Sharpening of Turán type inequalities for polar derivative of a polynomial

Abstract

Let p(z) be a polynomial of degree n. The polar derivative of p(z) with respect to a complex number $\alpha$ is defined by

$$\begin{aligned} D_{\alpha }p(z)=np(z)+(\alpha -z) p'(z). \end{aligned}$$

If $p(z)=z^s\displaystyle {\sum _{j=0}^{n-s}c_jz^j}$, $0\le s\le n,$ has all its zeros in $|z| \le k, k\ge 1$, then for $|\alpha |\ge k$, Kumar and Dhankhar [Bull, Math. Soc. Sci. Math., 63(4), 359-367 (2020)] proved

$$\begin{aligned}&\max _{ |z |=1} |D_{\alpha }p(z) |\ge \frac{n( |\alpha |-k)}{1+k^{n-s}}\left( 1+\frac{( |c_{n-s} |k^n- |c_0 |k^s)(k-1)}{2( |c_{n-s} |k^n+ |c_0 |k^{s+1})}\right) \max _{ |z |=1} |p(z) |. \end{aligned}$$

In this paper, we first improve the above inequality. Besides, we are able to prove an improvement of a result due to Govil and Mctume [Acta. Math. Hungar., 104, 115-126 (2004)] and also prove an inequality for a subclass of polynomials having no zero in $|z |< k, k\le 1$.

1 Introduction

Let $p(z)=\displaystyle {\sum _{j=0}^{n}c_jz^j}$ be a polynomial of degree n over the set of complex numbers. We will use q(z) to represent the polynomial $z^n\overline{p\left( \frac{1}{\bar{z}}\right) }$.

According to the famous Bernstein’s inequality [6],

$$\begin{aligned} \max _{ |z |=1}|p'(z)|\le n\max _{ |z |=1} |p(z) |. \end{aligned}$$

(1)

Equality in (1) holds for $p(z)=\alpha z^n, \alpha \ne 0$.

If we restrict the zeros of p(z), inequality (1) can be refined. In this direction, Erdös conjectured and later Lax [19, p. 1] proved that if p(z) is a polynomial of degree n having no zero in $|z |< 1$, then

$$\begin{aligned} \max _{ |z |=1} |p'(z) |\le \frac{n}{2}\max _{ |z |=1} |p(z) |. \end{aligned}$$

(2)

Inequality (2) is best possible for $p(z)=\alpha +\beta z^n$, where $|\alpha |= |\beta |$.

It was R. P. Boas who asked that if p(z) is a polynomial of degree n not vanishing in $|z |<k$, $k>0$, then how large

$$\begin{aligned} \left\{ \max _{ |z |=1} |p'(z) |\big | \bigg /\max _{ |z |=1} |p(z) |\right\} ~~~\text {can be ?} \end{aligned}$$

A partial answer to this problem was given by Malik [20, Theorem, p. 58], who proved that if p(z) is a polynomial of degree n having no zeros in $|z |<k$, $k\ge 1$, then

$$\begin{aligned} \max _{ |z |=1} |p'(z) |\le \frac{n}{1+k}\max _{ |z |=1} |p(z) |. \end{aligned}$$

(3)

In the literature, there exist generalizations and improvements of inequality (3), for brief understanding one can refer to: Chan and Malik [8], Qazi [21], Bidkham and Dewan [7], Aziz and Zargar [4], Chanam and Dewan [9], Aziz and Shah [3] etc.

On the other hand, for the class of polynomials p(z) such that $p(z)\ne 0$ for $|z |<k$, $k\le 1$, the precise estimate for maximum of $|p'(z) |$ on $|z |=1$ does not seem to be easily obtainable. For quit some time, it was believed that the inequality analogous to (3) for $p(z)\ne 0$ in $|z |<k$, $k\le 1$, should be

$$\begin{aligned} \max _{ |z |=1} |p'(z) |\le \frac{n}{1+k^n}\max _{ |z |=1} |p(z) |, \end{aligned}$$

(4)

till E. B. Saff gave the example $p(z)=\left( z-\frac{1}{2}\right) \left( z+\frac{1}{3}\right)$ to counter this belief.

With extra assumption inequality (4) could be satisfied. In this direction, Govil [11] proved that if p(z) is a polynomial of degree n having no zero in $|z |<k$, $k\le 1$, with additional hypothesis that $|p'(z) |$ and $|q'(z) |$ attain their maxima at the same point on $|z |=1$, then

$$\begin{aligned} \max _{ |z |=1} |p'(z) |\le \frac{n}{1+k^{n}}\max _{ |z |=1} |p(z) |. \end{aligned}$$

(5)

Under the same set of hypothesis, Kumar and Dhankar [18, Theorem 2] further improved inequality (5) by proving

$$\begin{aligned}{} & {} \max _{|z |=1} |p'(z) |\nonumber \\{} & {} \quad \le \frac{n}{1+k^{n}}\left\{ 1-\frac{k^n\left( |c_0 |- |c_n |k^n\right) (1-k)}{2\left( |c_0 |k+ |c_n |k^n\right) } \right\} \max _{ |z |=1} |p(z) |.\nonumber \\ \end{aligned}$$

