Paper29-Acta26-2011. 184KB Jun 04 2011 12:03:17 AM
Acta Universitatis Apulensis
ISSN: 1582-5329
No. 26/2011
pp. 267-288
SANDWICH THEOREMS FOR ANALYTIC FUNCTIONS
INVOLVING WRIGHT’S GENERALIZED HYPERGEOMETRIC
FUNCTION
M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian
Abstract. In this paper, we obtain some applications of first order differential
subordination and superordination results involving Wright’s generalized hypergeometric function defined by certain p-valent analytic functions.
2000 Mathematics Subject Classification: 30C45.
1.Introduction
Let A(p) denote the class of functions of the form:
p
f (z) = z +
∞
X
an z n (p ∈ N = {1, 2, ...}), 1.1
(1)
n=p+1
which are analytic and p-valent in the open unit disk U = {z : z ∈ C, |z| < 1} .
Let H(U ) be the class of analytic functions in U and let H[a, p] be the subclass of
H(U ) consisting of functions of the form:
f (z) = a + ap z p + ap+1 z p+1 ... (a ∈ C).
For simplicity, let H[a] = H[a, 1]. Also, let A(1) = A be the subclass of H(U )
consisting of functions of the form:
f (z) = z + a2 z 2 + ... .1.2
(2)
If f , g ∈ H(U ), we say that f is subordinate to g, written f (z) ≺ g(z) if there
exists a Schwarz function w(z), which (by definition) is analytic in U with w(0) = 0
and |w(z)| < 1 for all z ∈ U, such that f (z) = g(w(z)), z ∈ U. Furthermore, if the
function g(z) is univalent in U, then we have the following equivalence, (cf., e.g.,[6],
[19]; see also [20]):
f (z) ≺ g(z) ⇔ f (0) = g(0)andf (U ) ⊂ g(U ).
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
Let p, h ∈ H(U ) and let ϕ(r, s, t; z) : C 3 × U → C. If p and ϕ(p(z), zp′ (z),
are univalent and if p satisfies the second order superordination
z 2 p′′ (z); z)
h(z) ≺ ϕ(p(z), zp′ (z), z 2 p′′ (z); z), 1.3
(3)
then p is a solution of the differential superordination (1.3). Note that if f is subordinate to g, then g is superordinate to f. An analytic function q is called a subordinant
if q(z) ≺ p(z) for all p satisfying (1.3). A univalent subordinant qe that satisfies q ≺ qe
for all subordinants of (1.3) is called the best subordinant. Recently Miller and Mocanu [21] obtained conditions on the functions h, q and ϕ for which the following
implication holds:
h(z) ≺ ϕ(p(z), zp′ (z), z 2 p′′ (z); z) ⇒ q(z) ≺ p(z).1.4
(4)
Using the results of Miller and Mocanu [21], Bulboača [4] considered certain classes
of first order differential superordinations as well as superordination-preserving integral operators [5] many researshers ([1] and [29]) have obtained sufficient conditions
on normalized analytic functions f (z) by means of differential subordinations and
superordinations.
∞
∞
P
P
For analytic functions f (z) =
an z n and g (z) =
bn z n , the Hadamard
n=0
n=0
product (orconvolution) of f (z) and g (z), is defined by
(f ∗ g) (z) =
∞
X
an bn z n .1.5
(5)
n=0
Let α1 , A1 , ..., αq , Aq and β1 , B1 , ..., βs , Bs (q, s ∈ N) be positive and real parameters such that
1+
s
X
Bj −
j=1
q
X
Aj ≥ 0.
j=1
The Wright generalized hypergeometric function [31] (seealso [12] )
h
i
q Ψs [(α1 , A1 ) , ..., (αq , Aq ) ; (β1 , B1 ) , ..., (βs , Bs ) ; z] =q Ψs (αi , Ai )1,q ; (βi , Bi )1,s ; z
is defined by
q Ψs
h
(αi , Ai )1,q ; (βi , Bi )1,s ; z
i
q
Γ (αi
∞
=n=0 i=1
s
Q
+ nAi ) z n
.
n!
Γ (βi + nBi )
i=1
268
(z ∈ U ) .
M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
If Ai = 1(i = 1, ..., q) and Bi = 1 (i = 1, ..., s) , we have the relationship:
h
i
Ωq Ψs (αi , 1)1,q ; (βi , 1)1,s ; z = q Fs (α1 , ..., αq ; β1 , ..., βs ; z) ,
where q Fs (α1 , ..., αq ; β1 , ..., βs ; z) is the generalized hypergeometric function ( see for
details [9] , [10] , [11] , [16] ) and
Ω=
s
Q
i=1
q
Q
Γ (βi )
.1.6
(6)
Γ (αi )
i=1
The Wright generalized hypergeometric functions were invoked in the geometric
function theory ( see [8] , [9] , [10] , [11] , [25] , [26] and [27] ).
By using the generalized hypergeometric function Dziok and Srivastava [10] introduced a linear operator. In [8] Dziok and Raina and in [2] Aouf and Dziok
extended the linear operator by using
function.
h Wright generalized hypergeometric
i
p
First we define a function q Φs (αi , Ai )1,q ; (βi , Bi )1,s ; z by
p
q Φs
h
i
h
i
(αi , Ai )1,q ; (βi , Bi )1,s ; z = Ωz p q Ψs (αi , Ai )1,q ; (βi , Bi )1,s ; z
and consider the following linear operator
i
h
θp,q,s (αi , Ai )1,q ; (βi , Bi )1,s : A(p) → A(p), 1.7
(7)
defined by the convolution
h
i
h
i
θp,q,s (αi , Ai )1,q ; (βi , Bi )1,s f (z) =q Φps (αi , Ai )1,q ; (βi , Bi )1,s ; z ∗ f (z) .
We observe that, for a function f (z) of the form (1.1) , we have
i
h
p
θp,q,s (αi , Ai )1,q ; (βi , Bi )1,s f (z) = z +
∞
X
Ωσn,p (α1 ) an z n , 1.8
(8)
n=p+1
where Ω is given by (1.6) and σn,p (α1 ) is defined by
σn,p (α1 ) =
Γ (α1 + A1 (n − p)) ...Γ (αq + Aq (n − p))
.1.9
Γ (β1 + B1 (n − p)) ...Γ (βs + Bs (n − p)) (n − p)!
(9)
If, for convenience, we write
θp,q,s [α1 , A1 , B1 ] f (z) = θp,q,s [(α1 , A1 ) , ..., (αq , Aq ) ; (β1 , B1 ) , ..., (βs , Bs )] f (z) ,
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
then one can easily verify from (1.8) that
zA1 (θp,q,s [α1 , A1 , B1 ] f (z))′ = α1 θp,q,s [α1 + 1, A1 , B1 ] f (z)
−(α1 − pA1 )θp,q,s [α1 , A1 , B1 ] f (z) (A1 > 0).1.10
(10)
We note that for Ai = 1 (i = 1, 2, ..., q) and Bi = 1 (i = 1, 2, ..., s) , we obtain
θp,q,s [α1 , 1, 1] f (z) = Hp,q,s [α1 ]f (z) , which was introduced and studied by Dziok
and Srivastava [10] . Also for f (z) ∈ A, the linear operator θ1,q,s [α1 , A1 , B1 ] = θ [α1 ]
was introduced by Dziok and Raina [8] and studied by Aouf and Dziok [2].
We note that, for f (z) ∈ A(p), Ai = 1 (i = 1, ..., q) , Bi = 1 (i = 1, ..., s) , q = 2
and s = 1, we have:
(i) θp,2,1 [a, 1; c] f (z) = Lp (a, c) f (z) (a > 0, c > 0, p ∈ N) (seeSaitoh [28]);
(ii) θp,2,1 [µ + p, p; p] f (z) = Dµ+p−1 f (z) (µ > −p, p ∈ N) , where Dµ+p−1 f (z)
is the (µ + p − 1) −the order Ruscheweyh derivative of a function f (z) ∈ A(p)
(seeKumarandShukla [13] and [14]);
(µ,p)
(µ,p)
(iii) θp,2,1 [1 + p, 1; 1 + p − µ] f (z) = Ωz f (z) , where the operator Ωz
is
defined by ( see Srivastava and Aouf [30] )
f (z) =
Ω(µ,p)
z
Γ (1 + p − µ) µ µ
z Dz f (z) (0 ≤ µ < 1; p ∈ N) ,
Γ (1 + p)
where Dzµ is the fractional derivative operator (see, f ordetails, [23] and [24] ) ;
(iv) θp,2,1 [ν + p, 1; ν + p + 1] f (z) = Jν,p (f ) (z), where Jν,p (f ) (z) is the generalized Bernardi-Libera-Livingston-integral operator (see [3] , [15] and [18] ), defined
by
Z
ν + p z ν−1
Jν,p (f ) (z) =
t f (t) dt (ν > −p; p ∈ N) ;
zν
0
(v) θp,2,1 [p + 1, 1; n + p] f (z) = In,p f (z) (n ∈ Z, n > −p, p ∈ N) ,where the operator In,p was considered by Liu and Noor [17] ;
(vi) θp,2,1 [λ + p, c; a] f (z) = Ipλ (a, c) f (z) a, c ∈ R\Z̄0 , λ > −p, p ∈ N , where
Ipλ (a, c) is the Cho-Kwon-Srivastava operator [7] .
The main object of the present paper is to find sufficient condition for certain
normalized analytic functions f (z) in U such that θp,q,s [α1 , A1 , B1 ](f ∗Ψ) (z) 6= 0 and
f (z) to satisfy
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
q1 (z) ≺
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
≺ q2 (z),
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
where q1 , q2 are given univalent functions in U and Φ(z) = z p +
∞
P
λn z n , Ψ(z) =
n=p+1
zp +
∞
P
µn z n are analytic functions in U with λn ≥ 0, µn ≥ 0. Also, we obtain
n=p+1
number of known results as special cases.
2. Definitions and Preliminaries
In order to prove our results, we shall make use of the following definition and
lemmas.
