# A Study of a Certain Subclass of Hurwitz-Lerch-Zeta Function Related to a Linear Operator

Authors: F. Ghanim
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

By using a linear operator with Hurwitz-Lerch-Zeta function, which is defined
here by means of the Hadamard product (or convolution), the author investigates
interesting properties of certain subclasses of meromorphically univalent functions in the
punctured unit diskU*.


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## Body

## 1. Introduction

A meromorphic function is a single-valued function that is analytic in all but possibly a discrete subset of its domain, and at those singularities, it must go to infinity like a polynomial (i.e., these exceptional points must be poles and not essential singularities). A simpler definition states that a meromorphic functionf(z) is a function of the form
(1)f(z)=g(z)h(z),
where g(z) and h(z) are entire functions with h(z)≠0 (see [1, page 64]). A meromorphic function therefore may only have finite-order, isolated poles and zeros and no essential singularities in its domain. A meromorphic function with an infinite number of poles is exemplified by csc(1/z) on the punctured disk U*={z:0<|z|<1}. An equivalent definition of a meromorphic function is a complex analytic map to the Riemann sphere. Another definition for a meromorphic function is the following (see [2]).Definition 1.
A functionf on an open set Ω is meromorphic if there exists a discrete set of points S={z:z∈Ω} such that f is holomorphic on Ω-S and has poles at each z∈S. Furthermore, f is meromorphic in the extended complex plane if  F(z)=f(1/z) is either meromorphic or holomorphic at 0. In this case, we say that f has a pole or is holomorphic at infinity.Example 2.
The Gamma function is meromorphic in the whole complex plane.Example 3.
All rational functions such as(2)f(z)=z3-2z+10z5+3z-1
are meromorphic on the whole complex plane.Example 4.
The functionsf1(z)=ez/z and f2(z)=sinz/(1-z)2 are meromorphic on the whole complex plane.Example 5.
The functionf(z)=e1/z is defined in the whole complex plane except for the origin 0. However, 0 is not a pole of this function, rather an essential singularity. Thus, this function is not meromorphic in the whole complex plane.Example 6.
The complex logarithm functionf(z)=lnz is not meromorphic on the whole complex plane, as it cannot be defined on the whole complex plane except for an isolated set of points.
The aim of this paper is to investigate interesting properties of certain subclasses of meromorphically univalent functions with linear operator which are defined here by means of the Hadamard product (or convolution) in the punctured unit diskU*.


