Understanding better quadratic variation and fractional derivative I'm don't really understand what's the meaning of fractional derivative, neither where it apply in the nature. Nevertheless, I often see that formally for a Brownian motion, we use the notation $dB_t=(dt)^{1/2}$. 


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*Q1) Does it mean that despite the fact that the Brownian motion has no derivative, it has a $\frac{1}{2}-$derivative ?

*Q2) More generally, I know that if $F$ has bounded variation, then it's derivable a.e. So if $f$ has $p$-bounded variation (i.e. $$\lim_{n\to \infty }\sum_{a\leq t_0<t_1<...< t_n\leq b}|f(t_{i+1})-f(t_i)|^p<\infty,$$
but not $q-$bounded variation for all $q<p$, would it make sense to say that $df$ repreent the $\frac{1}{p}-$derivative of $f$ ? (i.e. $df=(dt)^{\frac{1}{p}}$).

*Q3) Do you have an example of function that has quadric variation but is not of bounded variation on a compact set ? (in determinist case, i.e. not the Brownian motion).
 A: I guess that in the definition of a function $f$ of bounded $p$-variation should be the condition $$\sup \sum_{a\leq t_0<t_1<...< t_n\leq b}|f(t_{i+1})-f(t_i)|^p<\infty.$$ At least so is for $p=1$. Also $f(t_i)$ should be defined for each $t_i$ in the sum. Below I’ll use this definition. 
Now about Q3. There exists a function $f$ continuous of $[0,1]$ such that variation of $f$ on $[0,1]$ is unbounded [LYB, p.338]. Namely, put $f(x)=x\cos\tfrac \pi{2x}$ for $0<x\le 1$ and $f(0)=0$. Given a natural $n$, put $$(t_0,t_1,\dots, t_k)=(0, \tfrac 1{2n},\tfrac 1{2n-1},\dots,\tfrac 12,1).$$ Then 
$$\sum|f(t_{i+1})-f(t_i)|=1+\frac 12+\dots+\frac 1n\to\infty.$$ 
On the other hand, I conjectured that $2$-variation of $f$ on $[0,1]$ is bounded. I tried to prove this, but I found no easy way to do this. Maybe somebody will find it for this or an other similar function with unbounded variation.
References 
[LYB] I. Lyashko, V. Yemel’yanov, O. Boyarchuk, Mathematical analysis, vol. 1, Kyiv, Vyshcha shkola, 1992 (in Ukrainian).
