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# An analogue of the Bernstein-Walsh lemma in Jordan regions of the complex plane

Journal of Inequalities and Applications20132013:570

https://doi.org/10.1186/1029-242X-2013-570

• Accepted: 7 November 2013
• Published:

## Abstract

In this paper we continue to study two-dimensional analogues of Bernstein-Walsh estimates for arbitrary Jordan domains.

MSC:Primary 30A10; 30C10; secondary 41A17.

## Keywords

• algebraic polynomials
• conformal mapping
• Bernstein lemma

## 1 Introduction and main results

Let $G\subset \mathbb{C}$ be a finite region, with $0\in G$, bounded by a Jordan curve $L:=\partial G$, $\mathrm{\Delta }:=\left\{w:|w|>1\right\}$, $\mathrm{\Omega }:=ext\overline{G}$ (with respect to $\overline{\mathbb{C}}$). Let $w=\mathrm{\Phi }\left(z\right)$ be the univalent conformal mapping of Ω onto the Δ normalized by $\mathrm{\Phi }\left(\mathrm{\infty }\right)=\mathrm{\infty }$, ${\mathrm{\Phi }}^{\prime }\left(\mathrm{\infty }\right)>0$, and $\mathrm{\Psi }:={\mathrm{\Phi }}^{-1}$.

Let ${\mathrm{\wp }}_{n}$ denote the class of arbitrary algebraic polynomials ${P}_{n}\left(z\right)$ of degree at most $n\in \mathbb{N}$.

Let ${A}_{p}\left(G\right)$, $p>0$, denote the class of functions f which are analytic in G and satisfy the condition
${\parallel f\parallel }_{{A}_{p}\left(G\right)}:={\left({\iint }_{G}|f\left(z\right){|}^{p}\phantom{\rule{0.2em}{0ex}}d{\sigma }_{z}\right)}^{1/p}<\mathrm{\infty },$

where σ denotes a two-dimensional Lebesgue measure.

When L is rectifiable, let ${\mathcal{L}}_{p}\left(L\right)$, $p>0$, denote the class of functions f which are integrable on L and satisfy the condition
${\parallel f\parallel }_{{\mathcal{L}}_{p}\left(L\right)}:={\left({\int }_{L}|f\left(z\right){|}^{p}|dz|\right)}^{1/p}<\mathrm{\infty }.$
From the well-known Bernstein-Walsh lemma [, p.101], we see that
$|{P}_{n}\left(z\right)|\le |\mathrm{\Phi }\left(z\right){|}^{n}{\parallel {P}_{n}\parallel }_{C\left(\overline{G}\right)},\phantom{\rule{1em}{0ex}}z\in \mathrm{\Omega }.$
(1.1)
For $R>1$, let us set ${L}_{R}:=\left\{z:|\mathrm{\Phi }\left(z\right)|=R\right\}$, ${G}_{R}:=int{L}_{R}$, ${\mathrm{\Omega }}_{R}:=ext{L}_{R}$. Then (1.1) can be written as follows:
${\parallel {P}_{n}\parallel }_{C\left({\overline{G}}_{R}\right)}\le {R}^{n}{\parallel {P}_{n}\parallel }_{C\left(\overline{G}\right)}.$
(1.2)

Hence, setting $R=1+\frac{1}{n}$, according to (1.2), we see that the C-norm of a polynomial ${P}_{n}\left(z\right)$ in ${\overline{G}}_{R}$ and $\overline{G}$ is equivalent, i.e., the norm ${\parallel {P}_{n}\parallel }_{C\left({\overline{G}}_{R}\right)}$ increases with no more than a constant with respect to ${\parallel {P}_{n}\parallel }_{C\left(\overline{G}\right)}$.

In the case when L is rectifiable, a similar estimate of (1.2) type in space ${\mathcal{L}}_{p}\left(L\right)$ was obtained in  as follows:
${\parallel {P}_{n}\parallel }_{{\mathcal{L}}_{p}\left({L}_{R}\right)}\le {R}^{n+\frac{1}{p}}{\parallel {P}_{n}\parallel }_{{\mathcal{L}}_{p}\left(L\right)},\phantom{\rule{1em}{0ex}}p>0.$
(1.3)
The Berstein-Walsh type estimation for regions with quasiconformal boundary [, p.97] in the space ${A}_{p}\left(G\right)$, $p>0$, is contained in :
${\parallel {P}_{n}\parallel }_{{}_{{A}_{p}\left({G}_{R}\right)}}\le {c}_{2}{R}^{{\ast }^{n+\frac{1}{p}}}{\parallel {P}_{n}\parallel }_{{}_{{A}_{p}\left(G\right)}},\phantom{\rule{1em}{0ex}}p>0,$
(1.4)

where ${R}^{\ast }:=1+{c}_{1}\left(R-1\right)$ and ${c}_{1}>0$, ${c}_{2}={c}_{2}\left({c}_{1},p,G\right)>0$ are constants. Therefore, if we choose $R=1+\frac{{c}_{3}}{n}$, then (1.4) we can see that the ${A}_{p}$-norm of polynomials ${P}_{n}\left(z\right)$ in ${G}_{R}$ and G is equivalent.

