# Characterising Hilbert spaces

I found this question:

Characterization of Hilbert spaces using orthogonal decomposition

It deals with the question of when pre-Hilbert spaces are Hilbert. It proves that (assuming separability)

A pre-Hilbert space $$H$$ is Hilbert if for all subspaces $$K$$ $$K^\perp = \{0\} \implies K\; \text{dense in}\; H$$

But I'm afraid I do not really understand the answer. In particular I would have thought that neither $$\eta$$ nor $$\xi$$ lie in $$K$$.

Additionally I would very much like a proof in the non-separable case. There is a comment saying it should be possible, but again I cannot fill in the details.

Let $$H$$ be a pre-Hilbert space, denote with $$\mathcal H$$ its completion which is Hilbert. For a subset $$K$$ of $$H$$ denote with $$K'$$ the orthogonal complement in $$H$$ and $$K^\perp$$ the orthogonal complement in $$\mathcal H$$. Its clear that $$K'\subseteq K^\perp$$ and even that $$K'=K^\perp \cap H$$.

Suppose there is a vector $$v\in \mathcal H - H$$ (ie that $$H$$ is not Hilbert), then the span $$V:=\Bbb C\cdot v$$ is a closed sub-space of $$\mathcal H$$ and define: $$K := V^\perp \cap H.$$ Note that since $$H$$ is dense in $$\mathcal H$$ you have that $$K$$ is dense in $$V^\perp$$ in $$\mathcal H$$. In particular you have that $$K^\perp = (\overline K)^\perp = (V^\perp)^\perp = V$$ since $$V$$ is closed. But $$K'= K^\perp\cap H = V\cap H = \{0\}$$. However $$K$$ cannot be dense in all of $$H$$, because then it must be dense in $$\mathcal H$$ and then $$V^\perp$$ must be dense in $$\mathcal H$$. But $$v$$ itself cannot be approximated by elements of $$V^\perp$$.

This shows $$[\text{H is not Hilbert}]\implies [\text{there exists a not dense subspace K with K\neq H and K^\perp = \{0\}}]$$ independently of any countability assumptions.

The other direction is elementary as $$(K^\perp)^\perp = \overline K$$ in a Hilbert space, and $$\{0\}^\perp$$ is always the entire space.

• Can you elaborate on why is $K$ dense in $V^\perp$? The intersection of a dense set with a subspace is sometimes not dense in that subspace.
– Ruy
Mar 20 at 13:45
• Thats true - that step was an oversight but the statement remains true because $V$ is one-dimensional. Let $w\in V^\perp$ and $x_n\in H$ with $x_n\to w$. Fix some $x\in H$ with $\langle x, v\rangle\neq 0$ then $x_n - \langle x_n, v\rangle \frac{x}{\langle x,v\rangle}$ which is in $V^\perp \cap H$ and converges to $w$. Mar 20 at 18:08
• Perfect! Thank you!
– Ruy
Mar 20 at 18:43