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I came upon this question - I am unsure if it is true.

Let $S \subseteq \mathbb{R}^n$, path connected, such that $p \in S \Rightarrow -p \in S$, then for any continuous function $f:S \rightarrow \mathbb{R}$ exists $q \in S$ such that $f(q)=f(-q)$. (We take usual topology for both spaces).

My thoughts:

Let $f' = f(-p)$ and $g = f - f'$. Take $s \in S$, then if $g(s)= 0$ we are done, otherwise wlog $g(s)>0$, so $g(-s)<0$. Take path $\gamma :[0,1] \rightarrow S$ where $\gamma(0)=s, \gamma(1)=-s$, then $g \circ \gamma :[0,1] \rightarrow R$ yields existence of $c \in S$ such that $g(c) = 0$ by Intermediate Value Theorem.

Is this right?

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  • $\begingroup$ It's worth writing another line to explain how you know $g(-s)<0$. $\endgroup$ – 5xum Jun 12 '17 at 10:25
  • $\begingroup$ The definition of f' is confusing. Do you mean f':S -> R, p -> f(-p)? $\endgroup$ – William Elliot Jun 12 '17 at 10:32
  • $\begingroup$ Why bother with f'? Simply define g(x) = f(x) - f(- x) for x in S. $\endgroup$ – William Elliot Jun 12 '17 at 10:38
  • $\begingroup$ it is right, and it is well-written. Anyone who cannot follow it is their own fault :) $\endgroup$ – Mirko Jun 12 '17 at 18:18
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The proof is generally correct. That is, if you were writing a test, I would give you full points. However, it jumps a couple of steps. In order to make it easier to read (say, if you were writing a paper) I would just advise writing out some details a bit more to make it clearer:

  • Explain why $g(-s)<0$
  • Explain that $(g\circ \gamma)(0) > 0$ and $(g\circ \gamma)(1)<0$ to make it clear that the conditions of IVT are met
  • Given that IVT gives a point $t'\in [0,1]$, explain how this point then gives you the point $c\in S$ such that $g(c)=0$.

I understand you know all the answers to the three points above, but making it easier to the reader goes a long way.

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g:S -> R, x -> f(x) - f(-x) is continuous.
Assume S is only connected and g is never zero.
Pick any p in S. Since g(-p) = -g(p), wlog, g(p) > 0, g(-p) < 0.
g^-1((-oo,0)) and g^-1((0,oo)) disconnect S, a contradiction.
Thus there is a p in S with g(p) = 0, ie f(p) = f(-p).

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