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Let $\Omega\subset\mathbb R^n$ open and $u\in\mathrm C^0(\Omega)$ subharmonic.

Why does $\Omega$ contain

$(i)$ a sequence of smooth subharmonic functions $u_i$ cum $u_i\rightarrow u$ locally uniformly convergent and

$(ii)$ $\forall i\in\mathbb N$ is $\Omega_i$ open cum $\bar\Omega_i\subset\Omega_{i+1}$ and $\Omega=\cup_i \ \Omega_i$

? Thanks for helping!

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For the second question try to work with the distance to the complement of $\Omega$. – Davide Giraudo May 31 '12 at 16:44
up vote 2 down vote accepted


$u$ is subharmonic in $U$ iff, for every $x\in U$ and $r>0$ such that $B_r(x) \subset U $ the following mean value inequality is true: $$u(x) \le \frac{1}{|B_r(x)|} \int_{B_r(x)} u(y) dy $$

($|.|$ denoting Lebesgue volume). Let $\phi_k(x) = \psi_k(||x||) \ge 0 $ a sequence of smooth mollifiers such that $\mbox{supp} \,\phi_k \subset B_{1/k}(0)$ and such that $\int \phi_k = 1$. Define $$u_k(y):=\int u(y-x) \phi_k(x) dx$$ It is known (and I assume you do know) that $u_k\rightarrow u$ uniformly and $u_k$ is smooth on all open domains $V$ such that $V+B_{1/k}(0)\subset U$ for large $k$ (this is the reason why you will need ii)). Now $$\frac{1}{|B_r(x)|} \int_{B_r(x)} u_k(y) dy = \frac{1}{|B_r(x)|} \int_{B_r(x)} \int u(y-z) \phi_k(z) dzdy$$ The inner integral may be taken over any open set the closure of which contains the support of $\phi_k$, that is, $B_{1/k}(0)$. In particular, for large $k$, you may take the ball of radius $r$ around $0$

Now transform (simply by translation) the inner integral so that you are integrating over $B_r(y)$, and use the mean value inequality in the inner integral to show that the expressions in this formula are $\ge u_k(y)$. That is, $u_k$ satisfies the mean value inequality and therefore $u_k$ is subharmonic. (There are still some gaps where the details need to be filled which I leave to you :-).

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