Gen. Math. Notes, Vol. 15, No. 1, March, 2013, pp. 72-83
ISSN 2219-7184; Copyright © ICSRS Publication, 2013
www.i-csrs.org
Available free online at http://www.geman.in
Differential Sandwich Theorems for Integral
Operator of Certain Analytic Functions
Abbas Kareem Wanas
Department of Mathematics
College of Computer Science and Mathematics
University of Al-Qadisiya
Diwaniya – Iraq
E-mail: k.abbaswmk@yahoo.com
(Received: 16-11-12 / Accepted: 3-2-13)
Abstract
In the present paper, we obtain some subordination and superordination
results involving the integral operator ℑ for certain normalized analytic
functions in the open unit disk. These results are applied to obtain sandwich
results.
Keywords: Analytic functions, Differential subordination, Superordination,
Sandwich theorems, Dominant, Subordinant, Integral operator.
1
Introduction
Let = be the class of analytic functions in the open unit disk = {
∈
ℂ ∶ |
| < 1}. For a positive integer and ∈ ℂ. Let , be the subclass of
consisting of functions of the form:
∈ ℂ.
= + + + ⋯
Also, let be the subclass of consisting of functions of the form:
1.1
Differential Sandwich Theorems for Integral…
= +
∞
!"
73
.
1.2
Let , $ ∈ . The function is said to be subordinate to $, or $ is said to be
superordinate to , if there exists a Schwarz function % analytic in with
%0 = 0 and |%
| < 1 ∈ such that = $%
. In such a case we
write ≺ $ or ≺ $
∈ . If $ is univalent in , then ≺ $ if and
only if 0 = $0 and ⊂ $.
Let ), ℎ ∈ and +,, -, .; : ℂ1 × ⟶ ℂ. If ) and +)
, )′ , " )′′ ; are univalent functions in and if ) satisfies the second-order differential
superordination
1.3
ℎ
≺ +)
, )′ , " )′′ ; ,
then ) is called a solution of the differential superordination (1.3). (If is
subordinate to $, then $ is superordinate to ). An analytic function 5 is called a
subordinant of (1.3),if 5 ≺ ) for all the functions ) satisfying (1.3). An univalent
subordinant 56 that satisfies 5 ≺ 56 for all the subordinants 5 of (1.3) is called the
best subordinant. Miller and Mocanu [6] have obtained conditions on the
functions ℎ, 5 and + for which the following implication holds:
ℎ
≺ +)
, )′ , " )′′ ; ⟹ 5
≺ )
.
1.4
Komatu [4] introduced and investigated a family of integral operator ℑ ∶ ⟶ , which is defined as follows:
ℑ =
+
∞
!"
>
9
;
;"
=
<
=
@AB$
C
=D=
Γ:
;
=
?
9
E
G 9+−1
We note from (1.5) that, we have
∈ , 9 > 0, : ≥ 0.
@ℑ C = 9ℑ − 9 − 1ℑ .
′
1.5
1.6
Ali et al. [1] obtained sufficient conditions for certain normalized analytic
functions to satisfy
′
5 ≺
≺ 5" ,
where 5 and 5" are given univalent functions in with 5 0 = 5" 0 = 1.
Also, Tuneski [9] obtained a sufficient conditions for star likeness of in terms of
74
Abbas Kareem Wanas
the quantity
L ′′ >L>
@L ′ >C
M
. Recently, Shanmugam et al. [7,8], Goyal et al. [3] also
obtained sandwich results for certain classes of analytic functions.
The main object of the present paper is to find sufficient conditions for certain
normalized analytic functions to satisfy
P
ℑ 5 ≺ N
O ≺ 5" ,
P
.ℑ + 1 − .ℑ 5 ≺ N
O ≺ 5" ,
