Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2007, Article ID 39176, 8 pages
doi:10.1155/2007/39176
Research Article
On Bloch-Type Functions with Hadamard Gaps
Stevo Stević
Received 2 May 2007; Accepted 20 August 2007
Recommended by Simeon Reich
We give some sufficient and necessary conditions
for an analytic function f on the unit
ball B with Hadamard gaps, that is, for f (z) = ∞
k=1 Pnk (z) (the homogeneous polynomial
expansion of f ) satisfying nk+1 /nk ≥ λ > 1 for all k ∈ N, to belong to the space Ꮾαp (B) =
α
{ f |sup 0<r<1 (1 − r 2 ) R fr p < ∞, f ∈ H(B)}, p = 1,2, ∞ as well as to the corresponding
little space. A remark on analytic functions with Hadamard gaps on mixed norm space
on the unit disk is also given.
Copyright © 2007 Stevo Stević. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction
Let B = {z ∈ Cn : |z| < 1} be the open unit ball of Cn , ∂B = {z ∈ Cn : |z| = 1} its boundary, D the unit disk in C, dv the normalized Lebesgue measure of B (i.e., v(B) = 1), and
dσ the normalized rotation invariant Lebesgue measure of S satisfying σ(∂B) = 1. We
denote the class of all holomorphic functions on the
unit ball by H(B).
For f ∈ H(B) with the Taylor expansion f (z) = |β|≥0 aβ zβ , let R f (z) = |β|≥0 |β|aβ zβ
β
β
1
n
be the radial derivative of f , where
β = (β1 ,β2 ,...,βn )
is a multi-index and zβ
= z1 · · · zn .
n
∞
∞
It is well known that R f (z) = j =1 z j (∂ f /∂z j )(z) = k=0 kPk (z), if f (z) = k=0 Pk (z).
As usual, we write
fr =
p
S
f (rζ) p dσ(ζ)
1/ p
if p ∈ (0, ∞), and where fr (ζ) = f (rζ). If p = ∞, then f ∞ = supz∈B | f (z)|.
(1.1)
2
Abstract and Applied Analysis
Let α > 0. The α-Bloch space Ꮾα = Ꮾα (B) is the space of all holomorphic functions f
on B such that
α bα ( f ) = sup 1 − |z|2 R f (z) < ∞.
(1.2)
z ∈B
It is clear that Ꮾα is a normed space under the norm f Ꮾα = | f (0)| + bα ( f ), and Ꮾα1 ⊂
Ꮾα2 for α1 < α2 . Let Ꮾα0 denote the subspace of Ꮾα consisting of those f ∈ Ꮾα for which
(1 − |z|2 )α |R f (z)| → 0 as |z| → 1. This space is called the little α-Bloch space. For α =
1, the α-Bloch space and the little α-Bloch space become Bloch space Ꮾ and the little
Bloch space Ꮾ0 . Some characterizations of these spaces can be found, for example, in the
following papers [1–6].
We say that an analytic function f on the unit disk D has Hadamard gaps if f (z) =
∞
nk
k=1 ak z where nk+1 /nk ≥ λ > 1, for all k ∈ N.
In [7], Yamashita proved the following result.
Theorem 1.1. Assume that f is an analytic function on D with Hadamard gaps. Then for
α > 0, the following two propositions hold:
(a) f ∈ Ꮾα (D) if and only if limsupk→∞ |ak |n1k−α < ∞;
(b) f ∈ Ꮾα0 (D) if and only if limk→∞ |ak |n1k−α = 0.
An analytic function on B with the homogeneous expansion f (z) = ∞
k=1 Pnk (z) (here,
Pnk is a homogeneous polynomial of degree nk ) is said to have Hadamard gaps if nk+1 /nk ≥
λ > 1, for all k ∈ N. In [8], among others, Choa generalizes the main result in [9], proving
the following result.
Theorem 1.2. Assume that p ∈ (0, ∞) and f (z) = ∞
k=1 Pnk (z) is an analytic function on
B with Hadamard gaps.
