Weak maximum principle of strictly elliptic equation with solution in Sobolev space ( Gilberg Trudiger theorem 8.1)

Question

\begin{equation}\label{eq:81} Lu=D_i(a^{ij}(x)D_ju+b^i(x)u)+c^i(x)D_iu+d(x)u \end{equation}

with

\begin{equation}\label{eq:88} \int_{\Omega}(dv-b^iD_iv)dx\leq 0\qquad \forall v\geq 0,v\in C_0^1(\Omega) \end{equation} Let $u\in W^{1,2}(\Omega)$ satisfy $Lu\geq 0(\leq 0)$ in $\Omega$. Then \begin{equation}\label{eq:89} \sup_{\Omega} u\leq \sup_{\partial \Omega}u^+\quad(\inf_{\Omega} u\geq \inf_{\partial \Omega}u^+) \end{equation} Proof: If $u\in W^{1,2}(\Omega)$ and $v\in W_0^{1,2}(\Omega)$ we have $uv\in W_0^{1,1}$ and $D(uv)=vDu+uDv$ .

$\mathfrak L(u,v)\leq 0$ $ \int \{(a^{ij}D_ju+b^iu)D_iv-(c^iD_iu+du)v)\}dx\leq 0$

\begin{equation} \int_{\Omega}\{a^{ij}D_juD_iv-(b^i+c^i)vD_iu\}dx\leq \int_{\Omega}\{duv-b^iD_i(uv)\}\leq 0 \end{equation} for all $v\geq 0$ such that $uv\geq 0$. last inequality.(Here we used coefficient of u negative).

Hence , by coefficient bounds ,we have \begin{equation}\label{eq:810} \int_{\Omega}a^{ij}D_jD_ivdx\leq 2\lambda\nu\int v|Du|dx \end{equation} for all $v\geq 0$ such that $uv\geq 0$.

In special case $b^i+c^i=0$, the proof is immediate by taking $v=\max \{u-l,0\}$ where $l=sup_{\partial \Omega}u^+$.

Suppose on contrary ,$\sup_{\Omega} u> \sup_{\partial \Omega}u^+$ For general case choose $k$ to satisfy $l\leq k<\sup_{\Omega}u,$ and we set $v=(u-k)^+.$( If no such $k$ exists then we are done.\ $v\in W_0^{1,2}(\Omega)$ and by chain rule \begin{align*} Dv = \left\{ \begin{array}{cc} Du & \hspace{5mm} u>k\qquad(v\neq 0) \\ 0 & \hspace{5mm} u\leq k\qquad(v=0)\\ \end{array} \right. \end{align*} Consequently, we obtain above \begin{equation*} \int_{\Omega}a^{ij}D_jD_ivdx\leq 2\lambda\nu\int_{\Gamma} v|Du|dx\qquad \Gamma=supp (Dv)\subset v \end{equation*} and Hence by Strict ellipticity of $L$, \begin{equation*} \int_{\Omega} |Du|^2dx\leq 2\nu\int_{\Gamma} v|Du|dx\leq 2\nu ||v||_{2,\Gamma}||Dv||_{2,\Gamma} \end{equation*} So that \begin{equation*} ||Dv||_2\leq 2\nu||v||_{2,\Gamma} \end{equation*} By theorem 7.10, $n\geq 3$ \begin{equation*} ||v||_{2n/(n-2}\leq C||Dv||_2. \end{equation*} Also by Schwartz's inequality \begin{equation*} 2\nu||v||_{2,\Gamma}\leq C|\Gamma|^{1/n}||v||_{2n/(n-2)}. \end{equation*} So \begin{equation*} ||v||_{2n/(n-2}\leq C|\Gamma|^{1/n}||v||_{2n/(n-2)}. \end{equation*} where $C=C(n,v)$ so that \begin{equation*} |\Gamma|\geq C^{-n}>0 \end{equation*} If $n=2$\ \begin{equation*} \sup_{\Omega}|u|\leq C|\Omega|^{1/2-1/p}||Du||_p \end{equation*} By replacing $2n/(n-2)$ by any number greater that 2 we get \begin{equation*} |\Gamma|\geq C^{-n}>0 \end{equation*} As above inequality is independent of $k$, we can take $k\to \sup_{\Omega}u$. $u $ attain its supremum in $\Omega$ on set of positive measure, where $Du=0$. This is contradiction to preceding inequality.

I do not understand how to come to contradiction.I understand everything except how contradiction came I don't know.

Any Help will be appreciated.

Answer 1

First note that $$ \vert \Gamma_k \vert = \int_{\text{supp}(\nabla u)} \mathbb{1}_{\{u \ge k\}}. $$ Choosing an increasing sequence $\{k_m\}_m$ such that $k_m \to \sup_\Omega u$, we find that $\{\mathbb{1}_{\{u \ge k_m\}}\}_m$ is monotone decreasing. However, $\vert \Omega \vert < \infty$, so we can apply the decreasing version of the monotone convergence theorem to see that $$ \lim_{m \to \infty} \int_{\text{supp}(\nabla u)} \mathbb{1}_{\{u \ge k_m\}} = \int_{\text{supp}(\nabla u)} \mathbb{1}_{\{u = \sup_\Omega u\}} =0. $$ The latter follows from the fact that $\nabla u =0$ a.e. on sets where $u$ is constant. Combining these shows that $C^{-n} \le \lim_{m \to \infty} \vert \Gamma_{k_m} \vert =0$, and this is the contradiction.