In this paper, we investigate the persistency of loglog singular structure for 2D Euler equation under the perturbation of continuous function. First, we prove that when loglog vortex is perturbed by the function with modulus of continuity $\mu(r)=\left( \log \frac{1}{r}\right)^{-\alpha} ~~ (0<\alpha<1)$, the loglog vortex is propagated while the perturbation term maintaining same modulus of continuity. We then show that this result is sharp: There exists a smooth function such that when the loglog vortex is perturbed by the smooth function, the norm of perturbation term corresponding to modulus of continuity $\mu(r)=\left( \log \frac{1}{r}\right)^{-\alpha} $ instantly blows up for all $\alpha>1$.

§ INTRODUCTION

§.§ Persistence of singular structure

Singularity, by which we mean a quantity is infinite, is one of the most important and interesting object in mathematical physics. For example, a fundamental question in fluid dynamics is whether a singularity could be formed from smooth initial data in finite time, e.g., for 3D Navier-Stokes equation and 3D Euler equation. Besides the singularity formation issues, one can alternatively start from singularity and investigate whether the singular structure persists. We illustrate the persistency of singular structure for 2D Euler equation in the vorticity form

\begin{equation} \label{EE}
\left\{ \begin{aligned}
& w_t+u \cdot \nabla w=0, \\
& w(\cdot,0)=w_0,
\end{aligned} \right.
\end{equation}

where the vorticity $w: \bbR^2 \rightarrow \bbR$ and velocity field $u: \bbR^2 \rightarrow \bbR^2$ are related through Biot-Savart law

\begin{equation} \label{Biot-Savart}
u(x,t) = \frac{1}{2\pi} \int_{\bbR^2} \frac{(x-y)^\perp}{\abs{x-y}^2}w(y) \, dy
\end{equation}

for $(a,b)^\perp \coloneqq (-b,a)$. The system (<ref>)-(<ref>) is known to be globally well-posded in several standard spaces, e.g., Sobolev space $H^s~~(s>1)$ [8], Hölder space $C^{k,\alpha}~~ (0<\alpha<1)$ by Wolibner [10], and $L^1 \cap L^\infty$ by Yudovich [11][If a function space is written without domain, its domain is automatically assumed to be $\bbR^2$.]. However, in many physical phenomena, vorticity is sharply concentrated in a small region so that it would be modeled through unbounded or even measure-valued function, which could be regarded as a singularity. One of the most important example of the singularity is so-called point vortex where the vorticity is Dirac measure concentrated in the origin

\begin{equation*}
w_0(x) = \Gamma \delta_0, \qquad v_0(x) = \frac{1}{2\pi} \frac{x^\perp}{\abs{x}^2}
\end{equation*}

for $\Gamma \in \bbR$. Due to its radial nature, the solution emanating from the point vortex is stationary. However, it is more physically reasonable to consider a point vortex with a perturbation term rather than the exact point vortex. Thus, consider initial vorticity of the following form

\begin{equation} \label{equ13}
w_0(x) = \Gamma \delta_0+g_0(x)
\end{equation}

for $g_0 \in L^1 \cap L^\infty$. Then, it is natural to ask whether the solution $w(x,t)$ of 2D Euler equation starting from initial vorticity (<ref>) maintains the same structure as (<ref>), i.e.,

\begin{equation*}
w(x,t) = \Gamma \delta_{\phi^*(t)}+g(x,t)
\end{equation*}

for continuous map $\phi^*:[0,\infty) \rightarrow \bbR^2$ and $g \in L^\infty_{\textnormal{loc}}([0,\infty); L^1 \cap L^\infty)$. The answer to the question, to the best of our knowledge, was shown to be positive under the additional assumption that $g_0$ is constant near the center of the vortex [7]. Actually, not to mention the solution structure, even the uniqueness for initial data of the form (<ref>) is not known due to the seriousness of Dirac singularity. Therefore, to treat the problem of persistency of singular structure rigorously, it is desirable to consider milder singularity for which global well-posdedness was verified. In fact, there has been several results which guaranteed global well-posedness for specific kinds of unbounded function, e.g., for function with mild $L^p$-norm growth with respect to $p$ by Yudovich [12] (see [3] for more elementary proof by Crippa–Stefani), for critical Besov space by Vishik [9], and for BMO-type space by Bernicot–Keraani [1]. Among these admissible class for global well-posedness, the loglog vortex is nearly the most severe singularity [5], and this is a consequence of Theorem 1.6 & Theorem 1.8 of [3] with its mild $L^p$-norm growth (see Lemma 1 in [4])

\begin{equation*}
\nrm{\log \log \frac{1}{\abs{\cdot}}}_{L^p(B(0,e^{-2}))} \leq M ( \log p+1).
\end{equation*}

Therefore, for a loglog-type singular vorticity $w_s:\bbR^2 \backslash \set{0} \rightarrow \bbR$ defined by

\begin{equation} \label{def of ws}
w_s(x) \coloneqq \begin{cases}
\log \log \frac{1}{\abs{x}}, & (0<\abs{x}<e^{-2}), \\
s(\abs{x}), & (e^{-2} \leq \abs{x} \leq e^{-1}), \\
0, & (\abs{x} > e^{-1}),
\end{cases}
\end{equation}

where $s:(0,\infty) \rightarrow \bbR$ could be any function which makes $w_s$ smooth function with bounded support on $\bbR^2$, we have the following global well-posedness of loglog vortex with a bounded perturbation.

Let $g$ be any bounded function with compact support. Then, there exists a unique global weak solution $($in the standard distribution sense$)$ for (<ref>)-(<ref>) starting from initial vorticity

\begin{equation*}
w_0(x) = w_s +g_0.
\end{equation*}

In addition, the solution $w$ is Lagrangian; The solution $w(x,t)$ could be written by

\begin{equation*}
\end{equation*}

where $\Phi_t^{-1}(x)$ is the inverse flow map of $u~(\coloneqq \nabla^\perp \Delta^{-1} w)$.

As we have global well-posedness for initial vorticity of the form

\begin{equation} \label{initial vorticity 1}
w_0(x) = w_s(x) + \text{(compactly supported bounded function)},
\end{equation}

we can safely return to the following precise question on the persistency of loglog singular structure:

For a Banach space $X \subset L^\infty$, consider the initial vorticity of the form

\begin{equation} \label{initial vorticity 2}
\end{equation}

for compactly supported $g_0 \in X$. Then, is it possible to write the solution $w(x,t)$ of 2D Euler equation as the form of

\begin{equation} \label{solution form vorticity}
\end{equation}

for continuous $\phi^*:[0,\infty) \rightarrow \bbR^2$ and $g \in L^\infty_{\textnormal{loc}}([0,\infty); X)$?

To the best of our knowledge, Question <ref> was first investigated by Drivas–Elgindi–La [4], where they gave positive answer to for $X=L^\infty$. In this paper, we aim to answer the Question <ref> for other spaces $X$ consisting of continuous function. We also provide a sharp regularity threshold which separate the answer to Question <ref>.

§.§ Main Result

We first define a function space induced from a modulus of continuity.

Let $\mu: [0,c] \rightarrow [0,\infty)$ be a modulus of continuity (MOC), i.e., $\mu$ is a non-decreasing continuous function such that $\mu(0)=0$. Then, the space $C^\mu$ is defined by

\begin{equation*}
C^\mu \coloneqq \Big\{ f \in L^\infty \, \Big\vert \, \nrm{f}_{C^\mu} \coloneqq \nrm{f}_{\infty}+ \sup_{0<\abs{x-y}<c} \frac{\abs{f(x)-f(y)}}{\mu(\abs{x-y})} < \infty \Big\}.
\end{equation*}

For example, for the modulus of continuities $\mu_1(r)= r \log \frac{1}{r} $ and $\mu_2(r) =r^\alpha$, the space $C^{\mu_1}$ and $C^{\mu_2}$ are the collection of log-Lipschitz function and Hölder continuous function respectively. In this paper, we fix the following three modulus of continuities:

\begin{equation} \label{MOCs}
\varphi_\alpha(r)=\left( \log \frac{1}{r} \right)^{-\alpha}, \qquad \psi_\alpha(r) = r \left( \log \frac{1}{r}\right)^{1-\alpha}, \qquad \phi_\beta(r) = \left( \log \frac{1}{r} \right)^{-\beta}
\end{equation}

for $\alpha \in (0,1), \beta \in (1,\infty)$.

We are now ready to state the first main result which corresponds to persistence of singular structure in the propagation of loglog vortex $w_s$.

Let $\alpha \in (0,1)$ be given and $g_0 \in C^{\varphi_\alpha}$ be a compactly supported function. Then, the unique solution $w(x,t)$ of 2D Euler equation starting from initial vorticity

\begin{equation*}
\end{equation*}

could be written as the form of

\begin{equation*}
\end{equation*}

for $g \in L^\infty_{\textnormal{loc}}([0,\infty), \cphi)$. Here, $\phi^*:[0,\infty) \rightarrow \bbR^2$ is the particle trajectory of the origin.

Theorem <ref> gives positive answer to Question <ref> for $X=\cphi$. While constant vorticity condition is necessary in the neighborhood of point vortex to prove the propagation of point vortex up to bounded correction, Theorem <ref> does not require any specific condition in the neighborhood of loglog vortex. Also, our result shows that loglog vortex is not strong enough to break continuity of perturbation term for wide class of continuous initial vorticity, beyond the boundedness of perturbation term which was shown in [4].

Theorem <ref> leads to the natural question of the existence of other function space $X$ in $\bbR^2$ which also gives positive answer to Question <ref>. However, the next main result shows that Theorem <ref> is sharp: The answer to Question <ref> is No for $X=C^{\phi_\beta}$ for any $\beta \in (1,\infty)$. More precisely:

There exists a compactly supported smooth function $g_0$ such that for any $\beta\in (1,\infty)$ and $T>0$, the unique solution $w(x,t)$ starting from the initial vorticity

\begin{equation*}
\end{equation*}

which, by Theorem <ref>, could be written in the form of

\begin{equation*}
\end{equation*}

for $g \in L^\infty([0,T];L^\infty)$ and trajectory of the origin $\phi^*:[0,\infty)\rightarrow \bbR^2$, we have

\begin{equation} \label{blowup}
\sup_{t \in [0,T]} \nrm{g(\cdot,t)}_{C^{\phi_\beta}}=\infty.
\end{equation}

Note that the initial perturbation $g_0$ chosen in Theorem <ref> is compactly supported smooth function. Furthermore, as can be seen in its precise definition (<ref>), $g_0$ is identically zero in a neighborhood of the origin in which the local $C^{\phi_\beta}$-norm blows up instantaneously in the sense of (<ref>). This implies that the blow-up (<ref>) does not come from a delicate effect of perturbation term, but essentially originates from the structure of loglog singularity. It is also noteworthy that one does not prevent blow-up even though the norm is estimated in lower regularity space whose regularity deteriorates continuously in time, i.e., $H^{s(t)}$ or $C^{0,\alpha(t)}$ (See, e.g., [2] for this kind of loss of regularity for logarithmically singular surface quasi-geostophic equation).

Since modulus of continuity $\mu(r)=r^\alpha ~~(0<\alpha<1)$ is regular than $\phi_\beta(r)=\left(\log \frac{1}{r}\right)^{-\beta} ~~(\beta>1)$, Theorem <ref> implies breakdown of Hölder continuity for the perturbation term.

In the definition of singular vortex $w_s$ (<ref>), $w_s(x)$ does not need to be equal to $\log \log \frac{1}{\abs{x}}$ exactly near the origin. Instead, any $C^2$ radial function on $\bbR^2 \backslash \set{0}$ which is equivalent to $\log \log \frac{1}{\abs{x}}$ up to second derivative and supported in bounded set suffices to replace $w_s$. More precisely, if a $C^2$ function $q:(0,\infty) \rightarrow \bbR$ satisfies

* $\supp(q)$ is bounded,

* There exist constants $R<e^{-2}, M_1, M_2>0$ such that

\begin{equation*}
M_1 \frac{d^k}{dx^k} \left( \log \log \frac{1}{x}\right) \leq \frac{d^k q}{dx^k} \leq M_2 \frac{d^k}{dx^k} \left( \log \log \frac{1}{x} \right)
\end{equation*}

for all $0<\abs{x}<R$ and $k=0, 1, 2$,

$w_s(x)$ could be replaced with $q(\abs{x})$. This is because, in the proof of our main theorem, we only use information of $w_s$ up to second derivatives.

If $g_0$ is smooth, something non-trivial could happen only along the center of radial function $w_s$, which was denoted by $\phi^*(t)$. More precisely, by the propagation of regularity for 2D Euler equation (Proposition 8.3 of [8]), we already know that $w(\cdot,t)$ is smooth at $\bbR^2 \backslash \{\phi^*(t)\}$[In the Proposition 8.3 of [8], there is $L^1 \cap L^\infty$ assumption on $w_0$. However, this $L^1 \cap L^\infty$ condition is only used to get log-Lipschitz flow. For the solution $w$ in (<ref>), even though $w_s(x-\phi^*(t))$ does not belong to $L^\infty$, it generates log-Lipschitz velocity field by Lemma <ref> so that the proof of Proposition 8.3 in [8] works without modification.]. Therefore, it suffices to focus on value of perturbation term $g$ in (<ref>) near the center of loglog vortex.

Note in Theorem <ref> that $g \notin L^\infty([0,T]; C^{\phi_\beta})$ does not imply $g(\cdot, t) \notin C^{\phi_\beta}$ for any fixed $t \in (0,\infty)$. We hope that the blow-up of $C^{\phi_\beta}$-norm for fixed time will be (dis-)proved in the future works.

§.§ Notations

In this subsection, we introduce some notations which will be used throughout this paper.

* For $x \in \bbR^2$ and $r>0$, $B(x,r)$ is a ball of radius $r$ centered at $x$.

* For $0<a<b$, the annulus $A(a,b)$ is defined by

\begin{equation*}
A(a,b) \coloneqq B(0,b) \backslash B(0,a). \vspace{2mm}
\end{equation*}

* By $C^\infty_c$, we mean the space of compactly supported smooth function.

