# On a Subclass of Analytic Functions Related to a Hyperbola

Authors: Jagannath Patel; Ashok Kumar Sahoo
Journal: International Journal of Mathematics and Mathematical Sciences
 (2014)

---

## Abstract

The object of the present investigation is to solve Fekete-Szegö problem and determine the sharp upper bound to the second Hankel determinant for a new classℛ
̃
(
a
,
c
,
ρ
) of analytic functions in the unit disk. We also obtain a sufficient condition for an analytic function to be in this class.


---

## Body

## 1. Introduction and Preliminaries

Let𝒜 be the class of functions f of the form:
(1)
f
(
z
)
=
z
+
∑
n
=
2
∞
a
n
z
n
which are analytic in the open unit disk 𝒰
=
{
z
∈
ℂ
:
|
z
|
<
1
}.A functionf
∈
𝒜 is said to be starlike function of order ρ and convex function of order ρ, respectively, if and only if Re
{
z
f
′
(
z
)
/
f
(
z
)
}
>
ρ and Re
{
1
+
(
z
f
′′
(
z
)
/
f
′
(
z
)
)
}
>
ρ, for 0
≤
ρ
<
1 and for all z
∈
𝒰. By usual notations, we denote these classes of functions by 𝒮
⋆
(
ρ
) and 𝒦
(
ρ
)
(
0
≤
ρ
<
1
), respectively. We write 𝒮
⋆
(
0
)
=
𝒮
⋆ and 𝒦
(
0
)
=
𝒦, the familiar subclasses of starlike functions and convex functions in 𝒰.Furthermore, a functionf
∈
𝒜 is said to in the class ℛ
(
ρ
), if it satisfies the inequality:
(2)
Re
{
f
′
(
z
)
}
>
ρ
(
0
≤
ρ
<
1
;
z
∈
𝒰
)
.
Note that ℛ
(
ρ
) is a subclass of close-to-convex functions of order ρ
(
0
≤
ρ
<
1
) in 𝒰.Let𝒫 denote the class of analytic functions of the form:
(3)
ϕ
(
z
)
=
1
+
p
1
z
+
p
2
z
2
+
⋯
(
z
∈
𝒰
)
satisfying the condition Re
{
ϕ
(
z
)
}
>
0 in 𝒰.Let the functionsf and g be analytic in 𝕌. We say that f is subordinate to g, written as f
≺
g or f
(
z
)
≺
g
(
z
)
(
z
∈
𝕌
), if there exists a Schwarz function ω, which (by definition) is analytic in 𝕌 with ω
(
0
)
=
0
,
|
ω
(
z
)
|
<
1 and f
(
z
)
=
g
(
ω
(
z
)
)
,
z
∈
𝕌. Furthermore, if the function g is univalent in 𝕌, then we have the following equivalence relation (cf., e.g., [1]):
(4)
f
(
z
)
≺
g
(
z
)
⟺
f
(
0
)
=
g
(
0
)
,
f
(
𝕌
)
⊂
g
(
𝕌
)
.For the functionsf
,
g analytic in 𝒰 and given by the power series
(5)
f
(
z
)
=
∑
n
=
0
∞
a
n
z
n
,
g
(
z
)
=
∑
n
=
0
∞
b
n
z
n
,
their Hadamard product (or convolution), denoted by f
⋆
g is defined as
(6)
(
f
⋆
g
)
(
z
)
=
∑
n
=
0
∞
a
n
b
n
z
n
=
(
g
⋆
f
)
(
z
)
(
z
∈
𝒰
)
.
Note that f
⋆
g is analytic in 𝒰.TheGauss hypergeometric function
2
F
1 is defined by the infinite series
(7)
2
F
1
(
a
,
b
;
c
;
z
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
z
n
n
!
(
a
,
b
,
c
∈
ℂ
,
c
∉
ℤ
0
-
=
{
0
,
-
1
,
-
2
,
…
}
;
z
∈
𝒰
)
,
where (
κ
)
n denotes thePochhammer symbol (orshifted factorial) given, in terms of the Gamma function Γ, by
(8)
(
κ
)
n
=
Γ
(
κ
+
n
)
Γ
(
κ
)
=
{
κ
(
κ
+
1
)
⋯
(
κ
+
n
-
1
)
,
n
∈
ℕ
1
,
n
=
0
.
We note that the series, given by (7), converges absolutely for z
∈
𝒰 and hence the function 2
F
1 represents an analytic function in the unit disc 𝒰 [2].We further observe that theGauss hypergeometric function
2
F
1 plays an important role in the study of various properties and characteristics of subclasses of univalent/multivalent functions in geometric function theory (cf., e.g. [3–5]). In our present investigation, we consider the incomplete beta function ψ, defined by
(9)
ψ
(
a
,
c
;
z
)
=
z
2
F
1
(
a
,
1
;
c
;
z
)
=
∑
n
=
0
∞
‍
(
a
)
n
(
c
)
n
z
n
+
1
hhhhhh
(
a
,
c
∈
ℂ
,
c
∉
ℤ
0
-
;
z
∈
𝒰
)
.By making use of the Hadamard product and the functionψ, Carlson and Shaffer [6] defined the linear operator ℒ
(
a
,
c
)
:
𝒜
→
𝒜 by
(10)
ℒ
(
a
,
c
)
f
(
z
)
=
ψ
(
a
,
c
;
z
)
⋆
f
(
z
)
(
f
∈
𝒜
;
z
∈
𝒰
)
.
If f
∈
𝒜 is given by (1), then it follows from (10) that
(11)
ℒ
(
a
,
c
)
f
(
z
)
=
z
+
∑
n
=
1
∞
‍
(
a
)
n
(
c
)
n
a
n
+
1
z
n
+
1
(
z
∈
𝒰
)
,
(12)
z
(
ℒ
(
a
,
c
)
f
)
′
(
z
)
=
a
ℒ
(
a
+
1
,
c
)
f
(
z
)
-
(
a
-
1
)
ℒ
(
a
,
c
)
f
(
z
)
(
z
∈
𝒰
)
.
