Maximal Convergence and Interpolation on Unconnected Sets

Abstract

A theorem of Grothmann states that interpolating polynomials to a holomorphic function on a compact set E is maximally convergent to f only if a subsequence of the interpolation points converges to the equilibrium distribution of E in the weak* sense. Grothmann’s proof applies only for connected sets E. The objective of this paper is to provide a new necessary condition for maximal convergence which is the crucial tool to prove Grothmann’s theorem for unconnected sets E.

1 Introduction

For $B \subset {\mathbb {C}}$, we denote by ${\overline{B}}$ its closure and by $\partial {B}$ the boundary of B and we use $\Vert \cdot \Vert _B$ for the supremum norm over B. Let ${\mathcal {A}}(B)$ be the class of functions that are holomorphic (i.e., analytic and single-valued) in a neighborhood of B, and we denote by ${\mathcal {P}}_n$ the set of algebraic polynomials of degree at most n.

Let E be a compact subset of the complex plane ${\mathbb {C}}$, and let ${\mathcal {M}}(E)$ be the collection of all probability measures supported on E, then the logarithmic potential of $\mu \in {\mathcal {M}}(E)$ is defined by

$$\begin{aligned} U^\mu (z) = \int \log \frac{1}{|z-t|} \; {\mathrm {d}} \mu (t) \end{aligned}$$

and the logarithmic energy $I(\mu )$ by

$$\begin{aligned} I(\mu ) := \int \int \log \frac{1}{|z-t|} \; {\mathrm {d}}\mu (t) \; {\mathrm {d}}\mu (z) = \int U^\mu (z) \; {\mathrm {d}}\mu (z). \end{aligned}$$

Let

$$\begin{aligned} V(E) := \mathrm{inf} \{ I(\mu ) : \mu \in {\mathcal {M}}(E)\}, \end{aligned}$$

then V(E) is either finite or $V(E) = +\infty$. The quantity

$$\begin{aligned} \mathrm{cap} \; E = e^{-V(E)} \end{aligned}$$

is called the logarithmic capacity or capacity of E.

Let E be compact in the complex plane ${\mathbb {C}}$ with connected complement $\Omega = {\overline{{\mathbb {C}}}} \setminus E$ in the extended plane ${\overline{{\mathbb {C}}}}$. The domain $\Omega$ is called regular if the Green function $G(z) = G(z,\infty )$ on $\Omega$ with pole at $\infty$ tends to 0 as $z\in \Omega$ tends to the boundary $\partial \Omega$ of $\Omega$. If $\Omega$ is regular, then $\mathrm{cap} \; E > 0$ and there exists a unique measure $\mu _E \in {\mathcal {M}}(E)$ such that

$$\begin{aligned} I(\mu _E) = - \log \mathrm{cap} \; E = V(E) \end{aligned}$$

and we have

$$\begin{aligned} U^{\mu _E}(z) = - G(z) - \log \mathrm{cap} \; E, \quad z\in \Omega . \end{aligned}$$

$\mu _E$ is called equilibrium measure of E.

In the following, let E be compact in ${\mathbb {C}}$ with regular connected complement $\Omega = {\overline{{\mathbb {C}}}} \setminus E$. Then, we define for $\sigma > 1$ the Green domains $E_\sigma$ by

$$\begin{aligned} E_\sigma := \{ z \in \Omega : G(z) < \log \sigma \} \cup E \end{aligned}$$

with boundary $\Gamma _\sigma := \partial E_\sigma$. Since $\Omega$ is regular, the Green domain $E_\sigma$ consists of a finite number of Jordan regions which are mutually exterior (cf. Walsh ([4], Chapter 4, section 4.1). Only in the case that E is connected, each $E_\sigma$ is a single Jordan region for any $\sigma > 1$.

If $f \in {\mathcal {A}}(E)$, then there exists $\rho > 1$ and polynomials $p_n \in {\mathcal {P}}_n$, $n \in {\mathbb {N}}$, such that

$$\begin{aligned} \underset{n \rightarrow \infty }{\lim \sup } \; \Vert f - p_n\Vert ^{1/n}_E \le \frac{1}{\rho }, \end{aligned}$$

due to a result of Walsh ([4]).

Let $\rho (f)$ denote the maximal parameter $\rho > 1$, $1< \rho \le \infty$, such that f is holomorphic on $E_\rho$. Then, there exist polynomials $p_n \in {\mathcal {P}}_n$ such that

$$\begin{aligned} \underset{n \rightarrow \infty }{\lim \sup } \; \Vert f - p_n\Vert ^{1/n}_E = \frac{1}{\rho (f)}. \end{aligned}$$

Such a sequence $p_n \in {\mathcal {P}}_n$ is called maximally convergent. Moreover, Walsh ([4], Sect 4.7, Theorem 7, Theorem 8 and its Corollary, pp. 79–81) proved that for such maximally convergent polynomials

$$\begin{aligned} \underset{n \rightarrow \infty }{\lim \sup } \; \Vert f - p_n\Vert ^{1/n}_{\Gamma _\sigma } = \frac{\sigma }{\rho (f)}, \quad 1< \sigma < \rho (f). \end{aligned}$$

(1.1)

Consider Lagrange–Hermite interpolation to f on point sets

$$\begin{aligned} Z_n : \quad z_{n,0},z_{n,1},\ldots ,z_{n,n} \in E \end{aligned}$$

by polynomials $p_n \in {\mathcal {P}}_n$. Then, it is known, due to Bernstein–Walsh, that the interpolation on such schemes $Z_n$ yields (1.1) if the normalized counting measures $\nu _n$, i.e.,

$$\begin{aligned} \nu _n(B) := \frac{\#\{z_{n,j} : z_{n,j} \in B\}}{n+1} \quad (B \subset {\mathbb {C}}), \end{aligned}$$

satisfy

$$\begin{aligned} {\widehat{\nu }}_n \;\overset{*}{\underset{n\rightarrow \infty }{\longrightarrow }} \; \mu _E \end{aligned}$$

in the weak* sense, where ${\widehat{\nu }}_n$ denotes the balayage measure of $\nu _n$ on the boundary of E, i.e., ${\widehat{\nu }}_n$ is the measure supported on the boundary of E such that

$$\begin{aligned} U^{{\widehat{\nu }}_n}(z) = U^{\nu }(z) \quad \text {for all} \; z\in {\overline{{\mathbb {C}}}} \setminus E. \end{aligned}$$

Conversely, Grothmann stated the following theorem.

Theorem

(Grothmann [1]): Let ${p_n}$ be the polynomial of interpolation on $Z_n \subset E$. If $f \in {\mathcal {A}}(E), 1< \rho (f)< \infty$, and if

$$\begin{aligned} \underset{n \rightarrow \infty }{\lim \sup } \; {\Vert f - p_n\Vert ^{1/n}_E = \frac{1}{\rho (f)}}, \end{aligned}$$

then $\mu _E$ is a weak* limit point of $\left\{ {\widehat{\nu }}_n\right\} _{n \in {\mathbb {N}}}$.

The proof given in ([1]) applies only if E is connected or, at least, if $E_{\rho (f)}$ is connected. Hence, one objective of this paper is to provide a proof of Grothmann’s theorem even for unconnected sets E. The crucial tool will be a new necessary condition for maximally convergent polynomials which seems to be interesting itself.

2 Maximal Convergence and Interpolation

Let $E \subset {\mathbb {C}}$ be compact with regular connected complement $\Omega = {\overline{{\mathbb {C}}}} \setminus E$ and let $p_n \in {\mathcal {P}}_n$, $n \in {\mathbb {N}}$, denote polynomials such that

$$\begin{aligned} \underset{n \rightarrow \infty }{\lim \sup } \; \Vert f - p_n\Vert ^{1/n}_E = \frac{1}{\rho (f)}, \end{aligned}$$

where $\rho (f)$ is the maximal parameter of holomorphy of f and $1< \rho (f) < \infty$.

The Green domains $E_r$, $1< r < \infty$, consist of a finite number of disjoint regions ${E_r}^{i}$,

$$\begin{aligned} E_r \; = \;E^{1}_{r} \cup E^{2}_{r} \cup \ldots \cup E^{l_r}_{r}, \quad l_r \in {\mathbb {N}}. \end{aligned}$$

(2.1)

Each $E^{i}_{r}$ is a Jordan region, and we write $\Gamma _r^{i}=\partial E^{i}_{r}$. Then, the boundary $\Gamma _r = \partial E_r$ is

$$\begin{aligned} \Gamma _r = \bigcup \limits ^{l_r}_{i=1}\Gamma ^{i}_{r} \end{aligned}$$

and we note that $\Gamma ^{i}_{r}$ and $\Gamma ^{j}_{r}$ may have points in common if $i \ne j$, but only a finite number of points (cf. Walsh ([4], chapter 4, section 4.1).

Our first result is a necessary condition for maximal convergence that is new if E is not connected.

Proposition 1

Let $f\in {\mathcal {A}}(E_\rho )$, $1< \rho < \infty$, and let $p_n \in {\mathcal {P}}_n, n \in {\mathbb {N}}$, be polynomials such that

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim \sup } \;\Vert f - p_n\Vert ^{1/n}_E \le \frac{1}{\rho }. \end{aligned}$$

(2.2)

If $1< \sigma < \rho$ and if

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim \sup } \;\underset{1 \le i \le l_\sigma }{\min }\;\Vert f - p_n\Vert ^{1/n}_{\Gamma ^{i}_{\sigma }} < \frac{\sigma }{\rho }, \end{aligned}$$

(2.3)

then the maximal parameter $\rho (f)$ of holomorphy of f satisfies $\rho (f) > \rho$.

As a consequence of Proposition 1, we get immediately

Theorem 1

Let $f\in {\mathcal {A}}(E)$, and let $1< \sigma< \rho (f) < \infty$. Then, the sequence $\left\{ p_n\right\} _{n\in {\mathbb {N}}}$ with $p_n\in {\mathcal {P}}_n$ is maximally convergent to f if and only if

$$\begin{aligned} \frac{\sigma }{\rho (f)} \;=\ \underset{n\rightarrow \infty }{\lim \sup }\; \Vert f - p_n\Vert ^{1/n}_{\Gamma _\sigma } \;=\; \underset{n\rightarrow \infty }{\lim \sup } \underset{1 \le i \le l_\sigma }{\min }\; \Vert f - p_n\Vert ^{1/n}_{\Gamma ^{i}_{\sigma }}. \end{aligned}$$

(2.4)

If E is connected, then $l_\sigma = 1$ for any $\sigma$ and Theorem 1 coincides with the well-known characterization of Bernstein–Walsh.

Next, we want to use the above results to characterize interpolating polynomials converging maximally to f by the distribution of the interpolation points.

Proposition 2

Let $f\in {\mathcal {A}}(E_\rho )$, $1< \rho < \infty$, and let $p_n \in {\mathcal {P}}_n$ be the interpolating polynomial to f on the point set $Z_n \subset E$, $n\in {\mathbb {N}}$, with

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim \sup }\; \Vert f - p_n\Vert ^{1/n}_{\Gamma _{\sigma ^{*}}} \;\le \;\frac{\sigma ^{*}}{\rho } \end{aligned}$$

(2.5)

for all $\sigma ^{*}$, $1< \sigma ^{*} <\rho$.