(6)

Another improvement of (5) was also recently obtained by Singh and Chanam [23, Theorem 3] by proving

$$\begin{aligned}{} & {} \max _{ |z |=1} |p'(z) |\nonumber \\{} & {} \quad \le \left[ \frac{n}{1+k^n}-\frac{\left( \sqrt{ |c_0 |}-k^{\frac{n}{2}}\sqrt{ |c_n |}\right) k^n}{(1+k^n)\sqrt{ |c_0 |}}\right] \max _{ |z |=1} |p(z) |.\nonumber \\ \end{aligned}$$

(7)

In 1939, Turán [26] provided a lower bound estimate of the derivative to the size of the polynomial by restricting its zeros, and proved that if p(z) has all its zeros in $|z |\le 1$, then

$$\begin{aligned} \max _{ |z |=1}|p'(z) |\ge \frac{n}{2}\max _{ |z |=1} |p(z) |. \end{aligned}$$

(8)

Aziz and Dawood [1, Theorem 4] further refined inequality (8) by involving $\displaystyle {\min _{ |z |=1} |p(z) |}$. In fact, they proved

$$\begin{aligned} \max _{ |z |=1}|p'(z) |\ge \frac{n}{2}\left\{ \max _{ |z |=1} |p(z) |+\min _{ |z |=1} |p(z) |\right\} . \end{aligned}$$

(9)

Both the inequalities (8) and (9) are best possible and equality holds if p(z) has all its zeros on $|z |=1$.

Inequalities (8) and (9) have been extended and generalized in different directions (see [3, 5, 12,13,14]). For polynomial p(z) having all its zeros in $|z |\le k, k\ge 1,$ Govil [12, Theorem, p. 544] proved that

$$\begin{aligned} \max _{ |z |=1}|p'(z) |\ge \frac{n}{1+k^n} \max _{ |z |=1} |p(z) |. \end{aligned}$$

(10)

Further, as an improvement of (10) and a generalization of (9), Govil [13, Theorem 2] proved

$$\begin{aligned} \max _{ |z |=1}|p'(z) |\ge \frac{n}{1+k^n} \max _{ |z |=1} |p(z) |+\frac{n}{1+k^n}\min _{ |z |=k} |p(z) |. \end{aligned}$$

(11)

Inequalities (10) and (11) are sharp and equality holds for $p(z)=z^n+k^n$.

The concept of ordinary derivative of a polynomial has been generalized to polar derivative of a polynomial as follows:

If p(z) is a polynomial of degree n and $\alpha$ be any real or complex number, the polar derivative of p(z) with respect to $\alpha$, denoted by $D_{\alpha }p(z)$, is defined as

$$\begin{aligned} D_{\alpha }p(z)=np(z)+(\alpha -z) p'(z). \end{aligned}$$

It is easy to see that $D_{\alpha }p(z)$ is a polynomial of degree at most $n-1$ and it generalizes the ordinary derivative in the sense that

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }\left[ \frac{D_{\alpha }p(z)}{\alpha }\right] = p'(z). \end{aligned}$$

Shah [22] extended inequality (8) to the polar derivative and proved that if p(z) is a polynomial of degree n having all its zeros in $|z |\le 1,$ then for any complex number $\alpha$ with $|\alpha |\ge 1$

$$\begin{aligned} \max _{ |z |=1} |D_{\alpha }p(z) |\ge \frac{n( |\alpha |-1)}{2}\max _{ |z |=1} |p(z) |. \end{aligned}$$

(12)

Recently, Gulzar et al. [17, Theorem 2.1] refined inequality (12) and proved that if $p(z)=\displaystyle {\sum _{j=0}^{n}c_jz^j}$ is a polynomial of degree n having all its zeros in $|z |\le 1,$ then for any complex number $\alpha$ with $|\alpha |\ge 1$ and $|z |=1$

$$\begin{aligned} |D_{\alpha }p(z) |\ge \frac{( |\alpha |-1)}{2}\left( n+\frac{\sqrt{ |c_n |}-\sqrt{ |c_0 |}}{\sqrt{ |c_n |}}\right) |p(z) |. \end{aligned}$$

(13)

In 1998, Aziz and Rather [2, Theorem 2] extended inequality (10) to polar derivative by proving that if p(z) is a polynomial of degree n having all its zeros in $|z |\le k, k\ge 1$, then for every complex number $\alpha$ with $|\alpha |\ge k$,

$$\begin{aligned} \max _{ |z |=1} |D_{\alpha }p(z) |\ge n\left( \frac{ |\alpha |-k}{1+k^n}\right) \max _{ |z |=1} |p(z) |. \end{aligned}$$

(14)

