Subordination of Ces$\grave{a}$ro Means of Convex Functions

Abstract

For all $n\in {\mathbb {N}},$ we find sharp constants $\mu _n^{(2)}$ and $\mu _n^{(3)}$ such that $\mu _n^{(2)}\sigma _n^{(2)}(z,f)\prec f(z)$ and $\mu _n^{(3)}\sigma _n^{(3)}(z,f)\prec f(z)$ in the open unit disc ${\mathbb {D}}$ for all f- normalized convex univalent functions in ${\mathbb {D}}$. Here $\sigma _n^{(\alpha )}(z,f)$ stands for nth Ces$\grave{a}$ro mean of order $\alpha ,\; \alpha \ge 0,$ of $f(z)=\sum _{k=1}^{\infty }a_k z^k$ defined by $\sigma _n^{(\alpha )}(z,f):=\left( {\begin{array}{c}n+\alpha -1\\ n-1\end{array}}\right) ^{-1}\sum _{k=1}^n\left( {\begin{array}{c}n+\alpha -k\\ n-k\end{array}}\right) a_k z^k$ and the symbol $'\prec '$ stands for subordination between two analytic functions. Among other things, a generalization of an earlier known result related to subordination is also presented.

1 Introduction

Let ${\mathbb {D}}=\{z\in {{\mathbb {C}}:|z|<1}\}$ denote the open unit disc in the complex plane and ${\mathscr {S}}$ the class of functions $f(z)=z+a_2z^2+...$ which are analytic and univalent in ${\mathbb {D}}$. Denote by ${\mathscr {S}}^*$ and ${\mathscr {K}}$ the usual subclasses of ${\mathscr {S}}$ consisting of functions which map ${\mathbb {D}}$ onto starlike (w.r.t. the origin) and convex domains, respectively. Let ${\mathscr {S}}^{*}(1/2)\subset {\mathscr {S}}$ be the class of functions which are starlike of order 1/2. It is known that ${\mathscr {K}}\subset {\mathscr {S}}^{*}(1/2).$ Further, by ${\mathscr {C}}$ we denote the class of functions $f\in {\mathscr {S}}$ for which there exist $g\in {\mathscr {K}}$ such that $\Re \left\{ f'(z)/g'(z)\right\} >0$, $z\in {\mathbb {D}}$. Functions in the class ${\mathscr {C}}$ are called close-to-convex.

The convolution or Hadamard product of two power series $f(z)=\sum _{n=0}^{\infty }a_nz^n$ and $g(z)=\sum _{n=0}^{\infty }b_nz^n$ is denoted by $(f*g)(z)$ and is defined as

$$\begin{aligned} (f*g)(z):=\sum _{n=0}^{\infty }a_n b_nz^n. \end{aligned}$$

A function f is said to be subordinate to a function g (in symbols $f(z)\prec g(z))$ in $|z|<r ~(0<r<1)$ if g is univalent in $|z|<r,f(0)=g(0)$ and $f(|z|<r)\subset g(|z|<r)$.

For a given function $f(z)=\sum _{n=1}^{\infty }a_nz^n$ and $n\in {{\mathbb {N}}}$, $\alpha \ge 0,$ let

$$\begin{aligned} s_n(z,f)= & {} \sum _{k=1}^na_kz^k,\\ \sigma _n^{(\alpha )}(z,f)= & {} \left( {\begin{array}{c}n+\alpha -1\\ n-1\end{array}}\right) ^{-1}\sum _{k=1}^n\left( {\begin{array}{c}n+\alpha -k\\ n-k\end{array}}\right) a_k z^k \end{aligned}$$

and

$$\begin{aligned} V_n(z,f)=\left( {\begin{array}{c}2n\\ n\end{array}}\right) ^{-1}\sum _{k=1}^n\left( {\begin{array}{c}2n\\ n+k\end{array}}\right) a_k z^k \end{aligned}$$

denote, the nth partial sum, the nth Ces$\grave{a}$ro mean of order $\alpha$ and the nth de la Vall$\acute{e}$e Poussin mean of f, respectively.

P$\acute{o}$lya and Schoenberg ( [7], Theorem 2, p.298) showed that the de la Vall$\acute{e}$e Poussin means $V_n(z,f)$ are convex (starlike) if and only if f is convex (starlike) and also, established the following fascinating result : $\square$

Theorem 1.1

If $f\in {{\mathscr {K}}}$, then $V_n(z,f)\prec f(z)$ in ${\mathbb {D}}.$

Robertson [8] further extended this result by proving that if f is univalent in ${\mathbb {D}}$, then converse of Theorem 1.1 is also true.

It has been a long tradition to study mapping properties of Ces$\grave{a}$ro means (for example, see [4, 2, 3, 6, 9, 11, 12]). In 1995, Singh and Singh [14] proved the following analogues of above theorem of P$\acute{o}$lya and Schoenberg for certain transformations of the nth partial sum, $s_n(z,f),$ and the nth Ces$\grave{a}$ro mean of first order, $\sigma _n^{(1)}(z,f),$ of $f\in {\mathscr {K}}$.

Theorem 1.2

If $f\in {{\mathscr {K}}}$, then

(i) $(1/z)\int _0^zs_n(t,f)dt\prec f(z)$ in ${\mathbb {D}}$ for every $n\in {\mathbb {N}}$.

(ii) $(n/(n+1))\sigma _n^{(1)}(z,f)\prec f(z)$ in ${\mathbb {D}}.$ This result is sharp for every $n\in {\mathbb {N}}$.

