Boundedness and pointwise convergence imply weak convergence in $\ell^p$

There is a more general result.

Theorem. Let $E$ be a normed space. Let $\{x_n:n\in\mathbb{N}\}\subset E$ and $x\in E$, then the following conditions are equivalent:

• $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x\in E$
• $\{x_n:n\in\mathbb{N}\}$ is bounded and for all $S\subset E^*$ such that $\mathrm{cl}(\mathrm{span} S)=E^*$ holds $\lim\limits_{n\to\infty} f(x_n)=f(x)$ for all $f\in S$.

Proof. Assume $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x$. Then it is well knonw that $\{x_n:\in\mathbb{N}\}$ is bounded. Moreover for all $f\in X^*$ we know that $\lim\limits_{n\to\infty}f(x_n)=f(x)$. Hence for all $S\subset X^*$ such that $\mathrm{cl}(\mathrm{span}S)=X^*$ we have $\lim\limits_{n\to\infty}f(x_n)=f(x)$ for all $f\in S$.

Assume that $\{x_n:n\in\mathbb{N}\}$ is bounded by some conatant $M>0$ and for all $S\subset X^*$ such that $\mathrm{cl}(\mathrm{span}S)=X^*$ holds $\lim\limits_{n\to\infty}f(x_n)=f(x)$ for all $f\in S$. Take arbitrary $g\in\mathrm{span}S$, then $g=\sum\limits_{k=1}^m \alpha_k f_k$. Then it is easy to check that $\lim\limits_{n\to\infty}g(x_n)=g(x)$. So for all $g\in\mathrm{span}S$ we have $\lim\limits_{n\to\infty}g(x_n)=g(x)$. Now take arbitrary $g\in X^*=\mathrm{cl}(\mathrm{span}S)$, then $g=\lim\limits_{k\to\infty} g_k$ for some $\{g_k:k\in\mathbb{N}\}\subset\mathrm{span}S$. Fix arbitrary $\varepsilon>0$. Since $g=\lim\limits_{k\to\infty} g_k$, then there exist $K\in\mathbb{N}$ such that $\Vert g-g_k\Vert\leq\varepsilon$ for all $k>K$. Then $$ \begin{align} |g(x_n)-g(x)| &\leq|g(x_n)-g_k(x_n)|+|g_k(x_n)-g_k(x)|+|g_k(x)-g(x)|\\ &\leq\Vert g-g_k\Vert\Vert x_n\Vert+|g_k(x_n)-g_k(x)|+\Vert g_k-g\Vert\Vert x\Vert\\ &\leq\varepsilon M+|g_k(x_n)-g_k(x)|+\varepsilon\Vert x\Vert \end{align} $$ Let's take a limit $n\to\infty$ in this inequality, then $$ \limsup\limits_{n\to\infty}|g(x_n)-g(x)|\leq \varepsilon M+\lim\limits_{n\to\infty}|g_k(x_n)-g_k(x)|+\varepsilon\Vert x\Vert $$ Since $g_k\in\mathrm{span}S$, then $\lim\limits_{n\to\infty}|g_k(x_n)-g_k(x)|=0$ and we get $$ \limsup\limits_{n\to\infty}|g(x_n)-g(x)|\leq \varepsilon M+\varepsilon\Vert x\Vert $$ Since $\varepsilon>0$ is arbitrary we conclude $\limsup\limits_{n\to\infty}|g(x_n)-g(x)|=0$. This is equivalent to $\lim\limits_{n\to\infty}|g(x_n)-g(x)|=0$ i.e. $\lim\limits_{n\to\infty}g(x_n)=g(x)$. Since $g\in X^*$ is arbitrary, then $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x$.