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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 $m>2$. According to Riesz decomposition theorem (Hayman and Kennedy, "subharmonic functions", vol. 1, pg 104) if $u$ is subharmonic on $D$, then there is a unique Borel measure $\mu$ such that for all compact $E$ in $D$ we have $$u(x)=\int_{E}K(x-\zeta)d\mu(\zeta)+h(x)$$ where $h$ is harmonic on the interior of $E$.

I have two questions:

1) Suppose $E$ is a compact of $D$ with no interior. Then what happens to the above formula? Does it still hold?

2) Is it true that the function $h$ is given by the integral of $u$ over the boundary of $\partial E$ of $E$ with respect to the harmonic measure?

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1) The formula is then meaningless: $h$ can be completely arbitrary (as there is no interior of $E$).

2) Yes, this is true. Consider a larger compact set $E'$ which is contained in $D$ and such that the interior of $E'$ contains $D$. Apply the decomposition theorem to $E'$ to get $$u(x) = \int_{E'} K(x - \xi) d\mu(\xi) + h'(x).$$ Then $$h(x) = h'(x) - \int_{E' \setminus E} K(x - \xi) d\mu(\xi).$$ Both $h'$ and $K(x - \xi)$ (for $\xi \in E' \setminus E$) are given as integrals with respect to the harmonic measure of $E$, and it remains to use Fubini's theorem.

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  • $\begingroup$ If $E$ is a compact with empty interior can we steel write formally $$u(x)=\int_{E}K(x-\zeta)d\mu{\zeta)+\int_{E}u(\zeta)d\omega_{x}(\zeta)$$ where $\omega_{x}$ is the harmonic measure at $x$? Of course it does not make sense to talk about harmonicity of the function defined by the harmonic mesure, but is such equation holds formally? $\endgroup$ – M. Rahmat Sep 11 '19 at 3:03
  • $\begingroup$ Yes, the integral not only makes sense, but it can well be non-zero. Just imagine that $\mu$ is the Lebesgue measure, and $K$ is a fat Cantor set. $\endgroup$ – Mateusz Kwaśnicki Sep 11 '19 at 5:52

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