Definition 1 [21] . Denote by Q, the set of all functions f that are analytic and
injective on U \ E(f ), where
E(f ) = {ξ ∈ ∂U : lim f (z) = ∞},
z→ξ
and are such that
f ′ (ξ)
6= 0 for ξ ∈ ∂U \ E(f ).
Lemma 1 [20] . Let q be univalent in the unit disk U and θ and ϕ be analytic in a
domain D containing q(U ) with ϕ(w) 6= 0 when w ∈ q(U ). Set
ψ(z) = zq ′ (z)ϕ(q(z)) and h(z) = θ(q(z)) + ψ(z).it2.1
(11)
Suppose that
(i) ψ(z)nis starlike
univalent in U,
o
zh′ (z)
(ii) Re ψ(z) > 0 for z ∈ U.
If p is analytic with p(0) = q(0), p(U ) ⊆ D and
θ(p(z)) + zp′ (z)ϕ(p(z)) ≺ θ(q(z)) + zq ′ (z)ϕ(q(z)), it2.2
(12)
then p(z) ≺ q(z) and q is the best dominant.
Lemma 2 [4] . Let q be convex univalent in U and ϑ and φ be analytic in a domain
D containing q(U ). Suppose that
(i) Re{ϑ′ (q(z))/φ(q(z))} > 0 for z ∈ U and
(ii) ψ(z) = zq ′ (z)φ(q(z)) is starlike univalent in U.
If p(z) ∈ H[q(0), 1] ∩ Q, with p(U ) ⊆ D, and ϑ(p(z)) + zp′ (z)φ(p(z)) is univalent
in U and
ϑ(q(z)) + zq ′ (z)φ(q(z)) ≺ ϑ(p(z)) + zp′ (z)φ(p(z)), it2.3
(13)
then q(z) ≺ p(z) and q is the best subordinant.
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
3. Subordination results for analytic functions
Unless otherwise mentioned, we shall assume in the reminder of this paper that,
α1 , A1 , ..., αq , Aq and β1 , B1 , ..., βs , Bs (q, s ∈ N) be positive real parameters, Φ, Ψ ∈
A(p), p ∈ N, ξ1 , ξ2 , ξ3 ∈ C and ξ4 ∈ C∗ = C\{0}.
Theorem 1. Let q be convex univalent in U with q(0) = 1. Further, assume that
zq ′′ (z)
ξ3 2ξ2
Re
+
q(z) + 1 + ′
> 0 (z ∈ U ).it3.1
(14)
ξ4
ξ4
q (z)
If f ∈ A(p) satisfies
̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.2
(15)
where
̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
ξ1 + ξ2
2
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
θp,q,s [α1 +1,A1 ,B1 ](f ∗Ψ)(z)
+ ξ3
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
θp,q,s [α1 +1,A1 ,B1 ](f ∗Ψ)(z)
θ
[α1 +1,A1 ,B1 ](f ∗Φ)(z)
θp,q,s [α1 +2,A1 ,B1 ](f ∗Ψ)(z)
+ξ4 α1 p,q,s
−
(α
+
1)
+
1
1
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
θp,q,s [α1 +1,A1 ,B1 ](f ∗Ψ)(z)
× θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z) ,
θp,q,s [α1 +1,A1 ,B1 ](f ∗Ψ)(z)
3.3
(16)
then
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
≺ q(z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
and q is the best dominant.
Proof. Define the function p(z) by
p(z) =
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
(z ∈ U ).3.4
(17)
Then the function p(z) is analytic in U and p(0) = 1, Therefore, by making use of
(3.4), we obtain
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
ξ1 + ξ2
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
2
+ ξ3
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 2, A1 , B1 ](f ∗ Ψ) (z)
+ξ4 α1
− (α1 + 1)
+1
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
×
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
= ξ1 + ξ2 p2 (z) + ξ3 p(z) + ξ4 zp′ (z).3.5
(18)
By using (3.2) in (3.5), we have
ξ1 + ξ2 p2 (z) + ξ3 p(z) + ξ4 zp′ (z) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z).
By setting
θ(w) = ξ1 + ξ2 w2 + ξ3 w
and
ϕ(w) = ξ4 ,
it can be easily observed that θ(w) is analytic in C, ϕ(w) is analytic in C∗ and that
ϕ(w) 6= 0, w ∈ C∗ , then by using Lemma 1 we complete the proof of Theorem 1.
zp
Putting Φ(z) = Ψ(z) = 1−z
(or, µn = λn = 1, n ≥ p + 1, p ∈ N) in Theorem 1,
we obtain the following corollary:
Corollary 1. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true. If
f ∈ A(p) and satisfies
Z(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.6
(19)
where
Z(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
ξ1 + ξ2
2
θp,q,s [α1 ,A1 ,B1 ]f (z)
θp,q,s [α1 +1,A1 ,B1 ]f (z)
+ ξ3
θp,q,s [α1 ,A1 ,B1 ]f (z)
θp,q,s [α1 +1,A1 ,B1 ]f (z)
θ
[α1 +1,A1 ,B1 ]f (z)
θp,q,s [α1 +2,A1 ,B1 ]f (z)
+ξ4 α1 p,q,s
−
(α
+
1)
+
1
1
θp,q,s [α1 ,A1 ,B1 ]f (z)
θp,q,s [α1 +1,A1 ,B1 ]f (z)
× θp,q,s [α1 ,A1 ,B1 ]f (z) ,
θp,q,s [α1 +1,A1 ,B1 ]f (z)
3.7
(20)
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
then
θp,q,s [α1 , A1 , B1 ]f (z)
≺ q(z)
θp,q,s [α1 + 1, A1 , B1 ]f (z)
and q is the best dominant.
Remark 1. Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s) and p = 1, in
Corollary 1,we obtain the result obtained by Murugusundaramoorthy and Magesh
[22, Corollary 2.6 ].
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2 , s = 1 and α1 =
α2 = β1 = 1 in Theorem 1, hence we obtain the next result:
Corollary 2. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true. If
f ∈ A(p) and satisfies
B(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.8
(21)
where
B(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
ξ1 + ξ2
+ξ4
(f ∗Φ)(z)
z(f ∗Ψ)′ (z)+(1−p)(f ∗Ψ)(z)
z(f ∗Φ)′ (z)
(f ∗Φ)(z)
−
2
+ ξ3
(f ∗Φ)(z)
z(f ∗Ψ)′ (z)+(1−p)(f ∗Ψ)(z)
z 2 (f ∗Ψ)′′ (z)+z(3−p)(f ∗Ψ)′ (z)+(1−p)(3−p)(f ∗Ψ)′ (z)
z(f ∗Ψ)′ (z)+(1−p)(f ∗Ψ)(z)
(f ∗Φ)(z)
×
′
z(f ∗Ψ) (z)+(1−p)(f ∗Ψ)(z) ,
+ (2 − p)
3.9
(22)
then
z(f ∗
(f ∗ Φ)(z)
≺ q(z)
+ (1 − p)(f ∗ Ψ)(z)
Ψ)′ (z)
and q is the best dominant.
Remark 2. (i) Putting p = 1 in Corollary 2, we obtain the result obtained by
Murugusundaramoorthy and Magesh [22, Corollary 2.7 ];
(ii) Putting p = 1 and Φ(z) = Ψ(z) in Corollary 2, we obtain the result obtained
by Murugusundaramoorthy and Magesh [22, Corollary 2.8 ];
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
z
(or, p = 1, µn = λn = 1 and n ≥ 2) in
(iii) Putting p = 1 and Φ(z) = Ψ(z) = 1−z
Corollary 2, we obtain the result obtained by Murugusundaramoorthy and Magesh
[22, Corollary 2.9 ].
Taking q(z) = 1 + Az1 + Bz (−1 ≤ A < B ≤ 1) in Theorem 1, we obtain the next
result:
Corollary 3. Let q be convex univalent in U with q(0) = 1 and
ξ3 2ξ2
1 − Bz
Re
+
(1 + Az1 + Bz) +
> 0 (z ∈ U ).
ξ4
ξ4
1 + Bz
If f ∈ A(p) , −1 ≤ A < B ≤ 1 and
̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 +ξ2 (1 + Az1 + Bz)2 (z)+ξ3 (1 + Az1 + Bz)+ξ4
z (A − B)
3.10
(1 + Bz)2
(23)
where ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined by (3.3), then
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
≺ 1 + Az1 + Bz3.11
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
(24)
and 1 + Az1 + Bz is the best dominant.
Remark 3. Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s) and p = 1, in
Corollary 3,we obtain the result obtained by Murugusundaramoorthy and Magesh [
22, Corollary 2.11 ].
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
a, α2 = 1 and β1 = c (a, c > 0, p ∈ N) in Theorem 1, we obtain the following
corollary:
Corollary 4. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true.
If f ∈ A(p) satisfies
K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.12
where
K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
2
Lp (a,c)(f ∗Φ)(z)
Lp (a,c)(f ∗Φ)(z)
+
ξ
ξ
+
ξ
3
1
2
Lp (a+1,c)(f ∗Ψ)(z)
Lp (a+1,c)(f ∗Ψ)(z)
L (a+2,c)(f ∗Ψ)(z)
L (a+1,c)(f ∗Φ)(z)
=
+ξ4 a Lp p (a,c)(f ∗Φ)(z) − (a + 1) Lpp (a+1,c)(f ∗Ψ)(z) + 1
× Lp (a,c)(f ∗Φ)(z) ,
3.13
Lp (a+1,c)(f ∗Ψ)(z)
(26)
then
Lp (a, c)(f ∗ Φ) (z)
≺ q(z)
Lp (a + 1, c)(f ∗ Ψ) (z)
and q is the best dominant of (3.12).
Remark 4. Putting p = 1 in Corollary 4, we obtain the result obtained by Murugusundaramoorthy and Magesh [22, Corollary 2.5 ].
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
µ+p (µ > −p, p ∈ N) , α2 = β1 = p in Theorem 1, we obtain the following corollary:
Corollary 5. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true.
