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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?

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See… for the sums. – Ricky Demer Oct 19 '10 at 2:19
To elaborate on Ricky's comment, this is unconditional convergence of the series (or unconditional summation) – Yemon Choi Oct 19 '10 at 2:23
up vote 1 down vote accepted

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.

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What is the character basis? – Ricky Demer Oct 22 '10 at 22:24
The trigonometric system--$1,e^{it},e^{-it),...$. – Bill Johnson Oct 22 '10 at 22:40

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 $n$?

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Can you give an example of an orthonormal basis $B$ for $L_2(0,1)$ so that no element of $B$ is in $L_{\infty}$? Also, where do you get your definition of Schauder basis? – Ricky Demer Oct 19 '10 at 20:25
Another question is how can it be shown that $B$ still satisfies the first condition with the $L_1$ norm. – Ricky Demer Oct 19 '10 at 20:40
@Alejandro: That's the one I knew of, but it does not require that the coordinate functions be continuous. – Ricky Demer Oct 19 '10 at 21:49
@Ricky Demer: To get such an ON basis, take $f$ in $L_2$ but not in $L_\infty$ s.t. $\int f h_n$ is not zero for every Haar function $h_n$ and apply Gram-Schmidt to $f, h_1, h_2, ...$. (For $f$ you can take any strictly decreasing positive $L_2$ function that is not bounded.) However, I do not see that such a sequence $g_n$ satisfies your first condition in the $L_1$ norm. It will if inf $\|g_n\|_1 >0$, but I don't see how to build $g_n$ to have this additional property. I'll think more about it when I have time. – Bill Johnson Oct 20 '10 at 23:41

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