# Composition operators on weighted Bergman spaces of the polydisc

Frédéric Bayart, Anne Dorval

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Frédéric Bayart, Anne Dorval. Composition operators on weighted Bergman spaces of the polydisc. 2026. hal-05477672

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Distributed under a Creative Commons CC BY 4.0 - Attribution - International License
COMPOSITION OPERATORS ON WEIGHTED BERGMAN SPACES OF
THE POLYDISC

FRÉDÉRIC BAYART AND ANNE DORVAL

ABSTRACT. We study composition operators between weighted Bergman spaces of the poly-
disc induced by smooth symbols. We prove a general result of continuity which only involves
the behaviour of the symbol on the polytorus. We deduce from this several consequences
about the automatic continuity of the induced operator. We study in depth the case of the
tridisc and exhibit several examples showing that a characterization of continuity using only
derivatives seems impossible.

1. INTRODUCTION

Let $\mathcal{U}$ be a domain in $\mathbb{C}^d$, let $X$ be a Banach space of holomorphic functions on $\mathcal{U}$ and let $\phi : \mathcal{U} \to \mathcal{U}$ be holomorphic (we will write $\phi \in \mathcal{O}(\mathcal{U}, \mathcal{U})$). The composition operator with symbol $\phi$ is defined by $C_\phi(f) = f \circ \phi, f \in X$. The first question to solve when studying composition operators is that of continuity: for which symbols $\phi$ does $C_\phi$ induce a bounded composition operator on $X$? When $X$ is the Hardy space or a weighted Bergman space of the unit disc $\mathbb{D}$, the answer is easy: by the Littlewood subordination principle, this is always the case.

The situation breaks down dramatically on the polydisc $\mathbb{D}^d$, $d \ge 2$, where there are simple examples of symbols $\phi$ such that $C_\phi$ does not map $H^2(\mathbb{D}^d)$ or any Bergman space $A_\beta^2(\mathbb{D}^d)$ into itself. The papers [1, 11, 13, 14] contain some general results as well as a detailed study of the case of affine symbols.

It turns out that studying the continuity of composition operators on Hardy and Bergman
spaces of the bidisc and the tridisc already becomes an issue. Regarding the bidisc, a character-
ization of the symbols $\phi : \mathbb{D}^2 \to \mathbb{D}^2$ which are regular up to the boundary inducing a bounded
composition operator on $A^2(\mathbb{D}^2)$ was given in [1]. This characterization was extended to $H^2(\mathbb{D}^2)$
and to all Bergman spaces $A_\beta^2(\mathbb{D}^2)$ in [12]. Precisely, let $\phi \in \mathcal{O}(\mathbb{D}^2, \mathbb{D}^2) \cap C^2(\bar{\mathbb{D}}^2)$ and let $\beta > -1$.
Then $C_\phi$ is bounded on $A_\beta^2(\mathbb{D}^2)$ (resp. on $H^2(\mathbb{D}^2)$) if and only if $d\phi(\xi)$ is invertible for all $\xi \in \mathbb{T}^2$
such that $\phi(\xi) \in \mathbb{T}^2$. Hence, to study the continuity of $C_\phi$ on Hardy or Bergman spaces of
the bidisc, it is sufficient to know conditions on the first order derivative for any $\xi \in \mathbb{T}^2$ such
that $\phi(\xi)$ belongs to the boundary of $\mathbb{D}^2$. Moreover, this shows that if $C_\phi$ is continuous on one
$A_\beta^2(\mathbb{D}^2)$, then it is continuous on all $A_\beta^2(\mathbb{D}^2)$. We also mention the recent papers [3, 4, 5, 7, 10]
on composition operators on the bidisc.

The problem of continuity is trickier on the tridisc. In [12], a characterization of holomorphic symbols $\phi : \mathbb{D}^3 \to \mathbb{D}^3$ which are $C^2$-smooth on $\bar{\mathbb{D}}^3$ inducing a bounded composition operator on $A^2(\mathbb{D}^3)$ is given. This characterization again only involves conditions on the first-order derivative of $\phi$. A similar characterization was given in [2] for the Hardy space $H^2(\mathbb{D}^3)$. It now involves conditions on the second-order derivatives and it is shown that first-order derivatives are not sufficient: for a map $\phi : \bar{\mathbb{D}}^d \to \bar{\mathbb{D}}^d$ we denote by $\mathcal{E}(\phi)$ the set $\{z \in \bar{\mathbb{D}}^d : \phi(z) \in \partial\bar{\mathbb{D}}^d\}$. Then one can construct two holomorphic functions $\phi, \psi \in \mathcal{O}(\mathbb{D}^3, \mathbb{D}^3) \cap C^\infty(\bar{\mathbb{D}}^3)$ such that

<footnote>The authors are partially supported by the grant ANR-24-CE40-0892-01 of the French National Research Agency ANR.</footnote>
• $\mathcal{E}(\phi) = \mathcal{E}(\psi)$.

• For all $\xi \in \mathcal{E}(\phi)$, $\phi(\xi) = \psi(\xi)$ and $\nabla\phi(\xi) = \nabla\psi(\xi)$.

• $C_\phi$ is continuous on $H^2(\mathbb{D}^3)$ whereas $C_\psi$ is not.

In this paper we investigate composition operators $C_\phi$ on general weighted Bergman spaces $A_\beta^2(\mathbb{D}^d)$ when the symbol $\phi$ is regular up to the boundary. Our first main result, Theorem 2.10, gives a necessary and sufficient condition for $C_\phi$ to map boundedly $A_{\beta_1}^2(\mathbb{D}^d)$ into $A_{\beta_2}^2(\mathbb{D}^d)$. This condition only involves boundary values of $\phi$, contrary to the usual condition based on Carleson measures. It has several interesting corollaries, like a stability result: if $C_\phi$ is continuous on $A_{\beta_1}^2(\mathbb{D}^d)$, then it is continuous on $A_{\beta_2}^2(\mathbb{D}^d)$ for all $\beta_2 \ge \beta_1$ (see Theorem 2.13 below).

Then one concentrates on Bergman spaces of the tridisc. In view of the results of [2] and [12], one could expect that conditions on the second-order derivatives of $\phi$ would be sufficient to characterize the boundedness of $C_\phi$ on any $A_\beta^2(\mathbb{D}^3)$. This conjecture is far from being true: for all $n \ge 1$, even the knowledge of the first $n$-th order derivatives will not be enough to characterize the continuity of $C_\phi$ on all $A_\beta^2(\mathbb{D}^3)$.

**Theorem 1.1.** Let $n \ge 1$. There exist $\phi, \psi \in \mathcal{O}(\mathbb{D}^3, \mathbb{D}^3) \cap C^\infty(\bar{\mathbb{D}}^3)$ and $\beta \in (-1, 0)$ such that

• $\mathcal{E}(\phi) = \mathcal{E}(\psi)$.

• For any $\xi \in \mathcal{E}(\phi)$, for any $\alpha \in \mathbb{Z}_+^3$ with $|\alpha| \le n$, $\partial^\alpha \phi(\xi) = \partial^\alpha \psi(\xi)$.

• $C_\phi$ is continuous on $A_\beta^2(\mathbb{D}^3)$ and $C_\psi$ is not continuous on $A_\beta^2(\mathbb{D}^3)$.

Moreover, two different values of $\beta \in [-1, 0]$ require two different characterizations of the boundedness of composition operators, as the following result indicates.

**Theorem 1.2.** Let $-1 \le \beta_1 < \beta_2 \le 0$. Then there exists $\phi \in \mathcal{O}(\mathbb{D}^3, \mathbb{D}^3) \cap C^1(\bar{\mathbb{D}}^3)$ such that $C_\phi$ is continuous on $A_{\beta_2}^2(\mathbb{D}^3)$ and $C_\phi$ is not continuous on $A_{\beta_1}^2(\mathbb{D}^3)$.

Hence it seems impossible to characterize in full generality the continuity of $C_\phi$ on $A_\beta^2(\mathbb{D}^3)$, $\beta \in (-1, 0)$. Nevertheless, we do in Section 5 an in-depth study to understand what can happen when we have a full knowledge of the second-order derivative. This allows us to give plenty of examples in Section 6. In Section 7, we come back to composition operators on the bidisc and we are interested in the following problem: let $\beta_1 > -1$; for which values of $\beta_2 \ge \beta_1$ does there exist $\phi \in \mathcal{O}(\mathbb{D}^2, \mathbb{D}^2) \cap C^1(\bar{\mathbb{D}}^2)$ such that $C_\phi$ maps $A_{\beta_1}(\mathbb{D}^2)$ into $A_\beta(\mathbb{D}^2)$ if and only if $\beta \ge \beta_2$. We show that this is the case if and only if $\beta_2 \in [\beta_1, \beta_1 + 1/2]$ or $\beta_2 = 2\beta_1 + 2$. This answers a question of [10].

## 2. SOME GENERAL RESULTS

2.1. **Notations.** Let us start by introducing some notation that will be used throughout this paper. The unit vector $(1, \dots, 1)$ will be denoted by $e$. For $\theta = (\theta_1, \dots, \theta_d) \in \mathbb{R}^d$, we will write $e^{i\theta}$ for $(e^{i\theta_1}, \dots, e^{i\theta_d})$. Any $z \in \bar{\mathbb{D}}^d$ such that $z_k \ne 0$ for all $k$ will be uniquely written $((1 - \rho_k)e^{i\theta_k})$ with $\theta_k \in [-\pi, \pi)$ and $\rho_k \in [0, 1)$.

For $f: \bar{\mathbb{D}}^d \to A$, $\theta \in \mathbb{R}^d$ and $\rho \in [0, 1]^d$, we write $f(\theta)$ for $f(e^{i\theta_1}, \dots, e^{i\theta_d})$ and $f(\rho, \theta)$ for $f(z)$ with $z = ((1 - \rho_k)e^{i\theta_k})$. If $f$ maps $A$ into $C^d$, the map $f_I$, where $I = \{i_1, \dots, i_p\} \subset \{1, \dots, d\}$, will denote $(f_{i_1}, \dots, f_{i_p}) : A \to \mathbb{C}^p$. For $f, g: A \to \mathbb{R}$, we shall write $f \lesssim g$ provided there exists $C > 0$ such that, for all $x \in A$, $f(x) \le Cg(x)$. When we write $f \simeq g$, we mean $f \lesssim g$ and $g \lesssim f$.

The Lebesgue measure on $\mathbb{R}^d$ will be denoted $\lambda_d$ and the normalized surface measure on $\mathbb{T}^d$ will be denoted $\sigma_d$. Let $dA$ denote the normalized area measure on $\mathbb{D}$. For $\beta > -1$, we put $dA_\beta(z) = (\beta + 1)(1 - |z|^2)^\beta dA(z)$. On the polydisc $\mathbb{D}^d$, we define

$$dV_\beta(z) = dA_\beta(z_1) \cdots dA_\beta(z_d), \quad z = (z_1, \dots, z_d) \in \mathbb{D}^d.$$
For $\beta > -1$, the Bergman space $A_{\beta}^{2}(\mathbb{D}^{d})$ is the function space defined by

$$A_{\beta}^{2}(\mathbb{D}^{d}) = \left\{ f \in \mathcal{O}(\mathbb{D}^{d}, \mathbb{D}^{d}) : \|f\|_{2} = \int_{\mathbb{D}^{d}} |f(z)|^{2} dV_{\beta}(z) < +\infty \right\},$$

and the Hardy space $H^2(\mathbb{D}^d)$ is defined by

$$H^2(\mathbb{D}^d) = \left\{ f \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}^d) : \|f\| = \sup_{0<r<1} \int_{\mathbb{T}^d} |f(re^{i\theta})|^2 d\sigma_d(z) < +\infty \right\}.$$

For simplicity, we will sometimes write $A_{-1}^2(\mathbb{D}^d)$ for $H^2(\mathbb{D}^d)$.

For $\eta \in \mathbb{T}^d$ and $\bar{\delta} \in (0, 2)^d$, the Carleson box is the set

$$S(\eta, \bar{\delta}) = \{z \in \mathbb{D}^d : |z_j - \eta_j| < \delta_j, j = 1, \dots, d\}.$$

If $I \subset \{1, \dots, d\}$, $S_I(\eta, \bar{\delta})$ will mean $\{z \in \mathbb{D}^d : |z_j - \eta_j| < \delta_j, j \in I\}$.

**2.2. Technical lemmas.** In this subsection, we collect several lemmas which will be used throughout the proofs. We start with two geometric lemmas. The first one is the parametrized Morse lemma (see [6, Section 4.44] for instance).

**Lemma 2.1.** Let $f : \mathbb{R}^n \times \mathbb{R}^{d-n} \to \mathbb{R}$ be a $C^3$-function such that $f(0) = 0$, $d_y f(0) = 0$ and $d_y^2 f(0)$ is non-singular. There exist a neighbourhood $U$ of $0$ in $\mathbb{R}^d$, a diffeomorphism $\Gamma : U \to \mathbb{R}^d$, $(x, y) \mapsto (x, \gamma(x, y))$ and a $C^1$-map $h : \mathbb{R}^n \to \mathbb{R}$ such that for any $(x, y) \in U$,

$$f(x, y) = \sum_{j=1}^{d-n} \pm \gamma_j(x, y)^2 + h(x).$$

The second one is a kind of Julia-Caratheodory theorem in several variables (see, in particular, [1, Lemma 2.7]).

**Lemma 2.2.** Let $\varphi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}) \cap C^1(\bar{\mathbb{D}}^d)$ and $\xi \in \mathbb{T}^d$. Assume that $\varphi(\xi) = 1$. Then for all $k \in \{1, \dots, d\}$, $\frac{\partial \varphi}{\partial z_k}(\xi) \ge 0$. Moreover, if $\frac{\partial \varphi}{\partial z_k}(\xi) = 0$, then $\varphi$ does not depend on $z_k$.

Our next lemma will help us to provide examples of self-maps of the polydisc with prescribed derivatives.

**Lemma 2.3.** Let $\psi$ be defined by $\psi(z) = 1 + \frac{z-1}{2} - \frac{(z-1)^2}{8} + \frac{3}{128}(z-1)^3$ for $z \in \mathbb{D}$. Then $\psi(\bar{\mathbb{D}}) \subset \bar{\mathbb{D}}$ and $|\psi(z)| = 1$ if and only if $z = 1$.

*Proof.* Indeed, let $z = e^{i\theta} \in \mathbb{T}$. After some computations, we find

$$\begin{aligned} |\psi(\theta)|^2 &= \frac{1}{16384} (2(135 \cos(3\theta) - 810 \cos(2\theta) + 2025 \cos(\theta)) + 13684) \\ &= \frac{1}{16384} (2(540 \cos^3(\theta) - 1620 \cos^2(\theta) + 1620 \cos(\theta)) + 15304) \end{aligned}$$

Then,

$$\begin{aligned} |\psi(\theta)|^2 \le 1 &\Leftrightarrow 2(540 \cos^3(\theta) - 1620 \cos^2(\theta) + 1620 \cos(\theta)) \le 1080 \\ &\Leftrightarrow \cos^3(\theta) - 3 \cos^2(\theta) + 3 \cos(\theta) \le 1 \\ &\Leftrightarrow (\cos(\theta) - 1)^3 \le 0. \end{aligned}$$

The last inequality is always satisfied with equality if and only if $\theta = 0$ or $\theta = 2\pi$. $\square$

<figure><img src="image_3.png" /></figure>

We will also need to estimate the volume of certain sets to characterize continuity. This is done in the following lemmas:
**Lemma 2.4.** Let $\beta > -1$. There exist $C_1, C_2 > 0$ such that, for all measurable subsets $A$ of $\mathbb{R}^d$ and for all $\bar{\delta} \in (0, 1)^d$,

$$V_{\beta}(\{( (1 - \rho_k)e^{i\theta_k})_{k=1,...,d} : 0 \le \rho_k \le \delta_k, \theta \in A \}) \simeq (\delta_1 \cdots \delta_d)^{(1+\beta)} \lambda_d(A).$$

*Proof.* First, let us consider

$$\int_{1-\delta_i}^{1} (\beta+1)(1-r^2)^\beta r dr = \left[ \frac{-(1-r^2)^{\beta+1}}{2} \right]_{1-\delta_i}^{1} = 2^\beta \delta_i^{\beta+1} \left(1 - \frac{\delta_i}{2}\right)^{\beta+1} \underset{0}{\sim} 2^\beta \delta_i^{\beta+1}.$$

Thus, there exist $c_1, c_2 > 0$ such that

$$c_1 \delta_i^{\beta+1} \le \int_{1-\delta_i}^{1} (\beta+1)(1-r^2)^\beta r dr \le c_2 \delta_i^{\beta+1}.$$

We conclude using Fubini's theorem.

□

*Remark.* A look at the proof shows that the involved constants are independent of $\beta$ provided $\beta \in (-1, 0]$ (see also [2, Lemma 3.6]).

**Lemma 2.5.** Let $O$ be a bounded open subset of $\mathbb{R}^2$. Then there exists $C > 0$ such that for all $a \in \mathbb{R}$ and all $\delta \in (0, e^{-1})$, we have

$$\lambda_2(\{(x, y) \in O : |x^2 - y^2 - a| < \delta\}) \le -C\delta \log(\delta).$$

*Proof.* We can assume that $a > 0$ and $O = [-M/2, M/2]^2$ for some $M > 1$. Let $(x, y) \in O$ and set $u = x - y$ and $v = x + y$. Then, $(u, v) \in [-M, M]^2$ and

$$|x^2 - y^2 - a| < \delta \Leftrightarrow |uv - a| < \delta.$$

In particular, if the set $\{(x, y) \in O : |x^2 - y^2 - a| < \delta\}$ is nonempty, then $a$ has to be smaller than $M^2 + \delta$, an assumption that we now make. By symmetry, one only has to estimate the measure of

$$\mathcal{E} = \{(u, v) \in [0, M] \times [-M, M] : |uv - a| < \delta\}.$$

Assume first that $a \ge \delta$ which implies $\mathcal{E} \subset [0, +\infty)^2$. Using $0 \le v \le M$ and $(a - \delta)/u \le v \le (a + \delta)/u$, we find that $\mathcal{E} \subset \mathcal{E}_1 \cup \mathcal{E}_2$ with

$$\mathcal{E}_1 = \left\{ (u, v) \in \mathbb{R}^2 : \frac{a - \delta}{M} \le u \le \frac{a + \delta}{M}, 0 \le v \le M \right\}$$

$$\mathcal{E}_2 = \left\{ (u, v) \in \mathbb{R}^2 : \frac{a + \delta}{M} \le u \le M + 2, \frac{a - \delta}{u} \le v \le \frac{a + \delta}{u} \right\}$$

(we have taken here $M + 2$ instead of $M$ to be sure that $M + 2 > (a + \delta)/M$). Therefore,

$$\begin{aligned} \lambda_2(\mathcal{E}) &\le 2\delta + \int_{(a+\delta)/M}^{M+2} \frac{2\delta}{u} du \\ &\le 2\delta + 2\delta \log(M+2) - 2\delta \log(a+\delta) + 2\delta \log(M) \\ &\le -C\delta \log(\delta). \end{aligned}$$

On the other hand, if $a < \delta$, then we write $\mathcal{E} \subset \mathcal{E}'_1 \cup \mathcal{E}'_2 \cup \mathcal{E}'_3$ with

$$\mathcal{E}'_1 = \left\{ (u, v) \in \mathbb{R}^2 : 0 \le u \le \frac{\delta - a}{M}, -M \le v \le M \right\}$$

$$\mathcal{E}'_2 = \left\{ (u, v) \in \mathbb{R}^2 : \frac{\delta - a}{M} \le u \le \frac{a + \delta}{M}, \frac{a - \delta}{u} \le v \le M \right\}$$

$$\mathcal{E}'_3 = \left\{ (u, v) \in \mathbb{R}^2 : \frac{a + \delta}{M} \le u \le M + 2, \frac{a - \delta}{u} \le v \le \frac{a + \delta}{u} \right\}$$
so that

$$
\begin{align*}
\lambda_2(\mathcal{E}) &\le 2(\delta - a) + \int_{\frac{\delta-a}{M}}^{\frac{\delta+a}{M}} \left(M + \frac{\delta-a}{u}\right) du + 2\delta \log(M+2) - 2\delta \log(a+\delta) + 2\delta \log(M) \\
&\le 2\delta + (\delta - a) \log\left(\frac{\delta+a}{\delta-a}\right) + 4\delta \log(M+2) - 2\delta \log(a+\delta) \\
&\le -C\delta \log(\delta)
\end{align*}
$$

since $x \mapsto -x \log(x)$ is decreasing on $(0, e^{-1})$ and $(\delta - a) \log(\delta + a) \le 0$. $\square$

<figure><img src="image_4.png" /></figure>

2.3. **How to prove continuity.** It is well known that the continuity of composition operators on Hardy and Bergman spaces of the polydisc is linked to volume estimates of the preimage of Carleson sets. As a consequence of [9, Theorem 2.5], for $\beta_1, \beta_2 > -1$, we know that $C_\phi$ maps boundedly $A_{\beta_1}^2(\mathbb{D}^d)$ into $A_{\beta_2}^2(\mathbb{D}^d)$ if and only if

$$
(1) \quad \exists C > 0, \forall \eta \in \mathbb{T}^d, \forall \bar{\delta} \in (0, 1)^d, V_{\beta_2}(\phi^{-1}(S(\eta, \bar{\delta}))) \leq C V_{\beta_1}(S(\eta, \bar{\delta})).
$$

This means that $V_{\beta_2} \circ \phi^{-1}$ is a Carleson measure for $A_{\beta_1}^2(\mathbb{D}^d)$. Recall that

$$
V_{\beta_1}(S(\eta, \bar{\delta})) \approx \prod_{i=1}^{d} \delta_i^{2+\beta_1}.
$$

This characterization extends to $\beta_2 = -1$ (that is, replacing $A_{\beta_2}(\mathbb{D}^d)$ by $H^2(\mathbb{D}^d)$), with the measure $\lambda_d \circ \phi^{-1}$ instead of $V_{\beta_2} \circ \phi^{-1}$. It becomes more complicated for $\beta_1 = -1$ since Carleson measures on Hardy spaces of the polydisc are difficult objects. The condition

$$
V_{\beta_2}(\phi^{-1}(S(\eta, \bar{\delta}))) \leq C \prod_{i=1}^{d} \delta_i
$$

is still necessary but not sufficient. To prove continuity, a strategy first devised in [1] is to prove
that $C_\phi$ maps continuously $A_{\beta_1}^2(\mathbb{D}^d)$ into $A_{\beta_2}^2(\mathbb{D}^d)$ with $\|C_\phi\|_{\mathcal{L}(A_{\beta_1}^2, A_{\beta_2}^2)} \le C$ with $C$ independent
of $\beta_1$ and to allow $\beta_1$ to $-1$. For this we will need the following result (see [1, Proposition 9.3]):

**Lemma 2.6.** There exists a constant $C(d)$ depending only on the dimension $d$ such that, for all $\phi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D})$, all $\beta_1 \in (-1, 0]$ and all $\beta_2 > -1$, if (1) is satisfied, then $\|C_\phi\|_{\mathcal{L}(A_{\beta_1}^2, A_{\beta_2}^2)} \le C(d) \cdot C$.

Our first main result is a characterization of the continuity of $C_\phi : A_{\beta_1}^2(\mathbb{D}^d) \to A_{\beta_2}^2(\mathbb{D}^d)$ using only the boundary values of $\phi$, provided that $\phi$ is sufficiently regular. We first need some definitions.

**Definition 2.7.** (1) Let $\varphi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}) \cap \mathcal{C}^1(\bar{\mathbb{D}}^d)$. Assume that there exists $\theta \in \mathbb{R}^d$ such that
$\varphi(\theta) \in \mathbb{T}$. We define

$$
P_{\varphi} = \left\{ k \in \{1, \dots, d\} : \frac{\partial \varphi(\theta)}{\partial z_k} \neq 0 \right\}.
$$

(2) Let $\phi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}) \cap \mathcal{C}^1(\bar{\mathbb{D}}^d)$ and let $I \subset \{1, \dots, d\}$. We say that $\phi_I$ touches the polycircle if $\phi_I(\mathbb{T}^d) \cap \mathbb{T}^{|I|} \neq \emptyset$.

(3) Let $\phi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}) \cap \mathcal{C}^1(\bar{\mathbb{D}}^d)$ and $I \subset \{1, \dots, d\}$ such that $\phi_I$ touches the boundary. For $j \in I$, we simply write $P_j = P_{\phi_j}$ and we set $P_I = \bigcup_{j \in I} P_j$.
(4) Let $\phi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}) \cap \mathcal{C}^1(\bar{\mathbb{D}}^d)$, let $I \subset \{1, \dots, d\}$ such that $\phi_I$ touches the boundary, let $\bar{\delta} \in (0, 1)^{|I|}$ and let $k \in \{1, \dots, d\}$. We set

$$ \omega_{I,k}(\bar{\delta}) = \begin{cases} 1 & \text{provided } k \notin \bigcup_{j \in I} P_j \\ \min\{\delta_j, j \in I, k \in P_j\} & \text{otherwise.} \end{cases} $$

We point out that $P_\varphi$ does not depend on $\theta$ since, by Lemma 2.2, for any $\theta, \theta' \in \mathbb{R}^d$ such that $\varphi(\theta), \varphi(\theta') \in \mathbb{T}^d$,

$$ \frac{\partial \varphi(\theta)}{\partial z_k} = 0 \iff \frac{\partial \varphi(\theta')}{\partial z_k} = 0 $$

and this is equivalent to saying that $\varphi$ does not depend on $z_k$. In other words, $P_\varphi$ denotes the set of (the indices of) the variables involved in $\varphi$.