If f ∈ A(p) satisfies
M (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.14
(27)
where
M (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
ξ1 + ξ2
Dµ+p−1 (f ∗Φ)(z)
Dµ+p (f ∗Ψ)(z)
2
+ ξ3
Dµ+p−1 (f ∗Φ)(z)
Dµ+p (f ∗Ψ)(z)
µ+p
Dµ+p+1 (f ∗Ψ)(z)
+ξ4 (µ + p) DDµ+p−1(f(f∗Φ)(z)
+
1
−
(µ
+
p
+
1)
Dµ+p (f ∗Ψ)(z)
∗Φ)(z)
× Dµ+p−1 (f ∗Φ)(z) ,
µ+p
D
(f ∗Ψ)(z)
3.15
(28)
then
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
Dµ+p−1 (f ∗ Φ) (z)
≺ q(z)
Dµ+p (f ∗ Ψ) (z)
and q is the best dominant of (3.14).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
1 + p, α2 = 1 and β1 = 1 + p − µ (0 ≤ µ < 1, p ∈ N) , in Theorem 1, we obtain the
following corollary:
Corollary 6. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true. If
f ∈ A(p) satisfies
N (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.16
(29)
where
N (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
2
(µ,p)
(µ,p)
(f ∗Φ)(z)
Ωz
Ωz
(f ∗Φ)(z)
+
ξ
ξ
+
ξ
3
1
2
(µ+1,p)
(µ+1,p)
Ωz
(f ∗Ψ)(z)
Ωz
(f ∗Ψ)(z)
(µ+2,p)
(µ+1,p)
Ωz
(f ∗Ψ)(z)
Ωz
(f ∗Φ)(z)
(1
+
p)
−
(p
+
2)
+
1
+ξ
4
(µ,p)
(µ+1,p)
Ωz
Ωz
(f ∗Ψ)(z)
(f ∗Φ)(z)
(µ,p)
(f ∗Φ)(z)
z
× Ω(µ+1,p)
,
Ωz
3.17
(f ∗Ψ)(z)
(30)
then
(µ,p)
Ωz
(f ∗ Φ) (z)
(µ+1,p)
Ωz
(f
∗ Ψ) (z)
≺ q(z)
and q is the best dominant of (3.16).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
ν + p (ν > −p, p ∈ N), α2 = 1 and β1 = ν + p + 1 in Theorem 1, we obtain the
following corollary:
Corollary 7. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true.
If f ∈ A(p) satisfies
L(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.18
277
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where
L(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
ξ1 + ξ2
2
Jν,p (f ∗Φ)(z)
Jν+1,p (f ∗Ψ)(z)
+ ξ3
Jν,p (f ∗Φ)(z)
Jν+1,p (f ∗Ψ)(z)
Jν+2,p (f ∗Ψ)(z)
J
(f ∗Φ)(z)
+ξ4 (ν + p) ν+1,p
−
(ν
+
p
+
1)
+
1
Jν,p (f ∗Φ)(z)
Jν+1,p (f ∗Ψ)(z)
× Jν,p (f ∗Φ)(z) ,
Jν+1,p (f ∗Ψ)(z)
3.19
(32)
then
Jν,p (f ∗ Φ) (z)
≺ q(z)
Jν+1,p (f ∗ Ψ) (z)
and q is the best dominant of (3.18).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
p + 1, α2 = 1 and β1 = n + p (n ∈ Z, n > −p, p ∈ N) in Theorem 1, we obtain the
following corollary:
Corollary 8. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true.
If f ∈ A(p) satisfies
R(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.20
(33)
where
R(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
2
In,p (f ∗Φ)(z)
In,p (f ∗Φ)(z)
ξ
+
ξ
+
ξ
1
2 In+1,p (f ∗Ψ)(z)
3 In+1,p (f ∗Ψ)(z)
I
(f ∗Φ)(z)
In+2,p (f ∗Ψ)(z)
=
+ξ4 (p + 1) n+1,p
−
(p
+
2)
+
1
In,p (f ∗Φ)(z)
In+1,p (f ∗Ψ)(z)
× In,p (f ∗Φ)(z) ,
3.21
In+1,p (f ∗Ψ)(z)
then
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In,p (f ∗ Φ) (z)
≺ q(z)
In+1,p (f ∗ Ψ) (z)
and q is the best dominant of (3.20).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ...,
s), q = 2, s = 1, α1 =
λ + p , α2 = c and β1 = a a, c ∈ R\Z̄0 , λ > −p, p ∈ N , in Theorem 1, we obtain
the following corollary:
Corollary 9. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true.
If f ∈ A(p) satisfies
G(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.22
(35)
where
G(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
ξ1 + ξ2
Ipλ (a,c)(f ∗Φ)(z)
Ipλ+1 (a,c)(f ∗Ψ)(z)
2
+ ξ3
Ipλ (a,c)(f ∗Φ)(z)
Ipλ+1 (a,c)(f ∗Ψ)(z)
I λ+2 (a,c)(f ∗Ψ)(z)
I λ+1 (a,c)(f ∗Φ)(z)
+1
+ξ4 (λ + p) pI λ (a,c)(f ∗Φ)(z) − (λ + p + 1) pλ+1
×
Ip
p
Ipλ (a,c)(f ∗Φ)(z)
Ipλ+1 (a,c)(f ∗Ψ)(z)
3.23
(a,c)(f ∗Ψ)(z)
,
(36)
then
Ipλ (a, c) (f ∗ Φ) (z)
Ipλ+1 (a, c) (f ∗ Ψ) (z)
≺ q(z)
and q is the best dominant of (3.22).
4. Superordination results for analytic functions
Now, by applying Lemma 2, we prove the following theorem.
Theorem 2. Let q be convex univalent in U with q(0) = 1 and
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
f ∈ A(p), θp,q,s
[α1 +1,A1 ,B1 ](f ∗Ψ)(z) ∈ H[q(0), 1] ∩ Q.Further, assume that
ξ3 2ξ2
+
q(z) > 04.1
Re
ξ4
ξ4
279
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and ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) be univalent in U and satisfy
ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z) ≺ ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ), 4.2
(38)
where ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is given by (3.3), then
q(z) ≺
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
and q is the best subordinant of (4.2).
Proof. Define the function p(z) by (3.4), with simple computation from (3.4), we
get
̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) = ξ1 + ξ2 p2 (z) + ξ3 p(z) + ξ4 zp′ (z),
then
ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z) ≺ ξ1 + ξ2 p2 (z) + ξ3 p(z) + ξ4 zp′ (z).4.3
(39)
By setting
υ(w) = ξ1 + ξ2 w2 + ξ3 w
and
ϕ(w) = ξ4 ,
it is easy observed that υ(w) is analytic in C, ϕ(w) is analytic in C∗ and ϕ(w) 6=
0, w ∈ C∗ . Since q is convex and univalent function, it follows that
ξ3 2ξ2
Re θ′ (q(z))ϕ(q(z)) = Re
+
q(z) > 0 (z ∈ U )
ξ4
ξ4
and then, by using Lemma 2 we complete the proof of Theorem 2.
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
a, α2 = 1 and β1 = c (a, c > 0) in Theorem 2, we obtain the following corollary:
Corollary 10. Let q be convex univalent in U with q(0) = 1, (4.1) holds true and
Lp (a,c)(f ∗Φ)(z)
f ∈ A(p), Lp (a+1,c)(f
∗Ψ)(z) ∈ H[q(0), 1]∩Q.Further, assume that K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) be
univalent in U which satisfying
ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z) ≺ K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ), 4.4
where K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is given by (3.13), then
280
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q(z) ≺
Lp (a, c)(f ∗ Φ) (z)
Lp (a + 1, c)(f ∗ Ψ) (z)
and q is the best subordinant of (4.4).
Remark 5. Putting p = 1 in Corollary 10, we obtain the result obtained by Murugusundaramoorthy and Magesh [22, Corollary 2.13 ].
Remark 6. (i) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), p = 1, q = 2
, s = 1, α1 = α2 = 1 and β1 = 1 in Theorem 2, we obtain the result obtained by
Murugusundaramoorthy and Magesh [22, Corollary 2.14 ];
(ii) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2 , s = p = α1 =
α2 = β1 = 1 and Φ(z) = Ψ(z) in Theorem 2, we obtain the result obtained by
Murugusundaramoorthy and Magesh [22, Corollary 2.15];
(iii) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2 , s = p = α1 =
z
in Theorem 2, we obtain the result obtained by
α2 = β1 = 1 and Φ(z) = Ψ(z) = 1−z
Murugusundaramoorthy and Magesh [22, Corollary 2.16].
Taking q(z) = 1 + Az1 + Bz (−1 ≤ A < B ≤ 1) in Theorem 2, we obtain the next
result:
Corollary 11. Let q be convex univalent in U with q(0) = 1, f ∈ A(p),
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
θp,q,s [α1 +1,A1 ,B1 ](f ∗Ψ)(z)
∈ H[q(0), 1] ∩ Q and
ξ3 2ξ2
Re
+
1 + Az1 + Bz > 0 (−1 ≤ A < B ≤ 1) .
ξ4
ξ4
Further, assume that ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) be univalent in U and satisfy
z (A − B)
≺ ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ), 4.5
(1 + Bz)2
(41)
where ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is given by (3.3), then
ξ1 +ξ2 (1 + Az1 + Bz)2 (z)+ξ3 (1 + Az1 + Bz)+ξ4
1 + Az1 + Bz ≺
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
and 1 + Az1 + Bz is the best subordinant (4.5).
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Remark 7. Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s) and p = 1, in
Corollary 11,we obtain the result obtained by Murugusundaramoorthy and Magesh
[22, Corollary 2.17 ].
