# Geometric Properties of Cesàro Averaging Operators

Authors: Priyanka Sangal; A. Swaminathan
Journal: Journal of Complex Analysis
 (2017)

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## Abstract

Using positivity of trigonometric cosine and sine sums whose coefficients are generalization of Vietoris numbers, we find the conditions on coefficient{ak} to characterize the geometric properties of the corresponding analytic function f(z)=z+∑k=2∞akzk in the unit disc D. As an application, we also find geometric properties of generalized Cesàro-type polynomials.


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## Body

## 1. Introduction

Inequalities involving trigonometric sums arise naturally in various problems of pure and applied mathematics. Inequalities that assure nonnegativity or boundedness of partial sums of trigonometric series are of particular interest and applications in various fields. For example, the positivity of trigonometric polynomials are studied in geometric function theory by Gluchoff and Hartmann [1] and Ruscheweyh and Salinas [2]. For a detailed application in signal processing, we refer to the monograph of Dumitrescu [3]. For other applications in this direction, we refer to Dimitrov and Merlo [4], Fernández-Durán [5], and Gasper [6]. The positive trigonometric polynomials played an important role in the proof of Bieberbach conjecture; see [7]. For the applications of positive trigonometric polynomials in Fourier series, approximation theory, function theory, and number theory, we refer to the work of Dimitrov [8] and references therein. For the study of extremal problems, we refer to the dissertation of Révész [9] wherein several applications are outlined.The problem of finding the behaviour of the coefficients to validate the positivity of trigonometric sum has been dealt by many researchers. Among them, the contributions of Vietoris [10] followed by Koumandos [11] are of interest to the present investigation. Precisely, Vietoris [10] gave sufficient conditions on the coefficient of a general class of sine and cosine sums that ensure their positivity in (0,π). For further details in this direction one can refer to [11–13] and the references therein. An account of recent results available in this direction is given in [13] and one of the main results in [13] is as follows.Theorem 1 (see [13]).
Suppose thatα≥0, β≥0, and λ,μ≥0 such that λ+μ≥1 then for b0=2, b1=1, and bk=1/(k+α)λ(k+β)μ, k≥2, we have (1)b02+∑k=1nbkcos⁡kθ>0,∑k=1nbksin⁡kθ>0,for 0<θ<π and n∈N.Using summation by parts, the following corollary of Theorem1 can be obtained.Corollary 2.
Forα≥0, β≥0 and λ≥0, μ≥0 such that λ+μ≥1. If there exists a sequence {ak} of positive real numbers, such that (2)k+1+αλk+1+βμak+1≤k+αλk+βμak≤⋯≤2+αλ2+βμa2≤a1≤a02,then, for n∈N, the following inequalities hold: (3)a02+∑k=1nakcos⁡kθ>0,∑k=1naksin⁡kθ>0,where0<θ<π.The main purpose of this note is to use Corollary2 to find certain geometric properties of analytic functions, in particular univalent functions. Let A0 be the subclass of the class of analytic functions f∈A with normalized conditions f(0)=0, f′(0)=1 in the unit disc D={z∈D,|z|<1}. The subclasses of A0 consisting of univalent function are denoted by S. Several subclasses of univalent functions play a prominent role in the theory of univalent functions. For 0≤γ<1, let S∗(γ) be the family of functions f starlike of order γ; that is, if f∈A0 satisfies the analytic characterization, (4)f∈S∗γ⟺Rezf′zfz>γ,forz∈D.For 0≤γ<1, let C(γ) be the family of functions f convex of order γ; that is, if f∈A0 satisfies the analytic characterization, (5)f∈Cγ⟺Re1+zf′′zf′z>γ,forz∈D.These two classes are related by the Alexander transform, f∈C(γ)⇔zf′∈S∗(γ). The usual classes of starlike functions (with respect to origin) and convex functions are denoted, respectively, by S∗(0)≡S∗ and C(0)≡C. An analytic function f is said to be close-to-convex of order γ, (0≤γ<1) with respect to a fixed starlike function g if and only if it satisfies the analytic characterization: (6)Reeiηzf′zgz-μ>0,z∈D,η∈-π2,π,2,g∈S∗.The family of all close-to-convex function of order μ with respect to g∈S∗ is denoted by Kg(μ). Further, for 0≤μ<1, each f∈Kg(μ) is also univalent in D. The proper inclusion between these classes is given by (7)C⫋S∗⫋K⫋S.Another important subclass is the class of typically real functions. A function f∈A0 is typically real if Im(z)Im(f(z))≥0 where z∈D. Its class is denoted by T. For several interesting geometric properties of these classes, one can refer to the standard monographs [14–16] on univalent functions.Remark 3.
