As a matter of fact, it is impossible to find a continuous $f$ such that $(f(z))^2=z$ for all $z$. ("Calculus Fourth Edition" by Michael Spivak.)

Question

As a matter of fact, it is impossible to find a continuous $f$ such that $(f(z))^2=z$ for all $z$. In fact, it is even impossible for $f(z)$ to be defined for all $z$ with $|z|=1$. To prove this by contradiction, we can assume that $f(1)=1$ (since we could always replace $f$ by $-f$). Then we claim that for all $\theta$ with $0\leq\theta <2\pi$ we have

$$f(\cos{\theta}+i\sin{\theta})=\cos{\frac{\theta}{2}}+i\sin{\frac{\theta}{2}}\tag{*}$$

The argument for this is left to you (it is a standard type of least upper bound argument). But (*) implies that

$$\lim\limits_{\theta\to 2\pi} f(\cos{\theta}+i\sin{\theta}) = \cos{\pi}+i\sin{\pi}$$

$$=-1$$

$$\neq f(1)$$

even though $\cos{\theta}+i\sin{\theta} \to 1$ as $\theta\to 2\pi$. Thus we have our contradiction.

Let $g(\theta):=\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}$.
Let $h(\theta):=-\cos\frac{\theta}{2}-i\sin\frac{\theta}{2}$.
$f(\cos\theta+i\sin\theta)=g(\theta)$ or $f(\cos\theta+i\sin\theta)=h(\theta)$ for all $\theta\in [0,2\pi)$.
How to prove $f(\cos\theta+i\sin\theta)=g(\theta)$ for all $\theta\in [0,2\pi)$?

My intuition:
By the author's assumption, $f(\cos 0+i\sin 0)=f(1)=1=\cos\frac{0}{2}+i\sin\frac{0}{2}=g(0)$.
Because $f$ is continuous, it is impossible $f(\cos\theta_1+i\sin\theta_1)=g(\theta_1)$ and $f(\cos\left(\theta_1+\Delta t\right)+i\sin\left(\theta_1+\Delta t\right))=h(\theta_1+\Delta t)$ for $\theta_1\in [0,2\pi)$ when $\Delta t$ is very small. ($f(\cos\theta+i\sin\theta)$ cannot jump. Please see the following figure.)

But I cannot prove this obvious fact.

Please give me a hint.

Answer 1

The question is about finding continuous solutions $w(z)$ of $w^2 = z$ on the unit circle. Exclude $z=1$, and take the Cayley transform $$\begin{align} w &= \frac{y + \mathrm{i}}{y - \mathrm{i}} & z = \frac{x + \mathrm{i}}{x - \mathrm{i}}\text{.} \end{align}$$ Then the possible $w(z)$ are in bijection with the solutions $y(x)$ on the real line of $$y^2-2xy-1 = 0\text{.}$$ Since $y$ cannot be zero, by the intermediate value theorem $y(x)$ cannot change sign: it must be positive for all $x$ or negative for all $x$. In trigonometric language, we must choose either $\cot \tfrac{\theta}{4}$ or $-1/(\cot \tfrac{\theta}{4})$, corresponding by the Cayley transform to your $g$ and $h$, respectively.

---

The rest of the argument can also be done without invoking trigonometric functions. The positive choice of $y(x)$ above satisfies $\lim_{x\to \infty}y(x) = \infty$ and $\lim_{x\to -\infty}y(x)=0$. Consequently, the corresponding limit $\lim_{z \to 1}w(z)$ does not exist.

Answer 2

$f(\cos\theta+i\sin\theta)=g(\theta)$ or $f(\cos\theta+i\sin\theta)=h(\theta)$ for each $\theta\in [0,2\pi)$.

Since $f(\cos\theta+i\sin\theta)$ is continuous, there exists $t_1>0$ such that $\theta\in [0,t_1)\implies |(f(\cos\theta+i\sin\theta)-f(\cos 0+i\sin 0)|<1$.

Since $h(\theta)$ is continuous, there exists $t_2>0$ such that $\theta\in [0,t_2)\implies |h(\theta)-h(0)|<1$.

Suppose $\theta\in [0,\min(t_1,t_2))$.
Then, $|f(\cos\theta+i\sin\theta)-1|<1$ and $|h(\theta)+1|<1$.

If $f(\cos\theta+i\sin\theta)=h(\theta)$, then $|h(\theta)-1|<1$.

On the other hand, $2=|1-(-1)|=|(1-h(\theta))+(h(\theta)-(-1))|\leq |1-h(\theta)|+|h(\theta)+1|$.
So, $1=2-1<2-|h(\theta)+1|\leq|h(\theta)-1|$.

This is a contradiction.

So, $f(\cos\theta+i\sin\theta)=g(\theta)$ for all $\theta\in [0,\min(t_1,t_2))$.

Let $S:=\{t\in (0,\infty):f(\cos\theta+i\sin\theta)=g(\theta)\text{ for all }\theta\in [0,t)\}$.

Then, $\min(t_1,t_2)\in S$.
So, $S\neq\varnothing$.

$f(\cos 2\pi+i\sin 2\pi)=f(1)=1\neq -1=\cos\pi+i\sin\pi=g(2\pi)$.
So, $2\pi\notin S$.
So, if $2\pi<t$, then $t\notin S$.
So, if $t\in S$, then $t\leq 2\pi$.
So, $S$ is bounded above.

$\sup S\leq 2\pi$ holds.

Suppose $\sup S<2\pi$.

Since $f(\cos\theta+i\sin\theta)$ is continuous, there exists $\delta_1>0$ such that $|\theta-\sup S|<\delta_1\implies |f(\cos\theta+i\sin\theta)-f(\cos\sup S+i\sin\sup S)|<\frac{1}{4}$.
Since $g(\theta)$ is continuous, there exists $\delta_2>0$ such that $|\theta-\sup S|<\delta_2\implies |g(\theta)-g(\sup S)|<\frac{1}{4}$.
Since $h(\theta)$ is continuous, there exists $\delta_3>0$ such that $|\theta-\sup S|<\delta_3\implies |h(\theta)-h(\sup S)|<\frac{1}{4}$.
Let $\delta:=\min(\sup S, 2\pi-\sup S,\delta_1,\delta_2,\delta_3)$.

Suppose $\sup S-\delta<\theta_1<\sup S$.
Then there exists $t_1$ such that $\theta_1<t_1\leq\sup S$ and $t_1\in S$.
So, $f(\cos\theta_1+i\sin\theta_1)=g(\theta_1)$ holds.

Suppose $\sup S<t_2<\sup S+\delta$.
Since $t_2\notin S$, there exists $\theta_2$ such that $\sup S\leq\theta_2<t_2$ and $f(\cos\theta_2+i\sin\theta_2)=h(\theta_2)$.

Therefore, $$2=|2g(\sup S)|=|g(\sup S)-h(\sup S)|\\\leq|g(\sup S)-f(\cos\theta_1+i\sin\theta_1)|+|f(\cos\theta_1+i\sin\theta_1)-f(\cos\sup S+i\sin\sup S)|+|f(\cos\sup S+i\sin\sup S)-f(\cos\theta_2+i\sin\theta_2)|+|f(\cos\theta_2+i\sin\theta_2)-h(\sup S)|\\\leq 4\cdot\frac{1}{4}=1$$.

This is a contradiction.

So, $\sup S=2\pi$.

Therefore $f(\cos\theta+i\sin\theta)=g(\theta)$ for all $\theta\in [0,2\pi)$.