Proof. Note first that $M\subset St(x)$ for every $x\in M$ implies that

\[ M\subset\bigcap_{x\in M} St(x). \]

To prove the converse, choose $z\in \bigcap_{x\in M} St(x)$. Then $\|z+x\|=2$ for all $x\in M$. Set two open convex sets

\[ S_1=M+\text{int} B_X \text{ and } S_2={-}z+\text{int} B_X, \]

where $\mbox {int}B_X$ denotes the interior points of $B_X$. Obviously, $S_1\cap S_2=\emptyset$. By the separation theorem, there is a linear functional $x^{*}\in S_{X^{*}}$ and a real number $c$ such that $x^{*}(z_1)>c>x^{*}(z_2)$ for every $z_1\in S_1$ and every $z_2\in S_2$. It follows that

\[ \sup x^{*}({-}z+ \mbox{int} B_X) \leq c\leq \inf x^{*}(x+\mbox{int} B_X) \]

for every $x\in M$. Consequently,

\[{-}x^{*}(z)+1\leq c \leq x^{*}(x)-1 \]

for every $x\in M$. This shows that $c=0$ and $x^{*}(z)=x^{*}(x)=1$ for every $x\in M$. By the maximality of $M$, we have $M=\{x\in S_X: x^{*}(x)=1\}$, and hence $z\in M$. The proof is complete.

Proof. That (1) implies (2) is obvious. To prove that (2) implies (3), suppose that there are two sets $M_1,\, M_2\in \mathfrak {M}(X)$ containing $x$. Clearly, $M_1\cup M_2 \subset St(x)$. Now, take an arbitrary $z\in M_1$. Since (2) holds, it follows that $[z,\, y]\subset S_X$ for every $y\in M_2$. Hence we have $M_2 \subset \mbox {co}(M_2,\, z) \subset S_X$. Then the maximality of $M_2$ ensures us that $\mbox {co}(M_2,\,z)=M_2$, and thus $z\in M_2$. It follows that $M_1\subset M_2$, and so $M_1=M_2$. Finally, (3) implies (1) following from the identity (2.1).

Proof. Since the inverse of $f$ denoted by $f^{-1}$ is also a surjective phase-isometry, we only need to prove that if $St(x)\in \mathfrak {M}(X)$ then $St(f(x))\in \mathfrak {M}(Y)$. Suppose that it is not true. Then Proposition 2.3 allows the existence of two distinct elements $y_1,\, y_2\in St(f(x))$ such that $\|y_1+y_2\|<2$. We set $z_1:=\tfrac {1}{2}(y_1+f(x))$ and $z_2:=\tfrac {1}{2}(y_2+f(x))$. It is easy to see that $z_1,\, z_2\in St(f(x))$ satisfy

\[ \|z_1+z_2\|<2 \text{and } \|z_2-z_1\|<2. \]

Since $f$ is surjective, there are $x_1,\,x_2\in S_X$ such that $f(x_1)=z_1$ and $f(x_2)=z_2$. Observe from the fact that $f$ is a phase-isometry that

\[ x_1, x_2\in St(x) \cup St({-}x). \]

Thus, $\|x_1+x_2\|=2$ or $\|x_1-x_2\|=2$, which leads to a contradiction that $\|z_1+z_2\|=2$ or $\|z_1-z_2\|=2$. The proof is complete.

Proof. By Lemma 3.2 and Remark 3.4, it is apparent that $St(f(x_0))\in \mathfrak {M}(Y)$ and $f_M(M)\subset St(f(x_0))$. For every $y\in St(f(x_0))$ with $f(x)=y$ for some $x\in S_X$. Since $f$ is a phase-isometry, we see that

\[ \|x+x_0\|=\|f(x)+f(x_0)\|=2\text{ or }\|x-x_0\|=\|f(x)+f(x_0)\|=2. \]

It follows that $x$ or $-x \in St(x_0)=M$ by Proposition 2.3. As a consequence, we get $y=f_M(x)$ with $x\in M$ or $y=f_M(-x)$ with $-x\in M$, which implies that $f_M(M)=St(f(x_0))$.

