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If $v \in M^\bot$, then $Pv=0$, since $P$ is the orthogonal projection on $M$.
Further, Proposition (3.1) shows that $M^\bot$ is invariant under each $\pi(x)$. Hence, $\pi(x)v \in M^\bot$. With the same argument as above, we get $P \pi(x)v=0$.
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If $v \in M^\bot$, then $Pv=0$, since $P$ is the orthogonal projection on $M$.
Further, Proposition (3.1) shows that $M^\bot$ is invariant under each $\pi(x)$. Hence, $\pi(x)v \in M^\bot$. With the same argument as above, we get $P \pi(x)v=0$.