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

## Abstract

For a polynomial $p\left(z\right)$ of degree n, having all zeros in $|z|â‰¤k$, where $kâ‰¤1$, Dewan et al. (Southeast Asian Bull. Math. 34:69-77, 2010) proved that for every $\mathrm{Î±}âˆˆ\mathbb{C}$ with $|\mathrm{Î±}|â‰¥k$ and for each $r>0$,

$n\left(|\mathrm{Î±}|âˆ’k\right){\left\{{âˆ«}_{0}^{2\mathrm{Ï€}}{|p\left({e}^{i\mathrm{Î¸}}\right)|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right\}}^{\frac{1}{r}}â‰¤{\left\{{âˆ«}_{0}^{2\mathrm{Ï€}}{|1+k{e}^{i\mathrm{Î¸}}|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right\}}^{\frac{1}{r}}\underset{|z|=1}{max}|{D}_{\mathrm{Î±}}p\left(z\right)|.$

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 $p\left(z\right)$ be a polynomial of degree n. Then according to Bernsteinâ€™s inequality [1] on the derivative of a polynomial, we have

$\underset{|z|=1}{max}|{p}^{â€²}\left(z\right)|â‰¤n\underset{|z|=1}{max}|p\left(z\right)|.$
(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 $|z|â‰¤1$, then it has been proved by TurÃ¡n [2] that

$\underset{|z|=1}{max}|{p}^{â€²}\left(z\right)|â‰¥\frac{n}{2}\underset{|z|=1}{max}|p\left(z\right)|.$
(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 [3] proved that if $p\left(z\right)$ has all zeros in $|z|â‰¤k$, where $kâ‰¤1$, then

$\underset{|z|=1}{max}|{p}^{â€²}\left(z\right)|â‰¥\frac{n}{1+k}\underset{|z|=1}{max}|p\left(z\right)|.$
(1.3)

This result is best possible and equality holds for $p\left(z\right)={\left(zâˆ’k\right)}^{n}$.

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 $p\left(z\right)$ on $|z|=1$. In fact he proved that if $p\left(z\right)$ has all its zeros in $|z|â‰¤1$, then for each $r>0$,

$n{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|p\left({e}^{i\mathrm{Î¸}}\right)|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}â‰¤{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|1+{e}^{i\mathrm{Î¸}}|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}\underset{|z|=1}{max}|{p}^{â€²}\left(z\right)|.$
(1.4)

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

$n{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|p\left({e}^{i\mathrm{Î¸}}\right)|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}â‰¤{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|1+k{e}^{i\mathrm{Î¸}}|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}\underset{|z|=1}{max}|{p}^{â€²}\left(z\right)|.$
(1.5)

As a generalization of (1.5), Aziz and Shah [6] proved that if $p\left(z\right)={a}_{n}{z}^{n}+{âˆ‘}_{\mathrm{Î½}=\mathrm{Î¼}}^{n}{a}_{nâˆ’\mathrm{Î½}}{z}^{nâˆ’\mathrm{Î½}}$, $1â‰¤\mathrm{Î¼}â‰¤n$, is a polynomial of degree n, having all its zeros in $|z|â‰¤kâ‰¤1$, then for each $r>0$,

$n{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|p\left({e}^{i\mathrm{Î¸}}\right)|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}â‰¤{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|1+{k}^{\mathrm{Î¼}}{e}^{i\mathrm{Î¸}}|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}\underset{|z|=1}{max}|{p}^{â€²}\left(z\right)|.$
(1.6)

Let ${D}_{\mathrm{Î±}}p\left(z\right)$ denote the polar derivative of the polynomial $p\left(z\right)$ of degree n with respect to $\mathrm{Î±}âˆˆ\mathbb{C}$. Then ${D}_{\mathrm{Î±}}p\left(z\right)=np\left(z\right)+\left(\mathrm{Î±}âˆ’z\right){p}^{â€²}\left(z\right)$. The polynomial ${D}_{\mathrm{Î±}}p\left(z\right)$ is of degree at most $nâˆ’1$ and it generalizes the ordinary derivative in the sense that

