Are there periodic functions without a smallest period?












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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?






analysis periodic-functions







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  • 4




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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.
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    – 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?
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    – abligh
    Nov 1 '14 at 9:30






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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.
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    – fonini
    Nov 3 '14 at 7:16






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    @Przemysław: I don't understand the motivation for the bounty. "This question has not received enough attention"?
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    – Rahul
    Dec 25 '14 at 4:16






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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".
    $endgroup$
    – Przemysław Scherwentke
    Dec 25 '14 at 4:19
















44












$begingroup$


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?






analysis periodic-functions







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    If you don't require the function to be continuous, every subgroup of $mathbb{R}$ is the period group of a periodic function.
    $endgroup$
    – Daniel Fischer♦
    Oct 31 '14 at 21:54






  • 2




    $begingroup$
    Perhaps more interesting: are there periodic continuous non-constant functions with no smallest period?
    $endgroup$
    – abligh
    Nov 1 '14 at 9:30






  • 2




    $begingroup$
    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.
    $endgroup$
    – fonini
    Nov 3 '14 at 7:16






  • 1




    $begingroup$
    @Przemysław: I don't understand the motivation for the bounty. "This question has not received enough attention"?
    $endgroup$
    – Rahul
    Dec 25 '14 at 4:16






  • 1




    $begingroup$
    @Rahul Alas, I can only choose one of the possibilities, not my own choice. And this would be "It is a Christmas gift".
    $endgroup$
    – Przemysław Scherwentke
    Dec 25 '14 at 4:19














44












44








44


14



$begingroup$


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?






analysis periodic-functions







share|cite|improve this question









$endgroup$




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?






analysis periodic-functions




analysis periodic-functions






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share|cite|improve this question











share|cite|improve this question




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asked Oct 31 '14 at 21:49









user28375028user28375028

1,0491714




1,0491714








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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.
    $endgroup$
    – Daniel Fischer♦
    Oct 31 '14 at 21:54






  • 2




    $begingroup$
    Perhaps more interesting: are there periodic continuous non-constant functions with no smallest period?
    $endgroup$
    – abligh
    Nov 1 '14 at 9:30






  • 2




    $begingroup$
    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.
    $endgroup$
    – fonini
    Nov 3 '14 at 7:16






  • 1




    $begingroup$
    @Przemysław: I don't understand the motivation for the bounty. "This question has not received enough attention"?
    $endgroup$
    – Rahul
    Dec 25 '14 at 4:16






  • 1




    $begingroup$
    @Rahul Alas, I can only choose one of the possibilities, not my own choice. And this would be "It is a Christmas gift".
    $endgroup$
    – Przemysław Scherwentke
    Dec 25 '14 at 4:19














  • 4




    $begingroup$
    If you don't require the function to be continuous, every subgroup of $mathbb{R}$ is the period group of a periodic function.
    $endgroup$
    – Daniel Fischer♦
    Oct 31 '14 at 21:54






  • 2




    $begingroup$
    Perhaps more interesting: are there periodic continuous non-constant functions with no smallest period?
    $endgroup$
    – abligh
    Nov 1 '14 at 9:30






  • 2




    $begingroup$
    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.
    $endgroup$
    – fonini
    Nov 3 '14 at 7:16






  • 1




    $begingroup$
    @Przemysław: I don't understand the motivation for the bounty. "This question has not received enough attention"?
    $endgroup$
    – Rahul
    Dec 25 '14 at 4:16






  • 1




    $begingroup$
    @Rahul Alas, I can only choose one of the possibilities, not my own choice. And this would be "It is a Christmas gift".
    $endgroup$
    – Przemysław Scherwentke
    Dec 25 '14 at 4:19








4




4




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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.
$endgroup$
– Daniel Fischer♦
Oct 31 '14 at 21:54




$begingroup$
If you don't require the function to be continuous, every subgroup of $mathbb{R}$ is the period group of a periodic function.
$endgroup$
– Daniel Fischer♦
Oct 31 '14 at 21:54