Recently, Kumar and Dhankhar [18, Theorem 3] obtained a generalization as well as improvement of (14) by establishing that if $p(z)=z^s\displaystyle {\sum _{j=0}^{n-s}c_jz^j},0\le s\le n,$ is a polynomial of degree n having all its zeros in $|z |\le k, k\ge 1,$ then for any complex number $\alpha$ with $|\alpha |\ge k,$

$$\begin{aligned}{} & {} \max _{ |z |=1} |D_{\alpha }p(z) |\nonumber \\ {}{} & {} \quad \ge \frac{n( |\alpha |-k)}{1+k^{n-s}}\left( 1+\frac{( |c_{n-s} |k^n- |c_0 |k^s)(k-1)}{2( |c_{n-s} |k^n+ |c_0 |k^{s+1})}\right) \nonumber \\{} & {} \qquad \times \max _{|z |=1} |p(z) |. \end{aligned}$$

(15)

With the same hypothesis, Singh and Chanam [23, Theorem 1] provided another improvement of (14) and a generalization of (13) and obtained

$$\begin{aligned}{} & {} \max _{ |z |=1} |D_{\alpha }p(z) |\nonumber \\ {}{} & {} \quad \ge \frac{\left( |\alpha |-k\right) }{1+k^n}\left( n+s+\frac{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}-\sqrt{ |c_0 |}}{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}}\right) \nonumber \\{} & {} \qquad \times \max _{ |z |=1} |p(z) |. \end{aligned}$$

(16)

Govil and Mctume [15, Theorem 3] extended inequality (11) to polar derivative and proved

$$\begin{aligned}{} & {} \max _{ |z |=1} |D_{\alpha }p(z) |\nonumber \\ {}{} & {} \quad \ge n\left( \frac{ |\alpha |-k}{1+k^n}\right) \max _{ |z |=1} |p(z) |+n\left( \frac{ |\alpha |-(1+k+k^n)}{1+k^n}\right) \nonumber \\{} & {} \qquad \times \min _{ |z |=k} |p(z) |, \end{aligned}$$

(17)

where $\alpha$ is any complex number with $|\alpha |\ge 1+k+k^n$.

Improvements of inequality (17) by involving leading coefficient and constant term of the polynomial can be seen in recent works of Singh and Chanam [23, Theorem 2] and Singh et al. [24, Theorem 4].

2 Main results

We begin by presenting the following refinement of inequality (15) and inequality (16).

Theorem 1

If $p(z)=z^s\displaystyle {\sum _{j=0}^{n-s}c_jz^j},0\le s\le n,$ is a polynomial of degree n having all its zeros in $|z |\le k, k\ge 1,$ then for any complex number $\alpha$ with $|\alpha |\ge k$,

$$\begin{aligned}{} & {} \max _{ |z |=1} |D_{\alpha }p(z) |\nonumber \\{} & {} \quad \ge \left( \frac{ |\alpha |-k}{1+k^{n-s}}\right) \left( n+s+\frac{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}-\sqrt{ |c_0 |}}{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}}\right) \nonumber \\{} & {} \qquad \times \left\{ 1+\frac{\left( |c_{n-s} |k^n- |c_0 |k^s\right) (k-1)}{2\left( |c_{n-s} |k^n+ |c_0 |k^{s+1}\right) }\right\} \max _{ |z |=1} |p(z) |. \end{aligned}$$

(18)

Remark 1

Since the polynomial $h(z)=\frac{p(z)}{z^s}=\displaystyle {\sum _{j=0}^{n-s}c_jz^j}$ has all its zeros in $|z |\le k, k\ge 1$, we have

$$\begin{aligned} |\frac{c_0}{c_{n-s}} |\le k^{n-s}, \end{aligned}$$

which is equivalent to

$$\begin{aligned} |c_0 |k^s\le |c_{n-s} |k^n, \end{aligned}$$

and

$$\begin{aligned} k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}\ge \sqrt{ |c_0 |}. \end{aligned}$$

Dividing both sides of (18) by $|\alpha |$ and taking limit as $|\alpha |\rightarrow \infty$, we get the following generalization and refinement of inequality (10) due to Govil [12].

Corollary 1

If $p(z)=z^s\displaystyle {\sum _{j=0}^{n-s}c_jz^j},0\le s\le n,$ is a polynomial of degree n having all its zeros in $|z |\le k, k\ge 1,$ then

$$\begin{aligned}{} & {} \max _{ |z |=1} |p'(z) |\ge \left( \frac{1}{1+k^{n-s}}\right) \left( n+s+\frac{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}-\sqrt{ |c_0 |}}{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}}\right) \nonumber \\{} & {} \quad \times \left\{ 1+\frac{\left( |c_{n-s} |k^n- |c_0 |k^s\right) (k-1)}{2\left( |c_{n-s} |k^n+ |c_0 |k^{s+1}\right) }\right\} \max _{ |z |=1} |p(z) |. \end{aligned}$$

(19)

When $s=0$, Theorem 1, in particular, gives the following improvement of inequality (14) proved by Aziz and Rather [2] and a generalization and an improvement of inequality (13) of Gulzar et al. [17].

