Proof Note that $\varphi (\mathbb {B}_N) \subset \mathbb {B}_N$ , $\varphi $ is continuous, and $K_k$ is a compact subset of $\mathbb {B}_N$ . Therefore, $\varphi (K_k)$ is also a compact subset of $\mathbb {B}_N$ , and there exists $j \in \mathbb {N}$ such that $\varphi (K_k) \subset K_j$ . Hence, for all $f \in \mathrm {Hol}(\mathbb {B}_N)$ ,

$$\begin{align*}\left\Vert C_\varphi(f)\right\Vert{}_{\infty, k} = \left\Vert f \circ \varphi\right\Vert{}_{\infty, k} = \sup_{z \in K_k} \left\lvert (f \circ \varphi)(z) \right\rvert \le \sup_{w \in K_j} \left\lvert f(w) \right\rvert = \left\Vert f\right\Vert{}_{\infty, j}.\\[-43pt] \end{align*}$$

Proof Let $x \in \mathbb {B}_N$ . It is immediate that $k_x$ is linear and multiplicative. Moreover, since $\{x\}$ is a compact subset of $\mathbb {B}_N$ , $k_x$ is continuous on $\mathrm {Hol}(\mathbb {B}_N)$ , so $k_x$ is a character.

Conversely, let k be a character on $\mathrm {Hol}(\mathbb {B}_N)$ . We set $x_0 = (k(e_1), \ldots , k(e_N))$ , where $e_p(z) = z_p$ . For $\vec {\jmath } = (j_1, \ldots , j_N) \in \mathbb {N}_0^N$ , denote $\left \lvert \vec {\jmath } \right \rvert = j_1 + \cdots + j_N$ . For all ${f \in \mathrm {Hol}(\mathbb {B}_N)}$ , if we consider the truncated Maclaurin series $(p_n)_{n \in \mathbb {N}_0}$ defined by

$$\begin{align*}p_n = \sum_{\vec{\jmath} \in \mathbb{N}_0^N, \; \left\lvert \vec{\jmath} \right\rvert \le n} a_{\vec{\jmath}} \prod_{\ell = 1}^N e_\ell^{j_\ell}, \end{align*}$$

then $(p_n)$ converges uniformly to f on all compact subsets of $\mathbb {B}_N$ . Finally, using the properties of k, it follows that

$$\begin{align*}k(f) = \lim_{n \to \infty} k(p_n) = \lim_{n \to \infty} \sum_{\left\lvert \vec{\jmath} \right\rvert \le n} a_{\vec{\jmath}} \prod_{\ell = 1}^N k \left( e_\ell \right)^{j_\ell} = \lim_{n \to \infty} \sum_{\left\lvert \vec{\jmath} \right\rvert \le n} a_{\vec{\jmath}} \prod_{\ell = 1}^N x_\ell^{j_\ell} = f(x_0).\\[-45pt] \end{align*}$$

Proof If $\varphi $ is an automorphism of $\mathbb {B}_N$ , then we can consider its inverse $\varphi ^{-1}$ , and it is clear that the operator $C_\varphi $ is invertible, with inverse $C_{\varphi ^{-1}}$ .

Conversely, suppose that $C_\varphi $ is invertible. Then there exists an operator T on $\mathrm {Hol}(\mathbb {B}_N)$ such that $C_\varphi \circ T = T \circ C_\varphi = \mathrm {Id}$ . Let $z \in \mathbb {B}_N$ . If $\psi = (Te_1, \ldots , Te_N)$ , we have

$$\begin{align*}\begin{aligned} \psi(\varphi(z)) & = (Te_1(\varphi(z)), \ldots, Te_N(\varphi(z))) \\ & = ([(C_\varphi \circ T)(e_1)](z), \ldots, [(C_\varphi \circ T)(e_N)](z)) = (z_1, \ldots, z_N) = z. \end{aligned} \end{align*}$$

We deduce that $\psi \circ \varphi = \mathrm {Id}$ , so $\varphi $ is univalent.

To show that $\varphi $ is onto, let $x \in \mathbb {B}_N$ , and k be a character on $\mathrm {Hol}(\mathbb {B}_N)$ . Since $C_\varphi $ is linear, continuous (by Proposition 2.1), and multiplicative, $k \circ C_\varphi $ is also a character on $\mathrm {Hol}(\mathbb {B}_N)$ . Moreover, we can choose k so that $k \circ C_\varphi = k_x$ , and by Lemma 2.2, there exists $y \in \mathbb {B}_N$ such that $k = k_y$ , so

$$\begin{align*}k_x = k \circ C_\varphi = k_y \circ C_\varphi = k_{\varphi(y)}. \end{align*}$$

Finally, $x = \varphi (y)$ , so $\varphi $ is onto, and bijective.

Proof Suppose that $\left \Vert \varphi \right \Vert {}_{\infty } = r_0 < 1$ . Then, the family $C_\varphi (V_{r_0, 1})$ is uniformly bounded by 1 on $\mathbb {B}_N$ . By Montel’s theorem [Reference Conway16], $C_\varphi (V_{r_0, 1})$ is relatively compact.

Conversely, if $\left \Vert \varphi \right \Vert {}_{\infty } = 1$ , let $\varepsilon , r \in (0, 1)$ . We show that $C_\varphi (V_{r, \varepsilon })$ is not relatively compact. There exists $z_0 \in \mathbb {B}_N$ such that $\left \lvert \varphi (z_0) \right \rvert> r$ , so let $w_0 = \varphi (z_0)$ and $\rho> 0$ small enough so that if $K = \overline {B(0, r)} \cup \{w_0\}$ and $\Omega = B(0, r+\rho ) \cup B(w_0, \rho )$ , then

• • K is a compact subset of $\mathbb {B}_N$ ,

• • $\Omega $ is an open set of $\mathbb {B}_N$ containing K,

• • $B(0, r+\rho ) \cap B(w_0, \rho ) = \varnothing $ .

Also, K is polynomially convex by the separation lemma of [Reference Kallin22], since $\overline {B(0, r)}$ and $\{w_0\}$ are convex, compact and disjoint sets.

Finally, for all $n \in \mathbb {N}_0$ , set . Then $h_n \in \mathrm {Hol}(\Omega )$ , and by Oka–Weil’s theorem, there exists a sequence $(p_n)$ of polynomials such that for all $n \in \mathbb {N}_0$ ,

$$\begin{align*}\left\Vert h_n - p_n\right\Vert{}_{\infty, K} < \varepsilon. \end{align*}$$

Therefore, we have $p_n \in \mathrm {Hol}(\mathbb {B}_N) \cap V_{r, \varepsilon }$ , so that $p_n \circ \varphi \in C_\varphi (V_{r, \varepsilon })$ , and also $\left \lvert (p_n \circ \varphi )(z_0) \right \rvert \ge n - \varepsilon \to \infty $ . Hence, no subsequence of $(p_n \circ \varphi )$ converges on $\mathrm {Hol}(\mathbb {B}_N)$ , so $C_\varphi (V_{r, \varepsilon })$ is not relatively compact.

Notation For $\vec {\alpha } = (\alpha _1, \cdots , \alpha _N) \in \mathbb {N}_0^N$ and $z \in \mathbb {C}^N$ , we write

$$\begin{align*}\left\lvert \vec{\alpha} \right\rvert = \sum_{k = 1}^N \alpha_k, \qquad \vec{\alpha} ! = \prod_{k = 1}^N \alpha_k !, \qquad z^{\vec{\alpha}} = \prod_{k = 1}^N z_k^{\alpha_k}, \qquad f^{(\vec{\alpha})} = \frac{\partial^{\left\lvert \vec \alpha \right\rvert} f}{\partial z^{\vec \alpha}}. \end{align*}$$

We also define a strict order $\prec $ on $\mathbb {N}_0^N$ . For $\vec {\imath }, \vec {\jmath } \in \mathbb {N}_0^N$ , $\vec {\imath } \prec \vec {\jmath }$ if

$$\begin{align*}\left\lvert \vec{\imath} \right\rvert < \left\lvert \vec{\jmath} \right\rvert ~ \text{ or } ~ \begin{cases} \left\lvert \vec{\imath} \right\rvert = \left\lvert \vec{\jmath} \right\rvert \\ \exists k \in \{1, \ldots, N\}, ~ i_1 = j_1, ~ \ldots, i_{k-1} = j_{k-1}, ~ i_k < j_k. \end{cases} \end{align*}$$

We write $\vec {\imath } \preccurlyeq \vec {\jmath }$ if $\vec {\imath } \prec \vec {\jmath }$ or $\vec {\imath } = \vec {\jmath }$ . The order $\preccurlyeq $ is a well-order. Hence, we can define the predecessor and the successor of an element $\vec {\jmath } \in \mathbb {N}_0^N$ , denoted respectively by $\vec {\jmath }_-$ and $\vec {\jmath }_+$ .

