Surjective isometries on a Banach space of analytic functions with bounded derivatives

TL;DR

This paper characterizes surjective isometries on a space of analytic functions with bounded derivatives, extending previous work on linear isometries in Hardy spaces to a broader class of isometries.

Contribution

It provides a complete description of surjective isometries on the space of analytic functions with bounded derivatives, generalizing known results for linear isometries.

Findings

Characterization of surjective isometries on the space of functions with bounded derivatives.

Extension of previous linear isometry results to non-linear isometries.

New insights into the structure of isometries in analytic function spaces.

Abstract

Let $H(\mathbb{D})$ be the linear space of all analytic functions on the open unit disc $\mathbb{D}$ and $H^p(\mathbb{D})$ the Hardy space on $\mathbb{D}$. The characterization of complex linear isometries on $\mathcal{S}^p=\{f\in H(\mathbb{D}):f'\in H^p(\mathbb{D})\}$ was given for $1\leq p<\infty$ by Novinger and Oberlin in 1985. Here, we characterize surjective, not necessarily linear, isometries on $\mathcal{S}^\infty$.

Equations

$$\Vert T(f)-T(g){\Vert }_N=\Vert f-g{\Vert }_N(f,g\in N).$$

$$T(f)(z)$$

$$T(f)(z)=cf(\lambda z),(f\in {\mathcal{S}}^{\infty },z\in \mathbb{D}).$$

$$T(f)(z)=c_{0}f(a)+{\int }_{[a,z]}{c}_1{f}^{\prime }(\rho (\zeta ))d\zeta ,(f\in {\mathcal{S}}^{\infty },z\in \mathbb{D}),$$

$${\mathcal{S}}^{\infty }=\{f\in H(\mathbb{D}):f^{\prime }\in H^{\infty }(\mathbb{D})\}$$

$${\Vert f\Vert }_{(\Sigma )}=\Vert f{\Vert }_{\mathbb{D}}+\Vert f^{\prime }{\Vert }_{\mathbb{D}},{\Vert f\Vert }_{(C)}=\underset{z\in \mathbb{D}}{\sup }(|f(z)|+|f^{\prime }(z)|)\text{mbox}and{\Vert f\Vert }_{(\sigma )}=|f(a)|+\Vert f^{\prime }{\Vert }_{\mathbb{D}}$$

$$|\widehat{u_{\eta_{1}}}(\eta)|=1\quad\mbox{if and only if}\quad\widehat{u_{\eta_{1}}}(\eta)=1\quad\mbox{for $\eta\in\mathcal{M}$}.$$

$${\mathfrak{D}}_{\Sigma }=\mathbb{T}\times \partial \mathcal{A},{\mathfrak{D}}_C=\{(\eta ,\eta ):\eta \in \partial \mathcal{A}\}\text{mbox}and{\mathfrak{D}}_{\sigma }=\{a\}\times \partial \mathcal{A}.$$

$${\mathcal{K}}_e={\mathfrak{D}}_e\times \mathbb{T}.$$

$$\tilde{f}(z,\eta ,w)=\hat{f}(z)+\widehat{f^{\prime }}(\eta )w(z,\eta ,w)\in {\mathcal{K}}_e:$$

$$B_e=\{\tilde{f}\in C({\mathcal{K}}_e):f\in {\mathcal{S}}^{\infty }\}$$

$$\Vert \tilde{f}{\Vert }_{{\mathcal{K}}_{\Sigma }}=\Vert \hat{f}{\Vert }_{\mathbb{T}}+\Vert \widehat{f^{\prime }}{\Vert }_{\partial \mathcal{A}}={\Vert f\Vert }_{(\Sigma )}\text{mbox}and\Vert \tilde{f}{\Vert }_{{\mathcal{K}}_{\sigma }}=|f(a)|+\Vert \widehat{f^{\prime }}{\Vert }_{\partial \mathcal{A}}={\Vert f\Vert }_{(\sigma )}$$

$${\Vert f\Vert }_{(\Sigma )}=\Vert \hat{f}{\Vert }_{\mathbb{T}}+\Vert \widehat{f^{\prime }}{\Vert }_{\partial \mathcal{A}}=|\hat{f}({\xi }_0)+\widehat{f^{\prime }}({\eta }_0)w_0)|\le \Vert \tilde{f}{\Vert }_{{\mathcal{K}}_{\Sigma }}\le \Vert \hat{f}{\Vert }_{\mathbb{T}}+\Vert \widehat{f^{\prime }}{\Vert }_{\partial \mathcal{A}}={\Vert f\Vert }_{(\Sigma )},$$

$$\Vert \tilde{f}{\Vert }_{{\mathcal{K}}_C}=\underset{(\eta ,\eta ,w)\in {\mathcal{K}}_C}{\sup }|\hat{f}(\eta )+\widehat{f^{\prime }}(\eta )w|={\Vert f\Vert }_{(C)}.$$

$$|\hat{f}(z)|+|\widehat{f^{\prime }}(z)|=|\hat{f}(z)+\widehat{f^{\prime }}(z)w_z|\le \underset{\eta \in \partial \mathcal{A}}{\sup }|\hat{f}(\eta )+\widehat{f^{\prime }}(\eta )w_z|\le \Vert \tilde{f}{\Vert }_{{\mathcal{K}}_C}.$$

$$\Vert \tilde{f}{\Vert }_{{\mathcal{K}}_C}=|\hat{f}({\eta }_0)+\widehat{f^{\prime }}({\eta }_0)w_0|=|\tilde{f}({\eta }_0,{\eta }_0,w_0)|\le |\tilde{f}(z_0,z_0,w_0)|+\epsilon =|\hat{f}(z_0)+\widehat{f^{\prime }}(z_0)w_0|+\epsilon ,$$

$${\delta }_x(\tilde{f})=\tilde{f}(x)(\tilde{f}\in B_e).$$

$$I_0(v)(z)={\int }_{[0,z]}v(\zeta )d\zeta (v\in H^{\infty }(\mathbb{D}),z\in \mathbb{D}),$$

$$\widehat{g_1}(z_1)\ne 0=\widehat{g_1}(z_2)=\widehat{g_1^{\prime }}(z_1)=\widehat{g_1^{\prime }}(z_2)\text{mbox}and\widehat{g_2}(z_1)\ne 0=\widehat{g_2}({\xi }_2)=\widehat{g_2^{\prime }}(z_1)=\widehat{g_2^{\prime }}({\xi }_2).$$

$$\Vert f{\Vert }_{\mathbb{D}}\le \epsilon ,\widehat{f^{\prime }}({\eta }_0)=1=\Vert \widehat{f^{\prime }}{\Vert }_{\mathcal{M}}\text{mbox}and|\widehat{f^{\prime }}|<\epsilon \text{mbox}on\mathcal{M}\setminus V.$$

$$|f(z_0)|=|I_0(v)(z_0)|=\left|{\int }_{[0,z_0]}v(\zeta )d\zeta \right|\le |z_0|\underset{\zeta \in [0,z_0]}{\sup }|v(\zeta )|<{\epsilon }_0.$$

$$|f(z_0)|$$

$$\le |f(z_r)|+|z_r-z_0|\underset{\zeta \in [z_r,z_0]}{\sup }|v(\zeta )|\le {\epsilon }_0+r{\epsilon }_0=(1+r){\epsilon }_0.$$

$$P_V={\pi }^{-1}(V)\times \mathbb{T}\text{mbox}and{P}_{V^c}={\pi }^{-1}(V^c)\times \mathbb{T}.$$

$$(z_0,{\eta }_0,w_0)\in P_V,P_V\cup P_{V^c}={\mathcal{K}}_e\text{mbox}and{P}_V\cap P_{V^c}=\varnothing .$$

$$\Vert f_0{\Vert }_{\mathbb{D}}\le \epsilon ,\widehat{f_0^{\prime }}({\eta }_0)=1=\Vert \widehat{f_0^{\prime }}{\Vert }_{\mathcal{M}}\text{mbox}and|\widehat{f_0^{\prime }}|<\epsilon \text{mbox}on{V}^c.$$

$$|\widetilde{f_0}|\le \epsilon +1\text{mbox}on{P}_V\text{mbox}and|\widetilde{f_0}|<2\epsilon \text{mbox}on{P}_{V^c}.$$

$${\int }_{{\mathcal{K}}_e}\widetilde{f_0}d\mu ={\delta }_{x_0}(\widetilde{f_0})=\widehat{f_0}(z_0)+\widehat{f_0^{\prime }}({\eta }_0)w_0=\widehat{f_0}(z_0)+w_0.$$

Full text

∗Corresponding author: miura@math.sc.niigata-u.ac.jp

The first author was supported by KAKENHI Grant Number 20K03650.

1. Introduction and Main results

Let $(N,\left\|\cdot\right\|_{N})$ be a real or complex normed linear space. A mapping $T\colon N\to N$ is said to be an isometry if

[TABLE]

Here we notice that isometries need not be linear. A linear map $S\colon N\to N$ is an isometry if it preserves the norm in the following sense: $\|S(f)\|_{N}=\|f\|_{N}$ for $f\in N$. Banach [1] gave the characterization of surjective, not necessarily linear, isometries on the Banach space of all continuous real valued functions $C_{\mathbb{R}}(K)$, $K$ a compact metric space, with the supremum norm. Many mathematicians have been investigating linear isometries on various subspaces of the linear space $H(\mathbb{D})$ of all analytic functions on the open unit disc $\mathbb{D}$. Nagasawa [21] characterized surjective complex linear isometries on uniform algebras. Let $H^{p}(\mathbb{D})$ be the Hardy spaces for $1\leq p\leq\infty$. By the characterization of Nagasawa, we see that every surjective complex linear isometry on $H^{\infty}(\mathbb{D})$ is a weighted composition operator. deLeeuw, Rudin and Wermer [6] obtained the form of surjective complex linear isometries on $H^{\infty}(\mathbb{D})$ and $H^{1}(\mathbb{D})$. Forelli [9] extended the result by deLeeuw, Rudin and Wermer to the characterization of complex linear isometries, which need not be surjective, on $H^{p}(\mathbb{D})$ for $1\leq p<\infty$ with $p\neq 2$. Results for isometries on other spaces of analytic functions were obtained by Botelho [2], Cima and Wogen [5], Hornor and Jamison [15] and Kolaski [16].

Let $f^{\prime}$ be the derivative of $f\in H(\mathbb{D})$. Novinger and Oberlin [22] gave the characterization of complex linear isometries on the space $\mathcal{S}^{p}=\{f\in H(\mathbb{D}):f^{\prime}\in H^{p}(\mathbb{D})\}$ for $1\leq p<\infty$, $p\neq 2$ with respect to the following norms: $|f(0)|+\left\|f^{\prime}\right\|_{H^{p}}$ and $\|f\|_{\infty}+\left\|f^{\prime}\right\|_{H^{p}}$, where $\left\|\cdot\right\|_{H^{p}}$ and $\left\|\cdot\right\|_{\infty}$ are the usual norm of $H^{p}(\mathbb{D})$ and the supremum norm on $\mathbb{D}$, respectively. In [22], the authors assume no surjectivity of isometries on $\mathcal{S}^{p}$, while they do assume that $p$ is finite. The purpose of this paper is to characterize all surjective, not necessarily linear, isometries on $\mathcal{S}^{\infty}$ with the norms $\|f\|_{\infty}+\|f^{\prime}\|_{\infty}$, $\sup_{z\in\mathbb{D}}(|f(z)|+|f^{\prime}(z)|)$ and $|f(a)|+\|f^{\prime}\|_{\infty}$ for $a\in\mathbb{D}$.

The following are the main results of this paper.

Theorem 1**.**

A surjective map $T\colon\mathcal{S}^{\infty}\to\mathcal{S}^{\infty}$ is an isometry with the norms $\left\|f\right\|_{(\Sigma)}=\sup_{z\in\mathbb{D}}|f(z)|+\sup_{w\in\mathbb{D}}|f^{\prime}(w)|$ or $\left\|f\right\|_{(C)}=\sup_{z\in\mathbb{D}}(|f(z)|+|f^{\prime}(z)|)$ for $f\in\mathcal{S}^{\infty}$, if and only if there exist constants $c,\lambda\in\mathbb{C}$ with $|c|=|\lambda|=1$ such that

[TABLE]

Theorem 2**.**

Let $a\in\mathbb{D}$. A surjective map $T\colon\mathcal{S}^{\infty}\to\mathcal{S}^{\infty}$ is an isometry with the norm $\left\|f\right\|_{(\sigma)}=|f(a)|+\sup_{w\in\mathbb{D}}|f^{\prime}(w)|$ for $f\in\mathcal{S}^{\infty}$, if and only if there exist constants $c_{0},c_{1},\lambda\in\mathbb{C}$ with $|c_{0}|=|c_{1}|=|\lambda|=1$ and $b\in\mathbb{D}$ such that, for a function $\displaystyle\rho(z)=\lambda\,\frac{z-b}{1-\overline{b}z}$ defined on $\mathbb{D}$,

[TABLE]

As direct consequences of Theorems 1 and 2, we have the following results: They give the characterizations of surjective complex linear isometries on $\mathcal{S}^{\infty}$ with respect to the norms $\left\|\cdot\right\|_{(\Sigma)}$ and $\left\|\cdot\right\|_{(\sigma)}$, which had been open since 1985 (see [22]).

Corollary 1**.**

If $T\colon\mathcal{S}^{\infty}\to\mathcal{S}^{\infty}$ is a surjective complex linear isometry with the norms $\left\|f\right\|_{(\Sigma)}$ or $\left\|f\right\|_{(C)}$, then there exist constants $c,\lambda\in\mathbb{C}$ with $|c|=|\lambda|=1$ such that

[TABLE]

Corollary 2**.**

Let $a\in\mathbb{D}$ and $T\colon\mathcal{S}^{\infty}\to\mathcal{S}^{\infty}$ a surjective complex linear isometry with the norm $\left\|f\right\|_{(\sigma)}$. There exist constants $c_{0},c_{1},\lambda\in\mathbb{C}$ with $|c_{0}|=|c_{1}|=|\lambda|=1$ and $b\in\mathbb{D}$ such that

[TABLE]

where $\displaystyle\rho(z)=\lambda\,\frac{z-b}{1-\overline{b}z}$ for $z\in\mathbb{D}$.

Theorem 1 states that every surjective isometry $T$ on $(\mathcal{S}^{\infty},\left\|\cdot\right\|_{(\Sigma)})$ or $(\mathcal{S}^{\infty},\left\|\cdot\right\|_{(C)})$ is a weighted composition operator, or a combination of such operator and the complex conjugate, provided that $T(0)=0$. The weighted composition operator is induced by constant functions of modulus $1$. If $T$ is a surjective isometry on $(\mathcal{S}^{\infty},\left\|\cdot\right\|_{(\sigma)})$ with $T(0)=0$, then it is represented by a combination of weighted composition operator, the complex conjugate and the integral operator. Here, the weighted composition operator is induced by a constant function of modulus $1$ and a Möbius transform on $\mathbb{D}$. These maps are trivial examples of surjective isometries on $\mathcal{S}^{\infty}$ with the norms $\left\|\cdot\right\|_{(\Sigma)}$, $\left\|\cdot\right\|_{(C)}$ and $\left\|\cdot\right\|_{(\sigma)}$. We see that there are no other surjective isometries on the space $\mathcal{S}^{\infty}$ by Theorems 1 and 2.

Though our proofs of Theorems 1 and 2 seem to be complicated, the main idea of them is quite simple. It is based on, what is so called, extreme point argument: More explicitly, let $C(X)$ be a Banach space of all continuous complex valued functions defined on a compact Hausdorff space $X$. Let $S$ be an surjective, complex linear isometry from a subspace $A$ of $C(X)$ onto itself. Then the adjoint operator $T^{*}$ defined on the dual space $A^{*}$ of $A$ preserves extreme points of the closed unit ball $A_{1}^{*}$ of $A^{*}$. It is well-known that each extreme point of $A_{1}^{*}$ is of the form $\lambda\delta_{x}$ for some $\lambda\in\mathbb{C}$ with $|\lambda|=1$ and $x\in\operatorname{Ch}(A)$, Choquet boundary for $A$. Here, $\delta_{x}$ denotes the linear functional on $A$, defined by $\delta_{x}(f)=f(x)$ for $f\in A$. Since $S^{*}(\delta_{x})$ is an extreme point of $A_{1}^{*}$ for $x\in\operatorname{Ch}(A)$, there exist $\mu\in\mathbb{C}$ with $|\mu|=1$ and $y\in\operatorname{Ch}(A)$ such that $S^{*}(\delta_{x})=\mu\delta_{y}$. We can define two mappings $\alpha$ and $\phi$ satisfying $S^{*}(\delta_{x})=\alpha(x)\delta_{\phi(x)}$, which yields $S(f)(x)=S^{*}(\delta_{x})(f)=\alpha(x)\delta_{\phi(x)}(f)=\alpha(x)f(\phi(x))$ for all $f\in A$ and $x\in\operatorname{Ch}(A)$. That is, $S$ is a weighted composition operator.

To apply the above arguments to a surjective isometry $T$ on $\mathcal{S}^{\infty}$ in Theorems 1 and 2, we first embed $\mathcal{S}^{\infty}$ into $C(K)$ for some compact Hausdorff space $K$, which depends on the norms of $\mathcal{S}^{\infty}$. In section 2, we define a compact Hausdorff space $K$ and a complex linear isometry $U\colon\mathcal{S}^{\infty}\to C(K)$. Suppose that $T$ is a surjective isometry on $\mathcal{S}^{\infty}$. By the Mazur-Ulam theorem, we may assume that $T$ is real linear. Setting $A=U(\mathcal{S}^{\infty})$, we define $S=U|_{A}\circ T\circ(U|_{A})^{-1}$, and thus $S$ is a surjective real linear isometry from $A$ onto itself. We introduce a kind of adjoint operator, $S_{*}$, between $A_{1}^{*}$; since $S$ is real linear, the adjoint operator $S^{*}$ is not a well defined map on $A_{1}^{*}$. Roughly speaking, we prove that $S_{*}$ preserves extreme points, and thus it is a weighted composition operator. In section 3, we investigate the map $S=U|_{A}\circ T\circ(U|_{A})^{-1}$. Then we have the form of $T=(U|_{A})^{-1}\circ S\circ U|_{A}$, which is unsatisfactory for our purpose. We give complete description of surjective isometry $T$ on $\mathcal{S}^{\infty}$ in section 4

2. Preliminaries and an embedding of $\mathcal{S}^{\infty}$ into $C(K)$

In this section, we embed the space $\mathcal{S}^{\infty}$ into $C(\mathcal{K})$ for some compact Hausdorff space $\mathcal{K}$. Let $B$ be the isometric image of $\mathcal{S}^{\infty}$ in $C(\mathcal{K})$. Then we investigate extreme points of the closed unit ball of the dual space of $B$. Such a space $\mathcal{K}$ depends on norm on $\mathcal{S}^{\infty}$, which is closely related to Shilov boundary for a uniform algebra. We prove, in particular, that Choquet boundary coincides with the Shilov boundary.

Let $H(\mathbb{D})$ be the algebra of all analytic functions on the open unit disc $\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}$ in the complex number field $\mathbb{C}$. The constant function in $H(\mathbb{D})$, which takes the value only $1$, is denoted by $\mathbf{1}$. We denote by $H^{\infty}(\mathbb{D})$ the commutative Banach algebra of all bounded analytic functions on $\mathbb{D}$ with the supremum norm $\|v\|_{\mathbb{D}}=\sup_{z\in\mathbb{D}}|v(z)|$ for $v\in H^{\infty}(\mathbb{D})$. We define a Banach space $\mathcal{S}^{\infty}$ by

[TABLE]

with norms $\left\|f\right\|_{(\Sigma)}$, $\left\|f\right\|_{(C)}$ and $\left\|f\right\|_{(\sigma)}$, given by

[TABLE]

for $f\in\mathcal{S}^{\infty}$, where $a\in\mathbb{D}$. Let $A(\overline{\mathbb{D}})$ be the disc algebra, that is, the commutative Banach algebra of all analytic functions on $\mathbb{D}$ which can be extended to continuous functions on the closed unit disc $\overline{\mathbb{D}}$. We see that $\mathcal{S}^{\infty}$ is a subset of $A(\overline{\mathbb{D}})$ by [7, Theorem 3.11].

We need some properties of $H^{\infty}(\mathbb{D})$. The maximal ideal space $\mathcal{M}$ of $H^{\infty}(\mathbb{D})$ is a compact Hausdorff space with the relative weak*-topology of the dual space of $H^{\infty}(\mathbb{D})$. The Gelfand transform $\widehat{v}$ of $v\in H^{\infty}(\mathbb{D})$ is a continuous function on $\mathcal{M}$, defined by $\widehat{v}(\eta)=\eta(v)$ for $\eta\in\mathcal{M}$. Then $\|\widehat{v}\|_{\mathcal{M}}=\sup_{\eta\in\mathcal{M}}|\widehat{v}(\eta)|=\|v\|_{\mathbb{D}}$ for every $v\in H^{\infty}(\mathbb{D})$. Let $C(\mathcal{M})$ be the commutative Banach algebra of all continuous complex valued functions on $\mathcal{M}$ with the supremum norm $\|\!\cdot\!\|_{\mathcal{M}}$. In the rest of this paper, we denote by $\mathcal{A}$ the uniform algebra $\{\widehat{v}\in C(\mathcal{M}):v\in H^{\infty}(\mathbb{D})\}$ on $\mathcal{M}$, i.e., $\mathcal{A}$ is a uniformly closed subalgebra of $C(\mathcal{M})$ that contains the constant functions and separates the points of $\mathcal{M}$. Shilov boundary $\partial\mathcal{A}$ for $\mathcal{A}$ is the smallest closed subset of $\mathcal{M}$ with the property that $\|\widehat{v}\|_{\partial\mathcal{A}}=\sup_{\eta\in\partial\mathcal{A}}|\widehat{v}(\eta)|=\|\widehat{v}\|_{\mathcal{M}}$ for all $\widehat{v}\in\mathcal{A}$. Let $\mathbb{T}$ be the unit circle in $\mathbb{C}$. Then $\mathbb{T}$ is the Shilov boundary for $A(\overline{\mathbb{D}})$. Hence, $\|\widehat{f}\|_{\mathbb{T}}=\sup_{z\in\mathbb{T}}|\widehat{f}(z)|=\|f\|_{\mathbb{D}}$ for all $f\in A(\overline{\mathbb{D}})$. The identity function on $\mathbb{D}$ is denoted by ${\rm id}$.

We shall introduce the Choquet boundary $\operatorname{Ch}(\mathcal{A})$ for $\mathcal{A}$. Let $\mathcal{A}^{*}_{1}$ be the closed unit ball of the dual space of $\mathcal{A}$. The Choquet boundary $\operatorname{Ch}(\mathcal{A})$ for $\mathcal{A}$ is the set of all points $\eta\in\mathcal{M}$ such that the point evaluation functional $\delta_{\eta}\colon\mathcal{A}\to\mathbb{C}$ at $\eta$ is an extreme point of $\mathcal{A}_{1}^{*}$. It is well-known that Choquet boundaries are dense in Shilov boundaries for uniform algebras. In fact, $\operatorname{Ch}(\mathcal{A})=\partial\mathcal{A}$; the authors of this paper believe that it is a well-known fact. Here, we give its proof for completeness.

Proposition 2.1**.**

The Choquet boundary for $\mathcal{A}$ is $\partial\mathcal{A}$.

