# Starlikeness of Functions Defined by Third-Order Differential Inequalities and Integral Operators

Authors: R. Chandrashekar; Rosihan M. Ali; K. G. Subramanian; A. Swaminathan
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Sufficient conditions are obtained to ensure starlikeness of positive order for analytic functions defined in the open unit disk satisfying certain third-order differential inequalities. As a consequence, conditions for starlikeness of functions defined by integral operators are obtained. Connections are also made to earlier known results.


---

## Body

## 1. Introduction

Letℋ denote the class of analytic functions f defined in the open unit disk U
:
=
{
z
∈
ℂ
:
|
z
|
<
1
}. For a
∈
ℂ and n a positive integer, let
(1)
ℋ
n
(
a
)
=
{
f
∈
ℋ
:
f
(
z
)
=
a
+
∑
k
=
n
∞
‍
a
k
z
k
}
,
and 𝒜
n
=
{
f
∈
ℋ
:
f
(
z
)
=
z
+
∑
k
=
n
+
1
∞
a
k
z
k
}, with 𝒜
1
:
=
𝒜. For β
∈
[
0,1
), denote by 𝒮
*
(
β
) the subclass of 𝒜 consisting of functions starlike of order β satisfying
(2)
Re
(
z
f
′
(
z
)
f
(
z
)
)
>
β
,
z
∈
U
.
The class 𝒮
*
:
=
𝒮
*
(
0
) is the well-known subclass of starlike functions studied widely in geometric function theory.In the sequel, we give emphasis to the class(3)
𝒮
1
(
β
)
=
{
f
∈
𝒜
:
|
z
f
′
(
z
)
f
(
z
)
-
1
|
<
1
-
β
,
z
∈
U
}
,
β
∈
[
0,1
), where 𝒮
1
:
=
𝒮
1
(
0
)
⊂
𝒮
*. Evidently 𝒮
1
(
β
)
⊂
𝒮
*
(
β
) for 0
≤
β
<
1. The class 𝒮
1
(
β
) was investigated by Silverman [1], who showed that 𝒮
1
(
β
) coincides with 𝒮
*
(
β
) for univalent functions with negative coefficients. This class has subsequently been studied in several other works (see, e.g., [2]).The problem of determining sufficient conditions to ensure starlikeness of functions has been widely investigated. These include conditions in terms of differential inequalities; see, for example, [2–11]. Miller and Mocanu [12], Kuroki and Owa [13], and, more recently, Ali et al. [14] determined conditions for starlikeness of functions defined by an integral operator of the form
(4)
f
(
z
)
=
∫
0
1
‍
W
(
r
,
z
)
d
r
,
or by the double integral operator
(5)
f
(
z
)
=
∬
0
1
W
(
r
,
s
,
z
)
d
r
d
s
.In this paper, conditions on certain third-order differential inequalities are found that would imply starlikeness of positive order. As a consequence, conditions on the kernel of certain integral operators are also obtained to ensure that the functions defined by these operators are starlike. Connections are also made to earlier known results.Recall that an analytic functionf issubordinate to an analytic function g in U, written as f
(
z
)
≺
g
(
z
), if there exists an analytic self-map w of U with w
(
0
)
=
0 satisfying f
(
z
)
=
g
(
w
(
z
)
).The following lemmas will be required in the sequel.Lemma 1 (see [15, Theorem 1, page 192] and see also [16, Theorem 3.1b, page 71]).
Leth be convex in U with h
(
0
)
=
a, γ
≠
0 and Re
γ
≥
0. If p
∈
ℋ
n
(
a
) and
(6)
p
(
z
)
+
z
p
′
(
z
)
γ
≺
h
(
z
)
,
then
(7)
p
(
z
)
≺
q
(
z
)
≺
h
(
z
)
,
where
(8)
q
(
z
)
=
γ
n
z
γ
/
n
∫
0
z
‍
h
(
t
)
t
(
γ
/
n
)
-
1
d
t
.
The function q is convex and is the best (
a
,
n
)-dominant.Lemma 2 (see [17] and see also [16, Theorem 3.1d, page 76]).