The real numbers $\omega_{I,k}(\bar{\delta})$ will be relevant to link the intersection of the preimage of a Carleson window with the polycircle to the whole preimage, as the following lemma indicates.

**Lemma 2.8.** Let $\phi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}^d) \cap \mathcal{C}^1(\bar{\mathbb{D}}^d)$.

a) For all $I \subset \{1, \dots, d\}$ such that $\phi_I$ touches the polycircle, there exists $D \ge 1$ such that, for all $\eta \in \mathbb{T}^{|I|}$, for all $\bar{\delta} \in (0, 1)^{|I|}$,

$$ \{((1 - \rho_k)e^{i\theta_k}) : 0 \le \rho_k \le \omega_{I,k}(\bar{\delta}), k \in P_I \text{ and } e^{i\theta} \in \phi_I^{-1}(S_I(\eta, \bar{\delta}))\} \subset \phi_I^{-1}(S_I(\eta, D\bar{\delta})). $$

b) For all $\xi \in \bar{\mathbb{D}}^d$, for all $I \subset \{1, \dots, d\}$ with $\phi_I(\xi) \in \mathbb{T}^{|I|}$, there exist $D \ge 1$ and a neighbourhood $\mathcal{U}$ of $\xi$ in $\bar{\mathbb{D}}^d$ such that, for all $\eta \in \mathbb{T}^{|I|}$, for all $\bar{\delta} \in (0, 1)^{|I|}$,

$$ \phi_I^{-1}(S_I(\eta, \bar{\delta})) \cap \mathcal{U} \subset \{((1 - \rho_k)e^{i\theta_k}) : 0 \le \rho_k \le D\omega_{I,k}(\bar{\delta}), k \in P_I \text{ and } e^{i\theta} \in \phi_I^{-1}(S_I(\eta, D\bar{\delta}))\}. $$

*Proof.* a) Let $I$ be as in the assumptions and let $\bar{\delta} \in (0, 1)^{|I|}$. Let also $\theta \in \mathbb{R}^d$ and $\rho \in [0, 1]^d$ satisfying $0 \le \rho_k \le \omega_{I,k}(\bar{\delta})$ for all $k \in P_I$. If we set $z = ((1 - \rho_k)e^{i\theta_k})$, the mean value inequality yields for any $j \in I$,

$$ |\phi_j(z) - \phi_j(\theta)| \le \left( \sup_l \|\nabla \phi_l\| \right) \sup_{k \in P_j} \|z_k - e^{i\theta_k}\| $$

(we can take the supremum over $P_j$ since $\phi_j$ does not depend on the other variables) so that

$$ |\phi_j(z) - \phi_j(\theta)| \le \left( \sup_l \|\nabla \phi_l\| \right) \sup_{k \in P_j} \omega_{I,k}(\bar{\delta}) \le \left( \sup_l \|\nabla \phi_l\| \right) \delta_j. $$

The conclusion of a) now follows from the triangle inequality, with $D = 1 + \left( \sup_l \|\nabla \phi_l\| \right)$

b) Without loss of generality, we assume that $P_I = \{1, \dots, q\}$. We first observe that

$$ \phi_I(\xi_1/|\xi_1|, \dots, \xi_d/|\xi_d|) = \phi_I(\xi_1, \dots, \xi_d) $$

where $\xi_k/|\xi_k| = 1$ if $\xi_k = 0$. Indeed, assume that $\xi_1 \notin \mathbb{T}$. Then for all $j \in I$, $z_1 \mapsto \phi_j(z_1, \xi_2, \dots, \xi_d)$ is holomorphic by Weierstrass' theorem, thus constant by the maximum principle. Therefore, the claimed result follows from an easy induction. Moreover, for all $k \in P_I$, $|\xi_k| = 1$. Indeed, let $j \in I$ be such that $k \in P_j$. Then $\phi_j(\xi_1/|\xi_1|, \dots, \xi_k, \dots, \xi_d/|\xi_d|) \in \mathbb{T}$. By Lemma 2.2 and by definition of $P_j$, the map $z_k \mapsto \phi_j(\xi_1/|\xi_1|, \dots, z_k, \dots, \xi_d/|\xi_d|)$ cannot be constant, hence cannot have modulus 1 at $\xi_k$ if $|\xi_k| < 1$.

To fix the notations, we assume $\xi_1 = \dots = \xi_q = 1$ and $\phi_I(\xi) = (1, \dots, 1)$. Let $j \in I$. A Taylor expansion of $\phi_j$ near $e$ gives

$$ |\phi_j(z)|^2 = 1 + \sum_{k \in P_j} \rho_k F_{j,k}(\rho, \theta) + G_j(\theta) $$
where $F_{j,k}(0,0) = -2\frac{\partial\phi_j(e)}{\partial z_k}$, $z_k = (1-\rho_k)e^{i\theta_k}$, $0 \le \rho_k \le 1$ and $\theta_k \in \mathbb{R}$. For $k \in P_j$, by Lemma 2.2, $F_{j,k}(0,0) < 0$. Let $\mathcal{V}$ be a neighbourhood of $(1, \dots, 1)$ in $\bar{\mathbb{D}}^q$ and let $c > 0$ be such that $F_{j,k}(\rho, \theta) \le -c$ for all $j \in I$, all $k \in P_j$, all $\rho \in (0, 1)^q$ and all $\theta \in \mathbb{R}^q$ such that $((1-\rho_k)e^{i\theta_k})_{k=1,\dots,q} \in \mathcal{V}$. Besides, if we choose $\rho_k = 0$ for all $k = 1, \dots, q$, then we get $G_j(\theta) \le 0$ for all $(\theta_1, \dots, \theta_q)$ sufficiently close to $(0, \dots, 0)$. Reducing $\mathcal{V}$ if necessary, we may assume that this holds when $((1-\rho_k)e^{i\theta_k})_{k=1,\dots,q} \in \mathcal{V}$. We finally set $\mathcal{U} = \mathcal{V} \times \bar{\mathbb{D}}^{d-q}$ which is a neighbourhood of $\xi$ in $\bar{\mathbb{D}}^d$.

Let $\eta \in \mathbb{T}^{|I|}$, $\bar{\delta} \in (0, 1)^{|I|}$ and pick $z \in \phi_I^{-1}(S_I(\eta, \bar{\delta})) \cap \mathcal{U}$. Write it $z = ((1-\rho_k)e^{i\theta_k})_{k=1,\dots,d}$ and consider $j \in I$. Since $|\phi_j(z)|^2 \ge 1 - 2\delta_j$ and $G_j(\theta) \le 0$, we get

$$ \sum_{k \in P_j} \rho_k F_{j,k}(\rho, \theta) \ge -2\delta_j $$

which yields $\rho_k \le 2\delta_j/c$ for $k \in P_j$. Taking the infimum over $j \in I$, we get, for all $k \in P_I$, $\rho_k \le 2\omega_{I,k}(\bar{\delta})/c$. Moreover, we also have

$$ \begin{aligned} |\phi_j(\theta) - \eta_j| &\le |\phi_j(\theta) - \phi_j(z)| + |\phi_j(z) - \eta_j| \\ &\le \left( \sup_l \|\nabla\phi_l\| \right) \max_{k \in P_j} \|z_k - e^{i\theta_k}\| + \delta_j \\ &\le \left( 1 + \frac{2 \sup_l \|\nabla\phi_l\|}{c} \right) \delta_j. \end{aligned} $$

This yields the result.

□

*Remark.* In both cases, a compactness argument could prove that one can choose $D$ independently of $I$ and $\theta_0$. However, we will not use this and a further compactness argument will be needed during the proof of Theorem 2.10.

The value of the $w_{I,k}(\bar{\delta})$, $k = 1, \dots, d$, is rather intricate. We will use several times the following result to estimate its product.

**Lemma 2.9.** Let $\phi \in \mathcal{O}(\bar{\mathbb{D}}^d, \mathbb{D}^d) \cap \mathcal{C}^1(\bar{\mathbb{D}}^d)$, $I \subset \{1, \dots, d\}$ be such that $\phi_I$ touches the polycircle. Then for all $\bar{\delta} \in (0, 1)^{|I|}$,

$$ (2) \qquad \prod_{k=1}^{d} w_{I,k}(\bar{\delta}) \le \prod_{j \in I} \delta_j^{\frac{|P_I|}{|I|}}. $$

*Proof.* We first observe that $\prod_{k=1}^d w_{I,k}(\bar{\delta}) = \prod_{k \in P_I} w_{I,k}(\bar{\delta})$. We shall do the proof by induction on $|I|$. If $|I| = 1$, this is obvious. Let now $q \in \{2, \dots, d\}$ and let us assume that (2) holds for all $I$ with $|I| = q-1$ and all $\bar{\delta} \in (0, 1)^{|I|}$. Let $I \subset \{1, \dots, d\}$ with $|I| = q$, for instance $I = \{1, \dots, q\}$ and let $\bar{\delta} \in (0, 1)^{|I|}$. Without loss of generality, we may assume that $\delta_1 \le \dots \le \delta_q$. If $P_1 = P_I$, then

$$ \prod_{k=1}^{d} \omega_{I,k}(\bar{\delta}) = \delta_1^{|P_I|} \le \prod_{j \in I} \delta_j^{\frac{|P_I|}{|I|}}. $$
On the contrary, if $P_1 \neq P_I$, let us set $I' = I \setminus \{1\}$. Notice that if $k \notin P_1$, then $\omega_{I,k}(\bar{\delta}) = \omega_{I',k}(\bar{\delta})$ and for $k \in P_1$ we have $\omega_{I,k}(\bar{\delta}) = \delta_1$. Thus,

$$ \prod_{k=1}^{d} \omega_{I,k}(\bar{\delta}) = \prod_{k \in P_I} \omega_{I,k}(\bar{\delta}) = \prod_{k \in P_1} \omega_{I,k}(\bar{\delta}) \prod_{k \in P_{I'}} \omega_{I',k}(\bar{\delta}) \leq \delta_1^{|P_1|} \prod_{j \in I'} \delta_j^{\frac{|P_I|-|P_1|}{q-1}} $$

where the last inequality comes from the induction hypothesis. Now,

$$ \begin{aligned} \delta_1^{|P_1|} \prod_{j \in I'} \delta_j^{\frac{|P_I|-|P_1|}{q-1}} &= \delta_1^{|P_1| - \frac{|P_I|-|P_1|}{q-1}} \prod_{j \in I} \delta_j^{\frac{|P_I|-|P_1|}{q-1}} \\ &= \left( \delta_1^{q|P_1|-|P_I|} \prod_{j \in I} \delta_j^{|P_I|-|P_1|} \right)^{\frac{1}{q-1}} \\ &\leq \left( \prod_{j \in I} \delta_j^{|P_1|-\frac{|P_I|}{q}} \prod_{j \in I} \delta_j^{|P_I|-|P_1|} \right)^{\frac{1}{q-1}} \\ &\leq \prod_{j \in I} \delta_j^{\frac{|P_I|}{q}}. \end{aligned} $$

□

We are now ready to prove our first main result.

**Theorem 2.10.** Let $\beta_1, \beta_2 \geq -1$. Let $\phi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}^d) \cap C^1(\overline{\mathbb{D}}^d)$. Then, the following assertions are equivalent:

(1) $C_\phi : A_{\beta_1}^2(\mathbb{D}^d) \to A_{\beta_2}^2(\mathbb{D}^d)$ is bounded.

(2) For all $I \subset \{1, \dots, d\}$ such that $\phi_I$ touches the polycircle, there exists $C > 0$ such that for all $\eta \in \mathbb{T}^{|I|}$ and all $\bar{\delta} \in (0, 1)^{|I|}$,

$$ \lambda_d(\{\theta \in [-\pi, \pi)^d : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \leq C \frac{\prod_{j \in I} \delta_j^{2+\beta_1}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta_2}}. $$

(3) For all $\theta_0 \in \mathbb{R}^d$, all $I \subset \{1, \dots, d\}$ such that $\phi_I(\theta_0) \in \mathbb{T}^{|I|}$ and $I$ is maximal with respect to this property, there exist a neighbourhood $O$ of $\theta_0$ in $\mathbb{R}^d$ and $C > 0$ such that for all $\eta \in \mathbb{T}^{|I|}$ and all $\bar{\delta} \in (0, 1)^{|I|}$,

$$ (3) \qquad \lambda_d(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \leq C \frac{\prod_{j \in I} \delta_j^{2+\beta_1}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta_2}}. $$

*Proof.* We first observe that 2) ⇒ 3) is immediate. We split the proof of the other implications in several cases, following $A_{\beta_1}^2$ (resp. $A_{\beta_2}^2$) is, or is not, the Hardy space.

*Case 1:* $C_\phi : A_{\beta_1}^2(\mathbb{D}^d) \to A_{\beta_2}^2(\mathbb{D}^d)$ with $\beta_1, \beta_2 > -1$.
1) ⇒ 2) Let $I \subset \{1, \dots, d\}$ be such that $\phi_I$ touches the polycircle. Let also $\eta \in \mathbb{T}^{|I|}$ and $\bar{\delta} \in (0, 1)^{|I|}$. By Lemma 2.8, there exists $D \ge 1$ such that

$$E := \left\{ \left( (1 - \rho_k)e^{i\theta_k} \right) : 0 \le \rho_k \le \omega_{I,k}(\bar{\delta}), k \in P_I \text{ and } e^{i\theta} \in \phi_I^{-1}(S_I(\eta, \bar{\delta})) \right\} \subset \phi_I^{-1}(S_I(\eta, D\bar{\delta})).$$

Let us now extend $\bar{\delta}$ to $(0, 1)^d$ by setting $\delta_j = 2$ if $j \notin I$ and $\eta$ to $\mathbb{T}^d$ by taking $\eta_j = 1$ if $j \notin I$. Then

$$\phi_I^{-1}(S_I(\eta, D\bar{\delta})) = \phi^{-1}(S(\eta, D\bar{\delta})).$$

We now use Lemma 2.4 and the characterization of continuity of composition operators using Carleson measures to get

$$\begin{align*}
\lambda_d(\{\theta \in [-\pi, \pi)^d : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta_2} &\lesssim V_{\beta_2}(E) \\
&\lesssim V_{\beta_2}(\phi_I^{-1}(S_I(\eta, D\bar{\delta}))) \\
&\lesssim V_{\beta_2}(\phi^{-1}(S(\eta, D\bar{\delta}))) \\
&\lesssim V_{\beta_1}(S(\eta, D\bar{\delta})) \\
&\lesssim \prod_{j \in I} \delta_j^{2+\beta_1}
\end{align*}$$

where the involved constants do not depend on $\eta$ and $\bar{\delta}$.

3) ⇒ 1) The proof will be divided into two steps. We first show that (3) implies a local Carleson type condition and then we conclude with a covering argument.

**Fact.** Let $\xi \in \bar{\mathbb{D}}^d$ and $I \subset \{1, \dots, d\}$ be such that $\phi_I(\xi) \in \mathbb{T}^{|I|}$ and $I$ is maximal with respect to this property. Then there exist a neighbourhood $\mathcal{U}$ of $\xi$ in $\bar{\mathbb{D}}^d$ and $D := D(\phi, I, \xi, \beta_1, \beta_2)$ such that, for all $\eta \in \mathbb{T}^{|I|}$ and all $\bar{\delta} \in (0, 1)^{|I|}$,

$$V_{\beta_2}(\phi_I^{-1}(S_I(\eta, \bar{\delta})) \cap \mathcal{U}) \le DV_{\beta_1}(S_I(\eta, \bar{\delta})),$$

where $D$ does not depend on $\beta_1$ (resp. $\beta_2$) if $\beta_1 \in (-1, 0]$ (resp. $\beta_2 \in (-1, 0]$).

*Proof of the fact.* Let us write $\xi = ((1 - \tilde{\rho}_k)e^{i\tilde{\theta}_k})$ and observe (as in the proof of Lemma 2.8) that $\phi_I(e^{i\tilde{\theta}}) \in \mathbb{T}^{|I|}$. Let $O$ be a neighbourhood of $\tilde{\theta}$ and $C > 0$ be given by 3) of Theorem 2.10. Let $D \ge 1$ and $\mathcal{U}_1$ be the neighbourhood of $\xi$ given by b) of Lemma 2.8. We set

$$\mathcal{U} = \mathcal{U}_1 \cap \left\{ \left( (1 - \rho_k)e^{i\theta_k} \right)_{k=1, \dots, d} : 0 \le \rho_k \le 3/2, e^{i\theta} \in O \right\}.$$

Then $\mathcal{U}$ is a neighbourhood of $\xi$ and

$$\phi_I^{-1}(S_I(\eta, \bar{\delta})) \cap \mathcal{U} \subset \left\{ \left( (1 - \rho_k)e^{i\theta_k} \right) : 0 \le \rho_k \le D\omega_{I,k}(\bar{\delta}), k \in P_I \text{ and } e^{i\theta} \in \phi_I^{-1}(S_I(\eta, D\bar{\delta})) \cap O \right\}.$$

We use Lemma 2.4 to get

$$\begin{align*}
V_{\beta_2}(\phi_I^{-1}(S_I(\eta, \bar{\delta})) \cap \mathcal{U}) &\le C_{\beta_2} \prod_{k \in P_I} \omega_{I,k}(D\bar{\delta})^{1+\beta_2} \lambda_d(\phi_I^{-1}(S_I(\eta, D\bar{\delta})) \cap O) \\
&\le C \cdot C_{\beta_2} \prod_{k \in P_I} \omega_{I,k}(D\bar{\delta})^{1+\beta_2} \frac{\prod_{j \in I} (D\delta_j)^{2+\beta_1}}{\prod_{k=1}^{d} \omega_{I,k}(D\bar{\delta})^{1+\beta_2}} \\
&\le C \cdot C_{\beta_2} \prod_{j \in I} (D\delta_j)^{2+\beta_1} \\
&\le C \cdot C_{\beta_2} D^{(2+\beta_1)d} C_{\beta_1} V_{\beta_1}(S_I(\eta, D\bar{\delta})).
\end{align*}$$
That $C_{\beta_1}$ and $C_{\beta_2}$ may be chosen independently of $\beta_1$ and $\beta_2$ when they belong to $(-1, 0]$ follows from the remark after Lemma 2.4 and from [1, Lemma 8.1]. $\square$

We are now ready for the second half of 3) $\implies 1)$, the covering argument. We introduce

$$A_1 = \{\xi \in \bar{\mathbb{D}}^d : \phi(\xi) \in \mathbb{T}^d\}.$$

Let $\xi \in A_1$. By the fact, there exist $\mathcal{U}_{1,\xi}$ a neighbourhood of $\xi$ in $\bar{\mathbb{D}}^d$ and $C_{1,\xi} > 0$ such that for all $\eta \in \mathbb{T}^d$, and all $\bar{\delta} \in (0, 1)^d$,

$$V_{\beta_2}(\phi^{-1}(S(\eta, \bar{\delta})) \cap \mathcal{U}_{1,\xi}) \le C_{1,\xi}(\beta_1, \beta_2)V_{\beta_1}(S(\eta, \bar{\delta})).$$

Then, the sets $(\mathcal{U}_{1,\xi})_\xi$ form an open covering of $A_1$ which is compact so we can extract a finite covering $(\mathcal{U}_{1,k})_{k=1,...,s_1}$. For all $\eta \in \mathbb{T}^d$ and all $\bar{\delta} \in (0, 2)^d$, we get

$$V_{\beta_2} \left( \phi^{-1}(S(\eta, \bar{\delta})) \cap \bigcup_{k=1}^{s_1} \mathcal{U}_{1,k} \right) \le \sum_{k=1}^{s_1} V_{\beta_2}(\phi^{-1}(S(\eta, \bar{\delta})) \cap \mathcal{U}_{1,k}) \le \left( \sum_{k=1}^{s_1} C_{1,k} \right) V_{\beta_1}(S(\eta, \bar{\delta})),$$

with $C_{1,k}$ independent of $\beta_i$ if $\beta_i \in (-1, 0]$. For notational convenience, we set $D_{1,k} = C_{1,k}$.

Then there exists $\varepsilon_1 \in (0, 1)$ such that for all $\xi \in B_1 := \bar{\mathbb{D}}^d \setminus \bigcup_{k=1}^{s_1} \mathcal{U}_{1,k}$, we have $|\phi_j(\xi)| < 1 - 2\varepsilon_1$ for some $j \in \{1, ..., d\}$. We introduce

$$A_2 = \{\xi \in B_1 : \exists I \subset \{1, ..., d\}, |I| = d - 1, \phi_I(\xi) \in \mathbb{T}^{d-1}\}.$$

Let $\xi \in A_2$. There exists $I \subset \{1, ..., d\}$ with $|I| = d - 1$ such that $\phi_I(\xi) \in \mathbb{T}^{d-1}$ and $|\phi_j(\xi)| < 1 - 2\varepsilon_1$ for the single $j \in \{1, ..., d\}$ which does not belong to $I$. So, applying again the fact, we get that there exist $\mathcal{U}_{2,\xi}$ a neighbourhood of $\xi$ in $\bar{\mathbb{D}}^d$ and $C_{2,\xi} > 0$ such that, for all $\eta \in \mathbb{T}^{|I|}$, for all $\bar{\delta} \in (0, 1)^{|I|}$ and for all $z \in \mathcal{U}_{2,\xi}$, we have

$$V_{\beta_2}(\phi_I^{-1}(S_I(\eta, \bar{\delta})) \cap \mathcal{U}_{2,z}) \le C_{2,\xi}V_{\beta_1}(S_I(\eta, \bar{\delta})) \text{ and } |\phi_j(z)| < 1 - \varepsilon_j.$$

Let now $\bar{\delta} \in (0, 1)^{|I|}$. Then $\phi_j^{-1}(S(\eta, \bar{\delta})) \cap \mathcal{U}_{2,\xi} = \emptyset$ if $\delta_j < \varepsilon_1$. It $\delta_j \ge \varepsilon_1$, then

$$\begin{align*}
V_{\beta_2}(\phi^{-1}(S(\eta, \bar{\delta})) \cap \mathcal{U}_{2,\xi}) &\le V_{\beta_2}(\phi_I^{-1}(S_I(\eta, \bar{\delta})) \cap \mathcal{U}_{2,\xi}) \\
&\le C_{2,\xi}V_{\beta_1}(S_I(\eta, \bar{\delta})) \\
&\le \frac{C_{2,\xi}}{\delta_j^{2+\beta_1}} V_{\beta_1}(S(\eta, \bar{\delta})) \\
&\le \frac{C_{2,\xi}}{\varepsilon_1^{2+\beta_1}} V_{\beta_1}(S(\eta, \bar{\delta})) \\
&= D_{2,\xi}V_{\beta_1}(S(\eta, \bar{\delta})).
\end{align*}$$

As $\frac{1}{\varepsilon_1^{2+\beta_1}} \le \frac{1}{\varepsilon_1^2}$ when $\beta_1 \in (-1, 0]$, we get that $D_{2,\xi}$ is independent of $\beta_i$ if $\beta_i \in (-1, 0]$. Moreover, the sets $(\mathcal{U}_{2,\xi})_\xi$ form an open covering of $A_2$ which is compact so we can extract a finite covering $(\mathcal{U}_{2,k})_{k=1,...,s_2}$ and we have

$$V_{\beta_2} \left( \phi^{-1}(S(\eta, \bar{\delta})) \cap \bigcup_{k=1}^{s_2} \mathcal{U}_{2,k} \right) \le \sum_{k=1}^{s_2} V_{\beta_2}(\phi^{-1}(S(\eta, \bar{\delta})) \cap \mathcal{U}_{2,k}) \le \left( \sum_{k=1}^{s_2} D_{2,k} \right) V_{\beta_1}(S(\eta, \bar{\delta})).$$

Then there exists $\varepsilon_2 \in (0, \varepsilon_1)$ such that for all $\xi \in B_2 := B_1 \setminus \bigcup_{k=1}^{s_2} \mathcal{U}_{2,k}$, there exist two distinct $j_1, j_2 \in \{1, ..., d\}$ such that $|\phi_{j_1}(\xi)| < 1 - 2\varepsilon_2$ and $|\phi_{j_2}(\xi)| < 1 - 2\varepsilon_2$. We repeat the reasoning with

$$A_3 = \{\xi \in B_2 : \exists I \subset \{1, ..., d\}, |I| = d - 2, \phi_I(\xi) \in \mathbb{T}^{d-2}\}.$$
and we define $B_3, A_4, \dots, B_{d-1}, A_d$. At the last step, there exists $\varepsilon_d \in (0, \varepsilon_{d-1})$ such that for all $\xi \in B_d := B_{d-1} \setminus \bigcup_{k=1}^{s_d} \mathcal{U}_{d,k}$, we have $|\phi_j(z)| < 1 - \varepsilon_d$ for all $j \in \{1, \dots, d\}$.

