A generalized inequality for the polar derivative of a polynomial

Abstract

In this paper, we extend a result recently proved by Liman et al. (Complex Anal. Oper. Theory 6:1199-1209, 2012. doi:10.1007/s11785-010-0120-3) to the polar derivative of a polynomial and thereby obtain some more general results for polynomials with restricted zeros.

MSC:30A10, 30C10, 30C15.

1 Introduction and statement of results

Let $P(z):={\sum }_{j=0}^n{a}_j{z}^j$ be a polynomial of degree n, and let $P^{\prime }(z)$ be its derivative, then

$$\underset{|z|=1}{max}|P^{\prime }(z)|\le n\underset{|z|=1}{max}|P(z)|.$$

(1)

Inequality (1) is a famous result due to Bernstein and is best possible with equality holding for the polynomial $P(z)=\lambda z^n$, where λ is a complex number.

If we restrict ourselves to a class of polynomials having no zeros in $|z|<1$, then the above inequality can be sharpened. In fact, Erdös conjectured and later Lax [1] proved that if $P(z)\ne 0$ in $|z|<1$, then

$$\underset{|z|=1}{max}|P^{\prime }(z)|\le \frac{n}{2}\underset{|z|=1}{max}|P(z)|.$$

(2)

As a refinement of (2), Aziz and Dawood [2] proved that if $P(z)$ is a polynomial of degree n having no zeros in $|z|<1$, then

$$\underset{|z|=1}{max}|P^{\prime }(z)|\le \frac{n}{2}\{\underset{|z|=1}{max}|P(z)|-\underset{|z|=1}{min}|P(z)|\}.$$

(3)

As an improvement of (3), Dewan and Hans [3] proved that if $P(z)$ is a polynomial of degree n having no zeros in $|z|<1$, then for any β with $|\beta |\le 1$ and $|z|=1$,

(4)

Let $D_{\alpha }P(z)$ denote the polar derivative of the polynomial $P(z)$ of degree n with respect to α, then

$$D_{\alpha }P(z)=nP(z)+(\alpha -z)P^{\prime }(z).$$

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

$$\underset{\alpha \to \mathrm{\infty }}{lim}\frac{D_{\alpha }P(z)}{\alpha }=P^{\prime }(z).$$

As an extension of (1) to the polar derivative, Aziz and Shah ([4], Theorem 4 with $k=1$) showed that if $P(z)$ is a polynomial of degree n, then for every complex number α with $|\alpha |>1$,

$$|D_{\alpha }P(z)|\le n|\alpha |\underset{|z|=1}{max}|P(z)|\text{for }|z|=1.$$

(5)

Inequality (5) becomes equality for $P(z)=a{z}^n$, $a\ne 0$.

If we divide the two sides of (5) by $|\alpha |$ and let $|\alpha |\to \mathrm{\infty }$, we get inequality (1).

Aziz and Shah [5] proved that if $P(z)$ is a polynomial of degree n that does not vanish in $|z|<1$, then for every complex number α with $|\alpha |\ge 1$,

$$\underset{|z|=1}{max}|D_{\alpha }P(z)|\le \frac{n}{2}\{(|\alpha |+1)\underset{|z|=1}{max}|P(z)|-(|\alpha |-1)\underset{|z|=1}{min}|P(z)|\}.$$

(6)

The estimate (6) is best possible with equality for $P(z)=z^n+1$. If we divide both sides of (6) by $|\alpha |$ and make $|\alpha |\to \mathrm{\infty }$, we get inequality (3).

As an improvement and generalization to (6) and (4), Liman et al. [6] recently proved the following theorem.

Theorem 1 If $P(z)$ is a polynomial of degree n that does not vanish in $|z|<1$, then for every complex number α, β with $|\alpha |\ge 1$, $|\beta |\le 1$ and $|z|=1$,

$$\begin{array}{rcl}|z{D}_{\alpha }P(z)+n\beta \frac{|\alpha |-1}{2}P(z)|& \le & \frac{n}{2}\{(|\alpha +\beta \frac{|\alpha |-1}{2}|+|z+\beta \frac{|\alpha |-1}{2}|)\underset{|z|=1}{max}|P(z)|\\ -(|\alpha +\beta \frac{|\alpha |-1}{2}|-|z+\beta \frac{|\alpha |-1}{2}|)\underset{|z|=1}{min}|P(z)|\}.\end{array}$$

(7)

In this paper, we prove the following more general result which is an extension as well as generalization of Theorem 1 and yields a number of known polynomial inequalities.

