Proof. Let \(X(t,x)\) be the trajectory map associated with \(u\), defined by \[\label{Eq46trajectory} \begin{align} \partial_t X(t,x) = u(t, X(t,x)), \quad X(0,x) = x. \end{align}\tag{12}\] Then the transport equation implies that \[\label{transol} \rho(t, X(t,x)) = \rho_0(x), \quad \text{i.e.,} \quad \rho(t,x) = \rho_0(X_t^{-1}(x)),\tag{13}\] where \(X_t : \mathbb{R}^3 \to \mathbb{R}^3\) is defined by \(X_t(x) = X(t,x)\).

Note that \[-\|\nabla u(t)\|_{L^\infty} |X(t,x) - X(t,y)| \leq \frac{d}{dt} |X(t,x) - X(t,y)| \leq \|\nabla u(t)\|_{L^\infty} |X(t,x) - X(t,y)|.\] By Grönwall’s lemma, we obtain \[\label{IFM} |X(s,x) - X(s,y)| \leq \exp\left( \int_s^t \|\nabla u(\tau)\|_{L^\infty} \, d\tau \right) |X(t,x) - X(t,y)|, \quad \forall s < t.\tag{14}\] Moreover, from the definition 5 of \(\|u\|_T\), we have \[\int_s^t \|\nabla u(\tau)\|_{L^\infty} \, d\tau \leq C\|u\|_T \log\left( \frac{t}{s} \right),\] which together with 14 yields \[\label{diffrho} |X(s,x) - X(s,y)| \leq \exp\Bigl( C\|u\|_T \log\bigl( \frac{t}{s} \bigr) \Bigr) |X(t,x) - X(t,y)|, \quad \forall s < t.\tag{15}\]

Now we use this inequality to derive the Hölder estimate of \(\rho\). Let \(x, y \in \mathbb{R}^3\). Then there exist \(\overline{x}, \overline{y} \in \mathbb{R}^3\) such that \[\begin{align} &\rho(t,x) = \rho_0(\overline{x}), \quad x = X(t, \overline{x}),\\ &\rho(t,y) = \rho_0(\overline{y}), \quad y = X(t, \overline{y}). \end{align}\] Consequently, for any \(\tau < t\), by 12 and 13 we have \[\begin{align} |\rho(t,x) - \rho(t,y)| &\leq \|\nabla \rho_0\|_{L^\infty} |\overline{x} - \overline{y}| \\ &\leq \|\nabla \rho_0\|_{L^\infty} |X(\tau, \overline{x}) - X(\tau, \overline{y})| + 2 \|\nabla \rho_0\|_{L^\infty} \int_0^\tau \|u(\tau)\|_{L^\infty} \, d\tau \\ &\leq \|\nabla \rho_0\|_{L^\infty} |X(\tau, \overline{x}) - X(\tau, \overline{y})| + 2 \|\nabla \rho_0\|_{L^\infty} \|u\|_T \tau^{\frac{1}{2}}. \end{align}\] For any \(\gamma \in (0,1)\) and \(t \in [0,T]\), we take \(\tau = t \frac{|x-y|^{2\gamma}}{1 + 2|x-y|^{2\gamma}}\). Then it follows from 15 and assumption 11 that \[\begin{align} |\rho(t,x) - \rho(t,y)| &\leq \|\nabla \rho_0\|_{L^\infty} \exp\left( \log\Bigl(2 + \frac{1}{|x-y|^{2\gamma}}\Bigr) \|u\|_T \right) |X(t, \overline{x}) - X(t, \overline{y})| \\ &\quad + 2\sqrt{t} \|\nabla \rho_0\|_{L^\infty} \|u\|_T |x-y|^\gamma \\ &\leq \|\nabla \rho_0\|_{L^\infty} \left( \exp\bigl(-C_0 \gamma \varepsilon_1 \log(|x-y|)\bigr) |x-y| + 2\sqrt{t} \varepsilon_1 |x-y|^\gamma \right) \\ &\lesssim \|\nabla \rho_0\|_{L^\infty} |x-y|^\gamma, \quad \forall |x-y| < 1/2, \end{align}\] for some universal \(C_0\), provided we choose \(\varepsilon_1\) sufficiently small such that \(1 - C_0 \gamma \varepsilon_1 \geq \gamma\). Note that a direct corollary of 13 is \[\label{lowerbd} \|\rho(t)\|_{L^\infty} \leq \|\rho_0\|_{L^\infty}, \quad \forall t \geq 0.\tag{16}\] This completes the proof ?? . ◻

Proof. Let \(\psi \in \mathcal{S}(\mathbb{R}^3)\) be a standard mollifier. Denote \(\psi_\varepsilon = \varepsilon^{-3} \psi(\cdot/\varepsilon)\) and \(\rho_\varepsilon = \rho \ast \psi_\varepsilon\). Then \[\begin{align} |D|^{\alpha'}\rho(t) - |D|^{\alpha'}\rho(s) = &\bigl( |D|^{\alpha'}\rho(t) - |D|^{\alpha'}\rho_\varepsilon(t) \bigr) \\ &- \bigl( |D|^{\alpha'}\rho(s) - |D|^{\alpha'}\rho_\varepsilon(s) \bigr) + \bigl( |D|^{\alpha'}\rho_\varepsilon(t) - |D|^{\alpha'}\rho_\varepsilon(s) \bigr). \end{align}\] For the mollification error, using standard inequalities, we obtain \[\begin{align} ||D|^{\alpha'}\rho(x) - |D|^{\alpha'}\rho_\varepsilon(x)| &= \left| \int \bigl( |D|^{\alpha'}\rho(x) - |D|^{\alpha'}\rho(y) \bigr) \psi_\varepsilon(x-y) \, dy \right| \\ &\lesssim \||D|^\alpha \rho\|_{L_T^\infty L^\infty} \| |\cdot|^{\alpha-\alpha'} \phi_\varepsilon(\cdot) \|_{L^1} \lesssim \||D|^\alpha \rho\|_{L_T^\infty L^\infty} \varepsilon^{\alpha-\alpha'}. \end{align}\] Moreover, the mollified function \(\rho_\varepsilon\) satisfies \[\partial_t \rho_\varepsilon + \psi_\varepsilon \ast (u \cdot \nabla \rho_\varepsilon) + \psi_\varepsilon \ast \nabla \cdot \bigl( u (\rho - \rho_\varepsilon) \bigr) = 0,\] hence \[\begin{align} ||D|^{\alpha'}\rho_\varepsilon(t) - |D|^{\alpha'}\rho_\varepsilon(s)| &\lesssim \Bigl ||D|^{\alpha'} \psi_\varepsilon \ast \int_s^t (\nabla \rho_\varepsilon \cdot u)(\tau) \, d\tau \Bigr| + \Bigl| |D|^{\alpha'} \nabla \psi_\varepsilon \ast \int_s^t (\rho - \rho_\varepsilon) u(\tau) \, d\tau \Bigr| \\ &\lesssim \||D|^\alpha \rho\|_{L^\infty} \|u\|_T \int_s^t \varepsilon^{\alpha-\alpha'-1} \tau^{-\frac{1}{2}} \, d\tau \\ &\lesssim \||D|^\alpha \rho\|_{L^\infty} \|u\|_T \, \varepsilon^{\alpha-\alpha'-1} \min\bigl\{ s^{-\frac{1}{2}} (t-s), (t-s)^{\frac{1}{2}} \bigr\}, \end{align}\] where we used the estimate \[|\nabla \rho_\varepsilon| \lesssim ||D|^\alpha \rho \ast \nabla |D|^{-\alpha} \psi_\varepsilon| \lesssim \||D|^\alpha \rho\|_{L^\infty} \varepsilon^{-1+\alpha}.\] Optimizing with respect to \(\varepsilon\) by taking \(\varepsilon = \min\bigl\{ s^{-\frac{1}{2}} (t-s), (t-s)^{\frac{1}{2}} \bigr\}\) yields the desired estimate. ◻

Proof. The estimate ?? follows from a classical decomposition into low and high frequencies; details can be found in [1]. For ?? , we observe that \[\delta_a^t \partial_t^{m} \mathcal{P}^{\eta} K_{x_0}(t,x) = \int_t^{t+a} \partial_t^{m+1} \mathcal{P}^{\eta} K_{x_0}(s,x) \, ds,\] and then ?? follows directly from ?? . ◻

Proof. We first deduce from 6 that \(\widehat{K}_{x_0}(t,\xi)=e^{-\frac{t}{\rho_0(x_0)}|\xi|^2},\) from which, we infer \[\label{S3eq5} \begin{align} \sup_{t \leq T} t^{\frac{1}{2}} \|u_{L,x_0}(t)\|_{L^\infty} =&\sqrt{\rho_0(x_0)}\sup_{t \leq T} \left( \Bigl(\frac{t }{\rho_0(x_0)}\Bigr)^{\frac{1}{2}}\bigl\|e^{\frac{t}{\rho_0(x_0)}\Delta}u_0\bigr\|_{L^\infty}\right)\\ \leq & \sqrt{\rho_0(x_0)}\|u_0\|_{\mathrm{BMO}_b^{-1}},\quad if\;\;T\leq \rho(x_0)b^2, \end{align}\tag{18}\] and \[\begin{align} \sup_{t \leq T}& t^{1+\frac{\kappa}{2}} \| (|D|^{1+\kappa} u_{L,x_0})(t) \|_{L^\infty} \\ =&\rho_0^{1+\frac{\kappa}{2}}(x_0)\sup_{t \leq T} \left(\Bigl(\frac{t}{\rho_0(x_0)}\Bigr)^{\frac{1}{2}} \Bigl(\frac{t}{\rho_0(x_0)}\Bigr)^{\frac{1+\kappa}{2}} \bigl\| (|D|^{1+\kappa}e^{\frac{t}{2\rho_0(x_0)}\Delta} e^{\frac{t}{2\rho_0(x_0)}\Delta}u_0\bigr\|_{L^\infty} \right)\\ \lesssim &\rho_0^{1+\frac{\kappa}{2}}(x_0)\sup_{t \leq T} \left(\Bigl(\frac{t}{\rho_0(x_0)}\Bigr)^{\frac{1}{2}} \bigl\|e^{\frac{t}{2\rho_0(x_0)}\Delta}u_0\bigr\|_{L^\infty} \right)\\ \lesssim &\rho_0^{1+\frac{\kappa}{2}}(x_0)\|u_0\|_{\mathrm{BMO}_b^{-1}},\quad if\;\;T\leq 2\rho(x_0)b^2. \end{align}\] Observe that for \(t\geq s\), if \(t-s\leq \frac{s}{2}\), then \[\label{S3eq4} \begin{align} \left\|e^{\frac{t}{\rho_0(x_0)}\Delta}u_0- e^{\frac{s}{\rho_0(x_0)}\Delta}u_0\right\|_{L^\infty}&= \Bigl\|\int_{\frac{s}{2}}^{t-\frac{s}{2}}\partial_\tau \left(e^{\frac{\tau}{\rho_0(x_0)}\Delta}\right)e^{\frac{s}{2\rho_0(x_0)}\Delta}u_0d\tau\Bigr\|_{L^\infty}\\ &\lesssim \frac{1}{\rho_0(x_0)}\int_{\frac{s}{2}}^{t-\frac{s}{2}}\bigl\|e^{\frac{\tau}{\rho_0(x_0)}\Delta}\Delta e^{\frac{s}{2\rho_0(x_0)}\Delta}u_0\bigr\|_{L^\infty}d\tau\\ &\lesssim s^{-1}(t-s)\|e^{\frac{s}{4\rho_0(x_0)}\Delta}u_0\|_{L^\infty}\\ &\lesssim \sqrt{\rho_0(x_0)}s^{-\frac{3}{2}}(t-s)\|u_0\|_{BMO^{-1}_b},\quad if\;\;T\leq 4\rho(x_0)b^2. \end{align}\tag{19}\] While it follows from 18 that \[\left\|e^{\frac{t}{\rho_0(x_0)}\Delta}u_0- e^{\frac{s}{\rho_0(x_0)}\Delta}u_0\right\|_{L^\infty}\lesssim \sqrt{\rho_0(x_0)}s^{-\frac{1}{2}}\|u_0\|_{BMO_b^{-1}},\quad if\;\;s\leq t\leq T\leq \rho(x_0)b^2.\] As a result, it comes out \[\label{S3eq6} \left\|e^{\frac{t}{\rho_0(x_0)}\Delta}u_0- e^{\frac{s}{\rho_0(x_0)}\Delta}u_0\right\|_{L^\infty}\lesssim \sqrt{\rho_0(x_0)}\min\{1,s^{-1}(t-s)\}s^{-\frac{1}{2}}\|u_0\|_{BMO_b^{-1}}.\tag{20}\] For the case \(t-s\geq \frac{s}{2}\), 20 follows from 18 .

By virtue of 20 , we deduce that \[\label{S3eq11} \begin{align} \sup_{s < t \leq T} s^{\frac{1}{2} + \varkappa_1} \frac{\|u_{L,x_0}(t) - u_{L,x_0}(s)\|_{L^\infty}}{(t-s)^{\varkappa_1}} \lesssim& \frac{\min\{1,s^{-1}(t-s)\}}{(t-s)^{\varkappa_1}} s^{\varkappa_1}\|u_0\|_{BMO_b^{-1}}\\ \lesssim &\|u_0\|_{BMO_b^{-1}}, \end{align}\tag{21}\] where we used the fact that if \(t-s\leq s,\) there holds \[\frac{\min\{1,s^{-1}(t-s)\}}{(t-s)^{\varkappa_1}} s^{\varkappa_1}\leq (t-s)^{-1+\varkappa_1} s^{1-\varkappa_1}\leq 1\quad \forall \varkappa_1\in (0,1),\] whereas if \(t-s\geq s,\) one has \[\frac{\min\{1,s^{-1}(t-s)\}}{(t-s)^{\varkappa_1}} s^{\varkappa_1}\leq (t-s)^{-\varkappa_1} s^{\varkappa_1}\leq 1\quad \forall \varkappa_1\in (0,1).\]

For \(\|u_{L,x_0}\|_{V_T^b},\) we get, by using changes of variable, that \[\begin{align} \|u_{L,x_0}\|_{V_T^b}= &R^{-\frac{3}{2}}\Bigl(\int_0^{\min\{T,R^2\}}\int_{B(x,R)}\bigl|e^{\frac{t}{\rho_0(x_0)}\Delta}u_0(y)\bigr|^2\,dy\,dt\Bigr)^{\frac{1}{2}}\\ \lesssim &\rho_0(x_0)R^{-\frac{3}{2}}\Bigl(\int_0^{\min\bigl\{\frac{T}{\rho_0(x_0)}, \frac{R^2}{\rho_0(x_0)}\bigr\}}\int_{B(x,R)}\bigl|e^{s\Delta}u_0(y)\bigr|\,dy\,ds\Bigr)^{\frac{1}{2}}\\\ \lesssim & C(\rho_0(x_0))R^{-\frac{3}{2}}\Bigl(\int_0^{\min\bigl\{\frac{T}{\rho_0(x_0)},\frac{R^2}{\rho_0(x_0)}\bigr\}}\int_{B\bigl(x,\frac{R}{\sqrt{\rho_0(x_0)}}\bigr)}\bigl|e^{s\Delta}u_0(y)\bigr|^2\,dy\,\Bigr)^{\frac{1}{2}}\\\ \lesssim &\|u_0\|_{BMO_b^{-1}} \end{align}\] where in the last step, we used the fact that the ball with radius \(R\) can be covered by finite balls with radius \(\frac{R}{\sqrt{\rho_0(x_0)}}\), and with a constant multiple to \(T,b\) with respect to \(\|\rho_0^{-1}\|_{L^\infty}\).

As a consequence, we deduce that, \[\label{S3eq1} \begin{align} &\sup_{x_0} \Bigl( \sup_{t \leq T} \left( t^{\frac{1}{2}} \|u_{L,x_0}(t)\|_{L^\infty} + t^{1+\frac{\kappa}{2}} \| (|D|^{1+\kappa} u_{L,x_0})(t) \|_{L^\infty} \right) + \|u_{L,x_0}\|_{V_T^b} \Bigr) \\ &\quad + \sup_{x_0} \sup_{s < t \leq T} s^{\frac{1}{2} + \varkappa_1} \frac{\|u_{L,x_0}(t) - u_{L,x_0}(s)\|_{L^\infty}}{(t-s)^{\varkappa_1}} \lesssim \|u_0\|_{\mathrm{BMO}_b^{-1}}. \end{align}\tag{22}\]

On the other hand, we observe that for any function \(K(t,s,x,z)\geq 0\), \[\begin{align} &\left(\int_{\mathbb{R}^3}\Bigl(\int_{\mathbb{R}^3}K(t,s,x-y,x)u_0(y)\,dy\Bigr)^2\,dx\right)^{\frac{1}{2}}\\ &\lesssim \left(\int_{\mathbb{R}^3}\Bigl(\Bigl(\int_{\mathbb{R}^3} K(t,s,x-y,x)|u_0(y)|^2\,dy \Bigr)^{\frac{1}{2}}\Bigl(\int_{\mathbb{R}^3}K(t,s,x-y,x)\,dy \Bigr)^{\frac{1}{2}}\Bigr)^2dx\right)^{\frac{1}{2}}\\ &\lesssim \Bigl(\sup_{z\in\mathbb{R}^3}\|K(t,s,\cdot,z)\|_{L^1}\Bigr)^{\frac{1}{2}}\left(\int_{\mathbb{R}^3}\int_{\mathbb{R}^3} K(t,s,x-y,x)|u_0(y)|^2\,dy dx\right)^{\frac{1}{2}}\\ &\lesssim \|\sup_{z\in\mathbb{R}^3}K(t,s,\cdot,z)\|_{L^1}\|u_0\|_{L^2}. \end{align}\] We now take \(K(t,s,x,z)=K(t,x,z)=\frac{\rho_0^{\frac{3}{2}}(z)}{(4\pi t)^{\frac{3}{2}}}\exp(-\frac{\rho_0(z)|x|^2}{4t}),\) then it follows from ?? that \[\sup_{z\in\mathop{\mathbb{R}\kern 0pt}\nolimits^3}K(t,s,x,z)\leq \frac{C_2^{\frac{3}{2}}}{(4\pi t)^{\frac{3}{2}}}\exp\Bigl(-\frac{C_1|x|^2}{4t}\Bigr) \quadand\quad\|\sup_{z\in\mathbb{R}^3}K(t,\cdot,z)\|_{L^1}\lesssim 1,\quad for any\;t\geq 0.\] Hence we obtain \[\label{S3eq9} \sup_{t\leq T}\|\tilde{u}_L(t)\|_{L^2}\lesssim \|u_0\|_{L^2}\tag{23}\] from which and 22 , we infer \[\label{S3eq3} \begin{align} \sup_{t \leq T} &t^{\frac{5}{12}} \left\| (K_{x_0} \ast u_0)|_{x_0=x}(t,x) \right\|_{L_x^{12}}\\ \leq &\Bigl(\sup_{x_0\in\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\sup_{t \leq T} t^{\frac{1}{2}} \left\| (K_{x_0} \ast u_0)(t,x) \right\|_{L_x^{\infty}}\Bigr)^{\frac{5}{6}}\Bigl(\sup_{x_0\in\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\sup_{t \leq T} \| \widetilde{u}_L(t,x)\|_{L_x^{2}}\Bigr)^{\frac{1}{6}}\\ \lesssim & \|u_0\|_{\mathrm{BMO}_b^{-1}}^{\frac{5}{6}} \|u_0\|_{L^2}^{\frac{1}{6}}.\end{align}\tag{24}\]

For the term involving time difference of \(\widetilde{u}_L,\) we take \[\begin{align} K(t,s,x,z)&=\frac{(\rho_0(z))^{\frac{3}{2}}}{(4\pi t)^{-\frac{3}{2}}}\exp\Bigl(-\frac{\rho_0(z)|x|^2}{4t}\Bigr)-\frac{(\rho_0(z))^{\frac{3}{2}}}{(4\pi s)^{-\frac{3}{2}}}\exp\Bigl(-\frac{\rho_0(z)|x|^2}{4s}\Bigr) \\ &=\int_s^t\partial_\tau \Bigl(\frac{(\rho_0(z))^{\frac{3}{2}}}{(4\pi \tau)^{-\frac{3}{2}}}\exp\Bigl(-\frac{\rho_0(z)|x|^2}{4\tau}\Bigr)\Bigr)\,d\tau, \end{align}\] from which and ?? , we infer \[\bigl\|\sup_{z\in\mathbb{R}^3}K(t,s,\cdot,z)\bigr\|_{L^1}\lesssim \min\bigl\{1,s^{-1}(t-s)\bigr\},\quad for any\;t\geq s\geq 0,\] so that we obtain \[\begin{align} &\|\tilde{u}_L(t)-\tilde{ u}_L(s)\|_{L^2}\lesssim \min\bigl\{1,s^{-1}(t-s)\bigr\}\|u_0\|_{L^2},\\ &\|\tilde{u}_L(t)-\tilde{ u}_L(s)\|_{L^\infty}\min\bigl\{1,s^{-1}(t-s)\bigr\}\|u_0\|_{BMO_b^{-1}}. \end{align}\] As a consequence, we deduce that \[\label{S3eq12} \begin{align} & \sup_{s<t<T}s^{\frac{5}{12}+\varkappa_2}\frac{\|\tilde{u}_L(t)-\tilde{ u}_L(s)\|_{L^{12}}}{(t-s)^{\varkappa_2}}\\ &\lesssim \sup_{s<t<T}\Bigl(\frac{s^{\varkappa_2}}{(t-s)^{\varkappa_2}} \bigl(s^{\frac{1}{2}}\|\tilde{u}_L(t)-\tilde{ u}_L(s)\|_{L^\infty}\bigr)^{\frac{5}{6}}\|\tilde{u}_L(t)-\tilde{ u}_L(s)\|_{L^2}^{\frac{1}{6}} \Bigr)\\ &\lesssim \frac{s^{\varkappa_2}}{(t-s)^{\varkappa_2}} {\min\{1,s^{-1}(t-s)\}}\|u_0\|_{BMO_b^{-1}}^{\frac{5}{6}}\|u_0\|_{L^2}^{\frac{1}{6}}\\ &\lesssim \|u_0\|_{BMO_b^{-1}}^{\frac{5}{6}}\|u_0\|_{L^2}^{\frac{1}{6}}, \end{align}\tag{25}\]

By summarizing the estimates 22 , 24 and 25 , we complete the proof of [uL]. ◻

Proof. Define the norms \[\begin{align} &\|f\|_{Z_T^{1,b}} := \sup_{t \leq T} t^{\frac{1}{2}} \|f(t)\|_{L^\infty} + \sup_{x_0 \in \mathbb{R}^3,\, R \leq b} R^{-\frac{3}{2}} \Bigl( \int_0^{\min\{T, R^2\}} \int_{B_{x_0}(R)} |f(\tau,y)|^2 \, dy \, d\tau \Bigr)^{\frac{1}{2}},\\[6pt] &\|f\|_{Z_T^{2,b}} := \sup_{t \leq T} t \|f(t)\|_{L^\infty} + \sup_{x_0 \in \mathbb{R}^3,\, R \leq b} R^{-3} \int_0^{\min\{T, R^2\}} \int_{B_{x_0}(R)} |f(\tau,y)| \, dy \, d\tau. \end{align}\]

Thanks to the approach of Koch and Tataru, we have the following lemma (see also [29]).

The proof of this lemma is postponed to the appendix.

It follows from Lemma 1 that \[\label{S3eq17} \sup_{x_0\in\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \sup_{t \leq T} t^{\frac{1}{2}} \|u_{N,x_0}(t)\|_{L^\infty} + \sup_{x_0\in\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \|u_{N,x_0}\|_{V_T^b} \lesssim \|\rho_0\|_{L^\infty} \|u\|_{X_T}^2.\tag{26}\]

Let us turn to the estimate of the remaining norms of \(u_{N,x_0}\) in the space \(X_T,\) which we will handle term by term below.

(i): The \(C^{1+\kappa}\) norm of \(u_{N,x_0}\).

