On some starlike mappings involving certain convolution operators

Lemma 1 ([1, 5]). Let u = u1 + u2i, v = v1 + v2i and ψ a complex-valued function satisfying: (a) ψ(u, v) is continuous in a domain Ωof C2, (b) (1, 0) ∈Ωand Re ψ(1, 0) > 0 and (c) Re ψ(u2i, v1) ≤0 when (u2i, v1) ∈Ω and 2v1 ≤−(1+u2 2). If p(z) = 1+c1z +c2z2 +... satisfies (p(z), zp′(z)) ∈Ω and Re ψ(p(z), zp′(z)) > 0 for z ∈E, then Re p(z) > 0 in E.

Lemma 2 ([4]). Let η and µ be complex constants and h(z) a convex uni- valent function in E satisfying h(0) = 1, and Re (ηh(z) + µ) > 0. Suppose p ∈P satisfies the differential subordination: (5) p(z) + zp′(z) ηp(z) + µ ≺h(z), z ∈E. If the differential equation: (6) q(z) + zq′(z) ηq(z) + µ = h(z), q(0) = 1 has univalent solution q(z) in E, then p(z) ≺q(z) ≺h(z) and q(z) is the

Lemma 3 ([6]). Let η ̸= 0 and µ be complex constants, and h(z) regular in E with h′(0) ̸= 0, then the solution q(z) of (6) (given by (7)) is univalent in E if (i) Re {G(z) = ηh(z) + µ} > 0 and (ii) Q(z) = zG′(z)/G(z) and R(z) = Q(z)/G(z) are both starlike in E. 3. Main results

Theorem 1. Let σ ≥n + 1 and h(z) a convex univalent function in E satisfying h(0) = 1, and Re (σ −(n + 1) + h(z)) > 0, z ∈E. Let f ∈A. If Lσ n+2f(z) Lσ n+1f(z) ≺h(z), then Lσ n+1f(z) Lσnf(z) ≺h(z).

Theorem 2. Let σ ≥n + 1 and h(z) a convex univalent function in E satisfying h(0) = 1, and Re (σ −(n + 1) + h(z)) > 0, z ∈E. Let f ∈A. If f ∈Sσ n+1, then Lσ n+1f(z) Lσnf(z) ≺q(z) where (9) q(z) = 1 + P∞ k=1 σ−n σ−n+k(k + 1)2zk

Theorem 3. Sσ n+1 ⊂Sσ n, n ∈N.

Lemma 1, we have Re z(Lσ n+1f(z))′ Lσ n+1f(z) > 0 implies Re z(Lσ nf(z))′ Lσnf(z) > 0. By Remark 1(b) Re Lσ n+1f(z) Lσnf(z) > σ −(n + 1) σ −n

Corollary 1. All functions in Sσ n are starlike univalent in E. Following from the inclusion relations, setting σ = 2 and n = 1, we have the following important univalence condition.

Corollary 2. Let f ∈A satisfy Re 2zf ′(z) + z2f ′′(z) f(z) + zf ′(z) > 0, z ∈E. Then f(z) is starlike univalent in E. Example 1. The functions fj(z) j = 1, 2, 3, 4., given by f1(z) = 2[1 −(1 −z)ez] z , f2(z) = 2[1 −(1 + z)e−z] z , f3(z) = −2[z + log(1 −z)] z

Theorem 4. Functions in Sσ n have integral representation: f(z) = lσ n  z. exp Z z 0 p(t) −1 t dt  for some p ∈P.

Theorem 5. The class Sσ n is closed under F.

Theorem 2, ψ satisfies all the conditions of Lemma 1, hence z(Lσ nf(z))′ Lσnf(z) > 0 implies Re z(Lσ nF (z))′ LσnF (z) > 0 and by Remark 1(b) we have Re Lσ n+1F(z) LσnF(z) > σ −(n + 1) σ −n as required. □

Theorem 6. Let f ∈Sσ n. Then we have the inequalities |ak| ≤ σ! (σ + k −1)! (σ + k −1 −n)! (σ −n)! k, k ≥2. The function kσ n(z), given by (10), show that the inequalities are sharp.