Integral kernel smooth Assume that $Tf(x):=\int_{\mathbb{R}^n} K(x,y)f(y) dy$ is an operator such that $T \in L( H^{-k}, H^k)$ is continuous for any $k$, where $H^k$ is the $k-th$ order Sobolev space on $\mathbb{R}^n$. 
How can one prove that $K(\cdot,\cdot)$ is in fact a $C^{\infty}$-function? 
 A: Your operator $T$ sends $H^{-\infty}=\cup_{s\in \mathbb R} H^s$ into 
$H^{+\infty}=\cap_{s\in \mathbb R}H^s$, and thus the Dirac mass at $y_0$ to a $C^\infty$ function of $x$, which means 
$$
\forall y_0,\quad x\mapsto K(x,y_0) \quad\text{is smooth}.
$$
Tha adjoint operator $T^*$ sends also $H^{-\infty}$ into $H^{+\infty}$ and its kernel is $K^*(x,y)=\overline{K(y,x)}$, proving also that 
$$\forall x_0,\quad
y\mapsto K(x_0,y)\quad\text{is smooth.}
$$
Using the derivatives of the Dirac mass, as suggested in the Christian Remling's comment leads to
$$\forall \alpha,
\forall \beta, \forall y_0,\quad x\mapsto \partial_x^\alpha\partial_y^\beta K(x,y_0) \text{ is smooth}\text{ and}\quad
\forall x_0,\quad y\mapsto \partial_x^\alpha\partial_y^\beta K(x_0,y)\text{ is smooth.}
$$
To obtain that $K$ is indeed a smooth function of $(x,y)$, it is enough to note that your continuity assumption implies that you have local bounds for these derivatives: for instance taking as $f$ the Dirac mass at $y_0$, for $y_0$ in a compact set multiplied by a smooth compactly supported function will provide bounds for $K(x,y)$ on a compact set. Same thing for the derivatives.
