AQFT: Can test functions obey the Klein-Gordon equation? In AQFT we can choose test functions with compact support. Can such functions obey a Klein-Gordon equation?
I start with a test function $g$ with compact support and I apply the Klein-Gordon operator: $\mathrm{KG}(g) = f$. I can use the advanced and retarded Green's functions to get solutions  $g_1$ and $g_2$ such that
$$\mathrm{KG}(g_1) = \mathrm{KG}(g_2) = f$$
and so $\mathrm{KG}(g_1 - g_2) = 0$.
If my $g$ (with compact support) was $g_1$, I get $g - g_2$ which obey the KG equation but what about the support of $g_2$?
 A: No, they cannot. There is no non-vanishing smooth KG solution with compact support. Let $\psi$ be a compactly supported solution (I assume $m>0$, the massless case is a bit more complicated) and $T_{ab}$ the associated stress energy tensor. The integral of $T_{00}\geq 0$ over a spacelike Cauchy surface at constant Minkowski time, $\Sigma$, does not depend on the chosen Cauchy surface. 
Since the support of $\psi$ is compact we can fix $\Sigma$ far away from the support of $\psi$, obtaining $\int_\Sigma T_{00} d^3x =0$ on $\Sigma$ which, in turn, implies that 
$T_{00}|_\Sigma=0$ (because $T_{00}\geq 0$) and thus
both $\psi$ and $\partial_0\psi$ vanishes on $\Sigma$ just in view of the form of $T_{00}$ ($m>0$). Since $\Sigma$ is a Cauchy surface, we have that $\psi=0$ everywhere in the spacetime.
Dealing with a generic globally hyperbolic spacetime you can rearrange this proof.
ADDENDUM. If $m=0$, the condition $T_{00}|_\Sigma =0$ implies $\partial_0\psi|_\Sigma=0$ and $\partial_k\psi|_\Sigma=0$ for $k=1,2,3$, referring to Minkowskian coordinates adapted to $\Sigma$ (thus corresponding to $t=t_0$).
Consequently $\psi$ is constant over $\Sigma$. Since the support of $\psi|_\Sigma$ is compact as well, it must be $\psi|_\Sigma=0$ reducing to the previously treated case.