## 2. Preliminaries

LetΣ denote the class of meromorphic functions f(z) normalized by
(3)f(z)=1z+∑n=1∞‍anzn,
which are analytic in the punctured unit disk U*. For 0≤β, we denote by S*(β) and k(β) the subclasses ofΣ consisting of all meromorphic functions which are, respectively, starlike of order β and convex of order β in U*.For functionsfj(z)(j=1;2) defined by
(4)fj(z)=1z+∑n=1∞‍an,jzn,
we denote the Hadamard product (or convolution) of f1(z) and f2(z) by
(5)(f1*f2)=1z+∑n=1∞‍an,1an,2zn.Let us define the functionϕ~(α,β;z) by
(6)ϕ~(α,β;z)=1z+∑n=0∞‍(α)n+1(β)n+1zn,
for β≠0,-1,-2,…, and α∈ℂ∖{0}, where (λ)n=λ(λ+1)n+1 is the Pochhammer symbol. We note that
(7)ϕ~(α,β;z)=1zr2F1(1,α,β;z),
where
(8)2F1(b,α,β;z)=∑n=0∞‍(b)n(α)n(β)nznn!
is the well-known Gaussian hypergeometric function.We recall here a general Hurwitz-Lerch-Zeta function, which is defined in [3, 4] by the following series:
(9)Φ(z,t,a)=1at+∑n=1∞‍zn(n+a)t(a∈ℂ∖ℤ0-,ℤ0-={0,-1,-2,…};t∈ℂwhenz∈U=U*⊂{0};ℜ(t)>1whenz∈∂U).Important special cases of the functionΦ(z,t,a) include, for example, the Riemann Zeta function ζ(t)=Φ(1,t,1), the Hurwitz Zeta function ζ(t,a)=Φ(1,t,a), the Lerch Zeta function lt(ζ)=Φ(exp2πiξ,t,1), (ξ∈ℝ,ℜ(t)>1), and the polylogarithm Lti(z)=zΦ(z,t,a). Recent results on Φ(z,t,a) can be found in the expositions [5, 6]. By making use of the following normalized function we define
(10)Gt,a(z)=(1+a)t[Φ(z,t,a)-at+1z(1+a)t]=1z+∑n=1∞‍(1+an+a)tzn(z∈U*).Corresponding to the functionsGt,a(z) and using the Hadamard product for f(z)∈Σ, we define a new linear operator Lt,a(α,β) on Σ by the following series:
(11)Lat(α,β)f(z)=ϕ(α,β;z)*Gt,a(z)=1z+∑n=1∞‍(α)n+1(β)n+1(1+an+a)tanzn(z∈U*).The meromorphic functions with the generalized hypergeometric functions were considered recently by many others; see, for example, [7–12].It follows from (11) that
(12)z(Lat(α,β)f(z))′=α(Lat(α+1,β)f(z))-(α+1)Lat(α,β)f(z).We denote by the classΣα,βa,t(γ,δ,μ,λ) of all functions f(z)∈Σ such that
(13)ℜ{+λ(Lat(α+1,β)f(z)Lat(α+1,β)g(z))·(Lat(α,β)f(z)Lat(α,β)g(z))μ-1(1-λ)(Lat(α,β)f(z)Lat(α,β)g(z))μHH+λ(Lat(α+1,β)f(z)Lat(α+1,β)g(z))·(Lat(α,β)f(z)Lat(α,β)g(z))μ-1}>γ,
where g(z)∈Σ satisfies the following condition:
(14)ℜ{Lat(α,β)g(z)Lat(α+1,β)g(z)}>δ(0≤δ<1,z∈U*),
where γ and μ are real numbers such that 0≤γ<1, μ>0, and λ∈ℂ with ℜ{λ}>0.To establish our main results, we need the following lemmas.Lemma 7 (see [13]).
LetΩ be a set in the complex plane ℂ and let the function Ψ:ℂ2→ℂ satisfy the condition Ψ(ir2,s1)∉Ω for all real r2,s1≤(1+r22)/2. If q(z) is analytic in U* with q(0)=1 and Ψ(q(z),zq′(z))∈Ω, z∈U*, then ℜ{q(z)}>0.Lemma 8 (see [14]).
Ifq(z) is analytic in U* with q(0)=1, and if λ∈ℂ∖{0} with ℜ{λ}>0, then ℜ{q(z)+λzq′(z)}>γ, (0≤γ<1) implies ℜ{q(z)}>γ+(1-γ)(2ε-1), where ε is given by
(15)ε=ε(ℜλ)=∫01‍(1+tℜλ)-1dt
which is increasing function of ℜ{λ} and 1/2≤ε<1. The estimate is sharp in the sense that the bound cannot be improved.
For real or complex numbersa, b, c, (c∉z0-), the Gauss hypergeometric function is defined by
(16)2F1(a,b,c;z)=1+abcz1!+a(a+1)b(b+1)c(c+1)z22!+⋯.
One notes that the above series converges absolutely for z∈U* and hence represents an analytic function in the unit disc U* (see, for details, [15, Chapter 14]).Each of the identities (asserted by Lemma9) is fairly well known (cf., e.g., [15, Chapter 14]).Lemma 9.
For real or complex parametersa, b, c, (c∉z0-), then
(17)∫01‍tb-1(1-t)c-b-1(1-tz)-adt=Γ(b)Γ(c-b)Γ(c)y2F1(a,b,c;z)(ℜ(c)>ℜ(b)>0),
(18)2F1(a,b,c;z)=(1-z)-a2F1(a,c-b,c;zz-1),2F1(a,b,c;z)=y2F1(b,c,c;z),2F1(1,1,2;12)=2ln2.


