# Differential Subordination Results for Certain Integrodifferential Operator and Its Applications

Authors: M. A. Kutbi; A. A. Attiya
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We introduce an integrodifferential operatorJs,b(f) which plays an important role in the Geometric Function Theory. Some theorems in differential subordination for Js,b(f) are used. Applications in Analytic Number Theory are also obtained which give new results for Hurwitz-Lerch Zeta function and Polylogarithmic function.


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## Body

## 1. Introduction

LetA denote the class of functions f(z) normalized by
(1.1)f(z)=z+∑k=2∞‍akzk
which are analytic in the open unit disc 𝕌={z∈ℂ:|z|<1}.Also, letμ denote the class of analytic functions in the form
(1.2)r(z)=1+∑k=1∞‍akzk.We begin by recalling that a general Hurwitz-Lerch Zeta functionΦ(z,s,b) defined by (cf., e.g., [1, P. 121 et seq.])
(1.3)Φ(z,s,b)=∑k=0∞‍zk(k+b)s,(b∈ℂ∖ℤ0-,ℤ0-=ℤ-∪{0}={0,-1,-2,…},  s∈ℂ when z∈𝕌, Re(s)>1 when |z|=1)which contains important functions ofAnalytic Number Theory, as the Polylogarithmic function:
(1.4)Lis(z)=∑k=1∞‍zkks=zΦ(z,s,1),(s∈ℂwhenz∈𝕌,Re(s)>1when|z|=1).Several properties ofΦ(z,s,b) can be found in the recent papers, for example Choi et al. [2], Ferreira and López [3], Gupta et al. [4], and Luo and Srivastava [5]. See, also [6–16].Recently, Srivastava and Attiya [8] introduced the operator Js,b(f) which makes a connection between Geometric Function Theory and Analytic Number Theory, defined by
(1.5)Js,b(f)(z)=Gs,b(z)*f(z),(z∈𝕌;f∈A;b∈ℂ∖ℤ0-;s∈ℂ),
where
(1.6)Gs,b(z)=(1+b)s[Φ(z,s,b)-b-s]
and * denotes the Hadamard product (or convolution).Furthermore, Srivastava and Attiya [8] showed that
(1.7)Js,b(f)(z)=z+∑k=2∞‍(1+bk+b)sakzk.
As special cases of Js,b(f), Srivastava and Attiya [8] introduced the following identities:
(1.8)J0,b(f)(z)=f(z),J1,0(f)(z)=∫0zf(t)tdt=A(f)(z),J1,1(f)(z)=2z∫0zf(t)dt=ℒ(f)(z),J1,γ(f)(z)=1+γzγ∫0zf(t)tγ-1dt=ℒγ(f)(z)(γreal;γ>-1),Jσ,1(f)(z)=2σzΓ(σ)∫0z(log(zt))σ-1f(t)dt=Iσ(f)(z)(σreal;σ>0),
where, the operators A(f) and ℒ(f) are the integral operators introduced earlier by Alexander [17] and Libera [18], respectively, ℒγ(f) is the generalized Bernardi operator, ℒγ(f)(γ∈ℕ={1,2,…}) introduced by Bernardi [19], and Iσ(f) is the Jung-Kim-Srivastava integral operator introduced by Jung et al. [20].Moreover, in [8], Srivastava and Attiya defined the operator Js,b(f) for b∈ℂ∖ℤ-, by using the following relationship:
(1.9)Js,0(f)(z)=limb→0Js,b(f)(z).Some applications of the operatorJs,b(f) to certain classes in Geometric Function Theory can be found in [21, 22].In our investigations we need the following definitions and lemma.Definition 1.1.
Letf(z) andF(z) be analytic functions. The function f(z) is said to be subordinate to F(z), written f(z)≺F(z), if there exists a function w(z) analytic in 𝕌, with w(0)=0 and |w(z)|≤1, and such that f(z)=F(w(z)). If F(z) is univalent, then f(z)≺F(z) if and only if  f(0)=F(0) and  f(𝕌)⊂F(𝕌).Definition 1.2.
LetΨ:ℂ2×𝕌→ℂ be analytic in domain 𝔻, and let h(z) be univalent in 𝕌. If p(z) is analytic in 𝕌 with (p(z),zp′(z))∈𝔻 when z∈𝕌, then we say that p(z) satisfies a first order differential subordination if
(1.10)Ψ(p(z),zp′(z);z)≺h(z)(z∈U).
The univalent function q(z) is called dominant of the differential subordination (1.10), if p(z)≺q(z) for all p(z) satisfying (1.10), if q~(z)≺q(z) for all dominant of (1.10), then we say that q~(z) is the best dominant of (1.10).Lemma 1.3 (see [8]).
Ifz∈𝕌,f∈A,b∈ℂ∖ℤ- and s∈ℂ, then
(1.11)zJs+1,b′(f)(z)=(1+b)Js,b(f)(z)-bJs+1,b(f)(z).The purpose of the present paper is to extend the use ofJs,b(f) as integrodifferential operator, and some theorems in differential subordination for Js,b(f) are used. Applications in Analytic Number Theory are also obtained which give new results for Hurwitz-Lerch Zeta function and Polylogarithmic function.


