Von Neumann-Bounded Set in Weak-* Topology of Normed Dual of Banach Space is Norm Bounded
Theorem
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,} }$.
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $E$ be a von Neumann-bounded subset of $\struct {X^\ast, w^\ast}$.
Then $E$ is a von Neumann-bounded subset of $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$.
Proof
Let $x \in X$.
Then from Open Sets in Weak-* Topology of Topological Vector Space:
- $V = \set {f \in X^\ast : \cmod {\map f x} < 1}$ is an open neighborhood of $\mathbf 0_{X^\ast}$ in $\struct {X^\ast, w^\ast}$.
Then there exists $r > 0$ such that:
- $E \subseteq r V$
Then for each $f \in E$, we have $f = r g$ for some $g \in V$.
Then $\cmod {\map f x} = r \cmod {\map g x} < r$.
So:
- $\ds \sup_{f \in E} \cmod {\map f x} < \infty$
for each $x \in X$.
Since $X$ is a Banach space, we may apply the Banach-Steinhaus Theorem to obtain:
- $\ds \sup_{f \in E} \norm f_{X^\ast} < \infty$
That is, there exists $M > 0$ such that:
- $\norm f_{X^\ast} \le M$ for all $f \in E$.
From Characterization of von Neumann-Boundedness in Normed Vector Space, we may conclude that $E$ is von Neumann-bounded in $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$.
$\blacksquare$