Infinite variation of Brownian motion and Continuity Let $C>0$ be a constant. Brownian motion is Hölder continuous for $\alpha=1/2$:
$$| B(t+h) - B(t) | \leq C \sqrt{h \log(1/h)} \leq C h^\alpha,$$
for every sufficiently small $h$.
But Brownian motion is of unbounded variation:
$$\sup \sum_{j}^k |B(t_j) - B(t_{j-1})| < \infty.$$
Consider a stochastic differential equation
$$dX_t = X_t dt + \alpha B_t,$$
where $\alpha>0$ and $B_t$ is Brownian motion.
The Euler-Maruyama simulation gives
$$X_{i+1} = X_{i} + X_{i}\Delta t + \alpha \cdot (B_{t_{i+1}} - B_{t_i}).$$
Is the diffusion term of the simulated equation $X_{i+1}$ then bounded w.r.t the Hölder continuity although Brownian motion is of infinite variation? Can we define an upper bound for the increment $B_{t_{i+1}} - B_{t_i}$?
 A: Thing is $B_{t_{i+1}} - B_{t_i}$ is a random variable with distribution $\mathscr N(0, t_{i+1} - t_i)$ and hence can take value above any finite bound with a small but positive probability. How does that relate to Hölder continuity? I guess you may feel a bit of contradiction here. Well, here is the point.
Brownian motion is almost surely only locally $\alpha$-Hölder continuous (also, it seems that $\alpha$ must be strictly less than $\frac12$). What it means is that with probability $1$ you will have a path $B(\omega)$ which is locally $\alpha$-Hölder continuous, no problem here. But at a given $t_i$ you do not know which $\varepsilon(t_i,\omega)$ you need to pick so that your path is $\alpha$-Hölder continuous in the interval $(t_i - \varepsilon(t_i, \omega), t_i + \varepsilon(t_i, \omega))$. For this reason, given a fixed $\Delta t$ even if you had some true realization of $B(\omega)$ you'd still could not get any a priori bounds on $B_{t_{i+1}} - B_{t_i}$. Not to mention that since you deal with numerical schemes, you do not have access to the full information of your realization of $B(\omega)$, on which $\varepsilon(t_i,\omega)$ crucially depends.
