I am trying to show that any radial tempered distribution can be approximated by radial Schwarz functions, where $T\in{S}'(\mathbb{R}^n)$ being radial means $\langle{T},\phi\circ{R^T}\rangle=\langle{T,\phi}\rangle$ for all $\phi\in{S(\mathbb{R}^n)}$, $R\in{SO(n)}$.
I have so far shown that $T$ is radial iff its Fourier transform is, and that if $T\in{S'}$, $\psi\in{S}$ are radial then so are $\psi{T}$ and $T\ast{\psi}$.
I want to approximate $T$ by something like $T\ast{\phi_{\epsilon}}$ for $\phi_{\epsilon}$ a mollifying sequence, but can't see why this would be identifiable with a Schwarz function.
Thanks for any help.
With $\phi\in C^\infty_c$ radial $\ge 0$ and $\int \phi = 1$ then
$\phi(x/k) (T\ast k^n\phi(kx))$ is radial and $C^\infty_c$ and it converges to $T$
A good exercice is to show that replacing $\phi$ by $e^{-\pi |x|^2}$ works.