Proof.

We need to show that $\partial\mathcal{A}\subset\operatorname{Ch}(\mathcal{A})$, since $\operatorname{Ch}(\mathcal{A})$ is dense in $\partial\mathcal{A}$. Choose $\eta_{0}\in\partial\mathcal{A}$ arbitrarily. Notice that the inner functions of $H^{\infty}(\mathbb{D})$ separate the points of $\mathcal{M}$ (cf. [11, Corollary 5.2 in Chapter X]). Let $\eta_{1}\in\mathcal{M}$ with $\eta_{1}\neq\eta_{0}$. There exists an inner function $f_{\eta_{1}}\in H^{\infty}(\mathbb{D})$ such that $\widehat{f_{\eta_{1}}}(\eta_{1})\neq\widehat{f_{\eta_{1}}}(\eta_{0})$. It follows from [11, Theorem 2.2 in Chapter V] that $|\widehat{f_{\eta_{1}}}|=1$ on $\partial\mathcal{A}$. Then $|\widehat{f_{\eta_{1}}}(\eta_{0})|=1=\|\widehat{f_{\eta_{1}}}\|_{\mathcal{M}}$, since $\partial\mathcal{A}$ is a boundary for $\mathcal{A}$. We set $\lambda_{0}=\widehat{f_{\eta_{1}}}(\eta_{0})\in\mathbb{T}$, and thus $\widehat{f_{\eta_{1}}}(\eta_{1})\neq\lambda_{0}$. Define $u_{\eta_{1}}(z)=(\overline{\lambda_{0}}f_{\eta_{1}}(z)+1)/2$ for $z\in\mathbb{D}$. Then $u_{\eta_{1}}\in H^{\infty}(\mathbb{D})$ with $\widehat{u_{\eta_{1}}}(\eta_{0})=1=\|\widehat{u_{\eta_{1}}}\|_{\mathcal{M}}$ and

[TABLE]

Therefore, $\widehat{u_{\eta_{1}}}$ is a peaking function for $\mathcal{A}$. The set $P_{\widehat{u_{\eta_{1}}}}=\{\eta\in\mathcal{M}:\widehat{u_{\eta_{1}}}(\eta)=1\}$ satisfies $\eta_{0}\in P_{\widehat{u_{\eta_{1}}}}$ but $\eta_{1}\not\in P_{\widehat{u_{\eta_{1}}}}$. Hence, $\cap_{\eta_{1}\in\mathcal{M}\setminus\{\eta_{0}\}}P_{\widehat{u_{\eta_{1}}}}=\{\eta_{0}\}$. We observe that $\eta_{0}\in\operatorname{Ch}(\mathcal{A})$ (see, for example, [12, Propositions 3.18 and 3.19]), and thus $\partial\mathcal{A}\subset\operatorname{Ch}(\mathcal{A})$, as is claimed. ∎

The Choquet boundary $\operatorname{Ch}(\mathcal{A})$ for $\mathcal{A}$ has the following property, which we shall use later:

For each $\varepsilon>0$, $\eta_{0}\in\operatorname{Ch}(\mathcal{A})$ and open neighborhood $V$ of $\eta_{0}$ in $\mathcal{M}$, there exists $\widehat{v}\in\mathcal{A}$ such that $\widehat{v}(\eta_{0})=1=\|\widehat{v}\|_{\mathcal{M}}$ and $|\widehat{v}|<\varepsilon$ on $\mathcal{M}\setminus V$.

We refer the reader to [3] for information about uniform algebras.

We may and do regard $\mathbb{D}$ as a subset of $\mathcal{M}$. Then $\mathbb{D}$ is dense in $\mathcal{M}$ by the Corona theorem [11, Theorem 1.8 in Chapter V, and Chapter VIII]. We define the fiber $\mathcal{M}_{\xi}=\{\eta\in\mathcal{M}:\widehat{{\rm id}}(\eta)=\xi\}$ for each $\xi\in\mathbb{T}$. Then we have $\mathcal{M}=\mathbb{D}\cup\bigcup_{\xi\in\mathbb{T}}\mathcal{M}_{\xi}$ (see [11, Section 1 of Chapter V]). If $f\in A(\overline{\mathbb{D}})$, then $\widehat{f}(\xi)=\widehat{f}(\eta)$ for all $\xi\in\mathbb{T}$ and $\eta\in\mathcal{M}_{\xi}$ by [14, Theorem, p.161] (see, also [11, Exercises and further results in Chapter V]). Recall that the Shilov boundary $\partial\mathcal{A}$ for $\mathcal{A}$ is the smallest closed subset of $\mathcal{M}$ so that $\|\widehat{v}\|_{\partial\mathcal{A}}=\|\widehat{v}\|_{\mathcal{M}}$ for all $\widehat{v}\in\mathcal{A}$. By the maximum modulus principle, we obtain $\|\widehat{v}\|_{\mathcal{M}}=\|\widehat{v}\|_{\mathcal{M}\setminus\mathbb{D}}$ for $v\in H^{\infty}(\mathbb{D})$. Hence, $\partial\mathcal{A}\subset\mathcal{M}\setminus\mathbb{D}$. If, in addition, $v\in A(\overline{\mathbb{D}})$, then $\|v\|_{\mathbb{D}}=\|\widehat{v}\|_{\mathcal{M}}=\|\widehat{v}\|_{\mathcal{M}\setminus\mathbb{D}}=\|\widehat{v}\|_{\mathbb{T}}$, since $\widehat{v}(\xi)=\widehat{v}(\eta)$ for $\xi\in\mathbb{T}$ and $\eta\in\mathcal{M}_{\xi}$. We refer the reader to the following books for information about the maximal ideal space $\mathcal{M}$ of $H^{\infty}(\mathbb{D})$, [11] and [14].

To embed $\mathcal{S}^{\infty}$ into $C(\mathcal{K}_{e})$ for some compact Hausdorff space $\mathcal{K}_{e}$, we introduce a compact Hausdorff space $\mathfrak{D}_{e}$, which depends on the norms $\left\|\cdot\right\|_{(\Sigma)}$, $\left\|\cdot\right\|_{(C)}$ and $\left\|\cdot\right\|_{(\sigma)}$. Let $a\in\mathbb{D}$, and we define $\mathfrak{D}_{\Sigma}$, $\mathfrak{D}_{C}$ and $\mathfrak{D}_{\sigma}$ by

[TABLE]

For $\mathfrak{D}_{e}\in\{\mathfrak{D}_{\Sigma},\mathfrak{D}_{C},\mathfrak{D}_{\sigma}\}$, we set

[TABLE]

Then $\mathfrak{D}_{e}$ and $\mathcal{K}_{e}$ are compact Hausdorff spaces with the product topology. For each $f\in\mathcal{S}^{\infty}$, we define $\widetilde{f}$ by

[TABLE]

For simplicity of notation, we shall write $(z,\eta,w)$ instead of $((z,\eta),w)$ for each element of $\mathcal{K}_{e}$. Here, we note that $f\in A(\overline{\mathbb{D}})$ and $\widehat{f^{\prime}}\in\mathcal{A}$ for $f\in\mathcal{S}^{\infty}$. We define the normed linear subspace $B_{e}$ of $C(\mathcal{K}_{e})$ by

[TABLE]

with the supremum norm $\|\widetilde{f}\|_{\mathcal{K}_{e}}=\sup_{x\in\mathcal{K}_{e}}|\widetilde{f}(x)|$ for $\widetilde{f}\in B_{e}$. Note that $\|v\|_{\mathbb{D}}=\|\widehat{v}\|_{\mathcal{M}}=\|\widehat{v}\|_{\partial\mathcal{A}}$ for $v\in H^{\infty}(\mathbb{D})$. Thus, we observe that

[TABLE]

for every $f\in\mathcal{S}^{\infty}$: In fact, since $f\in\mathcal{S}^{\infty}\subset A(\overline{\mathbb{D}})$, we have $\|f\|_{\mathbb{D}}=\|\widehat{f}\,\|_{\mathbb{T}}$ and $\|f^{\prime}\|_{\mathbb{D}}=\|\widehat{f^{\prime}}\|_{\partial\mathcal{A}}$. Hence, $\left\|f\right\|_{(\Sigma)}=\|f\|_{\mathbb{D}}+\|f^{\prime}\|_{\mathbb{D}}=\|\widehat{f}\,\|_{\mathbb{T}}+\|\widehat{f^{\prime}}\|_{\partial\mathcal{A}}$. Let $(\xi_{0},\eta_{0})\in\mathbb{T}\times\partial\mathcal{A}$ be such that $|\widehat{f}(\xi_{0})|=\|\widehat{f}\,\|_{\mathbb{T}}$ and $|\widehat{f^{\prime}}(\eta_{0})|=\|\widehat{f^{\prime}}\|_{\partial\mathcal{A}}$. Such a pair of points do exist, since $\widehat{f}$ and $\widehat{f^{\prime}}$ are continuous on the compact sets $\mathbb{T}$ and $\partial\mathcal{A}$, respectively. Choose $w_{0}\in\mathbb{T}$ so that $|\widehat{f}(\xi_{0})+\widehat{f^{\prime}}(\eta_{0})w_{0}|=|\widehat{f}(\xi_{0})|+|\widehat{f^{\prime}}(\eta_{0})|$. We obtain

[TABLE]

which prove the first equalities in (2.4). We can prove the second equalities in (2.4) by a similar argument to the above, and we omit it.

We shall prove that $\|\widetilde{f}\|_{\mathcal{K}_{C}}=\left\|f\right\|_{(C)}$ for every $f\in\mathcal{S}^{\infty}$.

Proposition 2.2**.**

For each $f\in\mathcal{S}^{\infty}$,

[TABLE]

Proof.

Let $f\in\mathcal{S}^{\infty}$ and $z\in\mathbb{D}$. There exists $w_{z}\in\mathbb{T}$ such that $|\widehat{f}(z)|+|\widehat{f^{\prime}}(z)|=|\widehat{f}(z)+\widehat{f^{\prime}}(z)w_{z})|$. Since $f,f^{\prime}w_{z}\in H^{\infty}(\mathbb{D})$, we have $\widehat{f+f^{\prime}w_{z}}=\widehat{f}+\widehat{f^{\prime}}w_{z}$ and

[TABLE]

Since $z\in\mathbb{D}$ is arbitrarily chosen, we obtain $\left\|f\right\|_{(C)}\leq\|\widetilde{f}\|_{\mathcal{K}_{C}}$.

By the continuity of $\widetilde{f}$, there exists $\eta_{0}\in\partial\mathcal{A}$ and $w_{0}\in\mathbb{T}$ such that $|\widehat{f}(\eta_{0})+\widehat{f^{\prime}}(\eta_{0})w_{0}|=\|\widetilde{f}\|_{\mathcal{K}_{C}}$. Since $\mathbb{D}$ is dense in $\mathcal{M}$ by the Corona theorem, for each $\varepsilon>0$ there exists $z_{0}\in\mathbb{D}$ such that $|\widetilde{f}(\eta_{0},\eta_{0},w_{0})-\widetilde{f}(z_{0},z_{0},w_{0})|<\varepsilon$. Then we obtain

[TABLE]

and therefore, $\|\widetilde{f}\|_{\mathcal{K}_{C}}\leq\left\|f\right\|_{(C)}+\varepsilon$. Since $\varepsilon>0$ is arbitrarily chosen, we get $\|\widetilde{f}\|_{\mathcal{K}_{C}}\leq\left\|f\right\|_{(C)}$, and consequently, $\|\widetilde{f}\|_{\mathcal{K}_{C}}=\left\|f\right\|_{(C)}$. ∎

Let $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{C},\mathcal{K}_{\sigma}\}$. For each $x\in\mathcal{K}_{e}$, we define the point evaluation functional $\delta_{x}\colon B_{e}\to\mathbb{C}$ at $x$ by

[TABLE]

Define the complex linear operator $I_{0}\colon H^{\infty}(\mathbb{D})\to H^{\infty}(\mathbb{D})$ by

[TABLE]

where $[0,z]$ is the straight line interval from [math] to $z$.

We shall show that $B_{e}$ separates the points of $\mathcal{K}_{e}$ in the following two propositions, in which the operator $I_{0}$ plays an important role.

Proposition 2.3**.**

Let $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{\sigma}\}$. For each pair of distinct points $x_{1},x_{2}\in\mathcal{K}_{e}$ there exists $\widetilde{f}\in B_{e}$ such that $\widetilde{f}(x_{1})\neq\widetilde{f}(x_{2})$.

Proof.

Let $x_{j}=(z_{j},\eta_{j},w_{j})\in\mathcal{K}_{e}$ for $j=1,2$ with $x_{1}\neq x_{2}$. Suppose first that $z_{1}\neq z_{2}$. We set $\xi_{j}=\widehat{{\rm id}}(\eta_{j})$ for $j=1,2$. Then $\xi_{j}\in\mathbb{T}$, since $\eta_{j}\in\partial\mathcal{A}$. We have three possible cases to consider.

Case 1. $\xi_{1},\xi_{2}\not\in\{z_{1},z_{2}\}$. Let $f(z)=(z-z_{2})(z-\xi_{1})^{2}(z-\xi_{2})^{2}\in\mathcal{S}^{\infty}$. Then we obtain $\widehat{f}(z_{1})\neq 0=\widehat{f}(z_{2})$ and $\widehat{f^{\prime}}(\xi_{1})=0=\widehat{f^{\prime}}(\xi_{2})$. Equality (2.2) shows that $\widetilde{f}(x_{1})=\widehat{f}(z_{1})\neq 0=\widetilde{f}(x_{2})$.

Case 2. $\xi_{j}\in\{z_{1},z_{2}\}$ and $\xi_{k}\not\in\{z_{1},z_{2}\}$ for $j,k\in\{1,2\}$ with $j\neq k$. Without loss of generality, we may and do assume that $\xi_{1}\in\{z_{1},z_{2}\}$ and $\xi_{2}\not\in\{z_{1},z_{2}\}$. For functions $g_{1}(z)=(2z-3z_{1}+z_{2})(z-z_{2})^{2}\in\mathcal{S}^{\infty}$ and $g_{2}(z)=(2z-3z_{1}+\xi_{2})(z-\xi_{2})^{2}\in\mathcal{S}^{\infty}$, we see that

[TABLE]

We set $g=g_{1}g_{2}\in\mathcal{S}^{\infty}$, and then $\widehat{g}(z_{1})\neq 0=\widehat{g}(z_{2})$. The multiplication law, $g^{\prime}={g_{1}}^{\prime}g_{2}+g_{1}{g_{2}}^{\prime}$, shows that $\widehat{g^{\prime}}(z_{1})=0=\widehat{g^{\prime}}(z_{2})=\widehat{g^{\prime}}(\xi_{2})$. Because $\xi_{1}\in\{z_{1},z_{2}\}$, we get $\widehat{g^{\prime}}(\xi_{1})=0$. Since $g^{\prime}\in A(\overline{\mathbb{D}})$, we have $\widehat{g^{\prime}}(\xi)=\widehat{g^{\prime}}(\eta)$ for all $\xi\in\mathbb{T}$ and $\eta\in\mathcal{M}_{\xi}$. Therefore, $\widehat{g^{\prime}}(\eta_{j})=\widehat{g^{\prime}}(\xi_{j})=0$ for $j=1,2$. It follows from (2.2) that $\widetilde{g}(x_{1})=\widehat{g}(z_{1})\neq 0=\widetilde{g}(x_{2})$.

Case 3. $\xi_{1},\xi_{2}\in\{z_{1},z_{2}\}$. Let $h(z)=(2z-3z_{1}+z_{2})(z-z_{2})^{2}\in\mathcal{S}^{\infty}$, and then we see that $\widehat{h}(z_{1})\neq 0=\widehat{h}(z_{2})$ and $\widehat{{h}^{\prime}}(z_{1})=0=\widehat{{h}^{\prime}}(z_{2})$. By the choice of $\xi_{j}$, we have $\widehat{{h}^{\prime}}(\xi_{j})=0$ for $j=1,2$. Equality (2.2) shows that $\widetilde{h}(x_{1})=\widehat{h}(z_{1})\neq 0=\widetilde{h}(x_{2})$.

Now we assume that $z_{1}=z_{2}$ and $w_{1}\neq w_{2}$. We have that $\widetilde{{\rm id}}(x_{1})=z_{1}+w_{1}\neq z_{2}+w_{2}=\widetilde{{\rm id}}(x_{2})$ by (2.2).

If $z_{1}=z_{2}$ and $w_{1}=w_{2}$, then $\eta_{1}\neq\eta_{2}$. Since the image $\mathcal{A}$ of the Gelfand transform of $H^{\infty}(\mathbb{D})$ separates the points of $\mathcal{M}$, there exists $v\in H^{\infty}(\mathbb{D})$ such that $\widehat{v}(\eta_{1})\neq\widehat{v}(\eta_{2})$. Because $z_{1}=z_{2}$ and $w_{1}=w_{2}$, we see that $\widetilde{I_{0}(v)}(x_{1})\neq\widetilde{I_{0}(v)}(x_{2})$ by (2.2) and (2.6).

Consequently, $\widetilde{f}(x_{1})\neq\widetilde{f}(x_{2})$ for some $\widetilde{f}\in B_{e}$, as is claimed. ∎

Proposition 2.4**.**

For each pair of distinct points $x_{1},x_{2}\in\mathcal{K}_{C}$ there exists $\widetilde{f}\in B_{C}$ such that $\widetilde{f}(x_{1})\neq\widetilde{f}(x_{2})$.

Proof.

Let $x_{j}=(\eta_{j},\eta_{j},w_{j})\in\mathcal{K}_{C}$ for $j=1,2$ with $x_{1}\neq x_{2}$. Suppose that $\eta_{1}\neq\eta_{2}$ and $\widehat{{\rm id}}(\eta_{1})=\widehat{{\rm id}}(\eta_{2})$. Since $\mathcal{A}$ separates the points of $\mathcal{M}$, we can choose $v_{0}\in H^{\infty}(\mathbb{D})$ so that $\widehat{v_{0}}(\eta_{1})=1$ and $\widehat{v_{0}}(\eta_{2})=0$. We set $f_{0}=I_{0}(v_{0})$, and then $f_{0}\in A(\overline{\mathbb{D}})$. Since $\widehat{{\rm id}}(\eta_{1})=\widehat{{\rm id}}(\eta_{2})$ and $f_{0}\in A(\overline{\mathbb{D}})$, we have $\widehat{f_{0}}(\eta_{1})=\widehat{f_{0}}(\eta_{2})$ with $\widehat{{f_{0}}^{\prime}}(\eta_{1})=1$ and $\widehat{{f_{0}}^{\prime}}(\eta_{2})=0$. Thus, $\widetilde{f_{0}}(x_{1})=\widehat{f_{0}}(\eta_{1})+w_{1}\neq\widehat{f_{0}}(\eta_{2})=\widetilde{f_{0}}(x_{2})$ by (2.2). We now assume that $\eta_{1}\neq\eta_{2}$ and $\widehat{{\rm id}}(\eta_{1})\neq\widehat{{\rm id}}(\eta_{2})$. We set $\xi_{j}=\widehat{{\rm id}}(\eta_{j})$ for $j=1,2$, and $f_{1}(z)=(2z-3\xi_{1}+\xi_{2})(z-\xi_{2})^{2}\in A(\overline{\mathbb{D}})$. Then we get $\widehat{f_{1}}(\xi_{1})\neq 0=\widehat{f_{1}}(\xi_{2})$ and $\widehat{f_{1}^{\prime}}(\xi_{1})=\widehat{f_{1}^{\prime}}(\xi_{2})=0$. Therefore, $\widetilde{f_{1}}(x_{1})\neq 0=\widetilde{f_{1}}(x_{2})$ by (2.2).

Suppose now that $\eta_{1}=\eta_{2}$, and then $w_{1}\neq w_{2}$ by the choice of $x_{j}$’s. If we take $f_{2}={\rm id}\in\mathcal{S}^{\infty}$, then we see that $\widetilde{f_{2}}(x_{1})\neq\widetilde{f_{2}}(x_{2})$. ∎

We shall investigate Choquet boundary for $B_{e}$, which is deeply connected with the set of all extreme points of the closed unit ball of the dual space of $B_{e}$. To this end, we need some topological properties of the maximal ideal space $\mathcal{M}$ of $H^{\infty}(\mathbb{D})$. Essential ideas of proof of Lemmas 2.5 and 2.6 below are due to Professor Osamu Hatori. The authors are grateful for his suggestions and comments on them.

Lemma 2.5**.**

Let $W_{0}=\{z\in\mathbb{D}:|z|<1-\varepsilon_{0}\}$ and $W_{1}=\{z\in\mathbb{D}:|z|>1-\varepsilon_{0}\}$ for each $\varepsilon_{0}$ with $0<\varepsilon_{0}<1$. Then $\mathcal{M}\setminus W_{0}$ is the weak-closure of $W_{1}$ in $\mathcal{M}$.*

Proof.

Let $\operatorname{cl}(W_{1})$ be the weak*-closure of $W_{1}$ in $\mathcal{M}$. Since $\mathcal{M}\setminus W_{0}$ is a closed subset of $\mathcal{M}$ containing $W_{1}$, we see that $\operatorname{cl}(W_{1})\subset\mathcal{M}\setminus W_{0}$.

Now we shall prove that $\mathcal{M}\setminus\mathbb{D}\subset\operatorname{cl}(W_{1})$. Let $\eta_{0}\in\mathcal{M}\setminus\mathbb{D}$. By the Corona theorem, the open unit disc $\mathbb{D}$ is dense in $\mathcal{M}$, and then there exists a net $\{z_{\nu}\}$ in $\mathbb{D}$ such that $\{z_{\nu}\}$ converges to $\eta_{0}$ with respect to the weak*-topology. We assert $\eta_{0}\in\operatorname{cl}(W_{1})$; in fact, if $\eta_{0}$ were not in $\operatorname{cl}(W_{1})$, then we would have that $\eta_{0}$ is in the open set $\mathcal{M}\setminus\operatorname{cl}(W_{1})$. Since $\{z_{\nu}\}$ converges to $\eta_{0}$, we may and do assume that $z_{\nu}\in\mathcal{M}\setminus\operatorname{cl}(W_{1})$ for all $\nu$. Then we get $z_{\nu}\in\mathbb{D}\setminus W_{1}$ for all $\nu$. Let $\mathcal{M}_{\xi}$ be the fiber over $\xi\in\mathbb{T}$. Because $\eta_{0}\in\mathcal{M}\setminus\mathbb{D}=\cup_{\xi\in\mathbb{T}}\mathcal{M}_{\xi}$, there exists a unique $\xi_{0}\in\mathbb{T}$ such that $\eta_{0}\in\mathcal{M}_{\xi_{0}}$. As $z_{\nu}\in\mathbb{D}\setminus W_{1}$, we obtain $|z_{\nu}|\leq 1-\varepsilon_{0}$, and then $|z_{\nu}-\xi_{0}|\geq\varepsilon_{0}$ for all $\nu$. Since $\{z_{\nu}\}$ converges to $\eta_{0}$ with weak*-topology, $z_{\nu}=\widehat{{\rm id}}(z_{\nu})$ tends to $\widehat{{\rm id}}(\eta_{0})=\xi_{0}$, which contradicts $|z_{\nu}-\xi_{0}|\geq\varepsilon_{0}$ for all $\nu$. This implies $\eta_{0}\in\operatorname{cl}(W_{1})$, and consequently $\mathcal{M}\setminus\mathbb{D}\subset\operatorname{cl}(W_{1})$.

