Proof. If $G=0$ , we have $(x_0, \xi , e) \in P_{G}$ for any $(x_0, \xi , e) \in X \times \beta (\widetilde {X} \times \mathbb {B}(E^{*})) \times \mathbb {B}(E)$ . Thus, let $G \in \operatorname {Lip}(X, E)$ with $G \neq 0$ . Since $\beta (\widetilde {X} \times \mathbb {B}(E^{*}))$ is compact, there exists $\xi \in \beta (\widetilde {X} \times \mathbb {B}(E^{*}))$ such that $|\widetilde {G}(\xi )|=\|\widetilde {G}\|_{\infty }=L(G)$ . There are $x_0 \in X$ such that $\|G(x_0)\|_{E}=\|G\|_{\infty }$ and $\alpha \in \mathbb {C}$ with $|\alpha |=1$ such that $\alpha \widetilde {G}(\xi )=\|\widetilde {G}\|_{\infty }=L(G)$ . We get

$$ \begin{align*} \|\Gamma(G)(x_0, \xi, \frac{\alpha}{\|G(x_0)\|_{E}}G(x_0))\|_{E}&=\|G(x_0)+\widetilde{G}(\xi) \frac{\alpha}{\|G(x_0)\|_{E}}G(x_0)\|_{E}\\ &=(1+L(G)\frac{1}{\|G\|_{\infty}})\|G\|_{\infty}=\|G\|_{\infty}+L(G)=\|G\|_{L}. \end{align*} $$

This implies that $(x_0, \xi , \frac {\alpha }{\|G(x_0)\|_{E}}G(x_0) )\in P_{G}$ .

Proof. Let $e\in E$ . If $e=0$ , then $T(1\otimes e)=T(0)=0=1\otimes 0 \in 1\otimes E$ . Thus, we assume that $0 \neq e \in \mathbb {B}(E)$ . Fix $x' \in X$ , $(x,y)\in \widetilde {X}$ and $e^{*}\in \mathbb {B}(E^{*})$ . Let $\theta \in [0, 2\pi )$ . We obtain

(3) $$ \begin{align} &{\Gamma(1 \otimes e)(x',((x,y),e^{i\theta}e^{*}),e)}\nonumber\\ &=(1 \otimes e)(x')+e^{i\theta}e^{*}\left(\frac{(1 \otimes e)(x)- (1\otimes e)(y)}{d(x,y)}\right) e=e+0 e =e. \end{align} $$

This implies that

$$\begin{align*}\|\Gamma(1 \otimes e)(x',((x,y),e^{i\theta}e^{*}),e)\|_{E}=\|1\otimes e\|_{L}. \end{align*}$$

Thus, we get $(x',((x,y),e^{i\theta }e^{*}),e) \in P_{1\otimes e}$ . Choose a choice function $\Psi _{\theta }: \operatorname {Lip}(X, E) \to X\times \beta (\widetilde {X} \times \mathbb {B}(E^{*})) \times \mathbb {B}(E)$ such that

$$\begin{align*}\Psi_{\theta}(1 \otimes e)=(x',((x,y),e^{i\theta}e^{*}),e) \end{align*}$$

and define a semi-inner product $[\cdot ,\cdot ]_{\Psi _{\theta }L}$ on $\operatorname {Lip}(X, E)$ in the manner as in (2). Since T is a hermitian operator, we have $[T(1 \otimes e), 1\otimes e]_{\Psi _{\theta }L} \in \mathbb {R}$ . By (3), it follows that

(4) $$ \begin{align} \mathbb{R}&\ni [T(1 \otimes e), 1\otimes e]_{\Psi_{\theta}L}\nonumber\\ &=[\Gamma(T(1 \otimes e))(\Psi_{\theta}(1 \otimes e)), \Gamma(1 \otimes e)(\Psi_{\theta}(1 \otimes e))]_{E}\nonumber\\ &=[T(1 \otimes e)(x')+e^{i\theta}e^{*}\left(\frac{T(1 \otimes e)(x)-T(1 \otimes e)(y)}{d(x,y)}\right) e, e]_{E}\nonumber\\ &=[T(1 \otimes e)(x'),e]_{E}+e^{i\theta}e^{*}\left(\frac{T(1 \otimes e)(x)-T(1 \otimes e)(y)}{d(x,y)}\right)\|e\|^2_{E}. \end{align} $$

As $e \neq 0$ , we see that $\|e\|^{2}_{E}>0$ . Since $\theta \in [0, 2\pi )$ is arbitrary, it must be

(5) $$ \begin{align} e^{*}\left(\frac{T(1 \otimes e)(x)-T(1 \otimes e)(y)}{d(x,y)}\right)=0 \end{align} $$

for any $e^{*} \in \mathbb {B}(E^{*})$ . This implies

$$\begin{align*}\frac{T(1 \otimes e)(x)-T(1 \otimes e)(y)}{d(x,y)}=0 \end{align*}$$

for any $(x,y) \in \widetilde {X}$ . Thus, we deduce $L(T(1\otimes e))=0$ . Therefore, there exists $e_0 \in E$ such that $T(1\otimes e)=1 \otimes e_0$ .