* For each term in (<ref>), we denote the velocity field generated by each term by

\begin{equation} \label{def of velocities}
u \coloneqq \nabla^\perp \Delta^{-1}w, \qquad u_s \coloneqq \nabla^\perp \Delta^{-1}w_s, \qquad u_g \coloneqq \nabla^\perp \Delta^{-1} g,
\end{equation}

so that we could get

\begin{equation} \label{solution form velocity}
\end{equation}

from (<ref>).

* When we write $M=M(\alpha_1, ..., \alpha_n)>0$, we mean a constant $M>0$ depends on variables $\alpha_1, ..., \alpha_n$. Thus, if we write just $M>0$, it is an absolute constant which does not depend on any other variables, or a constant which depends only on fixed solution of 2D Euler throughout this paper.

* Although a constant could be changed line by line in successive inequalities, we represent different constants with same symbol for simplicity.

§.§ Idea of the proof

To present the idea for our main results, suppose $g_0$ in (<ref>) is symmetric, i.e., $g_0(x)=g(-x)$, just for simplicity. Then, equation for $g$ could be written by

\begin{equation} \label{equ8}
\left\{ \begin{aligned}
& g_t+u \cdot \nabla g =-u_g \cdot \nabla w_s, \\
& g(\cdot,0)=g_0,
\end{aligned} \right.
\end{equation}

which is a forced transport equation. Here, $u$ is a unique solution of (<ref>)-(<ref>) starting from initial vorticity $w_0$ in (<ref>). By symmetry on $g_0$, we have $u_g(0)=0$ so that it offsets the singularity of $\nabla w_s$, which is equal to $-\frac{1}{\abs{x} \log \frac{1}{\abs{x}}}\frac{x}{\abs{x}}$ near the origin. The following Table <ref> shows the regularity of forcing term $-u_g \cdot \nabla w_s$ under the various regularity assumptions on $g$.

Case     MOC  for $g$     MOC for $u_g$     $u_g \cdot \nabla w_s (x)$

(i)     $L^\infty$     $r \log \frac{1}{r}$     $\approx 1$

(ii)     $r^\alpha ~~(0<\alpha<1)$     $r$     $\approx \left( \log \frac{1}{\abs{x}}\right)^{-1}$

(iii)     $\left( \log \frac{1}{r}\right)^{-\alpha} ~~ (0<\alpha<1)$     $r \left( \log \frac{1}{r} \right)^{1-\alpha}$     $\approx \left( \log \frac{1}{\abs{x}}\right)^{-\alpha}$

(vi)     $\left( \log \frac{1}{r} \right)^{-1}$     $r \log \log \frac{1}{r}$     $\approx \frac{\log \log \frac{1}{\abs{x}}}{\log \frac{1}{\abs{x}}}$

(v)     $\left( \log \frac{1}{r} \right)^{-\beta} ~~ (\beta>1)$     $r$     $\approx \left( \log \frac{1}{\abs{x}}\right)^{-1} $

Magnitude of forcing term under various modulus of continuity for $g$

In the case (i) and (iii), we can see that assumed modulus of continuity on $g$ (second column) and forcing term on the equation (<ref>) (fourth column) is consistent. This is the fundamental reason the Theorem <ref> holds. Having noticed this consistency, Theorem <ref> could be proven by considering the following standard iteration scheme: Set $g^{(0)}(x,t)=g_0(x)$ and define $g^{(n+1)}$ by the solution of

\begin{equation} \label{equ11}
\left\{ \begin{aligned}
& g^{(n+1)}_t+ u \cdot \nabla g^{n+1} = -u_{g^{(n)}} \cdot \nabla w_s, \\
& g^{(n+1)}(\cdot,0) = g_0.
\end{aligned} \right.
\end{equation}

Then, it will be shown that the sequence $\{ g^{(n)} \}_{n \in \bbN}$ is the Cauchy sequence in $L^\infty([0, T]; \cphi)$ for any $T \geq 0$ and $\alpha \in (0,1)$. Even if symmetric assumption on $g_0$ is eliminated, the same proof works since offset of singularity in $\nabla w_s(x-\phi^*(t))$ still occurs as can be seen in the general equation for $g$ (<ref>).

In contrast, in the case (ii) and (v), we can see that assumed modulus of continuity on $g$ and forcing term on the (<ref>) is inconsistent: Forcing term is far greater. This is the essential reason the Theorem <ref> and Remark <ref> holds. We briefly sketch the proof of Theorem <ref>. To show $\lim_{t \searrow 0} \nrm{g(\cdot,t)}_{C^{\phi_\beta}}=\infty$, we first construct symmetric initial perturbation $g_0 \in C^\infty_c$ as in <ref> so that $u_{g_0}(\coloneqq \nabla^\perp \Delta^{-1} g_0)$ is a hyperbolic Lipschitz velocity field satisfying (<ref>) and then investigate the value of $g$ along the trajectory $\phi_r(t)$, where $\phi_r: [0,\infty) \rightarrow \bbR^2$ is the unique solution of ODE

\begin{equation}
\left\{ \begin{aligned}
& \frac{d}{dt} \phi_r(t) = u(\phi_r(t),t), \\
& \phi_r(0)=(0,r) \quad (r>0).
\end{aligned} \right.
\end{equation}

From (<ref>) and the fact that $g_0$ vanishes near the origin, we have

\begin{equation} \label{equ9}
g(\phi_r(t),t) = -\int_0^t u_g(\phi_r(s),s) \cdot \nabla w_s (\phi_r(s)) \, ds
\end{equation}

for all small $r>0$. Note that when $s=0$, the integrand becomes

\begin{equation*}
u_g(\phi_r(0),0) \cdot \nabla w_s (\phi_r(0)) \underset{\eqref{def of ws} }{=} u_{g_0}(0,r) \cdot \frac{1}{\abs{(0,r)}\log \frac{1}{\abs{(0,r)}}} \frac{-(0,r)}{\abs{(0,r)}} \underset{\eqref{Lip initial} }{\geq} \frac{K}{\boldsymbol{\log \frac{1}{r}}},
\end{equation*}

which strongly suggests that the forcing term is strong enough to break modulus of continuity less severe than $\left( \log \frac{1}{r}\right)^{-1}$. Thus, to prove Theorem <ref>, it suffices to show the integrand does not deviate far from the value when $s=0$ up to time $t$, i.e.,

\begin{align}
u_g(\phi_r(s),s) \cdot \nabla w_s (\phi_r(s)) &\approx u_g(\phi_r(0),0) \cdot \nabla w_s (\phi_r(0)), \label{equ10} \\
\abs{\phi_r(s)}& \approx \abs{\phi_r(0)} \label{equ10-1}
\end{align}

for all small $r>0$ and $0 \leq s \leq t$ [Here, $A \approx B$ means there exists $c>1$ such that $\frac{1}{c}A \leq B \leq cA$].

However, The main obstacle is that we do not have Lipschitz regularity on velocity field $u$ a priori, even though information on $\nabla u$ is necessary to establish (<ref>). Furthermore, this lack of regularity could create infinitely small scale in the origin so that the particle trajectory starting at $(0,r)$ could get closer to the origin, a center of loglog singularity near which estimates become worse, far faster than exponential rate (i.e., $\abs{\phi_r(t)}\ll re^{-t}$). This made (<ref>) seem infeasible. To overcome this problems, in the earlier version of this paper, which proved breakdown of Hölder continuity mentioned in Remark <ref>, used proof by contradiction; the author made contradictory assumption that perturbation term $g(x,t)$ has uniform in time $C^{0,\alpha}$-norm by which they could avoid aforementioned issues.

In this paper, without adopting proof by contradiction, we use a posteriori estimates available from Section <ref>. Although results in Section <ref> still does not guarantee that $u$ is Lipschitz (precisely, the modulus of continuity for $u$ will be $\mu(r)=r \left( \log \frac{1}{r}\right)^{1-\alpha}$), we show that this regularity is enough to control $\abs{\phi_r(t)}$ for short time. To be more precise, we show that for any $\alpha \in (0,1)$, there exist $M>0$, $R=R(\alpha)>0$ and $T>0$ such that for all $0<r<R$ and $0 \leq t \leq T$,

\begin{equation} \label{equ10-2}
re^{-Mt\left( \log \frac{1}{r}\right)^{1-\alpha}}<\abs{\phi_r(t)} < re^{Mt\left( \log \frac{1}{r}\right)^{1-\alpha}},
\end{equation}

so that if we take time $t=t(r)$ sufficiently small than $\left( \log \frac{1}{r}\right)^{1-\alpha}$, we could get (<ref>). Also, combining (<ref>) with interior Schauder estimate, we could get estimate for $\abs{\nabla u(x)}$ for all $x\in \bbR^2 \backslash \set{0}$ (Proposition <ref>), and this leads to (<ref>) without a global Lipschitz estimate on $u$. Using (<ref>)-(<ref>), we get Theorem <ref> almost as a corollary.

§.§ Organization of the paper

We prove the conservation of modulus of continuity (Theorem <ref>) in Section <ref> and breakdown of modulus of continuity (Theorem <ref>) in Section <ref>.

In Section <ref>, we gather elementary lemmas to apply iteration scheme (<ref>). In Section <ref>, we prove Theorem <ref> by completing the iteration scheme.

In Section <ref>, we construct $g_0$ in (<ref>) and get some estimates for the solution $(w,u)$ starting from initial vorticity $w_0$ in (<ref>). In Section <ref>, we gather results in Section <ref> to prove the Key Lemma (Lemma <ref>). Then, we finish the proof of Theorem <ref> in Section <ref> as a corollary of Proposition <ref>.

Acknowledgements. The author thanks In-Jee Jeong for fruitful discussions. The author also acknowledges partial support by Samsung Science and Technology Foundation (SSTF-BA2002-04).

§ CONSERVATION OF MODULUS OF CONTINUITY

In this section, we prove Theorem <ref>. For simplicity, we prove the following theorem where we assume the center of loglog vortex is fixed at the origin due to symmetric property of $g_0$ in initial data (<ref>)

Let $\alpha \in (0,1)$ be given. Suppose compactly supported $g_0 \in C^{\varphi_\alpha}$ is symmetric with respect to the origin $($i.e., $g_0(-x)=g_0(x)$ for all $x \in \bbR^2$$)$ so that perturbation term $g(x,t)$ in (<ref>) is symmetric with respect to the origin and $u_g$ in (<ref>) is odd-symmetric with respect to the origin $($i.e., $u_g(x,t)=-u_g(x,t)$ for all $x\in \bbR^2, t \in [0,\infty)$$)$. Then, $ g \in L^\infty_{\textnormal{loc}}([0,\infty); C^{\varphi_\alpha})$.

To prove Theorem <ref>, we gather a few elementary lemmas.

§.§ Elementary Lemmas

Let $\alpha \in (0,1)$ be given. Then, there exists a constant $C=C(\alpha, \abs{\supp (w)})$ such that for compactly supported $w \in \cphi$, we have $u=\nabla^\perp \Delta^{-1}w \in \cpsi$ with

\begin{equation} \label{vor to vel}
\nrm{u}_{\cpsi} \leq C \nrm{w}_{\cphi}.
\end{equation}

The proof is almost same as the proof of the fact that velocity field generated by $L^1 \cap L^\infty$ vorticity is log-Lipschitz (see Lemma 8.1 of [8]). From the proof of Lemma 8.1 of [8], there exists $C>0$ such that for $\abs{x-y}<1$,

\begin{equation} \label{eq1}
\begin{aligned}
& \abs{u(x)-u(y)} \\
\underset{\substack{\text{Lem 8.1 } \text{in [8]}} }&{\leq} C ( \nrm{w}_1 + \nrm{w}_{\infty}) \abs{x-y}+ C\abs{x-y} \int_{B(x,1) \backslash B(x, 2\abs{x-y})} \frac{\abs{w(x)-w(z)}}{\abs{x-z}^2} \, dz \\
& \leq C ( \nrm{w}_1 + \nrm{w}_{\infty}) \abs{x-y}+ C\nrm{w}_{\cphi}\abs{x-y}  \int_{B(x,1) \backslash B(x, 2\abs{x-y})} \frac{\logalpha{x-z}}{\abs{x-z}^2} \, dz \\
&= C ( \nrm{w}_1 + \nrm{w}_{\infty}) \abs{x-y}+ \frac{C}{1-\alpha}  \nrm{w}_{\cphi} \abs{x-y} \left( \log \frac{1}{2\abs{x-y}}\right)^{1-\alpha} \\
&\leq C \nrm{w}_{\cphi} \abs{x-y}\left( \log \frac{1}{\abs{x-y}}\right)^{1-\alpha},
\end{aligned}
\end{equation}

where $C$ is a constant depending only on $\alpha$ and $\abs{\supp (w)}$. Combining (<ref>) with

\begin{equation*}
\nrm{u}_{\infty} \leq \nrm{w}_{1} + \nrm{w}_{\infty} \leq (\abs{\supp (w)}+1) \nrm{w}_{\cphi}
\end{equation*}

yields the inequality (<ref>).

Next, we point out the fact that under the log-Lipschitz vector field, the two points could get closer or farther at most double exponential rate.

Let $u$ be a log-Lipschitz vector field such that

\begin{equation*}
\frac{u(x,t)-u(y,t)}{\abs{x-y} \log \frac{1}{\abs{x-y}}} \leq N
\end{equation*}

for all $\abs{x-y} < e^{-1}, 0 \leq t \leq T$, and $\Phi(\cdot,t)$ be its flow map. Then, for $\abs{x-y}<e^{-NT}$,

\begin{equation} \label{access rate}
\abs{x-y}^{e^{Nt}} \leq \abs{\Phi(x,t)-\Phi(y,t)} \leq \abs{x-y}^{e^{-Nt}}.
\end{equation}

This follows directly from the comparison with ODE

\begin{equation*}
\left\{ \begin{aligned}
x'(t)&= \pm Nx(t)\log \frac{1}{x(t)}, & (0 \leq t \leq T), \\
x(0)&=x_0 ,& (x_0 < e^{-NT}),
\end{aligned} \right.
\end{equation*}

whose solution is $x(t)= x_0^{e^{\mp Nt}}$.

