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.$\square$
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