Caffarelli-Kohn-Nirenberg Theory

The first claim is clear:
\begin{equation*}
\begin{split}
\phi(\theta r,v)=& \theta r\cdot |Q_{\theta r} |^{-1}\int_{Q_{\theta r}}(v-v_{r})+\theta r v_r\\
\le & C'\theta r\cdot |Q_{\theta r} |^{1-\frac 13-1}\norm{v-v_r}_{L^3(Q_{\theta r})}+\theta rv_r\\
\le & C'\theta^{-\frac 23} \varphi(r,v)+\theta \phi(r,v)\\
\le & C'\epsilon_2 \theta^{-\frac 23}+\frac 12\le 1
\end{split}
\end{equation*}if $\epsilon_2\le \theta^{\frac 23} /(2C')$. As for the latter, we prove it by contradiction: if not, then $\forall C>0,\exists \beta\sl \gamma, \theta \in (0,\frac 12]$, $\forall \epsilon_2,r_2>0$(where we set both as $\dfrac 1n$), if $\exists r_n \le \dfrac 1n $, $(v_n,p_n)$ are SWS of \eqref{NS} in $Q_{r_n}$ such that
$$\psi(r_n, v_n)\le 1, \epsilon_n\coloneqq \varphi(r_n,v_n,p_n)+\psi(r_n, f_n) \le \frac 1n ,$$then $\varphi(\theta r_n,v_n,p_n)>C \theta^{1+\frac 23} \br{\varphi(r_n,v_n,p_n )+\psi(r_n,f_n) }=C \epsilon_n \theta^{1+\frac 23} .$
Before the contradiction, first we take a scaling by following formulation:
\begin{equation*}
(b_n,u_n,q_n,g_n)\coloneqq \br{(v_n^{r_n})_1, \epsilon_n^{-1} (v_n^{r_n}-(v_n^{r_n})_1 ), \epsilon_n^{-1}(p_n^{r_n}-(p_n^{r_n})_1 ), \epsilon_n^{-1}f^{r_n}_n}.
\end{equation*}Then the homogeneity of $\psi$ and $\varphi$ transform the above relations as¹
\begin{equation}
\label{Uniform Bounds}
|b_n|\le 1, \varphi(1, u_n,q_n)+ r_n^{-(\gamma-\beta)}\psi(1,g_n)=
\norm{u_n}_3+\norm{q_n}_\frac 32+ r_n^{-(\gamma-\beta)} \norm{g_n}_{2,2\lambda+1} = 1,
\end{equation}and the claim: $\forall C>0, \varphi(\theta, u_n,q_n)\ge C \theta^{1+\frac 23}\br{\varphi(1,u_n,q_n)+r_n^{-(\gamma-\beta)} \psi(1,g_n ) }=C \theta^{1+\frac 23}$. The remained work is to show $\varphi(\theta ,u_n), \varphi(\theta, q_n)\lesssim \theta^{1+\frac 23} $, which counter the claim immediately.

1. $\varphi(\theta, u_n)\lesssim \theta^{1+\frac 23}$: Notice $(u_n,q_n)$ is suitable weak solution of following system on $Q_1$:
   \begin{equation*}
   \begin{cases}
   \partial_t u_n-\Delta u_n+\br{b_n+\epsilon_n u_n }\cdot \nabla u_n+\nabla q_n=g_n,\\
   \nabla\cdot u_n=0.
