# Geometric Properties of a Class of Analytic Functions Defined by a Differential Inequality

Authors: Manpreet Kaur; Sushma Gupta; Sukhjit Singh
Journal: International Journal of Analysis
 (2015)

---

## Abstract

LetA be the class of analytic functions f defined in the open unit disk E and normalized by f
(
0
)
=
f
'
(
0
)
-
1
=
0. For f
(
z
)
/
z
≠
0 in E, let M
α
,
λ
:
=
{
f
∈
A
:
|
-
α
z
2
(
z
/
f
(
z
)
)
′
′
+
f
′
(
z
)
(
z
/
f
(
z
)
)
2
-
1
|
≤
λ
,
z
∈
E
}, where λ
>
0 and α
∈
R
∖
[
-
1
/
2,0
]. In the present paper, we find conditions under which functions in the class M
(
α
,
λ
) are starlike of order γ, 0
≤
γ
<
1.


---

## Body

## 1. Introduction

LetA denote the class of all functions f that are analytic in the open unit disk E
=
{
z
:
z
<
1
} and are normalized by the conditions f
(
0
)
=
f
′
(
0
)
-
1
=
0. We denote by S the subclass of A consisting of functions which are univalent in E and denote by S
∗
(
γ
), 0
≤
γ
<
1, the class of functions in S which are starlike of order γ in E. Analytically,(1)
S
∗
γ
=
f
∈
A
:
Re
⁡
z
f
′
z
f
z
>
γ
,
z
∈
E
.Note that S
∗
(
0
) is the usual class of starlike (with respect to the origin) functions in E and we denote it simply by S
∗. We say that f
∈
R
(
β
), 0
<
β
≤
1, if f
∈
A and a
r
g
f
′
z
<
β
π
/
2, for all z
∈
E. It is well known that the functions in R
(
1
) are close-to-convex and hence univalent in E [1, 2].In 1972, Ozaki and Nunokawa [3] studied the class U
(
λ
), which is defined as(2)
U
λ
=
f
∈
A
:
f
′
z
z
f
z
2
-
1
≤
λ
,
z
∈
E
,where f
(
z
)
/
z
≠
0 in E. They proved that U
(
λ
)
⊂
S for 0
≤
λ
≤
1. Several researchers studied this class (e.g., see [4–8]) and obtained many significant results.Recently, Obradovic and Ponnusamy [9] investigated another class (3)
M
λ
=
f
∈
A
:
z
2
z
f
z
′
′
+
f
′
z
z
f
z
2
-
1
≤
λ
,
z
∈
E
,where λ
>
0 and f
(
z
)
/
z
≠
0 in E. They obtained certain inclusion relations, characterization formula, and coefficient conditions. They also posed a question about the starlikeness of the functions in the class M
(
λ
). The purpose of the present paper is to answer this question. In fact, we study a more general class (4)
M
α
,
λ
=
f
∈
A
:
-
α
z
2
z
f
z
′
′
+
f
′
z
z
f
z
2
-
1
≤
λ
,
z
∈
E
,where f
(
z
)
/
z
≠
0 in E, λ
>
0, and α
∈
R
∖
[
-
1
/
2,0
]. We find conditions on α, λ, and f involved in the class M
(
α
,
λ
) under which members of M
(
α
,
λ
) are starlike of a given order γ, 0
≤
γ
<
1. We remark that the class M
(
α
,
λ
) follows essentially from the class S
(
α
,
β
,
λ
) studied by Baricz and Ponnusamy [10] by taking β
=
0; however, we will study those issues for the class M
(
α
,
λ
) which are not studied by the authors for S
(
α
,
β
,
λ
).


