Proposition 12.15.5.label Let $E, F, G$ be TVSs over $K \in \RC$, $E_{0} \subset E$ and $F_{0} \subset F$ be dense subspaces, and $\sigma \subset B(E_{0})$ and $\tau \subset B(F_{0})$ be ideals of bounded sets. Denote $\ol{\sigma}$ and $\ol \tau$ as the ideals generated by $\bracsn{\ol{S}|S \in \sigma}$ and $\bracsn{\ol{T}|T \in \tau}$, respectively. If
- (a)
$G$ is a complete Hausdorff TVS.
- (b)
$\ol\sigma$ covers $E$ and $\ol{\tau}$ covers $F$.
Then, for any $(\sigma, \tau)$-hypocontinuous bilinear map $\lambda: E_{0} \times F_{0} \to G$, there exists a unique $\Lambda: E \times F \to G$ such that:
- (1)
$\Lambda|_{E_0 \times F_0}= \lambda$.
- (2)
$\Lambda$ is bilinear and $(\ol\sigma, \ol\tau)$-hypocontinuous.
Proof. By (2) of Proposition 12.15.4, for each $S \in \sigma$ and $T \in \tau$, $\lambda|_{S \times T}$ is uniformly continuous. Thus (a) and Theorem 6.7.6 imply that there exists a unique continuous extension of $\lambda$ to $\ol S \times \ol T$. By (b) and the gluing lemma, there exists a unique $\Lambda: E \times F \to G$ such that:
- (1)
$\Lambda|_{E_0 \times F_0}= \lambda$.
- (2’)
For each $S \in \sigma$ and $T \in \tau$, $\Lambda|_{\ol S \times \ol T}$ is continuous.
so the extension is unique by (1) of Proposition 12.15.4.
It remains to show that $\Lambda$ is bilinear and $(\ol \sigma, \ol \tau)$-hypocontinuous. To this end, observe that for each $x \in E_{0}$, $\lambda(x, \cdot) \in L(F_{0}; G)$, and extends to a unique element of $L(F; G)$ by the linear extension theorem. For each $S \in \sigma$ and $T \in \tau$, $\Lambda|_{\ol S \times \ol T}$ is the unique continuous extension of $\lambda|_{S \times T}$. As such, $\Lambda(x, \cdot)$ is the unique continuous extension of $\lambda(x, \cdot)$ to an element of $L(F; G)$, so $\Lambda(x, \cdot) \in L(F; G)$ for all $x \in E_{0}$. By symmetry, $\Lambda(\cdot, y) \in L(E; G)$ for all $y \in F_{0}$.
Now, let $S \in \sigma$, then $\bracsn{\lambda(x, \cdot)|x \in S}$ is equicontinuous by the $\sigma$-hypocontinuity of $\lambda$. For any $U \in \cn_{0}(G)$, there exists $V \in \cn_{0}(F)$ such that $\bigcup_{x \in S}\lambda(x, V \cap F_{0}) \subset U$. By Proposition 5.5.3, $\bigcup_{x \in S}\Lambda(x, \ol V) \subset \ol U$. Thus Proposition 12.1.6 implies that $\bracsn{\Lambda(x, \cdot)|x \in S}$ is equicontinuous as well.
For each $x_{0} \in \ol S$ and $y_{0} \in F$, there exists $T \in \tau$ with $y_{0} \in \ol T$. As $\Lambda|_{\ol S \times \ol T}$ is the unique continuous extension of $\lambda|_{S \times T}$, $\Lambda(x_{0}, \cdot)$ is a pointwise limit of elements of $\bracsn{\lambda(x, \cdot)|x \in S}$. By the Arzelà-Ascoli Theorem,
- (1)
$\bracsn{\Lambda(x, \cdot)|x \in \ol S}\subset \ol{\bracsn{\Lambda(x, \cdot)|x \in S}}^{L_s(F; G)}$.
- (2)
$\bracsn{\Lambda(x, \cdot)|x \in \ol S}$ is also equicontinuous.
so $\Lambda$ is $\ol \sigma$-hypocontinuous. Therefore $\Lambda$ is $(\ol \sigma, \ol \tau)$-hypocontinuous by symmetry.$\square$
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