Weakly Continuous Linear Operators

# Weakly Continuous Linear Operators

 Definition: Let $(E, F)$ and $(G, H)$ be dual pairs and let $t : E \to G$ be a linear operator. $t$ is said to be Weakly Continuous if $t$ is continuous when $E$ is equipped with the $\sigma(E, F)$ topology and when $G$ is equipped with the $\sigma(G, H)$ topology.

The following proposition gives us criterion for when $t$ is weakly continuous.

 Proposition 1: Let $(E, F)$ and $(G, H)$ be dual pairs and let $t : E \to G$ be a linear operator with transpose $t' : H \to E^*$. Then $t'(H) \subseteq F$ if and only if $t$ is weakly continuous
• Proof: $\Rightarrow$ Suppose that $t'(H) \subseteq F$. Let $V$ be a $\sigma(G, H)$-neighbourhood of the origin of the form:
(1)
\begin{align} \quad V := \{ g : \sup_{1 \leq i \leq n} |\langle g, h_i \rangle| \leq 1 \} \end{align}
• where $h_1, h_2, ..., h_n \in H$. (Note that $\sigma(G, H)$-neighbourhoods of the form above form a base of neighbourhoods for the $\sigma(G, H)$-topology).
• Let:
(2)
\begin{align} \quad U := \{ e : \sup_{1 \leq i \leq n} |\langle e, t'(h_i) \rangle| \leq 1 \} \end{align}
• Then $U$ is a $\sigma(E, F)$-neighbourhood of the origin. For each $u \in U$, we have that $|\langle u, t'(h_i) \rangle| \leq 1$ for all $1 \leq i \leq n$, or equivalently, $|\langle t(u), h_i \rangle| \leq 1$ for all $1 \leq i \leq n$, so that:
(3)
\begin{align} \quad \sup_{1 \leq i \leq n} |\langle t(u), h_i \rangle| \leq 1 \end{align}
• So $t(u) \in V$. So $t(U) \subseteq V$. Thus, for every $\sigma(G, H)$-neighbourhood of the origin $W$, let $V$ be a $\sigma(G, H)$-neighbourhood of the origin of the form above such that $V \subseteq W$. Then there exists a $\sigma(E, F)$-neighbourhood of the origin such that $t(U) \subseteq V \subseteq W$. So $t$ is continuous when $E$ is equipped with $\sigma(E, F)$ and when $G$ is equipped with $\sigma(G, H)$. Hence $t$ is weakly continuous.
• $\Leftarrow$ Suppose that $t : E \to G$ is weakly continuous.
• For each fixed $h \in H$, observe that $t'(h) = \langle t(\cdot), h \rangle : E \to \mathbf{F}$ is a continuous linear form on $E$ since $t$ is continuous, and so $t'(h) \in E'$. So $t'(H) \subseteq E'$. However, since $E$ is equipped with the $\sigma(E, F)$-topology, we have by the theorem on The Topological Dual of E Equipped with σ(E, F) is F page that $E' = (E^{\sigma(E, F)})' = F$. Thus $t'(H) \subseteq F$. $\blacksquare$.