   \end{cases}
   \end{equation*}Moreover, the uniform bound \eqref{Uniform Bounds} implies
   $$b_n\xrightarrow{\R} b, u_n \overset{L^3}{\rightharpoonup} u, q_n\overset{L^\frac 32}{\rightharpoonup} q,g_n\xrightarrow{L^{2,2\gamma+1} } 0 .$$ Now we attempt to derive a strong $L^q$-convergence for $u_n$ by Aubin-Lion lemma: The local energy inequality implies²
   $$ \norm{u_n}_{L^{10/3}(Q_{7/8})}\lesssim \norm{u_n}_{L^\infty L^2\cap L^2\dot H^1(Q_{7/8}) }\lesssim 1 ,$$and following calculation implies $ \norm{\partial_t u_n }_{L^{\frac 43}(H^1_\sigma)' }\lesssim 1$:
   \begin{equation*}
   \begin{split}
   \abs{\int \partial_t u_n \cdot \zeta }\le & \abs{\int \nabla u_n\cdot \nabla \zeta }+\abs{\int \nabla u_n\cdot \zeta }+\abs{\int u_n u_n \nabla \zeta }\\
   \lesssim & \norm{ u_n }_{L^2\dot H^1}\norm{\zeta }_{\dot H^1}+\norm{u_n }_{L^2 \dot H^1}\norm{\zeta }_{L^2}+\norm{u_n }_{L^\infty L^2}^\frac 12\norm{u_n }_{L^2L^6}^{\frac 32} \norm{\nabla \zeta }_{L^4L^2}\\
   \lesssim & \norm{\zeta}_{L^4(H^1_\sigma )}.
   \end{split}
   \end{equation*}Since $H^1\overset{K}{\hookrightarrow} L^2\hookrightarrow (H^1_\sigma)'$, together with $ \norm{u_n }_{L^2H^1}\lesssim 1 $ and $\norm{\partial_t u_n }_{L^\frac 43(H^1_\sigma)' }\lesssim 1$, we see $\{u_n\}$ is precompact in $L^2L^2$ and thus $u_n\xrightarrow{L^2}u$. Moreover, since $\norm{u_n}_{L^{10/3}}\lesssim 1 $, so the Holder interpolation implies³ $u_n\xrightarrow{L^q(Q_{7/8})} u,\forall q\in [2,10/3)$.
   Consequently, $(u,q)$ solves⁴ following system in $Q_{\frac 78}$:
   \begin{equation*}
   \begin{cases}
   \partial_t u-\Delta u+b\cdot \nabla u+\nabla q=0,\\
   \nabla\cdot u=0.
   \end{cases}
   \end{equation*}Following we show that $u$ is Holder continuous:
   \begin{equation*}
   \begin{split}
   (\partial_t -\Delta+b\cdot \nabla )(\nabla \times u )=0\Longrightarrow \nabla \times u\in L^{\infty}(Q_{5/6})\xRightarrow{\nabla \cdot u=0 } \nabla u\in L^\infty L^{100}(Q_{4/5} ),\\
   q\in L^{\frac 32}, \Delta q(t)=0\Longrightarrow \nabla q\in L^{3/2}C^\infty \xRightarrow{\nabla u\in L^{\infty}L^{100}, \Delta u\in \qst} \partial_t u\in L^{3/2}L^\infty (Q_{4/5}).
   \end{split}
   \end{equation*}It comes $u\in C^{\alpha,\frac{\alpha}{2}}(\ol{Q_{3/4}} )$ for $\frac \alpha 2=1-\frac{1}{3/2}\Longrightarrow\alpha=\frac 23$. Then the Campanato characterization in parabolic version gives out following bound since $\theta \le 3/4$:
   \begin{equation*}
   \theta^{-3-2-3\alpha} \int_{Q_\theta} |u-(u)_{\theta} |^3\lesssim 1\xRightarrow{u_n\xrightarrow{L^3}u } \theta^{-3-2-3\alpha} \int_{Q_\theta} |u_n-(u_n)_{\theta} |^3\lesssim 1,
   \end{equation*}which implies $\varphi(\theta, u)\lesssim \theta^{1+\alpha}$.
2. $\varphi(\theta ,q_n)\lesssim 1:$ notice $\Delta q_n=\epsilon_n \nabla \cdot (u_n\cdot \nabla u_n )+\nabla \cdot g_n =\epsilon_n (\partial_i\partial_j(v^iv^j) )+\nabla \cdot g_n $ on $Q_{7/8}$, so we split $q_n=\tilde q_n+h_n $ where
   \begin{equation*}
   \begin{split}
   \Delta \tilde q_n=\epsilon_n \zeta(\partial_i\partial_j (v^i v^j))+\zeta \nabla\cdot g_n , & \zeta\in D(Q_\frac 78) \text{ cutoff on }Q_\frac 34;\\
   \Delta h_n= \Delta( q_n- \tilde q_n)=0 & \text{ on } Q_{\frac 34}.
