Can we approximate a $C_0(\mathbb R)$-function by $C_c^\infty(\mathbb R)$-functions?

Question

Let $C_0(\mathbb R)$ denote the closure of $C_c(\mathbb R)$ with respect to the supremum norm.

Are we able to show that the closure of $C_c^\infty(\mathbb R)$ with respect to the supremum norm is $C_0(\mathbb R)$?

Answer 1

Let $f \in C_0(\mathbb R)$ and $\epsilon >0$. There exists $M>0$ such that $|f(x)| <\epsilon $ for $|x|>M$. There exists a polynomial $p$ such that $|f(x)-p(x)| <\epsilon$ for $|x| \leq M+1$. There exists a function $g \in C_c(\mathbb R)$ such that $0\leq g(x)\leq 1$ for all $x$ , $g(x)=1$ if $|x| \leq M$ and $g(x)=0$ if $|x| >M+\epsilon$. Let $h=pg$. Then $h \in C_c(\mathbb R)$, $$|f(x)-p(x)g(x)|=|f(x)-p(x)|<\epsilon$$ for $|x| \leq M$. For $|x| >M+\epsilon$ we have $$|f(x)-p(x)g(x)|=|f(x)| <\epsilon$$. Finally, for $M\leq |x| \leq M+\epsilon$ we have $$|f(x)-p(x)g(x)|\leq \epsilon+ |p(x)-p(x)g(x)| \leq \epsilon+ |p(x)|$$ (because $0\leq 1-g(x) \leq 1$) and $|p(x)|<\epsilon +|f(x)| <2\epsilon$. Remark: I just realized that I could have replaced $(M,M+\epsilon) $ by $(M,M+1)$ but that is only a minor point).