I am 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 $U$ ?
[Math] Orthonormal basis in Hilbert space
functional-analysishilbert-spaces
Best Answer
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.