Theorem 2 Let $P(z)$ be a polynomial of degree n that does not vanish in $|z|<k$, $k\le 1$, then for all real or complex numbers ${\alpha }_i$ with $|{\alpha }_i|\ge k$, $k\le 1$, $i=1,2,\dots ,t$, $t\le n-1$ and for any real or complex number β with $|\beta |\le 1$ and for $|z|=1$,

(8)

Remark Theorem 1 is a special case of Theorem 2 when we take $t=k=1$.

If we take $t=1$ in Theorem 2, we get the following corollary.

Corollary 1 If $P(z)$ is a polynomial of degree n that does not vanish in $|z|<k$, $k\le 1$, then for all complex numbers α, β with $|\alpha |\ge k$, $k\le 1$, $|\beta |\le 1$, and for $|z|=1$,

$$\begin{array}{rcl}|z{D}_{\alpha }P(z)+n\beta \frac{|\alpha |-k}{1+k}P(z)|& \le & \frac{n}{2}\{(\frac{1}{k^n}|\alpha +\beta \frac{|\alpha |-k}{1+k}|+|z+\beta \frac{|\alpha |-k}{1+k}|)\underset{|z|=1}{max}|P(z)|\\ -(\frac{1}{k^n}|\alpha +\beta \frac{|\alpha |-k}{1+k}|-|z+\beta \frac{|\alpha |-k}{1+k}|)\underset{|z|=k}{min}|P(z)|\}.\end{array}$$

If we take $k=1$ in Theorem 2, we get the following result.

Corollary 2 Let $P(z)$ be a polynomial of degree n that does not vanish in $|z|\le 1$, then for all real or complex numbers ${\alpha }_i$ with $|{\alpha }_i|\ge 1$, $i=1,2,\dots ,t$, $t\le n-1$ and for any real or complex number β with $|\beta |\le 1$ and for $|z|=1$,

(9)

For $\beta =0$ and $t=1$ in Theorem 2, we get the following.

Corollary 3 Let $P(z)$ be a polynomial of degree n that does not vanish in $|z|<k$, $k\le 1$, then for any real or complex number α with $|\alpha |\ge k$, $k\le 1$,

$$\underset{|z|=1}{max}|D_{\alpha }P(z)|\le \frac{n}{2}\{(\frac{|\alpha |}{k^n}+1)\underset{|z|=1}{max}|P(z)|-(\frac{|\alpha |}{k^n}-1)\underset{|z|=k}{min}|P(z)|\}.$$

(10)

If we take $k=1$ in Corollary 3, then (10) reduces to (6).

By taking $t=1$ in (8), dividing both sides of (8) by $|\alpha |$ and letting $|\alpha |\to \mathrm{\infty }$, we have the following generalization of inequality (4).

Corollary 4 Let $P(z)$ be a polynomial of degree n that does not vanish in $|z|<k$, $k\le 1$, then for any real or complex number β with $|\beta |\le 1$and $|z|=1$,

$$\begin{array}{rcl}|z{P}^{\prime }(z)+\frac{n\beta }{1+k}P(z)|& \le & \frac{n}{2}\{(\frac{1}{k^n}|1+\frac{\beta }{1+k}|+\left|\frac{\beta }{1+k}\right|)\underset{|z|=1}{max}|P(z)|\\ -(\frac{1}{k^n}|1+\frac{\beta }{1+k}|-\left|\frac{\beta }{1+k}\right|)\underset{|z|=k}{min}|P(z)|\}.\end{array}$$

(11)

Taking $\beta =0$ and $k=1$ in Corollary 4, (11) reduces to (3).

2 Lemmas

We require the following lemmas. The first lemma follows from Laguerre’s theorem [[7], p.52] (see also [8]).

