For Normed Spaces E and F, t is Continuous IFF t' is Continuous

# For Normed Spaces E and F, t is Continuous IFF t' is Continuous

 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.

Recall that if $E$ is a normed space, then its topological dual $E'$ is also a normed space with norm defined for all $f \in E'$ by:

(1)
\begin{align} \quad \| f \| := \sup_{\| x \| \leq 1} |f(x)| \end{align}
• 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 and $t' : F' \to E^*$.
• Let $\mathcal A$ be the set of norm closed origin-centered balls in $E$, i.e.:
(2)
\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.:
(3)
\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:
(4)
\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:
(5)
\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:
(6)
\begin{align} \quad B^{\circ} \subseteq (t(A))^{\circ} \end{align}
• Therefore the norm topology on $F'$ is finer than the topology of $t(\mathcal A) := \{ t(A) : A \in \mathcal A \}$-convergence.
• $\Leftarrow$ Proven analogously. $\blacksquare$