Uniform continuity exchanges derivative I am trying to prove the following thing:
Let I = [a,b]. Let f and g be two functions from I to $\mathbb{R}$. 
Let $f_n$ a succession of functions  that is $C^1(I)$. 
I know that $f_n$ converges punctually to f. 
I also know that $f'_n$ converges uniformly to g. 
I want to prove the following things:
$f$ is $C^1$, $f_n$ converges uniformly to $f$ and  $f' = g$. 
Here it is my attempt:
I define $\widetilde{f} = f(x_0) + \int_{x}^{x_0}g(t)dt$. 
Now, if I can prove that $\widetilde{f} = f$ and that $f_n$ converges uniformly to $\widetilde{f}$, then $f' = g$ follows from Fundamental theorem of calculus. 
Now I write my definitions:
Punctual convergence: 
$\forall x \in [a,b]$, $\forall \epsilon > 0$, there exists $N(x,\epsilon)$ such that $\forall n>N(x,\epsilon$) we have $\vert f_n(x) - f(x)\vert < \epsilon$
Uniform convergence: 
$\forall x \in [a,b]$, $\forall \epsilon > 0$, there exists $N(\epsilon)$ such that $\forall n>N(\epsilon$) we have $\vert f'_n(x) - g(x)\vert < \epsilon$
Now however I struggle to keep going. Should I consider the epsilons and the N's in the two definitions different, right? 
Then how do I complete the proof?
 A: How I would do it:
Using that the $f_n$ are $C^1(I)$, write 
$$
f_n(x)=f_n(x_0)+\int_{x_0}^xf_n'(t)\mathrm dt
$$
for $x,x_0\in I$. Now taking limits and using that $f_n$ converges to $f$ pointwise we have 
$$
f(x)=f(x_0)+\lim_{n\to \infty}\int_{x_0}^xf_n'(t)\mathrm dt
$$
Finally, using uniformity of the limit of $f_n'\to g$, interchange limit and integral and find 
$$
f(x)=f(x_0)+\int_{x_0}^xg(t)\mathrm dt
$$
From here we see that $f$ is $C^1$ with derivative $g$.
For uniformity of $f_n\to f$, use our formula for $f_n$ to estimate
$$
|f(x)-f_n(x)|=\left|f(x_0)-f_n(x_0)+ \int_{x_0}^x g(t)-f_n'(t)\mathrm dt  \right|\\
\stackrel{\text{triangle ineq.}}{\leq} \left| \int_{x_0}^x |g(t)-f_n'(t)|\mathrm dt  \right|+|f(x_0)-f_n(x_0)|\\
\leq \left| \sup_{t\in [a,b]}|g(t)-f_n'(t)|\int_{x_0}^x \mathrm dt\right|+|f(x_0)-f_n(x_0)|\\
=|x-x_0|\sup_{t\in [a,b]}|g(t)-f_n'(t)|+|f(x_0)-f_n(x_0)|\\
\leq |a-b|\sup_{t\in [a,b]}|g(t)-f_n'(t)|+|f(x_0)-f_n(x_0)|\to 0
$$
as $n\to \infty$ independently of $x$ by assumption that $f_n'\to g$ uniformly. Uniform convergence of the $f_n$ then follows.
In answer to  your question, for any given epsilon, you can find an $N(\epsilon)$ making your estimate on the derivative hold, and then you can find $N_1(x,\epsilon)$ to make the estimate on the difference $|f_n(x)-f(x)|$ hold. These will in general be different natural numbers. 
It is worth learning the following criterion for uniform convergence: $f_n\to f$ uniformly on some domain $D$ if and only if 
$$
\sup_{x\in D}|f_n(x)-f(x)|\leq M_n
$$
where $M_n$ is a sequence tending to $0$. This avoids messing about with $\epsilon$'s and makes things cleaner.
A: Your definition of $\tilde f$ should be$$\tilde f(x)=f(x_0)+\int_{x_0}^xg(t)\,\mathrm dt.$$With this definition\begin{align}\left\lvert\tilde f(x)-f_n(x)\right\rvert&=\left\lvert f(x_0)+\int_{x_0}^xg(t)\,\mathrm dt-\left(f_n(x_0)+\int_{x_0}^xf_n'(t)\,\mathrm dt\right)\right\rvert\\&=\left\lvert f(x_0)-f_n(x_0)+\int_{x_0}^xg(t)-f_n'(t)\,\mathrm dt\right\rvert\\&\leqslant\left\lvert f(x_0)-f_n(x_0)\right\rvert+\left\lvert\int_{x_0}^xg(t)-f_n'(t)\,\mathrm dt\right\rvert\\&\leqslant\left\lvert f(x_0)-f_n(x_0)\right\rvert+\int_{x_0}^x\left\lvert g(t)-f_n'(t)\right\rvert\,\mathrm dt.\end{align}Since $\lim_{n\to\infty}f_n(x_0)=f(x_0)$ and since $(f_n')_{n\in\mathbb N}$ converges uniformly to $g$, it follows from this that $(f_n)_{n\in\mathbb N}$ converges uniformly to $\tilde f$.
