Proving that uniform convergence of a derivative series implies uniform convergence of the series In a past answer to a different thread the following theorem was used: 

Theorem 1: If $\sum_{n=1}^{\infty}f_n(x)$ converges at least at one point, and the series of derivatives $\sum_{n=1}^{\infty}f_n^{'}(x)$ converges uniformly in some interval $I$, then in that interval the series can be differentiated term by term, meaning that the sum of the derivatives converges to the derivative of the sum.

After I asked the proof to the author of the answer it was pointed to me this link: Uniform convergence of derivatives, Tao 14.2.7. where the following theorem is found:

Theorem 2Let $I = [a,b]$ and $f_n : I \to \mathbb{R}$ a sequence of differentiable functions such that the sequence of derivatives $(f_n')_n$ converges uniformly to a function $g : I \to \mathbb{R}$. Also $\exists x_0 \in I$ such that the sequence $(f_n(x_0))_n$ converges. Then $(f_n)_n$ converges uniformly to a differentiable function $f : I \to \mathbb{R}$ with $f' = g$.

I understand that I could treat $F_k=\sum_{n=1}^{\infty}f_n(x)$, since $d\frac{\sum_{n=1}^{\infty}f_n(x)}{dx}=\sum_{n=1}^{\infty}\frac{df_n(x)}{dx}$. However I am looking for a proof for Theorem 1.
Question:
Could someone help me prove Theorem 1?
Thanks in advance!
 A: I shall assume that each $f_n'$ is Riemann integrable, in which case $g$ is Riemann integrable too. Note that, for each $x\in[a,b]$,$$f_n(x)=f_n(x_0)+g_n(x)-f_n(x_0)=f_n(x_0)+\int_{x_0}^xf_n'(t)\,\mathrm dt.$$So, define$$f(x)=\lim_{n\to\infty}f_n(x_0)+\int_{x_0}^xg(t)\,\mathrm dt$$and then\begin{align}\bigl|f(x)-f_n(x)\bigr|&\leqslant\left|f(x_0)-\lim_{n\to\infty}f_n(x_0)\right|+\left|\int_{x_0}^xg(t)-f_n(t)\,\mathrm dt\right|\\&\leqslant\left|f(x_0)-\lim_{n\to\infty}f_n(x_0)\right|+\int_{x_0}^x\bigl|g(t)-f_n'(t)\bigr|\,\mathrm dt.\end{align}So, $(f_n)_{n\in\mathbb N}$ converges uniformly to $f$.
A: This is Theorem 7.17 of Rudin's Principles of mathematical analysis.  The statement and proof occupy pages 152-153 of the text.
The result and its proof also appear as Theorem 13.8 of my honors calculus notes.  Notice that just before discussing this result, I state and prove a similar result but with slightly stronger hypotheses: instead of differentiable I assume that the functions are continuously differentiable (and thus the derivative is Riemann integrable). The point is that this allows an easier argument using the Fundamental Theorem of Calculus -- the same one that appears in the answer of José Carlos Santos.