Let $\Lambda \subset {\mathbb {N}}$ and let $\mu _E$ be not a weak* limit point of ${\widehat{\nu }}_n$, $n \in \Lambda$. If $\sigma$ is fixed with $1< \sigma < \rho$, then for all $n \in \Lambda$ there exists an index s(n), $1 \le s(n) \le l_\sigma$, such that the strict inequality

$$\begin{aligned} \underset{n\in \Lambda , n\rightarrow \infty }{\lim \sup }\; \Vert f - p_n\Vert ^{1/n}_{\Gamma ^{s(n)}_{\sigma }} \;<\; \frac{\sigma }{\rho } \end{aligned}$$

(2.6)

holds.

Finally, we combine Proposition 2 with Proposition 1 to obtain a characterization of maximally converging interpolation polynomials.

Theorem 2

Let $f\in {\mathcal {A}}(E)$ with $1< \rho (f) < \infty$, and let $\left\{ p_n\right\} _{n\in {\mathbb {N}}}$ be maximally convergent to f on E. If $p_n$ interpolates f at the interpolation point set $Z_n\subset {E}$ with counting measure $\nu _n$ and balayage measure ${\widehat{\nu }}_n$ on $\partial E$, then the following holds:

• (a) $\mu _E$ is a weak* limit point of $\left\{ {\widehat{\nu }}_n\right\} _{n \in {\mathbb {N}}}$.

• (b) For every fixed $\sigma$, $1< \sigma < \rho (f)$, there exists a subset $\Lambda \subset {\mathbb {N}}$ such that

  $$\begin{aligned} \frac{\sigma }{\rho (f)}\; = \;\underset{n\rightarrow \infty }{\lim \sup }\; \Vert f - p_n\Vert ^{1/n}_{\Gamma _\sigma } \;=\; \underset{n\in \Lambda , n\rightarrow \infty }{\lim } \; \underset{1 \le i \le l_\sigma }{\min }\; \Vert f - p_n\Vert ^{1/n}_{\Gamma ^{i}_{\sigma }} \end{aligned}$$

  (2.7)

  and

  $$\begin{aligned} {{\widehat{\nu }}_n {\mathop {\rightarrow }\limits ^{*}}\mu _E \quad as \; { n \rightarrow \infty , n\in \Lambda }}. \end{aligned}$$

The first statement (a) is Grothmann’s theorem, even proved here for unconnected E. The second statement (b) describes the subsequences $\Lambda \in {\mathbb {N}}$ which lead to $\mu _E$ as a weak* limit point of ${\widehat{\nu }}_n, n \in {\mathbb {N}}$.

3 Proof of Proposition 1

The proof is based on constructing a telescoping series of f,

$$\begin{aligned} f = p_{n_1} + \sum \limits ^{\infty }_{j=1} (p_{n_{j+1}} - p_{n_j}), \end{aligned}$$

which is holomorphic in a neighborhood of ${\overline{E}}_\rho$.

Because of (2.2), the Lemma of Bernstein–Walsh induces that there exists for $\varepsilon > 0$ and $1 \le r < \rho$ a number $n_\varepsilon (r) \in {\mathbb {N}}$ such that

$$\begin{aligned} \frac{1}{n} \log \Vert f - p_n\Vert _{\Gamma _{r}}\; \le \; \log \frac{r}{\rho } + \varepsilon \end{aligned}$$

(3.1)

for $n \ge n_\varepsilon (r)$. Because of (2.3), there exist a map

$$\begin{aligned} s:\; {\mathbb {N}}\;\rightarrow \left\{ 1,2,\ldots ,l_\sigma \right\} \end{aligned}$$

and $\delta > 0$ and $n_1(\delta ) \in {\mathbb {N}}$ such that

$$\begin{aligned} \frac{1}{n} \log \Vert f - p_n\Vert _{\Gamma ^{s(n)}_{\sigma }}\; \le \;\log \frac{\sigma }{\rho } - \delta . \end{aligned}$$

(3.2)

for all $n \ge n_1(\delta )$.

3.1 The Starting Telescoping Series

Let us fix a parameter $\kappa > 1$, then the telescoping series

$$\begin{aligned} f = p_{m_1} + \sum \limits ^{\infty }_{i=1} \; (p_{m_{i+1}} - p_{m_i}) \end{aligned}$$

(3.3)

and the sequence $\Lambda _1(\kappa ) := \left\{ m_{i}\right\} ^{\infty }_{i=1}$ is defined recursively:

Set $m_1 := 1$. If $m_i$ is defined and if there exists $m > m_i$ with

$$\begin{aligned} s(m) = s(m_i) \quad \text {and} \quad m/{m_i} \le \kappa , \end{aligned}$$

then we define

$$\begin{aligned} m_{i+1} := m,\quad {{ otherwise}}\quad m_{i+1} := m_i + 1. \end{aligned}$$

Hence, $\Lambda _1(\kappa ) = \left\{ m_{i}\right\} ^{\infty }_{i=1}$ has the following properties:

$$\begin{aligned} m_{i+1} / m_i \le \kappa \quad \text {and}\quad s(m_{i+1}) = s(m_i) \end{aligned}$$

(3.4)

or

$$\begin{aligned} m_{i+1} = m_i + 1 \;\text {and} \;s(m) \ne s(m_i) \;\text {for}\; m_i+1 \le m \le \kappa \, m_i. \end{aligned}$$

(3.5)

Next, we decompose $\Lambda _1(\kappa )$ into

$$\begin{aligned} \Lambda _1(\kappa ) := \Lambda _{1,1}(\kappa ) \;\cup \; \Lambda _{1,2}(\kappa ) \end{aligned}$$

(3.6)

with

$$\begin{aligned} \Lambda _{1,1}(\kappa )\;=\; \left\{ \,m_i \in \Lambda _1(\kappa ) : \; m_i \;\text {satisfies}\;(3.4)\right\} \end{aligned}$$

(3.7)

and

$$\begin{aligned} \Lambda _{1,2}(\kappa )\; =\; \left\{ \,m_i \in \Lambda _1(\kappa ) : \; m_i \;\text {satisfies}\;(3.5)\right\} \; = \;\Lambda _1(\kappa ) \setminus \Lambda _{1,1}(\kappa ). \end{aligned}$$

(3.8)

The next lemma shows how to estimate the norm of the difference

$$\begin{aligned} p_{m_{i+1}} - p_{m_{i}} \quad \text {for}\quad m_i \in \Lambda _{1,1}(\kappa ) \end{aligned}$$

outside of $E_r$. We use as auxiliary tool the harmonic measure

$$\begin{aligned} h_r^{i}(z) = \omega ({z,\Gamma _r^{i}, {\overline{{\mathbb {C}}}} \setminus \overline{E_r}}), \quad 1 \le i \le l_r, \end{aligned}$$

where $\Gamma _r^{i} = \partial {E^{i}_{r}}$ is the boundary of the Jordan region $E_r^{i}$ in the decomposition of $E_r$ in (2.1), i.e., $h_r^{i}$ is the harmonic function in ${\overline{{\mathbb {C}}}} \setminus \overline{E_r}$ that satisfies the boundary conditions $h_r^{i} = 1$ on $\Gamma _r^{i}$ and $h_r^{i} = 0$ on $\Gamma _r \setminus \Gamma _r^{i}$, possibly except of a finite number of points (cf. [3], p. 111, section III, 17 or [2]). Then,

$$\begin{aligned} h_r^{i} > 0 \quad \text {for all}\; z \in {\overline{{\mathbb {C}}}} \setminus \overline{E_r} \end{aligned}$$

and if we define for $r^{*} > r$

$$\begin{aligned} \alpha _r(r^{*}) := \underset{1 \le i \le l_r}{\min } \;\underset{ z\in \Gamma _{r^{*}}}{\min }\; {h_r^{i}(z)}, \end{aligned}$$

we obtain

$$\begin{aligned} \alpha _r(r^{*}) > 0. \end{aligned}$$

Lemma 1

Let $n,m \in {\mathbb {N}}$ with $m < n$ and $n/m \le \kappa$, and let $1< r < \rho$ such that

$$\begin{aligned} \frac{1}{m}\; \log {\Vert f - p_m\Vert }_{\Gamma _r^{s(m)}}\;\le \; \log {\frac{r}{\rho } - \delta _r} \end{aligned}$$

(3.9)

and

$$\begin{aligned} \frac{1}{n}\; \log {\Vert f - p_n\Vert }_{\Gamma _r^{s(n)}}\;\le \; \log {\frac{r}{\rho } - \delta _r}. \end{aligned}$$

(3.10)

where $\delta _r > 0$. If $r^{*} > r$, $s(m)=s(n)$ and if

$$\begin{aligned} 1< \kappa < 1 + \frac{\alpha _r(r^{*})}{\log {(\rho /r)}}\;\delta _r, \end{aligned}$$

(3.11)

then there exists $\delta ^{*}_{r^{*}} > 0$ and $n^{*} = n^{*}(\kappa )\in {\mathbb {N}}$ such that

$$\begin{aligned} \frac{1}{n}\; \log {\Vert p_n - p_m\Vert }_{\Gamma _{r^{*}}}\;\le \;\log {\frac{r^{*}}{\rho }} - \delta ^{*}_{r^{*}} \end{aligned}$$

(3.12)

for $m \ge n^{*}$, where

$$\begin{aligned} \delta ^{*}_{r^{*}}\ge \frac{1}{2}\left( \frac{\delta _r}{\kappa }\alpha _r(r^{*}) - \left( 1-\frac{1}{\kappa }\right) \log \frac{\rho }{r}\right) . \end{aligned}$$

(3.13)

Proof

(3.9) and (3.10) imply

$$\begin{aligned} \Vert p_n - p_m\Vert _{\Gamma ^{s(n)}_{r}} \le \;2\left( \frac{r}{\rho }\;e^{-\delta _r}\right) ^{m}\; \le \;2\left( \frac{r}{\rho }\;e^{-\delta _r} \right) ^{n/\kappa } \end{aligned}$$

or

$$\begin{aligned} \frac{1}{n}\; \log \Vert p_n - p_m\Vert _{\Gamma ^{s(n)}_{r}}\le & {} \frac{\log 2}{n} + \left( \log \frac{r}{\rho }-\delta _r\right) \frac{1}{\kappa } \nonumber \\= & {} \log \frac{r}{\rho } - \frac{\delta _r}{\kappa } + \left( \frac{1}{\kappa } -1\right) \; \log \frac{r}{\rho } + \frac{\log 2}{n}. \end{aligned}$$

(3.14)

Fix $0< \varepsilon < \log {\rho /\sigma }$. Because of (3.1), there exists $n_\varepsilon (r) \in {\mathbb {N}}$ such that

$$\begin{aligned} \frac{1}{n} \log \Vert f - p_n\Vert _{\Gamma _{r}}\; \le \; \log \frac{r}{\rho } + \varepsilon \end{aligned}$$

for $n \ge n_\varepsilon (r)$. Then,

$$\begin{aligned} \Vert p_n - p_m\Vert _{\Gamma ^{s(n)}_{r}} \; \le 2\;\left( \log \frac{r}{\rho }\; e^{\varepsilon }\right) ^{m}\; \le 2 \left( \log \frac{r}{\rho }\; e^{\varepsilon }\right) ^{n/\kappa } \end{aligned}$$

or

$$\begin{aligned} \frac{1}{n}\; \log \Vert p_n - p_m\Vert _{\Gamma ^{s(n)}_{r}}\le & {} \frac{\log 2}{n} + \frac{1}{\kappa } \log \frac{r}{\rho } +\frac{\varepsilon }{\kappa } \nonumber \\\le & {} \log \frac{r}{\rho } + \left( \frac{1}{\kappa } - 1\right) \; \log \frac{r}{\rho } + \frac{\kappa \;\log 2}{n} + \frac{\varepsilon }{\kappa }. \end{aligned}$$