Proof of Theorem 2.3

Suppose that T is of the form described as (7) in the statement of Theorem 2.3. To prove that T is a hermitian operator, we apply the fact that T is a hermitian if and only if $e^{itT}$ is a surjective isometry for every $t \in \mathbb {R}$ ; see [Reference Fleming and Jamison6, Theorem 5.2.6]. Let $t \in \mathbb {R}$ . By the definition of T, we have

$$\begin{align*}e^{itT}(F)(x)=e^{it\phi}(F(x)) \end{align*}$$

for any $F \in \operatorname {Lip}(X, E)$ and $x \in X$ . Since $\phi $ is a hermitian on E, $e^{it\phi }$ is a surjective isometry. This implies that $\|e^{itT}(F)\|_{\infty }=\|F\|_{\infty }$ and $L(e^{itT}(F))=L(F).$ Thus, we deduce $\|e^{itT}(F)\|_{L}=\|F\|_{L}$ for any $F \in \operatorname {Lip}(X, E)$ . Since $e^{itT}$ is a surjective isometry for every $t \in \mathbb {R}$ , we conclude T is a hermitian operator. We prove the converse. Suppose that $T:\operatorname {Lip}(X, E) \to \operatorname {Lip}(X, E)$ is a hermitian operator. Let $\phi $ be the operator defined by (6). A similar argument as above yields an operator from $\operatorname {Lip}(X, E)$ into itself given by $F \mapsto \phi \circ F$ is a hermitian operator. Hence, we define a hermitian operator $T_0: \operatorname {Lip}(X, E) \to \operatorname {Lip}(X, E)$ by

$$\begin{align*}T_0 (F)(x)=T(F)(x)-\phi(F(x)) \end{align*}$$

for all $F \in \operatorname {Lip}(X, E)$ and $x \in X$ . We shall prove that $T_0=0$ on $\operatorname {Lip}(X, E)$ in two steps.

Note that the same idea with [Reference Oi14, Theorem 2.2] is valid even if we replace a finite dimensional Banach space E with a Banach space E.

By [Reference Fleming and Jamison6, p. 10], there is a semi-inner product $[\cdot ,\cdot ]_{E}$ on E compatible with the norm such that $[e_1,\lambda e_2]_{E}=\bar \lambda [e_1,e_2]_{E}$ for any $e_i \in E$ and $\lambda \in \mathbb {C}$ . Let $e \in \mathbb {S}(E)$ . We define a map $S_{e}: \operatorname {Lip}(X) \to \operatorname {Lip}(X)$ by

$$\begin{align*}S_{e}(f)(x)=[T_0 (f \otimes e)(x),e]_{E}, \quad f \in \operatorname{Lip}(X),\,\, x \in X. \end{align*}$$

By simple calculations, we have that $S_e$ is a bounded linear operator with $\|S_e\| \le \|T_0\|$ . Moreover, we shall prove that $S_e$ is a hermitian operator. Let $t \in \mathbb {R}$ . By the definition of $S_{e}$ , we get $(I+itS_{e})(1)(x)=1$ for any $x \in X$ . This implies that

(8) $$ \begin{align} 1 \le \|I+itS_{e}\|. \end{align} $$

However, let $f \in \operatorname {Lip}(X)$ . We obtain for any $x, y \in X$ ,

$$ \begin{align*} |(I+itS_{e})(f)(x)|\le \|(I+itT_0)(f \otimes e)\|_{\infty} \end{align*} $$

and

$$ \begin{align*} |(I+itS_{e})(f)(x)-(I+itS_{e})(f)(y)| \le L((I+itT_0)(f \otimes e))d(x,y). \end{align*} $$

Therefore, we get

$$ \begin{align*} \|(I+itS_{e})(f)\|_{L} &\le \|(I+itT_0)(f \otimes e)\|_{\infty}+L((I+itT_0)(f \otimes e))\\ &\le \|I+itT_0\|\|f\otimes e\|_{L}=\|I+itT_0\|\|f\|_{L} \end{align*} $$

for any $f \in \operatorname {Lip}(X)$ . We conclude that

(9) $$ \begin{align} \|I+itS_{e}\| \le \|I+itT_0\|. \end{align} $$

Since $T_0$ is a hermitian operator on $\operatorname {Lip}(X, E)$ , we have $\|I+itT_0\|=1+o(t)$ by [Reference Fleming and Jamison6, Theorem 5.2.6]. By (8) and (9), we see that

$$\begin{align*}1 \le \|I+itS_{e}\| \le \|I+itT_0\|=1+o(t). \end{align*}$$

This implies that $S_{e}: \operatorname {Lip}(X) \to \operatorname {Lip}(X)$ is a hermitian operator. By [Reference Botelho, Jamison, Jiménez-Vargas and Villegas-Vallecillos3, Theorem 3.1.], we have that $S_e$ is a real multiple of the identity. Since $S_{e}(1)(x)=[T_0(1\otimes e)(x), e]=0$ , we deduce $S_{e}(f)(x)=0f(x)=0$ for any $f \in \operatorname {Lip}(X)$ and $x \in X$ . This implies that $[T_0(f \otimes e)(x), e]_{E}=0$ for all $f \in \operatorname {Lip}(X)$ and $x \in X$ . As $e \in \mathbb {S}(E)$ is arbitrary, we obtain

(10) $$ \begin{align} [T_0(f \otimes e)(x), e]_{E}=0, \quad e \in E, \quad f \in \operatorname{Lip}(X), \quad x \in X. \end{align} $$

Let $f \in \operatorname {Lip}(X)$ and $x \in X$ . Then we define a map $S_{fx}: E \to E$ by $S_{fx}(e)=T_0(f \otimes e)(x)$ for any $e \in E$ . Since $T_0$ is a bounded linear operator, $S_{fx}$ is also a bounded linear operator with $\|S_{fx}\| \le \|T_0\|\|f\|_{L}$ . By (10), we have $[S_{fx}(e),e]_{E}=[T_0(f \otimes e)(x), e]_{E}=0$ for all $e \in E$ . Applying [Reference Lumer11, Theorem 5], we have $T_0(f \otimes e)(x)=S_{fx}(e)=0$ for any $e \in E$ . As $f \in \operatorname {Lip}(X)$ and $x \in X$ are arbitrary, we conclude step 1.

