Consider the differential operator
$$ A:=\sum_{|\alpha|\leq m}a_{\alpha}(x)D^{\alpha}, $$ where $\alpha$ is a mutiindex on $\mathbb{R^n}$.
It can be checked that the Schwartz kernel of $A$ is
$$ K(x,y)=\sum_{|\alpha|\leq m}a_{\alpha}(x)(D^{\alpha}\delta)(x-y). $$
I would like to prove that
$$ \text{supp}Au\subset\text{supp}u,\quad\text{and}\quad\text{sing supp}Au\subset\text{sing supp}u, $$
where $u\in\mathscr{S}'$
From Hintz's notes (Page10) https://people.math.ethz.ch/~hintzp/notes/micro.pdf, there states that:
From the perspective of the Schwartz kernel K of A, the first inclusion is really due to the fact that $K(x, y)$ is supported on the diagonal $x = y$, while the second inclusion is really due to the fact that $K(x, y)$ is smooth away from $x = y$. (That is, adding to $K$ an element of $\mathscr{S}(\mathbb{R}^{2n})$ preserves the second inclusion, but destroys the first inclusion.).
I don't really understand what this means, i.e., I don't know how to understand these two inclusions from perspective of Schwartz kernel. Can anyone please explain the idea behind it? Thanks a lot!
The main idea is working with the formula:
\begin{equation} Au(x)=\int K(x,y)u(y)dy \end{equation}
and the set operation for $A\subset X\times Y$, $B\subset Y$:
\begin{equation} A\circ B = \{x\in X : \exists y\in Y \text{ so that } (x,y)\in A,y\in B\} \end{equation}
(draw some pictures in $X=Y=\mathbb{R}^2$ for intuition).
The formula then basically tells you that:
\begin{equation} supp \;Au \subset supp \; K \; \circ \; supp \; u \end{equation} because the LHS can only be non-zero if the integrand in the RHS is at some point non-zero, but it could also happen that even with non-zero integrand, the integral vanishes (therefore $\subset$ and not $=$). Analogous reasoning with $singsupp$ instead of $supp$, since differential operators preserve smoothness and extend to distributions.
Look again at the form of $K$ and notice that $supp \; K = singsupp \; K \subset Diagonal(X\times X)$, since it is a linear combination of derivatives of delta functions in $(x-y)$, so only non-zero for $x=y$ (and always singular when non-zero).
This means $supp \; Au \subset Diagonal(X\times X) \;\circ supp \; u = supp \; u$ and same for $singsupp$. Notice that the last equality follows because in the case $X=Y$ the diagonal is a left identity for $\circ$ (check).