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.$\square$