KYUNGPOOK Math. J. 53(2013), 1-12

Lemma 1.1([4]). Suppose that the function H : C2 →C satisfies the condition: Re{H(is, t)} ≤0, for all real s and t ≤−n(1 + s2)/2, where n is a positive integer. If the function p(z) = 1 + pnzn + · · · is analytic in U and Re{H(p(z), zp′(z))} > 0 (z ∈U), then Re{p(z)} > 0 in U.

Lemma 1.2([5]). Let β, γ ∈C with β ̸= 0 and let h ∈H(U) with h(0) = c. If Re{βh(z) + γ} > 0 for z ∈U, then the solution of the differential equation q(z) + zq′(z) βq(z) + γ = h(z) (z ∈U; q(0) = c) is analytic in U and satisfies Re{βq(z) + γ} > 0 for z ∈U.

Lemma 1.3([6]). Let p ∈Q with p(0) = a and let q(z) = a+anzn +· · · be analytic in U with q(z) ̸≡a and n ≥1. If q is not subordinate to p, then there exist points z0 = r0eiθ ∈U and ζ0 ∈∂U\E(f), for which q(Ur0) ⊂p(U), p(z0) = q(ζ0) and z0p′(z0) = mζ0q′(ζ0) (m ≥n).

Lemma 1.4([7]). Let q ∈H[a, 1], let φ : C2 →C and set φ(q(z), zq′(z)) ≡h(z). If L(z, t) = φ(q(z), tzq′(z)) is a subordination chain and p ∈H[a, 1] ∩Q, then h(z) ≺φ(p(z), zp′(z)). implies that q(z) ≺p(z). Furthermore, if φ(q(z), zp′(z)) = h(z) has a univalent solution q ∈Q, then q is the best subordinant. A function L(z, t) defined on U × [0, ∞) is the subordination chain (or L¨owner chain) if L(·, t) is analytic and univalent in U for all t ∈[0, ∞), L(z, ·) is contin- uously differentiable on [0,

Lemma 1.5([12]). The function L(z, t) = a1(t)z + · · · , with a1(t) ̸= 0 and limt→∞|a1(t)| = ∞, is a subordination chain if and only if Re {z∂L(z, t)/∂z ∂L(z, t)/∂t } > 0 (z ∈U; 0 ≤t < ∞).

Theorem 2.1. Let f, g ∈Apand 0 ≤α < 1, −∞< λ < p. Suppose that (2.1) Re { 1 + zϕ′′(z) ϕ′(z) } > −δ ( z ∈U; ϕ(z) := (1 −α)Ω1+λ,p z g(z) + αΩλ,p z g(z) zp ) ,

Theorem 2.2. Let f, g ∈Ap and 0 ≤α < 1, −∞< λ < p. Suppose that Re { 1 + zϕ′′(z) ϕ′(z) } > −δ ( z ∈U; ϕ(z) := (1 −α)Ω1+λ,p z g(z) + αΩλ,p z g(z) zp ) ,

Lemma 1.4, we conclude that the superordination condition (2.13) must imply the superordination given by (2.15). Furthermore, since the differential equation (2.14) has the univalent solution G, it is the best subordinant of the given differential superordination. Therefore we complete the proof of Theorem 2.2. 2 If we combine this Theorem 2.1 and Theorem 2.2, then we obtain the following sandwich-type theorem.

Theorem 2.3. Let f, gk ∈Ap(k = 1, 2) and 0 ≤α < 1, −∞< λ < p. Suppose that Re { 1 + zϕ′′ k(z) ϕ′ k(z) } > −δ ( z ∈U; ϕk(z) := (1 −α)Ω1+λ,p z gk(z) + αΩλ,p z gk(z) zp

Corollary 2.1. Let f, gk ∈Ap(k = 1, 2) and 0 ≤α < 1, −∞< λ < p. Suppose that the condition (2.16) is satisfied and Re { 1 + zψ′′(z) ψ′(z) } > −δ ( z ∈U; ψk(z) := (1 −α)Ω1+λ,p z f(z) + αΩλ,p z f(z) zp ; f ∈Q

Corollary 2.1. 2 By setting p = 1 and λ = α = 1/2 in Theorem 2.3, so that δ = 1/2, we deduce the following consequence of Theorem 2.3.

Corollary 2.2. Let f, gk ∈A1(k = 1, 2). Suppose that Re { 1 + zϕ′′ k(z) ϕ′ k(z) } > −1 2 ( z ∈U; ϕk(z) := (1/2)Ω3/2,1 z gk(z) + (1/2)Ω1/2,1 z

Theorem 2.4. Let f, gk ∈Ap(k = 1, 2) and 0 ≤α < 1, λ < 0. Suppose that Re { 1 + zϕ′′ k(z) ϕ′ k(z) } > −δ ( z ∈U; ϕk(z) := (1 −α)Ω1+λ,p z gk(z) + αΩλ,p z gk(z) zp−1