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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