We split the time interval into two parts as \[\begin{align} |D|^\kappa \nabla u_{N,x_0}(t,x) =& \left( \int_0^{t/2} + \int_{t/2}^t \right) \int |D|^\kappa \mathbb{P} \nabla^2 K_{x_0}(t-\tau, x-y) (\rho u \otimes u)(\tau, y) \, dy \, d\tau\\ =:&u_{N1,x_0} + u_{N2,x_0}. \end{align}\] Note that by Proposition 1, if \(\|u\|_{X_T} < \varepsilon_0\) for some sufficiently small \(\varepsilon_0 > 0\), then \[\label{rhoreg} \|\rho\|_{C^\gamma} \lesssim C(\gamma, \varepsilon_0) \|\rho_0\|_{C^1}, \quad \forall \gamma < 1.\tag{27}\]

For \(u_{N1}\), we use a method similar to that in the Appendix 7.2. In order to do so, for all \((t,x) \in [0,+\infty) \times \mathbb{R}^3,\) we define \[\label{defnewker} \Gamma_R^1(t,x) := \mathbf{1}_{|x| > R} \frac{1}{|x|^{3+\alpha}}, \qquad \Gamma_R^2(t,x) := \mathbf{1}_{|x| < R} \frac{1}{(\sqrt{t} + |x|)^{3+\alpha}},\tag{28}\] for some \(\alpha>0\). It is easy to observe that \[||D|^\kappa \mathbb{P} \nabla^2 K_{x_0}(t, x)| \lesssim \Gamma_R^1(t,x) + \Gamma_R^2(t,x),\] with \(\alpha = 2 + \kappa\). Then \[\begin{align} |u_{N1,x_0}| \lesssim& \int_0^{t/2} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \Gamma_R^1(x-y) |\rho u \otimes u|(\tau, y) \, dy \, d\tau\\ &+ \int_0^{t/2} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \Gamma_R^2(t-\tau, x-y) |\rho u \otimes u|(\tau, y) \, dy \, d\tau := u_{N1,x_0}^1 + u_{N1,x_0}^2. \end{align}\] Taking \(R = \sqrt{t}\) in ?? of Lemma 6 gives \[\|u_{N1,x_0}^1(t)\|_{L^\infty} \lesssim t^{-1 - \frac{\kappa}{2}} \|\rho u \otimes u\|_{Z_T^{2,b}}.\] Note that \[\begin{align} \|u_{N1,x_0}^2(t)\|_{L^\infty} &\lesssim \int_0^{t/2} \int_{|y| \leq \sqrt{t}} \frac{1}{(\sqrt{t})^{5+\kappa}} |\rho u \otimes u|(\tau, y) \, d\tau \\ &\lesssim \frac{t^{\frac{3}{2}}}{t^{\frac{5+\kappa}{2}}} \|\rho u \otimes u\|_{Z_T^{2,b}} \lesssim t^{-1 - \frac{\kappa}{2}} \|\rho u \otimes u\|_{Z_T^{2,b}}. \end{align}\] Combining the estimates for \(u_{N1}^1\) and \(u_{N1}^2\), and noting that \(\|\rho u \otimes u\|_{Z_T^{2,b}} \lesssim \|u\|_{Z_T^{1,b}}^2 \|\rho_0\|_{L^\infty}\), we obtain \[\label{uN1} \sup_{t \leq T} t^{1 + \frac{\kappa}{2}} \|u_{N1,x_0}(t)\|_{L^\infty} \lesssim \|u\|_{X_T}^2 \|\rho_0\|_{L^\infty}.\tag{29}\]

For \(u_{N2}\), we have \[\begin{align} |u_{N2,x_0}(t,x)| &\lesssim \Bigl| \int_{t/2}^t \int_{\mathbb{R}^3} \mathbb{P} \nabla^2 |D|^\kappa K_{x_0}(t-\tau, x-y) \bigl( (\rho u \otimes u)(\tau, y) - (\rho u \otimes u)(\tau, x) \bigr) \, dy \, d\tau \Bigr| \\ &\lesssim \int_{t/2}^t \int_{\mathbb{R}^3} \left| \mathbb{P} \nabla^2 |D|^\kappa K_{x_0}(t-\tau, x-y) \right| |x-y|^{\kappa'} \|\rho u \otimes u(\tau)\|_{\dot{C}^{\kappa'}} \, dy \, d\tau \\ &\lesssim \int_{t/2}^t (t-\tau)^{-1 - \frac{\kappa}{2} + \frac{\kappa'}{2}} \|\rho u \otimes u(\tau)\|_{\dot{C}^{\kappa'}} \, d\tau. \end{align}\] for some \(\kappa' \in (\kappa, 1)\). Yet we get, by using interpolation inequality in Hölder space and the definition of \(\|\cdot\|_{X_T}\) given by 4 that \[\begin{align} \|u(t)\|_{\dot{C}^{\kappa'}}\lesssim \|u(t)\|_{L^\infty}^{1-\frac{\kappa'}{1+\kappa}}\|u(t)\|_{\dot{C}^{1+\kappa}}^{\frac{\kappa'}{1+\kappa}} \lesssim t^{\frac{1}{2}\left(1+\kappa'\right)}\|u\|_{X_T}, \end{align}\] from which and the law of product in Hölder space (see [29]), we infer \[\label{S3eq13} \begin{align} \|\rho u \otimes u(\tau)\|_{\dot{C}^{\kappa'}} \lesssim &\|\rho u\|_{L^\infty}\|u\|_{\dot{C}^{\kappa'}} +\|\rho u\|_{\dot{C}^{\kappa'}}\|u\|_{L^\infty}\\ \lesssim &\|\rho_0\|_{L^\infty}\|u\|_{L^\infty}\|u\|_{\dot{C}^{\kappa'}} +\|\rho\|_{\dot{C}^{\kappa'}}\|u\|_{L^\infty}^2 \lesssim \tau^{-1-\frac{\kappa'}{2}}\|u\|_{X_T}^2 \|\rho_0\|_{C^1}. \end{align}\tag{30}\] As a consequence, we obtain \[\label{uN2} |u_{N2,x_0}(t,x)|\\ \lesssim t^{-1 - \frac{\kappa}{2}} \|u\|_{X_T}^2 \|\rho_0\|_{C^1}\tag{31}\]

Thus, by 29 and 31 , we have proved \[\sup_{t \leq T} t^{1 + \frac{\kappa}{2}} \|\nabla |D|^\kappa u_{N,x_0}(t)\|_{L^\infty} \lesssim \|u\|_{X_T}^2 \|\rho_0\|_{C^1} .\] Note that all of the estimates above are uniform with respect to \(x_0\). Therefore, for \(T \sim b^2\), we have \[\sup_{t \leq T} t^{1 + \frac{\kappa}{2}} \| (\nabla |D|^\kappa u_{N,x_0})|_{x_0=x}(t) \|_{L^\infty} \lesssim \|u\|_{X_T}^2 \|\rho_0\|_{C^1}.\]

(ii): The non-critical \(L^{12}\) norm.

We first observe from Proposition 3 that \[\begin{align} |\nabla_x K_{x_0}(t,x)|\leq \frac{C(C_1,C_2)}{\left(\sqrt{t}+|x|\right)^4} \quadand\quad\|\nabla_x K_{x_0}(t,\cdot)\|_{L^1}\lesssim t^{-\frac{1}{2}}, \end{align}\] from which and 8 , we get, by direct computation, that \[\begin{align} |\tilde{u}_N(t,x)| &\lesssim \int_0^t \int \bigl| \nabla_y K_{x_0}(t-\tau, x, x-y) \bigr|^{\frac{11}{12}} \Bigl( \bigl| \nabla_y K_{x_0}(t-\tau, x, x-y) \bigr|^{\frac{1}{12}} |\rho u \otimes u(\tau, y)| \Bigr) dy \, d\tau \\ &\lesssim \int_0^t (t-\tau)^{-\frac{11}{24}} \Bigl( \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \bigl| \nabla_y K_{x_0}(t-\tau, x, x-y) \bigr| |\rho u \otimes u(\tau, y)|^{12} \, dy \Bigr)^{\frac{1}{12}} d\tau. \end{align}\] Taking the \(L^{12}\) norm with respect to \(x\), we obtain \[\begin{align} \|\tilde{u}_N(t)\|_{L^{12}}&\lesssim \int_0^t (t-\tau)^{-\frac{11}{24}} \Bigl( \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \bigl| \nabla_y K_{x_0}(t-\tau, x, x-y) \bigr| |\rho u \otimes u(\tau, y)|^{12} \, dy\,dx \Bigr)^{\frac{1}{12}} \,d\tau\\ &\lesssim \int_0^t (t-\tau)^{-\frac{1}{2}} \|\rho u \otimes u(\tau) \|_{L^{12}}\, d\tau \\ &\lesssim \int_0^t (t-\tau)^{-\frac{1}{2}} \tau^{-\frac{11}{12}} \, d\tau \, \|u\|_{X_T}^2 \|\rho\|_{L_T^\infty L^\infty} \\ &\lesssim t^{-\frac{5}{12}} \|u\|_{X_T}^2 \|\rho_0\|_{L^\infty}, \end{align}\] where we used the fact that \[\begin{align} \|u(\tau)\|_{L^{12}}^2\leq \|u(\tau)\|_{L^{\infty}}\|u(\tau)\|_{L^{12}}\lesssim \tau^{-\frac{11}{12}} \|u\|_{X_T}^2. \end{align}\]

(iii): The time difference for \(L^\infty\).

For the quantity \(\frac{\|u_{N,x_0}(t)-u_{N,x_0}(s)\|_{L^\infty}}{(t-s)^{\varkappa_1}}\), it suffices to consider the case \(t-s < \frac{s}{4}\), since otherwise, we have \[\begin{align} \sup_{s<t\leq T}s^{\frac{1}{2}+\varkappa_1}\frac{\|u_{N,x_0}(t)-u_{N,x_0}(s)\|_{L^\infty}}{(t-s)^{\varkappa_1}} \lesssim &\sup_{s<t\leq T}s^{\frac{1}{2}}{\|u_{N,x_0}(t)-u_{N,x_0}(s)\|_{L^\infty}}\\ \lesssim &\sup_{t<T} t^{\frac{1}{2}} \|u_{N,x_0}(t)\|_{L^\infty}. \end{align}\]

Let \(a := t-s\) and write \[\label{decopuN} \begin{align} u_{N,x_0}(t) - u_{N,x_0}(s) &= \int_s^t \mathbb{P} \nabla K_{x_0}(t-\tau) \ast (\rho u \otimes u)(\tau) \, d\tau \\ &\quad + \left( \int_0^{s/2} + \int_{s/2}^s \right) \delta_a^t \mathbb{P} \nabla K_{x_0}(s-\tau) \ast (\rho u \otimes u)(\tau) \, d\tau \\ &=: Du_{N,x_0}^1(t,s) + Du_{N,x_0}^2(t,s) + Du_{N,x_0}^3(t,s). \end{align}\tag{32}\]

By virtue of 30 , for any \(\eta\in (0,1),\) we get, by first taking \(\kappa'\in (1-\eta,1)\) and then using the interpolation inequality in Besov spaces, that \[\label{S3eq14} \begin{align} \|D^{1-\eta}(\rho u \otimes u)(\tau)\|_{L^\infty} \lesssim &\|(\rho u \otimes u)(\tau)\|_{\dot{B}^{1-\eta}_{\infty,1}} \lesssim \|(\rho u \otimes u)(\tau)\|_{L^\infty}^{1-\theta}\|\rho u \otimes u(\tau)\|_{\dot{C}^{\kappa'}}^{\theta}\\ \lesssim &\bigl(\|\rho_0\|_{L^\infty}\tau\|u(\tau)\|_{L^\infty}^2\bigr)^{1-\theta} \tau^{1-\theta}\times \tau^{-\left(1+\frac{\kappa'}{2}\right)\theta}\bigl(\|\rho_0\|_{C^1}\|u\|_{X_T}^2\bigr)^\theta\\ \lesssim &\tau^{-\frac{3-\eta}{2}}\|\rho_0\|_{C^1}\|u\|_{X_T}^2, \end{align}\tag{33}\] where \(\theta=\frac{1-\eta}{\kappa'}.\)

While for \(\Gamma_\eta(t) := \mathbb{P} \nabla |D|^{\eta-1} K_{x_0}(t)\) with some \(\eta \ll \kappa\) sufficiently small, it follows from Proposition 3 that \[\begin{align} |\Gamma_\eta(t,x)|\leq \frac{C(C_1,C_2)}{(\sqrt{t}+|x|)^{3+\eta}} \quadand\quad\|\Gamma_\eta(t,\cdot)\|_{L^1}\lesssim t^{-\frac{\eta}{2}}. \end{align}\]

As a consequence, we deduce that \[\|Du_{N,x_0}^1(t,s)\|_{L^\infty} \lesssim \int_s^t \|\Gamma_\eta(t-\tau)\|_{L^1} \|D^{1-\eta}(\rho u \otimes u)(\tau)\|_{L^\infty} \, d\tau \lesssim a^{1-\frac{\eta}{2}} s^{-\frac{3-\eta}{2}} \|u\|_{X_T}^2 \|\rho_0\|_{C^1}.\]

Similarly, for \(Du_{N,x_0}^3(t,s)\), by ?? we have \[\label{diffker} \|\delta_a^t \Gamma_\eta(t)\|_{L^1} \lesssim \min\left\{ t^{-\frac{\eta}{2}}, a t^{-1-\frac{\eta}{2}} \right\}.\tag{34}\] In particular, for \(0<\varkappa_1<1-\frac{\eta}{2},\) we deduce from 33 that \[\begin{align} \|Du_{N,x_0}^3(t,s)\|_{L^\infty} \lesssim & \int_{s/2}^s \|\delta_a^t \Gamma_\eta(s-\tau)\|_{L^1} \|D^{1-\eta}(\rho u \otimes u)(\tau)\|_{L^\infty} \, d\tau\\ \lesssim&a^{\varkappa_1}\int_{s/2}^s (s-\tau)^{-\frac{\eta}{2}-\varkappa_1}\tau^{-\frac{3-\eta}{2}}\,d\tau \|u\|_{X_T}^2 \|\rho_0\|_{C^1}\\ \lesssim &a^{\varkappa_1} s^{-\frac{1}{2}-\varkappa_1} \|u\|_{X_T}^2 \|\rho_0\|_{C^1} . \end{align}\]

For \(Du_{N,x_0}^2(t,s)\), we write \[Du_{N,x_0}^2(t,s) = \int_0^{s/2} \left( \int_s^t \partial_\gamma \mathbb{P} \nabla K_{x_0}(\gamma-\tau) \, d\gamma \right) \ast (\rho u \otimes u)(\tau) \, d\tau.\] Denote \(H(\tau) := \partial_\gamma \mathbb{P} \nabla K_{x_0}(\tau)\); then \(H(\gamma-\tau) \lesssim \frac{1}{(\sqrt{\gamma-\tau} + |y|)^6}\). Using the same decomposition as in Lemma 6, we write \(H = H^1 + H^2\), with \(H^1 \lesssim \mathbf{1}_{|y| \geq R} \frac{1}{|y|^6}\) and \(H^2 \lesssim \mathbf{1}_{|y| \leq R} \frac{1}{(\sqrt{\gamma-\tau} + |y|)^6}\). Take \(R = \sqrt{s} < b\). By ?? in Lemma 6 with \(\alpha = 3\), we obtain \[\left| \int_0^{s/2} \int_s^t H^1(\gamma-\tau) \ast (\rho u \otimes u)(\tau) \, d\gamma \, d\tau \right| \lesssim \frac{t-s}{R^3} \|u\|_{X_T}^2 \|\rho_0\|_{C^1}.\] Moreover, for the part concerning \(H^2,\) we have \[\begin{align} &\left| \int_0^{s/2} \int_s^t H^2(\gamma-\tau) \ast (\rho u \otimes u)(\tau) \, d\gamma \, d\tau \right| \\ &\quad \lesssim \int_0^{s/2} \int_s^t \int_{|y| \leq R} \frac{1}{(\gamma-\tau)^3} |\rho u \otimes u|(\tau, x-y) \, dy \, d\gamma \, d\tau \\ &\quad \lesssim \int_s^t \frac{R^3}{(\gamma - \frac{s}{2})^3} \, d\gamma \, \|u\|_{X_T}^2 \|\rho_0\|_{C^1} \lesssim \frac{(t-s) R^3}{s^3} \|u\|_{X_T}^2 \|\rho_0\|_{C^1}. \end{align}\] As a result, since \(R=\sqrt{s}\), it comes out \[\|Du_{N,x_0}^3(t,s)\|_{L^\infty} \lesssim \frac{t-s}{s^{3/2}} \|u\|_{X_T}^2 \|\rho_0\|_{C^1}.\] Since all the estimates above are uniform with respect to \(x_0\) and \(t-s < \frac{s}{4}\), we conclude \[\sup_{x_0} \sup_{s < t < T} s^{\frac{1}{2} + \varkappa_1} \frac{\|u_{N,x_0}(t) - u_{N,x_0}(s)\|_{L^\infty}}{(t-s)^{\varkappa_1}} \lesssim \|u\|_{X_T}^2 \|\rho_0\|_{C^1}.\]

(iv): The time difference for \(L^{12}\).

For any \(\eta\in (0,1),\) we get, by using 17 , that \[\begin{align} \left\||D|^{1-\eta}(\rho u\otimes u)\right\|_{L^{12}}&\lesssim\int_{\mathbb{R}^3}\frac{\left\|\delta_a(\rho u\otimes u)\right\|_{L^{12}}}{|\alpha|^{4-\eta}}d\alpha\\ &\lesssim \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\frac{\|\delta_\alpha \rho\|_{L^\infty}\|u\|_{L^\infty}\|u\|_{L^{12}}}{|\alpha|^{4-\eta}}d\alpha+\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\frac{\|\rho\|_{L^\infty}\|\delta_\alpha u\|_{L^\infty}\|u\|_{L^{12}}}{|\alpha|^{4-\eta}}d\alpha\\ &\qquad+\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\frac{\|\rho\|_{L^\infty}\|\delta_\alpha u\|_{L^\infty}\sup_{\alpha}\|u(\cdot-\alpha)\|_{L^{12}}}{|\alpha|^{4-\eta}}d\alpha \end{align}\] As a result, it comes out \[\label{S3eq18} \begin{align} \| |D|^{1-\eta}&(\rho u \otimes u)(t) \|_{L^{12}} \lesssim \|u(t)\|_{L^\infty} \|u(t)\|_{L^{12}} \|\rho(t)\|_{C^{1-\eta}} \\ &+ \|u(t)\|_{L^\infty}^{\eta}\|u(t)\|_{\dot{C}^1}^{1-\eta} \|u(t)\|_{L^{12}} \|\rho\|_{L^\infty} \lesssim \|\rho_0\|_{C^1}t^{-\left(\frac{17}{12}-\frac{\eta}{2}\right)}\|u\|_{X_T}^2. \end{align}\tag{35}\]

Let us now turn to the estimate of \(\sup_{s < t < T} s^{\frac{5}{12} + \varkappa_2} \frac{\|\tilde{u}_N(t) - \tilde{u}_N(s)\|_{L^{12}}}{(t-s)^{\varkappa_2}}.\) We again use the corresponding decomposition in 32 . Let \(a = t-s < \frac{s}{4}\). Then we deduce from ?? and 35 that \[\begin{align} \|D\tilde{u}_N^1(t,s)\|_{L^{12}} &\lesssim \int_s^t \| \tilde{\Gamma}_\eta(t-\tau, z, y)\|_{L^1_y(L^\infty_z)} \| |D|^{1-\eta}(\rho u \otimes u)(\tau) \|_{L^{12}} \, d\tau \\ &\lesssim a^{1-\frac{\eta}{2}} s^{-\frac{2-\eta}{2} - \frac{5}{12}} \|\rho_0\|_{C^1} \|u\|_{X_T}^2. \end{align}\] Here, we denote \(\tilde{\Gamma}_\eta(t, z, y) := (\nabla |D|^{\eta-1} K_{x_0}(t, y))|_{x_0=z}\) for \(0 < \eta \ll \kappa\) and \(T \leq 1\).

Similarly, we deduce from ?? and 35 that \[\begin{align} \|D\tilde{u}_N^3\|_{L^{12}} &\lesssim \int_{s/2}^s \| \delta_a^t \tilde{\Gamma}_\eta(s-\tau, z, y) \|_{L^1_y L^\infty_z} \| D^{1-\eta}(\rho u \otimes u) \|_{L^{12}} \, d\tau \\ &\lesssim a^{1-\theta} \int_{s/2}^s (s-\tau)^{-\theta-\frac{\eta}{2}}\tau^{-\left(\frac{17}{12}-\frac{\eta}{2}\right)}\,d\tau\|\rho_0\|_{C^1} \|u\|_{X_T}^2\\ &\lesssim a^{\theta} s^{-\theta- \frac{5}{12}} \|\rho_0\|_{C^1} \|u\|_{X_T}^2, \end{align}\] for any \(\theta<1-\frac{\eta}{2}.\)

For \(u_{N,x_0}^2\), using the fact that \(\|\rho u \otimes u(t)\|_{L^{12}} \lesssim t^{-\frac{11}{12}} \|u\|_{X_T}^2\) and 34 , we infer \[\begin{align} \|D\tilde{u}_N^2(t,s)\|_{L^{12}} \lesssim& \int_0^{s/2} \| \delta_a^t \tilde{\Gamma}_1(s-\tau, z, y) \|_{L^1_y (L^\infty_z)} \|\rho u \otimes u(\tau)\|_{L^{12}} \, d\tau\\ \lesssim&a^{\varkappa_2}\int_0^{s/2}(s-\tau)^{-\frac{1}{2}-\varkappa_2}\tau^{-\frac{11}{12}}\,d\tau\|u\|_{X_T}^2 \|\rho_0\|_{C^1}\\ \lesssim & a^{\varkappa_2} s^{-\varkappa_2 - \frac{5}{12}} \|u\|_{X_T}^2 \|\rho_0\|_{C^1}. \end{align}\] Thus we conclude that \[\sup_{s < t < T} s^{\frac{5}{12} + \varkappa_2} \frac{\|\tilde{u}_N(t) - \tilde{u}_N(s)\|_{L^{12}}}{(t-s)^{\varkappa_2}} \lesssim \|u\|_{X_T}^2 \|\rho_0\|_{C^1}.\]

Combining the results (i)–(iv) above with 26 , we complete the proof of Proposition 5. ◻