## 3. Main Results

Theorem 10.
Letf(z)∈Σα,βa,t(γ,δ,μ,λ), α∈ℝ∖{0} and λ≥0. Then
(19)ℜ(Lat(α,β)g(z)Lat(α,β)g(z))≥2αγμ+δλ2αμ+δλ,(0≤γ<1,μ>0,z∈U*), where the function g(z)∈Σ satisfies condition (14).Proof.
Letξ=(2αγμ+δλ)/(2αμ+δλ), and we define the function q(z) by
(20)q(z)=1(1-ξ){(Lat(α,β)f(z)Lat(α,β)g(z))μ-ξ}.
Then q(z) is analytic in U* and q(0)=1. If we set
(21)h(z)=Lat(α,β)g(z)Lat(α+1,β)g(z),
then by the hypothesis ℜ{h(z)}>δ. Differentiating (19) and using the identity (12), we have
(22)(1-λ)(Lat(α,β)f(z)Lat(α,β)g(z))μ+λLat(α+1,β)f(z)Lat(α+1,β)g(z)(Lat(α,β)f(z)Lat(α,β)g(z))μ-1=[ξ+(1-ξ)q(z)]+λ(1-ξ)αμh(z)zq′(z).
Let us define the function Ψ(r,s) by
(23)Ψ(r,s)=ξ+(1-ξ)r+λ(1-ξ)αμh(z)s.
Using (21) and the fact that f(z)∈Σα,βa,t(γ,δ,μ,λ), we obtain
(24){Ψ(q(z),zq′(z));z∈U*}⊂Ω={w∈C:ℜ(w)>γ}.
Now for all real r2,s1≤(1+r22)/2, we have
(25)ℜ{Ψ(ir2,s1)}=ξ+λ(1-ξ)αμℜ{h(z)}≤ξ-λδ(1-ξ)(1+r22)2αμ≤ξ-λδ(1-ξ)2αμ=γ.
Hence for each z∈U*, Ψ(ir2,s1)∉Ω. Thus by Lemma 7, we have ℜ{q(z)}>0, and hence
(26)ℜ(Lat(α,β)f(z)Lat(α,β)g(z))μ>ξ,z∈U*.
This proves Theorem 10.Corollary 11.
Let the functionsf(z) and g(z) be in Σ, and let g(z) satisfy condition (14). If α∈ℝ∖{0}, λ≥0, and
(27)ℜ{(1-λ)Lat(α,β)f(z)Lat(α,β)g(z)+λLat(α+1,β)f(z)Lat(α+1,β)g(z)}>γ,0≤γ<1, z∈U*, then
(28)ℜ{Lat(α+1,β)f(z)Lat(α+1,β)g(z)}>ξ=γ(2α+δ)+δ(λ-1)2α+λδ.Proof.
We have(29)λLat(α+1,β)f(z)Lat(α+1,β)g(z)=[(1-λ)Lat(α,β)f(z)Lat(α,β)g(z)+λLat(α+1,β)f(z)Lat(α+1,β)g(z)]+(λ-1)Lat(α,β)f(z)Lat(α,β)g(z).
Since λ>1 making use of (27) and (19) (for μ=1), we deduce that
(30)ℜ{Lat(α+1,β)f(z)Lat(α+1,β)g(z)}>ξ=γ(2α+δ)+δ(λ-1)2α+λδ.Corollary 12.
Letλ∈ℂ∖{0} with ℜ{λ}>0 and α∈ℝ∖{0}. If f(z)∈Σ satisfies the following condition:
(31)ℜ{(zLat(α,β)f(z))μ-1(1-λ)(zLat(α,β)f(z))μ+λ(zLat(α+1,β)f(z))hh·(zLat(α,β)f(z))μ-1}>γ(0≤γ<1,μ>0,z∈U*), then
(32)ℜ{zLat(α,β)f(z)}μ>2μαγ+ℜ{λ}2μα+ℜ{λ}.
Further, ifλ≥1, α∈ℝ∖{0}, and f(z)∈Σ satisfies
(33)ℜ{(1-λ)zLat(α,β)f(z)+λ(zLat(α+1,β)f(z))}>γ,
then
(34)ℜ{zLat(α+1,β)f(z)}>(2a+1)γ+λ-12a+λhhhhhhhhhhhhhh.hhh(0≤γ<1,z∈U*).Proof.
The results (19) and (20) follow by putting g(z)=1/z in Theorem 10 and Corollary 11, respectively.Remark 13.
(i) Puttingλ=1 and α,β>0 in Corollary 12, we have that
(35)ℜ{zLat(α+1,β)f(z)·(zLat(α,β)f(z))μ-1}>γ
implies
(36)ℜ{zLat(α,β)f(z)}μ>2αγ+ℜ{λ}2α+ℜ{λ},z∈U*.
