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I have the definition of a periodic function in my Real Analysis "textbook" given by my professor and below is a remark that states "A function can be periodic without its period being defined"

The definition given is : Let E ⊂ R and t ∈ R*+ (positive real numbers). A function of f : E → R is t-periodic if, for all x ∈ E, we have x + t ∈ E and f(x + t) = f(x). The smallest t > 0 with this property, if it exists, is called the period of f.

What does it mean for the period to be undefined? And what are some examples of periodic functions with an undefined period? (if the examples could be a bit elaborated that would really really help)

Thank you so much in advance and if you have any good sources for Real Analysis to help with the understanding of concepts that would be amazing. I'm taking my first real analysis course and having issues wrapping my head around certain definitions because I'm a visual learner and it's hard for me to learn from definitions.

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  • $\begingroup$ Could you please put the definition of a "periodic function" given by your professor? That would help. $\endgroup$ Dec 30, 2022 at 11:08
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    $\begingroup$ Would be nice if you could state the definition of periodic function used in your textbook. Without it, I can only guess that for example constant functions $f(t)=c$ are periodic but don't have a period. $\endgroup$
    – NeitherNor
    Dec 30, 2022 at 11:09
  • $\begingroup$ Yes, sorry. The definition given is : Let E ⊂ R and t ∈ R*+ (positive real numbers). A function of f : E → R is t-periodic if, for all x ∈ E, we have x + t ∈ E and f(x + t) = f(x). The smallest t > 0 with this property, if it exists, is called the period of f. $\endgroup$
    – Cup of Tea
    Dec 30, 2022 at 11:25
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    $\begingroup$ Presumably the issue is that the "period" has been defined as the minimal positive value $p$ such that $f(x+p)=f(x)$ and that minimum may not exist. As has been remarked, constants are certainly periodic but there is no minimal period. $\endgroup$
    – lulu
    Dec 30, 2022 at 11:26
  • $\begingroup$ @lulu Very good comment ! If there is no such smallest $p$ , we should not speak of a periodical function at all despite of the usual deifinition screening them as "periodical". $\endgroup$
    – Peter
    Dec 30, 2022 at 11:45

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Based on the standard definitions for periodic functions and period, a function $f:\mathbb{R}\to\mathbb{R}$ is said to be periodic if for some $p>0$, for every $x\in\mathbb{R}$, $$f(x+p) = f(x).$$ Among such $p>0$, if there is a smallest one, then we call the smallest $p$ as the period of $f$. But there can be no smallest such number. For example, let $f(x) = 0$. Then for any $p>0$ we have $f(x+p) = 0 = f(x)$. More complicated example will be the Dirichlet function, $$\chi_{\mathbb{Q}}(x) = \begin{cases} 1 &x\in\mathbb{Q}\\ 0 &x\notin\mathbb{Q}\end{cases}.$$ This is periodic since for any positive rational $p$, $f(x+p)=f(x)$(take case when $x$ is rational or irrational). So there's no smallest such $p$, and its period is not defined. This is the case your professor is talking about.

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  • $\begingroup$ THANK YOU SO MUCH!! $\endgroup$
    – Cup of Tea
    Dec 30, 2022 at 11:22
  • $\begingroup$ Another nice example are functions with empty domain $E$. Trivially, this function is t-periodic for all $t$, and we again have the "problem" that no smallest $t$ exists... $\endgroup$
    – NeitherNor
    Dec 30, 2022 at 11:35
  • $\begingroup$ @ZAhmed: But the question was explicitly about the statement "A function can be periodic without its period being defined", i.e. periodic functions, so talking about non periodic functions is kind of besides the point. $\endgroup$ Dec 30, 2022 at 12:15

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