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I am reviewing masters exams and can't recall the multivariable calculus one needs to prove that this is true. A reference would suffice. Thank you!

Suppose $x_1,x_2,x_3 \in \mathbb{R}^2$ are three points in the plane that do not lie on a line, and denote by $T$ the closed triangle with vertices $x_1,x_2,x_3$. Suppose $f : T \to \mathbb{R}$ is a continuous function which is differentiable on the interior of $T$ and which vanishes on the boundary of $T$.

Prove that there exists a point $x$ in the interior of $T$ such that the tangent plane to the graph of $f$ at the point $x$ is horizontal.

EDIT: I've been looking through the Wikipedia links, and I understand that for every line there is a point where the derivative must be zero, but I don't see how to the prove the existence of a point where the derivative along every line must be zero. So I may need to be walked through this a little more, if you have time.

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up vote 5 down vote accepted

I wouldn't use the mean value theorem here. Here are some hints:

  1. The triangle $T$ is compact.
  2. If the function is constant, the problem is void.
  3. If the function is not constant, the function will either have a global maximum or a global minimum in the interior of $T$.
  4. The tangent plane to the graph of $f$ at a local maximum or minimum in the interior is horizontal.
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Got it. So the general argument is: For 1, argue $T$ is compact as a closed, bounded subset of $\mathbb{R}^2$. For 3), the continuous function $f: T (\subset \mathbb{R}^2) \to \mathbb{R}$ maps into an ordered set, so use the Extreme Value Theorem to demonstrate existence of a global maximum and minimum. Then because $f$ is differentiable, the derivative at the global maximum/minimum must be zero. In particular, every directional derivative is zero, so the tangent plane must be horizontal. – Michael Chen Apr 25 '11 at 17:17
@Michael: That's exactly the argument I had in mind. Christian Blatter's answer yields another argument for 4) – t.b. Apr 25 '11 at 17:22

If $f$ is continuous on $T$ it assumes a maximum on $T$, i.e., there is a $\xi\in T$ with $\eta:=f(\xi)\geq f(x)$ for all $x\in T$. The horizontal plane $H:\ z=\eta$ is a supporting plane for the graph of $f$, which means it meets the graph in the point $(\xi,\eta)$, and no part of the graph lies above $H$. So you may call $H$ a "tangent plane" to the graph of $f$. If $f$ is $\equiv 0$ on $\partial T$ and positive in some interior points of $T$ the point $\xi$ will lie in the interior of $T$, and the question arises: How do we find $\xi\ $? Here calculus comes to our help. There is a lemma saying that in a point $\xi$ with such a supporting plane $H$ we must have $\nabla f(\xi)=0$; therefore we go and try to solve this equation for $\xi$.

To prove the lemma consider a point $a$ in the interior of $T$ where $\nabla f(a)=:p\ne 0$. Then for the auxiliary function $\phi(t):=f(a + t p)$ one has $\phi'(0)=|p|^2 > 0$ which says that along the line $t\mapsto a + t p$ the function $f$ is neither maximal nor minimal at $a$. This shows that points $a$ with $\nabla f(a)\ne 0$ do not admit such a supporting hyperplane.

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I like this answer. Two small things: 1. You can turn your argument around and say that at a local extremum $x$ in the interior the scalar function $t \mapsto f(x + te_i)$ will have a local extremum at $t = 0$ for $i = 1,2$ by Rolle's theorem, hence $\partial_i f (x) = 0$ and thus $\nabla f = 0$. 2. The condition $\nabla f = 0$ is necessary but not sufficient. – t.b. Apr 22 '11 at 21:12

William Wade's "Introduction to Analysis" got me through my multivariable calculus qualifying exams. It's good for reviewing single dimensional analysis as well.

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