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Let $f_1,f_2$ be two smooth odd functions, i.e. $f_i(-x)=-f_i(x)$. Prove that there exists $x\in S^2$ such that $f_1(x)=f_2(x)=0$

My attempt is to consider $(f_1,f_2):S^2\to \mathbb R^2$, and the image is symmetric about the origin. We only need to show the origin $0$ lies on the image. Another attempt is to consider $f_1^{-1}(0)$ and $f_2^{-1}(0)$. It suffices to show they intersect to each other.

What is more, I was wondering whether we can drop the smooth condition of $f_i$?

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  • $\begingroup$ Any odd function $f$ with $0$ in its domain has $f(0)=0$. Did I miss something? $\endgroup$
    – smcc
    Aug 22, 2016 at 14:07
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    $\begingroup$ @smcc $0\notin S^2$ $\endgroup$ Aug 22, 2016 at 14:36
  • $\begingroup$ Sorry, what is $S^2$ here? (OP did not say, but perhaps it's a standard notation I am unaware of.) Edit: Ah okay its a sphere. $\endgroup$
    – smcc
    Aug 22, 2016 at 14:48

1 Answer 1

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This is an application of Borsuk-Ulam theorem, and you can indeed drop the smoothness condition, just keep the continuity.

Consider the function $g: x \mapsto (f_1(x), f_2(x))$. $g$ is continuous, and, according to Borsuk-Ulam theorem, there exists $x \in S^2$ such that $g(x) = g(-x)$.

Hence, there exists $x$ such that $f_1(x) = f_1(-x)$ and $f_2(x) = f_2(-x)$. One concludes by observing that $f_1$ and $f_2$ are odd.

Edit: removed my edits. I need not to say anything on mondays.

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