   \end{split}
   \end{equation*}We estimate $\tilde q_n$ by Riesz potential, and $h_n$ by properties of harmonic function:⁵
   \begin{equation*}
   \begin{split}
   \norm{\tilde q_n }_{L^3(Q_{7/8} )}=& \norm{I^2 \br{ ( \epsilon_n\zeta(\partial_i\partial_j (u_n^i u_n^j) )+\zeta \nabla\cdot g_n)} }_{L^{3/2}(Q_{7/8})}\\
   \lesssim & \epsilon_n \norm{\partial^2 \Gamma * (u_n^iu_n^j) }_{L^{3/2}(Q_{7/8})}+ \norm{g_n}_{L^\frac 32 L^2(Q_{7/8})}\\
   \lesssim & \epsilon_n \norm{u_n}_{L^3(Q(7/8))}^2+\norm{g_n}_{L^{2,2\gamma+1}(Q_{7/8})}\lesssim \epsilon_n+ r_n^{\gamma-\beta};
   \end{split}
   \end{equation*}
   \begin{equation*}
   \begin{split}
   \int_{Q_\theta} |h_n-(h_n)_\theta |^\frac 32=& \int_{-\theta^2}^0\int_{B_{\theta}} |h_n-(h_n)_\theta |^\frac 32 =\int_{-\theta^2}^0\int_{B_{\theta}} (\theta\norm{h_n}_{lip(Q_\theta)} )^\frac 32\\
   \lesssim & \theta^{3+\frac 32} \int_{-9/16}^0 \norm{\partial h_n}_{L^\infty(B_{3/4})}^{\frac 32}\lesssim \theta^{3+\frac 32}\int^0_{-9/16}\norm{h_n}^{3/2}_{L^1(B_{3/4})}\\
   \lesssim &\theta^{3+\frac 32}\int^0_{-9/16}\norm{h_n}^{3/2}_{L^{3/2}(B_{3/4})}\lesssim \theta^{3+\frac 32}\int_{Q_{3/4}}|h_n|^{3/2}\\
   \lesssim & \theta^{3+\frac 32}\int_{Q_{3/4}}\br{ |q_n|^{3/2}+|\tilde q_n|^{3/2}}\lesssim \theta^{3+\frac 32 }.
   \end{split}
   \end{equation*}And then the estimate $q_n$ follows:
   \begin{equation*}
   \begin{split}
   \int_{Q_\theta}\abs{q_n-(q_n)_{\theta}}^\frac 32\lesssim & \int_{Q_\theta}\abs{h_n-(h_n)_{\theta}}^\frac 32+\int_{Q_\theta}\abs{\tilde q_n-(\tilde q_n)_{\theta}}^\frac 32\\
   \lesssim & \theta^{3+\frac 32}+\br{\epsilon_n +r_n^{\gamma-\beta}}^{3/2} \lesssim \theta^{3+\frac 32}
   \end{split}
   \end{equation*}if $\epsilon_n +r_n^{\gamma-\beta}\lesssim \theta^3$. Consequently, we have $\varphi(\theta, q_n)\lesssim \theta^{1+\frac 23} $.

Combine together, we get $\varphi(\theta, u_n,p_n)\lesssim \theta^{1+\frac 23}$.

First we set following relations to apply the above lemma:
\begin{equation*}
\begin{split}
& \beta=\frac 12(\alpha+\gamma ); \theta\in (0,\frac 12], C \theta^\frac 23+\theta^\beta\le \theta^\alpha;\\
& (\epsilon_2, r_2) \sim (\theta, \beta,\gamma)\sim (\gamma, \alpha ) ; \\ & r_1=\min \set{ r_2, \br{\frac{\epsilon_2 }{2\norm{f}_{L^{2,2\lambda+1}(Q_R)} }}^{\frac 1{1+\beta}} }.