Corollary 2

If $p(z)=\displaystyle \sum _{j=0}^{n}c_jz^j$ is a polynomial of degree n having all its zeros in $|z |\le k, k\ge 1$, then for any complex number $|\alpha |$ with $|\alpha |\ge k$

$$\begin{aligned}{} & {} \max _{ |z |=1} |D_{\alpha }p(z) |\ge \left( \frac{ |\alpha |-k}{1+k^{n}}\right) \left( n+\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{ |c_0 |}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\right) \nonumber \\{} & {} \quad \times \left\{ 1+\frac{\left( |c_n |k^n- |c_0 |\right) (k-1)}{2\left( |c_n |k^n+ |c_0 |k\right) }\right\} \max _{ |z |=1} |p(z) |. \end{aligned}$$

(20)

Dividing both sides of (20) by $|\alpha |$ and taking limit as $|\alpha |\rightarrow \infty$, we get the following refinement of inequality (10) due to Govil [12].

Corollary 3

If $p(z)=\displaystyle \sum _{j=0}^{n}c_jz^j$ is a polynomial of degree n having all its zeros in $|z |\le k, k\ge 1$, then

$$\begin{aligned}{} & {} \max _{ |z |=1} |p'(z) |\ge \left( \frac{1}{1+k^{n}}\right) \left( n+\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{ |c_0 |}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\right) \nonumber \\{} & {} \quad \times \left\{ 1+\frac{\left( |c_n |k^n- |c_0 |\right) (k-1)}{2\left( |c_n |k^n+ |c_0 |k\right) }\right\} \max _{ |z |=1} |p(z) |. \end{aligned}$$

(21)

The inequality (21) is best possible for $p(z)=z^n+k^n$.

Remark 2

Taking $k=1$ in Corollary 3, inequality (21) provides a refinement of inequality (8) due to Turán [26].

As an application of Theorem 1, we obtain the following result which is a refinement of inequality (17) due to Govil and Mctume [15] and a result recently proved by Singh and Chanam [23, Theorem 2].

Theorem 2

If $p(z)=\displaystyle \sum _{j=0}^{n}c_jz^j$ is a polynomial of degree n having all its zeros in $|z |\le k, k\ge 1$, then for any complex number $\alpha$ with $|\alpha |\ge 1+k+k^n$

$$\begin{aligned}{} & {} \max _{ |z |=1} |D_{\alpha }p(z) |\nonumber \\{} & {} \quad \ge \frac{\left( |\alpha |-k\right) }{1+k^n}\left( n+\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{ |c_0+ e^{i\theta _{0}}m |}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\right) \nonumber \\{} & {} \quad \times \left\{ 1+\frac{( |c_n |k^n- |c_0+ e^{i\theta _{0}}m |)(k-1)}{2( |c_n |k^n+ |c_0+ e^{i\theta _{0}}m |k)}\right\} \max _{ |z |=1} |p(z) |\nonumber \\{} & {} \quad + \Bigg [n\left( \frac{ |\alpha |-(1+k+k^n)}{1+k^n}\right) \nonumber \\{} & {} \quad +\frac{ |\alpha |-k}{1+k^n}\Bigg \{\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{ |c_0+ e^{i\theta _{0}}m |}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\nonumber \\{} & {} \quad +\frac{( |c_n |k^n- |c_0+ e^{i\theta _{0}}m |)(k-1)}{2( |c_n |k^n+ |c_0+ e^{i\theta _{0}}m |k)}\nonumber \\{} & {} \quad \left( n+\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{ |c_0+ e^{i\theta _{0}}m |}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\right) \Bigg \}\Bigg ]m, \end{aligned}$$

(22)

where $m=\displaystyle {\min _{ |z |=k} |p(z) |}$ and $\theta _{0}=\arg \left\{ p(e^{i\phi _{0}})\right\}$ such that $|p(e^{i\phi _{0}}) |=\displaystyle {\max _{ |z |=1} |p(z) |}$.

Remark 3

If $p(z)=\displaystyle \sum _{j=0}^{n}c_jz^j$ is a polynomial of degree n having all its zeros in $|z |\le k, k\ge 1,$ then for any complex number $|\lambda |e^{i\theta _{0}}$ with $|\lambda |<1$, by Rouche’s theorem it follows that the polynomial $p(z)+ |\lambda |e^{i\theta _{0}} m=(c_0+ |\lambda |e^{i\theta _{0}} m)+c_1z+\cdots +c_nz^n$ has all its zeros in $|z |\le k$, where $m=\displaystyle {\min _{ |z |=k} |p(z) |}$, then

$$\begin{aligned} k^n\ge \vert \frac{c_0+ |\lambda |e^{i\theta _{0}} m}{c_n}|, \end{aligned}$$

which implies that

$$\begin{aligned} k^{\frac{n}{2}}\sqrt{ |c_n |}\ge \sqrt{|c_0+ |\lambda |e^{i\theta _{0}} m|}. \end{aligned}$$

Taking $|\lambda |\rightarrow 1$, we get

$$\begin{aligned} k^{\frac{n}{2}}\sqrt{ |c_n |}\ge \sqrt{|c_0+e^{i\theta _{0}} m|}, \end{aligned}$$

and

$$\begin{aligned} k^n |c_n |\ge |c_0+e^{i\theta _{0}} m |. \end{aligned}$$

Remark 4

Dividing both sides of (22) by $|\alpha |$ and taking limit as $|\alpha |\rightarrow \infty$, we have the following refinement of inequality (11) due to Govil [13].

