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Let $f: (0,1)\rightarrow (0,1)$ be a given function. A real function $\phi:(0,1)\rightarrow \mathbb{R}$ is said to satisfy the Schroder functional equation, with parameter $s\neq 0, 1$, if $$\phi(f(x))=s\phi(x), \ \ \forall x\in (0,1).$$

I am particularly interested in the case where $s=2$ and on what conditions one must impose on $f$ so that there exists a nonzero continuous solution $\phi$ of the corresponding equation. According to "Kuczma - Functional equations in a single variable", pg. 136, if $f$ belongs in $R_\xi^0$, then there do not exist nonzero continuous solutions. The set $R_\xi^0$ contains the strictly increasing continuous functions with the property that $f(x)>x$ in $(0,\xi)$ and $f(x)<x$ in $(\xi, 1)$, where $\xi$ is the fixed point of $f$.

So this gives us a necessary condition that $f$ needs to satisfy so that the Schroder's equation has a nonzero solution. But is there a sufficient condition? In particular, if $f$ is a strictly increasing function satisfying $f(x)<x$ in $(0,\xi)$ and $f(x)>x$ in $(\xi, 1)$, is it true that the corresponding equation $$\phi(f(x))=2\phi(x),$$ always has a nonzero continuous solution?

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    $\begingroup$ Don't have that one, I have Iterative Functional Equations by Kuczma, Choczewski and Ger(1990). Probably in here somewhere, but it takes some time to interpret. First I will see if I can find your nonexistence result. $\endgroup$ – Will Jagy May 3 '17 at 23:05
  • $\begingroup$ requested your book from library, there now, see what is meant $\endgroup$ – Will Jagy May 6 '17 at 17:28
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Alright, got the book. It appears the main thing you missed is the freedom to consider $f^{-1}$ instead of $f.$ So, for example, in Theorem 6.1 on page 137, let $h(x) = f^{-1}(x)$ so that $h \in R^0_0.$ For now, demand $h$ to be $C^2.$ then we can solve $$ \psi(h(t)) = \frac{1}{2} \psi(t), $$ with $\psi'(0)$ whatever you prefer. Taking $t = f(x)$ gives $$ \psi(x) = \frac{1}{2} \psi(f(x)),$$ $$ 2\psi(x) = \psi(f(x)).$$

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