Dugundji extension theorem

Remarks

The proof covers $X \setminus A$ by small open balls whose radii are comparable to the distance to $A$. It then takes a locally finite open refinement, which exists because metric spaces are paracompact. For each set $W$ of the refinement, pick a point $a_W \in A$ close to $W$, and let $(\varphi_W)$ be a partition of unity subordinate to the refinement. Set $F(x) = \sum_W \varphi_W(x) f(a_W)$ for $x \notin A$ and $F = f$ on $A$. Continuity at points of $A$ uses that $a_W$ is near $x$ when $W$ is near $x$, together with the local convexity of $L$.

Example: with $L = \R$, the theorem gives the Tietze extension theorem for metric spaces, with the bound $\inf f \le F \le \sup f$.

The hypothesis that $A$ is closed is necessary. On $X = [0, 1]$, the map $f(x) = \sin(1/x)$ is continuous on $A = (0, 1]$, but it has no continuous extension to $X$, because $f$ has no limit at $0$.

Going deeper

For a student

The theorem extends the Tietze extension theorem from real-valued to vector-valued maps, at the price of assuming that $X$ is metric. Its main consequence is that every convex subset of a locally convex metrizable topological vector space is an absolute retract. The proof is a model use of partitions of unity on paracompact spaces, and the same technique appears in the Michael selection theorem.