# Global well-posedness of slightly supercritical SQG equation and exponential
gradient estimate

Hyungjun Choi
Department of Mathematics
Princeton University, Princeton NJ 08544, USA<EMAIL_ADDRESS>
(Date: October 19, 2022)

###### Abstract.

We prove the global regularity of smooth solutions for a dissipative surface
quasi-geostrophic equation with both velocity and dissipation logarithmically
supercritical compared to the critical equation. By this, we mean that a
symbol defined as a power of logarithm is added to both velocity and
dissipation terms to penalize the equation’s criticality. Our primary tool is
the nonlinear maximum principle which provides transparent proofs of global
regularity for nonlinear dissipative equations. In addition, we prove an
exponential gradient estimate for the critical surface quasi-geostrophic
equation which improves the previous double exponential bound [13].

###### Key words and phrases:

Surface quasi-geostrophic (SQG) equation, $L^{\infty}$-critical, slightly
supercritical, global well-posedness, active scalar, nonlinear maximum
principle, modulus of continuity

###### 2020 Mathematics Subject Classification:

35Q35, 76U05

##  1\. Introduction

The global well-posedness of the dissipative surface quasi-geostrophic (SQG)
equation (1.1) has been widely studied in recent literature.

$\left\\{\begin{array}[]{rl}\partial_{t}\theta+(u\cdot\nabla)\theta+\Lambda^{\alpha}\theta&=0,\\\
u&=\nabla^{\perp}\Lambda^{-1}\theta.\end{array}\right.$ (1.1)

The SQG equation first appeared in the mathematical literature in [2]. Due to
the positivity of the operator $\Lambda^{\alpha}$, it can be easily seen that
the $L^{p}$ norms of $\theta$ are nonincreasing in time under smooth
evolution. Indeed, the equation has $L^{\infty}$-maximum principle [7]. The
situation differs by whether the order of the diffusion term is
higher/identical/lower than the nonlinear transport term under
$L^{\infty}$-rescaling of a solution. The case $\alpha=1$ is termed “critical”
since the order is identical. The weak formulation and the global well-
posedness for the subcritical case ($\alpha>1$) were understood in the usual
sense [3, 16]. In addition, a small initial data case was studied in [4]. The
global well-posedness in the critical case was first proved in [1](in
$\mathbb{R}^{n}$) and [13, 14](in $\mathbb{T}^{n}$). In particular, [13]
analyzed a breakthrough scenario of a modulus of continuity, called the
nonlocal maximum principle, which is studied further in [15]. Shortly after
that, [6] presented the third proof. They introduced the nonlinear maximum
principle, which can be applied to well-posedness proofs for various critical
active scalar equations.

In the supercritical case ($\alpha<1$), the global well-posedness for generic
initial remains an open problem. There have been several results regarding
conditional regularity [5, 11](a Hölder continuous solution is smooth) and
eventual regularization of solutions [17, 8, 15]. The global regularity has
been studied if velocity or dissipation is slightly supercritical. In
particular, the slightly supercritical velocity was studied in [9], where the
veloctiy is obtained from $\theta$ by a Fourier multiplier with symbol
$i\zeta^{\perp}\left\lvert\zeta\right\rvert^{-1}m(\zeta)$. Here, $m(\zeta)$
grows slower than $\log\log\left\lvert\zeta\right\rvert$ as $\zeta\to\infty$.
Moreover, the slightly supercritical dissipation for several active scalar
equations was studied in [10], where the dissipation term is given by
multiplier with behavior
$P(\zeta)\sim\left\lvert\zeta\right\rvert(\log\left\lvert\zeta\right\rvert)^{-\alpha}$
($0\leq\alpha\leq 1$) for large $\zeta$. These proofs were based on the
nonlocal maximum principle introduced in [13, 15].

In this paper, we consider the SQG equation, which is slightly supercritical
in both velocity and dissipation:

$\left\\{\begin{array}[]{rl}\partial_{t}\theta+(u\cdot\nabla)\theta+\mathcal{L}\theta&=0,\\\
u&=\nabla^{\perp}\Lambda^{-1}m(\Lambda)\theta,\\\
\theta(t=0)&=\theta_{0}.\end{array}\right.$ (1.2)

The nonlocal operator $\mathcal{L}$ is defined by a convolutional kernel as
follows:

$\mathcal{L}\theta(x)=P.V.\int_{\mathbb{R}^{2}}(\theta(x)-\theta(y))\frac{k(x-y)}{\left\lvert
x-y\right\rvert^{2}}\,dy.$

Note that $\mathcal{L}=\Lambda$ when $k(x)=\frac{1}{2\pi\left\lvert
x\right\rvert}$. The main example we have in mind is

$k(r)=\frac{1}{r(\log(10+r^{-1}))^{\alpha_{1}}},\quad
m(\zeta)=(\log(10+\left\lvert\zeta\right\rvert))^{\alpha_{2}},$ (1.3)

for some $\alpha_{1},\alpha_{2}\geq 0$.

We prove the global regularity with the help of the nonlinear maximum
principle [6]. We show a conditional regularity result in the first step, then
we prove that desired regularity is conserved by evolution in the second step.
The main difference between our proof and [6] is that we provide a modulus of
continuity while [6] argues only with uniform continuity (the only small shock
condition). An explicit modulus of continuity allows us to construct an
exponential gradient estimate, Main Theorem 2.

In section 2, we list some estimates regarding the relationship between the
kernel and the Fourier multiplier. We provide suitable conditions for $k$ and
$m$ to behave similarly as the main example (1.3). We also collect some
elementary integral estimates for later use. In section 3, we provide
conditional regularity result for (1.2). If $\theta$ is a bounded weak
solution to (1.2) in time $[0,T]$ and $\theta(\cdot,t)$ admits a modulus of
continuity $\Omega$ for $t\in[0,T]$ which satisfies

$\lim_{r\to 0+}\frac{m(r^{-1})\Omega(r)}{rk(r)}=0,$ (1.4)

then $\theta$ is a smooth solution. In section 4, we prove that a modulus of
continuity $M\omega$ for $\theta(\cdot,t)$ is conserved by $\eqref{eq:1.2}$ if
a constant $M$ depends on $\theta_{0}$ is sufficiently large and

$\lim_{r\to 0+}\frac{m(r^{-1})}{k(r)}\frac{\omega^{\prime}(r)}{\omega(r)}=0.$
(1.5)

Combining two steps, we prove the following:

Main Theorem 1. Assume that $\theta_{0}\in\mathcal{S}(\mathbb{R}^{2})$.
Suppose there is a non-decreasing, continuous, concave function
$\omega:[0,\infty)\to[0,\infty)$ satisfying $\omega(0)=0$, (1.4), (1.5), and
(2.1). Then there exists a global smooth solution $\theta$ of the slightly
supercritical SQG equation (1.2).

######  Remark.

An initial $\theta_{0}\in W^{1,\infty}\cap H^{s}(\mathbb{R}^{2})$ with $s>1$
is enough to guarantee a smooth solution.

A typical example is a logarithmic multiplier (1.3).

######  Corollary 1.

Assume that $\theta_{0}\in\mathcal{S}(\mathbb{R}^{2})$. If

$\alpha_{1}+\alpha_{2}<1,$

then there exists a global smooth solution $\theta$ of

$\left\\{\begin{array}[]{rl}\partial_{t}\theta+(u\cdot\nabla)\theta+\Lambda(\log^{-\alpha_{1}}\Lambda)\theta&=0,\\\
u&=\nabla^{\perp}\Lambda^{-1}(\log^{\alpha_{2}}\Lambda)\theta,\\\
\theta(t=0)&=\theta_{0}.\end{array}\right.$ (1.6)

This result improves [9] which studied a slightly supercritical velocity,
where $m(\zeta)$ being growing slower than double logarithm. Moreover, we
prove the global regularity when both velocity and dissipation are slightly
supercritical. Hence this paper extends [10], which studied the slightly
supercritical dissipation.

In section 5, we prove an exponential gradient estimate for the critical SQG
equation:

$\left\\{\begin{array}[]{rl}\partial_{t}\theta+(u\cdot\nabla)\theta+\Lambda\theta&=0,\\\
u&=\nabla^{\perp}\Lambda^{-1}\theta,\\\
\theta(t=0)&=\theta_{0}\end{array}\right.$ (1.7)

Main Theorem 2. The critical SQG equation (1.7) with initial
$\theta_{0}\in\mathcal{S}(\mathbb{R}^{2})$ has a unique global smooth
solution. In addition, for any $\gamma>0$,

$\lVert\nabla\theta\rVert_{L^{\infty}_{t,x}}\leq
C\lVert\nabla\theta_{0}\rVert_{L^{\infty}}\exp(C\lVert\theta_{0}\rVert_{L^{\infty}}^{1+\gamma}),$
(1.8)

where constants depend only on $\gamma$.

Our result (1.8) improves the following double exponential estimate (1.9). In
[13], they used a modulus of continuity $\omega(r)$, which behaves like
$\log\log(r/\delta)$ for $\delta\ll r\lesssim 1$. While we use a modulus of
continuity $|\log r|^{-\beta}$ for $r\lesssim 1$ with some $\beta>0$.

######  Proposition 2 ([13]).

The critical SQG equation (1.7) with initial $\theta_{0}\in
C^{\infty}(\mathbb{T}^{2})$ has a unique global smooth solution. In addition,

$\lVert\nabla\theta\rVert_{L^{\infty}_{t,x}}\leq
C\lVert\nabla\theta_{0}\rVert_{L^{\infty}}\exp(\exp(C\lVert\theta_{0}\rVert_{L^{\infty}})).$
(1.9)

Lastly, these results can be easily reproduced for other $L^{\infty}$-critical
active scalar equations with nonlocal dissipation and a linear balance law
$u=T\theta$.

##  2\. Preliminaries

The local regularity result for the SQG-type equation is standard. It is well-
known that $\lVert\nabla\theta\rVert_{L^{1}_{t}L^{\infty}_{x}}$ is a blow-up
criterion.

######  Proposition 3 (local existence of smooth solution, [12]).

Suppose $\theta_{0}\in W^{1,\infty}\cap H^{s}(\mathbb{R}^{2})$ and $s>1$. Then
there exists $T>0$ and a solution $\theta\in
C^{\infty}((0,T]\times\mathbb{R}^{2})$ of (1.2). Moreover, the smooth solution
may be continued as long as

$\lVert\nabla\theta\rVert_{L^{1}_{t}L^{\infty}_{x}}=\int_{0}^{T}\lVert\nabla\theta(\cdot,t)\rVert_{L^{\infty}}<\infty.$

Next, we provide assumptions on $k$ and $m$ and arrange some estimates on the
kernel of the operator $\nabla\Lambda^{-1}m(\Lambda)$ and the kernel
$r^{-2}k(r)$. Throughout the paper, we assume on $k$ and $m$ as follows:

  * •

The kernel $k(x)=k(\left\lvert x\right\rvert)$ is radially symmetric,
nonnegative, and nonincreasing. In addition, $r^{2}k(r)$ is nondecreasing for
$r>0$.

  * •

(Slightly supercitical dissipation) For any $\epsilon>0$,

$r^{1-\epsilon}k(r)$

is nonincreasing for $0<r<r_{\epsilon}$ and tends to $\infty$ as $r\to 0$.

  * •

There exists a constant $c\geq 1$ such that

$k\left(\frac{r}{2}\right)\leq c\,k(r),\quad r>0.$

  * •

The Fourier multiplier $m(\zeta)=m(\left\lvert\zeta\right\rvert)\geq 1$ is
radially symmetric, nondecreasing, and satisfies the Hörmander-Mikhlin
condition:

$\left\lvert\zeta\right\rvert^{k}|\nabla^{k}m(\zeta)|\leq
C_{k}m(\zeta),\quad\zeta\neq 0,$

for some constant $C_{k}>0$.

  * •

(Slightly supercritical velocity) For any $\epsilon>0$,

$\left\lvert\zeta\right\rvert^{\epsilon}m(\left\lvert\zeta\right\rvert^{-1})$

is nondecreasing in $0<\left\lvert\zeta\right\rvert<r_{\epsilon}$ and tends to
$0$ as $\left\lvert\zeta\right\rvert\to 0$.

The following lemmas reveal relations between an operator’s kernel and the
Fourier multiplier. Lemma 4 deals with velocity $u$ and the Lemma 5 deals with
dissipation term $\mathcal{L}\theta$.

