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I'm trying to prove something about periodic functions and I'd need someone to tell me if what I wrote is right!

If $f$ is a periodic function with fundamental period $\tau$. Then, all periods of $f$ are in the form $k\tau$, with $k$ in $\mathbb{Z}$ Obviously, if $\tau$ is the fundamental period of $f$, $k\tau$ is too: \begin{equation*} f(x+k\tau)=f(x+\underbrace{\tau+\tau+\dots+\tau}_{k \text{ times}})=f(x+\underbrace{\tau+\tau+\dots+\tau}_{k-1 \text{ times}})=\dots=f(x+\tau)=f(x) \end{equation*} Viceversa, let's suppose that a period different from $k\tau$ exists, and let's call it $\beta>\tau$. Hypothesis tell us that $\frac{\beta}{\tau}\neq k$, and so $\frac{\beta}{\tau}$ is not an integer. We have two possibilities:

  • $\frac{\beta}{\tau}$ is rational, but not integer;
  • $\frac{\beta}{\tau}$ is irrational.

In the first case, it could be:

  • $\beta$ and $\tau$ are two irrational numbers that gives a rational when divided: $\beta=s\tau$ with $s\in\mathbb{Q}\verb|\| \mathbb{Z} $
  • $\beta$ and $\tau$ are two integers without factors in common.

Let's analyze case by case

If $\beta=s\tau$ with $s$ like I said, $\beta$ is not a period. In fact, even if $\tau$ is a period of $f$, $\beta=s\tau$ isn't anymore. We can consider $\sin(\pi+2\pi)=\sin(\pi)=0\neq\sin(\pi+\frac{1}{3}2\pi)$ for a counterexample.

If $\beta$ and $\tau$ are two integers without factors in common, we cand divide $\beta$ by $\tau$, finding the quotient $q$ and the remainder $r$ so that: \begin{equation*} \beta=q\cdot\tau+r \end{equation*} with $0<r<\tau$. Now, for hypothesis $\beta$ is a period, and so: \begin{equation*} f(x)=f(x+\beta)=f(x+q\cdot\tau+r)=f(x+r) \end{equation*} That means that $r$ too is a period, but this is absurd because $\tau$ is the minimum period and $r<\tau$.

If $\frac{\beta}{\tau}=t$ with $t$ irrational, $\beta=\tau t$ isn't a period anymore (it's possible to create a counterexample like done before). So, if $\beta\neq k\tau$, $\beta$ isn't a period, therefore $\beta=k\tau$, with $k\in\mathbb{Z}$

Now let's use this result to show that: If $f$ and $g$ are periodic functions with fundamental period $s$ and $t$ respectively, then if:

  • $i)$ $\frac{s}{t}$ is a rational number $\neq1$, $f+g$, $fg$, $f/g$ are periodic functions with period $mcm(s,t)$.
  • $ii)$ $s=t$, $f+g$, $fg$, $f/g$ are periodic functions and their period is $\leq s=t$;
  • $iii)$ $\frac{s}{t}$ is irrational, $f+g$, $fg$, $f/g$ are not periodic functions.

Here we extend the notion of $mcm$ to real number as it follow: \begin{equation*} z=mcm(\alpha,\beta) \iff \exists m,n \in \mathbb{Z} : \begin{cases} \alpha=m\cdot z\\ \beta=n\cdot z \end{cases} \end{equation*}

  • $i)$ If $\frac{s}{t}=k$ is rational, we have $\frac{s}{t}=\frac{m}{n} \implies sn=mt$, with $s$ and $m$ integers. $mcm(s,t)=sn=mt$.

