Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2013, Article ID 294378, 6 pages
http://dx.doi.org/10.1155/2013/294378
Research Article
On Certain Classes of Convex Functions
Young Jae Sim and Oh Sang Kwon
Department of Mathematics, Kyungsung University, Busan 608-736, Republic of Korea
Correspondence should be addressed to Oh Sang Kwon; oskwon@ks.ac.kr
Received 25 February 2013; Accepted 7 May 2013
Academic Editor: Heinrich Begehr
Copyright © 2013 Y. J. Sim and O. S. Kwon. 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.
For real numbers 𝛼 and 𝛽 such that 0 ≤ 𝛼 < 1 < 𝛽, we denote by K(𝛼, 𝛽) the class of normalized analytic functions which
󸀠󸀠
󸀠
satisfy the following two sided-inequality: 𝛼 < R{1 + (𝑧𝑓 (𝑧)/𝑓 (𝑧))} < 𝛽 (𝑧 ∈ U), where U denotes the open unit disk. We find
some relationships involving functions in the class K(𝛼, 𝛽). And we estimate the bounds of coefficients and solve the Fekete-Szegö
problem for functions in this class. Furthermore, we investigate the bounds of initial coefficients of inverse functions or biunivalent
functions.
1. Introduction
Let A denote the class of analytic functions in the unit disc
U = {𝑧 : 𝑧 ∈ C, |𝑧| < 1}
(1)
It is clear that K(𝛼, 𝛽) ⊂ K. And we remark that, for
given real numbers 𝛼 and 𝛽 (0 ≤ 𝛼 < 1 < 𝛽), 𝑓 ∈ K(𝛼, 𝛽) if
and only if 𝑓 satisfies each of the following two subordination
relationships:
which is normalized by
𝑓 (0) = 0,
󸀠
𝑓 (0) = 1.
(2)
Also let S denote the subclass of A which is composed of
functions which are univalent in U. And, as usual, we denote
by K the class of functions in A which are convex in U.
We say that 𝑓 is subordinate to 𝐹 in U, written as 𝑓 ≺
𝐹 (𝑧 ∈ U), if and only if 𝑓(𝑧) = 𝐹(𝑤(𝑧)) for some Schwarz
function 𝑤(𝑧) such that
𝑤 (0) = 0,
|𝑤 (𝑧)| < 1
(𝑧 ∈ U) .
(3)
If 𝐹 is univalent in U, then the subordination 𝑓 ≺ 𝐹 is
equivalent to
𝑓 (0) = 𝐹 (0) ,
𝑓 (U) ⊂ 𝐹 (U) .
(4)
Definition 1. Let 𝛼 and 𝛽 be real numbers such that 0 ≤ 𝛼 <
1 < 𝛽. The function 𝑓 ∈ A belongs to the class K(𝛼, 𝛽) if 𝑓
satisfies the following inequality:
𝛼 < R {1 +
𝑧𝑓󸀠󸀠 (𝑧)
}<𝛽
𝑓󸀠 (𝑧)
(𝑧 ∈ U) .
(5)
𝑧𝑓󸀠󸀠 (𝑧) 1 + (1 − 2𝛼) 𝑧
≺
𝑓󸀠 (𝑧)
1−𝑧
(𝑧 ∈ U) ,
𝑧𝑓󸀠󸀠 (𝑧) 1 + (1 − 2𝛽) 𝑧
≺
1+ 󸀠
𝑓 (𝑧)
1−𝑧
(𝑧 ∈ U) .
1+
(6)
Now, we define an analytic function 𝑝 : U → C by
𝑝 (𝑧) = 1 +
𝛽−𝛼
1 − 𝑒2𝜋𝑖((1−𝛼)/(𝛽−𝛼)) 𝑧
𝑖 log (
).
𝜋
1−𝑧
(7)
The above function 𝑝 was introduced by Kuroki and Owa [1],
and they proved that 𝑝 maps U onto a convex domain
Λ = {𝑤 : 𝛼 < R {𝑤} < 𝛽}
(8)
conformally. Using this fact and the definition of subordination, we can obtain the following lemma, directly.
2
International Journal of Mathematics and Mathematical Sciences
Lemma 2. Let 𝑓(𝑧) ∈ A and 0 ≤ 𝛼 < 1 < 𝛽. Then 𝑓 ∈
K(𝛼, 𝛽) if and only if
Theorem 4. Let 𝑓 ∈ A, 0 ≤ 𝛼 < 1 and
R {1 +
𝑧𝑓󸀠󸀠 (𝑧)
1+ 󸀠
𝑓 (𝑧)
(𝑧 ∈ U) .
(15)
Then
𝛽−𝛼
1 − 𝑒2𝜋𝑖((1−𝛼)/(𝛽−𝛼)) 𝑧
≺1+
𝑖 log (
)
𝜋
1−𝑧
(𝑧 ∈ U) .
