# A Study of New Class of Star-Like Functions Associated by Symmetricp,q-Calculus

Authors: Khalid Akbar; Rashid Murtaza; Adnan; Umar Khan; Ilyas Khan; Md. Fayz-Al-Asad
Journal: Journal of Mathematics
 (2021)

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## Abstract

As of late quantum calculus is broadly utilized in different parts of mathematics. Uniquely, the hypothesis of univalent functions can be newly portrayed by utilizingq-calculus. In this paper, we utilize our recently presented symmetric p,q-number m˜p,q to characterize new symmetric p,q-derivative Dp,q of analytic function f in the open unit disk U. Utilizing Dp,q, we introduce new class of analytic star-like functions and examine some fascinating results.


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## Body

## 1. Introduction

The mathematical study ofq-calculus has been a topic of great interest for researchers due to its wide applications in different fields. Some of the earlier work on the applications of q-calculus was introduced by Jackson [1]. Later on, q-calculus attained much popularity among the researchers. Recently, q-calculus has gained the attention of researchers because of its huge applications in mathematics and physics. The in-depth analysis of q-calculus was firstly discussed by Jackson [1, 2], where he defined q-integral and q-derivative in a very systematic way. Recently, authors are using these q-integral and q-derivative to define new subclasses of the class of univalent functions and obtained variety of new results. Extending the idea of q-number, which contains only one variable q, the p,q-number which contains two independent parameters p and q was independently considered by Chakrabarti and Jagannathan [3]. As q-calculus or quantum calculus originated by using q-number, similarly by using p,q-number, the p,q-calculus or postquantum calculus has been studied and discussed by several researchers (see, for example, Duran et al. [4] and references therein). Let f∈ℂ. Furthermore, f is normalized analytic, if f is single valued and differentiable for z∈U along f0=0,f′0=1 and is represented as(1)fz=z+∑m=2∞amzm.The symbolA is used for this type of functions. Let f∈A be given by (1). Then,(2)fis univalent⟺⋋1≠⋋2⟹f⋋1≠f⋋2∀⋋1,⋋2∈U.Assume thatp˜∈ℂ is analytic. Furthermore,(3)℘∈P⟺Re℘z⊂u>0along℘0=1,and is presented as(4)℘z=1+∑m=1∞cmzm.LetD⊂ℂ. If w0 is a constant in D and for all z0 in D, we have(5)w0and∀z0∈D⟹1−tw0+tz0∈D,0≤t≤1.At that point, we state thatD is a star-formed region. Geometrically, f∈A is star-like if fU is a star-shaped region. We meant by S∗ the family of these functions. Analytically, we attain(6)f∈S∗⟺zf′f∈P.Assume thatD⊂ℂ. If ζ1≠ζ2∈D, then(7)Dis convex⟺1−tζ1+tζ2∈D,0≤t≤1.Geometrically,(8)f∈Ais convex⟺fUis convex region.The family containing all the convex function is represented asℭ. Analytically f∈ℭ⟺zf′′/f′∈P,Note(9)f∈ℭ⟺zf′∈S∗.Letf be presented as in (1), and the convolution f∗g is characterized as(10)g∗fz=∑m=1∞ambmzm=f∗gz,z∈U,where(11)gz=z+b2z2+⋯=∑m=1∞bmzm.Leth1,h2 be two functions. Then, h1≺h2⟺∃, if there exist a Schwarz function ϖ analytic along ϖ0=0,ϖz<1, such that h1z=h2∘ϖz. It can be found in [5] that if h2∈S, then(12)h10=h20,h1U⊂h2U⟺h1≺h2,and for details, see [6].Kanas and Wisniowska [7] presented the conic regions Λk,k∈0,∞ by(13)Λk=w˜=μ˜+iν˜∈ℂ:μ˜>kμ˜−12+ν˜2,with 1∈Λk. Assume(14)∂Λk=w˜=μ˜+iν˜∈ℂ:μ˜2=k2μ˜−12+ν˜2.Therefore,(15)k=0⟹Λ0=μ˜>0,right half plane,k=1⟹Λ1=ν˜2<2μ˜−1,parabolic region,k∈0,1⟹Λk=μ˜>kμ˜−12+ν˜2,hyperbolic region.Identified with the regionΛk, the accompanying functions are extremal and pkU⊂Λk:(16)pkz=1+z1−z,k=0,1+2π2log1+z1−z2,k=1,1+21−k2sinh22πarccoskarctanhz,0<k<1.The familiesk−UℭB and k−SI are fascinating subfamilies of S and are characterized in [8] for all k≥0, z∈U, as(17)f belongs to k−UℭB⟺Re1+zf′zf′z>kzf″zf′z,f belongs to k−SI⟺Rezf′zfz>kzf′zfz−1.Note from [8] that(18)f∈k−UℭB⟺zf′∈k−SI.Geometrically,(19)f∈k−UℭB⟺zf′z′f′zU⊂Λk.


