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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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – user294388
    May 9, 2020 at 10:32
  • $\begingroup$ It is continuous by definition of distribution and so does its derivative $T'\ast \phi(kx)$ and so on $\endgroup$
    – reuns
    May 9, 2020 at 11:48
  • $\begingroup$ 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. $\endgroup$
    – user294388
    May 9, 2020 at 13:19
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    $\begingroup$ $\langle{T\ast{\phi}},\psi{\rangle}:=\langle{T},\phi\ast{\psi}\rangle$ And it is immediate that it is a function $T\ast \phi(a) = <T,\phi(a-.)>$ $\endgroup$
    – reuns
    May 9, 2020 at 13:21

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