## 2. Main Results

LetA
n denote the class of analytic functions p in E such that p
(
k
)
(
0
)
=
0 for k
=
0,1
,
2
,
…
,
n, where p
(
0
)
(
0
)
=
p
(
0
). With w
(
0
)
(
0
)
=
w
(
0
), we set B
n
=
{
w: w is analytic, w
z
≤
1 in E and w
(
k
)
(
0
)
=
0 for k
=
0,1
,
2
,
…
,
n
}. Functions in B
n are called Schwarz functions. Obviously, w
∈
B
k implies that |
w
(
z
)
|
≤
|
z
|
k
+
1 in E, for k
=
0,1
,
2,3
,
…
,
n.We begin with the following result.Lemma 1.
Letf
(
z
)
=
z
+
a
2
z
2
+
a
3
z
3
+
⋯ be in M
(
α
,
λ
). Then, R
e
f
(
z
)
/
z
≥
1
/
2 in E whenever a
2
≤
1
-
λ
/
2
α
+
1.Proof.
Asf
∈
M
(
α
,
λ
), there exists a Schwarz function w
∈
B
1 such that(5)
-
α
z
2
z
f
z
′
′
+
f
′
z
z
f
z
2
-
1
=
λ
w
z
. Since w
∈
B
1, w
(
0
)
=
w
′
(
0
)
=
0. If we set(6)
p
z
=
f
′
z
z
f
z
2
-
1
=
z
f
z
-
z
z
f
z
′
-
1
,where p is an analytic function in E with p
(
0
)
=
p
′
(
0
)
=
0, then (5) is equivalent to(7)
α
z
p
′
z
+
p
z
=
λ
w
z
,from which we get(8)
p
z
=
λ
α
∫
0
1
w
t
z
t
1
/
α
-
1
d
t
,or, equivalently,(9)
-
z
z
f
z
-
1
+
a
2
z
′
+
z
f
z
-
1
+
a
2
z
=
λ
α
∫
0
1
w
t
z
t
1
/
α
-
1
d
t
.Solving this equation for z
/
f
(
z
), we obtain(10)
z
f
z
=
1
-
a
2
z
-
λ
α
∫
0
1
∫
0
1
w
s
t
z
s
2
t
1
/
α
-
1
d
s
d
t
.Now, using the fact that w
s
t
z
≤
s
t
z
2, it follows that(11)
z
f
z
-
1
≤
a
2
z
+
λ
α
∫
0
1
∫
0
1
t
1
/
α
+
1
z
2
d
s
d
t
,
z
∈
E
,
≤
z
a
2
+
λ
z
2
α
+
1
,
z
∈
E
.
The inequality (11) is equivalent to(12)
f
z
z
-
1
1
-
z
2
a
2
+
λ
z
/
2
α
+
1
2
≤
z
a
2
+
λ
z
/
2
α
+
1
1
-
z
2
a
2
+
λ
z
/
2
α
+
1
2
, which gives(13)
R
e
f
z
z
≥
1
1
+
z
a
2
+
λ
z
/
2
α
+
1
>
1
1
+
a
2
+
λ
/
2
α
+
1
≥
1
2
,for all z
∈
E whenever a
2
≤
1
-
λ
/
2
α
+
1.In the next result, we find the range of values ofλ for which f
∈
M
(
α
,
λ
) implies that f
∈
R
(
β
), 0
<
β
≤
1.Theorem 2.
Letf
(
z
)
=
z
+
a
2
z
2
+
a
3
z
3
+
⋯ be in M
(
α
,
λ
). Then, f
∈
R
(
β
) for 0
<
λ
≤
λ
β, where λ
β satisfies the inequality(14)
1
-
λ
2
2
α
+
1
2
sin
⁡
β
π
2
-
2
a
2
+
λ
2
α
+
1
1
-
a
2
+
λ
2
α
+
1
2
≥
λ
2
α
+
1
cos
⁡
β
π
2
.Proof.
From (8), we get(15)
p
z
<
λ
2
α
+
1
,
z
∈
E
,where(16)
p
z
=
f
′
z
z
f
z
2
-
1
.Therefore,(17)
arg
⁡
f
′
z
z
f
z
2
≤
arcsin
⁡
λ
2
α
+
1
,
z
∈
E
.Also, in view of (11), we obtain(18)
arg
⁡
z
f
z
≤
arcsin
⁡
a
2
+
λ
2
α
+
1
,
z
∈
E
.Therefore,(19)
arg
⁡
f
′
z
≤
arg
⁡
f
′
z
z
f
z
2
+
2
arg
⁡
z
f
z
≤
arcsin
⁡
λ
2
α
+
1
+
2
arcsin
⁡
a
2
+
λ
2
α
+
1
=
arcsin
⁡
λ
2
α
+
1
+
arcsin
⁡
2
a
2
+
λ
2
α
+
1
1
-
a
2
+
λ
2
α
+
1
2
.Now, the desired result follows, if(20)
a
r
c
s
i
n
λ
2
α
+
1
+
a
r
c
s
i
n
2
a
2
+
λ
2
α
+
1
1
-
a
2
+
λ
2
α
+
1
2
≤
β
π
2
.By using a
r
c
s
i
n
x
+
a
r
c
s
i
n
y
=
a
r
c
s
i
n
x
1
-
y
2
+
y
1
-
x
2
,
x
,
y
∈
-
1,1, and x
2
+
y
2
≤
