Prove that $\exists x_0\in S^1$ such that $f(x_0)=f(-x_0)$

Question

Let $f:S^1\to \Bbb R$ be a mapping.

Prove that $\exists x_0\in S^1$ such that $f(x_0)=f(-x_0)$.

I am unable to figure out how to show this.

Let $f(x)\neq f(-x)$ forall $x\in S^1$. Take $A=\{x:f(x)>f(-x)\}$ and $B=\{x:f(x)<f(-x)\}$.

I know that $A\cup B$ is open.

But I need to show that both $A,B$ are open to arrive at a contradiction to the fact that $S^1$ is connected.

How to show this?Please help.

Answer 1

Take $g(x)=f(x)-f(-x)$, $A=g^{-1}(\{x:x>0\}$ and $B=g^{-1}(\{x:x<0\}$ since $g$ is continuous, $A$ and $B$ are open.

Answer 2

Consider $f:[0,2\pi]\to\mathbb{R}$ such that $f(0)=f(2\pi)$. Define $g(x)=f(x)-f(x+\pi)$. If $f(0)=f(\pi)$, you are done. Otherwise, let $f(0)>f(\pi)$. Then $g(0)>0$ and $g(\pi)<0$, so the intermediate value theorem implies there is a $\theta\in(0,1)$ such that $g(\theta)=0$. I think you can now quotient $[0,2\pi]$ to get the circle, and the result should be ok, if you use the universal property of quotients.

Answer 3

Define $F: S^1 \to \mathbb{R}$ by $F(x) = f(x) - f(-x)$.

If $f$ is continuous, so is $F$ (we can then write it as a composition of continuous functions ( e.g. $x \to (f(x), f(-x)) \in \mathbb{R}^2$ followed by $(x,y) \in \mathbb{R}^2 \to x-y \in \mathbb{R}$).

Take any $p \in S^1$. If $F(p) = 0$ this means that $f(p) - f(-p)= 0$ or $f(p) = f(-p)$. So a value where $F$ takes the value $0$ is required for the conclusion.

Because $F(-x) = f(-x) - f(--x) = f(-x) - f(x) = -(f(x) - f(-x)) = -F(x)$, if $F$ assumes a value $>0$ (at some $p$) it also assumes a value $<0$ (at $-p$) and vice versa. So $F[S^1]$ is a connected subset of $\mathbb{R}$ that contains two points of opposite sign. This implies that $F$ also assumes the value $0$ (fact: $A$ connected in the reals and $x,y \in A$ implies $[x,y] \subseteq A$, and $0 \in [-r,r]$ for any $r>0$). Done.