In this work, we study a problem similar to (1.4) in ${A}_{p}\left(G\right)$, $p>0$, for regions with arbitrary Jordan boundary.

Now we can state our new result.

Theorem 1.1 Let $p>0$; G be a Jordan region. Then, for any ${P}_{n}\in {\mathrm{\wp }}_{n}$, ${R}_{1}=1+\frac{1}{n}$ and arbitrary R, $R>{R}_{1}$, we have
${\parallel {P}_{n}\parallel }_{{}_{{A}_{p}\left({G}_{R}\right)}}\le {c}_{4}{R}^{{}^{n+\frac{2}{p}}}{\parallel {P}_{n}\parallel }_{{}_{{A}_{p}\left({G}_{{R}_{1}}\right)}},$
(1.5)

where ${c}_{4}={\left(\frac{2}{{e}^{p}-1}\right)}^{\frac{1}{p}}\left[1+O\left(\frac{1}{n}\right)\right]$, $n\to \mathrm{\infty }$.

The sharpness of (1.5) can be seen from the following remark:

Remark 1.1 For any $n=1,2,\dots$ , there exist a polynomial ${P}_{n}^{\ast }\in {\mathrm{\wp }}_{n}$, region ${G}^{\ast }\subset \mathbb{C}$ and number $R>{R}_{1}=1+\frac{1}{n}$ such that
${\parallel {P}_{n}^{\ast }\parallel }_{{}_{{A}_{p}\left({G}_{R}^{\ast }\right)}}\ge {\left(\frac{2}{{e}^{p}-1}\right)}^{\frac{1}{p}}{R}^{{}^{n+\frac{2}{p}}}{\parallel {P}_{n}^{\ast }\parallel }_{{}_{{A}_{p}\left({G}_{{R}_{1}}^{\ast }\right)}}.$
(1.6)

## 2 Some auxiliary results

Let $G\subset \mathbb{C}$ be a finite region bounded by the Jordan curve L. Let ${L}_{R}:=\left\{z:|\mathrm{\Phi }\left(z\right)|=R,R>1\right\}$, ${G}_{t}:=int{L}_{t}$, ${\mathrm{\Omega }}_{t}:=ext{L}_{t}$.

We note that, throughout this paper, ${c}_{1},{c}_{2},\dots$ (in general, different in different relations) are positive constants.

Lemma 2.1 Let $p>0$; f be an analytic function in $|z|>1$ and have a pole of degree at most n, $n\ge 1$ at $z=\mathrm{\infty }$. Then, for any ${R}_{1}$ and $R>{R}_{1}$, we have
${\parallel f\parallel }_{{A}_{p}\left({R}_{1}<|z|
(2.1)
Proof The function $g\left(z\right):=\frac{f\left(z\right)}{{z}^{n}}$ is analytic in $|z|>1$ and continuous in $|z|\ge 1$. Applying Hardy’s convexity theorem [, p.9: Th.1.5], for any arbitrary ${R}_{1}$ and R ($R>{R}_{1}$), and ρ, s such that ${R}_{1}\le \rho , $1, we can write
${\int }_{|z|=\rho }|\frac{f\left(z\right)}{{z}^{n+\frac{1}{p}}}{|}^{p}|dz|\le {\int }_{|z|={R}_{1}}|\frac{f\left(z\right)}{{z}^{n+\frac{1}{p}}}{|}^{p}|dz|,$
(2.2)
${\int }_{|z|={R}_{1}}|\frac{f\left(z\right)}{{z}^{n+\frac{1}{p}}}{|}^{p}|dz|\le {\int }_{|z|=s}|\frac{f\left(z\right)}{{z}^{n+\frac{1}{p}}}{|}^{p}|dz|,$
(2.3)
respectively. Thus,
${\int }_{|z|=\rho }|f\left(z\right){|}^{p}|dz|\le {\rho }^{np+1}{\int }_{|z|={R}_{1}}|\frac{f\left(z\right)}{{z}^{n+\frac{1}{p}}}{|}^{p}|dz|,$
(2.4)
${s}^{np+1}{\int }_{|z|={R}_{1}}|\frac{f\left(z\right)}{{z}^{n+\frac{1}{p}}}{|}^{p}|dz|\le {\int }_{|z|=s}|f\left(z\right){|}^{p}|dz|.$
(2.5)
Integrating (2.4) over ρ from ${R}_{1}$ to R, and (2.5) over s from 1 to ${R}_{1}$, we get
$\begin{array}{c}{\int }_{{R}_{1}}^{R}{\int }_{|z|=\rho }|f\left(z\right){|}^{p}|dz|\phantom{\rule{0.2em}{0ex}}d\rho \le \frac{1}{np+2}\left({R}^{np+2}-{R}_{1}^{np+2}\right){\int }_{|z|={R}_{1}}|\frac{f\left(z\right)}{{z}^{n+\frac{1}{p}}}{|}^{p}|dz|,\hfill \\ \frac{1}{np+2}\left({R}_{1}^{np+2}-1\right){\int }_{|z|={R}_{1}}|\frac{f\left(z\right)}{{z}^{n+\frac{1}{p}}}{|}^{p}|dz|\le {\int }_{1}^{{R}_{1}}{\int }_{|z|=s}|f\left(z\right){|}^{p}|dz|\phantom{\rule{0.2em}{0ex}}ds.\hfill \end{array}$
After calculation we have
${\iint }_{{R}_{1}<|z|
(2.6)

