Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2008, Article ID 404636, 9 pages
doi:10.1155/2008/404636
Research Article
Some Sufficient Conditions for Analytic Functions
to Belong to QK,0p, q Space
Xiaoge Meng
Department of Mathematics, Jia Ying University, Meizhou 514015, Guangdong, China
Correspondence should be addressed to Xiaoge Meng, mengxiaoge80@163.com
Received 31 May 2008; Accepted 7 June 2008
Recommended by Stevo Stevic
This paper gives some sufficient conditions for an analytic function to belong to the space consisting
q
of all analytic functions f on the unit disk such lim|a|→1 D |f z|p 1 − |z|2 Kgz, adAz 0.
Copyright q 2008 Xiaoge Meng. 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 D be the open unit disk in the complex plane C and HD the space of all analytic functions
in D. For a ∈ D, let
ϕa z a−z
1 − az
1.1
be the Möbius transformation of D and let
1 − az gz, a log a−z 1.2
be the Green’s function on D. Let Da, r denote the pseudo-hyperbolic metric disk centered at
a ∈ D with radius r ∈ 0, 1, that is, Da, r {z ∈ D : |ϕa z| < r}.
nk
It is said that an analytic function fz ∞
k1 ak z is defined by a lacunary series if
nk1
> 1.
k∈N nk
λ inf
For some results in the topic, see, for example, 1–6 and the references therein.
1.3
2
Abstract and Applied Analysis
Given a function K : 0, ∞ → 0, ∞, we consider the space QK p, q of all functions
f ∈ HD such that
p q p
fQK p,q sup f z 1 − |z|2 K gz, a dAz < ∞.
1.4
a∈D D
By QK,0 p, q we denote the space consisting of all f ∈ QK p, q such that
p q lim f z 1 − |z|2 K gz, a dAz 0,
|a| → 1 D
1.5
where 0 < p < ∞, −2 < q < ∞, and dAz 1/πdx dy 1/πr dr dθ.
For p 2, q 0, the space QK p, q is reduced to QK see, e.g., 7. If Kgz, a gz, as , 0 ≤ s < ∞, then QK p, q Fp, q, s see, e.g., 8, 9.
Throughout the paper, we assume that the condition holds see 7
1
q
1
< ∞,
1.6
1 − r 2 K log
r
0
so that the space QK p, q we study is nontrivial. We also assume that K as a nondecreasing
function. An important tool in the study of QK spaces is the auxiliary function ϕK defined by
see 10
Kst
,
0<t≤1 Kt
ϕK s sup
The following condition
∞
1
ϕK s
0 < s < ∞.
ds
<∞
s2
1.7
1.8
is crucial in this paper. It has played an important role in the study of QK spaces during the
last few years.
In this paper, we give some sufficient conditions for an analytic function f to belong to
the space QK,0 p, q.
The followings are our main results in this paper.
Theorem 1.1. Let f ∈ HD, 0 < p < ∞, −2 < q < ∞, and let ϕ be a monotone increasing function in
r on 0, 1 such that |f z| ≤ ϕr, for |z| r. If
1
q ϕp r 1 − r 2 K 1 − r 2 r dr < ∞,
1.9
0
then f ∈ QK,0 p, q.
Theorem 1.2. For 1 ≤ p < 2, 0 ≤ q < ∞, and 1 ≤ p − 2q < 3. If K satisfies condition 1.7 and is
nk
a function with the property that Kt K1 for t ≥ 1, then a lacunary series fz ∞
k1 ak z
belongs to QK,0 p, q if
∞
p
1
p−q−1 nk
< ∞.
1.10
ak K
n
k
k1
Throughout this paper, C stands for a positive constant, whose value may differ from
one occurrence to the other. The expression a ≈ b means that there is a positive constant C such
that C−1 a ≤ b ≤ Ca.
Xiaoge Meng
3
2. Main results and proofs
In this section, we give the proofs of Theorems 1.1 and 1.2. Before formulating the main results,
we give some lemmas which are used in the proofs.
