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I read in a book the following assertion:

Let $\Omega\subset\mathbb{R^n}$ be a open set, then any bounded and uniformly continuous function in $\Omega$ has a unique bounded continuous extension to $\overline{\Omega}$.

I can't see this readly, first I thought was a simple consequence of the Tietze extension theorem, but I didn't have any progress. Someone could help me?


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Tietze Extension is not helpful here. Rather, once you have answered your question, you can use TE to show that the function can be extended (not necessarily uniquely) to all of $\mathbb{R}^n$. – Pete L. Clark Aug 18 '13 at 15:30
up vote 4 down vote accepted

For any $x \in \overline{\Omega}$, find a sequence $(x_n) \in \Omega$ such that $x_n \to x$.

  1. Since $(x_n)$ is Cauchy, and $f$ is uniformly continuous, show that $(f(x_n))$ is Cauchy.
  2. Define $\overline{f}(x) = \lim_{n\to \infty} f(x_n)$
  3. Show that this definition is independent of the choice of the sequence $(x_n)$.

Now $\overline{f}$ is your extension (it coincides with $f$ because of #3)

Edit: Uniqueness holds because any continuous extension of $f$ must satisfy #2 : ie. if $g$ is another continuous extension of $f$, then for any $x_n$ as above $$ g(x) = \lim_{n\to \infty} g(x_n) = \lim_{n\to \infty} f(x_n) = \overline{f}(x) $$

As for boundedness, it again follows from #2 : If $|f(y)| \leq M$ for all $y \in \Omega$, then $|\overline{f}(x)| = \lim_{x_n\to x}|f(x_n)| \leq M$ as well.

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Note that this works in general for "Cauchy-continuous" functions on any metric space. – Anthony Carapetis Aug 18 '13 at 15:22
I upvoted this answer, since it treats the hardest part of the question (correctly!). The OP also asked about uniqueness and boundedness of the extension. Maybe you'd like to say a few words about that as well? – Pete L. Clark Aug 18 '13 at 15:28
And how about the boundedness and unicity? – DiegoMath Aug 18 '13 at 15:32
Just edited my earlier post. Hope this helps :) – Prahlad Vaidyanathan Aug 18 '13 at 16:21

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