Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / P / P3 / d

Set $u={\sum }_{k=1}^n⟨x,e_k⟩e_k$, with the convention that the inner product is linear in its first argument. Orthonormality gives $⟨x-u,e_j⟩=0$, hence $x-u\perp U_n$. For any $v\in U_n$ the Pythagorean identity yieldsThus this $u$ is the best approximation, and uniqueness gives $\boxed{P_{n}x=\sum _{k=1}^n⟨x,e_k⟩e_k}$. No completeness assumption on $H$ is necessary for this finite-dimensional construction.

Take the squared norm of the orthonormal expansion from (i). All mixed inner products vanish and every diagonal inner product is one, so

The decomposition $x=P_{n}x+(x-P_{n}x)$ is orthogonal by (i). Therefore $\Vert x{\Vert }^2=\Vert P_{n}x{\Vert }^2+\Vert x-P_{n}x{\Vert }^2$. Substitute (ii) and rearrange to obtainIn particular nonnegativity of the left side gives the finite Bessel inequality.