1 Introduction and statement of results

For an arbitrary entire function f(z), let \(M(f,r):=\max _{|z|=r}|f(z)|\). For a polynomial P(z) of degree n, it is known that

$$\begin{aligned} M(P,\rho )\le \rho ^{n}M(P,1), \quad \rho \ge 1. \end{aligned}$$
(1.1)

Inequality (1.1) is a simple consequence of Maximum Modulus Principle (see [4]). It was shown by Ankeny and Rivlin [1] that if \(P(z)\ne 0\) in \(|z|< 1\), then (1.1) can be replaced by

$$\begin{aligned} M(P,\rho )\le \frac{\rho ^{n}+1}{2}M(P,1), \quad \rho \ge 1. \end{aligned}$$
(1.2)

In 1988, Aziz and Dawood further improved the bound in (1.2) and proved under the same hypothesis that

$$\begin{aligned} M(P,\rho )\le \frac{\rho ^{n}+1}{2}M(P,1)-\frac{\rho ^n -1}{2}\min _{|z|=1}|P(z)|, \quad \rho \ge 1. \end{aligned}$$
(1.3)

Recently, Dubinin [3] obtained the following refinement of (1.2) by using the classical Schwarz lemma.

Theorem 1

If\(P(z)=\sum \nolimits _{j=0}^{n}c_jz^j\)is a polynomial of degree\(n\ge 2\)with no zeros in\(|z|< 1\), then for any\(\rho >1\),

$$\begin{aligned} M(P,\rho )\le \frac{(1+\rho ^n)(|c_0|+\rho |c_n|)}{(1+\rho )(|c_0|+|c_n|)}M(P,1). \end{aligned}$$
(1.4)

The result is best possible and equality holds in (1.4) for \(P(z)=\frac{\mu +\nu z^n}{2}\), \(|\mu |=|\nu |=1.\)

In this note, we prove the following generalization of (1.4) which sharpens the bounds in (1.2) and (1.3) as well.

Theorem 2

If\(P(z)=\sum \nolimits _{j=0}^{n}c_jz^j\)is a polynomial of degree\(n\ge 2\)with no zeros in\(|z|< 1\), then for any\(\rho >1\)and\(0\le t\le 1,\)

$$\begin{aligned}&M(P,\rho )\le \left( \frac{(1+\rho ^n)(|c_0|+\rho |c_n|-tm)}{(1+\rho )(|c_0|+|c_n|-tm)}\right) M(P,1)\\ {}&\nonumber \qquad \qquad \quad -\left( \frac{(1+\rho ^n)(|c_0|+\rho |c_n|-tm)}{(1+\rho )(|c_0|+|c_n|-tm)}-1\right) tm, \end{aligned}$$
(1.5)

where \(m=min_{|z|=1}|P(z)|.\)

The result is best possible and equality holds in (1.5) for \(P(z)=\frac{\mu +\nu z^n}{2}\), \(|\mu |=|\nu |=1.\)

Remark 1

Since if \(P(z)=\sum \nolimits _{j=0}^{n}c_jz^j\ne 0\) in \(|z|< 1,\) then \(|c_0|\ge |c_n|.\) Also, as in the proof of the Theorem 2 (given in the next section), we have for every \(\lambda \) with \(|\lambda |\le 1,\) the polynomial \(P(z)-\lambda m\) does not vanish in \(|z|<1,\) hence

$$\begin{aligned} |c_0-\lambda m|\ge |c_n|. \end{aligned}$$
(1.6)

If in (1.6), we choose the argument of \(\lambda \) suitably and note that \(|c_0|>m\) (from (2.2), proof of Theorem 2), we get

$$\begin{aligned} |c_0|-|\lambda |m\ge |c_n|. \end{aligned}$$
(1.7)

If we take \(|\lambda |=t\) in (1.7) so that \(0\le t\le 1\), we get \(tm+|c_n|\le |c_0|.\)

Remark 2

Here, we show that for \(\rho >1,\)

$$\begin{aligned} \frac{|c_0|+\rho |c_n|-tm}{|c_0|+|c_n|-tm}\le \frac{1+\rho }{2}, \end{aligned}$$
(1.8)

which is equivalent to showing

$$\begin{aligned} |c_0|+\rho |c_n|-tm\le |c_n|+\rho |c_0|-\rho tm, \end{aligned}$$

that is

$$\begin{aligned} |c_n|+tm\le |c_0|, \end{aligned}$$

which clearly holds by Remark 1. Also, the function \(xM(P,1)-(x-1)tm\) is a non-decreasing function of x. If we combine this fact with (1.8) according to which

$$\begin{aligned} \frac{(1+\rho ^n)(|c_0|+\rho |c_n|-tm)}{(1+\rho )(|c_0|+|c_n|-tm)}\le \frac{1+\rho ^n}{2}, \end{aligned}$$

it follows that the right hand side of (1.5) does not exceed \((\frac{1+\rho ^n}{2})M(P,1)-(\frac{\rho ^n-1}{2}) t m,\) we have a refinement of (1.3).

