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

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.

1 Introduction and main results

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

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

Let $A_p(G)$, $p>0$, denote the class of functions f which are analytic in G and satisfy the condition

$${\parallel f\parallel }_{A_p(G)}:={\left({\iint }_G\right|f(z){|}^{p}d{\sigma }_z)}^{1/p}<\mathrm{\infty },$$

where σ denotes a two-dimensional Lebesgue measure.

When L is rectifiable, let ${\mathcal{L}}_p(L)$, $p>0$, denote the class of functions f which are integrable on L and satisfy the condition

$${\parallel f\parallel }_{{\mathcal{L}}_p(L)}:={\left({\int }_L\right|f(z){|}^p|dz|)}^{1/p}<\mathrm{\infty }.$$

From the well-known Bernstein-Walsh lemma [[1], p.101], we see that

$$|P_n(z)|\le |\mathrm{\Phi }(z){|}^n{\parallel P_n\parallel }_{C(\overline{G})},z\in \mathrm{\Omega }.$$

(1.1)

For $R>1$, let us set $L_R:=\{z:|\mathrm{\Phi }(z)|=R\}$, $G_R:=int{L}_R$, ${\mathrm{\Omega }}_R:=ext{L}_R$. Then (1.1) can be written as follows:

$${\parallel P_n\parallel }_{C({\overline{G}}_R)}\le R^n{\parallel P_n\parallel }_{C(\overline{G})}.$$

(1.2)

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

In the case when L is rectifiable, a similar estimate of (1.2) type in space ${\mathcal{L}}_p(L)$ was obtained in [2] as follows:

$${\parallel P_n\parallel }_{{\mathcal{L}}_p(L_R)}\le R^{n+\frac{1}{p}}{\parallel P_n\parallel }_{{\mathcal{L}}_p(L)},p>0.$$

(1.3)

The Berstein-Walsh type estimation for regions with quasiconformal boundary [[3], p.97] in the space $A_p(G)$, $p>0$, is contained in [4]:

$${\parallel P_n\parallel }_{{}_{A_p(G_R)}}\le c_2{R}^{{\ast }^{n+\frac{1}{p}}}{\parallel P_n\parallel }_{{}_{A_p(G)}},p>0,$$

(1.4)

where $R^{\ast }:=1+c_1(R-1)$ and $c_1>0$, $c_2=c_2(c_1,p,G)>0$ 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 $P_n(z)$ in $G_R$ and G is equivalent.

In this work, we study a problem similar to (1.4) in $A_p(G)$, $p>0$, 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

$${\parallel P_n\parallel }_{{}_{A_p(G_R)}}\le c_4{R}^{{}^{n+\frac{2}{p}}}{\parallel P_n\parallel }_{{}_{A_p(G_{R_1})}},$$

(1.5)

where $c_4={(\frac{2}{e^p-1})}^{\frac{1}{p}}[1+O(\frac{1}{n})]$, $n\to \mathrm{\infty }$.

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

$${\parallel P_n^{\ast }\parallel }_{{}_{A_p(G_R^{\ast })}}\ge {\left(\frac{2}{e^p-1}\right)}^{\frac{1}{p}}{R}^{{}^{n+\frac{2}{p}}}{\parallel P_n^{\ast }\parallel }_{{}_{A_p(G_{R_1}^{\ast })}}.$$

(1.6)

2 Some auxiliary results

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

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

$${\parallel f\parallel }_{A_p(R_1<|z|<R)}\le {\left(\frac{R^{np+2}-R_1^{np+2}}{R_1^{np+2}-1}\right)}^{\frac{1}{p}}{\parallel f\parallel }_{A_p(1<|z|<R_1)}.$$

(2.1)

Proof The function $g(z):=\frac{f(z)}{z^n}$ is analytic in $|z|>1$ and continuous in $|z|\ge 1$. Applying Hardy’s convexity theorem [[5], p.9: Th.1.5], for any arbitrary $R_1$ and R ($R>R_1$), and ρ, s such that $R_1\le \rho <R$, $1<s\le R_1$, we can write

$${\int }_{|z|=\rho }|\frac{f(z)}{z^{n+\frac{1}{p}}}{|}^p|dz|\le {\int }_{|z|=R_1}|\frac{f(z)}{z^{n+\frac{1}{p}}}{|}^p|dz|,$$

(2.2)

$${\int }_{|z|=R_1}|\frac{f(z)}{z^{n+\frac{1}{p}}}{|}^p|dz|\le {\int }_{|z|=s}|\frac{f(z)}{z^{n+\frac{1}{p}}}{|}^p|dz|,$$

