Riesz decomposition theorem

Remarks

Proof idea: the measure $\mu$ restricted to $U$ is finite, so $p(x) = \int_U E(x - y)\, d\mu(y)$ is subharmonic on $\R^n$ and $\Delta p = \mu$ on $U$ in the sense of distributions. Then $\Delta(u - p) = 0$ on $U$ as a distribution, and Weyl's lemma makes $u - p$ equal, almost everywhere, to a harmonic function $h$. Both $u$ and $p + h$ are subharmonic, and two subharmonic functions that agree almost everywhere agree everywhere, since each is the limit of its ball averages.

Example: for $u(z) = \log\abs{z}$ on $\C$, $\mu = 2\pi\delta_0$, so $\int E(z - w)\, d\mu(w) = \log\abs{z}$ and $h = 0$. Let $f$ be a polynomial with leading coefficient $c$ and zeros $a_1, \dots, a_k$ listed with multiplicity. Then $\log\abs{f(z)} = \sum_j \log\abs{z - a_j} + \log\abs{c}$ on all of $\C$, which has the same form, with the constant $h = \log\abs{c}$ as harmonic part.

A global form holds on a domain $\Omega$ with a Green function $G$, normalised by $-\Delta G(\cdot, y) = \delta_y$. Let $u$ be a superharmonic function on $\Omega$ that has a subharmonic minorant. Then $u$ is the sum of the Green potential of the measure $-\Delta u$ and the greatest harmonic minorant of $u$.

Going deeper

For a student

The theorem reduces many questions about subharmonic functions to questions about measures and their potentials. For $\log\abs{f}$ with $f$ holomorphic it is the analytic form of factoring $f$ through its zeros, as in Jensen's formula and the Hadamard factorisation of entire functions. In the global form on Green domains, it identifies the positive superharmonic functions with zero greatest harmonic minorant, the potentials, with Green potentials of measures.