# Bounded Doubly Close-to-Convex Functions

Authors: Dorina Răducanu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We consider a new classC
C
(
α
,
β
) of bounded doubly close-to-convex functions. Coefficient bounds, distortion theorems, and radius of convexity for the class C
C
(
α
,
β
) are investigated. A corresponding class of doubly close-to-starlike functions S
*
S
(
α
,
β
) is also considered.


---

## Body

## 1. Introduction

LetA denote the class of functions of the form
(1)
f
(
z
)
=
z
+
∑
n
=
2
∞
‍
a
n
z
n
,
which are analytic in the unit disk U
=
{
z
∈
C
:
|
z
|
<
1
}.LetS
*, K, and C denote the well-known classes of starlike, convex, and close-to-convex functions, respectively.A functionf
∈
A is said to be in the class C
(
β
) of close-to-convex functions of order β
≥
0 (see [1]) if there exist g
∈
K and θ
∈
R such that
(2)
|
arg
(
e
i
θ
f
′
(
z
)
g
′
(
z
)
)
|
≤
β
π
2
(
z
∈
U
)
.
It is clear that C
(
0
)
=
K and C
(
1
)
=
C.Denote byB the class of analytic functions ω in U with ω
(
0
)
=
0 and such that |
ω
(
z
)
|
<
1 for all z
∈
U.Suppose thatf and g are two analytic functions in U. The function f is said to be subordinate to the function g, denoted by f
≺
g, if there exists a function ω
∈
B such that f
(
z
)
=
g
(
ω
(
z
)
), z
∈
U.LetP be the well-known class of analytic functions p normalized by p
(
0
)
=
1 and having positive real part in U.For a fixedα
>
1
/
2 let P
α denote the subclass of P defined by
(3)
P
α
=
{
p
∈
P
:
|
p
(
z
)
-
α
|
<
α
,
z
∈
U
}
.
The class P
α has been investigated by Goel [2] and also by Libera and Livingston [3].It is easy to observe that whenα
→
∞, the class P
α reduces to the class P.Forα
>
1
/
2, the function
(4)
p
α
(
z
)
=
1
+
z
1
-
(
1
-
(
1
/
α
)
)
z
(
z
∈
U
)
maps the unit disk U onto the domain D
α
=
{
z
∈
U
:
|
z
-
α
|
<
α
}. It follows that a function p is in the class P
α if and only if p
≺
p
α.If(5)
p
α
(
z
)
=
1
+
∑
n
=
1
∞
‍
P
n
z
n
,
then it is easy to check that
(6)
P
1
=
2
-
1
α
,
P
2
=
(
2
-
1
α
)
(
1
-
1
α
)
.Some properties of the functions belonging to the classP
α are listed in the next lemma.Lemma 1 (see [2, 3]).
Letp
(
z
)
=
1
+
∑
n
=
1
∞
p
n
z
n be in the class P
α  (
α
>
1
/
2
). Then
(7)
|
p
n
|
≤
2
-
1
α
(
n
∈
{
1,2
,
…
}
)
;
(8)
1
-
r
1
+
(
1
-
(
1
/
α
)
)
r
≤
|
p
(
z
)
|
≤
1
+
r
1
-
(
1
-
(
1
/
α
)
)
r
;
(9)
|
z
p
′
(
z
)
p
(
z
)
|
≤
(
2
-
(
1
/
α
)
)
r
1
-
(
1
/
α
)
r
-
(
1
-
(
1
/
α
)
)
r
2
,
for z
∈
U and |
z
|
=
r
<
1. All the inequalities are sharp.LetC
C
(
α
)  (
α
>
1
/
2
) denote the class of all functions f
∈
A for which there exists a function g
∈
K such that f
′
/
g
′
≺
p
α, where p
α is given by (4), or equivalently
(10)
|
f
′
(
z
)
g
′
(
z
)
-
α
|
<
α
(
z
∈
U
)
.It is easy to see that whenα
→
∞ the class C
C
(
α
) reduces to the class C of close-to-convex functions.A slightly different class than the classC
C
(
α
) was investigated in [2].Recently, in [4] the authors considered a new class of analytic functions defined in a similar way to the class C
(
β
). For fixed β
≥
0 and γ
≥
0 a function f
∈
A is called doubly close-to-convex if there exist g
∈
C
(
β
) and ϕ
∈
R such that
(11)
|
arg
(
e
i
ϕ
f
′
(
z
)
g
′
(
z
)
)
|
≤
γ
π
2
(
z
∈
U
)
.Motivated by the ideas from [4] we define a new class of bounded doubly close-to-convex functions.Definition 2.
Letα
>
1
/
2 and β
>
1
/
2 be fixed. A function f
∈
A is said to be in the class C
C
(
α
,
β
) if there exists a function g
∈
C
C
(
α
) such that f
′
/
g
′
≺
p
β, where p
β is given by (4) with β instead of α, or equivalently
(12)
|
f
′
(
z
)
g
′
(
z
)
-
β
|
<
β
(
z
∈
U
)
.From the above definition and the definition of the classC
C
(
α
), it follows that f
∈
C
C
(
α
,
β
) if there exist a function h
∈
K and a function g
∈
A such that g
′
/
h
′
≺
p
α and f
′
/
g
′
≺
p
β, or equivalently
(13)
|
g
′
(
z
)
h
′
(
z
)
-
α
|
<
α
,
|
f
′
(
z
)
g
′
(
z
)
-
β
|
<
β
(
z
∈
U
)
.In the next lemma we prove that the new classC
C
(
α
,
β
) is nonempty.Lemma 3.
Letα
>
1
/
2 and β
>
1
/
2. Then, there exists a function f
∈
C
C
(
α
,
β
).Proof.
Define the following three functions:(14)
f
(
z
)
=
∫
0
z
‍
(
1
+
ϵ
1
u
)
(
1
+
ϵ
2
u
)
mmm
×
(
(
1
+
ϵ
3
u
)
2
[
1
-
(
1
-
(
1
/
α
)
)
ϵ
1
u
]
mmmnm
×
[
1
-
(
1
-
(
1
/
β
)
)
ϵ
2
u
]