Let's see if $sn$ is a period for $f+g$, $fg$ e $f/g$. We have: \begin{equation*} \begin{aligned} &(f+g)(x+m)=f(x+m)+g(x+m)=f(x+n\cdot kt)+g(x+l\cdot t)=f(x)+g(x)=(f+g)(x)\\ &(fg)(x+sn)=f(x+sn)g(x+sn)=f(x+sn)g(x+mt)=f(x)g(x) \\ &\frac{f(x+sn)}{g(x+mt)}=\frac{f(x)}{g(x)} \end{aligned} \end{equation*} So $sn=mt$ is a period for $f+g$, $fg$, $f/g$. We have to show that $sn=mt$ is the minimum of the positive periods of $f$: suppose that $\alpha$ is the period of $f$, so $\alpha\leq sn$. By the precedent proposition, we know that $sn$ is in the form $\alpha k_1$, and so: \begin{equation*} k_1=\frac{sn}{\alpha} \end{equation*} $k_1$ is an integer, so $\frac{sn}{\alpha}$ must be an integer. That happens if $sn=mt=\alpha$ (in this case,we conclude), or $\frac{s}{\alpha}$ is integer: $s=k_2\cdot \alpha$.

But that means that $s$ is a period for $f+g, fg, f/g$. We can repeat the same with $mt$, and we would get that $t$ is a period for $f+g, fg, f/g$.

However, since $s=kt$; if $k$ is rational not integer, that isn't true (in fact $s$ wouldn't be a period for $g$), if $k$ is an integer, $t$ can't be a period for $f$ (otherwise it would be a positive period less than the fundamental period),and so it can't be a period for$f+g, fg, f/g$ too. That means that the only possibility is $sn=mt=\alpha$.

$ii)$ In this case it's obvious that $s=t$ is a period for the functions $f+g$, $fg$, $f/g$. However, we don't manage to say much on he period of these functions. The only thing we can say is that if $\alpha$ is \textbf{the} fundamental period, it is the minimum of the positive periods, and so it surely will be $\alpha \leq s=t$

  • $iii)$ If $\frac{s}{t}$ is irrational, then we can't find integers $m$, $n$ so that $ms=nt$.

Anyway, suppose that $f+g$ $fg$ $f/g$ are periodics with fundamental period $c$. $c$ can't be, at the same time, period of $f$ and period of $g$, because such number doesn't exist.

If $c$ was a period for $f$ ($c=kt$), we would have: \begin{equation*} f(x+kt)+g(x+kt)=f(x)+g(x+kt)\neq f(x)+g(x) \end{equation*} And so $kt$ is not the period of $f+g$,same thing if $c$ was a period for $g$. But that means that $\exists x\in X : f(x+c)\neq f(x)$ and $g(x+c)\neq g(x)$ and so $f(x)+g(x) \neq f(x+c)+g(x+c)$.

Therefore, $c$ is not a period for $f+g$, contraddiction: $f+g$ is not periodic. This can be done in the same way for the function $fg$ $f/g$, and we conclude.

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  • $\begingroup$ Quicker: If $\tau$ is the least positive period, and $\beta$ is some other period, then We consider $n=\lfloor \frac {\beta}{\tau}\rfloor$ and write $\beta =n\tau+\gamma$ for $0≤\gamma<\tau$. Easy to check that $\gamma $ is another period, hence must be $0$. $\endgroup$
    – lulu
    Commented Jan 7 at 19:03

3 Answers 3

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A few comments on what you wrote.

  • In general, assumptions like "Viceversa, let's suppose that a period different from $k \tau$ exists, and let's call it $\beta > \tau$" are a bit unwieldy and easily misleads students into almost-correct-but-very-wobbly proofs. Instead it is often much tidier to assume that $\beta$ is any period of the function, and prove that there exists an integer $k$ such that $\beta = k \tau$.

  • As an illustration of what I mean by "wobbly", consider this sentence that you have written: "So, if $\beta \neq k \tau$, $\beta$ isn't a period, therefore $\beta = k \tau$, with $k \in \mathbb Z$.". Can you honestly say that this sentence is meaningful and that you are 100% confident about its meaning?

  • An important issue is the total absence of explicit quantifiers. In general, you should never use a variable name without having introduced that variable name before. For instance, you write "let's suppose that a period different from $k \tau$ exists, and let's call it $\beta > \tau$", but no variable named $k$ has been introduced! Instead, you should say something like: "Let's suppose there exists a period $\beta > \tau$ such that for any integer $k$, $\beta \neq k \tau$" or alternatively "Let's suppose there exists a period $\beta > \tau$ such that there exists no integer $k$ such that $\beta = k \tau$". See how I always introduced $k$ with "there exists no integer $k$" or "for any integer $k$" before using $k$ in an equation? That's important.

