Let $S^2$ denotes the subset of $\mathbb{R^3}$ which includes the points $(x,y,z)$ s.t $x^2+y^2+z^2=1$
i.e the boundary of a unit sphere. Let $f$ be a continuous function from $S^2$ to $\mathbb{R}$ Prove there exists$ p=(x,y,z) $ s.t $f(p)=f(-p)$
Exam hint: The intermediate value theorem. I have deduced the following: If $p$ is in $S^2$ then $-p$ is also.
$S^2$ is connected since its path connected and its closed and bounded so by Heine Borel its compact.
Continuous functions preserve compactness and connectedness. We can use fact that the only sets in $\mathbb{R}$ that satisfy this are closed intervals. So the image of $f$ will always be a closed interval. The extreme value theorem also applies because of compactness so the endpoints of the interval will be attained. I am thinking maybe define a new function $g$ from the reals to the reals to somehow get the result. Otherwise I certainly can't use it on $f$. I think I might have most of the ideas I need but I don't know how to go further. Thanks
Think about the function $g(x)=f(x)-f(-x)$. Now the goal of the problem is to show that there is some $P\in S^2$ such that $g(P)=0$. if $g$ is identically $0$ then there is nothing to prove. Otherwise let $Q\in S^2$ be a point such that $g(Q)\geq 0$. then $g(-Q)=f(-Q)-f(Q)=-g(Q)$. Now you can apply the intermediate value theorem.
You should also consider the case where $g(Q)\leq 0$.