(9)
1
(16)
(𝑧 ∈ U) .
2−𝛼
Proof. First of all, we put 𝛾 = 1/(2 − 𝛼) and note that 1/2 ≤
𝛾 < 1 for 0 ≤ 𝛼 < 1. Let
R {√𝑓󸀠 (𝑧)} >
And we note that the function 𝑝, defined by (7), has the
form
∞
𝑝 (𝑧) = 1 + ∑ 𝐵𝑛 𝑧𝑛 ,
(10)
𝑛=1
𝑝 (𝑧) =
𝛽−𝛼
𝑖 (1 − 𝑒2𝑛𝜋𝑖((1−𝛼)/(𝛽−𝛼)) )
𝑛𝜋
1+
(𝑛 ∈ N) .
(11)
For given real numbers 𝛼 and 𝛽 such that 0 ≤ 𝛼 < 1 <
𝛽, we denote by K𝜎 (𝛼, 𝛽) the class of biunivalent functions
consisting of the functions in A such that
𝑓 ∈ K (𝛼, 𝛽) ,
𝑓−1 ∈ K (𝛼, 𝛽) ,
(12)
where 𝑓−1 is the inverse function of 𝑓.
In our present investigation, we first find some relationships for functions in bounded positive class K(𝛼, 𝛽). And
we solve several coefficient problems including Fekete-Szegö
problems for functions in the class. Furthermore, we estimate
the bounds of initial coefficients of inverse functions and biunivalent functions. For the coefficient bounds of functions
in special subclasses of S, the readers may be referred to the
works [2–4].
1
(√𝑓󸀠 (𝑧) − 𝛾) .
1−𝛾
2 (1 − 𝛾) 𝑧𝑝󸀠 (𝑧)
𝑧𝑓󸀠󸀠 (𝑧)
=
1
+
= 𝜓 (𝑝 (𝑧) , 𝑧𝑝󸀠 (𝑧)) ,
𝑓󸀠 (𝑧)
(1 − 𝛾) 𝑝 (𝑧) + 𝛾
(18)
where
𝜓 (𝑟, 𝑠) = 1 +
2 (1 − 𝛾) 𝑠
.
(1 − 𝛾) 𝑟 + 𝛾
(19)
Using (15), we have
{𝜓 (𝑝 (𝑧) , 𝑧𝑝󸀠 (𝑧)) : 𝑧 ∈ U} ⊂ {𝑤 ∈ C : R (𝑤) > 𝛼} := Ω𝛼 .
(20)
Now for all real 𝜌, 𝜎 ∈ R with 𝜎 ≤ −(1 + 𝜌2 )/2,
R {𝜓 (𝑖𝜌, 𝜎)} = 1 +
≤1−
2𝛾 (1 − 𝛾) 𝜎
2
(1 − 𝛾) 𝜌2 + 𝛾2
𝛾 (1 − 𝛾) (1 + 𝜌2 )
2
(1 − 𝛾) 𝜌2 + 𝛾2
(21)
.
Define a function 𝑔 : R → R by
𝑔 (𝜌) =
2. Relations Involving Bounds on
the Real Parts
(17)
Differentiating (17), we can obtain
where
𝐵𝑛 =
𝑧𝑓󸀠󸀠 (𝑧)
}>𝛼
𝑓󸀠 (𝑧)
1 + 𝜌2
2
(1 − 𝛾) 𝜌2 + 𝛾2
.
(22)
Then 𝑔 is a continuous even function and
In this section, we will find some relations involving the
functions in K(𝛼, 𝛽). And the following lemma will be
needed in finding the relations.
𝑔󸀠 (𝜌) =
2 (2𝛾 − 1) 𝜌
2
2
((1 − 𝛾) 𝜌2 + 𝛾2 )
.
(23)
Lemma 3 (see Miller and Mocanu [5, Theorem 2.3.i]). Let Ξ
be a set in the complex plane C and let 𝑏 be a complex number
such that R{𝑏} > 0. Suppose that a function 𝜓 : C2 × U → C
satisfies the condition
Hence 𝑔󸀠 (0) = 0 and 𝑔 is increasing on (0, ∞), since 1/2 ≤
𝛾 < 1. Hence 𝑔 satisfies that
𝜓 (𝑖𝜌, 𝜎; 𝑧) ∉ Ξ
for all 𝜌 ∈ R. Therefore, by combining (21) and (24), we can
get
(13)
for all real 𝜌, 𝜎 ≤ −|𝑏 − 𝑖𝜌|2 /(2R{𝑏}) and all 𝑧 ∈ U. If the
function 𝑝(𝑧) defined by 𝑝(𝑧) = 𝑏 + 𝑏1 𝑧 + 𝑏2 𝑧2 + ⋅ ⋅ ⋅ is analytic
in U and if
𝜓 (𝑝 (𝑧) , 𝑧𝑝󸀠 (𝑧) ; 𝑧) ∈ Ξ,
then R{𝑝(𝑧)} > 0 in U.
