Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 34 / 3 / a / i / Solution

For each sample point put $A_n={\max }_{0\le k<2^n}|X_{(k+1)2^{-n}}-X_{k{2}^{-n}}|$. First take $s<t$ in the dyadic rationals, set $h=t-s$, and choose $N\ge 0$ with $2^{-N}\le h<2^{1-N}$. The left grid approximations $s_N=2^{-N}\lfloor 2^{N}s\rfloor$ and $t_N=2^{-N}\lfloor 2^{N}t\rfloor$ are at most two grid steps apart, so $|X_{t_N}-X_{s_N}|\le 2A_N$.

For either endpoint $r$, successive left approximations $r_n,r_{n+1}$ are equal or one step apart on the level-$(n+1)$ grid. Since $r$ is dyadic these approximations eventually equal $r$ exactly. The triangle inequality therefore givesThe second inequality is valid also for $\alpha =0$. Thus the required bound holds simultaneously for every dyadic pair whenever $K_{\alpha }$ is finite; if it is infinite the inequality holds in the extended sense. For $s=t$ the left side is zero. This is dyadic increment chaining, with the stated constant two.