Constant function as periodic

Prove that every continuous and periodic function that has every $\frac{1}{n}$ as a period is constant.

Solution

We shall show that $f(x)=f(0) , \; \forall x >0$. (It does not change anything in case $x<0$). Suppose that this does not hold. Then there exists an $a>0$ such that $f(a) \neq 0$. Consider the infimum $m$ of $x>0$ with $f(x)=f(a)$. Due to continuity we have that $f(m)=f(a)$. Thus $m=0$, because if $m>0$ then $0<\frac{1}{n}<m$ for some $n$ or equivelantly $0<m-\frac{1}{n}<m$. (contradiction). But since $m=0$ we have that $f(x)=f(a)=f(a)$ which is again a contradiction.

Thus $f(x)=f(0)$.

Remarks

Here are some similar interesting results.

1. Every periodic polynomial function is constant.
2. If $f:\mathbb{R} \rightarrow \mathbb{Q}$ is continuous , then $f$ is constant.
3. If $f:\mathbb{R} \rightarrow \mathbb{R}$ is a continuous function such that:
   
   $$f \left( x+\frac{1}{n} \right) =f(x)$$
   
   for every rational $x$ and $n \in \mathbb{Z}^+$ , then $f$ is constant.