(14)
1
1
≤ 𝑔 (𝜌) <
,
2
2
𝛾
(1 − 𝛾)
R {𝜓 (𝑖𝜌, 𝜎)} ≤ 1 − 𝛾 (1 − 𝛾) 𝑔 (𝜌) ≤ 2 −
(24)
1
= 𝛼.
𝛾
(25)
And this shows that R{𝜓(𝑖𝜌, 𝜎)} ∉ Ω𝛼 for all 𝜌, 𝜎 ∈ R with
𝜎 ≤ −(1 + 𝜌2 )/2. By Lemma 3, we get R{𝑝(𝑧)} > 0 for all
𝑧 ∈ U, and this shows that the inequality (16) holds and the
proof of Theorem 4 is completed.
International Journal of Mathematics and Mathematical Sciences
Theorem 5. Let 𝑓 ∈ A, 1 < 𝛽 < 2 and
R {1 +
󸀠󸀠
3. Coefficient Problems Involving
Functions in K(𝛼,𝛽)
𝑧𝑓 (𝑧)
}<𝛽
𝑓󸀠 (𝑧)
(𝑧 ∈ U) .
(26)
1
2−𝛽
(𝑧 ∈ U) .
(27)
Then
R {√𝑓󸀠 (𝑧)} <
3
Proof. We put 𝛿 = 1/(2−𝛽) and note that 𝛿 > 1 for 1 < 𝛽 < 2.
And let
1
(√𝑓󸀠 (𝑧) − 𝛿) ,
𝑝 (𝑧) =
1−𝛿
(28)
2 (1 − 𝛿) 𝑠
𝜓 (𝑟, 𝑠) = 1 +
.
(1 − 𝛿) 𝑟 + 𝛿
In the present section, we will solve some coefficient problems
involving functions in the class K(𝛼, 𝛽). And our first
result on the coefficient estimates involves the function class
K(𝛼, 𝛽) and the following lemma will be needed.
Lemma 7 (see Rogosinski [6, Theorem 10]). Let 𝑝(𝑧) =
𝑛
∑∞
𝑛=1 𝐶𝑛 𝑧 be analytic and univalent in U and suppose that
𝑛
𝑝(𝑧) maps U onto a convex domain. If 𝑞(𝑧) = ∑∞
𝑛=1 𝐴 𝑛 𝑧 is
analytic in U and satisfies the following subordination:
𝑞 (𝑧) ≺ 𝑝 (𝑧)
{𝜓 (𝑝 (𝑧) , 𝑧𝑝 (𝑧)) : 𝑧 ∈ U} ⊂ {𝑤 ∈ C : R (𝑤) < 𝛽} := Ω𝛽 ,
(29)
by (26). And for all real 𝜌, 𝜎 with 𝜎 ≤ −(1 + 𝜌2 )/2,
R {𝜓 (𝑖𝜌, 𝜎)} = 1 +
≥1−
2𝛿 (1 − 𝛿) 𝜎
(1 − 𝛿)2 𝜌2 + 𝛿2
𝛿 (1 − 𝛿) (1 + 𝜌2 )
(1 −
𝛿)2 𝜌2
+
(30)
𝛿2
󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨
󵄨󵄨𝐴 𝑛 󵄨󵄨 ≤ 󵄨󵄨𝐶1 󵄨󵄨
1 󵄨󵄨 󵄨󵄨
󵄨𝐵 󵄨 ,
{
{
{ 2 󵄨 1󵄨
󵄨󵄨 󵄨󵄨 {
󵄨󵄨𝑎𝑛 󵄨󵄨 ≤ { 󵄨󵄨 󵄨󵄨 𝑛−2
{
{ 󵄨󵄨𝐵1 󵄨󵄨 ∏ (1 +
{
{ 𝑛 (𝑛 − 1) 𝑘=1
(1 − 𝛿)2 𝜌2 + 𝛿2
.
(31)
󵄨󵄨 󵄨󵄨
󵄨󵄨𝐵1 󵄨󵄨
) , 𝑛 = 3, 4, 5, . . . ,
𝑘
(32)
𝑝 (𝑧) = 1 +
for all 𝜌 ∈ R. Therefore,
R {𝜓 (𝑖𝜌, 𝜎)} ≥ 1 − 𝛿 (1 − 𝛿) 𝑔 (𝜌) ≥ 2 −
1
= 𝛽.
𝛿
(33)
And this shows that R{𝜓(𝑖𝜌, 𝜎)} ∉ Ω𝛽 for all 𝜌, 𝜎 ∈ R with
𝜎 ≤ −(1 + 𝜌2 )/2. By Lemma 3, we get R{𝑝(𝑧)} > 0 for all
𝑧 ∈ U, and this shows that the inequality (27) holds and the
proof of Theorem 5 is completed.