To prove Theorem <ref>, we now derive the equation for perturbation term $g$ in (<ref>). Since $(w,u)$ is a vorticity-velocity pair for 2D Euler equation, solution form (<ref>), (<ref>) give

\begin{equation*}
(\underbrace{w_s(x-\phi^*(t))+g}_{=w})_t + ( \underbrace{u_s(x-\phi^*(t))+u_g}_{=u}) \cdot \nabla (w_s(x-\phi^*(t))+g) = 0.
\end{equation*}

Then, we have

\begin{equation} \label{eq2}
g_t+u \cdot \nabla g = - \left( u_g - \frac{d \phi^*}{dt}(t) \right) \cdot \nabla w_s(x-\phi^*(t)) -u_s(x-\phi^*(t)) \cdot \nabla w_s(x-\phi^*(t)).
\end{equation}

Since $w_s$ is radial and $w_s(x)= \log \log \frac{1}{\abs{x}}$ for $0<\abs{x}<e^{-2}$, the explicit formula $u_s(x)= \left( \frac{1}{\abs{x}}\int_0^{\abs{x}}r \log \log \frac{1}{r}\, dr \right)\mathbf{e}_\theta$ gives

\begin{equation*}
\lim_{x \rightarrow 0} \frac{\abs{u_s(x)}}{\abs{x}  \log \log \frac{1}{\abs{x}}}=\frac{1}{2},
\end{equation*}

which yields $u_s(0)=0$. Therefore,

\begin{equation}\label{eq3}
\frac{d\phi^*}{dt}(t) = u(\phi^*(t),t) \underset{\eqref{solution form velocity}}{=} \underbrace{u_s(\phi^*(t)-\phi^*(t)) }_{=0}+u_g(\phi^*(t),t)=u_g(\phi^*(t),t).
\end{equation}

Since radial symmetry of $w_s$ makes $u_s$ and $\nabla w_s$ orthogonal, (<ref>) could be simplified to

\begin{equation} \label{equation for g original}
g_t+u \cdot \nabla g = -(u_g(x,t) -u_g(\phi^*(t),t)) \cdot \nabla w_s (x-\phi^*(t)).
\end{equation}

Here, just for simplicity, suppose initial perturbation term $g_0$ is symmetric with respect to the origin. Then, initial velocity field $u_0$ satisfies $u_0(-x)=-u_0(x)$, so does $u$ by a symmetric property of the Euler equation. As $u(0,t)=0$ for all $t \in [0,\infty)$, $\phi^*(t) \equiv 0$ for all $t \in [0,\infty)$ so that (<ref>) could be simplified to

\begin{equation} \label{equation for g}
g_t+u \cdot \nabla g = -u_g \cdot \nabla w_s.
\end{equation}

This is nothing but a transport equation with a forcing term. To prove <ref> using (<ref>), one have to a priori expect that for compactly supported symmetric $g\in \cphi$, forcing term on the right hand side also belongs to space $\cphi$. Heuristically, this is plausible as $\abs{u_g(x)} \approx \abs{x} \logonealpha{x}$ by (<ref>) and $\abs{\nabla w_s(x)} \approx \frac{1}{\abs{x} \log \frac{1}{\abs{x}}}$ near the origin[Even though we remove the assumption of symmetry on $g_0$, this cancellation of `zero times infinity' still occurs near $\phi^*(t)$ on the right hand side of (<ref>). This guarantees that similar proof goes on without assumption on symmetry.]. More precisely, we prove

Let $\alpha \in (0,1)$ be given. Suppose a vector field $v \in \cpsi$ satisfies $v(0)=0$. Then, there exists a constant $C>0$ such that the scalar function $v \cdot \nabla w_s \in \cphi$ with

\begin{equation*}
\nrm{v \cdot \nabla w_s}_{\cphi} \leq C \nrm{v}_{\cpsi}.
\end{equation*}

Suppose two points $x$ and $y$ are given with $\abs{x-y} \leq \frac{e^{-2}}{2}$. Without loss of generality, we assume $\abs{x} \leq \abs{y}$. We divide the case.

$\bullet$ Case I. $\abs{y} \geq e^{-2}$

Note that $\nabla w_s$ is compactly supported smooth function on $\bbR^2 \backslash B \left( 0, \frac{e^{-2}}{2} \right)$ so that

\begin{equation*}
\nrm{\nabla w_s}_{\cphi \left(\bbR^2 \backslash B \left( 0, \frac{e^{-2}}{2}\right)\right)} \coloneqq M< \infty.
\end{equation*}

Then, we have

\begin{align*}
&\frac{\abs{v(x) \cdot \nabla w_s(x) - v(y) \cdot \nabla w_s(y)}}{\logalpha{x-y}} \\
&\leq \frac{\abs{v(x)-v(y)} \cdot \abs{\nabla w_s(x)}}{\abs{x-y} \logonealpha{x-y}} \abs{x-y} \log \frac{1}{\abs{x-y}}+\abs{v(y)} \frac{\abs{\nabla w_s(x)-\nabla w_s(y)}}{\logalpha{x-y}} \\
&\leq \left( \frac{e^{-2}}{2} \log (2e^2 ) + M \right) \nrm{v}_{\cpsi}.
\end{align*}

$\bullet$ Case II. $\abs{y} \leq e^{-2}$

Note that for all $z \in B \left( 0, e^{-2} \right)$,

\begin{equation} \label{eq4}
\frac{\abs{v(z) \cdot \nabla w_s(z)}}{\logalpha{z}} \underset{\substack{\eqref{def of ws} }}{\leq} \frac{\nrm{v}_{\cpsi}\abs{z} \logonealpha{z} \cdot \frac{1}{\abs{z} \log \frac{1}{\abs{z}}}}{\logalpha{z}} = \nrm{v}_{\cpsi}.
\end{equation}

Now, we divide the Case II into two cases based on size of $\abs{x-y}$ and $\abs{y}$.

$\bullet$ Case II-(i). $\abs{y}\leq e^{-2},~~ \abs{x-y} \geq \frac{\abs{y}}{2}$

In this case, we estimate $\cphi$-norm of $v \cdot \nabla w_s$ as follows,

\begin{align*}
\frac{\abs{v(x) \cdot \nabla w_s(x) - v(y) \cdot \nabla w_s(y)}}{\logalpha{x-y}} &\leq \frac{\abs{v(x) \cdot \nabla w_s(x)}}{\logalpha{x-y}}+\frac{\abs{v(y) \cdot \nabla w_s(y)}}{\logalpha{x-y}} \\
&\leq \frac{\abs{v(x) \cdot \nabla w_s(x)}}{\logalpha{x}} \cdot \frac{\logalpha{x}}{\logalpha{x-y}} + \frac{\abs{v(y) \cdot \nabla w_s(y)}}{\logalpha{y}} \cdot \frac{\logalpha{y}}{\logalpha{x-y}}.
\end{align*}

Combining inequality (<ref>) with

\begin{equation*}
\frac{\logalpha{x}}{\logalpha{x-y}} \leq 2^\alpha , \qquad \frac{\logalpha{y}}{\logalpha{x-y}} \leq 2^\alpha,
\end{equation*}

which is implied by the condition $\abs{x-y}\geq \frac{\abs{y}}{2}$, we have

\begin{equation*}
\frac{\abs{v(x) \cdot \nabla w_s(x) - v(y) \cdot \nabla w_s(y)}}{\logalpha{x-y}} \leq 2^{\alpha+1} \nrm{v}_{\cpsi}.
\end{equation*}

$\bullet$ Case II-(ii). $\abs{y}\leq e^{-2},~~ \abs{x-y} \leq \frac{\abs{y}}{2}$

In this case, we have

\begin{equation*}
\frac{\abs{v(x) \cdot \nabla w_s(x) - v(y) \cdot \nabla w_s(y)}}{\logalpha{x-y}} \leq \frac{\abs{v(x)-v(y)}}{\logalpha{x-y}} \cdot \abs{\nabla w_s(x)} + \abs{v(y)} \cdot \frac{\abs{\nabla w_s(x) - \nabla w_s (y)}}{\logalpha{x-y}}.
\end{equation*}

Note that $\abs{x-y}\leq \frac{\abs{y}}{2}$ implies $\abs{x-y} \leq \abs{x}$ and $\abs{y} \leq 2\abs{x}$. Then,

\begin{equation*}
\frac{\abs{v(x)-v(y)}}{\logalpha{x-y}} \cdot \abs{\nabla w_s(x)} \underset{\eqref{def of ws} }{=} \frac{\abs{v(x)-v(y)}}{\abs{x-y} \logonealpha{x-y}} \cdot \frac{\abs{x-y} \log \frac{1}{\abs{x-y}}}{\abs{x} \log \frac{1}{\abs{x}}}\leq \nrm{v}_{\cpsi}.
\end{equation*}

\begin{align*}
\abs{v(y)} \cdot \frac{\abs{\nabla w_s(x) - \nabla w_s (y)}}{\logalpha{x-y}} &\leq \frac{\abs{v(y)}}{\logalpha{x-y}} \cdot \abs{x-y} \frac{2}{\abs{x}^2 \log \frac{1}{\abs{x}}} \\
&= \frac{2 \abs{v(y)}}{\abs{y} \logonealpha{y}}\cdot \frac{\abs{y} \logonealpha{y}}{\abs{x} \logonealpha{x}} \cdot \frac{\abs{x} \logalpha{x}}{\abs{x-y} \logalpha{x-y}}\\
&\leq 4 \nrm{v}_{\cpsi}.
\end{align*}

As we have calculated $\cphi$-seminorm so far, we now calculate $L^\infty$-norm of $v \cdot \nabla w_s$. For $0 < \abs{z} < e^{-2}$,

\begin{equation*}
\abs{v(z) \cdot \nabla w_s (z)} \underset{\eqref{eq4} }{\leq} \nrm{v}_{\cpsi} \sup_{z \in B (0, e^{-2})} \logalpha{z} \leq \nrm{v}_{\cpsi}.
\end{equation*}

For $\abs{z} \geq e^{-2}$,

\begin{equation*}
\abs{v(z) \cdot \nabla w_s(z)} \leq \nrm{v}_{\cpsi} \sup_{z \in \bbR^2 \backslash B(0,e^{-2})} \abs{\nabla w_s(z)}.
\end{equation*}

Combining above inequalities implies the Lemma <ref>.

Gathering above elementary lemmas, we can prove the following proposition. Here, $LL$ is the space of log-Lipschitz function, i.e.,

\begin{equation*}
LL \coloneqq \set{f \, \Big\vert \, \nrm{f}_{LL} \coloneqq \nrm{f}_{\infty}+\sup_{\abs{x-y}<e^{-1}}\frac{
\abs{f(x)-f(y)}}{\abs{x-y} \log \frac{1}{\abs{x-y}}}<\infty}.
\end{equation*}

Suppose $v \in L^\infty([0,T]; \cpsi)$ with $v(0,t)=0$ for $t\in [0,T]$ and $u \in L^\infty([0,T];LL)$. Then, there exists $C=C(\alpha, T,\nrm{u}_{L^\infty_T LL})>0$ such that the unique weak solution for forced transport equation

\begin{equation} \label{transport eq}
\left\{ \begin{aligned}
& g_t+u \cdot \nabla g = - v \cdot \nabla w_s, \\
& g(\cdot,0)=g_0
\end{aligned} \right.
\end{equation}

belongs to $L^\infty([0,T]; \cphi)$ with

\begin{equation} \label{consistency}
\nrm{g(\cdot,t)}_{\cphi} \leq C(\nrm{g_0}_{\cphi} + t \nrm{v}_{L^\infty_t\cpsi}).
\end{equation}

Let $\nrm{u}_{L^\infty_T LL} \eqqcolon N$ and take two points $X, Y$ with $\abs{X-Y}\leq e^{-NT}$. Also, fix a time $t \in [0,T]$. Then, there exists $x, y \in \bbR^2$ such that $\Phi(x,t)=X, \Phi(y,t)=Y$ with $\abs{x-y} \leq e^{-1}$, where $\Phi: \bbR^2 \times [0,T] \rightarrow \bbR^2$ is a flow map generated by $u$. Since the solution for (<ref>) could be explicitly written by

\begin{equation} \label{eq5}
g(\Phi(z,t),t) = g_0(z)-\int_0^t v(\Phi(z,s),s) \cdot \nabla w_s (\Phi(z,s),s) \, ds,
\end{equation}

we have

\begin{equation} \label{eq6}
\begin{aligned}
&\frac{\abs{g(\Phi(x,t),t)-g(\Phi(y,t),t)}}{\logalpha{\Phi(x,t)-\Phi(y,t)}} \leq \frac{\abs{g_0(x)-g_0(y)}}{\logalpha{x-y}} \cdot \frac{\logalpha{x-y}}{\logalpha{\Phi(x,t)-\Phi(y,t)}}\\ &+ \int_0^t \underbrace{\frac{\abs{v(\Phi(x,s),s) \cdot \nabla w_s (\Phi(x,s),s)-v(\Phi(y,s),s) \cdot \nabla w_s (\Phi(y,s),s)}}{\logalpha{\Phi(x,s)-\Phi(y,s)}}}_{
\underset{\text{Lem }\ref{forcing term}}{\leq} C \sup_{0 \leq s \leq t} \nrm{v(\cdot,s)}_{\cpsi}} \cdot \frac{\logalpha{\Phi(x,s)-\Phi(y,s)}}{\logalpha{\Phi(x,t)-\Phi(y,t)}} \, ds
\end{aligned}
\end{equation}

As we have

\begin{equation*}
\frac{\logalpha{x-y}}{\logalpha{\Phi(x,t)-\Phi(y,t)}} \leq e^{N \alpha t}, \qquad \frac{\logalpha{\Phi(x,s)-\Phi(y,s)}}{\logalpha{\Phi(x,t)-\Phi(y,t)}} \leq e^{2N\alpha t},
\end{equation*}

by (<ref>), (<ref>) leads to

\begin{equation} \label{eq7}
\sup_{\abs{X-Y}\leq e^{-NT}}\frac{\abs{g(X,t)-g(Y,t)}}{\logalpha{X-Y}} \leq e^{N \alpha t} \nrm{g_0}_{\cphi} + Cte^{2N\alpha t}\nrm{v}_{L^\infty_t \cpsi},
\end{equation}

where $C$ is the constant in Lemma <ref>.