Set $M_1:=f_{M}(M)$. It remains to verify that $f_M(C_M^{X})=C_{M_1}^{Y}$. By the definition of $f_{M}$, for every $0\neq x\in C_M^{X}$, we have

\[ \|f_{M}(x)+f_{M}(x_0)\|=\|x+x_0\|=\|x\|+\|x_0\|=\|f_{M}(x)\|+\|f_{M}(x_0)\|. \]

This allows us to conclude that

\[ \|\frac{f_{M}(x)}{\|f_M(x)\|}+f_{M}(x_0)\|=2. \]

An immediate conclusion from this is that $f_M(x)/\|f_M(x)\| \in St(f_M(x_0))=M_1$, and so $f_M(x)\in C_{M_1}^{Y}$. For every $0\neq y\in C_{M_1}^{Y}$ with $f(x)=y$ for some $x\in X$. Note that $\|f(x)+f(x_0)\|=\|f(x)\|+\|f(x_0)\|$, and since $f$ is a phase-isometry, we have either

\[ \|x\|+\|x_0\|=\|f(x)+f(x_0)\|=\|x+x_0\| \]

or

\[ \|x\|+\|x_0\|=\|f(x)+f(x_0)\|=\|x-x_0\|. \]

It follows that $\pm x/\|x\|\in St(x_0)=M$. As a consequence, $f(x)=f_{M}(x)$ with $x\in C_M^{X}$ or $f(x)=f_{M}(-x)$ with $-x\in C_M^{X}$. Both cases show that $y=f(x)\in f_{M}(C_M^{X})$. Thus, we finish the proof of $f_{M}(C_M^{X})=C_{M_1}^{Y}$.

Finally, as seen from the above for all $x,\,y\in C_M^{X}$, we have

\[ \|f_{M}(x)+f_{M}(y)\|=\|f_{M}(x)\|+\|f_{M}(y)\|=\|x\|+\|y\|=\|x+y\|. \]

That $f_{M}$ is a phase-isometry leads to

\[ \|f_{M}(x)-f_{M}(y)\|=\|x-y\| \]

as required, which finishes the proof.

Proof. We can assume that $x_0\in M$ is a star point. Then $f(x_0)$ is also a star point of $S_Y$ by Lemma 3.2. Since $f$ is a phase-isometry, we have

\[ 2=\|x_0+x_0'\|\in \{\|f(x_0)+f(x'_0)\|, \|f(x_0)-f(x'_0)\|\}. \]

If $\|f(x_0)+f(x'_0)\|=\|x_0+x'_0\|=2$, then $f(x'_0)\in St(f(x_0))$. We conclude from this and $f_M$'s definition that for every $0\neq x\in C_M^{X}$ and $\theta \in \{-1,\,1\}$, $f_M(x)=\theta f(x)$ implies that $f_M'(x)=\theta f(x)$. Namely, $f_M=f_M'$. Another possibility is that

\[ \|f(x_0)-f(x'_0)\|=\|x_0+x'_0\|=2>\|x_0-x'_0\|. \]

Applying a similar argument as above shows that $f_M'=-f_M$.

Proof. Let $X_1=[A]$ and $Y_1=[f(X_1)]$. Choose $\{x_n\}\subset X_1$ which is dense in $X_1$. To see that $Y_1$ is separable, it suffices to show that the set $\{f(x_n)\}\cup \{-f(x_n)\}$ is dense in $f(X_1)$. By Lemma 3.1, it is clear that $\{f(x_n)\}\cup \{-f(x_n)\} \subset f(X_1)$. Now for every $y\in f(X_1)$ with $f(x)=y$ for some $x\in X_1$, there is a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $x_{n_k}$ converges to $x$ in norm. For every $k\geq 1$, choose $y_k\in Y$ satisfying

\[ y_{k}= \left \{ \begin{array}{@{}ll} f(x_{n_k}) & \text{if } \|f(x_{n_k})-f(x)\|=\|x_{n_k}-x\|\neq\|x_{n_k}+x\| ; \\ -f(x_{n_k}) & \text{if }\|f(x_{n_k})-f(x)\|=\|x_{n_k}+x\|. \end{array} \right. \]

Thus, it is obvious that $\|y_k-y\|=\|y_k-f(x)\|=\|x_{n_k}-x\|.$ This means that $\{f(x_n)\}\cup \{-f(x_n)\}$ is dense in $f(X_1)$ as desired. Define inductively separable subspaces $X_n\subset X$ and $Y_n\subset Y$ such that for all $n\geq 2$,

\[ X_n=[f^{{-}1}(Y_{n-1})] \text{ and } Y_n=[f(X_{n})]. \]

Then it is easy to verify that $X_0=\bigcup_{n=1}^{\infty } X_n$ and $Y_0=\bigcup_{n=1}^{\infty } Y_n$ are just the subspaces that we need.

