# Certain Subclass ofm-Valent Functions Associated with a New Extended Ruscheweyh Operator Related to Conic Domains

Authors: Amnah E. Shammaky; Tamer M. Seoudy
Journal: Journal of Function Spaces
 (2021)

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## Abstract

The main object of the present paper is to introduce certain subclass ofm-valent functions associated with a new extended Ruscheweyh linear operator in the open unit disk. Also, we investigate a number of geometric properties including coefficient estimates and the Fekete–Szegö type inequalities for this subclass. Several known consequences of the main results are also pointed out.


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## Body

## 1. Introduction

LetAm denote the class of functions of the next form:(1)fξ=ξm+∑n=m+1∞anξnm∈ℕ=1,2,3,⋯,which are analytic and m-valent in the open unit disc D=ξ∈ℂ:ξ<1, and let A1=A. Also, let f,g be analytic in D, and the function fξ is said to be subordinate to gξ if there exists a function ωξ analytic in D with ω0=0 and ∣ωξ∣<1,ξ∈D, such that fξ=gωξ. In such a case, we write fξ≺gξ. If g is univalent function, then fξ≺gξ if and only if f0=g0 and fD⊂gD (see [1, 2] and [3]).For functionsfξ given by (1) and gξ is defined by(2)gξ=ξm+∑n=m+1∞bnξn,and the Hadamard product or convolution of fξ and gξ is defined by(3)f∗gξ=ξm+∑n=m+1∞anbnξn.Forv∈ℂ, k∈ℝ, and n∈ℕ, the Pochhammer k-symbol vn,k is given by (see [4])(4)vn,k=vv+kv+2k⋯v+n−1k=∏i=1nv+i−1k.We define the function ϕmδ,k;ξ by(5)ϕmδ,k;ξ=ξm1−ξδ+mk/k=ξm+∑n=m+1∞δ+mkn−m,kkn−m,kξnδ>−mk;k>0;ξ∈D.Corresponding to the functionϕmδ,k;ξ, we consider a linear operator Dδ+mk−k:Am⟶Amδ>−mk,k>0 which is defined by means of the following Hadamard product (or convolution):(6)Dδ+mk−kfξ=ϕmδ,k;ξ∗fξ=ξm+∑n=m+1∞δ+mkn−m,kkn−m,kanξnξ∈D.It is easily verified from (6) that(7)kξDδ+mk−kfξ′=δ+mkDδ+mkfξ−δDδ+mk−kfξk>0.We note that(1)
Fork=1, the operator Dδ+mk−kfξ reduced to the differential operator Dδ+m−1fξ introduced by Goel and Sohi [5] (see also [6, 7] and [8])(2)
