I would like to show the following result
Let $(f_n)_n$ be a sequence of functions $C^{1}$ of an interval $I$ of $\mathbb{R}$ with real or complex values. We suppose that we have the following two conditions
The sequence of derivatives $(f_{n}^{'})_n$ converges uniformly to a function $g$ on $I$.
There exists $x_0\in I$ such that the numerical sequence $(f_n(x_0))_n$ converges to a limit $l$.
Then the sequence $(f_n)_n$ simply converges on $I$ to a function $f$ which is $C^{1}$ and verifies $f'=g$.
What I have tried : We have for all $x\in I$:
$f_n(x)=f_n(x_0)+\int_{x_0}^{x}f_{n}^{'}(t)dt$
Then we apply the theorem that allows to invert limit and integral when we have uniform convergence:
$\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}f_n(x_0)+\lim_{n\to\infty}\int_{x_0}^{x}f_{n}^{'}(t)dt=l+\int_{x_0}^{x}\lim_{n\to\infty}f_{n}^{'}(t)dt=l+\int_{x_0}^{x}g(t)dt$
and we notice then, using the continuity of $g(t)$ and posing $d'(t)=g(t)$, that
$f'(x)=\frac{d}{dx}\int_{x_0}^{x}g(t)dt=\frac{d}{dx}[d(x)-d(x_0)]=g(x)$
Do you think it is correct ?