Let $f$ be a continuous function on the unit circle. Then there exists a fixed point $p$ such that $f(p) = p$.

Let $$S = \{p = (x, y) \in \mathbb R^2 : x^2 + y^2 = 1\}.$$ Let $$f : S \to S$$ be a continuous function. Then, there always exists $$p \in S$$ such that $$f(p) = p.$$

My try:- $$S = \{p = (x, y) \in \mathbb R^2 : x^2 + y^2 = 1\}.$$ is compact but not convex. So, we cannot apply Brouwer's fixed-point theorem. So, we can not conclude from here. Can you give any suggesion?

• How about $f(x,y)=(-x,-y)$? – Lord Shark the Unknown Jun 1 at 17:00
• This is an incorrect assertion. Please reword or correct as appropriate. – copper.hat Jun 1 at 17:18