Open Journal of Mathematical Analysis

On certain subclasses of p-valent functions with negative coefficients defined by a generalized differential operator

Bitrus Sambo1, Gideon Benjamin Meller
Department of Mathematics, Gombe State University, P.M.B.127, Gombe, Nigeria.; (B.S & G.B.M)
1Corresponding Author: bitrussambo3@gmail.com

Abstract

In this article, we introduce new subclasses of normalized analytic functions in the unit disk U, defined by a generalized Raducanu-Orhan differential Operator. Various results are driven including coefficient inequalities, growth and distortion theorem, closure property, δ$-neighborhoods, extreme points, radii of close-to-convexity, starlikeness and convexity for these subclasses.

Keywords:

Multivalent functions, Raducanu-Orhan differential operator, extreme points, coefficient inequality, closure properties.

1. Introduction

Let A denote the class of all functions of the form
f(z)=z+k=2akzk,
(1)
which are analytic in the open unit disk U={z:|z|<1}. For a function fA, Raducanu and Orhan [1] introduced the following operator: D0ανf(z)=f(z) D1ανf(z)=ανz2f(z)+(αν)zf(z)+(1α+ν)f(z)
Dnανf(z)=Dαν(Dn1ανf(z)),(0να1,nN).
(2)
If f is given by (1), then from the definition of the operator Dnανf, the Equation (2) can be rewritten as:
Dnανf(z)=z+k=2[1+(ανk+αν)(k1)]nakzk,
(3)
where (nN0=N{0}).

Remark 1.

  1. When α=1,ν=0, we get the Sălăgean differential operator introduced by Sălăgean in [2].
  2. When ν=0, we obtain differential operator defined by Al-Oboudi in [3].

Let Ap denote the class of functions of the form
f(z)=zp+k=p+1akzk,(p=1,2,3,...)
(4)
which are analytic and p-valent in the open unit disk U={z:|z|<1}. We can write the following equalities for the functions fAp: D0,pανf(z)=f(z)
D1,pανf(z)=ανpz2f(z)+1p[(1p)αν+αν]zf(z)+(1α+ν)f(z)
(5)
Dn,pανf(z)=Dαν(Dn1ανf(z)),(nN=1,2,3,...)
(6)
If f is given by Equation 4, then from Equation 5 and Equation 6, we see that
Dn,pανf(z)=zp+k=p+1[1+(ανk+αν)(kp1)]nakzk,(nN0=N{0},pN=1,2,3,).
(7)

Remark 2.

  1. If ν=0, Dn,pανf=Dnα,pf defined by Bulut in [4]
  2. If p=1, Dn,pανf=Dnανf introduced by Raducanu and Orhan in [1]
  3. If p=1,α=1,ν=0, Dn,pανf=Dnf defined by Sălăgean in [2]
  4. If p=1,ν=0, Dn,pανf=Dnαf defined by Al-Oboudi in [3].

Let Tp denote the subclass of Ap consisting of functions of the form
f(z)=zpk=p+1akzk,(ak0,p=1,2,3,...).
(8)
If f is given by Equation 8, then from Equation 5 and Equation 6, we get
Dn,pανf(z)=zpk=p+1[1+(ανk+αν)(kp1)]nakzk,(nN0)
(9)

Definition 1. A function fTp is in the class, Snp(ϑ,β,γ,φ) if and only if

|(Dn,pανf(z))pzp1ϑ(Dn,pανf(z))+(βγ)|<φ,(zU,nN0)
(10)
for 0να1,0ϑ<1,0γ<1,0<β1,0<φ<1,pN,Dn,pανf(z) as in 9.

In this paper, basic properties of the class Snp(ϑ,β,γ,φ) are studied such as: coefficient inequalities, growth and distortion theorem, closure property, δ-neighborhoods, extreme points, radii of close-to-convexity, starlikeness and convexity for these subclasses.

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 fTp 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|<1δ0...(zU,0δ0<1) for 0ϑ<1, 0γ<1, 0<β1, 0<φ<1.

Definition 3. For a function fTp, δ0, δ-neighborhood of f is defined as:

Npδ(f,g)={g:g=zpk=p+1bkzkTpandk=p+1k|akbk|δ},
(11)
in particular, for a function hTp, given by h(z)=zp (pN), we immediately have
Npδ(h,g)={g:g=zpk=p+1bkzkTp,andk=p+1k|bk|δ}.
(12)
The concept of neighborhoods was first introduced by Goodman [5] and generalized by Ruschewey [6] and Altintas [7] (see also [8, 9].