(2.3)

respectively. Thus,

$${\int }_{|z|=\rho }|f(z){|}^p|dz|\le {\rho }^{np+1}{\int }_{|z|=R_1}|\frac{f(z)}{z^{n+\frac{1}{p}}}{|}^p|dz|,$$

(2.4)

$$s^{np+1}{\int }_{|z|=R_1}|\frac{f(z)}{z^{n+\frac{1}{p}}}{|}^p|dz|\le {\int }_{|z|=s}|f(z){|}^p|dz|.$$

(2.5)

Integrating (2.4) over ρ from $R_1$ to R, and (2.5) over s from 1 to $R_1$, we get

$$\begin{array}{c}{\int }_{R_1}^R{\int }_{|z|=\rho }|f(z){|}^p|dz|d\rho \le \frac{1}{np+2}(R^{np+2}-R_1^{np+2}){\int }_{|z|=R_1}|\frac{f(z)}{z^{n+\frac{1}{p}}}{|}^p|dz|,\hfill \\ \frac{1}{np+2}(R_1^{np+2}-1){\int }_{|z|=R_1}|\frac{f(z)}{z^{n+\frac{1}{p}}}{|}^p|dz|\le {\int }_1^{R_1}{\int }_{|z|=s}|f(z){|}^p|dz|ds.\hfill \end{array}$$

After calculation we have

$${\iint }_{R_1<|z|<R}|f(z){|}^{p}d{\sigma }_z\le \frac{R^{np+2}-R_1^{np+2}}{R_1^{np+2}-1}{\iint }_{1<|z|<R_1}|f(z){|}^{p}d{\sigma }_z,$$

(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

$${\parallel f\parallel }_{A_p(R_1<|z|<R)}\le c_1{R}^{n+\frac{2}{p}}{\parallel f\parallel }_{A_p(1<|z|<R_1)},$$

(2.7)

where $c_1:=c_1(p,n)={(\frac{1}{e^p-1})}^{\frac{1}{p}}[1+O(\frac{1}{n})]$, $n\to \mathrm{\infty }$.

Proof Let us put

$$S^p:=S^p(R,R_1,n,p):=\frac{R^{np+2}-R_1^{np+2}}{R_1^{np+2}-1}=R^{np+2}\cdot \frac{1-{(\frac{R_1}{R})}^{np+2}}{R_1^{np+2}-1},$$

and taking $R_1=1+\frac{1}{n}$, we have

$$S^p=R^{np+2}\cdot \frac{1-{(\frac{R_1}{R})}^{np+2}}{{(1+\frac{1}{n})}^{np+2}-1}\le \frac{R^{np+2}}{{(1+\frac{1}{n})}^{np+2}-1}.$$

(2.8)

According to the right-hand side of the well-known estimation (see, for example, [[6], p.52 (Problem 170)])

$$\frac{e}{2n+2}<e-{(1+\frac{1}{n})}^n<\frac{e}{2n+1},n=1,2,\dots ,$$

(2.9)

we have

$${(1+\frac{1}{n})}^{np+2}\ge {(1+\frac{1}{n})}^{np}\ge {(e-\frac{e}{2n+1})}^p=e^p\cdot {(1-\frac{1}{2n+1})}^p\ge {({\epsilon }_n\cdot e)}^p,$$

where

$$\frac{2}{3}\le {\epsilon }_n:=1-\frac{1}{2n+1}\to 1,n\to \mathrm{\infty }.$$

(2.10)

Therefore

$$S^p\le \frac{1}{{({\epsilon }_{n}e)}^p-1}{R}^{np+2}=R^{np+2}\frac{1}{e^p-1}[1+O\left(\frac{1}{n}\right)],n\to \mathrm{\infty }.$$

(2.11)

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

Remark 2.1 For the polynomial $Q_n(z)=z^n$, $R_1=1+\frac{1}{n}$ and any $R>R_1$,

$${\parallel Q_n\parallel }_{A_p(R_1<|z|<R)}\ge c_2{R}^{n+\frac{2}{p}}{\parallel Q_n\parallel }_{A_p(1<|z|<R_1)},$$

(2.12)

where $c_2:=c_2(p,n):={(\frac{1}{e^p-1})}^{\frac{1}{p}}[1-O(\frac{1}{n})]$, $n\to \mathrm{\infty }$.

