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

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.