Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 29 / 2 / c / Solution

For $x\le 0$, $f_n(x)=|x|$; for $x>0$, only the interval $(0,1/n)$ contributes an error to the integral of the derivative. HenceBy the Itô formula, the continuous processesare nondecreasing, since $f_n^{\prime \prime }\ge 0$. Parts (a) and (b), together with the uniform approximation of absolute value, show that $C^{(n)}$ converges uniformly on compacts in probability toThe limit is continuous: the martingale integral is continuous, and the bounded-integrand finite-variation integral is continuous because $A$ is continuous. For any rational $s<t$, convergence in probability and $C_t^{(n)}-C_s^{(n)}\ge 0$ imply $Z_t-Z_s\ge 0$ almost surely. Intersect these countably many events and use continuity to extend to all real $s<t$. ThusSince $|X|={\mathrm{sgn}}_{-}(X)\cdot M+({\mathrm{sgn}}_{-}(X)\cdot A+Z)$ is a continuous local martingale plus continuous finite variation, $|X|$ is a continuous semimartingale. This is the smooth convex approximation proof of the Tanaka formula. With the specified sign value $-1$ at zero, $Z$ is the right local time at zero, rather than an unspecified symmetric normalization.

For an arbitrary continuous semimartingale starting at zero, use its continuous decomposition $X=M+A$, with both parts starting at zero, and letThe continuity and local finiteness of variation imply ${\rho }_k\uparrow \infty$ almost surely. Up to ${\rho }_k$ the required deterministic bound holds, and the stopped local martingale is bounded, hence a true $L^2$-bounded martingale. Apply the previous argument to $X^{{\rho }_k}$ for each integer $k$. Stopping the defining integrals gives precisely $Z_{t\wedge {\rho }_k}$. On the intersection of their full-probability events, each stopped $Z$ is nondecreasing; every finite interval is eventually contained before ${\rho }_k$, so $Z$ is nondecreasing globally. Therefore