ON AN INTEGRAL OPERATOR ON THE UNIT BALL IN
C
STEVO STEVI ´CReceived 23 December 2003
LetH(B) denote the space of all holomorphic functions on the unit ballB⊂Cn. In this paper, we investigate the integral operatorTg(f)(z)=
1
0 f(tz)g(tz)(dt/t), f ∈H(B), z∈B, whereg∈H(B) andg(z)=nj=1zj(∂g/∂zj)(z) is the radial derivative ofg. The operator can be considered as an extension of the Ces`aro operator on the unit disk. The boundedness of the operator ona-Bloch spaces is considered.
1. Introduction
LetUbe the unit disc in the complex planeCandH(U) the space of all analytic functions inU.
For each complexγwith Reγ >−1 andknonnegative integer, letAγkbe defined as the kth coefficient in the expression
1 (1−x)γ+1 =
∞
k=0
Aγkxk, (1.1)
so thatAγk=(γ+ 1)···(γ+k)/k!.
For an analytic function f(z)=∞n=0anznonU, the generalized Ces`aro operator is defined by
Ꮿγ(f)(z)=∞ n=0
1 Aγn+1
n
k=0 Aγn−kak
zn. (1.2)
Forγ=0 we obtain the Ces`aro operator onU. The boundedness of the operator on some spaces of analytic functions was considered by a number of authors, see, for exam-ple, [8,10,13], and the references therein.
The integral form ofᏯ0=Ꮿis
Ꮿ(f)(z)=1
z z
0 f(ζ) 1 (1−ζ)dζ=
1 z
z
0 f(ζ)
ln 1
(1−ζ)
dζ, (1.3)
82 On an integral operator on the unit ball inC
or, taking simply as a path the segment joining 0 andz,
Ꮿ(f)(z)=
1
0 f(tz)
ln 1
(1−ζ)
ζ=tz
dt. (1.4)
On most holomorphic function spaces the boundedness of the previous operator is equiv-alent to the boundedness of the operator
zᏯ(f)(z)=
z
0 f(ζ)
1−ζ dζ. (1.5)
Hence, Aleman and Siskakis [2] have introduced and investigated the following natural generalization of operator (1.5):
Tgf(z)=
z
0 f(ζ)g
(ζ)dζ. (1.6)
In [1,2,3] were investigated the boundedness and the compactness of the operator on Hardy and Bergman spaces. A natural question is to define a similar integral operator which acts onH(B) (the space of all holomorphic functions in the unit ballB).
Letz=(z1,. . .,zn) andw=(w1,. . .,wn) be points in complex vector spaceCnand
z,w =z1w¯1+···+znw¯n. (1.7)
LetdVNstand for the normalized Lebesgue measure onCn. For a holomorphic function
f we denote
∇f =
∂ f ∂z1,. . .,
∂ f ∂zn
. (1.8)
Letf(z)=n
j=1zj(∂ f /∂zj)(z) stand for the radial derivative of f ∈H(B) (see [7]). It is easy to see that if f ∈H(B), f(z)=αaαzα, whereαis a multi-index, then
f(z)= α
|α|aαzα. (1.9)
Leta >0. Thea-Bloch spaceᏮa=Ꮾa(B) is the space of all f ∈H(B) such that
ba(f)=sup z∈B
1− |z|2 af(z)<∞. (1.10)
The littlea-Bloch spaceᏮa0=Ꮾ0(a B) consists of all f ∈H(B) such that
lim |z|→1
1− |z|2 af(z)=0. (1.11)
OnᏮathe norm is introduced by
With this normᏮa is a Banach space andᏮa
0 is a closed subspace of Ꮾa. If a=1, we denoteᏮaandᏮa
0simply byᏮandᏮ0.
The aim of this paper is to investigate the boundedness of the following operator:
Tg(f)(z)=
1
0 f(tz)g(tz) dt
t , f ∈H(B),z∈B, (1.13)
whereg∈H(B), on thea-Bloch spaces. This operator can be considered as a natural ex-tension of operator (1.6) onH(B) (whenn=1 we indeed obtain (1.6)). Operator (1.13) has appeared, for the first time, in [6] where its boundedness and compactness are inves-tigated.
Closely related operators to the above mentioned on the unit polydisc were investi-gated in [4,5,9,11,12].
In this paper, we prove the following results.