Lemma 1 If all the zeros of the nth degree polynomial $P(z)$ lie in a circular region $\mathcal{C}$, and if ξ is any zero of

$$D_{\alpha }P(z)=nP(z)+(\alpha -z)P^{\prime }(z),$$

the polar derivative of $P(z)$, then both points ξ and α may not lie outside of $\mathcal{C}$.

By repeated applications of Lemma 1, we get the following result, when the circular region $\mathcal{C}$ is the circle $|z|\le r$.

Lemma 2 If all the zeros of the nth degree polynomial P(z) lie in $|z|\le r$ and if none of the points ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_t$ lie in $|z|\le r$, then each of the polar derivatives $D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}P(z)$, $t=1,2,\dots ,n-1$, has all its zeros in $|z|\le r$.

Lemma 3 If $P(z):=a_0+{\sum }_{j=\mu }^n{a}_j{z}^j$, $1\le \mu \le n$, is a polynomial of degree n having no zeros in the disk $|z|<k$, $k\ge 1$, then

$$k^{\mu }|P^{\prime }(z)|\le |Q^{\prime }(z)|\mathit{\text{for}}|z|=1,$$

where $Q(z)=z^n\overline{P(\frac{1}{\overline{z}})}$ and $\frac{\mu }{n}|\frac{a_{\mu }}{a_0}|k^{\mu }\le 1$.

The above lemma is due to Chan and Malik [9].

Lemma 4 If $P(z):=a_n{z}^n+{\sum }_{j=\mu }^n{a}_{n-j}{z}^{n-j}$ is a polynomial of degree n having all its zeros in the disk $|z|\le k\le 1$, then

$$|Q^{\prime }(z)|\le k^{\mu }|P^{\prime }(z)|\mathit{\text{for}}|z|=1,1\le \mu \le n,$$

(12)

where $Q(z)=z^n\overline{P(\frac{1}{\overline{z}})}$.

Proof of Lemma 4 Since all the zeros of $P(z)$ lie in $|z|\le k\le 1$, therefore all the zeros of $Q(z)=z^n\overline{P(\frac{1}{\overline{z}})}$ lie in $|z|\ge \frac{1}{k}\ge 1$. Hence applying Lemma 3 to the polynomial $Q(z):={\overline{a}}_n+{\sum }_{j=\mu }^n{\overline{a}}_{n-j}{z}^j$, we get

$$\frac{1}{K^{\mu }}|Q^{\prime }(z)|\le |P^{\prime }(z)|.$$

Hence, inequality (12) follows. □

Lemma 5 Let $P(z)=a_n{z}^n+{\mathrm{\Sigma }}_{j=\mu }^n{a}_{n-j}{z}^{n-j}$, $1\le \mu \le n$, be a polynomial of degree n having all its zeros in the disk $|z|\le k$, $k\le 1$, then for every real or complex number α with $|\alpha |\ge k$, $k\le 1$ and for $|z|=1$,

$$|D_{\alpha }P(z)|\ge n\left(\frac{|\alpha |-k^{\mu }}{1+k^{\mu }}\right)|P(z)|.$$

Proof of Lemma 5 Let $Q(z)=z^n\overline{P(\frac{1}{\overline{z}})}$, we have $P(z)=z^n\overline{Q(\frac{1}{\overline{z}})}$. Then it can be easily verified that

$$|Q^{\prime }(z)|=|nP(z)-z{P}^{\prime }(z)|\text{for }|z|=1.$$

(13)

Since $P(z)$ has all its zeros in $|z|\le k\le 1$, by Lemma 4, we get

$$|Q^{\prime }(z)|\le k^{\mu }|P^{\prime }(z)|\text{for }|z|=1.$$

This implies

$$|P^{\prime }(z)|+|Q^{\prime }(z)|\le (1+k^{\mu })|P^{\prime }(z)|.$$

(14)

Also, for $|z|=1$, by using (13), we have

$$n|P(z)|=|nP(z)+z{P}^{\prime }(z)-z{P}^{\prime }(z)|\le |nP(z)-z{P}^{\prime }(z)|+|P^{\prime }(z)|=|Q^{\prime }(z)|+|P^{\prime }(z)|.$$

