Equi-continuous sequence of functions from a continuous function having vanishing limit at infinity Let $f:[1, \infty) \rightarrow \mathbb{R}$ be a continuous function such that
$$
\lim _{x \rightarrow \infty}|f(x)|=0
$$
For $n \geq 1$, let $g_{n}:[1, \infty) \rightarrow \mathbb{R}$ be given by
$$
g_{n}(x)=f(n x)
$$
Show that $\{g_n\}_{n \geq 1}$ is equicontinuous on $[1, \infty)$.
How to make $|f(nx)-f(ny)|<\epsilon$ for all $n \geq 1$ while from the limit, we can have $|f(nx)-f(ny)|\leq|f(nx)|+|f(ny)|<\epsilon$ whenever $n \geq M$ for some $M>0$.
 A: For any $\epsilon >0~~~\exists~M>0$ such that $|f(x)|<\frac{\epsilon}{4}$ for all $x\geq M$. Without loss of generality we can let $M>1$.
$f$ is uniformly continuous on $[1,M]$ (since it is continuous there). For the above $\epsilon$ there exists $\delta >0$ such that  $x_1, x_2 \in [1,M]$ and $|x_1-x_2|<\delta$ implies $|f(x_1)-f(x_2)|<\frac{\epsilon}{2}$.
If $x_1, x_2 \geq M$ then $|f(x_1)-f(x_2)|\leq |f(x_1)|+|f(x_2)|<\frac{\epsilon}{2}<\epsilon$.  (No role of  $\delta$  here).
If  $|x_1-x_2|<\delta$ and  $x_1 <M$,  $x_2>M$  then  $|f(x_1)-f(x_2)|\leq |f(x_1)-f(M)|+|f(M)-f(x_2)| \leq \frac{\epsilon}{2}+\frac{\epsilon}{4}+\frac{\epsilon}{4}=\epsilon$   (because $|x_1-M|<\delta$).
Combining we get $f$ is uniformly continuous on $[1, \infty)$ and $|x_1-x_2|< \delta$ implies  $|f(x_1)-f(x_2)| \leq \epsilon$.
Now for $x_1, x_2 \in [1, \infty)$ and $n \in \mathbb{N}$, $nx_1, nx_2$ are just two points in $[1, \infty)$. So $|nx_1-nx_2|< \delta$ implies  $|f(nx_1)-f(nx_2)|\leq  \epsilon$, i.e. $|g_n(x_1)-g_n(x_2)|\leq \epsilon$ for all $n$.
