Results & Lemmas (14)
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Lemma 1
Lemma 1 (see [60]). Let p(z) ∈𝒫. Then, |ci| ≦2, i ∈ℕ. (19) Also, ⃓⃓ci – μcjci–j ⃓⃓≦2, i > j, μ ∈[0,1]. (20) Sharpness holds true for the…
Lemma 1 (see [60]). Let p(z) ∈𝒫. Then, |ci| ≦2, i ∈ℕ. (19) Also, ⃓⃓ci – μcjci–j ⃓⃓≦2, i > j, μ ∈[0,1]. (20) Sharpness holds true for the function g(z) given by g(z) = 1 + z 1 – z.
Lemma 2
Lemma 2 (See [50]). Let p ∈𝒫. Then, ⃓⃓ci+j – νcicj ⃓⃓≦2, for 0 ≦ν ≦1, (21) ⃓⃓⃓ci+2j – νcic2 j ⃓⃓⃓≦2(1 + 2ν), for ν ∈ℝ (22) and ⃓⃓c2 – νc2 1…
Lemma 2 (See [50]). Let p ∈𝒫. Then, ⃓⃓ci+j – νcicj ⃓⃓≦2, for 0 ≦ν ≦1, (21) ⃓⃓⃓ci+2j – νcic2 j ⃓⃓⃓≦2(1 + 2ν), for ν ∈ℝ (22) and ⃓⃓c2 – νc2 1 ⃓⃓≦2max{1,|2υ – 1| for ν ∈ℂ. (23)
Lemma 3
Lemma 3 (See [50]). Let the function p ∈𝒫. Then, ⃓⃓c3 – 2Tc1c2 + Dc3 1 ⃓⃓≦2, if 0 ≦T ≦1, and T(2T – 1) ≦D ≦T.
Lemma 3 (See [50]). Let the function p ∈𝒫. Then, ⃓⃓c3 – 2Tc1c2 + Dc3 1 ⃓⃓≦2, if 0 ≦T ≦1, and T(2T – 1) ≦D ≦T.
Lemma 4
Lemma 4 (See [61] and [60]). Let the function p ∈𝒫. Then, 2c2 = c2 1 + x(4 – c2 1) and 4c3 = c3 1 + 2(4 – c2 1)c1x – (4 – c2 1)c1x2 + 2(4 –…
Lemma 4 (See [61] and [60]). Let the function p ∈𝒫. Then, 2c2 = c2 1 + x(4 – c2 1) and 4c3 = c3 1 + 2(4 – c2 1)c1x – (4 – c2 1)c1x2 + 2(4 – c2 1)(1 – ⃓⃓x2⃓⃓)z, where x,z ∈ℂ, with |z| ≦1 and |x| ≦1.
Lemma 5
Lemma 5 (See [9]). Let p ∈𝒫, 0 < A < 1, 0 < α < 1 and 8A(1 – A) [︁ (αβ – 2λ)2 + (α (A + α) – β)2]︁ + α (1 – α)(β – 2Aα)2 ≦4α2A(1 – α)2 (1 –…
Lemma 5 (See [9]). Let p ∈𝒫, 0 < A < 1, 0 < α < 1 and 8A(1 – A) [︁ (αβ – 2λ)2 + (α (A + α) – β)2]︁ + α (1 – α)(β – 2Aα)2 ≦4α2A(1 – α)2 (1 – A). (24) Then, ⃓⃓⃓⃓λc4 1 + Ac2 2 + 2αc1c3 – 3 2βc2 1c2 – c4 ⃓⃓⃓⃓≦2. (25)
Theorem 7
Theorem 7 and in Theorem 8. 3 Main results
Theorem 7 and in Theorem 8. 3 Main results
Theorem 1
Theorem 1 Let g of the form (1) be in the class 𝒞G. Then, |d2| ≦1 4, |d3| ≦1 12, |d4| ≦1 24, |d5| ≦1 40. The above results are sharp.
Theorem 1 Let g of the form (1) be in the class 𝒞G. Then, |d2| ≦1 4, |d3| ≦1 12, |d4| ≦1 24, |d5| ≦1 40. The above results are sharp.
Theorem 2
Theorem 2 Let the function g of the form (1) belong to the class 𝒞G. Then, ⃓⃓d3 – γ d2 2 ⃓⃓≦1 12 max ︃ 1, ⃓⃓⃓⃓ 9γ – 2 6 ⃓⃓⃓⃓ ︃. (37) The…
Theorem 2 Let the function g of the form (1) belong to the class 𝒞G. Then, ⃓⃓d3 – γ d2 2 ⃓⃓≦1 12 max {︃ 1, ⃓⃓⃓⃓ 9γ – 2 6 ⃓⃓⃓⃓ }︃ . (37) The function g3 given by (34) provides a sharp result.
Theorem 3
Theorem 3 Let the function g of the form (1) belong to the class 𝒞G. Then, ⃓⃓d3 – d2 2 ⃓⃓≦1 12. (39) The function g3 given by (34) provides…
Theorem 3 Let the function g of the form (1) belong to the class 𝒞G. Then, ⃓⃓d3 – d2 2 ⃓⃓≦1 12. (39) The function g3 given by (34) provides a sharp result.
Theorem 4
Theorem 4 Let the function g of the form (1) be in the class 𝒞G. Then, |d2d3 – d4| ≦1 24. The function g4 given by (35) provides a sharp…
Theorem 4 Let the function g of the form (1) be in the class 𝒞G. Then, |d2d3 – d4| ≦1 24. The function g4 given by (35) provides a sharp result.
Theorem 5
Theorem 5 Let the function g of the form (1) be in the class 𝒞G. Then, ⃓⃓d2d4 – d2 3 ⃓⃓≦ 1 144. The function g3 given by (34) provides a…
Theorem 5 Let the function g of the form (1) be in the class 𝒞G. Then, ⃓⃓d2d4 – d2 3 ⃓⃓≦ 1 144. The function g3 given by (34) provides a sharp result.
Theorem 6
Theorem 6 Let the function g of the form (1) belong to the class CG. Then, ⃓⃓H3,1 (︁ g )︁⃓⃓≦ 1 480 + 1 576 + 1 1724 = 5461 1241280.
Theorem 6 Let the function g of the form (1) belong to the class CG. Then, ⃓⃓H3,1 (︁ g )︁⃓⃓≦ 1 480 + 1 576 + 1 1724 = 5461 1241280.
Theorem 7
Theorem 7 Let the function g of the form (1) belong to the class 𝒞G. Then, |β1| ≦1 8, |β2| ≦1 24, |β3| ≦1 48 (︃ |β4| ≦1 80 )︃.
Theorem 7 Let the function g of the form (1) belong to the class 𝒞G. Then, |β1| ≦1 8, |β2| ≦1 24, |β3| ≦1 48 (︃ |β4| ≦1 80 )︃ .
Theorem 8
Theorem 8 If the function g ∈CG has the following series representation: g–1(w) = w + A2w2 + A3w3 + ···. Then, |A2| ≦1 4 and |A3| ≦1 12.…
Theorem 8 If the function g ∈CG has the following series representation: g–1(w) = w + A2w2 + A3w3 + ··· . Then, |A2| ≦1 4 and |A3| ≦1 12. These bounds are sharp.
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