Updated the separable dual proposition.

This commit is contained in:
Bokuan Li
2026-06-15 14:16:32 -04:00
parent 9f05b9dabd
commit 35efec2d90
2 changed files with 19 additions and 6 deletions

View File

@@ -14,6 +14,15 @@
In the context of a dual system, $E$ and $F$ are identified as subspaces of each others' algebraic duals. In the context of a dual system, $E$ and $F$ are identified as subspaces of each others' algebraic duals.
\end{definition} \end{definition}
\begin{definition}[Norming Duality]
\label{definition:norming-duality}
Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality of normed vector spaces over $K$, then $\dpn{E, F}{\lambda}$ is \textbf{norming} if:
\begin{enumerate}
\item For each $x \in E$, $\norm{x}_E = \sup_{y \in F, \norm{y}_F \le 1}\dpn{x, y}{\lambda}$.
\item For each $y \in F$, $\norm{y}_F = \sup_{x \in E, \norm{x}_E \le 1}\dpn{x, y}{\lambda}$.
\end{enumerate}
\end{definition}
\begin{definition}[Weak Topology] \begin{definition}[Weak Topology]
\label{definition:duality-weak-topology} \label{definition:duality-weak-topology}
Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality over $K$, then the weak topology generated by $F$, denoted $\sigma(E, F)$, is the \textbf{weak topology} of the duality on $E$. Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality over $K$, then the weak topology generated by $F$, denoted $\sigma(E, F)$, is the \textbf{weak topology} of the duality on $E$.

View File

@@ -3,22 +3,26 @@
\begin{proposition} \begin{proposition}
\label{proposition:separable-dual} \label{proposition:separable-dual}
Let $E$ be a separable normed vector space, then $E^*$ is separable with respect to the weak*-topology. Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality of normed vector spaces over $K$ with $E$ being separable, then
\begin{enumerate}
\item The closed unit ball $S = \bracsn{y \in F|\ \norm{y}_{F} \le 1}$ is separable with respect to the $\sigma(F, E)$-topology.
\item If the duality is norming, then there exists $\seq{y_n} \subset F$ such that for each $x \in E$, $\norm{x}_E = \sup_{n \in \natp}\dpn{x, y_n}{\lambda}$.
\end{enumerate}
\end{proposition} \end{proposition}
\begin{proof} \begin{proof}
Let $\seq{x_n} \subset E$ be a dense subset and $S = \bracsn{\phi \in E^*| \norm{\phi}_{E^*} \le 1}$. For each $N \in \natp$, let Let $\seq{x_n} \subset E$ be a dense subset. For each $N \in \natp$, let
\[ \[
T_N: S \to \real^N \quad \phi \mapsto (\dpn{x_1, \phi}{E}, \cdots, \dpn{x_N, \phi}{E}) T_N: S \to \real^N \quad y \mapsto (\dpn{x_1, y}{E}, \cdots, \dpn{x_N, y}{E})
\] \]
Since $\real^N$ is separable, $T_N(S)$ is separable by \autoref{proposition:separable-metric-space}. Thus there exists $\bracs{\phi_{N, k}}_{k = 1}^\infty \subset S$ such that $\bracs{T_N\phi_{N, k}}_{k = 1}^\infty$ is dense in $T_N(S)$. Since $\real^N$ is separable, $T_N(S)$ is separable by \autoref{proposition:separable-metric-space}. Thus there exists $\bracs{y_{N, k}}_{k = 1}^\infty \subset S$ such that $\bracs{T_Ny_{N, k}}_{k = 1}^\infty$ is dense in $T_N(S)$.
Let $\phi \in S$, then for each $N \in \natp$, there exists $k_N \in \natp$ such that for each $1 \le n \le N$, Let $y \in S$, then for each $N \in \natp$, there exists $k_N \in \natp$ such that for each $1 \le n \le N$,
\[ \[
|\dpn{x_n, \phi_{N, k_N}}{E} - \dpn{x_n, \phi}{E}| \le \frac{1}{N} |\dpn{x_n, \phi_{N, k_N}}{E} - \dpn{x_n, \phi}{E}| \le \frac{1}{N}
\] \]
Thus for each $N \in \natp$, $\dpn{x_n, \phi_{N, k_N}}{E} \to \dpn{x_n, \phi}{E}$ as $N \to \infty$. Since $\phi_{N, k_N} \to \phi$ pointwise on a dense subset of $E$, and $\bracsn{\phi_{N, k_N}|N \in \natp} \subset S$ is uniformly equicontinuous, $\phi_{N, k_N} \to \phi$ in the weak*-topology by \autoref{proposition:strong-operator-dense}. Thus for each $N \in \natp$, $\dpn{x_n, y_{N, k_N}}{E} \to \dpn{x_n, y}{E}$ as $N \to \infty$. Since $\phi_{N, k_N} \to \phi$ pointwise on a dense subset of $E$, and $\bracsn{y_{N, k_N}|N \in \natp} \subset S$ is uniformly equicontinuous, $\phi_{N, k_N} \to \phi$ in the $\sigma(F, E)$-topology by \autoref{proposition:strong-operator-dense}.
\end{proof} \end{proof}
\begin{proposition} \begin{proposition}