Proof Faà di Bruno’s formula [Reference Ma26] gives the equation

$$ \begin{align*} (f \circ \varphi)^{(\vec{\jmath})}(0) = \vec{\jmath} ! \sum_{(s, p, M, r) \in \mathcal{D}_{\vec{\jmath}}} f^{(\vec{r})} (0) \prod_{k = 1}^s \prod_{\ell = 1}^N \frac{1}{m_{k\ell} !} \left( \frac{1}{\vec{p}_k !} \varphi_{\ell}^{(\vec{p}_k)} (0) \right)^{m_{k\ell}}, \end{align*} $$

where $(s, p, M, r) \in \mathcal {D}_{\vec {\jmath }}$ if and only if

$$\begin{align*}\vec{\jmath} = \sum_{k = 1}^s \left\lvert \vec{m}_k \right\rvert \vec{p}_k, \qquad r_\ell = \sum_{k = 1}^s m_{k\ell}, \qquad \vec{r} = (r_1, \cdots r_N). \end{align*}$$

However, if $\left \lvert \vec r \right \rvert < n$ , then $f^{(\vec {r})} (0) = 0$ . As a result, we focus on the terms with $\left \lvert \vec r \right \rvert = n$ .

If there exists $1 \le k_0 \le s$ such that $\left \lvert \vec {p}_{k_0} \right \rvert \ge 2$ , then

$$\begin{align*}n = \left\lvert \vec{\jmath} \right\rvert = \sum_{k = 1}^s \left\lvert \vec{m}_k \right\rvert \left\lvert \vec{p}_k \right\rvert \ge \sum_{k = 1}^s \left\lvert \vec{m}_k \right\rvert + \left\lvert \vec{m}_{k_0} \right\rvert = \left\lvert \vec r \right\rvert + \left\lvert \vec{m}_{k_0} \right\rvert> n. \end{align*}$$

Since it is impossible, we must have $\left \lvert \vec {p}_1 \right \rvert = \cdots = \left \lvert \vec {p}_s \right \rvert = 1$ . Also, even if we write some $\vec {m}_k = \vec {0}$ , we may assume that $s = N$ , , and $\left \lvert \vec {m}_k \right \rvert = j_k$ . Hence,

$$\begin{align*}(f \circ \varphi)^{(\vec{\jmath})}(0) = \vec{\jmath} ! \sum_{\left\lvert \vec{m}_1 \right\rvert = j_1} \cdots \sum_{\left\lvert \vec{m}_N \right\rvert = j_N} f^{(\vec{r})} (0) \prod_{k, \ell = 1}^N \frac{1}{m_{k\ell} !} \underbrace{\left( \frac{\partial \varphi_{\ell}}{\partial z_k} (0) \right)^{m_{k\ell}}}_{= 0 \text{ if } k> \ell}. \end{align*}$$

If there exists one $k> \ell $ such that $m_{k\ell }> 0$ , then the product is zero. Thus, there remain only the terms with $m_{k\ell } = 0$ for $k> \ell $ . Moreover, note that $r_1 = m_{11} < j_1$ , unless $m_{11} = j_1$ . In this case, $r_2 = m_{22} < j_2$ , unless $m_{22} = j_2$ . Iterating until the rank N, we get $\vec r \prec \vec {\jmath }$ , so $f^{(\vec r)}(0) = 0$ , unless , in which case $\vec r = \vec {\jmath }$ . Finally, there remains only one term, which is

$$\begin{align*}(f \circ \varphi)^{(\vec{\jmath})}(0) = \vec{\jmath} ! f^{(\vec{\jmath})}(0) \prod_{k = 1}^N \frac{1}{j_k !} \lambda_k^{j_k} = \lambda^{\vec{\jmath}} f^{(\vec{\jmath})}(0).\\[-45pt] \end{align*}$$

Proof For all $k \in \{1, \ldots , N\}$ , using Theorem 4.2 (in particular, the first coordinate of $\kappa $ ) and changes of basis, we can find maps $\eta _1, \ldots , \eta _N$ such that $\eta _k \circ \varphi = \lambda _k \eta _k$ . Thus, $\lambda _k \in \sigma _p(C_\varphi )$ . Lemma 4.1 concludes the inclusion $\supset $ .

Conversely, let $\mu \not \in \{ \lambda ^{\vec {\jmath }} : \vec {\jmath } \in \mathbb {N}_0^N\}$ . If $\mu \in \sigma _p(C_\varphi )$ , then there exists $f \in \mathrm {Hol}(\mathbb {B}_N)$ such that $f \not \equiv 0$ and $f \circ \varphi = \mu f$ . We show by induction that $f^{(\vec {\jmath })}(0) = 0$ for all ${\vec {\jmath } \in \mathbb {N}_0^N}$ .

• • First, since $f(\varphi (0)) = f(0) = \mu f(0)$ and $\mu \neq 1$ , we get $f(0) = 0$ .

• • Assume that for all $\vec {\imath } \prec \vec {\jmath }$ , we have $f^{(\vec {\imath })}(0) = 0$ . By Lemma 4.3, we get the equation (4.1):

  $$\begin{align*}(f \circ \varphi)^{(\vec{\jmath})}(0) = \lambda^{\vec{\jmath}} f^{(\vec{\jmath})}(0). \end{align*}$$
  But $f \circ \varphi = \mu f$ , so $\lambda ^{\vec {\jmath }} f^{(\vec {\jmath })}(0) = (f \circ \varphi )^{(\vec {\jmath })}(0) = \mu f^{(\vec {\jmath })} (0)$ . Since $\mu \neq \lambda ^{\vec {\jmath }}$ , we have $ f^{(\vec {\jmath })} (0) = 0$ .

We conclude that $f \equiv 0$ , which is impossible. Finally, $\mu \not \in \sigma _p(C_\varphi )$ .

Proof Note that for all $k \in \{1, \ldots , N\}$ , the Maclaurin series of $\kappa _k$ is

$$\begin{align*}\kappa_k(z) = \sum_{\ell = 1}^k \alpha_{k\ell} z_\ell + o(\left\lvert z \right\rvert), \quad \text{with} \quad \alpha_{k\ell} \in \mathbb{C}. \end{align*}$$

Hence, by the multinomial theorem,

$$ \begin{align*} \kappa^{\vec{\jmath}} (z) & = \prod_{k = 1}^N \left(\sum_{\ell = 1}^k \alpha_{k\ell} z_\ell \right)^{j_k} + o(\left\lvert z \right\rvert^{\left\lvert \vec{\jmath} \right\rvert}) \\ & = \prod_{k = 1}^N \left( \sum_{\left\lvert \vec{p}_k \right\rvert = j_k} \binom{j_k}{\vec{p}_k} (\alpha_{k} z)^{\vec{p}_k} \right) + o(\left\lvert z \right\rvert^{\left\lvert \vec{\jmath} \right\rvert}) = \sum_{\left\lvert \vec{p}_1 \right\rvert = j_1} \cdots \sum_{\left\lvert \vec{p}_N \right\rvert = j_N} \beta_p z^{\vec r} + o(\left\lvert z \right\rvert^{\left\lvert \vec{\jmath} \right\rvert}), \end{align*} $$

with $\beta _p \in \mathbb {C}$ , $p_{k\ell } = 0$ if $k < \ell $ , and

$$\begin{align*}\vec r = (p_{11} + \cdots + p_{N1}, p_{22} + \cdots + p_{N2}, \ldots, p_{NN}). \end{align*}$$

However, for $\vec {p_1}, \ldots , \vec {p_N}$ satisfying these assumptions, $\left \lvert \vec r \right \rvert = \left \lvert \vec {\jmath } \right \rvert $ . Also, note that $p_{11} = j_1$ , so $\vec r \succcurlyeq \vec {\jmath }$ . This gives the result for $\vec {\imath } \prec \vec {\jmath }$ . If $\vec {\imath } = \vec {\jmath }$ , the only way to get $\vec r = \vec {\jmath }$ is by taking . Hence, we conclude since $\beta _p = 1$ , because $\alpha _{kk} = 1$ for all k.

