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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.

Thanks in advance!

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Ruy
    Mar 20 at 13:45
  • $\begingroup$ 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 $w$. $\endgroup$
    – s.harp
    Mar 20 at 18:08
  • $\begingroup$ Perfect! Thank you! $\endgroup$
    – Ruy
    Mar 20 at 18:43

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