Goldstine's Theorem

Theorem
Let $\GF \in \set {\R, \C}$.

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

Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $X$.

Let $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ be the second normed dual of $X$.

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

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

Let $B_X^-$ be the closed unit ball of $X$.

Let $B_{X^{\ast \ast} }^-$ be the closed unit ball of $X^{\ast \ast}$.

Then:


 * $\map {\cl_{w^\ast} } {\iota B_X^-} = B_{X^{\ast \ast} }^-$

Proof
From Closed Unit Ball in Normed Dual Space is Weak-* Closed, $B_{X^{\ast \ast} }^-$ is closed in $\struct {X^{\ast \ast}, w^\ast}$.

In Normed Vector Space is Reflexive iff Closed Unit Ball in Original Space is Mapped to Closed Unit Ball in Second Dual, it is shown that $\iota B_X^- \subseteq B_{X^{\ast \ast} }$.

Hence, we have:


 * $\map {\cl_{w^\ast} } {\iota B_X^-} \subseteq \map {\cl_{w^\ast} } {B_{X^{\ast \ast} }^-}$

from Set Closure Preserves Set Inclusion.

From Set is Closed iff Equals Topological Closure, we have $\map {\cl_{w^\ast} } {B_{X^{\ast \ast} }^-} = B_{X^{\ast \ast} }^-$.

So we have $\map {\cl_{w^\ast} } {\iota B_X^-} \subseteq B_{X^{\ast \ast} }^-$.

Now suppose that $\Phi \in X^{\ast \ast} \setminus \map {\cl_{w^\ast} } {\iota B_X^-}$.

We aim to deduce that $\norm \Phi_{X^{\ast \ast} } > 1$ so that we have:


 * $\Phi \in X^{\ast \ast} \setminus B_{X^{\ast \ast} }^-$

This will give:


 * $X^{\ast \ast} \setminus \map {\cl_{w^\ast} } {\iota B_X^-} \subseteq X^{\ast \ast} \setminus B_{X^{\ast \ast} }^-$

and hence:


 * $B_{X^{\ast \ast} }^- \subseteq \map {\cl_{w^\ast} } {\iota B_X^-}$

from Set Complement inverts Subsets.

From Finite Topological Space is Compact, $\set \Phi$ is compact in $\struct {X^{\ast \ast}, w^\ast}$.

Further, $\set \Phi$ is convex from Singleton is Convex Set.

We have that $\map {\cl_{w^\ast} } {\iota B_X^-}$ is closed in $\struct {X^{\ast \ast}, w^\ast}$.

Further, $\map {\cl_{w^\ast} } {\iota B_X^-}$ is convex from Image of Convex Set under Linear Transformation is Convex and Closure of Convex Set in Topological Vector Space is Convex.

Let $A = \set \Phi$ and $B = \map {\cl_{w^\ast} } {\iota B_X^-}$.

We can then apply:


 * Hahn-Banach Separation Theorem: Hausdorff Locally Convex Space: Real Case: Compact Convex Set and Closed Convex Set in the case $\GF = \R$
 * Hahn-Banach Separation Theorem: Hausdorff Locally Convex Space: Complex Case: Compact Convex Set and Closed Convex Set

to obtain $\Psi \in \struct {X^{\ast \ast}, w^\ast}^\ast$ such that:


 * $\ds \sup_{x \in A} \map \Re {\map \Psi x} < \inf_{x \in B} \map \Re {\map \Psi x}$

From Characterization of Continuity of Linear Functional in Weak-* Topology, there exists $f \in X^\ast$ such that $\Psi = f^\wedge$.

Then we have:


 * $\ds \sup_{x \in \set \Phi} \map \Re {\map {f^\wedge} \Psi} < \inf_{x \in \map {\cl_{w^\ast} } {\iota B_X^-} } \map \Re {\map {f^\wedge} x}$

That is:


 * $\ds \map \Re {\map \Psi f} < \inf_{x \in \map {\cl_{w^\ast} } {\iota B_X^-} } \map \Re {\map {f^\wedge} x}$

So from Negative of Infimum is Supremum of Negatives we have:


 * $\ds \map \Re {\map \Psi g} > \sup_{x \in \map {\cl_{w^\ast} } {\iota B_X^-} } \map \Re {\map {g^\wedge} x}$

for $g = -f$.

First, we have:

We have that $\iota B_X^- \subseteq \map {\cl_{w^\ast} } {\iota B_X^-}$, so we have:

From Realification of Normed Dual is Isometrically Isomorphic to the Normed Dual of Realification, we have:


 * $\ds \sup_{x \in B_X^-} \map \Re {\map g x} = \norm g_{X^\ast}$

Hence we obtain:


 * $\norm \Psi_{X^{\ast \ast} } \norm g_{X^\ast} > \norm g_{X^\ast}$

Hence we obtain, since all terms are positive:


 * $\norm \Psi_{X^{\ast \ast} } > 1$

So we have:


 * $\Phi \in X^{\ast \ast} \setminus B_{X^{\ast \ast} }^-$

giving:


 * $X^{\ast \ast} \setminus \map {\cl_{w^\ast} } {\iota B_X^-} \subseteq X^{\ast \ast} \setminus B_{X^{\ast \ast} }^-$

and hence:


 * $B_{X^{\ast \ast} }^- \subseteq \map {\cl_{w^\ast} } {\iota B_X^-}$

So we obtain:


 * $\map {\cl_{w^\ast} } {\iota B_X^-} = B_{X^{\ast \ast} }^-$