The operator ℒ
(
a
,
c
) extends several operators introduced and studied by earlier researchers in geometric function theory.  For example, ℒ
(
m
+
1,1
)
f
(
z
)
=
𝒟
m
f
(
z
)
(
f
∈
𝒜
,
m
∈
ℤ
,
m
>
-
1
;
z
∈
𝒰
), the well-known Ruscheweyh derivative operator [7] of f and ℒ
(
2,2
-
λ
)
f
(
z
)
=
Ω
z
λ
f
(
z
)
(
f
∈
𝒜
,
0
≤
λ
<
1
;
z
∈
𝒰
), the familiar Owa-Srivastava fractional differential operator [8] of f.With the aid of the linear operatorℒ
(
a
,
c
), we introduce a subclass of 𝒜 as follows.Definition 1.
A functionf
∈
𝒜 is said to be in the class ℛ
~
(
a
,
c
,
ρ
), if it satisfies the following subordination relation:
(13)
ℒ
(
a
,
c
)
f
(
z
)
z
≺
{
1
+
(
1
-
2
ρ
)
z
1
-
z
}
1
/
2
(
0
≤
ρ
<
1
;
z
∈
𝒰
)
,
where the power in the right hand side of (13) indicates the principal branch. Note that if f
∈
ℛ
~
(
a
,
c
,
ρ
), then by (13)
(14)
ℒ
(
a
,
c
)
f
(
z
)
z
=
{
ρ
+
(
1
-
ρ
)
ϕ
(
z
)
}
1
/
2
(
ϕ
∈
𝒫
;
z
∈
𝒰
)
.We denote byℛ
~
(
2,1
,
ρ
)
=
ℛ
~
(
ρ
), the class of functions f
∈
𝒜 satisfying the subordination condition:
(15)
f
′
(
z
)
≺
{
1
+
(
1
-
2
ρ
)
z
1
-
z
}
1
/
2
(
0
≤
ρ
<
1
;
z
∈
𝒰
)
.
In fact, by suitably specializing the parameters a
,
c, and ρ in the class ℛ
~
(
a
,
c
,
ρ
), we can obtain several subclasses of 𝒜.Remark 2.
To bring out the geometrical significance of the classℛ
~
(
a
,
c
,
ρ
), we set
(16)
h
ρ
(
z
)
=
{
1
+
(
1
-
2
ρ
)
z
1
-
z
}
1
/
2
(
0
≤
ρ
<
1
;
z
∈
𝒰
)
and note that
(17)
ω
=
h
ρ
(
e
i
θ
)
=
1
+
(
1
-
2
ρ
)
e
i
θ
1
-
e
i
θ
(
0
≤
θ
≤
2
π
)
which gives e
i
θ
(
ω
2
-
2
ρ
+
1
)
=
ω
2
-
1 or |
ω
2
-
1
|
=
|
ω
2
+
1
-
2
ρ
|. Letting ω
=
u
+
i
v, we deduce that
(18)
1
-
(
u
2
-
v
2
)
2
+
4
u
2
v
2
=
(
u
2
-
v
2
+
1
-
2
ρ
)
2
+
4
u
2
v
2
,
which on simplification reduces to u
2
-
v
2
=
ρ. Thus, h
ρ
(
𝒰
) is the interior of the right half branch of the hyperbola u
2
-
v
2
=
ρ. Hence, if f
∈
ℛ
~
(
a
,
c
,
ρ
), then the set of values ℒ
(
a
,
c
)
f
(
z
)
/
z for z
∈
𝒰 lie in h
ρ
(
𝒰
), where h
ρ is given by (16).Fekete and Szegö [9] defined the Hankel determinant of a function f, given by (1) as
(19)
H
2
(
1
)
=
|
a
1
a
2
a
2
a
3
|
=
a
3
-
a
2
2
(
a
1
=
1
)
.
In our present investigation, we also consider the second Hankel determinant of f, given by
(20)
H
2
(
2
)
=
|
a
2
a
3
a
3
a
4
|
=
a
2
a
4
-
a
3
2
.
It is known [10] that if f given by (1) is analytic and univalent in 𝒰, then the sharp inequality H
2
(
1
)
=
|
a
3
-
a
2
2
|
≤
1 holds. For a family ℱ of functions in 𝒜 of the form (1), the more general problem of finding the sharp upper bounds for the functionals (
a
3
-
μ
a
2
2
)
(
μ
∈
ℝ
/
ℂ
) is popularly known as Fekete-Szegö problem for the class ℱ. The Fekete-Szegö problem for the known classes of univalent functions, starlike functions, convex functions, and close-to-convex functions has been completely settled [9, 11–18]. Recently, Janteng et al. [19, 20] have obtained the sharp upper bounds to the second Hankel determinant H
2
(
2
) for the family ℛ of functions in 𝒜 whose derivatives have positive real part in 𝒰. For initial work on the class ℛ, one may refer to the paper by MacGregor [21].Our objective in the present paper is to solve the Fekete-Szegö problem and also to determine the sharp upper bound to the second Hankel determinant for the classℛ
~
(
a
,
c
,
ρ
) by following the techniques devised by Libera and Złotkiewicz [22, 23]. The criteria for functions in 𝒜 to be in this class are also obtained.To establish our main results, we will need the following results about the functions belonging to the class𝒫.Lemma 3.
