Let $f:\mathbb R\to\mathbb R$ be a periodic function. We say $f$ is without minimum period if, $\forall t$ such that $f(x+t)=f(x)\forall x$, there is a $t'$ such that $0 Is there a periodic function without minimum period such that$P_f\cap\mathbb Q=\emptyset$? - ## closed as off topic by Qiaochu Yuan, Andres Caicedo, Anton Petrunin, Bill Johnson, Andy PutmanJan 21 2012 at 1:45 ## 2 Answers I guess what you can get as a set of periods is exactly any additive subgroup of the reals. Certainly the periods are closed under addition. On the other hand, for any subgroup$G$or$\mathbb R$, mimic the Dirichlet function by defining$f$to be the indicator function of$G$. Here the set of periods is exactly$G$itself. Another example would be the additive subgroup generated by$\sqrt 2$and$\sqrt 3$. - this is the complete question, sorry. Let$f:\mathbb R\to\mathbb R$be a periodic function. We say$f$is without minimum period if,$\forall t$such that$f(x+t)=f(x)\forall x$, there is a$t'$such that$0 < t' < t$and$f(x+t')=f(x)\forall x$. The easiest examples of such functions are constant functions. Dirichlet's function ($1$if$x\in\mathbb Q$and$0$if$x\not\in\mathbb Q$) too is a periodic function without minimum period, cause$\forall q\in\mathbb Q$it's true that$f(x+q)=f(x)$. Let's say that$P_f$is the set of all possible periods of$f$. (example:$P_{constant}=\mathbb R$,$P_{dirichlet's}=\mathbb Q$) Is there a periodic function without minimum period such that$P_f\cap\mathbb Q=\emptyset$? - How about$g(x)=f(ax)$where$f$is Dirichlet's function and$a$is irrational? – Anthony Quas Jan 19 2012 at 0:48 You actually want$P_f \cap \mathbb{Q} = \{ 0 \}\$. – Qiaochu Yuan Jan 19 2012 at 0:55
great, thanks a lot :) – alberto.bosia Jan 19 2012 at 2:05