Proof. For simplicity, we only prove ?? , since ?? can be proved along the same line. Denote that \(\partial_t K(t,x,y) = (\partial_t K_{x_0}(t,y))|_{x_0=x}\). First, for the \(L^p\) norm, we write \[\begin{align} \tilde{g}(t,x) =& \int_0^{t/2} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K(t-\tau, x, x-y) f(\tau, y) \, dy \, d\tau \\ &+ \int_{t/2}^t \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K(t-\tau, x, x-y) \bigl( f(\tau, y) - f(t, y) \bigr) \, dy \, d\tau \\ &+ f(t,x) - \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} K(t/2, x, x-y) f(t, y) \, dy. \end{align}\] Note that for any \(h(x)\) and \(m,n\geq 0\), if we denote \(\mathbf{h}(t,x) := \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_t^{m_1} \nabla_y^{m_2} K(t, x, x-y) h(y) \, dy\), then by Hölder’s inequality, \[\begin{align} |\mathbf{h}(t,x)| &\lesssim \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} |\partial_t^{m_1} \nabla_y^{m_2} K(t, x, x-y)|^{\frac{1}{p'}} |\partial_t^{m_1} \nabla_y^{m_2} K(t, x, x-y)|^{\frac{1}{p}} |h(y)| \, dy \\ &\lesssim \bigl\| \sup_z \partial_t^{m_1} \nabla_x^{m_2} K(t, z, x) \bigr\|_{L_x^1}^{\frac{1}{p'}} \left( \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} |\partial_t^{m_1} \nabla_y^{m_2} K(t, x, x-y)| |h(y)|^p \, dy \right)^{\frac{1}{p}}, \end{align}\] which leads to \[\label{refY} \begin{align} \|\mathbf{h}(t)\|_{L^p} &\lesssim \bigl\| \sup_z \partial_t^{m_1} \nabla_x^{m_2} K(t, z, x) \bigr\|_{L_x^1}^{\frac{1}{p'}} \left( \iint_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3\times\mathop{\mathbb{R}\kern 0pt}\nolimits^3} |\partial_t^{m_1} \nabla_y^{m_2} K(t, x, x-y)| |h(y)|^p \, dy \, dx \right)^{\frac{1}{p}} \\ &\lesssim \bigl\| \sup_z \partial_t^{m_1} \nabla_x^{m_2} K(t, z, x) \bigr\|_{L_x^1} \|h\|_{L^p}. \end{align}\tag{37}\] Note that, for any \(x_0\) and \(t\), one has \(\left\|K(t,x_0,x)\right\|_{L_x^1}=1,\) so that it follows from 37 that \[\left\|\int_{\mathbb{R}^d}K({t}/{2},x,x-y)f(t,y)\,dy\right\|_{L_x^p}\lesssim \|f(t,\cdot)\|_{L^p}\lesssim t^{-\eta}\sup_{t\leq T}t^\eta\|f(t)\|_{L^p}.\] By 37 and the pointwise estimate ?? , we find \[\label{S4eq1} \begin{align} \|\tilde{g}(t)\|_{L^p} &\lesssim \Bigl( \int_0^{t/2} (t-\tau)^{-1} \tau^{-\eta} \, d\tau + t^{-\eta} \Bigr) \sup_{t \leq T} t^{\eta} \|f(t)\|_{L^p} \\ &\quad + \int_{t/2}^t (t-\tau)^{-1+\sigma} \tau^{-\sigma-\eta} \, d\tau \sup_{s < t \leq T} s^{\eta+\sigma} \frac{\|f(t) - f(s)\|_{L^p}}{(t-s)^{\sigma}} \\ &\lesssim t^{-\eta} \Bigl( \sup_{t \leq T} t^{\eta} \|f(t)\|_{L^p} + \sup_{s < t \leq T} s^{\eta+\sigma} \frac{\|f(t) - f(s)\|_{L^p}}{(t-s)^{\sigma}} \Bigr). \end{align}\tag{38}\] For the time differences, let \(a:= t-s\). For the same reason as in the previous section, we only need to consider the case \(a < \frac{s}{4}\). We write \[\begin{align} \tilde{g}(t,x) &= \int_0^{t/2} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K(t-\tau, x, x-y) f(\tau, y) \, dy \, d\tau \\ &\quad + \int_{t/2}^t \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K(t-\tau, x, x-y) f(\tau, y) \, dy \, d\tau:= I(t,x) + II(t,x). \end{align}\] For \(I(t)\), since \[\begin{align} I(t) - I(s) = &\int_0^{s/2} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \delta_a^t \partial_\tau K(s-\tau, x, x-y) f(\tau, y) \, dy \, d\tau\\ & + \int_{s/2}^{t/2} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K(t-\tau, x, x-y) f(\tau, y) \, dy \, d\tau, \end{align}\] by ?? and 37 , one obtains \[\begin{align} {\|I(t) - I(s)\|_{L^p}} \lesssim& \Bigl(\int_0^{s/2}\min\bigl\{(s-\tau)^{-1}, a(s-\tau)^{-2}\big\}\tau^{-\eta}\,d\tau\\ &+\int_{s/2}^{t/2}(t-\tau)^{-1}\tau^{-\eta}\,d\tau\Bigr) \sup_{t \leq T} t^{\eta} \|f(t)\|_{L^p}\\ \lesssim & a^{\sigma}s^{-\sigma-\eta}\sup_{t \leq T} t^{\eta} \|f(t)\|_{L^p}, \end{align}\] where we used the fact that \[\begin{align} \int_{s/2}^{t/2}(t-\tau)^{-1}\tau^{-\eta}\,d\tau\lesssim a t^{-1}s^{-\eta} \lesssim a^{\sigma}s^{-\sigma-\eta} \Bigl(\frac{a}{t}\Bigr)^{1-\sigma}\lesssim a^\sigma s^{-\sigma-\eta}. \end{align}\] Furthermore, notice that \[\begin{align} II(t) - II(s) &= -\int_{s/2}^{t/2} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K(s-\tau, x, x-y) f(\tau, y) \, dy \, d\tau \\ &\quad + \int_{t/2}^{s} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \delta_a^t \partial_\tau K(s-\tau, x, x-y) f(\tau, y) \, dy \, d\tau \\ &\quad + \int_{s}^{t} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K(t-\tau, x, x-y) f(\tau, y) \, dy \, d\tau \\ &= -\int_{s/2}^{t/2} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K(s-\tau, x, x-y) f(\tau, y) \, dy \, d\tau \\ &\quad + \int_{t/2}^{s} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \delta_a^t \partial_\tau K(s-\tau, x, x-y) \left(f(\tau, y)-f(s,y)\right) \, dy \, d\tau \\ &\quad + \int_{s}^{t} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K(t-\tau, x, x-y) \left(f(\tau, y)-f(t,y)\right) \, dy \, d\tau \\ &\quad +\left(\int_{t/2}^{s} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \delta_a^t \partial_\tau K(s-\tau, x, x-y)f(s,y) \, dy \, d\tau\right.\\ &\left.\qquad+\int_{s}^{t} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K(t-\tau, x, x-y)f(t,y) \, dy \, d\tau\right)\\ &:= DII_1(t,s) + DII_2(t,s) + DII_3(t,s)+DII_4(t,s). \end{align}\] Observing that \[\sup_x\|\partial_\tau K(s-\tau,x,y)\|_{L_y^1}\lesssim (s-\tau)^{-1},\] and on the time interval \((s/2, t/2)\), there holds \((s-\tau)^{-1}\lesssim s^{-1}\), we deduce from 37 and \(a<\frac{s}{4}\), that \[\begin{align} \|DII_1(t,s)\|_{L^p}&\lesssim s^{-1}\int_{\frac{s}{2}}^{\frac{t}{2}}\|f(\tau)\|_{L^p}\,d\tau\lesssim s^{-1}\int_{\frac{s}{2}}^{\frac{t}{2}}\tau^{-\eta}\,d\tau\sup_{\tau\leq T}\tau^{\eta}\|f(\tau)\|_{L^p}\\ &\lesssim (t-s)s^{-1-\eta}\sup_{\tau\leq T}\tau^{\eta}\|f(\tau)\|_{L^p}\lesssim (t-s)^{\sigma}s^{-\sigma-\eta}\sup_{\tau\leq T}\tau^{\eta}\|f(\tau)\|_{L^p}. \end{align}\]

By ?? and 37 (pointwisely for each \(\tau\)) we have \[\begin{align} &\bigl\| \int_{t/2}^{s} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \delta_a^t \partial_\tau K(s-\tau, x, x-y) \bigl( f(\tau, y) - f(s, y) \bigr) \, dy \, d\tau\bigr\|_{L^p}\\ &\lesssim \int_{\frac{t}{2}}^s(s-\tau)^{-1+\sigma} \min\Bigl\{1,\frac{a}{s-\tau}\Bigr\}\,\tau^{-\sigma-\eta}\,d\tau \sup_{s < t \leq T} s^{\eta+\sigma} \frac{\|f(t) - f(s)\|_{L^p}}{(t-s)^{\sigma}}\\ &\lesssim a^{\sigma}s^{-\sigma-\eta}\sup_{s < t \leq T} s^{\eta+\sigma} \frac{\|f(t) - f(s)\|_{L^p}}{(t-s)^{\sigma}}, \end{align}\] where we used the fact that \[\begin{align} \int_{\frac{t}{2}}^s(s-\tau)^{-1+\sigma} \min\Bigl\{1,\frac{a}{s-\tau}\Bigr\}\,\tau^{-\sigma-\eta}\,d\tau = &a\int_{\frac{t}{2}}^{2s-t}(s-\tau)^{-2+\sigma} \tau^{-\sigma-\eta}\,d\tau \\ &+\int_{2s-t}^s(s-\tau)^{-1+\sigma} \tau^{-\sigma-\eta}\,d\tau \lesssim a^{\sigma}s^{-\sigma-\eta}. \end{align}\] Then along the same line to the proof of 38 , we find \[\begin{align} &\|DII_2(t,s)\|_{L^p} + \|DII_3(t,s)\|_{L^p}\\ &\quad \lesssim (t-s)^{\sigma} s^{-\eta-\sigma} \Bigl( \sup_{t \leq T} t^{\eta} \|f(t)\|_{L^p} + \sup_{s < t \leq T} s^{\eta+\sigma} \frac{\|f(t) - f(s)\|_{L^p}}{(t-s)^{\sigma}} \Bigr). \end{align}\] For the term \(DII_4(t,s)\), we write \[\begin{align} DII_4(t,s)=&\Bigl(\int_{\mathbb{R}^3}K(t-s,x,x-y)\bigl(f(s,y)dy-f(t,y)\bigr)\,dy\Bigr)\\ &+\Bigl(\int_{\mathbb{R}^3}K(s-\frac{t}{2},x,x-y)f(s,y)dy-\int_{\mathbb{R}^3}K(\frac{t}{2},x,x-y)f(s,y)dy\Bigr)\\ &+\left(f(t,x)-f(s,x)\right). \end{align}\] Then by 34 and 37 , we deduce that \[\begin{align} \|DII_4(t,s)\|_{L^p}&\lesssim \|f(t)-f(s)\|_{L^p}+\|f(s)\|_{L^p}\sup_{x}\left\|K(s-{t}/{2},x,y)-K({t}/{2},x,y)\right\|_{L_y^1}\\ &\lesssim a^\sigma s^{-\sigma-\eta}\Bigl(\sup_{t \leq T} t^{\eta} \|f(t)\|_{L^p} + \sup_{s < t \leq T} s^{\eta+\sigma} \frac{\|f(t) - f(s)\|_{L^p}}{(t-s)^{\sigma}} \Bigr). \end{align}\] This completes the proof of Lemma 2. ◻

Proof. We divide the proof into the following two steps:

i) Estimates for spatial integrability.

In view of 36 , we write \[\label{estR11n} \begin{align} u_{R1,x_0}(t,x) &= \int_0^t K_{x_0}(t-\tau) \ast \nabla \Delta^{-1} \partial_\tau \bigl( \nabla \rho_0 \cdot (u(\tau) - u(t)) \bigr) \, d\tau \\ &= \int_0^t \partial_t K_{x_0}(t-\tau) \ast \nabla \Delta^{-1} \bigl( \nabla \rho_0 \cdot (u(\tau) - u(t)) \bigr) \, d\tau \\ &\quad - K_{x_0}(t) \ast \nabla \Delta^{-1} \bigl( \nabla \rho_0 \cdot (u_0 - u(t)) \bigr)\\ &:=u_{R1,x_0}^1(t,x)+u_{R1,x_0}^2(t,x), \end{align}\tag{39}\] and correspondingly we define \(\tilde{u}_{R1}^i=(u_{R1,x_0}^i)|_{x_0=x}\) as in 8 . Furthermore, by the formula 6 , \(u_{R1,x_0}^1(t,x)\) satisfies \[u_{R1,x_0}^1(t,x)=\frac{1}{\rho_0(x_0)}\int_0^t \nabla K_{x_0}(t-\tau) \ast \bigl( \nabla \rho_0 \cdot (u(\tau) - u(t)) \bigr) \, d\tau.\] It follows from Lemmas 3, 4 and 5 that there holds ?? . Below we will frequently use the following interpolation inequalities: \[\label{intpoeq} \begin{align} &\|\nabla \Delta^{-1} u\|_{L^\infty} \lesssim \min\{\|u\|_{L^2}^{\frac{2}{3}} \|u\|_{L^\infty}^{\frac{1}{3}},\|u\|_{L^2}^{\frac{3}{5}}\|u\|_{L^{12}}^{\frac{2}{5}}\}, \quad &&\|\nabla \Delta^{-1}\nabla u\|_{L^\infty} \lesssim\|u\|_{L^2}^{\frac{2\kappa}{2\kappa+3}}\|u\|_{\dot{C}^{\kappa}}^{\frac{3}{2\kappa+3}} ,\\ & \|\nabla \Delta^{-1} u\|_{L^{12}} \lesssim \min\{\|u\|_{L^2}^{\frac{5}{6}} \|u\|_{L^\infty}^{\frac{1}{6}},\|u\|_{L^2}^{\frac{4}{5}}\|u\|_{L^{12}}^{\frac{1}{5}}\},\quad &&\|\nabla \Delta^{-1}\nabla u\|_{L^{12}} \lesssim \|u\|_{L^{12}}. \end{align}\tag{40}\]

First we consider the case \(\rho_0\in C^1\). By 39 and ?? , we find \[\begin{align} \|u_{R1,x_0}^1(t)\|_{L^\infty} \lesssim& \int_0^t (t-\tau)^{-\frac{1}{2}} \|\rho_0\|_{C^1} \|(u(\tau) - u(t))\|_{L^\infty} \, d\tau \lesssim \|\rho_0\|_{C^1}\|u\|_{X_T}\int_0^t(t-\tau)^{-\frac{1}{2}} \tau^{-\frac{1}{2}}\,d\tau \end{align}\] which implies \[\label{uS4eq20} \begin{align} \sup_{x_0} \|u_{R1,x_0}^1(t)\|_{L^\infty} \lesssim \mathbf{C}(u)\|\rho_0\|_{C^1}\|u\|_{X_T}. \end{align}\tag{41}\] Along the same line, we get, by using 37 , that \[\begin{align} \|\tilde{u}_{R1}^1(t)\|_{L^{12}} \lesssim& \int_0^t (t-\tau)^{-\frac{1}{2}} \|\rho_0\|_{C^1} \|u(\tau) - u(t)\|_{L^{12}} \, d\tau \lesssim\|\rho_0\|_{C^1}\|u\|_{X_T}\int_0^t(t-\tau)^{-\frac{1}{2}}\tau^{-\frac{5}{12}}\,d\tau \end{align}\] which leads to \[\label{S4eq21} \|\tilde{u}_{R1}^1(t)\|_{L^{12}}\lesssim t^{\frac{1}{12}} \mathbf{C}(u) \|\rho_0\|_{C^1}\|u\|_{X_T}.\tag{42}\]

Furthermore, for the \(C^{1+\kappa}\) norm, we have \[\label{S4eq22} \begin{align} \sup_{x_0} \| |D|^{1+\kappa} u_{R1,x_0}^1(t) \|_{L^\infty} \lesssim & \int_0^t (t-\tau)^{ -1- \frac{\kappa}{2}} \|\rho_0\|_{C^1}\|u(\tau) - u(t)\|_{L^\infty} \, d\tau \\ \lesssim &\int_0^t (t-\tau)^{-1 - \frac{\kappa}{2}} \tau^{-\frac{1}{2}}\min\{1,(t-\tau)^{\varkappa_1}\tau^{-\varkappa_1}\}d\tau\|\rho_0\|_{C^1}\|u\|_{X_T}\\ \lesssim& t^{-\frac{1 + \kappa}{2}} \mathbf{C}(u) \|\rho_0\|_{C^1}\|u\|_{X_T} . \end{align}\tag{43}\]

For \(u_{R1,x_0}^2\), by 39 and 40 , we have the \(L^\infty\) estimate \[\label{S4eq14} \begin{align} \|u_{R1,x_0}^2(t)\|_{L^\infty}&\lesssim \bigl\| K_{x_0}(t) \ast \bigl( \nabla \rho_0 (u_0 - u(t)) \bigr) \bigr\|_{L^2}^{\frac{2}{3}}\bigl\| K_{x_0}(t) \ast \bigl( \nabla \rho_0 (u_0 - u(t)) \bigr) \bigr\|_{L^\infty}^{\frac{1}{3}}\\ &\lesssim \|\rho_0\|_{C^1}\|u_0\|_{L^2}\|K_{x_0}(t)\|_{L^2}^{\frac{1}{3}}\lesssim t^{-\frac{1}{4}}\|\rho_0\|_{C^1}\mathbf{C}(u), \end{align}\tag{44}\] the corresponding \(L^{12}\) estimate \[\label{S4eq15} \begin{align} \|\tilde{u}_{R1}^2(t)\|_{L^{12}}&\lesssim \bigl\| \bigl((K_{x_0}(t,\cdot))\big|_{x_0=x} \ast (\nabla \rho_0 (u_0 - u(t))) \bigr) \bigr\|_{L_x^{2}}^{\frac{5}{6}} \bigl\| \bigl( (K_{x_0}(t,\cdot))|_{x_0=x} \ast (\nabla \rho_0 (u_0 - u(t))) \bigr) \bigr\|_{L_x^{\infty}}^{\frac{1}{6}}\\ &\lesssim\|\sup_{x_0\in\mathbb{R}^3}K_{x_0}(t,\cdot)\|_{L^2}^{\frac{1}{6}}\|\rho_0\|_{C^1}\|u_0\|_{L^2}\lesssim t^{-\frac{1}{8}} \|\rho_0\|_{C^1}\mathbf{C}(u), \end{align}\tag{45}\] and the \(C^{1+\kappa}\) estimate \[\label{S4eq16} \begin{align} \sup_{x_0}\||D|^{1+\kappa}u_{R1,x_0}^2(t)\|_{L^\infty}&\lesssim \|\sup_{x_0} \nabla \Delta^{-1} |D|^{1+\kappa} K_{x_0}(t) \|_{L^2} \| \nabla \rho_0 (u_0 - u(t)) \|_{L^2}\\ &\lesssim\|\sup_{x_0} \nabla \Delta^{-1} |D|^{1+\kappa} K_{x_0}(t) \|_{L^2} \| \nabla \rho_0 (u_0 - u(t)) \|_{L^2}\\ &\lesssim t^{-\frac{3 + 2\kappa}{4}}\|\rho_0\|_{C^1}\mathbf{C}(u). \end{align}\tag{46}\]

So combine 41 and 44 , we have \[\label{S4eq68} \sup_{x_0}\sup_{t\leq T}t^{\frac{1}{2}}\|u_{R1,x_0}(t)\|_{L^\infty}\lesssim T^{\frac{1}{4}}\|\rho_0\|_{C^1}\mathbf{C}(u).\tag{47}\] Combine 42 and 45 , we have \[\label{S4eq69} \sup_{t\leq T}t^{\frac{5}{12}}\|\tilde{u}_{R1}(t)\|_{L^{12}}\lesssim T^{\frac{7}{24}}\|\rho_0\|_{C^1}\mathbf{C}(u).\tag{48}\] Combine 43 and 46 , we have \[\label{S4eq70} \sup_{x_0}\sup_{t\leq T}t^{1+\frac{\kappa}{2}}\||D|^{1+\kappa}u_{R1,x_0}(t)\|_{L^\infty}\lesssim T^{\frac{1}{4}}\|\rho_0\|_{C^1}\mathbf{C}(u).\tag{49}\]

Note that \[\label{uR1bmo} \|u_{R1,x_0}\|_{V_T^b} \lesssim T^{\frac{1}{12}}\sup_{t \leq T} t^{\frac{5}{12}} \|u_{R1,x_0}\|_{L^\infty} \lesssim \sup_{t \leq T} t^{\frac{5}{12}} \|u_{R1,x_0}\|_{L^\infty}, \quad \forall T < 1.\tag{50}\] Hence, similar estimates as 71 also implies the estimate for \(\sup_{x_0} \|u_{R1,x_0}\|_{V_T^b}\).
ii) Estimates for time differences.
We still use 39 . For \(u_{R1,x_0}^1\) and \(s<t<T\), let \(a=t-s\), we denote \[\begin{align} u_{R1,x_0}^1(t,x)-u_{R1,x_0}^1(s,x)=&\frac{1}{\rho_0(x_0)}\int_s^t\nabla K_{x_0}(t-\tau)\ast (\nabla\rho_0 (u(\tau)-u(t)))d\tau\\ &+\frac{1}{\rho_0(x_0)}\int_0^s\nabla K_{x_0}(t-\tau)\ast (\nabla\rho_0 (u(t)-u(s)))d\tau\\ &+\frac{1}{\rho_0(x_0)}\int_0^s\nabla \delta_a^t K_{x_0}(s-\tau)\ast (\nabla\rho_0 (u(\tau)-u(s)))d\tau. \end{align}\] Then separately, we have the \(L^\infty\) estimate \[\begin{align} \|u_{R1,x_0}^1(t)-u_{R1,x_0}^1(s)\|_{L^\infty} \lesssim &\int_s^t(t-\tau)^{-\frac{1}{2}}(t-\tau)^{\varkappa_1}\tau^{-\frac{1}{2}-\varkappa_1}d\tau\|\rho_0\|_{C^1}\|u\|_{X_T}\\ & +\int_0^s(t-\tau)^{-\frac{1}{2}}(t-s)^{\varkappa_1}s^{-\frac{1}{2}-\varkappa_1}d\tau\|\rho_0\|_{C^1}\|u\|_{X_T}\\ &+\int_0^s(s-\tau)^{-\frac{1}{2}}\min\{1,a(s-\tau)^{-1}\}\\ &\qquad\times\tau^{-\frac{1}{2}}\min\{1,(s-\tau)^{\varkappa_1}\tau^{-\varkappa_1}\}d\tau\|\rho_0\|_{C^1}\|u\|_{X_T}\\ \lesssim &a^{\varkappa_1}s^{-\varkappa_1}\|\rho_0\|_{C^1}\|u\|_{X_T}, \end{align}\] and the \(L^{12}\) estimate \[\begin{align} \|\tilde{u}_{R1}^1(t)-\tilde{u}_{R1}^1(s)\|_{L^{12}} \lesssim &\int_s^t(t-\tau)^{-\frac{1}{2}}(t-\tau)^{\varkappa_2}\tau^{-\frac{5}{12}-\varkappa_2}d\tau\|\rho_0\|_{C^1}\|u\|_{X_T}\\ & +\int_0^s(t-\tau)^{-\frac{1}{2}}(t-s)^{\varkappa_2}s^{-\frac{5}{12}-\varkappa_2}d\tau\|\rho_0\|_{C^1}\|u\|_{X_T}\\ &+\int_0^s(s-\tau)^{-\frac{1}{2}}\min\{1,a(s-\tau)^{-1}\}\\ &\qquad\times\tau^{-\frac{5}{12}}\min\{1,(s-\tau)^{\varkappa_2}\tau^{-\varkappa_2}\}d\tau\|\rho_0\|_{C^1}\|u\|_{X_T}\\ \lesssim &a^{\varkappa_2}s^{\frac{1}{12}-\varkappa_2}\|\rho_0\|_{C^1}\|u\|_{X_T}. \end{align}\] Combine the two estimates above, we can obtain that \[\begin{align} &\sup_{x_0}\sup_{s\leq T\leq T}s^{\frac{1}{2}+\varkappa_1}\frac{\|u_{R1,x_0}^1(t)-u_{R1,x_0}^1(s)\|_{L^{\infty}}}{(t-s)^{\varkappa_1}}\lesssim T^{\frac{1}{2}}\|\rho_0\|_{C^1}\|u\|_{X_T},\label{S4eq41}\\ &\sup_{s\leq T\leq T}s^{\frac{5}{12}+\varkappa_2}\frac{\|\tilde{u}_{R1}^1(t)-u_{R1}^1(s)\|_{L^{12}}}{(t-s)^{\varkappa_2}}\lesssim T^{\frac{1}{2}}\|\rho_0\|_{C^1}\|u\|_{X_T}\nonumber \end{align}\tag{51}\]

For \(u_{R1,x_0}^2\), by 39 , we have \[\begin{align} u_{R1,x_0}^2(t,x)-u_{R1,x_0}^2(s,x)=&\left(K_{x_0}(t)-K_{x_0}(s)\right)\ast \nabla\Delta^{-1}\left(\nabla\rho_0(u_0-u(t))\right)(x)\label{S4eq40}\\ &+K_{x_0}(s)\ast \nabla\Delta^{-1}\left(\nabla\rho_0(u(t)-u(s))\right)(x)\nonumber\\ =&\frac{1}{\rho_0(x_0)}\int_s^t\nabla K_{x_0}(\gamma)d\gamma\ast \left(\nabla\rho_0(u_0-u(t))\right)(x)\nonumber\\ &+\nabla\Delta^{-1}K_{x_0}(s)\ast \left(\nabla\rho_0(u(t)-u(s))\right)(x).\nonumber \end{align}\tag{52}\] While it follows from ?? and Lemma 5 that for \(s<t,\) \[\begin{align} \|u(t) - u(s)\|_{L^2}\leq \int_s^t\|u_\tau(\tau)\|_{L^2}\,d\tau\leq \int_s^t\tau^{-1}\,d\tau\|\tau u_\tau(\tau)\|_{L^\infty_t(L^2)}\leq s^{-1}(t-s)\|u_0\|_{L^2}, \end{align}\] so that \[\label{S4eqp} \|u(t) - u(s)\|_{L^2}\lesssim \min\bigl\{1, s^{-1}(t-s)\bigr\}\|u_0\|_{L^2}.\tag{53}\] By 52 , 53 and Lemma 3, we have \[\begin{align} \|u_{R1,x_0}^2(t)-u_{R1,x_0}^2(s)\|_{L^\infty}\lesssim& \int_s^t\|\sup_{x_0}|\nabla K_{x_0}(\gamma)|\|_{L^2}d\gamma\|\rho_0\|_{C^1}\|u_0\|_{L^2}\\ &+\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(s)|\|_{L^2}\|\rho_0\|_{C^1}\|u_0\|_{L^2}\min\{1,(t-s)s^{-1}\}\\ \lesssim &s^{-\frac{1}{4}}\min\{1,(t-s)s^{-1}\}\|\rho_0\|_{C^1}\|u_0\|_{L^2},\\ \|\tilde{u}_{R1}^2(t)-\tilde{u}_{R1}^2(s)\|_{L^{12}}\lesssim &\int_s^t\|\sup_{x_0}|\nabla K_{x_0}(\gamma)|\|_{L^{\frac{12}{7}}}d\gamma\|\rho_0\|_{C^1}\|u_0\|_{L^2}\\ &+\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(s)|\|_{L^{\frac{12}{5}}}\|\rho_0\|_{C^1}\|u_0\|_{L^2}\min\{1,(t-s)s^{-1}\}\\ \lesssim& s^{-\frac{1}{8}}\min\{1,(t-s)s^{-1}\}\|\rho_0\|_{C^1}\|u_0\|_{L^2}. \end{align}\] These lead to \[\label{S4eq42} \begin{align} &\sup_{x_0}\sup_{s<t<T}s^{\frac{1}{2}+\varkappa_1}\frac{\|u_{R1,x_0}^2(t)-u_{R1,x_0}^2(s)\|_{L^\infty}}{(t-s)^{\varkappa_1}}\lesssim T^{\frac{1}{4}}\|\rho_0\|_{C^1}\|u_0\|_{C^2}\\ &\sup_{s<t<T}s^{\frac{5}{12}+\varkappa_2}\frac{\|\tilde{u}_{R1}^2(t)-\tilde{u}_{R1}^2(s)\|_{L^{12}}}{(t-s)^{\varkappa_2}}\lesssim T^{\frac{7}{24}}\|\rho_0\|_{C^1}\|u_0\|_{C^2} \end{align}\tag{54}\]

So 51 and 54 lead to \[\label{S4eq71} \begin{align} &\sup_{x_0}\sup_{s<t<T}s^{\frac{1}{2}+\varkappa_1}\frac{\|u_{R1,x_0}(t)-u_{R1,x_0}(s)\|_{L^\infty}}{(t-s)^{\varkappa_1}}\lesssim T^{\frac{1}{4}}\|\rho_0\|_{C^1}\mathbf{C}(u),\\ &\sup_{s<t<T}s^{\frac{5}{12}+\varkappa_2}\frac{\|\tilde{u}_{R1}(t)-\tilde{u}_{R1}(s)\|_{L^{12}}}{(t-s)^{\varkappa_2}}\lesssim T^{\frac{7}{24}}\|\rho_0\|_{C^1}\mathbf{C}(u). \end{align}\tag{55}\]

Now combine 47 , 48 , 49 , 50 and 55 , we have proved ?? . ◻

Proof. Note that from the decomposition 39 , the estimates of \(u_{R1,x_0}^1\) follows from 41 , 42 , 43 , 50 and 51 . We only need to get new estimates for \(u_{R1,x_0}^2\).

i) Estimates for spatial integrability.

