Zero-free approximants to derivatives of prestarlike functions

Abstract

For a prestarlike function f of nonnegative order α, $0\le \alpha <1$, and a close-to-convex function zg of order α, the convolution $g\ast f^{\prime }$ 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 $g\ast f^{\prime }$ 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 $f(z)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_n{z}^n$ in the unit disk $\mathbb{D}=\{z:|z|<1\}$ of the complex plane, and let be its subclass consisting of univalent functions. For $\mu <1$, let ${\mathcal{S}}^{\ast }(\mu )$ and $\mathcal{C}(\mu )$ be the subclasses of consisting respectively of starlike and convex functions of order μ defined analytically by

$$f\in {\mathcal{S}}^{\ast }(\mu )\iff Re\left(\frac{z{f}^{\prime }(z)}{f(z)}\right)>\mu ,\text{and}f\in \mathcal{C}(\mu )\iff Re(1+\frac{z{f}^{″}(z)}{f^{\prime }(z)})>\mu .$$

For brevity, denote $\mathcal{C}:=\mathcal{C}(0)$ and ${\mathcal{S}}^{\ast }={\mathcal{S}}^{\ast }(0)$. The closely-related class $\mathcal{K}(\mu )$ of close-to-convex functions of order μ consists of functions $f\in \mathcal{A}$ satisfying

$$Re\frac{z{f}^{\prime }(z)}{g(z)}>0$$

for some $g\in {\mathcal{S}}^{\ast }(\mu )$. Evidently, for $0\le \mu <1$, $\mathcal{C}(\mu )\subset {\mathcal{S}}^{\ast }(\mu )\subset \mathcal{K}(\mu )\subset \mathcal{K}:=\mathcal{K}(0)\subseteq \mathcal{S}$.

For $f(z)={\sum }_{k=0}^{\mathrm{\infty }}{a}_k{z}^k$ and $g(z)={\sum }_{k=0}^{\mathrm{\infty }}{b}_k{z}^k$ in , the convolution (or Hadamard product) $f\ast g$ is given by the series $(f\ast g)(z)={\sum }_{k=0}^{\mathrm{\infty }}{a}_k{b}_k{z}^k$. 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 ${\sigma }_n^{\beta }$ be the Cesàro polynomial of nonnegative order β defined by

$${\sigma }_n^{\beta }(z)=\frac{n!}{{(1+\beta )}_n}\sum _{k=0}^n\frac{{(1+\beta )}_{n-k}}{(n-k)!}{z}^k(n\in \mathbb{N}),$$

where ℕ is the set of positive integers. Here ${(a)}_k$ denotes the Pochhammer symbol given by ${(a)}_0=1$ and ${(a)}_k=a{(a+1)}_{k-1}$, $k\in \mathbb{N}$. The Cesàro means ${\sigma }_n^{\beta }(z,f)$ of order β for a function $f(z)={\sum }_{k=0}^{\mathrm{\infty }}{a}_k{z}^k$ is

$${\sigma }_n^{\beta }(z,f):={\sigma }_n^{\beta }(z)\ast f(z)=\frac{n!}{{(1+\beta )}_n}\sum _{k=0}^n\frac{{(1+\beta )}_{n-k}}{(n-k)!}{a}_k{z}^k(n\in \mathbb{N}).$$

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

A function f is said to be zero-free in if $f(z)\ne 0$ for all $z\in \mathbb{D}$. The outer functions, which play an important role in the theory of ${\mathcal{H}}^p$ spaces, are functions of the form

$$F(z)=e^{i\gamma }exp(\frac{1}{2\pi }{\int }_0^{2\pi }\frac{e^{it}+z}{e^{it}-z}log\psi (t)dt)(z\in \mathbb{D}),$$

where $\gamma \in \mathbb{R}$, $\psi (t)\ge 0$, $log\psi (t)\in L^1$ and $\psi (t)\in L^p$. 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 ${\sigma }_n^{\beta }$ and polynomial approximants associated with Jacobi polynomials for the derivative of a prestarlike function of a certain order. Prestarlike functions [7]${\mathcal{R}}_{\mu }$ of order μ, $\mu <1$, consists of functions $f\in \mathcal{A}$ satisfying $f\ast k_{\mu }\in {\mathcal{S}}^{\ast }(\mu )$, $k_{\mu }(z):=z/{(1-z)}^{2-2\mu }$, while ${\mathcal{R}}_1$ consists of functions $f\in \mathcal{A}$ satisfying $Re(f(z)/z)>1/2$. Evidently, ${\mathcal{R}}_{1/2}={\mathcal{S}}^{\ast }(1/2)$ and ${\mathcal{R}}_0=\mathcal{C}$. The works by [8–10] contained interesting exposition on prestarlike functions.

