Caffarelli-Kohn-Nirenberg Theory

The first claim is clear:
$$\begin{array}{rl}\varphi (\theta r,v)=& \theta r\cdot |Q_{\theta r}{|}^{-1}{\int }_{Q_{\theta r}}(v-v_r)+\theta r{v}_r\\ \le & C^{\prime }\theta r\cdot |Q_{\theta r}{|}^{1-\frac{1}{3}-1}{\Vert v-v_r\Vert }_{L^3(Q_{\theta r})}+\theta r{v}_r\\ \le & C^{\prime }{\theta }^{-\frac{2}{3}}\phi (r,v)+\theta \varphi (r,v)\\ \le & C^{\prime }{ϵ}_2{\theta }^{-\frac{2}{3}}+\frac{1}{2}\le 1\end{array}$$if ${ϵ}_2\le {\theta }^{\frac{2}{3}}/(2C^{\prime })$. As for the latter, we prove it by contradiction: if not, then $\mathrm{\forall }C>0,\mathrm{\exists }\beta <\gamma ,\theta \in (0,\frac{1}{2}]$, $\mathrm{\forall }{ϵ}_2,r_2>0$(where we set both as ${\displaystyle \frac{1}{n}}$), if $\mathrm{\exists }{r}_n\le {\displaystyle \frac{1}{n}}$, $(v_n,p_n)$ are SWS of $\text{(NS)}$ in $Q_{r_n}$ such that
$$\psi (r_n,v_n)\le 1,{ϵ}_n:=\phi (r_n,v_n,p_n)+\psi (r_n,f_n)\le \frac{1}{n},$$then $\phi (\theta r_n,v_n,p_n)>C{\theta }^{1+\frac{2}{3}}(\phi (r_n,v_n,p_n)+\psi (r_n,f_n))=C{ϵ}_n{\theta }^{1+\frac{2}{3}}.$
Before the contradiction, first we take a scaling by following formulation:
$$(b_n,u_n,q_n,g_n):=((v_n^{r_n}{)}_1,{ϵ}_n^{-1}(v_n^{r_n}-(v_n^{r_n}{)}_1),{ϵ}_n^{-1}(p_n^{r_n}-(p_n^{r_n}{)}_1),{ϵ}_n^{-1}{f}_n^{r_n}).$$Then the homogeneity of $\psi$ and $\phi$ transform the above relations as¹
$$\begin{array}{}\text{(2)}& |b_n|\le 1,\phi (1,u_n,q_n)+r_n^{-(\gamma -\beta )}\psi (1,g_n)={\Vert u_n\Vert }_3+{\Vert q_n\Vert }_{\frac{3}{2}}+r_n^{-(\gamma -\beta )}{\Vert g_n\Vert }_{2,2\lambda +1}=1,\end{array}$$and the claim: $\mathrm{\forall }C>0,\phi (\theta ,u_n,q_n)\ge C{\theta }^{1+\frac{2}{3}}(\phi (1,u_n,q_n)+r_n^{-(\gamma -\beta )}\psi (1,g_n))=C{\theta }^{1+\frac{2}{3}}$. The remained work is to show $\phi (\theta ,u_n),\phi (\theta ,q_n)\lesssim {\theta }^{1+\frac{2}{3}}$, which counter the claim immediately.