Then the following statements are equivalent:
(a) f X p = ( B |R f (z)| p (1 − |z|2 ) p−1 dv(z))1/ p < ∞;
p
(b) ∞
k =1 P n k p < ∞ .
This result motivates us to find some characterizations for certain function spaces of
analytic functions on the unit ball, in terms of the sequence (Pnk p )k∈N .
Now note that the quantity bα in the definition of the α-Bloch spaces can be written in
the following form:
bα ( f ) = sup 1 − r 2
0<r<1
α
α
sup R f (rζ) = sup 1 − r 2 M∞ (R f ,r).
ζ ∈S
(1.3)
0<r<1
On the other hand, the quantity bα can be considered as the limit case of the following
quantities:
α f Ꮾαp = sup 1 − r 2 R fr p ,
(1.4)
0<r<1
as p → ∞. Note that for every f ∈ H(B) and p ∈ (0, ∞),
α α sup 1 − r 2 R fr p ≤ sup 1 − r 2 R fr ∞ .
0<r<1
0<r<1
(1.5)
Stevo Stević 3
Hence, in this paper we also consider analytic functions with Hadamard gaps on the
following spaces:
α Ꮾαp = f | sup 1 − r 2 R fr p < ∞, f ∈ H(B) ,
Ꮾαp,0
=
0<r<1
f | lim(1 − r 2 )α R fr p = 0, f ∈ H(B) .
(1.6)
r →1
Motivated by Theorem 1.1 in this paper, we study analytic functions with Hadamard
gaps, which belong to Ꮾαp or Ꮾαp,0 space when p = 1,2, ∞. Some characterizations for
these classes of functions on the unit ball are given in terms of the sequence (Pnk p )k∈N .
The following are the main results.
Theorem 1.3. Assume that α > 0, p = 1,2, ∞, and f (z) = ∞
k=1 Pnk (z) is an analytic function on B with Hadamard gaps. Then the following statements are equivalent:
(a) f ∈ Ꮾαp ;
(b) limsupk→∞ Pnk p n1k−α < ∞.
Theorem 1.4. Assume that α > 0, p = 1,2, ∞, and f (z) = ∞
k=1 Pnk (z) is an analytic function on B with Hadamard gaps. Then the following statements are equivalent:
(a) f ∈ Ꮾαp,0 ;
(b) limk→∞ Pnk p n1k−α = 0.
Throughout this paper, constants are denoted by C, they are positive and may differ
from one occurrence to the other. The notation A B means that there is a positive
constant C such that B/C ≤ A ≤ CB.
2. Proof of main results
Before proving the main results of this paper we quote two auxiliary results which are
incorporated in the lemmas which follow (see [9, 10]).
Lemma 2.1. Assume that p ∈ (0, ∞). If (nk ) is an increasing sequence of positive integers
satisfying nk+1 /nk ≥ λ > 1, for all k, then there is a positive constant A depending only on p
and λ such that
⎛
∞
⎞1/2
1 ⎝
|ak |2 ⎠
A k =1
⎛
1
≤⎝
2π
p ⎞1/ p
⎛
⎞1/2
2π ∞
∞
in
θ
2
ak e k dθ ⎠ ≤ A ⎝ |ak | ⎠
0 k =1
k =1
(2.1)
for any number ak , k ∈ N.
Lemma 2.2. Assume that α > 0, p > 0, n ∈ N
0 , (an )n∈N0 is the sequence of nonnegative
n
numbers, In = {k | 2n ≤ k < 2n+1 , k ∈ N}, tn = k∈In ak , and g(x) = ∞
n=1 an x . Then there
is a positive constant K depending only on p and α such that
1
p
p
∞
∞
1 tn
tn
α −1 p
≤
(1
−
x)
g
(x)dx
≤
K
.
K n=0 2nα
2nα
0
n =0
(2.2)
4
Abstract and Applied Analysis
Proof of Theorem 1.3. (a)⇒(b) (Case p = 1). Let f ∈ Ꮾα1 . Let fζ (w) = f (ζw), ζ ∈ S, where
ζ is fixed and w ∈ D, be a slice function. By some calculation we see that
fζ (w) = ζ1
∂f
∂f
1
(wζ) + · · · + ζn
(wζ) = R f (wζ).