2




2




$begingroup$
Perhaps more interesting: are there periodic continuous non-constant functions with no smallest period?
$endgroup$
– abligh
Nov 1 '14 at 9:30




$begingroup$
Perhaps more interesting: are there periodic continuous non-constant functions with no smallest period?
$endgroup$
– abligh
Nov 1 '14 at 9:30




2




2




$begingroup$
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.
$endgroup$
– fonini
Nov 3 '14 at 7:16




$begingroup$
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.
$endgroup$
– fonini
Nov 3 '14 at 7:16




1




1




$begingroup$
@Przemysław: I don't understand the motivation for the bounty. "This question has not received enough attention"?
$endgroup$
– Rahul
Dec 25 '14 at 4:16




$begingroup$
@Przemysław: I don't understand the motivation for the bounty. "This question has not received enough attention"?
$endgroup$
– Rahul
Dec 25 '14 at 4:16




1




1




$begingroup$
@Rahul Alas, I can only choose one of the possibilities, not my own choice. And this would be "It is a Christmas gift".
$endgroup$
– Przemysław Scherwentke
Dec 25 '14 at 4:19




$begingroup$
@Rahul Alas, I can only choose one of the possibilities, not my own choice. And this would be "It is a Christmas gift".
$endgroup$
– Przemysław Scherwentke
Dec 25 '14 at 4:19










5 Answers
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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 $r$.






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  • $begingroup$
    I guess this is a consequence of the fact that there are irrationals(in fact infinitely many) between any 2 rationals?
    $endgroup$
    – 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 $r$.
    $endgroup$
    – Hagen von Eitzen
    Oct 31 '14 at 23:34










  • $begingroup$
    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 $n$.
    $endgroup$
    – kasperd
    Nov 2 '14 at 16:08






  • 1




    $begingroup$
    @MJD Your $f$ in the comment does not seem to be periodic, and in the next comment, you probably left "dense" out?
    $endgroup$
    – JiK
    Nov 3 '14 at 9:19






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    Thanks. meant rationals of the form $k2^{-n}$.
    $endgroup$
    – MJD
    Nov 3 '14 at 12:53



















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Yes, for example constant function.






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  • 13




    $begingroup$
    I cannot believe I overlooked that.
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    – user28375028
    Oct 31 '14 at 21:53






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    @PrzemysławScherwentke This appears to be the only continuous one, so it's certainly note-worthy! Although, I'm almost tempted to say that the period is 0 in this case.
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    – Cruncher
    Oct 31 '14 at 23:26








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    @Cruncher But the fundamental period is smallest positive period and every positive number is a period.
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    – Przemysław Scherwentke
    Oct 31 '14 at 23:30






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    #cruncher We don't consider zero to be a period because if we did, every function would be periodic, since $f(x+0) = f(x)$.
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    – MJD
    Nov 1 '14 at 3:00








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    @Kartik no, it should be a comment or an edit to the original question. Clearly the reply is "I mean besides constant" so it adds little value.
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    – djechlin
    Nov 2 '14 at 19:38



















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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.






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  • 6




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    Would be nice if you included a proof of the first statement.
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    – Mehrdad
    Nov 1 '14 at 19:16






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    @Mehrdad Did you read the second sentence?
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    – Najib Idrissi
    Nov 2 '14 at 14:11






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    @Mehrdad Left as an exercise for the reader. It's quite trivial.
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    – Bruno Joyal
    Nov 2 '14 at 21:17






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    Actually, you don't need to suppose $f$ is globally continuous. You only have to suppose $f$ is continuous at at least one point for it to be constant.
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    – fonini
    Nov 3 '14 at 7:26






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    @Mehrdad Check out my comment XD
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    – BCLC
    May 3 '18 at 11:20



















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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 $varepsilonto 0$, 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).