The functions(8)z,z1±z,z1±z2,z1±z2,z1±z+z2are the only nine starlike univalent functions having integer coefficients in D. It will be interesting to find f to be close-to-convex when the corresponding starlike function g takes one of the above forms.If we takeη=0 and g(z)=z/(1-z)2 then Re((1-z)2f′(z))>0 which implies zf′(z) is typically real function. A function f∈A0 is said to be typically real if Imf(z)Im(z)>0 whenever Im(z)≠0, z∈D. The function kγ(z)≔z/(1-z)2-2γ is the extremal function for the class of starlike function of order γ. Note that k0(z) is the well-known Koebe function and the function k1/2(z)=z/(1-z) is the extremal function for the class C. A function f(z) is said to be prestarlike of order γ, 0≤γ<1, if kγ(z)∗f(z)=z/(1-z)2∗f(z)∈S∗(γ) where “∗” is the convolution operator or Hadamard product. This class was introduced by Ruscheweyh [17]. For more details of this class see [18]. Here the Hadamard product or convolution is defined as follows: let f(z)=∑k=0∞‍akzk and g(z)=∑k=0∞‍bkzk, z∈D. Then, (9)f∗gz=∑k=0∞akbkzk,z∈D.Among all applications of positivity of trigonometric polynomials, the geometric properties of the subclasses of analytic functions are considered in this note. In this direction, Ruscheweyh [19] obtained some coefficient conditions for the class of starlike functions using the classical result of Vietoris [10]. So it would be interesting to find the geometric properties of function f(z) in which Corollary 2 plays a vital role.


## 2. Geometric Properties of an Analytic Function

In this section, we provide conditions on the Taylor coefficients of an analytic functionf to guarantee the admissibility of f in subclasses of S, using Corollary 2. The next lemma which is the generalization of [19, Lemma 2] is the crucial ingredient in the proof of the following theorem.Lemma 4 (see [20, Theorem 3.1]).
Let0≤γ<1 and f∈A be such that f′(z) and f′(z)-γf(z)/z are typically real in D. Further if Ref′(z)>0 and Re(f′(z)-γf(z)/z)>0, then f∈S∗(γ).Theorem 5.
Letα≥0, β≥0, λ≥0, μ≥0 such that λ+μ≥1; let {ak}k=1∞ be any sequence of positive real numbers such that a1=1. Let {ak} satisfy the following conditions: (1)
(2-γ)a2≤(1-γ)a1.(2)
(3-γ)a3≤1/(2+α)λ(2+β)μ(2-γ)a2.(3)
(k+2-γ)ak+2≤1+1/k+α-λ1+1/k+β-μ(k+1-γ)ak+1, ∀k≥2.
Then, for0≤γ<1, fn(z)=z+∑k=1n‍akzk and f(z)=limn→∞fn(z)=z+∑k=2∞‍akzk are starlike of order γ.Proof.
Letfn(z)=z+∑k=2n‍akzk, z∈D be the partial sum of f. Then fn′(z)=1+∑k=1n-1‍(k+1)ak+1zk.