Corollary 4

If $p(z)=\displaystyle \sum _{j=0}^{n}c_jz^j$ is a polynomial of degree n having all its zeros in $|z |\le k, k\ge 1$, then

$$\begin{aligned}{} & {} \max _{ |z |=1} |p'(z) |\nonumber \\{} & {} \quad \ge \frac{1}{1+k^n}\left( n+\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{ |c_0+ e^{i\theta _{0}}m |}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\right) \nonumber \\{} & {} \quad \times \left\{ 1+\frac{( |c_n |k^n- |c_0+ e^{i\theta _{0}}m |)(k-1)}{2( |c_n |k^n+ |c_0+ e^{i\theta _{0}}m |)}\right\} \max _{ |z |=1} |p(z) |\nonumber \\{} & {} \quad + \Bigg [\frac{n}{1+k^n}+\frac{1}{1+k^n}\Bigg \{\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{ |c_0+ e^{i\theta _{0}}m |}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\nonumber \\{} & {} \quad +\frac{( |c_n |k^n- |c_0+ e^{i\theta _{0}}m |)(k-1)}{2( |c_n |k^n+ |c_0+ e^{i\theta _{0}}m |)}\nonumber \\{} & {} \quad \left( n+\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{ |c_0+ e^{i\theta _{0}}m |}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\right) \Bigg \}\Bigg ]m, \end{aligned}$$

(23)

where $m=\displaystyle {\min _{ |z |=k}p(z)}$ and $\theta _{0}=\arg \left\{ p(e^{i\phi _{0}})\right\}$ such that $|p(e^{i\phi _{0}}) |=\displaystyle {\max _{ |z |=1} |p(z) |}$.

Inequality (23) is best possible for $p(z)=z^n+k^n$.

Remark 5

Taking $k=1$ in Corollary 4, inequality (23) reduces to a refinement of inequality (9) due to Aziz and Dawood [1].

Corollary 5

If $p(z)=\displaystyle \sum _{j=0}^{n}c_jz^j$ is a polynomial of degree n having all its zeros in $|z |\le 1$, then

$$\begin{aligned}{} & {} \max _{ |z |=1} |p'(z) |\nonumber \\{} & {} \quad \ge \frac{1}{2}\left( n+\frac{\sqrt{ |c_n |}-\sqrt{ |c_0+e^{i\theta _{0}} m |}}{\sqrt{ |c_n |}}\right) \max _{ |z |=1} |p(z) |\nonumber \\{} & {} \qquad + \frac{1}{2}\left[ n+\left( \frac{\sqrt{ |c_n |}-\sqrt{ |c_0+ e^{i\theta _{0}}m |}}{\sqrt{ |c_n |}}\right) \right] m, \end{aligned}$$

(24)

where $m=\displaystyle {\min _{ |z |=1}} |p(z) |$ and $\theta _{0}=\arg \left\{ p(e^{i\phi _{0}})\right\}$ such that $|p(e^{i\phi _{0}}) |=\displaystyle {\max _{ |z |=1} |p(z) |}$.

Further, we are able to prove an improvement of inequalities (6) and (7).

Theorem 3

If $p(z)=\displaystyle {\sum _{j=0}^{n}}c_jz^j$ is a polynomial of degree n having no zero in $|z |< k, k\le 1$. If $|p'(z) |$ and $|q'(z) |$ attain their maxima at the same point on $|z |=1$, then

$$\begin{aligned}{} & {} \max _{ |z |=1} |p'(z) |\le \frac{1}{1+k^n} \Bigg [n-k^n\Bigg \{\frac{\sqrt{ |c_0 |}-k^{\frac{n}{2}}\sqrt{ |c_n |}}{\sqrt{ |c_0 |}}\nonumber \\{} & {} \quad +\frac{( |c_0 |-k^n |c_n |)(1-k)}{2( |c_0 |k+k^n |c_n |)}\left( n+\frac{\sqrt{ |c_0 |}-k^{\frac{n}{2}}\sqrt{ |c_n |}}{\sqrt{ |c_0 |}}\right) \Bigg \}\Bigg ]\nonumber \\{} & {} \quad \max _{ |z |=1} |p(z) |. \end{aligned}$$

(25)

The result is sharp and equality in (25) holds for $p(z)=z^n+k^n$.

Remark 6

Since $p(z)=\displaystyle \sum _{j=0}^{n}c_jz^j$ has all its zeros in $|z |\ge k,k\le 1$, q(z) has all its zeros in $|z |\le \frac{1}{k}, \frac{1}{k}\ge 1$, then

$$\begin{aligned} |\frac{c_n}{c_{0}} |\le \frac{1}{k^{n}}, \end{aligned}$$

which equivalently gives

$$\begin{aligned} |c_0 |\ge |c_n |k^n, \end{aligned}$$

(26)

and

$$\begin{aligned} \sqrt{ |c_0 |}\ge k^{\frac{n}{2}}\sqrt{ |c_n |}. \end{aligned}$$

(27)

From inequalities (26) and (27), it is evident that the bound (25) improves both the bounds given by (6) and (7).

Remark 7

Taking $k=1$ in Theorem 3, we get the following improvement of (2) due to Erdös and Lax for a subclass of polynomials.