Lemma 2.1 see 7. Let 0 < p < ∞, −2 < q < ∞, f ∈ HD. Then, f ∈ QK,0 p, q if and only if
lim
|a| → 1 D
p f z 1 − |z|2 q K 1 − ϕa z2 dAz 0.
2.1
Lemma 2.2 see 5. Let K be a function with the property that Kt K1 for t ≥ 1. If K satisfies
condition 1.8, then there exists a constant C > 0 such that K2t ≈ Kt for t > 0.
Lemma 2.3 see 5. If K satisfies condition 1.8, then we can find another nonnegative function K ∗
such that QK QK∗ and the new function K ∗ has the following properties:
a K ∗ is nondecreasing on 0, ∞;
b K ∗ satisfies condition (∗);
c K ∗ 2t ≈ K ∗ t on 0, ∞;
d K ∗ is differentiable (up to any given order) on 0, ∞;
e K ∗ is concave on 0, ∞;
f K ∗ t K ∗ 1 for t ≥ 1;
g K ∗ t ≈ K1 on 0, 1.
Lemma 2.4 see 5. If K satisfies condition 1.8, then for any α ≥ 1 and 0 ≤ β < 1, one has
1
r
α−1
0
1
log
r
−β
1 − β 1−β
1−β
1
,
Klog dr ≈ Cβ
K
r
α
α
2.2
where Cβ is a constant depending on β alone.
Lemma 2.5 see 11. For 0 < p ≤ 1, a ∈ D, and z reiθ ∈ D,
2π
0
dθ
C
p ,
≤
1 − areiθ 2p
1 − |a|r
2.3
where C > 0 is a constant.
Proof of Theorem 1.1. Let z reiθ . By Lemma 2.3, we may also assume that K is concave, so that
the following inequality true holds
1
2π
2π
0
2 K 1 − ϕa reiθ dθ ≤ K
1
2π
2π
0
2 1 − ϕa reiθ dθ .
2.4
4
Abstract and Applied Analysis
From the definition of ϕK for 0 < s, t < 1, we have that Kst ≤ ϕK sKt. Using these facts
and polar coordinates, it follows that
p 2 q Ia f z 1 − |z|2 K 1 − ϕa z dAz
D
q 2 1 1 2π iθ p 1 − r 2 K 1 − ϕa reiθ r dr dθ
f re
2π 0 0
1
2π
iθ 2 1
p
2 q
K 1 − ϕa re
dθ r dr
≤ 2 ϕ r 1 − r
2π 0
0
1
2π
q
2 1
1 − ϕa reiθ dθ r dr
≤ 2 ϕp r 1 − r 2 K
2π 0
0
1
2π
q
1 − |a|2
1
2
r dr
2 ϕp r 1 − r 2 K
2 dθ 1 − r
2π 0 1 − areiθ 0
2
1
q
ϕ r 1 − r 2 K
p
0
≤2
1
2.5
1 − |a|2 2
r dr
1−r
1 − |a|2 r 2
q ϕp r 1 − r 2 K 1 − r 2 ϕK
0
1 − |a|2
r dr.
1 − |a|2 r 2
The last integral exists for every a ∈ D in view of 1.9, and ϕK 1 − |a|2 /1 − |a|2 r 2 ≤ 1.
Further, since
1 − |a|2
lim ϕK
ϕK 0 0,
2.6
|a| → 1
1 − |a|2 r 2
then the last integral tends to zero for every r ∈ 0, 1 as |a| → 1. By Lebesgue’s dominated
convergence theorem, we obtain lim|a| → 1 Ia 0. By Lemma 2.1, we get f ∈ QK,0 p, q.
From Theorem 1.1, we have the following corollary. Here, we give a different and
technical proof.
Corollary 2.6. Let f ∈ HD, 0 < p < ∞, −2 < q < ∞, 0 < s ≤ 1, and let ϕ be a monotone increasing
function of r in 0, 1 such that |f z| ≤ ϕr, for |z| r. If
1
qs
ϕp r 1 − r 2
r dr < ∞,
2.7
0
then f ∈ F0 p, q, s.
