# On the maximum modulus of a polynomial and its polar derivative

## Abstract

For a polynomial p(z) of degree n, having all zeros in |z| ≤ 1, Jain is shown that

$\begin{array}{r}\underset{|z|=1}{\mathrm{max}}|{D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)|\ge \frac{n\left(n-1\right)\cdots \left(n-t+1\right)}{{2}^{t}}×\left[\\ \left\{\left(|{\alpha }_{1}|-1\right)\cdots \left(|{\alpha }_{t}|-1\right)\right\}\underset{|z|=1}{\mathrm{max}}|p\left(z\right)|+\\ \left\{{2}^{t}\left(|{\alpha }_{1}|\cdots |{\alpha }_{t}|\right)-\left\{\left(|{\alpha }_{1}|-1\right)\cdots \left(|{\alpha }_{t}|-1\right)\right\}\right\}\underset{|z|=1}{\mathrm{min}}|p\left(z\right)|\right],\\ |{\alpha }_{1}|\ge 1,\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}|{\alpha }_{2}|\ge 1,\cdots |{\alpha }_{t}|\ge 1,\phantom{\rule{0.5em}{0ex}}\left(t

In this paper, the above inequality is extended for the polynomials having all zeros in |z| ≤ k, where k ≤ 1. Our result generalizes certain well-known polynomial inequalities.

(2010) Mathematics Subject Classification. Primary 30A10; Secondary 30C10, 30D15.

## 1. Introduction and statement of results

Let p(z) be a polynomial of degree n, then according to the well-known Bernstein's inequality  on the derivative of a polynomial, we have

$\underset{|z|=1}{max}\left|{p}^{\prime }\left(z\right)\right|\le n\underset{|z|=1}{max}\left|p\left(z\right)\right|.$
(1.1)

This result is best possible and equality holding for a polynomial that has all zeros at the origin.

If we restrict to the class of polynomials which have all zeros in |z| ≤ 1, then it has been proved by Turan  that

$\underset{|z|=1}{max}\left|{p}^{\prime }\left(z\right)\right|\ge \frac{n}{2}\underset{|z|=1}{max}\left|p\left(z\right)\right|.$
(1.2)

The inequality (1.2) is sharp and equality holds for a polynomial that has all zeros on |z| = 1.

As an extension to (1.2), Malik  proved that if p(z) has all zeros in |z| ≤ k, where k ≤ 1, then

$\underset{|z|=1}{max}\left|{p}^{\prime }\left(z\right)\right|\ge \frac{n}{1+k}\underset{|z|=1}{max}\left|p\left(z\right)\right|.$
(1.3)

This result is best possible and equality holds for p(z) = (z - k)n.

Aziz and Dawood  obtained the following refinement of the inequality (1.2) and proved that if p(z) has all zeros in |z| ≤ 1, then

$\underset{|z|=1}{max}\left|{p}^{\prime }\left(z\right)\right|\ge \frac{n}{2}\left\{\underset{|z|=1}{max}\left|p\left(z\right)\right|+\underset{|z|=1}{min}\left|p\left(z\right)\right|\right\}.$
(1.4)

This result is best possible and equality attains for a polynomial that has all zeros on |z| = 1.

Let D α p(z) denote the polar differentiation of the polynomial p(z) of degree n with respect to α . Then, D α p(z) = np(z) + (α - z)p'(z). The polynomial D α p(z) is of degree at most n - 1, and it generalizes the ordinary derivative in the sense that

$\underset{\alpha \to \infty }{lim}\left[\frac{{D}_{\alpha }p\left(z\right)}{\alpha }\right]={p}^{\prime }\left(z\right).$

Shah  extended (1.2) to the polar derivative of p(z) and proved that if all zeros of the polynomial p(z) lie in |z| ≤ 1, then for every α with |α| ≥ 1, we have

$\underset{|z|=1}{max}\left|{D}_{\alpha }p\left(z\right)\right|\ge \frac{n}{2}\left(\left|\alpha \right|-1\right)\underset{|z|=1}{max}\left|p\left(z\right)\right|.$
(1.5)

This result is best possible and equality holds as p(z) = (z - 1)nwith α ≥ 1.

