Are coordinate functions on topological vector spaces always continuous? Let $V$ be a 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 $\displaystyle\sum_{b\in B} \; c(b)\cdot b = 0$, 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?
 A: You ask whether an unconditional basis must be a Schauder basis.  The answer is no even when $V$ is a separable normed space.  Let $B$ be an orthonormal basis for $L_2(0,1)$ so that no element of $B$ is in $L_\infty$.  Then $B$ is an unconditional basis for the incomplete space  $L_2(0,1)$ with the $L_1$ norm, but none of the coordinate functionals are continuous.
EDIT 10/22/10.  Consider $f(t) = |t|^{-1/4}$ in $L_2(-\pi,\pi)$ with normalized Lebesgue measure.  Gram-Schmidt $f, e^{ix}, e^{-ix}, e^{2ix}, e^{-2ix}, ,,,$; this produces an orthonormal basis $g_n$ for $L_2(-\pi,\pi)$.  Since the Fourier coefficients of $f$ are all non zero, no $g_n$ is in $L_\infty$. Since successive Fourier coefficients of $f$ are comparable (they decay like $n^{-3/4}$), the $L_1$ norms $\|g_n\|_1$ are bounded away from zero. 
Suppose that $\sum a_n g_n$ unconditionally converges in $L_1$ to zero.  Since $L_1$ has cotype two, $\sum \|a_n g_n\|_1^2 <\infty$, so that 
 $\sum \|a_n|^2 <\infty$.  Thus $\sum a_n g_n$ converges in $L_2$ and the sum can be zero iff $a_n=0$ for all $n$.
I do not see that $g_n$ is a basis for $(L_2, \| \cdot\|_1)$.  That is, if
$\sum_{n=1}^\infty a_n g_n$ converges to zero in $L_1$, must $a_n=0$ for all $n$?
A: Here is a non orthogonal example that is however natural and probably simpler than the other one I gave.  Let $b_1, b_2,...$ be the character basis for $L_2(-\pi,\pi)$.  Let $b_0$ be an $L_1$ function whose Fourier series does not converge  in the $L_1$ norm.  Then $b_0,b_1,b_2,...$ is countably linearly independent because inner product with $b_1,b_2,...$ is continuous in the $L_1$ norm, and $b_0,b_1,b_2,...$ is thus an unconditional basis, in the $L_1$ norm,  for the linear span of $b_0$ and $L_2$.  The coordinate functional for $b_0$ is obviously discontinuous.  For $n\ge 1$, the coordinate functional for $b_n$ is discontinuous iff $\langle b_0, b_n \rangle \not= 0$.  You can guarantee that this happens for every $n$ by perturbing the original $b_0$ by an appropriate $L_2$ function.
