Boundedness of weakly convergent sequences

Let X be a Banach space. A net $\{x_\alpha\}$ in X is said to weakly converge to $x$ if $\ell(x_\alpha) \rightarrow \ell(x)$ for all $\ell \in X^*$. Clearly, every norm convergent net is weakly convergent (with the same limit). We know that a norm convergent sequence is necessarily bounded. It is natural to ask boundedness still hold for weakly convergent sequences.

Theorem. Every weakly convergent sequence in X is bounded.
Proof: Let $\{x_n\}$ be a weakly convergent sequence in X. Let $T_n \in X^{**}$ be defined by $T_n(\ell) = \ell(x_n)$ for all $\ell \in X^*$. Fix an $\ell \in X^*$. For any $n \in \mathbb{N}$, since the sequence $\{\ell(x_n)\}$ is convergent, $\{T_n(\ell)\}$ is a bounded set. By Uniform Boundedness Principle,
\[ \sup_{n \in \mathbb{N}} \|x_n\| = \sup_{n \in \mathbb{N}} \|T_n\| < \infty, \] i.e. $\{x_n\}$ is bounded.

However, the above result is no longer true when sequences are replaced by nets. To give a counterexample, let $X = L^1(\mathbb{R})$. Denote the characteristic function of a subset A by $\chi_A$. Let $\mathcal{O}$ be the collection of all open neighborhoods of 0 in $\mathbb{R}$ of finite Lebesgue measure with the ordering given by $U \prec V$ iff $U \supseteq V$. Consider the net $\{\chi_U: U \in \mathcal{O}\}$ in X. It is not bounded since $\|\chi_{(-n, n)}\|_1 = 2n \rightarrow \infty$ as $n \rightarrow \infty$. On the other hand, it converges weakly to 0. Fix any $f \in X^* = L^\infty(\mathbb{R})$. Choose M > 0 such that $|f| \leq M$ almost everywhere. For every $\epsilon > 0$, we can choose some $V \in \mathcal{O}$ with $m(V) < \epsilon/M$, then $|\int f\chi_U dm| \leq M m(U) \leq M m(V) < \epsilon$ whenever $U \succ V$.