# On measure solutions of the Keller-Segel equation in the subcritical and
critical regimes

Nicolas Fournier and Yoan Tardy  Sorbonne Université, LPSM-UMR 8001, Case
courrier 158, 75252 Paris Cedex 05, France.  nicolas.fournier@sorbonne-
universite.fr<EMAIL_ADDRESS>
###### Abstract.

We show global existence of a weak solution to the $2$-dimensional parabolic-
elliptic Keller-Segel equation when starting from any initial measure $f_{0}$
such that $f_{0}({\mathbb{R}}^{2})\leq 8\pi$ and
$\max_{x\in{\mathbb{R}}^{2}}f_{0}(\\{x\\})<8\pi$. The proof relies on a rather
simple two-particles moment computation.

###### Key words and phrases:

Keller-Segel equation, Chemotaxis, Existence of weak solutions.

###### 2010 Mathematics Subject Classification:

35K57, 35D30, 92C17

##  1\. Introduction

###  1.1. The model

We consider the classical parabolic-elliptic Keller-Segel model, also called
Patlak-Keller-Segel, of chemotaxis in ${\mathbb{R}}^{2}$, which writes

(1) $\displaystyle\partial_{t}f+\nabla\cdot(f\nabla c)=\Delta
f\qquad\hbox{and}\qquad\Delta c+f=0.$

The unknown $(f,c)$ is composed of two nonnegative functions $f_{t}(x)$ and
$c_{t}(x)$ of $t\geq 0$ and $x\in{\mathbb{R}}^{2}$, and the initial condition
$f_{0}$ is given.

This equation models the collective motion of a population of bacteria which
emit a chemical substance that attracts them. The quantity $f_{t}(x)$
represents the density of bacteria at position $x\in{\mathbb{R}}^{2}$ at time
$t\geq 0$, while $c_{t}(x)$ represents the concentration of chemical substance
at position $x\in{\mathbb{R}}^{2}$ at time $t\geq 0$. Note that in this model,
the speed of diffusion of the chemo-attractant is supposed to be infinite.
This equation has been introduced by Keller and Segel [16], see also Patlak
[18]. We refer to the recent book of Biler [3] and to the review paper of
Arumugam and Tyagi [1] for some complete descriptions of what is known about
this model.

We classically observe, see e.g. Blanchet-Dolbeault-Perthame [9, Page 4], that
necessarily $\nabla c_{t}=K\ast f_{t}$ for each $t\geq 0$, where

$\displaystyle K(x)=-\frac{x}{2\pi\|x\|^{2}}\quad\mbox{for
}x\in{\mathbb{R}}^{2}\setminus\\{0\\}\quad\mbox{ and (arbitrarily)}\quad
K(0)=0.$

Hence (1) may be rewritten as

(2) $\displaystyle\partial_{t}f+\nabla\cdot[f\;(K\ast f)]=\Delta f.$

###  1.2. Weak solutions

We will deal with weak measure solutions. For each $M>0$, we set

${\mathcal{M}}_{M}({\mathbb{R}}^{2})=\Big{\\{}\mu\mbox{ nonnegative measure on
}{\mathbb{R}}^{2}\mbox{ such that }\mu({\mathbb{R}}^{2})=M\Big{\\}}$

and we endow ${\mathcal{M}}_{M}({\mathbb{R}}^{2})$ with the weak convergence
topology, i.e. taking $C_{b}({\mathbb{R}}^{d})$, the set of continuous and
bounded functions, as set of test functions. We also denote by
$C_{b}^{2}({\mathbb{R}}^{d})$ the set of $C^{2}$-functions, bounded together
with all their derivatives. The following notion of weak solutions is
classical, see e.g. Blanchet-Dolbeault-Perthame [9, Page 5].

######  Definition 1.1.

Fix $M>0$. We say that $f\in
C([0,\infty),{\mathcal{M}}_{M}({\mathbb{R}}^{2}))$ is a weak solution of (2)
if for all $\varphi\in C^{2}_{b}({\mathbb{R}}^{2})$, all $t\geq 0$,

$\displaystyle\int_{{\mathbb{R}}^{2}}\varphi(x)f_{t}({\rm d}x)=$
$\displaystyle\int_{{\mathbb{R}}^{2}}\varphi(x)f_{0}({\rm
d}x)+\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\Delta\varphi(x)f_{s}({\rm d}x){\rm
d}s$
$\displaystyle+\frac{1}{2}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}K(x-y)\cdot(\nabla\varphi(x)-\nabla\varphi(y))f_{s}({\rm
d}x)f_{s}({\rm d}y){\rm d}s.$

All the terms in this equality are well-defined. In particular concerning the
last term, it holds that
$|K(x-y)\cdot(\nabla\varphi(x)-\nabla\varphi(y))|\leq\|\nabla^{2}\varphi\|_{\infty}/2\pi$.
However, $K(x-y)\cdot(\nabla\varphi(x)-\nabla\varphi(y))$, which equals $0$
when $x=y$ because we (arbitrarily) imposed that $K(0)=0$, is not continuous
near $x=y$. Hence a good weak solution has to verify that
$\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\hbox{\rm
1}{\hskip-2.8pt}\hbox{\rm I}_{\\{x=y\\}}f_{s}({\rm d}x)f_{s}({\rm d}y)=0$ for
a.e. $s\geq 0$.

###  1.3. Main result

Our goal is to prove the following global existence results.

######  Theorem 1.2.

(i) Fix $M\in(0,8\pi)$ and assume that
$f_{0}\in{\mathcal{M}}_{M}({\mathbb{R}}^{2})$ satisfies
$\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{0}({\rm d}x)<\infty$ for some
$\gamma\in(M/(4\pi),2)$. Then there exists a global weak solution $f$ to (2)
with initial condition $f_{0}$ and for all $t>0$, we have

(3) $\sup_{s\in[0,t]}\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{s}({\rm
d}x)+\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma-2}f_{s}({\rm
d}x)f_{s}({\rm d}y){\rm d}s<\infty.$

(ii) Assume that $f_{0}\in{\mathcal{M}}_{8\pi}({\mathbb{R}}^{2})$ satisfies
$\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm d}x)<\infty$ and
$\max_{x\in{\mathbb{R}}^{2}}f_{0}(\\{x\\})<8\pi$. Then there exists a global
weak solution $f$ to (2) with initial condition $f_{0}$ and for all $t>0$, we
have

(4) $\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{t}({\rm
d}x)=\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm
d}x)\quad\mbox{and}\quad\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\log\Big{(}1+\frac{1}{\|x-y\|}\Big{)}f_{s}({\rm
d}x)f_{s}({\rm d}y){\rm d}s<\infty.$

These solutions indeed satisfy that
$\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\hbox{\rm
1}{\hskip-2.8pt}\hbox{\rm I}_{\\{x=y\\}}f_{s}({\rm d}x)f_{s}({\rm d}y)=0$ for
a.e. $s\geq 0$. The proof of the following fact, which informally easily
follows from Definition 1.1, can be found in Blanchet-Dolbeaut-Perthame [9,
Page 5].

######  Remark 1.3.

Fix $M>0$ and consider $f_{0}\in{\mathcal{M}}_{M}({\mathbb{R}}^{2})$ such that
$\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm d}x)<\infty$. Then any (possibly
local) weak solution $f$ starting from $f_{0}$ and defined on a time interval
$[0,T]$ satisfies

$\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{t}({\rm
d}x)=\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm
d}x)+4M\Big{(}1-\frac{M}{8\pi}\Big{)}t\qquad\hbox{for all }t\in[0,T].$

According to this remark, Theorem 1.2 seems to be optimal, because

$\bullet$ if $M>8\pi$, there cannot be any global weak solution $f$, at least
if $\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm d}x)<\infty$, because else
$\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{t}({\rm d}x)$ would become negative in
finite time; $\bullet$ if $M=8\pi$ and if e.g. $f_{0}(\\{0\\})=8\pi$, then
$f_{0}=8\pi\delta_{0}$, so that Remark 1.3 would imply that
$f_{t}=8\pi\delta_{0}$ for all $t\geq 0$, but this is not a weak solution to
(2) because of the diffusion.

###  1.4. References and comments

This model was first introduced by Patlak [18] and Keller-Segel [16], as a
model for chemotaxis. For an exhaustive summary of the knowledge about this
equation and related models, we refer the reader to the review paper Arumugam-
Tyagi [1] and to the book of Biler [3].

The main difficulty of this model lies in the tight competition between
diffusion and attraction. Therefore it is not clear that a solution exists
because a blow-up could occur due to the emergence of a cluster, i.e. a Dirac
mass. Thus, the whole problem is about determining if the solution ends-up by
being concentrated in finite time or not.

As shown in Jäger-Luckhaus [15], this depends on the initial mass of the
solution $M=\int_{{\mathbb{R}}^{2}}f_{0}({\rm d}x)$, the solution globally
exists if $M$ is small enough and explodes in the other case. The fact that
solutions must explode in finite time if $M>8\pi$ is rather easy to show, as
recalled in Remark 1.3 above. But the fact that $8\pi$ is indeed the correct
thresold was much more difficult.

Biler-Karch-Laurençot-Nadzieja [4, 5] proved the global existence of a weak
solution in the subcritical case (first in the disc and later in
${\mathbb{R}}^{2}$) for every initial data which is a radially symmetric
measure such that $f_{0}(\\{0\\})=0$ and $f_{0}({\mathbb{R}}^{2})=M\leq 8\pi$
and satisfying a few other anodyne technical conditions. The proof relies on a
rather simple a priori estimate: to show that solutions do not explode, it
suffices to show that there is no blow up at $0$ since solutions are radial.
The argument is very similar to our approach, see Subsection 1.5, but is
simpler because of the radial symmetry. In spirit, for example in the
subcritical case $M<8\pi$, their computations more or less amount to look at
$({\rm d}/{\rm d}t)\int\|x\|^{\gamma}f_{t}({\rm d}x)$ and they conclude
something like $\int_{0}^{T}\int\|x\|^{\gamma-2}f_{t}({\rm d}x){\rm
d}t<\infty$.

At the same time, Blanchet-Dolbeault-Perthame [9] proved the existence of a
global weak free energy solution for initial data $f_{0}\in
L^{1}_{+}({\mathbb{R}}^{2})$ with mass $M<8\pi$, a finite moment of order $2$
and satisfying $\int f_{0}(x)\log(f_{0}(x)){\rm d}x<\infty$. The core of the
argument lies in the use of the logarithmic Hardy-Littlewood-Sobolev
inequality applied on a well chosen free-energy quantity. Something noticeable
is that the authors use this inequality with its optimal constant to get the
correct thresold $8\pi$.

In Bedrossian-Masmoudi [2], it is proven that under the condition that
$\max_{x\in{\mathbb{R}}^{2}}f_{0}(\\{x\\})<8\pi$, one can build mild solutions
even in the supercritical case, which are stronger solutions than weak
solutions, but these solutions are local in time. See also Biler-Zienkiewicz
[6] for much simpler approach where they build weaker solutions under the same
conditions. Wei [20] built global mild solutions in the subcritical and
critical cases and local mild solutions in the supercritical case for every
initial data $f_{0}\in L^{1}({\mathbb{R}}^{2})$, without any other condition.

