Past exam of the analysis course of the University of Oxford / 2010 / B / 6 / a

For finite subsets $F\subset \mathbb{Z}$, orthonormality givesFor $S_N={\sum }_{|n|\le N}{\alpha }_n{e}_n$, the squared norm of $S_N-S_M$ is the corresponding finite tail sum. Since ${\sum }_{n\in \mathbb{Z}}|{\alpha }_n{|}^2<\infty$, these tails tend to zero. Thus $(S_N)$ is a Cauchy sequence in the complete Hilbert space and converges.

The same estimate for arbitrary finite tails proves that the sum is unconditional, so the bilateral orthogonal series converges in $X$ independently of its ordering. This is the Riesz-Fischer theorem for an orthonormal system.

For $a_n=⟨x,e_n⟩$, the finite orthogonal projection $P_{N}x={\sum }_{|n|\le N}{a}_n{e}_n$ satisfiesThis is Bessel's inequality, so the coefficients are square summable. The preceding result gives a limit $s={\sum }_n{a}_n{e}_n$. For each fixed $k$, continuity of the inner product gives $⟨x-s,e_k⟩=0$. The completeness assumption on the orthonormal system forces $x-s=0$. HenceTaking squared norms also gives Parseval's identity, $\Vert x{\Vert }^2={\sum }_n|⟨x,e_n⟩{|}^2$.

Apply continuity of the inner product to the convergent expansion of $x$:The series is absolutely convergent because the Cauchy-Schwarz inequality for scalar sequences and Bessel's inequality giveThis is the polarized Parseval identity.