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?



Are coordinate functions on topological vector spaces always continuous?
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.