We will use the following formula for \(u_{R1,x_0}^2\). By 1 , \(u_{R1,x_0}^2(t,x)\) satisfies \[\begin{align} u_{R1,x_0}^2(t,x)=&-\nabla\Delta^{-1}K_{x_0}(t)\ast\left(\nabla\rho_0\cdot(\rho_0u_0-\rho(t)u(t))\right)+\nabla\Delta^{-1}K_{x_0}(t)\ast\left(\nabla\rho_0\cdot(\rho_0-1) u_0\right)\nonumber\\ &-\nabla\Delta^{-1}K_{x_0}(t)\ast\left(\nabla\rho_0\cdot(\rho(t)-1)u(t)\right)\nonumber\\ =&\,\nabla\Delta^{-1}K_{x_0}(t)\ast\left(\nabla\rho_0\cdot\int_0^t\left(\Delta u-\nabla\cdot(\rho u\otimes u)+\nabla P\right)(\tau,x)d\tau\right)\label{S4eq11} \\ &+\nabla\Delta^{-1}K_{x_0}(t)\ast\left(\nabla\rho_0\cdot(\rho_0-1)u_0\right)-\nabla\Delta^{-1}K_{x_0}(t)\ast\left(\nabla\rho_0\cdot(\rho(t)-1)u(t)\right)\nonumber\\ &:=\sum_{i=1}^3u_{R1,x_0}^{2i},\nonumber \end{align}\tag{56}\] and we denote \(\tilde{u}_{R1}^{2i}=u_{R1,x_0}^{2i}|_{x_0=x}\). Note that for the error terms come from density, by 13 and Lemma 3, we have \[\label{S4eq55} \begin{align} &\sup_{x_0}\left(\|u_{R1,x_0}^{22}\|_{L^\infty}+\|u_{R1,x_0}^{23}\|_{L^\infty}\right)\lesssim \|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}|(t)\|_{L^2}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}\\ &\qquad\lesssim t^{-\frac{1}{4}}\|u_0\|_{L^2}\|\rho_0-1\|_{L^\infty},\\ &\left(\|\tilde{u}_{R1}^{22}\|_{L^{12}}+\|\tilde{u}_{R1}^{23}\|_{L^{12}}\right)\lesssim \|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}|(t)\|_{L^{\frac{12}{7}}}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}\\ &\qquad\lesssim t^{-\frac{1}{8}}\|u_0\|_{L^2}\|\rho_0-1\|_{L^\infty},\\ &\sup_{x_0}\left(\||D|^{1+\kappa}u_{R1,x_0}^{22}\|_{L^\infty}+\||D|^{1+\kappa}u_{R1,x_0}^{23}\|_{L^\infty}\right)\lesssim \|\sup_{x_0}||D|^{1+\kappa}\nabla\Delta^{-1}K_{x_0}|(t)\|_{L^2}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}\\ &\qquad\lesssim t^{-\frac{3+2\kappa}{4}}\|u_0\|_{L^2}\|\rho_0-1\|_{L^\infty}. \end{align}\tag{57}\] For \(u_{R1,x_0}^{21}\), we further decompose \(u_{R1,x_0}^{21}=u_{R1,x_0}^{21,m}+u_{R1,x_0}^{21,P}\), with \[\begin{align} &u_{R1,x_0}^{21,m}=\nabla\Delta^{-1}K_{x_0}(t)\ast\left(\nabla\rho_0\cdot\int_0^t\left(\Delta u-\nabla\cdot(\rho u\otimes u)\right)(\tau,x)d\tau\right),\label{S4eq60}\\ &u_{R1,x_0}^{21,P}=\nabla\Delta^{-1}K_{x_0}(t)\ast\left(\nabla\rho_0\cdot\int_0^t\nabla P(\tau,x)d\tau\right).\nonumber \end{align}\tag{58}\] For \(u_{R1,x_0}^{21,m}\), using integration by parts and ?? , we have the \(L^\infty\) estimate \[\label{S4eq53} \begin{align} \sup_{x_0}\|u_{R1,x_0}^{21,m}\|_{L^\infty}&\lesssim \int_0^t\bigl\|\sup_{x_0}|\nabla\Delta^{-1}\nabla K_{x_0}(t)\|_{L^{\frac{4}{3}}} \|\nabla\rho_0\|_{L^\infty}\left\|\nabla u(\tau)-(\rho u\otimes u)(\tau)\right\|_{L^{4}}d\tau\\ &\quad+\int_0^t\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(t)|\|_{L^2}\|\rho_0\|_{C^2}\|\nabla u(\tau)-(\rho u\otimes u)(\tau)\|_{L^2}d\tau \\ &\lesssim \|\rho_0\|_{C^1}^2\int_0^tt^{-\frac{3}{8}}\tau^{-\frac{3}{4}}d\tau \mathbf{C}(u)^2\|u\|_{X_T}^{\frac{1}{2}}+\|\rho_0\|_{C^2}^2t^{-\frac{1}{4}}\int_0^t\tau^{-\frac{1}{2}}d\tau \mathbf{C}(u)^2\\ &\lesssim \|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(t^{-\frac{1}{8}}\|u\|_{X_T}^{\frac{1}{2}}+t^{\frac{1}{4}}\right), \end{align}\tag{59}\] the \(L^{12}\) estimate \[\label{S4eq18} \begin{align} \|\tilde{u}_{R1}^{21,m}(t)\|_{L^{12}}&\lesssim \int_0^t\bigl\|K_{x_0}(t)\ast \left(\nabla\rho_0\cdot\left(\nabla u(\tau)-(\rho u\otimes u)(\tau) \right)\right)\bigr\|_{L^{12}}d\tau\\ &\quad+\int_0^t\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(t)|\|_{L^{\frac{12}{7}}} \|\rho_0\|_{C^2}\|\nabla u(\tau)-(\rho u\otimes u)(\tau)\|_{L^2}d\tau \\ &\lesssim \|\rho_0\|_{C^1}^2\int_0^t\tau^{-\frac{11}{12}}d\tau \mathbf{C}^2(u)\|u\|_{X_T}^{\frac{5}{6}}+\|\rho_0\|_{C^2}^2\int_0^tt^{-\frac{1}{8}}\tau^{-\frac{1}{2}}d\tau\mathbf{C}^2(u)\\ &\lesssim \|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(t^{\frac{1}{12}}\|u\|_{X_T}^{\frac{5}{6}}+t^{\frac{3}{8}}\right), \end{align}\tag{60}\] and the \(C^{1+\kappa}\) estimate \[\begin{align} &\sup_{x_0}\||D|^{1+\kappa}u_{R1,x_0}^{21,m}(t)\|_{L^\infty \label{S4eq19}}\\ &\lesssim \int_0^t\|\sup_{x_0}\nabla\Delta^{-1}\nabla|D|^{1+\kappa}K_{x_0}(t)\|_{L^{\frac{4}{3}}}\|\rho_0\|_{C^1}^2\left(\|\nabla u(\tau)\|_{L^4}+\|(\rho u\otimes u)(\tau)\|_{L^4}\right)\nonumber\\ &\quad+\int_0^t\|\sup_{x_0}\nabla\Delta^{-1}|D|^{1+\kappa}K_{x_0}(t)\|_{L^{\frac{4}{3}}}\|\rho_0\|_{C^2}^2\left(\|\nabla u(\tau)\|_{L^4}+\|(\rho u\otimes u)(\tau)\|_{L^4}\right)\nonumber\\ &\lesssim \|\rho_0\|_{C^1}^2\int_0^tt^{-\frac{7+4\kappa}{8}}\tau^{-\frac{3}{4}}d\tau\mathbf{C}^2(u)\|u\|_{X_T}^{\frac{1}{2}}+\|\rho_0\|_{C^2}\int_0^tt^{-\frac{3+4\kappa}{8}}\tau^{-\frac{3}{4}}\mathbf{C}^2(u)\|u\|_{X_T}^{\frac{1}{2}}\nonumber\\ &\lesssim t^{-\frac{5+4\kappa}{8}}\|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\|u\|_{X_T}^{\frac{1}{2}}.\nonumber \end{align}\tag{61}\] For \(u_{R1,x_0}^{21,P}\), since by 1 and due to the divergence free condition of \(u\) and \(\partial_tu\), we have \[\label{S4eq61} \begin{align} -\int_0^t\nabla P(\tau)d\tau&=\int_0^t\left(\nabla\Delta^{-1}\nabla\cdot\left(\rho\partial_tu+\rho u\cdot\nabla u\right)\right)(\tau)d\tau\\ &=\int_0^t\nabla\Delta^{-1}\nabla\cdot\left((\rho-\rho_0)\partial_tu\right)(\tau)d\tau+\int_0^t\nabla\Delta^{-1}\nabla\cdot\left(\rho u\cdot\nabla u\right)(\tau)d\tau\\ &\qquad+\nabla\Delta^{-1}\nabla\cdot \big((\rho_0-1)(u(t)-u_0)\big)\\ &:=P_1(t)+P_2(t)+P_3(t). \end{align}\tag{62}\] For \(P_1(t)\), we use Proposition 2 and Lemma 5 to show that \[\begin{align} &\left\|\nabla\Delta^{-1}K_{x_0}(t)\ast \left(\nabla\rho_0 P_1(t)\right)\right\|_{L^\infty}\lesssim \left\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(t)| \right\|_{L^2}\int_0^t\|\rho(\tau)-\rho_0\|_{L^\infty}\|\partial_\tau u(\tau)\|_{L^2}d\tau\label{S4eq50}\\ &\qquad\lesssim t^{-\frac{1}{4}}\int_0^t\tau^{\frac{3}{8}}\tau^{-1}d\tau\|\rho_0\|_{C^1}\mathbf{C}^2(u)\lesssim t^{\frac{1}{8}}\|\rho_0\|_{C^1}\mathbf{C}^2(u),\nonumber\\ &\left\|\left(\nabla\Delta^{-1}K_{x_0}(t)\ast \left(\nabla\rho_0 P_1(t)\right)\right)\big|_{x_0=x}\right\|_{L^{12}}\lesssim \left\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(t)| \right\|_{L^\frac{12}{7}}\int_0^t\|\rho(\tau)-\rho_0\|_{L^\infty}\|\partial_\tau u(\tau)\|_{L^2}d\tau\nonumber\\ &\qquad\lesssim t^{-\frac{1}{8}}\int_0^t\tau^{\frac{3}{8}}\tau^{-1}d\tau\|\rho_0\|_{C^1}\mathbf{C}^2(u)\lesssim t^{\frac{1}{4}}\|\rho_0\|_{C^1}\mathbf{C}^2(u),\nonumber\\ &\left\||D|^{1+\kappa}\nabla\Delta^{-1}K_{x_0}(t)\ast \left(\nabla\rho_0 P_1(t)\right)\right\|_{L^\infty}\lesssim \left\|\sup_{x_0}||D|^{1+\kappa}\nabla\Delta^{-1}K_{x_0}(t)| \right\|_{L^2}\int_0^t\|\rho(\tau)-\rho_0\|_{L^\infty}\|\partial_\tau u(\tau)\|_{L^2}d\tau\nonumber\\ &\qquad\lesssim t^{-\frac{3+2\kappa}{4}}\int_0^t\tau^{\frac{3}{8}}\tau^{-1}d\tau\|\rho_0\|_{C^1}\mathbf{C}^2(u)\lesssim t^{-\frac{3+4\kappa}{8}}\|\rho_0\|_{C^1}\mathbf{C}^2(u).\nonumber \end{align}\tag{63}\] For \(P_3(t)\), by Lemma 3, we simply have \[\begin{align} &\left\|\nabla\Delta^{-1}K_{x_0}(t)\ast \left(\nabla\rho_0 P_3(t)\right)\right\|_{L^\infty}\lesssim \left\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(t)| \right\|_{L^2}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}\label{S4eq54}\\ &\qquad\lesssim t^{-\frac{1}{4}}\|\rho_0-1\|_{L^\infty}\mathbf{C}^2(u),\nonumber\\ &\left\|\left(\nabla\Delta^{-1}K_{x_0}(t)\ast \left(\nabla\rho_0 P_3(t)\right)\right)\big|_{x_0=x}\right\|_{L^{12}}\lesssim \left\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(t)| \right\|_{L^\frac{12}{7}}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}\nonumber\\ &\qquad\lesssim t^{-\frac{1}{8}}\|\rho_0-1\|_{L^\infty}\mathbf{C}^2(u),\nonumber\\ &\left\||D|^{1+\kappa}\nabla\Delta^{-1}K_{x_0}(t)\ast \left(\nabla\rho_0 P_3(t)\right)\right\|_{L^\infty}\lesssim \left\|\sup_{x_0}||D|^{1+\kappa}\nabla\Delta^{-1}K_{x_0}(t)| \right\|_{L^2}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}d\tau\nonumber\\ &\qquad\lesssim t^{-\frac{3+2\kappa}{4}}\|\rho_0-1\|_{L^\infty}\mathbf{C}^2(u).\nonumber \end{align}\tag{64}\] Finally, for \(P_2\), we use \(\rho_0\) to approximate \(\rho\), and write \[\label{S4eq63} P_2(t)=\int_0^t\nabla\Delta^{-1}\nabla\cdot((\rho(\tau)-\rho_0)u\cdot\nabla u)(\tau)d\tau+\int_0^t\nabla\Delta^{-1}\nabla\cdot(\rho_0 u\cdot\nabla u)(\tau)d\tau:=P_2^1(t)+P_2^2(t).\tag{65}\] For \(P_2^1\), since by Lemma 3 and 4, \(\|u\cdot\nabla u\|_{L^2}(\tau)\lesssim \tau^{-1}\), we can prove similar estimate as 63 , and obtain \[\begin{align} &\left\|\nabla\Delta^{-1}K_{x_0}(t)\ast \left(\nabla\rho_0 P_2^1(t)\right)\right\|_{L^\infty}\lesssim t^{\frac{1}{8}}\|\rho_0\|_{C^1}\mathbf{C}^2(u),\label{S4eq51}\\ &\left\|\left(\nabla\Delta^{-1}K_{x_0}(t)\ast \left(\nabla\rho_0 P_2^1(t)\right)\right)\big|_{x_0=x}\right\|_{L^{12}} \lesssim t^{\frac{1}{4}}\|\rho_0\|_{C^1}\mathbf{C}^2(u),\nonumber\\ &\left\||D|^{1+\kappa}\nabla\Delta^{-1}K_{x_0}(t)\ast \left(\nabla\rho_0 P_2^1(t)\right)\right\|_{L^\infty}\lesssim t^{-\frac{3+4\kappa}{8}}\|\rho_0\|_{C^1}\mathbf{C}^2(u).\nonumber \end{align}\tag{66}\] For \(P_2^2\), also by the divergence free condition of \(u\), we note that \[P_2^2(t)=\int_0^t\nabla\Delta^{-1}\nabla\cdot\operatorname{div}(\rho_0u\otimes u)(\tau)d\tau-\int_0^t\nabla\Delta^{-1}\nabla\cdot(u\nabla\rho_0\cdot u)(\tau)d\tau,\] which implies \[\begin{align} \nabla\Delta^{-1}K_{x_0}(t)\ast \nabla\rho_0 P_2^2(t)=&\nabla\Delta^{-1}\nabla K_{x_0}(t)\ast \left(\nabla\rho_0\int_0^t\Delta^{-1}\nabla\cdot\operatorname{div}(\rho_0 u\otimes u)(\tau)d\tau\right)\label{S4eq64}\\ &-\nabla\Delta^{-1} K_{x_0}(t)\ast \left(\nabla\rho_0\int_0^t\nabla\Delta^{-1}\nabla\cdot(u\nabla\rho_0\cdot u)(\tau)d\tau\right)\nonumber\\ &-\nabla\Delta^{-1} K_{x_0}(t)\ast \left(\nabla^2\rho_0\int_0^t\left(\Delta^{-1}\nabla\cdot\operatorname{div}(\rho_0 u\otimes u)d\tau\right)\right)\nonumber. \end{align}\tag{67}\] By this way, we can prove that \[\begin{align} &\left\|\nabla\Delta^{-1}K_{x_0}(t)\ast \left(\nabla\rho_0 P_2^2(t)\right)\right\|_{L^\infty}\lesssim \left\|\sup_{x_0}|\nabla^2\Delta^{-1}K_{x_0}(t)| \right\|_{L^2}\int_0^t\|\rho_0\|_{C^1}^2\left(\|u\otimes u(\tau)\|_{L^2} \right)d\tau\label{S4eq52}\\ &\qquad\qquad+ \left\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(t)| \right\|_{L^2}\|\rho_0\|_{C^2}^2\int_0^t\|u(\tau)\|_{L^\infty}\|u(\tau)\|_{L^2}d\tau \nonumber\\ &\qquad\lesssim t^{-\frac{3}{4}}\int_0^t\tau^{-\frac{1}{2}}d\tau\|\rho_0\|_{C^1}\mathbf{C}^2(u)\|u\|_{X_T}+t^{-\frac{1}{4}}\|\rho_0\|_{C^2}\int_0^t\tau^{-\frac{1}{2}}d\tau\mathbf{C}^2(u)\|u\|_{X_T}\nonumber\\ &\qquad\lesssim t^{-\frac{1}{4}}\|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\|u\|_{X_T},\nonumber\\ &\left\|\left(\nabla\Delta^{-1}K_{x_0}(t)\ast \left(\nabla\rho_0 P_2^2(t)\right)\right)\big|_{x_0=x}\right\|_{L^{12}}\lesssim \left\|\sup_{x_0}|\nabla^2\Delta^{-1}K_{x_0}(t)| \right\|_{L^\frac{12}{7}}\int_0^t\|\rho_0\|_{C^1}^2\|u\otimes u(\tau)\|_{L^2}d\tau\nonumber\\ &\qquad\qquad+ \left\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(t)| \right\|_{L^\frac{12}{7}}\|\rho_0\|_{C^2}^2\int_0^t\|u(\tau)\|_{L^\infty}\|u(\tau)\|_{L^2}d\tau\nonumber\\ &\qquad\lesssim t^{-\frac{5}{8}}\int_0^t\tau^{-\frac{1}{2}}d\tau\|\rho_0\|_{C^1}\mathbf{C}^2(u)\|u\|_{X_T}+t^{-\frac{1}{8}}\|\rho_0\|_{C^2}\int_0^t\tau^{-\frac{1}{2}}d\tau\mathbf{C}^2(u)\|u\|_{X_T}\nonumber\\ &\qquad\lesssim t^{-\frac{1}{8}}\|\rho_0\|_{C^1}\mathbf{C}^2(u)\|u\|_{X_T},\nonumber\\ &\left\||D|^{1+\kappa}\nabla\Delta^{-1}K_{x_0}(t)\ast \left(\nabla\rho_0 P_2^2(t)\right)\right\|_{L^\infty}\lesssim \left\|\sup_{x_0}||D|^{1+\kappa}\nabla^2\Delta^{-1}K_{x_0}(t)| \right\|_{L^2}\int_0^t\|\rho_0\|_{C^1}^2\|u\otimes u(\tau)\|_{L^2}d\tau\nonumber\\ &\qquad\qquad+\left\|\sup_{x_0}||D|^{1+\kappa}\nabla\Delta^{-1}K_{x_0}(t)| \right\|_{L^2}\|\rho_0\|_{C^2}^2\int_0^t\|u(\tau)\|_{L^\infty}\|u(\tau)\|_{L^2}d\tau\nonumber\\ &\qquad\lesssim t^{-\frac{5+2\kappa}{4}}\int_0^t\tau^{-\frac{1}{2}}d\tau\|\rho_0\|_{C^1}\mathbf{C}^2(u)\|u\|_{X_T}+t^{-\frac{3+2\kappa}{4}}\|\rho_0\|_{C^2}^2\int_0^t\tau^{-\frac{1}{2}}d\tau\mathbf{C}^2(u)\|u\|_{X_T}\nonumber\\ &\qquad\lesssim t^{-\frac{3+2\kappa}{4}}\|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\|u\|_{X_T}.\nonumber \end{align}\tag{68}\] Now we combine 63 , 64 , 66 and 68 , then we can obtain \[\label{S4eq56} \begin{align} &\sup_{x_0}\sup_{t\leq T}t^{\frac{1}{2}}\|u_{R1,x_0}^{21,P}(t)\|_{L^\infty}\lesssim \|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(T^{\frac{5}{8}}+T^{\frac{3}{8}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{1}{4}}\left(\|\rho_0-1\|_{L^\infty}+\|u\|_{X_T}\right) \right),\\ &\sup_{t\leq T}t^{\frac{5}{12}}\|\tilde{ u}_{R1}^{21,P}(t)\|_{L^{12}}\lesssim \|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(T^{\frac{2}{3}}+T^{\frac{5}{12}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{7}{24}}\left(\|\rho_0-1\|_{L^\infty}+\|u\|_{X_T}\right) \right),\\ &\sup_{x_0}\sup_{t\leq T}t^{\frac{2+\kappa}{2}}\||D|^{1+\kappa}u_{R1,x_0}^{21,P}(t)\|_{L^\infty}\lesssim\|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(T^{\frac{5}{8}}+T^{\frac{3}{8}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{1}{4}}\left(\|\rho_0-1\|_{L^\infty}+\|u\|_{X_T}\right) \right). \end{align}\tag{69}\] Finally, 57 , 59 , 60 , 61 and 69 imply that \[\label{S4eq57} \begin{align} &\sup_{x_0}\sup_{t\leq T}t^{\frac{1}{2}}\|u_{R1,x_0}^{2}(t)\|_{L^\infty}\lesssim \|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(T^{\frac{5}{8}}+T^{\frac{3}{8}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{1}{4}}\left(\|\rho_0-1\|_{L^\infty}+\|u\|_{X_T}\right) \right),\\ &\sup_{t\leq T}t^{\frac{5}{12}}\|\tilde{ u}_{R1}^{2}(t)\|_{L^{12}}\lesssim \|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(T^{\frac{2}{3}}+T^{\frac{5}{12}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{7}{24}}\left(\|\rho_0-1\|_{L^\infty}+\|u\|_{X_T}\right) \right),\\ &\sup_{x_0}\sup_{t\leq T}t^{\frac{2+\kappa}{2}}\||D|^{1+\kappa}u_{R1,x_0}^{2}(t)\|_{L^\infty}\lesssim\|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(T^{\frac{5}{8}}+T^{\frac{3}{8}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{1}{4}}\left(\|\rho_0-1\|_{L^\infty}+\|u\|_{X_T}\right) \right). \end{align}\tag{70}\]

So 41 and 70 show that \[\label{uR1Linf} \begin{align} &\sup_{x_0} \sup_{t\leq T}t^{\frac{1}{2}}\|u_{R1,x_0}(t)\|_{L^\infty}\\ &\lesssim \mathbf{C}^2(u)\|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(T^{\frac{5}{8}}+T^{\frac{3}{8}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{1}{4}}\left(\|\rho_0-1\|_{L^\infty}+\|u\|_{X_T}\right) \right), \end{align}\tag{71}\] 42 and 70 show that \[\label{uR1L12} \begin{align} &\sup_{t\leq T}t^{\frac{5}{12}}\|\tilde{u}_{R1}(t)\|_{L^{12}}\\ &\lesssim \mathbf{C}^2(u) \|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(T^{\frac{2}{3}}+T^{\frac{5}{12}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{7}{24}}\left(\|\rho_0-1\|_{L^\infty}+\|u\|_{X_T}\right) \right), \end{align}\tag{72}\] and 43 and 70 show that \[\label{uR1Hol} \begin{align} &\sup_{x_0} \sup_{t\leq T}t^{1+\frac{\kappa}{2}}\| |D|^{1+\kappa} u_{R1,x_0}(t) \|_{L^\infty} \\ &\lesssim \mathbf{C}^2(u) \|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(T^{\frac{5}{8}}+T^{\frac{3}{8}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{1}{4}}\left(\|\rho_0-1\|_{L^\infty}+\|u\|_{X_T}\right) \right). \end{align}\tag{73}\]

ii) Estimates for time differences.

