Acta Mathematica Academiae Paedagogicae Nyı́regyháziensis
25 (2009), 195–209
www.emis.de/journals
ISSN 1786-0091
ON DIFFERENTIAL SANDWICH THEOREMS FOR SOME
SUBCLASS OF MULTIVALENT ANALYTIC FUNCTIONS
T. N. SHANMUGAM, J. PANDURANGAN, AND M. P. JEYARAMAN
Abstract. In this present investigation we study certain application of
differential subordination and superordination for the class of multivalent
functions to be subordinated and superordinated by convex functions.
1. Introduction
Let H be the class of analytic functions in ∆ := {z ∈ C : |z| < 1} and
H[a, n] be a subclass of H consisting of functions of the form
f (z) = a + an z n + an+1 z n+1 + . . . .
Let Ap denote the class of functions of the form
∞
X
p
(1.1)
f (z) := z +
an z n
(z ∈ ∆),
n=p+1
and let A := A1 . Komatu [4] introduced the family of integral operators,
defined by
Z
(1 + a)σ z ³
z ´σ−1 a−1
σ
(1.2)
Ia f (z) := a
log
t f (t)dt,
z Γ(σ) 0
t
where a > −1, σ > 0 and f ∈ A. It can be easily observed that
¶
∞ µ
X
1+a
σ
(1.3)
Ia f (z) = z +
an z n .
n
+
a
n=2
From (1.2) and (1.3) it can be seen that
z(Iaσ+1 f (z))0 = (1 + a)Iaσ f (z) − aIaσ+1 f (z).
Let p, h ∈ H and let
φ(r, s, t; z) : C3 × ∆ → C.
2000 Mathematics Subject Classification. Primary 30C45, secondary 30C80.
Key words and phrases. Analytic functions, differential subordination, superordination,
Komatu operator.
195
196
T. N. SHANMUGAM, J. PANDURANGAN, AND M. P. JEYARAMAN
If p(z) and φ(p(z), zp0 (z), z 2 p00 (z); z) are univalent and if p(z) satisfy the second
order superordination
h(z) ≺ φ(p(z), zp0 (z), z 2 p00 (z); z),
(1.4)
then p is the solution of the differential superordination (1.4). (If f is subordinate to F , then we say F is superordinate to f ). An analytic function q is
called a subordinant if q ≺ p for all p satisfying (1.4). A univalent subordinant
qe that satisfy q ≺ qe for all subordinants q of (1.4) is said to be best subordinant. Recently Miller and Mocanu [6] obtained conditions on h, q and φ for
which the following implication holds:
h(z) ≺ φ(p(z), zp0 (z), z 2 p00 (z); z) ⇒ q(z) ≺ p(z).
Using the results of Miller and Mocanu [6], Bulboacă [3] have considered certain
classes of first order differential superordinations as well as superordination
preserving integral operators [2].
Over many years, several authors have studied the application of differential
0 (z)
00 (z)
subordination and superordination for functionals like zff (z)
, 1 + zff 0 (z)
, zff (z)
0 (z) .
Recently
³
´ Obradović and Owa [7] obtained some subordination result in terms
of
f (z)
z
µ
.
In this present investigation we give some applications of first order differential subordination and superordination to obtain sufficient conditions for
certain normalized analytic functions f to satisfy
µ
¶µ
1 f (z)
q1 (z) ≺
≺ q2 (z)
p
zp
where q1 and q2 are univalent in ∆. Interestingly various well known results
are special cases of our results.
2. Preliminaries
For the present investigation we need the following definition and results.
Definition 2.1. [6, Definition 2, p. 817] Let Q be the set of all functions f
¯ − E(f ), where
that are analytic and injective on ∆
½
¾
E(f ) = ζ ∈ ∂∆ : lim f (z) = ∞ ,
z→ζ
and are such that f 0 (ζ) 6= 0 for ζ ∈ ∂∆ − E(f ).
Theorem 2.1 ([5, Theorem 3.4h, p. 132 ]). Let q be univalent in the disk ∆
and θ and φ be analytic in a domain D containing q(∆) with φ(w) 6= 0 when
w ∈ q(∆).
Set Q(z) = zq 0 (z)φ(q(z)), h(z) = θ(q(z)) + Q(z). Suppose that
(1) Q is starlike univalent in ∆ and
0 (z)
> 0 for z ∈ ∆.