(
1
+
ϵ
3
u
)
2
[
1
-
(
1
-
(
1
/
α
)
)
ϵ
1
u
]
)
-
1
d
u
,
(15)
g
(
z
)
=
∫
0
z
‍
1
+
ϵ
1
u
(
1
+
ϵ
3
u
)
2
[
1
-
(
1
-
(
1
/
α
)
)
ϵ
1
u
]
d
u
,
h
(
z
)
=
z
1
+
ϵ
3
z
,
with z
∈
U and |
ϵ
k
|
=
1, k
∈
{
1,2
,
3
}. Since h
∈
K (see [5]) and
(16)
g
′
(
z
)
h
′
(
z
)
=
1
+
ϵ
1
z
1
-
(
1
-
(
1
/
α
)
)
ϵ
1
z
,
it follows that g
∈
C
C
(
α
). The equality
(17)
f
′
(
z
)
g
′
(
z
)
=
1
+
ϵ
2
z
1
-
(
1
-
(
1
/
β
)
)
ϵ
2
z
together with g
∈
C
C
(
α
) shows that the function f defined by (14) belongs to C
C
(
α
,
β
).In this paper we obtain distortion theorems, radius of convexity, and coefficient bounds for the classC
C
(
α
,
β
). In the last section of the paper a corresponding class of bounded doubly close-to-starlike functions S
*
S
(
α
,
β
) is also considered.