$\square$

In the same paper, i.e., [14], Singh and Singh also proved the following result :

Theorem 1.3

For every $f\in {\mathscr {S}}^{*}(1/2)$ and for every positive integer n, we have

$$\begin{aligned} \Re \frac{V_n(z,f)}{\sigma _n^{(2)}(z,f)}>0, \ \ \ z\in {\mathbb {D}}. \end{aligned}$$

$\square$

In the present note, we establish the analogues of Theorem 1.1 for certain transformations of the nth Ces$\grave{a}$ro means $\sigma _n^{(2)}(z,f)$ and $\sigma _n^{(3)}(z,f),$ of $f\in {\mathscr {K}}$. We also prove an analogue of Theorem 1.3 by replacing $\sigma _n^{(2)}(z,f)$ there with $\sigma _n^{(3)}(z,f)$. A generalization of a result of Singh and Singh [13] related to subordination between $V_2(z,f)$ and $\sigma _2^{(1)}(z,f)$ of $f\in {\mathscr {K}}$ is also presented.

2 Preliminaries

In this section, we collect following definition and results which shall be needed to prove our results in this paper.

Definition 2.1

A sequence $\{b_n\}_1^\infty$ of complex numbers is said to be a subordinating factor sequence if, whenever $f(z)= \sum _{n=1}^\infty a_n z^n$, $a_1 =1$, is univalent and convex in ${\mathbb {D}}$, we have

$$\begin{aligned} \sum _{n=1}^\infty a_nb_n z^n \prec f(z). \end{aligned}$$

$\square$

Lemma 2.2

[15] A sequence $\{b_n\}_1^\infty$ of complex numbers is a subordinating factor sequence if and only if

$$\begin{aligned} \Re \left[ 1+2 \sum \limits _{n=1}^\infty b_n z^n\right] >0,\ \ \ z \in {\mathbb {D}}. \end{aligned}$$

$\square$

Lemma 2.3

( [5], p.3) If $h(z)= nz+(n-1)z^2+ \ldots +z^n$, then

$$\begin{aligned} h(e^{i \theta })=-\frac{n+1}{2}+ \frac{\sin ^2(n+1)\theta /2}{2 \sin ^2 \theta /2}+ \frac{(n+1) \sin \theta -\sin (n+1) \theta }{4 \sin ^2 \theta /2}. \end{aligned}$$

$\square$

Lemma 2.4

[10] Let f and g belong to ${\mathscr {S}}^{*}(1/2)$. Then for each function F analytic in ${\mathbb {D}}$ and satisfying $\Re F(z)>0, z \in {\mathbb {D}}$, we have

$$\begin{aligned} {\Re } \frac{f(z)* F(z)g(z)}{f(z)*g(z)}>0, \ \ \ z \in {\mathbb {D}}. \end{aligned}$$

$\square$

Lemma 2.5

[1] Suppose that $b_0$, $b_1$, $b_2$ are complex numbers, $b_2\ne 0,$ and let $P(z)=b_0+b_1z+b_2z^2.$ Then the zeros of P(z) lie on $\overline{{\mathbb {D}}}=\{z\in {\mathbb {C}}:|z|\le 1\}$ if, and only if

(i) $|b_0|\le |b_2|$ and

(ii) $|b_0 {\bar{b}}_1 -b_1 {\bar{b}}_2| \le |b_2|^2-|b_1|^2.$

Lemma 2.6

[10] Let $\phi$ and $\psi$ be convex functions in ${\mathbb {D}}$ and suppose that f is subordinate to $\phi$. Then $f*\psi$ is subordinate to $\phi *\psi$ in ${\mathbb {D}}.$

$\square$

3 Main Results

Ces$\grave{a}$ro means not only play an important role in areas like approximation theory, summability, Fourier analysis etc., but also find many applications in geometric function theory. One of the striking properties of the polynomial approximations $\sigma _n^{(\alpha )}(z,f)$ is that they converge to f in the sense of compact convergence as $n\rightarrow \infty$. In the following two theorems we establish analogues of Theorem 1.1 for some transformations of $\sigma _n^{(\alpha )}(z,f),\alpha = 2,3.$

Theorem 3.1

For all elements $f \in {\mathscr {K}}$ and for all positive integers n, we have

$$\begin{aligned} \frac{n}{n+1} \sigma _n^{(2)}(z,f) \prec f(z) \end{aligned}$$

(1)

in ${\mathbb {D}}$. This result is sharp for every n.

$\square$

Proof

Let $f(z)= z+\sum \limits _{n=2}^\infty a_n z^n$ be any member of the class ${\mathscr {K}}$. Then

$$\begin{aligned} \frac{n}{n+1} \sigma _n^{(2)}(z,f)=\frac{n}{n+1} z+ \frac{n(n-1)}{(n+1)^2} a_2 z^2+ \frac{(n-1)(n-2)}{(n+1)^2} a_3 z^3 + \cdots + \frac{2.1}{(n+1)^2} a_n z^n. \end{aligned}$$

