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Let $V$ be a (Hausdorff) topological vector space, $W\subset V$ a closed linear subspace, $X\subset V $ a compact.

(Q): When there is a continuous map $P:X\to W$ such that $P(x)=x$ for every $x\in X\cap W$?

In general there is none (see below); to have a positive answer I'd like as mild conditions on $W$ as possible, while $X$, as a topological space, should possibly be any Hausdorff compact space, with no other assumptions.

Motivation: this is related to a previous question: for a Hausdorff compact space $X$, say that a closed subset $A$ of $X$ is nice (at least, to me) iff there exists an extension operator $E:C(A)\to C(X)$, that is a (bounded, linear) right inverse to the restriction operator $R:C(X)\to C(A)$ $-$maybe "the pair $(X,A)$ has the Dugundji Extension Property" sounds better. Since $x\mapsto \delta_x$ embeds $X$ as subspace of $C(X)^*$ with its $w^*$ topology, the composition $P:X\subset C(X)^*\xrightarrow{E^*}C(A)^*$ is an instance of the situation in (Q), for $P(\delta_x)=\delta_x$ for all $x\in A$. In other words, any extension operator has the form $Ef(x)=\mu_x(f)$ for a Baire measure $\mu_x$ continuously depending on $x\in X$ w.r.to the $w^*$ topology of $C(A)^*$, and $\mu_x=\delta_x$ for all $x\in A$.

May the abstract formulation (Q) shed some light on the problems listed in the previous linked question? Of course, a direct construction for the situation $X\subset V:=C(X)^*$, $W:=C(A)^*$, is most welcome.

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    $\begingroup$ "$X$, as a topological space, should possibly be any Hausdorff compact space, with no other assumptions." Do you still require that it has the topology induced from $V$? $\endgroup$
    – TaQ
    Commented Mar 3 at 13:23
  • $\begingroup$ Yes, sure, it has the induced topology. $\endgroup$ Commented Mar 3 at 13:25

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