MATEMATIQKI VESNIK
UDK 517.547
originalniresearch
nauqnipaper
rad
58 (2006), 119–124
ON CERTAIN NEW SUBCLASS OF
CLOSE-TO-CONVEX FUNCTIONS
Zhi-Gang Wang, Chun-Yi Gao and Shao-Mou Yuan
(k)
Abstract. In the present paper, the authors introduce a new subclass Ks (α, β) of close-toconvex functions. The subordination and inclusion relationship, and some coefficient inequalities
for this class are provided. The results presented here would provide extensions of those given in
earlier works.
1. Introduction
Let A denote the class of functions of the form
∞
P
f (z) = z +
an z n ,
n=2
which are analytic in the open unit disk U = {z ∈ C : |z| < 1}. Let S, S ∗ and K
denote the usual subclasses of A whose members are univalent, starlike and closeto-convex in U , respectively. Also let S ∗ (α) denote the class of starlike functions
of order α, 0 ≤ α < 1.
Sakaguchi [4] introduced a class Ss∗ of functions starlike with respect to symmetric points, consisting of functions f (z) ∈ S satisfying
½
¾
zf 0 (z)
<
> 0 (z ∈ U ).
f (z) − f (−z)
In a later paper, Gao and Zhou [1] discussed a class Ks of analytic functions related
to the starlike functions, that is the subclass of f (z) ∈ S satisfying the following
inequality
½ 2 0
¾
z f (z)
<
< 0 (z ∈ U ),
g(z)g(−z)
where g(z) ∈ S ∗ ( 12 ).
AMS Subject Classification: 30C45.
Keywords and phrases: Starlike functions, close-to-convex functions, differential subordination.
This work was supported by the Scientific Research Fund of Hunan Provincial Education
Department and the Hunan Provincial Natural Science Foundation (No. 05JJ30013) of People’s
Republic of China.
119
120
Zhi-Gang Wang, Chun-Yi Gao, Shao-Mou Yuan
More recently, Wang, Gao and Yuan [6] discussed a subclass Ks (α, β) of the
class Ks , that is the subclass of f (z) ∈ S satisfying the following inequality
¯ 2 0
¯
¯
¯
2 0
¯ z f (z)
¯
¯
¯
¯
¯ < β ¯ αz f (z) − 1¯ (z ∈ U ),
+
1
¯ g(z)g(−z)
¯
¯ g(z)g(−z)
¯
where 0 ≤ α ≤ 1, 0 < β ≤ 1 and g(z) ∈ S ∗ ( 12 ). Note that Ks (1, 1) = Ks , so
Ks (α, β) is a generalization of Ks .
Let f (z) and F (z) be analytic in U. Then we say that the function f (z)
is subordinate to F (z) in U, if there exists an analytic function ω(z) in U such
that |ω(z)| ≤ |z| and f (z) = F (ω(z)), denoted by f ≺ F or f (z) ≺ F (z). If
F (z) is univalent in U, then the subordination is equivalent to f (0) = F (0) and
f (U) ⊂ F (U) (see [3]).
In the present paper, we introduce and investigate the following class of analytic functions, and obtain some interesting results.
(k)
Definition 1. Let Ks (α, β) denote the class of functions in S satisfying the
inequality
¯ k 0
¯
¯ k 0
¯
¯ z f (z)
¯
¯ αz f (z)
¯
¯
¯
¯
¯
(1.1)
¯ gk (z) − 1¯ < β ¯ gk (z) + 1¯ (z ∈ U ),
where 0 ≤ α ≤ 1, 0 < β ≤ 1, g(z) ∈ S ∗ ( k−1
k ), k ≥ 1 is a fixed positive integer and
gk (z) is defined by the following equality
gk (z) =
k−1
Y
ε−ν g(εν z) (εk = 1).
(1.2)
ν=0
(k)
(2)
Note that Ks (α, β) = Ks (α, β), so Ks (α, β) is a generalization of Ks (α, β).
In the present paper, we shall provide the subordination and inclusion rela(k)
tionships, and some coefficient inequalities for the class Ks (α, β). The results
presented here would provide extensions of those given in earlier works.
2. Coefficient estimate
(k)
We first give two meaningful conclusions about the class Ks (α, β). The proof
of Theorem 1 below is much akin to that of Theorem 1 in [6], here we omit the
details.
(k)
Theorem 1. A function f (z) ∈ Ks (α, β) if and only if
z k f 0 (z)
1 + βz
≺
(z ∈ U).
gk (z)
1 − αβz
Remark 1. From Theorem 1, we know that
½
¾
zf 0 (z)
> 0 (z ∈ U ),
<
gk (z)/z k−1
½
¾
1 + βz
because of <
> 0 (z ∈ U ).
1 − αβz
(2.1)
(2.2)
121
On certain new subclass of close-to-convex functions
In order to prove our next theorem, we shall require the following lemma.