######  Lemma 4 (Lemma 4.1 of [9]).

Suppose $K$ is the kernel corresponding to the operator
$\partial_{j}\Lambda^{-1}m(\Lambda)$ and $m$ satisfies the prescribed
conditions. Then

$\left\lvert K(x)\right\rvert\leq C\left\lvert x\right\rvert^{-d}m(\left\lvert
x\right\rvert^{-1}),$

and

$\left\lvert\nabla K(x)\right\rvert\leq C\left\lvert
x\right\rvert^{-d-1}m(\left\lvert x\right\rvert^{-1}),$

for all $x\neq 0$.

######  Lemma 5 ([10, 18]).

Suppose $P(\zeta)=P(\left\lvert\zeta\right\rvert)$ is a radially symmetric
function that is smooth, nonnegative, nondecreasing from zero, $P(0)=0$, and
$P(\zeta)\to\infty$ as $\left\lvert\zeta\right\rvert\to\infty$. In addition,
assume the following for $P$:

  * •

There is a constant $c\geq 1$ so that $P(2\zeta)\leq cP(\zeta)$ for all
$\zeta\in\mathbb{R}^{d}$.

  * •

$P$ is of the Hörmander-Mikhlin type:

$\left\lvert\zeta\right\rvert^{k}|\nabla^{k}P(\zeta)|\leq
C_{k}P(\zeta),\quad\zeta\neq 0,$

for some constant $C_{k}$.

  * •

$P$ satisfies growth condition

$\int_{0}^{1}P(\left\lvert\zeta\right\rvert^{-1})\left\lvert\zeta\right\rvert\,d\zeta<\infty.$

Then, the corresponding radially symmetric kernel
$K={\mathchoice{{\ooalign{\hbox{\raise
7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\displaystyle\widehat{\vrule
width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=7.80904pt}$}}}}\cr\hbox{$\displaystyle
P$}}}}{{\ooalign{\hbox{\raise 7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\textstyle\widehat{\vrule width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=7.80904pt}$}}}}\cr\hbox{$\textstyle
P$}}}}{{\ooalign{\hbox{\raise 6.40926pt\hbox{\scalebox{1.0}[-1.0]{\lower
6.40926pt\hbox{$\scriptstyle\widehat{\vrule width=0.0pt,height=4.78333pt\vrule
height=0.0pt,width=5.46632pt}$}}}}\cr\hbox{$\scriptstyle
P$}}}}{{\ooalign{\hbox{\raise 5.9537pt\hbox{\scalebox{1.0}[-1.0]{\lower
5.9537pt\hbox{$\scriptscriptstyle\widehat{\vrule
width=0.0pt,height=3.41666pt\vrule
height=0.0pt,width=3.90451pt}$}}}}\cr\hbox{$\scriptscriptstyle P$}}}}}$
satisfies

$\left\lvert K(y)\right\rvert\leq C\left\lvert y\right\rvert^{-d}P(\left\lvert
y\right\rvert^{-1}),$

and

$\left\lvert\nabla K(y)\right\rvert\leq C\left\lvert
y\right\rvert^{-d-1}P(\left\lvert y\right\rvert^{-1}),$

for all $y\not=0\in\mathbb{R}^{d}$. Moreover, if $P$ satisfies

  * •

$\displaystyle(-\Delta)^{\frac{d}{2}+1}P(\zeta)\geq
c\left\lvert\zeta\right\rvert^{-d-2}P(\zeta)$

for some constant $c>0$. Then, $K$ is bounded below as

$K(y)\geq c\left\lvert y\right\rvert^{-d}P(\left\lvert y\right\rvert^{-1}),$

for all sufficiently small $y$.

######  Remark.

If we consider $\mathcal{L}=\Lambda(\log^{-\alpha}\Lambda)$, i.e.,

$P(\zeta)=\zeta\left(\log(10+\left\lvert\zeta\right\rvert)\right)^{-\alpha},\quad\alpha>0,$

then Lemma 5 shows that it is as same as considering

$k(r)=\frac{1}{r\left(\log(10+r^{-1})\right)^{\alpha}}.$

Lastly, we denote some technical integral estimates which we will use
frequently.

######  Lemma 6.

Suppose $k$ and $m$ satisfy the prescribed conditions and
$\omega:[0,\infty)\to[0,\infty)$ is a nondecreasing, continuous, concave
function with $\omega(0)=0$. In addition, assume that there exists
$0<\gamma<1$ such that

$\lim_{r\to
0+}\frac{\omega(r)}{r^{\gamma}}=\infty,\quad\text{and}\quad\lim_{r\to\infty}\frac{\omega(r)}{r^{\gamma}}=0.$
(2.1)

For sufficiently small $r$, namely $0<r<r_{0}$, the following inequalities
hold.

  1. (i)

$\displaystyle\int_{r}^{\infty}\frac{k(\rho)}{\rho}\,d\rho\geq\frac{k(r)}{2}$.

  2. (ii)

$\displaystyle\int_{r/2}^{r}\frac{\omega(\rho)k(\rho)}{\rho}\,d\rho\leq
C\omega(r)k(r)$ for a constant $\displaystyle C>0$.

  3. (iii)

$\displaystyle\int_{r}^{\infty}\omega(\rho)\left(-\frac{k^{\prime}(\rho)}{\rho}+\frac{2k(\rho)}{\rho^{2}}\right)\,d\rho\leq\frac{C\omega(r)k(r)}{r}$
for a constant $\displaystyle C>0$.

  4. (iv)

$\displaystyle\int_{0}^{r}\frac{m(\rho^{-1})^{2}}{\rho
k(\rho)}\,d\rho\leq\frac{4m(r^{-1})^{2}}{k(r)}$.

  5. (v)

$\displaystyle\int_{r}^{\infty}\frac{\omega(\rho)m(\rho^{-1})}{\rho^{2}}\,d\rho\leq\frac{C\omega(r)m(r^{-1})}{r}$
for a constant $\displaystyle C>0$.

###### Proof.

  1. (i)

$\displaystyle\int_{r}^{\infty}\frac{k(\rho)}{\rho}\,d\rho\geq
r^{2}k(r)\int_{r}^{\infty}\frac{\,d\rho}{\rho^{3}}=\frac{k(r)}{2}$.

  2. (ii)

$\displaystyle\int_{r/2}^{r}\frac{\omega(\rho)k(\rho)}{\rho}\,d\rho\leq\omega(r)k\left(\frac{r}{2}\right)\int_{r/2}^{r}\frac{\,d\rho}{\rho}\leq
c\log 2\cdot\omega(r)k(r)$.

  3. (iii)

$\displaystyle\int_{r}^{\infty}\omega(\rho)\left(-\frac{k^{\prime}(\rho)}{\rho}+\frac{2k(\rho)}{\rho^{2}}\right)\,d\rho\leq\int_{r}^{\infty}\frac{\omega(\rho)}{\rho^{\gamma}}\frac{d}{d\rho}\left(-\frac{2(1-\gamma)^{-1}k(\rho)}{\rho^{1-\gamma}}\right)\,d\rho\leq\frac{\omega(r)}{r^{\gamma}}\cdot\frac{2(1-\gamma)^{-1}k(r)}{r^{1-\gamma}}$.

  4. (iv)

$\displaystyle\int_{0}^{r}\frac{m(\rho^{-1})^{2}}{\rho
k(\rho)}\,d\rho\leq\int_{0}^{r}\frac{(\rho^{1/4}m(\rho^{-1}))^{2}}{\rho^{3/4}\big{(}\rho^{3/4}k(\rho)\big{)}}\,d\rho\leq\frac{(r^{1/4}m(r^{-1}))^{2}}{r^{3/4}k(r)}\int_{0}^{r}\frac{\,d\rho}{\rho^{3/4}}\leq\frac{4m(r^{-1})^{2}}{k(r)}$.

  5. (v)

$\displaystyle\int_{r}^{\infty}\frac{\omega(\rho)m(\rho^{-1})}{\rho^{2}}\,d\rho\leq
r^{-\gamma}\omega(r)m(r^{-1})\int_{r}^{\infty}\frac{\,d\rho}{\rho^{2-\gamma}}\leq\frac{(1-\gamma)^{-1}\omega(r)m(r^{-1})}{r}$.
∎

##  3\. Conditional regularity

In this section, we prove that some modulus of continuity for a solution of
(1.2) implies global regularity.

######  Theorem 7.

Suppose $\theta$ is a bounded weak solution of the slightly supercritical SQG
equation

$\left\\{\begin{array}[]{rl}\partial_{t}\theta+(u\cdot\nabla)\theta+\mathcal{L}\theta&=0,\\\
u&=\nabla^{\perp}\Lambda^{-1}m(\Lambda)\theta,\\\
\theta(t=0)&=\theta_{0}.\end{array}\right.$

with $k$ and $m$ satisfying the prescribed conditions. If $\theta$ admits a
modulus of continuity $\Omega$ for $[0,T]$ and $k$, $m$, $\Omega$ satisfy

$\lim_{r\to 0+}\frac{m(r^{-1})\Omega(r)}{rk(r)}=0,$

then there is a uniform-in-time estimate for
$\lVert\nabla\theta(\cdot,t)\rVert_{L^{\infty}}$ and $\theta$ is a smooth
solution on $(0,T]$.

###### Proof.

It suffices to prove the assertion for a smooth solution $\theta$. One may
consider regularized equation with the term $-\epsilon\Delta\theta$, get a
uniform estimate independent of $\epsilon$, and then take the inviscid limit
$\epsilon\to 0+$ to get the desired result.

Step 1. Evolution of $\left\lvert\nabla\theta\right\rvert^{2}$
We take gradient on both sides of the equation to get:

$\partial_{t}\nabla\theta+u\cdot\nabla^{2}\theta+\nabla
u\cdot\nabla\theta+\mathcal{L}\nabla\theta=0.$

Multiply $\nabla\theta$ to both sides:

$\nabla\theta\cdot\partial_{t}\nabla\theta+\nabla\theta\cdot
u\cdot\nabla^{2}\theta+\nabla\theta\cdot\nabla
u\cdot\nabla\theta+\nabla\theta\cdot\mathcal{L}\nabla\theta=0.$

Considering singular integral formulation for the nonlocal operator
$\mathcal{L}$ gives:

$\frac{1}{2}(\partial_{t}+u\cdot\nabla+\mathcal{L})\left\lvert\nabla\theta\right\rvert^{2}+\frac{1}{2}\underbrace{\int_{\mathbb{R}^{2}}\frac{\left\lvert\nabla\theta(x,t)-\nabla\theta(y,t)\right\rvert^{2}}{\left\lvert
x-y\right\rvert^{2}}k(x-y)\,dy}_{=:D(x,t)}=-\nabla\theta\cdot\nabla
u\cdot\nabla\theta.$ (3.1)

Due to the maximum principle for $\mathcal{L}$, it suffices to show that

$(\partial_{t}+u\cdot\nabla+\mathcal{L})\left\lvert\nabla\theta\right\rvert^{2}(x,t)<0$

whenever $\left\lvert\nabla\theta(x,t)\right\rvert$ is sufficiently large in
terms of $\lVert\theta_{0}\rVert_{L^{\infty}},k,m$, and $\Omega$.

