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 $\norm{\lambda}_{L^2(E, F; G)}= \norm{\Lambda_1}_{L^2(E^{**}, F^{**}; G^{**})}$.$\square$
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