Notice that $\{z\in\mathbb{D}:|z|\geq 1-\varepsilon_{0}\}\subset\operatorname{cl}(W_{1})$. As $\mathcal{M}\setminus\mathbb{D}\subset\operatorname{cl}(W_{1})$, we get $\mathcal{M}\setminus W_{0}=\{z\in\mathbb{D}:|z|\geq 1-\varepsilon_{0}\}\cup(\mathcal{M}\setminus\mathbb{D})\subset\operatorname{cl}(W_{1})$. Thus $\operatorname{cl}(W_{1})=\mathcal{M}\setminus W_{0}$ as is claimed. ∎

Lemma 2.6**.**

Let $W_{1}=\{z\in\mathbb{D}:|z|>1-\varepsilon_{0}\}$ for each $\varepsilon_{0}$ with $0<\varepsilon_{0}<1$. Then $\partial\mathcal{A}$ is contained in the interior of $\operatorname{cl}(W_{1})$.

Proof.

Let $W_{0}=\{z\in\mathbb{D}:|z|<1-\varepsilon_{0}\}$. Then $\operatorname{cl}(W_{1})=\mathcal{M}\setminus W_{0}$ by Lemma 2.5. Because $\partial\mathcal{A}\subset\mathcal{M}\setminus\mathbb{D}$, we deduce $\partial\mathcal{A}\subset\operatorname{cl}(W_{1})$. Let $\eta_{0}\in\partial\mathcal{A}$. We shall prove that $\eta_{0}$ is an interior point of $\operatorname{cl}(W_{1})$. If any open neighborhood $O$ of $\eta_{0}$ were not contained in $\operatorname{cl}(W_{1})$, there would exist a net $\{\eta_{\nu}\}$ in $\mathcal{M}\setminus\operatorname{cl}(W_{1})$ such that $\{\eta_{\nu}\}$ converges to $\eta_{0}$ with respect to the weak*-topology. Since $\operatorname{cl}(W_{1})=\mathcal{M}\setminus W_{0}$, we get $\eta_{\nu}\in W_{0}$, and hence $|\eta_{\nu}|<1-\varepsilon_{0}$ for all $\nu$. By the choice of $\{\eta_{\nu}\}$ and $\eta_{0}$, we obtain $\eta_{\nu}=\widehat{{\rm id}}(\eta_{\nu})\to\widehat{{\rm id}}(\eta_{0})$. Here, $|\widehat{{\rm id}}(\eta_{0})|=1$, since $\eta_{0}\in\partial\mathcal{A}\subset\mathcal{M}\setminus\mathbb{D}$. Therefore, $|\eta_{\nu}|\to 1$, which contradicts $|\eta_{\nu}|<1-\varepsilon_{0}$ for all $\nu$. Thus, $\eta_{0}$ is an interior point of $\operatorname{cl}(W_{1})$. ∎

In the next lemma, we shall prove the existence of “peaking functions” in $\mathcal{S}^{\infty}$, which will be significant to determine the set of all extreme points the closed unit ball of the dual space of $B_{e}$. We need topological properties of $\mathcal{M}$ as in Lemmas 2.5 and 2.6 to prove the next lemma.

Lemma 2.7**.**

Let $\varepsilon>0$, $\eta_{0}\in\partial\mathcal{A}$, and let $V$ be an open neighborhood of $\eta_{0}$ in $\mathcal{M}$. There exists $f\in\mathcal{S}^{\infty}$ such that

[TABLE]

Proof.

Let $\varepsilon>0$, $\eta_{0}\in\partial\mathcal{A}$ and $V$ an open neighborhood of $\eta_{0}$ in $\mathcal{M}$. Let $2\varepsilon_{0}=\min\{\varepsilon,1\}$, and we set $W_{0}=\{z\in\mathbb{D}:|z|<1-\varepsilon_{0}\}$ and $W_{1}=\{z\in\mathbb{D}:|z|>1-\varepsilon_{0}\}$. By Lemma 2.6, $\eta_{0}$ is an interior point of $\operatorname{cl}(W_{1})$. Thus, there exists an open neighborhood $O_{1}$ of $\eta_{0}$ in $\mathcal{M}$ such that $O_{1}\subset\operatorname{cl}(W_{1})$. We set $V_{1}=V\cap O_{1}$, and then $V_{1}$ is an open neighborhood of $\eta_{0}$ with $V_{1}\subset\operatorname{cl}(W_{1})$. Since $\operatorname{cl}(W_{1})=\mathcal{M}\setminus W_{0}$ by Lemma 2.5, we see that $V_{1}\cap\mathbb{D}\subset\{z\in\mathbb{D}:|z|\geq 1-\varepsilon_{0}\}$. Notice that $\eta_{0}\in\partial\mathcal{A}=\operatorname{Ch}(\mathcal{A})$, the Choquet boundary for $\mathcal{A}$, by Proposition 2.1. There exists $v\in H^{\infty}(\mathbb{D})$ such that $\widehat{v}(\eta_{0})=1=\|\widehat{v}\|_{\mathcal{M}}$ and $|\widehat{v}|<\varepsilon_{0}$ on $\mathcal{M}\setminus V_{1}$. Define $f\in\mathcal{S}^{\infty}$ by $f=I_{0}(v)$, and then $f^{\prime}=v$ on $\mathbb{D}$ and $\widehat{f^{\prime}}(\eta_{0})=1=\|\widehat{f^{\prime}}\|_{\mathcal{M}}$. Since $V_{1}\subset V$, we obtain $|\widehat{f^{\prime}}|<\varepsilon$ on $\mathcal{M}\setminus V$.

We show that $\|f\|_{\mathbb{D}}\leq\varepsilon$. Let $z_{0}$ be an arbitrary point of $\mathbb{D}$. If $z_{0}\in W_{0}=\{z\in\mathbb{D}:|z|<1-\varepsilon_{0}\}$, then the line interval $[0,z_{0}]$ from [math] to $z_{0}$ lies in $W_{0}$. Since $V_{1}\cap\mathbb{D}\subset\{z\in\mathbb{D}:|z|\geq 1-\varepsilon_{0}\}$, it follows that $|v|<\varepsilon_{0}$ on $[0,z_{0}]$. Thus,

[TABLE]

Hence, $|f(z_{0})|<\varepsilon_{0}$ for each $z_{0}\in W_{0}$.

If $z_{0}\in\mathbb{D}\setminus W_{0}$, then we set $z_{r}=(1-r\varepsilon_{0})z_{0}/|z_{0}|$ for each $r$ with $1<r<2$. Note that $\varepsilon_{0}<r\varepsilon_{0}<2\varepsilon_{0}\leq 1$. Because $|z_{r}|=1-r\varepsilon_{0}<1-\varepsilon_{0}$, we obtain $z_{r}\in W_{0}$. Then $|f(z_{r})|\leq\varepsilon_{0}$ by the fact proved in the last paragraph. By the choice of $z_{r}$ and $z_{0}$, we get $|z_{r}-z_{0}|<r\varepsilon_{0}$. It follows that

[TABLE]

Letting $r\searrow 1$, we obtain $|f(z_{0})|\leq 2\varepsilon_{0}\leq\varepsilon$ for $z_{0}\in\mathbb{D}\setminus W_{0}$.

We thus conclude that $|f|\leq\varepsilon$ on $\mathbb{D}$, and consequently $\|f\|_{\mathbb{D}}\leq\varepsilon$. ∎

Recall that $B_{e}$ is a normed linear subspace of $C(\mathcal{K}_{e})$, defined by (2.3). Let $\Lambda\colon B_{e}\to\mathbb{C}$ be a bounded linear functional with the operator norm $\|\Lambda\|$. By the Hahn-Banach theorem, $\Lambda$ is extended to a bounded linear functional $\Lambda_{1}$ on $C(\mathcal{K}_{e})$ with $\|\Lambda\|=\|\Lambda_{1}\|$. Thus, the Riesz representation theorem ensures that there exists a regular Borel measure $\mu$ on $\mathcal{K}_{e}$ such that $\displaystyle\Lambda(\widetilde{f})=\int_{\mathcal{K}_{e}}\widetilde{f}\,d\mu$ for every $\widetilde{f}\in B_{e}$ and $\|\Lambda\|=\|\mu\|$, where $\|\mu\|$ denotes the total variation of the measure $\mu$.

We investigate representing measures for the point evaluation functional $\delta_{x}$ with $x\in\mathcal{K}_{e}$ to determine the set of extreme points of the closed unit ball of $B_{e}$. We first show that any representing measure $\mu$ for $\delta_{x}$ with $x\in\mathcal{K}_{C}$ or $x\in\mathcal{K}_{\sigma}$ is a Dirac measure. To this end, we shall prove that any such measure $\mu$ is concentrated on some sets.

Here, we recall that $\mathfrak{D}_{\Sigma}=\mathbb{T}\times\partial\mathcal{A}$, $\mathfrak{D}_{C}=\{(\eta,\eta):\eta\in\partial\mathcal{A}\}$ and $\mathfrak{D}_{\sigma}=\{a\}\times\partial\mathcal{A}$, $a\in\mathbb{D}$ by (2.1). We denote by $\pi$ the natural projection from $\mathfrak{D}_{e}$ onto $\partial\mathcal{A}$, that is, $\pi(z,\eta)=\eta$ for $(z,\eta)\in\mathfrak{D}_{e}$.

Lemma 2.8**.**

Let $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{C},\mathcal{K}_{\sigma}\}$, $x_{0}=(z_{0},\eta_{0},w_{0})\in\mathcal{K}_{e}$ and let $V$ be an open neighborhood of $\eta_{0}$ in $\mathcal{M}$. Then $\mu(\pi^{-1}(V)\times\mathbb{T})=1$ for each representing measure $\mu$ for $\delta_{x_{0}}$.

Proof.

Let $x_{0}=(z_{0},\eta_{0},w_{0})$ be an arbitrary point of $\mathcal{K}_{e}$. Let $\mu$ be a representing measure for $\delta_{x_{0}}$. Since $\delta_{x_{0}}(\widetilde{\mathbf{1}})=1=\|\delta_{x_{0}}\|$, we see that $\mu$ is a probability measure (see, for example, [3, p. 81]).

Let $\varepsilon>0$ be an arbitrary positive real number and $V$ an open neighborhood of $\eta_{0}\in\partial\mathcal{A}$ in $\mathcal{M}$. We set $V^{c}=\partial\mathcal{A}\setminus V$,

[TABLE]

Then we obtain

[TABLE]

By Lemma 2.7, there exists $f_{0}\in\mathcal{S}^{\infty}$ such that

[TABLE]

Recall that $\widetilde{f_{0}}(z,\eta,w)=\widehat{f_{0}}(z)+\widehat{f_{0}^{\prime}}(\eta)w$ by (2.2). By the choice of $f_{0}$, we have

[TABLE]

Since $\mu$ is a representing measure for $\delta_{x_{0}}$, we get

[TABLE]

We derive from the above equalities with (2.7) that

[TABLE]

Since $\varepsilon>0$ is arbitrary, we get $1\leq\mu(P_{V})\leq\mu(\mathcal{K}_{e})=1$. Hence $1=\mu(P_{V})=\mu(\pi^{-1}(V)\times\mathbb{T})$. ∎

Lemma 2.9**.**

Let $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{C},\mathcal{K}_{\sigma}\}$ and $x_{0}=(z_{0},\eta_{0},w_{0})\in\mathcal{K}_{e}$. Then $\mu(\pi^{-1}(\{\eta_{0}\})\times\mathbb{T})=1$ for each representing measure $\mu$ for $\delta_{x_{0}}$.

Proof.

Let $P$ be an open set in $\mathcal{K}_{e}=\mathfrak{D}_{e}\times\mathbb{T}$ with $\pi^{-1}(\{\eta_{0}\})\times\mathbb{T}\subset P$. For each $(z,\eta_{0},w)\in\pi^{-1}(\{\eta_{0}\})\times\mathbb{T}$, there exist open sets $Z,V,W$ such that $(z,\eta_{0},w)\in Z\times V\times W\subset P$ by the definition of the product topology. Note that $\pi^{-1}(\{\eta_{0}\})\times\mathbb{T}$ is a compact set, and then there exist finitely many open sets $Z_{k},V_{k},W_{k}$ such that $\pi^{-1}(\{\eta_{0}\})\times\mathbb{T}\subset\cup_{k=1}^{n}(Z_{k}\times V_{k}\times W_{k})$ for some $n\in\mathbb{N}$. We now set $V_{0}=\cap_{k=1}^{n}V_{k}$. Then $V_{0}$ is an open neighborhood of $\eta_{0}$. We show that $\pi^{-1}(V_{0})\times\mathbb{T}\subset P$. In fact, if $(z_{1},\eta_{1},w_{1})\in\pi^{-1}(V_{0})\times\mathbb{T}$, then $(z_{1},\eta_{0},w_{1})\in\pi^{-1}(\{\eta_{0}\})\times\mathbb{T}$. By the choice of $Z_{k},V_{k},W_{k}$, we obtain $(z_{1},\eta_{0},w_{1})\in Z_{m}\times V_{m}\times W_{m}$ for some $m$ with $1\leq m\leq n$. We assure from $\eta_{1}\in V_{0}\subset V_{m}$ that $(z_{1},\eta_{1},w_{1})\subset Z_{m}\times V_{m}\times W_{m}\subset P$, which implies $\pi^{-1}(V_{0})\times\mathbb{T}\subset P$, as is claimed. We thus have that $1=\mu(\pi^{-1}(V_{0})\times\mathbb{T})\leq\mu(P)$ by Lemma 2.8. Since $\mu$ is a probability measure, we get $\mu(P)=1$. Since $P$ is an arbitrary open set with $\pi^{-1}(\{\eta_{0}\})\times\mathbb{T}\subset P$, the regular measure $\mu$ satisfies $\mu(\pi^{-1}(\{\eta_{0}\})\times\mathbb{T})=1$. ∎

Remark 2.1**.**

If $\mathfrak{D}_{e}=\mathfrak{D}_{C}=\{(\eta,\eta):\eta\in\partial\mathcal{A}\}$, then $\pi^{-1}(\{\eta_{0}\})=\{(\eta_{0},\eta_{0})\}$. If $\mathfrak{D}_{e}=\mathfrak{D}_{\sigma}=\{a\}\times\partial\mathcal{A}$, then $\pi^{-1}(\{\eta_{0}\})=\{(a,\eta_{0})\}$. Lemma 2.9 states that if $x_{0}=(z_{0},\eta_{0},w_{0})\in\mathcal{K}_{e}$ and if $\mathcal{K}_{e}\in\{\mathcal{K}_{C},\mathcal{K}_{\sigma}\}$, then $\mu(\{z_{0}\}\times\{\eta_{0}\}\times\mathbb{T})=1$ for every representing measure $\mu$ for $\delta_{x_{0}}$. The next lemma says that a similar result is true even if $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$.

Lemma 2.10**.**

If $x_{0}=(z_{0},\eta_{0},w_{0})\in\mathcal{K}_{\Sigma}=\mathbb{T}\times\partial\mathcal{A}\times\mathbb{T}$ with $\eta_{0}\not\in\mathcal{M}_{z_{0}}$ then $\mu(\{z_{0}\}\times\{\eta_{0}\}\times\mathbb{T})=1$ for each representing measure $\mu$ for $\delta_{x_{0}}$.

Proof.

Let $\varepsilon>0$ and $Z$ an open neighborhood of $z_{0}$ in $\mathbb{T}$. We set $Z^{c}=\mathbb{T}\setminus Z$, $Q_{Z}=Z\times\{\eta_{0}\}\times\mathbb{T}$ and $Q_{Z^{c}}=Z^{c}\times\{\eta_{0}\}\times\mathbb{T}$. Then we see that

[TABLE]

Notice that $\pi^{-1}(\{\eta_{0}\})=\mathbb{T}\times\{\eta_{0}\}$, and hence $\mu(Q_{Z}\cup Q_{Z^{c}})=1$ by Lemma 2.9. For each $m\in\mathbb{N}$, we define a function $g_{m}\in\mathcal{S}^{\infty}$ by $g_{m}(z)=\{(\overline{z_{0}}\,z+1)/2\}^{m}$ for $z\in\mathbb{D}$. We observe that

[TABLE]

Set $\xi_{0}=\widehat{{\rm id}}(\eta_{0})\in\mathbb{T}$. By the assumption, $\eta_{0}\not\in\mathcal{M}_{z_{0}}$. Thus $\xi_{0}\neq z_{0}$, and we get $|(\overline{z_{0}}\,\xi_{0}+1)/2|<1$. Therefore, $m\{(\overline{z_{0}}\,\xi_{0}+1)/2\}^{m-1}$ converges to [math] as $m\to\infty$. In addition, since $|\widehat{g_{m}}(z)|<1$ for all $z\in\mathbb{T}\setminus\{z_{0}\}$, we can find $m_{0}\in\mathbb{N}$ such that $|\widehat{g_{m_{0}}}|<\varepsilon$ on $Z^{c}$ and

[TABLE]

We set $f_{1}=g_{m_{0}}\in\mathcal{S}^{\infty}$, and hence

[TABLE]

Equality (2.2) with the above shows that $|\delta_{x_{0}}(\widetilde{f_{1}})|=|\widehat{f_{1}}(z_{0})+\widehat{f_{1}^{\prime}}(\eta_{0})w_{0}|\geq 1-\varepsilon$. We deduce from $\mu(Q_{Z}\cup Q_{Z^{c}})=1$ that

[TABLE]

Because $\varepsilon>0$ is arbitrarily chosen, we get $1\leq\mu(Q_{Z})$. Since $\mu$ is a probability measure, $1=\mu(Q_{Z})=\mu(Z\times\{\eta_{0}\}\times\mathbb{T})$ for all open neighborhood $Z$ of $z_{0}$. By a similar argument to proof of Lemma 2.9, we observe that $\mu(\{z_{0}\}\times\{\eta_{0}\}\times\mathbb{T})=1$. ∎

Now we are ready to characterize representing measure for $x_{0}\in\mathcal{K}_{e}$. Applying Lemmas 2.9 and 2.10, we shall prove that the representing measure for $\delta_{x_{0}}$ is concentrated on the point $x_{0}\in\mathcal{K}_{e}$.

Lemma 2.11**.**

Let $x_{0}=(z_{0},\eta_{0},w_{0})\in\mathcal{K}_{e}$ with $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{C},\mathcal{K}_{\sigma}\}.$ Assume further that $\eta_{0}\not\in\mathcal{M}_{z_{0}}$ if $x_{0}\in\mathcal{K}_{\Sigma}$. Then the Dirac measure concentrated at $x_{0}$ is the unique representing measure for $\delta_{x_{0}}$.

Proof.

Let $x_{0}=(z_{0},\eta_{0},w_{0})\in\mathcal{K}_{e}$. We assume that $\eta_{0}\not\in\mathcal{M}_{z_{0}}$ if $x_{0}\in\mathcal{K}_{\Sigma}$. Let $\mu$ be a representing measure for $\delta_{x_{0}}$. We shall prove that $\mu(\{x_{0}\})=1$. Set $R=\{z_{0}\}\times\{\eta_{0}\}\times\mathbb{T}$, and thus $\mu(R)=1$ by Lemma 2.9 (see, also Remark 2.1) and Lemma 2.10.

Let $f_{0}={\rm id}-\widehat{{\rm id}}(z_{0})\mathbf{1}\in\mathcal{S}^{\infty}$. Then $\widehat{f_{0}}(z_{0})=0$ and $\widehat{f_{0}^{\prime}}(\eta_{0})=1$. By (2.2), $\delta_{x_{0}}(\widetilde{f_{0}})=\widehat{f_{0}}(z_{0})+\widehat{f_{0}^{\prime}}(\eta_{0})w_{0}=w_{0}$. Since $\mu$ is a representing measure for $\delta_{x_{0}}$, the fact $\mu(R)=1$ shows that

[TABLE]

Hence, $\displaystyle\int_{R}(w-w_{0})\,d\mu=0$. Multiplying this equality by $-\overline{w_{0}}$, we have $\displaystyle\int_{R}(1-\overline{w_{0}}w)\,d\mu=0$. Since $\mu$ is a positive measure,

[TABLE]

where ${\rm Re}\,\zeta$ is the real part of a complex number $\zeta$. Here, we note that $1-{\rm Re}\,(\overline{w_{0}}w)>0$ on $R_{0}=\{z_{0}\}\times\{\eta_{0}\}\times(\mathbb{T}\setminus\{w_{0}\})\subset R$. We thus obtain $\mu(R_{0})=0$, and hence $\mu(\{z_{0}\}\times\{\eta_{0}\}\times\{w_{0}\})=\mu(R)=1$. This implies that $\mu(\{x_{0}\})=1$. We conclude that $\mu$ is a Dirac measure concentrated at $x_{0}=(z_{0},\eta_{0},w_{0})$, as is claimed. ∎

Let $\operatorname{Ch}(B_{e})$ be the Choquet boundary for $B_{e}$, that is, the set of all $x\in\mathcal{K}_{e}$ such that $\delta_{x}$ is an extreme point of $(B_{e})_{1}^{*}$, the closed unit ball of the dual space $(B_{e})^{*}$ of $(B_{e},\|\cdot\|_{\mathcal{K}_{e}})$. We denote by $\operatorname{ext}((B_{e})_{1}^{*})$ the set of all extreme points of $(B_{e})_{1}^{*}$. By the Arens-Kelley theorem, we see that $\operatorname{ext}((B_{e})_{1}^{*})=\{\lambda\delta_{x}:\lambda\in\mathbb{T},\ x\in\operatorname{Ch}(B_{e})\}$ (cf. [8, Corollary 2.3.6 and Theorem 2.3.8]).

The next lemma gives the characterization of $\operatorname{Ch}(B_{e})$. In other words, we can determine $\operatorname{ext}((B_{e})_{1}^{*})$, which is essential to our arguments below.

Lemma 2.12**.**

Let $x_{0}=(z_{0},\eta_{0},w_{0})\in\mathcal{K}_{e}$ with $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{C},\mathcal{K}_{\sigma}\}$. Assume that $\eta_{0}\not\in\mathcal{M}_{z_{0}}$ if $x_{0}\in\mathcal{K}_{\Sigma}$. Then $x_{0}\in\operatorname{Ch}(B_{e})$.

Proof.

