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Zero-free approximants to derivatives of prestarlike functions

Abstract

For a prestarlike function f of nonnegative order α, 0α<1, and a close-to-convex function zg of order α, the convolution g f is shown to be zero-free in the open unit disk. The result can be applied to a wide spectrum of interesting approximants, including those involving the Cesàro means and Jacobi polynomials. If zg is also prestarlike, then the range of g f is shown to be contained in a sector with opening angle strictly less than 2π.

MSC:30C45, 33C05, 40G05, 41A10.

Dedication

Dedicated to Professor Hari M Srivastava

1 Introduction

Let be the class of analytic functions f(z)=z+ n = 2 a n z n in the unit disk D={z:|z|<1} of the complex plane, and let be its subclass consisting of univalent functions. For μ<1, let S (μ) and C(μ) be the subclasses of consisting respectively of starlike and convex functions of order μ defined analytically by

f S (μ)Re ( z f ( z ) f ( z ) ) >μ,andfC(μ)Re ( 1 + z f ( z ) f ( z ) ) >μ.

For brevity, denote C:=C(0) and S = S (0). The closely-related class K(μ) of close-to-convex functions of order μ consists of functions fA satisfying

Re z f ( z ) g ( z ) >0

for some g S (μ). Evidently, for 0μ<1, C(μ) S (μ)K(μ)K:=K(0)S.

For f(z)= k = 0 a k z k and g(z)= k = 0 b k z k in , the convolution (or Hadamard product) fg is given by the series (fg)(z)= k = 0 a k b k z k . The Cesàro means of a given function is of special interest in this paper. It is the convolution between the function with the Cesàro polynomial. Specifically, let σ n β be the Cesàro polynomial of nonnegative order β defined by

σ n β (z)= n ! ( 1 + β ) n k = 0 n ( 1 + β ) n k ( n k ) ! z k (nN),

where is the set of positive integers. Here ( a ) k denotes the Pochhammer symbol given by ( a ) 0 =1 and ( a ) k =a ( a + 1 ) k 1 , kN. The Cesàro means σ n β (z,f) of order β for a function f(z)= k = 0 a k z k is

σ n β (z,f):= σ n β (z)f(z)= n ! ( 1 + β ) n k = 0 n ( 1 + β ) n k ( n k ) ! a k z k (nN).

The works of [1, 2] elucidated the geometric properties of the Cesàro polynomial.

A function f is said to be zero-free in if f(z)0 for all zD. The outer functions, which play an important role in the theory of H p spaces, are functions of the form

F(z)= e i γ exp ( 1 2 π 0 2 π e i t + z e i t z log ψ ( t ) d t ) (zD),

where γR, ψ(t)0, logψ(t) L 1 and ψ(t) L p . It is known [3, 4] that the derivatives of bounded convex functions are outer functions.

Taylor series or its partial sums are of course natural approximants to a given function. However, Barnard et al. [5] showed that the Taylor approximants of outer functions can vanish in , while the Cesàro means of order one for the derivative of convex functions are zero-free. It is therefore [5, 6] natural to investigate the problem of finding a suitable polynomial approximant for a given outer function f that retains the zero-free property of f.

Swaminathan [6] showed the zero-free property of the Cesàro means σ n β and polynomial approximants associated with Jacobi polynomials for the derivative of a prestarlike function of a certain order. Prestarlike functions [7] R μ of order μ, μ<1, consists of functions fA satisfying f k μ S (μ), k μ (z):=z/ ( 1 z ) 2 2 μ , while R 1 consists of functions fA satisfying Re(f(z)/z)>1/2. Evidently, R 1 / 2 = S (1/2) and R 0 =C. The works by [810] contained interesting exposition on prestarlike functions.

For prestarlike (and convex) functions f, the present work finds approximants derived from the convolution between f and g, where zg are close-to-convex of nonnegative order. This general result can be widely applied to include a range of interesting polynomial approximants, and thus connects with the earlier works by [5, 6, 11]. Section 3 gives examples of such applications. If zg is also prestarlike, then the range of g f is shown to be contained in a sector with opening angle strictly less than 2π.

The following two results will be required.

Lemma 1.1 [7]

  1. (i)

    If f,g R μ , μ1, then fg R μ .

  2. (ii)

    If μβ1, then R μ R β .

  3. (iii)

    If f S (μ) (or K(μ)) and g R μ , μ<1, then fg S (μ) (or K(μ)).

  4. (iv)

    R μ S if and only if μ1/2.

For 0α<1, let P(α) denote the class of all analytic functions p defined in satisfying p(0)=1 and Rep(z)>α. Also simply denote by P:=P(0). The result in [[9], Theorem 2.4, p.54] can be expressed in the following form.