𝑧𝑓󸀠󸀠 (𝑧)
,
𝑓󸀠 (𝑧)
𝛽−𝛼
1 − 𝑒2𝜋𝑖((1−𝛼)/(𝛽−𝛼)) 𝑧
𝑖 log (
).
𝜋
1−𝑧
(39)
(40)
(41)
Then, the subordination (9) can be written as follows:
𝑞 (𝑧) ≺ 𝑝 (𝑧)
(𝑧 ∈ U) .
(42)
Note that the function 𝑝(𝑧) defined by (41) is convex in U and
has the form
∞
𝑝 (𝑧) = 1 + ∑ 𝐵𝑛 𝑧𝑛 ,
By combining Theorems 4 and 5, we can obtain the
following result.
(43)
𝑛=1
where
Theorem 6. Let 𝑓 ∈ A, 0 ≤ 𝛼 < 1 < 𝛽 < 2 and
(𝑧 ∈ U) .
(38)
Proof. Let us define
𝑞 (𝑧) = 1 +
1
1
≤ 𝑔 (𝜌) <
,
𝛿2
(1 − 𝛿)2
𝐵𝑛 =
(34)
𝛽−𝛼
𝑖 (1 − 𝑒2𝑛𝜋𝑖((1−𝛼)/(𝛽−𝛼)) ) .
𝑛𝜋
(44)
If we let
Then
1
1
< R {√𝑓󸀠 (𝑧)} <
2−𝛼
2−𝛽
𝑛 = 2,
𝜋 (1 − 𝛼)
󵄨󵄨 󵄨󵄨 2 (𝛽 − 𝛼)
sin
.
󵄨󵄨𝐵1 󵄨󵄨 =
𝜋
𝛽−𝛼
Since 𝛿 > 1, 𝑔 satisfies the inequality
𝑧𝑓󸀠󸀠 (𝑧)
𝛼 < R {1 + 󸀠
}<𝛽
𝑓 (𝑧)
(37)
where |𝐵1 | is given by
where 𝑔(𝜌) is given by
1 + 𝜌2
(𝑛 = 1, 2, . . .) .
Theorem 8. Let 𝛼 and 𝛽 be real numbers such that 0 ≤ 𝛼 <
𝑛
1 < 𝛽. If the functions 𝑓(𝑧) = 𝑧 + ∑∞
𝑛=2 𝑎𝑛 𝑧 ∈ K(𝛼, 𝛽), then
= 1 − 𝛿 (1 − 𝛿) 𝑔 (𝜌) ,
𝑔 (𝜌) =
(36)
then
As in the proof of Theorem 4, we can get
󸀠
(𝑧 ∈ U) ,
∞
(𝑧 ∈ U) .
(35)
𝑞 (𝑧) = 1 + ∑ 𝐴 𝑛 𝑧𝑛 ,
𝑛=1
(45)
4
International Journal of Mathematics and Mathematical Sciences
then by Lemma 7, we see that the subordination (42) implies
that
󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨
󵄨󵄨𝐴 𝑛 󵄨󵄨 ≤ 󵄨󵄨𝐵1 󵄨󵄨
(𝑛 = 1, 2, . . .) ,
(46)
inequality (51) holds for 𝑛 = 𝑚. Then, some calculation gives
us that
󵄨
󵄨󵄨
󵄨󵄨𝑎𝑚+1 󵄨󵄨󵄨 ≤
where
≤
𝜋 (1 − 𝛼)
󵄨󵄨 󵄨󵄨 2 (𝛽 − 𝛼)
sin
.
󵄨󵄨𝐵1 󵄨󵄨 =
𝜋
𝛽−𝛼
(47)
(𝑧 ∈ U) .
=
(48)
1
(𝐴 + 2𝑎2 𝐴 𝑛−2 + 3𝑎3 𝐴 𝑛−3
𝑛 (𝑛 − 1) 𝑛−1
(49)
+ ⋅ ⋅ ⋅ + (𝑛 − 1) 𝑎𝑛−1 𝐴 1 ) .
A simple calculation combined with the inequality (46) yields
that
󵄨󵄨 󵄨󵄨
󵄨󵄨𝑎𝑛 󵄨󵄨 =
󵄨󵄨 󵄨󵄨 𝑛−2
󵄨𝐵 󵄨
≤ 󵄨 1 󵄨 ∏ (1 +
𝑛 (𝑛 − 1) 𝑘=1
󵄨󵄨 󵄨󵄨
󵄨󵄨𝐵1 󵄨󵄨
)
𝑘
(52)
󵄨󵄨 󵄨󵄨
󵄨󵄨𝐵1 󵄨󵄨
),
𝑘
󵄨󵄨 󵄨󵄨 𝑛−2
󵄨𝐵1 󵄨
󵄨󵄨 󵄨󵄨
󵄨󵄨𝑎𝑛 󵄨󵄨 ≤ 󵄨 󵄨 ∏ (1 +
𝑛 (𝑛 − 1) 𝑘=1
󵄨󵄨 󵄨󵄨
󵄨󵄨𝐵1 󵄨󵄨
)
𝑘
(𝑛 = 3, 4, 5, . . .) ,
(53)
where |𝐵1 | is given by (47). This completes the proof of
Theorem 8.
