Equicontinuous collection Suppose that $f: \mathbb{R} \to \mathbb{R}$ is a continuous function. For every integer $n \geq 1$ define $f_n(x) = f(nx)$ for all $x \in \mathbb{R}$ Assume the collection $\{f_n\}$ is equicontinuous.
What can you say about $f$?
 A: $f$ is constant on $[0,\infty)$ , $x,y\in[0,\infty)$ such that $x\neq y$, and suppose that $f(x)\neq f(y)$, let $\epsilon<|f(x)-f(y)|$, and let $\delta>0$ be such that  $|f(nt)-f(ns)|=|f_n((t)-f_n(s)|<\epsilon\forall t,s\in[0,1]$ such that $|t-s|<\delta$ and for all $n\ge 1$. There exists $N_1$ such that $\frac{|x-y|}{N_1}<\delta$ since both $x$, $y$ are greater than or equalto $0$ we see  that there exist $\frac{x}{N_2}\in [0,1],$ $\frac{y}{N_3}\in[0,1]$, If we choose $N>\{N_1,N_2,N_3\}$  then all three conditions holds simultaneously, but then we have $|f(x)-f(y)|=|f(Nx/N)-f(Ny/N)|=|f_N(x/N)-f_n(y/N)|<\epsilon$ (why?) but this contradicst the choice of $\epsilon$ to be smaller than $|f(x)-f(y)|$ hence we must have $f(x)=f(y)$
A: You have $|f(x)-f(y)|=|f_n(\frac{x}{n})-f_n(\frac{y}{n})|$. Equicontinuity says that $f_n$ behaves like f in a nerighborhood of every point (in the sense that the same delta can be chosen in the definition of continuity for $f_n$), let's say we choose $x_0=0$. This means that the second term tends to zero as $n \to \infty$ (because $f(x/n)-f(y/n)\to 0$), and so $f$ is constant.