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  • $begingroup$
    Both of your examples were already given in other answers.
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    – MJD
    Nov 5 '14 at 22:27






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    I know. I'm not claiming ownership of them, but I'm sorry if it looks like I am. My motivation for posting this answer was giving the proof that there exists an "only if" property for the functions of the kind mentioned in the question.
    $endgroup$
    – fonini
    Nov 6 '14 at 3:35






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    I edited it; hope it sounds better now.
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    – fonini
    Nov 9 '14 at 21:29



















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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 $f_alpha$.






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    5 Answers
    5






    active

    oldest

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    5 Answers
    5






    active

    oldest

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    active

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    active

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    78












    $begingroup$

    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 $r$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I guess this is a consequence of the fact that there are irrationals(in fact infinitely many) between any 2 rationals?
      $endgroup$
      – Cruncher
      Oct 31 '14 at 23:25






    • 25




      $begingroup$
      @Cruncher Actually no, rather it's because $alpha$ is irrational if and only if $alpha+r$ is, for any rational $r$.
      $endgroup$
      – Hagen von Eitzen
      Oct 31 '14 at 23:34










    • $begingroup$
      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 $n$.
      $endgroup$
      – kasperd
      Nov 2 '14 at 16:08






    • 1




      $begingroup$
      @MJD Your $f$ in the comment does not seem to be periodic, and in the next comment, you probably left "dense" out?
      $endgroup$
      – JiK
      Nov 3 '14 at 9:19






    • 1




      $begingroup$
      Thanks. meant rationals of the form $k2^{-n}$.
      $endgroup$
      – MJD
      Nov 3 '14 at 12:53
















    78












    $begingroup$

    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 $r$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I guess this is a consequence of the fact that there are irrationals(in fact infinitely many) between any 2 rationals?
      $endgroup$
      – Cruncher
      Oct 31 '14 at 23:25






    • 25




      $begingroup$
      @Cruncher Actually no, rather it's because $alpha$ is irrational if and only if $alpha+r$ is, for any rational $r$.
      $endgroup$
      – Hagen von Eitzen
      Oct 31 '14 at 23:34










    • $begingroup$
      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 $n$.
      $endgroup$
      – kasperd
      Nov 2 '14 at 16:08






    • 1




      $begingroup$
      @MJD Your $f$ in the comment does not seem to be periodic, and in the next comment, you probably left "dense" out?
      $endgroup$
      – JiK
      Nov 3 '14 at 9:19






    • 1




      $begingroup$
      Thanks. meant rationals of the form $k2^{-n}$.
      $endgroup$
      – MJD
      Nov 3 '14 at 12:53














    78












    78








    78





    $begingroup$

    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 $r$.






    share|cite|improve this answer











    $endgroup$



    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 $r$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    answered Oct 31 '14 at 21:55


























    community wiki





    MJD













    • $begingroup$
      I guess this is a consequence of the fact that there are irrationals(in fact infinitely many) between any 2 rationals?
      $endgroup$
      – Cruncher
      Oct 31 '14 at 23:25






    • 25




      $begingroup$
      @Cruncher Actually no, rather it's because $alpha$ is irrational if and only if $alpha+r$ is, for any rational $r$.
      $endgroup$
      – Hagen von Eitzen
      Oct 31 '14 at 23:34










    • $begingroup$
      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 $n$.
      $endgroup$
      – kasperd
      Nov 2 '14 at 16:08






    • 1




      $begingroup$
      @MJD Your $f$ in the comment does not seem to be periodic, and in the next comment, you probably left "dense" out?
      $endgroup$
      – JiK
      Nov 3 '14 at 9:19






    • 1




      $begingroup$
      Thanks. meant rationals of the form $k2^{-n}$.
      $endgroup$
      – MJD
      Nov 3 '14 at 12:53


















    • $begingroup$
      I guess this is a consequence of the fact that there are irrationals(in fact infinitely many) between any 2 rationals?
      $endgroup$
      – Cruncher
      Oct 31 '14 at 23:25