Step 2. Pointwise lower bound of $D(x,t)$
A smooth function $\varphi:[0,\infty)\to\mathbb{R}$ is a non-decreasing cutoff
function such that

$\varphi(x)=0\text{ on }x\leq\frac{1}{2},\quad\varphi(x)=1\text{ on }x\geq
1,\quad 0\leq\varphi^{\prime}\leq 4.$

For some sufficiently small $R=R(x,t)>0$ which will be determined later, we
have the following estimate on $D(x,t)$:

$\displaystyle\frac{D(x,t)}{2\pi}$
$\displaystyle\geq\frac{1}{2\pi}\int_{\mathbb{R}^{2}}\frac{\left\lvert\nabla\theta(x,t)-\nabla\theta(y,t)\right\rvert^{2}}{\left\lvert
x-y\right\rvert^{2}}k(x-y)\varphi\Big{(}\frac{\left\lvert
x-y\right\rvert}{R}\Big{)}\,dy$
$\displaystyle\geq\left\lvert\nabla\theta(x,t)\right\rvert^{2}\frac{1}{2\pi}\int_{\left\lvert
x-y\right\rvert\geq R}\frac{k(x-y)}{\left\lvert
x-y\right\rvert^{2}}\,dy-2\nabla\theta(x,t)\cdot\frac{1}{2\pi}\int_{\mathbb{R}^{2}}\nabla\theta(y,t)\frac{k(x-y)}{\left\lvert
x-y\right\rvert^{2}}\varphi\Big{(}\frac{\left\lvert
x-y\right\rvert}{R}\Big{)}\,dy$
$\displaystyle\geq\left\lvert\nabla\theta(x,t)\right\rvert^{2}\int_{R}^{\infty}\frac{k(\rho)}{\rho}\,d\rho-2\nabla\theta(x,t)\cdot\frac{1}{2\pi}\int_{\mathbb{R}^{2}}(\theta(y,t)-\theta(x,t))\left\\{\nabla_{v}\left(\frac{k(\left\lvert
v\right\rvert)}{\left\lvert v\right\rvert^{2}}\varphi\Big{(}\frac{\left\lvert
v\right\rvert}{R}\Big{)}\right)\right\\}_{v=x-y}\,dy$
$\displaystyle\geq\frac{1}{2}\left\lvert\nabla\theta(x,t)\right\rvert^{2}k(R)-2\left\lvert\nabla\theta(x,t)\right\rvert\int_{0}^{\infty}\rho\Omega(\rho)\left\lvert\frac{d}{d\rho}\Big{(}\frac{k(\rho)\varphi(\rho/R)}{\rho^{2}}\Big{)}\right\rvert\,d\rho.$

By Lemma 6(ii) and (iii),

$\displaystyle\int_{0}^{\infty}\rho\Omega(\rho)\left\lvert\frac{d}{d\rho}\Big{(}\frac{k(\rho)\varphi(\rho/R)}{\rho^{2}}\Big{)}\right\rvert\,d\rho$
$\displaystyle\qquad\leq\frac{4}{R}\int_{R/2}^{R}\frac{\Omega(\rho)k(\rho)}{\rho}\,d\rho+\int_{R/2}^{\infty}\Omega(\rho)\left(-\frac{k^{\prime}(\rho)}{\rho}+\frac{2k(\rho)}{\rho^{2}}\right)\,d\rho$
$\displaystyle\qquad\leq\frac{c_{1}}{8}\frac{\Omega(R)k(R)}{R}.$

Therefore,

$\frac{D(x,t)}{2\pi}\geq\frac{1}{2}\left\lvert\nabla\theta(x,t)\right\rvert^{2}k(R)-\left\lvert\nabla\theta(x,t)\right\rvert\frac{c_{1}\Omega(R)k(R)}{4R}=\frac{1}{2}\left\lvert\nabla\theta(x,t)\right\rvert^{2}k(R)\left(1-\frac{c_{1}}{2\left\lvert\nabla\theta(x,t)\right\rvert}\frac{\Omega(R)}{R}\right).$

Set $R=R(x,t)>0$ to satisfy

$\frac{\Omega(R)}{R}=\frac{\left\lvert\nabla\theta(x,t)\right\rvert}{c_{1}}.$
(3.2)

Then

$D(x,t)\geq 4c_{2}\left\lvert\nabla\theta(x,t)\right\rvert^{2}k(R).$ (3.3)

Note that $R(x,t)$ is arbitrarily small when
$\left\lvert\nabla\theta(x,t)\right\rvert$ is large enough.

Step 3. Estimate of $\nabla u$

$\nabla
u(x,t)=\nabla^{\perp}\Lambda^{-1}m(\Lambda)\nabla\theta(x,t)=P.V.\int_{\mathbb{R}^{2}}K(x-y)(\nabla\theta(x,t)-\nabla\theta(y,t))\,dy.$

We estimate $\nabla u$ by splitting $\mathbb{R}^{2}$ into two pieces; an inner
piece $\left\lvert x-y\right\rvert\leq r$ and an outer piece $\left\lvert
x-y\right\rvert>r$ for some $r=r(x,t)>0$.

For the inner piece, we use the Cauchy-Schwartz inequality:

$\displaystyle\left\lvert\nabla u_{\text{in}}(x,t)\right\rvert$
$\displaystyle\leq C\sqrt{\bigg{(}\int_{\left\lvert x-y\right\rvert\leq
r}\frac{\left\lvert\nabla\theta(x,t)-\nabla\theta(y,t)\right\rvert^{2}}{\left\lvert
x-y\right\rvert^{2}}k(x-y)\,dy\bigg{)}\bigg{(}\int_{\left\lvert
x-y\right\rvert\leq r}K(x-y)^{2}\frac{\left\lvert
x-y\right\rvert^{2}}{k(x-y)}\,dy\bigg{)}}$ $\displaystyle\leq
C\sqrt{D(x,t)\int_{0}^{r}\frac{m(\rho^{-1})^{2}}{\rho k(\rho)}\,d\rho}\leq
c_{3}\sqrt{D(x,t)\frac{m(r^{-1})^{2}}{k(r)}}.$ (Lemma 4 and 6(iv))

For the outer piece, we apply integration by parts and use the modulus of
continuity of $\theta$:

$\displaystyle\left\lvert\nabla u_{\text{out}}(x,t)\right\rvert$
$\displaystyle\leq\left\lvert\int_{\left\lvert
x-y\right\rvert>r}\big{(}\theta(x,t)-\theta(y,t)\big{)}\nabla
K(x-y)\,dy\right\rvert$ $\displaystyle\quad+\left\lvert\int_{\left\lvert
x-y\right\rvert=r}\big{(}\theta(x,t)-\theta(y,t)\big{)}K(x-y)\nu(y)\,d\sigma(y)\right\rvert$
$\displaystyle\leq
C\int_{r}^{\infty}\frac{\Omega(\rho)m(\rho^{-1})}{\rho^{2}}\,d\rho+C\frac{\Omega(r)m(r^{-1})}{r}$
(Lemma 4) $\displaystyle\leq c_{4}\frac{\Omega(r)m(r^{-1})}{r}.$ (Lemma 6(v))

Therefore,

$\displaystyle\begin{aligned} \left\lvert\nabla
u\right\rvert\left\lvert\nabla\theta\right\rvert^{2}&\leq\left\lvert\nabla
u_{\text{in}}\right\rvert\left\lvert\nabla\theta\right\rvert^{2}+\left\lvert\nabla
u_{\text{out}}\right\rvert\left\lvert\nabla\theta\right\rvert^{2}\\\
&\leq\frac{D}{4}+c_{3}^{2}\frac{m(r^{-1})^{2}}{k(r)}\left\lvert\nabla\theta\right\rvert^{4}+c_{4}\frac{\Omega(r)m(r^{-1})}{r}\left\lvert\nabla\theta\right\rvert^{2}.\end{aligned}$
(3.4)

Step 4. Maximum principle
Combining (3.1), (3.3), and (3.4),

$\displaystyle(\partial_{t}+u\cdot\nabla+\mathcal{L})\left\lvert\nabla\theta\right\rvert^{2}+\frac{D}{2}+2c_{2}k(R)\left\lvert\nabla\theta\right\rvert^{2}$
$\displaystyle\leq$
$\displaystyle\,(\partial_{t}+u\cdot\nabla+\mathcal{L})\left\lvert\nabla\theta\right\rvert^{2}+D$
$\displaystyle\leq$ $\displaystyle\,2\left\lvert\nabla
u\right\rvert\left\lvert\nabla\theta\right\rvert^{2}\leq\frac{D}{2}+2c_{3}^{2}\frac{m(r^{-1})^{2}}{k(r)}\left\lvert\nabla\theta\right\rvert^{4}+2c_{4}\frac{\Omega(r)m(r^{-1})}{r}\left\lvert\nabla\theta\right\rvert^{2}.$

Hence

$(\partial_{t}+u\cdot\nabla+\mathcal{L})\left\lvert\nabla\theta\right\rvert^{2}+2c_{2}k(R)\left\lvert\nabla\theta\right\rvert^{2}\leq
2c_{3}^{2}\frac{m(r^{-1})^{2}}{k(r)}\left\lvert\nabla\theta\right\rvert^{4}+2c_{4}\frac{\Omega(r)m(r^{-1})}{r}\left\lvert\nabla\theta\right\rvert^{2}.$

What we want to show are

$2c_{3}^{2}\frac{m(r^{-1})^{2}}{k(r)}\left\lvert\nabla\theta\right\rvert^{2}<c_{2}k(R),$
(3.5)

and

$2c_{4}\frac{\Omega(r)m(r^{-1})}{r}<c_{2}k(R),$ (3.6)

whenever $\left\lvert\nabla\theta\right\rvert$ is sufficiently large. Recall
(3.2) that

$\left\lvert\nabla\theta\right\rvert=c_{1}\frac{\Omega(R)}{R}.$

Hence (3.5) is equivalent to

$\frac{m(r^{-1})^{2}\Omega(R)^{2}}{R^{2}k(r)k(R)}<\frac{c_{2}}{2c_{1}^{2}c_{3}^{2}}.$

And (3.6) is equivalent to

$\frac{m(r^{-1})\Omega(r)}{rk(R)}<\frac{c_{2}}{2c_{4}}.$

Set $r=R=R(x,t)>0$ and then the above two inequalities hold for sufficiently
small $R$, since we assumed

$\lim_{r\to 0+}\frac{m(r^{-1})\Omega(r)}{rk(r)}=0.\qed$

Consider the main example

$k(r)\sim\frac{1}{r(-\log r)^{\alpha_{1}}},\quad
m(\zeta)\sim(\log\left\lvert\zeta\right\rvert)^{\alpha_{2}}.$

Then, a modulus of continuity

$\Omega(r)\sim\frac{1}{(-\log r)^{\beta}}$

satisfies the condition

$\frac{m(r^{-1})\Omega(r)}{rk(r)}\sim\frac{(-\log
r)^{\alpha_{1}+\alpha_{2}}}{(-\log r)^{\beta}}\to 0,\quad\text{as }r\to 0+,$

if $\beta>\alpha_{1}+\alpha_{2}$.

######  Corollary 8.

Suppose $\theta$ is a bounded weak solution of the slightly supercritical SQG
equation

$\left\\{\begin{array}[]{rl}\partial_{t}\theta+(u\cdot\nabla)\theta+\Lambda(\log^{-\alpha_{1}}\Lambda)\theta&=0,\\\
u&=\nabla^{\perp}\Lambda^{-1}(\log^{\alpha_{2}}\Lambda)\theta.\end{array}\right.$

and $\theta$ admits a modulus of continuity $\Omega$ for $[0,T]$. If

$\lim_{r\to 0+}\Omega(r)(-\log r)^{\alpha_{1}+\alpha_{2}}=0,$

then $\theta$ is a smooth solution on $(0,T]$.

##  4\. Conservation of modulus of continuity

In this section, we prove that some modulus of continuity of $\theta(\cdot,t)$
is conserved by (1.2).

######  Theorem 9.

Suppose $\theta$ is a bounded weak solution of the slightly supercritical SQG
equation

$\left\\{\begin{array}[]{rl}\partial_{t}\theta+(u\cdot\nabla)\theta+\mathcal{L}\theta&=0,\\\
u&=\nabla^{\perp}\Lambda^{-1}m(\Lambda)\theta,\\\
\theta(t=0)&=\theta_{0}.\end{array}\right.$

with $k$ and $m$ satisfying the prescribed conditions. Suppose a
$C^{1}$-function $\omega:[0,\infty)\to[0,\infty)$ with $\omega(0)=0$ is
nondecreasing, concave, and satisfies (2.1) and

$\lim_{r\to 0+}\frac{m(r^{-1})}{k(r)}\frac{\omega^{\prime}(r)}{\omega(r)}=0.$

Then there exists a constant $M>0$ depending on
$\lVert\theta_{0}\rVert_{L^{\infty}}$, $k$, $m$, and $\omega$ such that if
$\theta_{0}\in L^{p}\cap L^{\infty}$ satisfies the modulus of continuity
$\frac{M}{2}\omega$, then $\theta$ satisfies the modulus of continuity
$M\omega$ as long as the solution is defined.

###### Proof.

For notational convenience, define $\delta_{h}f(x)=f(x+h)-f(x)$ for a function
$f$. We consider the maximum principle for

$v(x,h;t):=\left(\frac{\delta_{h}\theta(x,t)}{\omega(\left\lvert
h\right\rvert)}\right)^{2}F(h).$

The function $F(h)=\exp(-G(|h|))$ is just for the decay of $v$ in $h$, so that
$v\in L^{p}\cap L^{\infty}(\mathbb{R}^{2}\times\mathbb{R}^{2})$ for some
$1\leq p<\infty$. We set $G$ being smooth, nonnegative, nondecreasing, and
$G(r)=0$ for $0\leq r\leq 1$.

