Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
For the second question try to work with the distance to the complement of $\Omega$. –  Davide Giraudo May 31 '12 at 16:44

1 Answer 1

up vote 1 down vote accepted

Sketch:

$u$ 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 :-).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.