Let $\Omega \subset \mathbb{R}^n$ open and let $u$ be a harmonic function
in $\Omega.$ If $K \subset \Omega$ is compact, then prove$$ \sup\limits_{x \in K} |u(x)| \le \frac{n}{\omega_n ~dist(K,\partial \Omega)^n} \int\limits_\Omega |u(x)| ~dx. $$
My attempt:
Consider the following theorem
Theorem (Mean-Value Property):
Let $\Omega \subset \mathbb{R}^n$ open and let $u \in C^2(\Omega)$ be a harmonic function. If $\overline{B(x_0, r)} \subset \Omega$, then
$$
u(x_0) = \frac{1}{|B(x_0, r)|}\int_{B(x_0)}u(x)~dx.
$$
There exists $x_0 \in \partial K$ such that $|u(x)| \leq |u(x_0)|$. At this early point, I'd like to use the mean-value theorem of a function, but I can't manage to prove one of its hypothesis, namely that $\overline{B(x_0, r)} \subset \Omega$.
Best Answer
This follows directly from mean-value property of harmonic functions. We will assume that $\Omega$ is open.
Since $K \subset \Omega$ is compact and $\Omega\subset \mathbb{R}^n$ is open then $r: = \mathrm{dist}(K ,\partial \Omega) > 0$ (notice that if $\Omega = \mathbb{R}^n$, then $r = \infty$, in which case we take $r$ as any positive number).
In view of the definition of $r$, for $x\in K$ we have $\overline{B(x,r - \varepsilon)} \subset \Omega$ for any fixed $\varepsilon > 0$ small enough. In view of the mean value property for $u$ we have $$ u(x) = \frac{n}{\omega_n (r - \varepsilon)^n} \int\limits_{B(x,r - \varepsilon)} u(y) dy, $$ hence $$ |u(x)| \leq \frac{n}{\omega_n (r - \varepsilon)^n} \int\limits_{B(x,r - \varepsilon)} |u(y)| dy \leq \frac{n}{\omega_n (r - \varepsilon)^n} \int\limits_{ \Omega } |u(y)| dy. $$ The right-hand side of the last inequality does not depend on $x$, and taking $\sup$ on the left-hand side over $x \in K$ we get $$ \sup\limits_{x \in K} |u(x)| \leq \frac{n}{\omega_n (r - \varepsilon)^n} \int\limits_{ \Omega } |u(y)| dy, $$ for any $\varepsilon > 0$ small. Taking $\varepsilon \to 0$ produces the desired inequality.