I don't know if it's true but I think in a finite dimensional veccter space, an orthonormal set which is complete becomes a basis. But in a Hilbert space, the books say that the set of finite linear combinations of the elements in a complete orthonormal set is only dense in the Hilbert space. So why aren't they equal? Does the problem lie in closedness or infinity of dimension? Would you please give me an example of a Hilbert space where a complete orthonormal set is not a basis?

Thanks to everyone.


For any infinite sequence $u_n$ of distinct members of your orthonormal set, you can take convergent infinite sums such as $\sum_{n=1}^\infty u_n/n$, i.e. $\lim_{N \to \infty} \sum_{n=1}^N u_n/n$ (convergent in the Hilbert space norm). The limit exists because the Hilbert space is a complete metric space. It's easy to prove that the limit is not a linear combination of finitely many members of the orthonormal set.

A complete orthonormal set in a Hilbert space is called an "orthonormal basis", but this use of the term "basis" is different from the ordinary vector space "basis".

  • $\begingroup$ Thank you very much. I don't know why is $\lim_{N \rightarrow \infty} \sum_{n=1}^N u_n/n$, but I think $ \lim_{N\rightarrow \infty} \sum_{n=1}^N u_n/n^2$ must be convergent. $\endgroup$ – ShinyaSakai Jan 2 '13 at 11:35
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    $\begingroup$ The limit exists because $\|\sum_{n=N}^M u_n/n\|^2 = \sum_{n=N}^M 1/n^2 < \sum_{n=N}^\infty 1/n^2$, and the latter goes to $0$ as $N \to \infty$. $\endgroup$ – Robert Israel Jan 2 '13 at 19:28
  • $\begingroup$ I see. Thank you very much. $\endgroup$ – ShinyaSakai Jan 4 '13 at 9:11

$\ell^2(\mathbb{Z})$ has a countable orthonormal basis in the Hilbert space sense but is a vector space of uncountable dimension in the ordinary sense. It is probably impossible to write down a basis in the ordinary sense in ZF, and this is a useless thing to do anyway. The whole point of working in infinite-dimensional Hilbert spaces is that the topology matters and helps you do useful things, and you shouldn't ignore it, so you should incorporate it into all of your definitions (including your definition of what constitutes a reasonable notion of basis).

Note that $\ell^2(\mathbb{Z})$ does not have an orthonormal basis in the ordinary sense (that is, a basis which consists of orthonormal vectors) at all (exercise).

  • $\begingroup$ I see. Thanks you very much. $\endgroup$ – ShinyaSakai Jan 2 '13 at 11:38

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