# A function $V : \mathbb{R}^2 \to \mathbb{R}$ is a (logarithmic) potential

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 !

• Do you insist that the function $V(x)$ is really defined for all $x$, or you want to treat $V(x)$ as a distribution? The problem is that there are signed Borel measures, even with compact support for which the RHS is not defined at all points but only defined almost everywhere (as an integrable distribution). This problem becomes easier if you restrict it to positive measures: then $V$ is defined everywhere if you allow $V(x)=-\infty$ at some points. – Alexandre Eremenko Oct 11 at 13:53
• @AlexandreEremenko Thank you for your comment. I should have given more information on $V$. It's an integrable function w.r.t. to some measure (not necessarily absolutely continuous w.r.t. the Lebesgue measure). As such, it could potential take the value $-\infty$. Now, I think I might be able to restrict my attention to positive measure, since I have some vague results which tell me that if $V$ 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 ? – Hermès Oct 11 at 14:51

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 $$\int_{S(r)}V(x)d\sigma-\mu(R^2)\log r\to 0,\quad r\to\infty.$$