Inequalities for the polar derivative of a polynomial

Abstract

Let P(z) be a polynomial of degree n having all its zeros in $|z|\le 1$, then according to Turan (Compositio Mathematica 7:89–95, 2004)

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

In this paper, we shall use polar derivative and establish a generalisation and an extension of this result. Our results also generalize variety of other results.

1 Introduction

Let $\mathcal {P}_n$ denote the class of all complex polynomials of degree at most n. Let $B=\{z;|z|=1\}$ denotes the unit disk and $B_-$ and $B_+$ denote the regions inside and outside the disk B respectively. If $P\in \mathcal {P}_n$, then according to the well known result of Bernstein [4]

$$\begin{aligned} \max \limits _{z\in B}|P'(z)|\le n \max \limits _{z\in B}|P(z)|. \end{aligned}$$

(1)

Inequality (1) is best possible and equality holds for the polynomial $P(z)=\lambda z^n,$ where $\lambda$ is a complex number. If we restrict ourselves to the class of polynomials having no zeros in $B\cup B_-$, then it was conjectured by Erdös and later on proved by Lax [6] that

$$\begin{aligned} \max \limits _{z\in B}|P'(z)|\le \frac{n}{2}\max \limits _{z\in B}|P(z)|, \end{aligned}$$

(2)

and if P has no zero in $B\cup B_+,$ then it was proved by Turan [8] that

$$\begin{aligned} \max \limits _{z\in B}|P'(z)|\ge \frac{n}{2}\max \limits _{z\in B}|P(z)|. \end{aligned}$$

(3)

The inequalities (2) and (3) are also best possible and equality holds for polynomials which have all zeros on B.

If P(z) is a polynomial of degree n and $\alpha$ a complex number, then the polar derivative of P(z) with respect to $\alpha$, denoted by $D_{\alpha }P(z)$ is defined by

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

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

$$\begin{aligned} \lim \limits _{\alpha \rightarrow \infty }\frac{D_{\alpha }P(z)}{\alpha }=P'(z). \end{aligned}$$

As an extension of (1), Aziz and Shah [3] used polar derivative and established that if P(z) is a polynomial of degree n, then for every real or complex number $\alpha$ with $|\alpha | >1$ and for $z\in B$,

$$\begin{aligned} |D_{\alpha }P(z)|\le n|\alpha |\max \limits _{z\in B}|P(z)|. \end{aligned}$$

(4)

Aziz [1] extended inequality (2) to the polar derivative and proved that if p is a polynomial of degree n having all zero in $z\in B\cup B_+$ then for $\alpha \in \mathbb {C}$ with $|\alpha |\ge 1$

$$\begin{aligned} \max \limits _{z\in B}|D_{\alpha }P(z)|\le \frac{n(|\alpha |+1)}{2}\max \limits _{z\in B}|P(z)|. \end{aligned}$$

(5)

If we divide the two sides of (4) and (5) by $|\alpha |$ and let $|\alpha |\rightarrow \infty$, we get inequalities (1) and (2) respectively.

Shah [7] extended (3) to the polar derivative and proved the following result:

Theorem 1.1

If $P\in \mathcal {P}_n$ and has all zeros in $z\in B\cup B_-$, then for $|\alpha |\ge 1$

$$\begin{aligned} \max \limits _{z\in B}|D_{\alpha }P(z)|\ge \frac{n(|\alpha |-1)}{2}\max \limits _{z\in B}|P(z)|. \end{aligned}$$

(6)

Theorem (1.1) generalizes (3) and to obtain (3), divide both sides of Theorem (1.1) by $|\alpha |$ and let $|\alpha |\rightarrow \infty$.

2 Main results

In this paper we obtain some more general results. First we prove the following generalization of Theorem (6).

Theorem 2.1

If $P\in \mathcal {P}_n$ and $P(z)=\sum \nolimits _{j=0}^{n}c_jz^j$ has all its zeros in $B\cup B_-$, then for $\alpha \in \mathbb {C}$ with $|\alpha |\ge 1$ and $z\in B$,

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

(7)

The result is sharp and equality holds for the polynomial $P(z)=c_n z^n +c_0$ with $|c_0|=|c_n|\ne 0.$

Remark 2.1

Since P(z) has all its zeros in $B\cup B_-$, therefore $|c_n|\ge |c_0|$, it follows that Theorem 2.1 is an improvement of inequality (6)

Remark 2.2

If we divide the two sides of Theorem 2.1 by $|\alpha |$ and let $|\alpha |\rightarrow \infty$, we get a result due to Dubinin [5].