Thus, in view of the Definition 2.1, the assertion (1) holds if and only if the sequence

$$\begin{aligned} \left\langle \frac{n}{n+1}, \frac{n(n-1)}{(n+1)^2}, \frac{(n-1)(n-2)}{(n+1)^2}, \cdots , \frac{2.1}{(n+1)^2},0,0,\cdots \right\rangle \end{aligned}$$

is a subordinating factor sequence. By Lemma 2.2, this is equivalent to

$$\begin{aligned} \Re F(z)>0, \ \ \ z \in {\mathbb {D}}. \end{aligned}$$

where

$$\begin{aligned} F(z)= & {} 1+ \frac{2}{(n+1)^2} \left\{ (n+1)nz+n(n-1)z^2+(n-1)(n-2)z^3+ \cdots +2.1 z^n \right\} . \nonumber \\= & {} 1+ \frac{2}{(n+1)^2} F_n(z)~ (say). \end{aligned}$$

(2)

Here

$$\begin{aligned} F_n(z)=(n+1)nz+n(n-1)z^2+(n-1)(n-2)z^3+ \cdots +2.1 z^n, \end{aligned}$$

(3)

from which we get,

$$\begin{aligned} (1-z)F_n(z)= & {} (n+1)n z-2n z^2-2(n-1)z^3-\cdots - 2 z^{n+1}. \nonumber \\= & {} (n+1)nz-2z \sum _{k=1}^{n} (n-k+1)z^k. \end{aligned}$$

It compiles to

$$\begin{aligned} F_n(z)=n(n+1)\frac{z}{1-z}-2\frac{z}{1-z}\left[ \sum _{k=1}^{n} (n-k+1)z^k \right] . \end{aligned}$$

Setting $z=e^{i\theta }, 0\le \theta <2\pi$ and making use of Lemma 2.3, we get

$$\begin{aligned} F_n(e^{i \theta })= & {} n(n+1)\left( \frac{e^{i\theta }}{1-e^{i\theta }}\right) -2\left( \frac{e^{i\theta }}{1-e^{i\theta }}\right) \cdot \left[ -\frac{n+1}{2} \right. \nonumber \\{} & {} \left. +\frac{\sin ^2(n+1)\theta /2}{2\sin ^2\theta /2}+i\frac{(n+1)\sin \theta -\sin (n+1)\theta }{4\sin ^2\theta /2}\right] . \end{aligned}$$

(4)

On substituting $\frac{e^{i\theta }}{1-e^{i\theta }}=\left( \frac{-1}{2}+\frac{i}{2} \cot \theta /2\right)$, we get

$$\begin{aligned} F_n(e^{i \theta })= & {} n(n+1)\left( \frac{-1}{2}+\frac{i}{2} \cot \theta /2\right) -2\left( \frac{-1}{2}+\frac{i}{2} \cot \theta /2\right) \cdot \left[ -\frac{n+1}{2} \right. \nonumber \\{} & {} \left. +\frac{\sin ^2(n+1)\theta /2}{2\sin ^2\theta /2}+i\frac{(n+1)\sin \theta -\sin (n+1)\theta }{4\sin ^2\theta /2}\right] . \end{aligned}$$

(5)

From (2), (3) and (5), we obtain

$$\begin{aligned} \Re (F(e^{i \theta }))= & {} 1+ \frac{2}{(n+1)^2}\left\{ \frac{-(n+1)^{2}}{2}+\left[ \frac{\sin ^2(n+1)\theta /2}{2 \sin ^2 \theta /2}\right. \right. \\{} & {} \left. \left. + \frac{(n+1) \sin \theta -\sin (n+1) \theta }{4 \sin ^2 \theta /2}{\cot \theta /2}\right] \right\} . \end{aligned}$$

or

$$\begin{aligned} \Re (F(e^{i \theta }))= \frac{2}{(n+1)^2}\left[ \frac{\sin ^2(n+1)\theta /2}{2 \sin ^2 \theta /2}+ \frac{(n+1) \sin \theta -\sin (n+1) \theta }{4 \sin ^2 \theta /2}{\cot \theta /2}\right] . \end{aligned}$$

We note that ${(n+1) \sin \theta -\sin (n+1) \theta }$ and ${\cot \theta /2}$ are both non-negative for $\theta \in [0,\pi ]$ and are both negative for $\theta \in (\pi , 2\pi ).$ Hence,

$$\begin{aligned} \Re (F(z))>0 \ \ \mathrm{for \ all } \ \ \theta \in [0,2\pi ). \end{aligned}$$

In order to prove the sharpness of our result, we consider the function $h(z)=z/(1-z)$ which is a member of the class ${\mathscr {K}}$. We have

$$\begin{aligned} \sigma _n^{(2)}(z,h)= & {} z+\frac{n(n-1)}{n(n+1)} z^2+ \frac{(n-1)(n-2)}{n(n+1)} z^3+\cdots +\frac{2.1}{n(n+1)}z^n.\nonumber \\= & {} \frac{1}{n(n+1)}F_n(z). \end{aligned}$$

In view of (5), taking $z=e^{i \theta }, 0\le \theta <2\pi$, we get

$$\begin{aligned}{} & {} \sigma _n^{(2)}(e^{i\theta },h)= \frac{1}{n(n+1)}\left\{ n(n+1)\left( \frac{-1}{2}+\frac{i}{2} \cot \theta /2\right) -2\left( \frac{-1}{2}+\frac{i}{2} \cot \theta /2\right) .\right. \\{} & {} \qquad \cdot \left. \left[ -\frac{n+1}{2}+\frac{\sin ^2(n+1)\theta /2}{2\sin ^2\theta /2} +i\frac{(n+1)\sin \theta -\sin (n+1)\theta }{4\sin ^2\theta /2}\right] \right\} . \end{aligned}$$

