TITLE: $f:S^1 \rightarrow R$ continuous means there exists a point $x$ such that $f(x) = f(-x)$.
QUESTION [5 upvotes]: Let $S^1$ be the unit circle in $\mathbb{R}^2$, then for a continuous map from $S^1$ into $\mathbb{R}$, there exists $x \in S^1$ such that $f(x) = f(-x)$.  
I know that $S^1$ is path connected and so connected.  And the intermediate value theorem.  For any choice of $-x, x$, and any $r$ between $f(-x)$ and $f(x)$ there is $c \in S^1$ such that $f(c) = r$.  Let $\psi : [0,2\pi] \rightarrow S^1, \ x \mapsto e^{i(a + x)}$ for any $a\in \mathbb{R}$ be a circular path in $S^1$ identified with the circle group as a subset of $\mathbb{C}$.
Let $g(x) = f(x) - f(-x)$.  Let $a\in S^1$ be any point such that there exist points in the image of $g$ to the right of $g(a)$ for at least a run of $\epsilon$.  Now translate everything to near the origin with $h(x) = g(x) - g(a) - \epsilon/2$
Can't figure this one out.

REPLY [7 votes]: We define $g(x)=f(x)−f(−x)$, this is a continuous map from $S^1$ to $\Bbb R$. Note that $g(x)=-g(-x).$ If $g(x)>0$ for some $x$, then $g(-x)<0$. Now apply the intermediate value theorem.