Abstract
A motivation comes from {\em M. Ismail and et al.: A generalization of starlike functions, Complex Variables Theory Appl., 14 (1990), 77--84} to study a generalization of close-to-convex functions by means of a $q$-analog of a difference operator acting on analytic functions in the unit disk $\mathbb{D}=\{z\in \mathbb{C}:\,|z|<1\}$. We use the terminology {\em $q$-close-to-convex functions} for the $q$-analog of close-to-convex functions. The $q$-theory has wide applications in special functions
Results & Lemmas (24)
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Lemma 2.1.
Lemma 2.1. [20, Lemma 1.1(1)] Let f(z) be of the form (2.1) and P∞ n=1 |Bn+1 −Bn| ≤1, with Bn = An(1 −qn)/(1 −q). Then f(z) ∈Kq with g(z) =…
Lemma 2.1. [20, Lemma 1.1(1)] Let f(z) be of the form (2.1) and P∞ n=1 |Bn+1 −Bn| ≤1, with Bn = An(1 −qn)/(1 −q). Then f(z) ∈Kq with g(z) = z/(1 −z). As a consequence of Lemma 2.1, we have
Theorem 2.2.
Theorem 2.2. Let An be a sequence of real numbers such that Bn = An(1 −qn)/(1 −q) for all n ≥1. Suppose that 1 ≥B2 ≥B3 ≥· · · ≥Bn ≥· · · ≥0…
Theorem 2.2. Let {An} be a sequence of real numbers such that Bn = An(1 −qn)/(1 −q) for all n ≥1. Suppose that 1 ≥B2 ≥B3 ≥· · · ≥Bn ≥· · · ≥0 or 1 ≤B2 ≤B3 ≤· · · ≤Bn ≤· · · ≤2. Then f(z) = z + P∞ n=2 Anzn ∈Kq with g(z) = z/(1 −z).
Lemma 2.1
Lemma 2.1, we prove the assertion of our theorem. □
Lemma 2.1, we prove the assertion of our theorem. □
Theorem 2.3.
Theorem 2.3. Let f be defined by (2.1) and suppose that ∞ X n=1 |Bn −Bn−1| ≤1, Bn = An+1(1 −qn+1) 1 −q −An(1 −qn) 1 −q. Then f ∈Kq with g(z)…
Theorem 2.3. Let f be defined by (2.1) and suppose that ∞ X n=1 |Bn −Bn−1| ≤1, Bn = An+1(1 −qn+1) 1 −q −An(1 −qn) 1 −q . Then f ∈Kq with g(z) = z/(1 −z)2.
Theorem 2.4.
Theorem 2.4. Let An be a sequence of real numbers such that A0 = 0, A1 = 1 and Bn = An+1(1 −qn+1) 1 −q −An(1 −qn) 1 −q. Suppose that 1 ≥B1…
Theorem 2.4. Let {An} be a sequence of real numbers such that A0 = 0, A1 = 1 and Bn = An+1(1 −qn+1) 1 −q −An(1 −qn) 1 −q . Suppose that 1 ≥B1 ≥B2 ≥· · · ≥Bn ≥· · · ≥0 or 1 ≤B1 ≤B2 ≤· · · ≤Bn ≤· · · ≤2. Then f(z) = z + P∞ n=2 Anzn ∈Kq with g(z) = z/(1 −z)2.
Theorem 2.5.
Theorem 2.5. Let f be defined by f(z) = z + P∞ n=2 A2n−1z2n−1 and suppose that ∞ X n=1 |B2n−1 −B2n+1| ≤1, Bn = An(1 −qn) 1 −q. Then f ∈Kq…
Theorem 2.5. Let f be defined by f(z) = z + P∞ n=2 A2n−1z2n−1 and suppose that ∞ X n=1 |B2n−1 −B2n+1| ≤1, Bn = An(1 −qn) 1 −q . Then f ∈Kq with g(z) = z/(1 −z2).
Theorem 2.6.
Theorem 2.6. Let An be a sequence of real numbers such that Bn = An(1 −qn)/(1 −q) for all n ≥1. Suppose that 1 ≥B3 ≥B5 ≥· · · ≥B2n−1 ≥· · ·…
Theorem 2.6. Let {An} be a sequence of real numbers such that Bn = An(1 −qn)/(1 −q) for all n ≥1. Suppose that 1 ≥B3 ≥B5 ≥· · · ≥B2n−1 ≥· · · ≥0 or 1 ≤B3 ≤B5 ≤· · · ≤B2n−1 ≤· · · ≤2. Then f(z) = z + P∞ n=2 A2n−1z2n−1 ∈Kq with g(z) = z/(1 −z2).
Lemma 2.7.
