Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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


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 $f(x)=\pm x\;\;\;(\forall x\in\mathbb{R})$.

share|cite|improve this answer

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 $c=\pm1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.