5_aouf.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 5, No. 2, 2012, 141-159

ISSN 1307-5543 – www.ejpam.com

Applications Of Differential Subordination To Certain Subclasses

Of Meromorphically Multivalent Functions Associated With

Generalized Hypergeometric Function

M. K. Aouf

Faculty of Science, Mansoura University, Mansoura 35516, Egypt.

Abstract. By making use of the principle of differential subordination, we investigate several inclu-

sion relationships and other interesting properties of certain subclasses of meromorphically multiva-

lent functions which are defined by certain linear operator involving the generalized hypergeometric

function.

2010 Mathematics Subject Classifications: 30C45

Key Words and Phrases: Differential subordination, Hadamard product, meromorphic function, Hy-

pergeometric function.

1. Introduction

For any integer m > −p, let
∑

p,m denote the class of all meromorphic functions f of the

form:

f (z) = z−p +

∞
∑

k=m

akzk (p ∈ N = {1,2, . . .}), (1)

which are analytic and p-valent in the punctured disc

U∗ = {z : z ∈ C and 0< |z| < 1}= U\{0}. For convenience, we write
∑

p,−p+1 =
∑

p. If f and

g are analytic in U , we say that f is subordinate to g, written symbolically as follows:

f ≺ g or f (z)≺ g(z),

if there exists a Schwarz function w, which (by definition) is analytic in U with w(0) = 0 and

|w(z)| < 1 (z ∈ U) such that f (z) = g(w(z)) (z ∈ U).

In particular, if the function g is univalent in U , we have the equivalence (cf., e. g., [7]; see

also [8, p. 4]):

f (z)≺ g(z)⇔ f (0) = g(0) and f (U)⊂ g(U).

Email address: mkaouf127�yahoo.
om (M. Aouf)

http://www.ejpam.com 141 c© 2012 EJPAM All rights reserved.



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 142

For functions f ∈∑p,m, given by (1), and g ∈∑p,m defined by

g(z) = z−p +

∞
∑

k=m

bkzk (m> −p; p ∈ N), (2)

then the Hadamard product (or convolution) of f and g is given by

( f ∗ g) = z−p +

∞
∑

k=m

ak bkzk = (g ∗ f )(z) (m> −p; p ∈ N). (3)

For complex parameters

α1, . . .αq β1, . . . ,β s (β j /∈ Z−0 = {0,−1,−2, . . .}; j = 1,2, . . . , s),

we now define the generalized hypergeometric function qFs(α1, . . . ,αq;β1, . . . ,β s; z) by (see,

for example, [14, p.19])

qFs(α1, . . . ,αq;β1, . . . ,β s; z) =

∞
∑

k=0

(α1)k . . . (αq)k

(β1)k . . . (β s)k
· z

k

k!

(q ≤ s+ 1; q, s ∈ N0 =N ∪ {0}; z ∈ U), (4)

where (θ)ν is the Pochhammer symbol defined, in terms of the Gamma function Γ, by

(θ )ν =
Γ(θ + ν)

Γ(θ)
=

¨

1 (ν = 0;θ ∈ C\{0}),
θ(θ − 1) . . . (θ + ν − 1) (ν ∈ N ;θ ∈ C).

(5)

Corresponding to the function hp(α1, . . . ,αq;β1, . . . ,β s; z), defined by

hp(α1, . . . ,αq;β1, . . . ,β s; z) = z−p
qFs(α1, . . . ,αq;β1, . . . ,β s; z), (6)

we consider a linear operator

Hp(α1, . . . ,αq;β1, . . . ,β s; z) : Σp→ Σp,

which is defined by the following Hadamard product (or convolution):

Hp(α1, . . . ,αq;β1, . . . ,β s) f (z) = hp(α1, . . . ,αq;β1, . . . ,β s; z) ∗ f (z). (7)

We observe that, for a function f (z) of the form (1), we have

Hp(α1, . . . ,αq;β1, . . . ,β s) f (z) = z−p +

∞
∑

k=m

(α1)k+p . . . (αq)k+p

(β1)k+p . . . (β s)k+p

· ak

(k+ p)!
zk. (8)

If, for convenience, we write

Hp,q,s(α1) = Hp(α1, . . . ,αq;β1, . . . ,β s), (9)



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 143

then one can easily verify from the definition (7) that

z(Hp,q,s(α1) f (z))
′
= α1Hp,q,s(α1+ 1) f (z)− (α1 + p)Hp,q,s(α1) f (z). (10)

For m = −p + 1 (p ∈ N), the linear operator Hp,q,s(α1) was investigated recently by Liu and

Srivastava [5] and Aouf [1].

In particular, for s = 1,q = 2,α1 > 0,β1 > 0 and α2 = 1, we obtain the linear operator

ℓp(α1,β1) f (z) = Hp(α1, 1;β1) f (z)( f ∈∑p), which was introduced and studied by Liu and

Srivastava [4].