Step 1. Evolution of $v$
Since

$\partial_{t}\theta(x)+u(x)\cdot\nabla\theta(x)+\mathcal{L}\theta(x)=0,$

and

$\partial_{t}\theta(x+h)+u(x+h)\cdot\nabla\theta(x+h)+\mathcal{L}\theta(x+h)=0.$

Subtracting two formulas gives:

$\displaystyle\partial_{t}(\theta(x+h)-\theta(x))$
$\displaystyle\qquad+u(x)\cdot\nabla(\theta(x+h)-\theta(x))+(u(x+h)-u(x))\cdot\nabla\theta(x+h)+(\mathcal{L}\theta(x+h)-\mathcal{L}\theta(x))=0,$

and thus the evolution of $\delta_{h}\theta$ is given by

$\big{(}\partial_{t}+u\cdot\nabla_{x}+(\delta_{h}u)\cdot\nabla_{h}+\mathcal{L}_{x}\big{)}\delta_{h}\theta=0.$

Multiply $\delta_{h}\theta$ to both sides:

$\frac{1}{2}\big{(}\partial_{t}+u\cdot\nabla_{x}+(\delta_{h}u)\cdot\nabla_{h}+\mathcal{L}_{x}\big{)}(\delta_{h}\theta)^{2}(x,h,t)+\frac{1}{2}\underbrace{\int_{\mathbb{R}^{2}}\frac{\left\lvert\delta_{h}\theta(x,t)-\delta_{h}\theta(y,t)\right\rvert^{2}}{\left\lvert
x-y\right\rvert^{2}}k(x-y)\,dy}_{=:D_{h}(x,t)}=0.$

Therefore

$\big{(}\partial_{t}+u\cdot\nabla_{x}+(\delta_{h}u)\cdot\nabla_{h}+\mathcal{L}_{x}\big{)}v(x,h,t)+D_{h}(x,t)\frac{F(h)}{\omega(\left\lvert
h\right\rvert)^{2}}=-(\delta_{h}u)\cdot\frac{h}{\left\lvert
h\right\rvert}\left(G^{\prime}(\left\lvert
h\right\rvert)+\frac{2\omega^{\prime}(\left\lvert
h\right\rvert)}{\omega(\left\lvert h\right\rvert)}\right)v.$ (4.1)

Step 2. Breakthrough moment and maximum principle
Fix $M>0$, which will be determined later. Suppose $\theta_{0}$ satisfies a
modulus of continuity $\frac{M}{2}\omega$ and $\theta(t)$ admits a modulus of
continuity $M\omega$ for $0\leq t\leq t^{\ast}$. However, at the time
$t=t^{\ast}$,

$\left\lvert\delta_{h}\theta(x,t^{\ast})\right\rvert\geq M\omega(\left\lvert
h\right\rvert)$ (4.2)

for some $x,h$. Suppose we prove that

$(\partial_{t}+u\cdot\nabla_{x}+(\delta_{h}u)\cdot\nabla_{h}+\mathcal{L}_{x})v(x,h,t^{\ast})<0,$
(4.3)

whenever the such event occurs, then we get contraction from the maximum
principle of $\mathcal{L}_{x}$. Hence such an event is impossible, and the
modulus of continuity $M\omega(\cdot)$ is conserved.

Therefore it suffices to show that

$(\partial_{t}+u\cdot\nabla_{x}+(\delta_{h}u)\cdot\nabla_{h}+\mathcal{L}_{x})v(x,h,t^{*})<0,$

whenever

$M\omega(\left\lvert
h\right\rvert)\leq\left\lvert\delta_{h}\theta(x,t^{*})\right\rvert\leq
2M\omega(\left\lvert h\right\rvert).$

If (4.2) is satisfed,

$M\omega(\left\lvert
h\right\rvert)\leq\left\lvert\delta_{h}\theta(x,t^{*})\right\rvert\leq
2\lVert\theta_{0}\rVert_{L^{\infty}}.$

Hence, we only need to consider $h$ with $M\omega(\left\lvert
h\right\rvert)\leq 2\lVert\theta_{0}\rVert_{L^{\infty}}$. Let $r_{0}>0$ be
$M\omega(r_{0})=2\lVert\theta_{0}\rVert_{L^{\infty}}$. We choose $M$ large so
that $r_{0}<1$ and thus $F(h)=1$, $G^{\prime}(\left\lvert h\right\rvert)=0$
for $\left\lvert h\right\rvert\leq r_{0}$. Due to (4.1), it suffices to show
that if $M$ is large enough, then

$D_{h}(x,t^{*})\geq 2\frac{\omega^{\prime}(\left\lvert
h\right\rvert)}{\omega(\left\lvert
h\right\rvert)}\left\lvert\delta_{h}u(x,t^{*})\right\rvert\left\lvert\delta_{h}\theta(x,t^{*})\right\rvert^{2},$
(4.4)

whenever $M\omega(\left\lvert
h\right\rvert)\leq\left\lvert\delta_{h}\theta(x,t^{*})\right\rvert\leq
2M\omega(\left\lvert h\right\rvert)$. Let us denote $t^{*}$ by simply $t$ from
here.

Step 3. Pointwise lower bound of $D_{h}$
A smooth function $\varphi:[0,\infty)\to\mathbb{R}$ is a non-decreasing cutoff
function such that

$\varphi(r)=0\text{ on }r\leq\frac{1}{2},\quad\varphi(r)=1\text{ on }r\geq
1,\quad 0\leq\varphi^{\prime}\leq 4.$

For some sufficiently small $R=R(x,h,t)\geq 6\left\lvert h\right\rvert$ which
will be determined later, we have the following estimate on $D_{h}(x,t)$:

$\displaystyle\frac{D_{h}(x,t)}{2\pi}$
$\displaystyle\geq\frac{1}{2\pi}\int_{\mathbb{R}^{2}}\frac{\left\lvert\delta_{h}\theta(x,t)-\delta_{h}\theta(y,t)\right\rvert^{2}}{\left\lvert
x-y\right\rvert^{2}}k(x-y)\varphi\Big{(}\frac{\left\lvert
x-y\right\rvert}{R}\Big{)}\,dy$
$\displaystyle\geq\left\lvert\delta_{h}\theta(x,t)\right\rvert^{2}\frac{1}{2\pi}\int_{\left\lvert
x-y\right\rvert\geq R}\frac{k(x-y)}{\left\lvert
x-y\right\rvert^{2}}\,dy-2\delta_{h}\theta(x,t)\cdot\frac{1}{2\pi}\int_{\mathbb{R}^{2}}\delta_{h}\theta(y,t)\frac{k(x-y)}{\left\lvert
x-y\right\rvert^{2}}\varphi\Big{(}\frac{\left\lvert
x-y\right\rvert}{R}\Big{)}\,dy$
$\displaystyle\geq\left\lvert\delta_{h}\theta(x,t)\right\rvert^{2}\int_{R}^{\infty}\frac{k(\rho)}{\rho}\,d\rho$
$\displaystyle\qquad-2\delta_{h}\theta(x,t)\cdot\frac{1}{2\pi}\int_{\mathbb{R}^{2}}(\theta(y,t)-\theta(x,t))\bigg{\\{}\delta_{h}\left(\frac{k(\left\lvert
v\right\rvert)}{\left\lvert v\right\rvert^{2}}\varphi\Big{(}\frac{\left\lvert
v\right\rvert}{R}\Big{)}\right)\bigg{\\}}_{v=x-y}\,dy.$

By the mean value theorem,

$\displaystyle\left\lvert\delta_{h}\left(\frac{k(\left\lvert
v\right\rvert)}{\left\lvert v\right\rvert^{2}}\varphi\Big{(}\frac{\left\lvert
v\right\rvert}{R}\Big{)}\right)\right\rvert_{v=x-y}\leq\left\lvert
h\right\rvert\left\lvert\nabla_{v}\left(\frac{k(\left\lvert
v\right\rvert)}{\left\lvert v\right\rvert^{2}}\varphi\Big{(}\frac{\left\lvert
v\right\rvert}{R}\Big{)}\right)\right\rvert_{v=x-y+h^{\ast}}$
$\displaystyle\hskip 56.9055pt\leq\left\lvert
h\right\rvert\left\\{\frac{4}{R}\frac{k(\rho^{\ast})}{(\rho^{\ast})^{2}}\mathbf{1}_{R\geq\rho^{\ast}\geq\frac{R}{2}}+\left(-\frac{k^{\prime}(\rho^{\ast})}{(\rho^{\ast})^{2}}+\frac{2k(\rho^{\ast})}{(\rho^{\ast})^{3}}\right)\mathbf{1}_{\rho^{\ast}\geq\frac{R}{2}}\right\\}_{\rho^{\ast}=\left\lvert
x-y+h^{\ast}\right\rvert}$

Since $\rho^{\ast}=\left\lvert x-y+h^{\ast}\right\rvert\geq\frac{R}{2}$ and
$\left\lvert h^{\ast}\right\rvert\leq\left\lvert
h\right\rvert\leq\frac{R}{6}$, it follows $\rho=\left\lvert
x-y\right\rvert\geq\frac{R}{3}$ and
$\rho^{\ast}\geq\rho-\frac{R}{6}\geq\frac{\rho}{2}$. Hence the above formula
is less than equal to the same formula where $\rho^{\ast}$ replaced by
$\rho/2$, $R\geq\rho^{*}\geq\frac{R}{2}$ replaced by
$\frac{7R}{6}\geq\rho\geq\frac{R}{3}$, and $\rho^{\ast}\geq\frac{R}{2}$
replaced by $\rho\geq\frac{R}{3}$. In addition, regarding the modulus of
continuity of $\theta(\cdot,t)$, we have

$\left\lvert\theta(y,t)-\theta(x,t)\right\rvert\leq M\omega(\left\lvert
x-y\right\rvert).$

Therefore, wrapping up the estimates with Lemma 6, we have the following
estimate for $D_{h}$ with a constant $c_{5}(>2)$.

$\displaystyle\frac{D_{h}(x,t)}{2\pi}$
$\displaystyle\geq\frac{1}{2}\left\lvert\delta_{h}\theta(x,t)\right\rvert^{2}k(R)-2\left\lvert\delta_{h}\theta(x,t)\right\rvert\left\lvert
h\right\rvert M\cdot c_{5}\frac{\omega(R)k(R)}{R}$
$\displaystyle\geq\frac{1}{2}\left\lvert\delta_{h}\theta(x,t)\right\rvert^{2}k(R)\left(1-\frac{4c_{5}\left\lvert
h\right\rvert
M}{\left\lvert\delta_{h}\theta(x,t)\right\rvert}\frac{\omega(R)}{R}\right).$

Set $R=R(x,h,t)>0$ to satisfy

$\frac{\omega(R)}{R}=\frac{1}{8c_{5}M}\frac{\left\lvert\delta_{h}\theta(x,t)\right\rvert}{\left\lvert
h\right\rvert}.$

Since

$\frac{\omega(R)}{R}=\frac{1}{8c_{5}M}\frac{\left\lvert\delta_{h}\theta(x,t)\right\rvert}{\left\lvert
h\right\rvert}\leq\frac{1}{4c_{5}}\frac{\omega(\left\lvert
h\right\rvert)}{\left\lvert h\right\rvert}<\frac{\omega(6\left\lvert
h\right\rvert)}{6\left\lvert h\right\rvert},$

we get $R\geq 6\left\lvert h\right\rvert$ as assumed. On the other hand,

$\frac{\omega(R)}{R}=\frac{1}{8c_{5}M}\frac{\left\lvert\delta_{h}\theta(x,t)\right\rvert}{\left\lvert
h\right\rvert}\geq\frac{1}{8c_{5}}\frac{\omega(\left\lvert
h\right\rvert)}{\left\lvert h\right\rvert},$

and thus

$8c_{5}\frac{\left\lvert
h\right\rvert^{1-\gamma}}{R^{1-\gamma}}\geq\frac{\omega(\left\lvert
h\right\rvert)\left\lvert h\right\rvert^{-\gamma}}{\omega(R)R^{-\gamma}}\geq
1.$

Hence $R\leq(8c_{5})^{1/(1-\gamma)}\left\lvert h\right\rvert$. Summing up,

$D_{h}(x,t)\geq ck(R)\left\lvert\delta_{h}\theta(x,t)\right\rvert^{2}\geq
8c_{6}k(\left\lvert
h\right\rvert)\left\lvert\delta_{h}\theta(x,t)\right\rvert^{2}.$ (4.5)

Step 4. Estimate of $\delta_{h}u$

$\delta_{h}u(x,t)=P.V.\int_{\mathbb{R}^{2}}K(x-y)(\delta_{h}\theta(x,t)-\delta_{h}\theta(y,t))\,dy.$

We estimate $\nabla u$ by splitting $\mathbb{R}^{2}$ into two pieces; an inner
piece $\left\lvert x-y\right\rvert\leq 3\left\lvert h\right\rvert$ and an
outer piece $\left\lvert x-y\right\rvert>3\left\lvert h\right\rvert$.