If $F \in \operatorname {Lip}(X) \otimes E$ , step 1 yields that $T_0(F)=0$ by the linearity of $T_0$ . Thus, it suffices to show $T_0(F)=0$ holds for any $F \in \operatorname {Lip}(X, E) \setminus \operatorname {Lip}(X) \otimes E$ . Let $F \in \operatorname {Lip}(X, E) \setminus \operatorname {Lip}(X) \otimes E$ with $F(x_0)=0$ . For any $e \in \mathbb {S}(E)$ , put

$$\begin{align*}G_e=(\|F\|_{\infty}-|F|)\otimes e +F, \end{align*}$$

where $|F|(x):=\|F(x)\|_{E}$ and $|F| \in \operatorname {Lip}(X)$ . Then we have

$$\begin{align*}G_{e}(x_0)=\|F\|_{\infty}e \end{align*}$$

and

$$ \begin{align*} &\|G_{e}(x)\|_{E}=\|(\|F\|_{\infty}-\|F(x)\|_{E})\otimes e +F(x)\|_{E} \\ &\le \|F\|_{\infty}-\|F(x)\|_{E}+\|F(x)\|_{E} = \|F\|_{\infty} \end{align*} $$

for any $x \in X$ . Thus, we obtain $\|G_e(x_0)\|_{E}=\|F\|_{\infty }=\|G_e\|_{\infty }$ . As $\beta (\widetilde {X} \times \mathbb {B}(E^{*})) $ is compact, there are $\xi \in \beta (\widetilde {X} \times \mathbb {B}(E^{*})) $ and $\alpha \in \mathbb {C}$ with $|\alpha |=1$ such that $\alpha \widetilde {G_e}(\xi )=L(G_e)$ . This implies that $(x_0, \xi , \alpha e) \in P_{G_e}$ . We choose a choice function $\Psi _{e}:\operatorname {Lip}(X, E) \to X \times \beta (\widetilde {X} \times \mathbb {B}(E^{*})) \times \mathbb {B}(E)$ such that $\Psi _e(G_e)=(x_0, \xi , \alpha e)$ and define a semi-inner product $[\cdot ,\cdot ]_{\Psi _e L}$ in the manner as in (2). Since $T_0: \operatorname {Lip}(X,E) \to \operatorname {Lip}(X,E)$ is a hermitian operator, we get

$$ \begin{align*} \mathbb{R} &\ni [T_0(G_e),G_e]_{\Psi_e L}= [T_0(F),G_e]_{\Psi_e L}\\ &\quad= [T_0(F)(x_0)+\alpha \widetilde{T_0(F)}(\xi)e,\|F\|_{\infty}e+L(G_e)e]_{E}\\ &\qquad=(e^{*}(T_0(F)(x_0))+\alpha \widetilde{T_0(F)}(\xi))\|G_{e}\|_{L}, \end{align*} $$

where $e^{*} \in \mathbb {B}(E^{*})$ with $e^{*}(e)=1$ for any $e \in \mathbb {S}(E)$ . We have

(11) $$ \begin{align} e^{*}(T_0(F)(x_0))+ \alpha \widetilde{T_0(F)}(\xi) \in \mathbb{R}. \end{align} $$

However, there exists $y_0 \in X$ such that $\|F(y_0)\|_{E}=\|F\|_{\infty }\neq 0$ , and there is $f_{y_0} \in \mathbb {S}(E)$ such that $F(y_0)=\|F\|_{\infty }f_{y_0}$ . We get

$$\begin{align*}G_e(y_0)=F(y_0)=\|F\|_{\infty}f_{y_0}. \end{align*}$$

This implies that $\|G_e(y_0)\|_{E}=\|F(y_0)\|_{E}=\|F\|_{\infty }=\|G_{e}\|_{\infty }$ . We have

$$ \begin{align*} &\|\Gamma (G_{e})(y_0,\xi,\alpha f_{y_0})\|_{E}=\|G_e(y_0)+\alpha \widetilde{G_e}(\xi)f_{y_0}\|_{E}\\ &\qquad=\|\|F\|_{\infty}f_{y_0}+L(G_e)f_{y_0}\|_{E}=\|F\|_{\infty}+L(G_e)=\|G_e\|_{L}. \end{align*} $$

Thus, we get $(y_0, \xi , \alpha f_{y_0}) \in P_{G_{e}}$ . In the same manner, there is a choice function $\Psi _{f_{y_0}}:\operatorname {Lip}(X, E) \to X \times \beta (\widetilde {X} \times \mathbb {B}(E^{*})) \times \mathbb {B}(E)$ such that $\Psi _{f_{y_0}}(G_e)=(y_0, \xi , \alpha f_{y_0}) $ , and we can define a semi-inner product $[\cdot ,\cdot ]_{\Psi _{f_{y_0}}L}$ on $\operatorname {Lip}(X, E)$ . It follows that

$$ \begin{align*} \mathbb{R} &\ni [T_0(G_e),G_e]_{\Psi_{f_{y_0}}L}= [T_0(F),G_e]_{\Psi_{f_{y_0}}L}\\ &\quad= [T_0(F)(y_0)+\alpha \widetilde{T_0(F)}(\xi)f_{y_0},\|G_{e}\|_{L}f_{y_0}]_{E}\\&\qquad=({f_{y_0}}^{*}(T_0(F)(y_0))+\alpha \widetilde{T_0(F)}(\xi))\|G_{e}\|_{L}, \end{align*} $$

where ${f_{y_0}}^{*} \in \mathbb {B}(E^{*})$ with ${f_{y_0}}^{*} (f_{y_0})=1$ . We obtain

(12) $$ \begin{align} {f_{y_0}}^{*}(T_0(F)(y_0))+\alpha \widetilde{T_0(F)}(\xi) \in \mathbb{R}. \end{align} $$

By (11) and (12), we get $e^{*}(T_0(F)(x_0))-{f_{y_0}}^{*}(T_0(F)(y_0)) \in \mathbb {R}$ . Since $e \in \mathbb {S}(E)$ is arbitrary, it follows that $T_0(F)(x_0)=0$ . Let $F \in \operatorname {Lip}(X, E) \setminus \operatorname {Lip}(X) \otimes E$ and $x \in X$ . We define $F_x=F-1 \otimes F(x)$ . Since $F_x(x)=0$ , we get

$$\begin{align*}0=T_0(F_x)(x)=T_0(F)(x)-T_0(1 \otimes F(x))(x)=T_0(F)(x). \end{align*}$$

Thus, we have $T_0(F)=0$ for any $F \in \operatorname {Lip}(X, E)$ and conclude step 2.

