$\mathcal{l}^1$ space and proving a graph is closed Let $X= \{\xi =(x_n)_{n \in \mathbb{N}} \in \mathcal{l}^1 : \sum_{n=1}^{\infty} n |x_n| < \infty \}$
And define the map $T: X \to \mathcal{l}^1$ by $(T \xi)_n := n x_n$  
I'm trying to proof that the graph of $T$ is closed.
We can't use the Closed Graph Theorem because $T$ is not continuous and $X$ is not a Banach Space.
When I just take a sequence $(\xi_m, T \xi_m)_m$ in the graph of $T$ and assume it converges, I don't know why the limit should be in the graph of $T$.  
Any ideas? Thanks.

Suppose $\xi_n \to \hat{\xi}$ and $T\xi_n \to \hat{y}$. You need to show that $T \hat{\xi} = y$.
Since $T\xi_n \to \hat{y}$, we have $\| T\xi_n - \hat{y} \| \to 0$. Since
$|[T\xi_n]_k - [\hat{y}]_k| \le \| T\xi_n - \hat{y} \| $, we see that 
$[T\xi_n]_k = k [\xi_n] \to [\hat{y}]_k$ for all $k$. Since $[\xi_n]_k \to [\hat{\xi}]_k$ for all $k$, this gives $k [\hat{\xi}]_k = [\hat{y}]_k$, or, in other words, $T \hat{\xi} = \hat{y}$.