Form=1, we obtain the k-Ruscheweyh derivative operator Dkδ ([9]), where(8)Dkδfξ=ξ1−ξδ+k/k∗fξ=ξ+∑n=2∞δ+kn−1,kkn−1,kanξn.(3)
Fork=m=1, the operator Dδ+mk−kfξ reduced tothe well-familiar Ruscheweyh operator Dδ ([10])(4)
Forδ=k−mk,we have D0fξ=fξ, and for δ=2k−mk, we get Dkfξ=ξmξ1−mfξ′By using the linear operatorDδ+mk−kfξ, we define the subclass β−STmδ,k,b of Am as follows:Definition 1.
Letβ≥0, δ>−mk, m∈ℕ, k>0, b∈ℂ∗=ℂ\0, and ξ∈D. A function f∈Am is in the class β−STmδ,k,b, if it satisfies(9)R1+1bξDδ+mk−kfξ′mDδ+mk−kfξ−1>β1bξDδ+mk−kfξ′mDδ+mk−kfξ−1.Geometrically, a functionf∈β−STmδ,k,b if and only if(10)1+1bξDδ+mk−kfξ′mDδ+mk−kfξ−1,takes all the values in the conic domainΩβ=ψβD, where(11)Ωβ=u+iv:u>βu−12+v2,or equivalently,(12)1+1bξDδ+mk−kfξ′mDδ+mk−kfξ−1≺ψβξ,Ωβ=ψβD.The boundary∂Ωβ of the above set becomes the imaginary axis when β=0, a hyperbola when 0<β<1, a parabola when β=1, and an ellipse when 1<β<∞. The functions ψβξ are defined by(13)ψβξ=1+ξ1−ξβ=0,1+11−β2cos2πcos−1βilog1+ξ1−ξ0<β<1,1+2π2log1+ξ1−ξ2β=1,1+1β2−1sinπ2Rt∫0uξ/tdx1−x21−t2x2+β2β2−11<β<∞,with uξ=ξ−t/1−tξ0<t<1,ξ∈D, where t is chosen such that k=coshπR′t/4Rt, and Rt is the Legendre’s complete elliptic integral of the first kind and R′t the complementary integral of Rt (see [11, 12] and [13]).By taking specific values to the parametersβ, m, δ, k, and b in the subclass β−STmδ,k,b, we obtain(1)
β−STmδ,k,1−α/mcosγe−iγ=β−STmγδ,k,α0≤α<m;γ<π/2=f∈Am:eiγξDδ+mk−kfξ′/Dδ+mk−kfξ≺m−αcosγψβξ+αcosγ+imsinγ and β−STmδ,k,1−α/m=β−STmδ,k,α0≤α<m=f∈Am:1/m−αξDδ+mk−kfξ′/Dδ+mk−kfξ−α≺ψβξ(2)
β−STmδ,1,b=β−STmδ,b=f∈Am:1+1/bξDδ+m−1fξ′/mDδ+m−1fξ−1≺ψβξ,β−STmδ,1,1−α/mcosγe−iγ=β−STmγδ,α0≤α<m;γ<π/2=f∈Am:eiγξDδ+m−1fξ′/Dδ+m−1fξ≺m−αcosγψβξ+αcosγ+imsinγ and β−STmδ,1,1−α/m=β−STmδ,α0≤α<m=f∈Am:1/m−αξDδ+m−1fξ′/Dδ+m−1fξ−α≺ψβξ(3)
β−ST1δ,1,b=β−STδ,b=f∈A:1+1/bξDδfξ′/Dδfξ−1≺ψβξ,β−ST1δ,1,1−αcosγe−iγ=β−STγδ,α0≤α<1;γ<π/2=f∈A:eiγξDδfξ′/Dδfξ≺1−αcosγψβξ+αcosγ+isinγ and β−ST1δ,1,1−α/m=β−STδ,α0≤α<1=f∈A:1/1−αξDδfξ′/Dδfξ−α≺ψβξ(4)
β−ST1δ,k,b=β−STδ,k,b=f∈A:1+1/bξDkδfξ′/Dkδfξ−1≺ψβξ,β−ST1δ,k,1−αcosγe−iγ=β−STγδ,k,α0≤α<1;γ<π/2=f∈A:eiγξDkδfξ′/Dkδfξ≺1−αcosγψβξ+αcosγ+isinγ and β−ST1δ,k,1−α/m=β−STδ,k,α0≤α<1=f∈A:1/1−αξDkδfξ′/Dkδfξ−α≺ψβξ(5)
