The Mazur--Ulam property in $\ell_\infty$-sum and $c_0$-sum of strictly convex Banach spaces

TL;DR

This paper investigates the Mazur--Ulam property in certain Banach spaces, proving it holds for specific sums and function spaces involving strictly convex spaces, with implications for renorming and function space structures.

Contribution

It establishes the Mazur--Ulam property for c0-sums and C0 spaces of strictly convex Banach spaces under various conditions, extending known results.

Findings

c0-sums of strictly convex spaces satisfy the Mazur--Ulam property

C0 spaces over totally disconnected spaces satisfy the Mazur--Ulam property

Renorming allows many Banach spaces to satisfy the Mazur--Ulam property

Abstract

In this paper we deal with those Banach spaces $Z$ which satisfy the Mazur--Ulam property, namely that every surjective isometry $\Delta$ from the unit sphere of $Z$ to the unit sphere of any Banach space $Y$ admits an unique extension to a surjective real-linear isometry from $Z$ to $Y$. We prove that for every countable set $\Gamma$ with $\vert \Gamma \vert \geq 2$, the Banach space $\bigoplus_{\gamma \in \Gamma}^{c_0} X_\gamma $ satisfies the Mazur--Ulam property, whenever the Banach space $X_\gamma $ is strictly convex with dim$((X_\gamma )_{\mathbb{R}})\geq 2$ for every $\gamma $. Moreover we prove that the Banach space $C_0(K,X)$ satisfies the Mazur--Ulam property whenever $K$ is a totally disconnected locally compact Hausdorff space with $\vert K\vert \geq 2$, and $X$ is a strictly convex separable Banach space with dim$(X_{\mathbb{R}})\geq 2$. As consequences, we obtain the following results: (1) Every weakly countably determined Banach space can be equivalently renormed so that it satisfies the Mazur--Ulam property. (2) If $X$ is a strictly convex Banach space with dim$(X_{\mathbb{R}}) \geq 2$, then $C(\mathfrak{C} ,X)$ satisfies the Mazur--Ulam property, where $ \mathfrak{C}$ denotes the Cantor set.

Equations

$$\Vert 2x-\frac{x}{\Vert x\Vert }\Vert <1$$

$$A({\gamma }_0,x_{{\gamma }_0}):=\{z\in S_Z:z({\gamma }_0)=x_{{\gamma }_0}\}.$$

$$\mathrm{supp}({\gamma }_0,x_{{\gamma }_0}):=\{\psi \in S_{Y^{\ast }}:{\psi }^{-1}(\{1\})\cap B_Y=\Delta (A({\gamma }_0,x_{{\gamma }_0}))\}$$

$$St(x):=\{y\in S_X:\Vert x+y\Vert =2\}.$$

$$p\in Ext(B_{C(K,X)})if\; and\; only\; if\; \Vert p(t)\Vert =1\; for\; allt\in K.$$

$$A(t_0,x_0):=\{f\in S_{C_0(K,X)}:f(t_0)=x_0\}.$$

$$\mathrm{supp}(t_0,x_0):=\{\psi \in S_{Y^{\ast }}:{\psi }^{-1}(\{1\})\cap B_Y=\Delta (A(t_0,x_0))\}$$

$$\Delta (-A(t_0,x_0))={\psi }^{-1}(-1)\cap B_{C_0(K,X)}=-\Delta (A(t_0,x_0))$$

$$2=\psi (\Delta (f_1))-\psi (\Delta (-f_2))\le \Vert \Delta (f_1)-\Delta (-f_2)\Vert =\Vert f_1+f_2\Vert =1,$$

$$2=\psi (\Delta (f_0))+\psi (\Delta (f_1))\le \Vert \Delta (f_0)+\Delta (f_1)\Vert =\Vert f_1+f_2\Vert =\Vert x_0+x^{\prime }\Vert <2,$$

$$O_{\epsilon }:=\{t\in K:\Vert f(t)\Vert <\epsilon \}.$$

$$|\psi (\Delta (f))\pm 1|=|\psi (\Delta (f))-\psi (\Delta (\pm g\otimes x_0))|\le \Vert \Delta (f)-\Delta (\pm g\otimes x_0)\Vert$$

$$\le \Vert f\pm g\otimes x_0\Vert \le 1+\epsilon .$$

$$C_{\epsilon }:=\{t\in K:\Vert f(t)\Vert \ge 1-\epsilon \}.$$

$$|\psi (\Delta (f))|=|\psi (\Delta (f))-\psi (\Delta (hf))|\le \Vert \Delta (f)-\Delta (hf)\Vert$$

$$=\Vert f-hf\Vert \le 1-\epsilon <1.$$

$$t_{0}\in\mathcal{O}\subseteq\Big{\{}s\in K:\|f(s)-f(t_{0})\|<\frac{\|f(t_{0})\|}{2}\Big{\}}.$$

$$\|f-h\|=\sup_{s\in\mathcal{O}}\big{\|}\frac{f(s)}{\|f(s)\|}-f(s)\big{\|}=\sup_{s\in\mathcal{O}}\|1-\|f(s)\|\|\leq 1-\frac{\|f(t_{0})\|}{2}.$$

$$1=\psi (\Delta (h))-\psi (\Delta (f))\le \Vert \Delta (h)-\Delta (f)\Vert =\Vert h-f\Vert \le 1-\Vert f(t_0)\Vert <1,$$

$$\psi \Delta (z)={\phi }_x\otimes P_{\gamma }(z)$$

$$A({\gamma }_1,p({\gamma }_1))=\{z\in S_Z:z({\gamma }_1)=p({\gamma }_1)\}=p_1+B_{Z_{{\gamma }_1}}.$$

$$\phi (z):=\psi \Delta (p_1+z)-\psi \Delta (p_1)$$

$$\tilde{\phi }(z)=\psi \Delta (p_{{\gamma }_1}+P_{\Gamma \setminus {\gamma }_1}(z))\text{mbox}foreveryz\in B_Z.$$

$$\tilde{\phi }(z)={\phi }_{x_0}(z({\gamma }_0))\text{mbox}foreveryz\in Z.$$

$$|\varphi (z({\gamma }_0))-{\phi }_{x_0}(z({\gamma }_0))|<\frac{\epsilon }{4}\text{mbox}forevery\varphi \in B_{X_{{\gamma }_0}^{\ast }}\text{mbox}with\varphi (x_0)>1-\rho .$$

$$\{{\phi }_x\otimes P_{\gamma }:\gamma \in \Gamma ,x\in Sm(X_{\gamma })\}\subseteq B_{Z^{\ast }}$$

$${\overline{co}}^{w^{\ast }}\{{\phi }_x\otimes P_{\gamma }:\gamma \in \Gamma ,x\in Sm(X_{\gamma })\}=B_{Z^{\ast }}.$$

$$\left|\tilde{\phi }(u)-\left(\sum _{i=1}^n{\lambda }_i{\phi }_{x_{{\gamma }_i}}\otimes P_{{\gamma }_i}\right)(u)\right|<\rho ,\text{mbox}foru\in \{z_0,z\}.$$

$$\left|1-\left(\sum _{i=1}^n{\lambda }_i{\phi }_{x_{{\gamma }_i}}\otimes P_{{\gamma }_i}\right)(z_0)\right|=\left|\left({\phi }_{x_0}-\sum _{i\in N}{\lambda }_i{\phi }_{x_i}\right)(x_0)\right|<\rho .$$

$$\left|\left({\phi }_{x_0}-\sum _{i\in N}{\lambda }_i{\phi }_{x_i}\right)z({\gamma }_0)\right|<\frac{\epsilon }{4}.$$

Full text

1. Introduction

Let $X$ and $Y$ be two Banach spaces over ${\mathbb{K}}$ (${\mathbb{R}}$ or ${\mathbb{C}}$) with unit spheres $S_{X}$ and $S_{Y}$, respectively. Then the clasical Mazur–Ulam theorem states that every surjective isometry $\Delta:X\to Y$ is affine. As usual, by a convex body of a normed space $X$ we mean a closed convex subset of $X$ with non-empty interior in $X$. In 1972, Mankiewicz [26] proved that every surjective isometry between convex bodies in two arbitrary normed spaces can be uniquely extended to an affine function between the spaces. Motivated by this observation, Tingley [43] raised the following extension problem:

Problem 1.1**.**

Suppose that $\Delta:S_{X}\to S_{Y}$ is a surjective isometry. Is $\Delta$ neccesarily the restriction of a surjective real-linear isometry from $X$ to $Y$?

This problem has been addressed in many papers, and has been answered affirmatively for particular choices of $X$ and $Y$ (see [12, 13, 14, 15, 16, 19, 20, 25, 37, 41, 44]).

Let us say that a Banach space $X$ satisfies the Mazur–Ulam property if, for any Banach space $Y$, every surjective isometry $\Delta:S_{X}\to S_{Y}$ admits an extension to a surjective real-linear isometry from $X$ to $Y$. The pioneering paper dealing with this property is that of Ding [16], who proves that the space $c_{0}({\mathbb{N}},{\mathbb{R}})$ of all null sequences of real numbers satisfies the Mazur-Ulam property. More examples of Banach spaces satisfying the Mazur–Ulam property are $c(\Gamma,{\mathbb{K}})$, $c_{0}(\Gamma,{\mathbb{K}})$, and $\ell_{\infty}(\Gamma,{\mathbb{K}})$ for any set $\Gamma$ (see [25] for ${\mathbb{K}}={\mathbb{R}}$ and [22, 33] for ${\mathbb{K}}={\mathbb{C}}$). More recently, it has been shown that this property is satisfied by unital complex C∗-algebras and real von Neumann algebras [30], by JBW∗-triples with rank one or rank bigger than or equal to three [4], and by the space $C(K,H)$ of all continuous functions from any compact Hausdorff space $K$ to a real or complex Hilbert space $H$ with $\dim(H_{\mathbb{R}})\geq 2$ [9]. In [24, Theorem 4.6], Li shows that, if $X_{1}$ and $X_{2}$ are strictly convex Banach spaces, then a surjective isometry $\Delta$ from the unit sphere of $X_{1}\bigoplus_{\infty}X_{2}$ to the unit sphere of any Banach space $Y$ admits an extension to a surjective real-linear isometry from $X_{1}\bigoplus_{\infty}X_{2}$ to $Y$ whenever ${\mathbb{R}}\Delta(S_{X_{1}})$ and ${\mathbb{R}}\Delta(S_{X_{2}})$ are subspace of $Y$. Other references dealing with the Mazur–Ulam property are [35, 37, 38, 39]. Anyway, as a matter of fact, Problem 1.1 remains unanswered even if the Banach spaces $X$ and $Y$ are two-dimensional [6, 23, 45].

The main aim of this paper is to provide the reader with new examples of Banach spaces satisfying the Mazur–Ulam property.

In Section 2, we revisit some previously known results, like [5], [21], [27, Corollary 9], and [32, Theorem 5], in order to establish the following:

1. (1)

If $\{X_{\gamma}\}_{\gamma\in\Gamma}$ is a family of strictly convex Banach spaces with dim$((X_{\gamma})_{{\mathbb{R}}})\geq 2$ and $|\Gamma|\geq 2$ (where $|\cdot|$ means cardinality), then the closed unit ball of $\bigoplus_{\gamma\in\Gamma}^{\ell_{\infty}}X_{\gamma}$ is the convex hull of its extreme points (Proposition 2.5). 2. (2)

If $X$ is a strictly convex Banach space with dim$(X_{{\mathbb{R}}})\geq 2$, and if $K$ is a totally disconnected compact Hausdorff space with $|K|\geq 2$, the the closed unit ball of $C(K,X)$ is the convex hull of its extreme points (Proposition 2.10).

