Is there a uniform Dugundji theorem

A theorem of Dugundji states that if $X$ is a separable metric space and $A \subseteq X$ is closed then any continuous function $f$ from $A$ to some normed linear space $L$ may be extended to a continuous map $\bar{f} \colon X \to L$. I got this formulation from van Mill's book "Infinite-Dimensional Topology". Does there exist a version of Dugundji's Theorem where we assume that $f$ is uniformly continuous an conclude the existence of a uniformly continuous extension? Since my main interest is when $L$ is a Banach space (or even a $C^*$-algebra), I would be very happy to add that assumption to get a positive answer.

Well, of course not. Take $X=\mathbb R$, $A=\cup [2n,2n+1]$ over $n\in \mathbb N$. Set $f(x)=n^2$ for $x\in [2n,2n+1]$. It is uniformly continuous but you cannot extend it to a uniformly continuous function on $\mathbb R$.

• Ahh, I see. Does the answer changes if one also asks that $f$ be bounded? Jun 21, 2012 at 13:31

Note that the case of real-valued functions is easy. A function $f:A\to\mathbb{R}$ has a uniformly continuous extension to any metric space $X\supset A$ iff it has a sub-linear modulus of continuity (that is, that verifies $\omega(t)\le a|t| + b$).

Rmk. Note that these general extensions problems may be approached as selection problems for multivalued functions. Precisely, given $f:A\to L$ and a concave modulus of continuity $\omega$, consider the multi $F:X\to 2^L$, taking $x\in X$ to the set $F(x)\subset L$ of all admissible values for an $\omega$-continuous extension of $f$ to $A\cup \{x\}$, that is $$F(x)=\cap _ {a\in A} \bar B\big(a,\omega(d(a,x))\big) \, .$$

Then, any extension $\tilde f$ of $f$ to $X$ has to be a selection of $F$, $f(x)\in F(x)$; note that $F(a)=\{a\}$ for $a\in A$. If $\omega$ is not too small and $L$ is compact enough, you may hope to prove that $F(x)$ is a not-empty bounded closed set, for all $x$. You may then correspondingly look for a modulus of continuity for $F$ seen as a map valued into non-empty, bounded, closed subsets of $L$ (even convex, if $L$ is a normed space) with respct to the Hausdorff distance. Finally, you may construct $\tilde f$ as a composition $c\circ F$, where $c$ is a map that picks a point $c(S)$ within every such subset $S$. This program can be achieded in a very satisfactory way in the case of the Kierszbraun theorem, where the Lipschitz constant is preserved: Hilbert spaces $L$ are a nice setting, because in this case there is a 1-Lipschitz function $c$, taking a bounded convex set $S$ to the center of the smallest closed ball containing $S$. Slightly more genreal, I guess, uniformly convex Banach spaces $L$ could work, but I guess the modulus of continuity of the extension in general will be larger. If $L$ is a metric space, then of course there are in general topological obstructions even for continuous extensions.

• Thanks. I didn't know about the modulus of continuity. If I understand it correct, we have that if $f \colon A \to \mathbb{R}$ is uniformly continuous and bounded then it has an extension to a uniformly continuous function, since its modulus of continuity is bounded. The wikipedia article you link to seems to indicate that a lot of research have been devoted to deciding when Lipschitz function extend, for instance the Kirszbraun theorem. Have similar research been done for (bounded) uniformly continuous functions? Jun 22, 2012 at 13:44
• I do think so, although I do not have references. I've edited and added some general hints. Jun 22, 2012 at 14:53
• Adam, the result on extending bounded uniformly continuous functions into $\Bbb R$ is called Katetov's theorem, and you can find references to several different proofs under 2.31 (in subsection 2.D) in arxiv.org/abs/1106.3249v3. In particular, Itzkowitz's proof is along the lines of a standard proof of the Tietze/Urysohn extesion theorem; older proofs are more intricate. Regarding extension of bounded u.c. maps into topological vector spaces, see the rest of the above-mentioned subsection 2.D, and subsection 4.A, especially Remark 4.5 with lots of references. Jun 23, 2012 at 0:04
• Sergey, thank you for this excellent reference. Jun 25, 2012 at 8:03
• What's up with all the recent edits and all the bumps to the front page? I count something like 15 within the past 30-45 minutes. Sep 5, 2013 at 16:04