TITLE MARCH 2012 - Pakistan Academy of Sciences

56 M.K. Aouf et al
and
1
Az
K
pa
b,
c;
K
1
Bz
a,
b,
c;
A,
B ( 1
B A 1).
,
,
,
p
Inclusion properties was investigated by
several authors (e.g. see [18], [19], [20] and [21]).
In this paper, we investigate several inclusion
properties of the classes ( a,
b,
c;
),
S, p
K p
( a,
b,
c;
) and C p
( a,
b,
c;
,
) associated
,
,
with the general integral operator ( a,
b,
)
I , p
c .
Some applications involving these and other
families of integral operators also considered.
2 . INCLUSION PROPERTIES INVOLVING
I ,
p
To establish our main results, we shall need the
following lemmas.
Lemma 1 [22]. Let h be convex univalent in
U with h ( 0) 1
and
Re
h(
z)
0 ( ,
C)
.
If q (z)
is analytic in U with q ( 0) 1, then
zq(
z)
q( z)
h(
z)
( z U)
q(
z)
implies that q( z)
h(
z)
( z U)
.
Lemma 2 [23]. Let h be convex in U with
h ( 0) 1. Suppose also that Q (z)
is analytic in U
with ReQ
( z)
0 ( zU)
. If q (z)
is analytic in
U with q ( 0) 1, then
q( z)
Q(
z)
zq(
z)
h(
z)
( zU)
implies that q( z)
h(
z)
( z U)
.
Theorem 1. Let p,
a p and p
. Then
, p
,
p
S ( a 1, b, c; ) S ( a, b, c; )
1,
p
S ( a, b, c; ) ( S).
Proof. First of all, we show that
,
p
S ( a 1, b, c; )
,
p
S ( a, b, c; ) ( S; p; a p; p N).
Let f ( z)
S
, p(
a 1,
b,
c;
)
and set
z I
pI
,
p
,
p
( a,
b,
c)
f ( z)
( a,
b,
c)
f ( z)
q(
z)
,
(2.1)
2
where q ( z)
1
q1z
q2z
...
is analytic in U
and q ( z)
0 for all z U . Using the identity
(1.14) in (2.1), we obtain
I,
p(
a 1,
b,
c)
f ( z)
a
pq(
z)
a p .
I ( a,
b,
c)
f ( z)
,
p
(2.2)
Differentiating (2.2) logarithmically with
respect to z , we have
I
( a 1,
b,
c)
f ( z)
zI
( a,
b,
c)
f ( z)
z
I
,
p
,
p
( a 1,
b,
c)
f ( z)
Since
I
,
p
,
p
( a,
b,
c)
f ( z)
zq(
z)
q(
z)
.
pq(
z)
a p
zq(
z)
pq(
z)
a p
(2.3)
a p, ( z)
S
, and f ( z)
S
, p(
a 1,
b,
c;
)
,
from (2.3) we see that
Re
p
( z)
a p 0 ( zU)
and
zq(
z)
q( z)
(
z)
( z U)
pq(
z)
a p
Thus, by using Lemma 1 and (2.1), we
observe that
q( z)
(
z)
( zU) ,
so that
f ( z)
S
,
( a,
b,
c;
) .
p
This implies that
S, p( a 1, b,
c;
)
S,
p(
a,
b,
c;
) .
To prove the second part, let
f ( z) S ( a, b, c; ) ( p; a p; p ) and
put
z I
pI
1,
p
1,
p
,
p
( a,
b,
c)
f ( z)
( a,
b,
c)
f ( z)
g(
z)
,
2
where g ( z)
1
d z d z ... is analytic in U
1 2