For a function $f:\mathbb{R}\rightarrow \mathbb{R}$ denote $f_0(x)=x$, $f_n(x)=f(f_{n-1}(x))$. Assume that $f$ satisfies a functional equation $$f_n(x)+a_{n-1} f_{n-1}(x)+\dots+a_0 f_0(x)\equiv 0$$ for some constant real coefficients $a_{0},a_1,\dots,a_{n-1}$. Assume also that $f$ is continuous. What are conditions for polynomial $t^n+a_{n-1}t^{n-1}+\dots +a_0$ which allow to conclude that $f$ is linear?
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asked Mar 1, 2016 at 7:35
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$\begingroup$ Out of curiosity, have you an example where a condition imposed on the polynomial is sufficient to guarantee that $f$ is linear? Or a condition implying that no linear $f$ can be solution? $\endgroup$– Loïc TeyssierCommented Mar 1, 2016 at 8:05
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3$\begingroup$ yes, for example equation $f(f(x))=f(x)+2x$ implies that $f(x)=-x$ or $f(x)=2x$ (this is a problem of forgotten origin which I eventually gave to children I am currently teaching). $\endgroup$– Fedor PetrovCommented Mar 1, 2016 at 8:32
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1$\begingroup$ In mathoverflow.net/questions/152834, the conditions for existence of solutions are discussed. THere are solutions if and only if the corresponding polynomial has a real root. $\endgroup$– Ilya BogdanovCommented Mar 2, 2016 at 17:20
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1$\begingroup$ If there are two roots of the same sign, then there are non-linear piecewise-linear solutions. Say, for $t^2-3t+2$, a function $f(x)=(3x+|x|)/2$ fits. $\endgroup$– Ilya BogdanovCommented Mar 2, 2016 at 17:24
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