Corollary 6

If $p(z)=\displaystyle {\sum _{j=0}^{n}}c_jz^j$ is a polynomial of degree n having no zero in $|z |< 1$. If $|p'(z) |$ and $|q'(z) |$ attain their maxima at the same point on $|z |=1$, then

$$\begin{aligned} \max _{ |z |=1} |p'(z) |\le \frac{1}{2}\left( n-\frac{\sqrt{ |c_0 |}-\sqrt{ |c_n |}}{\sqrt{ |c_0 |}}\right) \max _{ |z |=1} |p(z) |. \end{aligned}$$

(28)

3 Lemmas

We need the following lemmas to prove our theorems.

Lemma 1

If $p(z)=\displaystyle {\sum _{j=0}^{n}c_jz^j}$ is a polynomial of degree $n\ge 1$ having all its zeros in $|z |\le 1,$ then for all z on $|z |=1$ with $p(z)\ne 0$.

$$\begin{aligned} \Re \left( z\frac{ p'(z)}{p(z)}\right) \ge \frac{1}{2}\left( n+\frac{\sqrt{ |c_n |}-\sqrt{ |c_0 |}}{\sqrt{ |c_n |}}\right) . \end{aligned}$$

(29)

The above result is due to Dubin [10, Theorem 4]( also see Singh and Chanam [23, Lemma 3] and Wali and Shah [25, Inequality 9]).

Lemma 2

Let $p(z)=z^s\displaystyle {\sum _{j=0}^{n-s}c_jz^j},0\le s\le n$ be a polynomial of degree n having all its zeros in $|z |\le k,k\ge 1,$ then

$$\begin{aligned}{} & {} \max _{ |z |=k} |p(z) |\ge \frac{2k^n}{1+k^{n-s}}\left( 1+\frac{( |c_{n-s} |k^n- |c_0 |k^s)(k-1)}{2( |c_{n-s} |k^n+ |c_0 |k^{s+1})}\right) \nonumber \\{} & {} \quad \max _{ |z |=1} |p(z) |. \end{aligned}$$

(30)

The above result appears in Kumar and Dhankar [18, Lemma 4].

Lemma 3

If $p(z)=z^s\displaystyle {\sum _{j=0}^{n-s}c_jz^j},0\le s\le n$ is a polynomial of degree n having all its zeros in $|z |\le 1,$ with $s-$fold zeros at the origin, then for any complex number $\alpha$ with $|\alpha |\ge 1$ and on $|z |=1$

$$\begin{aligned} |D_{\alpha }p(z) |\ge \frac{\left( |\alpha |-1\right) }{2}\left( n+s+\frac{\sqrt{ |c_{n-s} |}-\sqrt{ |c_0 |}}{\sqrt{ |c_{n-s} |}}\right) |p(z) |.\nonumber \\ \end{aligned}$$

(31)

This result appears in Singh and Chanam [23, Lemma 5].

Lemma 4

If p(z) is a polynomial of degree n, then on $|z |=1$

$$\begin{aligned} |p'(z) |+ |q'(z) |\le n\max _{ |z |=1} |p(z) |. \end{aligned}$$

(32)

The above result is a particular case of a result [16, Inequality 3.2] due to Govil and Rahman.

4 Proofs of the theorems

Proof of Theorem 1

Since $p(z)=z^s\displaystyle {\sum _{j=0}^{n-s}c_jz^j}$ has all its zeros in $|z |\le k,k\ge 1,$ the polynomial $p(kz)=z^s\left( k^sc_0+k^{s+1}c_1z+\cdots k^{n}c_nz^{n-s}\right)$ has all its zeros in $|z |\le 1$. Using Lemma 3 to p(kz), we get for $|\frac{\alpha }{k} |\ge 1$

$$\begin{aligned}{} & {} \max _{ |z |=1} |D_{\frac{\alpha }{k}}p(kz) |\\{} & {} \quad \ge \frac{ |\alpha |-k}{2k}\left( n+s+\frac{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}-\sqrt{ |c_0 |}}{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}}\right) \\{} & {} \qquad \times \max _{ |z |=1}|p(kz)|, \end{aligned}$$

that is

$$\begin{aligned}{} & {} \max _{ |z |=1}|np(kz)+\left( \frac{\alpha }{k}-z\right) kp'(kz)|\nonumber \\{} & {} \quad \ge \frac{\left( |\alpha |-k\right) }{2k}\left( n+s+\frac{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}-\sqrt{ |c_0 |}}{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}}\right) \nonumber \\{} & {} \quad \max _{ |z |=k}|p(z)|. \end{aligned}$$

(33)