## 2. Making Use ofJs,b(f) as a Differential Operator

From the definition ofJs,b(f) in (1.5) and using (1.7), we obtain the following identities.Forz∈𝕌,f∈A,n∈ℕ0=ℕ∪{0} and b∈ℂ∖ℤ-, we have
(2.1)J-1,0(f)(z)=zf′(z),J-1,1(f)(z)=12{f(z)+zf′(z)},J-1,1/(1-λ)(f)(z)=λf(z)+(1-λ)zf′(z)(λ≠1),J-n,0(f)(z)=Dn(f)(z),J-n,(1/λ)-1(f)(z)=Dλn(f)(z)(λ≠0),J-n,λ(f)(z)=Iλn(f)(z)(λ>-1),J-n,1(f)(z)=In(f)(z),
where Dn(f) is the Sălăgean differential operator which introduced by Sălăgean [23], Dλn(f) is the generalized of operator, Dλn(f)(λ>0;real) introduced by Al-Oboudi [24], Iλn(f) was studied by Cho and Srivastava [25] and by Cho and Kim [26], and the operator In(f) was studied by Uralegaddi and Somanatha [27].Also, we note that(2.2)J-n,0(f)(z)=Li-n(z)*f(z)(n∈ℕ0;f∈A),J-n,1(f)(z)=Li-n(z)z*f(z)(n∈ℕ0;f∈A),
where Lis(z) is the Polylogarithmic function defined by (1.4).Now, we prove the following lemma.Lemma 2.1.
Ifz∈𝕌,f∈A,n∈ℕ0 and b∈ℂ∖ℤ-, then
(2.3)J-n,b(f)(z)=1(1+b)n(zD+b)nf(z)(D:=ddz),
where (zD+b)n=(zD+b)∘(zD+b)∘⋯∘(zD+b) to n-times, and ∘ denotes the composition (I∘J)(f)(z)=I(J(f(z))).Proof.
Puttings=-n(n∈ℕ0) in (1.11), we have
(2.4)(1+b)(J-n,b)(f)(z)=[zddzJ-n+1,b(f)(z)+bJ-n+1,b(f)(z)]=(zD+b)J-n+1,b(f)(z)(D:=ddz),
therefore,
(2.5)J-n,b(f)(z)=1(1+b)(zD+b)J-n+1,b(f)(z).
Noting that the relation (2.5) is a recurrence relation, by using mathematical induction, we get (2.3), which completes the proof of the lemma.Puttingf(z)=f0(z)=z/(1-z) in Lemma 2.1, we obtain the following properties for both Hurwitz-Lerch Zeta function Φ(z,s,b) and Polylogarithmic function Lis(z).Corollary 2.2.
LetΦ(z,s,b) and Lis(z) be the Hurwitz-Lerch Zeta function and Polylogarithmic function defined by (1.3) and (1.4), respectively, then we have
(2.6)Φ(z,-n,b)=bn+(zddz+b)n(z1-z)(|z|<1),Li-n(z)=z{1+(zddz+1)n(z1-z)}(|z|<1),
where b∈ℂ∖ℤ0- and n∈ℕ0.Example 2.3.
Using Corollary2.2, we have the following well known results for z(z∈ℂ;|z|<1).(i)
Φ(z,0,b)=1/(1-z).(ii)
Φ(z,-1,b)=b+((1+b)z-bz2)/(1-z)2.(iii)
Φ(z,-2,b)=b2+((1+b)2z+(1-2b-2b2)z2+b2z3)/(1-z)3.(iv)
Li0(z)=z/(1-z).(v)
Li-1(z)=z/(1-z)2.(vi)
Li-2(z)=z(1+z)/(1-z)3.


