Supose we have a sequence of real continuous functions on an interval $[0,b] \in \mathbb{R}$ $(f_n)_{n=0}^{\infty}$ (hence uniformly continuous). Moreover the sequence has the following properties:
- it is uniformly bounded, i.e. $f_n \leq \mu \in \mathbb{R}$ for every $n$,
- it is monotone (increasing),
- the derivatives are uniformly bounded $f_n' \leq M \in \mathbb{R}$ for every $n$.
How can i conclude using well knoknw theorems about convergence (which i don't remember) that the limit exists, it's Lipschitz continuous and the convergence is uniform?
From 1. and 2. i can get the pointwise convergence. Now using dominated convergence i can prove that $f_n$ is uniformly Cauchy in $\mathcal{C}[0,1]$ and so it converges uniformly to it's limit $f$: $$\sup_{x \in [0,b]}\lvert f_n(x)-f_m(x) \rvert \leq \lvert f_n(0)-f_m(0) \rvert + \int_0^b \lvert f_n'(t)-f_m'(t) \rvert dt$$
Is my proof correct? Does somebody know a shorter and more elegant proof that uses some theorem on uniform convergence? How can i conclude that the limit is Lupschitz continuous? (i could use that it has bounded derivative but how do i prove that the limit of the derivatives is the derivative of the limit and that it's bounded?)
Edit 1: I found a reasoning that might do the trick $$\lvert f(x)-f(y) \rvert = \lim\limits_{n \to \infty}\lvert f_n(x)-f_n(y) \rvert \leq M(x-y)$$ but i'm not 100% sure on the first equality (i'm a bit rusty on these topics). Is it because both quantities have finite limit?