Proof We argue by induction on the number of variables n.

For $n=1,$ the statement is Theorem 8.3 of [Reference Brudnyi and Mashreghi9]. Assume the result holds for dimension n, and consider $H^\infty (\mathbb {D}^{\,n+1},A)$ .

Let $(\widehat U_i)_{1\le i\le k}$ be an open cover of $[M(H^\infty (\mathbb {D}))]^{\,n+1}$ . By the product topology, there exist a finite open cover $(\widehat V_j)_{1\le j\le l}$ of $[M(H^\infty (\mathbb {D}))]^n$ and, for each j, a finite open cover $(\widehat W_{jm})_{1\le m\le l_j}$ of $M(H^\infty (\mathbb {D}))$ such that the cover $(\widehat V_j\times \widehat W_{jm})_{j,m}$ of $[M(H^\infty (\mathbb {D}))]^{\,n+1}$ refines $(\widehat U_i)_{1\le i\le k}$ . Set

$$\begin{align*}U_{jm} := V_j\times W_{jm},\qquad V_j = \mathbb{D}^n\cap \widehat V_j,\quad W_{jm}=\mathbb{D}\cap \widehat W_{jm}. \end{align*}$$

Then $(U_{jm})_{j,m}$ is a finite cover of $\mathbb {D}^{\,n+1}$ refining $(U_i)_{1\le i\le k}$ .

Suppose the Bézout equation

(3.1) $$ \begin{align} \sum_{i=1}^k f_i g_i = 1 \end{align} $$

admits local solutions $(g_{1;jm},\dots , g_{k;jm})\in [H^\infty (U_{jm},A)]^k$ on the cover $(U_{jm})_{j,m}$ . For each $i\in \{1,\dots , k\}$ , define slice functions on $W_{jm}$ ,

$$\begin{align*}g_{i;jm}^s(z)(z_1,\dots,z_n):=g_{i;jm}(z_1,\dots,z_n,z),\qquad (z_1,\dots, z_n)\in V_j,\ z\in W_{jm}. \end{align*}$$

These belong to $H^\infty (W_{jm},H^\infty (V_j,A))$ . Likewise, each $f_i$ defines

$$\begin{align*}f_{i;j}^s \in H^\infty(\mathbb D,H^\infty(V_j,A)). \end{align*}$$

Passing to slice functions in (3.1) yields that the $H^\infty (V_j,A)$ -valued Bézout equation on $\mathbb D$ ,

(3.2) $$ \begin{align} \sum_{i=1}^k f_{i;j}^s \,g_i=1, \end{align} $$

has local solutions $(g_{1;jm}^s,\dots , g_{k;jm}^s)\in [H^\infty (W_{jm},H^\infty (V_j,A))]^k$ on the cover $(W_{jm})$ of $\mathbb D$ .

Since $W_{jm}=\mathbb {D}\cap \widehat W_{jm}$ and $\mathcal W_j=(\widehat W_{jm})_{1\le m\le l_j}$ is an open cover of $M(H^\infty (\mathbb {D}))$ , applying [Reference Brudnyi and Mashreghi9, Theorem 8.3], we obtain global solutions $(h_{1;j},\dots , h_{k;j})\in [H^\infty (\mathbb D,H^\infty (V_j,A))]^k$ , $1\le j\le l$ , of (3.2).

Define

$$\begin{align*}g_{i;j}(z_1,\dots,z_{n+1}) := (h_{i;j}(z_{n+1}))(z_1,\dots,z_n),\qquad (z_1,\dots, z_{n+1})\in V_j\times\mathbb D. \end{align*}$$

Then $g_{i;j}\in H^\infty (V_j\times \mathbb D,A)$ and

$$\begin{align*}\sum_{i=1}^k f_i g_{i;j}=1 \quad\text{on } V_j\times \mathbb D. \end{align*}$$

Hence, we obtain local bounded holomorphic solutions of the original Bézout equation (3.1) on the cover $(V_j\times \mathbb D)_{1\le j\le l}$ of $\mathbb D^{\,n+1}$ .

Now view each $g_{i;j}$ as a slice function

$$\begin{align*}g_{i;j}^s \in H^\infty(V_j,H^\infty(\mathbb D,A)). \end{align*}$$

Similarly, $f_i$ induce slice functions

$$\begin{align*}f_i^s \in H^\infty(\mathbb D^n,H^\infty(\mathbb D,A)). \end{align*}$$

Then the solvability of the Bézout equation (3.1) on the cover $(V_j\times \mathbb D)$ transforms into the identities

$$\begin{align*}\sum_{i=1}^k f_i^s g_{i;j}^s = 1 \quad \text{in } H^\infty(\mathbb D,A)\ \text{for each } V_j, \end{align*}$$

that is, we obtain, with respect to the cover $(V_j)_{1\le j\le l}$ , local bounded holomorphic solutions of the Bézout equation on $\mathbb D^n$ ,

(3.3) $$ \begin{align} \sum_{i=1}^k f_i^s h_i = 1 \end{align} $$

with values in the algebra $H^\infty (\mathbb D,A)$ , over the cover $(V_j)_{1\le j\le l}$ of $\mathbb D^n$ , where $V_j = \mathbb {D}^n\cap \widehat V_j$ and $\mathcal V=(\widehat V_j)_{1\le j\le l}$ is a finite open cover of $[M(H^\infty (\mathbb {D}))]^n$ .

By the induction hypothesis, there exists a solution $(h_1,\dots , h_k) \in [H^\infty (\mathbb D^n, H^\infty (\mathbb D,A))]^k$ of (3.3).

Finally, set

$$\begin{align*}g_i(z_1,\dots,z_{n+1}) := (h_i(z_1,\dots,z_n))(z_{n+1}), \quad (z_1,\dots, z_{n+1})\in\mathbb D^{n+1}. \end{align*}$$

Then $g_i\in H^\infty (\mathbb D^{n+1},A)$ and

$$\begin{align*}\sum_{i=1}^k f_i g_i=1. \end{align*}$$

This completes the induction.

