Find all additive real valued functions such that $f(x^{2019})=f(x)^{2019}$

Question

The following is the final problem from this page:

Find all the functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $$f(x+y)=f(x)+f(y) \; \; \; \forall \,x,y\in \mathbb{R}$$ and also (this is the important part) $$f(x^{2019})=f(x)^{2019}\tag{$*$}$$

My idea is to prove that $f(x)=x \; \; \forall x \in \mathbb{R}$, $f(x)=-x \; \; \forall x \in \mathbb{R}$ or $f\equiv 0$.

If we change $2019$ for an even number this is easy because it implies that the image of a positive number is positive and from there $f$ is linear and hence the identity or zero.

If we change $2019$ by $3$ then this is related (although I don't know how to deal with the case $f(1)=0$ or $f(1)=-1$)

But in this case I don't know how to prove any type of regularity from $(*)$ to conclude that $f$ must be linear.

Answer 1

Let $f$ satisfy the premises. Then $f(ax)=af(x)$ for any $x\in\mathbb{R}$ and $a\in\mathbb{Q}$. Now $$f\big((a+x)^{2019}\big)=f(a+x)^{2019}$$ (with both sides expanded using the binomial formula and the above), being a polynomial identity in $a\in\mathbb{Q}$, implies $$f(x^k)=f(1)^{2019-k}f(x)^k\qquad(0\leqslant k\leqslant 2019).$$ Taking $k=2$, we get $f(x^2)=f(1)f(x)^2$. This reduces to the case you have worked out (after replacing $f$ by $-f$ if needed).

Answer 2

We can prove that $f$ is continuous at $\mathbb{R}$. Furthermore $f$ has the first derivative on $(0, 0)$. Using these properties leads to show that $f(x) = x$ for every $x \in \mathbb{R}$ or $f \equiv 0$.

Answer 3

The first equation is Cauchy's functional equation, and hence the existence of nonlinear solutions depends on the axiom of choice.

If we assume that $f$ is linear, then it must be of the form $f(x)=ax$ for some $a \in \mathbb{R}$. The second equation then says that $\forall x \in \mathbb{R} (ax)^{2019} = ax^{2019}$. Setting $x=1$, it follows that $a$ must be its own 2019th power. The only real numbers that are their own 2019th (or nth for any odd $n>1$) powers are $0$, $1$, and $-1$. Hence, the three linear solutions are $f(x)=0$, $f(x)=x$, and $f(x)=-x$.

If instead of considering solutions over $\mathbb{R}$, we had considered solutions over $\mathbb{C}$, then there would be 2016 more linear solutions, with one corresponding to each non-real 2018th root of unity.

For nonlinear solutions (assuming AC), we don't know.