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.
Best Answer
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