8.5 Continuous Linear Maps
Definition 8.5.1 (Continuous Linear Map). Let $E, F$ be TVSs over $K \in \RC$, and $T \in \hom({E, F})$ be a linear map, then the following are equivalent:
$T \in UC(E; F)$.
$T \in C(E; F)$.
$T$ is continuous at $0$.
If the above holds, then $T$ is a continuous linear map. The set $L(E; F)$ denotes the vector space of all continuous linear maps from $E$ to $F$.
Proof. $(1) \Rightarrow (2) \Rightarrow (3)$: By Proposition 5.2.2 and Definition 4.6.1.
$(3) \Rightarrow (1)$: Let $U$ be an entourage of $F$, there exists an entourage $V$ of $E$ such that $T(V(0)) \subset U(0)$. Using Proposition 8.1.6 and Lemma 8.1.5, assume without loss of generality that $U$ and $V$ are symmetric and translation-invariant.
For any $x, y \in V$, $x - y \in V(0)$, so $Tx - Ty \in U(0)$, $Ty \in U(Tx)$ by symmetry, and $(Tx, Ty) \in U$. Therefore $T$ is uniformly continuous.$\square$
Definition 8.5.2 (Continuous Multilinear Map). Let $\seqf{E}$, $F$ be TVSs over $K \in \RC$, then the set $L^{n}(E_{1}, \cdots, E_{n}; F) = L^{n}(\seqf{E_j}; F)$ is the space of all continuous $n$-linear maps from $\prod_{j = 1}^{n} E_{j}$ to $F$.
Proposition 8.5.3. Let $E, F$ be TVSs over $K \in \RC$ and $T \in L(E; F)$, then for any $B \subset E$ bounded, $T(B)$ is also bounded.
Proof. Let $U \in \cn_{F}(0)$, then $T^{-1}(U) \in \cn_{E}(0)$, so there exists $\lambda \in K$ such that $\lambda T^{-1}(U) = T^{-1}(\lambda U) \supset B$ and $\lambda U \supset T(B)$.$\square$
Definition 8.5.4 (Product Topology). Let $\seqi{E}$ be TVSs over $K \in \RC$ and $E = \prod_{i \in I}E_{i}$ be their product as a vector space, and $\fU$ be the initial uniformity generated by the projection maps, then
$E$ equipped with the topology induced by $\fU$ is a topological vector space.
For any TVS $F$ over $K$ and $\seqi{T}$ where $T_{i} \in L(F; E_{i})$ for each $i \in I$, there exists a unique $U \in L(F; E)$ such that the following diagram commutes
\[\xymatrix{ F \ar@{->}[rd]^{T_i} \ar@{->}[d]_{T} & \\ \prod_{i \in I}E_i \ar@{->}[r]_{\pi_i} & E_i }\]
The uniformity $\fU$ and its induced topology are the product uniformity/topology, and $E$ equipped with $\fU$ is the product TVS of $\seqi{E}$.