2. Coefficient inequalities

Theorem 4. A function fTp is in the class Snp(ϑ,β,γ,φ) if and only if

k=p+1k[1+(ανk+αν)(kp1)]n(1+φϑ)akφ(ϑp+βγ),
(13)
for 0να1,0ϑ<1, 0γ<1, 0<β1, 0<φ<1, nN0, pN. Furthermore, the result is sharp for the function given as f(z)=zpφ(ϑp+βγ)k[1+(ανk+αν)(kp1)]n(1+φϑ)ak,(kp+1).

Proof. Suppose that fSnp(ϑ,β,γ,φ), then from inequality 10, we have |(Dn,pανf(z))pzp1ϑ(Dn,pανf(z))+(βγ)|=|pzp1k=p+1k[1+(ανk+αν)(kp1)]nakzk1pzp1ϑ(pzp1k=p+1k[1+(ανk+αν)(kp1)]nakzk1)+(βγ)|=|k=p+1k[1+(ανk+αν)(kp1)]nakzk1ϑ(pzp1k=p+1k[1+(ανk+αν)(kp1)]nakzk1)+(βγ)|<φ,(zU,nN0) it is well known that z|z|, therefore, we obtain {K=p+1k[1+(ανk+αν)(kp1)]nakzk1ϑ(pzp1k=p+1k[1+(ανk+αν)(kp1)]nakzk1)+(βγ)}<φ. If we choose z real and let z1, then we get K=p+1k[1+(ανk+αν)(kp1)]nakφ{ϑ(pk=p+1k[1+(ανk+αν)(kp1)]nak)+(βγ)} which is precisely the assertion 13. On contrary, suppose that the inequality 13 hold true and let zδU={zC:|z|=1}. Then, from 10, we have |(Dn,pανf(z))pzp1|φ|ϑ(Dn,pανf(z))+(βγ)|k=p+1k[1+(ανk+αν)(kp1)]nak|z|k1φ(ϑp+βγ)+φϑk=p+1k[1+(ανk+αν)(kp1)]nak|z|k1=k=p+1k[1+(ανk+αν)(kp1)]nak|z|k1(1+φϑ)akφ(ϑp+βγ)0. By maximum modulus theorem, we have fSnp(ϑ,β,γ,φ).

Corollary 5. If fSnp(ϑ,β,γ,φ), then ap+1φ(ϑp+βγ)(p+1)[1+(αν(p+1)+αν)(1p)]n(1+φϑ).

3. Growth and distortion theorem

Theorem 6. For each f(z)Snp(ϑ,β,γ,φ), we have |z|pφ(ϑp+βγ)[1+(αν(p+1)+αν)(1p)]n(1+φϑ)(p+1)|z|p+1|f(z)||z|p+φ(ϑp+βγ)[1+(αν(p+1)+αν)(1p)]n(1+φϑ)(p+1)|z|p+1.

Proof. Let f(z)Snp(ϑ,β,γ,φ),zU, the bound on f(z) is given by

|f(z)||z|p+|z|p+1k=p+1ak,zU,
(14)
from Theorem 4, we have
k=p+1akφ(ϑp+βγ)(p+1)[1+(αν(p+1)+αν)(1p)]n(1+φϑ),
(15)
by using (15) in (14), we obtain
|f(z)||z|p+φ(ϑp+βγ)(p+1)[1+(αν(p+1)+αν)(1p)]n(1+φϑ)|z|p+1,
(16)
again using (15), we have
|f(z)||z|pφ(ϑp+βγ)(p+1)[1+(αν(p+1)+αν)(1p)]n(1+φϑ)|z|p+1.
(17)
Consequently, combining (16) and (17) we obtain the desired result.

Theorem 7. For each f(z)Snp(ϑ,β,γ,φ), we have p|z|p1φ(ϑp+βγ)[1+(αν(p+1)+αν)(1p)]n(1+φϑ)|z|p|f(z)|p|z|p1+φ(ϑp+βγ)[1+(αν(p+1)+αν)(1p)]n(1+φϑ)|z|p.

