# Hilbert subspaces of indefinite inner product spaces

Let $E$ be a real linear space, endowed with a non-degenerate symmetric bilinear form $(.,.)$.

Suppose that the [indefinite] inner product space $(E,(.,.))$ satisfies the following [sequential] properties:

(WSC) If the sequence { ${(x_{n},\ y)}$ } is Cauchy for each $y$ in $E$, then there exists some $x$ in $E$ such that $\left(x_{n}-x,y\right)$ $\rightarrow$ $0$ whenever $y$ in $E$.

(That is to say, $(E,(.,.))$ is weakly sequentially complete.)

and

(DPG) If $\left(x_{n},y\right)$ $\rightarrow$ $0$ for every $y$ in $E$, then $\left(x_{n},x_{n}\right)$ $\rightarrow$ $0$.

(That would be sort of "Dunford-Pettis & Grothendieck'' property for indefinite inner product spaces.)

Suppose also that $|E|$ is "big enough'' (at least $|E|$ $>|\mathbb{R}|$).

Conjecture. $(E,(.,.))$ contains an infinite-dimensional Hilbert subspace.

(That is, there exists a linear isometry from $(\ell^{2},<.,.>)$ into $(E,(.,.))$ .)

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Presumably there are supposed to be commas somewhere in the properties listed above. –  Steve Huntsman Jan 29 '10 at 23:40
Indeed. Now it's OK ? –  Ady Jan 30 '10 at 0:06
I'm curious - which condition fails if we take a Hilbert space with all norms multiplied by -1, or if we form an orthogonal direct sum of that with a positive definite copy of R? –  S. Carnahan Jan 30 '10 at 0:27
Correct me if I'm wrong, but wouldn't the Conjecture contradict (DPG)? –  Mark Meckes Jan 30 '10 at 1:05
@Scott Carnahan The (DPG) fails. –  Ady Jan 30 '10 at 1:22