10.12 Spaces of Linear Maps
Definition 10.12.1 (Space of Bounded Linear Maps).label Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^{E}$ be an ideal, and $k \in \nat$. The space $B_{\sigma}^{k}(E; F)$ is the set of all $k$-linear maps $T: E^{k} \to F$ with $T(S^{k}) \in \mathfrak{B}(F)$ for all $S \in \sigma$, equipped with the $\bracsn{S^k| S \in \sigma}$-uniform topology.
Let $\fB \subset 2^{E}$ be the collection of all bounded subsets of $E$, then $B_{\sigma}(E; F) = B(E; F)$ is the space of bounded linear maps from $E$ to $F$.
Proposition 10.12.2 ([III.3.3, SW99]).label Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^{E}$ be an ideal, and $A \subset B_{\sigma}(E; F)$, then the following are equivalent:
- (1)
$A \subset B_{\sigma}(E; F)$ is bounded with respect to the $\sigma$-uniform topology.
- (2)
For each $V \in \cn_{F}(0)$, $\bigcap_{T \in A}T^{-1}(V)$ absorbs every $S \in \sigma$.
- (3)
For every $S \in \sigma$, $\bigcup_{T \in A}T(A)$ is bounded in $F$.
Proposition 10.12.3.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 $k \in \natp$.$\square$
Definition 10.12.4 (Strong Operator Topology).label Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^{E}$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F^{E}$ is the strong operator topology.
The space $L_{s}(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology.
Proposition 10.12.5.label Let $E, F$ be TVSs over $K \in \RC$ and $\net{T}\subset L(E; F)$ and $T \in L_{s}(E; F)$. If
- (a)
There exists a dense subset $S \subset E$ such that $T_{\alpha} x \to Tx$ strongly for all $x \in S$.
- (b)
$\bracs{T_\alpha|\alpha \in A}$ is uniformly equicontinuous.
then $T_{\alpha} \to T$ in $L_{s}(E; F)$.
Proof. Let $x \in E$, $U \in \cn_{F}(Tx)$, and $V \in \cn_{F}(Tx)$ be balanced such that $V + V + V \subset U$. By (b), there exists a balanced neighbourhood $W \in \cn_{E}(0)$ such that $T(W) \cup \bigcup_{\alpha \in A}T_{\alpha}(W) \subset V$. By (a), there exists $y \in S \cap (x + W)$ and $\alpha_{0} \in A$ such that for all $\alpha \ge \alpha_{0}$, $T_{\alpha} y - Ty \in V$. In which case, for any $\alpha \ge \alpha_{0}$,
$\square$
Definition 10.12.6 (Weak Operator Topology).label Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^{E}$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F_{w}^{E}$ is the weak operator topology.
The space $L_{w}(E; F) = L_{s}(E; F_{w})$ denotes $L(E; F)$ equipped with the weak operator topology.
Definition 10.12.7 (Bounded Convergence Topology).label Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^{E}$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the topology of bounded convergence.
The space $L_{b}(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
Proposition 10.12.8.label Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:
- (1)
$\hom(E; F)$ is a closed subspace of $F^{E}$ with respect to the product topology.
- (2)
$B(E; F)$ is a closed subspace of $F^{E}$ with respect to the topology of bounded convergence. In particular, if $F$ is complete, then so is $B(E; F)$.
Proof. (1): For each $x, y \in E$ and $\lambda \in K$, the mappings
and
are continuous with respect to the product topology. Since
and $\bracs{0}$ is closed in $F$, $\hom(E; F)$ is a closed subspace of $F^{E}$.
(2): By Definition 10.11.2 and (1), the space of bounded functions and the space of linear functions from $E$ to $F$ are closed subspaces of $F^{E}$ with respect to the topology of bounded convergence. Therefore $B(E; F)$ is also a closed subspace.$\square$