## 2. Postquantum Calculus

Extending the idea ofq-number, the p,q-number with two variables p and q was independently considered in [3].For0<q<p≤1, the twin basic number or p,q-number is defined for m∈ℕ as(20)mp,q=pm−qmp−q=pm−1+pm−2q+⋯+pqm−2+qm−1,which is the normal speculation of q-number with the end goal that for p⟶1−, q⟶1− we have(21)mp,q=m1,q=mq,andmp,q=m1,1=m.Furthermore, note thatmp,q=mq,p.In 1991, Chakrabarti and Jagannathan [3] characterized p,q-derivative of f∈A as(22)Dp,qfz=fpz−fqzp−qz,z≠0,f′0,z=0.Furthermore, iffz=zm, then Dp,qfz=mp,qzm−1, and for (1), we have(23)Dp,qfz=1+∑m=2∞mp,qamzm−1,z∈U.We remark here as(24)p⟶1−⟹Dp,qfz=Dqfz,(25)p⟶1−,q⟶1−⟹mp,q=m,Dp,qfz=f′z.We now extend the idea in [9] and define symmetric p,q-number as follows:(26)m˜p,q=pm−q−mp−q−1,0<q≤p<1.Analogous top,q-derivative defined by (22), the symmetric p,q-derivative is characterized as(27)Dp,qfz=fpz−fq−1zp−q−1z,z≠0,f′0,z=0.If(28)fz=zm⟹Dp,qfz=m˜p,qzm−1,∀z∈U.In this manner for (1), we acquire(29)Dp,qfz=1+∑m=2∞m˜p,qamzm−1,z∈U.Equivalently, by using the same technique as in [9], it can be seen that(30)Dp,qfz+gz=Dp,qfz+Dp,qgz,Dp,qfzgz=gpzDp,qfz+fq−1zDp,qgz=gq−1zDp,qfz+fpzDp,qgz,Dp,qfzgz=gq−1zDp,qfz−fq−1zDp,qgzgpzgq−1z=gpzDp,qfz−fpzDp,qgzgpzgq−1z.Assume thatT denotes the subfamily of A with negative coefficients such as(31)fz=z−∑m=2∞amzm,z∈U.For further developments about classT and its subclasses, one can refer to a wide range of extraordinary articles written by famous mathematicians (see [10–13] and references therein).Presently, utilizing the symmetricp,q-derivative characterized by (27) of f, we present another subclass of A as follows.Definition 1.
Let0≤k≤∞,0≤η<1and0<q≤p<1. Assume thatf∈Acharacterized by (1). Furthermore,(32)f∈k−STp,q˜η⟺kzDp,qfzfz−1+η<RezDp,qfzfz,z∈U.
Geometrically,(33)zDp,qfzfzU⊂Λk,η⟺f∈k−STp,q˜η,where Λk,η is given by(34)Λk,η=1−η+ηΛk,and Λk is defined as in (13).
Utilizing the functionspk,η defined in [14], we have(35)zDp,qfzfz≺pk,η⟺f∈k−STp,q˜η,z∈U.Remark 1.
Likewise, we setk−SI˜p,q−η=k−SI˜p,qη∩T.
Note that on the off chance thatp⟶1−,q⟶1−, at that point(36)k−SI˜p,qη=k−SIη.


