Energy of a measure

Remarks

Equivalently, $I(\mu) = \int U^\mu \, d\mu$, where $U^\mu(x) = \int k(x - y)\, d\mu(y)$ is the potential of $\mu$. Some authors include the normalising constant of the fundamental solution in $k$, which multiplies the energy by a constant. The infimum of $I(\mu)$ over probability measures on a compact set $K$ determines the capacity of $K$. For instance, when $n \ge 3$ and with this kernel, the Newtonian capacity of $K$ is the reciprocal of that infimum.

Example: let $\sigma$ be the uniform probability measure on the unit sphere in $\R^3$. Its potential is $U^\sigma(x) = 1/\max(\abs{x}, 1)$, which equals $1$ on the sphere, so $I(\sigma) = 1$. In the plane, the uniform probability measure on the unit circle has potential $\log(1/\max(\abs{z}, 1))$, which is $0$ on the circle, so its energy is $0$.

Non-example: the Dirac measure $\delta_0$ is a positive Borel measure with compact support, but $I(\delta_0) = k(0) = +\infty$, so it does not have finite energy.