We note that, for any integer n> −p and f ∈∑p,m,

Hp,2,1(n+ p, 1; 1) f (z) = Dn+p−1 f (z) =
1

zp(1− z)n+p
∗ f (z)

where Dn+p−1 is the differential operator studied by Uralegaddi and Somanatha [17].

Making use of the principle of differential subordination as well as the linear operator

Hp,q,s(α1), we now introduce a subclass of the function class
∑

p,m as follows:

For fixed parameters A and B(−1 ≤ B < A≤ 1), we say that a function f ∈∑p,m is in the

class
∑m

p,q,s(α1; A, B), if it satisfies the following subordination condition:

−zp+1(Hp,q,s(α1) f (z))
′

p
≺ 1+ Az

1+ Bz
. (11)

In view of the definition of subordination, (11) is equivalent to the following condition:

�

�

�

�

�

zp+1(Hp,q,s(α1) f (z))
′
+ p

Bzp+1(Hp,q,s(α1) f (z))
′
+ pA

�

�

�

�

�

< 1 (z ∈ U).

For convenience, we write

Σm
p,q,s(α1; 1− 2ζ

p
, 1) = Σm

p,q,s(α1;ζ),

where Σm
p,q,s(α1;ζ) denotes the class of functions f (z) ∈ Σp,m satisfying the following inequal-

ity:

Re
¦

−zp+1(Hp,q,s(α1) f (z))
′©
> ζ (0≤ ζ < p; z ∈ U) .

We note that Σ
−p+1
p,q,s (α1; A+ (B − A)

ρ

p
, B) = Σp,q,s(α1,A, B,ρ), 0 ≤ ρ < p; p ∈ N), where the

class Σp,q,s(α1,A, B,ρ) was introduced and studied by Aouf [1]. We also observe that:

(i)
∑−p+1

p,2,1 (n+ p, 1; 1; A, B) = Cn,p(A, B) (n> −p; p ∈ N ;−1≤ B < A≤ 1), is the subclass of
∑

p studied by Uralegaddi and Somanatha [17];

(ii)
∑−p+1

p,2,1 (n+ p, 1; 1; 1− 2α

p
,−1) =
∑

n,p(α) (n> −p; p ∈ N ; 0≤ α < p), is the subclass of
∑

p studied by Cho and Nunokawa [2];



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 144

(iii) For q = 2, s = 1, α1 = a > 0, β1 = c > 0 and α2 = 1, we have

Σa,c(p; m,A, B) =

(

f (z) ∈ Σp,m :−zp+1(ℓp(a, c) f (z))
′

p
≺ 1+ Az

1+ Bz
,−1≤ B < A≤ 1, z ∈ U

)

,

(12)

where the class Σa,c(p; m,A, B) was studied by Patel and Cho [12].

2. Preliminary Lemmas

To establish our main results, we need the following lemmas.

Lemma 1 ([3]). Let the function h be analytic and convex (univalent) in U with h(0) = 1.

Suppose also that the function ϕ given by

ϕ(z) = 1+ cp+mzp+m+ cp+m+1zp+m+1 + . . . (13)

in analytic in U. If

ϕ(z) +
zϕ
′
(z)

γ
≺ h(z) (Re(γ)≥ 0;γ 6= 0), (14)

then

ϕ(z) ≺ψ(z) = γ

p+m
z
−γ

p+m

z
∫

0

t
γ

p+m
−1

h(t)d t ≺ h(z),

and ψ is the best dominant of (14).

With a view to starting a well-known result (Lemma 2 below), we denote by P(γ) the class

of functions ϕ given by

ϕ(z) = 1+ b1z + b2z2 + . . . , (15)

which are analytic in U and satisfy the following inequality:

Re
�

ϕ(z)
	

> γ (0≤ γ < 1; z ∈ U) .

Lemma 2 ([10]). Let the function ϕ, given by (15), be in the class P(γ). Then

Re
�

ϕ(z)
	 ≥ 2γ− 1+

2(1− γ)
1+ |z| (0≤ γ < 1; z ∈ U).

Lemma 3 ([16]). For 0≤ γ1,γ2 < 1, we have

P(γ1) ∗ P(γ2)⊂ P(γ3) (γ3 = 1− 2(1− γ1)(1− γ2)).

The result is the best possible.



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 145

For real or complex numbers a, b and c (c /∈ Z−0 ), the Gaussian hypergeometric function is

defined by

2F1(a, b; c; z) = 1+
ab

c
· z

1!
+

a(a+ 1)b(b+ 1)

c(c + 1)
· z

2

z!
+ . . . .

We note that the above series converges absolutely for z ∈ U and hence represents an analytic

function in U (see, for details [18, Chapter 14]).

Each of the identities (asserted by Lemma 4 below) is well-known (cf., e.g., [18, Chapter

14]).