∂z1
∂zn
w
From (2.3) and since fζ (w) =
S
∞
k=1 nk Pnk (ζ)w
n k −1 ,
(2.3)
we have that
1 η fζ (η) dη
dσ(ζ)
n +1
S 2πi ∂r D η k
R f (ζη)
1
dσ(ζ)|dη|
≤
2π ∂r D S ηnk +1 fr α
Ꮾ1
≤
,
(1 − r)α r nk
nk Pnk (ζ)dσ(ζ) =
(2.4)
which implies that
n k r n k P n k 1 ≤
f Ꮾα1
,
(1 − r)α
(2.5)
for every k ∈ N and r ∈ (0,1). Choosing r = 1 − (1/nk ), we obtain n1k−α Pnk 1 ≤ C, as
desired.
(b)⇒(a) (Case p = 1). Assume limsupk→∞ Pnk 1 n1k−α < ∞. We have that
α
f Ꮾα1 = sup 1 − r 2
0<r<1
S
R f (rζ)dσ(ζ)
∞
2 α
n
k
dσ(ζ)
= sup 1 − r
n
P
(ζ)r
k
n
k
S
0<r<1
k =1
≤ sup 1 − r
0<r<1
∞
2 α
k =1
≤ sup 1 − r
0<r<1
2 α+1
∞
n =1
∞
α+1 ≤ C sup 1 − r 2
0<r<1
n k P n k 1 r n k
n =1
n k P n k
n k ≤n
n k ≤n
1
(2.6)
rn
nαk r n
∞
α+1 ≤ C sup 1 − r 2
nα r n ≤ C,
0<r<1
n =1
where
we have used the fact that there is a positive constant C independent of n such
that nk ≤n nαk ≤ Cnα (here is used the assumption that nk+1 /nk ≥ λ > 1) and the following
well-known estimate:
∞
nα r n ≤ C(1 − r)−(α+1) ,
n =1
α > 0, r ∈ [0,1); see, for example, [11].
(2.7)
Stevo Stević 5
Case p = 2. Since
f
Ꮾα2
= sup 1 − r
2 α
0<r<1
∞
2
n2k Pnk 2 r 2nk
1/2
(2.8)
k =1
we have that
α
sup 1 − r 2 nk Pnk 2 r nk ≤ f Ꮾα2 ≤ sup 1 − r 2
0<r<1
∞
α 0<r<1
n k P n k 2 r n k ,
(2.9)
k =1
from which the result follows similar to the case p = 1.
Now we show that (a)⇔(b) for case p = ∞. As above, the function fζ (w) =
∞
nk
iθ
k=1 Pnk (ζ)w , where w = re , is a lacunary series in D and
α
1 − r 2 R f (rζ) = reiθ 1 − r 2
α
fζe −iθ (reiθ ),
(2.10)
from which by Theorem 1.1 the equivalence follows.
Proof of Theorem 1.4. (a)⇒(b) (Case p = 1). Let f ∈
δ > 0 such that
1 − r2
α
S
Ꮾα1,0 ,
then for every ε > 0 there is a
R f (rζ)dσ(ζ) < ε,
(2.11)
whenever δ < r < 1. From (2.4), (2.11), and rotational invariance of dσ, we have that
S
R f (ζη)
dσ(ζ)|dη|
2π ∂r D S ηnk +1 α 1 − r 2 R f (ζη)
1
α
≤
dσ(ζ)|dη|
2π ∂r D S
1 − r 2 r nk +1
1
nk Pnk (ζ)dσ(ζ) ≤
≤
(2.12)
ε
,
(1 − r)α r nk
which implies that
n k r n k P n k 1 ≤
ε
(1 − r)α
(2.13)
for every k ∈ N and r ∈ (δ,1). Choosing r = 1 − (1/nk ), we obtain
nk Pnk 1 ≤ Cεnαk ,
(2.14)
from which (b) follows in this case.
(b)⇒(a) (Case p = 1). Assume that limk→∞ Pnk 1 n1k−α = 0, then for every ε > 0 there
is a k0 ∈ N such that
Pn ≤ εnα−1 ,
k 1
k
for k ≥ k0 .