(3.15)

Define

$$\begin{aligned} A(\kappa ) := \left( 1 - \frac{1}{\kappa }\right) \;\log \frac{\rho }{r} \end{aligned}$$

and

$$\begin{aligned} H(z) := \frac{1}{n} \; \log |(p_n(z) - p_m(z))| - G(z) + \log {\rho }. \end{aligned}$$

Then, H(z) is subharmonic in ${\overline{{\mathbb {C}}}} \setminus {\overline{E}}_r$ and the harmonic function

$$\begin{aligned} - \frac{\delta _r}{\kappa } \; h^{s(n)}_{r}(z) + A(\kappa ) + \frac{\log 2}{n} + \frac{\varepsilon }{\kappa }\; \end{aligned}$$

is an upper bound of H(z) in ${\mathbb {C}}\setminus {{\overline{E}}_r}$, where we have taken into account (3.14) and (3.15) and the definition of $h^{i}_{r}$ for $i=s(n)=s(m)$. Inserting $z\in \Gamma _{r^{*}}$, we obtain

$$\begin{aligned} H(z) < -\frac{\delta _r}{\kappa }\; \alpha _r(r^{*})+ A(\kappa ) + \frac{\log 2}{n} + \frac{\varepsilon }{\kappa }\ \end{aligned}$$

and

$$\begin{aligned} \frac{1}{n}\; \log \Vert p_n - p_m\Vert _{\Gamma _{r^{*}}} \le \log \frac{r^{*}}{\rho } - \frac{\delta _r}{\kappa } \alpha _r(r^{*}) + A(\kappa ) + \frac{\log 2}{n} + \frac{\varepsilon }{\kappa }. \end{aligned}$$

(3.16)

If we choose $\kappa$ as in (3.11), we get with

$$\begin{aligned} -\frac{\delta _r}{\kappa }\; \alpha _r(r^{*}) + A(\kappa )= & {} -\frac{\delta _r}{\kappa }\; \alpha _r(r^{*}) + \log \frac{\rho }{r} - \frac{1}{\kappa }\log \frac{\rho }{r} \\= & {} -\frac{1}{\kappa }\left( \delta _r\; \alpha _r(r^{*}) + \log \frac{\rho }{r} \right) + \log \frac{\rho }{r} \\< & {} 0. \end{aligned}$$

Now, let us define

$$\begin{aligned} \delta ^{*}_{r^{*}} := \frac{1}{2} \left( \frac{\delta _r}{\kappa }\; \alpha _r(r^{*}) - A(\kappa )\right) , \end{aligned}$$

then $\delta ^{*}_{r^{*}} > 0$, and we can choose n large enough and $\varepsilon$ sufficiently small such that

$$\begin{aligned} \frac{\log 2}{n} + \frac{\varepsilon }{\kappa } < \delta ^{*}_{r^{*}}. \end{aligned}$$

Hence, we can find $n^{*} = n^{*}(\kappa ) \in {\mathbb {N}}$ such that by (3.16) we obtain

$$\begin{aligned} \frac{1}{n} \; \log \Vert p_n - p_m \Vert _{\Gamma _{r^{*}}} \le \log \frac{r^{*}}{\rho } - \delta ^{*}_{r^{*}} \end{aligned}$$

for $m \ge n^{*}$, and (3.12) and (3.13) are proven. $\square$

We know from (3.2) that

$$\begin{aligned} \frac{1}{n}\log \Vert f - p_n\Vert _{\Gamma ^{s(n)}_{\sigma }}\;\le \;\log \frac{\sigma }{\rho } - \delta , \quad \ n \ge n_1(\delta ). \end{aligned}$$

(3.17)

Let

$$\begin{aligned} \kappa ^{*}_{1}\; := \, 1 + \frac{\alpha _\sigma (\rho )}{\log {(\rho /\sigma )}}\;\delta \end{aligned}$$

(3.18)

and let $1< \kappa < \kappa _1^{*}$ in the definition of $\Lambda _1(\kappa )$ in (3.6)−(3.8), then we obtain by Lemma 1 and choosing $r = \sigma$ and $r^{*} = \rho$:

There exists $\delta ^{*}_{1}\ > 0$ and $n^{*}_{1} = n^{*}_{1}(\kappa ) \in {\mathbb {N}}$ such that

$$\begin{aligned} \frac{1}{m_{i+1}} \log \Vert p_{m_{i+1}} - p_{m_i}\Vert _{\Gamma _\rho } \;\le \; -\delta ^{*}_{1} \end{aligned}$$

(3.19)

for all $m_i \in \Lambda _{1,1}(\kappa )$, $m_i \ge n^{*}_{1}$, and

$$\begin{aligned} \delta ^{*}_{1} \ge \frac{1}{2}\left( \frac{\delta }{\kappa }\alpha _\sigma (\rho ) - \left( 1-\frac{1}{\kappa }\right) \log \frac{\rho }{\sigma }\right) . \end{aligned}$$

Corollary 1

Let $1< \kappa < \kappa ^{*}_{1}$ and assume that $\Lambda _{1,2}(\kappa )$ is a finite set in the decomposition of $\Lambda _1(\kappa )$ in (3.6). Then, the telescoping series (3.3) converges uniformly on compact sets of a neighborhood of $\overline{E_\rho }$, i.e., $\rho (f) > \rho .$

Proof

We apply the Bernstein–Walsh Lemma to the differences

$$\begin{aligned} p_{m_{i+1}} - p_{m_i} \end{aligned}$$

and use the inequality (3.19). $\square$

Hence, Proposition 1 is proved for this special situation.

3.2 The Auxiliary Parameter $1< \sigma _0 < \sigma$

To restrict $\kappa$ in the definition of $\Lambda _1(\kappa )$ completely, we start with the decomposition of $E_\sigma$ in (2.1),

$$\begin{aligned} E_{\sigma } \; = \;E^{1}_{\sigma } \cup E^{2}_{\sigma }\cup \ldots \cup E^{l_\sigma }_\sigma , \quad l_{\sigma } \in {\mathbb {N}}. \end{aligned}$$

Then, we can define a parameter $\sigma _0, 1< \sigma _0 < \sigma$, such that the decomposition of $E_{\sigma _0}$ into disjoint Jordan regions $E^{i}_{\sigma _0}$,

$$\begin{aligned} E_{\sigma _0} \; = \;E^{1}_{\sigma _0} \cup E^{2}_{\sigma _0}\cup \ldots \cup E^{l_{\sigma _0}}_{\sigma _0}, \quad l_{\sigma _0}\in {\mathbb {N}}, \end{aligned}$$

satisfies

$$\begin{aligned} l_{\sigma _0} = l_\sigma \quad \text {and}\quad \overline{E^{i}_{\sigma _0}}\subset E^{i}_{\sigma }\quad \text {for}\;1 \le i \le l_{\sigma _0}. \end{aligned}$$

This can be achieved by the strict monotonicity of $E_r$ with respect to r and the fact that the Green function G(z) of $\Omega$ has only a finite number of critical points in ${\mathbb {C}}\setminus E$ (cf. Walsh [4], chapter 4, section 4.1).

Lemma 2

Let $\delta > 0$ and let $1< \sigma _0 < \sigma$ such that according to (3.2)

$$\begin{aligned} \frac{1}{n}\log \Vert f - p_n\Vert _{\Gamma _\sigma ^{s(n)}}\;\le \;\log \frac{\sigma }{\rho } - \delta , \quad \ n \ge n_1(\delta ). \end{aligned}$$

Then, there exist $\delta _0 > 0$ and $n_0(\delta )$ such that

$$\begin{aligned} \frac{1}{n}\log \Vert f - p_n\Vert _{\Gamma _{\sigma _0}^{s(n)}}\;\le \;\log \frac{\sigma _0}{\rho } - \delta _0 \quad \text {for} \quad \ n \ge n_0(\delta ). \end{aligned}$$

(3.20)

Proof

Because of (3.1), there exists $n_\varepsilon (1)$ such that

$$\begin{aligned} \frac{1}{n}\log \Vert f - p_n\Vert _E\;\le \;\log \frac{1}{\rho } + \varepsilon , \quad n \ge n_\varepsilon (1). \end{aligned}$$

Let us consider the Dirichlet problem for the harmonic function $g_{i}(z)$ in the region

$$\begin{aligned} E^{i}_{\sigma } \setminus E, \quad 1\le i \le l_\sigma , \end{aligned}$$

with the boundary conditions

$$\begin{aligned} g_{i}(z)= -\delta \; \text {for} \; z\in \Gamma ^{i}_{\sigma } \quad \text {and}\quad g_{i}(z)= 0 \;\text {for}\;z\in {E^{i}_{\sigma }} \cap \Gamma , \end{aligned}$$

where $\Gamma = \partial E$. Then, $g_{i}(z) < 0$ for $z\in E^{i}_\sigma \setminus E$. Define

$$\begin{aligned} \beta _{i} := \underset{z\in \Gamma ^{i}_{\sigma _o}}{\max }\;g_{i}(z), \end{aligned}$$

then $\beta _{i} < 0$ and also

$$\begin{aligned} \beta := \underset{1\le i \le l_{\sigma }}{\max }\beta _{i} < 0. \end{aligned}$$

Moreover, the function

$$\begin{aligned} g_{s(n)}(z) + \varepsilon \end{aligned}$$

is a harmonic majorant of

$$\begin{aligned} \frac{1}{n}\, \log \, |f(z) - p_{n}(z)| - G(z) + \log \rho \quad \text {in} \quad E^{s(n)}_{\sigma } \setminus E. \end{aligned}$$

That leads to

$$\begin{aligned} \frac{1}{n}\, \log \Vert f - p_n\Vert _{\Gamma ^{s(n)}_{\sigma _0}} - \log \frac{\sigma _o}{\rho } \;\le \underset{z\in \Gamma ^{s(n)}_{\sigma _o}}{\max }\;g_{s(n)}(z) + \varepsilon \end{aligned}$$

or

$$\begin{aligned} \frac{1}{n}\, \log \Vert f - p_n\Vert _{\Gamma ^{s(n)}_{\sigma _0}} \le \log \frac{\sigma _o}{\rho } +\beta + \varepsilon . \end{aligned}$$

If we define $\varepsilon := - \beta /2$, then we get

$$\begin{aligned} \frac{1}{n}\, \log \Vert f - p_n\Vert _{\Gamma ^{s(n)}_{\sigma _0}} \le \log \frac{\sigma _0}{\rho } + \frac{\beta }{2} \end{aligned}$$

for all $n \ge n_\varepsilon (1)$. Therefore,

$$\begin{aligned} n_{0}(\delta ) := n_\varepsilon (1) \quad \text {and} \quad \delta _0 := - \beta /2 \end{aligned}$$

satisfy the statement of Lemma 2. $\square$

Lemma 3

Let $n,m \in {\mathbb {N}}$ with $m < n$ and $n/m \le \kappa$. Let $\delta _0 > 0$ and $n_{0}(\delta ) \in {\mathbb {N}}$ such that (3.20) holds according to Lemma 2. Moreover, let $s(m) = s(n)$ and let

$$\begin{aligned} 1< \kappa < \kappa ^{*}_{2} := 1 + \frac{\alpha _{\sigma _0}(\sigma )}{\log (\rho /\sigma _0)}\;\delta _0, \end{aligned}$$