Proof We prove most of the properties by induction. For $\vec {\jmath } \in \mathbb {N}_0^N$ , since $\preccurlyeq $ is a well-order, we can define the predecessor of the vector $\vec {\jmath }$ , that will be denoted as $\vec {\jmath }_-$ .

1. (1) First, we show that $W_{\vec {\jmath }} \cap \mathrm {Hol}_{\vec {\jmath }}(\mathbb {B}_N) = \{0\}$ . Indeed, by considering

   $$\begin{align*}f = \sum_{\vec{\imath} \preccurlyeq \vec{\jmath}} c_{\vec{\imath}} \kappa^{\vec{\imath}} \in \mathrm{Hol}_{\vec{\jmath}}(\mathbb{B}_N), \end{align*}$$
   we obtain $c_{\vec 0} = f(0) = 0$ . Let $\vec k \preccurlyeq \vec {\jmath }$ such that for $\vec {\imath } \prec \vec k$ , $c_{\vec {\imath }} = 0$ . By Lemma 4.5,
   $$\begin{align*}f^{(\vec k)} (0) = \sum_{\vec k \preccurlyeq \vec{\imath} \preccurlyeq \vec{\jmath}} c_{\vec{\imath}} \times (\kappa^{\vec{\imath}})^{(\vec k)} (0) = \vec k ! c_{\vec k}. \end{align*}$$

   Since $f \in \mathrm {Hol}_{\vec {\jmath }}(\mathbb {B}_N)$ , we get $c_{\vec k} = 0$ . Finally, $f \equiv 0$ .

   Now, we show that $W_{\vec {\jmath }} + \mathrm {Hol}_{\vec {\jmath }}(\mathbb {B}_N) = \mathrm {Hol}(\mathbb {B}_N)$ . The inclusion $\subset $ is trivial, so we focus on the other one. Note that $\mathrm {Im}(Q_{\vec {\jmath }}) \subset W_{\vec {\jmath }}$ . We use an induction argument.

   • • Let $f \in \mathrm {Hol}(\mathbb {B}_N)$ . Then, $(f - Q_{\vec 0}(f))(0) = (f - P_{\vec 0}(f))(0) = f(0) - f(0) = 0$ .

     Hence, $f - Q_{\vec 0}(f) \in \mathrm {Hol}_{\vec 0}(\mathbb {B}_N)$ , so $f \in W_{\vec 0} + \mathrm {Hol}_{\vec 0}(\mathbb {B}_N)$ .

   • • Suppose now that $f - Q_{\vec {\jmath }_-}(f) \in \mathrm {Hol}_{\vec {\jmath }_-}(\mathbb {B}_N)$ . Then,

     $$\begin{align*}Q_{\vec{\jmath}}(f) = Q_{\vec{\jmath}_-}(f) + P_{\vec{\jmath}}(f) = Q_{\vec{\jmath}_-}(f) + \frac{1}{\vec{\jmath} !} \left[ f - Q_{\vec{\jmath}_-}(f) \right]^{(\vec{\jmath})} (0) \times \kappa^{\vec{\jmath}}. \end{align*}$$

     Let $\vec {\imath } \preccurlyeq \vec {\jmath }$ . By definition of $Q_{\vec {\jmath }}$ ,

     $$\begin{align*}(f - Q_{\vec{\jmath}}(f))^{(\vec{\imath})}(0) = (f - Q_{\vec{\jmath}_-}(f))^{(\vec{\imath})}(0) - \frac{1}{\vec{\jmath} !} \left[ f - Q_{\vec{\jmath}_-}(f) \right]^{(\vec{\jmath})} (0) \times (\kappa^{\vec{\jmath}})^{(\vec{\imath})}(0). \end{align*}$$

     If $\vec {\imath } \prec \vec {\jmath }$ , we have $(f - Q_{\vec {\jmath }_-}(f))^{(\vec {\imath })}(0) = (\kappa ^{\vec {\jmath }})^{(\vec {\imath })} (0) = 0$ , using Lemma 4.5 and the inductive assumption. Hence, $(f - Q_{\vec {\jmath }}(f))^{(\vec {\imath })}(0) = 0$ . Otherwise, if $\vec {\imath } = \vec {\jmath }$ ,

     $$\begin{align*}(f - Q_{\vec{\jmath}}(f))^{(\vec{\jmath})}(0) = (f - Q_{\vec{\jmath}_-}(f))^{(\vec{\jmath})}(0) - \frac{1}{\vec{\jmath} !} [f - Q_{\vec{\jmath}_-}(f)]^{(\vec{\jmath})} (0) \times \vec{\jmath} ! = 0. \end{align*}$$

     Finally, $f - Q_{\vec {\jmath }}(f) \in \mathrm {Hol}_{\vec {\jmath }}(\mathbb {B}_N)$ , so $f \in W_{\vec {\jmath }} + \mathrm {Hol}_{\vec {\jmath }}(\mathbb {B}_N)$ .

2. (2) By $(1)$ , if $f \in \mathrm {Hol}(\mathbb {B}_N)$ , then f has the form $f = g+h$ , with $g \in W_{\vec {\jmath }}$ and $h \in \mathrm {Hol}_{\vec {\jmath }}(\mathbb {B}_N)$ uniquely determined. Moreover, $Q_{\vec {\jmath }}(f) = g$ , so $Q_{\vec {\jmath }}$ is indeed the projection on $W_{\vec {\jmath }}$ in parallel to $\mathrm {Hol}_{\vec {\jmath }}(\mathbb {B}_N)$ .

3. (3) First, we show that for all $\vec {\jmath } \in \mathbb {N}_0^N$ , $C_\varphi \circ P_{\vec {\jmath }} = \lambda ^{\vec {\jmath }} P_{\vec {\jmath }}$ . Indeed,

   $$ \begin{align*} (C_\varphi \circ P_{\vec{\jmath}})(f) & = \frac{\kappa^{\vec{\jmath}} \circ \varphi}{\vec{\jmath} !} \left[ f - Q_{\vec{\jmath}_-}(f) \right]^{(\vec{\jmath})} (0) \\ & = \frac{\lambda^{\vec{\jmath}} \kappa^{\vec{\jmath}}}{\vec{\jmath} !} \left[ f - Q_{\vec{\jmath}_-}(f) \right]^{(\vec{\jmath})} (0) = \lambda^{\vec{\jmath}} P_{\vec{\jmath}}(f). \end{align*} $$

   Now, we prove that for all $\vec {\jmath } \in \mathbb {N}_0^N$ , $Q_{\vec {\jmath }} \circ C_\varphi = C_\varphi \circ Q_{\vec {\jmath }}$ . To do so, it suffices to show that $W_{\vec {\jmath }}$ and $\mathrm {Hol}_{\vec {\jmath }}(\mathbb {B}_N)$ are invariant by $C_\varphi $ .

   • • Let $\vec {\imath } \preccurlyeq \vec {\jmath }$ . Since $\kappa ^{\vec {\imath }} \circ \varphi = \lambda ^{\vec {\imath }} \kappa ^{\vec {\imath }}$ , then $\kappa ^{\vec {\imath }} \circ \varphi \in W_{\vec {\jmath }}$ , so $C_\varphi (W_{\vec {\jmath }}) \subset W_{\vec {\jmath }}$ .

   • • Let $f \in \mathrm {Hol}_{\vec {\jmath }}(\mathbb {B}_N)$ . Then, $(f \circ \varphi )(0) = f(0) = 0$ . In addition, for $0 \prec \vec {\imath } \preccurlyeq \vec {\jmath }$ , by Lemma 4.3, $(f \circ \varphi )^{(\vec {\imath })}(0) = \lambda ^{\vec {\imath }} f^{(\vec {\imath })}(0) = 0$ . Hence, $f \circ \varphi \in \mathrm {Hol}_{\vec {\jmath }}(\mathbb {B}_N)$ .

