A certain q-Ruscheweyh type derivative operator and its applications involving multivalent functions

Lemma 1 (see [25]) Let ϕj ∈P(βj) be given by (1.1) for (0 ≦βj < 1;j = 1,2). Then ϕ1 ∗ϕ2 ∈P(β3), where β3 = 1 – (1 – β1)(1 – β2).

Lemma 2 (see [24]) Let the function ϕ, given by (1.4), be in the class P(β). Then ℜ  ϕ(z)  > 2β – 1 + 2(1 – β) 1 + |z| (0 ≦β < 1).

Lemma 3 (see [29]) The function given by (1 – z)γ = eγ log(1–z) (γ ̸= 0) is univalent in U if and only if γ is either in the closed disk |γ – 1| ≦1 or in the closed disk |γ + 1| ≦1.

Lemma 4 (see [3]) Let the function h(z) be analytic and convex univalent in U with h(0) = 1. Also, let the function g(z) given by g(z) = 1 + b1z + b2z2 + ··· be analytic in U. If g(z) + zDqg(z) c ≺h(z) (z ∈U;c ̸= 0), then, for ℜ(c) ≧0, the following subordination relation holds true: g(z) ≺c zc  z 0 tc–1h(t)dt.

Lemma 5 (see [3]) Let the function u(z) be univalent in U, and let the functions θ(w) and ϕ(w) be analytic in the domain D containing u(U) with ϕ(w) ̸= 0 when w ∈u(U). Set Q(z) = zDq  u(z)  ϕ  u(z)  and h(z) = θ  u(z) + Q(z) 

Theorem 1 Let λ > 0, α > 0, and –1 ≦B ≦A < 1. If f ∈A(p) satisfies the following subor- dination relation: (1 – α)Rλ+p–1 q f (z) zp + α Rλ+p q f (z) zp ≺h(A,B,z), then ℜ Rλ+p–1 q

Theorem 2 Let A = 1 – 2α, B = –1, α,λ > 1, n ≧1, and 0 ≦β < 1. If the function f ∈A(p) satisfies the following subordination condition: (1 – α)Rλ+p–1 q f (z) zp + α Rλ+p q f (z) zp ≺h(1 – 2α,–1,z),

Theorem 3 Let λ > 0 and 0 ≦ρ < 1. Also, let the parameter γ ∈C \ {0} satisfy either 2γ (1 – ρ) [λ + p]q αqλ[p]q – 1  ≦1 or 2γ (1 – ρ) [λ + p]q αqλ[p]q + 1  ≦1. If f ∈A(p) satisfies the following inequality: ℜ  Rλ+p

Theorem 4 Let λ > 0, α < 1, and –1 ≦Bj ≦Aj < 1 (j = 1,2). If each of the functions fj ∈A(p) (j = 1,2) satisfies the following subordination condition: (1 – α)Rλ+p–1 q fj(z) zp + α Rλ+p q fj(z) zp ≺h(Aj,Bj,z) (j = 1,2), then (1 – α)Rλ+p–1 q