function $u:\mR^{2d}\to\mR$, $$u_{\eps}(z):=u*\varrho_\eps(z)=\int_{\mR^{2d}}u(z')\varrho_\eps(z-z')\dif z'.$$ Let $\sP$ be an operator on the space of locally integrable functions. We define $$\begin{aligned}
\label{Def0}
[\varrho_\eps, \sP]u:=(\sP u)*\varrho_\eps-\sP(u*\varrho_\eps).\end{aligned}$$ We need the following commutator estimate results. \[Le24\]

(i) Let $p\in[1,\infty)$ and $q,r\in[p,\infty]$ with $\frac{1}{p}=\frac{1}{q}+\frac{1}{r}$. For any $b\in L^q(\mR^{2d})$ and $u\in H^{0,1}_r$, we have $$\begin{aligned}
    \label{ES9}
    \lim_{\eps\to 0}\|[\varrho_\eps, b\cdot\nabla_\v]u\|_p=0.\end{aligned}$$

(ii) Let $a:\mR^{2d}\to\mR^d\otimes\mR^d$ be a bounded measurable function. For any $p\in[1,\infty)$ and $u\in H^{0,2}_p$, we have $$\begin{aligned}
    \label{ES8}
    \lim_{\eps\to 0}\|[\varrho_\eps, \sL^a]u\|_p=0.\end{aligned}$$

\(i) It follows by [@Zh22 Lemma 4.2].

\(ii) By definition, we can write for $z=(x,\v)$, $$\begin{aligned}
[\varrho_\eps, \sL^a]u(z)&=(\sL^au)*\varrho_\eps(z)-\sL^a(u*\varrho_\eps)(z)\\
&=\int_{\mR^{2d}}\tr((a(z')-a(z))\cdot\nabla^2_\v u(z'))\varrho_\eps(z-z')\dif z'\\
&\quad+\int_{\mR^{2d}}(\v'-\v) \cdot\nabla_x u(z')\varrho_\eps(z-z')\dif z'\\
&=:I^\eps_1(t,z)+I^\eps_2(t,z).\end{aligned}$$ For $I^\eps_1(t,z)$, by Jensen’s inequality and the assumption, we have $$\begin{aligned}
|I^\eps_1(t,z)|^p&\leq \int_{\mR^{2d}}|\tr((a(z')-a(z))\cdot\nabla^2_\v u(z'))|^p\varrho_\eps(z-z')\dif z'\\
&\leq(2\|a\|_\infty)^p\int_{\mR^{2d}}\|\nabla^2_\v u(z')\|^p\varrho_\eps(z-z')\dif z'.\end{aligned}$$ For $I^\eps_2(t,z)$, by the integration by parts and Hölder’s inequality, we have $$\begin{aligned}
|I^\eps_2(t,z)|^p&=\left|\int_{\mR^{2d}}(\v'-\v) \cdot \nabla_x\varrho_\eps(z-z')u(z')\dif z'\right|^p\\
&\leq \eps^{p}\left(\int_{\mR^{2d}}|\nabla_x\varrho_\eps(z-z')|\cdot|u(z')|\dif z'\right)^p\\
&\leq \eps^{p}\left(\int_{\mR^{2d}}|\nabla_x\varrho_\eps(z)|\dif z\right)^{p-1}
\int_{\mR^{2d}}|\nabla_x\varrho_\eps(z-z')|\cdot|u(z')|^p\dif z'\\
&=\left(\int_{\mR^{2d}}|\nabla_x\varrho(z)|\dif z\right)^{p-1}
\int_{\mR^{2d}}|(\nabla_x\varrho)_\eps(z-z')|\cdot|u(z')|^p\dif z'.\end{aligned}$$ Combining the above calculations, we get $$\|[\varrho_\eps, \sL^a]u\|_p\leq C\|u\|_{0,2;p}.$$ Hence, for any $u\in H^{0,2}_p$, it is easy to see that $$\begin{aligned}
\label{GH0}
\lim_{\eps'\to 0}\sup_{\eps\in(0,1)}\|[\varrho_\eps, \sL^a](u_{\eps'}-u)\|_p\leq C\lim_{\eps'\to 0}\|u_{\eps'}-u\|_{0,2;p}=0.\end{aligned}$$ Moreover, for fixed $\eps'\in(0,1)$, since $u_{\eps'}\in H^{\infty,\infty}_p$, by [@Zh22 Lemma 4.2] we have $$\lim_{\eps\to 0}\|[\varrho_\eps, \sL^a]u_{\eps'}\|_p=0,$$ which together with implies .

Let $\sigma_t(x,\v)=\sigma_t$ be independent of $(x,\v)$. 