Abstract
Let $\mathcal{A}_n$ be the class of analytic functions $f(z)$ of the form $f(z)=z+\sum_{k=n+1}^\infty a_kz^k,n\in\mathbb{N}$ and let \begin{align*} Ω_n:=\left\{f\in\mathcal{A}_n:\left|zf'(z)-f(z)\right|<\frac{1}{2},\; z\in\mathbb{D}\right\}. \end{align*} We make use of differential subordination technique to obtain sufficient conditions for the class $Ω_n$, and then employ these conditions to construct functions which involve double integrals and members of $Ω_n$. We also consider a subclass $\w
Results & Lemmas (35)
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Lemma 1.1
Lemma 1.1 ( [14, Theorem 3.1b, p. 71]). Let h(z) be a convex function in D with h(0) = a, γ ̸= 0 and Reγ ≥0. If p(z) ∈Hn(a) and p(z) +…
Lemma 1.1 ( [14, Theorem 3.1b, p. 71]). Let h(z) be a convex function in D with h(0) = a, γ ̸= 0 and Reγ ≥0. If p(z) ∈Hn(a) and p(z) + γ−1zp′(z) ≺h(z), then p(z) ≺q(z) ≺h(z), where q(z) = γ nz γ n Z z 0 h(ξ)ξ γ
Theorem 2.1.
Theorem 2.1. Let n ∈N and γ ≥1. If f ∈An satisfies zf ′′(z) + (γ −1)
Theorem 2.1. Let n ∈N and γ ≥1. If f ∈An satisfies zf ′′(z) + (γ −1)
Lemma 1.1
Lemma 1.1 that p(z) ≺ 1 nz γ n Z z 0 n + γ 2 ξ ξ γ n−1dξ = 1
Lemma 1.1 that p(z) ≺ 1 nz γ n Z z 0 n + γ 2 ξ ξ γ n−1dξ = 1
Corollary 2.1.
Corollary 2.1. If f ∈An satisfies |zf ′′(z)| < n + 1 2, then f ∈Ωn. The result is sharp. Fixing n = 1, and then taking γ = 1 and γ = 2 in…
Corollary 2.1. If f ∈An satisfies |zf ′′(z)| < n + 1 2 , then f ∈Ωn. The result is sharp. Fixing n = 1, and then taking γ = 1 and γ = 2 in Theorem 2.1, respectively, we obtain the following sufficient conditions proved by Obradovi´c and Peng [15]
Corollary 2.2
Corollary 2.2 ( [15, Theorem 2]). If f ∈A satisfies |zf ′′(z)| < 1, then f ∈Ω. The number 1 is best possible.
Corollary 2.2 ( [15, Theorem 2]). If f ∈A satisfies |zf ′′(z)| < 1, then f ∈Ω. The number 1 is best possible.
Corollary 2.3
Corollary 2.3 ( [15, Theorem 3]). Let f ∈A. If z2f ′′(z) + zf ′(z) −f(z) < 3 2, then f ∈Ω. The number 3 2 is best possible. Note. For…
Corollary 2.3 ( [15, Theorem 3]). Let f ∈A. If z2f ′′(z) + zf ′(z) −f(z) < 3 2, then f ∈Ω. The number 3 2 is best possible. Note. For brevity, we fix bfn,µ(z) = z+ µ 2nzn+1 (|µ| = 1) and write bf1(z) = bf1,1(z) = z+ 1 2z2.
Theorem 2.2.
Theorem 2.2. Let γ ≥1, n ∈N, and let f(z) = z + ∞ X k=n+1 akzk ∈An, z ∈D.
Theorem 2.2. Let γ ≥1, n ∈N, and let f(z) = z + ∞ X k=n+1 akzk ∈An, z ∈D.
Corollary 2.4.
Corollary 2.4. If f(z) = z + P∞ k=n+1 akzk ∈An satisfies ∞ X k=n+1 k(k −1)|ak| ≤n + 1 2, then f ∈Ωn. The result is sharp. If we fix n = 1,…
Corollary 2.4. If f(z) = z + P∞ k=n+1 akzk ∈An satisfies ∞ X k=n+1 k(k −1)|ak| ≤n + 1 2 , then f ∈Ωn. The result is sharp. If we fix n = 1, and then take γ = 1 and γ = 2 in Theorem 2.2, respectively, we obtain the following sufficient conditions for the function class Ω.