Remark 8. (i) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q =
2, s = 1, α1 = µ + p (µ > −p, p ∈ N) , α2 = β1 = p in Theorem 2, we obtain the
superordination result for the operator Dµ+p−1 ;
(ii) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
1 + p, α2 = 1 and β1 = 1 + p − µ (0 ≤ µ < 1, p ∈ N) , in Theorem 2, we obtain the
(µ,p)
superordination result for the operator Ωz ;
(iii) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
ν + p (ν > −p, p ∈ N), α2 = 1 and β1 = ν + p + 1 in Theorem 2, we obtain the
superordination result for the operator Jν,p ;
(iv) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
p + 1, α2 = 1 and β1 = n + p (n ∈ Z, n > −p, p ∈ N) in Theorem 2, we obtain the
superordination result for the operator In,p ;
(v) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ...,
s), q = 2, s = 1, α1 =
λ + p , α2 = c and β1 = a a, c ∈ R\Z̄0 , λ > −p, p ∈ N , in Theorem 2, we obtain
the superordination result for the operator Ipλ (a, c) .
4. Sandwich results
Combining Theorem 1 with Theorem 2, we get the following sandwich theorem.
Theorem 3. Let q1 , q2 be convex univalent in U. Suppose q1 and
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
q2 satisfies (4.1) and (3.1), respectively. If f ∈ A(p), θp,q,s
[α1 +1,A1 ,B1 ](f ∗Ψ)(z) ∈
H[q(0), 1] ∩ Q and ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, which satisfying
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.1
(42)
where ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is given by (3.3), then
q1 (z) ≺
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
≺ q2 (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.1).
Taking q1 (z) = 1 + A1 z1 + B1 z (−1 ≤ A1 < B1 ≤ 1) and q2 (z) = 1 + A2 z1 + B2 z (−1 ≤ A2 < B2 ≤ 1)
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in Theorem 3, we obtain the following corollary:
Corollary 12. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
(4.1) and (3.1), respectively. If f ∈ A(p), θp,q,s
[α1 +1,A1 ,B1 ](f ∗Ψ)(z) ∈ H[q(0), 1] ∩
Q and ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, which satisfying
ξ1 +ξ2 (1 + A1 z1 + B1 z)2 (z)+ξ3 (1 + A1 z1 + B1 z)+ξ4
(A1 − B1 ) z
≺ ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
(1 + B1 z)2
≺ ξ1 + ξ2 (1 + A2 z1 + B2 z)2 + ξ3 (1 + A2 z1 + B2 z) + ξ4
(A2 − B2 ) z
, 5.2
(1 + B2 z)2
(43)
where ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined in (3.3), then
1 + A1 z1 + B1 z ≺
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
≺ 1 + A2 z1 + B2 z
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
and 1 + A1 z1 + B1 z, 1 + A2 z1 + B2 z are, respectively, the best subordinant and
dominant of (5.2).
Remark 9. Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2 and
s = p = α1 = α2 = β1 = 1 in Corollary 12, we obtain the result obtained by
Murugusundaramoorthy and Magesh [22, Corollary 3.2 ].
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
a, α2 = 1 and β1 = c (a, c > 0) in Theorem 3, we obtain the following corollary:
Corollary 13. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
Lp (a,c)(f ∗Φ)(z)
(4.1) and (3.1), respectively. If f ∈ A(p), Lp (a+1,c)(f
∗Ψ)(z) ∈ H[q(0), 1] ∩ Q and
K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U ,where K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined in
(3.13), then
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.3
implies
q1 (z) ≺
Lp (a, c)(f ∗ Φ) (z)
≺ q2 (z)
Lp (a + 1, c)(f ∗ Ψ) (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.3).
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Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
µ + p (µ > −p, p ∈ N) and α2 = β1 = p in Theorem 3, we obtain the following
corollary:
Corollary 14. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
µ+p−1 ∗Φ)(z)
(4.1) and (3.1), respectively. If f ∈ A(p), DDµ+p (f(f∗Ψ)(z)
∈ H[q(0), 1] ∩ Q and
M (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, where M (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined
in (3.15), then
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ M (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.4
(45)
implies
q1 (z) ≺
Dµ+p−1 (f ∗ Φ) (z)
≺ q2 (z)
Dµ+p (f ∗ Ψ) (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.4).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
1 + p, α2 = 1 and β1 = 1 + p − µ (0 ≤ µ < 1, p ∈ N) in Theorem 3, we obtain the
following corollary:
Corollary 15. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
(4.1) and (3.1), respectively. If f ∈ A(p),
(µ,p)
Ωz
(f ∗Φ)(z)
(µ+1,p)
Ωz
(f ∗Ψ)(z)
∈ H[q(0), 1] ∩ Q and
N (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, where N (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined in
(3.17), then
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ N (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.5
(46)
implies
(µ,p)
q1 (z) ≺
Ωz
(f ∗ Φ) (z)
(µ+1,p)
Ωz
(f
∗ Ψ) (z)
≺ q2 (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.5).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
ν + p (ν > −p, p ∈ N), α2 = 1 and β1 = ν + p + 1 in Theorem 3, we obtain the
following corollary:
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Corollary 16. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
Jν,p (f ∗Φ)(z)
(4.1) and (3.1), respectively. If f ∈ A(p), Jν+1,p
(f ∗Ψ)(z) ∈ H[q(0), 1] ∩ Q and
L(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, where L(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined in
(3.19), then
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ L(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.6
(47)
implies
q1 (z) ≺
Jν,p (f ∗ Φ) (z)
≺ q2 (z)
Jν+1,p (f ∗ Ψ) (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.6).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
p + 1, α2 = 1 and β1 = n + p (n > −p, p ∈ N) in Theorem 3, we obtain the following
corollary:
Corollary 17. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
In,p (f ∗Φ)(z)
(4.1) and (3.1), respectively. If f ∈ A(p), In+1,p
(f ∗Ψ)(z) ∈ H[q(0), 1] ∩ Q and
R(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, where R(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined in
(3.21), then
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ R(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.7
(48)
implies
q1 (z) ≺
In,p (f ∗ Φ) (z)
≺ q2 (z)
In+1,p (f ∗ Ψ) (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.7).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2,
s = 1, α1 =
λ + p (λ > −p, p ∈ N), α2 = c and β1 = a a, c ∈ R\Z̄0 , λ > −p in Theorem 3,
we obtain the following corollary:
Corollary 18. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
(4.1) and (3.1), respectively. If f ∈ A(p),
Ipλ (a,c)(f ∗Φ)(z)
λ+1
Ip (a,c)(f ∗Ψ)(z)
285
∈ H[q(0), 1] ∩ Q and
M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
G(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, where G(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined in
(3.23), then
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ G(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.8
(49)
implies
q1 (z) ≺
Ipλ (a, c) (f ∗ Φ) (z)
Ipλ+1 (a, c) (f ∗ Ψ) (z)
≺ q2 (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.8).
References
[1] R. M. Ali, V. Ravichandran , M. H. Khan and K. G. Subramanian,
Differential sandwich theorems for certain analytic functions, Far East
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[2] M.K. Aouf and J. Dziok, Certain class of analytic functions associated
with the Wright generalized hypergeometric function, J. Math. Appl., 30
(2008), 23-32.
[3] S.D. Bernardi, Convex and univalent functions, Trans. Amer. Math.
Soc., 135 (1996), 429-446.
[4] T. Bulboacă, Classes of first order differential superordinations, Demonstratio Math. 35 (2002), no. 2, 287-292.
[5] T. Bulboacă, A class of superordination-preserving integral operators,
Indag. Math. (N. S.), 13 (2002), no. 3, 301-311.
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properties for certain subclasses of multivalent functions associated with
a family of linear operators, J. Math. Anal. Appl., 292 (2004), 470-483.
[8] J. Dziok and R.K. Raina, Families of analytic functions associated
with the Wright generalized hypergeometric function, Demonstratio Math.,
37 (2004), no. 3, 533-542.
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[9] J. Dziok, R.K. Raina and H.M. Srivastava, Some classes of analytic
functions associated with operators on Hilbert space involving Wright hypergeometric function, Proc. Jangieon Math. Soc.,7 (2004), 43-55.
[10] J. Dziok and H.M. Srivastava, Classes of analytic functions with the
generalized hypergeometric function, Appl. Math. Comput., 103 (1999),
1-13.
[11] J. Dziok and H.M. Srivastava, Certain subclasses of analytic functions associated with the Wright generalized hypergeometric function, Integral Transform. Spec. Funct., 14 (2003), 7-18.
[12] P.W. Karlsson and H.M. Srivastava, Multiple Gaussian Hypergeometric Series, Halsted Press (Ellis Horwood Ltd, Chichester ), John
Wiley and Sons, New York, Chichester, Brisbane and London, 1985.
[13] V. Kumar and S.L. Shukla, Multivalent functions defined by Ruscheweyh
derivatives, Indian J. Pure Appl. Math., 15 (1984), no. 11, 1216-1227.
[14] V. Kumar and S.L. Shukla, Multivalent functions defined by Ruscheweyh
derivatives II, Indian J. Pure Appl. Math., 15 (1984), no. 11, 1228-1238.
[15] R.J. Libera, Some classes of regular univalent functions, Proc. Amer.
Math. Soc., 16 (1969), 755-758.
[16] J.-L. Liu, Strongly starlike functions associated with the Dziok- Srivastava operator, Tamkang J. Math., 35 (2004), no. 1, 37-42.
[17] J.-L. Liu and K.I. Noor, Some properties of Noor integral operator,
J. Natur. Geom., 21 (2002), 81-90.
[18] A. E. Livingston, On the radius of univalence of certain analytic
functions, Proc. Amer. Math. Soc., 17 (1966), 352-357.
[19] S. S. Miller and P. T. Mocanu, Differential subordinations and univalent functions, Michigan Math. J., 28 (1981), no. 2, 157-171.
[20] S. S. Miller and P. T. Mocanu, Differential Subordination : Theory
and Applications, Series on Monographs and Textbooks in Pure and
Applied Mathematics, Vol. 225, Marcel Dekker Inc., New York and
Basel, 2000.
[21] S. S. Miller and P. T. Mocanu, Subordinates of differential superordinations, Complex Variables, 48 (2003), no. 10, 815-826.
