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!