## 3. Main Results

Now in this segment, we will demonstrate our principle results. It is worthy to mention here that our main results are extension of results studied by Kanas et al. [9]. We utilize symmetric p,q-derivative operator to obtain these results.The accompanying lemmas are useful to prove main result.Lemma 1.
Assume thatr∈0,1andβ>0. Furthermore, fandgare analytic inUalongf≺g. Then,(37)∫02πfreiθβdθ≤∫02πgreiθβdθ,z=reiθ,and for details, see [15].Lemma 2.
For the sequencebmj=1∞and∀z∈U,(38)Re1+2∑m=1∞bmzm>0⟺bmm=1∞,is a subordinating factor sequence (see [15]).Now we will extend the existing results of [13] with the help of symmetric p,q-derivative.Theorem 1.
Assume that0<q≤p<1and letf∈Sbe defined by (1). If the inequality(39)∑m=2∞m˜p,qk+1−k+ηam≤1−η,holds true, then f∈k−STp,q˜η.Proof.
To prove (39), it is adequate to prove that(40)kzDp,qfzfz−1−RezDp,qfzfz−1<1−η.
Consider(41)kzDp,qfzfz−1−RezDp,qfzfz−1≤kzDp,qfzfz−1+zDp,qfzfz−1=k+1zDp,qfzfz−1.
Now using (1) and (29), we obtain(42)kzDp,qfzfz−1−RezDp,qfzfz−1≤∑m=2∞m˜p,qamzm−∑m=2∞azmz+∑m=2∞amzm≤k+1∑m=2∞m˜p,q−1amzm1+∑m=2∞amzm≤k+1∑m=2∞m˜p,q−1am1−∑m=2∞am≤1−η.
This can be composed as(43)k+1∑m=2∞m˜p,q−1am≤1−η1−∑m=2∞am.
From (43), we obtain (39) and the proof is completed.Remark 2.
In Theorem1, by putting p=q, we have the result presented in [9].Corollary 1.
Assume that0<q<1and letf∈Sbe presented as in (1). If(44)∑m=2∞m˜qk+1−k+ηam≤1−η,holds true for 0≤k<∞, 0≤η<1, then f∈k−STq˜η.Remark 3.
It can be seen that the quantity(45)m˜p,qk+1−k+η=pm−q−mp−q−1k+1−k+η>0,for all 0≤k<∞,0≤η<1and0<q≤p<1, unless otherwise mentioned.
Now using (39), we can find some special members of f∈k−STp,q˜η. One of them is the following.Corollary 2.
Let for0≤k<∞,0≤η<1and0<q≤p<1.If, forfz=z+amzm,the following inequality:(46)am≤1−ηm˜p,qk+1−k+η,m≥2,holds, then f∈k−STp,q˜η. Especially(47)fz=z+1−ηqpq+1k+1−k+ηqz2∈k−STp,q˜η,z∈U.Forp=q, we obtain the following known results (see [9]).Corollary 3.
Let0≤k<∞, 0≤η<1 and 0<q<1. If, forfz=z+amzm, the following inequality:(48)am≤1−ηm˜qk+1−k+η,m≥2,holds, then f∈k−STq˜η. Especially(49)fz=z+1−ηqq2+1k+1−k+ηqz2∈k−STq˜η,z∈U.Also, fork−STp,q˜−η, we obtain the following.Corollary 4.
Let0≤k<∞,0≤η<1and0<q≤p<1.An essential and adequate condition offz=z−a2z2−⋅⋅⋅,am≥0belongs tok−STp,q˜−η,such as(50)∑m=2∞m˜p,qk+1−k+ηam≤1−η.The accompanying function yields quality.(51)fz=z−1−ηm˜p,q,k+1−k+ηzm.Proof.
Proof immediately follows by using Theorem1.
On the off chance that we takep=q in Corollary 4, we have the accompanying corollary (see [9]).Corollary 5.
Let0≤k<∞,0≤η<1and0<q<1.A necessary and sufficient condition forfof the formfz=z−a2z2−⋅⋅⋅,am≥0, to be in the classk−STq˜−ηis that(52)∑m=2∞m˜qk+1−k+ηam≤1−η.The result is sharp and accompanying function yields quality.(53)fz=z−1−ηm˜qk+1−k+ηzm.Theorem 2.
Let0≤k<∞, 0≤η<1 and 0<q≤p<1. Assumefz=z−a2z2−⋅⋅⋅,am≥0belongs tok−STp,q˜−η.Furthermore, |z|=r<1yields(54)r−q1−ηpq+1k+1−qk+ηr2≤fz≤r+q1−ηpq+1k+1−qk+ηr2.Accompanying function yields quality for (54).(55)fz=z+q1−ηpq+1k+1−qk+ηz2,z∈U.Proof.
