The (2D) Radon transform $R$ is usually defined for functions in the Schwartz space $S(\mathbb{R}^2)$ or bump functions $C_c^\infty(\mathbb{R}^2)$ by \begin{align*} R\colon C_c^\infty(\mathbb{R}^2)&\to L^2(\mathbb{R}\times[0,\pi))\newline (Rf)(r,\varphi)&:=\int_{L(r,\varphi)}f(x) dx \end{align*} where $L(r,\varphi)$ is the line in the plane parameterized by the angle $\varphi$ and the directed distance from the origin $r$.
This definition ensures that the integral exists. It is easy to show (by estimating the operator norm) that the Radon transform is continuous when it is (for example) restricted to $C_c^\infty(\Omega)$, where $\Omega$ is the unit ball in $\mathbb{R}^2$.
Because $C_c^\infty(\Omega)$ is dense in $L^2(\Omega)$, the Radon transform then has a unique continuous extension to $L^2(\Omega)$.
In the literature however, I often see references to the Radon transform defined on $L^2(\mathbb{R}^2)$. But I do not understand how the Radon transform can be extended from the bump functions to $L^2(\mathbb{R}^2)$ since the proof for continuity only works for bounded sets.
Is the Radon transform continuous when defined on $C_c^\infty(\mathbb{R}^2)$?
If so, how can this be shown?
If not, how can it then be extended to $L^2(\mathbb{R}^2)$?