1. $\phi (\theta ,u_n)\lesssim {\theta }^{1+\frac{2}{3}}$: Notice $(u_n,q_n)$ is suitable weak solution of following system on $Q_1$:
   $$\{\begin{array}{l}{\partial }_t{u}_n-\mathrm{\Delta }{u}_n+(b_n+{ϵ}_n{u}_n)\cdot \mathrm{\nabla }{u}_n+\mathrm{\nabla }{q}_n=g_n,\\ \mathrm{\nabla }\cdot u_n=0.\end{array}$$Moreover, the uniform bound $\text{(2)}$ implies
   $$b_n\stackrel{\mathbb{R}}{\to }b,u_n\stackrel{L^3}{\rightharpoonup }u,q_n\stackrel{L^{\frac{3}{2}}}{\rightharpoonup }q,g_n\stackrel{L^{2,2\gamma +1}}{\to }0.$$ Now we attempt to derive a strong $L^q$-convergence for $u_n$ by Aubin-Lion lemma: The local energy inequality implies²
   $${\Vert u_n\Vert }_{L^{10/3}(Q_{7/8})}\lesssim {\Vert u_n\Vert }_{L^{\mathrm{\infty }}{L}^2\cap L^2{\dot{H}}^1(Q_{7/8})}\lesssim 1,$$and following calculation implies ${\Vert {\partial }_t{u}_n\Vert }_{L^{\frac{4}{3}}(H_{\sigma }^1{)}^{\prime }}\lesssim 1$:
   $$\begin{array}{rl}|\int {\partial }_t{u}_n\cdot \zeta |\le & |\int \mathrm{\nabla }{u}_n\cdot \mathrm{\nabla }\zeta |+|\int \mathrm{\nabla }{u}_n\cdot \zeta |+|\int u_n{u}_n\mathrm{\nabla }\zeta |\\ \lesssim & {\Vert u_n\Vert }_{L^2{\dot{H}}^1}{\Vert \zeta \Vert }_{{\dot{H}}^1}+{\Vert u_n\Vert }_{L^2{\dot{H}}^1}{\Vert \zeta \Vert }_{L^2}+{\Vert u_n\Vert }_{L^{\mathrm{\infty }}{L}^2}^{\frac{1}{2}}{\Vert u_n\Vert }_{L^2{L}^6}^{\frac{3}{2}}{\Vert \mathrm{\nabla }\zeta \Vert }_{L^4{L}^2}\\ \lesssim & {\Vert \zeta \Vert }_{L^4(H_{\sigma }^1)}.\end{array}$$Since $H^1\stackrel{K}{\hookrightarrow }{L}^2\hookrightarrow (H_{\sigma }^1{)}^{\prime }$, together with ${\Vert u_n\Vert }_{L^2{H}^1}\lesssim 1$ and ${\Vert {\partial }_t{u}_n\Vert }_{L^{\frac{4}{3}}(H_{\sigma }^1{)}^{\prime }}\lesssim 1$, we see $\{u_n\}$ is precompact in $L^2{L}^2$ and thus $u_n\stackrel{L^2}{\to }u$. Moreover, since ${\Vert u_n\Vert }_{L^{10/3}}\lesssim 1$, so the Holder interpolation implies³ $u_n\stackrel{L^q(Q_{7/8})}{\to }u,\mathrm{\forall }q\in [2,10/3)$.
   Consequently, $(u,q)$ solves⁴ following system in $Q_{\frac{7}{8}}$:
   $$\{\begin{array}{l}{\partial }_{t}u-\mathrm{\Delta }u+b\cdot \mathrm{\nabla }u+\mathrm{\nabla }q=0,\\ \mathrm{\nabla }\cdot u=0.\end{array}$$Following we show that $u$ is Holder continuous:
   $$\begin{array}{r}({\partial }_t-\mathrm{\Delta }+b\cdot \mathrm{\nabla })(\mathrm{\nabla }\times u)=0⟹\mathrm{\nabla }\times u\in L^{\mathrm{\infty }}(Q_{5/6})\stackrel{\mathrm{\nabla }\cdot u=0}{\Rightarrow }\mathrm{\nabla }u\in L^{\mathrm{\infty }}{L}^{100}(Q_{4/5}),\\ q\in L^{\frac{3}{2}},\mathrm{\Delta }q(t)=0⟹\mathrm{\nabla }q\in L^{3/2}{C}^{\mathrm{\infty }}\stackrel{\mathrm{\nabla }u\in L^{\mathrm{\infty }}{L}^{100},\mathrm{\Delta }u\in ?}{\Rightarrow }{\partial }_{t}u\in L^{3/2}{L}^{\mathrm{\infty }}(Q_{4/5}).\end{array}$$It comes $u\in C^{\alpha ,\frac{\alpha }{2}}(\overline{Q_{3/4}})$ for $\frac{\alpha }{2}=1-\frac{1}{3/2}⟹\alpha =\frac{2}{3}$. Then the Campanato characterization in parabolic version gives out following bound since $\theta \le 3/4$:
   $${\theta }^{-3-2-3\alpha }{\int }_{Q_{\theta }}|u-(u{)}_{\theta }{|}^3\lesssim 1\stackrel{u_n\stackrel{L^3}{\to }u}{\Rightarrow }{\theta }^{-3-2-3\alpha }{\int }_{Q_{\theta }}|u_n-(u_n{)}_{\theta }{|}^3\lesssim 1,$$which implies $\phi (\theta ,u)\lesssim {\theta }^{1+\alpha }$.
