DIFFERENTIAL SUBORDINATION IMPLICATIONS FOR CERTAIN

Lemma 2.1. [17, Theorem 3.4h, p.132] Let q : D →C be analytic, and ψ and v be analytic in a domain U ⊇q(D) with ψ(w) ̸= 0 whenever w ∈q(D). Set Q(z) := zq′(z)ψ(q(z)) and h(z) := v(q(z)) + Q(z), z ∈D. Suppose that (i) either h(z) is convex, or Q(z) is starlike univalent in D and (ii) Re  zh′(z) Q(z)  > 0, z ∈D. If p is analytic in D, with p(0) = q(0), p(D) ⊂U and v(p(z)) + zp′(z)ψ(p(z)) ≺v(q(z)) + zq′(z)ψ(q(z))

Theorem 2.2. Let Q(z) ∈P be defined by (1.1) and further L = Z 0 −1 eet−1 −1 t dt and U = Z 1 0 eet−1 −1 t dt. (2.1)

Corollary 2.3. Set M(z) := 1 −zf ′(z)/f(z) + zf ′′(z)/f ′(z). If the function f ∈A satisfies 1 + β zf′(z) f(z) M(z) ≺Q(z), then (a) f ∈S∗ q if β ≥ 1/ √ 2
U, (b) f ∈S∗ B if β ≥  1/(1 −e(e−1−1)) 

Theorem 2.4. Let U and L be given by (2.1) and Q(z) be given by (1.1). Let p be an analytic function in D with p(0) = 1. If Λβ(z, p(z)) ≺Q(z), then (a) p(z) ≺φq(z) for β ≥ 1 log(1+ √ 2)U ≈2.40301. (b) p(z) ≺Q(z) for β ≥ 1 e−1U ≈1.23260. (c) p(z) ≺φc(z) for β ≥ 1 log 3U ≈1.92784. (d) p(z) ≺φ0(z) for β ≥ 1

Theorem 2.5. Let U and L be given by (2.1) and Q(z) be given by (1.1). Assume p to be an analytic function in D with p(0) = 1. If Θβ(z, p(z)) ≺Q(z), then each of the following subordination holds: (a) p(z) ≺φq(z) for β ≥ 1 2− √ 2U ≈3.61556. (b) p(z) ≺Q(z) for β ≥ ee−1 ee−1−1U ≈2.58089. (c) p(z) ≺φc(z) for β ≥3 2U ≈3.17692. (d) p(z) ≺φ0(z) for β ≥2U ≈4.2359. (e) p(z) ≺φlim(z) for β ≥5+4

Theorem 2.6. Let φSG be given by (1.2) and further I−= Z 0 −1 et −1 t(et + 1)dt and I+ = Z 1 0 et −1 t(et + 1)dt. (2.4) Assume p to be an analytic function in D with p(0) = 1. If the subordination Ψβ(z, p(z)) ≺φSG(z)

Lemma 2.1, it follows that the subordination 1 + βzp′(z) ≺1 + βzq′ β(z) implies p(z) ≺

Corollary 2.7. Let f ∈A be analytic function which satisfies 1 + β zf ′(z) f(z) M(z) ≺φSG(z). Then, (a) f ∈S∗ q if β ≥ 1/(2 − √ 2)
I−, (b) f ∈S∗ c if β ≥(3/2) I−, (c) f ∈S∗ R if β ≥(3 + 2

Theorem 2.8. Let I+ and I−be given by 2.4 and φSG be given by (1.2). Assume p to be an analytic function in D with p(0) = 1. If Λβ(z, p(z)) ≺φSG(z), then each of the following holds. (a) p(z) ≺φq(z) for β ≥ 1 log(1+ √ 2)I−≈0.55242. (b) p(z) ≺φc(z) for β ≥ 1 log 3I−≈0.443185. (c) p(z) ≺φ0(z) for β ≥ 1 log 

Theorem 2.6, we conclude the result. □

Theorem 2.9. Let I+ and I−be given by (2.4). Assume p to be an analytic function in D with p(0) = 1. If Θβ(z, p(z)) ≺φSG(z), then (a) p(z) ≺φq(z) for β ≥ 1 2− √ 2I+ ≈0.83117. (b) p(z) ≺φc(z) for β ≥3 2I+ ≈0.73033. (c) p(z) ≺φ0(z) for β ≥(2 + 2 √ 2)I−≈2.35090. (d) p(z) ≺Q(z) for β ≥ ee−1 ee−1−1I+ ≈0.59331.

Theorem 2.10. Let p be an analytic function in D with p(0) = 1. Then each of the following subordination implies p(z) ≺Q(z) := eez−1: (a) Ψβ(z, p(z)) ≺φc(z) if β ≥ 1 1−e(e−1−1) ≈2.13430. (b) Λβ(z, p(z)) ≺φc(z) if β ≥ e e−1 ≈1.581976. (c) Θβ(z, p(z)) ≺φc(z) if β ≥ 5ee−1 3(ee−1−1) ≈2.030970. The bounds in each case are sharp.

Corollary 2.11. Let f ∈A be given by f(z) = z + ∞ P n=2 anzn. If one of the following subordinations holds (a) 1 + β zf′(z) f(z) M(z) ≺φc(z) for β ≥ 1 1−e(e−1−1), (b) 1 + βM(z) ≺φc(z) for β ≥ e e−1, (c) 1 + β 

Theorem 2.12. Let p be an analytic function in D with p(0) = 1. Then the following subordinations hold for p(z) ≺φSG(z) := 2/(1 + e−z). (a) Ψβ(z, p(z)) ≺φ0(z) if β ≥(e+1)(1− √ 2−2 log(2− √ 2)) e−1 ≈1.418226. (b) Λβ(z, p(z)) ≺φ0(z) if β ≥1− √ 2−2 log(2− √ 2) 1+log 2−log(1+e) ≈1.725221.

Theorem 2.13. Let p be an analytic function in D which satisfies p(0) = 1. Then each of the following subordination is sufficient for p(z) ≺φSG(z). (a) Ψβ(z, p(z)) ≺φc(z) if β ≥5(e+1) 3(e−1) ≈3.60659. (b) Λβ(z, p(z)) ≺φc(z) if β ≥ 5 3(1+log 2−log(1+e)) ≈4.387286. (c) Θβ(z, p(z)) ≺φc(z) if β ≥ 10e 3(e−1) ≈5.27326. The bounds on β in each case are sharp. References [1] O. P. Ahuja, S. Kumar and V. Ravichandran, Applications of first order differential subordination for functions with positive real part