I have a question about the proof a a certain version of the Itô-formula. First the author defines an Itô-process and states the formula:
My question is in regarding the proof. The proof uses Taylor expansion, and in the last line in the proof he says that the above argument shows that the remainder terms go to 0, but I can not see why this is the case:
Do you see why also the remainder term goes to 0? Why does that follow from the argument?
reference for the proof: http://th.if.uj.edu.pl/~gudowska/dydaktyka/Oksendal.pdf, chapter 4
Here is the idea (you need to fill in some details)
$R_j = o(|\Delta t_j|^2 + |\Delta X_j|^2)$ means $$R_j \le r(|\Delta t_j|^2 + |\Delta X_j|^2)(|\Delta t_j|^2 + |\Delta X_j|^2)$$, where $r(y)\to 0 \ as \ y\to 0$ (using the condition that $g$ is twice continuously differentiable, you can show such $r$ exists and does not depends on $j$). Now $$|\sum_j R_j| \le \max_j\ r(|\Delta t_j|^2 + |\Delta X_j|^2) \times \sum_j (|\Delta t_j|^2 + |\Delta X_j|^2)$$ By similar method used in the proof, you can show $\sum_j (|\Delta t_j|^2 + |\Delta X_j|^2) \to \int v^2 ds$ in $L_2$, and $\max_j\ r(|\Delta t_j|^2 + |\Delta X_j|^2) \to 0$, by the continuity of $X$