Proof. Let a ∈ D. We have
p q s
Ja f z 1 − |z|2 gz, a dAz
D
D\Da,1/2
Da,1/2
q
|f z|p 1 − |z|2
log
1
|ϕa z|
s
dAz I1 a I2 a.
2.8
Xiaoge Meng
5
For z ∈ D \ Da, 1/2, 1/|ϕa z| ≤ 2, hence
a − z 2
1 − az 1 − az 1 − |z|2 1 − |a|2
<
−1≤4 1−
4
.
log a−z a−z 1 − az 1 − az2
2.9
For 0 < s ≤ 1, by Lemma 2.5, we have
s
p f z 1 − |z|2 q log 1 dAz
ϕa z
D\Da,1/2
s
p qs 1 − |a|2
s
2
f z 1 − |z|
≤4
dAz
1 − az2s
D\Da,1/2
s 2π 1 − |a|2 dθ
4s 1 iθ p 2 qs
≤
1−r
f re
r dr
π 0
1 − areiθ 2s
0
I1 a 2s 4 s
≤
C
π
The last integral exists since
1
0
2.10
s
1 − |a|
2 qs
ϕ r 1 − r
s r dr.
1 − |a|r
0
1
p
ϕp r1 − r 2 qs r dr < ∞ and 1 − |a|/1 − |a|r ≤ 1. It is clear that
1 − |a|
lim 0,
|a| → 1 1 − |a|r
2.11
s
1 − |a|
2 qs
lim ϕ r 1 − r
s r dr 0.
|a| → 1 0
1 − |a|r
2.12
which implies
1
p
By Lebesgue’s dominated convergence theorem, we get
lim I1 a 0.
|a| → 1
2.13
Now, we consider the case z ∈ Da, 1/2. Note that
1 − az a−z >2
2.14
in the case. By a well-known inequality see, e.g., 12, page 3, we have that
|z| − |a| 1
≤
1 − |a|z|
2
2.15
and consequently, for z ∈ Da, 1/2, we have
|z| ≤
1 2|a|
.
2 |a|
2.16
6
Abstract and Applied Analysis
By the monotonicity of ϕ, we have
s
p 1
2 q
log
I2 a f z 1 − |z|
dAz
ϕa z
Da,1/2
≤ϕ
p
ϕp
ϕp
1 2|a|
2 |a|
1 2|a|
2 |a|
1 2|a|
2 |a|
Da,1/2
2 q
1 − |z|
1
log ϕa z
s
dAz
2
2 q
1 s 1 − |a|2 log
1 − ϕa w
dAw
|w| 1 − aw4
|w|<1/2
q2 q
1 s
1
log
dAw
1 − |a|2
1 − |w|2
|w| 1 − aw2q4
|w|<1/2
≤
q2 1/2 q
dθ
1 p 1 2|a| 1 s 2π
1 − t2
log
ϕ
1 − |a|2
2q4 t dt
π
2 |a|
t
0
0
1 − at
≤
1/2
q2
q
1 s
1
1 p 1 2|a| q2 log
2
2π
1 − |a|
1 − t2
ϕ
2q4 t dt
π
2 |a|
t
0
1 − at
≤ ϕp
q2 1/2 q
1 2|a| q3 2q4 1 s
2 2
log
t dt.
1 − |a|
1 − t2
2 |a|
t
0
This implies
q2
1 2|a| ,
1 − |a|
2 |a|
I2 a ≤ Cϕp
2.17
since the following integral exists
1/2
2 q
1−t
0
1
log
t
s
t dt.
2.18
Choosing s 1, for every r ∈ 0, 1, it follows that
1
1
q1
ϕp t 1 − t2
t dt ≥ ϕp r
r
r
1 − t2
q1
q2
1
1 .
1 − r2
t dt ϕp r
2
q2
2.19
This and
1
q1
ϕp r 1 − r 2
r dr < ∞
2.20
limϕp r1 − rq2 0
2.21
0
imply that
r→1
Xiaoge Meng
7
or
p
ϕ
1 2|a|
2 |a|
q2
1 − |a|
1 2|a| q2
p 1 2|a|
1−
ϕ
−→ 0,
2 |a|
2 |a| 2 |a|q2
2.22
as |a| → 1. Consequently,
lim I2 a ≤ C lim ϕp
|a| → 1
|a| → 1
q2
1 2|a| 0.
