# On Subordinations for Certain Multivalent Analytic Functions in the Right-Half Plane

Authors: Yi-Hui Xu; Jin-Lin Liu
Journal: Journal of Function Spaces
 (2016)

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## Abstract

The object of the present paper is to investigate some properties of multivalent analytic functions associated with the lemniscate of Bernoulli.


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## Body

## 1. Introduction

LetA
n
(
p
) denote the class of functions of the form(1)
f
z
=
z
p
+
∑
k
=
n
+
p
∞
a
k
z
k
p
,
n
∈
N
=
1,2
,
3
,
…
,which are analytic in the open unit disk U
=
{
z
:
|
z
|
<
1
}. A function f
∈
A
n
(
p
) is said to be starlike of order ρ in U if it satisfies(2)
R
e
z
f
′
z
f
z
>
ρ
0
≤
ρ
<
p
;
z
∈
U
.For functionsf and g analytic in U, we say that f is subordinate to g and write f
(
z
)
≺
g
(
z
)
(
z
∈
U
), if there exists an analytic function w
(
z
) in U such that (3)
w
z
≤
z
,
f
z
=
g
w
z
z
∈
U
.Furthermore, if the function g is univalent in U, then (4)
f
z
≺
g
z
z
∈
U
⟺
f
0
=
g
0
,
f
U
⊂
g
U
.LetS
L be the class of functions defined by(5)
S
L
≔
f
∈
A
1
1
:
z
f
′
z
f
z
2
-
1
<
1
.Thus a function f
∈
S
L if z
f
′
(
z
)
/
f
(
z
) lies in the region bounded by the right-half of the lemniscate of Bernoulli given by |
w
2
-
1
|
<
1. In terms of subordination, the class S
L consists of normalized analytic functions f satisfying z
f
′
(
z
)
/
f
(
z
)
≺
1
+
z. This class S
L was introduced by Sokól and Stankiewicz [1]. Several geometric properties had been investigated by Sokól and Stankiewicz [1]. They determined the radius of convexity for functions in the class S
L. They also obtained structural formula, as well as growth and distortion theorems for these functions. Estimates for the first few coefficients of functions in S
L were obtained in [2]. Recently, Sokól [3] determined various radii for functions belonging to the class S
L; these include the radii of convexity, starlikeness, and strong starlikeness of order ρ. The S
L-radii for certain well-known classes of functions including the Janowski starlike functions were obtained in [4]. General radii problems were also recently considered in [5, 6] wherein certain radii results for the class S
L were obtained as special cases.Motivated essentially by the above and some recent works [7], the main object of the present paper is to investigate some properties of multivalent analytic functions associated with the lemniscate of Bernoulli.


