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Since there was essentially no answers on my previous question, I will ask a partial case of it, which is very easy to state.

Let $\left(X,\left<\cdot,\cdot\right>\right)$ be an inner product (pre-Hilbert) space. Is it possible to describe the weak topology on $X$ explicetely in terms of $\left<\cdot,\cdot\right>$ as a function on $X\times X$? That means without mentioning such not-enough-constructive objects as "completion" and "the dual".

Thank you.

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Let $\bar X$ denote the completion. The weak topology $\sigma(X,X)$ on $X$ is strictly coarser than the weak topology $\sigma(X,\bar X)$; see [Schaefer, Topological Vector Spaces, (IV.1.2)].

However, both topologies have the same bounded subsets (which are the norm-bounded sets). To see this, note that the bounded sets in $\bar X$ for $\sigma(\bar X,X)$ are contained in the completions of the bounded sets in $X$, for $\sigma(X,X)$. Moreover the bounded sets in $\bar X$ for $\sigma(\bar X,\bar X)$ are the same as those for $\sigma(\bar X,X)$, by Schaefer (III, 3.4).

And each equicontinuous subset of $X$ for the $\sigma(X,X)$-topology (i.e., norm bounded by the uniform boundedness theorem) the topologies $\sigma(X,X)$ and $\sigma(X,\bar X)$ coincide.

EDIT: Correction: The finest locally convex topology on $X$ which coincides with $\sigma(X,X)$ on each norm-ball (each bounded set), is at least as fine as the topology of uniform convergence on precompact subsets, i.e., on compact subsets of $\bar X$. It might be the same.

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    $\begingroup$ The above answer contains two errors. The two topologies do not have the same bounded sets in general and the finest locally convex topology which agrees on the unit ball with the weak topology is, by definition, the bounded weak topology which, in the case of a Hilbert space, is complete, something which the weak topology is not. $\endgroup$
    – weather
    Commented Mar 4, 2015 at 10:23
  • $\begingroup$ @weather, but, still, despite the weak dual not being complete, it is quasi-complete, so all is not lost, etc. $\endgroup$
    – paul garrett
    Commented Jun 2, 2015 at 19:06

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