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.
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