A representation is continuous if $v \mapsto \pi(g)v$ and $g \mapsto \pi(g)v$ are each continuous Let $G$ be a locally compact topological group, and let $V$ be a complex Banach space.  Let $\operatorname{GL}(V)$ be the group of linear isomorphisms of $V$ as an abstract vector space.  A representation $\pi$ of $G$ is called continuous if the map $G \times V \rightarrow V, (g,v) \mapsto \pi(g)v$ is continuous, where $V$ is taken in the norm topology.
In order to show that a representation is continuous, does it suffice to show that $G \times V \rightarrow V$ is continuous in each variable?  
 A: Yes.  Suppose $G \times V \rightarrow V$ is continuous in each variable.  For $g \in G$, the condition that $v \mapsto \pi(g)v$ is continuous is equivalent to saying that $\pi(g)$ is a bounded linear operator.
Now let $U$ be any set in $G$ with compact closure.  For each $v \in V$, the continuity of $g \mapsto \pi(g)v$ implies that $||\pi(g)v|| : g \in U$ is bounded.  The uniform boundedness principle then tells us that the operator norms $||\pi(g)||$ for $g \in U$ are bounded as well.    
To show that $G \times V \rightarrow V$ is continuous, it suffices to show continuity at zero.  So given $\epsilon > 0$, and a pair $(g_0,v_0) \in G \times V$, we need to show there exist neighborhoods $U$ and $W$ of $g_0$ and $v_0$ such that $||\pi(g)v - \pi(g_0)v_0|| < \epsilon$ whenever $g \in U$ and $v \in W$.
By the continuity of $g \mapsto \pi(g)v_0$, and the fact that $G$ is locally compact, we can find a compact neighborhood $U'$ of $1_G$ such that $$||\pi(h)v_0 - v_0|| < \frac{\epsilon}{2||\pi(g_0)||}$$ for all $h \in U'$.  If we set $U = g_0U'$, then $U$ is a compact neighborhood of $g_0$, and we have
$$||\pi(g_0^{-1}g)v_0 - v_0|| < \frac{\epsilon}{2||\pi(g_0)||}$$
for all $g \in U$.  
By the uniform boundedness principle above, there exists an $M > 0$ such that $||\pi(g)|| < M$ for all $g \in U$.  Letting 
$$W = \{ v \in V : ||v - v_0|| < \frac{\epsilon}{2M} \}$$
we have for $g \in U$ and $v \in W$ that
$$||\pi(g)v - \pi(g_0)v_0|| \leq ||\pi(g)v - \pi(g)v_0|| + ||\pi(g)v_0 - \pi(g_0)v_0||$$
$$ \leq ||\pi(g)|| \cdot ||v - v_0|| + ||\pi(g_0)|| \cdot || \pi(g_0^{-1}g)v_0 - v_0|| < \epsilon$$