    • 25




      $begingroup$
      @Cruncher Actually no, rather it's because $alpha$ is irrational if and only if $alpha+r$ is, for any rational $r$.
      $endgroup$
      – Hagen von Eitzen
      Oct 31 '14 at 23:34










    • $begingroup$
      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 $n$.
      $endgroup$
      – kasperd
      Nov 2 '14 at 16:08






    • 1




      $begingroup$
      @MJD Your $f$ in the comment does not seem to be periodic, and in the next comment, you probably left "dense" out?
      $endgroup$
      – JiK
      Nov 3 '14 at 9:19






    • 1




      $begingroup$
      Thanks. meant rationals of the form $k2^{-n}$.
      $endgroup$
      – MJD
      Nov 3 '14 at 12:53
















    $begingroup$
    I guess this is a consequence of the fact that there are irrationals(in fact infinitely many) between any 2 rationals?
    $endgroup$
    – Cruncher
    Oct 31 '14 at 23:25




    $begingroup$
    I guess this is a consequence of the fact that there are irrationals(in fact infinitely many) between any 2 rationals?
    $endgroup$
    – Cruncher
    Oct 31 '14 at 23:25




    25




    25




    $begingroup$
    @Cruncher Actually no, rather it's because $alpha$ is irrational if and only if $alpha+r$ is, for any rational $r$.
    $endgroup$
    – Hagen von Eitzen
    Oct 31 '14 at 23:34




    $begingroup$
    @Cruncher Actually no, rather it's because $alpha$ is irrational if and only if $alpha+r$ is, for any rational $r$.
    $endgroup$
    – Hagen von Eitzen
    Oct 31 '14 at 23:34












    $begingroup$
    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 $n$.
    $endgroup$
    – kasperd
    Nov 2 '14 at 16:08




    $begingroup$
    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 $n$.
    $endgroup$
    – kasperd
    Nov 2 '14 at 16:08




    1




    1




    $begingroup$
    @MJD Your $f$ in the comment does not seem to be periodic, and in the next comment, you probably left "dense" out?
    $endgroup$
    – JiK
    Nov 3 '14 at 9:19




    $begingroup$
    @MJD Your $f$ in the comment does not seem to be periodic, and in the next comment, you probably left "dense" out?
    $endgroup$
    – JiK
    Nov 3 '14 at 9:19




    1




    1




    $begingroup$
    Thanks. meant rationals of the form $k2^{-n}$.
    $endgroup$
    – MJD
    Nov 3 '14 at 12:53




    $begingroup$
    Thanks. meant rationals of the form $k2^{-n}$.
    $endgroup$
    – MJD
    Nov 3 '14 at 12:53











    78












    $begingroup$

    Yes, for example constant function.






    share|cite|improve this answer









    $endgroup$









    • 13




      $begingroup$
      I cannot believe I overlooked that.
      $endgroup$
      – user28375028
      Oct 31 '14 at 21:53






    • 1




      $begingroup$
      @PrzemysławScherwentke This appears to be the only continuous one, so it's certainly note-worthy! Although, I'm almost tempted to say that the period is 0 in this case.
      $endgroup$
      – Cruncher
      Oct 31 '14 at 23:26








    • 1




      $begingroup$
      @Cruncher But the fundamental period is smallest positive period and every positive number is a period.
      $endgroup$
      – Przemysław Scherwentke
      Oct 31 '14 at 23:30






    • 14




      $begingroup$
      #cruncher We don't consider zero to be a period because if we did, every function would be periodic, since $f(x+0) = f(x)$.
      $endgroup$
      – MJD
      Nov 1 '14 at 3:00








    • 3




      $begingroup$
      @Kartik no, it should be a comment or an edit to the original question. Clearly the reply is "I mean besides constant" so it adds little value.
      $endgroup$
      – djechlin
      Nov 2 '14 at 19:38
















