## Example for deterministic function with unbounded total variation and bounded quadratic variation

It is well known that e.g. $sin(1/x)$ is of unbounded total variation (in the interval [0,1] assuming $f(0)=0$). (Preliminary numerical tests suggest that) it is also of unbounded quadratic variation. $x\ sin(1/x)$ is of unbounded total variation too but (preliminary numerical tests suggest that) it is of zero quadratic variation.

My question: How to construct a deterministic function with unbounded total variation and bounded (non zero) quadratic variation along these lines? I don't want to have a function which is defined by a sum of terms (like the Weierstrass function) but one which is defined straight forward like the two above mentioned examples. References (if available) would also be appreciated!

Addendum: If some of these conjectures are not true please tell me. And please tell me also if it is not possible to construct such a deterministic function (and why not).

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Function $x^a\sin(1/x)$ on $(0,1]$ has bounded variation iff $a>1$ and finite quadratic variation iff $a>1/2$. So for your example, take $1/2 < a \le 1$. It looks like more than numerical tests may be needed to decide questions like this...

I took "quadratic variation" to mean $$\sup \sum_{j=1}^n |f(x_j)-f(x_{j-1})|^2$$ with sup over all finite sequences $x_0\lt x_1\lt\cdots\lt x_n$ in $(0,1]$. Perhaps you meant something else?

On the other hand, maybe you mean $$\lim_{\delta \to 0}\sup \sum_{j=1}^n |f(x_j)-f(x_{j-1})|^2$$ with sup over all finite sequences $x_0\lt x_1\lt\cdots\lt x_n$ in $(0,1]$ such that $x_j-x_{j-1}<\delta$. In that case, on any inverval where $f$ is monotone it has quadratic variation $0$, so you won't find any such simple example.