Section 3 is devoted to proving our results on the Mazur–Ulam property in $c_{0}$- and $\ell_{\infty}$-sums of families of Banach spaces. We show that, if $\{X_{\gamma}\}_{\gamma\in\Gamma}$ is any family of strictly convex Banach spaces such that $|\Gamma|\geq 2$, and such that dim$((X_{\gamma})_{{\mathbb{R}}})\geq 2$ and the norm of $X_{\gamma}$ is Gâteaux differentiable in a dense subset of its unit sphere for every $\gamma$, then both $\bigoplus_{\gamma\in\Gamma}^{c_{0}}X_{\gamma}$ and $\bigoplus_{\gamma\in\Gamma}^{\ell_{\infty}}X_{\gamma}$ satisfy the Mazur–Ulam property (Theorem 3.3). In the particular cases of countable $c_{0}$-sums and finite $\ell_{\infty}$-sums, the hypothesis of Gâteaux differentiability of the norm can be removed. Indeed, we show that for a family $\{X_{n}\}_{n\in{\mathbb{N}}}$ of strictly convex Banach spaces with dim$(X_{n})\geq 2$, the Banach spaces $X_{1}\oplus_{\infty}\cdots\oplus_{\infty}X_{n}$ ($n\geq 2$) and $\bigoplus_{n\in{\mathbb{N}}}^{c_{0}}X_{n}$ satisfies the Mazur–Ulam property (Theorem 3.6). As a direct consequence, we obtain that every strictly convex Banach space can be equivalently renormed so that it satisfies the Mazur–Ulam property (Corollary 3.7). For instance, every separable Banach space, every reflexive Banach space, and more generally every weakly Lindelof Banach space (see [2] and [11]), can be equivalently renormed so that it satisfies the Mazur–Ulam property. A characterization in linear topological terms of the normed spaces which are strictly convex renormable can be found in [29].

The concluding Section 4 is devoted to studying the Mazur–Ulam property in the Banach space $C_{0}(K,X)$ of all continuous functions vanishing at infinity from a locally compact Hausdorff space $K$ to a Banach space $X$. As the most outstanding result in this setting, we show that, if $K$ is totally disconnected with $|K|\geq 2$, if $X$ is strictly convex with dim$(X_{{\mathbb{R}}})\geq 2$, and if the norm of $X$ is Gâteaux differentiable in a dense subset of its unit sphere, then $C_{0}(K,X)$ satisfies the Mazur–Ulam property (Theorem 4.4). In the case that $K$ is actually a metrizable compact space, the hypothesis of Gâteaux differentiability of the norm can be removed. As a consequence, if $X$ is a strictly convex Banach space with dim$(X_{{\mathbb{R}}})\geq 2$, then $C(\mathfrak{C},X)$ satisfies the Mazur–Ulam property, where $\mathfrak{C}$ denotes the Cantor set (Corollary 4.6).

Notation. Given a Banach space $X$, $B_{X}$, $S_{X}$, and $X^{*}$ shall stand for the closed unit ball, the unit sphere, and the dual of $X$, respectively. We denote by $Ext(B_{X})$ the set of all extreme points of $B_{X}$.

Given a family $\{X_{\gamma}\}_{\gamma\in\Gamma}$ of Banach spaces, we set $Z_{0}:=\bigoplus_{\gamma\in\Gamma}^{c_{0}}X_{\gamma}$ and $Z_{\infty}:=\bigoplus_{\gamma\in\Gamma}^{\ell_{\infty}}X_{\gamma}$. Given a subset $\mathcal{R}\subseteq\Gamma$, we denote by $P_{\mathcal{R}}$ the canonical projection from $\bigoplus_{\gamma\in\Gamma}X_{\gamma}$ to $\bigoplus_{\gamma\in\mathcal{R}}X_{\gamma}$. The symbol $Z$ shall stand for any of the spaces $Z_{0}$ or $Z_{\infty}$.

We recall that a Banach space $X$ is said to be strictly convex if every element of $S_{X}$ is an extreme point of $B_{X}$.

Given a compact Hausdorff space $K$, the symbol $C(K,X)$ shall stand for the Banach space of all continuous functions from $K$ to $X$ equipped with the sup norm. Given $f\in C(K)$ and $x\in X$, we denote by $f\otimes x$ the function in $C(K,X)$ defined by $(f\otimes x)(t):=f(t)x$ for all $t\in K$. If $K$ is a locally compact Hausdorff space, we denote by $C_{0}(K,X)$ the space of continuous $X$-valued functions on $K$ vanishing at infinity. Recall that $f:K\to X$ vanishes at infinity if for every $\varepsilon>0$ there exists a compact subset $K_{\varepsilon}$ of $K$ satisfying $\|f(t)\|<\varepsilon$ for all $t\in K\setminus K_{\varepsilon}$.

It is clear that $X$ satisfies the Mazur–Ulam property if and only if so does $X_{{\mathbb{R}}}$. According to this remark, throughout this paper we shall assume that all Banach spaces are real.

2. The strong Mankiewicz property: preliminary results

The result of Mankiewicz in [26] is one of the main tools applied in those papers devoted to explore new progress on Tingley’s problem and to determine new Banach spaces satisfying the Mazur–Ulam property.

In the recent paper [30], Mori and Ozawa introduce new techniques that are essential for our work. Following these authors, we shall say that a convex subset $C$ of a normed space $X$ satisfies the strong Mankiewicz property if every surjective isometry $\Delta$ from $C$ to an arbitrary convex subset $L$ in a normed space $Y$ is affine. In [30, Theorem 2] it is show that some of the hypothesis in Mankiewicz’s theorem can be somehow relaxed. The precise result reads as follows.

Theorem 2.1**.**

[30, Theorem 2]** Let $X$ be a Banach space such that the closed convex hull of $Ext(B_{X})$ has non-empty interior in $X$. Then, every convex body $K\subseteq X$ has the strong Mankiewicz property. $\hfill\Box$

Throughout this section we shall work with a family of nonzero Banach spaces $\{X_{\gamma}\}_{\gamma\in\Gamma}$.

It is known that $p\in Ext(B_{Z_{\infty}})$ implies $\|p(\gamma)\|_{X_{\gamma}}=1$ for all $\gamma\in\Gamma$. In the case that $X_{\gamma}$ is strictly convex for every $\gamma\in\Gamma$, the converse implication is true. Therefore we are provided with the following.

Lemma 2.2**.**

Suppose that $X_{\gamma}$ is strictly convex for every $\gamma\in\Gamma$. Then $p\in Ext(B_{Z_{\infty}})$ if and only if $\|p(\gamma)\|_{X_{\gamma}}=1$ for all $\gamma\in\Gamma$.

The following lemma is folklore.

Lemma 2.3**.**

Let $X$ be a Banach space with $\dim(X)\geq 2$. Then every element in $B_{X}$ can be expressed as a mean of two elements in $S_{X}$.

Proof.

Let $x$ be in $X$ with $0<\|x\|<1$. Then

[TABLE]

Therefore, since $S_{X}$ is connected, there exists $y\in S_{X}$ such that $\|2x-y\|=1$. Now $x=\frac{1}{2}(2x-y+y)$. ∎

As a straightforward consequence of the above lemma, we derive the following.

Corollary 2.4**.**

Suppose that $\dim(X_{\gamma})\geq 2$ for every $\gamma$. Then every element in $z\in B_{Z_{\infty}}$ can be written as $z=\frac{1}{2}(x+y)$ with $\|x(\gamma)\|_{X_{\gamma}}=1=\|y(\gamma)\|_{X_{\gamma}}$ for all $\gamma\in\Gamma$.

Now, combining Lemma 2.2 and Corollary 2.4, we obtain the following.

Proposition 2.5**.**

Suppose that $\dim(X_{\gamma})\geq 2$ and that $X_{\gamma}$ is strictly convex for every $\gamma\in\Gamma$. Then every element in $B_{Z_{\infty}}$ admits a expression as a mean of two elements in $Ext(B_{Z_{\infty}})$.

Proposition 2.6**.**

Suppose that $\dim(X_{\gamma})\geq 2$ and that $X_{\gamma}$ is strictly convex for every $\gamma\in\Gamma$. Then every convex body in $Z_{\infty}$ satisfies the strong Mankiewicz property.

Proof.

By Proposition 2.5, the closed unit ball of $Z_{\infty}$ is the convex hull of its extreme points, and hence the result follows from the Mori–Ozawa Theorem 2.1. ∎

For each $\gamma_{0}\in\Gamma$ and each $x_{\gamma_{0}}\in S_{X_{\gamma_{0}}}$ we set

[TABLE]

Lemma 2.7**.**

Suppose that $X_{\gamma}$ is strictly convex for every $\gamma\in\Gamma$. Then, for $\gamma_{0}\in\Gamma$ and $x_{\gamma_{0}}\in S_{X_{\gamma_{0}}}$, $A(\gamma_{0},x_{\gamma_{0}})$ is a maximal norm-closed proper face of $B_{Z}$, equivalently, a maximal convex subset of $S_{Z}$.

Proof.

We define $z_{0}\in S_{Z}$ by $z_{0}(\gamma):=0$ if $\gamma\neq\gamma_{0}$ and $z_{0}(\gamma_{0}):=x_{\gamma_{0}}$. Since $A(\gamma_{0},x_{\gamma_{0}})$ is a convex subset of $S_{Z}$, by Zorn’s lemma, there exists a maximal convex subset $C$ of $S_{Z}$ that contains $A(\gamma_{0},x_{\gamma_{0}})$. By [42, Lemma 3.1], there exists $z^{*}\in S_{Z^{*}}$ such that $C=\{z\in B_{Z}:z^{*}(z)=1\}$. Given $z\in C$ we have that $z^{*}(z+z_{0})=2=\|z+z_{0}\|$. This implies that $\|z(\gamma_{0})+x_{\gamma_{0}}\|_{X_{\gamma_{0}}}=2$, and hence $z(\gamma_{0})=x_{\gamma_{0}}$ because $X_{\gamma_{0}}$ is strictly convex. We conclude that $C=A(\gamma_{0},x_{\gamma_{0}})$.∎

The proof of the next result is an adaptation of [22, Lemma 2.1].

Lemma 2.8**.**

Suppose that $X_{\gamma}$ is strictly convex for every $\gamma\in\Gamma$. Let $Y$ be a Banach space, and let $\Delta:S_{Z}\to S_{Y}$ be a surjective isometry. Then, for $\gamma_{0}\in\Gamma$ and $x_{\gamma_{0}}\in S_{X_{\gamma_{0}}}$, the set

[TABLE]

is a non-empty weak∗-closed face of $B_{Y^{*}}$.

Proof.

By Lemma 2.7, the set $A(\gamma_{0},x_{\gamma_{0}})$ is a maximal convex subset of $B_{Z}$. Then, by [40, Lemma 3.5], $\Delta(A(\gamma_{0},x_{\gamma_{0}}))$ is a maximal convex subset of $B_{Y}$, and therefore, by [42, Lemma 3.1], there exists $\psi\in S_{Y^{*}}$ such that $\Delta(A(\gamma_{0},x_{\gamma_{0}}))=\{y\in B_{Y}:\psi(y)=1\}$. It is clear that given $\psi^{\prime}\in\overline{{\rm supp}(\gamma_{0},x_{\gamma_{0}})}^{w^{*}}$, we have $\Delta(A(\gamma_{0},x_{\gamma_{0}}))\subseteq\{y\in B_{Y}:\psi^{\prime}(y)=1\}$. Since $\Delta(A(\gamma_{0},x_{\gamma_{0}}))$ is a maximal convex subset of $B_{Y}$, $\psi^{\prime}\in{\rm supp}(\gamma_{0},x_{\gamma_{0}})$. ∎

Given a norm-one element $x$ in a Banach space $X$, the star of $x$ with respect to $S_{X}$, $St(x)$, is defined by

[TABLE]

It is known that $St(x)$ is precisely the union of all maximal convex subsets of $S_{X}$ containing $x$. The arguments in the proofs of [22, Lemmas 2.2 and 2.3, and Proposition 2.4] allow us to obtain the following result.

