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The Wikipedia page for periodic functions states that the smallest positive period $P$ of a function is called the fundamental period of the function (if it exists). I was intrigued by the condition that the function actually has a smallest period, so my question is, what properties of a function would cause it to be periodic but not have a smallest period?

user28375028
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    If you don't require the function to be continuous, every subgroup of $\mathbb{R}$ is the period group of a periodic function. – Daniel Fischer Oct 31 '14 at 21:54
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    Perhaps more interesting: are there periodic continuous non-constant functions with no smallest period? – abligh Nov 01 '14 at 09:30
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    No, there doens't exist periodic continuous non-constant functions with no smallest period. Actually, it doesn't end there: there doesn't exist periodic "somewhere continous" non-constant functions with no smallest period. (That is, if you want a periodic function to have no smallest period, the you must make it either constant or nowhere continuous). There is a proof below. – fonini Nov 03 '14 at 07:16
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    @Przemysław: I don't understand the motivation for the bounty. "This question has not received enough attention"? –  Dec 25 '14 at 04:16
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    @Rahul Alas, I can only choose one of the possibilities, not my own choice. And this would be "It is a Christmas gift". – Przemysław Scherwentke Dec 25 '14 at 04:19
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    @PrzemysławScherwentke I still don't understand your motivation. Can you please explain it? – fonini Dec 30 '14 at 05:49
  • I'm just going to leave a link to a related question on Quora. – PinkyWay Aug 21 '20 at 11:48

5 Answers5

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For a nontrivial example, consider the Dirichlet function, which has $$\delta(x) = \begin{cases}0 & \text{ if $x$ is rational}\\1 & \text{ if $x$ is irrational}\end{cases}$$

Then $\delta(x)$ is periodic with period $r$ for every rational number $r$.

MJD
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  • I guess this is a consequence of the fact that there are irrationals(in fact infinitely many) between any 2 rationals? – Cruncher Oct 31 '14 at 23:25
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    @Cruncher Actually no, rather it's because $\alpha$ is irrational if and only if $\alpha+r$ is, for any rational $r$. – Hagen von Eitzen Oct 31 '14 at 23:34
  • I noticed that this construction is constant, if you restrict the domain to only the rational numbers. I then wondered if there exist any construction, which is non-constant even if the domain is only the rational numbers. Turns out there does. One could define $f(x)$ to be $0$ if $x$ can be written with a finite number of decimals, and $1$ if $x$ requires infinitely many (repeating) decimals. This would be periodic with period $10^{-n}$ for any $n$. – kasperd Nov 02 '14 at 16:08
  • Yes, you can do a similar thing by letting $f(x) = 1$ for rationals of the form $2^{-n}$ and $f(x) = 0$ for rational numbers not of that form. – MJD Nov 02 '14 at 16:18
  • As other comments have pointed out, if $S$ is any additively-closed subset of $\Bbb R$ (that is, $a+b\in S$ whenever $a\in S$ and $b\in S$) then the characteristic function of $S$ has no smallest period. In my example in this post, $S=\Bbb Q$; in your example it is the set of terminating fractions. – MJD Nov 02 '14 at 21:30
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    @MJD Your $f$ in the comment does not seem to be periodic, and in the next comment, you probably left "dense" out? – JiK Nov 03 '14 at 09:19
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    Thanks. meant rationals of the form $k2^{-n}$. – MJD Nov 03 '14 at 12:53
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Yes, for example constant function.

Przemysław Scherwentke
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In fact, a continuous function of a real variable having arbitrarily small periods is necessarily a constant. Indeed, the set of periods is then a dense additive subgroup of the real line, and the function is constantly equal to its value at any point.

Bruno Joyal
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You can show that if a periodic function is continuous at at least one point, and doesn't have a minimum period, then it is constant.

In other words: in order for a function to have arbitrarily small periods, it must be either constant or everywhere discontinuous.

The reasoning is simple: if $f$ doesn't have a smallest period, then the image of $f$ must be the same in every open interval. Therefore, no matter how small is $\varepsilon$, the oscillation of $f$ in any $(a, a+\varepsilon) $ (defined as $\sup f - \inf f$ restricted to that interval) is going to be the same as the global oscillation. Since the oscillation is constant as $\varepsilon\to 0$, and we know that the oscillation should converge to $0$ if we want $f$ to be continuous at $a$, we reach the conclusion that $f$ is nowhere continuous (allowing that $f$ is constant in the case that this global constant oscillation is $0$).

It is easy to obtain "examples" of constant functions. As MJD's answer says, an example of a non-constant function with arbitrarily small periods is the indicator function of the rationals ($1$ at the rationals and $0$ at the irrationals). As noted by Hagen von Eitzen in MJD's post, this function doesn't accept a smallest period because a real number $x$ is rational if and only if $x+r$ is also rational for any rational $r$ (hence any rational number is a period of this function).

fonini
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Here's a large family of nontrivial examples with an uncountable set of periods. Select a Hamel basis $\{x_\alpha\}_{\alpha \in A}$ of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and let $f_\alpha: \mathbb{R} \to \mathbb{Q}$ be the extraction of the coefficient of some fixed $x_\alpha$. Then any rational multiple of any element of the Hamel basis other than $x_\alpha$ itself is a period of $f_\alpha$.

Connor Harris
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