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A function $f:[0,2] \to \mathbb R $ is given in this way :
$f(x)=\inf\{|x-\frac{1}{n}|:n \in \mathbb Z^+\}$

How can one find all the points where $f$ has derivation on them?

Note : I know that existence of the derivation is equivalent to the existence of a limit. But that's not an algorithm. I can't just put every single point of $[0,2]$ in the formula and calculate that limit. What should i do ?

Tina Ashtiani
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2 Answers2

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I would suggest you think about the behaviour of your function on closed intervals of the form $[\frac{1}{k+1}, \frac{1}{2}(\frac{1}{k}+\frac{1}{k+1})]$ and $[\frac{1}{2}(\frac{1}{k}+\frac{1}{k+1}), \frac{1}{k}]$, for a $k \in \mathbb{Z}^{+}$. I suspect that if you analyze the way the function behaves on these intervals then that will help you to solve your problem.

Rupert
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  • would you please explain it more? – Tina Ashtiani Dec 08 '16 at 12:06
  • Well, I mean, I haven't solved the problem, this is just my guess about a strategy that might work. Let's try it. If $x \in [\frac{1}{k+1}, \frac{1}{2}(\frac{1}{k}+\frac{1}{k+1})]$ then $f(x)=x-\frac{1}{k+1}$. So what can you say if $x$ is in the interior of the closed interval (not one of the two endpoints)? Does $f$ have a derivative for that value of $x$ if that is the case? What will you do to solve the question about the endpoints? – Rupert Dec 08 '16 at 12:09
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You should break it up into cases. I'll do the first one for you, may add more later if I have time. If $x \in [1, 2]$, then $x- \dfrac 1n$ is positive, and it achieves its minimum when $n=1$. Therefore on $[1, 2]$ your $f(x)=x-1$, which you know is differentiable on $(1, 2)$.

Ovi
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