A sequence converging weakly in $\ell^p$, for $p >1$ and failing to converge weakly for $p=1$ For $1 \le p < \infty$ and each index $n$, let $e_n \in \ell^p$ have $n$-th component 1 and all other componenets $0$. I want to show that $p>1 \Rightarrow \{e_n\} \to 0$ weakly in $\ell^p$ and that this is not the case in $\ell^1$. Any suggestions?
 A: The definition of 
$\ell^p, \ 1\leq p<+\infty$
 is 
$$\ell^p=\left\{x=(x(k))_{k\in\mathbb N}:\|x\|_p:=\sqrt[p]{\displaystyle\sum_{k=1}^\infty |x(k)|^p}<+\infty\right\}$$
and 
$$\ell^\infty=\left\{x=(x(k))_{k\in\mathbb N}:\|x\|_\infty:=\sup\{|x(k)|:k\in\mathbb N\}<+\infty\right\}.$$
A sequense 
$(x_n)_{n\in\mathbb N}$
 in 
$\ell^p, \ 1\leq p<+\infty$
, converges weakly to 
$x\in\ell^p$
 (denoted $x_n\rightharpoonup x$ )
if for any 
$y\in\ell^q$ 
(where $q$ is defined by $p+q=pq$ for $p\neq1$ and $q=+\infty$ for $p=1$):
$$\langle x_n,y \rangle \xrightarrow[n\to+\infty]{}\langle x,y \rangle,$$ 
where 
$\langle x,y \rangle :=\displaystyle\sum_{k=1}^{\infty}x(k)y(k)$
 for 
$x=(x(k))_{k\in\mathbb N}\in\ell^p,y=(y(k))_{k\in\mathbb N}\in\ell^q$.
We want to show that for 
$p>1, \ e_n\rightharpoonup 0$  in $\ell^p$.
Equivalently that $\displaystyle\sum_{k=1}^{\infty}e_n(k)y(k)\xrightarrow[n\to+\infty]{}0$ 
for all 
$(y(k))_{k\in\mathbb N}\in\ell^q$. 
Note that 
$e_n(k)y(k)=\delta_{nk}y(k) \ \forall n,k \in\mathbb N$;
 from $\displaystyle \sum_{k=1}^\infty|y(k)|<+\infty$ it follows that $$\displaystyle\sum_{k=1}^{\infty}e_n(k)y(k)=y(n)\xrightarrow[n\to\infty]{}0.$$
Now for 
$p=1$ 
find an element 
$(y(k))_{k\in\mathbb N}\in\ell^\infty$
s.t. $\displaystyle\sum_{k=1}^{\infty}e_n(k)y(k) \not\rightarrow0$ 
(hint: find one with $\displaystyle\sum_{k=1}^{\infty}e_n(k)y(k)=1, \ \forall n\in\mathbb N $).
A: Following the same steps as here, we can establish the following weak convergence conditions.

Let $1<p<\infty$ and $\{x^{(n)}\}\subset \ell^p$. This sequences converges weakly to $0$ in $\ell^p$ is and only if the following two conditions are satisfied:

*

*We have $\lim_{n\to +\infty}x^{(n)}_N= 0$ for all $N\in\Bbb N$.

*The sequence $\{\lVert x^{(n)}\rVert_p\}\subset\Bbb R$ is bounded.


To see that $\{e_n\}$ doesn't converge to $0$ in $\ell^1$, consider $L\colon\ell^1\to \Bbb R$, $L(x):=\sum_{j=0}^{+\infty}x_j$. Actually, strong and weak convergence of sequences are the same thing in $\ell^1$.
