 Solution

Since $\rho (M)=1$, ${\Pi }^{2}f=\rho (f)\rho (M)M=\Pi f$. Also $Lf=0$ exactly when $f=\rho (f)M$, which is exactly the range of $\Pi$. ThusA bounded self-adjoint idempotent is an orthogonal projection: if $u=\Pi u$ and $w=(I-\Pi )w$, then $⟨u,w⟩=⟨u,\Pi w⟩=0$. Here the orthogonal complement is precisely the zero-mass subspace $\rho (f)=0$.