How to show set of periods is dense in $\mathbb{R}$

Question

I just saw a result in my class notes of Real analysis, but I am unable to prove it.

Result: set of periods of any periodic function, which doesn't have fundamental period is dense in $\mathbb{R}$.

I had many examples which support above result. For eg: constant function, Dirichlet function etc.

But, couldn't able to prove the result. Though I had given try, as below:

Let $P$ be set of period of periodic function $f(x)$ which doesn't have the fundamental period.

and $P'$ be set of limit points of $P$.

Let $P'≠ \mathbb{R}$ then there exists $k$ in $\mathbb{R}$ such that $k$ is not limit point of $P$.

$→$ there exists $δ >0$ such that, $(k-δ, k+δ) ∩ P =∅ $

that is, in δ nbd of k, there does not exists any period of function $f(x)$.

From here I am unable to go further. Please help me.

Answer 1

We'll consider the following two cases separately.

• If all ratios of two periods are rational, let $\,T_1\,$ be a positive period. If all periods larger than $\,T_1\,$ are integer multiples of $\,T_1\,$ then $\,T_1\,$ is the fundamental period (since any positive period smaller than $\,T_1\,$ would have multiples which are not integer multiples of $\,T_1\,$). Given that there is no fundamental period, it follows that there must exist a positive period $\,T'_1 \gt T_1 \gt 0\,$ which is a rational non-integer multiple of $\,T_1\,$.
  
  Let $\,\,T_1' / T_1 = m_1/n_1 \gt 1\,$, $\,m_1 \in \mathbb Z\,$, $\,n_1 \in \mathbb N\,$, $\,n_1 \gt 1\,$, $\,\gcd(m_1, n_1) = 1\,$. By Bézout's identity there exist integers $\,a,b\,$ such that $\,am_1 + bn_1 = 1\,$. Dividing by $\,n_1\,$ and multiplying by $\,T_1\,$ gives $\,aT'_1 + bT_1=\frac{1}{n_1}T_1\,$, so $\,\frac{1}{n_1} T_1\,$ is also a period, and $\,T_2 = \frac{1}{n_1}T_1\le \frac{1}{2}T_1\,$.
  
  Repeating the steps produces a sequence of periods $\,T_k \le \frac{1}{2^k}T_1\,$ which tends to $\,0\,$. All that's left to prove is that for any real sequence $\,T_k \to 0\,$ the set $\,\{n T_k \mid n \in \mathbb Z, k \in \mathbb N\}\,$ is dense in $\,\mathbb R\,$, which is straightforward.

• Otherwise, there must exist two incommensurable periods $\,T_1 / T_2 = \alpha \in \mathbb{R} \setminus \mathbb{Q}\,$.
  
  By Dirichlet's approximation theorem there exist arbitrarily large integers $\,p, q\,$ such that $|\alpha - p/q| \lt 1 /q^2$ $\iff |q T_1 - p T_2| \lt T_2/q\,$. But $\,qT_1 - pT_2\,$ is also a period, so this implies that the function has arbitrarily small periods. Therefore $\,0\,$ is a limit point of the set of periods, and density follows from there.