(2) < zh
Q(z)
ON DIFFERENTIAL SANDWICH THEOREMS
197
If ξ is analytic in ∆ with ξ(∆) ⊆ D, and
θ(ξ(z)) + zξ 0 (z)φ(ξ(z)) ≺ θ(q(z)) + zq 0 (z)φ(q(z)),
(2.1)
then ξ ≺ q and q is the best dominant.
Theorem 2.2 ([3]). Let q be univalent in the unit disk ∆ and ϑ and φ be
analytic in a domain D containing q(∆) , suppose that
0
ϑ q(z)
(1) < ψ(q(z))
> 0 for all z ∈ ∆ and
0
(2) ξ(z) = zq (z)ψ(q(z)) is starlike univalent function in ∆.
If ξ ∈ H [q(0), 1] ∩ Q , with ξ(∆) ⊂ D, and ϑ(ξ(z)) + zξ 0 (z)ψ(ξ(z)) is univalent
in ∆ , and
(2.2)
ϑ(q(z)) + zq 0 (z)ψ(q(z)) ≺ ϑ(ξ(z)) + zξ 0 (z)ψ(ξ(z)),
then q ≺ ξ and q is the best subordinant.
Theorem 2.3 ([5, Lemma 1, p. 71]). Let h be convex univalent in ∆ with
h(0) = a and 0 6= γ ∈ C and <γ > 0. If p ∈ H[a, n] and
zp0 (z)
p(z) +
≺ h(z)
γ
then
p(z) ≺ q(z) ≺ h(z),
where
Z z
γ
γ
q(z) =
h(t)t n −1 dt.
γ
nz n 0
The function q is convex and is the best dominant.
3. Application of Differential Subordination
Theorem 3.1. Let α, β and γ be complex numbers with γ 6= 0. Let q be convex
univalent in ∆ with q(0) = 1 and satisfy
½
¾
α + 2βq(z)
(3.1)
<
> 0.
γ
Let f ∈ Ap and
(3.2)
α
ψ(z) :=
p
µ
f (z)
zp
¶µ
β
+ 2
p
µ
f (z)
zp
µ
¶2µ
+ γµ
f (z)
zp
If
ψ(z) ≺ αq(z) + βq 2 (z) + γzq 0 (z)
then
and q is the best dominant.
1
p
µ
f (z)
zp
¶µ
≺ q(z)
¶µ ·
¸
zf 0 (z)
−1 .
pf (z)
198
T. N. SHANMUGAM, J. PANDURANGAN, AND M. P. JEYARAMAN
Proof. Define the function ξ(z) by
1
ξ(z) :=
p
(3.3)
µ
f (z)
zp
¶µ
.
A computation using (3.3) shows that
zµf 0 (z)
zξ 0 (z)
=
− µp.
ξ(z)
f (z)
Also we find that
ψ(z) := αξ(z) + βξ 2 (z) + γzξ 0 (z)
µ
¸
¶µ
µ
¶2µ
¶µ · 0
µ
α f (z)
β f (z)
f (z)
zf (z)
=
−1 .
+ 2
+ γµ
p
zp
p
zp
zp
pf (z)
Since ψ(z) ≺ αq(z) + βq 2 (z) + γzq 0 (z), this can be written as (2.1), when
θ(w) := αw + βw2 and φ(w) := γ. Note that φ(w) 6= 0 and θ(w), φ(w) are
analytic in C. Set
Q(z) := γzq 0 (z)
h(z) := θ(q(z)) + Q(z)
= αq(z) + βq 2 (z) + γzq 0 (z).
In light of the hypothesis of Theorem 2.1, we see that Q is starlike and
½
¾
zh0 (z)
α + 2βq(z)
zq 00 (z)
<
=<
+ (1 + 0
) > 0.
Q(z)
γ
q (z)
Hence the result follows as an application of Theorem 2.1.
¤
Theorem 3.2.
(1) Let 0 6= δ ∈ C and q be convex univalent in ∆ with q(0) = 1 and satisfy
nµo
> 0.
<
δ
If f ∈ A satisfy
Ã
¶µ
µ
¶µ−1 !
µ
f (z)
f
(z)
δ
+ δ f 0 (z)
≺ q(z) + zq 0 (z)
(1 − δ)
z
z
µ
then
µ
(2) If f ∈ A satisfy
0
f (z)
then
µ
f (z)
z
f (z)
z
¶µ
≺ q(z).
