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Abstract
The present work is an attempt to give partial proofs of certain conjectures on the fifth
coefficient of certain normalized analytic functions. Further, bounds on the sixth and seventh
coefficients for the starlike functions related to a lune are also investigated. The non-sharp bound
on third and fourth Hankel determinants are also obtained.
2010 Mathematics Subject Classification: 30C45; 30C50; 30C80
Results & Lemmas (10)
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.
Theorem 1.
Theorem 1. Let f(z) = z+a2z2 +a3z3 +··· ∈U(λ). Then |a5| ≤1 2 2+2λ+5λ2 +6λ3 +4λ4. (2.1) We need the following lemma to prove our result.…
Theorem 1. Let f(z) = z+a2z2 +a3z3 +··· ∈U(λ). Then |a5| ≤1 2 2+2λ+5λ2 +6λ3 +4λ4 . (2.1) We need the following lemma to prove our result. Before proceeding further we recall that the class of functions p(z) = 1 + ∑∞ n=1 pnzn with ℜp(z) > 0,z ∈D is de- noted by P and they are related with the Schwarz function w : D →C by means of the relation w(z) = p(z)−1 p(z)+1.
Lemma 1
Lemma 1 ([25], Lemma 2.3, p. 507). Let p ∈P. Then for all n,m ∈N, |µpnpm −pm+n| ≤ 2, 0 ≤µ ≤1; 2|2µ−1|, elsewhere. If 0 < µ < 1, then the…
Lemma 1 ([25], Lemma 2.3, p. 507). Let p ∈P. Then for all n,m ∈N, |µpnpm −pm+n| ≤ 2, 0 ≤µ ≤1; 2|2µ−1|, elsewhere. If 0 < µ < 1, then the inequality is sharp for the function p(z) = (1+zm+n)/(1−zm+n). In the other cases, the inequality is sharp for the function ˆp0.
Theorem 2.
Theorem 2. Let f(z) = z+a2z2 +a3z3 +··· ∈S ∗ q. Then
Theorem 2. Let f(z) = z+a2z2 +a3z3 +··· ∈S ∗ q . Then
Lemma 2
Lemma 2 ([14], Theorem 4(b), p. 678). A function p ∈P if and only if ∞ ∑ j=0 2zj + ∞ ∑ k=1 pkzk+j
Lemma 2 ([14], Theorem 4(b), p. 678). A function p ∈P if and only if ∞ ∑ j=0 2zj + ∞ ∑ k=1 pkzk+j
Lemma 3
Lemma 3 ([20], Proposition 6, p. 7). Let D:= z ∈C: |z| ≦1. Also, for any real numbers a, b and c, let the quantity Y(a,b,c):= maxz∈D …
Lemma 3 ([20], Proposition 6, p. 7). Let D := {z ∈C : |z| ≦1}. Also, for any real numbers a, b and c, let the quantity Y(a,b,c) := maxz∈D |a+bz+cz2|+1−|z|2 . If ac ≧0, then Y(a,b,c) = |a|+|b|+|c|
Lemma 4
Lemma 4 ([10], Lemma 1). Let p(z) = 1+ p1z+ p2z2 + p3z3 +··· ∈P. Then, for any real number µ, µp3 −p3 1 ≤
Lemma 4 ([10], Lemma 1). Let p(z) = 1+ p1z+ p2z2 + p3z3 +··· ∈P. Then, for any real number µ, µp3 −p3 1 ≤
Lemma 5
Lemma 5 ([1], Corollary 1, p. 68). Let p ∈P. Then |p3 −(µ+1)p1p2 +µp3 1| ≤ 2, 0 ≤µ ≤1; 2|µ−1|, elsewhere.
Lemma 5 ([1], Corollary 1, p. 68). Let p ∈P. Then |p3 −(µ+1)p1p2 +µp3 1| ≤ 2, 0 ≤µ ≤1; 2|µ−1|, elsewhere.
Lemma 6
Lemma 6 ([15], Libera and Zlotkiewicz). Let p(z) = 1+ p1z+ p2z2 +··· ∈P with p1 ≥0. Then 2p2 = p2 1 +x(4−p2 1) (3.7) and 4p3 = p3 1…
Lemma 6 ([15], Libera and Zlotkiewicz). Let p(z) = 1+ p1z+ p2z2 +··· ∈P with p1 ≥0. Then 2p2 = p2 1 +x(4−p2 1) (3.7) and 4p3 = p3 1 +2p1(4−p2 1)x−p1(4−p2 1)x2 +2(4−p2 1)(1−|x|2)y (3.8) for some x and y such that |x| ≤1 and |y| ≤1. To prove (3.6), we consider (23/22)p1p3 −p2
Theorem 3.
Theorem 3. Let f(z) = z+a2z2 +a3z3 +··· ∈S ∗ q. Then |H3,1(f)| ≤2 3 ≈0.67.
Theorem 3. Let f(z) = z+a2z2 +a3z3 +··· ∈S ∗ q . Then |H3,1(f)| ≤2 3 ≈0.67.
Theorem 4.
Theorem 4. Let f(z) = z+a2z2 +a3z3 +··· ∈S ∗ q. Then |H4,1(f)| ≤1.93977. ACKNOWLEDGEMENT The first author was supported by the Basic Science…
Theorem 4. Let f(z) = z+a2z2 +a3z3 +··· ∈S ∗ q . Then |H4,1(f)| ≤1.93977. ACKNOWLEDGEMENT The first author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2019R1I1A3A01050861). The part of this work was completed during the second author’s visit to the Department of Applied Mathemat- ics, Pukyong National University (PKNU), Busan, South, Korea. The second author is thankful
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