Consider the norm $\|f\|=\sup_{x>0}|xf^\prime(x)|$ on the space $\mathcal{C}_c^\infty(0,\infty)$, the space of smooth functions with compact support in $(0,\infty)$.

I want to prove that the closure of $\mathcal{C}_c^\infty(0,\infty)$ in the norm $\|\cdot\|$ is separable.


If you want to show that the closure is separable, then it is enough to show that the set before was separable.

Every function in $C_c^\infty(0;\infty)$ has support contained in some $[1/n;n]$ where $n\in \mathbb{N}$. However, on $[1/n;n]$ we can approximate using polynomial functions (by the Stone-Weierstrass theorem). To render it countable, we can use polynomial function with rational coefficients. Now we just need to extend the polynomial functions to all of $(0;\infty)$. This we can do by multiplying with a cut-off function with suitable support. Hence, the space is separable and therefore also its closure.


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