Proof. We first conclude that this conclusion is true under the assumption that $X$ is a separable Banach space. By Example 2.4, $M$ has a star point. In other words, if the chosen $x_0\in M$ is not a star point, then by Lemma 3.6 we can replace it by a star point $x'_0\in M$. Proposition 3.5 yields the desired conclusion.

Now we handle the general case. For every finite subset $A\subset M$ with $x_0\in A$, we apply Lemma 3.7 to $F_M$ to obtain two separable subspaces $X_A\subset X$ and $Y_A\subset Y$ such that $A\subset X_A$ and $F_M(X_A)=Y_A$. Let $M_A\in \mathfrak {M}(X_A)$ such that $M\cap X_A\subset M_A$. The previous argument ensures us that

\[ f_{M_A}(M_A)\in \mathfrak{M}(Y_A), \]

where $f_{M_A}: C_{M_A}^{X_A} \rightarrow Y_A$ is the induced map with respect to $x_0$ and $M_A$. As $A\subset M\cap M_A$, we have $f_{M}(x)=f_{M_A}(x)$ for every $x\in A$. It follows that

\[ \mbox{co}(f_M(M))\subset S_Y. \]

So we can choose $M_1\in \mathfrak { M}(Y)$ such that $f_M(M)\subset M_1$. For the converse, since the inverse $f^{-1}$ of $f$ is also a surjective phase-isometry, there is a map $h_{M_1}:C_{M_1}^{Y}\rightarrow X$ with respect to $f(x_0)\in M_1$ and induced from $f^{-1}$. We can see that

\[ h_{M_1}\circ f_M(x)=x \]

for all $x\in C_M^{X}$. Using the above argument for $h_{M_1}$ and $M_1$, we also have

\[ M\subset h_{M_1}(M_1)\subset \mbox{co}(h_{M_1}(M_1)) \subset S_X. \]

This and the maximality of $M$ entail that $M=h_{M_1}(M_1)$, and so $f_M(M)=M_1$.

We also need to prove that $f_M(C_M^{X})=C_{M_1}^{Y}$, where $M_1=f_M(M)$. For all $0\neq z\in C_M^{X}$ and $x\in M$, set $B:=\{z,\, x,\, x_0\}$. Following a similar argument as above, we get two separable subspaces $X_B\subset X$ and $Y_B\subset Y$ such that $B\subset X_B$ and $F_M(X_B)=Y_B$. Let $f_{M_B}: C_{M_B}^{X_B} \rightarrow Y_B$ be the induced map with respect to $x_0$ and $M_B$. Then $f_{M_B}(M_B)\in \mathfrak {M}(Y_B)$ and $f_{M_B}(C_{M_B}^{X_B})\subset Y_B$ is a cone generated by $f_{M_B}(M_B)$. It follows that

\[ \|f_{M}(z)+f_{M}(x)\|=\|f_{M_B}(z)+f_{M_B}(x)\|=\|f_{M}(z)\|+\|f_{M}(x)\|. \]

Consequently, we obtain $f_{M}(z)/\|f_{M}(z)\|\in St (f_{M}(x))$ for every $x\in M$. Applying Proposition 2.1, we get $f_{M}(z)/\|f_{M}(z)\|\in M_1$, and so $f_M(C_M^{X})\subset C_{M_1}^{Y}$. Conversely, repeating the same argument to $h_{M_1}: C_{M_1}^{Y} \rightarrow X$, we obtain $h_{M_1}(C_{M_1}^{Y})\subset C_M^{X}$. Observe that the map $h_{M_1}$ is injective and

\[ h_{M_1}\circ f_M(x)=x \]

for all $x\in C_M^{X}$. We get $f_M(C_M^{X})=C_{M_1}^{Y}$ as desired. It is simply verified that $f_M: C_M^{X} \rightarrow Y$ is an isometry by noticing that

\[ \|f_M(x)+f_M(y)\|=\|x+y\|, \]

for all $x,\,y\in C_M^{X}$. The proof is complete.