Proof Recall that $A(F_\xi )$ is the uniform closure of $\widehat {H^\infty (\mathbb {D}^n)}|_{F_\xi }$ , hence $M\!\left (A(F_\xi )\right )$ is exactly the $H^\infty $ -hull of $F_\xi $ . Also recall that

$$\begin{align*}\pi_n \colon M\!\left(H^\infty(\mathbb{D}^n)\right)\longrightarrow \big[M(H^\infty(\mathbb{D}))\big]^{n} \end{align*}$$

is the transpose of the canonical embedding

$$\begin{align*}\widehat{\otimes}_{\varepsilon}^{\,n} H^\infty(\mathbb{D})\ \hookrightarrow\ H^\infty(\mathbb{D}^n), \end{align*}$$

and that the embedded algebra separates the fibers of $\pi _n$ .

Since the $H^\infty $ -hull of $\pi _n^{-1}(\xi )$ coincides with $\pi _n^{-1}(\xi )$ , we have $M\!\left (A(F_\xi )\right )\subset \pi _n^{-1}(\xi )$ .

For the reverse inclusion, suppose $\zeta \in \pi _n^{-1}(\xi )\setminus M\!\left (A(F_\xi )\right )$ . By the definition of the $H^\infty $ -hull, there exists $f\in H^\infty (\mathbb {D}^n)$ with $\hat f(\zeta )=1$ and

$$\begin{align*}m:=\max_{F_\xi}|\hat f|<1. \end{align*}$$

Set $g:=1-f$ . Then $\hat g(\zeta )=0$ and $|\hat g(x)|\ge 1-m>0$ for all $x\in F_\xi $ .

By continuity of $\hat g$ and compactness of $F_\xi $ , there exists an open neighborhood $\mathcal U$ of $F_\xi $ in $\mathrm {cl}(\mathbb D^n)$ on which $|\hat g|\ge \tfrac 12(1-m)$ . Since the sets

$$\begin{align*}\pi_n^{-1}(\widehat U)\cap\mathrm{cl}(\mathbb D^n), \quad \widehat U\subset\big[M(H^\infty(\mathbb D))\big]^n \end{align*}$$

with $\widehat U$ an open neighborhood of $\xi $ , form a neighborhood basis of $F_\xi $ , we may choose $\widehat U$ so that

$$\begin{align*}\pi_n^{-1}(\widehat U)\cap\mathrm{cl}(\mathbb D^n)\subset\mathcal U. \end{align*}$$

Set $U:=\widehat U\cap \mathbb D^n$ . Then

(4.1) $$ \begin{align} |g(z)|\ge \tfrac12(1-m)\quad\text{for all }z\in U. \end{align} $$

For each $x=(x_1,\dots ,x_n)\in \big [M(H^\infty (\mathbb D))\big ]^n\setminus \widehat U$ , choose an index $r=r(x)$ such that $x_r\ne \xi _r$ . Since $H^\infty (\mathbb D)$ separates points of its maximal ideal space [Reference Suárez38, Theorem 2.4], choose $\psi _x\in H^\infty (\mathbb D)$ with $\widehat {\psi _x}(x_r)\ne \widehat {\psi _x}(\xi _r)$ , and set

$$\begin{align*}\varphi_x:=\psi_x-\widehat{\psi_x}(\xi_r). \end{align*}$$

Then

$$\begin{align*}\widehat{\varphi_x}(\xi_r)=0, \quad \widehat{\varphi_x}(x_r)\ne0. \end{align*}$$

Define

$$\begin{align*}g_x(z_1,\dots,z_n):=\varphi_x(z_r). \end{align*}$$

Then $g_x\in \widehat \otimes _{\varepsilon }^{\,n}H^\infty (\mathbb D)$ , and if

$$\begin{align*}G_x(\eta_1,\dots,\eta_n):=\widehat{\varphi_x}(\eta_r), \end{align*}$$

then $\widehat {g_x}=G_x\circ \pi _n$ . In particular, $\widehat {g_x}$ vanishes on $\pi _n^{-1}(\xi )$ . By continuity of $G_x$ , there exist an open neighborhood $\widehat W_x$ of x and $\delta _x>0$ such that

$$\begin{align*}|\widehat{g_x}(y)|\ge\delta_x \quad\text{for all }y\in\pi_n^{-1}(\widehat W_x). \end{align*}$$

The compact set $\big [M(H^\infty (\mathbb D))\big ]^n\setminus \widehat U$ is covered by finitely many such neighborhoods $\widehat W_{x_1},\dots ,\widehat W_{x_q}$ . Write

$$\begin{align*}g_i:=g_{x_i},\quad \widehat W_i:=\widehat W_{x_i},\quad W_i:=\widehat W_i\cap\mathbb D^n,\quad \delta_i:=\delta_{x_i}. \end{align*}$$

Then

(4.2) $$ \begin{align} \widehat g_i\big|_{\pi_n^{-1}(\xi)}=0, \quad |g_i(z)|\ge\delta_i\quad\text{for all }z\in W_i, \quad 1\le i\le q. \end{align} $$

The set $\widehat U$ , together with the sets $\widehat W_i$ , $1\le i\le q$ , forms a finite open cover of $\big [M(H^\infty (\mathbb D))\big ]^n$ . On U, the tuple $(g,g_1,\dots ,g_q)$ has the bounded holomorphic local Bézout solution $(g^{-1},0,\dots ,0)$ , while on $W_i$ , it has the local solution whose $g_i$ -coordinate is $g_i^{-1}$ and whose other coordinates are zero. Therefore, Theorem 2.1, with $A=\mathbb C$ , yields $f_0,f_1,\dots ,f_q\in H^\infty (\mathbb D^n)$ such that

$$\begin{align*}gf_0+g_1f_1+\cdots+g_qf_q=1\quad\text{on }\mathbb D^n. \end{align*}$$

Taking Gelfand transforms and restricting to $\pi _n^{-1}(\xi )$ , equation (4.2) gives

$$\begin{align*}\widehat f_0(x)\,\widehat g(x)=1 \qquad\text{for all }x\in\pi_n^{-1}(\xi). \end{align*}$$

But $\hat g(\zeta )=0$ , a contradiction. Hence, $\pi _n^{-1}(\xi )\subset M\!\left (A(F_\xi )\right )$ , and therefore

$$\begin{align*}\pi_n^{-1}(\xi)=M\!\left(A(F_\xi)\right). \end{align*}$$

Finally,

$$\begin{align*}\mathrm{cl}(\mathbb{D}^n)=M\!\left(H^\infty(\mathbb{D}^n)\right) \end{align*}$$

(i.e., the corona theorem holds for $H^\infty (\mathbb {D}^n)$ ) if and only if

$$\begin{align*}M\!\left(A(F_\xi)\right)=F_\xi\qquad \text{for all }\xi\in \big[M(H^\infty(\mathbb{D}))\big]^n, \end{align*}$$

equivalently, the $H^\infty $ -hull of each fiber $F_\xi $ coincides with $F_\xi $ .

