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 $$0. Now, for hypothesis $$\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.

• 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 $0$.
– lulu
Commented Jan 7 at 19:03

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.

• 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! 😂
– m05
Commented Jan 7 at 18:54
• @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.
– Stef
Commented Jan 7 at 18:58
• 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 $frac(x)<1$)
– m05
Commented Jan 7 at 19:44
• @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 $\tau$.
– Stef
Commented Jan 7 at 20:14

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 $$0 satisfies \eqref{zero}, then $$\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 $$0. Hence, for any $$a\in\mathbb{R}$$, $$a=\lfloor \frac{a}{p}\rfloor p+ s$$ for some $$0\leq s 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 $$n\tau\leq np' Hence $$0 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.

• 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?
– Stef
Commented Jan 7 at 22:30
• @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. Commented Jan 7 at 23:32

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 $$\alpha=(k+1)\tau.$$