by the interpolation inequality and Young’s inequality, we get the claim.

Now by , we have $$\begin{aligned}
\label{ES6}
\begin{split}
&\|\nabla^2_\v u\|_{\mL^p(t,T)}
=\left(\int_{\mR^{2d}}\|(\nabla^2_\v u)\phi^{z^o_\cdot}_\delta\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}\\
&\quad\leq \left(\int_{\mR^{2d}}\|\nabla^2_\v (u\phi^{z^o_\cdot}_\delta)-(\nabla^2_\v u)\phi^{z^o_\cdot}_\delta\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}\\
&\qquad+\left(\int_{\mR^{2d}}\|\nabla^2_\v u^{z^o}_\delta\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}=:I_1+I_2.
\end{split}\end{aligned}$$ For $I_1$, by and the interpolation inequality, we have $$I_1\leq C_\delta\Big(\|\nabla_\v u\|_{\mL^p(t,T)}+\|u\|_{\mL^p(t,T)}\Big)\leq \omega_a(\delta)\|\nabla^2_\v u\|_{\mL^p(t,T)}+C_{\delta}\|u\|_{\mL^p(t,T)}.$$ For $I_2$, noticing that by and Duhamel’s formula, $$\begin{aligned}
\label{EQ1}
u^{z^o}_{\delta,t}(z)=\int^T_t\e^{\lambda(t-s)} P^{z^o}_{t,s}g^{z^o}_{\delta,s}(z)\dif s,\end{aligned}$$ where $P^{z^o}_{t,s}$ is defined by in terms of $\sigma^{z^o}_t:=(a^{z^o}_t)^{1/2}$, by and , we have $$\begin{aligned}
\begin{split}
I_2\leq C_0&\left(\int_{\mR^{2d}}\|g^{z^o}_{\delta}\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}
\leq C_0\omega_a(\delta)\|\nabla^2_\v u\|_{\mL^p(t,T)}\\
&\quad+C_\delta\Big(\|u\|_{\mL^p(t,T)}+C\|f\|_{\mL^p(t,T)}\Big).
\end{split}\end{aligned}$$ Substituting these two estimates into and by $\omega_a(\delta)\leq\frac{1}{2(C_0+1)}$, we get $$\begin{aligned}
\label{LK9}
\|\nabla^2_\v u\|_{\mL^p(t,T)}\leq C\Big(\|u\|_{\mL^p(t,T)}+\|f\|_{\mL^p(t,T)}\Big).\end{aligned}$$ Similarly, one can show that (see also step (c) below) $$\begin{aligned}
\label{LK8}
\|\Delta_x^{\frac{1}{3}}u\|_{\mL^p(t,T)}\leq C\Big(\|u\|_{\mL^p(t,T)}+\|f\|_{\mL^p(t,T)}\Big).\end{aligned}$$

[**(b)**]{} By and the contraction of operator $P^{z^o}_{t,s}$ in $L^p(\mR^{2d})$, we have $$\begin{aligned}
\|u_t\|_p^p&\stackrel{\eqref{LK1}}{=}\int_{\mR^{2d}}\|u^{z^o}_{\delta,t}\|^p_p\dif z^o\leq\int_{\mR^{2d}}\left(\int^T_t \|g^{z^o}_{\delta,s}\|_{p}\dif s\right)^p\dif z^o\\
&\leq \left(\int^T_t\e^{p\lambda(t-s)/(p-1)}\dif s\right)^{p-1}\int_{\mR^{2d}}\int^T_t\|g^{z^o}_{\delta,s}\|^p_{p}\dif s\dif z^o\\
&\stackrel{\eqref{LK3}}{\preceq} ((T-t)\wedge\lambda^{-1})^{p-1}\Big(\|\nabla^2_\v u\|_{\mL^p(t,T)}^p+\|u\|_{\mL^p(t,T)}^p+\|f\|_{\mL^p(t,T)}^p\Big)\\
&\stackrel{\eqref{LK9}}{\preceq}((T-t)\wedge\lambda^{-1})^{p-1}\left(\int^T_t\|u_s\|_{p}^p\dif s+\|f\|_{\mL^p(t,T)}^p\right),\end{aligned}$$ which yields by Gronwall’s inequality that $$\begin{aligned}
\label{ES77}
\|u_t\|^p_p\leq C((T-t)\wedge\lambda^{-1})^{p-1}\|f\|_{\mL^p(t,T)}^p.\end{aligned}$$ Substituting it into and , we obtain , and also by , $$\begin{aligned}
\label{LP2}
\int_{\mR^{2d}}\|g^{z^o}_{\delta}\|_{\mL^p(t,T)}^p\dif z^o\leq C\|f\|_{\mL^p(t,T)}^p,\ \ t\in[0,T].\end{aligned}$$

[**(c)**]{} Let $\alpha\in(0,\frac{2}{3})$. By , and Hölder’s inequality, we have $$\begin{aligned}
\|u^{z^o}_{\delta,t}\|_{\alpha,0; p}&\leq \int^T_t\e^{\lambda(t-s)} \|P^{z^o}_{t,s}g^{z^o}_{\delta,s}\|_{\alpha,0;p}\dif s
\preceq\int^T_t\e^{\lambda(t-s)}(s-t)^{-\frac{3\alpha}{2}}\|g^{z^o}_{\delta,s}\|_{p}\dif s\no\\
&\preceq \left(\int^T_t\e^{\lambda(t-s)} (s-t)^{-\frac{3p\alpha}{2(p-1)}}\dif s\right)^{1-\frac{1}{p}}\|g^{z^o}_{\delta}\|_{\mL^p(t,T)}\no\\
&\preceq( (T-t)\wedge\lambda^{-1})^{1-\frac{1}{p}-\frac{3\alpha}{2}}\|g^{z^o}_{\delta}\|_{\mL^p(t,T)}.\label{LP1}\end{aligned}$$ By again, we have $$\begin{aligned}
\label{LP4}