Lemma 2.9**.**

Suppose that $X_{\gamma}$ is strictly convex for every $\gamma\in\Gamma$. Let $Y$ be a Banach space and let $\Delta:S_{Z}\to S_{Y}$ be a surjective isometry. Pick $\gamma_{0}\in\Gamma$, $x_{\gamma_{0}}\in S_{X_{\gamma_{0}}}$ and define $z_{0}\in S_{Z}$ by $z_{0}(\gamma):=x_{\gamma_{0}}$ if $\gamma=\gamma_{0}$ and $z_{0}(\gamma):=0$ if $\gamma\neq\gamma_{0}$. Then the following assertions hold:

1. (1)

$St(\Delta(z_{0}))=\Delta(A(\gamma_{0},x_{\gamma_{0}}))$. 2. (2)

$\psi\Delta(z)=-1$* for all $z\in A(\gamma_{0},-x_{\gamma_{0}})$ and $\psi\in{\rm supp}(\gamma_{0},x_{\gamma_{0}})$.* 3. (3)

$\Delta(-A(\gamma_{0},x_{\gamma_{0}}))=-\Delta(A(\gamma_{0},x_{\gamma_{0}}))$. 4. (4)

${\rm supp}(\gamma_{0},x_{\gamma_{0}})\cap{\rm supp}(\gamma,x)=\varnothing$, for all $\gamma\neq\gamma_{0}$ and $x\in S_{X_{\gamma}}$. 5. (5)

${\rm supp}(\gamma_{0},x_{\gamma_{0}})\cap{\rm supp}(\gamma_{0},x^{\prime})=\varnothing$, for every $x^{\prime}\in S_{X_{\gamma_{0}}}$ with $x^{\prime}\neq x_{\gamma_{0}}$. 6. (6)

If $\psi\in{\rm supp}(\gamma_{0},x_{\gamma_{0}}),$ and if $z\in S_{Z}$ with $z(\gamma_{0})=0$, then $\psi\Delta(z)=0$. Furthermore, $|\psi(\Delta(z))|<1$, for all $\psi\in{\rm supp}(\gamma_{0},x_{0})$ and $z\in S_{Z}$ with $\|z(\gamma_{0})\|<1$. 7. (7)

If $z$ is in $S_{Z}$ such that $\psi\Delta(z)=0$ for all $x\in S_{X_{\gamma_{0}}}$ and $\psi\in{\rm supp}(\gamma_{0},x)$, then $z(\gamma_{0})=0$.

By [3, Lemma 1.4], if $p\in Ext(B_{C(K,X)})$, then $\|p(t)\|=1$ for all $t\in K$. The reciprocal implication is not true in general, however by [3, Remark 1.5], if $X$ is a strictly convex Banach space, then

[TABLE]

Given a compact Hausdorff space $K$, we denote by dim$(K)$ the covering dimension of $K$ [17, page 385]. We recall that a space $K$ has dim$(K)=0$ if and only if each point of $K$ has a neighborhoods base consisting of clopen sets [46, Definition 29.4]. We shall simply observe that, if $K$ is a totally disconnected compact Hausdorff space, then dim$(K)=0$ [46, Theorem 29.7, page 211].

Proposition 2.10**.**

Let $X$ be a strictly convex Banach space and let $K$ be a compact Hausdorff space satisfying one of the following conditions:

1. $(1)$

$X$* is infinite dimensional.* 2. $(2)$

$X$* is $n$-dimensional with $n\geq 2$ ($n\in\mathbb{N}$) and dim*$(K)\leq n-1$.

Then $B_{C(K,X)}$ is the convex hull of its extreme points, and hence every convex body in $C(K,X)$ satisfies the strong Mankiewicz property.

Proof.

Under any one of the conditions (1) or (2), we deduce from [27, Corollaries 8 and 9] that the closed unit ball of $C(K,X)$ is the convex hull of its extreme points. Therefore the desired conclusion is a consequence of Mori–Ozawa’s Theorem 2.1. ∎

In the remaining of this section, $K$ shall denote a locally compact Hausdorff space, and $X$ shall denote a strictly convex Banach space.

For each $t_{0}\in K$ and each $x_{0}\in S_{X}$ we set

[TABLE]

Lemma 2.11**.**

Let $t_{0}\in K$ and $x_{0}\in S_{X}$. Then $A(t_{0},x_{0})$ is a maximal norm-closed proper face of $B_{C_{0}(K,X)}$, equivalently, a maximal convex subset of $S_{C_{0}(K,X)}$.

Proof.

Let $U$ be an open set of $K$ with $t_{0}\in U$. Applying Urysohn’s lemma we find $h\in C_{0}(K)$ with $0\leq h\leq 1$, $h(t_{0})=1$ and $h\arrowvert_{K\setminus U}=0$. We consider $h\otimes x_{0}\in S_{C_{0}(K,X)}$. Since $A(t_{0},x_{0})$ is a convex set of $S_{C_{0}(K,X)}$, by Zorn’s lemma, there exists a maximal convex subset $C$ of $S_{C_{0}(K,X)}$ that contains $A(t_{0},x_{0})$. By [42, Lemma 3.1], there exists $\varphi\in S_{C_{0}(K,X)^{*}}$ such that $C=\{f\in B_{C_{0}(K,X)}:\varphi(f)=1\}$. Given $g\in C$ we have that $\varphi(h\otimes x_{0}+g)=2=\|h\otimes x_{0}+g\|$. This implies that there exists $t_{1}\in K$ such that $\|h(t_{1})x_{0}+g(t_{1})\|=2$. Since $\|h(t_{1})x_{0}\|=1=\|g(t_{1})\|$ and $0\leq h\leq 1$, we have that $h(t_{1})=1$. Now the strict convexity of $X$ yields $x_{0}=g(t_{1})$. We conclude that $C=A(t_{0},x_{0})$. ∎

The proof of the next result is similar to that of Lemma 2.8.

Lemma 2.12**.**

Let $Y$ be a Banach space and let $\Delta:S_{C_{0}(K,X)}\to S_{Y}$ be a surjective isometry. Then for each $t_{0}\in K$ and each $x_{0}\in S_{X}$ the set

[TABLE]

is a non-empty weak∗-closed face of $B_{Y^{*}}$.$\hfill\Box$

With the appropriate changes, the proof of the next lemma follows the arguments in the proofs of [8, Lemmas 2.4 and 2.5, and Propositions 2.6 and 3.1].

Lemma 2.13**.**

Let $Y$ be a Banach space and let $\Delta:S_{C_{0}(K,X)}\to S_{Y}$ be a surjective isometry. Pick $t_{0}\in K$ and $x_{0}\in S_{X}$. Then the following assertions hold:

1. (1)

$\psi\Delta(f)=-1$* for all $f\in A(t_{0},-x_{0})$ and $\psi\in{\rm supp}(t_{0},x_{{}_{0}})$.* 2. (2)

$\Delta(-A(t_{0},x_{0}))=-\Delta(A(t_{0},x_{0}))$. 3. (3)

${\rm supp}(t_{0},x_{0})\cap{\rm supp}(t,x)=\varnothing$, for all $t\neq t_{0}$ and $x\in S_{X}$. 4. (4)

${\rm supp}(t_{0},x_{0})\cap{\rm supp}(t_{0},x^{\prime})=\varnothing$, for every $x^{\prime}\in S_{X}$ with $x^{\prime}\neq x_{0}$. 5. (5)

If $\psi\in{\rm supp}(t_{0},x_{0}),$ and if $f\in S_{C_{0}(K,X)}$ with $f(t_{0})=0$, then $\psi\Delta(f)=0$. Furthermore, $\|\psi(\Delta(f))\|<1$ for all $f\in S_{C_{0}(K,X)}$ with $\|f(t_{0})\|<1$ and $\psi\in{\rm supp}(t_{0},x_{0})$. 6. (6)

If $K$ is totally disconnected, then $f(t_{0})=0$ whenever $f$ is in $S_{C_{0}(K,X)}$ with $\psi\Delta(f)=0$ for all $x\in S_{X}$ and $\psi\in{\rm supp}(t_{0},x)$.

Proof.

Let $O$ be an open set of $K$ with $t_{0}\in O$. Via Urysohn’s lemma, we find $h\in C_{0}(K)$ with $0\leq h\leq 1$, $h(t_{0})=1$ and $h|_{K\setminus O}=0$. Let us set $f_{0}=h\otimes x_{0}$.

First we see that $St(\Delta(f_{0}))=\Delta(A(t_{0},x_{0}))$. Let $y\in\Delta(A(t_{0},x_{0}))$. Then $\Delta^{-1}(y)\in A(t_{0},x_{0})$, and hence $\|\Delta^{-1}(y)+f_{0}\|=2$. By [18, Corollary 2.2], $\|y+\Delta(f_{0})\|=2$, and so $y\in St(\Delta(f_{0}))$. Conversely, let $y\in St(\Delta(f_{0}))$. Then $\|y+\Delta(f_{0})\|=2$ and hence, by [18, Corollary 2.2], $\|\Delta^{-1}(y)+f_{0}\|=2$. This implies that $\|\Delta^{-1}(y)(t_{0})+x_{0}\|=2$. Now, since $X$ is strictly convex, $\Delta^{-1}(y)(t_{0})=x_{0}$, and so $\Delta^{-1}(y)\in A(t_{0},x_{0})$.

(1) Let $f\in A(t_{0},-x_{0})$ and $\psi\in{\rm supp}(t_{0},x_{0})$. Since $-\Delta(f)\in S_{Y}$, there exists $h\in S_{C_{0}(K,X)}$ with $\Delta(h)=-\Delta(f)$. Let $g\in A(t_{0},x_{0})$. Then $2=\|g-f\|=\|\Delta(g)-\Delta(f)\|$, and hence $\Delta(h)=-\Delta(f)\in St(\Delta(g))$ and $\|\Delta(g)+\Delta(h)\|=2$. By [18, Corollary 2.2], $\|g+h\|=2$ for every $g\in A(t_{0},x_{0})$. Then there exists $t_{g}\in K$ such that $g(t_{g})=h(t_{g})$. Let $\mathcal{F}$ be the family of all open subset of $K$ containing $t_{0}$, and let $O\in\mathcal{F}$. Then, by Urysohn’s lemma, there exists $g_{O}\in C_{0}(K)$ with $0\leq g_{O}\leq 1$, $g_{O}(t_{0})=1$ and $g_{O}|_{K\setminus O}=0$. Since $g_{O}\otimes x_{0}\in A(t_{0},x_{0})$, there exists $t_{O}\in K$ such that $g_{O}(t_{O})x_{0}=h(t_{O})$ and $\|g_{O}(t_{O})x_{0}\|=\|h(t_{O})\|=1$. Since $g_{O}(t_{O})\in[0,1]$ and $\|g_{O}(t_{O})x_{0}\|=1$, we have $g_{O}(t_{O})=1$, hence $x_{0}=h(t_{O})$. Considering $\mathcal{F}$ as a directed set under the reverse inclusion, the net $(t_{O})_{O\in\mathcal{F}}$ converges to $t_{0}$. It follows from the continuity of $h$ that $h(t_{0})=x_{0}$ and $h\in A(t_{0},x_{0})$. Therefore $\psi(\Delta(h))=1=-\psi(\Delta(f))$.

(2) By (1), $\psi(\Delta(f))=-1$ whenever $f\in A(t_{0},-x_{0})$ and $\psi\in{\rm supp}(t_{0},x_{0})$. Since $A(t_{0},-x_{0})=-A(t_{0},x_{0})$ and $\psi^{-1}(1)\cap B_{C_{0}(K,X)}=\Delta(A(t_{0},x_{0}))$, we have that

[TABLE]

for every $\psi\in{\rm supp}(t_{0},x_{0})$.

