Function problem involving limits and periodicity

Question

Let $f: \Bbb R \to \Bbb R$ a continuous, non-constant and periodic function, such that $T \in \Bbb R-\Bbb Q$ is one of its periods. Show that the sequence $f(n)$ cannot be convergent, where $n \in \Bbb N$

Answer 1

Let $m=\min f$, and $M=\max f$ (which exist as we can simpy consider the compact interval $[0,T]$). By assumption, $M>m$, and we find $x_m,x_M$ and $\epsilon>0$ such that $f(x)>\frac{2M+m}3$ for $|x-x_M|<\epsilon$ and $f(x)<\frac{M+2m}3$ for $|x-x_m|<\epsilon$. As $\frac 1T$ is irrational, the set of fractional parts $\{\frac nT\}$ is dense in $[0,1]$. Therefore, there are infinitely many $n$ such that $\{\frac nT\}-\{\frac {x_m}T\}<\frac \epsilon {|T|}$ and hence $f(n)<\frac{M+2m}3$. Likewise, there are infinitely many $n$ such that $f(n)>\frac{2M+m}3$. This is not compatible with $f(n)$ converging.