Results & Lemmas (176)
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Theorem 1.
Theorem 1. Let f (z) be defined by (1) and p(z) by (19). Then, f ∈GS∗ F ψ(m, β) where m ∈[−1, 1] and β ≥1, if a1 s ≤
Theorem 1. Let f (z) be defined by (1) and p(z) by (19). Then, f ∈GS∗ F ψ(m, β) where m ∈[−1, 1] and β ≥1, if a1 s ≤
Theorem 2.
Theorem 2. Let f (z) be defined by (1) and γ(z) by (20). Then, f ∈GS∗ F ψ(m, β) where m ∈[−1, 1] and β ≥1 if the following condition holds…
Theorem 2. Let f (z) be defined by (1) and γ(z) by (20). Then, f ∈GS∗ F ψ(m, β) where m ∈[−1, 1] and β ≥1 if the following condition holds true
Theorem 3.
Theorem 3. Let f (z) be defined by (1) and γ(z) by (20). Then, f ∈GS∗ F ψ(m, β) where m ∈[−1, 1] and β ≥1, if a1 s ≤
Theorem 3. Let f (z) be defined by (1) and γ(z) by (20). Then, f ∈GS∗ F ψ(m, β) where m ∈[−1, 1] and β ≥1, if a1 s ≤
Theorem 4.
Theorem 4. Let f (z) be defined by (1) and Q(z) by (21). Then, f ∈GS∗ F ψ(m, β) where m ∈[−1, 1] and β ≥1 if the following condition holds…
Theorem 4. Let f (z) be defined by (1) and Q(z) by (21). Then, f ∈GS∗ F ψ(m, β) where m ∈[−1, 1] and β ≥1 if the following condition holds true
Theorem 5.
Theorem 5. Let f (z) be defined by (1) and Q(z) by (21). Then, f ∈GS∗ F ψ(m, β) where m ∈[−1, 1] and β ≥1, if a1 s ≤
Theorem 5. Let f (z) be defined by (1) and Q(z) by (21). Then, f ∈GS∗ F ψ(m, β) where m ∈[−1, 1] and β ≥1, if a1 s ≤
Theorem 6.
Theorem 6. Let f (z) be defined by (1) and Q(z) by (21). Then, f ∈GS∗ F ψ(m, β) where m ∈[−1, 1] and β ≥1 if the following condition holds…
Theorem 6. Let f (z) be defined by (1) and Q(z) by (21). Then, f ∈GS∗ F ψ(m, β) where m ∈[−1, 1] and β ≥1 if the following condition holds true
Theorem 1.
Theorem 1. Let f ∈Σ given by (5) belongs to the class Gt Σ(x, p, q, λ, β, γ). Then |a2| ≤ tx √ 2tx λee 1 2 (1−λ2) r h 5(1 + 2γ + 6β)(tx)2…
Theorem 1. Let f ∈Σ given by (5) belongs to the class Gt Σ(x, p, q, λ, β, γ). Then |a2| ≤ tx √ 2tx λee 1 2 (1−λ2) r h 5(1 + 2γ + 6β)(tx)2 −8ee(1−λ2)(1 + γ + 2β)2(ptx2 + aq) i , and
Theorem 2.
Theorem 2. Let f ∈Σ given by (5) belongs to the class Gt Σ(x, p, q, λ, β, γ). Then a3 −ηa2 2 ≤
Theorem 2. Let f ∈Σ given by (5) belongs to the class Gt Σ(x, p, q, λ, β, γ). Then a3 −ηa2 2 ≤
Corollary 1.
Corollary 1. Let f ∈Σ given by (5) belongs to the class Gt Σ(x, p, q, λ, γ). Then |a2| ≤ tx √ 2tx λee 1 2 (1−λ2) r h 5(1 + 2γ)(tx)2…
Corollary 1. Let f ∈Σ given by (5) belongs to the class Gt Σ(x, p, q, λ, γ). Then |a2| ≤ tx √ 2tx λee 1 2 (1−λ2) r h 5(1 + 2γ)(tx)2 −8ee(1−λ2)(1 + γ)2(ptx2 + aq) i , |a3| ≤
Corollary 2.
Corollary 2. Let f ∈Σ given by (5) belongs to the class Gt Σ(x, p, q, λ). Then |a2| ≤ tx √ 2tx λee 1 2 (1−λ2) r h 15(tx)2 −32ee(1−λ2)(ptx2…
Corollary 2. Let f ∈Σ given by (5) belongs to the class Gt Σ(x, p, q, λ). Then |a2| ≤ tx √ 2tx λee 1 2 (1−λ2) r h 15(tx)2 −32ee(1−λ2)(ptx2 + aq) i , |a3| ≤
Corollary 3.
Corollary 3. Let f ∈Σ given by (5) belongs to the class Gt Σ(x, p, q, λ, 0). Then |a2| ≤ tx √ 2tx λee 1 2 (1−λ2) r h 5(tx)2 −8ee(1−λ2)(ptx2…
Corollary 3. Let f ∈Σ given by (5) belongs to the class Gt Σ(x, p, q, λ, 0). Then |a2| ≤ tx √ 2tx λee 1 2 (1−λ2) r h 5(tx)2 −8ee(1−λ2)(ptx2 + aq) i , |a3| ≤
Corollary 4.
Corollary 4. Let f ∈Σ given by (5) belongs to the classGt Σ(x, p, q, β, γ). Then |a2| ≤ tx √ 2tx e r h 5(1 + 2γ + 6β)(tx)2 −8e(1 + γ +…
Corollary 4. Let f ∈Σ given by (5) belongs to the classGt Σ(x, p, q, β, γ). Then |a2| ≤ tx √ 2tx e r h 5(1 + 2γ + 6β)(tx)2 −8e(1 + γ + 2β)2(ptx2 + aq) i , |a3| ≤ t2x2 4e2(1 + γ + 2β)2 +
Lemma 1.
Lemma 1. (See [2].) Let f (z) = z|z|2βh(z)g(z) be a log-harmonic mapping on E, 0 /∈hg(E). Then, f ∈STLH(α) if and only if ϕ(z) = zh(z) g(z)…
Lemma 1. (See [2].) Let f (z) = z|z|2βh(z)g(z) be a log-harmonic mapping on E, 0 /∈hg(E). Then, f ∈STLH(α) if and only if ϕ(z) = zh(z) g(z) ∈S∗(α). In [2], the authors studied the class of α −spirallike functions and proved that, if f (z) = z|z|2βh(z)g(z) is a log-harmonic mapping on E, 0 /∈hg(E), then f is α −spirallike if Re e−iα z fz −z fz f > 0, 0 ≤α < 1 for all z ∈E. We remark that a simply connected domain Ωin C containing the origin is said to be α −spirallike, −π 2 < α < π
Lemma 2.
Lemma 2. ([2]) If f (z) = z|z|2βh(z)g(z) is log-harmonic on E and 0 /∈hg(E), with Reβ > −1 2, then f ∈Sα LH(ρ) if and only if ψ(z) = zh(z)…
Lemma 2. ([2]) If f (z) = z|z|2βh(z)g(z) is log-harmonic on E and 0 /∈hg(E), with Reβ > −1 2, then f ∈Sα LH(ρ) if and only if ψ(z) = zh(z) g(z)e2iα ∈Sα(ρ). In the celebrated paper [11], the authors introduce a new way of studying harmonic functions in Geometric Function Theory. Additionally, many authors investigated the linear combinations of harmonic functions in a plane; see, for example, [12–14]. In Section 2 of this paper, taking the convex-exponent product combination of two elements, a sp
Theorem 1.
Theorem 1. Let f (z) = zh(z)g(z) ∈STLH(ρ), (0 ≤ρ < 1) with respect to a ∈B0, φ ∈S∗(γ), (0 ≤γ < 1) and α, β be real numbers with α + β = 1.…
Theorem 1. Let f (z) = zh(z)g(z) ∈STLH(ρ), (0 ≤ρ < 1) with respect to a ∈B0, φ ∈S∗(γ), (0 ≤γ < 1) and α, β be real numbers with α + β = 1. Then, F(z) = f (z)αK(z)β is starlike log-harmonic mapping of order αρ + βγ with respect to a, where K(z) = φ(z) exp 2Re Z z 0 a(s) 1 −a(s) φ′(s) φ(s) ds .
Theorem 2.
Theorem 2. Let f (z) = zh(z)g(z) ∈Sβ LH(ρ) with respect to a ∈B0 and γ be a constant with Reγ > −1 2. Then, F(z) = f (z)| f (z)|2γ is an α…
Theorem 2. Let f (z) = zh(z)g(z) ∈Sβ LH(ρ) with respect to a ∈B0 and γ be a constant with Reγ > −1 2. Then, F(z) = f (z)| f (z)|2γ is an α −spirallike log-harmonic mapping of order ρ with respect to ˆa(z) = (1 + ¯γ)a(z) + ¯γ 1 + γ + γa(z) , where |β| < π 2 and α = tan−1 tan β+2Imγ 1+2Reγ . 30
Theorem 3.
Theorem 3. Let fk(z) = zhk(z)gk(z) ∈Sβ LH(ρ) with k = 1, 2 and with respect to the same a ∈B0 and γ be a constant with Reγ > −1 2.…
Theorem 3. Let fk(z) = zhk(z)gk(z) ∈Sβ LH(ρ) with k = 1, 2 and with respect to the same a ∈B0 and γ be a constant with Reγ > −1 2. Moreover, let F1(z) = f1(z)| f1(z)|2γ and F2(z) = f2(z)| f2(z)|2γ. Then, F(z) = Fλ 1 (z)F1−λ 2 (z) is an α-spirallike log-harmonic mapping of order ρ with respect to ˆa(z) = (1 + ¯γ)a(z) + ¯γ 1 + γ + γa(z) , where |β| < π
Theorem 4.
Theorem 4. Let fk(z) = zhk(z)gk(z) ∈Sβ LH(ρ) with respect to ak ∈B0(k = 1, 2). Moreover, suppose that Reγ > −1 2, F1(z) = f1(z)| f1(z)|2γ…
Theorem 4. Let fk(z) = zhk(z)gk(z) ∈Sβ LH(ρ) with respect to ak ∈B0(k = 1, 2). Moreover, suppose that Reγ > −1 2, F1(z) = f1(z)| f1(z)|2γ and F2(z) = f2(z)| f2(z)|2γ. If Re " (1 −a1(z)a2(z)) 1 + zh′ 1(z) h1(z) 1 + zh′
Theorem 5.
Theorem 5. Let fk(z) = zhk(z)gk(z) be univalent log-harmonic functions with respect to ak ∈B0(k = 1, 2) and Reγ > −1 2. Moreover, suppose…
Theorem 5. Let fk(z) = zhk(z)gk(z) be univalent log-harmonic functions with respect to ak ∈B0(k = 1, 2) and Reγ > −1 2. Moreover, suppose that zhkgk = φk(z), where φk(z) = zexp 2 Z z 0 ak(t) t(1 −ak(t))dt and F1(z) = f1(z)| f1(z)|2γ and F2(z) = f2(z)| f2(z)|2γ.
Theorem 6.
Theorem 6. Let fk(z) = zhk(z)gk(z)(k = 1, 2) be log-harmonic functions with respect to ak ∈B0. Moreover, suppose that zhkgk = z and F1(z) =…
Theorem 6. Let fk(z) = zhk(z)gk(z)(k = 1, 2) be log-harmonic functions with respect to ak ∈B0. Moreover, suppose that zhkgk = z and F1(z) = f1(z)| f1(z)|2γ and F2(z) = f2(z)| f2(z)|2γ. Then, F(z) = Fλ 1 (z)F1−λ 2 (z) ∈Sα LH(1), where 0 ≤λ ≤1 and α = tan−1 2Imγ (1+2Reγ)
Theorem 4
Theorem 4 implies that F(z) = Fλ 1 (z)F1−λ 2 (z) ∈Sα LH(1 2), where 0 ≤λ ≤1 and α = tan−1 2Imγ 1+2Reγ . The images in Example 2–4 are…
Theorem 4 implies that F(z) = Fλ 1 (z)F1−λ 2 (z) ∈Sα LH(1 2), where 0 ≤λ ≤1 and α = tan−1 2Imγ 1+2Reγ . The images in Example 2–4 are shown in Figures 2–4. Example 4. Let Reγ > 1 2, a1(z) = z, and h1(z) = g1(z) =
Theorem 6
Theorem 6 are satisfied. Hence, according to Theorem 6, by taking 38
Theorem 6 are satisfied. Hence, according to Theorem 6, by taking 38
Theorem 1.
