# Maximum principle proof clarification

Maximum Principle: Let $$\Omega$$ be a bounded connected region of $$\mathbb{R}^m$$ with $$u$$ defined and continuous in $$\bar{\Omega}, \, \Delta u = 0$$. Then $$u$$ achieves its maximum (and minimum) on $$\partial \Omega$$.

Proof:

$$\Delta u = 0, \, \Omega$$ $$\exists \, x^0: \max_{x \in \bar{\Omega}}u(x)=u(x^0)=M < +\infty$$

• If $$x^0 \in \partial \Omega$$: The principle holds true.

• If $$x^0 \in \Omega$$: Using the mean value property:

$$u(x^0) = \frac{1}{4 \pi \delta^2} \int_\limits{S(x^0,\delta)} u(x) \, ds = M$$ $$\Rightarrow u(x) = M, \, \forall x \in S(x^0,\delta) \Rightarrow u(x) = M, \, \forall x \in \Omega. \quad\square$$

I don't get the last line of this proof. Why $$u(x) = M, \, \forall x \in S(x^0,\delta)$$ follows from the mean value property and why $$u(x) = M$$ holds for every $$x$$ in $$\Omega$$ ?

Suppose that there exists $$x_0\in \Omega$$, with $$u(x_0)=M=\max_{\bar{\Omega}}u,$$ and let $$S=\{x\in \Omega: u(x)=M\}.$$ We will show that this set is non-empty and both open and closed in $$\Omega$$. If we can show this, then connectedness will imply that $$S$$ is all of $$\Omega.$$ Clearly, it is non-empty, as we assumed that $$x_0\in S.$$ Note $$S=u^{-1}\left(\{M\}\right)\cap\Omega$$, and since $$u$$ is continuous, this is relatively closed.
Next, take any $$x_0\in S.$$ By the MVP, we have that for any $$r>0$$ so that $$B(x_0,r)\subset\Omega$$ holds, we have that
$$M=u(x_0)=\frac{1}{|B(x_0,r)|} \int\limits_{B(x_0,r)}u\, dy\leq M,$$ since $$u\leq M$$ on $$\bar{\Omega}.$$
We claim that the above can only be equal if $$u=M$$ on all of $$B(x_0,r)$$. Suppose not. Then, there exists $$x_1\in B(x_0,r)$$ so that $$u(x_1) By the continuity of $$u$$, this holds on a neighborhood, which will contradict the equality given by the MVP. Thus, $$u(x)=M$$ for all $$x\in B(x_0,r),$$ which shows that $$B(x_0,r)\subset S.$$ So, $$S$$ is open in $$\Omega$$, as well.