Proof of (i)

Let $ \xi = (\xi _1, \dots , \xi _n) \in M_a^n $ . Since each coordinate $ \xi _j \in M(H^\infty (\mathbb {D})) $ lies in a nontrivial Gleason part, a classical result of Hoffman (see, e.g., [Reference Hoffman25, Theorem 5.5]) guarantees the existence of an interpolating sequence

$$\begin{align*}\zeta_j = \{ z_{j,k} \}_{k=1}^\infty \subset \mathbb{D} \end{align*}$$

whose closure in $ M(H^\infty (\mathbb {D})) $ contains $ \xi _j $ , for $ j = 1, \dots , n $ .

Define the product sequence

$$\begin{align*}\zeta := \zeta_1 \times \cdots \times \zeta_n = \left\{ z_k := (z_{1,k}, \dots, z_{n,k}) \right\}_{k=1}^\infty \subset \mathbb{D}^n, \end{align*}$$

whose closure in the product topology on $ [M(H^\infty (\mathbb {D}))]^n $ contains $ \xi $ .

Indeed, for each $ j $ and $ k $ , let $ \varphi _{j,k} \in H^\infty (\mathbb {D}) $ be bounded holomorphic functions satisfying

$$\begin{align*}\varphi_{j,k}(z_{j,\ell}) = \delta_{k,\ell}, \quad \sum_{k=1}^\infty |\varphi_{j,k}(z)| \le C_j, \quad \text{for all } z \in \mathbb{D}, \end{align*}$$

for some constants $ C_j = C_j(\zeta _j)> 0 $ (see [Reference Garnett19, Chapter VII, Theorem 2.1]). Define

$$\begin{align*}\psi_k(z) := \prod_{j=1}^n \varphi_{j,k}(z_j), \quad z = (z_1, \dots, z_n) \in \mathbb{D}^n, \quad k \in \mathbb{N}. \end{align*}$$

Then $ \psi _k \in H^\infty (\mathbb {D}^n) $ , and

$$\begin{align*}\psi_k(z_\ell) = \delta_{k,\ell}, \quad \forall\, k, \ell \in \mathbb{N}, \qquad \sum_{k=1}^\infty |\psi_k(z)| \le C := \prod_{j=1}^n C_j, \quad \forall\, z \in \mathbb{D}^n. \end{align*}$$

Thus, the map

$$\begin{align*}L \colon \ell^\infty(\zeta) \to H^\infty(\mathbb{D}^n), \qquad L\big( (c_k)_{k=1}^\infty \big) := \sum_{k=1}^\infty c_k \psi_k, \end{align*}$$

is a bounded linear operator of norm $ \le C $ , with $ L((c_k))|_\zeta = (c_k) $ for all $ (c_k) \in \ell ^\infty (\zeta ) $ . Hence, $ \zeta $ is interpolating for $H^\infty (\mathbb {D}^n)$ , as claimed.

It follows from Claim 1 that the restriction map

$$\begin{align*}H^\infty(\mathbb{D}^n) \to \ell^\infty(\zeta), \qquad f \mapsto f|_\zeta, \end{align*}$$

induces, by transposition, a continuous map of maximal ideal spaces

$$\begin{align*}i_\zeta \colon M\big( \ell^\infty(\zeta) \big) \longrightarrow M\big( H^\infty(\mathbb{D}^n) \big). \end{align*}$$

Because the restriction map is surjective, $i_\zeta $ is injective; since its domain is compact and its codomain is Hausdorff, $i_\zeta $ is a topological embedding. Since $ \ell ^\infty (\zeta ) $ is self-adjoint, its maximal ideal space $ M(\ell ^\infty (\zeta )) $ is homeomorphic to the Stone–Čech compactification $ \beta (\mathbb {N}) $ , and

$$\begin{align*}i_\zeta \big( M(\ell^\infty(\zeta)) \big) = \mathrm{cl}(\zeta) \subset \mathrm{cl}(\mathbb{D}^n) \end{align*}$$

is totally disconnected.

Let $ B_{\zeta _j} \in H^\infty (\mathbb {D}) $ be the Blaschke product vanishing precisely on $ \zeta _j $ , and denote by $ \hat {B}_{\zeta _j} \in C\big ( M(H^\infty (\mathbb {D})) \big )$ its Gelfand transform. Let $ p_j \colon [M(H^\infty (\mathbb {D}))]^n \to M(H^\infty (\mathbb {D})) $ be the projection onto the $ j $ th coordinate, and define

$$\begin{align*}\widetilde{B}_{\zeta_j} := \hat{B}_{\zeta_j} \circ p_j \in C\big( [M(H^\infty(\mathbb{D}))]^n \big). \end{align*}$$

By the product topology, the joint zero set of the functions $ \widetilde {B}_{\zeta _j} $ , $ 1 \le j \le n $ , coincides with the closure of the interpolating sequence $ \zeta $ in $ [M(H^\infty (\mathbb {D}))]^n $ and hence contains  $ \xi $ .

By definition, the pullbacks $ \pi _n^* \widetilde {B}_{\zeta _j} \in C\big (M(H^\infty (\mathbb {D}^n))\big ) $ are the Gelfand transforms of the functions $ (p_j|_{\mathbb {D}^n})^*(B_{\zeta _j}) \in H^\infty (\mathbb {D}^n) $ , given by

$$\begin{align*}(p_j|_{\mathbb{D}^n})^*(B_{\zeta_j})(z_1,\dots, z_n) = B_{\zeta_j}(z_j), \qquad 1 \le j \le n, \end{align*}$$

and they vanish on $F_\xi $ .

We next identify the joint zero set of these pullbacks.

