### Proof of Theorem 2.1

By Lemma 3.1, since \(F\in \mathcal{H}(E)\) and \(\mu \in \mathbf{Reg}_{1}^{*}(E)\),

$$ F(z):=\sum_{\nu =0}^{\infty } \langle F\rangle _{\nu }p_{\nu }(z)= \sum_{\nu =0}^{\infty } [F]_{\nu }\varPhi _{\nu }(z) $$

for all \(z\in D_{\rho _{0}(F)}\).

Assume that (a) holds. We will show that if *F* is an entire function, then *F* is a polynomial. Let

$$ Q_{n,m}^{\mu }(z):=\prod_{k=1}^{m_{n}}(z- \zeta _{n,k})=\sum_{j=0}^{m _{n}} q_{n,j} z^{j}. $$

Note that \(m_{n}\geq 1\) for all *n* sufficiently large and \(q_{n,m _{n}}=1\).

From the definition of \(Q_{n,m}^{\mu }\), we have, for all \(k=1,\ldots ,m\),

$$\begin{aligned} 0&=\bigl\langle Q_{n,m}^{\mu } F\bigr\rangle _{n+k}=\sum _{j=0}^{m_{n}} \sum _{ \nu =0}^{\infty } \langle F\rangle _{\nu }q_{n,j} \bigl\langle z^{j} p_{ \nu } \bigr\rangle _{n+k}= \sum _{j=0}^{m_{n}} \sum _{\nu =n+k-j}^{\infty } \langle F\rangle _{\nu }q_{n,j} \bigl\langle z^{j} p_{\nu } \bigr\rangle _{n+k} \\ &=\sum_{\nu =n+k-m_{n}}^{\infty } \langle F\rangle _{\nu }\bigl\langle z^{m _{n}} p_{\nu } \bigr\rangle _{n+k}+\sum_{j=0}^{m_{n}-1} \sum _{\nu =n+k-j} ^{\infty } \langle F\rangle _{\nu }q_{n,j} \bigl\langle z^{j} p_{\nu } \bigr\rangle _{n+k} \\ &=\frac{\kappa _{n+k-m_{n}}}{\kappa _{n+k}}\langle F\rangle _{n+k-m_{n}}+ \sum _{\nu =n+k-m_{n}+1}^{\infty } \langle F\rangle _{\nu } \bigl\langle z ^{m_{n}} p_{\nu } \bigr\rangle _{n+k}+\sum _{j=0}^{m_{n}-1}\sum_{\nu =n+k-j} ^{\infty } q_{n,j} \langle F\rangle _{\nu } \bigl\langle z^{j}p_{\nu } \bigr\rangle _{n+k}. \end{aligned}$$

(4)

Using the Vieta formulas, since

$$ \inf_{N \geq m} \sup_{n\geq N} \bigl\{ \vert \zeta \vert :\zeta \in \mathcal{P}_{n} ^{\mu }\bigr\} < \infty , $$

there exists \(c_{1}>0\) such that

$$ \sup \bigl\{ \vert q_{n,j} \vert : 0\leq j\leq m_{n}, n\geq n_{0}\bigr\} \leq c_{1}. $$

(5)

From the Cauchy–Schwarz inequality and the orthonormality of \(p_{\nu }\), for all \(n,\nu ,k\in \mathbb{N}_{0}\) and \(j=0,\ldots ,m\), there exists \(c_{2}>0\) such that

$$ \bigl\vert \bigl\langle z^{j}p_{\nu } \bigr\rangle _{n+k} \bigr\vert = \bigl\vert \bigl\langle z^{j}p_{\nu } , p_{n+k} \bigr\rangle _{\mu } \bigr\vert \leq c_{2}. $$

(6)