## 2. Main Results

The following lemma will be required.Lemma 1 (see [8]).
Letq be univalent in the unit disk U and let θ and φ be analytic in a domain D containing q
(
U
) with φ
(
w
)
≠
0 when w
∈
q
(
U
). Set Q
(
z
)
=
z
q
′
(
z
)
φ
(
q
(
z
)
), h
(
z
)
=
θ
(
q
(
z
)
)
+
Q
(
z
). Suppose that(1)
eitherh is convex or Q is a starlike univalent function in U,
(2)
R
e
z
h
′
(
z
)
/
Q
(
z
)
>
0 for z
∈
U.
Ifp is analytic in U, p
(
0
)
=
q
(
0
), and satisfies (6)
θ
p
z
+
z
p
′
z
φ
p
z
≺
θ
q
z
+
z
q
′
z
φ
q
z
,then p
(
z
)
≺
q
(
z
), and q is the best dominant.Theorem 2.
Letα
>
0. If f
∈
A
n
(
p
) satisfies(7)
f
z
z
p
1
+
α
z
f
′
z
f
z
-
p
≺
1
+
z
z
∈
U
,then f
(
z
)
/
z
p
≺
1
+
z.Proof.
Letp
(
z
)
=
f
(
z
)
/
z
p. Then p
(
z
) is analytic in U with p
(
0
)
=
1. Condition (7) becomes that(8)
f
z
z
p
1
+
α
z
f
′
z
f
z
-
p
=
p
z
+
α
z
p
′
z
≺
1
+
z
z
∈
U
.
Define the functionq
:
U
→
C by q
(
z
)
=
1
+
z with q
(
0
)
=
1. Since q
(
U
)
=
w
:
|
w
2
-
1
|
<
1 is the right-half of the lemniscate of Bernoulli, q
(
U
) is a convex set and hence q is a convex function. This shows that the function z
q
′
(
z
) is starlike with respect to the origin. We now show that p
(
z
)
≺
q
(
z
) if the function p satisfies the differential chain(9)
p
z
+
α
z
p
′
z
≺
q
z
+
α
z
q
′
z
.For this purpose, let the functions θ and φ be defined by θ
(
w
)
=
w and φ
(
w
)
=
α. Clearly the functions θ and φ are analytic in C and φ
(
w
)
≠
0. Also let Q and h be the functions defined by(10)
Q
z
=
z
q
′
z
φ
q
z
=
α
z
q
′
z
,
h
z
=
θ
q
z
+
Q
z
=
q
z
+
α
z
q
′
z
=
2
+
2
+
α
z
2
1
+
z
.It is obvious that Q is a starlike univalent function in U and(11)
R
e
z
h
′
z
Q
z
=
1
α
+
R
e
1
+
z
q
′
′
z
q
′
z
>
0
z
∈
Ufor α
>
0. By the lemma, it follows that p
≺
q
=
1
+
z. To complete the proof, it is left to show that(12)
q
z
=
1
+
z
≺
2
+
2
+
α
z
2
1
+
z
=
h
z
.Since q
-
1
(
w
)
=
w
2
-
1, it follows that(13)
q
-
1
h
z
=
z
1
+
α
+
α
2
4
1
+
z
.For z
=
e
i
t, t
∈
[
-
π
,
π
], clearly(14)
q
-
1
h
e
i
t
=
1
+
α
+
α
2
4
1
+
e
i
t
.A simple calculation shows that the minimum of the above expression is attained at t
=
0. Hence (15)
q
-
1
h
e
i
t
≥
1
+
α
+
α
2
8
>
1
.Thus q
-
1
(
h
(
U
)
)
⊃
U or h
(
U
)
⊃
q
(
U
). This shows that q
(
z
)
≺
h
(
z
) and completes the proof.Theorem 3.
Letf belong to the class A
n
(
p
) and satisfy(16)
f
z
z
p
≺
1
+
z
z
∈
U
.Then(17)
R
e
1
-
δ
f
z
z
p
2
+
δ
z
f
′
z
f
z
>
ρ
z
<
r
n
δ
,
ρ
,where 0
≤
ρ
<
p, ρ
/
p
<
δ
≤
1, and r
n
(
δ
,
ρ
) is the smallest root in (
0,1
) of the equation (18)
2
1
-
δ
r
2
n
-
2
2
-
ρ
+
δ
n
+
2
p
-
4
r
n
+
2
1
-
ρ
+
2
δ
p
-
1
=
0
.The result is sharp.Proof.
From (16) we can write(19)
f
z
z
p
2
=
1
+
z
n
φ
z
,where φ is analytic and |
φ
(
z
)
|
≤
1 in U. Differentiating both sides of (19) logarithmically, we arrive at(20)
z
f
′
z
f
z
=
p
+
n
z
n
φ
z
2
1
+
z
n
φ
z
+
z
n
