First, my definition of weak convergence in $X$ is that $x_n \rightharpoonup x$ if $\phi(x_n) \to \phi(x)$ for all $\phi \in X^*$.
I recently read the statement that $e_n \rightharpoonup 0$ in $\ell^p$, $p>1$, where $e_n$ is the canonical basis vector.
In $\ell^2$, this is clear to me, since the weak convergence $x_n \rightharpoonup x$ in $\ell^2$ (which is Hilbert) is equivalent to $\langle x_n, y \rangle \to \langle x, y \rangle$ for all $y \in \ell^2$.
But when $p \neq 2$, I struggle to prove this since I don't know the form of a general functional $\phi \in X^*$. Can you help me understand this?
Best Answer
Here is an approach without an explicit characterisation of the dual space of $\ell^p$. Let $\varphi$ be a linear continuous functional on $\ell^p$, $\left(a_n\right)_{n\geqslant 1}$ be an element of $\ell^p$. Let $c_n=a_n\operatorname{sgn}(a_n)\operatorname{sgn}\left(\varphi\left(e_n\right)\right)$, where $\operatorname{sgn}(x)=1$ if $x\gt0$, $-1$ if $x\lt 0$ and $0$ for $x=0$. Let $x_N:=\sum_{n=1}^Nc_ne_n$. Then the sequence $\left(x_N\right)_{N\geqslant 1}$ converges in $\ell^p$ to some $x$ (indeed, $\left\lVert x_{M+N}-x_N\right\rVert_p^p=\sum_{n=N+1}^{N+M}\lvert a_n\rvert^p$).
As a consequence, the sequence $\left(\varphi\left(x_N\right)\right)_{N\geqslant 1}$ is bounded and it follows that $\sum_{n\geqslant 1}a_n\left\lvert \varphi\left(e_n\right)\right\rvert$ is convergent (indeed, $\varphi(x_N)=\sum_{n=1}^Nc_n\varphi(e_n)=\sum_{n=1}^N\lvert a_n \varphi(e_n)\rvert$). In other words, we proved $$\tag{*} \left(a_n\right)_{n\geqslant 1}\in\ell^p\Rightarrow \sum_{n\geqslant 1}\left\lvert a_n\varphi\left(e_n\right)\right\rvert<+\infty. $$ This implies that the sequence $\left(\varphi\left(e_n\right)\right)_{n\geqslant 1}$ converges to $0$. Indeed, if not, there is a $\delta\gt 0$ and a sequence $(n_k)$ of integers growing to infinity such that $\left\lvert \varphi\left(e_{n_k}\right)\right\rvert\gt \delta$ for all $k$. Let $\left(b_k\right)\in \ell^p\setminus \ell^1$ and define $a_n$ by $a_{n_k}=b_k$ and $a_n=0$ if $n$ is not of the form $n_k$ for some $k$ to get a contradiction with $(*)$.