TITLE: Why must this function have a critical point inside the sphere?
QUESTION [7 upvotes]: 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?

REPLY [6 votes]: 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 $\nabla f = 0$.