Corollary 2.5.
Corollary 2.5. Let f(z) = z + P∞ k=2 akzk ∈A. If ∞ X k=2 k(k −1)|ak| ≤1, then f ∈Ωand the result is sharp.
Corollary 2.5. Let f(z) = z + P∞ k=2 akzk ∈A. If ∞ X k=2 k(k −1)|ak| ≤1, then f ∈Ωand the result is sharp.
Corollary 2.6.
Corollary 2.6. Let f(z) = z + P∞ k=2 akzk ∈A satisfies ∞ X k=2 (k2 −1)|ak| ≤3 2. Then f ∈Ωand the result is sharp.
Corollary 2.6. Let f(z) = z + P∞ k=2 akzk ∈A satisfies ∞ X k=2 (k2 −1)|ak| ≤3 2. Then f ∈Ωand the result is sharp.
Theorem 2.3.
Theorem 2.3. Let γ ≥1, n ∈N, and let J (z) be analytic in D such that |J (z)| ≤n + γ 2. (2.5) Then the function f(z) = z + zn+1 ZZ 1 0 J…
Theorem 2.3. Let γ ≥1, n ∈N, and let J (z) be analytic in D such that |J (z)| ≤n + γ 2 . (2.5) Then the function f(z) = z + zn+1 ZZ 1 0 J (stz)sn−1tn+γ−1dsdt (2.6) belongs to the class Ωn. Moreover, if equality holds in (2.5), then the function in (2.6) becomes bfn,µ ∈Ωn.
Corollary 2.7.
Corollary 2.7. Let J ∈H such that |J (z)| ≤1. Then the function f(z) = z + z2 ZZ 1 0 J (stz)tdsdt belongs to the class Ω. We conclude this…
Corollary 2.7. Let J ∈H such that |J (z)| ≤1. Then the function f(z) = z + z2 ZZ 1 0 J (stz)tdsdt belongs to the class Ω. We conclude this section by showing that for each n ∈N and for each µ ∈C with |µ| = 1, the function bfn,µ(z) is an extreme point of Ωn. We prove it by showing that bfn,µ(z) satisfies the condition established by Peng and Zhong [16, Theorem 3.14]. Observe that bfn,µ(z) bfn,µ(z) = z + µ 2nzn+1 = z + 1 2z Z z 0 φ(ξ), where φ(ξ) = µξn−1 satisfies
Lemma 3.1.
Lemma 3.1. The sequence sk ∞ k=1 is a subordinating factor sequence if and only if Re
Lemma 3.1. The sequence {sk}∞ k=1 is a subordinating factor sequence if and only if Re
Theorem 3.1.
Theorem 3.1. Let n ∈N, and let f(z) = z + P∞ k=n+1 akzk ∈bΩn. Then for every convex function ν(z) = z + P∞ k=2 ckzk in D, we have τ(f…
Theorem 3.1. Let n ∈N, and let f(z) = z + P∞ k=n+1 akzk ∈bΩn. Then for every convex function ν(z) = z + P∞ k=2 ckzk in D, we have τ(f ∗ν)(z) ≺ν(z) (3.3) and Re (f(z)) > −1 2τ , (3.4)
Corollary 3.1.
Corollary 3.1. Let f ∈bΩ. Then for every convex function ν(z) in D, we have 1 3(f ∗ν)(z) ≺ν(z) and Re(f(z)) > −3 2. (3.6) The sharpness of…
Corollary 3.1. Let f ∈bΩ. Then for every convex function ν(z) in D, we have 1 3(f ∗ν)(z) ≺ν(z) and Re(f(z)) > −3 2. (3.6) The sharpness of the estimate 1 3 is guaranteed by the function bf1,−1(z) = z −z2/2 ∈Ω (see Figure 2).
