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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.

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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.

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  • $\begingroup$ In $f(x^{2})=f(1)^{n-2}f(x)^{2}=f(1)f(x)^{2}$ when $f(1)\neq 0$ how does the second equality follow? Sorry if I am missing something obvious. $\endgroup$ – Kavi Rama Murthy Jul 16 '18 at 5:46
  • $\begingroup$ As $f(1)\in\{-1,0,1\}$ and $f(1)\neq 0$, we have $|f(1)|=1\implies f(1)^2=1$; since $n$ is odd, $$f(1)^{n-2}=\left(f(1)^2\right)^{\frac{n-3}{2}}f(1)=1^{\frac{n-3}{2}}f(1)=f(1).$$ $\endgroup$ – Carl Schildkraut Jul 16 '18 at 17:02
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    $\begingroup$ Nice and simpler than mine! $\endgroup$ – Sungjin Kim Jul 16 '18 at 18:27
  • $\begingroup$ @KaviRamaMurthy The second equality is by $f(y^2)=f(1)f(y)^2$. $\endgroup$ – Sungjin Kim Jul 16 '18 at 18:27
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This is to show that if $n=5$ (so that $f(x+y^5)=f(x)+f(y)^5$ for all $x,y\in\mathbb{R}$), then $f(x)=0$, $f(x)=x$ or $f(x)=-x$. I think you can generalize the argument, so I leave the general cases to you.

As you mentioned, $f(x+y)=f(x)+f(y)$ for all $x, y\in \mathbb{R}$. Additionally, we have $f(x^5)=f(x)^5$. Since $f$ is additive, $f(qx)=qf(x)$ for all $q\in\mathbb{Q}$, $x\in\mathbb{R}$.

Since $f( (x+y)^5)= f(x+y)^5 = ( f(x)+f(y))^5$, we have $$ \begin{align} f & (x^5+5x^4 y + 10x^3y^2 + 10x^2y^3+5xy^4+y^5)\\ &=f(x)^5+5f(x)^4 f(y) + 10 f(x)^3f(y)^2+10f(x)^2f(y)^3+5f(x)f(y)^4+f(y)^5. \end{align} $$ Then, the first and the last term cancel due to $f(x^5)=f(x)^5$, $f(y^5)=f(y)^5$.

Since $5x^4 y + 10x^3y^2 + 10x^2y^3+5xy^4=(x+y)(5x^3y+5x^2y^2+5xy^3)$, we have $$ \begin{align} f((x+y)(5x^3y+5x^2y^2+5xy^3))=(f(x)+f(y))(5f(x)^3f(y)+5f(x)^2f(y)^2+5f(x)f(y)^3) \end{align} $$ Now, impose $x+y=q\in\mathbb{Q}\backslash\{0\}$ and substitute $y=q-x$, then $$ f(5x^3y+5x^2y^2+5xy^3)=f(1)\left(5f(x)^3f(y)+5f(x)^2f(y)^2+5f(x)f(y)^3\right), $$ hence $$ \begin{align} f & (x^3(q-x)+x^2(q-x)^2+x(q-x)^3)\\ &=f(1)\left(f(x)^3(qf(1)-f(x))+f(x)^2 (qf(1)-f(x))^2+f(x)(qf(1)-f(x)^3)\right) \end{align} $$ This is a polynomial identity in $q$, and holds for infinitely many values of $q$. Thus, it should also hold for $q=0$. Thus, we obtain $$ f(-x^4)=-f(1)f(x)^4 $$ So, $f(x^4)=f(1)f(x)^4$.

Since $f(1)=f(1)^5$, $f(1)=0, 1, \mathrm{or} \ -1$. If $f(1)=0$, then $f(x)=0$. If $f(1)=1$, then $f$ is increasing, so $f(x)=x$. If $f(1)=-1$, then $f$ is decreasing, so $f(x)=-x$.

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  • $\begingroup$ Thanks for this wonderful argument. I am yet to convince myself that this works for all odd $n$. $\endgroup$ – Kavi Rama Murthy Jul 16 '18 at 6:22

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