    78












    $begingroup$

    Yes, for example constant function.






    share|cite|improve this answer









    $endgroup$









    • 13




      $begingroup$
      I cannot believe I overlooked that.
      $endgroup$
      – user28375028
      Oct 31 '14 at 21:53






    • 1




      $begingroup$
      @PrzemysławScherwentke This appears to be the only continuous one, so it's certainly note-worthy! Although, I'm almost tempted to say that the period is 0 in this case.
      $endgroup$
      – Cruncher
      Oct 31 '14 at 23:26








    • 1




      $begingroup$
      @Cruncher But the fundamental period is smallest positive period and every positive number is a period.
      $endgroup$
      – Przemysław Scherwentke
      Oct 31 '14 at 23:30






    • 14




      $begingroup$
      #cruncher We don't consider zero to be a period because if we did, every function would be periodic, since $f(x+0) = f(x)$.
      $endgroup$
      – MJD
      Nov 1 '14 at 3:00








    • 3




      $begingroup$
      @Kartik no, it should be a comment or an edit to the original question. Clearly the reply is "I mean besides constant" so it adds little value.
      $endgroup$
      – djechlin
      Nov 2 '14 at 19:38














    78












    78








    78





    $begingroup$

    Yes, for example constant function.






    share|cite|improve this answer









    $endgroup$



    Yes, for example constant function.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 31 '14 at 21:51









    Przemysław ScherwentkePrzemysław Scherwentke

    11.9k52751




    11.9k52751








    • 13




      $begingroup$
      I cannot believe I overlooked that.
      $endgroup$
      – user28375028
      Oct 31 '14 at 21:53






    • 1




      $begingroup$
      @PrzemysławScherwentke This appears to be the only continuous one, so it's certainly note-worthy! Although, I'm almost tempted to say that the period is 0 in this case.
      $endgroup$
      – Cruncher
      Oct 31 '14 at 23:26








    • 1




      $begingroup$
      @Cruncher But the fundamental period is smallest positive period and every positive number is a period.
      $endgroup$
      – Przemysław Scherwentke
      Oct 31 '14 at 23:30






    • 14




      $begingroup$
      #cruncher We don't consider zero to be a period because if we did, every function would be periodic, since $f(x+0) = f(x)$.
      $endgroup$
      – MJD
      Nov 1 '14 at 3:00








    • 3




      $begingroup$
      @Kartik no, it should be a comment or an edit to the original question. Clearly the reply is "I mean besides constant" so it adds little value.
      $endgroup$
      – djechlin
      Nov 2 '14 at 19:38














    • 13




      $begingroup$
      I cannot believe I overlooked that.
      $endgroup$
      – user28375028
      Oct 31 '14 at 21:53






    • 1




      $begingroup$
      @PrzemysławScherwentke This appears to be the only continuous one, so it's certainly note-worthy! Although, I'm almost tempted to say that the period is 0 in this case.
      $endgroup$
      – Cruncher
      Oct 31 '14 at 23:26








    • 1




      $begingroup$
      @Cruncher But the fundamental period is smallest positive period and every positive number is a period.
      $endgroup$
      – Przemysław Scherwentke
      Oct 31 '14 at 23:30






    • 14




      $begingroup$
      #cruncher We don't consider zero to be a period because if we did, every function would be periodic, since $f(x+0) = f(x)$.
      $endgroup$
      – MJD
      Nov 1 '14 at 3:00








    • 3




      $begingroup$
      @Kartik no, it should be a comment or an edit to the original question. Clearly the reply is "I mean besides constant" so it adds little value.
      $endgroup$
      – djechlin
      Nov 2 '14 at 19:38








    13




    13




    $begingroup$
    I cannot believe I overlooked that.
    $endgroup$
    – user28375028
    Oct 31 '14 at 21:53




    $begingroup$
    I cannot believe I overlooked that.
    $endgroup$
    – user28375028
    Oct 31 '14 at 21:53