Let $x_{0}=(z_{0},\eta_{0},w_{0})\in\mathcal{K}_{e}$, and $\eta_{0}\not\in\mathcal{M}_{z_{0}}$ if $x_{0}\in\mathcal{K}_{\Sigma}$. We prove that $\delta_{x_{0}}$ is an extreme point of $(B_{e})^{*}_{1}$. Let $\chi_{1},\chi_{2}\in(B_{e})_{1}^{*}$ be such that $\delta_{x_{0}}=(\chi_{1}+\chi_{2})/2$. Then $\chi_{1}(\widetilde{\mathbf{1}})+\chi_{2}(\widetilde{\mathbf{1}})=2\delta_{x_{0}}(\widetilde{\mathbf{1}})=2$ by (2.2). Since $\chi_{j}\in(B_{e})_{1}^{*}$, we get $|\chi_{j}(\widetilde{\mathbf{1}})|\leq 1$ and thus $\chi_{j}(\widetilde{\mathbf{1}})=1=\|\chi_{j}\|$ for $j=1,2$. Let $\tau_{j}$ be a representing measure for $\chi_{j}$, that is, $\displaystyle\chi_{j}(\widetilde{f})=\int_{\mathcal{K}_{e}}\widetilde{f}\,d\tau_{j}$ for $\widetilde{f}\in B_{e}$ with $\|\chi_{j}\|=\|\tau_{j}\|$, the total variation of $\tau_{j}$, for $j=1,2$. We see that $\tau_{j}$ is a probability measure on $\mathcal{K}_{e}$ (cf. [3, p. 81]). Then $(\tau_{1}+\tau_{2})/2$ is a representing measure for $\delta_{x_{0}}$. It follows from Lemma 2.11 that $(\tau_{1}+\tau_{2})/2$ is the Dirac measure $\gamma_{x_{0}}$ concentrated at $x_{0}$. Because $\tau_{j}$ is a positive measure, $\tau_{j}(E)=0$ for each Borel set $E$ with $x_{0}\not\in E$. Hence $\tau_{j}=\gamma_{x_{0}}$ for $j=1,2$, and consequently $\chi_{1}=\chi_{2}$. Therefore, $\delta_{x_{0}}$ is an extreme point of $(B_{e})_{1}^{*}$, and thus $x_{0}\in\operatorname{Ch}(B_{e})$. ∎

Remark 2.2**.**

By Lemma 2.12, we see that $\mathcal{K}_{C}=\operatorname{Ch}(B_{C})$ and $\mathcal{K}_{\sigma}=\operatorname{Ch}(B_{\sigma})$. The authors are not sure whether or not $\mathcal{K}_{\Sigma}=\operatorname{Ch}(B_{\Sigma})$ holds. But, the following result on $\mathcal{K}_{\Sigma}$ is enough for our proof.

Lemma 2.13**.**

Define $\mathcal{K}_{\Sigma}^{\circ}=\{(z,\eta,w)\in\mathcal{K}_{\Sigma}:\eta\not\in\mathcal{M}_{z}\}$. Then the set $\mathcal{K}_{\Sigma}^{\circ}$ is dense in $\mathcal{K}_{\Sigma}=\mathfrak{D}_{\Sigma}\times\mathbb{T}$.

Proof.

Let $x_{0}=(z_{0},\eta_{0},w_{0})\in\mathcal{K}_{\Sigma}$ and $O$ an open neighborhood of $x_{0}$ in $\mathcal{K}_{\Sigma}$. We show that $O\cap\mathcal{K}_{\Sigma}^{\circ}\neq\emptyset$. To this end, we need to consider the case when $x_{0}\not\in\mathcal{K}_{\Sigma}^{\circ}$, and hence $\eta_{0}\in\mathcal{M}_{z_{0}}$. By the definition of the product topology, we can choose open sets $Z,W\subset\mathbb{T}$ and $V\subset\partial\mathcal{A}$ such that $(z_{0},\eta_{0},w_{0})\in Z\times V\times W\subset O$. Choose $z_{1}\in Z$ with $z_{0}\neq z_{1}$, and then we obtain $(z_{1},\eta_{0},w_{0})\in O$ and $\widehat{{\rm id}}(\eta_{0})=z_{0}\neq z_{1}$. Hence $\eta_{0}\not\in\mathcal{M}_{z_{1}}$, and therefore, $(z_{1},\eta_{0},w_{0})\in O\cap\mathcal{K}_{\Sigma}^{\circ}$. This shows that $\mathcal{K}_{\Sigma}^{\circ}$ is dense in $\mathcal{K}_{\Sigma}$. ∎

We now define an isometry, $S$, on $B_{e}$ and investigate it. To do this, we introduce a kind of adjoint operator, $S_{*}$, on the dual space $(B_{e})^{*}$ of $B_{e}$. In the rest of this section, we shall prove that the map $S_{*}$ preserves extreme points of $(B_{e})^{*}_{1}$ in some sense.

Let $(\mathcal{K}_{e},\left\|\cdot\right\|_{(e)})$ be one of $(\mathcal{K}_{\Sigma},\left\|\cdot\right\|_{(\Sigma)})$ or $(\mathcal{K}_{C},\left\|\cdot\right\|_{(C)})$ or $(\mathcal{K}_{\sigma},\left\|\cdot\right\|_{(\sigma)})$. For a surjective, not necessarily linear, isometry $T$ from $(\mathcal{S}^{\infty},\left\|\cdot\right\|_{(e)})$ onto itself, define the mapping $T_{0}$ on $(\mathcal{S}^{\infty},\left\|\cdot\right\|_{(e)})$ by

[TABLE]

By the Mazur-Ulam theorem [17, 24], $T_{0}$ is a surjective, real linear isometry from $(\mathcal{S}^{\infty},\left\|\cdot\right\|_{(e)})$ onto itself. We define the map $U\colon(\mathcal{S}^{\infty},\left\|\cdot\right\|_{(e)})\to(B_{e},\left\|\cdot\right\|_{\mathcal{K}_{e}})$ by

[TABLE]

where $\widetilde{f}$ is defined as in (2.2). By (2.4) and (2.5), $U$ is a surjective complex linear isometry. Denote $UT_{0}U^{-1}$ by $S$; the mapping $S\colon B_{e}\to B_{e}$ is a well-defined surjective real linear isometry, since $U\colon(\mathcal{S}^{\infty},\left\|\cdot\right\|_{(e)})\to(B_{e},\left\|\cdot\right\|_{\mathcal{K}_{e}})$ is a surjective complex linear isometry.

[TABLE]

By the definition of the mapping $S$, we derive

[TABLE]

We define the mapping $S_{*}\colon(B_{e})^{*}\to(B_{e})^{*}$ by

[TABLE]

for $\chi\in(B_{e})^{*}$ and $\widetilde{f}\in B_{e}$. Here we note that $S_{*}$ coincides with the usual adjoint operator, provided that $S$ is complex linear. The mapping $S_{*}$ is a surjective real linear isometry with respect to the operator norm on $(B_{e})^{*}$ (see, for example, [23, Proposition 5.17]). This shows that $S_{*}$ preserves extreme points, that is, $S_{*}(\operatorname{ext}((B_{e})_{1}^{*}))=\operatorname{ext}((B_{e})_{1}^{*})$.

We define the topological subspace $\mathcal{E}_{e}$ of $(B_{e})^{*}_{1}$ with the weak*-topology by

[TABLE]

Now we consider the case when $\mathcal{K}_{e}\in\{\mathcal{K}_{C},\mathcal{K}_{\sigma}\}$. Then $\mathcal{K}_{e}=\operatorname{Ch}(B_{e})$ by Remark 2.2. Applying the Arens-Kelley theorem to the above, we obtain

[TABLE]

Since $S_{*}$ preserves extreme points, we get $S_{*}(\mathcal{E}_{e})=\mathcal{E}_{e}$. In the rest of this section, we shall prove that the equality is true even if $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$.

For $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{C},\mathcal{K}_{\sigma}\}$, we define the mapping $\mathbf{h}\colon\mathbb{T}\times\mathcal{K}_{e}\to\mathcal{E}_{e}$ by

[TABLE]

We shall prove that $\mathbf{h}$ is a homeomorphism. The property on $\mathbf{h}$ is quite important to show that $S_{*}$ preserves the set $\mathcal{E}_{e}$. The ideas of the following two lemmas are due to Professor Kazuhiro Kawamura. The authors are grateful for his suggestions and comments.

Lemma 2.14**.**

Let $\mathbb{T}\times\mathcal{K}_{e}$ be the compact Hausdorff space with the product topology. Then the mapping $\mathbf{h}\colon\mathbb{T}\times\mathcal{K}_{e}\to\mathcal{E}_{e}$ is a homeomorphism. In particular, $\mathbf{h}(\mathbb{T}\times\operatorname{Ch}(B_{e}))=\operatorname{ext}((B_{e})_{1}^{*})$.

Proof.

The map $\mathbf{h}$ is surjective by definition. Because $B_{e}$ contains the constant function $\widetilde{\mathbf{1}}$ and separates the points of $\mathcal{K}_{e}$ (see Propositions 2.3 and 2.4), we see that $\mathbf{h}$ is injective. By the definition of the weak*-topology, we observe that $\mathbf{h}$ is a continuous map from the compact space $\mathbb{T}\times\mathcal{K}_{e}$ onto the Hausdorff space $\mathcal{E}_{e}$. Hence, $\mathbf{h}\colon\mathbb{T}\times\mathcal{K}_{e}\to\mathcal{E}_{e}$ is a homeomorphism. By the Arens-Kelley theorem, $\operatorname{ext}((B_{e})_{1}^{*})=\{\lambda\delta_{x}:\lambda\in\mathbb{T},\,x\in\operatorname{Ch}(B_{e})\}$, and thus $\operatorname{ext}((B_{e})_{1}^{*})=\mathbf{h}(\mathbb{T}\times\operatorname{Ch}(B_{e}))$. ∎

The next lemma is a generalization of the well-known fact that the adjoint operator of a surjective complex linear isometry on $(\mathcal{S}^{\infty},\left\|\cdot\right\|_{(e)})$ preserves the set $\operatorname{ext}((B_{e})_{1}^{*})$ of all extreme points of $(B_{e})_{1}^{*}$. In fact, if $\mathcal{K}_{e}=\mathcal{K}_{C}$ or $\mathcal{K}_{e}=\mathcal{K}_{\sigma}$, then $\mathcal{E}_{e}=\operatorname{ext}((B_{e})_{1}^{*})$, and thus $S_{*}(\mathcal{E}_{e})=\mathcal{E}_{e}$, as mentioned above. We shall prove that $S_{*}(\mathcal{E}_{e})=\mathcal{E}_{e}$ holds even if $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$

Lemma 2.15**.**

The mapping $S_{*}$ preserves $\mathcal{E}_{e}$, that is, $S_{*}(\mathcal{E}_{e})=\mathcal{E}_{e}$.

Proof.

We need to consider the case when $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$, and then $\mathcal{E}_{e}=\mathcal{E}_{\Sigma}$. Let $\mathbf{h}$ be the homeomorphism as in (2.11). Lemma 2.14 shows that

[TABLE]

By Lemma 2.12, $\mathcal{K}_{\Sigma}^{\circ}\subset\operatorname{Ch}(B_{\Sigma})\subset\mathcal{K}_{\Sigma}$, and hence we observe that

[TABLE]

By the definition of the map $\mathbf{h}$, we get $\mathbf{h}(\mathbb{T}\times\mathcal{K}_{\Sigma})=\mathcal{E}_{\Sigma}$, and hence $S_{*}(\mathbf{h}(\mathbb{T}\times\mathcal{K}_{\Sigma}^{\circ}))\subset\mathcal{E}_{\Sigma}$. We denote by $\operatorname{cl}_{M}(L)$ the closure of a subset $L$ in a topological space $M$. By Lemma 2.13, $\mathcal{K}_{\Sigma}^{\circ}$ is dense in $\mathcal{K}_{\Sigma}$, and thus $\operatorname{cl}_{\mathcal{K}_{\Sigma}}(\mathcal{K}_{\Sigma}^{\circ})=\mathcal{K}_{\Sigma}$. Since the map $\mathbf{h}$ is a homeomorphism, we obtain

[TABLE]

and thus $S_{*}(\mathcal{E}_{\Sigma})=S_{*}(\operatorname{cl}_{\mathcal{E}_{\Sigma}}(\mathbf{h}(\mathbb{T}\times\mathcal{K}_{\Sigma}^{\circ})))$. Because $S_{*}\colon(B_{\Sigma})^{*}\to(B_{\Sigma})^{*}$ is a surjective isometry with the operator norm, it is bijective. By the definition of $S_{*}$, it is continuous with respect to the weak*-topology on $(B_{\Sigma})^{*}$. In the same way, we can show that $S_{*}^{-1}$ is continuous on $(B_{\Sigma})^{*}$ as well. Hence $S_{*}\colon(B_{\Sigma})^{*}\to(B_{\Sigma})^{*}$ is a homeomorphism with respect to the weak*-topology. Since $S_{*}(\mathbf{h}(\mathbb{T}\times\mathcal{K}_{\Sigma}^{\circ}))\subset\mathcal{E}_{\Sigma}$, we obtain

[TABLE]

which implies $S_{*}(\mathcal{E}_{\Sigma})\subset\mathcal{E}_{\Sigma}$. By the same arguments, applied to $S_{*}^{-1}$ instead of $S_{*}$, we have $S_{*}^{-1}(\mathcal{E}_{\Sigma})\subset\mathcal{E}_{\Sigma}$. This shows that $S_{*}(\mathcal{E}_{\Sigma})=\mathcal{E}_{\Sigma}$. ∎

3. Characterization of the induced map $S$

In this section, we use the same notation as in the previous section. We assume that $S\colon B_{e}\to B_{e}$ is a surjective real linear isometry, which satisfies (2.9), and $S_{*}\colon(B_{e})^{*}\to(B_{e})^{*}$ is a surjective real linear isometry defined by (2.10). The mapping $\mathbf{h}\colon\mathbb{T}\times\mathcal{K}_{e}\to\mathcal{E}_{e}$ is a homeomorphism as in (2.11).

We shall analyze the isometry $S_{*}$ on $(B_{e})^{*}$, which leads to the characterization of the map $S=UT_{0}U^{-1}$: The idea of the arguments is a generalization of a well-known proof of the Banach-Stone theorem. In fact, if $S$ is complex linear, then the fact that the adjoint operator $S^{*}$ preserves extreme points implies that $S$ is a weighted composition operator. Our argument below is a kind of real linear version of the above.

Definition 3.1**.**

Let $p_{1}\colon\mathbb{T}\times\mathcal{K}_{e}\to\mathbb{T}$ and $p_{2}\colon\mathbb{T}\times\mathcal{K}_{e}\to\mathcal{K}_{e}$ be the natural projections from $\mathbb{T}\times\mathcal{K}_{e}$ onto the first and second coordinates, respectively. We define maps $\alpha\colon\mathbb{T}\times\mathcal{K}_{e}\to\mathbb{T}$ and $\Phi\colon\mathbb{T}\times\mathcal{K}_{e}\to\mathcal{K}_{e}$ by $\alpha=p_{1}\circ\mathbf{h}^{-1}\circ S_{*}\circ\mathbf{h}$ and $\Phi=p_{2}\circ\mathbf{h}^{-1}\circ S_{*}\circ\mathbf{h}$.

[TABLE]

Recall that $S_{*}$ is a real linear isometry with $S_{*}(\mathcal{E}_{e})=\mathcal{E}_{e}$ by Lemma 2.15, and thus it is bijective. By the definition of $S_{*}$, it is continuous with respect to the weak*-topology. Therefore, $S_{*}|_{\mathcal{E}_{e}}\colon\mathcal{E}_{e}\to\mathcal{E}_{e}$ is a homeomorphism with respect to the relative weak*-topology. Hence $\alpha$ and $\Phi$ are both well-defined, surjective continuous maps.

We get $(\mathbf{h}^{-1}\circ S_{*}\circ\mathbf{h})(\lambda,x)=(\alpha(\lambda,x),\Phi(\lambda,x))$ for $(\lambda,x)\in\mathbb{T}\times\mathcal{K}_{e}$ by the definition of $\alpha$ and $\Phi$. Hence $(S_{*}\circ\mathbf{h})(\lambda,x)=\mathbf{h}(\alpha(\lambda,x),\Phi(\lambda,x))$, which implies $S_{*}(\lambda\delta_{x})=\alpha(\lambda,x)\delta_{\Phi(\lambda,x)}$ for every $(\lambda,x)\in\mathbb{T}\times\mathcal{K}_{e}$. Set, for each $\lambda\in\mathbb{T}$, $\alpha_{\lambda}(x)=\alpha(\lambda,x)$ and $\Phi_{\lambda}(x)=\Phi(\lambda,x)$ for $x\in\mathcal{K}_{e}$. Then $\alpha_{\lambda}$ and $\Phi_{\lambda}$ satisfy

[TABLE]

for every $\lambda\in\mathbb{T}$ and $x\in\mathcal{K}_{e}$.

We first investigate the map $\alpha$. In fact, since $S_{*}$ is a real linear isometry, we can prove that $\alpha(\cdot,x)$ is a real linear isometry on $\mathbb{T}$ for each $x\in\mathcal{K}_{e}$ as well.

Lemma 3.1**.**

For each $x\in\mathcal{K}_{e}$, $\alpha_{i}(x)=i\alpha_{1}(x)$ or $\alpha_{i}(x)=-i\alpha_{1}(x)$.

Proof.

Let $x\in\mathcal{K}_{e}$, and we set $\lambda_{0}=(1+i)/\sqrt{2}\in\mathbb{T}$. By the real linearity of $S_{*}$, we obtain

[TABLE]

Evaluating these equalities at $\widetilde{\mathbf{1}}\in B_{e}$, we get $\sqrt{2}\,\alpha_{\lambda_{0}}(x)=\alpha_{1}(x)+\alpha_{i}(x)$. Since $|\alpha_{\lambda}(x)|=1$ for all $\lambda\in\mathbb{T}$, we have $\sqrt{2}=|\alpha_{1}(x)+\alpha_{i}(x)|=|1+\alpha_{i}(x)\overline{\alpha_{1}(x)}|$. Then we see that $\alpha_{i}(x)\overline{\alpha_{1}(x)}=i$ or $\alpha_{i}(x)\overline{\alpha_{1}(x)}=-i$. We thus conclude $\alpha_{i}(x)=i\alpha_{1}(x)$ or $\alpha_{i}(x)=-i\alpha_{1}(x)$. ∎

Now we can describe a connection between $\alpha_{1}$ and $\alpha_{i}$.

Lemma 3.2**.**

There exists a continuous function $s_{0}\colon\mathcal{K}_{e}\to\{\pm 1\}$ such that $S_{*}(i\delta_{x})=is_{0}(x)\alpha_{1}(x)\delta_{\Phi(i,x)}$ for every $x\in\mathcal{K}_{e}$.

Proof.

For each $x\in\mathcal{K}_{e}$, we obtain $\alpha_{i}(x)=i\alpha_{1}(x)$ or $\alpha_{i}(x)=-i\alpha_{1}(x)$ by Lemma 3.1. We define $E_{+}$ and $E_{-}$ by

[TABLE]

Then we have $\mathcal{K}_{e}=E_{+}\cup E_{-}$ and $E_{+}\cap E_{-}=\emptyset$. By the continuity of the functions $\alpha_{1}=\alpha(1,\cdot)$ and $\alpha_{i}=\alpha(i,\cdot)$ on $\mathcal{K}_{e}$, we see that $E_{+}$ and $E_{-}$ are both closed subsets of $\mathcal{K}_{e}$. Hence, the function $s_{0}\colon\mathcal{K}_{e}\to\{\pm 1\}$, defined by

[TABLE]

is continuous on $\mathcal{K}_{e}$. Then we obtain $\alpha_{i}(x)=is_{0}(x)\alpha_{1}(x)$ for every $x\in\mathcal{K}_{e}$. This shows $S_{*}(i\delta_{x})=is_{0}(x)\alpha_{1}(x)\delta_{\Phi(i,x)}$ for all $x\in\mathcal{K}_{e}$. ∎

We are ready to prove that $\alpha(\cdot,x)$ is a real linear isometry on $\mathbb{T}$ for each $x\in\mathcal{K}_{e}$; more explicitly, we have the following identity, which says that $\Phi_{\lambda}$ depends on $\Phi_{1}$ and $\Phi_{i}$.

Lemma 3.3**.**

Let $s_{0}$ be the continuous function as in Lemma 3.2. For each $\lambda=r+it\in\mathbb{T}$ with $r,t\in\mathbb{R}$ and $x\in\mathcal{K}_{e}$,

[TABLE]

for all $\widetilde{f}\in B_{e}$.

Proof.

Let $\lambda=r+it\in\mathbb{T}$ with $r,t\in\mathbb{R}$ and $x\in\mathcal{K}_{e}$. Recall that $S_{*}(\delta_{x})=\alpha_{1}(x)\delta_{\Phi(1,x)}$, and $S_{*}(i\delta_{x})=is_{0}(x)\alpha_{1}(x)\delta_{\Phi(i,x)}$ by Lemma 3.2. Because $S_{*}$ is real linear,

[TABLE]

The evaluation of these equalities at $\widetilde{\mathbf{1}}\in B_{e}$ shows that $\alpha_{\lambda}(x)=\alpha_{1}(x)(r+its_{0}(x))$. Since $r+its_{0}(x)=\lambda^{s_{0}(x)}$, then $\alpha_{\lambda}(x)=\lambda^{s_{0}(x)}\alpha_{1}(x)$. In particular, $\alpha(\cdot,x)$ is a real linear isometry on $\mathbb{T}$. We thus obtain $\lambda^{s_{0}(x)}\delta_{\Phi(\lambda,x)}=r\delta_{\Phi(1,x)}+its_{0}(x)\delta_{\Phi(i,x)}$, which implies that $\lambda^{s_{0}(x)}\widetilde{f}(\Phi_{\lambda}(x))=r\widetilde{f}(\Phi_{1}(x))+its_{0}(x)\widetilde{f}(\Phi_{i}(x))$ for all $\widetilde{f}\in B_{e}$. ∎

We next analyze the map $\Phi$, which satisfies $S_{*}(\lambda\delta_{x})=\alpha_{\lambda}(x)\delta_{\Phi_{\lambda}(x)}$ for every $\lambda\in\mathbb{T}$ and $x\in\mathcal{K}_{e}$. Roughly speaking, we prove that $\Phi_{\lambda}(x)$ is independent from $\lambda\in\mathbb{T}$.

Definition 3.2**.**

Recall that $\mathcal{K}_{e}=\mathfrak{D}_{e}\times\mathbb{T}$, where

[TABLE]

We denote by $q_{j}$ the projection from $\mathcal{K}_{e}$ onto the $j$-th coordinate of $\mathcal{K}_{e}$ for $j=1,2,3$: If $\mathcal{K}_{e}=\mathcal{K}_{C}$, then we define $q_{j}(\eta,\eta,w)=\eta$ for $j=1,2$ and $(\eta,\eta,w)\in\mathcal{K}_{C}$. Let $\Phi\colon\mathbb{T}\times\mathcal{K}_{e}\to\mathcal{K}_{e}$ be the map as in Definition 3.1. Define $\phi\colon\mathbb{T}\times\mathcal{K}_{e}\to q_{1}(\mathcal{K}_{e})$, $\psi\colon\mathbb{T}\times\mathcal{K}_{e}\to\partial\mathcal{A}$ and $\varphi\colon\mathbb{T}\times\mathcal{K}_{e}\to\mathbb{T}$ by

[TABLE]

respectively. For each $\lambda\in\mathbb{T}$, we also define $\phi_{\lambda}$, $\psi_{\lambda}$ and $\varphi_{\lambda}$ by $\phi_{\lambda}(x)=\phi(\lambda,x)$, $\psi_{\lambda}(x)=\psi(\lambda,x)$ and $\varphi_{\lambda}(x)=\varphi(\lambda,x)$ for all $x\in\mathcal{K}_{e}$.

By the definition of $\phi$, $\psi$ and $\varphi$, we get $\Phi_{\lambda}(x)=(\phi_{\lambda}(x),\psi_{\lambda}(x),\varphi_{\lambda}(x))$ for every $\lambda\in\mathbb{T}$ and $x\in\mathcal{K}_{e}$; if $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$ then $\phi_{\lambda}(x)\in\mathbb{T}$, if $\mathcal{K}_{e}=\mathcal{K}_{C}$ then $\phi_{\lambda}(x)=\psi_{\lambda}(x)\in\partial\mathcal{A}$, and if $\mathcal{K}_{e}=\mathcal{K}_{\sigma}$ then $\phi_{\lambda}(x)=a$. By (2.2),

[TABLE]

for all $f\in\mathcal{S}^{\infty}$ and $(\lambda,x)\in\mathbb{T}\times\mathcal{K}_{e}$. Note that $\phi$, $\psi$ and $\varphi$ are surjective and continuous, since so is $\Phi$ (see Definition 3.1).