µ
¶µ−1
−
µ
f (z)
z
f (z)
z
¶µ
¶µ
≺ q(z)
≺
1 0
zq (z)
µ
ON DIFFERENTIAL SANDWICH THEOREMS
199
and q is the best dominant.
Proof. Proof of the first part follows from Theorem 3.1, by taking α = p =
1, β = 0 and γ = µδ .
The proof of the second part follows from Theorem 3.1, by taking α = β =
0, p = 1 and γ = µ1 .
¤
£
¤
By taking δ = µ = n and q(z) = β + (1 − β) −1 − z2 log(1 − z) in first part
of Theorem 3.2, we get the following result of Ponnusamy [8].
Corollary 3.3. Let f ∈ A. Then for a positive integer n, we have
(
µ
¶n
µ
¶n−1 )
f (z)
f
(z)
< (1 − n)
+ nf 0 (z)
>β
z
z
implies
µ
f (z)
z
¶n
≺ β + (1 − β)(−1 −
2
log(1 − z))
z
and β + (1 − β)(−1 − z2 log(1 − z)) is the best dominant.
¡ A ¢
By taking µ = 1 and q(z) = 1 + 1+δ
z in Theorem 3.2 and µ = δ = 1 and
¡
¢
A
A log(1+Bz)
q(z) = B + 1 − B
where δ, A and B are non zero complex numbers
Bz
and <δ > 0 and −1 ≤ B < A ≤ 1 in Theorem 3.2 we get the following result
of Ponnusamy and Juneja [9].
Corollary 3.4. Let f ∈ A. Let δ be a complex number with <δ ≥ 0 and
−1 ≤ B < A ≤ 1. Then
(1 − δ)
f (z)
+ δf 0 (z) ≺ 1 + Az
z
implies
and the function 1 +
¡
A
1+δ
¢
f (z)
≺1+
z
and the function
A
B
A
1+δ
¶
z
z is the best dominant. Also
f 0 (z) ≺
implies
µ
1 + Az
1 + Bz
µ
¶
A
A log(1 + Bz)
f (z)
≺ + 1−
z
B
B
Bz
¡
¢
A log(1+Bz)
+ 1− B
is the best dominant.
Bz
By taking α = p = 1, β = 0 and γ =
result:
1
µ
in Theorem 3.1, we have the following
200
T. N. SHANMUGAM, J. PANDURANGAN, AND M. P. JEYARAMAN
Corollary 3.5. If f ∈ A satisfy
µ
¶µ−1
f (z)
zq 0 (z)
0
f (z)
≺ q(z) +
z
µ
implies
µ
f (z)
z
¶µ
≺ q(z)
and q is the best dominant.
Theorem 3.6. Let α, β, γ ∈ C with γ 6= 0. Let q be convex univalent in ∆
0 (z)
be starlike univalent in ∆. Further assume that
and zqq(z)
½
¾
βq(z) zq 0 (z)
(3.4)
<
−
> 0.
γ
q(z)
Let f ∈ Ap and if
µ
¶µ
· 0
¸
zf (z)
β f (z)
γzq 0 (z)
α+
+
γµ
−
p
≺
α
+
βq(z)
+
p
zp
f (z)
q(z)
then
1
p
µ
f (z)
zp
¶µ
≺ q(z)
where q is the best dominant.
Proof. Let θ(w) := α + βw and φ(w) := wγ . Note that θ(w) and φ(w) are
analytic in C³\ {0}.
´µHence the result follows as an application of Theorem 2.1
1 f (z)
for ξ(z) := p z
.
¤
By taking α = p = 1, β = 0, γ = µ1 and q(z) = eλAz , in Theorem 3.6 we get
the following result obtained by Obradović and Owa [7].
Corollary 3.7. Let f ∈ A. If
zf 0 (z)
≺ 1 + Az
f (z)
then
µ
f (z)
z
¶µ
≺ eλAz ,
where eλAz is the best dominant.
We remark here that q(z) = eλAz is univalent if and only if |λA| < π.
1
For a special case when q(z) = (1−z)
2b where b ∈ C \ {0}, and α = µ = p =
1, β = 0 and γ = 1b in Theorem 3.6, we have the following result obtained by
the Srivatsava and Lashin [10].
