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

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

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.

• Strictly speaking, if you have a function, you already have the domain.
– zhw.
Aug 6, 2020 at 20:41

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.