# Infinite convergent sum of differentiable functions always differentiable?

Let $\{f_i\}_{i \in \mathbb{N}}$ be a sequence of differentiable real-valued functions in $\mathbb{R}$.

Assume that for any $x \in \mathbb{R}$, the sum $N(x) = \sum\limits_{i=0}^\infty f_i(x)$ is convergent. Is there a general theorem that states that $N(x)$ is also a differentiable function?

• Definitely not without more assumptions, any continuous function on an interval can be approximated arbitrarily well by polynomials (Weierstrass theorem). May 14, 2018 at 21:32

Yes: if the series $\sum_{i=0}^\infty f_i'(x)$ converges uniformly to $g$, then $N$ is differentiable and $N'=g$.

• Thanks. It means then that the general statement is not true, but only under the assumption of uniform convergence of $\sum\limits_{i=0}^{\infty} f'_i(x)$ it is true? May 14, 2018 at 21:38
• No, it is not true. It is sufficient that that series converges uniformly, but it is not necessary. May 14, 2018 at 21:39

In general the sum is not differentiable even if $\sum_{n=1}^\infty f_n$ converges uniformly.

Example from @zhw. here:

Consider the smooth functions $f_n : \mathbb{R} \to \mathbb{R}$ defined as $f_n(x) = \sqrt {x^2+\frac1n}$.

For any $x\in \mathbb{R}$ we have

$$0\le f_n(x) - |x| =\sqrt { x^2+\frac1n }-\sqrt {x^2} =\frac{1/n}{ \sqrt {x^2+\frac1n}+\sqrt {x^2} } \le \frac{\frac1n}{\frac1{\sqrt n}} = \frac{1}{\sqrt n}$$

so $f_n \xrightarrow{n\to\infty} |\cdot|$ uniformly on $\mathbb{R}$. Every sequence can be turned into a series so we have:

$$f_1 + \sum_{n=1}^{\infty}(f_{n+1}-f_n) \xrightarrow{n\to\infty} |\cdot|$$

however the absolute value $|\cdot|$ is clearly not differentiable at $0$.

• But if $\sum f_n(x)$ converges of some $x_0\in I$ AND $\sum f'_n(x)$ converges uniformly for $x\in I$, then $\frac{d}{dx}\sum f_n(x)= \sum f'_n(x)$ for $x\in I$. But those conditions are only sufficient ones. May 14, 2018 at 21:49