Letf∈k−STp,q˜−η. By using Theorem 1, we have(56)2˜p,qk+1−k+η∑m=2∞am≤∑m=2∞m˜p,qk+1−k+η≤1−η.
This means that(57)∑m=2∞am≤1−η2˜p,qk+1−k+η.
Furthermore if|z|=r<1,(58)fz≤z+∑m=2∞amzm≤r+q1−ηpq+11−ηk+1−qk+ηr2,and(59)fz≥z−∑m=2∞amzm≥r−q1−ηpq+11−ηk+1−qk+ηr2.
Combining (58) and (59), we obtain (54).Theorem 3.
Suppose thatf∈k−SI˜p,q−η. Moreover,(60)f2z=z−1−η2˜p,qk+1−k+ηz2.If(61)∀z=reiθ,0<r<1⟹∫02πfzηdθ≤∫02πf2zηdθ.Proof.
Note that to prove (61) for fz=z−∑m=2∞amzm, it is equivalent to prove(62)∫02π1−∑m=2∞amzm−1ηdθ≤∫02π1−1−η2˜p,qk+1−k+ηηdθ.
By Lemma1, it is sufficient to show that(63)1−∑m=2∞amzm−1≺1−1−η2˜p,qk+1−k+η.
Set(64)1−∑m=2∞amzm−1=1−1−η2˜p,qk+1−k+ηwz.
This means that(65)wz=∑m=2∞2˜p,qk+1−k+η1−ηamzm−1.
Now using (39), we get wz≤|z|<1 and this finishes the proof.Theorem 4.
Letf∈k−SI˜p,qηalong∈ℭ. Then,(66)2˜p,qk+1−k+η21−η+2˜p,qk+1−k+ηf∗gz≺gz,that is,(67)Refz>−1−η+2˜p,qk+1−k+η2˜p,qk+1−k+η>0,z∈U.The constant2˜p,qk+1−k+η/21−η+2˜p,qk+1−k+η cannot be increased.Proof.
Letf∈k−SI˜p,qηandg∈ℭ.
Then, consider(68)2˜p,qk+1−k+η21−η+2˜p,qk+1−k+ηf∗gz=2˜p,qk+1−k+η21−η+2˜p,qk+1−k+ηz+∑m=2∞ambmzm.Therefore, by using the definition of subordinating factor sequence, (66) holds true, if(69)2˜p,qk+1−k+η21−η+2˜p,qk+1−k+ηamn=1∞,is a subordinating factor sequence with a1=1. Equivalently, by Lemma 2, we have(70)Re1+∑m=1∞2˜p,qk+1−k+η1−η+2˜p,qk+1−k+ηamzm>0,z∈U.Now forz=r<1, we can write(71)Re1+2˜p,qk+1−k+η1−η+2˜p,qk+1−k+η∑m=1∞amzm=Re1+2˜p,qk+1−k+η1−η+2˜p,qk+1−k+ηz+2˜p,qk+1−k+η∑m=2∞amzm1−η+2˜p,qk+1−k+η>1−2˜p,qk+1−k+η1−η+2˜p,qk+1−k+ηr−∑m=2∞m˜p,qk+1−k+ηamrj1−η+2˜p,qk+1−k+η.It can be seen that the function defined by(72)ψp,qn,k,η=m˜p,qk+1−k+η,is increasing for n≥2; furthermore, by utilizing affirmation (39) of Theorem 1, we have(73)Re1+2˜p,qk+1−k+η1−η+2˜p,qk+1−k+η∑m=1∞amzm>1−2˜p,qk+1−k+η1−η+2˜p,qk+1−k+ηr−1−η1−η+2˜p,qk+1−k+ηr>0.This equivalently proves the subordination condition given by (66). The inequality given by (67) can be obtained from (66) by taking gz=z/1−z∈ℭ. Now from (47), we obtained the following function:(74)Fz=z+1−η2˜p,qk+1−k+η∈k−SI˜p,qη.Function (66) becomes(75)2˜p,qk+1−k+η21−η+2˜p,qk+1−k+ηFz≺z1−z.It is easily verified that(76)minRe2˜p,qk+1−k+η21−η+2˜p,qk+1−k+ηFz=−12,z∈U.This implies that the constant2˜p,qk+1−k+η/21−η+2˜p,qk+1−k+η is possible.Theorem 5.
Let0≤k<∞,0≤η<1and0<q≤p<1.If we set(77)f1z=zfmz=z−1−ηm˜p,qk+1−k+ηzm,then(78)f∈k−SI˜p,q−η⟺fz=∑m=1∞δmfmz,∑m=1∞δm=1.Proof.
Suppose that(79)fz=∑m=1∞δmfmz=δ1f1z+∑m=2∞δmfmz.
Utilizing (77), we get(80)fz=δ1f1z+∑m=2∞δmz−1−ηm˜p,qk+1−k+ηzm=∑m=1∞δmz−∑m=2∞δm1−ηm˜p,qk+1−k+ηzm=z−∑m=2∞δm1−ηm˜p,qk+1−k+ηzm.
Since∑m=1∞δm=1, this means that ∑m=2∞δm=∑m=1∞δm−δ1=1−δ1≤1, and using the assertion (51) of Corollary 4, we obtain f∈k−SI˜p,q−η.
Conversely, suppose thatf∈k−SI˜p,q−η. Making use of (39), we may set(81)δm=m˜p,qk+1−k+η1−ηam,∑m=1∞δm=1.
Then,(82)fz=z−∑m=2∞amzm=z−∑m=2∞δm1−ηm˜p,qk+1−k+ηzm=z−∑m=2∞δmz−fmz=1−∑m=2∞δmz+∑m=2∞δmfmz=δ1z+∑m=2∞δmfmz,δ1=1−∑m=2∞δm=∑m=1∞δmfmz,z∈U,which is required.



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