Q:

How to prove that if $-\nabla^2 \Phi = \rho$ and $\rho$ is bounded then $\Phi$ is bounded.

This problem is obviously inspired by the study of electrostatics, and requires a little more detail than in the title. Precisely, suppose that two functions defined in 3-$d$ space, $\Phi(\mathbf{x})$ and $\rho(\mathbf{x})$, that are related through Poisson's equation:
$$-\nabla^2 \Phi = \rho.$$
If it is given that $\rho$ is bounded from above and below ($R_L \le \rho \le R_M$) in some finite volume $V$, is it possible to prove that $\Phi$ is bounded in that same volume?
I can argue that the presence of point charges and line charges produces divergences in $\Phi$, but surface charges do not. The pattern is that charge densities with support on sets with dimension $0$ and $1$ produce divergent values of $\Phi$, but those on sets of dimension $2$ do not, so $3$ must not either. Is that sufficient? Is there a more rigorous argument that can be made based on Green's theorem with Dirichlet boundary conditions?
$$\Phi(\mathbf{x}) = \int_V \frac{\rho(\mathbf{x} - \mathbf{x}')}{4\pi|\mathbf{x} - \mathbf{x}'|} \operatorname{d}^3\mathbf{x} + \Phi_0(\mathbf{x}),$$
where $\Phi_0(\mathbf{x})$ is the field produced by the boundary conditions/charges external to $V$ and is bounded in $V$ (it satisfies $\nabla^2 \Phi_0 = 0$ in $V$).

A:

Because the integral kernel $\frac{1}{4\pi \lvert \mathbf{x} \rvert}$ is positive, the analysis is quite straightforward for this: if we have $ R_L \leq \rho \leq R_M $, then positivity of the integral ($\int f>0$ if $f>0$) implies that
$$ R_L\int_V \frac{1}{4\pi \lvert \mathbf{x}-\mathbf{x}' \rvert} \, dV \leq \int_V \frac{\rho(\mathbf{x}-\mathbf{x}')}{4\pi \lvert \mathbf{x}-\mathbf{x}' \rvert} \, dV \leq R_M\int_V \frac{1}{4\pi \lvert \mathbf{x}-\mathbf{x}' \rvert} \, dV $$
To show that the integral $\int_V \frac{1}{4\pi \lvert \mathbf{x}-\mathbf{x}' \rvert} \, dV$ is bounded, if $V$ is bounded, we can find a large ball $B$ of radius $R$ centred on $\mathbf{x}$ so that $V$ is contained in this ball whenever $\mathbf{x} \in V$. But again since $\frac{1}{4\pi \lvert \mathbf{x} \rvert}>0$, we have
$$ \int_V \frac{1}{4\pi \lvert \mathbf{x}-\mathbf{x}' \rvert} \, dV < \int_B \frac{1}{4\pi \lvert \mathbf{x}-\mathbf{x}' \rvert} \, dV = \frac{1}{4\pi}\int_0^R \frac{1}{r} 4\pi r^2 \, dr = \int_0^R r \, dr = \frac{1}{2}R^2,  $$
using polar coordinates centred at $\mathbf{x}$.