Theorem 1. Assume that the function f (z) is an entire function of q−1-exponential growth of order 1 and a finite type α less than ξ1, or it…
Theorem 1. Assume that the function f (z) is an entire function of q−1-exponential growth of order 1 and a finite type α less than ξ1, or it is an entire function of q−1-exponential growth of an order of less than 1. Then, f (z) has a convergent q-Lidstone representation f (z) = ∞ ∑ n=0 h D2n q−1 f (1)An(z) −D2n q−1 f (0)Bn(z) i , (3) where (An)n and (Bn)n are the q-Lidstone polynomials defined, respectively, by the generating
Lemma 1.
Lemma 1. For any x ∈[0, 1], we have Sinqξ1x ≤ξ1x. (11)
Lemma 1. For any x ∈[0, 1], we have Sinqξ1x ≤ξ1x. (11)
Proposition 1
Proposition 1 ([16]). Let ξk k∈N be the sequence of the positive zeros of Sinq(x) and m ∈N0. Then, (−1)n−1An(x) = 2Sinq(ξ1x) ξ2n+1 1 Sin′…
Proposition 1 ([16]). Let {ξk}k∈N be the sequence of the positive zeros of Sinq(x) and m ∈N0. Then, (−1)n−1An(x) = 2Sinq(ξ1x) ξ2n+1 1 Sin′ q(ξ1) + O(ξ−(2n+1) 2 ); (12) (−1)n−1Bn(x) = Sinq(ξ1x)Cosq(ξ1)
Proposition 2
Proposition 2 ([17]). If f ∈C2n q ([0, 1]), then f (x) = n−1 ∑ m=0 h D2m q−1 f (1)Am(x) −D2m q−1 f (0)Bm(x) i + Z 1 0 Gn(x, qt)D2n q−1 f…
Proposition 2 ([17]). If f ∈C2n q ([0, 1]), then f (x) = n−1 ∑ m=0 h D2m q−1 f (1)Am(x) −D2m q−1 f (0)Bm(x) i + Z 1 0 Gn(x, qt)D2n q−1 f (q2t) dqt,
Proposition 3.
Proposition 3. Let ξ1 be the smallest positive zero of Sinq(x). Then, there exist some constants M1 and M2 and a positive integer n0 such…
Proposition 3. Let ξ1 be the smallest positive zero of Sinq(x). Then, there exist some constants M1 and M2 and a positive integer n0 such that the following inequalities hold 0 ≤(−1)nAn(x) ≤M1 ξ2n 1 ; (19) 0 ≤(−1)n−1Bn(x) ≤M2 ξ2n 1 , (20) for all x ∈[0, 1] and n ≥n0.
Proposition 4.
Proposition 4. There exists a constant M such that 0 ≤ Z 1 0 (−1)n Gn(x, qt) dqt ≤M ξ2n 1.
Proposition 4. There exists a constant M such that 0 ≤ Z 1 0 (−1)n Gn(x, qt) dqt ≤M ξ2n 1 .
Proposition 5.
Proposition 5. For any fixed point x0 ∈(0, 1) and sufficiently large n, there exist some constants M1 and M2 such that (−1)nAn(x0) ≥M1 ξ2n 1;…
Proposition 5. For any fixed point x0 ∈(0, 1) and sufficiently large n, there exist some constants M1 and M2 such that (−1)nAn(x0) ≥M1 ξ2n 1 ; (24) (−1)n−1Bn(x0) ≥M2 ξ2n 1 . (25)
Theorem 2.
Theorem 2. If the series S = a0A0(x) + b0B0(x) + a1A1(x) + b1B1(x) +... (26) converges for a single value x0 ∈(0, 1), then the series ∑∞…
Theorem 2. If the series S = a0A0(x) + b0B0(x) + a1A1(x) + b1B1(x) + . . . (26) converges for a single value x0 ∈(0, 1), then the series ∑∞ n=0(−1)nh an+bn ξ2n 1 i is absolutely convergent.
Proposition 6.
Proposition 6. If a function f ∈C∞ q [0, a] is q-completely convex, then (−1)nD2n q−1 f (0) ≥0 (n ∈N0). (30)
Proposition 6. If a function f ∈C∞ q [0, a] is q-completely convex, then (−1)nD2n q−1 f (0) ≥0 (n ∈N0). (30)
Proposition 7.
Proposition 7. Let f ∈C∞ q (0, 1) be a q-completely convex function on [0, 1]. Then, for a suffi- ciently large n, we have D2n q−1 f (0) =…
Proposition 7. Let f ∈C∞ q (0, 1) be a q-completely convex function on [0, 1]. Then, for a suffi- ciently large n, we have D2n q−1 f (0) = O(ξ2n 1 ); (31) D2n q−1 f (1) = O(ξ2n 1 ). (32)
Proposition 8.
Proposition 8. Let f be a q-completely convex function on [0, 1]. Then, there exists a positive constant C such that for all x ∈Aq 0…
Proposition 8. Let f be a q-completely convex function on [0, 1]. Then, there exists a positive constant C such that for all x ∈Aq 0 ≤(−1)nD2n q−1 f (x) ≤C ξ1 x 2n , (35) where ξ1 is the smallest positive zero of Sinq(x).
Lemma 2.
Lemma 2. Let f (x) and −D2 q−1 f (x) be non-negative on A∗ q, and continuous at 0. Assume that there exists a number x0 ∈Aq such that f…
Lemma 2. Let f (x) and −D2 q−1 f (x) be non-negative on A∗ q, and continuous at 0. Assume that there exists a number x0 ∈Aq such that f (x0) ≤α (α ∈R). Then, f (x) ≤(1 + q)α (1 −q)x0 , for all x ∈A∗ q.
Corollary 1.
Corollary 1. If f ∈C∞ q [0, 1] is a q-completely convex function, then there exists a positive constant M such that 0 ≤(−1)nD2n q−1 f (x)…
Corollary 1. If f ∈C∞ q [0, 1] is a q-completely convex function, then there exists a positive constant M such that 0 ≤(−1)nD2n q−1 f (x) ≤Mξ2n 1 (n ∈N0, x ∈A∗ q). (41)
Lemma 3.
Lemma 3. If f ∈C∞ q [0, 1] is a q-completely convex function on [0, 1], then there exists a constant K > 0 such that |Dn q−1 f (x)| ≤Kξn 1…
Lemma 3. If f ∈C∞ q [0, 1] is a q-completely convex function on [0, 1], then there exists a constant K > 0 such that |Dn q−1 f (x)| ≤Kξn 1 (x ∈A∗ q), (42) where ξ1 is the smallest positive zero of Sinq(z).
Theorem 3.
Theorem 3. Let f ∈C∞ q [0, 1] be a q-completely convex on [0, 1]. If f is analytic at zero, then the following q-Lidstone series expansion…
Theorem 3. Let f ∈C∞ q [0, 1] be a q-completely convex on [0, 1]. If f is analytic at zero, then the following q-Lidstone series expansion holds for all x ∈[0, 1]. f (x) = ∞ ∑ n=0 h D2n q−1 f (1)An(x) −D2n q−1 f (0)Bn(x) i . (48) Moreover, f (x) is the restriction of an entire function of q−1-exponential growth of order 1
Theorem 4.
Theorem 4. Let n ∈N0, (an)n and (bn)n be two sequences of non-negative integers. Assume that the series ∞ ∑ n=0 h (−1)nan An(x) −(−1)nbn…
Theorem 4. Let n ∈N0, (an)n and (bn)n be two sequences of non-negative integers. Assume that the series ∞ ∑ n=0 h (−1)nan An(x) −(−1)nbn Bn(x) i converges to a function f (x), 0 ≤x ≤1. Then, f (x) is a minimal q-completely convex on the interval [0, 1].
Theorem 5.
Theorem 5. If f (x) is a minimal q-completely convex function on [0, 1], then it can be expanded into a convergent q-Lidstone series: f (x)…
Theorem 5. If f (x) is a minimal q-completely convex function on [0, 1], then it can be expanded into a convergent q-Lidstone series: f (x) = f (1)A0(x) −f (0)B0(x) + D2 q−1 f (1)A1(x) −D2 q−1 f (0)B1(x) + . . . . (54)
Theorem 6.
Theorem 6. A real function f (x) can be represented by an absolutely convergent q-Lidstone series if and only if it is the difference of…
Theorem 6. A real function f (x) can be represented by an absolutely convergent q-Lidstone series if and only if it is the difference of two minimal q-completely convex functions on [0, 1].
Theorem 4
Theorem 4, g(x) and h(x) are minimal q-completely convex functions on [0, 1]. Since f (x) = h(x) −g(x), the proof is complete. 6.…
Theorem 4, g(x) and h(x) are minimal q-completely convex functions on [0, 1]. Since f (x) = h(x) −g(x), the proof is complete. 6. Conclusions We introduced the class of q-completely convex functions in the interval [0, a], with the functions satisfying the inequality (−1)nD2n q−1 f (aqk) ≥0 ({n, k} ⊂N0)). This class of functions is a generalization of the class of completely convex functions introduced by Widder [10]. First, we presented some properties of a q-completely convex function, and the
Theorem 1.
Theorem 1. Suppose that 0 ≤η ≤1, 0 ≤λ ≤1 and 0 ≤δ ≤1. If f ∈Σ of the style (1) be an element of class WΣ(η, δ, λ, σ, θ, α, β, p, q; h),…
Theorem 1. Suppose that 0 ≤η ≤1, 0 ≤λ ≤1 and 0 ≤δ ≤1. If f ∈Σ of the style (1) be an element of class WΣ(η, δ, λ, σ, θ, α, β, p, q; h), with h(z) = 1 + e1z + e2z2 + · · · , then |a2| ≤ (η + λ(δ + 1))[Ψ2(σ, α, β)]θ p,q|e1| [Ψ1(σ, α, β)]θp,q = |e1| Ω 57
Corollary 1.
Corollary 1. If f ∈Σ given by style (1) is in the family WΣ(η, δ, λ, σ, θ, α, β, p, q; Hγ(τ, z)), then |a2| ≤ (η + λ(δ + 1))[Ψ2(σ, α, β)]θ…
Corollary 1. If f ∈Σ given by style (1) is in the family WΣ(η, δ, λ, σ, θ, α, β, p, q; Hγ(τ, z)), then |a2| ≤ (η + λ(δ + 1))[Ψ2(σ, α, β)]θ p,q|1 + γ −τ| [Ψ1(σ, α, β)]θp,q = |1 + γ −τ| Ω and |a3| ≤ min ( max (
Theorem 2.
Theorem 2. Suppose that 0 ≤ξ ≤1 and 0 ≤ρ < 1. If f ∈Σ of the style (1) be an element of the class KΣ(ξ, ρ, σ, θ, α, β, p, q; h), with h(z)…
Theorem 2. Suppose that 0 ≤ξ ≤1 and 0 ≤ρ < 1. If f ∈Σ of the style (1) be an element of the class KΣ(ξ, ρ, σ, θ, α, β, p, q; h), with h(z) = 1 + e1z + e2z2 + · · · , then |a2| ≤ (ξ + 1)(1 −ρ)[Ψ2(σ, α, β)]θ p,q|e1| [Ψ1(σ, α, β)]θp,q = |e1| Υ and |a3| ≤min ( max ( e1 Φ ,
Corollary 2.
Corollary 2. If f ∈Σ of the style (1) be an element of the class KΣ(ξ, ρ, σ, θ, α, β, p, q; Hγ(τ, z)), then |a2| ≤ (ξ + 1)(1 −ρ)[Ψ2(σ, α,…
Corollary 2. If f ∈Σ of the style (1) be an element of the class KΣ(ξ, ρ, σ, θ, α, β, p, q; Hγ(τ, z)), then |a2| ≤ (ξ + 1)(1 −ρ)[Ψ2(σ, α, β)]θ p,q|1 + γ −τ| [Ψ1(σ, α, β)]θp,q = |1 + γ −τ| Υ and |a3| ≤ min ( max (
Theorem 3.
