TITLE: Prove an analog of Rolle's theorem for several variables
QUESTION [7 upvotes]: On p. 135 of Buck's Advanced Calculus, he asks the reader to prove an analog of Rolle's theorem for functions of two variables (I suspect the number two is arbitrary). The hint is to assume that $f=0$ on the boundary of a bounded open set, and I belive what he's looking for is that if $f\in $  $C^1(U)$ for $U$ bounded in $\mathbb{R}^{n}$, and such that $f=0$ on $\partial U$, then $\exists x_0 \in U$ such that $\mathbf{D}f(x_0)= \mathbf{0}$. 
This occurs in the section after the introduction of the derivative on $\mathbb{R}^n$, which suggests that methods like Taylor's theorem, etc. are not required.  
My intuition is to draw lines parallel to each axis and note that by Rolle's theorem in one dimension, there must be a point on each line where $\frac{\partial f}{\partial x_i}=0$, and then somehow argue that there must be a point where all are zero. It seems to me that the points where say $\frac{\partial f}{\partial x_1}=0$ must be something like an $(n-1)$-manifold sitting in $U$, since each line parallel to $e_i$ through $U$ must intersect this set in some point. And same for those points where $\frac{\partial f}{\partial x_2}=0$, but I don't necessarily see why these two hypersurfaces would intersect.
I think I am overcomplicating it. Can anyone think of a simpler argument?

REPLY [5 votes]: Assume that $\overline{U}$ is compact. If $f$ is constant on $\overline{U}$, then we are done. Otherwise, $f$ attains a global maximum or minimum at some $x\in\overline{U}$ such that $f(x)\ne0$, and in fact $x\in U$ since $f=0$ on $\partial U$. Now use the fact that the derivative is zero at a local minimum or maximum.