2

$f(x)$ is a function defined on $\mathbb{R}$, with fundamental period 1. Prove that for all $f$, $g(x) = f(x^2)$ is not periodic, or give a counterexample.

If the condition "fundamental" is removed, $f(x) = \begin{cases} 1, \ x \in \mathbb{A}, \\ 0, \ \mathrm{otherwise} \end{cases}$ ($\mathbb{A}$ denotes algebraic numbers) would be a counterexample. However, it doesn't have a fundamental period.

One suggests constructing $f$ that satisfies the periodicity condition, and then prove that $f(x)$ has a fundamental period:

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 $\alpha$.

But I have no idea what to do next.

AlumKal
  • 87
  • $T$ is fundamental if it is the smallest positive satisfying the identity $f(x+T)=f(x)$? – AlvinL May 28 '23 at 11:26
  • @AlvinL You are right. – AlumKal May 28 '23 at 11:28
  • 1
    Why not just pick something like a function $f(x) = 1$ on $[2k, 2k+1)$ for $k \in \mathbb{Z}$ and $0$ elsewhere? $f$ would have fundamental period $1$. Clearly, we would have that $f(x^2)$ could not be periodic, because there is no such $T> 0$ such that $g(x +T) = g(x)$ for all $x$. We can argue this using Archimedean property of reals. – mildboson May 28 '23 at 12:00
  • 2
    I think the objective is to either prove that $f(x^2)$ cannot be (fundamentally?) periodic or find an example where both $f(x)$ and $f(x^2)$ are periodic. – AlvinL May 28 '23 at 12:01
  • Two other posts asked this question before, though there's no definitive answers there it would seem: https://math.stackexchange.com/q/2436486/1104384 and https://math.stackexchange.com/q/2434582/1104384 – Bruno B May 28 '23 at 13:57
  • If $f$ is differentiable, then $f(x^2)'$ is unbounded, so $f(x^2)$ can't be periodic. So we may assume $f$ is not differentiable. – AlvinL May 28 '23 at 14:11

0 Answers0