Lemma 9 (see Keogh and Merkes [7]). Let 𝑝(𝑧) = 1 + 𝑐1 𝑧 +
𝑐2 𝑧2 + ⋅ ⋅ ⋅ be a function with positive real part in U. Then, for
any complex number ],
󵄨󵄨
󵄨
󵄨󵄨𝑐2 − ]𝑐12 󵄨󵄨󵄨 ≤ 2 max {1, |1 − 2]|} .
󵄨
󵄨
(50)
(54)
Now, the following result holds for the coefficient of 𝑓 ∈
K(𝛼, 𝛽).
Theorem 10. Let 0 ≤ 𝛼 < 1 < 𝛽 and let the function 𝑓(𝑧)
𝑛
given by 𝑓(𝑧) = 𝑧 + ∑∞
𝑛=2 𝑎𝑛 𝑧 be in the class K(𝛼, 𝛽). Then,
for a complex number 𝜇,
where |𝐵1 | is given by (47) and |𝑎1 | = 1. Hence, we have |𝑎2 | ≤
|𝐵1 |/2. To prove the assertion of the theorem, we need to show
that
󵄨󵄨 󵄨󵄨 𝑛
󵄨𝐵1 󵄨
󵄨
󵄨󵄨 󵄨󵄨
󵄨
󵄨󵄨𝑎𝑛 󵄨󵄨 ≤ 󵄨 󵄨 ∑ (𝑘 − 1) 󵄨󵄨󵄨𝑎𝑘−1 󵄨󵄨󵄨
𝑛 (𝑛 − 1) 𝑘=2
󵄨󵄨 󵄨󵄨 𝑚−1
󵄨󵄨𝐵1 󵄨󵄨
∏ (1 +
(𝑚 + 1) 𝑚 𝑘=1
󵄨󵄨 󵄨󵄨
󵄨󵄨𝐵1 󵄨󵄨
)
𝑘
And now, we will solve the Fekete-Szegö problem for 𝑓 ∈
K(𝛼, 𝛽), and we will need the following lemma.
1
𝑛 (𝑛 − 1)
󵄨
× 󵄨󵄨󵄨𝐴 𝑛−1 + 2𝑎2 𝐴 𝑛−2 + 3𝑎3 𝐴 𝑛−3
󵄨
+ ⋅ ⋅ ⋅ + (𝑛 − 1) 𝑎𝑛−1 𝐴 1 󵄨󵄨󵄨
1
≤
𝑛 (𝑛 − 1)
󵄨 󵄨󵄨
󵄨 󵄨󵄨
󵄨
󵄨
󵄨
󵄨
× (󵄨󵄨󵄨𝐴 𝑛−1 󵄨󵄨󵄨 + 2 󵄨󵄨󵄨𝑎2 󵄨󵄨󵄨 󵄨󵄨󵄨𝐴 𝑛−2 󵄨󵄨󵄨 + 3 󵄨󵄨󵄨𝑎3 󵄨󵄨󵄨 󵄨󵄨󵄨𝐴 𝑛−3 󵄨󵄨󵄨
󵄨
󵄨󵄨 󵄨
+ ⋅ ⋅ ⋅ + (𝑛 − 1) 󵄨󵄨󵄨𝑎𝑛−1 󵄨󵄨󵄨 󵄨󵄨󵄨𝐴 1 󵄨󵄨󵄨)
󵄨󵄨 󵄨󵄨 𝑛
󵄨󵄨𝐵1 󵄨󵄨
󵄨
󵄨
≤
∑ (𝑘 − 1) 󵄨󵄨󵄨𝑎𝑘−1 󵄨󵄨󵄨 ,
𝑛 (𝑛 − 1) 𝑘=2
󵄨󵄨 󵄨󵄨
󵄨󵄨𝐵1 󵄨󵄨
)
𝑘
which implies that the inequality (51) is true for 𝑛 = 𝑚 + 1.
