Which closed subsets $Y$ of a compact space $X$ admit a linear extensor $C(Y)\to C(X)$?

Question

In the following $X$ is a Hausdorff compact topological space. Let $Y$ be a closed subset of $X$. The restriction operator $R_Y:C(X)\to C(Y)$ is surjective (Tietze), so it admits a continuous right inverse (by Bartle-Graves Selection Theorem) but it may fail to have a bounded linear right inverse (an "extension operator"). Equivalently, $\text{ker}R_Y=\{f\in C(X): f_{|Y}=0\}$ may fail to be complemented in $C(X)$.

The possibly simpler counterexample is mentioned here, with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$.

Notably, here the closed set $Y$ is not metrisable. In fact (Lutzer and Martin, 1974) a metrisable $G_\delta$ subset $Y$ of $X$ does admit an extension operator --this generalises the classical Borsuk-Dugundji theorem, where $X$ is a metric space.

Metrisability of $Y$ is certainly not necessary, in general (e.g. in the quite trivial case of a clopen subset $Y$). Being a $G_\delta$ subset, that is, the zero set of a continuous function on $X$, is also not necessary, in general (e.g. in the quite trivial case of $Y=\{x_0\}$, where $x_0\in\ X$ does not have a countable basis of neighborhoods).

Question 1. Can we weaken the metrisability condition from Lutzer-Martin theorem, that is, what more general conditions on a closed set $Y$ ensure the existence of an extension operator?

The reason why in the above example with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$ the restriction operator $C(X)\to C(Y)$ is not a left-inverse is that $C(X)\sim\ell_\infty$ has a $w^*$-separable dual, whereas $C(Y)\sim \ell_\infty/c_0$ has not. This seems quite difficult to translate into more direct topological properties of $(X,Y)$. So it may be difficult to find a topological condition that guarantees the existence of an extension operator.

Question 2. Given a nbd $U$ of a closed set $Y$, can we find a closed set $Y'$ that admits an extension operator, $Y\subset Y'\subset U$?

Question 3. Suppose $Y_1$ and $Y_2$ admits an extension operator. Does $Y_1\cap Y_2$ admit an extension operator too?

Rmk. An extension operator $E:C(Y)\to C(X)$ is always of the form $Ef(x)= \langle Px,f\rangle$, where $P:X\to C(Y)^*$ is continuous w.r.to the $w^*$ topology of $C(Y)^*$ and $Py=\delta_y$ for all $y\in Y$.

Answer 1

This is an extended comment expanding on the remarks I left in the comments. It's far from a definitive answer. To make the discussion less technical, I'll restrict to compact Hausdorff spaces so that there is no distinction between bounded and unbounded functions. Continuity of functions regarding $C(X)$ will always refer to the norm topology (and I hope I got these aspects correct).

Theorem Let $X$ be a compact Hausdorff space and $A\subseteq X$ a closed subspace. Then the pair $(X,A)$ admits both a linear extender and a continuous extender. It may fail to admit a continuous linear extender.

To produce a linear extender $C(A)\rightarrow C(X)$ it suffices to define it on the members of a linear basis of $C(A)$. Due to the Tietze Extension Theorem, this can always be done. The result is almost never continuous. On the other hand, continuous, nonlinear extenders have been produced by Lutzer and Przymusiński [5].

Now, here is a positive result which slightly weakens the metrisability condition. It's an old observation due to Carlos Borges.

Theorem (Borges) Let $X$ be compact Hausdorff. Then for any closed subspace $A\subseteq X$ which has metrisable boundary, the pair $(X,A)$ admits a continuous linear extender.

The trick is to use the assumption to produce an extender $\gamma:C(\partial A)\rightarrow C(X\setminus A^\circ)$ and then define $\psi:C(A)\rightarrow C(X)$ by $$\psi(f)(x)=\begin{cases}\gamma(f|_{\partial A})(x)&x\in X\setminus A^\circ\\f(x)&x\in A^\circ.\end{cases}$$

I'll refer to Borges's paper [1] for the details, where the assumption that $A$ is a zero set in $X$ is not used.

There is a class of spaces in which continuous linear extenders always exist. These are the generalised ordered spaces.

Theorem (Heath, Lutzer [3]) If $X$ is a compact GO space, then a continuous linear extender exists for any closed subspace.

Finding a complete characterisation of those spaces $X$ for which every closed subset $A\subseteq X$ admits a continuous linear extender is difficult. Clearly such a space must be normal. van Douwen in the paper [2] linked by KP Hart in the comments showed that if the operator norms of the extenders are required to be suitably uniformly bounded, then $X$ must even be hereditarily collectionwise normal.

While every compact Hausdorff space is normal, they are generally not hereditarily so. Still, van Douwen does produce in [2] an example of a basically disconnected compact Hausdorff space admitting continuous linear extenders for each closed subspace, which does fail to be hereditarily collectionwise normal.

A few no-go results appear in the literature. One of the most useful is as follows.

Proposition If $X$ is compact Hausdorff and $A\subseteq X$ is a closed subspace with $c(A)>d(X)$, then the pair $(X,A)$ admits no continuous linear extender.

This is proved, for instance in the survey paper [4]. Here $c(A)$ is the cellularity, or Souslin number, of $A$, defined to be the supremum of all cardinalities of pairwise-disjoint families of open sets of $A$. The density $d(X)$ is the least cardinality of any dense subset of $X$.

Since $c(\beta\mathbb{N}\setminus\mathbb{N})=2^{\aleph_0}$, we obtain a topological explanation for the following.

Corollary The pair $(\beta\mathbb{N},\beta\mathbb{N}\setminus\mathbb{N})$ admits no continuous linear extender.

Finally, I want to ask if extenders exist for monotonically normal compacta. Monotone normality is one of the strongest normality properties. It's hereditary, and it implies both collectionwise normality and countable paracompactness. Every GO space is monotonically normal, and there is the result of Mary Ellen Rudin that the monotonically normal compacta are exactly the images of compact linearly ordered spaces.

It's known that every monotonically normal space admits monotone extenders for each closed subspace. On the other hand, there are examples of monotonically normal spaces which do not admit continuous linear extenders for their closed subspaces. I don't know of any compact examples appearing in the literature.

Question Is there a monotonically normal compactum $X$ and a closed a closed subspace $A\subseteq X$ without a continuous linear extender?

Edit: Thanks to KP Hart for a correction in the comments.

References

[1] C. Borges, Another Generalization of the Dugundji Extension Theorem, Colloquium Mathematicae 56 (2) 2 (1988), 263-265.

[2] E. van Douwen, Simultaneous linear extension of Continuous functions, Gen. Top. and App. 5 (4) 4 (1975), 297-319.

[3] R. Heath, D. Lutzer, Dugundji Extension Theorems for Linearly Ordered Spaces, Pacific J. Math. 55 (1974), 419-425.

[4] D. Lutzer, A survey of Dugundji Extension Theory, Seminar Uniform Spaces (1978), 24-42.

[5] D. Lutzer, T. Przymusiński, Continuous Extenders in Normal and Collectionwise Normal Spaces, Fundamenta Math. 102 (1979), 165-171.