14.6 The Arens Product
Definition 14.6.1 (Arens Extension).label Let $E, F, G$ be normed vector spaces over $K \in \RC$ and $\lambda \in L^{2}(E, F; G)$ be a continuous bilinear mapping, then there exists a unique bilinear mapping $\Lambda_{1}: E^{**}\times F^{**}\to G^{**}$ such that:
- (1)
For each $(x, y) \in E \times F$, $\Lambda_{1}(x, y) = \lambda(x, y)$.
- (2)
For each $x \in E$, $\Lambda_{1}(x, \cdot)$ is weak*-continuous.
- (3)
For each $y \in F^{**}$, $\Lambda_{1}(\cdot, y)$ is weak*-continuous.
Similarly, there exists a unique bilinear mapping $\Lambda_{2}: E^{**}\times F^{**}\to G^{**}$ such that:
- (1)
For each $(x, y) \in E \times F$, $\Lambda_{2}(x, y) = \lambda(x, y)$.
- (2’)
For each $x \in E^{**}$, $\Lambda_{2}(x, \cdot)$ is weak*-continuous.
- (3’)
For each $y \in F$, $\Lambda_{2}(\cdot, y)$ is weak*-continuous.
The mappings $\Lambda_{1}, \Lambda_{2}: E^{**}\times F^{**}\to G^{**}$ are the first and second Arens extensions of $\lambda$, respectively.
Proof, [Section 1, Theorem 3.2, Are51]. For each $x \in E$, the mapping $\lambda(x, \cdot) \in L(F; G)$ admits an adjoint $\lambda^{*}(x, \cdot) \in L(G^{*}; F^{*})$, which induces an adjoint of the bilinear map as follows
Applying the above operation again yields a second adjoint
and finally, applying the adjoint operation a third time gives
(1): Let $(x, y) \in E \times F$, then for each $\phi \in G^{*}$,
By the Hahn-Banach Theorem, $\Lambda_{1}$ is an extension of $\lambda$.
(2): Fix $x \in E$ and $\phi \in G^{*}$, then for each $y \in F^{**}$,
Since $\lambda^{*}(x, \phi) \in F^{*}$, $\Lambda_{1}(x, \cdot)$ is weak*-continuous.
(3): Fix $y \in F^{**}$ and $\phi \in G^{*}$, then for each $x \in E^{**}$, $\dpn{\phi, \Lambda_1(x, y)}{G^*}= \dpn{\lambda^{**}(y, \phi), x}{E^*}$. Since $\lambda^{**}(y, \phi) \in E^{*}$, $\Lambda_{1}(\cdot, y)$ is weak*-continuous.
(Uniqueness): By Goldstine’s Theorem, $E$ is weak*-dense in $E^{**}$, and $F$ is weak*-dense in $F^{**}$, so the extension is uniquely determined.$\square$
Proposition 14.6.2.label Let $E, F, G$ be normed vector spaces over $K \in \RC$, $\lambda \in L^{2}(E, F; G)$ be a continuous bilinear mapping, then $\Lambda_{1}, \Lambda_{2} \in L^{2}(E^{**}, F^{**}; G^{**})$, where
Proof. Assume without loss of generality that $\norm{\lambda}_{L^2(E, F; G)}= 1$. Fix $x \in \ol{B_{E}(0, 1)}$, then
Since for any $z \in G^{**}$, $\norm{z}_{G^{**}}= \sup_{\phi \in G^*, \norm{\phi}_{G^*} \le 1}\dpn{z, \phi}{G^*}$, the norm on $G^{**}$ is weak*-lower semicontinuous, and $\ol{B_{G^{**}}(0, 1)}$ is weak*-closed. By Goldstine’s Theorem, $\ol{B_{F}(0, 1)}$ is weak*-dense in $\ol{B_{F^{**}}(0, 1)}$. As such, weak*-continuity of $\Lambda_{1}(x, \cdot)$ and Proposition 5.5.3 implies that $\Lambda_{1}(x, \ol{B_{F^{**}}(0, 1)}) \subset \ol{B_{G^{**}}(0, 1)}$ as well.
