Let, \begin{align*} \alpha(x) &= 1 \\ \beta(x) &= x\\ s_{n,k}(x) &= \max\{1-|2(2^nx-k)-1|, 0\} \text{ for } 0 \le n \text{ and } 0 \le k \le 2^{n}-1 \end{align*} We define the Faber-Schauder system \begin{equation*} S := \{ \alpha, \beta \} \cup \{ s_{n,k} \}_{n \ge 0, 2^{n}-1 \ge k \ge 0 } \end{equation*}
$s_{n,k}$ denotes the hat function supported on the interval $[\frac{k}{2^n},\frac{k+1}{2^n}]$ and whose peak is 1 and attained at $\frac{2k+1}{2^{n+1}}$. S is a basis for $C[0,1]$ with the supremum norm. It is known that any basis for $C[0,1]$ is conditional. I am trying to find a concrete example of a series of functions which converges conditionally. My strategy is to use some sort of linear combinations of the basis functions i.e, $g_M = \sum_{n = 0}^{M} \sum_{k = 0}^{2^{n}-1} c_{n,k} s_{n,k}$. I will show that $g_M$ uniformly converges(show that it is uniformly cauchy). Then I show that $(-1)^n g_M$ doesn't uniformly converge(ideally goes to infinity). Vice versa is fine too, i.e, $(-1)^n g_M$ uniformly converges and $(-1)^n g_M$ doesn't uniformly converge. This shows that $g_M$(or $(-1)^n g_M$) is conditionally convergent. Or maybe, I could use another characterization of conditional convergence.
I tried couple of linear combinatons to construct $g_M$. For example, $g_M = \sum_{n = 0}^{M} s_{n,0}$ which didn't work. Another example is constructed by firs considering interval $[0,1]$ and its corresponding function $t_0 = s_{0,0}$ then diveding it by half and taking the left part,$[0,\frac{1}{2}]$, which corresponds to $t_1 = s_{1,0}$ then divide half and take the right part, $[\frac{1}{4},\frac{1}{2}]$, which correspond to $t_2 = s_{2,1}$ and so on. We always halve the current interval and alternate between taking the left part and right part with its corresponding function. $g_M = \sum_{n=0}^{M} t_n$.In this case, $g_M$ goes to infinity which is good. But I don't think $(-1)^ng_{M}$ uniformly converges.
I am looking forward for your ideas.
Take $$ f_n(x) = \frac{1}{n} s_{n,2^n-2}(x)\qquad \mathrm{for}\ n = 1, 2, 3, \ldots $$ and the sum $ \sum f_n $ is conditionally convergent. Let's prove it. First, we have $ \sum \|f_n\| = \sum{1/n} = \infty $ so we know it's not absolutely convergent.
What about conditional convergence? Easy. The functions $ f_n $ were chosen with disjoint supports. So their pointwise convergence is trivial. The limit of the sum is depicted with this cartoon which is clearly continuous (even at $ x = 1 $!)