ON DIFFERENTIAL SANDWICH THEOREMS
201
Corollary 3.8. Let 0 6= b ∈ C. If f ∈ A and
·
¸
1 zf 0 (z)
1+z
−1 ≺
1+
b f (z)
1−z
then
f (z)
1
≺
,
z
(1 − z)2b
1
where (1−z)
2b is the best dominant.
λ(A−B)
By taking q(z) = (1+Bz) B , α = p = 1, β = 0 and γ = µ1 in Theorem 3.6,
then we have the following result of Obradović and Owa [7].
Corollary
3.9. Let
¯
¯ −1 ≤¯ B < A ≤ ¯1. Let µ, A and B satisfy the relation
¯ λ(A−B)
¯
¯
¯
either ¯ B − 1¯ ≤ 1 or ¯ λ(A−B)
+ 1¯ ≤ 1. If f ∈ A and
B
zf 0 (z)
1 + Az
≺
f (z)
1 + Bz
then
and (1 + Bz)
µ
λ(A−B)
B
f (z)
z
¶µ
≺ (1 + Bz)
λ(A−B)
B
is the best dominant.
Theorem 3.10. Let α, β and γ be complex numbers and γ 6= 0. Let q(z) be
univalent in ∆ with q(0) = 1. Let f ∈ Ap satisfy (3.1). Let
(3.5)
µ p ¶µ
·µ p ¶µ
µ p ¶µ ¸
µ p ¶2µ
α
z
z
z
β
z
1 zf 0 (z)
ψ(z) :=
+ 2
+ γµ
−
.
p f (z)
p f (z)
f (z)
p f (z) f (z)
If
ψ(z) ≺ αq(z) + βq 2 (z) + γzq 0 (z)
then
µ p ¶µ
1
z
≺ q(z)
p f (z)
and q is the best dominant.
Proof. The proof is a straight forward application of Theorem 2.1.
By putting α = p = 1, β = 0 and γ =
following result:
λ
µ
¤
in Theorem 3.10 we have the
Corollary 3.11. If f (z) ∈ A and
µ
µ
¶µ
¶µ+1
z
z
λ
0
(1 + λ)
− λf (z)
≺ q(z) + zq 0 (z)
f (z)
f (z)
µ
then
µ
¶µ
z
≺ q(z).
f (z)
202
T. N. SHANMUGAM, J. PANDURANGAN, AND M. P. JEYARAMAN
By taking λ = −1 in Corollary 3.11 we get the following result.
Corollary 3.12. If f ∈ A and
µ
¶µ+1
z
zq 0 (z)
0
f (z)
≺ q(z) −
f (z)
µ
implies
¶µ
µ
z
≺ q(z)
f (z)
and q is the best dominant.
4. Application of Superordination
Theorem 4.1. Let α, β and γ be complex numbers and γ 6= 0. Let q be convex
univalent in ∆ with q(0) = 1 and satisfies
½µ
¶¾
α + 2βq(z)
(4.1)
<
> 0.
γ
Let
µ
¸
¶µ
µ
¶2µ
µ
¶µ · 0
α f (z)
f (z)
zf (z)
β f (z)
ψ(z) :=
−1
+ 2
+ γµ
p
zp
p
zp
zp
f (z)
³
´µ
and is univalent in ∆. If f ∈ Ap , 0 6= p1 fz(z)
∈ H[1, 1] ∩ Q then
p
αq(z) + βq 2 (z) + γzq 0 (z) ≺ ψ(z)
implies
1
q(z) ≺
p
where q is the best subordinant.
Proof. Define the function ξ(z) by
1
ξ(z) :=
p
(4.2)
µ
µ
f (z)
zp
f (z)
zp
¶µ
¶µ
.
A computation using (4.2) shows that
zµf 0 (z)
zξ 0 (z)
=
− pµ.
ξ(z)
f (z)
Note that
µ
¶µ
µ
¶2µ
¶µ · 0
¸
µ
β f (z)
f (z)
α f (z)
zf (z)
+ 2
+ γµ
ψ(z) :=
−1
p
zp
p
zp
zp
f (z)
= αξ(z) + βξ 2 (z) + γzξ 0 (z).