Theorem 3. If f ∈Σ of the style (1) be an element of family WΣ(η, δ, λ, σ, θ, α, β, p, q; h), then a3 −ζa2 2 ≤|e1| ∆min max 1,
Theorem 3. If f ∈Σ of the style (1) be an element of family WΣ(η, δ, λ, σ, θ, α, β, p, q; h), then a3 −ζa2 2 ≤|e1| ∆min max 1,
Corollary 3.
Corollary 3. If f ∈Σ of the style (1) be an element of WΣ(η, δ, λ, σ, θ, α, β, p, q; Hγ(τ, z)), then a3 −ζa2 2
Corollary 3. If f ∈Σ of the style (1) be an element of WΣ(η, δ, λ, σ, θ, α, β, p, q; Hγ(τ, z)), then a3 −ζa2 2
Theorem 4.
Theorem 4. If f ∈Σ of the style (1) is in the family KΣ(ξ, ρ, σ, θ, α, β, p, q; h), then a3 −ζa2 2 ≤|e1| Φ min max 1,
Theorem 4. If f ∈Σ of the style (1) is in the family KΣ(ξ, ρ, σ, θ, α, β, p, q; h), then a3 −ζa2 2 ≤|e1| Φ min max 1,
Corollary 4.
Corollary 4. If f ∈Σ of the style (1) be an element of KΣ(ξ, ρ, σ, θ, α, β, p, q; Hγ(τ, z)), then a3 −ζa2 2
Corollary 4. If f ∈Σ of the style (1) be an element of KΣ(ξ, ρ, σ, θ, α, β, p, q; Hγ(τ, z)), then a3 −ζa2 2
Lemma 1
Lemma 1 ([52]). Let the Schwarz function ω(η) be given by ω(η) = ω1η + ω2η2 + ω3η3 +... + ωkηk +... (η ∈∇); then |ω1| ≤1, |ω2| ≤1 −|ω1|2,…
Lemma 1 ([52]). Let the Schwarz function ω(η) be given by ω(η) = ω1η + ω2η2 + ω3η3 + . . . + ωkηk + . . . (η ∈∇); then |ω1| ≤1, |ω2| ≤1 −|ω1|2, |ω2 −tω2 1| ≤1 + (|t| −1)|ω1|2. (10) 71
Theorem 1.
Theorem 1. Let Φ be given by (1). For χ ≥1, 0 ≤α < 1, (µ, β, γ, δ ≥0), γ > δ, β > µ and k ∈N0. If Φ ∈Dℓ Σq(χ, δ, γ, µ; ϕ) and ϱm = 0; m =…
Theorem 1. Let Φ be given by (1). For χ ≥1, 0 ≤α < 1, (µ, β, γ, δ ≥0), γ > δ, β > µ and k ∈N0. If Φ ∈Dℓ Σq(χ, δ, γ, µ; ϕ) and ϱm = 0; m = 2, . . . , k −1, then |ϱk| ≤ 2(1 −q) 1 + (q −qk)χ
Theorem 2.
Theorem 2. For χ ≥1, 0 ≤α < 1, (µ, β, γ, δ ≥0), γ > δ, β > µ and k ∈k0, if Φ ∈Dℓ Σq(χ, δ, γ, µ; ϕ) where Φ(η) is given by (1), then we have…
Theorem 2. For χ ≥1, 0 ≤α < 1, (µ, β, γ, δ ≥0), γ > δ, β > µ and k ∈k0, if Φ ∈Dℓ Σq(χ, δ, γ, µ; ϕ) where Φ(η) is given by (1), then we have the following consequence |ϱ2| ≤min 2 (1 + χq)Ωℓ 2 , 2 q (1 + χ(q2 + q))Ωℓ 3
Corollary 1
Corollary 1 ([37]). Let χ ≥1. A bi-univalent function Φ given by (1) belongs to the class DΣ(q, χ; ϕ) (χ ≥1). If ϱm = 0; m = 2,..., k −1.…
Corollary 1 ([37]). Let χ ≥1. A bi-univalent function Φ given by (1) belongs to the class DΣ(q, χ; ϕ) (χ ≥1). If ϱm = 0; m = 2, . . . , k −1. Then |ϱk| ≤ 2(1 −q) 1 −q + (q −qk)χ (n ≥4). 75
Corollary 2
Corollary 2 ([37]). Let χ ≥1. A bi-univalent function Φ given by (1) belongs to the class Rσ(χ, ϕ)(χ ≥1). If ϱm = 0; m = 2,..., k −1. Then…
Corollary 2 ([37]). Let χ ≥1. A bi-univalent function Φ given by (1) belongs to the class Rσ(χ, ϕ)(χ ≥1). If ϱm = 0; m = 2, . . . , k −1. Then |ϱk| ≤ 2 1 + χ(k −1) (n ≥4). For k = 0 in Theorem 2, we obtain the following corollary.
Corollary 3
Corollary 3 ([37]). Let χ ≥1. A bi-univalent function Φ given by (1) belongs to the class DΣ(q; χ, ϕ). Then (1) |ϱ2| ≤ 2 1+qχ, (2) |ϱ3| ≤ 4…
Corollary 3 ([37]). Let χ ≥1. A bi-univalent function Φ given by (1) belongs to the class DΣ(q; χ, ϕ). Then (1) |ϱ2| ≤ 2 1+qχ, (2) |ϱ3| ≤ 4 (1+qχ)2 + 2 1+(q2+q)χ, (3) |2ϱ2 2 −ϱ3| ≤
Corollary 4.
Corollary 4. A bi-univalent function Φ given by (1) belongs to the class Rσ(χ, ϕ)(χ ≥1). Then (1) |ϱ2| ≤ 2 1+χ, (2) |ϱ3| ≤ 4 (1+3χ)2 + 2…
Corollary 4. A bi-univalent function Φ given by (1) belongs to the class Rσ(χ, ϕ)(χ ≥1). Then (1) |ϱ2| ≤ 2 1+χ, (2) |ϱ3| ≤ 4 (1+3χ)2 + 2 1+2χ. 6. Conclusions This article investigated a novel subclass of bi-univalent functions, DℓΣq(χ, δ, γ, µ; ϕ), on the symmetry disk ∇. For functions belonging to each of these three classes of bi- univalent functions, we calculated estimates for the upper bound of the Taylor–Maclaurin
Lemma 1
Lemma 1 (see Pescar [33]). Let α and β be complex number such that ℜ(α) > 0 and |β| ≦1(β ̸= −1). 82
Lemma 1 (see Pescar [33]). Let α and β be complex number such that ℜ(α) > 0 and |β| ≦1(β ̸= −1). 82
Lemma 2
Lemma 2 (see Pascu [34]). Let ϖ ∈C such that ℜ(ϖ) > 0. If h ∈A satisfies the following inequality:
Lemma 2 (see Pascu [34]). Let ϖ ∈C such that ℜ(ϖ) > 0. If h ∈A satisfies the following inequality:
Lemma 3.
Lemma 3. Let ξ > −1 and ϱ > 0. Then, for ∀ρ ∈D, the function Eξ,ϱ defined by (4) provides the following inequalities: E′ ξ,ϱ(ρ) −Eξ,ϱ(ρ) ρ ≦…
Lemma 3. Let ξ > −1 and ϱ > 0. Then, for ∀ρ ∈D, the function Eξ,ϱ defined by (4) provides the following inequalities: E′ ξ,ϱ(ρ) −Eξ,ϱ(ρ) ρ ≦ ϱ(ξ + 1) (ξ −ϱ + 1)2 (ϱ −1 < ξ), (9)
Theorem 1.
Theorem 1. Let v = 1, 2,..., n, ξv > −1, ϱ > 0 and 2 + √ 2 ϱ −1 < ξv. Also, let γ, β, ηv and ζv be in C such that ℜ(γ) > 0, |β| ≤1(β ̸=…
Theorem 1. Let v = 1, 2, . . . , n, ξv > −1, ϱ > 0 and 2 + √ 2 ϱ −1 < ξv. Also, let γ, β, ηv and ζv be in C such that ℜ(γ) > 0, |β| ≤1(β ̸= −1). Assume that these numbers satisfy the following inequality: 84
Theorem 2.
Theorem 2. Let the parameters ϱ, γ, ηv, ξv and ζv (v = 1, 2,..., n) be as in Theorem 1. Suppose that ξ = min ξ1, ξ2,..., ξn and that the…
Theorem 2. Let the parameters ϱ, γ, ηv, ξv and ζv (v = 1, 2, . . . , n) be as in Theorem 1. Suppose that ξ = min{ξ1, ξ2, . . . , ξn} and that the following inequality holds true: ℜ(γ) ≧ n ∑ v=1 |ηv| (ξ −ϱ + 1)3 + 2ϱ(ξ + 1)2 (ξ −ϱ + 1)2 + (ϱ −1)2 −(2ξϱ + 1) + n ∑ v=1 |ζv| ϱ(ξ + 1)
Theorem 3.
Theorem 3. Let the parameters ϱ, ηv, ξv and ζv (v = 1, 2,..., n) be as in Theorem 1. Suppose that ξ = min ξ1, ξ2,..., ξn and that the…
Theorem 3. Let the parameters ϱ, ηv, ξv and ζv (v = 1, 2, . . . , n) be as in Theorem 1. Suppose that ξ = min{ξ1, ξ2, . . . , ξn} and that the following inequality holds true: 0 < n ∑ v=1
Corollary 1.
Corollary 1. Let γ, β, η and ζ be in C such that ℜ(γ) > 0 and |β| ≤1(β ̸= −1). If the inequality |β| + 1 |γ| 59 4 |η| + 2 3|ζ| ≦1 holds…
Corollary 1. Let γ, β, η and ζ be in C such that ℜ(γ) > 0 and |β| ≤1(β ̸= −1). If the inequality |β| + 1 |γ| 59 4 |η| + 2 3|ζ| ≦1 holds true, then the function γ Z ρ 0 t−ζ+γ−1 et/2η 2et/2 −2
Corollary 2.
Corollary 2. Let γ, β, η and ζ be in C such that ℜ(γ) > 0 and |β| ≤1(β ̸= −1). If the inequality |β| + 1 |γ| 59 2 |η| + 2 3|ζ| ≦1 holds,…
Corollary 2. Let γ, β, η and ζ be in C such that ℜ(γ) > 0 and |β| ≤1(β ̸= −1). If the inequality |β| + 1 |γ| 59 2 |η| + 2 3|ζ| ≦1 holds, then the function 6η3ζγ Z ρ 0 t−3η−3ζ+γ−1 tet −2et + t + 2 η 2et −t2 −2t −2
Corollary 3.
Corollary 3. Let η and ζ be complex numbers such that 59 4 |η| + 2 3|ζ| ≦1. Then the function Z ρ 0 t−ζ et/2η 2et/2 −2 ζ dt is convex…
Corollary 3. Let η and ζ be complex numbers such that 59 4 |η| + 2 3|ζ| ≦1. Then the function Z ρ 0 t−ζ et/2η 2et/2 −2 ζ dt is convex of order δ given by δ = 1 −59 4 |η| −2 3|ζ|.
Corollary 4.
Corollary 4. Let η and ζ be complex numbers such that 0 < 59 2 |η| + 2 3|ζ| ≦1. Then the function 6η3ζ Z ρ 0 t−3η−3ζ tet −2et + t + 2 η…
Corollary 4. Let η and ζ be complex numbers such that 0 < 59 2 |η| + 2 3|ζ| ≦1. Then the function 6η3ζ Z ρ 0 t−3η−3ζ tet −2et + t + 2 η 2et −t2 −2t −2 ζ dt is convex of order δ given by δ = 1 −59 2 |η| −2
Theorem 4.
