---
tag: JTCKJWTFEN
kind: definition
name: Strong dual space
msc: [46A20]
defines: RG52DZ1A8Y
uses: [4A82S5FVRT, 3BAJ8ZTVPX, ZVFXQ7S6FY]
sources:
  - url: https://en.wikipedia.org/wiki/Strong_dual_space
---

## Definition

Let $\mathbb{K}$ be $\R$ or $\C$, let $E$ be a topological vector space over $\mathbb{K}$, and let $E'$ be its [[continuous dual space|4A82S5FVRT]].
For each [[von Neumann bounded set|3BAJ8ZTVPX]] $B \subseteq E$, put $p_B(f) = \sup_{x \in B} \abs{f(x)}$ for $f \in E'$.
The *strong dual space* $E'_b$ of $E$ is $E'$ with the [[topology generated by the seminorms|ZVFXQ7S6FY]] $p_B$, for all bounded subsets $B$ of $E$.

## 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.