Therefore, we obtain $T(F)(x)=\phi (F(x))$ for any $F \in \operatorname {Lip}(X, E)$ . This completes the proof.

Proof. We conclude this Lemma by the Hahn $-$ Banach theorem immediately.

Proof. It follows from the Mazur $-$ Ulam theorem that every surjective isometry U between two normed spaces with $U(0)=0$ is a real linear isometry. By the maximality of T-sets and surjectivity of U, we conclude that U preserves T-sets.

Proof. For any $F_1,\cdots ,F_n \in S(x,\mathbb {U},\mathbb {T})$ , we have $F_i(x) \in \mathbb {U}$ and $\|F_{i}(x)\|_{E}=\|F_i\|_{\infty }$ for any $i=1,\cdots ,n$ . We get

$$\begin{align*}\|\Sigma^{n}_{i=1} F_i\|_{\infty} \le \Sigma^{n}_{i=1} \|F_i\|_{\infty} =\Sigma^{n}_{i=1} \|F_i(x)\|_{E}=\|\Sigma^{n}_{i=1} F_i(x)\|_{E} \le \|\Sigma^{n}_{i=1} F_i\|_{\infty}. \end{align*}$$

This implies that $\|\Sigma ^{n}_{i=1} F_i\|_{\infty }=\Sigma ^{n}_{i=1}\|F_i\|_{\infty }$ . Since $F_i \in \mathbb {T}$ for any $i=1,\cdots ,n$ , we also get $L(\Sigma ^{n}_{i=1} F_i)=\Sigma ^{n}_{i=1} L(F_i)$ . This implies that $\|\Sigma ^{n}_{i=1} F_i\|_{L}=\Sigma ^{n}_{i=1}\|F_i\|_{L}$ .

Proof. For any $F \in \mathcal {S}$ , we write $P(F):=\{x \in X : \|F(x)\|_{E}=\|F\|_{\infty }\}$ . We shall show that $\bigcap _{F \in \mathcal {S}}P(F)\neq \emptyset $ . For any finite collection $F_1,\cdots ,F_n \in \mathcal {S}$ , we get $\|\Sigma ^{n}_{i=1} F_i\|_{L}=\Sigma ^{n}_{i=1}\|F_i\|_{L}$ . As $\|\Sigma ^{n}_{i=1} F_i\|_{\infty } \le \Sigma ^{n}_{i=1} \|F_i\|_{\infty }$ and $ L(\Sigma ^{n}_{i=1} F_i) \le \Sigma ^{n}_{i=1} L(F_i)$ , we have $\|\Sigma ^{n}_{i=1} F_i\|_{\infty }=\Sigma ^{n}_{i=1}\|F_i\|_{\infty }$ . Since $\Sigma ^{n}_{i=1} F_i \in \operatorname {Lip}(X, E)$ , there is $x \in X$ such that $\|(\Sigma ^{n}_{i=1} F_i)(x)\|_{E}=\|\Sigma ^{n}_{i=1} F_i\|_{\infty }$ . Thus, we get

$$\begin{align*}\Sigma^{n}_{i=1} \|F_i\|_{\infty}=\|\Sigma^{n}_{i=1} F_i\|_{\infty}=\|(\Sigma^{n}_{i=1} F_i)(x)\|_{E} \le \Sigma^{n}_{i=1}\|F_i(x)\|_{E} \le \Sigma^{n}_{i=1} \|F_i\|_{\infty}. \end{align*}$$

This implies that $\|F_i\|_{\infty }=\|F_i(x)\|_{E}$ for any $i=1,\cdots ,n$ and $x \in \bigcap ^{n}_{i=1}P(F_i)$ . Since X is compact and $P(F)$ is a closed set for each $F \in \mathcal {S}$ , we have $\bigcap _{F \in \mathcal {S}}P(F) \neq \emptyset $ by the finite intersection property.

Let $x \in \bigcap _{F \in \mathcal {S}}P(F) $ . We consider the set

$$\begin{align*}R_{x}(\mathcal{S}):=\{F(x) \in E : F \in \mathcal{S}\}. \end{align*}$$

Choose any finite collection $F_1(x),\cdots ,F_n(x) \in R_{x}(\mathcal {S})$ . Since $\Sigma ^{n}_{i=1} F_i \in \mathcal {S}$ , we have $x \in P(\Sigma ^{n}_{i=1} F_i )$ . This implies

$$\begin{align*}\Sigma^{n}_{i=1} \|F_i\|_{\infty}=\|\Sigma^{n}_{i=1} F_i\|_{\infty}=\|\Sigma^{n}_{i=1} F_i(x)\|_{E} \le \Sigma^{n}_{i=1}\|F_i(x)\|_{E}=\Sigma^{n}_{i=1} \|F_i\|_{\infty}. \end{align*}$$

Thus, we have $\|\Sigma ^{n}_{i=1} F_i(x)\|_{E} = \Sigma ^{n}_{i=1}\|F_i(x)\|_{E}$ , which means that there is a T-set $\mathbb {U}$ of E such that $R_{x}(\mathcal {S}) \subset \mathbb {U}$ . Therefore, for any $F \in \mathcal {S}$ , we have $F(x) \in \mathbb {U}$ and $\|F(x)\|_{E}=\|F\|_{\infty }$ .

Since $L(\Sigma ^{n}_{i=1} F_i)=\Sigma ^{n}_{i=1} L(F_i)$ for any finite collection $F_1,\cdots ,F_n \in \mathcal {S}$ , there exists a T-set $\mathbb {T}$ of $\operatorname {Lip}(X, E)$ with respect to $L(\cdot )$ such that $\mathcal {S} \subset \mathbb {T}$ . This implies that $\mathcal {S} \subset S(x,\mathbb {U},\mathbb {T})$ . By Lemma 3.4 and maximality of $\mathcal {S}$ , we conclude that $\mathcal {S}=S(x,\mathbb {U},\mathbb {T})$ .

