A period of a function f : ℝ → X is a T ∈ ℝ such that for every t ∈ ℝ, we have f(t + T) = f(t). Note that X may be any set; the definition does not refer to any structure on X other than its equality relation.
For every function f : ℝ → X, the set of periods of f is an additive subgroup of ℝ:
- For any two periods T1 and T2 of f and every t ∈ ℝ, we have
f(t + T1 + T2) = f(t + T1) = f(t),
so T1 + T2 is a period of f. - For every period T of f and every t ∈ ℝ, we have
f(t − T) = f((t − T) + T) = f(t),
so − T is a period of f.
Therefore, we call it the group of periods of f.
The converse also holds: every additive subgroup G of ℝ is the group of periods of a function, namely the canonical projection f : ℝ → ℝ/G (which takes every t ∈ ℝ to its equivalence class wrt the equivalence relation {(t1, t2) : t2 − t1 ∈ G}):
- For every T ∈ G and every t ∈ ℝ, we have (t + T) − t = T ∈ G and hence f(t + T) = f(t), so T is a period of f.
- Conversely, for every period T of f, we have f(T) = f(0) and hence T − 0 = T ∈ G.
Henceforth, we shall always refer to additive subgroups of ℝ as groups of periods.
It is worth noting explicitly that, in addition to the two closure properties proven above, the fact that the groups of periods are groups implies that they are closed under integer multiples, and in particular contain 0.
A function is aperiodic iff its group of periods is trivial, i.e. 0 is its only period. Otherwise, it is periodic.
If a group G of periods is nontrivial and cyclic, then it has a nonzero generator a. Since the set of the generators of G is closed under negation, we can choose a to be positive. This a is then the least positive element of G, because for every positive x ∈ G, there is an n ∈ ℤ such that x = na where n ∈ ℤ, and this n has to be positive (since x and a are positive), so that x ≥ a. The logic here works for any ordered group, not just groups of periods. It also works for any positive generator of G—so the least positive element of G is the only positive generator of G.
Conversely, if a group G of periods has a least positive element a, then a generates G. To see this, suppose x ∈ G and let n be the floor of x/a, so that n ≤ x/a < n + 1, i.e. na ≤ x < (n + 1)a, i.e. 0 ≤ x − na < a. Then x − na is an element of G less than a, so it cannot be positive; it must be 0, so that x = na. (The logic here doesn’t work for any ordered group, because it relies on the Archimedean property of the real numbers.)
So a nontrivial group G of periods has a least positive element a iff it is cyclic, in which case a is the unique positive generator of G. In this situation, we call a the fundamental period of G, or more loosely just the period of G.
The “canonical” examples of periodic functions have fundamental periods. For example, the sine and cosine functions have a fundamental period of 2π. However, constant functions are also technically periodic, and they do not have fundamental periods: a function f : ℝ → X is constant iff its group of periods is ℝ. There are more obscure examples. Consider the indicator function 1ℚ of ℚ. For every T ∈ ℚ and every t ∈ ℝ, we have t + T ∈ ℚ iff t ∈ ℚ and hence f(t + T) = f(t); so T is a period of f. On the other hand, for every irrational T ∈ ℝ, we have ( − T) + T = 0 ∈ ℚ but − T ∉ ℚ, and hence f(( − T) + T) = 1 but f( − T) = 0; so there is a t ∈ ℝ such that f(t + T) ≠ f(t), and hence T is not a period of f. It follows that the group of periods of 1ℚ is ℚ.
There is a topological criterion which can be used to determine whether a periodic function has a fundamental period: a group G of periods is cyclic iff it is not dense in ℝ. To see this, observe first that if G is dense in ℝ, then it has elements arbitrarily close to 0, symmetrically on both the positive and negative sides of the number line (since G is closed under negation). Therefore G does not have a least positive element and hence is acyclic.
Conversely, suppose G is acyclic and hence has no least positive element. Then for every positive x ∈ G, there is a positive y ∈ G less than x. The difference x − y is also a positive element of G less than x, and at least one of y and x − y is less than or equal to x/2 (otherwise their sum would be greater than x). So in fact there is a positive y ∈ G less than or equal to x/2. By repeatedly applying this we can see that for every n ∈ ℕ, there is a positive y ∈ G less than or equal to x/2n. So G has positive elements arbitrarily close to 0. It follows that for every a ∈ ℝ and every positive ε ∈ ℝ, there is a positive x ∈ G less than ε. The interval (a − ε, a + ε), being of length 2ε > x, must contain nx for some n ∈ ℤ. And since G is closed under integer multiples, we have nx ∈ G. So G has elements arbitrarily close to a. This holds for every a ∈ ℝ, so G is dense in ℝ.
This criterion is simplified for continuous periodic functions f : ℝ → X, where X is an accessible topological space (a topological space X is accessible iff any two distinct points in X have neighbourhoods not containing the other point). This is because the group G of periods of such a function f is closed. To see this, suppose T0 is a limit point of G; it will suffice to prove that T0 ∈ G. To see that, suppose t ∈ ℝ; it will suffice to prove that f(t + T0) = f(t). To see that, suppose N is a neighbourhood of f(t + T0); it will suffice to prove that f(t) ∈ N. Since f is continuous, so is its right-composition with adding t. Therefore there is a neighbourhood M of T0 such that f(t + M) ⊆ N. And since T0 is a limit point of G, there is a T ∈ M ∩ G. Since T ∈ M, we have f(t + T) ∈ N. Since T ∈ G, we have f(t + T) = f(t) and hence f(t) ∈ N.
Since G is closed, it is equal to its closure. So it can only be dense in ℝ (i.e. have its closure equal to ℝ) iff it is equal to ℝ. Therefore, given that f is a continuous periodic function with an accessible codomain, it has a fundamental period iff it is nonconstant.