Taking $\theta = 2 \pi /(n+1)$, for any positive real number $\rho$, we have

$$\begin{aligned} \Re \rho \ \sigma _n^{(2)}(e^{i \theta },h)= & {} -\frac{\rho }{2n}\left\{ (n+1)+\left( 1-\frac{1}{\sin ^{2}(\frac{\pi }{n+1})}\right) \right\} \\\ge & {} -\frac{\rho (n+1)}{2n}, \end{aligned}$$

because for all $n\in {\mathbb {N}},$

$$\begin{aligned} 1-\frac{1}{\sin ^2(\pi /(n+1))}\le 0. \end{aligned}$$

It follows that if $\rho > n/(n+1)$, then $\Re \rho \sigma _n^{(2)}(z,h)< -1/2$ and hence $\rho \sigma _n^{(2)}(z,h)$ will not be subordinate to h in ${\mathbb {D}}$ as h maps ${\mathbb {D}}$ onto the right half plane $\Re w > -1/2$.

This completes the proof.

Fig. 1

Graphs of $h(\partial \mathbb {D})$ and $(n/(n + 1))\sigma _n^{(2)}(\partial {\mathbb {D}},h),$ for $n=2,3,4$

Fig. 2

Enlargement of the critical portion of Fig.1.

For the sake of illustration, graphs of $h(\partial {\mathbb {D}})$ and $(n/(n+1))\sigma _n^{(2)}(\partial {\mathbb {D}},h), n=2,3,4$ are plotted in Fig. 1, and Fig. 2 is an enlargement of the critical portion of Fig. 1. $\square$

Theorem 3.2

For all $f \in {\mathscr {K}}$ and for all positive integers n, we have

$$\begin{aligned} \frac{2n(n+2)}{(2n+1)(n+3)} \sigma _n^{(3)}(z,f) \prec f(z) \end{aligned}$$

(6)

in ${\mathbb {D}}$. This result is sharp for every n.

$\square$

Proof

For $f(z)= z+\sum \limits _{n=2}^\infty a_n z^n$, we have

$$\begin{aligned} \frac{2n(n+2)}{(2n+1)(n+3)} \sigma _n^{(3)}(z,f)= & {} \frac{2}{(2n+1)(n+3)(n+1)}\\{} & {} \left[ (n+2)(n+1) nz +(n+1) n (n-1)z^2\right. \\{} & {} \left. + n(n-1)(n-2)z^3+\ldots + 3.2.1 z^n\right] . \end{aligned}$$

The assertion (6) will hold, if

$$\begin{aligned}{} & {} \left\langle \frac{2n(n+1)(n+2)}{(2n+1)(n+3)(n+1)}, \frac{2n(n+1)(n-1)}{(2n+1)(n+3)(n+1)}, \frac{2n(n-1)(n-2)}{(2n+1)(n+1)(n+3)},\cdots ,\right. \\{} & {} \quad \left. \frac{3.2.1}{(2n+1)(n+1)(n+3)},0,0,\cdots \right\rangle \end{aligned}$$

is a subordinating factor sequence. By Lemma 2.2, it is equivalent to show that

$$\begin{aligned} \Re G(z) >0 \end{aligned}$$

for all z in ${\mathbb {D}},$ where

$$\begin{aligned} G(z)= & {} 1+\frac{4}{(2n+1)(n+3)(n+1)} \left\{ (n+2)(n+1) nz +(n+1) n (n-1)z^2\right. \nonumber \\{} & {} \left. +(n+1) n (n-1)z^2+ n(n-1)(n-2)z^3+\ldots + 3.2.1 z^n\right\} . \end{aligned}$$

(7)

If we write

$$\begin{aligned} G_n(z)= (n+2)(n+1)nz+(n+1)n(n-1)z^2+n(n-1)(n-2)z^3+ \cdots +3.2.1 z^n , \end{aligned}$$

then

$$\begin{aligned} (1-z)G_n(z)= & {} (n+2)(n+1) nz-3(n+1)nz^2-3n(n-1)z^3- \ldots -3.2.1z^{n+1} \\= & {} (n+2)(n+1)nz-3z \left[ (n+1)nz+n(n-1)z^2+ \ldots + 2.1z^n \right] . \end{aligned}$$

Thus, we obtain

$$\begin{aligned} G(z)=1+ \frac{4}{(n+1)(n+3)(2n+1)}\{n(n+1)(n+2)\frac{z}{1-z} -3\frac{z}{1-z}F_n(z)\}, \end{aligned}$$

where $F_n$ is as in (3). Setting $z=e^{i\theta },~ 0\le \theta <2\pi$ and using (4), we get

$$\begin{aligned}{} & {} G(e^{i\theta })=1+ \frac{4}{(n+1)(n+3)(2n+1)}\left\{ n(n+1)(n+2)\frac{e^{i\theta }}{1-e^{i\theta }}\right. \\{} & {} \quad \left. -3\frac{e^{i\theta }}{1-e^{i\theta }}\left\{ n(n+1)\left( \frac{e^{i\theta }}{1-e^{i\theta }}\right) \right. \right. \\{} & {} \quad \left. \left. -2\left( \frac{e^{i\theta }}{1-e^{i\theta }}\right) \left[ -\frac{n+1}{2}+\frac{\sin ^2 (n+1)\theta /2}{2 \sin ^2 \theta /2}+i \frac{(n+1) \sin \theta -\sin (n+1) \theta }{4 \sin ^2 \theta /2}\right] \right\} \right\} . \end{aligned}$$

