I would like to prove that Brownian motion, denoted $B_t$, is not of bounded variation using the fact that its quadratic variation is finite.
Here is my attempt:
Consider $[t,s]\subset[0,+\infty)$ and a sequence $(\pi_n)_n$ of partitions such that the diameter $\lvert \pi_n\rvert\to 0$ as $n\to\infty$. We know that
$$
\| T_{[s,t]}^{\pi_n}(B) – (s-t)\|_{L^2}\to_{n\to\infty} 0\quad\text{where}\quad T_{[s,t]}^{\pi_n}(B) = \sum_{i=0}^{n-1}(B_{t_{i+1}} – B_{t_{i}})^{2}
$$
The convergence in $L^2$ implies the existence of a subsequence $T_{[s,t]}^{\pi_{n_k}}(B)$ that converges almost surely to $s-t$, we denote $A$ this set. Now we notice that the Brownian motion is almost surely continuous in its trajectory, denote $B$ the set associated to this. Then we have that $\mathbb{P}(C)=\mathbb{P}(A\cap B) = 1$.
Now take $\omega\in C$, we know that $B_t(\omega)$ is uniformly continuous on $[t,s]$. Take $\epsilon>0$, there exists $\delta>0$ such that for all $p,q\in [t,s]$ satisfying $\lvert p-q\rvert\leq\delta$ we have $\lvert B_p(\omega) – B_q(\omega)\rvert\leq\epsilon$. Moreover, for the subsequence $\pi_{n_k}$ there exists a rank $N\in\mathbb{N}$ such that for all $n_k\geq N$ we have $\lvert\pi_{n_k}\rvert\leq\delta$. Thus we have
$$
T_{[s,t]}^{\pi_{n_k}}(B(\omega))= \sum_{i=0}^{n-1}(B_{t_{i+1}}(\omega) – B_{t_{i}}(\omega))^{2} = \epsilon\sum_{i=0}^{n-1}\lvert B_{t_{i+1}}(\omega) – B_{t_{i}}(\omega)\rvert\leq \epsilon V_{t,s}(B(\omega))
$$
But we know that $T_{[s,t]}^{\pi_{n_k}}(B(\omega)$ converges to $s-t$ which is strictly positive so this implies that the constant $V_{t,s}(B(\omega))$ which is the variation of $B_t$ on the interval $[t,s]$ is equal to $\infty$.
I would like first to know if my reasoning is correct and if you have another way to prove this I would be glad to see this.
Best Answer
Minor nitpick: you meant (before your edit) for all $|p-q| < \delta$.
Another way to see this is to prove the more general result that a continuous martingale $M_t$ of bounded variation is in fact constant. Say, WLOG, that $M_0=0$ and observe that: $$\mathbb{E}(M_T^2) = \mathbb{E} \left[ \sum_{i\in \pi_n} (M_{t_{i+1}} -M_{t_i})^2\right] \leq V_{[0,T]}(M) \mathbb{E} \left[ \sup_{i\in\pi_n} |M_{t_{i+1}} - M_{t_i}| \right] \to 0$$ where the limit is due to uniform continuity of the paths of $M$ on $[0,T]$ and the fact $V_{[0,T]}(M) < \infty$. This implies $M_t = 0$ for all $t$.
Since Brownian motion is a non-constant martingale, it must be of unbounded variation.