functional equations - Solutions of $f(x+y^{n})=f(x)+[f(y)]^{n}$.

Consider the functional equation $f(x+y^{n})=f(x)+[f(y)]^{n}$ where $f:\mathbb R \to \mathbb R$ and $n$ is given integer $>1$. This equation was discussed yesterday and it was shown that $f$ is necessarily additive. Assuming continuity it was concluded that $f(x)\equiv cx$ for some $c$. [ Necessarily $c$ is an n-th root of unity]. If $n$ is even then the given functional equation gives $f(x+y^{n}) \geq f(x)$ which easily leads to the conclusion that $f$ is an increasing function. It follows that $f$ is Borel measurable; since any Borel measurable additive function if of the type $f(x)\equiv cx$ the assumption that $f$ is continuous is not necessary. My question is what can be said for $n$ odd? Can one use some trick to prove that $f$ is necessarily Borel measurable? Or is there a counter-example? Discontinuous additive functions are constructed using Hamel basis but I am unable to use this method to construct a counter-example. I would appreciate receiving any ideas about this question.

Answer

Here's a generalization of i707107's argument that is actually a bit simpler, as long as I didn't make any mistakes:

You have

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

and

$$\begin{align} \sum_{i=0}^n \binom{n}{i}f(x^iy^{n-i}) &=f((x+y)^n)\\ &=f(x+y)^n\\ &=(f(x)+f(y))^n\\ &=\sum_{i=0}^n \binom{n}{i}f(x)^if(y)^{n-i}. \end{align}$$

Taking $y$ rational, we have $f(x^iy^{n-i})=y^{n-i}f(x^i)$ and $f(y)=yf(1)$, so

$$\sum_{i=0}^n\binom{n}{i}y^{n-i}\left[f(x^i)-f(1)^{n-i}f(x)^i\right]=0$$

As this is a polynomial of degree $n$ that is $0$ for all rationals, it is identically $0$, so

$$f(x^i)=f(1)^{n-i}f(x)^i$$

for all $0\leq i\leq n$. Originally, we had $f(1)=f(1)^n$, so $f(1)\in\{-1,0,1\}$. If $f(1)=0$, we have $f(x^i)=0$, so $\boxed{f(x)\equiv 0}$. Otherwise, we have

$$f(x^2)=f(1)^{n-2}f(x)^2=f(1)f(x)^2$$

$$f(x+y^2)=f(x)+f(y^2)=f(x)+f(1)f(y)^2.$$

If $f(1)=1$, this means $f$ is increasing, and if $f(1)=-1$ this means $f$ is decreasing. Either way, $f$ is not everywhere dense, so $f(x)=cx$ for some $c$ and all $x$. The observation that $f(1)=\pm 1$ means $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ are our only other solutions.