Because \(\mu \in \mathbf{Reg}_{1}^{*}(E)\), there exists \(c_{3}>0\) such that

$$ \frac{\kappa _{n+k-m_{n}}}{\kappa _{n+k}}\geq c_{3}, \quad n\geq n_{0}, $$

(7)

where \(c_{3}\) does not depend on *k* and \(m_{n}\). Combining (5), (6), and (7), it is easy to check that (4) implies that, for all \(k=1,\ldots ,m\), and for all \(n\geq n_{0}\),

$$ \bigl\vert \langle F\rangle _{n+k-m_{n}} \bigr\vert \leq c_{4} \sum_{\nu =n+k-m_{n}+1}^{ \infty } \bigl\vert \langle F \rangle _{\nu } \bigr\vert , $$

where \(c_{4}\) is a positive constant that does not depend on *n*, *k* and \(m_{n}\). For each \(n \geq n_{0}\), we choose \(k=m_{n}\) in the previous inequality and we obtain, for all \(n \geq n_{0}\),

$$ \bigl\vert \langle F\rangle _{n} \bigr\vert \leq c_{4} \sum_{\nu =n+1}^{\infty } \bigl\vert \langle F \rangle _{\nu } \bigr\vert . $$

Using Lemma 3.3 when \(A_{n}=\langle F\rangle _{n}\), because the above inequality holds and

$$ \lim_{n \rightarrow \infty } \vert A_{n} \vert ^{1/n}=0, $$

\(\langle F\rangle _{n}=0\) for all sufficiently large *n* and *F* is a polynomial as desired.

Now, we assume that (b) holds. We will follow the same plan by proving that if *F* is an entire function, then *F* is a polynomial. Let

$$ Q_{n,m}^{E}(z):=\prod_{k=1}^{m_{n}}(z- \zeta _{n,k})=\sum_{j=0}^{m_{n}} q_{n,j} z^{j}, $$

where \(q_{n,m_{n}}=1\). Arguments analogous to those used to derive (4) show that, for \(k=1,\ldots ,m\),

$$\begin{aligned} 0={}&\bigl(\operatorname{cap}(E)\bigr)^{m_{n}}[F]_{n+k-m_{n}} \\ &{}+ \sum_{\nu =n+k-m_{n}+1}^{\infty } [F]_{\nu }\bigl[ z^{m_{n}} \varPhi _{\nu } \bigr]_{n+k}+ \sum_{j=0}^{m_{n}-1}\sum _{\nu =n+k-j}^{\infty } q_{n,j} [F]_{\nu } \bigl[ z ^{j}\varPhi _{\nu } \bigr]_{n+k}. \end{aligned}$$

(8)

Moreover, there exists \(c_{5}>0\) such that

$$ \sup \bigl\{ \vert q_{n,j} \vert : 0\leq j\leq m_{n}, n\geq n_{0}\bigr\} \leq c_{5}. $$

(9)

Take \(\rho >1\). Using Lemma 3.2, for \(j=0,1,\ldots ,m\), \(k=1,\ldots ,m\), and \(n,\nu \in \mathbb{N}_{0}\), we obtain

$$ \bigl\vert \bigl[ z^{j} \varPhi _{\nu } \bigr]_{n+k} \bigr\vert = \biggl\vert \frac{1}{2\pi i} \int _{\varGamma _{\rho }} \frac{z^{j} \varPhi _{\nu }(z) \varPhi '(z)}{ \varPhi ^{n+k+1}(z)} \,dz \biggr\vert \leq c_{6} \frac{\rho ^{\nu }}{\rho ^{n}}. $$

(10)

Combining (8), (9), and (10), it is easy to check that, for all \(k=1,\ldots ,m\), and for all \(n\geq n_{0}\),

$$ \bigl\vert [F]_{n+k-m_{n}} \bigr\vert \rho ^{n}\leq c_{7} \sum_{\nu =n+k-m_{n}+1}^{\infty } \bigl\vert [F]_{\nu } \bigr\vert \rho ^{\nu }, $$

where \(c_{7}\) is a positive constant that does not depend on *n*, *k* and \(m_{n}\). For each \(n \geq n_{0}\), we choose \(k=m_{n}\) in the previous inequality and we obtain, for all \(n \geq n_{0}\),

$$ \bigl\vert [F]_{n} \bigr\vert \rho ^{n} \leq c_{7} \sum_{\nu =n+1}^{\infty } \bigl\vert [F]_{\nu } \bigr\vert \rho ^{\nu }. $$

(11)

Using Lemma 3.3 by setting \(A_{n}=[F]_{n}\rho ^{n}\), it follows that \([F]_{n}=0\). Consequently, *F* is a polynomial.

