3 Nowhere boundedness of summable periodic functions
Lemma 3.1 If \(\Delta_q f\) is nowhere bounded, then \(f\) is nowhere bounded.
Proof. Suppose for contradiction that \(f\) is bounded on some open interval \((a,b)\). Then for any \(x \in (a, b-q)\) (assuming \(q > 0\); the case \(q < 0\) is analogous), we have \(\Delta_q f(x) = f(x+q) - f(x)\), where both \(f(x+q)\) and \(f(x)\) are bounded on the interval \((a, \min(b, b-q))\). Hence \(\Delta_q f\) is bounded on a nonempty open interval, contradicting the assumption that \(\Delta_q f\) is nowhere bounded.
Lemma 3.2 Let \(f\) be a periodic function and let \(q \notin P_f\). If \(q \in P_{\Delta_q f}\), then \(q \notin \operatorname{span}_\mathbb{Q}(P_f)\). In other words, no rational multiple of a period of \(f\) that is not itself a period of \(f\) can be a period of \(\Delta_q f\).
Proof. Suppose \(q \in P_{\Delta_q f}\), so that \(\Delta_q f(x + q) = \Delta_q f(x)\) for all \(x\). This means \(\Delta_q f\) is \(q\)-periodic, i.e., \(\Delta_q^2 f = 0\). Now suppose for contradiction that \(q \in \operatorname{span}_\mathbb{Q}(P_f) \setminus P_f\). Then there exist a period \(p \in P_f\) and integers \(m, n\) with \(n \neq 0\) such that \(nq = mp\). Since \(mp \in P_f\), we have \(f(x + nq) = f(x)\) for all \(x\).
On the other hand, since \(\Delta_q^2 f = 0\), the function \(\Delta_q f\) is \(q\)-periodic. Together with the fact that \(P_f \subseteq P_{\Delta_q f}\), we conclude that \(\Delta_q f\) is constant on each coset of \(P_f \oplus q\mathbb{Z}\) — but we must first verify that this is indeed a direct sum. However, since \(nq = mp \in P_f\), the submodule \(P_f + q\mathbb{Z}\) is not a direct sum; rather, \(q\mathbb{Z}\) collapses modulo \(P_f\) into a finite cyclic group of order dividing \(n\).
More directly: since \(\Delta_q^2 f = 0\), we have \(\Delta_q f(x) = \Delta_q f(x + q)\) for all \(x\), so by induction \(\Delta_q f(x) = \Delta_q f(x + kq)\) for all integers \(k\). In particular, \(\Delta_q f(x) = \Delta_q f(x + nq) = \Delta_q f(x + mp) = \Delta_q f(x)\), which is trivially true and gives no information. Instead, we use the telescoping identity: \[f(x + nq) = f(x) + \sum_{k=0}^{n-1} \Delta_q f(x + kq) = f(x) + n \cdot \Delta_q f(x)\]
where the last equality follows from the \(q\)-periodicity of \(\Delta_q f\). Since \(f(x + nq) = f(x + mp) = f(x)\), we obtain \(n \cdot \Delta_q f(x) = 0\) for all \(x\), and thus \(\Delta_q f \equiv 0\). But this means \(q \in P_f\), contradicting our assumption.
Theorem 3.1 Let \(f\) be a periodic function and let \(q \notin P_f\). If there exists a positive integer \(N\) such that \(\Delta_q^N f = 0\), then \(f\) is nowhere bounded (and in particular, \(\mathcal{C}(f) = \emptyset\)).
Proof. We proceed by induction on the smallest \(N\) such that \(\Delta_q^N f = 0\).
Base case (\(N = 2\)). Suppose \(\Delta_q^2 f = 0\), meaning \(\Delta_q f\) is \(q\)-periodic. Since \(q \notin P_f\), Lemma 3.2 tells us that \(q \notin \operatorname{span}_\mathbb{Q}(P_f)\). Therefore the submodule \(G := P_f \oplus q\mathbb{Z}\) is a genuine direct sum and has rank at least 2, so by Proposition 1.3 it is dense in \(\mathbb{R}\).
Since \(P_f \subseteq P_{\Delta_q f}\) and \(q \in P_{\Delta_q f}\), the function \(\Delta_q f\) is periodic with respect to all of \(G\). Being periodic with respect to a dense submodule, \(\Delta_q f\) is either identically zero or nowhere continuous (by Proposition 1.1). Since \(q \notin P_f\), it is not identically zero, so \(\Delta_q f\) is nowhere continuous and in particular not everywhere zero.
Now, \(\Delta_q f\) is constant on each coset of \(G\): for any \(g \in G\), we have \(\Delta_q f(x + g) = \Delta_q f(x)\). Since \(\Delta_q f\) is not identically zero, there exists \(y \in \mathbb{R}\) such that \(\Delta_q f(y) = c \neq 0\). By the constancy on cosets, \(\Delta_q f(x) = c\) for all \(x \in y + G\). By the telescoping identity and the \(q\)-periodicity of \(\Delta_q f\):
\[f(y + p + nq) = f(y) + nc \quad \text{for all } p \in P_f, \, n \in \mathbb{Z}.\]
Now let \(I\) be any open interval. Since \(y + G\) is dense in \(\mathbb{R}\), the set \(I \cap (y + G)\) is infinite. Each element of this set can be written uniquely as \(y + p + nq\) for some \(p \in P_f\) and \(n \in \mathbb{Z}\), with \(f(y + p + nq) = f(y) + nc\). It remains to show that \(n\) cannot be bounded on \(I \cap (y + G)\).
Suppose for contradiction that \(|n| \leq n_0\) for all \(y + p + nq \in I\). For any \(\epsilon > 0\), by the density of \(G\) near the origin, there exists \(r \in P_f\) and an integer \(M\) with \(|M| > n_0\) such that \(|r + Mq| < \epsilon\). Choosing \(\epsilon\) small enough and \(x = y + p + nq \in I\) appropriately, the point \(x + r + Mq = y + (p + r) + (n + M)q\) also lies in \(I \cap (y + G)\) but has \(|n + M| > n_0 - n_0 = 0\) — and by choosing \(|M|\) large enough, we can make \(|n + M|\) exceed \(n_0\), a contradiction.
Therefore \(f\) is unbounded on every open interval, i.e., \(f\) is nowhere bounded.
Inductive step. Suppose the result holds for all smaller values of \(N\), and let \(N \geq 3\) be the smallest positive integer such that \(\Delta_q^N f = 0\). Write \(h = \Delta_q^{N-2} f\). Then \(\Delta_q^2 h = \Delta_q^N f = 0\), and \(q \notin P_h\) (since if \(q \in P_h\) then \(\Delta_q h = \Delta_q^{N-1} f = 0\), contradicting the minimality of \(N\)). By the base case, \(h = \Delta_q^{N-2} f\) is nowhere bounded. Applying Lemma 3.1 repeatedly (\(N - 2\) times), we conclude that \(f\) is nowhere bounded.