Let $\bar{\delta} \in (0,2)^d$. Assume that at least one of the $\delta_j$ is less than $\varepsilon_d$, then $\phi^{-1}(S(\eta, \bar{\delta})) \cap B_d = \emptyset$ so that

$$ \phi^{-1}(S(\eta, \bar{\delta})) = \phi^{-1}(S(\eta, \bar{\delta})) \cap \bigcup_{l,k} \mathcal{U}_{l,k}. $$

Hence

$$ \begin{aligned} V_{\beta_2}(\phi^{-1}(S(\eta, \bar{\delta}))) &\le \sum_l V_{\beta_2} \left( \phi^{-1}(S(\eta, \bar{\delta})) \cap \bigcup_k \mathcal{U}_{l,k} \right) \\ &\le \left( \sum_{k,l} D_{l,k} \right) V_{\beta_1}(S(\eta, \bar{\delta})) \\ &= DV_{\beta_1}(S(\eta, \bar{\delta})), \end{aligned} $$

with $D > 0$ independent of $\beta_i$ if $\beta_i \in (-1,0]$. Multiplying eventually $D$ with some power of $\varepsilon_d$, as above, this remains true if all $\delta_j$ are bigger than $\varepsilon_d$. In conclusion, we have shown that there exists $D(\beta_1, \beta_2) > 0$ such that for all $\eta \in \mathbb{T}^d$ and all $\bar{\delta} \in (0,2)^d$, we have

$$ V_{\beta_2}(\phi^{-1}(S(\eta, \bar{\delta}))) \le D(\beta_1, \beta_2) V_{\beta_1}(S(\eta, \bar{\delta})), $$

that is $C_\phi$ maps continuously $A_{\beta_1}^2(\mathbb{D}^d)$ into $A_{\beta_2}^2(\mathbb{D}^d)$ and $\|C_\phi\| \le F(C, \beta_1, \beta_2)$ with $F(C, \beta_1, \beta_2)$ independent of $\beta_i$ if $\beta_i \in (-1,0]$, $i=1,2$ (see Lemma 2.6).

Case 2: $C_\phi: A_{\beta_1}^2(\mathbb{D}^d) \to H^2(\mathbb{D})$ with $\beta_1 > -1$. The proof is identical since the characterization of continuity extends to $\beta_2 = -1$. We replace everywhere $V_{\beta_2}$ by $\lambda_d$.

Case 3: $C_\phi: H^2(\mathbb{D}^d) \to A_{\beta_2}^2(\mathbb{D}^d)$ with $\beta_2 > -1$. The proof of 1) $\Rightarrow$ 2) is again identical. For the proof of 3) $\Rightarrow$ 1), we are going to prove that $C_\phi: A_{\beta_1}^2(\mathbb{D}^d) \to A_{\beta_2'}^2(\mathbb{D}^d)$ is bounded with $\|C_\phi\| \le C$, where $C > 0$ is independent of $\beta_1 \in (-1,0]$ and $\beta_2' \to \beta_2$ when $\beta_1 \to -1$. Thus, let us consider $\beta_1 \in (-1,0]$ and let us set $\beta_2' = \beta_2 + d(1+\beta_1)$. By assumption, we know that for all $\theta_0 \in \mathbb{R}^d$, all $I \subset \{1, \dots, d\}$ such that $\phi_I(\theta_0) \in \mathbb{T}^{|I|}$ and $I$ is maximal, there exist $O$ a neighbourhood of $\theta_0$ in $\mathbb{R}^d$ and $C > 0$ such that for all $\eta \in \mathbb{T}^{|I|}$ and all $\bar{\delta} \in (0,1)^{|I|}$,

$$ \lambda_d(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C \frac{\prod_{j \in I} \delta_j}{\prod_{k=1}^d \omega_{I,k}(\bar{\delta})^{1+\beta_2}}. $$

Now,

$$ \begin{aligned} \frac{\prod_{j \in I} \delta_j}{\prod_{k=1}^d \omega_{I,k}(\bar{\delta})^{1+\beta_2}} &\le \frac{\prod_{j \in I} \delta_j}{\prod_{k=1}^d \omega_{I,k}(\bar{\delta})^{1+\beta_2'}} \left( \prod_{k=1}^d \omega_{I,k}(\bar{\delta}) \right)^{d(1+\beta_1)} \\ &\le \frac{\prod_{j \in I} \delta_j}{\prod_{k=1}^d \omega_{I,k}(\bar{\delta})^{1+\beta_2'}} \left( \prod_{j \in I} \delta_j \right)^{1+\beta_1} \end{aligned} $$

by Lemma 2.9. Hence, for all $\theta_0 \in \mathbb{R}^d$, all $I \subset \{1, \dots, d\}$ such that $\phi_I(\theta_0) \in \mathbb{T}^{|I|}$ and $I$ is maximal, there exist $O$ a neighbourhood of $\theta_0$ in $\mathbb{R}^d$ and $C > 0$ such that for all $\eta \in \mathbb{T}^{|I|}$ and
all $\bar{\delta} \in (0, 1)^{|I|}$,

$$ \lambda_d(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C \frac{\prod_{j \in I} \delta_j^{2+\beta_1}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta'_2}}. $$

By Case 1 of the present theorem, we find that $C_\phi : A_{\beta_1}^2(\mathbb{D}^d) \to A_{\beta_2}^2(\mathbb{D}^d)$ is bounded with $\|C_\phi\| \le D$ where $D$ is independent from $\beta_1 \in (-1, 0]$. Thus, $C_\phi : H^2(\mathbb{D}^d) \to A_{\beta_2}^2(\mathbb{D}^d)$ is bounded.

**Case 4 :** $C_\phi : H^2(\mathbb{D}^d) \to H^2(\mathbb{D}^d)$. We prove that 1) $\Rightarrow$ 2) as above. For 3) $\Rightarrow$ 1), let us consider $\beta \in (-1, -d/(1+d)]$ and let us set $\beta_1 = \beta$, $\beta_2 = \beta + d(1+\beta)$. Arguing as in Case 3, we find that $C_\phi : A_{\beta_1}^2(\mathbb{D}^d) \to A_{\beta_2}^2(\mathbb{D}^d)$ is bounded with $\|C_\phi\| \le D$ where $D$ is independent of $\beta \in (-1, 0]$. When $\beta \to -1$, we have that $\beta_1 \to -1$ and $\beta_2 \to -1$, thus, we get that $C_\phi : H^2(\mathbb{D}^d) \to H^2(\mathbb{D}^d)$ is bounded. $\square$

It can be difficult to apply Theorem 2.10 in practice because of the difficulty to compute $\omega_{I,k}(\bar{\delta})$ in general. We shall use sometimes the following corollaries which only involve $\bar{\delta}$

**Corollary 2.11.** Let $\beta_1, \beta_2 \ge -1$. Let $\phi \in O(\mathbb{D}^d, \mathbb{D}^d) \cap C^1(\overline{\mathbb{D}}^d)$.

(a) Let $\theta_0 \in \mathbb{R}^d$ and $I \subset \{1, \dots, d\}$ be such that $\phi_I(e^{i\theta_0}) \in \mathbb{T}^{|I|}$ and there exist $C > 0$ and $O$ a neighbourhood of $\theta_0$ in $\mathbb{R}^d$ such that for all $\eta \in \mathbb{T}^{|I|}$ and all $\bar{\delta} \in (0, 1)^{|I|}$,

$$ (4) \qquad \lambda_d(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C \left( \prod_{j \in I} \delta_j \right)^{\alpha}, $$
with $\alpha + \frac{|P_I|}{|I|} (1 + \beta_2) \ge 2 + \beta_1$.

Then

$$ \lambda_d(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C \frac{\prod_{j \in I} \delta_j^{2+\beta_1}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta_2}}. $$

(b) Assume that for all $\theta_0 \in \mathbb{R}^d$ and all $I \subset \{1, \dots, d\}$ such that $\phi_I(e^{i\theta_0}) \in \mathbb{T}^{|I|}$ and $I$ is maximal with respect to this property, there exist $C > 0$ and $O$ a neighbourhood of $\theta_0$ in $\mathbb{R}^d$ such that for all $\eta \in \mathbb{T}^{|I|}$ and all $\bar{\delta} \in (0, 1)^{|I|}$,

$$ (5) \qquad \lambda_d(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C \left( \prod_{j \in I} \delta_j \right)^{\alpha}, $$
with $\alpha + \frac{|P_I|}{|I|} (1 + \beta_2) \ge 2 + \beta_1$. Then $C_\phi : A_{\beta_1}^2(\mathbb{D}^d) \to A_{\beta_2}^2(\mathbb{D}^d)$ is bounded.

*Proof.* We need to prove that under the assumptions of the corollary,

$$ \left( \prod_{j \in I} \delta_j \right)^{\alpha} \le \frac{\prod_{j \in I} \delta_j^{2+\beta_1}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta_2}}. $$
Taking into account our assumption on $\alpha$, it suffices to prove that

$$ \prod_{k=1}^{d} \omega_{I,k}(\bar{\delta}) \leq \prod_{j=1}^{d} \delta_j^{\frac{|P_I|}{|I|}}. $$

This is precisely the content of Lemma 2.9. The second point now follows from the first one and an application of Theorem 2.10. $\square$

**Corollary 2.12.** Let $\beta_1, \beta_2 \geq -1$. Let $\phi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}^d) \cap \mathcal{C}^1(\bar{\mathbb{D}}^d)$. Assume that there exist $c > 0$, $I \subset \{1, \dots, d\}$ such that $\phi_I$ touches the polycircle and $\eta \in \mathbb{T}^{|I|}$ such that for all $\delta \in (0, 1)$,

$$ \lambda_d(\{\theta \in [-\pi, \pi)^d : |\phi_j(\theta) - \eta_j| < \delta, j \in I\}) \geq cf(\delta) $$

where

$$ \lim_{\delta \to 0} \frac{f(\delta)}{\delta^{|I|(2+\beta_1)-|P_I|(1+\beta_2)}} = +\infty. $$

Then $C_\phi: A_{\beta_1}^2(\mathbb{D}^d) \to A_{\beta_2}^2(\mathbb{D}^d)$ is not bounded. This holds in particular with $f(\delta) = \delta^{|I|}$ when $\alpha + \frac{|P_I|}{|I|}(1 + \beta_2) < 2 + \beta_1$.

**Proof.** When $\delta_1 = \dots = \delta_d = \delta$, inequality (3) reads

$$ \lambda_d(\{\theta \in [-\pi, \pi)^d : |\phi_j(\theta) - \eta_j| < \delta, j \in I\}) \leq C \delta^{|I|(2+\beta_1)-|P_I|(1+\beta_2)}. $$

Therefore Corollary 2.12 follows from the contrapositive statement of 1) $\implies$ 2) in Theorem 2.10. $\square$

## 2.4. Applications.
We now develop some applications of Theorem 2.10 and its corollaries. We start with a stability result about the continuity of composition operators:

**Theorem 2.13.** Let $\phi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}^d) \cap \mathcal{C}^1(\bar{\mathbb{D}}^d)$ and $\beta_1, \beta_2 \geq -1$. Assume that $C_\phi$ maps boundedly $A_{\beta_1}^2(\mathbb{D}^d)$ into $A_{\beta_2}^2(\mathbb{D}^d)$. Then for all $\beta'_1 \geq \beta_1$, $C_\phi$ maps boundedly $A_{\beta'_1}^2(\mathbb{D}^d)$ into $A_{\beta'_2}^2(\mathbb{D}^d)$ where

$$ \beta'_2 = \frac{\beta'_1(\beta_2 + 2) + 2(\beta_2 - \beta_1)}{\beta_1 + 2}. $$

In particular, if $C_\phi$ is bounded on $A_{\beta_1}^2(\mathbb{D}^d)$, then it is bounded on $A_{\beta'_1}^2(\mathbb{D}^d)$.

In order to prove Theorem 2.13, we first need a lemma.

**Lemma 2.14.** Let $\phi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}^d) \cap \mathcal{C}^1(\bar{\mathbb{D}}^d)$ and let $\beta_1, \beta_2 \geq -1$. Assume that $C_\phi$ maps boundedly $A_{\beta_1}^2(\mathbb{D}^d)$ into $A_{\beta_2}^2(\mathbb{D}^d)$. Then there exists $D > 0$ such that, for all $I \subset \{1, \dots, d\}$ such that $\phi_I$ touches the polycircle,

$$ \prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{2+\beta_2} \leq D \prod_{j \in I} \delta_j^{2+\beta_1}. $$

**Proof.** Assume first that $\beta_2 > -1$. Let $I \subset \{1, \dots, d\}$ and $\theta_0 \in \mathbb{R}^d$ with $\phi_I(e^{i\theta_0}) \in \mathbb{T}^{|I|}$. We may assume $e^{i\theta_0} = e$. We first show that for all $\bar{\delta} \in (0, 1)^{|I|}$,

$$ E := \{((1 - \rho_k)e^{i\theta_k}) : 0 \leq \rho_k \leq \omega_{I,k}(\bar{\delta}), |\theta_k| \leq \omega_{I,k}(\bar{\delta}), k \in P_I\} \subset \phi_I^{-1}(S_I(\phi_I(e), C_1\bar{\delta})) $$

with $C_1 = 2 \sup_l \|\nabla \phi_l\|$. Indeed, for $j \in I$,

$$ \begin{aligned} |\phi_j(z) - \phi_j(e)| &\leq \|\nabla \phi_j\| \max_{k \in P_j} |z_k - 1| \\ &\leq (2 \sup_l \|\nabla \phi_l\|) \delta_j \end{aligned} $$
by the mean value inequality. We extend $\bar{\delta}$ to $(0, 2]^d$ by setting $\delta_j = 2$ if $j \notin I$, so that

$$E \subset \phi^{-1}(S(\phi(e), C_1\bar{\delta})).$$

Since $C_\phi$ maps boundedly $A_{\beta_1}^2(\mathbb{D}^d)$ into $A_{\beta_2}^2(\mathbb{D}^d)$, we know that there exists $C_2 > 0$ such that

$$V_{\beta_2}(\phi^{-1}(S(\phi(e), C_1\bar{\delta}))) \le C_2 \prod_{j \in I} \delta_j^{2+\beta_1}.$$

We get the result since Lemma 2.4 implies

$$V_{\beta_2}(E) \gtrsim \left( \prod_{k=1}^{d} \omega_{I,k}(\bar{\delta}) \right)^{2+\beta_2}.$$

The result remains true for $\beta_2 = -1$ by replacing $V_{\beta_2}$ by $\lambda_d$. $\square$

*Proof of Theorem 2.13.* In view of Theorem 2.10, we only have to show that for all $I \subset \{1, \dots, d\}$ such that $\phi_I$ touches the polycircle, there exists $C > 0$ such that, for all $\bar{\delta} \in (0, 1)^{|I|}$,

$$\frac{\prod_{j \in I} \delta_j^{2+\beta_1}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta_2}} \le C \frac{\prod_{j \in I} \delta_j^{2+\beta'_1}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta'_2}}.$$

Let $D > 0$ be given by the previous lemma. Then we write, setting $t = (\beta'_1 + 2)/(\beta_1 + 2) \ge 1$,

$$\begin{align*}
\frac{\prod_{j \in I} \delta_j^{2+\beta_1}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta_2}} &= \prod_{k=1}^{d} \omega_{I,k}(\bar{\delta}) \times D^{-1} \times \left( \frac{D \prod_{j \in I} \delta_j^{2+\beta_1}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{2+\beta_2}} \right) \\
&\le \prod_{k=1}^{d} \omega_{I,k}(\bar{\delta}) \times D^{-1} \times \left( \frac{D \prod_{j \in I} \delta_j^{2+\beta_1}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{2+\beta_2}} \right)^t \\
&\le D^{t-1} \frac{\prod_{j \in I} \delta_j^{2t+\beta_1}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{2t-1+t\beta_2}} \\
&\le D^{t-1} \frac{\prod_{j \in I} \delta_j^{2+\beta'_1}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta'_2}}.
\end{align*}$$

$\square$

Part of the previous result was announced in [8, Corollary 6], without any assumption of regularity at the boundary of $\phi$. However, the proof of [8] is not correct, since it uses Theorem C of [9] with a measure which does not satisfy the assumptions of this last theorem.

In the particular case of the tridisc, Theorem 2.13 can be used in conjunction with Theorem 10 of [12] to lead to the following result.

**Theorem 2.15.** Let $\phi \in \mathcal{O}(\mathbb{D}^3, \mathbb{D}^3) \cap \mathcal{C}^2(\bar{\mathbb{D}}^3)$. The following propositions are equivalent:

(1) $C_\phi$ is bounded on $A^2(\mathbb{D}^3)$.
(2) For all $\beta \ge 0$, $C_\phi$ is bounded on $A_\beta^2(\mathbb{D}^3)$.

(3) There exists $\beta \ge 0$ such that $C_\phi$ is bounded on $A_\beta^2(\mathbb{D}^3)$.

(4) For any $\xi \in \mathbb{T}^3$ such that $\phi(\xi) \in \mathbb{T}^3$, $d\phi(\xi)$ is invertible and for any $1 \le i_1 < i_2 \le 3$ and any $\xi \in \mathbb{T}^3$ such that $(\phi_{i_1}(\xi), \phi_{i_2}(\xi)) \in \mathbb{T}^2$, either

(a) $\nabla\phi_{i_1}(\xi), \nabla\phi_{i_2}(\xi)$ are linearly independent, or

(b) $\frac{\partial\phi_{i_k}}{\partial z_j}(\xi) \neq 0$ for $j = 1, 2, 3$ and $k = 1, 2$.

Proof. 1) ⇒ 2) This is exactly the content of Theorem 2.13.

2) ⇒ 3) Obvious.

3) ⇒ 4) To prove this implication, we just need to check if what was done in [12] can be extended to $\beta \ge 0$. Let $\beta \ge 0$ and assume that $C_\phi$ is bounded on $A_\beta^2(\mathbb{D}^3)$. Let $\xi \in \mathbb{T}^3$. If $\phi(\xi) \in \mathbb{T}^3$, the necessary condition comes from [11]. If $\phi_I(\xi) \in \mathbb{T}^2$ for some $I \subset \{1, 2, 3\}$ with $|I| = 2$, and if the necessary condition is not satisfied, the work of [12] shows that one can assume that there exist $C > 0$, $\eta \in \mathbb{T}^3$ and $\alpha_1, \alpha_2 \in \mathbb{C}$ such that, for all $\delta \in (0, 1)$, setting $\bar{\delta} = (\delta, \delta, 1)$,

$$ \{z \in \mathbb{D}^3 : |\alpha_1(z_1 - 1) + \alpha_2(z_2 - 1)| < C\delta\} \subset \phi^{-1}(S(\eta, \bar{\delta})). $$

Now, $V_\beta(\{z \in \mathbb{D}^3 : |\alpha_1(z_1 - 1) + \alpha_2(z_2 - 1)| < C\delta\}) \gtrsim \delta^{2\beta+7/2}$ by [12, Proposition 5] whereas $V_\beta(S(\eta, \bar{\delta})) \simeq \delta^{2\beta+4}$. Hence, $C_\phi$ cannot be bounded on $A_\beta^2(\mathbb{D}^3)$.

4) ⇒ 1) This is Theorem 10 of [12]. □

As another application of Theorem 2.10, we give an enhancement of [13, Theorem A] under the supplementary assumption that $\phi \in C^1(\bar{\mathbb{D}}^d)$.

**Theorem 2.16.** Let $\phi \in O(\mathbb{D}^d, \mathbb{D}) \cap C^1(\bar{\mathbb{D}}^d)$ such that $\phi(\bar{\mathbb{D}}^d) \cap \partial\mathbb{D}^2 \neq \emptyset$, let $\beta \ge -1$ and let $d_\phi = \max(|I| : I \subset \{1, \dots, d\}, \phi_I(\mathbb{T}^d) \cap \mathbb{T}^{|I|} \neq \emptyset)$. Then $C_\phi$ maps boundedly $A_\beta^2(\mathbb{D}^d)$ into $A_{d_\phi(\beta+2)-2}^2(\mathbb{D}^d)$. In particular, $C_\phi$ maps boundedly $A_\beta^2(\mathbb{D}^d)$ into $A_{d_\phi(\beta+2)-2}^2(\mathbb{D}^d)$.

The proof will need the following easy lemma.

**Lemma 2.17.** Let $\varphi \in O(\mathbb{D}^d, \mathbb{D}) \cap C^1(\bar{\mathbb{D}}^d)$ and let $\theta_0 \in \mathbb{R}^d$ be such that $\frac{\partial\varphi}{\partial z_k}(\theta_0) \neq 0$ for some $k \in \{1, \dots, d\}$. There exist a neighbourhood $O$ of $\theta_0$ in $\mathbb{R}^d$ and $C > 0$ such that, for all $\eta \in \mathbb{C}$ and all $\delta > 0$,

$$ \lambda_d(\{\theta \in O : |\varphi(\theta) - \eta| < \delta\}) \le C\delta. $$

Proof. To fix the notations we assume that $k = 1$ so that $\frac{\partial\varphi}{\partial z_1}(\theta_0) > 0$. There exist a bounded neighbourhood $O = O_1 \times O_1'$ with $O_1 \subset \mathbb{R}$ of $\theta_0$ and $c > 0$ such that, for all $\theta \in O$, $\frac{\partial\varphi}{\partial z_1}(\theta) \ge c$. This implies that, for all $(\theta_2, \dots, \theta_d) \in O_1'$ and all $\theta_1, \theta_1' \in O_1$,

$$ |\varphi(\theta_1, \theta_2, \dots, \theta_d) - \varphi(\theta_1', \theta_2, \dots, \theta_d)| \ge c|\theta_1 - \theta_1'|. $$

This yields that, given $(\theta_2, \dots, \theta_d) \in O_1'$, $\{\theta_1 \in O_1 : |\varphi(\theta_1, \dots, \theta_d) - \eta| < \delta\}$ is contained in an interval of length $4\delta/c$. The lemma then follows from an application of Fubini's theorem. □

**Proof of Theorem 2.16.** We show that Condition 3) of Theorem 2.10 is satisfied. Let $\theta_0 \in \mathbb{R}^d$ and $I \subset \{1, \dots, d\}$ maximal be such that $\phi_I(\theta_0) \in \mathbb{T}^{|I|}$. We assume that $I = \{1, \dots, q\}$ with $q \le d_\phi$ and let $O, C > 0$ be given by Lemma 2.17 and working for $\phi_1, \dots, \phi_q$. Let $\eta \in \mathbb{T}^{|I|}$, $\bar{\delta} \in (0, 1)^{|I|}$ and let us assume that $\delta_1 \le \dots \le \delta_q$. Then since $P_1 \neq \emptyset$, at least one $\omega_{I,k}(\bar{\delta})$ is less
than or equal to $\delta_1$ so that

$$ \begin{aligned} \frac{\prod_{j \in I} \delta_j^{2+\beta}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{d_{\phi}(\beta+2)-1}} &\ge \frac{\prod_{j \in I} \delta_j^{2+\beta}}{\delta_1^{d_{\phi}(\beta+2)-1}} \\ &\ge \delta_1^{(q-d_{\phi})(\beta+2)+1} \\ &\ge \delta_1. \end{aligned} $$

On the other hand,

$$ \begin{aligned} \lambda_d(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) &\le \lambda_d(\{\theta \in O : |\phi_1(\theta) - \eta_1| < \delta_1\}) \\ &\le C\delta_1. \end{aligned} $$

This shows that $C_{\phi}$ maps boundedly $A_{\beta}^2(\mathbb{D}^d)$ into $A_{d_{\phi}(\beta+2)-2}^2(\mathbb{D}^d)$. $\square$

Notice that the index $\beta' = d_{\phi}(\beta + 2) - 2$ for $C_{\phi}: A_{\beta}^2(\mathbb{D}^d) \to A_{\beta'}^2(\mathbb{D}^d)$ to be bounded is optimal. Indeed, consider the function $\phi \in O(\mathbb{D}^d, \mathbb{D}^d) \cap C^1(\bar{\mathbb{D}}^d)$ defined by $\phi(z) = (z_1, \dots, z_1, 0, \dots, 0)$. For $j = 1, \dots, d_{\phi}$, we can write $\phi_j(\theta) = 1 + i\theta_1 + O(\theta_1^2)$. Thus, for $\delta > 0$ small enough,

$$ [-\delta/2, \delta/2] \times [-\pi, \pi)^{d-1} \subset \{\theta \in [-\pi, \pi]^d : |\phi_j(\theta) - 1| < \delta, j = 1, \dots, d_{\phi}\} $$

and $\lambda_d([-\delta, \delta] \times [-\pi, \pi)^{d-1}) \gtrsim \delta$. Applying Corollary 2.12, we find that $C_{\phi}: A_{\beta}^2(\mathbb{D}^d) \to A_{\beta'}^2(\mathbb{D}^d)$ is not bounded for $\beta' < d_{\phi}(2 + \beta) - 2$.

We finish this subsection with a class of symbols which will be used several times throughout the paper.