Remark 3.1**.**

We notice that $\Phi_{\mu}(x)=\Phi_{-\mu}(x)$ for $\mu\in\{1,i\}$ and $x\in\mathcal{K}_{e}$: In fact, equality (3.1) shows that

[TABLE]

for all $\widetilde{f}\in B_{e}$ and $x\in\mathcal{K}_{e}$. Since $B_{e}$ separates the points of $\mathcal{K}_{e}$ by Propositions 2.3 and 2.4, we obtain $\Phi_{1}(x)=\Phi_{-1}(x)$ and $\Phi_{i}(x)=\Phi_{-i}(x)$ for all $x\in\mathcal{K}_{e}$. Consequently, $\phi_{1}=\phi_{-1}$, $\psi_{1}=\psi_{-1}$ and $\varphi_{1}=\varphi_{-1}$.

We shall prove that $\phi_{\lambda}(x)$, $\psi_{\lambda}(x)$ and $\varphi_{\lambda}(x)$ are independent from $\lambda\in\mathbb{T}$, which is a consequence of (3.1) and (3.2). We need the following elementary properties of $\mathcal{S}^{\infty}$ to demonstrate it. In fact, we shall prove that $\mathcal{S}^{\infty}$ separates the points of $\overline{\mathbb{D}}$ and $\partial\mathcal{A}$ in the next two lemmas.

Proposition 3.4**.**

Let $z_{0},z_{1},z_{2}\in\overline{\mathbb{D}}$ with $z_{0}\not\in\{z_{1},z_{2}\}$, and let $\eta_{1},\eta_{2},\eta_{3}\in\partial\mathcal{A}$. There exists $f_{0}\in\mathcal{S}^{\infty}$ such that

[TABLE]

Proof.

We set $z_{3}=\widehat{{\rm id}}(\eta_{1})$, $z_{4}=\widehat{{\rm id}}(\eta_{2})$ and $z_{5}=\widehat{{\rm id}}(\eta_{3})$. Then $z_{j}\in\mathbb{T}$ for $j=3,4,5$.

Case 1. Suppose that $z_{0}\not\in\{z_{3},z_{4},z_{5}\}$. We set

[TABLE]

and then $\widehat{g}(z_{0})\neq 0$, $\widehat{g}(z_{1})=0=\widehat{g}(z_{2})$ and $\widehat{g^{\prime}}(z_{j})=0$ for $j=3,4,5$. If we set $f_{1}=g/\widehat{g}(z_{0})\in\mathcal{S}^{\infty}$, then we obtain

[TABLE]

Case 2. If $z_{0}\in\{z_{3},z_{4},z_{5}\}$, then we define $J=\{j\in\{3,4,5\}:z_{j}\neq z_{0}\}$ and $J_{0}=\{1,2\}\cup J$. Let $j\in J_{0}$, and set

[TABLE]

Here, we notice that $z_{0}\neq z_{j}$. We see that $\widehat{g_{j}}(z_{0})\neq 0$ and $\widehat{g_{j}^{\prime}}(z_{0})=0=\widehat{g_{j}}(z_{j})=\widehat{g_{j}^{\prime}}(z_{j})$. The function $f_{j}=g_{j}/\widehat{g_{j}}(z_{0})\in\mathcal{S}^{\infty}$ satisfies $\widehat{f_{j}}(z_{0})=1$ and $\widehat{f_{j}^{\prime}}(z_{0})=0=\widehat{f_{j}}(z_{j})=\widehat{f_{j}^{\prime}}(z_{j})$. We set $f=\Pi_{k\in J_{0}}f_{k}\in\mathcal{S}^{\infty}$, and then $\widehat{f}(z_{0})=1$ and $\widehat{f}(z_{j})=0$. In particular, $\widehat{f}(z_{1})=0=\widehat{f}(z_{2})$. We note that $f^{\prime}=f_{j}^{\prime}\,\Pi_{k\in J_{0}\setminus\{j\}}f_{k}+f_{j}\,(\Pi_{k\in J_{0}\setminus\{j\}}f_{k})^{\prime}$ by the multiplication law. Since $\widehat{f_{j}}(z_{j})=0=\widehat{f_{j}^{\prime}}(z_{j})$, we obtain $\widehat{f^{\prime}}(z_{j})=0$ for each $j\in J_{0}$. If $l\not\in J_{0}$, then $z_{l}=z_{0}$. Set $J_{0}=\{k_{1},\dots,k_{m}\}$, a finite set. Since $\widehat{f_{k_{1}}}(z_{0})=1$ and $\widehat{f_{k_{1}}^{\prime}}(z_{0})=0$, the multiplication law shows that

[TABLE]

By the same reasonings, we have $\widehat{f^{\prime}}(z_{l})=\left\{\left(\Pi_{k\in J_{0}\setminus\{k_{1},\cdots,k_{m-1}\}}f_{k}\right)^{\prime}\right\}\widehat{\mathstrut}\,(z_{0})=\widehat{f_{k_{m}}^{\prime}}(z_{0})=0.$ Hence $f$ is a desired function. ∎

Proposition 3.5**.**

For each $\eta_{0},\eta_{1},\eta_{2}\in\partial\mathcal{A}$ with $\eta_{0}\not\in\{\eta_{1},\eta_{2}\}$, there exists $f_{0}\in\mathcal{S}^{\infty}$ and $j_{0}\in\{0,1,2\}$ such that

[TABLE]

Proof.

Set $z_{j}=\widehat{{\rm id}}(\eta_{j})$ for $j=0,1,2$.

Case 1. Suppose that $z_{0}=z_{1}=z_{2}$. Since $\eta_{0}\not\in\{\eta_{1},\eta_{2}\}$, there exists $v_{1}\in H^{\infty}(\mathbb{D})$ such that $\widehat{v_{1}}(\eta_{0})=1$ and $\widehat{v_{1}}(\eta_{1})=0=\widehat{v_{1}}(\eta_{2})$. We set $g_{1}=I_{0}(v_{1})\in\mathcal{S}^{\infty}$, and then $\widehat{g_{1}^{\prime}}(\eta_{0})=1$ and $\widehat{g_{1}^{\prime}}(\eta_{1})=0=\widehat{g_{1}^{\prime}}(\eta_{2})$ (see (2.6)). Because $\widehat{g_{1}}$ is continuous on $\mathbb{D}$, we have that $\widehat{g_{1}}(\xi)=\widehat{g}(\eta)$ for each $\xi\in\mathbb{T}$ and $\eta\in\mathcal{M}_{\xi}$, the fiber over $\xi$. This implies that $\widehat{g_{1}}(\eta_{0})=\widehat{g_{1}}(\eta_{1})=\widehat{g_{1}}(\eta_{2})$, since $z_{0}=z_{1}=z_{2}$. Then the function $f_{1}=g_{1}-\widehat{g_{1}}(\eta_{0})\mathbf{1}\in\mathcal{S}^{\infty}$ satisfies

[TABLE]

Case 2. Suppose that $z_{0}=z_{1}\neq z_{2}$. Setting $g_{2}(z)=(z-z_{0})^{2}(z-z_{2})\in\mathcal{S}^{\infty}$, we see that $\widehat{g_{2}}(z_{0})=0=\widehat{g}(z_{2})$, $\widehat{g_{2}^{\prime}}(z_{0})=0$ and $\widehat{g_{2}^{\prime}}(z_{2})\neq 0$. Then $f_{2}=g_{2}/\widehat{g_{2}^{\prime}}(z_{2})\in\mathcal{S}^{\infty}$ satisfies

[TABLE]

Case 3. Suppose that $z_{0}=z_{2}\neq z_{1}$. By a quite similar argument to Case 2, we can find $f_{3}\in\mathcal{S}^{\infty}$ with

[TABLE]

Case 4. Now we suppose that $z_{0}\neq z_{1}$ and $z_{0}\neq z_{2}$. Set

[TABLE]

We see that $\widehat{g_{4}^{\prime}}(z_{0})\neq 0$, $\widehat{g_{4}^{\prime}}(z_{1})=0=\widehat{g_{4}^{\prime}}(z_{2})$ and $\widehat{g_{4}}(z_{j})=0$ for $j=0,1,2$. Set $f_{4}=g_{4}/\widehat{g_{4}^{\prime}}(z_{0})\in\mathcal{S}^{\infty}$, and then

[TABLE]

The proof is complete. ∎

Before proving that $\phi_{\lambda}(x)$ is independent of $\lambda\in\mathbb{T}$, we prepare one lemma, which is an essential part of it.

Lemma 3.6**.**

Let $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{C},\mathcal{K}_{\sigma}\}$. Then $\phi_{\lambda}(x)\in\{\phi_{1}(x),\phi_{i}(x)\}$ for all $\lambda\in\mathbb{T}$ and $x\in\mathcal{K}_{e}$.

Proof.

Note first that $\phi$ is a map from $\mathbb{T}\times\mathcal{K}_{e}$ to $q_{1}(\mathcal{K}_{e})$. If $\mathcal{K}_{e}=\mathcal{K}_{\sigma}$, then $q_{1}(\mathcal{K}_{e})=\{a\}$. Thus $\phi_{\lambda}(x)=\phi_{1}(x)$ for all $\lambda\in\mathbb{T}$ and $x\in\mathcal{K}_{\sigma}$.

Let $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{C}\}$ and $x_{0}\in\mathcal{K}_{e}$. Suppose, on the contrary, that $\phi_{\lambda_{0}}(x_{0})\not\in\{\phi_{1}(x_{0}),\phi_{i}(x_{0})\}$ for some $\lambda_{0}\in\mathbb{T}\setminus\{\pm 1,\pm i\}$ (see Remark 3.1).

If $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$, then by Proposition 3.4, there exists $f_{0}\in\mathcal{S}^{\infty}$ such that

[TABLE]

By (3.2), we have $\widetilde{f_{0}}(\Phi_{\lambda_{0}}(x_{0}))=1$ and $\widetilde{f_{0}}(\Phi_{1}(x_{0}))=0=\widetilde{f_{0}}(\Phi_{i}(x_{0}))$. Applying these equalities to (3.1) to get $\lambda_{0}^{s_{0}(x_{0})}=0$, which contradicts $\lambda_{0}\in\mathbb{T}$.

If $\mathcal{K}_{e}=\mathcal{K}_{C}$, then $\phi=\psi$ by definition. Proposition 3.5 shows that there exist $f_{1}\in\mathcal{S}^{\infty}$ and $\mu_{0}\in\{\lambda_{0},1,i\}$ such that

[TABLE]

Equality (3.2) shows that $\widetilde{f_{1}}(\Phi_{\mu_{0}}(x_{0}))=\varphi_{\mu_{0}}(x_{0})$ and $\widetilde{f_{1}}(\Phi_{\nu}(x_{0}))=0$ for $\nu\in\{\lambda_{0},1,i\}\setminus\{\mu_{0}\}$. If $\mu_{0}=\lambda_{0}$, then equality (3.1) yields $\lambda_{0}^{s_{0}(x_{0})}\varphi_{\lambda_{0}}(x_{0})=0$. This contradicts $\lambda_{0}\varphi_{\lambda_{0}}(x_{0})\in\mathbb{T}$. Hence $\mu_{0}\in\{1,i\}$. Let $\lambda_{0}=r_{0}+it_{0}$ with $r_{0},t_{0}\in\mathbb{R}$. If $\mu_{0}=1$, then $0=r_{0}\varphi_{1}(x_{0})$ by (3.1), which implies $r_{0}=0$. Thus $it_{0}=\lambda_{0}\in\mathbb{T}\setminus\{\pm 1,\pm i\}$, a contradiction. If $\mu_{0}=i$, we will lead a similar contradiction by the same reasonings.

From the above arguments, we conclude that $\phi_{\lambda}(x_{0})\in\{\phi_{1}(x_{0}),\phi_{i}(x_{0})\}$ for all $\lambda\in\mathbb{T}$. Since $x_{0}\in\mathcal{K}_{e}$ was arbitrary, the proof is complete. ∎

Now we are in a position to show that $\phi_{\lambda}$ does not depend on $\lambda\in\mathbb{T}$. We first prove it for $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$ and $\mathcal{K}_{e}=\mathcal{K}_{\sigma}$.

Lemma 3.7**.**

Let $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{\sigma}\}$. Then $\phi_{1}\colon\mathcal{K}_{e}\to q_{1}(\mathcal{K}_{e})$ is a surjective and continuous function with $\phi_{1}(x)=\phi_{\lambda}(x)$ for all $x\in\mathcal{K}_{e}$ and $\lambda\in\mathbb{T}$.

Proof.

If $\mathcal{K}_{e}=\mathcal{K}_{\sigma}$, then $\phi$ is a surjective and continuous function from $\mathbb{T}\times\mathcal{K}_{\sigma}$ onto $q_{1}(\mathcal{K}_{\sigma})=\{a\}$. Hence $\phi_{1}\colon\mathcal{K}_{\sigma}\to\{a\}$ is a surjective and continuous function satisfying $\phi_{1}(x)=\phi_{\lambda}(x)$ for all $x\in\mathcal{K}_{\sigma}$ and $\lambda\in\mathbb{T}$.

Let $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$ and $x_{0}\in\mathcal{K}_{\Sigma}$. We shall prove that $\phi_{1}(x_{0})=\phi_{i}(x_{0})$. Since $\phi$ is a continuous function from $\mathbb{T}\times\mathcal{K}_{\Sigma}$ onto $q_{1}(\mathcal{K}_{\Sigma})=\mathbb{T}$, the function $\phi(\cdot,x_{0})\colon\mathbb{T}\to\mathbb{T}$, which maps $\lambda\in\mathbb{T}$ to $\phi(\lambda,x_{0})$, is continuous on $\mathbb{T}$ as well. This implies that the image $\phi(\mathbb{T},x_{0})$ of $\mathbb{T}$ under the function $\phi(\cdot,x_{0})$ is a connected subset of $\mathbb{T}$. Because $\phi(\mathbb{T},x_{0})\in\{\phi_{1}(x_{0}),\phi_{i}(x_{0})\}$ by Lemma 3.6, we conclude that $\phi_{1}(x_{0})=\phi_{i}(x_{0})$. We thus obtain $\phi_{\lambda}(x_{0})=\phi_{1}(x_{0})$ for all $\lambda\in\mathbb{T}$.

We show that $\phi_{1}\colon\mathcal{K}_{\Sigma}\to\mathbb{T}$ is surjective. Since $\phi$ is surjective, for each $\xi\in\mathbb{T}$ there exists $(\lambda_{1},x_{1})\in\mathbb{T}\times\mathcal{K}_{\Sigma}$ such that $\phi(\lambda_{1},x_{1})=\xi$. Then we have that $\phi_{1}(x_{1})=\phi_{\lambda_{1}}(x_{1})=\phi(\lambda_{1},x_{1})=\xi$. This shows that $\phi_{1}$ is surjective. ∎

We now prove that the above result holds for $\mathcal{K}_{e}=\mathcal{K}_{C}$.

Lemma 3.8**.**

The map $\phi_{1}\colon\mathcal{K}_{C}\to\partial\mathcal{A}$ is surjective and continuous. In addition, $\phi_{1}(x)=\phi_{\lambda}(x)$ holds for all $x\in\mathcal{K}_{C}$ and $\lambda\in\mathbb{T}$.

Proof.

Note that $\phi=\psi$ on $\mathbb{T}\times\mathcal{K}_{C}$ by definition. Let $x_{0}\in\mathcal{K}_{C}$. We shall prove that $\phi_{1}(x_{0})=\phi_{i}(x_{0})$. To this end, suppose not, and then $\phi_{1}(x_{0})\neq\phi_{i}(x_{0})$. Let $\lambda_{0}=(1+i)/\sqrt{2}\in\mathbb{T}$, and set $\eta_{0}=\phi_{\lambda_{0}}(x_{0})$, $\eta_{1}=\phi_{1}(x_{0})$ and $\eta_{2}=\phi_{i}(x_{0})$. We see that $\eta_{0}\in\{\eta_{1},\eta_{2}\}$ by Lemma 3.6. If $\eta_{0}=\eta_{1}\neq\eta_{2}$, Proposition 3.5 shows that there exist $f_{0}\in\mathcal{S}^{\infty}$ and $j_{0}\in\{0,1,2\}$ such that

[TABLE]

Substituting these equalities into (3.2), we get $\widetilde{f_{0}}(\Phi_{\mu_{0}}(x_{0}))=\varphi_{\mu_{0}}(x_{0})$ for some $\mu_{0}\in\{\lambda_{0},1,i\}$ and $\widetilde{f_{0}}(\Phi_{\nu}(x_{0}))=0$ for $\nu\in\{\lambda_{0},1,i\}\setminus\{\mu_{0}\}$. We can lead a contradiction by a similar argument to Proof of Lemma 3.6. If $\eta_{0}=\eta_{2}\neq\eta_{1}$, we will arrive at a contradiction by the same reasonings. Thus, we have $\phi_{1}(x_{0})=\phi_{i}(x_{0})$, and therefore, $\phi_{\lambda}(x_{0})=\phi_{1}(x_{0})$ for all $\lambda\in\mathbb{T}$. Since $x_{0}\in\mathcal{K}_{C}$ is arbitrarily chosen, we conclude that $\phi_{1}(x)=\phi_{\lambda}(x)$ for all $x\in\mathcal{K}_{C}$ and $\lambda\in\mathbb{T}$.

We show that $\phi_{1}$ is surjective. For each $\eta\in q_{1}(\mathcal{K}_{C})$, we can choose $(\lambda_{1},x_{1})\in\mathbb{T}\times\mathcal{K}_{C}$ such that $\phi(\lambda_{1},x_{1})=\eta$, since $\phi$ is surjective. Then $\phi_{1}(x_{1})=\phi_{\lambda_{1}}(x_{1})=\phi(\lambda_{1},x_{1})=\eta$. Hence, $\phi_{1}$ is surjective. ∎

We shall prove that $\psi_{\lambda}(x)$ is independent of $\lambda\in\mathbb{T}$. The proof below is quite similar to that of Lemmas 3.7.

Lemma 3.9**.**

Let $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{C},\mathcal{K}_{\sigma}\}$. Then $\psi_{1}\colon\mathcal{K}_{e}\to\partial\mathcal{A}$ is a surjective and continuous map with $\psi_{1}(x)=\psi_{\lambda}(x)$ for all $x\in\mathcal{K}_{e}$ and $\lambda\in\mathbb{T}$.

Proof.

If $\mathcal{K}_{e}=\mathcal{K}_{C}$, then Lemma 3.8 yields the conclusion, since $\phi=\psi$ for $\mathcal{K}_{C}$. We will consider the case when $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{\sigma}\}$. Let $x_{0}\in\mathcal{K}_{e}$. By Lemma 3.7, $\Phi_{\lambda}(x_{0})=(\phi_{1}(x_{0}),\psi_{\lambda}(x_{0}),\varphi_{\lambda}(x_{0}))$ for $\lambda\in\mathbb{T}$. Equality (3.2) is reduced to

[TABLE]

for all $f\in\mathcal{S}^{\infty}$ and $\lambda\in\mathbb{T}$.

First, we show that $\psi_{\lambda}(x_{0})\in\{\psi_{1}(x_{0}),\psi_{i}(x_{0})\}$ for all $\lambda\in\mathbb{T}$. Suppose, on the contrary, that $\psi_{\lambda_{0}}(x_{0})\not\in\{\psi_{1}(x_{0}),\psi_{i}(x_{0})\}$ for some $\lambda_{0}\in\mathbb{T}\setminus\{1,i\}$. Set $\eta_{0}=\psi_{\lambda_{0}}(x_{0})$, $\eta_{1}=\psi_{1}(x_{0})$ and $\eta_{2}=\psi_{i}(x_{0})$. Then $\eta_{j}\in\partial\mathcal{A}$ for $j=0,1,2$ and $\eta_{0}\not\in\{\eta_{1},\eta_{2}\}$. There exists $v_{0}\in H^{\infty}(\mathbb{D})$ such that $\widehat{v_{0}}(\eta_{0})=1$ and $\widehat{v_{0}}(\eta_{1})=0=\widehat{v_{0}}(\eta_{2})$. The function $f_{0}\in\mathcal{S}^{\infty}$, defined by $f_{0}=I_{0}(v_{0})-\widehat{I_{0}(v_{0})}(\phi_{1}(x_{0}))\mathbf{1}$, satisfies that $\widehat{f_{0}}(\phi_{1}(x_{0}))=0$, $\widehat{f_{0}^{\prime}}(\eta_{0})=1$ and $\widehat{f_{0}^{\prime}}(\eta_{1})=0=\widehat{f_{0}^{\prime}}(\eta_{2})$. By (3.3), $\widetilde{f_{0}}(\Phi_{\lambda_{0}}(x_{0}))=\varphi_{\lambda_{0}}(x_{0})$ and $\widetilde{f_{0}}(\Phi_{1}(x_{0}))=0=\widetilde{f_{0}}(\Phi_{i}(x_{0}))$. If we substitute these equalities into (3.1), we obtain $\lambda_{0}^{s_{0}(x_{0})}\varphi_{\lambda_{0}}(x_{0})=0$, which contradicts $\lambda_{0},\varphi_{\lambda_{0}}(x_{0})\in\mathbb{T}$. Consequently, $\psi_{\lambda}(x_{0})\in\{\psi_{1}(x_{0}),\psi_{i}(x_{0})\}$ for all $\lambda\in\mathbb{T}$.

We next prove that $\psi_{1}(x_{0})=\psi_{i}(x_{0})$. Suppose that $\psi_{1}(x_{0})\neq\psi_{i}(x_{0})$. Let $\lambda_{1}=(1+i)/\sqrt{2}\in\mathbb{T}$. Then $\psi_{\lambda_{1}}(x_{0})\in\{\psi_{1}(x_{0}),\psi_{i}(x_{0})\}$ as proved above. If $\psi_{\lambda_{1}}(x_{0})=\psi_{1}(x_{0})\neq\psi_{i}(x_{0})$, then we can choose $f_{1}\in\mathcal{S}^{\infty}$ so that

[TABLE]

by the same arguments as in the previous paragraph. Applying these equalities to (3.3), we have $\widetilde{f}(\Phi_{i}(x_{0}))=\varphi_{i}(x_{0})$ and $\widetilde{f}(\Phi_{1}(x_{0}))=0=\widetilde{f}(\Phi_{\lambda_{1}}(x_{0}))$, since $\psi_{\lambda_{1}}(x_{0})=\psi_{1}(x_{0})$. By (3.1), we get $0=is_{0}(x_{0})\varphi_{i}(x_{0})$, which is impossible. If $\psi_{\lambda_{1}}(x_{0})=\psi_{i}(x_{0})\neq\psi_{1}(x_{0})$, then we will reach a contradiction by a quite similar argument. Therefore, we conclude that $\psi_{1}(x_{0})=\psi_{i}(x_{0})$. Consequently $\psi_{\lambda}(x)=\psi_{1}(x)$ for all $\lambda\in\mathbb{T}$ and $x\in\mathcal{K}_{e}$.

