Recently, someone asked me the following question:
Let $f:S^1 \to \mathbb{R}^1$ be a continuous function and $g:S^1 \times [0,2\pi) \to S^1, g(x,\varphi) = x \cdot \mathrm{e}^{\mathrm{i}\varphi}$ be the rotation of $x$ on the unit circle by $\varphi$ degrees (we identify $\mathbb{C}$ with $\mathbb{R}^2$). Is it possible to find an $x \in S^1$ such that $f(x)=f(g(x,\varphi))$ for a fixed $\varphi \in [0,2\pi)$? If not, can we restrict $f$ to a certain property such that this equation holds?
If we set $\varphi=\pi$, we have Borsuk-Ulam. I am sure that the statement is not true for arbitrary $\varphi$, but cannot think of an example. I am therefore more interested in knowing if one can restrict the "continuous function" condition a bit more so that the statement holds.
Does anyone have an idea? Thanks in advance!
We can actually show, that this holds for every continuous function $g:S^1 \to S^1$.
Let $h(x):=f(x)-f(g(x))$. So we are looking for $h(x)=0$.
Now let $x_{\text{max}}\in S^1$ s.t. $\forall x\in S^1 : f(x_{\text{max}}) \geq f(x)$
and $x_{\text{min}}\in S^1$ s.t. $\forall x\in S^1 : f(x_{\text{min}}) \leq f(x)$
But then $h(x_{\text{max}}) \geq 0$ and $h(x_{\text{min}}) \leq 0$.
If one is equal to $0$, we found a solution, if not, the intermediate value theorem states, that there must be at least one solution for $h(x)=0$ for every path between $x_{\text{max}}$ and $x_{\text{min}}$.