Let $G$ be a locally profinite group.
A smooth representation is a complex representation ($V,\rho$) of $G$ such that the stabiliser of any $v \in V$ is open.
One can show that (as $\text{GL}_n(\mathbb{C})$ is a lie group and has NSS), a (finite dimensional) representation of $G$ is continuous if and only if $\ker(\rho)$ is open.
Therefore, in finite dimensions, continuous representations are smooth.
Furthermore, as $$ \ker(\rho) = \cap_v \text{Stab}_G(v), $$ and the intersection on the right can be taken as finite for finite dimensional $V$, smooth also implies continuous. So these are equivalent for finite dimensions.
What about infinite dimensions? Does either imply the other?
What is the reason for this terminology? I only ask because I am conditioned to thinking that these implications must be smooth implies continuous, and not necessarily the other way around!
I suppose continuous here means the map $P:G \times V \rightarrow V$ is continuous, given V the discrete topology. Then smooth certainly implies continuous, literally by definition ( check the inverse image of a single vector under P is open)
But I don’t think the other side is right as it should depend on the group.