By 56 , for \(s<t<T\), \[\begin{align} u_{R1,x_0}^2(t,x)-u_{R1,x_0}^2(s,x)=\sum_{i=1}^3\left(u_{R1,x_0}^{2i}(t,x)-u_{R1,x_0}^{2i}(s,x)\right). \end{align}\] For the latter two terms, since by Proposition 2 we have \[\|\rho(t)-\rho(s)\|_{L^\infty}\lesssim_\eta \|\rho_0\|_{C^1}\mathbf{C}^2(u)(t-s)^{1-\eta}s^{-\frac{1-\eta}{2}},\qquad\forall \eta\in(0,1).\] By the estimate above and 53 , for \(a=t-s\leq \frac{s}{4}\), we have \[\begin{align} &\|u_{R1,x_0}^{22}(t)-u_{R1,x_0}^{22}(s)\|_{L^\infty}\lesssim \int_s^t\|\nabla\Delta^{-1}\partial_\tau K_{x_0}(\tau)\|_{L^2}d\tau\|\nabla\rho_0\|_{L^\infty}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}\label{S4eq58}\\ &\qquad\lesssim s^{-\frac{1}{4}}\min\{1,as^{-1}\}\|\nabla\rho_0\|_{L^\infty}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2},\nonumber\\ &\|\tilde{u}_{R1}^{22}(t)-\tilde{u}_{R1}^{22}(s)\|_{L^{12}}\lesssim \int_s^t\|\nabla\Delta^{-1}\partial_\tau K_{x_0}(\tau)\|_{L^\frac{12}{7}}d\tau\|\nabla\rho_0\|_{L^\infty}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}\nonumber\\ &\qquad\lesssim s^{-\frac{1}{8}}\min\{1,as^{-1}\}\|\nabla\rho_0\|_{L^\infty}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2},\nonumber\\ &\|u_{R1,x_0}^{23}(t)-u_{R1,x_0}^{23}(s)\|_{L^\infty}\lesssim \int_s^t\|\nabla\Delta^{-1}\partial_\tau K_{x_0}(\tau)\|_{L^2}d\tau\|\nabla\rho_0\|_{L^\infty}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}\nonumber\\ &\qquad\qquad+\left\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(s)|\right\|_{L^2 }\|\nabla\rho_0\|_{L^\infty}\left(\|\rho(t)-\rho(s)\|_{L^\infty}\|u_0\|_{L^2}+\|\rho_0-1\|_{L^\infty}\|u(t)-u(s)\|_{L^2}\right) \nonumber \\ &\qquad\lesssim a^{\varkappa_1}\|\nabla\rho_0\|_{L^\infty}\|u_0\|_{L^2}\left(s^{-\frac{1}{4}-\varkappa_1}\|\rho_0-1\|_{L^\infty}+s^{\frac{1-\eta}{2}-\varkappa_1-\frac{1}{4}}\right),\nonumber\\ &\|\tilde{u}_{R1}^{23}(t)-\tilde{u}_{R1}^{23}(s)\|_{L^{12}}\lesssim \int_s^t\|\nabla\Delta^{-1}\partial_\tau K_{x_0}(\tau)\|_{L^{\frac{12}{7}}}d\tau\|\nabla\rho_0\|_{L^\infty}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}\nonumber\\ &\qquad\qquad+\left\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(s)|\right\|_{L^{\frac{12}{7}} }\|\nabla\rho_0\|_{L^\infty}\left(\|\rho(t)-\rho(s)\|_{L^\infty}\|u_0\|_{L^2}+\|\rho_0-1\|_{L^\infty}\|u(t)-u(s)\|_{L^2}\right) \nonumber \\ &\qquad\lesssim a^{\varkappa_2}\|\rho_0\|_{C^1}^2\|u_0\|_{L^2}\left(s^{-\frac{1}{8}-\varkappa_2}\|\rho_0-1\|_{L^\infty}+s^{\frac{1-\eta}{2}-\varkappa_1-\frac{1}{8}}\right),\nonumber \end{align}\tag{74}\] for some \(\eta\in(0,1)\). So we can take \(\eta=\frac{1}{4}\) to get the following difference estimate: \[\label{S4eq59} \begin{align} &\sup_{x_0}\sup_{s<t<T}s^{\frac{1}{2}+\varkappa_1}\frac{\left\|u_{R1,x_0}^{22}(t)-u_{R1,x_0}^{22}(s)\right\|_{L^\infty}}{(t-s)^{\varkappa_1}}+\sup_{x_0}\sup_{s<t<T}s^{\frac{1}{2}+\varkappa_1}\frac{\left\|u_{R1,x_0}^{23}(t)-u_{R1,x_0}^{23}(s)\right\|_{L^\infty}}{(t-s)^{\varkappa_1}}\\ &+\sup_{s<t<T}s^{\frac{5}{12}+\varkappa_2}\frac{\left\|\tilde{u}_{R1}^{22}(t)-\tilde{u}_{R1,}^{22}(s)\right\|_{L^{12}}}{(t-s)^{\varkappa_2}}+\sup_{s<t<T}s^{\frac{5}{12}+\varkappa_2}\frac{\left\|\tilde{u}_{R1}^{23}(t)-\tilde{u}_{R1}^{23}(s)\right\|_{L^{12}}}{(t-s)^{\varkappa_2}}\\ &\qquad\lesssim \|\rho_0\|_{C^1}^2\|u_0\|_{L^2}\left(T^{\frac{5}{8}}+T^{\frac{1}{4}}\|\rho_0-1\|_{L^\infty}\right) \end{align}\tag{75}\] For \(u_{R1,x_0}^{21}\), we still apply the decomposition 58 . For \(u_{R1,x_0}^{21,m}\), we have \[\begin{align} u_{R1,x_0}^{21,m}(t,x)-u_{R1,x_0}^{21,m}(s,x)=&\int_s^t\nabla\Delta^{-1}\nabla\partial_\gamma K_{x_0}(\gamma)d\gamma\ast \int_0^t\nabla\rho_0\left(\nabla u(\tau)-(\rho u\otimes u)(\tau)\right)d\tau\\ &-\int_s^t\nabla\Delta^{-1}\partial_\gamma K_{x_0}(\gamma)d\gamma\ast \int_0^t\nabla^2\rho_0\left(\nabla u(\tau)-(\rho u\otimes u)(\tau)\right)d\tau\\ &+\nabla\Delta^{-1}\nabla K_{x_0}(s)\ast \int_s^t\nabla\rho_0\left(\nabla u(\tau)-(\rho u\otimes u)(\tau)\right)d\tau\\ &-\nabla\Delta^{-1} K_{x_0}(s)\ast \int_s^t\nabla^2\rho_0\left(\nabla u(\tau)-(\rho u\otimes u)(\tau)\right)d\tau \end{align}\] So by ?? , Lemma 3, 4 and 5 we separately have the \(L^\infty\) estimate \[\begin{align} \|u_{R1,x_0}^{21,m}(t)-u_{R1,x_0}^{21,m}(s)\|_{L^\infty}\lesssim &\int_s^t\|\nabla^2K_{x_0}(\gamma)\|_{L^{\frac{4}{3}}}d\gamma\int_0^t\|\rho_0\|_{C^1}^2\left(\|\nabla u(\tau)\|_{L^4}+\|u(\tau)\|_{L^\infty}\|u(\tau)\|_{L^4}\right)d\tau\\ &+\int_s^t\|\nabla K_{x_0}(\gamma)\|_{L^{\frac{4}{3}}}d\gamma\int_0^t\|\rho_0\|_{C^2}^2\left(\|\nabla u(\tau)\|_{L^4}+\|u(\tau)\|_{L^\infty}\|u(\tau)\|_{L^4}\right)d\tau\\ &+\|\rho_0\|_{C^1}^2\|\nabla\Delta^{-1}\nabla K_{x_0}(s)\|_{L^{\frac{4}{3}}}\int_s^t\left(\|\nabla u(\tau)\|_{L^4}+\|u(\tau)\|_{L^\infty}\|u(\tau)\|_{L^4}\right)d\tau\\ &+\|\rho_0\|_{C^2}^2\|\nabla\Delta^{-1}K_{x_0}(s)\|_{L^{2}}\int_s^t\|\nabla u(\tau)-\mathbb{P}(u\otimes u)(\tau)\|_{L^{2}}d\tau\\ \lesssim &\|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(\int_s^t\gamma^{-1-\frac{3}{8}}d\gamma\int_0^t\tau^{-\frac{3}{4}}d\tau +\int_s^t\gamma^{-\frac{1}{2}-\frac{3}{8}}d\gamma \int_0^t\tau^{-\frac{3}{4}}d\tau\right)\|u\|_{X_T}^{\frac{1}{2}}\\ &+\|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(s^{-\frac{3}{8}}\int_0^t\tau^{-\frac{3}{4}}d\tau\|u\|_{X_T}^{\frac{1}{2}}+s^{-{\frac{1}{4}}}\int_s^t\tau^{-\frac{1}{2}}d\tau\|u_0\|_{L^2} \right)\\ \lesssim&\|\rho_0\|_{C^2}^2(t-s)^{\varkappa_1} \mathbf{C}^2(u)\left(s^{-\frac{1}{8}-\varkappa_1}\|u\|_{X_T}^{\frac{1}{2}}+s^{\frac{1}{4}-\varkappa_1}\right), \end{align}\] and the \(L^{12}\) estimate \[\begin{align} \|\tilde{u}_{R1}^{21,m}(t)-\tilde{u}_{R1}^{21,m}(s)\|_{L^{12}}\lesssim &\int_s^t\|\sup_{x_0}|\nabla^2K_{x_0}(\gamma)|\|_{L^{\frac{6}{5}}}d\gamma\int_0^t\|\rho_0\|_{C^1}^2\left(\|\nabla u(\tau)\|_{L^4}+\|u(\tau)\|_{L^\infty}\|u(\tau)\|_{L^4}\right)d\tau\\ &+\int_s^t\|\sup_{x_0}|\nabla K_{x_0}(\gamma)|\|_{L^{\frac{6}{5}}}d\gamma\int_0^t\|\rho_0\|_{C^2}^2\left(\|\nabla u(\tau)\|_{L^4}+\|u(\tau)\|_{L^\infty}\|u(\tau)\|_{L^4}\right)d\tau\\ &+\|\rho_0\|_{C^1}^2\|\sup_{x_0}|\nabla\Delta^{-1}\nabla K_{x_0}(s)|\|_{L^{\frac{6}{5}}}\int_s^t\left(\|\nabla u(\tau)\|_{L^4}+\|u(\tau)\|_{L^\infty}\|u(\tau)\|_{L^4}\right)d\tau\\ &+\|\rho_0\|_{C^2}^2\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(s)|\|_{L^\frac{12}{7}}\int_s^t\|\nabla u(\tau)-(u\otimes u)(\tau)\|_{L^{2}}d\tau\\ \lesssim &\|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(\int_s^t\gamma^{-1-\frac{1}{4}}d\gamma\int_0^t\tau^{-\frac{3}{4}}d\tau +\int_s^t\gamma^{-\frac{1}{2}-\frac{1}{4}}d\gamma \int_0^t\tau^{-\frac{3}{4}}d\tau\right)\|u\|_{X_T}^{\frac{1}{2}}\\ &+\|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(s^{-\frac{1}{4}}\int_0^t\tau^{-\frac{3}{4}}d\tau\|u\|_{X_T}^{\frac{1}{2}}+s^{-{\frac{1}{8}}}\int_s^t\tau^{-\frac{1}{2}}d\tau \right)\\ \lesssim&\|\rho_0\|_{C^2}^2(t-s)^{\varkappa_2} \mathbf{C}^2(u)\left(s^{-\varkappa_2}\|u\|_{X_T}^{\frac{1}{2}}+s^{\frac{3}{8}-\varkappa_2}\right), \end{align}\] So we conclude that \[\label{S4eq43} \begin{align} &\sup_{x_0}\sup_{s<t<T}s^{\frac{1}{2}+\varkappa_1}\frac{\|u_{R1,x_0}^{21,m}(t)-u_{R1,x_0}^{21,m}(s)\|_{L^\infty}}{(t-s)^{\varkappa_1}}\lesssim \|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(T^{\frac{3}{8}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{3}{4}} \right)\\ &\sup_{s<t<T}s^{\frac{5}{12}+\varkappa_2}\frac{\|\tilde{u}_{R1}^{21,m}(t)-\tilde{u}_{R1}^{21,m}(s)\|_{L^{12}}}{(t-s)^{\varkappa_2}}\lesssim \|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\left(T^{\frac{5}{12}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{19}{24}} \right) \end{align}\tag{76}\] For \(u_{R1,x_0}^{21,P}\), we apply the decomposition of \(\nabla P\) as in 62 and 65 . Note that by 53 , Proposition 2 and Lemma 5, for any \(\eta\in(0,1)\) and \(s<t\), we have \[\begin{align} &\left\|P_1(t)\right\|_{L^2}\lesssim \int_0^t \tau^{-1+\frac{1-\eta}{2}}d\tau\|\rho_0\|_{C^1}\mathbf{C}^2(u)\lesssim t^{\frac{1-\eta}{2}}\|\rho_0\|_{C^1}\mathbf{C}^2(u),\\ & \left\|P_1(t)-P_1(s) \right\|_{L^2}\lesssim \int_s^t\tau^{-1+\frac{1-\eta}{2}}d\tau\|\rho_0\|_{C^1}\mathbf{C}^2(u)\lesssim s^{-\frac{1+\eta}{2}}\min\{1,(t-s)s^{-1}\}\|\rho_0\|_{C^1}\mathbf{C}^2(u),\\ &\left\|P_2^1(t)\right\|_{L^2}\lesssim \int_0^t \tau^{-1+\frac{1-\eta}{2}}d\tau\|\rho_0\|_{C^1}\mathbf{C}^3(u)\lesssim t^{\frac{1-\eta}{2}}\|\rho_0\|_{C^1}\mathbf{C}^3(u),\\ & \left\|P_2^1(t)-P_2^1(s) \right\|_{L^2}\lesssim \int_s^t\tau^{-1+\frac{1-\eta}{2}}d\tau\|\rho_0\|_{C^1}\mathbf{C}^3(u)\lesssim s^{-\frac{1+\eta}{2}}\min\{1,(t-s)s^{-1}\}\|\rho_0\|_{C^1}\mathbf{C}^3(u),\\ &\left\|P_3(t)\right\|_{L^2}\lesssim \|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2},\\ & \left\|P_3(t)-P_3(s) \right\|_{L^2}\lesssim \|\rho_0-1\|_{L^\infty}\|u(t)-u(s)\|_{L^2}\lesssim \|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}\min\{1,(t-s)s^{-1}\}. \end{align}\] Then we denote the components in \(u_{R1,x_0}^{21,P}\) related to \(P_1,P_2^1, P_2^2,P_3\) by \(u_{R1,x_0}^{21,P1},u_{R1,x_0}^{21,P21}, u_{R1,x_0}^{P22},u_{R1,x_0}^{21,P3}\) separately, and by ?? we can show that \[\begin{align} &\left\|u_{R1,x_0}^{21,P1}(t)-u_{R1,x_0}^{21,P1}(s)\right\|_{L^\infty}\lesssim \|\rho_0\|_{C^1}\int_s^t\left\|\sup_{x_0}\partial_\tau\nabla\Delta^{-1}K_{x_0}(\tau)\right\|_{L^2}d\tau \|P_1(t)\|_{L^2} \label{S4eq62}\\ &\qquad\qquad+\|\rho_0\|_{C^1}\left\|\sup_{x_0}\nabla\Delta^{-1}K_{x_0}(s)\right\|_{L^2}\|P_1(t)-P_1(s)\|_{L^2} \nonumber\\ &\qquad\lesssim (t-s)^{\varkappa_1}s^{\frac{1}{8}-\varkappa_1}\|\rho_0\|_{C^1}^2\mathbf{C}^2(u),\nonumber\\ &\left\|u_{R1,x_0}^{21,P21}(t)-u_{R1,x_0}^{21,P21}(s)\right\|_{L^\infty}\lesssim \|\rho_0\|_{C^1}\int_s^t\left\|\sup_{x_0}\partial_\tau\nabla\Delta^{-1}K_{x_0}(\tau)\right\|_{L^2}d\tau \|P_2^1(t)\|_{L^2} \nonumber\\ &\qquad\qquad+\|\rho_0\|_{C^1}\left\|\sup_{x_0}\nabla\Delta^{-1}K_{x_0}(s)\right\|_{L^2}\|P_2^1(t)-P_2^1(s)\|_{L^2} \nonumber\\ &\qquad\lesssim (t-s)^{\varkappa_1}s^{\frac{1}{8}-\varkappa_1}\|\rho_0\|_{C^1}^2\mathbf{C}^3(u)\nonumber\\ &\left\|u_{R1,x_0}^{21,P3}(t)-u_{R1,x_0}^{21,P3}(s)\right\|_{L^\infty}\lesssim \|\rho_0\|_{C^1}\int_s^t\left\|\sup_{x_0}\partial_\tau\nabla\Delta^{-1}K_{x_0}(\tau)\right\|_{L^2}d\tau \|P_3(t)\|_{L^2} \nonumber\\ &\qquad\qquad+\|\rho_0\|_{C^1}\left\|\sup_{x_0}\nabla\Delta^{-1}K_{x_0}(s)\right\|_{L^2}\|P_3(t)-P_3(s)\|_{L^2} \nonumber\\ &\qquad\lesssim (t-s)^{\varkappa_1}s^{-\frac{1}{4}-\varkappa_1}\|\rho_0\|_{C^1}^2\|\rho_0-1\|_{L^\infty}\mathbf{C}^2(u),\nonumber\\ &\left\|\tilde{u}_{R1}^{21,P1}(t)-\tilde{u}_{R1}^{21,P1}(s)\right\|_{L^{12}}\lesssim \|\rho_0\|_{C^1}\int_s^t\left\|\sup_{x_0}\partial_\tau\nabla\Delta^{-1}K_{x_0}(\tau)\right\|_{L^{\frac{12}{7}}}d\tau \|P_1(t)\|_{L^2}\nonumber \\ &\qquad\qquad+\|\rho_0\|_{C^1}\left\|\sup_{x_0}\nabla\Delta^{-1}K_{x_0}(s)\right\|_{L^{\frac{12}{7}}}\|P_1(t)-P_1(s)\|_{L^2}\nonumber \\ &\qquad\lesssim (t-s)^{\varkappa_2}s^{\frac{1}{4}-\varkappa_2}\|\rho_0\|_{C^1}^2\mathbf{C}^2(u),\nonumber\\ &\left\|\tilde{u}_{R1}^{21,P21}(t)-\tilde{u}_{R1}^{21,P21}(s)\right\|_{L^{12}}\lesssim \|\rho_0\|_{C^1}\int_s^t\left\|\sup_{x_0}\partial_\tau\nabla\Delta^{-1}K_{x_0}(\tau)\right\|_{L^{\frac{12}{7}}}d\tau \|P_2^1(t)\|_{L^2}\nonumber \\ &\qquad\qquad+\|\rho_0\|_{C^1}\left\|\sup_{x_0}\nabla\Delta^{-1}K_{x_0}(s)\right\|_{L^{\frac{12}{7}}}\|P_2^1(t)-P_2^1(s)\|_{L^2}\nonumber \\ &\qquad\lesssim (t-s)^{\varkappa_2}s^{\frac{1}{4}-\varkappa_2}\|\rho_0\|_{C^1}^2\mathbf{C}^3(u),\nonumber\\ &\left\|\tilde{u}_{R1}^{21,P3}(t)-\tilde{u}_{R1}^{21,P3}(s)\right\|_{L^{12}}\lesssim \|\rho_0\|_{C^1}\int_s^t\left\|\sup_{x_0}\partial_\tau\nabla\Delta^{-1}K_{x_0}(\tau)\right\|_{L^{\frac{12}{7}}}d\tau \|P_3(t)\|_{L^2}\nonumber \\ &\qquad\qquad+\|\rho_0\|_{C^1}\left\|\sup_{x_0}\nabla\Delta^{-1}K_{x_0}(s)\right\|_{L^{\frac{12}{7}}}\|P_3(t)-P_3(s)\|_{L^2}\nonumber \\ &\qquad\lesssim (t-s)^{\varkappa_2}s^{-\frac{1}{8}-\varkappa_2}\|\rho_0\|_{C^1}^2\|\rho_0-1\|_{L^\infty}\mathbf{C}^2(u).\nonumber \end{align}\tag{77}\] Finally, by 67 , and the fact that \[\begin{align} &\|\int_0^t(u\otimes u)(\tau)d\tau\|_{L^2}\lesssim t^{\frac{1}{2}}\|u\|_{X_T}\mathbf{C}(u),\\ &\|\int_s^t(u\otimes u)(\tau)d\tau\|_{L^2}\lesssim s^{\frac{1}{2}}\min\{1,(t-s)s^{-1}\}\|u\|_{X_T}\mathbf{C}(u), \end{align}\] together with ?? , we have \[\begin{align} &\left\|u_{R1,x_0}^{21,P22}(t)-u_{R1,x_0}^{21,P22}(s) \right\|_{L^\infty}\lesssim \int_s^t\left\|\sup_{x_0}|\partial_\tau\nabla^2\Delta^{-1}K_{x_0}(\tau)|\right\|_{L^2}d\tau\|\rho_0\|_{C^1}^2t^{\frac{1}{2}}\|u\|_{X_T}\mathbf{C}(u)\label{S4eq65}\\ &\qquad\qquad+\left\|\sup_{x_0}|\nabla^2\Delta^{-1}K_{x_0}(s)| \right\|_{L^2}s^{\frac{1}{2}}\|\rho_0\|_{C^1}\min\{1,(t-s)s^{-1}\}\|u\|_{X_T}\mathbf{C}(u)\nonumber\\ &\qquad\qquad+\int_s^t\left\|\sup_{x_0}|\partial_\tau\nabla\Delta^{-1}K_{x_0}(\tau)|\right\|_{L^2}d\tau\|\rho_0\|_{C^2}^2t^{\frac{1}{2}}\|u\|_{X_T}\mathbf{C}(u)\nonumber\\ &\qquad\qquad+\left\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(s)| \right\|_{L^2}s^{\frac{1}{2}}\|\rho_0\|_{C^2}^2\min\{1,(t-s)s^{-1}\}\|u\|_{X_T}\mathbf{C}(u)\nonumber\\ &\qquad\lesssim (t-s)^{\varkappa_1}s^{-\frac{1}{4}-\varkappa_1}\|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\|u\|_{X_T},\nonumber\\ &\left\|\tilde{u}_{R1}^{21,P22}(t)-\tilde{u}_{R1}^{21,P22}(s) \right\|_{L^{12}}\lesssim \int_s^t\left\|\sup_{x_0}|\partial_\tau\nabla^2\Delta^{-1}K_{x_0}(\tau)|\right\|_{L^{\frac{12}{7}}}d\tau\|\rho_0\|_{C^1}^2t^{\frac{1}{2}}\|u\|_{X_T}\mathbf{C}(u)\nonumber\\ &\qquad\qquad+\left\|\sup_{x_0}|\nabla^2\Delta^{-1}K_{x_0}(s)| \right\|_{L^{\frac{12}{7}}}s^{\frac{1}{2}}\|\rho_0\|_{C^1}\min\{1,(t-s)s^{-1}\}\|u\|_{X_T}\mathbf{C}(u)\nonumber\\ &\qquad\qquad+\int_s^t\left\|\sup_{x_0}|\partial_\tau\nabla\Delta^{-1}K_{x_0}(\tau)|\right\|_{L^{\frac{12}{7}}}d\tau\|\rho_0\|_{C^2}^2t^{\frac{1}{2}}\|u\|_{X_T}\mathbf{C}(u)\nonumber\\ &\qquad\qquad+\left\|\sup_{x_0}|\nabla\Delta^{-1}K_{x_0}(s)| \right\|_{L^{\frac{12}{7}}}s^{\frac{1}{2}}\|\rho_0\|_{C^2}^2\min\{1,(t-s)s^{-1}\}\|u\|_{X_T}\mathbf{C}(u)\nonumber\\ &\qquad\lesssim (t-s)^{\varkappa_1}s^{-\frac{1}{8}-\varkappa_1}\|\rho_0\|_{C^2}^2\mathbf{C}^2(u)\|u\|_{X_T}.\nonumber \end{align}\tag{78}\]

So by 75 , 76 , 77 and 78 , we have \[\label{S4eq27} \begin{align} &\sup_{x_0} \sup_{s < t \leq T} s^{\frac{1}{2} + \varkappa_1} \frac{\|u_{R1,x_0}^2(t) - u_{R1,x_0}^2(s)\|_{L^\infty}}{(t-s)^{\varkappa_1}}+\sup_{x_0} \sup_{s < t \leq T} s^{\frac{5}{12} + \varkappa_2} \frac{\|u_{R1,x_0}^2(t) - u_{R1,x_0}^2(s)\|_{L^{12}}}{(t-s)^{\varkappa_2}} \\ &\qquad\lesssim \mathbf{C}^3(u)\|\rho_0\|_{C^2}^2\left(T^{\frac{1}{4}}(\|u\|_{X_T}+\|\rho_0-1\|_{L^\infty})+T^{\frac{3}{8}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{5}{8}}\right) . \end{align}\tag{79}\]

By summarizing the estimates 71 , 72 , 73 , 50 and 79 , we conclude the proof of ?? . ◻

Proof. We divide the proof into the following two steps:

(i) Estimates for spatial integrability.