$\underset{\mathrm{Î±}â†’\mathrm{âˆž}}{lim}\left[\frac{{D}_{\mathrm{Î±}}p\left(z\right)}{\mathrm{Î±}}\right]={p}^{â€²}\left(z\right).$

Shah [7] extended (1.2) to the polar derivative of $p\left(z\right)$ and proved that if all zeros of the polynomial $p\left(z\right)$ lie in $|z|â‰¤1$, then for every Î± with $|\mathrm{Î±}|â‰¥1$, we have

$\underset{|z|=1}{max}|{D}_{\mathrm{Î±}}p\left(z\right)|â‰¥\frac{n}{2}\left(|\mathrm{Î±}|âˆ’1\right)\underset{|z|=1}{max}|p\left(z\right)|.$
(1.7)

This result is best possible and equality holds for $p\left(z\right)={\left(zâˆ’1\right)}^{n}$ with $\mathrm{Î±}â‰¥1$.

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 $p\left(z\right)$ lie in $|z|â‰¤k$, $kâ‰¤1$, then for every Î± with $|\mathrm{Î±}|â‰¥k$, we get

$\underset{|z|=1}{max}|{D}_{\mathrm{Î±}}p\left(z\right)|â‰¥\frac{n}{1+k}\left(|\mathrm{Î±}|âˆ’k\right)\underset{|z|=1}{max}|p\left(z\right)|.$
(1.8)

This result is best possible and equality holds for $p\left(z\right)={\left(zâˆ’k\right)}^{n}$ with $\mathrm{Î±}â‰¥k$.

Recently Dewan et al. [9] generalized the inequalities (1.5) and (1.8). They proved that if $p\left(z\right)$ has all its zeros in $|z|â‰¤kâ‰¤1$, then for every $\mathrm{Î±}âˆˆ\mathbb{C}$ with $|\mathrm{Î±}|â‰¥k$ and for each $r>0$,

$n\left(|\mathrm{Î±}|âˆ’k\right){\left\{{âˆ«}_{0}^{2\mathrm{Ï€}}{|p\left({e}^{i\mathrm{Î¸}}\right)|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right\}}^{\frac{1}{r}}â‰¤{\left\{{âˆ«}_{0}^{2\mathrm{Ï€}}{|1+k{e}^{i\mathrm{Î¸}}|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right\}}^{\frac{1}{r}}\underset{|z|=1}{max}|{D}_{\mathrm{Î±}}p\left(z\right)|.$
(1.9)

In the limiting case, when $râ†’\mathrm{âˆž}$, the above inequality is sharp and equality holds for the polynomial $p\left(z\right)={\left(zâˆ’k\right)}^{n}$ with $\mathrm{Î±}â‰¥k$.

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 $p\left(z\right)={a}_{n}{z}^{n}+{âˆ‘}_{\mathrm{Î½}=\mathrm{Î¼}}^{n}{a}_{nâˆ’\mathrm{Î½}}{z}^{nâˆ’\mathrm{Î½}}$, $1â‰¤\mathrm{Î¼}â‰¤n$, is a polynomial of degree n, having all its zeros in $|z|â‰¤kâ‰¤1$ and $m={min}_{|z|=k}|p\left(z\right)|$, then for $\mathrm{Î»},\mathrm{Î±}âˆˆ\mathbb{C}$ with $|\mathrm{Î»}|â‰¤1$, $|\mathrm{Î±}|â‰¥{s}_{\mathrm{Î¼}}$ and $r>0$, $d>1$, $q>1$, with $\frac{1}{d}+\frac{1}{q}=1$, we have