Define(10)gnz≔fn′z-γfnzz=b02+∑k=1n-1bkzk,z∈D,where b0=2(1-γ) and bk=(k+1-γ)ak+1, ∀k≥1. Consider, (11)b02-b1=1-γa1-2-γa2≥0,b12+αλ2+βμ-b2=12+αλ2+βμ2-γa2-3-γa3≥0.Now, for k≥2, (12)k+αλk+βμbk-k+1+αλk+1+βμbk+1=k+1+αλk+1+βμ×1+1k+α-λ1+1k+β-μk+1-γak+1-k+2-γak+2≥0.So by the given hypothesis, {bk} satisfy the conditions of Corollary 2 which implies Regn(z)>0 and Imgn(z)>0 if Im(z)>0. By reflection principle Imgn(z)<0 if Im(z)<0. So gn(z) is typically real function. In order to prove the theorem it is remaining to show that Refn′(z)>0 and fn′(z) is typically real. In this case bk=(k+1)ak+1 and b0=2. So such bk also satisfy given hypothesis because (k+1-γ)/(k+2-γ)<(k+1)/(k+2), for all k≥0. So Refn′(z)>0 and again using reflection principle we get that fn′(z) is typically real in D.
Applying Lemma4, we get that fn(z)∈S∗(γ). Since limn→∞fn(z)=f(z) and the family of starlike functions is normal [21, p. 217], we get f(z)=limn→∞fn(z) is also starlike of order γ.Remark 6.
Ifγ=0 in Theorem 5, then we get Re(fn′(z))>0 which implies fn(z) is close-to-convex with respect to z and fn′(z) is typically real also and with (13)Re1-zfn′z=Re1-zRefn′z+ImzImfn′z>0this yields fn(z) that is close-to-convex with respect to starlike function z/(1-z).Example 7.
Consider the sequence{ak} as a1=1, a2=1/2, and ak=1/k2 for k≥3; then by Theorem 5, the function (14)fz=z+z22+∑k=3∞zkk2,z∈Dis starlike univalent. But [22, Theorem 2.1] fails to include this function. Hence Theorem 5 is better than [22, Theorem 2.1] in the sense that it is likely to include more cases.By proving thatzfn′(z) is typically real function in the similar fashion, we obtain the next result.Theorem 8.
Letα≥0, β≥0 and λ≥0, μ≥0 such that λ+μ≥1; let {ak}k=1∞ be any sequence of positive real numbers such that a1=1, if {ak} satisfy the following conditions: (15)k+1+αλk+1+βμk+1ak+1≤k+αλk+βμkak≤⋯≤2+αλ2+βμ2a2≤1,fork≥2.Then fn(z)=z+∑k=2n‍akzk and f(z)=z+∑k=2∞‍akzk are close-to-convex with respect to starlike function z/(1-z2).Note that Theorem8 provides close-to-convexity of f with respect to the function z/(1-z2). Results for the close-to-convexity of f with respect to other four starlike functions given in Remark 3 are of considerable interest, and the authors have considered some of these cases separately elsewhere. The next result provides the coefficient conditions for f to be in the class of prestarlike functions of order γ, 0≤γ<1.Theorem 9.
Letα≥0, β≥0, λ≥0, μ≥0 such that λ+μ≥1; let {ak}k=1∞ be any sequence of positive real numbers such that a1=1. Let {ak} satisfy the following conditions: (1)
(2+α)λ(2+β)μ(3-γ)(3-2γ)a3≤2(2-γ)a2≤a1.(2)
(k+1+α)λ(k+1+β)μ(k+2-γ)(k+2-2γ)ak+2≤(k+α)λ(k+β)μ(k+1-γ)(k+1)ak+1, ∀k≥2.
Then for0≤γ<1, fn(z)=z+∑k=2n‍akzk is prestarlike of order γ. Moreover, f(z)=z+∑k=2∞‍akzk is prestarlike of order γ.Proof.