(3.21)

then there exists $\delta ^{*}_{0} > 0$ and $n^{*}_{2}= n^{*}_{2}(\kappa ) \in {\mathbb {N}}$ such that

$$\begin{aligned} \frac{1}{n} \log \Vert p_n -p_m\Vert _{\Gamma _\sigma } \le \log \frac{\sigma }{\rho }- \delta ^{*}_{0} \end{aligned}$$

(3.22)

for $m \ge n^{*}_{2}$ and

$$\begin{aligned} \delta _0^{*} \ge \frac{1}{2}\left( \frac{\delta _0}{\kappa }\alpha _{\sigma _0}(\sigma ) - \left( 1-\frac{1}{\kappa }\right) \log \frac{\rho }{\sigma _0}\right) . \end{aligned}$$

(3.23)

Proof

Because of Lemma 2, there exists $n_0(\delta )$ such that

$$\begin{aligned} \frac{1}{m} \log \Vert f - p_m\Vert _{\Gamma ^{s(m)}_{\sigma _0}} \le \log \frac{\sigma _0}{\rho }-\delta _0 \end{aligned}$$

and

$$\begin{aligned} \frac{1}{n} \log \Vert f - p_n\Vert _{\Gamma ^{s(n)}_{\sigma _0}} \le \log \frac{\sigma _0}{\rho }-\delta _0 \end{aligned}$$

for $m < n$ and $m \ge n_{0}(\delta )$. Since $s(m) = s(n)$ and $\kappa$ satisfies (3.21), then Lemma 1 yields that there exists $n^{*}_{2} = n^{*}_{2}(\kappa )\in {\mathbb {N}}$ and $\delta ^{*}_{0} > 0$ such that

$$\begin{aligned} \frac{1}{n} \log \Vert p_m - p_n\Vert _{\Gamma _\sigma } \le \log \frac{\sigma }{\rho }-\delta ^{*}_{0} \end{aligned}$$

for $m \ge n^{*}_{2}$ and $\delta ^{*}_{0}$ satisfies (3.23). $\square$

3.3 The Final Telescoping Series

We start with the telescoping series associated with

$$\begin{aligned} \Lambda _1(\kappa ) = \left\{ m_i\right\} ^{\infty }_{i=1} \end{aligned}$$

satisfying (3.4) and (3.5) and choosing the parameter $\kappa$ such that

$$\begin{aligned} 1< \kappa < \min \left( \kappa ^{*}_{1}, \kappa ^{*}_{2}, \kappa ^{*}_{3}\right) . \end{aligned}$$

$\kappa ^{*}_{1}$ is defined by (3.18), i.e.,

$$\begin{aligned} \kappa ^{*}_{1}\; = \; 1 + \frac{\alpha _\sigma (\rho )}{\log {(\rho /\sigma )}}\;\delta \end{aligned}$$

and $\delta$ satisfies (3.17). $\kappa ^{*}_{2}$ is defined by (3.21), i.e.,

$$\begin{aligned} \kappa ^{*}_{2}\; =\; 1 + \frac{\alpha _{\sigma _0}(\sigma )}{\log {(\rho /\sigma _0)}}\;\delta _0, \end{aligned}$$

and $\delta _0$ satisfies (3.20). $\kappa ^{*}_{3}$ will be defined by

$$\begin{aligned} \kappa ^{*}_{3}\; := \, 1 + \frac{\alpha _\sigma (\rho )}{\log {(\rho /\sigma )}}\;\frac{\delta ^{*}_{0}}{2} \end{aligned}$$

(3.24)

and $\delta ^{*}_{0}$ satisfies (3.23). The role of $\kappa ^{*}_{3}$ will be seen in the proof of Lemma 5. As above, we use the decomposition

$$\begin{aligned} \Lambda _1(\kappa ) := \Lambda _{1,1}(\kappa ) \;\cup \; \Lambda _{1,2}(\kappa ) \end{aligned}$$

Hence, by (3.19) there exist $\delta ^{*}_{1} > 0$ and $n^{*}_{1} = n^{*}_{1}(\kappa ) \in {\mathbb {N}}$ such that

$$\begin{aligned} \frac{1}{m_{i+1}}\log \;\Vert p_{m_{i+1}} - p_{m_i}\Vert _{\Gamma _\rho } \le - \delta ^{*}_{1} \end{aligned}$$

(3.25)

for all $m_i \in \Lambda _{1,1}(\kappa ), \; m_i \ge n^{*}_{1}$. So, as critical differences in the telescoping series with respect to $\Lambda _1(\kappa )$ remain $\;p_{m_{i+1}} - p_{m_i}$, where

$$\begin{aligned} m_{i+1} = m_i + 1 \; \text {and} \; s(m_{i+1}) \ne s(m_i). \end{aligned}$$

In Corollary 1, we have given already the proof of Proposition 1 for the case that $\Lambda _{1,2}(\kappa )$ is a finite sequence. Therefore, we assume henceforth that

$$\begin{aligned} \Lambda _{1,2}(\kappa )\; = \;\left\{ \lambda _1< \lambda _2< \lambda _3 <\; \;\ldots \; \right\} \end{aligned}$$

(3.26)

is an infinite sequence.

In the following, we use a real parameter $c \in {\mathbb {R}}$, $0< c < 1$.

Lemma 4

Let $\lambda _k \in \Lambda _{1,2}(\kappa )$ be fixed. Then, there exist at most $l_\sigma$ elements of $\Lambda _{1,2}(\kappa )$ in the interval

$$\begin{aligned} \left( \lambda _k\; , \;\kappa \lambda _k\right] . \end{aligned}$$

Moreover, let the parameter $c \in {\mathbb {R}}$ , $0< c < 1$, be fixed and let the semi-open intervals $I(\lambda _k, j)$ be defined by

$$\begin{aligned} I(\lambda _k, j) := \left( \lambda _k\;\left( 1 + \left( \frac{c}{1+c}\right) ^{j+1} (\kappa -1)\right) ,\ \lambda _k\;\left( 1 + \left( \frac{c}{1+c}\right) ^{j} (\kappa -1)\right) \right] \; \end{aligned}$$

for $0 \le j \le l_{\sigma }-1$. Then, there exists $\widetilde{l_k}$, $0 \le \widetilde{l_k} \le l_{\sigma }-1$, such that

$$\begin{aligned} I(\lambda _k,\widetilde{l_k} )\; \cap \; \Lambda _{1,2}(\kappa ) = \emptyset . \end{aligned}$$

Proof

Let us assume that there exist at least $l_\sigma$ elements of $\Lambda _{1,2}(\kappa )$ in the interval $\left( \lambda _k,\kappa \lambda _k\right]$. Then, the definition of $\Lambda _1(\kappa )$, resp. $\Lambda _{1,2}(\kappa )$, implies that the values of the function s at the points

$$\begin{aligned} \lambda _k, \lambda _{k+1}, \ldots , \lambda _{k+l_\sigma } \end{aligned}$$

are all different, which contradicts the definition of $l_\sigma$.

Let us assume that the second part of the Lemma is false. Then, in each

$$\begin{aligned} I(\lambda _k,j),\;0 \le j \le l_{\sigma }-1, \end{aligned}$$

there exists at least one element of $\Lambda _{1,2}(\kappa )$. Hence, the interval

$$\begin{aligned} \left( \lambda _k\left( 1 + \left( \frac{c}{1+c}\right) ^{l_\sigma }(\kappa - 1)\right) \;, \;\lambda _\kappa \right] \end{aligned}$$

contains at least $l_\sigma$ elements of $\Lambda _{1,2}(\kappa )$, contradicting the first part of the lemma. $\square$

3.3.1 The Telescoping Defining Sequence $\Lambda (\kappa ,c)$

Let

$$\begin{aligned} \Lambda _1(\kappa ) = \left\{ m_i\right\} ^{\infty }_{i=1} \end{aligned}$$

satisfies (3.4) and (3.5) with a parameter $\kappa$ where

$$\begin{aligned} 1< \kappa < \min \left( \kappa ^{*}_{1}, \kappa ^{*}_{2}, \kappa ^{*}_{3}\right) . \end{aligned}$$

$\kappa ^{*}_{1}$ is defined by (3.18), $\kappa ^{*}_{2}$ by (3.21), $\kappa ^{*}_{3}$ by (3.24). As in (3.6) - (3.8), we decompose

$$\begin{aligned} \Lambda _1(\kappa ) := \Lambda _{1,1}(\kappa ) \;\cup \; \Lambda _{1,2}(\kappa ). \end{aligned}$$

Then, we define the sequence

$$\begin{aligned} \Lambda (\kappa , c) = \left\{ n_j\right\} ^{\infty }_{j=1} \end{aligned}$$

as follows: If $\Lambda _{1,2}(\kappa )$ is a finite sequence, then $\Lambda (\kappa ,c) := \Lambda _1(\kappa )$. If $\Lambda _{1,2}(\kappa )$ is an infinite sequence, we define

$$\begin{aligned} \gamma := \left\lfloor \frac{1}{\left( \frac{c}{1+c}\right) ^{l_\sigma } \left( \kappa -1\right) } \right\rfloor + 1 \end{aligned}$$

(3.27)

and we set

$$\begin{aligned} M := \min \left\{ m_i\in \Lambda _1(\kappa ): \; m_i > \gamma \right\} . \end{aligned}$$

Then, we define $n_j := m_j$ for $1\le m_j \le M$. The remaining elements $n_j \in \Lambda (\kappa ,c), \;n_j > M$, will be defined recursively:

If $n_j \ge M$ is already constructed, we note that we obtain by (3.27) for $0 \le \widetilde{l_{j}} \le l_\sigma -1$

$$\begin{aligned} n_j\left( \frac{c}{1+c}\right) ^{\widetilde{l_{j}}+1}(\kappa -1) \ge n_j\left( \frac{c}{1+c}\right) ^{l_{\sigma }}(\kappa -1)> 1. \end{aligned}$$

(3.28)

Then, we fix

$$\begin{aligned} m:= \min \left\{ m_i\in \Lambda _1(\kappa ): \; m_i > n_j \right\} \end{aligned}$$

and distinguish 2 cases:

• (i) If $s(m) = s(n_j)$, then $n_{j+1} := m$.

