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It is easy to see that $f$ is uniformly continuous (basically by definition of a contraction). Now I'm trying to understand why this implies that $f$ extends uniquely to a continuous $\tilde{f}: \tilde{M} \to \tilde{M}$ where $\tilde M$ is the completion of M. Also, is $\tilde{f}$ a contraction as well? I think it is, but I'm having a hard time trying to justify it.

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that is indeed true –  Aden Dong Feb 22 '13 at 0:55
you may use the definition of completion of $M$, for each point in $\tilde{M}$, find sequence to approximate it. –  Yimin Feb 22 '13 at 0:56

1 Answer 1

up vote 3 down vote accepted

I assume $M$ is a metric space.

We can see $M$ as a dense subspace of its completion $\widetilde{M}$.

Then $f$ takes values in $\widetilde{M}$, a fortiori, which is complete: $$ f:M\longrightarrow \widetilde{M}. $$

Since $f$ is uniformly continuous, it extends uniquely to a uniformly continuous function from $\overline{M}=\widetilde{M}$ to itself.

The basic idea is that $f$ takes Cauchy sequences to Cauchy sequences.

So now you can read this for the proof and more motivation: http://drexel28.wordpress.com/2010/11/03/extending-uniformly-continuous-functions/. Note this great write-up is from the blog of a contributor of this website.

To answer your second question, now. If there exists $K>0$ such that $$ d(f(x),f(y))\leq Kd(x,y) $$ for all $x, y\in M$, then the extension $\tilde{f}$ satisifies $$ d(\tilde{f}(x),\tilde{f}(y))\leq Kd(x,y) $$ for all $x,y\in\widetilde{M}$ by continuity and density of $M$ in $\widetilde{M}$.

Indeed, for all $x,y\in\widetilde{M}$ there exist sequences $x_n$, $y_n$ in $M$ which converge to $x$ and $y$ respectively, and for which the inequality is satisfied. Then passing to the limit and using the continuity of $f$, we get the same estimate for $\tilde{f}$.

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