   Finally, for all $f \in \mathrm {Hol}(\mathbb {B}_N)$ , there exist $g, h \in \mathrm {Hol}_{\vec {\jmath }}(\mathbb {B}_N)$ such that $f = Q_{\vec {\jmath }}(f) + g$ and $f \circ \varphi = Q_{\vec {\jmath }}(f \circ \varphi ) + h$ . Hence, $(Q_{\vec {\jmath }}(f) \circ \varphi ) + (g \circ \varphi ) = Q_{\vec {\jmath }}(f \circ \varphi ) + h$ . But f can only be written in a unique way on $W_{\vec {\jmath }} \oplus \mathrm {Hol}_{\vec {\jmath }}(\mathbb {B}_N)$ , so

   $$\begin{align*}Q_{\vec{\jmath}}(f) \circ \varphi = Q_{\vec{\jmath}}(f \circ \varphi). \end{align*}$$

   Finally, we show that $P_{\vec {\jmath }} \circ C_\varphi = C_\varphi \circ P_{\vec {\jmath }}$ . Since $P_{\vec 0} = Q_{\vec 0}$ , we have $P_{\vec 0} \circ C_\varphi = C_\varphi \circ P_{\vec 0}$ . Let $\vec {\jmath } \in \mathbb {N}_0^N \backslash \{\vec 0\}$ . Using the equations above,

   $$\begin{align*}P_{\vec{\jmath}} \circ C_\varphi = (Q_{\vec{\jmath}} - Q_{\vec{\jmath}_-}) \circ C_\varphi = C_\varphi \circ (Q_{\vec{\jmath}} - Q_{\vec{\jmath}_-}) = C_\varphi \circ P_{\vec{\jmath}}.\\[-37pt] \end{align*}$$

Proof The inclusion $\supset $ is trivial, since $\varphi $ is not one-to-one. We focus on the other inclusion. Let $\mu \not \in \{0\} \cup \sigma _p(C_\varphi )$ . We want to show that $C_\varphi - \mu \mathrm {Id}$ is invertible. By Lemma 4.8, there exist $0 < \varepsilon < 1$ and $p \in \mathbb {N}_0$ such that for all $g \in \mathrm {Hol}_p(\mathbb {B}_N)$ ,

$$\begin{align*}\sum_{n \ge 0} \frac{g \circ \varphi^{[n]}}{\mu^n} \text{ converges uniformly on } \overline{B(0, \varepsilon)}. \end{align*}$$

Moreover, by Theorem 4.6, $\mathrm {Hol}(\mathbb {B}_N) = W_p \oplus \mathrm {Hol}_p(\mathbb {B}_N)$ , with $W_p$ and $\mathrm {Hol}_p(\mathbb {B}_N)$ both invariant by $C_\varphi $ . Hence, we can write $C_\varphi = \mathrm {diag}(S, T)$ , where $S \in \mathcal {L}(W_p)$ and $T \in \mathcal {L}(\mathrm {Hol}_p(\mathbb {B}_N))$ .

• • First, with respect to the basis $(\kappa ^{\vec {\jmath }} : \left \lvert \vec {\jmath } \right \rvert \le p)$ of $W_{\vec {\jmath }}$ , $S = \mathrm {diag}(\lambda ^{\vec {\jmath }} : \left \lvert \vec {\jmath } \right \rvert \le p)$ . However, $\mu \not \in \sigma _p(C_\varphi ) = \{\lambda ^{\vec {\jmath }} : \left \lvert \vec {\jmath } \right \rvert \le p\}$ , so $S - \mu \mathrm {Id}$ is invertible.

• • Consider the function

  $$\begin{align*}h = -\frac{1}{\mu} \sum_{n \ge 0} \frac{g \circ \varphi^{[n]}}{\mu^n} \in \mathrm{Hol}(B(0, \varepsilon)). \end{align*}$$
  Then, on $B(0, \varepsilon )$ ,
  $$\begin{align*}(h \circ \varphi) - \mu h = \sum_{n \ge 0} \frac{g \circ \varphi^{[n]}}{\mu^n} - \sum_{n \ge 0} \frac{g \circ \varphi^{[n+1]}}{\mu^{n+1}} = g \circ \varphi^{[0]} = g. \end{align*}$$
  Hence, by Lemma 4.7, there exists a function $\tilde h \in \mathrm {Hol}(\mathbb {B}_N)$ such that $(\tilde h \circ \varphi ) - \mu \tilde h = g$ . This shows that $T - \mu \mathrm {Id}$ is onto. In addition, $T - \mu \mathrm {Id}$ is univalent since $\mu \not \in \sigma _p(C_\varphi )$ and $\mathbb {C} \backslash \sigma _p(C_\varphi ) \subset \mathbb {C} \backslash \sigma _p(T)$ . We conclude that $T - \mu \mathrm {Id}$ is invertible on $\mathcal {L}(\mathrm {Hol}_p(\mathbb {B}_N))$ .

Finally, $C_\varphi - \mu \mathrm {Id}$ is invertible, so $\mu \not \in \sigma (C_\varphi )$ .

Proof Denote by the constant map equal to 1. Then, , so $1 \in \sigma _p(C_\varphi )$ . Moreover, we may have $0 \in \sigma _p(C_\varphi )$ , for instance if the range of $\varphi $ is contained in $\mathbb {C}^M$ for some $M < N$ , by considering a nonzero function f such that $f = 0$ on $\text {ran}(\varphi )$ .

Conversely, let $\mu \in \mathbb {C} \backslash \{0, 1\}$ . The goal is to prove that the operator $C_\varphi - \mu \mathrm {Id}$ is invertible. Note that since the diagonal of $\varphi '(0)$ only has zeroes, there exists $n_0 \in \mathbb {N}$ such that $(\varphi ^{[n_0]})'(0) = (\varphi '(0))^{n_0} = 0$ . We consider two cases.

• • For $c \in \mathbb {C}^*$ , we set $f_c(z) = c/(1 - \mu )$ . Then, for all $z \in \mathbb {B}_N$ ,

  $$\begin{align*}f_c(\varphi(z)) - \mu f_c(z) = \frac{c}{1 - \mu} - \mu \frac{c}{1-\mu} = c. \end{align*}$$

• • Let $g \in \mathrm {Hol}(\mathbb {B}_N)$ such that $g(0) = 0$ . We know that there exists $0 < \varepsilon < 1$ and $d> 1$ such that for all $\left \lvert z \right \rvert \le \varepsilon $ , we have $\left \lvert g(z) \right \rvert \le d\left \lvert z \right \rvert $ , $\left \lvert \varphi (z) \right \rvert \le d\left \lvert z \right \rvert $ and $\left \lvert \varphi ^{[n_0]}(z) \right \rvert \le d\left \lvert z \right \rvert ^2$ . Using Schwarz’s lemma [Reference Abate1, Proposition 2.2.12], for all $k \in \mathbb {N}$ , $k \ge 3$ and $\left \lvert z \right \rvert \le \varepsilon $ ,

  $$\begin{align*}\left\lvert \varphi^{[kn_0]}(z) \right\rvert \le d^k \left\lvert z \right\rvert^{2^k} \le d^k \left\lvert z \right\rvert^{2k+2}. \end{align*}$$
  For $n \in \mathbb {N}$ , if $n \ge 3 n_0$ , we can write $n = kn_0 + p$ , with $k \ge 3$ and $0 \le p < n_0$ . Therefore, $k = (n-p)/n_0 \ge n/n_0 - 1$ , and
  $$ \begin{align*} \qquad \left\lvert \frac{g(\varphi^{[n]}(z))}{\mu^n} \right\rvert & \le \frac{d \left\lvert \varphi^{[n]}(z) \right\rvert}{\left\lvert \mu \right\rvert^n} \le \frac{d^{p+1} \left\lvert \varphi^{[kn_0]}(z) \right\rvert}{\left\lvert \mu \right\rvert^n} \\ & \le \frac{d^{k+p+1} \left\lvert z \right\rvert^{2k+2}}{\left\lvert \mu \right\rvert^n} \le d \frac{d^n \left\lvert z \right\rvert^{2n/n_0}}{\left\lvert \mu \right\rvert^n} = d \left( \frac{d \left\lvert z \right\rvert^{2/n_0}}{\left\lvert \mu \right\rvert} \right)^n. \end{align*} $$
  Now, we choose $\tilde \varepsilon < \varepsilon $ small enough to have $d \tilde \varepsilon ^{2/n_0} < \left \lvert \mu \right \rvert $ , so that
  $$\begin{align*}h = - \frac 1 \mu \sum_{n \ge 0} \frac{g(\varphi^{[n]}(z))}{\mu^n} \in \mathrm{Hol}(B(0, \tilde \varepsilon)). \end{align*}$$
  Moreover, as in the proof of Proposition 4.9, we show that $h \circ \varphi - \mu h = g$ . By Lemma 4.7, there exists $\tilde h \in \mathrm {Hol}(\mathbb {B}_N)$ such that $\tilde h \circ \varphi - \mu \tilde h = g$ .