Lemma 1. Let ψi (z) ∈ S ∗ (αi ), where 0 ≤ αi < 1 (i = 0, 1, . . . , k − 1). Then
k−1
P
for k − 1 ≤
αi < k, we have
i=0
Qk−1
i=0 ψi (z)
z k−1
∈ S∗
µk−1
P
i=0
¶
αi − (k − 1) .
Proof. Since ψi (z) ∈ S ∗ (αi ) (i = 0, 1, . . . , k − 1), we have
¾
½ 0
¾
½ 0
¾
½ 0
zψk−1 (z)
zψ1 (z)
zψ0 (z)
> α0 , <
> α1 , . . . , <
> αk−1 .
<
ψ0 (z)
ψ1 (z)
ψk−1 (z)
(2.3)
We now let
ψ0 (z)ψ1 (z) · · · ψk−1 (z)
.
z k−1
Differentiating (2.4) logarithmically, we have
F (z) =
(2.4)
0
zψk−1
(z)
zψ00 (z) zψ10 (z)
zF 0 (z)
=
+
+ ··· +
− (k − 1),
F (z)
ψ0 (z)
ψ1 (z)
ψk−1 (z)
(2.5)
from (2.5) together with (2.3), we can get
½ 0 ¾
½ 0
¾
½ 0
¾
½ 0
¾
zψk−1 (z)
zF (z)
zψ0 (z)
zψ1 (z)
<
=<
+<
+ ··· + <
− (k − 1)
F (z)
ψ0 (z)
ψ1 (z)
ψk−1 (z)
k−1
P
>
αi − (k − 1).
i=0
Thus, if 0 ≤
k−1
P
i=0
αi − (k − 1) < 1, we know that
Qk−1
F (z) =
i=0 ψi (z)
z k−1
∈ S∗
∞
P
Theorem 2. Let g(z) = z +
n=2
µk−1
P
i=0
¶
αi − (k − 1) .
bn z n ∈ S ∗
¡ k−1 ¢
, then gk (z)/z k−1 ∈ S ∗ ⊂ S.
k
Proof. From (1.2), we know
Qk−1 −ν ν
Qk−1 −ν ν
P∞
g(ε z)
[ε z + n=2 bn (εν z)n ]
gk (z)
ν=0 ε
ν=0 ε
=
=
z k−1
z k−1
z k−1
¤
P∞
Qk−1 £
z + n=2 bn ε(n−1)ν z n
= ν=0
.
z k−1
Now, suppose that
g(z) = z +
∞
P
n=2
µ
bn z n ∈ S ∗
k−1
k
¶
.
(2.6)
122
Zhi-Gang Wang, Chun-Yi Gao, Shao-Mou Yuan
Then, by Lemma 1 and equality (2.6), we can get the assertion of Theorem 2
easily.
Remark 2. From Theorem 2 and inequality (2.2), we know that if f (z) ∈
(k)
(k)
Ks (α, β), then f (z) is a close-to-convex function. So Ks (α, β) is a subclass of
the class of close-to-convex functions.
(k)
In order to give the coefficient estimate of f (z) ∈ Ks (α, β), we shall require
the following lemma.
P∞
P∞
Lemma 2 [5] Let f (z) = z + n=2 an z n ∈ S, g(z) = z + n=2 bn z n ∈ S, and
satisfy the inequality
¯ 0
¯
¯
¯
¯ zf (z)
¯ αzf 0 (z)
¯
¯
¯
¯
¯
¯
¯ g(z) − 1¯ < β ¯ g(z) + 1¯ (z ∈ U ),
where 0 ≤ α ≤ 1, 0 < β ≤ 1. Then for n ≥ 2, we have
2
|nan − bn | ≤ 2(1 + αβ 2 )
n−1
P
k |ak | |bk | (|a1 | = |b1 | = 1).
(2.7)
k=1
We now give the following theorem.
P∞
P∞
Theorem 3. Let f (z) = z + n=2 an z n ∈ S, g(z) = z + n=2 bn z n ∈ S, and
satisfy the inequality (1.1). Then for n ≥ 2, we have
2
|nan − Bn | ≤ 2(1 + αβ 2 )
n−1
P
k |ak | |Bk | (|a1 | = |B1 | = 1),
(2.8)
k=1
where Bn is given by (2.11).
Proof. Note that inequality (1.1) can be written as
¯
¯
¯
¯
0
¯ zf 0 (z)
¯
¯
¯
¯
¯ < β ¯ αzf (z) + 1¯ .
−
1
¯ gk (z)/z k−1
¯
¯
¯ gk (z)/z k−1
(2.9)
At the same time, we know that equality (2.6) can be written as
∞
P
gk (z)
=
z
+
Bn z n ∈ S ∗ ⊂ S.
(2.10)
z k−1
n=2
P∞
Now, suppose that f (z) = z + n=2 an z n ∈ S and satisfy (2.9). So f (z) and
gk (z)/z k−1 satisfy the condition of Lemma 2. Thus, from (2.7), we can get (2.8)
easily.
3. Inclusion relationship
In order to give the inclusion relationship for the class Ks (λ, α, β), we shall
require the following lemma.
