Integral mean estimates for the polar derivative of polynomials whose zeros are within a circle

Abstract

For a polynomial $p(z)$ of degree n, having all zeros in $|z|\le k$, where $k\le 1$, Dewan et al. (Southeast Asian Bull. Math. 34:69-77, 2010) proved that for every $\alpha \in \mathbb{C}$ with $|\alpha |\ge k$ and for each $r>0$,

$$n(|\alpha |-k){\left\{{\int }_0^{2\pi }{\left|p\left(e^{i\theta }\right)\right|}^{r}d\theta \right\}}^{\frac{1}{r}}\le {\left\{{\int }_0^{2\pi }{|1+k{e}^{i\theta }|}^{r}d\theta \right\}}^{\frac{1}{r}}\underset{|z|=1}{max}|D_{\alpha }p(z)|.$$

In this paper we improve and extend the above inequality. Our result generalizes certain well-known polynomial inequalities.

MSC:30A10, 30C10, 30D15.

1 Introduction and statement of results

Let $p(z)$ be a polynomial of degree n. Then according to Bernstein’s inequality [1] on the derivative of a polynomial, we have

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

(1.1)

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

If we restrict to the class of polynomials which have all zeros in $|z|\le 1$, then it has been proved by Turán [2] that

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

(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 [3] proved that if $p(z)$ has all zeros in $|z|\le k$, where $k\le 1$, then

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

(1.3)

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

On the other hand, Malik [4] obtained a generalization of (1.2) in the sense that the right-hand side of (1.2) is replaced by a factor involving the integral mean of $p(z)$ on $|z|=1$. In fact he proved that if $p(z)$ has all its zeros in $|z|\le 1$, then for each $r>0$,

$$n{\left[{\int }_0^{2\pi }{\left|p\left(e^{i\theta }\right)\right|}^{r}d\theta \right]}^{\frac{1}{r}}\le {\left[{\int }_0^{2\pi }{|1+e^{i\theta }|}^{r}d\theta \right]}^{\frac{1}{r}}\underset{|z|=1}{max}|p^{\prime }(z)|.$$

(1.4)

As an extension of (1.4), Aziz [5] proved that if $p(z)$ has all its zeros in $|z|\le k\le 1$, then for each $r>0$,

$$n{\left[{\int }_0^{2\pi }{\left|p\left(e^{i\theta }\right)\right|}^{r}d\theta \right]}^{\frac{1}{r}}\le {\left[{\int }_0^{2\pi }{|1+k{e}^{i\theta }|}^{r}d\theta \right]}^{\frac{1}{r}}\underset{|z|=1}{max}|p^{\prime }(z)|.$$

(1.5)

As a generalization of (1.5), Aziz and Shah [6] proved that if $p(z)=a_n{z}^n+{\sum }_{\nu =\mu }^n{a}_{n-\nu }{z}^{n-\nu }$, $1\le \mu \le n$, is a polynomial of degree n, having all its zeros in $|z|\le k\le 1$, then for each $r>0$,

$$n{\left[{\int }_0^{2\pi }{\left|p\left(e^{i\theta }\right)\right|}^{r}d\theta \right]}^{\frac{1}{r}}\le {\left[{\int }_0^{2\pi }{|1+k^{\mu }{e}^{i\theta }|}^{r}d\theta \right]}^{\frac{1}{r}}\underset{|z|=1}{max}|p^{\prime }(z)|.$$

(1.6)

Let $D_{\alpha }p(z)$ denote the polar derivative of the polynomial $p(z)$ of degree n with respect to $\alpha \in \mathbb{C}$. 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

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

Shah [7] extended (1.2) to the polar derivative of $p(z)$ and proved that if all zeros of the polynomial $p(z)$ lie in $|z|\le 1$, then for every α with $|\alpha |\ge 1$, we have

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

(1.7)

This result is best possible and equality holds for $p(z)={(z-1)}^n$ with $\alpha \ge 1$.

Aziz and Rather [8] extended the inequality (1.3) to the polar derivative of a polynomial. In fact, they proved that if all zeros of $p(z)$ lie in $|z|\le k$, $k\le 1$, then for every α with $|\alpha |\ge k$, we get

$$\underset{|z|=1}{max}|D_{\alpha }p(z)|\ge \frac{n}{1+k}(|\alpha |-k)\underset{|z|=1}{max}|p(z)|.$$

(1.8)

This result is best possible and equality holds for $p(z)={(z-k)}^n$ with $\alpha \ge k$.