2. $\phi (\theta ,q_n)\lesssim 1:$ notice $\mathrm{\Delta }{q}_n={ϵ}_n\mathrm{\nabla }\cdot (u_n\cdot \mathrm{\nabla }{u}_n)+\mathrm{\nabla }\cdot g_n={ϵ}_n({\partial }_i{\partial }_j(v^i{v}^j))+\mathrm{\nabla }\cdot g_n$ on $Q_{7/8}$, so we split $q_n={\tilde{q}}_n+h_n$ where
   $$\begin{array}{rl}\mathrm{\Delta }{\tilde{q}}_n={ϵ}_n\zeta ({\partial }_i{\partial }_j(v^i{v}^j))+\zeta \mathrm{\nabla }\cdot g_n,& \zeta \in D(Q_{\frac{7}{8}})\text{ cutoff on }{Q}_{\frac{3}{4}};\\ \mathrm{\Delta }{h}_n=\mathrm{\Delta }(q_n-{\tilde{q}}_n)=0& \text{ on }{Q}_{\frac{3}{4}}.\end{array}$$We estimate ${\tilde{q}}_n$ by Riesz potential, and $h_n$ by properties of harmonic function:⁵
   $$\begin{array}{rl}{\Vert {\tilde{q}}_n\Vert }_{L^3(Q_{7/8})}=& {\Vert I^2(({ϵ}_n\zeta ({\partial }_i{\partial }_j(u_n^i{u}_n^j))+\zeta \mathrm{\nabla }\cdot g_n))\Vert }_{L^{3/2}(Q_{7/8})}\\ \lesssim & {ϵ}_n{\Vert {\partial }^2\mathrm{\Gamma }\ast (u_n^i{u}_n^j)\Vert }_{L^{3/2}(Q_{7/8})}+{\Vert g_n\Vert }_{L^{\frac{3}{2}}{L}^2(Q_{7/8})}\\ \lesssim & {ϵ}_n{\Vert u_n\Vert }_{L^3(Q(7/8))}^2+{\Vert g_n\Vert }_{L^{2,2\gamma +1}(Q_{7/8})}\lesssim {ϵ}_n+r_n^{\gamma -\beta };\end{array}$$
   $$\begin{array}{rl}{\int }_{Q_{\theta }}|h_n-(h_n{)}_{\theta }{|}^{\frac{3}{2}}=& {\int }_{-{\theta }^2}^0{\int }_{B_{\theta }}|h_n-(h_n{)}_{\theta }{|}^{\frac{3}{2}}={\int }_{-{\theta }^2}^0{\int }_{B_{\theta }}(\theta {\Vert h_n\Vert }_{lip(Q_{\theta })}{)}^{\frac{3}{2}}\\ \lesssim & {\theta }^{3+\frac{3}{2}}{\int }_{-9/16}^0{\Vert \partial h_n\Vert }_{L^{\mathrm{\infty }}(B_{3/4})}^{\frac{3}{2}}\lesssim {\theta }^{3+\frac{3}{2}}{\int }_{-9/16}^0{\Vert h_n\Vert }_{L^1(B_{3/4})}^{3/2}\\ \lesssim & {\theta }^{3+\frac{3}{2}}{\int }_{-9/16}^0{\Vert h_n\Vert }_{L^{3/2}(B_{3/4})}^{3/2}\lesssim {\theta }^{3+\frac{3}{2}}{\int }_{Q_{3/4}}|h_n{|}^{3/2}\\ \lesssim & {\theta }^{3+\frac{3}{2}}{\int }_{Q_{3/4}}(|q_n{|}^{3/2}+|{\tilde{q}}_n{|}^{3/2})\lesssim {\theta }^{3+\frac{3}{2}}.\end{array}$$And then the estimate $q_n$ follows:
   $$\begin{array}{rl}{\int }_{Q_{\theta }}{|q_n-(q_n{)}_{\theta }|}^{\frac{3}{2}}\lesssim & {\int }_{Q_{\theta }}{|h_n-(h_n{)}_{\theta }|}^{\frac{3}{2}}+{\int }_{Q_{\theta }}{|{\tilde{q}}_n-({\tilde{q}}_n{)}_{\theta }|}^{\frac{3}{2}}\\ \lesssim & {\theta }^{3+\frac{3}{2}}+{({ϵ}_n+r_n^{\gamma -\beta })}^{3/2}\lesssim {\theta }^{3+\frac{3}{2}}\end{array}$$if ${ϵ}_n+r_n^{\gamma -\beta }\lesssim {\theta }^3$. Consequently, we have $\phi (\theta ,q_n)\lesssim {\theta }^{1+\frac{2}{3}}$.