Finally, since $\psi$ is surjective, for each $\eta\in\partial\mathcal{A}$ there exists $(\lambda,x)\in\mathbb{T}\times\mathcal{K}_{e}$ such that $\psi(\lambda,x)=\eta$. We thus obtain $\eta=\psi_{\lambda}(x)=\psi_{1}(x)$, which shows the surjectivity of $\psi_{1}$. ∎

We investigate the function $\varphi_{\lambda}$. To be more explicit, we shall prove, for each $x\in\mathcal{K}_{e}$, that $\varphi_{i}(x)$ is $\varphi_{1}(x)$ or $-\varphi_{1}(x)$.

Lemma 3.10**.**

Let $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{C},\mathcal{K}_{\sigma}\}$. There exists a continuous function $s_{1}\colon\mathcal{K}_{e}\to\{\pm 1\}$ such that $\varphi_{i}(x)=s_{1}(x)\varphi_{1}(x)$ for all $x\in\mathcal{K}_{e}$.

Proof.

Fix an arbitrary $x_{0}\in\mathcal{K}_{e}$. According to Lemmas 3.7, 3.8 and 3.9, we can write $\Phi_{\lambda}(x_{0})=(\phi_{1}(x_{0}),\psi_{1}(x_{0}),\varphi_{\lambda}(x_{0}))$ for all $\lambda\in\mathbb{T}$. Let $\lambda_{0}=(1+i)/\sqrt{2}\in\mathbb{T}$ and $f_{0}(z)=z-\phi_{1}(x_{0})\mathbf{1}\in\mathcal{S}^{\infty}$. Then $\widehat{f_{0}}(\phi_{1}(x_{0}))=0$ and $\widehat{f_{0}^{\prime}}=1$ on $\partial\mathcal{A}$. By (3.3), $\widetilde{f_{0}}(\Phi_{\mu}(x_{0}))=\varphi_{\mu}(x_{0})$ for $\mu=\lambda_{0},1,i$. If we apply these equalities to (3.1), then we obtain $\sqrt{2}\,\lambda_{0}^{s_{0}(x_{0})}\varphi_{\lambda_{0}}(x_{0})=\varphi_{1}(x_{0})+is_{0}(x_{0})\varphi_{i}(x_{0})$. As $\varphi_{\lambda}(x_{0})\in\mathbb{T}$ for all $\lambda\in\mathbb{T}$, we have

[TABLE]

Then we get $is_{0}(x_{0})\varphi_{i}(x_{0})\overline{\varphi_{1}(x_{0})}=i$ or $is_{0}(x_{0})\varphi_{i}(x_{0})\overline{\varphi_{1}(x_{0})}=-i$. Thus, for each $x\in\mathcal{K}_{e}$, we obtain $\varphi_{i}(x)=s_{0}(x)\varphi_{1}(x)$ or $\varphi_{i}(x)=-s_{0}(x)\varphi_{1}(x)$. By the continuity of $\varphi_{1}$, $\varphi_{i}$ and $s_{0}$, there exists a continuous function $s_{1}\colon\mathcal{K}_{e}\to\{\pm 1\}$ such that $\varphi_{i}(x)=s_{1}(x)\varphi_{1}(x)$ for all $x\in\mathcal{K}_{e}$. ∎

In the rest of this paper, we denote $a+ibs$ by $[a+ib]^{s}$ for $a,b\in\mathbb{R}$ and $s\in\{\pm 1\}$. Thus, for each $\lambda\in\mathbb{C}$, $[\lambda]^{s}=\lambda$ if $s=1$ and $[\lambda]^{s}=\overline{\lambda}$ if $s=-1$. Therefore, $[\lambda\mu]^{s}=[\lambda]^{s}[\mu]^{s}$ for all $\lambda,\mu\in\mathbb{C}$. If, in addition, $\lambda\in\mathbb{T}$, then $[\lambda]^{s}=\lambda^{s}$ for $s\in\{\pm 1\}$.

From the above arguments, we have a partial form of the isometry $S$ on $B_{e}$. This is an important part of the isometry $T_{0}=U^{-1}SU$ on $\mathcal{S}^{\infty}$ (see (2.9)).

Lemma 3.11**.**

Let $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{C},\mathcal{K}_{\sigma}\}$. For each $f\in\mathcal{S}^{\infty}$ and $x\in\mathcal{K}_{e}$,

[TABLE]

Proof.

Let $f\in\mathcal{S}^{\infty}$ and $x\in\mathcal{K}_{e}$. By (2.10), ${\rm Re}\,\{S_{*}(\chi)(\widetilde{f})\}={\rm Re}\,\{\chi(S(\widetilde{f}))\}$ for every $\chi\in(B_{e})^{*}$. Taking $\chi=\delta_{x}$ and $\chi=i\delta_{x}$ into the last equality, we get

[TABLE]

respectively, where ${\rm Im}\,z$ is the imaginary part of a complex number $z$. Therefore,

[TABLE]

Recall that $S_{*}(\delta_{x})=\alpha_{1}(x)\delta_{\Phi(1,x)}$, and $S_{*}(i\delta_{x})=is_{0}(x)\alpha_{1}(x)\delta_{\Phi(i,x)}$ by Lemma 3.2. Substitute these two equalities into (3.5) to obtain

[TABLE]

Lemmas 3.7, 3.8, 3.9 and 3.10 imply that

[TABLE]

Applying these two equalities to the above formula of $S(\widetilde{f})(x)$, we assure from (2.2) that

[TABLE]

This completes the proof. ∎

4. Proof of the main results

We recall that $S(\widetilde{f})=\widetilde{T_{0}(f)}$ for $f\in\mathcal{S}^{\infty}$ by (2.9). Applying (2.2), we can rewrite equality (3.4) as

[TABLE]

for all $f\in\mathcal{S}^{\infty}$ and $x=(z,\eta,w)\in\mathcal{K}_{e}=\mathfrak{D}_{e}\times\mathbb{T}$.

In this section, we shall derive from (4.1) that $\widehat{T_{0}(f)}(z)=[\alpha_{1}(x)\widehat{f}(\phi_{1}(x))]^{s_{0}(x)}$ and $\widehat{T_{0}(f)^{\prime}}(\eta)w=[\alpha_{1}(x)\widehat{f^{\prime}}(\psi_{1}(x))\varphi_{1}(x)]^{s_{0}(x)s_{1}(x)}$; the main idea of its proof is to show that $\alpha_{1}(x)$, $\phi_{1}(x)$ and $s_{0}(x)$ are all independent of $w$ for $x=(z,\eta,w)\in\mathcal{K}_{e}$. Once we obtain it, then the first term of the right-hand side of (4.1) is constant with respect to $w$, and therefore, we get $\widehat{T_{0}(f)}(z)=[\alpha_{1}(x)\widehat{f}(\phi_{1}(x))]^{s_{0}(x)}$. For this end, we first show that $\phi_{1}(z,\eta,w)$ does not depend on $w\in\mathbb{T}$.

In Lemmas 4.1 and 4.2, we assume that $\mathfrak{D}_{e}$ is one of $\mathfrak{D}_{\Sigma}=\mathbb{T}\times\partial\mathcal{A}$, $\mathfrak{D}_{C}=\{(\eta,\eta):\eta\in\partial\mathcal{A}\}$ and $\mathfrak{D}_{\sigma}=\{a\}\times\partial\mathcal{A}$ with $a\in\mathbb{D}$.

Lemma 4.1**.**

For each $(z,\eta)\in\mathfrak{D}_{e}$, the value $\phi_{1}(z,\eta,w)$ is independent of $w\in\mathbb{T}$.

Proof.

Take any $(z,\eta)\in\mathfrak{D}_{e}$ and $w_{1},w_{2}\in\mathbb{T}$ with $w_{1}\neq w_{2}$. We shall show that $\phi_{1}(z,\eta,w)\in\{\phi_{1}(z,\eta,w_{1}),\phi_{1}(z,\eta,w_{2})\}$ for all $w\in\mathbb{T}$. If $\mathcal{K}_{e}=\mathcal{K}_{\sigma}$, then $\phi_{1}$ is a constant function with the value $a$ on $\mathcal{K}_{\sigma}$. We need to consider the case when $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{C}\}$. Suppose that there exists $w_{0}\in\mathbb{T}$ such that $\phi_{1}(z,\eta,w_{0})\not\in\{\phi_{1}(z,\eta,w_{1}),\phi_{1}(z,\eta,w_{2})\}$. We set $x_{j}=(z,\eta,w_{j})$, $z_{j}=\phi_{1}(x_{j})$ and $\eta_{j}=\psi_{1}(x_{j})$ for $j=0,1,2$. Then $z_{0}\not\in\{z_{1},z_{2}\}$ by hypothesis. If $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$, then $z_{j}\in\mathbb{T}$ for $j=0,1,2$. There exists $f_{0}\in\mathcal{S}^{\infty}$ such that

[TABLE]

by Proposition 3.4. Substituting these equalities into (4.1), we obtain

[TABLE]

Since $w_{1}\neq w_{2}$, the above equalities yield $\widehat{T_{0}(f_{0})^{\prime}}(\eta)=0=\widehat{T_{0}(f_{0})}(z)$. Therefore, $[\alpha_{1}(x_{0})]^{s_{0}(x_{0})}=0$, which is impossible, since $\alpha_{1}(x_{0})\in\mathbb{T}$. Hence, $\phi_{1}(z,\eta,w)\in\{\phi_{1}(z,\eta,w_{1}),\phi_{1}(z,\eta,w_{2})\}$ for all $w\in\mathbb{T}$.

If $\mathcal{K}_{e}=\mathcal{K}_{C}$, then $z_{j}=\eta_{j}\in\partial\mathcal{A}$ for $j=0,1,2$. By Proposition 3.5, there exist $f_{1}\in\mathcal{S}^{\infty}$ and $j_{0}\in\{0,1,2\}$ such that

[TABLE]

If we apply these equalities to (4.1), we get

[TABLE]

We notice that $w_{0},w_{1}$ and $w_{2}$ are mutually distinct by the choice of them. Then the above equalities show that $\widehat{T_{0}(f_{1})^{\prime}}(\eta)=0=\widehat{T_{0}(f_{1})}(z)$. Therefore, $[\alpha_{1}(x_{j_{0}})\varphi_{1}(x_{j_{0}})]^{s_{0}(x_{j_{0}})s_{1}(x_{j_{0}})}=0$, which contradicts $\alpha_{1}(x_{j_{0}}),\varphi_{1}(x_{j_{0}})\in\mathbb{T}$. Hence, $\phi_{1}(z,\eta,w)\in\{\phi_{1}(z,\eta,w_{1}),\phi_{1}(z,\eta,w_{2})\}$ for all $w\in\mathbb{T}$.

We now prove that the value $\phi_{1}(z,\eta,w)$ is independent of $w\in\mathbb{T}$. Since $\phi$ is continuous, so is the function $\phi_{1}(z,\eta,\cdot)\colon\mathbb{T}\to q_{1}(\mathcal{K}_{e})$, which maps $w\in\mathbb{T}$ to $\phi_{1}(z,\eta,w)$. Thus, the set $\phi_{1}(z,\eta,\mathbb{T})$ is connected. By the previous paragraph, we see that $\phi_{1}(z,\eta,\mathbb{T})\subset\{\phi_{1}(z,\eta,w_{1}),\phi_{1}(z,\eta,w_{2})\}$. We thus obtain $\phi_{1}(z,\eta,w_{1})=\phi_{1}(z,\eta,w_{2})$. This implies that the value $\phi_{1}(z,\eta,w)$ is independent of $w\in\mathbb{T}$. ∎

We next prove that $\psi_{1}(z,\eta,w)$ does not depend on $w\in\mathbb{T}$; proof of the result is quite similar to that of Lemma 4.1.

Lemma 4.2**.**

For each $(z,\eta)\in\mathfrak{D}_{e}$, the value $\psi_{1}(z,\eta,w)$ is independent of $w\in\mathbb{T}$.

Proof.

Fix arbitrary $(z,\eta)\in\mathfrak{D}_{e}$ and $w_{1},w_{2}\in\mathbb{T}$ satisfying $w_{1}\neq w_{2}$. We shall prove that $\psi_{1}(z,\eta,w)\in\{\psi_{1}(z,\eta,w_{1}),\psi_{1}(z,\eta,w_{2})\}$ for all $w\in\mathbb{T}$. Suppose, on the contrary, that there exists $w_{0}\in\mathbb{T}$ such that $\psi_{1}(z,\eta,w_{0})\not\in\{\psi_{1}(z,\eta,w_{1}),\psi_{1}(z,\eta,w_{2})\}$. We set $x_{j}=(z,\eta,w_{j})$ and $\eta_{j}=\psi_{1}(x_{j})$ for $j=0,1,2$. Then $\eta_{0}\not\in\{\eta_{1},\eta_{2}\}$ by the hypothesis. Choose $v_{0}\in H^{\infty}(\mathbb{D})$ so that $\widehat{v_{0}}(\eta_{0})=1$ and $\widehat{v_{0}}(\eta_{1})=0=\widehat{v_{0}}(\eta_{2})$. Recall that $\phi_{1}(x_{j})$ is independent of $j$ by Lemma 4.1. If we set $f_{0}=I_{0}(v_{0})-\widehat{I_{0}(v_{0})}(\phi_{1}(x_{0}))\mathbf{1}\in\mathcal{S}^{\infty}$, then we observe

[TABLE]

Equality (4.1) with the above implies that

[TABLE]

Because $w_{1}\neq w_{2}$, we assure from the above equalities that $\widehat{T_{0}(f_{0})^{\prime}}(\eta)=0=\widehat{T_{0}(f_{0})}(z)$. Hence, $[\alpha_{1}(x_{0})\varphi_{1}(x_{0})]^{s_{0}(x_{0})s_{1}(x_{0})}=0$, which contradicts $\alpha_{1}(x_{0})\varphi_{1}(x_{0})\in\mathbb{T}$. We thus conclude that $\psi_{1}(z,\eta,w)\in\{\psi_{1}(z,\eta,w_{1}),\psi_{1}(z,\eta,w_{2})\}$ for all $w\in\mathbb{T}$.

We show that the value $\psi_{1}(z,\eta,w)$ is independent of $w\in\mathbb{T}$. The set $\psi_{1}(z,\eta,\mathbb{T})=\{\psi_{1}(z,\eta,w):w\in\mathbb{T}\}$ is connected, since $\psi_{1}$ is continuous. By the last paragraph, we see that $\psi_{1}(z,\eta,w_{1})=\psi_{1}(z,\eta,w_{2})$. Hence, the value $\psi_{1}(x,\eta,w)$ does not depend on $w\in\mathbb{T}$, as is claimed. ∎

Now we shall prove that both $s_{0}(z,\eta,w)$ and $s_{1}(z,\eta,w)$ are constant functions with respect to $w\in\mathbb{T}$.

Lemma 4.3**.**

1. (1)

If $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{\sigma}\}$, then for $x=(z,\eta,w)\in\mathcal{K}_{e}$, the values $s_{0}(x)$ and $s_{1}(x)$ are independent of $z\in q_{1}(\mathcal{K}_{e})$ and $w\in\mathbb{T}$. 2. (2)

If $\mathcal{K}_{e}=\mathcal{K}_{C}$, then for $x=(\eta,\eta,w)\in\mathcal{K}_{e}$, the values $s_{0}(x)$ and $s_{1}(x)$ are independent of $w\in\mathbb{T}$.

Proof.

Let $k=0,1$ and $\eta\in\partial\mathcal{A}$.

(1) Suppose that $\mathcal{K}_{e}\in\{\mathcal{K}_{\Sigma},\mathcal{K}_{\sigma}\}$. The function $s_{k}(\cdot,\eta,\cdot)$, which sends $(z,w)\in q_{1}(\mathcal{K}_{e})\times\mathbb{T}$ to $s_{k}(z,\eta,w)$, is continuous on $q_{1}(\mathcal{K}_{e})\times\mathbb{T}$. Since $q_{1}(\mathcal{K}_{e})=\mathbb{T}$ if $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$, and $q_{1}(\mathcal{K}_{e})=\{a\}$ if $\mathcal{K}_{e}=\mathcal{K}_{\sigma}$, the product space $q_{1}(\mathcal{K}_{e})\times\mathbb{T}$ is connected. Then the image $s_{k}(q_{1}(\mathcal{K}_{e}),\eta,\mathbb{T})$ of $q_{1}(\mathcal{K}_{e})\times\mathbb{T}$ under the continuous mapping $s_{k}(\cdot,\eta,\cdot)$ is a connected subset of $\{\pm 1\}$. Then we deduce that the value $s_{k}(z,\eta,w)$ does not depend on $z\in q_{1}(\mathcal{K}_{e})$ and $w\in\mathbb{T}$.

(2) The function $s_{k}(\eta,\eta,\cdot)$, which sends $w\in\mathbb{T}$ to $s_{k}(\eta,\eta,w)$, is continuous on $\mathbb{T}$. The image $s_{k}(\eta,\eta,\mathbb{T})$ of $\mathbb{T}$ under the mapping $s_{k}(\eta,\eta,\cdot)$ is a connected subset of $\{\pm 1\}$. Hence, the value $s_{k}(\eta,\eta,w)$ does not depend on $w\in\mathbb{T}$. ∎

By Lemmas 4.1, 4.2 and 4.3, we may write $\phi_{1}(z,\eta,w)=\phi_{1}(z,\eta)$, $\psi_{1}(z,\eta,w)=\psi_{1}(z,\eta)$ and, for $k=0,1$, $s_{k}(z,\eta,w)=s_{k}(\eta)$ for $(z,\eta,w)\in\mathcal{K}_{e}$. Then we can rewrite equality (4.1) as follows:

[TABLE]

for $f\in\mathcal{S}^{\infty}$ and $x=(z,\eta,w)\in\mathcal{K}_{e}$.

We need to show that $\alpha_{1}(z,\eta,w)$ is independent of $w\in\mathbb{T}$. We use the following elementary result to prove it.

Proposition 4.4**.**

Let $\lambda,\mu\in\mathbb{C}$. If $|\lambda+\mu w|=1$ for all $w\in\mathbb{T}$, then $\lambda\mu=0$ and $|\lambda|+|\mu|=1$.

Proof.

Suppose, on the contrary, that $\lambda\mu\neq 0$. Choose $w_{1}\in\mathbb{T}$ so that $\mu\,w_{1}=\lambda|\mu||\lambda|^{-1}$, and set $w_{2}=-w_{1}$. By hypothesis, $|\lambda+\mu\,w_{1}|=1=|\lambda+\mu\,w_{2}|$, that is,

[TABLE]

These equalities show that |\lambda|+|\mu|=\bigl{|}|\lambda|-|\mu|\bigr{|}. This implies that $\lambda=0$ or $\mu=0$, which contradicts the hypothesis that $\lambda\mu\neq 0$. Thus we obtain $\lambda\mu=0$, and then $|\lambda|+|\mu|=1$. ∎

4.1. Proof of Theorem 1

In this subsection, we give proof of Theorem 1; to this end, we consider the case when $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}=\mathfrak{D}_{\Sigma}\times\mathbb{T}$ and $\mathcal{K}_{e}=\mathcal{K}_{C}=\mathfrak{D}_{C}\times\mathbb{T}$, where $\mathfrak{D}_{\Sigma}=\mathbb{T}\times\partial\mathcal{A}$ and $\mathfrak{D}_{C}=\{(\eta,\eta):\eta\in\partial\mathcal{A}\}$.

Lemma 4.5**.**

1. (1)

The function $s_{0}$ as in Lemma 3.2 is constant on $\partial\mathcal{A}$; we will write $s_{0}$ instead of $s_{0}(\eta)$. 2. (2)

There exists a constant $c\in\mathbb{T}$ such that

1. (a)

$\widehat{T_{0}(\mathbf{1})}(z)=c^{\,s_{0}}$* for all $z\in\overline{\mathbb{D}}$,* 2. (b)

$\widehat{T_{0}(i\mathbf{1})}(z)=is_{0}\widehat{T_{0}(\mathbf{1})}(z)$* for all $z\in\overline{\mathbb{D}}$,* 3. (c)

$\alpha_{1}(x)=c$* for all $x\in\mathcal{K}_{e}$.*

Proof.

Let $\lambda\in\{1,i\}$. If we substitute $f=\lambda\mathbf{1}\in\mathcal{S}^{\infty}$ into (4.2), then

[TABLE]

for all $x=(z,\eta,w)\in\mathcal{K}_{e}$. We shall show that $\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}=0$ on $\mathcal{M}$ for $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$ and $\mathcal{K}_{e}=\mathcal{K}_{C}$.

(i) Let $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$. Suppose, on the contrary, that there exists $\eta_{0}\in\partial\mathcal{A}$ such that $\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}(\eta_{0})\neq 0$. Let $z\in\mathbb{T}$. By (4.3), $|\widehat{T_{0}(\lambda\mathbf{1})}(z)+\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}(\eta_{0})w|=1$ for all $w\in\mathbb{T}$. Because $\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}(\eta_{0})\neq 0$, Proposition 4.4 shows that $\widehat{T_{0}(\lambda\mathbf{1})}(z)=0$. Since $z\in\mathbb{T}$ is arbitrary, we deduce that $\widehat{T_{0}(\lambda\mathbf{1})}=0$ on $\mathbb{T}$. Since $T_{0}(\lambda\mathbf{1})$ is an analytic function on $\mathbb{D}$, it follows that $T_{0}(\lambda\mathbf{1})$ is identically zero on $\mathbb{D}$. This implies that $\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}=0$ on $\mathcal{M}$, which contradicts $\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}(\eta_{0})\neq 0$. Hence, $\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}(\eta)=0$ for all $\eta\in\partial\mathcal{A}$. Since $\partial\mathcal{A}$ is a boundary for $\mathcal{A}$, we get $\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}=0$ on $\mathcal{M}$.

(ii) Now, we suppose that $\mathcal{K}_{e}=\mathcal{K}_{C}$. Choose $\eta_{0}\in\partial\mathcal{A}$ arbitrarily. Equality (4.3) shows that $|\widehat{T_{0}(\lambda\mathbf{1})}(\eta_{0})+\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}(\eta_{0})w|=1$ for all $w\in\mathbb{T}$. By Proposition 4.4,

[TABLE]

Because $\eta_{0}\in\partial\mathcal{A}$ is arbitrarily chosen, we deduce that $\widehat{T_{0}(\lambda\mathbf{1})}\,\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}=0$ and $|\widehat{T_{0}(\lambda\mathbf{1})}|+|\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}|=1$ on $\partial\mathcal{A}$. Since $\partial\mathcal{A}$ is a boundary for $\mathcal{A}$, we see that $\widehat{T_{0}(\lambda\mathbf{1})}\,\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}=0$ on $\mathcal{M}$. Hence $T_{0}(\lambda\mathbf{1})T_{0}(\lambda\mathbf{1})^{\prime}=0$ on $\mathbb{D}$. Both $T_{0}(\lambda\mathbf{1})$ and $T_{0}(\lambda\mathbf{1})^{\prime}$ are analytic on the connected set $\mathbb{D}$, and thus $T_{0}(\lambda\mathbf{1})=0$ on $\mathbb{D}$, or $T_{0}(\lambda\mathbf{1})^{\prime}=0$ on $\mathbb{D}$. If $T_{0}(\lambda\mathbf{1})=0$ on $\mathbb{D}$, then we would have $T_{0}(\lambda\mathbf{1})^{\prime}=0$ on $\mathbb{D}$, and hence $\widehat{T_{0}(\lambda\mathbf{1})}=\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}=0$ on $\mathcal{M}$. This contradicts $|\widehat{T_{0}(\lambda\mathbf{1})}|+|\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}|=1$ on $\partial\mathcal{A}$, and consequently we obtain $T_{0}(\lambda\mathbf{1})^{\prime}=0$ on $\mathbb{D}$. This implies that $\widehat{T_{0}(\lambda\mathbf{1})^{\prime}}$ is identically zero on $\mathcal{M}$, as is claimed.

