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Define a smooth function $f(x,y)$ on $\mathbb{R}^2$,which satisfies the following condition:$f(x,y)=f(x+1,y);\quad f(x,y)=f(x,y+1)$ (i.e.$f$ can be defined on a torus)then hotw to proof that there are at least 4 critical points($f_{x}=f_{y}=0$) on the unit cube $[0,1]\times[0,1]$ (obviously,it has at least 2 critical point)?

I'm also like to know other examples of thhis kind(the topology implies some properties of the function defined on that space)

*Edit*I think there are 2 different proffs of this question,one only need some use of calculus techniques(which I tried but failed),another one may include some topological statements like Neal said.

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    I tried to use some basic calculas to prove it(construc some auxiliary function)but failed.BTW,I don't have much backgroud on topology,but I'm very glad if you can show me that way2012-09-20

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