Lemma 2.7. [20, Lemma 1.1(4)] Let f be defined by (2.1) and suppose that ∞ X n=2 |Bn −Bn−2| ≤1, Bn = An(1 −qn) (1 −q). Then f ∈Kq with g(z)…
Lemma 2.7. [20, Lemma 1.1(4)] Let f be defined by (2.1) and suppose that ∞ X n=2 |Bn −Bn−2| ≤1, Bn = An(1 −qn) (1 −q) . Then f ∈Kq with g(z) = z/(1 −z2).
Lemma 2.7
Lemma 2.7 leads the following sufficient conditions for functions to be in Kq.
Lemma 2.7 leads the following sufficient conditions for functions to be in Kq.
Theorem 2.8.
Theorem 2.8. Let An be a sequence of real numbers such that A1 = 1 and Bn = An(1 −qn) 1 −q for all n ≥1. Suppose that 1 ≥B1 + B2 ≥· · ·…
Theorem 2.8. Let {An} be a sequence of real numbers such that A1 = 1 and Bn = An(1 −qn) 1 −q for all n ≥1. Suppose that 1 ≥B1 + B2 ≥· · · ≥Bn−1 + Bn ≥· · · ≥0 or 1 ≤B1 + B2 ≤· · · ≤Bn−1 + Bn ≤· · · ≤2. Then f(z) = z + P∞ n=2 Anzn ∈Kq with g(z) = z/(1 −z2).
Theorem 2.7
Theorem 2.7, we complete the proof. □ As a consequence of Theorem 2.8, one can obtain the following new criteria for functions to be in the…
Theorem 2.7, we complete the proof. □ As a consequence of Theorem 2.8, one can obtain the following new criteria for functions to be in the close-to-convex family.
Theorem 2.9.
Theorem 2.9. Let an be a sequence of real numbers such that a1 = 1 and bn = nan for all n ≥1. Suppose that 1 ≥b1 + b2 ≥· · · ≥bn−1 + bn ≥·…
Theorem 2.9. Let {an} be a sequence of real numbers such that a1 = 1 and bn = nan for all n ≥1. Suppose that 1 ≥b1 + b2 ≥· · · ≥bn−1 + bn ≥· · · ≥0 or 1 ≤b1 + b2 ≤· · · ≤bn−1 + bn ≤· · · ≤2. Then f(z) = z + P∞ n=2 anzn is close-to-convex with g(z) = z/(1 −z2).
Lemma 2.10.
Lemma 2.10. [20, Lemma 1.1(2)] Let f be defined by (2.1) and suppose that ∞ X n=1 |Bn−1 −Bn + Bn+1| ≤1, Bn = An(1 −qn) 1 −q. Then f ∈Kq with…
Lemma 2.10. [20, Lemma 1.1(2)] Let f be defined by (2.1) and suppose that ∞ X n=1 |Bn−1 −Bn + Bn+1| ≤1, Bn = An(1 −qn) 1 −q . Then f ∈Kq with g(z) = z/(1 −z + z2).
Lemma 2.10
Lemma 2.10 yields the following sufficient condition.
Lemma 2.10 yields the following sufficient condition.
Theorem 2.11.
Theorem 2.11. Let An be a sequence of real numbers such that A1 = 1 and Bn = An(1 −qn) 1 −q for all n ≥1. Suppose that 0 ≥B2 −B1 ≥B3 ≥B2…
Theorem 2.11. Let {An} be a sequence of real numbers such that A1 = 1 and Bn = An(1 −qn) 1 −q for all n ≥1. Suppose that 0 ≥B2 −B1 ≥B3 ≥B2 +B4 ≥B2 +B3 +B5 ≥· · · ≥B2 +B3 +B4 +· · ·+Bn−1 +Bn+1 ≥−1 or 0 ≤B2 −B1 ≤B3 ≤B2 + B4 ≤B2 + B3 + B5 ≤· · · ≤B2 + B3 + B4 + · · · + Bn−1 + Bn+1 ≤1 holds. Then f(z) = z + P∞ n=2 Anzn ∈Kq with g(z) = z/(1 −z + z2).
Theorem 2.12.
Theorem 2.12. Let an be a sequence of real numbers such that a1 = 1 and bn = nan, for all n ≥1. Suppose that 0 ≥b2 −b1 ≥b3 ≥b2 + b4 ≥b2 +…
Theorem 2.12. Let {an} be a sequence of real numbers such that a1 = 1 and bn = nan, for all n ≥1. Suppose that 0 ≥b2 −b1 ≥b3 ≥b2 + b4 ≥b2 + b3 + b5 ≥· · · ≥b2 + b3 + b4 + · · · + bn−1 + bn+1 ≥−1 or 0 ≤b2 −b1 ≤b3 ≤b2 + b4 ≤b2 + b3 + b5 ≤· · · ≤b2 + b3 + b4 + · · · + bn−1 + bn+1 ≤1. Then f(z) = z+ ∞ X n=2 anzn is in the close-to-convex family with respect to g(z) = z/(1−z+z2). 3. The Bieberbach-de Branges Theorem for Kq A necessary and sufficient condition for functions to be in S∗ q was obtained in
Lemma 3.1.
