orthonormal basis of infinite dimensional Hilbert space H is not a basis of H as vector space?

Question

Apparently the orthonormal basis $(e_n)_{n\in \mathbb{N}}$ of the Hilbert space $H$ (in special case, infinitly dimensional) is not a basis of $H$ as a vectorspace. Is there a way to prove this?

Answer 1

A basis means that for everu $u\in H$ you have $u=a_{i_1}e_{i_1}+...+a_{i_n}e_{i_n}$ for $n$ finite, and orthogonal basis means that $u=\sum_{n\geq 0}a_ne_n$ where the sequence $\sum_{i=0}^{i=n}a_ie_i$ converges, so for orthogonal basis, all the coefficients may not be zero.

Answer 2

Every vector space basis of an infinite dimensional Banach or Hilbert space is necessarily uncountable. This is actually a non-trivial result and follows from the Baire category theorem, see e.g. Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.

Answer 3

First show or recall that every $x\in H$ has a unique representation $$x=\sum a_ne_n,$$with $\sum|a_n|^2<\infty$. So if $$x=\sum_{n=1}^\infty e_n/n$$then the uniqueness shows that $x$ is not a (finite) linear combination of the $e_n$.