Hence, by the mathematical induction, we prove that
Then, the coefficients of 𝑧𝑛−1 in both sides lead to
𝑎𝑛 =
󵄨󵄨 󵄨󵄨 𝑚−2
󵄨󵄨𝐵1 󵄨󵄨
∏ (1 +
(𝑚 + 1) 𝑚 𝑘=1
󵄨󵄨 󵄨󵄨2
𝑚−2
󵄨󵄨𝐵1 󵄨󵄨
+
∏ (1 +
𝑚 (𝑚 + 1) (𝑚 − 1) 𝑘=1
Now, equality (40) implies that
𝑓󸀠 (𝑧) + 𝑧𝑓󸀠󸀠 (𝑧) = 𝑞 (𝑧) 𝑓󸀠 (𝑧)
󵄨󵄨 󵄨󵄨 𝑚+1
󵄨󵄨𝐵1 󵄨󵄨
󵄨
󵄨
∑ (𝑘 − 1) 󵄨󵄨󵄨𝑎𝑘−1 󵄨󵄨󵄨
(𝑚 + 1) 𝑚 𝑘=2
󵄨
󵄨󵄨
󵄨󵄨𝑎3 − 𝜇𝑎22 󵄨󵄨󵄨
󵄨
󵄨
≤
𝛽−𝛼
1−𝛼
sin (
𝜋)
3𝜋
𝛽−𝛼
󵄨󵄨󵄨 1
𝛽−𝛼
3
⋅ max {1; 󵄨󵄨󵄨 + (1 − 𝜇)
𝑖
2
𝜋
󵄨󵄨 2
𝛽−𝛼
3
1
𝑖)
+ ( − (1 − 𝜇)
2
2
𝜋
󵄨󵄨
󵄨
× 𝑒2𝜋𝑖((1−𝛼)/(𝛽−𝛼)) 󵄨󵄨󵄨} .
󵄨󵄨
(51)
(𝑛 = 3, 4, 5, . . .) .
We now use the mathematical induction for the proof of the
theorem. For the case 𝑛 = 3, it is clear. We assume that the
The result is sharp.
(55)
International Journal of Mathematics and Mathematical Sciences
Proof. Let us consider a function 𝑞(𝑧) given by 𝑞(𝑧) = 1 +
𝑧𝑓󸀠󸀠 (𝑧)/𝑓󸀠 (𝑧). Then, since 𝑓 ∈ K(𝛼, 𝛽), we have 𝑞(𝑧) ≺
𝑝(𝑧)(𝑧 ∈ U), where
𝑝 (𝑧) = 1 +
𝛽−𝛼
1 − 𝑒2𝜋𝑖((1−𝛼)/(𝛽−𝛼)) 𝑧
𝑖 log (
)
𝜋
1−𝑧
∞
5
Corollary 11. Let 0 ≤ 𝛼 < 1 < 𝛽 and let the function 𝑓, given
𝑛
by 𝑓(𝑧) = 𝑧 + ∑∞
𝑛=2 𝑎𝑛 𝑧 , be in the class K(𝛼, 𝛽). Also let the
function 𝑓−1 , defined by
𝑓−1 (𝑓 (𝑧)) = 𝑧 = 𝑓 (𝑓−1 (𝑧)) ,
(56)
be the inverse of 𝑓. If
= 1 + ∑ 𝐵𝑛 𝑧𝑛 ,
∞
𝑓−1 (𝑤) = 𝑤 + ∑ 𝑏𝑛 𝑤𝑛
𝑛=1
𝑛=2
where 𝐵𝑛 is given by (11). Let
1 + 𝑝−1 (𝑞 (𝑧))
= 1 + ℎ1 𝑧 + ℎ2 𝑧2 + ⋅ ⋅ ⋅
ℎ (𝑧) =
1 − 𝑝−1 (𝑞 (𝑧))
(𝑧 ∈ U) .
Then ℎ is analytic and has positive real part in the open unit
disk U. We also have
ℎ (𝑧) − 1
)
ℎ (𝑧) + 1
(𝑧 ∈ U) .
(58)
1
𝑎2 = 𝐵1 ℎ1 ,
4
1
1
1
1
𝐵 ℎ − 𝐵 ℎ2 + 𝐵 ℎ2 + 𝐵2 ℎ2 ,
12 1 2 24 1 1 24 2 1 24 1 1
(59)
1
𝐵 (ℎ − ]ℎ12 ) ,
12 1 2
𝐵
3
1
(1 − 2 − 𝐵1 + 𝜇𝐵1 ) .
2
𝐵1
2
(61)
Applying Lemma 9, we can obtain
1 󵄨󵄨 󵄨󵄨 󵄨󵄨
󵄨󵄨
󵄨
2 󵄨󵄨
󵄨󵄨𝑎3 − 𝜇𝑎22 󵄨󵄨󵄨 =
󵄨
󵄨
󵄨
󵄨 12 󵄨󵄨𝐵1 󵄨󵄨 󵄨󵄨ℎ2 − ]ℎ1 󵄨󵄨
≤
1 󵄨󵄨 󵄨󵄨
󵄨𝐵 󵄨 ⋅ max {1; |1 − 2]|} .
6 󵄨 1󵄨
(62)
𝛽−𝛼
𝐵1 =
𝑖 (1 − 𝑒2𝜋𝑖((1−𝛼)/(𝛽−𝛼)) ) ,
𝜋
(63)
𝛽−𝛼
𝑖 (1 − 𝑒4𝜋𝑖((1−𝛼)/(𝛽−𝛼)) )
2𝜋
(64)
in (62), we can obtain the result as asserted. The estimate is
sharp for the function 𝑓 : U → C defined by
𝑧
𝜁
0
0
𝑓 (𝑧) = ∫ {exp (∫
𝑝 (𝜉) − 1
𝑑𝜉)} 𝑑𝜁,
𝜉
(65)
where the function 𝑝 is given by (7). Hence the proof of
Theorem 10 is completed.
