# weak convergence of nth unit vector in $\ell^p (\mathbb N)$

I would like to know,
Why is it true that $e_n$, the nth unit vector in $\ell^p(\mathbb N)$ converges weakly to $0$. $1 < p < \infty$ According to Mazur's lemma, which says that $y_n$ is convex combination of $e_n$ converges to $0$ in norm topology. I am trying to construct an explicit convex combination but i failed.

How do i construct it, any example would be nice. Thanks.

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You need $1<p<\infty$, here. Take an element $x\in\ell_q$. Note its coordinates converge to $0$. Then, what must $x(e_i)$ converge to? For the convex combinations, consider $y_n={1\over n}\sum_{j=1}^n e_j$. – David Mitra Dec 13 '12 at 17:09
@DavidMitra : Its will give co-ordinate wise convergence to $0$, at every co-ordinate. But i am getting confused because for some $x \in \ell_q$, nth co-ordinate ie 1 will not be zero ? Where is the flaw in what i am thinking. – Theorem Dec 13 '12 at 17:22
You get $\lim_i x(e_i)=\lim_i x_i=0$. This is true for all $x\in\ell_q$, so $(e_i)$ is weakly convergent to $0$ for $1<p<\infty$. When $p=1$, $(e_i)$ is not weakly convergent and any convex combination of them will have norm 1. I'm not sure what happens when $p=\infty$. The $(y_n)$ I defined previously do tend to $0$ in norm. I'm not sure if $(e_i)$ converges weakly or not. – David Mitra Dec 13 '12 at 17:31
@David Mitra : Thank you :) – Theorem Dec 13 '12 at 17:32

I suppose here $1<p<\infty, and \frac{1}{p}+\frac{1}{q}=1$.
Note that by definition, $e_n\overset{w}{\to}0$ iff $f(e_k)\to 0, \forall f\in (l^p)'=l^q$, so for any $f=(a_k)\in l^q$, we check that $|f(e_k)|=|a_k|\to 0.$
For the second part, just take $y_n=\sum_{k=n+1}^{2n}\frac{1}{n}e_k$, then calculation shows that $||y_n||^p=\sum_{k=n+1}^{2n}\frac{1}{n^p}=\frac{1}{n^{p-1}}\to 0$, since $p>1$.
$a_k \to 0$ is because of that fact that if it belongs to $\ell_q$ then it has to converge to $0$ if it is a infinite sequence , and if it is a finite sequence then its obvious . – Theorem Dec 13 '12 at 17:31