To prove this, suppose, to the contrary, that there exists a point $ \eta \in \mathrm {cl}(\mathbb {D}^n) \setminus \mathrm {cl}(\zeta ) $ such that

$$\begin{align*}\pi_n^* \widetilde{B}_{\zeta_j}(\eta) = 0 \quad \text{for all } j = 1, \dots, n. \end{align*}$$

Let $ U \subset M(H^\infty (\mathbb {D}^n)) $ be an open neighborhood of $ \eta $ whose closure is disjoint from $ \mathrm {cl}(\zeta ) $ . For each $ \varepsilon \in (0,1) $ , consider the open set

$$\begin{align*}U_\varepsilon := U \cap \big\{ \theta \in M(H^\infty(\mathbb{D}^n)) : |\pi_n^* \widetilde{B}_{\zeta_j}(\theta)| < \varepsilon \text{ for all } 1 \le j \le n \big\}. \end{align*}$$

Define the pseudohyperbolic metric on $ \mathbb {D}^n $ by

$$\begin{align*}\rho_n(z, w) := \max_{1 \le j \le n} \rho(z_j, w_j), \qquad z, w \in \mathbb{D}^n, \end{align*}$$

where $ \rho $ is the pseudohyperbolic metric on $ \mathbb {D} $ (see (2.5)). Since $ \mathbb {D}^n $ is dense in $ \mathrm {cl}(\mathbb {D}^n) $ , we may select a point

$$\begin{align*}z_\varepsilon \in U_\varepsilon \cap \mathbb{D}^n. \end{align*}$$

By a result of Hoffman (see, e.g., [Reference Garnett19, Chapter X, Lemma 1.4]), we have

(5.1) $$ \begin{align} \lim_{\varepsilon \to 0} \mathrm{dist}(z_\varepsilon, \zeta) = 0, \end{align} $$

where, for $ z \in \mathbb {D}^n $ and $ Z \subset \mathbb {D}^n $ ,

$$\begin{align*}\mathrm{dist}(z, Z) := \inf_{y \in Z} \rho_n(z, y). \end{align*}$$

Let $ \nu \in \mathrm {cl}(\mathbb {D}^n) $ be a limit point of the sequence $ \{ z_{1/k} \}_{k \in \mathbb {N}} $ . By the Schwarz–Pick lemma (see, e.g., [Reference Garnett19, Chapter I, Lemma 1.2]) and (5.1), we obtain $ \nu \in \mathrm {cl}(\zeta ) $ . On the other hand, since $ \nu \in \mathrm {cl}(U) $ and $ \mathrm {cl}(U) \cap \mathrm {cl}(\zeta ) = \varnothing $ , a contradiction, proving Claim 2 .

Claim 2 implies that

$$\begin{align*}F_\xi := \pi_n^{-1}(\xi) \cap \mathrm{cl}(\mathbb{D}^n) \subset \left\{ \nu\in \mathrm{cl}(\mathbb{D}^n) : \pi_n^* \widetilde{B}_{\zeta_j}(\nu) = 0,\ 1 \le j \le n \right\} \subset \mathrm{cl}(\zeta). \end{align*}$$

Since $ \zeta $ is an interpolating sequence for $ H^\infty (\mathbb {D}^n) $ , the restriction of the algebra $ \mathcal {O}\big (M(H^\infty (\mathbb {D}^n))\big ) $ to $ \mathrm {cl}(\zeta ) $ coincides with the full algebra $ C\big (\mathrm {cl}(\zeta )\big ) $ . By the Tietze–Urysohn extension theorem, the same property holds for any closed subset of $ \mathrm {cl}(\zeta ) $ , and in particular for the compact set $ F_\xi \subset \mathrm {cl}(\zeta ) $ . Hence, $ A(F_\xi ) = C(F_\xi ) $ , and therefore the maximal ideal space $ M(A(F_\xi )) $ coincides with $ F_\xi $ . By Theorem 2.2, we conclude that

$$\begin{align*}\pi_n^{-1}(\xi) = F_\xi \subset \mathrm{cl}(\zeta). \end{align*}$$

As $ \mathrm {cl}(\zeta ) $ is totally disconnected, the same is true for $ F_\xi $ .

This completes the proof of part (i) of the theorem.

Proof of (ii) If exactly one coordinate of $ \xi $ lies in $ M_s $ and the remaining $ n-1 $ coordinates lie in $ M_a $ , then the natural action of the symmetric group on coordinates of $ \mathbb {D}^n $ preserves both algebras $ \widehat {\otimes }_\varepsilon ^n H^\infty (\mathbb {D}) $ and $ H^\infty (\mathbb {D}^n) $ . Consequently, it extends to actions on their maximal ideal spaces $ \big [M\big (H^\infty (\mathbb {D})\big )\big ]^n$ and $ M\big (H^\infty (\mathbb {D}^n)\big ) $ , and these actions commute with the projection

$$\begin{align*}\pi_n \colon M\big(H^\infty(\mathbb{D}^n)\big) \to \big[M\big(H^\infty(\mathbb{D})\big)\big]^n. \end{align*}$$

In particular, the fibers of $ \pi _n $ corresponding to points in the same orbit are homeomorphic and have identical analytic structure. Hence, without loss of generality, we may and will assume that

$$\begin{align*}\xi \in M_a^{n-1} \times M_s. \end{align*}$$

We first show that

$$\begin{align*}\pi_n^{-1}(\xi) = F_\xi := \pi_n^{-1}(\xi)\cap \mathrm{cl}(\mathbb{D}^n) \end{align*}$$

(see Lemma 5.1 below).

Let $\xi = (\xi ^{\,n-1},\xi _n)$ , where $\xi ^{\,n-1} \in M_a^{\,n-1}$ and $\xi _n \in M_s$ . By Claim 1 in the proof of part (i) of the theorem, there exists an interpolating sequence $\zeta = \{z_k\} \subset \mathbb {D}^{\,n-1}$ for $H^\infty (\mathbb {D}^{\,n-1})$ , which is the Cartesian product of interpolating sequences for $H^\infty (\mathbb {D})$ in the coordinate disks, such that the closure of $\zeta $ in $\big [M(H^\infty (\mathbb {D}))\big ]^{n-1}$ contains the point $\xi ^{\,n-1}$ . Moreover, there exist functions $\psi _k \in H^\infty (\mathbb {D}^{\,n-1})$ , $k \in \mathbb {N}$ , such that

$$\begin{align*}\psi_k(z_i) = \delta_{k,i}, \quad \forall\, k,i \in \mathbb{N}, \qquad \sum_{k=1}^{\infty} \lvert \psi_k(z) \rvert \le C = C(\zeta), \quad \forall\, z \in \mathbb{D}^{\,n-1}. \end{align*}$$

