n), respectively, (see [5]). It was shown that f is a multiplier of ℬ if and only if f∈H∞ and f∈ℒℬ in [6].Anf∈H(𝔹n) is said to belong to the logarithmic-type space Hlog∞, if
(1.5)∥f∥Hlog∞=supz∈𝔹n|f(z)|ln(e/(1-|z|2))<∞.
It is easy to see that Hlog∞ becomes a Banach space under the norm ∥·∥Hlog∞, and that the inclusions H∞⊂ℬ⊂Hlog∞, hold. For some information of the space Hlog∞ see [7, 8].Letg∈H(𝔹n). The extended Cesàro operator on H(𝔹n) is defined by
(1.6)Tgf(z)=∫01f(tz)ℜg(tz)dtt‍,f∈H(𝔹n),z∈𝔹n.
This operator is a natural extension of a one-dimensional operator defined in [9]. Some other results on the one-dimensional operator can be found, for example, in [10, 11] (see also the references therein). For some extensions of operator (1.6) on the unit disk see [12–16]. On related operators on the unit polydisk see, for example, [17–21] and references therein. The boundedness and compactness of operator (1.6) between various spaces of holomorphic functions has been extensively studied recently, see, [17, 22–36]. For some integral operators on spaces of harmonic functions see, for example, [37] as well as the references therein. A new extension of operator (1.6) in the unit ball case have been recently introduced by Stević in [38] (see also [5, 39]).In this paper, we study the extended Cesàro operator fromHlog∞ to Bloch-type spaces ℬμ and ℬμ,0. Sufficient and necessary conditions for the extended Cesàro operator Tg to be bounded and compact are given.Throughout the paper, constants are denoted byC, they are positive and may not be the same in every occurrence.


## 2. Main Results and Proofs

In this section, we give our main results and their proofs. Before stating these results, we need some auxiliary results, which are incorporated in the lemmas which follows.Lemma 2.1.
Assume thatg∈H(𝔹n) and μ:𝔹n→[0,∞) are normal. Then Tg:Hlog∞→ℬμ is compact if and only if Tg:Hlog∞→ℬμ is bounded and for any bounded sequence (fk)k∈ℕinHlog∞ which converges to zero uniformly on compact subsets of 𝔹nask→∞, one has ∥Tgfk∥ℬμ→0 as k→∞.The proof of Lemma2.1 follows by standard arguments (see, e.g., Lemmas 3 in [20, 21, 29]). Hence, we omit the details.Lemma 2.2.
Assume thatμ:𝔹n→[0,∞) is normal. A closed set K in ℬμ,0 is compact if and only if it is bounded and satisfies
(2.1)lim|z|→1supf∈Kμ(z)|ℜf(z)|=0.This lemma can be found in [5], and its proof is similar to the proof of Lemma 1 in [40]. Hence, it will be omitted.The following result was proved in [8].Lemma 2.3.
There exist two functionsf1,f2∈Hlog∞(𝔹1) such that
(2.2)|f1(z)|+|f2(z)|≥ln11-|z|,z∈𝔹1.Now we are in a position to state and prove our main results.Theorem 2.4.
Assume thatg∈H(𝔹n) and μ:𝔹n→[0,∞) is normal. Then Tg:Hlog∞→ℬμ is bounded if and only if
(2.3)M=supz∈𝔹nμ(z)|ℜg(z)|lne1-|z|2<∞.
Moreover, if Tg:Hlog∞→ℬμ is bounded then the following asymptotic relation holds
(2.4)∥Tg∥Hlog∞→ℬμ≍supz∈𝔹nμ(z)|ℜg(z)|lne1-|z|2<∞.Proof.
Assume that (2.3) holds. Then, for any f∈Hlog∞, we have
(2.5)μ(z)|ℜ(Tgf)(z)|=μ(z)|ℜg(z)||f(z)|≤μ(z)|ℜg(z)|(lne1-|z|2)∥f∥Hlog∞.
In addition, it is easy to see that Tgf(0)=0. Therefore we have
(2.6)∥Tgf∥ℬμ=supz∈𝔹nμ(z)|ℜ(Tgf)(z)|≤M∥f∥Hlog∞
as desired.