\end{split}
\end{equation*}Then for any $z\in Q_{\theta R}, r\in [\theta (1-\theta)R, (1-\theta)R]$, denote $\Psi(z,r)\coloneqq \varphi(z,r,v,p)+\psi(z,r,f)$, we attempt to show
\begin{equation}
\label{Psi iteration}
\Psi(z, \theta^k r)\le \theta^{(1+\alpha)k}\epsilon_2, \forall k\ge 0,
\end{equation}which means for any $\rho\in (0, (1-\theta)R](\rho=\theta^k r)$,
$$\varphi(z,\rho, v)\le \Psi(z, \rho)\le \br{\frac \rho r }^{1+\alpha}\epsilon_2\le \br{\frac \rho {\theta r_0} }^{1+\alpha}\epsilon_2\lesssim \rho^{1+\alpha}.$$That is say, $v\in C^{\alpha, \frac \alpha 2}(\ol{ Q_{\theta R}} )$ by Campanato characterization. Then the remained work is to verify \eqref{Psi iteration}. Indeed, the condition implies $k=0$ case:
\begin{equation*}
\begin{split}
\phi(z,r)\le & 1 ;\qst \\
\Psi(z,r)= & \varphi(z,r,v,p)+\psi(z,r,f)\le c \psi(z,r,v,p)+ \psi(z, r, f)\\
\le & c\br{ R/r }^\frac{4}{3}\psi(R,v,p)+ (r/r_0)^{1+\beta}\psi(z, r_0, f) \\
\le & c_{\theta}\epsilon_1+ \frac{\epsilon_2}2 \le \epsilon_2
\end{split}
\end{equation*}for $\epsilon_1$ small enough. Moreover, the iteration follows by the lemma:
\begin{equation*}
\begin{split}
\phi(z,\theta^k r)=& 1; \\
\Psi(z,\theta^{k}r)= & \varphi(z ,\theta^k r,v,p )+\psi (z, \theta^kr, f )\\
\le & C \theta^{1+\frac 23}\br{ \varphi (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} \Psi(z, \theta^{k-1}r)\le \cdots \le \theta^{(1+\alpha)k}\epsilon_2.
\end{split}
\end{equation*}Thus the claim finished.

It is convenient to denote homogenous quantities:
\begin{equation*}
\begin{split}
& A(r)=\frac 1r\norm{v }_{L^\infty L^2(Q_r)}^2, B(r)=\frac 1r\norm{\nabla v }_{L^2L^2(Q_r)}^2;\\
& C(r)= \frac 1{r^2}\norm{v }_{L^3(Q_r)}^3, \tilde C(r)=\frac 1{r^2}\norm{v-(v)_{B_r} }_{L^3(Q_r) }^3,D(r)=\frac 1{r^2}\norm{p }_{L^\frac 32(Q_r) }^\frac 32;\\
& G_1(r)= \frac{1}{r}\norm{v-(v)_{B_r} }_{L^sL^q}(Q_r), G_2(r)=\frac 1r\norm{\nabla v }_{L^sL^q(Q_r)}.
\end{split}
\end{equation*}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: $\forall r\le \frac \rho 2$,
\begin{equation}
\label{Iteration ABCD}
\begin{split}
& C(r)\lesssim \br{\frac r\rho }C(\rho)+\br{\frac \rho r }^2 \tilde C(\rho ),
D(r)\lesssim \br{\frac r \rho }D(\rho)+\br{\frac \rho r }^2 \tilde C(\rho);\\
& \tilde C(r)\lesssim A^{\frac 1s}(r)E^{1-\frac 1s}(r)G(r), (A+E)(r) \lesssim 1+(C+D)(2r).