and we see that (2.1) is true. □

Corollary 2.2 Under the assumptions of Lemma  2.1 for ${R}_{1}=1+\frac{1}{n}$, we have
${\parallel f\parallel }_{{A}_{p}\left({R}_{1}<|z|
(2.7)

where ${c}_{1}:={c}_{1}\left(p,n\right)={\left(\frac{1}{{e}^{p}-1}\right)}^{\frac{1}{p}}\left[1+O\left(\frac{1}{n}\right)\right]$, $n\to \mathrm{\infty }$.

Proof Let us put
${S}^{p}:={S}^{p}\left(R,{R}_{1},n,p\right):=\frac{{R}^{np+2}-{R}_{1}^{np+2}}{{R}_{1}^{np+2}-1}={R}^{np+2}\cdot \frac{1-{\left(\frac{{R}_{1}}{R}\right)}^{np+2}}{{R}_{1}^{np+2}-1},$
and taking ${R}_{1}=1+\frac{1}{n}$, we have
${S}^{p}={R}^{np+2}\cdot \frac{1-{\left(\frac{{R}_{1}}{R}\right)}^{np+2}}{{\left(1+\frac{1}{n}\right)}^{np+2}-1}\le \frac{{R}^{np+2}}{{\left(1+\frac{1}{n}\right)}^{np+2}-1}.$
(2.8)
According to the right-hand side of the well-known estimation (see, for example, [, p.52 (Problem 170)])
$\frac{e}{2n+2}
(2.9)
we have
${\left(1+\frac{1}{n}\right)}^{np+2}\ge {\left(1+\frac{1}{n}\right)}^{np}\ge {\left(e-\frac{e}{2n+1}\right)}^{p}={e}^{p}\cdot {\left(1-\frac{1}{2n+1}\right)}^{p}\ge {\left({\epsilon }_{n}\cdot e\right)}^{p},$
where
$\frac{2}{3}\le {\epsilon }_{n}:=1-\frac{1}{2n+1}\to 1,\phantom{\rule{1em}{0ex}}n\to \mathrm{\infty }.$
(2.10)
Therefore
${S}^{p}\le \frac{1}{{\left({\epsilon }_{n}e\right)}^{p}-1}{R}^{np+2}={R}^{np+2}\frac{1}{{e}^{p}-1}\left[1+O\left(\frac{1}{n}\right)\right],\phantom{\rule{1em}{0ex}}n\to \mathrm{\infty }.$
(2.11)

From (2.8) and (2.11) we complete the proof. □

Remark 2.1 For the polynomial ${Q}_{n}\left(z\right)={z}^{n}$, ${R}_{1}=1+\frac{1}{n}$ and any $R>{R}_{1}$,
${\parallel {Q}_{n}\parallel }_{{A}_{p}\left({R}_{1}<|z|
(2.12)

where ${c}_{2}:={c}_{2}\left(p,n\right):={\left(\frac{1}{{e}^{p}-1}\right)}^{\frac{1}{p}}\left[1-O\left(\frac{1}{n}\right)\right]$, $n\to \mathrm{\infty }$.

