No one answered last week's problem. You can find the solution below.
[sp]We will prove a more general result.
Let $E$ be a normed space. Let $\{x_n:n\in\mathbb{N}\}\subset E$ and $x\in E$, then TFAE:
- $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x\in E$.
- $\{x_n:n\in\mathbb{N}\}$ is bounded and for all $S\subset E^{\ast}$ such that $\mathrm{cl}(\mathrm{span}\,S)=E^{\ast}$ holds, $\displaystyle\lim_{n\to\infty} f(x_n) = f(x)$ for all $f\in S$.
Proof: Assume $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x$. Then it is well known that $\{x_n:n\in\mathbb{N}\}$ is bounded. Moreover for all $f\in X^{\ast}$ we know that $\displaystyle\lim_{n\to\infty} f(x_n) = f(x)$. Hence for all $S\subset X^{\ast}$ such that $\mathrm{cl}(\mathrm{span}\,S)=X^{\ast}$, we have $\displaystyle\lim_{n\to\infty} f(x_n)=f(x)$ for all $f\in S$.
Assume that $\{x_n:n\in\mathbb{N}\}$ is bounded by some constant $M>0$ and for all $S\subset X^{\ast}$ such that $\mathrm{cl}(\mathrm{span}\,S)=X^{\ast}$ holds, $\displaystyle\lim_{n\to\infty}f(x_n)=f(x)$ for all $f\in S$. Take arbitrary $g\in\mathrm{span}\,S$, then $g=\displaystyle\sum_{k=1}^m\alpha_kf_k$. Then it is easy to check that $\displaystyle\lim_{n\to\infty} g(x_n)=g(x)$. So for all $g\in\mathrm{span}\,S$ we have $\displaystyle\lim_{n\to\infty}g(x_n)=g(x)$. Now take arbitrary $g\in X^{\ast}=\mathrm{cl}(\mathrm{span}\,S)$, then $g=\displaystyle\lim_{k\to\infty}g_k$ for some $\{g_k:k\in\mathbb{N}\}\subset\mathrm{span}\,S$. Fix arbitrary $\varepsilon>0$. Since $g=\displaystyle\lim_{k\to\infty}g_k$, then there exist $K\in \mathbb{N}$ such that $\|g−g_k\|\leq\varepsilon$ for all $k>K$. Then
\[\begin{aligned}
|g(x_n)-g(x)|
&\leq|g(x_n)-g_k(x_n)|+|g_k(x_n)-g_k(x)|+|g_k(x)-g(x)|\\
&\leq\| g-g_k\|\| x_n\|+|g_k(x_n)-g_k(x)|+\| g_k-g\|\| x\|\\
&\leq\varepsilon M+|g_k(x_n)-g_k(x)|+\varepsilon\| x\|
\end{aligned}\]Let's take a limit $n\to\infty$ in this inequality, then
\[\limsup\limits_{n\to\infty}|g(x_n)-g(x)|\leq\varepsilon M+\lim\limits_{n\to\infty}|g_k(x_n)-g_k(x)|+\varepsilon\| x\|\]
Since $g_k\in\mathrm{span}\,S$, then $\displaystyle lim_{n\to\infty}|g_k(x_n)−g_k(x)|=0$ and we get
\[\limsup\limits_{n\to\infty}|g(x_n)-g(x)|\leq\varepsilon M+\varepsilon\| x\|\]
Since $\varepsilon>0$ is arbitrary we conclude $\displaystyle \limsup_{n\to\infty}|g(x_n)−g(x)|=0$. This is equivalent to $\displaystyle\lim_{n\to\infty}|g(x_n)−g(x)|=0$ i.e. $\displaystyle\lim_{n\to\infty}g(x_n)=g(x)$. Since $g\in X^{\ast}$ is arbitrary, then $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x$.$\hspace{.25in}\blacksquare$[/sp]