Show $\exists x$ such that $Df(x) = 0$ for $f = 0$

Question

Let $f:\mathbb{R}^n \to \mathbb{R}$ be continuously differentiable, $\Omega \subseteq \mathbb{R}^n$ an open and bounded set and $f = 0$ on $\partial \Omega$. Show that then there exists a $x \in \Omega$ such that $Df(x) = 0$.

I dont't understand what to show, isn't $Df(x) = 0$ a direct consequence from $f = 0$ ?

An attempt:

Because $x$ lies in the interior, $\exists \epsilon \gt 0$ such that the partial function $$f(x_1,...,x_{j-1},x_j,x_{j+1},...,x_d) := ]x_j - \epsilon,x_j - \epsilon[ \to \mathbb{R}, x \mapsto f(x_1,...,x_{j-1},a,x_{j+1},...,x_d)$$ is well defined.

It has an extremum ar $x_j$, so $\frac{\partial f}{\partial a_j}(x) = 0$. This is the case for all $j \in \{1,...,d\}$ and by assumption $$Df(x) = \left( \frac{\partial f}{\partial a_1}(x) \frac{\partial f}{\partial a_2}(x) ... \frac{\partial f}{\partial a_d}(x)\right) = 0$$

Is this attempt correct?

$f=0$ only on the boundary, you need to show $Df=0$ in the interior. — vadim123, May 04 '14 at 19:53
Compare it with the Rolle's theorem, it's a generalization to multiple dimensions. — dtldarek, May 04 '14 at 19:56
@dtldarek you're right I did not see that. I don't understand why $f = 0$ on $\partial \Omega$. How am I supposed to use that information? — sj134, May 04 '14 at 20:04
Note that $\Omega\cup\partial\Omega$ is closed and bounded, aka. compact. — Hagen von Eitzen, May 04 '14 at 20:18
@sj134 For Rolle's theorem you have $\partial\Omega = {a,b}$ and $f(a) = f(b)$ (read it as, $f$ is constant on $\partial\Omega$); it wouldn't change anything if you were to assume that $f(a) = 0 = f(b)$. Similarly, in your case the constant is zero, but any real number would do, i.e. $f|_{\partial\Omega} = c$ for some $c \in \mathbb{R}$. — dtldarek, May 04 '14 at 20:26
You don't want to consider the 'partial functions'- just consider the whole function itself from the outset. — Alex Zorn, May 04 '14 at 21:08
@sj134 you need to show that the extremum is not on the boundary but that's the right track — Thomas, May 05 '14 at 20:33

triple_sec · Accepted Answer · 2014-05-05T02:43:49.093

Since $\Omega$ is open and bounded, its closure is bounded and hence compact. Now, Weierstrass's theorem implies that $f$ attains both its maximum and minimum on $\operatorname{cl}\Omega=\Omega\cup\partial\Omega$.

If both the maximum and the minimum occur on the boundary, then, since $f|\partial\Omega$ is identically zero, it follows that $f$ is identically zero on $\Omega$ as well, from which the claim readily follows. (Derivative of a constant function vanishes everywhere.)

If it is not true that both the maximum and the minimum lie on the boundary, then assume, without loss of generality, that no maximum point lies on the boundary. In this case, there exists some $x^*\in\Omega$ such that $f(x^*)>0$ and $f(x^*)\geq f(x)$ for any $x\in \Omega\cup\partial\Omega$. Hence, $x^*$ is a maximal point not only over $\operatorname{cl}\Omega$ but a fortiori also over $\Omega$, which is open, implying that the derivative of $f$ must vanish at $x^*$.

(Note: One ought to assume that $\Omega$ is non-empty at the outset.)

score 1 · Answer 2 · answered May 04 '14 at 23:06

1

Hint: The closure $\overline{\Omega}$ is closed and bounded, and thus compact. What does that imply for the existence of extrema of $f$?

answered May 04 '14 at 23:06

Daniel Robert-Nicoud

29,795

Show $\exists x$ such that $Df(x) = 0$ for $f = 0$

2 Answers2