(3) We argue by contradiction, and hence we assume that there exist $t_{1}\neq t_{0}$, $x_{1}\in S_{X}$, and $\psi\in{\rm supp}(t_{0},x_{0})\cap{\rm supp}(t_{1},x_{1})$. By Urysohn’s lemma, there are $u,v\in C_{0}(K)$ with $0\leq u,v\leq 1$, $u(t_{0})=1=v(t_{1})$ and $u(t)v(t)=0$ for every $t\in K$. Define the elements $f_{1}:=u\otimes x_{0}\in A(t_{0},x_{0})$ and $f_{2}:=v\otimes x_{1}\in A(t_{1},x_{1})$. By (1), $\psi(\Delta(-f_{2}))=-1$, and hence

[TABLE]

the desired contradiction.

(4) Let $u\in C_{0}(K)$ and $x^{\prime}\in S_{X}$ be such that $u(t_{0})=1$ and $x_{0}\neq x^{\prime}$. Then $f_{0}:=u\otimes x_{0}\in A(t_{0},x_{0})$ and $f_{1}:=u\otimes x^{\prime}\in A(t_{0},x^{\prime})$. If $\psi\in{\rm supp}(t_{0},x_{0})\cap{\rm supp}(t_{0},x^{\prime})$, then

[TABLE]

a contradiction. Therefore ${\rm supp}(t_{0},x_{0})\cap{\rm supp}(t_{0},x^{\prime})=\varnothing$, as desired.

(5) Let $f$ be in $S_{C_{0}(K,X)}$, and let $\psi$ be in ${\rm supp}(t_{0},x_{0})$. Suppose that $f(t_{0})=0$. Let $0<\varepsilon<1$, and define the open set

[TABLE]

By Urysohn’s lemma, there exists $g\in C_{0}(K)$ with $0\leq g\leq 1$, $g(t_{0})=1$ and $g(K\setminus O_{\varepsilon})=0$. Then $\|f\pm g\otimes x_{0}\|\leq 1+\varepsilon$. By (2), $\psi(\Delta(-g\otimes x_{0}))=-1$ and, by definition, $\psi(\Delta(g\otimes x_{0}))=1$. It follows

[TABLE]

[TABLE]

By letting $\varepsilon\to 0$, we obtain $|\psi(\Delta(f))\pm 1|\leq 1$. This implies $\psi(\Delta(f))=0$. Now suppose merely that $\|f(t_{0})\|<1$. Pick $0<\varepsilon<1$ with $\|f(t_{0})\|<1-\varepsilon$ and consider the closed set

[TABLE]

By Urysohn’s lemma, there exists $h\in C_{0}(K)$ such that $0\leq h\leq 1$, $h(t_{0})=0$ and $h(C_{\varepsilon})=1$. The function $hf$ in $S_{C_{0}(K,X)}$ satisfies $(hf)(t_{0})=0$. Consequently, $\psi(\Delta(hf))=0$. We conclude that

[TABLE]

[TABLE]

(6) Suppose that $K$ is totally disconnected. Let $f$ be in $S_{C_{0}(K,X)}$ such that $\psi(\Delta(f))=0$ for all $x\in S_{X}$ and $\psi\in{\rm supp}(t_{0},x)$. Assume towards a contradiction that $f(t_{0})\neq 0$. If $\|f(t_{0})\|=1$, then $\psi(\Delta(f))=1$ whenever $\psi$ lies in ${\rm supp}(t_{0},f(t_{0}))$, and this is impossible. Therefore $\|f(t_{0})\|<1$. It is known that a totally disconnected locally compact Hausdorff space has a basis of its topology consisting of compact open sets. We can always find a compact open subset $\mathcal{O}$ satisfying that

[TABLE]

Since $0<\frac{1}{2}\|f(t_{0})\|<\|f(s)\|$ for every $s\in\mathcal{O}$, the function $h$ defined on $K$ by $h(s):=f(s)$ if $s\in K\setminus\mathcal{O}$ and $h(s):=\frac{f(s)}{\|f(s)\|}$ if $s\in\mathcal{O}$ belongs to $S_{C_{0}(K,X)}$, and $\|h(t_{0})\|=1$. Therefore,

[TABLE]

Now, $h\in A(t_{0},\frac{f(t_{0})}{\|f(t_{0})\|})$, and for $\psi$ in ${\rm supp}(t_{0},\frac{f(t_{0})}{\|f(t_{0})\|})$ it follows that

[TABLE]

the desired contradiction, hence $f(t_{0})=0$.∎

3. The Mazur–Ulam property in $\bigoplus_{\gamma\in\Gamma}^{c_{0}}X_{\gamma}$ and $\bigoplus_{\gamma\in\Gamma}^{\ell_{\infty}}X_{\gamma}$

We recall that an element $x\in S_{X}$ is a smooth point of $X$ if there is a unique $f\in S_{X^{*}}$ such that $f(x)=1$. We denote by $\it{Sm}(X)\subseteq S_{X}$ the set of all smooth points of $X$ and, given $x\in\it{Sm}(X)$, we denote by $\varphi_{x}\in S_{X^{*}}$ the unique functional such that $\varphi_{x}(x)=1$. It is known that $x$ is a smooth point of $X$ if and only if the norm of $X$ is Gâteaux differentiable at $x$ [11, Corollary 1.5]. We refer to [11, §I.1] for the basic results on Gâteaux differentiability.

We note that the class of those Banach spaces $X$ such that $\it{Sm}(X)$ is norm dense in $S_{X}$ contains all separable Banach spaces [36, Proposition 9.4.3] and all Asplund Banach spaces [11, §I.1].

Proposition 3.1**.**

Let $\{X_{\gamma}\}_{\gamma\in\Gamma}$ be a family of strictly convex Banach spaces such that $|\Gamma|\geq 2$, and such that dim$(X_{\gamma})\geq 2$ and the set $\it{Sm}(X_{\gamma})$ is norm dense in $S_{X_{\gamma}}$ for every $\gamma$. Let $Y$ be a Banach space, and let $\Delta:S_{Z}\to S_{Y}$ be a surjective isometry such that $\Delta\arrowvert_{A(\gamma,x)}$ is an affine map whenever $\gamma\in\Gamma$ and $x\in S_{X_{\gamma}}$. Then the equality

[TABLE]

holds for all $\gamma\in\Gamma$, $x\in S_{X_{\gamma}}$, $\psi\in{\rm supp}(\gamma,x)$ and $z\in S_{Z}$ with $\|z(\gamma^{\prime})\|=1$ for some $\gamma^{\prime}\neq\gamma$.

Proof.

Let us fix $\gamma_{0},\gamma_{1}\in\Gamma$ with $\gamma_{1}\neq\gamma_{0}$, $x_{0}\in\it{Sm}(X_{\gamma_{0}})$, $\psi\in{\rm supp}(\gamma_{0},x_{0})$, and $p\in S_{Z}$ with $\|p(\gamma_{1})\|=1$. We set $Z_{\gamma_{1}}:=P_{\Gamma\setminus{\gamma_{1}}}(Z)$ and define $p_{1}\in Z$ by $p_{1}(\gamma_{1}):=p(\gamma_{1})$ and $p_{1}(\gamma):=0$ if $\gamma\neq\gamma_{1}$. It follows that

[TABLE]

Since $\Delta:A(\gamma_{1},p(\gamma_{1}))\to\Delta(A(\gamma_{1},p(\gamma_{1})))$ is affine, the mapping $\Delta_{p_{1}}:B_{Z_{\gamma_{1}}}\to\Delta(A(\gamma_{1},p(\gamma_{1})))$ defined by $\Delta_{p_{1}}(z):=\Delta(p_{1}+z)$ is affine. Now, the mapping $\varphi$ from $B_{Z_{\gamma_{1}}}$ to $\mathbb{R}$ defined by

[TABLE]

is affine with $|\varphi(z)|\leq 1$ for all $z\in B_{Z_{\gamma_{1}}}$. We can regard $\varphi$ as the restriction to $B_{Z_{\gamma_{1}}}$ of a functional $\varphi^{\prime}$ in $B_{Z_{\gamma_{1}}^{*}}$. Let $\widetilde{\varphi}$ be the function from $Z$ to ${\mathbb{R}}$ defined by $\widetilde{\varphi}(z):=\varphi^{\prime}(P_{\Gamma\setminus\gamma_{1}}(z))$ for every $z\in Z$. It follows that $\widetilde{\varphi}$ belongs to $B_{Z^{*}}$. By Lemma 2.9 (6), $\psi\Delta(p_{1})=0$ since $p_{1}(\gamma_{0})=0$ and $\psi\in{\rm supp}(\gamma_{0},x_{0})$, and hence

[TABLE]

We define $z_{0}\in S_{Z_{\gamma_{1}}}\subseteq S_{Z}$ by $z_{0}(\gamma_{0}):=x_{0}$ and $z_{0}(\gamma):=0$ if $\gamma\neq\gamma_{0}$. It is clear that $p_{1}+z_{0}\in A(\gamma_{0},x_{0})$, and since $\psi\in{\rm supp}(\gamma_{0},x_{0})$, we get $\widetilde{\varphi}(z_{0})=\psi\Delta(p_{1}+z_{0})=1,$ and so $\|\widetilde{\varphi}\|=1$ in $Z^{*}$.

We claim that

[TABLE]

Let us fix $\varepsilon>0$ and $z\in B_{Z}$. Since $X_{\gamma_{0}}$ is smooth at $x_{0}$, it follows from [11, Theorem I.1.4] that there exists $0<\rho<\frac{\varepsilon}{8}$ with the following property:

[TABLE]

Since $\it{Sm}(X_{\gamma})$ is norm dense in $S_{X_{\gamma}}$, for each $\gamma\in\Gamma$, the set

[TABLE]

is a norming set, and therefore, by the Hahn–Banach theorem,

[TABLE]

By (4), there exist $\lambda_{1},\ldots,\lambda_{n}\in(0,1]$ with $\sum_{i=1}^{n}\lambda_{i}=1$, $\gamma_{i}\in\Gamma$, and $x_{\gamma_{i}}\in\it{Sm}(X_{\gamma_{i}})$ such that

[TABLE]

Set $N:=\{i\in\{1,\ldots,n\}:\gamma_{i}=\gamma_{0}\}$. Since $\widetilde{\varphi}(z_{0})=1$, we have that

[TABLE]

It follows from (3) that

[TABLE]

On the other hand, we have $1-\sum_{i\in N}\lambda_{i}\ <{\rho},$ and so $\sum_{i\notin N}\lambda_{i}<\rho$ because $\sum_{i=1}^{n}\lambda_{i}=1$ . Now, for each $i\in N$, we set $\displaystyle\mu_{i}:=\frac{\lambda_{i}}{\sum_{j\in N}\lambda_{j}}$.

It is not hard to check that

[TABLE]

and hence

[TABLE]

Keeping in mind the above inequalities we obtain

[TABLE]

[TABLE]

[TABLE]

[TABLE]

We therefore have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

By letting $\varepsilon\to 0$, we realize that $\widetilde{\varphi}(z)=\varphi_{x_{0}}\otimes P_{\gamma_{0}}(z)$, which shows that (2) holds. Consequently,

[TABLE]

as desired. ∎

Proposition 3.2**.**

Let $\{X_{\gamma}\}_{\gamma\in\Gamma}$ be a family of strictly convex Banach spaces such that $|\Gamma|\geq 2$, and such that dim$(X_{\gamma})\geq 2$ and the set $\it{Sm}(X_{\gamma})$ is norm dense in $S_{X_{\gamma}}$ for every $\gamma$. Let $Y$ be a Banach space, and let $\Delta:S_{Z}\to S_{Y}$ be a surjective isometry. Then, for each $\gamma\in\Gamma$ and each $x\in\it{Sm}(X_{\gamma})$, the equality

[TABLE]

holds for all $\psi\in{\rm supp}(\gamma,x)$ and $z\in S_{Z}$.

