\begin{split}
\|\Delta_x^{\frac{\alpha}{2}}u_t\|^p_{p}&=\int_{\mR^{2d}}\|(\Delta_x^{\frac{\alpha}{2}} u_t)\phi^{z^o_t}_\delta\|^p_{p}\dif z^o 
\preceq\int_{\mR^{2d}}\|\Delta_x^{\frac{\alpha}{2}} u^{z^o}_{\delta,t}\|^p_{p}\dif z^o \\
&\qquad+\int_{\mR^{2d}}\|\Delta_x^{\frac{\alpha}{2}} (u_t\phi^{z^o_t}_\delta)-(\Delta_x^{\frac{\alpha}{2}} u_t)\phi^{z^o_t}_\delta\|^p_{p}\dif z^o.
\end{split}\end{aligned}$$ By and , we have $$\begin{aligned}
\begin{split}
\int_{\mR^{2d}}\|\Delta_x^{\frac{\alpha}{2}}u^{z^o}_{\delta,t}\|^p_{p}\dif z^o 
&\preceq((T-t)\wedge\lambda^{-1})^{p-1-\frac{3p\alpha}{2}}\int_{\mR^{2d}}\|g^{z^o}_{\delta}\|_{\mL^p(t,T)}^p\dif z^o\\
&\preceq((T-t)\wedge\lambda^{-1})^{p-1-\frac{3p\alpha}{2}}\|f\|^p_{\mL^p(t,T)}.
\end{split}\label{LP5}\end{aligned}$$ On the other hand, noticing that by definition , $$\Delta_x^{\frac{\alpha}{2}} (u_t\phi^{z^o_t}_\delta)-(\Delta_x^{\frac{\alpha}{2}} u_t)\phi^{z^o_t}_\delta
=u_t\cdot\Delta_x^{\frac{\alpha}{2}}\phi^{z^o_t}_\delta+\int_{\mR^d}\delta_{(y,0)}u_t\cdot\delta_{(y,0)}\phi^{z^o_t}_\delta|y|^{-d-\alpha}\dif y,$$ and $$\sup_z\int_{\mR^{2d}}|\Delta_x^{\frac{\alpha}{2}}\phi_\delta(z^o_t-z)|^p\dif z^o\leq C_\delta,$$ $$\sup_z\left(\int_{\mR^{2d}}|\delta_{(y,0)}\phi^{z^o_t}_\delta(z)|^p\dif z^o\right)^{1/p}\leq C_\delta(|y|\wedge 1),$$ by Minkovskii’s inequality, we have $$\begin{aligned}
\label{LP6}
\begin{split}
&\int_{\mR^{2d}}\|\Delta_x^{\frac{\alpha}{2}} (u_t\phi^{z^o_t}_\delta)-(\Delta_x^{\frac{\alpha}{2}} u_t)\phi^{z^o_t}_\delta\|_p^p\dif z^o
\preceq\int_{\mR^{2d}}\|u_t\cdot\Delta_x^{\frac{\alpha}{2}}\phi^{z^o_t}_\delta\|_p^p\dif z^o\\
&\qquad+\int_{\mR^{2d}}\left(\int_{\mR^d}|\delta_{(y,0)}u_t(z)|\cdot\left(\int_{\mR^{2d}}|\delta_{(y,0)}\phi^{z^o_t}_\delta(z)|^p\dif z^o\right)^{1/p}|y|^{-d-\alpha}\dif y\right)^p\dif z\\
&\qquad\preceq \|u_t\|^p_p+\int_{\mR^{2d}}\left(\int_{\mR^d}|\delta_{(y,0)}u_t(z)|\cdot(|y|\wedge 1)|y|^{-d-\alpha}\dif y\right)^p\dif z\\
&\qquad\preceq \|u_t\|^p_p+\left(\int_{\mR^d}\|\delta_{(y,0)}u_t\|_p\cdot(|y|\wedge 1)|y|^{-d-\alpha}\dif y\right)^p\preceq\|u_t\|^p_p.
\end{split}\end{aligned}$$ Combining , and with , we arrive at $$\|u_t\|^p_{\alpha,0; p}\preceq((T-t)\wedge\lambda^{-1})^{p-1-\frac{3p\alpha}{2}}\|f\|^p_{\mL^p(t,T)}.$$ Similarly, for any $\beta\in(0,2)$, one can show that $$\|u_t\|^p_{0,\beta; p}\preceq((T-t)\wedge\lambda^{-1})^{p-1-\frac{p\beta}{2}}\|f\|^p_{\mL^p(t,T)}.$$ Combining the above two estimates, we obtain .

Below we assume that $u\in\mH^{0,2}_p(T)$ is a solution of in the sense of Definition \[Def31\]. Let $\varrho_\eps$ be defined by and set $$u_\eps:=u*\varrho_\eps,\ \ f_\eps:=f*\varrho_\eps.$$ Taking $\varphi=\varrho_\eps(z-\cdot)$ in , we obtain $$\p_t u_\eps+\sL^a_t u_\eps-\lambda u_\eps+[\varrho_\eps,\sL^a_t]u_t+f_\eps=0,\ \ u_{\eps,T}=0,$$ where $\sL^a_t:=\sL^{a,0}_t$ and $[\varrho_\eps,\sL^a_t]$ is defined by . By what we have proved, it holds that $$\|u_\eps\|_{\mH^{2/3,2}_p(T)}\leq C_1(\|f_\eps\|_{\mL^p(T)}+\|[\varrho_\eps,\sL^{a}]\|_{\mL^p(T)}).$$ Since $\nabla^2_\v u\in\mL^p(T)$, by the property of convolutions and (ii) of Lemma \[Le24\], we get by taking limits. Similarly, we also have .

Finally, we use the standard continuity argument to show the existence of a solution (see [@Kr0]). Consider the 