Theorem 4. Let v = 1, 2,..., n, ξv > −1, ϱ > 0 and 2ϱ −1 < ξv. Also, let β, ηv and ζv be in C such that |β| ≤1(β ̸= −1) and ℜ
Theorem 4. Let v = 1, 2, . . . , n, ξv > −1, ϱ > 0 and 2ϱ −1 < ξv. Also, let β, ηv and ζv be in C such that |β| ≤1(β ̸= −1) and ℜ
Theorem 5.
Theorem 5. Let the parameters ϱ, ηv, ξv and ζv (v = 1, 2,..., n) be as in Theorem 4. Suppose that ξ = min ξ1, ξ2,..., ξn and that the…
Theorem 5. Let the parameters ϱ, ηv, ξv and ζv (v = 1, 2, . . . , n) be as in Theorem 4. Suppose that ξ = min{ξ1, ξ2, . . . , ξn} and that the following inequality holds true: n ∑ v=1 " |ηv| (ξ + 1)2 (ξ −ϱ + 1)(ξ −2ϱ + 1) + |ζv| ξ + 1 ξ −ϱ + 1 2# ≦1. Then the function Kξ1,ξ2,...,ξn;ϱ
Theorem 6.
Theorem 6. Let v = 1, 2,..., n, ξv > −1, ϱ > 0 and 2ϱ −1 < ξv. Also, let ηv and ζv be in C such that ℜ
Theorem 6. Let v = 1, 2, . . . , n, ξv > −1, ϱ > 0 and 2ϱ −1 < ξv. Also, let ηv and ζv be in C such that ℜ
Corollary 5.
Corollary 5. Let β, η and ζ be in C such that ℜ(1 + η) > 0 and |β| ≤1(β ̸= −1). If these numbers satisfy the inequality: |β| + 1 |1 + η| 9…
Corollary 5. Let β, η and ζ be in C such that ℜ(1 + η) > 0 and |β| ≤1(β ̸= −1). If these numbers satisfy the inequality: |β| + 1 |1 + η| 9 5|η| + 36 25|ζ| ≦1, 90
Corollary 6.
Corollary 6. Let β, η and ζ be in C such that ℜ(1 + η) > 0 and |β| ≤1(β ̸= −1). If these numbers satisfy the follwing inequality |β| + 1 |1…
Corollary 6. Let β, η and ζ be in C such that ℜ(1 + η) > 0 and |β| ≤1(β ̸= −1). If these numbers satisfy the follwing inequality |β| + 1 |1 + η| 8 3|η| + 16 9 |ζ| ≦1, then the function " 3η(1 + η) Z ρ 0
Corollary 7.
Corollary 7. Let η and ζ be a complex numbers such that ℜ(1 + η) > 0 and 0 < 9 5|η| + 36 25|ζ| ≦1. 91
Corollary 7. Let η and ζ be a complex numbers such that ℜ(1 + η) > 0 and 0 < 9 5|η| + 36 25|ζ| ≦1. 91
Corollary 8.
Corollary 8. Let η and ζ be a complex numbers such that ℜ(1 + η) > 0 and 0 < 8 3|η| + 16 9 |ζ| ≦1. Then the function defined by (31) is…
Corollary 8. Let η and ζ be a complex numbers such that ℜ(1 + η) > 0 and 0 < 8 3|η| + 16 9 |ζ| ≦1. Then the function defined by (31) is convex of order δ given by δ = 1 −8 3|η| −16 9 |ζ|. 5. Univalence and Convexity Conditions for the Integral Operator in (7) In this section, we derive the univalence and convexity results for the integral operator defined by (7).
Theorem 7.
Theorem 7. Let v = 1, 2,..., n, ξv > −1, ϱ > 0 and 2 + √ 2 ϱ −1 < ξv. Also, let γ, β, ηv and ζv be in C such that ℜ(γ) > 0, |β| ≤1(β ̸=…
Theorem 7. Let v = 1, 2, . . . , n, ξv > −1, ϱ > 0 and 2 + √ 2 ϱ −1 < ξv. Also, let γ, β, ηv and ζv be in C such that ℜ(γ) > 0, |β| ≤1(β ̸= −1). Assume that these numbers satisfy the following inequality: |β| + 1 |γ| ( n ∑
Theorem 8.
Theorem 8. Let the parameters ϱ, γ, ηv, ξv and ζv (v = 1, 2,..., n) be as in Theorem 7. Suppose that ξ = min ξ1, ξ2,..., ξn and that the…
Theorem 8. Let the parameters ϱ, γ, ηv, ξv and ζv (v = 1, 2, . . . , n) be as in Theorem 7. Suppose that ξ = min{ξ1, ξ2, . . . , ξn} and that the following inequality holds true: ℜ(γ) ≧ ( n ∑ v=1 |ηv| (ξ −ϱ + 1)3 + 2ϱ(ξ + 1)2 (ξ −ϱ + 1)2 + (ϱ −1)2 −(2ξϱ + 1) + n ∑ v=1 |ζv|
Theorem 9.
Theorem 9. Let v = 1, 2,..., n, ξv > −1, ϱ > 0 and 2 + √ 2 ϱ −1 < ξv. Also, let γ, ηv and ζv be in C such that ℜ(γ) > 0. Assume that…
Theorem 9. Let v = 1, 2, . . . , n, ξv > −1, ϱ > 0 and 2 + √ 2 ϱ −1 < ξv. Also, let γ, ηv and ζv be in C such that ℜ(γ) > 0. Assume that these numbers satisfy the following inequality: 0 < n ∑ v=1 |ηv| (ξ −ϱ + 1)3 + 2ϱ(ξ + 1)2 (ξ −ϱ + 1)2 + (ϱ −1)2 −(2ξϱ + 1)
Corollary 9.
Corollary 9. Let γ, β, η and ζ be in C such that ℜ(γ) > 0, |β| ≤1(β ̸= −1). If these numbers satisfy the inequality: |β| + 1 |γ| 59 4 |η|…
Corollary 9. Let γ, β, η and ζ be in C such that ℜ(γ) > 0, |β| ≤1(β ̸= −1). If these numbers satisfy the inequality: |β| + 1 |γ| 59 4 |η| + 16 9 |ζ| ≦1 then the function γ Z ρ 0 tγ−1 et/2η
Corollary 10.
Corollary 10. Let γ, β, η and ζ be in C such that ℜ(γ) > 0 and β ̸= −1. If these numbers satisfy the inequality |β| + 1 |γ| 59 2 |η| + 16…
Corollary 10. Let γ, β, η and ζ be in C such that ℜ(γ) > 0 and β ̸= −1. If these numbers satisfy the inequality |β| + 1 |γ| 59 2 |η| + 16 9 |ζ| ≦1 then the function 6ηγ Z ρ 0 t−3η+γ−1 tet −2et + t + 2 ηe3ζ
Corollary 11.
Corollary 11. Let η and ζ be complex numbers such that 59 4 |η| + 16 9 |ζ| ≦1. Then the function Z ρ 0 et/2η e2et/2−2ζ dt 94
Corollary 11. Let η and ζ be complex numbers such that 59 4 |η| + 16 9 |ζ| ≦1. Then the function Z ρ 0 et/2η e2et/2−2ζ dt 94
Corollary 12.
Corollary 12. Let η and ζ be complex numbers such that 59 2 |η| + 16 9 |ζ| ≦1. Then the function 6η Z ρ 0 t−3η tet −2et + t + 2 ηe3ζ …
Corollary 12. Let η and ζ be complex numbers such that 59 2 |η| + 16 9 |ζ| ≦1. Then the function 6η Z ρ 0 t−3η tet −2et + t + 2 ηe3ζ 2et−t2−2t−2 t2 dt is convex of order δ given by
Lemma 1
Lemma 1 (see [58]). If p is a function with a positive real part and p(z) = 1 + ∞ ∑ l=1 clzl, then |cl| ≦2. The problem of finding bounds…
Lemma 1 (see [58]). If p is a function with a positive real part and p(z) = 1 + ∞ ∑ l=1 clzl, then |cl| ≦2. The problem of finding bounds for the coefficients has always been a key concern in geometric function theory. The size of their coefficients can determine a number of properties of analytic functions, including univalency, rate of growth and distortion. Many scholars have used a variety of methods to overcome the aforementioned issues. Similar to univalent functions, bi-univalent function co
Theorem 1.
Theorem 1. If h has the series representation stated in (1) and belongs to the class KΣ(q, α, ϱ), and if ai = 0 and 2 ≦i ≦l −1, then |al|…
Theorem 1. If h has the series representation stated in (1) and belongs to the class KΣ(q, α, ϱ), and if ai = 0 and 2 ≦i ≦l −1, then |al| ≦Γq(2)Γq(l + 1 −ϱ)(2(1 −α) + l) [l]qΓq(2 −ϱ)Γq(l + 1) (l ≧3).
Corollary 1.
Corollary 1. If the function h has the series representation stated in (1) and belongs to the class KΣ(q, 0, 1), and if ai = 0 (2 ≦i ≦l…
Corollary 1. If the function h has the series representation stated in (1) and belongs to the class KΣ(q, 0, 1), and if ai = 0 (2 ≦i ≦l −1), then |al| ≦Γq(2)Γq(l)(2 + l) [l]qΓq(l + 1) (l ≧3).
Corollary 2.
Corollary 2. If the function h has the series representation stated in (1) and belongs to KΣ(q, α, 1), and if ai = 0 (2 ≦i ≦l −1), then…
Corollary 2. If the function h has the series representation stated in (1) and belongs to KΣ(q, α, 1), and if ai = 0 (2 ≦i ≦l −1), then |al| ≦Γq(2)Γq(l)(2(1 −α) + l) [l]qΓq(l + 1) (l ≧3). 106
Corollary 3.
Corollary 3. If the function h has the series representation stated in (1) and belongs to the class KΣ(q →1−, α, ϱ), and if ai = 0 (2 ≦i ≦l…
Corollary 3. If the function h has the series representation stated in (1) and belongs to the class KΣ(q →1−, α, ϱ), and if ai = 0 (2 ≦i ≦l −1), then |al| ≦Γ(2)Γ(l + 1 −ϱ)(2(1 −α) + l) lΓ(2 −ϱ)Γ(l + 1) (l ≧3).
Corollary 4.
Corollary 4. If the h has the series representation stated in (1) and belongs to the class KΣ(q → 1−, α, 1), and if ai = 0 (2 ≦i ≦l −1),…
Corollary 4. If the h has the series representation stated in (1) and belongs to the class KΣ(q → 1−, α, 1), and if ai = 0 (2 ≦i ≦l −1), then |al| ≦Γ(2)Γ(l)(2(1 −α) + l) lΓ(l + 1) (l ≧3). The following known consequence of Theorem 1 for ϱ = 0 and q →1−was demon- strated in [27].
Corollary 5
Corollary 5 (see [27]). Let h ∈KΣ(α). If ai+1 = 0 (1 ≦i ≦l), then |al| ≦1 + 2(1 −α) l (l ≧3).
Corollary 5 (see [27]). Let h ∈KΣ(α). If ai+1 = 0 (1 ≦i ≦l), then |al| ≦1 + 2(1 −α) l (l ≧3).
Corollary 6
Corollary 6 (see [56]). If the function h has the series representation stated in (1) and belongs to the class KΣ(q →1−, 0, ϱ), and if ai =…
Corollary 6 (see [56]). If the function h has the series representation stated in (1) and belongs to the class KΣ(q →1−, 0, ϱ), and if ai = 0 (2 ≦i ≦l −1), then |al| ≦(2 + l)Γ(l + 1 −ϱ) lΓ(2 −ϱ)Γ(l + 1) (l ≧3). As a special form of Theorem 1, our next result (Theorem 2 below) provides estimates for the initial coefficients |a2| and |a3|, and also for the Fekete–Szegö-type functional involved in a3 −a2 2 for functions in the class KΣ(m, α, q).
Theorem 2.
Theorem 2. Let the function h ∈KΣ(q, α, ϱ) be given by (1). Then, |a2| ≦
Theorem 2. Let the function h ∈KΣ(q, α, ϱ) be given by (1). Then, |a2| ≦
Corollary 7.