Lemma 4.1
Lemma 4.1 ( [16, Corollary 3.12]). If f(z) = z + ∞ X k=2 akzk is in Ω, then |ak| ≤ 1 2(k −1), k ≥2. (4.1)
Lemma 4.1 ( [16, Corollary 3.12]). If f(z) = z + ∞ X k=2 akzk is in Ω, then |ak| ≤ 1 2(k −1), k ≥2. (4.1)
Lemma 4.2
Lemma 4.2 ( [17, Theorem 3]). The function fk(z) = z + akzk is in Sp if and only if |ak| ≤ 1 (2k −1), k ≥2.
Lemma 4.2 ( [17, Theorem 3]). The function fk(z) = z + akzk is in Sp if and only if |ak| ≤ 1 (2k −1), k ≥2.
Lemma 4.3
Lemma 4.3 ( [9, Corollary 2.4]). Let f(z) = z + ∞ X k=2 akzk ∈A. If ∞ X k=2 (2k −1)|ak| ≤1, then f ∈Sp.
Lemma 4.3 ( [9, Corollary 2.4]). Let f(z) = z + ∞ X k=2 akzk ∈A. If ∞ X k=2 (2k −1)|ak| ≤1, then f ∈Sp.
Lemma 4.4.
Lemma 4.4. The function fk(z) = z + akzk is in Ωif and only if |ak| ≤ 1 2(k −1), k ≥2. (4.2)
Lemma 4.4. The function fk(z) = z + akzk is in Ωif and only if |ak| ≤ 1 2(k −1), k ≥2. (4.2)
Theorem 4.1.
Theorem 4.1. If fk(z) = z + akzk belongs to Sp, then fk ∈Ωfor every k ≥2.
Theorem 4.1. If fk(z) = z + akzk belongs to Sp, then fk ∈Ωfor every k ≥2.
Theorem 4.2.
Theorem 4.2. Let f(z) = z + ∞ X k=3 akzk ∈bΩ(i.e., a2 = 0). Then f ∈Sp.
Theorem 4.2. Let f(z) = z + ∞ X k=3 akzk ∈bΩ(i.e., a2 = 0). Then f ∈Sp.
Lemma 4.5
Lemma 4.5 ( [13, Theorem 5]). If |ak| ≤ s k + 1 2k3, k ≥2, then the function fk(z) = z + akzk is in UST. Using Lemma 4.5, we prove the…
Lemma 4.5 ( [13, Theorem 5]). If |ak| ≤ s k + 1 2k3 , k ≥2, then the function fk(z) = z + akzk is in UST. Using Lemma 4.5, we prove the following theorem.
Theorem 4.3.
Theorem 4.3. Let fk(z) = z + akzk be in Ω. Then, for all k ≥3, f(z) is in UST.
Theorem 4.3. Let fk(z) = z + akzk be in Ω. Then, for all k ≥3, f(z) is in UST.
Lemma 5.1
Lemma 5.1 ( [16, Theorem 3.1]). If f ∈Ω, then |z| −1 2|z|2 ≤|f(z)| ≤|z| + 1 2|z|2, (5.1) and 1 −|z| ≤|f ′(z)| ≤1 + |z|. (5.2) Further, for…
Lemma 5.1 ( [16, Theorem 3.1]). If f ∈Ω, then |z| −1 2|z|2 ≤|f(z)| ≤|z| + 1 2|z|2, (5.1) and 1 −|z| ≤|f ′(z)| ≤1 + |z|. (5.2) Further, for each 0 ̸= z ∈D, equality occurs in both the estimates if and only if f(z) = bf1,µ(z) = z + µ 2 z2 with |µ| = 1. (5.3)
Lemma 5.2.
Lemma 5.2. Let f ∈Ω. Then for |z| = r < 1, we have the sharp estimate
Lemma 5.2. Let f ∈Ω. Then for |z| = r < 1, we have the sharp estimate
Theorem 5.1.
Theorem 5.1. The S∗(α)-radius for the class Ωis r(α) = 2(1−α) 2−α, where 0 ≤α < 1.
Theorem 5.1. The S∗(α)-radius for the class Ωis r(α) = 2(1−α) 2−α , where 0 ≤α < 1.
Theorem 5.2.