Since αq(z) + βq 2 (z) + γzq 0 (z) ≺ ψ(z), this can be written as (2.2), when
θ(w) := αw + βw2 and φ(w) := γ. Hence the result follows as an application
of Theorem 2.2.
¤
ON DIFFERENTIAL SANDWICH THEOREMS
203
Corollary 4.2. Let f ∈ A and δ be complex number with <δ > 0 and −1 ≤
B < A ≤ 1.
(i) If (1 − δ) f (z)
+ δzf 0 (z) is univalent in ∆, then
z
1 + Az ≺ (1 − δ)
f (z)
A
f (z)
+ δzf 0 (z) ⇒ 1 +
z≺
z
1+δ
z
A
and 1+δ
z is the best subordinant.
(ii) If f 0 (z) is univalent in ∆ then
1 + Az
A
A log(1 + Bz)
f (z)
≺ f 0 (z) ⇒ + (1 − )
≺
1 + Bz
B
B
Bz
z
and
A
B
A log(1+Bz)
+ (1 − B
) Bz
is the best subordinant.
Proof. Proof of first part follows from Theorem 4.1 by taking α = p = 1, β = 0,
A
γ = δ, µ = 1 and q(z) := 1 + 1+δ
z.
Proof of the second part follows from Theorem 4.1 by taking α = p = 1,
A
A log(1+Bz)
β = 0, γ = δ = 1, µ = 1 and q(z) := B
+ (1 − B
) Bz .
¤
Theorem 4.3. Let α, β and γ be complex numbers and γ 6= 0. Let q be convex
0 (z)
univalent in ∆ and zqq(z)
be starlike univalent in ∆. Further assume that
½
¾
βq(z)
(4.3)
<
> 0.
γ
³
´µ
h 0
i
³
´µ
zf (z)
1 f (z)
+
γµ
−
p
is
univalent
in
∆.
If
f
∈
A
,
∈
Let α + βp fz(z)
p p
p
f (z)
zp
H[1, 1] ∩ Q then
µ
¶µ
· 0
¸
zf (z)
β f (z)
γzq 0 (z)
α + βq(z) +
≺α+
+ γµ
−p
q(z)
p
zp
f (z)
implies
1
q(z) ≺
p
µ
f (z)
zp
¶µ
where q is the best subordinant.
Proof. Let θ(w) := α + βw and φ(w) := wγ . Note that θ(w) and φ(w) are
analytic in C \ {0}.
the result follows as an application of Theorem 2.2,
³ Hence
´
when ξ(z) :=
1
p
f (z)
zp
µ
.
¤
Note that by taking α = p = 1, β = 0, γ = µ1 and q(z) := eλAz we get
the corresponding superordination result of Corollary 3.7. Also by taking
1
α = µ = p = 1, β = 0, γ = 1b and q(z) := (1−z)
2b we obtain the superordination
result of Corollary 3.8
204
T. N. SHANMUGAM, J. PANDURANGAN, AND M. P. JEYARAMAN
Theorem 4.4. Let α, β, γ ∈ C and γ 6= 0. Let q be convex univalent in ∆
with q(0) = 1 and satisfy
½µ
¶¾
α + 2βq(z)
<
> 0.
γ
³ p ´µ
Let ψ(z) as defined by (3.5) be univalent in ∆. If f ∈ Ap and 0 6= p1 fz(z) ∈
H[1, 1] ∩ Q then
αq(z) + βq 2 (z) + γzq 0 (z) ≺ ψ(z)
implies
1
q(z) ≺
p
where q is the best subordinant.
By letting α = p = 1, β = 0 and γ =
result:
µ
λ
µ
zp
f (z)
¶µ
in Theorem 4.4, we get the following
Corollary 4.5. Let 0 6= λ ∈ C. Let q be convex univalent in ∆ with q(0) = 1
and satisfy
nµ
o
<
q 0 (z) > 0.
λ
Let
µ
¶µ
µ
¶µ+1
z
z
0
ψ1 (z) := (1 + λ)
− λf (z)
f (z)
f (z)
³
´µ
z
be univalent in ∆. If f ∈ A and f (z)
∈ H[1, 1] ∩ Q then
q(z) +
λ 0
q (z) ≺ ψ1 (z)
µ
implies that
µ
q(z) ≺
z
f (z)
¶µ
and q is the best subordinant.
