Positive definiteness of the energy

Remarks

Proof idea for $n \ge 3$: the Fourier transform of $\abs{x}^{2-n}$, as a tempered distribution, is $c_n \abs{\xi}^{-2}$ with a constant $c_n > 0$. For $\lambda = \mu - \nu$ this gives $$I(\lambda) = c\, \int_{\R^n} \frac{\abs{\hat{\lambda}(\xi)}^2}{\abs{\xi}^2}\, d\xi$$ with a constant $c > 0$, which is nonnegative and vanishes only when $\hat\lambda = 0$, that is, $\mu = \nu$. In the plane the logarithmic kernel has a similar representation on measures of total mass $0$, which is why the masses must agree.

The mass condition is necessary in the plane. Let $\mu$ be the uniform probability measure on the circle $\abs{z} = R$ with $R > 1$, and $\nu = 0$. Then $I(\mu - \nu) = I(\mu) = \log(1/R) < 0$.

For $n \ge 3$ the theorem implies the Cauchy–Schwarz inequality $I(\mu, \nu)^2 \le I(\mu) I(\nu)$, since $I(\mu - t\nu) = I(\mu) - 2t\, I(\mu, \nu) + t^2 I(\nu) \ge 0$ for every $t > 0$ and $I(\mu, \nu) \ge 0$; take $t = I(\mu, \nu)/I(\nu)$ when $I(\nu) > 0$.

Going deeper

For a student

This is the key fact of the energy method in potential theory. It makes $\sqrt{I(\mu - \nu)}$ a distance on measures of finite energy, and a theorem of Cartan says that positive measures of finite energy are complete for this distance when $n \ge 3$. Uniqueness of the equilibrium measure follows: if $\mu$ and $\nu$ both minimise the energy among probability measures on a compact set, then the energy of $\frac12(\mu + \nu)$ forces $I(\mu - \nu) = 0$, so $\mu = \nu$.