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Let $f : \mathbb{R}^3 \to \mathbb{R}$ be a differentiable function and it has a maximum in $S^2$. Show that exists a point $p \in S^2=\left \{ (x,y,z) \in \mathbb{R} : x^2+y^2+z^2=1\right \}$ in which gradient $\triangledown_{f}(p)$ and $p$ are linearly independent.

HINT (that is given in exercise): The function $f$ has a maximum in unit sphere $S^2$.

Melina
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    Yes. Did you find this result in a book? I think it is wrong, because one should be able to construct a differentiable function that is constant in a neighborhood of $S^2$ and decreases rapidly to zero outside this neighborhood. Then $\nabla f(p)$ would be zero on $S^2$ and there is no chance that $\nabla f(p)$ and $p$ would be linearly independent. – Derived Cats Aug 16 '16 at 15:21
  • No, I couldn't find it. :( – Melina Aug 16 '16 at 15:58
  • I added to my answer below. – zhw. Sep 01 '16 at 15:00

2 Answers2

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Let $f(x) = \cos (1-|x|^2).$ Then $f$ attains a maximum at each point of $S.$ But $\nabla f (x) = 2\sin(1-|x|^2)x$ for all $x.$ Thus $x, \nabla f (x),$ are never independent at any point of $\mathbb R^3.$

Added later: Dominique asked the question: What if the maximum is isolated? That's a good question, and we can obtain the result for this case. Suppose $f$ has an isolated maximum at $p_0\in S.$ Choose a smooth injective curve $p(t):(-1,1) \to S$ such that $p(0) = p_0.$ (For example, let $u$ be a unit vector perpendicular to $p_0.$ Then $p(t) = (\cos t)p_0 + (\sin t)u$ would do nicely.) Because $p(t)$ is a curve in $S,$ $p'(t)\perp p(t)=0$ for all $t.$

Set $g(t)= f(p(t)).$ Then $g$ has an isolated maximum at $t=0.$ Therefore $g$ is not constant, hence $g'(s)\ne 0$ for some $s\in (-1,1).$ But note

$$g'(t) = \nabla f (p(t))\cdot p'(t),\, t\in (-1,1).$$

So at $s,$ $\nabla f (p(s))$ is not perpendicular to $p'(s),$ while $p(s)$ is perpendicular to $p'(s).$ Thus $\nabla f (p(s))$ is not a scalar multiple of $p(s)$ and we conclude $\nabla f (p(s))$ and $p(s)$ are linearly independent.

zhw.
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  • What if the maximum is isolated? I have a feeling that this is what the OP has in mind... though I might be wrong. – Derived Cats Aug 30 '16 at 22:40
  • I don't know how to do this rigorously right now, but here are some thoughts about this. If the maximum (at the point $x$, say) is isolated, in a sufficiently small neighborhood around $x$ the gradient will be non-zero (except at $x$, of course) and will be pointing roughly in the direction of $x$. For points of this neighborhood that lie on the sphere, this roughly means that the gradient points more in the tangential direction than in the radial direction. So we should be able to find a point where $p$ and $\nabla f (p)$ are linearly independent. – Derived Cats Aug 31 '16 at 18:58
  • @DominiqueR.F. Good question, and I added to my post to answer it. – zhw. Sep 01 '16 at 14:58
  • the gradient of $f$ is a 3 dimensional vector, you are talking about something else – marmouset Sep 01 '16 at 15:00
  • @marmouset I don't understand your comment. – zhw. Sep 01 '16 at 16:02
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Suppose that the conclusion is wrong so that for every $p\in S^2$ we have $\nabla f(p) =c(p) \vec{p}$ for some function $c(p)$. Let $\gamma(t)\in S^2$, $0\geq t\geq 1$ be a $C^1$ path in $S^2$. Then:

$$ \frac{d}{dt} f(\gamma(t))=\nabla f(\gamma(t)) \cdot \gamma'(t) = c(p) p \cdot \gamma'(t)= 0$$ because $\gamma'(t)$ is tangential to $S^2$, whence orthogonal to $p$. As $S^2$ is connected $f$ must be constant on $S^2$. Conversely, if f is constant on $S^2$, then $\nabla f(p)$ must be perpendicular to all tangents to $S^2$ whence linearly dependent with $p$. Whether or not $f$ has a maximum on $S^2$ is not of importance here.

H. H. Rugh
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