# Can this function be of bounded variation?

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. –  Robert Israel Jun 7 '11 at 0:35
@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. –  Aaron Jun 7 '11 at 0:53
@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? –  Did Jun 7 '11 at 6:58
@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. –  Aaron Jun 7 '11 at 7:14
@Aaron: Yes.  –  Did Jun 7 '11 at 8:25

The function will be of bounded variation whenever $a>1$ because even without knowing where the function is switching direction (it will depend on how points of the form $1/n$ and $\frac{m^{-a}+(m+1)^{-a}}{2}$ are interspersed), you can still say that the variation of the function will be less than $2\sum n^{-a}$, which is a convergent series.

The function is also clearly not of bounded variation when $a=1$, because there is a zero between every peak.

However, when $a<1$, the estimate above cannot show that the function is of bounded variation, and without estimates on how often points of them form $\frac{m^{-a}+(m+1)^{-a}}{2}$ are peaks of the function, I can't say that the function will definitely not be of bounded variation. For example, if for some choice of $a$, we only had peaks for $n\approx 2^k$ $k\in \mathbb N$, then the variation would be given by an almost geometric series. Of course, this is very unlikely, but the problem does not seem amenable to any standard techniques.

My guess is that the function is never of bounded variation for $0<a<1$, but I cannot prove the claim.

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