## 2. Distortion Theorems

In this section distortion theorems for the classC
C
(
α
,
β
) are obtained.Theorem 4.
Letα
>
1
/
2 and β
>
1
/
2. If f
∈
C
C
(
α
,
β
), then
(18)
(
1
-
r
)
2
(
1
+
r
)
2
[
1
+
(
1
-
(
1
/
α
)
)
r
]
[
1
+
(
1
-
(
1
/
β
)
)
r
]
≤
|
f
′
(
z
)
|
≤
(
1
+
r
)
2
(
1
-
r
)
2
[
1
-
(
1
-
(
1
/
α
)
)
r
]
[
1
-
(
1
-
(
1
/
β
)
)
r
]
for z
∈
U and |
z
|
=
r
<
1.Proof.
Letf
∈
C
C
(
α
,
β
). Then there exists g
∈
C
C
(
α
) such that f
′
(
z
)
=
g
′
(
z
)
q
(
z
), where q belongs to the class P
β defined by (3) with β instead of α. Since q
∈
P
β, making use of inequalities (8) from Lemma 1, we obtain
(19)
1
-
r
1
+
(
1
-
(
1
/
β
)
)
r
≤
|
f
′
(
z
)
g
′
(
z
)
|
≤
1
+
r
1
-
(
1
-
(
1
/
β
)
)
r
for z
∈
U and |
z
|
=
r
<
1.
The functiong belongs to the class C
C
(
α
) and thus, there exists a function h
∈
K such that g
′
(
z
)
=
h
′
(
z
)
p
(
z
), where p
∈
P
α. Using once more the inequalities (8) from Lemma 1, we have
(20)
1
-
r
1
+
(
1
-
(
1
/
α
)
)
r
≤
|
g
′
(
z
)
h
′
(
z
)
|
≤
1
+
r
1
-
(
1
-
(
1
/
α
)
)
r
for z
∈
U and |
z
|
=
r
<
1.
Moreover,h
∈
K implies that (see [5, 6])
(21)
1
(
1
+
r
)
2
≤
|
h
′
(
z
)
|
≤
1
(
1
-
r
)
2
for z
∈
U and |
z
|
=
r
<
1.
Combining the inequalities (19), (20), and (21), we obtain the desired inequality (18).Theorem 5.
Letα
>
1
/
2 and β
>
1
/
2. If f
∈
C
C
(
α
,
β
), then
(22)
|
f
(
z
)
|
≤
4
α
β
[
r
1
-
r
+
(
α
+
β
-
1
)
log
⁡
(
1
-
r
)
]
-
α
β
α
-
β
{
[
1
-
(
1
-
1
β
)
r
]
(
2
α
-
1
)
2
log
⁡
[
1
-
(
1
-
1
α
)
r
]
mmmnmmm
-
(
2
β
-
1
)
2
log
⁡
[
1
-
(
1
-
1
β
)
r
]
}
,
(23)
|
f
(
z
)
|
≥
4
α
β
[
r
1
+
r
-
(
α
+
β
-
1
)
log
⁡
⁡
(
1
+
r
)
]
+
α
β
α
-
β
{
[
1
+
(
1
-
1
β
)
r
]
(
2
α
-
1
)
2
log
⁡
[
1
+
(
1
-
1
α
)
r
]
mmmmmm
-
(
2
β
-
1
)
2
log
⁡
[
1
+
(
1
-
1
β
)
r
]
}
if α
≠
β and
(24)
|
f
(
z
)
|
≤
4
α
2
[
r
1
-
r
+
(
2
α
-
1
)
log
⁡
(
1
-
r
)
]
+
(
2
α
-
1
)
{
r
1
-
(
1
-
(
1
/
α
)
)
r
mmmmmmm
-
4
α
2
log
⁡
[
1
-
(
1
-
1
α
)
r
]
r
1
-
(
1
-
(
1
/
α
)
)
r
}
,
(25)
|
f
(
z
)
|
≥
4
α
2
[
r
1
+
r
-
(
2
α
-
1
)
log
⁡
(
1
+
r
)
]
+
(
2
α
-
1
)
{
r
1
+
(
1
-
(
1
/
α
)
)
r
m
m
m
m
m
i
m
m
+
4
α
2
log
⁡
[
1
+
(
1
-
1
α
)
r
]
r
1
-
(
1
-
(
1
/
α
)
)
r
}
if α
=
β.Proof.
Letf
∈
C
C
(
α
,
β
). Integrating along the straight line segment from origin to z
=
r
e
i
θ  (
0
<
r
<
1
) the right-hand side of inequality (18) we obtain
(26)
|
f
(
z
)
|
≤
∫
0
r
‍
|
f
′
(
ρ
e
i
θ
)
|
d
ρ
≤
∫
0
r
‍
(
1
+
ρ
)
2
(
1
-
ρ
)
2
[
1
-
(
1
-
(
1
/
α
)
)
ρ
]
[
1
-
(
1
-
(
1
/
β
)
)
ρ
]
d
ρ
,
which leads to inequalities (22) and (24).
To prove the lower bound of|
f
(
z
)
| we proceed in the following way. Let δ
>
0 be the radius of the open disk contained entirely in f
(
U
). Consider z
0 with |
z
0
|
=
r
<
1 such that |
f
(
z
0
)
|
=
min
⁡
|
z
|
=
r
|
f
(
z
)
|. The minimum increases with r and is less than δ. Hence, the linear segment Γ which connects the origin with the point f
(
z
0
) will be covered entirely by the values of f
(
z
). Denote by γ the arc in U which is mapped by w
=
f
(
z
) in Γ. Making use of the left-hand side of inequality (18) we get
(27)
|
f
(
z
)
|
≥
|
f
(
z
0
)
|
=
∫
Γ
‍
d
w
≥
∫
γ
‍
|
f
′
(
z
)
|
|
d
z
|
≥
∫
0
r
‍
(
1
-
ρ
)
2
(
1
+
ρ
)
2
[
1
+
(
1
-
(
1
/
α
)
)
ρ
]
[
1
+
(
1
-
(
1
/
β
)
)
ρ
]
d
ρ
.
After simple calculations we obtain the inequalities (23) and (25). Thus, the proof of our theorem is completed.