For the inner piece, we use the Cauchy-Schwartz inequality:

$\left\lvert\delta_{h}u_{\text{in}}(x,t)\right\rvert\leq
c_{7}\sqrt{D_{h}(x,t)\frac{m(\left\lvert
h\right\rvert^{-1})^{2}}{k(\left\lvert h\right\rvert)}}.$ (4.6)

For the outer piece:

$\displaystyle\left\lvert\delta_{h}u_{\text{out}}(x,t)\right\rvert$
$\displaystyle=\left\lvert\int_{\left\lvert x-y\right\rvert>3\left\lvert
h\right\rvert}K(x-y)(\theta(y,t)-\theta(y+h,t))\,dy\right\rvert$
$\displaystyle\leq\left\lvert\int_{\left\lvert x-y\right\rvert>3\left\lvert
h\right\rvert}(K(x-y)-K(x-y+h))\theta(y,t)\,dy\right\rvert$
$\displaystyle\qquad+\left\lvert\int_{\begin{subarray}{c}\\{y:\left\lvert
x-y\right\rvert>3\left\lvert h\right\rvert\\}\triangle\\\ \\{y:\left\lvert
x-y+h\right\rvert>3\left\lvert
h\right\rvert\\}\end{subarray}}K(x-y+h)\theta(y,t)\,dy\right\rvert$

The first term is estimated as follows:

$\displaystyle\leq\lVert\theta_{0}\rVert_{L^{\infty}}\int_{\left\lvert
x-y\right\rvert>3\left\lvert h\right\rvert}\left\lvert
h\right\rvert\left\lvert\nabla K(x-y+h^{\ast})\right\rvert\,dy$ ($\left\lvert
h^{\ast}\right\rvert\leq\left\lvert h\right\rvert<\frac{1}{3}\left\lvert
x-y\right\rvert$) $\displaystyle\leq
c\lVert\theta_{0}\rVert_{L^{\infty}}\left\lvert h\right\rvert\int_{\left\lvert
x-y\right\rvert>3\left\lvert
h\right\rvert}\left\lvert\frac{2}{3}(x-y)\right\rvert^{-3}m\big{(}\left\lvert{\textstyle\frac{2}{3}}(x-y)\right\rvert^{-1}\big{)}\,dy$
$\displaystyle\leq c\lVert\theta_{0}\rVert_{L^{\infty}}\left\lvert
h\right\rvert\int_{\left\lvert
h\right\rvert}^{\infty}\frac{m\big{(}\rho^{-1}\big{)}}{\rho^{2}}\,d\rho\leq
c_{3}\lVert\theta_{0}\rVert_{L^{\infty}}m\big{(}\left\lvert
h\right\rvert^{-1}\big{)}.$

And the second term is estimated as follows:

$\displaystyle\leq\lVert\theta_{0}\rVert_{L^{\infty}}\int_{2\left\lvert
h\right\rvert<\left\lvert x-y+h\right\rvert\leq 4\left\lvert
h\right\rvert}\left\lvert K(x-y+h)\right\rvert\,dy\leq
C\lVert\theta_{0}\rVert_{L^{\infty}}\int_{2\left\lvert
h\right\rvert}^{4\left\lvert
h\right\rvert}\frac{m(\rho^{-1})}{\rho}\,d\rho\leq
C\lVert\theta_{0}\rVert_{L^{\infty}}m(\left\lvert h\right\rvert^{-1}).$

Therefore,

$\left\lvert\delta_{h}u_{\text{out}}(x,t)\right\rvert\leq
c_{8}\lVert\theta_{0}\rVert_{L^{\infty}}m(\left\lvert h\right\rvert^{-1}).$
(4.7)

Step 5. Proof of (4.4)
By the estimates (4.5), (4.6), and (4.7), it is enough to show that

$\frac{D_{h}}{4}+2c_{6}k(\left\lvert
h\right\rvert)\left\lvert\delta_{h}\theta\right\rvert^{2}\geq\left(c_{7}\sqrt{D_{h}(x,t)\frac{m(\left\lvert
h\right\rvert^{-1})^{2}}{k(\left\lvert
h\right\rvert)}}+c_{8}\lVert\theta_{0}\rVert_{L^{\infty}}m(\left\lvert
h\right\rvert^{-1})\right)\frac{\omega^{\prime}(\left\lvert
h\right\rvert)}{\omega(\left\lvert
h\right\rvert)}\left\lvert\delta_{h}\theta\right\rvert^{2}.$ (4.8)

By Young’s inequality,

R.H.S. of (4.8) $\displaystyle\leq\frac{D_{h}}{4}+c_{7}^{2}\frac{m(\left\lvert
h\right\rvert^{-1})^{2}}{k(\left\lvert
h\right\rvert)}\left(\frac{\omega^{\prime}(\left\lvert
h\right\rvert)}{\omega(\left\lvert
h\right\rvert)}\right)^{2}\left\lvert\delta_{h}\theta\right\rvert^{4}+c_{8}\lVert\theta_{0}\rVert_{L^{\infty}}m(\left\lvert
h\right\rvert^{-1})\left(\frac{\omega^{\prime}(\left\lvert
h\right\rvert)}{\omega(\left\lvert
h\right\rvert)}\right)\left\lvert\delta_{h}\theta\right\rvert^{2}.$

Our goal (4.3) is verified if

$c_{7}^{2}\frac{m(\left\lvert h\right\rvert^{-1})^{2}}{k(\left\lvert
h\right\rvert)}\left(\frac{\omega^{\prime}(\left\lvert
h\right\rvert)}{\omega(\left\lvert
h\right\rvert)}\right)^{2}\left\lvert\delta_{h}\theta\right\rvert^{2}\leq
c_{6}k(\left\lvert h\right\rvert),$ (4.9)

and

$c_{8}\lVert\theta_{0}\rVert_{L^{\infty}}m(\left\lvert
h\right\rvert^{-1})\left(\frac{\omega^{\prime}(\left\lvert
h\right\rvert)}{\omega(\left\lvert h\right\rvert)}\right)\leq
c_{6}k(\left\lvert h\right\rvert).$ (4.10)

Since $\left\lvert\delta_{h}\theta\right\rvert\leq
2\lVert\theta_{0}\rVert_{L^{\infty}}$, it is enough to show

$\left(\frac{m(\left\lvert h\right\rvert^{-1})}{k(\left\lvert
h\right\rvert)}\frac{\omega^{\prime}(\left\lvert
h\right\rvert)}{\omega(\left\lvert
h\right\rvert)}\right)^{2}\leq\frac{c_{6}}{4c_{7}^{2}\lVert\theta_{0}\rVert_{L^{\infty}}^{2}},$

instead of (4.9). And (4.10) is equivalent to

$\frac{m(\left\lvert h\right\rvert^{-1})}{k(\left\lvert
h\right\rvert)}\frac{\omega^{\prime}(\left\lvert
h\right\rvert)}{\omega(\left\lvert
h\right\rvert)}\leq\frac{c_{6}}{c_{8}\lVert\theta_{0}\rVert_{L^{\infty}}}.$
(4.11)

The above two inequalities hold for sufficiently small $\left\lvert
h\right\rvert$ by our assumption on $k,m,$ and $\omega$. It completes the
proof that a modulus of continuity $M\omega$ is conserved for sufficiently
large $M$. ∎

Consider the main example

$k(r)\sim\frac{1}{r(-\log r)^{\alpha_{1}}},\quad
m(\zeta)\sim(\log\left\lvert\zeta\right\rvert)^{\alpha_{2}}.$

Then, a modulus of continuity

$\omega(r)\sim\frac{1}{(-\log r)^{\beta}}$

satisfies the condition

$\frac{m(r^{-1})}{k(r)}\frac{\omega^{\prime}(r)}{\omega(r)}\sim\frac{r(-\log
r)^{\alpha_{1}+\alpha_{2}}}{r(-\log r)}\to 0\quad\text{as }r\to 0+,$

if $\alpha_{1}+\alpha_{2}<1$. Combining this with Corollary 8, then we obtain
Corollary 1 that (1.6) admits a global smooth solution if
$\alpha_{1}+\alpha_{2}<1$.

##  5\. Exponential Gradient Estimate

In this section, we prove the estimate (1.8) for the critical SQG equation
(1.7). Note that the critical SQG equation is a special case in the previous
discussion that

$k(r)=\frac{1}{2\pi r},\quad m(\zeta)=1.$

Refer to the Appendix for self-contained proof. Here, we present it as a
consequence of previous discussions.

Proof of Main Theorem 2. A solution $\theta$ to the critical SQG equation
exhibits the following rescaling property:

$\theta_{\lambda}(x,t)=\theta(\lambda x,\lambda t).$

Since this rescaling changes $\lVert\nabla\theta\rVert_{L^{\infty}}$ by
$\lambda$ times and remains $\lVert\theta\rVert_{L^{\infty}}$ the same, we may
assume that $\lVert\nabla\theta_{0}\rVert_{L^{\infty}}=1$. In addition, it
suffices to show our estimate for $\lVert\theta_{0}\rVert_{L^{\infty}}\geq 1$
as we already have (1.9).

For a fixed $\beta>0$, let $\omega(r)=(-\log r)^{-\beta}$ for small $r$ and
set $\Omega(r)=\min\\{M\omega(r),2\lVert\theta_{0}\rVert_{L^{\infty}}\\}$ with
a constant $M>0$ which will be determined later. Define $r_{0}\in(0,1)$ to be

$M\omega(r_{0})=2\lVert\theta_{0}\rVert_{L^{\infty}}.$

Due to our assumptions on $\theta_{0}$, it automatically admits $\Omega$ as a
modulus of continuity. If $M$ is large enough so that (4.11) is satisfied for
$\left\lvert h\right\rvert\leq r_{0}$, then the modulus of continuity $\Omega$
is conserved. In the case of critical SQG,

$\frac{r\omega^{\prime}(r)}{\omega(r)}=\beta(-\log
r)^{-1}\leq\frac{1}{c\lVert\theta_{0}\rVert_{L^{\infty}}},\quad\text{for
}r\leq r_{0}.$

Take $-\log r_{0}=c\beta\lVert\theta_{0}\rVert_{L^{\infty}}$ and
$M=2(c\beta)^{\beta}\lVert\theta_{0}\rVert_{L^{\infty}}^{1+\beta}$, which
satisfy the above conditions.

Let $r_{1}\in(0,1)$ be

$\Omega(r_{1})=\min\bigg{\\{}\frac{c_{2}^{\frac{1}{2}}}{2^{\frac{1}{2}}c_{1}c_{3}},\frac{c_{2}}{2c_{4}},1\bigg{\\}}=:2(c^{\prime})^{-1},$

where $c_{1},c_{2},c_{3},c_{4}$ are numeric constants in the proof of Theorem
7. According to the proof of Theorem 7, in particular the last part, we get

$\lVert\nabla\theta\rVert_{L^{\infty}}\leq
c_{1}\frac{\Omega(r_{1})}{r_{1}}\leq
c_{1}\exp\Big{(}c(c^{\prime})^{\frac{1}{\beta}}\beta\lVert\theta_{0}\rVert_{L^{\infty}}^{1+\frac{1}{\beta}}\Big{)}.$

######  Remark.

Since there is no control for constants as $\beta^{-1}\to 0$, we do not know
how to improve it to $C\exp(C\lVert\theta_{0}\rVert_{L^{\infty}})$.

### Acknowledgment

The author thanks In-Jee Jeong for suggesting the topic and providing valuable
discussions.