Corollary 7. Let the function h ∈KΣ(q, α, 1) be given by (1). Then, |a2| ≦
Corollary 7. Let the function h ∈KΣ(q, α, 1) be given by (1). Then, |a2| ≦
Corollary 8.
Corollary 8. Let h ∈KΣ(q, 0, 1) be given by (1). Then, |a2| ≦
Corollary 8. Let h ∈KΣ(q, 0, 1) be given by (1). Then, |a2| ≦
Corollary 9.
Corollary 9. Let h ∈KΣ(q →1−, α, ϱ) be given by (1). Then, |a2| ≦
Corollary 9. Let h ∈KΣ(q →1−, α, ϱ) be given by (1). Then, |a2| ≦
Corollary 10
Corollary 10 (see [27]). Let h ∈KΣ(q →1−, α, 0). Then, |a2| ≦ p 2(1 −α) 0 ≦α < 1 2
Corollary 10 (see [27]). Let h ∈KΣ(q →1−, α, 0). Then, |a2| ≦ p 2(1 −α) 0 ≦α < 1 2
Corollary 11
Corollary 11 (see [56]). Let h ∈KΣ(q →1−, 0, ϱ) be given by (1). Then, |a2| ≦min s 2Γ(3 −ϱ)Γ(4 −ϱ) Γ(2 −ϱ) 3Γ(4)Γ(3 −ϱ)…
Corollary 11 (see [56]). Let h ∈KΣ(q →1−, 0, ϱ) be given by (1). Then, |a2| ≦min s 2Γ(3 −ϱ)Γ(4 −ϱ) Γ(2 −ϱ){3Γ(4)Γ(3 −ϱ) −2Γ(3)Γ(4 −ϱ)}, 2Γ(2)Γ(3 −ϱ) 2Γ(2 −ϱ)Γ(3) −Γ(2)Γ(3 −ϱ)
Lemma 1
Lemma 1 ([19]). Let ℑ= φ + ψ where φ and ψ are given by (1) and suppose that ϱ ≥0, 0 ≤γ < 1 and ∞ ∑ ν=2 ν[1 + ϱ ν2 −1 ]|aν| + ∞ ∑ ν=1…
Lemma 1 ([19]). Let ℑ= φ + ψ where φ and ψ are given by (1) and suppose that ϱ ≥0, 0 ≤γ < 1 and ∞ ∑ ν=2 ν[1 + ϱ ν2 −1 ]|aν| + ∞ ∑ ν=1 ν[1 + ϱ
Lemma 2
Lemma 2 ([6]). Let ℑ= φ + ψ where φ and ψ are given by (2) and suppose that 0 ≤τ < 1. Then ℑ∈T NHF(τ) if and only if ∞ ∑ ν=2 ν|aν| + ∞ ∑…
Lemma 2 ([6]). Let ℑ= φ + ψ where φ and ψ are given by (2) and suppose that 0 ≤τ < 1. Then ℑ∈T NHF(τ) if and only if ∞ ∑ ν=2 ν|aν| + ∞ ∑ ν=1 ν|bν| ≤1 −τ. (9) Moreover, if ℑ∈T NHF(τ), then |aν| ≤1 −τ ν , ν ≥2,
Lemma 3
Lemma 3 ([18]). Let ℑ= φ + ψ where φ and ψ are given by (2), and suppose that 0 ≤τ < 1. Then ℑ∈T RHF(τ) if and only if ∞ ∑ ν=2 ν2|aν| + ∞ ∑…
Lemma 3 ([18]). Let ℑ= φ + ψ where φ and ψ are given by (2), and suppose that 0 ≤τ < 1. Then ℑ∈T RHF(τ) if and only if ∞ ∑ ν=2 ν2|aν| + ∞ ∑ ν=1 ν2|bν| ≤1 −τ. (12) Moreover, if ℑ∈T RHF(τ), then |aν| ≤1 −τ ν2 , ν ≥2
Lemma 4
Lemma 4 ([5]). If ℑ= φ + ψ ∈S∗ HF where φ and ψ are given by (1) with b1 = 0, then |aν| ≤(2ν + 1)(ν + 1) 6 and |bν| ≤(2ν −1)(ν −1) 6. (15)
Lemma 4 ([5]). If ℑ= φ + ψ ∈S∗ HF where φ and ψ are given by (1) with b1 = 0, then |aν| ≤(2ν + 1)(ν + 1) 6 and |bν| ≤(2ν −1)(ν −1) 6 . (15)
Lemma 5
Lemma 5 ([5]). If ℑ= φ + ψ ∈KHF where φ and ψ are given by (1) with b1 = 0, then |aν| ≤ν + 1 2 and |bν| ≤ν −1 2. (16) 119
Lemma 5 ([5]). If ℑ= φ + ψ ∈KHF where φ and ψ are given by (1) with b1 = 0, then |aν| ≤ν + 1 2 and |bν| ≤ν −1 2 . (16) 119
Theorem 1.
Theorem 1. Let ϱ ≥0, γ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0. If h 2ϱ χ (5) ρ,ϵ(1) + χ (5) η,ϵ(1) + 23ϱχ (4) ρ,ϵ(1) + (67ϱ + 2)χ (3)
Theorem 1. Let ϱ ≥0, γ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0 . If h 2ϱ χ (5) ρ,ϵ(1) + χ (5) η,ϵ(1) + 23ϱχ (4) ρ,ϵ(1) + (67ϱ + 2)χ (3)
Theorem 2.
Theorem 2. Let ϱ ≥0, γ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0. If [ϱχ (4) ρ,ϵ(1) + 7ϱχ (3) ρ,ϵ(1) + (9ϱ + 1)χ (2) ρ,ϵ(1) + 2χ′ ρ,ϵ(1) + ϱχ (4) η,ϵ +…
Theorem 2. Let ϱ ≥0, γ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0 . If [ϱχ (4) ρ,ϵ(1) + 7ϱχ (3) ρ,ϵ(1) + (9ϱ + 1)χ (2) ρ,ϵ(1) + 2χ′ ρ,ϵ(1) + ϱχ (4) η,ϵ + 5ϱχ (3) η,ϵ(1) + (5ϱ −1)χ
Theorem 3.
Theorem 3. Let ϱ ≥0, γ, τ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0. If (1 −τ) h ϱ χ (2) ρ,ϵ(1) + χ (2) η,ϵ(1) + ϱ χ′
Theorem 3. Let ϱ ≥0, γ, τ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0 . If (1 −τ) h ϱ χ (2) ρ,ϵ(1) + χ (2) η,ϵ(1) + ϱ χ′
Theorem 4.
Theorem 4. Let ϱ ≥0, γ, τ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0. If (1 −τ) ϱ χ′ ρ,ϵ(1) + χ′ η,ϵ(1) + 1 Z 0 χρ,ϵ(s)
Theorem 4. Let ϱ ≥0, γ, τ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0 . If (1 −τ) ϱ χ′ ρ,ϵ(1) + χ′ η,ϵ(1) + 1 Z 0 χρ,ϵ(s)
Theorem 5.
Theorem 5. Let ϱ ≥0, γ, τ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0. If χρ,ϵ(1) + χη,ϵ(1) ≤3 −|b1| 1 −γ then Θ(HF(ϱ, γ)) ⊂HF(ϱ, γ).
Theorem 5. Let ϱ ≥0, γ, τ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0 . If χρ,ϵ(1) + χη,ϵ(1) ≤3 −|b1| 1 −γ then Θ(HF(ϱ, γ)) ⊂HF(ϱ, γ).
Corollary 1.
Corollary 1. Let γ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0. If 2 χ (3) ρ,ϵ(1) + χ (3) η,ϵ(1) + 9χ (2) ρ,ϵ(1) + 6χ ′ ρ,ϵ(1) + 3χ
Corollary 1. Let γ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0 . If 2 χ (3) ρ,ϵ(1) + χ (3) η,ϵ(1) + 9χ (2) ρ,ϵ(1) + 6χ ′ ρ,ϵ(1) + 3χ
Corollary 2.
Corollary 2. Let γ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0. If [χ (2) ρ,ϵ(1) −χ (2) η,ϵ(1) + 2 χ′ ρ,ϵ(1) −χ′ η,ϵ(1) ] ≤2(1 −γ), (31)
Corollary 2. Let γ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0 . If [χ (2) ρ,ϵ(1) −χ (2) η,ϵ(1) + 2 χ′ ρ,ϵ(1) −χ′ η,ϵ(1) ] ≤2(1 −γ), (31)
Corollary 3.
Corollary 3. Let γ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0. If (1 −τ) χρ,ϵ(1) + χη,ϵ(1) −2 ≤1 −γ −|b1|, then Θ(T NHF(τ)) ⊂HF(γ).
Corollary 3. Let γ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0 . If (1 −τ) χρ,ϵ(1) + χη,ϵ(1) −2 ≤1 −γ −|b1|, then Θ(T NHF(τ)) ⊂HF(γ).
Corollary 4.
Corollary 4. Let γ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0. If (1 −τ) 1 Z 0 χρ,ϵ(s) s dt + 1 Z 0 χη,ϵ(s)
Corollary 4. Let γ ∈[0, 1) and ρ, ϵ, η, δ ̸∈Z− 0 . If (1 −τ) 1 Z 0 χρ,ϵ(s) s dt + 1 Z 0 χη,ϵ(s)
Lemma 1.
Lemma 1. The function ϕBS(ξ) = ξ eξ−1 2 satisfies min |ξ|=ℓRe ϕBS(ξ) = ϕBS(ℓ) = min |ξ|=ℓ|ϕBS(ξ)| max |ξ|=ℓRe ϕBS(ξ) = ϕBS(−ℓ) = max
Lemma 1. The function ϕBS(ξ) = ξ eξ−1 2 satisfies min |ξ|=ℓRe ϕBS(ξ) = ϕBS(ℓ) = min |ξ|=ℓ|ϕBS(ξ)| max |ξ|=ℓRe ϕBS(ξ) = ϕBS(−ℓ) = max
Theorem 1.
Theorem 1. The class S∗ BS satisfies the following inclusion relations: 1. If 0 ≤λ ≤ 1 (e−1)2, then S∗ BS ⊂S∗(λ). 2. If β ≥( e e−1)2, then…
Theorem 1. The class S∗ BS satisfies the following inclusion relations: 1. If 0 ≤λ ≤ 1 (e−1)2 , then S∗ BS ⊂S∗(λ). 2. If β ≥( e e−1)2, then S∗ BS ⊂RS∗(1/β) ⊂M(β). 3. S∗ BS ⊂SS∗(β), where β0 ≤β ≤1, wherein β0 = 2h(y2)/π ≈0.6454469651m and h is
Lemma 2
Lemma 2 ([25]). If p ∈Pn(λ), then for |ξ| = ℓ,
Lemma 2 ([25]). If p ∈Pn(λ), then for |ξ| = ℓ,
Lemma 3
Lemma 3 ([26]). Let p ∈P. Then, |jp3 1 −kp1p2 + lp3| ≤2|j| + 2|k −2j| + 2|j −k + l|.
Lemma 3 ([26]). Let p ∈P. Then, |jp3 1 −kp1p2 + lp3| ≤2|j| + 2|k −2j| + 2|j −k + l|.
Lemma 4
Lemma 4 ([27]). If p ∈Pn[a, b], then for |ξ| = ℓ, p(ξ) −1 −abℓ2n 1 −b2ℓ2n ≤(a −b)ℓn 1 −b2ℓ2n. If p ∈Pn(λ), then for |ξ| = ℓ, p(ξ) −(1 + (1…
Lemma 4 ([27]). If p ∈Pn[a, b], then for |ξ| = ℓ, p(ξ) −1 −abℓ2n 1 −b2ℓ2n ≤(a −b)ℓn 1 −b2ℓ2n . If p ∈Pn(λ), then for |ξ| = ℓ, p(ξ) −(1 + (1 −2λ))ℓ2n 1 −ℓ2n ≤2(1 −λ)ℓn 1 −ℓ2n . In the following lemmas, we find disks centered at (ν, 0) and (1, 0) of the largest and the smallest radii, respectively, such that ℧BS := ϕBS(E) lies in the disk with the smallest radius and contains the largest disk.
