Every period of function is multiple of fundamental period

Question

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 $x \in D_f$.

Answer 1

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.

Answer 2

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

$T=nT_0+r, r<T_0$, $f(x+r)=f(x+T-nT_0)=f(x)$ implies that $r=0$ since $r$ is a period and $r<T_0$.

Answer 3

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.

Answer 4

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 $T-T_0<T_0$ 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 $T=nT_0$.