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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!

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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.

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  • $\begingroup$ Thank you! And now I'm also embarrassed by this question, because of how easy this was... $\endgroup$
    – btfm
    Commented Mar 31, 2021 at 23:39

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