Then for every uniformly bounded sequence of functions $\{f_k\} \subset H^\infty (\mathbb {D})$ , there exists $f \in H^\infty (\mathbb {D}^n)$ such that

$$\begin{align*}f(z_k, z) = f_k(z) \quad \forall\, z \in \mathbb{D},\ \forall\, k \in \mathbb{N}. \end{align*}$$

Indeed, set

$$\begin{align*}f(w,z) := \sum_{k=1}^{\infty} \psi_k(w)\, f_k(z), \qquad (w,z) \in \mathbb{D}^{\,n-1} \times \mathbb{D}. \end{align*}$$

Since $\sum _{k=1}^{\infty } \lvert \psi _k(w) \rvert \le C$ for all $w \in \mathbb {D}^{\,n-1}$ and $\sup _k \lVert f_k \rVert _\infty < \infty $ , the series defining f converges uniformly on compact subsets of $\mathbb {D}^{\,n-1} \times \mathbb {D}$ , and

$$\begin{align*}\lVert f \rVert_\infty \le C\, \sup_{k} \lVert f_k \rVert_\infty. \end{align*}$$

Uniform convergence on compacta shows that $f \in H^\infty (\mathbb {D}^n)$ . Moreover, for each $i \in \mathbb {N}$ and every $z \in \mathbb {D}$ ,

$$\begin{align*}f(z_i, z) = \sum_{k=1}^{\infty} \psi_k(z_i)\, f_k(z) = \sum_{k=1}^{\infty} \delta_{k,i}\, f_k(z) = f_i(z), \end{align*}$$

as required.

Thus, the restriction map

(5.2) $$ \begin{align} R \colon H^\infty(\mathbb{D}^n) \to H^\infty(\mathbb{D}\times\mathbb{N}), \quad R(f)(k) := f \big|_{\{z_k\}\times \mathbb{D}}, \end{align} $$

is a surjective homomorphism of Banach algebras, where $H^\infty (\mathbb {D}\times \mathbb {N})$ is equipped with the norm

$$\begin{align*}\lVert (g_k)_{k\in\mathbb{N}} \rVert := \sup_{k\in\mathbb{N}} \lVert g_k \rVert_\infty. \end{align*}$$

In particular, the induced transpose map on maximal ideal spaces

$$\begin{align*}R^* \colon M\big(H^\infty(\mathbb{D}\times\mathbb{N})\big) \longrightarrow M\big(H^\infty(\mathbb{D}^n)\big), \qquad R^*(\varphi) := \varphi \circ R, \end{align*}$$

is a continuous embedding, and its image is $H^\infty $ -convex; that is, for every character $\eta \in M\big (H^\infty (\mathbb {D}^n)\big ) \setminus R^*\!\big (M(H^\infty (\mathbb {D}\times \mathbb {N}))\big )$ , there exists $f \in H^\infty (\mathbb {D}^n)$ such that

$$\begin{align*}\lvert \hat{f}(\eta) \rvert> \sup_{ R^*(M(H^\infty(\mathbb{D}\times\mathbb{N}))) } \lvert \hat{f} \rvert. \end{align*}$$

Lemma 5.1 We have

$$\begin{align*}R^*\big(M(H^\infty(\mathbb{D}\times\mathbb{N}))\big) \supset \pi_n^{-1}(\xi). \end{align*}$$

Moreover,

$$\begin{align*}\pi_n^{-1}(\xi)=F_\xi. \end{align*}$$

Proof Let us first prove that $\pi _n^{-1}(\xi ) \subset \mathrm {im}\ R^*$ . Set

$$\begin{align*}I := \ker R = \big\{ f \in H^\infty(\mathbb{D}^n) : f(z_k, z) = 0 \ \forall\,k\in\mathbb{N},\ \forall z\in\mathbb{D} \big\}. \end{align*}$$

Let U be an open neighborhood of

$$\begin{align*}F_\xi := \pi_n^{-1}(\xi) \cap \mathrm{cl}(\mathbb{D}^n) \end{align*}$$

in $M\big (H^\infty (\mathbb {D}^n)\big )$ . As observed above, we may assume, without loss of generality, that

$$\begin{align*}U=\pi_n^{-1}(V)\cap \mathrm{cl}(\mathbb{D}^n) \end{align*}$$

for some open neighborhood V of $\xi $ in $\big [M(H^\infty (\mathbb {D}))\big ]^n$ . Since the closure of $\zeta \times \mathbb {D}$ in $\big [M(H^\infty (\mathbb {D}))\big ]^n$ contains $\xi $ , there exists a subsequence $\zeta _U\subset \zeta $ such that $\zeta _U\times \mathbb {D}\subset V$ . Varying U over a basis of open neighborhoods of $F_\xi $ , we conclude that

$$\begin{align*}F_\xi \subset \mathrm{cl}(\zeta\times\mathbb{D}) \qquad \text{(recall that } \mathrm{cl} = \mathrm{cl}_{M(H^\infty(\mathbb{D}^n))}). \end{align*}$$

Thus, the Gelfand transform of every $f\in I$ vanishes on $F_\xi $ , and hence on its $H^\infty $ -hull. By Theorem 2.2, this hull equals $\pi _n^{-1}(\xi )$ . Therefore, every $\eta \in \pi _n^{-1}(\xi )$ annihilates I and thus factors through

$$\begin{align*}H^\infty(\mathbb{D}^n)/I\simeq H^\infty(\mathbb{D}\times\mathbb{N}), \end{align*}$$

that is,

$$\begin{align*}\eta=R^*(\varphi) \qquad\text{for some }\varphi\in M\!\big(H^\infty(\mathbb{D}\times\mathbb{N})\big). \end{align*}$$

This proves $\pi _n^{-1}(\xi )\subset \mathrm {im}\ R^*$ . Using this fact, let us prove that $\pi _n^{-1}(\xi )=F_\xi $ .

