TITLE MARCH 2012 - Pakistan Academy of Sciences

Inclusion Properties of Certain Operators 57
and g ( z)
0 for all z U . Then, by using
arguments similar to those detailed above with the
identity (1.13), it follows that
g( z)
(
z)
( zU) ,
which implies that f z)
S ( a,
b,
c;
) . Hence
we conclude that
(
1,
p
S, p( a 1,
b,
c;
)
S,
p(
a,
b,
c;
)
S
1,
p(
a,
b,
c;
)
,
which completes the proof of Theorem 1.
1z Putting n, c a,
b p 1
and ( z)
1 z
( zU)
in
Theorem 1, we obtain the following corollary.
Corollary 1. Let
S
S
n p1
n
p
.
n p
and p
. Then
Remark 1. Putting p 1
in Corollary 1, we
obtain the result obtained by Noor [15].
Theorem 2. Let p,
a p and p
. Then
C ( a 1, b, c; ) C ( a, b, c; )
, p
,
p
C ( a, b, c; ) ( S).
1,
p
Proof. Applying (1.15) and Theorem 1, we
observe that
f ( z) C ,
p
( a 1, b, c; )
I
,
p
( a 1, b, c) f ( z) K p( )
z ( I
,
( a 1, b , c ) f ( z )) S
p p
p( )
zf
()
z
I
, p
( a 1, b, c) p S
p( )
zf ()
z
p S
,
p
( a 1, b, c; )
zf ()
z
p S
,
p
( a, b, c; )
zf
()
z
I
,
( a, b, c) S
p p p ( )
z
p I
, p
( a , b , c ) f ( z ) S
p( )
I
,
p
( a, b, c) f ( z) K p( )
f ( z) C ,
p
( a, b, c; )
and
f ( z) K ( a, b, c; )
1,
p
,
p
zf ()
z
p
zf ()
z
p
1,
p
,
p
S ( a, b, c; )
1,
p
S ( a, b, c; )
z
I ( a, b, c) f ( z) S ( )
p
I ( a, b, c) f ( z) K ( )
f ( z) K ( a, b, c; ),
1,
p
which evidently proves Theorem 2.
Taking
1
Az
( z)
( 1
B A 1;
z U)
1
Bz
in Theorem 1 and 2, we have
p
Corollary 2. Let p , a p,
p
and
1
B A 1.
Then
1, , ; , p , , ; ,
a, b, c; A,
B
, p
,
S a b c A B S a b c A B
S
1,
p
and
K a 1, b, c; A, B K a, b, c; A,
B
p
a, b, c; A, B.
, p
,
K
1,
p
Theorem 3. Let p,
a p and p
. Then
C ( a 1, b, c; , ) C ( a, b, c; , )
, p
,
p
C ( a, b, c; , ) ( , S).
1,
p
Proof. We begin by proving that
C ( a 1, b, c; , ) C ( a, b, c; , )
, p
,
p
( p; a p; p ; , S).
Let f ( z)
C
, p(
a 1,
b,
c;
,
) . Then, in view of
(1.7), there exists a function h ( z)
S ( )
such
that
z I
(
,
p
( a 1,
b,
c)
f ( z))
( z)
ph(
z)
p
p
( z U) .
Choose the function g(z)
such that