Finite Coulomb energy implies potential level sets enclose finite volume? Suppose $d\in \{3,4,...\}$ and $\rho \in L_{loc.}^1(\mathbb{R}^d,[0,+\infty])$ is so that the Coulomb energy
$$E(\rho):=\int_{\mathbb{R}^d\times \mathbb{R}^d}d\lambda(x,y)\frac{\rho(x)\rho(y)}{|x-y|^{d-2}}<+\infty.$$
Q: Is it true that for every $\phi^*\in (0,+\infty)$, we have
$$\lambda\left(\left\{x\in \mathbb{R}^d\left|\,\phi^*<\int_{\mathbb{R}^d}d\lambda(y)\,\frac{\rho(y)}{|x-y|^{d-2}}=:\phi(x)\right.\right\}\right)<+\infty?$$

*

*If $\rho$ is (globally) bounded above, I might know a proof. To be more precise I suspect that the statement "$\phi$ does not tend to zero at infinity and $\rho$ globally bounded above implies that $E(\rho)=+\infty$" holds. The would-be proof would use the convexity and rotational invariance of the Coulomb energy to take spherical averages of $\rho$ and $\phi$ (averaging over spheres centered on a sequence of centers $x_n\to +\infty$ s.t. $\phi(x_n)$ does not go to zero) and would then borrow a bit here and there from Newton's shell theorem.

*I thought of attacking the problem via the Parseval identity (Below, $\rho$ is taken as a compactly supported approximation to the $\rho$ from above)
$$\int_{\mathbb{R}^d\times \mathbb{R}^d}d\lambda(x,y)\frac{\rho(x)\rho(y)}{|x-y|^{d-2}}=\int_{\mathbb{R}^d}d\lambda(x)\,\rho(x)\phi(x)=C\int_{\mathbb{R}^d}d\lambda(k)\,\overline{\hat{\phi}(k)}\hat{\rho}(k)=C\int_{\mathbb{R}^d}d\lambda(k)\frac{|\hat{\rho}(k)|^2}{|k|^2},$$
thinking the integrand might sufficiently blow up at the origin $k=0$. This consideration alone can however never suffice to finish the job (it's easy to conjure up a pair of non-negative $L^2$-functions with a very large norm but which are mutually orthogonal).

 A: The conjecture seems to be true. To see why, recall (consult e.g. Lieb & Loss' analysis book) that the Coulomb energy $E$ is a non-degenerate quadratic form and hence induces an inner product
$$D(\rho_1,\rho_2):= \int_{\mathbb{R}^d\times \mathbb{R}^d}d\lambda(x,y)\frac{\rho_1(x)\rho_2(y)}{|x-y|^{d-2}},\qquad (E(\rho)=D(\rho,\rho))$$
with in particular the following Cauchy-Schwarz inequality:
$$[D(\rho_1,\rho_2)]^2\leq D(\rho_1,\rho_1)D(\rho_2,\rho_2).$$
Now take an arbitrary $\rho \in L_{loc.}^1(\mathbb{R}^d,[0,+\infty])$, $\phi^*\in (0,+\infty)$ and let (for arbitrary $R\in (0,+\infty)$) $$f_R(.):=\mathbf{1}_{V_R}(.)\\
V_R:=B(0,R)\cap\left\{x\in \mathbb{R}^d\left|\,\phi^*<\int_{\mathbb{R}^d}d\lambda(y)\frac{\rho(y)}{|x-y|^{d-2}}=:\phi_{\rho}(x)\right.\right\}$$
Then
$$\left[\phi^* \lambda(V_R)\right]^2\leq [D(\rho,f_R)]^2 \leq D(\rho,\rho)D(f_R,f_R)=E(\rho)E(f_R).\qquad(*)$$
Riesz' rearrangement inequality and a standard calculation concerning the Coulomb energy of uniformly charged ball implies that
$$E(f_R)\leq E(f_R^*)= C^{st.} \frac{\|f_R^*\|_1^2}{[\lambda(V_R^*)]^{\frac{d-2}{d}}}= C^{st.} [\lambda(V_R^*)]^{\frac{d+2}{d}}=C^{st.} [\lambda(V_R)]^{\frac{d+2}{d}}.\qquad (**)$$
($C^{st.}\in (0,+\infty)$ is a constant depending only on the dimension $d$.)
Inserting (**) into (*) yields
$$[\lambda(V_R)]^{\frac{d-2}{d}}\leq C^{st.} E(\rho)$$
implying that if $E(\rho)<+\infty$, then also
$$\lambda\left(\left\{x\in \mathbb{R}^d\left|\,\phi^*<\phi_{\rho}(x)\right.\right\}\right)=\sup_{R\in (0,+\infty)} \lambda(V_R)<+\infty.$$
