Problem on a sequence of functions, Rudin PMA, page 168

Question

Suppose $f$ is real continuous function on $\mathbb{R}$, $f_{n}(t)=f(nt)$ for $n=1, 2, 3,.....,$ and {$f_{n}$} is equicontinuous on $[0,1].$ What conclusion can you draw about $f$?

Answer 1

$f$ must be constant, suppose not and let $f(a)\neq f(b)$, let $\epsilon=|f(a)-f(b)|$, there must be a $\delta>0$ such that $|f_n(x)-f_n(y)|<\epsilon$ for all $n$ whenever $|x-y|<\delta$.

Pick $N$ such that $|a-b|/n<\delta$, we immediately reach a contradiction by noticing $|a/N-b/N|<\delta$ while $|f_N(a/N)-f_N(b/N)|=|f(a)-f(b)|=\epsilon$.

The contradiction comes from assuming there exists $a$ and $b$ with $f(a)\neq f(b)$. So $f$ must be constant, and clearly all constant functions satisfy that the sequence $f_n$ is equicontinuous.