Let's consider $\ell_1$ space with the standard unconditional Schauder basis $(e_n)_{n=1}^\infty$ and let's define the following sequence:
$$x_1:=e_1$$
and
$$x_n := e_{n-1} – e_n$$
for $n >1$.
I found this sequence also mentioned on this site, as being an example of a conditional Schauder basis for the space $\ell_1$. So this is what I wanted to check.
Let's fix $x \in \ell_1$. Then we can write
$$x=\sum_{n=1}^\infty a_n e_n$$
since $(e_n)_{n=1}^\infty$ is a basis. Then, by doing some series manipulation and working on $x_n$ we can define the following sequence:
$$b_1:=0$$
and
$$b_n:=\sum_{k=1}^{n-1} a_k.$$
Then we have that
$$x=\sum_{n=1}^\infty b_n x_n$$
so this shows that $(x_n)_{n=1}^\infty$ is a Schauder basis for $\ell^1$ (hopefully here I didn't make any mistakes). Now, in order to show that it's conditional I wanted to use one of the characterizations, namely either find a sequence $(\varepsilon_n)_{n=1}^\infty \subset \{-1,1\}$ such that $\sum_{n=1}^\infty \varepsilon_n x_n$ doesn't converge in $\ell_1$, or equivalently $(\lambda_n)_{n=1}^\infty \subset \mathbb{K}$ such that $\sup_n |\lambda_n| < \infty$ and again, such that $\sum_{n=1}^\infty \lambda_n x_n$ doesn't converge in $\ell_1$ but I'm not sure what kind of a sequence we can choose here. Could anyone share some insight how we can choose such a sequence?
Thank you!
Best Answer
Just take $\epsilon_n=(-1)^{n}$ and compute the partial sums of $\sum \epsilon_n x_n$. You wil see easily that the series does not converge.