I am struggling with the proof of the following theorem: Let $(f_n)$ be a sequence of differentiable functions in a closed and bounded set $[a,b]$ s.t. $(f_n(x))$ is convergent for each $x\in [a,b]$ and, for a constant $M\geq0$, we have
$|f_n^{'}(x)|\leq M$ $\qquad$ $\forall n\in\Bbb{N}$ , $\forall x\in [a,b]$
Prove that $(f_n)$ is uniformly convergent in $[a,b]$...
I tried with this: as a consequence of the Lagrange Theorem we have that if $f_n$ is continuous and differentiable in a closed and bounded set and $f_n'$ is bounded then $f_n$ is Lipschitz. Then $|f_n(x_1)-f_n(x_2)|\leq |f_n^{'}(x) ||x_1-x_2|\leq M(x_1-x_2)<\epsilon$ $\quad$ whenever $|x_1-x_2|<\delta$. Thus $(f_n)$ is uniformly continuous.
Is this right? and from this how can I prove that $f_n$ is uniformly convergent?
It will be useful to define for $m\in \mathbb{N}$ the following points $$x_j^{(m)}:=a+ j(b-a)/m \text{ for } j\in \{ 0, 1, \dots, m \}.$$ Now note that $(f_n(x_j^{(m)}))_{n\geq 1}$ converges. For $m$ sufficiently large we find for any $x\in [a,b]$ some $j\in \{ 0, 1, \dots, m \}$ such that $\vert x - x_j^{(m)} \vert \leq 1/(2m).$ Then you can use your approach to get $$ \vert f_n(x) - f_\ell(x) \vert \leq \vert f_n(x) - f_n(x_j^{(m)}) \vert + \vert f_n(x_j^{(m)}) - f_\ell(x_j^{(m)}) \vert + \vert f_\ell(x_j^{(m)}) - f_\ell(x) \vert \leq 2M \vert x- x_j^{(m)} \vert + \vert f_n(x_j^{(m)}) - f_\ell(x_j^{(m)}) \vert \leq M/m + \vert f_n(x_j^{(m)}) - f_\ell(x_j^{(m)}) \vert. $$ Use this to show that for every $\varepsilon>0$ there exists $N\in \mathbb{N}$ such that for all $n,\ell \geq N$ holds $$ \sup_{x\in [a,b]} \vert f_n(x) - f_\ell(x) \vert < \varepsilon. $$ Now $(f_n)_{n\in \mathbb{N}}$ is a Cauchy sequence with respect to the supremum norm and thus converges uniformly.