Lemma 5.
Lemma 5. Let 1 e−1 2 ≤ν ≤ e e−1 2. Then, w ∈C: |w −ν| < ℓν ⊂℧BS ⊂ ( w ∈C: |w −1| < e e −1
Lemma 5. Let 1 e−1 2 ≤ν ≤ e e−1 2. Then, {w ∈C : |w −ν| < ℓν} ⊂℧BS ⊂ ( w ∈C : |w −1| < e e −1
Theorem 2.
Theorem 2. The sharp RS∗ BS,n for Sn is RS∗ BS,n(Sn) =
Theorem 2. The sharp RS∗ BS,n for Sn is RS∗ BS,n(Sn) =
Theorem 3.
Theorem 3. Let R1 =
Theorem 3. Let R1 =
Theorem 4.
Theorem 4. The S∗ BS,n-radius for S∗ n[a, b] is RS∗ BS,n(S∗ n[a, b]) = min 1; ℓ1, −1 ≤b ≤0 < a ≤1, min 1; ℓ2, 0 < b < a ≤1, where ℓ1 =
Theorem 4. The S∗ BS,n-radius for S∗ n[a, b] is RS∗ BS,n(S∗ n[a, b]) = min{1; ℓ1}, −1 ≤b ≤0 < a ≤1, min{1; ℓ2}, 0 < b < a ≤1, where ℓ1 =
Theorem 5.
Theorem 5. Let −1 < b < a ≤1. If either (a) (1 −b) ≤(e −1)2(1 −a) and 2(1 −b2) ≤(1 −ab)(e −1)2 < (1 −b2)(1 + e2) or if (b) (a + 1)(e −1)2…
Theorem 5. Let −1 < b < a ≤1. If either (a) (1 −b) ≤(e −1)2(1 −a) and 2(1 −b2) ≤(1 −ab)(e −1)2 < (1 −b2)(1 + e2) or if (b) (a + 1)(e −1)2 ≤e2(1 + b) and (1 −b2) 1 + e2 ≤2(1 −ab)(e −1)2 ≤2e2(1 −b2) hold, then S∗ n[a, b] ⊂S∗ BS,n. 135
Theorem 6.
Theorem 6. The sharp radii for S∗ L, S∗ RL, S∗ e, and S∗ lim are RS∗ BS(S∗ L) = (e−1)4−1 (e−1)4 ≈0.889, RS∗ BS(S∗ RL) = (5+4 √ 2)(e−1)4+(−6
Theorem 6. The sharp radii for S∗ L, S∗ RL, S∗ e, and S∗ lim are RS∗ BS(S∗ L) = (e−1)4−1 (e−1)4 ≈0.889, RS∗ BS(S∗ RL) = (5+4 √ 2)(e−1)4+(−6
Theorem 7.
Theorem 7. The sharp radii for functions in the families F1, F2, and F3 respectively, are: RS∗ BS,n(F1) = e(e−2) 2n(e−1)2+√…
Theorem 7. The sharp radii for functions in the families F1, F2, and F3 respectively, are: RS∗ BS,n(F1) = e(e−2) 2n(e−1)2+√ 1+(4n2+1)(e−1)4−2(e−1)2 1/n , RS∗ BS,n(F2) = 2e(e−2) 3n(e−1)2+√ (9n2+4n+4)(e−1)4−4(n+2)(e−1)2+4
Lemma 6
Lemma 6 ([5]). Let p ∈P and be of the form (6). Then for v, a complex number |p2 −vp2 1| ≤2 max(1, |2v −1|).
Lemma 6 ([5]). Let p ∈P and be of the form (6). Then for v, a complex number |p2 −vp2 1| ≤2 max(1, |2v −1|).
Lemma 7
Lemma 7 ([29,30]). Let p ∈P and be of the form (6) such that |ρ| ≤1, and |η| ≤1. Then, 2p2 = p2 1 + ρ(4 −p2 1), (10) 4p3 = p3 1 + 2(4 −p2…
Lemma 7 ([29,30]). Let p ∈P and be of the form (6) such that |ρ| ≤1, and |η| ≤1. Then, 2p2 = p2 1 + ρ(4 −p2 1), (10) 4p3 = p3 1 + 2(4 −p2 1)p1ρ −(4 −p2 1)p2 1ρ + 2(4 −p2 1)(1 −|ρ|2)η,
Lemma 8
Lemma 8 ([31]). Let ϖ ∈B be given by ϖ(z) = ∞ ∑ n=0 cnξn, and thus ψ(u, v) = c3 + µ1c1c2 + µ2c3 1. Then, ψ(u, v) ≤|ν| if (u, v) ∈D6, where…
Lemma 8 ([31]). Let ϖ ∈B be given by ϖ(z) = ∞ ∑ n=0 cnξn, and thus ψ(u, v) = c3 + µ1c1c2 + µ2c3 1 . Then, ψ(u, v) ≤|ν| if (u, v) ∈D6, where D6 = (u, v) : 2 ≤|µ| ≤4, ν ≥1 12
Lemma 9
Lemma 9 ([32]). Let E:= ρ ∈C: |ρ| ≤1, and, for j, k, and l ∈R, let Y(j, k, l):= max n |j + kρ + lρ2| + 1 −|ρ|2: ρ ∈E o. (12) If jl ≥0, then…
Lemma 9 ([32]). Let E := {ρ ∈C : |ρ| ≤1}, and, for j, k, and l ∈R, let Y(j, k, l) := max n |j + kρ + lρ2| + 1 −|ρ|2 : ρ ∈E o . (12) If jl ≥0, then Y(j, k, l) = |j| + |k| + |l|, |k| ≥2(1 −|l|), 1 + |j| +
Theorem 8.
Theorem 8. Let ϝ ∈S∗ BS and be of the form (2). Then, |d2| ≤1, |d3| ≤17 24, |d4| ≤29 72. These bounds are the best possible.
Theorem 8. Let ϝ ∈S∗ BS and be of the form (2). Then, |d2| ≤1, |d3| ≤17 24, |d4| ≤29 72. These bounds are the best possible.
Theorem 9.
Theorem 9. Let ϝ ∈S∗ BS and have the series representation given in (2). Then, |d3 −d2 2| ≤1 2. (20)
Theorem 9. Let ϝ ∈S∗ BS and have the series representation given in (2). Then, |d3 −d2 2| ≤1 2. (20)
Theorem 10.
Theorem 10. Let ϝ ∈S∗ BS and have the series representation given in (2). Then, H2,2(ϝ) = |d2d4 −d2 3| ≤521 576. (21) The equality is…
Theorem 10. Let ϝ ∈S∗ BS and have the series representation given in (2). Then, H2,2(ϝ) = |d2d4 −d2 3| ≤521 576. (21) The equality is obtained by the ϝ1 given in (19).
Lemma 1
Lemma 1 (see [42]). If h ∈P, then |ck| ≤2 for each k, where P is the family of all functions h, analytic in U, for which ℜ h(z) > 0 (z ∈U),…
Lemma 1 (see [42]). If h ∈P, then |ck| ≤2 for each k, where P is the family of all functions h, analytic in U, for which ℜ{h(z)} > 0 (z ∈U), where h(z) = 1 + c1z + c2z2 + · · · (z ∈U). We begin by estimating the coefficients |a2| and |a3| for functions in the class Bσ,m;℘ j,δ,Σ (λ, Ξ). Let P(z) be defined by P(z) : = 1 + ϖ(z) 1 −ϖ(z) = 1 + c1z + c2z2 + · · · . It is evident that
Theorem 1.
Theorem 1. Let assume that the f function is as in (1) and in the class Bσ,m;℘ j,δ,Σ (λ, Ξ). Then |a2| ≤min 1 (λ+1)V2, q 2 |…
Theorem 1. Let assume that the f function is as in (1) and in the class Bσ,m;℘ j,δ,Σ (λ, Ξ). Then |a2| ≤min 1 (λ+1)V2 , q 2 |{(λ−1)(λ+2)−(℘−1)(λ+1)2}V2 2 +2(λ+2)V3| (22) and |a3| ≤min
Corollary 1.
Corollary 1. Let assume that the f function is as in (1) and in the class Sσ,m;℘ j,δ,Σ (Ξ). Then |a2| ≤min ( 1 V2, q 2 |4V3−(℘+1)V2 2 |…
Corollary 1. Let assume that the f function is as in (1) and in the class Sσ,m;℘ j,δ,Σ (Ξ). Then |a2| ≤min ( 1 V2 , q 2 |4V3−(℘+1)V2 2 | (42) and |a3| ≤min
Corollary 2.
Corollary 2. Let assume that the f function is as in (1) and in the class Rσ,m;℘ j,δ,Σ (Ξ). Then |a2| ≤min ( 1 2V2, q 1 3V3−2(℘−1)V2 2 (44)…
Corollary 2. Let assume that the f function is as in (1) and in the class Rσ,m;℘ j,δ,Σ (Ξ). Then |a2| ≤min ( 1 2V2 , q 1 3V3−2(℘−1)V2 2 (44) and |a3| ≤min
Theorem 2.
Theorem 2. Let assume that the f function is as in (1) and f ∈Mσ,m;℘ j,δ,Σ (τ, Ξ), τ ≥1. Then |a2| ≤min 1 2(2τ−1)V2, 1 q…
Theorem 2. Let assume that the f function is as in (1) and f ∈Mσ,m;℘ j,δ,Σ (τ, Ξ), τ ≥1. Then |a2| ≤min 1 2(2τ−1)V2 , 1 q |(1+τ)−2(2τ−1)2(℘−1)|V2 2 (46) and |a3| ≤min
Corollary 3.
Corollary 3. Let assume that the f function is as in (1) and f ∈Kσ,m;℘ j,δ,Σ (Ξ). Then |a2| ≤min ( 1 2V2, 1 √ |2−2(℘−1)|V2 2 (61) and |a3|…
Corollary 3. Let assume that the f function is as in (1) and f ∈Kσ,m;℘ j,δ,Σ (Ξ). Then |a2| ≤min ( 1 2V2 , 1 √ |2−2(℘−1)|V2 2 (61) and |a3| ≤min
Lemma 2
Lemma 2 ([44]). Let k ∈R and z1, z2 ∈C. If |z1| < R and |z2| < R then |(k + 1)z1 + (k −1)z2| ≤ 2|k|R, |k| ≥1 2R |k| ≤1. (63)
Lemma 2 ([44]). Let k ∈R and z1, z2 ∈C. If |z1| < R and |z2| < R then |(k + 1)z1 + (k −1)z2| ≤ 2|k|R, |k| ≥1 2R |k| ≤1. (63)
Lemma 3
Lemma 3 ([44]). Let k, l ∈R and z1, z2 ∈C. If |z1| < R and |z2| < R then |(k + l)z1 + (k −l)z2| ≤ 2|k|R, |k| ≥|l| 2|l|R |k| ≤|l|.…
Lemma 3 ([44]). Let k, l ∈R and z1, z2 ∈C. If |z1| < R and |z2| < R then |(k + l)z1 + (k −l)z2| ≤ 2|k|R, |k| ≥|l| 2|l|R |k| ≤|l|. (64) Now, we obtain Fekete-Szegö inequalities for f ∈Bσ,m;℘ j,δ,Σ (λ, Ξ) :
Theorem 3.
Theorem 3. For ℵ∈R, let assume that the f function is as in (1) and f ∈Bσ,m;℘ j,δ,Σ (λ, Ξ), then a3 −ℵa2 2 ≤ ( 1 (2+λ)V3; 0 ≤|h(ℵ)| ≤ 1…
Theorem 3. For ℵ∈R, let assume that the f function is as in (1) and f ∈Bσ,m;℘ j,δ,Σ (λ, Ξ), then a3 −ℵa2 2 ≤ ( 1 (2+λ)V3 ; 0 ≤|h(ℵ)| ≤ 1 4(2+λ)V3 4|h(ℵ)| ; |h(ℵ)| ≥
Corollary 4.