Using Theorem 10, we can get the following result.
Proof. The relations (66) and (67) give
𝑏3 = 2𝑎22 − 𝑎3 .
(69)
1−𝛼
󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 𝛽 − 𝛼
sin (
𝜋) ,
󵄨󵄨𝑏2 󵄨󵄨 = 󵄨󵄨𝑎2 󵄨󵄨 ≤ 󵄨󵄨𝐵1 󵄨󵄨 =
2
𝜋
𝛽−𝛼
(70)
immediately. Furthermore, an application of Theorem 10
(with 𝜇 = 2) gives the estimates for |𝑏3 |; hence, the proof of
Corollary 11 is completed.
Finally, we will estimate some initial coefficients for the
bi-univalent functions 𝑓 ∈ K𝜎 (𝛼, 𝛽).
And substituting
𝐵2 =
1−𝛼
󵄨󵄨 󵄨󵄨 𝛽 − 𝛼
sin (
𝜋)
󵄨󵄨𝑏3 󵄨󵄨 ≤
3𝜋
𝛽−𝛼
󵄨󵄨󵄨 1 2 (𝛽 − 𝛼)
⋅ max {1; 󵄨󵄨󵄨󵄨 −
𝑖
𝜋
󵄨󵄨 2
Thus, we can get the estimate for |𝑏2 | by
(60)
where
]=
1−𝛼
󵄨󵄨 󵄨󵄨 𝛽 − 𝛼
sin (
𝜋) ,
󵄨󵄨𝑏2 󵄨󵄨 ≤
𝜋
𝛽−𝛼
𝑏2 = −𝑎2 ,
which imply that
𝑎3 − 𝜇𝑎22 =
(67)
󵄨󵄨
1 2 (𝛽 − 𝛼)
󵄨
+( +
𝑖) 𝑒2𝜋𝑖((1−𝛼)/(𝛽−𝛼)) 󵄨󵄨󵄨󵄨} .
2
𝜋
󵄨󵄨
(68)
We find from (58) that
𝑎3 =
1
(|𝑤| < 𝑟0 ; 𝑟0 > ) ,
4
then
(57)
𝑞 (𝑧) = 𝑝 (
(66)
Theorem 12. For given 𝛼 and 𝛽 such that 0 ≤ 𝛼 < 1 < 𝛽, let 𝑓
𝑛
given by 𝑓(𝑧) = 𝑧 + ∑∞
𝑛=2 𝑎𝑛 𝑧 , be in the class K𝜎 (𝛼, 𝛽). Then
󵄨󵄨 󵄨󵄨
󵄨󵄨𝑎2 󵄨󵄨 ≤
󵄨󵄨 󵄨󵄨 √󵄨󵄨 󵄨󵄨
󵄨󵄨𝐵1 󵄨󵄨 󵄨󵄨𝐵1 󵄨󵄨
,
(71)
󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 󵄨󵄨
󵄨
󵄨󵄨𝑎3 󵄨󵄨 ≤ (󵄨󵄨𝐵1 󵄨󵄨 + 󵄨󵄨𝐵2 − 𝐵1 󵄨󵄨󵄨) ,
2
(72)
√2 󵄨󵄨󵄨󵄨𝐵12 − 2𝐵1 + 2𝐵2 󵄨󵄨󵄨󵄨
where 𝐵1 and 𝐵2 are given by (63) and (64).
Proof. If 𝑓 ∈ K𝜎 (𝛼, 𝛽), then 𝑓 ∈ K(𝛼, 𝛽) and 𝑔 ∈ K(𝛼, 𝛽),
where 𝑔 = 𝑓−1 . Hence
𝑧𝑓󸀠󸀠 (𝑧)
≺ 𝑝 (𝑧)
𝑓󸀠 (𝑧)
(𝑧 ∈ U) ,
𝑧𝑔󸀠󸀠 (𝑧)
𝐿 (𝑧) := 1 + 󸀠
≺ 𝑝 (𝑧)
𝑔 (𝑧)
(𝑧 ∈ U) ,
𝑄 (𝑧) := 1 +
(73)
6
International Journal of Mathematics and Mathematical Sciences
where 𝑝(𝑧) is given by (7). Let
Acknowledgment
1 + 𝑝−1 (𝑄 (𝑧))
1 − 𝑝−1 (𝑄 (𝑧))
ℎ (𝑧) =
= 1 + ℎ1 𝑧 + ℎ2 𝑧2 + ⋅ ⋅ ⋅
The research was supported by Kyungsung University Research Grants in 2013.
