Between Tietze's and Dugundji's extension theorems

Question

The celebrated Tietze extension theorem asserts that any continuous real-valued function defined on a closed subset of a normal space, can be extended to a continuous function on the whole space. Seen as a property of the target space $\mathbb{R}$, this leads to the important notion of absolute neighborhood retract, or AR(normal); in Dugundji's notation, the Tietze extension theorem can thus be rephrased saying that $\mathbb{R}$ is an AR(normal) space.

If in the Tietze theorem we restrict the class of domains from normal to metric spaces, by the Dugundji extension theorem, at least all locally convex topological vector spaces are suitable codomains: any continuous LCTVS-valued function on a closed subset of a metric space can be extended to a continuous function on the whole space.

Of course, this situation in principle allows a wide variety of intermediate situations. The first natural questions, that I would be glad to learn an answer of, are:

Q1. Does Dugundji's theorem hold true for normal spaces, namely, can any continuous LCTVS-valued function on a closed subset of a normal topological space be extended to a continuous function on the whole space?

I guess the answer is no, but I can't imagine a counterexample. In case of a negative (or not known) answer:

Q2. Are Banach spaces absolute retract for Hausdorff compact spaces, namely, can any continuous Banach-valued function on a closed subset of a Hausdorff compact space be extended to a continuous function on the whole space?

edit After Bill Johnson's answer to question 2, and the other useful comments, I would like to focus on the following question, that should have some good reference in the (wide) literature.

Q3. Let $X$ be a Hausdorff compact topological space, $Y\subset X$ a closed set, $E$ a Banach space. Does there always exist a bounded linear extension operator $C(Y,E)\to C(X,E)$?

Answer 1

I think that question Q3 has nothing to do with Banach space valued functions: If $Y$ is a closed subset of a compact space $X$ such that there is a continuous linear extension operator $F:C(Y)\to C(X)$ then one can use the injective tensor product to get an extension operator $$F\otimes id_E: C(Y,E) \cong C(Y)\hat\otimes_\varepsilon E \to C(X)\hat\otimes_\varepsilon E \cong C(X,E)$$ even for every complete locally convex space $E$. The isomorphism $C(X)\hat\otimes_\varepsilon E\cong C(X,E)$ is desribed, e.g., in Jarchow's book Locally Convex Spaces, chapter 16.

I would be very surprised if it would be unknown whether the restriction operator $C(X)\to C(Y)$ always has a continuous linear right inverse.

Edit. I just learned from Tomasz Kania's answer to this question Nonseparable counterexamples in analysis that an example is $X=\beta\mathbb N$, the Stone-Cech compactification of $\mathbb N$, and $Y=\beta\mathbb N\setminus\mathbb N$: If there were an extension operator $C(Y)\to C(X)$ then the kernel of the restriction operator would be complemented in $C(X)=C(\beta\mathbb N)=\ell^\infty(\mathbb N)$ but for the remainder $Y$ this kernel is $c_0$ which is not complemented (the theorem of Phillips from 1940).

Answer 2

Here is an answer to Q2.

Since compact subsets of Banach spaces are separable, WLOG the target Banach space is separable.

Since all separable infinite dimensional Banach spaces are homeomorphic, WLOG the target Banach space is $c_0$.

Since $c_0$ is a Lipschitz retract of $\ell_\infty$, WLOG the target Banach space is $\ell_\infty$.

The space $\ell_\infty$ clearly has the desired property.

Sorry, Pietro; this being the day after April 1, I could not resist giving this answer.

Answer 3

The following is a rather well-known theorem of Haydon that might be useful for your purpose.

Theorem: The following are equivalent for a compact Hausdorff space $Y$:

1. ($Y$ is a Dugundji space): For every compact $X$ and embedding $e:Y \to X $ there is a bounded linear extension operator $T:C(Y) \to C(X)$ such that $T(1)=1$ and $T(f)\geq0$ whenever $f\geq0$.

2. ($Y$ is $\mathrm{AE}(0)$): For every compact zero-dimensional $Z$, every closed $A \subseteq Z$ and every map $f:A \to Y$ there is an extension $f:Z \to Y$.

3. $Y$ is the inverse limit of a Haydon system.

See for example Todorcevic´s book "Topics in Topology" for a proof of this theorem (and the definition of Haydon system).

For example any product of compact metrizable spaces is Dugundji and compact topological groups are Dugundji.

Answer 4

This is a comment but the system doesn't regard me as worthy. An important ingredient of Dugundji's theorem is that an extension can be found with values in the closed convex hull of the range. Without this, the fact that the values of the function to be extended are in a lcs is rather artificial as is also evident in Bill Johnson's answer and comment. As regards the stronger version of Q2 mentioned in the comments to the OP, the existence of a continuous, linear extension operator was investigated in some detail (for the scalar case). One could check the publications of Pelczyński, Corson, Lindenstrauss, Semadeni for some results of this nature.

Answer 5

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

1. $x \in W(n,x)$ for each $n$,
2. $W(n,x) \subset W(n+1,x)$ for each $x$, and
3. 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).