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?

  • $\begingroup$ 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

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.


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.