Theorem 13.12.6.label Let $E, F$ be locally convex spaces over $K \in \RC$, then $(E \wh \otimes_{\eps} F)^{*} = I(E, F)$.

Proof. Let $\lambda \in (E \wh \otimes_{\eps} F)^{*}$, then there exists equicontinuous subsets $S \subset E^{*}$ and $T \subset F^{*}$ such that for each $x \in E$ and $y \in F$,

\[|\lambda(x, y)| \le \sup_{\phi \in S}\sup_{\psi \in T}|\dpn{x, \phi}{E}\dpn{y, \psi}{F}|\]

For any $(x, y) \in E \times F$ and $(\phi, \psi) \in S \times T$, let $f_{xy}(\phi, \psi) = \dpn{x, \phi}{E}\dpn{y, \psi}{F}$. By the Hahn-Banach Theorem, there exists $\Lambda \in C(S \times T; K)^{*}$ such that the following diagram commutes:

\[\xymatrix{ & C(S \times T; K) \ar@{->}[rd]^{\Lambda} & \\ E \times F \ar@{->}[ru]^{{(x, y) \mapsto f_{xy}}} \ar@{->}[rr]_{\lambda} & & K }\]

Now, since $S$ and $T$ are equicontinuous, using the Banach-Alaoglu Theorem, assume without loss of generality that $S$ and $T$ are weak*-compact. In which case, by the Riesz Representation Theorem, there exists $\mu \in M_{R}(S \times T; K)$ such that $\Lambda(f) = \int_{S \times T}f d\mu$ for all $f \in C(S \times T; K)$. Therefore

\[\lambda(x, y) = \Lambda(f_{xy}) = \int_{S \times T}f_{xy}d\mu = \int_{S \times T}\dpn{x, \phi}{E}\dpn{y, \psi}{F}d\mu\]

for all $(x, y) \in E \times F$, and $\lambda \in I(E; F)$.$\square$

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