Conversely, assume thatTg:Hlog∞→ℬμ is bounded. For a∈𝔹n, set
(2.7)fa(z)=lne1-〈z,a〉.
It is easy to see that fa∈Hlog∞ and supa∈𝔹n∥fa∥Hlog∞<∞.
For anyb∈𝔹n, we have
(2.8)∞>∥Tgfb∥ℬμ=supz∈𝔹nμ(z)|ℜ(Tgfb)(z)|=supz∈𝔹nμ(z)|ℜg(z)||fb(z)|≥μ(b)|ℜg(b)|lne1-|b|2,
from which (2.3) follows, moreover
(2.9)supz∈𝔹nμ(z)|ℜg(z)|lne1-|z|2≤C∥Tg∥Hlog∞→ℬμ.
From (2.6) and (2.9), we see that (2.4) holds. The proof is completed.Theorem 2.5.
Assume thatg∈H(𝔹n) and μ:𝔹n→[0,∞) is normal. Then Tg:Hlog∞→ℬμ is compact if and only if
(2.10)lim|z|→1μ(z)|ℜg(z)|lne1-|z|2=0.Proof.
Suppose thatTg:Hlog∞→ℬμ is compact. Let (zk)k∈ℕ be a sequence in 𝔹n such that limk⇀∞|zk|=1. Set
(2.11)fk(z)=(lne1-〈z,zk〉)2(lne1-|zk|2)-1,k∈ℕ.
It is easy to see that supk∈ℕ∥fk∥Hlog∞<∞. Moreover fk→0 uniformly on compact subsets of 𝔹n as k→∞. By Lemma 2.1,
(2.12)limk→∞∥Tgfk∥ℬμ=0.
In addition,
(2.13)∥Tgfk∥ℬμ=supz∈𝔹nμ(z)|ℜg(z)fk(z)|≥μ(zk)|ℜg(zk)|lne1-|zk|2,
which together with (2.12) implies that
(2.14)limk→∞μ(zk)|ℜg(zk)|lne1-|zk|2=0.
From the above inequality we see that (2.10) holds.
Conversely, assume that (2.10) holds. From Theorem 2.4 we see that Tg:Hlog∞→ℬμ is bounded. In order to prove that Tg:Hlog∞→ℬμ is compact, according to Lemma 2.1, it suffices to show that if (fk)k∈ℕ is a bounded sequence in Hlog∞ converging to 0 uniformly on compact subsets of 𝔹n, then
(2.15)limk→∞∥Tgfk∥ℬμ=0.
Let(fk)k∈ℕ be a bounded sequence in Hlog∞ such that fk→0 uniformly on compact subsets of 𝔹n as k→∞. By (2.10) we have that for any ε>0, there is a constant δ∈(0,1), such that
(2.16)μ(z)|ℜg(z)|lne1-|zk|2<ε
whenever δ<|z|<1. Let K={z∈𝔹n:|z|≤δ}. From (2.10) we see that g∈ℬμ. Equality (2.16) along with the fact that g∈ℬμ implies
(2.17)∥Tgfk∥ℬμ=supz∈𝔹nμ(z)|ℜ(Tgfk)(z)|=supz∈𝔹nμ(z)|ℜg(z)fk(z)|≤(sup{z∈𝔹n:|z|≤δ}+sup{z∈𝔹n:δ≤|z|<1})μ(z)|ℜg(z)||fk(z)|≤∥g∥ℬμsupz∈K|fk(z)|+sup{z∈𝔹n:δ≤|z|<1}μ(z)|g(z)|(lne1-|z|2)∥fk∥Hlog∞≤∥g∥ℬμsupz∈K|fk(z)|+Cε.
Observe that K is a compact subset of 𝔹n, so that
(2.18)limk→∞supz∈K|fk(z)|=0.
Therefore
(2.19)limsupk→∞∥Tgfk∥ℬμ≤Cε.
Since ε>0 is an arbitrary positive number it follows that the last limit is equal to zero. Therefore, Tg:Hlog∞→ℬμ is compact. The proof is completed.Remark 2.6.
