This question stems from this one, where if $f$ is continuous and $f(x) = f(x+1) = f(x+\pi)$, then $f$ must be constant. The above is shown using the density of $\mathbb{Z}+\pi\mathbb{Z}$ in $\mathbb{R}$ and the continuity of $f$. However, what jumps out at me from the question is the (seeming) incongruity of a period that is both rational and irrational.

Specifically, if $f$ is non-constant and has a least positive period, $T$, then it cannot be that $f(x)=f(x+1)=f(x+\pi)$, since that would imply that $\pi$ is rational. But, as was pointed out to me, this cannot be used to prove the above because not every periodic function has a least positive period.

So my question is, what determines if a function has a least positive period? Are there certain classes of functions that, if periodic, must have a least positive period? For instance, continuous periodic functions?

Also, what other examples are there of non-constant periodic functions that do not have a least positive period? The example given to me (and the only one given on the Wikipedia page) is the indicator function of rational numbers.


3 Answers 3


For a given function $f$, consider the set $P = \{ p \in \mathbb{R} \mid \forall x \in \mathbb{R} : f(x+p) = f(x) \}$. Clearly $0 \in P$, and if $p, q \in P$ also $p - q \in P$. That means $P$ is a subgroup of the additive group of the real numbers. Several cases can arise

  1. $P = \{0\}$. In that case $f$ is not periodic at all.
  2. $P = p\mathbb{Z}$ for some $p > 0$. Then $p$ is the least period of $f$.
  3. $P = \mathbb{R}$. This is the case for constant functions.
  4. $P$ is a dense proper subgroup of $\mathbb{R}$, as in your example of the indicator function of the rationals, where $P = \mathbb{Q}$.

No other subgroups exist. To construct an arbitrary periodic function, you can take any nontrivial subgroup $P$ and define an arbitrary function on the quotient $\mathbb{R}/P$. This fixes the whole function, since it must be constant on every coset of $P$. In case 2 the quotient can be represented as the interval $[0, p)$; for case 4 this tends to be trickier, but a slightly more interesting example would be $$ f(x) = \begin{cases} p & \text{for}\, x = q + \sqrt{p},\; q \in \mathbb{Q}, p \,\text{prime}, \\ 0 & \text{otherwise}. \end{cases} $$

  • $\begingroup$ Great answer. +1 $\endgroup$
    – Pedro
    Jul 10, 2013 at 16:46
  • $\begingroup$ So if there are both rational and irrational periods, the subgroup must be of either the third or fourth type. Such a subgroup of periods of a continuous function could not be of the fourth type, so it must be of the third, which would then imply that such a function is constant. $\endgroup$ Jul 10, 2013 at 22:28
  • 1
    $\begingroup$ That's correct. Functions of the fourth type are necessarily nowhere continuous. $\endgroup$ Jul 11, 2013 at 20:53
  • $\begingroup$ Thanks. This is very cool. $\endgroup$ Jul 13, 2013 at 4:19
  • $\begingroup$ Sorry for the bump, but are there known specific hypotheses on $f$ that grant 2. to hold? $\endgroup$ May 8, 2014 at 20:37

The other answers are very good. Here is a 1915 theorem of Burstin that also seems relevant:

If a Lebesgue measurable function $f: \mathbb{R} \rightarrow \mathbb{R}$ has arbitrarily small periods, then $f$ is constant almost everywhere.

A nice proof is given in this one page MONTHLY note of J.M. Henle. A comment at the bottom claims that Burstin's original proof was faulty.

  • $\begingroup$ When should I expect a response? $\endgroup$
    – Pedro
    Aug 17, 2013 at 23:58
  • $\begingroup$ Let me remark that Burstin's proof was incorrect. This theorem was discovered independently by Łomnicki: A. Łomnicki, O wielookresowych funkcjach jednoznacznych zmiennej rzeczywistej, Sprawozdania z Posiedzeń Towarzystwa Naukowego Warszawskiego, Wydział III Nauk Matematycznych i Przyrodniczych 11 (1918), 807–846. and sometimes it is called the Burstin–Łomnicki theorem. $\endgroup$ Dec 19, 2014 at 8:12

It seems that a nonconstant continuous periodic function must have a least positive period. First, if there are periods arbitarily close to zero, then pick your favorite point $x$ and value $f(z)$. Since the periods are getting smaller and smaller, there must be a sequence of $z_n \rightarrow x$ with $f(z_i) = f(z)$ for all $i$. Hence $f(z) = f(x)$ by continuity, violating that $f$ is nonconstant. So there is some positive lower bound $L$ to the set of periods.

If there is a sequence of periods $p_i \neq L$ converging to $L$, then $f(x + L) = f(x) = f(x + L + p_i)$ and so we have a sequence of periods converging to zero. So $L$ must be a period.


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