By taking λ = −1 in Corollary 4.5 we get the following result:
Corollary 4.6. If f ∈ A and
q 0 (z)
q(z) −
≺ f 0 (z)
µ
implies
µ
q(z) ≺
and q is the best subordinant.
z
f (z)
µ
z
f (z)
¶µ
¶µ
ON DIFFERENTIAL SANDWICH THEOREMS
205
5. Sandwich Results
By combining Theorem 3.1 and Theorem 4.1 we get the following sandwich
type result.
Theorem 5.1. Let q1 and q2 be convex univalent in ∆, satisfying (4.1) and
(3.1)
Let ψ(z) as given by (3.2) be univalent in ∆. If 0 6=
´µ
³ respectively.
1 f (z)
∈ H[1, 1] ∩ Q then
p
zp
αq1 (z) + βq12 (z) + γzq10 (z) ≺ ψ(z) ≺ αq2 (z) + βq22 (z) + γzq20 (z)
implies
µ
¶µ
1 f (z)
q1 (z) ≺
≺ q2 (z)
p
zp
where q1 and q2 are respectively the best subordinant and best dominant.
Now by combining Theorem 3.6 and Theorem 4.3 with p = 1 we have the
following result.
Theorem 5.2. Let q1 and q2 be convex univalent in ∆, satisfying (4.3) and
zq 0 (z)
(3.4) respectively. Suppose qii(z) be starlike univalent in ∆ for i = 1, 2. Let
µ
¶µ
µ 0
¶
f (z)
zf (z)
η(z) := α + β
+ γµ
−1
z
f (z)
³
´µ
be univalent in ∆. If 0 6= f (z)
∈ H[1, 1] ∩ Q then
z
α + βq1 (z) + γ
zq10 (z)
zq 0 (z)
≺ η(z) ≺ α + βq2 (z) + γ 2
q1 (z)
q2 (z)
implies
µ
¶µ
f (z)
q1 (z) ≺
≺ q2 (z)
z
where q1 and q2 are respectively best subordinant and best dominant.
Theorem 5.3. Let q³1 and
´µq2 be convex univalent satisfying (4.1) and (3.1)
f (z)
respectively. Let 0 6= z
∈ H[1, 1] ∩ Q.
³
´µ
³
´µ−1
f (z)
0
(i) Let f ∈ A, and (1 − δ) f (z)
+
δf
(z)
is univalent in ∆ then
z
z
1 + (1 − 2β1 )z
≺ (1 − δ)
1−z
µ
f (z)
z
implies
2
1 + (−1 − β1 )(1 − log(1 − z) ≺
z
µ
¶µ
0
+ δf (z)
µ
f (z)
z
f (z)
z
¶µ−1
≺
1 + (1 − 2β2 )z
1−z
¶µ
≺ 1 + (−1 − β2 )(1 −
2
log(1 − z)
z
206
T. N. SHANMUGAM, J. PANDURANGAN, AND M. P. JEYARAMAN
where 1 + (−1 − β1 )(1 − z2 log(1 − z) and 1 + (−1 − β2 )(1 − z2 log(1 − z) are
respectively the best subordinant and best dominant.
(ii) If (1 − δ) f (z)
+ δf 0 (z) is univalent in ∆, then
z
1 + A1 z ≺ (1 − δ)
f (z)
+ δf 0 (z) ≺ 1 + A2 z
z
implies
A1
f (z)
A2
1+
z≺
≺1+
z
1+δ
z
1+δ
¡A ¢
¡A ¢
2
1
where 1 + 1+δ
z and 1 + 1+δ
z are respectively the best subordinant and best
dominant.
Proof. The proof of the first part follows from Theorem 5.1 by taking
qi (z) = 1+(1−βi )(−1−2 log(1−z) for i = 1, 2 and by taking α = p = 1,¡ β =¢ 0
Ai
and γ = µδ and the proof of second part follows by taking qi (z) = 1 + 1+δ
z
δ
¤
for i = 1, 2 and by taking α = µ = p = 1, β = 0 and γ = µ .
In a similar manner we may obtain the sandwich result by combining Theorem 4.4 and Theorem 3.10.
