Superharmonic potential

Remarks

Since $u \ge 0$, the function $0$ is a harmonic minorant, so the greatest harmonic minorant exists; it is $-h_0$, where $h_0$ is the least harmonic majorant of $-u$. Let $\Omega$ have a Green function $G$. Then the Riesz decomposition theorem shows that the potentials are exactly the functions $x \mapsto \int G(x, y)\, d\mu(y)$. Here $\mu$ runs over the positive measures for which this function is not identically $+\infty$. Every nonnegative superharmonic function is uniquely the sum of a potential and a nonnegative harmonic function.

Example: for $n \ge 3$ let $u(x) = \abs{x}^{2-n}$ on $\Omega = \R^n$, with $u(0) = +\infty$. It is superharmonic and positive. Let $h$ be harmonic with $h \le u$, and fix $a$. By the mean value property, $h(a)$ is the mean of $h$ over the sphere $S(a, R)$, which is at most the mean of $u$ there, namely $\max(\abs{a}, R)^{2-n}$. Letting $R \to \infty$ gives $h(a) \le 0$, so $u$ is a potential.

Non-example: on a bounded domain the constant $u = 1$ is superharmonic and nonnegative, but $h = 1$ is a harmonic minorant with $h > 0$, so clause 3 fails.