In a locally compact Hausdorff space, given a compact set $K$ can we construct a continuous linear function which vanishes outside K? Specifically I would like to show that $C_0(M)$, the space of real valued bounded functions vanishing at infinity defined on a LCH space, separates points. So I can separate two points by disjoint compact sets, but can I guarantee the existence of a function (something like the indicator function, but cts) that is supported on a given compact set? If LCH implies normal I can use Urysohn's lemma, but I think LCH is only regular (please verify).
 A: Indeed, unlike compact spaces, a  LCH space $M$  is not always normal so Uryshon's Theorem cannot be applied indiscriminately.
That is, if $F$ and $G$ are closed, disjoint subsets of $M$, you cannot guarantee the existence of a continuous, real-valued
function $f$ such that $f|_F\equiv 1$, and $f|_G\equiv 0$.
However there is a trick that works in case $F$ (or $G$) is compact.  Considering the one-point compactification
$$
  M^+=M\cup \{\infty \},
  $$
observe that $F$ (the compact guy) is still closed in
$M^+$.  The same cannot be said of $G$ (in case it is not compact), but the closure of $G$ in $M^+$ can only add $\infty $ to
it, so it is still disjoint from $F$.  You may then apply Uryshon for $M^+$ (which is compact,  hence normal) relative to the subsets $F$ and
$$
  G'=G\cup \{\infty \}
  $$
(which is closed), and
then restrict the resulting function $f$ to $M$!
Even better, since $f$ vanishes on $G'$,  it also vanishes on $\infty $, so the restriction of $f$ to $M$ lies in $C_0(M)$.
