Open Journal of Mathematical Analysis
ISSN: 2616-8111 (Online) 2616-8103 (Print)
DOI: 10.30538/psrp-oma2019.0037
On certain subclasses of p-valent functions with negative coefficients defined by a generalized differential operator
Department of Mathematics, Gombe State University, P.M.B.127, Gombe, Nigeria.; (B.S & G.B.M)
1Corresponding Author: bitrussambo3@gmail.com
Abstract
Keywords:
1. Introduction
Let A denote the class of all functions of the formRemark 1.
- When α=1,ν=0, we get the Sălăgean differential operator introduced by Sălăgean in [2].
- When ν=0, we obtain differential operator defined by Al-Oboudi in [3].
Remark 2.
- If ν=0, Dn,pανf=Dnα,pf defined by Bulut in [4]
- If p=1, Dn,pανf=Dnανf introduced by Raducanu and Orhan in [1]
- If p=1,α=1,ν=0, Dn,pανf=Dnf defined by Sălăgean in [2]
- If p=1,ν=0, Dn,pανf=Dnαf defined by Al-Oboudi in [3].
Definition 1. A function f∈Tp is in the class, Snp(ϑ,β,γ,φ) if and only if
Remark 3. If ν=0, ϑ=α, φ=μ, the class Snp(ϑ,β,γ,φ) reduces to the class Rnp(α,β,γ,μ) investigated by Bulut [4]
Definition 2. A function f∈Tp is in the class Sn,(δ0)p(ϑ,β,γ,φ), if there exists a function g(z)∈Snp(ϑ,β,γ,φ) such that |f(z)g(z)−1|<1−δ0...(z∈U,0≤δ0<1) for 0≤ϑ<1, 0≤γ<1, 0<β≤1, 0<φ<1.
Definition 3. For a function f∈Tp, δ≥0, δ-neighborhood of f is defined as:
2. Coefficient inequalities
Theorem 4. A function f∈Tp is in the class Snp(ϑ,β,γ,φ) if and only if
Proof. Suppose that f∈Snp(ϑ,β,γ,φ), then from inequality 10, we have |(Dn,pανf(z))′−pzp−1ϑ(Dn,pανf(z))′+(β−γ)|=|pzp−1−∑∞k=p+1k[1+(ανk+α−ν)(kp−1)]nakzk−1−pzp−1ϑ(pzp−1−∑∞k=p+1k[1+(ανk+α−ν)(kp−1)]nakzk−1)+(β−γ)|=|∑∞k=p+1k[1+(ανk+α−ν)(kp−1)]nakzk−1ϑ(pzp−1−∑∞k=p+1k[1+(ανk+α−ν)(kp−1)]nakzk−1)+(β−γ)|<φ,(z∈U,n∈N0) it is well known that ℜz≤|z|, therefore, we obtain ℜ{∑∞K=p+1k[1+(ανk+α−ν)(kp−1)]nakzk−1ϑ(pzp−1−∑∞k=p+1k[1+(ανk+α−ν)(kp−1)]nakzk−1)+(β−γ)}<φ. If we choose z real and let z→1−, then we get ∞∑K=p+1k[1+(ανk+α−ν)(kp−1)]nak≤φ{ϑ(p−∞∑k=p+1k[1+(ανk+α−ν)(kp−1)]nak)+(β−γ)} which is precisely the assertion 13. On contrary, suppose that the inequality 13 hold true and let z∈δU={z∈C:|z|=1}. Then, from 10, we have |(Dn,pανf(z))′−pzp−1|−φ|ϑ(Dn,pανf(z))′+(β−γ)|≤∞∑k=p+1k[1+(ανk+α−ν)(kp−1)]nak|z|k−1−φ(ϑp+β−γ)+φϑ∞∑k=p+1k[1+(ανk+α−ν)(kp−1)]nak|z|k−1=∞∑k=p+1k[1+(ανk+α−ν)(kp−1)]nak|z|k−1(1+φϑ)ak−φ(ϑp+β−γ)≤0. By maximum modulus theorem, we have f∈Snp(ϑ,β,γ,φ).
