Let $\mathfrak{X}$ be a Banach space. As a standard corollary of the Principle of Uniform Boundedness, any weak-* convergent sequence in $\mathfrak{X}^*$ must be (norm) bounded. A weak-* convergent net need not be bounded in general, but must it be eventually bounded?
It seems like the following should prove that the answer is yes: If $\{y_\nu\}$ is a net in $\mathfrak{X}^*$, suppose it's not eventually bounded. Then we can recursively construct an unbounded subsequence: since the net is not bounded, there exists some $\nu_1$ with
$\|y_{\nu_1}\| > 1$. By hypothesis the tail subnet $\{y_\nu \mid \nu \geq \nu_1\}$ is not bounded, so there exists some $\nu_2 \geq \nu_1$ with $\|y_{\nu_2}\| > 2$, and so on. If the original net were weak-* convergent, so would this unbounded subsequence, contradicting PUB.
It would then follow that weakly convergent nets in $\mathfrak{X}$ are bounded as well, because the image in $\mathfrak{X}^{**}$ would be weak-* convergent.
Question: This is legit, right? I'm still not quite comfortable enough with nets or with the weak-* topology to entirely trust myself here, and I'd like to know the answer since I seem to be bumping into this question a lot recently.
Best Answer
Here's a counterexample. Let $X$ be any infinite-dimensional Banach space.
Let $\mathcal{U}$ be the collection of all weak-* open neighborhoods of $0 \in X^*$. One can show that each $U \in \mathcal{U}$ contains a line. (Think about what the basic open sets are. Indeed, $U$ contains a vector subspace of finite codimension.) For each $U$, let $f_U$ be a nonzero point on such a line, so that $\mathbb{R} f_U \subset U$.
Set $I = \mathcal{U} \times \mathbb{N}$ with the preorder $\preceq$ defined by $$(U,n) \preceq (V,m) \text{ iff } V \subset U.$$ This makes $I$ into a directed set. Let $f_{(U,n)} = n f_U$; this defines a net indexed by $I$. This net converges to 0, since for any weak-* neighborhood $V$ of 0 we have $$\{ f_{(U,n)} : (U,n) \succeq (V,0) \} \subset V.$$ But for any $(V,m)$ the segment $\{f_{(U,n)} : (U,n) \succeq (V,m)\}$ is unbounded, since in particular it contains all the $f_{(V,k)} = k f_V$ for $k \in \mathbb{N}$.