Proof.

We first show the result in the case $Z=Z_{\infty}$.

Let us fix $\gamma\in\Gamma$ and $x\in\it{Sm}(X_{\gamma})$. Since the set

[TABLE]

is dense in $S_{Z_{\infty}}$, to prove the equality

[TABLE]

it is enough to verify it in the case that $z$ lies in $B$, and hence there exists $\gamma_{0}\in\Gamma$ such that $\|z(\gamma_{0})\|_{X_{\gamma_{0}}}=1$. We define $z_{0}\in S_{Z_{\infty}}$, by $z_{0}(\gamma_{0})=z(\gamma_{0})$ and $z_{0}(\gamma)=0$ for $\gamma\neq\gamma_{0}$, and $Z_{\gamma_{0}}:=\bigoplus_{\gamma\in\Gamma\setminus\gamma_{0}}^{\ell_{\infty}}X_{\gamma}$. By Lemma 2.7, $A(\gamma_{0},z(\gamma_{0}))$ is a maximal convex subset of $S_{Z_{\infty}}$. Then, by [7, Lemma 5.1], $\Delta(A(\gamma_{0},z(\gamma_{0})))$ is a maximal convex subset of $Y$. Furthermore, the mapping $\Delta_{z_{0}}:B_{Z_{\gamma_{0}}}\to\Delta(A(\gamma_{0},p(\gamma_{0})))$ defined by $\Delta_{z_{0}}(z):=\Delta(z_{0}+z)$ is a surjective isometry. It follows from Proposition 2.6 and Theorem 2.1 that $\Delta_{z_{0}}$ is an affine map. Then the restriction of $\Delta$ to $A(\gamma_{0},z(\gamma_{0}))$, regarded as a mapping onto $\Delta(A(\gamma_{0},z(\gamma_{0})))$, is affine. By Lemma 2.2, $\|p(\gamma)\|_{X_{\gamma}}=1$ for all $p\in Ext(B_{Z_{\infty}})$ and $\gamma\in\Gamma$, and hence, by Proposition 3.1, the equality

[TABLE]

is true for every $\psi\in{\rm supp}(\gamma,x)$. By Proposition 2.5, there exists $p,q\in Ext(B_{Z_{\infty}})$ such that $z=\frac{1}{2}(p+q)$. Since $X_{\gamma_{0}}$ is strictly convex and $z(\gamma_{0})=\frac{1}{2}(p(\gamma_{0})+q(\gamma_{0}))$, it follows that $p,q\in A(\gamma_{0},z(\gamma_{0}))$. Since $\Delta(z)=\frac{1}{2}(\Delta(p)+\Delta(q))$, we have that

[TABLE]

[TABLE]

for every $\psi\in{\rm supp}(\gamma,x).$

Now we show the result in the case $Z=Z_{0}$. We begin by showing that, given $\gamma_{0}\in\Gamma$ and $z\in S_{Z_{0}}$ with $\|z(\gamma_{0})\|_{X_{\gamma_{0}}}\!\!=1$, the restriction of $\Delta$ to $A(\gamma_{0},z(\gamma_{0}))$, regarded as a mapping onto $\Delta(A(\gamma_{0},z(\gamma_{0})))$, is affine.

By Lemma 2.7, $A(\gamma_{0},z(\gamma_{0}))$ is a maximal convex subset of $S_{Z_{0}}$. Then, by [7, Lemma 5.1], $\Delta(A(\gamma_{0},z(\gamma_{0})))$ is a maximal convex subset of $Y$. Let us take $\varepsilon>0$, $z_{1},z_{2}\in A(\gamma_{0},z(\gamma_{0}))$ and $\alpha\in(0,1)$. Then there exists a finite subset $\Gamma_{0}\subseteq\Gamma$ with $|\Gamma_{0}|\geq 2$, such that $\gamma_{0}\in\Gamma_{0}$ and $\|z_{i}-z_{i}^{\prime}\|<\varepsilon$ where $z_{i}^{\prime}:=P_{\Gamma_{0}}(z_{i})$ ($i=1,2$). We set

[TABLE]

and define $z_{0}\in S_{Z_{0}}$ by $z_{0}(\gamma_{0}):=z(\gamma_{0})$ and $z_{0}(\gamma):=0$ for every $\gamma\in\Gamma\setminus\gamma_{0}$. It is clear that

[TABLE]

and that $z_{1}^{\prime},z_{2}^{\prime}\in A_{\Gamma_{0}}(\gamma_{0},z(\gamma_{0}))$.

Let us verify that the set $\Delta(A_{\Gamma_{0}}(\gamma_{0},z(\gamma_{0})))$ is a convex subset of $S_{Y}$. Let $a,b\in\Delta(A_{\Gamma_{0}}(\gamma_{0},z(\gamma_{0})))\subseteq\Delta(A(\gamma_{0},z(\gamma_{0})))$ and $\beta\in(0,1)$. Then $\beta a+(1-\beta)b\in\Delta(A(\gamma_{0},z(\gamma_{0})))$. Take $u,v\in A_{\Gamma_{0}}(\gamma_{0},z(\gamma_{0}))$ and $w\in A(\gamma_{0},z(\gamma_{0}))$ such that $\Delta(u)=a$, $\Delta(v)=b$, and $\Delta(w)=\beta a+(1-\beta)b$. By Lemma 2.9 (6), for all $\gamma\in\Gamma\setminus\Gamma_{0}$ and $x\in S_{X_{\gamma}}$ we have $\psi\Delta(u)=0=\psi\Delta(v)$ for every $\psi\in{\rm supp}(\gamma,x)$. This implies that $\psi\Delta(w)=0$ for all $\gamma\in\Gamma\setminus\Gamma_{0}$, $x\in S_{X_{\gamma}}$, and $\psi\in{\rm supp}(\gamma,x)$. By Lemma 2.9 (7), $w(\gamma)=0$ for every $\gamma\in\Gamma\setminus\Gamma_{0}$. It follows that $w\in A_{\Gamma_{0}}(\gamma_{0},z(\gamma_{0}))$, and hence $\Delta(A_{\Gamma_{0}}(\gamma_{0},z(\gamma_{0})))$ is a convex subset of $S_{Y}$, as desired. Furthermore, the mapping $\Delta_{z_{0}}:B_{\bigoplus_{\gamma\in\Gamma_{0}\setminus\gamma_{0}}^{\ell_{\infty}}X_{\gamma}}\to\Delta(A_{\Gamma_{0}}(\gamma_{0},z(\gamma_{0})))$ defined by $\Delta_{z_{0}}(z):=\Delta(z_{0}+z)$ is a surjective isometry. Therefore, by Proposition 2.6 and Theorem 2.1, $\Delta_{z_{0}}$ is an affine map, and so the restriction of $\Delta$ to $A_{\Gamma_{0}}(\gamma_{0},z(\gamma_{0}))$, regarded as a mapping onto $\Delta(A_{\Gamma_{0}}(\gamma_{0},z(\gamma_{0})))$, is affine. This fact, together with the inequalities

[TABLE]

and

[TABLE]

allows us to obtain

[TABLE]

By letting $\varepsilon\to 0$, we realize that the restriction of $\Delta$ to $A(\gamma_{0},z(\gamma_{0}))$, regarded as a mapping onto $\Delta(A(\gamma_{0},z(\gamma_{0})))$, is affine.

Now the proof shall be concluded by showing that, for $x_{0}\in\it{Sm}(X_{\gamma_{0}})$, the equality

[TABLE]

is true for all $\psi\in{\rm supp}(\gamma_{0},x_{0})$ and $z\in S_{Z_{0}}$.

Let us fix $z\in S_{Z_{0}}$ and $\gamma_{1}\in\Gamma$ such that $\|z(\gamma_{1})\|_{X_{\gamma_{1}}}=1$. Pick $\varepsilon>0$, $\gamma_{2}\in\Gamma\setminus\{\gamma_{0}\}$ with $\|z(\gamma_{2})\|_{X_{\gamma_{2}}}<\varepsilon$, and $x_{\gamma_{2}}\in S_{X_{\gamma_{2}}}$. For $i=1,2$, define $p_{i}\in S_{Z_{0}}$ by $p_{i}(\gamma):=z(\gamma)$ if $\gamma\neq\gamma_{2}$ and $p_{i}(\gamma_{2}):=(-1)^{i}x_{\gamma_{2}}$. Then $p_{i}\in A(\gamma_{1},z(\gamma_{1}))$ ($i=1,2$) and $\|z-\frac{p_{1}+p_{2}}{2}\|<\varepsilon$. By Proposition 3.1,

[TABLE]

for every $\psi\in{\rm supp}(\gamma_{0},x_{0})$. Since $\Delta\arrowvert_{A(\gamma_{1},z(\gamma_{1}))}$ is an affine map, we have that

[TABLE]

Then

[TABLE]

[TABLE]

By letting $\varepsilon\to 0$, we obtain

[TABLE]

We can now establish the main result of this section.

Theorem 3.3**.**

Let $\{X_{\gamma}\}_{\gamma\in\Gamma}$ be a family of strictly convex Banach spaces such that $|\Gamma|\geq 2$, and such that dim$(X_{\gamma})\geq 2$ and the set $\it{Sm}(X_{\gamma})$ is norm dense in $S_{X_{\gamma}}$ for all $\gamma$. Then $Z_{0}$ and $Z_{\infty}$ satisfy the Mazur–Ulam property.

Proof.

Let $Y$ be a Banach space, and let $\Delta:S_{Z}\to S_{Y}$ be a surjective isometry. By Proposition 3.2, the equality

[TABLE]

is true for all $\gamma\in\Gamma$, $x\in\it{Sm}(X_{\gamma})$, $\psi\in{\rm supp}(\gamma,x)$, and $z\in S_{Z}$. Now, the set

[TABLE]

is a norming set for $Z$, and hence the set

[TABLE]

is a norming set for $Y$. It follows that

[TABLE]

[TABLE]

[TABLE]

for all $\lambda>0$ and $z,\tilde{z}\in S_{Z}$. By [18, Lemma 2.1], $\Delta$ can be extended to a real-linear isometry of $Z$ onto $Y$.∎

Now, let us see that the hypothesis about the Gâteaux differentiability of the norm in Theorem 3.3 can be removed in the particular cases of countable $c_{0}$-sums and finite $\ell_{\infty}$-sums.

Lemma 3.4**.**

Let $X_{1}$, $X_{2}$, and $Y$ be Banach spaces, and let $A$ be a separable subset of $X_{1}\oplus_{\infty}X_{2}$. Suppose that $\Delta:S_{X_{1}\oplus_{\infty}X_{2}}\to S_{Y}$ is a surjective isometry. Then there exist separable subspaces $M_{1}\subseteq X_{1}$, $M_{2}\subseteq X_{2}$ and $N\subseteq Y$ such that $A\subseteq M_{1}\oplus_{\infty}M_{2}$ and $\Delta(S_{M_{1}\oplus_{\infty}M_{2}})=S_{N}$.

Proof.