Lemma 4 ([18]). For real or complex parameters a, b and c (c /∈ Z−0 ),

1
∫

0

t b−1(1− t)c−b−1(1− zt)−ad t =
Γ(b)Γ(c− b)

Γ(c)
2Γ1(a, b; c; z) (Re(c) > Re(b)> 0); (16)

2F1(a, b; c; z) = (1− z)−a
2F1(a, b; c;

z

z − 1
); (17)

2F1(a, b; c; z) = 2F1(a, b− 1; c; z) +
az

c
2F1(a+ 1, b; c + 1; z); (18)

2F1(a, b;
a+ b+ 1

2
;

1

2
) =

p
πΓ( a+b+1

2
)

Γ( a+1

2
)Γ( b+1

2
)
. (19)

Lemma 5 ([13]). Let Φ be analytic in U with

Φ(0) = 1 and Re {Φ(z)} > 1

2
(z ∈ U).

Then, for any function F analytic in U, (Φ ∗ F) (U) is contained in the convex hull of F(U).

3. Main Results

Remark 1. Throughout our present paper, we assume that:

−1≤ B < A≤ 1,λ > 0, p ∈ N and α1 ∈ C\{0}.

Theorem 1. Let the function f defined by (1) satisfying the following subordination condition:

− (1−λ)z
p+1(Hp,q,s(α1) f (z))

′
+λzp+1(Hp,q,s(α1 + 1) f (z))

′

p
≺ 1+ Az

1+ Bz
.

Then

−zp+1(Hp,q,s(α1) f (z))
′

p
≺ Q(z) ≺ 1+ Az

1+ Bz
, (20)



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 146

where the function Q given by

Q(z) =







A

B
+ (1− A

B
)(1+ Bz)−1

2F1(1,1;
α1

λ(p+m)
+ 1; Bz

1+Bz
) (B 6= 0)

1+
α1A

λ(p+m)+α1
z (B = 0)

is the best dominant of (20). Furthermore,

Re

(

−zp+1(Hp,q,s(α1) f (z))
′

p

)

> ξ (z ∈ U), (21)

where

ξ =







A

B
+ (1− A

B
)(1− B)−1

2F1(1,1;
α1

λ(p+m)
+ 1; B

B−1
) (B 6= 0)

1− α1A

λ(p+m)+α1
(B = 0).

The estimate in (21) is the best possible.

Proof. Consider the function ϕ defined by

ϕ(z) = −zp+1(Hp,q,s(α1) f (z))
′

p
(z ∈ U). (22)

Then ϕ is of the form (13) and is analytic in U . Differentiating (22) with respect to z and

using (10), we obtain

− (1−λ)z
p+1(Hp,q,s(α1) f (z))

′
+λzp+1(Hp,q,s(α1) f (z))

′

p

= ϕ(z) +
λ

α1

zϕ
′
(z)≺ 1+ Az

1+ Bz
(z ∈ U).

Now, by using Lemma 1 for β =
α1

λ
, we obtain

−zp+1(Hp,q,s(α1) f (z))
′

p
≺ Q(z)

=
α1

λ(p+m)
z
− α1
λ(p+m)

z
∫

0

t
α1

λ(p+m)
−1

�

1+At

1+ Bt

�

d t

=







A

B
+ (1− A

B
)(1+ Bz)−1

2F1(1,1;
α1

λ(p+m)
+ 1; Bz

1+Bz
) (B 6= 0)

1+
α1A

λ(p+m)+α1
z (B = 0),



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 147

by change of variables followed by the use of the identities (16), (17) and (18) (with a = 1,

c = b+ 1, b =
α1

λ(p+m)
). This proves the assertion (20) of Theorem 1.

Next, in order to prove the assertion (21) of Theorem 1, it suffices to show that

inf
|z|<1
{Re(Q(z))}= Q(−1). (23)

Indeed we have, for |z| ≤ r < 1,

Re

�

1+ Az

1+ Bz

�

≥ 1− Ar

1− Br
.

Upon setting

g(ζ, z) =
1+ Aζz

1+ Bζz
and dν(ζ) =

α1

λ(p+m)
ζ

α1
λ(p+m)

−1
dζ (0≤ ζ≤ 1),

which is a positive measure on the closed interval [0,1], we get

Q(z) =

1
∫

0

g(ζ, z)dν(ζ),

so that

Re {Q(z)} ≥
1
∫

0

�

1− Aζr

1− Bζr

�

dν(ζ) = Q(−r) (|z| ≤ r < 1) .

Letting r → 1− in the above inequality, we obtain the assertion (21) of Theorem 1.

Finally, the estimate in (21) is the best possible as the function Q is the best dominant of

(20).

Taking λ = 1, A= 1− 2σ

p
(0 ≤ σ < p) and B = −1 in Theorem 1, we obtain the following

corollary.

