Let $D\subset\mathbb R$ and let $T\in(0,+\infty)$. A function $f\colon D\longrightarrow\mathbb R$ is called a periodic function with period $T$ if, for each $x\in D$, $x+T\in D$ and $f(x+T)=f(x)$.

If $D\subset\mathbb R$ and $f\colon D\longrightarrow\mathbb R$ is continuous and periodic, must there be, among all periods of $f$, a minimal one?

Questions like this one have been posted here before, but in each case, as far as I can see, the domain of $f$ was $\mathbb R$, which implies that the set $P$ of periods, together with $0$ and $-P$, is a subgroup of $(\mathbb{R},+)$. Using that (together with continuity), it is easy to see that a minimal period must exist indeed. But I don't know whether it is true or not in the general case.

  • $\begingroup$ Maybe I misunderstand, but I feel like the argument carries over pretty cleanly. Suppose there is no minimal one. Because periods get arbitrarily small, $D$ must be dense in a positive ray (starting at any $x \in D$), and in fact $f$ is constant on this dense set. Thus $f$ is constant. $\endgroup$ – Mees de Vries Oct 11 '18 at 10:37
  • $\begingroup$ @MeesdeVries How do you go from “periods get arbitrarily small” to “$D$ must be dense in a positive ray”? Doesn't that assume that if $T$ and $T^\star$ are periods, with $T>T^\star$, then $T-T^\star$ is a period too? Is that obvious in this context? $\endgroup$ – José Carlos Santos Oct 11 '18 at 10:40
  • $\begingroup$ If $T_n$ is a sequence of periods of $f$ which tends to zero, and $x \in D$, then the set of points $x'$ where $f(x) = f(x')$ -- so in particular $x' \in D$ -- includes at least $\{x + kT_n \mid k, n \in \mathbb N\}$, which is certainly dense in $[x, \infty)$. Is this wrong? $\endgroup$ – Mees de Vries Oct 11 '18 at 10:44
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    $\begingroup$ @MeesdeVries It looks right, but you are assuming that “there is no minimal period” is equivalent to “there is a sequence of periods which converges to $0$”. Why do you think so? $\endgroup$ – José Carlos Santos Oct 11 '18 at 10:47
  • $\begingroup$ Ah, thank you for the correction. I see where the difficulty lies. Interesting question! $\endgroup$ – Mees de Vries Oct 11 '18 at 10:48

This has the air of exploiting a hole left in the question parameters, but here comes.

Let $D=\{0\}\cup(1,\infty)$ and let $f(x)$ be a constant function. Then $T$ will be a period if and only if $T+D\subseteq D$. In particular:

  • every $T>1$ is a period, but
  • $T=1$ is not a period (and there cannot be smaller periods $\le 1$),
  • so there is no smallest period.

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