Using Lemma 2 and the fact that $\displaystyle {\max _{ |z |=1}}|np(kz)+\left( \frac{\alpha }{k}-z\right) kp'(kz)|=\displaystyle {\max _{ |z |=k} |D_{\alpha }p(z) |}$, inequality (33) implies

$$\begin{aligned}{} & {} \max _{ |z |=k} |D_{\alpha }p(z) |\nonumber \\{} & {} \quad \ge \frac{\left( |\alpha |-k\right) }{2k}\left( n+s+\frac{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}-\sqrt{ |c_0 |}}{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}}\right) \nonumber \\{} & {} \quad \times \frac{2k^n}{1+k^{n-s}}\left\{ 1+\frac{( |c_{n-s} |k^n- |c_0 |k^s)(k-1)}{2( |c_{n-s} |k^n+ |c_0 |k^{s+1})}\right\} \nonumber \\ {}{} & {} \qquad \times \max _{ |z |=1} |p(z) |. \end{aligned}$$

(34)

As we can see that $D_{\alpha }p(z)$ is a polynomial of degree at most $n-1$ and $k\ge 1$, it is well-known that

$\displaystyle {\max _{ |z |=k} |D_{\alpha }p(z) |\le k^{n-1}\max _{ |z |=1} |D_{\alpha }p(z) |}$. Using this fact, inequality (34) gives

$$\begin{aligned}{} & {} k^{n-1}\max _{ |z |=1} |D_{\alpha }p(z) |\\{} & {} \quad \ge \left( |\alpha |-k\right) \left( n+s+\frac{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}-\sqrt{ |c_0 |}}{k^{\frac{n-s}{2}}\sqrt{ |c_{n-s} |}}\right) \\{} & {} \qquad \times \frac{k^{n-1}}{1+k^{n-s}}\left\{ 1+\frac{( |c_{n-s} |k^n- |c_0 |k^s)(k-1)}{2( |c_{n-s} |k^n+ |c_0 |k^{s+1})}\right\} \nonumber \\ {}{} & {} \qquad \times \max _{ |z |=1} |p(z) |. \end{aligned}$$

which gives inequality (18), and the proof of Theorem 1 is complete. $\square$

Proof of Theorem 2

If p(z) has a zero on $|z |=k,$ then $m=0$ and the result follows trivially from Theorem 1. So, without loss of generality, let us assume that p(z) has all its zeros in $|z |<k,k\ge 1,$ then it follows by Rouche’s theorem that for any complex number $\lambda$ with $|\lambda |<1$, the polynomial $p(z)+\lambda m=(c_{0}+\lambda m)+c_1z+\cdots +c_nz^n$ has all its zeros in $|z |<k,k\ge 1$. Therefore, applying Theorem 1 to $p(z)+\lambda m$ with $s=0$, we get for $|\alpha |\ge 1+k+k^n$

$$\begin{aligned}{} & {} \max _{ |z |=1}|D_{\alpha }\left[ p(z)+\lambda m\right] |\nonumber \\{} & {} \quad \ge \left( \frac{ |\alpha |-k}{1+k^{n}}\right) \left( n+\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{|c_0+\lambda m|}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\right) \nonumber \\{} & {} \qquad \times \left\{ 1+\frac{\left( |c_n |k^n-|c_0+\lambda m|\right) (k-1)}{2\left( |c_n |k^n+|c_0+\lambda m|k\right) }\right\} \nonumber \\{} & {} \qquad \times \max _{ |z |=1}|p(z)+\lambda m|. \end{aligned}$$

(35)

Let $0\le \phi _0<2\pi$, be such that $|p(e^{i\phi _0})|=\displaystyle {\max _{ |z |=1} |p(z) |}$. Then, inequality (35) takes

$$\begin{aligned}{} & {} \max _{ |z |=1}|D_{\alpha }p(z)+n\lambda m|\nonumber \\{} & {} \quad \ge \left( \frac{ |\alpha |-k}{1+k^{n}}\right) \left( n+\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{|c_0+\lambda m|}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\right) \end{aligned}$$

(36)

$$\begin{aligned}{} & {} \quad \times \left\{ 1+\frac{\left( |c_n |k^n-|c_0+\lambda m|\right) (k-1)}{2\left( |c_n |k^n+|c_0+\lambda m|k\right) }\right\} |p(e^{i\phi _0})+\lambda m|.\nonumber \\ \end{aligned}$$

(37)

Now,

$$\begin{aligned} |p(e^{i\phi _0})+\lambda m|= & {} ||p(e^{i\phi _{0}})|e^{i\theta _{0}}+|\lambda |e^{i\phi } m|\\= & {} ||p(e^{i\phi _{0}})|+|\lambda |e^{i(\phi -\theta _{0})} m|. \end{aligned}$$

Setting the argument $\phi$ such that $\phi =\theta _{0}$, then

$$\begin{aligned} |p(e^{i\phi _{0}})+\lambda m|=|p(e^{i\phi _{0}})|+|\lambda |m. \end{aligned}$$

(38)

