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. 
 A: 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.
A: 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 $\mathbb R$.