Now, fix $y \in \ol{B_{F^{**}}(0, 1)}$, then $\Lambda_{1}(\ol{B_E(0, 1)}, y) \subset \ol{B_{G^{**}}(0, 1)}$. By Goldstine’s Theorem, $\ol{B_{E}(0, 1)}$ is weak*-dense in $\ol{B_{E^{**}}(0, 1)}$. Thus the weak*-continuity of $\Lambda_{1}(\cdot, y)$ and Proposition 5.5.3 implies that $\Lambda_{1}(\ol{B_{E^{**}}(0, 1)}, y) \subset \ol{B_{G^{**}}(0, 1)}$. Therefore
and $\norm{\lambda}_{L^2(E, F; G)}= \norm{\Lambda_1}_{L^2(E^{**}, F^{**}; G^{**})}$.$\square$
Definition 14.6.3 (Arens Regularity).label Let $E, F, G$ be normed vector spaces over $K \in \RC$, $\lambda \in L^{2}(E, F; G)$ be a continuous bilinear mapping, and $\Lambda_{1}, \Lambda_{2}: E^{**}\times F^{**}\to G^{**}$ be its first and second Arens extensions, respectively, then the following are equivalent:
- (1)
$\Lambda_{1} = \Lambda_{2}$.
- (2)
There exists an extension $\Lambda: E^{**}\times F^{**}\to G^{**}$ of $\lambda$ that is separately weak*-continuous.
- (3)
There exists an extension $\Lambda: E^{**}\times F^{**}\to G^{**}$ of $\lambda$ that is separately weak*-continuous when restricted to $\ol{B_{E^{**}}(0, 1)}\times \ol{B_{F^{**}}(0, 1)}$.
If the above holds, then $\lambda$ is an Arens regular bilinear map, and $\Lambda = \Lambda_{1} = \Lambda_{2}$ is the Arens extension of $\lambda$.
Proof, [Theorem 3.3, Are51]. (3) $\Rightarrow$ (1): By Goldstine’s Theorem, $\ol{B_{E}(0, 1)}$ is weak*-dense in $\ol{B_{E^{**}}(0, 1)}$, and $\ol{B_{F}(0, 1)}$ is weak*-dense in $\ol{B_{F^{**}}(0, 1)}$. Thus the restrictions of $\Lambda_{1}$ and $\Lambda_{2}$ to $\ol{B_{E^{**}}(0, 1)}\times \ol{B_{F^{**}}(0, 1)}$ are uniquely determined by the value of $\lambda$ on $\ol{B_{E}(0, 1)}\times \ol{B_{F}(0, 1)}$, in the following sense:
- (i)
$\Lambda_{1}|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}}$ is the unique extension of $\lambda|_{\ol{B_{E}(0, 1)} \times \ol{B_{F}(0, 1)}}$ such that
- (a)
For each $x \in \ol{B_E(0, 1)}$, $\Lambda_{1}(x, \cdot)$ is weak*-continuous.
- (b)
For each $y \in \ol{B_{F^{**}}(0, 1)}$, $\Lambda_{1}(\cdot, y)$ is weak*-continuous.
- (ii)
$\Lambda_{2}|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}}$ is the unique extension of $\lambda|_{\ol{B_{E}(0, 1)} \times \ol{B_{F}(0, 1)}}$ such that
- (a)
For each $x \in \ol{B_{E^{**}}(0, 1)}$, $\Lambda_{2}(x, \cdot)$ is weak*-continuous.
- (b)
For each $y \in \ol{B_{F}(0, 1)}$, $\Lambda_{2}(\cdot, y)$ is weak*-continuous.
Since the given extension $\Lambda$ satisfies (i.a), (i.b), (ii.a), and (ii.b),
As $\Lambda_{1}, \Lambda_{2}$ are bilinear, the above implies that $\Lambda_{1} = \Lambda_{2}$.$\square$
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