If $(e_1,\ldots)$ is an orthonormal basis and $x$ is orthogonal to the $(e_i)$, must $x$ be $0$?

Question

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.

Answer 1

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 $||x||=0$.