Let $T$ be a distribution, $f$ be smooth. It is known that
$T(f(x - \cdot) ) \in C ^ \infty$ .
On the other hand, for $\phi$ a test function
$( T \ast f ) (\phi) = T( f( - \cdot ) \ast \phi ) = T( \int f( x - \cdot ) \phi( \cdot ) )$,
because the convolution is still a distribution. However, with regard to the first result we expect
$ T( \int f( x - \cdot ) \phi( \cdot ) ) = \int T ( f( x - \cdot ) \phi( \cdot ) )$.
This intuitively makes sense, as $T$ is a distribution in the $x$-variable, whereas we integrate over the unnamed second variable.
Nevertheless, I don't know a proof of the last equality, and I am not familiar enough with the topologies of the smooth functions, test functions and distribution, to use approximative arguments. Can you help?
Thanks.