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 $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$.