Q:

Equicontinuity with decaying oscillators

Suppose $D \subset \mathbb{R}$, $f : D \to \mathbb{R}$ is a bounded continuous function, $a_n$ is a sequence of positive numbers tending to $+\infty$, and $b_n$ is a sequence of positive numbers tending to zero. Define $f_n(x)=b_n f(a_n x)$. Is the family $f_n$ equicontinuous?
I think the answer is yes, but I am not sure my construction is legitimate. Here is my idea. Define $M=\sup f - \inf f$; assume $M>0$, since the case $M=0$ is trivial. Let $\epsilon > 0$. Choose $N$ large enough that if $n > N$ then $b_n<\epsilon/M$. Let $n > N,x \in D,y \in D$. Then 
$$|f_n(x)-f_n(y)|=b_n|f(a_n x)-f(a_n y)| \leq \epsilon M/M = \epsilon.$$
Now the family $f_1,f_2,\dots,f_N$ is automatically equicontinuous, because it is finite and each function is separately continuous. So by defining the equi-modulus of continuity for the whole family at $\epsilon$ to be the equi-modulus of continuity of these first $N$ functions, we conclude that the entire family is equicontinuous. Is this last step legitimate? If not, is it legitimate if $D$ is assumed compact?
The title comes because this result is justified by considering problems like "is $f_n(x)=\frac{1}{n} \sin(n^n x)$ equicontinuous?"

A:

You need to assume $f$ is uniformly continuous, otherwise it's a nonstarter. Then the answer is yes: Each $f_n$ is then uniformly continuous and $f_n\to 0$ uniformly. So given $\epsilon>0,$ choose $N$ such that $n>N$ implies $|f_n| <\epsilon/2$ on $D.$ It's clear there exists $\delta > 0$ that works for $f_1,\dots , f_N.$ It also works for the rest of the $f_n$'s because for these $n,$
$$|f_n(y) - f_n(x)| \le |f_n(y)| + |f_n(x)| <\epsilon.$$