**Theorem 2.18.** Let $\beta_1, \beta_2 \ge -1$. Let $\varphi \in O(\mathbb{D}, \mathbb{D}) \cap C^1(\bar{\mathbb{D}})$. Assume that, for any $z \in \bar{\mathbb{D}}$, $|\varphi(z)| = 1$ if and only if $z = 1$ and that there exist $\kappa > 1$, $c > 0$ such that

$$ |\varphi(e^{i\theta})| = 1 - c|\theta|^{\kappa} + o(|\theta|^{\kappa}). $$

Let $d \ge k \ge 1$ and let us set

$$ \phi(z) = (\varphi(z_1) \cdots \varphi(z_k)z_{k+1} \cdots z_d, \dots, \varphi(z_1) \cdots \varphi(z_k)z_{k+1} \cdots z_d, 0, \dots, 0) $$

where the first $q$ coordinates, $q \in \{1, \dots, d\}$, are identical. Then $C_{\phi}$ maps boundedly $A_{\beta_1}^2(\mathbb{D}^d)$ into $A_{\beta_2}^2(\mathbb{D}^d)$ if and only if

$$ d\beta_2 \ge \left( 2q - d - 1 - \frac{k}{\kappa} \right) + q\beta_1. $$

*Proof.* By definition of $\phi$, for $j \in \{1, \dots, q\}$ and $e^{i\theta} \in \mathbb{T}^d$, we know that

$$ \phi_j(e^{i\theta}) = e^{i(\theta_{k+1} + \dots + \theta_d)} \varphi(e^{i\theta_1}) \dots \varphi(e^{i\theta_k}). $$

Doing the linear change of variables $u_d = \theta_{k+1} + \dots + \theta_d$, $u_1 = \theta_1, \dots, u_{d-1} = \theta_{d-1}$, we get $\phi_j(e^{iu_j}) = e^{iu_d}\varphi(e^{iu_1}) \dots \varphi(e^{iu_k})$. We will apply the second point of Corollary 2.11 and Corollary 2.12 with $|I| = q$ and $|P_I| = d$. We shall estimate the $\lambda_d$-volumes with respect to the new coordinates $u_1, \dots, u_d$ which does not change the conclusion.

First, we focus on the discontinuity of the operator. Let $\delta \in (0, 1)$. Let $u_j$ be such that $|u_j|^{\kappa} \le \frac{\delta}{c}$, $j = 1, \dots, k$. Then, provided $\delta$ is small enough, $|\varphi(e^{iu_j})| \ge 1 - 2\delta$. We write $\varphi(e^{iu_1}) \cdots \varphi(e^{iu_k}) = \rho e^{i\alpha}$ with $\rho \ge 1 - 2d\delta$ and $\alpha \in [-\pi, \pi)$. Now, consider $I_{u_1, \dots, u_k} = [-\alpha -
$\delta, -\alpha + \delta] \cap [-\pi, \pi)$. For $u_d \in I_{u_1, \dots, u_k}$, we write $u_d = -\alpha + \gamma$, with $|\gamma| < \delta$ and by the triangle inequality, we get

$$|\phi_j(e^{iu}) - 1| = |e^{i(-\alpha+\gamma)} \rho e^{i\alpha} - 1| \le |\rho e^{i\gamma} - \rho| + |\rho - 1| \le (2d+1)\delta.$$

Hence, $E := \{u \in [-\pi, \pi)^d : u_1, \dots, u_k \in [-(\frac{\delta}{c})^{1/\kappa}, (\frac{\delta}{c})^{1/\kappa}], u_d \in I_{u_1, \dots, u_k}, (u_{k+1}, \dots, u_{d-1}) \in [-\pi, \pi)^{d-1-k}\}$ is contained in $\{u \in [-\pi, \pi)^d : |\phi_j(e^{iu}) - 1| < (2d+1)\delta \text{ for all } j = 1, \dots, q\}$ so that

$$\lambda_d(E) \gtrsim \delta^{1+k/\kappa}.$$

Thus, applying Corollary 2.12, we get that $C_\phi$ is not bounded if

$$\frac{1}{q} \left(1 + \frac{k}{\kappa}\right) + \frac{d}{q} (1 + \beta_2) < 2 + \beta_1 \Leftrightarrow d\beta_2 < \left(2q - d - 1 - \frac{k}{\kappa}\right) + q\beta_1.$$

Conversely, let us study the continuity of the operator. The only maximal $I$ for which $\phi_I(e^{i\theta_0}) \in T^{|I|}$ for some $\theta_0 \in \mathbb{R}^d$ is $I = \{1, \dots, q\}$. Let $\eta \in T^q$ and $\bar{\delta} \in (0, 1)^q$. Let $j \in \{1, \dots, q\}$. We can notice that $|\phi_j(e^{iu}) - \eta_j| < \delta_j$ implies that for $l = 1, \dots, k$,

$$1 - c|u_l|^\kappa + o(|u_l|^\kappa) = |\varphi(e^{iu_l})| \ge 1 - \delta_j.$$

Hence there exists a neighbourhood $O$ of $0$ in $\mathbb{R}^d$ such that, if $u \in O$ and $e^{iu} \in \phi_I^{-1}(S(\eta, \bar{\delta}))$, then $|u_l| \lesssim \min(\delta_j)^{1/\kappa}$, $l = 1, \dots, k$. Now, fix $u_1, \dots, u_k$ such that $|u_l| \lesssim \min(\delta_j)^{1/\kappa}$, $l = 1, \dots, k$, and write $\varphi(e^{iu_1}) \cdots \varphi(e^{iu_k}) = \rho e^{i\alpha}$. We have

$$\begin{aligned} |\phi_j(e^{iu}) - \eta_j| < \delta_j &\Rightarrow |\rho e^{i(u_d+\alpha)} - \eta_j| < \delta_j \\ &\Rightarrow \left|e^{iu_d} - \frac{\eta_j e^{-i\alpha}}{\rho}\right| < \frac{\delta_j}{\rho} \\ &\Rightarrow \left|\sin(u_d) - a(\eta_j, u_1, \dots, u_k)\right| < 2\delta_j \end{aligned}$$

provided $\delta_j \le 1/2$ which implies $\rho \ge 1/2$. Shrinking $O$ if necessary, we may assume that for all $u \in O$, $(\sin)'(u_d) = \cos(u_d) \ge 1/2$. Hence, by the triangle inequality, $u_d$ belongs to an interval $I(\eta_j, u_1, \dots, u_k)$ whose length is less than $8\delta_j$. Finally, we have found a neighbourhood $O$ of $0$ in $\mathbb{R}^d$ such that

$$\lambda_d(\{u \in O : |\phi_j(e^{iu}) - \eta_j| < \delta_j, j \in I\}) \lesssim \min(\delta_j)^{1+\frac{k}{\kappa}} \le \left(\prod_{j \in I} \delta_j\right)^{\frac{1}{q}\left(\frac{k}{\kappa}+1\right)}.$$

Applying the second point of Corollary 2.11, we get that $C_\phi$ maps continuously $A_{\beta_1}^2(\mathbb{D}^d)$ into $A_{\beta_2}^2(\mathbb{D}^d)$ as soon as

$$d\beta_2 \ge 2q - d - 1 - \frac{k}{\kappa} + q\beta_1.$$

□

The tridisc case for $q = 2$ and $k = 1$ writes:

**Corollary 2.19.** Let $\varphi \in \mathcal{O}(\mathbb{D}, \mathbb{D}) \cap \mathcal{C}^1(\bar{\mathbb{D}})$. Assume that, for any $z \in \bar{\mathbb{D}}$, $|\varphi(z)| = 1$ if and only if $z = 1$ and that there exist $\kappa > 1$, $c > 0$ such that

$$|\varphi(e^{i\theta})| = 1 - c|\theta|^\kappa + o(|\theta|^\kappa).$$

We set $\phi(z) = (\varphi(z_1)z_2z_3, \varphi(z_1)z_2z_3, 0)$. Then $C_\phi$ is bounded on $A_\beta^2(\mathbb{D}^3)$ if and only if $\beta \ge -1/\kappa$.
### 3. N-TH ORDER DERIVATIVES ARE NOT ENOUGH

In this section, we show that the knowledge of the N-th order derivatives at the boundary points of the symbol is not sufficient to know if the induced composition operator is continuous on some $A_{\beta}^2(\mathbb{D}^d)$. We first need to exhibit two functions which have the same first derivatives and which have a different behaviour. We fix $n \ge 1$ and we consider the two functions $g_n$ and $h_n$ defined by

$$ 
\begin{aligned}
g_n(z) &= z^n - \frac{1}{2^n} \left( z - \left( \frac{1+z^2}{2} \right) \right)^n \\
h_n(z) &= z^{2n} - \frac{z^n}{2^n} \left( z - \left( \frac{1+z^2}{2} \right) \right)^n .
\end{aligned}
$$

**Fact 1.** $g_n(\bar{\mathbb{D}}) \subset \bar{\mathbb{D}}$, $h_n(\bar{\mathbb{D}}) \subset \bar{\mathbb{D}}$ and $|g_n(z)| = 1 \Leftrightarrow |h_n(z)| = 1 \Leftrightarrow z = 1$.

*Proof.* Let $\theta \in \mathbb{R}$. We have

$$ g_n(e^{i\theta}) = e^{in\theta} - \frac{1}{2^n} \left( e^{i\theta} - \frac{1+e^{2i\theta}}{2} \right)^n = e^{in\theta} \left( 1 - \left( \frac{1-\cos(\theta)}{2} \right)^n \right). $$

Thus, $|g_n(e^{i\theta})| = \left| 1 - \left( \frac{1-\cos(\theta)}{2} \right)^n \right|$. However, $0 \le \frac{1-\cos(\theta)}{2} \le 1$ so that $|g_n(e^{i\theta})| \le 1$. Moreover,

$$ |g_n(e^{i\theta})| = 1 \Leftrightarrow \frac{1-\cos(\theta)}{2} = 0 \Leftrightarrow \theta = 0 \quad [2\pi]. $$

Finally, $g_n(\bar{\mathbb{D}}) \subset \bar{\mathbb{D}}$ and $|g_n(z)| = 1 \Leftrightarrow z = 1$. We deduce the results for $h_n$ if we observe that $h_n(z) = z^n g_n(z)$. $\square$

**Fact 2.** For all $k \in \{1, \dots, n\}$, we have $g_{2n}^{(k)}(1) = h_n^{(k)}(1) = 2n \cdots (2n-k+1)$.

*Proof.* Setting $u_n(z) = \left(z - \left(\frac{1+z^2}{2}\right)^n\right)^n$, this fact easily follows from Leibniz's formula if we are able to prove that $u_n^{(k)}(1) = 0$ for all $k = 0, \dots, n$. This last property can be proved by induction, using $u_{n+1}(z) = u_n(z)u_1(z)$. $\square$

We now prove a variant of Theorem 1.1 where we fix $\beta$ and we allow the dimension to change.

**Theorem 3.1.** Let $\beta \ge -1$ and $n \in \mathbb{N}$. There exist $d \ge 3$, $\phi, \psi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}^d) \cap \mathcal{C}^\infty(\bar{\mathbb{D}}^d)$ such that

* $\mathcal{E}(\phi) = \mathcal{E}(\psi)$.

* For any $\xi \in \mathcal{E}(\phi)$, for any $\alpha \in \mathbb{Z}_+^d$ with $|\alpha| \le n$, $\partial^\alpha \phi(\xi) = \partial^\alpha \psi(\xi)$.

* $C_\phi$ is continuous on $A_\beta^2(\mathbb{D}^d)$ whereas $C_\psi$ is not.

*Proof.* Let us consider $(a, b) \in \mathbb{N} \times \mathbb{Z}$ such that

$$ 0 \le \beta - \frac{b}{a} < \frac{b}{a} $$

and $a > 4n$. We set $d = 2a + b + 3$, $q = a + b + 3$, $k = 4n$ and

$$ \phi(z) = (h_n(z_1) \cdots h_n(z_k)z_{k+1} \cdots z_d, \dots, h_n(z_1) \cdots h_n(z_k)z_{k+1} \cdots z_d, 0, \dots, 0) $$

$$ \psi(z) = (g_{2n}(z_1) \cdots g_{2n}(z_k)z_{k+1} \cdots z_d, \dots, g_{2n}(z_1) \cdots g_{2n}(z_k)z_{k+1} \cdots z_d, 0, \dots, 0) $$

where the first $q$ coordinates are identical. Observe that $d > k$ since $b \ge -a$. By Fact 1, $\phi$ and $\psi$ belong to $\mathcal{O}(\mathbb{D}^d, \mathbb{D}^d) \cap \mathcal{C}^\infty(\bar{\mathbb{D}}^d)$ and $\mathcal{E}(\phi) = \mathcal{E}(\psi) = \{(1, \dots, 1)\} \times \mathbb{T}^{d-k}$. Moreover, Fact 2 yields
$\partial^{\alpha}\phi(\xi) = \partial^{\alpha}\psi(\xi)$ for all $|\alpha| \le n$ and all $\xi \in \mathcal{E}(\phi)$.

To discuss the continuity of $C_{\phi}$ and $C_{\psi}$, we observe that

$$|h_n(e^{i\theta})| = \left| 1 - \left( \frac{1 - \cos(\theta)}{2} \right)^n \right| = 1 - \frac{\theta^{2n}}{4^n} + O(\theta^{2n})$$

and

$$|g_{2n}(e^{i\theta})| = \left| 1 - \left( \frac{1 - \cos(\theta)}{2} \right)^{2n} \right| = 1 - \frac{\theta^{4n}}{4^{2n}} + O(\theta^{4n}).$$

Now we get

$$(d-q)\beta - \left(2q - d - 1 - \frac{4n}{2n}\right) = a\beta - b \ge 0$$

whereas

$$(d-q)\beta - \left(2q - d - 1 - \frac{4n}{4n}\right) = a\beta - b - 1 < 0.$$

Therefore Theorem 2.18 yields the continuity of $C_{\phi}$ on $A_{\beta}^{2}(\mathbb{D}^{d})$ and the discontinuity of $C_{\psi}$ on $A_{\beta}^{2}(\mathbb{D}^{d})$. $\square$

The proof of Theorem 1.1 exactly follows the same lines. We consider $\phi$ and $\psi$ defined for all $z \in \mathbb{D}^3$ by

$$\phi(z) = (h_n(z_1)z_2z_3, h_n(z_1)z_2z_3, 0) \quad \text{and} \quad \psi(z) = (g_{2n}(z_1)z_2z_3, g_{2n}(z_1)z_2z_3, 0).$$

Applying Corollary 2.19, we get that $C_{\phi}$ is bounded on $A_{\beta}^{2}(\mathbb{D}^{3})$ if and only if $\beta \ge \frac{-1}{2n}$ and $C_{\psi}$ is bounded on $A_{\beta}^{2}(\mathbb{D}^{3})$ if and only if $\beta \ge \frac{-1}{4n}$. In particular, for $\beta \in [\frac{-1}{2n}, \frac{-1}{4n})$, we have that $C_{\phi}$ is bounded on $A_{\beta}^{2}(\mathbb{D}^{3})$ whereas $C_{\psi}$ is not.

Theorems 1.1 and 3.1 lead to the following natural question.

**Question 3.2.** Let $d \ge 3$ and $n \in \mathbb{N}$. For which values of $\beta \ge -1$ do there exist $\phi, \psi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}^d) \cap \mathcal{C}^n(\overline{\mathbb{D}}^d)$ such that

* $\mathcal{E}(\phi) = \mathcal{E}(\psi)$,

* for any $\xi \in \mathcal{E}(\phi)$, for any $\alpha \in \mathbb{Z}_+^d$ with $|\alpha| \le n$, $\partial^{\alpha}\phi(\xi) = \partial^{\alpha}\psi(\xi)$,

* $C_{\phi}$ is continuous on $A_{\beta}^{2}(\mathbb{D}^{d})$ whereas $C_{\psi}$ is not?

We do not know the answer to this question but we can at least prove that this holds for

$$\beta \in \bigcup_{q=1}^{d-1} \left[ \frac{1}{d-q} \left( 2q - 1 - d - \frac{d-1}{2n} \right), \frac{1}{d-q} (2q - 1 - d) \right)$$

Indeed, for $n \in \mathbb{N}$ and $p \ge 1$ we consider

$$H_n(z) = z^{n(p+1)} + \frac{z^{pn}}{2^n} \left( z + \left( \frac{1+z}{2} \right)^2 \right)^n = z^{pn} g_n(z).$$

Then, as it was done for $h_n$, we can show that $|H_n(z)| \le 1$ for all $z \in \mathbb{D}$ and $|H_n(z)| = 1$ if and only if $z = 1$. Moreover, in the same way as the proof of Fact 2, we can show that $H_n^{(k)}(1) = g_{(p+1)n}^{(k)}(1) = (p+1)n \cdots ((p+1)n - k + 1)$ for all $k \in \{1, \dots, n\}$. Then, consider

$$\phi(z) = (H_n(z_1) \cdots H_n(z_{d-1})z_d, \dots, H_n(z_1) \cdots H_n(z_{d-1})z_d, 0, \dots, 0)$$

$$\psi(z) = (g_{(p+1)n}(z_1) \cdots g_{(p+1)n}(z_{d-1})z_d, \dots, g_{(p+1)n}(z_1) \cdots g_{(p+1)n}(z_{d-1})z_d, 0, \dots, 0)$$
where the $q$-first coordinates are identical. Then, $\phi, \psi \in O(\mathbb{D}^d, \mathbb{D}^d) \cap C^n(\bar{\mathbb{D}}^d)$ with $\mathcal{E}(\phi) = \mathcal{E}(\psi)$ and $\partial^\alpha \phi(\xi) = \partial^\alpha \psi(\xi)$ for all $\xi \in \mathcal{E}(\phi)$ and all $\alpha \in \mathbb{Z}_+$ such that $|\alpha| \le n$. To study the continuity of $C_\phi$ and $C_\psi$, observe that

$$ 
\begin{aligned}
|H_n(e^{i\theta})| &= 1 - \frac{\theta^{2n}}{4^n} + O(\theta^{2n}) \\
|g_{(p+1)n}(e^{i\theta})| &= 1 - \frac{\theta^{2(p+1)n}}{4(p+1)n} + O(\theta^{2(p+1)n}).
\end{aligned}
$$

Applying Theorem 2.18, we get that $C_\phi : A_\beta^2(\mathbb{D}^d) \to A_\beta^2(\mathbb{D}^d)$ is bounded if and only if

$$ \beta \ge \frac{1}{d-q} \left( 2q - d - 1 - \frac{d-1}{2n} \right) $$

and $C_\psi : A_\beta^2(\mathbb{D}^d) \to A_\beta^2(\mathbb{D}^d)$ if and only if

$$ \beta \ge \frac{1}{d-q} \left( 2q - d - 1 - \frac{d-1}{2n(p+1)} \right). $$

Letting $p$ to $+\infty$, we find that for $\beta \in \bigcup_{q=1}^{d-1} \left[ \frac{1}{d-q} \left( 2q - d - 1 - \frac{d-1}{2n} \right), \frac{1}{d-q} (2q - d - 1) \right)$, there exist $\phi, \psi \in O(\mathbb{D}^d, \mathbb{D}^d) \cap C^n(\bar{\mathbb{D}}^d)$ with $\mathcal{E}(\phi) = \mathcal{E}(\psi)$ and $\partial^\alpha \phi(\xi) = \partial^\alpha \psi(\xi)$ for all $\xi \in \mathcal{E}(\phi)$ and all $\alpha \in \mathbb{Z}_+$ with $|\alpha| \le n$ such that $C_\phi$ is bounded on $A_\beta^2(\mathbb{D}^d)$ while $C_\psi$ is not.

## 4. TWO DIFFERENT WEIGHTS

In this section, we shall prove the following general version of Theorem 1.2.

**Theorem 4.1.** Let $d \ge 3$ and let $-1 \le \beta_1 < \beta_2 \le d-3$. Then there exist $\phi \in O(\mathbb{D}^d, \mathbb{D}^d) \cap C^1(\bar{\mathbb{D}}^d)$ such that $C_\phi$ is continuous on $A_{\beta_2}^2(\mathbb{D}^d)$ and $C_\phi$ is not continuous on $A_{\beta_1}^2(\mathbb{D}^d)$.

*Proof.* Assume first that $\beta_2 > d-4$. Let $\kappa > 1$ be such that $\beta_1 < d-3 - \frac{1}{\kappa} < \beta_2$. We set $\alpha = \kappa - \lfloor \kappa \rfloor$. Let $f : \mathbb{R} \to [1/2, 1]$ be such that

* $f$ is $C^{\lfloor \kappa \rfloor, \alpha}$-smooth on $\mathbb{R}$ and $2\pi$-periodic.

* $f(\theta) = 1 \Leftrightarrow \theta = 0$ $[2\pi]$.

* $f(\theta) \sim_0 |\theta|^\kappa$.

Now, consider $\Gamma$ the Jordan curve defined by the following equation

$$ r(\theta) = 1 - f(\theta). $$

We have $\Gamma \subset \bar{\mathbb{D}}$. Applying Riemann's and Kellogg-Warschawski's theorems, we get that there exists a $C^1$ function $\varphi : \bar{\mathbb{D}} \to \bar{\mathbb{D}}$ such that $\varphi : \mathbb{D} \to \mathbb{D}$ is holomorphic, $\varphi$ is a conformal map that maps $\mathbb{D}$ onto $\Delta$, where $\Delta$ is the interior of $\Gamma$, and $|\varphi(z)| = 1$ if and only if $z = 1$. For all $\theta \in \mathbb{R}$, we have $\varphi(e^{i\theta}) \in \Gamma$ so

$$ 1 - |\varphi(e^{i\theta})| = f(\arg(\varphi(e^{i\theta}))). $$

But, for $\theta$ close to 0, $\arg(\varphi(e^{i\theta})) = \arctan\left(\frac{\operatorname{Im}(\varphi(e^{i\theta}))}{\operatorname{Re}(\varphi(e^{i\theta}))}\right)$ and

$$ \varphi(e^{i\theta}) = 1 + \varphi'(1)(e^{i\theta} - 1) + o(\theta) = 1 + i\varphi'(1)\theta + o(\theta). $$

Thus, $\arg(\varphi(e^{i\theta})) = \arctan(\varphi'(1)\theta + o(\theta)) = \varphi'(1)\theta + o(\theta)$. We deduce that

$$ 1 - |\varphi(e^{i\theta})| \sim_0 \varphi'(1)|\theta|^\kappa \Rightarrow |\varphi(e^{i\theta})| = 1 - \varphi'(1)|\theta|^\kappa + o(|\theta|^\kappa). $$
Now, we consider $\phi : \mathbb{D}^d \to \mathbb{D}^d$ defined as in Theorem 2.18 with $q = d-1$, $k = 1$ and this specific $\varphi$. Since

$$d\beta_2 \geq 2q - d - 1 - \frac{1}{\kappa} + q\beta_2$$

and

$$d\beta_1 \leq 2q - d - 1 - \frac{1}{\kappa} + q\beta_1,$$

$C_\phi$ is the composition operator we were looking for.

If $\beta_2 \leq d - 4$, let $q \in [3, d - 1]$ be such that $q - 4 < \beta_2 \leq q - 3$. There exists $\psi \in O(\mathbb{D}^q, \mathbb{D}^q) \cap C^1(\bar{\mathbb{D}}^q)$ such that $C_\psi$ is bounded on $A_{\beta_2}^2(\mathbb{D}^q)$ and $C_\psi$ is not bounded on $A_{\beta_1}^2(\mathbb{D}^q)$. We come back to $\mathbb{D}^d$ by setting

$$\phi(z_1, \dots, z_d) = (\psi(z_1, \dots, z_q), \dots, \psi(z_1, \dots, z_q), 0, \dots, 0)$$

where the first $q$ coordinates are identical. $\square$

In particular, for $-1 \leq \beta_1 < \beta_2$, we can always distinguish continuity on $A_{\beta_1}^2(\mathbb{D}^d)$ and $A_{\beta_2}^2(\mathbb{D}^d)$ provided we allow the dimension to be large enough. It is interesting to know the lowest dimension $d$ for which this is possible. For $d = 3$, the bound $d - 3$ on $\beta_2$ is optimal by Theorem 2.15. This leads to the following question.

**Question 4.2.** Let $d \geq 4$ and let $\phi \in O(\mathbb{D}^d, \mathbb{D}^d) \cap C^1(\bar{\mathbb{D}}^d)$. Is it true that if $C_\phi$ is continuous on $A_\beta^2(\mathbb{D}^d)$ for some $\beta \geq d - 3$, then it is continuous on $A_\beta^2(\mathbb{D}^d)$ for all $\beta \geq d - 3$?

We can observe that if $\beta_1 < d - 3 \leq \beta_2$, there exists a composition operator $C_\phi$ which is bounded on $A_{\beta_2}^2(\mathbb{D}^d)$ and which is unbounded on $A_{\beta_1}^2(\mathbb{D}^d)$ (see [12, Corollary 6]), given by $\phi(z) = (z_1 \cdots z_d, \dots, z_1 \cdots z_d, 0)$.

## 5. ABOUT CONTINUITY ON THE TRIDISC

In this section, we focus on the continuity of the composition operator $C_\phi$ on the Bergman space $A_\beta^2(\mathbb{D}^3)$. We provide necessary and sufficient conditions for $C_\phi$ to be bounded on $A_\beta^2(\mathbb{D}^3)$. We begin by recalling the terminology of [2]. Let $\phi \in O(\mathbb{D}^3, \mathbb{D}^3) \cap C^3(\bar{\mathbb{D}}^3)$. Assume that there exist $I \subset \{1, 2, 3\}$ with $|I| = 2$, $I = \{i_1, i_2\}$ and $\xi \in \mathbb{T}^3$ such that $\phi_I(\xi) \in \mathbb{T}^2$ and such that $\nabla\phi_{i_1}(\xi)$, $\nabla\phi_{i_2}(\xi)$ are linearly dependent.

