Uniform convergence of derivatives, Tao 14.2.7. This is ex. 14.2.7. from Terence Tao's Analysis II book. 
Let $I:=[a,b]$ be an interval and $f_n:I \rightarrow \mathbb R$ differentiable functions with $f_n'$ converges uniform to a function $g:I \rightarrow \mathbb R$. Suppose $\exists x_0 \in I: \lim \limits_{n \rightarrow \infty} f_n(x_0) = L \in \mathbb R$. Then the $f_n$ converge uniformly to a differentiable function $f:I \rightarrow \mathbb R$ with $f' = g$.
We are not given that the $f_n'$ are continuous but he gives the hint that
$$
d_{\infty}(f_n',f_m') \leq \epsilon \Rightarrow |(f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0))| \leq \epsilon |x-x_0|
$$ This can be shown by the mean value theorem. My question is : How does this help me to prove the theorem ?
 A: Since $\{f_n(x_0)\}$ converges, for each $\epsilon > 0$ and $n, m$ large enough we have
$$
\begin{align}
\lvert f_n(x) - f_m(x) \rvert &\leq \left\lvert (f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0)) \right\rvert + \left\lvert f_n(x_0) - f_m(x_0) \right\rvert \\ 
&\leq \epsilon \left\lvert x - x_0 \right\rvert + \epsilon \\
&\leq \epsilon (b - a) + \epsilon
\end{align}
$$
Hence $f_n$ converges uniformly on $I$ to a function $f$, moreover for each $\epsilon > 0$ and $m, n$ large enough, the inequality
$$
\left\lvert \frac {f_n(y) - f_n(x)} {y - x} - \frac {f_m(y) - f_m(x)} {y - x} \right\rvert \leq \epsilon
$$
holds for each $x\neq y\in I$. (It is the same inequality of the hint but now we can assume it holds for generic $y\in I$, because we showed $f_n(y)$ converges for all $y \in I$)
The above relation implies that $\frac {f_n(y) - f_n(x)} {y - x}$ converges uniformly to $\frac {f(y) - f(x)} {y - x}$.
Now we can write
$$
\left\lvert\frac {f(y) - f(x)} {y - x} - g(x) \right\rvert \leq \\
\left\lvert\frac {f(y) - f(x)} {y - x} - \frac {f_n(y) - f_n(x)} {y - x} \right\rvert + \left\lvert \frac {f_n(y) - f_n(x)} {y - x}  - f_n'(x)\right\rvert + \left\lvert f_n'(x) - g(x) \right\rvert
$$
For each $\epsilon > 0$ and $n$ large enough we get
$$
\left\lvert\frac {f(y) - f(x)} {y - x} - g(x) \right\rvert \leq 2\frac \epsilon 3 + \left\lvert \frac {f_n(y) - f_n(x)} {y - x}  - f_n'(x)\right\rvert
$$
and for $y$ close enough to $x$
$$
\left\lvert\frac {f(y) - f(x)} {y - x} - g(x) \right\rvert \leq \epsilon 
$$
So $f'(x)$ exists and is equal to $g(x)$.
Edit
To clarify the point raised by @DavidC.Ullrich.
Since ${f'_n}$ converges uniformly, there exists $N \in \mathbb N$ such that $\lVert f'_n - f'_m \rVert_\infty < \epsilon$ for all $n, m > N$, that is
$$
|f'_n(x) - f'_m(x)| < \epsilon \qquad \forall m,n > N, \forall x\in I
$$
So, by means of the mean value theorem, for each $m,n > N$ and for each $x \neq y\in I$ we can write
$$
\left\lvert \frac {f_n(y) - f_n(x)} {y - x} - \frac {f_m(y) - f_m(x)} {y - x} \right\rvert = \\
\left\lvert \frac {f_n(y) - f_m(y)} {y - x} - \frac {f_n(x) - f_m(x)} {y - x} \right\rvert = \\
\left\lvert \frac {(f_n - f_m)(y)- (f_n - f_m)(x)} {y - x}\right\rvert = \\
\lvert (f_n - f_m)'(\xi) \rvert = \\
\lvert f_n'(\xi) - f_m'(\xi)\rvert < \epsilon
$$
A: Due to the uniform convergence of the $f'_n$ you can find an $N$ for every $\epsilon$ such that $(f'_n(x) - g(x)) \leq \epsilon$ for all $n \geq N$, which is equivalent to $d_\infty(f'_n,g) \leq \epsilon$. Thus, $d_\infty(f_n',g) \to0 $ as $n\to\infty$.
Now, $\int_{a}^x f_n'(y) - g(y) dy \leq d_\infty(f_n',g)|x-a| \leq d_\infty(f_n',g)|b-a|$. Since $d_\infty(f_n',g) \to0 $ as $n\to\infty$ you get that $\int_a^x f_n'(y)dy$ converges uniformly to $\int_a^x g(y)dy$.
In general, that won't transfer to $f_n(x) = c_n + \int_a^x f_n'(y)dy$ because the $c_n$ could be chosen maliociously. But if $\lim_{n\to\infty}f_n(x) = \lim_{n\to\infty} c_n + \int_a^x f_n'(y)dy$ converges for one $x$, then $\lim_{n\to\infty}c_n$ must converge, since the second term converges too (every uniformly!).
Which in turn means the limit must actually converge for all $x$, ecause $\lim_{n\to\infty}c_n$ doesn't actually depend on $x$. And for the same reason (and because the other term converges uniformly), the convergence is even uniform.
