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?



Ultraproduct of metric spaces
No, it's not an isometry in general. Let the index set be $\mathbb{N}$ and let each $X\_n = \mathbb{N}$ with its usual metric. Let $\mathcal{U}$ be any free ultrafilter on $\mathbb{N}$. For each $n$ define $f\_n(i) = 0$ for $i < n$ and $1$ for $i \geq n$. Then the ultraproduct of the $f\_n$ is not zero (the distance from each $f\_n$ to the zero function is $1$) but it is taken to zero by your map. (So I wouldn't call your map an "embedding".)


Incidentally, nowadays the more usual notation for $X^{\#}$ is ${\rm Lip}\_0(X)$.


