Show every closed Ideal in $C_0(X)$ for a locally compact Hausdorff space X is of the form $K_F=\{f\in C_0(X)\colon f\lvert_F \equiv 0\}$ for some closed subspace $F\subseteq X$.
Here is what I have so far:
Let $I$ be a closed ideal.
I tried to use the Gelfand Transform.
Consider the evaluation map $ev_x: C_0(X) \rightarrow \mathbb{C}$ defined by $ev_x(f)= f(x)$ for all $x \in F$. Then we have $Ker (ev_x)= K_F$
From here, I'm pretty much lost what to do to get the result desired.
Also, I thought about using Urysohn's Lemma here: https://www.math.ksu.edu/~nagy/real-an/1-05-top-loc-comp.pdf . But still didn't anywhere.
Any help will be appreciated.
Best Answer
Given a closed ideal $I\trianglelefteq C_0(X)$, let $$ F=\{x\in X: f(x)=0, \text{ for all } f\in I\}, $$ and let us prove that $I=K_F$.
The inclusion $I\subseteq K_F$ holds for a pretty elementary reason (which nevertheless sounds a bit like a tongue-twister): every function in $I$ vanishes on any point where all functions in $I$ vanish.
The hard part is to prove the reverse inclusion $K_F\subseteq I$. For this pick $f\in K_f$. Fixing $\varepsilon >0$, recall that the set $$ C=\{x\in X: |f(x)|\geq \varepsilon \} $$ is compact.
Given $x$ in $C$, we have that $f(x)$ is nonzero, so clearly $x$ is not in $F$. Consequently not all functions in $I$ vanish at $x$, and hence we may choose some element $a_x\in I$, such that $a_x(x)\neq 0$. The set $$ U_x=\{y\in X: a_x(y)\neq 0\} $$ is then an open neighborhood of $x$, and hence $\{U_x\}_{x\in C}$ is an open cover of $C$. By compactness we may pick a finite subcover $\{U_{x_1},U_{x_2},\ldots ,U_{x_n}\}$. Next observe that $$ g:= \sum_{i=1}^n\overline {a}_{x_i}a_{x_i} $$ lies in $I$, and does not vanish on any point of $C$.
There are multiple endings for this story, and perpahs the slickest way is as follows: besides an ideal, $I$ is a C*-algebra and $g$ is a non-negative element in $I$, so $g^{1/p}$ lies in $I$ for every positive integer $p$. Moreover the sequence $\{g^{1/p}\}_{p\in {\mathbb N}}$ converges uniformly to 1 on $C$. It is also easy to see that $$ \|g^{1/p}\| = \|g\|^{1/p} \leq 2, $$ for large enough $p$. Therefore we may find $p$ such that $$ |f(x)-g(x)^{1/p}f(x)| < \varepsilon , \quad\forall x\in C, $$ and $$ |f(x)-g(x)^{1/p}f(x)| \leq \big (1+|g(x)^{1/p}|\big ) |f(x)|\leq (1+2)\varepsilon , $$ for every $x\in X\setminus C$. Consequently $\|f - g^{1/p}f\|\leq 3\varepsilon $, so we deduce that $f$ lies in the closure of $I$, and hence also in $I$ because $I$ is closed.
EDIT. Here is another approach, assuming we already know the answer in case $X$ is compact.
Assuming that $X$ is not compact, let $\hat X=X\cup \{\infty \}$ be the one-point compactification of $X$, and let $$ f\in C_0(X)\mapsto \hat f\in C(\hat X) $$ be the natural identification. One may then prove that $\hat I$ is a closed ideal in $C(\hat X)$, and hence, by the compact case, $$ \hat I=\{g\in C(\hat X) : g|_G=0\}, $$ where $G$ is a closed subset of $\hat X$.
Notice that necessarily $\infty \in G$, since otherwise one could use Urysohn to produce some $g\in C(\hat X)$, vanishing on $G$, with $g(\infty )\neq 0$, but this would imply that $g$ lies in $\hat I\setminus C_0(X)$, which is impossible.
Setting $F=G\cap X$, we have that $F$ is a closed subset of $X$. So, upon identifying $C_0(X)$ as the subalgebra of $C(\hat X)$ formed by the functions vanishing at $\infty $, we have: $$ \hat I=\{g\in C(\hat X) : g|_G=0\} = $$$$ = \{g\in C(\hat X) : g|_F=0\} \cap \{g\in C(\hat X) : g(\infty )=0\} = $$$$ = \{g\in C(\hat X) : g|_F=0\} \cap C_0(X) = $$$$ = \{g\in C_0(X) : g|_F=0\} = K_F. $$