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Let $f$ be a periodic function such that it has a (non-zero) fundamental period (Smallest nonzero period). Can $f(x^2)$ also be periodic?

So the constant functions and Dirichlet function are not examples we want here.

If $f$ is continuous, then it is impossible, because $f(x^2)$ fails to be uniformly continuous.

Ice sea
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    I believe the same topic is being discussed at https://math.stackexchange.com/questions/2434582/a-problem-about-periodic-functions/2434646#2434646 – An aedonist Sep 19 '17 at 20:10
  • I believe the answer to be no, and I think there might be a "geometrical" argument. If $f(x)$ has fundamental period $p$, we can see it as a map $S^1(p):=\mathbb{R}/p\mathbb{Z}\to\mathbb{R}$. If $f(x^2)$ has fundamental period $q$, then it would be a map $S^1(q)\to\mathbb{R}$. But the map $x\mapsto x^2$ doesn't give a well defined map $S^1(p)\to S^1(q)$ and this should imply some kind of contradiction. I couldn't make this argument formal yet. – Daniel Robert-Nicoud Sep 19 '17 at 20:11

1 Answers1

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I think the answer is yes. Choose a transcendental constant $\alpha$. Consider the equivalence relation $\sim$ on $\mathbb R$ generated (as a transitive closure) by $x \sim x \pm 1$ and $x^2 \sim (x\pm \alpha)^2$. Note that all equivalence classes are countable. Take $f(x) = 1$ if $x \sim 0$ and $f(x) = 0$ otherwise. Then $f$ has period $1$ and $f(x^2)$ has period $\alpha$. I'm pretty sure that there is no non-integer rational $r$ such that $0 \sim r$, implying that $1$ is the fundamental period of $f$, as any chain of equivalences $x \sim x \pm 1$ or $x^2 \sim (x \pm \alpha)^2$ starting with $0$ and ending at $r$ would imply a polynomial identity for $\alpha$.

Robert Israel
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  • Sorry. I didn't follow. You need to show that this equivalent classes is not trivial. Could you give more details? Thank you – Ice sea Sep 19 '17 at 20:35
  • I think $f(x^2)$ is not periodic – Ice sea Sep 19 '17 at 21:14
  • Very nice idea with using trancendental $\alpha$. But you only excluded rational fundamental periods $<1$. There can still be irrational ones. – M. Winter Sep 19 '17 at 21:17
  • @Icesea Yes it is. $$f(x^2)=1\Leftrightarrow x^2\sim 0\Leftrightarrow (x\pm\alpha)^2\sim 0\Leftrightarrow f((x\pm\alpha)^2)=1.$$ – M. Winter Sep 19 '17 at 21:20
  • @M.Winter I'm confused.Why $f(x^2)=1$ implies $x^2\sim 0$. Is not it that $x^2=n$ for some integer? – Ice sea Sep 19 '17 at 21:37
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    @Icesea Thats how $f$ was defined. Robert chose to define $f(x)$ to be one whenever $x$ and zero are in the same equivalence class. – M. Winter Sep 20 '17 at 06:27
  • @M.Winter Could you explain more about the equivalent classes. I'm quite confuse. Does $x\sim x\pm 1$ mean that $x\sim y$ if and only if $x-y=n $ and What does $x^2 \sim (x\pm \alpha)^2$ mean? Does it mean $x\sim y$ if and only if $x^2=(y+n\alpha)^2$. In this case it doesn't work. If it means $y \sim x$ if and only if $x=(y+n\alpha)^2$. I don't think it is an equivalent class at all. – Ice sea Sep 20 '17 at 10:38
  • Let us continue this discussion in chat. – M. Winter Sep 20 '17 at 10:40
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    To elaborate on the "I'm pretty sure" part: Suppose $0\sim r$, i.e., there is a finite sequence $x_0=0, \ldots, x_n=r$ with $x_{k+1}=x_k\pm1$ or $x_{k+1}^2=(x_k\pm \alpha)^2$. By induction on $n$, the steps give us a polynomial $f\in \Bbb Z[X,Y,Z]$ with $f(x_0,x_n,\alpha)=0$ such that neither $f(x_0,Y,\alpha)$ nor $f(X,x_n,\alpha)$ is trivial. If $r\in\Bbb Q$, then $f(0,r,Z)\in\Bbb Q[Z]$ is a rational polynomial with root $\alpha$, hence is the zero polynomial. We conclude that everything works also with $\alpha=0$. But then $x_{k+1}=\x_k\pm1$ or $x_{k+1}=\pm x_k$ and so $r=x_n\in\Bbb Z$. – Hagen von Eitzen Sep 23 '17 at 16:35