Similarly, from (<ref>),

\begin{equation} \label{eq8}
\nrm{g(\cdot,t)}_{\infty} \leq \nrm{g_0}_{\infty} + t \nrm{v \cdot \nabla w_s}_{L^\infty_t L^\infty} \underset{\text{Lem }\ref{forcing term} }{\leq} \nrm{g_0}_{\cphi}+Ct \nrm{v}_{L^\infty_t \cpsi}.
\end{equation}

Combining (<ref>) and (<ref>) implies Proposition <ref>.

§.§ Proof of the Main Theorem: Conservation Part

In this subsection, we prove Theorem <ref> and Theorem <ref> using the results in Section <ref>. The proof is quite standard.

Proof of Theorem <ref>.

Let $T<\infty$ be given. Define a sequence $\set{g^{(n)}}_{n \in \bbN}$ inductively by setting $g^{(0)}(\cdot,t) \equiv g_0$ and $g^{(n+1)}$ being a solution of forced transport equation

\begin{equation} \label{iteration scheme}
\left\{ \begin{aligned}
& g^{(n+1)}_t + u \cdot \nabla g^{(n+1)}=-u_{g^{(n)}} \cdot \nabla w_s, \\
& g^{(n+1)}(x,0)=g_0(x),
\end{aligned} \right.
\end{equation}

where $u \in L^\infty ([0,T]; LL)$ is the fixed velocity field which is the solution of 2D Euler equation

\begin{equation*}
\left\{ \begin{aligned}
& w_t+ u \cdot \nabla w=0, \\
& w(\cdot,0)=w_s + g_0.
\end{aligned} \right.
\end{equation*}

Note that if $g^{(n)}\in \cphi$ is symmetric with respect to the origin for all $t \in [0,T]$, then $g^{(n+1)}$ is well-defined through Lemma <ref> and Proposition <ref>, and symmetric with respect to the origin. Therefore,  $\set{g^{(n)}}_{n \in \bbN}$ is well-defined for all $n \in \bbN$. Since $g^{(n+1)} - g^{(n)}$ satisfies the forced transport equation

\begin{equation*}
\left\{ \begin{aligned}
& (g^{(n+1)} - g^{(n)})_t + u \cdot \nabla (g^{(n+1)} - g^{(n)}) = -(u_{g^{(n+1)}}-u_{g^{(n)}}) \cdot \nabla w_s, \\
& (g^{(n+1)} - g^{(n)})(x,0) =0.
\end{aligned} \right.
\end{equation*}

Proposition <ref> and Lemma <ref> implies that there exists a constant $C=C(\alpha, T, \nrm{u}_{L^\infty_T LL}, \abs{\supp g_0})$ such that

\begin{equation} \label{eq9}
\nrm{g^{(n+1)} - g^{(n)}}_{L^\infty_t \cphi} \underset{\text{Lem } \ref{consistency lemma} }{\leq}Ct \nrm{u_{g^{(n)}}-u_{g^{(n-1)}}}_{L^\infty_t \cpsi} \underset{\eqref{vor to vel} }{\leq} Ct \nrm{g^{(n+1)} - g^{(n)}}_{L^\infty_t \cphi}.
\end{equation}

If we take $T_0>0$ sufficiently small so that $CT_0 \leq \frac{1}{2}$, (<ref>) implies that  $\set{g^{(n)}}_{n \in \bbN}$ is a Cauchy sequence in $L^\infty_{T_0}\cphi$. Therefore, there exists $h \in L^\infty_{T_0}\cphi$ so that $g^{(n)} \rightarrow h$ in $L^\infty_{T_0}\cphi$. Since $u \in L^\infty_T L^\infty$ and $w_s$ has compact support regardless of $n \in \bbN$, we can iterate this scheme so that maximal convergence time could be extended up to $T$ like in the proof of classical Cauchy-Lipschitz theorem.

Also, since

\begin{equation*}
g^{(n)} \rightarrow h \qquad \text{and} \qquad u_{g^{(n)}} \cdot \nabla w_s \xrightarrow[\text{Lem }\ref{forcing term} ]{} u_h \cdot \nabla w_s \qquad \text{in }~~ L^\infty_T L^\infty,
\end{equation*}

it is straightforward to show that $h$ is the weak solution of (<ref>) by taking $n \rightarrow \infty$ in the weak formulation of (<ref>)

\begin{equation*}
\int_{\bbR^2} \left( \phi(x,T) g^{(n+1)}(x, T) - \phi(x,0) g_0 \right)\, dx + \int_0^T \int_{\bbR^2} \left( -\frac{D \phi}{Dt}g^{(n+1)} + (u_{g^{(n)}}\cdot \nabla w_s)\phi \right)\, dxdt=0
\end{equation*}

for a test function $\phi \in C^\infty_c (\bbR^2 \times [0,T])$.

Lastly, we conclude our proof by showing that the unique solution $g$ of (<ref>) whose regularity is guaranteed to be at most $L^\infty_TL^\infty$ at this point, is equal to $h \in L^\infty_T\cphi$. Since both $g$ and $h$ are weak solutions of (<ref>), $g-h$ satisfies the following transport equation

\begin{equation*}
\left\{ \begin{aligned}
&(g-h)_t + u \cdot \nabla (g-h) = - (u_g-u_h) \cdot \nabla w_s, \\
& (g-h)(\cdot,0)=0.
\end{aligned} \right.
\end{equation*}

Thus, we have $L^\infty$- estimate

\begin{equation} \label{eq10}
\nrm{g-h}_{L^\infty_t L^\infty} \leq \int_0^t \nrm{(u_g-u_h)(\cdot,s) \cdot \nabla w_s(\cdot,s)}_{L^\infty} \, ds.
\end{equation}

Since $u_g$ and $u_h$ are log-Lipschitz and $u_g(0)=u_h(0)=0$ by symmetry, we have for $z \in B(0,e^{-2})$,

\begin{equation*}
\abs{(u_g-u_h)(z) \cdot \nabla w_s(z))} \underset{\eqref{def of ws}}{\leq} C \nrm{g-h}_\infty \abs{z} \log \frac{1}{\abs{z}} \cdot \frac{1}{\abs{z} \log \frac{1}{\abs{z}}} = C \nrm{g-h}_{\infty},
\end{equation*}

where $C$ is a constant depending only on $\abs{\supp (g-h)}$ which is already known to be bounded. Also, for $z \in \bbR^2 \backslash B(0,e^{-2})$,

\begin{equation*}
\abs{(u_g-u_h)(z) \cdot \nabla w_s(z)} \leq C \nrm{g-h}_\infty \sup_{z \in \bbR^2 \backslash B(0,e^{-2})} \abs{\nabla w_s(z)}.
\end{equation*}

Therefore, there exists a absolute constant $C>0$ such that

\begin{equation} \label{eq11}
\nrm{(u_g-u_h)\cdot \nabla w_s}_\infty \leq C \nrm{g-h}_{\infty}.
\end{equation}

Combining (<ref>), (<ref>) with Grönwall inequality leads to $g=h$ on $[0,T]$. $\square$

Proof of Theorem <ref>.

In the proof of Theorem <ref>, the reason we impose symmetry condition on $g_0$ is to apply Lemma <ref> and Proposition <ref>. The vanishing velocity condition in Lemma <ref> is used to cancel singularity of $\nabla w_s$ at the origin. However, investigating general equation for $g$ (<ref>), where there is no symmetry condition on $g$, it is immediate to see that vanishing velocity condition is redundancy since $u_g(x,t) - u_g(\phi^*(t),t)$ automatically has zero velocity at the singular point of $\nabla w_s(x-\phi^*(t))$. For this reason, similar proof goes on with obvious modification to Theorem <ref> without any symmetry condition.$\hfill \square$

§ BREAKDOWN OF MODULUS OF CONTINUITY

In this section, we prove Theorem <ref>. The first thing we have to do is to clarify initial perturbation $g_0$.

§.§ Construction of Initial Data and Basic Properties of the Solution

Let $\eta \in C^\infty_c$ be a standard mollifier and $\eta_\varepsilon \coloneqq \frac{1}{\varepsilon^2}\eta \left( \frac{\cdot}{\varepsilon} \right)$. For a function $c_0$ defined by

\begin{equation*}
c_0(x)= \begin{cases}
1, & (x\in [21,23] \times [21,23]  ~~ \cup ~~ [-23, -21] \times [-23, -21]), \\
-1, & (x\in [-23,-21] \times [21,23]  ~~ \cup ~~ [21,23] \times [-23,-21]), \\
0, & (\text{otherwise}),
\end{cases}
\end{equation*}

which is symmetric with respect to the origin, define $g_0$ by

\begin{equation} \label{initial data}
g_0 \coloneqq \eta_\varepsilon * c_0.
\end{equation}

Then, an elementary calculation shows that we can choose small $\varepsilon>0$ such that there exist $K, \delta >0$ so that $u_{g_0}^2$, the second component of $u_{g_0}$, satisfies

\begin{equation} \label{Lip initial}
-(K+1)r\leq u_{g_0}^2(0,r) \leq -Kr
\end{equation}

for all $0<r<\delta$.

Let $w$ be the unique weak solution of 2D Euler equation

\begin{equation} \label{equ1}
\left\{ \begin{aligned}
& w_t+u \cdot \nabla w=0, \\
& w(\cdot,0)=w_s(x) +g_0.
\end{aligned} \right.
\end{equation}

Then, by Theorem 3 of [4], the solution $w(x,t)$ could be written as the form of

\begin{equation*}
\end{equation*}

where $\phi^*(t)$ is the particle trajectory of the origin and $g$  belongs to $ L^\infty_{\textnormal{loc}} ([0,\infty), L^\infty)$. From now on, by $w, u$ and $g$, we always mean above fixed solution, not a general solution.

The velocity field $u(x,t)$ satisfies $u(-x,t)=-u(x,t)$ for all $x \in \bbR^2, t \in [0,\infty)$ so that $u(0,t)=0$ for all $t \in [0,\infty)$. Therefore, $\phi^*(t) \equiv 0$ so that the solution can be written in the form of

\begin{equation*}
w(x,t)=w_s(x) + g(x,t).
\end{equation*}

Clear as both $w_s(x)$ and $g_0$ are symmetric with respect to the origin.

Now, we derive an estimate for velocity field $u_s, u_g$ and $u$ defined as in (<ref>).

There exists $R>0$ such that for all $0<\abs{x}<R$,

\begin{equation} \label{us}
\abs{u_s(x)} \leq \abs{x} \log \log \frac{1}{\abs{x}},
\end{equation}

\begin{equation} \label{del us}
\abs{\nabla u_s(x)} \leq 3 \log \log \frac{1}{\abs{x}}.
\end{equation}

\begin{equation*}
\nrm{u_s}_{\infty} + \sup_{\abs{x-y}<e^{-3}}\frac{\abs{u_s(x)-u_s(y)}}{\abs{x-y}\log \log \frac{1}{\abs{x-y}}} < \infty.
\end{equation*}

Let $\mathbf{e}_r= \begin{pmatrix}
\frac{x_1}{\abs{x}} \\
\frac{x_2}{\abs{x}}
\end{pmatrix}, \mathbf{e}_\theta=\begin{pmatrix}
-\frac{x_2}{\abs{x}} \\
\frac{x_1}{\abs{x}}
\end{pmatrix}$. Remembering the definition of $w_s$ (<ref>), direct calculation (e.g., see Example 2.1 of [8]) gives us

\begin{align*}
u_s(x)&= \left( \frac{1}{\abs{x}} \int_0^{\abs{x}} r w_s(r) \, dr \right) \mathbf{e}_r \eqqcolon G(\abs{x}) \mathbf{e}_\theta , \\
\nabla u_s (x) &= \frac{G(\abs{x})}{\abs{x}} \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix} + \left( G'(\abs{x}) - \frac{G(\abs{x})}{\abs{x}}\right) \mathbf{e}_\theta \mathbf{e}_r^t,
\end{align*}

from which the result is obvious[Since $w_s$ is radial, $w_s(r)$ is well-defined without confusion.].

Now, we give more refined rate of access of two points than Lemma <ref> using upgraded regularity of $g$ obtained from Theorem <ref>.

