On page $210$ of Howe and Tan's Non-Abelian Harmonic Analysis there is the following proposition and proof:
Let $(\rho, V) $ be a unitary representation of $SL(n,\mathbb{R})$. The following are equivalent:
(a) weak closure of $\rho(SL(n,\mathbb{R})) = \rho(SL(n,\mathbb{R})) \cup \{0\}$
(b) $ \phi_{u,v}(g) = \langle u, \rho(g) v \rangle $ vanishes at $\infty$ for all $u, v \in V$.
Proof: Let $\{\rho(g_n)\}$ be a sequence of operators with $g_n \in SL(n,\mathbb{R})$. They are bounded by unitarity. By passing to a subsequence, we can assume that this sequence has a weak limit $T$, that is, $$ \phi_{u,v}(g_n) = \langle u, \rho(g_n)v \rangle \to \langle u, Tv \rangle$$ for every $u, v \in V$. This means that $T \neq 0$, if and only if $\phi_{u,v}(g_n) \not\to 0 $ for some $u, v$, and this shows that $(a)$ holds if and only if $(b)$ holds.
This seems to work for showing that $(b)$ implies $(a)$, but the converse is not clear to me. If we assume $0$ is in the weak closure, then we have the existence of a $g_n \to \infty$ such that $\phi_{u,v}(g_n) \to 0$ for every $u, v$, but we need to show that this holds for any $g_n$ which goes to $\infty$.