"Basis" for $\ell^{1}(\mathbb{N})$?

Question

This may be a really trivial question. Anyway, suppose that $x=(x_{1},x_{2},\ldots)\in\ell^{1}$ and let $e_{n}\in\ell^{1}$ denote the standard $n$-th coordinate sequence. Is it then true that $$x=\sum_{n\in\mathbb{N}}x_{n}e_{n}?$$ Here is what I know:

It is clear that $$\sum_{n\in\mathbb{N}}\|x_{n}e_{n}\|_{1}=\sum_{n\in\mathbb{N}}|x_{n}|=\|x\|_{1}<\infty.$$ So because $\ell^{1}$ is complete, it follows that the series $\sum_{n\in\mathbb{N}}x_{n}e_{n}$ indeed converges in $\ell^{1}$.

But how can I conclude that the series is actually equal to $x$? And does this imply that $(e_{1},e_{2},\ldots)$ is some sort of basis for $\ell^{1}$?

It is not an orthonormal basis since we have no inner-product that induces he norm $\|\cdot\|_{1}$ on $\ell^{1}$.

Answer 1

If $s_n$'s are the partial sums then $\|s_n-x\|= \sum\limits_{k=n+1}^{\infty} |x_k|$ and this last sum tends to $0$.

Answer 2

For $k \in \mathbb N$ let

$$s_k:=\sum_{j=1}^kx_j e_j.$$

Then we have

$$||x-s_k||_1= ||\sum_{j=k+1}^{\infty}x_j e_j||_1 \le \sum_{j=k+1}^{\infty}|x_j|.$$

Since $\sum_{j=1}^{\infty}|x_j|$ is convergent, we have $\sum_{j=k+1}^{\infty}|x_j| \to 0$ as $k \to \infty.$

This shows that

$$||x-s_k||_1 \to 0$$

as $k \to \infty.$

$(e_{1},e_{2},\ldots)$ is a Schauder basis of $\ell^1$.

https://en.wikipedia.org/wiki/Schauder_basis