Abstract
For analytic functions f in the unit disk D normalized by f (0) = 0 and f ′(0) = 1 satisfying
in D respectively the conditions Re{(1 −z) f ′(z)} > 0, Re{(1 −z2) f ′(z)} > 0, Re{(1 −
z + z2) f ′(z)} > 0, Re{(1 −z)2 f ′(z)} > 0, the sharp upper bound of the third logarithmic
coefficient in case when f ′′(0) is real was computed.
Results & Lemmas (7)
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 ∈A be of the form (1). Then 1. if f ∈F1 and 1 ≤a2 ≤3/2, then |γ3| ≤ 1 288(11 + 15 √ 30) = 0.323466...; 123
Theorem 1 Let f ∈A be of the form (1). Then 1. if f ∈F1 and 1 ≤a2 ≤3/2, then |γ3| ≤ 1 288(11 + 15 √ 30) = 0.323466 . . . ; 123
Lemma 1
Lemma 1 If p ∈P is of the form (8), then c1 = 2ζ1, (9) c2 = 2ζ 2 1 + 2(1 −|ζ1|2)ζ2 (10) and c3 = 2ζ 3 1 + 4(1 −|ζ1|2)ζ1ζ2 −2(1 −|ζ1|2)ζ1ζ 2…
Lemma 1 If p ∈P is of the form (8), then c1 = 2ζ1, (9) c2 = 2ζ 2 1 + 2(1 −|ζ1|2)ζ2 (10) and c3 = 2ζ 3 1 + 4(1 −|ζ1|2)ζ1ζ2 −2(1 −|ζ1|2)ζ1ζ 2 2 + 2(1 −|ζ1|2)(1 −|ζ2|2)ζ3 (11) for some ζi ∈D, i ∈{1, 2, 3}. For ζ1 ∈T, there is a unique function p ∈P with c1 as in (9), namely, p(z) = 1 + ζ1z 1 −ζ1z ,
Lemma 2
Lemma 2 [5] If ac ≥0, then Y(a, b, c) = ⎧ ⎨ ⎩ |a| + |b| + |c|, |b| ≥2(1 −|c|), 1 + |a| + b2 4(1 −|c|), |b| < 2(1 −|c|). If ac < 0, then…
Lemma 2 [5] If ac ≥0, then Y(a, b, c) = ⎧ ⎨ ⎩ |a| + |b| + |c|, |b| ≥2(1 −|c|), 1 + |a| + b2 4(1 −|c|), |b| < 2(1 −|c|). If ac < 0, then Y(a, b, c) = ⎧
Theorem 2
Theorem 2 If f ∈F1 is of the form (1) with a2 ∈R, then |γ3| ≤ 1 288(11 + 15 √ 30) = 0.323466... (15) The inequality is sharp with the…
Theorem 2 If f ∈F1 is of the form (1) with a2 ∈R, then |γ3| ≤ 1 288(11 + 15 √ 30) = 0.323466 . . . (15) The inequality is sharp with the extremal function f (z) = z 0 p(t) 1 −t dt, z ∈D, (16)
Theorem 3
Theorem 3 If f ∈F2 is of the form (1) with a2 ∈R, then |γ3| ≤ 1 972(95 + 23 √ 46) = 0.258223... (32) The inequality is sharp with the…
Theorem 3 If f ∈F2 is of the form (1) with a2 ∈R, then |γ3| ≤ 1 972(95 + 23 √ 46) = 0.258223 . . . (32) The inequality is sharp with the extremal function f (z) = z 0 p(t) 1 −t2 dt, z ∈D, (33)
Theorem 4
Theorem 4 If f ∈F3 is of the form (1) with a2 ∈R, then |γ3| ≤ 1 7776(743 + 131 √ 262) = 0.368238... (44) This result is sharp.
Theorem 4 If f ∈F3 is of the form (1) with a2 ∈R, then |γ3| ≤ 1 7776(743 + 131 √ 262) = 0.368238 . . . (44) This result is sharp.
Theorem 5
Theorem 5 If f ∈F4 is of the form (1) with a2 ∈R, then |γ3| ≤ 1 243(28 + 19 √ 19) = 0.456045... (57) The inequality is sharp with the…
Theorem 5 If f ∈F4 is of the form (1) with a2 ∈R, then |γ3| ≤ 1 243(28 + 19 √ 19) = 0.456045 . . . (57) The inequality is sharp with the extremal function f (z) = z 0 p(t) (1 −t)2 dt, z ∈D, (58)
Function classes studied:
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