 Solution

Let $d={\inf }_{v\in V}\Vert f-v\Vert$ and choose $v_n\in V$ with $\Vert f-v_n\Vert \to d$. The parallelogram identity givesCompleteness and closedness give $v_n\to v\in V$, attaining the distance. Put $w=f-v$. For $u\in V$ and real $t$, minimality gives $\Vert w-tu{\Vert }^2\ge \Vert w{\Vert }^2$, so $\mathrm{Re}⟨w,u⟩=0$. In a complex Hilbert space, also replace $u$ with $iu$ to get the imaginary part zero. ThereforeThe decomposition is unique since $V\cap V^{\perp }=\{0\}$. This proves the required orthogonal projection result rather than assuming it before the averaging argument.