Sequentially closure on Hilbert separable space

Question

Let $H$ a Hilbert separable space with a Hilbert basis $(e_n)_{n\in\mathbb{N}}$. Let $$ F = \{e_m + m e_n \, : \, n,m \geq 1\} $$

1. Show that 0 is not in the sequentially weak closure of F.
2. Show that 0 in sequentially weak closure of sequentially weak closure of F

Any hint of suggestion?

I tried by contradiction but I'm stuck

Answer 1

Assume the sequence $x_k=e_{m_k}+m_ke_{n_k}$ is weakly convergent to $x.$ Then $x_k$ is bounded say by $N,$ by the uniform boundedness theorem. Therefore the sequence $m_k$ is bounded, because $N\ge \|x_k\|\ge m_k-1.$ Thus there is an infinite constant subsequence $m_{k_l}=:m.$ Hence $$x_{k_l}=e_m+me_{n_{k_l}}\ {\rm and}\ \langle x_{k_l},e_m\rangle \ge 1$$ Therefore $$\langle x,e_m\rangle =\lim_{l\to\infty}\langle x_{k_l},e_m\rangle \ge 1$$ which implies $x\neq 0.$

For every fixed $m$ the sequence $\{e_m+me_n\}_{n=1}^\infty$ tends weakly to $e_m.$ Hence $e_m$ belongs to the sequential weak closure of $F.$ In turn the sequence $\{e_m\}_{m=1}^\infty $ tends weakly to $0.$ This shows $2.$