β−STmk−mk,k,b=β−STmb=f∈Am:1+1/bξf′ξ/mfξ−1≺ψβξβ−STmk−mk,k,1−α/mcosγe−iγ=β−STmγα0≤α<m;γ<π/2=f∈Am:eiγξf′ξ/fξ≺m−αcosγψβξ+αcosγ+imsinγ and (see [14])(6)
β−STmk−mk,k,1−α/m=β−STmα0≤α<m=f∈Am:1/m−αξf′ξ/fξ−α≺ψβξ;β−ST10,k,b=β−STb=f∈A:1+1/bξf′ξ/fξ−1≺ψβξ,β−ST10,k,1−αcosγe−iγ=β−STγα0≤α<1;γ<π/2=f∈Am:eiγξf′ξ/fξ≺1−αcosγψβξ+αcosγ+isinγ and (see [15]) β−ST10,k,1−α=β−STα0≤α<1=f∈A:1/1−αξf′ξ/fξ−α≺ψβξ(7)
0−STmδ,k,b=Smδ,k,b=f∈Am:Rm+1/bξDδ+mk−kfξ′/Dδ+mk−kfξ−m>0,0−STmδ,k,1−α/mcosγe−iγ=Smγδ,k,α0≤α<m;γ<π/2=f∈Am:ReiγξDδ+mk−kfξ′/Dδ+mk−kfξ>αcosγ, (see [16] and [17]) 0−STmk−mk,k,b=Smγα0≤α<m;γ<π/2=f∈Am:Reiγξf′ξ/fξ>αcosγ,(see [18]) 0−STmk−mk,k,b=Smb=f∈Am:Rm+1/bξf′ξ/fξ−m>0 and S1b=Sb (see [19, 20])In order to establish our main results, we need the following lemmas.Lemma 2 [21].
Letψβξ0≤β<∞ be defined by (13). If(14)ψβξ=1+L1ξ+L2ξ2+⋯,then(15)L1=2A21−β20≤β<1,8π2β=1,π24tβ2−1R2t1+t1<β<∞,(16)L2=A2+23L10≤β<1,23L1β=1,4R2tt2+6t+1−π224tR2t1+tL11≤β<∞,where(17)A=2cos−1βπ,and t∈0,1 is chosen such that β=coshπR′t/Rt, and Rt is the Legendre’s complete elliptic integral of the first kind.Lemma 3 [22].
Lethξ=1+∑n=1∞cnξn∈P, i.e., let h be analytic in D and satisfy Rhξ>0 for ξ in D; then, the following sharp estimate holds(18)c2−vc12≤2max1,2v−1forallv∈ℂ.
The result is sharp for the functions given by(19)gξ=1+ξ21−ξ2orgξ=1+ξ1−ξ.Lemma 4 [22].
Ifhξ=1+∑n=1∞cnξn∈P, then(20)c2−νc12≤−4ν+2ifν≤0,2if0≤ν≤1,4ν−2ifν≥1,and when υ<0 or ν>1, the equality holds if and only if hξ=1+ξ/1−ξ or one of its rotations. If 0<ν<1, then the equality holds if and only if hξ=1+ξ2/1−ξ2 or one of its rotations. If ν=0, the equality holds if and only if(21)hξ=1+λ21+ξ1−ξ+1−λ21−ξ1+ξ0≤λ≤1,or one of its rotations. If ν=1, the equality holds if and only if g is the reciprocal of one of the functions such that equality holds in the case of ν=0.Also, the above upper bound is sharp, and it can be improved as follows when0<ν<1:(22)c2−νc12+νc12≤20≤ν≤12,c2−νc12+1−νc12≤212≤ν≤1.In this paper, we investigate a coefficient estimates and the familiar Fekete–Szegö type inequalities for the subclassβ−STmδ,k,b.


