Abstract
The theory of first-order differential subordination developed by Miller and Mocanu was recently extended to functions with fixed initial coefficient by R. M. Ali, S. Nagpal and V. Ravichandran [Second-order differential subordination for analytic functions with fixed initial coefficient, Bull. Malays. Math. Sci. Soc. (2) 34 (2011), 611--629] and applied to obtain several generalization of classical results in geometric function theory. In this paper, further applications of this subordination t
Results & Lemmas (10)
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Theorem 1.1.
Theorem 1.1. [4, Theorem 3.4] Let p ∈Hµ,n with 0 < µ ≤2. Let ψ ∈Ψn,µ with associated domain D. If (p(z), zp′(z)) ∈D and Re ψ(p(z), zp′(z))…
Theorem 1.1. [4, Theorem 3.4] Let p ∈Hµ,n with 0 < µ ≤2. Let ψ ∈Ψn,µ with associated domain D. If (p(z), zp′(z)) ∈D and Re ψ(p(z), zp′(z)) > 0, then Re p(z) > 0 for z ∈D. For α ̸= 1, let S∗(α) := f ∈A : zf ′(z) f(z) ≺1 + (1 −2α)z 1 −z . The function pα(z) := (1 + (1 −2α)z)/(1 −z) maps D onto {w ∈C : Re w > α} for α < 1 and onto {w ∈C : Re w < α} for α > 1. Therefore, for α < 1, S∗(α) is the class of starlike functions of order α consisting of functions f ∈A for which
Theorem 2.1.
Theorem 2.1. Let α ≥0, β ̸= 1, and 0 ≤µ = nb ≤2. Let δ1, δ2, δ3 and δ4 be given by δ1 = −α 2 (1 −β) n + 2 −µ 2 + µ + (1 −α)β + αβ2, δ2…
Theorem 2.1. Let α ≥0, β ̸= 1, and 0 ≤µ = nb ≤2. Let δ1, δ2, δ3 and δ4 be given by δ1 = −α 2 (1 −β) n + 2 −µ 2 + µ + (1 −α)β + αβ2, δ2 = −1 2(1 −β) n + 2 −µ 2 + µ
Theorem 2.2.
Theorem 2.2. Let α ≥0, β ̸= 1, and 0 ≤µ = −(n + 1)b ≤2. Let δ1, δ2, δ3 and δ4 be given by δ1 = α 2 (1 −β) n + 2 −µ 2 + µ + (1 −α)β +…
Theorem 2.2. Let α ≥0, β ̸= 1, and 0 ≤µ = −(n + 1)b ≤2. Let δ1, δ2, δ3 and δ4 be given by δ1 = α 2 (1 −β) n + 2 −µ 2 + µ + (1 −α)β + αβ2, δ2 = 1 2(1 −β) n + 2 −µ 2 + µ
Theorem 2.3.
Theorem 2.3. Let α ≥0, β ̸= 1, and 0 ≤µ = (n + 1)b ≤2. Let δ1, δ2, δ3 and δ4 be given as in Theorem 2.1. If f ∈An,b satisfies one of the…
Theorem 2.3. Let α ≥0, β ̸= 1, and 0 ≤µ = (n + 1)b ≤2. Let δ1, δ2, δ3 and δ4 be given as in Theorem 2.1. If f ∈An,b satisfies one of the following subordinations f ′(z) α zf ′′(z) f ′(z) + f ′(z) −1 + 1 ≺1 + (1 −2δ1)z 1 −z , (2.9) f ′(z) + zf ′′(z) ≺1 + (1 −2δ2)z
Lemma 2.1.
Lemma 2.1. Let α ≥0, β ̸= 1, γ > 0, and 0 ≤µ ≤2. For function p ∈Hµ,n and δ:= −γ 2(1 −β) n + 2 −µ 2 + µ + (1 −α)β + αβ2, if p satisfies…
Lemma 2.1. Let α ≥0, β ̸= 1, γ > 0, and 0 ≤µ ≤2. For function p ∈Hµ,n and δ := −γ 2(1 −β) n + 2 −µ 2 + µ + (1 −α)β + αβ2, if p satisfies (1 −α)p(z) + αp2(z) + γzp′(z) ≺1 + (1 −2δ)z 1 −z (2.13) then p(z) ≺1 + (1 −2β)z 1 −z
Lemma 2.2.
Lemma 2.2. Let β ̸= 1, γ > 0, and 0 ≤µ ≤2. For function p ∈Hµ,n and δ:= −γ 2(1 −β) n + 2 −µ 2 + µ + β if p satisfies p(z) + γzp′(z) ≺1 +…
Lemma 2.2. Let β ̸= 1, γ > 0, and 0 ≤µ ≤2. For function p ∈Hµ,n and δ := −γ 2(1 −β) n + 2 −µ 2 + µ + β if p satisfies p(z) + γzp′(z) ≺1 + (1 −2δ)z 1 −z then p(z) ≺1 + (1 −2β)z 1 −z .
Lemma 2.3.
Lemma 2.3. Let α > 0, β ̸= 1 and 0 ≤µ ≤2. Let δ = −αβ 2(1−β) n + 2−µ 2+µ + β:= δ2, if β ≤1 2,
Lemma 2.3. Let α > 0, β ̸= 1 and 0 ≤µ ≤2. Let δ = −αβ 2(1−β) n + 2−µ 2+µ + β := δ2, if β ≤1 2,
Lemma 2.4.
Lemma 2.4. Let β ̸= 1 and 0 ≤µ ≤2. Let δ = −β 2(1−β) n + 2−µ 2+µ , if β < 1 2,
Lemma 2.4. Let β ̸= 1 and 0 ≤µ ≤2. Let δ = −β 2(1−β) n + 2−µ 2+µ , if β < 1 2,
Lemma 2.5.
Lemma 2.5. Let α > 0, β ̸= 1, and 0 ≤µ ≤2. Let δ = −1 2 (1−β) (αβ+γ) n + 2−µ 2+µ + β, if
Lemma 2.5. Let α > 0, β ̸= 1, and 0 ≤µ ≤2. Let δ = −1 2 (1−β) (αβ+γ) n + 2−µ 2+µ + β, if
Lemma 2.6.
Lemma 2.6. Let β ̸= 1, γ > 0, and 0 ≤µ ≤2. If the function p ∈Hµ,n satisfies p2(z) + γzp′(z) ≺1 + (1 −2δ)z 1 −z (z ∈D) (2.18) where δ:= −γ…
Lemma 2.6. Let β ̸= 1, γ > 0, and 0 ≤µ ≤2. If the function p ∈Hµ,n satisfies p2(z) + γzp′(z) ≺1 + (1 −2δ)z 1 −z (z ∈D) (2.18) where δ := −γ 2(1 −β) n + 2 −µ 2 + µ + β2 then p(z) ≺1 + (1 −2β)z
Function classes studied:
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