Past exam of the options II course of the University of Oxford / 2010 / P / P3 / d / i

Let $v={\sum }_{k=1}^n⟨x,e_k⟩e_k$, using the convention that the inner product is linear in its first argument. Orthonormality givesfor every $j$. Hence $x-v$ is orthogonal to $U_n$. For any $u\in U_n$, the Pythagorean identity gives $\Vert x-u{\Vert }^2=\Vert x-v{\Vert }^2+\Vert v-u{\Vert }^2$, so $v$ is the unique best approximation. Therefore