On substituting $\frac{e^{i\theta }}{1-e^{i\theta }}=\left( -\frac{1}{2}+\frac{i}{2}\cot \theta /2\right)$ and $(\frac{e^{i\theta }}{1-e^{i\theta }})^{2}= \frac{-e^{i\theta }}{4\sin ^{2}\theta /2}$, we get

$$\begin{aligned}{} & {} G(e^{i\theta })=1+ \frac{4}{(n+1)(n+3)(2n+1)}\left\{ n(n+1)(n+2)\left( \frac{-1}{2}+\frac{i}{2} \cot \theta /2 \right) \right. \\{} & {} \quad \left. +3(n+1)^2\left( \frac{e^{i \theta }}{4 \sin ^2 \theta /2}\right) \right. \\{} & {} \quad \left. -6\left( \frac{e^{i \theta }}{4 \sin ^2 \theta /2}\right) \frac{\sin ^2 (n+1)\theta /2}{2 \sin ^2 \theta /2}-6\left( i\frac{e^{i \theta }}{4 \sin ^2 \theta /2}\right) \frac{(n+1) \sin \theta -\sin (n+1)\theta }{4 \sin ^2 \theta /2}\right\} . \end{aligned}$$

Writing the real part, we have

$$\begin{aligned} \Re G(e^{i\theta })= & {} 1+ \frac{4}{(n+1)(n+3)(2n+1)}\left\{ \frac{-n(n+1)(n+2)}{2} +\frac{3}{4}(n+1)^2\frac{\cos \theta }{\sin ^{2}\theta /2}\right. \\{} & {} \left. -\frac{6}{8}\frac{\cos \theta \sin ^{2}(n+1)\theta /2}{\sin ^{4}\theta /2} +\frac{6}{16}\frac{\sin \theta [(n+1)\sin \theta -\sin (n+1)\theta ]}{\sin ^{4}\theta /2}\right\} . \end{aligned}$$

Writing $\sin ^2(n+1) \theta /2=(1-\cos (n+1) \theta )/2$ and regrouping the terms, we have

$$\begin{aligned} \Re G(e^{i\theta })= & {} 1+ \frac{4}{(n+1)(n+3)(2n+1)}\left\{ \frac{-n(n+1)(n+2)}{2}+ \frac{3}{4} (n+1)^2 (-1+ \cot ^2 \theta /2)\right. \\{} & {} \left. +\frac{6}{16} \frac{\cos (n+2)\theta }{\sin ^4 \theta /2}-\frac{6}{16} \frac{\cos \theta }{\sin ^4 \theta /2}+\frac{6}{4} (n+1) \cot ^2 \theta /2 \right\} , \end{aligned}$$

i.e.,

$$\begin{aligned} \Re G(e^{i\theta })= & {} 1+ \frac{4}{(n+1)(n+3)(2n+1)}\left\{ \frac{-(n+1)(n+3)(2n+1)}{4}\right. \\{} & {} \left. +\frac{3}{4}(n+1)(n+3)\cot ^{2}\frac{\theta }{2}+\frac{3}{8}\frac{\cos (n+2)\theta -\cos \theta }{\sin ^{4}\theta /2}\right\} , \end{aligned}$$

or

$$\begin{aligned} \Re G(e^{i\theta })= & {} 1+ \frac{4}{(n+1)(n+3)(2n+1)}\left\{ \frac{-(n+1)(2n+1)(n+3)}{4}\right. \\{} & {} \left. +\frac{3}{2}\cot ^{2}\frac{\theta }{2}\left( \frac{(n+3)(n+1)}{2} +\frac{\cos (n+2)\theta -\cos \theta }{\sin ^{2}\theta }\right) \right\} , \end{aligned}$$

which further simplifies to

$$\begin{aligned} \Re G(e^{i\theta })= & {} \frac{4}{(n+1)(n+3)(2n+1)}\left\{ \frac{3}{2}\cot ^{2}\frac{\theta }{2} \left( \frac{(n+3)(n+1)}{2}\right. \right. \\{} & {} \left. \left. +2\frac{\cos ^{2}(n+2)\theta /2+\sin ^{2}\theta /2}{\sin ^{2}\theta }\right) \right\} . \end{aligned}$$

All terms on the right-hand side of above expression are positive. So, we have

$$\begin{aligned} \Re G(z)>0 \end{aligned}$$

for all z in ${\mathbb {D}}.$

Thus, the desired result holds.

To prove the sharpness of our result, we consider the function $h(z)=z/(1-z)$ which is a member of the class ${\mathscr {K}}$. We have

$$\begin{aligned} \sigma _n^{(3)}(z,h)=z+\frac{(n+1)n(n-1)}{n(n+1)(n+2)} z^2+ \frac{n(n-1)(n-2)}{n(n+1)(n+2)} z^3+\cdots \frac{3.2.1}{n(n+1)(n+2)}z^n. \end{aligned}$$

Setting $z=e^{i \theta }$, we get

$$\begin{aligned} \sigma _n^{(3)}(e^{i \theta },h)= & {} \frac{1}{n(n+1)(n+2)}[(n+2)(n+1)ne^{i \theta }+(n+1)n(n-1) e^{2i \theta }\\{} & {} + n(n-1)(n-2)e^{3i \theta }+\cdots 3.2.1e^{ni \theta }]. \end{aligned}$$

