Evaluation Linear Transformation on Normed Vector Space is Weak to Weak-* Continuous Embedding into Second Normed Dual

Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $X$ be a normed vector space over $\Bbb F$.

Let $X^\ast$ be the normed dual of $X$.

Let $X^{\ast \ast}$ be the second normed dual of $X$.

Let $w$ be the weak topology on $X$.

Let $w^\ast$ be the weak-$\ast$ topology on $X^{\ast \ast}$.

Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation.

Then $\iota$ is $\struct {w, w^\ast}$-continuous.

Proof

From the definition of the initial topology, $w^\ast$ is generated by the mappings $f^\wedge : X^{\ast \ast} \to \Bbb F$ defined by:

$\map {f^{\wedge} } \Phi = \map \Phi f$

for each $f \in X^\ast$ and $\Phi \in X^{\ast \ast}$.

From Continuity in Initial Topology, it therefore suffices to verify that $f^\wedge \circ \iota : \struct {X, w} \to \Bbb F$ is continuous for each $f \in X^\ast$.

From Characterization of Continuity of Linear Functional in Weak Topology, we can equivalently show that $f^\wedge \circ \iota \in X^\ast$ for each $f \in X^\ast$.

For each $x \in X$ and $f \in X^\ast$, we have:

\(\ds \map {\paren {f^\wedge \circ \iota} } x\)

\(=\)

\(\ds \map {f^\wedge} {x^\wedge}\)

\(\ds \)

\(=\)

\(\ds \map {x^\wedge} f\)

\(\ds \)

\(=\)

\(\ds \map f x\)

Hence we have:

$f^\wedge \circ \iota = f \in X^\ast$

and we are done.

$\blacksquare$