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In my work I came up with a continuous function $f(x)$ on $[0, 1]$ with the following properties:

  1. $f(0) = 0$,
  2. $f\left(\frac{1}{n}\right) = 0$ for all natural $n$,
  3. $\displaystyle{f\left(\frac{1}{2}\left(\frac{1}{n^a} + \frac{1}{(n + 1)^a}\right)\right) = \frac{1}{n^a} }$, where $a$ is some positive real parameter, and
  4. in all other intervals the function $f(x)$ is linear.

Given that, for what values of parameter $a$ is the function $f(x)$ of bounded variation on $[0, 1]$?

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    Are you sure it's $f(1/n) = 0$ rather than $f(1/n^a) = 0$? The way you have it, there's no guarantee that you can't have $\frac{1}{m} = \frac{1}{2} \left( \frac{1}{n^a} + \frac{1}{(n+1)^a} \right)$ for some $m$ and $n$, producing a contradiction.2011-06-07
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    @Robert Israel We can add the condition that $a$ is irrational to make sure that the function is well defined. I wouldn't be surprised if there were no conflicts for "most" rational choices of $a$ as well, though I would not place money on it. If $f(n^{-a})=0$ instead, it trivializes the problem.2011-06-07
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    @Aaron: I understand you are saying that if $a$ is irrational, then the equation $2/m=1/n^a+1/(n+1)^a$ has no integer solution. Why is that so?2011-06-07
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    @Didier Piau I was not thinking carefully when I wrote that. In retrospect, the argument that I had does not work. In fact, for fixed $n$, the right hand side is a smooth, decreasing function of $a$, and for most values of $m$ there will be a solution for some $n$. However, this means that there are only countably many $a$ such that there is a solution $(m,n)$ to the equation, and so for generic choice of $a$, there is no solution.2011-06-07
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    @Aaron: Yes. $ $2011-06-07

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