Theorem1.1. Letg∈H(B)anda∈(0, 1). Then the following statements are equivalent: (a)Tgis bounded onᏮa;
(b) supz∈B|g(z)|(1− |z|2)a<∞. MoreoverTg supz∈B|g(z)|(1− |z|2)a.
Theorem1.2. Letg∈H(B). Then the following statements are equivalent: (a)Tgis bounded onᏮ;
(b)Tgis bounded onᏮ0;
(c) supz∈B|g(z)|(1− |z|2) ln 1/(1− |z|2)<∞;
and the relationshipTg supz∈B|g(z)|(1− |z|2) ln 1/(1− |z|2)holds.
Theorem1.3. Letg∈H(B)anda >1. Then the following statements are equivalent: (a)Tgis bounded onᏮa;
(b) supz∈B|g(z)|(1− |z|2)<∞. MoreoverTg supz∈B|g(z)|(1− |z|2).
2. Auxiliary results
In order to prove our results, we need some auxiliary results which are incorporated in the following lemmas.
Lemma2.1. For every f,g∈H(B), it holds that
Tg(f)(z)=f(z)g(z). (2.1)
Proof. Assume that the holomorphic function fghas the expansionαaαzα. Then
Tg(f)
(z)=
1
0
α
aα(tz)αdtt =
α
aα |α|zα
=
α
aαzα, (2.2)
84 On an integral operator on the unit ball inC
Lemma2.2. Let f ∈Ꮾa(B),0< a <∞. Then
f(z)≤C
f(0)+fᏮa, a∈(0, 1),
f(0)+fᏮaln e
1− |z|2, a=1, f(0)+ fᏮa
1− |z|2 a−1, a >1,
(2.3)
for someC >0independent off.
Proof. Let|z|>1/2,z=rζ, andζ∈∂B. We have
f(z)−f
rζ
2
=
1
1/2
∇f(tz),zdt≤
1
1/2
ft(tz)dt
≤4fᏮa
1
0
|z|dt
1−t2|z|2 a.
(2.4)
LetIa=
1
0(|z|dt/(1−t2|z|2)a). Ifa∈(0, 1), then
Ia≤
1
0
|z|dt
1−t|z| a =
1−1− |z| 1−a
1−a ≤
1
1−a. (2.5)
Ifa=1, then
1
0
|z|dt
1−t2|z|2 a = 1 2ln
1 +|z|
1− |z|≤
1 2ln
4
1− |z|2. (2.6)
Finally, ifa >1, then
Ia≤
1
0
|z|dt
1−t|z| a =
1 a−1
1
1− |z| a−1−1
≤ 2a−1
(a−1)1− |z|2 a−1. (2.7)
From all of the above we have
f(z)≤ M 1 2
+4fᏮa
1−a , a∈(0, 1),
M 1
2
+ 2fᏮaln 4
1− |z|2, a=1, M
1
2
+ 2
a+1f Ꮾa
(a−1)1− |z|2 a−1, a >1,
(2.8)
Let|z| ≤1/2, then, by the mean value property of the function f(z)−f(0) (see [7]), and Jensen’s inequality, we obtain
max |z|≤1/2
f(z)−f(0)2
≤4n
|z|≤3/4
f(w)−f(0)2
dVN(w)
≤4n
|z|≤3/4
f(w)2
dVN(w)
≤3n max
|z|≤3/4
f(z)2 .
(2.9)
The second inequality can be easily proved by using the homogeneous expansion of f. Hence,
M
1 2
≤f(0)+√3 n max |z|≤3/4
f(z)≤f(0)+24a√3 n
7a fᏮa. (2.10)
From (2.8) and (2.10), the result follows easily whena=1. Ifa=1, then we have
f(z)≤f(0)+16√3 n
7 fᏮ+ 2fᏮln 4 1− |z|2
≤
16√
3 n 7 + ln 16
f(0)+fᏮln1− |ez|2
,
(2.11)
thus finishing the proof.
3. Proofs of the main results
Proof ofTheorem 1.1. Assume thatTgis bounded onᏮa. Choosef0(z)≡1. It is clear that
f0∈Ꮾa0and thatf0Ꮾa=1. The boundedness ofTgimplies
1− |z|2 aT g
f0 (z)=
1− |z|2 ag(z)≤T
gf0Ꮾa=Tg<∞. (3.1) Henceg∈Ꮾa, as desired.
Assume now thatg∈Ꮾa. Then, byLemma 2.2we have
1− |z|2 aT
g(f)(z)=1− |z|2 af(z)g(z) ≤ gᏮaCf(0)+fᏮa
≤2CgᏮafᏮa.