Proof. We may assume that $\theta =1$ to simplify the notation, that is $f_{M_1}(x_0)=f_{M_2}(x_0)$. Suppose, on the contrary, that there is a nonzero element $z_0\in C_{M_1}^{X} \cap C_{M_2}^{X}$ such that $f_{{M_1}}(z_0)=-f_{{M_2}}(z_0).$ We observe first from Proposition 3.8 that

\[ \|x_0-z_0\|=\|f_{M_1}(x_0)-f_{M_1}(z_0)\|=\|f_{M_2}(x_0)+f_{M_2}(z_0)\|=\|x_0\|+\|z_0\|=\|x_0+z_0\|. \]

Let $u_0$ be the mid-point of $x_0$ and $z_0$, i.e., $u_0=(x_0+z_0)/2$. Obviously, $u_0\in C_{M_1}^{X}\cap C_{M_2}^{X}$. If $f_{M_1}(u_0)=-f_{M_2}(u_0)$, then

\[ \|u_0\|=\|x_0-u_0\|=\|f_{M_1}(x_0)-f_{M_1}(u_0)\|=\|f_{M_2}(x_0)+f_{M_2}(u_0)\|=\|x_0\|+\|u_0\|; \]

If $f_{M_1}(u_0)=f_{M_2}(u_0)$, then

\[ \|u_0\|=\|u_0-z_0\|=\|f_{M_1}(u_0)-f_{M_1}(z_0)\|=\|f_{M_2}(u_0)+f_{M_2}(z_0)\|=\|u_0\|+\|z_0\|. \]

Either case leads to a contradiction. Thus, the proof is complete.

Theorem 4.2 Let $X$ be an $n$-dimensional polyhedral Banach space. Then $X$ has the Wigner property.

Proof. Let $Y$ be a Banach space, and let $f: X\rightarrow Y$ be a surjective isometry. Given $M_0\in \mathfrak {M}(X)$, let $f_{M_0}$ be defined as in Definition 3.3. Since $X$ is a polyhedral Banach space of finite dimension, the cone $C_{M_0}^{X}$ generated by $M_0$ is a convex body of $X$. Proposition 3.8 guarantees that $f_{M_0}(C_{M_0}^{X})$ is also a convex body in $Y$. P. Mankiewicz [Reference Mankiewicz13] has proved that every isometry between convex bodies is the restriction of an affine onto isometry between the corresponding spaces. By this result, there is a linear isometry $g_{M_0}: X\rightarrow Y$ such that its restriction on the cone $C_{M_0}^{X}$ is $f_{M_0}$.

We now apply Lemma 4.1 to obtain that for every $w^{*}$-exposed point $x^{*}$ of $B_{X^{*}}$, there exists a linear functional $\varphi \in S_{Y^{*}}$ such that

(4.1)\begin{equation} x^{*}(x)={\pm}\varphi(f(x))={\pm}\varphi(f_M(x))\quad (M\in\mathfrak{M}(X),x\in C_M^{X}). \end{equation}

We claim that for every $M\in \mathfrak {M}(X)$, either $x^{*}(x)=\varphi (f_M(x))$ for all $x\in C_M^{X}$ or $x^{*}(x)=-\varphi (f_M(x))$ for all $x\in C_M^{X}$ holds. It remains to show that the third possibility, that is $x^{*}(x)=\varphi (f_M(x))\neq 0$ and $x^{*}(y)=-\varphi (f_M(y))\neq 0$ for some $x,\, y \in C_M^{X}$, leads to a contradiction. Indeed, it follows from (4.1) that

\[ x^{*}(x+y)={\pm}\varphi(f_M(x+y))={\pm}\varphi(f_M(x)+f_M(y)). \]