Indeed, the corona theorem for $H^\infty (\mathbb {D}\times \mathbb {N})$ (which follows from uniform bounds in the proof of Carleson’s original theorem, see, e.g., [Reference Garnett19]) implies that $\mathbb {D}\times \mathbb {N}$ is an open dense subset of $M(H^\infty (\mathbb {D}\times \mathbb {N}))$ . By construction of $R^*$ ,

$$\begin{align*}R^*(\mathbb{D}\times\mathbb{N}) = \zeta\times\mathbb{D}. \end{align*}$$

Since $R^*$ is continuous and injective, it follows that

$$\begin{align*}R^*\big(M(H^\infty(\mathbb{D}\times\mathbb{N}))\big) = \mathrm{cl}\big(\zeta\times\mathbb{D}\big) \subset \mathrm{cl}(\mathbb{D}^n). \end{align*}$$

Hence,

$$\begin{align*}\pi_n^{-1}(\xi) = \pi_n^{-1}(\xi)\cap R^*\big(M(H^\infty(\mathbb{D}\times\mathbb{N}))\big) \subset \pi_n^{-1}(\xi)\cap \mathrm{cl}(\mathbb{D}^n) = F_\xi. \end{align*}$$

Thus, $\pi _n^{-1}(\xi )=F_\xi $ , completing the proof of the lemma.

By Lemma 5.1 and Theorem 2.2, we conclude that $M(A(F_\xi )) = F_\xi $ .

We now use the known structure of $M\big (H^\infty (\mathbb {D}\times \mathbb {N})\big )$ to analyze the internal geometry of $F_\xi $ . As shown in [Reference Brudnyi8], every Gleason part in $M\big (H^\infty (\mathbb {D}\times \mathbb {N})\big )$ is either a singleton or an analytic disk; the set of nontrivial parts is open and dense, the set of trivial parts is totally disconnected, and

$$\begin{align*}\dim M\big(H^\infty(\mathbb{D}\times\mathbb{N})\big)=2. \end{align*}$$

We relate this structure to the fibers of the projection $T^*$ introduced below.

The Banach algebra homomorphism

(5.3) $$ \begin{align} T \colon \ell^\infty \longrightarrow H^\infty(\mathbb{D}\times\mathbb{N}), \qquad T\big((c_k)_{k\in\mathbb{N}}\big) := (g_k)_{k\in\mathbb{N}}, \quad g_k(z) \equiv c_k,\ z\in\mathbb{D}, \end{align} $$

is an isometric embedding. Identifying $M(\ell ^\infty )$ with the Stone–Čech compactification $\beta \mathbb {N}$ , its transpose

$$\begin{align*}T^* \colon M\big(H^\infty(\mathbb{D}\times\mathbb{N})\big) \longrightarrow \beta\mathbb{N} \end{align*}$$

is surjective. For characters corresponding to points $(z,n)\in \mathbb {D}\times \mathbb {N}$ , the map $T^*$ assigns evaluation at the index n; in particular,

$$\begin{align*}T^*\big(\mathbb{D}\times\mathbb{N}\big)=\mathbb{N}\subset \beta\mathbb{N}. \end{align*}$$

Fix $\chi \in \beta \mathbb {N}\setminus \mathbb {N}$ . If $G \subset M \big (H^\infty (\mathbb {D}\times \mathbb {N})\big )$ is an analytic disk with $T^*(G)$ containing $\chi $ , then, since $T^*$ is continuous and $\beta \mathbb {N}$ is totally disconnected, the connected set $T^*(G)$ must be a singleton. Hence, $T^*(G)=\{\chi \}$ , so $G \subset (T^*)^{-1}(\chi )$ . Consequently, for every $\chi \in \beta \mathbb {N}\setminus \mathbb {N}$ , the fiber $(T^*)^{-1}(\chi )$ is a disjoint union of analytic disks and points belonging to trivial Gleason parts of $M \big (H^\infty (\mathbb {D}\times \mathbb {N})\big )$ (each such part is a singleton).

Lemma 5.2 Let G be an analytic disk in $M\big (H^\infty (\mathbb {D}\times \mathbb {N})\big )$ such that $R^*(G)$ contains a point $\eta \in F_{\xi } = \pi _n^{-1}(\xi )$ , where $\xi = (\xi ^{\,n-1},\xi _n) \in M_a^{n-1}\times M_s \subset \big [M(H^\infty (\mathbb {D}))\big ]^n$ . Then

$$\begin{align*}R^*(G) \subset F_\xi. \end{align*}$$

Proof Let

$$\begin{align*}p^{\,n-1}\colon \mathbb{D}^n \to \mathbb{D}^{\,n-1} \end{align*}$$

be the projection onto the first $n-1$ coordinates. It induces the following algebra homomorphisms. First, the pullback

(5.4) $$ \begin{align} (p^{\,n-1})^*\colon H^\infty(\mathbb{D}^{\,n-1}) \longrightarrow H^\infty(\mathbb{D}^n),\qquad (p^{\,n-1})^*(f)=f\circ p^{\,n-1}, \end{align} $$

whose transpose is the map

$$\begin{align*}\hat{p}^{\,n-1}\colon M\big(H^\infty(\mathbb{D}^n)\big) \longrightarrow M\big(H^\infty(\mathbb{D}^{\,n-1})\big),\qquad \hat{p}^{\,n-1}(\varphi)=\varphi\circ (p^{\,n-1})^*, \end{align*}$$

and $\hat {p}^{\,n-1}|_{\mathbb {D}^n}=p^{\,n-1}$ .

Second, the restriction

$$\begin{align*}(p^{\,n-1})^*\big|_{\widehat{\otimes}_\varepsilon^{\,n-1} H^\infty(\mathbb{D})} \;\colon\; \widehat{\otimes}_\varepsilon^{\,n-1} H^\infty(\mathbb{D}) \longrightarrow \widehat{\otimes}_\varepsilon^{\,n} H^\infty(\mathbb{D}) \end{align*}$$

whose transpose

$$\begin{align*}\bar{p}^{\,n-1}\colon \big[M\big(H^\infty(\mathbb{D})\big)\big]^n \longrightarrow \big[M\big(H^\infty(\mathbb{D})\big)\big]^{\,n-1}, \end{align*}$$

coincides with the projection onto the first $n-1$ coordinates.

