Proceedings of paspk.org Volume42-4 - Pakistan Academy of ...

Lemma 9 [6]Theorem 8Proof:Let f , ( α )Rk. Then G = f * φ∈ R k( α)where φ is convex in EααLet F ∈ P [ A, B]and φ is convex. Then F* φ∈ P [ A, B].αLet F, P k[ A,B]kk. Then F (z)=P(z) g(z) , where g belongs to R (α ) and P(z),P[A,B]. ItF * φfollows from the Lemma 9 that g*φ , Rk(α ) .Then PAB [ , ]g * φ ∈ .RemarkAs an application of Theorem 8, we have the followingα(1) The family P [ A, B]is invariant under the following operators.kzf ( ξ )F( f ) = dξ= ( f * φ )( z)∫1 1ξ02F ( f ) = f ( ξ) dξ = ( f * φ )( z)2 2z0zf ( ζ) − f ( xζ)F3( f ) = ∫dζ, | x | ≤1 , x ≠1ζ − x ζ0z∫= ( f * φ )( z)3z1+c c 14( ) = ξ − ( ξ) ξ , Re 0c∫>0F f f d cwhere F( f ( z)) = ( f ∗φ)( z)and φ ( i =1,2,3,4 )iiiare convex univalent functions which satisfyk∞1 nφ1( z) = ∑ z =−log(1 −z) ,nn = 1∞2 n − 2[ z + log(1 −z)]φ2( z) = ∑ z =,n + 1zn = 1∞ n1−xn 1 1−xzφ3( z) = ∑ z = log ,| x | ≤1, x ≠1,n(1 −x) 1−x 1−zn = 1∞1+c nφ4( z) = ∑ z ,Rec> 0.n + cn = 117