1. Introduction and Preliminaries
In the previous post, we motivated a series of questions surrounding the sums of periodic functions and the existence of functions with incommensurable periods. Now, we are ready to dive in. We will see that many of the things we intuitively hold to be true (or that I at least held to be true) about periodic functions do indeed hold for contiuous functions in particular. We will also start to see, however, that the structural richness of sets of periods leaves much room for pathological behaviour deviating starkly from intuitions based on the continuous functions.
To begin, let us review the basic definitions and facts about periodic functions that set the stage for our further explorations, revealing the beautiful interplay of algebra and topology in the world of periodicity.
Definition 1 (Periodic Function) A function \(f:\mathbb{R}\to\mathbb{R}\) is said to be periodic if there exists a non-zero real number \(p\) such that \(f(x+p)=f(x)\) for all \(x\in\mathbb{R}\). If there exists a smallest positive such \(p\), it is called the fundamental period of \(f\), and we say \(f\) is strictly periodic.
Remark: Note that a function can be periodic without being strictly periodic; for instance, the constant functions are periodic with any real period, but they are not strictly periodic. In this series, we will examine pathological examples of functions that fail to be strictly periodic but are also not constant. Some are even constant only on sets of measure zero, which is as far from being constant as one can get. We will also see pathological functions that actually are strictly periodic.
Definition 2 (The Period Module) For any function \(f:\mathbb{R}\to\mathbb{R}\), we denote by \(P_f\) the set of all periods of \(f\), i.e., \(P_f := \{p\in\mathbb{R}: f(x+p)=f(x) \text{ for all } x\in\mathbb{R}\}\). We call \(P_f\) the period module of \(f\). Also, we say two periodic functions \(f\) and \(g\) are incommensurable if \(P_f\cap P_g = \{0\}\), and they are commensurable otherwise.
(Note: Constant functions are trivially commensurable with any periodic function. Non-periodic functions still have a period module, but it is just \(\{0\}\)).
Definition 3 (\(\mathbb{Z}\)-submodule and its Rank) A subset \(M\) of \(\mathbb{R}\) is called a \(\mathbb{Z}\)-submodule if it is closed under addition and under multiplication by integers. The proper algebraic measure of its size is its rank, which is the cardinality of a maximal linearly independent subset of \(M\) over \(\mathbb{Z}\).
We immediately find some basic facts about periodic functions and about the algebra and topology of their period modules:
- Fact 0: If \(f, g\) are incommensurable functions, any (nonzero) periods must be incommensurable real numbers. If \(p\in P_f\) and \(q\in P_g\) are such that \(p/q\in\mathbb{Q}\), then there exist integers \(n, m\) such that \(np = mq\), and thus \(np\in P_f\cap P_g\), showing that \(f\) and \(g\) are not incommensurable functions.
- Fact 1: If \(f\) is periodic with period \(p\), then it is also periodic with respect to any integer linear combinations of periods of \(f\). This means \(P_f\) is a \(\mathbb{Z}\)-submodule of \(\mathbb{R}\).
- Fact 2: If \(f\) is strictly periodic with fundamental period \(p\), then the set of all periods of \(f\) is precisely the \(\mathbb{Z}\)-module \(p\mathbb{Z}:= \{np : n\in\mathbb{Z}\}\). I.e., \(P_f\) is necessarily a submodule of rank 1.
- Fact 3: If \(P_f\) is dense, then \(f\) is either constant or else it is nowhere continuous. (To see this, suppose \(f\) is not constant and let \(x \in \mathbb{R}\). For any open interval \((a,b)\) containing \(x\), the density of \(P_f\) guarantees the existence of periods \(p, q \in P_f\) such that \(a < p < q < b\). Since \(P_f\) is a \(\mathbb{Z}\)-submodule, the difference \(r = q-p\) is a strictly positive period. The closed interval \([p, q] \subset (a,b)\) has length exactly \(r\), meaning it spans a full period of \(f\). Consequently, the image of \([p, q]\) under \(f\) is the entire range of \(f\). Because \(f\) is not constant, its range contains some value distinct from \(f(x)\), guaranteeing there is some \(z \in [p, q] \subset (a,b)\) such that \(f(z) \neq f(x)\). Thus, \(f\) cannot be continuous at \(x\)).
We can start to see how closely the nature of a periodic function relates to the topological and algebraic features of its period module. The following propositions make this even clearer:
Proposition 1 \(P_f\) is dense if and only if \(f\) is not strictly periodic. Thus, if a periodic \(f\) is non-constant and continuous, it must be strictly periodic.
Proof. If \(P_f\) is dense, then the infimum of positive periods must be zero, and thus no fundamental period can exist. Conversely, if no fundamental period exists, the infimum \(p\) of positive periods must be zero; for otherwise, we could find two distinct periods arbitrarily close to \(p\) such that their positive difference would be a positive period smaller than \(p\), violating the infimum. Thus, any interval around \(0\) must contain elements of \(P_f\). Now take any interval \((a, b)\) with \(0\not\in (a, b)\). By the foregoing, we can find \(x\in P_f\) with \(0<|x|<\min(\{|b|, |a|, |b-a|\})\). There thus must be some integer \(n\) so that \(nx\in (a, b)\), for otherwise we could find an \(n\) so that \((n-1)x \le a < b \le nx\), which contradicts \(|x| < |b-a|\). But \(nx\in P_f\), showing \(P_f\) is dense in \(\mathbb{R}\). The last part is an immediate consequence of this density and Fact 3.
