Von Neumann-Bounded Set in Weak Topology is Norm Bounded

Theorem

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.

Let $w$ be the weak topology on $\struct {X, \norm {\, \cdot \,} }$.

Let $B$ be a von Neumann-bounded subset of $\struct {X, w}$.

Then $B$ is a von Neumann-bounded subset of $\struct {X, \norm {\, \cdot \,} }$

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

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

Let $f \in X^\ast$.

$V = \set {y \in X : \cmod {\map f y} < 1}$ is an open neighborhood of $\mathbf 0_X$ in $\struct {X, w}$.

So there exists $r > 0$ such that:

$B \subseteq r V$

That is, if $x \in B$ we have $x = r y$ for some $y \in V$.

That is:

$\cmod {\map f x} = r \cmod {\map f y} < r$

giving:

$\ds \cmod {\map f x} < r$

for all $x \in B$.

Hence:

$\ds \sup_{x \mathop \in B} \cmod {\map f x} < \infty$

for each $f \in X^\ast$.

That is:

$\ds \sup_{x \mathop \in B} \cmod {\map {x^\wedge} f} < \infty$

for each $f \in X^\ast$.

Since $X^\ast$ is a Banach space, we can apply the Banach-Steinhaus Theorem to the maps $\family {x^\wedge : X^\ast \to \Bbb F}_{x \in X}$.

Doing that, we obtain:

$\ds \sup_{x \mathop \in B} \norm {x^\wedge}_{X^{\ast \ast} } < \infty$

$\ds \sup_{x \mathop \in B} \norm x < \infty$

So there exists $M \in \R_{>0}$ such that:

$\norm x \le M$ for each $x \in B$.

$\blacksquare$