Combine together, we get $\phi (\theta ,u_n,p_n)\lesssim {\theta }^{1+\frac{2}{3}}$.

First we set following relations to apply the above lemma:
$$\begin{array}{rl}& \beta =\frac{1}{2}(\alpha +\gamma );\theta \in (0,\frac{1}{2}],C{\theta }^{\frac{2}{3}}+{\theta }^{\beta }\le {\theta }^{\alpha };\\ & ({ϵ}_2,r_2)\sim (\theta ,\beta ,\gamma )\sim (\gamma ,\alpha );\\ & r_1=min\{r_2,{\left(\frac{{ϵ}_2}{2{\Vert f\Vert }_{L^{2,2\lambda +1}(Q_R)}}\right)}^{\frac{1}{1+\beta }}\}.\end{array}$$Then for any $z\in Q_{\theta R},r\in [\theta (1-\theta )R,(1-\theta )R]$, denote $\mathrm{\Psi }(z,r):=\phi (z,r,v,p)+\psi (z,r,f)$, we attempt to show
$$\begin{array}{}\text{(3)}& \mathrm{\Psi }(z,{\theta }^{k}r)\le {\theta }^{(1+\alpha )k}{ϵ}_2,\mathrm{\forall }k\ge 0,\end{array}$$which means for any $\rho \in (0,(1-\theta )R](\rho ={\theta }^{k}r)$,
$$\phi (z,\rho ,v)\le \mathrm{\Psi }(z,\rho )\le {\left(\frac{\rho }{r}\right)}^{1+\alpha }{ϵ}_2\le {\left(\frac{\rho }{\theta r_0}\right)}^{1+\alpha }{ϵ}_2\lesssim {\rho }^{1+\alpha }.$$That is say, $v\in C^{\alpha ,\frac{\alpha }{2}}(\overline{Q_{\theta R}})$ by Campanato characterization. Then the remained work is to verify $\text{(3)}$. Indeed, the condition implies $k=0$ case:
$$\begin{array}{rl}\varphi (z,r)\le & 1;?\\ \mathrm{\Psi }(z,r)=& \phi (z,r,v,p)+\psi (z,r,f)\le c\psi (z,r,v,p)+\psi (z,r,f)\\ \le & c{\left(R/r\right)}^{\frac{4}{3}}\psi (R,v,p)+(r/r_0{)}^{1+\beta }\psi (z,r_0,f)\\ \le & c_{\theta }{ϵ}_1+\frac{{ϵ}_2}{2}\le {ϵ}_2\end{array}$$for ${ϵ}_1$ small enough. Moreover, the iteration follows by the lemma:
$$\begin{array}{rl}\varphi (z,{\theta }^{k}r)=& 1;\\ \mathrm{\Psi }(z,{\theta }^{k}r)=& \phi (z,{\theta }^{k}r,v,p)+\psi (z,{\theta }^{k}r,f)\\ \le & C{\theta }^{1+\frac{2}{3}}(\phi (z,{\theta }^{k-1}r,v,p)+\psi (z,{\theta }^{k-1}r,f))+{\theta }^{1+\beta }\psi (z,{\theta }^{k-1}r,f)\\ \le & {\theta }^{1+\alpha }\mathrm{\Psi }(z,{\theta }^{k-1}r)\le \cdots \le {\theta }^{(1+\alpha )k}{ϵ}_2.\end{array}$$Thus the claim finished.