    1




    1




    $begingroup$
    @PrzemysławScherwentke This appears to be the only continuous one, so it's certainly note-worthy! Although, I'm almost tempted to say that the period is 0 in this case.
    $endgroup$
    – Cruncher
    Oct 31 '14 at 23:26






    $begingroup$
    @PrzemysławScherwentke This appears to be the only continuous one, so it's certainly note-worthy! Although, I'm almost tempted to say that the period is 0 in this case.
    $endgroup$
    – Cruncher
    Oct 31 '14 at 23:26






    1




    1




    $begingroup$
    @Cruncher But the fundamental period is smallest positive period and every positive number is a period.
    $endgroup$
    – Przemysław Scherwentke
    Oct 31 '14 at 23:30




    $begingroup$
    @Cruncher But the fundamental period is smallest positive period and every positive number is a period.
    $endgroup$
    – Przemysław Scherwentke
    Oct 31 '14 at 23:30




    14




    14




    $begingroup$
    #cruncher We don't consider zero to be a period because if we did, every function would be periodic, since $f(x+0) = f(x)$.
    $endgroup$
    – MJD
    Nov 1 '14 at 3:00






    $begingroup$
    #cruncher We don't consider zero to be a period because if we did, every function would be periodic, since $f(x+0) = f(x)$.
    $endgroup$
    – MJD
    Nov 1 '14 at 3:00






    3




    3




    $begingroup$
    @Kartik no, it should be a comment or an edit to the original question. Clearly the reply is "I mean besides constant" so it adds little value.
    $endgroup$
    – djechlin
    Nov 2 '14 at 19:38




    $begingroup$
    @Kartik no, it should be a comment or an edit to the original question. Clearly the reply is "I mean besides constant" so it adds little value.
    $endgroup$
    – djechlin
    Nov 2 '14 at 19:38











    33





    +100







    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$









    • 6




      $begingroup$
      Would be nice if you included a proof of the first statement.
      $endgroup$
      – Mehrdad
      Nov 1 '14 at 19:16






    • 4




      $begingroup$
      @Mehrdad Did you read the second sentence?
      $endgroup$
      – Najib Idrissi
      Nov 2 '14 at 14:11






    • 3




      $begingroup$
      @Mehrdad Left as an exercise for the reader. It's quite trivial.
      $endgroup$
      – Bruno Joyal
      Nov 2 '14 at 21:17






    • 1




      $begingroup$
      Actually, you don't need to suppose $f$ is globally continuous. You only have to suppose $f$ is continuous at at least one point for it to be constant.
      $endgroup$
      – fonini
      Nov 3 '14 at 7:26






    • 1




      $begingroup$
      @Mehrdad Check out my comment XD
      $endgroup$
      – BCLC
      May 3 '18 at 11:20
















    33





    +100







    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$









    • 6




      $begingroup$
      Would be nice if you included a proof of the first statement.
      $endgroup$
      – Mehrdad
      Nov 1 '14 at 19:16






    • 4




      $begingroup$
      @Mehrdad Did you read the second sentence?
      $endgroup$
      – Najib Idrissi
      Nov 2 '14 at 14:11






    • 3




      $begingroup$
      @Mehrdad Left as an exercise for the reader. It's quite trivial.
      $endgroup$
      – Bruno Joyal
      Nov 2 '14 at 21:17






    • 1




      $begingroup$
      Actually, you don't need to suppose $f$ is globally continuous. You only have to suppose $f$ is continuous at at least one point for it to be constant.
      $endgroup$
      – fonini
      Nov 3 '14 at 7:26






    • 1




      $begingroup$
      @Mehrdad Check out my comment XD
      $endgroup$
      – BCLC
      May 3 '18 at 11:20