Let us finally mention [13], where global weak solutions were built for any
measure initial condition $f_{0}$ such that $f_{0}({\mathbb{R}}^{2})<2\pi$,
with a light additional moment condition. This work was inspired by the work
of Osada [17] on vortices. Our work consists in refining this approach, and
surprisingly, this allows us to treat the whole subcritical case and, working
more, the critical case.

We show in this paper the existence of global weak solutions in both
subcritical and critical cases for all initial data $f_{0}$ which is a measure
such that $\max_{x\in 8\pi}f_{0}(\\{x\\})<8\pi$ and a moment condition, at
worst $\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm d}x)<\infty$. This moment
condition is anodyne, even if other references such as [2, 20] get rid of
these assumption. Our result seems to be stronger than all the preceding
references, and actually optimal as explained just after Remark 1.3, from the
point of view of the regularity of the initial data. To our knowledge, the
only reference that deals with global solutions for measure initial conditions
is Biler-Karch-Laurençot-Nadzieja [4, 5] but the initial measure is supposed
to be radially symmetric with no concentration of mass at $0$: $f_{0}$ is not
allowed to have some atoms. Nevertheless, we build a weak solution and the
estimate that quantifies the fact that the solution do not concentrate too
much is kind of weak.

Moreover, the proof relies on a simple (in the subcritical case at least)
computation which involves a well-chosen two-particles moment. This seems
rather simple and does not use a sophisticated functional inequality with some
optimal constant.

Informally, our result shows that one needs to have a cluster of mass of at
least $8\pi$ to see a blow-up occurring, whereas starting from one (or more)
cluster of mass strictly less than $8\pi$, particles in the cluster spread out
instantaneously. This implies that in the subcritical case, i.e if the total
mass is strictly less than $8\pi$, there cannot be any blow-up. We also show
that in the critical case, i.e if the total mass equals $8\pi$, if we have
$\max_{x\in{\mathbb{R}}^{2}}f_{0}(\\{x\\})<8\pi$ then there is no blow-up.
This means that possible initial clusters immediately spread out, and no
cluster may appear later. Moreover, if the initial mass strictly exceeds
$8\pi$, Remark 1.3 seems to indicate the emergence of a cluster in finite
time, and Suzuki [19] has indeed shown that explosion is due to the formation
of some Dirac of mass precisely equal to $8\pi$. We refer the reader to [14]
to have a probabilistic precise description of this phenomenon in the case of
a $N$-particle system closely related to the Keller-Segel equation, but the
arguments are not uniform in $N$.

Let us mention a few other papers: Carillo-Chen-Liu-Wang [11], with a result
and a proof slightly different than in Blanchet-Dolbeault-Perthame [9],
Carlen-Figalli [10] who proved a bound of the convergence of weak solutions to
a steady state of the Keller-Segel equation in the critical case, Egaña-
Mischler [12] concerning uniqueness of global weak solutions in both
subcritical and critical cases and of local weak solutions in the
supercritical case under a condition on solutions slightly less stringent than
boundedness, Blanchet-Carlen-Carrillo [7] for the convergence to steady states
in the critical case, etc.

###  1.5. Main a priori estimate in the subcritical case

The difficulty to build a global weak solution is to find some a priori
estimate, such as (3), showing that the solution cannot concentrate too much,
nor lose mass at infinity, in finite time. We consider a solution $f$ to (2)
starting from $f_{0}\in{\mathcal{M}}_{M}({\mathbb{R}}^{2})$, with
$M\in(0,8\pi)$, satisfying $\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{0}({\rm
d}x)<\infty$ for some $\gamma\in(M/(4\pi),2)$. The first estimate, saying that
the solution cannot lose mass at infinity because

(5) $\hbox{for all $t\geq
0$,}\qquad\sup_{s\in[0,t]}\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{s}({\rm
d}x)<\infty,$

is not difficult: it is informally easier than for the heat equation, since
the additional drift term is attractive. The second estimate is more tricky.
We set
$S_{t}=\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma}f_{t}({\rm
d}x)f_{t}({\rm d}y)$. Using (2) and proceeding to some integration by parts,
one finds that $S_{t}=S_{0}+I^{1}_{t}+I^{2}_{t}$, where

$\displaystyle
I^{1}_{t}=2\gamma^{2}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma-2}f_{s}({\rm
d}x)f_{s}({\rm d}y){\rm d}s,$ $\displaystyle
I^{2}_{t}=\gamma\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}(K(x-z)-K(y-z))\cdot(\|x-y\|^{\gamma-2}(x-y))f_{s}({\rm
d}x)f_{s}({\rm d}y)f_{s}({\rm d}z){\rm d}s.$

Symmetrizing and recalling the expression of $K$, we get

$\displaystyle
I^{2}_{t}=\frac{\gamma}{6\pi}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}F(x,y,z)f_{s}({\rm
d}x)f_{s}({\rm d}y)f_{s}({\rm d}z){\rm d}s,$

where, setting $X=x-y$, $Y=y-z$ and $Z=z-x$, $F(x,y,z)$ equals

$\displaystyle\Big{(}\frac{Z}{\|Z\|^{2}}+\frac{Y}{\|Y\|^{2}}\Big{)}\cdot\|X\|^{\gamma-2}X+\Big{(}\frac{X}{\|X\|^{2}}+\frac{Z}{\|Z\|^{2}}\Big{)}\cdot\|Y\|^{\gamma-2}Y+\Big{(}\frac{Y}{\|Y\|^{2}}+\frac{X}{\|X\|^{2}}\Big{)}\cdot\|Z\|^{\gamma-2}Z.$

We now define

$G(x,y,z)=\|X\|^{\gamma-2}+\|Y\|^{\gamma-2}+\|Z\|^{\gamma-2}=\|X\|^{\gamma-2}X\cdot\frac{X}{\|X\|^{2}}+\|Y\|^{\gamma-2}Y\cdot\frac{Y}{\|Y\|^{2}}+\|Z\|^{\gamma-2}Z\cdot\frac{Z}{\|Z\|^{2}}$

and observe that

$\displaystyle
F(x,y,z)+G(x,y,z)=\Big{(}\frac{X}{\|X\|^{2}}+\frac{Y}{\|Y\|^{2}}+\frac{Z}{\|Z\|^{2}}\Big{)}\cdot(\|X\|^{\gamma-2}+\|Y\|^{\gamma-2}Y+\|Z\|^{\gamma-2}Z),$

which is nonnegative according to Lemma 4.1 below, of which the proof is
rather quick. This implies that $F\geq-G$, whence

$I^{2}_{t}\geq-\frac{\gamma}{6\pi}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}G(x,y,z)f_{s}({\rm
d}x)f_{s}({\rm d}y)f_{s}({\rm d}z){\rm d}s.$

De-symmetrizing and using the conservation of mass, i.e. that
$\int_{{\mathbb{R}}^{2}}f_{t}({\rm d}z)=M$, we conclude that

$I^{2}_{t}\geq-\frac{\gamma
M}{2\pi}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma-2}f_{s}({\rm
d}x)f_{s}({\rm d}y){\rm d}s.$

All this proves that a priori,

$\displaystyle S_{t}\geq
S_{0}+2\gamma\Big{(}\gamma-\frac{M}{4\pi}\Big{)}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma-2}f_{s}({\rm
d}x)f_{s}({\rm d}y){\rm d}s.$

Since now $S_{t}$ is finite by (5), since $S_{0}\geq 0$ and since
$\gamma-M/4\pi>0$, we conclude that

$\hbox{for all $t\geq
0$,}\qquad\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma-2}f_{s}({\rm
d}x)f_{s}({\rm d}y){\rm d}s<\infty.$

This indeed provides some (very weak) non-concentration estimate because
$\gamma<2$. As we will see, this informal argument is sufficient to build
rigorously some global weak solutions. The critical case where $M=8\pi$ (but
$\max_{x\in{\mathbb{R}}^{2}}f_{0}(\\{x\\})<8\pi$) is more intricate.

###  1.6. Plan of the paper

In Section 2, we introduce a regularized version of (2), for which there
obviously exists a global classical solution. In Section 3, we quickly show
the compactness of the family of solutions to these regularized problems. In
Section 4, we conclude the existence proof in the subcritical case, i.e. we
prove Theorem 1.2-(i). Finally, we treat the critical case, i.e. we prove
Theorem 1.2-(ii), in Section 5, which is more tedious.

##  2\. A regularized model

We fix $M>0$ and $f_{0}\in{\mathcal{M}}_{M}({\mathbb{R}}^{2})$. For
$\varepsilon\in(0,1]$, we introduce the following regularized versions

$K_{\varepsilon}(x)=-\frac{x}{2\pi(\|x\|^{2}+\varepsilon)}\quad\mbox{and}\quad
f_{0}^{\varepsilon}(x)=\frac{1}{2\pi\varepsilon}\int_{{\mathbb{R}}^{2}}e^{-\|x-y\|^{2}/(2\varepsilon)}f_{0}({\rm
d}y).$

of $K$ and $f_{0}$. Since $K_{\varepsilon}$ and $f_{0}^{\varepsilon}$ are
smooth, the equation

(1)
$\partial_{t}f^{\varepsilon}+\nabla\cdot[f^{\varepsilon}(K_{\varepsilon}\ast
f^{\varepsilon})]=\Delta f^{\varepsilon}$

starting from $f^{\varepsilon}_{0}$ has a unique classical solution
$(f^{\varepsilon}_{t}(x))_{t\geq 0,x\in{\mathbb{R}}^{2}}$. This solution
preserves mass, i.e

(2) $\displaystyle\int_{{\mathbb{R}}^{2}}f_{t}^{\varepsilon}({\rm
d}x)=\int_{{\mathbb{R}}^{2}}f_{0}^{\varepsilon}({\rm
d}x)=\int_{{\mathbb{R}}^{2}}f_{0}({\rm d}x)=M\qquad\hbox{for all $t\geq 0$}.$

Here and below, we write $f^{\varepsilon}_{t}({\rm
d}x)=f^{\varepsilon}_{t}(x){\rm d}x$.