Corollary 1. The following inclusion property holds true for the function class Σm
p,q,s(α1;σ):

Σm
p,q,s(α1 + 1;σ)⊂ Σm

p,q,s(α1;β(p, m,α1,σ))⊂ Σm
p,q,s(α1;σ),

where

β(p, m,α1,σ) = σ+ (p−σ)
�

2F1(1,1;
α1

p+m
+ 1;

1

2
)− 1

�

.

The result is the best possible.

Taking λ= 1 and m= 1− p (p ∈ N) in Theorem 1, we obtain the following corollary.



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 148

Corollary 2. The following inclusion property holds true for the function class Σp,q,s(α1; A, B):

Σp,q,s(α1 + 1; A, B) ⊂ Σp,q,s(α1; 1− 2σ

p
,−1)⊂ Σp,q,s(α1; A, B),

where

σ =







A

B
+ (1− A

B
)(1+ B)−1

2F1(1,1;α1 + 1; B

B−1
) (B 6= 0)

1− α1A

1+α1
(B = 0).

The result is the best possible.

Remark 2.

(i) Taking λ = 1, q = 2, s = 1, α1 = a, β1 = c (a > 0; c > 0) and α2 = 1 in Theorem 1, we

obtain the result obtained by Patel and Cho [12, Theorem 1];

(ii) Taking m = −p+ 1, λ = 1, q = 2, s = 1, α1 = n+ p(n > −p), α2 = β1 = 1 in Theorem

1, we obtain the result obtained by Patel and Cho [12, Corollary 2] which improves the

corresponding result obtained by Uralegaddi and Somanatha [17];

(iii) Taking q = 2, s = 1, α1 = a > 0, β1 = c > 0 and α2 = 1 in Corollary 2, we obtain the

result obtained by Patel and Cho [12, Corollary 1].

Theorem 2. If f ∈ Σm
p,q,s(α1;θ) (0≤ θ < p), then

Re
¦

−zp+1
�

(1−λ)(Hp,q,s(α1) f (z))
′
+λ(Hp,q,s(α1 + 1) f (z))

′�©
> θ (|z| < R), (24)

where

R=







p

α2
1 +λ

2(p+m)2 −λ(p+m)

α1







1

p+m

.

The result is the best possible.

Proof. Since f ∈ Σm
p,q,s(α1;θ), we write

−zp+1(Hp,q,s(α1) f (z))
′
= θ + (p− θ)u(z) (z ∈ U). (25)

Then, clearly, u is of the form (13), is analytic in U , and has a positive real part in U . Differ-

entiating (25) with respect to z and using (10), we obtain

−
zp+1
�

(1−λ)(Hp,q,s(α1) f (z))
′
+λ(Hp,q,s(α1 + 1) f (z))

′�
+ θ

p− θ = u(z) +
λ

α1

zu
′
(z). (26)



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 149

Now, by applying the well-known estimate [6]

�

�

�zu
′
(z)

�

�

�

Re{u(z)} ≤
2(p+m)r p+m

1− r2(p+m)
(|z| = r < 1)

in (26), we obtain

Re

(

−
zp+1
�

(1−λ)(Hp,q,s(α1) f (z))
′
+λ(Hp,q,s(α1 + 1) f (z))

′�
+ θ

p− θ

)

≥ Re{u(z)} ·
�

1− 2λ(p+m)r p+m

α1(1− r2(p+m))

�

. (27)

It is easily seen that the right-hand side of (27) is positive provided that r < R, where R is

given as in Theorem 2. This proves the assertion (24) of Theorem 2.

In order to show that the bound R is the best possible, we consider the function f ∈ Σp,m

defined by

−zp+1(Hp,q,s(α1) f (z))
′
= θ + (p− θ )1+ zp+m

1− zp+m
(0≤ θ < p; p ∈ N ; z ∈ U).

Noting that

−
zp+1
�

(1−λ)(Hp,q,s(α1) f (z))
′
+λ(Hp,q,s(α1 + 1) f (z))

′�
+ θ

p− θ

=
α1−α1z2(p+m) + 2λ(p+m)zp+m

α1(1− zp+m)2
= 0

for z = R
1

p+m exp
�

iπ

p+m

�

, we complete the proof of Theorem 2.

Putting λ= 1 in Theorem 2, we obtain the following result.

Corollary 3. If f ∈ Σm
p,q,s(α1;θ) (0 ≤ θ < p; p ∈ N), then f ∈ Σm

p,q,s(α1 + 1;θ) for |z| < R∗,
where

R∗ =







p

α2
1
+ (p+m)2 − (p+m)

α1







1

p+m

.

The result is the best possible.

Remark 3. Taking s = 1, q = 2, α1 = a and β1 = c (a > 0; c > 0) and α2 = 1 in Corollary 3,

we obtain the result obtained by Patel and Cho [12, Theorem 2].