## 3. Radius of Convexity

In this section we obtain the radius of the disk which is mapped onto a convex domain by the functions belonging toC
C
(
α
,
β
).Theorem 6.
Letα
>
1
/
2 and β
>
1
/
2. Suppose that f
∈
C
C
(
α
,
β
). Then, the function f maps the disk {
z
∈
C
:
|
z
|
<
r
0
<
1
} onto a convex domain, where r
0 is the smallest positive root of the equation
(28)
(
α
+
β
-
α
β
-
1
)
r
4
+
4
(
α
β
-
α
-
β
+
1
)
r
3
+
(
10
α
β
-
5
α
-
5
β
+
1
)
r
2
+
4
α
β
r
-
α
β
=
0
.Proof.
Letf
∈
C
C
(
α
,
β
). Then, there exists g
∈
C
C
(
α
) such that
(29)
f
′
(
z
)
=
q
(
z
)
g
′
(
z
)
(
z
∈
U
)
,
where q
∈
P
β. Since g
∈
C
C
(
α
), there exists a function h
∈
K such that
(30)
g
′
(
z
)
=
p
(
z
)
h
′
(
z
)
(
z
∈
U
)
,
where p
∈
P
α. Moreover, since h
∈
K it follows that there exists a function k
∈
S
* (see [5, 6]) such that
(31)
k
(
z
)
=
z
h
′
(
z
)
(
z
∈
U
)
.
Combining the equalities (29), (30), and (31), we get
(32)
z
f
′
(
z
)
=
q
(
z
)
p
(
z
)
k
(
z
)
(
z
∈
U
)
.
By taking logarithmic derivative in (32), we obtain
(33)
1
+
z
f
′′
(
z
)
f
′
(
z
)
=
z
q
′
(
z
)
q
(
z
)
+
z
p
′
(
z
)
p
(
z
)
+
z
k
′
(
z
)
k
(
z
)
which leads to
(34)
R
(
1
+
z
f
′′
(
z
)
f
′
(
z
)
)
≥
min
⁡
R
(
z
k
′
(
z
)
k
(
z
)
)
-
max
⁡
|
z
q
′
(
z
)
q
(
z
)
|
-
max
⁡
|
z
p
′
(
z
)
p
(
z
)
|
.
For k
∈
S
*, we have (see [7])
(35)
R
(
z
k
′
(
z
)
k
(
z
)
)
≥
1
-
r
1
+
r
with z
∈
U and |
z
|
=
r
<
1.
Sincep
∈
P
α and q
∈
P
β, making use of inequality (9) from Lemma 1, we obtain
(36)
|
z
p
′
(
z
)
p
(
z
)
|
≤
(
2
-
(
1
/
α
)
)
r
1
-
(
1
/
α
)
r
-
(
1
-
(
1
/
α
)
)
r
2
,
|
z
q
′
(
z
)
q
(
z
)
|
≤
(
2
-
(
1
/
β
)
)
r
1
-
(
1
/
β
)
r
-
(
1
-
(
1
/
β
)
)
r
2
with z
∈
U and |
z
|
=
r
<
1.
Substituting (35) and (36) in (34) we have
(37)
R
(
1
+
z
f
′′
(
z
)
f
′
(
z
)
)
≥
1
-
r
1
+
r
-
(
2
-
(
1
/
α
)
)
r
1
-
(
1
/
α
)
r
-
(
1
-
(
1
/
α
)
)
r
2
-
(
2
-
(
1
/
β
)
)
r
1
-
(
1
/
β
)
r
-
(
1
-
(
1
/
β
)
)
r
2
.
It follows that the functionf is convex whenever the expression in the right-hand side of (37) is positive. The numerator of this expression can be written as P
(
r
)
=
(
1
-
r
)
Q
(
r
), where
(38)
Q
(
r
)
=
(
α
+
β
-
α
β
-
1
)
r
4
+
4
(
α
β
-
α
-
β
+
1
)
r
3
+
(
10
α
β
-
5
α
-
5
β
+
1
)
r
2
+
4
α
β
r
-
α
β
.
We observe that Q
(
0
)
=
-
α
β
<
0 and Q
(
1
)
=
4
(
2
α
-
1
)
(
2
β
-
1
)
>
0. It follows that the smallest root r
0 of Q
(
r
)
=
0 and also of P
(
r
)
=
0 lies between 0 and 1 and, thus, the theorem is proved.


