I'm looking for references given some sort of *inverse problem* in logarithmic potential theory. That is, given a function $V : \mathbb{R}^2 \to \mathbb{R}$, what is a sufficient (and perhaps necessary) condition for $V$ to be a (logarithmic) potential, that is that there exists a (signed) Borel measure $\mu$ such that
$$
V(x) = -\int_{\mathbb{R}^2} \log|x-y|{\rm d}\mu(y).
$$
This question seems rather natural, but I haven't come across anything similar in the literature for the moment. Note that I'm looking for solutions to this problem that might make sense only in a weak sense (i.e. distributions etc.). In fact, I'm also considering the Coulomb potential in higher dimension $d \geq 3$, so that references for this case (if any) are welcomed !

There are two kinds of conditions:

a) the local one: distributional Laplacian of $V$ must be a signed measure (difference of two non-negative distributions). I do not think that there is a simpler restatement of this condition.

b) the first global one. Once you know that the distributional Laplacian is a signed measure $\mu$, you want to know that the integral $P(z)=\int\log|z-\zeta|d\mu$ converges in some sense, at least for almost all $z$. Convergence of this integral almost everywhere implies convergence quasi-everywhere, so $P$ is defined on the spheres a. e. with respect to the surface measure.

c) the second global one: if a) and b) are satisfied you want to know that the difference $V-P$ is zero (rather than some harmonic function). The easiest way to ensure this is to check that $$\int_{S_r}|V(z)-P(z)|d\sigma\to 0,$$ where $S_r=\{ z:|z|=r\}$ and $d\sigma$ is the normalized surface measure on the sphere. (This simplifies when $n\geq 3$, you can remove $P(z)$ from the integral.)

Everything simplifies if you restrict your class to potentials of positive measures. Then for a) you can simply check that $$V(z)\leq \int_{S(r,z)}V(\zeta)d\sigma,$$ for all $z$, and all $r>0$ where $S(r,z)$ is the sphere of radius $a$ centered at $z$ and for $n\geq 3$ check that $$\int_{S(r)}V(z)d\sigma\to 0,\quad r\to\infty.$$ b) can be skipped in this case. These conditions will ensure that $V=P$ almost everywhere. If you want everywhere, you need to add the condition that $V$ is upper semi-continuous.

When $n=2$ it is slightly more comlicated. You can find $\mu(R^2)$ by the formula $$\mu(R^2)=\lim_{r\to\infty}r\frac{d}{dr}\int_{S(r)} V(x)d\sigma.$$ and then $c$ becomes $$\int_{S(r)}V(x)d\sigma-\mu(R^2)\log r\to 0,\quad r\to\infty.$$

vagueresults which tell me thatif$V$ is $C^2$, then it is superharmonic. As such, the measure which generates it (if any) must be positive ! So, in this case (positive measures), what results are known considering this question ? $\endgroup$ – Hermès Oct 11 at 14:51