Dugundji extension theorem

Let ${\displaystyle X}$ be a metrizable space, ${\displaystyle A}$ a closed subset of X, and ${\displaystyle L}$ a locally convex topological vector space. Then:

• Every continuous map ${\displaystyle f:A\to L}$ admits a continuous extension ${\displaystyle F:X\to L}$ such that the image ${\displaystyle F(X)}$ is contained in the convex hull of ${\displaystyle f(A).}$^[1]

or, equivalently:

• Every continuous map from ${\displaystyle A}$ into a convex subset ${\displaystyle K}$ of ${\displaystyle L}$ admits a continuous extension from ${\displaystyle X}$ into ${\displaystyle K.}$^[2]

The first version of the Tietze extension theorem corresponds to the special case of the above theorem where the target space L is the real line ℝ. Urysohn generalized this to replacing the domain being a metric space by an arbitrary normal space. The Dugundji extension theorem is a transverse generalization, replacing the target ℝ by an arbitrary locally convex space. There is another generalization of the Tietze theorem assuming that the domain X is paracompact and the target L is a Banach space.

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Fix some metric ${\displaystyle d}$ on ${\displaystyle X.}$ Consider the open cover of ${\displaystyle X-A}$ that consists of the open balls ${\displaystyle B(x,d(x,A)/2)}$ for ${\displaystyle x\in X-A.}$ Since every metric space is paracompact, there exists a locally finite open cover ${\displaystyle \{U_{i}\}_{i\in I}}$ of ${\displaystyle X-A}$ such that each ${\displaystyle U_{i}}$ is contained in one of those balls. Choose a partition of unity ${\displaystyle \varphi _{i}}$ subordinate to this cover. For each ${\displaystyle i}$, pick a point ${\displaystyle a_{i}\in A}$ satisfying

${\displaystyle d(x_{i},a_{i})\leq 2d(x_{i},A),}$

which is possible since for each ${\displaystyle \epsilon >0}$, there is an ${\displaystyle a\in A}$ with ${\displaystyle d(x,a)\leq d(x,A)+\epsilon }$. Define the extension ${\displaystyle F}$ on ${\displaystyle X}$ by:

$${\displaystyle F(x)={\begin{cases}f(x)&{\text{if }}x\in A,\\\sum _{i}\varphi _{i}(x)f(a_{i})&{\text{else.}}\end{cases}}}$$

The map ${\displaystyle F}$ is clearly continuous on ${\displaystyle X-A}$. We shall then show it is continuous at each point ${\displaystyle a}$ in ${\displaystyle A}$ as well. For each ${\displaystyle x}$ in ${\displaystyle U_{i}}$, we have: ${\displaystyle 2d(x,x_{i})\leq d(x_{i},A)}$ or

${\displaystyle d(x,x_{i})+2d(x_{i},A)\leq 5d(x_{i},A)-5d(x,x_{i}).}$

Thus, we have:

${\displaystyle d(x,a_{i})\leq 5d(x,A)}$

and then

${\displaystyle d(a,a_{i})\leq 6d(a,x).}$

Now, let a convex neighborhood C of ${\displaystyle F(a)}$ be given. Then, since ${\displaystyle f}$ is continuous, there is some ${\displaystyle \delta >0}$ such that ${\displaystyle f(B(a,\delta )\cap A)\subset C}$. Then we have ${\displaystyle F(B(a,\delta /6))\subset C}$ by the above inequality, completing the proof of the continuity. ${\displaystyle \square }$