######  Lemma 2.1.

(i) For all $\varepsilon\in(0,1]$, all $t\geq 0$ and all $\varphi\in
C^{2}_{b}({\mathbb{R}}^{2})$, we have

(3) $\displaystyle\int_{{\mathbb{R}}^{2}}\varphi(x)f_{t}^{\varepsilon}({\rm
d}x)=$ $\displaystyle\int_{{\mathbb{R}}^{2}}\varphi(x)f_{0}^{\varepsilon}({\rm
d}x)+\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\Delta\varphi(x)f_{s}^{\varepsilon}({\rm
d}x){\rm d}s$
$\displaystyle+\frac{1}{2}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}K_{\varepsilon}(x-y)\cdot[\nabla\varphi(x)-\nabla\varphi(y)]f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s.$

(ii) For all $\varepsilon\in(0,1]$, all $t\geq 0$ and all $\varphi\in
C^{2}_{c}({\mathbb{R}}_{+})$, we have

$\displaystyle\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}$
$\displaystyle\varphi(\|x-y\|^{2})f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm
d}y)=\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\varphi(\|x-y\|^{2})f_{0}^{\varepsilon}({\rm
d}x)f_{0}^{\varepsilon}({\rm d}y)$
$\displaystyle+8\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}[\varphi^{\prime}(\|x-y\|^{2})+\|x-y\|^{2}\varphi^{\prime\prime}(\|x-y\|^{2})]f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s$
$\displaystyle+2\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}[K_{\varepsilon}(x-z)-K_{\varepsilon}(y-z)]\cdot(x-y)\varphi^{\prime}(\|x-y\|^{2})f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y)f_{s}^{\varepsilon}({\rm d}z){\rm d}s.$

###### Proof.

For (i), we first consider $\varphi\in C^{2}_{c}({\mathbb{R}}^{2})$, we
multiply (1) by $\varphi$, integrate on $[0,t]\times{\mathbb{R}}^{2}$, and
proceed to some integrations by parts. This leads to

(4) $\displaystyle\int_{{\mathbb{R}}^{2}}\varphi(x)f_{t}^{\varepsilon}({\rm
d}x)=$ $\displaystyle\int_{{\mathbb{R}}^{2}}\varphi(x)f_{0}^{\varepsilon}({\rm
d}x)+\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\Delta\varphi(x)f_{s}^{\varepsilon}({\rm
d}x){\rm d}s$
$\displaystyle+\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}K_{\varepsilon}(x-y)\cdot\nabla\varphi(x)f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s.$

This formula easily extends to any $\varphi\in C^{2}_{b}({\mathbb{R}}^{2})$ by
approximation and dominated convergence, since $K_{\varepsilon}$ is bounded.
Finally, a symmetry argument allows us to conclude.

For (ii), we first consider $\psi\in
C^{2}_{b}({\mathbb{R}}^{2}\times{\mathbb{R}}^{2})$ and write

$\displaystyle\frac{{\rm d}}{{\rm
d}t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\psi(x,y)f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)$ $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{2}}\Big{(}\frac{{\rm d}}{{\rm
d}t}\int_{{\mathbb{R}}^{2}}\psi(x,y)f_{t}^{\varepsilon}({\rm
d}x)\Big{)}f_{t}^{\varepsilon}({\rm
d}y)+\int_{{\mathbb{R}}^{2}}\Big{(}\frac{{\rm d}}{{\rm
d}t}\int_{{\mathbb{R}}^{2}}\psi(x,y)f_{t}^{\varepsilon}({\rm
d}y)\Big{)}f_{t}^{\varepsilon}({\rm d}x)$ $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}(\Delta_{x}\psi(x,y)+\Delta_{y}\psi(x,y))f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)$ (5)
$\displaystyle+\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\Big{(}K_{\varepsilon}(x-z)\cdot\nabla_{x}\psi(x,y)+K_{\varepsilon}(y-z)\cdot\nabla_{y}\psi(x,y)\Big{)}f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)f_{t}^{\varepsilon}({\rm d}z)$

by (4). For $\varphi\in C^{2}_{c}({\mathbb{R}}_{+})$, it holds that
$\psi(x,y)=\varphi(\|x-y\|^{2})$ belongs to
$C^{2}_{b}({\mathbb{R}}^{2}\times{\mathbb{R}}^{2})$, and

$\displaystyle\nabla_{x}\psi(x,y)=-\nabla_{y}\psi(x,y)=2(x-y)\varphi^{\prime}(\|x-y\|^{2}),$
$\displaystyle\Delta_{x}\psi(x,y)=\Delta_{y}\psi(x,y)=4\varphi^{\prime}(\|x-y\|^{2})+4\|x-y\|^{2}\varphi^{\prime\prime}(\|x-y\|^{2}).$

Inserting these formulas into (5) completes the proof. ∎

##  3\. Compactness

We fix $M>0$, $f_{0}\in{\mathcal{M}}_{M}({\mathbb{R}}^{2})$ and consider the
corresponding family $(f^{\varepsilon})_{\varepsilon\in(0,1]}$ of solutions to
the regularized problems introduced in Section 2. The aim of this section is
to prove the following.

######  Proposition 3.1.

Assume that $\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{0}({\rm d}x)<\infty$ for
some $\gamma\in(0,2]$.

(i) There is a constant $C_{\gamma}$, depending only on $\gamma$, such that
for all $\varepsilon\in(0,1]$, all $t\geq 0$,

$\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{t}^{\varepsilon}({\rm d}x)\leq
2\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{0}({\rm
d}x)+C_{\gamma}(M+M^{2})(1+t).$

(ii) If $\gamma\geq 1$, then for all $\varepsilon\in(0,1]$, all $t\geq 0$,
$\int_{{\mathbb{R}}^{2}}xf_{t}^{\varepsilon}({\rm
d}x)=\int_{{\mathbb{R}}^{2}}xf_{0}({\rm d}x)$.

(iii) The family $(f^{\varepsilon})_{\varepsilon\in(0,1]}$ is relatively
compact in $C([0,\infty),{\mathcal{M}}_{M}({\mathbb{R}}^{2}))$, endowed with
the uniform convergence on compact time intervals,
${\mathcal{M}}_{M}({\mathbb{R}}^{2})$ being endowed with the weak convergence
topology.

###### Proof.

We start with (i). We introduce a smooth function
$\chi:{\mathbb{R}}_{+}\to[0,1]$ such that $\chi(r)=1$ for $r\in[0,1]$ and
$\chi(r)=0$ for $r\geq 2$. For $a\in(0,1]$, we consider
$\varphi_{a}(x)=(1+\|x\|^{2})^{\gamma/2}\chi(a\|x\|^{2})$, which belongs to
$C^{2}_{c}({\mathbb{R}}^{2})$. One easily checks, using that $\gamma\in(0,2]$,
that $A_{\gamma}=\sup_{a\in(0,1]}\|\nabla^{2}\varphi_{a}\|_{\infty}<\infty$.
Applying (3), we conclude that for all $\varepsilon\in(0,1]$, all $a\in(0,1]$,
all $t\geq 0$,

$\int_{{\mathbb{R}}^{2}}\varphi_{a}(x)f_{t}^{\varepsilon}({\rm
d}x)=I^{\varepsilon}_{a,1}+I^{\varepsilon}_{a,2}(t)+I^{\varepsilon}_{a,3}(t),$

with

$\displaystyle
I^{\varepsilon}_{a,1}=\int_{{\mathbb{R}}^{2}}\varphi_{a}(x)f_{0}^{\varepsilon}({\rm
d}x),\qquad
I^{\varepsilon}_{a,2}(t)=\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\Delta\varphi_{a}(x)f_{s}^{\varepsilon}({\rm
d}x){\rm d}s,$ $\displaystyle
I^{\varepsilon}_{a,3}(t)=\frac{1}{2}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}K_{\varepsilon}(x-y)\cdot[\nabla\varphi_{a}(x)-\nabla\varphi_{a}(y)]f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s.$

First,
$I^{\varepsilon}_{a,1}\leq\int_{{\mathbb{R}}^{2}}(1+\|x\|^{2})^{\gamma/2}f_{0}^{\varepsilon}({\rm
d}x)\leq\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{0}^{\varepsilon}({\rm
d}x)+M\leq 2\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{0}({\rm d}x)+4M$. Next, we
obviously have $I^{\varepsilon}_{a,2}(t)\leq 2MA_{\gamma}t$ by (2), while

$\displaystyle I^{\varepsilon}_{a,3}(t)\leq$
$\displaystyle\frac{A_{\gamma}}{2}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|K_{\varepsilon}(x-y)\|\,\|x-y\|f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s\leq\frac{M^{2}A_{\gamma}t}{4\pi},$

since $\|z\|\,\|K_{\varepsilon}(z)\|\leq 1/(2\pi)$ and by (2) again. All in
all, we found a constant $C_{\gamma}$ such that
$\int_{{\mathbb{R}}^{2}}\varphi_{a}(x)f_{t}^{\varepsilon}({\rm d}x)\leq
2\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{0}({\rm
d}x)+C_{\gamma}(M+M^{2})(1+t)$. We easily conclude by monotone convergence,
since $\varphi_{a}(x)$ increases to
$(1+\|x\|^{2})^{\gamma/2}\geq\|x\|^{\gamma}$ as $a\to 0$.

If $\gamma\geq 1$, we easily deduce from point (i) that (3) can be extended to
any $\varphi\in C^{2}({\mathbb{R}}^{2})$ with bounded derivatives of order $1$
and $2$. In particular, we may apply (3) with $\varphi(x)=x$ (or rather with
$\varphi(x)=x_{1}$ and $\varphi(x)=x_{2}$). This gives
$\int_{{\mathbb{R}}^{2}}xf_{t}^{\varepsilon}({\rm
d}x)=\int_{{\mathbb{R}}^{2}}xf_{0}^{\varepsilon}({\rm d}x)$, which equals
$\int_{{\mathbb{R}}^{2}}xf_{0}({\rm d}x)$.

Concerning (iii), we recall that we may find a family $(\varphi_{n})_{n\geq
0}$ of elements of $C^{2}_{b}({\mathbb{R}}^{2})$ satisfying

$\|\varphi_{n}\|_{\infty}+\|\nabla\varphi_{n}\|_{\infty}+\|\nabla^{2}\varphi_{n}\|_{\infty}\leq
1\qquad\hbox{for all $n\geq 0$}$

and such that the distance $\delta$ on ${\mathcal{M}}_{M}({\mathbb{R}}^{2})$
defined through

$\delta(f,g)=\sum_{n\geq
0}2^{-n}\Big{|}\int_{{\mathbb{R}}^{2}}\varphi_{n}(x)f({\rm
d}x)-\int_{{\mathbb{R}}^{2}}\varphi_{n}(x)g({\rm d}x)\Big{|}$

metrizes the weak convergence topology on
${\mathcal{M}}_{M}({\mathbb{R}}^{2})$. We now check that we can apply the
Arzela-Ascoli theorem. First, for each $t\geq 0$,
$(f_{t}^{\varepsilon})_{\varepsilon\in(0,1]}$ is tight by (i) (since
$\gamma>0$) and thus relatively compact in
${\mathcal{M}}_{M}({\mathbb{R}}^{2})$ by the Prokhorov theorem. To conclude,
it is enough to show that $f^{\varepsilon}$ is uniformly Lipschitz continuous
in time, in that there exists a constant $C>0$ such that for all
$\varepsilon\in(0,1]$, all $t\geq s\geq 0$,
$\delta(f_{t}^{\varepsilon},f_{s}^{\varepsilon})\leq C|t-s|$. But using (3),
for all $n\geq 0$,

$\displaystyle\Big{|}\int_{{\mathbb{R}}^{2}}\varphi_{n}(x)f_{t}^{\varepsilon}({\rm
d}x)-\int_{{\mathbb{R}}^{2}}\varphi_{n}(x)f_{s}^{\varepsilon}({\rm
d}x)\Big{|}$ $\displaystyle=$
$\displaystyle\Big{|}\int_{s}^{t}\int_{{\mathbb{R}}^{2}}\Delta\varphi_{n}(x)f_{u}^{\varepsilon}({\rm
d}x){\rm
d}u+\frac{1}{2}\int_{s}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}K_{\varepsilon}(x-y)\cdot[\nabla\varphi_{n}(x)-\nabla\varphi_{n}(y)]f_{u}^{\varepsilon}({\rm
d}x)f_{u}^{\varepsilon}({\rm d}y){\rm d}u\Big{|}$ $\displaystyle\leq$
$\displaystyle(2M+M^{2}/(4\pi))(t-s),$

by (2), since $\|\nabla^{2}\varphi_{n}\|_{\infty}\leq 1$ and since
$\|z\|\,\|K_{\varepsilon}(z)\|\leq 1/(2\pi)$. We conclude that

$\delta(f_{t}^{\varepsilon},f_{s}^{\varepsilon})\leq\sum_{n\geq
0}2^{-n}(2M+M^{2}/(4\pi))(t-s)=2(2M+M^{2}/(4\pi))(t-s)$

as desired. ∎

##  4\. The subcritical case

In this section we prove Theorem 1.2-(i). We fix $M\in(0,8\pi)$ and
$f_{0}\in{\mathcal{M}}_{M}({\mathbb{R}}^{2})$ such that
$\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{0}({\rm d}x)<\infty$ for some
$\gamma\in(M/(4\pi),2)$ and we consider the corresponding family
$(f^{\varepsilon})_{\varepsilon\in(0,1]}$ introduced in Section 2. We begin
with a simple geometrical result which is crucial for our purpose.

