Remark 2

To say that $U$ is continuous is equivalent to the following set of three conditions:

a) for every $s \in G$, $U(s)$ is continuous; b) there exists a neighborhood $V$ of $e$ such that $U(V)$ is equicontinuous; c) there exists a total set $D$ in $E$ such that, for every $x \in D$, the mapping $s \mapsto U(s)x$ is continuous.

These conditions are obviously necessary. Conversely, suppose that the conditions a), b), c) are satisfied. On $U(V)$, the topology of pointwise convergence is identical to the topology of pointwise convergence in $D$ (TVS, III, §3, No. 4, Prop. 5). Therefore the mapping $(s, x) \mapsto U(s)x$ of $V \times E$ into $E$ is continuous (GT, X, §2, No. 1, Cor. 3 of Prop. 1). Since $U(s_0s)x = U(s_0)(U(s)x)$ for all $s_0 \in G$, $s \in G$, $x \in E$, one sees that $U$ is continuous.

When $G$ is locally compact, the conditions a) and b) are equivalent to the condition: $a')$ for every compact subset $K$ of $G$, $U(K)$ is equicontinuous.