MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

(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?

share|cite|improve this question
The sums are unconditional, as explained at…;. – Ricky Demer Dec 9 '10 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 '10 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 '10 at 0:43
up vote 2 down vote accepted

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 $e_0^*$.

share|cite|improve this answer
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 '10 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 '10 at 3:32
I will accept this answer, I'm just hoping fedja responds to my comment first. – Ricky Demer Dec 12 '10 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 '10 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 '12 at 13:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.