How can I prove that for any smooth function $f\colon S^n\longrightarrow \mathbb{R}$ always exist $x,y\in S^n$ such that $df_x=df_y=0$?
I have tried it by induction, but I don't know to prove it even with $n=1$.
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The continuous image of a compact set is compact, and $S^n$ is compact. The continuous image of a connected set is connected, and $S^n$ is connected for $n>0$. Therefore the image $f(S^n)$ is a compact connected set; i.e. a closed interval say $[a,b]$. So $f$ achieves its minimum at some point $x$ such that $f(x)=a$, and its maximum at some point $y$ such that $f(y)=b$. Then by Fermat's theorem the derivative vanishes at the extremal points $x$ and $y$.
Note that this theorem applies not just for maps $\mathbb{R}\to\mathbb{R}$, but also for maps $f\colon M\to \mathbb{R}$ for any smooth manifold $M$: for any tangent vector $v$ at $x$ we have a curve $\exp_v(t)$ through $x$ with derivative $v$, defined on some interval. Then apply the 1d Fermat theorem to the map $t\mapsto f(\exp(t))$.
ziggurism
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What do you mean by "by Fermat"? – Carlos Ramírez Nov 28 '16 at 01:03
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@CarlosRamírez fermats theorem says that if a function achieves a max or min then its derivative vanishes – ziggurism Nov 28 '16 at 01:05
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But the Fermat theorem is for intervals of $\mathbb{R}$ to intervals of $\mathbb{R}$, why this makes correct the proof? – Carlos Ramírez Nov 28 '16 at 01:13
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1@CarlosRamírez actually Fermat applies to maps from manifolds of any dimension. Just consider $f$ evaluated along curves in the manifold. See my edit, is it clear? – ziggurism Nov 28 '16 at 01:22
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yes, now is totally clear. – Carlos Ramírez Nov 28 '16 at 01:22