# Prove that $f(x) = \sum_{n=1}^{\infty} 2^{-n} |x-1/n|$ is differentiable on $x_0 \notin \{1, 1/2, 1/3, ... \}$

Prove that $$f(x) = \sum_{n=1}^{\infty} 2^{-n} |x-1/n|$$ is differentiable on $$x_0 \notin \{1, 1/2, 1/3, ...\}$$

I think the relevant theorem is the term-by-term differentiation theorem which states:

Let $$f_n$$ be differentiable functions on A and assume $$\sum_{n=1}^{\infty} f_n'(x)$$ converges uniformly and that $$\exists x_0 \in A$$ where $$\sum_{n=1}^{\infty} f_n(x)$$ converges. Then $$f'(x) =\sum_{n=1}^{\infty} f_n'(x)$$ on A.

Here I know that $$\sum_{n=1}^{\infty} f_n(x)$$ converges for all R. So what I have left to show is that

1. Each $$f_n$$ is differentiable on $$\{1, 1/2, 1/3, ...\}^c$$
2. $$\sum_{n=1}^{\infty} f_n'(x)$$ converges uniformly on $$\{1, 1/2, 1/3, ...\}^c$$

But I'm stuck on showing 1. Once I write the limit definition I'm stuck.

• Think about the function $f(x) = |x-a|$, where $a \in \mathbb{R}$. Is it differentiable everywhere? Where is the critical point? Dec 12, 2020 at 18:15

Take $$n\in\Bbb N$$. Then $$f_n(x)=2^{-n}\left(x-\frac1n\right)$$ when $$x>\frac1n$$, and therefore $$f_n'(x)=2^{-n}$$ then. And $$f_n(x)=-2^{-n}\left(x-\frac1n\right)$$ when $$x<\frac1n$$, and therefore $$f_n'(x)=-2^{-n}$$ then.

This also proves that $$(\forall n\in\Bbb N)\left(\forall x\in\Bbb R\setminus\left\{1,\frac12,\frac13,\ldots\right\}\right):|f_n'(x)|=2^{-n}$$. Therefore, the series $$\sum_{n=1}^\infty f_n'(x)$$ is uniformly convergent.

• Don't I have to show first that $f_n$ is differentiable before jumping to prove 2.?
– CHTM
Dec 12, 2020 at 19:18
• Didn't I do just that? Dec 12, 2020 at 19:22
• Oh. I thought I had to use the limit definition to show that each $f_n$ is differentiable and then show $\sum_{1}^{\infty} f'_n(x)$ converges uniformly.
– CHTM
Dec 12, 2020 at 19:34
• Do you think that I have answered your question? Dec 12, 2020 at 20:22
• Yea sorry I was just trying to see why $f$ is not differentiable at 1
– CHTM
Dec 12, 2020 at 20:47