Abstract
This paper introduces a new class T γ
α,β,k(η) of analytic functions which
is defined by means of a linear operator involving generalized Mittag-Leffler func-
tion Hγ
α,β,k(f). The results investigated in this paper include, an inclusion rela-
tion for functions in the class T γ
α,β,k(η) and also some subordination results of
the linear operator Hγ
α,β,k(f). Several consequences of our results are also pointed
out.
Mathematics Subject Classification (2010): 33E12, 30C45.
Results & Lemmas (11)
Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.
Lemma 1.2.
Lemma 1.2. ([5]) If p(z) = 1 + p1z + p2z2 + · · · is analytic in U and h(z) is convex function in U with h(0) = 1 and µ is a complex…
Lemma 1.2. ([5]) If p(z) = 1 + p1z + p2z2 + · · · is analytic in U and h(z) is convex function in U with h(0) = 1 and µ is a complex constant such that Reµ > 0, then p(z) + zp′(z) µ ≺h(z), (1.7) implies p(z) ≺q(z) ≺h(z), where q(z) = µ zµ zZ 0 h(t)tµ−1dt, and q(z) is the best dominant.
Lemma 1.3.
Lemma 1.3. ([10]) Let q be a convex function in U and let h(z) = q(z) + αzq′(z), where α > 0. If p(z) = q(0) + p1z + · · · and p(z) +…
Lemma 1.3. ([10]) Let q be a convex function in U and let h(z) = q(z) + αzq′(z), where α > 0. If p(z) = q(0) + p1z + · · · and p(z) + αzp′(z) ≺h(z), then p(z) ≺q(z), and this result is sharp. 2. Inclusion relation We begin by showing the following inclusion relation.
Theorem 2.1.
Theorem 2.1. If η ∈[0, 1), then T γ+1 α,β,k(η) ⊂T γ α,β,k(δ), (2.1) where δ = δ(η, γ, k) = 2η −1 + 2(1 −η)(γ + k) k B γ + k k , (2.2)
Theorem 2.1. If η ∈[0, 1), then T γ+1 α,β,k(η) ⊂T γ α,β,k(δ), (2.1) where δ = δ(η, γ, k) = 2η −1 + 2(1 −η)(γ + k) k B γ + k k , (2.2)
Theorem 3.1.
Theorem 3.1. Let q(z) be convex univalent in U with q(0) = 1 and let h be a function such that h(z) = q(z) + k γ + k zq′(z). (3.1) If f ∈A…
Theorem 3.1. Let q(z) be convex univalent in U with q(0) = 1 and let h be a function such that h(z) = q(z) + k γ + k zq′(z). (3.1) If f ∈A and verifies the differential subordination Hγ+1 α,β,k(f)(z) ′ ≺h(z), (3.2) then
Theorem 3.2.
Theorem 3.2. Let h ∈A with h(0) = 1 and h′(0) ̸= 0, which verifies the inequality Re 1 + zh′′(z) h′(z) > −1 2, (z ∈U). (3.4) If f ∈A and…
Theorem 3.2. Let h ∈A with h(0) = 1 and h′(0) ̸= 0, which verifies the inequality Re 1 + zh′′(z) h′(z) > −1 2, (z ∈U). (3.4) If f ∈A and verifies the differential subordination Hγ+1 α,β,k(f)(z) ′
Theorem 3.3.
Theorem 3.3. Let q(z) be convex univalent in U with q(0) = 1. And let h be a function such that h(z) = q(z) + zq′(z), (z ∈U). (3.7) If f ∈A…
Theorem 3.3. Let q(z) be convex univalent in U with q(0) = 1. And let h be a function such that h(z) = q(z) + zq′(z), (z ∈U). (3.7) If f ∈A and verifies the differential subordination Hγ α,β,k(f)(z) ′ ≺h(z), (3.8) then Hγ α,β,k(f)(z)
Theorem 3.4.
Theorem 3.4. Let h ∈A with h(0) = 1 and h′(0) ̸= 0, which verifies the inequality (3.4). If f ∈A and verifies the differential subordination …
Theorem 3.4. Let h ∈A with h(0) = 1 and h′(0) ̸= 0, which verifies the inequality (3.4). If f ∈A and verifies the differential subordination Hγ α,β,k(f)(z) ′ ≺h(z), (z ∈U), (3.12)
Corollary 3.5.
Corollary 3.5. Let q(z) be convex univalent in U with q(0) = 1 and let h be a function such that h(z) = q(z) + 1 2zq′(z). (3.14) If f ∈A…
Corollary 3.5. Let q(z) be convex univalent in U with q(0) = 1 and let h be a function such that h(z) = q(z) + 1 2zq′(z). (3.14) If f ∈A and verifies the differential subordination f ′(z) + 1 2zf ′′(z) ≺h(z), (3.15) then f ′(z) ≺q(z), (3.16) and the result is sharp.
Corollary 3.6.
Corollary 3.6. Let h ∈A with h(0) = 1 and h′(0) ̸= 0, which verifies the inequality (3.4). If f ∈A and verifies the differential subordination…
Corollary 3.6. Let h ∈A with h(0) = 1 and h′(0) ̸= 0, which verifies the inequality (3.4). If f ∈A and verifies the differential subordination f ′(z) + 1 2zf ′′(z) ≺h(z), (3.17) then f ′(z) ≺q(z), (3.18) where q(z) = 2 z2 zZ 0 h(t)tdt. The function q is convex and is the best dominant.
Corollary 3.7.
Corollary 3.7. Let q(z) be convex univalent in U with q(0) = 1 and let h be a function such that h(z) = q(z) + zq′(z), (z ∈U). (3.19) If f…
Corollary 3.7. Let q(z) be convex univalent in U with q(0) = 1 and let h be a function such that h(z) = q(z) + zq′(z), (z ∈U). (3.19) If f ∈A and verifies the differential subordination f ′(z) ≺h(z), (3.20) then f(z) z ≺q(z), (3.21) and the result is sharp.
Corollary 3.8.
Corollary 3.8. Let h ∈A with h(0) = 1 and h′(0) ̸= 0, which verifies the inequality (3.4). If f ∈A and verifies the differential subordination…
Corollary 3.8. Let h ∈A with h(0) = 1 and h′(0) ̸= 0, which verifies the inequality (3.4). If f ∈A and verifies the differential subordination f ′(z) ≺h(z), (z ∈U), (3.22) then f(z) z ≺q(z), (z ∈U, z ̸= 0), (3.23) where q(z) = 1 z zZ
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