DIFFERENTIAL SUBORDINATIONS FOR CLASSES OF MEROMORPHIC p-VALENT FUNCTIONS DEFINED BY MULTIPLIER TRANSFORMATIONS

LEMMA 2.1 [5]. Let the function h be convex (univalent) in U, with h(0) = 1. Suppose also that the function ϕ given by ϕ(z) = 1 + cp+nz p+n + cp+n+1z p+n+1 + · · · (2.1) is analytic in U. Then ϕ(z) + zϕ′(z) δ ≺h(z), Re δ ≥0, δ ̸= 0, implies that ϕ(z) ≺ψ(z) = δ p + n z−δ/(p+n) Z z

LEMMA 2.3 [14]. Let the function ϕ given by (2.3) be in the class P(γ ). Then Re ϕ(z) ≥2γ −1 + 2(1 −γ ) 1 + |z| , z ∈U (0 ≤γ < 1).

LEMMA 2.4 [17]. For 0 ≤γ1 < γ2 < 1, the inclusion P(γ1) ∗P(γ2) ⊂P(γ3) where γ3 = 1 −2(1 −γ1)(1 −γ2), holds and the result is the best possible. The symbol ‘∗’ stands for the previous mentioned Hadamard product of the power series.

LEMMA 2.5 [15]. Let 8 be an analytic function in U, with 8(0) = 1 and Re 8(z) > 1/2, z ∈U. Then, for any function F analytic in U, the set (8 ∗F)(U) is contained in the convex hull of F(U), that is, (8 ∗F)(U) ⊂co F(U).

LEMMA 2.6 [19]. For all real or complex numbers α1, α2, β1, where β1 /∈Z− 0 = {0, −1, −2, . . .}, Z 1 0 tα2−1(1 −t)β1−α2−1(1 −zt)−α1 dt = 0(α2)0(β1 −α2) 0(β1) 2F1(α1, α2, β1; z) for Re β1 > Re α2 > 0, (2.4) 2F1(α1, α2, β1; z) = 2F1(α2, α1, β1; z), (2.5) 2F1(α1, α2, β1; z) = (1 −z)−α1 2F1 

THEOREM 3.1. If the function f ∈P p,n satisfies the subordination condition − (1 −β)z p+1(I m p (n; λ, l) f (z))′ + βz p+1(I m+1 p (n; λ, l) f (z))′ p ≺1 + Az 1 + Bz , then − z p+1(I m p (n; λ, l) f (z))′ p

COROLLARY 3.3. The inclusions Rm+1,p(A, B) ⊂Rm,p(A, B) ⊂Rm,p(1 −2ρ, −1) hold, where ρ =    A B +  1 −A B  (1 −B)−12F1 

Theorem 3.1, and using (2.7), we get the following result.

COROLLARY 3.4. If the function f ∈P p,−p+2 satisfies the inequality Re{−z p+1[(p + 2) f ′(z) + zf ′′(z)]} > α, z ∈U (0 ≤α < p), then Re[−z p+1 f ′(z)] > α + (p −α) π 2 −1  , z ∈U, and the result is the best possible. REMARK 3.5. Taking α = −p(π −2)/(4 −π) in the above corollary, we obtain that if the function f ∈P p,−p+2 satisfies

THEOREM 3.6. If the function f ∈Pm p,n(λ, l; α), 0 ≤α < p, then Re{−z p+1[(1 −β)(I m p (n; λ, l) f (z))′ + β(I m+1 p (n; λ, l) f (z))′]} > α, for |z| < R, where R = s 1 + βλ l 2 (p + n)2 −βλ l (p + n)

COROLLARY 3.8. If the function f ∈Pm p,n(λ, l; α), 0 ≤α < p, then f ∈Pm+1 p,n (λ, l; α) for |z| < eR, where eR = s 1 + λ l 2

THEOREM 3.9. Let f ∈Pm p,n(λ, l; A, B), and let Fp,c( f )(z) = c zc+p Z z 0 tc+p−1 f (t) dt, c > 0. (3.8) Then − z p+1(I m p (n; λ, l)Fp,c( f )(z))′ p

Theorem 2]. If c > 0 and f ∈Rm,p(A, B), then Fp,c(Rm,p(A, B)) ⊂Rm,p(1 −2ζ, −1) ⊂Rm,p(A, B), where ζ =    A B +  1 −A B  (1 −B)−12F1 

COROLLARY 3.11. If c > 0 and if f ∈P p,n satisfies the inequality Re[−z p+1 f ′(z)] > α, z ∈U (0 ≤α < p), then Re  −c zc Z z 0 tc+p f ′(t) dt  > α + (p −α) 

THEOREM 3.12. Let the function f ∈P p,n, and suppose that g ∈P p,n satisfies the inequality Re[z pI m p (n; λ, l)g(z)] > 0, z ∈U. If

THEOREM 3.13. Let −1 ≤Bi < Ai ≤1, i = 1, 2, and suppose that each of the functions fi ∈P p satisfies the subordination condition (1 −β)z pI m p (λ, l) fi(z) + βz pI m+1 p (λ, l) fi(z) ≺1 + Aiz 1 + Biz , i = 1, 2, (3.17)

COROLLARY 3.14. If the functions fi ∈P p, i = 1, 2, satisfy the inequality Re{(1 + βp)z p fi(z) + βz p+1 f ′ i (z)} > αi, z ∈U (0 ≤αi < 1, i = 1, 2), (3.21) then Re{(1 + βp)z p( f1 ∗f2)(z) + βz p+1( f1 ∗f2)(z)} > η0, z ∈U, where η0 = 1 −4(1 −α1)(1 −α2)  1 −1 2 2F1 

THEOREM 3.15. If the function f ∈P p,n satisfies the subordination condition (1 −β)z pI m p (n; λ, l) f (z) + βz pI m+1 p (n; λ, l) f (z) ≺1 + Az 1 + Bz , then Re[z pI m p (n; λ, l) f (z)]1/q > ρ1/q, z ∈U (q ∈N), where ρ = Q(−1) is given as in Theorem 3.1. The result is the best possible.

COROLLARY 3.16. Let the functions fi ∈P p (i = 1, 2), satisfy inequality (3.21). Then Re[z p( f1 ∗f2)(z)] > η0 + (1 −η0)  2F1  1, 1, 1 β + 1; 1 2  −1  , z ∈U,

THEOREM 3.17. If the function g ∈P p,n satisfies the inequality Re[z pg(z)] > 1 2, z ∈U, (3.23) then, for any function f ∈Pm p,n(λ, l, A; B), we have f ∗g ∈ m X p,n (λ, l; A, B). PROOF. It is easy to check that −