\end{split}
\end{equation}And all above estimates give out: $\forall r\le \dfrac {\rho}{4},$
\begin{equation*}
\begin{split}
\tilde C\br{\frac \rho 2 }\lesssim & A^{\frac 1s}\br{\frac \rho 2 }E^{1-\frac 1s}\br{\frac \rho 2 }G\br{\frac \rho 2 }\lesssim \br{\frac 1s A\br{\frac \rho 2 }+\br{1-\frac 1s}E\br{\frac \rho 2 }}G\br{\frac \rho 2 } \\
\lesssim & \br{A+E}\br{\frac \rho 2 }G\br{\frac \rho 2 }\lesssim (1+(C+D)(\rho))G(\rho); \\
(C+D)(r)\lesssim & \br{\frac r\rho }(C+D)\br{ \frac \rho 2 }+\br{\frac \rho r }^2 \tilde C\br{ \frac \rho 2 };\\
\lesssim & \br{\frac r\rho }(C+D )(\rho)+\br{\frac \rho r }^2\br{1+C(\rho)+D(\rho) }G(\rho).
\end{split}
\end{equation*}Set the constant as $c>0$, then we choose $\theta\in (0,1/4)$ so that $c\theta \sl 1/4$. By assumption, there is $\displaystyle r_0>0, \forall r\le r_0, G(r)\sl \frac{\theta^2 \epsilon_1 }{1+8c } $. Then the estimate indicates:
\begin{equation*}
\begin{split}
(C+D)(\theta r )\le & \frac 12 (C+D)(r)+\theta^{-2} \br{1+(C+D)(r) }G(r)\\
\le & \frac 12(C+D)(r)+\frac {\epsilon_1 }{4}\qst,\\
\Longrightarrow (C+D)(\theta^k r )\le & \frac 1{2^k} (C+D)(r)+\frac {\epsilon_1} 2,\forall r\sl r_0.\\
\end{split}
\end{equation*}Then for $k$ big enough(depends on $r$), we have $(C+D)(\theta^k r)\le \epsilon_1 $. Let $R=\theta^k r$, then the above criterion shows $v$ is regular near $0$. Following are the verification of \eqref{Iteration ABCD}:

1. $C(r)\lesssim \br{\frac r\rho }C(\rho)+\br{\frac \rho r }^2 \tilde C(\rho )$:
   \begin{equation*}
   \begin{split}
   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 & \br{\frac{\rho}{r} }^2\br{\frac 1{\rho^2}\int_{Q_\rho} |v-(v)_{B_\rho}|^3}+\frac{r}{\rho}\br{\frac 1{\rho^2}\int_{Q_{\rho}}|v|^3 } \\
   \lesssim & \br{\frac{\rho}{r} }^2\tilde C(\rho)+\frac{r}{\rho}C(\rho);
   \end{split}
   \end{equation*}⁶
2. $D(r)\lesssim \br{\frac r \rho }D(\rho)+\br{\frac \rho r }^2 \tilde C(\rho):$ Since $\Delta p= \partial_i\partial_j(v^iv^j )$, then we decompose $p=\tilde p+h$, where $\Delta \tilde p=\zeta(\partial_i\partial_j(v^iv^j ) ), \zeta$ cutoff on $B_{\rho /2}$. Then
   \begin{equation*}
   \begin{split}
   \norm{\tilde p }_{L^\frac 32(B_\rho)}\lesssim \norm{\Gamma*(\partial_i \partial_j (v^i-(v)^i_{B_\rho})(v^j-(v)_{B_\rho}^j ) ) }\lesssim \norm{v-(v)_{B_\rho}}_{L^3(B_\rho)},
   \end{split}
   \end{equation*}and since $|h|^\frac 32$ is sub-harmonic\footnote{Sub-harmonicity is preserved under convex function.}, it comes
   \begin{equation*}
   \begin{split}
   D(r)\lesssim & \frac{1}{r^2}\int_{Q_r}|\tilde p |^\frac 32+\frac 1{r^2}\int_{Q_r}|h|^\frac 32
   \lesssim \frac{1}{r^2}\int_{Q_\rho}|\tilde p |^\frac 32+\frac{r}{\rho^3}\int_{B_\rho} |h|^{\frac 32}\\
   \lesssim &\br{\frac r\rho }\br{\frac 1{\rho^2} \int_{B_\rho}|p|^\frac 32}+\br{\frac r{\rho^3}+\frac 1{r^2} }\br{\int_{B_\rho} |\tilde p |^\frac 32 }\\
   \lesssim & \br{\frac{r }{\rho } }D(\rho)+\br{\frac{\rho }{r} }^2\tilde C(\rho).