For prestarlike (and convex) functions f, the present work finds approximants derived from the convolution between $f^{\prime }$ 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 $g\ast f^{\prime }$ 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]

• (i) If $f,g\in {\mathcal{R}}_{\mu }$, $\mu \le 1$, then $f\ast g\in {\mathcal{R}}_{\mu }$.

• (ii) If $\mu \le \beta \le 1$, then ${\mathcal{R}}_{\mu }\subset {\mathcal{R}}_{\beta }$.

• (iii) If $f\in {\mathcal{S}}^{\ast }(\mu )$ (or $\mathcal{K}(\mu )$) and $g\in {\mathcal{R}}_{\mu }$, $\mu <1$, then $f\ast g\in {\mathcal{S}}^{\ast }(\mu )$ (or $\mathcal{K}(\mu )$).

• (iv) ${\mathcal{R}}_{\mu }\subset \mathcal{S}$ if and only if $\mu \le 1/2$.

For $0\le \alpha <1$, let $\mathcal{P}(\alpha )$ denote the class of all analytic functions p defined in satisfying $p(0)=1$ and $Rep(z)>\alpha$. Also simply denote by $\mathcal{P}:=\mathcal{P}(0)$. 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 $\alpha <1$, and $0\le \beta <1$. If $f\in {\mathcal{R}}_{\alpha }$, $g\in {\mathcal{S}}^{\ast }(\alpha )$ and $p\in \mathcal{P}(\beta )$, then there exists $p_1\in \mathcal{P}(\beta )$ such that $f\ast gp=(f\ast g)p_1$.

2 Main results

Theorem 2.1 Let $0\le \alpha <1$. If $f\in {\mathcal{R}}_{\alpha }$ and $zg\in \mathcal{K}(\alpha )$, then $g\ast f^{\prime }$ is zero-free in .

Proof It is sufficient to show that $g\ast f^{\prime }$ is a product of two zero-free functions in . Rewrite $z(g\ast f^{\prime })(z)$ as

$$z(g\ast f^{\prime })(z)=z{(zg)}^{\prime }(z)\ast f(z).$$

(1)

Since $zg\in \mathcal{K}(\alpha )$, there exists a function $h\in {\mathcal{S}}^{\ast }(\alpha )$ and $p\in \mathcal{P}$ such that $z{(zg)}^{\prime }(z)=h(z)p(z)$. Therefore, the expression on the right side of (1) can be written as

$$z{(zg)}^{\prime }(z)\ast f(z)=((hp)\ast f)(z).$$

Since $f\in {\mathcal{R}}_{\alpha }$, and $h\in {\mathcal{S}}^{\ast }(\alpha )$, Lemma 1.2 yields a $p_1\in \mathcal{P}$ such that

$$((hp)\ast f)(z)=(h\ast f)(z)p_1(z).$$

Therefore, (1) implies that

$$(g\ast f^{\prime })(z)=\frac{(h\ast f)(z)}{z}{p}_1(z).$$

(2)

It also follows from Lemma 1.1(iii) that $h\ast f\in {\mathcal{S}}^{\ast }(\alpha )$. Since ${\mathcal{S}}^{\ast }(\alpha )\subset \mathcal{S}$ for $0\le \alpha <1$, $(h\ast f)(z)=0$ if and only if $z=0$. Therefore, $(h(z)\ast f(z))/z$ is zero-free in . Further, as $Re{p}_1(z)>0$, (2) implies that $g\ast f^{\prime }$ is a product of two zero-free functions, and, hence, it is also zero-free in . □

Lewis [1] proved that $z{\sigma }_n^{\beta }\in \mathcal{K}$ for $\beta \ge 1$. Since ${\mathcal{R}}_0=\mathcal{C}$, 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 $f\in \mathcal{C}$, then the function ${\sigma }_n^{\beta }(z,f^{\prime })=({\sigma }_n^{\beta }\ast f^{\prime })(z)$ is zero-free in for $\beta \ge 1$.