Letgn(z)≔fn(z)∗z/(1-z)2-2γ, z∈D, 0≤γ<1. To prove required theorem, it is sufficient to prove that gn(z)∈S∗(γ):(16)gnz=fnz∗z1-z2-2γ=z+∑k=2n2-2γk-1k-1!akzk,z∈D.We prove that gn(z) satisfy the conditions of Lemma 4. For this, define (17)hnz≔gn′z-γgnzz=b02+∑k=1n-1bkzk,z∈D,where b0=2(1-γ) and bk=(k+1-γ)(2-2γ)k/k!ak+1 for k≥1. Using simple calculations, along with the hypothesis, {bk} satisfy the conditions of Corollary 2. Continuing the same argument as earlier, we get the desired result.Remark 10.
Note thatR(1/2)=S∗(1/2). It can be easily verified that all the conditions of Theorem 9 for R(1/2) coincide with the conditions of Theorem 5 for S∗(1/2).Forγ=0, R∗(0)≡C and the following result is immediate.Corollary 11.
Forα≥0, β≥0, λ≥0, μ≥0 such that λ+μ≥1, let {ak}k=1∞ be any sequence of positive real numbers such that a1=1. Let {ak} satisfy the following condition: (18)k+1+αλk+1+βμk+22ak+2≤k+αλk+βμk+12ak+1≤⋯≤2+αλ2+βμ9a3≤4a2≤a1.Then fn(z)=z+∑k=2n‍akzk is convex function. In particular f(z)=z+∑k=2∞‍akzk is convex univalent.Example 12.
Letf(z)=z+z2/4+∑k=3∞‍zk/(k-1+α)λ(k-1+β)μk2 is convex univalent.
In particular ifα=β=1 and λ=μ=1/2, we get that z+z2/4+∑k=3∞‍zk/k3 is convex.


## 3. Application to Cesàro Mean of Type(b-1,c)

Thenth Cesàro mean of type (b-1,c) of f(z)∈A0 is given by (19)snb-1,cz,f≔z+∑k=2nBn-kBn-1akzk=snb-1,cz∗fz,n∈N,where b and c are real numbers such that b+1>c>0 and B0=1 and Bk=(1+b-c)/b(b)k/(c)k for k≥1. Here by (α)k, k≥N, which is the well-known Pochhammer symbol, we mean the following: (20)αk=αα+1k-1withα0=1.For b=1+δ and c=1, it follows that (21)snδ,1f,z=snδf,z=z+∑k=2n1+δn-kn-k!n-1!1+δn-1akzk,which is the Cesàro mean of order δ for δ>-1. Since (19) is one type of generalization of the well-known Cesàro mean [23], we call these Cesàro means of type (b-1;c) as generalized Cesàro operators. The coefficients given in (19) were considered in [13] while finding positivity of trigonometric polynomials. Using (19), generalized Cesàro averaging operators were studied in [24] which are generalization of the Cesàro operator given by Stempak [25]. The geometric properties of snδ(z) are well-known. For details, see [23, 26, 27]. Lewis [28] proved that snδ(z) is close-to-convex and hence univalent for δ≥1. Ruscheweyh [23] proved that it is prestarlike of order (3-δ)/2. Hence it would be interesting to see if the geometric properties of snδ(z) can be extended to sn(b-1,c)(f,z). Such investigations are possible by various known methods in geometric function theory. In particular, the positivity techniques used in Koumandos [11] or Mondal and Swaminathan [20] can be applied to sn(b-1,c)(z) as well. However, in view of Example 7, we are interested in using the results available in Section 2 to obtain the geometric properties of sn(b-1,c)(z).Theorem 13.
