For interactions via a Newtonian gravitational potential, the virial theorem states that $$ 2 \langle T \rangle = -\langle V \rangle $$ i.e. the time-averaged total kinetic energy and total potential energy are proportional, with a constant factor. In (for example) the Wikipedia article on the virial theorem, this is derived via an equation (there) called Lagrange’s identity $$ \frac{\mathrm{d}^2 I}{\mathrm{d} t^2} = 2T + V$$ where $I$ is the total moment of inertia.
When applied to astrophysical objects (e. g. dark matter halos, galaxy clusters), it is often said that these objects “virialize” after collapsing. Unfortunately (in my experience), the explanations invariably become very hand-waving at this point, and somehow the time averages from the virial theorem are replaced by the original quantities (e. g. in the same Wikipedia article), i. e. $$ 2T = -V, \quad E = T + V = -T = \frac{1}{2} V = \mathrm{const.} $$ Comparing with Lagrange’s identity above, this is equivalent to the condition $\frac{\mathrm{d}^2 I}{\mathrm{d} t^2} = 0$. When this holds, the system is said to be “virialized” or “in virial equilibrium”, and it’s generally assumed that objects like dark matter halos are in this state.
However, I have never really found a justification for this assumption of virial equilibrium other than “empirical evidence”. This answer has an intuitive explanation of why a system would evolve towards $\frac{\mathrm{d}^2 I}{\mathrm{d} t^2} = 0$ over time, but it’s conceivable (and even explicitly mentioned there) that the system might not settle in a state with with constant $T$ and $V$:
If $\Sigma$ is negligible (which is assumed in the form you give), then a system which has $2T < -U$ will have a dynamical evolution that drives an increase in $I$, or in other words the system contracts, and vice-versa for the opposite case. The system can also respond in other ways, for instance it might shed some $T$ via radiation.
If $2T\sim-U$, then $\frac{{\rm d}^2 I}{{\rm d}t^2}=0$, so the expansion/contraction is neither accelerating nor decelerating. This is a necessary condition for equilirbium, but it is not the same as $\frac{{\rm d}I}{{\rm d}t}=0$: the system might still be expanding/contracting and be about to 'overshoot' its equilibrium point. For such an overshoot to happen requires $\frac{{\rm d}T}{{\rm d}t}$ and $\frac{{\rm d}U}{{\rm d}t}\neq 0$. A zero time average implies that these derivatives are zero and that the system is actually 'stuck' on its equilibrium (or oscillating around it if the time average is over a long enough window, which could sloppily be called an equilibrium system; this might be fine, depending on context).
This question asking how long it takes for a system to settle into virial equilibrium is related, but unfortunately the answer doesn’t address under which conditions this happens in the first place.
To summarize my question: Do all self-gravitating systems eventually settle in virial equilibrium ($2T = -V$)? If yes, how can this be proven? If not, under what conditions does it happen? I’m imagining that there could be some kind of statistical/ergodic argument, but I’ve never encountered anything to this effect.
