Ma-Minda φ-classes studied in this paper:
Abstract
Let -1\leq B<A\leq 1. Condition on β, is determined so that 1+βzp'(z)/p^k(z)\prec(1+Az)/(1+Bz)\;(-1<k\leq3) implies p(z)\prec \sqrt{1+z}. Similarly, condition on βis determined so that 1+βzp'(z)/p^n(z) or p(z)+βzp'(z)/p^n(z)\prec\sqrt{1+z}\;(n=0, 1, 2) implies p(z)\prec(1+Az)/(1+Bz) or \sqrt{1+z}. In addition to that condition on βis derived so that p(z)\prec(1+Az)/(1+Bz) when p(z)+βzp'(z)/p(z)\prec\sqrt{1+z}. Few more problems of the similar flavor are also considered.
Results & Lemmas (14)
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Lemma 1.1.
Lemma 1.1. [6, Corollary 3.4h, p.135] Let q be univalent in D, and let ϕ be analytic in a domain D containing q(D). Let zq′(z)ϕ(q(z)) be…
Lemma 1.1. [6, Corollary 3.4h, p.135] Let q be univalent in D, and let ϕ be analytic in a domain D 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. The following is a more general form of the above lemma:
Lemma 1.2.
Lemma 1.2. [6, Corollary 3.4i, p.134] Let q be univalent in D, and ϕ and ν be analytic in a domain D containing q(D) with ϕ(w) ̸= 0 when w…
Lemma 1.2. [6, Corollary 3.4i, p.134] Let q be univalent in D, and ϕ and ν be analytic in a domain D containing q(D) with ϕ(w) ̸= 0 when w ∈q(D). Set Q(z) := zq′(z)ϕ(q(z)), h(z) := ν(q(z)) + Q(z). Suppose that (1) h is convex or Q(z) is starlike univalent in D and (2) Re zh′(z) Q(z) > 0 for z ∈D. If (1.1) ν(p(z)) + zp′(z)ϕ(p(z)) ≺ν(q(z)) + zq′(z)ϕ(q(z)),
Lemma 1.3.
Lemma 1.3. [6, Corollary 3.4a, p.120] Let q be analytic in D, let φ be analytic in a domain D containing q(D) and suppose (1) Re φ[q(z)] >…
Lemma 1.3. [6, Corollary 3.4a, p.120] Let q be analytic in D, let φ be analytic in a domain D containing q(D) and suppose (1) Re φ[q(z)] > 0 and either (2) q is convex, or (3) Q(z) = zq′(z).φ[q(z)] is starlike. If p is analytic in D, with p(0) = q(0), p(D) ⊂D and p(z) + zp′(z)φ[p(z)] ≺q(z), then p(z) ≺q(z). 2. Results Associated with Lemniscate of Bernoulli In the first result condition on β is obtained so that the subordination 1 + β zp′(z) pk(z) ≺1 + Az 1 + Bz (−1 < B < A ≤1) implies p(z) ≺√1 +
Lemma 2.1.
Lemma 2.1. Let |β| ≥2(k+3)/2(A −B) + |Bβ|, −1 < k ≤3. Let p be an analytic function defined on D with p(0) = 1 satisfies 1 + β zp′(z) pk(z)…
Lemma 2.1. Let |β| ≥2(k+3)/2(A −B) + |Bβ|, −1 < k ≤3. Let p be an analytic function defined on D with p(0) = 1 satisfies 1 + β zp′(z) pk(z) ≺1 + Az 1 + Bz (−1 < B < A ≤1), then p(z) ≺√1 + z
Lemma 2.2.
Lemma 2.2. Let (A−B)β ≥ √ 2(1+|B|)2 +(1−B)2 and −1 ≤B < A ≤1. Let p be an analytic function defined on D with p(0) = 1 satisfies 1 + βzp′(z)…
Lemma 2.2. Let (A−B)β ≥ √ 2(1+|B|)2 +(1−B)2 and −1 ≤B < A ≤1. Let p be an analytic function defined on D with p(0) = 1 satisfies 1 + βzp′(z) ≺ √ 1 + z, then p(z) ≺1+Az 1+Bz.
Lemma 2.3.
Lemma 2.3. Let (A −B)β ≥( √ 2 −1)(1 + |A|)(1 + |B|) and −1 ≤B < A ≤1. Let p be an analytic function defined on D with p(0) = 1 satisfies 1 +…
Lemma 2.3. Let (A −B)β ≥( √ 2 −1)(1 + |A|)(1 + |B|) and −1 ≤B < A ≤1. Let p be an analytic function defined on D with p(0) = 1 satisfies 1 + β zp′(z) p(z) ≺ √ 1 + z, then p(z) ≺1+Az 1+Bz.
