Representation of $C_0(X)$ on $\ell^2(X)$ - density of the image

Question

Suppose $X$ is a locally compact (Hausdorff) space and consider the Hilbert space $\ell^2(X)$ of square summable functions $\phi:X\rightarrow \mathbb{C}$ that vanish outside a countable subset of $X$.

Define a representation $\pi:C_0(X)\rightarrow B(\ell^2(X))$ by $\pi(f)(\phi):=f\phi$.

Why $\pi(C_0(X))\ell^2(X)$ have to be dense in $\ell^2(X)$ ? Maybe it is obvious, but I do not see now any simple argument.

Answer 1

It will suffice to show that $\pi(C_0(X))\ell^2(X)$ contains $c_c(X)$, the space of functions $X \rightarrow \mathbb{C}$ vanishing outside a finite subset of $X$. Indeed, given a $\phi\in c_c(X)$ we may find a $C_c(X)$ function, $f$, that is $1$ on the support of $\phi$. Then $\phi=f \phi=\pi(f)\phi\in\pi(C_0(X))\ell^2(X)$.