Let the functionϕ, given by (3), be a member of the class 𝒫. Then
(21)
|
p
k
|
≤
2
(
k
≥
1
)
,
(22)
|
p
2
-
ν
p
1
2
|
≤
2
max
⁡
{
1
,
|
2
ν
-
1
|
}
(
ν
∈
ℂ
)
,
(23)
p
2
=
1
2
{
p
1
2
+
(
4
-
p
1
2
)
x
}
,
(24)
p
3
=
1
4
{
p
1
3
+
2
(
4
-
p
1
2
)
p
1
x
-
(
4
-
p
1
2
)
p
1
x
2
j
j
j
j
j
j
j
+
2
(
4
-
p
1
2
)
(
1
-
|
x
|
2
)
z
}
for some complex numbers x
,
z satisfying |
x
|
≤
1 and |
z
|
≤
1. The estimates in (21) and (22) are sharp.We note that the estimate (21) is contained in [10]; the estimate (22) is obtained by Ma and Minda [24]; the results in (23) and (24) are due to Libera and Złotkiewicz [23] (see also [22]).


## 2. Main Results

Unless otherwise mentioned, we assume throughout the sequel that(25)
a
>
0
,
c
>
0
,
0
≤
ρ
<
1
.Now, we determine the sharp upper bound for the functional|
a
3
-
μ
a
2
2
|
(
μ
∈
ℂ
) for functions of the form (1) belonging to the class ℛ
~
(
a
,
c
,
ρ
).Theorem 4.
Leta
>
0 and c
>
0. If the function f, given by (1), belongs to the class ℛ
~
(
a
,
c
,
ρ
), then for any μ
∈
ℂ
(26)
|
a
3
-
μ
a
2
2
|
≤
(
c
)
2
(
1
-
ρ
)
(
a
)
2
×
max
⁡
{
1
,
|
2
(
a
+
1
)
c
(
1
-
ρ
)
μ
-
a
(
c
+
1
)
(
1
+
ρ
)
|
2
a
(
c
+
1
)
}
.
The estimate in (26) is sharp.Proof.
Sincef
∈
ℛ
~
(
a
,
c
,
ρ
), by (14) we have
(27)
ℒ
(
a
,
c
)
f
(
z
)
z
=
{
ρ
+
(
1
-
ρ
)
ϕ
(
z
)
}
1
/
2
(
z
∈
𝒰
)
,
where ϕ
∈
𝒫 is given by (3). It is easily seen that
(28)
{
ρ
+
(
1
-
ρ
)
ϕ
(
z
)
}
1
/
2
=
1
+
1
2
(
1
-
ρ
)
p
1
z
+
1
-
ρ
2
{
p
2
-
(
1
-
ρ
)
4
p
1
2
}
z
2
+
1
-
ρ
2
{
p
3
-
(
1
-
ρ
)
2
p
1
p
2
+
(
1
-
ρ
)
2
8
p
1
3
}
z
3
+
⋯
hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
(
z
∈
𝒰
)
.
Writing the series expansion of ℒ
(
a
,
c
)
f
(
z
) given by (11), {
ρ
+
(
1
-
ρ
)
ϕ
(
z
)
}
1
/
2, in (27) and equating the coefficients of z
,
z
2
,
z
3 in the resulting equation, we obtain
(29)
a
2
=
(
1
-
ρ
)
c
2
a
p
1
,
(30)
a
3
=
(
1
-
ρ
)
(
c
)
2
2
(
a
)
2
{
p
2
-
(
1
-
ρ
)
4
p
1
2
}
,
(31)
a
4
=
(
1
-
ρ
)
(
c
)
3
2
(
a
)
3
{
p
3
-
(
1
-
ρ
)
2
p
1
p
2
+
(
1
-
ρ
)
2
8
p
1
3
}
.
Thus for any μ
∈
ℂ,
(32)
|
a
3
-
μ
a
2
2
|
=
(
c
)
2
(
1
-
ρ
)
2
(
a
)
2
|
p
2
-
(
1
-
ρ
)
{
a
(
c
+
1
)
+
2
(
a
+
1
)
c
μ
}
4
a
(
c
+
1
)
p
1
2
|
and by using (22) in the above expression, we get
(33)
|
a
3
-
μ
a
2
2
|
≤
(
c
)
2
(
1
-
ρ
)
(
a
)
2
×
max
⁡
{
1
,
|
(
1
-
ρ
)
2
a
(
c
+
1
)
{
a
(
c
+
1
)
+
2
(
a
+
1
)
c
μ
}
-
1
|
}
which, upon simplification, gives the required assertion of Theorem 4.
Equality in (26) holds for the function f
0 defined in 𝒰 by
(34)
f
0
(
z
)
=
{
ψ
(
c
,
a
;
z
)
⋆
z
h
ρ
(
z
)
,
|
2
(
a
+
1
)
c
(
1
-
ρ
)
μ
-
a
(
c
+
1
)
(
1
+
ρ
)
|
2
a
(
c
+
1
)
≤
1
,
ψ
(
c
,
a
;
z
)
⋆
z
h
ρ
(
z
2
)
,
|
2
(
a
+
1
)
c
(
1
-
ρ
)
μ
-
a
(
c
+
1
)
(
1
+
ρ
)
|
2
a
(
c
+
1
)
>
1
,
where the function h
ρ is given by (16). This completes the proof of Theorem 4.Theorem 5.
Leta
>
0
,
c
>
0 and μ
∈
ℝ. If the function f, given by (1), belongs to the class ℛ
~
(
a
,
c
,
ρ
), then
(35)
|
a
3
-
μ
a
2
2
|
≤
{
{
a
(
c
+
1
)
(
1
+
ρ
)
-
2
(
a
+
1
)
c
(
1
-
ρ
)
μ
}
c
(
1
-
ρ
)
2
a
(
a
)
2
,
μ
<
-
a
(
c
+
1
)
2
(
a
+
1
)
c
(
c
)
2
(
1
-
ρ
)
(
a
)
2
,
-
a
(
c
+
1
)
2
(
a
+
1
)
c
≤
μ
≤
a
(
c
+
1
)
(
3
+
ρ
)
2
(
a
+
1
)
c
(
1
-
ρ
)
{
2
(
a
+
1
)
c
(
1
-
ρ
)
μ
-
a
(
c
+
1
)
(
1
+
ρ
)
}
c
(
1
-
ρ
)
2
a
(
a
)
2
,
μ
>
a
(
c
+
1
)
(
3
+
ρ
)
2
(
a
+
1
)
c
(
1
-
ρ
)
.