Thus, if $\eta \in \mathrm {Hol}(\mathbb {B}_N)$ , then $g = \eta - \eta (0)$ satisfies $g(0) = 0$ , and

$$\begin{align*}(f_{\eta(0)} + \tilde h) \circ \varphi - \mu (f_{\eta(0)} + \tilde h) = \eta(0) + g = \eta. \end{align*}$$

Finally, $C_\varphi - \mu \mathrm {Id}$ is invertible, so $\mu \not \in \sigma (C_\varphi )$ , and $\sigma (C_\varphi ) \subset \{0, 1\}$ . To conclude, since $\varphi $ is not one-to-one, Proposition 2.3 implies that $0 \in \sigma (C_\varphi )$ .

Proof The fact that $\{1\} \subset \sigma _p(C_\varphi ) \subset \{0, 1\}$ comes from the exact same argument as in the beginning of the proof of Proposition 5.1.

To finish the proof, we only have to show that if $\mu \not \in \{\lambda ^{\vec {\jmath }} : \vec {\jmath } \in \mathbb {N}_0^N\}$ , then $\mu \not \in \sigma _p(C_\varphi )$ . To do so, assume that there exists $f \in \mathrm {Hol}(\mathbb {B}_N) \backslash \{0\}$ such that $f \circ \varphi = \mu f$ . Then, the same inductive argument as in Proposition 4.4 gives Equation (4.1), and $f^{(\vec {\jmath })}(0) = 0$ for all $\vec {\jmath } \in \mathbb {N}_0^N$ . Thus, $f \equiv 0$ , and this finishes the proof.

Proof Since $\varphi $ is not one-to-one, by Proposition 2.3, the operator $C_\varphi $ is not invertible. Hence, $0 \in \sigma (C_\varphi )$ . Moreover, $1 \in \sigma _p(C_\varphi ) \subset \sigma (C_\varphi )$ by Proposition 5.3.

To obtain the second inclusion, let $\mu \in \mathbb {C}$ such that $\left \lvert \mu \right \rvert> \left \Vert \varphi '(0)\right \Vert $ and $\mu \neq 1$ . Since $\mu \not \in \sigma _p(C_\varphi )$ , we only have to show that $C_\varphi - \mu \mathrm {Id}$ is onto. Similarly to Lemma 4.8, there exists $\varepsilon> 0$ such that for all $h \in \mathrm {Hol}_0(\mathbb {B}_N)$ , the series $\sum (h \circ \varphi ^{[n]})/\mu ^n$ converges uniformly on $\overline {B(0, \varepsilon )}$ . For $g \in \mathrm {Hol}(\mathbb {B}_N)$ , we define the functions h and f by

$$\begin{align*}h(z) = g(z) - g(0), \qquad f(z) = \frac{g(0)}{1-\mu} - \frac{1}{\mu} \sum_{n \ge 0} \frac{h \circ \varphi^{[n]}}{\mu^n}. \end{align*}$$

Hence, $f \in \mathrm {Hol}(B(0, \varepsilon ))$ , $h \in \mathrm {Hol}_0(\mathbb {B}_N)$ , and on $B(0, \varepsilon )$ ,

$$\begin{align*}f \circ \varphi - \mu f = (1 - \mu) \frac{g(0)}{1-\mu} + \sum_{n \ge 0} \frac{h \circ \varphi^{[n]}}{\mu^n} - \sum_{n \ge 0} \frac{h \circ \varphi^{[n+1]}}{\mu^{n+1}} = g(0) + h = g. \end{align*}$$

By Lemma 4.7, there exists a function $\tilde f \in \mathrm {Hol}(\mathbb {B}_N)$ such that $\tilde f \circ \varphi - \mu \tilde f = g$ . This finishes the proof.

Proof For all $k \in \{1, \ldots , N\}$ , consider $f_k(z) = z_k$ . Then

$$\begin{align*}(f_k \circ D)(z) = [Dz]_k = e^{i\theta_k} z_k = e^{i\theta_k} f_k(z). \end{align*}$$

Hence, $e^{i\theta _k} \in \sigma _p(C_\varphi )$ . Lemma 4.1 concludes for the inclusion $\supset $ .

For the converse, set $e = (e^{i\theta _1}, \ldots , e^{i\theta _N})$ . Let $\mu \not \in \{ e^{\vec {\jmath }} : \vec {\jmath } \in \mathbb {N}_0^N\}$ . If $\mu \in \sigma _p(C_D)$ , there exists $f \in \mathrm {Hol}(\mathbb {B}_N) \backslash \{0\}$ such that $f \circ D = \mu f$ . The same inductive proof as in Proposition 4.4, using Equation (4.1), gives $f^{(\vec {\jmath })}(0) = 0$ for all $\vec {\jmath } \in \mathbb {N}_0^N$ . Thus, $f \equiv 0$ , and $\mu \not \in \sigma _p(C_D)$ .

Proof The first inclusion is immediate by Proposition 6.1.

To prove the other one, let $\mu \in \mathbb {C} \backslash \mathbb {T}$ and $g \in \mathrm {Hol}(\mathbb {B}_N)$ . We write

$$\begin{align*}g = \sum_{\vec{\jmath} \in \mathbb{N}_0^N} a_{\vec{\jmath}} z^{\vec{\jmath}}. \end{align*}$$

If $f = \sum b_{\vec {\jmath }} z^{\vec {\jmath }}$ satisfies $f \circ D - \mu f = g$ , then for all $\vec {\jmath } \in \mathbb {N}_0^N$ , $b_{\vec {\jmath }} (e^{\vec {\jmath }} - \mu ) = a_{\vec {\jmath }}$ . Hence,

$$\begin{align*}b_{\vec{\jmath}} = \frac{a_{\vec{\jmath}}}{e^{\vec{\jmath}} - \mu} \quad \Longrightarrow \quad \frac{1}{1 + \left\lvert \mu \right\rvert} \le \left\lvert \frac{b_{\vec{\jmath}}}{a_{\vec{\jmath}}} \right\rvert = \frac{1}{\left\lvert e^{\vec{\jmath}} - \mu \right\rvert} \le \frac{1}{\left\lvert 1 - \left\lvert \mu \right\rvert \right\rvert}. \end{align*}$$

Finally, $f = \sum b_{\vec {\jmath }} z^{\vec {\jmath }} \in \mathrm {Hol}(\mathbb {B}_N)$ , so $C_\varphi - \mu \mathrm {Id}$ is invertible.

Proof Let . Since p is a multiple of $p_k$ and $p_k\theta _k \in 2\pi \mathbb {Z}$ , then $p \theta _k \in 2\pi \mathbb {Z}$ . Hence, $e^{ip\theta _k} = 1$ , so

$$\begin{align*}D^p = \mathrm{diag}(e^{ip\theta_1}, \ldots, e^{ip\theta_N}) = \mathrm{diag}(1, \ldots, 1) = \mathrm{Id}. \end{align*}$$

Now, we show the minimality of p. To do so, let $q \in \mathbb {N}$ such that $D^q = \mathrm {Id}$ . Then, 1 is the unique eigenvalue of $D^q$ , so for all $k \in \{1, \ldots , N\}$ , $e^{iq\theta _k} = 1$ , that is $q \theta _k \in 2\pi \mathbb {Z}$ . Hence, for all k, q is a multiple of $p_k$ , so $q \ge lcm(p_1, \ldots , p_N) = p$ .

Proof Note that since $D^p = \mathrm {Id}$ , by polynomial functional calculus [Reference Bourbaki12, Reference Dunford, Schwartz, Bade and Bartle19] and multiplicativity of $C_D$ , we have $\sigma (C_D)^p = \sigma (C_D^p) = \sigma (C_{D^p}) = \sigma (C_{\mathrm {Id}}) = \{1\}$ . Hence, if $\lambda \in \sigma (C_D)$ , then $\lambda ^p = 1$ .

Conversely, let $\lambda \in \mathbb {C}$ such that $\lambda ^p = 1$ . Set $e = (e^{i\theta _1}, \ldots , e^{i\theta _N})$ .