The embeddings $\widehat {\otimes }_\varepsilon ^{\,k} H^\infty (\mathbb {D}) \hookrightarrow H^\infty (\mathbb {D}^k)$ are compatible with the pullback $(p^{\,n-1})^*$ . Passing to transposes (see (2.2)) yields

(5.5) $$ \begin{align} \bar{p}^{\,n-1}\circ \pi_{n} \;=\; \pi_{\,n-1}\circ \hat{p}^{\,n-1}. \end{align} $$

Let $\zeta = \{z_k\} \subset \mathbb {D}^{\,n-1}$ be the interpolating sequence fixed at the beginning of the proof, whose closure in $\big [M(H^\infty (\mathbb {D}))\big ]^{n-1}$ contains $\xi ^{\,n-1}$ . Define the restriction homomorphism

$$\begin{align*}R_\zeta\colon H^\infty(\mathbb{D}^{\,n-1}) \longrightarrow \ell^\infty(\mathbb{N}),\qquad R_\zeta(f)=\big(f(z_k)\big)_{k\in\mathbb{N}}. \end{align*}$$

By the definitions of R, T, and $(p^{\,n-1})^*$ , see (5.2)–(5.4), we have the commutative relation

$$\begin{align*}T\circ R_\zeta \;=\; R\circ (p^{\,n-1})^*. \end{align*}$$

Passing to transposes and combining with (5.5) gives the commutative diagram:

(5.6)

Since $R^*(G)\subset M\big (H^\infty (\mathbb {D}^n)\big )$ meets the fiber $F_\xi =\pi _n^{-1}(\xi )$ and $T^*$ is constant on G, the diagram (5.6) gives, for $\eta \in R^*(G)\cap F_\xi $ ,

(5.7) $$ \begin{align} (\pi_{\,n-1}\circ \hat{p}^{\,n-1})(\eta)=\xi^{\,n-1}. \end{align} $$

Let $T^*(G)=\{\omega \}\subset \beta \mathbb N$ . Then the top square in (5.6) yields

$$\begin{align*}\hat{p}^{\,n-1}\big(R^*(G)\big)=\{(R_\zeta)^*(\omega)\}. \end{align*}$$

Applying $\pi _{\,n-1}$ and using (5.7), we obtain

$$\begin{align*}(\pi_{\,n-1}\circ \hat{p}^{\,n-1})\big(R^*(G)\big)=\{\xi^{\,n-1}\}. \end{align*}$$

For the last coordinate, let

$$\begin{align*}p_n:\big[M(H^\infty(\mathbb{D}))\big]^n\longrightarrow M(H^\infty(\mathbb{D})) \end{align*}$$

be the projection onto the nth factor. Then $(p_n\circ \pi _n)\big (R^*(G)\big )$ is an analytic disk in $M(H^\infty (\mathbb {D}))$ passing through $\xi _n$ . Since $\xi _n\in M_s$ is a trivial Gleason part, Hoffman’s results [Reference Hoffman25, Theorems 4.3, 5.5, and 5.6] imply that this image is a singleton:

$$\begin{align*}(p_n\circ \pi_n)\big(R^*(G)\big)=\{\xi_n\}. \end{align*}$$

Combining the two displays gives

$$\begin{align*}\pi_n\big(R^*(G)\big)=\{(\xi^{\,n-1},\xi_n)\}=\{\xi\}, \end{align*}$$

hence $R^*(G)\subset \pi _n^{-1}(\xi )=F_\xi $ , as claimed.

Combining Lemmas 5.1 and 5.2 with the structure of $M\big (H^\infty (\mathbb {D}\times \mathbb {N})\big )$ established in [Reference Brudnyi8], we obtain that $F_\xi $ is a disjoint union of its Gleason parts, each of which is either a single point or an analytic disk. Moreover, since $F_\xi $ embeds into $M\big (H^\infty (\mathbb {D}\times \mathbb {N})\big )$ , its covering dimension satisfies

$$\begin{align*}\dim F_\xi \le 2. \end{align*}$$

To complete the proof of part (ii), it remains to show that the subset of trivial Gleason parts in $F_\xi $ is nonempty. Equivalently, if the fiber $F_\xi $ contains a nontrivial Gleason part (an analytic disk), then it must also contain at least one trivial Gleason part.

Proof It was shown in [Reference Behrens4, Lemma 2] that the algebra $H^\infty (\mathbb {D}\times \mathbb {N})$ is logmodular on $\mathbb {D} \times \mathbb {N}$ . This, together with the general result of Hoffman [Reference Hoffman24], implies that the Gleason parts of $H^\infty (\mathbb {D}\times \mathbb {N})$ are either points or analytic disks. The latter means that there exists a continuous injective map

$$\begin{align*}\pi:\mathbb{D}\longrightarrow M\big(H^\infty(\mathbb{D}\times\mathbb{N})\big) \end{align*}$$

onto this Gleason part such that $\hat f\circ \pi \in H^\infty (\mathbb {D})$ for all $f\in H^\infty (\mathbb {D}\times \mathbb {N})$ .

Suppose, to the contrary, that there exists a nontrivial Gleason part $G \subset M\big (H^\infty (\mathbb {D}\times \mathbb {N})\big )$ whose closure $\overline {G}$ consists entirely of nontrivial Gleason parts, so that $\overline {G}$ is a disjoint union of analytic disks.

For $f\in H^\infty (\mathbb {D}\times \mathbb {N})$ , let $\mathcal {C}_{\hat f}$ denote the set of all analytic disks in $\overline {G}$ on which the Gelfand transform $\hat f$ is constant.

Claim Each $\mathcal {C}_{\hat f}$ is a nonempty compact subset of $\overline {G}$ .

Indeed, the function $|\hat f|$ , being continuous on the compact set $\overline {G}$ , attains its maximum at some point $\eta $ . Let $G_\eta $ be the nontrivial Gleason part containing $\eta $ , and let $\pi :\mathbb {D}\to G_\eta $ be its analytic parametrization (as in the above definition). Then the function $|\hat f \circ \pi |$ is subharmonic on $\mathbb {D}$ and attains its maximum at the interior point $\pi ^{-1}(\eta )$ . By the maximum modulus principle, it must be constant, and hence $\hat f \circ \pi $ , and therefore $\hat f$ , is constant on $G_\eta $ . Thus, $G_\eta \subset \mathcal {C}_{\hat f}$ , showing that $\mathcal {C}_{\hat f}$ is nonempty.