where 5 and 5" are given univalent functions in with 5 0 = 5" 0 = 1.
and
2
Preliminaries
In order to prove our subordination and superordination results, we need the
following definition and lemmas.
Definition 2.1 [5]: Denote by Q the set of all functions that are analytic and
injective on ∖ S, where
E = U V ∈ W: lim = ∞]
>→\
and are such that ′ V ≠ 0 for V ∈ W\S.
2.1
Lemma 2.1 [5]: Let 5 be univalent in the unit disk and let ` and a be analytic
in a domain b containing 5 with a% ≠ 0 when % ∈ 5 .Set Q
=
5 c ad5
e and ℎ
= `d5
e + Q
. Suppose that
(i) Q
is starlike univalent in ,
>i′ >
(ii) fg h j> k > 0 for ∈ .
If ) is analytic in , with )0 = 50, ) ⊂ b and
`d)
e + )′ ad)
e ≺ `d5
e + 5 ′ ad5
e,
then ) ≺ 5 and 5 is the best dominant of (2.2).
2.2
Lemma 2.2 [6]: Let 5 be a convex univalent function in and let l ∈ ℂ , m ∈ ℂ ∖
{0} with
Differential Sandwich Theorems for Integral…
5 ′′ l
fg n1 + ′
o > max U0, −fg ].
5 m
If ) is analytic in and
l)
+ m
)′ ≺ l5
+ m
5 ′ ,
then ) ≺ 5 and 5 is the best dominant of (2.3).
75
2.3
Lemma 2.3 [6]: Let 5 be convex univalent in and let m ∈ ℂ. Further assume
that fgm > 0. If ) ∈ 50,1 ∩ Q and )
+ m
)c is univalent in ,
then
5
+ m
5 ′ ≺ )
+ m
)′ ,
2.4
which implies that 5 ≺ ) and 5 is the best subordinant of (2.4).
Lemma 2.4 [2]: Let 5 be convex univalent in the unit disk and let ` and a be
analytic in a domain b containing 5. Suppose that
(i) fg U udt>e ] > 0 for ∈ ,
s′ dt>e
(ii) Q
= 5 ′ ad5
e is starlike univalent in .
If ) ∈ 50,1 ∩ Q, with ) ⊂ b, `d)
e + )′ ad)
e is univalent in
and
`d5
e + 5 ′ ad5
e ≺ `d)
e + )′ ad)
e,
2.5
then 5 ≺ ) and 5 is the best subordinant of (2.5).
3
Subordination Results
Theorem 3.1: Let 5 be convex univalent in with 50 = 1, 0 ≠ v ∈ ℂ, w > 0
and suppose that 5 satisfies
5 ′′ w
fg n1 + ′
o > max U0, −fg E G].
5 v
If ∈ satisfies the subordination
P
P
3.1
ℑ ℑ ℑ v
1 − 9v N
O + 9v N
O N O ≺ 5
+ 5 ′ ,
w
ℑ 3.2
76
Abbas Kareem Wanas
P
ℑ N
O ≺ 5
then
and 5 is the best dominant of (3.2).
Proof: Define the function ) by
P
ℑ )
= N
O .
Differentiating (3.4) with respect to logarithmically, we get
@ℑ C
)′ = wx
− 1y.
)
ℑ ′
3.3
3.4
3.5
Now, in view of (1.6), we obtain the following subordination
Therefore,
ℑ )′ = w9 N − 1O.
)
ℑ P
ℑ ℑ )′ = 9N
O N − 1O.
w
ℑ The subordination (3.2) from the hypothesis becomes
v
v
)
+ )′ ≺ 5
+ 5 ′ .
w
w
An application of Lemma 2.2 with m = P and l = 1, we obtain (3.3).
z
> {
Putting 5
= @;>C 0 < | ≤ 1 in Theorem 3.1, we obtain the following
corollary:
Corollary 3.1: Let 0 < | ≤ 1, 0 ≠ v ∈ ℂ, w > 0 and
fg n
1 + 2|
+ "
w
o > max U0, −fg E G].
"
1−
v
If ∈ satisfies the subordination
Differential Sandwich Theorems for Integral…
77
P
then
P
ℑ ℑ ℑ 1 − 9v N
O + 9v N
O N O
ℑ 2v|
1+
{
≺ E1 +
G
E
G ,
w1 − " 1 − P
ℑ 1+
{
N
O ≺E
G
1−
> {
and 5
= @;>C is the best dominant.
Theorem 3.2: Let 5 be convex univalent in with 50 = 1, 5
≠ 0 ∈ and assume that 5 satisfies
fg n1 +
~ € + 1
5 ′ 5 ′′ +
5
+  − 1
+ ′
o > 0,
v
v
5
5 3.6
where ~, €,  ∈ ℂ, v ∈ ℂ ∖ {0} and ∈ .
Suppose that d5
e
; ′
5
is starlike univalent in . If ∈ satisfies

