How to prove $x_0\in S^1$ s.t., $f(x_0)=f(-x_0)$

Question

Let $f:S^1 \to \mathbb{R}$ continuous function , then there exists $x_0\in S^1$ s.t., $f(x_0)=f(-x_0)$

I tried to calculate by taking $x:=(\cos{t}, \sin{t})$, but $f$ is not specific, I can't.

Answer 1

We need one point $x_0$ on circle such that $f(x_0)=f(-x_0)$.

Let g be an odd real-valued continuous function on a circle;Pick an arbitrary $x$.

if $g(x)$=0 then we are done otherwise $g(x)>0$ without loss of generality now since g is odd function $g(-x)<0$ so by Intermediate value theorem there is $y$ s.t $g(y)$=0 between x and -x and so $g(y)$=0.

Now only question is what odd function we can choose so take $g(x)$=$f(x)$ $-$ $ f(-x)$ this function is clearly odd and continuous since f is continuous.So this function work for us now if g(x) is even then it is trivially true that $f(x)$ $=$ $f(-x)$ hence we consider odd function $g(x)$.

Also there is generalization of this known as Borsuk–Ulam theorem.

Answer 2

Consider the function $g(x)=f(x)-f(-x)$ has no zeros.

Check how $g(x)$ relates to $g(-x)$.

Now assume there is no such $x_0$ as in the question.

Can you finish from here?