Theorem 2.2

If $P\in \mathcal {P}_n$ and $P(z)=\sum \nolimits _{j=0}^{n}c_jz^j$ has all its zeros in $B\cup B_-$, then for $\alpha \in \mathbb {C}$ with $|\alpha |\ge 1$, $0\le l<1$ and $z\in B$,

$$\begin{aligned} \begin{aligned} \max \limits _{z\in B}|D_\alpha P(z)|\ge {}&\frac{n}{2}\left\{ \left( |\alpha |-1\right) \max \limits _{z\in B}|P(z)|+\left( |\alpha |+1 \right) lm\right\} \\&+\frac{\left( |\alpha |-1 \right) }{2}\left\{ \frac{\sqrt{|c_n|-l m}-\sqrt{|c_0|}}{\sqrt{|c_n|-l m}} \right\} \left\{ \max \limits _{z\in B}|P(z)|-lm \right\} , \end{aligned} \end{aligned}$$

(8)

where $m= \min \nolimits _{z\in B} |P(z)|$.

Dividing both sides of (8) by $|\alpha |$ and let $|\alpha |\rightarrow \infty$, we get the following result:

Corollary 2.1

If $P\in \mathcal {P}_n$ and $P(z)=\sum \nolimits _{j=0}^{n}c_jz^j$ has all its zeros in $B\cup B_-$, then for $0\le l<1$ and $z\in B$,

$$\begin{aligned} \begin{aligned} |P'(z)|\ge {}&\frac{1}{2}\left\{ n + \frac{\sqrt{|c_n|-l m}-\sqrt{|c_0|}}{\sqrt{|c_n|-lm}} \right\} \max \limits _{z\in B}|P(z)|\\&+\frac{1}{2}\left\{ n - \frac{\sqrt{|c_n|-l m}-\sqrt{|c_0|}}{\sqrt{|c_n|-lm}} \right\} lm. \end{aligned} \end{aligned}$$

(9)

3 Lemmas

For the proof of above Theorems, we need the following lemmas.

Lemma 3.1

If $P\in \mathcal {P}_n$ and P(z) has all its zeros in $B \cup B_-$ and $Q(z)=z^n\overline{P}(\frac{1}{\bar{z}})$, then for $z \in B$,

$$\begin{aligned} |Q'(z)|\le |P'(z)|. \end{aligned}$$

Lemma 3.1 is a special case of a result due to Aziz and Rather [2].

We also need the following result which is due to Dubinin [5].

Lemma 3.2

If $P\in \mathcal {P}_n$ and P(z) has all zeros in $B\cup B_-$, then

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

(10)

Inequality (10) is sharp and equality holds for polynomials which have all zeros on B.

4 Proofs of the theorems

Proof of Theorem (2.1)

If $Q(z)=z^n\overline{P} \left(\frac{1}{\bar{z}}\right)$, it can be easily seen that $|Q'(z)|=|nP(z)-zP'(z)|$ for $z\in B$. Also P(z) has all its zeros in $z\in B\cup B_-$ so by Lemma 3.1, we have

$$\begin{aligned} \begin{aligned} |P'(z)|{}&\ge |Q'(z)|\\&=|nP(z)-zP'(z)| \quad \text {for} \quad z \in B. \end{aligned} \end{aligned}$$

(11)

Now for every complex $\alpha$ with $|\alpha |\ge 1,$ we have for $z\in B,$

$$\begin{aligned} \begin{aligned} |D_\alpha P(z)|{}&=|nP(z)+(\alpha -z)P'(z)|\\&\ge |\alpha ||P'(z)|-|nP(z)-zP'(z)|. \end{aligned} \end{aligned}$$

This gives with the help of (11) that

$$\begin{aligned} |D_\alpha P(z)|\ge (|\alpha |-1)P'(z). \end{aligned}$$

(12)

By Lemma 3.2, we have for each z on B at which P(z) does not vanish,

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

This gives

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

(13)

Combining (12) and (13), we get for $z\in B$,

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

(14)

That is

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

(15)

This completes proof of Theorem 2.1.

Proof of Theorem (2.2)

Since $P\in P_n$ and by hypothesis P(z) has all its zeros in $B\cup B_-$, if P(z) has a zero on B, then $m=\min \limits _{|z|=1}|P(z)|=0$ and the result follows from Theorem 2.1. So, assume that all the zeros of P(z) lie in $B_-$ so that $m>0$. Now $m \le |P(z)|$ for $z\in B$.

If $\lambda$ is any complex number such that $|\lambda | <1$, then $|m\lambda z^n|<|P(z)|$ for $z\in B$. Since all zeros of P(z) lie in $B_-$, it follows by Rouche’s Theorem that all the zeros of $F(z)=P(z)-\lambda m z^n$ also lie in $B_-$.

