Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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: 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 $\tilde{f}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.