It's from Brownian Motion and Stochastic Calculus by I. Karatzas chapter 1, problem 5.7 property (iv) trying to show an inequality wrt total variation and cross variation of martingales.
$$\check{\xi}_t - \check{\xi}_s\le \frac{1}{2}[\langle X\rangle_t+\langle Y\rangle_t-(\langle X\rangle_s+\langle Y\rangle_s)]$$ where $X$ and $Y$ are two martingales and $\check{\xi}_t$ is the total variation of the cross variation $\langle X,Y\rangle_t$ on $[0,t]$.
For now, I know how to show that $\check{\xi}_t\le \frac{1}{2}[\langle X\rangle_t+\langle Y\rangle_t]$ using $\langle X-Y\rangle\ge0$ but as for the $s$-term, I don't know how to deal with it.
It is said to be the final step to verify that cross variation is a bilinear function but I really don't understand why aren't the first three properties sufficient to show the result. I'll appreciate help of any kind. Thank you so much!
Properties (i) and (ii) show that $\langle\cdot,\cdot\rangle$ is a bilinear form. Properties (iii) and (iv) seem to be extra properties.
Let $0\le s< t$.
Observation 1. Suppose $h$ is given by the difference between two nondecreasing functions $f$ and $g$: $h=f-g$. Then $\check h(t)-\check h(s)\le f(t)-f(s)$.
Proof. $|f(t_2)-g(t_2)-f(t_1)+g(t_1)|\le f(t_2)-f(t_1)$, so $\sum_i|h(t_{i+1})-h(t_i)|\le f(t)-f(s)$ for any partition.
Observation 2. For any function $f$ of bounded variation, $f(t)-f(s)\le \check f(t)-\check f(s)$.
Now Property (iv): \begin{align} \check\xi_t-\check\xi_s &\le \frac{1}{4}\left(\langle X+Y\rangle_t-\langle X+Y\rangle_s\right),\quad\text{Observation 1},\\ &=\frac{1}{4}( \langle X\rangle_t-\langle X\rangle_s+\langle Y\rangle_t-\langle Y\rangle_s +2\langle X,Y\rangle_t-2\langle X,Y\rangle_s ),\quad\text{Properties (i) and (ii)},\\ &\le \frac{1}{4}( \langle X\rangle_t-\langle X\rangle_s+\langle Y\rangle_t-\langle Y\rangle_s +2(\check\xi_t-\check\xi_s) ),\quad\text{Observation 2}, \end{align} which, rearranged, yields the desired inequality.