For a positive real number $\rho$, we have

$$\begin{aligned} \rho \sigma _n^{(3)}(e^{i\theta },h)= & {} \frac{\rho }{n(n+1)(n+2)}\left\{ n(n+1)(n+2) \frac{e^{i\theta }}{1-e^{i\theta }}-3\frac{e^{i\theta }}{1-e^{i\theta }} \left\{ (n+1)^2\frac{e^{i\theta }}{1-e^{i\theta }}\right. \right. \\{} & {} \left. \left. -2\frac{e^{i\theta }}{1-e^{i\theta }}\left( \frac{\sin ^2(n+1)\theta /2}{2 \sin ^2 \theta /2}+i \frac{(n+1) \sin \theta -\sin (n+1) \theta }{4 \sin ^2 \theta /2}\right) \right\} \right\} . \end{aligned}$$

Simplifying, we obtain

$$\begin{aligned} \Re \rho \sigma _n^{(3)}(e^{i\theta },h)= & {} \frac{\rho }{n(n+1)(n+2)}\left\{ \frac{-(n+1)(n+3)(2n+1)}{4}\right. \\{} & {} \left. +\frac{3}{4}(n+1)(n+3)\cot ^{2}\frac{\theta }{2}+\frac{3}{8}\frac{\cos (n+2)\theta -\cos \theta }{\sin ^{4}\theta /2}\right\} . \end{aligned}$$

For $\theta = \pi$, we get

$$\begin{aligned} \Re \rho \ \sigma _n^{(3)}(e^{i\theta },h)= & {} \frac{\rho }{n(n+1)(n+2)} \left\{ \frac{-(n+1)(n+3)(2n+1)}{4}\right. \nonumber \\{} & {} \left. +\frac{3}{8}(1+\cos (n+2)\pi )\right\} \end{aligned}$$

(8)

$$\begin{aligned}\ge & {} -\frac{\rho (n+3)(2n+1)}{n(n+2)}. \end{aligned}$$

(9)

It follows that if $\rho > 2n(n+2)/((n+3)(2n+1))$, then $\displaystyle \Re \rho \sigma _n^{(3)}(z,h)< -1/2$ and hence $\displaystyle \rho \sigma _n^{(3)}(z,h)$ will not be subordinate to h in $\displaystyle {\mathbb {D}}$ as h maps $\displaystyle {\mathbb {D}}$ onto the right half plane $\displaystyle \Re w > -1/2$.

This completes the proof.

Fig. 3

Graphs of $h(\partial \mathbb {D}) \text {\,\,and\,\,} 2n(n + 2)/((2n + 1)(n + 3))\sigma _n^{(3)}(\partial {\mathbb {D}},h), for~n=2,3,4$

Fig. 4

Enlargement of the critical portion of Fig. 3.

For visual illustration, graphs of $h(\partial {\mathbb {D}})$ and $2n(n+2)/((2n+1)(n+3))\sigma _n^{(3)}(\partial {\mathbb {D}},h)$, $n=2,3,4,$ are plotted in Fig. 3, and Fig. 4 is an enlargement of the critical portion of Fig. 3. $\square$

Ruscheweyh [11] proved that $\displaystyle \sigma _n^{(3)}(z,z/(1-z))\in {\mathscr {K}}\subset {\mathscr {S}}^*(1/2)$. Since the class $\displaystyle {\mathscr {S}}^*(1/2)$ is closed under convolution (see [10], Theorem 3.1), so $\displaystyle \sigma _n^{(3)}(z,f)= \sigma _n^{(3)}(z,z/(1-z))*f\in {\mathscr {S}}^{*}(1/2)$ for every $\displaystyle f\in {\mathscr {S}}^*(1/2).$ In the next theorem we establish that $\sigma _n^{(2)}(z,f)$ in Theorem 1.3 can be replaced with $\sigma _n^{(3)}(z,f)$.

Theorem 3.3

Let $\displaystyle f(z)=z+ \sum _{n=2}^\infty a_n z^n$ be a member of the class $\displaystyle {\mathscr {S}}^{*}(1/2)$. Then for every positive integer n and each $\displaystyle z \in {\mathbb {D}}$, we have

$$\begin{aligned} {\Re } \frac{V_{n}(z,f)}{\sigma ^{(3)}_{n}(z,f)} >0. \end{aligned}$$

$\square$

Proof

Consider the function

$$\begin{aligned} F_{n}(z)= & {} (1-z)\left[ \frac{n}{(n+1)}+\frac{n}{(n+1)}z+\frac{n}{(n+3)}z^{2}\right. \nonumber \\{} & {} \left. +\frac{n^{2}}{(n+3)(n+4)}z^{3} +\frac{n^{2}(n-1)}{(n+3)(n+4)(n+5)}z^{4}\right. \nonumber \\{} & {} \left. +...+\frac{n^{2}(n-1)(n-2)(n-3)...4}{(n+3)(n+4)(n+5)...(2n)}z^{n-1}\right] . \end{aligned}$$

(10)

Then

$$\begin{aligned} \sigma _n^{(3)}(z,f)* \frac{z}{1-z} F_n(z)= & {} \left\{ z+ \frac{n-1}{n+2} z^2+ \frac{(n-1)(n-2)}{(n+2)(n+1)}z^3+ \ldots \right. \\{} & {} \left. +\frac{3.2.1}{(n+2)(n+1)n} z^n\right\} \\{} & {} * \left\{ \frac{n}{(n+1)}z+\frac{n}{(n+1)}z^2+\frac{n}{(n+3)}z^{3}\right. \\{} & {} \left. +\frac{n^{2}}{(n+3)(n+4)}z^{4} + \ldots +\frac{n^{2}(n-1)(n-2)(n-3)...4}{(n+3)(n+4)(n+5)...(2n)}z^n \right\} \\{} & {} =\frac{n}{n+1}z+\frac{n(n-1)}{(n+1)(n+2)} z^2+ \frac{n(n-1)(n-2)}{(n+1)(n+2)(n+3)}z^3\\{} & {} + \ldots \frac{n(n-1)(n-2) \ldots 4.3.2.1}{(n+1)(n+2) \ldots (2n)} z^n.\\{} & {} = V_n(z,f). \end{aligned}$$

