summing basis of $c_0$ is a conditional basis The standard unit vector basis $(e_n)_{n=1}^\infty$ is an unconditional basis of
$c_0$ and $l^p$ for $1 \leq p < \infty.$ An
example of a  Schauder basis that is normalized conditional (i.e., not unconditional) is
the summing basis of $c_0$, defined as
$$f_n=e_1+...+e_n, n \in \mathbb{N}.$$
To see that $(f_n)_{n=1}^\infty$ is a basis for $c_0$ we prove that for each $\xi= (\xi(n))_{n=1}^\infty \in c_0$ we have $\xi = \sum_{n=1}^\infty f_n^*(\xi)f_n,$ where $f_n^* = e_n^*-e_{n+1}^*$ are the biorthogonal functionals of $(f_n)_{n=1}^\infty.$
Given $N \in \mathbb{N},$
\begin{eqnarray}
\nonumber \displaystyle\sum_{n=1}^N
 f_n^*(\xi)f_n&=&\displaystyle\sum_{n=1}^N(e_n^*(\xi)-e_{n+1}^*(\xi))f_n\\
&=&\nonumber\displaystyle\sum_{n=1}^N(\xi(n)-\xi(n+1))f_n\\
&=&\nonumber\displaystyle\sum_{n=1}^N\xi(n)f_n-\displaystyle\sum_{n=2}^{N+1}\xi(n)f_{n-1}\\
&=&\nonumber\displaystyle\sum_{n=1}^N\xi(n)(f_n-f_{n-1})-\xi(N+1)f_N\\
&=&\nonumber(\displaystyle\sum_{n=1}^N\xi(n)e_n)-\xi(N+1)f_N,
\end{eqnarray}
where we have used the convention that $f_0=0.$ Therefore,
\begin{eqnarray} \nonumber
\|\xi-\displaystyle\sum_{n=1}^Nf_n^*(\xi)f_n\|_\infty&=&  \|\xi - \sum_{n=1}^N (e_n^*(\xi)- e_{n+1}^*(\xi))f_n\|_\infty\\
\nonumber&=&\|\displaystyle\sum_{n=1}^\infty e_n^*(\xi)e_n-\displaystyle\sum_{n=1}^N(\xi(n)-\xi(n+1))f_n\|_\infty\\
\nonumber &=& \|\sum_{n=1}^\infty \xi(n)e_n -[(\xi(1)-\xi(2))f_1 + (\xi(2)-\xi(3))f_2 + ... + \\
\nonumber &(\xi(N)&-\xi(N+1))f_N]\|_\infty\\
\nonumber &=& \|\sum_{n=1}^\infty \xi(n)e_n - [\xi(1)(f_1-f_0) + \xi(2)(f_2-f_1)+ \xi(3)(f_3-f_2) + ... \\
\nonumber &+& \xi(N)(f_N - f_{N-1}) - \xi(N+1)f_N]\|_\infty \\
\nonumber&=&\|\displaystyle\sum_{n=1}^\infty
\nonumber \xi(n)e_n-\displaystyle\sum_{n=1}^N\xi(n)e_n+\xi(N+1)f_N\|_\infty\\
\nonumber&=&\|\displaystyle\sum_{N+1}^\infty\xi(n)e_n+\xi(N+1)f_N\|_\infty\\
\nonumber&\leq&\|\displaystyle\sum_{N+1}^\infty\xi(n)e_n\|_\infty+|\xi(N+1)|\|f_N\|_\infty
%\overrightarrow
\end{eqnarray}
This tends to zero as $N\rightarrow\infty$ because $\xi \in c_0$ and $\|f_N\|_\infty=1 \forall N.$
So, $(f_n)_{n=1}^\infty$ is a basis.
Now we will identify the set $S$ of coefficients
$(\alpha_n)_{n=1}^\infty$ such that the series $\sum_{n=1}^\infty
\alpha_n f_n$ converges. In fact, we have that
$(\alpha_n)_{n=1}^\infty \in S$ if and only if there exists
$\xi=(\xi(n))_{n=1}^\infty \in c_0 $such that
$\alpha_n=\xi(n)-\xi(n+1)$ for all $n.$ Then, clearly, unless the
series $\sum_{n=1}^\infty \alpha_n$ converges absolutely, the convergence of $\sum_{n=1}^\infty \alpha_n f_n$ in $c_0$ is not equivalent to the convergence of $\sum_{n=1}^\infty \epsilon_n \alpha_n f_n$ for all choices of signs $(\epsilon_n)_{n=1}^\infty.$
In fact, we have
\begin{eqnarray}
\nonumber \sum_{i=m}^{m+p} \alpha_i f_i &=& \sum_{i=m}^{m+p} \alpha_i \sum_{j=1}^i e_j  \\
\nonumber &=& \biggl (\sum_{i=m}^{m+p} \alpha_i \biggr ) \biggl (\sum_{j=1}^m e_j \biggr ) + \biggl (\sum_{i = m+1}^{m+p} \alpha_i \biggr) e_{m+1} + ... + \alpha_{m+p} e_{m + p},
\end{eqnarray}
whence
$$\left\lVert  \displaystyle \sum_{i=m}^{m+p} \alpha_i f_i \right\rVert = \displaystyle \sup_{m \leq k \leq m+p} \left \vert \displaystyle \sum_{i=k}^{m+p} \alpha_i \right \vert.$$
Consequently, $\displaystyle \sum_{i=1}^\infty \alpha_i f_i$ converges if and only if $\displaystyle \sum_{i=1}^\infty \alpha_i$ converges.
