# 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.

1. 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 $$n$$.
1. 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}