[22] G. Murugusundaramoorthy and N. Magesh, Differential subordinations and superordinations for analytic functions defined by convolution
structure, Studia Univ. ”Babes-Bolyai”, Math., 56 (2009), no. 2, 83-96.
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[23] S. Owa, On the distortion theorems. I, Kyungpook Math. J., 18
(1978), 53-59.
[24] S. Owa and H.M. Srivastava, Univalent and starlike generalized hypergeometric functions, Canad. J. Math., 39 (1987), 1057-1077.
[25] R.K. Raina and T.S. Naher, A note on boundedness properties of
Wright’s generalized hypergeometric function, Ann. Math. Blaise Pascal,
4 (1997), 83-95.
[26] R.K. Raina and T.S. Naher, On characterization of certain Wright’s
generalized hypergeometric functions involving certain subclasses of analytic functions, Informatica, 10 (1999), 219-230.
[27] R.K. Raina and T.S. Naher, On univalent and starlike Wright’s
hypergeometric functions, Rend. Sem. Math. Univ. Pavoda, 95 (1996),
11-22.
[28] H. Saitoh, A linear operator and its applications of first order differential subordinations, Math. Japon., 44 (1996), 31-38.
[29] T. N. Shanmugam, V. Ravichandran and S. Sivasubramanian, Differential sandwich theorems for some subclasses of analytic functions, J.
Austr.Math. Anal. Appl., 3 (2006), no. 1, Art. 8, 1-11.
[30] H.M. Srivastava and M.K. Aouf, A certain fractional derivative operator and its application to a new class of analytic and multivalent functions with negative coefficients I and II, J. Math. Anal. Appl., 171
(1992), 1-13; ibid. 192 (1995), 673-688.
[31] E.M. Wright, The asymptotic expansion of the generalized hypergeometric function, Proc. London Math. Soc., 46 (1946), 389-408.
M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian
Department of Mathematics
Faculty of Science
Mansoura University
Mansoura 35516, Egypt.
email: [email protected], [email protected], [email protected],
samar [email protected]
288
ISSN: 1582-5329
No. 26/2011
pp. 267-288
SANDWICH THEOREMS FOR ANALYTIC FUNCTIONS
INVOLVING WRIGHT’S GENERALIZED HYPERGEOMETRIC
FUNCTION
M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian
Abstract. In this paper, we obtain some applications of first order differential
subordination and superordination results involving Wright’s generalized hypergeometric function defined by certain p-valent analytic functions.
2000 Mathematics Subject Classification: 30C45.
1.Introduction
Let A(p) denote the class of functions of the form:
p
f (z) = z +
∞
X
an z n (p ∈ N = {1, 2, ...}), 1.1
(1)
n=p+1
which are analytic and p-valent in the open unit disk U = {z : z ∈ C, |z| < 1} .
Let H(U ) be the class of analytic functions in U and let H[a, p] be the subclass of
H(U ) consisting of functions of the form:
f (z) = a + ap z p + ap+1 z p+1 ... (a ∈ C).
For simplicity, let H[a] = H[a, 1]. Also, let A(1) = A be the subclass of H(U )
consisting of functions of the form:
f (z) = z + a2 z 2 + ... .1.2
(2)
If f , g ∈ H(U ), we say that f is subordinate to g, written f (z) ≺ g(z) if there
exists a Schwarz function w(z), which (by definition) is analytic in U with w(0) = 0
and |w(z)| < 1 for all z ∈ U, such that f (z) = g(w(z)), z ∈ U. Furthermore, if the
function g(z) is univalent in U, then we have the following equivalence, (cf., e.g.,[6],
[19]; see also [20]):
f (z) ≺ g(z) ⇔ f (0) = g(0)andf (U ) ⊂ g(U ).
267
M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
Let p, h ∈ H(U ) and let ϕ(r, s, t; z) : C 3 × U → C. If p and ϕ(p(z), zp′ (z),
are univalent and if p satisfies the second order superordination
z 2 p′′ (z); z)
h(z) ≺ ϕ(p(z), zp′ (z), z 2 p′′ (z); z), 1.3
(3)
then p is a solution of the differential superordination (1.3). Note that if f is subordinate to g, then g is superordinate to f. An analytic function q is called a subordinant
if q(z) ≺ p(z) for all p satisfying (1.3). A univalent subordinant qe that satisfies q ≺ qe
for all subordinants of (1.3) is called the best subordinant. Recently Miller and Mocanu [21] obtained conditions on the functions h, q and ϕ for which the following
implication holds:
h(z) ≺ ϕ(p(z), zp′ (z), z 2 p′′ (z); z) ⇒ q(z) ≺ p(z).1.4
(4)
Using the results of Miller and Mocanu [21], Bulboača [4] considered certain classes
of first order differential superordinations as well as superordination-preserving integral operators [5] many researshers ([1] and [29]) have obtained sufficient conditions
on normalized analytic functions f (z) by means of differential subordinations and
superordinations.
∞
∞
P
P
For analytic functions f (z) =
an z n and g (z) =
bn z n , the Hadamard
n=0
n=0
product (orconvolution) of f (z) and g (z), is defined by
(f ∗ g) (z) =
∞
X
an bn z n .1.5
(5)
n=0
Let α1 , A1 , ..., αq , Aq and β1 , B1 , ..., βs , Bs (q, s ∈ N) be positive and real parameters such that
1+
s
X
Bj −
j=1
q
X
Aj ≥ 0.
j=1
The Wright generalized hypergeometric function [31] (seealso [12] )
h
i
q Ψs [(α1 , A1 ) , ..., (αq , Aq ) ; (β1 , B1 ) , ..., (βs , Bs ) ; z] =q Ψs (αi , Ai )1,q ; (βi , Bi )1,s ; z
is defined by
q Ψs
h
(αi , Ai )1,q ; (βi , Bi )1,s ; z
i
q
Γ (αi
∞
=n=0 i=1
s
Q
+ nAi ) z n
.
n!
Γ (βi + nBi )
i=1
268
(z ∈ U ) .
M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
If Ai = 1(i = 1, ..., q) and Bi = 1 (i = 1, ..., s) , we have the relationship:
h
i
Ωq Ψs (αi , 1)1,q ; (βi , 1)1,s ; z = q Fs (α1 , ..., αq ; β1 , ..., βs ; z) ,
where q Fs (α1 , ..., αq ; β1 , ..., βs ; z) is the generalized hypergeometric function ( see for
details [9] , [10] , [11] , [16] ) and
Ω=
s
Q
i=1
q
Q
Γ (βi )
.1.6
(6)
Γ (αi )
i=1
The Wright generalized hypergeometric functions were invoked in the geometric
function theory ( see [8] , [9] , [10] , [11] , [25] , [26] and [27] ).
By using the generalized hypergeometric function Dziok and Srivastava [10] introduced a linear operator. In [8] Dziok and Raina and in [2] Aouf and Dziok
extended the linear operator by using
function.
h Wright generalized hypergeometric
i
p
First we define a function q Φs (αi , Ai )1,q ; (βi , Bi )1,s ; z by
p
q Φs
h
i
h
i
(αi , Ai )1,q ; (βi , Bi )1,s ; z = Ωz p q Ψs (αi , Ai )1,q ; (βi , Bi )1,s ; z
and consider the following linear operator
i
h
θp,q,s (αi , Ai )1,q ; (βi , Bi )1,s : A(p) → A(p), 1.7
(7)
defined by the convolution
h
i
h
i
θp,q,s (αi , Ai )1,q ; (βi , Bi )1,s f (z) =q Φps (αi , Ai )1,q ; (βi , Bi )1,s ; z ∗ f (z) .
We observe that, for a function f (z) of the form (1.1) , we have
i
h
p
θp,q,s (αi , Ai )1,q ; (βi , Bi )1,s f (z) = z +
∞
X
Ωσn,p (α1 ) an z n , 1.8
(8)
n=p+1
where Ω is given by (1.6) and σn,p (α1 ) is defined by
σn,p (α1 ) =
Γ (α1 + A1 (n − p)) ...Γ (αq + Aq (n − p))
.1.9
Γ (β1 + B1 (n − p)) ...Γ (βs + Bs (n − p)) (n − p)!
(9)
If, for convenience, we write
θp,q,s [α1 , A1 , B1 ] f (z) = θp,q,s [(α1 , A1 ) , ..., (αq , Aq ) ; (β1 , B1 ) , ..., (βs , Bs )] f (z) ,
269
M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
then one can easily verify from (1.8) that
zA1 (θp,q,s [α1 , A1 , B1 ] f (z))′ = α1 θp,q,s [α1 + 1, A1 , B1 ] f (z)
−(α1 − pA1 )θp,q,s [α1 , A1 , B1 ] f (z) (A1 > 0).1.10
(10)
We note that for Ai = 1 (i = 1, 2, ..., q) and Bi = 1 (i = 1, 2, ..., s) , we obtain
θp,q,s [α1 , 1, 1] f (z) = Hp,q,s [α1 ]f (z) , which was introduced and studied by Dziok
and Srivastava [10] . Also for f (z) ∈ A, the linear operator θ1,q,s [α1 , A1 , B1 ] = θ [α1 ]
was introduced by Dziok and Raina [8] and studied by Aouf and Dziok [2].