Let $\alpha \in (0,1)$ be given and $\Phi: \bbR^2 \times [0,1] \rightarrow \bbR^2$ be a flow map of $u$. Then, there exist $M>0$[Originally, the constant depends on $\cpsi$- norm of the solution $u$. However, we declared in Section <ref> that as we fix $u$ as the specific solution of (<ref>), dependence on $u$ would not be mentioned.], $ 0<R=R(\alpha)<e^{-3}$ such that for all $0<\abs{x-y}<R$ and $0 \leq t \leq 1$,

\begin{equation} \label{access rate alpha}
\abs{x-y} e^{-Mt \logonealpha{x-y}} \leq \abs{\Phi(x,t)-\Phi(y,t)} \leq \abs{x-y} e^{Mt \logonealpha{x-y}}.
\end{equation}

Since $g_0$ is compactly supported smooth function, $g_0$ also belongs to $\cphi$ regardless of $\alpha$. Then, $g\in L^\infty([0,1]; \cphi)$ by Theorem <ref> and $u_g \in L^\infty([0,1];  \cpsi)$ by Lemma <ref>. Thus $u \in L^\infty([0,1]; \cpsi)$ and let

\begin{equation*}
\sup_{0 \leq t \leq 1} \nrm{u(\cdot,t)}_{\cpsi} \eqqcolon N.
\end{equation*}

Then, comparison with ODE

\begin{equation*}
\left\{ \begin{aligned}
& x'(t)= \pm N x(t) \left( \log \frac{1}{x(t)} \right)^{1-\alpha}, \\
& x(0)=x_0 \quad (0<x_0<e^{-3(N+1)}),
\end{aligned} \right.
\end{equation*}

whose solution is

\begin{equation*}
x(t)=e^{-\left(\mp \alpha Nt+ \left( \log \frac{1}{x_0}\right)^\alpha\right)^{1/\alpha}},
\end{equation*}

gives us

\begin{equation} \label{eq12}
e^{-\left( \alpha Nt+ \left( \log \frac{1}{\abs{x-y}}\right)^\alpha\right)^{1/\alpha}} \leq \abs{\Phi(x,t)-\Phi(y,t)} \leq e^{-\left(- \alpha Nt+ \left( \log \frac{1}{\abs{x-y}}\right)^\alpha\right)^{1/\alpha}}
\end{equation}

for $0<\abs{x-y}<e^{-3(N+1)}$. Applying elementary inequalities

\begin{equation*}
(a+b)^k \leq a^k +bk(a+b)^{k-1}, \quad (a-b)^k  \geq a^k -bka^{k-1} \quad (a,b,k>0)
\end{equation*}

to (<ref>), we have

\begin{equation} \label{eq13}
e^{-\log \frac{1}{\abs{x-y}} -Nt \left( \alpha Nt+ \left( \log \frac{1}{\abs{x-y}} \right)^\alpha \right)^{1/\alpha -1}} \leq \abs{\Phi(x,t)-\Phi(y,t)} \leq e^{-\log \frac{1}{\abs{x-y}}+Nt \logonealpha{x-y}}.
\end{equation}

If $\abs{x-y}<e^{-\left( \frac{\alpha N}{2^\alpha -1} \right)^{1/\alpha}}$, then $\alpha Nt \leq (2^\alpha-1) \left( \log \frac{1}{\abs{x-y}}\right)^\alpha$ so that (<ref>) could be simplified to

\begin{equation*}
\abs{x-y} e^{-2Nt \logonealpha{x-y}} \leq \abs{\Phi(x,t)-\Phi(y,t)} \leq \abs{x-y} e^{Nt \logonealpha{x-y}}.
\end{equation*}

Now, we estimate the value of $g$. The inequality (<ref>) implies that there exists a constant $C=C(\alpha)$ such that for $0 \leq t \leq 1$,

\begin{equation} \label{eq14}
\nrm{g(\cdot,t)}_{\cphi} \leq \nrm{g_0}_{\cphi} + Ct.
\end{equation}

However, we need to get sharper time-dependent local estimates near the origin.

Let $\alpha \in (0,1)$ be given. Then, there exist $R>0, M=M(\alpha)>0$ and $0<T\leq 1$ such that for all $0<\abs{x}<R$ and $0 \leq t \leq T$,

\begin{equation} \label{g}
\abs{g(x,t)} \leq M \logalpha{x} t,
\end{equation}

\begin{equation} \label{ug}
\abs{u_g(x,t)} \leq M\abs{x} \left( 1+ \logonealpha{x}t \right).
\end{equation}

Since $g\in \cphi$ and $u_g(0,t)=0$ due to symmetry, Proposition <ref> could be applied. As $g$ satisfies (<ref>), we have

\begin{equation} \label{eq15}
g(x,t) = \underbrace{g_0(\Phi_t^{-1}(x))}_{\eqqcolon F(x,t)} - \underbrace{\int_0^t u_g(\Phi(\Phi_t^{-1}(x),s)) \cdot \nabla w_s(\Phi(\Phi_t^{-1}(x),s))\, ds}_{\eqqcolon G(x,t)},
\end{equation}

where $\Phi: \bbR^2 \times [0,1] \rightarrow \bbR^2$ is the flow map generated by $u$ and $\Phi_t^{-1}$ is the inverse flow map. Since $\sup_{0 \leq t \leq 1} \nrm{u}_{LL}<\infty$, there exits $0<T<1$ such that if $0\leq t \leq T$, then $\abs{\Phi(x,t)-x}\leq 1, \abs{\Phi_t^{-1}(x)-x}\leq 1$ for all $x \in \bbR^2$. Then, considering the fact that $g_0$ is supported in $\bbR^2 \backslash B(0,21)$ and $w_s$ is supported in $B(0,1)$ (see (<ref>) and (<ref>)),

\begin{equation*}
\left\{ \begin{aligned}
& \supp(F(\cdot,t) ) \subset \bbR^2 \backslash B(0,20), \\
& \supp (G(\cdot,t) ) \subset B(0,2),
\end{aligned} \right.
\end{equation*}

for $0 \leq t \leq T$. Using same method in the proof of Proposition <ref>, there exists $M=M(\alpha)>0$ such that

\begin{equation*}
\nrm{ u_g(\Phi(\Phi_t^{-1}(x),s)) \cdot \nabla w_s(\Phi(\Phi_t^{-1}(x),s)) }_{\cphi} \leq M \nrm{u_g(\cdot,t)}_{\cpsi}
\underset{\eqref{vor to vel} }{\leq} M \nrm{g(\cdot,t)}_{\cphi}
\underset{\eqref{eq14} }{\leq} \nrm{g_0}_{\cphi}+Ct.
\end{equation*}

Therefore, there exists a constant $M=M(\alpha)>0$ such that

\begin{equation} \label{equ12}
\nrm{G(\cdot,t)}_{\cphi} \leq Mt,\
\end{equation}

for $0 \leq t \leq T$, which implies inequality (<ref>). Then, we could get (<ref>) through the decomposition (<ref>) by applying (<ref>) with Lemma <ref> and the fact that $F(\cdot,t)$ generates Lipschitz velocity field near the origin.

Using Lemma <ref>, we give an estimate for $\nabla u$.

Let $\alpha \in (0,1)$ be given. Then, there exist $0<R=R(\alpha)<e^{-3} , M>0$, and $0 \leq T \leq 1$ such that for all $0< \abs{x} <R$ and $0 \leq t \leq T$,

\begin{equation} \label{del u}
\abs{\nabla u(x,t)} \leq M \left( \log \log \frac{1}{\abs{x}}+ \logonealpha{x}t + e^{Mt \logonealpha{x}}   \right).
\end{equation}

Application of interior Schauder estimate (see, e.g., Theorem 2.14 of [6]) to relation

\begin{equation*}
\Delta \psi =w \quad \text{in} \quad B\left(x,\frac{\abs{x}}{2}\right),
\end{equation*}

where $\psi:\bbR^2 \rightarrow \bbR$ is a stream function such that $\nabla^\perp \psi =u$, and $\psi(0)=0$, yields that there exists $C>0$ such that

\begin{equation} \label{Shauder}
\nrm{\nabla u(\cdot,t)}_{L^\infty \left(B \left(x, \frac{\abs{x}}{4}\right)\right)} \leq C  \left(  \frac{1}{\abs{x}^2} \nrm{\psi(\cdot,t)}_{L^\infty \left(B \left(x, \frac{\abs{x}}{2}\right)\right)} + \nrm{w(\cdot,t)}_{L^\infty \left(B \left(x, \frac{\abs{x}}{2}\right)\right)} + \abs{x}^\alpha [w(\cdot,t)]_{C^{0,\alpha} \left(B \left(x, \frac{\abs{x}}{2}\right)\right)}  \right).
\end{equation}

We first estimate $\nrm{\psi(\cdot,t)}_{L^\infty \left(B \left(x, \frac{\abs{x}}{2}\right)\right)}$. By (<ref>) and (<ref>), there exists $0 < R < e^{-4}, M=M(\alpha)>0$, and $T>0$ such that for all $0<\abs{x}<R$ and $0 \leq t \leq T$,

\begin{equation} \label{u}
\abs{u(x,t)} \leq \abs{u_s(x)}+ \abs{u_g(x,t)} \underset{\eqref{us}, \eqref{ug} }{\leq} \abs{x} \log \log \frac{1}{\abs{x}} + M \abs{x} \logonealpha{x}t.
\end{equation}

As $u=\nabla^\perp \psi$ and $\psi(0)=0$, the Fundamental Theorem of Calculus implies that

\begin{equation} \label{eq16}
\begin{aligned}
\nrm{\psi(\cdot,t)}_{L^\infty \left(B \left(x, \frac{\abs{x}}{2}\right)\right)} & \leq \frac{3}{2} \abs{x} \sup_{0 \leq \abs{z} \leq \frac{3}{2}\abs{x}} \abs{u(z,t)} \\
\underset{\eqref{us}, \eqref{ug} }&{\leq} M \abs{x} \left( \abs{x} \log \log \frac{1}{\abs{x}} + M \abs{x} \logonealpha{x}t \right).
\end{aligned}
\end{equation}

Also, by (<ref>) and (<ref>), there exist $0<R=R(\alpha) < e^{-4},M=M(\alpha)>0$ and $T>0$ such that for all $0<\abs{x}<R$ and $0 \leq t \leq T \leq 1$ (we can choose $R=R(\alpha)$ so that $M\left( \log \frac{1}{R}\right)^{-\alpha} \leq \log \log \frac{1}{R}$ holds),

\begin{equation} \label{eq17}
\abs{w(x,t)} \underset{\text{Lem } \ref{symmetry lemma}}{\leq} \abs{w_s(x)}+\abs{g(x,t)} \underset{\eqref{def of ws}, \eqref{g} }{\leq} \log \log \frac{1}{\abs{x}} + M\logalpha{x}t \leq 2 \log \log \frac{1}{\abs{x}}.
\end{equation}

Lastly, we calculate Hölder seminorm $[w(\cdot,t)]_{C^{0,\alpha} \left(B \left(x, \frac{\abs{x}}{2}\right)\right)}$. Suppose two points $y, z \in B\left( x, \frac{\abs{x}}{2} \right)$ are given. Then, for all $\abs{x}<R$ and $0 \leq t \leq T$, where $R$ is a constant in Lemma <ref> and $T$ is the constant in Lemma <ref>,

\begin{equation*}
\begin{aligned}
\frac{\abs{w(y,t)-w(z,t)}}{\abs{y-z}^\alpha} &= \frac{\abs{w_0(\Phi_t^{-1}(y))-w_0(\Phi_t^{-1}(z))}}{\abs{y-z}^\alpha} \\
\underset{\text{Lem } \ref{symmetry lemma}}&{\leq} \frac{|w_s(\Phi_t^{-1}(y))-w_s(\Phi_t^{-1}(z))+\overbrace{g_0(\Phi_t^{-1}(y))-g_0(\Phi_t^{-1}(z))}^{=0 ~~(\because \, \supp(g_0) \subset \bbR^2 \backslash B(0,2))}|}{\abs{y-z}^\alpha} \\
& = \frac{|w_s(\Phi_t^{-1}(y))-w_s(\Phi_t^{-1}(z))|}{\abs{\Phi_t^{-1}(y) - \Phi_t^{-1}(z)}} \cdot \frac{\abs{\Phi_t^{-1}(y) - \Phi_t^{-1}(z)}}{\abs{y-z}^\alpha}.
\end{aligned}
\end{equation*}

By mean value theorem,

\begin{equation} \label{equ2}
\begin{aligned}
\sup_{y,z\in B\left( x, \frac{\abs{x}}{2}\right)} \frac{|w_s(\Phi_t^{-1}(y))-w_s(\Phi_t^{-1}(z))|}{\abs{\Phi_t^{-1}(y) - \Phi_t^{-1}(z)}} &\leq \sup_{p \in \Phi_t^{-1}\left( B \left( x, \frac{\abs{x}}{2}\right)\right)} \abs{\nabla w_s(p)} \\
\underset{\text{Lem } \ref{access rate alpha lemma} }&{\leq} \abs{\nabla w_s \left( \frac{\abs{x}}{2} e^{-Mt \left( \log \frac{1}{\abs{x}/2}\right)^{1-\alpha}}\right)} \\
& \leq \frac{2e^{2^{1-\alpha}Mt \logonealpha{x}}}{\abs{x} \log \frac{1}{\abs{x}}}.
\end{aligned}
\end{equation}

Also, by Lemma <ref>,

\begin{equation*}
\frac{\abs{\Phi_t^{-1}(y) - \Phi_t^{-1}(z)}}{\abs{y-z}^\alpha} \leq \abs{y-z}^{1-\alpha} e^{Mt \logonealpha{y-z}}.
\end{equation*}

Note that for one-variable function $f(r)=r^{1-\alpha}e^{Mt\left( \log \frac{1}{r}\right)^{1-\alpha}}$,

\begin{equation*}
f'(r)=(1-\alpha)r^{-\alpha}e^{Mt \left( \log \frac{1}{r}\right)^{1-\alpha}} \left( 1-Mt \left( \log \frac{1}{r} \right)^{-\alpha} \right).
\end{equation*}

This means that the supremum of $\abs{y-z}^{1-\alpha} e^{Mt \logonealpha{y-z}}$ for $y, z \in B \left( x, \frac{\abs{x}}{2} \right)$ is achieved when $\abs{y-z}=\abs{x}$ for sufficiently small $x$ (precisely, $0<\abs{x}<R$ suffices for $R$ satisfying $M \left( \log \frac{1}{R}\right)^{-\alpha} <\frac{1}{2}$). Then, we have

\begin{align*}
\sup_{y,z\in B\left( x, \frac{\abs{x}}{2}\right)} \frac{\abs{\Phi_t^{-1}(y) - \Phi_t^{-1}(z)}}{\abs{y-z}^\alpha} & \leq \sup_{y,z\in B\left( x, \frac{\abs{x}}{2}\right)}  \abs{y-z}^{1-\alpha} e^{Mt \logonealpha{y-z}} \\
& = \abs{x}^{1-\alpha} e^{Mt \logonealpha{x}}.
\end{align*}

