I am reading Francois Treves' *Introduction to pseduodifferential and Fourier integral operators, vol. I.* I am having trouble understanding the proof of Lemma 2.1, which is stated as follows.

Let $\Omega \subseteq \mathbb{R}^d$ be open. Suppose $K : C^\infty_c(\Omega) \to C^\infty(\Omega)$ is sequentially continuous (this in particular means that we can extend to $K: \mathcal{E}'(\Omega) \to \mathcal{D}'(\Omega)$ by transposition). Also, suppose that the corresponding kernel $k \in \mathcal{D}'(\Omega \times \Omega)$ is very regular (meaning that it is $C^\infty$ in the complement of the diagonal of $\Omega \times \Omega$). Then for any $u \in \mathcal{E}'(\Omega)$, the singular support of $Ku$ is contained in the singular support of $u$ .

The proof strategy is to take arbitrary $u \in \mathcal{E}'(\Omega)$. Suppose that $U \subseteq \Omega$ is open with $u \in C^\infty(U)$. Take $V$ to be an arbitrary open set such that $V \subset \subset U$.Choose $\phi \in C_c^\infty(U)$ such that $\phi \equiv 1$ on $V$. And then look at

$$Ku = K(\phi u) + K((1 - \phi)u).$$

Since $\phi u \in C_c^\infty(\Omega)$, we have $K(\phi u) \in C^\infty(\Omega)$. I am having trouble showing that the second term is $C^\infty$. My attempt so far is to take $\psi \in C_c^\infty(V)$ and then calculate

$$K((1-\phi)u)(\psi) = ((1 - \phi) u)(K\psi) = \cdots $$ But I am not sure what to do with the calculation from here. I know I will somehow need to use the fact that $k$ is smooth away from the diagonal, but I can't figure out how to get $k$ to show up somewhere. If $(1 - \phi)u$ were $C_c^\infty(\Omega)$, this would be no problem (just use the statement of the Schwartz kernel theorem). But I don't have this since $u$ is just a compact supported distribution.

Hints or solutions are greatly appreciated!