# Comparison of the gradients of two harmonic functions near the boundary

Let $\Omega$ an open bounded domain in $R^n$. Let $u,v$ be nonconstant smooth functions in the interior of $\Omega$ and harmonic in $\Omega$. Suppose that $u,v \in C(\overline{\Omega})$ and $u \geq v$ in $\Omega$. Suppose that exists $x_0 \in \partial \Omega$ with $u(x_0) = v(x_0)$ and $|\nabla u| \leq 1$ in $\Omega$. (I took $1$ for convenience). Is the following inequality true?

$$\limsup_{y \rightarrow x_0} |\nabla u (y)| \geq \limsup_{y \rightarrow x_0} |\nabla v (y)|$$

I dont know if is true, but if yes, it will help me to understand a passage of a paper. I tried to prove, but no success. Someone could give me a help?

This question is related to the last inequality of the case i) (and the case ii)) of lemma 2.3 of the article Existence of classical solutions to a free boundary problem for the p-Laplace operator, (I) by A. Henrot and H. Shahgholian

(the article is in the right side of the page)

• Is it possible in these assumptions that $u$ is constant? Just trying to plausibility check this in my head. Jul 31 '14 at 20:00
• ops, assume that the functions is non constant . I forget to write this. sorry Jul 31 '14 at 20:05
• If we change $u,v$ with $-u,-v$ the inequality changes "direction" so if it is true it implies equality Jul 31 '14 at 20:45
• Saying it's about "a passage of a paper" but not telling us which one is not so helpful.
– user147263
Jul 31 '14 at 20:52
• I'll elaborate on the previous comment. You see claim X in a paper. You think that it may be true because Y holds. So you ask: is Y true? Others reply: no, it isn't. Everyone's time has been wasted. Just ask about X.
– user147263
Jul 31 '14 at 21:03

No, this is not true. Let $\Omega$ be unit disk in the plane, and let $u(x_1,x_2) = x_1$. Define $v$ on the boundary so that $v(x_1,x_2) = x_1 - M|x_2|$, where $M$ is large. Then $\partial v/\partial \theta$ is the harmonic function which takes large values on the boundary near $(1,0)$, approximately $\pm M$. Therefore, it will take on large values near $(1,0)$ inside the domain.

• i updated the question Jul 31 '14 at 21:16