## Bad News

The answer to Q3 as stated is no. Let $X$ be the Michael line, and let $Y$ be the closed subset consisting of all the rationals. Then, there is no bounded linear extension $C(Y,\mathbb{R}) \to C(X,\mathbb{R})$. A proof (I'm not sure if this is the first place where it appeared) may be found in Example 3.3 of the paper

RW Heath and DJ Lutzer, *Dugundji extension theorems for linearly ordered spaces*, Pacific Journal of Math, 55(2), 419-425 (1974).

On the other hand, the authors provide an extension theorem for what they call *linearly ordered spaces*. There is a zoo of such conditions which are related to -- but strictly stronger than -- normality under which you can find a simultaneous extension. It would be quite a painful task to try and list all of them anywhere. There was some work on the class of *linearly stratifiable spaces*, I think going back to work of CJR Borges in the mid-70s, but people found some gaps and some counterexamples, so I'm not sure where things stand with all that right now.

## Good News

Here is one rather typical example where Dugundji extension certainly works - I'll call this assumption $N^+$. A space $X$ satisfies $N^+$ if it admits a distinguished collection of open sets $\{W(n,x) \subset X ~ \mid ~ x \in X \text{ and } n \in \omega\}$ so that

- $x \in W(n,x)$ for each $n$,
- $W(n,x) \subset W(n+1,x)$ for each $x$, and
- For any open $U \subset X$ with $x \in U$, there exists an open $V \subset X$ so that for each $y \in V$ there is some $n$ with $x \in W(n,y) \subset U$.

By the way, if $X$ is $T_1$ then this condition is equivalent to metrizability. Here is the result you want:

**Theorem** Let $Y$ be a closed subspace of a topological space $X$ satisfying $N^+$. If $L$ is any locally convex topological vector space, then there exists a linear map $C(Y,L) \to C(X,L)$ which produces Dugundji extensions (and in particular, satisfies the convex hull condition).

You can find the proof in a nice and short paper:

IC Starc, *Concerning the Dugundji extension property*, Topology and its Applications 63(2), 165–172 (1995).

celebrated", but the theorem is by Urysohn, not by Tietze. $\endgroup$ – Włodzimierz Holsztyński Apr 3 '15 at 19:20