satisfies for Lebesgue almost all $t\in[0,T]$. For given $p\geq 1$, let $\phi$ be a nonnegative symmetric smooth function on $\mR^{2d}$ with support in the unit ball and $$\int_{\mR^{2d}}|\phi(z)|^p\dif z=1.$$ Let $\delta\in(0,1)$ be as in [**(H$^{\delta,p}_K$)**]{} and set $$\phi_\delta(z):=\delta^{-2d/p}\phi(z/\delta),$$ and for $z^o=(x^o,\v^o)$ and $t\in[0,T]$, define $$z^o_t:=(x^o-t\v^o,\v^o), \  \phi^{z^o_t}_\delta(z):=\phi_\delta(z^o_t-z).$$ By definition, it is easy to see that $$\begin{aligned}
\label{LK1}
\int_{\mR^{2d}}|\phi^{z^o_t}_\delta(z)|^p\dif z^o=1,\ \ t\in[0,T],\ z\in\mR^{2d},\end{aligned}$$ and for $j=1,2$, $$\begin{aligned}
\label{LK2}
\sup_{t\in[0,T]}\sup_{z\in\mR^{2d}}\int_{\mR^{2d}}|\nabla^j_\v\phi^{z^o_t}_\delta(z)|^p\dif z^o\leq C_\delta.\end{aligned}$$ Define the freezing functions at point $z^o=(x^o,\v^o)$ as follows: $$a^{z^o}_t:=a_{t}(z^o_t),\ \ u^{z^o}_{\delta,t}(z):=u_t(z)\phi^{z^o_t}_\delta(z).$$ By and easy calculations, one sees that $$\begin{aligned}
\label{LK88}
\p_t u^{z^o}_\delta+\mathrm{tr}(a^{z^o}_t\cdot\nabla^2_{\v} u^{z^o}_\delta)+\v\cdot \nabla_x u^{z^o}_\delta-\lambda u^{z^o}_\delta=g^{z^o}_{\delta},\end{aligned}$$ where for $z=(x,\v)$, $$\begin{aligned}
g^{z^o}_{\delta,t}(z)&:=\mathrm{tr}(a^{z^o}_t\cdot\nabla^2_{\v} u^{z^o}_{\delta,t})(z)-\mathrm{tr}(a_t\cdot\nabla^2_\v u_t)(z)\phi_\delta^{z^o_t}(z)\\
&\quad+(\v-\v^o)\cdot \nabla_x \phi_\delta^{z^o_t}(z) u_t(z)+f_t(z)\phi_\delta^{z^o_t}(z).\end{aligned}$$ We have the following [*claim:*]{} $$\begin{aligned}
\label{LK3}
\begin{split}
&\left(\int_{\mR^{2d}}\|g^{z^o}_{\delta}\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}
\leq \omega_a(\delta)\|\nabla^2_\v u\|_{\mL^p(t,T)}\\
&\quad+C_\delta\Big(\|u\|_{\mL^p(t,T)}+\|f\|_{\mL^p(t,T)}\Big),\ \ t\in[0,T].
\end{split}\end{aligned}$$ [*Proof of the claim:*]{} Observe that $$\begin{aligned}
g^{z^o}_{\delta,t}(z)&=\mathrm{tr}((a^{z^o}_t-a_t)\cdot\nabla^2_\v u_t)(z)\phi_\delta^{z^o_t}(z)
+\mathrm{tr}(a^{z^o}_t\cdot(\nabla_\v u_t\otimes\nabla_\v\phi^{z^o_t}_\delta))(z)\\
&\quad+[\mathrm{tr}(a^{z^o}_t\cdot\nabla^2_\v\phi^{z^o_t}_\delta)(z)+(\v-\v^o)\cdot \nabla_x \phi_\delta^{z^o_t}(z)] u_t(z)
+f_t(z)\phi_\delta^{z^o_t}(z)\\
&=:I^\delta_1(t,z,z^o)+I^\delta_2(t,z,z^o)+I^\delta_3(t,z,z^o)+I^\delta_4(t,z,z^o).\end{aligned}$$ For $I^\delta_1(t,z,z^o)$, since the support of $\phi_\delta$ is in $B_\delta:=\{z\in\mR^{2d}: |z|\leq\delta\}$, by the definition of $\omega_a(\delta)$ and , we have $$\begin{aligned}
\left(\int_{\mR^{2d}}\|I^\delta_1(\cdot,\cdot,z^o)\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}
&\leq \omega_a(\delta)\left(\int_{\mR^{2d}}\|\nabla^2_\v u\cdot\phi^{z^o_\cdot}_\delta\|^p_{\mL^p(t,T)}\dif z^o\right)^{1/p}\\
&=\omega_a(\delta)\|\nabla^2_\v u\|_{\mL^p(t,T)}.\end{aligned}$$ For $I^\delta_2(t,z,z^o)$, by we have $$\left(\int_{\mR^{2d}}\|I^\delta_2(\cdot,\cdot,z^o)\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}\leq C_\delta\|\nabla_\v u\|_{\mL^p(t,T)}.$$ For $I^\delta_3(t,z,z^o)$, we similarly have $$\left(\int_{\mR^{2d}}\|I^\delta_3(\cdot,\cdot,z^o)\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}\leq C_\delta\|u\|_{\mL^p(t,T)}.$$ For $I^\delta_4(t,z,z^o)$, by we have $$\left(\int_{\mR^{2d}}\|I^\delta_4(\cdot,\cdot,z^o)\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}=\|f\|_{\mL^p(t,T)}.$$ Combining the above calculations, and 