Proof Really, from (2.6) we get
${S}^{p}={R}^{np+2}\cdot \frac{1-{\left(\frac{{R}_{1}}{R}\right)}^{np+2}}{{\left(1+\frac{1}{n}\right)}^{np+2}-1}={R}^{np+2}\cdot \frac{1-{\delta }_{n}}{{\left(1+\frac{1}{n}\right)}^{np+2}-1},$
(2.13)
where
${\delta }_{n}:={\left(\frac{{R}_{1}}{R}\right)}^{np+2}\to 0,\phantom{\rule{1em}{0ex}}n\to \mathrm{\infty }.$
(2.14)
According to the left-hand side of (2.9), we obtain
$\begin{array}{rl}{\left(1+\frac{1}{n}\right)}^{np+2}& ={\left(1+\frac{1}{n}\right)}^{np}{\left(1+\frac{1}{n}\right)}^{2}\le {\left(e-\frac{e}{2n+2}\right)}^{p}{\eta }_{n}\\ ={e}^{p}\cdot {\left(1-\frac{1}{2n+2}\right)}^{p}{\eta }_{n}\le {e}^{p}\cdot {\eta }_{n},\end{array}$
where
${\eta }_{n}:={\left(1+\frac{1}{n}\right)}^{2}\to 1,\phantom{\rule{1em}{0ex}}n\to \mathrm{\infty }.$
Therefore,
$\begin{array}{rcl}{S}^{p}& \ge & {R}^{np+2}\cdot \frac{1-{\delta }_{n}}{{\eta }_{n}{e}^{p}-1}\\ =& {R}^{np+2}\cdot \left[\frac{1}{{\eta }_{n}{e}^{p}-1}-\frac{{\delta }_{n}}{{\eta }_{n}{e}^{p}-1}\right]\\ =& {R}^{np+2}\cdot \left\{\frac{1}{{e}^{p}-1}\left[1-O\left(\frac{1}{n}\right)\right]-O\left({\delta }_{n}\right)\right\}\\ =& {R}^{np+2}\cdot \frac{1}{{e}^{p}-1}\left[1-O\left(\frac{1}{n}\right)\right],\phantom{\rule{1em}{0ex}}n\to \mathrm{\infty }.\end{array}$

□

Corollary 2.3 For $f\equiv {P}_{n}$, we have
${\parallel {P}_{n}\parallel }_{{A}_{p}\left(|z|

where ${c}_{3}:={c}_{3}\left(p,n\right):={\left(\frac{2}{{e}^{p}-1}\right)}^{\frac{1}{p}}\left[1+O\left(\frac{1}{n}\right)\right]$, $n\to \mathrm{\infty }$.

Proof Really, (2.1) implies, for any $f\equiv {P}_{n}$,
${\parallel {P}_{n}\parallel }_{{A}_{p}\left({R}_{1}<|z|
Adding ${\parallel {P}_{n}\parallel }_{{A}_{p}\left(|z|<{R}_{1}\right)}^{p}$ to the both sides, we obtain
$\begin{array}{rcl}{\parallel {P}_{n}\parallel }_{{A}_{p}\left(|z|
Passing to the limit as ${R}_{1}=1+\frac{1}{n}\to 1$, from (2.11) we obtain
${\parallel {P}_{n}\parallel }_{{A}_{p}\left(|z|

□

## 3 Proof of the theorem

Proof First of all, let us convince ourselves that for the proof of (1.5) it is sufficient to show the fulfilment of estimation
${\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{R}\mathrm{\setminus }{G}_{{R}_{1}}\right)}\le c{R}^{n+\frac{2}{p}}{\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{{R}_{1}}\mathrm{\setminus }G\right)}$
(3.1)
for some constant $c=c\left(p,{R}_{1}\right)>0$ independent of R and n. Really, let (3.1) be true. Then
${\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{R}\mathrm{\setminus }{G}_{{R}_{1}}\right)}^{p}\le {c}^{p}{R}^{np+2}{\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{{R}_{1}}\mathrm{\setminus }G\right)}^{p}.$
(3.2)
Now, we will add to both sides ${\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{{R}_{1}}\right)}^{p}$:
$\begin{array}{rcl}{\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{R}\right)}^{p}& \le & {c}^{p}{R}^{np+2}{\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{{R}_{1}}\mathrm{\setminus }G\right)}^{p}+{\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{{R}_{1}}\right)}^{p}\\ \le & {c}^{p}{R}^{np+2}{\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{{R}_{1}}\mathrm{\setminus }G\right)}^{p}+{c}^{p}{R}^{np+2}{\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{{R}_{1}}\right)}^{p}\\ =& 2{c}^{p}{R}^{np+2}{\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{{R}_{1}}\right)}^{p}.\end{array}$
(3.3)
Therefore,
${\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{R}\right)}\le {2}^{\frac{1}{p}}c{R}^{n+\frac{2}{p}}{\parallel {P}_{n}\parallel }_{{A}_{p}\left({G}_{{R}_{1}}\right)}.$

Now, let us make a proof of (3.1).

