# On the Fekete-Szegö Problem for a Class of Analytic Functions

Authors: Zhigang Peng
Journal: ISRN Mathematical Analysis
 (2014)

---

## Abstract

Let𝒜 denote the class of functions which are analytic in the unit disk D
=
{
z
:
|
z
|
<
1
} and given by the power series f
(
z
)
=
z
+
∑
n
=
2
∞
‍
a
n
z
n. Let C be the class of convex functions. In this paper, we give the upper bounds of |
a
3
-
μ
a
2
2
| for all real number μ and for any f
(
z
) in the family 𝒱
=
{
f
(
z
)
:
f
∈
𝒜, Re(
f
(
z
)
/
g
(
z
)
)
>
0 for  some g
∈
C
}.


---

## Body

## 1. Introduction

Let𝒜 denote the class of functions which are analytic in the unit disk D
=
{
z
:
|
z
|
<
1
} and satisfy f
(
0
)
=
f
′
(
0
)
-
1
=
0. The set of all functions f
∈
𝒜 that are univalent will be denoted by S. Let C
,
S
*
(
β
) and K be the classes of convex, starlike of order β and close-to-convex functions, respectively. Fekete and Szegö [1] proved that
(1)
|
a
3
-
μ
a
2
2
|
≤
{
4
μ
-
3
,
μ
≥
1
,
1
+
2
exp
⁡
[
-
2
μ
(
1
-
μ
)
]
,
0
≤
μ
<
1
,
3
-
4
μ
,
μ
<
0
holds for any f
(
z
)
=
z
+
∑
n
=
2
∞
‍
a
n
z
n
∈
S and that this inequality is sharp. The coefficient functional Λ
μ
(
f
)
=
a
3
-
μ
a
2
2 on f in 𝒜 plays an important role in function theory. For example, a
3
-
a
2
2
=
S
f
(
0
)
/
6, where S
f is the Schwarzian derivative. The problem of maximizing the absolute value of the functional Λ
μ
(
f
) is called the Fekete-Szegö problem. In the literature, there exist a large number of results about the Fekete-Szegö problem (see, for instance, [2–11]).For0
≤
α
<
1 and 0
≤
β
<
1, let u
α
β denote the class of functions f satisfying f
∈
𝒜 and
(2)
Re
{
α
z
2
f
′′
(
z
)
g
(
z
)
+
z
f
′
(
z
)
g
(
z
)
}
>
0
for some g
∈
S
*
(
β
). Al-Abbadi and Darus [7] investigated the Fekete-Szegö problem on the class u
α
β.LetC
1
(
β
) be the class of functions f in 𝒜 satisfying the inequality
(3)
Re
{
z
f
′
(
z
)
ϕ
(
z
)
e
i
β
}
>
0
(
|
z
|
<
1
,
-
π
2
<
β
<
π
2
)
for some function ϕ
∈
C. In [11], Srivastara et al. studied the Fekete-Szegö problem on the class
(4)
C
1
=
⋃
β
‍
C
1
(
β
)
(
-
π
2
<
β
<
π
2
)
for 0
≤
μ
≤
1 by proving that
(5)
|
a
3
-
μ
a
2
2
|
≤
{
5
3
-
9
4
μ
,
0
≤
μ
≤
2
9
,
2
3
+
1
9
μ
,
2
9
≤
μ
≤
2
3
,
5
6
,
2
3
<
μ
≤
1
.
Srivastara et al. held that the inequality (5) was sharp. However, the extremal function given in [11] did not exist in the case of 2
/
3
<
μ
≤
1.In this paper, we solve the Fekete-Szegö problem for the family(6)
𝒱
=
{
f
(
z
)
:
f
∈
𝒜
,
Re
f
(
z
)
g
(
z
)
>
0
for
some
g
∈
C
}
.
As a corollary of the main result, we find the sharp upper bounds for absolute value of the Fekete-Szegö functional for the class v
α defined by
(7)
v
α
=
{
f
(
z
)
:
f
∈
𝒜
,
Re
{
α
z
2
f
′′
(
z
)
g
(
z
)
+
z
f
′
(
z
)
g
(
z
)
}
>
0
for
some
g
∈
C
{
α
z
2
f
′′
(
z
)
g
(
z
)
+
z
f
′
(
z
)
g
(
z
)
}
}
.
Clearly, v
α is a subclass of u
α
0. In the case of α
=
0, we get sharp estimation of the absolute value of the Fekete-Szegö functional for the class C
1
(
0
) and for all real number μ, which prove that the inequality (5) is not sharp actually when 2
/
3
<
μ
≤
1.