$\begin{array}{r}n\left(|\mathrm{Î±}|âˆ’{s}_{\mathrm{Î¼}}\right){\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|p\left({e}^{i\mathrm{Î¸}}\right)+\mathrm{Î»}m|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}\\ \phantom{\rule{1em}{0ex}}â‰¤{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|1+{s}_{\mathrm{Î¼}}{e}^{i\mathrm{Î¸}}|}^{dr}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{dr}}{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|{D}_{\mathrm{Î±}}p\left({e}^{i\mathrm{Î¸}}\right)|}^{qr}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{qr}},\end{array}$
(1.10)

where ${s}_{\mathrm{Î¼}}=\frac{n|{a}_{n}|{k}^{2\mathrm{Î¼}}+\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|{k}^{\mathrm{Î¼}âˆ’1}}{n|{a}_{n}|{k}^{\mathrm{Î¼}âˆ’1}+\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|}$. In the limiting case, when $râ†’\mathrm{âˆž}$, the above inequality is sharp and equality holds for the polynomial $p\left(z\right)={\left(zâˆ’k\right)}^{n}$ with $\mathrm{Î±}â‰¥k$.

Letting $qâ†’\mathrm{âˆž}$ (so that $dâ†’1$) in (1.10), we have the following.

Corollary 1.2 If $p\left(z\right)={a}_{n}{z}^{n}+{âˆ‘}_{\mathrm{Î½}=\mathrm{Î¼}}^{n}{a}_{nâˆ’\mathrm{Î½}}{z}^{nâˆ’\mathrm{Î½}}$, $1â‰¤\mathrm{Î¼}â‰¤n$, is a polynomial of degree n, having all its zeros in $|z|â‰¤kâ‰¤1$ and $m={min}_{|z|=k}|p\left(z\right)|$, then for $\mathrm{Î»},\mathrm{Î±}âˆˆ\mathbb{C}$ with $|\mathrm{Î»}|â‰¤1$, $|\mathrm{Î±}|â‰¥{s}_{\mathrm{Î¼}}$ and $r>0$,

$n\left(|\mathrm{Î±}|âˆ’{s}_{\mathrm{Î¼}}\right){\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|p\left({e}^{i\mathrm{Î¸}}\right)+\mathrm{Î»}m|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}â‰¤{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|1+{s}_{\mathrm{Î¼}}{e}^{i\mathrm{Î¸}}|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}\underset{|z|=1}{max}|{D}_{\mathrm{Î±}}p\left(z\right)|,$
(1.11)

where ${s}_{\mathrm{Î¼}}$ is defined as in Theorem  1.1.

Remark 1.3 Since by Lemma 2.3, ${s}_{\mathrm{Î¼}}â‰¤k$, 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 $|\mathrm{Î±}|$ and make $|\mathrm{Î±}|â†’\mathrm{âˆž}$, we obtain the following refinement and generalization of the inequality (1.6).

Corollary 1.4 If $p\left(z\right)={a}_{n}{z}^{n}+{âˆ‘}_{\mathrm{Î½}=\mathrm{Î¼}}^{n}{a}_{nâˆ’\mathrm{Î½}}{z}^{nâˆ’\mathrm{Î½}}$, $1â‰¤\mathrm{Î¼}â‰¤n$, is a polynomial of degree n, having all its zeros in $|z|â‰¤kâ‰¤1$ and $m={min}_{|z|=k}|p\left(z\right)|$, then for every $\mathrm{Î»}âˆˆ\mathbb{C}$ with $|\mathrm{Î»}|â‰¤1$ and $r>0$,

$n{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|p\left({e}^{i\mathrm{Î¸}}\right)+\mathrm{Î»}m|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}â‰¤{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|1+{s}_{\mathrm{Î¼}}{e}^{i\mathrm{Î¸}}|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}\underset{|z|=1}{max}|{p}^{â€²}\left(z\right)|,$
(1.12)

where ${s}_{\mathrm{Î¼}}$ is defined as in Theorem  1.1.

Letting $râ†’\mathrm{âˆž}$ in (1.10) and choosing the argument of Î» suitably with $|\mathrm{Î»}|=1$, we have the following result.