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 150

Theorem 3. Let f ∈ Σm
p,q,s(α1; A, B) and let

Fδ,p( f )(z) =
δ

zδ+p

z
∫

0

tδ+p−1 f (t)d t (δ > 0; z ∈ U). (28)

Then

−zp+1(Hp,q,s(α1)Fδ,p f (z))
′

p
≺ Φ(z)≺ 1+ Az

1+ Bz
, (29)

where the function Φ given by

Φ(z) =







A

B
+ (1− A

B
)(1+ Bz)−1

2F1(1,1; δ

p+m
+ 1; Bz

Bz+1
) (B 6= 0)

1+ δ

δ+p+m
Az (B = 0),

is the best dominant of (29). Furthermore,

Re

(

−zp+1(Hp,q,s(α1)Fδ,p( f )(z))
′

p

)

> ξ∗ (z ∈ U), (30)

where

ξ∗ =







A

B
+ (1− A

B
)(1− B)−1

2F1(1,1; δ

p+m
+ 1; B

B−1
) (B 6= 0)

1− δ

δ+p+m
A (B = 0).

The result is the best possible.

Proof. Defining the function ϕ by

ϕ(z) = −zp+1(Hp,q,s(α1)Fδ,p( f )(z))
′

p
(z ∈ U), (31)

we note that ϕ is of the form (13) and is analytic in U . Using the following operator identity:

z(Hp,q,s(α1)Fδ,p( f )(z))
′
= δHp,q,s(α1) f (z)− (δ+ p)Hp,q,s(α1)Fδ,p( f )(z) (32)

in (31) and differentiating the resulting equation with respect to z, we find that

−zp+1(Hp,q,s(α1) f (z))
′

p
≺ ϕ(z) + zϕ

′
(z)

δ
≺ 1+ Az

1+ Bz
.

Now the remaining part of Theorem 3 follows by employing the techniques that we used in

proving Theorem 1 above.

Putting m= 1− p (p ∈ N) in Theorem 3, we obtain the following corollary.



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 151

Corollary 4. If δ > 0 and f ∈ Σp,q,s(α1; A, B), then

Fδ,p( f )(z) ∈ Σp,q,s(α1; 1− 2ξ

p
,−1)⊂ Σp,q,s(α1; A, B),

where

ξ=







A

B
+ (1− A

B
)(1+ B)−1

2F1(1,1;δ+ 1; B

B−1
) (B 6= 0)

1− δ

δ+1
A (B = 0).

The result is the best possible.

Remark 4. By observing that

zp+1(Hp,q,s(α1)Fδ,p( f )(z))
′
=
δ

zδ

z
∫

0

tδ+p(Hp,q,s(α1) f (t))
′
d t ( f ∈ Σp,m; z ∈ U), (33)

Corollary 4 can be restated as follows:

If δ > 0 and f ∈ Σp,q,s(α1; A, B), then

Re







− δ

pzδ

z
∫

0

tδ+p(Hp,q,s(α1) f (t))
′
d t







> ξ (z ∈ U).

where ξ is given as in Corollary 4.

In view of (33), Theorem 3 for A= 1− 2θ

p
(0≤ θ < p; p ∈ N) and B = −1 yields

Corollary 5. If δ > 0 and if f ∈ Σp,m satisfies the following inequality:

Re
¦

−zp+1(Hp,q,s(α1) f (z))
′©
> θ (0≤ θ < p; p ∈ N ; z ∈ U),

then

Re







−δ
zδ

z
∫

0

(Hp,q,s(α1) f (t))
′
d t







> θ + (p− θ)
�

2F1(1,1;
δ

p+m
+ 1;

1

2
)− 1

�

(z ∈ U).

The result is the best possible.

Remark 5. Putting s = 1, q = 2, α1 = a, β1 = c (a > 0; c > 0) and α2 = 1 in Theorem 3, we

obtain the result obtained by Patel and Cho [12, Theorem 3].



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 152

Theorem 4. Let f ∈ Σp,m. Suppose also that g ∈ Σp,m satisfies the following inequality:

Re
¦

zp(Hp,q,s(α1)g(z))
©

> 0 (z ∈ U).

If
�

�

�

�

�

Hp,q,s(α1) f (z)

Hp,q,s(α1)g(z)
− 1

�

�

�

�

�

< 1 (z ∈ U),

then

Re

(

−z(Hp,q,s(α1) f (z))
′

Hp,q,s(α1) f (z)

)

> 0 (|z| < R0),

where

R0 =

p

g(p+m)2 + 4p(2p+m)− 3(p+m)

2(2p+m)
.

