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Let $\Omega \subset \mathbb{R}^N$ be a bounded, connected, open, and regular set. Let $u \in C^\infty(\Omega)$, such that $$u=0, \mbox{ on }\partial \Omega.$$ Let us suppose that as a consequence of an application of the strong maximum principle for the operator $-\Delta + M \mbox{ Id}$, where $M\geq 0$, we have $$u>0 \mbox{ in }\Omega,\;\;\; \frac{\partial u}{\partial n}<0\mbox{ on } \partial \Omega.$$
(to be more precise, u is a nontrivial supersolution for the operator $-\Delta+M\mbox{ Id}$).

Prove that there exists $C_1,C_2>0$ such that $$0<C_1 d(x,\partial \Omega)\leq u(x)\leq C_2 d(x,\partial \Omega),\; \forall x\in \Omega.$$

This result is trivial if we consider a compact set $K\subset \Omega$. My problem is how to prove this result near the boundary.

Charlie
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1 Answers1

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I'll deal with $u(x)\ge C_1 d(x,\partial \Omega)$. Suppose this statement fails. Then there is a sequence $x_n$ such that $d(x_n,\partial \Omega)\to0$ and $$u(x_n)\le \frac{1}{n}d(x_n,\partial \Omega) \tag{1}$$ We may assume that $x_n$ converge to some $y\in \partial \Omega$. Let $y_n$ be a point of $x_n$ such that $|x_n-y_n|=d(x_n,\partial \Omega)$. Using (1) and the mean value theorem, conclude that there is a point $z_n$ on the line segment from $x_n$ to $y_n$ such that $$D_{\xi_n} u (z_n)\ge -\frac{1}{n}\tag{2}$$ where $ \xi_n= \frac{y_n-x_n}{|y_n-x_n|}$.

Let's recap: $x_n\to y$, $y_n\to y$, hence $z_n\to y$. Also, since $\xi_n$ is orthogonal to the boundary at $y_n$, the vectors $\xi_n$ converge to the outward normal $\xi$ at $y$. From (2), $D_{\xi} u (y)\ge 0$, a contradiction.

The argument for the other part is similar, except (1) is replaced with $$u(x_n)\ge {n}d(x_n,\partial \Omega) \tag{1$'$}$$ and so forth.

Post No Bulls
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