Tstats

Proposition 1: Let $(E, F)$ and $(G, H)$ be dual pairs, $\mathcal A$ a collection of $\sigma(E, F)$-weakly bounded sets, and let $t : E \to G$ be a linear mapping. Then if $t$ is weakly continuous then its transpose $t'$ is continuous when $F$ is equipped with the polar topology of uniform convergence of the sets of $\mathcal A$ and $H$ is equipped with the polar topology of uniform convergence of the sets of $t(\mathcal A) := \{ t(A) : A \in \mathcal A \}$. |

*Recall that $t : E \to G$ being weakly continuous simply means that $t$ is continuous when $E$ is equipped with the $\sigma(E, F)$-weak topology and when $G$ is equipped with the $\sigma(G, H)$-weak topology.*

*Also recall that if $t : E \to G$ then $t' : H \to E^*$. But recall that $t$ is weakly continuous if and only if $t'(H) \subseteq F$ (see Weakly Continuous Linear Operators). Thus if $t$ is weakly continuous, then $t'$ maps $H$ to a subset of $F$.*

**Proof:**Let $t$ be weakly continuous so that $t'(H) \subseteq F$ and so that $t' : H \to F$.

- Since $F$ is equipped with the polar topology of uniform convergence of the sets of $\mathcal A$, we have that $\{ A^{\circ} : A \in \mathcal A \}$ is a base of neighbourhoods of $o_F$. So let $V$ be any neighbourhood in $F$. Then there exists an $A \in \mathcal A$ such that $A^{\circ} \subseteq F$ and thus:

\begin{align} \quad t'^{-1}(A^{\circ}) \subseteq t'^{-1}(V) \end{align}

- But since $(E, F)$, $(G, H)$ are dual pairs and $t : E \to G$ is weakly continuous, we have by the proposition on the For Dual Pairs (E, F), (G, H), and a Weakly Continuous Linear Operator t from E to G, (t(A))° = t'-1(A°) page that $t'^{-1}(A^{\circ}) = (t(A))^{\circ}$, and so from the inclusion above:

\begin{align} \quad (t(A))^{\circ} \subseteq t'^{-1}(V) \end{align}

- But $(t(A))^{\circ}$ is a neighbourhood of $o_H$ in $H$, with $H$ equipped with the topology of uniform convergence of the sets $t(\mathcal A) := \{ t(A) : A \in \mathcal A \}$.

- So for every neighbourhood $V$ of $o_F$ in $F$, there exists a neighbourhood $(t(A))^{\circ}$ of $o_H$ in $H$ such that $(t(A))^{\circ} \subseteq t'^{-1}(V)$. Thus $t'$ is continuous when $F$ and $H$ are given the respective topologies above. $\blacksquare$

Proposition 1: Let $E$ and $F$ be normed spaces, let $E'$ and $F'$ denote their topological duals and equip them with their normed topologies. Let $t : E \to F$ be a linear operator with transpose $t' : F ' \to E^*$. Then $t$ is continuous if and only if $t'$ is continuous. |

**Proof:**Since $E$ and $F$ are normed spaces, they are Hausdorff and locally convex topological vector spaces so that $(E, E')$ and $(F, F')$ are dual pairs so that $t' : F' \to E^*$.

- Let $\mathcal A$ be the set of norm closed origin-centered balls in $E$, i.e.:

\begin{align} \quad \mathcal A := \{ \overline{B_E}(x, \epsilon) : \epsilon > 0 \} \end{align}

- Similarly, let $\mathcal B$ be the set of norm closed origin-centered balls in $F$, i.e.,:

\begin{align} \quad \mathcal B := \{ \overline{B_F}(y, \epsilon) : \epsilon > 0 \} \end{align}

- It should be remarked that a base of neighbourhoods for the norm topology on $E'$ is given by:

\begin{align} \quad \mathcal A^{\circ} := [\overline{B_E}(x, \epsilon)]^{\circ} : \epsilon > 0 \} \end{align}

- Indeed, observe that $f \in [\overline{B_E}(x, \epsilon)]^{\circ}$ is equivalent to saying that $f \in E'$ is such that:

\begin{align} \quad \sup \{ |\langle x_1, f \rangle | : x_1 \in \overline{B_E}(x, \epsilon) \} \leq 1 \end{align}

- or equivalently, $f \in E'$ is such that:

\begin{align} \quad \sup_{\| x \| \leq 1} |f(x)| \leq \epsilon \end{align}

- which tells us that $\| f \| \leq \epsilon$. So $[\overline{B_E}(x, \epsilon)]^{\circ} = B_{E'}(f, \epsilon)$. But $\{ [\overline{B_E}(x, \epsilon)]^{\circ} : \epsilon > 0 \} = \{ \overline{B_{E'}}(f, \epsilon) : \epsilon > 0 \}$ is a base of neighbourhoods of the origin in $E'$.

- A similar discussion leads us to see that $\{ [\overline{B_F}(y, \epsilon)]^{\circ} : \epsilon > 0 \} = \{ \overline{B_{F'}}(g, \epsilon) : \epsilon > 0 \}$ is a base of neighbourhoods of the origin in $F'$.

- $\Rightarrow$ Suppose that $t : E \to F$ is continuous. Then for each $B \in \mathcal B$ there exists an $A \in \mathcal A$ for which $t(A) \subseteq B$ (as $\mathcal A$ forms a base of neighbourhoods of the origin for $E$, and similarly, $\mathcal B$ forms a base of neighbourhoods of the origin for $F$). Taking polars (in $F'$) and we get that:

\begin{align} \quad B^{\circ} \subseteq (t(A))^{\circ} \end{align}