######  Lemma 4.1.

For all pair of nonincreasing functions
$\varphi,\psi:(0,\infty)\to(0,\infty)$, for all $X,Y,Z\in{\mathbb{R}}^{2}$
such that $X+Y+Z=0$, we have

$\Delta=[\varphi(\|X\|)X+\varphi(\|Y\|)Y+\varphi(\|Z\|)Z]\cdot[\psi(\|X\|)X+\psi(\|Y\|)Y+\psi(\|Z\|)Z]\geq
0.$

More precisely, if e.g. $\|X\|\leq\|Y\|\leq\|Z\|$, we have

$\Delta\geq[\varphi(\|X\|)-\varphi(\|Y\|)][\psi(\|X\|)-\psi(\|Y\|)]\|X\|^{2}.$

###### Proof.

We may study only the case where $\|X\|\leq\|Y\|\leq\|Z\|$. Since $Y=-X-Z$,

$\displaystyle\varphi(\|X\|)X+\varphi(\|Y\|)Y+\varphi(\|Z\|)Z=\lambda X-\mu
Z,$
$\displaystyle\psi(\|X\|)X+\psi(\|Y\|)Y+\psi(\|Z\|)Z=\lambda^{\prime}X-\mu^{\prime}Z,$

where $\lambda=\varphi(\|X\|)-\varphi(\|Y\|)\geq 0$,
$\mu=\varphi(\|Y\|)-\varphi(\|Z\|)\geq 0$,
$\lambda^{\prime}=\psi(\|X\|)-\psi(\|Y\|)\geq 0$ and
$\mu^{\prime}=\psi(\|Y\|)-\psi(\|Z\|)\geq 0$. Therefore,

$\Delta=\lambda\lambda^{\prime}\|X\|^{2}+\mu\mu^{\prime}\|Z\|^{2}-(\lambda\mu^{\prime}+\lambda^{\prime}\mu)X\cdot
Z\geq\lambda\lambda^{\prime}\|X\|^{2}$

as desired, because $X\cdot Z\leq 0$. Indeed, if $X\cdot Z>0$, then
$\|Y\|^{2}=\|Z+X\|^{2}=\|Z\|^{2}+\|X\|^{2}+2X\cdot Z>\|Z\|^{2}\geq\|Y\|^{2}$,
which is absurd. ∎

The following computation is the core of the paper, at least in the
subcritical case.

######  Proposition 4.2.

Recall that $M\in(0,8\pi)$, that $f_{0}\in{\mathcal{M}}_{M}({\mathbb{R}}^{2})$
and that $\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{0}({\rm d}x)<\infty$ for
some $\gamma\in(M/(4\pi),2)$. For all $\varepsilon\in(0,1]$, all $t>0$,

$\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma-2}f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm
d}s\leq\frac{4M}{2\gamma(\gamma-M/(4\pi))}\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{t}^{\varepsilon}({\rm
d}x).$

###### Proof.

We fix $\varepsilon\in(0,1]$ and divide the proof into four steps. Step 1. For
$a>0$, we apply Lemma 2.1-(ii) with the function
$\varphi_{a}(r)=(a+r)^{\gamma/2}$, whence

$\varphi_{a}^{\prime}(r)=\gamma(a+r)^{\gamma/2-1}/2\qquad\hbox{and}\qquad\varphi_{a}^{\prime\prime}(r)=\gamma(\gamma-2)(a+r)^{\gamma/2-2}/4.$

We find
$I_{a}^{\varepsilon}(t)=I_{a}^{\varepsilon}(0)+\int_{0}^{t}J_{a}^{\varepsilon}(s){\rm
d}s+\int_{0}^{t}S_{a}^{\varepsilon}(s){\rm d}s$, with

$\displaystyle
I_{a}^{\varepsilon}(t)=\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}(a+\|x-y\|^{2})^{\gamma/2}f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y),$ $\displaystyle
J_{a}^{\varepsilon}(t)=\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\Big{(}4\gamma(a+\|x-y\|^{2})^{\gamma/2-1}+2\gamma(\gamma-2)\|x-y\|^{2}(a+\|x-y\|^{2})^{\gamma/2-2}\Big{)}f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y),$ $\displaystyle
S_{a}^{\varepsilon}(t)=\gamma\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}[K_{\varepsilon}(x-z)-K_{\varepsilon}(y-z)]\cdot(x-y)(a+\|x-y\|^{2})^{\gamma/2-1}f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)f_{t}^{\varepsilon}({\rm d}z).$

The use of Lemma 2.1-(ii) is not licit, however, it suffices to approximate
$\varphi_{a}$ by a suitable sequence of compactly supported functions and to
pass to the limit by dominated convergence using that
$\sup_{[0,t]}\int_{{\mathbb{R}}^{2}}(1+\|x\|^{\gamma})f_{s}^{\varepsilon}({\rm
d}x)<\infty$ by Proposition 3.1-(i) and (2), since the integrand $\bullet$ in
$I_{a}^{\varepsilon}(t)$ is bounded by a constant times
$1+\|x\|^{\gamma}+\|y\|^{\gamma}$,

$\bullet$ in $J_{a}^{\varepsilon}(t)$ is bounded (since $\gamma<2$),

$\bullet$ in $S_{a}^{\varepsilon}(t)$ is bounded by a constant times
$\|K_{\varepsilon}\|_{\infty}(1+\|x-y\|^{2})^{\gamma/2-1/2}\leq\|K_{\varepsilon}\|_{\infty}(1+\|x\|^{\gamma}+\|y\|^{\gamma})$.

Step 2. It holds that

$J_{a}^{\varepsilon}(t)\geq
2\gamma^{2}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}(a+\|x-y\|^{2})^{\gamma/2-1}f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y).$

Indeed, it suffices to observe that for all $r\geq 0$, using that $2>\gamma$
for the last inequality,

$4\gamma(a+r)^{\gamma/2-1}+2\gamma(\gamma-2)r(a+r)^{\gamma/2-2}=2\gamma(a+r)^{\gamma/2-2}[2a+\gamma
r]\geq 2\gamma^{2}(a+r)^{\gamma/2-1}.$

Step 3. We next verify that

$S^{\varepsilon}_{a}(t)\geq-\frac{\gamma
M}{2\pi}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}(a+\|x-y\|^{2})^{\gamma/2-1}f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y).$

By symmetry,

(1)
$S^{\varepsilon}_{a}(t)=\frac{\gamma}{3}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}F_{a}^{\varepsilon}(x,y,z)f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)f_{t}^{\varepsilon}({\rm d}z),$

with

$\displaystyle F_{a}^{\varepsilon}(x,y,z)=$
$\displaystyle[K_{\varepsilon}(x-z)-K_{\varepsilon}(y-z)]\cdot(x-y)(a+\|x-y\|^{2})^{\gamma/2-1}$
$\displaystyle+[K_{\varepsilon}(y-x)-K_{\varepsilon}(z-x)]\cdot(y-z)(a+\|y-z\|^{2})^{\gamma/2-1}$
$\displaystyle+[K_{\varepsilon}(z-y)-K_{\varepsilon}(x-y)]\cdot(z-x)(a+\|z-x\|^{2})^{\gamma/2-1}.$

Introducing $X=x-y$, $Y=y-z$ and $Z=z-x$ and recalling that
$K_{\varepsilon}(X)=\frac{-X}{2\pi(\|X\|^{2}+\varepsilon)}$, we find

$\displaystyle 2\pi F_{a}^{\varepsilon}(x,y,z)=$
$\displaystyle\Big{[}\frac{Z}{\|Z\|^{2}+\varepsilon}+\frac{Y}{\|Y\|^{2}+\varepsilon}\Big{]}\cdot
X(a+\|X\|^{2})^{\gamma/2-1}$
$\displaystyle+\Big{[}\frac{X}{\|X\|^{2}+\varepsilon}+\frac{Z}{\|Z\|^{2}+\varepsilon}\Big{]}\cdot
Y(a+\|Y\|^{2})^{\gamma/2-1}$
$\displaystyle+\Big{[}\frac{Y}{\|Y\|^{2}+\varepsilon}+\frac{X}{\|X\|^{2}+\varepsilon}\Big{]}\cdot
Z(a+\|Z\|^{2})^{\gamma/2-1}.$

We now introduce
$G_{a}(x,y,z)=(a+\|X\|^{2})^{\gamma/2}+(a+\|Y\|^{2})^{\gamma/2}+(a+\|Z\|^{2})^{\gamma/2}$
and note that

$G_{a}(x,y,z)\geq
X\cdot\frac{X}{\|X\|^{2}+\varepsilon}(a+\|X\|^{2})^{\gamma/2}+Y\cdot\frac{Y}{\|Y\|^{2}+\varepsilon}(a+\|Y\|^{2})^{\gamma/2}+Z\cdot\frac{Z}{\|Z\|^{2}+\varepsilon}(a+\|Z\|^{2})^{\gamma/2}.$

Hence $G_{a}(x,y,z)+2\pi F_{a}^{\varepsilon}(x,y,z)$ is larger than

$\displaystyle\Big{(}\frac{X}{\|X\|^{2}\\!+\\!\varepsilon}\\!+\\!\frac{Y}{\|Y\|^{2}\\!+\\!\varepsilon}\\!+\\!\frac{Z}{\|Z\|^{2}\\!+\\!\varepsilon}$
$\displaystyle\Big{)}\cdot\Big{(}X(a\\!+\\!\|X\|^{2})^{\gamma/2-1}\\!+\\!Y(a\\!+\\!\|Y\|^{2})^{\gamma/2-1}\\!+\\!Z(a\\!+\\!\|Z\|^{2})^{\gamma/2-1}\Big{)},$

which is nonnegative according to Lemma 4.1, since $r\to
1/(r^{2}+\varepsilon)$ and $r\to(a+r^{2})^{\gamma/2-1}$ are both nonincreasing
on $(0,\infty)$ and since $X+Y+Z=0$. Thus $F_{a}^{\varepsilon}(x,y,z)\geq-
G_{a}(x,y,z)/(2\pi)$. Recalling (1), using a symmetry argument and (2), we
conclude that

