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.
Best Answer
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.