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(I'm just adding the completeness condition to $V$ from this 2 month old question of mine, because I realized it's relevant to whether Bill Johnson's answer to this 4 month old question of mine actually answers the question.)

Let $V$ be a complete Hausdorff locally convex topological vector space over the field $\mathbb{K}$.
Let $B$ be a subset of $V$ such that

$\;$ for all functions $c : B\to \mathbb{K}$, if $\displaystyle\sum_{b\in B} \; c(b)\cdot b = \textbf{0}$, then $c$ is identically zero

and $f : B\times V \to \mathbb{K}$ be a function such that

$\;$ for all vectors $v$ in $V$, $\; \displaystyle\sum_{b\in B} \; f(b,v)\cdot b = v$.

Let $b$ be a member of $B$, and $g : V \to \mathbb{K}$ be given by $g(v) := f(b,v)$. Does it follow that $g$ is continuous?

flag	asked Dec 9 2010 at 4:32 Ricky Demer 6,421●8●31

	The sums are unconditional, as explained at en.wikipedia.org/wiki/…. – Ricky Demer Dec 9 2010 at 4:34
	For Banach (or Frechet) spaces the answer is "Yes", but I think you know that youself since you posted the question in such generality. – fedja Dec 11 2010 at 18:00
	I didn't actually know that (and I still don't know a proof or reference, do you know one?), I was just more interested in the general case. – Ricky Demer Dec 12 2010 at 0:43

Bill Johnson · Accepted Answer · 2010-12-11 23:28:06Z UTC

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Consider $\ell_1:=\ell_1(\Bbb{N}\cup \{0\})$ as the dual of $c$, the space of convergent sequences indexed by $\Bbb{N}$, where the action is given by $e_0^*(x)=\lim_n x(n)$ and $e_n^*(x)=x(n)$ for $n\ge 1$. Put the bw$^*$ topology on $\ell_1$ under this pairing, which is the largest locally convex topology that agrees with the weak $^*$ topology on bounded sets. IIRC, this is a complete topology (while the weak$^*$ topology is only boundedly complete). This is an unconditional basis that is not a Schauder basis because $e_n^*$ converges weak $^*$ to $e_0^*$.

answered Dec 11 2010 at 23:28

Bill Johnson
14.4k●1●33●60

	Does the bw* topology have a more concrete description in this instance? Also, what definition of Schauder basis are you using? – Ricky Demer Dec 12 2010 at 2:37
	Schauder means all of the coordinate functionals are continuous. I never thought much about the bw$^$ topology, although I know some books on functional analysis do treat it. If you only care about sequential completeness or that bounded Cauchy nets converge, use the weak$^$ topology instead. – Bill Johnson Dec 12 2010 at 3:32
	I will accept this answer, I'm just hoping fedja responds to my comment first. – Ricky Demer Dec 12 2010 at 21:29
	@Ricky: Look at almost any book on Banach space theory. It is the first theorem mentioned in Lindenstrauss-Tzafriri (without proof) and the first theorem mentioned in Albiac-Kalton (with proof). – Bill Johnson Dec 13 2010 at 1:32
	@Ricky. Regarding your query about a more concrete description of the topology in question, it is the topology of uniform convergence on compact subsets. This works for the dual of eny Banach space (even, in a suitable form, of a Frechet space). It is even the finest topology with the above property (not just locally convex topology), a fact which is often useful. This is all part of the circle of ideas surrounding the theorem of Banach-Dieudonne. – jbc Nov 20 at 13:06

Are coordinate functionals on complete vector spaces always continuous?

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