‚~, €, w, :, ., , 9, v; ≺ d~ + €5
ed5
e + v
d5
e
where
‚~, €, w, :, ., , 9, v; .ℑ + 1 − .ℑ = ~N
O
P
P
3.7
P
.ℑ + 1 − .ℑ; N − 1O,
.ℑ + 1 − .ℑ 0 ≤ . ≤ 1, w > 0, ∈ ,
P
.ℑ + 1 − .ℑ N
O ≺ 5
and 5 is the best dominant of (3.7).
Proof: Define the function ) by
5
.ℑ + 1 − .ℑ +€N
O
.ℑ + 1 − .ℑ +vw9 N
O
then
; ′
,
P
.ℑ + 1 − .ℑ )
= N
O .
3.8
3.9
3.10
78
Abbas Kareem Wanas
By setting
`% = ~ + €%%  D a% = v% ; , % ≠ 0,
we see that `% is analytic in ℂ , a% is analytic in ℂ ∖ {0} and that a% ≠
0, % ∈ ℂ ∖ {0}. Also, we get
and
Q
= 5 ′ ad5
e = v
d5
e

; ′
5
ℎ
= `d5
e + Q
= d~ + €5
ed5
e + v
d5
e
It is clear that Q
is starlike univalent in ,
; ′
5
.
ℎ′ ~ € + 1
5 ′ 5 ′′ fg n
o = fg n1 +
+
5
+  − 1
+ ′
o > 0.
Q
v
v
5
5 By a straightforward computation, we obtain

; ′
d~ + €)
ed)
e + v
d)
e
) = ‚~, €, w, :, ., , 9, v; , 3.11
where ‚~, €, w, :, ., , 9, v; is given by (3.8).
From (3.7) and (3.11), we have

; ′
d~ + €)
ed)
e + v
d)
e
) 
≺ d~ + €5
ed5
e + v
d5
e
; ′
.
5
3.12
Therefore, by Lemma 2.1, we get )
≺ 5
. By using (3.10), we obtain the
result.
Putting 5
= ‡> −1 ≤ ˆ < ‰ ≤ 1 in Theorem 3.2, we obtain the following
corollary:
†>
Corollary 3.2: Let −1 ≤ ˆ < ‰ ≤ 1 and
~ € + 11 + ‰
1 + ‰ − ˆ
− ‰ˆ
"
fg n
+
+
o> 0,
1 + ‰
1 + ˆ
v
v1 + ˆ
where ~, €,  ∈ ℂ, v ∈ ℂ ∖ {0} and ∈ . If ∈ satisfies
‚~, €, w, :, ., , 9, v; 1 + ‰
1 + ‰
 v‰ − ˆ1 + ‰
; ≺ N~ + € E
GO E
G +
,
1 + ˆ
1 + ˆ
1 + ˆ

where ‚~, €, w, :, ., , 9, v; is given by (3.8),
then
Differential Sandwich Theorems for Integral…
79
P
and 5
=
4
†>
‡>
.ℑ + 1 − .ℑ 1 + ‰
N
O ≺
1 + ˆ
is the best dominant.
Superordination Results
Theorem 4.1: Let 5 be convex univalent in with 50 = 1, w > 0 and fg{v} >
0. Let ∈ satisfies
P
ℑ N
O ∈ 50,1 ∩ Q
and
P
P
ℑ ℑ ℑ 1 − 9v N
O + 9v N
O N O
ℑ be univalent in . If
P
P
ℑ ℑ ℑ v ′
5
+ 5 ≺ 1 − 9v N
O + 9v N
O N O,
w
ℑ 4.1
then
P
ℑ 5
≺ N
O
4.2
and 5 is the best subordinant of (4.1).
Proof: Define the function ) by
P
ℑ )
= N
O .
Differentiating (4.3) with respect to logarithmically, we get
@ℑ C
)′ = wx
− 1y.
)
ℑ ′
4.3
4.4
After some computations and using (1.6), from (4.4), we obtain
P
P
ℑ ℑ ℑ v
1 − 9v N
O + 9v N
O N O = )
+ )′ ,
w
ℑ and now, by using Lemma 2.3, we get the desired result.
80
Abbas Kareem Wanas
> {
Putting 5
= @ C 0 < | ≤ 1 in Theorem 4.1, we obtain the following
;>
corollary:
Corollary 4.1: Let 0 < | ≤ 1, w > 0 and fg{v} > 0. If ∈ satisfies
P
and
ℑ N
O ∈ 50,1 ∩ Q
P
P
ℑ ℑ ℑ 1 − 9v N
O + 9v N
O N O
ℑ be univalent in . If
2v|
1+
{
E1 +
GE
G
w1 − " 1 − P
then
P
ℑ ℑ ℑ ≺ 1 − 9v N
O + 9v N
O N O,
ℑ P
ℑ 1+
{
E
G ≺N
O
1−
> {
and 5
= @;>C is the best subordinant.
Theorem 4.2: Let 5 be convex univalent in with 50 = 1, and assume that 5
satisfies
~ ′
€ + 1
4.5
fg n
5 +
5
5 ′ o > 0,
v
v
where ~, €,  ∈ ℂ, v ∈ ℂ ∖ {0} and ∈ .
;
Suppose that d5
e
5 ′ is starlike univalent in . Let ∈ satisfies
P
.ℑ + 1 − .ℑ N
O ∈ 50,1 ∩ Q
and ‚~, €, w, :, ., , 9, v; is univalent in , where ‚~, €, w, :, ., , 9, v; is
given by (3.8). If

