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

Table of Contents

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.

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.

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