# $f(x+f(y))=f(x)+y^n$

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

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

$f(0)=0\;\;\;(1)$.

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

$f(f(y))=y^n\;\;\;(2).$

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

$f(x+f(y))=f(x)+y^n=f(x)+f(f(y))\;\;\;(3).$

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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It can be shown that instead of monotonicity, we can assume other properties for $f$ and get the same result. For this, we may first find some properties of $f$ without assuming monotonicity and then apply the extra assumption.

Suppose that we have: $$f(x+f(y))=f(x)+y^n\qquad\bf(1)$$ Letting $x=y=0$ in (1) we get: $$f(f(0))=f(0)+0^n=f(0)$$ $$\therefore f(f(f(0)))=f(f(0))=f(0)$$ Again, letting $x=0$ and $y=f(0)$ we get: $$f(f(f(0)))=f(0)+f(0)^n$$ $$\therefore f(0)=f(0)+f(0)^n$$ $$\therefore f(0)=0$$ Now, letting $x=0$ in (1) we get: $$f(f(y))=y^n\qquad\bf(2)$$ So by applying $f$ on the both sides of (2) we have: $$f(f(f(y)))=f(y^n)$$ On the other hand, substituting $f(y)$ for $y$ in (2) we get: $$f(f(f(y)))=f(y)^n$$ Combining the last two results, we find that: $$f(y^n)=f(y)^n\qquad\bf(3)$$ Now, by (1), (2) and (3) we have: $$f(f(x+f(y)))=f(y^n+f(x))$$ $$\therefore(x+f(y))^n=f(y^n)+x^n=f(y)^n+x^n$$ $$\therefore(x+f(f(y)))^n=f(f(y))^n+x^n$$ $$\therefore(x+y^n)^n=x^n+y^{n^2}$$ Letting $x=y=1$ in the last equation we get $2^n=2$ and so $n=1$. Thus, substituting $f(y)$ for $y$ in (1), by (2) we have: $$f(x+y)=f(x)+f(y)\qquad\bf(4)$$ $$f(f(x))=x\qquad\bf(5)$$ So, $f$ satisfies (1) iff $n=1$ and $f$ satisfies (4) and (5).

It's well-known that given a Hamel basis, one can construct wild functions satisfying (4) and (5). On the other hand, it's also well-known that if $f$ satisfies (4) and it's also continuous, monotone, lebesgue measurable or bounded when restricted to a bounded interval, then there is a constant $c$ such that for every real number $x$ we have $f(x)=cx$. For example, see Overview of basic facts about Cauchy functional equation. By (5) you can see that $c=\pm1$.

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