Let h be a starlike function withh
(
0
)
=
0. If p
∈
ℋ
n
(
a
) satisfies
(9)
z
p
′
(
z
)
≺
h
(
z
)
,
then
(10)
p
(
z
)
≺
q
(
z
)
=
a
+
1
n
∫
0
z
‍
h
(
t
)
t
d
t
.
The function q is convex and is the best (
a
,
n
)-dominant.


## 2. Main Results

The following two results are easily obtained by simple adaptations of Theorem 2.1 and Theorem 2.6 in [13]. The proofs are therefore omitted.Lemma 3.
Letf
∈
𝒜
n, 0
≤
α
<
n
δ, and 0
≤
β
<
1. If
(11)
|
δ
z
f
′′
(
z
)
-
α
(
f
′
(
z
)
-
1
)
|
<
(
n
+
1
)
(
1
-
β
)
(
n
δ
-
α
)
n
+
1
-
β
,
then f
∈
𝒮
*
(
β
) with an extremal function f
(
z
)
=
z
+
(
1
-
β
)
z
n
+
1
/
(
n
+
1
-
β
).Lemma 4.
Let0
≤
α
<
n
δ, 0
≤
β
<
1, and g
∈
ℋ. If
(12)
|
g
(
z
)
|
<
(
n
+
1
)
(
1
-
β
)
(
n
δ
-
α
)
n
+
1
-
β
,
then
(13)
f
(
z
)
=
z
+
z
n
+
1
δ
∬
0
1
g
(
r
s
z
)
r
[
(
n
-
1
)
δ
-
α
]
/
δ
s
n
d
r
d
s
is a starlike function of order β.Remark 5.
Even though the conditions given in Lemmas3 and 4 are sufficient to deduce f
∈
𝒮
*
(
β
), they are in fact sufficient to imply f
∈
𝒮
1
(
β
)
⊂
𝒮
*
(
β
).The above two lemmas are next used to obtain conditions in terms of a third-order differential inequality and a third-order integral operator to deduce starlikeness off of order β.Theorem 6.
Letf
∈
𝒜
n, 0
<
α
<
n
ν, δ
>
α
≥
γ
≥
0, and 0
≤
β
<
1. Further let μ and ν satisfy
(14)
ν
-
α
μ
=
δ
-
γ
,
ν
μ
=
γ
.
If
(15)
|
γ
z
2
f
′′′
(
z
)
+
δ
z
f
′′
(
z
)
-
α
(
f
′
(
z
)
-
1
)
|
<
(
1
+
n
μ
)
(
n
+
1
)
(
1
-
β
)
(
n
ν
-
α
)
n
+
1
-
β
,
then f
∈
𝒮
1
(
β
)
⊂
𝒮
*
(
β
). Equality is attained for f
(
z
)
=
z
+
(
1
-
β
)
z
n
+
1
/
(
n
+
1
-
β
).Proof.
Let(16)
p
(
z
)
=
ν
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
1
)
.
A brief computation shows that
(17)
p
(
z
)
+
μ
z
p
′
(
z
)
=
γ
z
2
f
′
′
′
(
z
)
+
δ
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
1
)
.
Hence, (15) can be written in the subordination form as
(18)
p
(
z
)
+
μ
z
p
′
(
z
)
≺
(
1
+
n
μ
)
(
n
+
1
)
(
1
-
β
)
(
n
ν
-
α
)
n
+
1
-
β
z
.
It follows from Lemma 1 that
(19)
p
(
z
)
≺
1
μ
n
z
1
/
μ
n
∫
0
z
‍
(
1
+
μ
n
)
(
n
+
1
)
(
1
-
β
)
(
n
ν
-
α
)
n
+
1
-
β
t
1
/
μ
n
d
t
=
(
n
+
1
)
(
1
-
β
)
(
n
ν
-
α
)
n
+
1
-
β
z
,
which implies
(20)
|
ν
z
f
′′
(
z
)
-
α
(
f
′
(
z
)
-
1
)
|
≤
(
n
+
1
)
(
1
-
β
)
(
n
ν
-
α
)
n
+
1
-
β
.