We can assume that $\xi = e = (1, 1, 1)$, $I = \{1, 2\}$ and $\phi_I(e) = (1, 1)$. For $z \in \mathbb{T}^3$, we write $z = (e^{i\theta_1}, e^{i\theta_2}, e^{i\theta_3})$ with $\theta_i \in \mathbb{R}$, $i = 1, 2, 3$. Thanks to a second-order Taylor expansion at $e$ for $\phi_j$, $j = 1, 2$, we obtain:

$$\phi_j(z) = 1 + \sum_{k=1}^{3} a_{j,k}(z_k - 1) + \sum_{k,l=1}^{3} \beta_{j,k,l}(z_k - 1)(z_l - 1) + O(|z_1 - 1|^3 + |z_2 - 1|^3 + |z_3 - 1|^3).$$

By Julia-Caratheodory's theorem, we know that $a_{j,k} \geq 0$ for $k = 1, 2, 3$ and $j = 1, 2$. At fixed $j$, all $a_{j,k}$ cannot be simultaneously equal to zero otherwise $\phi_j$ is constant which is not the case. Finally, by a Taylor expansion of $e^{i\theta_k}$, we get:

$$\begin{aligned} \mathrm{Re}\phi_j(\theta) &= 1 - \sum_{k=1}^{3} a_{j,k} \frac{\theta_k^2}{2} - \sum_{k,l=1}^{3} \mathrm{Re}(\beta_{j,k,l})\theta_k\theta_l + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3) \\ &= 1 - Q_j(\theta) + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3) \end{aligned}$$
where $Q_j$ is a quadratic form and

$$ \mathrm{Im}\phi_j(\theta) = \sum_{k=1}^{3} a_{j,k}\theta_k - \sum_{k,l=1}^{3} \mathrm{Im}(\beta_{j,k,l})\theta_k\theta_l + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3). $$

As $\nabla\phi_1(e)$ and $\nabla\phi_2(e)$ are linearly dependent and nonzero, we can write

$$ \mathrm{Im}\phi_j(\theta) = \kappa_j L(\theta) - \sum_{k,l=1}^{3} \mathrm{Im}(\beta_{j,k,l})\theta_k\theta_l + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3) $$

where $L$ is a fixed nonzero linear form and $\kappa_j \in \mathbb{R}^*$, $j = 1, 2$. Each $Q_j$ is a quadratic form of signature $(n_j, 0)$ so it can be written $Q_j = \sum_{k=1}^{n_j} L_{j,k}^2$ with $\mathrm{sign}(Q_1 + Q_2) = (m, 0)$. Moreover,

$$ \mathrm{span}(L_{j,k}, j = 1, 2, k = 1, \dots, n_j)^\perp = \ker(Q_1 + Q_2), $$

thus $\mathrm{dim}(\mathrm{span})^\perp = 3 - m$. We then consider $(L_1, \dots, L_m)$ a basis of $\mathrm{span}(L_{j,k})$. The linear form $L$ belongs to $\mathrm{span}(L_1, \dots, L_m)$. Indeed, if $L \notin \mathrm{span}(L_{j,k})$, then there exists $\theta \in \mathbb{R}^3$ such that $L(\theta) \neq 0$ but $Q_j(\theta) = 0$ for $j = 1, 2$. Therefore, for $\varepsilon$ small enough, we have

$$ \begin{aligned} |\phi_j(\varepsilon\theta)|^2 &= 1 - 2\varepsilon^2 Q_j(\theta) + \kappa_j^2\varepsilon^2 L(\theta)^2 + O(\varepsilon^3) \\ &= 1 + \kappa_j^2\varepsilon^2 L(\theta)^2 + O(\varepsilon^3) \\ &> 1, \end{aligned} $$

a contradiction. Then we complete $(L_1, \dots, L_m)$ to have $(L_1, L_2, L_3)$ a basis of $(\mathbb{R}^3)^*$ and we can write

$$ \mathrm{Im}\phi_j = \kappa_j \left( L + \sum_{\substack{1 \le k \le m \\ k \le l \le 3}} b_{j,k,l} L_k L_l + \sum_{m+1 \le k \le l \le 3} b_{j,k,l} L_k L_l \right) + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3). $$

We define the quadratic form $R$ by

$$ R = \kappa_2 \sum_{m+1 \le k \le l \le 3} b_{1,k,l} L_k L_l - \kappa_1 \sum_{m+1 \le k \le l \le 3} b_{2,k,l} L_k L_l. $$

Finally we set

$$ s(\phi, I, \xi) = m \text{ and } r(\phi, I, \xi) = \mathrm{sign}(R). $$

It is shown in [2] that $r(\phi, I, \xi)$ does not depend on the way we define the linear forms $L_j$. One can also notice that if we consider $\phi$ defined by $\phi = (\phi_1, \phi_1, 0)$ then $R = 0$, thus $r(\phi, I, \xi) = (0, 0)$.

The notations $s(\phi, I, \xi)$ and $r(\phi, I, \xi)$ were introduced in [2] in order to give a characterization of bounded composition operators on $H^2(\mathbb{D}^3)$. In this section we investigate how their knowledge helps us to determine whether $C_\phi$ is continuous on $A_\beta^2(\mathbb{D}^3)$ or not. Before stating the main result of this section, let us start by some results of boundedness and non-boundedness.

**Theorem 5.1.** Let $\phi \in \mathcal{O}(\mathbb{D}^3, \mathbb{D}^3) \cap \mathcal{C}^3(\bar{\mathbb{D}}^3)$. Assume that there exist $I \subset \{1, 2, 3\}$, with $|I| = 2$, $I = \{i_1, i_2\}$ and $\xi \in \mathbb{T}^3$ such that $\phi_I(\xi) \in \mathbb{T}^2$ and $\nabla\phi_{i_1}(\xi), \nabla\phi_{i_2}(\xi)$ are linearly dependent.

* If $|P_I| = 3$, then

(a) If $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (0, 0)$ then $C_\phi$ is not bounded on $A_\beta^2(\mathbb{D}^3)$ for $\beta < \frac{-2}{3}$,
(b) If $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (1,0)$ or $(0,1)$ then $C_\phi$ is not bounded on $A_\beta^2(\mathbb{D}^3)$
for $\beta < -\frac{5}{6}$.

• If $|P_I| \le 2$, then $C_\phi : A_\beta^2(\mathbb{D}^3) \to A_\beta^2(\mathbb{D}^3)$ is never bounded for any $\beta \ge -1$.

*Proof.* We intend to apply Corollary 2.12. Without loss of generality, we may assume that $I = \{1,2\}$, $\xi = (1,1,1)$ and $\phi_I(\xi) = (1,1)$. Because it only depends on what happens on the unit polycircle, making linear changes of variables on the coordinates $(\theta_1, \theta_2, \theta_3)$ will not change the volume estimates. Thus, in everything that follows, we will always assume that $L(\theta) = \theta_1$. We start by assuming that $|P_I| = 3$. First, suppose that $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (0,0)$. We can assume that $Q_j(\theta) = \gamma_j \theta_1^2$ with $\gamma_j > 0$, $j = 1,2$ because $\text{sign}(Q_1(\theta) + Q_2(\theta)) = 1$ and since, $L(\theta) = \theta_1$ and $s(\phi, I, \xi) = 1$, we can choose $L_2(\theta) = \theta_2$ and $L_3(\theta) = \theta_3$. We write for $j = 1,2$,

$$ \operatorname{Re}\phi_j(\theta) = 1 - \gamma_j \theta_1^2 + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3) $$

$$ \operatorname{Im}\phi_j(\theta) = \kappa_j (\theta_1 - a_j \theta_2^2 - b_j \theta_2 \theta_3 - c_j \theta_3^2) + O(\theta_1^2 + |\theta_1 \theta_2| + |\theta_1 \theta_3| + |\theta_2|^3 + |\theta_3|^3). $$

However, as

$$ R(\theta_2, \theta_3) = \kappa_2 \kappa_1 (-a_1 \theta_2^2 - b_1 \theta_2 \theta_3 - c_1 \theta_3^2) - \kappa_1 \kappa_2 (-a_2 \theta_2^2 - b_2 \theta_2 \theta_3 - c_2 \theta_3^2) = 0 $$

and $\kappa_2 \kappa_1 \neq 0$, we have

$$ a_1 = a_2 = a, \quad b_1 = b_2 = b \quad \text{and} \quad c_1 = c_2 = c. $$

Thus,

$$ \operatorname{Im}\phi_j(\theta) = \kappa_1(\theta_1 - a\theta_2^2 - b\theta_2\theta_3 - c\theta_3^2) + O(\theta_1^2 + |\theta_1\theta_2| + |\theta_1\theta_3| + |\theta_2|^3 + |\theta_3|^3) $$

$$ \operatorname{Re}\phi_j(\theta) = 1 - \gamma_j \theta_1^2 + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3). $$

Let $\delta \in (0,1)$ and let us consider $(\theta_2, \theta_3) \in [-\delta^{1/3}, \delta^{1/3}]^2$. For those $(\theta_2, \theta_3)$, we set $I_{\theta_2, \theta_3} := \{\theta_1: |\theta_1 - a\theta_2^2 - b\theta_2\theta_3 - c\theta_3^2| \le \delta\}$. For $\theta_1 \in I_{\theta_2, \theta_3}$, we can easily check that $|\theta_1| \lesssim \delta^{2/3}$ and we have

$$ |\operatorname{Im}\phi_j(\theta)| \lesssim \delta + O(\delta^{4/3} + \delta + \delta + \delta + \delta) \lesssim \delta $$

$$ |\operatorname{Re}\phi_j(\theta) - 1| \lesssim \delta^{4/3} + O(\delta^2 + \delta + \delta) \lesssim \delta. $$

Since

$$ \lambda_3 \left( \left\{ \theta \in \mathbb{R}^3 : (\theta_2, \theta_3) \in [-\delta^{1/3}, \delta^{1/3}]^2, \theta_1 \in I_{\theta_2, \theta_3} \right\} \right) \gtrsim \delta^{1+2/3}, $$

applying Corollary 2.12 with $|I| = 2$, $|P_I| = 3$ and $\alpha = \frac{5}{6}$, we get that $C_\phi$ is not bounded on $A_\beta^2(\mathbb{D}^3)$ for $\beta < -\frac{2}{3}$.

Now, suppose that $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (1,0)$ or $(0,1)$. $R$ is a quadratic form with signature equal to $(1,0)$ or $(0,1)$ so, doing a linear change of variables, we may assume that $R(\theta_2, \theta_3) = \pm \theta_3^2$. Thus, we write for $j = 1,2$,

$$ \operatorname{Re}\phi_j(\theta) = 1 - \gamma_j \theta_1^2 + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3) $$

$$ \operatorname{Im}\phi_j(\theta) = \kappa_j(\theta_1 - h_j \theta_1 \theta_2 - l_j \theta_1 \theta_3 - a_j \theta_2^2 - b_j \theta_2 \theta_3 - c_j \theta_3^2) + O(\theta_1^2 + |\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3). $$

However, as

$$ R(\theta_2, \theta_3) = \kappa_2 \kappa_1 (-a_1 \theta_2^2 - b_1 \theta_2 \theta_3 - c_1 \theta_3^2) - \kappa_1 \kappa_2 (-a_2 \theta_2^2 - b_2 \theta_2 \theta_3 - c_2 \theta_3^2) = \pm \theta_3^2 $$

and $\kappa_2 \kappa_1 \neq 0$, we have

$$ a_1 = a_2 = a, \quad b_1 = b_2 = b \quad \text{and} \quad c_1 - c_2 \neq 0. $$
Thus,

$$ \mathrm{Im}\phi_j(\theta) = \kappa_j(\theta_1 - h_j\theta_1\theta_2 - l_j\theta_1\theta_3 - a\theta_2^2 - b\theta_2\theta_3 - c_j\theta_3^2) + O(\theta_1^2 + |\theta_2|^3 + |\theta_3|^3) $$

$$ \mathrm{Re}\phi_j(\theta) = 1 - \gamma_j\theta_1^2 + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3). $$

Let $\delta \in (0, 1)$. We aim to prove that there exist $C > 0$ and $O$ a neighbourhood of 0 in $\mathbb{R}^3$ such that

$$ \lambda_3(\{\theta \in O : |\phi_j(\theta) - 1| \le \delta, j = 1, 2\}) \ge C\delta^{1+5/6}. $$

Let $U$ be defined on $\mathbb{R}^3$ by $U(\theta) = \left(\frac{\mathrm{Im}\phi_1}{\kappa_1}, \theta_2, \theta_3\right)$. There exist $O_1$ and $O_2$ two neighbourhoods of 0 in $\mathbb{R}^3$ and $C > 0$ such that $U$ is a diffeomorphism from $O_1$ into $O_2$ and

$$ (6) \qquad \left\{ \begin{array}{ll} |\det(dU)| & \le C \quad \text{on } O_1 \\ |\det(dU^{-1})| & \le C \quad \text{on } O_2. \end{array} \right. $$

As $dU(0) = I_3$, we have

$$ \mathrm{Im}\phi_1 \circ U^{-1}(u) = \kappa_1 u_1 $$

$$ \mathrm{Im}\phi_2 \circ U^{-1}(u) = \kappa_2(u_1 + \tilde{h}_2 u_1 u_2 + \tilde{l}_2 u_1 u_3 + c_2 u_3^2) + O(|u_1|^3 + |u_2|^3 + |u_3|^3) $$

$$ \mathrm{Re}\phi_j \circ U^{-1}(u) = 1 - \gamma_j u_1^2 + O(|u_1|^3 + |u_2|^3 + |u_3|^3). $$

Then, if $|u_1| \le \delta$, $|u_2| \le \delta^{1/3}$ and $|u_3| \le \delta^{1/2}$, we have for $j = 1, 2$,

$$ |\mathrm{Im}\phi_j \circ U^{-1}(u)| \lesssim \delta \quad \text{and} \quad |\mathrm{Re}\phi_j \circ U^{-1}(u) - 1| \lesssim \delta. $$

Thus, because of (6),

$$ \lambda_3(\{\theta \in O_1 : |\phi_j(\theta) - 1| < \delta, j \in I\}) \ge \lambda_3 ([-\delta, \delta] \times [-\delta^{1/3}, \delta^{1/3}] \times [-\delta^{1/2}, \delta^{1/2}]) \ge \delta^{1+5/6}. $$

So, applying Corollary 2.12, we get that $C_\phi$ is not bounded on $A_\beta^2(\mathbb{D}^3)$ for $\beta < -\frac{5}{6}$.

Now, assume that $|P_I| = 2$. We can write

$$ \mathrm{Re}\phi_j(\theta) = 1 - \gamma_j\theta_1^2 - \varepsilon_j\theta_2^2 - \Delta_j\theta_1\theta_2 + O(|\theta_1|^3 + |\theta_2|^3) $$

$$ \mathrm{Im}\phi_j(\theta) = \kappa_j(\theta_1 - g_j\theta_1^2 - h_j\theta_1\theta_2 - a_j\theta_2^2) + O(|\theta_1|^3 + |\theta_2|^3). $$

Let $\delta \in (0, 1)$. Consider $\theta_2 \in [-\delta^{1/2}, \delta^{1/2}]$ and $\theta_1 \in [-\delta, \delta]$. Then

$$ \lambda_3(\{\theta \in [-\pi, \pi)^3 : |\phi_j(\theta) - 1| < \delta, j \in I\}) \ge \lambda_3 ([-\delta, \delta] \times [-\delta^{1/2}, \delta^{1/2}] \times [-\pi, \pi)) \ge \delta^{1+1/2}. $$

Applying Corollary 2.12, we get that $C_\phi : A_\beta^2(\mathbb{D}^3) \to A_\beta^2(\mathbb{D}^3)$ is never bounded for any $\beta \ge -1$.

Finally, consider the case $|P_I| = 1$. We can write

$$ \mathrm{Re}\phi_j(\theta) = 1 - \gamma_j\theta_1^2 + O(|\theta_1|^3) \quad \text{and} \quad \mathrm{Im}\phi_j(\theta) = \kappa_j\theta_1 + O(\theta_1^2). $$

Let $\delta \in (0, 1)$. Then $[-\delta, \delta] \times [-\pi, \pi)^2 \subset \{\theta \in [-\pi, \pi)^3 : |\phi_j(\theta) - 1| < \delta, j = 1, 2\}$ and $\lambda_3([-\delta, \delta] \times [-\pi, \pi)^2) \ge \delta$. Applying Corollary 2.12, we get that $C_\phi$ is not bounded on $A_\beta^2(\mathbb{D}^3)$ for any $\beta \ge -1$. $\square$

**Theorem 5.2.** Let $\phi \in O(\mathbb{D}^3, \mathbb{D}^3) \cap C^3(\bar{\mathbb{D}}^3)$. Let $\theta_0 \in \mathbb{R}^3$ and $I \subset \{1, 2, 3\}$ with $|I| = 2$ and $I = \{i_1, i_2\}$ be such that $\phi_I(\theta_0) \in \mathbb{T}^2$ and $\nabla\phi_{i_1}(\theta_0), \nabla\phi_{i_2}(\theta_0)$ are linearly dependent. If $|P_I| = 3$, then there exist $C > 0$ and $O$ a neighbourhood of $\theta_0$ in $\mathbb{R}^3$ such that for all $\eta \in \mathbb{T}^2$ and all $\bar{\delta} \in (0, 1)^2$, we have

$$ \lambda_3(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C(\delta_{i_1} \delta_{i_2})^\alpha $$

with
(a) $\alpha = \frac{3}{4}$ if $s(\phi, I, e^{i\theta_0}) = 1$, $r(\phi, I, e^{i\theta_0}) \in \{(0,1), (1,0)\}$ or $s(\phi, I, e^{i\theta_0}) = 2$, $r(\phi, I, e^{i\theta_0}) = (0,0)$

(b) $0 < \alpha < 1$ if $s(\phi, I, e^{i\theta_0}) = 1$, $r(\phi, I, e^{i\theta_0}) = (1,1)$

(c) $\alpha = \frac{1}{2}$ if $s(\phi, I, e^{i\theta_0}) = 1$, $r(\phi, I, e^{i\theta_0}) = (0,0)$.

*Proof.* Without loss of generality, we can assume that $I = \{1,2\}$, $\theta_0 = (0,0,0)$ and $\phi_I(\theta_0) = (1,1)$. First, let us consider the case $s(\phi, I, e) = 1$ and $r(\phi, I, e) = (0,1)$ or $(1,0)$. We may argue as in the proof of Theorem 5.1 to write, for $j = 1,2$,

$$ \mathrm{Im}\phi_j(\theta) = \kappa_j(\theta_1 - g_j\theta_1^2 - h_j\theta_1\theta_2 - l_j\theta_1\theta_3 - a\theta_2^2 - b\theta_2\theta_3 - c_j\theta_3^2) + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3) $$

$$ \mathrm{Re}\phi_j(\theta) = 1 - \gamma_j\theta_1^2 + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3), $$

with $c_1 \neq c_2$. Let

$$ \begin{aligned} f(\theta) &= \kappa_2 \mathrm{Im}\phi_1(\theta) - \kappa_1 \mathrm{Im}\phi_2(\theta) \\ &= \alpha\theta_1^2 + \beta\theta_1\theta_2 + \gamma\theta_1\theta_3 + \kappa_1\kappa_2(c_2 - c_1)\theta_3^2 + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3). \end{aligned} $$

If $\delta_1 \le \delta_2$, let $U$ be defined on $\mathbb{R}^3$ by $U(\theta) = \left(\frac{\mathrm{Im}\phi_1}{\kappa_1}, \theta_2, \theta_3\right)$, let $O_1$ and $O_2$ be two neighbourhoods of $0$ in $\mathbb{R}^3$ and let $C > 0$ be such that $U$ is a diffeomorphism from $O_1$ into $O_2$ and (6) is satisfied. As $dU(0) = I_3$, we have

$$ \mathrm{Im}\phi_1 \circ U^{-1}(u) = \kappa_1 u_1 $$

$$ f \circ U^{-1}(u) = \tilde{\alpha}u_1^2 + \tilde{\beta}u_1u_2 + \tilde{\gamma}u_1u_3 + \kappa_1\kappa_2(c_2 - c_1)u_3^2 + O(|u_1|^3 + |u_2|^3 + |u_3|^3). $$

We observe that $f(0) = 0$, $d_{u_3}f(0) = 0$ and $d_{u_3}^2f(0) = 2\kappa_1\kappa_2(c_2 - c_1) \neq 0$ as $c_1 \neq c_2$. Hence, by the parametrized Morse's lemma 2.1, there exist a neighborhood $O'_2 \subset O_2$ of $0$ in $\mathbb{R}^3$, a diffeomorphism $V: O'_2 \to V(O'_2)$ defined by $V(u_1, u_2, u_3) = (u_1, u_2, v_3(u_1, u_2, u_3))$, a $C^1$-map $h: \mathbb{R}^2 \to \mathbb{R}$ and $C_1 > 0$ such that

$$ f \circ (V \circ U)^{-1}(v) = \pm v_3^2 + h(v_1, v_2) $$

and

$$ (7) \qquad \left\{ \begin{array}{l} |\det(dV)| \le C_1 \text{ on } O'_2 \\ |\det(dV^{-1})| \le C_1 \text{ on } V(O'_2). \end{array} \right. $$

By the definition of $V$, we still have $\mathrm{Im}\phi_1 \circ (V \circ U)^{-1}(v) = \kappa_1 v_1$. Let $O = U^{-1}(O'_2)$. For $\theta \in O \cap \phi_I^{-1}(S_I(\eta, \bar{\delta}))$, considering $v = V \circ U(\theta)$, we have

$$ |\kappa_1 v_1 - \mathrm{Im}\eta_1| \le \delta_1 \quad \text{and} \quad |\pm v_3^2 + h(v_1, v_2) - (\kappa_2 \mathrm{Im}\eta_1 - \kappa_1 \mathrm{Im}\eta_2)| \le \max(\delta_1, \delta_2) = \delta_2. $$

So, by (6), (7) and Fubini's theorem,

$$ \lambda_3(\phi_I^{-1}(S_I(\eta, \bar{\delta})) \cap O) \lesssim \lambda_3(V \circ U(O \cap \phi_I^{-1}(S_I(\eta, \bar{\delta})))) \lesssim \delta_1 \delta_2^{1/2} \lesssim (\delta_1 \delta_2)^{3/4}. $$

If $\delta_2 \le \delta_1$, we work with $U(\theta) = \left(\frac{\mathrm{Im}\phi_2}{\kappa_2}, \theta_2, \theta_3\right)$ and in the same way, we prove that

$$ \lambda_3(V \circ U(O \cap \phi_I^{-1}(S_I(\eta, \bar{\delta})))) \lesssim \delta_2 \delta_1^{1/2} \lesssim (\delta_1 \delta_2)^{3/4}. $$

Now, let us consider the case $s(\phi, I, e) = 2$ and $r(\phi, I, e) = (0,0)$. We write

$$ \mathrm{Re}\phi_j(\theta) = 1 - Q_j(\theta) + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3) \text{ and } \mathrm{Im}\phi_j(\theta) = \kappa_j L(\theta) + O(|\theta_1|^2 + |\theta_2|^2 + |\theta_3|^2) $$

where $Q_j$ is a quadratic form and $L$ is a nonzero linear form.

Let us consider

$$ f(\theta) = \mathrm{Re}\phi_1(\theta) + \mathrm{Re}\phi_2(\theta) = 2 - Q(\theta) + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3), $$
with $Q$ a quadratic form of signature $(2,0)$. Doing a linear change of variables, we may assume

$$ \mathrm{Im}\phi_1(\theta) = a_1\theta_1 + a_2\theta_2 + O(|\theta_1|^2 + |\theta_2|^2 + |\theta_3|^2) $$

$$ f(\theta) = 2 - \theta_1^2 - \theta_2^2 + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3). $$

If $\delta_1 \le \delta_2$, let $U(\theta_1, \theta_2, \theta_3) = (\mathrm{Im}\phi_1(\theta), \theta_2, \theta_3)$ and let $O_1, O_2$ and $C > 0$ be as above. Doing this change of variables, we get

$$ \mathrm{Im}\phi_1 \circ U^{-1}(u) = u_1 \quad \text{and} \quad f \circ U^{-1}(u) = 2 - \tilde{Q}(u_1, u_2) + O(|u_1|^3 + |u_2|^3 + |u_3|^3) $$

with $\mathrm{sign}(\tilde{Q}) = (2,0)$. In particular, $\frac{\partial^2 \tilde{Q}(0,0)}{\partial u_2^2} > 0$. Thus, applying Morse's lemma to $f$ regarding $u_2$, we get that there exist $O'_2, O_3$ two neighbourhoods of $0$ in $\mathbb{R}^3$, a diffeomorphism $V: O'_2 \to O_3$ defined by $V(u_1, u_2, u_3) = (u_1, v_2(u_1, u_2, u_3), u_3)$, a $C^1$-map $h: \mathbb{R}^2 \to \mathbb{R}$ and $C_1 > 0$ such that (7) is satisfied and

$$ \mathrm{Im}\phi_1 \circ (V \circ U)^{-1}(v) = v_1 \quad \text{and} \quad f \circ (V \circ U)^{-1}(v) = 2 - v_2^2 + h(v_1, v_3). $$

Thus, if $O = U^{-1}(O'_2)$ and $\theta \in \phi_I^{-1}(S_I(\eta, \bar{\delta})) \cap O$, considering $v = V \circ U(\theta)$ we have

$$ |v_1 - \mathrm{Im}\eta_1| \le \delta_1 \quad \text{and} \quad |v_2^2 - (\mathrm{Re}\eta_1 + \mathrm{Re}\eta_2)| \le \max(\delta_1, \delta_2) = \delta_2. $$

Therefore, we have $\lambda_3(\phi_I^{-1}(S_I(\eta, \bar{\delta}))) \cap O \lesssim \delta_1\delta_2^{1/2} \lesssim (\delta_1\delta_2)^{3/4}$.

