# Calderón splitting and weak solutions for Navier-Stokes equations with initial data in weighted Lp spaces

Pierre Gilles Lemarié-Rieusset

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Pierre Gilles Lemarié-Rieusset. Calderón splitting and weak solutions for Navier-Stokes equations with initial data in weighted Lp spaces. 2025. hal-05406704

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Preprint submitted on 9 Dec 2025

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Distributed under a Creative Commons Attribution 4.0 International License
Calderón splitting and weak solutions for
Navier–Stokes equations with initial data in
weighted $L^p$ spaces.

Pierre Gilles Lemarié-Rieusset*‡

**Abstract**

We show the existence of global weak solutions of the 3D Navier–Stokes equations with initial velocity in the weighted spaces $L_{\Phi_\gamma}^p = L^p(\mathbb{R}^3, \Phi_\gamma(x) dx)$, where $2 < p < +\infty$, $0 < \gamma < 2$ and $\Phi_\gamma(x) = \frac{1}{(1+|x|^2)^{\frac{\gamma}{2}}}$, using Calderón splitting $L_{\Phi_\gamma}^p \subset L_{\Phi_2}^2 + L^r$ (with some $r \in (3, +\infty)$) and energy controls in $L_{\Phi_2}^2$.

**Keywords :** Navier–Stokes equations, weighted spaces, weak solutions, energy controls, Calderón’s splitting

**AMS classification :** 35Q30, 76D05.

**Introduction**

We consider the Navier–Stokes equations on $(0, T) \times \mathbb{R}^3$

$$ egin{cases} 
\partial_t u = \Delta u - u \cdot \nabla u - \nabla p \\ 
\text{div } u = 0 \\ 
\lim_{t \to 0} u(t, \cdot) = u_0 
\end{cases} 
\quad (1) $$

when the non-linearity $u \cdot \nabla u$ is rewritten as $\text{div}(u \otimes u)$ and the pressure $p$ is eliminated due to the Leray projection operator $\mathbb{P}$, rewritting $u \cdot \nabla u + \nabla p$ as $\mathbb{P}\text{div}(u \otimes u)$.

<footnote>*LaMME, Univ Evry, CNRS, Université Paris-Saclay, 91025, Evry, France</footnote>

<footnote>†emergitus professor, (Univ Evry)</footnote>

<footnote>‡e-mail : pierregilles.lemarierieusset@univ-evry.fr</footnote>
We thus shall consider distribution $\mathbf{u}$ such that $\mathbf{u} \in L^2((0, T), L^2(\mathbb{R}^3, \frac{dx}{(1+|x|^2)^2}))$, we first prove that $\mathbb{P} \operatorname{div}(\mathbf{u} \otimes \mathbf{u})$ is well defined as a distribution on $(0, T) \times \mathbb{R}^3$; assuming moreover that $\mathbf{u}$ satisfies the Navier-Stokes equations

$$ \left\{ \begin{array}{l} \partial_t \mathbf{u} = \Delta \mathbf{u} - \mathbb{P} \operatorname{div}(\mathbf{u} \otimes \mathbf{u}) \\ \operatorname{div} \mathbf{u} = 0 \end{array} \right. \qquad (2) $$

we shall prove that $\mathbf{u}$ may be defined as a continuous in time distribution $t \in [0, T] \mapsto \mathbf{u}(t, \cdot) \in S'(\mathbb{R}^3)$, (more precisely, that $t \mapsto \frac{1}{(1+|x|^2)^2} \mathbf{u}(t, \cdot)$ is continuous from $[0, T]$ to $H^{-4}(\mathbb{R}^3)$, so that the initial value condition

$$ \lim_{t \to 0} \mathbf{u}(t, \cdot) = \mathbf{u}_0 \qquad (3) $$

is meaningful. Those results are described in Proposition 1 and Theorem 2.

There are many examples of solutions such that $\mathbf{u} \in L^2((0, T), L^2(\mathbb{R}^3, \frac{dx}{(1+|x|^2)^2}))$. A special subclass of such solutions are the solutions such that

$$ \sup_{x_0 \in \mathbb{R}^3} \iint_{(0,T) \times B(x_0, 1)} |\mathbf{u}(s, y)|^2 ds dy < +\infty. \qquad (4) $$

In this subclass, one can find

* Kato's mild solutions in Lebesgue spaces $\mathbf{u} \in L^\infty((0, T), L^p)$ with $3 \le p \le +\infty$ [5] corresponding to the Cauchy initial value problem with $\mathbf{u}_0 \in L^p$

* more generally, Koch and Tataru's mild solutions [6] corresponding to the Cauchy initial value problem with $\mathbf{u}_0 \in \text{bmo}^{-1}$

* Leray's weak solutions $\mathbf{u} \in L^\infty((0, T), L^2)$ [9] corresponding to the Cauchy initial value problem with $\mathbf{u}_0 \in L^2$

* Calderón's weak solutions $\mathbf{u} \in L^\infty((0, T), L^2 + L^\infty)$ [2] corresponding to the Cauchy initial value problem with $\mathbf{u}_0 \in L^p$ with $p \ge 2$

* Lemarié-Rieusset's weak solutions $\mathbf{u} \in L^\infty((0, T), L_{\text{uloc}}^2)$ [7] corresponding to the Cauchy initial value problem with $\mathbf{u}_0 \in L_{\text{uloc}}^2$

More recently, Fernández-Dalgo & Lemarié-Rieusset [3] and Bradshaw, Kucavica & Tsai [1] considered the Cauchy initial value problem with $\mathbf{u}_0 \in L^2(\mathbb{R}^3, \frac{dx}{1+|x|^2})$. Their solutions belong to $L^\infty((0, T), L^2(\frac{dx}{1+|x|^2}))$ and don't satisfy condition (4).