Corollary 1.5 If $p\left(z\right)={a}_{n}{z}^{n}+{âˆ‘}_{\mathrm{Î½}=\mathrm{Î¼}}^{n}{a}_{nâˆ’\mathrm{Î½}}{z}^{nâˆ’\mathrm{Î½}}$, $1â‰¤\mathrm{Î¼}â‰¤n$, is a polynomial of degree n, having all its zeros in $|z|â‰¤kâ‰¤1$, then for every $\mathrm{Î±}âˆˆ\mathbb{C}$ with $|\mathrm{Î±}|â‰¥{s}_{\mathrm{Î¼}}$,

$\frac{n\left(|\mathrm{Î±}|âˆ’{s}_{\mathrm{Î¼}}\right)}{1+{s}_{\mathrm{Î¼}}}\left[\underset{|z|=1}{max}|p\left(z\right)|+\underset{|z|=k}{min}|p\left(z\right)|\right]â‰¤\underset{|z|=1}{max}|{D}_{\mathrm{Î±}}p\left(z\right)|,$
(1.13)

where ${s}_{\mathrm{Î¼}}$ 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 $p\left(z\right)$ lie in a circular region C and w is any zero of ${D}_{\mathrm{Î±}}p\left(z\right)$, then at most one of the points w and Î± may lie outside C.

Lemma 2.2 If $p\left(z\right)={a}_{n}{z}^{n}+{âˆ‘}_{\mathrm{Î½}=\mathrm{Î¼}}^{n}{a}_{nâˆ’\mathrm{Î½}}{z}^{nâˆ’\mathrm{Î½}}$; $1â‰¤\mathrm{Î¼}â‰¤n$, is a polynomial of degree n having all its zeros in $|z|â‰¤kâ‰¤1$ and $q\left(z\right)={z}^{n}\stackrel{Â¯}{p\left(\frac{1}{\stackrel{Â¯}{z}}\right)}$, then on $|z|=1$

$|{q}^{â€²}\left(z\right)|â‰¤{s}_{\mathrm{Î¼}}|{p}^{â€²}\left(z\right)|$
(2.1)

and

$\frac{\mathrm{Î¼}}{n}|\frac{{a}_{nâˆ’\mathrm{Î¼}}}{{a}_{n}}|â‰¤{k}^{\mathrm{Î¼}},$
(2.2)

where ${s}_{\mathrm{Î¼}}=\frac{n|{a}_{n}|{k}^{2\mathrm{Î¼}}+\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|{k}^{\mathrm{Î¼}âˆ’1}}{n|{a}_{n}|{k}^{\mathrm{Î¼}âˆ’1}+\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|}$.

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

Lemma 2.3 If $p\left(z\right)={a}_{n}{z}^{n}+{âˆ‘}_{\mathrm{Î½}=\mathrm{Î¼}}^{n}{a}_{nâˆ’\mathrm{Î½}}{z}^{nâˆ’\mathrm{Î½}}$, $1â‰¤\mathrm{Î¼}â‰¤n$, has all its zeros in $|z|â‰¤k$, $kâ‰¤1$, then

${s}_{\mathrm{Î¼}}â‰¤{k}^{\mathrm{Î¼}},$
(2.3)

where ${s}_{\mathrm{Î¼}}$ is same as above.

Proof By using Lemma 2.2, we have

$\frac{\mathrm{Î¼}}{n}|\frac{{a}_{nâˆ’\mathrm{Î¼}}}{{a}_{n}}|â‰¤{k}^{\mathrm{Î¼}},$
(2.4)

or

$\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|â‰¤n|{a}_{n}|{k}^{\mathrm{Î¼}},$

or equivalently,

$\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|âˆ’n|{a}_{n}|{k}^{\mathrm{Î¼}}â‰¤0.$