Recently Dewan et al. [9] generalized the inequalities (1.5) and (1.8). They proved that if $p(z)$ has all its zeros in $|z|\le k\le 1$, then for every $\alpha \in \mathbb{C}$ with $|\alpha |\ge k$ and for each $r>0$,

$$n(|\alpha |-k){\left\{{\int }_0^{2\pi }{\left|p\left(e^{i\theta }\right)\right|}^{r}d\theta \right\}}^{\frac{1}{r}}\le {\left\{{\int }_0^{2\pi }{|1+k{e}^{i\theta }|}^{r}d\theta \right\}}^{\frac{1}{r}}\underset{|z|=1}{max}|D_{\alpha }p(z)|.$$

(1.9)

In the limiting case, when $r\to \mathrm{\infty }$, the above inequality is sharp and equality holds for the polynomial $p(z)={(z-k)}^n$ with $\alpha \ge k$.

The following result which we prove is a generalization as well as a refinement of inequalities (1.9) and (1.8). In a precise set up, we have the following.

Theorem 1.1 If $p(z)=a_n{z}^n+{\sum }_{\nu =\mu }^n{a}_{n-\nu }{z}^{n-\nu }$, $1\le \mu \le n$, is a polynomial of degree n, having all its zeros in $|z|\le k\le 1$ and $m={min}_{|z|=k}|p(z)|$, then for $\lambda ,\alpha \in \mathbb{C}$ with $|\lambda |\le 1$, $|\alpha |\ge s_{\mu }$ and $r>0$, $d>1$, $q>1$, with $\frac{1}{d}+\frac{1}{q}=1$, we have

$$\begin{array}{r}n(|\alpha |-s_{\mu }){\left[{\int }_0^{2\pi }{|p\left(e^{i\theta }\right)+\lambda m|}^{r}d\theta \right]}^{\frac{1}{r}}\\ \le {\left[{\int }_0^{2\pi }{|1+s_{\mu }{e}^{i\theta }|}^{dr}d\theta \right]}^{\frac{1}{dr}}{\left[{\int }_0^{2\pi }{\left|D_{\alpha }p\left(e^{i\theta }\right)\right|}^{qr}d\theta \right]}^{\frac{1}{qr}},\end{array}$$

(1.10)

where $s_{\mu }=\frac{n|a_n|k^{2\mu }+\mu |a_{n-\mu }|k^{\mu -1}}{n|a_n|k^{\mu -1}+\mu |a_{n-\mu }|}$. In the limiting case, when $r\to \mathrm{\infty }$, the above inequality is sharp and equality holds for the polynomial $p(z)={(z-k)}^n$ with $\alpha \ge k$.

Letting $q\to \mathrm{\infty }$ (so that $d\to 1$) in (1.10), we have the following.

Corollary 1.2 If $p(z)=a_n{z}^n+{\sum }_{\nu =\mu }^n{a}_{n-\nu }{z}^{n-\nu }$, $1\le \mu \le n$, is a polynomial of degree n, having all its zeros in $|z|\le k\le 1$ and $m={min}_{|z|=k}|p(z)|$, then for $\lambda ,\alpha \in \mathbb{C}$ with $|\lambda |\le 1$, $|\alpha |\ge s_{\mu }$ and $r>0$,

$$n(|\alpha |-s_{\mu }){\left[{\int }_0^{2\pi }{|p\left(e^{i\theta }\right)+\lambda m|}^{r}d\theta \right]}^{\frac{1}{r}}\le {\left[{\int }_0^{2\pi }{|1+s_{\mu }{e}^{i\theta }|}^{r}d\theta \right]}^{\frac{1}{r}}\underset{|z|=1}{max}|D_{\alpha }p(z)|,$$

(1.11)

where $s_{\mu }$ is defined as in Theorem  1.1.

Remark 1.3 Since by Lemma 2.3, $s_{\mu }\le k$, the inequality (1.11) provides a refinement and generalization of the inequality (1.9).

If we divide both sides of the inequality (1.11) by $|\alpha |$ and make $|\alpha |\to \mathrm{\infty }$, we obtain the following refinement and generalization of the inequality (1.6).

