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.

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

1 Answer 1


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.

  • $\begingroup$ Don't I have to show first that $f_n$ is differentiable before jumping to prove 2.? $\endgroup$
    – CHTM
    Dec 12, 2020 at 19:18
  • $\begingroup$ Didn't I do just that? $\endgroup$ Dec 12, 2020 at 19:22
  • $\begingroup$ 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. $\endgroup$
    – CHTM
    Dec 12, 2020 at 19:34
  • $\begingroup$ Do you think that I have answered your question? $\endgroup$ Dec 12, 2020 at 20:22
  • $\begingroup$ Yea sorry I was just trying to see why $f$ is not differentiable at 1 $\endgroup$
    – CHTM
    Dec 12, 2020 at 20:47

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