Proof. Let f(z)Snp(ϑ,β,γ,φ),zU, the bound on the derivative of f(z) is given by |f(z)|p|z|p1+(p+1)|z|pk=p+1ak,zU, and, in the same way as above, we get our desired result.

4. Closure properties

Theorem 8. Let the functions f(z)=zpk=p+1akzk,(ak0)g(z)=zpk=p+1bkzk,(bk0), be in the class Snp(ϑ,β,γ,φ). Then for 0λ1, the function h is defined as h(z)=(1λ)f(z)+λg(z)=zpk=p+1ckzk, where ck:=(1λ)ak+λbk0, 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+αν)(kp1)]n(1+φϑ)ck=(1λ)k=p+1k[1+(ανk+αν)(kp1)]n(1+φϑ)ak+λk=p+1k[1+(ανk+αν)(kp1)]n(1+φϑ)bk(1λ)φ(ϑp+βγ)+λφ(ϑp+βγ)=φ(ϑp+βγ), which completes the proof.

5. δ-Neighborhoods

Theorem 9. If

δ:=φ(ϑp+βγ)[1+(αν(p+1)+αν)(1p)]n(1+φϑ),
(18)
then Snp(ϑ,β,γ,φ)Npδ(h,g).

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)+αν)(1p)]n(1+φϑ)k=p+1akφ(ϑp+βγ), therefore,

k=p+1akφ(ϑp+βγ)(p+1)[1+(αν(p+1)+αν)(1p)]n(1+φϑ).
(19)
On the other hand, we also find from (13) that
k=p+1kakφ(ϑp+βγ)[1+(αν(p+1)+αν)(1p)]n(1+φϑ),
(20)
that is
k=p+1kakφ(ϑp+βγ)[1+(αν(p+1)+αν)(1p)]n(1+φϑ):=δ,
(21)
which completes the proof.

Theorem 10. If g(z)Snp(ϑ,β,γ,φ) and

δ0=1δp+1(p+1)[1+(αν(p+1)+αν)(1p)]n(1+φϑ)(p+1)[1+(αν(p+1)+αν)(1p)]n(1+φϑ)φ(ϑp+βγ),
(22)
then Npδ(f,g)Sn,(δ0)p(ϑ,β,γ,φ).

Proof. Suppose that fNpδ(f,g), then by Definition 3, we have k=p+1k|akbk|δ, which readily implies the coefficient inequality given by k=p+1|akbk|δp+1(pN). Next, since gSnp(ϑ,β,γ,φ), we have from inequality (13) that k=p+1bkφ(ϑp+βγ)(p+1)[1+(αν(p+1)+αν)(1p)]n(1+φϑ), so from the definition of the class, we have |f(z)g(z)1|<k=p+1|akbk|1k=p+1bkδp+1(p+1)[1+(αν(p+1)+αν)(1p)]n(1+φϑ)(p+1)[1+(αν(p+1)+αν)(1p)]n(1+φϑ)φ(ϑp+βγ)=1δ0, provided that δ0 is given precisely by (22). Thus, by the definition, fSn,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)=zpφ(ϑp+βγ)k[1+(ανk+αν)(kp1)]n(1+φϑ)zk(kp+1) then, fSnp(ϑ,β,γ,φ) if and only if it can be expressed in the form f(z)=λpfp(z)+k=p+1λkfk(z), where λk0 and λp=1k=p+1λk.

Proof. Assume that f(z)=λpfp(z)+k=p+1λkfk(z), then f(z)=(1k=p+1λk)zp+k=p+1λk{zpφ(ϑp+βγ)k[1+(ανk+αν)(kp1)]n(1+φϑ)zk}=zpk=p+1λk{φ(ϑp+βγ)k[1+(ανk+αν)(kp1)]n(1+φϑ)zk}. Thus, k=p+1k[1+(ανk+αν)(kp1)]n(1+φϑ)λkφ(ϑp+βγ)k[1+(ανk+αν)(kp1)]n(1+φϑ)=φ(ϑp+βγ)k=p+1λk=φ(ϑp+βγ)(1λp)φ(ϑp+βγ), which shows that f satisfies condition (13) and therefore,fSnp(ϑ,β,γ,φ). Conversely, suppose that fSnp(ϑ,β,γ,φ), since akφ(ϑp+βγ)k[1+(ανk+αν)(kp1)]n(1+φϑ),(kp+1), we may set λk=k[1+(ανk+αν)(kp1)]n(1+φϑ)φ(ϑp+βγ)ak, and λp=1k=p+1λk, then we obtain from f(z)=zpk=p+1akzk=(λp+k=p+1λk)zpk=p+1λkφ(ϑp+βγ)k[1+(ανk+αν)(kp1)]n(1+φϑ)zk=λpzp+k=p+1λk(zpφ(ϑp+βγ)k[1+(ανk+αν)(kp1)]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)=zpφ(ϑp+βγ)k[1+(ανk+αν)(kp1)]n(1+φϑ)zk(kp+1)

