# 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? Jul 23, 2018 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, 2018 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, 2018 at 22:15
• You're right. I've added that as an explicit assumption. Jul 23, 2018 at 22:19
• Thanks for your answer! The proof regarded the theorem 2 that concerns sequences not series. How would it work with series? Jul 24, 2018 at 12:02
• @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 $f'=g$. Jul 24, 2018 at 12:07
• There is one thing that theorem 1 is confusing me. You say in theorem 2 that when the derivative of a sequence converges uniformly then the sequence itself converges uniformly. But on theorem 1(regarding the series) it is said that the derivative of each element of the series can only be taken if the series converge to a point in the interval considered and if the the derivative of the series converges uniformly. How can theorem 1 relate to theorem 2? Thanks in advance! Jul 24, 2018 at 13:15

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.