# Non-Coinciding Rotations on the Unit Circle

Let $$C$$ be the unit circle and $$F, G: C \to C$$ be continuous functions. Does there exist a continuous function $$\alpha: C \to \mathbb{R}$$ such that there exists no $$x\in C$$ with $$F_{\alpha}=e^{i\alpha (x)}F(x) = G(x)?$$ If not, under what conditions we can expect this.

For the case $$F,G : [a,b]\rightarrow C$$ I have the following idea, but does not seem to work when the domain is $$C$$ instead of [a,b].

Let $$F, G: [a, b] \to C$$ be continuous functions where $$C$$ is the unit circle. We want to prove that there exists a continuous bounded function $$\alpha: [a, b] \to \mathbb{R}$$ such that for all $$x \in [a, b]$$, the function $$F_{\alpha}(x)$$ (which represents the counterclockwise rotation of $$F(x)$$ by $$\alpha(x)$$ degrees) is not equal to $$G(x)$$.

Since $$F$$ and $$G$$ are continuous on the compact interval $$[a, b]$$, they are uniformly continuous. Therefore, for every $$\epsilon > 0$$, there exists a $$\delta > 0$$ such that for all $$x, y \in [a, b]$$, if $$|x - y| < \delta$$, then $$|F(x) - F(y)| < \epsilon$$ and $$|G(x) - G(y)| < \epsilon$$.

Choose $$\epsilon$$ small enough such that the arcs subtended by the images of the intervals of length $$\delta$$ in $$C$$ are less than the length corresponding to an angle of $$\pi/2$$ radians. Since $$[a, b]$$ is compact, we can cover it with finitely many such intervals $$\{[x_{i-1}, x_{i}]\}_{i=1}^{n}$$ where $$x_0 = a$$ and $$x_n = b$$.

We will now define $$\alpha$$ on each $$[x_{i-1}, x_i]$$. For the first interval $$[x_0, x_1]$$, choose any $$\alpha(x)$$ such that the rotation $$F_{\alpha}(x)$$ does not coincide with $$G(x)$$ for $$x \in [x_0, x_1]$$. This is possible because the images under $$F$$ and $$G$$ are less than $$\pi/2$$ apart, providing us with at least $$\pi/2$$ radians (or $$90$$ degrees) of "wiggle room" to avoid overlap.

For $$[x_1, x_2]$$, we choose $$\alpha$$ such that $$\alpha(x_1)$$ from the previous step is equal to $$\alpha(x_1)$$ defined for $$[x_1, x_2]$$ to maintain continuity. Again, we can rotate $$F(x)$$ for $$x \in [x_1, x_2]$$ to avoid coincidence with $$G(x)$$ while ensuring that $$|\alpha(x) - \alpha(x_1)| < \pi/2$$ to maintain the bounded rotation. This process is repeated for each subsequent interval.

By construction, $$\alpha$$ is continuous on each interval $$[x_{i-1}, x_i]$$ and matches at the endpoints, ensuring continuity on the entire interval $$[a, b]$$. Furthermore, $$\alpha$$ is bounded because the rotation at each step is less than $$\pi/2$$, and these bounded rotations accumulate to a finite bound over the finite number of intervals.

Therefore, we have constructed a continuous bounded function $$\alpha: [a, b] \to \mathbb{R}$$ such that $$F_{\alpha}(x) \neq G(x)$$ for all $$x \in [a, b]$$. This completes the proof.

• It is a potentially interesting question, but: (1) Why do you say "continuous bounded"? (Continuous functions on a compact space are always bounded.) (2) What is the origin of this problem and what are your own thoughts about it? What did you try to solve it, etc? Lastly, a concise description of $F(z,\alpha(z))$ is $e^{i\alpha(z)} F(z)$. Apr 2 at 14:43
• @MoisheKohan Thank you for your comment. You are absolutely about the boundedness assumption. I revised the question. If the domain of the functions $F,G$ is an interval $[a,b]$ instead of $C$ this could be done by dividing the interval into small subintervals and doing a continuous shift in each of the subintervals, but my methods breaks down if the domain is a periodic region like $C$ instead of $[a,b]$. I want to see if this is just a technical issue or something deep. Apr 2 at 15:01
• @MoisheKohan I just added my argument for $[a,b]$. Apr 2 at 15:30
• Hint: think about winding numbers.
– Karl
Apr 2 at 16:07

Expanding on my hints in the comments, let $$w(\cdot)$$ denote the winding number of maps $$C\to C$$. We have:
• $$w(fg)=w(f)+w(g)$$, where $$fg$$ is the pointwise product.
• $$e^{i\alpha(x)}$$ has winding number $$0$$.
• If $$\frac fg$$ is surjective then $$\exists x[f(x)=g(x)]$$.
Putting these things together, we get that a necessary condition for the desired $$\alpha$$ to exist is $$w(F)=w(G)$$. It's easy to find simple examples where this is the case. The condition is also sufficient because if $$w(F/G)=0$$, we can define $$\alpha$$ so that $$F_\alpha/G$$ is constant (and never hits $$1$$).