A sequence converging weakly in $\ell^p$, for $p >1$ and failing to converge weakly for $p=1$

Question

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?

Answer 1

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 $p=1$): $$<x_n,y>\xrightarrow[n\to+\infty]{}<x,y>,$$ where $<x,y>:=\displaystyle\sum_{k=1}^{\infty}x(k)y(k)$ 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 $).

Answer 2

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:

1. We have $\lim_{n\to +\infty}x^{(n)}_N= 0$ for all $N\in\Bbb N$.
2. 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 $\ell^1$.