• (ii) If $s(m) \ne s(n_j)$, then we apply Lemma 4. Hence, there exists $k_0 \in {\mathbb {N}}$ and $0 \le \widetilde{l_{j}} \le l_\sigma -1$, such that

  $$\begin{aligned} \lambda _{k_0} \le n_j\left( 1 + \left( \frac{c}{1+c}\right) ^{\widetilde{l_{j}}+1}(\kappa -1)\right) \end{aligned}$$

  and

  $$\begin{aligned} \lambda _{k_{0+1}} > n_j\left( 1 + \left( \frac{c}{1+c}\right) ^{\widetilde{l_{j}}} (\kappa -1)\right) , \end{aligned}$$

  where we have used (3.28) and the enumeration of $\Lambda _{1,2}(\kappa )$ as in (3.26). Then, we define

  $$\begin{aligned} n_{j+1} := \lambda _{k_0}. \end{aligned}$$

Properties of $\Lambda (\kappa ,c)$ We have always   $n_{j+1} \le \kappa \, n_j$. If $s(n_{j+1}) \ne s(n_j)$, then

$$\begin{aligned} n_{j+1} - n_j \;\le \; n_j \left( \frac{c}{1+c}\right) ^{\widetilde{l_j}+1} (\kappa - 1) \end{aligned}$$

(3.29)

and

$$\begin{aligned} \underset{\lambda \in \Lambda _{1,2}(\kappa )}{\min }\left\{ \lambda : \lambda> n_{j+1}\right\} > n_j \left( 1 + \left( \frac{c}{1+c}\right) ^{\widetilde{l_j}}(\kappa -1)\right) , \end{aligned}$$

(3.30)

where $0 \le \widetilde{l_j} \le l_\sigma - 1$. Moreover, $s(m) = s(n_{j+1})$ for

$$\begin{aligned} m \in \Lambda _1(\kappa ),\; \text {where} \; n_{j+1} \le m < n_j\left( 1 + \left( \frac{c}{1+c}\right) ^{\widetilde{l_j}} (\kappa -1)\right) . \end{aligned}$$

(3.31)

In the following, we use the decomposition

$$\begin{aligned} \Lambda (\kappa ,c) := \Lambda _1(\kappa ,c) \;\cup \; \Lambda _2(\kappa ,c), \end{aligned}$$

where

$$\begin{aligned} \Lambda _1(\kappa ,c) = \left\{ \,n_j \in \Lambda (\kappa ,c) : n_{j+1}/n_j \le \kappa \; \text {and} \; s(n_{j+1})=s(n_j) \right\} \end{aligned}$$

and

$$\begin{aligned} \Lambda _2(\kappa ,c) = \left\{ \,n_j \in \Lambda (\kappa ,c) : n_{j+1}/n_j \le \kappa \; \text {and} \; s(n_{j+1})\ne s(n_j). \right\} \end{aligned}$$

Lemma 5

Let $n_j \in \Lambda _2(\kappa ,c)$, then there exists $\delta _{2,j} > 0$ and $n^{*}_{3} = n^{*}_{3}(\kappa ) \in {\mathbb {N}}$ such that

$$\begin{aligned} \frac{1}{n_{j+1}} \log \Vert f - p_{n_{j+1}}\Vert _{\Gamma _\sigma } \; \le \;\log \frac{\sigma }{\rho } - \delta _{2,j} \quad \text {for} \; n_j \ge n^{*}_{3}. \end{aligned}$$

(3.32)

Moreover, $\delta _{2,j}$ can be chosen in such a way that

$$\begin{aligned} \delta _{2,j} \;\ge \; \frac{1}{2} \min \left( \delta ^{*}_{0},\; \frac{1}{2\kappa }\left( \frac{c}{1+c}\right) ^{\widetilde{l_j}}\frac{\kappa -1}{1+c} \log \frac{\rho }{\sigma }\right) \end{aligned}$$

(3.33)

with $\delta ^{*}_{0}$ satisfying (3.22) and (3.23) of Lemma 3.

Proof

We consider the telescoping series

$$\begin{aligned} f = p_{n_{j+1}} + \sum \limits ^{\infty }_{k=1} \; (p_{n_{j+k+1}} - p_{n_{j+k}}) \end{aligned}$$

and define

$$\begin{aligned} k_j := \sup \left\{ k : s(n_{j+1}) = s(n_{j+2}) = \ldots = s(n_{j+k})\right\} . \end{aligned}$$

(3.34)

Because of (3.30), we have

$$\begin{aligned} n_{j+k_j} \ge n_j \left( 1 + \left( \frac{c}{1+c}\right) ^{\widetilde{l_j}}(\kappa - 1)\right) - 1 , \end{aligned}$$

(3.35)

keeping in mind that $n_{j+k_j} \in {\mathbb {N}}$. Now, we write

$$\begin{aligned} \sum \limits ^{\infty }_{k=1} \; (p_{n_{j+k+1}} - p_{n_{j+k}}) = \; A_j +B_j, \end{aligned}$$

where

$$\begin{aligned} A_j = \sum \limits ^{k_j-1}_{k=1} \; (p_{n_{j+k+1}} - p_{n_{j+k}}) \end{aligned}$$

and

$$\begin{aligned} B_j = \sum \limits ^{\infty }_{k=k_j} \; (p_{n_{j+k+1}} - p_{n_{j+k}}). \end{aligned}$$

Estimation of $A_j$ on $\Gamma _\sigma$ Because of the definition of $k_j$ in (3.34), we use (3.31) and apply Lemma 3 for all differences

$$\begin{aligned} p_{n_{j+k+1}} - p_{n_{j+k}} \end{aligned}$$

occurring in $A_j$. We obtain with $\delta ^{*}_{0} > 0$ and $n^{*}_{2} = n^{*}(\kappa )\in {\mathbb {N}}$ that

$$\begin{aligned} \frac{1}{n_{j+k+1}} \; \log \;\Vert p_{n_{j+k+1}} - p_{n_{j+k}}\Vert _{\Gamma _\sigma } \; \le \; \log \frac{\sigma }{\rho } - \delta ^{*}_{0},\quad n_{j+k}\ge n^{*}_{2}, \end{aligned}$$

where $\delta ^{*}_{0}$ satisfies the inequality (3.23), since

$$\begin{aligned} 1< \kappa < \kappa ^{*}_{2} = 1 + \frac{\alpha _{\sigma _0}(\sigma )}{\log (\rho /{\sigma _0})}\; \delta _0 \end{aligned}$$

and $\delta _0$ is defined by Lemma 2 in (3.20). Then,

$$\begin{aligned} \Vert A_j\Vert _{\Gamma _\sigma }\le & {} \sum ^{k_j-1}_{k=1} \Vert p_{n_{j+k+1}} - p_{n_{j+k}}\Vert _{\Gamma _\sigma } \nonumber \\\le & {} \sum ^{\infty }_{k=1}\;\left( \frac{\sigma }{\rho } e^{-\delta ^{*}_{0}}\right) ^{n_{j+k+1}} \nonumber \\= & {} \beta _1 \left( \frac{\sigma }{\rho } e^{-{\delta }^{*}_{0}}\right) ^{n_{j+1}} \end{aligned}$$

(3.36)

for all j with $n_j \ge n^{*}_{2}$ and $\beta _1$ is a constant independent of j.

Estimation of $B_j$ on $\Gamma _\sigma$ Let us define

$$\begin{aligned} \gamma _j := 1 + \left( \frac{c}{1+c}\right) ^{\widetilde{l_j}}(\kappa - 1). \end{aligned}$$

(3.37)

Because of (3.1), there exists $n_\varepsilon (\sigma )$ such that for $n \ge n_\varepsilon (\sigma )$

$$\begin{aligned} \frac{1}{n}\log \Vert f - p_n\Vert _{\Gamma _\sigma } \le \log {\frac{\sigma }{\rho }} + \varepsilon , \end{aligned}$$

where $0< \varepsilon < \log (\rho /\sigma )$ is fixed. Then, for $n_j \ge n_\varepsilon (\sigma )$

$$\begin{aligned} \Vert p_{n_j+k_j+1} - p_{n_j+k_j}\Vert _{\Gamma _\sigma } \le 2\left( \frac{\sigma }{\rho } \;e^{\varepsilon } \right) ^{n_j+k_j} \end{aligned}$$

and with (3.35) and (3.37) we obtain

$$\begin{aligned} \begin{aligned} \Vert p_{n_j+k_j+1} - p_{n_j+k_j}\Vert _{\Gamma _\sigma }&\le \ 2\left( \frac{\sigma }{\rho } \; e^{\varepsilon }\right) ^{\gamma _j n_j} \left( \frac{\sigma }{\rho }\; e^{\varepsilon }\right) ^{-1} \\&\le \ \beta _2\left( \frac{\sigma }{\rho }\; e^{\varepsilon } \right) ^{\gamma _j n_j}\\ \end{aligned} \end{aligned}$$

with

$$\begin{aligned} \beta _2 = 2\left( \frac{\sigma }{\rho }\; e^{\varepsilon }\right) ^{-1}. \end{aligned}$$

Analogously,

$$\begin{aligned} \Vert p_{n_j+k+1} - p_{n_j+k}\Vert _{\Gamma _\sigma } \; \le \; \beta _2\left( \frac{\sigma }{\rho }\; e^{\varepsilon } \right) ^{\gamma _{j}n_j+k-k_j} \end{aligned}$$

for all $k \ge k_j$ and $n_j \ge n_\varepsilon (\sigma )$. Hence, for such $n_j$

$$\begin{aligned} \begin{aligned} \Vert B_j\Vert _{\Gamma _\sigma } \quad&\le \quad \sum \limits ^{\infty }_{k=k_j} \Vert p_{n_{j+k+1}} - p_{n_{j+k}}\Vert _{\Gamma _\sigma } \\&\le \quad \beta _2\left( \frac{\sigma }{\rho } e \,^{\varepsilon }\right) ^{\gamma _{j} n_j}\sum \limits ^{\infty }_{\nu =0} \left( \frac{\sigma }{\rho } \,e^{\varepsilon }\right) ^{\nu } \\&= \quad \beta _2 \;\frac{\rho }{\rho - \sigma \,e^{\varepsilon }} \left( \frac{\sigma }{\rho }\; e^{\varepsilon }\right) ^{\gamma _{j} n_j} \\&= \quad \beta _3 \left( \frac{\sigma }{\rho } \;e^{\varepsilon } \right) ^{\gamma _{j} n_j}, \end{aligned} \end{aligned}$$

(3.38)

where

$$\begin{aligned} \beta _3 \; = \; \beta _2 \;\frac{\rho }{\rho - \sigma \, e^{\varepsilon }}. \end{aligned}$$

Because of (3.29),

$$\begin{aligned} 1 \; \le \; \frac{n_j}{n_{j+1}} \left( 1 + \left( \frac{c}{1+c}\right) ^{\widetilde{l_j}+1}(\kappa -1)\right) \end{aligned}$$

and therefore

$$\begin{aligned} \begin{aligned} \frac{{\gamma _j}n_j}{n_{j+1}} \quad&\ge \quad \frac{1 + \left( \frac{c}{1+c}\right) ^{\widetilde{l_j}}(\kappa -1)}{1+\left( \frac{c}{1+c}\right) ^{\widetilde{l_j}+1}(\kappa -1)} \\ \quad&= \quad 1 + \frac{\left( \frac{c}{1+c}\right) ^{\widetilde{l_j}}\frac{\kappa -1}{1+c}}{1+\left( \frac{c}{1+c}\right) ^{\widetilde{l_j}+1}(\kappa -1)}. \end{aligned} \end{aligned}$$

(3.39)

For abbreviation, we define

$$\begin{aligned} \widetilde{\delta _j} \; := \;\frac{\left( \frac{c}{1+c}\right) ^{\widetilde{l_j}}\frac{\kappa -1}{1+c}}{1+\left( \frac{c}{1+c}\right) ^{\widetilde{l_j}+1}(\kappa -1)} \end{aligned}$$

and note that

$$\begin{aligned} \frac{1}{\kappa }\left( \frac{c}{1+c}\right) ^{\widetilde{l_j}} \frac{\kappa -1}{1+c}< \widetilde{\delta _j} < \kappa - 1. \end{aligned}$$