Proof. Suppose that $U(F)(y_0)\neq 0$ . Put $a=U(F)(y_0) / \|U(F)(y_0)\|_{E_2}$ . The map from $\mathbb {S}(E_{2})$ to $\mathbb {R}$ defined by

$$\begin{align*}e \mapsto \|U(F)(y_0)+(\|U(F)\|_{\infty}+1)e\|_{E_2} \end{align*}$$

is continuous. Since $\|U(F)(y_0)\|_{E_2} \neq 0$ , we have

$$ \begin{align*} &\|U(F)(y_0)+(\|U(F)\|_{\infty}+1)a\|_{E_2}=\left \|\frac{U(F)(y_0)}{\|U(F)(y_0)\|_{E_2}}(\|U(F)(y_0)\|_{E_2}+\|U(F)\|_{\infty}+1) \right\|_{E_2}\\&\quad= \|U(F)(y_0)\|_{E_2}+\|U(F)\|_{\infty}+1>\|U(F)\|_{\infty}+1. \end{align*} $$

There exists $\delta>0$ such that if $e \in \mathbb {S}(E_2)$ with $\|a-e\|_{E_2}<\delta $ , then $\|U(F)(y_0)+(\|U(F)\|_{\infty }+1)e\|_{E_2}>\|U(F)\|_{\infty }+1$ . We choose $\theta \in (0,2\pi )$ such that $|e^{i\theta }-1|<\delta $ . We write $e_{\theta }:=\psi ^{-1}(e^{i\theta }a)$ . This implies that

(13) $$ \begin{align} \|U(F)(y_0)+(\|U(F)\|_{\infty}+1)\psi(e_{\theta})\|_{E_2}>\|U(F)\|_{\infty}+1. \end{align} $$

For any $n \in \mathbb {N}$ , we define $g_n \in \operatorname {Lip}(X_1)$ by

$$\begin{align*}g_n(x)=(\|U(F)\|_{\infty}+1)\max\{1-nL(F)d(x,x_0),0\}, \quad x \in X_1. \end{align*}$$

By Zorn’s lemma, there is $\mathcal {S}_n$ which is a T-set of $\operatorname {Lip}(X_1, E_1)$ with respect to $\|\cdot \|_{L}$ such that $F+g_n\otimes e_{\theta } \in \mathcal {S}_n$ . We have

$$\begin{align*}(F+g_n\otimes e_{\theta})(x_0)=(\|U(F)\|_{\infty}+1)e_{\theta}. \end{align*}$$

When $x \neq x_0$ and $1-nL(F)d(x,x_0)\ge 0$ , we have

$$ \begin{align*} &\|(F+g_n\otimes e_{\theta})(x)\|_{E_1}=\|F(x)-F(x_0)+(g_n\otimes e_{\theta})(x)\|_{E_1}\\ &\quad\le L(F)d(x,x_0)+(\|U(F)\|_{\infty}+1)(1-nL(F)d(x,x_0))\\&\qquad\quad=(1-n(\|U(F)\|_{\infty}+1)) L(F)d(x,x_0)+\|U(F)\|_{\infty}+1< \|U(F)\|_{\infty}+1. \end{align*} $$

When $1-nL(F)d(x,x_0)\le 0$ , we have

$$\begin{align*}\|(F+g_n\otimes e_{\theta})(x)\|_{E_1}=\|F(x)\|_{E_1}\le 1< \|U(F)\|_{\infty}+1. \end{align*}$$

Thus, we obtain $P(F+g_n\otimes e_{\theta } )=\{x_0\}$ . By Proposition 3.5, there are T-set $\mathbb {U}_n \subset E_1$ and T-set $\mathbb {T}_n \subset \operatorname {Lip}(X_1, E_1)$ such that $ F+g_n\otimes e_{\theta } \in \mathcal {S}_n=S(x_0,\mathbb {U}_n,\mathbb {T}_n)$ . In particular, we have

$$\begin{align*}(\|U(F)\|_{\infty}+1)e_{\theta}=(F+g_n\otimes e_{\theta})(x_0) \in \mathbb{U}_n. \end{align*}$$

By Lemma 3.2, $e_{\theta } \in \mathbb {U}_n$ . Since U is a surjective isometry with $U(0)=0$ , Lemma 3.3 shows that there are $y_n \in X_2$ , T-set $\mathbb {V}_n \subset E_2$ and T-set $\mathbb {T'}_n \subset \operatorname {Lip}(X_2, E_2)$ with respect to $L(\cdot )$ such that $U(S(x_0,\mathbb {U}_n,\mathbb {T}_n))=S(y_n,\mathbb {V}_n,\mathbb {T'}_n)$ . Since $e_{\theta } \in \mathbb {U}_n$ , we have $1 \otimes e_{\theta } \in S(x_0,\mathbb {U}_n,\mathbb {T}_n) $ . By the assumption, we have $U(1 \otimes e_{\theta })=1 \otimes \psi (e_{\theta }) \in S(y_n,\mathbb {V}_n,\mathbb {T'}_n)$ . It implies that $\psi (e_{\theta }) \in \mathbb {V}_n$ for any $n \in \mathbb {N}$ . For any $y \in X_2$ ,

$$ \begin{align*} &U(F+g_n \otimes e_{\theta})(y)=U(F)(y)+\psi(g_n(\varphi(y))e_{\theta})\\&\quad=U(F)(y)+(\|U(F)\|_{\infty}+1)\max\{1-nL(F)d(\varphi(y),x_0),0\}\psi(e_{\theta}). \end{align*} $$