$\underline {\text {Step 1}}$ : We show that there exists $\vec {\jmath } \in \mathbb {N}_0^N$ such that p is the smallest positive integer verifying $(e^{\vec {\jmath }})^p = 1$ . First, note that for all $k \in \{1, \ldots , N\}$ , we may write $\theta _k = 2\pi q_k/p_k$ , with $p_k \wedge q_k = 1$ and $p = lcm(p_1, \ldots , p_N)$ . We show that

$$\begin{align*}\left\{\exp\left(\frac{2i\pi q_k j}{p_k}\right) : j \in \mathbb{N}_0 \right\} = \left\{\exp\left(\frac{2i\pi j}{p_k}\right) : j \in \mathbb{N}_0 \right\}. \end{align*}$$

To do so, we only have to show the inclusion $\supset $ . Let $j \in \mathbb {N}_0$ . By Bézout’s identity, there exist $\ell \in \mathbb {Z}$ and $m \in \mathbb {N}$ such that $\ell p_k + mq_k = 1$ . Hence,

$$\begin{align*}\exp\left(i\frac{2\pi j}{p_k}\right) = \exp\left(i\frac{2\pi (j\ell p_k + jm q_k)}{p_k}\right) = \exp\left(i\frac{2\pi (jm) q_k}{p_k}\right). \end{align*}$$

Therefore, for all $k \in \{1, \ldots , N\}$ , there exists $m_k \in \mathbb {N}$ such that

$$\begin{align*}\exp\left( \frac{2i\pi m_k q_k}{p_k} \right) = \exp\left( \frac{2i\pi}{p_k} \right). \end{align*}$$

We show now that by setting $r_k = p/p_k$ , the $r_k$ are setwise coprime. Indeed, if all the $r_k$ were divisible by a certain $r \ge 2$ , then r divides p, and $p/r$ is a common multiple of $p_1, \ldots , p_N$ , which is impossible. Thus, there exists $\ell _1, \ldots , \ell _N \in \mathbb {Z}$ such that

$$\begin{align*}\sum_{k = 1}^N \ell_k r_k = 1, \end{align*}$$

and we obtain

$$ \begin{align*} \exp\left( 2i\pi \sum_{k = 1}^N \frac{q_k \ell_k m_k}{p_k} \right) & = \exp\left( 2i\pi \sum_{k = 1}^N \frac{\ell_k}{p_k} \right) \\ & = \exp\left( \frac{2i\pi}{p} \sum_{k = 1}^N r_k \ell_k \right) = \exp\left( \frac{2i\pi}{p} \right). \end{align*} $$

By setting $j_k = \ell _k m_k$ (even if we add a multiple of $p_k$ before, so that $j_k \ge 0$ ), we have $e^{\vec {\jmath }} = \exp (2i\pi /p)$ , so p is the smallest positive integer verifying $(e^{\vec {\jmath }})^p = 1$ .

$\underline {\text {Step 2}}$ : Since $\lambda ^p = 1$ , $\lambda ^p \in \sigma (\mathrm {Id}) = \sigma (C_D^p) = \sigma (C_D)^p$ . Therefore, there exists $\gamma \in \sigma (C_D)$ such that $(\lambda /\gamma )^p = 1$ . If $\lambda /\gamma = e^{i\alpha }$ , then

$$\begin{align*}e^{i\alpha p} = 1 = (e^{\vec{\jmath}})^p. \end{align*}$$

Hence, there exist $\ell , q \in \{0, \ldots , p-1\}$ such that

$$\begin{align*}\alpha = \sum_{k = 1}^{N} j_k \theta_k + \frac{2q\pi}{p} \quad \text{and} \quad \sum_{k = 1}^{N} j_k \theta_k = \frac{2\ell \pi}{p}, \end{align*}$$

with $\ell \wedge p = 1$ . By Bézout’s theorem, there exist $s \in \mathbb {N}$ , $t \in \mathbb {Z}$ such that $\ell s + tp = 1$ . Thus,

$$ \begin{align*} \frac{\lambda}{\gamma} = e^{i\alpha} & = \exp\left(i\left[ \sum_{k = 1}^N j_k \theta_k + \frac{2q\pi}{p} \right]\right) \\ & = \exp\left(i\left[ \sum_{k = 1}^N j_k \theta_k + \frac{2\ell sq\pi}{p} + 2tq\pi \right]\right) \\ & = \exp\left(i(sq+1)\left[ \sum_{k = 1}^N j_k \theta_k \right]\right) = (e^{\vec{\jmath}})^{sq+1}. \end{align*} $$

$\underline {\text {Step 3}}$ : Consider the map $\Gamma _{\vec {\jmath }} : \mathrm {Hol}(\mathbb {B}_N) \to z^{\vec {\jmath }} \mathrm {Hol}(\mathbb {B}_N)$ defined by

$$\begin{align*}[\Gamma_{\vec{\jmath}}(f)](z) = z^{\vec{\jmath}} f(z). \end{align*}$$

Hence, $\Gamma _{\vec {\jmath }}$ is bijective. Set $T = (C_D)_{|z^{\vec {\jmath }} \mathrm {Hol}(\mathbb {B}_N)}$ . Thus, for $f \in \mathrm {Hol}(\mathbb {B}_N)$ ,

$$\begin{align*}[(\Gamma_{\vec{\jmath}}^{-1} \circ T \circ \Gamma_{\vec{\jmath}})(f)](z) = \frac{(z^{\vec{\jmath}}f \circ D)(z)}{z^{\vec{\jmath}}} = \frac{D(z)^{\vec{\jmath}} f(D(z))}{z^{\vec{\jmath}}} = e^{\vec{\jmath}} f(D(z)). \end{align*}$$

Therefore, T is similar to $e^{\vec {\jmath }} C_D$ . Moreover, if the operator $C_D - \beta \mathrm {Id}$ is bijective, then $\beta \not \in \{e^{\vec {\imath }} : \vec {\imath } \in \mathbb {N}_0^N\}$ , and for all $g \in z^{\vec {\jmath }} \mathrm {Hol}(\mathbb {B}_N)$ , there exists $f \in \mathrm {Hol}(\mathbb {B}_N)$ such that $f \circ D - \beta f = g$ .

Denote respectively by $(a_{\vec {\imath }})$ and $(b_{\vec {\imath }})$ the Maclaurin coefficients of f and g. Hence, for all $\vec {\imath } \in \mathbb {N}_0^N$ , $a_{\vec {\imath }} = b_{\vec {\imath }}/(e^{\vec {\imath }} - \beta )$ , so $a_{\vec {\imath }} = 0$ if and only if $b_{\vec {\imath }} = 0$ . Therefore, $f \in z^{\vec {\jmath }} \mathrm {Hol}(\mathbb {B}_N)$ , and $T - \beta \mathrm {Id}$ is bijective. So $\sigma (T) \subset \sigma (C_D)$ .

Finally, since T is similar to $e^{\vec {\jmath }} C_D$ , we get $\sigma (T) = e^{\vec {\jmath }} \sigma (C_D)$ . Hence, $e^{\vec {\jmath }} \sigma (C_D) \subset \sigma (C_D)$ . Since $\lambda = (e^{\vec {\jmath }})^{qs+1} \gamma $ , with $\gamma \in \sigma (C_D)$ , we conclude that $\lambda \in \sigma (C_D)$ .

Proof Let $\lambda \in \mathbb {C}$ such that there exists $f \in \mathrm {Hol}(\mathbb {B}_N) \backslash \{0\}$ satisfying

$$\begin{align*}f \circ \varphi = \lambda f. \end{align*}$$

Iterating this equality, we obtain, for all $n \in \mathbb {N}$ , $f \circ \varphi ^{[n]} = \lambda ^n f$ . If $\left \lvert \lambda \right \rvert> 1$ , then the right hand side diverges to $+\infty $ at least at one point of $\mathbb {B}_N$ . This is impossible.

Moreover, as in the previous proofs, since , we get $1 \in \sigma _p(C_\varphi )$ .

Proof We prove $(1)$ by induction.

$\underline {m = 1}$ : We have to show that $\mathrm {Hol}(\mathbb {B}_2) = W_1 \oplus X_1$ .

Let $f \in \mathrm {Hol}(\mathbb {B}_2)$ . Using the Maclaurin coefficients of f, we may write

$$\begin{align*}f(z) = \sum_{j \ge 0} \sum_{k \ge 0} a_{jk} z_1^j z_2^k = \underbrace{\sum_{k \ge 0} a_{0k} z_2^k}_{\in W_1} + \underbrace{z_1 \sum_{j \ge 1} \sum_{k \ge 0} a_{jk} z_1^{j-1} z_2^k}_{\in X_1}. \end{align*}$$

Hence, $\mathrm {Hol}(\mathbb {B}_2) = W_1 + X_1$ .

Now, assume that $f \in W_1 \cap X_1$ . We can write

$$\begin{align*}f(z) = f_0(z_2) = z_1 \tilde f (z), \end{align*}$$

with $f_0 \in \mathrm {Hol}(\mathbb {D})$ and $\tilde f \in \mathrm {Hol}(\mathbb {B}_2)$ . Considering $z = (0, z_2)$ , we get $f_0(z_2) = 0$ for all $z_2 \in \mathbb {D}$ . Thus, $f \equiv 0$ , so $\mathrm {Hol}(\mathbb {B}_2) = W_1 \oplus X_1$ .