Next, we show that $\mathcal {C}_{\hat f}$ is compact. Let $\theta $ be a limit point of $\mathcal {C}_{\hat f}$ . By the hypothesis on $\overline {G}$ , the point $\theta $ belongs to some maximal analytic disk, that is, to the Gleason part of $\theta $ . By [Reference Brudnyi8, Theorems 2.5(b) and (d)], there exists an open neighborhood U of $\theta $ in the set of nontrivial Gleason parts of $H^\infty (\mathbb {D}\times \mathbb {N})$ , and a homeomorphism

$$\begin{align*}h : S\times \mathbb{D} \longrightarrow U, \end{align*}$$

where S is a compact subset of the Stone–Čech compactification $\beta \mathbb {N}$ , such that each slice $h(\{\omega \}\times \mathbb {D})$ lies in a nontrivial Gleason part of $H^\infty (\mathbb {D}\times \mathbb {N})$ , and for every $f\in H^\infty (\mathbb {D}\times \mathbb {N}),$ the pullback $\hat f\circ h(\omega ,\cdot )$ belongs to $H^\infty (\mathbb {D})$ .

Since $\theta $ is a limit point of $\mathcal {C}_{\hat f}$ , there exists a net $(\theta _\alpha )\subset \mathcal {C}_{\hat f}\cap U$ converging to $\theta $ . If $\theta _\alpha $ lies in the image under h of the slice corresponding to $\omega _\alpha $ , then $\hat f$ is constant on this image. Passing to the limit within U shows that $\hat f$ is constant on the image under h of the slice containing $\theta $ . By analyticity, $\hat f$ is constant on the entire Gleason part containing $\theta $ ; hence, $\theta \in \mathcal {C}_{\hat f}$ .

Thus, $\mathcal {C}_{\hat f}$ is closed in $\overline {G}$ , and compactness follows since $\overline {G}$ is compact.

Further, given any finite collection $f_1,\dots ,f_k \in H^\infty (\mathbb {D}\times \mathbb {N})$ , consider the continuous nonnegative function

$$\begin{align*}\widehat F := \sum_{j=1}^k |\widehat{f_j}|^2 \end{align*}$$

on $\overline G$ . It attains its maximum at some point $\eta $ . Let $G_\eta $ be the nontrivial Gleason part containing $\eta $ , and let $\pi :\mathbb D\to G_\eta $ be its analytic parametrization. Then

$$\begin{align*}\widehat F\circ\pi =\sum_{j=1}^k|\widehat f_j\circ\pi|^2 \end{align*}$$

is a continuous nonnegative subharmonic function on $\mathbb D$ and attains its maximum at the interior point $\pi ^{-1}(\eta )$ . By the maximum principle for subharmonic functions, it is constant. Hence,

$$\begin{align*}0=\Delta(\widehat F\circ\pi) =4\sum_{j=1}^k\big|(\widehat f_j\circ\pi)'\big|^2, \end{align*}$$

so each $\widehat f_j\circ \pi $ is constant. Thus, each $\widehat f_j$ is constant on $G_\eta $ , and consequently

$$\begin{align*}\mathcal{C}_{\hat{f_1}} \cap \cdots \cap \mathcal{C}_{\hat{f_k}} \neq \varnothing. \end{align*}$$

Since each $\mathcal {C}_{\hat f}$ is compact and all finite intersections are nonempty, the finite intersection property implies that

$$\begin{align*}\bigcap_{f\in H^\infty(\mathbb{D}\times\mathbb{N})} \mathcal{C}_{\hat f} \neq \varnothing. \end{align*}$$

Hence, there exists an analytic disk in $\overline {G}$ on which $\hat f$ is constant for every $f \in H^\infty (\mathbb {D}\times \mathbb {N})$ . This yields a contradiction, since the Gelfand transforms of functions from the algebra $H^\infty (\mathbb {D}\times \mathbb {N})$ separate points of its maximal ideal space. Therefore, $\overline {G}$ cannot consist solely of analytic disks, that is, it must contain a trivial Gleason part, as claimed.

Since $F_\xi $ is contained in the homeomorphic image of $M\big (H^\infty (\mathbb {D}\times \mathbb {N})\big )$ under $R^*$ , which preserves the analytic structure, the subset of trivial Gleason parts in $F_\xi $ is nonempty. This completes the proof of part (ii) of Theorem 2.3.

Proof We assume that H is nonconstant; otherwise, both statements are trivial.

(1) $\Rightarrow $ (2). Choose radii $0<r_1<r_2<\cdots \nearrow 1$ and consider the exhaustion of $\mathbb D^n$ by open polydisks $\mathbb D_{r_j}^n$ of radii $r_j$ . By compactness of the closure $\overline {\mathbb D}_{r_j}^{\,n}$ , there exist $x_1,\dots ,x_N$ with

$$\begin{align*}\overline{\mathbb D}_{r_j}^{\,n} \subset U_{x_1}\cup\cdots\cup U_{x_N}, \end{align*}$$

hence

$$\begin{align*}H(\mathbb D_{r_j}^n)\subset \Sigma_j:=\Sigma_{x_1}\cup\cdots\cup\Sigma_{x_N}, \end{align*}$$

a one-dimensional analytic set.

Let $\nu _j:\Sigma _j^\nu \to \Sigma _j$ be the normalization [Reference Grauert and Remmert21, Chapter 8]. By its universal property, the holomorphic map $H|_{\mathbb D_{r_j}^n}$ lifts to a holomorphic map

$$\begin{align*}H_j^\nu:\mathbb D_{r_j}^n \longrightarrow \Sigma_j^\nu, \qquad H = \nu_j \circ H_j^\nu. \end{align*}$$

Since $H_j^\nu $ is nonconstant and $\mathbb D_{r_j}^n$ is connected, the open mapping theorem implies that

$$\begin{align*}R_j := H_j^\nu(\mathbb D_{r_j}^n) \end{align*}$$

is a connected open Riemann surface.

Set

$$\begin{align*}h_j:=H_j^\nu:\mathbb D_{r_j}^n\to R_j,\qquad \widetilde H_j:=\nu_j|_{R_j}:R_j\to\mathbb C^m, \end{align*}$$

so $H|_{\mathbb D_{r_j}^n}=\widetilde H_j\circ h_j$ . If $r_i<r_j$ , uniqueness of normalization factorizations yields a unique holomorphic map $\varphi _{ij}:R_i\to R_j$ such that $h_j=\varphi _{ij}\circ h_i$ on $\mathbb D_{r_i}^n$ . Thus, $\{R_j,\varphi _{ij}\}$ forms a direct system of Riemann surfaces.