@Jochen I need indeed that for each $y$, $f(⋅,y)$ is a distribution on $\mathbb R^m_x$, then I can differentiate with the above rule. Using the $\overline{\partial}$ operator as suggested in my answer, you can prove a weak analyticity result. I doubt that much more could be proven: think about a smooth non-analytic function $f$ depending only on $x$. Sorry for the wrong reference for the variation on this topic: it is Theorem 2.1.3. – Bazin 4 hours ago

I guess that the top left $-\beta$ cannot be changed into $-\beta-\Delta$: that would make your matrix Toeplitz, a class on which much is known.

@BR I guess the cyrillic letter for S reads C. About the Schwartz space $\mathscr S(\mathbb R^n)$: Laurent Schwartz defined that space in the mid-forties and the letter S was not after his own name, but did stand at the time for ``Spherical functions'' which are smooth functions defined on the sphere with $n$ dimensions and that are flat at the north pole (all derivatives vanish there). By stereographic projection, you recover what is now known as the Schwartz space.

Federico Poloni is absolutely right.

Theorem 18.1.8 and its proof in H\"ormander's third volume of ALPDO.

This is true in the bornology sense defined in my answer.

It is true in the bornology sense defined in my answer.

The commutator $[A_1,A_2]$ of operators $A_j\in \Psi^{m_j}$ indeed belongs to $\Psi^{m_1+m_2-1}$. This is a key classical -but nontrivial- point in the theory of pseudodifferential operators. Just to support that result, it is easy to check directly that the Poisson bracket {$a_1,a_2$} of symbols $a_j\in S^{m_j}_{1,0}$ belongs to $S^{m_1+m_2-1}_{1,0}$.