$\underline {m \rightarrow m+1}$ : Assume that $\mathrm {Hol}(\mathbb {B}_2) = W_m \oplus X_m$ .

Let $f \in \mathrm {Hol}(\mathbb {B}_2)$ . By induction hypothesis,

$$\begin{align*}f(z) = \sum_{\ell=0}^{m-1} \kappa_1^\ell(z_1) f_\ell(z_2) + z_1^m \tilde f (z), \end{align*}$$

with $f_0, \ldots , f_{m-1} \in \mathrm {Hol}(\mathbb {D})$ and $\tilde f \in \mathrm {Hol}(\mathbb {B}_2)$ . Denote

$$\begin{align*}\tilde f (z) = \sum_{j \ge 0} \sum_{k \ge 0} b_{jk} z_1^j z_2^k. \end{align*}$$

Hence, there exists $g \in \mathrm {Hol}(\mathbb {B}_2)$ such that

$$\begin{align*}z_1^m \tilde f (z) = z_1^m f_m(z_2) + z_1^{m+1} g(z), \qquad f_m(z) = \sum_{k \ge 0} b_{0k} z_2^k. \end{align*}$$

Moreover, by the properties of $\kappa _1$ , there exists $h \in \mathrm {Hol}(\mathbb {D})$ such that

$$\begin{align*}\kappa_1^m (z_1) = z_1^m + z_1^{m+1} h(z_1). \end{align*}$$

Therefore,

$$\begin{align*}z_1^m \tilde f (z) = \kappa_1^m(z_1) f_m(z_2) + z_1^{m+1} \left( g(z) - h(z_1) f_m(z_2) \right). \end{align*}$$

Finally, we have

$$\begin{align*}f(z) = \underbrace{\sum_{\ell=0}^{m-1} \kappa_1^\ell(z_1) f_\ell(z_2) + \kappa_1^m(z_1) f_m(z_2)}_{\in W_{m+1}} + \underbrace{z_1^{m+1} \left( g(z) - h(z_1) f_m(z_2) \right) \vphantom{\sum_a}}_{\in X_{m+1}}, \end{align*}$$

so $\mathrm {Hol}(\mathbb {B}_2) = W_{m+1} + X_{m+1}$ .

Now, assume that $f \in W_{m+1} \cap X_{m+1}$ . Then,

$$\begin{align*}f(z) = \sum_{\ell = 0}^{m} \kappa_1^\ell (z_1) f_\ell(z_2) = z_1^{m+1} \tilde f (z), \end{align*}$$

with $f_0, \ldots , f_m \in \mathrm {Hol}(\mathbb {D})$ and $\tilde f \in \mathrm {Hol}(\mathbb {B}_2)$ . Hence,

$$ \begin{align*} \sum_{\ell = 0}^{m-1} \kappa_1^\ell (z_1) f_\ell(z_2) & = z_1^{m+1} \tilde f (z) - \kappa_1^m(z_1) f_m(z_2) \\ & = z_1^m (z_1 \tilde f (z) - f_m(z_2) - z_1 h(z_1) f_m(z_2)) \in W_m \cap X_m = \{0\}. \end{align*} $$

Thus, $f(z) = \kappa _1^m(z_1) f_m(z_2) = z_1^{m+1} \tilde f (z)$ , so

$$\begin{align*}z_1^m f_m(z_2) = z_1^{m+1} (\tilde f(z) - h(z_1)f_m(z_2)). \end{align*}$$

Denote by $(c_{jk})$ the Maclaurin coefficients of the map $z \mapsto z_1^m f_m(z_2)$ , and $(d_{jk})$ those of $z \mapsto z_1^{m+1} (\tilde f(z) - h(z_1)f_m(z_2))$ . Then,

$$\begin{align*}j \neq m \Longrightarrow c_{jk} = 0 \quad \text{and} \quad j = m \Longrightarrow d_{jk} = 0. \end{align*}$$

Finally, since $c_{jk} = d_{jk}$ for all $j, k \in \mathbb {N}_0$ , we obtain

$$\begin{align*}z_1^m f_m(z_2) = z_1^{m+1} (\tilde f(z) - h(z_1)f_m(z_2)) = 0. \end{align*}$$

Hence, $f_m \equiv 0$ , so $f \equiv 0$ . We proved that $\mathrm {Hol}(\mathbb {B}_2) = W_m \oplus X_m$ .

Now, let us prove $(2)$ . If $f \in W_m$ , then

$$\begin{align*}f = \sum_{\ell = 0}^{m-1} \kappa_1^\ell (z_1) f_\ell(z_2), \end{align*}$$

with $f_0, \ldots , f_{m-1} \in \mathrm {Hol}(\mathbb {D})$ . Thus,

$$\begin{align*}(f \circ \varphi)(z) = \sum_{\ell = 0}^{m-1} \varphi_1'(0)^\ell \kappa_1^\ell(z_1) f_\ell(\beta_2 z_2) = \sum_{\ell = 0}^{m-1} \kappa_1^\ell(z_1) \tilde f_\ell(z_2), \end{align*}$$

where $\tilde f_\ell (z_2) = \varphi _1'(0)^\ell f_\ell (\beta _2 z_2)$ . Thus, $f \circ \varphi \in W_m$ .

If $f \in X_m$ , then $f(z) = z_1^m g(z)$ , with $g \in \mathrm {Hol}(\mathbb {B}_2)$ . Hence,

$$\begin{align*}(f \circ \varphi)(z) = \varphi_1(z_1)^m g(\varphi_1(z_1), \beta_2 z_2). \end{align*}$$

However, since $\varphi _1(0) = 0$ , we may write $\varphi _1(z_1) = z_1 \psi (z_1)$ , with $\psi \in \mathrm {Hol}(\mathbb {D})$ . Therefore,

$$\begin{align*}(f \circ \varphi)(z) = z_1^m \psi(z_1)^m g(\varphi_1(z_1), \beta_2 z_2) \in X_m.\\[-33pt] \end{align*}$$

Proof of Theorem 6.8

Let $g \in \mathrm {Hol}(\mathbb {B}_N)$ . Denote $z = (z_1, \tilde z)$ . Then, using multi-index for $\tilde z$ instead of index for $z_2$ , in a same way as in Proposition 6.9, we may write for all $m \in \mathbb {N}$

$$\begin{align*}g(z) = \sum_{\ell = 0}^{m-1} \kappa_1^\ell (z_1) g_\ell (\tilde z) + z_1^{m} G_1(z), \end{align*}$$

with $g_0, \ldots , g_{m-1} \in \mathrm {Hol}(\mathbb {B}_{N-1})$ and $G_1 \in \mathrm {Hol}(\mathbb {B}_N)$ . If we do the same decomposition for each $g_\ell $ , with $\tilde z = (z_2, w)$ , we obtain for all $(m_1, m_2) \in \mathbb {N}^2$

$$\begin{align*}g(z) = \sum_{\ell_1 = 0}^{m_1 - 1} \sum_{\ell_2 = 0}^{m_2 - 1} \kappa_1^{\ell_1} (z_1) \kappa_2^{\ell_2} (z_2) g_{\ell_1, \ell_2} (z_3, \ldots, z_N) + z_2^{m_2} G_2(z) + z_1^{m_1} G_1(z), \end{align*}$$

with $g_{\ell _1, \ell _2} \in \mathrm {Hol}(\mathbb {B}_{N-2})$ and $G_1, G_2 \in \mathrm {Hol}(\mathbb {B}_N)$ .

Iterating this p times, and we get for all $(m_1, \ldots , m_p) \in \mathbb {N}^p$

$$\begin{align*}g(z) = \sum_{\ell_1 = 0}^{m_1 - 1} \cdots \sum_{\ell_p = 0}^{m_p - 1} \left( \prod_{k = 1}^p \kappa_k^{\ell_k} (z_k) \right) g_{\ell_1, \ldots, \ell_p} (z_{p+1}, \ldots, z_N) + \sum_{k = 1}^p z_k^{m_k} G_k(z), \end{align*}$$

with $g_{\ell _1, \ldots , \ell _p} \in \mathrm {Hol}(\mathbb {B}_{N-p})$ and $G_k \in \mathrm {Hol}(\mathbb {B}_N)$ . In the following, we will denote $\tilde G_k(z) = z_k^{m_k} G_k(z)$ .

