I'm doing Ex 3.17.2 in Brezis's book of Functional Analysis:
Let $E := \ell^p$ with $1 < p \le \infty$. Let $x=(x_{1}, x_{2}, \ldots) \in E$ and $(x^{n})$ with $x^{n}=(x_{1}^{n}, x_{2}^{n}, \ldots)$ be a sequence in $E$. Assume $(x^{n})$ is bounded and $x_{m}^{n} \overset{n \to \infty}{\longrightarrow} x_{m}$ for all $m$. Then $x^{n} \rightharpoonup x$ in $\sigma(E, E')$.
I just want to add a screenshot to make sure that I did not incorrectly quote the exercise.
I solved the case $p \in (1, \infty)$ by using the following 2 lemmas:
Lemma 1: Let $E := \ell^p$ with $1 < p < \infty$. Let $\pi_n: E \to \mathbb R, x \mapsto x_n$ be the canonical projection. Clearly, $\pi_n \in E'$. Let $G := \{\pi_1, \pi_2, \ldots\}$. Then $\operatorname{span} G$ is dense in $E'$. [A proof is given here]
In this comment, @Bazyli said that Lemma 1 is not true if $p \in \{1, \infty\}$ because $E'$ is not separable in those cases.
Lemma 2: Let $E$ be a normed space. Let $(x_n) \subset E$ and $x\in E$. Then $(x_n)$ weakly converges to $x\in E$ if and only if $(x_n)$ is bounded and $\exists S\subset E'$ such that $\mathrm{span} S$ is dense in $E'$ and $f(x_n) \to f(x)$ for all $f\in S$. [A proof is given here]
Could you elaborate on the proof in case $ p = \infty$?
Best Answer
This result is not true as stated. By the Hahn-Banach theorem, there exists a bounded linear functional $\phi$ on $\ell^\infty$ such that $\phi(x)=\lim_{k\to\infty} x(k)$ whenever this limit exists.
Now let $$ x_n(k)=\begin{cases}1&\text{if }k\leq n,\\0&\text{otherwise}.\end{cases} $$ Clearly, $(x_n)$ is bounded in $\ell^\infty$ and $x_n(k)\to 1$ for every $k\in\mathbb N$. However, $\phi(x_n)=0$ and $\phi(x)=1$.