ABSTRACT
This problem was solved by R. Singh and S. Singh [ 389 ] in the cases when
A is real and y = 0 or y = 1. V. Selinger [ 371 ] extended these results for
following result.
THEOREM 5.4a. Let I be the integral operator defined by (5.4-1) with
where 2 Fj is the hypergeometric function given in ( 1. 2-12) . If It E C satisfies
(5.4-3) 1 y
290 SPECIAL DIFFERENTIAL SUBORDINATIONS
Proof. Without loss of generality we can assume thatf is analytic and convex
R zF' (z) > Y + 1 _ " . e F(z) - ' 2F1(1;1;y+2;!)
Using identity (1.2-16) we can rewrite this result as
e F(z) ;::: o(y),
If F is not subordinate to f, then according to Lemma 2.2d there exist points
Zo E U and ~0 E JU, and an m ;::: 1, SUCh that
Next we introduce the constant Q and use (5.4-7) and (5.4-6) to obtain
If we set w = F(z0 )jz0 F'(z0 ), then condition (5.4-4) at the point z0 is equivalent to
Using this condition, combined with (5.4-8) and (5.4-3) we deduce
0(0) = 1 = 1 2· 2 FJ.(1,1,2;-1) 2ln2 ·
Using this result in Theorem 5.4a leads to the following corollary.