(3.40)

Since $\varepsilon < \log ({\rho /\sigma })$, multiplication of (3.39) by $\log ({\sigma /\rho }) + \varepsilon$ yields

$$\begin{aligned} \frac{{\gamma _j}n_j}{n_{j+1}}\;\left( \log {\frac{\sigma }{\rho } + \varepsilon }\right) \;\le \;\log \frac{\sigma }{\rho } + \widetilde{\delta _j} \log {\frac{\sigma }{\rho }} + \varepsilon \left( 1+ \widetilde{\delta _j}\right) . \end{aligned}$$

Hence, the upper bound in (3.40) leads to

$$\begin{aligned} \frac{{\gamma _j}n_j}{n_{j+1}}\;\left( \log {\frac{\sigma }{\rho } + \varepsilon }\right) < \;\log {\frac{\sigma }{\rho }} - \widetilde{\delta _j} \log {\frac{\rho }{\sigma }} + \varepsilon \,\kappa . \end{aligned}$$

Next, we define

$$\begin{aligned} \varepsilon \;:= \;\frac{1}{2\,{\kappa }^{2}}\left( {\left( \frac{c}{1+c}\right) ^{\widetilde{l_j}}\frac{\kappa -1}{1+c} \log {\frac{\rho }{\sigma }}}\right) . \end{aligned}$$

(3.41)

Then the general condition $\varepsilon < \log (\rho /\sigma )$ is satisfied and the lower bound of (3.40) yields

$$\begin{aligned}&- \widetilde{\delta _j} \log {\frac{\rho }{\sigma }} + \varepsilon \,\kappa \; < \;- \widetilde{\delta _j} \log {\frac{\rho }{\sigma }} + \frac{1}{2}\;\widetilde{\delta _j} \log {\frac{\rho }{\sigma }} \\&\quad = \; - \frac{1}{2}\;\widetilde{\delta _j} \log {\frac{\rho }{\sigma }} . \end{aligned}$$

Therefore, for such $\varepsilon$ we obtain by (3.38) for $n_j \in \Lambda _2(\kappa ,c)$ and $n_j \ge n_\varepsilon (\sigma )$

$$\begin{aligned} \Vert B_j\Vert _{\Gamma _\sigma } \; < \; \beta _3 \left( \frac{\sigma }{\rho } e^{ - \delta ^{*}_{2.j}}\right) ^{n_{j+1}}, \end{aligned}$$

(3.42)

where $\delta ^{*}_{2.j}$ is defined by

$$\begin{aligned} \delta ^{*}_{2.j} \; := \; \frac{\widetilde{\delta _j}}{2} \log {\frac{\rho }{\sigma }} \end{aligned}$$

and $\varepsilon$ is defined by (3.41).

Summarizing, by (3.36) and (3.42) we have got for $n_j \in \Lambda _2(\kappa ,c)$ and $n_j \ge \; \max \,(n^{*}_{2}, n_\varepsilon (\sigma ))$

$$\begin{aligned} \begin{aligned} \Vert f - n_{j+1}\Vert _{\Gamma _\sigma }&\le \quad \Vert A_j\Vert _{\Gamma _\sigma } + \Vert B_j\Vert _{\Gamma _\sigma } \\ \\&\quad \le \quad \beta _1\left( \frac{\sigma }{\rho } e^{-\delta ^{*}_{0}}\right) ^{n_{j+1}} + \beta _3\left( \frac{\sigma }{\rho } e^{-\delta ^{*}_{2,j}}\right) ^{n_{j+1}}. \end{aligned} \end{aligned}$$

$\beta _1$ and $\beta _3$ are constants, independent of $n_j$. Hence, if we define

$$\begin{aligned} \delta _{2,j} \; := \; \frac{1}{2} \min \left( \delta ^{*}_{0},\; \frac{1}{2\kappa }\left( \frac{c}{1+c}\right) ^{\widetilde{l_j}} \frac{\kappa -1}{1+c}\log \frac{\rho }{\sigma }\right) \end{aligned}$$

and if we use the lower bound in (3.40), then there exists $n^{*}_{3} = n^{*}_{3}(\kappa ,c)$ such that

$$\begin{aligned} \frac{1}{n_{j+1}} \log \Vert f - p_{n_{j+1}}\Vert {\Gamma }_{\sigma }\; \le \; \log \frac{\sigma }{\rho } - \delta _{2,j}, \quad n_j \ge n^{*}_{3}, \end{aligned}$$

and (3.32) and (3.33) of Lemma 5 are proven. $\square$

3.3.2 Fixing the Parameter c in $\Lambda (\kappa ,c)$

In the case that $n_j \in \Lambda _2(\kappa ,c)$, we have by Lemma 5: There exists $\delta _{2,j} > 0$ and $n^{*}_{3} = n^{*}_{3}(\kappa ,c)$ such that

$$\begin{aligned} \frac{1}{n_{j+1}} \log \Vert f - p_{n_{j+1}}\Vert _{\Gamma _\sigma }\; \le \; \log {\frac{\sigma }{\rho }} - \delta _{2,j} \end{aligned}$$

(3.43)

for all $n_j \ge n^{*}_{3}(\kappa )$. Moreover,

$$\begin{aligned} \delta _{2,j} \ge \frac{1}{2} \; \min \;\left( \delta ^{*}_{0},\; \frac{1}{2\,\kappa }\,\left( \frac{c}{1+c}\right) ^{\widetilde{l_j}} \frac{\kappa -1}{1+c} \log \frac{\rho }{\sigma } \right) \end{aligned}$$

(3.44)

with $\delta ^{*}_{0} > 0$ as in Lemma 3. Because of (3.43), we have a fortiori

$$\begin{aligned} \frac{1}{n_{j+1}} \log \Vert f - p_{n_{j+1}}\Vert _{\Gamma ^{s(n_j)}_{\sigma }} \; \le \; \log {\frac{\sigma }{\rho }} - \delta _{2,j}. \end{aligned}$$

(3.45)

On the other hand, we have by (3.2)

$$\begin{aligned} \frac{1}{n_j} \log \Vert f - p_{n_j}\Vert _{\Gamma ^{s(n_j)}_{\sigma }} \;\le \; \log {\frac{\sigma }{\rho }} - \delta \end{aligned}$$

(3.46)

for all $n_j \ge n_1(\delta )$ with $\delta > 0$. Now, we can apply Lemma 1 by taking into account (3.45) and (3.46): There exists $n^{*}_{4} = n^{*}_{4}(\kappa )$ such that

$$\begin{aligned} \frac{1}{n_{j+1}} \log \Vert p_{n_{j+1}} - p_{n_j}\Vert _{\Gamma _\rho } \; \le \; - \delta ^{*}_{3}, \quad n_j \ge n^{*}_{4}, \end{aligned}$$

(3.47)

where

$$\begin{aligned} \delta ^{*}_{3} \; \ge \; \frac{1}{2} \, \left( \frac{\alpha _{\sigma } (\rho )}{\kappa } {\text {min}}\, (\delta , \, \delta _{2,j}) - \left( 1 - {\frac{1}{\kappa }}\right) \, \text {log} {\frac{\rho }{\sigma }}\right) > 0, \end{aligned}$$

if we can achieve, i.e., if we can arrange c with 0 < c <1 such that

$$\begin{aligned} \kappa _j := \frac{n_{j+1}}{n_j} \;\le \; \kappa \; < \; 1 + \frac{\alpha _\sigma (\rho )}{\log {(\rho /\sigma )}} \min \,(\delta , \delta _{2,j}). \end{aligned}$$

(3.48)

Since

$$\begin{aligned} \kappa _j = \frac{n_{j+1}}{n_j} \;\le \; \kappa \; < \;\kappa ^{*}_{1}\; =\; 1 + \frac{\alpha _\sigma (\rho )}{\log {(\rho /\sigma )}}\, \delta , \end{aligned}$$

the inequalities (3.48) are fulfilled if

$$\begin{aligned} \kappa _j = \frac{n_{j+1}}{n_j} \;< \;1 + \frac{\alpha _\sigma (\rho )}{\log {(\rho /\sigma )}} \, \delta _{2,j}. \end{aligned}$$

Taking into account (3.44) and

$$\begin{aligned} \kappa \;<\; \kappa ^{*}_{3} \;=\; 1 + \frac{\alpha _\sigma (\rho )}{\log {(\rho /\sigma )}} \, \frac{\delta ^{*}_{0}}{2}, \end{aligned}$$

the inequality (3.48) is satisfied if

$$\begin{aligned} \kappa _j \;\le \;\kappa \;<\; 1\;+\; \frac{\alpha _\sigma (\rho )}{\log (\rho /\sigma )}\; \frac{1}{4\,\kappa }\; \left( \frac{c}{1+c}\right) ^{\widetilde{l_j}} \frac{\kappa -1}{1+c} \;\log \frac{\rho }{\sigma }. \end{aligned}$$

Because of (3.29), we know that

$$\begin{aligned} \kappa _j = \frac{n_{j+1}}{n_j} \;\le \; 1 + \left( \frac{c}{1+c}\right) ^{\widetilde{l_j}+1}\left( \kappa -1\right) . \end{aligned}$$

Therefore, (3.48) is guaranteed if

$$\begin{aligned} c \; < \; \frac{1}{4\,\kappa }\; \alpha _\sigma (\rho ). \end{aligned}$$

(3.49)

3.4 Conclusions

We consider the telescoping series

$$\begin{aligned} f \;=\; p_{n_1} + \sum ^{\infty }_{j=1} \;\left( p_{n_{j+1}} - p_{n_j}\right) , \end{aligned}$$

associated with the sequence

$$\begin{aligned} \Lambda (\kappa ,c) = \left\{ n_j\right\} ^{\infty }_{j=1} = \Lambda _1(\kappa ,c) \cup \Lambda _2(\kappa ,c). \end{aligned}$$

The parameter $\kappa$ satisfies

$$\begin{aligned} 1< \kappa < \min \,(\kappa ^{*}_{1},\kappa ^{*}_{2},\kappa ^{*}_{3}), \end{aligned}$$

and we fix a parameter c such that

$$\begin{aligned} 0< c < \frac{1}{4\,\kappa } \, \alpha _\sigma (\rho ), \end{aligned}$$

where $\kappa ^{*}_{1}$, $\kappa ^{*}_{2}$, $\kappa ^{*}_{3}$ are defined by (3.18), (3.21), (3.24). If $n_j \in \Lambda _1(\kappa ,c)$, then according to (3.25)

$$\begin{aligned} \frac{1}{n_{j+1}} \log \; \Vert p_{n_{j+1}} - p_{n_j}\Vert _{\Gamma _\rho } \le - \delta ^{*}_{1} \end{aligned}$$

for all $n_j \in \Lambda _1(\kappa ,c)$, $n_j \ge n^{*}_{1}(\kappa )$. If $n_j \in \Lambda _2(\kappa ,c)$, then according to (3.47)

$$\begin{aligned} \frac{1}{n_{j+1}} \log \,\Vert p_{n_{j+1}} - p_{n_j}\Vert _{\Gamma _\rho } \le - \delta ^{*}_{3}. \end{aligned}$$

for all $n_j \in \Lambda _2(\kappa ,c)$, $n_j \ge n^{*}_{4}(\kappa )$, since c satisfies (3.49). Therefore,

$$\begin{aligned} \frac{1}{n_{j+1}} \log \,\Vert p_{n_{j+1}} - p_{n_j}\Vert _{\Gamma _\rho } \le - \min \, (\delta ^{*}_{1},\delta ^{*}_{3}) < 0 \end{aligned}$$

for all $n_j \in \Lambda (\kappa ,c)$ with $n_j \ge \max \,\left( n^{*}_{1}(\kappa ),n^{*}_{4}(\kappa )\right)$. Finally, the Lemma of Bernstein–Walsh implies that f is holomorphic in a neighborhood of $\overline{E_\rho }$, i.e.,

$$\begin{aligned} \rho (f) \;>\; \rho , \end{aligned}$$

and Proposition 1 is proven.

