1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $||x||=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.