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:*

\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.:

\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_{\| 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}

- Therefore the norm topology on $F'$ is finer than the topology of $t(\mathcal A) := \{ t(A) : A \in \mathcal A \}$-convergence.

- Now since $t$ is continuous and since normed spaces are Hausdorff and locally convex, we have by the result on the For Hausdorff LCTVS E and F, if t from E to F is Continuous then t is Weakly Continuous page that $t'$ is weakly continuous. Therefore, by the result on the Weak Continuity Criterion of t for the Continuity of t' page we have that $t'$ being weakly continuous implies that $t'$ is continuous when $E'$ is equipped with the topology of $\mathcal A$-convergence (which is norm convergence) and $F'$ is equipped with the topology of $t(\mathcal A)$-convergence. Thus $t'$ is continuous when $E'$ is equipped its norm topology and when $F'$ is equipped with its norm topology.

- $\Leftarrow$ Proven analogously. $\blacksquare$