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.

  • $\begingroup$ 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 $e^{i\alpha(z)} F(z)$. $\endgroup$ Apr 2 at 14:43
  • $\begingroup$ @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. $\endgroup$ Apr 2 at 15:01
  • 1
    $\begingroup$ Please include your thoughts in the revision of your question instead of replying in comments. $\endgroup$ Apr 2 at 15:12
  • $\begingroup$ @MoisheKohan I just added my argument for $[a,b]$. $\endgroup$ Apr 2 at 15:30
  • $\begingroup$ Hint: think about winding numbers. $\endgroup$
    – Karl
    Apr 2 at 16:07

1 Answer 1


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 $0$.
  • A map with nonzero winding number is surjective.
  • 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 $1$).


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