$\displaystyle S^{\varepsilon}_{a}(t)$
$\displaystyle\geq-\frac{\gamma}{6\pi}\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\\!G_{a}(x,y,z)f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)f_{t}^{\varepsilon}({\rm d}z)=-\frac{\gamma
M}{2\pi}\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\\!(a+\|X\|^{2})^{\gamma/2}f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y).$

Step 4. Combining Steps 1, 2 and 3 and using that $I_{a}^{\varepsilon}(0)\geq
0$, we have proved that

$\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}(a+\|x-y\|^{2})^{\gamma/2}f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)\geq\Big{(}2\gamma^{2}-\frac{\gamma
M}{2\pi}\Big{)}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}(a+\|x-y\|^{2})^{\gamma/2-1}f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s.$

Letting $a\to 0$ by dominated convergence (due to Proposition 3.1-(i)) for the
LHS and monotone convergence for the RHS, we get

$\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma}f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)\geq\Big{(}2\gamma^{2}-\frac{\gamma
M}{2\pi}\Big{)}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma-2}f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s.$

Since finally
$\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma}f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)\leq
4M\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{t}^{\varepsilon}({\rm d}x)$ because
$(r+u)^{\gamma}\leq 2r^{\gamma}+2u^{\gamma}$, the conclusion follows. ∎

We now give the

###### Proof of Theorem 1.2-(i).

Recall that $M\in(0,8\pi)$, that $f_{0}\in{\mathcal{M}}_{M}({\mathbb{R}}^{2})$
and $\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{0}({\rm d}x)<\infty$ for some
$\gamma\in(M/(4\pi),2)$, and that $(f^{\varepsilon})_{\varepsilon\in(0,1]}$ is
the corresponding family introduced in Section 2.

Step 1. By Proposition 3.1-(iii), we can find $(\varepsilon_{k})_{k\geq 0}$
and $f\in C([0,\infty),{\mathcal{M}}_{M}({\mathbb{R}}^{2}))$ such that
$\lim_{k}\varepsilon_{k}=0$ and $\lim_{k}f^{\varepsilon_{k}}=f$ in
$C([0,\infty),{\mathcal{M}}_{M}({\mathbb{R}}^{2}))$, endowed with the uniform
convergence on compact time intervals, ${\mathcal{M}}_{M}({\mathbb{R}}^{2})$
being endowed with the weak convergence topology. By definition of
$f_{0}^{\varepsilon}$, we obviously have $f|_{t=0}=f_{0}$. By Propositions
3.1-(i) and 4.2 and the Fatou lemma, we have

$\hbox{for all
$t>0$,}\qquad\sup_{s\in[0,t]}\int_{{\mathbb{R}}^{2}}\|x\|^{\gamma}f_{s}({\rm
d}x)+\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma-2}f_{s}({\rm
d}x)f_{s}({\rm d}y){\rm d}s<\infty.$

Step 2. It only remains to check that $f$ is a weak solution to (2). We fix
$\varphi\in C^{2}_{b}({\mathbb{R}}^{2})$ and use (3) to write
$\int_{{\mathbb{R}}^{2}}\varphi(x)f_{t}^{\varepsilon_{k}}({\rm
d}x)=I_{k}(t)+J_{k}(t)$, where

$\displaystyle
I_{k}(t)=\int_{{\mathbb{R}}^{2}}\varphi(x)f_{0}^{\varepsilon_{k}}({\rm
d}x)+\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\Delta\varphi(x)f_{s}^{\varepsilon_{k}}({\rm
d}x){\rm d}s,$ $\displaystyle
J_{k}(t)=\frac{1}{2}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}K_{\varepsilon_{k}}(x-y)\cdot[\nabla\varphi(x)-\nabla\varphi(y)]f_{s}^{\varepsilon_{k}}({\rm
d}x)f_{s}^{\varepsilon_{k}}({\rm d}y){\rm d}s.$

Since $\varphi$ and $\Delta\varphi$ are continuous and bounded, we immediately
conclude that

$\lim_{k}\int_{{\mathbb{R}}^{2}}\\!\varphi(x)f_{t}^{\varepsilon_{k}}({\rm
d}x)=\int_{{\mathbb{R}}^{2}}\\!\varphi(x)f_{t}({\rm
d}x)\quad\hbox{and}\quad\lim_{k}I_{k}(t)=\int_{{\mathbb{R}}^{2}}\\!\varphi(x)f_{0}({\rm
d}x)+\int_{0}^{t}\\!\int_{{\mathbb{R}}^{2}}\\!\Delta\varphi(x)f_{s}({\rm
d}x){\rm d}s.$

It only remains to check that for
$J(t)=\frac{1}{2}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}K(x-y)\cdot[\nabla\varphi(x)-\nabla\varphi(y)]f_{s}({\rm
d}x)f_{s}({\rm d}y){\rm d}s$, we have $\lim_{k}J_{k}(t)=J(t)$. To this end, we
write $J_{k}(t)=J_{k}^{1}(t)+J_{k}^{2}(t)$, where

$\displaystyle
J_{k}^{1}(t)=\frac{1}{2}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}K(x-y)\cdot[\nabla\varphi(x)-\nabla\varphi(y)]f_{s}^{\varepsilon_{k}}({\rm
d}x)f_{s}^{\varepsilon_{k}}({\rm d}y){\rm d}s,$ $\displaystyle
J_{k}^{2}(t)=\frac{1}{2}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}[K_{\varepsilon_{k}}(x-y)-K(x-y)]\cdot[\nabla\varphi(x)-\nabla\varphi(y)]f_{s}^{\varepsilon_{k}}({\rm
d}x)f_{s}^{\varepsilon_{k}}({\rm d}y){\rm d}s.$

Recalling the expression of $K$ and that $\varphi\in
C^{2}_{b}({\mathbb{R}}^{2})$, we see that
$g(x,y)=K(x-y)\cdot[\nabla\varphi(x)-\nabla\varphi(y)]$ is bounded and
continuous on the set ${\mathbb{R}}^{2}\setminus D$, where
$D=\\{(x,y)\in{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}:x=y\\}$. Since
$f^{\varepsilon_{k}}_{s}\otimes f^{\varepsilon_{k}}_{s}$ goes weakly to
$f_{s}\otimes f_{s}$ for each $s\geq 0$ and since $(f_{s}\otimes f_{s})(D)=0$
for a.e. $s\geq 0$ by Step 1, we conclude that
$\lim_{k}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}g(x,y)f_{s}^{\varepsilon_{k}}({\rm
d}x)f_{s}^{\varepsilon_{k}}({\rm
d}y)=\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}g(x,y)f_{s}({\rm
d}x)f_{s}({\rm d}y)$ for a.e. $s\geq 0$, whence
$\lim_{k}J^{1}_{k}(t)=J_{k}(t)$ by dominated convergence.

We finally have to verify that $\lim_{k}J^{2}_{k}(t)=0$. But

$\|z\|\,\|K(z)-K_{\varepsilon}(z)\|=\frac{\varepsilon}{2\pi(\varepsilon+\|z\|^{2})}\leq\min(1,\varepsilon\|z\|^{-2})\leq(\varepsilon\|z\|^{-2})^{1-\gamma/2}=\varepsilon^{1-\gamma/2}\|z\|^{\gamma-2},$

where we used that $\gamma\in(0,2)$. Hence

$\displaystyle|J^{2}_{k}(t)|\leq$
$\displaystyle\|\nabla^{2}\varphi\|_{\infty}\varepsilon_{k}^{1-\gamma/2}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma-2}f_{s}^{\varepsilon_{k}}({\rm
d}x)f_{s}^{\varepsilon_{k}}({\rm d}y){\rm d}s,$

which tends to $0$ as desired since
$\sup_{\varepsilon\in(0,1]}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|^{\gamma-2}f_{s}^{\varepsilon_{k}}({\rm
d}x)f_{s}^{\varepsilon_{k}}({\rm d}y){\rm d}s<\infty$ by Propositions 4.2 and
3.1-(i). ∎

##  5\. The critical case

Here we prove Theorem 1.2-(ii). We fix
$f_{0}\in{\mathcal{M}}_{8\pi}({\mathbb{R}}^{2})$ such that
$\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm d}x)<\infty$ and
$\max_{x\in{\mathbb{R}}^{2}}f_{0}(\\{x\\})<8\pi$. We consider the
corresponding family $(f^{\varepsilon})_{\varepsilon\in(0,1]}$ introduced in
Section 2. We will need the following lemma, which shows that
$f^{\varepsilon}_{t}$ cannot be too concentrated on one single point. We use
the standard notation $B(x,r)=\\{z\in{\mathbb{R}}^{2}:\|z-x\|<r\\}$.

######  Lemma 5.1.

Recall that $f_{0}\in{\mathcal{M}}_{8\pi}({\mathbb{R}}^{2})$, that
$\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm d}x)<\infty$ and that
$\max_{x\in{\mathbb{R}}^{2}}f_{0}(\\{x\\})<8\pi$. For all $T>0$, there exist
$\nu\in(0,1]$ and $\zeta>0$ such that for all $\varepsilon\in(0,1]$, all
$t\in[0,T]$, all $x_{0}\in{\mathbb{R}}^{2}$,

$f_{t}^{\varepsilon}(B(x_{0},\nu)^{c})\geq\zeta.$

###### Proof.

We divide the proof into $3$ steps. Step 1. The goal of this step is to prove
that

$\displaystyle\alpha:=\inf_{\varepsilon\in(0,1]}\inf_{t\geq
0}\inf_{x_{0}\in{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-x_{0}\|^{2}f_{t}^{\varepsilon}({\rm
d}x)>0.$

First of all, we have that for all $\varepsilon\in(0,1]$, all $t\geq 0$,

$\displaystyle\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{t}^{\varepsilon}({\rm
d}x)\geq\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}^{\varepsilon}({\rm d}x).$

Indeed, this inequality follows from (3) with $\varphi(x)=\|x\|^{2}$, noting
that for all $s\geq 0$,

$\displaystyle\int_{{\mathbb{R}}^{2}}\Delta\varphi(x)f_{s}^{\varepsilon}({\rm
d}x)$
$\displaystyle+\frac{1}{2}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}K_{\varepsilon}(x-y)\cdot[\nabla\varphi(x)-\nabla\varphi(y)]f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y)$ $\displaystyle=$ $\displaystyle 4\times
8\pi-\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\frac{\|x-y\|^{2}}{2\pi(\|x-y\|^{2}+\varepsilon)}f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y)\geq 32\pi-\frac{(8\pi)^{2}}{2\pi}=0.$

We used the expression of $K_{\varepsilon}$ and that
$f_{s}\in{\mathcal{M}}_{8\pi}({\mathbb{R}}^{2})$. To be perfectly rigorous,
one rather has to use a compactly supported smooth approximation of
$\varphi(x)=\|x\|^{2}$.

