Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

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 $f$ $$ f(x)=\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}\alpha+\beta x_n=\alpha+\beta x._\blacksquare$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.