## 4. Coefficient Estimates

In order to find coefficient estimates for the classC
C
(
α
,
β
), we will find first the coefficient estimates for the class C
C
(
α
).Theorem 7.
Letα
>
1
/
2. If the function f given by (1) is in the class C
C
(
α
), then
(39)
|
a
n
|
≤
1
+
(
2
α
-
1
)
(
n
-
1
)
2
α
(
n
∈
{
2,3
,
…
}
)
.Proof.
Sincef
∈
C
C
(
α
) we have
(40)
f
′
(
z
)
=
p
(
z
)
g
′
(
z
)
(
z
∈
U
)
,
where
(41)
g
(
z
)
=
z
+
∑
n
=
2
∞
‍
b
n
z
n
∈
K
,
p
(
z
)
=
1
+
∑
n
=
1
∞
‍
p
n
z
n
∈
P
α
.
Equating the coefficients of z
n on both sides of (40), we find the following relation between the coefficients:
(42)
n
a
n
=
n
b
n
+
p
n
-
1
+
2
b
2
p
n
-
2
+
⋯
+
(
n
-
1
)
b
n
-
1
p
1
(
n
∈
{
2,3
,
…
}
)
.
For g
∈
K we have |
b
k
|
≤
1, k
≥
2 (see [5, 6]). In virtue of inequality (7) of Lemma 1 we have |
p
k
|
≤
2
-
(
1
/
α
), k
∈
{
1,2
,
…
}.
Making use of (42), we find
(43)
n
|
a
n
|
≤
n
+
(
2
-
1
α
)
∑
k
=
1
n
-
1
‍
k
=
n
+
(
2
-
1
α
)
n
(
n
-
1
)
2
(
n
∈
{
2,3
,
…
}
)
and, thus, we get the desired inequality (39).Whenα
→
∞ we find the well-known coefficient estimates of close-to-convex functions (see [5, 6]).In the next theorem we obtain the coefficient estimates for the classC
C
(
α
,
β
).Theorem 8.
Letα
>
1
/
2 and β
>
1
/
2. If the function f given by (1) is in the class C
C
(
α
,
β
), then
(44)
|
a
n
|
≤
1
+
(
4
-
1
α
-
1
β
)
n
-
1
2
+
(
2
-
1
α
)
(
2
-
1
β
)
(
n
-
1
)
(
n
-
2
)
6
(
n
∈
{
2,3
,
…
}
)
.Proof.
Letf
∈
C
C
(
α
,
β
). Then, there exist
(45)
g
(
z
)
=
z
+
∑
n
=
2
∞
‍
b
n
z
n
∈
C
C
(
α
)
,
q
(
z
)
=
1
+
∑
n
=
1
∞
‍
q
n
z
n
∈
P
β
such that
(46)
f
′
(
z
)
=
q
(
z
)
g
′
(
z
)
(
z
∈
U
)
.
Comparing the coefficients of z
n on both sides of the above equality, we obtain the next relation:
(47)
n
a
n
=
n
b
n
+
q
n
-
1
+
2
b
2
q
n
-
2
+
⋯
+
(
n
-
1
)
b
n
-
1
q
1
(
n
∈
{
2,3
,
…
}
)
.
Since g
∈
C
C
(
α
) and q
∈
P
β, from (39) and (7), we get
(48)
|
b
k
|
≤
1
+
(
2
α
-
1
)
(
k
-
1
)
2
α
(
k
∈
{
2,3
,
…
}
)
,
|
q
k
|
≤
2
-
1
β
(
k
∈
{
1,2
,
…
}
)
.
From (47) in connection with (48), we obtain
(49)
|
a
n
|
≤
1
+
(
2
α
-
1
)
(
n
-
1
)
2
α
+
1
n
(
2
-
1
β
)
[
1
+
∑
k
=
2
n
-
1
‍
k
(
1
+
(
2
α
-
1
)
(
k
-
1
)
2
α
)
]
=
1
+
(
2
α
-
1
)
(
n
-
1
)
2
α
+
1
n
[

(
2
-
1
β
)

(
1
+
∑
k
=
2
n
-
1
‍
k
)

+
(
2
-
1
β
)

2
α
-
1
2
α
∑
k
=
2
n
-
1
‍
k
(
k
-
1
)

]
=
1
+
(
2
α
-
1
)
(
n
-
1
)
2
α
+
2
β
-
1
2
β
(
n
-
1
)
+
(
2
α
-
1
)
(
2
β
-
1
)
6
α
β
(
n
-
1
)
(
n
-
2
)
which leads to inequality (44).


