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Here is the problem:

Fix $n\in\mathbb{N}$. Find all monotonic solutions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+f(y))=f(x)+y^n$.

I've tried to show that $f(0)=0$ and derive some properties from that but have been unable to do so.

A solution would be appreciated.

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Since $f$ is monotonic and $f(x+f(0)) = f(x)$ for all $x$, we know that $f(0) = 0$. –  Jim Belk Jun 18 '11 at 4:11

1 Answer 1

If $|f(0)|>0$, f is a monotonic periodic function and thus is constant; however, the functional equation admits no constant solutions so we must have


Now, substituting $x=0$ into the functional equation gives us


Note that (2) allows to write the functional equation as


Also note that by the functional equation and (1), the range of $f$ contains all non-negative numbers (fix $x=0$ and vary y). Hence any non-negative number may be substituted for f(y).

Thus the functional equation can be written as

$f(x+y)=f(x)+f(y),x\in\mathbb{R}, y\geq0\;\;\;(4)$

Since f is monotonic, we must have that for $x\geq0, f(x)=Cx$ for some constant C.

Substituting this partial solution into the original functional equation, we see that this is only possible if $n=1$; that is, if $n\neq1$ the functional equation does not admit solutions.

If $n=1$, f is clearly surjective by the original functional equation and thus (4) holds $\forall x,y\in\mathbb{R}.$

It follows that the only potential solutions are of the form $f(x)=Cx$ for some constant C (monotonic additive functions are linear).

Substituting into the functional equation, we find that the only solutions are $f(x)=\pm x\;\;\;(\forall x\in\mathbb{R})$.

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