On certain subclasses of analytic functions

Lemma 1 [8]. If −1 ≤B < A ≤1, β > 0 and the complex number γ satisfy Re(γ) ≥−β(1 −A)/(1 −B), then the differential equation q(z) + zq′(z) βq(z) + γ = 1 + Az 1 + Bz has a univalent solution in E given by (2.1) q(z) =      

Lemma 2 [16]. Let µ be a positive measure on the unit interval [0, 1]. Let g(t, z) be an analytic function in E for each t ∈[0, 1], and integrable in t for each z ∈E and for almost all t ∈[0, 1], and suppose that Re{g(t, z)} > 0 on E, g(t, −r) is real and Re{1/g(t, z)} ≥1/g(t, −r) for |z| ≤r and t ∈[0, 1]. If g(z) = z 0 g(t, z) dµ(t), then Re{1/g(z)} ≥1/g(−r) for |z| ≤r. For real or complex numbers α1, α2 and β1 (β1 ̸= 0, −1, −2, . . .), the hypergeometric function 2F1(z) is defined by (2.2) 2F1

Lemma 3. For real or complex α1, α2 and β1 (β1 ̸= 0, −1, −2, . . .), we have (2.3) 1 0 tα2−1(1 −t)β1−α2−1(1 −tz)−α1 dt = Γ(α2)Γ(β1 −α2) Γ(β1) 2F1(α1, α2; β1; z) (Re α1 > Re α2 > 0), (2.4) 2F1(α1, α2; β1; z) = 2F1(α2, α1; β1; z), (2.5) 2F1(α1, α2; β1; z) = (1 −z)−α12F1(α1, β1 −α2; β1; z).

Lemma 4. Let p(z) be analytic in E with p(0) = 1 and p(z) ̸= 0 for 0 < |z| < 1, and let −1 ≤B < A ≤1. (i) Let B ̸= 0 and µ be a complex number with µ ̸= 0. Let A, B and µ satisfy either (2.6) µ A −B B −1 ≤1 or µ A −B B + 1 ≤1. If p(z) satisfies

Theorem 1. If B < A ≤(c −B)/(c −A), then: (i) S(a, c, A, B) ⊂S(a, c + 1, A∗, B) where A∗= {(c −1)A + B}/c. Furthermore, if f ∈S(a, c, A, B) then L(a, c)f(z) L(a, c + 1)f(z) ≺ 1 cQ(z) = eq(z) (z ∈E)

Corollary 1. If f ∈S(λ, α, β), then Re  Jλ z f(z) Jλ−1 z f(z)  >  2F1  1, 2(1 −λ)(1 −α); 3 −λ; β β + 1

Corollary 2. If f ∈S∗(α), then Re n zf(z)  z 0 f(t) dt −1o > 2(2F1(1, 2(1 −α); 3; 1/2))−1 (z ∈E). The result is best possible.

Theorem 2. Let f ∈S(a, c, A, B), where −1 ≤B < A ≤1 (B ̸= 0). If either (c −1) A −B B −1 ≤1 or (c −1) A −B B + 1 ≤1 then (3.9) L(a, c)f(z) z

Corollary 3. Under the hypotheses of Theorem 2, we have, for |z| = r < 1, (3.11) |L(a, c)f(z)| ≤  r(1 + Br)(c−1)(A−B)/B, B ̸= 0, r exp((c −1)Ar), B = 0, and (3.12) |L(a, c)f(z)| ≥  r(1 −Br)(c−1)(A−B)/B, B ̸= 0,

Corollary 4. If f ∈S(λ, α, β), then for |z| = r < 1, r (1 + βr)2(1−λ)(1−α) ≤|Jλ z (z)| ≤ r (1 −βr)2(1−λ)(1−α) . The bounds are sharp.

Corollary 5. If f ∈S(λ, α, β), then Re Jλ z (z) z  > (1 + β)−2(1−λ)(1−α) (z ∈E). The result is sharp.

Theorem 3. Let δ be a real number satisfying (3.13) B < A ≤B + (1 −B)(δ + 1) c −1 .

Corollary 6. Let δ be a real number satisfying δ ≥{(1 −λ)(1 −2α) − (1 + βλ)}/(1 + β). (i) If f ∈S(λ, α, β), then the function Fδ defined by (1.2) belongs to the class S(λ, α, β). Furthermore, J1+λ z Fδ(z) Jλ z Fδ(z) ≺ 1 1 −λ  1 Q(z) −(δ + λ)

Theorem 4. Let f ∈A be given by (1.1) and −1 ≤B < 0. If (3.19) ∞ X n=2 {(1 −B)(n −1) + (A −B)(c −1)}(a)n (c −1)n |an| ≤A −B then f ∈S(a, c, A, B). The result is sharp.

Corollary 7. Let f ∈A be given by (1.1). If ∞ X n=2 Γ(n + 1)Γ(1 −λ){(1 + β)(n −1) + 2β(1 −α)} Γ(n + 1 −λ) |an| ≤2β(1 −α) then f ∈S(λ, α, β). The result is sharp.

Theorem 5. If f given by (1.1) belongs to S(a, c, A, B), then (3.20) |an| ≤(A −B)(c −1)n (n −1)(a)n−1 n−1 Y j=2  1 + (A −B)(c −1) j −1  (n ≥2). The result is sharp.

Corollary 8. If f, given by (1.1), belongs to the class S(λ, α, β), then |an| ≤ 2β(1 −α)Γ(n + 1 −λ) (n −1)Γ(n + 1)Γ(1 −λ) n−1 Y j=2  1 + 2β(1 −α)(1 −λ) j −1  (n ≥2). The result is sharp.

Theorem 6. Let f, given by (1.1), belong to the class S(a, c, A, B) and µ be any complex number. Then |a3 −µa2 2| ≤(A −B)(c −1)3 2(a)2 × max  1, {B −(A −B)(c −1)} + µ2(A −B)(a + 1)(c −1)2 a(c + 1)

Corollary 9. If f, given by (1.1), belongs to the class S(λ, α, β), then for any complex number µ |a3 −µa2 2| ≤β(1 −α)Γ(4 −λ) 3! Γ(1 −λ) × max  1,

Definition 1. The fractional integral of order λ is defined, for a func- tion f(z), by (1.3) D−λ z f(z) = 1 Γ(λ) z 0 f(ζ)

Definition 2. Under the hypotheses of Definition 1, the fractional deri- vative of order n + λ is defined by Dn+λ z f(z) = dn dzn (Dλ z f(z)) (0 ≤λ < 1, n ∈N0 = {0, 1, . . .}). Let φ(a, c; z) =

Definition 3. A function f ∈A is said to be in the class S(a, c, A, B) if it satisfies (1.6) L(a, c −1)f(z) L(a, c)f(z) ≺1 + Az 1 + Bz (z ∈E) for some a > 0, c > 1, and −1 ≤B < A ≤1. By the definition of subordi- nation, it follows that