1 and carrying out some simplifications, we conclude that (20) is equivalent to (14).Ifa
2
=
0 and β
=
1, then Theorem 2 gives the following.Corollary 3.
Iff
(
z
)
=
z
+
a
3
z
3
+
⋯ is in M
(
α
,
λ
), and then R
e
f
′
(
z
)
>
0 in E, whenever 0
<
λ
≤
2
α
+
1
/
2.Settinga
2
=
0, β
=
1, and α
=
-
1 in Theorem 2, we obtain the following.Corollary 4.
Iff
(
z
)
=
z
+
a
3
z
3
+
⋯ is in M
(
-
1
,
λ
), and then R
e
f
′
(
z
)
>
0 in E, whenever 0
<
λ
≤
1
/
2.In the following theorem, we find conditions under which functions in the classM
(
α
,
λ
) belong to S
∗
(
γ
), 0
≤
γ
<
1.Theorem 5.
Letf
(
z
)
=
z
+
a
2
z
2
+
a
3
z
3
+
⋯ be in M
(
α
,
λ
) and let f
′
′
(
0
)
=
0. Then, for 0
≤
γ
<
1, f
∈
S
∗
(
γ
) provided 0
<
λ
≤
1
-
γ
2
α
+
1
/
(
2
+
γ
).Proof.
Asa
2
=
f
′
′
(
0
)
/
2
=
0, from (8) and (10), we get(21)
z
f
′
z
f
z
=
f
′
z
z
/
f
z
2
z
/
f
z
=
1
+
λ
/
α
∫
0
1
w
t
z
t
1
/
α
-
1
d
t
1
-
λ
/
α
∫
0
1
∫
0
1
w
s
t
z
/
s
2
t
1
/
α
-
1
d
s
d
t
.Now, R
e
z
f
′
(
z
)
/
f
(
z
)
>
γ is equivalent to(22)
z
f
′
z
/
f
z
-
γ
1
-
γ
≠
i
T
,
T
∈
R
.Using (21) in (22), we get(23)
λ
α
∫
0
1
w
t
z
t
1
/
α
-
1
d
t
-
γ
1
-
i
T
1
-
λ
α
∫
0
1
∫
0
1
w
s
t
z
s
2
t
1
/
α
-
1
d
s
d
t
+
λ
α
i
T
∫
0
1
∫
0
1
w
s
t
z
s
2
t
1
/
α
-
1
d
s
d
t
≠
-
1
-
i
T
,or, equivalently,(24)
λ
2
α
∫
0
1
w
t
z
t
1
/
α
-
1
d
t
-
∫
0
1
∫
0
1
w
s
t
z
s
2
t
1
/
α
-
1
d
s
d
t
-
γ
1
-
λ
α
∫
0
1
∫
0
1
w
s
t
z
s
2
t
1
/
α
-
1
d
s
d
t
+
λ
2
α
1
+
i
T
1
-
i
T
∫
0
1
w
t
z
t
1
/
α
-
1
d
t
+
∫
0
1
∫
0
1
w
s
t
z
s
2
t
1
/
α
-
1
d
s
d
t
≠
-
1
.If we denote the left-hand side of (24) by H
(
w
,
T
,
z
) and let(25)
M
=
sup
T
∈
R
,
w
∈
B
1
,
z
∈
E
⁡
H
w
,
T
,
z
, then, in view of the rotation invariance property of the set B
1, we obtain that (22) holds if M
≤
1.
A simple calculation gives(26)
M
≤
sup
w
∈
B
1
,
z
∈
E
⁡
λ
2
α
∫
0
1
w
t
z
t
1
/
α
-
1
d
t
-
∫
0
1
∫
0
1
w
s
t
z
s
2
t
1
/
α
-
1
d
s
d
t
+
λ
2
α
∫
0
1
w
t
z
t
1
/
α
-
1
d
t
+
∫
0
1
∫
0
1
w
s
t
z
s
2
t
1
/
α
-
1
d
s
d
t
+
γ
1
-
λ
α
∫
0
1
∫
0
1
w
s
t
z
s
2
t
1
/
α
-
1
d
s
d
t
. Now, by the parallelogram Law, z
1
+
z
2
+
z
1
-
z
2
≤
2
z
1
2
+
z
2
2, we have(27)
M
≤
λ
α
sup
w
∈
B
1
,
z
∈
E
⁡
∫
0
1
w
t
z
t
1
/
α
-
1
d
t
2
+
∫
0
1
∫
0
1
w
s
t
z
s
2
t
1
/
α
-
1
d
s
d
t
2
+
γ
+
λ
γ
α
∫
0
1
∫
0
1
w
s
t
z
s
2
t
1
/
α
-
1
d
s
d
t
. Using the fact that w
z
≤
z
2 in E, we get(28)
M
≤
λ
α
sup
z
∈
E
⁡
∫
0
1
z
2
t
1
/
α
+
1
d
t
2
+
∫
0
1
∫
0
1
z
2
t
1
/
α
+
1
d
t
2
+
γ
+
λ
γ
α
∫
0
1
∫
0
1
z
2
t
1
/
α
+
1
d
t
=
λ
2
+
γ
2
α
+
1
+
γ
.Thus, M
≤
1, if λ
≤
1
-
γ
2
α
+
1
/
(
2
+
γ
).Takingγ
=
0, in Theorem 5, we get the following result.Corollary 6.
Letf be as in Theorem 5. Then, f
∈
S
∗ provided 0
<
λ
≤
2
α
+
1
/
2.



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*Source: 185635-2015-10-04.xml*