4
$\begingroup$

I have this problem at the moment which the strong topology $\beta (E;E^* )$ is defined, when $E$ is a locally convex space. This topology is generated by the basic open sets: $$U=\{x \in E : \sup_{f \in B} |\langle f,x \rangle|<\varepsilon\},$$

where $B\subset E^* $ is bounded. In this way, we say that $B$ is bounded if for all $x\in E, \ \sup_{f\in B} |\langle f,x \rangle|<\infty,$ which is equivalent to the $\omega^* -$boundless. If we consider the strong topology $\beta(E^* ; E),$ now in $E^* , $ the basic open sets are $$V=\{f \in E^* : \sup_{x \in A} |\langle f,x \rangle|<\varepsilon\},$$ where $A\subset E$ is bounded. It is well known that a set $A$ is bounded if, and only if, is weakly bounded, and because of that, we have $\sup_{x \in A}|\langle f,x \rangle|<\infty, \ \forall f \in E^* . $ So, my question is: when we say that $B\subset E^* $ is bounded, do we mean that it is bounded in the strong topology or in the weak* topology? Or are they equivalent?

$\endgroup$

1 Answer 1

8
$\begingroup$

In general, $\sigma(E^*,E)$-bounded sets need not be $\beta(E^*,E)$-bounded. For an example, let $E$ be the set of scalar sequences with only finitely many non-zero terms endowed with the norm $\|x\|_\infty=\sup\{|x_n|:n\in\mathbb N\}$. For the evaluations $\delta_n(x)=x_n$, the set $B=\{n\delta_n:n\in\mathbb N\}$ is $\sigma(E^*,E)$-bounded but not $\beta(E^*,E)$-bounded.

A sufficient condition for the coincidence of weak$^*$- and strongly bounded sets is barrelledness of the locally convex space $E$ since then $\sigma(E^*,E)$-bounded sets are even equi-continuous.

I am not sure whether there is a standard what is meant by just boundedness and I would always specify to pointwise boundedness or uniform boundedness on $E$-bounded sets.

$\endgroup$

This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .