# Product of weakly convergent sequence in $c_0$ and weakly/weakly* convergent sequence in $l^1$

we had in the class about the convergence of the product of a weakly convergent and a strongly convergent sequence.

But what about the product of weakly convergent sequence and a weakly/weakly* convergent one?

I mean:

1.) Let $(x_n)_{n \ge1} \in c_0$ be any sequence converging weakly to $0$ and $(y_n)_{n \ge1} \in l^1$ be any sequence converging to 0 in the weak*-topology.

2.) Let $(x_n)_{n \ge1} \in c_0$ be any sequence converging weakly to $0$ and $(y_n)_{n \ge1} \in l^1$ be any sequence converging weakly to 0.

What would happen in those two cases to the product $<x_n,y_n>$ ? Can anyone explain it to me?

Thanks!

## 1 Answer

For your first question, consider $x_n=y_n=e_n$: then $x_n$ converges weakly to $0$ in $c_0$, and $y_n$ converges weakly-* to $0$ in $l_1$, but $\left<x_n,y_n\right>=1$ for all $n\in\mathbb N$, which does not converge to $0$.

The second question is different: if $y_n\in l_1$ and $y_n$ converges weakly to $0$, then it converges strongly to $0$, since $l_1$ has Schur's property. Therefore, if $x_n\to 0$ weakly in $c_0$, then $\left<x_n,y_n\right>\to 0$.

• Ah that makes sense! Thanks a lot for explaining! :) – TigerLa Nov 13 '16 at 22:03
• Oh sorry, I totally forgot it. Done! – TigerLa Nov 13 '16 at 22:08