Normed Vector Space is Reflexive iff Weak and Weak-* Topologies on Normed Dual coincide

Theorem

Let $X$ be a normed vector space.

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

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

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

Then $X$ is reflexive if and only if $\struct {X^\ast, w} = \struct {X^\ast, w^\ast}$.

Proof

Necessary Condition

Suppose that $X$ is reflexive.

Then for each $\Phi \in X^{\ast \ast}$ there exists $x \in X$ such that $\Phi = x^\wedge$.

Conversely, by Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have $x^\wedge \in X^{\ast \ast}$ for each $x \in X$.

Hence, we have:

$\set {x^\wedge : x \in X} = X^{\ast \ast}$

By definition, $w$ is the initial topology on $X^\ast$ generated by $X^{\ast \ast}$, and $w^\ast$ is the initial topology on $X^\ast$ generated by the family $\set {x^\wedge : x \in X}$.

It follows that $w = w^\ast$.

$\Box$

Sufficient Condition

Suppose that $\struct {X^\ast, w} = \struct {X^\ast, w^\ast}$.

Then $\struct {X^\ast, w}^\ast = \struct {X^\ast, w^\ast}^\ast$.

From Characterization of Continuity of Linear Functional in Weak Topology, we have $\struct {X^\ast, w}^\ast = X^{\ast \ast}$.

From Characterization of Continuity of Linear Functional in Weak-* Topology , we have $\struct {X^\ast, w^\ast}^\ast = \iota X$.

So we have:

$\iota X = X^{\ast \ast}$

From Normed Vector Space is Reflexive iff Surjective Evaluation Linear Transformation, we have that $X$ is reflexive.

$\blacksquare$