## References

  * [1] L. A. Caffarelli and A. Vasseur. “Drift diffusion equations with fractional diffusion and the quasi-geostrophic equation.” Annals of Mathematics 171.3 (2010): 1903-1930. arXiv:math/0608447
  * [2] P. Constantin, A. J. Majda, and E. Tabak. “Formation of strong fronts in the 2-D quasigeostrophic thermal active scalar.” Nonlinearity 7.6 (1994): 1495-1533.
  * [3] P. Constantin and J. Wu. “Behavior of solutions of 2D quasi-geostrophic equations.” SIAM journal on mathematical analysis 30.5 (1999): 937-948.
  * [4] P. Constantin, D. Córdoba, and J. Wu. “On the critical dissipative quasi-geostrophic equation.” Indiana University Mathematics Journal (2001): 97-107.
  * [5] P. Constantin and J. Wu. “Regularity of Hölder continuous solutions of the supercritical quasi-geostrophic equation.” Annales de l’Institut Henri Poincaré C, Analyse non linéaire. 25.6. (2008): 1103-1110. arXiv:math/0701592
  * [6] P. Constantin and V. Vicol. “Nonlinear maximum principles for dissipative linear nonlocal operators and applications.” Geometric And Functional Analysis 22.5 (2012): 1289-1321. arXiv:1110.0179
  * [7] A. Córdoba and D. Córdoba. “A maximum principle applied to quasi-geostrophic equations.” Communications in Mathematical Physics 249.3 (2004): 511-528.
  * [8] M. Dabkowski. “Eventual regularity of the solutions to the supercritical dissipative quasi-geostrophic equation.” Geometric and Functional Analysis 1.21 (2011): 1-13. arXiv:1007.2970
  * [9] M. Dabkowski, A. Kiselev, and V. Vicol. “Global well-posedness for a slightly supercritical surface quasi-geostrophic equation.” Nonlinearity 25.5 (2012): 1525. arXiv:1106.2137
  * [10] M. Dabkowski, A. Kiselev, L. Silvestre, and V. Vicol. “Global well-posedness of slightly supercritical active scalar equations.” Analysis & PDE 7.1 (2014): 43-72. arXiv:1203.6302
  * [11] H. Dong and N. Pavlović. “A regularity criterion for the dissipative quasi-geostrophic equations.” Annales de l’IHP Analyse non linéaire. 26.5 (2009): 1607-1619. arXiv:0710.5201
  * [12] H. Dong. “Dissipative quasi-geostrophic equations in critical Sobolev spaces: smoothing effect and global well-posedness.” Dicrete Contin. Dyn. Syst. 26.4 (2010): 1197-1211. arXiv:math/0701826
  * [13] A. Kiselev, F. Nazarov, and A. Volberg. “Global well-posedness for the critical 2D dissipative quasi-geostrophic equation.” Inventiones mathematicae 167.3 (2007): 445-453. arXiv:math/0604185
  * [14] A. Kiselev and F. Nazarov. “Variation on a theme of Caffarelli and Vasseur.” Journal of Mathematical Sciences 166.1 (2010): 31-39. arXiv:0908.0923
  * [15] A. Kiselev. “Nonlocal maximum principles for active scalars.” Advances in Mathematics 227.5 (2011): 1806-1826. arXiv:1009.0542
  * [16] S. G. Resnick. Dynamical problems in non-linear advective partial differential equations. 1996, Ph.D. Thesis, University of Chicago.
  * [17] L. Silvestre. “Eventual regularization for the slightly supercritical quasi-geostrophic equation.” Annales de l’Institut Henri Poincaré C, Analyse non linéaire. 27.2 (2010): 693-704. arXiv:0812.4901
  * [18] E. M. Stein. Singular Integrals and Differentiability Properties of Functions (PMS-30). Vol. 30. Princeton University Press, 1970.

## Appendix

This is supplementary to section 5. We repeat discussion on section 3 and 4,
in the sprit of the critical SQG, which gives a self-contained proof for Main
Theorem 2.

Main Theorem 2. The critical SQG equation

$\left\\{\begin{array}[]{rl}\partial_{t}\theta+(u\cdot\nabla)\theta+\Lambda\theta&=0,\\\
u&=\nabla^{\perp}\Lambda^{-1}\theta,\\\
\theta(t=0)&=\theta_{0}\end{array}\right.$ (5.0)

with initial $\theta_{0}\in\mathcal{S}(\mathbb{R}^{2})$ has a unique global
smooth solution. In addition, for any $\gamma>0$,

$\lVert\nabla\theta\rVert_{L^{\infty}_{t,x}}\leq
C\lVert\nabla\theta_{0}\rVert_{L^{\infty}}\exp(C\lVert\theta_{0}\rVert_{L^{\infty}}^{1+\gamma}),$
(5.1)

where constants depend only on $\gamma$.

A solution $\theta$ to the critical SQG equation exhibits the following
rescaling property:

$\theta_{\lambda}(x,t)=\theta(\lambda x,\lambda t).$

Since this rescaling changes $\lVert\nabla\theta\rVert_{L^{\infty}}$ by
$\lambda$ times and remains $\lVert\theta\rVert_{L^{\infty}}$ the same, we may
assume that $\lVert\nabla\theta_{0}\rVert_{L^{\infty}}=1$. In addition, it
suffices to show our estimate for $\lVert\theta_{0}\rVert_{L^{\infty}}\geq 1$
as we already have a double exponential estimate.

Let $\omega(r)=(-\log r)^{-\beta}$ for some $\beta>0$ and let
$\Omega(r)=\min\\{M\omega(r),2\lVert\theta_{0}\rVert_{L^{\infty}}\\}$ with a
constant $M$ which will be determined later. Note that $\theta_{0}$ admits
$\min\\{\frac{M}{2}\omega(r),2\lVert\theta_{0}\rVert_{L^{\infty}}\\}$ as a
modulus of continuity. We split the proof into two parts:

######  Proposition 10.

Suppose $\theta\in L^{\infty}([0,T];L^{p}\cap L^{\infty})$ is a weak solution
of the critical SQG equation (5.0) with an initial $\theta_{0}$ satisfying
prescribed assumptions. Then there exists constant
$M=C_{\beta}\lVert\theta_{0}\rVert_{L^{\infty}}^{1+\beta}$ so that
$\theta(t,\cdot)$ admits modulus of continuity $\Omega$ as long as the
solution is defined.

######  Proposition 11.

Suppose $\theta$ is a bounded weak solution of the critical SQG equation
(5.0). If $\theta$ admits a modulus of continuity $\Omega$ for $[0,T]$, then

$\lVert\nabla\theta\rVert_{L^{\infty}_{t,x}}\leq
C\lVert\nabla\theta_{0}\rVert_{L^{\infty}}\exp(CM^{\frac{1}{\beta}}),$

where constants depend on $\beta$. Thus $\theta$ is a smooth solution on
$(0,T]$.

###### Proof of Proposition 10.

Let $r_{0}\in(0,1)$ be $M\omega(r_{0})=2\lVert\theta\rVert_{L^{\infty}}$.
Since $L^{\infty}$-norm is conserved, it suffices to show the modulus of
continuity for distance less than $r_{0}$.

For notational convenience, define $\delta_{h}f(x)=f(x+h)-f(x)$ for a function
$f$. We consider the maximum principle for

$v(x,h;t):=\left(\frac{\delta_{h}\theta(x,t)}{\omega(\left\lvert
h\right\rvert)}\right)^{2}F(h).$

The function $F(h)=\exp(-G(|h|))$ is just for the decay of $v$ in $h$, so that
$v\in L^{p}\cap L^{\infty}(\mathbb{R}^{2}\times\mathbb{R}^{2})$ for some
$1\leq p<\infty$. We set $G$ being smooth, nonnegative, nondecreasing, and
$G(r)=0$ for $0\leq r\leq 1$.

Step 1. Evolution of $v$
Since

$\partial_{t}\theta(x)+u(x)\cdot\nabla\theta(x)+\Lambda\theta(x)=0,$

and

$\partial_{t}\theta(x+h)+u(x+h)\cdot\nabla\theta(x+h)+\Lambda\theta(x+h)=0.$

Subtracting two formulas gives:

$\displaystyle\partial_{t}(\theta(x+h)-\theta(x))$
$\displaystyle\qquad+u(x)\cdot\nabla(\theta(x+h)-\theta(x))+(u(x+h)-u(x))\cdot\nabla\theta(x+h)+(\Lambda\theta(x+h)-\Lambda\theta(x))=0,$

and thus the evolution of $\delta_{h}\theta$ is given by

$\big{(}\partial_{t}+u\cdot\nabla_{x}+(\delta_{h}u)\cdot\nabla_{h}+\Lambda_{x}\big{)}\delta_{h}\theta=0.$

Multiply $\delta_{h}\theta$ to both sides:

$\frac{1}{2}\big{(}\partial_{t}+u\cdot\nabla_{x}+(\delta_{h}u)\cdot\nabla_{h}+\Lambda_{x}\big{)}(\delta_{h}\theta)^{2}(x,h,t)+\frac{1}{2}\underbrace{\frac{1}{2\pi}\int_{\mathbb{R}^{2}}\frac{\left\lvert\delta_{h}\theta(x,t)-\delta_{h}\theta(y,t)\right\rvert^{2}}{\left\lvert
x-y\right\rvert^{3}}\,dy}_{=:D_{h}(x,t)}=0.$

Therefore

$\big{(}\partial_{t}+u\cdot\nabla_{x}+(\delta_{h}u)\cdot\nabla_{h}+\Lambda_{x}\big{)}v(x,h,t)+D_{h}(x,t)\frac{F(h)}{\omega(\left\lvert
h\right\rvert)^{2}}=-(\delta_{h}u)\cdot\frac{h}{\left\lvert
h\right\rvert}\left(G^{\prime}(\left\lvert
h\right\rvert)+\frac{2\omega^{\prime}(\left\lvert
h\right\rvert)}{\omega(\left\lvert h\right\rvert)}\right)v.$ (5.2)

Step 2. Breakthrough moment and maximum principle
Fix $M>0$, which will be determined later. Suppose $\theta_{0}$ satisfies a
modulus of continuity $\frac{M}{2}\omega$ and $\theta(t)$ admits a modulus of
continuity $M\omega$ for $0\leq t\leq t^{\ast}$. However, at the time
$t=t^{\ast}$,

$\left\lvert\delta_{h}\theta(x,t^{\ast})\right\rvert\geq M\omega(\left\lvert
h\right\rvert)$ (5.3)

for some $x,h$. Suppose we prove that

$(\partial_{t}+u\cdot\nabla_{x}+(\delta_{h}u)\cdot\nabla_{h}+\Lambda_{x})v(x,h,t^{\ast})<0,$
(5.4)

whenever the such event occurs, then we get contraction from the maximum
principle of $\Lambda_{x}$. Hence such an event is impossible, and the modulus
of continuity $M\omega(\cdot)$ is conserved.

Therefore it suffices to show that

$(\partial_{t}+u\cdot\nabla_{x}+(\delta_{h}u)\cdot\nabla_{h}+\Lambda_{x})v(x,h,t^{*})<0,$

whenever

$M\omega(\left\lvert
h\right\rvert)\leq\left\lvert\delta_{h}\theta(x,t^{*})\right\rvert\leq
2M\omega(\left\lvert h\right\rvert).$

If (5.3) is satisfed,

$M\omega(\left\lvert
h\right\rvert)\leq\left\lvert\delta_{h}\theta(x,t^{*})\right\rvert\leq
2\lVert\theta_{0}\rVert_{L^{\infty}}.$

Hence, we only need to consider $\left\lvert h\right\rvert\leq r_{0}$. Due to
(5.2), it suffices to show that

$\frac{D_{h}(x,t^{*})}{2\beta}\geq\frac{1}{\left\lvert
h\right\rvert(-\log\left\lvert
h\right\rvert)}\left\lvert\delta_{h}u(x,t^{*})\right\rvert\left\lvert\delta_{h}\theta(x,t^{*})\right\rvert^{2},$
(5.5)

whenever $M\omega(\left\lvert
h\right\rvert)\leq\left\lvert\delta_{h}\theta(x,t^{*})\right\rvert\leq
2M\omega(\left\lvert h\right\rvert)$, $\left\lvert h\right\rvert\leq r_{0}$.
Let us denote $t^{*}$ by simply $t$ from here.