We shall show that the sequence $\{y_n\}$ converges $y_0$ as $n \to \infty $ . Suppose that there exists $n \in \mathbb {N}$ such that $1-nL(F)d(\varphi (y_n),x_0) < 0$ . Since $U(F+g_n\otimes e_{\theta }) \in S(y_n,\mathbb {V}_n,\mathbb {T'}_n)$ , we have

(14) $$ \begin{align} \|U(F)\|_{\infty} &\ge \|U(F)(y_n)\|_{E_2}\nonumber\\&=\|(U(F)+U(g_n\otimes e_{\theta}))(y_n)\|_{E_2}=\|U(F)+U(g_n\otimes e_{\theta})\|_{\infty}. \end{align} $$

Moreover, we get $g_n(x_0)=(\|U(F)\|_{\infty }+1)\max \{1-nL(F)d(x_0,x_0),0\}=\|U(F)\|_{\infty }+1$ . Since $\varphi (y_0)=x_0$ , we have $U(g_n\otimes e_{\theta })(y_0)=\psi (g_n(\varphi (y_0)) e_{\theta })=(\|U(F)\|_{\infty }+1)\psi (e_{\theta })$ . This implies that

(15) $$ \begin{align} \|U(F)+U(g_n\otimes e_{\theta})\|_{\infty} &\ge \|U(F)(y_0)+(\|U(F)\|_{\infty}+1)\psi(e_{\theta})\|_{E_2}\nonumber\\&>\|U(F)\|_{\infty}+1, \end{align} $$

where the last inequality follows by (13). By (14) and (15), we have $\|U(F)\|_{\infty }>\|U(F)\|_{\infty }+1$ . This is a contradiction. Thus, for every $n \in \mathbb {N}$ , we have

$$\begin{align*}1-nL(F)d(\varphi(y_n),x_0) \ge 0. \end{align*}$$

Thus, we get $1/ nL(F)>d(\varphi (y_n),x_0)=d(\varphi (y_n),\varphi (y_0))=d(y_n,y_0)$ . This implies that $y_n \to y_0$ as $n \to \infty $ . Since $U(F)\in \operatorname {Lip}(X_2, E_2)$ , we get $U(F)(y_n) \to U(F)(y_0)$ .

Because we obtain $0\le 1-nL(F)d(\varphi (y_n),x_0) \le 1$ , the sequence $\{ 1-nL(F)d(\varphi (y_n),x_0)\}$ has a convergent subsequence. Without loss of generality, we can assume that the sequence converges to $\beta \in [0,1]$ as $n \to \infty $ . We write

$$\begin{align*}c_n:=U(F+g_n\otimes e_{\theta})(y_n)=U(F)(y_n)+(\|U(F)\|_{\infty}+1)(1-nL(F)d(\varphi(y_n),x_0))\psi(e_{\theta}) \end{align*}$$

and

(16) $$ \begin{align} c_0:=U(F)(y_0)+(\|U(F)\|_{\infty}+1)\beta \psi(e_{\theta}). \end{align} $$

We obtain that

(17) $$ \begin{align} \|c_n-c_0\|_{E_2} \to 0 \ \quad \text{if} \quad n \to \infty. \end{align} $$

As $U(F+g_n\otimes e_{\theta }) \in S(y_n,\mathbb {V}_n,\mathbb {T'}_n)$ , we get $c_n \in \mathbb {V}_n$ . Since $\psi (e_{\theta }) \in \mathbb {V}_n$ , we have $\|c_n+\psi (e_{\theta })\|_{E_2}=\|c_n\|_{E_2}+\|\psi (e_{\theta })\|_{E_2}$ . By (17), we get $\|c_0+\psi (e_{\theta })\|_{E_2}=\|c_0\|_{E_2}+\|\psi (e_{\theta })\|_{E_2}$ . As $\psi (e_{\theta })=e^{i\theta }a$ , we obtain

$$\begin{align*}\|e^{-i\theta}c_0+a\|_{E_2}=\|e^{-i\theta}c_0\|_{E_2}+\|a\|_{E_2}. \end{align*}$$

Thus, there is $\tau \in E_2^{*}$ such that $\|\tau \|=1$ , $\tau (e^{-i\theta }c_0)=\|c_0\|_{E_2}$ and $\tau (a)=\|a\|_{E_2}=1$ . By (16) and $a=U(F)(y_0) / \|U(F)(y_0)\|_{E_2}$ , we have

$$ \begin{align*} &e^{i\theta}\|c_0\|_{E_2}=\tau(c_0)=\tau(U(F)(y_0))+\tau((\|U(F)\|_{\infty}+1)\beta e^{i\theta}a)\\ &\quad=\|U(F)(y_0)\|_{E_2}+e^{i\theta}(\|U(F)\|_{\infty}+1)\beta. \end{align*} $$

We obtain that $\|U(F)(y_0)\|_{E_2}=e^{i\theta }(\|c_0\|_{E_2}-(\|U(F)\|_{\infty }+1)\beta )$ . As $\theta \in (0,2\pi )$ and $\|c_0\|_{E_2}-(\|U(F)\|_{\infty }+1)\beta \in \mathbb {R}$ , we conclude $U(F)(y_0)=0$ .

Proof of Proposition 3.6

By the assumption, it suffices to show that $U(F)(y)=\psi (F(\varphi (y)))$ holds for any $F \in \operatorname {Lip}(X_1, E_1)$ in which F is not a constant map. For any $x \in X_1$ , we define $G:=F-1 \otimes F(x)$ . Then we have $G(x)=0$ . As $G \neq 0$ , without loss of generality, we assume that $\|G\|_{\infty }=1$ . By Lemma 3.7, we obtain $U(G)(\varphi ^{-1}(x))=0$ . This implies that $U(F)(\varphi ^{-1}(x))=U(1 \otimes F(x))(\varphi ^{-1}(x))=\psi (1(x)F(x))=\psi (F(x))$ .