Since $A_{1}:=A$ is a separable subset of $X_{1}\oplus_{\infty}X_{2}$, there exists a countable subset of $A_{1}$ (say $\{a_{n}:n\in{\mathbb{N}}\}$) dense in $A_{1}$. For each $n\in{\mathbb{N}}$, we have $a_{n}=a_{n}^{1}+a_{n}^{2}$ with $a_{n}^{1}\in X_{1}$ and $a_{n}^{2}\in X_{2}$. Now, for $i=1,2$, consider the separable closed subspace $B_{(i,1)}\subseteq X_{i}$ defined by $B_{(i,1)}:=\overline{Lin}\{a_{n}^{i}\}_{n\in{\mathbb{N}}}$. It is clear that $A_{1}\subseteq B_{(1,1)}\oplus_{\infty}B_{(2,1)}\subseteq X_{1}\oplus_{\infty}X_{2}.$ Now, $B_{(1,1)}\oplus_{\infty}B_{(2,1)}$ is a separable closed subspace of $X_{1}\oplus_{\infty}X_{2}$, hence $\Delta(S_{B_{(1,1)}\oplus_{\infty}B_{(2,1)}})$ is a separable subset of $S_{Y}$. Define $N_{1}:=\overline{Lin}(\Delta(S_{B_{(1,1)}\oplus_{\infty}B_{(2,1)}}))$. Since $S_{N_{1}}$ is a separable subset of $S_{Y}$, we have that $A_{2}:=\Delta^{-1}(S_{N_{1}})$ is a separable subset of $S_{X_{1}\oplus_{\infty}X_{2}}$. Consider $B_{(i,2)}\subseteq X_{i}$ such that $B_{(i,1)}\subseteq B_{(i,2)}$ ($i=1,2$) and $A_{2}\subseteq B_{(1,2)}\oplus_{\infty}B_{(2,2)}$ and $N_{2}=\overline{Lin}(\Delta(S_{B_{(1,2)}\oplus_{\infty}B_{(2,2)}}))$. Define the separable closed subspaces $B_{(i,k)}\subseteq X_{i}$ ($i=1,2$) and $N_{k}\subseteq Y$ inductively by $\Delta^{-1}(S_{N_{k}})\subseteq S_{B_{(1,k+1)}\oplus_{\infty}B_{(2,k+1)}}$, $B_{(i,k)}\subseteq B_{(i,k+1)}$ ($i=1,2$) and $N_{k}:=\overline{Lin}(\Delta(S_{B_{(1,k)}\oplus_{\infty}B_{(2,k)}}))$ for every $k\geq 2$. We consider the separable closed subspaces $M_{i}:=\overline{\bigcup_{k\in{\mathbb{N}}}B_{(i,k)}}$ ($i=1,2$) and $N:=\overline{\bigcup_{k\in{\mathbb{N}}}N_{k}}$. The inclusion $\Delta(S_{M_{1}\oplus_{\infty}M_{2}})\subseteq S_{N}$ is clear. Let $y$ be in $\bigcup_{k\in{\mathbb{N}}}N_{k}$ with $\|y\|=1$. Then there exists $k\in{\mathbb{N}}$ such that $y\in S_{N_{k}}$, and hence $\Delta^{-1}(y)\in S_{B_{(1,k+1)}\oplus_{\infty}B_{(2,k+1)}}\subseteq S_{M_{1}\oplus_{\infty}M_{2}}$. Now let $y$ be arbitrary in $S_{N}$. Then we can find a sequence $\{y_{n}\}$ of norm-one elements in $\bigcup_{k\in{\mathbb{N}}}N_{k}$ such that $\{y_{n}\}\rightarrow y$. It is clear that $\{\Delta^{-1}(y_{n})\}$ is a Cauchy sequence in $S_{M_{1}\oplus_{\infty}M_{2}}$, hence there exists $z\in S_{M_{1}\oplus_{\infty}M_{2}}$ such that $\{\Delta^{-1}(y_{n})\}\rightarrow z$. We conclude that $\Delta(z)=y$. Therefore $\Delta(S_{M_{1}\oplus_{\infty}M_{2}})\supseteq S_{N}$. ∎

Lemma 3.5**.**

Let $\{X_{n}\}_{n\in{\mathbb{N}}}$ be a family of Banach spaces and let $A$ be a separable subset of $Z_{0}$. Let $Y$ be a Banach space, and let $\Delta:S_{Z_{0}}\to S_{Y}$ be a surjective isometry. Then there exist separable subspaces $M_{n}\subseteq X_{n}$ ($n\in{\mathbb{N}}$), and $N\subseteq Y$ such that $A\subseteq\bigoplus_{n\in{\mathbb{N}}}^{c_{0}}M_{n}$ and $\Delta(S_{\bigoplus_{n\in{\mathbb{N}}}^{c_{0}}M_{n}})=S_{N}$.

Proof.

Since $A$ is a separable subset of $\bigoplus_{n\in{\mathbb{N}}}^{c_{0}}X_{n}$, there exists a countable subset of $A$ (say $\{a_{k}:k\in{\mathbb{N}}\}$) dense in $A$. For each $n\in{\mathbb{N}}$, $P_{n}(\{a_{k}:k\in{\mathbb{N}}\})$ is a countable subset of $X_{n}$, and hence

[TABLE]

is a separable subspace of $X_{n}$. It is clear that $A\subseteq\bigoplus_{n\in{\mathbb{N}}}^{c_{0}}B_{n}$. For $n\in{\mathbb{N}}$, let us fix a dense countable subset $D_{n}$ of $B_{n}$ containing [math]. Then the subset $\mathcal{F}$ of $\bigoplus_{n\in{\mathbb{N}}}^{c_{0}}B_{n}$ consisting of all finitely supported vectors whose $n$-th coordinate is in $D_{n}$ for each $n$, is countable and dense in $\bigoplus_{n\in{\mathbb{N}}}^{c_{0}}B_{n}$. Therefore $\bigoplus_{n\in{\mathbb{N}}}^{c_{0}}B_{n}$ is a separable subspace of $\bigoplus_{n\in{\mathbb{N}}}^{c_{0}}X_{n}$. We conclude the proof arguing as in the proof of the previous lemma. ∎

Now, we are ready to prove the following.

Theorem 3.6**.**

Let $\{X_{n}\}_{n\in{\mathbb{N}}}$ be a family of strictly convex Banach spaces such that dim$(X_{n})\geq 2$ for all $n$. Then the Banach spaces $Z_{n}:=X_{1}\oplus_{\infty}\cdots\oplus_{\infty}X_{n}$ ($$n\geq 2$$) and $\bigoplus_{n\in{\mathbb{N}}}^{c_{0}}X_{n}$ satisfy the Mazur–Ulam property.

Proof.

Let $Y$ be a Banach space, and let $\Delta:S_{Z_{n}}\to S_{Y}$ be a surjective isometry. Let us take $z,\tilde{z}\in S_{Z_{n}}$ and $\lambda>0$. By Lemma 3.4, there exist separable subspaces $M_{i}\subseteq X_{i}$ for all $i=1,\cdots,n$ and $N\subseteq Y$ such that $z,\tilde{z}\in M_{1}\oplus_{\infty}\cdots\oplus_{\infty}M_{n}$ and

[TABLE]

Since $X_{i}$ is a strictly convex with dim$(X_{i})\geq 2$, we have that $M_{i}$ is a strictly convex with dim$(M_{i})\geq 2$ for all $i=1,\ldots,n$. According to Mazur’s theorem [36, Proposition 9.4.3], $\it{Sm}(M_{1}\oplus_{\infty}\cdots\oplus_{\infty}M_{n})$ is norm dense in $S_{M_{1}\oplus_{\infty}\cdots\oplus_{\infty}M_{n}}$. By Theorem 3.3, the restriction of $\Delta$ to $S_{M_{1}\oplus_{\infty}\cdots\oplus_{\infty}M_{n}}$, regarded as a mapping onto $S_{N}$, admits a unique extension to a surjective real-linear isometry $\Phi$ from $M_{1}\oplus_{\infty}\cdots\oplus_{\infty}M_{n}$ to $N$, and so

[TABLE]

Since $z,\tilde{z}\in S_{Z_{n}}$ and $\lambda>0$ are arbitrary, it follows from [18, Lemma 2.1] that $\Delta$ can be extended to a real-linear isometry from $Z_{n}$ to $Y$. By replacing Lemma 3.4 with Lemma 3.5, the proof in the case $\bigoplus_{\gamma\in\Gamma}^{c_{0}}X_{\gamma}$ is similar. ∎

Using the fact that $c_{0}(\Gamma)$ admits a strictly convex equivalent norm, one can realize that a Banach space $X$ is strictly convex renormable as soon as there exists a one-to-one bounded linear operator $T:X\to c_{0}(\Gamma)$. In this case, the equivalent strictly convex norm in $X$ is given by

[TABLE]

This fact implies that every weakly compactly generated space (in particular every separable space) admits a strictly convex equivalent renorming (see [11], [29] and [2]). Let us note that there are strictly convex renormable Banach spaces which cannot be linearly and continuously imbedded into $c_{0}(\Gamma)$ for any $\Gamma$ [10].

Corollary 3.7**.**

Let $X$ be a strictly convex Banach space. Then $X$ has a equivalent norm with the Mazur–Ulam property.

Proof.

The result is clear for finite-dimensional Banach spaces since every Hilbert space satisfies the Mazur–Ulam property [4, Proposition 4.5]. So that we can assume that $X$ is an infinite-dimensional Banach space. Let $M$ be a finite dimensional subspace of $X$ with dim$(M)\geq 2$. Since $M$ is a complemented subspace, there exists a closed subspace $N$ of $X$ such that $X=M\oplus N$. Since $X$ is strictly convex, then $M,N$ are strictly convex. We conclude the proof applying Theorem 3.6 to $X$ with the equivalent norm given by $X=M\oplus_{\infty}N$. ∎

Corollary 3.8**.**

Let $X$ be a Banach space such that $X^{*}$ is strictly convex. Then $X$ has a equivalent norm whose dual norm has the Mazur–Ulam property.

Proof.

We can assume that $X$ is infinite-dimensional. Let $M$ be a finite dimensional subspace of $X$ with dim$(M)\geq 2$. We consider the equivalen norm in $X$ given by $X=M\oplus_{1}N$. Then $X^{*}=M^{*}\oplus_{\infty}N^{*}$. Since $N^{*}$ and $M^{*}$ are strictly convex, it follows from Theorem 3.6 that $X^{*}$ has the Mazur–Ulam property. ∎

We recall that, if $X$ is a weakly compactly generated Banach space, then there exists $\|.\|_{1}$ and $\|.\|_{2}$ equivalent norms in $X$ such that $(X,\|.\|_{1})$ and $(X,\|.\|_{2})^{*}$ are strictely convex [11, Theorems VI.2.1 and VII.1.6].

Corollary 3.9**.**

Let $X$ be a weakly compactly generated Banach space. Then there exists $\|.\|_{1}$ and $\|.\|_{2}$ equivalent norms in $X$ such that $(X,\|.\|_{1})$ and $(X,\|.\|_{2})^{*}$ satisfy the Mazur–Ulam property.

4. The Mazur–Ulam property in $C_{0}(K,X)$

Let $X$ be a Banach space, and let $K$ be a locally compact Hausdorff space. Given $p\in S_{C_{0}(K,X)}$ and a compact open subset $\mathcal{O}$ of $K$ such that $\|p(t)\|=1$ for all $t\in\mathcal{O}$, we define the intersection face of $p$ relatively to $\mathcal{O}$ by

[TABLE]

Proposition 4.1**.**

Let $X$ be a strictly convex Banach space such that dim$(X)\geq 2$ and the set $\it{Sm}(X)$ is norm dense in $S_{X}$, let $K$ be a locally compact Hausdorff space with $|K|\geq 2$. Let $Y$ be a Banach space, and let $\Delta:S_{C_{0}(K,X)}\to S_{Y}$ be a surjective isometry such that $\Delta\arrowvert_{\mathcal{F}(p,\mathcal{O})}$ is an affine map where $p\in S_{C_{0}(K,X)}$ and $\mathcal{O}$ is a compact open subset of $K$ with $\|p(t)\|=1$ for every $t\in\mathcal{O}$ and $K\setminus\mathcal{O}\neq\emptyset$. Then the equality

[TABLE]

is true for all $t\in K\setminus\mathcal{O}$, $x\in\it{Sm}(X)$ and $\psi\in{\rm supp}(t,x)$.

Proof.