(3.2)
Taking supremumz∈Bin (3.2), we obtain Tg(f)
Ꮾa≤2CgᏮafᏮa. (3.3)
Hence
Tg≤2CgᏮa, (3.4)
86 On an integral operator on the unit ball inC
Proof ofTheorem 1.2. First, assume thatTgis bounded onᏮ. From the proof which fol-lows we will see that we also consider the case whenTgis bounded onᏮ0. Forw∈B, put
fw(z)=ln 1/(1− z,w). Since
1− |z|2 f w(z)≤
1− |z|2 ∇f w(z)=
1− |z|2
1− wz,w
≤
1− |z|2 1− z,w≤2,
(3.5)
we havefwᏮ≤2, for eachw∈B. On the other hand, we have
1− |z|2 f w(z)≤
1− |z|2 1− z,w≤
1− |z|2
1− |w| −→0, (3.6)
as|z| →1. Hence fw∈Ꮾ0, for eachw∈B. ByLemma 2.1we have
1− |w|2 g(w)ln 1
1− |w|2=fw(w)g(w)1− |w| 2
=Tgfw (w)1− |w|2 ≤TgfwᏮ≤2Tg.
(3.7)
Taking supremum in (3.7) overw∈B, we obtain that conditions (a) and (b) imply (c). Assume that (c) holds. Since|f(0)| ≤ fᏮ, and byLemma 2.2, we have
f(z)≤CfᏮ1 + ln 1 1− |z|2
, (3.8)
for someC >0. Hence
Tg(f)
(z)1− |z|2 =f(z)g(z)1− |z|2
≤CfᏮ
1 + ln 1 1− |z|2
g(z)1− |z|2
≤CfᏮ sup |z|≤1/2
1 + ln 1 1− |z|2
g(z)1− |z|2
+CfᏮ sup 1/2<|z|<1
1 + ln 1 1− |z|2
g(z)1− |z|2
≤C1fᏮsup z∈B
ln 1
1− |z|2g(z)1− |z| 2 ,
(3.9)
since (c) implies
sup z∈B
g(z)1− |z|2 <∞. (3.10)
We now prove that (c) implies (b). Since ln 1/(1− |z|)→ ∞as|z| →1, we have that g∈Ꮾ0. Hence, byLemma 2.1, we have that for each polynomialp(z),
1− |z|2 T g(p)
(z)=1− |z|2 p(z)g(z)≤M p
1− |z|2 g(z), (3.11)
whereMp=supz∈B|p(z)|. SinceMp<∞andg∈Ꮾ0, we obtain that for each polynomial
p,Tg(p)∈Ꮾ0. The set of polynomials is dense inᏮ0, thus for every f ∈Ꮾ0 there is a sequence of polynomials (pn) such thatpn−fᏮ→0. Hence
Tgpn−Tgf
Ꮾ≤Tgpn−fᏮ−→0, asn−→ ∞, (3.12)
since the operatorTgis bounded. HenceTg(Ꮾ0)⊂Ꮾ0, sinceᏮ0is closed subset ofᏮ. Finally, from (3.7) and (3.9) it follows that
Tgsup
z∈B
g(z)1− |z|2 ln 1
1− |z|2. (3.13)
Proof ofTheorem 1.3. LetTgbe bounded onᏮa. Letw∈B, and fw(z)=1/(1− z,w)a−1. It is clear that fw∈Ꮾaand thatfwᏮa≤(a−1)2a. The boundedness ofTgimplies
1− |w|2 aT g
fw (w)=
1− |w|2 ag(w)f w(w) =1− |w|2 g(w)
≤TgfwᏮa
=(a−1)2aT g<∞.
(3.14)
Hence supw∈B(1− |w|2)|g(w)|<∞, as desired.
Assume now thatg∈Ꮾ. Then, byLemma 2.2we have
1− |z|2 aT
g(f)(z)=1− |z|2 af(z)g(z)
≤b1(g)1− |z|2 a−1C
f(0)+ fᏮa
1− |z|2 a−1
≤2Cb1(g)fᏮa.
(3.15)
Hence
Tg(f)
Ꮾa≤2Cb1(g)fᏮa, (3.16)
and consequentlyTg ≤2Cb1(g), as desired.
88 On an integral operator on the unit ball inC
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Stevo Stevi´c: Mathematical Institute of Serbian Academy of Sciences and Arts, Knez Mihailova 35/I, 11000 Beograd, Serbia