## 5. Maximum Value of|
a
3
-
(
2
/
3
)
a
2
2
|

The problem of finding sharp upper bounds for the functional|
a
3
-
μ
a
2
2
| for a family of analytic functions is known as the Fekete-Szegö problem. For the classes K and C, the following estimates are known (see, e.g., [8–11]):
(50)
max
⁡
f
∈
K
|
a
3
-
μ
a
2
2
|
=
max
⁡
{
1
3
,
|
μ
-
1
|
}
,
(51)
max
⁡
f
∈
C
|
a
3
-
μ
a
2
2
|
=
{
3
-
4
μ
,
0
≤
μ
≤
1
3
1
3
+
4
9
μ
,
1
3
≤
μ
≤
2
3
1
,
2
3
≤
μ
≤
1
.In this section, the caseμ
=
2
/
3 of the Fekete-Szegö problem will be considered, first for the class C
C
(
α
) and then, for the class C
C
(
α
,
β
).In order to prove our results we need the following lemma due to Keogh and Merkes [9].Lemma 9.
Letω
(
z
)
=
e
1
z
+
e
2
z
2
+
⋯ be in the class B and let λ
∈
C. Then
(52)
|
e
2
-
λ
e
1
2
|
≤
max
⁡
{
1
,
|
λ
|
}
.
Equality may be attained for ω
(
z
)
=
z
2 and ω
(
z
)
=
z.Theorem 10.
Letα
>
1
/
2. If f
∈
C
C
(
α
) is of the form (1), then
(53)
|
a
3
-
2
3
a
2
2
|
≤
1
-
1
3
α
.Proof.
Letf
∈
C
C
(
α
). Then there exists g
∈
K such that f
′
(
z
)
/
g
′
(
z
)
≺
p
α
(
z
), where p
α
(
z
) is given by (4). Let g
(
z
)
=
z
+
b
2
z
2
+
b
3
z
3
+
⋯. Define
(54)
p
(
z
)
=
f
′
(
z
)
g
′
(
z
)
=
1
+
p
1
z
+
p
2
z
2
+
⋯
(
z
∈
U
)
.
Since p
≺
p
α, there exists ω
(
z
)
=
e
1
z
+
e
2
z
2
+
⋯
∈
B such that
(55)
p
(
z
)
=
p
α
(
ω
(
z
)
)
=
1
+
P
1
e
1
z
+
(
P
1
e
2
+
P
2
e
1
2
)
z
2
+
⋯
(
z
∈
U
)
,
where P
1 and P
2 are given by (6). Combining (54) and (55), after simple calculations, we get
(56)
a
2
=
b
2
+
p
1
2
,
a
3
=
b
3
+
p
2
3
+
2
3
b
2
p
1
,
(57)
p
1
=
P
1
e
1
,
p
2
=
P
1
e
2
+
P
2
e
1
2
.
Substituting (6) and (57) in (56) we obtain
(58)
a
2
=
b
2
+
(
1
-
1
2
α
)
e
1
,
a
3
=
b
3
+
2
3
(
2
-
1
α
)
b
2
e
1
+
2
3
(
2
-
1
α
)
[
e
2
+
(
1
-
1
α
)
e
1
2
]
so that
(59)
a
3
-
2
3
a
2
2
=
(
b
3
-
2
3
b
2
2
)
+
1
3
(
2
-
1
α
)
(
e
2
-
1
2
α
e
1
2
)
.
Since g
∈
K, making use of (50) with μ
=
2
/
3, we have
(60)
|
b
3
-
2
3
b
2
2
|
≤
1
3
.
In virtue of Lemma 9 and taking into account that α
>
1
/
2, we get
(61)
|
e
2
-
1
2
α
e
1
2
|
≤
1
.
Combining (59), (60), and (61), we obtain
(62)
|
a
3
-
2
3
a
2
2
|
≤
|
b
3
-
2
3
b
2
2
|
+
1
3
(
2
-
1
α
)
|
e
2
-
1
2
α
e
1
2
|
≤
1
3
+
1
3
(
2
-
1
α
)
=
1
-
1
3
α
and, thus, the proof is completed.It is easy to observe that whenα
→
∞ inequality (53) reduces to |
a
3
-
(
2
/
3
)
a
2
2
|
≤
1 which is the same with (51) for μ
=
2
/
3.Theorem 11.
Letα
>
1
/
2 and β
>
1
/
2. If f of the form (1) belongs to the class C
C
(
α
,
β
), then
(63)
|
a
3
-
2
3
a
2
2
|
≤
1
3
(
5
-
1
α
-
1
β
)
.Proof.
Sincef
∈
C
C
(
α
,
β
), there exists g
∈
C
C
(
α
) such that f
′
(
z
)
/
g
′
(
z
)
≺
p
β
(
z
), where p
β
(
z
) is given by (4) with β instead of α. Let g
(
z
)
=
z
+
b
2
z
2
+
b
3
z
3
+
⋯ and p
(
z
) defined by
(64)
p
(
z
)
=
f
′
(
z
)
g
′
(
z
)
=
1
+
p
1
z
+
p
2
z
2
+
⋯
(
z
∈
U
)
.
From p
≺
p
β it follows that there exists ω
(
z
)
=
e
1
z
+
e
2
z
2
+
⋯
∈
B such that p
(
z
)
=
p
β
(
ω
(
z
)
).
Using the same method as in the proof of Theorem10, we obtain
(65)
a
3
-
2
3
a
2
2
=
(
b
3
-
2
3
b
2
2
)
+
1
3
(
2
-
1
β
)
(
e
2
-
1
2
β
e
2
2
)
.
Since g
∈
C
C
(
α
), from (53), we have
(66)
|
b
3
-
2
3
b
2
2
|
≤
1
-
1
3
α
.
Moreover, for β
>
1
/
2, we get from Lemma 9 that
(67)
|
e
2
-
1
2
β
e
2
2
|
≤
1
.
Combining (65), (66), and (67), the inequality (63) follows.