Let $G(z)=z^n\overline{F}(\frac{1}{\bar{z}})$, it can be easily seen that

$$\begin{aligned} |G'(z)|=|nF(z)-zF'(z)| \quad for \quad z \in B. \end{aligned}$$

Also F(z) has all its zeros in $z \in B_-$, so by Lemma 3.1 ,we have

$$\begin{aligned} \begin{aligned} |F'(z)|{}&\ge |G'(z)|\\&=|nF(z)-zF'(z)| \quad for \quad z \in B. \end{aligned} \end{aligned}$$

(16)

Now for every complex $\alpha$ with $|\alpha |\ge 1,$ we have for $z\in B$,

$$\begin{aligned} \begin{aligned} |D_\alpha F(z)|{}&=|nF(z)+(\alpha -z)F'(z)|\\&\ge |\alpha ||F'(z)|-|nF(z)-zF'(z)|. \end{aligned} \end{aligned}$$

This gives with the help of (16) that

$$\begin{aligned} |D_\alpha F(z)|\ge (|\alpha |-1)F'(z). \end{aligned}$$

(17)

Since the polynomial $F(z)=c_0+c_1z+c_2z^2+\cdots +c_{n-1}z^{n-1}+(c_n-\lambda m)z^n$ does not vanish in $|z|<1,$ we have by Lemma 3.2

$$\begin{aligned} Re\frac{zF'(z)}{F(z)}\ge \frac{n+1}{2}- \frac{1}{2}\frac{\sqrt{|c_0|}}{\sqrt{|c_n-\lambda m|}}. \end{aligned}$$

This gives

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

(18)

Combining (17) and (18), we get for $|z|=1$,

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

(19)

That is

$$\begin{aligned}&|D_\alpha (P(z)-\lambda m z^n)|\ge \frac{(|\alpha |-1)}{2}\left[ n+ \frac{\sqrt{|c_n-\lambda m|}-\sqrt{|c_0|}}{\sqrt{|c_n-\lambda m|}}\right] |P(z)-\lambda m z^n|. \end{aligned}$$

(20)

$$\begin{aligned}&|D_\alpha P(z)-\lambda mn\alpha z^{n-1}|\ge \frac{(|\alpha |-1)}{2}\left[ n+ \frac{\sqrt{|c_n|-|\lambda | m}-\sqrt{|c_0|}}{\sqrt{|c_n|-|\lambda | m}}\right] |P(z)-\lambda m z^n|. \end{aligned}$$

(21)

It follows by a simple consequence of Laguerre Theorem on the polar derivative of a polynomial that for every $\alpha$ with $|\alpha |\ge 1$ ,the polynomial

$$\begin{aligned} D_\alpha (P(z)-\lambda m z^n)=D_\alpha P(z)-\lambda mn\alpha z^{n-1} \end{aligned}$$

(22)

has all its zeros in $B_-$. Thus, we have

$$\begin{aligned} |D_\alpha (P(z)|\ge \lambda mn|\alpha | |z|^{n-1} \quad for \quad |z|\ge 1. \end{aligned}$$

(23)

Now choosing the argument of $\lambda$ suitably in the left hand side of (21) such that

$$\begin{aligned} |D_\alpha P(z)-\lambda mn\alpha z^{n-1}|=|D_\alpha P(z)|-mn|\lambda ||\alpha | |z|^{n-1} \end{aligned}$$

which is possible by (23), we get for $z\in B$

$$\begin{aligned} |D_\alpha P(z)|-mn|\alpha ||\lambda | \ge \frac{(|\alpha |-1)}{2}\left[ n+ \frac{\sqrt{|c_n|-|\lambda | m}-\sqrt{|c_0|}}{\sqrt{|c_n|-|\lambda | m}}\right] \left\{ |P(z)|-|\lambda | m\right\}. \end{aligned}$$

(24)

From (24), one can easily obtain for $z\in B$ and for any $\alpha \in \mathbb {C}$ with $|\alpha |\ge 1$ that

$$\begin{aligned} \begin{aligned} \max \limits _{z\in B}|D_\alpha P(z)|\ge {}&\frac{n}{2}\left\{ \left( |\alpha |-1\right) \max \limits _{z\in B}|P(z)|+\left( |\alpha |+1 \right) lm\right\} \\&+\frac{\left( |\alpha |-1 \right) }{2}\left\{ \frac{\sqrt{|c_n|-lm}-\sqrt{|c_0|}}{\sqrt{|c_n|-lm}} \right\} \left\{ \max \limits _{z\in B}|P(z)|-lm \right\}, \end{aligned} \end{aligned}$$

(25)

where $0\le l<1$. That completes proof of Theorem 2.2.