Let \(\sum x_{i}\) be a series in a Banach space \(X, x \in X\). Show that the
following are equivalent:
(i) For every \(\varepsilon>0\), there is a finite set \(F \subset \mathbf{N}\)
such that \(\left\|x-\sum_{i \in F^{\prime}} x_{i}\right\|<\varepsilon\)
whenever \(F^{\prime}\) is a finite set in \(\mathbf{N}\) satisfying \(F^{\prime}
\supset F\).
(ii) If \(\pi\) is any permutation of \(\mathbf{N}\), then \(\sum x_{\pi(i)}=x\). If
these conditions hold, we say that the series is unconditionally convergent to
\(x .\) As in the real case, a series \(\sum x_{i}\) is unconditionally convergent
if it is absolutely convergent: \(\left\|\sum_{i \in G} x_{i}\right\| \leq
\sum_{i=n}^{\infty}\left\|x_{i}\right\|\) for \(G \subset \mathbf{N}\) with \(n
\leq \min (G)\).
Note that the unconditional convergence of a series does not in general imply
its absolute convergence; consider \(\sum \frac{1}{i} e_{i}\) in \(\ell_{2}\).
Hint: (i) \Longrightarrow (ii): For every \(\varepsilon>0\), get a finite \(F\)
such that \(\left\|\sum_{F^{\prime}} x_{i}-x\right\|<\varepsilon\)
whenever \(F^{\prime}\) is a finite set in \(\mathbf{N}\) such that \(F^{\prime}
\supset F\). For some \(n_{0}\), we get \(\left\\{\pi(1), \pi(2), \ldots,
\pi\left(n_{0}\right)\right\\} \supset F .\) Thus \(\left\|x-\sum_{i=1}^{n}
x_{\pi(i)}\right\|<\varepsilon\) for \(n \geq n_{0}\)
(ii) \(\Longrightarrow\) (i): By contradiction, get by induction a sequence
\(n_{k}\) such that \(\left\|\sum_{i=1}^{n_{k}} x_{i}-x\right\|<\frac{1}{k}\) and
a sequence of finite sets \(M_{k}\) such that \(n_{k+1}>\) \(\max
\left(M_{k}\right) \geq \min \left(M_{k}\right)>n_{k}\) in such a way that
\(\left\|\sum_{i=1}^{n_{k}} x_{i}+\sum_{M_{k}} x_{i}-x\right\| \geq
\varepsilon\)
Indeed, since (i) fails for \(F=\left\\{1, \ldots, n_{k}\right\\}\), we find
\(F^{\prime}\) and set \(M_{k}=F^{\prime} \backslash F\) Find a permutation \(\pi\)
such that \(\pi\left(\left\\{1, \ldots,
n_{k}+\left|M_{k}\right|\right\\}\right)=\left\\{1, \ldots, n_{k}\right\\}
\cup M_{k}\)
then \(\sum x_{\pi(i)}\) does not converge. To see that \(\sum \frac{1}{i} e_{i}\)
converges unconditionally in \(\ell_{2}\), note that \(\left\|\sum_{G}
\frac{1}{i} e_{i}\right\|\) is small if \(\min (G)\) is large enough.
