I haven't seen the following result stated as such, so I'm wondering if my proof is correct. I'd appreciate feedback.
Let $H$ be an infinite dimensional separable Hilbert space, and $D$ a proper dense subspace. Then:
- Any orthonormal basis (ONB) for $D$ is an orthonormal basis for $H$
- There is at least one ONB for $H$ which is an ONB for $D$.
- There is at least one ONB for $H$ which is not an ONB for $D$.
Proof:
- If $E$ is an ONB for $D$ then it is an orthonormal set in $H$ and so can be extended to an ONB for $H$. But one of the properties of a dense subset is that its orthogonal complement = $\{0\}$. So, $E$ is a maximal orthonormal set in $H$ and therefore an ONB for $H$.
- By Zorn's lemma, $D$ has some orthonormal basis $E$. By (1) this is also an ONB for $H$. Viewing this the other way round gives the result.
- Since $D$ is a proper subspace there is some $0 \ne e \in H \setminus D$. By linearity then $e/||e|| \in H \setminus D$. Then $\{e/||e|| \}$ is an orthonormal set (of one element) which can be expanded to an ONB, $E$, for $H$. But this cannot be an ONB for $D$ since $e/||e|| \not \in D$.
As a further thoughts
- this would seem to apply just as well to a non-separable space (i.e. where dense subspaces are not countable), with the term ONB replaced by complete orthonormal set ?
- there's no dependency on completeness that I can see, so it would seem to be true in any inner product space ?