Lemma 2.4.
Lemma 2.4. Let (A −B)β ≥( √ 2 −1)(1 + |A|)2 + (1 −A)2 and −1 ≤B < A ≤1. Let p be an analytic function defined on D with p(0) = 1 satisfies 1…
Lemma 2.4. Let (A −B)β ≥( √ 2 −1)(1 + |A|)2 + (1 −A)2 and −1 ≤B < A ≤1. Let p be an analytic function defined on D with p(0) = 1 satisfies 1 + β zp′(z) p2(z) ≺ √ 1 + z, then p(z) ≺1+Az 1+Bz.
Lemma 2.5.
Lemma 2.5. Let p be an analytic function defined on D with p(0) = 1 satisfies p(z) + βzp′(z) ≺ √1 + z, β > 0. Then p(z) ≺√1 + z.
Lemma 2.5. Let p be an analytic function defined on D with p(0) = 1 satisfies p(z) + βzp′(z) ≺ √1 + z, β > 0. Then p(z) ≺√1 + z.
Lemma 2.6.
Lemma 2.6. Let p be an analytic function defined on D with p(0) = 1 satisfies p(z) + β zp′(z) p(z) ≺ √ 1 + z, β > 0. Then p(z) ≺√1 + z.
Lemma 2.6. Let p be an analytic function defined on D with p(0) = 1 satisfies p(z) + β zp′(z) p(z) ≺ √ 1 + z, β > 0. Then p(z) ≺√1 + z.
Lemma 2.7.
Lemma 2.7. Let p be an analytic function defined on D with p(0) = 1 satisfies p(z) + β zp′(z) p2(z) ≺ √ 1 + z, β > 0. Then p(z) ≺√1 + z.
Lemma 2.7. Let p be an analytic function defined on D with p(0) = 1 satisfies p(z) + β zp′(z) p2(z) ≺ √ 1 + z, β > 0. Then p(z) ≺√1 + z.
Lemma 2.8.
Lemma 2.8. Let −1 ≤B < A ≤1, (A −B)β ≥ √ 2(1 + |A|)(1 + |B|) + |A|2 −1 and 1 β ≥max 0, A −B (1 + |A|)(1 + |B|) −1 −|B| 1 + |B| . Let p…
Lemma 2.8. Let −1 ≤B < A ≤1, (A −B)β ≥ √ 2(1 + |A|)(1 + |B|) + |A|2 −1 and 1 β ≥max 0, A −B (1 + |A|)(1 + |B|) −1 −|B| 1 + |B| . Let p be an analytic function defined on D with p(0) = 1 satisfies p(z) + β zp′(z) p(z) ≺
Lemma 3.1.
Lemma 3.1. Assume that −1 ≤B < A ≤1, −1 ≤E < D ≤1 and β(A −B) ≥(D −E)(1 + B2) + |2B(D −E) −Eβ(A −B)|. Let p be an analytic function defined…
Lemma 3.1. Assume that −1 ≤B < A ≤1, −1 ≤E < D ≤1 and β(A −B) ≥(D −E)(1 + B2) + |2B(D −E) −Eβ(A −B)|. Let p be an analytic function defined on D with p(0) = 1 satisfies 1 + βzp′(z) ≺1 + Dz 1 + Ez , β ̸= 0. Then p(z) ≺1+Az 1+Bz.
Lemma 3.2.
Lemma 3.2. Assume that −1 ≤B < A ≤1, −1 ≤E < D ≤1 and β(A −B) ≥(D −E)(1 + |AB|) + |(A + B)(D −E) −Eβ(A −B)|. Let p be an analytic function…
Lemma 3.2. Assume that −1 ≤B < A ≤1, −1 ≤E < D ≤1 and β(A −B) ≥(D −E)(1 + |AB|) + |(A + B)(D −E) −Eβ(A −B)|. Let p be an analytic function defined on D with p(0) = 1 satisfies 1 + β zp′(z) p(z) ≺1 + Dz 1 + Ez , β ̸= 0. Then p(z) ≺1+Az 1+Bz.
Lemma 3.3.
Lemma 3.3. Assume that −1 ≤B < A ≤1, −1 ≤E < D ≤1 and |β|(A −B) ≥(D −E)(1 + A2) + |2A(D −E) −Eβ(A −B)|. Let p be an analytic function…
Lemma 3.3. Assume that −1 ≤B < A ≤1, −1 ≤E < D ≤1 and |β|(A −B) ≥(D −E)(1 + A2) + |2A(D −E) −Eβ(A −B)|. Let p be an analytic function defined on D with p(0) = 1 satisfies 1 + β zp′(z) p2(z) ≺1 + Dz 1 + Ez . Then p(z) ≺1+Az 1+Bz.
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