Using (14) in the above inequality, we get

$$n|P(z)|\le (1+k^{\mu })|P^{\prime }(z)|,$$

or

$$|P^{\prime }(z)|\ge \frac{n}{(1+k^{\mu })}|P(z)|\text{for }|z|=1.$$

(15)

For every real or complex number α with $|\alpha |\ge k$, $k\le 1$, we have

$$|D_{\alpha }P(z)|=|nP(z)+(\alpha -z)P^{\prime }(z)|.$$

Now, by using Lemma 4 and (15), we have

$$\begin{array}{rcl}|D_{\alpha }P(z)|& =& |nP(z)+(\alpha -z)P^{\prime }(z)|\\ =& |\alpha P^{\prime }(z)+nP(z)-z{P}^{\prime }(z)|\\ \ge & |\alpha ||P^{\prime }(z)|-|nP(z)-z{P}^{\prime }(z)|\\ =& |\alpha ||P^{\prime }(z)|-|Q^{\prime }(z)|\text{for }|z|=1\\ \ge & |\alpha ||P^{\prime }(z)|-k^{\mu }|P^{\prime }(z)|\\ =& (|\alpha |-k^{\mu })|P^{\prime }(z)|\\ \ge & n\left(\frac{|\alpha |-k^{\mu }}{1+k^{\mu }}\right)|P^{\prime }(z)|\text{for }|z|=1.\end{array}$$

This completes the proof of Lemma 5. □

Lemma 6 Let $P(z)=a_n{z}^n+{\sum }_{j=\mu }^n{a}_{n-j}{z}^{n-j}$, $1\le \mu \le n$, be a polynomial of degree n having all its zeros in the disk $|z|\le k$, $k\le 1$, then for every real or complex number ${\alpha }_i$ with $|{\alpha }_i|\ge k$, $k\le 1$, $i=1,2,\dots ,t$, $t\le n-1$ and for $|z|=1$,

(16)

Proof of Lemma 6 If $|{\alpha }_i|=k$ for at least one i, $1\le i\le t$, then inequality (16) is trivial. Thus, we assume that $|{\alpha }_i|>k$, $k\le 1$, for all $1\le i\le t$. We proceed by the principle of mathematical induction. The result is true for $t=1$ by Lemma 5, that is, if $|{\alpha }_1|>k$, then

$$|D_{{\alpha }_1}P(z)|\ge n\left(\frac{|{\alpha }_1|-k^{\mu }}{1+k^{\mu }}\right)|P(z)|.$$

(17)

Now, for $t=2$ and for $|{\alpha }_1|>k$, $D_{{\alpha }_1}P(z)$ will be a polynomial of degree at most $n-1$. Since all the zeros of $P(z)$ in $|z|\le k$, $k\le 1$, therefore, by applying Lemma 1, all the zeros of $D_{{\alpha }_1}P(z)$ lie in $|z|\le k$, $k\le 1$, then using Lemma 5 for the polynomial $D_{{\alpha }_1}P(z)$ of degree at most $n-1$, and for $|{\alpha }_2|>k$, we have

$$\left|D_{{\alpha }_2}\{D_{{\alpha }_1}P(z)\}\right|\ge (n-1)\left(\frac{|{\alpha }_2|-k^{\mu }}{1+k^{\mu }}\right)|D_{{\alpha }_1}P(z)|.$$

(18)

Combining (17) and (18), we get

$$|D_{{\alpha }_2}{D}_{{\alpha }_1}P(z)|\ge n(n-1)\left\{\frac{(|{\alpha }_1|-k^{\mu })(|{\alpha }_2|-k^{\mu })}{{(1+k^{\mu })}^2}\right\}|P(z)|.$$

So, the result is true for $t=2$. Now, we assume that the result is true for $t=\upsilon <n$; that is, for $|z|=1$,

(19)

We need to show that the result is true for $t=\upsilon +1$.

