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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 $(\Omega,\mathcal A,\operatorname P)$
  • $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ be open
  • $\tilde C^1(\Lambda):=\left\{f:\Lambda\times\Lambda\to\mathbb R\mid\partial_x^\alpha\partial_y^\alpha f\text{ exists and is continuous for all }|\alpha|,|\beta|\le1\right\}$
  • $M:\Omega\times\overline I\times\Lambda\to\mathbb R$ such that $M(x)=M(\;\cdot\;,\;\cdot\;,x)$ is a continuous local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ for all $x\in\Lambda$ and $$A(x,y):=[M(x),M(y)]\;\;\;\text{for }x,y\in\Lambda,$$ where $[\;\cdot\;,\;\cdot\;]$ denotes the covariation bracket

Assume $$A_t\in\tilde C^1(\Lambda)\;\;\;\text{almost surely for all }t\in\overline I\tag1.$$ Let $\varepsilon>0$ and $$\Lambda_\varepsilon:=\left\{x\in\mathbb R^d:\rho(x,\Lambda^c)>\varepsilon\right\},$$ where $\rho$ denotes the Euclidean metric on $\mathbb R^d$. Now, let $i\in\left\{1,\ldots,d\right\}$ and $$N(x,\theta):=\frac{M(x+\theta e_i)-M(x)}\theta\;\;\;\text{for }x\in\Lambda_\varepsilon\text{ and }\theta\in(-\varepsilon,\varepsilon)\setminus\left\{0\right\}.$$ Note that $$[N(x,\theta),N(y,\vartheta)]=\frac{A(x+\theta e_i,y+\vartheta e_i)-A(x+\theta e_i,y)-A(x,y+\vartheta e_i)+A(x,y)}{\theta\vartheta}\tag2$$ almost surely for all $x,y\in\Lambda_\varepsilon$ and $\theta,\vartheta\in(-\varepsilon,\varepsilon)\setminus\left\{0\right\}$ By $(1)$, $$\frac{A_t(x+\theta e_i,y+\vartheta e_i)-A_t(x+\theta e_i,y)-A_t(x,y+\vartheta e_i)+A_t(x,y)}{\theta\vartheta}=\\\int_0^1\int_0^1\frac{\partial^2A_t}{\partial x_i\partial y_i}(x+\sigma\theta e_i,y+\tau\vartheta e_i)\:{\rm d}\tau\:{\rm d}\sigma\tag3$$ almost surely for all $t\in\overline I$, $x,y\in\Lambda_\varepsilon$ and $\theta,\vartheta\in(-\varepsilon,\varepsilon)\setminus\left\{0\right\}$. Clearly, the right-hand side of $(3)$ is defined even when $\theta=\vartheta=0$.

How can we conclude that $N_t(x,\theta)$ converges uniformly in $(x,t)$ as $\theta\to 0$ and the limit $N(x,0)$ is a continuous local martingale?

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