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

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.

• You have tagged this as "contest-math". Can you please provide a link to the contest that this problem comes from? – Xander Henderson Aug 16 '19 at 19:14
• @XanderHenderson Oh sorry, but I don't know if it comes from a contest. My source is that it is problem 10 of the proposed problems of this blog entry how-did-i-get-here.com/61 – A123 Aug 16 '19 at 19:51
• I have edited your question to include the link (in the future, you should provide such references yourself---these provide valuable context). Given that this problem is not from a contest, is there any particular reason that you have chosen to tag it as a contest problem? – Xander Henderson Aug 16 '19 at 23:29
• @XanderHenderson Similar problems (e.g. replacing 2019) have appeared as training exercises before. I can't link one off the top of my head, but I think the tag is perfectly fine here, even if it does not come from an explicit contest. – user574848 Aug 17 '19 at 0:35

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).

• That's really cool, thanks!! – A123 Aug 16 '19 at 21:33
• @metamorphy I still seem to be missing something. Why cannot $f$ satisfy something like e.g., $f(e)=e^2$? [where $e$ is the base of the natural logarithm]. – Mike Aug 17 '19 at 1:53
• @Mike If $f(x+y)=f(x)+f(y)$ and $x\geqslant 0\implies f(x)\geqslant 0$, then $f$ is (necessarily) linear. The Wikipedia article suggests this as an easy consequence of the fact that nonlinear solutions must have graphs (everywhere) dense in $\mathbb{R}^2$. – metamorphy Aug 17 '19 at 9:07
• I think I see it now: the condition $f(x^2)=f(1)(f(x)^2$ implies that for all positive $y$, the sign of $f(y)$ has to be the same as the sign of $f(1)$. Which implies that there is at least one quadrant of $\mathbb{R}^2$ not covered by the graph of $f$, which implies (via the reasoning in the link you posted) that $f$ has to be linear. Thank you for explaining @metamorphy ! – Mike Aug 17 '19 at 16:01

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.

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 $$f \equiv 0$$.

• Can you please show the continuity? – Virtuoz Aug 16 '19 at 18:49
• I agree that there is more work to be done. If $f$ were a function on $\mathbb{Q}$ instead of $\mathbb{R}$ then this would be much easier. But e.g., just because one has $f(e)$ does not mean that one has $f(e^2)$. – Mike Aug 16 '19 at 19:48
• For example, let $X$ be the subset of $\mathbb{R}$ of the form $\{a +be+ce^2\}$; $a,b,c \in \mathbb{Q}$. Consider $f$ defined as follows: $f(1)=1$; $f(e)=0$; $f(e^2)=100000$. Then this could be extended to an additive real-valued function on $X$. Now there may very well be "something else" happening for the entire real line $\mathbb{R}$ but if so what is it. – Mike Aug 16 '19 at 19:51