From [24] and Theorems 2.4 and 2.5, we see that Tg:Hlog∞→ℬ is bounded if and only if Tg:ℬ→ℬ is bounded; Tg:Hlog∞→ℬ is compact if and only if Tg:ℬ→ℬ is compact.Theorem 2.7.
Assume thatg∈H(𝔹n) and μ:𝔹n→[0,∞) is normal. Then the following statements are equivalent:(a)
Tg:Hlog∞→ℬμ,0 is bounded;(b)
Tg:Hlog∞→ℬμ,0 is compact;(c)
(2.20)lim|z|→1μ(z)|ℜg(z)|lne1-|z|2=0.Proof.
(b)⇒(a). This implication is obvious.
(a)⇒(c). Assume that Tg:Hlog∞→ℬμ,0 is bounded. Now we prove that (2.20) holds. Note that (2.20) is equivalent with
(2.21)lim|z|→1μ(z)|ℜg(z)|ln11-|z|=0.
Hence we only need to show that (2.21) holds. This can be done by contradiction. Now assume that the condition (2.21) does not hold. If it was, then it would exist ε0>0 and a sequence (z(j))j∈ℕ∈𝔹n, such that z(j)→∂𝔹n, and
(2.22)μ(z(j))|ℜg(z(j))|ln11-|z(j)|≥ε0>0
for sufficiently large j. We may assume that limj→∞z(j)=(1,0,…,0) and also
(2.23)1-|z(j)|>1-|z1(j)|2,j∈ℕ.
According to Lemma 2.3 we know that there exist two functions h1,h2∈Hlog∞(𝔹1) such that
(2.24)|h1(z1)|+|h2(z1)|≥ln11-|z1|,z1∈𝔹1.
Let
(2.25)F1(z)=h1(z1),F2(z)=h2(z1),z=(z1,…,zn)∈𝔹n.
Then clearly F1,F2∈Hlog∞. By the boundedness of Tg:Hlog∞→ℬμ,0 we have
(2.26)TgF1,TgF2∈ℬμ,0.
On the other hand,
(2.27)μ(z(j))(|ℜ(TgF1)(z(j))|+|ℜ(TgF2)(z(j))|)=μ(z(j))|ℜg(z(j))|(|F1(z(j))|+|F2(z(j))|)=μ(z(j))|ℜg(z(j))|(|h1(z1(j))|+|h2(z1(j))|)≥μ(z(j))|ℜg(z(j))|ln11-|z1(j)|≥Cμ(z(j))|ℜg(z(j))|ln11-|z(j)|≥Cε0>0
for sufficiently large j. Since z(j)→∂𝔹n, from the above inequality we obtain that TgF1,TgF2∉ℬμ,0, which is a contradiction.
(c)⇒(b). From (1.5) we have that
(2.28)μ(z)|ℜ(Tgf)(z)|≤μ(z)|ℜg(z)|(lne1-|z|2)∥f∥Hlog∞.
Taking the supremum in the above inequality over all f∈Hlog∞ such that ∥f∥Hlog∞≤1, then letting |z|→1, by (2.20) we arrive at
(2.29)lim|z|→1sup∥f∥Hlog∞≤1μ(z)|ℜ(Tg(f))(z)|=0.
From this and by employing Lemma 2.2, we see that Tg:Hlog∞→ℬμ,0 is compact. The proof is completed.From Theorems2.4, 2.5, and 2.7, we have the following corollary.Corollary 2.8.
Letg∈H(𝔹n). Then(1)
Tg:Hlog∞→ℒℬ is bounded if and only if
(2.30)supz∈𝔹n(1-|z|2)|ℜg(z)|(lne1-|z|2)2<∞.(2)
Tg:Hlog∞→ℒℬ is compact if and only if Tg:Hlog∞→ℒℬ0 is bounded if and only if Tg:Hlog∞→ℒℬ0 is compact if and only if
(2.31)lim|z|→1(1-|z|2)|ℜg(z)|(lne1-|z|2)2=0.



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