I have a general question about sequence of functions $(f_n)$ from $[0,1]$ to reals which are non-uniformly Lipschitz (i.e., each has a Lipschitz constant $D_n$ but we don't know if it is bounded) and converge uniformly to a function with Lipschitz constant $D$. I have been thinking whether if this was sufficient to say that the set $\{D_n, n \in \mathbb{N}\}$ is bounded. I have seen here examples which give examples of non-Lipschitz functions converging to Lipschitz ones so it seems this idea may not be correct. And indeed I was able to build a counter example which is
$f_n(x) = nxe^{-n^nx}$
The derivative of these functions at $0$ is $n$, so there are elements of this sequence with arbitrarily large Lipschitz constant (and each $f_n$ is indeed Lipschitz since their derivative is continuous in $[0,1]$). Moreover for these functions the maximum is achieved at $x=\frac{1}{n^n}$ with value $\frac{1}{en^{n-1}}$ so they uniformly converge to $0$.
Thus this builds a counterexample. Now my question is could one add some more assumptions to such a sequence so that the Lipschitz constants would be bounded (of course no easy assumptions like the functions being differentiable and their derivatives converging etc, it should be a topological property).
Thanks