If $x^{*}(x+y)=\varphi (f_M(x)+f_M(y))$, this shows that $x^{*}(y)=-\varphi (f_M(y))=0$, a contradiction. If $x^{*}(x+y)=-\varphi (f_M(x)+f_M(y))$, then obviously $x^{*}(x)=-\varphi (f_M(x))=0$. This is a contradiction which proves the claim. For every $w^{*}$-exposed point $z^{*}$ of $B_{X^{*}}$, we can choose $\varphi _{z^{*}}\in S_{Y^{*}}$ such that

\[ z^{*}(z)=\varphi_{z^{*}}(g_{M_0}(z)) \ (z\in C_{M_0}^{X}) \text{ and } z^{*}(x)={\pm} \varphi_{z^{*}}(g_{M_0}(x)) \ (x\in X/C_{M_0}^{X}). \]

To see our conclusion, we shall prove that $f$ is phase equivalent to $g_{M_0}$. Now for every $M\in \mathfrak {M}(X)$, although it may occur that $M\cap M_0=\emptyset$, there are $M_1,\,\cdots ,\, M_m\subset \mathfrak {M}(X)$ such that $M_i\cap M_{i+1}\neq \emptyset$ for $i=0,\,1,\,\cdots ,\,m$, where we use the notation $M_{m+1}=M$. We deduce from Lemma 3.11 and $M_1\cap M_0\neq \emptyset$ that there is $\theta _1\in \{-1,\,1\}$ such that $g_{M_0}(x)=\theta _1 f_{M_1}(x)$ for all $x\in C_{M_0}^{X}\cap C_{M_1}^{X}$. This and the claim show that

\[ \varphi_{z^{*}}(g_{M_0}(y))=\varphi_{z^{*}}(\theta_1 f_{M_1}(y)) \]

for all $y\in C_{M_1}^{X}$. Inductively, there are $\{\theta _i\}_{i=2}^{m+1}\subset \{-1,\,1\}$ such that

\[ \varphi_{z^{*}}(g_{M_0}(y))=\varphi_{z^{*}}(\theta_1\ldots\theta_{m+1}f_{M_{m+1}}(y))=\varphi_{z^{*}}(\theta_1\cdots\theta_{m+1}f_{M}(y)) \]

for all $y\in C_{M}^{X}$. Set $\theta _M=\theta _1\cdots \theta _{m+1}$. Then

(4.2)\begin{equation} \varphi_{z^{*}}(g_{M_0}(y))=\varphi_{z^{*}}(\theta_M f_M(y)) \end{equation}

for all $y\in C_M^{X}$. Let $W$ be the set consisting of all $w^{*}$-exposed points of $B_{X^{*}}$. By Proposition 3.8, the set $\{\varphi _{z^{*}}\}_{z^{*}\in W}$ is a norming set of $Y$. So (4.2) ensures us that $g_{M_0}=\theta _M f_M$ on $C_{M}^{X}$. This completes the proof.

Proof. If $M_1=M_2$, then the desired conclusion is obviously true. Now we assume that $M_1\neq M_2$. Choose an element $z\in M_1$ which is not in $M_2\cup -M_2$. Since $B_X=\mbox {co}(M_2\cup -M_2)$, there are $x_1\in M_2,\, x_2\in -M_2$ and $\lambda \in (0,\,1)$ such that

\[ z=\lambda x_1+(1-\lambda)x_2. \]

The identity $x_{M_1}^{*}(z)=1$ implies that $x_1,\, x_2\in M_1$. We conclude from this that $x_1\in M_1\cap M_2\neq \emptyset$.

Proof. We may assume that $f_{M_1}(x_0)=f_{{M_2}}(x_0)$ for some $x_0\in M_1\cap {M_2}$. Namely, the value of $\theta$ is chosen to be $1$. Moreover, by Corollary 3.10, we can assume that $f_{M_1}$ and $f_{M_2}$ are defined with respect to $x_0$. If $z\in C_{M_2}^{X}\cap C_{M_1}^{X}$, then by Lemma 3.11, we have $f_{M_2}(z)=f_{M_1}(z)$. This gives the desired conclusion in this case. Note that $f(-x)=-f(x)$ for every $x\in X$. Then in the case of $z\in C_{M_2}^{X}\cap -C_{M_1}^{X}$, we have $f_{M_2}(z)=-f_{M_1}(-z)$ following from $\|f(z)+f(x_0)\|=\|z+x_0\|$ or $\|f(z)-f(x_0)\|=\|z+x_0\|$. This entails the desired result.