Hence f
∈
𝒮
1
(
β
)
⊂
𝒮
*
(
β
) on using Lemma 3.
For sharpness, it is evident that the functionf
(
z
)
=
z
+
(
1
-
β
)
z
n
+
1
/
(
n
+
1
-
β
) satisfies
(21)
|
γ
z
2
f
′′′
(
z
)
+
δ
z
f
′′
(
z
)
-
α
(
f
′
(
z
)
-
1
)
|
=
(
1
+
n
μ
)
(
n
+
1
)
(
1
-
β
)
(
n
ν
-
α
)
n
+
1
-
β
.
Thus,
(22)
|
z
f
′
(
z
)
f
(
z
)
-
1
|
<
1
-
β
,
z
∈
U
.Theorem 7.
Let0
<
α
<
n
μ, δ
>
α
≥
γ
≥
0, 0
≤
β
<
1, and g
∈
ℋ. If
(23)
|
g
(
z
)
|
<
(
1
+
μ
n
)
(
n
+
1
)
(
1
-
β
)
(
n
ν
-
α
)
n
+
1
-
β
,
where
(24)
ν
-
α
μ
=
δ
-
γ
,
ν
μ
=
γ
,
then
(25)
f
(
z
)
=
z
+
z
n
+
1
γ
∭
0
1
g
(
r
s
t
z
)
r
n
-
1
-
α
/
ν
s
n
t
n
-
1
+
1
/
μ
d
r
d
s
d
t
satisfies f
∈
𝒮
1
(
β
)
⊂
𝒮
*
(
β
).Proof.
Letf
∈
𝒜
n satisfy
(26)
γ
z
2
f
′
′
′
(
z
)
+
δ
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
1
)
=
z
n
g
(
z
)
.
From Theorem 6, the solution of (26) belongs to the class 𝒮
1
(
β
)
⊂
𝒮
*
(
β
). Now (26) has the form
(27)
p
(
z
)
+
μ
z
p
′
(
z
)
=
z
n
g
(
z
)
,
where
(28)
p
(
z
)
=
ν
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
1
)
.
Equation (27) has a solution
(29)
p
(
z
)
=
z
-
1
/
μ
μ
∫
0
z
‍
g
(
ξ
)
ξ
n
-
1
+
1
/
μ
d
ξ
=
z
n
μ
∫
0
1
‍
g
(
t
z
)
t
n
-
1
+
1
/
μ
d
t
=
z
n
ϕ
(
z
)
,
with
(30)
ϕ
(
z
)
=
1
μ
∫
0
1
‍
g
(
t
z
)
t
n
-
1
+
1
/
μ
d
t
.
Note that the functionf in Lemma 4 satisfies δ
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
1
)
=
z
n
g
(
z
). Thus replacing the appropriate parameters in the equation
(31)
ν
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
1
)
=
z
n
ϕ
(
z
)
yields a solution
(32)
f
(
z
)
=
z
+
z
n
+
1
ν
∬
0
1
ϕ
(
r
s
z
)
r
n
-
1
-
α
/
ν
s
n
d
r
d
s
=
z
+
z
n
+
1
γ
∭
0
1
g
(
r
s
t
z
)
r
n
-
1
-
α
/
ν
s
n
t
n
-
1
+
1
/
μ
d
r
d
s
d
t
.
This completes the proof.The next result provides a sufficient condition for starlikeness of orderβ involving a second-order differential inequality.Lemma 8.
Letf
∈
𝒜
n, and 0
≤
α
<
δ with 0
≤
β
<
1. If
(33)
|
δ
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
f
(
z
)
z
)
|
<
n
(
1
-
β
)
(
δ
(
n
+
1
)
-
α
)
n
+
1
-
β
,
then f
∈
𝒮
1
(
β
)
⊂
𝒮
*
(
β
) with an extremal function f
(
z
)
=
z
+
(
1
-
β
)
z
n
+
1
/
(
n
+
1
-
β
).Proof.
Inequality (33) can be expressed in the subordination form
(34)
δ
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
f
(
z
)
z
)
≺
n
(
1
-
β
)
(
δ
(
n
+
1
)
-
α
)
n
+
1
-
β
z
,
z
∈
U
.
