Pointwise Convergence of Lipschitz Functions is Uniform Say we have $f_n:[a,b]\to \mathbb{R}$ , such that for all $n$ we have $|f_n(x)-f_n(y)|\leq L|x-y|$ and $f_n \to f$ pointwise.
Is the convergence uniform?
I started with an attempt to prove it by showing Cauchy Criterion for uniform convergence : for any $c\in[a,b]$
$$|f_n(x)-f_m(x)|=|f_n(x)-f_n(c)+f_n(c)-f_m(c)+f_m(c)-f_m(x)|\\\leq|f_n(x)-f_n(c)|+|f_n(c)-f_m(c)|+|f_m(c)-f_m(x)|\leq 2L|x-c|+|f_m(c)-f_n(c)|.$$
Now let $\epsilon>0$, there exists $n,m$ such that $|f_m(c)-f_n(c)|<\frac \epsilon 2$ from pointwise convergence.
Also, if we take $x\in(c-\frac \epsilon {4L},c+\frac \epsilon {4L})$, we get $|f_n(x)-f_m(x)|<2L\frac \epsilon {4L}+\frac \epsilon 2=\epsilon$.
So we have uniform convergence in $(c-\frac \epsilon {4L},c+\frac \epsilon {4L})$ for all $c\in[a,b]$. We can get finite cover of these covers, where we have uniform convergence, and take the maximum $N$ from all of those intervals to get uniform convergence in $[a,b]$.
However, My friend presented me with a possible counter-example : $f_n(x)=nxe^{-nx}$ in [0,1], where the derivative is bounded, and the convergence is only pointwise to 0, and not uniform.
I couldn't find where my proof fails, and I`d be glad if someone can point it out for me.
 A: The counter-example is wrong: the sequence $f_n(x)=nxe^{-nx}$ is not uniformly Lipschitz in $[0,1]$ since, as $n\to +\infty$,
$$\frac{f_n(1/n)-f_n(0)}{1/n-0}=ne^{-1}\to +\infty.$$
Moreover your proof is correct. See also Given sequence of $L-$Lipschitz functions which converges pointwise, prove uniform convergence
A: +1 for Robert's response...
Let $\varepsilon>0$ be given, and set $\delta=\min\left[\frac{\varepsilon}{4}, \frac{\varepsilon}{4L}\right]$. Since the collection of open balls $\mathcal{B}: = \{B(\, x, \delta) : x \in [a,b] \}$ is a cover for $[a,b]$, we may refine $\mathcal{B}$ to a finite subcover, say $\{B(\,x_1, \delta),  \, \ldots, \, B(\,x_K, \delta)\}$ (Heine-Borel Theorem).  Since $f_n$ converges pointwise on $[a,b]$, for each point $x_j \: \left(\,j=1,\ldots, K \right)$ there exists a positive integer $N_j$ so that 
\begin{equation} \left|\, f_n(x_j) -
 f_m(x_j) \right| <  \frac{\varepsilon}{4}  \text{ whenever } n, m \geq N_j \,.
\end{equation}
We set $N = \max [N_1,  \ldots, N_K]$. Now, given $x \in [a,b]$ we know there exists a positive integer $M_x \geq N$ so that $|\,f_{M_x}(x)-f(x)| < \frac{1}{4}\varepsilon$ (since $f_n \to f$ pointwise) and that $|x-x_j| < \delta$ for some $j \in \{1, \ldots, K\}$ (definition of a cover). We combine to notice that if $x \in [a,b]$ and $n \geq N$, then
\begin{aligned}
\left|\,f_n(x)- f(x) \right| & \leq \left| \,f_n (x)- f_n(x_j) \right| + \left| \,f_n (x_j)- f_{M_x}(x_j) \right| + \left|\, f_{M_x} (x_j)- f_{M_x}(x) \right| + \left|\, f_{M_x}(x)- f(x) \right| \\
& < \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} = \varepsilon \, . 
\end{aligned}
Therefore, $f_n \to f$ uniformly on $[a,b]$.