Since $kâ‰¤1$ and $\mathrm{Î¼}â‰¥1$, the above inequality implies

$\left({k}^{\mathrm{Î¼}âˆ’1}âˆ’{k}^{\mathrm{Î¼}}\right)\left(\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|âˆ’n|{a}_{n}|{k}^{\mathrm{Î¼}}\right)â‰¤0,$

that is,

$\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|{k}^{\mathrm{Î¼}âˆ’1}âˆ’n|{a}_{n}|{k}^{\mathrm{Î¼}}{k}^{\mathrm{Î¼}âˆ’1}âˆ’\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|{k}^{\mathrm{Î¼}}+n|{a}_{n}|{k}^{2\mathrm{Î¼}}â‰¤0,$

which is equivalent to

$\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|{k}^{\mathrm{Î¼}âˆ’1}+n|{a}_{n}|{k}^{2\mathrm{Î¼}}â‰¤\left(n|{a}_{n}|{k}^{\mathrm{Î¼}âˆ’1}+\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|\right){k}^{\mathrm{Î¼}},$

which implies

${s}_{\mathrm{Î¼}}=\frac{n|{a}_{n}|{k}^{2\mathrm{Î¼}}+\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|{k}^{\mathrm{Î¼}âˆ’1}}{n|{a}_{n}|{k}^{\mathrm{Î¼}âˆ’1}+\mathrm{Î¼}|{a}_{nâˆ’\mathrm{Î¼}}|}â‰¤{k}^{\mathrm{Î¼}}.$

â€ƒâ–¡

Lemma 2.4 If $p\left(z\right)={a}_{n}{z}^{n}+{âˆ‘}_{\mathrm{Î½}=\mathrm{Î¼}}^{n}{a}_{nâˆ’\mathrm{Î½}}{z}^{nâˆ’\mathrm{Î½}}$, $1â‰¤\mathrm{Î¼}â‰¤n$, is a polynomial of degree n, having all zeros in the closed disk $|z|â‰¤k$, $kâ‰¤1$, then for every real or complex number Î± with $|\mathrm{Î±}|â‰¥{s}_{\mathrm{Î¼}}$ and $|z|=1$, we have that

$|{D}_{\mathrm{Î±}}p\left(z\right)|â‰¥\left(|\mathrm{Î±}|âˆ’{s}_{\mathrm{Î¼}}\right)|{p}^{â€²}\left(z\right)|,$
(2.5)

where ${s}_{\mathrm{Î¼}}$ is same as above.

Proof Let $q\left(z\right)={z}^{n}\stackrel{Â¯}{p\left(1/\stackrel{Â¯}{z}\right)}$, then $|{q}^{â€²}\left(z\right)|=|np\left(z\right)âˆ’z{p}^{â€²}\left(z\right)|$ on $|z|=1$. Thus on $|z|=1$, we get

$\begin{array}{rl}|{D}_{\mathrm{Î±}}p\left(z\right)|& =|np\left(z\right)+\left(\mathrm{Î±}âˆ’z\right){p}^{â€²}\left(z\right)|=|\mathrm{Î±}{p}^{â€²}\left(z\right)+np\left(z\right)âˆ’z{p}^{â€²}\left(z\right)|\\ â‰¥|\mathrm{Î±}{p}^{â€²}\left(z\right)|âˆ’|np\left(z\right)âˆ’z{p}^{â€²}\left(z\right)|,\end{array}$

which implies

$|{D}_{\mathrm{Î±}}p\left(z\right)|â‰¥|\mathrm{Î±}||{p}^{â€²}\left(z\right)|âˆ’|{q}^{â€²}\left(z\right)|.$
(2.6)

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

$|{D}_{\mathrm{Î±}}p\left(z\right)|â‰¥\left(|\mathrm{Î±}|âˆ’{s}_{\mathrm{Î¼}}\right)|{p}^{â€²}\left(z\right)|.$

â€ƒâ–¡

## 3 Proof of the theorem

Proof of Theorem 1.1 If $k=0$, then $p\left(z\right)$ has all its zeros at the origin, therefore $p\left(z\right)={a}_{n}{z}^{n}$. In this case $m=0$, ${s}_{\mathrm{Î¼}}=0$ and ${D}_{\mathrm{Î±}}P\left(z\right)=n\mathrm{Î±}{a}_{n}{z}^{nâˆ’1}$, therefore on the left-hand side of (1.10), we have

