Let
- $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space
- $T>0$
- $I:=(0,T]$
- $(\mathcal F_t)_{t\in\overline I}$ be a complete and right-continuous filtration on $(\Omega,\mathcal A,\operatorname P)$
- $d\in\mathbb N$
- $M:\Omega\times\overline I\times\mathbb R^d\to\mathbb R$ such that $M(x)$ is a continuous local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ for all $x\in\mathbb R^d$
Moreover, let $$A(x,y):=[M(x),M(y)]\;\;\;\text{for }x,y\in\Lambda$$ and assume that $$A_t\in C^2(\mathbb R^{2d})\;\;\;\text{almost surely for all }t\in\overline I\tag1.$$ Now, let $i\in\left\{1,\ldots,d\right\}$ and $$N(x,\theta):=\frac{M(x+\theta e_i)-M(x)}\theta\;\;\;\text{for }x\in\mathbb R^d\text{ and }\theta\in\mathbb R\setminus\left\{0\right\}.$$
How can we show that $$N_t(x,\theta)\xrightarrow{\theta\to0}X\;\;\;\text{for all }x\in\mathbb R^d\text{ almost surely for all }t\in\overline I\tag2$$ for some continuous local $\mathcal F$-martingale $X$?
My idea is as follows: Let $x\in\mathbb R^d$, $(\theta_n)_{n\in\mathbb N}\subseteq\mathbb R\setminus\left\{0\right\}$ with $\theta_n\xrightarrow{n\to\infty}0$ and $$N^n:=N(x,\theta_n)\;\;\;\text{for }n\in\mathbb N.$$ We would obtain $(2)$, if we could show that $$\sup_{t\in\overline I}\left|N^m_t-N^n_t\right|\xrightarrow{m,\:n\:\to\:\infty}0\;\;\;\text{in probability}\tag3;$$ see this question. Now, I know that $$[Y^n]_T\xrightarrow{n\to\infty}0\Leftrightarrow\sup_{t\in\overline I}|Y_t^n|\xrightarrow{n\to\infty}0\tag4$$ for any sequence $(Y^n)_{n\in\mathbb N}$ of continuous local martingales, where the convergence is in probability. By $(1)$, we know that $$[N(x,\theta),N(y,\vartheta]_t=\int_0^1\int_0^1\frac{\partial ^2A_t}{\partial x_i,\partial y_i}(x+\sigma\theta e_i,y+\tau\vartheta e_i)\:{\rm d}\tau\:{\rm d}\sigma\tag5$$ almost surely for all $t\in\overline I$, $x,y\in\mathbb R^d$ and $\theta,\vartheta\in\mathbb R\setminus\left\{0\right\}$.
How can we conclude?