Quadratic Variation of Wiener's process

Question

I know I'm wrong, but I still don't understand why can't this operation be performed:

$$ \sum_{j=0}^{n-1}(W(t_{j+1})-W(t_j))^2\le \max[W(t_{j+1})-W(t_j)]*(W(T)-W(0) )$$

which would have a limit of $0$, as $\max[t_{j+1}-t_j]\to0$, because $W(t)$ is continuous.

Answer 1

You're missing some absolute values. The correct inequality would be $$\sum \limits_{j = 0}^{n - 1} (W(t_{j + 1}) - W(t_j))^2 \le \max(|W(t_{j + 1}) - W(t_j)|) \sum \limits_{j = 0}^{n - 1} |W(t_{j + 1}) - W(t_j)|.$$

But now you can't apply your argumentation, as the Brownian motion doesn't have a bounded variation.