4 Proof of Proposition 2

We choose r and R such that

$$\begin{aligned} 1< r< \sigma< R < \rho \end{aligned}$$

under the additional condition that in the decomposition of $E_R$, resp. $E_\rho$, analogous to (2.1), the numbers $l_R$ and $l_\rho$ satisfy $l_R = l_\rho$. For abbreviation, we define

$$\begin{aligned} l := l_R = l_\rho . \end{aligned}$$

Now, for all $z\in \Omega ={\overline{{\mathbb {C}}}}\setminus {E}$ we have

$$\begin{aligned} \left( U^{\nu _n} - U^{\mu _E}\right) (z) \; =\; U^{\nu _n}(z)\; + \,G(z)\; + \,\log \mathrm{cap} \; E \end{aligned}$$

and therefore

$$\begin{aligned} \begin{aligned} \underset{z\in \Gamma _r}{\max } \; \left( U^{\nu _n} - U^{\mu _E}\right) (z) \quad&= \quad \underset{z\in \Gamma _r}{\max }\; U^{\nu _n}(z)\, +\, \log \, r \,+\,\log \mathrm{cap} \; E \\ \quad&= \quad \underset{z\in \Gamma _r}{\max }\; U^{\nu _n}(z)\, +\, \log \mathrm{cap} \; \overline{E_r}. \end{aligned} \end{aligned}$$

Hence, the uniqueness of the equilibrium measure of $\overline{E_r}$ implies

$$\begin{aligned} \delta _n \; := \; \underset{z\in \Gamma _r}{\max } \; \left( U^{\nu _n} - U^{\mu _E}\right) (z) \; > \; 0 . \end{aligned}$$

(4.1)

Next, we fix $z_n\in \Gamma _r$ such that

$$\begin{aligned} \delta _n \;=\; \left( U^{\nu _n} - U^{\mu _E}\right) (z_n) \end{aligned}$$

(4.2)

and we choose $s^{*}(n)\in {\mathbb {N}}$ such that

$$\begin{aligned} 1 \le s^{*}(n) \le l \; \text {and} \; z_n \in E^{s^{*}(n)}_{R}. \end{aligned}$$

(4.3)

Consider

$$\begin{aligned} D^{s^{*}(n)}_{R,r} \; := \; E^{s^{*}(n)}_{R}\setminus {\overline{E_r}}, \end{aligned}$$

then $D^{s^{*}(n)}_{R,r}$ is a region with boundary

$$\begin{aligned} \Gamma ^{s^{*}(n)}_{R} \; \cup \; \left( \Gamma _r \cap E^{s^{*}(n)}_{R}\right) , \end{aligned}$$

where

$$\begin{aligned} \Gamma ^{s^{*}(n)}_{R} \; \cap \; \left( \Gamma _r \cap E^{s^{*}(n)}_{R}\right) = \emptyset . \end{aligned}$$

The Lagrange–Hermite formula for the error $f - p_n$ at $z\in \Gamma _r$ is

$$\begin{aligned} f(z) - p_n(z) \; = \; \frac{1}{2\pi i}\;\int _{\Gamma _R}\; \frac{w_n(z)}{w_n(t)}\;\frac{f(t)}{t-z} \,d(t) \end{aligned}$$

with

$$\begin{aligned} w_n(t) \;= \; \prod ^{n}_{i=0}\,\left( t - z_{n,i}\right) , \quad t\in {\mathbb {C}}, \end{aligned}$$

where $z_{n,i},\; 0\le i\le n,$ are the interpolation points of $Z_n$. Moreover, we can write

$$\begin{aligned} f(z) - p_n(z) \; = \; \frac{1}{2\pi i}\;\int _{\Gamma _R}\; \frac{w_n(z)}{w_n(t)}\;\frac{f(t)-p_n(t)}{t-z} \,d(t) \end{aligned}$$

for $z\in \Gamma _r$. If $z\in \Gamma _r \cap E^{s^{*}(n)}_{R}$, we may reduce the path of integration to $\Gamma ^{s^{*}(n)}_{R}$, hence

$$\begin{aligned} f(z) - p_n(z) \; = \; \frac{1}{2\pi i}\;\int _{\Gamma ^{s^{*}(n)}_{R}}\; \frac{w_n(z)}{w_n(t)}\;\frac{f(t)-p_n(t)}{t-z} \,d(t). \end{aligned}$$

(4.4)

Let $\varepsilon > 0$, then (2.5) implies that there exists $n_0(\varepsilon )$ such that

$$\begin{aligned} \frac{1}{n} \log \, \Vert f - p_n\Vert _{\Gamma _{R}} \;\le \; \log \,\frac{R}{\rho } + \varepsilon \end{aligned}$$

and

$$\begin{aligned} \frac{1}{n} \log \, \Vert f - p_n\Vert _{\Gamma _{r}} \;\le \; \log \,\frac{r}{\rho } + \varepsilon \end{aligned}$$

for all $n \ge n_0(\varepsilon )$. Using (4.4), we may choose $n_0(\varepsilon )$ in such a way that for all $z\in \Gamma ^{s^{*}(n)}_{r}$ and $n\ge n_0(\varepsilon )$

$$\begin{aligned} \begin{aligned} \frac{1}{n}\, \log |f(z) - p_n(z)|&\le -U^{\nu _n}(z) + \underset{t\in \Gamma _R}{\max }\; U^{\nu _n}(t) + \frac{1}{n} \log \Vert f-p_n\Vert _{\Gamma _{R}} + \varepsilon \\&\le -U^{\nu _n}(z) + \underset{t\in \Gamma _{R}}{\max } \; U^{\nu _n}(t) + \log \frac{R}{\rho } + 2\,\varepsilon . \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} U^{\mu _E}(t) = U^{\mu _E}(z) + \log \frac{r}{R} \quad \text {for}\; t\in \Gamma _R \quad \text {and}\; z\in \Gamma _r, \end{aligned}$$

we get for $z\in \Gamma ^{s^{*}(n)}_{r}$

$$\begin{aligned} \begin{aligned} \frac{1}{n} \log |f(z) - p_n(z)| \;&\le \;\left( U^{\mu _E} - U^{\nu _n}\right) (z) + \underset{t\in {\Gamma _R}}{\max }\; \left( U^{\nu _n} - U^{\mu _E}\right) (t) \\&\quad \; + \log \frac{r}{\rho } + 2\,\varepsilon . \end{aligned} \end{aligned}$$

(4.5)

Now, let us consider the difference

$$\begin{aligned} U^{\nu _n}(z) - U^{\mu _E}(z), \quad z\in \Omega , \end{aligned}$$

which is a harmonic function in $\Omega$. Then, the maximum of this difference on the level curve $\Gamma _{\sigma ^{*}}$ is increasing with decreasing $\sigma ^{*},\; 1< \sigma ^{*} < \infty$. Consequently,

$$\begin{aligned} \underset{t\in {\Gamma _r}}{\max }\,\left( U^{\nu _n} - U^{\mu _E}\right) (t) \;> \; \underset{t\in {\Gamma _R}}{\max }\,\left( U^{\nu _n} - U^{\mu _E}\right) (t). \end{aligned}$$

(4.6)

We note for further applications that (4.6) holds also if we replace $\nu _n$ by any probability measure $\nu \ne \mu _E$ with support in E.

Because of (4.1) and (4.2) and the choice of $s^{*}(n)$, we have

$$\begin{aligned} \begin{aligned} \underset{z\in \Gamma ^{s^{*}(n)}_{r}}{\max } \left( U^{\nu _n} - U^{\mu _E}\right) (z)&\; = \;\underset{z\in {\Gamma _r}}{\max } \left( U^{\nu _n} - U^{\mu _E}\right) (z) \\&\; > \;\underset{t\in {\Gamma _R}}{\max } \left( U^{\nu _n} - U^{\mu _E}\right) (t). \end{aligned} \end{aligned}$$

Next, we define the Dirichlet problem for the harmonic function $\Phi _n(z)$ in the region

$$\begin{aligned} D^{s^{*}(n)}_{R,r} = E^{s^{*}(n)}_{R}\setminus {\overline{E_r}} \end{aligned}$$

with the boundary conditions

$$\begin{aligned} \Phi _n(z)\, =\,0, \; z\in \Gamma ^{s^{*}(n)}_{R} \end{aligned}$$

and

$$\begin{aligned} \Phi _n(z) = \min {(0, \left( U^{\mu _E} - U^{\nu _n}\right) (z) + c(\nu _n;\Gamma _R))}, \; z\in \Gamma _r\cap E^{s^{*}(n)}_{R}, \end{aligned}$$

(4.7)

where

$$\begin{aligned} c(\nu _n;\Gamma _R) \;:=\; \underset{t\in \Gamma _R}{\max }\, \left( U^{\nu _n} - U^{\mu _E}\right) (t). \end{aligned}$$

Because of (4.3) and (4.6), $\Phi _n(z_n) < 0$ and therefore $\Phi _n(z) < 0$ for all $z\in D^{s^{*}(n)}_{R,r}$. Thus, if we define

$$\begin{aligned} \Phi ^{\sigma }_{n}\;:=\; \underset{t\in \Gamma _\sigma \cap E^{s^{*}(n)}_{R}}{\max } \Phi _n(t), \quad r< \sigma < R, \end{aligned}$$

then

$$\begin{aligned} \Phi ^{\sigma }_{n} < 0 \quad \text {for all} \; n\in \Lambda . \end{aligned}$$

Moreover, the maximum principle for harmonic functions, together with (4.5), implies that the harmonic function

$$\begin{aligned} \Phi _n(z) + G(z) - \log {\rho } + 2\,\varepsilon , \end{aligned}$$

is an upper bound for the subharmonic function

$$\begin{aligned} \frac{1}{n} \log |f(z) - p_n(z)|, \end{aligned}$$

i.e.,

$$\begin{aligned} \frac{1}{n} \log |f(z) - p_n(z)| \;\le \; \Phi _n(z) + G(z) - \log {\rho } + 2\,\varepsilon \end{aligned}$$

for all $z\in D^{s^{*}(n)}_{R,r}$. Hence, we obtain

$$\begin{aligned} \frac{1}{n} \log |f(z) - p_n(z)| \;\le \; \log {\frac{\sigma }{\rho }} + \Phi ^{\sigma }_{n} + 2\,\varepsilon \end{aligned}$$

(4.8)

for all $z\in \Gamma _\sigma \cap E^{s^{*}(n)}_{R}$ and all $n \ge n_0(\varepsilon )$.

