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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 $P_f\cap\mathbb Q=\emptyset$?

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closed as off topic by Qiaochu Yuan, Andres Caicedo, Anton Petrunin, Bill Johnson, Andy Putman Jan 21 '12 at 1:45

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2 Answers 2

up vote 7 down vote accepted

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 $\sqrt 3$.

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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 $P_f\cap\mathbb Q=\emptyset$?

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2  
How about $g(x)=f(ax)$ where $f$ is Dirichlet's function and $a$ is irrational? –  Anthony Quas Jan 19 '12 at 0:48
    
You actually want $P_f \cap \mathbb{Q} = \{ 0 \}$. –  Qiaochu Yuan Jan 19 '12 at 0:55
    
great, thanks a lot :) –  alberto.bosia Jan 19 '12 at 2:05
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