Proof. Letting

w(z) =
Hp,q,s(α1) f (z)

Hp,q,s(α1)g(z)
− 1= tp+mzp+m+ tp+m+1zp+m+1 + . . . (34)

we note that w is analytic in U , with w(0) = 0 and |w(z)| ≤ |z|p+m (z ∈ U). Then, by applying

the familiar Schwarz lemma [9], we obtain

w(z) = zp+mΨ(z),

where the functions Ψ is analytic in U and |Ψ(z)| ≤ 1 (z ∈ U). Therefore, (34) leads us to

Hp,q,s(α1) f (z) = Hp,q,s(α1)g(z) (1+ zp+mΨ(z)) (z ∈ U). (35)

Differentiating (35) logarithmically with respect to z, we obtain

z(Hp,q,s(α1) f (z))
′

Hp,q,s(α1) f (z)
=

z(Hp,q,s(α1)g(z))
′

Hp,q,s(α1)g(z)
+

zp+m
¦

(p+m)Ψ(z) + zΨ
′
(z)
©

1+ zp+mΨ(z)
. (36)

Putting ϕ(z) = zpHp,q,s(α1)g(z), we see that the function ϕ is of the form (13), is analytic in

U , Re{ϕ(z)} > 0 (z ∈ U) and

z(Hp,q,s(α1)g(z))
′

Hp,q,s(α1)g(z)
=

zϕ
′
(z)

ϕ(z)
− p,

so that we find from (36) that

Re

(

−z(Hp,q,s(α1) f (z))
′

Hp,q,s(α1) f (z)

)

≥ p−
�

�

�

�

�

zϕ
′
(z)

ϕ(z)

�

�

�

�

�

−
�

�

�

�

�

zp+m
¦

(p+m)Ψ(z) + zΨ
′
(z)
©

1+ zp+mΨ(z)

�

�

�

�

�

(z ∈ U). (37)



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 153

Now, by using the following known estimates [11] (see also [6]):
�

�

�

�

�

ϕ
′
(z)

ϕ(z)

�

�

�

�

�

≤ 2(p+m)r p+m−1

1− r2(p+m)
(|z| = r < 1)

and
�

�

�

�

�

(p+m)Ψ(z) + zΨ
′
(z)

1+ zp+mΨ(z)

�

�

�

�

�

≤ (p+m)

1− r p+m
(|z| = r < 1)

in (37), we obtain

Re

(

−z(Hp,q,s(α1) f (z))
′

Hp,q,s(α1) f (z)

)

≥ p− 3(p+m)r p+m− (2p+m)r2(p+m)

1− r2(p+m)
(|z|= r < 1) ,

which is certainly positive, provided that r < R0, R0 being given as in Theorem 4.

Theorem 5. Let −1 ≤ B j < A j ≤ 1 ( j = 1,2). If each of the functions f j ∈ Σp satisfies the

following subordination condition:

(1−λ)zpHp,q,s(α1) f j(z) +λzpHp,q,s(α1 + 1) f j(z) ≺
1+A jz

1+ B jz
( j = 1,2; z ∈ U), (38)

then

(1−λ)zpHp,q,s(α1)G(z) +λzpHp,q,s(α1 + 1)G(z)≺ 1+ (1− 2ζ)z

1− z
(z ∈ U), (39)

where

G(z) = Hp,q,s(α1) ( f1 ∗ f2)(z)

and

ζ = 1− 4(A1− B1)(A2− B2)

(1− B1)(1− B2)

�

1− 1

2
2F1(1,1;

α1

λ
+ 1;

1

2
)

�

.

The result is the best possible when B1 = B2 = −1.

Proof. Suppose that each of the functions f j ∈ Σp ( j = 1,2) satisfies the condition (38).

Then, by letting

φ j(z) = (1−λ)zpHp,q,s(α1) f j(z) +λzpHp,q,s(α1 + 1) f j(z) ( j = 1,2), (40)

we have

ϕ j(z) ∈ P(γ j) (γ j =
1− A j

1− B j

; j = 1,2).

Using the identity (10) in (40), we observe that

Hp,q,s(α1) f j(z) =
α1

λ
z−p− α1

λ

z
∫

0

t
α1
λ
−1φ j(t)d t ( j = 1,2),



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 154

which, in view of the definition of G given already with (39), yields

Hp,q,s(α1)G(z) =
α1

λ
z−p− α1

λ

z
∫

0

t
α1
λ
−1ϕ0(t)d t, (41)

where, for convenience,

φ0(z) = (1−λ)zpHp,q,s(α1)G(z) +λzpHp,q,s(α1 + 1)G(z)

=
α1

λ
z−

α1
λ

z
∫

0

t
α1
λ
−1 (ϕ1 ∗ϕ2)(t)d t. (42)

Since ϕ1 ∈ P(γ1) and ϕ2 ∈ P(γ2), it follows from Lemma 3 that

(ϕ1 ∗ϕ2) ∈ P(γ3) (γ3 = 1− 2(1− γ1)(1− γ2)). (43)

