Topologies of the Dual Pair (E, F)

# Topologies of the Dual Pair (E, F)

Definition: If $(E, F)$ is a dual pair, then a locally convex topology $\tau$ on $E$ with $(E^{\tau})' = F$ is called a **Topology of the Dual Pair $(E, F)$. |

For example, if $E$ is a locally convex topological vector space that is Hausdorff, then $(E, E')$ is a dual pair, and from the theorem on The Topological Dual of E Equipped with σ(E, F) is F page, $\sigma(E, E') $ is topology of the dual pair [[$ (E, E')$.

In the following proposition we will see that if $(E, F)$ is a dual pair and if $A \subseteq E$ then the closure of $A$ in any of the topologies of the dual pair $(E, F)$ is the same.

Proposition 1: If $(E, F)$ is a dual pair and if $A$ is a convex subset of $E$, then the closure of $A$ is the same subset of $E$ when $E$ is equipped with any topology of the dual pair $(E, F)$. |

**Proof:**We will prove that the closure of $A$ under any topology $\xi$ of the dual pair $(E, F)$ is the same as the closure of $A$ under $\sigma(E, F)$.

- First observe that since $\sigma(E, F)$ is the coarsest topology of the dual pair $(E, F)$, we must have that:

\begin{align} \overline{A}^{\xi} \subseteq \overline{A}^{\sigma(E, F)} \end{align}

- Now suppose that $a \not \in \overline{A}^{\xi}$. Since $A$ is convex and since $\xi$ is a locally convex topology, we have by one of the corollaries on The Hahn-Banach Separation Theorem page that since $a \not \in \overline{A}^{\xi}$ there exists a $\xi$-continuous linear form, call it $x'$ on $E$ such that:

\begin{align} \langle a, x' \rangle \not \in \overline{\langle A, x' \rangle} \end{align}

- (Note that since $x'$ is $\xi$-continuous, we have that $x' \in (E^{\xi})'$. However, since $\xi$ is topology of the dual pair $(E, F)$ we have that $(E^{\xi})' = F$, and so we may regard $x' \in F$ so that $\langle a, x' \rangle$ makes sense.)

- But $\overline{\langle A, x' \rangle}$ is a convex set of real or complex numbers (as $A$ is convex and $x'$ is linear) and since $\langle a, x' \rangle \not \in \overline{\langle A, x' \rangle}$, there exists a $ \delta > 0 $ such that:

\begin{align} |\langle a, x' \rangle - \langle x, x' \rangle| = | \langle a - x, x' \rangle | \geq \delta \end{align}

- for all $x \in A$. Let:

\begin{align} U := \{ x : |\langle x, x' \rangle | < \delta \} \end{align}

- Then $U$ is a $\sigma (E, F)$-neighbourhood of the origin, and so $a + U$ is a $\sigma (E, F)$-neighbourhood of $a$, and has the form:

\begin{align} a + U = a + \{ x : |\langle x, x'| \rangle < \delta \} = \{ a + x : |\langle x, x' \rangle | < \delta \} = \{ y : |\langle y - a, x' \rangle | < \delta \} = \{ x : |\langle x - a, x' \rangle| < \delta \} \end{align}

- But $|\langle x - a, x' \rangle| \geq \delta$ for all $x \in A$, and so $(a + U) \cap A = \emptyset$. Thus $a \not \in \overline{A^{\sigma(E, F)}}$.

- So $(\overline{A}^{\xi})^C \subseteq (\overline{A}^{\sigma(E,F)})^C$, and thus:

\begin{align} \overline{A}^{\sigma(E, F)} \subseteq \overline{A}^{\xi} \end{align}

- So indeed, we have that $\overline{A}^{\xi} = \overline{A}^{\sigma(E, F)}$. Since this is true for any topology $\xi$ of the dual pair $(E, F)$, we must have that the closure of a convex set $A$ is the same for every topology of the dual pair $(E, F)$. $\blacksquare$