Show that a very regular kernel $k(x,y)$ has operator $K : \mathcal{E}'(\Omega) \to \mathcal{D}'(\Omega)$ which is pseudo-local 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!
 A: I encountered the same problem when I read Taylor's book on pseudo differential operators recently. I think we can prove this in the following way:
Choose $U,V,\phi$ in the same way as you did. Take $W\subset\subset V$ and fix any $\psi\in C_c^\infty(W)$. Take some $\gamma\in C^\infty(\Omega)$ such that $\gamma\equiv 1$ outsides $V$, $\gamma\equiv 0$ in $W$ and $\text{dist}(W,\text{supp }\gamma)>0$. Then \begin{aligned}\langle K((1-\phi)u),\psi\rangle&=\langle (1-\phi)u,K\psi\rangle\\&=\langle (1-\phi)u,\gamma K\psi\rangle\\&=\langle (1-\phi)u,\gamma(\cdot)\int k(\cdot,y)\psi(y)\ dy\rangle\\&=\int \psi(y)\langle (1-\phi)u,\gamma(\cdot) k(\cdot,y)\rangle\ dy.\end{aligned}The second equality holds since $\phi\equiv 1$ in $V$; the third equality holds since $\text{dist}(W,\text{supp }\gamma)>0$; the last equality holds from the continuity of distributions. Thus, \begin{aligned}K((1-\phi)u)(y)=\langle (1-\phi)u,\gamma(\cdot) k(\cdot,y)\rangle\end{aligned} for all $y\in W$, which is smooth in $W$.
