1. Introduction
Let
A
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 closetoconvex 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
Let
A
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.
Let
f
(
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.
As
f
∈
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 of
λ
for which
f
∈
M
(
α
,
λ
)
implies that
f
∈
R
(
β
)
,
0
<
β
≤
1
.
Theorem 2.
Let
f
(
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).
If
a
2
=
0
and
β
=
1
, then Theorem 2 gives the following.
Corollary 3.
If
f
(
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
.
Setting
a
2
=
0
,
β
=
1
, and
α
=

1
in Theorem 2, we obtain the following.
Corollary 4.
If
f
(
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 class
M
(
α
,
λ
)
belong to
S
∗
(
γ
)
,
0
≤
γ
<
1
.
Theorem 5.
Let
f
(
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.
As
a
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 lefthand 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
)
.
Taking
γ
=
0
, in Theorem 5, we get the following result.
Corollary 6.
Let
f
be as in Theorem 5. Then,
f
∈
S
∗
provided
0
<
λ
≤
2
α
+
1
/
2
.