Let us fix $t_{0}\in K\setminus\mathcal{O}$, $x_{0}\in\it{Sm}(X)$ and $\psi\in{\rm supp}(t_{0},x_{0})$. For $j=1,2$, we set $p_{j}=\chi_{{}_{\mathcal{O}_{j}}}p$ where $\mathcal{O}_{1}=K\setminus\mathcal{O}$ and $\mathcal{O}_{2}=\mathcal{O}$. It follows that

[TABLE]

Since $\Delta\arrowvert_{\mathcal{F}(p,\mathcal{O}_{2})}$ is affine, the mapping $\Delta_{p_{2}}$ from $B_{C_{0}(\mathcal{O}_{1},X)}$ to $\Delta(\mathcal{F}(p,\mathcal{O}_{2}))$ defined by $\Delta_{p_{2}}(g):=\Delta(p_{2}+g)$ for every $g\in C_{0}(\mathcal{O}_{1},X)$ is affine. Now, the mapping $\varphi$ from $B_{C_{0}(\mathcal{O}_{1},X)}$ to $\mathbb{R}$ defined by

[TABLE]

is affine with $|\varphi(g)|\leq 1$ for every $g\in B_{C_{0}(\mathcal{O}_{1},X)}$. We can regard $\varphi$ as the restriction to $B_{C_{0}(\mathcal{O}_{1},X)}$ of a functional $\varphi^{\prime}$ in $B_{C_{0}(\mathcal{O}_{1},X)^{*}}$. Let $\widetilde{\varphi}$ be the function from $C_{0}(K,X)$ to ${\mathbb{R}}$ defined by $\widetilde{\varphi}(g):=\varphi^{\prime}(g|_{\mathcal{O}_{1}})$ for every $g\in C_{0}(K,X)$. It follows that $\widetilde{\varphi}$ belongs to $B_{C_{0}(K,X)^{*}}$. By Lemma 2.13 $(5)$, $\psi\Delta(p_{2})=0$ since $p_{2}(t_{0})=0$ and $\psi\in{\rm supp}(t_{0},x_{0})$, and hence $\widetilde{\varphi}(g)=\psi\Delta(g|_{\mathcal{O}_{1}}+p_{2})$ for all $g\in C_{0}(K,X)$. We set $f_{0}=x_{0}\otimes\chi_{{}_{\mathcal{O}_{1}}}$. It is clear that $f_{0}+p_{2}\in A(t_{0},x_{0})$, and since $\psi\in{\rm supp}(t_{0},x_{0})$, we get $\widetilde{\varphi}(f_{0})=\psi\Delta(f_{0}+p_{2})=1,$ and so $\widetilde{\varphi}\in S_{C_{0}(K,X)^{*}}$.

Since $\psi\Delta(p)=\widetilde{\varphi}(p)$, the proof of the proposition is concluded if we prove that

[TABLE]

Actually, we shall show more, namely that

[TABLE]

Let us fix $\varepsilon>0$ and $g\in C_{0}(K,X)$. Since $x_{0}\in\it{Sm}(X)$, it follows from [11, Theorem I.1.4] that there exists $0<\rho<\frac{\varepsilon}{4}$ with the following property:

[TABLE]

Since $\it{Sm}(X)$ is norm dense in $S_{X}$, the set

[TABLE]

is a norming set, and therefore, by the Hahn-Banach theorem,

[TABLE]

Let $\mathcal{O}_{3}$ denote the open set $\{t\in\mathcal{O}_{1}:\|g(t)-g(t_{0})\|<{\rho}\}.$ By Urysohn’s lemma, there exists $h_{0}\in C_{0}(K)$ such that $0\leq h_{0}\leq 1$, $h_{0}(t_{0})=1$ and $h_{0}|_{K\backslash\mathcal{O}_{3}}=0$. We set $\widetilde{f}_{0}=x_{0}\otimes h_{0}$. By (9), there exist $\lambda_{1},\ldots,\lambda_{n}\in(0,1]$ with $\sum_{i=1}^{n}\lambda_{i}$, $t_{1},\ldots,t_{n}\in K$, and $x_{1},\ldots,x_{n}\in\it{Sm}(X)$ such that

[TABLE]

It is clear that $\widetilde{f}_{0}\in A(t_{0},x_{0})$, and since $\widetilde{f}_{0}|_{\mathcal{O}_{2}}=0$, we get $\widetilde{f}_{0}+p_{2}\in A(t_{0},x_{0})$, and so

[TABLE]

Now we follow the argument in the proof of the Proposition 3.1, and we check that (7) is true. ∎

Proposition 4.2**.**

Let $X$ be a strictly convex Banach space such that dim$(X)\geq 2$, let $K$ be a totally disconnected locally compact Hausdorff space with $|K|\geq 2$. Let $Y$ be a Banach space, and let $\Delta:S_{C_{0}(K,X)}\to S_{Y}$ be a surjective isometry. Then the restriction of $\Delta$ to $\mathcal{F}(p,\mathcal{O})$, regarded as a mapping onto $\Delta(\mathcal{F}(p,\mathcal{O}))$, is an affine map for every $p\in S_{C_{0}(K,X)}$ and every compact open subset $\mathcal{O}$ of $K$ with $\|p(t)\|=1$ for every $t\in\mathcal{O}$ and $K\setminus\mathcal{O}\neq\emptyset$.

Proof.

Let $p$ be in $S_{C_{0}(K,X)}$, and let $\mathcal{O}$ be a compact open subset with $\|p(t)\|=1$ for every $t\in\mathcal{O}$ and $K\setminus\mathcal{O}\neq\emptyset$. By [30, Lemma 8], $\Delta(\mathcal{F}(p,\mathcal{O}))$ is an intersection face, and hence a non-empty convex subset of $S_{Y}$, since $\displaystyle\mathcal{F}(p,\mathcal{O})=\bigcap_{t\in\mathcal{O}}A(t,p(t))$ is a intersection face. Fix $z_{1},z_{2}\in\mathcal{F}(p,\mathcal{O})$, $\alpha\in(0,1)$, and $\varepsilon>0$. There exist $a_{1},a_{2}\in C_{0}(K\setminus\mathcal{O},X)$ with $\|(1-\chi_{\mathcal{O}})z_{i}-a_{i}\|<\varepsilon$ and $\overline{\{t\in K:a_{i}(t)\neq 0\}}$ is a non-empty compact subset of $K\setminus\mathcal{O}$ ($i=1,2$). For $i=1,2$, we define $z_{i}^{\prime}(t):=z_{i}(t)$ if $t\in\mathcal{O}$ and $z_{i}^{\prime}(t):=a_{i}(t)$ if $t\in K\setminus\mathcal{O}$. Then $\|z_{i}-z_{i}^{\prime}\|<\varepsilon$ and $\overline{\{t\in K:z_{i}^{\prime}(t)\neq 0\}}$ is a compact subset of $K$ ($i=1,2$). Since $K$ has a basis of its topology consisting of compact open sets, there exist $\mathcal{O}_{1},\cdots,\mathcal{O}_{n}$ compact open subsets of $K$ with

[TABLE]

Now, $K_{1}:=\cup_{i=1}^{n}\mathcal{O}_{i}$ is a totally disconnected compact Hausdorff subset of $K$, $z_{1}^{\prime},z_{2}^{\prime}\in C(K_{1},X)$, and $C_{0}(K,X)=C(K_{1},X)\oplus_{\infty}C_{0}(K_{2},X)$. We set

[TABLE]

It is clear that $z_{1}^{\prime},z_{2}^{\prime}\in\mathcal{F}_{K_{1}}(p,\mathcal{O})\subseteq\mathcal{F}(p,\mathcal{O})$ and that $\mathcal{F}_{K_{1}}(p,\mathcal{O})=\chi_{\mathcal{O}}p+B_{C(K_{1}\setminus\mathcal{O},X)}$.

Let us verify that the set $\Delta(\mathcal{F}_{K_{1}}(p,\mathcal{O}))$ is a convex subset of $S_{Y}$. Let $a,b\in\Delta(\mathcal{F}_{K_{1}}(p,\mathcal{O}))\subseteq\Delta(\mathcal{F}(p,\mathcal{O}))$ and $\beta\in(0,1)$. Then $\beta a+(1-\beta)b\in\Delta(\mathcal{F}(p,\mathcal{O}))$. Take $u,v\in\mathcal{F}_{K_{1}}(p,\mathcal{O})$ and $w\in\mathcal{F}(p,\mathcal{O})$ such that $\Delta(u)=a$, $\Delta(v)=b$ and $\Delta(w)=\beta a+(1-\beta)b$. By Lemma 2.13 (5), for all $t\in K\setminus K_{1}$ and $x\in S_{X}$, we have $\psi\Delta(u)=0=\psi\Delta(v)$ for every $\psi\in{\rm supp}(t,x)$. This implies that $\psi\Delta(w)=0$ for every $\psi\in{\rm supp}(t,x)$. By Lemma 2.13(6), $w(t)=0$ for $t\in K\setminus K_{1}$. It follows that $w\in\mathcal{F}_{K_{1}}(p,\mathcal{O})$, and hence $\Delta(\mathcal{F}_{K_{1}}(p,\mathcal{O}))$ is a convex subset of $S_{Y}$, as desired. Furthermore, the mapping $\Delta_{\chi_{\mathcal{O}}p}:B_{C(K_{1}\setminus\mathcal{O},X)}\to\Delta(\mathcal{F}_{K_{1}}(p,\mathcal{O}))$ defined by $\Delta_{\chi_{\mathcal{O}}p}(z):=\Delta(\chi_{\mathcal{O}}p+z)$ is a surjective isometry. Therefore, by Proposition 2.6 ($K_{1}\setminus\mathcal{O}$ is totally disconnected compact Hausdorff space) and Theorem 2.1, $\Delta_{\chi_{\mathcal{O}}p}$ is an affine map, and so the restriction of $\Delta$ to $\mathcal{F}_{K_{1}}(p,\mathcal{O})$, regarded as a mapping onto $\Delta(\mathcal{F}_{K_{1}}(p,\mathcal{O}))$, is affine. This fact, together with the inequalities

[TABLE]

and

[TABLE]

allows us to obtain

[TABLE]

By letting $\varepsilon\to 0$, we realize that the restriction of $\Delta$ to $\mathcal{F}_{K_{1}}(p,\mathcal{O})$, regarded as a mapping onto $\Delta(\mathcal{F}_{K_{1}}(p,\mathcal{O}))$, is affine.∎

Proposition 4.3**.**

Let $X$ be a strictly convex Banach space such that dim$(X)\geq 2$ and such that the set $\it{Sm}(X)$ is norm dense in $S_{X}$, and let $K$ be a totally disconnected compact Hausdorff space with $|K|\geq 2$. Then $C(K,X)$ satisfies the Mazur–Ulam property.

Proof.

