Solution
= Solution
Since $\rho(M)=1$, $\Pi^2f=\rho(f)\rho(M)M=\Pi f$. Also $Lf=0$ exactly when $f=\rho(f)M$, which is exactly the range of $\Pi$. Thus
$$
\boxed{\Pi^2=\Pi,\qquad\ker L=\operatorname{im}\Pi=\operatorname{span}\{M\}}.
$$
A bounded <self-adjoint> idempotent is an <orthogonal projection>: if $u=\Pi u$ and $w=(I-\Pi)w$, then $\langle u,w\rangle=\langle u,\Pi w\rangle=0$. Here the <orthogonal complement> is precisely the zero-mass subspace $\rho(f)=0$.