Explicit example of an additive map which is not R-linear

Question

Is there an explicit example of an additive map $\mathbb{R}^n \rightarrow \mathbb{R}^m$ which is not $\mathbb{R}$-linear? (I have mostly thought about the question when $m = n = 1$, and I don't think the general case is any easier.) I know that something like $f: \mathbb{C} \rightarrow \mathbb{C}$ which sends $f: z \mapsto \text{Real}(z)$ would be additive but not $\mathbb{C}$-linear. I also know that since $\mathbb{R}$ is a $\mathbb{Q}$-vector space, I can find some example where $1 \mapsto 1$ and $\sqrt{2} \mapsto 0$. Is there an explicit example? By explicit, I mean, given an element in the domain, there would be some procedure to decide where it maps. Thank you!

Answer 1

No.

It's not too hard to show that if $f:\Bbb R\to\Bbb R$ is additive and measurable then $f$ is $\Bbb R$-linear. And it's well known that ZF (that is, standard set theory minus AC) does not prove that there is a non-measurable subset of $\Bbb R$. (That's well known to people who know things like that, for me it's just hearsay.) Hence ZF does not suffice to show that there exists a non-linear additive $f$; the existence of such a thing requires at least some form of AC. (We haven't given an explicit definition of "explicit", but presumably a "construction" that uses AC is not "explicit"...)

Edit: There are two major gaps above; here's a proof for the one that I "should" be able to prove:

If $f:\Bbb R\to\Bbb R$ is additive and measurable then $f$ is $\Bbb R$-linear.

Proof: My first thoughts on this don't work unless we know that $f$ is locally integrable. So we start with a trick to get something not too big: Let $$g(x)=f\left(\frac x{2\pi}\right)-\frac x{2\pi}f(1).$$We're done if we can show $g=0$. Since $g$ is additive annd $g(2\pi)=0$, $g$ has period $2\pi$; hence $\chi$ also has period $2\pi$, if $$\chi(t)=e^{ig(t)}.$$

The Plan: Looking at the Fourier coefficients of $\chi$ will show that $\chi=1$, from which it is not hard to show that $g=0$.

Suppose then that $n$ is a non-zero integer. Then $ng(2\pi/n)=g(2\pi)=0$, so $g(2\pi/n)=0$, hence $$\chi(t+2\pi/n)=\chi(t).$$ Hence $$\chi(t+\pi/n)=\chi(t+2\pi/(2n))=\chi(t).$$

Now the fact that $\chi$ has period $2\pi$ shows that $$\int_0^{2\pi}\chi(t)e^{-int}=\int_{\pi/n}^{\pi/n+2\pi}\chi(t)e^{-int},$$while a change of variable $t\mapsto t-\pi/n$ shows that $$\int_{\pi/n}^{\pi/n+2\pi}\chi(t)e^{-int}=-\int_0^{2\pi}\chi(t)e^{-int},$$so $$\widehat\chi(n)=0\quad(n\in\Bbb Z, n\ne0).$$So elementary Fourier series blah blah shows that there exists a constant $c$ with $\chi=c$ almost everywhere.

Now if again $n$ is a non-zero integer then $\chi(t/n)=c$ almost everywhere. But $\chi(t/n)^n=\chi(t)$. So $c^n=c$ for every non-zero integer $n$, and since $|c|=1$ this shows that $c=1$.

Now let $G=\{t:\chi(t)=1\}$. Since $G$ has positive measure, $G+G$ contains an interval $I$. But $G+G\subset G$, so $\chi=1$ on $I$. Since $\chi$ has arbitrarily small periods $2\pi/n$ this shows that $$\chi=1.$$

So $$g(\Bbb R)\subset 2\pi\Bbb Z.$$Since $g(t/n)=g(t)/n$ it follows that $$g(t)=0.$$(Because if $g(t)=2\pi k$ with $k\ne0$ then $g(t/(2k))=\pi\notin 2\pi\Bbb Z$.)