I need a reference for a theorem that states: Let $D$ be a domain of $\mathbb{R}^{m}$ and let $K(x)= \log|x|$ if $m=2$, and $K(x)=|x|^{2-m}$ if $m>2$. Let $u$ be a subharmonic function on $D$. Then there is a unique Borel measure $\mu$ such that for all compact $E$ in $D$ we have $$u(x)=\int_{\partial E}u(\zeta)d\omega_{x}(\zeta)-\int_{E}K(x-\zeta)d\mu(\zeta),$$ where $\omega_{x}$ is the harmonic measure at $x$.

In Hayman and Kenney's book ("subharmonic functions", vol. 1 Theorem 3.9, pg 104) this theorem is given but instead of the first integral, they give a function $h$ harmonic in the interior of $E$ only. Also, on pg 120, the explicit form of the first integral is given but the integrand of the second integral is the green function $g$.

Same things in Armitage and Gardiner's book ("Classical potential theory"). I would appreciate if someone could give me a reference or explain how to go from one form to the other.

  • $\begingroup$ You need some assumptions on $E$ to define $\omega_x(\zeta)$. And more assumptions to make your formula true. $\endgroup$ Sep 13, 2019 at 12:08

1 Answer 1


Perhaps I misunderstood the question: you are right, there are two expressions available for $u$.

  1. We have $$ u(x) = \int_{\partial E} u(z) \omega_E^x(dz) - \int_E G_E(x, y) \mu(dy) ,$$ where $\omega^E_x$ is the harmonic measure, $G_E$ is the Green's function, and $\mu = -\Delta u$ in the sense of distributions.

  2. We also have $$ u(x) = h(x) - \int_E K(y - x) \mu(dy) ,$$ where $K$ is the Newtonian kernel (the Green's function for $\mathbb{R}^m$) and $\mu$ is as above.

These two expressions are of course closely related: since $$ G_E(x, y) = K(y - x) - \int_{\partial E} K(z - y) \omega^x_E(dz) ,$$ we have $$ h(x) = \int_{\partial E} \biggl( u(z) + \int_E K(z - y) \mu(dy) \biggr) \omega^x_E(dz) . $$

  • $\begingroup$ Thanks. Is the first expression with Green's function formally valid if $E$ is a compact with empty interior? $\endgroup$
    – M. Rahmat
    Sep 13, 2019 at 9:18
  • 1
    $\begingroup$ I think so (although I am used to thinking $E$ is an open set). The probabilistic argument works for general sets: if $T_E$ is the first exit time, then $G_E$ is the density function of the mean occupation measure, $\omega_E$ is the distribution at $T_E$, and the desired expression is a consequence of the strong Markov property of Brownian motion. $\endgroup$ Sep 13, 2019 at 9:26

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