Now, we claim: There exists $\delta > 0$ such that

$$\begin{aligned} \Phi ^{\sigma }_{n} \;\le \; - \, \delta \quad \text {for all}\;n\in \Lambda . \end{aligned}$$

(4.9)

Let us assume that the claim is false:

Then, there exists a subsequence $\Lambda _1 \subset \Lambda$ such that

$$\begin{aligned} \underset{n\in \Lambda _1,n\rightarrow \infty }{\lim } \;\Phi ^{\sigma }_{n} \;= \;0. \end{aligned}$$

By Helly’s theorem, there exists a subsequence $\Lambda _2\subset \Lambda _1$ such that

$$\begin{aligned} \underset{n\in \Lambda _2, n\rightarrow \infty }{\lim }\; {\widehat{\nu }}_n \; =\; \nu \end{aligned}$$

with supp$(\nu )\subset E$ and $\nu \ne \mu _E$. Since there are only l different sets

$$\begin{aligned} \Gamma _\sigma \cap E^{i}_{R}, \quad 1 \le i \le l, \end{aligned}$$

we can finally choose $\Lambda _2$ such that the sets

$$\begin{aligned} \Gamma _\sigma \cap E^{s^{*}(n)}_{R} \end{aligned}$$

are fixed for all $n\in \Lambda _2$, i.e., $s^{*}(n) = j_0$ is fixed for all $n\in \Lambda _2$.

Because of

$$\begin{aligned} {\widehat{\nu }}_n \; \overset{*}{\underset{n\in \Lambda _2,n\rightarrow \infty }{\longrightarrow }} \; \nu , \end{aligned}$$

there exists $n_1(\varepsilon ) \ge n_0(\varepsilon )$ such that

$$\begin{aligned} |U^{\nu }(z) - U^{\nu _n}(z)| \;<\; \varepsilon , \quad z\in \Gamma _r \cup \Gamma _R, \end{aligned}$$

for all $n\in \Lambda _2$, $n \ge n_1(\varepsilon )$. Then, for $z\in \Gamma _r$ and $n \ge n_1(\varepsilon )$

$$\begin{aligned} \begin{aligned}&\left( U^{\mu _E} - U^{\nu _n}\right) (z) + c(\nu _n;\Gamma _R) \\&\quad = \;\left( U^{\mu _E} - U^{\nu _n}\right) (z) + \underset{t\in \Gamma _R}{\max } \,\left( U^{\nu _n} - U^{\mu _E}\right) (t) \\&= \;\left( U^{\mu _E} - U^{\nu }\right) (z) + \left( U^{\nu } - U^{\nu _n}\right) (z) \\&\qquad + \underset{t\in \Gamma _R}{\max } \,\left( \left( U^{\nu _n} - U^{\nu }\right) (t) + \left( U^{\nu } - U^{\mu _E}\right) (t)\right) \\&\le \; \left( U^{\mu _E} - U^{\nu }\right) (z) + c(\nu ;\Gamma _R) + 2\,\varepsilon , \end{aligned} \end{aligned}$$

where we have defined

$$\begin{aligned} c(\nu ;\Gamma _R) := \underset{t\in \Gamma _R}{\max } \,\left( U^{\nu } - U^{\mu _E}\right) (t). \end{aligned}$$

(4.10)

Therefore, the boundary condition (4.7) can be estimated by

$$\begin{aligned} \begin{aligned}&\min \,(0, \left( U^{\mu _E} - U^{\nu _n}\right) (z) + c(\nu _n;\Gamma _R)) \\&\quad \le \,\min \,(0, \left( U^{\mu _E} - U^{\nu }\right) (z) + c(\nu ;\Gamma _R) + 2\,\varepsilon ) \end{aligned} \end{aligned}$$

(4.11)

for $z\in \Gamma _r\cap E^{j_0}_{R}.$

Now, we consider the Dirichlet Problem for the function $\Phi (z)$ in the region $D^{j_0}_{R,r} = E^{j_0}_{R} \setminus \overline{E_r}$ with the boundary conditions

$$\begin{aligned} \Phi (z) = 0 \;\text {for}\; z\in \Gamma ^{j_0}_{R} \end{aligned}$$

(4.12)

and

$$\begin{aligned} \Phi (z) = \min \left( 0, \left( U^{\mu _E} - U^{\nu }\right) (z) + c(\nu ;\Gamma _R) +2\,\varepsilon \right) , \; z\in {\Gamma _r\cap E^{j_0}_{R}} \end{aligned}$$

(4.13)

where $c(\nu ;\Gamma _R)$ is defined by (4.10).The continuous functions $U^{\mu _E} - U^{\nu _n}$ converge in $\Omega$ uniformly on compact sets, especially on $\Gamma _r \cup \Gamma _R$, as $n\in \Lambda _2, n\rightarrow \infty$. Hence, by (4.1) and (4.2)

$$\begin{aligned} \begin{aligned} \underset{z\in {\Gamma _r\cap E^{j_0}_{R}}}{\max } \left( U^{\nu } - U^{\mu _E}\right) (z) \quad&=\quad \underset{n\in \Lambda _2,n\rightarrow \infty }{\lim }\; \underset{z\in {\Gamma _r\cap E^{j_0}_{R}}}{\max } (U^{\nu _n} - U^{\mu _E})(z) \\&=\quad \underset{n\in \Lambda _2,n\rightarrow \infty }{\lim } \quad \underset{z\in \Gamma _r}{\max }\; (U^{\nu _n} - U^{\mu _E})(z) \\&=\quad \underset{z\in \Gamma _r}{\max } \,(U^{\nu } - U^{\mu _E})(z) \\&>\quad \underset{t\in \Gamma _R}{\max } \,(U^{\nu } - U^{\mu _E})(t). \\&=\quad c(\nu ;\Gamma _R). \end{aligned} \end{aligned}$$

The last inequality follows from $\nu \ne \mu _E$, mentioned in the remark following (4.6).

Next, we choose $\varepsilon > 0$ such that

$$\begin{aligned} \underset{z\in {\Gamma _r\cap E^{j_0}_{R}}}{\max } \left( U^{\nu } - U^{\mu _E}\right) (z) -2\,\varepsilon \; >\; c(\nu ;\Gamma _R). \end{aligned}$$

Hence, the boundary conditions for the harmonic function $\Phi (z)$ in (4.12) and (4.13) read as $\Phi (z) \le 0$, but $\Phi (z)$ is not identically 0 on $\Gamma _r \cap E^{j_0}_{R}$. Then, the maximum principle for harmonic functions yields

$$\begin{aligned} \Phi (t) < 0 \quad \text {for} \quad t \in D^{j_0}_{R,r} = E^{j_0}_{R,r}\setminus \overline{E_r}. \end{aligned}$$

If we compare the Dirichlet problems for $\Phi _n$ and $\Phi$, then by (4.11)

$$\begin{aligned} \Phi _n(t) \le \Phi (t) \quad \text {for} \; t\in D^{j_0}_{R,r} \quad \text {and for all}\; n\in \Lambda _2, n \ge n_1(\varepsilon ). \end{aligned}$$

Therefore,

$$\begin{aligned} \Phi ^{\sigma }_{n} = \underset{t\in \Gamma _\sigma \cap E^{j_0}_{R}}{\max } \Phi _n(t) \le \underset{t\in \Gamma _\sigma \cap E^{j_0}_{R}}{\max }\Phi (t) < 0 \end{aligned}$$

for $n\in \Lambda _2, n \ge n_1(\varepsilon )$, contradicting our assumption that (4.9) is not true.

Hence, (4.8) and (4.9) imply that

$$\begin{aligned} \frac{1}{n}\,|f(z) -p_n(z)| \;\le \;\log \frac{\sigma }{\rho } - \delta + 2\,\varepsilon \end{aligned}$$

for all $z\in \Gamma _\sigma \cap E^{s^{*}(n)}_{R}$ and $n \ge n_0(\varepsilon ), n\in \Lambda$. If we choose $\varepsilon = \delta /4$, then we finally get

$$\begin{aligned} \frac{1}{n}\,|f(z) -p_n(z)| \;\le \;\log \frac{\sigma }{\rho } - \frac{\delta }{2} \end{aligned}$$

(4.14)

for $z\in \Gamma _\sigma \cap E^{s^{*}(n)}_{R}$ and $n \ge n_0(\varepsilon ), n\in \Lambda$.

We note that each $\Gamma _\sigma \cap E^{s^{*}(n)}_{R}, 1 \le s^{*}(n) \le l$, consists of a finite number of connected components of $\Gamma _\sigma$. Therefore, because of (4.14) we can define for each $n\in \Lambda$ a number $s(n), 1 \le s(n) \le l_\sigma$ such that

$$\begin{aligned} \underset{n\in \Lambda ,n\rightarrow \infty }{\lim \sup } \,\Vert f - p_n\Vert ^{1/n}_{\Gamma ^{s(n)}_{\sigma }}\; < \; \frac{\sigma }{\rho }. \end{aligned}$$

Hence, (2.6) of Proposition 2 is proven.

5 Proof of the Theorems

We have already mentioned that Theorem 1 is a direct consequence of Proposition 1. More precisely, if the condition (2.4) of Theorem 1 is true, then the sequence $p_n\in {\mathcal {P}}_n$ is maximally convergent to f, due to Bernstein–Walsh. Conversely, if the condition (2.4) is not true for some $\sigma , \; 1< \sigma< \rho (f) < \infty$, i.e.,

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim \sup } \underset{1 \le i \le l_\sigma }{\min }\; \Vert f - p_n\Vert ^{1/n}_{\Gamma ^{i}_{\sigma }} \; <\frac{\sigma }{\rho (f)} \; = \; \underset{n\rightarrow \infty }{\lim \sup }\; \Vert f - p_n\Vert ^{1/n}_{\Gamma _\sigma }, \end{aligned}$$

then Proposition 1 shows that $\rho (f)$ is not the maximal parameter of holomorphy of f, which is a contradiction.

Concerning part (a) of Theorem 2, let us assume that $\mu _E$ is not a weak* limit point of ${\widehat{\nu }}_n, n\in {\mathbb {N}}$. Then, Proposition 2 yields—using $\Lambda = {\mathbb {N}}$—that there exist parameter $s(n), 1 \le s(n) \le l_\sigma$, such that

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim \sup } \; \Vert f-p_n\Vert ^{1/n}_{\Gamma ^{s(n)}_{\sigma }} \; < \;\frac{\sigma }{\rho (f)}. \end{aligned}$$

But according to Theorem 1, then $\rho (f)$ could not be the maximal parameter of holomorphy of f. This is a contradiction to the maximal convergence of $\left\{ p_n\right\} ^{\infty }_{n\in {\mathbb {N}}}$.

Concerning part (b), we know already that there exists a subsequence $\Lambda \in {\mathbb {N}}$ such that (2.7) holds. Let us assume that $\mu _E$ is not a weak* limit point of ${\widehat{\nu }}_n, \; n\in \Lambda$. Then, Proposition 2 implies that there exist

$$\begin{aligned} s(n),\; 1 \le s(n) \le l_\sigma , \end{aligned}$$

such that

$$\begin{aligned} \underset{n\in \Lambda ,n\rightarrow \infty }{\lim \sup } \;\Vert f - p_n\Vert ^{1/n}_{\Gamma ^{s(n)}_{\sigma }} \;<\; \frac{\sigma }{\rho (f)}. \end{aligned}$$

This contradicts (2.7) and Theorem 2 is proven.