  • Your handling of the case $s \in \mathbb Q \setminus \mathbb Z$ is not rigorous at all! You basically found one example and argued that because there is one example, it must always be true. You wrote: "If $\beta = s \tau$ with $s$ like I said, $\beta$ is not a period. In fact, even if $\tau$ is a period of $f$, $\beta = s \tau$ isn't anymore. We can consider $\sin(\pi+2 \pi)=\sin(\pi)=0\neq \sin(\pi+\frac 1 3 2\pi)$ for a counterexample."

  • In fact, the disjunction of cases that you made is not necessary. You can handle all cases at the same time. The idea is always the same: let $k = \left\lfloor \frac \beta \tau \right\rfloor$ (or in other words, let $k \in \mathbb Z$ be the largest integer such that $k \tau \leq \beta$). Then you can prove that either: $\beta = k \tau$; or $\beta - k \tau > 0$ is a period, but $0 \leq \beta - k \tau < \tau$, which would contradict the fact that $\tau$ is minimal.

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    $\begingroup$ Ok, I think I got It! I can't thank you enough! I'm a bit embarrassed cause I felt genius when I thought about the euclidean division! 😂 $\endgroup$
    – m05
    Commented Jan 7 at 18:54
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    $\begingroup$ @m05 It is genius! But $\beta$ and $\tau$ are not assumed to be integers, so instead of saying "Let $k, r$ be the quotients and remainder of the Euclidean division of $\beta$ by $\tau$", we have to say "Let $k$ be the largest integer such that $k \tau \leq \beta$, and let $r = \beta - k \tau$". Then we have to prove that $0 \leq r < \tau$, which is done exactly the same way as you proved this about the remainder when you proved the Euclidean division theorem. $\endgroup$
    – Stef
    Commented Jan 7 at 18:58
  • $\begingroup$ Is it something like this? Every real number can be written as $x=\lfloor x\rfloor+frac(x)$ where $0 \leq frac(x)<1$, so we can say: $ \beta=\tau\cdot \lfloor \frac{\beta}{\tau}\rfloor+\tau\cdot frac(x)$ and so, $\forall x \in X $ (domain of $f$), we have $f(x)=f(x+\beta)=f(x+\tau\cdot\lfloor\frac{\beta}{\tau}\rfloor+\tau\cdot frac(x))=f(x+\tau\cdot frac(x))$ and that means $\tau \cdot frac(x)$ is a period for $f$, but $\tau \cdot frac(x) < \tau$ (cause $frac(x)<1$) $\endgroup$
    – m05
    Commented Jan 7 at 19:44
  • $\begingroup$ @m05 Almost, except if you read your last comment you should notice that one of the $x$ is not in its place. It should be $\frac \beta \tau = \left\lfloor \frac \beta \tau \right \rfloor + \operatorname{frac}(\frac \beta \tau)$. Then multiply both sides by $\tau$. $\endgroup$
    – Stef
    Commented Jan 7 at 20:14
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The OP's argument needs some simplification. First a precise definition of what a periodic function $f$ with fundamental period $\tau$ is.

Here is a standard definition:

Definition: A function $f$ on $\mathbb{R}$ is periodic if there is $t\in\mathbb{R}$ such that \begin{align} f(x+t)=f(x),\qquad x\in\mathbb{R}\tag{0}\label{zero} \end{align} Any $t\in\mathbb{R}$ that satisfies \eqref{zero} is called a period of $f$. If there exist $\tau>0$ which satisfies \eqref{zero}, and no other $0<t<\tau$ satisfies \eqref{zero}, then $\tau$ is called the fundamental period of $f$, and that $f$ is $\tau$-periodic.

  • Suppose $f$ is periodic and let $\mathcal{P}$ be the set of all periods of $f$, that is $$\mathcal{P}=\{p\in\mathbb{R}: f(x+p)=f(x),\forall x\in \mathbb{R}\}$$ It is easy to check that if $p_1,p_2\in\mathcal{P}$ and $m\in\mathbb{Z}$, then $p_1+m p_2\in\mathcal{P}$, that is, $\mathcal{P}$ is an additive subgroup of $\mathbb{R}$.