We note that, for f (z) ∈ A(p), Ai = 1 (i = 1, ..., q) , Bi = 1 (i = 1, ..., s) , q = 2
and s = 1, we have:
(i) θp,2,1 [a, 1; c] f (z) = Lp (a, c) f (z) (a > 0, c > 0, p ∈ N) (seeSaitoh [28]);
(ii) θp,2,1 [µ + p, p; p] f (z) = Dµ+p−1 f (z) (µ > −p, p ∈ N) , where Dµ+p−1 f (z)
is the (µ + p − 1) −the order Ruscheweyh derivative of a function f (z) ∈ A(p)
(seeKumarandShukla [13] and [14]);
(µ,p)
(µ,p)
(iii) θp,2,1 [1 + p, 1; 1 + p − µ] f (z) = Ωz f (z) , where the operator Ωz
is
defined by ( see Srivastava and Aouf [30] )
f (z) =
Ω(µ,p)
z
Γ (1 + p − µ) µ µ
z Dz f (z) (0 ≤ µ < 1; p ∈ N) ,
Γ (1 + p)
where Dzµ is the fractional derivative operator (see, f ordetails, [23] and [24] ) ;
(iv) θp,2,1 [ν + p, 1; ν + p + 1] f (z) = Jν,p (f ) (z), where Jν,p (f ) (z) is the generalized Bernardi-Libera-Livingston-integral operator (see [3] , [15] and [18] ), defined
by
Z
ν + p z ν−1
Jν,p (f ) (z) =
t f (t) dt (ν > −p; p ∈ N) ;
zν
0
(v) θp,2,1 [p + 1, 1; n + p] f (z) = In,p f (z) (n ∈ Z, n > −p, p ∈ N) ,where the operator In,p was considered by Liu and Noor [17] ;
(vi) θp,2,1 [λ + p, c; a] f (z) = Ipλ (a, c) f (z) a, c ∈ R\Z̄0 , λ > −p, p ∈ N , where
Ipλ (a, c) is the Cho-Kwon-Srivastava operator [7] .
The main object of the present paper is to find sufficient condition for certain
normalized analytic functions f (z) in U such that θp,q,s [α1 , A1 , B1 ](f ∗Ψ) (z) 6= 0 and
f (z) to satisfy
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
q1 (z) ≺
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
≺ q2 (z),
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
where q1 , q2 are given univalent functions in U and Φ(z) = z p +
∞
P
λn z n , Ψ(z) =
n=p+1
zp +
∞
P
µn z n are analytic functions in U with λn ≥ 0, µn ≥ 0. Also, we obtain
n=p+1
number of known results as special cases.
2. Definitions and Preliminaries
In order to prove our results, we shall make use of the following definition and
lemmas.
Definition 1 [21] . Denote by Q, the set of all functions f that are analytic and
injective on U \ E(f ), where
E(f ) = {ξ ∈ ∂U : lim f (z) = ∞},
z→ξ
and are such that
f ′ (ξ)
6= 0 for ξ ∈ ∂U \ E(f ).
Lemma 1 [20] . Let q be univalent in the unit disk U and θ and ϕ be analytic in a
domain D containing q(U ) with ϕ(w) 6= 0 when w ∈ q(U ). Set
ψ(z) = zq ′ (z)ϕ(q(z)) and h(z) = θ(q(z)) + ψ(z).it2.1
(11)
Suppose that
(i) ψ(z)nis starlike
univalent in U,
o
zh′ (z)
(ii) Re ψ(z) > 0 for z ∈ U.
If p is analytic with p(0) = q(0), p(U ) ⊆ D and
θ(p(z)) + zp′ (z)ϕ(p(z)) ≺ θ(q(z)) + zq ′ (z)ϕ(q(z)), it2.2
(12)
then p(z) ≺ q(z) and q is the best dominant.
Lemma 2 [4] . Let q be convex univalent in U and ϑ and φ be analytic in a domain
D containing q(U ). Suppose that
(i) Re{ϑ′ (q(z))/φ(q(z))} > 0 for z ∈ U and
(ii) ψ(z) = zq ′ (z)φ(q(z)) is starlike univalent in U.
If p(z) ∈ H[q(0), 1] ∩ Q, with p(U ) ⊆ D, and ϑ(p(z)) + zp′ (z)φ(p(z)) is univalent
in U and
ϑ(q(z)) + zq ′ (z)φ(q(z)) ≺ ϑ(p(z)) + zp′ (z)φ(p(z)), it2.3
(13)
then q(z) ≺ p(z) and q is the best subordinant.
271
M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
3. Subordination results for analytic functions
Unless otherwise mentioned, we shall assume in the reminder of this paper that,
α1 , A1 , ..., αq , Aq and β1 , B1 , ..., βs , Bs (q, s ∈ N) be positive real parameters, Φ, Ψ ∈
A(p), p ∈ N, ξ1 , ξ2 , ξ3 ∈ C and ξ4 ∈ C∗ = C\{0}.
Theorem 1. Let q be convex univalent in U with q(0) = 1. Further, assume that
zq ′′ (z)
ξ3 2ξ2
Re
+
q(z) + 1 + ′
> 0 (z ∈ U ).it3.1
(14)
ξ4
ξ4
q (z)
If f ∈ A(p) satisfies
̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.2
(15)
where
̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
ξ1 + ξ2
2
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
θp,q,s [α1 +1,A1 ,B1 ](f ∗Ψ)(z)
+ ξ3
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
θp,q,s [α1 +1,A1 ,B1 ](f ∗Ψ)(z)
θ
[α1 +1,A1 ,B1 ](f ∗Φ)(z)
θp,q,s [α1 +2,A1 ,B1 ](f ∗Ψ)(z)
+ξ4 α1 p,q,s
−
(α
+
1)
+
1
1
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
θp,q,s [α1 +1,A1 ,B1 ](f ∗Ψ)(z)
× θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z) ,
θp,q,s [α1 +1,A1 ,B1 ](f ∗Ψ)(z)
3.3
(16)
then
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
≺ q(z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
and q is the best dominant.
Proof. Define the function p(z) by
p(z) =
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
(z ∈ U ).3.4
(17)
Then the function p(z) is analytic in U and p(0) = 1, Therefore, by making use of
(3.4), we obtain
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
ξ1 + ξ2
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
2
+ ξ3
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 2, A1 , B1 ](f ∗ Ψ) (z)
+ξ4 α1
− (α1 + 1)
+1
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
×
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
= ξ1 + ξ2 p2 (z) + ξ3 p(z) + ξ4 zp′ (z).3.5
(18)
By using (3.2) in (3.5), we have
ξ1 + ξ2 p2 (z) + ξ3 p(z) + ξ4 zp′ (z) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z).
By setting
θ(w) = ξ1 + ξ2 w2 + ξ3 w
and
ϕ(w) = ξ4 ,
it can be easily observed that θ(w) is analytic in C, ϕ(w) is analytic in C∗ and that
ϕ(w) 6= 0, w ∈ C∗ , then by using Lemma 1 we complete the proof of Theorem 1.
zp
Putting Φ(z) = Ψ(z) = 1−z
(or, µn = λn = 1, n ≥ p + 1, p ∈ N) in Theorem 1,
we obtain the following corollary:
Corollary 1. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true. If
f ∈ A(p) and satisfies
Z(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.6
(19)
where
Z(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
ξ1 + ξ2
2
θp,q,s [α1 ,A1 ,B1 ]f (z)
θp,q,s [α1 +1,A1 ,B1 ]f (z)
+ ξ3
θp,q,s [α1 ,A1 ,B1 ]f (z)
θp,q,s [α1 +1,A1 ,B1 ]f (z)
θ
[α1 +1,A1 ,B1 ]f (z)
θp,q,s [α1 +2,A1 ,B1 ]f (z)
+ξ4 α1 p,q,s
−
(α
+
1)
+
1
1
θp,q,s [α1 ,A1 ,B1 ]f (z)
θp,q,s [α1 +1,A1 ,B1 ]f (z)
× θp,q,s [α1 ,A1 ,B1 ]f (z) ,
θp,q,s [α1 +1,A1 ,B1 ]f (z)
3.7
(20)
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
then
θp,q,s [α1 , A1 , B1 ]f (z)
≺ q(z)
θp,q,s [α1 + 1, A1 , B1 ]f (z)
and q is the best dominant.
Remark 1. Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s) and p = 1, in
Corollary 1,we obtain the result obtained by Murugusundaramoorthy and Magesh
[22, Corollary 2.6 ].
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2 , s = 1 and α1 =
α2 = β1 = 1 in Theorem 1, hence we obtain the next result:
Corollary 2. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true. If
f ∈ A(p) and satisfies
B(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.8
(21)
where
B(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
ξ1 + ξ2
+ξ4
(f ∗Φ)(z)
z(f ∗Ψ)′ (z)+(1−p)(f ∗Ψ)(z)
z(f ∗Φ)′ (z)
(f ∗Φ)(z)
−
2
+ ξ3
(f ∗Φ)(z)
z(f ∗Ψ)′ (z)+(1−p)(f ∗Ψ)(z)
z 2 (f ∗Ψ)′′ (z)+z(3−p)(f ∗Ψ)′ (z)+(1−p)(3−p)(f ∗Ψ)′ (z)
z(f ∗Ψ)′ (z)+(1−p)(f ∗Ψ)(z)
(f ∗Φ)(z)
×
′
z(f ∗Ψ) (z)+(1−p)(f ∗Ψ)(z) ,
+ (2 − p)
3.9
(22)
then
z(f ∗
(f ∗ Φ)(z)
≺ q(z)
+ (1 − p)(f ∗ Ψ)(z)
Ψ)′ (z)
and q is the best dominant.
Remark 2. (i) Putting p = 1 in Corollary 2, we obtain the result obtained by
Murugusundaramoorthy and Magesh [22, Corollary 2.7 ];
(ii) Putting p = 1 and Φ(z) = Ψ(z) in Corollary 2, we obtain the result obtained
by Murugusundaramoorthy and Magesh [22, Corollary 2.8 ];
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
z
(or, p = 1, µn = λn = 1 and n ≥ 2) in
(iii) Putting p = 1 and Φ(z) = Ψ(z) = 1−z
Corollary 2, we obtain the result obtained by Murugusundaramoorthy and Magesh
[22, Corollary 2.9 ].
Taking q(z) = 1 + Az1 + Bz (−1 ≤ A < B ≤ 1) in Theorem 1, we obtain the next
result:
Corollary 3. Let q be convex univalent in U with q(0) = 1 and
ξ3 2ξ2
1 − Bz
Re
+
(1 + Az1 + Bz) +
> 0 (z ∈ U ).