The estimates are sharp.Proof.
First, we assume thatμ
<
-
{
a
(
c
+
1
)
}
/
2
(
a
+
1
)
c. Then
(36)
2
(
a
+
1
)
c
(
1
-
ρ
)
μ
-
a
(
c
+
1
)
(
1
+
ρ
)
2
a
(
c
+
1
)
<
-
1
so that by (26), we obtain
(37)
|
a
3
-
μ
a
2
2
|
≤
{
a
(
c
+
1
)
(
1
+
ρ
)
-
2
(
a
+
1
)
c
(
1
-
ρ
)
μ
}
c
(
1
-
ρ
)
2
a
(
a
)
2
.
Next, let
(38)
-
a
(
c
+
1
)
2
(
a
+
1
)
c
≤
μ
≤
a
(
c
+
1
)
(
3
+
ρ
)
2
(
a
+
1
)
c
(
1
-
ρ
)
.
Then, a routine calculation yields
(39)
|
2
(
a
+
1
)
c
(
1
-
ρ
)
μ
-
a
(
c
+
1
)
(
1
+
ρ
)
|
2
a
(
c
+
1
)
≤
1
and by using (26) again, we get
(40)
|
a
3
-
μ
a
2
2
|
≤
(
c
)
2
(
1
-
ρ
)
(
a
)
2
.
Finally, if {
a
(
c
+
1
)
(
3
+
ρ
)
}
/
2
(
a
+
1
)
c
(
1
-
ρ
)
>
1, then
(41)
2
(
a
+
1
)
c
(
1
-
ρ
)
μ
-
a
(
c
+
1
)
(
1
+
ρ
)
2
a
(
c
+
1
)
>
1
.
Thus, by (26), we have
(42)
|
a
3
-
μ
a
2
2
|
≤
{
2
(
a
+
1
)
c
(
1
-
ρ
)
μ
-
a
(
c
+
1
)
(
1
+
ρ
)
}
c
(
1
-
ρ
)
2
a
(
a
)
2
.
The estimates are sharp for the functionf
0 defined in 𝒰 by
(43)
f
0
(
z
)
=
{
ψ
(
c
,
a
;
z
)
⋆
z
h
ρ
(
z
)
,
μ
<
-
a
(
c
+
1
)
2
(
a
+
1
)
c
or
μ
>
a
(
c
+
1
)
(
3
+
ρ
)
2
(
a
+
1
)
c
(
1
-
ρ
)
ψ
(
c
,
a
;
z
)
⋆
z
h
ρ
(
z
2
)
,
-
a
(
c
+
1
)
2
(
a
+
1
)
c
≤
μ
≤
a
(
c
+
1
)
(
3
+
ρ
)
2
(
a
+
1
)
c
(
1
-
ρ
)
,
where the function h
ρ is given by (16) and the proof of Theorem 5 is completed.Using (21) in (29) and putting μ
=
0 and μ
=
1, respectively, in Theorem 5, we get the following.Corollary 6.
Leta
≥
c
>
0. If the function f, given by (1), belongs to the class ℛ
~
(
a
,
c
,
ρ
), then
(44)
|
a
2
|
≤
c
(
1
-
ρ
)
a
,
(45)
|
a
3
|
≤
(
c
)
2
(
1
-
ρ
)
(
a
)
2
,
(46)
|
a
3
-
a
2
2
|
≤
(
c
)
2
(
1
-
ρ
)
(
a
)
2
.
The estimates in (44) and (46) are sharp for the function f
0 defined by
(47)
f
0
(
z
)
=
ψ
(
c
,
a
;
z
)
⋆
z
h
ρ
(
z
)
(
z
∈
𝒰
)
,
whereas the estimate in (45) is sharp for the function f
0 given by
(48)
f
0
(
z
)
=
ψ
(
c
,
a
;
z
)
⋆
z
h
ρ
(
z
2
)
(
z
∈
𝒰
)
,
where the function h
ρ is given by (16).Lettinga
=
2 and c
=
1 in Theorem 8, we obtain the following.Corollary 7.
If the functionf, given by (1), belongs to the class ℛ
~
(
ρ
), then
(49)
|
a
3
-
μ
a
2
2
|
≤
{
{
2
(
1
+
ρ
)
-
3
μ
(
1
-
ρ
)
}
(
1
-
ρ
)
12
,
μ
<
-
2
3
(
1
-
ρ
)
3
,
-
2
3
≤
μ
≤
2
(
3
+
ρ
)
3
(
1
-
ρ
)
{
3
μ
(
1
-
ρ
)
-
2
(
1
+
ρ
)
}
(
1
-
ρ
)
12
,
μ
>
2
(
3
+
ρ
)
3
(
1
-
ρ
)
.
The estimates are sharp for the function f
0 defined in 𝒰 by
(50)
f
0
(
z
)
=
{
∫
0
z
‍
d
t
1
-
t
⋆
z
h
ρ
(
z
)
,
μ
<
-
2
3
o
r
μ
>
2
(
3
+
ρ
)
3
(
1
-
ρ
)
∫
0
z
‍
d
t
1
-
t
⋆
z
h
ρ
(
z
2
)
,
-
2
3
≤
μ
≤
2
(
3
+
ρ
)
3
(
1
-
ρ
)
,
where h
ρ is given by (16).Next, we find the sharp upper bound for the fourth coefficient of functions in the classℛ
~
(
a
,
c
,
ρ
).Theorem 8.
Let the functionf, given by (1), belong to the class ℛ
~
(
a
,
c
,
ρ
). Then
(51)
|
a
4
|
≤
(
c
)
3
(
1
-
ρ
)
(
a
)
3
and the estimate in (51) is sharp.Proof.
From (31), we have
(52)
|
a
4
|
=
(
c
)
3
(
1
-
ρ
)
2
(
a
)
3
|
p
3
-
1
-
ρ
2
p
1
p
2
+
(
1
-
ρ
)
2
8
p
1
3
|
.
