Jerry's Digital Garden

BibliographyComments
/Part 3: Functional Analysis/Chapter 17: Duality

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

  1. ($S_{1}$)

    For any $x_{0} \in E$, if $\lambda(x_{0}, y) = 0$ for all $y \in F$, then $x_{0} = 0$.

  2. ($S_{2}$)

    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. (1)

    For each $x \in E$, $\norm{x}_{E} = \sup_{y \in F, \norm{y}_F \le 1}\dpn{x, y}{\lambda}$.

  2. (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 $x \in E$,

\[|\dpn{x, \phi}{\lambda}| \le \sum_{k = 1}^{n} |\dpn{x, y_k}{\lambda}|\]

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

\[\xymatrix{ E \ar@{->}[rd]_{\phi} \ar@{->}[r]^{{(y_1, \cdots, y_n)}} & K^n \ar@{->}[d]^{\Phi} \\ & K }\]

For each $1 \le k \le n$, let $e_{k}$ be the $k$-th standard basis vector in $K^{n}$, then for any $x \in E$,

\[\dpn{x, \phi}{E}= \sum_{k = 1}^{n} \Phi(e_{k}) \dpn{x, y_k}{\lambda}\]

$\square$

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. (1)

    $\dpn{E, F_0}{\lambda}$ is a duality.

  2. (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. (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

\[\phi: E_{0} \to K \quad x \mapsto \dpn{x, y_0}{\lambda}\]

Since $\dpn{E, F_0}{\lambda}$ is a duality, there exists $\seqf{y_j}\subset F_{0}$ such that for all $x \in E_{0}$,

\[|\dpn{x, \phi}{E_0}| \le \sum_{j = 1}^{n} |\dpn{x, y_j}{\lambda}|\]

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 $x = 0$.$\square$

Theorem 17.1.6 (Goldstine).label 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 $B$.

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 Bipolar Theorem 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 $E_{A}^{*}$,

\[\ol{B \cap F}^{\sigma(E_A^*, E_A)}= (B \cap F)^{\bullet\bullet}= A^{\circ\bullet\bullet}= A^{\bullet} = B\]

$\square$

Corollary 17.1.7.label Let $E$ be a normed space over $K \in \RC$, then $E \cap \ol{B_{E^{**}}(0, 1)}$ is $\sigma(E^{**}, E^{*})$-dense in $\ol{B_{E^{**}}(0, 1)}$.

Proof. By the Banach-Alaoglu Theorem, $\ol{B_{E^*}(0, 1)}$ is convex, circled, and $\sigma(E^{*}, E)$-compact. By Goldstine’s Theorem, $E \cap \ol{B_{E^{**}}(0, 1)}$ is $\sigma(E^{**}, E^{*})$-dense in $\ol{B_{E^{**}}(0, 1)}$.$\square$

Post a Comment

Name:Email:
Please enter the tag of the current page (NZ) to post the comment.
Tag:

Direct References

Jerry's Digital Garden

BibliographyComments

Direct References