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I an reading a book about functional analysis and there is one thing i really don't understand. Let $\mathcal{H}$ be a Hilbert space. And $U \subset \mathcal{H}$ a closed subspace. Is it possible to choose an orthonormal basis $\{e_{i}\}_{i=1}^{\infty}$ such that there exists a subsequence of the $e_{i}$'s that span $U$ ?

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You've been watching too much Colbert Report :-) – joriki Oct 25 '11 at 21:01
my question is: if there exists $B = \{e_{i_{k}}\}_{k=1}^{\infty}$ such that $B$ is an orthonormal basis of $U$? – kim Oct 25 '11 at 21:08

As abatkai points out, this is true when $\mathcal{H}$ is separable.

A more general fact: if $A \subset \mathcal{H}$ is any orthonormal set, there exists an orthonormal basis $B$ which contains $A$. The proof is along the same lines as the proof that an orthonormal basis exists: consider the collection of all orthonormal sets which contain $A$, and use Zorn's lemma to find one which is maximal with respect to inclusion.

This implies your desired result, by taking $A$ to be an orthonormal basis of $U$. When $\mathcal{H}$ is separable, $B$ (and hence also $A$) will be countable, and so you can write $B$ as a sequence and $A$ as a subsequence.

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Assuming that your $\mathcal{H}$ is separable, the answer is yes. Then every closed subspace is a separable Hilbert space itself, meaning that both $U$ and $U^{\bot}$ have orthonormal basis $\{f_i\}$ and $\{g_i\}$. You just have to comb the two sequences together, then the odd subsequence will give you the basis of your subspace.

Added after remarks: The arguments above work if both $U$ and $U^{\bot}$ are infinite dimensional. Clearly, if $U$ is finite dimensional, then the answer is no (but you get a finite sequence of as an ONB}, and if $U^{\bot}$ is finite dimensional, then you can choose your subsequence by dropping the finitely many elements.

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Not quite: $U$ or $U^\perp$ could be finite-dimensional. – Robert Israel Oct 25 '11 at 22:33
Correct. But then, the question is easier... – abatkai Oct 25 '11 at 22:58

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