Ma-Minda φ-classes studied in this paper:
Abstract
Let p(z) be an analytic function defined on the open unit disk D and p(0) = 1.
Condition β in terms of complex numbers D and real E with −1 < E < 1 and |D| ≤1
is determined such that 1 + βzp′(z) ≺
1+Dz
1+Ez implies p(z) ≺√1 + z. Furthermore, the
expression 1 + βzp′(z)
p(z)
and 1 + βzp′(z)
p2(z)
are considered in obtaining similar results.
Results & Lemmas (10)
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Lemma 2.1
Lemma 2.1([2], p. 135. Let q be univalent in D and let ϕ be analytic in a domain containing q(D). Let zq′(z)ϕ[q(z)] be starlike. If p is…
Lemma 2.1([2], p. 135. Let q be univalent in D and let ϕ be analytic in a domain containing q(D) . Let zq′(z)ϕ[q(z)] be starlike. If p is analytic in D, p(0) = q(0) and satisfies zp′(z)ϕ[p(z)] ≺zq′(z)ϕ[q(z)] then p ≺q and q is the best dominant. Our first result is as follows:
Theorem 2.1.
Theorem 2.1. Let p be an analytic function on D and p(0) = 1. Let β ≥β0, β0 = 2 √ 2|D−E| (1−|E|) where −1 < E < 1 and |D| ≤1. If 1 +…
Theorem 2.1. Let p be an analytic function on D and p(0) = 1. Let β ≥β0, β0 = 2 √ 2|D−E| (1−|E|) where −1 < E < 1 and |D| ≤1. If 1 + βzp′(z) ≺1 + Dz 1 + Ez , then p(z) ≺ √ 1 + z .
Lemma 2.1
Lemma 2.1 suggests 1 + βzp′(z) ≺1 + βzq′(z) ⇒p(z) ≺q(z), so to prove our result, it is suffice to show s(z) = 1 + Dz 1 + Ez ≺1 + βzq′(z) = 1…
Lemma 2.1 suggests 1 + βzp′(z) ≺1 + βzq′(z) ⇒p(z) ≺q(z), so to prove our result, it is suffice to show s(z) = 1 + Dz 1 + Ez ≺1 + βzq′(z) = 1 + βz 2√1 + z = h(z). Since s−1(w) = w−1 D−Ew, then s−1[h(z)] = βz 2√1 + z(D −E) −βEz . For z = eiθ, θ ∈[−π, π] ,
Corollary 2.1.
Corollary 2.1. Let β ≥β0, β0 = 2 √ 2|D−E| (1−|E|) where −1 < E < 1, |D| ≤1,and f ∈A. i) If f satisfies the following 1 + β zf ′(z) f(z)
Corollary 2.1. Let β ≥β0, β0 = 2 √ 2|D−E| (1−|E|) where −1 < E < 1, |D| ≤1,and f ∈A. i) If f satisfies the following 1 + β zf ′(z) f(z)
Theorem 2.2.
Theorem 2.2. Let p be an analytic function in D and p(0) = 1. Let β ≥β0, β0 = 4|D−E| (1−|E|), −1 < E < 1 and |D| ≤1. 1 + β zp′(z) p(z) ≺1 +…
Theorem 2.2. Let p be an analytic function in D and p(0) = 1. Let β ≥β0, β0 = 4|D−E| (1−|E|) , −1 < E < 1 and |D| ≤1. 1 + β zp′(z) p(z) ≺1 + Dz 1 + Ez ⇒p(z) ≺ √ 1 + z .
Corollary 2.2.
Corollary 2.2. Let β ≥β0, β0 = 4|D−E| (1−|E|), −1 < E < 1 and |D| ≤1, i) 1 + β " 1 + zf ′′(z) f ′(z) −zf ′(z) f(z) # ≺1 + Dz 1 + Ez ⇒f…
Corollary 2.2. Let β ≥β0, β0 = 4|D−E| (1−|E|) , −1 < E < 1 and |D| ≤1, i) 1 + β " 1 + zf ′′(z) f ′(z) −zf ′(z) f(z) # ≺1 + Dz 1 + Ez ⇒f ∈SL⋆. ii) 1 + β "
Theorem 2.3.
Theorem 2.3. Let β ≥β0, β0 = 4 √ 2|D−E| (1−|E|), −1 < E < 1 and |D| ≤1. 1 + β zp′(z) p2(z) ≺1 + Dz 1 + Ez ⇒p(z) ≺ √ 1 + z.
Theorem 2.3. Let β ≥β0, β0 = 4 √ 2|D−E| (1−|E|) , −1 < E < 1 and |D| ≤1. 1 + β zp′(z) p2(z) ≺1 + Dz 1 + Ez ⇒p(z) ≺ √ 1 + z .
Corollary 2.3.
Corollary 2.3. Let β ≥β0, β0 = 4 √ 2|D−E| (1−|E|), −1 < E < 1, |D| ≤1 and f ∈A, 1 −β + β 1 + zf ′′(z) f ′(z) zf ′(z) f(z) ≺1 + Dz 1 +…
Corollary 2.3. Let β ≥β0, β0 = 4 √ 2|D−E| (1−|E|) , −1 < E < 1, |D| ≤1 and f ∈A , 1 −β + β 1 + zf ′′(z) f ′(z) zf ′(z) f(z) ≺1 + Dz 1 + Ez ⇒f ∈SL⋆.
Theorem 2.4.
Theorem 2.4. Let p be an analytic function in D and p(0) = 1. Let β ≥β0, 0 < α ≤1, β0 = |1+A||1+B||D−E| α|A−B|(1−|E|), −1 < E < 1, |D| ≤1…
Theorem 2.4. Let p be an analytic function in D and p(0) = 1. Let β ≥β0 , 0 < α ≤1, β0 = |1+A||1+B||D−E| α|A−B|(1−|E|) , −1 < E < 1, |D| ≤1 and −1 ≤B < A ≤1. 1 + β zp′(z) p(z) ≺1 + Dz 1 + Ez ⇒p(z) ≺ 1 + Az 1 + Bz α .
Corollary 2.4.
Corollary 2.4. Let β0 = |1+A||1+B||D−E| α|A−B|(1−|E|), −1 < E < 1, |D| ≤1 and −1 ≤B < A ≤1. 1 + β " 1 + zf ′′(z) f ′(z) −zf ′(z) f(z) # ≺1…
Corollary 2.4. Let β0 = |1+A||1+B||D−E| α|A−B|(1−|E|) , −1 < E < 1, |D| ≤1 and −1 ≤B < A ≤1. 1 + β " 1 + zf ′′(z) f ′(z) −zf ′(z) f(z) # ≺1 + Dz 1 + Ez ⇒zf ′(z) f(z) ≺ 1 + Az
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