## 3. Applications of Differential Subordination forJs,b(f)

To prove our results, we need the following lemmas due to Hallenbeck and Ruscheweyh [28] and Miller and Mocanu [29], respectively, see also Miller and Mocanu [30].Lemma 3.1.
Leth(z) be convex univalent in 𝕌, with h(0)=1,γ≠0 and  Re(γ)≥0.If q(z)∈μ and
(3.1)q(z)+zq′(z)γ≺h(z),
then
(3.2)q(z)≺S(z)≺h(z),
where
(3.3)S(z)=γzγ∫0zh(t)tγ-1dt.
The functionS(z) is convex univalent and is the best  dominant.Lemma 3.2.
Letλ>0, and let β0=β0(λ) be the root of the equation as follows:
(3.4)βπ=3π2-tan-1(λβ).
In addition, letα=α(β,λ)=β+(2/π)tan-1(λπ), for 0<β≤β0.
Ifp(z)∈μ and
(3.5)p(z)+λzp′(z)≺[1+z1-z]α
then
(3.6)p(z)≺[1+z1-z]β.Now, we define the functionL(f)(z):=L(s,b,λ)(f)(z) as the following:
(3.7)L(f)(z)=(1-λ-λb)Js,b(f)(z)+λ(1+b)Js-1,b(f)(z)(z∈𝕌),(z∈𝕌;f∈A;b∈ℂ∖ℤ-;{s,λ∈ℂ;λ≠0;Reλ≥0}).Theorem 3.3.
Let the functionL(f)(z) defined by (3.7) and for some α(0≤α<1). If
(3.8)Re{L(f)(z)z}>α,
then
(3.9)Re{Js,b(f)(z)z}>(2α-1)+2(1-α)2F1(1,1λ;1λ+1,-1).The constant(2α-1)+2(1-α)2F1(1,1/λ;(1/λ)+1,-1)  is the best estimate.Proof.
Defining the functionq(z)=Js,b(f)(z)/z, then we have q(z)∈μ.
If we takeγ=1/λ, and the convex univalent function h(z) defined by
(3.10)h(z)=1+(2α-1)z1+z,0≤α<1,
then, we have
(3.11)q(z)+zq′(z)γ=(1-λ)Js,b(f)(z)z+λJs,b′(f)(z).
Using Lemma 1.3 and (3.7), therefore (3.11) can be written as
(3.12)q(z)+zq′(z)γ=L(f)(z)z,
then,
(3.13)q(z)+zq′(z)γ≺h(z),
where h(z) is defined by (3.10) satisfying h(0)=1.
Applying Lemma3.1, we obtain that Js,b(f)(z)/z≺S(z), where the convex univalent function S(z) defined by
(3.14)S(z)=1λz1/λ∫0z1+(2α-1)t1+tt((1/λ)-1)dt.
Since Re{h(z)}>0 and S(z)≺h(z), we have Re{S(z)}>0.
This implies that(3.15)infz∈𝕌Re{S(z)}=S(1)=(2α-1)+2λ(1-α)∫01u((1/λ)-1)1+udu=(2α-1)+2(1-α)∫01dt1+tλ=(2α-1)+2(1-α)2F1(1,1λ;1λ+1,-1).
Hence, the constant (2α-1)+2(1-α)2F1(1,1/λ;(1/λ)+1,-1) cannot be replace by any larger one.
This completes the proof of Theorem3.3.Theorem 3.4.
Let the functionL(f)(z) with λ>0; real, defined by (3.7), and let β0 satisfy the following equation:
(3.16)β0π+tan-1(β02)=3π2.
If(3.17)L(f)(z)z≺[1+z1-z]β+(2/π)tan-1(λβ),
then
(3.18)Js,b(f)(z)z≺[1+z1-z]β(0<β≤β0).Proof.
Defining the functionp(z)=Js,b(f)(z)/z∈μ, then we have
(3.19)p(z)+λzp′(z)=(1-λ)Js,b(f)(z)z+λJs,b′(f)(z).
Using Lemma 1.3 and (3.7), therefore (3.11) can be written as
(3.20)p(z)+λzp′(z)=L(f)(z)z.
This completes the proof of Theorem 3.4 after applying Lemma 3.2