First in view of 36 , we can rewrite \(u_{R2,x_0}\) as \[\label{S4eq5} u_{R2,x_0} = \int_0^t \partial_\tau \mathbb{P} K_{x_0}(t-\tau) \ast \bigl( (\varrho u)(t) - (\varrho u)(\tau) \bigr) \, d\tau + K_{x_0}(t) \ast \mathbb{P} (\varrho u)(t) = u_{R2,x_0}^1 + u_{R2,x_0}^2.\tag{80}\]

It follows from the proof of Proposition 2 that for any \(\kappa\in (0,1),\) \[\begin{align} \|\rho(t)-\rho_0\|_{L^\infty}\lesssim& \||D|^\kappa\rho\|_{L^\infty_T(L^\infty)}\bigl(\varepsilon^{\kappa}+\|u\|_{X_T}\varepsilon^{\kappa-1}t^{\frac{1}{2}}\bigr)\\ \lesssim& \|\rho_0\|_{C^1}\bigl(1+\|u\|_{T}\bigr)t^{\frac{\kappa}{2}},\quadif we take \;\;\varepsilon=t^{\frac{1}{2}}, \end{align}\] from which, Propositions 1 and 2, we deduce that for any \(0 < \epsilon' < 1\) and for all \(s < t < T\), \[\label{rhouse} \begin{align} &\|\varrho(t)\|_{L^\infty} \lesssim t^{\frac{1-\epsilon'}{2}} \|\rho_0\|_{C^1}\bigl(1+\|u\|_{T}\bigr), \\ &\|\varrho(t) - \varrho(s)\|_{L^\infty} \lesssim \min\bigl\{ (t-s)^{\frac{1-\epsilon'}{2}}, s^{-\frac{1-\epsilon'}{2}} (t-s)^{1-\epsilon'} \bigr\} \|\rho_0\|_{C^1}\bigl(1+\|u\|_{T}\bigr). \end{align}\tag{81}\]

By ?? , 37 , ?? , and 81 , we infer \[\begin{align} \|u^1_{R2,x_0}(t)\|_{L^\infty} \lesssim & \int_0^t \|\partial_t \mathbb{P} K_{x_0}(t-\tau)\|_{L^1} \left( \|\varrho(\tau)(u(t)-u(\tau))\|_{L^\infty} + \|(\varrho(t)-\varrho(\tau)) u(t)\|_{L^\infty} \right) d\tau \\ \leq &C \int_0^t (t-\tau)^{-1} \bigl( \tau^{\frac{1-\epsilon'}{2}} (t-\tau)^{\varkappa_1} \tau^{-\frac{1}{2}-\varkappa_1} + (t-\tau)^{\frac{1-\epsilon'}{2}} t^{-1/2} \bigr) d\tau \mathbf{C}(u) \|\rho_0\|_{C^1} \|u\|_{T}\\ \leq & C\mathbf{C}(u) \|\rho_0\|_{C^1}\|u\|_{T}t^{-\frac{1}{12}}, \quad (if we take\;\epsilon' = 1/24), \end{align}\] and \[\begin{align} &\|\nabla|D|^{\kappa} u^1_{R2,x_0}(t)\|_{L^\infty} \\ &\lesssim \int_0^t \|\partial_t |D|^{1+\kappa} \mathbb{P} K_{x_0}(t-\tau)\|_{L^1} \left( \|\varrho(\tau)(u(t)-u(\tau))\|_{L^\infty} + \|(\varrho(t)-\varrho(\tau)) u(t)\|_{L^\infty} \right) d\tau \\ &\leq C\int_0^t (t-\tau)^{-1-\frac{1+\kappa}{2}} \bigl( \tau^{\frac{1-\epsilon'}{2}} (t-\tau)^{\varkappa_1} \tau^{-1/2-\varkappa_1} + (t-\tau)^{1-\epsilon'} \tau^{-\frac{1-\epsilon'}{2}} t^{-1/2} \bigr) d\tau \mathbf{C}(u) \|\rho_0\|_{C^1}\|u\|_{T}\\ &\leq C\mathbf{C}(u) \|\rho_0\|_{C^1}\|u\|_{T}t^{-\frac{\epsilon' + 1 + \kappa}{2}} \\ &\leq C\mathbf{C}(u) \|\rho_0\|_{C^1}\|u\|_{T} t^{-\frac{3}{4}}, \quad (if we take\;\epsilon' = 1/2 - \kappa), \end{align}\] and \[\begin{align} \|\tilde{u}^1_{R2}(t)\|_{L^{12}} \lesssim &\int_0^t \bigl\| \sup_{x_0} \partial_t \mathbb{P} K_{x_0}(t-\tau) \bigr\|_{L^1} \left( \|\varrho(\tau)(u(t)-u(\tau))\|_{L^{12}} + \|(\varrho(t)-\varrho(\tau)) u(t)\|_{L^{12}} \right) d\tau \\ \leq & C \int_0^t (t-\tau)^{-1} \bigl( \tau^{\frac{1-\epsilon'}{2}} (t-\tau)^{\varkappa_2} \tau^{-\frac{5}{12}-\varkappa_2} + t^{-\frac{5}{12}} (t-\tau)^{1-\epsilon'} \tau^{-\frac{1-\epsilon'}{2}} \bigr) d\tau \mathbf{C}(u) \|\rho_0\|_{C^1}\|u\|_{T}\\ \leq & C \mathbf{C}(u) \|\rho_0\|_{C^1}\|u\|_{T}t^{\frac{1}{12} - \frac{\epsilon'}{2}} \\ \leq &\mathbf{C}(u) \|\rho_0\|_{C^1}\|u\|_{T}, \quad (if we take\;\epsilon' = 1/6). \end{align}\]

Similarly, we deduce from ?? , 37 and ?? that \[\begin{align} \|u_{R2,x_0}^2(t)\|_{L^\infty}\lesssim& \|\mathbb{P} K_{x_0}(t)\|_{L^\frac{12}{11}} \|(\varrho u)(t)\|_{L^{12}} \leq Ct^{-\frac{1}{8}} t^{\frac{1-\epsilon'}{2}}t^{-\frac{5}{12}} \mathbf{C}(u)\|u\|_{T} \|\rho_0\|_{C^1} \\ \leq &C \mathbf{C}(u) \|u\|_{T}\|\rho_0\|_{C^1} t^{-\frac{1}{6}}, \quad (if we take\;\epsilon' = 1/4), \end{align}\] and \[\begin{align} \|\nabla|D|^{\kappa} u_{R2,x_0}^2\|_{L^\infty} \lesssim& \|\nabla|D|^{\kappa} K_{x_0}(t)\|_{L^1} \|(\varrho u)(t)\|_{L^\infty} \leq Ct^{-\frac{1+\epsilon}{2}} t^{\frac{1-\kappa}{2}} t^{-1/2} \mathbf{C}(u)\|u\|_{T} \|\rho_0\|_{C^1} \\ \leq & C\mathbf{C}(u)\|u\|_{T}\|\rho_0\|_{C^1} t^{-\frac{7}{12}}, \quad (if we take\;\epsilon' = 1/6 - \epsilon), \end{align}\] and \[\begin{align} \|\tilde{u}^2_{R2}(t)\|_{L^{12}} \lesssim& \|\varrho u(t)\|_{L^{12}} \leq C t^{\frac{1-\epsilon'}{2}} t^{-\frac{5}{12}} \|u\|_{T}\mathbf{C}(u) \|\rho_0\|_{C^1} \\ \leq& C\mathbf{C}(u)\|u\|_{T}\|\rho_0\|_{C^1}, \qquad (if we take\;\epsilon' = 1/6) \end{align}\]

Thanks to 50 , we get, by summarizing the above estimates, that \[\label{uR2spa} \begin{align} \sup_{x_0} \Bigl( & \sup_{t \leq T} \Bigl( t^{\frac{1}{2}} \|u_{R2,x_0}(t)\|_{L^\infty} + t^{1+\frac{\kappa}{2}} \|\nabla|D|^{\kappa} u_{R2,x_0}(t)\|_{L^\infty} + t^{\frac{5}{12}} \|\tilde{u}_{R2}(t)\|_{L^{12}} \Bigr) \\ &+\|u_{R2,x_0}\|_{V_T^b} \Bigr) \leq C T^{\frac{3}{8}} \mathbf{C}(u)\|u\|_{T} \|\rho_0\|_{C^1}. \end{align}\tag{82}\]

(ii) Estimates for time differences.

We first consider the \(L^{12}\) case. Note from 36 that \[\label{S4eq7} u_{R2,x_0} = \int_0^t \partial_t K_{x_0}(t-\tau) \ast \mathbb{P} (\varrho u)(\tau) \, d\tau - \mathbb{P} (\varrho u)(t).\tag{83}\] By 81 and the definition of the \(\|\cdot\|_{X_T^2}\) norm, we have \[\label{uR2diffL12} \begin{align} \|\mathbb{P} (\varrho u)(t) - \mathbb{P} (\varrho u)(s)\|_{L^{12}} \leq &\| \varrho(t) - \varrho(s)\|_{L^\infty}\|u(t)\|_{L^{12}}+\|\varrho(s)\|_{L^\infty}\|u(t)-u(s)\|_{L^{12}}\\ \lesssim & s^{-\left(\frac{\varkappa_2}{2}+\frac{5}{12}\right)}(t-s)^{\varkappa_2}\|\rho_0\|_{C^1}\|u\|_{X_T} \mathbf{C}(u) \quad \forall s < t, \end{align}\tag{84}\] from which, Lemma 2 and 83 , we infer \[\begin{align} &\sup_{s < t \leq T} s^{\frac{5}{12} + \varkappa_2} \frac{\|\tilde{u}_{R2}(t) - \tilde{u}_{R2}(s)\|_{L^{12}}}{(t-s)^{\varkappa_2}} \\ &\lesssim \sup_{t\leq T}t^{\frac{5}{12}}\|\mathbb{P} (\varrho u)(t)\|_{L^{12}}+\sup_{s < t \leq T}s^{\frac{5}{12} + \varkappa_2}\frac{\|\mathbb{P} (\varrho u)(t) - \mathbb{P} (\varrho u)(s)\|_{L^{12}}}{(t-s)^{\varkappa_2}}\\ &\lesssim T^{\frac{3}{8}}\mathbf{C}(u)\|u\|_{X_T} \|\rho_0\|_{C^1}. \end{align}\]

For the \(L^\infty\) case, we have for any \(\sigma>0,\) \[\|\mathbb{P} f\|_{L^\infty} \lesssim \|f\|_{L^{12}}^{\frac{4\sigma}{1+4\sigma}} \|f\|_{\dot{C}^{\sigma}}^{\frac{1}{1+4\sigma}}.\] Taking \(\sigma = \frac{9}{14}\) in the above inequality, and using 81 , 84 , and the fact that \(\varkappa_1 = \frac{18}{25} \varkappa_2\), we obtain \[\label{uR2diffLinf} \begin{align} &\sup_{s<t\leq T}s^{\frac{1}{2}+\varkappa_1}\frac{\|\mathbb{P} (\varrho u)(t) - \mathbb{P} (\varrho u)(s)\|_{L^\infty}}{(t-s)^{\varkappa_1}}\\ & \lesssim \Bigl(\sup_{s<t\leq T}s^{\frac{5}{12}+\varkappa_2}\frac{\|\varrho u(t) - \varrho u(s)\|_{L^{12}}}{(t-s)^{\varkappa_1}} \Bigr)^{\frac{18}{25}} \bigl(s^{\frac{5}{7}} \|\varrho u(s)\|_{\dot{C}^{\frac{9}{14}}} + t^{\frac{5}{7}}\|\varrho u(t)\|_{\dot{C}^{\frac{9}{14}}} \bigr)^{\frac{7}{25}}\\ & \lesssim T^{\frac{3}{8}} \mathbf{C}(u)\|u\|_{X_T} \|\rho_0\|_{C^1}, \end{align}\tag{85}\] where we used the fact that \(\|u(s)\|_{\dot{C}^{\frac{9}{14}}}\lesssim s^{-\frac{25}{28}}\|u\|_{X_T}\) and \[\begin{align} \|\varrho u(s)\|_{\dot{C}^{\frac{9}{14}}} \lesssim &\|\varrho(s)\|_{L^\infty}\| u(s)\|_{\dot{C}^{\frac{9}{14}}} +\|\varrho(s)\|_{\dot{C}^{\frac{9}{14}}}\|u(s)\|_{L^\infty}\\ \lesssim &\|u\|_{X_T}\|\rho_0\|_{C^1}s^{-\frac{1}{2}}. \end{align}\] Then it follows from Lemma 2, 83 and 85 that \[\begin{align} \sup_{x_0} \Bigl(\sup_{s < t < T} s^{\frac{1}{2} + \varkappa_1} \frac{\|u_{R2,x_0}(t) - u_{R2,x_0}(s)\|_{L^\infty}}{(t-s)^{\varkappa_1}} \Bigr) \lesssim T^{\frac{3}{8}}\mathbf{C}(u)\|u\|_{X_T} \|\rho_0\|_{C^1}. \end{align}\] This together with 82 completes the proof of ?? . ◻

Proof. We divide the proof into the following three steps:

(i) Estimates for spatial integrability.

We first observe from ?? that \[\label{S4eq31} \| \sup_{x_0} (|D|^{\eta} \partial_t^i K_{x_0}(t,y) |y|) \|_{L^p_y} \lesssim t^{-i + \frac{1-\eta}{2} -\frac{3}{2}(1-\frac{1}{p})},\tag{86}\] for \(\eta = 0, 1+\kappa\) and \(i = 0, 1\).

We have the formula of \(u_{R3,x_0}\) from 36 as \[\label{S4eq33} \begin{align} u_{R3,x_0}&=-\int_0^t\int_{\mathbb{R}^3}\partial_\tau K_{x_0}(t-\tau,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)u(\tau,y)dyd\tau\\ &\qquad+\left(\rho_0(x_0)-\rho_0(x)\right)u(t,x)-\int_{\mathbb{R}^3}K_{x_0}(t,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)u_0(y)dy\\ &:=\sum_{i=1}^3u_{R3,x_0}^i. \end{align}\tag{87}\] When we fix \(x_0=x\), the second term \(u_{R3,x_0}^2\) vanishes. The third term can be written as \[\label{S4eq34} \begin{align} &u_{R3,x_0}^3=\int_{\mathbb{R}^3}K_{x_0}(t,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)u_0(y)dy\\ &\quad=\int_{\mathbb{R}^3}K_{x_0}(t,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)\left(\left(\rho_0u_0(y)-(\rho u)(t,y)\right)-(\rho_0-1)u_0(y)+(\rho u)(t,y)\right)dy\\ &\quad=\int_{\mathbb{R}^3}K_{x_0}(t,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)\int_0^t\left(\Delta u(\tau,y)-\nabla\cdot(\rho u\otimes u)(\tau,y)+\nabla P\right)(\tau)d\tau dy\\ &\qquad+\int_{\mathbb{R}^3}K_{x_0}(t,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)\left((1-\rho_0)u_0(y)+(\rho u)(t,y)\right)dy\\ &\quad=\int_{\mathbb{R}^3}\nabla\left(K_{x_0}(t,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)\right)\cdot\int_0^t\left(\nabla u(\tau,y)-(\rho u\otimes u)(\tau,y)\right)(\tau)d\tau dy\\ &\qquad+\int_{\mathbb{R}^3}K_{x_0}(t,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)\int_0^t\nabla P(\tau)d\tau dy\\ &\qquad+\int_{\mathbb{R}^3}K_{x_0}(t,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)\left((1-\rho_0)u_0(y)+(\rho u)(t,y)\right)dy\\ &\quad:=\sum_{i=1}^3u_{R3,x_0}^{3i}.\\ \end{align}\tag{88}\] By 86 , it is easy to prove \[\begin{align} &\sup_{t\leq T}t^{\frac{1}{2}}\left(\|\tilde{u}_{R3}^1(t)\|_{L^\infty}+\|\tilde{u}_{R3}^{31}(t)\|_{L^\infty}+\|\tilde{u}_{R3}^{33}(t)\|_{L^\infty}\right)\label{S4eq32}\\ &\qquad\qquad+\sup_{t\leq T}t^{\frac{5}{12}}\left(\|\tilde{u}_{R3}^1(t)\|_{L^{12}}+\|\tilde{u}_{R3}^{31}(t)\|_{L^{12}}+\|\tilde{u}_{R3}^{33}(t)\|_{L^{12}}\right)\nonumber\\ &\quad\lesssim \sup_{t\leq T}t^{\frac{1}{2}}\int_0^t(t-\tau)^{-\frac{1}{2}}\tau^{-\frac{1}{2}}d\tau\|u\|_{X_T}+\sup_{t\leq T}t^{\frac{5}{12}}\int_0^t(t-\tau)^{-\frac{1}{2}}\tau^{-\frac{5}{12}}d\tau\|u\|_{X_T}\nonumber\\ &\qquad+\sup_{t\leq T}t^{\frac{1}{2}}\int_0^t\|\rho_0\|_{C^1}\bigl\|\sup_{x_0}\left|\left(\nabla K_{x_0}(t,y)|y|+K_{x_0}(t,y)\right) \right|\bigr\|_{L^{\frac{4}{3}}_y}\left(\|\nabla u(\tau)\|_{L^4}d\tau+\|u(\tau)\|_{L^\infty}\|u(\tau)\|_{L^4} \right)d\tau\nonumber\\ &\qquad+\sup_{t\leq T}t^{\frac{5}{12}}\int_0^t\|\rho_0\|_{C^1}\bigl\|\sup_{x_0}\left|\left(\nabla K_{x_0}(t,y)|y|+K_{x_0}(t,y)\right)\right|\bigr\|_{ L^{\frac{6}{5}}_y}\left(\|\nabla u(\tau)\|_{L^4}d\tau+\|u(\tau)\|_{L^\infty}\|u(\tau)\|_{L^4} \right)d\tau\nonumber\\ &\qquad+\sup_{t\leq T}t^{\frac{1}{2}}\|\sup_{x_0}\left(K_{x_0}(t,y)|y|\right) \bigr\|_{L^{2}_y}\|\rho_0\|_{C^1}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}\nonumber\\ &\qquad+\sup_{t\leq T}t^{\frac{5}{12}}\|\sup_{x_0}\left(K_{x_0}(t,y)|y|\right) \bigr\|_{L^{\frac{12}{7}}_y}\|\rho_0\|_{C^1}\|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2}\nonumber\\ &\qquad+\sup_{t\leq T}t^{\frac{1}{2}}\|\sup_{x_0}\left(K_{x_0}(t,y)|y|\right) \bigr\|_{L^{1}_y}\|\rho_0\|_{C^1}\|u(t)\|_{L^\infty}\nonumber\\ &\qquad+\sup_{t\leq T}t^{\frac{5}{12}}\|\sup_{x_0}\left(K_{x_0}(t,y)|y|\right) \bigr\|_{L^{1}_y}\|\rho_0\|_{C^1}\|u(t)\|_{L^{12}}\nonumber\\ &\quad\lesssim T^{\frac{3}{8}}\mathbf{C}^2(u)\|\rho\|_{C^1}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{1}{4}}\mathbf{C}^3(u)\|\rho\|_{C^1}\|\rho_0-1\|_{L^\infty}\nonumber. \end{align}\tag{89}\] For the most subtle term \(u_{R3,x_0}^{32}\), we use the decomposition 62 . Note that \[\begin{align} &\|P_1(t)\|_{L^2}\lesssim \|\rho_0\|_{C^1}\int_0^t\tau^{-1+\frac{1-\eta}{2}}d\tau\mathbf{C}^2(u)\|u_0\|_{L^2}\lesssim t^{\frac{1-\eta}{2}}\|\rho_0\|_{C^1}\mathbf{C}^3(u),\label{S4eq67}\\ &\|P_2(t)\|_{L^{\frac{12}{7}}}\lesssim \|\rho\|_{L^\infty}\int_0^t\|u(\tau)\|_{L^{12}}\|\nabla u(\tau)\|_{L^2}d\tau\lesssim t^{\frac{1}{12}}\|\rho_0\|_{C^1}\mathbf{C}(u)\|u\|_{X_T},\nonumber\\ &\|P_3(t)\|_{L^2}\lesssim \|\rho_0-1\|_{L^\infty}\|u_0\|_{L^2},\nonumber \end{align}\tag{90}\] which implies \[\begin{align} &\sup_{t\leq T}t^{\frac{1}{2}}\left(\|\tilde{u}_{R3}^{32}(t)\|_{L^\infty}\right)+\sup_{t\leq T}t^{\frac{5}{12}}\left(\|\tilde{u}_{R3}^{32}(t)\|_{L^{12}}\right)\label{S4eq66}\\ &\quad\lesssim \sup_{t\leq T}t^{\frac{1}{2}}\|\sup_{x_0}\left(K_{x_0}(t,y)|y|\right) \bigr\|_{L^{2}_y}\left(\|P_1(t)\|_{L^2}+\|P_3(t)\|_{L^2}\right)\nonumber\\ &\qquad+\sup_{t\leq T}t^{\frac{1}{2}}\|\sup_{x_0}\left(K_{x_0}(t,y)|y|\right) \bigr\|_{L^{\frac{12}{5}}_y}\|P_2(t)\|_{L^\frac{12}{7}}\nonumber\\ &\qquad+\sup_{t\leq T}t^{\frac{5}{12}}\|\sup_{x_0}\left(K_{x_0}(t,y)|y|\right) \bigr\|_{L^{\frac{12}{7}}_y}\left(\|P_1(t)\|_{L^2}+\|P_3(t)\|_{L^2}\right)\nonumber\\ &\qquad+\sup_{t\leq T}t^{\frac{1}{2}}\|\sup_{x_0}\left(K_{x_0}(t,y)|y|\right) \bigr\|_{L^{2}_y}\|P_2(t)\|_{L^\frac{12}{7}}\nonumber\\ &\quad\lesssim \|\rho_0\|_{C^1}\mathbf{C}^2(u)\left(T^{\frac{3}{8}}\|u\|_{X_T}+T^{\frac{1}{4}}\|\rho_0-1\|_{L^\infty}\right).\nonumber \end{align}\tag{91}\]

For \(C^{1+\kappa}\) estimate, we have \[\begin{align} &|D|^{1+\kappa}u_{R3,x_0}=-\int_0^t\int_{\mathbb{R}^3}\partial_\tau |D|^{1+\kappa}K_{x_0}(t-\tau,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)\left(u(\tau,y)-u(t,y)\right)dyd\tau\label{S4eq35}\\ &\qquad-\int_{\mathbb{R}^3}|D|^{1+\kappa}K_{x_0}(t,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)\left(u_0(y)-u(t,y)\right)dy.\nonumber \end{align}\tag{92}\] Note that the boundary term \(\tau=t\) vanishes. We use similar methods as 88 for the second term (but change the \(\nabla\left(K_{x_0}(t,x-y)(\rho_0(x)-\rho_0(y))\right)\) to \(\nabla\left(|D|^{1+\kappa}K_{x_0}(t,x-y)(\rho_0(x)-\rho_0(y))\right)\)), then we have \[\begin{align} \sup_{t\leq T}t^{1+\frac{\kappa}{2}}\||D|^{1+\kappa}u_{R3,x_0}(t)|_{x_0=x}\|_{L^\infty}\lesssim \|\rho_0\|_{C^1}\mathbf{C}^3(u)\left(T^{\frac{3}{8}}\|u\|_{X_T}^\frac{1}{2}+T^{\frac{1}{4}}\|\rho_0-1\|_{L^\infty}\right). \end{align}\] As a consequence of the estimate above and 89 , 91 , we obtain \[\label{uR3spa} \begin{align} &\sup_{t \leq T} \Bigl( t^{\frac{1}{2}} \|\tilde{u}_{R3}(t)\|_{L^\infty} + \sup_{t \leq T} t^{1 + \frac{\kappa}{2}} \| (\nabla|D|^{\kappa} u_{R3,x_0})(t)|_{x_0=x} \|_{L^{\infty}} + t^{\frac{5}{12}} \|\tilde{u}_{R3}(t)\|_{L^{12}} \Bigr) \\ &\quad \leq C\|\rho_0\|_{C^1}\mathbf{C}^3(u)\left(T^{\frac{3}{8}}\|u\|_{X_T}^\frac{1}{2}+T^{\frac{1}{4}}\|\rho_0-1\|_{L^\infty}\right). \end{align}\tag{93}\] (ii) Estimates for time differences.

