This arose from Sandeep's problem Do you know its property - 9
Starting question. If two functions and are periodic, is 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 , we have .
Claim 1: If the fundamental period of is and the fundamental period of is , then the fundamental period of is the LCM of ?
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 is and the period of is , and is rational, then a period of is the LCM of .
That leaves us to consider the irrational case.
Claim 2: If the period of is and the period of is , and is irrational, then is never periodic.
It is easy to find examples where is not periodic, like . 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 and are both continuous, periodic, non-constant functions with irrational ratios, then is not periodic.
Hint 3: Find a function that is not constant, whose periods is dense in but the set of periods is not . Note: The function is not continuous.
Hint 4: Now find and with irrational ratio of periods, such that is periodic (and non-constant).
Note: I used the Axiom of Choice (which I believe in).
Proof of hint 1. Let be a continuous, periodic function with period . Suppose that is not bounded. That means that for any , there exists an such that . WLOG, we may assume that , as we translate it to the first period.
Since the points lie in the closed interval , there exists an accumulation point, which means that there is a certain subsequence which converges to . Since is a continuous function, we have . But the RHS tends to infinity, which contradicts the assumption that is finite. Hence, the assumption is false, and is bounded.
Proof of hint 2. We first prove the following claim:
Claim If a function is continuous, periodic, and it's periods are dense in , then it is the constant function.
This follows immediately by considering , where is a period of . It is dense in , and their values are all equal. By continuity, it is the constant function.
Now, on to the proof. Suppose that and are continuous, periodic, non-constant functions with irrational period ratio, and is periodic. If is the constant function, then we can show that and must have the same period, which contradicts the assumption. Hence, we may assume that has a smallest period .
We are given that . Define , which is a continuous function (since it is the difference of 2 continuous functions).
For any pair of integers , observe that
This implies that the function is periodic with period . But since the ratio is irrational, this implies that is dense in , and thus must be a constant, which we will denote by .
If , then which is an unbounded sequence, contradicting Hint 1.
If , then and are both periodic with period . Now, at least one of or is irrational (WLOG, let it be the first ). Once again, this implies that is dense in , and thus that is a constant function. However, this contradicts the assumption.
Hence, what we can conclude at this point is
Fact: If is a continuous periodic function, then either is constant, or has a fundamental period (smallest).
Fact: If and are continuous, periodic, non-constant functions with an irrational ratio of fundamental periods, then 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
This function is periodic, and the set of periods is all rational numbers. This is dense in , but is not .
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 are rational-linearly independent. What this means is that if , where are rational numbers, then .
Define the rational span of to be the set of all numbers of the form .
Let be the characteristic function of the rational span of , and and similarly defined.
Then, set and . Then, we have is a period of , is a period of and is a period of .
Thus, we have 2 functions with irrational ratio of periods, but sum to give a periodic function.