Using this fact in inequality (37), we have

$$\begin{aligned}{} & {} \max _{ |z |=1} |D_{\alpha }p(z) |+n |\lambda |m\\{} & {} \quad \ge \left( \frac{ |\alpha |-k}{1+k^{n}}\right) \left( n+\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{|c_0+ |\lambda |e^{i\theta _0} m|}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\right) \\{} & {} \quad \times \left\{ 1+\frac{\left( |c_n |k^n-|c_0+ |\lambda |e^{i\theta _0} m|\right) (k-1)}{2\left( |c_n |k^n+|c_0+ |\lambda |e^{i\theta _0} m|k\right) }\right\} \\{} & {} \quad \left( |p(e^{i\phi _0})|+ |\lambda |m\right) . \end{aligned}$$

which is equivalent to

$$\begin{aligned}{} & {} \max _{ |z |=1} |D_{\alpha }p(z) |\\{} & {} \quad \ge \frac{\left( |\alpha |-k\right) }{1+k^n}\left( n+\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{|c_0+ |\lambda |e^{i\theta _{0}}m|}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\right) \\{} & {} \quad \times \left\{ 1+\frac{( |c_n |k^n-|c_0+ |\lambda |e^{i\theta _0}m|)(k-1)}{2( |c_n |k^n+|c_0+ |\lambda |e^{i\theta _0}m|k)}\right\} \max _{ |z |=1} |p(z) |\\{} & {} \quad + |\lambda |\Bigg [n\left( \frac{ |\alpha |-(1+k+k^n)}{1+k^n}\right) \\{} & {} \quad +\frac{ |\alpha |-k}{1+k^n}\Bigg \{\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{|c_0+ |\lambda |e^{i\theta _{0}}m|}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\\{} & {} \quad +\frac{( |c_n |k^n-|c_0+ |\lambda |e^{i\theta _0}m|)(k-1)}{2( |c_n |k^n+|c_0+ |\lambda |e^{i\theta _0}m|k)}\\{} & {} \quad \left( n+\frac{k^{\frac{n}{2}}\sqrt{ |c_n |}-\sqrt{|c_0+ |\lambda |e^{i\theta _{0}}m|}}{k^{\frac{n}{2}}\sqrt{ |c_n |}}\right) \Bigg \}\Bigg ]m, \end{aligned}$$

Taking $|\lambda |\rightarrow 1$, the above inequality reduces to (22). This completes the proof of Theorem 2. $\square$

Proof of Theorem 3

Since p(z) has all its zeros in $|z |\ge k,k\le 1$, q(z) has all its zeros in $|z |\le \frac{1}{k}, \frac{1}{k}\ge 1$. Then applying Corollary 3 to q(z), we have

$$\begin{aligned}{} & {} \max _{ |z |=1}|q'(z)|\nonumber \\ {}{} & {} \quad \ge \left( \frac{k^n}{1+k^{n}}\right) \left( n+\frac{\left( \frac{1}{k}\right) ^{\frac{n}{2}}\sqrt{ |c_0 |}-\sqrt{ |c_n |}}{\left( \frac{1}{k}\right) ^{\frac{n}{2}}\sqrt{ |c_0 |}}\right) \nonumber \\{} & {} \quad \times \left\{ 1+\frac{\left( |c_0 |\left( \frac{1}{k}\right) ^n- |c_n |\right) (\frac{1}{k}-1)}{2\left( |c_0 |\left( \frac{1}{k}\right) ^n+ |c_n |\frac{1}{k}\right) }\right\} \max _{ |z |=1} |p(z) |. \end{aligned}$$

(39)

By Lemma 4, we have on $|z |=1$,

$$\begin{aligned} |p'(z) |+ |q'(z) |\le n\max _{ |z |=1} |p(z) |. \end{aligned}$$

(40)

Since $|p'(z) |$ and $|q'(z) |$ attain their maxima at the same point on $|z |=1$, then

$$\begin{aligned} \max _{ |z |=1}\left\{ |p'(z) |+ |q'(z) |\right\} = \max _{ |z |=1} |p'(z) |+\max _{ |z |=1} |q'(z) |. \end{aligned}$$

(41)

Combining (39), (40) and (41), we have

$$\begin{aligned}{} & {} n\max _{ |z |=1} |p(z) |\\ {}{} & {} \quad \ge \left( \frac{k^n}{1+k^{n}}\right) \left( n+\frac{\sqrt{ |c_0 |}-k^{\frac{n}{2}}\sqrt{ |c_n |}}{\sqrt{ |c_0 |}}\right) \\{} & {} \quad \times \left\{ 1+\frac{\left( |c_0 |-k^n |c_n |\right) (1-k)}{2\left( |c_0 |k+k^n |c_n |\right) }\right\} \max _{ |z |=1} |p(z) |+ \max _{ |z |=1} |p'(z) |, \end{aligned}$$

which is equivalent to

$$\begin{aligned}{} & {} \max _{ |z |=1} |p'(z) |\\{} & {} \quad \le \frac{1}{1+k^n} \Bigg [n-k^n\Bigg \{\frac{\sqrt{ |c_0 |}-k^{\frac{n}{2}}\sqrt{ |c_n |}}{\sqrt{ |c_0 |}}\\{} & {} \quad +\frac{( |c_0 |-k^n |c_n |)(1-k)}{2( |c_0 |k+k^n |c_n |)}\left( n+\frac{\sqrt{ |c_0 |}-k^{\frac{n}{2}}\sqrt{ |c_n |}}{\sqrt{ |c_0 |}}\right) \Bigg \}\Bigg ]\\{} & {} \quad \max _{ |z |=1} |p(z) |, \end{aligned}$$

$\square$