Now, by using (43) in (42) and then appealing to Lemma 2 and Lemma 4, we obtain

Re{ϕ0(z)} =
α1

λ

1
∫

0

u
α1
λ
−1 Re{ϕ1 ∗ϕ2}(uz)du

≥ α1

λ

1
∫

0

u
α1
λ
−1(2γ3− 1+

2(1− γ3)

1+ u|z| )du

>
α1

λ

1
∫

0

u
α1
λ
−1(2γ3− 1+

2(1− γ3)

1+ u
)du

= 1− 4(A1− B1)(A2 − B2)

(1− B1)(1− B2)
(1− α1

λ

1
∫

0

u
α1
λ
−1(1+ u)−1du)

= 1− 4(A1− B1)(A2 − B2)

(1− B1)(1− B2)

�

1− 1

2
2F1(1,1;

α1

λ
+ 1;

1

2
)

�

= ζ (z ∈ U).

When B1 = B2 = −1, we consider the functions f j ∈ Σp ( j = 1,2), which satisfy the hypothesis

(38) of Theorem 5 and are defined by

Hp,q,s(α1) f j(z) =
α1

λ
z−

α1
λ

z
∫

0

t
α1
λ
−1(

1+A j t

1− t
)d t ( j = 1,2).



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 155

Thus it follows from (42) and Lemma 4 that

ϕ0(z) =
α1

λ

1
∫

0

u
α1
λ
−1

�

1− (1+ A1)(1+ A2) +
(1+ A1)(1+ A2)

1− uz

�

du

= 1− (1+ A1)(1+A2) + (1+ A1)(1+ A2)(1− z)−1. 2F1(1,1;
α1

λ
+ 1;

z

z − 1
)

→ 1− (1+A1)(1+ A2) +
1

2
(1+ A1)(1+ A2) ·2 F1(1,1;

α1

λ
+ 1;

1

2
) as z→−1,

which evidently completes the proof of Theorem 5.

Putting A j = 1− 2θ j, B j = −1 ( j = 1,2; 0≤ θ j < 1), s = 1, q = 2, α1 = a > 0, β1 = c > 0

and α2 = 1 in Theorem 5, we obtain the following corollary.

Corollary 6. If the functions f j ∈ Σp ( j = 1,2) satisfy the following inequality:

Re
¦

(1+λp)zpℓp(a, c) f j(z) +λzp+1 (ℓp(a, c) f j(z))
′©

> θ j (0≤ θ j < 1; j = 1,2; z ∈ U), (44)

then

Re
¦

(1+λp)zpℓp(a, c)( f1 ∗ f2)(z) +λzp+1(ℓp(a, c)( f1 ∗ f2)(z))
′©
> η0 (z ∈ U),

where

η0 = 1− 4(1− θ1)(1− θ2)

�

1− 1

2
2F1(1,1;

a

λ
+ 1;

1

2
)

�

.

The result is the best possible.

Choosing A j = 1− 2φ j , B j = 1 ( j = 1,2; 0≤ φ j < 1), q = s+ 1, α1 = β1 = p,

α j = 1 ( j = 2,3, . . . , s + 1) and β j = 1 ( j = 2,3, . . . , s) in Theorem 5, we obtain the following

result which refines the work of Yang [19, Theorem 4] and the work of Srivastava and Patel

[15, Corollary 6].

Corollary 7. If the functions f j ∈ Σp ( j = 1,2) satisfy the following inequality:

Re

�

(1+λ)zp f j(z) +
λ

p
zp+1 f

′
j (z)

�

> φ j (0≤ φ j < 1; j = 1,2; z ∈ U), (45)

then

Re

�

(1+λ)zp( f1 ∗ f2)(z) +
λ

p
zp+1( f1 ∗ f2)(z))

′
�

> ρ0 (z ∈ U),

where

ρ0 = 1− 4(1−φ1)(1−φ2)

�

1− 1

2
2F1(1,1;

p

λ
+ 1;

1

2
)

�

.

The result is the best possible.



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 156

Theorem 6. If f ∈ Σp,m satisfies the following subordination condition:

(1−λ)zpHp,q,s(α1) f (z) +λzpHp,q,s(α1 + 1) f (z)≺ 1+ Az

1+ Bz
,

then

Re
¦

zpHp,q,s(α1) f (z)
©

1

d > ξ
1

d (d ∈ N ; z ∈ U),

where ξ is given as in Theorem 1. The result is the best possible.

Proof. Defining the function ϕ by

ϕ(z) = zpHp,q,s(α1) f (z) ( f ∈ Σp,m; z ∈ U), (46)

we see that the function ϕ is of the form (13) and is analytic in U . Differentiating (46) with

respect to z and using the identity (10), we obtain

(1−λ)zpHp,q,s(α1) f (z) +λzpHp,q,s(α1 + 1) f (z) = ϕ(z) +
λ

α1

zϕ
′
(z)≺ 1+ Az

1+ Bz
.