Aziz and Rather  generalized (1.5) by extending (1.3) to the polar derivative of a polynomial. In fact, they proved that if all zeros of p(z) lie in |z| ≤ k, where k ≤ 1, then for every α with |α| ≥ k, we get

$\underset{|z|=1}{max}\left|{D}_{\alpha }p\left(z\right)\right|\ge \frac{n}{1+k}\left(\left|\alpha \right|-k\right)\underset{|z|=1}{max}\left|p\left(z\right)\right|.$
(1.6)

This result is best possible and equality holds for p(z) = (z - k)nwith αk.

In the same paper, Aziz and Rather  sharpened the inequality (1.5) by proving that if all the zeros of p(z) lie in |z| ≤ 1, then for every α with |α| ≥ 1, we would obtain

$\underset{|z|=1}{max}\left|{D}_{\alpha }p\left(z\right)\right|\ge \frac{n}{2}\left\{\left(\left|\alpha \right|-1\right)\underset{|z|=1}{max}\left|p\left(z\right)\right|+\left(\left|\alpha \right|-1\right)\underset{|z|=1}{min}\left|p\left(z\right)\right|\right\}.$
(1.7)

This result is best possible and equality attains for p(z) = (z - 1)nwith α ≥ 1.

As an extension to the inequality (1.7), Jain  proved that if p(z) has all zeros in |z| ≤ 1, then for all α1,... α t with |α1| ≥ 1, |α2| ≥ 1, ..., |α t | ≥ 1, (1 ≤ t < n), we have

$\begin{array}{r}\underset{|z|=1}{\mathrm{max}}|{D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)|\ge \frac{n\left(n-1\right)\cdots \left(n-t+1\right)}{{2}^{t}}\left[\\ \left\{\left(|{\alpha }_{1}|-1\right)\cdots \left(|{\alpha }_{t}|-1\right)\right\}\underset{|z|=1}{\mathrm{max}}|p\left(z\right)|+\\ \left\{{2}^{t}\left(|{\alpha }_{1}|\cdots |{\alpha }_{t}|\right)-\left\{\left(|{\alpha }_{1}|-1\right)\cdots \left(|{\alpha }_{t}|-1\right)\right\}\right\}\underset{|z|=1}{\mathrm{min}}|p\left(z\right)|\right],\end{array}$
(1.8)

where

$\begin{array}{c}{D}_{{\alpha }_{j}}{D}_{{\alpha }_{j-1}}\cdots {D}_{{\alpha }_{1}}p\left(z\right)={p}_{j}\left(z\right)=\\ \left(n-j+1\right){p}_{j-1}\left(z\right)+\left({\alpha }_{j}-z\right){p}_{j-{1}^{\prime }}\left(z\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}j=1,2,\cdots \phantom{\rule{0.3em}{0ex}},t,\\ {p}_{0}\left(z\right)=p\left(z\right).\end{array}$

This result is best possible and equality holds as p(z) = (z - 1)nwith α1 ≥ 1, α2 ≥ 1,..., α t ≥ 1.

The following result proposes an extension to (1.8). In a precise set up, we have

Theorem 1.1. Let p(z) be a polynomial of degree n having all zeros in |z| ≤ k, where k ≤ 1, then for all α1, ... α t with |α1| ≥ k, |α2| ≥ k,..., |α t | ≥ k, (1 ≤ t < n),

$\begin{array}{r}\underset{|z|=1}{\mathrm{max}}|{D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)|\ge \frac{n\left(n-1\right)\cdots \left(n-t+1\right)}{{\left(1+k\right)}^{t}}\left[\\ \left\{\left(|{\alpha }_{1}|-k\right)\cdots \left(|{\alpha }_{t}|-k\right)\right\}\underset{|z|=1}{\mathrm{max}}|p\left(z\right)|+\\ \left\{{\left(1+k\right)}^{t}\left(|{\alpha }_{1}|\cdots |{\alpha }_{t}|\right)-\left\{\left(|{\alpha }_{1}|-k\right)\cdots \left(|{\alpha }_{t}|-k\right)\right\}\right\}{k}^{-n}\underset{|z|=k}{\mathrm{min}}|p\left(z\right)|\right].\end{array}$
(1.9)

This result is best possible and equality holds for p(z) = (z - k)nwith α1k, α2k,..., α t k.

If we take k = 1 in Theorem 1.1, then inequality (1.9) reduces to inequality (1.8).

If we take t = 1 in Theorem 1.1, the following refinement of inequality (1.6) can be obtained.