1 − |a|
2 |a|
2.23
Combining 2.8, 2.13 with 2.23 we see that
lim
|a| → 1 D
p f z 1 − |z|2 q gz, a s dAz 0,
2.24
which means f ∈ F0 p, q, s. The proof is complete.
Proof of Theorem 1.2. Consider the monotone increasing function
∞
nk ak r nk −1 ,
0 < r < 1.
2.25
∞
iθ n
−1
f re nk ak r k ≤ ϕr,
k1
2.26
ϕr k1
For every θ ∈ 0, 2π, we have
By Theorem 1.1, we only need to prove
p
1 ∞
n −1 q k
I
nk ak r
1 − r 2 K 1 − r 2 r dr < ∞.
0
2.27
k1
By the inequality 1 − r 2 ≤ 2 log1/r, r ∈ 0, 1, and Lemma 2.2, there exists a constant C such
that
1
1
2
K 1 − r ≤ K 2 log
≈ CK log
.
2.28
r
r
Then for 0 ≤ q < ∞, we have
p
1 ∞
n
1 q
1
k
dr.
I≤C
nk ak r
r 1−p log
K log
r
r
0
k1
2.29
For p ≥ 1, assume In {k: 2n ≤ k< 2n1 , k ∈ N}. Since
∞
∞ 2n1
n
2n/2 r 2 ≤ 21/2
t−1/2 r t/2 dt ≤ 21/2
n0
n
n0 2
∞
0
1
1 −1/2
,
log
2
r
t−1/2 r t/2 dt 2Γ
2.30
8
Abstract and Applied Analysis
which together with Hölder’s inequality gives
p p p
∞
∞ ∞ n
n
2n
k
k
nk ak r
nk ak r
≤
nk ak r
n0 nk ∈In
k1
n0 nk ∈In
∞
n/2 2n 1−1/p 2n 1−pn/2 1/p nk ak 2 r
r 2
nk ∈In
n0
≤
∞
r 2 21−p/2n
n
1
≤ C log
r
Hence,
I≤C
∞
nk ak ∞
p−1
2.31
n
2n/2 r 2
n0
−p−1/2 ∞
n
r 2 21−p/2n
nk ak p
.
nk ∈In
n0
p
1
1 q−p−1/2
1
n
dr.
nk ak 21−p/2n r 2 1−p log
K log
r
r
0
nk ∈In
n0
p nk ∈In
n0
p
2.32
For 1 ≤ p ≤ 2, 1 ≤ p − 2q < 3, by Lemma 2.4, choosing α 2n 2 − p, β p − 2q − 1/2, we obtain
∞
p 1/22q − p 3 2q−p3/2
1/22q − p 3
1−p/2n
I≤C
nk ak
2
K
2n 2 − p
2n 2 − p
n0 nk ∈In
≤C
∞
n0
C
p nk ∈In
∞
n0
nk ak nk ∈In
nk ak p 1
2n
1
2n
2q−p3/2 1
2n
p−1/2 1
K n
2
2.33
1q 1
K n .
2
If nk ∈ In , then nk < 2n1 . The assumption that K is nondecreasing and Lemma 2.2 give
1q 1q
1
1
1
1
1
1
−1q
.
K n ≤
K n1 < nk
K
1q
2n
2
2n
n
2
2
k
2.34
Since fz is a lacunary series, the Taylor series of f has most logλ 2 1 terms ak znk such that
nk ∈ In . Combining this with the last inequality and Hölder’s inequality, we obtain
p
∞
1−1q/p 1
1/p
ak K
nk
I≤C
n
k
n0 nk ∈In
∞
p−1 p−q−1 p
1
≤C
nk
logλ 2 1
ak K
n
k
n0
nk ∈In
∞
p−1 p−q−1 ak p K
nk
C logλ 2 1
k1
This shows that f ∈ QK,0 p, q.
1
nk
< ∞.
2.35
Xiaoge Meng
9
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