Corollary 1.4 If $p(z)=a_n{z}^n+{\sum }_{\nu =\mu }^n{a}_{n-\nu }{z}^{n-\nu }$, $1\le \mu \le n$, is a polynomial of degree n, having all its zeros in $|z|\le k\le 1$ and $m={min}_{|z|=k}|p(z)|$, then for every $\lambda \in \mathbb{C}$ with $|\lambda |\le 1$ and $r>0$,

$$n{\left[{\int }_0^{2\pi }{|p\left(e^{i\theta }\right)+\lambda m|}^{r}d\theta \right]}^{\frac{1}{r}}\le {\left[{\int }_0^{2\pi }{|1+s_{\mu }{e}^{i\theta }|}^{r}d\theta \right]}^{\frac{1}{r}}\underset{|z|=1}{max}|p^{\prime }(z)|,$$

(1.12)

where $s_{\mu }$ is defined as in Theorem  1.1.

Letting $r\to \mathrm{\infty }$ in (1.10) and choosing the argument of λ suitably with $|\lambda |=1$, we have the following result.

Corollary 1.5 If $p(z)=a_n{z}^n+{\sum }_{\nu =\mu }^n{a}_{n-\nu }{z}^{n-\nu }$, $1\le \mu \le n$, is a polynomial of degree n, having all its zeros in $|z|\le k\le 1$, then for every $\alpha \in \mathbb{C}$ with $|\alpha |\ge s_{\mu }$,

$$\frac{n(|\alpha |-s_{\mu })}{1+s_{\mu }}[\underset{|z|=1}{max}|p(z)|+\underset{|z|=k}{min}|p(z)|]\le \underset{|z|=1}{max}|D_{\alpha }p(z)|,$$

(1.13)

where $s_{\mu }$ is defined as in Theorem  1.1.

2 Lemmas

For the proof of the theorem, the following lemmas are needed. The first lemma is due to Laguerre [10].

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 $D_{\alpha }p(z)$, then at most one of the points w and α may lie outside C.

Lemma 2.2 If $p(z)=a_n{z}^n+{\sum }_{\nu =\mu }^n{a}_{n-\nu }{z}^{n-\nu }$; $1\le \mu \le n$, is a polynomial of degree n having all its zeros in $|z|\le k\le 1$ and $q(z)=z^n\overline{p(\frac{1}{\overline{z}})}$, then on $|z|=1$

$$|q^{\prime }(z)|\le s_{\mu }|p^{\prime }(z)|$$

(2.1)

and

$$\frac{\mu }{n}\left|\frac{a_{n-\mu }}{a_n}\right|\le k^{\mu },$$

(2.2)

where $s_{\mu }=\frac{n|a_n|k^{2\mu }+\mu |a_{n-\mu }|k^{\mu -1}}{n|a_n|k^{\mu -1}+\mu |a_{n-\mu }|}$.

The above lemma is due to Aziz and Rather [8].

Lemma 2.3 If $p(z)=a_n{z}^n+{\sum }_{\nu =\mu }^n{a}_{n-\nu }{z}^{n-\nu }$, $1\le \mu \le n$, has all its zeros in $|z|\le k$, $k\le 1$, then

$$s_{\mu }\le k^{\mu },$$

(2.3)

where $s_{\mu }$ is same as above.

Proof By using Lemma 2.2, we have

$$\frac{\mu }{n}\left|\frac{a_{n-\mu }}{a_n}\right|\le k^{\mu },$$

(2.4)

or

$$\mu |a_{n-\mu }|\le n|a_n|k^{\mu },$$

or equivalently,

$$\mu |a_{n-\mu }|-n|a_n|k^{\mu }\le 0.$$

Since $k\le 1$ and $\mu \ge 1$, the above inequality implies

$$(k^{\mu -1}-k^{\mu })(\mu |a_{n-\mu }|-n|a_n|k^{\mu })\le 0,$$

that is,

$$\mu |a_{n-\mu }|k^{\mu -1}-n|a_n|k^{\mu }{k}^{\mu -1}-\mu |a_{n-\mu }|k^{\mu }+n|a_n|k^{2\mu }\le 0,$$

which is equivalent to

$$\mu |a_{n-\mu }|k^{\mu -1}+n|a_n|k^{2\mu }\le (n|a_n|k^{\mu -1}+\mu |a_{n-\mu }|)k^{\mu },$$

which implies

$$s_{\mu }=\frac{n|a_n|k^{2\mu }+\mu |a_{n-\mu }|k^{\mu -1}}{n|a_n|k^{\mu -1}+\mu |a_{n-\mu }|}\le k^{\mu }.$$

