= Solution
The <orthogonal projection> satisfies $\Pi L=\Pi(\Pi-I)=0$, hence $d\Pi f_t/dt=0$ and $\Pi f_t=\Pi f_0=\rho(f_0)M$. Set $h_t=f_t-\Pi f_0$. Then $\Pi h_t=0$ and $\dot h_t=Lh_t$. Crucially, applying the energy identity of part (e) to $h_t$ gives
$$
\frac{d}{dt}\|h_t\|_H^2=2\operatorname{Re}\langle Lh_t,h_t\rangle=-2\|h_t-\Pi h_t\|_H^2=-2\|h_t\|_H^2.
$$
Solving this scalar equation and taking square roots proves
$$
\boxed{\|f_t-\rho(f_0)M\|_H=e^{-t}\|f_0-\rho(f_0)M\|_H}.
$$
This argument uses (e) explicitly, rather than bypassing the requested <energy method> with an explicit solution formula.
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