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 $f\colon \mathbb R^+\to\mathbb R$.

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

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