Since the functions ϕ
(
z
) and ϕ
(
e
i
θ
z
)
(
θ
∈
ℝ
) are in the class 𝒫 simultaneously, we assume without loss of generality that p
1
>
0. For convenience of notation, we write p
1
=
p
(
0
≤
p
≤
2
). Now, by using (23) and (24) in (52), we deduce that
(53)
|
a
4
|
=
(
c
)
3
(
1
-
ρ
)
2
(
a
)
3
|
1
+
ρ
2
8
p
3
+
1
+
ρ
4
(
4
-
p
2
)
p
x
-
1
4
(
4
-
p
2
)
p
x
2
+
1
2
(
4
-
p
2
)
(
1
-
|
x
|
2
)
z
1
+
ρ
2
8
p
3
+
1
+
ρ
4
|
for some complex numbers x
(
|
x
|
≤
1
) and z
(
|
z
|
≤
1
).
Applying the triangle inequality in the above expression followed by the replacement of|
x
| with y in the resulting equation, we obtain
(54)
|
a
4
|
≤
(
c
)
3
(
1
-
ρ
)
2
(
a
)
3
{
1
+
ρ
2
8
p
3
+
1
+
ρ
4
(
4
-
p
2
)
p
y
+
1
4
(
4
-
p
2
)
(
p
-
2
)
y
2
+
1
2
(
4
-
p
2
)
1
+
ρ
2
8
p
3
+
1
+
ρ
4
}
=
G
(
p
,
y
)
(
0
≤
p
≤
2,0
≤
y
≤
1
)
(
say
)
.
We next maximize the functionG
(
p
,
y
) on the closed rectangle [
0,2
]
×
[
0,1
]. Since
(55)
∂
G
∂
y
=
1
4
(
4
-
p
2
)
{
p
(
1
+
ρ
)
-
2
(
2
-
p
)
y
}
,
we have ∂
G
/
∂
y
<
0 for 0
<
p
<
2 and 0
<
y
<
1. Thus, G
(
p
,
y
) cannot have a maximum in the interior on the closed rectangle [
0,2
]
×
[
0,1
]. Therefore, for fixed p
∈
[
0,2
]
(56)
max
⁡
0
≤
y
≤
1
G
(
p
,
y
)
=
G
(
p
,
0
)
=
F
(
p
)
(
say
)
,
where
(57)
F
(
p
)
=
1
+
ρ
2
8
p
3
+
1
2
(
4
-
p
2
)
(
0
≤
p
≤
2
)
.
A routine calculation yields
(58)
F
′
(
p
)
=
3
(
1
+
ρ
2
)
8
p
2
-
p
=
0
for p
=
0 or p
=
8
/
{
3
(
1
+
ρ
2
)
}. Since F
′′
(
0
)
=
-
1
<
0 and F
′′
(
8
/
{
3
(
1
+
ρ
2
)
}
)
=
1
>
0, we conclude that the maximum of F is attained at p
=
0. Thus, the upper bound of the function G corresponds to p
=
y
=
0. Putting p
=
y
=
0 in (54), we get our desired estimate (51).
Equality in (51) holds for the function f
0 defined by
(59)
f
0
(
z
)
=
ψ
(
c
,
a
;
z
)
⋆
z
h
ρ
(
z
3
)
(
0
≤
ρ
<
1
;
z
∈
𝒰
)
,
where h
ρ is given by (16).In the following theorem, we find the sharp upper bound to the second Hankel determinant for the classℛ
~
(
a
,
c
,
ρ
).Theorem 9.
Leta
≥
c
>
0 and (
a
+
2
)
(
c
+
1
)
-
3
(
a
-
c
)
>
0. If the function f, given by (1), belongs to the class ℛ
~
(
a
,
c
,
ρ
), then
(60)
|
a
2
a
4
-
a
3
2
|
≤
{
(
c
)
2
(
1
-
ρ
)
(
a
)
2
}
2
.
The estimate in (60) is sharp.Proof.
From (29), (30), and (31), we deduce that
(61)
|
a
2
a
4
-
a
3
2
|
=
c
(
c
)
2
(
1
-
ρ
)
2
4
a
(
a
)
2
×
|
(
c
+
2
a
+
2
)
p
1
p
3
-
(
c
+
1
a
+
1
)
p
2
2
-
(
a
-
c
)
(
1
-
ρ
)
2
(
a
+
1
)
(
a
+
2
)
p
1
2
p
2
+
{
(
c
+
2
)
(
a
+
1
)
+
(
a
-
c
)
}
(
1
-
ρ
)
2
16
(
a
+
1
)
(
a
+
2
)
p
1
4
|
.
As in Theorem 8, we assume without loss of generality that p
1
>
0 and for convenience of notation, we write p
1
=
p
(
0
≤
p
≤
2
). By using (23) and (24) in (61), we get
(62)
|
a
2
a
4
-
a
3
2
|
=
c
(
c
)
2
(
1
-
ρ
)
2
4
a
(
a
)
2
×
|
4
(
a
-
c
)
ρ
+
{
(
c
+
2
)
(
a
+
1
)
+
(
a
-
c
)
}
(
1
-
ρ
)
2
16
(
a
+
1
)
(
a
+
2
)
p
4
+
(
a
-
c
)
(
1
+
ρ
)
4
(
a
+
1
)
(
a
+
2
)
(
4
-
p
2
)
p
2
x
-
{
4
(
a
+
2
)
(
c
+
1
)
+
(
a
-
c
)
p
2
}
4
(
a
+
1
)
(
a
+
2
)
(
4
-
p
2
)
x
2
+
c
+
2
2
(
a
+
2
)
(
4
-
p
2
)
p
(
1
-
|
x
|
2
)
z
4
(
a
-
c
)
ρ
+
{
(
c
+
2
)
(
a
+
1
)
+
(
a
-
c
)
}
(
1
-
ρ
)
2
16
(
a
+
1
)
(
a
+
2
)
|
.
