Added Goldstine's Theorem.
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Bokuan Li
2026-07-17 18:49:11 -04:00
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commit 25d3f4e8d2
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@@ -78,4 +78,38 @@
(3) $\Rightarrow$ (1): Let $x \in E$ such that $\dpn{x, y}{\lambda} = 0$ for all $y \in F_0$, then since $F_0$ is $\sigma(F, E)$-dense in $F$, $\dpn{x, y}{\lambda} = 0$ for all $y \in F$. Hence $x = 0$.
\end{proof}
\begin{theorem}[Goldstine]
\label{theorem:goldstine-weak}
Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, $A \subset E$ be non-empty, convex, circled, and $\sigma(E, F)$-compact, and $B$ be the closed unit ball of $E_A^*$, then $B \cap F$ is $\sigma(E_A^*, E_A)$-dense in $B$.
\end{theorem}
\begin{proof}
For any $S \subset E$ or $S \subset F$, denote $S^\circ$ as the polar of $S$ with respect to $\dpn{E, F}{\lambda}$. For any $S \subset E_A$ or $S \subset E_A^*$, denote $S^\bullet$ as the polar of $S$ with respect to $\dpn{E_A, E_A^*}{E_A}$.
Since $B_{E_A}(0, 1)$ is circled and $A = \ol{B_{E_A}(0, 1)}^{E_A}$ is compact in $E$,
\[
B \cap F = \bracsn{\phi \in F| \text{Re}\dpn{x, \phi}{\lambda} \le 1 \forall x \in A} = A^\circ
\]
is the polar of $A$ with respect to $\dpn{E, F}{\lambda}$. Now, as $A$ is convex, circled, and compact, the \hyperref[Bipolar Theorem]{theorem:bipolar} implies that
\begin{align*}
A^{\circ\bullet} &= \bracsn{x \in E_A|\text{Re}\dpn{x, \phi}{\lambda} \le 1 \forall \phi \in A^\circ} \\
&= E_A \cap \bracsn{x \in E|\text{Re}\dpn{x, \phi}{\lambda} \le 1 \forall \phi \in A^\circ} = E_A \cap A^{\circ\circ} = A
\end{align*}
Given that $B \cap F$ is a convex and circled subset of $E_A^*$,
\[
\ol{B \cap F}^{\sigma(E_A^*, E_A)} = (B \cap F)^{\bullet\bullet} = A^{\circ\bullet\bullet} = A^\bullet = B
\]
\end{proof}
\begin{corollary}
\label{corollary:weak-dense-unit-ball}
Let $E$ be a normed space over $K \in \RC$, then $E \cap \ol{B_{E^{**}}(0, 1)}$ is $\sigma(E^{**}, E^*)$-dense in $\ol{B_{E^{**}}(0, 1)}$.
\end{corollary}
\begin{proof}
By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, $\ol{B_{E^*}(0, 1)}$ is convex, circled, and $\sigma(E^*, E)$-compact. By \hyperref[Goldstine's Theorem]{theorem:goldstine-weak}, $E \cap \ol{B_{E^{**}}(0, 1)}$ is $\sigma(E^{**}, E^*)$-dense in $\ol{B_{E^{**}}(0, 1)}$.
\end{proof}

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@@ -66,6 +66,22 @@
On the other hand, let $\mathcal{T} \subset 2^E$ be a locally convex topology consistent with $\dpn{E, F}{\lambda}$. By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, every $\mathcal{T}$-equicontinuous set is relatively $\sigma(F, E)$-compact. Therefore $\mathcal{T}$ is coarser than the topology of uniform convergence on relatively $\sigma(E, F)$-compact, convex, and circled sets.
\end{proof}
\begin{corollary}
\label{corollary:mackey-bounded}
Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, $\topo$ be a topology on $E$ consistent with $\dpn{E, F}{\lambda}$, and $B \subset E$, then $B$ is bounded with respect to $\topo$ if and only if $B$ is bounded with respect to $\sigma(E, F)$.
\end{corollary}
\begin{proof}
Assume without loss of generality that $\topo = \tau(E, F)$. Suppose that $B$ is $\sigma(E, F)$-bounded. Let $A \subset F$ be convex, circled, and $\sigma(F, E)$-compact. By continuity of dual pairing,
\[
A = \bracsn{\phi \in F|\ |\dpn{x, \phi}{\lambda}| \le 1 \forall x \in B}
\]
is a convex, circled, and closed subset. Given that $B$ is $\sigma(E, F)$-bounded, $\sup_{x \in B}|\dpn{x, \phi}{\lambda}| < \infty$ for all $x \in E$. Thus $A$ is absorbing and hence a barrel.
Since $B$ is compact, the auxiliary space $E_B$ is a Banach space. In particular, $E_B$ is barreled by \autoref{proposition:baire-barrel}. By continuity of the inclusion map, $A \cap E_B$ is a barrel in $E_B$. Therefore there exists $\lambda > 0$ such that $B \subset \lambda A \cap E_B \subset \lambda A$.
\end{proof}
\begin{definition}[Mackey Space]
\label{definition:mackey-space}
Let $E$ be a separated locally convex space over $K \in \RC$, then $E$ is a \textbf{Mackey space} if $E$ is equipped with the Mackey topology of $\dpn{E, E^*}{E}$.

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@@ -134,6 +134,13 @@
Let $B = \ol{\aconv}(\seq{\lambda_n x_n})$, then by \hyperref[Mazur's Theorem]{theorem:convex-hull-complete}, $B$ is a compact, convex, and circled subset of $E$ with $\seq{x_n} \subset B$. In addition, $\seq{x_n} \subset E_B$ with $\norm{x_n}_{E_B} \le \lambda_n^{-1}$ for all $n \in \natp$. Thus $\seq{x_n}$ is a null sequence in $E_B$ as well.
Now, let $A'$ be the closed convex hull of $\seq{x_n}$ with respect to $E_B$. Since the inclusion $E_B \to E$ is continuous, $A'$ is a compact convex set in $E$ by \autoref{proposition:compact-extensions}. As such, $A' = A$ by \autoref{proposition:closure-of-image}. Therefore $A$ is a compact subset of $E_B$.
Now, let $A'$ be the closed convex hull of $\seq{x_n}$ with respect to $E_B$. Since the inclusion $E_B \to E$ is continuous, $A'$ is a compact convex set in $E$ by \autoref{proposition:compact-extensions}. As such, $A' = A$ by \autoref{proposition:closure-of-image} and \autoref{proposition:compact-closed}. Therefore $A$ is a compact subset of $E_B$.
\end{proof}
\begin{lemma}
\label{lemma:auxiliary-weak-dense}
Let $E$ be a separated locally convex space over $K \in \RC$, $A \subset E$ be compact, convex, and circled, and $
\end{lemma}