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Let $E$ be a complex Banach space. A consequence of the uniform boundedness principle is the following. If $(\lambda_n)_{n\geq 1}$, $\lambda$, are elements of $E^*$ such that $$ \lambda_n(x) \rightarrow \lambda(x) \; \forall x\in E, $$ then $\lambda_n$ are uniformly bounded. This can be seen by considering the family $\mathcal{A} = \{\lambda_n\}_{n\geq 1}$ which must satisfy one of the following:

(i) $\mathcal{A}$ is uniformly bounded

(ii) There exists dense $S\subset E$ such that $$ \sup\limits_{n} |\lambda_n(x)| = \infty \; \forall x\in S $$

Option (ii) cannot happen, since for any $x$, $(\lambda_n(x))_{n\geq 1}$ is a convergent sequence. The last step seems to rely on the fact that we are dealing with sequences (indexed over $\mathbb{N}$).

My question is, if we now replace $(\lambda_n)$ by a net $(\lambda_\alpha)$, can we make the same conclusion? For instance, if the indexing set is $\mathbb{R}$, perhaps there could be a blow up of the net $\lambda_\alpha(x)$ at some finite $\alpha$, even though $\lambda_\alpha(x)$ converges as $\alpha \rightarrow \infty$.

This question came up because I was trying to show that a linear map $\Phi:(E^*$,weak-$*)\rightarrow \mathbb{C}$ is continuous. To do this, I could show that $\Phi(\lambda_\alpha)$ is convergent for every weak-$\ast$ convegent net $\lambda_\alpha$. A second question I have is, are there any properties of $(E^*$,weak-$*)$ that would allow me to show continuity by using just sequences indexed by $\mathbb{N}$?

Martin Argerami
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Matthew C
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Unbounded, convergent nets are constructed in this answer. As for what you are trying to prove, I don't think it holds. For instance let $E=c_0(\mathbb N)$, then $E^*=\ell^1(\mathbb N)$. Consider elements $f_n\in E^*$ given by $$ f_n(k)=\begin{cases}1,&\ k=n\\ 0,&\ k\ne n\end{cases} $$ and let $\tau:E^*\to\mathbb C$ be $\tau(f)=\sum_{k=1}^\infty f(k)$. Note that $f_n\to0$ weak$^*$, since for any $y\in c_0(\mathbb N)$ we have $$ \langle f_n,y\rangle=\sum_{k}f_n(k)y(k)=y(n)\to0. $$ But $\tau(f_n)=1$ for all $n\in\mathbb N$. So $\tau$ is a linear functional on $E^*$ that is not weak$^*$-continuous.

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Martin Argerami
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