12.15 Bilinear Mappings
Theorem 12.15.1.label Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg$ be separately continuous bilinear maps from $E \times F$ to $G$. If one of the following holds:
- (B)
$E$ is Baire.
- (B’)
$E$ is barrelled and $G$ is locally convex.
and that
- (M)
$E$ and $F$ are both metrisable.
- (E)
For each $x \in E$, $\bracsn{\lambda(x, \cdot)|\lambda \in \alg}\subset L(F; G)$ is equicontinuous.
then $\alg$ is equicontinuous.
Proof, [III.5.1, SW99]. Let $\seq{(x_n, y_n)}\subset E \times F$ and $\seq{\lambda_n}\subset \alg$ such that $(x_{n}, y_{n}) \to 0$ as $n \to \infty$. Since $\seq{y_n}$ is convergent, for each $n \in \natp$ and $x \in E$, $\bracsn{\lambda_n(x, y_n)|n \in \natp}$ is bounded by (E) and Proposition 6.5.2. By (B) or (B’) and the Banach-Steinhaus Theorem, $\bracsn{\lambda_n(\cdot, y_n)|n \in \natp}$ is equicontinuous, and $\lambda_{n}(x_{n}, y_{n}) \to 0$ as $n \to \infty$ by Proposition 6.5.2. By (M) and Proposition 6.5.2, $\alg$ is equicontinuous at $0$, and hence equicontinuous by Lemma 12.14.5.$\square$
Definition 12.15.2 (Hypocontinuity).label Let $E, F, G$ be TVSs over $K \in \RC$, $\sigma \subset 2^{E}$ be an ideal of bounded sets, and $\lambda: E \times F \to G$ be a separately continuous bilinear map, then the following are equivalent:
- (1)
For each $S \in \sigma$ and $V \in \cn_{G}(0)$, there exists $U \in \cn_{F}(0)$ such that $\lambda(S \times U) \subset V$.
- (2)
For each $S \in \sigma$, $\bracs{\lambda(x, \cdot)|x \in S}\subset F^{*}$ is equicontinuous.
If the above holds, then $\lambda$ is $\sigma$-hypocontinuous.
For any ideal $\tau \subset 2^{F}$ of bounded sets, $\lambda$ if $(\sigma, \tau)$-hypocontinuous if $\lambda$ is $\sigma$-hypocontinuous and $\tau$-hypocontinuous.
Proposition 12.15.3.label Let $E, F, G$ be TVSs over $K \in \RC$, and $\lambda: E \times F \to G$ be a separately continuous bilinear map. If one of the following holds:
- (B)
$E$ is Baire.
- (B’)
$E$ is barrelled and $G$ is locally convex.
then $\lambda$ is $B(E)$-hypocontinuous.
Proof. Since $\lambda$ is separately continuous, the mapping
is continuous with respect to the strong operator topology on $L(F; G)$. As such, for each $B \subset E$ bounded, $\bracs{\lambda(x, \cdot)|x \in B}$ is bounded in $L_{s}(F; G)$. By the Banach-Steinhaus Theorem, $\bracs{\lambda(x, \cdot)|x \in B}$ is equicontinuous.$\square$
Proposition 12.15.4.label Let $E, F, G$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be ideals of bounded sets, and $\lambda: E \times F \to G$ be a bilinear mapping.
- (1)
If $\lambda$ is $\sigma$-hypocontinuous, then $\lambda$ is continuous on $S \times F$ for all $S \in \sigma$.
- (2)
If $\lambda$ is $(\sigma, \tau)$-hypocontinuous, then $\lambda$ is uniformly continuous on $S \times T$ for all $S \in \sigma$ and $T \in \tau$.
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.$\square$
Proposition 12.15.6.label Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^{E}$ be a covering ideal, and $k \in \natp$, then
- (1)
The map
\[I: B_{\sigma}^{k}(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F)\]defined by
\[(IT)(x_{1}, \cdots, x_{k+1}) = T(x_{1}, \cdots, x_{k})(x_{k+1})\]is an isomorphism.
- (2)
The map
\[I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}}\to B^{k}_{\sigma}(E; F)\]defined by
\[IT(x_{1}, \cdots, x_{k}) = T(x_{1})\cdots (x_{k})\]is an isomorphism.
which allows the identification
under the map $I$ in (2).
Proof. (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and
Let $(x_{1}, \cdots, x_{k}) \in E^{k}$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_{1}^{k} \subset S$. In which case,
by assumption. Thus $I^{-1}T(x_{1}, \cdots, x_{k}) \in B_{\sigma}(E; F)$.
In addition, for any $S_{1} \in \sigma$ and entourage $E(S_{2}, U)$ of $B_{\sigma}(E; F)$ where $S_{2} \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_{1} \cup S_{2}$. Given that $T(S^{k+1}) \in \mathfrak{B}(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^{k}) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^{k}) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^{k}_{\sigma}(E; B_{\sigma}(E; F))$.
It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^{k}, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
On the other hand, let $S_{1} \in \sigma$ and $E(S_{2}, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_{2} \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_{1} \cup S_{2}$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
(2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
Thus (2) holds for all $k \in \natp$.$\square$
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