Schröder's equation

Remarks

Example: $f(x) = x^2 + 2x$ on $X = (-1, \infty)$, with $\lambda = 2$. The fixed point is $0$, and $f'(0) = 2$. The function $\Psi(x) = \log(1 + x)$ is a solution with $\Psi(0) = 0$ and $\Psi'(0) = 1$, since $\Psi(f(x)) = \log\big((1+x)^2\big) = 2 \log(1 + x)$.

Non-example: for the same $f$, the function $\Psi(x) = x$ is not a solution, since $f(x) = x^2 + 2x \ne 2x$ for $x \ne 0$.

If $\Psi$ is invertible, the equation says $f = \Psi^{-1} \circ (\lambda \Psi)$, so $f^n = \Psi^{-1} \circ (\lambda^n \Psi)$ for every $n \ge 0$. The function $\Psi$ is called a Schröder function of $f$. If $\Psi$ is a solution, so is $c \Psi$ for every constant $c$; the condition $\Psi'(0) = 1$ fixes this freedom.