## 6. Bounded Doubly Close-to-Starlike Functions

Letα
>
1
/
2. Consider S
*
S
(
α
) the class of all functions f
∈
A for which there exists a function g
∈
S
* such that f
/
g
≺
p
α, with p
α given by (4), or equivalently
(68)
|
f
(
z
)
g
(
z
)
-
α
|
<
α
(
z
∈
U
)
.It is easy to observe that whenα
→
∞ the class S
*
S
(
α
) reduces to the class of close-to-starlike functions defined by Reade [12].Forα
>
1
/
2 and β
>
1
/
2 we denote by S
*
S
(
α
,
β
) the class of functions f
∈
A for which there exists a function g
∈
S
*
S
(
α
) such that f
/
g
≺
p
β, with p
β given by (4), or equivalently
(69)
|
f
(
z
)
g
(
z
)
-
β
|
<
β
(
z
∈
U
)
.Connections between the classesS
*
S
(
α
) and C
C
(
α
) and also between S
*
S
(
α
,
β
) and C
C
(
α
,
β
) are given in the next theorem.Theorem 12.
Letα
>
1
/
2 and β
>
1
/
2. Then, the following relationships hold:
(70)
f
(
z
)
∈
C
C
(
α
)
i
f
f
z
f
′
(
z
)
∈
S
*
S
(
α
)
,
(71)
f
(
z
)
∈
C
C
(
α
,
β
)
i
f
f
z
f
′
(
z
)
∈
S
*
S
(
α
,
β
)
,
(72)
f
(
z
)
∈
S
*
S
(
α
)
i
f
f
∫
0
z
‍
f
(
t
)
t
d
t
∈
C
C
(
α
)
,
(73)
f
(
z
)
∈
S
*
S
(
α
,
β
)
i
f
f
∫
0
z
‍
f
(
t
)
t
d
t
∈
C
C
(
α
,
β
)
.Proof.
It is well known that a functiong
(
z
)
∈
K if and only if z
g
′
(
z
)
∈
S
*.
The definition of the classC
C
(
α
) implies that f
∈
C
C
(
α
) if and only if there exists g
∈
K such that f
′
(
z
)
/
g
′
(
z
)
≺
p
α
(
z
).
The relation (70) follows from
(74)
z
f
′
(
z
)
z
g
′
(
z
)
=
f
′
(
z
)
g
′
(
z
)
≺
p
α
(
z
)
,
z
g
′
(
z
)
∈
S
*
.
In the same way,f
∈
C
C
(
α
,
β
) if and only if there exists g
∈
C
C
(
α
) such that f
′
(
z
)
/
g
′
(
z
)
≺
p
β
(
z
). We have
(75)
z
f
′
(
z
)
z
g
′
(
z
)
=
f
′
(
z
)
g
′
(
z
)
≺
p
β
(
z
)
and taking into account (70), the relation (71) follows.
The proofs of (72) and (73) are similar and will be omitted.The condition (71) of Theorem 12 together with Lemma 3 and (14) shows that the function
(76)
g
(
z
)
=
z
(
1
+
ϵ
1
z
)
(
1
+
ϵ
2
z
)
(
1
+
ϵ
3
z
)
2
[
1
-
(
1
-
(
1
/
α
)
)
ϵ
1
z
]
[
1
-
(
1
-
(
1
/
β
)
)
ϵ
2
z
]
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
(
z
∈
U
)
,
where |
ϵ
k
|