If $\delta_2 \le \delta_1$, we replace $U$ by $U'$ defined by $U'(\theta_1, \theta_2, \theta_3) = (\mathrm{Im}\phi_2(\theta), \theta_2, \theta_3)$ and similarly, we get

$$ \lambda_3(\phi_I^{-1}(S_I(\eta, \bar{\delta}))) \cap O \lesssim (\delta_1\delta_2)^{3/4}. $$

Now, let us consider the case $s(\phi, I, e) = 1$ and $r(\phi, I, e) = (1,1)$. We write for $j=1,2$,

$$ \mathrm{Im}\phi_j(\theta) = \kappa_j(\theta_1 - g_j\theta_1^2 - h_j\theta_1\theta_2 - l_j\theta_1\theta_3 - a_j\theta_2^2 - b_j\theta_2\theta_3 - c_j\theta_3^2) + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3) $$

$$ \mathrm{Re}\phi_j(\theta) = 1 - \gamma_j\theta_1^2 + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3). $$

Since

$$ R(\theta) = \kappa_1\kappa_2((a_2 - a_1)\theta_2^2 + (b_2 - b_1)\theta_2\theta_3 + (c_2 - c_1)\theta_3^2) $$

has signature $(1,1)$, we may assume, doing a linear change of variables, that it is given by $R(\theta) = \kappa_1\kappa_2(\theta_2^2 - \theta_3^2)$. Hence, if set $f = \kappa_2\mathrm{Im}(\phi_1) - \kappa_1\mathrm{Im}(\phi_2)$, we get that we may write

$$ f(\theta) = \alpha\theta_1^2 + \beta\theta_1\theta_2 + \gamma\theta_1\theta_3 + \kappa_1\kappa_2(\theta_2^2 - \theta_3^2) + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3). $$

If $\delta_1 \le \delta_2$, we do the change of variables $U(\theta) = \left(\frac{\mathrm{Im}\phi_1}{\kappa_1}, \theta_2, \theta_3\right)$ where $U: O_1 \to O_2$ is a diffeomorphism, $O_1$ and $O_2$ are two neighbourhoods of $0$ in $\mathbb{R}^3$ and $C > 0$ is such that (6) is satisfied. Then, we have

$$ \mathrm{Im}\phi_1 \circ U^{-1}(u) = \kappa_1 u_1 $$

$$ f \circ U^{-1}(u) = \tilde{\alpha}u_1^2 + \tilde{\beta}u_1u_2 + \tilde{\gamma}u_1u_3 + \kappa_1\kappa_2(u_2^2 - u_3^2) + O(|u_1|^3 + |u_2|^3 + |u_3|^3). $$

We check that $f(0) = 0$, $d_{u_2,u_3}f(0) = (0,0)$ and $d_{u_2,u_3}^2 f(0) = \begin{pmatrix} 2\kappa_1\kappa_2 & 0 \\ 0 & -2\kappa_1\kappa_2 \end{pmatrix}$ with $\det(d_{u_2,u_3}f(0)) = -4(\kappa_1\kappa_2)^2 \neq 0$. Therefore, by Morse's lemma, there exist $O'_2, O_3$ two neighbourhoods of $0$ in $\mathbb{R}^3$, a diffeomorphism $V: O'_2 \to O_3$ defined by

$$ V(u_1, u_2, u_3) = (u_1, v_2(u_1, u_2, u_3), v_3(u_1, u_2, u_3)), $$

a $C^1$-map $h: \mathbb{R} \to \mathbb{R}$ and $C_1 > 0$ such that $f \circ (V \circ U)^{-1}(v) = v_2^2 - v_3^2 + h(v_1)$ and (7) is satisfied. By the definition of $V$, we still have $\mathrm{Im}\phi_1 \circ (V \circ U)^{-1}(v) = \kappa_1 v_1$. We set $O = U^{-1}(O'_2)$. Then, for $\theta \in O \cap \phi_I^{-1}(S_I(\eta, \bar{\delta}))$, considering $v = V \circ U(\theta)$, we get

$$ |\kappa_1 v_1 - \mathrm{Im}\eta_1| \le \delta_1 \quad \text{and} \quad |v_2^2 - v_3^2 + h(v_1) - (\kappa_2 \mathrm{Im}\eta_1 - \kappa_1 \mathrm{Im}\eta_2)| \le \max(\delta_1, \delta_2) = \delta_2. $$
Then, by Lemma 2.5, we deduce

$$ \lambda_3(\{\theta \in O : |\kappa_1 v_1 - \operatorname{Im}\eta_1| \le \delta_1, |v_2^2 - v_3^2 + h(v_1) - (\kappa_2 \operatorname{Im}\eta_1 - \kappa_1 \operatorname{Im}\eta_2)| \le \delta_2\}) \le -C\delta_1\delta_2 \log(\delta_2) \\ \lesssim (\delta_1\delta_2)^\alpha, $$

for all $0 < \alpha < 1$. If $\delta_2 \le \delta_1$, we work with $U'(\theta) = \left(\frac{\operatorname{Im}\phi_2}{\kappa_2}, \theta_2, \theta_3\right)$ and we get a similar estimate.

Finally, let us consider the case $s(\phi, I, e) = 1$ and $r(\phi, I, e) = (0, 0)$. Arguing as in the proof of Theorem 5.1, we write for $j = 1, 2$,

$$ \operatorname{Im}\phi_j(\theta) = \kappa_j(\theta_1 - g_j\theta_1^2 - h_j\theta_1\theta_2 - l_j\theta_1\theta_3 - a\theta_2^2 - b\theta_2\theta_3 - c\theta_3^2) + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3) $$

$$ \operatorname{Re}\phi_j(\theta) = 1 - \gamma_j\theta_1^2 + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3). $$

If $\delta_1 \le \delta_2$, we do the change of variables $U(\theta) = \left(\frac{\operatorname{Im}\phi_1}{\kappa_1}, \theta_2, \theta_3\right)$ where $U : O_1 \to O_2$ is a diffeomorphism with $O_1, O_2$ two neighbourhoods of 0 in $\mathbb{R}^3$ and $C > 0$ is such that (6) is satisfied. We then have

$$ \operatorname{Im}\phi_1 \circ U^{-1}(u) = \kappa_1 u_1. $$

For $\theta \in O_1 \cap \phi_I^{-1}(S_I(\eta, \bar{\delta}))$, considering $u = U(\theta)$, we have $|\kappa_1 u_1 - \operatorname{Im}\eta_1| \le \delta_1$. Thus,

$$ \lambda_3(U(O_1 \cap \phi_I^{-1}(S_I(\eta, \bar{\delta})))) \lesssim \delta_1 \le (\delta_1\delta_2)^{1/2}. $$

If $\delta_1 \ge \delta_2$, we work with $U'(\theta) = \left(\frac{\operatorname{Im}\phi_2}{\kappa_2}, \theta_2, \theta_3\right)$ and we also get

$$ \lambda_3(\{\theta \in O_1 : |\kappa_2 u_2 - \operatorname{Im}\eta_2| \le \delta_2\}) \lesssim (\delta_1\delta_2)^{1/2}. $$

□

*Remark.* The last case also follows from Lemma 2.17.

Before stating the main result of this section, we need one more technical lemma that links the fact that the derivatives $\nabla\phi_{i_k}(\xi)$ are independent for $i_k \in I$ such that $\phi_I(\xi) \in \mathbb{T}^{|I|}$ and the measure of the boundary values of $\phi$ for $\phi$ regular enough.

**Lemma 5.3.** Let $\phi \in \mathcal{O}(\mathbb{D}^d, \mathbb{D}^d) \cap C^1(\bar{\mathbb{D}}^d)$. Let $\beta \ge -1$. Let $\xi = e^{i\theta_0} \in \mathbb{T}^d$, $I = \{i_1, \dots, i_p\} \subset \{1, \dots, d\}$, $I \ne \emptyset$ such that $\phi_I(\xi) \in \mathbb{T}^{|I|}$ and $\nabla\phi_{i_1}(\xi), \dots, \nabla\phi_{i_p}(\xi)$ are linearly independent. Then there exists a neighbourhood $O$ of $\theta_0$ in $\mathbb{R}^d$ such that, for all $\eta \in \mathbb{T}^{|I|}$ and all $\bar{\delta} \in (0, 2)^{|I|}$, we have

$$ \lambda_d(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \lesssim \frac{\prod_{j \in I} \delta_j^{2+\beta}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta}}. $$

*Proof.* Let $\xi \in \mathbb{T}^d$ be as in the assumptions. Then, applying Lemma 12 of [12] (or Theorem 5.2 of [2]), we get that there exist $C > 0$ and $\mathcal{U}$ a neighbourhood of $\xi$ in $\bar{\mathbb{D}}^d$ such that for all $\eta \in \mathbb{T}^{|I|}$ and all $\bar{\delta} \in (0, 2)^{|I|}$,

$$ V_{\beta}(\{z \in \mathbb{D}^d \cap \mathcal{U} : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C \prod_{j \in I} \delta_j^{2+\beta}. $$

Then by Lemma 2.8, there exists $D \ge 1$ such that

$$ E := \{((1 - \rho_k)e^{i\theta_k}) : 0 \le \rho_k \le \omega_{I,k}(\bar{\delta}), k \in P_I \text{ and } e^{i\theta} \in \phi_I^{-1}(S_I(\eta, \bar{\delta}))\} \subset \phi_I^{-1}(S(\eta, D\bar{\delta})). $$
Let also $O$ be a neighbourhood of $\theta_0$ and $r > 0$ such that $((1 - \rho_k)e^{i\theta_k}) \in \mathcal{U}$ provided $0 \le \rho_k < r$ and $\theta \in O$. Then by Lemma 2.4, we have

$$ \prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta} \lambda_d(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \lesssim V_{\beta}(\phi_I^{-1}(S_I(\eta, D\bar{\delta}))) \cap \mathcal{U}) \\ \lesssim \prod_{j \in I} \delta_j^{2+\beta}. $$

Thus,

$$ \lambda_d(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \lesssim \frac{\prod_{j \in I} \delta_j^{2+\beta}}{\prod_{k=1}^{d} \omega_{I,k}(\bar{\delta})^{1+\beta}}. $$

□

Now, let us present the result of boundedness on $A_{\beta}^2(\mathbb{D}^3)$.

**Theorem 5.4.** Let $\phi \in O(\mathbb{D}^3, \mathbb{D}^3) \cap C^3(\bar{\mathbb{D}}^3)$ with $\phi(\bar{\mathbb{D}}^3) \cap \partial\mathbb{D}^3 \neq \emptyset$. Let $\beta \ge -1$. For $\xi \in \mathbb{T}^3$, we define $J_c(\phi, \xi)$ and $J_d(\phi, \xi)$ in the following way:

(1) $\phi(\xi) \in \mathbb{T}^3$

(a) If $d\phi(\xi)$ is invertible, then $J_c(\phi, \xi) = [-1, +\infty)$ and $J_d(\phi, \xi) = \emptyset$,

(b) If $d\phi(\xi)$ is not invertible, then $J_c(\phi, \xi) = \emptyset$ and $J_d(\phi, \xi) = [-1, +\infty)$.

(2) $\phi(\xi) \in \mathbb{T}^2 \times \mathbb{D} \cup \mathbb{T} \times \mathbb{D} \times \mathbb{T} \cup \mathbb{D} \times \mathbb{T}^2$. Let $1 \le i_1 < i_2 \le 3$ be such that $(\phi_{i_1}(\xi), \phi_{i_2}(\xi)) \in \mathbb{T}^2$ and $I = \{i_1, i_2\}$.

(a) If $(\nabla\phi_{i_1}(\xi), \nabla\phi_{i_2}(\xi))$ are linearly independent, then $J_c(\phi, \xi) = [-1, +\infty)$ and $J_d(\phi, \xi) = \emptyset$,

(b) If $(\nabla\phi_{i_1}(\xi), \nabla\phi_{i_2}(\xi))$ are linearly dependent and $|P_I| = 3$ then:

(α) if $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (0, 0)$, then $J_c(\phi, \xi) = [0, +\infty)$ and $J_d(\phi, \xi) = [-1, -\frac{2}{3})$,

(β) if $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (1, 0)$ or $(0, 1)$, then $J_c(\phi, \xi) = [-\frac{1}{2}, +\infty)$ and $J_d(\phi, \xi) = [-1, -\frac{5}{6})$,

(γ) if $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (1, 1)$, then $J_c(\phi, \xi) = (-1, +\infty)$ and $J_d(\phi, \xi) = \{-1\}$,

(δ) if $s(\phi, I, \xi) = 2$ and $r(\phi, I, \xi) = (0, 0)$, then $J_c(\phi, \xi) = [-\frac{1}{2}, +\infty)$ and $J_d(\phi, \xi) = \{-1\}$,

(ε) if $s(\phi, I, \xi) = 3$ or $s(\phi, I, \xi) = 2$ and $r(\phi, I, \xi) \in \{(0, 1), (1, 0)\}$ or $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) \in \{(0, 2), (2, 0)\}$, then $J_c(\phi, \xi) = [-1, +\infty)$ and $J_d(\phi, \xi) = \emptyset$.

(c) If $(\nabla\phi_{i_1}(\xi), \nabla\phi_{i_2}(\xi))$ are linearly dependent and $|P_I| \le 2$, then $J_c(\phi, \xi) = \emptyset$ and $J_d(\phi, \xi) = [-1, +\infty)$.

(3) $\phi(\xi) \in \mathbb{T} \times \mathbb{D}^2 \cup \mathbb{D} \times \mathbb{T} \times \mathbb{D} \cup \mathbb{D}^2 \times \mathbb{T}$, then $J_c(\phi, \xi) = [-1, +\infty)$ and $J_d(\phi, \xi) = \emptyset$.

We define $J_{\text{cont}}(\phi) = \bigcap_{\xi \in \mathbb{T}^3, \phi(\xi) \in \partial\mathbb{D}^3} J_c(\phi, \xi)$ and $J_{\text{discont}}(\phi) = \bigcup_{\xi \in \mathbb{T}^3, \phi(\xi) \in \partial\mathbb{D}^3} J_d(\phi, \xi)$. Then, for $\beta \in J_{\text{cont}}(\phi)$, $C_{\phi}$ is bounded on $A_{\beta}^2(\mathbb{D}^3)$ and for $\beta \in J_{\text{discount}}(\phi)$, $C_{\phi}$ is not bounded on $A_{\beta}^2(\mathbb{D}^3)$.

*Proof.* Let $\xi = e^{i\theta_0} \in \mathbb{T}^3$ and $\beta \in J_{\text{cont}}(\phi)$. We intend to apply Theorem 2.10 and to do that, we need to satisfy inequality (3). There are several cases to consider. The first case is $\phi(\xi) \in \mathbb{T}^3$. Applying Lemma 5.3, we get that there exist $C > 0$ and a neighbourhood $O$ of $\theta_0$ such that for all $\eta \in \mathbb{T}^3$ and all $\bar{\delta} \in (0, 2)^3$,

$$ \lambda_3(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j = 1, 2, 3\}) \le C \frac{\prod_{j=1}^{3} \delta_j^{2+\beta}}{\prod_{k=1}^{3} \omega_{\{1,2,3\},k}(\bar{\delta})^{1+\beta}}. $$
The second case is $\phi(\xi) \in \mathbb{T}^2 \times \mathbb{D} \cup \mathbb{T} \times \mathbb{D} \times \mathbb{T} \cup \mathbb{D} \times \mathbb{T}^2$. There exists $I = \{i_1, i_2\}$ such that $\phi_I(\xi) \in \mathbb{T}^2$. If $\nabla\phi_{i_1}(\xi), \nabla\phi_{i_2}(\xi)$ are linearly independent then applying Lemma 5.3, there exist $O$ a neighbourhood of $\theta_0$ and $C > 0$ such that for all $\eta \in \mathbb{T}^{|I|}$ and all $\bar{\delta} \in (0, 2)^{|I|}$,

$$ \lambda_3(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C \frac{\prod_{j \in I} \delta_j^{2+\beta}}{\prod_{k=1}^{3} \omega_{I,k}(\bar{\delta})^{1+\beta}}. $$

Now, if $\nabla\phi_{i_1}(\xi), \nabla\phi_{i_2}(\xi)$ are linearly dependent and $|P_I| = 3$, there are also several cases to consider. Firstly, if $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (0,0)$ or $r(\phi, I, \xi) = (1,0)$ or $(0,1)$ or $r(\phi, I, \xi) = (1,1)$ or if $s(\phi, I, \xi) = 2$ and $r(\phi, I, \xi) = (0,0)$, then, applying Theorem 5.2, we get that there exist $O$ a neighbourhood of $\theta_0$ and $C > 0$ such that for all $\eta \in \mathbb{T}^{|I|}$ and all $\bar{\delta} \in (0, 1)^{|I|}$,

$$ \lambda_3(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C(\delta_{i_1} \delta_{i_2})^{\alpha} $$

with $\alpha + \frac{|P_I|}{|I|}(1+\beta) \ge 2+\beta$ (this inequality on $\alpha$ holds because $\beta \in J_{cont}(\phi) \subset J_c(\phi, \xi)$). Then applying the first point of Corollary 2.11, we deduce that

$$ \lambda_3(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C \frac{\prod_{j \in I} \delta_j^{2+\beta_1}}{\prod_{k=1}^{3} \omega_{I,k}(\bar{\delta})^{1+\beta_2}}. $$

(It is the application of Corollary 2.11 that gives the bounds of the different $J_c(\phi, \xi)$ for such $\xi$). Secondly, if $s(\phi, I, \xi) = 3$ and $r(\phi, I, \xi) = (0,0)$ or $s(\phi, I, \xi) = 2$ and $r(\phi, I, \xi) = (1,0)$ or $(0,1)$ or $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (2,0)$ or $(0,2)$, then, following the proof of the fact of Theorem 5.1 of [2], there exist $O$ a neighbourhood of $\theta_0$ in $\mathbb{R}^3$ and $C > 0$ such that for all $\eta \in \mathbb{T}^{|I|}$ and all $\bar{\delta} \in (0, 2)^{|I|}$,

$$ \lambda_3(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C \delta_{i_1} \delta_{i_2}, $$

with $1 + \frac{|P_I|}{|I|}(1+\beta) \ge 2+\beta$. Then, applying the first point of Corollary 2.11, we get

$$ \lambda_3(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C \frac{\prod_{j \in I} \delta_j^{2+\beta}}{\prod_{k=1}^{3} \omega_{I,k}(\bar{\delta})^{1+\beta}}. $$

Finally, if $\phi(\xi) \in \mathbb{D}^2 \times \mathbb{T} \cup \mathbb{D} \times \mathbb{T} \times \mathbb{D} \cup \mathbb{T} \times \mathbb{D}^2$, then there exists $i_1 \in \{1, 2, 3\}$ such that $\phi_{i_1}(\xi) \in \mathbb{T}$. As $\nabla\phi_{i_1}(\xi) \ne 0$, applying Lemma 5.3, we get that there exist $O$ a neighbourhood of $\theta_0$ and $C > 0$ such that for all $\eta \in \mathbb{T}$ and all $\bar{\delta} \in (0, 1)$,

$$ \lambda_3(\{\theta \in O : |\phi_{i_1}(\theta) - \eta_{i_1}| < \delta\}) \le C \frac{\delta}{\prod_{k=1}^{3} \omega_{\{i_1\},k}(\bar{\delta})^{1+\beta}}. $$

Thus, we have shown that for all $\theta_0 \in \mathbb{R}^d$, all $I \subset \{1, \dots, d\}$ such that $\phi_I(\theta_0) \in \mathbb{T}^{|I|}$ and $I$ is maximal with respect to this property, there exist a neighbourhood $O$ of $\theta_0$ in $\mathbb{R}^d$ and $C > 0$ such that for all $\eta \in \mathbb{T}^{|I|}$ and all $\bar{\delta} \in (0, 1)^{|I|}$,

$$ \lambda_d(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j \in I\}) \le C \frac{\prod_{j \in I} \delta_j^{2+\beta}}{\prod_{k=1}^{3} \omega_{I,k}(\bar{\delta})^{1+\beta}}. $$
Applying Theorem 2.10 we get that $C_{\phi}: A_{\beta}^{2}(\mathbb{D}^{3}) \to A_{\beta}^{2}(\mathbb{D}^{3})$ is bounded.

Now, let $\beta \in J_{discont}(\phi)$. There exists $\xi \in \mathbb{T}^3$ such that $\beta \in J_d(\phi, \xi)$. There are several cases to consider. First, if $\phi(\xi) \in \mathbb{T}^3$, then $d\phi(\xi)$ is not invertible and the necessity from Theorem 1 of [11] gives that $C_{\phi}: A_{\beta}^{2}(\mathbb{D}^{3}) \to A_{\beta}^{2}(\mathbb{D}^{3})$ is not bounded. Now, if $\phi(\xi) \in \mathbb{T}^2 \times \mathbb{D} \cup \mathbb{T} \times \mathbb{D} \times \mathbb{T} \cup \mathbb{D} \times \mathbb{T}^2$, let $I = \{i_1, i_2\}$ be such that $\phi_I(\xi) \in \mathbb{T}^{|I|}$ and $\nabla\phi_{i_1}(\xi), \nabla\phi_{i_2}(\xi)$ are linearly dependent. Suppose first that $|P_I| = 3$. If $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (0,0)$ or $r(\phi, I, \xi) = (1,0)$ or $r(\phi, I, \xi) = (0,1)$, then applying Theorem 5.1, we get that $C_{\phi}: A_{\beta}^{2}(\mathbb{D}^{3}) \to A_{\beta}^{2}(\mathbb{D}^{3})$ is not bounded. If $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (1,1)$ or if $s(\phi, I, \xi) = 2$ and $r(\phi, I, \xi) = (0,0)$, then applying Theorem 4.1 of [2], we get that $C_{\phi}: A_{\beta}^{2}(\mathbb{D}^{3}) \to A_{\beta}^{2}(\mathbb{D}^{3})$ is not bounded. Now, if $|P_I| \le 2$ and $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (0,0)$ or $r(\phi, I, \xi) = (1,0)$ or $r(\phi, I, \xi) = (0,1)$ or $s(\phi, I, \xi) = 2$ and $r(\phi, I, \xi) = (0,0)$, then applying Theorem 5.1, we get that $C_{\phi}: A_{\beta}^{2}(\mathbb{D}^{3}) \to A_{\beta}^{2}(\mathbb{D}^{3})$ is not bounded. $\square$

## 6. EXAMPLES

We are going to give examples of the cases 2) $b)(\alpha), \beta), \gamma), \delta$) and $\varepsilon$) of Theorem 5.4 and to discuss the optimality of $J_c(\phi, \cdot)$ and $J_d(\phi, \cdot)$ in these cases.

Case 1: $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (0,0)$.

**Example 1.** Let $\phi: \mathbb{D}^3 \to \mathbb{D}^3$ be defined by $\phi(z_1, z_2, z_3) = (z_1z_2z_3, z_1z_2z_3, 0)$.

By Proposition 5 of [12], we know that $C_{\phi}$ is bounded on $A_{\beta}^{2}(\mathbb{D}^{3})$ if and only if $\beta \ge 0$. In particular, the interval $[0, +\infty)$ is optimal for $J_c(\phi)$ in this case.

**Example 2.** Let $\varphi: \mathbb{D} \to \mathbb{D}$ be defined by $\varphi(z) = \frac{3+6z-z^2}{8} = 1+\frac{z-1}{2}-\frac{(z-1)^2}{8}$. We then consider

$$ \phi: \begin{array}{rcl} \mathbb{D}^3 & \to & \mathbb{D}^3 \\ (z_1, z_2, z_3) & \mapsto & (\varphi(z_1)\varphi(z_2)\varphi(z_3), \varphi(z_1)\varphi(z_2)\varphi(z_3), 0). \end{array} $$

Doing a fourth-order Taylor expansion and the change of variables $u_1 = \theta_1 + \theta_2 + \theta_3$, $u_2 = \theta_2$ and $u_3 = \theta_3$, we get:

$$ \operatorname{Im}\phi_j \circ U^{-1}(u) = \frac{1}{2}u_1 + \frac{1}{24}u_1^3 - \frac{3}{16}(u_1^2(u_2+u_3) - u_1(u_2^2+u_3^2) + u_2u_3(u_2+u_3)) + \frac{3}{8}u_1u_2u_3 + O(|u_1|^4 + |u_2|^4 + |u_3|^4) $$

$$ \operatorname{Re}\phi_j \circ U^{-1}(u) = 1 - \frac{1}{8}u_1^2 + O(|u_1|^4 + |u_2|^4 + |u_3|^4) $$

Thus, $s(\phi, I, e) = 1$ and $R(u_2, u_3) = 0$ so $r(\phi, I, e) = (0,0)$ and we are indeed in Case 1.