Therefore, there exists $R=R(\alpha)>0, T>0$ and $M>0$ such that for $0<\abs{x}<R$ and $0 \leq t \leq T$,

\begin{equation} \label{eq18}
\begin{aligned}
\sup_{y,z\in B\left( x, \frac{\abs{x}}{2}\right)} \frac{\abs{w(y,t)-w(z,t)}}{\abs{y-z}^\alpha} & \leq \sup_{y,z\in B\left( x, \frac{\abs{x}}{2}\right)} \frac{|w_s(\Phi_t^{-1}(y))-w_s(\Phi_t^{-1}(z))|}{\abs{\Phi_t^{-1}(y) - \Phi_t^{-1}(z)}} \cdot \sup_{y,z\in B\left( x, \frac{\abs{x}}{2}\right)} \frac{\abs{\Phi_t^{-1}(y) - \Phi_t^{-1}(z)}}{\abs{y-z}^\alpha} \\
& \leq 2 \frac{e^{2^{1-\alpha}Mt\logonealpha{x}}}{\abs{x} \log \frac{1}{\abs{x}}} \abs{x}^{1-\alpha} e^{Mt\logonealpha{x}} \\
& \leq \frac{2e^{3Mt\logonealpha{x}}}{\abs{x}^\alpha \log \frac{1}{\abs{x}}}.
\end{aligned}
\end{equation}

Combining (<ref>), (<ref>), (<ref>) with (<ref>), we conclude that there exists $R=R(\alpha)>0, T>0$ and $M>0$ such that for $0<\abs{x}<R$ and $0 \leq t \leq T$,

\begin{equation*}
\abs{\nabla u(x,t)} \leq \nrm{\nabla u (\cdot,t)}_{L^\infty \left( B \left( x, \frac{\abs{x}}{4} \right)\right)} \underset{\substack{\eqref{Shauder}, \eqref{eq16} \\ \eqref{eq17}, \eqref{eq18}  }}{\leq} M  \left( \log \log \frac{1}{\abs{x}} + \logonealpha{x}t + e^{Mt \logonealpha{x}}\right).
\end{equation*}

§.§ The Key Lemma

Remember that the evolution equation for $g \in L^\infty_{\textnormal{loc}}([0,\infty); L^\infty)$ is given by (<ref>)

\begin{equation*}
g_t + u \cdot \nabla g = -u_g \cdot \nabla w_s.
\end{equation*}

Since $w_s(x)=\log \log \frac{1}{\abs{x}}$ for $0<\abs{x} <e^{-2}$, the equation for $g$ near the origin could be written by

\begin{equation} \label{equation for g near the origin}
g_t+u \cdot \nabla g = \frac{u_g \cdot \frac{x}{\abs{x}}}{\abs{x} \log \frac{1}{\abs{x}}}.
\end{equation}

Note that when time is zero, (<ref>) implies that there exists $K, \delta>0$ such that for $x=(0,r)$ with $0<r<\delta$,

\begin{equation*}
\frac{\abs{u_g(x,0) \cdot \frac{x}{\abs{x}}}}{\abs{x} \log \frac{1}{\abs{x}}} = \frac{\abs{u_{g_0}(0,r) \cdot (0,1) }}{r \log \frac{1}{r}} \geq \frac{K}{\boldsymbol{\log \frac{1}{r}}}.
\end{equation*}

This implies that forcing term on the right hand side of (<ref>) is so strong enough that one could expect that $g$ might not belong to $L^\infty([0,1]; C^{\phi_\beta})$ for any $\beta>1$ (remember $\phi_\beta(r) = \left( \log \frac{1}{r}\right)^{-\beta}$). To make this argument rigorous, what we have to do is to remove the qualifier `when time is zero' so that this large forcing maintains for finite length of time interval. This is what our Key Lemma exactly tells about.

A map $\phi_r:[0,\infty) \rightarrow\bbR^2$ is defined by particle trajectory of $(0,r)$, i.e., $\phi_r$ is a unique solution of ODE

\begin{equation*}
\left\{ \begin{aligned}
& \frac{d}{dt} \phi_r(t)=u(\phi_r(t),t), \\
\end{aligned} \right.
\end{equation*}

where $u$ is fixed solution of (<ref>).

By the definition of $w_s$ (<ref>) and Lemma <ref>, there exist $R=R(\alpha)>0$ and $0 \leq t \leq T$ such that forced transport equation (<ref>) could be written along particle trajectory $\phi_r$

\begin{equation} \label{equation for g integral}
g(\phi_r(t),t) = \int_0^t \frac{u_g(\phi_r(s),s) \cdot \frac{\phi_r(s)}{\abs{\phi_r(s)}}}{\abs{\phi_r(s)} \log \frac{1}{\abs{\phi_r(s)}}} \, ds
\end{equation}

for all $0<r<R$ and $0 \leq t \leq T$. We now state the Key Lemma.

Let $\alpha \in (0,1)$ be given. Then, for all $\varepsilon, \delta>0$, there exists $R=R(\alpha, \varepsilon)>0$ such that for $0<r<R$ and $0 \leq t \leq \frac{1}{\left( \log \frac{1}{r} \right) ^{1-\alpha +\varepsilon}}$,

\begin{equation*}
\abs{u_g(\phi_r(t),t)-u_g(\phi_r(0),0)} \leq \delta r.
\end{equation*}

Since $g_0$ is compactly supported smooth function, $g_0 \in \cphi$ for given $\alpha \in (0,1)$ so that results in the Section <ref> could be applied. From

\begin{equation*}
w_s(\Phi_t^{-1}(x))+g_0(\Phi_t^{-1}(x)) \underset{\eqref{equ1}}{=}w_0(\Phi_t^{-1}(x)) \underset{\eqref{equ1}}{=} w(x,t) \underset{\text{Lem }\ref{symmetry lemma} }{=} w_s(x)+g(x,t),
\end{equation*}

$g(x,t)$ could be written as

\begin{equation} \label{eq19}
\end{equation}

Considering the support of $w_s$ and $g_0$ with $\nrm{u}_{LL}<\infty$, there exists $T>0$ such that for all $0 \leq t \leq T$,

\begin{equation} \label{supp condition}
\left\{ \begin{aligned}
& \supp(w_s(\Phi_t^{-1}(\cdot))-w_s(\cdot)) \subset B(0,2), \\
& \supp(g_0(\Phi_t^{-1}(\cdot)) \subset A(20,40).
\end{aligned} \right.
\end{equation}

From now on, we always take all constraint on $t$ less than this $T$ so that this support condition (<ref>) holds. Using this decomposition, we can divide $\abs{u_g(\phi_r(t),t)-u_g(\phi_r(0),0)}$ into three parts

\begin{equation} \label{eq20}
\begin{aligned}
\abs{u_g(\phi_r(t),t)-u_g(\phi_r(0),0)} & \leq \abs{u_g(\phi_r(t),t)-u_g(\phi_r(0),t)}+\abs{u_g(\phi_r(0),t)-u_g(\phi_r(0),0)} \\
&=	\abs{u_g(\phi_r(t),t)-u_g(\phi_r(0),t)}+ \frac{1}{2\pi} \abs{\int_{\bbR^2} \frac{((0,r)-y)^\perp}{\abs{(0,r)-y}^2} (g(y,t)-g_0(y)) \, dy} \\
\underset{\eqref{eq19} }&{=} \underbrace{\abs{u_g(\phi_r(t),t)-u_g(\phi_r(0),t)}}_{\eqqcolon I_1(r,t)} \\
& \qquad + \frac{1}{2\pi}  \underbrace{\bigg\lvert\int_{B(0,2)} \frac{((0,r)-y)^\perp}{\abs{(0,r)-y}^2} (w_s(\Phi_t^{-1}(y))-w_s(y)) \, dy\bigg\rvert}_{\eqqcolon I_2(r,t)}  \\
& \qquad + \frac{1}{2\pi}  \underbrace{\bigg\lvert\int_{\bbR^2 \backslash B(0,20)} \frac{((0,r)-y)^\perp}{\abs{(0,r)-y}^2} (g_0(\Phi_t^{-1}(y))-g_0(y)) \, dy\bigg\rvert}_{\eqqcolon I_3(r,t)} .
\end{aligned}
\end{equation}

We now estimate each of $I_1, I_2$ and $I_3$.

Let $\alpha \in (0,1)$ be given.  Then, there exists $R=R(\alpha)>0, 0<T\leq 1$ and $M>0$ such that for all $0<r<R$ and $0 \leq t \leq T$,

\begin{equation*}
I_1(r,t)=\abs{u_g(\phi_r(t),t)-u_g(\phi_r(0),t)} \leq Mtr \left( \log \frac{1}{r} \right)^{1-\alpha}e^{Mt\left( \log \frac{1}{r}\right)^{1-\alpha}}.
\end{equation*}

As in the proof of Lemma <ref>, we can decompose $u_g$ into

\begin{equation*}
u_g = \underbrace{\nabla^\perp \Delta^{-1}F}_{\eqqcolon v_1}+\underbrace{\nabla^\perp \Delta^{-1}G}_{\eqqcolon v_2},
\end{equation*}

where $F$ and $G$ are defined same as in the proof of Lemma <ref>. Recall that $v_1 \in L^\infty([0,1];Lip(B(0,1)))$ and $v_2 \in L^\infty([0,1]; \cpsi)$ with

\begin{equation*}
\nrm{v_2(\cdot,t)}_{\cpsi} \leq Mt
\end{equation*}

for some constant $M=M(\alpha)>0$.

Take $0 < T \leq 1$ sufficiently small so that $\abs{\phi_r(t)-\phi_r(0)} < e^{-3}$ for all $0 \leq t \leq T$. Also, take $R>0$ sufficiently small so that $re^{M \left( \log \frac{1}{r}\right)^{1-\alpha}}<e^{-3}$ for all $0 < r \leq R$, where $M>0$ is a constant in Lemma <ref>. Then,

\begin{align*}
& \abs{u_g(\phi_r(t),t)-u_g(\phi_r(0),t)} \\
& \leq \abs{v_1(\phi_r(t),t)-v_1(\phi_r(0),t)}+ \abs{v_2 (\phi_r(t),t)} + \abs{v_2 (\phi_r(0),t)} \\
&\leq \nrm{v_1}_{L^\infty_T Lip(B(0,1))} \abs{\phi_r(t)-\phi_r(0)} +Mt \left( \abs{\phi_r(t)} \left( \log \frac{1}{\abs{\phi_r(t)}}\right)^{1-\alpha}+r \left( \log \frac{1}{r}\right)^{1-\alpha}  \right) \\
\underset{\text{Lem }\ref{access rate alpha} }&{\leq} \nrm{v_1}_{L^\infty_T Lip(B(0,1))} \abs{\phi_r(t)-\phi_r(0)} +Mt \left( r\left( \log \frac{1}{r} \right)^{1-\alpha}e^{Mt\left( \log \frac{1}{r}\right)^{1-\alpha}} +r \left( \log \frac{1}{r}\right)^{1-\alpha}  \right).
\end{align*}

Here, we can estimate $\abs{\phi_r(t)-\phi_r(0)}$ by

\begin{equation} \label{equ5}
\begin{aligned}
\abs{\phi_r(t)-\phi_r(0)} &= \abs{\int_0^t  \frac{d}{ds} \phi_r(s)\, ds} \\
& \leq \int_0^t \abs{u(\phi_r(s),s)}\, ds \\
\underset{\text{Lem } \ref{access rate alpha lemma} }&{\leq} t \cdot \sup_{x \in B ( 0,re^{Mt\left(\log \frac{1}{r}\right)^{1-\alpha}})} \abs{u(x)} \\
\underset{\eqref{u} }&{\leq} t \left( re^{Mt\left(\log \frac{1}{r}\right)^{1-\alpha}} \log \log \frac{1}{r} + M r\left( \log \frac{1}{r}\right)^{1-\alpha} e^{Mt\left(\log \frac{1}{r}\right)^{1-\alpha}}\right).
\end{aligned}
\end{equation}

Combining these estimates leads to Proposition <ref>.

Before estimating $I_2$, which is the hardest part, we give the estimate for $I_3$.

Let $\alpha \in (0,1)$ be given. Then, there exist $ 0<T\leq 1$ and $M>0$ such that for all $0<r<e^{-2}$ and $0 \leq t \leq T$,

\begin{equation*}
I_3(r,t)=\bigg\lvert\int_{\bbR^2 \backslash B(0,20)} \frac{((0,r)-y)^\perp}{\abs{(0,r)-y}^2} (g_0(\Phi_t^{-1}(y))-g_0(y)) \, dy\bigg\rvert \leq Mtr.
\end{equation*}

For convenience, let

\begin{equation*}
g_0(\Phi_t^{-1}(y))-g_0(y) \eqqcolon h(y,t).
\end{equation*}

Then, there exists $T>0$ such that for all $0 \leq t \leq T$, $\supp (h(\cdot,t)) \subset A(20,40)$. Also, since $g_0$ is Lipschitz,

\begin{equation} \label{eq21}
\abs{h(y,t)} \leq \nrm{g_0}_{Lip} \abs{\Phi_t^{-1}(y)-y} \leq \nrm{g_0}_{Lip} \nrm{u}_{L^\infty_T L^\infty} t
\end{equation}

for all $0 \leq t \leq T$. Then, as $h(y,t)=h(-y,t)$, we have

\begin{align*}
I_3(r,t) &=\bigg\lvert\int_{\bbR^2 \backslash B(0,20)} \frac{((0,r)-y)^\perp}{\abs{(0,r)-y}^2} h(y,t) \, dy\bigg\rvert \\
&=\bigg\lvert\int_{A(20,40)} \frac{((0,r)-y)^\perp}{\abs{(0,r)-y}^2} h(y,t) \, dy\bigg\rvert \\
&=  \Bigg\vert \int_{A(20,40) \cap \set{y \geq 0} } \frac{((0,r)-y)^\perp}{\abs{(0,r)-y}^2} h(y, t) \, dy+\int_{A(20, 40) \cap \set{y \leq 0}} \frac{((0,r)-y)^\perp}{\abs{(0,r)-y}^2} h(y, t) \, dy \Bigg\vert \\
&= \abs{\int_{A(20, 40) \cap \set{y \geq 0} } \frac{((0,r)-y)^\perp}{\abs{(0,r)-y}^2} h(y, t) \, dy+\int_{A(20, 40) \cap \set{y \geq 0}} \frac{((0,r)+y)^\perp}{\abs{(0,r)+y}^2} h(-y, t) \, dy} \\
&=\abs{\int_{A(20, 40) \cap \set{y \geq 0} } \left(\frac{((0,r)+y)^\perp}{\abs{(0,r)+y}^2}-\frac{(-(0,r)+y)^\perp}{\abs{(0,r)-y}^2}\right) h(y, t) \, dy} \\
&\leq \text{Vol}(A(20,40)) \cdot \sup_{y \in A(20, 40)} \left(\frac{((0,r)+y)^\perp}{\abs{(0,r)+y}^2}-\frac{(-(0,r)+y)^\perp}{\abs{(0,r)-y}^2}\right)  \cdot \sup_{\substack{y \in A(20, 40) \\ 0 \leq t \leq T}} \abs{h(y,t)} \\
\underset{\eqref{eq21} }&{\leq} \Bigg( \sup_{y \in A(20,40)} \frac{1}{\pi} \frac{2r}{\abs{(0,r)+y} \abs{(0,r)-y}} \Bigg) \cdot Mt \\
&= Mtr.
\end{align*}

