Suppose $G$ is a Lie group, and $(\pi, E)$ is a continuous representation of $G$ on a Fréchet space $E$. Let $E^\infty$ be the set of smooth vectors of $E$, which is known to be dense in $E$. In particular $E^\infty$ is non-empty if $E$ is.
Suppose now that $V\subseteq E$ is a non-empty closed invariant subspace. So, $V$ is also Fréchet, and $\pi$ restricts to a continuous representation $(\pi|_V,V)$. We can then consider the set $V^\infty$ of smooth vectors of this representation. Again, $V^\infty\neq \emptyset$.
It seems not hard to prove that $V^\infty=V\cap E^\infty$. Moreover, $E^\infty$ and $V^\infty$ can be given topologies from the natural injections $E^\infty\to C^\infty(G,E)$ and $V^\infty\to C^\infty(G,V)$. From $V^\infty=V\cap E^\infty$ we have $V^\infty\subseteq E^\infty$ and I believe that the above topology on $V^\infty$ coincides with the subspace topology induced from $E^\infty$. (All these facts follow from the natural injection $C^\infty(G,V)\to C^\infty(G,E)$ being well-defined and continuous.)
A simple consequence of (a subset of) these statements is that $V\cap E^\infty\neq \emptyset$ for any non-empty closed invariant subspace $V$ of $E$.
So my question is: am I making some simple mistake here? The motivation for this question comes from this paper (MR3215570). Their Theorem 5.17' and proposition 5.19' (in Introduction) seem to give an example of a unitary representation $E$ with a non-empty closed invariant subspace $V$ such that $V\cap E^\infty=\emptyset$. This is for $G=\widetilde{\mathrm{SU}(1,1)}$ (the universal cover).
EDIT: In response to Paul Garrett's comment, let me try to explain the argument that I have in mind, without equivalence of topologies. It is kind of trivial, so to compensate for this, let me write it out in an excessive amount of detail.
Let $\pi$ be any representation of $G$ on a topological vector space $E$, and let $V$ be any invariant subspace of $E$, endowed with subspace topology. Let $v\in V$, and consider the function $f: G\to V$ defined by $f(g) = \pi(g)v$. By definition, $v\in V^\infty$ iff $f$ is smooth. Similarly, $v\in E^\infty$ iff $f$ is smooth when considered as valued in $E$. By the below argument, $f$ is smooth when valued in $E$ iff it is smooth when valued in $V$. So $v\in V$ is in $E^\infty$ iff it is in $V^\infty$. This is the same as saying $V^\infty = E^\infty \cap V$ (as sets).
The below is a painfully trivial proof of the following statement: let $M$ be a manifold, $E$ a topological vector space, and $V$ a vector subspace of $E$ with subspace topology, $i:\,V\to E$ the continuous inclusion. Then $f:M\to V$ is smooth iff $i\circ f:M\to E$ is.
Let $U\subseteq \mathbb{R}^n$ be an open set and let $f:U\to E$ be a function valued in a locally convex topological vector space $E$. We say $f$ is differentiable at a point $x_0\in U$ if there exists $f'(x_0)\in \mathrm{Hom}(\mathbb{R}^n, E)$ such that $$\label{lim} \lim_{x\to x_0} \frac{f(x)-f(x_0)-f'(x_0)(x-x_0)}{|x-x_0|}=0.\tag{1} $$ Let $f$ be differentiable at all $x\in U$. Then we have a function $f':U\to E'$ where $E'\equiv \mathrm{Hom}(\mathbb{R}^n, E)$. (Sorry about this notation...) We give $E'$ the topology with a base consisting of sets $U(N,M)=\{h\in E': h(N)\subseteq M\}$ for all finite subsets of $\mathbb{R}^n$ and all open sets of $M$. (Pointwise convergence.) We say $f$ is continuous differentiable if $f'$ is continuous. We say that $f$ is smooth if it is continuously differentiable with derivative $f'$, which is continuously differentiable with derivative $f''$, ... (not sure if continuously is actually independent).
Now, suppose $V$ is a vector subspace of $E$. If $V$ is given the subspace topology from $E$, then the topology on $V'$ coincides with the subspace topology from $E'$. This is because $U(N,M)\cap V'=\{h\in V': h(N)\subseteq M\cap V\}$, but $M\cap V$ runs precisely through all open sets of $V$ as $M$ runs through open sets of $E$.
Let $f: \mathbb{R}^n \to V$ be some function. Consider $V$ on its own, with the subspace topology, and suppose $f$ is differentiable at $x_0\in U$. Then then $f$ is differentiable when valued in $E$, with the same derivative, because if the limit \eqref{lim} holds in $V$, it holds in $E$.
Suppose now we know instead that $f$ is differentiable at $x_0$ as a function valued in $E$. A priori, $f'(x_0)\in E'$. But it is easy to show that $f'(x_0)\in V'$. Indeed, let $t\in \mathbb{R}^n$, $t\neq 0$. Then, in $E$, $$ \lim_{\lambda\to 0} \frac{f(x_0+\lambda v)-f(x_0)-\lambda f'(x_0)(t)}{\lambda}=0.\tag{2} $$ This uniquely determines $f'(x_0)(t)$ in terms of $f$, which is valued in $V$. So $f'(x_0)(t)\in V$, and thus $f'(x_0)\in V'$. But then the expression under the limit in \eqref{lim} is valued in $V$, and then if the limit holds in $E$, it holds in $V$ with subpace topology.
Suppose $f:U\to V$ is differentiable in $U$ in any one of the two ways (as valued in $V$ or as valued in $E$). Then it is differentiable in the other way. If derivative $f'$ is continuous as valued in $V'$, it is continuous as valued in $E'$ and vice versa. Repeating infinitely many times, $f$ is smooth when valued in $V$ iff it is smooth when valued in $E$.
We can the translate this from $U\in \mathbb{R}^n$ to manifolds, and so to Lie groups.