   \end{split}
   \end{equation*}
3. $\tilde C(r)\lesssim A^{\frac 1s}(r)E^{1-\frac 1s}(r)G(r)$: Apply Holder interpolation:
   \begin{equation*}
   \begin{split}
   \tilde C(r)=& \frac{1}{r^2}\norm{v-(v)_{B_r} }_{L^3(Q_r)}^3\lesssim \frac 1{r^2}\norm{v-(v)_{B_r}}_{L^\infty L^2}^\frac 23\norm{v-(v)_{B_r} }_{L^2L^6}^{2-\frac 2s}\norm{v-(v)_{r}}_{L^sL^q}\\
   \lesssim & \br{\frac 1r\norm{v}_{L^\infty L^2}^2}^{\frac 1s }\br{\frac 1r \norm{\nabla v }_{L^2L^2}^2}^{1-\frac 1s }\norm{v-(v)_{B_r}}_{L^sL^q}\simeq A^\frac 1s(r)E^{1-\frac 1s}(r)G(r);
   \end{split}
   \end{equation*}The second condition is similar: For $(q,s)=(3,1)$,
   \begin{equation*}
   \begin{split}
   \norm{v-(v)_{B_r} }_{L^3(B_r)}^3\lesssim & \norm{v-(v)_{B_r} }_{L^3(B_r)}^2\norm{\nabla v }_{L^3(B_r)}+r^{-\frac 32}\norm{v-(v)_{B_r} }_{L^2(B_r)}^3 \\
   \le & \norm{v}^2_{L^2(B_r)}\norm{\nabla v }_{L^3(B_r)}.
   \end{split}
   \end{equation*}Integrating in time and apply Holder interpolation :
   \begin{equation*}
   \begin{split}
   \tilde C(r)=& r^{-2} \norm{\norm{v-(v)_{B_r} }^3_{L^3(B_r)}}_{L^1(-r^2,0)}\\
   \lesssim & r^{-2} \norm{\norm{v}^2_{L^2(B_r)}\norm{\nabla v }_{L^3(B_r)}}_{L^1(-r^2,0)}\\
   \lesssim &r^{-2}\norm{v}_{L^\infty L^2(Q_r)}^2\norm{\nabla v}_{L^1L^3(Q_r) }=A(r)G_2(r).
   \end{split}
   \end{equation*}
4. $(A+E)(r)\lesssim 1+(C+D)(2r):$ We take $\phi\in \Di(Q_{2r})$ cutoff on $Q_r$, then the local energy inequality gives:
   \begin{equation*}
   \begin{split}
   \norm{v}_{L^\infty L^2(Q_r)}^2+\norm{\nabla v}_{L^2(Q_r)}^2\le & \norm{v_0}_{L^2(B_r)}+ \int_{Q_{2r}} |v|^2(\partial_t \phi+\Delta \phi )+(|v|^2+2p )|v|\nabla \phi\\
   \lesssim & 1+r^{-2}\norm{v}_{L^2(Q_{2r})}^2+ r^{-2}\norm{v }_{L^3(Q_{2r})}^3+r^{-1}\norm{vp}_{L^1(Q_{2r})}.
   \end{split}
   \end{equation*}Take $r^{-1}$ at both hand, we get that
   \begin{equation*}
   \begin{split}
   A(r)+B(r)\lesssim & 1+ (r^{-2}\norm{v}^3_{L^3(Q_{2r})} )^\frac 32+r^{-2}\norm{v}^3_{L^3(Q_{2r})}\\
   & +\br{r^{-2}\norm{v}^3_{L^3(Q_{2r})}}^\frac 13 \br{r^{-2}\norm{p}_{L^\frac 32(Q_{2r})}^\frac 32 }^\frac 23\\
   \lesssim & 1+ C(2r)+D(2r).
   \end{split}
   \end{equation*}⁷