# Some Periodic Functions

Written by: Willie Wong

Primary Source: QNLW, 08/29/19.

Theorem 1
If is a non-constant, continuous periodic function on , then there exists a smallest positive real number satisfying for all .
Proof:

By definition of periodicity, there exists a subset satisfying for every and . Necessarily is closed under addition, and hence is a subgroup of , therefore is either equal to or is dense in . (See this proofwiki entry for justification of this assertion.)

By continuity of , has to be a closed set: if and , then . This means that if were dense, it is equal to . But in this case would be constant, which contradicts our assumption. So is discrete and we can take to be the smallest positive element.

Corollary 2

If is non-constant, continuous, and periodic, and are periods of , then is rational. The theorem is no longer true if you don’t assume continuity.
Example 3

Let be irrational. Define the equivalence relation on by setting whenever where . Since is irrational, the mapping is injective. Now notice that the equivalence classes of all have countably many elements, and hence has uncountably many classes. Let be any non-constant function. Then lifts to a function . By definition is both -periodic and -periodic, and has no smallest positive period.

## Sums of periodic functions

Here’s another “calculus fact”

Theorem 4

Given two continuous, non-constant, periodic functions on with periods respectively. Then
Proof:

The implication is immediate: if then there exists such that with . Thus is a common period of and and hence is periodic with period . (Note that this does not require continuity.)

For the reverse implication , we prove by contrapositive. First, without loss of generality we can assume are the smallest positive periods. Suppose now is irrational, and let be a period of . Then necessarily at least one of and is irrational. Without loss of generality assume it is .

If were rational, than we can assume without loss of generality that is a multiple of . This implies then that is -periodic, and hence so is . But then must be constant (based on Theorem 1 above).

If were irrational, we can assume that (otherwise switch and ). Consider and . By construction is constant, is a superposition of -periodic, and is -periodic. However, as is constant, we have that must also be -periodic, and thus is both and periodic and hence is constant, which implies that has to also be periodic (by the fundamental theorem of calculus). And since is irrational we must have is constant.

Dropping continuity we have the following theorem

Theorem 5

Given any , there exists non-constant functions with smallest positive periods respectively, such that is periodic.
Proof:

If is rational, then the theorem is trivial. So let us assume that is irrational.

Choose such that are linearly independent over the rationals. Observe that the mapping is therefore injective, denote by the image of this mapping. Notice that if and , then and .

Define and such that away from , and on , set These functions are well-defined by the injectivity property discussed above. By definition is periodic, and is periodic. But is periodic.

The following two tabs change content below.

#### Willie Wong

I am an assistant professor of mathematics at Michigan State University. My research combines analytic and geometric techniques to study nonlinear evolutionary partial differential equations that arise in physical and geometric contexts. I am particularly interested in situations where wave-like phenomena arise, such as general relativity, fluids, and the physics of elastic membranes and solids.