Lemma 3.1. A function f ∈Kq if and only if there exists g ∈S∗such that |g(z) + f(qz) −f(z)| |g(z)| ≤1 for all z ∈D.
Lemma 3.1. A function f ∈Kq if and only if there exists g ∈S∗such that |g(z) + f(qz) −f(z)| |g(z)| ≤1 for all z ∈D.
Theorem 3.2
Theorem 3.2 (Bieberbach-de Branges Theorem for Kq). If f ∈Kq, then |an| ≤1 −q 1 −qn n + n(n −1) 2 (1 + q) for all n ≥2.
Theorem 3.2 (Bieberbach-de Branges Theorem for Kq). If f ∈Kq, then |an| ≤1 −q 1 −qn n + n(n −1) 2 (1 + q) for all n ≥2.
Corollary 3.3.
Corollary 3.3. If f ∈Kq with the Koebe function g(z) = z/(1 −z)2, then for all n ≥2 we have |an| ≤1 −q 1 −qn n + (1 + q) n(n −1) 2 . If…
Corollary 3.3. If f ∈Kq with the Koebe function g(z) = z/(1 −z)2, then for all n ≥2 we have |an| ≤1 −q 1 −qn n + (1 + q) n(n −1) 2 . If f ∈K with g(z) = z, then for all n ≥2 it is well-known that |an| ≤2/n. As a generalization, we have the following:
Corollary 3.4.
Corollary 3.4. If f ∈Kq with g(z) = z, then for all n ≥2 we have |an| ≤(1 −q2)/(1 −qn). Here we note that the series z + P∞ n=2(1 −q2)/(1…
Corollary 3.4. If f ∈Kq with g(z) = z, then for all n ≥2 we have |an| ≤(1 −q2)/(1 −qn). Here we note that the series z + P∞ n=2(1 −q2)/(1 −qn)zn converges to the Heine hyperge- ometric function (z + qz)Φ[q, q; q2; q, z] −qz = z + z2Φ[q2, q2; q3; q2, z] and it follows from [24, 3.2.2, pp. 91]. If f ∈K with g(z) = z/(1 −z), then for all n ≥2 it is known that |an| ≤(2n −1)/n. We find the following analogous result:
Corollary 3.5.
Corollary 3.5. If f ∈Kq with g(z) = z/(1 −z), then for all n ≥2 we have |an| ≤1 −q 1 −qn [n + q(n −1)]. One can similarly verify that the…
Corollary 3.5. If f ∈Kq with g(z) = z/(1 −z), then for all n ≥2 we have |an| ≤1 −q 1 −qn [n + q(n −1)]. One can similarly verify that the series z + P∞ n=2 1−q 1−qn[n + q(n −1)] converges to the function z(1 + q) d dzΨ(q; z) −qΨ(q; z), where Ψ(q; z) := zΦ[q, q; q2; q, z] represents its Heine hypergeometric function. If f ∈K with g(z) = z/(1 −z2), then for all m ≥1 it is known that |an| ≤ 1, if n = 2m −1;
Corollary 3.6.
Corollary 3.6. If f ∈Kq with g(z) = z/(1 −z2), then for all m ≥1 we have |an| ≤ 1 −q 1 −qn n 2(1 + q) + 1 2(1 −q)
Corollary 3.6. If f ∈Kq with g(z) = z/(1 −z2), then for all m ≥1 we have |an| ≤ 1 −q 1 −qn n 2(1 + q) + 1 2(1 −q)
Corollary 3.7.
Corollary 3.7. If f ∈Kq with g(z) = z/(1 −z + z2), then for all m ≥1 we have |an| ≤
Corollary 3.7. If f ∈Kq with g(z) = z/(1 −z + z2), then for all m ≥1 we have |an| ≤
Theorem 4.1
Theorem 4.1 (The Bieberbach-de Branges Theorem for S∗ q ). If f ∈S∗ q, then for all n ≥2 we have (4.1) |an| ≤ 1 −q2 q −qn n−1 Y k=2 1…
Theorem 4.1 (The Bieberbach-de Branges Theorem for S∗ q ). If f ∈S∗ q , then for all n ≥2 we have (4.1) |an| ≤ 1 −q2 q −qn n−1 Y k=2 1 + 1 −q2 q −qk
Function classes studied:
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