For the $p>0$, let us set
${f}_{n}\left(w\right):={P}_{n}\left(\mathrm{\Psi }\left(w\right)\right){\left[{\mathrm{\Psi }}^{\prime }\left(w\right)\right]}^{\frac{2}{p}},\phantom{\rule{1em}{0ex}}w=\mathrm{\Phi }\left(z\right).$
The function ${f}_{n}$ is analytic in Δ and has a pole of degree at most n at $w=\mathrm{\infty }$. Then, according to Lemma 2.1, we have
${\parallel {f}_{n}\parallel }_{{A}_{p}\left({R}_{1}<|w|
where
${S}^{p}:=\frac{{R}^{np+2}-{R}_{1}^{np+2}}{{R}_{1}^{np+2}-1}={R}^{np+2}\cdot \frac{1-{\left(\frac{{R}_{1}}{R}\right)}^{np+2}}{{R}_{1}^{np+2}-1}.$
Then
$\begin{array}{rcl}{\iint }_{{G}_{R}\mathrm{\setminus }{G}_{{R}_{1}}}|{P}_{n}\left(z\right){|}^{p}\phantom{\rule{0.2em}{0ex}}d{\sigma }_{z}& =& {\iint }_{{R}_{1}<|w|
Therefore,
${\iint }_{{G}_{R}}|{P}_{n}\left(z\right){|}^{p}\phantom{\rule{0.2em}{0ex}}d{\sigma }_{z}\le 2{R}^{np+2}\cdot \frac{1}{{R}_{1}^{np+2}-1}{\iint }_{{G}_{{R}_{1}}}|{P}_{n}\left(z\right){|}^{p}\phantom{\rule{0.2em}{0ex}}d{\sigma }_{z}.$
(3.4)
Taking ${R}_{1}=1+\frac{1}{n}$, from (2.9) and (2.11) we get
$\frac{1}{{R}_{1}^{np+2}-1}=\frac{1}{{e}^{p}-1}\left[1+O\left(\frac{1}{n}\right)\right],\phantom{\rule{1em}{0ex}}n\to \mathrm{\infty }.$
(3.5)

Now, from (3.4) and (3.5) we complete the proof. □

### 3.1 Proof of the remark

Proof Let ${P}_{n}^{\ast }={z}^{n}$, ${G}^{\ast }=B:=\left\{z:|z|<1\right\}$ and $R\le \frac{8{e}^{p}}{{e}^{p}-1}$. Then
$\begin{array}{rcl}{}_{\parallel {P}_{n}^{\ast }\parallel }^{{}_{{A}_{p}\left({G}_{R}^{\ast }\right)}p}& =& {\iint }_{|z|
(3.6)
For ${R}_{1}=1+\frac{1}{n}$, from (2.9) we obtain
$\begin{array}{rcl}{\left(1+\frac{1}{n}\right)}^{np}& \le & {\left(e-\frac{e}{2n+2}\right)}^{p}\le {e}^{p},\\ {\left(1+\frac{1}{n}\right)}^{2}& \le & 4.\end{array}$
Then
$\frac{R}{{R}_{1}^{np+2}}\ge \frac{R}{4{e}^{p}}$
and
${}_{\parallel {P}_{n}^{\ast }\parallel }^{{}_{{A}_{p}\left({G}_{R}^{\ast }\right)}p}\ge \frac{R}{4{e}^{p}}\cdot {R}^{np+2}{}_{\parallel {P}_{n}^{\ast }\parallel }^{{}_{{A}_{p}\left({G}_{{R}_{1}}^{\ast }\right)}p}.$
In particular, for $R=\frac{8{e}^{p}}{{e}^{p}-1}$ we have
${}_{\parallel {P}_{n}^{\ast }\parallel }^{{}_{{A}_{p}\left({G}_{R}^{\ast }\right)}p}\ge \frac{2}{{e}^{p}-1}\cdot {R}^{np+2}{}_{\parallel {P}_{n}^{\ast }\parallel }^{{}_{{A}_{p}\left({G}_{{R}_{1}}^{\ast }\right)}p}.$

□

## Authors’ Affiliations

(1)
Faculty of Arts and Science, Department of Mathematics, Mersin University, Mersin, 33343, Turkey

## References
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