# Interesting question in analysis

I am trying to prove this :

Consider $\Omega \subset R^n$ ( $n \geq 2$) a bounded and open set and $u$ a smooth function defined in $\overline{\Omega}$. Suppose that $u(y) = 0$ for $y \in \partial \Omega$ and suppose that exists a $\alpha >0$ such that $|\nabla u (x)| = \sqrt{\displaystyle\sum_{i=1}^{n} (\frac{\partial u}{ \partial x_i }(x)} )^2\geq \alpha >0$ for all $x \in \Omega$, then

$$|u(x)| \geq \alpha |x-y|$$

for all $x \in \Omega$ and for all $y \in \partial \Omega$.

drawing a picture is easy to see the affirmation.. i am trying to prove this. but nothing ... My professor said that this is true....

Someone can give me a hint ?

• Something is funny.. on the unit disk in $R^2$, if I start at one boundary point $y$ and let $x$ tend toward another boundary point, $u(x)$ should go towards $0$ yet the absolute value is increasing as I go farther from $y$. Is there any missing condition (such as $x$ being a distance $h$ from the boundary?) – Evan Aug 28 '13 at 0:15
• @Evan , My professor stated the affirmation like i writed . maybe he did a mistake.. – math student Aug 28 '13 at 0:22
• If you reverse the inequality on the gradient and on $u(x)$ then there is a simple proof using Fundamental Theorem, but I'm guessing this is trying to illustrate something else. – Evan Aug 28 '13 at 0:23

At first sight one tends to believe the inequality because of the Mean Value Theorem, $$|u(x)|=|u(x)-u(y)|=|\nabla u(c)\cdot(x-y)|$$ for all $y$ in the boundary.

But there is an inner product involved, and so one cannot guarantee that the right-hand-side above is bounded below.

Indeed, let $\Omega$ be the unit disc in $\mathbb R^2$ and $$u(x)=2y(1-x^2-y^2).$$ Being a polynomial, $u$ is smooth. It is also zero at the boundary. We have $$\tag{1} |\nabla u(x,y)|^2=16x^2y^2+4(1-x^2-3y^2)^2,$$ which is never zero in the closed disk. As the expression in $(1)$ is continuous on a compact set, it achieves its minimum and so there exists $\alpha>0$ with $|\nabla u(x,y)|>\alpha$ for all $(x,y)\in\overline\Omega$.

Finally, $|u(1/2,0)|=0<\alpha\,|(1/2,0)-(1,0)|$

In hindsight, the assertion becomes suspicious when you notice that the condition on the gradient is only requiring for the gradient to be nonzero.

Let $x_0,y_0 \in \partial \Omega$. Let $x \in \Omega$. Then you have $$|u(x)| \ge \alpha |x - y_0|.$$ If $x \to x_0$ then by continuity you obtain $$|u(x_0)| \ge \alpha |x_0 - y_0|.$$ That is, $$0 \ge \alpha |x_0 - y_0|.$$

The problem is misstated.

I think there is something wrong here.

If $u$ is zero in $\partial \Omega$, then $u$ has a minimum or a maximum in $\Omega$. So there is a point $P$ in $\Omega$ such that $\nabla u(P)=0$.

• Should this have been added as a comment rather than an answer? – Simon Hayward Aug 28 '13 at 15:40