## 2. Main Results

We will assume throughout our discussion, unless otherwise stated, that0≤β<∞, m∈ℕ, δ>−mk, k>0, b∈ℂ∗, L1 is given by (15), L2 is given by (16), and ξ∈D.Theorem 5.
Letf∈Am be given by (1). If the inequality(23)∑n=m+1∞β+1n−m+mbδ+mkn−m,kkn−m,kan≤mb,holds, then the function f∈β−STmδ,k,b.Proof.
Suppose the inequality (23) holds. Also, let us assume(24)Hξ=1+1bξDδ+mk−kfξ′mDδ+mk−kfξ−1.
We have(25)Hξ−1=1bξDδ+mk−kfξ′mDδ+mk−kfξ−1=1mb∑n=m+1∞n−mδ+mkn−m,k/kn−m,kanξn−m1+∑n=m+1∞δ+mkn−m,k/kn−m,kanξn−m≤∑n=m+1∞n−mδ+mkn−m,k/kn−m,kanmb1−∑n=m+1∞δ+mkn−m,k/kn−m,kan.
Now consider(26)βHξ−1−RHξ−1≤β+1Hξ−1<β+1∑n=m+1∞n−mδ+mkn−m,k/kn−m,kanmb1−∑n=m+1∞δ+mkn−m,k/kn−m,kan.
The last expression is bounded by1 if (23) holds. This completes the proof of Theorem 5.☐Corollary 6.
Iffξ∈β−STmδ,k,b, then(27)an≤mbkn−m,kβ+1n−m+mbδ+mkn−m,kn≥m+1.
The result is sharp for the function(28)fξ=ξm+mbkn−m,kβ+1n−m+mbδ+mkn−m,kξnn≥m+1.Puttingm=1 in Theorem 5, we obtain the following corollary.Corollary 7.
Letf∈A be given by (1) with m=1. If the inequality(29)∑n=2∞β+1n−1+bδ+kn−1,kkn−1,kan≤b,holds, then the function f∈β−STδ,k,b.Theorem 8.
Iff∈β−STmδ,k,b, then(30)am+1≤mbL1kδ+mk,and for all n=3,4,5,⋯,(31)an+m−1≤mbL1kn−1,kn−1δ+mkn−1,k∏j=1n−21+mbL1j.Proof.
Let(32)gξ=1+1bξDδ+mk−kfξ′mDδ+mk−kfξ−1,where(33)gξ=1+∑n=1∞cnξn,is analytic function in D, and it can be written as(34)∑n=m+1∞n−mδ+mkn−m,kkn−m,kanξn=mbDδ+mk−kfξ∑n=1∞cnξn.
Comparing the coefficients ofξn+m−1 on both sides(35)n−1δ+mkn−1,kkn−1,kan+m−1=mbcn−1+δ+mk1,kk1,kcn−2am+1+⋯+δ+mkn−2,kkn−2,kc1an+m−2.
Taking absolute on both sides and then applying the coefficient estimatescn≤L1 (see [13]), we have(36)an+m−1≤mbL1kn−1,kn−1δ+mkn−1,k1+δ+mk1,kk1,kam+1+⋯+δ+mkn−2,kkn−2,kan+m−2.
We apply the mathematical induction on (36) so for n=2,(37)am+1≤mbL1k1,kδ+mk1,k=mbL1kδ+mk,and this shows that result is true for n=2. Now for n=3,(38)am+2≤mbL1k2,k2δ+mk2,k1+δ+mk1,kk1,kam+1,and using (37), we obtain(39)am+2≤mbL1k2,k2δ+mk2,k1+mbL1,which is true for n=3. Let us assume that (31) is true for n=t, that is,(40)at+m−1≤mbL1kt−1,kt−1δ+mkt−1,k∏j=1n−21+mbL1j.
Consider(41)at+m≤mbL1kt,ktδ+mkt,k1+δ+mk1,kk1,kam+1+⋯+δ+mkt−1,kkt−1,kat+m−1≤mbL1kt,ktδ+mkt,k1+mbL1+⋯+δ+mkt−1,kkt−1,kat+m−1≤mbL1kt,ktδ+mkt,k1+mbL1+mbL121+mbL1+⋯+mbL1t−11+mbL11+mbL12⋯1+mbL1t−2=mbL1kt,ktδ+mkt,k1+mbL11+mbL12⋯1+mbL1t−1=mbL1kt,ktδ+mkt,k∏j=1t−11+mbL1j.
Therefore, the result is true forn=t+1. Consequently, using mathematical induction, we proved that the result holds true for all nn≥2. This completes the proof of Theorem 8.☐Puttingm=1 in Theorem 8, we obtain the following corollary.Corollary 9.
Iff∈β−STδ,k,b, then(42)a2≤bL1kδ+k,and for all n=3,4,5,⋯,(43)an≤bL1kn−1,kn−1δ+kn−1,k∏j=1n−21+bL1j.Theorem 10.
Letf∈β−STmδ,k,b. Then., fD contains an open disk of radius(44)r=δ+mkm+1δ+mk+mbL1k.Proof.
Letw0≠0 be a complex number such that fξ≠w0 for ξ∈D. Then, f1ξ=w0fξ/w0−fξ=ξm+am+1+1/w0ξm+1+⋯.
