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Let $H,\langle\rangle$ be a pre-Hilbert space and $(e_1,\ldots,e_n,\ldots)$ an orthonormal basis of $H$.

Suppose that there is some $x$ such that $\forall i, \langle x,e_i\rangle=0$

Must $x$ be $0$ ?

This is obviously true when $\operatorname{Span}(e_i)$ is closed, or when $H$ is finite-dimensional. Otherwise, $\operatorname{Span}(e_i)^\bot$ needs not be in direct sum with $\operatorname{Span}(e_i)$.

I think a counter-example must exist, but I haven't found one.

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  • $\begingroup$ Are you aware of Parseval's identity? $\endgroup$ – Najib Idrissi Sep 28 '15 at 14:28
  • $\begingroup$ If $(e_1,e_2,...,e_n,...)$ is an orthonormal basis of $H$, this means $x=\sum\limits_{i=1}^{\infty}{a_i e_i}$ for some coefficients $a_i$. So $\langle x,e_i\rangle=\langle \lim\limits_{n\to \infty}{\sum\limits_{j=1}^{n}{a_j e_j}},e_i\rangle=\lim\limits_{n\to\infty}{\langle \sum\limits_{j=1}^{n}{a_j e_j},e_i\rangle}=a_i=0$. So all $a_i$ are $0$. Am I right ? $\endgroup$ – Svetoslav Sep 28 '15 at 14:28
  • $\begingroup$ @NajibIdrissi Yes I realized that. Why can I not freely delete a dumb question of mine? $\endgroup$ – Gabriel Romon Sep 29 '15 at 21:32
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    $\begingroup$ Because someone took the time to write an answer to your question, and your question could help future users. $\endgroup$ – Najib Idrissi Sep 30 '15 at 7:16
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I take it a pre-Hilbert space is just an inner product space? There are various definitions of "orthonormal basis" that are equivalent in a Hilbert space but not in a pre-Hilbert space; I'm assuming the $e_n$ are orthonormal and have dense span.

Then yes, $\langle x,e_n\rangle=0$ for all $n$ does imply $x=0$. Say $\epsilon>0$. Say $y$ is a linear combination of the $e_n$ and $||x-y||<\epsilon$. Then $$||x||^2=|\langle x-y,x\rangle|<\epsilon ||x||;$$hence $||x||=0$.

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