Now corresponding to an n th degree polynomial $P(z)$ whose all zeros lie in the disk $|z|\le k$, $k\le 1$, we construct $D_{{\alpha }_{\upsilon }}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}P(z)$ a polynomial of degree at most $n-\upsilon$ for all real or complex numbers ${\alpha }_i$ with $|{\alpha }_i|\ge k$, $k\le 1$, $i=1,2,\dots ,\upsilon$ ($\upsilon <n$) whose all zeros lie in $|z|\le k$. Therefore, for $|{\alpha }_{\upsilon +1}|>k$, by applying Lemma 5 to $D_{{\alpha }_{\upsilon }}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}P(z)$, we get

(20)

Combining (19) and (20), we obtain

This implies that the result is true for $t=\upsilon +1$ and this completes the proof of Lemma 6. □

Lemma 7 Let $P(z)=a_n{z}^n+{\mathrm{\Sigma }}_{j=\mu }^n{a}_{n-j}{z}^{n-j}$, $1\le \mu \le n$, be a polynomial of degree n having all its zeros in the disk $|z|\le k$, $k\le 1$. Then for every real or complex number ${\alpha }_i$ with $|{\alpha }_i|\ge k$, $k\le 1$, $i=1,2,\dots ,t$, $t\le n-1$, and for any real or complex number β with $|\beta |\le 1$ and for $|z|=1$,

(21)

Proof of Lemma 7 The result is clear if $P(z)$ has a zero on $|z|=k$, then $m={min}_{|z|=k}|P(z)|=0$. We now suppose that all the zeros of $P(z)$ lie in $|z|<k$, then $m>0$, and we have $m\le |P(z)|$ for $|z|=k$. Hence, for every λ with $|\lambda |<1$, we have $|P(z)|>|m\lambda {(\frac{z}{k})}^n|$ for $|z|=k$. Therefore, it follows by Rouche’s theorem that the polynomial $G(z)=P(z)-m\lambda {(\frac{z}{k})}^n$ has all its zeros in $|z|<k$, $k\le 1$. As ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_t$ are complex numbers with $|{\alpha }_i|\ge k$, $k\le 1$, $i=1,2,\dots ,t$, $t\le n-1$, then by Lemma 2 all the zeros of

lie in $|z|<k$. Applying Lemma 6 to the polynomial $G(z)$, we get

(22)

Since $z^t{D}_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}G(z)$ has all its zeros in $|z|<k$, $k\le 1$, therefore by Rouche’s theorem, it follows from inequality (22) that the polynomial

has all its zeros in $|z|<1$, where $|\beta |<1$.

By substituting for $G(z)$, we conclude that the polynomial

$$\begin{array}{rcl}T(z)& =& z^t{D}_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}P(z)+n(n-1)\cdots (n-t+1)\\ \times \beta \left(\frac{(|{\alpha }_1|-k^{\mu })(|{\alpha }_2|-k^{\mu })\cdots (|{\alpha }_t|-k^{\mu })}{{(1+k^{\mu })}^t}\right)P(z)\\ -\lambda m{\left(\frac{z}{k}\right)}^{n}n(n-1)\cdots (n-t+1)\\ \times \{{\alpha }_1\cdot {\alpha }_2\cdots {\alpha }_t+\beta \left(\frac{(|{\alpha }_1|-k^{\mu })(|{\alpha }_2|-k^{\mu })\cdots (|{\alpha }_t|-k^{\mu })}{{(1+k^{\mu })}^t}\right)\}\end{array}$$

(23)

will have no zeros in $|z|\ge 1$. This implies that for every β with $|\beta |<1$ and $|z|\ge 1$,

(24)

If (24) is not true, then there exists a point $z=\omega$ with $|\omega |\ge 1$ such that

(25)