(ii) Forλ∈ℂ∖{0} with ℜ{λ}>0, μ=1, and α,β>0 in Corollary 12, we have that
(37)ℜ{(1-λ)zLat(α,β)f(z)h..+λ(zLat(α+1,β)f(z))}>γ(0≤γ<1) implies
(38)ℜ{zLat(α,β)f(z)}>2αγ+ℜ{λ}2α+ℜ{λ},z∈U*.
Choosingt=0 and α=β=1 appropriately in Corollary 12, we obtain the following results.
(iii) Forλ=1, t=0, and α=β=1 in Corollary 12, we have that
(39)ℜ{zf′(z)f(z)(zf(z))μ}>γ,hhhhh...(0≤γ<1,μ>0)
implies
(40)ℜ{zf(z)}μ>2γμ+12μ+1,z∈U*.
(iv) Forλ∈ℂ∖{0} with ℜ{λ}>0, μ=1, t=0, and α=β=1 in Corollary 12, we have that
(41)ℜ{(1-λ)zf(z)+λ(z2f′(z))}>γ(0≤γ<1,μ>0) implies
(42)ℜ{zf(z)}>2γ+ℜ{λ}2+ℜ{λ},z∈U*.
(v) Replacingf(z) by -zf′(z) in the result (ii), we have that
(43)-ℜ{(1-λ)z2f(z)+λ(z3f′′(z))}>γ
implies
(44)-ℜ{z2f′(z)}>2γ+ℜ{λ}2+ℜ{λ},z∈U*.
(vi) Forλ∈ℝ with λ≥1, μ=1, t=0, and α=β=1 in Corollary 12, we have that
(45)ℜ{(1-λ)zf(z)+λ(z2f′(z))}>γ
implies
(46)ℜ{zf(z)}>3γ+λ-12+λ.Theorem 14.
Letλ∈ℂ∖{0} with ℜ{λ}>0 and α∈ℝ∖{0}. If f(z)∈Σ satisfies the following condition:
(47)ℜ{(zLat(α,β)f(z))μ-1(1-λ)(zLat(α,β)f(z))μ.h+λ(zLat(α+1,β)f(z))·(zLat(α,β)f(z))μ-1}>γhhhhhhhhhhhhhhhhhhh(0≤γ<1,μ>0,z∈U*),
then
(48)ℜ{zLat(α,β)f(z)}μ>γ+(1-γ)(2ρ-1),
where
(49)ρ=12h2F1(1,1,μaℜ{λ},12),z∈U*.Proof.
Let(50)q(z)=(zLat(α,β)f(z))μ.
Then q(z) is analytic in U* with q(0)=1. Differentiating q(z) with respect to z and using the identity (12), we obtain
(51)(1-λ)(zLat(α,β)f(z))μ+λ(zLat(α+1,β)f(z))·(zLat(α,β)f(z))μ-1=q(z)+λzq′(z)μa,
so that by the hypothesis (47), we have
(52)ℜ{q(z)+λzq′(z)μa}>γ.
In view Lemma 8, this implies that
(53)ℜ{q(z)}>γ+(1-γ)(2ρ-1),
where
(54)ρ=ρ(ℜ{λ})=∫01‍(1+tℜ{λ}/μa)-1dt.
Putting ℜ{λ}=λ1>0, we have
(55)ρ=∫01‍(1+tℜ{λ}/μα)-1dt=μαλ1∫01‍u(μα/λ1)-1(1+u)-1du.
Using (17)-(18), we obtain
(56)ρ=h2F1(1,μαλ1;μαλ1+1;-1)ρ*=12j2F1(1,1,αλ+1,12).
Thus the proof of Theorem14 is complete.Corollary 15.
Letλ∈ℝ, μ=1 with λ>1. If f(z)∈Σ satisfies
(57)ℜ{(1-λ)(zLat(α,β)f(z))hh+λ(zLat(α+1,β)f(z))}>γ(0≤γ<1,α∈ℝ∖{0},z∈U*),  then
(58)ℜ{(zLat(α+1,β)f(z))}h>γ+(1-γ)(2ρ*-1)(1-λ-1),
where
(59)ρ*=12j2F1(1,1,αλ+1,12).Proof.
The result follows by using the identity(60)λzLat(α+1,β)f(z)=[(1-λ)(zLat(α,β)f(z))+λ(zLat(α+1,β)f(z))]hhhh+(λ-1)(zLat(α,β)f(z)).Remark 16.
(i) We note that ifλ,μ>0, α=β=1, and t=0 in Corollary 15, that is,
(61)ℜ{(1-λ)(zf(z))λ+λz2f′(z)(zf(z))λ-1}>γ,
then (32) implies that
