\label{E28}
\|\nabla_\v f\|_{\alpha(\beta-1)/\beta,\beta-1;p}\leq C\|f\|_{\alpha,\beta;p}.\end{aligned}$$ (ii) Let $\alpha,\beta\geq 0$ and $p>1$ with $d\not=\frac{p\alpha\beta}{\alpha+\beta}$. Set $$p^*:=
\left\{
\begin{aligned}
&dp/(d-\tfrac{p\alpha\beta}{\alpha+\beta}),&\ \ d>\tfrac{p\alpha\beta}{\alpha+\beta};\\
&\infty,&\ \ d<\tfrac{p\alpha\beta}{\alpha+\beta}. 
\end{aligned}
\right.$$ For any $q\in[p,p^*]$, there is a constant $C=C(\alpha,\beta,p,q,d)>0$ such that $$\begin{aligned}
\label{E29}
\|f\|_{q}\leq C\|f\|_{\alpha,\beta;p}.\end{aligned}$$

\(i) It follows by the Mihlin multiplier theorem and .

\(ii) For , by , and with $\theta=\frac{\beta}{\alpha+\beta}$, we have $$\begin{aligned}
\|f\|^q_q&=\int_{\mR^d}\|f(\cdot,\v)\|_q^q\dif \v \preceq
\int_{\mR^d}\|(\mI-\Delta_x)^{\frac{\theta\alpha}{2}}f(\cdot,\v)\|_p^q\dif \v\\
&\leq\left(\int_{\mR^d}\left(\int_{\mR^d}|(\mI-\Delta_x)^{\frac{\theta\alpha}{2}}f(x,\v)|^q\dif\v\right)^{p/q}\dif x\right)^{q/p}\\
&\preceq\left(\int_{\mR^d}\|(\mI-\Delta_\v)^{\frac{(1-\theta)\beta}{2}}(\mI-\Delta_x)^{\frac{\theta\alpha}{2}}f(x,\cdot)\|^p_{p}\dif x\right)^{q/p}
\preceq\|f\|^q_{\alpha,\beta;p},\end{aligned}$$ where the second inequality is due to Minkovskii’s inequality.

Let $a:\mR^{2d}\to\mR^d\otimes\mR^d$ be a measurable function. Write $$\sL^a u:=\mathrm{tr}(a\cdot\nabla^2_\v u)+\v\cdot\nabla_x u.$$ We have \[Le22\] Let $\alpha\in(0,1)$ and $p>d/\alpha$. Suppose that $$\kappa_0:=\sup_{\v}\|\Delta_x^{\frac{\alpha}{2}}a(\cdot,\v)\|_p+\|a\|_\infty<\infty.$$ For any $\eps\in(0,1)$, there is a constant $C_\eps=C_\eps(p,d,\alpha,\kappa_0)>0$ such that for all $u\in H^{\infty,\infty}_p$, $$\begin{aligned}
\label{E210}
\|[\Delta_x^{\frac{\alpha}{2}},\sL^a]u\|_p\leq\eps\|\Delta_x^{\frac{\alpha}{2}}\nabla^2_\v u\|_{p}+C_\eps \|\nabla^2_\v u\|_{p},\end{aligned}$$ where $[\Delta_x^{\frac{\alpha}{2}},\sL^a]u:=\Delta_x^{\frac{\alpha}{2}}(\sL^a u)-\sL^a(\Delta_x^{\frac{\alpha}{2}}u)$.

Notice that by definition , $$[\Delta_x^{\frac{\alpha}{2}},\sL^a]u=\tr\big(\Delta_x^{\frac{\alpha}{2}} a\cdot\nabla^2_\v u\big)
+\int_{\mR^d}\tr\big(\delta_{(y,0)}a\cdot\nabla^2_\v\delta_{(y,0)}u\big)|y|^{-d-\alpha}\dif y.$$ Hence, $$\|[\Delta_x^{\frac{\alpha}{2}},\sL^a]u\|_p\leq\|\tr(\Delta_x^{\frac{\alpha}{2}} a\cdot\nabla^2_\v u)\|_p
+\int_{\mR^d}\big\|\tr\big(\delta_{(y,0)}a\cdot\nabla^2_\v\delta_{(y,0)}u\big)\big\|_p|y|^{-d-\alpha}\dif y.$$ Let $\beta\in(\frac{d}{p},\alpha)$. By , we have $$\begin{aligned}
\|\tr(\Delta_x^{\frac{\alpha}{2}} a\cdot\nabla^2_\v u)\|_p^p
&\preceq\int_{\mR^d}\|\Delta_x^{\frac{\alpha}{2}} a(\cdot,\v)\|^p_{p}\|\nabla^2_\v u(\cdot,\v)\|_\infty^p\dif \v\\
&\preceq\sup_{\v}\|\Delta_x^{\frac{\alpha}{2}}a(\cdot,\v)\|^p_p\|\nabla^2_\v u\|_{\beta,0; p}^p,\end{aligned}$$ and for $\gamma\in(0,\beta-\frac{d}{p})$, $$\begin{aligned}
&\big\|\tr\big(\delta_{(y,0)}a\cdot\nabla^2_\v\delta_{(y,0)}u\big)\big\|_p^p
\preceq\int_{\mR^d}\|\delta_y a(\cdot,\v)\|_{p}^p\cdot\|\nabla^2_\v\delta_y u(\cdot,\v)\|_{\infty}^p\dif \v\\
&\qquad\stackrel{\eqref{LI0}}{\preceq}|y|^{(\alpha+\gamma)p}\int_{\mR^d}\|\Delta_x^{\frac{\alpha}{2}}a(\cdot,\v)\|_p^p\|\nabla^2_\v u(\cdot,\v)\|_{\beta,p}^p\dif\v\\
&\qquad\ \preceq |y|^{(\alpha+\gamma)p}\sup_{\v}\|\Delta_x^{\frac{\alpha}{2}}a(\cdot,\v)\|^p_p\|\nabla^2_\v u\|_{\beta,0;p}^p.\end{aligned}$$ Moreover, it is easy to see that $$\big\|\tr\big(\delta_{(y,0)}a\cdot\nabla^2_\v\delta_{(y,0)}u\big)\big\|_p\preceq\|a\|_\infty\|\nabla^2_\v u\|_p.$$ Therefore, $$\big\|\tr\big(\delta_{(y,0)}a\cdot\nabla^2_\v\delta_{(y,0)}u\big)\big\|_p\preceq \kappa_0(|y|^{\alpha+\gamma}\wedge 1)\|\nabla^2_\v u\|_{\beta,0;p}.$$ Combining the above calculations, we get for some $C=C(p,d,\alpha,\beta)>0$, $$\begin{aligned}
\label{LJ1}
\|[\Delta_x^{\frac{\alpha}{2}},\sL^a]u\|_p\leq C\kappa_0\|\nabla^2_\v u\|_{\beta,0;p},\end{aligned}$$ On the 