Corollary 5. If f∈Snp(ϑ,β,γ,φ), then ap+1≤φ(ϑp+β−γ)(p+1)[1+(αν(p+1)+α−ν)(1p)]n(1+φϑ).
3. Growth and distortion theorem
Theorem 6. For each f(z)∈Snp(ϑ,β,γ,φ), we have |z|p−φ(ϑp+β−γ)[1+(αν(p+1)+α−ν)(1p)]n(1+φϑ)(p+1)|z|p+1≤|f(z)|≤|z|p+φ(ϑp+β−γ)[1+(αν(p+1)+α−ν)(1p)]n(1+φϑ)(p+1)|z|p+1.
Proof. Let f(z)∈Snp(ϑ,β,γ,φ),z∈U, the bound on f(z) is given by
Theorem 7. For each f(z)∈Snp(ϑ,β,γ,φ), we have p|z|p−1−φ(ϑp+β−γ)[1+(αν(p+1)+α−ν)(1p)]n(1+φϑ)|z|p≤|f′(z)|≤p|z|p−1+φ(ϑp+β−γ)[1+(αν(p+1)+α−ν)(1p)]n(1+φϑ)|z|p.
Proof. Let f(z)∈Snp(ϑ,β,γ,φ),z∈U, the bound on the derivative of f(z) is given by |f′(z)|≤p|z|p−1+(p+1)|z|p∞∑k=p+1ak,z∈U, and, in the same way as above, we get our desired result.
4. Closure properties
Theorem 8. Let the functions f(z)=zp−∞∑k=p+1akzk,(ak≥0)g(z)=zp−∞∑k=p+1bkzk,(bk≥0), be in the class Snp(ϑ,β,γ,φ). Then for 0≤λ≤1, the function h is defined as h(z)=(1−λ)f(z)+λg(z)=zp−∞∑k=p+1ckzk, where ck:=(1−λ)ak+λbk≥0, is also in Snp(ϑ,β,γ,φ).
Proof. Suppose that each of the functions f and g is in the class Snp(ϑ,β,γ,φ). Then making use of inequality (13), we have ∞∑k=p+1k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)ck=(1−λ)∞∑k=p+1k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)ak+λ∞∑k=p+1k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)bk≤(1−λ)φ(ϑp+β−γ)+λφ(ϑp+β−γ)=φ(ϑp+β−γ), which completes the proof.
5. δ-Neighborhoods
Theorem 9. If
Proof. For a function f(z)∈Snp(ϑ,β,γ,φ) of the form (8), Theorem 4 immediately yields (p+1)[1+(αν(p+1)+α−ν)(1p)]n(1+φϑ)∞∑k=p+1ak≤φ(ϑp+β−γ), therefore,
Theorem 10. If g(z)∈Snp(ϑ,β,γ,φ) and
Proof. Suppose that f∈Npδ(f,g), then by Definition 3, we have ∞∑k=p+1k|ak−bk|≤δ, which readily implies the coefficient inequality given by ∞∑k=p+1|ak−bk|≤δp+1(p∈N). Next, since g∈Snp(ϑ,β,γ,φ), we have from inequality (13) that ∞∑k=p+1bk≤φ(ϑp+β−γ)(p+1)[1+(αν(p+1)+α−ν)(1p)]n(1+φϑ), so from the definition of the class, we have |f(z)g(z)−1|<∑∞k=p+1|ak−bk|1−∑∞k=p+1bk≤δp+1(p+1)[1+(αν(p+1)+α−ν)(1p)]n(1+φϑ)(p+1)[1+(αν(p+1)+α−ν)(1p)]n(1+φϑ)−φ(ϑp+β−γ)=1−δ0, provided that δ0 is given precisely by (22). Thus, by the definition, f∈Sn,δ0p(ϑ,β,γ,φ) for δ0 given by (22), this completes our proof.
6. Extreme points
Theorem 11. If fp(z)=zp,fk(z)=zp−φ(ϑp+β−γ)k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)zk(k≥p+1) then, f∈Snp(ϑ,β,γ,φ) if and only if it can be expressed in the form f(z)=λpfp(z)+∑∞k=p+1λkfk(z), where λk≥0 and λp=1−∑∞k=p+1λk.