(62)ℜ{zf(z)}λ>2γ+12+1(z∈U*)
whereas if f(z)∈Σ satisfies condition (61); then, by using Theorem 14, we have
(63)ℜ{zf(z)}λ>2(1-ln2)γ+(2ln2-1)
which is better than (61).
(ii) We observe that ifλ∈ℝ satisfying λ≥0 and
(64)k(z)=Lat(α+1,β)f(z)Lat(α+1,β)g(z)+(1λ-1)Lat(α,β)f(z)Lat(α,β)g(z).
Then, from Theorem 10 (for μ=1), we have that
(65)ℜ{k(z)}>γλ
implies
(66)ℜ{Lat(α,β)f(z)Lat(α,β)g(z)}>2αγ+λδ2α+λδ,
whenever
(67)ℜ{Lat(α,β)g(z)Lat(α+1,β)g(z)}>δ,0≤δ<1.
Let λ→+∞; then, from (66), we have that ℜ{k(z)}>0 implies
(68)ℜ{Lat(α,β)f(z)Lat(α+1,β)g(z)}>1,z∈U*,
whenever
(69)ℜ{Lat(α,β)g(z)Lat(α+1,β)g(z)}>δ(0≤δ<1,z∈U*).
In the following theorem, we will extend the above results as follows.Theorem 17.
Suppose that the functionsf(z) and g(z) are in Σ, and suppose that g(z) satisfies condition (14). If
(70)ℜ{Lat(α+1,β)f(z)Lat(α+1,β)g(z)-Lat(α,β)f(z)Lat(α,β)g(z)}>(1-γ)δ2α(0≤γ<1,α∈ℝ∖{0},0≤δ<1,z∈U*),  then
(71)ℜ{Lat(α,β)f(z)Lat(α+1,β)g(z)}>γ,(72)ℜ{Lat(α+1,β)f(z)Lat(α+1,β)g(z)}>(2α+1+δ)γ-δ2α.Proof.
Let(73)q(z)=1(1-γ){Lat(α,β)f(z)Lat(α,β)g(z)-γ}.
Then q(z) is analytic in U* with q(0)=1. Putting
(74)Φ(z)=Lat(α,β)g(z)Lat(α+1,β)g(z),
we observe that by hypothesis ℜ{Φ(z)}>δ, in U*. A simple computation shows that
(75)(1-γ)zq′(z)ℜ{Φ}α=Lat(α+1,β)f(z)Lat(α+1,β)g(z)-Lat(α,β)f(z)Lat(α,β)g(z)=Ψ(q(z),zq′(z)),
where
(76)Ψ(r,s)=(1-γ)Φ(z)sα,α∈ℝ∖{0}.
Using the hypothesis (70), we obtain
(77)Ψ(q(z),zq′(z);z∈U*)⊂Ω={w∈ℂ:ℜ(w)>-δ(1-γ)2α}.
Now, for all real r2,s1≤-((1+r22)/2), we have
(78)ℜ{Ψ(ir2,s1)}=s1(1-γ)ℜ{Φ}α≤-δ(1-γ)(1-r22)2α≤-δ(1-γ)2α.
This shows that ℜ{Ψ(ir2,s1)}∉Ω for each z∈U*.
Hence by Lemma7, we have that this proves (71). The proof of (72) follows by using (71) and (72) in the identity:
(79)ℜ{Lat(α+1,β)f(z)Lat(α+1,β)g(z)}={Lat(α+1,β)f(z)Lat(α+1,β)g(z)-Lat(α,β)f(z)Lat(α,β)g(z)}+Lat(α,β)f(z)Lat(α,β)g(z).
This completes the proof of Theorem 17.Remark 18.
(i) Puttingα=β=1, t=0, and g(z)=1/z in Theorem 17 for all z∈U*, we obtain that
(80)ℜ{zf(z)+z2f′(z)}>-δ(1-γ)2
implies ℜ{zf(z)}>γ and
(81)ℜ{2zf(z)+z2f′(z)}>(2+δ)γ-δ2.
(ii) Forα=2, β=1, t=0, and g(z)=1/z in Theorem 17 for all z∈U*, we have that
(82)ℜ{zLat(3,1)f(z)-zLat(2,1)f(z)}>-(1-γ)δ4(0≤γ<1,0≤δ<1) implies
(83)ℜ{zLat(2,1)f(z)}>γ,ℜ{zLat(3,1)f(z)}>(4+δ)γ-δ4.



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