3 Examples of approximants

For applications of Theorem 2.1, this section looks at several interesting examples of approximants. For $\beta \ge 0$ and $\alpha \in [0,1)$, define the polynomial

$${\mathcal{G}}_{\alpha ,\beta }(z):=1+\frac{n!}{{(1+\beta )}_n}\sum _{k=1}^n\frac{{(1+\beta )}_{n-k}{(2-2\alpha )}_k}{(n-k)!(k+1)!}{z}^k={\sigma }_n^{\beta }(z,h),$$

(3)

where

$$h(z):=\{\begin{array}{cc}-\frac{log(1-z)}{z},\hfill & \alpha =1/2,\hfill \\ \frac{{(1-z)}^{2\alpha -1}-1}{z(1-2\alpha )},\hfill & \alpha \ne 1/2.\hfill \end{array}$$

(4)

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

Lemma 3.1 [[13], Theorem 4.2]

Let $n\in \mathbb{N}$. If $f\in \mathcal{C}(\lambda )$, $1/2\le \lambda <1$, and $\beta \ge 0$, then $z{\sigma }_{n-1}^{\beta }\ast f\in \mathcal{K}(\lambda )$.

Corollary 3.1 Let $1/2\le \alpha <1$, $\beta \ge 0$ and ${\mathcal{G}}_{\alpha ,\beta }$ be given by (3). If $f\in {\mathcal{R}}_{\alpha }$, then ${\mathcal{G}}_{\alpha ,\beta }\ast f^{\prime }$ is zero-free in .

Proof We show that $z{\mathcal{G}}_{\alpha ,\beta }\in \mathcal{K}(\alpha )$. It follows from (3) that

$$\begin{array}{rcl}z{\mathcal{G}}_{\alpha ,\beta }(z)& =& z+\frac{n!}{{(1+\beta )}_n}\sum _{k=1}^n\frac{{(1+\beta )}_{n-k}{(2-2\alpha )}_k}{(n-k)!(k+1)!}{z}^{k+1}\\ =& \left(\sum _{k=1}^{n+1}\frac{{(1+\beta )}_{n-k+1}}{(n-k+1)!}\frac{n!}{{(1+\beta )}_n}{z}^k\right)\ast \left(\sum _{k=1}^{\mathrm{\infty }}\frac{{(2-2\alpha )}_{k-1}}{k!}{z}^k\right)\\ :=& (z{\sigma }_n^{\beta }\ast {\tau }_{\alpha })(z).\end{array}$$

Since $z{\tau }_{\alpha }^{\prime }(z)=z{(1-z)}^{-(2-2\alpha )}\in {\mathcal{S}}^{\ast }(\alpha )$, Alexander’s theorem implies that ${\tau }_{\alpha }\in \mathcal{C}(\alpha )$, and hence Lemma 3.1 yields $z{\mathcal{G}}_{\alpha ,\beta }\in \mathcal{K}(\alpha )$. From Theorem 2.1, we deduce that ${\mathcal{G}}_{\alpha ,\beta }\ast f^{\prime }$ is zero-free in . □

Remark 3.1 For $\alpha =1/2$, simple computations show that $({\mathcal{G}}_{1/2,\beta }\ast f^{\prime })(z)=({\sigma }_n^{\beta }\ast f/z)(z)={\sigma }_n^{\beta }(z,f/z)$. If $f\in {\mathcal{R}}_{1/2}={\mathcal{S}}^{\ast }(1/2)$, it follows from Corollary 3.1 that ${\sigma }_n^{\beta }(z,f/z)\ne 0$ in . This is a result of Ruscheweyh [11].

The next example relates to the Lerch transcendental function $\mathrm{\Phi }(z,s,a)$ [14–16] given by

$$\mathrm{\Phi }(z,s,a)=\sum _{k=0}^{\mathrm{\infty }}\frac{z^k}{{(k+a)}^s},$$

$z\in \mathbb{D}$, $Res>0$ and $a\in \mathbb{C}\mathrm{\setminus }\{0,-1,-2,\dots \}$. For $Res<0$, the summand ${(k+a)}^{-s}$ can be continuously extended to $a=-k$, and in this case, $\mathrm{\Phi }(z,s,a)$ is defined for all $a\in \mathbb{C}$.