## 4. Applications in Analytic Number Theory

Puttingf(z)=f0(z)=z/(1-z) in Theorem 3.3, then we have the following property of Hurwitz-Lerch Zeta function.Corollary 4.1.
Let the functionGs,b(z) defined by (1.6). If
(4.1)Re{(1-λ-λb)Gs,b(z)+λ(1+b)Gs-1,b(z)z}>α,
then
(4.2)Re{Gs,b(z)z}>(2α-1)+2(1-α)2F1(1,1λ;1λ+1,-1),
where z∈𝕌,0≤α<1,b∈ℂ∖ℤ- and {s,λ∈ℂ;λ≠0;Reλ≥0}.The constant(2α-1)+2(1-α)2F1(1,1/λ;(1/λ)+1,-1)  is the best estimate.Puttingf(z)=f0(z)=z/(1-z) in Theorem 3.4, then we have another property of Hurwitz-Lerch Zeta function.Corollary 4.2.
Let the functionGs,b(z) defined by (1.6), and let β0 satisfy the following equation:
(4.3)β0π+tan-1(λβ0)=3π2.
If(4.4)(1-λ-λb)Gs,b(z)+λ(1+b)Gs-1,b(z)z≺[1+z1-z]β+(2/π)tan-1(λβ),
then
(4.5)Gs,b(z)z≺[1+z1-z]β(0<β≤β0),
where z∈𝕌,b∈ℂ∖ℤ-,s∈ℂ and λ>0; real.Puttingf(z)=f0(z)=z/(1-z) and b=1 in Theorem 3.3, then we have the following property of Polylogarithmic function.Corollary 4.3.
Let the functionHs(z) defined by
(4.6)Hs(z)=2s[Lis(z)z-1].
If(4.7)Re{(1-2λ)Hs(z)+2λHs-1(z)z}>α,
then
(4.8)Re{Hs(z)z}>(2α-1)+2(1-α)2F1(1,1λ;1λ+1,-1),
where z∈𝕌,0≤α<1 and {s,λ∈ℂ;λ≠0;Reλ≥0}.
The constant(2α-1)+2(1-α)2F1(1,1/λ;(1/λ)+1,-1)  is the best estimate.Puttingf(z)=f0(z)=z/(1-z) and b=1 in Theorem 3.4, then we have the following property of Polylogarithmic function.Corollary 4.4.
Let the functionsGs,b(z) and Hs(z) defined by (1.6) and (4.6), respectively, and let β0 satisfy the following:
(4.9)β0π+tan-1(λβ0)=3π2.
If(4.10)(1-2λ)Hs(z)+2λHs-1(z)z≺[1+z1-z]β+(2/π)tan-1(λβ),
then
(4.11)Gs,b(z)z≺[1+z1-z]β(0<β≤β0),
where z∈𝕌,s∈ℂ and λ>0; real.Settingf(z)=f0(z)=z/(1-z), b=1 and λ=1/2 in Theorem 3.3, then we have the following property of Polylogarithmic function.Corollary 4.5.
Let the functionHs(z) defined by (4.6).
If(4.12)Re{Hs-1(z)z}>α,
then
(4.13)Re{Hs(z)z}>2(2ln2-1)α+(3-4ln2),
where z∈𝕌,0≤α<1 and s∈ℂ.
The constant2(2ln2-1)α+(3-4ln2)  is the best estimate.Takingf(z)=f0(z)=z/(1-z), b=1 and λ=1/2 in Theorem 3.4, then we have the following property of polylogarithmic function.Corollary 4.6.
Let the functionHs(z) defined by (4.6).
If(4.14)Hs-1(z)z≺[1+z1-z]β+(2/π)tan-1(β),
then
(4.15)Hs(z)z≺[1+z1-z]β(0<β≤1.3148754023…),
where z∈𝕌 and s∈ℂ.Corollary 4.7.
Let the functionHs(z) defined by (4.6) as follows:
If(4.16)Hs-1(z)z≺[1+z1-z]3/2,
then
(4.17)Re{Hs+n(z)z}>1-(4ln2-2)n(n∈ℕ0),
where z∈𝕌 and s∈ℂ.Proof.
LetHs-1(z) satisfy the condition (4.16). Also, putting f(z)=f0(z)=z/(1-z), b=1, λ=1/2 and β=1 in Theorem 3.4.
Using (4.16), then we have
(4.18)Hs(z)z≺[1+z1-z],
therefore
(4.19)Re{Hs(z)z}>0.
Corollary 4.5, gives
(4.20)Re{Hs+1(z)z}>3-4ln2.
Applied (4.11) again and to n-times, which gives (4.17). This completes the proof of Corollary 4.7.Finally, we can put Corollary4.7 in the following form.Corollary 4.8.
Let the functionHs(z) defined by (4.6).
If(4.21)|Arg(Hs-1(z)z)|<3π4,
then
(4.22)Re{Hs+n(z)z}>1-(4ln2-2)n(n∈ℕ0),
where z∈𝕌 and s∈ℂ.



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