Does a bi-Lipschitz map from a space to itself extend continuously to its completion?

Question

If I have a metric space $X$ and a function $f$ from $X$ to $X$ that is bi-Lipschitz, then will $f$ extend continuously to a function from the completion of $X$ to itself?

I think that bi-Lipschitzness (or at least Lipschitz) is necessary here.

Answer 1

All you need is uniform continuity. We have the following:

Lemma: Suppose that $X,Y$ are metric spaces, where $Y$ is complete. Let $A \subset X$. If $f: A \to Y$ is uniformly continuous, then $f$ extends uniquely to a map from $\overline{A} \to Y$.

This lemma implies your statement, because you have that $X$ is a dense subspace of $X_{comp}$, and hence any Lipchitz map $f: X \to X_{comp}$ extends uniquely to a map $X_{comp} \to X_{comp}$.

---

Proof of Lemma: Let $x \in \overline{A}$. We can find a sequence of points $x_n \in A$ with $x_n \to x$. Fix $\epsilon>0$. By uniform continuity, there exists $\delta>0$ such that $d_X(a,b)<\delta$ implies $d_Y(f(a),f(b))<\epsilon$, for all $a,b \in A$. Choose $N$ such that $m,n \geq N$ implies $d_X(x_n,x_m)< \delta$ (we can do this since convergent sequences are Cauchy). Then $m,n \geq N$ implies $d_Y(f(x_n),f(x_m))<\epsilon$. Hence $(f(x_n))_n$ is a Cauchy sequence in $Y$, so it converges to some value that we will call $f^*(x)$. I will leave it to you to check that $f^*$ is well defined (it doesn't depend on the sequence $(x_n)$ chosen), and that it continuously extends $f$.

Answer 2

I think that $f$ being uniformly continuous is already enough here. This is the standard extension property for metric completions (see the section on completion here, e.g.). We consider $f$ from $X$ to itself as a map from $X$ to $\hat{X}$, the completion of $X$, in a trivial way and apply the extension property for the domain. So just Lipschitz is enough, as this already implies uniform continuity, but even weaker will do.