(2.15)
6
Abstract and Applied Analysis
We may assume that k0 = 1. From this and by the proof of Theorem 1.3, (b)⇒(a) (Case
p = 1), we have that
1−r
2 α
∞
α+1 R fr 1 ≤ sup 1 − r 2
0<r<1
n =1
≤ Cε sup 1 − r
k
n k ≤n
∞
2 α+1
0<r<1
n k P n n k ≤n
n =1
1
rn
nαk
rn
(2.16)
∞
α+1 ≤ Cε sup 1 − r 2
nα r n ≤ Cε,
0<r<1
n =1
from which the implication follows.
Case p = 2. By using (2.9) the result follows similar to the Case p = 1. The proof is omitted.
Finally, in view of (2.10) and employing Theorem 1.1(b) it is easy to see that (a)⇔(b)
for case p = ∞.
3. The case of mixed norm space
In this section, we give a note concerning analytic functions with Hadamard gaps on the
mixed norm space. The mixed norm space H p,q,α (B), p, q > 0, and α ∈ (−1, ∞), consists
of all f ∈ H(B) such that
f p,q,α =
1
0
f (rζ)q (1 − r)α dr
1/q
< ∞.
p
From [12, Theorem 4] the following result holds.
(3.1)
nk
Theorem 3.1. Assume that p ∈ (0, ∞), α > −1 and f (z) = ∞
k=1 ak z is an analytic func
qm−α−1
tion on D with Hadamard gaps. Then f (m) ∈ H p,q,α (D) if and only if ∞
|ak |q <
k =0 n k
∞.
Proof. First we consider the case m = 0. Similar to the proof of [12, Theorem 4] and by
Lemmas 2.1 and 2.2, we have that
q
f H p,q,α
p ⎞q/ p
2π ∞
ak r nk eink θ dθ ⎠ (1 − r)α dr
=
0
0 k =1
⎛
⎞q/2
1 ∞
⎝ ak 2 r 2nk ⎠ (1 − r)α dr
1
⎛
⎝ 1
2π
0
1
0
k =1
(3.2)
⎛
⎞q/2
∞
2 n
⎝ ak ρ k ⎠ (1 − ρ)α dρ
∞
k =0
k =1
1
2(α+1)k
⎛
⎞q/2
∞ ak q
2
am ⎠ ⎝
α+1 ,
m∈Ik
from which the result follows in this case.
k =0
nk
Stevo Stević 7
nk −m , it
Since f has Hadamard gaps and f (m) (z) = ∞
k=1 ak nk (nk − 1) · · · (nk − m + 1)z
(m)
has Hadamard gaps too. Applying the just proved result to the function
follows that f
f (m) , we obtain that f (m) ∈ H p,q,α (D) if and only if
∞ n k n k − 1 · · · n k − m + 1 ak q
k =0
finishing the proof.
nα+1
k
∞ ak q
k =0
α+1−mq
nk
< ∞,
(3.3)
Remark 3.2. Motivated
by [12, Theorems 3 and 4], we can conjecture that if p ∈ (0, ∞),
α > −1, and f (z) = ∞
k=1 Pnk (z) is an analytic function on B with Hadamard gaps, then
qm−α−1
q
(m)
R f ∈ H p,q,α (B) if and only if ∞
Pnk p < ∞. Note that the result is true for
k =0 n k
the case of the weighted Bergman space, that is, when p = q, see [12, Corollary 1]. It is
also expected that Theorems 1.3 and 1.4 hold for every p ∈ [1; ∞] (for the case n = 1, see
[13]).
Acknowledgment
The author would like to express his sincere thanks to the referees whose comments considerably improved the paper, in particular, for correcting a gap in the original versions
of Theorems 1.3 and 1.4.
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Abstract and Applied Analysis
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Stevo Stević: Mathematical Institute of the Serbian Academy of Science, Knez Mihailova 35/I,
11000 Beograd, Serbia
Email addresses: sstevic@ptt.yu; sstevo@matf.bg.ac.yu