Lemma: Compact time-slice test

For every compact smooth test $r$ and every $\mu \in \{0,m\}$, there is $r_N\in {C_c}^{\infty }(N)$ such that $E_{\mu }{r}_N=E_{\mu }r$. Moreover

Proof. Choose $\chi (t)$ equal to zero below a transition interval inside $(a,b)$ and one above it. Put

The causal solution $E_{\mu }r$ is spatially compact, so $r_N$ has compact support in the transition band. The compact function

satisfies $r-r_N=(\square +{\mu }^2)q$ by (eq:cutoff). Therefore $E_{\mu }{r}_N=E_{\mu }r$.

If $r$ is supported in $N$ and $E_{\mu }r=0$, then ${E_{\mu }}^{\mathrm{R}}r={E_{\mu }}^{\mathrm{A}}r=q$. Its support lies in $J^{+}(\mathrm{supp}r)\cap J^{-}(\mathrm{supp}r)$, a compact subset of $N$. Hence $r=(\square +{\mu }^2)q$ with $q\in {C_c}^{\infty }(N)$. The reverse inclusion is immediate.