Now, by following the lines of the proof of Theorem 1 mutates mutandis, and using the

elementary inequality:

Re
�

w
1

d

�

≥ (Re w)
1

d (Re(w)> 0; d ∈ N),

we arrive at the result asserted by Theorem 6.

Putting

A=

�

2F1(1,1;
α1

λ(p+m)
+ 1;

1

2
)− 1

�

·
�

2− 2F1(1,1;
α1

λ(p+m)
+ 1;

1

2
)

�−1

,

B = −1, s = 1, q = 2, α1 = a > 0,β1 = c > 0, α2 = 1 and d = 1 in Theorem 6, we obtain the

following corollary.

Corollary 8. If f ∈ Σp,m satisfies the following inequality:

Re
¦

(1+λp)zpℓp(a, c) f (z) +λzp+1(ℓp(a, c) f (z))
′©
>

3− 2 2F1(1,1; a

λ(p+m)
+ 1; 1

2
)

2
h

2− 2F1(1,1; a

λ(p+m)
+ 1; 1

2
)
i z ∈ U),

(47)

then

Re
¦

zpℓp(a, c) f (z)
©

>
1

2
(z ∈ U).

The result is the best possible.

From Corollary 6 and Theorem 6 (for m = −p+ 1, A= 1− 2η0, B = −1 and d = 1), we

obtain the following result.



M. Aouf / Eur. J. Pure Appl. Math, 5 (2012), 141-159 157

Corollary 9. If the function f j ∈ Σp ( j = 1,2) satisfy the inequality (44), then

Re
¦

zpℓp(a, c)( f1 ∗ f2)(z)
©

> η0 + (1−η0)

�

2F1(1,1;
a

λ
+ 1;

1

2
)− 1

�

(z ∈ U),

where η0 is given as in Corollary 6. The result is the best possible.

Putting

A=

�

2F1(1,1;
p

λ(p+m)
+ 1;

1

2
)− 1

�

·
�

2− 2F1(1,1;
p

λ(p+m)
+ 1;

1

2
)

�−1

,

B = −1, q = s+ 1, α1 = β1 = p, α j = 1 ( j = 2,3, . . . , s+ 1), β j = 1 ( j = 2,2, . . . , s) and d = 1

in Theorem 6, we obtain the following result which refines the work of Srivastava and Patel

[15, Corollary 7].

Corollary 10. If f ∈ Σp,m satisfies the following inequality:

Re

�

(1+λ)zp f (z) +
λ

p
zp+1 f

′
(z)

�

>
3− 2 2F1(1,1;

p

λ(p+m)
+ 1; 1

2
)

2
h

2− 2F1(1,1;
p

λ(p+m)
+ 1; 1

2
)
i (z ∈ U), (48)

then

Re
�

zp f (z)
	

>
1

2
(z ∈ U).

The result is the best possible.

From Corollary 7 and Theorem 6 (for m= −p+ 1, A= 1− 2η0, B = −1, d = 1, q = s+ 1,

α1 = β1 = p, α j = 1 ( j = 2,3, . . . , s+ 1) and β j = 1 ( j = 2,3, . . . , s)), we deduce the following

result.

Corollary 11. If the functions f j ∈ Σp ( j = 1,2) satisfy the inequality (45), then

Re
�

zp( f1 ∗ f2)(z)
�

> ρ0 + (1−ρ0)

�

2F1

�

1,1;
p

λ
+ 1;

1

2

�

− 1

�

(z ∈ U),

where ρ0 is given as in Corollary 7. The result is the best possible.

Theorem 7. Let f ∈ Σm
p,q,s(α1; A, B) and let g ∈ Σp,m satisfy the following inequality:

Re
�

zp g(z)
	

>
1

2
(z ∈ U).

Then

( f ∗ g) ∈ Σm
p,q,s(α1; A, B) .



REFERENCES 158

Proof. We have

−zp+1(Hp,q,s(α1)( f ∗ g)(z))
′

p
= −zp+1(Hp,q,s(α1) f (z))

′

p
∗ zp g(z) (z ∈ U).

Since

Re
�

zp g(z)
	

>
1

2
(z ∈ U)

and the function
1+ Az

1+ Bz

is convex (univalent) in U , it follows from (11) and Lemma 5 that ( f ∗ g)(z) ∈ Σm
p,q,s(α1; A, B).

This completes the proof of Theorem 7.

In view of Corollary 10 and Theorem 7, we have Corollary 11 below.

Corollary 12. If f ∈ Σm
p,q,s(α1; A, B) and the function g ∈ Σp,m satisfies the inequality (48), then

( f ∗ g) ∈ Σm
p,q,s(α1; A, B).

ACKNOWLEDGEMENTS The author is thankful to the referee for his comments and sugges-

tions.

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