Show that $f_n(\cdot)$ is uniformly convergent. 
Let $f_n : [0,1] \to \mathbb{R}$ be a sequence of continuously differentiable functions such that
  $$f_n(0) = 0, \ \ |f'_n(x)| \leq 1, \ \ \forall n \in 1, x \in (0,1).$$
  Suppose further that $f_n(\cdot)$ is convergent pointwise to some function $f(\cdot)$. Show that $f_n(\cdot)$ converges to $f(\cdot)$ uniformly.

We must show that $\forall \epsilon > 0$ there exists an index $N(\epsilon)$ independent of $x$ such that
$$|f_n(x) - f(x) | < \epsilon \ \forall n \geq N(\epsilon).$$
We know that $\forall \epsilon > 0$ there exists an index $N(\epsilon, x)$ dependent on $x$ such that
$$|f_n(x) - f(x) | < \epsilon \ \forall n \geq N(\epsilon,x).$$
We also know that since the functions are continuously differentiable, they are continuous. Since they are continuous, each function also achieves a maximum and minimum, and are bounded. 
I'm not sure where to go with all of this information. How can the bounded derivative help me? 
 A: Hint: By the mean value theorem, $|f_n(y)-f_n(x)| \le |y-x|$ for each $n.$ This implies, just from pointwise convergence, that $|f(y)-f(x)| \le |y-x|$ as well. Given $\epsilon >0$ think about a partition of $[0,1]$ into subintervals of length less than $\epsilon$ and use the pointwise convergence at the partition points.
A: A direct proof:
Let $\epsilon>0$. Take $(x_{l})_{l=1}^{m}\subseteq [0,1]$ such that $\bigcup_{l=1}^{m}\{x\in[0,1]:|x-x_{l}|<\epsilon\}=[0,1]$. If $n$ is large enough, then $|f(x_{l})-f(x_{l})|<\epsilon$, for all $l\in [m]$. Take $x\in [0,1]$ and $j\geq n$ such that $|f_{j}(x)-f(x)|<\epsilon$. We have:
$$ |f(x)-f_{n}(x)| \leq |f(x)-f_{j}(x)|+|f_{j}(x)-f_{j}(x_{l})|+|f_{j}(x_{l})-f_{n}(x_{l})|+|f_{n}(x_{l})-f_{n}(x)|$$
Then,
$$|f(x)-f_{n}(x)|<\epsilon +\|f'_{j}\||x-x_{l}|+2\epsilon+\|f'_{n}\||x-x_{l}|$$
Since $\|f'_{k}\|<1$, for all $k$, if we choose $x_{l}$ close to $x$, we finish the proof.
Edit: $\|f'_{k}\|=\sup_{x\in[0,1]}|f'_{k}(x)|$
A: Hints:


*

*Show that $f$ need to be Lipschitz continuous with constant $1$. Thus, you may assume without loss of generality that $f=0$ (if you assume $f_n$ has Lipschitz constant 2).

*Fix $m$. Let $x_0=0$ and let $x_k = k/m$ for $k=1,\dotsc,m$. Then, for $x\in [x_{k-1}, x_k]$ it follows
$$ |f_n(x)| \le |f_n(x_k)|  + |f_n(x_k) - f_n(x)| \le |f_n(x_k)| + \frac2m. $$
Show that the right hand side converges to $0$.
It is basically the proof of Arzela Ascoli. 
