"Standard reference" for $C_c^\infty(\mathbb R)$ is dense in $C_c(\mathbb R)$

Question

$C_c^\infty(\mathbb R)$ is dense in $C_c(\mathbb R)$. This can be shown by mollification. This is a well-known, widely used fact. However, I wasn't able to find any book which I could point in a reference to. Is there any kind of "standard reference" with a readable proof?

Answer 1

(I am assuming you are equipping $C_c^{\infty}(\mathbb R)$ and $C_c(\mathbb R)$ with the sup-norm). One can use the Stone-Weierstrass Theorem for locally compact Hausdorff spaces to show the result (for references to the Stone-Weierstrass Theorem, see Willard's General Topology Section 44 or Folland Chapter 4). In fact, the Stone-Weierstrass Theorem yields a stronger result: $C_c^{\infty}(\mathbb R^n)$ is dense in $C_0(\mathbb R^n)$ when both spaces are given the topology of uniform convergence. The sum, product, scalar multiple, and complex conjugate of smooth compactly supported functions is easily verified to also be smooth and compactly supported.

The fact that $C_c^{\infty}(\mathbb R^n)$ separates points and vanishes nowhere follows from the following theorem in Folland:

Theorem (Folland, 8.18). Let $K \subseteq \mathbb R^n$ be nonempty and compact, and let $U$ be an open set with $U \supseteq K$. Then, there is $f \in C_c^{\infty}(\mathbb R^n) $ such that $0 \leq f(x) \leq 1 $ for all $x \in \mathbb R^n$, $f(K)=\{1\}$, and $\text{supp}(f) \subseteq U$.

Theorem 8.17 in Folland also proves this in a different way using an approximation of the identity.

Answer 2

I don't have a reference at hand. But one can prove this without too much trouble using the Weierstrass approximation theorem.

Suppose $f\in C_c$ with support contained in $[a,b].$ Let $\epsilon>0.$ Choose $a'<a$ and $b'>b.$ Then there exists a polynomial $p$ such that $|p-f|<\epsilon$ on $[a',b'].$

Now there exists $g\in C^\infty_c(\mathbb R)$ with support in $[a',b']$ such that $0\le g\le 1$ everywhere, and $g=1$ on $[a,b].$ We then have $gp\in C^\infty_c(\mathbb R),$ and $|gp -f|<\epsilon$ everywhere. I'll leave the vefication of the last line to the reader; ask questions if you like.