Lemma 1.2 [[6], Lemma 3, p.120]

Let α<1, and 0β<1. If f R α , g S (α) and pP(β), then there exists p 1 P(β) such that fgp=(fg) p 1 .

2 Main results

Theorem 2.1 Let 0α<1. If f R α and zgK(α), then g f is zero-free in .

Proof It is sufficient to show that g f is a product of two zero-free functions in . Rewrite z(g f )(z) as

z ( g f ) (z)=z ( z g ) (z)f(z).
(1)

Since zgK(α), there exists a function h S (α) and pP such that z ( z g ) (z)=h(z)p(z). Therefore, the expression on the right side of (1) can be written as

z ( z g ) (z)f(z)= ( ( h p ) f ) (z).

Since f R α , and h S (α), Lemma 1.2 yields a p 1 P such that

( ( h p ) f ) (z)=(hf)(z) p 1 (z).

Therefore, (1) implies that

( g f ) (z)= ( h f ) ( z ) z p 1 (z).
(2)

It also follows from Lemma 1.1(iii) that hf S (α). Since S (α)S for 0α<1, (hf)(z)=0 if and only if z=0. Therefore, (h(z)f(z))/z is zero-free in . Further, as Re p 1 (z)>0, (2) implies that g f is a product of two zero-free functions, and, hence, it is also zero-free in . □

Lewis [1] proved that z σ n β K for β1. Since R 0 =C, Theorem 2.1 readily yields the following result on the Cesàro means of the derivative of convex functions.

Corollary 2.1 [[6], Theorem 2, p.120]

If fC, then the function σ n β (z, f )=( σ n β f )(z) is zero-free in for β1.

3 Examples of approximants

For applications of Theorem 2.1, this section looks at several interesting examples of approximants. For β0 and α[0,1), define the polynomial

G α , β (z):=1+ n ! ( 1 + β ) n k = 1 n ( 1 + β ) n k ( 2 2 α ) k ( n k ) ! ( k + 1 ) ! z k = σ n β (z,h),
(3)

where

h(z):={ log ( 1 z ) z , α = 1 / 2 , ( 1 z ) 2 α 1 1 z ( 1 2 α ) , α 1 / 2 .
(4)

The function zh is known to be extremal (see [12]) for many problems in the class C(α). The following result on Cesàro means for convex function of nonnegative order will be required.

Lemma 3.1 [[13], Theorem 4.2]

Let nN. If fC(λ), 1/2λ<1, and β0, then z σ n 1 β fK(λ).

Corollary 3.1 Let 1/2α<1, β0 and G α , β be given by (3). If f R α , then G α , β f is zero-free in .

Proof We show that z G α , β K(α). It follows from (3) that

z G α , β ( z ) = z + n ! ( 1 + β ) n k = 1 n ( 1 + β ) n k ( 2 2 α ) k ( n k ) ! ( k + 1 ) ! z k + 1 = ( k = 1 n + 1 ( 1 + β ) n k + 1 ( n k + 1 ) ! n ! ( 1 + β ) n z k ) ( k = 1 ( 2 2 α ) k 1 k ! z k ) : = ( z σ n β τ α ) ( z ) .

Since z τ α (z)=z ( 1 z ) ( 2 2 α ) S (α), Alexander’s theorem implies that τ α C(α), and hence Lemma 3.1 yields z G α , β K(α). From Theorem 2.1, we deduce that G α , β f is zero-free in . □

Remark 3.1 For α=1/2, simple computations show that ( G 1 / 2 , β f )(z)=( σ n β f/z)(z)= σ n β (z,f/z). If f R 1 / 2 = S (1/2), it follows from Corollary 3.1 that σ n β (z,f/z)0 in . This is a result of Ruscheweyh [11].

The next example relates to the Lerch transcendental function Φ(z,s,a) [1416] given by

Φ(z,s,a)= k = 0 z k ( k + a ) s ,

zD, Res>0 and aC{0,1,2,}. For Res<0, the summand ( k + a ) s can be continuously extended to a=k, and in this case, Φ(z,s,a) is defined for all aC.

Lemma 3.2 [[13], Theorem 5.5]

Let fC(α), 1/2α<1, and

Q ( z ) = 1 ( n + 1 ) γ k = 0 n z k ( n + 1 k ) γ = ( 1 ) γ ( n + 1 ) γ ( Φ ( z , γ , n 1 ) z n + 1 Φ ( z , γ , 0 ) ) ,
(5)

nN, γ0. Then zQfK(α).

The following result is evident from Lemma 3.2 and Theorem 2.1, and the details are therefore omitted.