ξ4
ξ4
1 + Bz
If f ∈ A(p) , −1 ≤ A < B ≤ 1 and
̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 +ξ2 (1 + Az1 + Bz)2 (z)+ξ3 (1 + Az1 + Bz)+ξ4
z (A − B)
3.10
(1 + Bz)2
(23)
where ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined by (3.3), then
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
≺ 1 + Az1 + Bz3.11
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
(24)
and 1 + Az1 + Bz is the best dominant.
Remark 3. Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s) and p = 1, in
Corollary 3,we obtain the result obtained by Murugusundaramoorthy and Magesh [
22, Corollary 2.11 ].
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
a, α2 = 1 and β1 = c (a, c > 0, p ∈ N) in Theorem 1, we obtain the following
corollary:
Corollary 4. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true.
If f ∈ A(p) satisfies
K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.12
where
K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
275
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
2
Lp (a,c)(f ∗Φ)(z)
Lp (a,c)(f ∗Φ)(z)
+
ξ
ξ
+
ξ
3
1
2
Lp (a+1,c)(f ∗Ψ)(z)
Lp (a+1,c)(f ∗Ψ)(z)
L (a+2,c)(f ∗Ψ)(z)
L (a+1,c)(f ∗Φ)(z)
=
+ξ4 a Lp p (a,c)(f ∗Φ)(z) − (a + 1) Lpp (a+1,c)(f ∗Ψ)(z) + 1
× Lp (a,c)(f ∗Φ)(z) ,
3.13
Lp (a+1,c)(f ∗Ψ)(z)
(26)
then
Lp (a, c)(f ∗ Φ) (z)
≺ q(z)
Lp (a + 1, c)(f ∗ Ψ) (z)
and q is the best dominant of (3.12).
Remark 4. Putting p = 1 in Corollary 4, we obtain the result obtained by Murugusundaramoorthy and Magesh [22, Corollary 2.5 ].
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
µ+p (µ > −p, p ∈ N) , α2 = β1 = p in Theorem 1, we obtain the following corollary:
Corollary 5. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true.
If f ∈ A(p) satisfies
M (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.14
(27)
where
M (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
ξ1 + ξ2
Dµ+p−1 (f ∗Φ)(z)
Dµ+p (f ∗Ψ)(z)
2
+ ξ3
Dµ+p−1 (f ∗Φ)(z)
Dµ+p (f ∗Ψ)(z)
µ+p
Dµ+p+1 (f ∗Ψ)(z)
+ξ4 (µ + p) DDµ+p−1(f(f∗Φ)(z)
+
1
−
(µ
+
p
+
1)
Dµ+p (f ∗Ψ)(z)
∗Φ)(z)
× Dµ+p−1 (f ∗Φ)(z) ,
µ+p
D
(f ∗Ψ)(z)
3.15
(28)
then
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
Dµ+p−1 (f ∗ Φ) (z)
≺ q(z)
Dµ+p (f ∗ Ψ) (z)
and q is the best dominant of (3.14).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
1 + p, α2 = 1 and β1 = 1 + p − µ (0 ≤ µ < 1, p ∈ N) , in Theorem 1, we obtain the
following corollary:
Corollary 6. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true. If
f ∈ A(p) satisfies
N (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.16
(29)
where
N (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
2
(µ,p)
(µ,p)
(f ∗Φ)(z)
Ωz
Ωz
(f ∗Φ)(z)
+
ξ
ξ
+
ξ
3
1
2
(µ+1,p)
(µ+1,p)
Ωz
(f ∗Ψ)(z)
Ωz
(f ∗Ψ)(z)
(µ+2,p)
(µ+1,p)
Ωz
(f ∗Ψ)(z)
Ωz
(f ∗Φ)(z)
(1
+
p)
−
(p
+
2)
+
1
+ξ
4
(µ,p)
(µ+1,p)
Ωz
Ωz
(f ∗Ψ)(z)
(f ∗Φ)(z)
(µ,p)
(f ∗Φ)(z)
z
× Ω(µ+1,p)
,
Ωz
3.17
(f ∗Ψ)(z)
(30)
then
(µ,p)
Ωz
(f ∗ Φ) (z)
(µ+1,p)
Ωz
(f
∗ Ψ) (z)
≺ q(z)
and q is the best dominant of (3.16).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
ν + p (ν > −p, p ∈ N), α2 = 1 and β1 = ν + p + 1 in Theorem 1, we obtain the
following corollary:
Corollary 7. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true.
If f ∈ A(p) satisfies
L(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.18
277
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
where
L(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
ξ1 + ξ2
2
Jν,p (f ∗Φ)(z)
Jν+1,p (f ∗Ψ)(z)
+ ξ3
Jν,p (f ∗Φ)(z)
Jν+1,p (f ∗Ψ)(z)
Jν+2,p (f ∗Ψ)(z)
J
(f ∗Φ)(z)
+ξ4 (ν + p) ν+1,p
−
(ν
+
p
+
1)
+
1
Jν,p (f ∗Φ)(z)
Jν+1,p (f ∗Ψ)(z)
× Jν,p (f ∗Φ)(z) ,
Jν+1,p (f ∗Ψ)(z)
3.19
(32)
then
Jν,p (f ∗ Φ) (z)
≺ q(z)
Jν+1,p (f ∗ Ψ) (z)
and q is the best dominant of (3.18).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
p + 1, α2 = 1 and β1 = n + p (n ∈ Z, n > −p, p ∈ N) in Theorem 1, we obtain the
following corollary:
Corollary 8. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true.
If f ∈ A(p) satisfies
R(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.20
(33)
where
R(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
2
In,p (f ∗Φ)(z)
In,p (f ∗Φ)(z)
ξ
+
ξ
+
ξ
1
2 In+1,p (f ∗Ψ)(z)
3 In+1,p (f ∗Ψ)(z)
I
(f ∗Φ)(z)
In+2,p (f ∗Ψ)(z)
=
+ξ4 (p + 1) n+1,p
−
(p
+
2)
+
1
In,p (f ∗Φ)(z)
In+1,p (f ∗Ψ)(z)
× In,p (f ∗Φ)(z) ,
3.21
In+1,p (f ∗Ψ)(z)
then
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
In,p (f ∗ Φ) (z)
≺ q(z)
In+1,p (f ∗ Ψ) (z)
and q is the best dominant of (3.20).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ...,
s), q = 2, s = 1, α1 =
λ + p , α2 = c and β1 = a a, c ∈ R\Z̄0 , λ > −p, p ∈ N , in Theorem 1, we obtain
the following corollary:
Corollary 9. Let q be convex univalent in U with q(0) = 1 and (3.1) holds true.
If f ∈ A(p) satisfies
G(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) ≺ ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z), 3.22
(35)
where
G(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
=
ξ1 + ξ2
Ipλ (a,c)(f ∗Φ)(z)
Ipλ+1 (a,c)(f ∗Ψ)(z)
2
+ ξ3
Ipλ (a,c)(f ∗Φ)(z)
Ipλ+1 (a,c)(f ∗Ψ)(z)
I λ+2 (a,c)(f ∗Ψ)(z)
I λ+1 (a,c)(f ∗Φ)(z)
+1
+ξ4 (λ + p) pI λ (a,c)(f ∗Φ)(z) − (λ + p + 1) pλ+1
×
Ip
p
Ipλ (a,c)(f ∗Φ)(z)
Ipλ+1 (a,c)(f ∗Ψ)(z)
3.23
(a,c)(f ∗Ψ)(z)
,
(36)
then
Ipλ (a, c) (f ∗ Φ) (z)
Ipλ+1 (a, c) (f ∗ Ψ) (z)
≺ q(z)
and q is the best dominant of (3.22).
4. Superordination results for analytic functions
Now, by applying Lemma 2, we prove the following theorem.
Theorem 2. Let q be convex univalent in U with q(0) = 1 and
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
f ∈ A(p), θp,q,s
[α1 +1,A1 ,B1 ](f ∗Ψ)(z) ∈ H[q(0), 1] ∩ Q.Further, assume that
ξ3 2ξ2
+
q(z) > 04.1
Re
ξ4
ξ4
279
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
and ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) be univalent in U and satisfy
ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z) ≺ ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ), 4.2
(38)
where ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is given by (3.3), then
q(z) ≺
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
and q is the best subordinant of (4.2).
Proof. Define the function p(z) by (3.4), with simple computation from (3.4), we
get
̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) = ξ1 + ξ2 p2 (z) + ξ3 p(z) + ξ4 zp′ (z),
then
ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z) ≺ ξ1 + ξ2 p2 (z) + ξ3 p(z) + ξ4 zp′ (z).4.3
(39)
By setting
υ(w) = ξ1 + ξ2 w2 + ξ3 w
and
ϕ(w) = ξ4 ,
it is easy observed that υ(w) is analytic in C, ϕ(w) is analytic in C∗ and ϕ(w) 6=
0, w ∈ C∗ . Since q is convex and univalent function, it follows that
ξ3 2ξ2
Re θ′ (q(z))ϕ(q(z)) = Re
+
q(z) > 0 (z ∈ U )
ξ4
ξ4
and then, by using Lemma 2 we complete the proof of Theorem 2.
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
a, α2 = 1 and β1 = c (a, c > 0) in Theorem 2, we obtain the following corollary:
Corollary 10. Let q be convex univalent in U with q(0) = 1, (4.1) holds true and
Lp (a,c)(f ∗Φ)(z)
f ∈ A(p), Lp (a+1,c)(f
∗Ψ)(z) ∈ H[q(0), 1]∩Q.Further, assume that K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) be
univalent in U which satisfying
ξ1 + ξ2 q 2 (z) + ξ3 q(z) + ξ4 zq ′ (z) ≺ K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ), 4.4
where K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is given by (3.13), then
280
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
q(z) ≺
Lp (a, c)(f ∗ Φ) (z)
Lp (a + 1, c)(f ∗ Ψ) (z)
and q is the best subordinant of (4.4).
Remark 5. Putting p = 1 in Corollary 10, we obtain the result obtained by Murugusundaramoorthy and Magesh [22, Corollary 2.13 ].
