Conditions for convergence of a derivative, given the function itself is convergent Suppose $\{f_n\}_{n \in \mathbb{N}}$ is a family of bounded, differentiable, monotone increasing functions on $[0,1]$, which converge uniformly to a limit $f$.  Also, suppose we know that $f_n'$ is $\alpha$-Lipschitz continuous for some constant $\alpha$ (not depending on $n$).   I want to analyze the differentiability of $f$ and the convergence of $f_n'$, if this is even possible.
Of course, the monotone increasing property of $f_n$ implies that $f$ is also monotone increasing, so it is differentiable Lebesgue almost surely on $[0,1]$.  Is it possible to say that $f_n'$ converges to $f'$ pointwise, wherever $f'$ exists (even if only on a subsequence)?
 A: The answer is positive, and weaker assumptions suffice.
Claim: Suppose $\{f_n\}_{n \in \mathbb{N}}$ is a family of bounded, differentiable,  functions on $[0,1]$, which converge pointwise  to a limit $f$.  Also, suppose we know that $\{f_n'\}$ are a uniformly equicontinuous$^{(*)}$ family of functions.
Then $f$ is differentiable on $[0,1]$ and $f_n' \to f$ uniformly.
$(*)$ (Equicontinuity, defined in [1],  certainly holds if $f_n'$ are $\alpha$-Lipschitz continuous for some constant $\alpha$  not depending on $n$. In fact,  Holder continuity with uniform constants also suffices.).
Proof of claim: Since $f_n$ are uniformly bounded and $\{f_n'\}$ are  uniformly equicontinuous, it follows$^{\bf (**)}$ that the derivatives $f'_n$ are uniformly bounded. The Ascoli-Arzela theorem [1] implies that $f_n'$ has a subsequence $f'_{n(k)}$ that converges uniformly to some $g \in C[0,1]$.
Therefore, for all $t$ in $[0,1]$,
$$f(t)-f(0)=\lim_k \int_0^t f'_{n(k)}(x) \,dx = \int_0^t g(x) \,dx \,,$$
so $f'=g$ in $[0,1]$ by the fundamental theorem of calculus.
If $f'_n$ do not  converge  uniformly to $g=f'$, then there exists some  $\epsilon>0$ and  another subsequence $f'_{m(k)}$ of $f_n'$, such that  $\|f'_{m(k)}-f'\|_\infty >\epsilon$ for all $k$.
But $f'_{m(k)}$ must also have a uniformly convergent subsequence by [1],
and the argument above shows that the limit of this subsequence must be $f'$, a contradiction.
Thus $f'_n \to f'$ uniformly in $[0,1]$.
$(**)$ Addendum: Since  there exists $M$ such that $|f_n| \le M$ in $[0,1]$ for all $n$,    the mean value theorem implies that for each $n$, there is some $t_n \in (0,1)$ such that $|f_n'(t_n)|=|f_n(1)-f_n(0)| \le 2M$. Since $f_n'$ are uniformly equicontinuous, there is some $\delta_1<1 $ so that $|x-y|<\delta_1$ implies $|f_n(x)-f_n(y)|<1$ for all $n$. Thus for all $n \ge 1$ and all $x \in [0,1]$, we have
$|f_n'(x)-f_n'(t_n)| \le 1+1/\delta_1$, so $|f_n'(x)| \le 2M+1+1/\delta_1$.
[1] https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem
A: Yes.  I will prove this via functional analysis; I do not know if there is a simpler method.

*

*Note that $f_n$ are uniformly bounded, since they converge uniformly.

*Suppose $m\leq f_n\leq M$ for any $n$.  By the intermediate value theorem, there is some point $a_n\in[0,1]$ and $b_n\in[0,M-m]$ such that $f_n'(a_n)=b_n$.

*Each $f_n'$ is uniformly bounded in $L^{\infty}$, since $\{f_n'\}_n$ are equi-Lipschitz on a compact domain, and attain the similar values of the $b_n$.  The domain $[0,1]$ has finite measure; thus $f_n'$ are uniformly bounded in any $L^p$ for $p\in[1,\infty]$.

*For any $p\in(1,\infty)$, the space $L^p$ is reflexive.  Pick your favorite such $p$ (mine is $p=2$); then there is a subsequence (which I will also call $f_n'$) that converges weakly, by separable Banach-Alaoglu.  Call the limit $g$, but note that $g$ is only well-defined up to a.e. equivalence.

*Since $[0,1]$ has finite measure, if $\phi\in L^\infty$, then $\phi\in L^2$.  Thus $$\int{\phi f_n'}\to\int{\phi g}$$ by weak convergence in $L^2$.  So actually $f_n'\rightharpoonup g$ in $L^1$.

*Fix $\epsilon$ and let $U(\epsilon)_n=\{x:f_n'(x)-g(x)\geq\epsilon\}$ and $V(\epsilon)_n=\{x:g(x)-f_n'(x)\geq\epsilon\}$.  Then $$0\leq\epsilon|U(\epsilon)_n|\leq\int{1_{U(\epsilon)_n}(f_n'-g)}\to0$$  Likewise $|V(\epsilon)_n|\to0$, and so $f_n'\to g$ in measure as well as weakly.

*Convergence in measure implies convergence a.e. up to a subsequence.  Again, I will call that subsequence $f_n'$.

*Let $U$ be a dense set on which $f_n'\to g$ pointwise.  A pointwise limit of uniformly Lipschitz functions is also Lipschitz (exercise); thus any representative of the a.e.-equivalence class $g$ is uniformly continuous on $U$.  A uniformly continuous function on a dense set has a unique continuous extension to that set's closure; thus $g$ has representatives that are continuous on $[0,1]$.  Pick one such, and also call it $g$.

*Pick $x\in[0,1]$.  Then there exists $\{x_t\}_{t=0}^\infty\in U^\omega$ converging to $x$.  For any $t$, \begin{align*}
\limsup_{n\to\infty}{|g(x)-f_n'(x)|}&\leq\limsup_{n\to\infty}{|g(x)-g(x_t)|+|g(x_t)-f_n'(x_t)|+|f_n'(x_t)-f_n'(x)|} \\
&\leq|g(x)-g(x_t)|+0+\alpha|x_t-x| \\
&\to0
\end{align*} as $t\to\infty$.  Thus $f_n'\to g$ pointwise everywhere.

*For any set $[0,x]$, we have $$\int_0^x{g(x)\,dx}=\int{1_{[0,x]}g}=\lim_{n\to\infty}{\int{1_{[0,x]}f_n'}}=\lim_{n\to\infty}{(f_n(x)-f_n(0))}=f(x)-f(0)$$

*Since $g$ is continuous (everywhere), $f'=g$ everywhere by Lebesgue's differentiation theorem.