Lemma 3.2 [[13], Theorem 5.5]

Let $f\in \mathcal{C}(\alpha )$, $1/2\le \alpha <1$, and

$$\begin{array}{rcl}Q(z)& =& \frac{1}{{(n+1)}^{\gamma }}\sum _{k=0}^n\frac{z^k}{{(n+1-k)}^{-\gamma }}\\ =& \frac{{(-1)}^{\gamma }}{{(n+1)}^{\gamma }}(\mathrm{\Phi }(z,-\gamma ,-n-1)-z^{n+1}\mathrm{\Phi }(z,-\gamma ,0)),\end{array}$$

(5)

$n\in \mathbb{N}$, $\gamma \ge 0$. Then $zQ\ast f\in \mathcal{K}(\alpha )$.

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

Corollary 3.2 Let $f\in {\mathcal{R}}_{\alpha }$, $1/2\le \alpha <1$. For $\gamma \ge 0$, let

$${\mathcal{H}}_{\alpha ,\gamma }(z):=\frac{1}{{(n+1)}^{\gamma }}\sum _{k=0}^n\frac{{(n+1-k)}^{\gamma }{(2-2\alpha )}_k}{(k+1)!}{z}^k=(Q\ast h)(z),$$

where h is given by (4) and Q by (5). Then ${\mathcal{H}}_{\alpha ,\gamma }\ast f^{\prime }$ is zero-free in .

Remark 3.2 Now let $a_k:={(n+1-k)}^{\gamma }/{(n+1)}^{\gamma }$, $\gamma \ge 0$, $k=0,1,\dots ,n$. A computation shows that $1=a_0\ge a_1\ge \cdots \ge a_n=1/{(n+1)}^{\gamma }>0$. For $\alpha =1/2$, Corollary 3.2 yields

$$({\mathcal{H}}_{1/2,\gamma }\ast f^{\prime })(z)=\left(\sum _{k=0}^n{a}_k{z}^k\right)\ast \left(\frac{f(z)}{z}\right)\ne 0.$$

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

$$b_k:=\frac{{(n+1-k)}^{\gamma }}{{(n+1)}^{\gamma }}\frac{{(2-2\alpha )}_k}{k!},\gamma \ge 0,\alpha \in (1/2,1),$$

then $1=b_0\ge b_1\ge \cdots \ge b_n={(2-2\alpha )}_n/(n!{(n+1)}^{\gamma })>0$ and

$$({\mathcal{H}}_{\alpha ,\gamma }\ast f^{\prime })(z)=\left(\sum _{k=0}^n{b}_k{z}^k\right)\ast \left(\frac{f(z)}{z}\right)\ne 0.$$

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

For $\alpha \le 1/2$, Lewis [[1], Lemma 3, p.1118] proved that

$$q_n^{\alpha }(z)=\frac{n!}{{(2-2\alpha )}_n}\sum _{k=0}^n\frac{{(2-2\alpha )}_k}{k!}\frac{{(2-2\alpha )}_{n-k}}{(n-k)!}{z}^k$$

is the derivative of a function in $\mathcal{K}(\alpha )$. The polynomial $q_n^{\alpha }$ is related [[1], p.1118] to the Jacobi polynomial $P_n^{a,b}(x):={\frac{{(1+a)}_n}{n!}}_2{F}_1(-n,n+a+b+1;1+a;(1-x)/2)$, $x\in [-1,1]$, by

$$q_n^{\alpha }\left(e^{i\theta }\right)=\frac{n!{(4-4\alpha )}_n}{{(2-2\alpha )}_n{(5/2-2\alpha )}_n}{e}^{in\theta /2}{P}_n^{(3/2-2\alpha ,3/2-2\alpha )}(cos(\theta /2)),0\le \theta \le 2\pi .$$

Here ${}_{2}F_1$ is the Gaussian hypergeometric function [17].

Consider now the polynomial

$${\mathcal{Q}}_{n,\alpha }(z):=\frac{n!}{{(2-2\alpha )}_n}\sum _{k=0}^n\frac{{(2-2\alpha )}_k}{k!}\frac{{(2-2\alpha )}_{n-k}}{(n-k)!}\frac{z^k}{k+1}.$$

(6)

A computation gives ${(z{\mathcal{Q}}_{n,\alpha })}^{\prime }=q_n^{\alpha }$, and, thus, $z{\mathcal{Q}}_{n,\alpha }\in \mathcal{K}(\alpha )$, $\alpha \le 1/2$. The following result is now easily derived from Theorem 2.1.