Since $\mathbb{D}$ is connected, we see that $T_{0}(\lambda\mathbf{1})$ is constant on $\mathbb{D}$. We set $c_{\lambda}=T_{0}(\lambda\mathbf{1})$, and then $c_{\lambda}=[\lambda\alpha_{1}(x)]^{s_{0}(\eta)}$ for all $x=(z,\eta,w)\in\mathcal{K}_{e}$ by (4.3). Then we obtain $|c_{\lambda}|=1$. It follows that $c_{i}=[i\alpha_{1}(x)]^{s_{0}(\eta)}=[i]^{s_{0}(\eta)}[\alpha_{1}(x)]^{s_{0}(\eta)}=is_{0}(\eta)c_{1}$, and hence $c_{i}=is_{0}(\eta)c_{1}$ for all $\eta\in\partial\mathcal{A}$. This shows that $s_{0}$ is a constant function on $\partial\mathcal{A}$. We set $c=c_{1}^{\,s_{0}}$, and then $\widehat{T_{0}(\mathbf{1})}=c_{1}=c^{\,s_{0}}$ and $\widehat{T_{0}(i\mathbf{1})}=c_{i}=is_{0}c^{\,s_{0}}$ on $\partial\mathcal{A}$. Furthermore, $\alpha_{1}=c_{1}^{\,s_{0}}=c$ on $\mathcal{K}_{e}$. Since $\partial\mathcal{A}$ is a boundary for $\mathcal{A}$, we have $\widehat{T_{0}(\mathbf{1})}(z)=c^{\,s_{0}}$ and $\widehat{T_{0}(i\mathbf{1})}(z)=is_{0}\widehat{T_{0}(\mathbf{1})}(z)$ for $z\in\overline{\mathbb{D}}$. ∎

By Lemma 4.5, equality (4.2) is reduced to

[TABLE]

for every $f\in\mathcal{S}^{\infty}$ and $x=(z,\eta,w)\in\mathcal{K}_{e}$. Then we show that $\phi_{1}(z,\eta)$ is constant for $\eta\in\partial\mathcal{A}$.

Lemma 4.6**.**

Let $c\in\mathbb{T}$ be the constant as in Lemma 4.5. Then $\widehat{T_{0}({\rm id})}(z)=[c\phi_{1}(z,\eta)]^{s_{0}}$ for all $(z,\eta)\in\mathfrak{D}_{e}$ with $\mathfrak{D}_{e}=\mathfrak{D}_{\Sigma}$ or $\mathfrak{D}_{e}=\mathfrak{D}_{C}$.

Proof.

Let $(z_{0},\eta_{0})\in\mathfrak{D}_{e}$. We set $\xi_{0}=\phi_{1}(z_{0},\eta_{0})$ and $g={\rm id}-\xi_{0}\mathbf{1}\in\mathcal{S}^{\infty}$. Then $\widehat{g}(\xi_{0})=0$ and $g^{\prime}=\mathbf{1}$, and thus, by (4.4),

[TABLE]

for all $w\in\mathbb{T}$. Hence $|\widehat{T_{0}(g)}(z_{0})+\widehat{T_{0}(g)^{\prime}}(\eta_{0})w|=1$ for all $w\in\mathbb{T}$. By Proposition 4.4, we obtain $\widehat{T_{0}(g)}(z_{0})\widehat{T_{0}(g)^{\prime}}(\eta_{0})=0$. We shall prove that $\widehat{T_{0}(g)}(z_{0})=0$. Suppose, on the contrary, that $\widehat{T_{0}(g)}(z_{0})\neq 0$, and then $\widehat{T_{0}(g)^{\prime}}(\eta_{0})=0$. By (4.5), we have

[TABLE]

for all $w\in\mathbb{T}$. By the surjectivity of $T_{0}$, there exists $h\in\mathcal{S}^{\infty}$ such that $\widehat{T_{0}(h)}(z_{0})=0$ and $\widehat{T_{0}(h)^{\prime}}(\eta_{0})=1$. Applying these two equalities to (4.4), we get

[TABLE]

for all $w\in\mathbb{T}$. The rightmost hand side of the above equalities is independent of $w$, and thus we arrive at a contradiction. Consequently, $\widehat{T_{0}(g)}(z_{0})=0$ as is claimed.

Since $T_{0}$ is real linear, $0=\widehat{T_{0}(g)}(z_{0})=\widehat{T_{0}({\rm id})}(z_{0})-\widehat{T_{0}(\xi_{0}\mathbf{1})}(z_{0})$, and hence $\widehat{T_{0}({\rm id})}(z_{0})=\widehat{T_{0}(\xi_{0}\mathbf{1})}(z_{0})$. Here, we note that $\widehat{T_{0}(\mathbf{1})}=c^{\,s_{0}}=[c]^{s_{0}}$ and $\widehat{T_{0}(i\mathbf{1})}=is_{0}\,\widehat{T_{0}(\mathbf{1})}$ by Lemma 4.5. These imply that

[TABLE]

where we have used the real linearity of $T_{0}$. Since $(z_{0},\eta_{0})\in\mathfrak{D}_{e}$ is arbitrarily chosen, we get $\widehat{T_{0}({\rm id})}(z)=[c\phi_{1}(z,\eta)]^{s_{0}}$ for all $(z,\eta)\in\mathfrak{D}_{e}$. ∎

Finally, we investigate the function $\varphi_{1}$ on $\mathcal{K}_{e}$. In fact, we prove that $\varphi_{1}(z,\eta,w)$ is constant for $z$. More explicitly, we have the following result on $\varphi_{1}$.

Lemma 4.7**.**

For each $(z,\eta,w)\in\mathcal{K}_{e}$,

[TABLE]

Proof.

Let $(z,\eta,w)\in\mathcal{K}_{e}$. Then $\widehat{T_{0}({\rm id})}(z)=[c\phi_{1}(z,\eta)]^{s_{0}}$ by Lemma 4.6. Substituting this equality and $f={\rm id}$ into (4.4), we obtain $\widehat{T_{0}({\rm id})^{\prime}}(\eta)w=[c\varphi_{1}(z,\eta,w)]^{s_{0}s_{1}(\eta)}$. If we take $w=1$ in the last equality, then $\widehat{T_{0}({\rm id})^{\prime}}(\eta)=[c\varphi_{1}(z,\eta,1)]^{s_{0}s_{1}(\eta)}$. Note that $\widehat{T_{0}({\rm id})^{\prime}}(\eta)\neq 0$, since $c,\varphi_{1}(z,\eta,1)\in\mathbb{T}$ (see Definition 3.2 and Lemma 4.5). We have

[TABLE]

and hence $[\varphi_{1}(z,\eta,w)]^{s_{0}s_{1}(\eta)}=w[\varphi_{1}(z,\eta,1)]^{s_{0}s_{1}(\eta)}$. We conclude $\varphi_{1}(z,\eta,w)=w^{s_{0}s_{1}(\eta)}\varphi_{1}(z,\eta,1)$ for all $(z,\eta,w)\in\mathcal{K}_{e}$. ∎

Now we are ready to prove Theorem 1 for the characterizations of the isometry $T_{0}$ on $\mathcal{S}^{\infty}$ with the norms $\left\|\cdot\right\|_{(\Sigma)}$ and $\left\|\cdot\right\|_{(C)}$. Then $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}=\mathfrak{D}_{\Sigma}\times\mathbb{T}$ and $\mathcal{K}_{e}=\mathcal{K}_{C}=\mathfrak{D}_{C}\times\mathbb{T}$ with $\mathfrak{D}_{\Sigma}=\mathbb{T}\times\partial\mathcal{A}$ and $\mathfrak{D}_{C}=\{(\eta,\eta):\eta\in\partial\mathcal{A}\}$.

Proof of Theorem 1.

Fix arbitrary $f\in\mathcal{S}^{\infty}$ and $(z_{0},\eta_{0})\in\mathfrak{D}_{e}$. We have $\varphi_{1}(z_{0},\eta_{0},w)=w^{s_{0}s_{1}(\eta_{0})}\varphi_{1}(z_{0},\eta_{0},1)$ for all $w\in\mathbb{T}$ by Lemma 4.7. Applying Lemmas 4.6 and 4.7, we may write $\phi_{1}(z,\eta)=\phi_{1}(z)$ and $\varphi_{1}(z,\eta,w)=w^{s_{0}s_{1}(\eta)}\varphi_{1}(\eta)$ for all $(z,\eta,w)\in\mathcal{K}_{e}$. Now equality (4.4) is reduced to

[TABLE]

for all $w\in\mathbb{T}$. The above equality holds for every $w\in\mathbb{T}$, and then $\widehat{T_{0}(f)}(z_{0})=[c\widehat{f}(\phi_{1}(z_{0}))]^{s_{0}}$ and $\widehat{T_{0}(f)^{\prime}}(\eta_{0})=[c\varphi_{1}(\eta_{0})]^{s_{0}s_{1}(\eta_{0})}[\widehat{f^{\prime}}(\psi_{1}(z_{0},\eta_{0}))]^{s_{0}s_{1}(\eta_{0})}$. Since $(z_{0},\eta_{0})\in\mathfrak{D}_{e}$ is arbitrary, we get

[TABLE]

for all $(z,\eta)\in\mathfrak{D}_{e}$. Set $\rho=c^{-s_{0}}\,T_{0}({\rm id})\in A(\overline{\mathbb{D}})$, and then $c^{\,s_{0}}\widehat{\rho}(z)=\widehat{T_{0}({\rm id})}(z)=[c\phi_{1}(z)]^{s_{0}}$ for all $z\in q_{1}(\mathcal{K}_{e})$ by (4.6). This shows that

[TABLE]

for all $z\in q_{1}(\mathcal{K}_{e})$. Consequently, by (4.6) and (4.8),

[TABLE]

for all $z\in q_{1}(\mathcal{K}_{e})$.

If $\mathcal{K}_{e}=\mathcal{K}_{\Sigma}$, then $q_{1}(\mathcal{K}_{e})=\mathbb{T}$. Since $\rho$ is an analytic function on $\mathbb{D}$, the maximum modulus principle shows that equality (4.9) holds for all $z\in\overline{\mathbb{D}}$. Note that $\rho\in A(\overline{\mathbb{D}})$ satisfies $|\widehat{\rho}(z)|=|\phi_{1}(z)|=1$ for all $z\in q_{1}(\mathcal{K}_{e})=\mathbb{T}$ by (4.8), and thus $\widehat{\rho}(\mathbb{D})\subset\overline{\mathbb{D}}$ by the maximum modulus principle.

If $\mathcal{K}_{e}=\mathcal{K}_{C}$, then $q_{1}(\mathcal{K}_{e})=\partial\mathcal{A}$. Since $\partial\mathcal{A}$ is a boundary for $\mathcal{A}$, we see that (4.9) holds for all $\eta\in\mathcal{M}$. In particular, (4.9) is valid for all $z\in\mathbb{D}$. By the continuity of $\widehat{T_{0}(f)}$, $\widehat{f}$ and $\widehat{\rho}$ on $\overline{\mathbb{D}}$, we observe that equality (4.9) is true for all $z\in\overline{\mathbb{D}}$. Note that $\rho\in A(\overline{\mathbb{D}})$ satisfies $|\widehat{\rho}(\eta)|=|\phi_{1}(\eta)|=1$ for all $\eta\in\partial\mathcal{A}$, and thus $\widehat{\rho}(\mathcal{M})\subset\overline{\mathbb{D}}$. Thus, we have $\widehat{\rho}(\mathbb{D})\subset\overline{\mathbb{D}}$.

We now prove that $\widehat{\rho}$ is injective on $\mathbb{D}$. Suppose that $\widehat{\rho}(z_{1})=\widehat{\rho}(z_{2})$ for $z_{1},z_{2}\in\mathbb{D}$. Since $T_{0}$ is surjective, there exists $f_{0}\in\mathcal{S}^{\infty}$ such that $T_{0}(f_{0})={\rm id}\in\mathcal{S}^{\infty}$. Since (4.9) holds for $z\in\mathbb{D}$, we obtain

[TABLE]

and thus $\widehat{\rho}$ is injective on $\mathbb{D}$, as is claimed. Then $\widehat{\rho}$ is a non-constant function on $\mathbb{D}$. By the maximum modulus principle, we see that $\widehat{\rho}(\mathbb{D})\subset\mathbb{D}$.

Since $T_{0}^{-1}$ is a surjective real linear isometry on $\mathcal{S}^{\infty}$ as well, the above arguments can be applied to $T_{0}^{-1}$. Then there exist $d\in\mathbb{T}$, $\theta\in A(\overline{\mathbb{D}})$ with $\widehat{\theta}(\mathbb{D})\subset\mathbb{D}$ and $s_{2}\in\{\pm 1\}$ such that

[TABLE]

for all $f\in\mathcal{S}^{\infty}$ and $z\in\mathbb{D}$. Let $z\in\mathbb{D}$. If we substitute $f=T_{0}(\mathbf{1})$ into (4.10), then we have $1=[d\widehat{T_{0}(\mathbf{1})}([\widehat{\theta}(z)]^{s_{2}})]^{s_{2}}$. As $\widehat{T_{0}(\mathbf{1})}=c^{\,s_{0}}$ by (4.9), we obtain $1=[dc^{\,s_{0}}]^{s_{2}}$. Substituting $f=T_{0}({\rm id})$ into (4.10) to get $z=[d\widehat{T_{0}({\rm id})}([\widehat{\theta}(z)]^{s_{2}})]^{s_{2}}$. By (4.9), $\widehat{T_{0}({\rm id})}(z)=c^{\,s_{0}}\widehat{\rho}(z)$, and hence

[TABLE]

where we have used that $1=[dc^{\,s_{0}}]^{s_{2}}$. Because $z\in\mathbb{D}$ is arbitrarily chosen, we get $z=[\widehat{\rho}([\widehat{\theta}(z)]^{s_{2}})]^{s_{2}}$ for all $z\in\mathbb{D}$. This shows that $\mathbb{D}\subset\widehat{\rho}(\mathbb{D})$, and therefore, $\widehat{\rho}(\mathbb{D})=\mathbb{D}$.

From the above arguments, we see that $\rho=c^{-s_{0}}\,T_{0}({\rm id})\in A(\overline{\mathbb{D}})$ is an analytic function on $\mathbb{D}$, which is a homeomorphism on $\mathbb{D}$ as well. It is well-known (cf. [23, Theorem 12.6]) that for such a function $\rho$ there exist $\lambda\in\mathbb{T}$ and $b\in\mathbb{D}$ such that

[TABLE]

Because $\widehat{T_{0}({\rm id})}(z)=c^{\,s_{0}}\widehat{\rho}(z)$ for $z\in\mathbb{D}$, we have $T_{0}({\rm id})^{\prime}(z)=c^{\,s_{0}}\rho^{\prime}(z)$ for $z\in\mathbb{D}$. Differentiating both sides of (4.11), we see that $\rho^{\prime}(z)=\lambda(1-|b|^{2})/(1-\overline{b}z)^{2}$ for $z\in\mathbb{D}$. Hence $c^{-s_{0}}T_{0}({\rm id})^{\prime}=\rho^{\prime}$ belongs to $A(\overline{\mathbb{D}})$. Therefore, $\widehat{\rho^{\prime}}(\xi)=\widehat{\rho^{\prime}}(\eta)$ for all $\xi\in\mathbb{T}$ and $\eta\in\mathcal{M}_{\xi}$. Substituting $f={\rm id}$ into (4.7), we get $|\widehat{T_{0}({\rm id})^{\prime}}|=|c\varphi_{1}|=1$ on $\partial\mathcal{A}$. Hence, $|\widehat{\rho^{\prime}}|=1$ on $\partial\mathcal{A}$. We see that $\partial\mathcal{A}\cap\mathcal{M}_{\xi}\neq\emptyset$ for all $\xi\in\mathbb{T}$. In fact, let $\xi_{0}\in\mathbb{T}$, and set $u(z)=(\overline{\xi_{0}}\,z+1)/2\in A(\overline{\mathbb{D}})$ for $z\in\mathbb{D}$. Then $\widehat{u}(\eta)=\widehat{u}(\xi)$ for all $\xi\in\mathbb{T}$ and $\eta\in\mathcal{M}_{\xi}$. Note that $\widehat{u}(\xi_{0})=1$ and $|\widehat{u}(\xi)|<1$ for all $\xi\in\mathbb{T}\setminus\{\xi_{0}\}$. Since $\partial\mathcal{A}$ is a boundary for $\mathcal{A}$, we obtain $\partial\mathcal{A}\cap\mathcal{M}_{\xi_{0}}\neq\emptyset$. This proves $\partial\mathcal{A}\cap\mathcal{M}_{\xi}\neq\emptyset$ for all $\xi\in\mathbb{T}$. For each $z\in\mathbb{T}$, we can find $\eta\in\partial\mathcal{A}\cap\mathcal{M}_{z}$. Then, $|\widehat{\rho^{\prime}}(z)|=|\widehat{\rho^{\prime}}(\eta)|=1$, since $\rho^{\prime}\in A(\overline{\mathbb{D}})$. Consequently, $|\widehat{\rho^{\prime}}(z)|=1$ for all $z\in\mathbb{T}$, that is, $1-|b|^{2}=|1-\overline{b}z|^{2}$ for all $z\in\mathbb{T}$. This implies that $b=0$: In fact, if $b\neq 0$, then we substitute $z=\pm b/|b|$ into the equality $1-|b|^{2}=|1-\overline{b}z|^{2}$ to obtain

[TABLE]

which is impossible. Therefore, $b=0$ and consequently $\widehat{\rho}(z)=\lambda z$ for all $z\in\mathbb{D}$. By (4.9) we conclude that $T_{0}(f)(z)=cf(\lambda z)$ for all $f\in\mathcal{S}^{\infty}$ and $z\in\mathbb{D}$, or $T_{0}(f)(z)=\overline{cf(\overline{\lambda z})}$ for all $f\in\mathcal{S}^{\infty}$ and $z\in\mathbb{D}$. By (2.8), $T_{0}=T-T(0)$, and thus $T(f)(z)=T(0)(z)+cf(\lambda z)$ for all $f\in\mathcal{S}^{\infty}$ and $z\in\mathbb{D}$, or $T(f)(z)=T(0)(z)+\overline{cf(\overline{\lambda z})}$ for all $f\in\mathcal{S}^{\infty}$ and $z\in\mathbb{D}$.

Conversely, if $T$ is one of the above forms, then it is easy to check that $T$ is a surjective isometry on $\mathcal{S}^{\infty}$. ∎

4.2. Proof of Theorem 2

In this subsection, we consider the case when $\mathcal{K}_{e}=\mathcal{K}_{\sigma}=\{a\}\times\partial\mathcal{A}\times\mathbb{T}$ for $a\in\mathbb{D}$. Recall that

[TABLE]

for all $x=(z,\eta,w)\in\mathcal{K}_{\sigma}$ by (4.4). By definition, $\phi_{1}$ is a map from $\mathcal{K}_{\sigma}$ to $q_{1}(\mathcal{K}_{\sigma})=\{a\}$. Hence $\phi_{1}$ is the constant function which takes the value only $a$. In addition, since $q_{1}(\mathcal{K}_{\sigma})=\{a\}$, we may write

[TABLE]

for $(z,\eta)\in\{a\}\times\partial\mathcal{A}$ and $w\in\mathbb{T}$. Now we obtain the following equality

[TABLE]

for all $f\in\mathcal{S}^{\infty}$ and $x=(a,\eta,w)\in\{a\}\times\partial\mathcal{A}\times\mathbb{T}$.

We need to prove that $\widehat{T_{0}(f)^{\prime}}(\eta)w=[\alpha_{1}(x)\widehat{f^{\prime}}(\psi_{1}(\eta))\varphi_{1}(\eta,w)]^{s_{0}(\eta)s_{1}(\eta)}$. We first show that $\alpha_{1}(a,\eta,w)$ is independent of $w\in\mathbb{T}$.

Lemma 4.8**.**

There exists $c_{0}\in\mathbb{T}$ such that $\widehat{T_{0}(\mathbf{1})}=c_{0}$ on $\overline{\mathbb{D}}$ and $\alpha_{1}(a,\eta,w)=[c_{0}]^{s_{0}(\eta)}=\alpha_{1}(a,\eta,1)$ for all $(\eta,w)\in\partial\mathcal{A}\times\mathbb{T}$.

Proof.

Let $g={\rm id}-a\mathbf{1}$, and then $g^{\prime}=\mathbf{1}$. If we apply $f=\mathbf{1},g\in\mathcal{S}^{\infty}$ to (4.13), then

[TABLE]

for all $(\eta,w)\in\partial\mathcal{A}\times\mathbb{T}$. We shall prove that $T_{0}(\mathbf{1})(a)\neq 0$. To this end, we assume that $T_{0}(\mathbf{1})(a)=0$, and then $\widehat{T_{0}(\mathbf{1})^{\prime}}(\eta)w=[\alpha_{1}(a,\eta,w)]^{s_{0}(\eta)}$ for all $(\eta,w)\in\partial\mathcal{A}\times\mathbb{T}$. Substituting this equality and (4.15) into (4.13) to obtain

[TABLE]

for all $f\in\mathcal{S}^{\infty}$ and $(\eta,w)\in\partial\mathcal{A}\times\mathbb{T}$. Since $w\in\mathbb{T}$ is arbitrary, we get

[TABLE]

for all $f\in\mathcal{S}^{\infty}$ and $\eta\in\partial\mathcal{A}$. Taking $f={\rm id}^{2}\in\mathcal{S}^{\infty}$ in (4.16), we have

[TABLE]

for all $\eta\in\partial\mathcal{A}$. Let $j\in\{\pm 1\}$. Since $\partial\mathcal{A}\cap\mathcal{M}_{j}\neq\emptyset$, we can choose $\eta_{j}\in\partial\mathcal{A}\cap\mathcal{M}_{j}$. Then $\widehat{{\rm id}}(\eta_{j})=j$. By Lemma 3.9 (see, also Lemma 4.2), $\psi_{1}\colon\partial\mathcal{A}\to\partial\mathcal{A}$ is surjective, and thus, there exists $\zeta_{j}\in\partial\mathcal{A}$ such that $\psi_{1}(\zeta_{j})=\eta_{j}$. We thus obtain $\widehat{{\rm id}}(\psi_{1}(\zeta_{j}))=j$. Substituting $\eta=\zeta_{j}$ into (4.17) to get $T_{0}({\rm id}^{2})(a)=2jT_{0}(g)(a)$, which yields $T_{0}(g)(a)=0$. We obtain $T_{0}(f)(a)=0$ for all $f\in\mathcal{S}^{\infty}$ by (4.16). This is impossible since $T_{0}$ is surjective. Consequently, $T_{0}(\mathbf{1})(a)\neq 0$, as is claimed.