We recall that the subset $\{\varphi_{x}\otimes\delta_{t}:t\in K,\ \ x\in\it{Sm}(X)\}$ of $C(K,X)^{*}$ is a norming set. Now, keeping in mind the proof of Theorem 3.3, the proof shall be concluded by showing that the equality

[TABLE]

holds for all $\psi\in{\rm supp}(t,x)$, $t\in K,\ \ x\in\it{Sm}(X)$, and $g\in S_{C(K,X)}$. Let $\varepsilon>0$, and take $g\in S_{C(K,X)}$ and $t_{0}\in K$ with $\|g(t_{0})\|=1$. There exists a clopen subset $\mathcal{O}_{1}$ of $K$ such that

[TABLE]

Define $h\in S_{C(K,X)}$ by $h:=\chi_{{}_{\mathcal{O}_{1}}}g(t_{0})+\chi_{{}_{\mathcal{O}_{2}}}g$ where $\mathcal{O}_{2}=K\setminus\mathcal{O}_{1}$. Then $\|h-g\|<\varepsilon$. By Proposition 2.10, there exist $\lambda_{1},\cdots,\lambda_{n}$ in $(0,1]$ with $\sum_{i=1}^{n}\lambda_{i}=1$, and $p_{1},\cdots,p_{n}$ in $Ext(B_{C(K,X)})$ such that $h=\sum_{i=1}^{n}\lambda_{i}p_{i}$. Therefore, $g(t_{0})=\sum_{i=1}^{n}\lambda_{i}p_{i}(t)$ for all $t\in\mathcal{O}_{1}$, and hence $g(t_{0})=p_{i}(t)$ for all $t\in\mathcal{O}_{1}$ and $i\in\{1,\ldots,n\}$, since $g(t_{0})$ is a extreme point of $B_{X}$. For each $i\in\{1,\ldots,n\}$, we denote $p_{i}^{j}:=p_{i}\rvert_{\mathcal{O}_{j}}$ ($j=1,2$). Then

[TABLE]

We recall that

[TABLE]

is an intersection face, by [30, Lemma 8], $\Delta(\mathcal{F}(h,\mathcal{O}_{1})$ is a non-empty convex subset of $S_{Y}$. Furthermore, the mapping

[TABLE]

defined by $\Delta_{h}(z):=\Delta(\chi_{{}_{\mathcal{O}_{1}}}g(t_{0})+z)$ for $z\in C(\mathcal{O}_{2},X)$ is a surjective isometry. By Proposition 2.10 and Theorem 2.1, $\Delta_{h}$ is affine. Then $\Delta|_{\mathcal{F}(h,\mathcal{O}_{1})}$ is an affine map, and so $\Delta(h)=\sum_{i=1}^{n}\lambda_{i}\Delta(\chi_{\mathcal{O}_{1}}g(t_{0})+p_{i}^{2})$. For all $x\in\it{Sm}(X)$, $t\in K$, and $\psi\in{\rm supp}(t,x)$, we have that $\psi\Delta(h)=\sum_{i=1}^{n}\lambda_{i}\psi\Delta(\chi_{\mathcal{O}_{1}}g(t_{0})+p_{i}^{2})$. Now, note that $\|(\chi_{\mathcal{O}_{1}}g(t_{0})+p_{i}^{2})(t)\|=1$ for all $t\in K$ and $i\in\{1,\cdots,n\}$, and hence, by Proposition 4.1, for all $x\in\it{Sm}(X)$, $t\in K$, and $\psi\in{\rm supp}(t,x)$, it follows

[TABLE]

and hence

[TABLE]

By letting $\varepsilon\to 0$, we realize that the equality

[TABLE]

is true for all $x\in\it{Sm}(X)$, $t\in K$, $\psi\in{\rm supp}(t,x)$, and $g\in S_{C(K,X)}$.∎

We can now establish the main result of this section.

Theorem 4.4**.**

Let $X$ be a strictly convex Banach space such that dim$(X)\geq 2$ and such that the set $\it{Sm}(X)$ is norm dense in $S_{X}$, and let $K$ be a totally disconnected locally compact Hausdorff space with $|K|\geq 2$. Then $C_{0}(K,X)$ satisfies the Mazur–Ulam property.

Proof.

We can assume that $K$ is not a compact space, since in other case we can apply Proposition 4.5. Let $Y$ be a Banach space, and let $\Delta:S_{C_{0}(K,X)}\to S_{Y}$ be a surjective isometry. We recall that the subset $\{\varphi_{x}\otimes\delta_{t}:t\in K,\ \ x\in\it{Sm}(X)\}$ of $C_{0}(K,X)^{*}$ is a norming set. Now, keeping in mind the proof of Theorem 3.3, the proof shall be concluded by showing that the equality

[TABLE]

is true for all $x\in\it{Sm}(X)$, $t\in K$, $\psi\in{\rm supp}(t,x)$, and $g\in S_{C(K,X)}$. Let $0<\varepsilon<\frac{1}{3}$, and take $g\in S_{C_{0}(K,X)}$. There exists $t_{0}\in K$ with $\|g(t_{0})\|=1$. Since $K$ is a totally disconnected locally compact Hausdorff space, there exists a compact open subset $\mathcal{O}_{1}$ of $K$ such that

[TABLE]

Define $h:=\chi_{\mathcal{O}_{1}}g(t_{0})+\chi_{\mathcal{O}_{2}}g$ where $\mathcal{O}_{2}=K\setminus\mathcal{O}_{1}$. We have that $\|h-g\|<\varepsilon$ and that $h\in S_{C_{0}(K,X)}$. By Proposition 4.2, the restriction of $\Delta$ to $\mathcal{F}(h,\mathcal{O}_{1})$, regarded as a mapping onto $\Delta(\mathcal{F}(h,\mathcal{O}_{1}))$, is an affine map. Since $h\in S_{C_{0}(K,X)}$ and $K$ is not a compact space, pick a compact open subset $\mathcal{O}_{3}$ of $K$ with $\|h(t)\|<\varepsilon$ for every $t\in\mathcal{O}_{3}$ and $x_{3}\in S_{X}$. For $i=1,2$, define the functions $p_{i}:=\chi_{\mathcal{O}_{3}}(-1)^{i}x_{3}+\chi_{\mathcal{O}_{4}}h$ where $\mathcal{O}_{4}=K\setminus\mathcal{O}_{3}$. Then $\mathcal{O}_{3}\cap\mathcal{O}_{1}=\emptyset$, $p_{1},p_{2}\in\mathcal{F}(h,\mathcal{O}_{1})$ and $\|h-\frac{1}{2}(p_{1}+p_{2})\|<\varepsilon$. By Proposition 4.1,

[TABLE]

for all $x\in\it{Sm}(X)$, $t\in K$, and $\psi\in{\rm supp}(t,x)$. Since $\Delta|_{\mathcal{F}(h,\mathcal{O}_{1})}$ is an affine map, we have that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Since $\|g-h\|<\varepsilon$,

[TABLE]

and

[TABLE]

It follows that $|\psi\Delta(h)-\varphi_{x}\otimes\delta_{t}(h)|<4\varepsilon.$ By letting $\varepsilon\to 0$, we obtain that

[TABLE]

for all $x\in\it{Sm}(X)$, $t\in K$, and $\psi\in{\rm supp}(t,x)$.∎

We conclude this paper obtaining a refinement of the previous theorem for a wide range of locally compact Hausdorff spaces. By [1, Theorem 3.44, p. 92], the one-point compactification $K^{*}$ of a noncompact locally compact Hausdorff space $K$ is metrizable if and only if $K$ is second countable. By [1, Lemma 3.99, p. 125], $C(K^{*})$ is separable if and only if $K^{*}$ is metrizable. It follows that a locally compact Hausdorff space $K$ is second countable if and only if $C_{0}(K)$ is a separable Banach space.

Proposition 4.5**.**

Let $X$ be a strictly convex Banach space such that dim$(X)\geq 2$, and let $K$ be a totally disconnected locally compact Hausdorff space such that $|K|\geq 2$ and such that $K$ is second countable. Then $C_{0}(K,X)$ satisfies the Mazur–Ulam property.

Proof.

Let $Y$ be a Banach space, and let $\Delta:S_{C_{0}(K,X)}\to S_{Y}$ be a surjective isometry. Let $A$ be a separable subset of $S_{C_{0}(K,X)}$. First, we check that there exists a separable closed subspace $M$ of $X$ such that

[TABLE]

Fix a countable subset $\{a_{k}:k\in{\mathbb{N}}\ \ a_{k}\in A\}$ which is dense in $A$. Take $k,n\in{\mathbb{N}}$. There exists $a_{k}^{\prime}\in C_{0}(K,X)$ such that $\|a_{k}-a_{k}^{\prime}\|<\frac{1}{n}$ and $\overline{\{t\in K:a_{k}^{\prime}(t)\neq 0\}}$ is a compact subset of $K$. There exist $\mathcal{O}_{1},\cdots,\mathcal{O}_{m}$ compact open subsets of $K$ with $\overline{\{t\in K:a_{k}^{\prime}(t)\neq 0\}}\subseteq\cup_{i=1}^{m}\mathcal{O}_{i}$. Then we have $C_{0}(K,X)=C(K_{1},X)\oplus_{\infty}C_{0}(K_{2},X)$ where $K_{1}:=\cup_{i=1}^{m}\mathcal{O}_{i}$ and $K_{2}:=K\setminus K_{1}$. It is clear that $a_{k}^{\prime}\in C(K_{1},X)$ and $K_{1}$ is a totally disconnected compact Hausdorff space. For each $t\in K_{1}$, define the open subset of $K_{1}$

[TABLE]

There exists a clopen subset $\mathcal{O}_{t}$ of $K_{1}$ such that $t\in\mathcal{O}_{t}\subseteq V_{t}$ for every $t\in K_{1}$. Since $K_{1}$ is compact and $K_{1}=\cup_{t\in K_{1}}\mathcal{O}_{t}$, there exists a finite subset $C_{(n,k)}$ of $K_{1}$ such that $K_{1}=\cup_{t\in C_{(n,k)}}\mathcal{O}_{t}$. We can assume that $\mathcal{O}_{t}\cap\mathcal{O}_{t^{\prime}}=\emptyset$ whenever $t\neq t^{\prime}$ and $t,t^{\prime}\in C_{(n,k)}$. We define the function $h_{(n,k)}:=\sum_{t\in C_{(n,k)}}\chi_{O_{t}}a_{k}^{\prime}(t)$. For $s\in K_{1}$, there exists $t\in C_{(n,k)}$ such that $s\in\mathcal{O}_{t}$, and hence

[TABLE]

Then $\|a_{k}^{\prime}-h_{(n,k)}\|_{C(K_{1},X)}=\|a_{k}^{\prime}-h_{(n,k)}\|\leq\frac{1}{n}$. This implies that $\|a_{k}-h_{(n,k)}\|\leq\frac{2}{n}$. We consider the separable closed subspace

[TABLE]

It follows that $C_{0}(K,M_{0})$ is a closed separable subspace of $C_{0}(K,X)$, and that $h_{(n,k)}\in C_{0}(K,M_{0})$ for all $n,k\in{\mathbb{N}}$, and hence $A\subseteq S_{C_{0}(K,M_{0})}$. We follow the argument in the proof of Lemma 3.4 to derive the existence of separable closed subspaces $M\subseteq X$ and $N\subseteq Y$ such that $A\subseteq S_{C_{0}(K,M)}$ and $\Delta(S_{C_{0}(K,M)})=S_{N}$. Let us take $z,\tilde{z}\in S_{C_{0}(K,X)}$ and $\lambda>0$. By the previous paragraph, there exist separable closed subspaces $M\subseteq X$ and $N\subseteq Y$ such that $z,\tilde{z}\subseteq S_{C_{0}(K,M)}$ and $\Delta(S_{C_{0}(K,M)})=S_{N}$. Since $X$ is a strictly convex, $M$ is a strictly convex and by Mazur’s theorem [36, Proposition 9.4.3]), the set of all smooth points of $M$ is dense in $S_{M}$. By Proposition 4.3, the restriction of $\Delta$ to $S_{C_{0}(K,M)}$, regarded as a mapping onto $S_{N}$, admits an extension to a surjective real-linear isometry $\Phi$ from $C_{0}(K,M)$ to $N$. It follows that

[TABLE]

and hence by [18, Lemma 2.1], $\Delta$ can be extended to a real-linear isometry of $C_{0}(K,X)$ onto $Y$. ∎

A 1910 theorem of Brouwer characterizes the Cantor set as the unique totally disconnected, compact metric space without isolated points.

Corollary 4.6**.**

Let $X$ be a strictly convex Banach space such that dim$(X)\geq 2$, and let $\mathfrak{C}$ be the Cantor set. Then $C(\mathfrak{C},X)$ satisfies the Mazur–Ulam property.

Acknowledgements

The author is very grateful to M. Cabrera and A. Rodríguez Palacios for fruitful remarks concerning the matter of the paper. The author is partially supported by the Junta de Andalucía grant FQM199.