Corollary 4. For ℵ∈R, let assume that the f function is as in (1) and f ∈Sσ,m;℘ j,δ,Σ (Ξ), then a3 −ℵa2 2 ≤ ( 1 2V3; 0 ≤|h(ℵ)| ≤ 1 8V3…
Corollary 4. For ℵ∈R, let assume that the f function is as in (1) and f ∈Sσ,m;℘ j,δ,Σ (Ξ), then a3 −ℵa2 2 ≤ ( 1 2V3 ; 0 ≤|h(ℵ)| ≤ 1 8V3 4|h(ℵ)| ; |h(ℵ)| ≥
Corollary 5.
Corollary 5. For ℵ∈R, let assume that the f function is as in (1) and f ∈Rσ,m;℘ j,δ,Σ (Ξ), then a3 −ℵa2 2 ≤ ( 1 3V3; 0 ≤|h(ℵ)| ≤ 1 12V3…
Corollary 5. For ℵ∈R, let assume that the f function is as in (1) and f ∈Rσ,m;℘ j,δ,Σ (Ξ), then a3 −ℵa2 2 ≤ ( 1 3V3 ; 0 ≤|h(ℵ)| ≤ 1 12V3 4|h(ℵ)| ; |h(ℵ)| ≥
Theorem 4.
Theorem 4. For ν ∈R, let assume that the f function is as in (1) and f ∈Mσ,m;℘ j,δ,Σ (τ, Ξ), then a3 −νa2 2 ≤ 2 3(3τ −1)V3; 0…
Theorem 4. For ν ∈R, let assume that the f function is as in (1) and f ∈Mσ,m;℘ j,δ,Σ (τ, Ξ), then a3 −νa2 2 ≤ 2 3(3τ −1)V3 ; 0 ≤|h(ν)| ≤ 1 6(3τ −1)V3
Corollary 6.
Corollary 6. For ν ∈R, let assume that the f function is as in (1) and f ∈Kσ,m;℘ j,δ,Σ (Ξ), then a3 −νa2 2 ≤ 2 6V3; 0 ≤|h(ν)| ≤ 1…
Corollary 6. For ν ∈R, let assume that the f function is as in (1) and f ∈Kσ,m;℘ j,δ,Σ (Ξ), then a3 −νa2 2 ≤ 2 6V3 ; 0 ≤|h(ν)| ≤ 1 12V3
Theorem 1.
Theorem 1. If the function s given by (3) belongs to the family Pτ σκ(η, ν, ϕ) and δ ∈R, then |dκ+1| ≤ |η|B1 √2B1 √ |…
Theorem 1. If the function s given by (3) belongs to the family Pτ σκ(η, ν, ϕ) and δ ∈R, then |dκ+1| ≤ |η|B1 √2B1 √ |{M(1+κ)+[Nτ(τ−1)+(1−(1+κ)τ)2ν](1+κ)2}ηB2 1−2L2B2|+2L2B1 , (9) |d2κ+1| ≤
Corollary 1.
Corollary 1. Let δ ∈R and let the function s given by (3) be in the family Iσκ(η, ν, ϕ). Then, |dκ+1| ≤ |η|B1 √2B1 √ | (1+κ)M1−2νκ(1+κ)2…
Corollary 1. Let δ ∈R and let the function s given by (3) be in the family Iσκ(η, ν, ϕ). Then, |dκ+1| ≤ |η|B1 √2B1 √ |{(1+κ)M1−2νκ(1+κ)2}ηB2 1−2L2 1B2|+2L2 1B1 , |d2κ+1| ≤
Corollary 2.
Corollary 2. If s ∈Pτ σ1(η, ν, ϕ) is given by (1) and δ ∈R, then |d2| ≤ |η|B1 √B1 √ | M2+2(N2τ(τ−1)+2ν(1−2τ)) ηB2 1−L2 2B2|+L2 2B1, |d3| ≤…
Corollary 2. If s ∈Pτ σ1(η, ν, ϕ) is given by (1) and δ ∈R, then |d2| ≤ |η|B1 √B1 √ |{M2+2(N2τ(τ−1)+2ν(1−2τ))}ηB2 1−L2 2B2|+L2 2B1 , |d3| ≤
Corollary 3.
Corollary 3. If s ∈P1 σ1(1, ν, ϕ) is given by (1) and δ ∈R, then |d2| ≤ B1 √B1 √ |(3−ν)B2 1−4B2|+4B1, |d3| ≤ B1 3(ν+1); 0 < B1 <
Corollary 3. If s ∈P1 σ1(1, ν, ϕ) is given by (1) and δ ∈R, then |d2| ≤ B1 √B1 √ |(3−ν)B2 1−4B2|+4B1 , |d3| ≤ B1 3(ν+1) ; 0 < B1 <
Corollary 4.
Corollary 4. If the function s given by (3) belongs to the family ∈Qτ σκ(η, ν, ϱ) and δ ∈R, then |dκ+1| ≤ 2ϱ|η| √ ϱ|…
Corollary 4. If the function s given by (3) belongs to the family ∈Qτ σκ(η, ν, ϱ) and δ ∈R, then |dκ+1| ≤ 2ϱ|η| √ ϱ|{M(1+κ)+[Nτ(τ−1)+(1−(1+κ)τ)2ν](1+κ)2}η−L2|+L2 , |d2κ+1| ≤ 2ϱ|η| M ; 0 < ϱ < L2 M(1+κ)|η|
Corollary 5.
Corollary 5. If the function s given by (3) belongs to the family Dσκ(η, ν, ϱ) and δ ∈R, then |dκ+1| ≤ 2ϱ|η| √ ϱ| M1(1+κ)−2νκ(1+κ)2 η−L2…
Corollary 5. If the function s given by (3) belongs to the family Dσκ(η, ν, ϱ) and δ ∈R, then |dκ+1| ≤ 2ϱ|η| √ ϱ|{M1(1+κ)−2νκ(1+κ)2}η−L2 1|+L2 1 , |d2κ+1| ≤ 2ϱ|η|
Corollary 4
Corollary 4 yields the following if κ = 1:
Corollary 4 yields the following if κ = 1:
Corollary 6.
Corollary 6. If s ∈Qτ σ1(η, ν, ϱ) is given by (1) and δ ∈R, then |d2| ≤ 2|η|ϱ √ ϱ| 2M2+4(N2τ(τ−1)+2ν(1−2τ)) η−L2 2|+L2 2, |d3| ≤
Corollary 6. If s ∈Qτ σ1(η, ν, ϱ) is given by (1) and δ ∈R, then |d2| ≤ 2|η|ϱ √ ϱ|{2M2+4(N2τ(τ−1)+2ν(1−2τ))}η−L2 2|+L2 2 , |d3| ≤
Corollary 6
Corollary 6 would yield the following if η = τ = 1.
Corollary 6 would yield the following if η = τ = 1.
Corollary 7.
Corollary 7. If the function s of the form (1) ∈Q1 σ1(1, ν, ϕ) and δ ∈R, then |d2| ≤ ϱ √ 2 √ ϱ(1−ν)+2, |d3| ≤ 2ϱ 3(ν+1); 0 < ϱ <
Corollary 7. If the function s of the form (1) ∈Q1 σ1(1, ν, ϕ) and δ ∈R, then |d2| ≤ ϱ √ 2 √ ϱ(1−ν)+2, |d3| ≤ 2ϱ 3(ν+1) ; 0 < ϱ <
Corollary 8.
Corollary 8. Let the function s of the form (3) belong to the class Xτ σκ(η, ν, ξ) and δ ∈R. Then, |dκ+1| ≤ (1 −ξ)2|η| p | (1 + κ)M + (1 +…
Corollary 8. Let the function s of the form (3) belong to the class Xτ σκ(η, ν, ξ) and δ ∈R. Then, |dκ+1| ≤ (1 −ξ)2|η| p |{(1 + κ)M + (1 + κ)2[Nτ(τ −1) + (1 −(1 + κ)τ)2ν]}(1 −ξ)η −L2| + L2 , |d2κ+1| ≤ (1−ξ)2|η|
Corollary 9.
Corollary 9. Let the function s of the form (3) belong to the class F τ σκ(η, ν, ξ) and δ ∈R. Then, |dκ+1| ≤ (1−ξ)2|η| √ |…
Corollary 9. Let the function s of the form (3) belong to the class F τ σκ(η, ν, ξ) and δ ∈R. Then, |dκ+1| ≤ (1−ξ)2|η| √ |{(1+κ)M1−2νκ(1+κ)2}(1−ξ)η−L2 1|+L2 1 , |d2κ+1| ≤
Corollary 10.
Corollary 10. Let the function s of the form (1) belong to the class Xτ σ1(η, ν, ξ) and δ ∈R. Then, |d2| ≤ 2(1−ξ)|η| √ |…
Corollary 10. Let the function s of the form (1) belong to the class Xτ σ1(η, ν, ξ) and δ ∈R. Then, |d2| ≤ 2(1−ξ)|η| √ |{2M2+4(N2τ(τ−1)+2ν(1−2τ))}(1−ξ)η−L2 2|+L2 2 , |d3| ≤
Corollary 11.
Corollary 11. If s ∈X1 σ1(1, ν, ϕ) is of the form (1) and δ ∈R, then |d2| ≤ √ 2(1−ξ) √ |(3−ν)(1−ξ)−2|+2, |d3| ≤ 2(1−ξ) 3(1+ν); 3ν+1
Corollary 11. If s ∈X1 σ1(1, ν, ϕ) is of the form (1) and δ ∈R, then |d2| ≤ √ 2(1−ξ) √ |(3−ν)(1−ξ)−2|+2, |d3| ≤ 2(1−ξ) 3(1+ν) ; 3ν+1
Lemma 1
Lemma 1 ([28,29]). Let f ∈S be given by (1). Then, the coefficients of its inverse map g = f −1 are given in terms of the Faber polynomials…
Lemma 1 ([28,29]). Let f ∈S be given by (1). Then, the coefficients of its inverse map g = f −1 are given in terms of the Faber polynomials of f with g(w) = f −1(w) = w + ∞ ∑ n=2 1 nK−n n−1(a2, a3, . . . , an)wn, where K−n n−1 = (−n)! (−2n + 1)!(n −1)! an−1 2
Lemma 2
Lemma 2 ([30]). Let f (z) = z + ∞ ∑ k=n akzk, n ≥2 be a univalent function in D and f −1(w) = w + ∞ ∑ k=n bkwk |w| < r0( f ), r0( f ) ≥1…
Lemma 2 ([30]). Let f (z) = z + ∞ ∑ k=n akzk, n ≥2 be a univalent function in D and f −1(w) = w + ∞ ∑ k=n bkwk |w| < r0( f ), r0( f ) ≥1 4 .
Lemma 3
Lemma 3 ([31] Exercise 9, p. 172). Assume that ϖ(z) = ∞ ∑ j=1 pjzj ∈B. Then, |pn| ≤1, n ≥2. This lemma represents a special case of the…
Lemma 3 ([31] Exercise 9, p. 172). Assume that ϖ(z) = ∞ ∑ j=1 pjzj ∈B. Then, |pn| ≤1, n ≥2. This lemma represents a special case of the result in [31] [Exercise 9, p. 172] obtained from this exercise for p0 = 0. 2. Main Results First, we prove the next lemma.
Lemma 4.
Lemma 4. Let u(z) = u1z + u2z2 + u3z3 + · · · ∈B and s be a complex number. Then, for all n ∈N, the following inequality holds: u2n −su2 n…
Lemma 4. Let u(z) = u1z + u2z2 + u3z3 + · · · ∈B and s be a complex number. Then, for all n ∈N, the following inequality holds: u2n −su2 n ≤1 + (|s| −1) u2 n ≤max{1; |s|}. Moreover, the functions u(z) = z and u(z) = z2 prove that the above inequality is sharp for |s| ≥1 and for |s| < 1, respectively.
Theorem 1.
