summing basis of $c_0$ is a conditional basis

Question

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.

Answer 1

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 $n$.

2. 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}