Writing(35)
P
(
z
)
=
f
′
(
z
)
-
f
(
z
)
z
=
P
n
z
n
+
⋯
,
it follows that
(36)
δ
z
P
′
(
z
)
+
(
δ
-
α
)
P
(
z
)
=
δ
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
f
(
z
)
z
)
≺
n
(
1
-
β
)
(
δ
(
n
+
1
)
-
α
)
n
+
1
-
β
z
.
Now Lemma 1 with γ
=
1
-
α
/
δ yields
(37)
P
(
z
)
≺
γ
n
z
γ
/
n
∫
0
z
‍
(
n
(
1
-
β
)
(
δ
(
n
+
1
)
-
α
)
(
δ
-
α
)
(
n
+
1
-
β
)
t
)
t
γ
/
n
-
1
d
t
=
n
(
1
-
β
)
n
+
1
-
β
z
,
which implies
(38)
|
f
′
(
z
)
-
f
(
z
)
z
|
<
n
(
1
-
β
)
n
+
1
-
β
.
Let(39)
p
(
z
)
=
f
(
z
)
z
=
1
+
p
n
z
n
+
⋯
.
Since
(40)
z
p
′
(
z
)
=
f
′
(
z
)
-
f
(
z
)
z
≺
n
(
1
-
β
)
n
+
1
-
β
z
,
an application of Lemma 2 shows that
(41)
p
(
z
)
≺
1
+
1
n
∫
0
z
‍
(
n
(
1
-
β
)
n
+
1
-
β
t
)
d
t
t
=
1
+
(
1
-
β
)
n
+
1
-
β
z
.
Therefore,
(42)
|
p
(
z
)
|
=
|
f
(
z
)
z
|
>
1
-
1
-
β
n
+
1
-
β
=
n
n
+
1
-
β
.
Combining (38) and (42) yields
(43)
n
n
+
1
-
β
|
z
f
′
(
z
)
f
(
z
)
-
1
|
=
n
n
+
1
-
β
|
z
f
(
z
)
|
|
f
′
(
z
)
-
f
(
z
)
z
|
<
n
(
1
-
β
)
n
+
1
-
β
,
which means |
z
f
′
(
z
)
/
f
(
z
)
-
1
|
<
1
-
β, whence f
∈
𝒮
1
(
β
)
⊂
𝒮
*
(
β
).Remark 9.
Forδ
=
1 and β
=
0, Lemma 8 reduces to [12, Lemma 2.2].The following result gives starlikeness for a function given by a double integral operator associated with Lemma4. The proof is analogous to Theorem 2.2 of [12] and is omitted.Lemma 10.
Let0
≤
α
<
n
δ, δ
>
α, 0
≤
β
<
1, and g
∈
ℋ. If
(44)
|
g
(
z
)
|
<
n
(
1
-
β
)
(
δ
(
n
+
1
)
-
α
)
n
+
1
-
β
,
then
(45)
f
(
z
)
=
z
+
z
n
+
1
δ
∬
0
1
g
(
r
s
z
)
r
(
n
δ
-
α
)
/
δ
s
n
-
1
d
r
d
s
satisfies f
∈
𝒮
1
(
β
)
⊂
𝒮
*
(
β
).An application of Lemma8 yields the following sufficient condition for starlikeness in terms of a third-order differential inequality.Theorem 11.
Letf
∈
𝒜
n, 0
≤
α
<
(
1
-
μ
)
(
n
+
1
)
ν, α
≥
γ
≥
0, δ
>
γ
+
α, and 0
≤
β
<
1. Further let
(46)
ν
-
α
μ
1
-
μ
=
δ
-
γ
,
ν
μ
=
γ
.
If
(47)
|
γ
z
2
f
′
′
′
(
z
)
+
δ
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
f
(
z
)
z
)
|
<
n
[
(
n
+
1
)
(
ν
-
γ
)
-
α
]
(
1
-
β
)
(
1
+
n
μ
)
(
n
+
1
-
β
)
(
1
-
μ
)
,
then f
∈
𝒮
1
(
β
)
⊂
𝒮
*
(
β
). Equality is attained for f
(
z
)
=
z
+
(
1
-
β
)
z
n
+
1
/
(
n
+
1
-
β
).Proof.
