# Every period of function is multiple of fundamental period

Suppose $$f : \mathbb{R} \to \mathbb{R}$$ is periodic with $$T$$. Is it necessarily true that $$T$$ is a multiple of fundamental period $$T_0$$? Obviously every multiple of $$T_0$$ is a period. Will the other way hold? Intuietvily it seems to hold but don't know how to prove that.

A function is said to be periodic if there exists $$T\gt0$$ such that $$f(x+T) = f(x)$$ for every $$x \in D_f$$.

No, not necessarily, not without further assumptions. A constant is periodic with any period at all, and the infamous function which is $$1$$ for rational argument and $$0$$ otherwise is periodic with any rational period. A primitive period exists for continuous and non-constant periodic functions, but the proof of that is not trivial.

• What's the further assumptions that we should take? Commented Apr 12, 2020 at 19:38
• I think if we assume $T_0$ exists then that statement is true, right? Commented Apr 12, 2020 at 19:46
• Does anybody know where I could find the proof for the statement that a primitive period exists for continuous, non-constant functions? Commented Aug 7, 2022 at 22:01

Yes, if the the fundamental beriod is the smallest period, divides $$T$$ by $$T_0$$

$$T=nT_0+r, r, $$f(x+r)=f(x+T-nT_0)=f(x)$$ implies that $$r=0$$ since $$r$$ is a period and $$r.

Yes. By definition, the fundamental period is the smallest period $$T_0$$ such that $$f$$ is $$T_0$$-periodic. So any period $$T$$ must satisfy $$T\geq T_0$$. For some period $$T$$, let $$T=n T_0+(T\text{ mod }T_0)$$. Since $$f$$ is $$T$$-periodic and $$T_0$$-periodic, we have that $$f(0)=f(T)=f(n T_0+(T\text{ mod }T_0))=f(T\text{ mod }T_0)$$. Then $$T\text{ mod }T_0$$ is a period of $$f$$ which is less than $$T_0$$ and $$T_0$$ cannot be the fundamental frequency, which is a contradiction.

We can prove this statement:

If $$f$$ is a closed curve (~a continuous, non constant and $$0\neq T$$-periodic curve) with smallest positive period $$T_0$$, then every positive period is a multiple of $$T_0$$.

Proof

We suppose the existence of $$T_0$$ (some Real Analysis can help). Let $$T>T_0$$ (if $$T=T_0$$, there is nothing to prove); the integer linear combination $$T-T_0$$ is a period of $$f$$. If $$T-T_0=T_0$$, we have finished, because $$T=2T_0$$. Otherwise $$T-T_0 or $$T-T_0>T_0$$. The first is impossible, for $$T_0$$ is the smallest positive period. Thus $$T-T_0>T_0$$ or $$T_0<\frac{T}{2}$$. Continue similarly and get $$\forall n\in\Bbb N: 0\leq T_0<\frac{T}{n}\rightarrow0$$, so $$T_0=0$$, which is a contradiction. Consequently, there exists a $$n\in\Bbb N$$ such that $$T=nT_0$$.