  • If $f$ is periodic and has fundamental period $\tau$, then $\mathcal{P}=\tau\mathbb{Z}$. Indeed, suppose $p\in\mathcal{P}$. Then $$p=\lfloor p/\tau\rfloor \tau +r, \qquad \text{for some}\quad 0\leq r<\tau$$ Consequently $$f(x+r)=f(x+p-\lfloor p/\tau\rfloor \tau)=f(x),\qquad \forall x\in\mathbb{R}$$ This means that $r\in\mathcal{P}$. By definition $\tau\in \mathcal{P}$ and no other $t\in(0,\tau)$ is in $\mathcal{P}$; hence, $r=0$ and so $p\in\{m\tau:m\in\mathbb{Z}\}$.

  • Not every periodic function has a fundamental period. For example $f(x)=\mathbb{1}_{\mathbb{Q}}(x)$ is periodic and any $p\in\mathbb{Q}$ is a period of $f$.

  • It is not difficult to check that if $f$ is continuous on $\mathbb{R}$, $f$ is not constant, and $f$ is periodic, then $f$ has a fundamental period $\tau>0$.

We have the following result:

Proposition: If $f$ is a periodic function on $\mathbb{R}$, then either $\mathcal{P}$ (the set of periods of $f$) is a dense additive subgroup of $\mathbb{R}$, or there exists $\tau>0$ such that $\mathcal{P}=\tau\mathbb{Z}$. In the later case, $f$ has fundamental period $\tau$.

Proof: Define $\tau_f:=\inf\{p\in \mathcal{P}: p>0\}$.

If $\tau=0$, then for any $\varepsilon>0$, there is $p\in\mathcal{P}$ with $0<p<\varepsilon$. Hence, for any $a\in\mathbb{R}$, $a=\lfloor \frac{a}{p}\rfloor p+ s$ for some $0\leq s<p$ ans so, $\lfloor \frac{a}{p}\rfloor p\in\mathcal{P}\cap(a-\varepsilon,a+\varepsilon)$. Consequently, $\mathcal{P}$ is dense.

Suppose then that $\tau=\tau_f>0$. If $p\in \mathcal{P}\cap(0,\infty)$, $$p=\lfloor \frac{p}{\tau}\rfloor \tau + s,\quad\text{for some}\quad 0\leq s<\tau$$ If $s>0$, then by definition of $\tau$, $n:=\lfloor \frac{p}{\tau}\rfloor\geq1$ and there is $p'\in\mathcal{P}\cap[\tau,\tau+\tfrac{s}{n})$. It follows that $$n\tau\leq np'<n\tau+s=p<(n+1)\tau$$ Hence $0<p-np'\in\mathcal{P}$ and $p-np'<\tau$ in contradiction to the definition of $\tau$. Thus, $s=0$ and so, $p\in\tau\mathbb{N}$. It follows from this that $\tau\in\mathcal{P}$ and that $\mathcal{P}=\tau\mathbb{Z}$. $\blacksquare$


Now we consider the following result:

Theorem: If $f,g$ are non constant continuous periodic functions and $h=f+g$ is not a constant function, then $h$ is periodic iff $\tau_f/\tau_g\in\mathbb{Q}\setminus\{0\}$, where $\tau_f$ and $\tau_g$ are the fundamental periods of $f$ and $g$ respectively. In such case, $\tau_h=\min \tau_f\mathbb{N} \cap \tau_g\mathbb{N}$.

Proof: Suppose first that $a\tau_f-b\tau_g=0$ for some $a,b\in\mathbb{N}$, with $g.c.d(a, b)=1$ (here $g.c.d$ stands for greatest common divisor). Then $p=a\tau_f= b\tau_g$ is is a period for $h$: $$ h(x+p)=f(x+p)+g(x+p)=f(x+a\tau_f)+g(x+b\tau_g)=f(x)+g(x),\quad\forall x\in\mathbb{R}$$ Hence $h$ is periodic. The assumption that $h$ is not constant, and the continuity of $f$ and $g$ imply that $h$ has a fundamental period $\tau_h>0$.

