on $T,\sigma,p,q,d, \|b\|_{\mL^q(T)}$ such that for any $(r,z)\in[0,T)\times\mR^{2d}$, any weak solution $((\tilde Z,\tilde W); (\tilde \Omega,\tilde \sF,\tilde \mP);(\tilde \sF_t)_{t\geq 0})$ of SDE with starting point $(r,z)$, and $r\leq t_0<t_1\leq T$ and $f\in\mL^p(T)$, $$\begin{aligned}
\label{BB9}
\tilde\mE\left(\int^{t_1}_{t_0} f_s(\tilde Z_s)\dif s\Big|\tilde \sF_{t_0}\right)
\leq C(t_1-t_0)^{\frac{1}{2d+1}-\frac{1}{p}}\|f\|_{\mL^p(t_0,t_1)},\end{aligned}$$ where $C>0$ is increasing in $T,\omega_\sigma(\delta)$ and $\|b\|_{\mL^q(T)}$.

Without loss of generality, we assume $(r,z)=(0,0)$ and drop the tilde in the definition of weak solutions for simplicity. We divide the proof into four steps.

[**(a)**]{} Let $a_t(z):=\frac{1}{2}(\sigma_t\sigma^*_t)(z)$ and $p\in(2(2d+1),q]$. For any $0\leq t_0<t_1\leq T$, $\lambda\geq 1$ and $f\in\mL^p(t_0,t_1)$, let $u=\cR^{a,b}_{\lambda,t_1}(f)$ be the solution of PDE with terminal time $T=t_1$. By and with $\alpha=\frac{4d}{3(2d+1)}$ and $\beta=\frac{4d}{2d+1}$, $$\begin{aligned}
\label{44}
\begin{split}
&\|\cR^{a,b}_{\lambda,t_1}(f)\|_{\mL^\infty(t_0,t_1)}
\preceq \big\|\cR^{a,b}_{\lambda,t_1}(f)\big\|_{L^\infty([t_0,t_1];H^{\alpha,\beta}_p)}\\
&\quad\preceq ((t_1-t_0)\wedge\lambda^{-1})^{\frac{1}{2d+1}-\frac{1}{p}}\|f\|_{\mL^p(t_0,t_1)},
\end{split}\end{aligned}$$ and by , , with $\alpha=\frac{4d+1}{3(2d+1)}$ and $\beta=\frac{4d+1}{2d+1}$, $$\begin{aligned}
\label{404}
\begin{split}
&\|\cR^{a,0}_{\lambda,t_1}(f)\|_{\mL^\infty(t_0,t_1)}\preceq 
\big\|\cR^{a,0}_{\lambda,t_1}(f)\big\|_{L^\infty([t_0,t_1];H^{\alpha,\beta}_p)}\\
&\quad\preceq ((t_1-t_0)\wedge\lambda^{-1})^{\frac{1}{2(2d+1)}-\frac{1}{p}}\|f\|_{\mL^p(t_0,t_1)},
\end{split}\end{aligned}$$ where the $C$ in the above $\preceq$ only depend on $d,p,q,K,T,\omega_a(\delta)$ and $\|b\|_{\mL^q(T)}$.

Now, for any $R>0$, define a stopping time $$\begin{aligned}
\label{TR}
\tau_R:=\Bigg\{t\geq 0: \int^t_0 |b_s(Z_s)|\dif s>R\Bigg\}.\end{aligned}$$ Let $\varrho_\eps$ be as in . We introduce a $d\times d$ matrix-valued function, which is crucial for us below. For $t\geq 0$ and $z\in\mR^{2d}$, let $$a^{\eps,R}_{t}(z):=
\left\{
\begin{array}{ll}
\tfrac{\mE\big(a_t(Z_t)\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)}{\mE\big(\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)},\ 
& \mathrm{ if }\ \ \mE\big(\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)\not=0,\\
a_t(z),\ & \mathrm{ if }\ \ 