But $\int_{{\mathbb{R}}^{2}}xf_{t}^{\varepsilon}({\rm d}x)$ and
$\int_{{\mathbb{R}}^{2}}f^{\varepsilon}_{t}({\rm d}x)$ are constant by
Proposition 3.1-(ii) and (2), so that for all $\varepsilon\in(0,1]$, all
$x_{0}\in{\mathbb{R}}^{2}$, all $t\geq 0$,

$\displaystyle\int_{{\mathbb{R}}^{2}}\|x-x_{0}\|^{2}f_{t}^{\varepsilon}({\rm
d}x)\geq\int_{{\mathbb{R}}^{2}}\|x-x_{0}\|^{2}f_{0}^{\varepsilon}({\rm d}x).$

We finally observe that, setting
$z_{0}=\int_{{\mathbb{R}}^{2}}xf_{0}^{\varepsilon}({\rm
d}x)=\int_{{\mathbb{R}}^{2}}xf_{0}({\rm d}x)$, it holds that

$\displaystyle\min_{x_{0}\in{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-x_{0}\|^{2}f_{0}^{\varepsilon}({\rm
d}x)=\int_{{\mathbb{R}}^{2}}\|x-z_{0}\|^{2}f_{0}^{\varepsilon}({\rm
d}x)\geq\int_{{\mathbb{R}}^{2}}\|x-z_{0}\|^{2}f_{0}({\rm d}x)$

by definition of $f_{0}^{\varepsilon}$. This last quantity is positive, since
$f_{0}(\\{z_{0}\\})<8\pi$ by assumption.

Step 2. Here we show that there exists some nondecreasing $\psi\in
C({\mathbb{R}}_{+})$ such that $\psi(0)=1$ and
$\lim_{r\to\infty}\psi(r)=\infty$ such that for all $T\geq 0$,

$\displaystyle
C_{T}:=\sup_{t\in[0,T]}\sup_{\varepsilon\in(0,1]}\int_{{\mathbb{R}}^{2}}\|x\|^{2}\psi(\|x\|^{2})f_{t}^{\varepsilon}({\rm
d}x)<\infty.$

Since $\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm d}x)<\infty$, by the de la
Vallée Poussin theorem, there exists $\psi\in C({\mathbb{R}}_{+})$ satisfying
the above assumptions and such that
$\int_{{\mathbb{R}}^{2}}\|x\|^{2}\psi(\|x\|^{2})f_{0}({\rm d}x)<\infty$.
Moreover, it is classically possible to choose $\psi$ in a such a way that
$\varphi(x):=\|x\|^{2}\psi(\|x\|^{2})$ is $C^{2}$, convex and e.g. such that
$\|\nabla^{2}\varphi(x)\|\leq C(1+\|x\|^{2})$. Besides, this implies that
$\sup_{\varepsilon\in(0,1]}\int_{{\mathbb{R}}^{2}}\|x\|^{2}\psi(\|x\|^{2})f_{0}^{\varepsilon}({\rm
d}x)<\infty$.

Next, we apply (3) with $\varphi(x):=\|x\|^{2}\psi(\|x\|^{2})$ and find that

$\int_{{\mathbb{R}}^{2}}\|x\|^{2}\psi(\|x\|^{2})f_{t}^{\varepsilon}({\rm
d}x)=\int_{{\mathbb{R}}^{2}}\|x\|^{2}\psi(\|x\|^{2})f_{0}^{\varepsilon}({\rm
d}x)+I^{\varepsilon}_{t}+J^{\varepsilon}_{t},$

where
$I^{\varepsilon}_{t}=\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\Delta\varphi(x)f^{\varepsilon}_{s}({\rm
d}x){\rm d}s\leq
C\int_{0}^{t}\int_{{\mathbb{R}}^{2}}(1+\|x\|^{2})f^{\varepsilon}_{s}({\rm
d}x){\rm d}s\leq C(1+t^{2})$ by Proposition 3.1-(i), for some constant $C>0$
depending only on $f_{0}$; and where the drift term $J^{\varepsilon}_{t}$ is
nonpositive, since by convexity of $\varphi$,

$K_{\varepsilon}(x-y)\cdot[\nabla\varphi(x)-\nabla\varphi(y)]=-\frac{x-y}{2\pi(\|x-y\|^{2}+\varepsilon)}\cdot[\nabla\varphi(x)-\nabla\varphi(y)]\leq
0.$

Step 3. We now fix $T>0$ and conclude. We set

$A=\Big{(}\frac{1}{4\pi}\sup_{t\in[0,T]}\sup_{\varepsilon\in(0,1]}\int_{{\mathbb{R}}^{2}}\|x\|^{2}f^{\varepsilon}_{t}({\rm
d}x)\Big{)}^{1/2},$

which is finite by Proposition 3.1-(i). First, for all $t\in[0,T]$, all
$x_{0}\in B(0,A+1)^{c}$, using that
$f^{\varepsilon}_{t}({\mathbb{R}}^{2})=8\pi$,

(1) $\displaystyle f_{t}^{\varepsilon}(B(x_{0},1)^{c})\geq
f_{t}^{\varepsilon}(B(0,A))\geq 8\pi-f_{t}^{\varepsilon}(B(0,A)^{c})\geq
8\pi-A^{-2}\int_{{\mathbb{R}}^{2}}\|x\|^{2}f^{\varepsilon}_{t}({\rm d}x)\geq
4\pi.$

Next, we fix $x_{0}\in B(0,A+1)$ and $t\in[0,T]$. By Step 1, we have, for any
$\nu\in(0,1]$,

$\alpha\leq\int_{{\mathbb{R}}^{2}}\|x-x_{0}\|^{2}f_{t}^{\varepsilon}({\rm
d}x)\leq
8\pi\nu^{2}+\int_{\|x-x_{0}\|>\nu}\hskip-9.0pt\|x-x_{0}\|^{2}f_{t}^{\varepsilon}({\rm
d}x)\leq
8\pi\nu^{2}+2\int_{\|x-x_{0}\|>\nu}\hskip-9.0pt(\|x\|^{2}+\|x_{0}\|^{2})f_{t}^{\varepsilon}({\rm
d}x).$

Hence,

$\displaystyle\alpha\leq$ $\displaystyle
8\pi\nu^{2}+2\|x_{0}\|^{2}f_{t}^{\varepsilon}(B(x_{0},\nu)^{c})+2\int_{\|x-x_{0}\|>\nu}\|x\|^{2}f_{t}^{\varepsilon}({\rm
d}x)$ $\displaystyle\leq$ $\displaystyle
8\pi\nu^{2}+2(\|x_{0}\|^{2}+B^{2})f_{t}^{\varepsilon}(B(x_{0},\nu)^{c})+2\int_{\|x-x_{0}\|>\nu,\|x\|\geq
B}\|x\|^{2}f_{t}^{\varepsilon}({\rm d}x)$ $\displaystyle\leq$ $\displaystyle
8\pi\nu^{2}+2((A+1)^{2}+B^{2})f_{t}^{\varepsilon}(B(x_{0},\nu)^{c})+2\frac{C_{T}}{\psi(B^{2})},$

where $\psi$ and $C_{T}$ were defined in Step 2. We now take $B$ large enough
so that $2C_{T}/\psi(B^{2})\leq\alpha/4$ and $\nu\in(0,1]$ small enough so
that $8\pi\nu^{2}\leq\alpha/4$. We get that for all $t\in[0,T]$, all $x_{0}\in
B(0,A+1)$,

(2) $\displaystyle
2((A+1)^{2}+B^{2})f_{t}^{\varepsilon}(B(x_{0},\nu)^{c})\geq\frac{\alpha}{2}.$

One easily concludes, gathering (1) and (2). ∎

We next proceed to a computation similar to Proposition 4.2 in the subcritical
case. For $\varepsilon\in(0,1]$ and $x,y,z\in{\mathbb{R}}^{2}$, we set
$X=x-y$, $Y=y-z$, $Z=z-x$ and

$\displaystyle
G_{\varepsilon}(x,y,z)=\Big{(}L(\|X\|^{2})X+L(\|Y\|^{2})Y+L(\|Z\|^{2})Z\Big{)}\cdot\Big{(}\frac{X}{\|X\|^{2}+\varepsilon}+\frac{Y}{\|Y\|^{2}+\varepsilon}+\frac{Z}{\|Z\|^{2}+\varepsilon}\Big{)},$

where $L(r)=\log(1+1/r)$ for all $r>0$. Observe that $G_{\varepsilon}$ is
nonnegative by Lemma 4.1.

######  Lemma 5.2.

If $f_{0}\in{\mathcal{M}}_{8\pi}({\mathbb{R}}^{2})$ satisfies
$\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm d}x)<\infty$, then for all $t>0$,

(3)
$\displaystyle\sup_{\varepsilon\in(0,1]}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}G_{\varepsilon}(x,y,z)f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y)f_{s}^{\varepsilon}({\rm d}z){\rm
d}s<\infty,$ (4)
$\displaystyle\sup_{\varepsilon\in(0,1]}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}L(\|x-y\|^{2})\frac{\varepsilon}{\|x-y\|^{2}+\varepsilon}f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s<\infty.$

###### Proof.

We fix $\varepsilon\in(0,1]$ and $\eta\in(0,1]$ and we introduce
$L_{\eta}(r)=\log(1+1/(\eta+r))$ for all $r\geq 0$. We apply Lemma 2.1-(ii)
with $\varphi_{\eta}(r)=(r+\eta)\log(1+1/(\eta+r))+\log(1+\eta+r)$ whence
$\varphi^{\prime}_{\eta}(r)=L_{\eta}(r)$ and
$\varphi_{\eta}^{\prime\prime}(r)=-1/(\eta+r+(\eta+r)^{2})$. We find that

(5)
$S^{\varepsilon,\eta}_{t}:=\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\varphi_{\eta}(\|x-y\|^{2})f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm
d}y)=S^{\varepsilon,\eta}_{0}+\int_{0}^{t}(A^{\varepsilon,\eta}_{s}+B^{\varepsilon,\eta}_{s}){\rm
d}s,$

where, setting $h_{\eta}(r)=r/[\eta+r+(\eta+r)^{2}]$,

$\displaystyle A^{\varepsilon,\eta}_{t}=$ $\displaystyle
8\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\Big{(}L_{\eta}(\|x-y\|^{2})-h_{\eta}(\|x-y\|^{2})\Big{)}f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)$ $\displaystyle\geq$
$\displaystyle\frac{1}{\pi}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}L_{\eta}(\|x-y\|^{2})f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)f_{t}^{\varepsilon}({\rm
d}z)-8\times(8\pi)^{2}$

because $M=8\pi$ and $h_{\eta}$ is bounded by $1$, and where

$B^{\varepsilon,\eta}_{t}=2\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}(K_{\varepsilon}(x-z)+K_{\varepsilon}(z-y))\cdot(x-y)L_{\eta}(\|x-y\|^{2})f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)f_{t}^{\varepsilon}({\rm d}z).$

