# 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?

• Which of the theorems you stated is Theorem 1? – Larara Jul 23 '18 at 21:50
• Are you interested in proving the theorem for sequences or series? If you want to prove it for series, You can just define $S_n(x)=\sum_0^nf_n(x)$ and use the second theorem. – Sar Jul 23 '18 at 21:54

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 $f$.
• There's no reason to think that $f_n'$ is Riemann integrable. – zhw. Jul 23 '18 at 22:15
• @PedroGomes Series are a particular type of sequences. In that case, the statement would be: if $\sum_{n=1}^\infty$ converges at some $x_0\in I$, if each $f_n$ is differentiable and $f_n'$ is Riemann-integrable, and if $\sum_{n=1}^\infty f_n'$ converges uniformly to $g$, then $\sum_{n=1}^\infty f_n$ converges pointwise to some function $f$. Furthermore, $f$ is differentiable and $f'=g$. – José Carlos Santos Jul 24 '18 at 12:07