17.1 Dual Systems
Definition 17.1.1 (Duality).label Let $K$ be a field, $E, F$ be vector spaces over $K$, and $\lambda: E \times F \to K$ be a bilinear map, then the triple $(E, F, \lambda)$ is a dual system/duality over $K$ if
- ($S_{1}$)
For any $x_{0} \in E$, if $\lambda(x_{0}, y) = 0$ for all $y \in F$, then $x_{0} = 0$.
- ($S_{2}$)
For any $y_{0} \in E$, if $\lambda(x, y_{0}) = 0$ for all $x \in E$, then $y_{0} = 0$.
The mapping $\lambda: E \times F \to K$ is the canonical bilinear form of the duality, denoted $(x, y) \mapsto \dpn{x, y}{\lambda}$, and the duality $(E, F, \lambda)$ is denoted $\dpn{E, F}{\lambda}$.
In the context of a dual system, $E$ and $F$ are identified as subspaces of each others’ algebraic duals.
Definition 17.1.2 (Norming Duality).label Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality of normed vector spaces over $K$, then $\dpn{E, F}{\lambda}$ is norming if:
- (1)
For each $x \in E$, $\norm{x}_{E} = \sup_{y \in F, \norm{y}_F \le 1}\dpn{x, y}{\lambda}$.
- (2)
For each $y \in F$, $\norm{y}_{F} = \sup_{x \in E, \norm{x}_E \le 1}\dpn{x, y}{\lambda}$.
Definition 17.1.3 (Weak Topology).label 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 weak topology of the duality on $E$.
Lemma 17.1.4.label Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality over $K$, then the dual of $(E, \sigma(E, F))$ is $F$. In other words, for any $\phi \in L(E, \sigma(E, F); K)$, there exists a unique $y \in F$ such that $\dpn{x, \phi}{E}= \dpn{x, y}{\lambda}$ for all $x \in E$.
Proof, [IV.1.2, SW99]. Since $\phi$ is continuous, there exists $\seqf{y_k}\subset F$ such that for all $x \in E$,
Assume without loss of generality that $\seqf{y_k}$ is linearly independent, then by the First Isomorphism Theorem, there exists $\Phi \in L(K^{n}; K)$ such that the following diagram commutes
For each $1 \le k \le n$, let $e_{k}$ be the $k$-th standard basis vector in $K^{n}$, then for any $x \in E$,
$\square$
Lemma 17.1.5.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 17.1.4, 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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