7. Radii of close-to-convexity,starlikeness and convexity

A function fTp is said to be p-valently close-to-convex of order ρ if it satisfies {f(z)}>ρ for some ρ(0ρ<p) and for all zU. Also, a function fTp is said to be p-valently starlike of order ρ if it satisfies {zf(z)f(z)}>ρ, for some ρ(0ρ<p) and for all zU. Further, a function fTp is said to be p-valently convex of order ρ if it satisfies {1+zf(z)f(z)}>ρ, for some ρ(0ρ<p) and for all zU.

Theorem 13. If fSnp(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+αν)(kp1)]n(1+φϑ)ak(pρ)φ(ϑp+βγ)}1kpkp+1.

Proof. It is sufficient to show that |f(z)zp1p|<pρ. Since |pzp1k=p+1kaKzk1zp1p|<pρ, which implies that |f(z)zp1p|k=p+1kak|z|kp<pρ, implies

k=p+1kak|z|kppρ<1,
(23)
and by applying the result of Theorem 4, we get k=p+1akφ(ϑp+βγ)k[1+(ανk+αν)(kp1)]n(1+φϑ)ak. Hence,(23) is true if
k|z|kppρk[1+(ανk+αν)(kp1)]n(1+φϑ)φ(ϑp+βγ),
(24)
solving (24) for z we obtain |z|{[1+(ανk+αν)(kp1)]n(1+φϑ)(pρ)φ(ϑp+βγ)}1kp which completes the proof.

Theorem 14. If fSnp(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+αν)(kp1)]n(1+φϑ)(pρ)φ(ϑp+βγ)(kρ)}1kpkp+1.

Proof. In order to prove, it suffices to show that |zf(z)f(z)p|<pρ.

|zf(z)f(z)p|=|zf(z)pf(z)f(z)|=|z(pzp1k=p+1kakzk1)p(zpk=p+1akzk)zpk=p+1akzk|=|k=p+1(kp)akzkp1k=p+1akzkp|k=p+1(kp)ak|z|kp1k=p+1ak|z|kp<pρ,
(25)
and by using inequality (13), we get k=p+1akφ(ϑp+βγ)k[1+(ανk+αν)(kp1)]n(1+φϑ)ak, so, (25) holds true if (kρ)|z|kρpρk[1+(ανk+αν)(kp1)]n(1+φϑ)φ(ϑp+βγ), and then f is starlike of order ρ.

Theorem 15. If fSnp(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+αν)(kp1)]n(1+φϑ)p(pρ)φ(ϑp+βγ)(kρ)}1kpkp+1.

Proof. To prove this, it suffices to show that |1+zf(z)f(z)p|<pρ. Since

|1+zf(z)f(z)p|=|f(z)+zf(z)pf(z)f(z)|=|pzp1k=p+1kakzk1+z(p(p1)zp2k=p+1k(k1)akzk2)p(pzp1k=p+1kakzk1)pzp1k=p+1kakzk1|
(26)
it implies that |1+zf(z)f(z)p|=|k=p+1k(kp)akzkppk=p+1kakzkp|k=p+1k(kp)ak|z|kppk=p+1kak|z|kp<pρ and by applying the result in Theorem 4, we get k=p+1akφ(ϑp+βγ)k[1+(ανk+αν)(kp1)]n(1+φϑ)ak so, (26) holds true if k(kρ)|z|kpp(pρ)k[1+(ανk+αν)(kp1)]n(1+φϑ)φ(ϑp+βγ) and then f is convex of order ρ.

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.

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