## 2. Main Result

LetB
0 be the class of functions ϕ
(
z
) that are analytic in D and satisfy |
ϕ
(
z
)
|
≤
|
z
| for all |
z
|
<
1. The following two lemmas can be found in [2].Lemma 1 (see [2]).
Ifϕ
(
z
)
=
∑
n
=
1
∞
‍
α
n
z
n is in the class B
0, then, for any complex number s, one has |
α
2
-
s
α
1
2
|
≤
1
+
(
|
s
|
-
1
)
|
α
1
2
|
≤
max
⁡
{
1
,
|
s
|
}. The inequality is sharp.Lemma 2 (see [2]).
Ifg
(
z
)
=
z
+
∑
n
=
2
∞
‍
c
n
z
n is in the class C and μ is a complex number, then |
c
3
-
μ
c
2
2
|
≤
1
/
3
+
(
|
μ
-
1
|
-
1
/
3
)
|
c
2
2
|
≤
max
⁡
{
1
/
3
,
|
μ
-
1
|
}. The inequality is sharp.Theorem 3.
Iff
(
z
)
=
z
+
∑
n
=
2
∞
‍
a
n
z
n is in the class 𝒱 and μ is a real number, then
(8)
|
a
3
-
μ
a
2
2
|
≤
{
5
-
9
μ
,
w
h
e
n
μ
≤
1
6
,
2
+
1
4
μ
,
w
h
e
n
1
6
≤
μ
≤
1
2
,
3
-
2
μ
+
1
4
(
1
-
μ
)
,
w
h
e
n
1
2
<
μ
≤
5
6
,
7
μ
-
3
,
w
h
e
n
5
6
<
μ
≤
1
,
9
μ
-
5
,
w
h
e
n
1
≤
μ
.Proof.
By definition,f
(
z
)
=
z
+
∑
n
=
2
∞
‍
a
n
z
n is in the class 𝒱 if and only if there exists a function g
(
z
)
=
z
+
∑
n
=
2
∞
‍
c
n
z
n
∈
C such that ϕ
(
z
)
=
[
f
(
z
)
-
g
(
z
)
]
/
[
f
(
z
)
+
g
(
z
)
]
=
∑
n
=
1
∞
‍
α
n
z
n
∈
B
0. A simple computation shows a
2
=
c
2
+
2
α
1
,
a
3
=
c
3
+
2
(
α
1
c
2
+
α
2
+
α
1
2
). Thus,
(9)
a
3
-
μ
a
2
2
=
c
3
-
μ
c
2
2
+
2
[
α
2
+
(
1
-
2
μ
)
α
1
2
]
+
2
(
1
-
2
μ
)
α
1
c
2
.
So, by Lemmas 1 and 2, we have
(10)
|
a
3
-
μ
a
2
2
|
≤
|
c
3
-
μ
c
2
2
|
+
2
|
α
2
+
(
1
-
2
μ
)
α
1
2
|
+
2
|
1
-
2
μ
∥
α
1
∥
c
2
|
≤
1
3
+
[
|
μ
-
1
|
-
1
3
]
|
c
2
|
2
+
2
[
1
+
(
|
1
-
2
μ
|
-
1
)
|
α
1
|
2
]
+
2
|
1
-
2
μ
∥
α
1
∥
c
2
|
.
Putting |
α
1
|
=
x and |
c
2
|
=
y, we get from (10) that |
a
3
-
μ
a
2
2
|
≤
F
(
x
,
y
), where
(11)
F
(
x
,
y
)
=
1
3
+
[
|
μ
-
1
|
-
1
3
]
y
2
+
2
[
1
+
(
|
1
-
2
μ
|
-
1
)
x
2
]
+
2
|
1
-
2
μ
|
x
y
.
Since |
α
1
|
≤
1 and |
c
2
|
≤
1, we will calculate the maximum value of F
(
x
,
y
) for (
x
,
y
)
∈
[
0,1
]
×
[
0,1
].
Case
1. Suppose μ
≤
1
/
2. Then it follows from (11) that
(12)
F
(
x
,
y
)
=
7
3
+
(
2
3
-
μ
)
y
2
-
4
μ
x
2
+
2
(
