Let $p = \sum_\alpha c_\alpha D^\alpha \in \mathbb{R}[X_1,\dots,X_n]$ and $\mathscr{L} = p(D) = \sum_\alpha c_\alpha D^\alpha \newcommand{\L}{\mathscr{L}} \DeclareMathOperator{\diff}{Diff}, \DeclareMathOperator{\supp}{supp}$ a linear partial differential operator.
I'm trying to understand the following result,
Let $\varphi \in C_c^\infty(\mathbb{R}^n)$. If $\L \varphi = 0$ in an open semispace $H$, then $\varphi = 0$ in $H$.
The reference I am reading proves it first for $H_0 = \{x_1 > 0\}$. This I have understood. The claim then is that the general result follows by composing $\varphi$ with translations and rotations, which seems rather straight forward but I'm having a hard time formalizing it.
If $\L \varphi = 0$ in $H$, we want to see that $\varphi = 0$ in $H$. Taking $\psi \in \diff(\mathbb{R}^n)$ as a rotation composed with a translation this is equivalent to showing that $\varphi \psi$ vanishes on $H_0$, and so it suffices to show that $\L (\varphi \psi)$ vanishes on $H_0$.
Rephrasing the original hypothesis, we know that $(\L\varphi) \circ \psi = 0$ on $H_0$, and so I would like to show that if $(\L \varphi) \circ \psi$ vanishes, then so does $\L (\varphi\psi)$. In other terms, is the inclusion
$$ \supp \L (\varphi\psi) \subset \sup (\L\varphi) \circ \psi $$
true, and if so, how could one go about showing this?