Taking $u_2, u_3 \in [-\delta^{1/4}, \delta^{1/4}]$ and $u_1 \in I_{u_2, u_3} := \{u_1: \left|\frac{u_1}{2} - \frac{3}{16}u_2u_3(u_2+u_3)\right| \le \delta\}$, it is easy to check that $|u_1| \lesssim \delta^{3/4}$ and we get:

$$ \begin{aligned} |\operatorname{Im}\phi_j \circ U^{-1}(u)| & \lesssim \delta + \delta^{9/4} + \delta^{3/2}(\delta^{1/4} + \delta^{1/4}) + \delta^{3/4}(\delta^{1/2} + \delta^{1/2}) + \delta^{5/4} + O(\delta^3 + \delta + \delta) \\ & \lesssim \delta \end{aligned} $$

$$ |\operatorname{Re}\phi_j \circ U^{-1}(u) - 1| \lesssim \delta^{3/2} + O(\delta^3 + \delta + \delta) \lesssim \delta. $$

Thus,

$$ \lambda_3 (\{u \in [-\pi, \pi]^3: |\phi_j(u) - 1| \lesssim \delta, j = 1, 2\}) \gtrsim \delta^{1+1/2}. $$

Applying Corollary 2.12 with $\alpha = \frac{3}{4}$, $|P_I| = 3$ and $|I| = 2$, we get that $C_{\phi}$ is not bounded on $A_{\beta}^{2}(\mathbb{D}^{3})$ for $\beta < \frac{1}{2}$.

Now, let us discuss the continuity of $C_{\phi}$ on $A_{\beta}^{2}(\mathbb{D}^{3})$. The only maximal $I$ for which $\phi_I(e^{i\theta_0}) \in \mathbb{T}^{|I|}$
for some $\theta_0 \in \mathbb{R}^3$ is $I = \{1, 2\}$ and this holds only for $\theta_0 = (0, 0, 0)$ (see Lemma 3.11 of [2]).
Thus, we intend to prove that there exist $C > 0$ and a neighbourhood $O$ of $0$ in $\mathbb{R}^3$ such that
for all $\bar{\delta} \in (0, 1)^2$ and all $\eta \in \mathbb{T}^2$,

$$ \lambda_3(\{\theta \in O : |\phi_j(\theta) - \eta_j| \le \delta_j, j = 1, 2\}) \le C(\delta_1 \delta_2)^{3/4}. $$

Note that for $j = 1, 2$,

$$ |\varphi(z_1)\varphi(z_2)\varphi(z_3) - \eta_j| < \delta_j \Rightarrow |\varphi(z_i)| \ge 1 - \delta_j \Rightarrow |\varphi(z_i)|^2 \ge 1 - 2\delta_j. $$

But, doing a fourth-order Taylor expansion of $|\varphi|^2$ gives

$$ |\varphi(e^{i\theta})|^2 = 1 - \frac{3}{64}\theta^4 + O(|\theta|^5). $$

So, $|\theta_k| \le c\delta_j^{1/4}$ if $|\phi_j(\theta) - \eta_j| \le \delta_j$, $j = 1, 2$, $k = 1, 2, 3$. Now, fix $\theta_2, \theta_3$ such that $|\theta_k| \lesssim \min(\delta_1, \delta_2)^{1/4}$. Then for $j = 1, 2$, the inequality $|\varphi(e^{i\theta_1})\varphi(e^{i\theta_2})\varphi(e^{i\theta_3}) - \eta_j| \le \delta_j$ implies that $|\varphi(e^{i\theta_1}) - h_{j,\theta_2,\theta_3}| \le c'\delta_j$ for some $h_{j,\theta_2,\theta_3}$ and $c' > 0$ independent of $\bar{\delta}$. Arguing as in the proof of Lemma 2.17, we find that $\theta_1$ belongs to some interval of length $\lesssim \min(\delta_j)$. Therefore, we have found $O$ a neighbourhood of $0$ in $\mathbb{R}^3$ such that for all $\eta \in \mathbb{T}^2$ and all $\bar{\delta} \in (0, 1)^2$,

$$ \lambda_3(\{\theta \in O : |\phi_j(\theta) - \eta_j| < \delta_j, j = 1, 2\}) \lesssim \min(\delta_j)^{3/2} \le (\delta_1 \delta_2)^{3/4}. $$

Then, applying the second point of Corollary 2.11, we deduce that $C_\phi$ is bounded on $A_\beta^2(\mathbb{D}^3)$ as soon as $\beta \ge -\frac{1}{2}$. To conclude, $C_\phi$ is bounded on $A_\beta^2(\mathbb{D}^3)$ if and only if $\beta \ge -\frac{1}{2}$.

It should be observed that these previous examples illustrate that the knowledge of $s$ and $r$ is not sufficient enough to study the continuity on any $A_\beta^2(\mathbb{D}^3)$, contrary to what happens for the Hardy space. We have also proved that in the case $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (0, 0)$, the best (=smaller) value for $J_d(\phi, \xi)$ that we can expect is $[-1, -1/2)$.

Case 2: $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (1, 0)$ or $(0, 1)$.

**Example.** For $\varepsilon \ge 0$, we define the map $F_\varepsilon$ on $\mathbb{C}$ by

$$ F_{\varepsilon}(z) = \frac{3 + 6z - z^2}{8} + 2i\varepsilon(z - 1)^2 - i\varepsilon(z - 1)^3. $$

It is shown in [2] that there exists $\varepsilon_1 > 0$ such that $F_\varepsilon \in O(\mathbb{D}, \mathbb{D})$ provided $\varepsilon \in [0, \varepsilon_1)$ and $|F_\varepsilon(z)| = 1$ if and only if $z = 1$. We then consider

$$ \phi: \begin{array}{rcl} \mathbb{D}^3 & \to & \mathbb{D}^3 \\ (z_1, z_2, z_3) & \mapsto & (F_0(z_1)F_b(z_2)F_0(z_3), F_0(z_1)F_b(z_2)F_c(z_3), 0) \end{array} $$

with $b, c \in (0, \varepsilon_1)$. We write $c_j = \begin{cases} 0 & \text{if } j = 1 \\ c & \text{if } j = 2 \end{cases}$. Doing a third-order Taylor expansion and then doing the linear change of variables $u_1 = \theta_1 + \theta_2 + \theta_3$, $u_2 = \theta_2$ and $u_3 = \theta_3$, we get

$$ \mathrm{Re}\phi_j \circ U^{-1}(u) = 1 - \frac{1}{8}u_1^2 + bu_2^2u_1 + c_ju_3^2u_1 + O(u_1^4 + u_2^4 + u_3^4) $$

$$ \mathrm{Im}\phi_j \circ U^{-1}(u) = \frac{1}{2}u_1 - 2(bu_2^2 + c_ju_3^2) + \frac{1}{24}u_1^3 - \frac{3}{16}(u_1^2u_2 + u_1^2u_3 - u_1u_2^2 - u_1u_3^2 + u_2^2u_3^2) \\ + \frac{3}{8}u_1u_2u_3 + O(u_1^4 + u_2^4 + u_3^4). $$

Hence, $s(\phi, I, e) = 1$ and $R(u_2, u_3) = (-2bu_2^2) - (-2bu_2^2 - 2cu_3^2) = 2cu_3^2$ so $r(\phi, I, e) = (1, 0)$ and we are indeed in Case 2.

We consider $u_2 \in [-\delta^{1/4}, \delta^{1/4}]$ and $u_3 \in [-\delta^{1/2}, \delta^{1/2}]$. For those $u_2, u_3$, we choose
$u_1 \in I_{u_2, u_3} := \{ u_1 : \left| \frac{u_1}{2} - 2bu_2^2 - \frac{3}{16}(u_2^2 u_3 + u_3^2 u_2) \right| \le \delta \}$. It is easy to check that $|u_1| \lesssim \delta^{1/2}$.

So,

$$|\mathrm{Re}\phi_1 \circ U^{-1}(u) - 1| \lesssim \delta + \delta + O(\delta^2 + \delta + \delta^2) \lesssim \delta$$

$$|\mathrm{Re}\phi_2 \circ U^{-1}(u) - 1| \lesssim \delta + \delta + \delta^{3/2} + O(\delta^2 + \delta + \delta^2) \lesssim \delta$$

$$|\mathrm{Im}\phi_1 \circ U^{-1}(u)| \lesssim \delta + \delta^{3/2} + \delta^{1+1/4} + \delta^{1+1/2} + \delta + \delta^{1+1/2} + \delta^{1+1/2+1/4} + O(\delta) \lesssim \delta$$

$$|\mathrm{Im}\phi_2 \circ U^{-1}(u)| \lesssim \delta + \delta + O(\delta) \lesssim \delta.$$

Thus,

$$\lambda_3(\{u \in [-\pi, \pi]^3 : |\phi_j(u) - 1| \lesssim \delta, j = 1, 2\}) \gtrsim \delta^{1+3/4}.$$

Applying Corollary 2.12, we get that $C_\phi$ is not bounded on $A_\beta^2(\mathbb{D}^3)$ for $\beta < -\frac{3}{4}$.

Now, about the continuity. The only maximal $I$ for which $\phi_I(e^{i\theta_0}) \in \mathbb{T}^{|I|}$ for some $\theta_0 \in \mathbb{R}^d$ is $I = \{1, 2\}$ and this only holds for $\theta_0 = (0, 0, 0)$ (see Lemma 3.12 of [2]). As before, we will be able to conclude if we prove that there exist $C > 0$ and a neighbourhood $O$ of $0$ in $\mathbb{R}^3$ such that for all $\bar{\delta} \in (0, 1)^2$ and all $\eta \in \mathbb{T}^2$, we have

$$\lambda_3(\{\theta \in O : |\phi_j(\theta) - \eta_j| \le \delta_j, j = 1, 2\}) \le C(\delta_1\delta_2)^{7/8}.$$

Note that

$$|\phi_j(z) - \eta_j| < \delta_j \Rightarrow \left\{ \begin{array}{ll} |F_0(z_1)|^2 & \ge 1 - 2\delta_j \\ |F_b(z_2)|^2 & \ge 1 - 2\delta_j \\ |F_c(z_3)|^2 & \ge 1 - 2\delta_j \end{array} \right. .$$

But, doing a fourth-order Taylor expansion of $|F_0|^2, |F_b|^2$ and $|F_c|^2$ gives

$$|F_0(e^{i\theta_1})|^2 = 1 - \frac{3}{64}\theta_1^4 + O(|\theta_1|^5)$$

$$|F_b(e^{i\theta_2})|^2 = 1 + \left(4b^2 - \frac{3}{64}\right)\theta_2^4 + O(|\theta_2|^5)$$

$$|F_c(e^{i\theta_3})|^2 = 1 + \left(4c^2 - \frac{3}{64}\right)\theta_3^4 + O(|\theta_3|^5).$$

So, $|\theta_i| \lesssim \delta_j^{1/4}$ if $|\phi_j(\theta) - \eta_j| \le \delta_j$. Then, we define $f(\theta) = \kappa_2\mathrm{Im}\phi_1(\theta) - \kappa_1\mathrm{Im}\phi_2(\theta)$. If $\delta_1 \le \delta_2$, we argue as above by considering the diffeomorphism $U : O_1 \to O_2$ defined by $U(\theta) = \left( \frac{\mathrm{Im}\phi_1}{\kappa_1}, \theta_2, \theta_3 \right)$. As $dU(0) = I_3$, we get

$$f \circ U^{-1}(u) = cu_3^2 + O(|u_1|^3 + |u_2|^3 + |u_3|^3).$$

As $c \ne 0$, $\frac{\partial^2 f(0)}{\partial u_3^2} \ne 0$ and applying Morse's lemma to $f$ regarding $u_3$, we get $V : O'_2 \subset O_2 \to O_3$ a diffeomorphism defined by $V(u_1, u_2, u_3) = (u_1, u_2, v_3(u_1, u_2, u_3))$ and a $\mathcal{C}^1$-map $h : \mathbb{R}^2 \to \mathbb{R}$ such that $f \circ (V \circ U)^{-1}(v) = cv_3^2 + h(v_1, v_2)$. Now, set $O = U^{-1}(O'_2)$. For $\theta \in O \cap \phi_I^{-1}(S_I(\eta, \bar{\delta}))$, considering $v = V \circ U(\theta)$, we have

$$|\kappa_1 v_1 - \mathrm{Im}\eta_1| \le \delta_1, \quad |v_3^2 \pm h(v_1, v_2) - (\kappa_2 \mathrm{Im}\eta_1 - \kappa_1 \mathrm{Im}\eta_2)| \le \delta_2 \quad \text{and} \quad |v_2| \lesssim \delta_1^{1/4}.$$

So,

$$\lambda_3(V \circ U(O \cap \phi_I^{-1}(S_I(\eta, \bar{\delta})))) \lesssim \delta_1^{5/4} \delta_2^{1/2} \le (\delta_1 \delta_2)^{7/8}.$$
If $\delta_2 \le \delta_1$, we replace $U$ by $U'$ defined by $U'(\theta) = \left(\frac{\mathrm{Im}\phi_2}{\kappa_2}, \theta_2, \theta_3\right)$ to get the same result.

In conclusion, $C_\phi$ is bounded on $A_\beta^2(\mathbb{D}^3)$ if and only if $\beta \ge -\frac{3}{4}$. In particular, in this case, one cannot expect a better (=bigger) set than $[-3/4, +\infty)$ for $J_c(\phi, \cdot)$ and a better set than $[-1, -3/4)$ for $J_d(\phi, \cdot)$.

Case 3: $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (1, 1)$.

**Example.** We consider

$$ \phi: \begin{array}{ccc} \mathbb{D}^3 & \to & \mathbb{D}^3 \\ (z_1, z_2, z_3) & \mapsto & (F_0(z_1)F_b(z_2)F_0(z_3), F_0(z_1)F_0(z_2)F_c(z_3), 0) \end{array} $$

with $b, c \in (0, \varepsilon_1)$. Doing the change of variables $u_1 = \theta_1 + \theta_2 + \theta_3$, $u_2 = \theta_2$, and $u_3 = \theta_3$, we get:

$$ \begin{aligned} \mathrm{Re}\phi_j \circ U^{-1}(u) &= 1 - \frac{1}{8}u_1^2 + O(|u_1|^3 + |u_2|^3 + |u_3|^3) \\ \mathrm{Im}\phi_1 \circ U^{-1}(u) &= \frac{1}{2}u_1 - 2bu_2^2 + O(|u_1|^3 + |u_2|^3 + |u_3|^3) \\ \mathrm{Im}\phi_2 \circ U^{-1}(u) &= \frac{1}{2}u_1 - 2cu_3^3 + O(|u_1|^3 + |u_2|^3 + |u_3|^3). \end{aligned} $$

Thus, $s(\phi, I, e) = 1$ and $R(u_2, u_3) = -2bu_2^2 + 2cu_3^2$ so $r(\phi, I, 0) = (1, 1)$. We are in Case 3. Applying Theorem 5.4, we get that $C_\phi : A_\beta^2(\mathbb{D}^3) \to A_\beta^2(\mathbb{D}^3)$ is bounded if and only if $\beta > -1$.

Case 4: $s(\phi, I, \xi) = 2$ and $r(\phi, I, \xi) = (0, 0)$.

$$ \mathbb{D}^3 \to \mathbb{D}^3 $$

**Example 1.** Let $\phi : (z_1, z_2, z_3) \mapsto \left( \frac{1}{2} \left( z_1z_2 + \frac{1+z_3}{2} \right), \frac{1}{2} \left( z_1z_2 + \frac{1+z_3}{2} \right), 0 \right)$. Let $\delta \in (0, 1)$. We start by doing the change of variables $u_1 = \theta_1 + \theta_2$, $u_2 = \theta_2$ and $u_3 = \theta_3$. We then have

$$ \phi(e^{iu}u) = \left( \frac{1}{2} \left( e^{iu_1} + \frac{1+e^{iu_3}}{2} \right), \frac{1}{2} \left( e^{iu_1} + \frac{1+e^{iu_3}}{2} \right), 0 \right). $$

By doing a third-order Taylor expansion and the linear change of variables $v_1 = u_1 + u_3/2$ and $v_3 = u_3$, we get:

$$ \begin{aligned} \mathrm{Re}\phi_j \circ (V \circ U)^{-1}(v) &= 1 - \frac{1}{4} \left( v_1 - \frac{1}{2}v_3 \right)^2 - \frac{2}{16}v_3^2 + O(v_1^4 + v_3^4) \\ \mathrm{Im}\phi_j \circ (V \circ U)^{-1}(v) &= \frac{1}{2}v_1 + \frac{1}{8}v_1^2v_3 - \frac{1}{16}v_1v_3^2 - \frac{1}{32}v_3^3 + O(v_1^4 + v_3^4). \end{aligned} $$

Hence, $s(\phi, I, e) = 2$ and $R(v_2, v_3) = 0$ so $r(\phi, I, e) = (0, 0)$ and we are indeed in Case 4. Considering $|v_1| \le \delta$ and $|v_3| \le \delta^{1/2}$, we have

$$ \begin{aligned} |\mathrm{Re}\phi_j \circ (V \circ U)^{-1}(v) - 1| &\lesssim \delta^2 + \delta^{3/2} + \delta + O(\delta^4 + \delta^2) \lesssim \delta \\ |\mathrm{Im}\phi_j \circ (V \circ U)^{-1}(v)| &\lesssim \delta + \delta^{2+1/2} + \delta^2 + \delta^{3/2} + O(\delta^4 + \delta^2) \lesssim \delta \end{aligned} $$

Thus,

$$ [-\delta, \delta] \times [-\pi, \pi) \times [-\delta^{1/2}, \delta^{1/2}] \subset \{u \in [-\pi, \pi)^3 : |\phi_j(u) - 1| \lesssim \delta, j = 1, 2\} $$

and

$$ \lambda_3 \left( [-\delta, \delta] \times [-\pi, \pi) \times [-\delta^{1/2}, \delta^{1/2}] \right) \gtrsim \delta^{1+1/2}. $$
Then, applying Corollary 2.12, we get that $C_\phi$ is not bounded for $\beta < \frac{-1}{2}$. Finally, by Theorem 5.4, $C_\phi$ is bounded on $A_\beta^2(\mathbb{D}^3)$ if and only if $\beta \ge \frac{-1}{2}$. In particular, the interval $[-\frac{1}{2}, +\infty)$ is optimal for $J_c(\phi, \cdot)$ in this case and we cannot expect something better than $[-1, -1/2)$ for $J_d(\phi, \cdot)$.

**Example 2.** Let $\psi : \mathbb{D} \to \mathbb{D}$ be defined by $\psi(z) = 1 + \frac{z-1}{2} - \frac{(z-1)^2}{8} + \frac{3}{128}(z-1)^3$. We then consider

$$ \phi: \begin{array}{ccc} \mathbb{D}^3 & \to & \mathbb{D}^3 \\ (z_1, z_2, z_3) & \mapsto & (z_1 z_2 \psi(z_3), z_1 z_2 \psi(z_3), 0) \end{array} $$

Doing the change of variables, $t_1 = \theta_1 + \theta_2$, $t_2 = \theta_2$ and $t_3 = \theta_3$ then a Taylor expansion and the change of variables $u_1 = 2t_1 + \frac{1}{2}t_3$, $u_2 = t_2$ and $u_3 = t_3$, we get, for $j = 1, 2$,

$$ \phi_j(u) = 1 + i u_1 - \frac{1}{2} u_1^2 - \frac{1}{6} u_3^2 + O(|u_1|^3 + |u_2|^3 + |u_3|^3). $$

Thus, $s(\phi, I, e) = 2$ and $r(\phi, I, e) = (0, 0)$ and we, indeed, are in Case 4. By Lemma 2.3, we know that $|\psi(z)| = 1$ if and only if $z = 1$. Moreover, by considering $z_3 = e^{i\theta_3}$ with $\theta_3 \in \mathbb{R}$ and doing a sixth-order Taylor expansion of $|\psi|^2$, we get

$$ |\psi(e^{i\theta_3})|^2 = 1 - \frac{135}{16384} \theta_3^6 + O(|\theta_3|^6). $$

Thus, applying Corollary 2.19, we get that $C_\phi$ is bounded on $A_\beta^2(\mathbb{D}^3)$ if and only if $\beta \ge \frac{-1}{6}$. Again, we observe that the knowledge of $r$ and $s$ is not sufficient to determine the set of $\beta$ such that $C_\phi$ is bounded on $A_\beta^2(\mathbb{D}^3)$.

Case 5: $s(\phi, I, \xi) = 3$ and $r(\phi, I, \xi) = (0, 0)$.

**Example.** Let $\phi : \begin{array}{ccc} \mathbb{D}^3 & \to & \mathbb{D}^3 \\ (z_1, z_2, z_3) & \mapsto & \left( \frac{z_1 + z_2 + z_3}{3}, \frac{z_1 + z_2 + z_3}{3}, 0 \right) \end{array}$. Doing a Taylor expansion at $e$ gives

$$ \phi_j(\theta) = 1 + \frac{i}{3}(\theta_1 + \theta_2 + \theta_3) - \frac{1}{6}(\theta_1^2 + \theta_2^2 + \theta_3^2) + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3). $$

Thus, $s(\phi, I, e) = 3$ and $r(\phi, I, e) = (0, 0)$ and, we are in fact in Case 5. Applying Theorem 5.4, we deduce that $C_\phi : A_\beta^2(\mathbb{D}^3) \to A_\beta^2(\mathbb{D}^3)$ is bounded for all $\beta \ge -1$.

Case 6: $s(\phi, I, \xi) = 2$ and $r(\phi, I, \xi) = (0, 1)$ or $(1, 0)$.

**Example.** Let $\varepsilon > 0$ be sufficiently small and let

$$ \phi: \begin{array}{ccc} \mathbb{D}^3 & \to & \mathbb{D}^3 \\ (z_1, z_2, z_3) & \mapsto & \left( \frac{1}{2} \left( z_1 z_2 + \frac{1+z_3}{2} + i \varepsilon (z_3-1)^3 \right), \frac{1}{2} \left( z_1 z_2 + \frac{1+z_3}{2} \right), 0 \right) \end{array}. $$

By [2, Example 7.2], we know that $\phi(\mathbb{D}^3) \subset \mathbb{D}^3$. Doing a Taylor expansion at $e$ we obtain

$$ \operatorname{Im} \phi_1(\theta) = \frac{1}{4}(\theta_1 + \theta_2 + \theta_3) - \varepsilon \frac{\theta_3^2}{2} + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3) $$

$$ \operatorname{Im} \phi_2(\theta) = \frac{1}{4}(\theta_1 + \theta_2 + \theta_3) + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3) $$

$$ \operatorname{Re} \phi_1(\theta) = 1 - \frac{1}{4}(\theta_1 + \theta_2)^2 - \frac{1}{8}\theta_3^2 + O(|\theta_1|^3 + |\theta_2|^3 + |\theta_3|^3) $$
Thus, $s(\phi, I, e) = 2$ and $R(\theta_2, \theta_3) = -2\epsilon\theta_3^2$ so $r(\phi, I, e) = (0,1)$ or $(1,0)$. By Theorem 5.4, $C_\phi : A_\beta^2(\mathbb{D}^3) \to A_\beta^2(\mathbb{D}^3)$ is bounded for any $\beta \ge -1$.

Case 7: $s(\phi, I, \xi) = 1$ and $r(\phi, I, \xi) = (0,2)$ or $(2,0)$.

**Example.** We consider

$$ \phi: \begin{array}{ccc} \mathbb{D}^3 & \to & \mathbb{D}^3 \\ (z_1, z_2, z_3) & \mapsto & (F_0(z_1)F_a(z_2)F_a(z_3), F_0(z_1)F_0(z_2)F_0(z_3), 0) \end{array} $$

with $a \ne 0$ small enough. Doing the change of variables $u_1 = \theta_1 + \theta_2 + \theta_3$, $u_2 = \theta_2$, and $u_3 = \theta_3$ we get:

$$ \mathrm{Re}\phi_j \circ U^{-1}(u) = 1 - \frac{1}{8}u_1^2 + O(|u_1|^3 + |u_2|^3 + |u_3|^3) $$

$$ \mathrm{Im}\phi_1 \circ U^{-1}(u) = \frac{1}{2}u_1 - 2a(u_2^2 + u_3^2) + O(|u_1|^3 + |u_2|^3 + |u_3|^3) $$

$$ \mathrm{Im}\phi_2 \circ U^{-1}(u) = \frac{1}{2}u_1 + O(|u_1|^3 + |u_2|^3 + |u_3|^3). $$

Thus, $s(\phi, I, e) = 1$ and $R(u_2, u_3) = -4a(u_2^2 + u_3^2)$ so $r(\phi, I, e) = (2,0)$ or $(2,0)$. We are in Case 7. Applying Theorem 5.4, we get that $C_\phi : A_\beta^2(\mathbb{D}^3) \to A_\beta^2(\mathbb{D}^3)$ is bounded for any $\beta \ge -1$.