Also, the function $F_{n}$ defined above is regular (in fact, an entire function) in ${\mathbb {D}}$ and can be written in the form

$$\begin{aligned} F_{n}(z)= & {} \left[ \frac{n}{(n+1)}-\frac{n}{(n+1)}\left( 1-\frac{n+1}{n+3}\right) z^{2} -\frac{n}{(n+3)}\left( 1-\frac{n}{n+4}\right) z^{3}\right. \nonumber \\{} & {} \left. -\frac{n^{2}}{(n+3)(n+4)} \left( 1-\frac{n-1}{n+5}\right) z^{4}-\right. \nonumber \\{} & {} \left. ...-\frac{n^{2}(n-1)(n-2)...5}{(n+3)(n+4)...(2n-1)}\left( 1-\frac{4}{2n}\right) z^{n-1} -\frac{n^{2}(n-1)(n-2)...4}{(n+3)(n+4)...(2n)}z^{n}\right] .\nonumber \\ \end{aligned}$$

(11)

In view of (10) and (11), it is clear that in ${{\mathbb {D}}}$, we have

$$\begin{aligned} {\Re } F_{n}(z) \ge F_{n}(|z|) > F_{n}(1)=0. \end{aligned}$$

Taking $\displaystyle f(z)=\sigma ^{(3)}_{n}(z,f), \ g(z)=z/(1-z)$ and $F(z)=F_{n}(z)$ in Lemma 2.4, we immediately get

$$\begin{aligned} {\Re }\frac{\sigma ^{(3)}_{n}(z,f)* z/(1-z) F_n(z)}{\sigma ^{(3)}_{n}(z,f)* z/(1-z)}= \ {\Re } \ \frac{V_{n}(z,f)}{\sigma ^{(3)}_{n}(z,f)} >0,\ z\in {\mathbb {D}}, \end{aligned}$$

(12)

as $\Re F_n(z) >0$ in ${\mathbb {D}}$.

This completes the proof. $\square$

It is known (see [6]) that for $\alpha \ge 1$ and $n\in {\mathbb {N}}$, $\sigma ^{(\alpha )}_{n}(z, z/(1-z))\in {\mathscr {C}}$. Then, using the fact that the class ${\mathscr {C}}$ is closed under convolution with convex functions (see [10], Theorem 2.2), we immediately get that for $\alpha \ge 1$ and $n\in {\mathbb {N}}$, $\sigma ^{(\alpha )}_{n}(z,f)\in {\mathscr {C}}$ for all $f\in {\mathscr {K}}.$ Singh and Singh [13] proved that if $f\in {\mathscr {K}}$, then $z/2\prec V_2(z,f)\prec \sigma _2^{(1)}(z,f)$ in ${\mathbb {D}}$. The theorem below generalizes this result of Singh and Singh [13] in the sense that the superordinate function $\sigma _2^{(1)}(z,f)$ can be replaced with $\sigma _2^{(\alpha )}(z,f)$, where $\alpha ,$ $\alpha \ge 1,$ is any real number.

Theorem 3.4

If $f \in {\mathscr {K}}$, then for all real numbers $\alpha$, $\alpha \ge 1$,

$$\begin{aligned} z/2\prec V_{2}(z,f) \prec \sigma _{2}^{(\alpha )}(z,f) \end{aligned}$$

in ${\mathbb {D}}$.

$\square$

Proof

We note that for $f(z)=z+\sum _{n=2}^{\infty }a_nz^n$,

$$\begin{aligned} V_2(z,f)=\frac{2}{3}z+\frac{1}{6}a_2z^2 \end{aligned}$$

and

$$\begin{aligned} \sigma _2^{(\alpha )}(z,f)=z+\frac{1}{1+\alpha }a_2z^2. \end{aligned}$$

For every $f\in {\mathscr {K}}$, the relation, $z/2\prec V_{2}(z,f)$, is well known. As $\sigma _{2}^{(\alpha )}(z,f)$ is univalent in ${\mathbb {D}}$ for $\alpha \ge 1$ and $V_2(0,f)=\sigma _2^{(\alpha )}(0,f)$, we need to show only that for all $f\in {\mathscr {K}}$,

$$\begin{aligned} V_{2}({\mathbb {D}},f) \subset \sigma _{2}^{(\alpha )}({\mathbb {D}},f). \end{aligned}$$

But this is equivalent to showing that for each real $\theta$, the polynomial

$$\begin{aligned} R(z)= z+\frac{1}{1+\alpha }a_2 z^2-\frac{2}{3}e^{i\theta }-\frac{1}{6}a_{2}e^{2i\theta } \end{aligned}$$

has a zero on $\overline{{\mathbb {D}}}$. Suppose that for some $\theta ,$ R(z) has no zero in ${\mathbb {D}}.$ Then the polynomial

$$\begin{aligned} T(z)= \left( \frac{2}{3}e^{i\theta }+\frac{1}{6}a_{2}e^{2i\theta }\right) z^2 -z-\frac{a_2}{1+\alpha } \end{aligned}$$

has both zeros on $\overline{{\mathbb {D}}}$. Hence by Lemma 2.5, condition (i), we must have

$$\begin{aligned} \left| \frac{a_2}{1+\alpha }\right| \le \frac{2}{3}\left| 1+\frac{a_2 e^{i\theta }}{4}\right| \end{aligned}$$