Furthermore, if we denote \(a = t-s < \frac{s}{4}\), we use 87 and have the decomposition \[\begin{align} &\delta_a^tu_{R3,x_0}(t,s,x)=-\int_s^t\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\partial_\tau K_{x_0}(t-\tau,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)u(\tau,y)dyd\tau\\ &\quad-\int_0^s\int_{\mathbb{R}^3}\delta_a^t\partial_\tau K_{x_0}(s-\tau,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)\left(u(\tau,y)-u(s,y)\right)dyd\tau\\ &\quad-\int_{\mathbb{R}^3}K_{x_0}(t-s,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)u(s,y)dy+\left(\rho_0(x_0)-\rho_0(x)\right)u(s,x)\\ &\quad+\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\delta_a^tK_{x_0}(s,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)u(s,y)dy+\left(\rho_0(x_0)-\rho_0(x)\right)\delta_a^tu(s,x)\\ &\quad-\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\delta_a^tK_{x_0}(s,x-y)\left(\rho_0(x_0)-\rho_0(y)\right)u_0(y)dy\\ &:=\sum_{i=1}^7Du_{R3,x_0}^i, \end{align}\] and we will also denote \(D\tilde{u}_{R3}^i(t,s,x)=(Du_{R3,x_0}^i)(t,s,x)|_{x_0=x}\). Easy to see that \(D\tilde{u}_{R3}^4(t,s,x)=D\tilde{u}_{R3}^6(t,s,x)=0\).
Observing that \[\begin{align} \left|\partial_t K_{x_0}(t,y) \right|&=\left|\Bigl(\frac{\rho_0(x_0)}{4\pi t}\Bigr)^{\frac{3}{2}}\exp\left(-\frac{\rho_0(x_0)|y|^2}{4t}\right)\left(-\frac{3}{2}t^{-1}+\frac{\rho_0(x_0)|y|^2}{4t^2}\right)\right|\\ &\lesssim \frac{C_2^{\frac{3}{2}}}{t^{\frac{5}{2}}}\exp\left(-\frac{C_1|y|^2}{8t}\right), \end{align}\] and \[\delta_a^t K_{x_0}(s,y)=\int_s^t\partial_\tau K_{x_0}(\tau,y)d\tau,\] we find \[\label{S4eq9} \begin{align} \| \sup_{x_0} |\delta_a^t K_{x_0}(s,y)| |y| \|_{L^p_y} \lesssim s^{\frac{1}{2}-\frac{3}{2}(1-\frac{1}{p})}\min\{1, a s^{-1} \}. \end{align}\tag{94}\] Similarly, observing that \[\begin{align} |\delta_a^t\partial_sK_{x_0}(s,y)|=\left|\int_s^t \partial_\tau^2 K_{x_0}(\tau,y)d\tau\right|\lesssim \frac{C^{\frac{3}{2}}}{s^{-\frac{7}{2}}}\exp\Bigl(-\frac{c|y|^2}{8s}\Bigr) \end{align}\] for any \(x_0\). We get, by a similar derivation of 94 , that \[\label{S4eq10} \begin{align} &\| \sup_{x_0} |\partial_s\delta_a^t K_{x_0}(s,y)| |y| \|_{L^2_y} \lesssim s^{-1}\min\bigl\{ s^{-\frac{1}{4}}, a^{\frac{3}{4}} s^{-1} \bigr\},\\ & \| \sup_{x_0} |\partial_s\delta_a^t K_{x_0}(s,y)| |y| \|_{L^{12/7}_y} \lesssim s^{-1}\min\bigl\{ s^{-\frac{1}{8}}, a^{\frac{7}{8}} s^{-1} \bigr\}. \end{align}\tag{95}\] So by 95 we have \[\begin{align} \|D\tilde{u}_{R3}^2(t,s)\|_{L^\infty}&\lesssim \int_s^t(s-\tau)^{-\frac{1}{2}}\min\{1,a(s-\tau)^{-1}\}\min\{1,(s-\tau)^{\varkappa_1}\tau^{-\varkappa_1}\}\tau^{-\frac{1}{2}}\|u\|_{X_T}\\ &\lesssim s^{-\varkappa_1}a^{\varkappa_1}\|u\|_{X_T},\\ \|D\tilde{u}_{R3}^2(t,s)\|_{L^{12}}&\lesssim \int_s^t(s-\tau)^{-\frac{5}{12}}\min\{1,a(s-\tau)^{-1}\}\min\{1,(s-\tau)^{\varkappa_2}\tau^{-\varkappa_2}\}\tau^{-\frac{5}{12}}\|u\|_{X_T}\\ &\lesssim s^{-\varkappa_2}a^{\varkappa_2}\|u\|_{X_T}. \end{align}\] Note that \[D\tilde{ u}_{R3}^3=\int_s^t\partial_\tau K_{x_0}(t-\tau,x-y)|_{x_0=x}\left(\rho_0(x)-\rho_0(y) \right)u(s,y)dyd\tau,\] we have \[D\tilde{u}_{R3}^1+D\tilde{u}_{R3}^3=-\int_s^t\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\partial_\tau K_{x_0}(t-\tau,x-y)|_{x_0=x}\left(\rho_0(x)-\rho_0(y)\right)\left(u(\tau,y)-u(s,y)\right)dyd\tau,\] so \[\begin{align} \|D\tilde{u}_{R3}^1(t,s)+D\tilde{u}_{R3}^3(t,s)\|_{L^\infty}&\lesssim \|\rho_0\|_{C^1}\int_s^t(t-\tau)^{-\frac{1}{2}}(s-\tau)^{\varkappa_1}s^{-\frac{1}{2}-\varkappa_1}d\tau\|u\|_{X_T}\\ &\lesssim s^{-\varkappa_1}a^{\varkappa_1}\|u\|_{X_T}\|\rho_0\|_{C^1},\\ \|D\tilde{u}_{R3}^1(t,s)+D\tilde{u}_{R3}^3(t,s)\|_{L^{12}}&\lesssim \|\rho_0\|_{C^1}\int_s^t(t-\tau)^{-\frac{1}{2}}(s-\tau)^{\varkappa_2}s^{-\frac{5}{12}-\varkappa_2}d\tau\|u\|_{X_T}\\ &\lesssim s^{\frac{1}{12}-\varkappa_2}a^{\varkappa_2}\|u\|_{X_T}\|\rho_0\|_{C^1}. \end{align}\] Finally, \(Du_{R3,x_0}^5\) and \(Du_{R3,x_0}^7\) have the following cancellation property, \[\begin{align} &D\tilde{u}_{R3}^5+D\tilde{u}_{R3}^7=\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\nabla_y\big(\delta_a^tK_{x_0}(s,x-y)\big)\big|_{x_0=x}\left(\rho_0(x)-\rho_0(y)\right)\int_0^s\left(\nabla u-\rho u\otimes u\right)(\tau)d\tau\\ &\qquad+\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\big(\delta_a^tK_{x_0}(s,x-y)\big)\big|_{x_0=x}\left(\rho_0(x)-\rho_0(y)\right)\int_0^s\nabla P(\tau)d\tau\\ &\qquad-\int_{\mathbb{R}^3}\big(\delta_a^tK_{x_0}(s,x-y)\left(\rho_0(x)-\rho_0(y) \right)\big)\big|_{x_0=x}(\rho(s,y)-1)u(s,y)dy\\ &\qquad+\int_{\mathbb{R}^3}\big(\delta_a^tK_{x_0}(s,x-y)\left(\rho_0(x)-\rho_0(y) \right)\big)\big|_{x_0=x}(\rho_0(y)-1)u_0(y)dy. \end{align}\] So by ?? and 90 , via s similar procedure as 89 and 91 , we have \[\begin{align} &\sup_{s<t<T}s^{\frac{1}{2}+\varkappa_1}\frac{\|D\tilde{u}_{R3}^5(t,x)+D\tilde{u}_{R3}^7(t,s)\|_{L^\infty}}{(t-s)^{\varkappa_1}}+\|D\tilde{u}_{R3}^5(t,x)+\sup_{s<t<T}s^{\frac{5}{12}+\varkappa_2}\frac{D\tilde{u}_{R3}^7(t,s)\|_{L^{12}}}{(t-s)^{\varkappa_2}}\\ &\quad\lesssim \|\rho_0\|_{C^1}^2\mathbf{C}^3(u)\left(T^{\frac{3}{8}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{1}{4}}\|\rho_0-1\|_{L^\infty}\right). \end{align}\]

By summarizing the above estimates, we arrive at \[\label{uR3diff} \begin{align} &\sup_{s < t \leq T} s^{\frac{1}{2} + \varkappa_1} \frac{\| \tilde{u}_{R3}(t) - \tilde{u}_{R3}(s) \|_{L^\infty}}{(t-s)^{\varkappa_1}} + \sup_{s < t \leq T} s^{\frac{5}{12} + \varkappa_2} \frac{\| \tilde{u}_{R3}(t) - \tilde{u}_{R3}(s) \|_{L^{12}}}{(t-s)^{\varkappa_2}} \\ &\qquad \lesssim \|\rho_0\|_{C^1}^2\mathbf{C}^3(u)\left(T^{\frac{3}{8}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{1}{4}}\|\rho_0-1\|_{L^\infty}\right) \end{align}\tag{96}\]

(iii) Estimates for the \(\mathrm{BMO}^{-1}\) norm.

Finally let us turn to the \(\mathrm{BMO}^{-1}\) type estimates. For any \(x_0 \in \mathbb{R}^3\), by virtue of 36 , we write \[\begin{align} u_{R3,x_0} =& \int_0^t \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K_{x_0}(t-\tau, x-y) (\rho_0(x_0)-\rho_0(y)) u(\tau,y) \, dy \, d\tau \\ &+ (\rho_0(x_0)-\rho_0(x)) u(t,x) - \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} K_{x_0}(t, x-y) (\rho_0(x_0)-\rho_0(y)) u_0(y) \, dy \\ =& (\rho_0(x_0)-\rho_0(x)) \Bigl( \int_0^t \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K_{x_0}(t-\tau, x-y) u(\tau,y) \, dy \, d\tau + u(x) - K_{x_0}(t) \ast u_0(x) \Bigr) \\ &+ \left(\int_0^t \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_\tau K_{x_0}(t-\tau, x-y) (\rho_0(x)-\rho_0(y)) u(\tau,y) \, dy \, d\tau \right.\\ &\left. - \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} K_{x_0}(t, x-y) (\rho_0(x)-\rho_0(y)) u_0(y) \, dy\right) \\ =:& (\rho_0(x_0)-\rho_0(x)) \sum_{i=1}^3 u_{R3,x_0}^{1i} + u_{R3,x_0}^2. \end{align}\] Since we still can write \[\begin{align} &\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} K_{x_0}(t, x-y) (\rho_0(x)-\rho_0(y)) u_0(y) \, dy\\ &=\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}K_{x_0}(t,x-y)(\rho_0(x_0)-\rho_0(y))\left((\rho_0u_0(y)-(\rho u)(t,y))+(\rho u)(t,y)\right)dy\\ &\quad+\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}K_{x_0}(t,x-y)(\rho_0(x_0)-\rho_0(y))\left((1-\rho_0)u_0(y)\right)dy\\ &=\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\nabla_y\left(K_{x_0}(t,x-y)(\rho_0(x_0)-\rho_0(y))\right)\cdot\int_0^t\left(\nabla u-\rho u\otimes u\right)(\tau,y)d\tau dy\\ &\quad+\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}\left(K_{x_0}(t,x-y)(\rho_0(x_0)-\rho_0(y))\right)\cdot\int_0^t\nabla P(\tau,y)d\tau dy\\ &\quad+\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3}K_{x_0}(t,x-y)(\rho_0(x_0)-\rho_0(y))\left((\rho u)(t,y)+(1-\rho_0)u_0(y)\right)dy, \end{align}\] we can get by a similar derivation of 93 that \[\sup_{t \leq T} t^{\frac{5}{12}} \| u_{R3,x_0}^2(t) \|_{L^\infty} \lesssim \mathbf{C}^2(u)\|\rho_0\|_{C^1}^2\left(T^{\frac{3}{8}} \|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{1}{4}}\|\rho_0-1\|_{L^\infty}\right).\] It remains to handle the estimate of the first term. Note that for any \(z \in \mathbb{R}^3\) and \(x \in B(z,R)\), we have \[|\rho_0(z) - \rho_0(x)| \lesssim R \|\nabla \rho_0\|_{L^\infty}.\] By the definitions of \(u_{R3,z}^{12}\), \(u_{R3,z}^{13}\) and the \(X_T\) norm, we have \[R^{-\frac{3}{2}} \Bigl( \int_0^{R^2} \int_{B(z,R)} | (\rho_0(z)-\rho_0(x)) (u_{R3,z}^{12} + u_{R3,z}^{13})(t,x) |^2 \, dx \, dt \Bigr)^{\frac{1}{2}} \lesssim R^{\frac{1}{2}} \|\rho_0\|_{C^1} \mathbf{C}(u).\] By Lemma 2 and Lemma 3, 5, for \(u_{R3,z}^{11}\), we have \[\begin{align} & \sup_{t\leq T}t^{\frac{1}{2}}\|u_{R3,z}^{11}(t)\|_{L^\infty}\lesssim \|u\|_{X_T},\\ & \sup_{t\leq T}\|u_{R3,z}^{11}(t)\|_{L^2}\lesssim \sup_{t\leq T}\|u(t)\|_{L^2}+\sup_{s<t\leq T}s^{\eta}\frac{\|u(t)-u(s)\|_{L^2}}{(t-s)^\eta}\lesssim \mathbf{C}^3(u) , \end{align}\] for some \(\eta\in(0,1)\). Then for \(x\in B(z,R)\), we have \[\begin{align} &R^{-\frac{3}{2}} \Bigl( \int_0^{R^2} \int_{B(z,R)} | (\rho_0(z)-\rho_0(x)) u_{R3,z}^{11}(t,x) |^2 \, dx \, dt \Bigr)^{\frac{1}{2}} \\ &\quad \lesssim R^{\frac{1}{4}} \|\rho_0\|_{C^1} \Bigl( \int_0^{R^2} \| u_{R3,z}^{11}(t) \|_{L^4}^2 \, dt \Bigr)^{\frac{1}{2}} \\ &\quad \lesssim R^{\frac{3}{4}} \|\rho_0\|_{C^1} \mathbf{C}^3(u)\|u\|_{X_T}^{\frac{1}{2}}. \end{align}\] Since \(R < b\), for \(T \sim b^2\) we obtain the following estimate for the \(\mathrm{BMO}^{-1}\) norm: \[\sup_{R \leq b} \sup_{z \in \mathbb{R}^3} R^{-\frac{3}{2}} \Bigl( \int_0^{R^2} \int_{B(z,R)} | u_{R3,z}(t,x) |^2 \, dt \, dx \Bigr)^{\frac{1}{2}} \lesssim CT^{\frac{3}{8}} \|\rho_0\|_{C^1} \mathbf{C}^3(u)\|u\|_{X_T}^{\frac{1}{2}}.\] Combining this with 93 and 96 completes the proof. ◻

Proof. Throughout all of our proof, the constant will depend on \(C_1\) and \(C_2\), and for simplicity we omit them. In view of 7 , we get, by applying 9 and Propositions 4, 5, 6, 8 and 9, that for any \(T\leq 1\), \[\begin{align} \|u\|_{X_T}\leq C\|\rho_0\|_{C^1}\Bigl(&\|u_0\|_{\mathrm{BMO}_b^{-1}} + \|u_0\|_{\mathrm{BMO}_b^{-1}}^{\frac{5}{6}} \|u_0\|_{L^2}^{\frac{1}{6}}+\|u\|_{X_T}^2\label{S5eq6}\\ &+T^{\frac{1}{4}}\bigl(1+\|u_0\|_{L^2}+\|u\|_{X_T}\bigr)^3\Bigr).\nonumber \end{align}\tag{97}\] We define \[\label{tdkmwyaz} \begin{align} \varepsilon_0=\min\left\{(100C(1+\|\rho_0\|_{C^1}))^{-1}(1+\|u_0\|_{L^2})^{-1},\|u_0\|_{L^2}\right\}. \end{align}\tag{98}\] Since \(u_0\in VMO^{-1}\), there exists \(b_*\) such that \(\|u_0\|_{BMO_{b_*}^{-1}}\leq \varepsilon_0\). Take \[\begin{align} &b=\min\left\{(\frac{1}{10}(2+\|u_0\|_{L^2})^{-3}\varepsilon_0)^{10},\frac{b_*}{2}\right\},\\ &T^*=4C_2b^2. \end{align}\] Then we define \[T_0:=\sup_{t\leq T^*}\{\|u\|_{X_T}\leq 8C\varepsilon_0^{\frac{5}{6}}\|u_0\|_{L^2}^{\frac{1}{6}}\|\rho_0\|_{C^1}\}\] We want to prove that \(T_0=T^*\). If \(T_0<T^*\), then by 97 and ?? , we have \[\|u\|_{X_{T_0}}\leq 2C\|\rho_0\|_{C^1}\varepsilon_0^{\frac{5}{6}}\|u_0\|_{L^2}^{\frac{1}{6}}+64C^2\|\rho_0\|_{C^1}^2\varepsilon_0^{\frac{5}{3}}\|u_0\|_{L^2}^{\frac{1}{3}}+\frac{1}{10}C\|\rho_0\|_{C^1}\varepsilon_0\leq 4C\varepsilon_0^{\frac{5}{6}}\|u_0\|_{L^2}^{\frac{1}{6}}\|\rho_0\|_{C^1}.\] This implies that \(T_0=T^*\), so for any \(T\leq T^*\), if \(\|u_0\|_{BMO^{-1}_b}\leq \varepsilon_0\) with \(\varepsilon_0,T^*,b\) above, there holds \[\begin{align} \label{S5eq2} \|u\|_{X_T}\leq 8C\|\rho_0\|_{C^1}\varepsilon_0^{\frac{5}{6}}\|u_0\|_{L^2}^{\frac{1}{6}}, \end{align}\tag{99}\] from which and ?? , we conclude the proof of Proposition 10. ◻

Proof. First, we take smooth approximating sequences \((\rho_{0,\varepsilon}, u_{0,\varepsilon})\) such that as \(\varepsilon \to 0\), \[\begin{align} &u_{0,\varepsilon} \to u_0 \quad \text{in } L^2 \cap \mathrm{VMO}^{-1}, \qquad \rho_{0,\varepsilon} \rightharpoonup \rho_0 \quad \text{in } L^\infty \text{ weak-}* \quadand\quad\\&\rho_{0,\varepsilon} \to \rho_0 \quad \text{in } L^p_{\rm{loc}}(\mathop{\mathbb{R}\kern 0pt}\nolimits^3)\quad for any\;p<\infty. \end{align}\] Then, by the classical well-posedness theory for the inhomogeneous Navier–Stokes equations, there exists a unique solution \((\rho^\varepsilon, u^\varepsilon)\) on \([0, T^\varepsilon]\). Furthermore, we deduce from Proposition 10 that \(T^\varepsilon\) has a uniform lower bound, which we denote by \(T^\star\), so that ?? holds on \([0,T^\star]\) for the approximate solutions \((\rho^\varepsilon, u^\varepsilon).\)

By the a priori estimates ?? and ?? for \((\rho^\varepsilon, u^\varepsilon)\), there exists a subsequence (still denoted by \((\rho^\varepsilon, u^\varepsilon)\)) such that \((\rho^\varepsilon, u^\varepsilon)\) converges to \((\rho, u)\) in the weak-\(*\) sense on \([0,T^\star]\). Then we deduce from the proof of Theorem 2.1 of [2] that thus obtained \((\rho,u)\) is indeed a weak solution of the system 1 . Furthermore, the solution satisfy ?? and ?? . This completes the proof of Theorem 1. ◻

Proof of Theorem 2. By applying Proposition 4, 7–9, we can prove that there exists a constant \(C_0<1\), such that for any \(T,b < C_0\), there holds \[\begin{align} \|u\|_{X_T}\leq C\|\rho_0\|_{C^2}\Bigl(&\|u_0\|_{\mathrm{BMO}^{-1}} + \|u_0\|_{\mathrm{BMO}^{-1}}^{\frac{5}{6}} \|u_0\|_{L^2}^{\frac{1}{6}}+\|u\|_{X_T}^2\label{S6eq2}\\ &+C\|\rho_0\|_{C^2} ^2\mathbf{C}^3(u)\left(T^{\frac{1}{4}}\left(\|u\|_{X_T}+\|\rho_0-1\|_{L^\infty}\right)+T^{\frac{3}{8}}\|u\|_{X_T}^{\frac{1}{2}}+T^{\frac{5}{8}}\right).\nonumber \end{align}\tag{100}\] We define \[\begin{align} \varepsilon_0=\min\left\{(100C(1+\|\rho_0\|_{C^2}))^{-1}(1+\|u_0\|_{L^2})^{-1},\|u_0\|_{L^2}\right\}. \end{align}\] Then, for any \(\varepsilon<\varepsilon_0\), define \[T_\varepsilon:=\varepsilon^{\frac{4}{3}}\|u_0\|_{L^2}^{\frac{2}{9}}\left(10C(2+\|\rho_0\|_{C^2}+\|u_0\|_{L^2})\right)^{-10}.\] Let \[T_\varepsilon^0=\sup\left\{T\leq T_\varepsilon: \|u\|_{X_T}\leq 8C\|\rho_0\|_{C^2}\varepsilon^{\frac{5}{6}}\|u_0\|_{L^2}^{\frac{1}{6}}\right\}.\] By 100 and the definition of \(T_\varepsilon\), \(\varepsilon_0\), if \(\|\rho_0-1\|_{L^\infty}+\|u_0\|_{BMO^{-1}}\leq \varepsilon<\varepsilon_0\), \(T_\varepsilon^0<T_\varepsilon\), then \[\begin{align} \|u\|_{X_{T_\varepsilon^0}}&\leq 2C\|\rho_0\|_{C^2}\varepsilon^{\frac{5}{6}}\|u_0\|_{L^2}^{\frac{1}{6}}+\frac{1}{100}\|u\|_{X_{T_\varepsilon^0}}+3C\|\rho_0\|_{C^2}\varepsilon^{\frac{5}{6}}\|u_0\|_{L^2}^{\frac{1}{6}}\\ &\leq 7C\|\rho_0\|_{C^2}\varepsilon^{\frac{5}{6}}\|u_0\|_{L^2}^{\frac{1}{6}}. \end{align}\] This implies \(T_\varepsilon^0=T_\varepsilon\). Then by definition of \(\|\cdot\|_{X_T}\), we have \[\label{S6eq1} \|u(\frac{T_\varepsilon}{2})\|_{L^\infty}\leq (\frac{T_\varepsilon}{2})^{-\frac{1}{2}} \|u\|_{X_{T_\varepsilon}}\lesssim \varepsilon^{\frac{5}{6}-\frac{2}{3}}\|u_0\|_{L^2}^{\frac{1}{6}-\frac{1}{9}}\|\rho_0\|_{C^2},\tag{101}\] which can be arbitrarily small if we take \(\varepsilon\) small.

