How does general relativity resolve the fact that energy is not positive definite in Newtonian gravity? In Newtonian gravity, the energy of the gravitational field $\vec{g}$ is
$$ U = -\frac{1}{8\pi G}\int |\vec{g}|^{2} d^{3}x $$
(assuming we don't have any point masses that lead to singularities and the gravitational field $\vec{g}(\vec{x})$ drops off rapidly enough as $\vec{x}$ goes to infinity).
This is always a nonpositive quantity and it is not bounded below, so unlike in the case of electromagnetism, any system involving Newtonian gravity + particles with positive kinetic energy does not have positive definite energy.
Does the story change in general relativity? That is, is energy bounded below in GR?  If so, how does it resolve the issue that gravity does not appear to have positive definite energy?
Based on the comments, I see that gravity carries positive energy but it can't be written in terms of an energy density in a unique and/or covariant manner. But this only seems to further the confusion. I have a hard time understanding how GR can say gravity carries positive energy, but approximations to GR say that gravity has negative energy. How is this not a contradiction?
 A: Suppose you have two masses interacting gravitationally in the Newtonian limit.  The mechanical energy is given by
$$
E = \frac12\mu\dot r^2 + \frac12 \frac{\ell^2}{\mu r^2} - \frac kr
$$
where $\frac1\mu = \frac1m + \frac1M$ describes the reduced mass, $r$ the distance between the two objects, $\ell = \mu r^2\dot\theta = \text{constant}$ is the angular momentum, and $k=GmM = G\mu(m+M)$ gives the strength of the gravitational attraction.  For a circular orbit, $\dot r=0$, we can show that $E = -\frac{k}{2r}$.
Your homework assignment is to construct a system where the relativistic energy $E + (m+M)c^2$ is negative, and ask whether the Newtonian approximation is still appropriate.
For example, we might take $m\ll M$ and set the circular orbital distance to $r_S = 2 GM/c^2$.  In that case our total energy becomes
$$
(M+m)c^2 - \frac{GmM}{2r_S}
=
(M+m)c^2 - \frac14 mc^2
$$
which is clearly always positive.  However, it is a famous result that $r_S$ is a coordinate singularity in the Schwarzchild metric, the location of an “event horizon,” and the Newtonian approximation is no good there.  For large distances $r\gg r_S$, where the Schwarzchild metric is approximately Newtonian, the negative $k/r$ term shrinks and the total energy becomes more positive.