Proceeding similarly as in the proof of Lemma8, inequality (47) can be written as
(48)
γ
z
2
f
′
′
′
(
z
)
+
δ
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
f
(
z
)
z
)
≺
n
[
(
n
+
1
)
(
ν
-
γ
)
-
α
]
(
1
-
β
)
(
1
+
n
μ
)
(
n
+
1
-
β
)
(
1
-
μ
)
z
.
Let
(49)
p
(
z
)
=
ν
z
f
′′
(
z
)
-
α
1
-
μ
(
f
′
(
z
)
-
f
(
z
)
z
)
.
Then a computation yields
(50)
p
′
(
z
)
=
ν
z
f
′
′
′
(
z
)
+
ν
f
′
′
(
z
)
-
α
1
-
μ
f
′
′
(
z
)
+
α
1
-
μ
(
f
′
(
z
)
z
-
f
(
z
)
z
2
)
,
so that
(51)
p
(
z
)
+
μ
z
p
′
(
z
)
=
γ
z
2
f
′′′
(
z
)
+
δ
z
f
′′
(
z
)
-
α
(
f
′
(
z
)
-
f
(
z
)
z
)
.
Hence
(52)
p
(
z
)
+
μ
z
p
′
(
z
)
≺
n
[
(
n
+
1
)
(
ν
-
γ
)
-
α
]
(
1
-
β
)
(
1
+
n
μ
)
(
n
+
1
-
β
)
(
1
-
μ
)
z
,
z
∈
U
.
Applying Lemma1 yields
(53)
p
(
z
)
≺
1
μ
n
z
1
/
n
μ
∫
0
z
‍
(
(
(
n
+
1
-
β
)
(
1
-
μ
)
)
-
1
(
n
[
(
n
+
1
)
(
ν
-
γ
)
-
α
]
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
1
-
β
)
(
1
+
n
μ
)
)
k
k
k
k
k
k
k
k
k
k
k
k
×
(
(
n
+
1
-
β
)
(
1
-
μ
)
)
-
1
)
t
1
/
n
μ
d
t
.
This implies that
(54)
ν
z
f
′
′
(
z
)
-
α
1
-
μ
(
f
′
(
z
)
-
f
(
z
)
z
)
≺
n
(
1
-
β
)
[
(
n
+
1
)
(
ν
-
γ
)
-
α
]
(
n
+
1
-
β
)
(
1
-
μ
)
z
,
and thus
(55)
|
ν
z
f
′′
(
z
)
-
α
1
-
μ
(
f
′
(
z
)
-
f
(
z
)
z
)
|
≤
n
(
1
-
β
)
[
(
n
+
1
)
(
ν
-
γ
)
-
α
]
(
n
+
1
-
β
)
(
1
-
μ
)
=
n
(
1
-
β
)
[
ν
(
n
+
1
)
-
α
/
(
1
-
μ
)
]
n
+
1
-
β
,
which, in comparison with Lemma 8, gives the required result.
Further the result is sharp forf
(
z
)
=
z
+
(
1
-
β
)
z
n
+
1
/
(
n
+
1
-
β
) which satisfies
(56)
|
γ
z
2
f
′
′
′
(
z
)
+
δ
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
f
(
z
)
z
)
|
=
n
[
(
n
+
1
)
(
ν
-
γ
)
-
α
]
(
1
-
β
)
(
1
+
n
μ
)
(
n
+
1
-
β
)
(
1
-
μ
)
.Remark 12.
Forf
∈
𝒜
n, the choice α
=
0
=
γ in Theorem 11 results in
(57)
|
δ
z
f
′
′
(
z
)
|
<
n
[
δ
(
n
+
1
)
]
(
1
-
β
)
n
+
1
-
β
⟹
f
∈
𝒮
1
(
β
)
⊂
𝒮
*
(
β
)
.