Proof Really, from (2.6) we get

$$S^p=R^{np+2}\cdot \frac{1-{(\frac{R_1}{R})}^{np+2}}{{(1+\frac{1}{n})}^{np+2}-1}=R^{np+2}\cdot \frac{1-{\delta }_n}{{(1+\frac{1}{n})}^{np+2}-1},$$

(2.13)

where

$${\delta }_n:={\left(\frac{R_1}{R}\right)}^{np+2}\to 0,n\to \mathrm{\infty }.$$

(2.14)

According to the left-hand side of (2.9), we obtain

$$\begin{array}{rl}{(1+\frac{1}{n})}^{np+2}& ={(1+\frac{1}{n})}^{np}{(1+\frac{1}{n})}^2\le {(e-\frac{e}{2n+2})}^p{\eta }_n\\ =e^p\cdot {(1-\frac{1}{2n+2})}^p{\eta }_n\le e^p\cdot {\eta }_n,\end{array}$$

where

$${\eta }_n:={(1+\frac{1}{n})}^2\to 1,n\to \mathrm{\infty }.$$

Therefore,

$$\begin{array}{rcl}{S}^p& \ge & R^{np+2}\cdot \frac{1-{\delta }_n}{{\eta }_n{e}^p-1}\\ =& R^{np+2}\cdot [\frac{1}{{\eta }_n{e}^p-1}-\frac{{\delta }_n}{{\eta }_n{e}^p-1}]\\ =& R^{np+2}\cdot \{\frac{1}{e^p-1}[1-O\left(\frac{1}{n}\right)]-O({\delta }_n)\}\\ =& R^{np+2}\cdot \frac{1}{e^p-1}[1-O\left(\frac{1}{n}\right)],n\to \mathrm{\infty }.\end{array}$$

 □

Corollary 2.3 For $f\equiv P_n$, we have

$${\parallel P_n\parallel }_{A_p(|z|<R)}\le c_3{R}^{n+\frac{2}{p}}{\parallel P_n\parallel }_{A_p(|z|<R_1)},$$

where $c_3:=c_3(p,n):={(\frac{2}{e^p-1})}^{\frac{1}{p}}[1+O(\frac{1}{n})]$, $n\to \mathrm{\infty }$.

Proof Really, (2.1) implies, for any $f\equiv P_n$,

$${\parallel P_n\parallel }_{A_p(R_1<|z|<R)}^p\le S^p\cdot {\parallel P_n\parallel }_{A_p(1<|z|<R_1)}^p.$$

Adding ${\parallel P_n\parallel }_{A_p(|z|<R_1)}^p$ to the both sides, we obtain

$$\begin{array}{rcl}{\parallel P_n\parallel }_{A_p(|z|<R)}^p& \le & S^p\cdot {\parallel P_n\parallel }_{A_p(1<|z|<R_1)}^p+{\parallel P_n\parallel }_{A_p(|z|<R_1)}^p\\ \le & 2max\{S^p,1\}\cdot {\parallel P_n\parallel }_{A_p(|z|<R_1)}^p.\end{array}$$

Passing to the limit as $R_1=1+\frac{1}{n}\to 1$, from (2.11) we obtain

$${\parallel P_n\parallel }_{A_p(|z|<R)}^p\le \frac{2}{e^p-1}[1+O\left(\frac{1}{n}\right)]\cdot R^{np+2}{\parallel P_n\parallel }_{A_p(|z|<R_1)}^p.$$

 □

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

$${\parallel P_n\parallel }_{A_p(G_R\mathrm{\setminus }{G}_{R_1})}\le c{R}^{n+\frac{2}{p}}{\parallel P_n\parallel }_{A_p(G_{R_1}\mathrm{\setminus }G)}$$

(3.1)

for some constant $c=c(p,R_1)>0$ independent of R and n. Really, let (3.1) be true. Then

$${\parallel P_n\parallel }_{A_p(G_R\mathrm{\setminus }{G}_{R_1})}^p\le c^p{R}^{np+2}{\parallel P_n\parallel }_{A_p(G_{R_1}\mathrm{\setminus }G)}^p.$$

(3.2)

Now, we will add to both sides ${\parallel P_n\parallel }_{A_p(G_{R_1})}^p$:

$$\begin{array}{rcl}{\parallel P_n\parallel }_{A_p(G_R)}^p& \le & c^p{R}^{np+2}{\parallel P_n\parallel }_{A_p(G_{R_1}\mathrm{\setminus }G)}^p+{\parallel P_n\parallel }_{A_p(G_{R_1})}^p\\ \le & c^p{R}^{np+2}{\parallel P_n\parallel }_{A_p(G_{R_1}\mathrm{\setminus }G)}^p+c^p{R}^{np+2}{\parallel P_n\parallel }_{A_p(G_{R_1})}^p\\ =& 2c^p{R}^{np+2}{\parallel P_n\parallel }_{A_p(G_{R_1})}^p.\end{array}$$