Proof. Assume that f(z)=λpfp(z)+∑∞k=p+1λkfk(z), then f(z)=(1−∞∑k=p+1λk)zp+∞∑k=p+1λk{zp−φ(ϑp+β−γ)k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)zk}=zp−∞∑k=p+1λk{φ(ϑp+β−γ)k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)zk}. Thus, ∞∑k=p+1k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)λkφ(ϑp+β−γ)k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)=φ(ϑp+β−γ)∞∑k=p+1λk=φ(ϑp+β−γ)(1−λp)≤φ(ϑp+β−γ), which shows that f satisfies condition (13) and therefore,f∈Snp(ϑ,β,γ,φ). Conversely, suppose that f∈Snp(ϑ,β,γ,φ), since ak≤φ(ϑp+β−γ)k[1+(ανk+α−ν)(kp−1)]n(1+φϑ),(k≥p+1), we may set λk=k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)φ(ϑp+β−γ)ak, and λp=1−∞∑k=p+1λk, then we obtain from f(z)=zp−∞∑k=p+1akzk=(λp+∞∑k=p+1λk)zp−∞∑k=p+1λkφ(ϑp+β−γ)k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)zk=λpzp+∞∑k=p+1λk(zp−φ(ϑp+β−γ)k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)zk)=λpzp+∞∑k=p+1λkfk(z), which completes the proof.
Corollary 12. The extreme points of Snp(ϑ,β,γ,φ) are given by fp(z)=zp,fk(z)=zp−φ(ϑp+β−γ)k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)zk(k≥p+1)
7. Radii of close-to-convexity,starlikeness and convexity
A function f∈Tp is said to be p-valently close-to-convex of order ρ if it satisfies ℜ{f′(z)}>ρ for some ρ(0≤ρ<p) and for all z∈U. Also, a function f∈Tp is said to be p-valently starlike of order ρ if it satisfies ℜ{zf′(z)f(z)}>ρ, for some ρ(0≤ρ<p) and for all z∈U. Further, a function f∈Tp is said to be p-valently convex of order ρ if it satisfies ℜ{1+zf″(z)f′(z)}>ρ, for some ρ(0≤ρ<p) and for all z∈U.Theorem 13. If f∈Snp(ϑ,β,γ,φ) then f is p-valently close-to-convex of order ρ in |z|<r1(ϑ,β,γ,φ,ρ), where r1(ϑ,β,γ,φ,ρ)=infk{[1+(ανk+α−ν)(kp−1)]n(1+φϑ)ak(p−ρ)φ(ϑp+β−γ)}1k−pk≥p+1.
Proof. It is sufficient to show that |f′(z)zp−1−p|<p−ρ. Since |pzp−1−∑∞k=p+1kaKzk−1zp−1−p|<p−ρ, which implies that |f′(z)zp−1−p|≤∞∑k=p+1kak|z|k−p<p−ρ, implies
Theorem 14. If f∈Snp(ϑ,β,γ,φ) then f is p-valently starlike of order ρ in |z|<r2(ϑ,β,γ,φ,ρ), where r2(ϑ,β,γ,φ,ρ)=infk{k[1+(ανk+α−ν)(kp−1)]n(1+φϑ)(p−ρ)φ(ϑp+β−γ)(k−ρ)}1k−pk≥p+1.
Proof. In order to prove, it suffices to show that |zf′(z)f(z)−p|<p−ρ.
Theorem 15. If f∈Snp(ϑ,β,γ,φ), then f is p-valently convex of order ρ in |z|<r3(ϑ,β,γ,φ,ρ), where r3(ϑ,β,γ,φ,ρ)=infk{[1+(ανk+α−ν)(kp−1)]n(1+φϑ)p(p−ρ)φ(ϑp+β−γ)(k−ρ)}1k−pk≥p+1.
Proof. To prove this, it suffices to show that |1+zf″(z)f′(z)−p|<p−ρ. Since
Author Contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.Competing Interests
The author(s) do not have any competing interests in the manuscript.References
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