# Tietze extension theorem in LCH spaces

Let $X$ be a locally compact Hausdorff space and $K$ be a compact subset of $X$. Then any function $f\in C(K)$ can be extended to a function in $C(X)$ which vanishes outside a compact set.

I have searched several books for the proof. It seems that the authors believe it is just an exercise-level proposition. Since $X$ is an LCH space, we can always find open set $V$ containing $K$ with compact closure $\overline{V}$ and we can extend $f$ to $\overline{V}$ by classical Tietze extension theorem. I also know the fact that $f\in C(K)$ implies that the the range of $f$ is contained in a closed interval $[a,b]$, but how do we define a $F \in C(X)$ satisfying the requirements?

• What statement of the Tietze extension theorem are we working from? E.g. the Wikipedia version for instance? Mar 13, 2015 at 20:44
• Yes.Let $X$ be a normal space and $K$ be a closed subspace of $X$. Then any function $f\in C(K)$ can be extended to a function $F \in C(X)$. Furthermore, if $f(K) \subset [a,b]$, then so is $F(X)$ Mar 13, 2015 at 20:51

Extend $f$ to $\hat f:\operatorname{cl}V\to\Bbb R$ just as you’ve already done. If $V$ is clopen in $X$, let

$$f^*:X\to\Bbb R:x\mapsto\begin{cases} \hat f(x),&\text{if }x\in V\\ 0,&\text{otherwise}\;. \end{cases}$$

Otherwise, apply Uryson’s lemma to the normal space $\operatorname{cl}V$ to get a continuous $g:\operatorname{cl}V\to[0,1]$ such that $g(x)=1$ for all $x\in K$, and $g(x)=0$ for all $x\in(\operatorname{cl}V)\setminus V$. Then define

$$f^*:X\to\Bbb R:x\mapsto\begin{cases} g(x)\hat f(x),&\text{if }x\in\operatorname{cl}V\\ 0,&\text{otherwise}\;. \end{cases}$$

• Sir, what about this extension that the target space is Banach? Jun 24, 2020 at 22:04

This can be done in few steps:

1. First extend $f$ to $K\cup \partial V$, putting $0$ on $\partial V$.
2. Then use Tietze to extend this to whole $\overline V$ (using the fact that it is compact Hausdorff, so normal).
3. Finally put $0$ outside of $\overline V$, obtaining $\bar f\colon X\to {\bf R}$.

Check continuity by the definition (preimage of a closed set containing zero is the preimage by $\bar f$ restricted to $\overline V$ plus the entire $V^c$, while the preimage of any other closed set is just the preimage by $\bar f$ restricted to $\overline V$).