Let $I$ be a set and $\mathcal{U}$ be an ultrafilter on $I$. Suppose that $(X_{i}, d_{i})_{i\in I}$ is a family of pointed metric spaces with a distinguished point $e_{i}$ for each $i\in I$. We set $$(X_{i})_{\mathcal{U}}:=\{(x_{i})_{i}\in \prod X_{i}:\sup_{i\in I}d_{i}(x_{i},e_{i})<\infty\}/\sim, $$ where the equivalent equation $\sim$ is defined as: $(x_{i})_{i}\sim (x'_{i})_{i}$ if and only if $\lim_{\mathcal{U}}d_{i}(x_{i},x'_{i})=0$.

A natural metric $d$ is defined on $(X_{i})_{\mathcal{U}}$ as: $d((x_{i})_{\mathcal{U}},(x'_{i})_{\mathcal{U}})=\lim_{\mathcal{U}}d_{i}(x_{i},x'_{i}).$

A canonical embedding $J: (X^{\#}_{i})_{\mathcal{U}}\rightarrow ((X_{i})_{\mathcal{U}})^{\#}$ is defined by $$\langle J((f_{i})_{\mathcal{U}}), (x_{i})_{\mathcal{U}}\rangle=\lim_{\mathcal{U}}f_{i}(x_{i}).$$

Where, for a pointed metric space $X$, $X^{\#}$ denotes the space of all real-valued Lipschitz functions defined on $X$ which vanish at a distinguished point $0$, with Lipschitz norm.

Question: Is the canonical embedding $J$ an into isometry?