10.5 Continuous Linear Maps
Definition 10.5.1 (Continuous Linear Map).label Let $E, F$ be TVSs over $K \in \RC$, and $T \in \hom({E, F})$ be a linear map, then the following are equivalent:
- (1)
$T \in UC(E; F)$.
- (2)
$T \in C(E; F)$.
- (3)
$T$ 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 6.2.2 and Definition 5.6.1.
$(3) \Rightarrow (1)$: Let $U$ be an entourage of $F$, there exists an entourage $V$ of $E$ such that $T(V(0)) \subset U(0)$. Using Proposition 10.1.6 and Lemma 10.1.5, 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 10.5.2 (Continuous Multilinear Map).label 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 10.5.3.label 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 10.5.4 (Product Topology).label 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
- (1)
$E$ equipped with the topology induced by $\fU$ is a topological vector space.
- (U)
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 $\seqi{E}$.
Theorem 10.5.5 (Linear Extension Theorem (TVS)).label Let $E$ be a TVS over $K \in \RC$, $F$ be a complete Hausdorff TVS over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then:
- (1)
There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_{D} = T$.
- (U)
For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.
Proof. By (3) of Definition 10.5.1, $T \in UC(D; F)$. By Theorem 6.5.6, there exists a unique $\ol T \in C(E; F)$ such that $\ol T|_{D} = T$.
It remains to show that $\ol T$ is linear. Since $\ol T$ is continuous, the maps
and
are continuous. Thus $\bracs{A = 0}\supset D \times D$ and $\bracs{M = 0}\supset K \times D$ are both closed. By density of $D$, $\bracs{A = 0}= E \times E$ and $\bracs{M = 0}= K \times E$. Therefore $T$ is linear.$\square$