Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 6 / 3 / c / Solution

The weighted inner product satisfies $⟨f,M⟩=\int f=\rho (f)$. ThereforeThe identity operator and $L$ are linear, bounded and self-adjoint, hence so is $\Pi$. Directly, $\Vert \Pi f{\Vert }_H=|\rho (f)|\le \Vert f{\Vert }_H$; equality holds at $f=M$. Thus $\Vert \Pi \Vert =1$.