$n\left(|\mathrm{Î±}|âˆ’{s}_{\mathrm{Î¼}}\right){\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|p\left({e}^{i\mathrm{Î¸}}\right)+\mathrm{Î»}m|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}=n|\mathrm{Î±}||{a}_{n}|{\left(2\mathrm{Ï€}\right)}^{\frac{1}{r}},$

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

$\begin{array}{r}{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|1+{s}_{\mathrm{Î¼}}{e}^{i\mathrm{Î¸}}|}^{dr}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{dr}}{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|{D}_{\mathrm{Î±}}p\left({e}^{i\mathrm{Î¸}}\right)|}^{qr}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{qr}}\\ \phantom{\rule{1em}{0ex}}={\left(2\mathrm{Ï€}\right)}^{\frac{1}{dr}}n|\mathrm{Î±}||{a}_{n}|{\left(2\mathrm{Ï€}\right)}^{\frac{1}{qr}}=n|\mathrm{Î±}||{a}_{n}|{\left(2\mathrm{Ï€}\right)}^{\frac{1}{r}\left(\frac{1}{d}+\frac{1}{q}\right)}=n|\mathrm{Î±}||{a}_{n}|{\left(2\mathrm{Ï€}\right)}^{\frac{1}{r}}.\end{array}$

Therefore, in the case $k=0$, Theorem 1.1 is true. So, we suppose that $k>0$, which implies ${s}_{\mathrm{Î¼}}>0$. Let $q\left(z\right)={z}^{n}\stackrel{Â¯}{p\left(\frac{1}{\stackrel{Â¯}{z}}\right)}$, then on $|z|=1$, we have

$|{p}^{â€²}\left(z\right)|=|nq\left(z\right)âˆ’z{q}^{â€²}\left(z\right)|.$
(3.1)

Let $m={min}_{|z|=k}|p\left(z\right)|$. Now $mâ‰¤|p\left(z\right)|$ for $|z|=k$, therefore, if Î» is any real or complex number such that $|\mathrm{Î»}|<1$, then

Since all the zeros of $p\left(z\right)$ lie in $|z|â‰¤k$, it follows by Roucheâ€™s theorem that all the zeros of

$F\left(z\right)=p\left(z\right)âˆ’\mathrm{Î»}m$

also lie in $|z|â‰¤k$. If $G\left(z\right)={z}^{n}\stackrel{Â¯}{F\left(\frac{1}{\stackrel{Â¯}{z}}\right)}=q\left(z\right)+\stackrel{Â¯}{\mathrm{Î»}}m{z}^{n}$, then by applying Lemma 2.2 to $F\left(z\right)$, we have

(3.2)

that is,

$|{q}^{â€²}\left(z\right)+\stackrel{Â¯}{\mathrm{Î»}}nm{z}^{nâˆ’1}|â‰¤{s}_{\mathrm{Î¼}}|{p}^{â€²}\left(z\right)|.$

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

$|{q}^{â€²}\left(z\right)+\stackrel{Â¯}{\mathrm{Î»}}nm{z}^{nâˆ’1}|â‰¤{s}_{\mathrm{Î¼}}|nq\left(z\right)âˆ’z{q}^{â€²}\left(z\right)|.$
(3.3)

Since $p\left(z\right)$ has all its zeros in $|z|â‰¤kâ‰¤1$, by the Gauss-Lucas theorem all the zeros of ${p}^{â€²}\left(z\right)$ also lie in $|z|â‰¤kâ‰¤1$. This implies that the polynomial

${z}^{nâˆ’1}\stackrel{Â¯}{{p}^{â€²}\left(\frac{1}{\stackrel{Â¯}{z}}\right)}=nq\left(z\right)âˆ’z{q}^{â€²}\left(z\right)$
(3.4)

has all its zeros in $|z|â‰¥\frac{1}{k}â‰¥1$.

