The sum operator is not weakly compact It's my first question in this forum. Perhaps I don't respect all the required rules. So all remarks ares wellcome to improve my next contributions.
Let $\sigma : l_1 \rightarrow l_\infty$ be the sum aoperator : $ \sigma((a_i)_{i=1}^\infty) = (\Sigma_{k=1}^n a_k)_{n=1}^\infty$
How can I prove that it's not weakly compact ?
I suppose I have to find a sequence $(y_n) \subset l_\infty$ which has no convergent sub-sequence. 
Thanks for any help
 A: Let's try to prove that the image by $\sigma$ of the canonical basis $(e_n)$ of $l_1$ has no w-convergent subsequence. 
Let $(\sigma(e_{n_k})) \subset l_\infty$ be a busbsequence. $l_\infty$ is isometric to the Stone-Cech space $C(\beta \mathbb N)$. So $(\sigma(e_{n_k}))$ can be viewed as a sequence $(f_n) \subset C(\beta \mathbb N)$. 
Hence it's enough to prove that there exists a functional $\varphi \in C(\beta \mathbb N)^*$ such that $<\varphi, f_n>$ doesn't converge.
If $f_n$ is convergent we can suppose without any loss that it's limit is $f = 0$.
Let's choose $b \in C(\beta \mathbb N) \backslash \mathbb N$ and let $(b_\gamma) \subset \mathbb N$ be a net converging to $b$. let $\delta_b \in C(\beta \mathbb N)^*$ be the linear form defined by : $\delta_b(f) = f(b)$.
We have $<\delta_b, f_n> = f_n(b) = f_n(lim _\gamma b_\gamma) = lim _\gamma f_n(b_\gamma) = 1$. That's true for each $n$. Hence $<\delta_b, f_n> \rightarrow 1$. So $f(b) = 1$ which is a contradiction with $f = 0$.
This proves that $(\sigma(e_n))$ has no w-convergent subsequence. Therefore $\sigma$ isn't w-compact.