We take

$$\lambda =\frac{{\omega }^t{D}_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}P(\omega )+n(n-1)\cdots (n-t+1)\beta \{\frac{(|{\alpha }_1|-k^{\mu })(|{\alpha }_2|-k^{\mu })\cdots (|{\alpha }_t|-k^{\mu })}{{(1+k^{\mu })}^t}\}P(\omega )}{n(n-1)\cdots (n-t+1)m{(\frac{\omega }{k})}^n({\alpha }_1\cdot {\alpha }_2\cdots {\alpha }_t+\beta \{\frac{(|{\alpha }_1|-k^{\mu })(|{\alpha }_2|-k^{\mu })\cdots (|{\alpha }_t|-k^{\mu })}{{(1+k^{\mu })}^t}\})}$$

so that $|\lambda |<1$, and with this choice of λ, we have $T(\omega )=0$ for $|\omega |\ge 1$ from (23). But this contradicts the fact that $T(z)\ne 0$ for $|z|\ge 1$. For β with $|\beta |=1$, (24) follows by continuity. This completes the proof of Lemma 7. □

Lemma 8 Let $P(z)=a_n{z}^n+{\mathrm{\Sigma }}_{j=\mu }^n{a}_{n-j}{z}^{n-j}$, $1\le \mu \le n$, be a polynomial of degree n. Then for all real or complex numbers ${\alpha }_i$ with $|{\alpha }_i|\ge k$, $k\le 1$, $i=1,2,\dots ,t$, $t\le n-1$, and for any real or complex number β with $|\beta |\le 1$ and for $|z|=1$,

Proof of Lemma 8 Let $M={max}_{|z|=k}|P(z)|$, if $|\lambda |<1$, then $|\lambda P(z)|<|M{(\frac{z}{k})}^n|$ for $|z|=k$. Therefore, it follows by Rouche’s theorem that the polynomial $G(z)=M{(\frac{z}{k})}^n-\lambda P(z)$ has all its zeros in $|z|<k$, $k\le 1$. As ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_t$ are complex numbers with $|{\alpha }_i|\ge k$, $k\le 1$, $i=1,2,\dots ,t$, $t\le n-1$, then by Lemma 2 all the zeros of

$$D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}G(z)=\frac{M}{k^n}\{n(n-1)\cdots (n-t+1){\alpha }_1\cdot {\alpha }_2\cdots {\alpha }_t\}{z}^{n-t}-\lambda D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}P(z)$$

lie in $|z|<k$. Applying Lemma 6 to the polynomial $G(z)$, we have

 □

Now, if we proceed similarly as in Lemma 7, the result follows.

Lemma 9 Let $P(z)$ be a polynomial of degree n, then for all real or complex numbers ${\alpha }_i$ with $|{\alpha }_i|\ge k$, $i=1,2,\dots ,t$, and for any real or complex number β with $|\beta |\le 1$,

where

$$Q(z):={\left(\frac{z}{k}\right)}^n\overline{P\left(\frac{k^2}{\overline{z}}\right)}.$$

Proof of Lemma 9 Let $M={max}_{|z|=k}|P(z)|$, then $|P(z)|\le M$ for $|z|\le k$. If λ is any real or complex number with $|\lambda |>1$, then by Rouche’s theorem the polynomial $G(z)=P(z)-\lambda M$ does not vanish in $|z|<k$. Consequently, the polynomial

$$H(z):={\left(\frac{z}{k}\right)}^n\overline{G\left(\frac{k^2}{\overline{z}}\right)}$$

has all zeros in $|z|\le k$ and $|G(z)|=|H(z)|$ for $|z|=k$. Since all the zeros of $H(z)$ lie in $|z|\le k$, $k\le 1$, therefore, for δ with $|\delta |>1$, by Rouche’s theorem, all the zeros of $G(z)+\delta H(z)$ lie in $|z|\le k$. Hence, by taking $\mu =1$ in Lemma 6 and using it, for every real or complex number ${\alpha }_i$ with $|{\alpha }_i|\ge k$, $i=1,2,\dots ,t$ ($1\le t<n$), $k\le 1$, and $|z|=1$, we have

Also, by Lemma 2, all the zeros $D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}(G(z)+\delta H(z))$ lie in $|z|<k\le 1$, where $|{\alpha }_i|\ge k$, $i=1,2,\dots ,t$ ($1\le t<n$). Therefore, for any β with $|\beta |<1$, Rouche’s theorem implies that all the zeros of

lie in $|z|<1$.