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

$w\left(z\right)=\frac{z\left({q}^{â€²}\left(z\right)+\stackrel{Â¯}{\mathrm{Î»}}nm{z}^{nâˆ’1}\right)}{{s}_{\mathrm{Î¼}}\left(nq\left(z\right)âˆ’z{q}^{â€²}\left(z\right)\right)}$
(3.5)

is analytic for $|z|â‰¤1$, and $|w\left(z\right)|â‰¤1$ for $|z|â‰¤1$. Furthermore, $w\left(0\right)=0$. Thus the function

$1+{s}_{\mathrm{Î¼}}w\left(z\right)$

is subordinate to the function

$1+{s}_{\mathrm{Î¼}}z$

for $|z|â‰¤1$.

Hence by a well-known property of subordination [11], we have for each $r>0$ and $0â‰¤\mathrm{Î¸}â‰¤2\mathrm{Ï€}$,

${âˆ«}_{0}^{2\mathrm{Ï€}}{|1+{s}_{\mathrm{Î¼}}w\left({e}^{i\mathrm{Î¸}}\right)|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}â‰¤{âˆ«}_{0}^{2\mathrm{Ï€}}{|1+{s}_{\mathrm{Î¼}}{e}^{i\mathrm{Î¸}}|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}.$
(3.6)

Also from (3.5), we have

$1+{s}_{\mathrm{Î¼}}w\left(z\right)=\frac{n\left(q\left(z\right)+\stackrel{Â¯}{\mathrm{Î»}}m{z}^{n}\right)}{nq\left(z\right)âˆ’z{q}^{â€²}\left(z\right)}.$

Therefore

$n|q\left(z\right)+\stackrel{Â¯}{\mathrm{Î»}}m{z}^{n}|=|1+{s}_{\mathrm{Î¼}}w\left(z\right)||nq\left(z\right)âˆ’z{q}^{â€²}\left(z\right)|.$
(3.7)

Since $|q\left(z\right)+\stackrel{Â¯}{\mathrm{Î»}}m{z}^{n}|=|p\left(z\right)+\mathrm{Î»}m|$ for $|z|=1$, we get from (3.7) and (3.1)

(3.8)

From (2.5) and (3.8), we have

(3.9)

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

$\begin{array}{r}{\left(n\left(|\mathrm{Î±}|âˆ’{s}_{\mathrm{Î¼}}\right)\right)}^{r}{âˆ«}_{0}^{2\mathrm{Ï€}}{|p\left({e}^{i\mathrm{Î¸}}\right)+\mathrm{Î»}m|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\\ \phantom{\rule{1em}{0ex}}â‰¤{âˆ«}_{0}^{2\mathrm{Ï€}}{|1+{s}_{\mathrm{Î¼}}{e}^{i\mathrm{Î¸}}|}^{r}{|{D}_{\mathrm{Î±}}p\left({e}^{i\mathrm{Î¸}}\right)|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}.\end{array}$
(3.10)

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

$\begin{array}{r}n\left(|\mathrm{Î±}|âˆ’{s}_{\mathrm{Î¼}}\right){\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|p\left({e}^{i\mathrm{Î¸}}\right)+\mathrm{Î»}m|}^{r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{r}}\\ \phantom{\rule{1em}{0ex}}â‰¤{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|1+{s}_{\mathrm{Î¼}}{e}^{i\mathrm{Î¸}}|}^{dr}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{dr}}{\left[{âˆ«}_{0}^{2\mathrm{Ï€}}{|{D}_{\mathrm{Î±}}p\left({e}^{i\mathrm{Î¸}}\right)|}^{qr}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right]}^{\frac{1}{qr}},\end{array}$
(3.11)

which is the desired result.â€ƒâ–¡

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## Acknowledgements

The authors are grateful to the referees for the careful reading of the paper and for the helpful suggestions and comments.

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The authors declare that they have no competing interests. All authors read and approved the final manuscript.

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Zireh, A., Khojastehnezhad, E. & Musawi, S.R. Integral mean estimates for the polar derivative of polynomials whose zeros are within a circle. J Inequal Appl 2013, 307 (2013). https://doi.org/10.1186/1029-242X-2013-307