Corollary 1.2. Let p(z) be a polynomial of degree n, having all zeros in |z| ≤ k, where k ≤ 1, then for every α with |α| ≥ k,

$\underset{|z|=1}{max}\left|{D}_{\alpha }p\left(z\right)\right|\ge \frac{n}{1+k}\left\{\left(\left|\alpha \right|-k\right)\underset{|z|=1}{max}\left|p\left(z\right)\right|+\left(\left|\alpha \right|+1\right){k}^{-\left(n-1\right)}\underset{|z|=k}{min}\left|p\left(z\right)\right|\right\}.$
(1.10)

This result is best possible and equality occurs if p(z) = (z - k)nwith αk.

If we divide both sides of the above inequality in (1.10) by |α| and make |α| → ∞, we obtain a result proved by Govil .

## 2. Lemmas

For proof of the theorem, the following lemmas are needed. The first lemma is due to Laguerre .

Lemma 2.1. If all the zeros of an nth degree polynomial p(z) lie in a circular region C and w is any zero of D α p(z), then at most one of the points w and α may lie outside C.

Lemma 2.2. If p(z) is a polynomial of degree n, having all zeros in the closed disk |z| ≤ k, k ≤ 1, then on |z| = 1,

$\left|{p}^{\prime }\left(z\right)\right|\ge \frac{n}{1+k}\left|p\left(z\right)\right|.$
(2.1)

This lemma is due to Govil .

Lemma 2.3. If p(z) is a polynomial of degree n, having no zeros in |z| < k, k ≥ 1, then on |z| = 1,

$k\left|{p}^{\prime }\left(z\right)\right|\le \left|{q}^{\prime }\left(z\right)\right|,$
(2.2)

where$q\left(z\right)={z}^{n}\overline{p\left(1∕\stackrel{̄}{z}\right)}$.

The above lemma is due to Chan and Malik .

Lemma 2.4. If p(z) is a polynomial of degree n, having all zeros in the closed disk |z| ≤ k, k ≤ 1, then on |z| = 1,

$\left|{q}^{\prime }\left(z\right)\right|\le k\left|{p}^{\prime }\left(z\right)\right|,$
(2.3)

where$q\left(z\right)={z}^{n}\overline{p\left(1∕\stackrel{̄}{z}\right)}$.

Proof. Since p(z) has all its zeros in |z| ≤ k, k ≤ 1, therefore q(z) has no zero in |z| < 1/k, 1/k ≥ 1. Now applying Lemma 2.3 to the polynomial q(z) and the result follows.

Lemma 2.5. If p(z) is a polynomial of degree n, having all zeros in the closed disk |z| ≤ k, k ≤ 1, then for every real or complex number α with |α| ≥ k and |z| = 1, we have

$\left|{D}_{\alpha }p\left(z\right)\right|\ge \frac{n}{1+k}\left(\left|\alpha \right|-k\right)\left|p\left(z\right)\right|.$
(2.4)

Proof. Let $q\left(z\right)={z}^{n}\overline{p\left(1∕\stackrel{̄}{z}\right)}$, then |q'(z)| = |np(z) - zp'(z)| on |z| = 1. Thus, on |z| = 1, we get

$\begin{array}{c}\left|{D}_{\alpha }p\left(z\right)\right|=\left|np\left(z\right)+\left(\alpha -z\right){p}^{\prime }\left(z\right)\right|=\left|\alpha {p}^{\prime }\left(z\right)+np\left(z\right)-z{p}^{\prime }\left(z\right)\right|\ge \\ \left|\alpha {p}^{\prime }\left(z\right)-|np\left(z\right)-z{p}^{\prime }\left(z\right)\right|,\end{array}$

that implies

$\left|{D}_{\alpha }p\left(z\right)\right|\ge \left|\alpha \right|\left|{p}^{\prime }\left(z\right)\right|-\left|{q}^{\prime }\left(z\right)\right|.$
(2.5)

By combining (2.3) and (2.5), we obtain

$\left|{D}_{\alpha }p\left(z\right)\right|\ge \left(\left|\alpha \right|-k\right)\left|{p}^{\prime }\left(z\right)\right|.$

that along Lemma 2.2, yields

$\left|{D}_{\alpha }p\left(z\right)\right|\ge \frac{n}{1+k}\left(\left|\alpha \right|-k\right)\left|p\left(z\right)\right|.$

Lemma 2.6. If$p\left(z\right)={a}_{0}+{a}_{1}z+{\sum }_{i=2}^{n}{a}_{i}{z}^{i}$is a polynomial of degree n, having no zeros in |z| < k, k ≥ 1, then

$\frac{k\left|{a}_{1}\right|}{\left|{a}_{0}\right|}\le n.$
(2.6)

The above lemma is due to Gardner et al. .

