Hausdorff LCTVS: Continuity Implies Weak Continuity
For Hausdorff LCTVS E and F, if t from E to F is Continuous then t is Weakly Continuous
| Proposition 1: Let $E$ and $F$ be Hausdorff locally convex topological vector spaces so that $(E, E')$ and $(F, F')$ are dual pairs, and let $t : E \to F$. If $t$ is continuous then $t$ is weakly continuous. |
- Proof: Suppose that $t : E \to F$ is continuous (where $E$ and $F$ are equipped with their given Hausdorff and locally convex topologies). Then for each $f \in F'$, we have that the linear operator $\langle t(\cdot), f \rangle : E \to \mathbf{F}$ is a continuous linear form on $E$, since $t$ and $f$ are both continuous, and $\langle (t \cdot), f \rangle = f(t(\cdot))$ is just the composition of $f$ and $t$. Therefore:
\begin{align} \quad t'(H) \subseteq E' \end{align}
- So by the proposition on the Weakly Continuous Linear Operators page, $t$ is weakly continuous. $\blacksquare$
It should be remarked that the converse of Proposition 1 above is NOT true in general. Indeed, let $E$ be a vector space and let $\tau_1$ and $\tau_2$ be Hausdorff and locally convex topologies on $E$ for which $\tau_1 \subset \tau_2$ (strict inclusion).
Let $t : (E, \tau_1) \to (E, \tau_2)$ be the identity function. Then $t$ is not continuous, for given an $\tau_2$-open set $U \in \tau_2 \setminus \tau_1$, $t^{-1}(U) = U \not \in \tau_1$ is not $\tau_1$-open.
However $t$ is still weakly continuous