It is convenient to denote homogenous quantities:
$$\begin{array}{rl}& A(r)=\frac{1}{r}{\Vert v\Vert }_{L^{\mathrm{\infty }}{L}^2(Q_r)}^2,B(r)=\frac{1}{r}{\Vert \mathrm{\nabla }v\Vert }_{L^2{L}^2(Q_r)}^2;\\ & C(r)=\frac{1}{r^2}{\Vert v\Vert }_{L^3(Q_r)}^3,\tilde{C}(r)=\frac{1}{r^2}{\Vert v-(v{)}_{B_r}\Vert }_{L^3(Q_r)}^3,D(r)=\frac{1}{r^2}{\Vert p\Vert }_{L^{\frac{3}{2}}(Q_r)}^{\frac{3}{2}};\\ & G_1(r)=\frac{1}{r}{\Vert v-(v{)}_{B_r}\Vert }_{L^s{L}^q}(Q_r),G_2(r)=\frac{1}{r}{\Vert \mathrm{\nabla }v\Vert }_{L^s{L}^q(Q_r)}.\end{array}$$For above quantities, there are rough dominations for $k\ge 1,$
$$A/B/C/\tilde{C}/D/G(r)\le c_{k}A/B/C/\tilde{C}/D/G(kr).$$To process further, we'd like to show the following estimates for iteration: $\mathrm{\forall }r\le \frac{\rho }{2}$,
$$\begin{array}{}\text{(4)}& \begin{array}{rl}& C(r)\lesssim \left(\frac{r}{\rho }\right)C(\rho )+{\left(\frac{\rho }{r}\right)}^2\tilde{C}(\rho ),D(r)\lesssim \left(\frac{r}{\rho }\right)D(\rho )+{\left(\frac{\rho }{r}\right)}^2\tilde{C}(\rho );\\ & \tilde{C}(r)\lesssim A^{\frac{1}{s}}(r)E^{1-\frac{1}{s}}(r)G(r),(A+E)(r)\lesssim 1+(C+D)(2r).\end{array}\end{array}$$And all above estimates give out: $\mathrm{\forall }r\le {\displaystyle \frac{\rho }{4}},$
$$\begin{array}{rl}\tilde{C}\left(\frac{\rho }{2}\right)\lesssim & A^{\frac{1}{s}}\left(\frac{\rho }{2}\right)E^{1-\frac{1}{s}}\left(\frac{\rho }{2}\right)G\left(\frac{\rho }{2}\right)\lesssim (\frac{1}{s}A\left(\frac{\rho }{2}\right)+(1-\frac{1}{s})E\left(\frac{\rho }{2}\right))G\left(\frac{\rho }{2}\right)\\ \lesssim & (A+E)\left(\frac{\rho }{2}\right)G\left(\frac{\rho }{2}\right)\lesssim (1+(C+D)(\rho ))G(\rho );\\ (C+D)(r)\lesssim & \left(\frac{r}{\rho }\right)(C+D)\left(\frac{\rho }{2}\right)+{\left(\frac{\rho }{r}\right)}^2\tilde{C}\left(\frac{\rho }{2}\right);\\ \lesssim & \left(\frac{r}{\rho }\right)(C+D)(\rho )+{\left(\frac{\rho }{r}\right)}^2(1+C(\rho )+D(\rho ))G(\rho ).\end{array}$$Set the constant as $c>0$, then we choose $\theta \in (0,1/4)$ so that $c\theta <1/4$. By assumption, there is ${\displaystyle r_0>0,\mathrm{\forall }r\le r_0,G(r)<\frac{{\theta }^2{ϵ}_1}{1+8c}}$. Then the estimate indicates:
$$\begin{array}{rl}(C+D)(\theta r)\le & \frac{1}{2}(C+D)(r)+{\theta }^{-2}(1+(C+D)(r))G(r)\\ \le & \frac{1}{2}(C+D)(r)+\frac{{ϵ}_1}{4}?,\\ ⟹(C+D)({\theta }^{k}r)\le & \frac{1}{2^k}(C+D)(r)+\frac{{ϵ}_1}{2},\mathrm{\forall }r<r_0.\end{array}$$Then for $k$ big enough(depends on $r$), we have $(C+D)({\theta }^{k}r)\le {ϵ}_1$. Let $R={\theta }^{k}r$, then the above criterion shows $v$ is regular near $0$. Following are the verification of $\text{(4)}$:

1. $C(r)\lesssim \left(\frac{r}{\rho }\right)C(\rho )+{\left(\frac{\rho }{r}\right)}^2\tilde{C}(\rho )$:
   $$\begin{array}{rl}C(r)\lesssim & \frac{1}{r^2}{\int }_{Q_r}|v-(v{)}_{B_{\rho }}{|}^3+\frac{1}{r^2}{\int }_{Q_r}|(v{)}_{B_{\rho }}{|}^3\\ \lesssim & {\left(\frac{\rho }{r}\right)}^2(\frac{1}{{\rho }^2}{\int }_{Q_{\rho }}|v-(v{)}_{B_{\rho }}{|}^3)+\frac{r}{\rho }\left(\frac{1}{{\rho }^2}{\int }_{Q_{\rho }}|v{|}^3\right)\\ \lesssim & {\left(\frac{\rho }{r}\right)}^2\tilde{C}(\rho )+\frac{r}{\rho }C(\rho );\end{array}$$⁶
2. $D(r)\lesssim \left(\frac{r}{\rho }\right)D(\rho )+{\left(\frac{\rho }{r}\right)}^2\tilde{C}(\rho ):$ Since $\mathrm{\Delta }p={\partial }_i{\partial }_j(v^i{v}^j)$, then we decompose $p=\tilde{p}+h$, where $\mathrm{\Delta }\tilde{p}=\zeta ({\partial }_i{\partial }_j(v^i{v}^j)),\zeta$ cutoff on $B_{\rho /2}$. Then
   $$\begin{array}{r}{\Vert \tilde{p}\Vert }_{L^{\frac{3}{2}}(B_{\rho })}\lesssim \Vert \mathrm{\Gamma }\ast ({\partial }_i{\partial }_j(v^i-(v{)}_{B_{\rho }}^i)(v^j-(v{)}_{B_{\rho }}^j))\Vert \lesssim {\Vert v-(v{)}_{B_{\rho }}\Vert }_{L^3(B_{\rho })},\end{array}$$and since $|h{|}^{\frac{3}{2}}$ is sub-harmonic\footnote{Sub-harmonicity is preserved under convex function.}, it comes
   $$\begin{array}{rl}D(r)\lesssim & \frac{1}{r^2}{\int }_{Q_r}|\tilde{p}{|}^{\frac{3}{2}}+\frac{1}{r^2}{\int }_{Q_r}|h{|}^{\frac{3}{2}}\lesssim \frac{1}{r^2}{\int }_{Q_{\rho }}|\tilde{p}{|}^{\frac{3}{2}}+\frac{r}{{\rho }^3}{\int }_{B_{\rho }}|h{|}^{\frac{3}{2}}\\ \lesssim & \left(\frac{r}{\rho }\right)\left(\frac{1}{{\rho }^2}{\int }_{B_{\rho }}|p{|}^{\frac{3}{2}}\right)+(\frac{r}{{\rho }^3}+\frac{1}{r^2})\left({\int }_{B_{\rho }}|\tilde{p}{|}^{\frac{3}{2}}\right)\\ \lesssim & \left(\frac{r}{\rho }\right)D(\rho )+{\left(\frac{\rho }{r}\right)}^2\tilde{C}(\rho ).\end{array}$$
3. $\tilde{C}(r)\lesssim A^{\frac{1}{s}}(r)E^{1-\frac{1}{s}}(r)G(r)$: Apply Holder interpolation:
