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i know the solutions of the well known Cauchy-functional-equation

$f(x+y)=f(x)+f(y)$

But what does it change if i have the following form

$f(x+g(y))=f(x)+f(g(y))$

? what can i say about g?

thanks

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There are lots of possibilities. For example, you might have $g(y) = 1$ for all $y$, in which case $f(x+1) = f(x) + f(1)$ just says that $f(x) = f(1) x + p(x)$ where $p(x)$ is periodic with period $1$ and $p(1) = p(0) = 0$. On the other hand, if $g$ is surjective, the equation reduces to Cauchy's.

EDIT: For any possible nonzero value $v$ of $g$, we have $$f(x + v) = f(x) + f(v)$$ and then $$f(x) = p(x) + \dfrac{f(v)}{v} x$$ where $p(x) = f(x) - \dfrac{f(v)}{v} x$ is periodic with period $v$. Nonlinear cases with two or more incommensurate $v$'s are going to be quite exotic (nonmeasurable, I would expect, just like nonlinear solutions to the Cauchy equation).

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  • $\begingroup$ thank you. is it possible to say that the solutions are of the form: $f(x)=f(g(y))x+p(x)$ with $p(x+kg(y))=p(x)$ for an integer k? or do i loose some solutions? $\endgroup$ – William Sep 10 '14 at 6:15
  • $\begingroup$ oh sorry i meant for p the condition $p(x+g(y))=p(x)+p(g(y))$ $\endgroup$ – William Sep 10 '14 at 17:00

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