    33





    +100







    33





    +100



    33




    +100



    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 31 '14 at 21:56









    Bruno JoyalBruno Joyal

    42.8k695186




    42.8k695186








    • 6




      $begingroup$
      Would be nice if you included a proof of the first statement.
      $endgroup$
      – Mehrdad
      Nov 1 '14 at 19:16






    • 4




      $begingroup$
      @Mehrdad Did you read the second sentence?
      $endgroup$
      – Najib Idrissi
      Nov 2 '14 at 14:11






    • 3




      $begingroup$
      @Mehrdad Left as an exercise for the reader. It's quite trivial.
      $endgroup$
      – Bruno Joyal
      Nov 2 '14 at 21:17






    • 1




      $begingroup$
      Actually, you don't need to suppose $f$ is globally continuous. You only have to suppose $f$ is continuous at at least one point for it to be constant.
      $endgroup$
      – fonini
      Nov 3 '14 at 7:26






    • 1




      $begingroup$
      @Mehrdad Check out my comment XD
      $endgroup$
      – BCLC
      May 3 '18 at 11:20














    • 6




      $begingroup$
      Would be nice if you included a proof of the first statement.
      $endgroup$
      – Mehrdad
      Nov 1 '14 at 19:16






    • 4




      $begingroup$
      @Mehrdad Did you read the second sentence?
      $endgroup$
      – Najib Idrissi
      Nov 2 '14 at 14:11






    • 3




      $begingroup$
      @Mehrdad Left as an exercise for the reader. It's quite trivial.
      $endgroup$
      – Bruno Joyal
      Nov 2 '14 at 21:17






    • 1




      $begingroup$
      Actually, you don't need to suppose $f$ is globally continuous. You only have to suppose $f$ is continuous at at least one point for it to be constant.
      $endgroup$
      – fonini
      Nov 3 '14 at 7:26






    • 1




      $begingroup$
      @Mehrdad Check out my comment XD
      $endgroup$
      – BCLC
      May 3 '18 at 11:20








    6




    6




    $begingroup$
    Would be nice if you included a proof of the first statement.
    $endgroup$
    – Mehrdad
    Nov 1 '14 at 19:16




    $begingroup$
    Would be nice if you included a proof of the first statement.
    $endgroup$
    – Mehrdad
    Nov 1 '14 at 19:16




    4




    4




    $begingroup$
    @Mehrdad Did you read the second sentence?
    $endgroup$
    – Najib Idrissi
    Nov 2 '14 at 14:11




    $begingroup$
    @Mehrdad Did you read the second sentence?
    $endgroup$
    – Najib Idrissi
    Nov 2 '14 at 14:11




    3




    3




    $begingroup$
    @Mehrdad Left as an exercise for the reader. It's quite trivial.
    $endgroup$
    – Bruno Joyal
    Nov 2 '14 at 21:17




    $begingroup$
    @Mehrdad Left as an exercise for the reader. It's quite trivial.
    $endgroup$
    – Bruno Joyal
    Nov 2 '14 at 21:17




    1




    1




    $begingroup$
    Actually, you don't need to suppose $f$ is globally continuous. You only have to suppose $f$ is continuous at at least one point for it to be constant.
    $endgroup$
    – fonini
    Nov 3 '14 at 7:26




    $begingroup$
    Actually, you don't need to suppose $f$ is globally continuous. You only have to suppose $f$ is continuous at at least one point for it to be constant.
    $endgroup$
    – fonini
    Nov 3 '14 at 7:26




    1




    1




    $begingroup$
    @Mehrdad Check out my comment XD
    $endgroup$
    – BCLC
    May 3 '18 at 11:20




    $begingroup$
    @Mehrdad Check out my comment XD
    $endgroup$
    – BCLC
    May 3 '18 at 11:20











    11












    $begingroup$

    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 $varepsilonto 0$, 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).