Now, by applying the triangle inequality in (62) and replacing |
x
| by y in the resulting equation, we get
(63)
|
a
2
a
4
-
a
3
2
|
≤
c
(
c
)
2
(
1
-
ρ
)
2
4
a
(
a
)
2
×
{
4
(
a
-
c
)
ρ
+
{
(
a
+
1
)
(
c
+
2
)
+
(
a
-
c
)
}
(
1
-
ρ
)
2
16
(
a
+
1
)
(
c
+
2
)
p
4
+
c
+
2
2
(
a
+
2
)
(
4
-
p
2
)
p
+
(
a
-
c
)
(
1
+
ρ
)
4
(
a
+
1
)
(
a
+
2
)
(
4
-
p
2
)
p
2
y
+
{
(
a
-
c
)
p
2
-
2
(
a
+
1
)
(
c
+
2
)
p
+
4
(
a
+
2
)
(
c
+
1
)
}
4
(
a
+
1
)
(
a
+
2
)
×
(
4
-
p
2
)
y
2
4
(
a
-
c
)
ρ
+
{
(
a
+
1
)
(
c
+
2
)
+
(
a
-
c
)
}
(
1
-
ρ
)
2
16
(
a
+
1
)
(
c
+
2
)
}
=
𝒢
(
p
,
y
)
(
0
≤
p
≤
2,0
≤
y
≤
1
)
(
say
)
.
We next maximize the function𝒢
(
p
,
y
) on the closed rectangle [
0,2
]
×
[
0,1
]. Since
(64)
∂
𝒢
∂
y
=
(
a
-
c
)
(
1
+
ρ
)
4
(
a
+
1
)
(
a
+
2
)
(
4
-
p
2
)
p
2
+
{
2
(
a
+
1
)
(
c
+
2
)
-
(
2
+
p
)
(
a
-
c
)
}
2
(
a
+
1
)
(
a
+
2
)
×
(
4
-
p
2
)
(
2
-
p
)
y
>
0
for 0
<
p
<
2 and 0
<
y
<
1, it follows that 𝒢
(
p
,
y
) cannot have a maximum in the interior on the closed rectangle [
0,2
]
×
[
0,1
]. Thus, for fixed p
∈
[
0,2
]
(65)
max
⁡
0
≤
y
≤
1
𝒢
(
p
,
y
)
=
𝒢
(
p
,
1
)
=
ℱ
(
p
)
(
say
)
,
where
(66)
ℱ
(
p
)
=
c
(
c
)
2
(
1
-
ρ
)
2
4
a
(
a
)
2
×
{
4
(
a
-
c
)
ρ
+
{
(
a
+
1
)
(
c
+
2
)
+
(
a
-
c
)
}
(
1
-
ρ
)
2
16
(
a
+
1
)
(
c
+
2
)
p
4
+
c
+
2
2
(
a
+
2
)
(
4
-
p
2
)
p
+
(
a
-
c
)
(
1
+
ρ
)
4
(
a
+
1
)
(
a
+
2
)
(
4
-
p
2
)
p
2
+
{
(
a
-
c
)
p
2
-
2
(
a
+
1
)
(
c
+
2
)
p
+
4
(
a
+
2
)
(
c
+
1
)
}
4
(
a
+
1
)
(
a
+
2
)
×
(
4
-
p
2
)
4
(
a
-
c
)
ρ
+
{
(
a
+
1
)
(
c
+
2
)
+
(
a
-
c
)
}
(
1
-
ρ
)
2
16
(
a
+
1
)
(
c
+
2
)
}
,
0
≤
ρ
<
1. and 0
≤
p
≤
2. Differentiating ℱ with respect to p, we deduce that
(67)
ℱ
′
(
p
)
=
c
(
c
)
2
(
1
-
ρ
)
2
4
a
(
a
)
2
×
[
{
(
a
+
1
)
(
c
+
2
)
+
(
a
-
c
)
}
(
1
-
ρ
)
2
-
8
(
a
-
c
)
4
(
a
+
1
)
(
a
+
2
)
p
3
-
2
{
(
a
+
2
)
(
c
+
1
)
-
(
a
-
c
)
(
2
+
ρ
)
}
(
a
+
1
)
(
a
+
2
)
p
]
=
0
for p
=
0 or
(68)
p
2
=
8
{
(
a
+
2
)
(
c
+
1
)
-
(
a
-
c
)
(
2
+
ρ
)
}
{
(
a
+
1
)
(
c
+
2
)
+
(
a
-
c
)
}
(
1
-
ρ
)
2
-
8
(
a
-
c
)
.
Since p
2
>
4 and
(69)
ℱ
′′
(
0
)
=
2
{
(
a
+
2
)
(
c
+
1
)
-
(
a
-
c
)
(
2
+
ρ
)
}
(
a
+
1
)
(
a
+
2
)
<
0
by the hypothesis, we conclude that the maximum value of ℱ is attained at p
=
0 so that the upper bound of the function 𝒢 corresponds to p
=
0 and y
=
1. Thus, by letting p
=
0 and y
=
1 in (63), we get the estimate (60).
The estimate in (60) is sharp for the function f
0 given by (48). This completes the proof of Theorem 9.Puttinga
=
2 and c
=
1 in Theorem 9, we get the following.Corollary 10.
If the functionf, given by (1) belongs to the class ℛ
~
(
ρ
), then
(70)
|
a
2
a
4
-
a
3
2
|
≤
(
1
-
ρ
)
2
9
and the estimate is sharp for the function f
0 defined by
(71)
f
0
(
z
)
=
∫
0
z
d
t
1
-
t
⋆
z
h
ρ
(
z
2
)
(
z
∈
𝒰
)
,
where the function h
ρ is given by (16).Theorem 11.
