= Solution
For an isolated <collisionless stellar system>, define the scalar second mass moment and total <kinetic energy> by
$$
I(t)=\int\rho(\boldsymbol r,t)r^2\,d^3r,
\qquad T=\frac12\int f(\boldsymbol r,\boldsymbol v,t)v^2\,d^3r\,d^3v.
$$
Take moments of the <Collisionless Boltzmann equation>, with vanishing surface fluxes. For any time-independent phase-space function $A$, <integration by parts> gives
$$
\frac d{dt}\int fA\,d^3r\,d^3v
=\int f\left(\boldsymbol v\cdot\nabla_r A-\nabla\phi\cdot\nabla_v A\right)d^3r\,d^3v.
$$
Apply this first to $A=r^2$ and then to $A=\boldsymbol r\cdot\boldsymbol v$. The result is
$$
\dot I=2\int f\boldsymbol r\cdot\boldsymbol v\,d^3r\,d^3v,
\qquad\frac12\ddot I=2T-\int\rho\boldsymbol r\cdot\nabla\phi\,d^3r.
$$
The potential is self-consistent Newtonian gravity, so
$$
\phi(\boldsymbol r)=-G\int\frac{\rho(\boldsymbol r')}{|\boldsymbol r-\boldsymbol r'|}\,d^3r',
\qquad W=-\frac G2\iint\frac{\rho(\boldsymbol r)\rho(\boldsymbol r')}{|\boldsymbol r-\boldsymbol r'|}\,d^3r\,d^3r'.
$$
Symmetrize the force integral under $\boldsymbol r\leftrightarrow\boldsymbol r'$:
$$
\begin{aligned}
\int\rho\boldsymbol r\cdot\nabla\phi\,d^3r
&=\frac G2\iint\rho\rho'\frac{(\boldsymbol r-\boldsymbol r')\cdot(\boldsymbol r-\boldsymbol r')}{|\boldsymbol r-\boldsymbol r'|^3}\,d^3r\,d^3r'\\
&=\frac G2\iint\frac{\rho\rho'}{|\boldsymbol r-\boldsymbol r'|}\,d^3r\,d^3r'=-W.
\end{aligned}
$$
Thus the dynamical <virial theorem> is $\ddot I/2=2T+W$. For a steady system, $\ddot I=0$, and consequently
$$
\boxed{2T+W=0.}
$$
For bounded statistically steady motion, averaging instead gives $2\langle T\rangle+\langle W\rangle=0$ provided $[\dot I(\tau)-\dot I(0)]/\tau\to0$. Collisionlessness alone does not imply instantaneous <virial equilibrium>: a collapsing system retains the $\ddot I$ term. The displayed derivation assumes a finite second mass moment, or a limiting version with vanishing boundary terms, and no confining boundary-pressure contribution.
In <virial equilibrium>, the total <energy> is $E_{\rm tot}=T+W=-T$. Define the kinetic <temperature> $\Theta$ at fixed particle number $N$ by $T=3Nk_B\Theta/2$. Along an equilibrium sequence,
$$
\boxed{\frac{dE_{\rm tot}}{d\Theta}=-\frac32Nk_B<0.}
$$
This is the <negative heat capacity of a virialized gravitational system>: removing total energy makes the kinetic temperature rise. A <collisionless stellar system> need not be in thermal equilibrium, so without that additional assumption this is a kinetic-temperature interpretation of <heat capacity>, rather than an assertion that every stellar distribution has a thermodynamic temperature.
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