1
-
2
μ
)
x
y
.
Since
(13)
F
x
(
x
,
y
)
=
-
8
μ
x
+
2
(
1
-
2
μ
)
y
,
F
y
(
x
,
y
)
=
2
(
1
-
2
μ
)
x
+
(
4
3
-
2
μ
)
y
,
|
-
8
μ
2
(
1
-
2
μ
)
2
(
1
-
2
μ
)
4
3
-
2
μ
|
=
16
3
μ
-
4
<
0
,
F
(
x
,
y
) does not have a local maximum at any point of the open rectangle (
0,1
)
×
(
0,1
). Hence, F
(
x
,
y
) must attain its maximum at a boundary point. Since F
(
0
,
y
)
≤
3
-
μ
,
F
(
1
,
y
)
≤
F
(
1,1
)
=
5
-
9
μ for μ
≤
1
/
2 and
(14)
F
(
x
,
0
)
≤
{
7
3
-
4
μ
,
if
μ
<
0
,
7
3
,
if
0
≤
μ
≤
1
2
,
F
(
x
,
1
)
≤
{
F
(
1,1
)
=
5
-
9
μ
,
if
μ
<
1
6
,
F
(
1
-
2
μ
4
μ
,
1
)
=
1
4
μ
+
2
,
if
1
6
≤
μ
≤
1
2
,
we have
(15)
|
a
3
-
μ
a
2
2
|
≤
F
(
x
,
y
)
≤
{
5
-
9
μ
,
when
μ
≤
1
6
,
2
+
1
4
μ
,
when
1
6
<
μ
≤
1
2
.
Case
2. Suppose 1
/
2
<
μ
≤
1. Then, we get from (11) that
(16)
F
(
x
,
y
)
=
7
3
+
(
2
3
-
μ
)
y
2
+
4
(
μ
-
1
)
x
2
+
2
(
2
μ
-
1
)
x
y
.
Since
(17)
F
x
(
x
,
y
)
=
8
(
μ
-
1
)
x
+
2
(
2
μ
-
1
)
y
,
F
y
(
x
,
y
)
=
2
(
2
μ
-
1
)
x
+
(
4
3
-
2
μ
)
y
,
|
8
(
μ
-
1
)
2
(
2
μ
-
1
)
2
(
2
μ
-
1
)
4
3
-
2
μ
|
=
4
(
-
8
μ
2
+
32
3
μ
-
11
3
)
<
0
,
F
(
x
,
y
) must attain its maximum at a boundary point of the rectangle [
0,1
]
×
[
0,1
].
Since
(18)
F
(
x
,
0
)
≤
7
3
,
if
1
2
<
μ
≤
1
,
F
(
x
,
1
)
≤
{
F
(
1
-
2
μ
4
(
μ
-
1
)
,
1
)
=
3
-
2
μ
+
1
4
(
1
-
μ
)
,
if
1
2
<
μ
≤
5
6
,
F
(
1,1
)
=
7
μ
-
3
,
if
5
6
<
μ
≤
1
,
F
(
0
,
y
)
≤
{
3
-
μ
,
if
1
2
<
μ
≤
2
3
,
7
3
,
if
2
3
<
μ
≤
1
,
F
(
1
,
y
)
≤
F
(
1,1
)
=
7
μ
-
3
,
if
1
2
<
μ
≤
1
,
we get
(19)
|
a
3
-
μ
a
2
2
|
≤
F
(
x
,
y
)
≤
{
3
-
2
μ
+
1
4
(
1
-
μ
)
,
when
1
2
<
μ
≤
5
6
,
7
μ
-
3
,
when
5
6
<
μ
≤
1
.
Case
3. Suppose μ
>
1. Then, (11) gives
(20)
F
(
x
,
y
)
=
7
3
+
(
μ
-
4
3
)
y
2
+
4
(
μ
-
1
)
x
2
+
2
(
2
μ
-
1
)
x
y
.
Since(21)
F
x
(
x
,
y
)
=
8
(
μ
-
1
)
x
+
2
(
2
μ
-
1
)
y
,
F
y
(
x
,
y
)
=
2
(
2
μ
-
1
)
x
+
(
2
μ
-
8
3
)
y
,
|
8
(
μ
-
1
)
2
(
2
μ
-
1
)
2
(
2
μ
-
1
)
2
μ
-
8
3
|
=
4
3
(
13
-
16
μ
)
<
0
,
F
(
x
,
y
) must attain its maximum at a boundary point of the rectangle [
0,1
]
×
[
0,1
]. Since
(22)
F
(
x
,
0
)
≤
4
μ
-
5
3
,
if