Letγ
>
0
,
a
≥
1
/
(
2
γ
)
,
c
>
0 and 1
/
2
≤
ρ
<
1. If f
∈
𝒜 satisfies the following inequality
(72)
R
e
{
ℒ
(
a
+
1
,
c
)
f
(
z
)
ℒ
(
a
,
c
)
f
(
z
)
}
>
(
2
a
γ
+
1
)
ρ
-
1
2
a
γ
ρ
(
z
∈
𝒰
)
,
then
(73)
ℒ
(
a
,
c
)
f
(
z
)
z
≺
{
1
+
(
1
-
2
ρ
)
z
1
-
z
}
1
/
γ
(
z
∈
𝒰
)
.
The result is the best possible.Proof.
We define the functionw by
(74)
ℒ
(
a
,
c
)
f
(
z
)
z
=
{
1
+
(
1
-
2
ρ
)
w
(
z
)
1
-
w
(
z
)
}
1
/
γ
(
z
∈
𝒰
)
.
Choosing the principal branch in the right hand side in (74), we note that w is analytic in 𝒰 with w
(
0
)
=
0. Furthermore, logarithmically differentiating (74) and using the identity (12) in the resulting equation, we find that
(75)
ℒ
(
a
+
1
,
c
)
f
(
z
)
ℒ
(
a
,
c
)
f
(
z
)
=
1
+
1
-
2
ρ
a
γ
z
w
′
(
z
)
1
+
(
1
-
2
ρ
)
w
(
z
)
+
1
a
γ
z
w
′
(
z
)
1
-
w
(
z
)
(
z
∈
𝒰
)
.
We claim that|
w
(
z
)
|
<
1 for all z
∈
𝒰. If not, then there exists a point z
0
∈
𝒰 such that
(76)
max
⁡
{
|
w
(
z
)
|
:
|
z
|
≤
|
z
0
|
}
=
|
w
(
z
0
)
|
=
1
(
w
(
z
0
)
≠
1
)
,
and let w
(
z
0
)
=
e
i
θ. Now, by applying Jack's lemma [25], we have
(77)
z
0
w
′
(
z
0
)
=
k
w
(
z
0
)
(
k
≥
1
)
.
From (75) and (77), we obtain
(78)
Re
{
ℒ
(
a
+
1
,
c
)
f
(
z
)
ℒ
(
a
,
c
)
f
(
z
)
}
=
1
+
k
a
γ
Re
(
e
i
θ
1
-
e
i
θ
)
+
(
1
-
2
ρ
)
k
a
γ
Re
{
e
i
θ
1
+
(
1
-
2
ρ
)
e
i
θ
}
=
1
-
c
2
a
γ
+
(
1
-
2
ρ
)
k
a
γ
×
1
-
2
ρ
+
cos
⁡
θ
1
+
2
(
1
-
2
ρ
)
cos
⁡
θ
+
(
1
-
2
ρ
)
2
≤
1
-
c
2
a
γ
+
(
2
ρ
-
1
)
k
2
a
γ
ρ
≤
(
2
a
γ
+
1
)
ρ
-
1
2
a
γ
ρ
,
which contradicts the hypothesis (72). Thus, we conclude that |
w
(
z
)
|
<
1 for all z
∈
𝒰 and (74) yields the required subordination relation (73).
To see that the result is the best possible, we consider the functionf
0
∈
𝒜 defined by
(79)
f
0
(
z
)
=
ψ
(
c
,
a
;
z
)
⋆
z
{
1
+
(
1
-
2
ρ
)
z
1
-
z
}
1
/
γ
(
γ
>
0
,
a
≥
1
2
γ
,
c
>
0
,
1
2
≤
ρ
<
1
;
z
∈
𝒰
)
from which it follows that
(80)
ℒ
(
a
,
c
)
f
0
(
z
)
z
=
{
1
+
(
1
-
2
ρ
)
z
1
-
z
}
1
/
γ
(
z
∈
𝒰
)
.
Thus, f
0 satisfies the subordination relation (73). On differentiating the expression in (80) followed by the use of the identity (12) in the resulting equation, we deduce that
(81)
ℒ
(
a
+
1
,
c
)
f
0
(
z
)
ℒ
(
a
,
c
)
f
0
(
z
)
=
1
+
(
1
-
2
ρ
a
γ
)
z
1
+
(
1
-
2
ρ
)
z
+
(
1
a
γ
)
z
1
-
z
(
z
∈
𝒰
)
.
This implies that
(82)
ℒ
(
a
+
1
,
c
)
f
0
(
z
)
ℒ
(
a
,
c
)
f
0
(
z
)
⟶
(
2
a
γ
+
1
)
ρ
-
1
2
a
γ
ρ
as
z
⟶
-
1
.
and the proof of Theorem 11 is completed.In the special caseγ
=
2, we get the following sufficient condition for the class ℛ
~
(
a
,
c
,
ρ
).Corollary 12.
Leta
≥
1
/
4
,
c
>
0 and 1
/
2
≤
ρ
<
1. If f
∈
𝒜 satisfies the following inequality:
(83)
R
e
{
ℒ
(
a
+
1
,
c
)
f
(
z
)
ℒ
(
a
,
c
)
f
(
z
)
}
>
(
4
a
+
1
)
ρ
-
1
4
a
(
1
2
≤
ρ
<
1
;
z
∈
𝒰
)
,
then f
∈
ℛ
~
(
a
,
c
,
ρ
). The result is the best possible for the function f
0 given by (47).Lettinga
=
2
,
c
=
1 and γ
=
2 in Theorem 11, we obtain the following.Corollary 13.
If1
/
2
≤
ρ
<
1 and f
∈
𝒜 satisfies
(84)
R
e
{
1
+
f
′′
(
z
)
f
′
(
z
)
}
>
5
ρ
-
1
4
ρ
(
1
2
≤
ρ
<
1
;
z
∈
𝒰
)
,
then f
∈
ℛ
~
(
ρ
). The result is the best possible for the function f
0 defined by
(85)
f
0
(
z
)
=
∫
0
z
d
t
1
-
t
⋆
z
h
ρ
(
z
)
(
1
2
≤
ρ
<
1
;
z
∈
𝒰
)
,
where the function h
ρ is given by (16).Theorem 14.
