Playing with a functional equation

Question

I was playing with a functional equation and proved the result below: Let $f$ be such that $$f(f(z))=z$$ If $f^{-1}$ exists then $$f(z)=f^{-1}(z)$$ If $f'$ exists then as $$(f^{-1}(z))'=\frac{1}{f'(f^{-1}(z))}=(f(z))'$$ then $(z)'=\frac{1}{f'(z)}$ so $f'(z)=1$ then $f(z)=z+C\rightarrow (z+C)+C=z\rightarrow C=0$ so $f(z)=z$. Any error? Did I assume innecesary things?

Answer 1

If $g:\mathbb R^{\geq 0}\to \mathbb R^{\geq 0}$ is such that $g(0)=0$, continuous, differentiable, $1-1$ and onto, and $g'(0)=1$, then you can define a function:

$$f(z) = -g(z) \text{ if } z\geq 0, g^{-1}(-z) \text{ if }z<0$$

Then it turns out that $f(f(z))=z$ for all $z$ and $f$ is differentiable everywhere. (The condition that $g'(0)=1$ is required differentiability.)

So there are a large number of such functions, most of them non-linear.

For example, if $g(z)=z^2+z$, then $g(0)=0$ and $g'(0)=1$ and $h(z)=\frac{-1+\sqrt{1+4z}}{2}$ is the inverse, and we can define $f(z)=-g(z)$ when $z\geq 0$ and $h(-z)$ when $z<0$.

The general non-trivial solution is to pick an arbitrary $C\in\mathbb R$ and $g$ as above. Then define

$$f(z)=\begin{cases}C-g(z-C)&z\geq C\\ C+g^{-1}(C-z)&z<C\end{cases}$$

In particular, $f(z)=C-z$ is an example.

Answer 2

I think the error lies on $(z)' = \frac{1}{f'(z)}$.

All I can see is $1 = (z)' = f(f(z))' = f'(f(z))f'(z)$, so $f'(f(z)) = f'(z)$.