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

A best approximation to $x$ from $U_n$ is a vector $u_{\ast }\in U_n$ such thatThe definition requires attainment of the infimum. This finite-dimensional subspace does have such an approximation, as the explicit projection below shows; completeness of the whole inner-product space is unnecessary.

Expand both squared norms using the inner product:Adding cancels the cross terms, giving the parallelogram lawThe real part may be omitted in a real inner-product space.

Suppose $u,v\in U_n$ both minimize the distance, with common value $d$. Apply the parallelogram identity to $x-u$ and $x-v$:The midpoint belongs to the linear subspace $U_n$, so its distance from $x$ is at least $d$. It follows that $\Vert u-v{\Vert }^2\le 0$, hence $u=v$. Thus the best approximation is unique.

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

Expand the norm of the orthonormal-basis expression. All cross terms vanish and $\Vert e_k\Vert =1$, so

The residual $x-P_{n}x$ is orthogonal to $P_{n}x\in U_n$. Thus the Pythagorean theorem gives $\Vert x{\Vert }^2=\Vert x-P_{n}x{\Vert }^2+\Vert P_{n}x{\Vert }^2$. Substitute part (ii) and rearrange:Nonnegativity of the left-hand side also gives the finite form of Bessel's inequality.