Lemma 2.7. If$p\left(z\right)={\sum }_{i=0}^{n}{a}_{i}{z}^{i}$is a polynomial of degree n, having all zeros in |z| ≤ k, k ≤ 1, then

$\frac{\left|{a}_{n-1}\right|}{\left|{a}_{n}\right|}\le nk.$
(2.7)

Proof. Since p(z) has all zeros in |z| ≤ k, k ≤ 1, therefore

$q\left(z\right)={z}^{n}\overline{p\left(1∕\stackrel{̄}{z}\right)}=\overline{{a}_{n}}+\overline{{a}_{n-1}}z+\cdots +\overline{{a}_{1}}{z}^{n-1}+\overline{{a}_{0}}{z}^{n},$

is a polynomial of degree at most n, which does not vanish in |z| < 1/k, 1/k ≥ 1. By applying Lemma 2.6 for q(z), we get

$\frac{\frac{1}{k}\left|{a}_{n-1}\right|}{\left|{a}_{n}\right|}\le \mathsf{\text{degree}}\left\{q\left(z\right)\right\}\le n,$

which completes the proof.

Lemma 2.8. If p(z) is a polynomial of degree n having all zeros in |z| ≤ k, k ≤ 1, then for all α1, ... α t with |α1| ≥ k, |α2| ≥ k,..., |α t | ≥ k, (1 ≤ t < n), and |z| = 1 we have

$\begin{array}{c}\left|{D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)\right|\ge \frac{n\left(n-1\right)\cdots \left(n-t+1\right)}{{\left(1+k\right)}^{t}}×\\ \left\{\left(\left|{\alpha }_{1}\right|-k\right)\cdots \left(\left|{\alpha }_{t}\right|-k\right)\right\}\left|p\left(z\right)\right|.\end{array}$
(2.8)

Proof. If |α j | = k for at least one j; 1 ≤ jt, then inequality (2.8) is trivial. Therefore, we assume that |α j | > k for all j; 1 ≤ jt.

In the rest, we proceed by mathematical induction. The result is true for t = 1, by Lemma 2.5, that means if |α1| > k then

$\left|{D}_{{\alpha }_{1}}p\left(z\right)\right|\ge \frac{n}{1+k}\left(\left|{\alpha }_{1}\right|-k\right)\left|p\left(z\right)\right|.$
(2.9)

Now for t = 2, since ${D}_{{\alpha }_{1}}p\left(z\right)=\left(n{a}_{n}{\alpha }_{1}+{a}_{n-1}\right){z}^{n-1}+\cdots +\left(n{a}_{0}+{\alpha }_{1}{a}_{1}\right)$, and |α1| > k, then ${D}_{{\alpha }_{1}}p\left(z\right)$ will be a polynomial of degree (n - 1). If it is not true, then the coefficient of zn-1must be equal to zero, which implies

$n{a}_{n}{\alpha }_{1}+{a}_{n-1}=0,$

i.e,

$\left|{\alpha }_{1}\right|=\frac{\left|{a}_{n-1}\right|}{n\left|{a}_{n}\right|}.$

Applying Lemma 2.7, we get

$\left|{\alpha }_{1}\right|=\frac{\left|{a}_{n-1}\right|}{n\left|{a}_{n}\right|}\le k.$

But this result contradicts the fact that |α1| > k. Hence, the polynomial ${D}_{{\alpha }_{1}}p\left(z\right)$ must be of degree (n - 1).

On the other hand, since all the zeros of p(z) lie in |z| ≤ k, therefore by applying Lemma 2.1, all the zeros of ${D}_{{\alpha }_{1}}p\left(z\right)$ lie in |z| ≤ k, then using Lemma 2.5 for the polynomial ${D}_{{\alpha }_{1}}p\left(z\right)$ of degree n - 1, and |α2 | > k, it concludes that

$\left|{D}_{{\alpha }_{2}}\left\{{D}_{{\alpha }_{1}}p\left(z\right)\right\}\right|\ge \frac{\left(n-1\right)}{1+k}\left(\left|{\alpha }_{2}\right|-k\right)\left|{D}_{{\alpha }_{1}}p\left(z\right)\right|.$

Substituting the term ${D}_{{\alpha }_{1}}p\left(z\right)$ from (2.9) in the above inequality, we obtain

$\left|{D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)\right|\ge \frac{n\left(n-1\right)}{{\left(1+k\right)}^{2}}\left(\left|{\alpha }_{1}\right|-k\right)\left(\left|{\alpha }_{2}\right|-k\right)\left|p\left(z\right)\right|.$

This implies result is true for t = 2.