So, the polynomial

$$\begin{array}{rcl}T(z)& =& z^t{D}_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}G(z)+\beta n(n-1)\cdots (n-t+1)\\ \times \left\{\frac{(|{\alpha }_1|-k)(|{\alpha }_2|-k)\cdots (|{\alpha }_t|-k)}{{(1+k)}^t}\right\}G(z)\\ +\delta (z^t{D}_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}H(z)+\beta n(n-1)\cdots (n-t+1)\\ \times \left\{\frac{(|{\alpha }_1|-k)(|{\alpha }_2|-k)\cdots (|{\alpha }_t|-k)}{{(1+k)}^t}\right\}H(z))\end{array}$$

will have no zeros in $|z|\ge 1$. Now, using a similar argument as that in the proof of Lemma 7 and taking $\mu =1$, we get for $|z|\ge 1$,

(26)

Now,

$$H(z):={\left(\frac{z}{k}\right)}^n\overline{G\left(\frac{k^2}{\overline{z}}\right)}={\left(\frac{z}{k}\right)}^n\overline{P\left(\frac{k^2}{\overline{z}}\right)}-\overline{\lambda }{\left(\frac{z}{k}\right)}^{n}M=Q(z)-\overline{\lambda }{\left(\frac{z}{k}\right)}^{n}M.$$

On substituting G(z) and H(z) in (26), we obtain the following:

This implies that

(27)

As $|P(z)|=|Q(z)|$ for $|z|=k$, that is, $M={max}_{|z|=k}|P(z)|={max}_{|z|=k}|Q(z)|$, by taking $\mu =1$ in Lemma 8 and using it to the polynomial $Q(z)$, we obtain the following:

Thus, taking the argument of λ suitably, we obtain

(28)

Using (28) in (27), we get for $|z|=1$ and $|\beta |<1$,

That is,

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

(29)

Then, by applying the maximum modulus principle for the polynomial $P(z)$ when $|k|\le 1$,

$$\underset{|z|=k}{max}|P(z)|\le \underset{|z|=1}{max}|P(z)|.$$

This in conjunction with (29) and the argument of continuity gives the result. □

Lemma 10 Let $H(z)$ be a polynomial of degree n having all its zeros in $|z|\le k$, $k\le 1$, and let $G(z)$ be a polynomial of degree not exceeding that of $H(z)$. If $|G(z)|\le |H(z)|$ for $|z|=k$, $k\le 1$, then for all real or complex numbers ${\alpha }_i$ with $|{\alpha }_i|\ge k$, $i=1,2,\dots ,t$, and for any real or complex number β with $|\beta |\le 1$, and $|z|=1$, we have

Proof of Lemma 10 For any real or complex number λ with $|\lambda |<1$, we have $|\lambda G(z)|<|G(z)|\le |H(z)|$ for $|z|=k$, $k\le 1$; therefore, by Rouche’s theorem $H(z)-\lambda G(z)$ and $H(z)$ have the same number of zeros in $|z|<k$. Also, as $|G(z)|\le |H(z)|$ for $|z|=k$, $k\le 1$, any zero of $H(z)$ that lies on $|z|=k$ is also a zero of $G(z)$. Therefore, $H(z)-\lambda G(z)$ has all its zeros in the closed disk $|z|\le k$. Therefore, using Lemma 6 with $\mu =1$, we have for every real or complex number ${\alpha }_i$ with $|{\alpha }_i|\ge k$, $i=1,2,\dots ,t$, $k\le 1$ $(1\le t<n)$ and $|z|=1$,

Now, by a similar argument as that used in the proof of Lemma 7, for any real or complex number β with $|\beta |<1$, we get

for $|z|=1$, which implies

$$\begin{array}{rcl}T(z)& =& z^t{D}_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}H(z)-\lambda z^t{D}_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}G(z)\\ +n(n-1)\cdots (n-t+1)\beta \left\{\frac{(|{\alpha }_1|-k)(|{\alpha }_2|-k)\cdots (|{\alpha }_t|-k)}{{(1+k)}^t}\right\}\\ \times (H(z)-\lambda G(z))\ne 0\end{array}$$

for $|z|=1$.