Theorem 1. Let the function f (z) = z + ∞ ∑ k=n0 akzk ∈NΣ(λ, δ, h), n0 ≥2. Then, |an0| ≤min |B1| 1 + (n0 −1)(λ + n0δ); v u
Theorem 1. Let the function f (z) = z + ∞ ∑ k=n0 akzk ∈NΣ(λ, δ, h), n0 ≥2. Then, |an0| ≤min |B1| 1 + (n0 −1)(λ + n0δ); v u
Lemma 1
Lemma 1 ([1]). Let Ur0 = z: |z| < r0, with 0 < r0 < 1. Let p(z) = b + bnzn + bn+1zn+1 +... be analytic in U with n ≥2 and p(z) ̸≡b, and let…
Lemma 1 ([1]). Let Ur0 = {z : |z| < r0}, with 0 < r0 < 1. Let p(z) = b + bnzn + bn+1zn+1 + . . . be analytic in U with n ≥2 and p(z) ̸≡b, and let q ∈Q(b). If there exist points z0 = r0eiθ0 ∈U and ξ0 ∈∂U\E(q) such that p(z0) = q(ξ0), p(U r0) ⊂q(U), ℜξ0q′′(ξ0) q′(ξ0) ≥0, and (9)
Lemma 2
Lemma 2 ([1]). Let Ur0 = z: |z| < r0, with 0 < r0 < 1. Suppose q given in (14) and p(z) = b + bnzn + bn+1zn+1 +... is analytic in U with n…
Lemma 2 ([1]). Let Ur0 = {z : |z| < r0}, with 0 < r0 < 1. Suppose q given in (14) and p(z) = b + bnzn + bn+1zn+1 + . . . is analytic in U with n ≥2 and p(z) ̸≡b . If there exist points z0 = r0eiθ0 ∈UM and w0 ∈∂U such that p(z0) = q(w0), p(Ur0) ⊂UM and |zp′(z)||[M + ¯beiθ]|2 ≤nM[M2 −|b|2] (15) when z ∈U r0 and θ ∈[0, 2π], then z0p′(z0) = nq(w0) |q(w0) −b|2 |q(w0)|2 −|b|2 , ℜ z0p′′(z0) p′(z0) + 1 ≥n |q(w0) −b|2
Theorem 1.
Theorem 1. Consider p ∈H[b, n] and q ∈Q(b) fulfills ℜξq′′(ξ) q′(ξ) ≥0 and
Theorem 1. Consider p ∈H[b, n] and q ∈Q(b) fulfills ℜξq′′(ξ) q′(ξ) ≥0 and
Corollary 1.
Corollary 1. Suppose q is univalent in U, q(0) = b and set qρ(z) ≡q(ρz) for ρ ∈(0, 1). Consider that p ∈H[b, n] and qρ fulfill ℜ ξq′′ ρ(ξ)…
Corollary 1. Suppose q is univalent in U, q(0) = b and set qρ(z) ≡q(ρz) for ρ ∈(0, 1). Consider that p ∈H[b, n] and qρ fulfill ℜ ξq′′ ρ(ξ) q′ρ(ξ) ≥0 and
Theorem 2.
Theorem 2. Consider p ∈H[b, n] and q ∈Q(b) and that they fulfill ℜξq′′(ξ) q′(ξ) ≥0 and
Theorem 2. Consider p ∈H[b, n] and q ∈Q(b) and that they fulfill ℜξq′′(ξ) q′(ξ) ≥0 and
Corollary 2.
Corollary 2. Suppose q is univalent in U, with q(0) = b, and set qρ(z) ≡q(ρz) for ρ ∈(0, 1). Consider that p ∈H[b, n] and qρ fulfill ℜ ξq′′…
Corollary 2. Suppose q is univalent in U, with q(0) = b, and set qρ(z) ≡q(ρz) for ρ ∈(0, 1). Consider that p ∈H[b, n] and qρ fulfill ℜ ξq′′ ρ(ξ) q′ρ(ξ) ≥0 and
Theorem 3.
Theorem 3. Consider p ∈H[b, n], Ξ: C4 × U × U −→C and that Ξ(p(z), zp′(z), z2p′′(z), z3p′′′(z); z, ζ) is analytic in U. Suppose h is…
Theorem 3. Consider p ∈H[b, n], Ξ : C4 × U × U −→C and that Ξ(p(z), zp′(z), z2p′′(z), z3p′′′(z); z, ζ) is analytic in U. Suppose h is univalent in U and the differential equation Ξ(p(z), zp′(z), z2p′′(z), z3p′′′(z); z, ζ) = h(z) (19) has a solution q ∈Q(b) and ℜξq′′(ξ) q′(ξ) ≥0 and
Theorem 4.
Theorem 4. Consider that the q given in (14) and p ∈H[b, n] satisfy |zp′(z)||M + ¯beiθ|2 ≤Mn h M2 −|b|2i, where z ∈U and 0 ≤θ ≤2π. If Ξ…
Theorem 4. Consider that the q given in (14) and p ∈H[b, n] satisfy |zp′(z)||M + ¯beiθ|2 ≤Mn h M2 −|b|2i , where z ∈U and 0 ≤θ ≤2π. If Ξ ∈Ξn[Π, M, b], then Ξ(p(z), zp′(z), z2p′′(z), z3p′′′(z); z, ζ) ⊂Π implies p(z) ≺q(z). Next, we obtain the following corollary when b = 0 in Theorem 4.
Corollary 3.
Corollary 3. Consider that q(w) = Mw and p ∈H[0, n] fulfill |zp′(z)| ≤Mn when z ∈U. If Π is a set in C and Ξ ∈Ξn[Π, M, 0] as characterized…
Corollary 3. Consider that q(w) = Mw and p ∈H[0, n] fulfill |zp′(z)| ≤Mn when z ∈U. If Π is a set in C and Ξ ∈Ξn[Π, M, 0] as characterized by (22), then Ξ(p(z), zp′(z), z2p′′(z), z3p′′′(z); z, ζ) ⊂Π implies p(z) ≺Mz. In this particular case, Theorem 4 becomes
Theorem 5.
Theorem 5. Consider that the q given in (14) and p ∈H[b, n] satisfy (17). If Π is a set in C and (i) Ξ ∈Ξn[Π, M, b], then Ξ(p(z), zp′(z),…
Theorem 5. Consider that the q given in (14) and p ∈H[b, n] satisfy (17). If Π is a set in C and (i) Ξ ∈Ξn[Π, M, b], then Ξ(p(z), zp′(z), z2p′′(z), z3p′′′(z); z, ζ) ⊂Π =⇒|p(z)| < M. (ii) If Ξ ∈Ξn[M, b], then |Ξ(p(z), zp′(z), z2p′′(z), z3p′′′(z); z, ζ)| < M =⇒|p(z)| < M. 197
Theorem 6.
Theorem 6. Consider Iλ p,δ f (z) ∈H[0, p] with p ≥2, q ∈Q(0) and that they satisfy ℜξq′′(ξ) q′(ξ) ≥0 and
Theorem 6. Consider Iλ p,δ f (z) ∈H[0, p] with p ≥2, q ∈Q(0) and that they satisfy ℜξq′′(ξ) q′(ξ) ≥0 and
Corollary 4.
Corollary 4. Consider q to be univalent in U, with q(0) = 0, and set qρ(z) ≡q(ρz) for ρ ∈(0, 1). Let Iλ p,δ f (z) ∈H[0, p] for p ≥2 and let…
Corollary 4. Consider q to be univalent in U, with q(0) = 0, and set qρ(z) ≡q(ρz) for ρ ∈(0, 1). Let Iλ p,δ f (z) ∈H[0, p] for p ≥2 and let Iλ p,δ f (z) and qρ satisfy (23). If Π is a set in C and Θ ∈ΘI[Π, qρ] and f (z) ∈A(p) fulfill Θ(Iλ p,δ f (z), Iλ−1 p,δ f (z), Iλ−2 p,δ f (z), Iλ−3 p,δ f (z); z, ζ) ⊂Π, then Iλ p,δ f (z) ≺q(z).
Theorem 7.
Theorem 7. Consider that Iλ p,δ f (z) ∈H[0, p] with p ≥2 and q ∈Q(0) and that they satisfy (23). If Π is a set in C, Θ ∈ΘI[Π, q], f (z)…
Theorem 7. Consider that Iλ p,δ f (z) ∈H[0, p] with p ≥2 and q ∈Q(0) and that they satisfy (23). If Π is a set in C, Θ ∈ΘI[Π, q], f (z) ∈A(p) and Θ(Iλ p,δ f (z), Iλ−1 p,δ f (z), Iλ−2 p,δ f (z), Iλ−3 p,δ f (z); z, ζ) is analytic in U, then Θ(Iλ p,δ f (z), Iλ−1 p,δ f (z), Iλ−2 p,δ f (z), Iλ−3 p,δ f (z); z, ζ) ≺≺h(z) implies
Corollary 5.
Corollary 5. Consider q to be univalent in U, with q(0) = 0, and set qρ(z) ≡q(ρz) for ρ ∈(0, 1). Let Iλ p,δ f (z) ∈H[0, p] for p ≥2 and let…
Corollary 5. Consider q to be univalent in U, with q(0) = 0, and set qρ(z) ≡q(ρz) for ρ ∈(0, 1). Let Iλ p,δ f (z) ∈H[0, p] for p ≥2 and let Iλ p,δ f (z) and qρ satisfy (23). If Π is a set in C, Θ ∈ΘI[Π, qρ] ,f (z) ∈A(p) and Θ(Iλ p,δ f (z), Iλ−1 p,δ f (z), Iλ−2 p,δ f (z), Iλ−3 p,δ f (z); z, ζ) is analytic in U, then Θ(Iλ p,δ f (z), Iλ−1 p,δ f (z), Iλ−2 p,δ f (z), Iλ−3
Theorem 8.
Theorem 8. Consider that Iλ p,δ f (z) ∈H[0, p] with p ≥2, Θ: C4 × U × U −→C and that Θ(Iλ p,δ f (z), Iλ−1 p,δ f (z), Iλ−2 p,δ f (z), Iλ−3…
Theorem 8. Consider that Iλ p,δ f (z) ∈H[0, p] with p ≥2, Θ : C4 × U × U −→C and that Θ(Iλ p,δ f (z), Iλ−1 p,δ f (z), Iλ−2 p,δ f (z), Iλ−3 p,δ f (z); z, ζ) is analytic in U. Suppose h is univalent in U and q ∈Q(0) is a solution of the following differential equation Θ(Iλ p,δ f (z), Iλ−1 p,δ f (z), Iλ−2 p,δ f (z), Iλ−3 p,δ f (z); z, ζ) = h(z) (31)
Corollary 6.
Corollary 6. Consider q(z) = Mz and Iλ p,δ f (z) ∈H[0, p] with p ≥2 to satisfy |z(Iλ p,δ f (z))′| ≤Mk, when z ∈U and k ≥p. If Θ ∈ΘI[Π, M],…
Corollary 6. Consider q(z) = Mz and Iλ p,δ f (z) ∈H[0, p] with p ≥2 to satisfy |z(Iλ p,δ f (z))′| ≤Mk, when z ∈U and k ≥p. If Θ ∈ΘI[Π, M], f (z) ∈A(p) satisfies Θ(Iλ p,δ f (z), Iλ−1 p,δ f (z), Iλ−2 p,δ f (z), Iλ−3 p,δ f (z); z, ζ) ⊂Π, then Iλ p,δ f (z) ≺q(z).
Corollary 7.
Corollary 7. Consider q(z) = Mz and Iλ p,δ f (z) ∈H[0, p] with p ≥2. If Π is a set in C and (i) Θ ∈ΘI[Π, M], f (z) ∈A(p) satisfies Θ(Iλ p,δ…
Corollary 7. Consider q(z) = Mz and Iλ p,δ f (z) ∈H[0, p] with p ≥2. If Π is a set in C and (i) Θ ∈ΘI[Π, M] , f (z) ∈A(p) satisfies Θ(Iλ p,δ f (z), Iλ−1 p,δ f (z), Iλ−2 p,δ f (z), Iλ−3 p,δ f (z); z, ζ) ⊂Π =⇒|p(z)| < M. (ii) If f (z) ∈A(p) and Θ ∈ΘI[M], it satisfies |Θ(Iλ p,δ f (z), Iλ−1 p,δ f (z), Iλ−2 p,δ f (z), Iλ−3 p,δ f (z); z, ζ)| < M =⇒|p(z)| < M. 4. Conclusions
Function classes studied:
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