Next, suppose that (c) holds. Our plan is to prove that if *F* is an entire function, then *F* is a polynomial. Let

$$ \tilde{Q}_{n,m}^{\mu }(z):=\prod_{k=1}^{m_{n}}(z- \zeta _{n,k})=\sum_{j=0}^{m_{n}} q_{n,j} z^{j}, $$

where \(q_{n,m_{n}}=1\). From the definition of \(\tilde{Q}_{n,m}^{ \mu }\), we have, for all \(k=0,\ldots ,m-1\),

$$\begin{aligned} 0={}&\bigl\langle z^{k}\tilde{Q}_{n,m}^{\mu } F\bigr\rangle _{n+1}=\sum_{j=0}^{m _{n}} \sum _{\nu =0}^{\infty }\langle F\rangle _{\nu } q_{n,j} \bigl\langle z^{k+j} p_{\nu } \bigr\rangle _{n+1}=\sum_{j=0}^{m_{n}} \sum _{\nu =n+1-k-j} ^{\infty }\langle F\rangle _{\nu } q_{n,j} \bigl\langle z^{k+j} p_{\nu } \bigr\rangle _{n+1} \\ ={}&\sum_{\nu =n+1-k-m_{n}}^{\infty }\langle F\rangle _{\nu } \bigl\langle z ^{k+m_{n}} p_{\nu } \bigr\rangle _{n+1} +\sum_{j=0}^{m_{n}-1} \sum _{\nu =n+1-k-j}^{\infty }\langle F\rangle _{\nu } q_{n,j} \bigl\langle z^{k+j} p_{\nu } \bigr\rangle _{n+1} \\ ={}&\frac{\kappa _{n+1-k-m_{n}}}{\kappa _{n+1}}\langle F\rangle _{n+1-k-m _{n}}+\sum _{\nu =n-k-m_{n}+2}^{\infty } \langle F\rangle _{\nu }\bigl\langle z^{k+m_{n}} p_{\nu } \bigr\rangle _{n+1} \\ &{}+\sum_{j=0}^{m_{n}-1} \sum _{\nu =n+1-k-j}^{\infty } \langle F \rangle _{\nu }q_{n,j} \bigl\langle z^{k+j}p_{\nu } \bigr\rangle _{n+1}. \end{aligned}$$

(12)

Applying exactly the same arguments as in (5), (6), and (7), there exists \(c_{8}>0\) such that

$$ \sup \bigl\{ \vert q_{n,j} \vert : 0\leq j\leq m_{n}, n\geq n_{0}\bigr\} \leq c_{8}, $$

(13)

there exists \(c_{9}>0\) such that, for all \(k=0,\ldots ,m-1\), \(j=0,\ldots ,m\), and \(n,\nu \in \mathbb{N}_{0}\),

$$ \bigl\vert \bigl\langle z^{k+j}p_{\nu } \bigr\rangle _{n+1} \bigr\vert = \bigl\vert \bigl\langle z^{j}p_{\nu } , p _{n+k}\bigr\rangle _{\mu } \bigr\vert \leq c_{9}, $$

(14)

there exists \(c_{10}>0\) such that, for all \(k=0,\ldots ,m-1\), \(m_{n}=1,\ldots ,m\), and \(n\geq n_{0}\),

$$ \frac{\kappa _{n+1-k-m_{n}}}{\kappa _{n+1}}\geq c_{10}. $$

(15)

Using (12), (13), (14), and (15), we have, for all \(k=0,\ldots ,m-1\) and \(n\geq n_{0}\),

$$ \bigl\vert \langle F\rangle _{n-k-m_{n}+1} \bigr\vert \leq c_{11} \sum_{\nu =n-k-m_{n}+2} ^{\infty } \bigl\vert \langle F \rangle _{\nu } \bigr\vert . $$