(3.3)

Therefore,

$${\parallel P_n\parallel }_{A_p(G_R)}\le 2^{\frac{1}{p}}c{R}^{n+\frac{2}{p}}{\parallel P_n\parallel }_{A_p(G_{R_1})}.$$

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

For the $p>0$, let us set

$$f_n(w):=P_n(\mathrm{\Psi }(w)){[{\mathrm{\Psi }}^{\prime }(w)]}^{\frac{2}{p}},w=\mathrm{\Phi }(z).$$

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

$${\parallel f_n\parallel }_{A_p(R_1<|w|<R)}^p\le S(R,R_1,n,p){\parallel f_n\parallel }_{A_p(1<|w|<R_1)}^p,$$

where

$$S^p:=\frac{R^{np+2}-R_1^{np+2}}{R_1^{np+2}-1}=R^{np+2}\cdot \frac{1-{(\frac{R_1}{R})}^{np+2}}{R_1^{np+2}-1}.$$

Then

$$\begin{array}{rcl}{\iint }_{G_R\mathrm{\setminus }{G}_{R_1}}|P_n(z){|}^{p}d{\sigma }_z& =& {\iint }_{R_1<|w|<R}|f_n(w){|}^{p}d{\sigma }_w\\ \le & S^p{\iint }_{1<|w|<R_1}|f_n(w){|}^{p}d{\sigma }_w\\ \le & R^{np+2}\cdot \frac{1}{R_1^{np+2}-1}{\iint }_{G_{R_1}\mathrm{\setminus }G}|P_n(z){|}^{p}d{\sigma }_z.\end{array}$$

Therefore,

$${\iint }_{G_R}|P_n(z){|}^{p}d{\sigma }_z\le 2R^{np+2}\cdot \frac{1}{R_1^{np+2}-1}{\iint }_{G_{R_1}}|P_n(z){|}^{p}d{\sigma }_z.$$

(3.4)

Taking $R_1=1+\frac{1}{n}$, from (2.9) and (2.11) we get

$$\frac{1}{R_1^{np+2}-1}=\frac{1}{e^p-1}[1+O\left(\frac{1}{n}\right)],n\to \mathrm{\infty }.$$

(3.5)

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

3.1 Proof of the remark

Proof Let $P_n^{\ast }=z^n$, $G^{\ast }=B:=\{z:|z|<1\}$ and $R\le \frac{8e^p}{e^p-1}$. Then

$$\begin{array}{rcl}{}_{\parallel P_n^{\ast }\parallel }^{{}_{A_p(G_R^{\ast })}p}& =& {\iint }_{|z|<R}|z^n{|}^{p}d{\sigma }_z\\ =& R^{np+2}\cdot R_1^{-(np+2)}{}_{\parallel P_n^{\ast }\parallel }^{{}_{A_p(G_{R_1}^{\ast })}p}\\ =& \frac{R}{R_1^2\cdot R_1^{np}}\cdot R^{np+2}{}_{\parallel P_n^{\ast }\parallel }^{{}_{A_p(G_{R_1}^{\ast })}p}.\end{array}$$

(3.6)

For $R_1=1+\frac{1}{n}$, from (2.9) we obtain

$$\begin{array}{rcl}{(1+\frac{1}{n})}^{np}& \le & {(e-\frac{e}{2n+2})}^p\le e^p,\\ {(1+\frac{1}{n})}^2& \le & 4.\end{array}$$

Then

$$\frac{R}{R_1^{np+2}}\ge \frac{R}{4e^p}$$

and

$${}_{\parallel P_n^{\ast }\parallel }^{{}_{A_p(G_R^{\ast })}p}\ge \frac{R}{4e^p}\cdot R^{np+2}{}_{\parallel P_n^{\ast }\parallel }^{{}_{A_p(G_{R_1}^{\ast })}p}.$$

In particular, for $R=\frac{8e^p}{e^p-1}$ we have

$${}_{\parallel P_n^{\ast }\parallel }^{{}_{A_p(G_R^{\ast })}p}\ge \frac{2}{e^p-1}\cdot R^{np+2}{}_{\parallel P_n^{\ast }\parallel }^{{}_{A_p(G_{R_1}^{\ast })}p}.$$

 □