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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?

Bruno Joyal
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ulilaka
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3 Answers3

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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?!

Bruno Joyal
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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.

Ted Shifrin
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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 $\nabla f = 0$.

Neal
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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 03:49
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    +1, I love it! By the way, I like to call this the "Poincaré-Hopf-Riemann-Roch" theorem. :) – Bruno Joyal Oct 18 '13 at 03:51
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    (It comes right before the "Poincaré-Hopf-Grothendieck-Riemann-Roch-Atiyah-Singer" theorem.) – Bruno Joyal Oct 18 '13 at 03:53
  • 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 03:55
  • @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 03:58
  • @Mike Did you mean the typo "nzeroes"? I originally typed "nondegenerate" there but decided that isolated was fine. (Also, if the vector field is not transverse to $0$, then the OP's question is already answered.) – Neal Oct 18 '13 at 04:00
  • @Ted & Neal, I just meant this: if you're needing to perturb the vector field to isolate its zeros, then obviously it already has zeros so there is no need to use this theorem. The only time you're needing to do anything is when the vector field has no zeros. In this case, the zeros are isolated (vacuously) so the theorem applies and gives a contradiction as Neals shows above. – Mike F Oct 18 '13 at 04:13
  • I haven't studied this theorem yet, but coincidently I'll have to write an essay about it in an extra discipline I'm taking this semester, so I'll keep this answer until I can understand it. – ulilaka Oct 18 '13 at 17:14