## 7. APPLICATIONS TO COMPOSITION OPERATORS ON THE BIDISC

### 7.1. On the range of composition operators.
In this final section, we use our results to get new informations on composition operators on the bidisc. Fix $\phi: \mathbb{D}^2 \to \mathbb{D}^2$ a holomorphic map which belongs to $C^1(\bar{\mathbb{D}}^2)$. It is proved in [13] that $C_\phi$ always maps $A_\beta^2(\mathbb{D}^2)$ into $A_{2\beta+2}^2(\mathbb{D}^2)$ (see also Theorem 2.16). Moreover, a jump phenomenon occurs at both sides of the interval $[\beta, 2\beta + 2]$:

* in [11], it is shown that if $C_\phi$ does not map $A_\beta^2(\mathbb{D}^2)$ into itself (namely if there exists $\xi \in \mathbb{T}^2$ such that $\phi(\xi) \in \mathbb{T}^2$ and $d\phi(\xi)$ is not invertible), then $C_\phi$ does not map $A_\beta^2(\mathbb{D}^2)$ into $A_{\beta+\frac{1}{4}-\epsilon}^2(\mathbb{D}^2)$ for any $\epsilon > 0$ provided $\phi \in C^2(\bar{\mathbb{D}}^2)$.

* in [10], it is proved that if $C_\phi$ maps $A_\beta^2(\mathbb{D}^2)$ into $A_{2\beta+2-\epsilon}^2(\mathbb{D}^2)$ for some $\epsilon > 0$, then it automatically maps $A_\beta^2(\mathbb{D}^2)$ into $A_{\beta+\frac{1}{2}}^2(\mathbb{D}^2)$.

Hence, this leads to introduce, for $\beta_1 \ge -1$, the set $\Lambda_{\beta_1}$ of real numbers $\beta_2 \ge \beta_1$ such that there exists $\phi \in O(\mathbb{D}^2, \mathbb{D}^2) \cap C^1(\bar{\mathbb{D}}^2)$ such that $C_\phi$ maps $A_{\beta_1}^2(\mathbb{D}^2)$ into $A_\beta^2(\mathbb{D}^2)$ if and only if $\beta \ge \beta_2$. The above results show that $\beta_1 + \frac{1}{4} \in \Lambda_{\beta_1}$ and that $((\beta_1 + \frac{1}{2}, 2\beta_1 + 2) \cup (2\beta_1 + 2, +\infty)) \cap \Lambda_{\beta_1} = \emptyset$. In [10], it is shown that $\beta_1 + 3/8 \in \Lambda_{\beta_1}$ and it is made the following conjecture: for any $m \in \mathbb{N}$, $\beta_1 + \frac{1}{2} - \frac{1}{4m}$ belongs to $\Lambda_{\beta_1}$.

We are able to completely describe $\Lambda_{\beta_1}$ and thus to prove this conjecture.

**Theorem 7.1.** Let $\beta_1 \ge -1$. Then $\Lambda_{\beta_1} = [\beta_1, \beta_1 + 1/2] \cup \{2\beta + 2\}$.

*Proof.* For $\beta_2 = 2\beta_1 + 2$ (resp. $\beta_2 = \beta_1 + 1/2$), the map $\phi(z) = (z_1, z_1)$ (resp. $\phi(z) = (z_1z_2, z_1z_2)$) does the job (see [10, Examples 7.1]). For $\beta_2 = \beta_1$, the map $\phi(z) = ((z_1 + z_2)/2, (2z_2 + z_3)/3)$ works, as any map $\phi$ such that $\mathcal{E}(\phi) = \{e\}$ and $d\phi(e)$ is invertible. For $\beta_2 \in (\beta_1, \beta_1 + 1/2)$, write it $\beta_2 = \beta_1 + \frac{1}{2} - \frac{1}{2\kappa}$ with $\kappa > 1$. Let $\varphi \in O(\mathbb{D}, \mathbb{D}) \cap C^1(\bar{\mathbb{D}})$ be such that for any $z \in \mathbb{T}$, $|\varphi(z)| = 1$ if and only if $z = 1$ and satisfying

$$ |\varphi(e^{i\theta})| = 1 - c|\theta|^\kappa + o(|\theta|^\kappa) $$

for some $c > 0$. Define $\phi(z) = (z_1\varphi(z_2), z_1\varphi(z_2))$. Then by Theorem 2.18, $C_\phi$ is continuous from $A_{\beta_1}^2(\mathbb{D}^2)$ into $A_{\beta_2}^2(\mathbb{D}^2)$ if and only if $\beta \ge \beta_2$.
For the converse inclusion, we use the quoted results of [13] if $\beta_2 > 2\beta_1 + 2$ and of [10] if $\beta \in (\beta_1 + 1/2, 2\beta_1 + 2)$. $\square$

*Remark.* If $\beta_2 = \beta_1 + \frac{1}{2} - \frac{1}{4n}$ for some $n \in \mathbb{N}$, then we can choose $\phi \in C^\infty(\overline{\mathbb{D}}^2)$ by taking for $\varphi$ either the function $h_n$ of Section 3.

**7.2. A characterization of the extreme cases.** A natural problem arises: for $\phi \in C^\infty(\overline{\mathbb{D}}^2)$ and $n \in \mathbb{N} \cup \{\infty\}$, is it possible to characterize when $C_\phi$ maps $A_\beta^2(\mathbb{D}^2)$ into $A_{\beta+\frac{1}{2}-\frac{1}{4n}}^2(\mathbb{D}^2)$? In view of the examples of Section 3, this seems a very difficult task. Indeed, if in the proof of Corollary 2.19, we choose respectively $\varphi = g_{2n}$ and $\varphi = h_n$, and if we call $\phi$ and $\psi$ the induced self-maps of $\mathbb{D}^2$, then we observe that for any $\xi \in \partial\mathbb{D}^2$, $\phi(\xi) \in \partial\mathbb{D}^2$ if and only if $\psi(\xi) \in \partial\mathbb{D}^2$ and that at these points $\phi$ and $\psi$ share the same first $(n-1)$-th derivatives. However, $C_\phi$ (resp. $C_\psi$) maps $A_{\beta_1}^2(\mathbb{D}^2)$ into $A_{\beta_2}^2(\mathbb{D}^2)$ if and only if $\beta_2 \ge \beta_1 + \frac{1}{2} - \frac{1}{8n}$ (resp. $\beta_2 \ge \beta_1 + \frac{1}{2} - \frac{1}{4n}$).

However, for the extreme cases, this is doable and this was done in [10]. In this section, we shall provide other characterizations which are maybe easier to state (and for the second one, to prove). Let us start with the characterization of when $C_\phi$ maps $A_\beta^2(\mathbb{D}^2)$ into $A_{\beta+\frac{1}{2}}^2(\mathbb{D}^2)$. It was shown in [10] that this holds if and only if $|J(\phi)(\xi)| + |D(\phi)(\xi)| > 0$ for all $\xi \in \phi^{-1}(\mathbb{T}^2) \cap \mathbb{T}^2$ where

$$J(\phi)(\xi) = \begin{vmatrix} \partial_1\phi_1(\xi) & \partial_2\phi_1(\xi) \\ \partial_1\phi_2(\xi) & \partial_2\phi_2(\xi) \end{vmatrix} \quad \text{and} \quad D(\phi)(\xi) = \begin{vmatrix} \partial_1\phi_1(\xi) & \partial_2\phi_1(\xi) \\ -\partial_1\phi_2(\xi) & \partial_2\phi_2(\xi) \end{vmatrix}.$$

Let $\xi \in \phi^{-1}(\mathbb{T}^2) \cap \mathbb{T}^2$ such that $J(\phi)(\xi) = D(\phi)(\xi) = 0$. We may assume that $\xi = (1, 1)$. Then it is easy to show that either $\partial_1\phi_1(\xi) = 0$ or $\partial_2\phi_1(\xi) = 0$. Assume $\partial_1\phi_1(\xi) = 0$. Then, since $J\phi(\xi) = 0$, we also have $\partial_1\phi_2(\xi) = 0$. By the Julia-Caratheodory theorem (namely Lemma 2.2), we then deduce that $\phi_1(z) = \phi_1(z_2)$ and $\phi_2(z) = \phi_2(z_2)$ with $\phi_1, \phi_2 : \mathbb{D} \to \mathbb{D}$. Conversely, it is easy to see that if $\phi(z) = (\phi_1(z_2), \phi_2(z_2))$, then $J(\phi)(\xi) = D(\phi)(\xi) = 0$ for all $\xi \in \overline{\mathbb{D}}^2$. Hence we obtain the following result:

**Theorem 7.2.** Let $\phi \in O(\mathbb{D}^2, \mathbb{D}^2) \cap C^1(\overline{\mathbb{D}}^2)$. Then $C_\phi$ maps $A_\beta^2(\mathbb{D}^2)$ into $A_{\beta+1/2}^2(\mathbb{D}^2)$ except if there exist $\varphi_1, \varphi_2 \in O(\mathbb{D}, \mathbb{D}) \cap C^1(\overline{\mathbb{D}})$ such that $\sup_\theta |\varphi_1(e^{i\theta})\varphi_2(e^{i\theta})| = 1$ and either $\phi(z) = (\varphi_1(z_1), \varphi_2(z_1))$ or $\phi(z) = (\varphi_1(z_2), \varphi_2(z_2))$.

In other words, $C_\phi$ maps $A_\beta^2(\mathbb{D}^2)$ into $A_{\beta+1/2}^2(\mathbb{D}^2)$ except if it depends on only one variable and if $\phi(\mathbb{T}^2) \cap \mathbb{T}^2 \ne \emptyset$.

Let us now turn to the characterization of symbols $\phi$ such that $C_\phi$ maps $A_\beta^2(\mathbb{D}^2)$ into $A_{\beta+\frac{1}{4}}^2(\mathbb{D}^2)$. They are characterized in [10] but the characterization is hard to state and hard to prove. We provide a different characterization, in the spirit of the work of this paper and of [2]. Let $\phi : \mathbb{D}^2 \to \mathbb{D}^2$ be holomorphic which extends to a $\mathbb{C}^3$-smooth function on $\overline{\mathbb{D}}^2$. Let $\xi \in \mathbb{T}^2$ be such that $\phi(\xi) \in \mathbb{T}^2$ and assume that $\nabla\phi_1(\xi)$ and $\nabla\phi_2(\xi)$ are linearly dependent (namely $J(\phi)(\xi) = 0$). We assume that $\xi = e$, $\phi(\xi) = e$ and we write, for $j = 1, 2$,

$$\mathrm{Im}\phi_j(\theta) = \kappa_j L(\theta) + o(|\theta_1| + |\theta_2|)$$

$$\mathrm{Re}\phi_j(\theta) = 1 - Q_j(\theta) + O(|\theta_1|^3 + |\theta_2|^3)$$

where $L$ is a nonzero linear form, $\kappa_j$ are nonzero real numbers and $Q_j$ are quadratic forms. As for the tridisc, the signature of each $Q_j$ should be $(n_j, 0)$ with $n_j \in \{0, 1, 2\}$ and there exists $s = s(\phi, \xi) \in \{1, 2\}$ such that the signature of $Q_1+Q_2$ is equal to $(s, 0)$. Taking $L_1, \dots, L_s$ a basis of $L_{j,k}$, where each $Q_j$ is written $Q_j = \sum_{k=1}^{n_j} L_{j,k}^2$, we know that $L$ belongs to $\mathrm{span}(L_1, \dots, L_s)$.
If $s = 1$, we complete $(L_1)$ into a basis $(L_1, L_2)$ of $(\mathbb{R}^2)^*$ and we write

$$ \mathrm{Im}\phi_j(\theta) = \kappa_j(L(\theta) + a_j L_1^2(\theta) + b_j L_1(\theta)L_2(\theta) + c_j L_2^2(\theta)) + O(|\theta_1|^3 + |\theta_2|^3). $$

We define $r(\phi, \xi) = 1$ if $c_1 \neq c_2$, $r(\phi, \xi) = 0$ if $c_1 = c_2$.

**Theorem 7.3.** Let $\phi \in O(\mathbb{D}^2, \mathbb{D}^2) \cap C^3(\bar{\mathbb{D}}^2)$ and let $\beta > -1$. Then $C_\phi$ maps $A_\beta^2(\mathbb{D}^2)$ into $A_{\beta+\frac{1}{4}}^2(\mathbb{D}^2)$ if and only if, for all $\xi \in \mathbb{T}^2$ such that $\phi(\xi) \in \mathbb{T}^2$,

* either $J(\phi)(\xi) \neq 0$;

* or $J(\phi)(\xi) = 0$ and $s(\phi, \xi) = 2$;

* or $J(\phi)(\xi) = 0$ and $s(\phi, \xi) = r(\phi, \xi) = 1$.

*Proof.* We first show that if there exists $\xi \in \mathbb{T}^2$ such that $\phi(\xi) \in \mathbb{T}^2$ with $s(\phi, \xi) = 1$ and $r(\phi, \xi) = 0$, then $C_\phi$ does not map $A_\beta^2(\mathbb{D}^2)$ into $A_{\beta+\frac{1}{4}}^2(\mathbb{D}^2)$. Indeed, assume that $\phi(e) = e$ with $s(\phi, e) = 1$ and $r(\phi, e) = 0$. First, let us consider the case $|P_I| = 2$ with $I = \{1, 2\}$. Doing a linear change of variables, we may assume that $L(\theta) = \theta_1$ and we obtain

$$ \mathrm{Im}\phi_j(\theta) = \kappa_j(\theta_1 + b_j\theta_1\theta_2 + c\theta_2^2) + o(\theta_1^2 + |\theta_2|^3) $$

$$ \mathrm{Re}\phi_j(\theta) = 1 + O(|\theta_1|^2 + |\theta_2|^3). $$

Let $\delta > 0$, set $\bar{\delta} = (\delta, \delta)$ and let us choose $\theta_2 \in [-\delta^{1/3}, \delta^{1/3}]$ and $\theta_1$ such that $|\theta_1 + c\theta_2^2| \le \delta$. It is obvious that $|\theta_1| \lesssim \delta^{2/3}$ so that, after an easy computation,

$$ |\mathrm{Im}\phi_j(\theta)| \lesssim \delta \text{ and } |\mathrm{Re}\phi_j(\theta) - 1| \lesssim \delta. $$

Hence,

$$ \lambda_2(\{\theta \in [-\pi, \pi]^2 : \phi(\theta) \in S(e, \bar{\delta})\}) \gtrsim \delta^{1+\frac{1}{3}}. $$

By Corollary 2.12, this prevents $C_\phi$ to map $A_\beta^2(\mathbb{D}^2)$ into $A_{\beta'}^2(\mathbb{D}^2)$ when $\beta' < \beta + \frac{1}{3}$. Now, if $|P_I| = 1$, then $\phi$ only depends on one variable and the result follows from Theorem 7.2.

Assume now that for all $\xi \in \mathbb{T}^2$ with $\phi(\xi) \in \mathbb{T}^2$, the conditions of Theorem 7.3 are satisfied. Let $\bar{\delta} = (\delta_1, \delta_2) \in (0, +\infty)^2$ and consider $\xi \in \mathbb{T}^2$ with $\phi(\xi) \in \partial\mathbb{D}^2$. If $\phi(\xi) \in \partial\mathbb{D}^2 \setminus \mathbb{T}^2$, let $I = \{j\}$ be such that $\phi_j(\xi) \in \mathbb{T}$. By Lemma 2.17, there exist $O$ a neighbourhood of $\xi$ in $\mathbb{T}^2$ and $C > 0$ such that

$$ \lambda_2(O \cap \phi_I^{-1}(S_I(\eta, \bar{\delta}))) \le C\delta_j $$

and observe that $1 + 2(1 + \beta + \frac{1}{4}) \ge 2 + \beta$.

If $\phi(\xi) \in \mathbb{T}^2$, we intend to show that there exist a neighbourhood $O$ of $\xi$ in $\mathbb{T}^2$ and $C > 0$ such that, for all $\eta \in \mathbb{T}^2$,

$$ (8) \qquad \lambda_2(O \cap \phi^{-1}(S(\eta, \bar{\delta}))) \le C\delta_1^{\frac{3}{4}}\delta_2^{\frac{3}{4}}. $$

Since $\frac{3}{4} + (1 + \beta + \frac{3}{4}) \ge 2 + \beta$, we will conclude with Corollary 2.11. We assume that $\xi = \phi(\xi) = e$ and without loss of generality, we assume that $\delta_1 \le \delta_2$. If $J(\phi)(e) \ne 0$, then (8) is well-known (see [12]) and we can even improve $\delta_1^{3/4}\delta_2^{3/4}$ into $\delta_1\delta_2$. Assume now that $J(\phi)(e) = 0$ and $s(\phi, e) = 2$. Doing a linear change of variables, we may assume that $L(\theta) = \theta_1$ and we write

$$ \mathrm{Im}\phi_j(\theta) = \kappa_j(\theta_1 + G_j(\theta)) $$

with $G_j(0) = 0$, $\nabla G_j(0) = 0$. We set $U(\theta) = (\theta_1 + G_1(\theta), \theta_2)$ and observe that there exist two neighbourhoods $O_1$ and $O_2$ of $0$ and $C_1 > 0$ such that $U$ is a diffeomorphism from $O_1$ onto $O_2$ and

$$ \left\{ \begin{array}{ll} |\det(dU)| & \le C_1 \quad \text{on } O_1 \\ |\det(dU^{-1})| & \le C_1 \quad \text{on } O_2. \end{array} \right. $$

We then write

$$ \mathrm{Re}\phi_j(\theta) = 1 - Q_j(\theta) + H_j(\theta) $$
with $H_j(\theta) = O(|\theta_1|^3 + |\theta_2|^3)$. Because $dU(0) = I_2$, for $u \in O_2$,
$$ \mathrm{Re}\phi_j \circ U^{-1}(u) = 1 - Q_j(u) + K_j(u) $$

with $K_j(u) = O(|u_1|^3 + |u_2|^3)$. Let us now define
$$ E(\bar{\delta}) = \{\theta \in \mathbb{R}^2 : |\phi_j(\theta) - \eta_j| \le \delta_j, j = 1, 2\} $$

and let us write $\eta_j = x_j + iy_j$. Provided $u \in U(E(\bar{\delta}) \cap O_1)$,
$$ \begin{aligned} & |-Q_1(u) - Q_2(u) + K_1(u) + K_2(u) + 2 - x_1 - x_2| \\ & \le |\mathrm{Re}\phi_1 \circ U^{-1}(u) - x_1| + |\mathrm{Re}\phi_2 \circ U^{-1}(u) - x_2| \\ & \le 2\delta_2. \end{aligned} $$

We set $F(u) = -Q_1(u) - Q_2(u) + K_1(u) + K_2(u)$. Since the signature of $Q_1 + Q_2$ is (2,0), $\frac{\partial^2 F}{\partial u_2^2}(0) < 0$. Hence we may apply the parametrized Morse lemma: there exist two neighborhoods $O'_2 \subset O_2$ and $O_3$ of 0, a constant $C_2 > 0$, a diffeomorphism $V : O'_2 \to O_3$, $V(u_1, u_2) = (u_1, V_2(u_1, u_2))$ and a $C^1$-map $h : \mathbb{R} \to \mathbb{R}$ such that, for all $v \in O_3$,

$$ F \circ V^{-1}(v) = -v_2^2 + h(v_1) \quad \text{and} \quad \left\{ \begin{array}{ll} |\det(dV)| & \le C_2 \quad \text{on } O'_2 \\ |\det(dV^{-1})| & \le C_2 \quad \text{on } O_3. \end{array} \right. $$

We finally set $O = U^{-1}(O'_2) \subset O_1$. Therefore, provided $v \in V \circ U(E(\bar{\delta}) \cap O)$, we infer from $|\mathrm{Im}\phi_1 - \mathrm{Im}\eta_1| \le \delta_1$ on $E(\bar{\delta})$ and from (9) that

$$ \left\{ \begin{array}{l} |v_1 - y_1| \le \delta_1/\kappa_1 \\ |-v_2^2 + h(v_1) + 2 - x_1 - x_2| \le 2\delta_2. \end{array} \right. $$

It then follows from Fubini's theorem that

$$ \lambda_2(V \circ U(E(\bar{\delta}) \cap O)) \lesssim \delta_1 \delta_2^{1/2} \le \delta_1^{3/4} \delta_2^{3/4} $$

which in turn yields (8).

Assume finally $s(\phi, e) = 1$ and $r(\phi, e) = 1$. We begin by arguing as above and we write

$$ \mathrm{Im}\phi_j(\theta) = \kappa_j(\theta_1 + a_j\theta_1^2 + b_j\theta_1\theta_2 + c_j\theta_2^2) + o(|\theta_1|^3 + |\theta_2|^3) $$

with $c_1 \neq c_2$. Doing the same change of variables and using $dU(0) = I_2$, we get

$$ \begin{aligned} \mathrm{Im}\phi_1 \circ U^{-1}(u) &= \kappa_1 u_1 \\ \mathrm{Im}\phi_2 \circ U^{-1}(u) &= \kappa_2(u_1 + a'_2 u_1^2 + (b_2 - b_1)u_1 u_2 + (c_2 - c_1)u_2^2) + K(u_1, u_2) \end{aligned} $$

with $K(u_1, u_2) = O(|u_1|^3 + |u_2|^3)$. We apply the parametrized Morse lemma to $\mathrm{Im}\phi_2 \circ U^{-1}$ to get a diffeomorphism $V : O'_2 \to O_3$, $V(u_1, u_2) = (u_1, V_2(u_1, u_2))$ and a $C^1$-map $h : \mathbb{R} \to \mathbb{R}$ such that, for all $v \in O_3$,

$$ \mathrm{Im}\phi_2 \circ (V \circ U)^{-1}(v) = \kappa_2(v_2^2 + h(v_1)) $$

$$ \mathrm{Im}\phi_1 \circ (V \circ U)^{-1}(v) = \kappa_1 v_1. $$

We then conclude as above, using only the imaginary parts $\mathrm{Im}\phi_1$ and $\mathrm{Im}\phi_2$, that

$$ \lambda_2(V \circ U(E(\bar{\delta}) \cap O)) \lesssim \delta_1 \delta_2^{1/2} \le \delta_1^{3/4} \delta_2^{3/4} $$

□

Let us finish this subsection with two examples that illustrate our results.

**Example 7.4.** Let $\phi(z) = ((z_1 + z_2)/2, (z_1 + z_2)/2)$. Then for all $\beta \ge -1$, $C_\phi$ maps $A_\beta^2(\mathbb{D}^2)$ into $A_{\beta+1/4}(\mathbb{D}^2)$.
*Proof.* Indeed, let $\xi \in \mathbb{T}^2$ be such that $\phi(\xi) \in \mathbb{T}^2$. By symmetry we may assume $\xi = e$. Then for $j = 1, 2$,

$$ 
\begin{aligned}
\mathrm{Im}(\phi_j) &= \frac{1}{2}(\theta_1 + \theta_2) + O(\theta_1^2 + \theta_2^2) \\
\mathrm{Re}(\phi_j) &= 1 - \frac{1}{4}(\theta_1^2 + \theta_2^2) + O(|\theta_1|^3 + |\theta_2|^3).
\end{aligned}
$$

Therefore, $s(\phi, e) = 2$ and the result follows from Theorem 7.3. □

For our second example, we recall that, for $\varepsilon \ge 0$, the map $F_\varepsilon$ defined on $\mathbb{C}$ by

$$ F_{\varepsilon}(z) = \frac{3+6z-z^2}{8} + 2i\varepsilon(z-1)^2 - i\varepsilon(z-1)^2 $$

is a holomorphic self-map of $\mathbb{D}$ provided $\varepsilon \in [0, \varepsilon_1)$ for some $\varepsilon_1 > 0$.

**Example 7.5.** Let $\beta \ge -1$, $a_1, a_2 \in [0, \varepsilon_1)$ and let

$$ \phi_{a_1, a_2}(z) = (F_{a_1}(z_1)F_{a_1}(z_1), F_{a_2}(z_2)F_{a_2}(z_2)). $$

Then $C_{\phi_{a_1, a_2}}$ maps $A_\beta^2(\mathbb{D}^2)$ into $A_{\beta+\frac{1}{4}}^2(\mathbb{D}^2)$ if and only if $a_1 \neq a_2$.

*Proof.* Observe first that for $\xi \in \mathbb{T}^2$, $\phi_{a_1, a_2}(\xi) \in \mathbb{T}^2 \iff \xi = e$. We may write, for $j \in \{1, 2\}$,

$$ 
\begin{aligned}
\mathrm{Im}\phi_j(\theta) &= \frac{1}{2}(\theta_1 + \theta_2) - a_j(\theta_1^2 + \theta_2^2) + O(|\theta_1|^3 + |\theta_2|^3) \\
\mathrm{Re}\phi_j(\theta) &= 1 - \frac{1}{8}(\theta_1 + \theta_2)^2 + O(|\theta_1|^3 + |\theta_2|^3).
\end{aligned}
$$

Therefore, $s(\phi_{a_1, a_2}, e) = 1$ and it is easy to check that $r(\phi_{a_1, a_2}, e) = 1$ if and only if $a_1 \neq a_2$. □

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LABORATOIRE DE MATHÉMATIQUES BLAISE PASCAL UMR 6620 CNRS, UNIVERSITÉ CLERMONT AUVERGNE, CAMPUS UNIVERSITAIRE DES CÉZEAUX, 3 PLACE VASARELY, 63178 AUBIÈRE CEDEX, FRANCE.

*Email address: frederic.bayart@uca.fr*

LABORATOIRE DE MATHÉMATIQUES BLAISE PASCAL UMR 6620 CNRS, UNIVERSITÉ CLERMONT AUVERGNE, CAMPUS UNIVERSITAIRE DES CÉZEAUX, 3 PLACE VASARELY, 63178 AUBIÈRE CEDEX, FRANCE.

*Email address: anne.dorval@uca.fr*