🧭 New here?
Take a guided tour of the site.
← Back to Papers

Results & Lemmas (8)

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.

Lemma 1 Lemma 1 If p ∈P is od the form (12), then c1 = 2ζ1, (13) c2 = 2ζ2 1 + 2(1 −|ζ1|2)ζ2 (14) and c3 = 2ζ3 1 + 2(1 −|ζ1|2)(2ζ1 −ζ1ζ2)ζ2 + 2(1…
Lemma 1 If p ∈P is od the form (12), then c1 = 2ζ1, (13) c2 = 2ζ2 1 + 2(1 −|ζ1|2)ζ2 (14) and c3 = 2ζ3 1 + 2(1 −|ζ1|2)(2ζ1 −ζ1ζ2)ζ2 + 2(1 −|ζ1|2)(1 −|ζ2|2)ζ3 (15) for some ζ1, ζ2, ζ3 ∈D := {z ∈C : |z| ≤1}. For ζ1 ∈T, there is a unique function p ∈P with c1 as in (13), namely, p(z) = 1 + ζ1z 1 −ζ1z , z ∈D.
Lemma 2 Lemma 2 ([4]) For real numbers A, B, C, let Y (A, B, C):= max  |A + Bz + Cz2| + 1 −|z|2: z ∈D
Lemma 2 ([4]) For real numbers A, B, C, let Y (A, B, C) := max  |A + Bz + Cz2| + 1 −|z|2 : z ∈D
Theorem 1 Theorem 1 If f ∈S∗, then |Γ1Γ3 −(Γ2)2| ≤13 12. (20) The inequality is sharp.
Theorem 1 If f ∈S∗, then |Γ1Γ3 −(Γ2)2| ≤13 12. (20) The inequality is sharp.
Lemma 2 Lemma 2 for 0 < ζ1 ≤ζ0 1 we get Γ1Γ3 −(Γ2)2 ≤1 3ζ1(1 −ζ2 1) (−|A| + |B| + |C|) = ρ(ζ1), where ρ(t):= 1 12(−24t4 + 8t2 + 3), t ∈[0, 1].…
Lemma 2 for 0 < ζ1 ≤ζ0 1 we get Γ1Γ3 −(Γ2)2 ≤1 3ζ1(1 −ζ2 1) (−|A| + |B| + |C|) = ρ(ζ1), where ρ(t) := 1 12(−24t4 + 8t2 + 3), t ∈[0, 1]. Since ρ′(t) = 0 for t ∈(0, 1) holds only for t0 = √ 6/6 > ζ0 1, we see that the function ρ is increasing in [0, ζ0 1] and therefore
Theorem 2 Theorem 2 If f ∈C, then Γ1Γ3 −(Γ2)2 ≤1 33. (26) The inequality is sharp.
Theorem 2 If f ∈C, then Γ1Γ3 −(Γ2)2 ≤1 33. (26) The inequality is sharp.
Lemma 2 Lemma 2 for 0 < ζ1 < ζ0 1 we get Γ1Γ3 −(Γ2)2 ≤1 24ζ1(1 −ζ2 1) (−|A| + |B| + |C|) = ρ(ζ1), where ρ(t):= 1 144(−11t4 + 4t2 + 4), t ∈[0, 1].
Lemma 2 for 0 < ζ1 < ζ0 1 we get Γ1Γ3 −(Γ2)2 ≤1 24ζ1(1 −ζ2 1) (−|A| + |B| + |C|) = ρ(ζ1), where ρ(t) := 1 144(−11t4 + 4t2 + 4), t ∈[0, 1].
Theorem 3 Theorem 3 If f ∈P′, then Γ1Γ3 −(Γ2)2 ≤17 144. (32) The inequality is sharp.
Theorem 3 If f ∈P′, then Γ1Γ3 −(Γ2)2 ≤17 144. (32) The inequality is sharp.
Theorem 4 Theorem 4 If f ∈T, then Γ1Γ3 −(Γ2)2 ≤7 3. (38) The inequality is sharp.
Theorem 4 If f ∈T , then Γ1Γ3 −(Γ2)2 ≤7 3. (38) The inequality is sharp.

Related Papers

An evolutionary approach to the coefficient problems in the class of starlike fu
2023
Successive Logarithmic Coefficients of Univalent Functions
2023
The sharp bound of the third Hankel determinant for Convex functions of order -1
2023
Sharp Bounds of the Hermitian Toeplitz Determinants for Certain Close-to-Star Fu
2022
Sharp inequalities for Hermitian Toeplitz determinants for strongly starlike and
2021
↑↓ navigate openesc close
✦ You're explorer #5,037 to wander the registry - thanks for stopping by. Tell us what you'd like to see →
💬 Feedback