Leta
>
0
,
c
>
0 and γ
>
0. If f
∈
𝒜 satisfies the following subordination relation:
(86)
ℒ
(
a
,
c
)
f
(
z
)
z
≺
{
1
+
(
1
-
2
ρ
)
z
1
-
z
}
1
/
γ
(
1
2
≤
ρ
<
1
;
z
∈
𝒰
)
,
then
(87)
R
e
(
ℒ
(
a
+
1
,
c
)
f
(
z
)
ℒ
(
a
,
c
)
f
(
z
)
)
>
ρ
(
|
z
|
<
r
0
(
a
,
γ
,
ρ
)
)
,
where
(88)
r
0
(
a
,
γ
,
ρ
)
=
{
(
1
+
a
γ
ρ
)
-
(
1
+
a
γ
ρ
)
2
-
(
a
γ
)
2
(
2
ρ
-
1
)
a
γ
(
2
ρ
-
1
)
,
1
2
<
ρ
<
1
a
γ
2
+
a
γ
,
ρ
=
1
2
.
The bound r
0
(
a
,
γ
,
ρ
) in (88) is the best possible.Proof.
From (86), we get
(89)
(
ℒ
(
a
,
c
)
f
(
z
)
z
)
=
{
ρ
+
(
1
-
ρ
)
ϕ
(
z
)
}
1
/
γ
(
ϕ
∈
𝒫
;
z
∈
𝒰
)
,
where we choose the principal branch in (89). Taking logarithmic differentiation in (89) and using the identity (12) in the resulting equation, we deduce that
(90)
Re
{
ℒ
(
a
+
1
,
c
)
f
(
z
)
ℒ
(
a
,
c
)
f
(
z
)
}
-
ρ
≥
(
1
-
ρ
)
[
1
-
|
z
ϕ
′
(
z
)
|
a
γ
{
|
ρ
+
(
1
-
ρ
)
ϕ
(
z
)
|
}
]
(
z
∈
𝒰
)
.
Using the following well-known estimates [21]
(91)
|
z
ϕ
′
(
z
)
|
Re
{
ϕ
(
z
)
}
≤
2
r
1
-
r
2
,
|
ϕ
(
z
)
|
≤
1
+
r
1
-
r
(
|
z
|
=
r
<
1
)
in (90), we get
(92)
Re
{
ℒ
(
a
+
1
,
c
)
f
(
z
)
ℒ
(
a
,
c
)
f
(
z
)
}
-
ρ
≥
(
1
-
ρ
)
[
1
-
2
r
a
γ
{
ρ
(
1
-
r
)
2
+
(
1
-
ρ
)
(
1
-
r
2
)
}
]
≥
(
1
-
ρ
)
[
1
-
2
r
a
γ
{
(
2
ρ
-
1
)
r
2
-
2
ρ
r
+
1
}
]
which is certainly positive for |
z
|
<
r
0
(
a
,
γ
,
ρ
), where r
0
(
a
,
γ
,
ρ
) is given by (88).
To show that the result is the best possible, we consider the functionf
0 defined by
(93)
f
0
(
z
)
=
ψ
(
c
,
a
;
z
)
⋆
z
{
ρ
+
(
1
-
ρ
)
1
+
z
1
-
z
}
1
/
γ
h
h
h
h
h
h
h
h
h
h
h
h
h
(
1
2
≤
ρ
<
1,0
<
γ
;
z
∈
𝒰
)
.
Noting that
(94)
{
ℒ
(
a
+
1
,
c
)
f
0
(
z
)
ℒ
(
a
,
c
)
f
0
(
z
)
}
-
ρ
=
(
1
-
ρ
)
[
1
+
2
z
a
γ
{
ρ
(
1
-
z
)
2
+
(
1
-
ρ
)
(
1
-
z
2
)
}
]
=
0
for z
=
-
r
0
(
a
,
γ
,
ρ
), we conclude that the bound is the best possible. This proves Theorem 14.Takingγ
=
2 in Theorem 14, we get the following.Corollary 15.
Ifa
>
0
,
c
>
0
,
1
/
2
≤
ρ
<
1 and f
∈
ℛ
~
(
a
,
c
,
ρ
), then
(95)
R
e
(
ℒ
(
a
+
1
,
c
)
f
(
z
)
ℒ
(
a
,
c
)
f
(
z
)
)
>
ρ
(
|
z
|
<
κ
(
a
,
ρ
)
)
,
where
(96)
κ
(
a
,
ρ
)
=
{
(
1
+
2
a
ρ
)
-
4
a
2
(
1
-
ρ
)
2
+
4
a
ρ
+
1
2
a
(
2
ρ
-
1
)
,
1
2
<
ρ
<
1
a
1
+
a
,
ρ
=
1
2
.
The bound κ
(
a
,
ρ
) is the best possible for the function f
0, given by (47).Settinga
=
2
,
c
=
1 and γ
=
1 in Theorem 14, we get the following.Corollary 16.
Iff
∈
𝒜 satisfies
(97)
R
e
{
f
′
(
z
)
}
>
ρ
(
1
2
≤
ρ
<
1
;
z
∈
𝒰
)
,
then
(98)
R
e
{
1
+
z
f
′′
(
z
)
f
′
(
z
)
}
>
2
ρ
-
1
(
|
z
|
<
ϰ
(
ρ
)
)
,
where
(99)
ϰ
(
ρ
)
=
{
(
1
+
2
ρ
)
-
4
ρ
2
-
4
ρ
+
5
2
(
2
ρ
-
1
)
,
1
2
<
ρ
<
1
1
2
,
ρ
=
1
2
.
The bound ϰ
(
ρ
) is the best possible for the function f
0, given in Corollary 13.



---

*Source: 570361-2014-05-07.xml*