That is,

$$\begin{array}{rcl}T(z)& =& z^t{D}_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}H(z)+n(n-1)\cdots (n-t+1)\\ \times \beta \left\{\frac{(|{\alpha }_1|-k)(|{\alpha }_2|-k)\cdots (|{\alpha }_t|-k)}{{(1+k)}^t}\right\}H(z)\\ -\lambda [z^t{D}_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}G(z)+n(n-1)\cdots (n-t+1)\\ \times \beta \left\{\frac{(|{\alpha }_1|-k)(|{\alpha }_2|-k)\cdots (|{\alpha }_t|-k)}{{(1+k)}^t}\right\}G(z)]\ne 0\end{array}$$

(30)

for $|z|=1$.

So, we conclude that

(31)

for $|z|=1$.

If (31) is not true, then there exists a point $z=\omega$ with $|\omega |=1$ such that

Take

$$\lambda =\frac{{\omega }^t{D}_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}H(\omega )+n(n-1)\cdots (n-t+1)\beta \{\frac{(|{\alpha }_1|-k)(|{\alpha }_2|-k)\cdots (|{\alpha }_t|-k)}{{(1+k)}^t}\}H(\omega )}{{\omega }^t{D}_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}G(\omega )+n(n-1)\cdots (n-t+1)\beta \{\frac{(|{\alpha }_1|-k)(|{\alpha }_2|-k)\cdots (|{\alpha }_t|-k)}{{(1+k)}^t}\}G(\omega )},$$

then $|\lambda |<1$, and with this choice of λ, we have from (30) $T(\omega )=0$ for $|\omega |=1$. But this contradicts the fact that $T(z)\ne 0$ for $|z|=1$. For β with $|\beta |=1$, (31) follows by continuity. This completes the proof of Lemma 10. □

3 Proof of the theorem

Since the polynomial $P(z)$ has no zeros in the disk $|z|<k$, and therefore, if $m={min}_{|z|=k}|P(z)|$, then $m\le |P(z)|$ for $|z|\le k$. If λ is any real or complex number with $|\lambda |<1$, we have

$$|\lambda m|<m\le |P(z)|\text{for }|z|=k.$$

Thus, by Rouche’s theorem, the polynomial $G(z)=P(z)-\lambda m$ does not vanish in $|z|<k$. Consequently, the polynomial

$$H(z):={\left(\frac{z}{k}\right)}^n\overline{G\left(\frac{k^2}{\overline{z}}\right)}=Q(z)-\overline{\lambda }m{\left(\frac{z}{k}\right)}^n$$

has all its zeros in $|z|\le k$, where $Q(z)={(\frac{z}{k})}^n\overline{P(\frac{k^2}{\overline{z}})}$, and also we have $|G(z)|=|H(z)|$ for $|z|=k$.

Applying Lemma 10 to the polynomials $H(z)$ and $G(z)$, we have

for every real or complex number ${\alpha }_i$ with $|{\alpha }_i|\ge k$, $i=1,2,\dots ,t$, and for any real or complex number β with $|\beta |\le 1$ and $|z|=1$.

On substituting $G(z)$ and $H(z)$ in the above inequality, we obtain the following for every real or complex number ${\alpha }_i$ with $|{\alpha }_i|\ge k$, $i=1,2,\dots ,t$, and for any real or complex number β with $|\beta |\le 1$ and $|z|=1$,

(32)

Since all the zeros of $Q(z)$ lie in $|z|\le k$ and $|P(z)|=|Q(z)|$ for $|z|=k$, therefore, by applying Lemma 7 to $Q(z)$, we have

Then, for an appropriate choice of the argument of λ, we get

(33)

for $|z|=1$.

Then combining the right-hand side of (32) and (33), we rewrite (32) as

Equivalently,

As $|\lambda |\to 1$, we have

It implies that for every real or complex number β with $|\beta |\le 1$ and $|z|=1$,

This in conjunction with Lemma 9 gives, for $|\beta |\le 1$ and $|z|=1$,

This completes the proof of the theorem.