For δ
=
1, this coincides with Lemma 8 at α
=
0, which was also exhibited in [13, Corollary 2.4]. Further, for n
=
1, (57) gives
(58)
|
δ
z
f
′
′
(
z
)
|
<
2
δ
(
1
-
β
)
2
-
β
⟹
f
∈
𝒮
1
(
β
)
⊂
𝒮
*
(
β
)
,
which for δ
=
1 and β
=
0 is the result given in [8, Theorem 1].Corresponding to Theorem11, a sufficient condition for starlikeness of order β for functions defined by a triple integral operator is obtained in the following result.Theorem 13.
Let0
≤
α
<
(
1
-
μ
)
(
n
+
1
)
ν, α
≥
γ
≥
0, δ
>
γ
+
α, 0
≤
β
<
1, and g
∈
ℋ. Further let
(59)
ν
-
α
μ
1
-
μ
=
δ
-
γ
,
ν
μ
=
γ
.
If
(60)
|
g
(
z
)
|
<
n
[
(
n
+
1
)
(
ν
-
γ
)
-
α
]
(
1
-
β
)
(
1
+
n
μ
)
(
n
+
1
-
β
)
(
1
-
μ
)
,
then
(61)
f
(
z
)
=
z
+
z
n
+
1
γ
∭
0
1
g
(
r
s
t
z
)
r
(
1
/
ν
)
[
n
ν
-
α
/
(
1
-
μ
)
]
k
k
k
k
k
k
k
k
k
k
k
k
×
s
n
t
n
-
1
+
1
/
μ
d
r
d
s
d
t
satisfies f
∈
𝒮
1
(
β
)
⊂
𝒮
*
(
β
).Proof.
Letf
∈
𝒜
n satisfy
(62)
γ
z
2
f
′
′
′
(
z
)
+
δ
z
f
′
′
(
z
)
-
α
(
f
′
(
z
)
-
f
(
z
)
z
)
=
z
n
g
(
z
)
.
From Theorem 11, we find that the solution of (62) lies in 𝒮
*
(
β
). Now (62) becomes
(63)
p
(
z
)
+
μ
z
p
′
(
z
)
=
z
n
g
(
z
)
,
where
(64)
p
(
z
)
=
ν
z
f
′
′
(
z
)
-
α
1
-
μ
(
f
′
(
z
)
-
f
(
z
)
z
)
.
Equation (63) has a solution
(65)
p
(
z
)
=
z
-
1
/
μ
μ
∫
0
z
‍
g
(
ξ
)
ξ
n
+
1
/
μ
-
1
d
ξ
=
z
n
μ
∫
0
1
‍
g
(
t
z
)
t
n
+
1
/
μ
-
1
d
t
=
z
n
ϕ
(
z
)
,
with
(66)
ϕ
(
z
)
=
1
μ
∫
0
1
‍
g
(
t
z
)
t
n
+
1
/
μ
-
1
d
t
.
In view of Lemma10, the equation
(67)
ν
z
f
′′
(
z
)
-
α
1
-
μ
(
f
′
(
z
)
-
f
(
z
)
z
)
=
z
n
ϕ
(
z
)
has a solution
(68)
f
(
z
)
=
z
+
z
n
+
1
ν
∬
0
1
ϕ
(
r
s
z
)
r
(
1
/
ν
)
[
n
ν
-
α
/
(
1
-
μ
)
]
s
n
-
1
d
r
d
s
=
z
+
z
n
+
1
μ
ν
∭
0
1
g
(
r
s
t
z
)
r
(
1
/
ν
)
[
n
ν
-
α
/
(
1
-
μ
)
]
k
k
k
k
k
k
k
k
k
k
k
k
×
s
n
-
1
t
n
-
1
+
1
/
μ
d
r
d
s
d
t
=
z
+
z
n
+
1
γ
∭
0
1
g
(
r
s
t
z
)
r
(
1
/
ν
)
[
n
ν
-
α
/
(
1
-
μ
)
]
k
k
k
k
k
k
k
k
k
k
k
k
×
s
n
-
1
t
n
-
1
+
1
/
μ
d
r
d
s
d
t
.
This completes the proof.



---

*Source: 723097-2014-04-03.xml*