=
1, k
∈
{
1,2
,
3
} belongs to the class S
*
S
(
α
,
β
) and, thus, this class is nonempty.Combining Theorem12 with Theorems 4 and 8 the next properties of the class S
*
S
(
α
,
β
) can be easily obtained.Corollary 13.
Letα
>
1
/
2 and β
>
1
/
2. If f
∈
S
*
S
(
α
,
β
), then
(77)
r
(
1
-
r
)
2
(
1
+
r
)
2
[
1
+
(
1
-
(
1
/
α
)
)
r
]
[
1
+
(
1
-
(
1
/
β
)
)
r
]
≤
|
f
(
z
)
|
≤
r
(
1
+
r
)
2
(
1
-
r
)
2
[
1
-
(
1
-
(
1
/
α
)
)
r
]
[
1
-
(
1
-
(
1
/
β
)
)
r
]
for z
∈
U and |
z
|
=
r
<
1.Corollary 14.
Letα
>
1
/
2 and β
>
1
/
2. If f
∈
S
*
S
(
α
,
β
) is given by (1), then
(78)
|
a
n
|
≤
n
+
(
4
-
1
α
-
1
β
)
n
(
n
-
1
)
2
+
(
2
-
1
α
)
(
2
-
1
β
)
n
(
n
-
1
)
(
n
-
2
)
6
m
m
m
m
m
m
m
m
m
m
m
m
(
n
∈
{
2,3
,
…
}
)
.Making use of Theorem12, we can also obtain an upper bound of |
a
3
-
(
1
/
2
)
a
2
2
| for functions in the class S
*
S
(
α
,
β
).Corollary 15.
Letα
>
1
/
2 and β
>
1
/
2. If f
∈
S
*
S
(
α
,
β
) is given by (1), then
(79)
|
a
3
-
1
2
a
2
2
|
≤
5
-
1
α
-
1
β
.Proof.
Letf
∈
S
*
S
(
α
,
β
). Then, from (73), the function F
(
z
)
=
z
+
b
2
z
2
+
b
3
z
3
+
⋯ given by
(80)
F
(
z
)
=
∫
0
z
‍
f
(
t
)
t
d
t
belongs to the class C
C
(
α
,
β
). Comparing the coefficients of z
2 and z
3 on both sides of the above equality, we obtain
(81)
a
2
=
2
b
2
,
a
3
=
3
b
3
so that
(82)
|
a
3
-
1
2
a
2
2
|
=
|
3
b
3
-
2
b
2
2
|
=
3
|
b
3
-
2
3
b
2
2
|
.
Now, the inequality (79) follows as an application of Theorem 11.Once again making use of Theorem12, we have that f
∈
S
*
S
(
α
,
β
) if and only if
(83)
z
f
′
(
z
)
f
(
z
)
=
1
+
z
F
′′
(
z
)
F
′
(
z
)
(
z
∈
U
)
for some F
∈
C
C
(
α
,
β
). Therefore, a radius of convexity for C
C
(
α
,
β
) will correspond to a radius of starlikeness for S
*
S
(
α
,
β
).The next result follows easily from Theorem6.Corollary 16.
Letα
>
1
/
2 and β
>
1
/
2. Suppose that f
∈
S
*
S
(
α
,
β
). Then, the function f maps the disk {
z
∈
C
:
|
z
|
<
r
0
<
1
} onto a starlike domain, where r
0 is the smallest positive root of (28) in Theorem 6.



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*Source: 804095-2014-08-18.xml*