$f:\Bbb R\to \Bbb R$ is continous and nonconstant function. Let $p$ be a positive real number such that $f(x+p)=f(x) $ for all $x \in \Bbb R$ . Then there exitst $n\in\Bbb N$ such that ${p \over n }=min\{a>0|f(x+a)=f(x), \forall x \in \Bbb R\}$.
Is this statement is true?
This is a subtle question, although our intuition seems to give a correct answer: yes, for non-constant $f$ the collection of $p\in\mathbb R$ such that $f(x+p)=f(x)$ for all $x\in\mathbb R$ is of the form $\mathbb Z\cdot \mu$ for some (minimal, non-zero) $\mu\in \mathbb R$.
Probably there is a more-elementary way to see this, but elementariness is not always a synonym for clarity... The way that I understand this is that the condition $f(x+p)=f(x)$ on $p$ (for fixed $f$) for a single $x\in\mathbb R$ is a closed condition, so the set of such $p$ is a closed subset of $\mathbb R$.
The conjunction of this requirement over all $x\in \mathbb R$ takes an intersection of those closed sets, so it is closed.
It is also a subgroup of $\mathbb R$!
Then one "classifies" the closed subgroups of $\mathbb R$: they are the trivial $\{0\}$, the whole $\mathbb R$, and the only intermediate cases are "lattices", namely subgroups of the form $\mathbb Z\cdot \mu$ for some $\mu>0$.
The hypotheses of the situation put us in the latter situation ...
Sorry, don't have enough points for a comment:
Note that $\frac {p}{n}+\frac{p}{n}+....+\frac{p}{n}$ ($n$ times)=$p$
If $p$ is the smallest period, then $n=1$ . If not, there is a smaller period $p'$. But, since $p$ itself is a period, then there is an integer $r$ with $rp$= $p+p+...+p$ ($r$ times) =$p$