Writing $a_2=\rho e^{i\phi }(\rho \le 1)$ and $\phi +\theta =\psi$, this is equivalent to

$$\begin{aligned} \frac{\rho }{1+\alpha } \le \frac{2}{3}\left| 1+\frac{\rho e^{i\psi }}{4}\right| \end{aligned}$$

(13)

From condition (ii) of Lemma 2.5, we must have

$$\begin{aligned} \left| \frac{a_2}{1+\alpha }+\frac{2}{3}e^{-i\theta }+\frac{{\bar{a}}_2}{6}e^{-2i\theta }\right| \le \left| \frac{2}{3}e^{i\theta }+\frac{{\bar{a}}_2}{6}e^{2i\theta } \right| ^2 - \left| \frac{a_2}{1+\alpha } \right| ^2. \end{aligned}$$

Again, writing $a_2=\rho e^{i\phi }$, $\rho \le 1,$ and $\phi +\theta =\psi$, this is equivalent to

$$\begin{aligned} 6\left| \frac{6\rho }{1+\alpha }e^{i\psi }+4+\rho e^{-i\psi } \right| \le \left| 4+\rho e^{i\psi } \right| ^2-\frac{36\rho ^2}{(1+\alpha )^2} \end{aligned}$$

or,

$$\begin{aligned} 6\left| \frac{6\rho }{1+\alpha }e^{i\psi }+4+\rho e^{-i\psi } \right| \le \left( \left| 4+\rho e^{i\psi } \right| -\frac{6\rho }{1+\alpha }\right) \left( \left| 4+\rho e^{i\psi } \right| +\frac{6\rho }{1+\alpha }\right) .\nonumber \\ \end{aligned}$$

(14)

As

$$\begin{aligned} \left| \frac{6\rho }{1+\alpha }e^{i\psi }+4+\rho e^{-i\psi } \right| \ge \left| 4+\rho e^{-i\psi }\right| -\left| \frac{6\rho }{1+\alpha }e^{i\psi }\right| =\left| 4+\rho e^{i\psi }\right| -\frac{6\rho }{1+\alpha }, \end{aligned}$$

therefore, if (14) holds, we must have

$$\begin{aligned} \left( \left| 4+\rho e^{i\psi } \right| -\frac{6\rho }{1+\alpha }\right) \left( \left| 4+\rho e^{i\psi } \right| +\frac{6\rho }{1+\alpha }-6\right) \ge 0. \end{aligned}$$

(15)

Obviously, minimum of left-hand side of (15) occurs at $\psi =\pi$; so, we must have

$$\begin{aligned} \left( \rho -\frac{4(1+\alpha )}{7+\alpha }\right) \left( \rho -\frac{2(1+\alpha )}{5-\alpha }\right) \le 0. \end{aligned}$$

(16)

Now, at $\psi =\pi$, (13) gives: $\rho \le 4(1+\alpha )/(7+\alpha ).$ But $4(1+\alpha )/(7+\alpha )\ge 1$ for $\alpha \ge 1$ and also for $1\le \alpha \le 5$, $2(1+\alpha )/(5-\alpha )\ge 1$. As $\rho \le 1$, (16) gives a contradiction (and therefore, R(z) has a zero in ${\mathbb {D}})$ except if $\rho = 1 ~(\alpha =1), \psi =\pi$ and $\alpha >5.$ When $\rho =1$ and $\psi =\pi$, we easily verify that $-e^{-i\phi }$ is a zero of R(z) on $\overline{{\mathbb {D}}}$. For $\alpha >5$, we proceed as follows. If $I(z)=z/(1-z)$, then

$$\begin{aligned} \max _{\theta }\left| V_2(e^{i\theta },I)\right| =\max _{\theta }\sqrt{\frac{4}{9}+\frac{1}{36}+\frac{2 \cos \theta }{9}}=\frac{5}{6}, \end{aligned}$$

and

$$\begin{aligned} \min _{\theta } |\sigma _2^{(\alpha )}(e^{i\theta },I)|=\min _{\theta }\sqrt{1+\frac{1}{(1+\alpha )^2}+\frac{2 \cos \theta }{1+\alpha }} =1-\frac{1}{(1+\alpha )}>\frac{5}{6} \end{aligned}$$

as $\alpha > 5.$ Thus, $V_2({\mathbb {D}},I)$ is contained in the disc $\{z:|z|\le 5/6\}$ and for $\alpha >5$, $\sigma _2^{(\alpha )}({\mathbb {D}},I)$ contains the disc $\{z:|z|\le 5/6\}.$ Therefore, $V_2(z,I)\prec \sigma _2^{(\alpha )}(z,I),\;\;\alpha >5.$ As, $\sigma _2^{(\alpha )}(z,I)$ is convex for $\alpha >5$ (infact, for $\alpha \ge 3$, see [11]), using Lemma 2.6, we immediately conclude that $V_2(z,f)\prec \sigma _2^{(\alpha )}(z,f)$ in ${\mathbb {D}},$ for all $\alpha >5$ and $f\in {\mathscr {K}}.$

This completes the proof. $\square$