At this stage, we assume that the result is true for t = s < n; it means that for |z| = 1, we have

$\begin{array}{c}\left|{D}_{{\alpha }_{s}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)\right|\ge \frac{n\left(n-1\right)\cdots \left(n-s+1\right)}{{\left(1+k\right)}^{s}}×\\ \left\{\left(\left|{\alpha }_{1}\right|-k\right)\cdots \left(\left|{\alpha }_{s}\right|-k\right)\right\}\left|p\left(z\right)\right|,\end{array}$
(2.10)

and we will prove that the result is true for t = s + 1 < n.

According to the above procedure, using Lemmas 2.7 and 2.1, the polynomial ${D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)$ must be of degree (n - 2) for |α1| > k, |α2| > k, and has all zeros in |z| ≤ k. One can continue that ${D}_{{\alpha }_{s}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)$ will be a polynomial of degree (n - s) for all α1,... α s with |α1| ≥ k, |α2| ≥ k,..., |α s | ≥ k, (s < n), and has all zeros in |z| ≤ k. Therefore, for |αs+1| > k, by applying Lemma 2.5 to ${D}_{{\alpha }_{s}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)$, we get

$\left|{D}_{{\alpha }_{s+1}}\left\{{D}_{{\alpha }_{s}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)\right\}\right|\ge \frac{\left(n-s\right)}{1+k}\left(\left|{\alpha }_{s+1}\right|-k\right)\left|{D}_{{\alpha }_{s}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)\right|.$
(2.11)

By combining the terms (2.10) and (2.11), we obtain

$\begin{array}{c}\left|{D}_{{\alpha }_{s+1}}{D}_{{\alpha }_{s}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)\right|\ge \frac{n\left(n-1\right)\cdots \left(n-s\right)}{{\left(1+k\right)}^{s+1}}×\\ \left\{\left(\left|{\alpha }_{1}\right|-k\right)\cdots \left(\left|{\alpha }_{s+1}\right|-k\right)\right\}\left|p\left(z\right)\right|.\end{array}$

This implies that the result is true for t = s + 1. The proof is complete.

Lemma 2.9. If$p\left(z\right)={\sum }_{i=0}^{n}{a}_{i}{z}^{i}$ is a polynomial of degree n, p(z) ≠ 0 in |z| < k, then m < |p(z)| for |z| < k, and in particular m < |a0|, where m = min|z|=k|p(z)|.

The above lemma is due to Gardner et al. .

Lemma 2.10. If$p\left(z\right)={\sum }_{i=0}^{n}{a}_{i}{z}^{i}$is a polynomial of degree n having all zeros in |z| ≤ k, then

$m\le {k}^{n}\left|{a}_{n}\right|,$
(2.12)

where m = min|z|=k|p(z)|.

Proof. If k = 0, then inequality (2.12) is trivial. Now we suppose that k > 0. Since the polynomial $p\left(z\right)={\sum }_{i=0}^{n}{a}_{i}{z}^{i}$ has all zeros in |z| ≤ k, the polynomial q(z) = znp(1/z) = a n + + a0znhas no zero in $\left|z\right|<\frac{1}{k}$. Thus, by applying Lemma 2.9 for the polynomial q(z), we get

$\underset{\left|z\right|=\frac{1}{k}}{min}\left|q\left(z\right)\right|<\left|{a}_{n}\right|.$
(2.13)

Since ${min}_{\left|z\right|=\frac{1}{k}}\left|q\left(z\right)\right|=\frac{1}{{k}^{n}}{min}_{\left|z\right|=k}\left|p\left(z\right)\right|$, (2.13) implies that $\frac{m}{{k}^{n}}<\left|{a}_{n}\right|$.

