Complete orthonormal bases give norm-convergent projections and Parseval

Fix $x\in H$ and $N$. Orthonormality gives $⟨x-P_{N}x,e_j{⟩}_H=0$ for each $j\le N$, so $x-P_{N}x$ is orthogonal to both $P_{N}x$ and every vector in the first-$N$ span. By [F1], $\mathrm{\parallel }x{\mathrm{\parallel }}_H^2=\mathrm{\parallel }{P}_{N}x{\mathrm{\parallel }}_H^2+\mathrm{\parallel }x-P_{N}x{\mathrm{\parallel }}_H^2,\mathrm{\parallel }{P}_{N}x{\mathrm{\parallel }}_H^2={\sum }_{k\le N}\mathrm{\mid }⟨x,e_k{⟩}_H{\mathrm{\mid }}^2\mathrm{.}$ For any $y$ in that span, $P_{N}x-y$ also lies in the span and is orthogonal to $x-P_{N}x$, hence $\mathrm{\parallel }x-y{\mathrm{\parallel }}_H^2=\mathrm{\parallel }x-P_{N}x{\mathrm{\parallel }}_H^2+\mathrm{\parallel }{P}_{N}x-y{\mathrm{\parallel }}_H^2\ge \mathrm{\parallel }x-P_{N}x{\mathrm{\parallel }}_H^2\mathrm{.}$ Thus $P_{N}x$ is a closest vector to $x$ in the first-$N$ span.

Let $\epsilon >0$. By density choose a finite linear combination $y$ of the $e_k$ with $\mathrm{\parallel }x-y{\mathrm{\parallel }}_H<\epsilon$, and choose $N$ at least as large as every index used in $y$. Then $y$ lies in the first-$N$ span, so step 1.1 gives $\mathrm{\parallel }x-P_{N}x{\mathrm{\parallel }}_H\le \mathrm{\parallel }x-y{\mathrm{\parallel }}_H<\epsilon$. Hence $P_{N}x\to x$; the first identity in step 1.1 then gives the asserted Parseval limit. For a finite family, $P_{N}x=x$ once all its vectors are included.