When to exchange integration and application of a distribution 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.
 A: I'll do it for when $f(x)$ has compact support; the general Schwartz function case is a modification of this. I use the fact that the Schwatrz functions are sequentially dense in the distributions, meaning that there is a sequence of Schwartz functions $g_n$ such that $\int g_n(x) \alpha(x)$ converges to $T(\alpha)$ as $n$ goes to infinity for any Schwartz function $\alpha(x)$. 
If you replace $T$ by $g_n(x)$, then your identity holds by Fubini's theorem. So one takes limits now as $n$ goes to infinity. The left hand side converges to $T \int f(x - \circ)\phi(\circ) d\circ$ immediately.
For the right-hand side, for each $x$ one has that $g_n(f(x - \circ) \phi(\circ))$ converges to $T(f(x - \circ) \phi(\circ))$. Since the $g_n(f(x - \circ) \phi(\circ))$ will be uniformly bounded (the construction of the $g_n$ is quite explicit and readily implies this) and since $f(x - \circ) \phi(\circ)$ is the zero distribution for $x$ outside a compact set, the dominated convergence theorem gives that the right hand side converges to $\int T (f(x - \circ) \phi(\circ)) dx$ as needed.