Theorem 5.2. The Sp-radius for the class Ωis 2 3. Figure 3. Sharpness of Sp-radius: bF1(z) = z bf ′ 1(z)/ bf1(z) with bf1(z) = z + z2/2.
Theorem 5.2. The Sp-radius for the class Ωis 2 3. Figure 3. Sharpness of Sp-radius: bF1(z) = z bf ′ 1(z)/ bf1(z) with bf1(z) = z + z2/2.
Lemma 5.3
Lemma 5.3 ( [12, Lemma 2.2]). For 1/e < a < e, let ra be given by ra = a −1 e, 1 e < a ≤1 2(e + 1 e) e −a, 1 2(e + 1 e) ≤a < e.
Lemma 5.3 ( [12, Lemma 2.2]). For 1/e < a < e, let ra be given by ra = a −1 e, 1 e < a ≤1 2(e + 1 e) e −a, 1 2(e + 1 e) ≤a < e.
Theorem 5.3.
Theorem 5.3. The S∗ e-radius for the class Ωis re = 2(1−e−1) 2−e−1 ≈0.77460032643.
Theorem 5.3. The S∗ e-radius for the class Ωis re = 2(1−e−1) 2−e−1 ≈0.77460032643.
Lemma 5.4
Lemma 5.4 ( [18, Lemma 2.5]). For 1/3 < a < 3, let ra be given by ra = 3a−1 3, 1 3 < a ≤5 3 3 −a, 5 3 ≤a < 3. Then w ∈C: |w −a| < ra…
Lemma 5.4 ( [18, Lemma 2.5]). For 1/3 < a < 3, let ra be given by ra = 3a−1 3 , 1 3 < a ≤5 3 3 −a, 5 3 ≤a < 3. Then {w ∈C : |w −a| < ra} ⊂RC.
Theorem 5.4.
Theorem 5.4. The S∗ C-radius for the class Ωis 4 5.
Theorem 5.4. The S∗ C-radius for the class Ωis 4 5.
Lemma 5.5
Lemma 5.5 ( [4, Lemma 3.3]). For 1 −sin 1 ≤a ≤1 + sin 1, let ra be given by ra = sin 1 −|a −1|. Then w ∈C: |w −a| < ra ⊂RS. Using Lemma…
Lemma 5.5 ( [4, Lemma 3.3]). For 1 −sin 1 ≤a ≤1 + sin 1, let ra be given by ra = sin 1 −|a −1|. Then {w ∈C : |w −a| < ra} ⊂RS. Using Lemma 5.5, the following theorem can be proved.
Theorem 5.5.
Theorem 5.5. The S∗ S-radius for the class Ωis rS = 2 sin 1 1+sin 1 ≈0.91391174962 (see Fig- ure 6). Figure 6. Sharpness of S∗ S-radius.…
Theorem 5.5. The S∗ S-radius for the class Ωis rS = 2 sin 1 1+sin 1 ≈0.91391174962 (see Fig- ure 6). Figure 6. Sharpness of S∗ S-radius. γS : Boundary curve of ϕS(D), where ϕS(z) = 1 + sin z. DS : bF1 (|z| < rS) , bF1(z) = zb f′ 1(z) b f1(z) with bf1(z) = z + z2
Theorem 5.6. · radius
Theorem 5.6. The S∗ SG-radius for the class Ωis rSG = e−1 e ≈0.6321205588285577. 6. Concluding Remarks and Some Open Problems Since the…
Theorem 5.6. The S∗ SG-radius for the class Ωis rSG = e−1 e ≈0.6321205588285577. 6. Concluding Remarks and Some Open Problems Since the members of Ωare starlike, it easily follows from the Alexander’s theorem that the members of the class Υ := f ∈A : z2f ′′(z) < 1 2 .
Corollary 6.1.
Corollary 6.1. If f(z) = z + P∞ k=2 akzk ∈A satisfies ∞ X k=2 k(k −1)|ak| ≤1 2, (6.1) then f(z) is convex.
Corollary 6.1. If f(z) = z + P∞ k=2 akzk ∈A satisfies ∞ X k=2 k(k −1)|ak| ≤1 2, (6.1) then f(z) is convex.
Function classes studied:
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