 □

Lemma 2.4 If $p(z)=a_n{z}^n+{\sum }_{\nu =\mu }^n{a}_{n-\nu }{z}^{n-\nu }$, $1\le \mu \le n$, is a polynomial of degree n, having all zeros in the closed disk $|z|\le k$, $k\le 1$, then for every real or complex number α with $|\alpha |\ge s_{\mu }$ and $|z|=1$, we have that

$$|D_{\alpha }p(z)|\ge (|\alpha |-s_{\mu })|p^{\prime }(z)|,$$

(2.5)

where $s_{\mu }$ is same as above.

Proof Let $q(z)=z^n\overline{p(1/\overline{z})}$, then $|q^{\prime }(z)|=|np(z)-z{p}^{\prime }(z)|$ on $|z|=1$. Thus on $|z|=1$, we get

$$\begin{array}{rl}|D_{\alpha }p(z)|& =|np(z)+(\alpha -z)p^{\prime }(z)|=|\alpha p^{\prime }(z)+np(z)-z{p}^{\prime }(z)|\\ \ge |\alpha p^{\prime }(z)|-|np(z)-z{p}^{\prime }(z)|,\end{array}$$

which implies

$$|D_{\alpha }p(z)|\ge |\alpha ||p^{\prime }(z)|-|q^{\prime }(z)|.$$

(2.6)

By combining (2.1) and (2.6), we obtain

$$|D_{\alpha }p(z)|\ge (|\alpha |-s_{\mu })|p^{\prime }(z)|.$$

 □

3 Proof of the theorem

Proof of Theorem 1.1 If $k=0$, then $p(z)$ has all its zeros at the origin, therefore $p(z)=a_n{z}^n$. In this case $m=0$, $s_{\mu }=0$ and $D_{\alpha }P(z)=n\alpha a_n{z}^{n-1}$, therefore on the left-hand side of (1.10), we have

$$n(|\alpha |-s_{\mu }){\left[{\int }_0^{2\pi }{|p\left(e^{i\theta }\right)+\lambda m|}^{r}d\theta \right]}^{\frac{1}{r}}=n|\alpha ||a_n|{(2\pi )}^{\frac{1}{r}},$$

and on the right-hand side of (1.10) we have

$$\begin{array}{r}{\left[{\int }_0^{2\pi }{|1+s_{\mu }{e}^{i\theta }|}^{dr}d\theta \right]}^{\frac{1}{dr}}{\left[{\int }_0^{2\pi }{\left|D_{\alpha }p\left(e^{i\theta }\right)\right|}^{qr}d\theta \right]}^{\frac{1}{qr}}\\ ={(2\pi )}^{\frac{1}{dr}}n|\alpha ||a_n|{(2\pi )}^{\frac{1}{qr}}=n|\alpha ||a_n|{(2\pi )}^{\frac{1}{r}(\frac{1}{d}+\frac{1}{q})}=n|\alpha ||a_n|{(2\pi )}^{\frac{1}{r}}.\end{array}$$

Therefore, in the case $k=0$, Theorem 1.1 is true. So, we suppose that $k>0$, which implies $s_{\mu }>0$. Let $q(z)=z^n\overline{p(\frac{1}{\overline{z}})}$, then on $|z|=1$, we have

$$|p^{\prime }(z)|=|nq(z)-z{q}^{\prime }(z)|.$$

(3.1)

Let $m={min}_{|z|=k}|p(z)|$. Now $m\le |p(z)|$ for $|z|=k$, therefore, if λ is any real or complex number such that $|\lambda |<1$, then