(𝑧 ∈ U) ,
1 + 𝑝−1 (𝐿 (𝑧))
1 − 𝑝−1 (𝐿 (𝑧))
𝑘 (𝑧) =
= 1 + 𝑘1 𝑧 + 𝑘2 𝑧2 + ⋅ ⋅ ⋅
(74)
(𝑧 ∈ U) .
Then ℎ and 𝑘 are analytic and have positive real part in U.
Also, we have
ℎ (𝑧) − 1
)
ℎ (𝑧) + 1
(𝑧 ∈ U) ,
𝑘 (𝑧) − 1
)
𝐿 (𝑧) = 𝑝 (
𝑘 (𝑧) + 1
(𝑧 ∈ U) .
𝑄 (𝑧) = 𝑝 (
(75)
By suitably comparing coefficients, we get
1
𝑎2 = 𝐵1 ℎ1 ,
4
(76)
1
1
1
3𝑎3 − 2𝑎22 = 𝐵1 ℎ2 − 𝐵1 ℎ12 + 𝐵2 ℎ12 ,
4
8
8
(77)
1
𝑏2 = 𝐵1 𝑘1 ,
4
(78)
1
1
1
3𝑏3 − 2𝑏22 = 𝐵1 𝑘2 − 𝐵1 𝑘12 + 𝐵2 𝑘12 ,
4
8
8
(79)
where 𝐵1 and 𝐵2 are given by (63) and (64), respectively. Now,
considering (76) and (78), we get
ℎ1 = −𝑘1 .
(80)
Also, from (77), (78), (79), and (80), we find that
𝑎22 =
𝐵13 (ℎ2 + 𝑘2 )
.
8 (𝐵12 − 2𝐵1 + 2𝐵2 )
(81)
Therefore, we have
󵄨󵄨 󵄨󵄨3
󵄨𝐵 󵄨
󵄨󵄨 2 󵄨󵄨
󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨
󵄨󵄨𝑎2 󵄨󵄨 ≤ 󵄨 2 󵄨 1 󵄨
󵄨 󵄨 8 󵄨󵄨𝐵 − 2𝐵 + 2𝐵 󵄨󵄨󵄨 (󵄨󵄨ℎ2 󵄨󵄨 + 󵄨󵄨𝑘2 󵄨󵄨)
1
2󵄨
󵄨 1
󵄨󵄨󵄨𝐵 󵄨󵄨󵄨3
≤ 󵄨󵄨 2 󵄨 1 󵄨
󵄨.
2 󵄨󵄨𝐵1 − 2𝐵1 + 2𝐵2 󵄨󵄨󵄨
(82)
This gives the bound on |𝑎2 | as asserted in (71). Now, further
computations from (77), (79), (80), and (81) lead to
𝑎3 =
1
1
𝐵 (2ℎ2 + 𝑘2 ) + ℎ12 (𝐵2 − 𝐵1 ) .
12 1
8
(83)
This equation, together with the well-known estimates
󵄨󵄨󵄨ℎ2 󵄨󵄨󵄨 ≤ 2,
󵄨󵄨󵄨𝑘2 󵄨󵄨󵄨 ≤ 2,
󵄨󵄨󵄨ℎ1 󵄨󵄨󵄨 ≤ 2,
(84)
󵄨 󵄨
󵄨 󵄨
󵄨 󵄨
leads us to the inequality (72). Therefore, the proof of
Theorem 12 is completed.
References
[1] K. Kuroki and S. Owa, “Notes on new class for certain analytic
functions,” RIMS Kokyuroku, vol. 1772, pp. 21–25, 2011.
[2] R. M. Ali, S. K. Lee, V. Ravichandran, and S. Supramaniam,
“Coefficient estimates for bi-univalent Ma-Minda starlike and
convex functions,” Applied Mathematics Letters, vol. 25, no. 3,
pp. 344–351, 2012.
[3] H. M. Srivastava, A. K. Mishra, and P. Gochhayat, “Certain
subclasses of analytic and bi-univalent functions,” Applied
Mathematics Letters, vol. 23, no. 10, pp. 1188–1192, 2010.
[4] Q.-H. Xu, Y.-C. Gui, and H. M. Srivastava, “Coefficient estimates
for a certain subclass of analytic and bi-univalent functions,”
Applied Mathematics Letters, vol. 25, no. 6, pp. 990–994, 2012.
[5] S. S. Miller and P. T. Mocanu, Differential Subordinations, Theory
and Applications, Marcel Dekker, New York, NY, USA, 2000.
[6] W. Rogosinski, “On the coefficients of subordinate functions,”
Proceedings of the London Mathematical Society, vol. 48, pp. 48–
82, 1943.
[7] F. R. Keogh and E. P. Merkes, “A coefficient inequality for certain
classes of analytic functions,” Proceedings of the American
Mathematical Society, vol. 20, pp. 8–12, 1969.
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