# Sets closed by sum and solutions to the Cauchy functional equation

Let $$f:\mathbb{R}\rightarrow\mathbb{R}$$ be a solution to the Cauchy functional equation $$f(a+b)=f(a)+f(b),\quad\forall a,b\in\mathbb{R}.$$

Observe that $$A:=\{a\in\mathbb{R}:f(a)\geq 0\},\quad B:=\{b\in\mathbb{R}:f(b)< 0\}$$ provide a partition of $$\mathbb{R}$$ into two closed by sum subsets.

I would like to know if the contrary holds true. So my question is:

Given a partition of $$\mathbb{R}$$ into two subsets $$A,B$$, both closed by sum, is it true that there exists a solution $$f$$ to the Cauchy functional equation such that $$A=\{a\in\mathbb{R}:f(a)\geq 0\}$$ and $$B=\{b\in\mathbb{R}:f(b)< 0\}$$?

Suppose we have a decomposition $$\mathbb R=A\cup B$$ as described in the question and that $$0\in A$$. (Clearly, $$0\in B$$ is impossible, so we have to denote by $$A$$ the set which contains zero.)

Now, let us denote $$A_0=\{x\in A; -x\in A\} \qquad\text{and}\qquad A_1=\{x\in A; -x\in B\}.$$ It's not difficult to see that $$A=A_0\cup A_1$$ and both $$A_0$$ and $$A_1$$ are closed under addition.

Moreover, notice that we have $$-x\in A_1$$ for any $$x\in B$$. (For $$x\in B$$ we have $$-x\in A$$, otherwise we would get $$x+(-x)=0\in B$$, a contradiction.) Thus we get that $$B=-A_1=\{-x; x\in A_1\}$$.

We can also check relatively easily that any of these three sets is closed under multiplication by a positive rational number. (Let $$X$$ be one of these three sets. It is easy to see that $$x\in X$$ and $$n\in\mathbb Z^+$$ we have $$nx\in X$$. Now for any rational number $$\frac pq$$ with $$p,q>0$$ and $$y=\frac xq$$ we get that $$qy\in X$$. This also implies $$y\in X$$; if $$y$$ belonged to some of the other too sets, so would $$qy$$. Consequently $$\frac pq \cdot x = py \in X$$.)

Now we can see that both $$A_0$$ and $$\{0\}\cup A_1\cup B$$ are subspaces of $$\mathbb R$$ considered as a $$\mathbb Q$$-vector space. Then we can choose basis of these subspaces such that $$B_0\subseteq A_0$$ and $$B_1\subseteq A_1$$.

Now if we prescribe $$f[B_0]=\{0\}$$ and $$f[B_1]=\{1\}$$, this gives us an additive function $$f\colon\mathbb R\to\mathbb R$$ such that $$A_0=\{x\in\mathbb R; f(x)=0\}$$, $$A_1=\{x\in\mathbb R; f(x)>0\}$$ and $$B=-A_1=\{x\in\mathbb R; f(x)<0\}$$.

It seems that similar approach would work for $$\mathbb R^+=A\cup B$$ and $$f\colon \mathbb R^+\to\mathbb R$$.

• Very clever, thank you! – Capublanca Sep 18 '19 at 15:45