I try to solve the problem 106 of the scottich book. I know the set of all rearranged sums is convex. Let $f_{j,k}$ the indicator function of the interval $(\frac{j}{2^k},\frac{j+1}{2^k}$). k = 0,1,2,$\cdots$ and j = 0..$2^k-1$. I have proofed, that $\sum_{k=0}^{\infty} x_k = f_{0,0} - f_{0,0} + \cdots$ -> 0. I also have proofed, that the rearrangement : $f_{0,0} + f_{0,1} + f_{1,1} - f_{0,0} + \cdots$ converges to the indicator function $1$.
In the scottish it is asked for a solution in infinite-dimensional room. The book of Werner - "Funktionalanalysis" presents in chapter V6.26 an idea. Proof: The set {x $ \in L²[0,1]:$ there exist a reorder with x = $\sum_{i=1}^{\infty} x_\pi(i)$} is not convex. The $x_n$ are from type $\mathbb{Z}$
And this proof is my problem. I don't want the solution, just a tip or breadcrumbs :).
Hint: For every $a \in \mathbb{Z}$ try to combine $f$'s in such an order that as a sum you will get constant function over interval $(0;1)$ equal $a$. After that try to think of a rearrangement of other terms which converges to $0$.