4
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ $f=0$ only on the boundary, you need to show $Df=0$ in the interior. $\endgroup$ – vadim123 May 4 '14 at 19:53
  • 1
    $\begingroup$ $\Omega$ should probably be bounded? $\endgroup$ – Alex Zorn May 4 '14 at 19:54
  • 1
    $\begingroup$ Compare it with the Rolle's theorem, it's a generalization to multiple dimensions. $\endgroup$ – dtldarek May 4 '14 at 19:56
  • 2
    $\begingroup$ @sj134 use my hint and you will see where you use it $\endgroup$ – Thomas May 4 '14 at 20:11
  • 1
    $\begingroup$ You don't want to consider the 'partial functions'- just consider the whole function itself from the outset. $\endgroup$ – Alex Zorn May 4 '14 at 21:08
2
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.