Symmetrizing, we get

(6) $\displaystyle
A^{\varepsilon,\eta}_{t}+B^{\varepsilon,\eta}_{t}\geq\frac{1}{3\pi}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}F_{\varepsilon,\eta}(x,y,z)f_{t}^{\varepsilon}({\rm
d}x)f_{t}^{\varepsilon}({\rm d}y)f_{t}^{\varepsilon}({\rm d}z)-512\pi^{2},$

where

$\displaystyle F_{\varepsilon,\eta}(x,y,z)=$ $\displaystyle
L_{\eta}(\|x-y\|^{2})[1+2\pi(x-y)\cdot(K_{\varepsilon}(x-z)+K_{\varepsilon}(z-y))]$
$\displaystyle+L_{\eta}(\|y-z\|^{2})[1+2\pi(y-z)\cdot(K_{\varepsilon}(y-x)+K_{\varepsilon}(x-z))]$
$\displaystyle+L_{\eta}(\|z-x\|^{2})[1+2\pi(z-x)\cdot(K_{\varepsilon}(z-y)+K_{\varepsilon}(y-x))].$

Setting $X=x-y$, $Y=y-z$, $Z=z-x$ and recalling that $2\pi
K_{\varepsilon}(X)=-\frac{X}{\|X\|^{2}+\varepsilon}$ we find

$\displaystyle F_{\varepsilon,\eta}(x,y,z)=$ $\displaystyle
L_{\eta}(\|X\|^{2})\Big{[}1+X\cdot\Big{(}\frac{Z}{\|Z\|^{2}+\varepsilon}+\frac{Y}{\|Y\|^{2}+\varepsilon}\Big{)}\Big{]}$
$\displaystyle
L_{\eta}(\|Y\|^{2})\Big{[}1+Y\cdot\Big{(}\frac{X}{\|X\|^{2}+\varepsilon}+\frac{Z}{\|Z\|^{2}+\varepsilon}\Big{)}\Big{]}$
$\displaystyle+L_{\eta}(\|Z\|^{2})\Big{[}1+Z\cdot\Big{(}\frac{Y}{\|Y\|^{2}+\varepsilon}+\frac{X}{\|X\|^{2}+\varepsilon}\Big{)}\Big{]}$
$\displaystyle=$ $\displaystyle
L_{\eta}(\|X\|^{2})X\cdot\Big{(}\frac{X}{\|X\|^{2}+\varepsilon}+\frac{Y}{\|Y\|^{2}+\varepsilon}+\frac{Z}{\|Z\|^{2}+\varepsilon}\Big{)}+L_{\eta}(\|X\|^{2})\frac{\varepsilon}{\|X\|^{2}+\varepsilon}$
$\displaystyle+L_{\eta}(\|Y\|^{2})Y\cdot\Big{(}\frac{X}{\|X\|^{2}+\varepsilon}+\frac{Y}{\|Y\|^{2}+\varepsilon}+\frac{Z}{\|Z\|^{2}+\varepsilon}\Big{)}+L_{\eta}(\|Y\|^{2})\frac{\varepsilon}{\|Y\|^{2}+\varepsilon}$
$\displaystyle+L_{\eta}(\|Z\|^{2})Z\cdot\Big{(}\frac{X}{\|X\|^{2}+\varepsilon}+\frac{Y}{\|Y\|^{2}+\varepsilon}+\frac{Z}{\|Z\|^{2}+\varepsilon}\Big{)}+L_{\eta}(\|Z\|^{2})\frac{\varepsilon}{\|Z\|^{2}+\varepsilon}$
$\displaystyle=$ $\displaystyle
G_{\varepsilon,\eta}(x,y,z)+H_{\varepsilon,\eta}(x,y,z),$

where we have put

$\displaystyle
G_{\varepsilon,\eta}(x,y,z)=\Big{(}L_{\eta}(\|X\|^{2})X+L_{\eta}(\|Y\|^{2})Y+L_{\eta}(\|Z\|^{2})Z\Big{)}\cdot\Big{(}\frac{X}{\|X\|^{2}+\varepsilon}+\frac{Y}{\|Y\|^{2}+\varepsilon}+\frac{Z}{\|Z\|^{2}+\varepsilon}\Big{)},$
$\displaystyle
H_{\varepsilon,\eta}(x,y,z)(x,y,z)=L_{\eta}(\|X\|^{2})\frac{\varepsilon}{\|X\|^{2}+\varepsilon}+L_{\eta}(\|Y\|^{2})\frac{\varepsilon}{\|Y\|^{2}+\varepsilon}+L_{\eta}(\|Z\|^{2})\frac{\varepsilon}{\|Z\|^{2}+\varepsilon}.$

Inserting this into (5) and (6), we find

$\displaystyle S^{\varepsilon,\eta}_{t}\geq$ $\displaystyle
S^{\varepsilon,\eta}_{0}-512\pi^{2}t+\frac{1}{3\pi}\\!\int_{0}^{t}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!G_{\varepsilon,\eta}(x,y,z)f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y)f_{s}^{\varepsilon}({\rm d}z){\rm d}s$
$\displaystyle+\frac{1}{3\pi}\\!\int_{0}^{t}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!H_{\varepsilon,\eta}(x,y,z)f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y)f_{s}^{\varepsilon}({\rm d}z){\rm d}s.$

Observing that $0\leq\varphi_{\eta}(r)\leq C(1+r)$ for some constant $C>0$, we
deduce from Proposition 3.1-(i), with $\gamma=2$, that
$\sup_{\varepsilon\in(0,1],\eta\in(0,1]}S_{t}^{\varepsilon,\eta}<\infty$.
Since moreover $S_{0}^{\varepsilon,\eta}\geq 0$, we conclude that

(7)
$\sup_{\varepsilon\in(0,1],\eta\in(0,1]}\int_{0}^{t}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\![G_{\varepsilon,\eta}(x,y,z)+H_{\varepsilon,\eta}(x,y,z)]f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y)f_{s}^{\varepsilon}({\rm d}z){\rm
d}s<\infty.$

Recall that both $G_{\varepsilon,\eta}$ and $H_{\varepsilon,\eta}$ are
nonnegative. Since

$\int_{0}^{t}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!H_{\varepsilon,\eta}(x,y,z)f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y)f_{s}^{\varepsilon}({\rm d}z){\rm d}s=3\times
8\pi\int_{0}^{t}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\frac{\varepsilon
L_{\eta}(\|x-y\|^{2})}{\|x-y\|^{2}+\varepsilon}f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s$

by de-symmetrization, we deduce (4) from (7) by monotone convergence, since
$L_{\eta}(r)$ increases to $L(r)$ as $\eta\to 0$. Finally, one easily shows,
using Lemma 4.1 and that $r\mapsto L_{\eta}(r)-L_{\eta^{\prime}}(r)$ is
nonincreasing and nonnegative if $0<\eta<\eta^{\prime}$, that $\eta\to
G_{\varepsilon,\eta}(x,y,z)$ increases to $G_{\varepsilon}(x,y,z)$ as $\eta$
decreases to $0$. We thus also deduce (3) from (7) by monotone convergence. ∎

We can now prove the main estimate in the critical case.

######  Proposition 5.3.

If $f_{0}\in{\mathcal{M}}_{8\pi}({\mathbb{R}}^{2})$ verifies
$\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm d}x)<\infty$ and
$\max_{x\in{\mathbb{R}}^{2}}f_{0}(\\{x\\})<8\pi$, then

$\displaystyle\hbox{for all
$t>0$,}\quad\sup_{\varepsilon\in(0,1]}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\log(1+\|x-y\|^{-2})f^{\varepsilon}_{s}({\rm
d}x)f^{\varepsilon}_{s}({\rm d}y){\rm d}s<\infty.$

###### Proof.

We fix $t\geq 0$ and we introduce

$I^{\varepsilon}_{t}:=\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}G_{\varepsilon}(x,y,z)f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y)f_{s}^{\varepsilon}({\rm d}z){\rm d}s.$

Recalling the definition of $G_{\varepsilon}$ and using Lemma 4.1, we see that
for all $x,y,z\in{\mathbb{R}}^{2}$, setting $X=x-y$, $Y=y-z$ and $Z=z-x$ as
usual and recalling that $L(r)=\log(1+1/r)$,

$G_{\varepsilon}(x,y,z)\geq\hbox{\rm 1}{\hskip-2.8pt}\hbox{\rm
I}_{\\{\|X\|\leq\|Y\|\leq\|Z\|\\}}\Big{(}L(\|X\|^{2})-L(\|Y\|^{2})\Big{)}\Big{(}\frac{1}{\|X\|^{2}+\varepsilon}-\frac{1}{\|Y\|^{2}+\varepsilon}\Big{)}\|X\|^{2}.$

We next consider $\nu>0$ and $\zeta>0$ introduced in Lemma 5.1 (with $T=t$),
i.e. such that
$\inf_{s\in[0,t],x_{0}\in{\mathbb{R}}^{2}}f_{s}(B(x_{0},\nu)^{c})\geq\zeta$,
and we write

$\displaystyle G_{\varepsilon}(x,y,z)\geq$ $\displaystyle\hbox{\rm
1}{\hskip-2.8pt}\hbox{\rm
I}_{\\{\|X\|\leq\nu/3,\nu/2\leq\|Y\|\leq\|Z\|\\}}\Big{(}L(\|X\|^{2})-L(\nu^{2}/4)\Big{)}\frac{\|X\|^{2}}{\|X\|^{2}+\varepsilon}\Big{(}1-\frac{\|X\|^{2}+\varepsilon}{\|Y\|^{2}+\varepsilon}\Big{)}$
$\displaystyle\geq$ $\displaystyle c\hbox{\rm 1}{\hskip-2.8pt}\hbox{\rm
I}_{\\{\|X\|\leq\nu/3,\nu/2\leq\|Y\|\leq\|Z\|\\}}\Big{(}L(\|X\|^{2})-L(\nu^{2}/4)\Big{)}\frac{\|X\|^{2}}{\|X\|^{2}+\varepsilon},$

where $c=1-[\nu^{2}/9+1]/[\nu^{2}/4+1]>0$. We used that
$1-[\nu^{2}/9+\varepsilon]/[\nu^{2}/4+\varepsilon]\geq c$ for all
$\varepsilon\in(0,1]$. We deduce that

$\displaystyle G_{\varepsilon}(x,y,z)\geq$ $\displaystyle c\hbox{\rm
1}{\hskip-2.8pt}\hbox{\rm
I}_{\\{\|X\|\leq\nu/3,\nu/2\leq\|Y\|\leq\|Z\|\\}}L(\|X\|^{2})\frac{\|X\|^{2}}{\|X\|^{2}+\varepsilon}-cL(\nu^{2}/4).$

All in all,

$\displaystyle I^{\varepsilon}_{t}+c(8\pi)^{3}L(\nu^{2}/4)t$
$\displaystyle\geq$ $\displaystyle
c\int_{0}^{t}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\hbox{\rm
1}{\hskip-2.8pt}\hbox{\rm
I}_{\\{\|x-y\|\leq\nu/3,\nu/2\leq\|y-z\|\leq\|z-x\|\\}}L(\|x-y\|^{2})\frac{\|x-y\|^{2}}{\|x-y\|^{2}+\varepsilon}f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y)f_{s}^{\varepsilon}({\rm d}z){\rm d}s$
$\displaystyle=$ $\displaystyle
c\int_{0}^{t}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\hbox{\rm
1}{\hskip-2.8pt}\hbox{\rm
I}_{\\{\|x-y\|\leq\nu/3\\}}L(\|x-y\|^{2})\frac{\|x-y\|^{2}}{\|x-y\|^{2}+\varepsilon}f_{s}^{\varepsilon}(\Gamma_{x,y})f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s,$

where $\Gamma_{x,y}=\\{z\in{\mathbb{R}}^{2}:\nu/2\leq\|z-y\|\leq\|z-x\|\\}$.
Using a symmetry argument, we find that

