About convolution of a distribution with test function I am fighting with the following problem: Given a a distribution T, I want to prove that $F_T:D(R^n)→C^\infty(R^n)$ defined by
$$F_T(ϕ)=T∗ϕ$$
(convolution of a distribution with a test function) is linear and continuous. I have proved that FT is well defined and is linear. How can I prove the continuity?
Any help or tip would be appreciated. Thanks in advanced.
 A: The topology of $\mathscr D(\mathbb R^n)$ is the locally convex inductive limit $\lim\limits_\to \mathscr D(K_k)$ where $K_k$ form a compact exhaustion of $\mathbb R^n$ (e.g., balls with center $0$ and radius $k$) and $\mathscr D(K_k)$ is the Frechet space of all test functions with support in $K_k$. Hence the continuity of the linear map $F_T$ is equivalent to the continuity of all restrictions to $\mathscr D(K_k)$. The latter can be shown either by direct estimations (the partial derivatives of $F_T(\varphi)$ are
$\partial^\alpha (T\ast \varphi)=T \ast \partial^\alpha \varphi$) or by the closed graph theorem (then does not need to deal with derivatives).
EDIT.
To see the continuity of $T_F$ on $\mathscr D(K_k)$ you have to check that it has closed graph and for this it is enough to see that $\varphi_m\to 0$ in $\mathscr D(K_k)$ implies $F_T(\varphi_m)(x)\to 0$ for each $x\in \mathbb R^n$. But $F_T(\varphi_m)(x)= (T\ast \varphi_m)(x)= T(\varphi_m(x-\cdot))$ and since the test function $\varphi_m(x-\cdot)$ again converge to $0$ (in $\mathscr D(K_k+x)$ and hence in $\mathscr D(\mathbb R^n)$) the continuity of $T$ implies $T(\varphi_m(x-\cdot))\to 0$.
