TITLE: $f : S^1 \to\mathbb R$ is continuous then $f(x)=f(-x)$ for some $x\in S^1$
QUESTION [12 upvotes]: Question is to prove :
$f : S^1 \to \mathbb R$ is continuous then $f(x)=f(-x)$ for some $x\in S^1$
I guess it would be helpful to use intermediate value theorem
Assuming $f(x)\neq f(-x)$ then given any $p\in (f(-x),f(x))$ (assuming $f(-x)<f(x)$) there exists $y\in (-x, x)$ such that $f(y)=p$
I am not very sure of how to use this.. 
It would be helpful if someone can give some hint which would help me to solve this..
Thank you.

REPLY [18 votes]: Define $g:[0,1]\to \mathbb{R}$ as $g(t):=f(e^{2\pi it})-f(-e^{2\pi it})$. 
Clearly $g$ is also continuous. 
We claim that $g$ is zero at some point. 
If that is true, say $g(t_0)=0$, then it means $f(e^{2\pi it_0})=f(-e^{2\pi it_0})$, hence we get what we wanted.
Suppose $g(t)\neq 0\; \forall\; t\in S^1$, in particular $g(0)\neq 0$. 
Observe that $g(0)=f(1)-f(-1)$ and $g(\frac{1}{2})=f(-1)-f(1)=-g(0)$.
Hence if $g(0)>0$, then $g(\frac{1}{2})<0$, and similarly, if $g(0)<0$, then $g(\frac{1}{2})>0$. 
Thus by the intermediate value theorem, there must be a point say $t_0\in (0,\frac{1}{2})$ such that $g(t_0)=0$, which gives a contradiction.
Observe that in the same way you can obtain not just a $t_0\in (0,\frac{1}{2})$ such that $g(t_0)=0$, but another point $t_1\in (\frac{1}{2},1)$ such that $g(t_1)=0$, because $g(1)=g(0)$.
Note: Instead of defining $g$, you may define, $h:S^1\to \mathbb{R}$, as $h(x)=f(x)-f(-x)$, and follow the same steps as above. The intermediate value theorem can be applied to the function $h$ also, because $S^1$ is a path-connected space.