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?

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    $\begingroup$ What is $X^\#$ for a metric space? $\endgroup$ – Joel David Hamkins Sep 10 '17 at 2:18
  • $\begingroup$ I add the definition of $X^{\#}$ for a metric space in my question. $\endgroup$ – Dongyang Chen Sep 10 '17 at 2:29

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 ${\rm Lip}_0(X)$.

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    $\begingroup$ You can also do it with each $X_n = [0,1]$. Take $f_n(t) =t$ for $t < 1/n$ and $1/n$ for $t \geq 1/n$. $\endgroup$ – Nik Weaver Sep 10 '17 at 5:04
  • $\begingroup$ I think that the question is true if each $X_{i}$ has finite diameter. Indeed, for each $(f_{i})_{\mathcal{U}}$ in $(X^{\#}_{i})_{\mathcal{U}}$ and every $\epsilon>0$. We choose $x_{i},x'_{i}\in X_{i}$ so that $|f_{i}(x_{i})-f_{i}(x'_{i})|/d_{i}(x_{i},x'_{i})>Lip(f_{i})-\epsilon$. Since each $X_{i}$ has finite diameter, $(x_{i})_{\mathcal{U}}$ and $(x'_{i})_{\mathcal{U}}$ are in $(X_{i})_{\mathcal{U}}$. It is easy to see that $\|J((f_{i})_{\mathcal{U}})\|\geq \|(f_{i})_{\mathcal{U}}\|-\epsilon.$ $\endgroup$ – Dongyang Chen Sep 10 '17 at 8:35
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    $\begingroup$ That is not right. The distance between the two sequences of points in the ultraproduct could be 0. My first comment gave a counterexample. $\endgroup$ – Nik Weaver Sep 10 '17 at 13:27
  • $\begingroup$ In your first comment, the diameter of each $X_{n}$ is infinity. In my opinion, the question is true if $\sup_{n} diam (X_{n})<\infty$. So I think that it is true if each $X_{n}=[0,1]$. $\endgroup$ – Dongyang Chen Sep 10 '17 at 15:29
  • $\begingroup$ In my first comment, each $X_n = [0,1]$, whose diameter is $1$. The claim is not true if each $X_n = [0,1]$. Do you need me to spell out the details of my example? $\endgroup$ – Nik Weaver Sep 10 '17 at 16:31

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