Lemma 16.1.4.label Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality over $K$, then for any subspace $F_{0} \subset F$, the following are equivalent:
- (1)
$\dpn{E, F_0}{\lambda}$ is a duality.
- (2)
For any $y_{0} \in F$, $\seqf{x_j}\subset E$, and $\eps > 0$, there exists $y \in F_{0}$ such that for each $1 \le j \le n$, $\dpn{x_j, y}{\lambda}= \dpn{x_j, y_0}{\lambda}$.
- (3)
$F_{0}$ is $\sigma(F, E)$-dense in $F$.
Proof. (1) $\Rightarrow$ (2): Let $E_{0} = \text{span}\bracs{x_j|1 \le j \le n}$, and
Since $\dpn{E, F_0}{\lambda}$ is a duality, there exists $\seqf{y_j}\subset F_{0}$ such that for all $x \in E_{0}$,
Hence $\phi \in L(E_{0}, \sigma(E_{0}, F_{0}); K)$. By the Hahn-Banach Theorem, there exists $\Phi \in L(E, \sigma(E, F_{0}); K)$ such that $\Phi|_{E_0}= \phi$. By Lemma 16.1.3, there exists $y \in F_{0}$ such that $\dpn{x_j, y}{\lambda}= \dpn{x_j, y_0}{\lambda}$ for all $1 \le j \le n$.
(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 $x = 0$.$\square$
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