# non-affine functions

it is obvious that if $f$ is an affine function, then $f$ has this property: there exist two function $g$ and $h$ such that $f(t+s)=g(t)+h(s)$ for all $t,s \in\mathbb{R}$. My question is: is there any non-affine function which has this property. In other words, is this property a characterisation of affine functions or not?

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We can consider $\mathbb R$ as a vector space over the field $\mathbb Q$. (Assuming the axiom of choice,) we can extend the $\mathbb Q$-lienar independent set $\{1, \sqrt 2\}$ to a $\mathbb Q$-basis of $\mathbb R$ and find a $\mathbb Q$-linear function $f\colon\mathbb R\to\mathbb R$ with $f(1)=1$ and $f(\sqrt2)=0$. Letting $g=h=f$, we see that $f(t+s)=f(t)+f(s) = g(t)+h(s)$ for all $s,t\in \mathbb R$ by $\mathbb Q$-linearity of $f$. Of course, $f$ is not affine, as $0=f(\sqrt2)-f(0)\ne\sqrt 2\cdot(f(1)-f(0))=\sqrt2$.

On the other hand, if we reduce to rationals, we have the following:

Proposition: Let $f,g,h\colon \mathbb Q\to \mathbb R$ be functions with $f(t+s)=g(t)+h(s)$ for all $s,t\in\mathbb Q$. Then there are $\alpha,\beta\in\mathbb R$ such that $f(x)=\alpha+\beta x$ for all $x\in\mathbb Q$.

Proof: Let $\alpha=f(0)$, $\beta=f(1)-f(0)$. For $a,b\in \mathbb Q$ we have $$\tag1f(a+b)=g(a)+h(b)=(f(a+0)-h(0))+(f(0+b)-g(0))\\=f(a)+f(b)-f(0).$$ Let $S=\{x\in\mathbb Q\mid f(x)=\alpha+\beta x\}$. By induction on $n$, we see from $(1)$ that $$\tag2f(a+nb) = f(a)+n\cdot(f(b)-f(0))$$ for $a,b\in\mathbb Q$, $n\in \mathbb N_0$, hence $\mathbb N_0\subseteq S$. Next, for a positive fraction $\frac mn$ with $m,n\in\mathbb N$, we see $f(m)=f(0+n\cdot\frac mn)=f(0)+n\cdot(f(\frac mn)-f(0))$, hence $\alpha+\beta m=\alpha+n\cdot(f(\frac mn)-\alpha)$ because $m,n\in S$ and finally $f(\frac mn)=\alpha+\frac mn \beta$, i.e. $\mathbb Q_{\ge0}\subseteq S$. As last step, if $a\in \mathbb Q$ is negative, then $b:=-a$ is positive, hence $\in S$ and $(1)$ yields $f(0)=f(a)+f(b)-f(0)$, i.e. $f(a)=2f(0)-f(b)=2\alpha-(\alpha-a\beta)=\alpha+a\beta$ and ultimately $S=\mathbb Q$, i.e. $f(x)=\alpha+\beta x$ for all $x\in \mathbb Q$.$_\blacksquare$

Corollary: Let $f,g,h\colon \mathbb R\to \mathbb R$ be functions with $f(t+s)=g(t)+h(s)$ for all $s,t\in\mathbb R$. Also assume that $f$ is continuous. Then there are $\alpha,\beta\in\mathbb R$ such that $f(x)=\alpha+\beta x$ for all $x\in\mathbb R$.

Proof: Observe that $(1)$ holds for all $a,b\in\mathbb R$ just as in the proof of the proposition. By theproposition, we have $f(x)=\alpha+\beta x$ at least for $x\in \mathbb Q$. Let $x\in \mathbb R$ be arbitrary and select a sequence $x_n\to x$ with $x_n\in \mathbb Q$. Then by continuity of $f$ $$f(x)=\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}\alpha+\beta x_n=\alpha+\beta x._\blacksquare$$

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