Remark 6. (i) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), p = 1, q = 2
, s = 1, α1 = α2 = 1 and β1 = 1 in Theorem 2, we obtain the result obtained by
Murugusundaramoorthy and Magesh [22, Corollary 2.14 ];
(ii) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2 , s = p = α1 =
α2 = β1 = 1 and Φ(z) = Ψ(z) in Theorem 2, we obtain the result obtained by
Murugusundaramoorthy and Magesh [22, Corollary 2.15];
(iii) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2 , s = p = α1 =
z
in Theorem 2, we obtain the result obtained by
α2 = β1 = 1 and Φ(z) = Ψ(z) = 1−z
Murugusundaramoorthy and Magesh [22, Corollary 2.16].
Taking q(z) = 1 + Az1 + Bz (−1 ≤ A < B ≤ 1) in Theorem 2, we obtain the next
result:
Corollary 11. Let q be convex univalent in U with q(0) = 1, f ∈ A(p),
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
θp,q,s [α1 +1,A1 ,B1 ](f ∗Ψ)(z)
∈ H[q(0), 1] ∩ Q and
ξ3 2ξ2
Re
+
1 + Az1 + Bz > 0 (−1 ≤ A < B ≤ 1) .
ξ4
ξ4
Further, assume that ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) be univalent in U and satisfy
z (A − B)
≺ ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ), 4.5
(1 + Bz)2
(41)
where ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is given by (3.3), then
ξ1 +ξ2 (1 + Az1 + Bz)2 (z)+ξ3 (1 + Az1 + Bz)+ξ4
1 + Az1 + Bz ≺
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
and 1 + Az1 + Bz is the best subordinant (4.5).
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
Remark 7. Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s) and p = 1, in
Corollary 11,we obtain the result obtained by Murugusundaramoorthy and Magesh
[22, Corollary 2.17 ].
Remark 8. (i) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q =
2, s = 1, α1 = µ + p (µ > −p, p ∈ N) , α2 = β1 = p in Theorem 2, we obtain the
superordination result for the operator Dµ+p−1 ;
(ii) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
1 + p, α2 = 1 and β1 = 1 + p − µ (0 ≤ µ < 1, p ∈ N) , in Theorem 2, we obtain the
(µ,p)
superordination result for the operator Ωz ;
(iii) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
ν + p (ν > −p, p ∈ N), α2 = 1 and β1 = ν + p + 1 in Theorem 2, we obtain the
superordination result for the operator Jν,p ;
(iv) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
p + 1, α2 = 1 and β1 = n + p (n ∈ Z, n > −p, p ∈ N) in Theorem 2, we obtain the
superordination result for the operator In,p ;
(v) Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ...,
s), q = 2, s = 1, α1 =
λ + p , α2 = c and β1 = a a, c ∈ R\Z̄0 , λ > −p, p ∈ N , in Theorem 2, we obtain
the superordination result for the operator Ipλ (a, c) .
4. Sandwich results
Combining Theorem 1 with Theorem 2, we get the following sandwich theorem.
Theorem 3. Let q1 , q2 be convex univalent in U. Suppose q1 and
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
q2 satisfies (4.1) and (3.1), respectively. If f ∈ A(p), θp,q,s
[α1 +1,A1 ,B1 ](f ∗Ψ)(z) ∈
H[q(0), 1] ∩ Q and ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, which satisfying
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.1
(42)
where ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is given by (3.3), then
q1 (z) ≺
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
≺ q2 (z)
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.1).
Taking q1 (z) = 1 + A1 z1 + B1 z (−1 ≤ A1 < B1 ≤ 1) and q2 (z) = 1 + A2 z1 + B2 z (−1 ≤ A2 < B2 ≤ 1)
282
M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
in Theorem 3, we obtain the following corollary:
Corollary 12. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
θp,q,s [α1 ,A1 ,B1 ](f ∗Φ)(z)
(4.1) and (3.1), respectively. If f ∈ A(p), θp,q,s
[α1 +1,A1 ,B1 ](f ∗Ψ)(z) ∈ H[q(0), 1] ∩
Q and ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, which satisfying
ξ1 +ξ2 (1 + A1 z1 + B1 z)2 (z)+ξ3 (1 + A1 z1 + B1 z)+ξ4
(A1 − B1 ) z
≺ ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
(1 + B1 z)2
≺ ξ1 + ξ2 (1 + A2 z1 + B2 z)2 + ξ3 (1 + A2 z1 + B2 z) + ξ4
(A2 − B2 ) z
, 5.2
(1 + B2 z)2
(43)
where ̥(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined in (3.3), then
1 + A1 z1 + B1 z ≺
θp,q,s [α1 , A1 , B1 ](f ∗ Φ) (z)
≺ 1 + A2 z1 + B2 z
θp,q,s [α1 + 1, A1 , B1 ](f ∗ Ψ) (z)
and 1 + A1 z1 + B1 z, 1 + A2 z1 + B2 z are, respectively, the best subordinant and
dominant of (5.2).
Remark 9. Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2 and
s = p = α1 = α2 = β1 = 1 in Corollary 12, we obtain the result obtained by
Murugusundaramoorthy and Magesh [22, Corollary 3.2 ].
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
a, α2 = 1 and β1 = c (a, c > 0) in Theorem 3, we obtain the following corollary:
Corollary 13. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
Lp (a,c)(f ∗Φ)(z)
(4.1) and (3.1), respectively. If f ∈ A(p), Lp (a+1,c)(f
∗Ψ)(z) ∈ H[q(0), 1] ∩ Q and
K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U ,where K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined in
(3.13), then
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ K(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.3
implies
q1 (z) ≺
Lp (a, c)(f ∗ Φ) (z)
≺ q2 (z)
Lp (a + 1, c)(f ∗ Ψ) (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.3).
283
(44)
M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
µ + p (µ > −p, p ∈ N) and α2 = β1 = p in Theorem 3, we obtain the following
corollary:
Corollary 14. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
µ+p−1 ∗Φ)(z)
(4.1) and (3.1), respectively. If f ∈ A(p), DDµ+p (f(f∗Ψ)(z)
∈ H[q(0), 1] ∩ Q and
M (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, where M (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined
in (3.15), then
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ M (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.4
(45)
implies
q1 (z) ≺
Dµ+p−1 (f ∗ Φ) (z)
≺ q2 (z)
Dµ+p (f ∗ Ψ) (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.4).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
1 + p, α2 = 1 and β1 = 1 + p − µ (0 ≤ µ < 1, p ∈ N) in Theorem 3, we obtain the
following corollary:
Corollary 15. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
(4.1) and (3.1), respectively. If f ∈ A(p),
(µ,p)
Ωz
(f ∗Φ)(z)
(µ+1,p)
Ωz
(f ∗Ψ)(z)
∈ H[q(0), 1] ∩ Q and
N (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, where N (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined in
(3.17), then
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ N (f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.5
(46)
implies
(µ,p)
q1 (z) ≺
Ωz
(f ∗ Φ) (z)
(µ+1,p)
Ωz
(f
∗ Ψ) (z)
≺ q2 (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.5).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
ν + p (ν > −p, p ∈ N), α2 = 1 and β1 = ν + p + 1 in Theorem 3, we obtain the
following corollary:
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
Corollary 16. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
Jν,p (f ∗Φ)(z)
(4.1) and (3.1), respectively. If f ∈ A(p), Jν+1,p
(f ∗Ψ)(z) ∈ H[q(0), 1] ∩ Q and
L(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, where L(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined in
(3.19), then
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ L(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.6
(47)
implies
q1 (z) ≺
Jν,p (f ∗ Φ) (z)
≺ q2 (z)
Jν+1,p (f ∗ Ψ) (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.6).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2, s = 1, α1 =
p + 1, α2 = 1 and β1 = n + p (n > −p, p ∈ N) in Theorem 3, we obtain the following
corollary:
Corollary 17. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
In,p (f ∗Φ)(z)
(4.1) and (3.1), respectively. If f ∈ A(p), In+1,p
(f ∗Ψ)(z) ∈ H[q(0), 1] ∩ Q and
R(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, where R(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined in
(3.21), then
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ R(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.7
(48)
implies
q1 (z) ≺
In,p (f ∗ Φ) (z)
≺ q2 (z)
In+1,p (f ∗ Ψ) (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.7).
Putting Ai = 1 (i = 1, 2, ..., q), Bj = 1 (j = 1, 2, ..., s), q = 2,
s = 1, α1 =
λ + p (λ > −p, p ∈ N), α2 = c and β1 = a a, c ∈ R\Z̄0 , λ > −p in Theorem 3,
we obtain the following corollary:
Corollary 18. Let q1 , q2 be convex univalent in U. Suppose q1 and q2 satisfies
(4.1) and (3.1), respectively. If f ∈ A(p),
Ipλ (a,c)(f ∗Φ)(z)
λ+1
Ip (a,c)(f ∗Ψ)(z)
285
∈ H[q(0), 1] ∩ Q and
M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian - Sandwich Theorems...
G(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is univalent in U, where G(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 ) is defined in
(3.23), then
ξ1 + ξ2 q12 (z) + ξ3 q1 (z) + ξ4 zq1′ (z) ≺ G(f, Φ, Ψ, ξ1 , ξ2 , ξ3 , ξ4 )
≺ ξ1 + ξ2 q22 (z) + ξ3 q2 (z) + ξ4 zq2′ (z), 5.8
(49)
implies
q1 (z) ≺
Ipλ (a, c) (f ∗ Φ) (z)
Ipλ+1 (a, c) (f ∗ Ψ) (z)
≺ q2 (z)
and q1 , q2 are, respectively, the best subordinant and dominant of (5.8).
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M. K. Aouf, A. Shamandy, A. O. Mostafa, S. M. Madian
Department of Mathematics
Faculty of Science
Mansoura University
Mansoura 35516, Egypt.
email: [email protected], [email protected], [email protected],
samar [email protected]
288