μ
≥
1
,
F
(
x
,
1
)
≤
F
(
1,1
)
=
9
μ
-
5
,
if
μ
≥
1
,
F
(
0
,
y
)
≤
{
7
3
,
if
1
<
μ
≤
4
3
,
1
+
μ
,
if
μ
>
4
3
,
F
(
1
,
y
)
≤
F
(
1,1
)
=
9
μ
-
5
,
if
μ
≥
1
,
it follows that
(23)
|
a
3
-
μ
a
2
2
|
≤
F
(
x
,
y
)
≤
9
μ
-
5
,
when
1
<
μ
.
Combining (15), (19) with (23), we get (8). Since inequalities in Lemmas 1 and 2 are sharp, it follows that inequality (8) is also sharp. The proof is completed.Sincef
∈
v
α if and only if F
(
z
)
=
α
z
2
f
′′
(
z
)
+
z
f
′
(
z
)
∈
𝒱, by a simple calculation, we have the following.Corollary 4.
Iff
(
z
)
=
z
+
∑
n
=
2
∞
‍
a
n
z
n
∈
v
α, then
(24)
|
a
3
-
μ
a
2
2
|
≤
{
5
3
(
1
+
2
α
)
-
9
μ
4
(
1
+
α
)
2
,
μ
≤
2
(
1
+
α
)
2
9
(
1
+
2
α
)
,
2
3
(
1
+
2
α
)
+
(
1
+
α
)
2
9
(
1
+
2
α
)
2
μ
,
2
(
1
+
α
)
2
9
(
1
+
2
α
)
≤
μ
≤
2
(
1
+
α
)
2
3
(
1
+
2
α
)
,
1
1
+
2
α
-
μ
2
(
1
+
α
)
2
+
(
1
+
α
)
2
12
(
1
+
2
α
)
(
1
+
α
)
2
-
9
(
1
+
2
α
)
2
μ
,
2
(
1
+
α
)
2
3
(
1
+
2
α
)
≤
μ
≤
10
(
1
+
α
)
2
9
(
1
+
2
α
)
,
7
4
(
1
+
α
)
2
μ
-
1
1
+
2
α
,
10
(
1
+
α
)
2
9
(
1
+
2
α
)
≤
μ
≤
4
(
1
+
α
)
2
3
(
1
+
2
α
)
,
9
4
(
1
+
α
)
2
μ
-
5
3
(
1
+
2
α
)
,
4
(
1
+
α
)
2
3
(
1
+
2
α
)
≤
μ
.Lettingα
=
0 in (24), we get the following.Corollary 5.
Iff
(
z
)
=
z
+
∑
n
=
2
∞
‍
a
n
z
n
∈
C
1
(
0
), then
(25)
|
a
3
-
μ
a
2
2
|
≤
{
5
3
-
9
4
μ
,
w
h
e
n
μ
≤
2
9
,
2
3
+
1
9
μ
,
w
h
e
n
2
9
≤
μ
≤
2
3
,
1
-
1
2
μ
+
1
12
-
9
μ
,
w
h
e
n
2
3
<
μ
≤
10
9
,
7
4
μ
-
1
,
w
h
e
n
10
9
<
μ
≤
4
3
,
9
4
μ
-
5
3
,
w
h
e
n
4
3
≤
μ
.Remark 6.
In [11], Srivastava et al. gave a function f
(
z
)
=
z
+
∑
n
=
2
∞
‍
a
n
z
n
∈
𝒜 satisfying ϕ
(
z
)
=
[
f
(
z
)
-
g
(
z
)
]
/
[
f
(
z
)
+
g
(
z
)
]
=
∑
n
=
1
∞
‍
α
n
z
n
∈
B
0, where g
(
z
)
=
z
+
z
2
+
z
3
+
⋯
∈
C
,
α
2
=
1
-
α
1
2 and
(26)
α
1
=
(
2
-
3
λ
)
±
i
6
μ
-
4
6
μ
.
Srivastava held that f
∈
C
1
(
0
) and |
a
3
-
μ
a
2
2
|
=
5
/
6 when 2
/
3
≤
μ
≤
1. But ϕ
∈
B
0 implies that |
α
2
|
≤
1
-
|
α
1
2
|. So ϕ
(
z
) satisfying the above conditions does not exist.



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*Source: 861671-2014-03-04.xml*