Must a weakly or weak-* convergent net be eventually bounded?

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.

Solution 1:

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 $k \in \mathbb{N}$.

Solution 2:

Nate Eldredge has done the hard work by giving a counterexample to the conjecture; here’s a brief explanation of what’s wrong with the argument given in the question.

A net $\psi:J\to X$ is a subnet of a net $\varphi:I\to X$ iff for each $i\in I$ there is a $j\in J$ such that $$\big\{\psi(j\,'):j\le j\,'\big\}\subseteq\big\{\varphi(i\,'):i\le i\,'\big\}\;.$$ Equivalently, if $\varphi$ is eventually in a set $A$, so is $\psi$.

Taking $D$ as the directed set underlying your net, there’s no reason to think that your sequence $\langle y_{\nu_k}:k\in\Bbb N\rangle$ is actually a subnet of $\langle y_\nu:\nu\in D\rangle$: there may well be a $\nu_0\in D$ such that $$\{y_{\nu_k}:k\in\Bbb N\}\setminus\{y_\nu:\nu_0\preceq\nu\}$$ is infinite. This is the case with Nate’s net, for instance.