Now we handle the remaining case of $z\in C_{M_2}^{X}$ and $z \notin C_{M_1}^{X}\cup -C_{M_1}^{X}$. Since $X$ is a CL-space, there are $x_1\in M_1,\, x_2\in -M_1$ and $\lambda \in (0,\,1)$ such that

\[ \frac{z}{\|z\|}=\lambda x_1+(1-\lambda)x_2. \]

It follows that $x_1,\, x_2\in {M_2}$, and therefore

\[ f_{{M_2}}(\|z\|x_1)=f_{M_1}(\|z\|x_1) \text{ and } f_{{M_2}}(\|z\|x_2)={-}f_{M_1}(-\|z\|x_2). \]

This together with the choice of $y^{*}_{M_1}$ yields

\begin{align*} 2\|z\|&= y^{*}_{M_1}(f_{{M_2}}(\|z\|x_1)-f_{{M_2}}(\|z\|x_2))\\ &=y^{*}_{M_1}(f_{{M_2}}(\|z\|x_1)-f_{M_2}(z)) + y^{*}_{M_1}(f_{{M_2}}(z)-f_{{M_2}}(\|z\|x_2))\\ &\leq \|f_{{M_2}}(\|z\|x_1)-f_{M_2}(z)\|+\|f_{{M_2}}(z)-f_{{M_2}}(\|z\|x_2)\|\\ &=\big\|\|z\|x_1-z\big\|+\big\|z-\|z\|x_2\big\|=2\|z\|. \end{align*}

Thus, equality holds in the above, i.e., we have

\begin{align*} &y^{*}_{M_1}(f_{{M_2}}(\|z\|x_1)-f_{M_2}(z))=\|f_{{M_2}}(\|z\|x_1)-f_{{M_2}}(z)\|=\big\|\|z\|x_1-z\big\|,\\ &y^{*}_{M_1}(f_{{M_2}}(z)-f_{{M_2}}(\|z\|x_2))=\|f_{{M_2}}(z)-f_{{M_2}}(\|z\|x_2)\|=\big\|z-\|z\|x_2\big\|. \end{align*}

On the other hand, we get

\begin{align*} 2\|z\| &= x^{*}_{M_1}(\|z\|x_1-\|z\|x_2)=x^{*}_{M_1}(\|z\|x_1-z)+x^{*}_{M_1}(z-\|z\|x_2) \\ &\leq \big\|\|z\|x_1-z\big\|+\big\|z-\|z\|x_2\big\|=2\|z\|. \end{align*}

This allows us to assert that

\[ x^{*}_{M_1}(\|z\|x_1-z)=\big\|\|z\|x_1-z\big\|=y^{*}_{M_1}(f_{{M_2}}(\|z\|x_1)-f_{{M_2}}(z)). \]

Since we have $y^{*}_{M_1}(f_{{M_2}}(\|z\|x_1))=x^{*}_{M_1}(\|z\|x_1)=\|z\|$, it follows that $y^{*}_{M_1} (f_{{M_2}}(z))=x^{*}_{M_1}(z)$ as expected. The proof is complete.

Theorem 4.5 Let $X$ be a CL-space. Then $X$ has the Wigner property.

Proof. Note that $\mathfrak {M}(X)$ is the set consisting of all maximal convex subsets of $S_X$. One can easily see that

\[ X=\bigcup_{M\in \mathfrak{M}(X)}C_M^{X}. \]

By the axiom of choice, there is a set $\mathfrak {M}(X)^{+}\subset \mathfrak {M}(X)$ such that for every $M\in \mathfrak {M}(X)$ either $M$ or $-M$ belongs to $\mathfrak {M}(X)^{+}$ and $M_1\cap M_2\neq \emptyset$ for every $M_1,\, M_2\in \mathfrak {M}(X)^{+}$ by Lemma 4.3. Fix $M_0\in \mathfrak {M}(X)^{+}$. Lemma 3.11 guarantees that there is a set $\{\theta _M\}_{M\in \mathfrak {M}(X)^{+}}$ of signs such that $f_{{M_0}}(x_0)=\theta _M f_{M}(x_0)$ for some $x_0\in M\cap M_0$ and each $\theta _M$ does not depend on the choices $x_0$ in $M\cap M_0$. So we can define a phase-isometry $F: X\rightarrow Y$ given by

