other hand, by the interpolation inequality and Young’s inequality, we have for any $\eps\in(0,1)$, $$\|\nabla^2_\v u\|_{\beta,0;p}\preceq \|\nabla^2_\v u\|^\frac{\beta}{\alpha}_{\alpha,0;p}\|\nabla^2_\v u\|^\frac{\alpha-\beta}{\alpha}_{p}
\leq\eps\|\nabla^2_\v u\|_{\alpha,0;p}+C_\eps\|\nabla^2_\v u\|_{p}.$$ Estimate now follows by .

\[Le23\] For any $\alpha,\beta\in(0,1)$ and $p>(\alpha+\beta)d/(\alpha\beta)$, there is a constant $C=C(\alpha,\beta,p,d)>0$ such that for all $b\in H^{\alpha,0}_p$ and $u\in H^{\infty,\infty}_p$, $$\|b\cdot\nabla_\v u\|_{\alpha,0; p}\leq C\|b\|_{\alpha,0;p}\Big(\|\Delta_x^{\frac{\alpha}{2}}\nabla_\v u\|_{0,\beta;p}+\|\nabla_\v u\|_{0,\beta; p}\Big).$$

Notice that by , $$\begin{aligned}
\|b\cdot\nabla_\v u\|_{\alpha,0;p}\preceq\|b\cdot\nabla_\v u\|_p+\|\Delta_x^{\frac{\alpha}{2}}(b\cdot\nabla_\v u)\|_p.\end{aligned}$$ By definition , we have $$\begin{aligned}
&\|\Delta_x^{\frac{\alpha}{2}}(b\cdot\nabla_\v u)\|_p\leq\|(\Delta_x^{\frac{\alpha}{2}} b)\cdot\nabla_\v u\|_p
+\|b\cdot\nabla_\v \Delta_x^{\frac{\alpha}{2}}u\|_p\\
&\qquad+\int_{\mR^d}\big\|\delta_{(y,0)}b\cdot\nabla_\v\delta_{(y,0)}u\big\|_p|y|^{-d-\alpha}\dif y=:I_1+I_2+I_3.\end{aligned}$$ For $I_1$, since $p>(\alpha+\beta)d/(\alpha\beta)$, by with $q=\infty$, we have $$\begin{aligned}
I_1\preceq \|\Delta_x^{\frac{\alpha}{2}} b\|_p\|\nabla_\v u\|_\infty\preceq\|b\|_{\alpha,0; p}\|\nabla_\v u\|_{\alpha,\beta;p}.\end{aligned}$$ For $I_2$, since $p\alpha>d$, by we have $$\begin{aligned}
I_2^p&=\int_{\mR^d}\|b(\cdot,\v)\cdot\nabla_\v\Delta_x^{\frac{\alpha}{2}} u(\cdot,\v)\|_p^p\dif \v\\
&\preceq\int_{\mR^d}\|b(\cdot,\v)\|_\infty^p\|\nabla_\v\Delta_x^{\frac{\alpha}{2}} u(\cdot,\v)\|_p^p\dif \v\\
&\preceq\int_{\mR^d}\|b(\cdot,\v)\|_{\alpha,p}^p\dif \v\sup_\v\|\nabla_\v\Delta_x^{\frac{\alpha}{2}} u(\cdot,\v)\|_p^p.\end{aligned}$$ For $I_3$, by and again, we have for any $\gamma\in(0,\alpha-\frac{d}{p})$, $$\begin{aligned}
&\int_{\mR^d}\|\delta_yb(\cdot,\v)\cdot\nabla_\v\delta_yu(\cdot,\v)\|_p^p\dif \v
\preceq\int_{\mR^d}\|\delta_y b(\cdot,\v)\|_p^p\|\delta_y\nabla_\v u(\cdot,\v)\|_\infty^p\dif \v\\
&\qquad\preceq\int_{\mR^d}\|b(\cdot,\v)\|_{\alpha,p}^p(|y|^{\alpha p}\wedge 1)\|\nabla_\v u(\cdot,\v)\|_{\alpha,p}^p(|y|^{\gamma p}\wedge 1)\dif \v\\
&\qquad\preceq\left(\int_{\mR^d}\|b(\cdot,\v)\|_{\alpha,p}^p\dif \v\right)\sup_\v\|\nabla_\v u(\cdot,\v)\|_{\alpha,p}^p(|y|^{(\alpha+\gamma)p}\wedge 1).\end{aligned}$$ On the other hand, notice that by $p\beta>d$ and , $$\begin{aligned}
\sup_\v\|\nabla_\v\Delta_x^{\frac{\alpha}{2}} u(\cdot,\v)\|_p^p
\leq\int_{\mR^d}\sup_\v |\nabla_\v\Delta_x^{\frac{\alpha}{2}} u(x,\v)|^p\dif x
\preceq\int_{\mR^d}\|\nabla_\v\Delta_x^{\frac{\alpha}{2}} u(x,\cdot)\|^p_{\beta, p}\dif x,\end{aligned}$$ and similarly, $$\|b\cdot\nabla_\v u\|^p_p\preceq\int_{\mR^d} \|b(\cdot,\v)\|^p_{\alpha,p}\|\nabla_\v u(\cdot,\v)\|^p_{p}\dif\v\preceq\|b\|^p_{\alpha,0;p}\|\nabla_\v u\|^p_{0,\beta;p}.$$ Combining the above calculations, we obtain the desired estimate.

Let $\varrho:\mR^{2d}\to[0,\infty)$ be a smooth function with support in the unit ball and $\int\varrho=1$. Define $$\begin{aligned}
\label{Rho}
\varrho_\eps(z):=\eps^{-2d}\varrho(\eps^{-1}z),\ \ \eps\in(0,1),\end{aligned}$$ and for a locally integrable 