$\displaystyle I^{\varepsilon}_{t}+c(8\pi)^{3}L(\nu^{2}/4)t$
$\displaystyle\geq$
$\displaystyle\frac{c}{2}\int_{0}^{t}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\hbox{\rm
1}{\hskip-2.8pt}\hbox{\rm
I}_{\\{\|x-y\|\leq\nu/3\\}}L(\|x-y\|^{2})\frac{\|x-y\|^{2}}{\|x-y\|^{2}+\varepsilon}f_{s}^{\varepsilon}(\Gamma_{x,y}\cup\Gamma_{y,x})f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s.$

But if $\|x-y\|\leq\nu/3$, we have $\Gamma_{x,y}\cup\Gamma_{y,x}\supset
B(x,\nu)^{c}$ because if $z\in B(x,\nu)^{c}$, then $\bullet$ if
$\|z-x\|\leq\|z-y\|$, then $z\in\Gamma_{y,x}$ because
$\|z-x\|\geq\nu\geq\nu/2$, $\bullet$ if $\|z-y\|\leq\|z-x\|$, then
$z\in\Gamma_{x,y}$ because $\|z-y\|\geq\|z-x\|-\|x-y\|\geq\nu-\nu/3\geq\nu/2$.

As a consequence, $f_{s}^{\varepsilon}(\Gamma_{x,y}\cup\Gamma_{y,x})\geq\zeta$
for all $s\in[0,t]$ (if $\|x-y\|\leq\nu/3$), and we end with

$\displaystyle\frac{c\zeta}{2}\int_{0}^{t}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\hbox{\rm
1}{\hskip-2.8pt}\hbox{\rm
I}_{\\{\|x-y\|\leq\nu/3\\}}L(\|x-y\|^{2})\frac{\|x-y\|^{2}}{\|x-y\|^{2}+\varepsilon}f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s\leq
I^{\varepsilon}_{t}+c(8\pi)^{3}L(\nu^{2}/4)t$

Moreover, we know from Lemma 5.2 that
$\sup_{\varepsilon\in(0,1]}I^{\varepsilon}_{t}<\infty$. All this shows that
there is a constant $C$ (depending on $t$) such that for all
$\varepsilon\in(0,1]$,

$\displaystyle\int_{0}^{t}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!\hbox{\rm
1}{\hskip-2.8pt}\hbox{\rm
I}_{\\{\|x-y\|\leq\nu/3\\}}L(\|x-y\|^{2})\frac{\|x-y\|^{2}}{\|x-y\|^{2}+\varepsilon}f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s\leq C.$

Since $L(u^{2})\leq\log(1+9/\nu^{2})$ for all $u\geq\nu/3$, we conclude that
for all $\varepsilon\in(0,1]$,

$\displaystyle\int_{0}^{t}\\!\int_{{\mathbb{R}}^{2}}\\!\int_{{\mathbb{R}}^{2}}\\!L(\|x-y\|^{2})\frac{\|x-y\|^{2}}{\|x-y\|^{2}+\varepsilon}f_{s}^{\varepsilon}({\rm
d}x)f_{s}^{\varepsilon}({\rm d}y){\rm d}s\leq C+t(8\pi)^{2}\log(1+9/\nu^{2}).$

We conclude using now (4). ∎

We are now ready to give the

###### Proof of Theorem 1.2-(ii).

Recall that $f_{0}\in{\mathcal{M}}_{8\pi}({\mathbb{R}}^{2})$ is such that
$\max_{x_{0}\in{\mathbb{R}}^{2}}f_{0}(\\{x_{0}\\})<8\pi$ and
$\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm d}x)<\infty$ and that the
corresponding family $(f^{\varepsilon})_{\varepsilon\in(0,1]}$ was introduced
in Section 2.

Step 1. By Proposition 3.1-(iii), we can find $(\varepsilon_{k})_{k\geq 0}$
and $f\in C([0,\infty),{\mathcal{M}}_{8\pi}({\mathbb{R}}^{2}))$ such that
$\lim_{k}\varepsilon_{k}=0$ and $\lim_{k}f^{\varepsilon_{k}}=f$ in
$C([0,\infty),{\mathcal{M}}_{8\pi}({\mathbb{R}}^{2}))$, endowed with the
uniform convergence on compact time intervals,
${\mathcal{M}}_{M}({\mathbb{R}}^{2})$ being endowed with the weak convergence
topology. By definition of $f_{0}^{\varepsilon}$, we obviously have
$f|_{t=0}=f_{0}$. By Proposition 5.3 and the Fatou Lemma, we directly get

$\hbox{for all $t\geq
0$,}\qquad\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\log(1+\|x-y\|^{-2})f_{s}({\rm
d}x)f_{s}({\rm d}y){\rm d}s<\infty.$

Step 2. It remains to show that $f$ is a weak solution to (2). Observe that
this automatically implies that $\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{t}({\rm
d}x)=\int_{{\mathbb{R}}^{2}}\|x\|^{2}f_{0}({\rm d}x)$ for all $t\geq 0$ by
Remark 1.3. It suffices to apply the very same arguments as in Step 3 of the
proof of Theorem 1.2-(i), except when showing that $\lim_{k}J_{k}^{2}(t)=0$,
which we now do. As already seen,
$\|z\|\,\|K(z)-K_{\varepsilon}(z)\|\leq\min(1,\varepsilon\|z\|^{-2})$, whence

$\|z\|\,\|K(z)-K_{\varepsilon}(z)\|\leq\min(1,\varepsilon\|z\|^{-2})\leq\frac{\log(1+\|z\|^{-2})}{\log(1+\varepsilon^{-1})},$

which is clear if $\|z\|^{2}\leq\varepsilon$, and which follows from the fact
that $r\mapsto r\log(1+r^{-1})$ is nondecreasing on ${\mathbb{R}}_{+}$ when
$\|z\|^{2}>\varepsilon$. Hence

$\displaystyle|J_{k}^{2}(t)|\leq$
$\displaystyle\|\nabla^{2}\varphi\|_{\infty}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\|x-y\|\|K(x-y)-K_{\varepsilon_{k}}(x-y)\|f_{s}^{\varepsilon_{k}}({\rm
d}x)f_{s}^{\varepsilon_{k}}({\rm d}y){\rm d}s$ $\displaystyle\leq$
$\displaystyle\frac{\|\nabla^{2}\varphi\|_{\infty}}{\log(1+\varepsilon_{k}^{-1})}\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}\log(1+\|x-y\|^{-2})f_{s}^{\varepsilon_{k}}({\rm
d}x)f_{s}^{\varepsilon_{k}}({\rm d}y){\rm d}s,$

which tends to $0$ by Proposition 5.3. ∎

## References

  * [1] G. Arumugam, J. Tyagi, Keller-Segel chemotaxis models: a review, Acta Appl. Math. 171, (2021), Paper 6, 82 pp.
  * [2] J. Bedrossian, N. Masmoudi, Existence, uniqueness and Lipschitz dependence for Patlak-Keller-Segel and Navier-Stokes in ${\mathbb{R}}^{2}$ with measure-valued initial data, Arch. Ration. Mech. Anal. 214, (2014), 717–801.
  * [3] P. Biler, Singularities of solutions to chemotaxis systems, De Gruyter Series in Mathematics and Life Sciences, De Gruyter, Berlin, 2020.
  * [4] P. Biler, G. Karch, P. Laurençot, T. Nadzieja, The $8\pi$-problem for radially symmetric solutions of a chemotaxis model in a disc, Topol. Methods Nonlinear Anal. 27, (2006), 133–147.
  * [5] P. Biler, G. Karch, P. Laurençot, T. Nadzieja, The $8\pi$-problem for radially symmetric solutions of a chemotaxis model in the plane, Math. Methods Appl. Sci. 29, (2006), 1563–1583.
  * [6] P. Biler, J. Zienkiewicz, Existence of solutions for the Keller-Segel model of chemotaxis with measures as initial data, Bull. Pol. Acad. Sci. Math. 63, (2015), 41–51.
  * [7] A. Blanchet, E. Carlen, J. Carillo, Functional inequalities, thick tails and asymptotics for the critical mass Patlak-Keller-Segel model, J. Funct. Anal. 262, (2012), 2142–2230.
  * [8] A. Blanchet, J. Carrillo, N. Masmoudi, Infinite time aggregation for the critical Patlak-Keller-Segel model in ${\mathbb{R}}^{2}$, Comm. Pure Appl. Math. 61, (2008), 1449–1481.
  * [9] A. Blanchet, J. Dolbeault, B. Perthame, Two-dimensional Keller-Segel model: optimal critical mass and qualitative properties of the solutions, Electron. J. Differential Equations 44 (2006), 32 pp.
  * [10] E. Carlen, A.Figalli, Stability for a GNS inequality and the log-HLS inequality, with application to the critical mass Keller-Segel equation, Duke Math. J. 162, (2013), 579–625.
  * [11] J. Carrillo, L. Chen, J. Liu, J. Wang, A note on the subcritical two dimensional Keller-Segel system, Acta Appl. Math. 119, (2012), 43–55.
  * [12] F. Egaña, S. Mischler, Uniqueness and long time asymptotics for the parabolic-parabolic Keller-Segel equation, Comm. Partial Differential Equations 42, (2017), 291–345.
  * [13] N. Fournier, B. Jourdain, Stochastic particle approximation of the Keller-Segel equation and two-dimensional generalization of Bessel processes, Ann. Appl. Probab. 27, (2017), 2807–2861.
  * [14] N. Fournier, Y. Tardy, Collisions of the supercritical Keller-Segel particle system, arXiv:2110.08490.
  * [15] W. Jäger, S. Luckaus, On explosions of solutions to a system of partial differential equations modelling chemotaxis, Trans. Amer. Math. Soc. 329, (1992), 819–824.
  * [16] E. Keller, L. Segel, Initiation of slime mold aggregation viewed as an instability, J. Theoret. Biol. 26, (1970), 399–415.
  * [17] H. Osada, A stochastic differential equation arising from the vortex problem, Proc. Japan Acad. Ser. A Math. Sci. 61, (1985), 333–336.
  * [18] C.S. Patlak, Random walk with persistence and external bias, Bull. Math. Biophys. 15, (1953), 311–338.
  * [19] T. Suzuki, Exclusion of boundary blowup for 2D chemotaxis system provided with Dirichlet boundary condition for the Poisson part, J. Math. Pures Appl., 100 (2013), 347–367.
  * [20] D. Wei, Global well-posedness and blow-up for the 2-D Patlak-Keller-Segel equation, J. Funct. Anal. 274, (2018), 388–401.