Proposition 2 If \(p, q\) are incommensurable (\(p/q\not\in \mathbb{Q}\)), then the \(\mathbb{Z}\)-module \(\mathbb{Z}_{p,q}:=\{np + mq : n,m\in\mathbb{Z}\}\) is dense in \(\mathbb{R}\). In particular, if a function \(f\) has two incommensurable periods \(p\) and \(q\), then \(\mathbb{Z}_{p,q}\subseteq P_f\), meaning its period module is dense and \(f\) is not strictly periodic. Therefore, non-constant functions with incommensurable periods are nowhere continuous.
Proof. To see the first part of the proposition, we make use of the fact that irrationals can be approximated by rationals very nicely: by Dirichlet’s approximation theorem, we know that for any \(\epsilon>0\), we can find integers \(-n, m\) such that \(|mp/q -(-n)|<\epsilon/|q|\), and thus \(|mp +nq|<\epsilon\). This shows all intervals containing the origin have elements of \(\mathbb{Z}_{p,q}\). From here, the density of \(\mathbb{Z}_{p,q}\) follows in the same way as the density of \(P_f\) in the previous proposition. The second part of the proposition then follows immediately from everything discussed previously.
2. The Difference Operator and Periodic Sums
Now, we revisit the question of sums of incommensurable functions, and in particular we show that in the continuous case, such sums are never periodic.
For this and for the future posts in this series, we introduce the following operator which is ubiquitously used in discussions of the periodic decomposition problem:
Definition 4 (Difference Operator) For any function \(f:\mathbb{R}\to\mathbb{R}\) and any \(p\in\mathbb{R}\), we define the difference operator in p \(\Delta_p\) by \(\Delta_p f(x) := f(x+p) - f(x)\) for all \(x\in\mathbb{R}\). As the reader can verify, these are linear operators for all \(p\), and they commute with each other, i.e., \(\Delta_p\Delta_q = \Delta_q\Delta_p\) for all \(p, q\in\mathbb{R}\).
This operator measures how much \(f\) changes when we translate by \(p\). In particular, \(\Delta_p f\) is identically zero if and only if \(p\) is a period of \(f\). A few more key facts about difference operators and sums of functions:
- For any \(f:\mathbb{R}\to\mathbb{R}\) and any \(q\in\mathbb{R}\), we have \(P_f\subseteq P_{\Delta_q f}\). That is, all periods of \(f\) are also periods of \(\Delta_q f\).
- Given two functions \(f,g\), their sum is periodic if and only if there is an \(r\neq 0\) such that \(\Delta_r f +\Delta_r g=0\). Such an \(r\) is then a period of \(f+g\).
- Thus, if \(f+g\) is periodic, then for all \(p\in P_f\) and all \(q\in P_g\) we have: \[0=\Delta_r \Delta_q f = \Delta_r\Delta_p g\] which implies: \[P_{\Delta_q f} \supseteq P_f + r\mathbb{Z} \quad \text{and} \quad P_{\Delta_p g} \supseteq P_g + r\mathbb{Z}\]
- Hence, if \(f+g\) is periodic, then \(\Delta_q f\) and \(\Delta_p g\) are commensurable for all \(p\in P_f\) and all \(q\in P_g\), even if \(f\) and \(g\) are incommensurable. This however is not a sufficient condition for \(f+g\) to be periodic, as the reader can verify (we will also see this in the next installment).
- When \(f, g\) are incommensurable and their sum is periodic, then the \(r\) above must be incommensurable with all periods of \(f\) and \(g\), and so we actually have: \[P_{\Delta_q f} \supseteq P_f \oplus r\mathbb{Z} \quad \text{and} \quad P_{\Delta_p g} \supseteq P_g \oplus r\mathbb{Z}\]
- If \(f, g\) are themselves periodic and incommensurable with \(f+g\) periodic, the above then implies \(\Delta_q f\) and \(\Delta_p g\) must have dense period modules, and thus be either constant or nowhere continuous.
This leads us to the final result of this installment:
Proposition 3 Let \(f, g\) be continuous periodic functions. Then \(f+g\) is periodic if and only if \(f, g\) are commensurable. Thus, no set of continuous periodic functions containing two or more incommensurable functions can be a vector subspace of \(\mathbb{R}^\mathbb{R}\).
Proof. We leave the details to the reader. As a hint, observe that if \(f\) is continuous, so is \(\Delta_q f\) for any \(q\), and thus if there are incommensurable numbers \(r, q\) such that \(\Delta_r\Delta_q f=0\), then \(\Delta_q f\) must be constant. Then use the boundedness of continuous periodic functions to conclude \(\Delta_q f = 0\) and finish the proof of the harder direction.
3. Beyond Continuity: A Preview of Pathology
With what we have established so far, a clear structural dichotomy emerges: any periodic function with a dense (and thus higher-rank) period module, and any pair of incommensurable functions whose sum manages to remain periodic, must be severely pathological—specifically, they must be discontinuous everywhere.
The requirement of continuity acts as a rigid cage, forcing period modules to remain of rank 1 over \(\mathbb{Z}\) and preventing sums of incommensurable functions from being periodic. But what happens when we remove this cage?
In the next installment of Pathological Periodicity, we will dive directly into the abyss of these nowhere-continuous functions. We will construct and dissect a variety of pathological examples, including some highly unintuitive functions that are constant only on sets of Lebesgue measure zero (though the functions themselves are non-measurable)—a construction where the Axiom of Choice becomes absolutely indispensable. Stay tuned!