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.

  • $\begingroup$ Thank you. So this should work also for $p=1$ i guess, but not for $p=\infty$ because its dual is not $\ell_1$? $\endgroup$ – balestrav Nov 9 '14 at 21:05
  • $\begingroup$ Actually it does not work for $\ell^1$ because in this case weak convergence is equivalent to strong convergence. $\endgroup$ – Davide Giraudo Nov 9 '14 at 21:07
  • $\begingroup$ Ok, but which part of the proof fails? And what about $\ell_\infty$? $\endgroup$ – balestrav Nov 9 '14 at 21:22
  • $\begingroup$ 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. $\endgroup$ – Davide Giraudo Nov 9 '14 at 21:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.