Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 25 / 3 / c / Solution

For $n>m$, define $r_j=\lfloor j{2}^{m-n}\rfloor 2^{-m}$ and $h_j=X_{j{2}^{-n}}-X_{r_j}$. Each $h_j$ is measurable at time $j{2}^{-n}$, so the summands in the given difference formula are orthogonal martingale transforms. ThereforeLetPath continuity on the compact interval gives ${\omega }_m\to 0$ almost surely, and ${\omega }_m\le 2C$. Since $0\le j{2}^{-n}-r_j<2^{-m}$, $|h_j|\le {\omega }_m$. Thus, by the Cauchy-Schwarz inequality and part (b),The last step is the dominated convergence theorem. The bound is uniform in $n>m$, and the other ordering follows by symmetry. Hence the terminal martingale transforms are Cauchy in $L^2$.