Let $f_n:[0,1]\to \mathbb{R}$ be a sequence of continuously differentiable functions such that
$$f_n(0)=0,\:\: |f_n'(x)|\leq 1, \text{for all }n\geq 1, x\in (0,1).$$ Suppose further that $f_n(.)$ is convergent to some function $f(.)$. Show that $f_n(.)$ converges to $f(.)$ uniformly.
I tired to prove this problem, but I'm lost and I cannot find a correct approach.
Firstly, I don't know if $f(.)$ is continuous. There must be some way to show $f$ is continuous, but I can't and I'm not sure if I need it. Secondly, I've obtained that $(f_n(.))$ are uniformly bounded and this because of $$\text{for some }z\in(0,x)\subset(0,1),\:\:|f'_n(z)|=|\frac{f_n(x)-f_n(0)}{x-0}|\leq 1\to |f_n(x)|< 1.$$
I tried to show $(f_n(.)) $ is equicontinuous. Here is my work.
Let $x\in (0,1),$ and let $\epsilon>0$. As $|f_n'(x)|\leq 1$, we can write $\lim_{y\to x}|\frac{f_n(x)-f_n(y)}{y-x}|\leq 1$, hence $|f_n(x)-f_n(y)|\leq |x-y|$. So if we take $\delta=\epsilon$,then for any $y$ that $|x-y|<\delta$, we have $|f_n(x)-f_n(y)|<\epsilon.$ So $(f_n)$ is equicontinuous on $(0,1)$.
Also we know $f_n(0)=0$, so this sequence is equicontinuous on $x=0$. If I'm correct in each of these steps, probably I'm done with the proof by showing that (f_n(.)) is equicontinuous on $x=1$.
Also in my proof, I didn't use this fact that $(f_n())$ are continuously differnetiable. I also saw the following proof, for this quetsion, but honestly, I think, it's not totally correct, since they use the continuity of $f(.)$ without proving it. here is the link "Show that $f_n(\cdot)$ is uniformly convergent."