Corollary 3.3 [[6], Theorem 4, p.122]

Let $f\in {\mathcal{R}}_{\alpha }$, $\alpha \le 1/2$, and ${\mathcal{Q}}_{n,\alpha }$ be given by (6). Then ${\mathcal{Q}}_{n,\alpha }\ast f^{\prime }$ is zero-free in .

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

$${\mathcal{I}}_n^{\beta }(z):=1+\sum _{k=1}^n\prod _{j=0}^{k-1}\frac{n-j}{\beta +n+j}{z}^k{=}_2{F}_1(1,-n;n+\beta ;-z),$$

and conjectured that $z{\mathcal{I}}_n^{\beta }\in {\mathcal{R}}_{(3-\beta )/2}$, whenever $\beta \ge 1$ and $n\in \mathbb{N}$. Ruscheweyh and Salinas [20] resolved the conjecture with the following more general result.

Lemma 3.3 [[20], Theorem 3, p.550]

Let $\lambda \ge -1$ and $\beta \ge max\{1,-2\lambda \}$. Then $z{\mathcal{I}}_{\lambda }^{\beta }(z)=z_2{F}_1(1,-\lambda ;\lambda +\beta ;-z)\in {\mathcal{R}}_{(3-\beta )/2}$.

A consequence of Lemma 3.3 is that $z{\mathcal{I}}_{\lambda -1}^3=((\lambda +1)/\lambda )V_{\lambda }\in {\mathcal{R}}_0=\mathcal{C}$ for $\lambda \ge -1/2$, where $V_{\lambda }(z):=\lambda /(\lambda +1)z_2{F}_1(1,1-\lambda ;\lambda +2;-z)$, $\lambda >0$ is a continuous extension (see [21]) of the de la Vallée Poussin means. Lemma 3.3 and Lemma 1.1(ii) together imply that $z{\mathcal{I}}_{\lambda -1}^{\beta +2}\in \mathcal{C}\subset \mathcal{K}$ for $\lambda >0$ and $\beta \ge 1$. Theorem 2.1 now gives a non-polynomial approximant for outer functions.

Corollary 3.4 If $f\in \mathcal{C}$, then ${\mathcal{I}}_{\lambda -1}^{\beta +2}(z,f^{\prime })=({\mathcal{I}}_{\lambda -1}^{\beta +2}\ast f^{\prime })(z)$ is zero-free in for all $\lambda >0$ and $\beta \ge 1$.

Remark 3.3 From [20], it is known that ${lim}_{\lambda \to \mathrm{\infty }}{\mathcal{I}}_{\lambda }^{\beta }(z)\ast f(z)=f(z)$. So if $f\in \mathcal{C}$ is bounded, then Corollary 3.4 implies that ${\mathcal{I}}_{\lambda -1}^{\beta +2}(z)$ is an approximant to the outer function $f^{\prime }$. 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 $a,b\in \mathbb{R}$ satisfy $2b+1\ge |2a+1|$. Then

$$z_2{F}_1(1,1+a,1+b,z)\in {\mathcal{R}}_{\frac{1-a-b}{2}}.$$

Theorem 3.1 Let $b\ge 1/2$ and $-b\le a\le 1-b$. Then the Cesàro means of order $(a+b)$ for the function ${}_{2}F_1(1+a+b,1+a;1+b;z)$ is zero-free in .

Proof Let $\alpha =(1-a-b)/2$. Under the given hypothesis, it is evident that $0\le \alpha \le 1/2$. The Cesàro means of order $a+b$ for the function ${}_{2}F_1(1+a+b,1+a;1+b;z)$ can be expressed in the form

$$\begin{array}{c}{\sigma }_n^{a+b}(z){\ast }_2{F}_1(1+a+b,1+a;1+b;z)\hfill \\ =\left(\sum _{k=0}^n\frac{n!}{{(2-2\alpha )}_n}\frac{{(2-2\alpha )}_{n-k}}{(n-k)!}\frac{{(2-2\alpha )}_k}{k!}\frac{z^k}{k+1}\right)\ast (\sum _{k=0}^{\mathrm{\infty }}\frac{{(1+a)}_k}{{(1+b)}_k}(k+1)z^k)\hfill \\ =(g\ast f^{\prime })(z),\hfill \end{array}$$

where $g(z)={\mathcal{Q}}_{n,\alpha }(z)$ is given by (6) and $f(z)=z_2{F}_1(1,1+a;1+b,z)$. It is known [1] that $zg=z{\mathcal{Q}}_{n,\alpha }\in \mathcal{K}(\alpha )$ for $\alpha \le 1/2$. Straightforward computations show that $2b+1\ge |2a+1|$, and, thus, Lemma 3.4 yields $f\in {\mathcal{R}}_{\alpha }$. Therefore, it follows from Theorem 2.1 that $g\ast f^{\prime }\ne 0$ in . □

Example 3.1

• (1) Choosing $a=b=1/2$, Theorem 3.1 yields ${\sigma }_n^1(z,{(1-z)}^{-2})$ is zero-free in .

