If, for every $a \in \partial U$, $\lim_{x \to a}f(x) = 0$, prove that there is at least a critical point of $f$ in $U$.

Question

Let $f: U \to \mathbb{R}$ be a differentiable function in the open $U \subset \mathbb{R}^{n}$. If, for every $a \in \partial U$, $\lim_{x \to a}f(x) = 0$, prove that there is at least a critical point of $f$ in $U$.

I don't know what to do. I cannot use $\lim_{x \to a}f(x) = f(a)$ because $a \in \partial U$. I know that $a = \lim x_{n}$ with $x_{n} \in U$ and I'm trying to figure out how to use it. Before, I tried to show that $U$ has a maximum or minimum point, but I couldnot develop. Can someone help me?

Answer 1

This is not true. Take $U=\{(x,y)\,|\,x>0\}$ and let $f(x,y)=x$. Then $\partial U=\{(0,y)\,|\,y\in\mathbb{R}\}$ and$$(\forall a\in\partial U):\lim_{X\to a}f(X)=0.$$However, $f$ has no critical point.

Answer 2

If $U$ is bounded, then it is possible to extend $f$ to the boundary, using the condition, so that the extension is still continuous (define it as $0$ on the boundary). The domain of the extended function is still bounded, so the extended function attains its maximum $M$ and minimum $m$. If $M=m=0$, then $f=0$, so it has critical points (every point of $U$). So WLOG assume that $M>0$. Then the maximum is attained at a point of $U$, so it has to be a critical point of $f$.