; ′
d~ + €5
ed5
e + v
d5
e
5 ≺ ‚~, €, w, :, ., , 9, v; , 4.6
then
P
.ℑ + 1 − .ℑ 5
≺ N
O
4.7
Differential Sandwich Theorems for Integral…
81
and 5 is the best subordinant of (4.6).
Proof: Define the function ) by
By setting
P
.ℑ + 1 − .ℑ )
= N
O .
`% = ~ + €%%  D a% = v% ; , % ≠ 0,
4.8
we see that `% is analytic in ℂ, a% is analytic in ℂ ∖ {0} and that a% ≠
0, % ∈ ℂ ∖ {0}. Also, we get
Q
= 5 ′ ad5
e = v
d5
e
It is clear that Q
is starlike univalent in ,
fg n
` ′ d5
e
ad5
e
o = fg n
; ′
.
5
~ ′
€ + 1
5 +
5
5′ o > 0.
v
v
By a straightforward computation, we obtain

; ′
‚~, €, w, :, ., , 9, v; = d~ + €)
ed)
e + v
d)
e
) , 4.9
where ‚~, €, w, :, ., , 9, v; is given by (3.8).
From (4.6) and (4.9), we have

; ′
d~ + €5
ed5
e + v
d5
e
5 
≺ d~ + €)
ed)
e + v
d)
e
; ′
.
)
4.10
Therefore, by Lemma 2.4, we get 5
≺ )
. By using (4.8), we obtain the
result.
5
Sandwich Results
Concluding the results of differential subordination and superordination, we arrive
at the following "sandwich results".
Theorem 5.1: Let 5 be convex univalent in with 5 0 = 1, fg{v} > 0 and let
5" be univalent in , 5" 0 = 1 and satisfies (3.1). Let ∈ satisfies
P
and
ℑ N
O ∈ 1,1 ∩ Q
82
Abbas Kareem Wanas
P
P
ℑ ℑ ℑ 1 − 9v N
O + 9v N
O N O
ℑ be univalent in . If
P
P
ℑ ℑ ℑ v
′
5 + 5 ≺ 1 − 9v N
O + 9v N
O N O
w
ℑ v
≺ 5" + 5" ′ ,
w
then
P
ℑ 5 ≺ N
O ≺ 5" and 5 and 5" are, respectively, the best subordinant and the best dominant.
Theorem 5.2: Let 5 be convex univalent in with 5 0 = 1 and satisfies (4.5)
and let 5" be univalent in , 5" 0 = 1 and satisfies (3.6). Let ∈ satisfies
P
.ℑ + 1 − .ℑ N
O ∈ 1,1 ∩ Q
and ‚~, €, w, :, ., , 9, v; is univalent in , where ‚~, €, w, :, ., , 9, v; is
given by (3.8). If

d~ + €5 ed5 e + v
d5 e
;

5 ′ ≺ ‚~, €, w, :, ., , 9, v; ≺ d~ + €5" ed5" e + v
d5" e
;
P
5" ′ ,
.ℑ + 1 − .ℑ 5 ≺ N
O ≺ 5" and 5 and 5" are, respectively, the best subordinant and the best dominant.
then
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[2]
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[4]
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