Dense Subspace: ONB

Question

This might be a duplicate. If so, then please let me know. Thanks!

Given a Hilbert space $\mathcal{H}$.

Consider a dense subspace $\overline{Z}=\mathcal{H}$.

Then it provides an ONB: $\mathcal{S}\subseteq Z$

(I guess it can be shown by slightly adjusting the usual proof via Zorn's lemma...)

This is not true if $\mathcal{H}$ is not assumed to be separable. See here http://math.stackexchange.com/a/201149/151552. — PhoemueX, Oct 20 '14 at 20:05
@PhoemueX: Oh pretty bad; then does it still hold that $x\in\mathcal{H}_{ac}\iff\nu_x\ll\lambda$? — C-star-W-star, Oct 20 '14 at 20:10
I am sorry, but: What is $\mathcal{H}_{ac}$? What are $\nu_x$ and $\lambda$? — PhoemueX, Oct 20 '14 at 20:18
@PhoemueX: No problem, it wasn't clear from the context anyway. Actually I meant $\mathcal{H}_{pp}:=\overline{\mathrm{span}{\phi\in\mathcal{H}:\exists\lambda\in\mathbb{C}:T\phi=\lambda\phi}}$ for a normal operator $T:\mathcal{D}(T)\to\mathcal{H}$ and $\nu_x(A):=|E(A)\phi|^2$ for its associated spectral measure $E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$ — C-star-W-star, Oct 20 '14 at 23:19
@PhoemueX: Ah and $\lambda$ was meant to be the Lebesgue measure but not needed here as now it is $\nu_\phi$ discrete iff $\phi\in\mathcal{H}_{pp}$. — C-star-W-star, Oct 20 '14 at 23:35

score 0 · Accepted Answer · answered Oct 21 '14 at 16:39

I just realized that this cannot be true for the following reason:

Assume every dense subspace would provide an ONB for the Hilbert space: $$\overline{\mathcal{S}}=\mathcal{H}\quad(\mathcal{S}\subseteq Z)$$ Then, it would serve as well as an ONB for the subspace: $$\overline{\mathcal{S}}\supseteq Z$$ But there are preHilbert spaces which do not admit any ONB (see Bourbaki or Robert Isreal).

Dense Subspace: ONB

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