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

A best approximation to $x$ from $U_n$ is an element $u_{\ast }\in U_n$ satisfying $\boxed{\Vert x-u_{\ast }\Vert =\underset{u\in U_n}{\inf }\Vert x-u\Vert }$. This is a minimum over the specified approximation subspace, not over all of $H$.

Expand using the inner product. The two cross terms cancel:Adding proves $\boxed{\Vert x-y{\Vert }^2+\Vert x+y{\Vert }^2=2(\Vert x{\Vert }^2+\Vert y{\Vert }^2)}$, for real or complex inner products.

Suppose $u,v\in U_n$ both attain the minimum distance $d$. Apply (b) to $x-u$ and $x-v$:The midpoint belongs to $U_n$, so its squared distance is at least $d^2$. Thus $\Vert u-v{\Vert }^2\le 0$, forcing $\boxed{u=v}$. Existence will also follow constructively in (d)(i).

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.