Brownian motion has bounded quadratic variation, however for almost every sample path, the $p$-variation is infinite for any $p>1/2$, where the $p$ variation takes the supremum over all possible partitions for a given path.
How can I prove this is the case $p=2$?
Ie: $lim_{\delta \rightarrow 0} (sup_{\pi:\delta(\pi)=\delta} \sum_1^{N(\pi)} (B(t_i)-B(t_{i-1}))^{2} ) = \infty$, where $\delta(\pi)$ is the mesh of the partition and the $sup$ ranges over partitions of $[0,T]$ for any $T>0$.
Let’s see conclusion first: a Brownian motion’s paths are a.s. with infinite 2-variation.
We consider a Brownian motion $B_t, t \in [0, 2]$, with triple $(\Omega, \mathcal{F}, P)$, and use a notation $\mathbb{B}_{s,t} := B_t - B_s, 0 \le s \le t \le 1$. We only focus the bounded interval $[0,1 ]$, but $B$ can value from more than $[0, 1]$ for convenience.
A key of proof is the law of iterated logarithms, which says when we fix a $t \in [0, 1]$, with probability 1, we have $$\limsup_{h\downarrow 0} \frac{\mathbb{B}_{t,t+h}} {\psi(h)} =c, $$ where $\psi(h)=\sqrt{hloglog1/h}$ ,and $c$ is a strictly positive real number(actually $c=\sqrt{2}$).
$\psi(h)$ is a strictly increasing function when $h$ is small enough, so it has an inverse $\phi(x)$. Intuitively we may have $$\limsup_{h\downarrow 0}\frac{\phi(\mathbb{B}_{t,t+h})}{h}>c’$$ for some $c’>0$. In fact $\bar{\phi}(x) = x^2/({loglog1/x})$ behaves like $\phi$, we have $$\limsup_{h\downarrow 0}\frac{\bar{\phi}(\mathbb{B}_{t,t+h})}{h}>c’’$$ for some $c’’>0$.
Let $\pi$ be an arbitrary partition of interval [0, 1], and denote $|\pi|$ is the mesh of $\pi $. If we have $\limsup\limits_{|\pi|\downarrow 0}\sum\limits_{i}\bar{\phi}(\mathbb{B}_{t_i,t_{i+1}})>c’’$, then by the fact that $\lim\limits_{x\downarrow 0}x^2/ \bar{\phi}(x)=\infty$, we get $\limsup\limits_{|\pi|\downarrow 0}\sum_{i}(\mathbb{B}_{t_i, t_{i+1}})^2=\infty$.
Vitality covering will be helpful. Consider process $X^{\delta}_t = \sup\limits_{0< h< \delta}\bar{\phi}(\mathbb{B}_{t,t+h})/h, $ with fixed $\delta$. $X^{\delta}_t$ can also be written as $X_t^{\delta} = \sup\limits_{0< h< \delta, h \in \mathbb{Q}}\bar{\phi}(\mathbb{B}_{t,t+h})/h$, where $\bar{\phi}(\mathbb{B}_{t,t+h})/h$ with variable $(t,\omega)$ is $[0,1]\times \Omega$ measurable due to its continuity with respect to $t$. So $X^{\delta}$ is also $[0, 1]\times \Omega$ measurable, and $X:= \lim\limits_{\delta\downarrow 0}X^{\delta}$ is also $[0, 1]\times \Omega$ measurable. Then by Fubini’s theorem and iterated logarithms law we get $$P[\mu\{X_{t}(\omega) > c’’\}]=\mu\{P[X_{t}(\omega) > c’’]\}= 1,$$ where $\mu$ is the Lebesgue measure on $[0, 1]$. This equation means with probability 1, Lebesgue measure of the set $E:= \{t \in [0, 1]: \limsup\limits_{\delta\downarrow 0}\bar{\phi}(\mathbb{B}_{t,t+\delta})/\delta>c’’\}$ is 1. Denote $I^{\delta}:= \{interval [t,t+h] \subset [0, 1]: \bar{\phi}(\mathbb{B}_{t,t+h})/h>c’’, h<\delta\}$. Then with probability 1, $I^{\delta}$ is a Vitali cover of $E$. By Vitali’s theorem, in these paths, we can choose disjoint intervals $[r_i, s_i], i=1, 2, … , n,$ so that $\sum\limits_{i}s_i - r_i > \mu\{ E \}-\varepsilon= 1- \varepsilon$, and$ \sum\limits_{i}\bar{\phi}(\mathbb{B}_{r_i,s_i})> c’’(1-\varepsilon)> c’’/2>0 $. And we can complete a partition $\pi$ based on $[r_i, s_i]$ so that it’s mesh is less than $\delta$, and the sum $\sum\limits_{\pi}\bar{\phi}(\mathbb{B}_{t_i, t_{i+1}})$ becomes larger. Then by the fact $\lim\limits_{x\downarrow 0}x^2/ \bar{\phi}(x)=\infty$,and let $\delta$ small enough, we get with probability 1, $$\limsup_{|\pi|\downarrow 0}\sum\limits_{i}(\mathbb{B}_{t_i, t_{i+1}})^2=\infty$$, which is we want.
For more details, see chapter 13.9 of “Multidimensional Stochastic Processes as Rough Paths Theory and Applications”, written by Peter K. Friz and Nicolas B. Victoir, ISBN 9780521876070.