## 3. Proof of the theorem

Proof of Theorem 1.1. Let m = min|z|=k|p(z)|. If p(z) has a zero on |z| = k, then m = 0 and the result follows from Lemma 2.8. Henceforth, we suppose that all the zeros of p(z) lie in |z| < k, so that m > 0. Now m ≤ |p(z)| for |z| = k, therefore if λ is any real or complex number such that |λ| < 1, then $\left|\lambda m{\left(\frac{z}{k}\right)}^{n}\right|<\left|p\left(z\right)\right|$ for |z| = k. Since all zeros of p(z) lie in |z| < k, by Rouche's theorem we can deduce that all zeros of the polynomial $G\left(z\right)=p\left(z\right)-\lambda m{\left(\frac{z}{k}\right)}^{n}$ lie in |z| < k. Also it follows from Lemma 2.10, that $\left|\lambda \right|\frac{m}{{k}^{n}}<\left|{a}_{n}\right|$, hence the polynomial $G\left(z\right)=p\left(z\right)-\lambda \left(\frac{m}{{k}^{n}}\right){z}^{n}$ is of degree n. Now we can apply Lemma 2.8 for the polynomial G(z) of degree n which has all zeros in |z| ≤ k. This implies that for all α1,... α t with |α1| ≥ k, |α2| ≥ k, ..., |α t | ≥ k, (t < n), on |z| = 1,

$\begin{array}{c}\left|{D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}G\left(z\right)\right|\ge \frac{n\left(n-1\right)\cdots \left(n-t+1\right)}{{\left(1+k\right)}^{t}}×\\ \left\{\left(\left|{\alpha }_{1}\right|-k\right)\cdots \left(\left|{\alpha }_{t}\right|-k\right)\right\}\left|G\left(z\right)\right|.\end{array}$

Equivalently

$\begin{array}{c}\left|{D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)-\lambda \frac{m}{{k}^{n}}\left\{n\left(n-1\right)\cdots \left(n-t+1\right){\alpha }_{1}{\alpha }_{2}\cdots {\alpha }_{1}\right\}{z}^{n-t}\right|\ge \\ \frac{n\left(n-1\right)\cdots \left(n-t+1\right)}{{\left(1+k\right)}^{t}}\left\{\left(\left|{\alpha }_{1}\right|-k\right)\cdots \left(\left|{\alpha }_{t}\right|-k\right)\right\}\left|p\left(z\right)-\lambda m{\left(\frac{z}{k}\right)}^{n}\right|.\end{array}$
(3.1)

But by Lemma 2.1, the polynomial $T\left(z\right)={D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}G\left(z\right)$ has all zeros in |z| ≤ k. That is,

$T\left(z\right)={D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}G\left(z\right)\ne 0,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}for\phantom{\rule{2.77695pt}{0ex}}\left|z\right|>k.$

Then, substituting G(z) in the above, we conclude that for every λ with |λ| < 1, and |z| > k,

$\begin{array}{c}T\left(z\right)={D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)-\\ \lambda \frac{m}{{k}^{n}}\left\{n\left(n-1\right)\cdots \left(n-t+1\right){\alpha }_{1}{\alpha }_{2}\cdots {\alpha }_{t}\right\}{z}^{n-t}\ne 0.\end{array}$
(3.2)

Thus, for |z| > k,

$\left|{D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)\right|\ge \frac{m}{{k}^{n}}\left\{n\left(n-1\right)\cdots \left(n-t+1\right)\left|{\alpha }_{1}\right|\left|{\alpha }_{2}\right|\cdots \left|{\alpha }_{t}\right|\right\}\left|{z}^{n-t}\right|.$
(3.3)

If the inequality (3.3) is not true, then there is a point z = z0 with |z0| > k such that

$\left|{D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left({z}_{0}\right)\right|<\frac{m}{{k}^{n}}\left\{n\left(n-1\right)\cdots \left(n-t+1\right)\left|{\alpha }_{1}\right|\left|{\alpha }_{2}\right|\cdots \left|{\alpha }_{t}\right|\right\}\left|{z}_{0}^{n-t}\right|.$

Now take

$\lambda =\frac{{D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left({z}_{0}\right)}{\frac{m}{{k}^{n}}\left\{n\left(n-1\right)\cdots \left(n-t+1\right){\alpha }_{1}{\alpha }_{2}\cdots {\alpha }_{t}\right\}{z}_{0}^{n-t}},$

then |λ| < 1 and with this choice of λ, we have, T(z0) = 0 for |z0| > k, from (3.2). But it contradicts the fact that T(z) ≠ 0 for |z| > k. Hence, for |z| > k, we have

