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Results & Lemmas (11)

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. A function f ∈Σ is said to be a member of the class S∗ a,s (A, B, α, β) if it satisfies the following inequality: ∞ X k=1 [k (1 +…
Theorem 1. A function f ∈Σ is said to be a member of the class S∗ a,s (A, B, α, β) if it satisfies the following inequality: ∞ X k=1 [k (1 + B) + (1 + A)]
Theorem 2. Theorem 2. A function f ∈Σ is said to be a member of the class ΣKa,s (A, B, α, β) if it satisfies the following inequality: ∞ X k=1 k (k (1…
Theorem 2. A function f ∈Σ is said to be a member of the class ΣKa,s (A, B, α, β) if it satisfies the following inequality: ∞ X k=1 k (k (1 + B) + (1 + A))
Theorem 3. Theorem 3. If f of the form (1) satisfies (13), then ℜ  Ls a(α, β)f(z) Ls a(α, β)fm(z)  ≧ 2(m + 1 + A) 2m + 2 + A + B
Theorem 3. If f of the form (1) satisfies (13), then ℜ  Ls a(α, β)f(z) Ls a(α, β)fm(z)  ≧ 2(m + 1 + A) 2m + 2 + A + B
Theorem 4. Theorem 4. If f of the form (1) satisfies (16), then ℜ  Ls a(α, β)f(z) Ls a(α, β)fm(z)  ≧ (2m + 2) (m + 1) + m (A + B) + 2A (m + 1) (2m +…
Theorem 4. If f of the form (1) satisfies (16), then ℜ  Ls a(α, β)f(z) Ls a(α, β)fm(z)  ≧ (2m + 2) (m + 1) + m (A + B) + 2A (m + 1) (2m + 2 + A + B)
Theorem 5. Theorem 5. (I) If f of the form (1) satisfies the condition (13), then ℜ Ls a(α, β)fm(z) Ls a(α, β)f(z)  ≧(2m + 2 + A + B) 2 (m + 1 + B)
Theorem 5. (I) If f of the form (1) satisfies the condition (13), then ℜ Ls a(α, β)fm(z) Ls a(α, β)f(z)  ≧(2m + 2 + A + B) 2 (m + 1 + B)
Theorem 6. Theorem 6. If f of the form (1) satisfies condition (13) with A = −B, then (I) ℜ  Ls a(α, β)f ′(z) Ls a(α, β)f ′ m(z)  ≧0 and (II) ℜ Ls…
Theorem 6. If f of the form (1) satisfies condition (13) with A = −B, then (I) ℜ  Ls a(α, β)f ′(z) Ls a(α, β)f ′ m(z)  ≧0 and (II) ℜ Ls a(α, β)f ′
Theorem 7. Theorem 7. If f of the form (1) satisfies condition (16), then (I) ℜ  Ls a(α, β)f ′(z) Ls a(α, β)f ′ m(z)  ≧ 2 (m + 1 + A) (2m + 2 + A + B)
Theorem 7. If f of the form (1) satisfies condition (16), then (I) ℜ  Ls a(α, β)f ′(z) Ls a(α, β)f ′ m(z)  ≧ 2 (m + 1 + A) (2m + 2 + A + B)
Lemma 1. Lemma 1. For 0 ≦θ ≦π, 1 2 + m X k=1 cos (kθ) k + 1 ≧0.
Lemma 1. For 0 ≦θ ≦π, 1 2 + m X k=1 cos (kθ) k + 1 ≧0.
Lemma 2. Lemma 2. Let P be analytic in U with P(0) = 1 and ℜ P(z) > 1 2 be in U. For any function Q which is analytic in U, the function P ∗Q takes…
Lemma 2. Let P be analytic in U with P(0) = 1 and ℜ{P(z)} > 1 2 be in U. For any function Q which is analytic in U, the function P ∗Q takes values in the convex hull of the image on U under Q.
Lemma 1 Lemma 1 is due to Rogosinski and Szego [21] and Lemma 2 is a well known result (c.f. [[1] and [10]]) that can be derived from the Herglotz…
Lemma 1 is due to Rogosinski and Szego [21] and Lemma 2 is a well known result (c.f. [[1] and [10]]) that can be derived from the Herglotz representation for P. After giving the above lemmas, we can proceed to the proof of our last result.
Theorem 8. Theorem 8. If f ∈ΣKa,s (A, B, α, β), then Fm ∈ΣKa,s (A, B, α, β).
Theorem 8. If f ∈ΣKa,s (A, B, α, β), then Fm ∈ΣKa,s (A, B, α, β).
Function classes studied:

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