Firstly, we should note that $\Omega$ is a bounded domain in $\mathbb R^n$ which you omit there.
It’s the weak minimum principle, we have two different ways to prove it.
First way :
Firstly, we consider the case $-\Delta u>0$ in $\Omega $. If there is a $x_0\in\Omega$ such that $$u(x_0)=\min_{\bar\Omega}u(x).$$
By necessity for extreme values, we know that the hessian matrix at $x_0$ is semi-positive definite, which means
$$-\Delta u(x_0)\leq 0.$$
It’s a contradiction. That is $$\displaystyle\min_{\bar{\Omega}}u = \displaystyle\min_{\partial{\Omega}}u.$$
Generally, for the case $-\Delta u\geq 0$ we define for $\forall \epsilon>0$
$$v(x)=u(x)-\epsilon|x|^2\quad in\,\Omega.$$
Hence, we obtain
$$-\Delta v=-\Delta(u-\epsilon|x|^2)=-\Delta u+2n\epsilon>0.$$
Thus
$$\displaystyle\min_{\bar{\Omega}}v= \displaystyle\min_{\partial{\Omega}}v.$$
Note that
$$\min_{\bar\Omega}u\geq \min_{\bar\Omega}v=\min_{\partial\Omega}v\geq\min_{\partial\Omega}u -\epsilon \max_{\partial\Omega}|x|^2.$$
Let $\epsilon\to 0$ yields
$$\min_{\bar\Omega}u\geq\min_{\partial\Omega}u$$
since $\Omega$ is bounded. And it’s clear that
$$\min_{\bar\Omega}u\leq\min_{\partial\Omega}u$$
which gives us
$$\min_{\bar\Omega}u =\min_{\partial\Omega}u.$$
Second way :
In this part, we will use mean value inequality.
Let $m:=\min_\limits{\bar\Omega}u$ and $D:=\{x\in\Omega: u(x)=m\}$. It’s clear that D is relative closed to $\Omega$. Next, we will prove D is relative open.
For $x_0\in D$, by mean value inequality we know
$$m= \displaystyle u(x_0) \geq \frac{1}{|\partial{B(x_0,r)|}}\int_{\partial{B(x_0,r)}} u(y) \ dS(y) \geq m$$for any $0<r< dist(x_0, \partial{\Omega})$,
which yields $u(y) = u(x_0)= m$ for all $y \in B(x_0,r)$.
Thus $B(x_0,r)\subset D$, that is D is relative open.And since $\Omega$ is connected, we get $D=\Omega$ or $D=\phi$.
Hence, we know
$$\min_{\bar\Omega} u=\min_{\partial\Omega} u.$$
If you want to learn more about maximal principle, we refer you to Han Qing and Lin Fanghua’s elliptic pde.
The first equation is a special case of the second one. Pick any x in $B⁰(0, r)$. Clearly $u(x) \leq \sup_V u(\cdot)$ holds and, due to Harnack’s inequality, $u(x) \leq \sup_V u(\cdot) \leq C \inf_V u(\cdot)$ is true. Since $\inf_V u(\cdot) \leq u(y)$ holds for any $y$, we can conclude $u(x) \leq C u(0)$. If we interchange the roles of $x$ and $0$, we can $u(0) \leq C u(x)$ or $\frac{1}{C} u(0) \leq u(x)$. Putting this together, we get $\frac{1}{C} u(0) \leq u(x) \leq C u(0)$. This is the first statement.
Addition: This is not exactly the first statement since the constants aren’t reciprocal to each other. The first equation can therefore be regarded as a more sophisticated variant. But I think the idea should be clear from the above.
Best Answer
Harnack's inequality concerns about the non-negative function which is harmonic in $U$, by the choice of $r$, $x$, and $y$, we know that $B(y,r)\subset B(x,2r) \subset U$, hence the inequality holds.