Let{ak} be any sequence of positive real numbers such that a1=1 and (b+n-2)a1≥2(c+n-2)a2. Let b≥c>0, 0≤α≤6/λ+4, 0≤β≤6/μ+4, and λ,μ≥0 such that λ+μ≥1 and 1≤λ+μ<2 and satisfy the following conditions: (i)
(2-αλ)(2-βμ)(b+n-3)a2≥2λ+μ+1(c+n-3)3a3.(ii)
(k-1+α-λ)(k-1+β-μ)(b+n-k-1)kak≥(k-1+α)(k-1+β)(c+n-k-1)(k+1)ak+1 for 3≤k≤n-3.(iii)
(n-2+α-λ)(n-2+β-μ)(1+b-c)(n-1)an-1≥(n-2+α)(n-2+β)cnan.
Thensn(b-1,c)(f,z) is close-to-convex with respect to z and z/(1-z) where f(z)=z+∑k=2∞‍akzk. Further for the same condition sn(b-1,c)(f,z) is starlike univalent.Proof.
Letsn(b-1,c)(f,z)=z+∑k=2n‍Bn-k/Bn-1akzk. Then, (22)snb-1,cf,z′=1+∑k=1n-1Bn-k-1Bn-1k+1ak+1zk.For 0≤r<1 and 0≤θ≤2π, (23)Resnb-1,cf,z′=b02+∑k=1n-1bkrkcos⁡kθ,Imsnb-1,cf,z′=∑k=1n-1bkrksin⁡kθ,where b0=2 and bk=Bn-k-1/Bn-1(k+1)ak+1 for k≥1. Hence bk and bk+1 can be related as follows: (24)bk+1=c+n-k-2k+2ak+2b+n-k-2k+1ak+1bk,for1≤k≤n-3,bn-1=c1+b-cnann-1an-1bn-2.For the sequence {bk}, our aim is to prove that b0/2+∑k=1n-1‍bkrkcos⁡kθ>0 and ∑k=1n-1‍bkrksin⁡kθ>0. Note that (25)b02-b1=1b+n-2b+n-2a1-2c+n-2a2≥0.For a given α and β, we can easily get (26)12+αλ2+βμ=12λ+μ1+α2-λ1+β2-μ=12λ+μ1-αλ2+λλ+12!α222-λλ+1λ+23!α323+λλ+1λ+2λ+34!α424-⋯×1-βμ2+μμ+12!β222-μμ+1μ+23!β323+μμ+1μ+2μ+34!β424-⋯=12λ+μ+22-αλ+λλ+1α2221-λ+26α+λλ+1λ+2λ+33·26α41-λ+410α+⋯×2-βμ+μμ+1β2221-μ+26β+μμ+1μ+2μ+33·26β41-μ+410β+⋯≥2-αλ2-βμ2λ+μ+2if0≤α≤6λ+2,0≤β≤6μ+2.Hence we see that (27)b12+αλ2+βμ-b2≥2-αλ2-βμb12λ+μ+2-c+n-33a3b+n-32a2b1≥0.For the other condition (k+α)λ(k+β)μbk≥(k+1+α)λ(k+1+β)μbk+1 to be satisfied, first we find (28)1+1k+α-λ1+1k+β-μ=1-λk+α+λλ+12!k+α21-2+λ3k+α+λλ+1λ+2λ+34!k+α41-λ+45k+α+⋯×1-μk+β+μμ+12!k+β21-2+μ3k+β+μμ+1μ+2μ+34!k+β41-μ+45k+β+⋯≥1-λk+α·1-μk+β,if2+λ3k+α≤1,2+μ3k+β≤1fork≥2.Clearly, (29)1+1k+α-λ1+1k+β-μbk-bk+1≥1-λk+α1-μk+βbk-c+n-k-2k+2ak+2b+n-k-2k+1ak+1bk,for2≤k≤n-3=bk-11-λk-1+α1-μk-1+β-c+n-k-1k+1ak+1b+n-k-1kak≥0for3≤k≤n-2.For k=n-2, consider (30)11+1/n-2+αλ1+1/n-2+βμbn-2-bn-1≥1-λn-2+α1-μn-2+β1+b-ccn-1an-1-nan≥0.We proved that b0/2+∑k=1n-1‍bkcos⁡kθ>0 and ∑k=1n-1‍bksin⁡kθ>0 for 0<θ<π. By the minimum principle for harmonic functions, b0/2+∑k=1n-1‍bkrkcos⁡kθ>0, 0≤r<1 and 0<θ<π and ∑k=1n-1‍bkrksin⁡kθ>0 for 0<θ<π and 0≤r<1. Using reflection principle, ∑k=1n-1‍bkrksin⁡kθ<0 for π<θ<2π and 0≤r<1. Note that sn(b-1,c)(f,z) is close-to-convex with respect to z if Resn(b-1,c)(f,z)>0 and sn(b-1,c)(f,z) is close-to-convex with respect to z/(1-z) if Re[1-zsnb-1,cf,z′]>0. Now (31)Re1-zsnb-1,cf,z′=Re1-zResnb-1,cf,z′-Im1-zImsnb-1,cf,z′=Re1-zResnb-1,cf,z′+ImzImsnb-1,cf,z′>0.Forb=1+δ, c=1, Theorem 13 leads to the following example.Example 14.