We see that $\widehat{T_{0}(\mathbf{1})^{\prime}}(\eta)=0$ for all $\eta\in\partial\mathcal{A}$ by equality (4.14) with Proposition 4.4. Since $\partial\mathcal{A}$ is a boundary for $\mathcal{A}$, we have $\widehat{T_{0}(\mathbf{1})^{\prime}}=0$ on $\mathcal{M}$. Then $T_{0}(\mathbf{1})$ is constant on $\mathbb{D}$, say $c_{0}\in\mathbb{C}$. Since $\mathbb{D}$ is dense in $\mathcal{M}$, we obtain $\widehat{T_{0}(\mathbf{1})}=c_{0}$ on $\mathcal{M}$. In particular, $\widehat{T_{0}(\mathbf{1})}=c_{0}$ on $\overline{\mathbb{D}}$. Substituting $\widehat{T_{0}(\mathbf{1})^{\prime}}(\eta)=0$ into (4.14) to get $c_{0}=T_{0}(\mathbf{1})(a)=[\alpha_{1}(a,\eta,w)]^{s_{0}(\eta)}$ for all $(\eta,w)\in\partial\mathcal{A}\times\mathbb{T}$, and thus $c_{0}\in\mathbb{T}$. Hence $\alpha_{1}(a,\eta,w)=[c_{0}]^{s_{0}(\eta)}=\alpha_{1}(a,\eta,1)$. ∎

By Lemma 4.8, $\alpha_{1}(a,\eta,w)=[c_{0}]^{s_{0}(\eta)}$ for all $(\eta,w)\in\partial\mathcal{A}\times\mathbb{T}$. Equality (4.13) is reduced to

[TABLE]

for every $f\in\mathcal{S}^{\infty}$ and $(\eta,w)\in\partial\mathcal{A}\times\mathbb{T}$.

In the next lemma, we determine $\varphi_{1}(\eta,w)$ as a function of variable $w\in\mathbb{T}$. Then we derive the form of $\widehat{T_{0}(f)^{\prime}}(\eta)$ from (4.18).

Lemma 4.9**.**

1. (1)

$\widehat{T_{0}({\rm id}_{a})^{\prime}}(\eta)=[c_{0}]^{s_{1}(\eta)}[\varphi_{1}(\eta,1)]^{s_{0}(\eta)s_{1}(\eta)}$* for all $\eta\in\partial\mathcal{A}$, where ${\rm id}_{a}={\rm id}-a\mathbf{1}$.* 2. (2)

$\varphi_{1}(\eta,w)=\varphi_{1}(\eta,1)w^{s_{0}(\eta)s_{1}(\eta)}$* for all $(\eta,w)\in\partial\mathcal{A}\times\mathbb{T}$.*

Proof.

Let $\eta_{0}\in\partial\mathcal{A}$. Taking $f={\rm id}_{a}$ in (4.18), we obtain

[TABLE]

for every $w\in\mathbb{T}$. Then $|T_{0}({\rm id}_{a})(a)+\widehat{T_{0}({\rm id}_{a})^{\prime}}(\eta_{0})w|=1$ for all $w\in\mathbb{T}$. We shall prove that $T_{0}({\rm id}_{a})(a)=0$. Suppose, on the contrary, that $T_{0}({\rm id}_{a})(a)\neq 0$. Proposition 4.4 shows that $\widehat{T_{0}({\rm id}_{a})^{\prime}}(\eta_{0})=0$. Then, from (4.19) we have $T_{0}({\rm id}_{a})(a)=[c_{0}]^{s_{1}(\eta_{0})}[\varphi_{1}(\eta_{0},w)]^{s_{0}(\eta_{0})s_{1}(\eta_{0})}$ for all $w\in\mathbb{T}$. Since $T_{0}$ is surjective, there exists $g\in\mathcal{S}^{\infty}$ such that $T_{0}(g)(a)=0$ and $\widehat{T_{0}(g)^{\prime}}(\eta_{0})=1$. Applying these equalities and $T_{0}({\rm id}_{a})(a)=[c_{0}]^{s_{1}(\eta_{0})}[\varphi_{1}(\eta_{0},w)]^{s_{0}(\eta_{0})s_{1}(\eta_{0})}$ to (4.18), we get

[TABLE]

for all $w\in\mathbb{T}$. This is impossible since the rightmost hand side of the above equalities is independent of $w\in\mathbb{T}$. Consequently, $T_{0}({\rm id}_{a})(a)=0$. Because $\eta_{0}\in\partial\mathcal{A}$ is arbitrarily chosen, it follows from (4.19) that $\widehat{T_{0}({\rm id}_{a})^{\prime}}(\eta)w=[c_{0}]^{s_{1}(\eta)}[\varphi_{1}(\eta,w)]^{s_{0}(\eta)s_{1}(\eta)}$ for all $(\eta,w)\in\partial\mathcal{A}\times\mathbb{T}$. Taking $w=1$ in the last equality, we get $\widehat{T_{0}({\rm id}_{a})^{\prime}}(\eta)=[c_{0}]^{s_{1}(\eta)}[\varphi_{1}(\eta,1)]^{s_{0}(\eta)s_{1}(\eta)}$ for all $\eta\in\partial\mathcal{A}$.

Note that $\widehat{T_{0}({\rm id}_{a})^{\prime}}(\eta)\neq 0$ for $\eta\in\mathbb{T}$. We thus obtain

[TABLE]

and therefore, $\varphi_{1}(\eta,w)=w^{s_{0}(\eta)s_{1}(\eta)}\varphi_{1}(\eta,1)$ for all $(\eta,w)\in\partial\mathcal{A}\times\mathbb{T}$. ∎

We are in a position to prove Theorem 2. We need to investigate $\psi_{1}$ to characterize the isometry $T_{0}$ on $\mathcal{S}^{\infty}$ for $\mathcal{K}_{\sigma}=\{a\}\times\partial\mathcal{A}\times\mathbb{T}$. The difficulty is to extend the map $\psi_{1}\colon\partial\mathcal{A}\to\partial\mathcal{A}$ to a homeomorphism defined on $\mathcal{M}$. The main idea of its proof is to induce a surjective real linear isometry on the uniform algebra $\mathcal{A}$. We then induce a real algebra isometric isomorphism on $\mathcal{A}$. It is known that such an isometric isomorphism is represented as a combination of a composition operator induced by a homeomorphism on $\mathcal{M}$ and the complex conjugate. Then we prove that $\psi_{1}$ is extended to $\mathcal{M}$.

Proof of Theorem 2.

For simplicity of notation, we write $\varphi_{1}(\eta,1)=\varphi_{1}(\eta)$. By Lemma 4.9, equality (4.18) is reduced to

[TABLE]

for all $f\in\mathcal{S}^{\infty}$ and $(\eta,w)\in\partial\mathcal{A}\times\mathbb{T}$. Since $w\in\mathbb{T}$ is arbitrary, we obtain

[TABLE]

and $T_{0}(f)(a)=c_{0}[f(a)]^{s_{0}(\eta)}$ for all $f\in\mathcal{S}^{\infty}$ and $\eta\in\partial\mathcal{A}$. Substituting $f=i\mathbf{1}\in\mathcal{S}^{\infty}$ into the last equality, we get $T_{0}(i\mathbf{1})(a)=c_{0}[i]^{s_{0}(\eta)}=is_{0}(\eta)c_{0}$ for all $\eta\in\partial\mathcal{A}$. This shows that $s_{0}$ is a constant function; we may and do write $s_{0}(\eta)=s_{0}$. Then we have

[TABLE]

for all $f\in\mathcal{S}^{\infty}$. For each $v\in H^{\infty}(\mathbb{D})$, we define $I_{a}(v)$ by

[TABLE]

where $[a,z]$ denotes the straight line interval from $a$ to $z$. Then $I_{a}(v)$ belongs to $H^{\infty}(\mathbb{D})$ with $I_{a}(v)(a)=0$ and

[TABLE]

Hence $I_{a}(v)\in\mathcal{S}^{\infty}$ and $\widehat{I_{a}(v)^{\prime}}=\widehat{v}$ on $\mathcal{M}$. We may regard $I_{a}$ as a complex linear operator from $H^{\infty}(\mathbb{D})$ to $\mathcal{S}^{\infty}$.

Recall that $\mathcal{A}=\{\widehat{v}\in C(\mathcal{M}):v\in H^{\infty}(\mathbb{D})\}$. Define $\mathcal{W}\colon\mathcal{A}\to\mathcal{A}$ by

[TABLE]

Since the Gelfand transform from $H^{\infty}(\mathbb{D})$ onto $\mathcal{A}$ is an isometric isomorphism, we see that the mapping $\mathcal{W}$ is well-defined.

[TABLE]

Because $T_{0}\colon\mathcal{S}^{\infty}\to\mathcal{S}^{\infty}$ is real linear, so is $\mathcal{W}\colon\mathcal{A}\to\mathcal{A}$. Recall that $\widehat{I_{a}(v)^{\prime}}=\widehat{v}$ on $\mathcal{M}$ for $v\in H^{\infty}(\mathbb{D})$. Substituting $f=I_{a}(v)$ into (4.20) to obtain $\widehat{\{T_{0}(I_{a}(v))\}^{\prime}}(\eta)=[c_{0}]^{s_{1}(\eta)}[\varphi_{1}(\eta)\widehat{v}(\psi_{1}(\eta))]^{s_{0}s_{1}(\eta)}$ for all $v\in H^{\infty}(\mathbb{D})$ and $\eta\in\partial\mathcal{A}$, where we have used that $s_{0}$ is a constant function. The last equality, combined with (4.23), shows that

[TABLE]

for $\widehat{v}\in\mathcal{A}$ and $\eta\in\partial\mathcal{A}$. Note that $c_{0},\varphi_{1}(\eta)\in\mathbb{T}$ by definition (see Lemma 4.8 and Definition 3.2), and thus $|\mathcal{W}(\widehat{v})(\eta)|=|\widehat{v}(\psi_{1}(\eta))|$ for all $\widehat{v}\in\mathcal{A}$ and $\eta\in\partial\mathcal{A}$. Since $\psi_{1}\colon\partial\mathcal{A}\to\partial\mathcal{A}$ is surjective by Lemma 3.9 and (4.12), we see that $\|\mathcal{W}(\widehat{v})\|_{\mathcal{M}}=\|\widehat{v}\|_{\mathcal{M}}$ for all $\widehat{v}\in\mathcal{A}$. Thus $\mathcal{W}$ is a real linear isometry on $(\mathcal{A},\left\|\cdot\right\|_{\mathcal{M}})$.

We shall prove that

[TABLE]

on $\mathcal{M}$ for each $f\in\mathcal{S}^{\infty}$. Let $f\in\mathcal{S}^{\infty}$, and then $\{I_{a}(f^{\prime})\}^{\prime}=f^{\prime}$ on $\mathbb{D}$ by (4.22). Thus $I_{a}(f^{\prime})-f$ is constant on $\mathbb{D}$. By the definition of $I_{a}$, we have $I_{a}(f^{\prime})(a)=0$, and hence

[TABLE]

Because $T_{0}$ is real linear, we obtain $T_{0}(I_{a}(f^{\prime}))=T_{0}(f)-T_{0}(f(a)\mathbf{1})$ on $\mathbb{D}$. Therefore, $\widehat{\{T_{0}(I_{a}(f^{\prime}))\}^{\prime}}=\widehat{T_{0}(f)^{\prime}}-\widehat{T_{0}(f(a)\mathbf{1})^{\prime}}$ on $\mathcal{M}$. Equality (4.20) shows that $\widehat{T_{0}(f(a)\mathbf{1})^{\prime}}=0$ on $\partial\mathcal{A}$, and thus $\widehat{T_{0}(f(a)\mathbf{1})^{\prime}}=0$ on $\mathcal{M}$, since $\partial\mathcal{A}$ is a boundary for $\mathcal{A}$. We deduce that $\widehat{\{T_{0}(I_{a}(f^{\prime}))\}^{\prime}}=\widehat{T_{0}(f)^{\prime}}$ on $\mathcal{M}$.

We show that $\mathcal{W}$ is surjective. Let $\widehat{v_{0}}\in\mathcal{A}$. By the surjectivity of $T_{0}\colon\mathcal{S}^{\infty}\to\mathcal{S}^{\infty}$, there exists $g\in\mathcal{S}^{\infty}$ such that $T_{0}(g)=I_{a}(v_{0})$ on $\mathbb{D}$, and hence $\widehat{T_{0}(g)^{\prime}}=\widehat{I_{a}(v_{0})^{\prime}}=\widehat{v_{0}}$ on $\mathcal{M}$ by (4.22). Equality (4.24) shows that $\widehat{\{T_{0}(I_{a}(g^{\prime}))\}^{\prime}}=\widehat{T_{0}(g)^{\prime}}=\widehat{v_{0}}$ on $\mathcal{M}$. By (4.23), $\mathcal{W}(\widehat{g^{\prime}})=\widehat{\{T_{0}(I_{a}(g^{\prime}))\}^{\prime}}=\widehat{v_{0}}$ on $\mathcal{M}$, which shows the surjectivity of $\mathcal{W}\colon\mathcal{A}\to\mathcal{A}$. Hence $\mathcal{W}$ is a surjective real linear isometry on the uniform algebra $(\mathcal{A},\left\|\cdot\right\|_{\mathcal{M}})$. By [18, Theorem 1.1] (see also [13, Theorem 3.3]), there exist a continuous function $u_{1}\colon\partial\mathcal{A}\to\mathbb{T}$, a closed and open set $G_{0}\subset\partial\mathcal{A}$ and a homeomorphism $\varrho\colon\partial\mathcal{A}\to\partial\mathcal{A}$ such that

[TABLE]

for all $\widehat{v}\in\mathcal{A}$. Then $\mathcal{W}(\widehat{\mathbf{1}})=u_{1}$ on $\partial\mathcal{A}$.

By the surjectivity of $\mathcal{W}$, there exists $v_{1}\in H^{\infty}(\mathbb{D})$ such that $\mathcal{W}(\widehat{v_{1}})=\widehat{\mathbf{1}}$. Then $\|\widehat{v_{1}}\|_{\mathcal{M}}=\|\mathcal{W}(\widehat{v_{1}})\|_{\mathcal{M}}=1$, since $\mathcal{W}$ is a real linear isometry. Equality (4.26) implies that $\mathcal{W}(\widehat{\mathbf{1}})\mathcal{W}(\widehat{v_{1}}^{2})=u_{1}\mathcal{W}(\widehat{v_{1}}^{2})=\mathcal{W}(\widehat{v_{1}})^{2}$ on $\partial\mathcal{A}$, and thus $\mathcal{W}(\widehat{\mathbf{1}})\mathcal{W}(\widehat{v_{1}}^{2})=\widehat{\mathbf{1}}$ on $\partial\mathcal{A}$. Since $\partial\mathcal{A}$ is a boundary for $\mathcal{A}$, we have $\mathcal{W}(\widehat{\mathbf{1}})\mathcal{W}(\widehat{v_{1}}^{2})=\widehat{\mathbf{1}}$ on $\mathcal{M}$. Note that $|\mathcal{W}(\widehat{v_{1}}^{2})(\eta)|\leq 1$ for all $\eta\in\mathcal{M}$, since $\|\mathcal{W}(\widehat{v_{1}}^{2})\|_{\mathcal{M}}=\|\widehat{v_{1}}^{2}\|_{\mathcal{M}}=1$. Since $\|\mathcal{W}(\widehat{\mathbf{1}})\|_{\mathcal{M}}=\|\widehat{\mathbf{1}}\|_{\mathcal{M}}=1$, we obatin $|\mathcal{W}(\widehat{\mathbf{1}})(\eta)|\leq 1$ for $\eta\in\mathcal{M}$. Combining these two inequalities with the equality $\mathcal{W}(\widehat{\mathbf{1}})(\eta)\mathcal{W}(\widehat{v_{1}}^{2})(\eta)=1$ for $\eta\in\mathcal{M}$, we deduce $|\mathcal{W}(\widehat{\mathbf{1}})|=1$ on $\mathcal{M}$. Since $\mathcal{W}(\widehat{\mathbf{1}})$ is the Gelfand transform of some analytic function on $\mathbb{D}\subset\mathcal{M}$, the maximum modulus principle shows that $\mathcal{W}(\widehat{\mathbf{1}})$ is a constant function of modulus one. Hence, $\mathcal{W}(\widehat{\mathbf{1}})=c_{1}$ on $\mathcal{M}$ for some $c_{1}\in\mathbb{T}$.

Since $\mathcal{W}$ is bijective and real linear, so is the mapping $\overline{c_{1}}\mathcal{W}\colon\mathcal{A}\to\mathcal{A}$. By (4.26), we see that $\overline{c_{1}}\mathcal{W}(\widehat{v_{1}}\widehat{v_{2}})=\overline{c_{1}}\mathcal{W}(\widehat{v_{1}})\overline{c_{1}}\mathcal{W}(\widehat{v_{2}})$ on $\partial\mathcal{A}$ for all $\widehat{v_{1}},\widehat{v_{2}}\in\mathcal{A}$. Because $\partial\mathcal{A}$ is a boundary for $\mathcal{A}$, we see that $\overline{c_{1}}\mathcal{W}$ is multiplicative, and hence it is a bijective real-algebra automorphism on $\mathcal{A}$. By [13, Themrem 2.1], there exist a homeomorphism $\rho_{1}\colon\mathcal{M}\to\mathcal{M}$ and a closed and open set $G_{1}$ of $\mathcal{M}$ such that

[TABLE]

for all $\widehat{v}\in\mathcal{A}$. By the connectedness of $\mathcal{M}$, we deduce that $G_{1}=\mathcal{M}$ or $G_{1}=\emptyset$, and hence there exists $s_{2}\in\{\pm 1\}$ such that

[TABLE]

for all $\widehat{v}\in\mathcal{A}$ and $\eta\in\mathcal{M}$.

Since $\mathcal{W}(\widehat{{\rm id}})\in\mathcal{A}$, there exists $h_{1}\in H^{\infty}(\mathbb{D})$ such that $\mathcal{W}(\widehat{{\rm id}})=\widehat{h_{1}}$. Then $|\widehat{h_{1}}|=|\widehat{{\rm id}}\circ\rho_{1}|$ on $\mathcal{M}$ by (4.27). Thus $\|\widehat{h_{1}}\|_{\mathcal{M}}=\|\widehat{{\rm id}}\circ\rho_{1}\|_{\mathcal{M}}=1$, which implies $h_{1}(\mathbb{D})\subset\overline{\mathbb{D}}$. Because $\rho_{1}$ is a homeomorphism on $\mathcal{M}$, we observe that $h_{1}$ is a non-constant analytic function on $\mathbb{D}$. The open mapping theorem yields that $h_{1}(\mathbb{D})$ is an open set, and hence $h_{1}(\mathbb{D})\subset\mathbb{D}$. This implies that $|\widehat{h_{1}}|<1$ on $\mathbb{D}$. Since $|\widehat{{\rm id}}|=1$ on $\mathcal{M}\setminus\mathbb{D}$ (see [14, p. 161]), the equality $|\widehat{{\rm id}}\circ\rho_{1}|=|\widehat{h_{1}}|$ shows that $\rho_{1}(\mathbb{D})\subset\mathbb{D}$.

The above arguments can be applied to the surjective real linear isometry $\mathcal{W}^{-1}$. Then there exist a homeomorphism $\rho_{-1}\colon\mathcal{M}\to\mathcal{M}$ with $\rho_{-1}(\mathbb{D})\subset\mathbb{D}$ and a constant $s_{3}\in\{\pm 1\}$ such that

[TABLE]

for all $\widehat{v}\in\mathcal{A}$ and $\eta\in\mathcal{M}$. By (4.27) and (4.28), the equality $\widehat{v}=\mathcal{W}^{-1}(\mathcal{W}(\widehat{v}))$ shows that

[TABLE]

for all $\widehat{v}\in\mathcal{A}$ and $\eta\in\mathcal{M}$, where $[\widehat{v}\circ\rho_{1}]^{s_{2}}(\eta)=[\widehat{v}(\rho_{1}(\eta))]^{s_{2}}$. For $\widehat{v}=\widehat{\mathbf{1}}$, the above equalities show that $1=c_{-1}[c_{1}]^{s_{3}}$, and hence $\widehat{v}=[\widehat{v}\circ\rho_{1}\circ\rho_{-1}]^{s_{2}s_{3}}$ for all $\widehat{v}\in\mathcal{A}$. Substituting $\widehat{v}=\widehat{{\rm id}}$ into this equality to obtain $\widehat{{\rm id}}=[\widehat{{\rm id}}\circ\rho_{1}\circ\rho_{-1}]^{s_{2}s_{3}}$ on $\mathcal{M}$. Since $\rho_{j}(\mathbb{D})\subset\mathbb{D}$ for $j=\pm 1$, we obtain $z=[\rho_{1}(\rho_{-1}(z))]^{s_{2}s_{3}}$ for all $z\in\mathbb{D}$. This shows that $\mathbb{D}\subset\rho_{1}(\mathbb{D})$, and therefore $\rho_{1}(\mathbb{D})=\mathbb{D}$. Hence $\rho_{1}|_{\mathbb{D}}\colon\mathbb{D}\to\mathbb{D}$ is a homeomorphism. For each $z\in\mathbb{D}$, $c_{1}[\rho_{1}(z)]^{s_{2}}=c_{1}[\widehat{{\rm id}}(\rho_{1}(z))]^{s_{2}}=\widehat{h_{1}}(z)=h_{1}(z)$ by (4.27) with $\mathcal{W}(\widehat{{\rm id}})=\widehat{h_{1}}$, and thus $[\rho_{1}|_{\mathbb{D}}]^{s_{2}}=\overline{c_{1}}h_{1}$ is analytic on $\mathbb{D}$. Since $\rho_{1}|_{\mathbb{D}}\colon\mathbb{D}\to\mathbb{D}$ is a homeomorphism, there exist $\lambda\in\mathbb{T}$ and $b\in\mathbb{D}$ such that

[TABLE]

(see, for example, [23, Theorem 12.6]).

Let $f\in\mathcal{S}^{\infty}$. Equality (4.24), combined with (4.23) and (4.27), shows that

[TABLE]

for every $\eta\in\mathcal{M}$. In particular, since $\rho_{1}(\mathbb{D})=\mathbb{D}$,

[TABLE]

Equality (4.25), applied to $T_{0}(f)$ instead of $f$, shows that $I_{a}(T_{0}(f)^{\prime})=T_{0}(f)-T_{0}(f)(a)\mathbf{1}$ on $\mathbb{D}$, that is,

[TABLE]

for all $z\in\mathbb{D}$. Recall that $T_{0}(f)(a)=c_{0}[f(a)]^{s_{0}}$ by (4.21), and that $[\rho_{1}(z)]^{s_{2}}=\lambda(z-b)/(1-\bar{b}z)$ for $z\in\mathbb{D}$ by (4.29). By the definition of $I_{a}$ with $T_{0}(f)^{\prime}=c_{1}[f^{\prime}\circ\rho_{1}]^{s_{2}}$, we get

[TABLE]

for all $f\in\mathcal{S}^{\infty}$ and $z\in\mathbb{D}$. Since $T_{0}=T-T(0)$ by (2.8), we obtain the forms displayed in Theorem 2.

Conversely, if $T$ is one of the above forms, then we can check that it is a surjective isometry on $\mathcal{S}^{\infty}$. This completes the proof. ∎

On behalf of all authors, the corresponding author states that there is no conflict of interest.

Acknowledgement**.**

The authors of the paper are deeply grateful to the referees for their valuable comments and suggestions that significantly improve the former manuscript.