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

Since all the zeros of $p(z)$ lie in $|z|\le k$, it follows by Rouche’s theorem that all the zeros of

$$F(z)=p(z)-\lambda m$$

also lie in $|z|\le k$. If $G(z)=z^n\overline{F(\frac{1}{\overline{z}})}=q(z)+\overline{\lambda }m{z}^n$, then by applying Lemma 2.2 to $F(z)$, we have

$$|G^{\prime }(z)|\le s_{\mu }|F^{\prime }(z)|\text{for }|z|=1,$$

(3.2)

that is,

$$|q^{\prime }(z)+\overline{\lambda }nm{z}^{n-1}|\le s_{\mu }|p^{\prime }(z)|.$$

Now using (3.1) in the above inequality, we get

$$|q^{\prime }(z)+\overline{\lambda }nm{z}^{n-1}|\le s_{\mu }|nq(z)-z{q}^{\prime }(z)|.$$

(3.3)

Since $p(z)$ has all its zeros in $|z|\le k\le 1$, by the Gauss-Lucas theorem all the zeros of $p^{\prime }(z)$ also lie in $|z|\le k\le 1$. This implies that the polynomial

$$z^{n-1}\overline{p^{\prime }\left(\frac{1}{\overline{z}}\right)}=nq(z)-z{q}^{\prime }(z)$$

(3.4)

has all its zeros in $|z|\ge \frac{1}{k}\ge 1$.

Therefore, it follows from (3.3) and (3.4) that the function

$$w(z)=\frac{z(q^{\prime }(z)+\overline{\lambda }nm{z}^{n-1})}{s_{\mu }(nq(z)-z{q}^{\prime }(z))}$$

(3.5)

is analytic for $|z|\le 1$, and $|w(z)|\le 1$ for $|z|\le 1$. Furthermore, $w(0)=0$. Thus the function

$$1+s_{\mu }w(z)$$

is subordinate to the function

$$1+s_{\mu }z$$

for $|z|\le 1$.

Hence by a well-known property of subordination [11], we have for each $r>0$ and $0\le \theta \le 2\pi$,

$${\int }_0^{2\pi }{|1+s_{\mu }w\left(e^{i\theta }\right)|}^{r}d\theta \le {\int }_0^{2\pi }{|1+s_{\mu }{e}^{i\theta }|}^{r}d\theta .$$

(3.6)

Also from (3.5), we have

$$1+s_{\mu }w(z)=\frac{n(q(z)+\overline{\lambda }m{z}^n)}{nq(z)-z{q}^{\prime }(z)}.$$

Therefore

$$n|q(z)+\overline{\lambda }m{z}^n|=|1+s_{\mu }w(z)||nq(z)-z{q}^{\prime }(z)|.$$

(3.7)

Since $|q(z)+\overline{\lambda }m{z}^n|=|p(z)+\lambda m|$ for $|z|=1$, we get from (3.7) and (3.1)

$$n|p(z)+\lambda m|=|1+s_{\mu }w(z)||p^{\prime }(z)|\text{for }|z|=1.$$

(3.8)

From (2.5) and (3.8), we have

$$n(|\alpha |-s_{\mu })|p(z)+\lambda m|\le |1+s_{\mu }w(z)||D_{\alpha }p(z)|\text{for }|z|=1.$$

(3.9)

By combining (3.6) and (3.9), for each $r>0$, we get

$$\begin{array}{r}{\left(n(|\alpha |-s_{\mu })\right)}^r{\int }_0^{2\pi }{|p\left(e^{i\theta }\right)+\lambda m|}^{r}d\theta \\ \le {\int }_0^{2\pi }{|1+s_{\mu }{e}^{i\theta }|}^r{\left|D_{\alpha }p\left(e^{i\theta }\right)\right|}^{r}d\theta .\end{array}$$

(3.10)

Now applying Holder’s inequality for $d>1$, $q>1$, with $\frac{1}{d}+\frac{1}{q}=1$ to (3.10), we get

$$\begin{array}{r}n(|\alpha |-s_{\mu }){\left[{\int }_0^{2\pi }{|p\left(e^{i\theta }\right)+\lambda m|}^{r}d\theta \right]}^{\frac{1}{r}}\\ \le {\left[{\int }_0^{2\pi }{|1+s_{\mu }{e}^{i\theta }|}^{dr}d\theta \right]}^{\frac{1}{dr}}{\left[{\int }_0^{2\pi }{\left|D_{\alpha }p\left(e^{i\theta }\right)\right|}^{qr}d\theta \right]}^{\frac{1}{qr}},\end{array}$$

(3.11)

which is the desired result. □