Letλ≥0, μ≥0 such that 1≤λ+μ<2, 0≤α≤6/(λ+4), and 0≤β≤6/(β+4); then (32)δ≥maxn≥10,n-22λ+μ+22-αλ2-βμ-1,n-32λ+μ+αμ+βλ+λμ2+α-λ2+β-μ.Then snδ(-log⁡(1-z),z) is close-to-convex with respect to z and z/(1-z). Further for the same condition it is also starlike univalent.Remark 15.
If we takeα=β=1 and λ=μ=1/2 then, for 1≤n≤3, snδ(-log⁡(1-z),z) is close-to-convex with respect to z and z/(1-z) for δ≥δ′ where 0<δ′<3. This conclusion cannot be obtained from [20, Corollary 4.2].Theorem 16.
Let{ak} be a sequence of positive real numbers with a1=1 and satisfy the hypothesis of Theorem 13. Then sn(b-1,c)(f,z)∈R(γ), γ≥0 where (33)γ≤1-c+n-2b+n-22a2,R(γ)={f∈A:Ref′(z)>γ} and f(z)=z+∑k=2∞‍akzk, z∈D.Proof.
Letsn(b-1,c)(f,z)=z+∑k=2n‍Bn-k/Bn-1akzk where B0=1 and Bk=(b)k/(c)k(1+b-c)/b, k≥1: (34)snb-1,cf,z′=1+∑k=1n-1Bn-k-1Bn-1k+1ak+1zk.We consider (35)snb-1,cf,z′-γ1-γ=b02+∑k=1n-1bkzk,where b0=2 and bk=Bn-k-1/Bn-1·(k+1)ak+1/(1-γ) for k≥1. Then bk and bk+1 are related by (36)bk+1=c+n-k-2k+2ak+2b+n-k-2k+1ak+1bkfor1≤k≤n-3,and, for k=n-2, (37)bn-1=1+b-ccn-1an-1nanbn-2.Using hypothesis, we can easily get (38)b02-b1=1-c+n-2b+n-22a21-γ≥0.The relation between the coefficients bk and bk+1 is the same as in the Theorem 13. So such bk also satisfy the conditions of Theorem 13 and from Corollary 2 we have the required result that (39)b02+∑k=1n-1bkcos⁡kθ>0for0<θ<π.From the minimum principle for harmonic functions for 0≤r<1 and 0<θ<2π we have (40)Resnb-1,cf,z′-γ1-γ=b02+∑k=1n-1bkrkcos⁡kθ>0.So, sn(b-1,c)(f,z)∈R(γ).It can be clearly seen that, forγ=0, Theorem 16 coincides with Theorem 13 for the case g(z)=z.Theorem 17.