Now, let $\lambda \in \mathbb {C}^*$ and $\zeta _1, \ldots , \zeta _p < 1$ such that

$$\begin{align*}\left\lvert \lambda \right\rvert \not\in \left\{ \prod_{k = 1}^p \left\lvert \varphi_k'(0) \right\rvert^{n_k} : n_1, \ldots, n_p \in \mathbb{N}_0 \right\} \quad \text{and} \quad \left\lvert \varphi_k'(0) \right\rvert < \zeta_k, \quad k = 1, \ldots, p. \end{align*}$$

Consider $m_1, \ldots , m_p \in \mathbb {N}$ such that for all $k = 1, \ldots , p$ , $\zeta _k^{m_k} < \left \lvert \lambda \right \rvert $ .

• • For all $k \in \{1, \ldots , p\}$ , there exists $\varepsilon _k> 0$ such that if $\left \lvert z_k \right \rvert < \varepsilon _k$ , then $\left \lvert \varphi _k(z_k) \right \rvert \le \zeta _k \left \lvert z_k \right \rvert $ . Using Schwarz’s lemma, we obtain for all $n \in \mathbb {N}_0$ $|\varphi _k^{[n]}(z_k)| \le \zeta _k^n \left \lvert z_k \right \rvert $ . Hence,

  $$\begin{align*}\left\lvert \frac{(\tilde G_k \circ \varphi^{[n]})(z)}{\lambda^n} \right\rvert \le C_k \frac{|\varphi_k^{[n]}(z_k)|^{m_k}}{\left\lvert \lambda \right\rvert^n} \le C_k \left\lvert z_k \right\rvert^{m_k} \left(\frac{\zeta_k^{m_k}}{\left\lvert \lambda \right\rvert}\right)^n. \end{align*}$$
  Therefore, the series $\sum (\tilde G_k \circ \varphi ^{[n]})(z)/\lambda ^n$ converges uniformly for $\left \lvert z_k \right \rvert < \varepsilon _k$ . Consider  $\tilde F_k$ the map defined by
  $$\begin{align*}\tilde F_k(z) = -\frac{1}{\lambda} \sum_{n \ge 0} \frac{\tilde G_k \circ \varphi^{[n]}(z)}{\lambda^n}. \end{align*}$$
  Then, $\tilde F_k$ is holomorphic on $B(0, \varepsilon _k)$ , and we obtain
  $$ \begin{align*} (\tilde F_k \circ \varphi)(z) - \lambda \tilde F_k(z) & = \sum_{n \ge 0} \frac{\tilde G_k \circ \varphi^{[n]}(z)}{\lambda^n} - \sum_{n \ge 0} \frac{\tilde G_k \circ \varphi^{[n+1]}(z)}{\lambda^{n+1}} \\ & = \tilde G_k \circ \varphi^{[0]} = \tilde G_k. \end{align*} $$
  Hence, for $\left \lvert z \right \rvert < \varepsilon = \min (\varepsilon _k : 1 \le k \le p)$ , we get
  $$\begin{align*}\left(\sum_{k = 1}^p \tilde F_k\right) \circ \varphi - \lambda \sum_{k = 1}^p \tilde F_k = \sum_{k = 1}^p \tilde G_k = \sum_{k = 1}^p z_k^{m_k} G_k. \end{align*}$$
  The last thing to do is extending the map $\tilde F_k$ to a map $F_k \in \mathrm {Hol}(\mathbb {B}_N)$ , satisfying
  $$\begin{align*}\left(\sum_{k = 1}^p F_k\right) \circ \varphi - \lambda \sum_{k = 1}^p F_k = \sum_{k = 1}^p z_k^{m_k} G_k, \end{align*}$$
  thanks to Lemma 4.7.

• • Let $\ell _1 \in \{0, \ldots , m_1 - 1\}$ , $\ldots $ , $\ell _p \in \{0, \ldots , m_p - 1\}$ . We now look at the map

  $$\begin{align*}\tilde g_{\ell_1, \ldots, \ell_p} = \left( \prod_{k = 1}^p \kappa_k^{\ell_k} (z_k) \right) g_{\ell_1, \ldots, \ell_p} (z_{p+1}, \ldots, z_N). \end{align*}$$
  Consider $\tilde f_{\ell _1, \ldots , \ell _p} \in \mathrm {Hol}(\mathbb {B}_N)$ of the form
  $$\begin{align*}\tilde f_{\ell_1, \ldots, \ell_p} = \left( \prod_{k = 1}^p \kappa_k^{\ell_k} (z_k) \right) f_{\ell_1, \ldots, \ell_p} (z_{p+1}, \ldots, z_N). \end{align*}$$
  Note that in this case, if we denote $\phi (z_{p+1}, \ldots , z_N) = (\beta _{p+1} z_{p+1}, \ldots , \beta _N z_N)$ ,
  $$ \begin{align*} & (\tilde f_{\ell_1, \ldots, \ell_p} \circ \varphi) - \lambda \tilde f_{\ell_1, \ldots, \ell_p} \\ = \; & \left( \prod_{k = 1}^p \kappa_k^{\ell_k} (z_k) \right) \left[ \prod_{k = 1}^p \varphi_k'(0)^{\ell_k} (f_{\ell_1, \ldots, \ell_p} \circ \phi) - \lambda f_{\ell_1, \ldots, \ell_p} \right]. \end{align*} $$
  Remark that $\phi $ is bijective. Since $0 < \left \lvert \varphi _k'(0) \right \rvert < 1$ for all k, and $\left \lvert \lambda \right \rvert \neq \prod _{k=1}^p \left \lvert \varphi _k'(0) \right \rvert ^{\ell _k}$ , by Proposition 6.2, there exists $f_{\ell _1, \ldots , \ell _p} \in \mathrm {Hol}(\mathbb {B}_p)$ such that
  (6.1) $$ \begin{align} \prod_{k = 1}^p \varphi_k'(0)^{\ell_k} (f_{\ell_1, \ldots, \ell_p} \circ \phi) - \lambda f_{\ell_1, \ldots, \ell_p} = g_{\ell_1, \ldots, \ell_p}. \end{align} $$
  Hence, we obtain $(\tilde f_{\ell _1, \ldots , \ell _p} \circ \varphi ) - \lambda \tilde f_{\ell _1, \ldots , \ell _p} = \tilde g_{\ell _1, \ldots , \ell _p}$ .

Finally, we may find $f \in \mathrm {Hol}(\mathbb {B}_N)$ such that $f \circ \varphi - \lambda f = g$ . It is defined as

$$\begin{align*}f(z) = \sum_{\ell_1 = 0}^{m_1 - 1} \cdots \sum_{\ell_p = 0}^{m_p - 1} \tilde f_{\ell_1, \ldots, \ell_p} (z) + \sum_{k = 1}^p F_k(z). \end{align*}$$

Hence, the operator $C_\varphi - \lambda \mathrm {Id}$ is bijective, so $\lambda \not \in \sigma (C_\varphi )$ .

Conversely, let $f_1, \ldots , f_N : \mathbb {D} \to \mathbb {C}$ holomorphic maps such that for all $1 \le k \le N$ ,

$$\begin{align*}f_k \circ \varphi_k = \varphi_k'(0) f_k \quad (1 \le k \le p), \qquad f_k \circ \varphi_k = \beta_k f_k \quad (p+1 \le k \le N). \end{align*}$$

Such maps exist using the results in one variable [Reference Arendt, Bernard, Célariès and Chalendar4, Reference Arendt, Célariès and Chalendar5], since $\varphi $ has separable variables. For $n_1, \ldots , n_N \in \mathbb {N}_0$ , consider $f : \mathbb {B}_N \to \mathbb {C}$ defined by $f(z) = \prod _{k = 1}^N f_k(z_k)^{n_k}$ . Then,

$$ \begin{align*} (f \circ \varphi)(z) & = \prod_{k = 1}^p f_k(\varphi_k(z_k))^{n_k} \times \prod_{k = p+1}^N f_k(\beta_k z_k)^{n_k} \\ & = \prod_{k = 1}^p \varphi_k'(0)^{n_k} f_k(z_k)^{n_k} \times \prod_{k = p+1}^N \beta_k^{n_k} f_k(z_k)^{n_k} \\ & = \left( \prod_{k = 1}^p \varphi_k'(0)^{n_k} \times \prod_{k = p+1}^N \beta_k^{n_k} \right) f(z). \end{align*} $$

Therefore, $\prod _{k = 1}^p \varphi _k'(0)^{n_k} \times \prod _{k = p+1}^N \beta _k^{n_k} \in \sigma _p(C_\varphi ) \subset \sigma (C_\varphi )$ .