Are coordinate functionals on complete vector spaces always continuous?

2010-12-09T04:32:17Z

(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?

Answer by Bill Johnson for Are coordinate functionals on complete vector spaces always continuous?

2010-12-11T23:28:06Z

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 $e_0^*$.

Are coordinate functionals on complete vector spaces always continuous? - MathOverflow

Are coordinate functionals on complete vector spaces always continuous?

Answer by Bill Johnson for Are coordinate functionals on complete vector spaces always continuous?