# Weak convergence in $\mathcal{l}_p$ and coordinatewise convergence

Let $x^n=(x^n_1, x^n_2,...)$ be a bounded sequence in $\mathcal{l}_p$ for $1<p<\infty$ and such that $x^n_i$ converges to $x_i$ for all $i\in\mathbb{N}$. I'm trying to prove that $x=(x_1,x_2,...)$ belongs to $\mathcal{l}_p$ and $x^n$ converges weakly to $x$.

Since $x_n$ is bounded in a reflexive space there exists a subsequence $x^{n_k}$ weakly convergent to some $y\in \mathcal{l}_p$. Testing with $e^n=(e^n_i)_i$, $e^n_i=\delta_{n,i}$ i get that the subsequence converges indeed to $x$, so $x\in \mathcal{l}_p$. I'm stuck for the second part. Since the dual of $\mathcal{l}_p$ is $\mathcal{l}_q$ I'm trying to show that for evry $z\in\mathcal{l}_q$ , $z(x^n)-z(x)=\sum_iz_i(x^n_i-x_i)$ converges to zero. $$|z(x^n)-z(x)|\leq \sum_i|z_i(x^n_i-x^{n_k}_i)|+\left|\sum_iz_i(x^{n_k}_i-x_i)\right|.$$ The second term goes to zero, but i don't know how to estimate the first sum.I don't know if this is the wrong approach, any help would be greatly appreciated.

For the first part, we can actually write $$\sum_{j=1}^N|x_j|^p=\lim_{n\to \infty}\sum_{j=1}^N|x_j^n|^p\leqslant \sup_l\lVert x^l\rVert_p^p.$$
As $N$ is arbitrary, we get that $x$ belongs to $\ell^p$.
For the second part, we approximate the element $z$ by the sequence whose $N$ first terms are the corresponding to $z$, and the other ones are $0$. Call this vector $z_N$. Then $$|z(x^n)-z(x)|\leqslant |z_N(x^n)-z_N(x)|+|(z-z_N)x^n|+|(z-z_N)x|.$$ Notice that $|(z-z_N)x^n|\leqslant \lVert z-z_N\rVert_q\lVert x^n\rVert_p\leqslant \lVert z-z_N\rVert_q\sup_l\lVert x^l\rVert_p$ and $|(z-z_N)x|\leqslant \lVert z-z_N\rVert_q\lVert x\rVert_p$, hence $$|z(x^n)-z(x)|\leqslant |z_N(x^n)-z_N(x)|+2\lVert z-z_N\rVert_q\sup_l\lVert x^l\rVert_p.$$ For a fixed $N$, the first terms goes to $0$ because of the coordinatewise convergence.
• Thank you. So this should work also for $p=1$ i guess, but not for $p=\infty$ because its dual is not $\ell_1$? – balestrav Nov 9 '14 at 21:05
• Actually it does not work for $\ell^1$ because in this case weak convergence is equivalent to strong convergence. – Davide Giraudo Nov 9 '14 at 21:07
• Ok, but which part of the proof fails? And what about $\ell_\infty$? – balestrav Nov 9 '14 at 21:22
• A linear functional on $\ell^1$ can be represented by an element of $\ell^\infty$, but not better, hence the approximation argument with $z_N$ does not work anymore. – Davide Giraudo Nov 9 '14 at 21:53