• (2) Since ${}_{2}F_1(1+b,1;1+b;z)={(1-z)}^{-1}$, with $a=0$, it follows that ${\sigma }_n^b(z)\ne 0$ for $b\in [1/2,1]$ and $z\in \mathbb{D}$.

• (3) If $b=-a=1/2$, Theorem 3.1 shows that the n th partial sum of the Taylor series of $arctanh(\sqrt{z})/\sqrt{z}$ 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 $f\in {\mathcal{R}}_{\alpha }$ and $zg\in {\mathcal{R}}_{\mu }$ with $(zg\ast f)/z$ bounded in , $\alpha ,\mu \le 1/2$. Then the range of $g\ast f^{\prime }$ is contained in a sector (from 0) with the opening $2\gamma \pi$ for some $\gamma <1$.

Proof Let $f\in {\mathcal{R}}_{\alpha }$, $\alpha \le 1/2$. By Lemma 1.1(ii), $f\in {\mathcal{R}}_{\alpha }\subset {\mathcal{R}}_{1/2}={\mathcal{S}}^{\ast }(1/2)$. Rewrite $g\ast f^{\prime }$ as

$$\begin{array}{rcl}(g\ast f^{\prime })(z)& =& \frac{1}{z}(z{(zg)}^{\prime }\ast f)(z)\\ =& \frac{1}{z}(zg(z)\frac{z{(zg)}^{\prime }(z)}{zg(z)}\ast f(z)).\end{array}$$

Since $zg\in {\mathcal{R}}_{\mu }\subset {\mathcal{R}}_{1/2}={\mathcal{S}}^{\ast }(1/2)$, there exists a function $p\in \mathcal{P}(1/2)$ satisfying

$$\frac{z{(zg)}^{\prime }(z)}{zg(z)}=p(z).$$

From Lemma 1.2, there exists a function $p_1\in \mathcal{P}(1/2)$ such that

$$(g\ast f^{\prime })(z)=\frac{(zg\ast f)(z)}{z}{p}_1(z).$$

Since $zg\in {\mathcal{S}}^{\ast }(1/2)$, and $f\in {\mathcal{S}}^{\ast }(1/2)$, Lemma 1.1(i) implies that $zg\ast f\in {\mathcal{S}}^{\ast }(1/2)$.

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

$$\frac{1}{z}(zg\ast f)(z)\in \mathcal{P}(1/2).$$

Since $(zg\ast f)/z$ is bounded in , there exists a ${\gamma }_1$ such that

$$|arg\frac{(zg\ast f)(z)}{z}|<\frac{{\gamma }_1\pi }{2},{\gamma }_1<1.$$

Therefore,

$$\begin{array}{rcl}|arg(g\ast f^{\prime })(z)|& =& |arg\{\frac{(zg\ast f)(z)}{z}{p}_1(z)\}|\\ \le & \frac{{\gamma }_1\pi }{2}+\frac{\pi }{2}=\gamma \pi ,\gamma :=({\gamma }_1+1)/2<1.\end{array}$$

 □

Example 3.2 Let g be either ${\mathcal{I}}_n^{\beta +1}$ or ${\sigma }_n^{\beta +1}$. The polynomial $(zg\ast f)/z$ is bounded in . Hence for $\beta \ge 1$, Theorem 3.2 implies that the range of both ${\mathcal{I}}_n^{\beta +1}(z,f^{\prime })$ and ${\sigma }_n^{\beta +1}(z,f^{\prime })$ are contained in a sector with opening $2\gamma \pi$, $\gamma <1$.

Remark 3.4 Example 3.2 reduces to a result of Swaminathan [[6], Theorem 3, p.121] in the case $f\in {\mathcal{R}}_0=\mathcal{C}$ and $g(z)={\sigma }_n^{\beta +1}(z)$.