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, $$\sup_{\partial \Omega} u = \inf\{k\in\mathbb{R} : u\le k\text{ on }\partial \Omega\}.$$


1 Answer 1


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 $C^{-n}$.


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