123
On certain new subclass of close-to-convex functions
Lemma 3. [2] Let −1 ≤ B2 ≤ B1 < A1 ≤ A2 ≤ 1. Then
1 + A1 z
1 + A2 z
≺
.
1 + B1 z
1 + B2 z
Theorem 4. Let 0 ≤ α1 ≤ α2 ≤ 1 and 0 < β1 ≤ β2 ≤ 1. Then we have
Ks(k) (α1 , β1 ) ⊂ Ks(k) (α2 , β2 ).
(k)
Proof. Suppose that f (z) ∈ Ks (α1 , β1 ). By Theorem 1 we have
z k f 0 (z)
1 + β1 z
≺
.
gk (z)
1 − α1 β1 z
Since 0 ≤ α1 ≤ α2 ≤ 1 and 0 < β1 ≤ β2 ≤ 1, we have
−1 ≤ −α2 β2 ≤ −α1 β1 < β1 ≤ β2 ≤ 1.
Thus, by Lemma 3, we have
z k f 0 (z)
1 + β2 z
≺
,
gk (z)
1 − α2 β2 z
(k)
(k)
(k)
that is f (z) ∈ Ks (α2 , β2 ). This means that Ks (α1 , β1 ) ⊂ Ks (α2 , β2 ). Hence
the proof is complete.
Taking k = 2 in Theorem 4, we have
Corollary 1. Let 0 ≤ α1 ≤ α2 ≤ 1 and 0 < β1 ≤ β2 ≤ 1. Then we have
Ks (α1 , β1 ) ⊂ Ks (α2 , β2 ).
4. Sufficient condition
At last, we give the sufficient condition for functions belonging to the class
(k)
Ks (α, β).
P∞
P∞
Theorem 5. Let f (z) = z + n=2 an z n , g(z) = z + n=2 bn z n be analytic in
U. If for 0 ≤ α ≤ 1, 0 < β ≤ 1, we have
∞
P
n=2
n(1 + αβ) |an | +
∞
P
n=2
(1 + β) |Bn | ≤ (1 + α)β,
(4.1)
(k)
where Bn is given by (2.10), then f (z) ∈ Ks (α, β).
P∞
Proof. Suppose that f (z) = z + n=2 an z n . In view of (2.9) and (2.10), let M
be denoted by
¯
¯
¯
¯
¯
¯
gk (z) ¯
gk (z) ¯
M = ¯¯zf 0 (z) − k−1 ¯¯ − β ¯¯αzf 0 (z) + k−1 ¯¯
z
z
¯
¯
¯
¯
∞
∞
∞
∞
¯
¯
¯
¯
P
P
P
P
= ¯¯z +
nan z n − z −
Bn z n ¯¯ − β ¯¯αz +
nαan z n + z +
Bn z n ¯¯ .
n=2
n=2
n=2
n=2
124
Zhi-Gang Wang, Chun-Yi Gao, Shao-Mou Yuan
Thus, for |z| = r < 1, we have
·
¸
∞
∞
P
P
|Bn | rn − β (1 + α)r −
nα |an | rn −
|Bn | rn
n=2
n=2
n=2
n=2
·
¸
∞
∞
P
P
< −(1 + α)β +
n(1 + αβ) |an | +
(1 + β) |Bn | r.
M≤
∞
P
n |an | rn +
∞
P
n=2
n=2
From inequality (4.1), we know that M < 0. Thus, we have
¯ k 0
¯ k 0
¯
¯
¯ αz f (z)
¯
¯
¯ z f (z)
¯
¯
¯
¯
¯ gk (z) − 1¯ < β ¯ gk (z) + 1¯ (z ∈ U ),
(k)
that is f (z) ∈ Ks (α, β). This completes the proof of Theorem 5.
REFERENCES
[1] C.-Y. Gao and S.-Q. Zhou, On a class of analytic functions related to the starlike functions,
Kyungpook Math. J. 45 (2005), 123–130.
[2] M.-S. Liu, On a subclass of p-valent close-to-convex functions of order β and type α, J. Math.
Study 30 (1997), 102–104.
[3] C. Pommerenke, Univalent Functions, Vandenhoeck and Ruprecht, Göttingen, 1975.
[4] K. Sakaguchi, On certain univalent mapping, J. Math. Soc. Japan 11 (1959), 72–75.
[5] T.V. Sudharsan, P. Balasubrahmanyam and K.G. Subramanian, On functions starlike with
respect to symmetric and conjugate points, Taiwanese J. Math. 2 (1998), 57–68.
[6] Z.-G. Wang, C.-Y. Gao and S.-M. Yuan, On certain subclass of close-to-convex functions,
Acta Math. Acad. Paedagog. Nyházi (N.S.) 22 (2006), 171–177.
(received 01.06.2005)
School of Mathematics and Computing Science, Changsha University of Science and Technology,
Changsha, 410076 Hunan, People’s Republic of China
E-Mail: zhigwang@163.com