Strong dual space

Remarks

Each $p_B$ is finite, because a continuous linear form maps bounded sets to bounded subsets of $\mathbb{K}$. A net $(f_i)$ converges to $f$ in $E'_b$ if and only if $f_i \to f$ uniformly on every bounded subset of $E$. The topology is also written $\beta(E', E)$. Example: let $E$ be a normed space with closed unit ball $B_1$. Every bounded set lies in some $r B_1$, and $p_{r B_1}(f) = r \norm{f}$, so the strong dual topology is the topology of the dual norm. Non-example: for an infinite-dimensional Banach space $E$, the weak-* topology $\sigma(E', E)$ uses only the finite sets $B$. It is not the strong topology, since its neighbourhoods of $0$ are not norm bounded.