$\left|{D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)\right|\ge \frac{m}{{k}^{n}}\left\{n\left(n-1\right)\cdots \left(n-t+1\right)\left|{\alpha }_{1}\right|\left|{\alpha }_{2}\right|\cdots \left|{\alpha }_{t}\right|\right\}\left|{z}^{n-t}\right|.$

Taking a relevant choice of argument of λ, arg $\lambda =arg\left\{{D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)\right\}-arg\left\{{\alpha }_{1}{\alpha }_{2}\cdots {\alpha }_{t}{z}^{n-t}\right\}$, we have

where |z| = 1.

Therefore, we can rewrite (3.1) as

where |z| = 1.

In an equivalent way

$\begin{array}{c}|{D}_{{\alpha }_{t}}\cdots {D}_{{\alpha }_{2}}{D}_{{\alpha }_{1}}p\left(z\right)|\ge \frac{n\left(n-1\right)\cdots \left(n-t+1\right)}{{\left(1+k\right)}^{t}}\left[\\ \left\{\left(|{\alpha }_{1}|-k\right)\cdots \left(|{\alpha }_{t}|-k\right)|p\left(z\right)|\right\}+\\ |\text{λ}|\left\{{\left(1+k\right)}^{t}\left(|{\alpha }_{1}||{\alpha }_{2}|\cdots |{\alpha }_{t}|\right)-\left\{\left(|{\alpha }_{1}|-k\right)\cdots \left(|{\alpha }_{t}|-k\right)\right\}\right\}\frac{m}{{k}^{n}}\right].\end{array}$

Making |λ| → 1, Theorem 1.1 follows.

## References

1. Bernstein S: Leons sur les Proprits Extrmales et la Meilleure Approximation des Fonctions Analytiques dune Variable relle. Gauthier Villars, Paris; 1926.

2. Turan P: Uber die ableitung von Polynomen. Compositio Math 1939, 7: 89–95.

3. Malik MA: On the derivative of a polynomial. J Lond Math Soc 1969, 1: 57–60. 10.1112/jlms/s2-1.1.57

4. Aziz A, Dawood QM: Inequalities for a polynomial and its derivative. J Approx Theory 1988, 54: 306–313. 10.1016/0021-9045(88)90006-8

5. Shah WM: A generalization of a theorem of Paul Turan. J Ramanujan Math Soc 1996, 1: 67–72.

6. Aziz A, Rather NA: A refinement of a theorem of Paul Turan concerning polynomials. Math Ineq Appl 1998, 1: 231–238.

7. Jain VK: Generalization of an inequality involving maximum moduli of a polynomial and its polar derivative. Bull Math Soc Sci Math Roum Tome 2007, 98: 67–74.

8. Govil NK: Some inequalities for derivative of polynomials. J Approx Theory 1991, 66: 29–35. 10.1016/0021-9045(91)90052-C

9. Laguerre E: OEuvres. Vol. Volume 1. 2nd edition. Chelsea, New York; 48–66.

10. Govil NK: On the derivative of a polynomial. Proc Am Math Soc 1973, 41: 543–546. 10.1090/S0002-9939-1973-0325932-8

11. Chan TN, Malik MA: On Erdoös-Lax Theorem. Proc. Indian Acad Sci 1983, 92: 191–193. 10.1007/BF02876763

12. Gardner RB, Govil NK, Weems A: Some results concerning rate of growth of polynomials. East J on Approx 2004, 10: 301–312.

13. Gardner RB, Govil NK, Musukula SR: Rate of growth of polynomials not vanishing inside a circle. J Inequal Pure Appl Math 2005, 6: 1–9.

## Acknowledgements

The author is grateful to the referees, for the careful reading of the paper and for the helpful suggestions and comments. This research was supported by Shahrood University of Technology.

## Author information

Authors

### Competing interests

The author declares that they have no competing interests.

## Rights and permissions

Reprints and Permissions

Zireh, A. On the maximum modulus of a polynomial and its polar derivative. J Inequal Appl 2011, 111 (2011). https://doi.org/10.1186/1029-242X-2011-111

• Accepted:

• Published:

• DOI: https://doi.org/10.1186/1029-242X-2011-111

### Keywords

• Polar derivative
• Polynomial
• Inequality
• Maximum modulus
• Zeros 