Remark 3

For \(t=0,\) (1.5) reduces to (1.4).

2 Proof of Theorem

Proof of Theorem 2. Since \(P(z)=\sum \nolimits _{j=0}^{n}c_jz^j\) has all its zeros in \(|z|\ge 1\) and \(m=\min _{|z|=1}|P(z)|\), therefore

$$\begin{aligned} m\le |P(z)|,~~ for~ |z|=1. \end{aligned}$$
(2.1)

It follows by the Maximum and Minimum Modulus Principles that the strict inequality

$$\begin{aligned} m< |P(z)|<M(P,1), \end{aligned}$$
(2.2)

holds for \(|z|<1\).

We show that for every complex \(\alpha \) with \(|\alpha |\le 1\), the polynomial \(F(z)=P(z)-\alpha m\) does not vanish in \(|z|< 1\). For if \(F(z)=P(z)-\alpha m\) has a zero in \(|z|<1,\) say at \(z=z_1\) with \(|z_1|<1,\) then

$$\begin{aligned} F(z_{1})=P(z_{1})-\alpha m=0. \end{aligned}$$

This gives,

$$\begin{aligned} |P(z_{1})|=|\alpha | m\le m, \end{aligned}$$

where \(|z_{1}|< 1\), which contradicts (2.2).

Hence, we conclude that the polynomial F(z) does not vanish in \(|z|< 1\). Applying Theorem 1 to the polynomial \(F(z)=P(z)-\alpha m=(c_{0}-\alpha m)+\sum \nolimits _{j=1}^{n}c_{j}z^{j}\), we get for every complex \(\alpha \) with \(|\alpha |\le 1\) and \(\rho > 1\),

$$\begin{aligned} \max _{|z|=\rho }\big |P(z)-\alpha m\big |\le \bigg (\frac{\rho ^n+1}{\rho +1}\bigg )\bigg (\frac{|c_0-\alpha m|+\rho |c_n|}{|c_0-\alpha m|+|c_n|}\bigg )\max _{|z|=1}\big |P(z)-\alpha m\big |. \end{aligned}$$
(2.3)

For every \(\alpha \) with \(|\alpha |\le 1,\) we have

$$\begin{aligned} |c_{0}-\alpha m|\ge \big ||c_{0}|-|\alpha |m\big |=|c_0|-|\alpha |m, \end{aligned}$$

since \(|\alpha |m \le m < |P(0)| = |c_0|,\) by (2.2).

Further, the function \(\big (\frac{x+\rho |c_n|}{x+|c_n|}\big )\) is decreasing on \(\{x:x>-|c_n|\}\cup \{x:x<-|c_n|\}\) for every \(\rho >1,\) it follows from (2.3) that for every \(\alpha \) with \(|\alpha |\le 1\) and for every \(\rho >1,\)

$$\begin{aligned} M(P,\rho )-|\alpha |m\le \bigg (\frac{\rho ^n+1}{\rho +1}\bigg )\bigg (\frac{|c_0|-|\alpha | m+\rho |c_n|}{|c_0|-|\alpha | m+|c_n|}\bigg )\big |P(z_0)-\alpha m\big |, \end{aligned}$$
(2.4)

where \(z_0\) is a point on \(|z|=1\) such that \(|P(z_0)|=M(P,1).\) Also by (2.1) and (2.2), we have

$$\begin{aligned} m\le |P(z)|\,\, for\,\, |z|\le 1, \end{aligned}$$
(2.5)

we take in particular \(z=z_0\) in (2.5) and get

$$\begin{aligned} m\le |P(z_0)|. \end{aligned}$$
(2.6)

Choosing the argument of \(\alpha \) with \(|\alpha |\le 1\) on the right hand side of (2.4) such that

$$\begin{aligned} |P(z_0)-\alpha m|=|P(z_0)|-|\alpha |m, \end{aligned}$$

which is possible by (2.6), we obtain from (2.4) that

$$\begin{aligned} M(P,\rho )-|\alpha |m\le \bigg (\frac{\rho ^n+1}{\rho +1}\bigg )\bigg (\frac{|c_0|-|\alpha | m+\rho |c_n|}{|c_0|-|\alpha | m+|c_n|}\bigg )(\big |P(z_0)|-|\alpha | m), \end{aligned}$$

for every \(\alpha \) with \(|\alpha |\le 1\) and for every \(\rho >1\).

The above inequality is equivalent to (1.5) and this completes the proof of Theorem 2.