Dugundji's extension theorem in its linear form

Let $X$ be a metric space, $A\subseteq X$ a nonempty closed subset and $L$ a locally convex topological vector space. Then every continuous $f\text{\hspace{0.17em}⁣}:A\to L$ extends to a continuous $F\text{\hspace{0.17em}⁣}:X\to L$ whose image lies in the convex hull of $f(A)$.

Moreover the extension can be produced by an extension operator: a single map $u\text{\hspace{0.17em}⁣}:C(A,L)\to C(X,L)$ with $u(f){\mathrm{\mid }}_A=f$ which is linear in $f$ and continuous for the topology of uniform convergence on compact sets. The values of $L$ are not restricted to an interval, and the extension respects convexity of the target.

Not proved in this library. The scalar and metric Tietze theorem is in scope and will be proved; the vector-valued statement with a linear extender is what is deferred.

What would prove it. A canonical locally finite open cover of $X\setminus A$ by sets whose diameters shrink as they approach $A$, a partition of unity subordinate to it, and an averaging formula $F(x)={\sum }_i{\phi }_i(x)f(a_i)$ with $a_i\in A$ chosen near the $i$-th cover element. Linearity in $f$ is visible in that formula, which is why the extender is linear. The construction needs paracompactness of metric spaces, A. H. Stone's theorem, which is itself choice-sensitive: if ZF is consistent, it is not provable in ZF + DC (Good, Tree and Watson, 1998) and is not implied by the Boolean prime ideal theorem (Corson, 2020). Both halves are relative-consistency results and neither is available unconditionally.

Why it matters here. The linearity of the extension operator, not the extension itself, is what makes the theorem a tool: it lets one extend a whole family of maps coherently, which is what retract theory and the theory of absolute neighbourhood retracts require. It is also the reason that Tietze in the scalar case looks elementary while the general case sits behind both a covering theorem and a choice principle.