\section{Continuous Linear Maps}
\label{section:tvs-linear-maps}


\begin{definition}[Continuous Linear Map]
\label{definition:continuous-linear}
    Let $E, F$ be TVSs over $K \in \RC$, and $T \in \hom({E, F})$ be a linear map, then the following are equivalent:
    \begin{enumerate}
        \item $T \in UC(E; F)$.
        \item $T \in C(E; F)$.
        \item $T$ is continuous at $0$.
    \end{enumerate}
    If the above holds, then $T$ is a \textbf{continuous linear map}. The set $L(E; F)$ denotes the vector space of all continuous linear maps from $E$ to $F$.
\end{definition}
\begin{proof}
    $(1) \Rightarrow (2) \Rightarrow (3)$: By \autoref{proposition:uniform-continuous} and \autoref{definition:continuity}.

    $(3) \Rightarrow (1)$: Let $U$ be an entourage of $F$, there exists an entourage $V$ of $E$ such that $T(V(0)) \subset U(0)$.  Using \autoref{proposition:tvs-uniform} and \autoref{lemma:translation-invariant-symmetric}, 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.
\end{proof}

\begin{definition}[Continuous Multilinear Map]
\label{definition:continuous-multilinear}
    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 $F$.
\end{definition}


\begin{proposition}
\label{proposition:continuous-bounded}
    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.
\end{proposition}
\begin{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 $\lambda U \supset T(B)$.
\end{proof}

\begin{definition}[Product Topology]
\label{definition:tvs-product}
    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
    \begin{enumerate}
        \item $E$ equipped with the topology induced by $\fU$ is a topological vector space.
        \item[(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

        \[
        \xymatrix{
        F \ar@{->}[rd]^{T_i} \ar@{->}[d]_{T} &  \\
        \prod_{i \in I}E_i \ar@{->}[r]_{\pi_i} & E_i
        }
        \]

    \end{enumerate}
    The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$.
\end{definition}

\begin{theorem}[Linear Extension Theorem (TVS)]
\label{theorem:linear-extension-theorem-tvs}
    Let $E$ be a TVS over $K \in \RC$, $F$ be a complete separated TVS over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then:
    \begin{enumerate}
        \item There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_D = T$.
        \item[(U)] For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.
    \end{enumerate}

\end{theorem}
\begin{proof}
    By (3) of \autoref{definition:continuous-linear}, $T \in UC(D; F)$. By \autoref{theorem:uniform-continuous-extension}, 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
    \[
    A: E \times E \to E \quad (x, y) \mapsto \ol Tx + \ol Ty - \ol T(x + y)
    \]

    and
    \[
    M: K \times E \to E \quad (\lambda x) \mapsto \lambda \ol Tx - Tx
    \]

    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. 
\end{proof}