This, together with the fact that \(\|\rho(t) - 1\|_{L^\infty} = \|\rho_0 - 1\|_{L^\infty}\) and \(\|u(t)\|_{L^2}\leq \|u_0\|_{L^2},\) implies that for any \(3 < p < \infty,\) \[\label{smallness} \left\| \mathbb{P}\!\left( \rho\!\left( \frac{T^\ast}{2} \right) u\!\left( \frac{T^\ast}{2} \right) \right) \right\|_{\dot{B}_{p,\infty}^{-1 + \frac{3}{p}}} \leq C \left\| u\!\left( \frac{T^\ast}{2} \right) \right\|_{L^3} \leq C\|u_0\|_{L^2}^{\frac{2}{3}} \left\| u\!\left( \frac{T^\ast}{2} \right) \right\|_{L^\infty}^{\frac{1}{3}}.\tag{102}\] Therefore, we can take \(\varepsilon_2\) sufficiently small in ?? , such that if \(\varepsilon<\varepsilon_2\), then 101 and 102 lead to \(\left\| \mathbb{P}\!\left( \rho\!\left( \frac{T^\ast}{2} \right) u\!\left( \frac{T^\ast}{2} \right) \right) \right\|_{\dot{B}_{p,\infty}^{-1 + \frac{3}{p}}} \leq \varepsilon_1\) in Theorem 3, then the system 1 with initial data at \(T^\ast/2\) generates a global solution. This completes the proof of Theorem 2. ◻

Proof. Taking the \(L^2\) inner product of the momentum equation with \(u\) and using \(\nabla \cdot u = 0\), we obtain \[\frac{1}{2} \frac{d}{dt} \left( \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \rho(t,x) |u(t,x)|^2 \, dx \right) + \|\nabla u(t)\|_{L^2}^2 = 0.\] Integrating in time and applying 103 yields the desired estimate. ◻

Proof. Taking the \(L^2\) inner product of the momentum equation with \(\partial_t u\), we obtain \[\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \rho |\partial_t u|^2 \, dx + \frac{1}{2} \frac{d}{dt } \|\nabla u(t)\|_{L^2}^2 = -\int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \rho \operatorname{div}(u \otimes u) \cdot \partial_t u \, dx.\] Multiplying both sides by \(t\) and applying Hölder’s inequality yields \[\begin{align} \frac{d}{dt}\bigl( t \|\nabla u(t)\|_{L^2}^2 \bigr) + t \|\sqrt{\rho} \partial_t u\|_{L^2}^2 \lesssim & \|\nabla u\|_{L^2}^2 + t \|\rho u \cdot \nabla u\|_{L^2}^2\\ \lesssim &\|\nabla u\|_{L^2}^2 +C_1^2 t \|u(t)\|_{L^\infty}^2 \|\nabla u\|_{L^2}^2. \end{align}\] Integrating over \((0, T)\) and using 103 together with Lemma 3 completes the proof. ◻

Proof. Differentiating 1 in time gives \[\rho \partial_{tt} u - \partial_t \Delta u = -\partial_t \rho \, \partial_t u - \partial_t \rho \, \nabla \cdot (u \otimes u) - \rho \partial_t u \cdot \nabla u - \rho u \cdot \nabla \partial_t u + \nabla \partial_t P.\] Taking the \(L^2\) inner product with \(\partial_t u\) and multiplying the resulting inequality by \(t^2\), we obtain \[\label{ener} \begin{align} &\frac{1}{2} \frac{d}{dt} \bigl( t^2 \|\sqrt{\rho} \partial_t u(t)\|_{L^2}^2 \bigr) + t^2 \|\partial_t \nabla u\|_{L^2}^2 = t \|\sqrt{\rho} \partial_t u(t)\|_{L^2}^2 \\ &\quad- t^2 \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_t \rho(t,x) |\partial_t u(t,x)|^2 \, dx - t^2 \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \partial_t \rho(t,x) (u \otimes \partial_t u)(t,x) : \nabla u(t,x) \, dx \\ &\quad - t^2 \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \rho(t,x) (\partial_t u \otimes \partial_t u)(t,x) : \nabla u(t,x) \, dx \\ &:= I(t) + II(t) + III(t) + IV(t). \end{align}\tag{104}\] By Lemma 3, Lemma 4, and 103 , we have \[\label{eng33} \begin{align} \int_0^T \bigl( |I(t)| + |IV(t)| \bigr) \, dt \lesssim & \|t^{\frac{1}{2}}\sqrt{\rho} \partial_t u(t)\|_{L^2_TL^2}^2 +C_1\|t\nabla u(t)\|_{L^\infty}\|t^{\frac{1}{2}}\sqrt{\rho} \partial_t u(t)\|_{L^2_TL^2}^2\\ \leq &C(1+ M)^{2} \|u_0\|_{L^2}^2, \end{align}\tag{105}\] where we used 103 and interpolation inequality so that \[\|t\nabla u(t)\|_{L^\infty}\lesssim \|t^{\frac{1}{2}}u(t)\|_{L^\infty}^{\frac{\eta{1+\eta}}{}}\|t^{1+\frac{\eta}{2}}u(t)\|_{\dot{C}^{1+\eta}}^{\frac{1}{1+\eta}}\lesssim M.}\]

For \(II\), note that by the transport equation for \(\rho\), the Cauchy–Schwarz inequality, and 103 , we obtain \[\begin{align} |II(t)| &= \left| 2t^2 \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \rho(t,x) (u \otimes \partial_t u)(t,x) : \partial_t \nabla u(t,x) \, dx \right| \\ &\leq C_2t^{\frac{1}{2}}\|u(t)\|_{L^\infty}\|t^{\frac{1}{2}}\partial_tu(t)\|_{L^2}\|t\nabla\partial_t u(t)\|_{L^2}\\ &\leq \frac{1}{8} t^2 \|\partial_t \nabla u(t)\|_{L^2}^2 + C M^2 t \|\partial_t u(t)\|_{L^2}^2, \end{align}\] which together with Lemma 4 yields \[\label{eng34} \begin{align} \int_0^T |II(t)| \, dt \leq \frac{1}{8} \| t \partial_t \nabla u(t) \|_{L_T^2 L^2}^2 + C (1+M)^2M^{2} \|u_0\|_{L^2}^2. \end{align}\tag{106}\]

For \(III\), using the transport equation for \(\rho\) in 1 , we have \[\begin{align} III &= t^2 \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \operatorname{div}(\rho u) (u \otimes \partial_t u) : \nabla u \, dx \\ &= -t^2 \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \rho (u \cdot \nabla u \otimes \partial_t u) : \nabla u \, dx - t^2 \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \rho (u \otimes u \cdot \nabla \partial_t u) : \nabla u \, dx \\ &\quad - t^2 \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \rho (u \otimes \partial_t u) : (u \cdot \nabla (\nabla u)) \, dx \\ &:= III_1 + III_2 + III_3. \end{align}\] By 103 , we deduce \[\begin{align} |III_1(t)| &\lesssim \|\sqrt{t} u(t)\|_{L^\infty} \|t \nabla u(t)\|_{L^\infty} \|\sqrt{t} \partial_t u(t)\|_{L^2} \|\nabla u(t)\|_{L^2} \\ &\lesssim M^2 \|\sqrt{t} \partial_t u\|_{L^2} \|\nabla u\|_{L^2}. \end{align}\] Then Lemma 3 and Lemma 4 lead to \[\label{III1} \int_0^T |III_1(t)| \, dt \lesssim (1+M)^2M^{2} \|u_0\|_{L^2}^2.\tag{107}\] By the Cauchy–Schwarz inequality, 103 , and Lemma 3, we obtain \[\label{III2} \int_0^T |III_2(t)| \, dt \leq C M^2 \int_0^T \|\nabla u\|_{L^2} \|t \nabla \partial_t u\|_{L^2} \, dt \leq \frac{1}{16} \| t \nabla \partial_t u \|_{L_T^2 L^2}^2 + C M^{4} \|u_0\|_{L^2}^2.\tag{108}\] For \(III_3\), we first need an estimate for \(\|\nabla^2 u(t)\|_{L^2}\). From 1 , we have \[-\Delta u = -\mathbb{P} (\rho \partial_t u + \rho u \cdot \nabla u).\] Thus, by singular integral theory, together with 103 , Lemma 3, and Lemma 4, we obtain \[\begin{align} \|\sqrt{t} \nabla^2 u\|_{L_T^2 L^2} &\lesssim \|\sqrt{t} \partial_t u\|_{L_T^2 L^2} + \|\sqrt{t} u\|_{L_T^\infty L^\infty} \|\nabla u\|_{L_T^2 L^2} \\ &\lesssim (1+M) \|u_0\|_{L^2}^2, \end{align}\] which implies \[\label{III3} \begin{align} \int_0^T |III_3(t)| \, dt &\lesssim \|\sqrt{t} u\|_{L_T^\infty L^\infty}^2 \|\sqrt{t} \partial_t u\|_{L_T^2 L^2} \|\sqrt{t} \nabla^2 u\|_{L_T^2 L^2} \\ &\lesssim (1+M)^2M^{2} \|u_0\|_{L^2}^2. \end{align}\tag{109}\] It follows from 107 , 108 , and 109 that \[\label{eng35} \begin{align} \int_0^T |III(t)| \, dt \leq \frac{1}{16} \| t \nabla \partial_t u \|_{L_T^2 L^2}^2 + C (1+M)^2M^{2} \|u_0\|_{L^2}^2. \end{align}\tag{110}\] Substitutining 105 , 106 , and 110 into 104 yields the desired result. ◻

Proof. First, for any \(R < b\), we split the convolution integral into annuli \(C(0, 2^p R, 2^{p+1} R)\), which yields \[\begin{align} |\Gamma_R^1 \ast f(t,x)| &\leq \sum_{p=0}^\infty \int_0^t \int_{C(0, 2^p R, 2^{p+1} R)} \frac{1}{|y|^{3+\alpha}} |f(\tau, x-y)| \, dy \, d\tau. \end{align}\] Note that \(C(0, 2^p R, 2^{p+1} R)\) can be covered by balls \(B(x', R)\) with center \(x'\) and radius \(R\), and the number of such \(x'\) is approximately \(2^{3p}\) for any \(t \leq R^2\). Consequently, \[|\Gamma_R^1 \ast f(t,x)| \leq \sum_{p=0}^\infty (2^p R)^{-3-\alpha} 2^{3p} \sup_{x'} \int_0^t \int_{B(x', R)} |f(\tau, x-y)| \, dy \lesssim \frac{C}{R^\alpha} \|f\|_{Z_T^{2,b}}.\]

Second, we write \[\begin{align} |\Gamma_R^2 \ast f(t,x)| &\lesssim \left( \int_{\min\{t/2, R^2\}}^t + \int_0^{\min\{R^2, t/2\}} \right) \int_{B_0(R)} \frac{1}{(\sqrt{t-\tau} + |x-y|)^{3+\alpha}} |f(\tau, y)| \, dy \, d\tau \\ &:= \Gamma_R^{21}(f) + \Gamma_R^{22}(f). \end{align}\] For \(\Gamma_R^{22}(f)\), since in this regime \(\frac{1}{(\sqrt{t-\tau} + |x-y|)^{3+\alpha}} \lesssim \frac{1}{t^{\frac{3+\alpha}{2}}}\), we obtain \[|\Gamma_R^{22}(f)| \lesssim \frac{R^3}{t^{\frac{3+\alpha}{2}}} \left( \frac{1}{R^3} \int_0^{R^2} \int_{B_0(R)} |f(\tau, y)| \, dy \, d\tau \right) \lesssim \frac{R^3}{t^{\frac{3+\alpha}{2}}} \|f\|_{Z_T^{2,b}} \lesssim \frac{1}{R^\alpha} \|f\|_{Z_T^{2,b}}, \quad \forall t \geq R^2.\] For the remaining part, note that \[\int_{B_0(R)} \frac{1}{(a + |y|)^{3+\alpha}} \, dy \lesssim \frac{1}{a^\alpha} \int_{\mathbb{R}^3} \frac{1}{(1 + |y|)^{3+\alpha}} \, dy.\] Hence, \[\begin{align} |\Gamma_R^{21}(f)| &\lesssim \int_{\min\{t/2, R^2\}}^t \int_{B_0(R)} \frac{1}{(\sqrt{t-\tau} + |x-y|)^{3+\alpha}} \|f(\tau)\|_{L^\infty} \, dy \, d\tau \\ &\lesssim \|f\|_{Z_T^{2,b}} \left( \int_{R^2}^t \frac{1}{(t-\tau)^{\frac{\alpha}{2}}} \frac{d\tau}{\tau} + \int_{t/2}^t \frac{R^3}{t^{\frac{3+\alpha}{2}}} \frac{d\tau}{\tau} \right) \\ &\lesssim \frac{1}{R^\alpha} \|f\|_{Z_T^{2,b}}, \quad \forall t \geq R^2,\; 0 < \alpha < 2. \end{align}\] This completes the proof. ◻

Proof. Without loss of generality, we may assume that \(x \in B_0(R)\). For any \(R < b\), we have \[\label{conbdd} \begin{align} |f_\theta(t,x)| &\lesssim t^{-\frac{3}{2}} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \theta\!\left(\frac{x-y}{\sqrt{t}}\right) |\mathbf{1}_{B_0(2R)} f(t,y)| \, dy \\ &\quad + t^{-\frac{3}{2}} \int_{\mathop{\mathbb{R}\kern 0pt}\nolimits^3} \frac{1}{(1 + |x-y|/\sqrt{t})^4} \frac{t}{R^2} |f(t,y)| \, dy \\ &\lesssim t^{-\frac{3}{2}} \left| \theta\!\left(\frac{\cdot}{\sqrt{t}}\right) \ast \mathbf{1}_{B_0(2R)} f(t,\cdot) \right| + \frac{1}{R^2} \sup_{t \leq b^2} t \|f(t)\|_{L^\infty}. \end{align}\tag{112}\] Integrating over \([0,R^2] \times B_0(R)\), we obtain \[\frac{1}{R^3} \|f_\theta\|_{L^1([0,R^2] \times B_0(R))} \lesssim \|f\|_{Z_T^{2,b}}.\] The \(L^\infty\) estimate also follows from 112 for any \(t < b^2\). ◻

Proof. We decompose \(f\) into low and high frequencies. Precisely, let \(\theta\) be a function whose Fourier transform \(\widehat{\theta} \in C_c^\infty(\mathbb{R}^3)\) satisfies \(\widehat{\theta}|_{B_0(1)} \equiv 1\). Write \[f = f^l + f^h, \qquad f^l = \mathcal{F}^{-1}\bigl( \widehat{\theta}(t^{1/2} \cdot) \widehat{f}(t,\cdot) \bigr)(x).\] Then \[\begin{align} \|f^h\|_{L^2([0,R^2]; \dot{H}^{-1})}^2 &\lesssim \int_0^{R^2}\int_{ \mathbb{R}^3} \frac{|1 - \widehat{\theta}(\sqrt{t} \xi)|^2}{t |\xi|^2} \, t |\widehat{f}(t,\xi)|^2 \, d\xi\, dt \\ &\lesssim \int_0^{R^2} t \|f(t)\|_{L^2}^2 \, dt \lesssim \|f\|_{L^1([0,R^2] \times \mathbb{R}^3)} \sup_{t < R^2} t \|f(t)\|_{L^\infty}, \end{align}\] and then the classical energy estimate for the Stokes equations yields \[\label{4461} \|L f^h\|_{L^2([0,R^2] \times \mathbb{R}^3)}^2 \lesssim R^3 \|f\|_{Z_T^{2,b}}^2.\tag{113}\] The hardest part is \(L f^l\). Observe that if we define \[\widetilde{f^l}(t,x) = \mathcal{F}^{-1}\bigl( e^{t|\cdot|^2} \widehat{\theta}(\sqrt{t} |\cdot|) \widehat{f}(t,\cdot) \bigr)(x),\] then \[L f^l = \nabla \int_0^t e^{(t-\tau)\Delta} \Bbb{P}f^l(\tau) \, d\tau = \nabla e^{t\Delta} \int_0^t \Bbb{P}\widetilde{f^l}(\tau) \, d\tau,\] where \(\Bbb{P}\) denotes the Leray projection operator to the divergence free vector field space. Since \(\widehat{\theta}\) is compactly supported, there exists a Schwartz function \(\widetilde{\theta}\) such that \[\label{formulalow} \widetilde{f^l} = t^{-\frac{3}{2}} \widetilde{\theta}\!\left(\frac{\cdot}{\sqrt{t}}\right) \ast f(t,\cdot).\tag{114}\] Direct calculation gives \[\label{4462} \begin{align} \mathcal{L}f &:= \|L f^l\|_{L^2([0,R^2] \times \mathbb{R}^3)}^2 \\ &= 2 \int_0^{R^2} \int_0^t \int_0^s \bigl\langle \nabla e^{t\Delta}\Bbb{P}\widetilde{f^l}(s), \nabla e^{t\Delta} \Bbb{P}\widetilde{f^l}(\tau) \bigr\rangle \, d\tau \, ds \, dt \\ &= -2 \int_0^{R^2} \int_0^t \int_0^s \bigl\langle \Delta e^{2t\Delta} \Bbb{P} \widetilde{f^l}(\tau),\widetilde{f^l}(s) \bigr\rangle \, d\tau \, ds \, dt \\ &= -\int_0^{R^2} \int_0^s \left\langle \left( \int_s^{R^2} \frac{d}{dt} e^{2t\Delta} \, dt \right) \Bbb{P} \widetilde{f^l}(\tau), \widetilde{f^l}(s) \right\rangle \, d\tau \, ds \\ &= \int_0^{R^2} \left\langle (e^{2s\Delta} - e^{2R^2\Delta}) \int_0^s \Bbb{P}\widetilde{f^l}(\tau) \, d\tau, \widetilde{f^l}(s) \right\rangle \, ds \\ &\lesssim \| \widetilde{f^l} \|_{L_T^1([0,R^2] \times \mathbb{R}^3)} \sup_{s < R^2} \left\| (e^{2s\Delta} - e^{2R^2\Delta}) \Bbb{P} \int_0^s \widetilde{f^l}(\tau) \, d\tau \right\|_{L^\infty}. \end{align}\tag{115}\] A simple consequence of 114 are that \[\begin{align} \| \widetilde{f^l} \|_{L_T^1([0,R^2] \times \mathbb{R}^3)} \lesssim \| {f^l} \|_{L_T^1([0,R^2] \times \mathbb{R}^3)} \lesssim \|f\|_{Z_T^{2,b}}, \end{align}\] and \[\begin{align} \label{4463} \left\| e^{2R^2\Delta} \int_0^s \widetilde{f^l}(\tau) \, d\tau \right\|_{L^\infty} \lesssim &R^{-3} \left\| e^{2R^2\Delta} \int_0^s \widetilde{f^l}(\tau) \, d\tau \right\|_{L^1}\\ \lesssim & R^{-3} \|f\|_{L^1([0,R^2] \times \mathbb{R}^3)} \lesssim \|f\|_{Z_T^{2,b}}. \end{align}\tag{116}\] For the remaining term, note that \[\label{4464} \left| e^{2s\Delta} \int_0^s \widetilde{f^l}(\tau, x) \, d\tau \right| \leq s^{-\frac{3}{2}} \sum_{{\boldsymbol{n}} \in \mathbb{Z}^3} \int_{B({\boldsymbol{n}}\sqrt{s},\sqrt{s})} \int_0^s e^{-\frac{|x-y|^2}{2s}} \mathbf{1}(y) \left| \widetilde{f^l}(\tau, y) \right| \, d\tau \, dy,\tag{117}\] where \(B({\boldsymbol{n}}\sqrt{s}, \sqrt{s})\) denotes the ball with center \({\boldsymbol{n}}\sqrt{s}\) and radius \(\sqrt{s}\). Using translation invariance, we may assume without loss of generality that \(x = 0\). Then \[\label{4465} \begin{align} \left| e^{2s\Delta} \int_0^s \widetilde{f^l}(\tau) \, d\tau (0)\right| \lesssim &\sum_{|{\boldsymbol{n}}| > 2} e^{-\frac{|{\boldsymbol{n}}|^2}{4}} s^{-\frac{3}{2}} \int_0^s \int_{B({\boldsymbol{n}}\sqrt{s},\sqrt{s})} |\widetilde{f^l}(\tau, y)| \, dy \, d\tau \\ &+ \sum_{|{\boldsymbol{n}}| \leq 2} s^{-\frac{3}{2}} \int_0^s \int_{B({\boldsymbol{n}}\sqrt{s}, \sqrt{s})} |\widetilde{f^l}(\tau, y)| \, dy \, d\tau \\ &\lesssim \Bigl(1 + \sum_{{\boldsymbol{n}} \in \mathbb{Z}^3} e^{-\frac{|n|^2}{4}}\Bigr) \| \widetilde{f^l} \|_{Z_T^{2,b}} \lesssim \|f\|_{Z_T^{2,b}}, \end{align}\tag{118}\] where the last inequality follows from Lemma 7. Here we also used \(\sqrt{s} < R < b\), and the fact that \(\int_0^s \int_{B{(\boldsymbol{n}}\sqrt{s},\sqrt{s})} |\widetilde{f^l}(\tau, y)| \, dy \, d\tau \lesssim s^{\frac{3}{2}} \| \widetilde{f^l} \|_{Z_T^{2,b}}\) for any \(n\). Combining 115 –118 , we obtain \[\|L f^l\|_{L^2([0,R^2] \times \mathbb{R}^3)}^2 \lesssim R^3 \|f\|_{Z_T^{2,b}}^2,\] which together with 113 completes the proof of Lemma 8. ◻

Proof of Lemma 1. Obviously, it suffices to prove that \(L\) (defined in 111 ) maps \(Z_T^{2,b}\) continuously into \(Z_T^{1,b}\). By the Duhamel principle, \(Lf\) admits the explicit representation (see [29] for instance) \[(L^m f)^k(t,x) = \sum_{\ell=1}^3 \int_0^t \int \Gamma_{m,\ell}^k(t-\tau, x-y) f^\ell(\tau, y) \, dy \, d\tau,\] where \(\Gamma_{j,\ell}^k\) is a kernel satisfying \[|\Gamma_{j,\ell}^k(t,x)| \lesssim \Gamma_R^1(t,x) + \Gamma_R^2(t,x), \quad \forall R > 0,\] with \(\Gamma_R^1\) and \(\Gamma_R^2\) defined in 28 . By Lemma 6 with \(\alpha = 1\) and \(R = \sqrt{t}\), we obtain \[\|Lf(t)\|_{L^\infty} \lesssim t^{-\frac{1}{2}} \|f\|_{Z_T^{2,b}}, \quad \forall t < b^2.\] Furthermore, we need to estimate the local \(L^2\) norm. Without loss of generality, we may restrict our attention to a neighborhood of the origin. Note that \[Lf = L(\mathbf{1}_{B(0,2R)} f) + L(\mathbf{1}_{B^c(0,2R)} f).\] For the exterior part, note that when \(|x| < R\) and \(|y| > 2R\), we have \(\frac{1}{(\sqrt{t} + |x-y|)^4} \lesssim \frac{1}{(\sqrt{t} + |y|)^4}\). The first estimate in Lemma 6 then yields \[R^{-\frac{3}{2}} \| L(\mathbf{1}_{B^c(0,2R)} f) \|_{L^2([0,R^2] \times B(0,R))} \lesssim R \| L(\mathbf{1}_{B^c(0,2R)} f) \|_{L^\infty([0,R^2] \times B(0,R))} \lesssim \|f\|_{Z_T^{2,b}}, \quad \forall R < b.\] The corresponding estimate for \(L(\mathbf{1}_{B_0(2R)} f)\) follows from Lemma 8. We thus complete the proof of Lemma 1. ◻