Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that whenever $a$, $b \in \mathbb{R}$ are such that $a-b \in \mathbb{Q}$, then also $f(a)-f(b)\in \mathbb{Q}$. Prove that $f$ is of the form $f(x) = qx + r$, and that $q$ is rational.
Update (16/04/2008): Solution posted! (show solution)
Consider the function $g_1(x) = f(x+1)-f(x)$, and note that it is continuous, and it only allows rational values. Thus, it has to be a constant $g_1(x) = K_1$ (because of the mean-value theorem, and that you can always find an irrational number between two rationals). We can also consider $g_{1/2} = f(x+{1/2)-f(x)$, this must also be a constant $g_{1/2}(x) = K_{1/2}$, and note that
$$K_1 = f(x+1)-f(x) =$$ $$\;= f(x+1)-f(x+1/2) + f(x+1/2)-f(x) =$$ $$\;= K_{1/2}+K_{1/2} =$$ $$\;= 2K_{1/2}$$ And we can repeat this with $K_{1/4}, K_{1/8}, ..., K_{2^{-n}}$, and get that $K_1 = 2^nK_{2^{-n}}$. So, for every diadic rational number $p2^{-n}$, we have that $f(p2^{-n}) - f(0) = pK_{2^{-n}} = p2^{-n}K_1$, ie. that $f(x)-f(0)$ is linear on diadic rationals, and by continuity it is linear everywhere. Thus $f(x) = qx+r$, and of course $f(1)-f(0) = q\in\mathbb{Q}$.
No comments:
Post a Comment