Let $\alpha \in (0,1)$ and $\varepsilon>0$ be given. Then, there exist $R=R(\alpha, \varepsilon)>0$ and $M=M(\alpha,\varepsilon)>0$ such that for all $0<r<R$ and $0 \leq t \leq \frac{1}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}}$,

\begin{equation*}
I_2(r,t) = \bigg\vert\int_{B(0,2)} \frac{( (0, r) -y)^\perp}{\abs{(0,r)-y }^2} (w_s(\Phi_t^{-1}(y))-w_s(y)) \, dy\bigg\vert \leq Mtr \log \log \frac{1}{r}.
\end{equation*}

Remember that we only consider sufficiently small $t$ so that $\supp(w_s(\Phi_t^{-1}(\cdot))-w_s(\cdot)) \subset B(0,2)$. Thus, for all $0<r<1$, we can decompose $I_2$ into two parts so that $I_2 \leq I_{21}+ I_{22}$ as following:

\begin{align*}
I_2(r,t)&=\bigg\vert \int_{B(0,2)} \frac{( (0, r) -y)^\perp}{\abs{(0,r)-y }^2} (w_s(\Phi_t^{-1}(y))-w_s(y)) \, dy\bigg\vert \\
&= \bigg\vert \int_{B((0,r),3)} \frac{( (0, r) -y)^\perp}{\abs{(0,r)-y }^2} (w_s(\Phi_t^{-1}(y))-w_s(y)) \, dy\bigg\vert \\
&\leq \underbrace{\bigg\vert \int_{B((0,r),2r)} \frac{( (0, r) -y)^\perp}{\abs{(0,r)-y }^2} (w_s(\Phi_t^{-1}(y))-w_s(y)) \, dy\bigg\vert}_{\eqqcolon I_{21}(r,t)} \\
& \qquad + \underbrace{\bigg\vert \int_{B((0,r),3)\, \backslash \,B((0,r),2r)} \frac{( (0, r) -y)^\perp}{\abs{(0,r)-y }^2} (w_s(\Phi_t^{-1}(y))-w_s(y)) \, dy\bigg\vert}_{\coloneqq I_{22}(r,t)}.
\end{align*}

$\bullet$ Estimation of $I_{21}$

\begin{align*}
I_{21}(r,t) &= \bigg\vert \int_{B((0,r),2r)} \frac{( (0, r) -y)^\perp}{\abs{(0,r)-y }^2} (w_s(\Phi_t^{-1}(y))-w_s(y)) \, dy\bigg\vert \\
& \leq \sup_{y \in B((0,r),2r)} \vert w_s(\Phi_t^{-1}(y))-w_s(y)\vert \cdot \int_{B((0,r),2r)} \frac{( (0, r) -y)^\perp}{\abs{(0,r)-y }^2} \, dy.
\end{align*}

Combining the equality (<ref>) and the support information on $g_0(\Phi_t^{-1}(\cdot))$ (<ref>) with Lemma <ref>, there exist constants $R>0, M=M(\alpha)>0$ and $T>0$ such that for all $0<r<R$ and $0 \leq t \leq T$,

\begin{align*}
\sup_{y \in B((0,r),2r)} \abs{w_s(\Phi_t^{-1}(y))-w_s(y)} \underset{\eqref{eq19} }&{=} \sup_{y \in B((0,r),2r)} \abs{g(x,t)-g_0(\Phi_t^{-1}(x))} \\
\underset{\eqref{supp condition} }&{\leq} \sup_{y \in B((0,r),2r)} \abs{g(x,t)} \\
\underset{\eqref{g} }&{\leq}  Mt \left( \log \frac{1}{3r} \right)^{-\alpha} \\
&\leq Mt \left( \log \frac{1}{r} \right)^{-\alpha}.
\end{align*}

Therefore, as $\int_{B((0,r),2r)} \frac{( (0, r) -y)^\perp}{\abs{(0,r)-y }^2} \, dy=4\pi r,$ there exists $R>0$ and $T>0$ such that for all $0<r<R$ and $0 \leq t \leq T$,

\begin{equation} \label{equ4}
I_{21}(r,t) \leq Mtr \left( \log \frac{1}{r} \right)^{-\alpha}.
\end{equation}

(End of the estimation of $I_{21}$)

$\bullet$ Estimation of $I_{22}$

We also divide $I_{22}$ into two parts so that $I_{22} \leq I_{221}+I_{222}$.

\begin{equation} \label{equ6}
\begin{aligned}
&I_{22}(r,t) \\
&= \bigg\vert\int_{B((0,r),3) \backslash B((0,r),2r)} \frac{((0,r)-y)^\perp}{\abs{(0,r)-y}^2} (w_s(\Phi_t^{-1}(y))-w_s(y)) \, dy\bigg\vert \\
&=\bigg\vert\int_{A(2r,3)} \frac{(-y)^\perp}{\abs{-y}^2} (w_s(\Phi_t^{-1}((0,r)+y))-w_s((0,r)+y)) \, dy\bigg\vert \\
&= \bigg\vert\int_{A(2r,3) \cap \set{y \geq 0}} \frac{(-y)^\perp}{\abs{-y}^2} (w_s(\Phi_t^{-1}((0,r)+y))-w_s((0,r)+y)) \, dy \\
& \qquad +\int_{A(2r,3)\cap \set{y \leq 0}} \frac{(-y)^\perp}{\abs{-y}^2} (w_s(\Phi_t^{-1}((0,r)+y))-w_s((0,r)+y)) \, dy\bigg\vert \\
&= \bigg\vert\int_{A(2r,3) \cap \set{y \geq 0}} \frac{(-y)^\perp}{\abs{-y}^2} (w_s(\Phi_t^{-1}((0,r)+y))-w_s((0,r)+y)) \, dy \\
& \qquad +\int_{A(2r,3)\cap \set{y \geq 0}} \frac{y^\perp}{\abs{y}^2} (\underbrace{w_s(\Phi_t^{-1}((0,r)-y))-w_s((0,r)-y)}_{w_s(\Phi_t^{-1}(y-(0,r))) - w_s(y-(0,r))}) \, dy\bigg\vert \\
&=\bigg\vert \int_{A(2r,3) \cap \set{y \geq 0}} \frac{y^\perp}{\abs{y}^2} \bigg( \Big( w_s(\Phi_t^{-1}(y-(0,r)))-w_s(y-(0,r)) \Big)- \Big(w_s(\Phi_t^{-1}(y+(0,r)))-w_s(y+(0,r)) \Big)  \bigg) \, dy \bigg\vert \\
& \leq  \underbrace{\bigg\vert \int_{\substack{A(2r,R) \cap  \set{y \geq 0}}}\frac{y^\perp}{\abs{y}^2} \bigg( \Big( w_s(\Phi_t^{-1}(y-(0,r)))-w_s(y-(0,r)) \Big)- \Big(w_s(\Phi_t^{-1}(y+(0,r)))-w_s(y+(0,r)) \Big)  \bigg) \, dy \bigg\vert}_{\eqqcolon I_{221}(r,t)}  \\
&~~+  \underbrace{\bigg\vert \int_{A(R,3) \cap \set{y \geq 0}} \frac{y^\perp}{\abs{y}^2} \bigg( \Big( w_s(\Phi_t^{-1}(y-(0,r)))-w_s(y-(0,r)) \Big)- \Big(w_s(\Phi_t^{-1}(y+(0,r)))-w_s(y+(0,r)) \Big)  \bigg) \, dy \bigg\vert}_{\eqqcolon I_{222}(r,t)}.
\end{aligned}
\end{equation}

$\bullet$ Estimation of $I_{221}$

To bound $I_{221}$ using mean value theorem, we first investigate the function

\begin{equation*}
q(x,t) \coloneqq w_s(\Phi_t^{-1}(x))-w_s(x).
\end{equation*}

By chain rule, we have

\begin{equation} \label{equ3}
\begin{aligned}
\nabla q(x,t) &= \nabla w_s(\Phi_t^{-1}(x)) \cdot \nabla \Phi_t^{-1}(x) - \nabla w_s (x) \\
&= \nabla w_s (\Phi_t^{-1}(x)) (\underbrace{\nabla \Phi_t^{-1}(x)-I}_{\eqqcolon J_1(x,t)}) + \underbrace{\nabla w_s (\Phi_t^{-1}(x)) - \nabla w_s(x)}_{\eqqcolon J_2(x,t)}.
\end{aligned}
\end{equation}

$\bullet$ Estimation of $J_1$

By differentiating an identity

\begin{equation*}
\Phi_t^{-1}(\Phi(x,t)) =x,
\end{equation*}

with respect to space variable, we have

\begin{equation*}
\nabla \Phi_t^{-1}(\Phi(x,t)) \cdot \nabla \Phi(x,t) =I.
\end{equation*}

Replacing $x$ with $\Phi_t^{-1}(x)$ leads to

\begin{equation*}
\nabla \Phi_t^{-1}(x)= (\nabla \Phi (\Phi_t^{-1}(x),t))^{-1}.
\end{equation*}

\begin{equation} \label{eq22}
J_1(x,t)= \nabla \Phi^{-1}_t (x) - I = \left( \nabla \Phi ( \Phi^{-1}_t (x), t) \right) ^{-1} -I=\left( \nabla \Phi ( \Phi^{-1}_t (x), t) \right)^{-1} (I- \nabla \Phi (\Phi^{-1}_t (x),t) ).
\end{equation}

We now derive an equation for $\nabla \Phi (\Phi_t^{-1}(x),t)$. Replacing $x$ with $\Phi_t^{-1}(x)$ after taking nabla on both sides of the identity

\begin{equation*}
\frac{d}{dt} \Phi (x,t) = u( \Phi (x,t),t),
\end{equation*}

we get

\begin{equation} \label{eq23}
\frac{d}{dt} \nabla \Phi ( \Phi^{-1}_t (x), t) = \nabla u(x,t) \cdot \nabla \Phi (\Phi^{-1}_t (x),t).
\end{equation}

Applying Grönwall inequality to (<ref>) yields

\begin{equation} \label{eq24}
\begin{aligned}
\abs{\nabla (\Phi_t^{-1}(x),t)} & \leq e^{\int_0^t \nabla u(x,s) \, ds} \\
\underset{\eqref{del u}}&{\leq} e^{Mt\left( \log \log \frac{1}{\abs{x}}+ \left( \log \frac{1}{\abs{x}} \right)^{1-\alpha}t+e^{Mt\left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}  \right) }.
\end{aligned}
\end{equation}

Then, we have

\begin{equation} \label{eq25}
\begin{aligned}
\abs{(\nabla \Phi (\Phi_t^{-1}(x),t))^{-1}} & = \bigg\vert \frac{1}{\det (\nabla \Phi (\Phi_t^{-1}(x),t))} \cdot \text{adj}\,(\nabla \Phi (\Phi_t^{-1}(x),t)) \bigg\vert\\
& = \abs{\nabla \Phi (\Phi_t^{-1}(x),t)} \\
\underset{\eqref{eq24}}&{\leq} e^{Mt\left( \log \log \frac{1}{\abs{x}}+ \left( \log \frac{1}{\abs{x}} \right)^{1-\alpha}t+e^{Mt\left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}  \right)}.
\end{aligned}
\end{equation}

In the second equality, we use incompressibility of velocity field.

\begin{equation} \label{eq26}
\begin{aligned}
& \abs{I-\Phi (\Phi_t^{-1}(x),t)} \\
&= \bigg\vert \int_0^t \frac{d}{ds} \Phi (\Phi_s^{-1}(x),s)\, ds  \bigg\vert \\
\underset{\eqref{eq23} }&{=} \bigg\vert \int_0^t \nabla u(x,s) \cdot \nabla \Phi (\Phi_s^{-1}(x),s) \, ds  \bigg\vert \\
\underset{ \eqref{del u}, \eqref{eq24} }&{\leq} Mt\left( \log \log \frac{1}{\abs{x}}+ \left( \log \frac{1}{\abs{x}} \right)^{1-\alpha}t+e^{Mt\left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}  \right)e^{Mt\left( \log \log \frac{1}{\abs{x}}+ \left( \log \frac{1}{\abs{x}} \right)^{1-\alpha}t+e^{Mt\left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}  \right)}.
\end{aligned}
\end{equation}

Applying estimates (<ref>), (<ref>) to (<ref>) implies that there exist $R=R(\alpha)>0, 0 < T \leq 1$ and $M>0$ such that for all $0< \abs{x} < R$ and $0 \leq t \leq T$,

\begin{equation} \label{eq27}
\begin{aligned}
\abs{J_1(x,t)} \underset{\eqref{eq22} }&{\leq} \abs{(\nabla \Phi(\Phi_t^{-1}(x),t))^{-1}} \cdot \abs{I-\nabla \Phi (\Phi_t^{-1}(x),t)} \\
\underset{\eqref{eq25}, \eqref{eq26} }&{\leq} Mt\left( \log \log \frac{1}{\abs{x}}+ \left( \log \frac{1}{\abs{x}} \right)^{1-\alpha}t+e^{Mt\left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}  \right)e^{Mt\left( \log \log \frac{1}{\abs{x}}+ \left( \log \frac{1}{\abs{x}} \right)^{1-\alpha}t+e^{Mt\left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}  \right)}.
\end{aligned}
\end{equation}

(End of the estimation of $J_1$)

$\bullet$ Estimation of $J_2$

Let $M>0$ be a constant in Lemma <ref> and take $0<R<e^{-3}$ such that $Re^{Mt\left( \log \frac{1}{R}\right)^{1-\alpha}}\leq \min \left( e^{-3},R' \right)$ for all $0 \leq t \leq T$, where $R'>0$ is a constant which makes inequality (<ref>) hold. then, by mean value theorem and Lemma <ref>,

\begin{equation} \label{eq28}
\begin{aligned}
\abs{J_2(x,t)} &= \abs{\nabla w_s (\Phi_t^{-1}(x)-\nabla w_s(x)} \\
\underset{\text{Lem } \ref{access rate alpha lemma} }&{\leq} \sup_{\abs{x}e^{-Mt \left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}<\abs{y}<\abs{x}e^{Mt \left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}} \abs{\nabla^2 w_s(y)} \cdot t \sup_{\abs{x}e^{-Mt \left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}<\abs{y}<\abs{x}e^{Mt \left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}} \abs{u(y,t)} \\
\underset{\eqref{def of ws}, \eqref{u}}&{\leq} \frac{e^{2Mt \left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}}{\abs{x}^2 \log \frac{1}{\abs{x}}} \left( Mt \abs{x}e^{Mt \left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}\left( \log \log \frac{1}{\abs{x}} +  \left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}t\right)\right) \\
&=\frac{Mte^{Mt\left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}}{\abs{x}\log \frac{1}{\abs{x}}} \left( \log \log \frac{1}{\abs{x}} +  \left( \log \frac{1}{\abs{x}}\right)^{1-\alpha} t\right)
\end{aligned}
\end{equation}

for all $0<\abs{x} < R$.

