# $cl(C_c(\Omega))$ is a subset of $C_0(\Omega)$

I am trying to show that $cl(C_c(\Omega))$ is dense in $C_0(\Omega)$ where $C_c(\Omega)$ is the set of compactly supported continuous functions $f: \Omega \rightarrow R$ and $C_0(\Omega)$ is the set of continuous functions $f: \Omega \rightarrow R$ that can be continuously extended to $\partial \Omega$ and $f |_{\partial \Omega} = 0$. As a first step I want to show that it's a subset.

Edit: $\Omega \subset R$ is a bounded open domain.

I think I have the idea but I don't know how to write it down, maybe someone can help me. Here is the idea:

Let $f \in cl(C_c(\Omega))$. Then $$\forall \varepsilon > 0 \exists g \in C_0(\Omega): || f - g ||_\infty < \varepsilon$$

Assume there doesn't exist an $f^\prime$ such that $f^\prime |_{\partial \Omega} = 0$ and $f^\prime |_\Omega = f$. Say, for all such $f^\prime$, $f^\prime \geq K$ on $\partial \Omega$.

Now I want to say that at the point where $\Omega$ ends and $\partial \Omega$ starts, $g$ has to be continuous and is $0$ on $\partial \Omega$ . But $f^\prime \neq 0$ and arbitrarily close to $g$, at least on $\Omega$, therefore there would have to be a point of discontinuity on $g$ or $f^\prime$ has to be zero on $\partial \Omega$.

Assuming my reasoning is correct, how can I write this properly? Many thanks for your help.

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Either you want to prove that $C_c(\Omega)$ is dense in $C_0(\Omega)$ or you want to prove that $\operatorname{cl}(C_c(\Omega)) = C_0(\Omega)$... Anyway, what is $\partial \Omega$ - this makes no sense unless $\Omega$ is a subspace of some other space (I assume that $\Omega$ is a locally compact space from your previous homework)? Use the definition: Given $f \in C_0(\Omega)$ and $\varepsilon \gt 0$ there is a compact set $K$ outside of which $|f| \lt \varepsilon$. Restrict $f|_K$ and use Tietze's extension theorem to find a continuous and compactly supported $g$ uniformly close to $f$. –  t.b. Oct 2 '11 at 14:14
What's $\Omega$? –  Dylan Moreland Oct 2 '11 at 14:14
@t.b.: Thanks, I forgot to say what $\Omega$ is. But no, this is a different question unrelated to my previous homework question. –  Rudy the Reindeer Oct 2 '11 at 14:21

If $g$ is in the closure of $C_c(\Omega)$, you want to show two things: $g$ is continuous, and $g = 0$ on $\partial \Omega$.
The second one is even easier: take some $x \in \partial \Omega$ and any $\epsilon > 0$. There exists $f \in C_c(\Omega)$ with $||g-f||_\infty < \epsilon$; in particular $|g(x) - f(x)| < \epsilon$. But $f(x)=0$ so $|g(x)|<\epsilon$. This is true for any $\epsilon$ so $g(x) = 0$.
You could also think in terms of sequences: a sequence of functions converging uniformly also converges pointwise, so $g(x)$ is the limit of a sequence that's identically zero.
Thanks, Nate. Yes, the first one is the proof of the uniform limit theorem with the three $\varepsilon / 3$ –  Rudy the Reindeer Oct 4 '11 at 20:01