Let{ak}k=1∞ be a sequence of positive real numbers such that a1=1. If for λ≥0, μ≥0 such that 1≤λ+μ<2 and 0≤α≤6/(λ+4), 0≤β≤6/(μ+4), ak satisfy the following conditions: (1)
(3-2λ-2μ)(b+n-2)a1≥(5-2λ-2μ)(c+n-2)a2.(2)
(2-αλ)(2-βμ)(5-2λ-2μ)(b+n-3)a2≥2λ+μ+2(7-2λ-2μ)(c+n-3)a3.(3)
(2k+1-2λ-2μ)(k-1+α-λ)(k-1+β-μ)(b+n-k-1)ak≥(2k+3-2λ-2μ)(k-1+α)(k-1+β)(c+n-k-1)ak+1 for 3≤k≤n-2.(4)
(n-2+α-λ)(n-2+β-μ)(2n+1-2λ-2μ)(1+b-c)an-1≥(n-2+α)(n-2+β)(2n+3-2λ-2μ)can.
Then,sn(b-1,c)(f,z)∈S∗(λ+μ-1/2), where f(z)=z+∑k=2∞‍akzk, z∈D.Proof.
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k, where b1=1 and bk=Bn-k/Bn-1ak for k≥2. Then, (41)bk+1=c+n-k-1b+n-k-1ak+1akbk,for2≤k≤n-2,and for k=n-1, bn=c/(1+b-c)an/an-1bn-1. It is enough to prove that {bk} satisfy the conditions of Theorem 5. For the sake of convenience we substitute γ=λ+μ-1/2. By a simple calculation we can get that (1-γ)b1-(2-γ)b2≥0. Now (42)12+αλ2+βμ2-γb2-3-γb3≥2-αλ2-βμ2λ+μ+25-2λ-2μb2-7-2λ-2μc+n-3b+n-3a3a2b2≥0.Now, for 2≤k≤n-3, (43)1+1k+α-λ1+1k+β-μk+1-γbk+1-k+2-γbk+2≥1-λk+α1-μk+β2k+3-2λ-2μbk+1-2k+5-2λ-2μc+n-k-2b+n-k-2ak+2ak+1bk+1≥0.And, for k=n-2, using the hypothesis, we obtain, (44)1+1n-2+α-λ1+1n-2+β-μ2n+1-2λ-2μbn-1-2n+3-2λ-2μbn≥0.From Theorem 5 the desired result follows.Theorem 18.
Letb≥c>0, 0≤α≤6/(λ+4), 0≤β≤6/(μ+4) and λ,μ≥0 such that 1≤λ+μ<2 and satisfies the following conditions: (1)
2(2-γ)(c+n-2)≤(b+n-2),(2)
(2+α)λ(2+β)μ(3-γ)(3-2γ)(c+n-3)≤2(2-γ)(b+n-3),(3)
(k+1+α)λ(k+1+β)μ(k+2-γ)(k+2-2γ)(c+n-k-2)≤(k+α)λ(k+β)μ(k+1-γ)(k+1)(b+n-k-2) for 2≤k≤n-3,(4)
(n-1+α)λ(n-1+β)μ(n-γ)(n-2γ)c≤(n-2+α)λ(n-2+β)μ(n-1-γ)(n-1)(1+b-c).
Thensn(b-1,c)(z) is prestarlike of order γ, where 0≤γ<1.Proof.
It is given thatsn(b-1,c)(z)=z+∑k=2n‍Bn-k/Bn-1zk=z+∑k=2n‍akzk, z∈D.
Then usingak=Bn-k/Bn-1 for k≥1 in Theorem 5 and following the same procedure the result can be proved.Ifγ=0 then sn(b-1,c)(z)∈R∗(0)=C. Further if we substitute b=1+δ and c=1 in Theorem 18, we have the following example.Example 19.
