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?



Weak* bounded and strong bounded are the same?
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*.