6. Application to Komatu operator
Theorem 6.1. Let h ∈ H, h(0) = 1, h0 (0) 6= 0 and satisfy
½
¾
zh00 (z)
1
< 1+ 0
>−
(z ∈ ∆).
h (z)
2
If f ∈ Am satisfy the differential subordination
Iaσ f (z)
≺ h(z)
z
then
Iaσ+1 f (z)
≺ g(z)
z
(6.1)
where
g(z) :=
1+a
1+a
m
Z
z
h(t)t
1+a
−1
m
mz
0
The function g is convex and is the best dominant.
dt.
Proof. Let the function p(z) be defined by
Iaσ+1 f (z)
p(z) :=
.
z
A simple computation shows that
¸
· σ+1
z(Ia f (z))0
zp0 (z)
−1 .
=
p(z)
Iaσ+1 f (z)
ON DIFFERENTIAL SANDWICH THEOREMS
207
By using the identity
z(Iaσ+1 f (z))0 = (1 + a)Iaσ f (z) − aIaσ+1 f (z),
we have
·
¸
(1 + a)Iaσ f (z)
zp0 (z)
=
− (a + 1) ,
p(z)
Iaσ+1 f (z)
and hence
zp0 (z)
Iaσ f (z)
p(z) +
=
.
a+1
z
The assertion (6.1) of Theorem 6.1 follows by an application of Theorem 2.3.
¤
Theorem 6.2. Let the function q(z) be convex univalent in ∆ and q(z) 6= 0.
0 (z)
Suppose that zqq(z)
is starlike univalent in ∆ and q(z) satisfy
½
¾
β
2γ 2
zq 0 (z)
(6.2)
<
q(z) + q (z) −
>0
δ
δ
q(z)
and let
(6.3) χ(z) :=
µ σ+1
µ σ+1
· σ
¶µ
¶2µ
¸
Ia f (z)
Ia f (z)
Ia f (z)
α+β
+γ
+ δµ(1 + a) σ+1
−1 .
z
z
Ia f (z)
If
(6.4)
δzq 0 (z)
χ(z) ≺ α + βq(z) + γq (z) +
q(z)
2
then
µ
Iaσ+1 f (z)
z
¶µ
≺ q(z)
and q is the best dominant.
Proof. Define the function p(z) by
p(z) :=
µ
Iaσ+1 f (z)
z
¶µ
.
Note the function p(z) is analytic in ∆. By a straight forward computation we
have
¸
µ σ+1
¶µ
µ σ+1
¶2µ
· σ
Ia f (z)
Ia f (z)
Ia f (z)
−1
χ(z) := α + β
+γ
+ δµ(1 + a) σ+1
z
z
Ia f (z)
(6.5)
= α + βp(z) + γp2 (z) +
δzp0 (z)
.
p(z)
208
T. N. SHANMUGAM, J. PANDURANGAN, AND M. P. JEYARAMAN
By using (6.5) in subordination (6.4), we have
(6.6)
α + βp(z) + γp2 (z) +
δzp0 (z)
δzq 0 (z)
≺ α + βq(z) + γq 2 (z) +
.
p(z)
q(z)
The subordination (6.6) is same as (2.1) with θ(w) := α + βw + γw2 and
φ(w) := wδ . Clearly θ(w) and φ(w) are analytic in C \ {0}. Hence the result
follows as an application of Theorem 2.1.
¤
0
(z)
be starlike univalent
Theorem 6.3. Let q(z) be convex univalent in ∆ and zqq(z)
in ∆. Further assume that
½
¾
2γ 2
β
<
q (z) + q(z) > 0.
δ
δ
³ σ+1 ´µ
Let χ(z) as defined by (6.3), is univalent in ∆. If f (z) ∈ A, 0 6= Ia zf (z) ∈
H[1, 1] ∩ Q then
α + βq(z) + γq 2 (z) +
implies
µ
q(z) ≺
δzq 0 (z)
≺ χ(z)
q(z)
Iaσ+1 f (z)
z
¶µ
and q is the best subordinant.
By combining Theorem 6.2 and Theorem 6.3 we obtain the sandwich result,
however the details are omitted.
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Received on October 12, 2006; revised on June 8, 2009; accepted on June 13, 2009
Department of Mathematics
Anna University
Chennai - 600 025 India
E-mail address: shan@annauni.edu
Department of Mathematics
Anna University, M I T campus
Chennai - 600 044 India
Department of Mathematics
Easwari Engineering College
Ramapuram, Chennai - 600089 India
E-mail address: jeyaraman mp@yahoo.co.in