A fundamental period claim

This arose from Sandeep's problem Do you know its property - 9

Starting question. If two functions \(f\) and \(g\) are periodic, is \( f+g \) also periodic?

Be warned that this note requires significant mathematical maturity, which is why I set it at level 5.

Note: The LCM of 2 real numbers is defined in the same way. Find the smallest positive number (if it exists) such that for some integers $n,m$, we have $n \alpha = m \beta$.

Claim 1: If the fundamental period of $f$ is $\alpha$ and the fundamental period of $g$ is $\beta$, then the fundamental period of $f + g$ is the LCM of $\alpha, \beta$?

Is this claim true? No, not necessarily. We only know that their LCM must be a period, but it need not be the fundamental period. There could be a smaller period due to cancellation (see Krishna's comment). What is true, is the following:

Fact: If the period of $f$ is $\alpha$ and the period of $g$ is $\beta$, and $\frac{\alpha} { \beta }$ is rational, then a period of $f + g$ is the LCM of $\alpha, \beta$.

That leaves us to consider the irrational case.

Claim 2: If the period of $f$ is $\alpha$ and the period of $g$ is $\beta$, and $\frac{\alpha} { \beta }$ is irrational, then $f+g$ is never periodic.

It is easy to find examples where $f + g$ is not periodic, like $\sin x + \sin \pi x$. However, must this statement be always true?

We know that there is no obvious candidate for a period, but can't there be some "weird" functions which cancel out, like in the rational case? The following 4 hints will guide us through the rest of this investigation. You should try working on these hints first, before reading further

Hint 1: Prove that a continuous periodic function is bounded.

Hint 2: Prove that if $f$ and $g$ are both continuous, periodic, non-constant functions with irrational ratios, then $f+g$ is not periodic.

Hint 3: Find a function that is not constant, whose periods is dense in $\mathbb{R}$ but the set of periods is not $\mathbb{R}$. Note: The function is not continuous.

Hint 4: Now find $f$ and $g$ with irrational ratio of periods, such that $f+g$ is periodic (and non-constant).

Note: I used the Axiom of Choice (which I believe in).

Proof of hint 1. Let $f$ be a continuous, periodic function with period $\alpha$. Suppose that $f$ is not bounded. That means that for any $N$, there exists an $x_N$ such that $| f ( x_N )| > N$. WLOG, we may assume that $0 \leq x_N \leq \alpha$, as we translate it to the first period.

Since the points $x_N$ lie in the closed interval $[0, \alpha ]$, there exists an accumulation point, which means that there is a certain subsequence $x_{N_k}$ which converges to $x^*$. Since $f$ is a continuous function, we have $f( x^*) = \lim_{k\rightarrow \infty} f(x_{N_k} )$. But the RHS tends to infinity, which contradicts the assumption that $f(x^*)$ is finite. Hence, the assumption is false, and $f$ is bounded. $_\square$

Proof of hint 2. We first prove the following claim:

Claim If a function $f$ is continuous, periodic, and it's periods are dense in $\mathbb{R}$, then it is the constant function.

This follows immediately by considering $f( \alpha )$, where $\alpha$ is a period of $f$. It is dense in $\mathbb{R}$, and their values are all equal. By continuity, it is the constant function. $_\square$

Now, on to the proof. Suppose that $f$ and $g$ are continuous, periodic, non-constant functions with irrational period ratio, and $f+ g$ is periodic. If $f+g$ is the constant function, then we can show that $f$ and $g$ must have the same period, which contradicts the assumption. Hence, we may assume that $f + g$ has a smallest period $\gamma > 0$.

We are given that $f( x+ \gamma) + g( x + \gamma ) = f(x) + g(x)$. Define $h(x) = f( x + \gamma) - f(x) = g(x) - g( x + \gamma)$, which is a continuous function (since it is the difference of 2 continuous functions).

For any pair of integers $n, m$, observe that $\begin{aligned} & h ( x + n \alpha + m \beta) \\ = & f( x + n \alpha + m\beta + \gamma) - f( x + n \alpha + m \beta ) \\ = & f ( x + m \beta + \gamma) - f ( x + m \beta ) = h ( x + m \beta ) \\ = & g ( x + m \beta ) - g ( x + m\beta + \gamma)\\ =& g ( x ) - g ( x + \gamma ) \\ =& h ( x ) \end{aligned}$

This implies that the function $h (x)$ is periodic with period $n \alpha + m \beta$. But since the ratio $\frac{ \alpha } { \beta }$ is irrational, this implies that $\mathbb{Z} \alpha + \mathbb{Z} \beta$ is dense in $\mathbb{R}$, and thus $h(x)$ must be a constant, which we will denote by $h$.

If $h \neq 0$, then $f( x + n \gamma) = f(x) + n h$ which is an unbounded sequence, contradicting Hint 1.

If $h = 0$, then $f$ and $g$ are both periodic with period $gamma$. Now, at least one of $\frac{ \alpha } { \gamma }$ or $\frac { \beta } { \gamma}$ is irrational (WLOG, let it be the first ). Once again, this implies that $\mathbb{Z} \alpha + \mathbb{Z} \gamma$ is dense in $\mathbb{R}$, and thus that $f$ is a constant function. However, this contradicts the assumption. $_\square$

Hence, what we can conclude at this point is

Fact: If $f$ is a continuous periodic function, then either $f$ is constant, or $f$ has a fundamental period (smallest).

Fact: If $f$ and $g$ are continuous, periodic, non-constant functions with an irrational ratio of fundamental periods, then $f + g$ is not periodic.

The claim is true, if we restrict out attention to only continuous functions.

Now, we move on to the fun and interesting part. The world is extremely well-behaved (relatively) when functions are continuous. Let's observe what happens when things go crazy.

Proof of Hint 3. Consider the function $f(x) = \begin{cases} 0 & x \in \mathbb{Q} \\ 1 & x \not \in \mathbb{Q} \\ \end{cases}.$

This function is periodic, and the set of periods is all rational numbers. This is dense in $\mathbb{R}$, but is not $\mathbb{R}$. $_\square$

In particular, this shows us that a non-continuous, non-constant periodic function need not have a fundamental period.

Proof of Hint 4. We know that $1, e, \pi$ are rational-linearly independent. What this means is that if $q_1 \times 1 + q_2 \times e + q_3 \times \pi = 0$, where $q_i$ are rational numbers, then $q _ i = 0$.

Define the rational span of $(a,b)$ to be the set of all numbers of the form $\mathbb{Q} a + \mathbb{Q} b$.

Let $\chi_{1, e }$ be the characteristic function of the rational span of $(1,e)$, and $\chi_{e, \pi}$ and $\chi_{1, \pi}$ similarly defined.

Then, set $f = \chi_{ 1, e } + \chi_{e, \pi}$ and $g = - \chi{ e, \pi} + \chi { 1 , \pi }$. Then, we have $e$ is a period of $f$, $\pi$ is a period of $g$ and $1$ is a period of $f + g$. $_ \square$

Thus, we have 2 functions with irrational ratio of periods, but sum to give a periodic function.