Step 3. Pointwise lower bound of $D_{h}$
A smooth function $\varphi:[0,\infty)\to\mathbb{R}$ is a non-decreasing cutoff
function such that

$\varphi(r)=0\text{ on }r\leq\frac{1}{2},\quad\varphi(r)=1\text{ on }r\geq
1,\quad 0\leq\varphi^{\prime}\leq 4.$

For some sufficiently small $R=R(x,h,t)\geq 6\left\lvert h\right\rvert$ which
will be determined later, we have the following estimate on $D_{h}(x,t)$:

$\displaystyle D_{h}(x,t)$
$\displaystyle\geq\frac{1}{2\pi}\int_{\mathbb{R}^{2}}\frac{\left\lvert\delta_{h}\theta(x,t)-\delta_{h}\theta(y,t)\right\rvert^{2}}{\left\lvert
x-y\right\rvert^{3}}\varphi\Big{(}\frac{\left\lvert
x-y\right\rvert}{R}\Big{)}\,dy$
$\displaystyle\geq\left\lvert\delta_{h}\theta(x,t)\right\rvert^{2}\frac{1}{2\pi}\int_{\left\lvert
x-y\right\rvert\geq R}\frac{1}{\left\lvert
x-y\right\rvert^{3}}\,dy-2\delta_{h}\theta(x,t)\cdot\frac{1}{2\pi}\int_{\mathbb{R}^{2}}\delta_{h}\theta(y,t)\frac{1}{\left\lvert
x-y\right\rvert^{3}}\varphi\Big{(}\frac{\left\lvert
x-y\right\rvert}{R}\Big{)}\,dy$
$\displaystyle\geq\left\lvert\delta_{h}\theta(x,t)\right\rvert^{2}\int_{R}^{\infty}\frac{1}{\rho^{2}}\,d\rho$
$\displaystyle\qquad-2\delta_{h}\theta(x,t)\cdot\frac{1}{2\pi}\int_{\mathbb{R}^{2}}(\theta(y,t)-\theta(x,t))\bigg{\\{}\delta_{h}\left(\frac{1}{\left\lvert
v\right\rvert^{3}}\varphi\Big{(}\frac{\left\lvert
v\right\rvert}{R}\Big{)}\right)\bigg{\\}}_{v=x-y}\,dy.$

By the mean value theorem,

$\displaystyle\left\lvert\delta_{h}\left(\frac{1}{\left\lvert
v\right\rvert^{3}}\varphi\Big{(}\frac{\left\lvert
v\right\rvert}{R}\Big{)}\right)\right\rvert_{v=x-y}\leq\left\lvert
h\right\rvert\left\lvert\nabla_{v}\left(\frac{1}{\left\lvert
v\right\rvert^{3}}\varphi\Big{(}\frac{\left\lvert
v\right\rvert}{R}\Big{)}\right)\right\rvert_{v=x-y+h^{\ast}}$
$\displaystyle\hskip 56.9055pt\leq\left\lvert
h\right\rvert\left\\{\frac{4}{R}\frac{1}{(\rho^{\ast})^{3}}\mathbf{1}_{R\geq\rho^{\ast}\geq\frac{R}{2}}+\frac{3}{(\rho^{*})^{4}}\mathbf{1}_{\rho^{\ast}\geq\frac{R}{2}}\right\\}_{\rho^{\ast}=\left\lvert
x-y+h^{\ast}\right\rvert}$

Since $\rho^{\ast}=\left\lvert x-y+h^{\ast}\right\rvert\geq\frac{R}{2}$ and
$\left\lvert h^{\ast}\right\rvert\leq\left\lvert
h\right\rvert\leq\frac{R}{6}$, it follows $\rho=\left\lvert
x-y\right\rvert\geq\frac{R}{3}$ and
$\rho^{\ast}\geq\rho-\frac{R}{6}\geq\frac{\rho}{2}$. Hence the above formula
is less than equal to the same formula where $\rho^{\ast}$ replaced by
$\rho/2$, $R\geq\rho^{*}\geq\frac{R}{2}$ replaced by
$\frac{7R}{6}\geq\rho\geq\frac{R}{3}$, and $\rho^{\ast}\geq\frac{R}{2}$
replaced by $\rho\geq\frac{R}{3}$. In addition, regarding the modulus of
continuity of $\theta(\cdot,t)$, we have

$\left\lvert\theta(y,t)-\theta(x,t)\right\rvert\leq M\omega(\left\lvert
x-y\right\rvert).$

Therefore, we have the following estimate for $D_{h}$ with a constant
$c_{5}(>3)$.

$\displaystyle D_{h}(x,t)$
$\displaystyle\geq\frac{\left\lvert\delta_{h}\theta(x,t)\right\rvert^{2}}{R}-2\left\lvert\delta_{h}\theta(x,t)\right\rvert\left\lvert
h\right\rvert M\cdot c_{5}\frac{\omega(R)}{R^{2}}$
$\displaystyle\geq\frac{\left\lvert\delta_{h}\theta(x,t)\right\rvert^{2}}{R}\left(1-\frac{2c_{5}\left\lvert
h\right\rvert
M}{\left\lvert\delta_{h}\theta(x,t)\right\rvert}\frac{\omega(R)}{R}\right).$

Set $R=R(x,h,t)>0$ to satisfy

$\frac{\omega(R)}{R}=\frac{1}{4c_{5}M}\frac{\left\lvert\delta_{h}\theta(x,t)\right\rvert}{\left\lvert
h\right\rvert}.$

Since

$\frac{\omega(R)}{R}=\frac{1}{4c_{5}M}\frac{\left\lvert\delta_{h}\theta(x,t)\right\rvert}{\left\lvert
h\right\rvert}\leq\frac{1}{2c_{5}}\frac{\omega(\left\lvert
h\right\rvert)}{\left\lvert h\right\rvert}<\frac{\omega(6\left\lvert
h\right\rvert)}{6\left\lvert h\right\rvert},$

we get $R\geq 6\left\lvert h\right\rvert$ as assumed. On the other hand,

$\frac{\omega(R)}{R}=\frac{1}{4c_{5}M}\frac{\left\lvert\delta_{h}\theta(x,t)\right\rvert}{\left\lvert
h\right\rvert}\geq\frac{1}{4c_{5}}\frac{\omega(\left\lvert
h\right\rvert)}{\left\lvert h\right\rvert},$

and thus

$4c_{5}\frac{\left\lvert
h\right\rvert^{1-\gamma}}{R^{1-\gamma}}\geq\frac{\omega(\left\lvert
h\right\rvert)\left\lvert h\right\rvert^{-\gamma}}{\omega(R)R^{-\gamma}}\geq
1.$

Hence $R\leq(4c_{5})^{1/(1-\gamma)}\left\lvert h\right\rvert$. Summing up,

$D_{h}(x,t)\geq\frac{\left\lvert\delta_{h}\theta(x,t)\right\rvert^{2}}{2R}\geq
8c_{6}\frac{\left\lvert\delta_{h}\theta(x,t)\right\rvert^{2}}{\left\lvert
h\right\rvert}.$ (5.6)

Step 4. Estimate of $\delta_{h}u$

$\delta_{h}u(x,t)=P.V.\int_{\mathbb{R}^{2}}\frac{(x-y)^{\perp}}{\left\lvert
x-y\right\rvert^{3}}(\delta_{h}\theta(x,t)-\delta_{h}\theta(y,t))\,dy.$

Name the kernel by $K(z)=z^{\perp}/\left\lvert z\right\rvert^{3}$. We estimate
$\nabla u$ by splitting $\mathbb{R}^{2}$ into two pieces; an inner piece
$\left\lvert x-y\right\rvert\leq 3\left\lvert h\right\rvert$ and an outer
piece $\left\lvert x-y\right\rvert>3\left\lvert h\right\rvert$.

For the inner piece, we use the Cauchy-Schwartz inequality:

$\displaystyle\begin{split}\left\lvert\delta_{h}u_{\text{in}}(x,t)\right\rvert&\leq
C\sqrt{\bigg{(}\int_{\left\lvert x-y\right\rvert\leq 3\left\lvert
h\right\rvert}\frac{\left\lvert\delta_{h}\theta(x,t)-\delta_{h}\theta(y,t)\right\rvert^{2}}{\left\lvert
x-y\right\rvert^{3}}\,dy\bigg{)}\bigg{(}\int_{\left\lvert x-y\right\rvert\leq
3\left\lvert h\right\rvert}\frac{1}{\left\lvert
x-y\right\rvert}\,dy\bigg{)}}\\\ &=c_{7}\sqrt{D_{h}(x,t)\left\lvert
h\right\rvert}.\end{split}$ (5.7)

For the outer piece:

$\displaystyle\left\lvert\delta_{h}u_{\text{out}}(x,t)\right\rvert$
$\displaystyle=\left\lvert\int_{\left\lvert x-y\right\rvert>3\left\lvert
h\right\rvert}K(x-y)(\theta(y,t)-\theta(y+h,t))\,dy\right\rvert$
$\displaystyle\leq\left\lvert\int_{\left\lvert x-y\right\rvert>3\left\lvert
h\right\rvert}(K(x-y)-K(x-y+h))\theta(y,t)\,dy\right\rvert$
$\displaystyle\qquad+\left\lvert\int_{\begin{subarray}{c}\\{y:\left\lvert
x-y\right\rvert>3\left\lvert h\right\rvert\\}\triangle\\\ \\{y:\left\lvert
x-y+h\right\rvert>3\left\lvert
h\right\rvert\\}\end{subarray}}K(x-y+h)\theta(y,t)\,dy\right\rvert$

The first term is estimated as follows:

$\displaystyle\leq\lVert\theta_{0}\rVert_{L^{\infty}}\int_{\left\lvert
x-y\right\rvert>3\left\lvert h\right\rvert}\left\lvert
h\right\rvert\left\lvert\nabla K(x-y+h^{\ast})\right\rvert\,dy$ ($\left\lvert
h^{\ast}\right\rvert\leq\left\lvert h\right\rvert<\frac{1}{3}\left\lvert
x-y\right\rvert$) $\displaystyle\leq
c\lVert\theta_{0}\rVert_{L^{\infty}}\left\lvert h\right\rvert\int_{\left\lvert
x-y\right\rvert>3\left\lvert h\right\rvert}\left\lvert
x-y\right\rvert^{-3}\,dy=c_{3}\lVert\theta_{0}\rVert_{L^{\infty}}.$

And the second term is estimated as follows:

$\displaystyle\leq\lVert\theta_{0}\rVert_{L^{\infty}}\int_{2\left\lvert
h\right\rvert<\left\lvert x-y+h\right\rvert\leq 4\left\lvert
h\right\rvert}\left\lvert K(x-y+h)\right\rvert\,dy\leq
C\lVert\theta_{0}\rVert_{L^{\infty}}\int_{2\left\lvert
h\right\rvert}^{4\left\lvert h\right\rvert}\frac{1}{\rho}\,d\rho\leq
C\lVert\theta_{0}\rVert_{L^{\infty}}.$

Therefore,

$\left\lvert\delta_{h}u_{\text{out}}(x,t)\right\rvert\leq
c_{8}\lVert\theta_{0}\rVert_{L^{\infty}}.$ (5.8)

Step 5. Proof of (5.5)
By the estimates (5.6), (5.7), and (5.8), it is enough to show that

$\frac{D_{h}}{4\beta}+\frac{2c_{6}}{\beta}\frac{\left\lvert\delta_{h}\theta\right\rvert^{2}}{\left\lvert
h\right\rvert}\geq\left(c_{7}\sqrt{D_{h}(x,t)\left\lvert
h\right\rvert}+c_{8}\lVert\theta_{0}\rVert_{L^{\infty}}\right)\frac{1}{\left\lvert
h\right\rvert(-\log\left\lvert
h\right\rvert)}\left\lvert\delta_{h}\theta\right\rvert^{2}.$ (5.9)

By Young’s inequality,

R.H.S. of (5.9) $\displaystyle\leq\frac{D_{h}}{4\beta}+\beta
c_{7}^{2}\left\lvert h\right\rvert\left(\frac{1}{\left\lvert
h\right\rvert(-\log\left\lvert
h\right\rvert)}\right)^{2}\left\lvert\delta_{h}\theta\right\rvert^{4}+c_{8}\lVert\theta_{0}\rVert_{L^{\infty}}\left(\frac{1}{\left\lvert
h\right\rvert(-\log\left\lvert
h\right\rvert)}\right)\left\lvert\delta_{h}\theta\right\rvert^{2}.$

If we verify that

$\beta c_{7}\left\lvert\delta_{h}\theta\right\rvert\left(-\log\left\lvert
h\right\rvert\right)^{-1}\leq c_{6},$ (5.10)

and

$\beta c_{8}\lVert\theta_{0}\rVert_{L^{\infty}}(-\log\left\lvert
h\right\rvert)^{-1}\leq c_{6},$ (5.11)

then (5.4) holds. Since $\left\lvert\delta_{h}\theta\right\rvert\leq
2\lVert\theta_{0}\rVert_{L^{\infty}}$, it suffices to show

$-\log\left\lvert h\right\rvert\geq\beta
c_{9}\lVert\theta_{0}\rVert_{L^{\infty}},\quad\text{for all }\left\lvert
h\right\rvert\leq r_{0},$

instead of (5.10) and (5.11). We completes the proof by setting $M=2(\beta
c_{9})^{\beta}\lVert\theta_{0}\rVert_{L^{\infty}}^{1+\beta}$. ∎

###### Proof of Proposition 11.

It suffices to prove the assertion for a smooth solution $\theta$. One may
consider regularized equation with the term $-\epsilon\Delta\theta$, get a
uniform estimate independent of $\epsilon$, and then take the inviscid limit
$\epsilon\to 0+$ to get the desired result.