So, we can find a series
$\sum_{n=1}^\infty \alpha_n f_n$ converges in $c_0$ while the convergence of $\sum_{n=1}^\infty \epsilon_n
\alpha_n f_n$ for all choices of signs $(\epsilon_n)_{n=1}^\infty$ fails.
Let $x = (x_n)$ be any sequence of reals such that $x_n \rightarrow zero$ but $\sum |x_n - x_{n+1}| = \infty$ (for example, $$x_n = a + \sum_{j=1}^n (-1)^j \frac{1}{j}$$ for a suitable value of $a$)
Now $$x = \sum (x_n - x_{n+1}) f_n.$$
But there exist $\epsilon_n = \pm{1}$ such that
$$\sum \epsilon_n (x_n - x_{n+1})f_n = \sum |x_n - x_{n+1}|f_n,$$
and that last sum does not converge in $c_0.$ Hence $(f_n)_{n=1}^\infty$ cannot be unconditional
Q1 Does the sentenses ""Now we will identify the set $S$ of coefficients
$(\alpha_n)_{n=1}^\infty$ such that the series $\sum_{n=1}^\infty
\alpha_n f_n$ converges. In fact, we have that
$(\alpha_n)_{n=1}^\infty \in S$ if and only if there exists
$\xi=(\xi(n))_{n=1}^\infty \in c_0 $such that
$\alpha_n=\xi(n)-\xi(n+1)$ for all $n.$""
means that the coefficients are unique so the basis is Schauder basis?? and how i can prove that??
Q2 * How $$\left\lVert  \displaystyle \sum_{i=m}^{m+p} \alpha_i f_i \right\rVert = \displaystyle \sup_{m \leq k \leq m+p} \left \vert \displaystyle \sum_{i=k}^{m+p} \alpha_i \right \vert.$$ imply that $\displaystyle \sum_{i=1}^\infty \alpha_i f_i$ converges if and only if $\displaystyle \sum_{i=1}^\infty \alpha_i$ converges.
*Can we say that $ | \displaystyle \sum_{i=m}^{m+p} \alpha_i | \leq \displaystyle \sup_{m \leq k \leq m+p} | \displaystyle \sum_{i=k}^{m+p} \alpha_i |.$ So, then the sequence of partial sums $\{\sum_{i=1}^n \alpha_i\}_{n=1}^\infty$ forms a cauchy sequence.
*Can we say that $\sum_{n=1}^\infty (\xi(n) - \xi(n+1))$ always converges for any $\xi \in c_0.$
Q3 is there a separate proof to prove that $\sum |x_n - x_{n+1}|f_n$ diverges for the example i mentioned.
 A: *

*Yes. Suppose that $\sum_j\alpha_jf_j=0$. Fix $\def\e{\varepsilon}$ $\e>0$. Then there exists $n_0$ such that for all $n_1\geq n_0$ we have
$$\tag1
\Big\|\sum_{n> n_1}\alpha_jf_j\Big\|<\e.
$$
Then
$$
\max\Big\{\Big|\sum_{j=1}^n\alpha_j\Big|:\ 1\leq n\leq n_1\Big\}=\Big\|\sum_{n\leq n_1}\alpha_jf_j\Big\|=\Big\|\sum_{n> n_1}\alpha_jf_j\Big\|<\e.
$$
It follows that
$$
\Big|\sum_{j=1}^n\alpha_j\Big|<\e,\qquad\qquad n\in\mathbb N.
$$
Then
$$
|\alpha_n|=\Big|\sum_{j=1}^n\alpha_j-\sum_{j=1}^{n-1}\alpha_j\Big|\leq2\e.
$$
As this can be done for any $\e>0$, we get that $\alpha_n=0$ for all $n$.



*The series $\sum_{j> n_1}\alpha_jf_j$ converges if and only if the sequence of partial sums converges. A sequence converges if and only if it is Cauchy. Hence
\begin{align}
\sum_{n> n_1}\alpha_jf_j\ \text{exists}
&\iff \forall\e>0,\ \exists m_0, \forall n>m\geq m_0,\ \Big\|\sum_{j=1 }^n\alpha_jf_j-\sum_{j=1}^m\alpha_jf_j\Big\|<\e\\[0.3cm]
&\iff \forall\e>0,\ \exists m_0, \forall n>m\geq m_0,\ \Big\|\sum_{j=m+1 }^n\alpha_jf_j\Big\|<\e\\[0.3cm]
&\iff \forall\e>0,\ \exists m_0, \forall n>m\geq m_0,\ \!\!\!\sup_{j=m+1,\ldots,n}\Big\{\Big|\sum_{n=m+1 }^n\alpha_j\Big|\Big\}<\e\\[0.3cm]
&\iff \sum_j\alpha_j\ \text{ converges}.
\end{align}
