$$\begin{aligned}
    \label{E36}
    \|\Delta_x^{\frac{1}{3}} u_t\|_{\alpha,\beta;p}\leq C_4((T-t)\wedge\lambda^{-1})^{1-\frac{1}{p}-\frac{(3\alpha)\vee\beta}{2}}\|f\|_{\mH^{2/3,0}_p(T)}.\end{aligned}$$

Here $C_3=C_3(d,\delta,K,\kappa_1, p,T)$ and $C_4=C_4(d,\delta,K,\kappa_1, \alpha,\beta,p,T)$ are increasing with respect to $T$.

As in the proof of Theorem \[Th33\], it suffices to show the apriori estimates and . Notice that using with $\alpha=\frac{2}{3}$, $\beta=2$ and by , $$\begin{aligned}
\label{ES00}
\|\Delta_x^{\frac{1}{6}}\nabla_\v u\|_{\mL^p(t,T)}\preceq\|u\|_{\mH^{2/3,2}_p(t,T)}\preceq\|f\|_{\mL^p(t,T)}.\end{aligned}$$ Assume $u\in C([0,T];H^{\infty,\infty}_p)$ and let $w_t(x,\v):=\Delta_x^{\frac{1}{3}}u_t(x,\v)$. By we have $$\p_t w+\sL^a_t w+[\Delta_x^{\frac{1}{3}},\sL^a_t]u+\Delta_x^{\frac{1}{3}}f=0, w_T=0.$$ By definition and the assumptions, it is easy to see that $$\sup_{(t,\v)\in[0,T]\times\mR^d}\|\Delta_x^{\frac{1}{3}}a_t(\cdot,\v)\|_p<\infty.$$ Hence, by , and , we have for any $\eps>0$, $$\begin{aligned}
&\|\nabla_x\nabla_\v u\|_{\mL^p(T)}+\|\Delta_x^{\frac{2}{3}}u\|_{\mL^p(T)}+\|\Delta_x^{\frac{1}{3}}\nabla^2_\v u\|_{\mL^p(T)}\\
&\quad\stackrel{\eqref{Riesz}}{\preceq}\|\Delta_x^{\frac{1}{6}}\nabla_\v w\|_{\mL^p(T)}+\|\Delta_x^{\frac{1}{3}} w\|_{\mL^p(T)}+\|\nabla^2_\v w\|_{\mL^p(T)}\\
&\quad\preceq \|[\Delta_x^{\frac{1}{3}},\sL^a_t]u+\Delta_x^{\frac{1}{3}}f\|_{\mL^p(T)}\\
&\quad\leq\|[\Delta_x^{\frac{1}{3}},\sL^a_t]u\|_{\mL^p(T)}+\|f\|_{\mH^{2/3,0}_p(T)}\\
&\quad\preceq\eps\|\Delta_x^{\frac{1}{3}}\nabla^2_\v u\|_{\mL^p(T)}+C_\eps\big(\|\nabla^2_\v u\|_{\mL^p(T)}+\|f\|_{\mH^{2/3,0}_p(T)}\big)\\
&\quad\preceq\eps\|\Delta_x^{\frac{1}{3}}\nabla^2_\v u\|_{\mL^p(T)}+C_\eps\|f\|_{\mH^{2/3,0}_p(T)},\end{aligned}$$ which implies the desired estimate by letting $\eps$ be small enough. As for , it follows by applying to $w$ and using the above estimate.

Proof of Theorem \[Th32\]
-------------------------

By a standard fixed point argument or Picard’s iteration, it suffices to prove the apriori estimate . Let $\alpha\in(0,\frac{2}{3})$, $\beta\in(1,2)$ and $p>\frac{2}{(2-3\alpha)\wedge(2-\beta)}$ be not equal to $\frac{d(\alpha+\beta)}{\alpha(\beta-1)}$.

\(i) Let $q\in[p\vee \frac{d(\alpha+\beta)}{\alpha(\beta-1)},\infty]$ and $r\in[p,\infty]$ with $\frac{1}{p}=\frac{1}{q}+\frac{1}{r}$. Notice that by Hölder’s inequality and , and using $(\alpha(\beta-1)/\beta, \beta-1)$ in place of $(\alpha,\beta)$ in , $$\|b\cdot\nabla_\v u\|^p_{\mL^p(t,T)}\leq \int^T_t\|b_s\|^p_{q}\|\nabla_\v u_s\|^p_{r}\dif s\preceq \int^T_t\|b_s\|^p_{q}\|u_s\|^p_{\alpha,\beta; p}\dif s.$$ Thus, by we have $$\begin{aligned}
\|u_t\|_{\alpha,\beta;p}^p&\preceq ((T-t)\wedge\lambda^{-1})^{p-1-\frac{p((3\alpha)\vee\beta)}{2}}\|b\cdot\nabla_\v u+f\|^p_{\mL^p(t,T)}\\
&\preceq((T-t)\wedge\lambda^{-1})^{p-1-\frac{p((3\alpha)\vee\beta)}{2}}\|f\|^p_{\mL^p(t,T)}+\int^T_t\|b_s\|^p_{q}\|u_s\|^p_{\alpha,\beta;p}\dif s,\end{aligned}$$ which implies by 