Suppose $f$ is a 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 $f$?
I read the proof for this problem, but don't fully understand it. The following is the proof with my questions.
Proof: The function $f(t)$ must be constant on $[0,\infty)$. (Before proving in details, how to tell this function will probably be constant? and how is this extended to $\infty$?)
For if $f(x)\neq f(y)$ and $0\leq x<y<\infty$, say $|f(x)-f(y)|=\epsilon>0$, it follows that $|f_n(\frac{x}{n})-f_n(\frac{y}{n})|=\epsilon$ for all $n$. Since $\frac{x-y}{n}\rightarrow 0$, ,it follows that the family $\{f_n\}$ cannot be equicontinuous on $[0,1]$, or, indeed, on any neighborhood of $0$". (I understand the contradiction on equicontinuity, but I'm confused about the difference of those intervals, $[0,1]$ and $[0,\infty)$, and "on any neighborhood of $0$. Could someone help me fill in the details?)
Let $0 < x < y$, and suppose $|f(x) - f(y)| = \epsilon > 0$. Now, since the functions are uniformly equicontinuous (because they are equicontinuous in a compact set), there exists a distance $\delta > 0$ s.t. if $z,w \in [0,1]$ and $\ |z - w| < \delta$, $$ |f_n(z) - f_n(w)| < \epsilon \ (\forall n) $$
now take $m \in \mathbb{N}$ such that $|\frac{x}{m} - \frac{y}{m}| < \delta$ and $\frac{x}{m} , \frac{y}{m} \in [0,1]$, then $$ |f_m(\frac{x}{m}) - f_m(\frac{y}{m})| = |f(x) - f(y)| < \epsilon = |f(x) - f(y)| $$
This is a contradiction, so $f(x) = f(y)$.
Edit: as for your second question, if we have equicontinuity on any neighbourhood of $0$, we can take a closed interval inside it which is compact. Now, the argument will work the same, except we will need $\frac{x}{m} , \frac{y}{m}$ to be in the neighbourhood, and that we can always do by taking a sufficiently large $m$.