   $$\begin{array}{rl}\tilde{C}(r)=& \frac{1}{r^2}{\Vert v-(v{)}_{B_r}\Vert }_{L^3(Q_r)}^3\lesssim \frac{1}{r^2}{\Vert v-(v{)}_{B_r}\Vert }_{L^{\mathrm{\infty }}{L}^2}^{\frac{2}{3}}{\Vert v-(v{)}_{B_r}\Vert }_{L^2{L}^6}^{2-\frac{2}{s}}{\Vert v-(v{)}_r\Vert }_{L^s{L}^q}\\ \lesssim & {\left(\frac{1}{r}{\Vert v\Vert }_{L^{\mathrm{\infty }}{L}^2}^2\right)}^{\frac{1}{s}}{\left(\frac{1}{r}{\Vert \mathrm{\nabla }v\Vert }_{L^2{L}^2}^2\right)}^{1-\frac{1}{s}}{\Vert v-(v{)}_{B_r}\Vert }_{L^s{L}^q}\simeq A^{\frac{1}{s}}(r)E^{1-\frac{1}{s}}(r)G(r);\end{array}$$The second condition is similar: For $(q,s)=(3,1)$,
   $$\begin{array}{rl}{\Vert v-(v{)}_{B_r}\Vert }_{L^3(B_r)}^3\lesssim & {\Vert v-(v{)}_{B_r}\Vert }_{L^3(B_r)}^2{\Vert \mathrm{\nabla }v\Vert }_{L^3(B_r)}+r^{-\frac{3}{2}}{\Vert v-(v{)}_{B_r}\Vert }_{L^2(B_r)}^3\\ \le & {\Vert v\Vert }_{L^2(B_r)}^2{\Vert \mathrm{\nabla }v\Vert }_{L^3(B_r)}.\end{array}$$Integrating in time and apply Holder interpolation :
   $$\begin{array}{rl}\tilde{C}(r)=& r^{-2}{\Vert {\Vert v-(v{)}_{B_r}\Vert }_{L^3(B_r)}^3\Vert }_{L^1(-r^2,0)}\\ \lesssim & r^{-2}{\Vert {\Vert v\Vert }_{L^2(B_r)}^2{\Vert \mathrm{\nabla }v\Vert }_{L^3(B_r)}\Vert }_{L^1(-r^2,0)}\\ \lesssim & r^{-2}{\Vert v\Vert }_{L^{\mathrm{\infty }}{L}^2(Q_r)}^2{\Vert \mathrm{\nabla }v\Vert }_{L^1{L}^3(Q_r)}=A(r)G_2(r).\end{array}$$
4. $(A+E)(r)\lesssim 1+(C+D)(2r):$ We take $\varphi \in \mathcal{D}(Q_{2r})$ cutoff on $Q_r$, then the local energy inequality gives:
   $$\begin{array}{rl}{\Vert v\Vert }_{L^{\mathrm{\infty }}{L}^2(Q_r)}^2+{\Vert \mathrm{\nabla }v\Vert }_{L^2(Q_r)}^2\le & {\Vert v_0\Vert }_{L^2(B_r)}+{\int }_{Q_{2r}}|v{|}^2({\partial }_t\varphi +\mathrm{\Delta }\varphi )+(|v{|}^2+2p)|v|\mathrm{\nabla }\varphi \\ \lesssim & 1+r^{-2}{\Vert v\Vert }_{L^2(Q_{2r})}^2+r^{-2}{\Vert v\Vert }_{L^3(Q_{2r})}^3+r^{-1}{\Vert vp\Vert }_{L^1(Q_{2r})}.\end{array}$$Take $r^{-1}$ at both hand, we get that
   $$\begin{array}{rl}A(r)+B(r)\lesssim & 1+(r^{-2}{\Vert v\Vert }_{L^3(Q_{2r})}^3{)}^{\frac{3}{2}}+r^{-2}{\Vert v\Vert }_{L^3(Q_{2r})}^3\\ & +{\left(r^{-2}{\Vert v\Vert }_{L^3(Q_{2r})}^3\right)}^{\frac{1}{3}}{\left(r^{-2}{\Vert p\Vert }_{L^{\frac{3}{2}}(Q_{2r})}^{\frac{3}{2}}\right)}^{\frac{2}{3}}\\ \lesssim & 1+C(2r)+D(2r).\end{array}$$⁷