+
1
φ
′
z
2
1
+
z
n
φ
z
z
∈
U
.Put |
z
|
=
r
<
1 and f
(
z
)
/
z
p
2
=
u
+
i
v
(
u
,
v
∈
R
). Then (19) implies that(21)
z
n
φ
z
=
u
-
1
+
i
v
,
1
-
r
n
≤
u
≤
1
+
r
n
.With the help of the Carathéodory inequality (22)
φ
′
z
≤
1
-
φ
z
2
1
-
r
2
,it follows from (21) that(23)
R
e
1
-
δ
f
z
z
p
2
+
δ
z
f
′
z
f
z
≥
1
-
δ
u
+
p
δ
+
n
δ
2
R
e
z
n
φ
z
1
+
z
n
φ
z
-
δ
2
z
n
+
1
φ
′
z
1
+
z
n
φ
z
≥
1
-
δ
u
+
p
δ
+
n
δ
2
1
-
u
u
2
+
v
2
+
δ
2
u
-
1
2
+
v
2
-
r
2
n
r
n
-
1
1
-
r
2
u
2
+
v
2
1
/
2
=
F
n
u
,
v
,
(24)
∂
F
n
u
,
v
∂
v
=
δ
v
G
n
u
,
v
,where 0
<
r
<
1, 0
<
δ
≤
1, and(25)
G
n
u
,
v
=
n
u
u
2
+
v
2
2
+
1
r
n
-
1
1
-
r
2
u
2
+
v
2
1
/
2
+
r
2
n
-
u
-
1
2
+
v
2
2
r
n
-
1
1
-
r
2
u
2
+
v
2
3
/
2
>
0because of (21). Since F
n
(
u
,
v
) is a dual function of v, from (23), (24), and (25), we see that(26)
F
n
u
,
v
≥
F
n
u
,
0
=
1
-
δ
u
+
p
δ
+
n
δ
2
1
-
1
u
+
δ
u
-
1
2
-
r
2
n
2
r
n
-
1
1
-
r
2
u
.
Now let us calculate the minimum value ofF
n
(
u
,
0
) on the closed interval [
1
-
r
n
,
1
+
r
n
]. From (26) we deduce that(27)
d
d
u
F
u
,
0
=
1
-
δ
+
δ
2
u
2
n
+
r
2
n
+
u
2
-
1
r
n
-
1
1
-
r
2
≥
1
-
δ
+
δ
2
1
+
r
n
2
n
-
2
r
1
-
r
n
1
-
r
2
=
1
-
δ
+
δ
2
1
+
r
n
2
I
n
r
,where (28)
I
n
r
=
n
-
2
r
1
-
r
n
1
-
r
2
.Note that I
1
(
r
)
=
1
-
r
/
1
+
r
>
0. Suppose that I
n
(
r
)
>
0. Then(29)
I
n
+
1
r
=
n
+
1
-
2
r
1
-
r
n
+
1
1
-
r
2
=
I
n
r
+
1
-
r
n
+
1
+
r
1
-
r
n
1
+
r
>
0
.Hence, by virtue of the mathematical induction, we have I
n
(
r
)
>
0 for all n
∈
N and 0
≤
r
<
1. This implies that(30)
d
d
u
F
u
,
0
>
0
1
-
r
n
≤
u
≤
1
+
r
n
.Further it follows from (23), (26), and (30) that(31)
R
e
1
-
δ
f
z
z
p
2
+
δ
z
f
′
z
f
z
-
ρ
≥
F
n
1
-
r
n
,
0
-
ρ
=
1
-
δ
1
-
r
n
+
p
δ
-
n
δ
r
n
2
1
-
r
n
-
ρ
=
2
1
-
δ
r
2
n
-
2
2
-
ρ
+
δ
n
+
2
p
-
4
r
n
+
2
1
-
ρ
+
2
δ
p
-
1
2
1
-
r
n
=
J
n
r
2
1
-
r
n
,where (32)
J
n
r
=
2
1
-
δ
r
2
n
-
2
2
-
ρ
+
δ
n
+
2
p
-
4
r
n
+
2
1
-
ρ
+
2
δ
p
-
1
.Note that J
n
(
0
)
=
2
(
1
-
ρ
)
+
2
δ
(
p
-
1
)
>
0 and J
n
(
1
)
=
-
n
δ
<
0. If we let r
n
(
δ
,
ρ
) denote the root in (
0,1
) of the equation J
n
(
r
)
=
0, then (31) yields the desired result (17).
To see that the boundr
n
(
δ
,
ρ
) is the best possible, we consider the function(33)
f
z
=
z
p
1
-
z
n
1
/
2
.Clearly the function f defined by (33) satisfies condition (16). Also it is obvious that, for z
=
r
∈
(
r
n
(
δ
,
ρ
)
,
1
), (34)
1
-
δ
f
r
r
p
2
+
δ
r
f
′
r
f
r
-
ρ
=
J
n
r
2
1
-
r
n
<
0
,which shows that the bound r
n
(
δ
,
ρ
) can not be increased. The proof of the theorem is completed.Settingδ
=
1, Theorem 3 reduces to the following.Corollary 4.
Letf belong to the class A
n
(
p
) and satisfy condition (16). Then f is starlike of order ρ  (
0
≤
ρ
<
p
) in (35)
z
<
2
p
-
ρ
2
p
-
ρ
+
n
1
/
n
.The result is sharp.



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*Source: 1782916-2016-05-17.xml*