Step 1. Evolution of $\left\lvert\nabla\theta\right\rvert^{2}$
We take gradient on both sides of the equation to get:

$\partial_{t}\nabla\theta+u\cdot\nabla^{2}\theta+\nabla
u\cdot\nabla\theta+\Lambda\nabla\theta=0.$

Multiply $\nabla\theta$ to both sides:

$\nabla\theta\cdot\partial_{t}\nabla\theta+\nabla\theta\cdot
u\cdot\nabla^{2}\theta+\nabla\theta\cdot\nabla
u\cdot\nabla\theta+\nabla\theta\cdot\Lambda\nabla\theta=0.$

Considering singular integral formulation for the nonlocal operator $\Lambda$
gives:

$\frac{1}{2}(\partial_{t}+u\cdot\nabla+\Lambda)\left\lvert\nabla\theta\right\rvert^{2}+\frac{1}{2}\underbrace{\frac{1}{2\pi}\int_{\mathbb{R}^{2}}\frac{\left\lvert\nabla\theta(x,t)-\nabla\theta(y,t)\right\rvert^{2}}{\left\lvert
x-y\right\rvert^{3}}\,dy}_{=:D(x,t)}=-\nabla\theta\cdot\nabla
u\cdot\nabla\theta.$ (5.12)

Our goal here is to prove

$(\partial_{t}+u\cdot\nabla+\Lambda)\left\lvert\nabla\theta\right\rvert^{2}(x,t)<0$

whenever $\left\lvert\nabla\theta(x,t)\right\rvert$ is sufficiently large in
terms of $M$.

Step 2. Pointwise lower bound of $D(x,t)$
A smooth function $\varphi:[0,\infty)\to\mathbb{R}$ is a non-decreasing cutoff
function such that

$\varphi(x)=0\text{ on }x\leq\frac{1}{2},\quad\varphi(x)=1\text{ on }x\geq
1,\quad 0\leq\varphi^{\prime}\leq 4.$

For some sufficiently small $R=R(x,t)>0$ which will be determined later, we
have the following estimate on $D(x,t)$:

$\displaystyle D(x,t)$
$\displaystyle\geq\frac{1}{2\pi}\int_{\mathbb{R}^{2}}\frac{\left\lvert\nabla\theta(x,t)-\nabla\theta(y,t)\right\rvert^{2}}{\left\lvert
x-y\right\rvert^{3}}\varphi\Big{(}\frac{\left\lvert
x-y\right\rvert}{R}\Big{)}\,dy$
$\displaystyle\geq\left\lvert\nabla\theta(x,t)\right\rvert^{2}\frac{1}{2\pi}\int_{\left\lvert
x-y\right\rvert\geq R}\frac{1}{\left\lvert
x-y\right\rvert^{3}}\,dy-2\nabla\theta(x,t)\cdot\frac{1}{2\pi}\int_{\mathbb{R}^{2}}\nabla\theta(y,t)\frac{1}{\left\lvert
x-y\right\rvert^{3}}\varphi\Big{(}\frac{\left\lvert
x-y\right\rvert}{R}\Big{)}\,dy$
$\displaystyle\geq\left\lvert\nabla\theta(x,t)\right\rvert^{2}\int_{R}^{\infty}\frac{1}{\rho^{2}}\,d\rho-2\nabla\theta(x,t)\cdot\frac{1}{2\pi}\int_{\mathbb{R}^{2}}(\theta(y,t)-\theta(x,t))\left\\{\nabla_{v}\left(\frac{1}{\left\lvert
v\right\rvert^{3}}\varphi\Big{(}\frac{\left\lvert
v\right\rvert}{R}\Big{)}\right)\right\\}_{v=x-y}\,dy$
$\displaystyle\geq\frac{\left\lvert\nabla\theta(x,t)\right\rvert^{2}}{\rho}-2\left\lvert\nabla\theta(x,t)\right\rvert\int_{0}^{\infty}\rho\Omega(\rho)\left\lvert\frac{d}{d\rho}\Big{(}\frac{\varphi(\rho/R)}{\rho^{3}}\Big{)}\right\rvert\,d\rho.$

The second term:

$\int_{0}^{\infty}\rho\Omega(\rho)\left\lvert\frac{d}{d\rho}\Big{(}\frac{\varphi(\rho/R)}{\rho^{3}}\Big{)}\right\rvert\,d\rho\leq\frac{4}{R}\int_{R/2}^{R}\frac{\Omega(\rho)}{\rho^{2}}\,d\rho+2\int_{R/2}^{\infty}\frac{\Omega(\rho)}{\rho^{3}}\,d\rho\leq\frac{c_{1}}{4}\frac{\Omega(R)}{R^{2}}.$

Therefore,

$D(x,t)\geq\frac{\left\lvert\nabla\theta(x,t)\right\rvert^{2}}{R}-\left\lvert\nabla\theta(x,t)\right\rvert\frac{c_{1}\Omega(R)}{2R^{2}}=\frac{\left\lvert\nabla\theta(x,t)\right\rvert^{2}}{R}\left(1-\frac{c_{1}}{2\left\lvert\nabla\theta(x,t)\right\rvert}\frac{\Omega(R)}{R}\right).$

Set $R=R(x,t)>0$ to satisfy

$\frac{\Omega(R)}{R}=\frac{\left\lvert\nabla\theta(x,t)\right\rvert}{c_{1}}.$
(5.13)

Then

$D(x,t)\geq\frac{\left\lvert\nabla\theta(x,t)\right\rvert^{2}}{2R}.$ (5.14)

Note that $R(x,t)$ is arbitrarily small when
$\left\lvert\nabla\theta(x,t)\right\rvert$ is large enough.

Step 3. Estimate of $\nabla u$

$\nabla
u(x,t)=\nabla^{\perp}\Lambda^{-1}\nabla\theta(x,t)=P.V.\int_{\mathbb{R}^{2}}K(x-y)(\nabla\theta(x,t)-\nabla\theta(y,t))\,dy,$

where $K(z)=z^{\perp}/\left\lvert z\right\rvert^{3}$. We estimate $\nabla u$
by splitting $\mathbb{R}^{2}$ into two pieces; an inner piece $\left\lvert
x-y\right\rvert\leq R$ and an outer piece $\left\lvert x-y\right\rvert>R$.

For the inner piece, we use the Cauchy-Schwartz inequality:

$\displaystyle\left\lvert\nabla u_{\text{in}}(x,t)\right\rvert$
$\displaystyle\leq C\sqrt{\bigg{(}\int_{\left\lvert x-y\right\rvert\leq
R}\frac{\left\lvert\nabla\theta(x,t)-\nabla\theta(y,t)\right\rvert^{2}}{\left\lvert
x-y\right\rvert^{3}}\,dy\bigg{)}\bigg{(}\int_{\left\lvert x-y\right\rvert\leq
R}\frac{1}{\left\lvert x-y\right\rvert}\,dy\bigg{)}}$
$\displaystyle=c_{3}\sqrt{D(x,t)R}.$

For the outer piece, we apply integration by parts and use the modulus of
continuity of $\theta$:

$\displaystyle\left\lvert\nabla u_{\text{out}}(x,t)\right\rvert$
$\displaystyle\leq\left\lvert\int_{\left\lvert
x-y\right\rvert>R}\big{(}\theta(x,t)-\theta(y,t)\big{)}\nabla
K(x-y)\,dy\right\rvert$ $\displaystyle\quad+\left\lvert\int_{\left\lvert
x-y\right\rvert=R}\big{(}\theta(x,t)-\theta(y,t)\big{)}K(x-y)\nu(y)\,d\sigma(y)\right\rvert$
$\displaystyle\leq
C\int_{R}^{\infty}\frac{\Omega(\rho)}{\rho^{2}}\,d\rho+C\frac{\Omega(R)}{R}\leq
c_{4}\frac{\Omega(R)}{R}.$

Therefore,

$\displaystyle\begin{aligned} \left\lvert\nabla
u\right\rvert\left\lvert\nabla\theta\right\rvert^{2}&\leq\left\lvert\nabla
u_{\text{in}}\right\rvert\left\lvert\nabla\theta\right\rvert^{2}+\left\lvert\nabla
u_{\text{out}}\right\rvert\left\lvert\nabla\theta\right\rvert^{2}\\\
&\leq\frac{D}{4}+c_{3}^{2}R\left\lvert\nabla\theta\right\rvert^{4}+c_{4}\frac{\Omega(R)}{R}\left\lvert\nabla\theta\right\rvert^{2}.\end{aligned}$
(5.15)

Step 4. Maximum principle
Combining (5.12), (5.14), and (5.15),

$\displaystyle(\partial_{t}+u\cdot\nabla+\Lambda)\left\lvert\nabla\theta\right\rvert^{2}+\frac{D}{2}+\frac{\left\lvert\nabla\theta\right\rvert^{2}}{2R}$
$\displaystyle\leq$
$\displaystyle\,(\partial_{t}+u\cdot\nabla+\Lambda)\left\lvert\nabla\theta\right\rvert^{2}+D$
$\displaystyle\leq$ $\displaystyle\,2\left\lvert\nabla
u\right\rvert\left\lvert\nabla\theta\right\rvert^{2}\leq\frac{D}{2}+2c_{3}^{2}R\left\lvert\nabla\theta\right\rvert^{4}+2c_{4}\frac{\Omega(R)}{R}\left\lvert\nabla\theta\right\rvert^{2}.$

Hence

$(\partial_{t}+u\cdot\nabla+\Lambda)\left\lvert\nabla\theta\right\rvert^{2}+\frac{\left\lvert\nabla\theta\right\rvert^{2}}{2R}\leq
2c_{3}^{2}R\left\lvert\nabla\theta\right\rvert^{4}+2c_{4}\frac{\Omega(R)}{R}\left\lvert\nabla\theta\right\rvert^{2}.$

Recall that (5.13) that

$\left\lvert\nabla\theta\right\rvert=c_{1}\frac{\Omega(R)}{R}.$

And let $r_{1}\in(0,1)$ be

$M(-\log
r_{1})^{-\beta}=\Omega(r_{1})=\min\Big{\\{}\frac{1}{8^{1/2}c_{1}c_{3}},\frac{1}{8c_{4}},1\Big{\\}}=:c^{\prime}.$

If $\left\lvert\nabla\theta\right\rvert>c_{1}\frac{\Omega(r_{1})}{r_{1}}$,
then $R<r_{1}$,

$2c_{3}^{2}R\bigg{(}c_{1}\frac{\Omega(R)}{R}\bigg{)}^{2}<\frac{1}{4R},\quad\text{and}\quad
2c_{4}\frac{\Omega(R)}{R}<\frac{1}{4R}.$

Hence,

$(\partial_{t}+u\cdot\nabla+\Lambda)\left\lvert\nabla\theta\right\rvert^{2}(x,t)<0.$

Due to the maximum principle,

$\lVert\nabla\theta\rVert_{L^{\infty}_{t,x}}\leq\max\Big{\\{}c_{1}\frac{\Omega(r_{1})}{r_{1}},1\Big{\\}}\leq\frac{C}{r_{1}}\leq
C\exp\big{(}(c^{\prime})^{-\frac{1}{\beta}}M^{\frac{1}{\beta}}\big{)}.\qed$

Since $M=C\lVert\theta_{0}\rVert_{L^{\infty}}^{1+\beta}$ according to
Proposition 10, we get (5.1):

$\lVert\nabla\theta\rVert_{L^{\infty}_{t,x}}\leq
C\lVert\nabla\theta_{0}\rVert_{L^{\infty}}\exp(C\lVert\theta_{0}\rVert_{L^{\infty}}^{1+\frac{1}{\beta}}).$