TITLE MARCH 2012 - Pakistan Academy of Sciences

50 M.K. Aouf et al
Proof. To prove the first inclusion, let
fz MC 1 p,q,s 1 , A 1 ,B 1 ;,;,. Then,
1
from the definition of MC ( , A , B ; , ; , ),
p, q, s 1 1 1
there exists a function gz MS 1 p,q,s 1 ,A 1 ,B 1 ;;
such that
1
p
Let
z p,q,s
1 1 ,A 1 ,B 1 fz
1 p,q,s 1 ,A 1 ,B 1 gz
z( , A , B f ( z))
q z z U
z.
1
p, q, s 1 1 1
( )
(
p
p, q, s
1, A1 , B1
g( z)
),
(2.3)
where qz is analytic function in U with
q(0) 1 . Using (1.10), we have
p, q, s 1 1 1
p, q, s 1 1 1
[ ( p ) q( z) ] , A , B g( z)
( + p) , A , B f ( z)
, , ( ). (2.4)
1
p, q, s 1
A1 B1
f z
Differentiating (2.4) with respect to
multiplying by z , we obtain
p, q, s 1 1 1
( p ) zq( z) , A , B g( z)
p, q, s 1 1 1
[ ( p ) q( z) ] z( , A , B g( z))
1
p, q, s 1 1 1
z( , A , B f ( z))
p, q, s 1 1 1
( + p) z( , A , B f ( z)) .
Since
1 1 1
z and
g( z) MS ( , A , B ; ; ) MS
1
p, q, s 1 1 1 p, q,
s
( , A, B; ; ), by Theorem 1, we set
1
p, q, s 1 1 1
( z)
,
p
p, q, s
1, A1 , B1
g( z)
z( , A , B g( z))
(2.5)
where ( z) ( z)
in U with the assumption
S . Then, by using (2.3), (2.4) and (2.5), we
have
1
1
z( p, q, s
1, A1 , B1
f ( z))
1
p
p, q, s
1, A1 , B1
g( z)
zq ( z)
q( z) ( z).
( p ) ( z)
p
(2.6)
Since 0 and ( z) ( z)
in U
p
with maxRe{ ( z)} , then
zU
p
Rep z p 0 z U.
Hence, by taking
z 1
p z p ,
in (2.6) and applying Lemma 2, we have
q( z) ( z)
in U , so that
f ( z) MC ( , A , B ; , ; , ). The second
p, q, s 1 1 1
inclusion can be proved by using arguments
similar to those detailed above with (1.9). This
compelets the proof of Theorem 3.
Theorem 4. Let , S with
1
1
p
maxRe{ ( )} min , A p
z
zU
1
( , A
0, 0 , p).
1
Then
MC
1
p, q, s 1 A1 B1
( , , ; , ; , )
p, q, s 1 A1 B1
p, q, s 1 A1 B1
p
MC ( , , ; , ; , )
MC ( 1, , ; , ; , ).
p
Proof. Just as we derived Theorem 2 as a
consequence of Theorem 1 by using the
equivalence (1.11), we can also prove Theorem 4
by using Theorem 3 in conjunction with the
equivalence (1.12).
3. PROPERTIES FOR THE INTEGRAL
OPERATOR
F , p
Let F , p
be the integral operator defined by (see
[14] and [27]):
z
p1
,
p
( )( )
( )
p
z
0
F f z t f t dt
p k
( z z ) f ( z)
k
p
k
1 p
(3.1)