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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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up vote 3 down vote accepted

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?

addition
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.

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Super! I was looking for something like that! Thank you! Is it ">1" and ">1/2" or ">=1" and ">=1/2"? Could you give me some reference where I could find more about these very interesting functions? –  vonjd Apr 30 '10 at 14:09
    
What "a" should I choose to get a quadratic variation of exactly "1" - how do I determine the a? –  vonjd Apr 30 '10 at 14:23
    
I meant the definition of quadratic variation that is normally used in the context of stochastic processes, e.g. that Brownian motion has unbounded variation but finite quadratic variation. Which of the two is that? –  vonjd Apr 30 '10 at 14:24
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Actually, quadratic variation of BM is neither of those definitions ( they both give infinity). For BM you have convergence in probability for a sequence of deterministic partitions with mesh tending to zero. You also have almost sure convergence if each sequence is a refinement of the previous one. –  George Lowther Apr 30 '10 at 21:08

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