Proof. Let $h \in H(\mathcal {A}_1)$ . We apply Proposition 4.3 to obtain $M_{1 \otimes h}$ is a hermitian operator on $\operatorname {Lip}(X_1,\mathcal {A}_1)$ . It follows from Proposition 4.4 that $UM_{1 \otimes h}U^{-1}$ is a hermitian operator on $\operatorname {Lip}(X_2,\mathcal {A}_2)$ . By applying Proposition 4.3 again, there exists $h' \in H(\mathcal {A}_2)$ and a $*$ -derivation D on $\mathcal {A}_2$ such that

(20) $$ \begin{align} UM_{1 \otimes h}U^{-1}=M_{1 \otimes h'}+i\widehat{D}. \end{align} $$

For any $y \in X_2$ , we have

$$\begin{align*}(UM_{1 \otimes h}U^{-1})(1)(y)=UM_{1 \otimes h}(1)(y)=U(1 \otimes h)(y) \end{align*}$$

and

$$\begin{align*}M_{1 \otimes h'}(1)(y)+i\widehat{D}(1)(y)=(1\otimes h')(y)+iD(1(y))=h'+i0=h'. \end{align*}$$

This implies that $U(1\otimes h)=1\otimes h'$ .

Proof. For any $h_2 \in H(\mathcal {A}_2)$ , by Proposition 4.4, we have that $U^{-1}M_{1 \otimes h_2}U$ is a hermitian operator on $\operatorname {Lip}(X_1,\mathcal {A}_1)$ . By Proposition 4.3, there are $h_1 \in H(\mathcal {A}_1)$ and a $*$ -derivation $D_1$ on $\mathcal {A}_1$ such that

$$\begin{align*}U^{-1}M_{1 \otimes h_2}U=M_{1 \otimes h_1}+i\widehat{D_1}. \end{align*}$$

Since we have $M_{1\otimes h_1}=U^{-1}M_{1 \otimes h_2}U-i\widehat {D_1}$ , we get

$$ \begin{align*} UM_{1\otimes h_1}U^{-1}(1)&= U(U^{-1}M_{1 \otimes h_2}U-i\widehat{D_1})U^{-1}(1)\\ &=M_{1 \otimes h_2}(1)-U(i\widehat{D_1}(1)) =1 \otimes h_2-iU(0) =1 \otimes h_2. \end{align*} $$

We obtain $U(1 \otimes h_1)=1 \otimes h_2$ and $\psi _0(h_1)=h_2$ . It follows that $\psi _0$ is surjective. For any $h \in H(\mathcal {A}_1)$ , we get $\|\psi _0(h)\|_{\mathcal {A}_2}=\|1 \otimes \psi _0(h)\|_{L}=\|U(1 \otimes h)\|_{L} =\|1 \otimes h \|_{L}=\|h\|_{\mathcal {A}_1}$ . Thus, we have $\psi _0$ is an isometry. Since U is a linear map, it is easy to see that $\psi _0$ is real linear. Moreover, $U(1)=1$ ; we get $\psi _0(1)=1$ .

Proof. By (22), we have $\psi $ is a complex linear isometry with $\psi (1)=1$ . Therefore, it suffices to show $\psi $ is surjective. For any $a \in \mathcal {A}_2$ , there exists $h_1$ , $h_2 \in H(\mathcal {A}_2)$ such that $a=h_1+ih_2$ . Since Lemma 4.6 shows that $\psi _0: H(\mathcal {A}_1) \to H(\mathcal {A}_2)$ is surjective, there are $h^{'}_1, h^{'}_2 \in H(\mathcal {A}_1)$ such that $\psi _0(h^{'}_1)=h_1$ and $\psi _0(h^{'}_2)=h_2$ . Then we get $a'=h^{'}_1+ih^{'}_2 \in \mathcal {A}_1$ . This implies that

$$\begin{align*}\psi(a')=\psi_0(h^{'}_1)+i\psi_0(h^{'}_2)=h_1+ih_2=a. \end{align*}$$

This completes the proof.

Proof. For any $b \in \mathcal {A}_2$ with $b^{*}=-b$ , we define a $*$ -derivation D on $\mathcal {A}_2$ by

$$\begin{align*}D(a)=ba-ab, \quad a \in \mathcal{A}_2. \end{align*}$$

Note that Proposition 4.3 shows that the map $i\widehat {D}: \operatorname {Lip}(X_2,\mathcal {A}_2) \to \operatorname {Lip}(X_2,\mathcal {A}_2)$ defined by

$$\begin{align*}(i\widehat{D})(F)(y)=iD(F(y)) \quad F \in \operatorname{Lip}(X_2,\mathcal{A}_2), \quad y \in X_2 \end{align*}$$

is a hermitian operator on $\operatorname {Lip}(X_2,\mathcal {A}_2)$ . Since the map U is an isometry, $U^{-1}i\widehat {D}U$ is a hermitian operator on $\operatorname {Lip}(X_1,\mathcal {A}_1)$ . By Proposition 4.3, there exists $h \in H(\mathcal {A}_1)$ and $*$ -derivation $D'$ on $\mathcal {A}_1$ such that

$$\begin{align*}U^{-1}i\widehat{D}U=M_{1 \otimes h}+i\widehat{D'}. \end{align*}$$

As $U(1)=1$ , we get

$$\begin{align*}(U^{-1}i\widehat{D}U)(1)=i(U^{-1}\widehat{D}U)(1)=iU^{-1}\widehat{D}(1)=iU^{-1}(0)=0. \end{align*}$$

This implies that

$$ \begin{align*} 0=(U^{-1}i\widehat{D}U)(1)=(M_{1 \otimes h}&+i\widehat{D'})(1)\\ &=1 \otimes h+i\widehat{D'}(1)=1\otimes h +i0=1\otimes h. \end{align*} $$

