# Weak maximum principle for elliptic operators without classical smoothness assumptions

I am currently reading chapter 8 in Gilbarg-Trudinger's Elliptic PDE of the Second order.

Let $$L$$ denote a strictly elliptic operator (with some additional assumptions on the coefficients). $$\Omega\subseteq \mathbb{R}^n$$ is a domain.

In Theorem 8.1, we are given $$u\in W^{1,2}(\Omega)$$ satisfying $$Lu \ge 0$$ in the weak sense. Suppose by way of contradiction that $$\sup_{\Omega} u > \sup_{\partial\Omega} u^+$$ Then, we are able to show that, given $$k\in [\sup_{\partial\Omega} u^+, \sup_{\Omega} u)$$, the function $$v = (u-k)^+\in W_0^{1,2}(\Omega)$$ satisfies $$\lvert\operatorname{supp}Dv\rvert \ge C^{-n}$$ where $$C$$ is a constant independent of $$u, k, v$$. The proof is then concluded with the following paragraph:

That is, the function $$u$$ must attain its supremum in $$\Omega$$ on a set of positive measure, where at the same time $$Du = 0$$ (by Lemma 7.7). This contradiction of the preceding inequality implies $$\sup_{\Omega} u \le \sup_{\partial\Omega} u^+$$.

The last paragraph loses me. At first, I found it intuitive. However, with my inability to formalize the argument, I had to come to terms with my subpar understanding of it. If someone understands the argument better than I, any explanation is appreciated.

###### Notes on notation:

$$\sup_\Omega u$$ is the essential supremum of $$u$$ on $$\Omega$$.

For $$u \in W^{1,2}(\Omega)$$, say that $$u \le v$$ on $$\partial\Omega$$ provided $$(u-v)^+ = \max\{u-v, 0\} \in W_0^{1,2}$$. Then, $$\sup_{\partial \Omega} u = \inf\{k\in\mathbb{R} : u\le k\text{ on }\partial \Omega\}.$$

Let $$m$$ be the essential supremum of $$u$$ and write $$v_k=(u-k)_+$$. Since $$u\le m$$ a.e., you have that $$v_m=0$$ a.e. and so its gradient is zero. On the other hand, if you call $$E_k$$ the support of the gradient of $$v_k$$, by the chain rule you have that the sets $$E_k$$ decrease as $$k$$ approaches $$m$$. If you have a decreasing sequence of sets of finite measure, their Lebesgue measure converges to the Lebesgue measure of their intersection, which is the set $$E_m$$. But this set is empty while the inequality in display says that the measures of the sets $$E_k$$ is bounded from below by $$C^{-n}$$.