Integral of test function is a test function? (Application in tensor theory of distributions) Let $f:\mathbb{R}^m\to\mathbb{C}$ and $g:\mathbb{R}^n\to\mathbb{C}$, then we can define the tensor product $f\otimes g$ of $f$ and $g$ as the function $\mathbb{R}^{m+n}\to\mathbb{C}$ for which $f\otimes g(x,y)=f(x)g(y)$.
If $f$ and $g$ are locally integrable on resp. $\mathbb{R}^m$ and $\mathbb{R}^n$, then so is $f\otimes g$ on $\mathbb{R}^{m+n}$. We can therefore consider the regular distribution $T_f$,$T_g$ and $T_{f\otimes g}$ associated with each of these functions. The latter one is defined as $$ \langle T_{f\otimes g},\phi(x,y)\rangle = \int_{\mathbb{R}^{m+n}} f\otimes g(x,y)\phi(x,y)dxdy = \int_{\mathbb{R}^m} f(x) \left(\int_{\mathbb{R}^n} g(y)\phi(x,y)dy\right) dx, \text{ for } \phi\in \mathcal{D}(\mathbb{R}^{m+n}).$$ For this last integral to be well-defined, we would need that $\psi(x) := \int_{\mathbb{R}^n} g(y)\phi(x,y)dy$ is a test function over $\mathbb{R}^m$, i.e. $\psi \in \mathcal{D}(\mathbb{R}^m)$. I don't see how I could show this rigorously. Does continuity of $\psi$ follow directly? Can I be sure that the integral will be continuous? As for the support, is taking the intersection of $[\phi]$ with a compact $K_1\subset \mathbb{R}^m$ for which $[\phi]\subset K_1\times K_2$ (and $K_2\subset \mathbb{R}^n$) enough?
Thanks in advance.
 A: First we should note that you really don't need this to show that $f\otimes g$ defines a distribution. You can just note that $f\otimes g$ is locally integrable, so $\langle T_{f\otimes g},\phi(x,y)\rangle = \int_{\mathbb{R}^{m+n}} f\otimes g(x,y)\phi(x,y)dxdy$ defines a distribution with no  problem.
But in fact yes, $\psi$ is a test function. You can use Leibniz' rule for differentiating under the integral sign to show $\psi$ is smooth. Choose $R>0$ so $\phi(x,y)=0$ for all $y$  if $|x|>R$. Now it's clear that $\psi(x)=0$ for $|x|>R$.
A: David already gave a nice self-contained answer, but I would like to add that this is just a special case of Fubini's Theorem for Schwartz distributions. See
https://mathoverflow.net/questions/72450/can-distribution-theory-be-developed-riemann-free/351028#351028
There I discussed the temperate case but the same holds with the change of spaces
$$
\mathscr{S}\rightarrow \mathscr{D}=C_{\rm c}^{\infty},\ 
\mathscr{S}'\rightarrow \mathscr{D}',\ 
\mathscr{O}_{\rm M}\rightarrow \mathscr{E}=C^{\infty}. 
$$