Sincef1 is univalent, so(45)am+1+1w0≤m+1.
Now using Theorem8, we have(46)1w0≤m+1+mbL1kδ+mk,and hence(47)w0≥δ+mkm+1δ+mk+mbL1k.
This completes the proof of Theorem10.☐Puttingm=1 in Theorem 10, we obtain the following corollary.Corollary 11.
Letf∈β−STδ,k,b. Then, fD contains an open disk of radius(48)r1=δ+k2δ+k+bL1k.Theorem 12.
Letf∈β−STmδ,k,b with the form (1). Then, for a complex number μ, we have(49)am+2−μam+12≤mbL1k2δ+mkδ+mk+kmax1,L2L1+mbL11−δ+mk+kδ+mkμ.Proof.
Iff∈β−STmδ,k,b, then there exists a Schwarz function w, with w0=0 and wξ<1 such that(50)1+1bξDδ+mk−kfξ′mDδ+mk−kfξ−1=ψβwξξ∈D.
Leth∈P be a function defined by(51)hξ=1+wξ1−wξ=1+c1ξ+c2ξ2+⋯ξ∈D.
This gives(52)wξ=c12ξ+12c2−c122ξ2+⋯,(53)ψβwξ=1+12c1L1ξ+12c12L22+c2−c122L1ξ2+⋯.
Using (53) in (50), we obtain(54)am+1=mbc1L1k2δ+mk,am+2=mbk22δ+mkδ+mk+kc12L22+c2−c122L1+mbc12L122.
For any complex numberμ, we have(55)am+2−μam+12=mbk22δ+mkδ+mk+kc12L22+c2−c122L1+mbc12L122−μm2b2c12L12k24δ+mk2.
Then (55) can be written as(56)am+2−μam+12=mbL1k22δ+mkδ+mk+kc2−vc12,where(57)v=121−L2L1−mbL11−δ+mk+kδ+mkμ.
Now, taking absolute value on both sides and using Lemma3, we obtain the required result.☐Puttingm=1 in Theorem 12, we obtain the following corollary.Corollary 13.
Letf∈β−STδ,k,b with the form (1). Then, for a complex number μ, we have(58)a3−μa22≤bL1k2δ+kδ+2kmax1,L2L1+bL11−δ+2kδ+kμ.Theorem 14.
Let(59)σ1=mbL12+L2−L1δ+mkmbδ+mk+kL12,σ2=mbL12+L2+L1δ+mkmbδ+mk+kL12,σ3=mbL12+L2δ+mkmbδ+mk+kL12.
Iff given by (1) belongs to β−STmδ,k,b with b>0, then(60)am+2−μam+12≤mbL1k2δ+mkδ+mk+kL2L1+mbL11−δ+mk+kδ+mkμμ≤σ1,mbL1k2δ+mkδ+mk+kσ1≤μ≤σ2,−mbL1k2δ+mkδ+mk+kL2L1+mbL11−δ+mk+kδ+mkμμ≥σ2.
Further, ifσ1≤μ≤σ3, then(61)am+2−μam+12+δ+mkmbδ+mk+kL11−L2L1−mbL11−δ+mk+kδ+mkμam+12≤mbL1k2δ+mkδ+mk+k.
Ifσ3≤μ≤σ2, then(62)am+2−μam+12+δ+mkmbδ+mk+kL11+L2L1+mbL11−δ+mk+kδ+mkμam+12≤mbL1k2δ+mkδ+mk+k.Proof.
Applying Lemma4 to (56) and (57), respectively, we can obtain our results asserted by Theorem 14.☐Puttingm=1 in Theorem 14, we obtainCorollary 15.
Let(63)σ4=bL12+L2−L1δ+kbδ+2kL12,σ5=bL12+L2+L1δ+kbδ+2kL12,σ6=bL12+L2δ+kbδ+2kL12.
Iff given by (1) belongs to β−STδ,k,b with b>0, then(64)a3−μa22≤bL1k2δ+kδ+2kL2L1+bL11−δ+2kδ+kμμ≤σ4,bL1k2δ+kδ+2kσ4≤μ≤σ5,−bL1k2δ+kδ+2kL2L1+bL11−δ+2kδ+kμμ≥σ5.
Further, ifσ4≤μ≤σ6, then(65)a3−μa22+δ+kbδ+2kL11−L2L1−bL11−δ+2kδ+kμa22≤bL1k2δ+kδ+2k.
Ifσ6≤μ≤σ5, then(66)a3−μa22+δ+kbδ+2kL11+L2L1+bL11−δ+2kδ+kμa22≤bL1k2δ+kδ+2k.Remark 16.
For different choices of the parametersβ, m, δ, k, and b in the above theorems, we can obtain the corresponding results for each of the following subclasses β−STmγδ,k,α, β−STmδ,k,α, β−STmδ,b, β−STmγδ,α, β−STmδ,α, β−STδ,b, β−STγδ,α, β−STδ,α, β−STγδ,k,α, β−STδ,k,α, β−STmb, β−STmγα, β−STmα, β−STb, β−STγα, β−STα, Smδ,k,b, Smγδ,k,α, Smγα, Smb, and Sb which are defined in Section 1.



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