Corollary 3.2 Let f R α , 1/2α<1. For γ0, let

H α , γ (z):= 1 ( n + 1 ) γ k = 0 n ( n + 1 k ) γ ( 2 2 α ) k ( k + 1 ) ! z k =(Qh)(z),

where h is given by (4) and Q by (5). Then H α , γ f is zero-free in .

Remark 3.2 Now let a k := ( n + 1 k ) γ / ( n + 1 ) γ , γ0, k=0,1,,n. A computation shows that 1= a 0 a 1 a n =1/ ( n + 1 ) γ >0. For α=1/2, Corollary 3.2 yields

( H 1 / 2 , γ f ) (z)= ( k = 0 n a k z k ) ( f ( z ) z ) 0.

This is Ruscheweyh result [[11], Theorem 1, p.682], obtained in his work on the extension of the classical Kakeya-Eneström theorem. For α>1/2, Corollary 3.2 asserts more. If now

b k := ( n + 1 k ) γ ( n + 1 ) γ ( 2 2 α ) k k ! ,γ0,α(1/2,1),

then 1= b 0 b 1 b n = ( 2 2 α ) n /(n! ( n + 1 ) γ )>0 and

( H α , γ f ) (z)= ( k = 0 n b k z k ) ( f ( z ) z ) 0.

Thus, the approximant is zero-free in in spite of the fact that f may not be univalent (see Lemma 1.1(iv)).

For α1/2, Lewis [[1], Lemma 3, p.1118] proved that

q n α (z)= n ! ( 2 2 α ) n k = 0 n ( 2 2 α ) k k ! ( 2 2 α ) n k ( n k ) ! z k

is the derivative of a function in K(α). The polynomial q n α is related [[1], p.1118] to the Jacobi polynomial P n a , b (x):= ( 1 + a ) n n ! 2 F 1 (n,n+a+b+1;1+a;(1x)/2), x[1,1], by

q n α ( e i θ ) = n ! ( 4 4 α ) n ( 2 2 α ) n ( 5 / 2 2 α ) n e i n θ / 2 P n ( 3 / 2 2 α , 3 / 2 2 α ) ( cos ( θ / 2 ) ) ,0θ2π.

Here F 1 2 is the Gaussian hypergeometric function [17].

Consider now the polynomial

Q n , α (z):= n ! ( 2 2 α ) n k = 0 n ( 2 2 α ) k k ! ( 2 2 α ) n k ( n k ) ! z k k + 1 .
(6)

A computation gives ( z Q n , α ) = q n α , and, thus, z Q n , α K(α), α1/2. The following result is now easily derived from Theorem 2.1.

Corollary 3.3 [[6], Theorem 4, p.122]

Let f R α , α1/2, and Q n , α be given by (6). Then Q n , α f is zero-free in .

We next turn to consider zero-free non-polynomial approximants. Robinson [18] (also see [[19], p.301]) introduced the polynomial

I n β (z):=1+ k = 1 n j = 0 k 1 n j β + n + j z k = 2 F 1 (1,n;n+β;z),

and conjectured that z I n β R ( 3 β ) / 2 , whenever β1 and nN. Ruscheweyh and Salinas [20] resolved the conjecture with the following more general result.

Lemma 3.3 [[20], Theorem 3, p.550]

Let λ1 and βmax{1,2λ}. Then z I λ β (z)= z 2 F 1 (1,λ;λ+β;z) R ( 3 β ) / 2 .

A consequence of Lemma 3.3 is that z I λ 1 3 =((λ+1)/λ) V λ R 0 =C for λ1/2, where V λ (z):=λ/(λ+1) z 2 F 1 (1,1λ;λ+2;z), λ>0 is a continuous extension (see [21]) of the de la Vallée Poussin means. Lemma 3.3 and Lemma 1.1(ii) together imply that z I λ 1 β + 2 CK for λ>0 and β1. Theorem 2.1 now gives a non-polynomial approximant for outer functions.

Corollary 3.4 If fC, then I λ 1 β + 2 (z, f )=( I λ 1 β + 2 f )(z) is zero-free in for all λ>0 and β1.

Remark 3.3 From [20], it is known that lim λ I λ β (z)f(z)=f(z). So if fC is bounded, then Corollary 3.4 implies that I λ 1 β + 2 (z) is an approximant to the outer function f . Thus, outer functions could also have zero-free non-polynomial approximants.

The following result on the prestarlikeness of functions, connected to the Gaussian hypergeometric function, will be required to prove the next theorem.