(End of the estimation of $J_2$)

Applying inequalities (<ref>), (<ref>), and (<ref>) to (<ref>), we have

\begin{equation} \label{eq29}
\begin{aligned}
&\abs{\nabla q(x,t)} \\
\underset{\eqref{equ3} }&{\leq }  \abs{\nabla w_s (\Phi_t^{-1}(x))} \cdot \abs{\nabla \Phi_t^{-1}(x)-I} + \abs{\nabla w_s (\Phi_t^{-1}(x)) - \nabla w_s(x)} \\
\underset{\substack{\eqref{equ2}, \eqref{eq27}\\ \eqref{eq28} }}&{\leq} \frac{Mt}{\abs{x} \log \frac{1}{\abs{x}}}\left( \log \log \frac{1}{\abs{x}}+ \left( \log \frac{1}{\abs{x}} \right)^{1-\alpha}t+e^{Mt\left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}  \right) e^{Mt\left( \log \log \frac{1}{\abs{x}}+ \left( \log \frac{1}{\abs{x}} \right)^{1-\alpha}+e^{Mt\left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}}  \right)},
\end{aligned}
\end{equation}

whenever $0 < \abs{x} < R$ and $0 \leq t \leq T$ for some constant $R=R(\alpha)>0, M>0$ and $T>0$. In particular, for any $\varepsilon>0$, we can take $R=R(\alpha,\varepsilon)<e^{-3}$ such that for all $0<\abs{x}<2R$,

\begin{equation*}
\frac{1}{\left( \log \frac{1}{\abs{x}}\right)^{1-\alpha+\varepsilon }} \left( \log \log \frac{1}{\abs{x}} + \left( \log \frac{1}{\abs{x}}\right)^{1-\alpha}\right) < \frac{1}{100(M+1)^2}
\end{equation*}

holds. Then, for all $0<\abs{x}<2R$ and $0 \leq t \leq \frac{1}{\left( \log \frac{1}{\abs{x}}\right)^{1-\alpha+\varepsilon}}$, we have

\begin{equation} \label{eq30}
\abs{\nabla q(x,t)} \leq \frac{3Mt}{\abs{x} \log \frac{1}{\abs{x}}} \log \log \frac{1}{\abs{x}}.
\end{equation}

Then, for all $0<r<\frac{R}{2}$ and $0 \leq t \leq \frac{1}{\left( \log \frac{1}{r} \right)^{1-\alpha+\varepsilon} }$,

\begin{equation} \label{eq31}
\begin{aligned}
&=\bigg\vert \int_{A(2r,R) \cap \set{y \geq 0}}\frac{y^\perp}{\abs{y}^2} \bigg( \Big( w_s(\Phi_t^{-1}(y-(0,r)))-w_s(y-(0,r)) \Big)- \Big(w_s(\Phi_t^{-1}(y+(0,r)))-w_s(y+(0,r)) \Big)  \bigg) \, dy \bigg\vert \\
&= \bigg\vert \int_{A(2r,R) \cap \set{y \geq 0}}\frac{y^\perp}{\abs{y}^2} \left(q(y-(0,r),t)-q(t+(0,r),t) \right) \, dy \bigg\vert \\
& \leq \int_{A(2r,R) \cap \set{y \geq 0}} \frac{1}{\abs{y}} \cdot 2r \cdot \sup_{0 \leq s \leq 2} \abs{\nabla q (y-(0,(1-s)r),t) } \, dy.
\end{aligned}
\end{equation}

Considering the fact that $\set{y-(0,(1-s)r) \, \vert \, y \in A(2r,R), 0 \leq s \leq 2} \subset B(r,2R)$ and $\abs{y-(0,(1-s)r)} \geq \frac{\abs{y}}{2} \text{ for } y \in A(2r,R)$, inequality (<ref>) leads to

\begin{align*}
I_{221}(r,t) & \leq 2r  \int_{A(2r,R) \cap \set{y \geq 0}} \frac{1}{\abs{y}} \frac{3Mt}{\frac{\abs{y}}{2} \log \frac{1}{\abs{\frac{y}{2}}}} \log \log \frac{1}{\abs{\frac{y}{2}}} \, dy \\
& \leq 24tr  \int_{A(2r,R) \cap \set{y \geq 0}} \frac{\log \log \frac{1}{\abs{y}}}{\abs{y}^2 \log \frac{1}{\abs{y}}} \, dy \\
&= 24 \pi tr \int_{2r}^R \frac{\log \log \frac{1}{\rho}}{\rho \log \rho} \, d\rho \\
& \leq 12 \pi t r \log \log \frac{1}{r}.
\end{align*}

(End of the estimation of $I_{221}$)

$\bullet$ Estimation of $I_{222}$

\begin{equation*}
I_{222(r,t)}=\bigg\vert \int_{A(R,3) \cap \set{y \geq 0}} \frac{y^\perp}{\abs{y}^2} \bigg( \Big( w_s(\Phi_t^{-1}(y-(0,r)))-w_s(y-(0,r)) \Big)- \Big(w_s(\Phi_t^{-1}(y+(0,r)))-w_s(y+(0,r)) \Big)  \bigg) \, dy \bigg\vert,
\end{equation*}

we can see that the integrand is same as $I_{221}$, but the integral range is more affirmative: It strictly avoids the origin where the integrand becomes singular. Therefore, it is far more easier to bound $I_{222}$ than $I_{221}$, using the fact that $w_s$ is compactly supported smooth function on $\overline{A(R,3)}$. Thus, we only provide the result to avoid redundancy: There exists a constant $M=M(R)$, where $R$ is the constant appearing in the bound of the term $I_{222}$, such that

\begin{equation} \label{eq32}
I_{222}(r,t) \leq Mtr.
\end{equation}

(End of the estimation of $I_{222}$)

Combining two estimates for $I_{221}$ and $I_{222}$ leads to the bound of $I_{22}$: There exist $R=R(\alpha, \varepsilon)>0$ and $M=M(\alpha,\varepsilon)>0$ such that for all $0<r<R$ and $0 \leq t \leq \frac{1}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}}$,

\begin{equation} \label{eq33}
I_{22} \underset{\eqref{equ6} }{\leq} I_{221}+I_{222} \underset{\eqref{eq31}, \eqref{eq32} }{\leq} Mtr\log \log \frac{1}{r}.
\end{equation}

(End of the estimation of $I_{22}$)

Finally, the Proposition <ref> is deduced by combining (<ref>) with the estimate for $I_{21}$ (<ref>).

Proof of the Key Lemma $($Lemma <ref>$)$.

Suppose $\alpha\in (0,1)$ and $\varepsilon, \delta>0$ are given. By Proposition <ref>, Proposition <ref>, and Proposition <ref>, there exist $R=R(\alpha,\varepsilon)>0$ and $M=M(\alpha,\varepsilon)>0$ such that for all $0<r<R$ and $0 \leq t \leq \frac{1}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}}$,

\begin{equation*}
\begin{aligned}
\abs{u_g(\phi_r(t),t)-u_g(\phi_r(0),0)} \underset{\eqref{eq20}}&{\leq} I_1(r,t)+I_2(r,t)+I_3(r,t) \\
\underset{\substack{\text{Prop }\ref{estimation of I1}, \text{ Prop }\ref{estimation of I3} \\ \text{Prop }\ref{estimation of I2} }}&{\leq} r \left( Mt+Mt \left( \log \frac{1}{r}\right)^{1-\alpha} e^{Mt\left(\log \frac{1}{r}\right)^{1-\alpha}}+ Mt \log \log \frac{1}{r} \right).
\end{aligned}
\end{equation*}

Then, the elementary fact

\begin{equation*}
\lim_{r \searrow 0} \frac{M}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}} =\lim_{r \searrow 0} \frac{M \left( \log \frac{1}{r}\right)^{1-\alpha}}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}} = \lim_{r \searrow 0} \frac{\log \log \frac{1}{r}}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}} =0
\end{equation*}

implies the Proposition <ref>.

§.§ Proof of the Main Theorem: Breakdown Part

To prove main theorem, Theorem <ref>, we point out one more elementary ingredient.

Let $\alpha \in (0,1)$ be given. Then, for any $\varepsilon, \delta>0$, there exists $R=R(\alpha, \varepsilon)>0$ such that for all $0<r<R$ and $0 \leq t \leq \frac{1}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}}$,

\begin{align*}
\abs{\phi_r(t)-(0, r)} &< \delta , \\
\abs{\frac{\phi_r (t)}{\abs{\phi_r(t)}}-(0,1)} &< \delta.
\end{align*}

This is just a corollary of (<ref>).

Proof of Theorem <ref>.

It suffices to consider the case $\beta \in (1,2)$. Suppose $\beta \in (0,1)$ and $T_0>0$ are given. Define $\alpha, \varepsilon>0$ by

\begin{equation} \label{eq34}
\alpha \coloneqq\frac{5-\beta}{4}, \qquad \varepsilon \coloneqq \frac{\beta-1}{4}, \qquad \delta \coloneqq \min \left( \frac{K}{4}, \frac{1}{10}\right),
\end{equation}

where $K>0$ is the constant in (<ref>). Then, for any $r$ with $0<r<R$, where $R>0$ is the constant in Proposition <ref>,

\begin{align*}
&\sup_{0 \leq t \leq T_0} \nrm{g(\cdot,t)}_{\phi_\beta} \\
&\leq  \sup_{0 \leq t \leq T_0} \sup_{0<\abs{x}<e^{-3}} \left( \log \frac{1}{\abs{x}}\right)^{\beta} \vert g(x,t)-\overbrace{g(0,t)}^{\overset{\eqref{g} }{=}0}\vert \\
& \geq \Bigg( \log \frac{1}{\abs{\phi_r \left(\frac{1}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}}\right)}}\Bigg)^{\beta} \abs{g \left( \phi_r \left( \frac{1}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}} \right), \frac{1}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}}\right)} \\
\underset{\eqref{equation for g integral} }&{=}\Bigg( \log \frac{1}{\abs{\phi_r \left(\frac{1}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}}\right)}}\Bigg)^{\beta} \bigg\vert \int_0^{\frac{1}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}}} \frac{u_g(\phi_r(s),s) \cdot \frac{\phi_r(s)}{\abs{\phi_r(s)}}}{\abs{\phi_r(s)}\log \frac{1}{\abs{\phi_r(s)}}} \, ds \bigg\vert \\
&= \Bigg( \log \frac{1}{\underbrace{\abs{\phi_r \left(\frac{1}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}}\right)}}_{\underset{\text{Lem } \ref{phirt estimate lemma} }{\geq}\frac{9}{10}r}}\Bigg)^{\beta} \cdot \\
& ~~ \bigg\vert \int_0^{\frac{1}{\left( \log \frac{1}{r}\right)^{1-\alpha+\varepsilon}}}  \frac{\overbrace{(u_g(\phi_r(s),s) - u_g(\phi_r(0),0))\cdot \frac{\phi_r(s)}{\abs{\phi_r(s)}}}^{\underset{\text{Lem } \ref{key lemma} }{\leq} \frac{K}{4}r}+\overbrace{u_g(\phi_r(0),0) \cdot (0,1)}^{\underset{\eqref{Lip initial} }{\leq} -Kr} + \overbrace{u_g(\phi_r(0),0) \cdot \left( \frac{\phi_r(s)}{\abs{\phi_r(s)}}-(0,1)\right)}^{\underset{\text{Lem } \ref{phirt estimate lemma} }{\leq}\frac{K}{4}r}}{\underbrace{\abs{\phi_r(s)}\log \frac{1}{\abs{\phi_r(s)}}}_{\underset{\text{Lem } \ref{phirt estimate lemma}}{\geq}\frac{9}{10}r \log \frac{1}{r}}}    \, ds \bigg\vert \\
& \geq \left( \frac{10}{9}\right)^{1+\beta} \frac{K}{2} \left( \log \frac{1}{r}\right)^{\beta-2+\alpha-\varepsilon} \\
\underset{\eqref{eq34} }&{=} \left( \frac{10}{9}\right)^{1+\beta} \frac{K}{2} \left( \log \frac{1}{r}\right)^{\frac{\beta-1}{2}} \xrightarrow[\quad  r \searrow 0 \quad ]{} \boldsymbol{\infty}.
\end{align*}

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