Let a function $f(z)$ defined for all $z\in\mathbb{C}$ (without any further restrictions) satisfy the equation: $$ f(f(z))=f(z-c)+c, $$ where $c\ne0$ is a constant. It can be easily demonstrated that the only analytic function satisfying the equation is $f(z)=z$ and I have strong (but possibly wrong) impression that it is in fact the only possible solution of the equation. Can it be proved or disproved? In the latter case what is the general form of solution?

I never dealt with functional equations, so any hint is appreciated.

  • $\begingroup$ If you are willing to consider matrices, the matrix equation $A^2 = A$ has plenty of solutions. This for the case $c=0$ of course. $\endgroup$ – Nigel Overmars Feb 14 '18 at 20:44
  • $\begingroup$ At the moment I would like to know the answer for the case when both function and argument are scalars. But of course the case of vectors is also interesting. In any case $c\ne0$ is the condition. Otherwise there is a plenty of solutions also in scalar case. $\endgroup$ – user Feb 14 '18 at 20:58
  • $\begingroup$ My guess is that the most general solution for $f(f(z))=f(z)$ is the following. Let ${\cal D}$ be a set of points $z$ such that $f(z)=z$. It suffices if the set contains a single point. Then in the domain $\mathbb{C}\setminus{\cal D}$ the function can take any value $f(z)\in{\cal D}$. $\endgroup$ – user Feb 14 '18 at 21:17

For a simple counterexample, consider $f(z)=\operatorname{Re}(z)$, which works for any $c\in\mathbb{R}$. More generally, given any decomposition $\mathbb{C}=A\oplus B$ of $\mathbb{C}$ as a group under addition with $c\in A$, you could take $f$ to be the projection onto $A$. I suspect there are other solutions besides these as well, and there is probably no nice way to describe all the solutions.

  • $\begingroup$ For integers with $c=2$, for instance, you could define $f(x)=x$ if $x$ is even and $f(x)=x-1$ if $x$ is odd. $\endgroup$ – Eric Wofsey Feb 15 '18 at 16:55
  • $\begingroup$ I have meanwhile realized this. I hope for your comments in the new thread. $\endgroup$ – user Feb 15 '18 at 17:06

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