# A simple clarification on Riesz decomposition theorem

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 $$E$$.

Here is my question: is this equation valid for all $$x\in D$$ or all $$x\in E$$?

• It holds only for $x\in E$. Since the measure is not restricted to $E$, the function does not have to be harmonic outside $E$. – Alexandre Eremenko Sep 21 '20 at 23:29
• It depends on the convention. You can also understand the claim as $u$ can be represented in the above form in $D$ with $h$ that is harmonic in the interior of $E$ (and subharmonic in the rest of the domain). The key point is that for every compact subset of $D$, the Riesz measure is finite, so the potential is well-defined but this may be not so in the entire domain $D$. – fedja Sep 22 '20 at 1:57
• But why do you say that h is subharmonic in the rest of the domain? – M. Rahmat Sep 22 '20 at 15:31

(You need a minus sign in front of $$\int_E K(x-\zeta)\,d\mu(\zeta)$$.)
The integral $$\,-\!\!\int_E K(x-\zeta)\,d\mu(\zeta)$$ is subharmonic throughout $$D$$ and harmonic in $$D\setminus E$$. If $$u$$ admits a harmonic majorant in $$D$$, hence a least such majorant (call it $$k$$), then $$h$$ (which depends on $$E$$) can be expressed as $$\,-\!\!\int_{E^c} K(x-\zeta)\,d\mu(\zeta)+k(x)$$ for $$x\in D$$.