Ifα, β, λ, and μ satisfy the conditions of Theorem 18 and if (45)δ≥max⁡n-13-2γ,n-22+αλ2+βμ3-γ3-2γ22-γ-1,n-33+αλ3+βμ4-γ4-2γ2+αλ2+βμ3-γ3-1.Then snδ(z) is prestarlike of order γ, where γ∈[0,1).It can be noted that if we takeα=β=0 and λ+μ=1 in Example 19, then, for δ≥(n-1)(3-2γ), snδ(z)∈R∗(γ). Similar type of result had been found in [23, Theorem 1]. From [26, Theorem 2.1], we deduce the following corollary.Corollary 20.
Ifα, β, λ, μ, and γ satisfy the hypothesis of Theorem 18 then for b≥c, sn(b-1,c)∈R∗(γ). Then for any zg∈K(γ)⇒g∗snb-1,cz′ is zero free in D.The following Lemma, which is the extension of the well-known Polya-Schoenberg Theorem, is ingredient to our next result.Lemma 21 (see [17, p. 499]).
Iff∈K(γ), g∈R∗(γ), 0≤γ<1 then f∗g∈K(γ).Clearly,sn(b-1,c)(g,z)∈K(γ) if g∈K(γ).Theorem 22.
Let{ak} be a sequence of positive real numbers such that a1=1. Then, for 0≤α≤6/λ+4, 0≤β≤6/μ+4 and λ≥0, μ≥0 such that 1≤λ+μ<2, if {ak} satisfy the following conditions: (1)
(2-αλ)(2-βμ)(b+n-2)a1≥(c+n-2)2λ+μ+3a2.(2)
k(k+α-λ)(k+β-μ)(b+n-k-1)ak≥(k+α)(k+β)(c+n-k-1)(k+1)ak+1, for all 2≤k≤n-2.(3)
(n-1+α-λ)(n-1+β-μ)(1+b-c)(n-1)an-1≥c(n-1+α)(n-1+β)nan.
Thensn(b-1,c)(f,z) is close-to-convex with respect to starlike function z/(1-z2) where f(z)=z+∑k=2∞‍akzk, z∈D.Proof.
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k is close-to-convex with respect to z/(1-z2) if zsnb-1,cf,z′ is typically real function. Consider (46)zsnb-1,cf,z′=z+∑k=2nBn-kBn-1kakzk=b1z+∑k=2nbkzk,where b1=1 and bk=Bn-k/Bn-1kak for k≥2. Clearly (47)bk+1=Bn-k-1Bn-1k+1ak+1⟹bk+1=c+n-k-1b+n-k-1k+1ak+1kakbk,∀1≤k≤n-2;bn=c1+b-cnann-1an-1bn-1,k=n-1.Now, (48)12+αλ2+βμb1-b2≥12λ+μ+22-αλ2-βμb1-c+n-2b+n-22a2a1b1=b12λ+μ+22-αλ2-βμ-c+n-2b+n-22λ+μ+3a2a1≥0.Further, for 2≤k≤n-2, (49)k+αλk+βμk+1+αλk+1+βμbk-bk+1≥1-λk+α1-μk+βbk-c+n-k-1b+n-k-1k+1ak+1kakbk=bkk+αk+βk+α-λk+β-μ-k+αk+βc+n-k-1b+n-k-1k+1ak+1kak≥0.For k=n-1, (50)1+1n-1+α-λ1+1n-1+β-μbn-1-bn≥1-λn-1+α1-μn-1+βbn-1-c1+b-cnann-1an-1bn-1,which is nonnegative. Following the same argument as in Theorem 5, zsn(b-1,c)(f,z) is typically real which completes the proof.Remark 23.
Note that we have no result for the close-to-convexity ofsn(b-1,c)(f,z) with respect to the starlike functions z/(1-z)2 and z/(1-z+z2). Although there are not many results in the literature for close-to-convexity with respect to z/(1-z+z2), it will be interesting if one can find the results in this direction.



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*Source: 6584584-2017-11-28.xml*