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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.

Andrés E. Caicedo
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Akash Patalwanshi
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1 Answers1

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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.

dxiv
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  • Thanks, please could you explain these points in your answer, $qT_1 - pT_1$ is also a period How? – Akash Patalwanshi Jan 05 '18 at 02:04
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    @AkashPatalwanshi Any linear combination of periods with integer coefficients is also a period: $f\big(x + (qT_1-pT_2)\big) = f\big((x + qT_1)-pT_2\big)=f\big(x+qT_1)$ because $T_2$ is a period, and $f\big(x+qT_1)=f(x)$ because $T_1$ is a period. – dxiv Jan 05 '18 at 02:08
  • Further, density follows because there exists infinitely many periods in the neighborhood of $0$ is , am I correct? – Akash Patalwanshi Jan 05 '18 at 02:09
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    @AkashPatalwanshi Correct. Otherwise said, with the notation from your question, choose a small enough period $T$ such that $|T| \lt \delta,$, then every interval $(k-\delta,k+\delta)$ will contain an integer multiple of $T$, which is itself a period. – dxiv Jan 05 '18 at 02:12
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    Thanks for your help sir. Got it. – Akash Patalwanshi Jan 05 '18 at 02:13
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    @dxiv: Can you explain why a periodic function without a fundamental period must have two periods which are incommensurate? – Leonidas Feb 29 '24 at 19:10
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    @Leonidas Thank you for catching that. The assertion is clearly false as stated, and the Dirichlet function is an immediate counterexample, as the OP mentioned. The case with all commensurate periods was more obvious, or so I must have thought when I left it out of the answer. It is getting late in my timezone now, will post an update and fill in the blanks tomorrow. – dxiv Mar 01 '24 at 07:22
  • @Leonidas Should be fixed now. See also Characterizing Dense Subgroups of the Reals, which is relevant because the set of periods is an additive subgroup of the reals. – dxiv Mar 01 '24 at 22:38