Lemma 3.4 [[9], Theorem 2.12]

Let a,bR satisfy 2b+1|2a+1|. Then

z 2 F 1 (1,1+a,1+b,z) R 1 a b 2 .

Theorem 3.1 Let b1/2 and ba1b. Then the Cesàro means of order (a+b) for the function F 1 2 (1+a+b,1+a;1+b;z) is zero-free in .

Proof Let α=(1ab)/2. Under the given hypothesis, it is evident that 0α1/2. The Cesàro means of order a+b for the function F 1 2 (1+a+b,1+a;1+b;z) can be expressed in the form

σ n a + b ( z ) 2 F 1 ( 1 + a + b , 1 + a ; 1 + b ; z ) = ( k = 0 n n ! ( 2 2 α ) n ( 2 2 α ) n k ( n k ) ! ( 2 2 α ) k k ! z k k + 1 ) ( k = 0 ( 1 + a ) k ( 1 + b ) k ( k + 1 ) z k ) = ( g f ) ( z ) ,

where g(z)= Q n , α (z) is given by (6) and f(z)= z 2 F 1 (1,1+a;1+b,z). It is known [1] that zg=z Q n , α K(α) for α1/2. Straightforward computations show that 2b+1|2a+1|, and, thus, Lemma 3.4 yields f R α . Therefore, it follows from Theorem 2.1 that g f 0 in . □

Example 3.1

  1. (1)

    Choosing a=b=1/2, Theorem 3.1 yields σ n 1 (z, ( 1 z ) 2 ) is zero-free in .

  2. (2)

    Since F 1 2 (1+b,1;1+b;z)= ( 1 z ) 1 , with a=0, it follows that σ n b (z)0 for b[1/2,1] and zD.

  3. (3)

    If b=a=1/2, Theorem 3.1 shows that the n th partial sum of the Taylor series of arctanh( z )/ z is zero-free in .

When both the source functions f and the approximant are prestarlike of certain order, the result below shows that the range of the approximant satisfies a sector-like condition on the boundary.

Theorem 3.2 Let f R α and zg R μ with (zgf)/z bounded in , α,μ1/2. Then the range of g f is contained in a sector (from 0) with the opening 2γπ for some γ<1.

Proof Let f R α , α1/2. By Lemma 1.1(ii), f R α R 1 / 2 = S (1/2). Rewrite g f as

( g f ) ( z ) = 1 z ( z ( z g ) f ) ( z ) = 1 z ( z g ( z ) z ( z g ) ( z ) z g ( z ) f ( z ) ) .

Since zg R μ R 1 / 2 = S (1/2), there exists a function pP(1/2) satisfying

z ( z g ) ( z ) z g ( z ) =p(z).

From Lemma 1.2, there exists a function p 1 P(1/2) such that

( g f ) (z)= ( z g f ) ( z ) z p 1 (z).

Since zg S (1/2), and f S (1/2), Lemma 1.1(i) implies that zgf S (1/2).

A result in [[22], Theorem 2.6a, p.57] shows that

1 z (zgf)(z)P(1/2).

Since (zgf)/z is bounded in , there exists a γ 1 such that

| arg ( z g f ) ( z ) z | < γ 1 π 2 , γ 1 <1.

Therefore,

| arg ( g f ) ( z ) | = | arg { ( z g f ) ( z ) z p 1 ( z ) } | γ 1 π 2 + π 2 = γ π , γ : = ( γ 1 + 1 ) / 2 < 1 .

 □

Example 3.2 Let g be either I n β + 1 or σ n β + 1 . The polynomial (zgf)/z is bounded in . Hence for β1, Theorem 3.2 implies that the range of both I n β + 1 (z, f ) and σ n β + 1 (z, f ) are contained in a sector with opening 2γπ, γ<1.

Remark 3.4 Example 3.2 reduces to a result of Swaminathan [[6], Theorem 3, p.121] in the case f R 0 =C and g(z)= σ n β + 1 (z).

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Acknowledgements

The work presented here was supported in part by a research university grant from Universiti Sains Malaysia.

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This work was completed when the second author was a postdoctoral fellow at Universiti Sains Malaysia (USM), and the third author was visiting USM. The research was funded by a grant from USM. The study was conceived and planned by all authors. Every author participated in the discussions of tackling the problem, and the directions of the proofs of the results. All authors read and approved the final manuscript.

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Ali, R.M., Mondal, S.R. & Ravichandran, V. Zero-free approximants to derivatives of prestarlike functions. J Inequal Appl 2013, 401 (2013). https://doi.org/10.1186/1029-242X-2013-401

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Keywords

  • Convex Function
  • Unit Disk
  • Jacobi Polynomial
  • Continuous Extension
  • Transcendental Function