Thus, we have $U^{-1}i\widehat {D}U=i\widehat {D'}$ . This implies that for any $f \in \operatorname {Lip}(X_1)$ , we have $(U^{-1}i\widehat {D}U)(f\otimes 1)(x)=i\widehat {D'}(f \otimes 1)(x)=0$ for all $x \in X_1$ . In addition, by the definition of D, we get

(23) $$ \begin{align} (U^{-1}i\widehat{D}U)(f\otimes 1)&=U^{-1}(i \widehat{D}U(f \otimes1))\nonumber\\ &=iU^{-1}(1 \otimes b U(f \otimes 1)-U(f \otimes 1)1 \otimes b). \end{align} $$

Therefore, we have

$$\begin{align*}U^{-1}(1 \otimes bU(f \otimes 1)-U(f \otimes 1)1 \otimes b)=0. \end{align*}$$

Since U is surjective, we have

(24) $$ \begin{align} 1 \otimes bU(f \otimes 1)=U(f \otimes 1)1 \otimes b. \end{align} $$

Note we choose $b \in \mathcal {A}_2$ with $b^{*}=-b$ arbitrary. For each $a \in \mathcal {A}_2$ , there exist unique elements $b_1, b_2 \in \mathcal {A}_2$ such that $b_i^{*}=-b_i$ for $i=1,2$ and $a=-ib_1+b_2$ . By applying (24), we have

$$\begin{align*}aU(f \otimes 1)(y)=U(f \otimes1)(y)a \end{align*}$$

for any $a \in \mathcal {A}_2$ and $y \in X_2$ . We get $U(f \otimes 1)(y) \in \{ b \in \mathcal {A}_2 \mid ab=ba \ \text {for all } a \in \mathcal {A}_2 \}=\mathbb {C}1$ . Thus, there is $g(y) \in \mathbb {C}$ such that $U(f \otimes 1)(y)=g(y)1$ . Since $U(f \otimes 1) \in \operatorname {Lip}(X_2,\mathcal {A}_2)$ , we get $g \in \operatorname {Lip}(X_2)$ and

$$\begin{align*}U(f\otimes 1)=g\otimes 1. \end{align*}$$

Thus, we can define a map $P_U: \operatorname {Lip}(X_1) \to \operatorname {Lip}(X_2)$ by

$$\begin{align*}U(f\otimes 1)=P_U(f) \otimes 1, \quad f \in \operatorname{Lip}(X_1). \end{align*}$$

It is easy to see that $P_U$ is a surjective complex linear isometry. Applying [Reference Hatori and Oi9, Corollary 15], there is a surjective isometry $\varphi : X_2 \to X_1$ such that

$$\begin{align*}U(f \otimes 1)(y)=P_U(f)(y)\otimes 1=f(\varphi(y)) \otimes 1, \quad f \in \operatorname{Lip}(X_1),\,\, y \in X_2.\\[-34pt] \end{align*}$$

Proof of Theorem 1.2

A simple calculation shows that the map U from $\operatorname {Lip}(X_1,\mathcal {A}_1)$ onto $\operatorname {Lip}(X_2,\mathcal {A}_2)$ , which has the form of the theorem, is a unital surjective linear isometry. We show the converse. Let us recall (21). Thus, for any $h \in H(\mathcal {A}_1)$ , there exists $\psi _0(h) \in H(\mathcal {A}_2)$ and $*$ -derivation D on $\mathcal {A}_2$ such that

$$\begin{align*}UM_{1 \otimes h}U^{-1}=M_{1 \otimes \psi_0(h)}+i \widehat{D}. \end{align*}$$

Let $f \in \operatorname {Lip}(X_1)$ . By Lemma 4.8, there is a surjective isometry $\varphi :X_2 \to X_1$ such that $U(f \otimes 1)=(f \circ \varphi ) \otimes 1$ . Thus, for any $y \in X_2$ , we obtain

$$\begin{align*}U(f \otimes 1)(y)=f (\varphi(y)) 1 \end{align*}$$

and

$$\begin{align*}\widehat{D}(U(f \otimes 1))(y)=D(U(f\otimes1)(y))=D(f (\varphi(y)) 1)=0. \end{align*}$$

We have

$$ \begin{align*} U(f \otimes h)(y)&=U(M_{1 \otimes h}(f \otimes 1))(y)=UM_{1 \otimes h}U^{-1}U(f \otimes 1)(y)\\ &=(M_{1 \otimes \psi_0(h)}+i \widehat{D})(U(f \otimes 1))(y)\\ &=M_{1 \otimes \psi_0(h)}(U(f \otimes 1))(y)+i \widehat{D}(U(f \otimes 1))(y)\\ &=\psi_0(h)U(f \otimes 1)(y)+0=f(\varphi(y))\psi_0(h). \end{align*} $$

For any $a \in \mathcal {A}_1$ , there exist $h_1,h_2 \in H(\mathcal {A}_1)$ such that $a=h_1+ih_2$ . Let us note that we define $\psi :\mathcal {A}_1 \to \mathcal {A}_2$ by $\psi (a)=\psi _0(h_1)+i\psi (h_2)$ . We get

$$ \begin{align*} U(f \otimes a)(y)&=U(f \otimes (h_1+ih_2))(y) =U(f \otimes h_1)(y)+iU(f \otimes h_2)(y)\\ &=f(\varphi(y))\psi_0(h_1)+if(\varphi(y))\psi_0(h_2)\\ &=f(\varphi(y))\psi(a) =\psi((f \otimes a)(\varphi(y)))=\psi(f(\varphi(y)) a) \end{align*} $$

for any $f \in \operatorname {Lip}(X_1)$ and $a \in \mathcal {A}_1$ . By Lemma 4.7, we recall that $\psi :\mathcal {A}_1 \to \mathcal {A}_2$ is a surjective complex linear isometry. Applying Proposition 3.6, we obtain

$$\begin{align*}U(F)(y)=\psi(F(\varphi(y))), \quad F \in \operatorname{Lip}(X_1,\mathcal{A}_1), y \in X_2.\\[-34pt] \end{align*}$$