Let $A:\mathcal{S}(\mathbb{R}^n) \to \mathcal{S}'(\mathbb{R}^n)$ be a continuous map and denote by $K \in \mathcal{S}'(\mathbb{R}^n)$ its Schwartz kernel. Show that $K \in \mathcal{S}(\mathbb{R}^2n)$ if and only if $A$ maps $\mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)$.
This is an exercise in Peter Hintz' lecture notes on microlocal analysis which can be found at https://people.math.ethz.ch/~hintzp/notes/micro.pdf. I am using them for exam preparation.
My ideas:\
$\Rightarrow$: Assume that $K \in \mathcal{S}(\mathbb{R}^{2n})$. Then: \begin{equation} \langle K, v \otimes u \rangle = \int_{\mathbb{R}^{2n}} K(x,y)v(x)u(y) dy = \int_{\mathbb{R}^n}\underbrace{(\int_{\mathbb{R}^n}K(x,y)u(y)dy)}_{Au(x)}v(x) dx \end{equation}
and now we can argue that for multiindices \begin{equation} \alpha, \beta: \text{sup}_{x \in \mathbb{R}^n}|x^{\alpha}D^{\beta} Au(x)| \end{equation}
can be written as: \begin{equation} \text{sup}_{x \in \mathbb{R}^n}|x^{\alpha}D^{\beta} \int_{\mathbb{R}^n}K(x,y)u(y)dy| < \infty \end{equation}
where in the last bound, we used that $K$ is a Schwartz-function in $x$, via dominated convergence.
Idea for the other direction:\
Maybe we can write $K(x,y) = \langle A \delta_y, \delta_x \rangle$