Results & Lemmas (8)
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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.
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