\[ F(x)= \left \{ \begin{array}{@{}ll} \theta_M f_{M}(x) & \text{if }x\in C_{M}^{X}\\ -\theta_M f_{{M}}({-}x) & \text{if } - x\in C_{M}^{X} \end{array} \right. \]

for all $M\in \mathfrak {M}(X)^{+}$. To see that $F$ is well defined, it remains to check that

\[ \theta_{M_1}f_{M_1}(x)= \theta_{M_2}f_{{M_2}}(x) \]

for all $x\in C_{M_1}^{X}\cap C_{M_2}^{X}$, where $M_1,\, M_2\in \mathfrak {M}(X)^{+}$. By Lemma 3.11, it suffices to show that there exists some $z_0\in M_1\cap M_2$ such that

\[ \theta_{M_1}f_{M_1}(z_0)= \theta_{M_2}f_{{M_2}}(z_0). \]

Lemma 4.4 ensures us the existence of a corresponding $y_{M_0}^{*}\in S_{Y^{*}}$ such that

\[ y_{M_0}^{*}(f_{M_1}(x))=\theta_{M_1}x_{M_0}^{*}(x)\text{ and } y_{M_0}^{*}(f_{{M_2}}(x))=\theta_{M_2}x_{M_0}^{*}(x), \]

for all $x\in M_1\cap M_2$. Consequently,

(4.3)\begin{equation} y_{M_0}^{*}(\theta_{M_1} f_{M_1}(x))= y_{M_0}^{*}(\theta_{M_2} f_{M_2}(x))=x_{M_0}^{*}(x) \end{equation}

for all $x\in M_1\cap M_2$. Choose $z_0\in M_1 \cap M_2$. If $x_{M_0}^{*}(z_0)\neq 0$, then (4.3) and $f_{M_1}(z_0)=\pm f_{{M_2}}(z_0)$ entail that $\theta _{M_1}f_{M_1}(z_0)= \theta _{M_2}f_{{M_2}}(z_0)$. If $x_{M_0}^{*}(z_0)=0$, then there are $z_1\in M_0,\, z_2\in -M_0$ and $\lambda \in (0,\,1)$ such that

\[ z_0=\lambda z_1+(1-\lambda)z_2. \]

We observe that $z_1\in M_1\cap M_2 \cap M_0$ is the desired element. Clearly, $F$ is a surjective phase-isometry, and it is phase-equivalent to $f$. By Lemma 3.1 and Proposition 3.8, we see that $F$ is bijective, $F(M)\in \mathfrak {M}(Y)$ and $F(C_M^{X})$ is a cone in $Y$ generated by $F(M)$.

We claim that for every $M\in \mathfrak {M}(X)$, the relation $y^{*}_M \circ F=\theta _{M} x^{*}_M$ holds on $X$. We can assume that $M\in \mathfrak {M}(X)^{+}$. By the above observation, we only need to check that

\[ y^{*}_M \circ F(z)= \theta_M x^{*}_M(z) \]

for every $z\in C_{M'}^{X}$ with $M'\in \mathfrak {M}(X)^{+}$. Note that $M\cap M'\neq \emptyset$ and $\theta _M f_M(v)=\theta _{M'}f_{M'}(v)$ for all $v\in M\cap M'$. Lemma 4.4 guarantees that

\[ y^{*}_M \circ F(z)= y_{M}^{*}(\theta_{M'} f_{M'}(z))=\theta_{M} x_{M}^{*}(z) \]

for all $z\in C_{M'}^{X}$. This finishes the claim.

We now prove that $F:X\rightarrow Y$ is a surjective isometry, and so $F$ is just the desired linear isometry by Mazur–Ulam theorem. Indeed, for all $u,\, v\in X$, it is easily seen that there is an $M\in \mathfrak {M}(X)$ such that $x^{*}_M(u-v)=\|u-v\|$. The claim implies that

\[ \|F(u)-F(v)\|\geq |y^{*}_M \circ F(u)-y^{*}_M \circ F(v)|=\|u-v\|. \]

Similarly, we have

\[ \|F(u)+F(v)\|\geq\|u+v\|. \]

Since $F$ is a phase-isometry, the previous two inequalities should be equalities. Namely $F:X\rightarrow Y$ is an isometry. The proof is complete.