# Weak* bounded and strong bounded are the same?

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?

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.