# Radial Schwarz functions are dense in radial tempered distributions

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.
• Thanks. How does one know that $T\ast{\phi(kx)}$ is a smooth function? The only definition I have seen for the convolution of a tempered distribution with a Schwarz function gives another tempered distribution as opposed to a function. Commented May 9, 2020 at 10:32
• It is continuous by definition of distribution and so does its derivative $T'\ast \phi(kx)$ and so on Commented May 9, 2020 at 11:48
• I think maybe I have a different definition of the convolution, to me $T\ast{\phi}$ is a tempered distribution itself, defined by $\langle{T\ast{\phi}},\psi{\rangle}:=\langle{T},\tilde{\phi}\ast{\psi}\rangle$, but perhaps this is uncommon. Commented May 9, 2020 at 13:19
• $\langle{T\ast{\phi}},\psi{\rangle}:=\langle{T},\phi\ast{\psi}\rangle$ And it is immediate that it is a function $T\ast \phi(a) = <T,\phi(a-.)>$ Commented May 9, 2020 at 13:21