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Both of your examples were already given in other answers.
      $endgroup$
      – MJD
      Nov 5 '14 at 22:27






    • 1




      $begingroup$
      I know. I'm not claiming ownership of them, but I'm sorry if it looks like I am. My motivation for posting this answer was giving the proof that there exists an "only if" property for the functions of the kind mentioned in the question.
      $endgroup$
      – fonini
      Nov 6 '14 at 3:35






    • 1




      $begingroup$
      I edited it; hope it sounds better now.
      $endgroup$
      – fonini
      Nov 9 '14 at 21:29
















    11












    $begingroup$

    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 $varepsilonto 0$, 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).






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Both of your examples were already given in other answers.
      $endgroup$
      – MJD
      Nov 5 '14 at 22:27






    • 1




      $begingroup$
      I know. I'm not claiming ownership of them, but I'm sorry if it looks like I am. My motivation for posting this answer was giving the proof that there exists an "only if" property for the functions of the kind mentioned in the question.
      $endgroup$
      – fonini
      Nov 6 '14 at 3:35






    • 1




      $begingroup$
      I edited it; hope it sounds better now.
      $endgroup$
      – fonini
      Nov 9 '14 at 21:29














    11












    11








    11





    $begingroup$

    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 $varepsilonto 0$, 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).






    share|cite|improve this answer











    $endgroup$



    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 $varepsilonto 0$, 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).







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 6 '14 at 5:14

























    answered Nov 3 '14 at 6:47









    foninifonini

    1,78911038




    1,78911038












    • $begingroup$
      Both of your examples were already given in other answers.
      $endgroup$
      – MJD
      Nov 5 '14 at 22:27






    • 1




      $begingroup$
      I know. I'm not claiming ownership of them, but I'm sorry if it looks like I am. My motivation for posting this answer was giving the proof that there exists an "only if" property for the functions of the kind mentioned in the question.
      $endgroup$
      – fonini
      Nov 6 '14 at 3:35






    • 1




      $begingroup$
      I edited it; hope it sounds better now.
      $endgroup$
      – fonini
      Nov 9 '14 at 21:29


















    • $begingroup$
      Both of your examples were already given in other answers.
      $endgroup$
      – MJD
      Nov 5 '14 at 22:27






    • 1




      $begingroup$
      I know. I'm not claiming ownership of them, but I'm sorry if it looks like I am. My motivation for posting this answer was giving the proof that there exists an "only if" property for the functions of the kind mentioned in the question.
      $endgroup$
      – fonini
      Nov 6 '14 at 3:35






    • 1




      $begingroup$
      I edited it; hope it sounds better now.
      $endgroup$
      – fonini
      Nov 9 '14 at 21:29
















    $begingroup$
    Both of your examples were already given in other answers.
    $endgroup$
    – MJD
    Nov 5 '14 at 22:27




    $begingroup$
    Both of your examples were already given in other answers.
    $endgroup$
    – MJD
    Nov 5 '14 at 22:27




    1




    1




    $begingroup$
    I know. I'm not claiming ownership of them, but I'm sorry if it looks like I am. My motivation for posting this answer was giving the proof that there exists an "only if" property for the functions of the kind mentioned in the question.
    $endgroup$
    – fonini
    Nov 6 '14 at 3:35




    $begingroup$
    I know. I'm not claiming ownership of them, but I'm sorry if it looks like I am. My motivation for posting this answer was giving the proof that there exists an "only if" property for the functions of the kind mentioned in the question.
    $endgroup$
    – fonini
    Nov 6 '14 at 3:35




    1




    1




    $begingroup$
    I edited it; hope it sounds better now.
    $endgroup$
    – fonini
    Nov 9 '14 at 21:29




    $begingroup$
    I edited it; hope it sounds better now.
    $endgroup$
    – fonini
    Nov 9 '14 at 21:29











    1












    $begingroup$

    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 $f_alpha$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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 $f_alpha$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $f_alpha$.






        share|cite|improve this answer









        $endgroup$



        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 $f_alpha$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 10 '18 at 14:21









        Connor HarrisConnor Harris

        4,430724




        4,430724






























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