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:
| \(\ds \map \Re {\map \Psi g}\) | \(\le\) | \(\ds \cmod {\map \Psi g}\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \norm \Psi_{X^{\ast \ast} } \norm g_{X^\ast}\) | Fundamental Property of Norm on Bounded Linear Functional |
We have that $\iota B_X^- \subseteq \map {\cl_{w^\ast} } {\iota B_X^-}$, so we have:
| \(\ds \sup_{x \in \map {\cl_{w^\ast} } {\iota B_X^-} } \map \Re {\map {g^\wedge} x}\) | \(\ge\) | \(\ds \sup_{x \in \iota B_X^-} \map \Re {\map {g^\wedge} x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{x \in B_X^-} \map \Re {\map {g^\wedge} {x^\wedge} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{x \in B_X^-} \map \Re {\map {x^\wedge} g}\) | Definition of Evaluation Linear Transformation | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{x \in B_X^-} \map \Re {\map g x}\) | Definition of Evaluation Linear Transformation |
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:
- $\map {\cl_{w^\ast} } {\iota B_X^-} = B_{X^{\ast \ast} }^-$
$\blacksquare$
Source of Name
This entry was named for Herman Heine Goldstine.