# Is an infinite sequence of orthogonal functions in $H$ closed in $H$?

Consider some countably infinite sequence of elements $$f_n$$, each belonging to an infinite dimensional Hilbert space $$H$$, that are all orthogonal to every other member of the sequence. Is this set closed on $$H$$?

A comment: For a Hilbert space with finite dimension $$k$$, it appears straightforward to show that an orthogonal sequence of elements of size $$k$$ is closed in that space. I was wondering if the infinite-dimensional nature of some Hilbert spaces throws a wrench in that.

$$f_n$$ could very well converge to zero (without $$0$$ being one of its elements): consider $$e_n$$ the canonical Hilbert basis of $$\ell^2$$ and $$f_n=2^{-n}e_n$$.
Added: With the additional condition that $$\inf\limits_{n\in\Bbb N}\lVert f_n\rVert>0$$, the support of the sequence is indeed closed: notice that $$\lVert f_n-f_m\rVert=\sqrt{\lVert f_n\rVert^2+\lVert f_m\rVert^2-2\left\langle f_n;f_m\right\rangle}=\sqrt{\lVert f_n\rVert^2+\lVert f_m\rVert^2}\ge \sqrt2 \inf\limits_{n\in\Bbb N}\lVert f_n\rVert$$ Therefore the sequence itself has no Cauchy subsequences, by which it follows that its range must be a discrete closed subset.
• @aghostinthefigures No, because there are not: the subsequent discussion contains basically all the ingredients to prove that either $\overline{\{f_n\,:\,n\in\Bbb N\}}=\{f_n\,:\,n\in\Bbb N\}$ or $\overline{\{f_n\,:\,n\in\Bbb N\}}=\{f_n\,:\,n\in\Bbb N\}\cup\{0\}$. – Gae. S. Sep 8 '19 at 21:32