For each \(n \geq n_{0}\), we choose \(k=m-m_{n}\) in the previous inequality and we obtain, for all \(n \geq n_{0}\),

$$ \bigl\vert \langle F\rangle _{n-m+1} \bigr\vert \leq c_{11} \sum_{\nu =n-m+2}^{\infty } \bigl\vert \langle F\rangle _{\nu } \bigr\vert . $$

Setting \(N=n-m+1\), we have

$$ \bigl\vert \langle F\rangle _{N} \bigr\vert \leq c_{11} \sum_{\nu =N+1}^{\infty } \bigl\vert \langle F \rangle _{\nu } \bigr\vert , \quad N\geq N_{0}. $$

Applying \(A_{N}=\langle F\rangle _{N}\) in Lemma 3.3, we can conclude that *F* is a polynomial.

Finally, we suppose that (d) holds. Following the same plan by assuming that *F* is an entire function, we want to show that *F* is a polynomial. Let

$$ \tilde{Q}_{n,m}^{E}(z):=\prod_{k=1}^{m_{n}}(z- \zeta _{n,k})=\sum_{j=0} ^{m_{n}} q_{n,j} z^{j}, $$

where \(q_{n,m_{n}}=1\). Arguments analogous to those used to derive (12) show that, for all \(k=0,\ldots ,m-1\),

$$\begin{aligned} 0={}&\bigl(\operatorname{cap}(E)\bigr)^{m_{m}+k} [F]_{n+1-k-m_{n}}+\sum _{\nu =n-k-m_{n}+2} ^{\infty } [F]_{\nu }\bigl[ z^{k+m_{n}} \varPhi _{\nu } \bigr]_{n+1} \\ &{}+\sum_{j=0}^{m_{n}-1} \sum _{\nu =n+1-k-j}^{\infty } [ F]_{\nu }q_{n,j} \bigl[ z^{k+j}\varPhi _{\nu }\bigr]_{n+1}. \end{aligned}$$

(16)

Moreover, there exists \(c_{12}>0\) such that

$$ \sup \bigl\{ \vert q_{n,j} \vert : 0\leq j\leq m_{n}, n\geq n_{0}\bigr\} \leq c_{12}. $$

(17)

Take \(\rho >1\). Using Lemma 3.2, we obtain, for all \(j=0,1,\ldots ,m\), \(k=0,\ldots ,m-1\), and \(n,\nu \in \mathbb{N}_{0}\),

$$ \bigl\vert \bigl[ z^{k+j} \varPhi _{\nu } \bigr]_{n+1} \bigr\vert = \biggl\vert \frac{1}{2\pi i} \int _{\varGamma _{\rho }} \frac{z^{k+j} \varPhi _{\nu }(z) \varPhi '(z)}{ \varPhi ^{n+2}(z)} \,dz \biggr\vert \leq c_{13} \frac{\rho ^{\nu }}{\rho ^{n}}. $$

(18)

By (16), (17), and (18), we have, for all \(k=0,\ldots ,m-1\) and \(n\geq n_{0}\),

$$ \bigl\vert [F]_{n+1-k-m_{n}} \bigr\vert \rho ^{n-m+1} \leq c_{14} \sum_{\nu =n-k-m_{n}+2} ^{\infty } \bigl\vert [F]_{\nu }\bigr\vert \rho ^{\nu }. $$

For each \(n \geq n_{0}\), we choose \(k=m-m_{n}\) in the previous inequality and we obtain, for all \(n \geq n_{0}\),

$$ \bigl\vert [F]_{n-m+1} \bigr\vert \rho ^{n-m+1}\leq c_{14} \sum_{\nu =n-m+2}^{\infty } \bigl\vert [ F]_{ \nu } \bigr\vert \rho ^{\nu }. $$

Setting \(N=n-m+1\) and \(A_{N}=[F]_{N} \rho ^{N}\) in Lemma 3.3, we arrive at the same conclusion. □