Suppose now that $h$ has a fundamental period $\tau_h>0$. Then $$\phi(x):=f(x+\tau_h)-f(x)=g(x+\tau_h)-g(x),\qquad\forall x\in\mathbb{R}$$ For any $m,n\in\mathbb{Z}$ and all $x\in\mathbb{R}$ \begin{align} \phi(x+m\tau_g+n\tau_g)&=f(x+m\tau_f+n\tau_g+\tau_h)-f(x+m\tau_f+n\tau_g)\\ &=f(x+n\tau_g+\tau_h)-f(x+n\tau_g)\\ &=g(x+n\tau_g+\tau_h)-g(x+n\tau_g)\\ &=g(x+\tau_h)-g(x)=\phi(x) \end{align} It follows that $\phi$ is periodic. If $\tau_f/\tau_g\notin\mathbb{Q}$, then either $\tau_f$ or $\tau_g$ (or both) are irrational and so, $\tau_f\mathbb{Z}+\tau_g\mathbb{Z}$ is dense in $\mathbb{R}$. As $\phi$ is continuous, it follows that $\phi$ is a constant function, say $k$. Then $$f(x)+g(x)=f(x+\tau_h)+g(x+\tau_h)=f(x)+g(x)+2k$$ which means that $k=0$. This means that $\tau_h\in \tau_f\mathbb{Z}\cap\tau_g\mathbb{Z}$, which in turn implies that $\tau_f/\tau_g\in\mathbb{Q}$ which is a contradiction. Therefore, $\tau_f/\tau_g\in\mathbb{Q}$ and $\tau_h=m.c.m(\tau_f,\tau_g)$ (here $m.c.m$ stands for minimal common multiple). $\blacksquare$

Under the continuity assumption, similar results can be obtain for $f\cdot g$.


If the assumption of continuity is relaxed, to say, measurability, we have the following results:

Theorem (Burstin): If $f$ is a measurable function on $\mathbb{R}$, periodic, and non-constant a.s. (for any $c\in\mathbb{R}$, $\lambda(\{x:f(x)\neq0\})>0$ where $\lambda$ is Lebesgue's measure), then $f$ has a fundamental period $\tau_f>0$ and so, the set $\mathcal{P}$ of all periods of $f$ is $\tau\mathbb{Z}$.

A proof of this using Lebesgue differentiation is given here.

Theorem: If $f$ and $g$ are measurable, periodic with fundamental periods $\tau_f,\tau_g>0$, and $h=f+g$ is not constant, then $h$ is periodic with fundamental period $\tau_h>0$ iff $\frac{\tau_f}{\tau_g}\in\mathbb{Q}$.

See Mirotin, A.R., and Mirotin, E.A., On Sums and Products of Periodic Functions, Real Analysis Exchange, Vol. 34(2), 2008/2009, pp. 1–12 for details of this and more general results.

If measurability is dropped, then I suspect the statement about sums does not hold in general. Analysis of this requires some set theoretic arguments beyond the scope of the posting. There might be a posting regarding this in MSE, in any event, here is an interesting link.

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  • $\begingroup$ Hi. In many places in this answer, you introduce variables somewhat implicitly, ie without an explicit "let..." or an explicit quantifier. For instance, $p_1, p_2$ and $r$ are introduced implicitly. Given that the OP was struggling a lot with the meaning of variables and quantifiers, may I suggest making these things more explicit in your answer? $\endgroup$
    – Stef
    Commented Jan 7 at 22:30
  • $\begingroup$ @Stef: I do not see any problem with my definitions and quantifiers; the $p$’s and the r are clearly defined in my opinion. I don’t see a particular need the extensive use of “let this and that”. But thanks for your comment. $\endgroup$
    – Mittens
    Commented Jan 7 at 23:32
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Let $\alpha> \tau $ be a period of $f.$ Let $k$ be the greatest positive integer such that $\alpha>k\tau.$ Then $\alpha_0=\alpha -k\tau $ is a period and $0< \alpha_0\le \tau.$ Indeed $$f(x+\alpha_0)=f(x+\alpha-k\tau)=f(x+\alpha)=f(x)$$ Thus $\alpha_0=\tau$ and $\alpha=(k+1)\tau.$

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