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Suppose we have $f: \mathbf{R}^{3} \to \mathbf{R}$ with the following property: $\langle \nabla f(x), x \rangle > 0$ for every $x \in S^{2}$, that is, it's gradient points outwards the unit sphere. It's asserted that there must a point $p$ inside the sphere with the property $\nabla f (p) = 0$.

Here's what I've done so far: suppose there's no such point in $B(0;1)$. Since $f$ is real valued and defined in the compact $\bar{B}(0;1)$, it must attain maximum and minimum. Since $f$ has no critical points in the interior, then these points must lie in $S^{2}$, say $x_{0}$ is the maximum and $y_{0}$ is the minimum. Now the problem seems to be that the gradient cannot point outwards in the minimum point, and from that we could derive a contradiction. But I don't know how to write that down -- using the directional derivative along the line joining the extrema points perhaps?

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up vote 10 down vote accepted

There is a topological argument. Consider the map

$$\sigma : S^2 \to S^2 : x \mapsto \frac{\nabla f}{\|\nabla f\|}.$$

By the condition on $f$ at the boundary, $\sigma$ is homotopic to the identity map of $S^2$ (because there is a unique geodesic going from $x$ to $\sigma(x)$). Now suppose that $\|\nabla f\| \neq 0$ for each $x \in \overline{B}(0,1)$. Consider the map

$$\overline{B}(0,1) \to S^2 : x \mapsto \frac{\nabla f}{\|\nabla f\|}.$$

Since $\overline{B}(0,1)$ is contractible, this map is homotopic to a constant map. But that is absurd, since its restriction to the boundary $S^2$ is $\sigma$, which is not homotopic to a constant map.

Cool, huh?!

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Do you mean because the ball is contractible? – Ted Shifrin Oct 18 '13 at 3:07
@TedShifrin Yes of course, thank you! – Bruno Joyal Oct 18 '13 at 3:08
That's indeed a very elegant answer. And also, lovely gif! Although the claim was made in a course in differential geometry in which homotopy wasn't even mentioned, I'll stick to it. – ulilaka Oct 18 '13 at 17:11
Dear @ulilaka, thank you for the compliment. Regards, – Bruno Joyal Oct 21 '13 at 16:46

You have it. At each point $x$ of the sphere, the directional derivative of $f$ at $x$ in the normal direction is positive, which says that $f$ increases as you approach $x$ from inside the ball. This means that the global minimum of $f$ on the closed ball cannot occur at a boundary point.

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I guess one needs to know that the gradient is continuous, which, if the OP is in a calculus class, is a reasonable assumption. – treble Oct 18 '13 at 3:14
I think $C^1$ is an automatic assumption unless pathology is made clear to start with! – Ted Shifrin Oct 18 '13 at 3:39
Yes, I should have mentioned it, but the function may be assumed $C^{\infty}$. – ulilaka Oct 18 '13 at 17:12

Here is a proof that appeals to the Poincare-Hopf theorem:

If $M$ is a manifold with boundary and $X$ is a vector field with isolated zeroes on $M$ and pointing outward on $\partial M$, then $$\sum_{\mbox{zeroes }m_i} \operatorname{index}_{m_i}(X) = \chi(M).$$

Since $\nabla f$ is a smooth vector field on the closed unit ball and the Euler characteristic of the closed unit ball is $1$, there must be at least one point in the interior of $M$ for which $\nabla f = 0$.

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I love fancy proofs, but now we're really using a cannon to kill a fly! – Ted Shifrin Oct 18 '13 at 3:49
+1, I love it! By the way, I like to call this the "Poincaré-Hopf-Riemann-Roch" theorem. :) – Bruno Joyal Oct 18 '13 at 3:51
(It comes right before the "Poincaré-Hopf-Grothendieck-Riemann-Roch-Atiyah-Singer" theorem.) – Bruno Joyal Oct 18 '13 at 3:53
I like how applying the theorem here seems to require observing that a nonvanishing vector field has isolated zeros. – Mike F Oct 18 '13 at 3:55
@Mike: What do you mean? Nonvanishing means none. But in general, you can wiggle a vector field to insure it has isolated zeroes. – Ted Shifrin Oct 18 '13 at 3:58

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