Other weak solutions which don't satisfy condition (4) are the statistically homogeneous solutions of Fursikov and Višik [4], which are proved to belong to $L^\infty((0, T), L^2(\frac{dx}{(1+|x|^2)^{\gamma/2}}))$ for every $\gamma > 3$.

Our main goal in this paper is to prove the following result:
**Theorem 1.** Let $2 < p < +\infty$ and $\gamma \in (0, 2)$. Let $u_0 \in L^p(\mathbb{R}^3, \frac{dx}{1+|x|^2})$ with $\text{div } u_0 = 0$. Then the Navier-Stokes equations

$$ egin{cases} 
\partial_t u = \Delta u - \mathbb{P} \text{div}(u \otimes u) \\ 
\text{div } u = 0 \\ 
\lim_{t \to 0} u(t, \cdot) = u_0 
\end{cases} 
\qquad (5) $$

have a solution on $(0, +\infty) \times \mathbb{R}^3$ with, for every $0 < T < +\infty$, $u \in L^\infty((0, T), L^2(\frac{dx}{1+|x|^2})) + L^\infty((0, T), L^r)$ for some $r \in (3, +\infty)$.

The limit case $L^\infty((0, T), L^2(\frac{dx}{1+|x|^2}))$ corresponds to the results of Fernández-Dalgo & Lemarié-Rieusset [3] and Bradshaw, Kucavica & Tsai [1]. The limit case $L^\infty((0, T), L^r)$ corresponds to Kato's mild solutions in $L^r$ [5] (existence of mild solutions is known only for a finite time $T \approx \frac{1}{\|u_0\|_r^{\frac{2r}{r-3}}}$). In order to deal with the case $2 < p < +\infty$, we shall use Calderón's method [2] and split the initial value in a sum of two vector fields corresponding to the limit cases which we know how to deal with.

# 1 Weighted Lebesgue and Sobolev spaces

Throughout the paper, we shall deal with weights $\Phi_\gamma(x) = \Phi(x)^\gamma$, $\gamma \in \mathbb{R}$, where

$$ \Phi(x) = \frac{1}{(1 + |x|^2)^{1/2}} \approx \frac{1}{1 + |x|}. $$

We will work in weighted spaces:

* weighted Lebesgue spaces $L^p(\Phi_\gamma dx)$ ($1 \le p \le +\infty$) with

$$ \|u\|_{L^p(\Phi_\gamma dx)} = \|\Phi_p^\gamma u\|_p. $$

* weighted Sobolev spaces $H^s(\Phi_\gamma dx)$ ($s \in \mathbb{R}$) with

$$ \|u\|_{H^s(\Phi_\gamma dx)} = \|\Phi^{\frac{\gamma}{2}} u\|_{H^s}. $$

$\mathcal{D}(\mathbb{R}^3)$ is dense in $L^p(\Phi_\gamma dx)$ for $1 \le p < +\infty$ and in $H^s(\Phi_\gamma dx)$. We have, for $u, v \in \mathcal{D}$ (and $p > 1$)

$$ |\int uv \, dx| \le \|u\|_{L^p(\Phi_\gamma dx)} \|v\|_{L^{\frac{p}{p-1}}(\Phi_{-\frac{\gamma}{p-1}} dx)} $$
and

$$ \left| \int uv \, dx \right| \le \|u\|_{H^s(\Phi_\gamma dx)} \|v\|_{H^{-s}(\Phi_{-\gamma} dx)}. $$

For $\gamma \le \delta$ and $\sigma \le s$, we have the following obvious continuous embeddings:

$$ L^p(\Phi_\gamma dx) \subset L^q(\Phi_\delta dx) \text{ and } H^s(\Phi_\gamma dx) \subset H^\sigma(\Phi_\gamma dx). $$

Using the Hölder inequality, we find as well the following embedding

$$ L^p(\Phi_\gamma dx) \subset L^q(\Phi_\delta dx) \text{ for } q < p \text{ and } \frac{\delta}{q} > \frac{\gamma}{p} + \frac{3(p-q)}{pq}. $$

Using the Sobolev inequalities, we find that, for $0 \le s < \frac{3}{2}$, $\frac{1}{q} = \frac{1}{2} - \frac{s}{3}$ and $\frac{1}{r} = \frac{1}{2} + \frac{s}{3}$, we have

$$ H^s(\Phi_\gamma dx) \subset L^q(\Phi_{\frac{r}{2}\gamma} dx) \text{ and } L^r(\Phi_{\frac{r}{2}\gamma} dx) \subset H^{-s}(\Phi_\gamma dx). $$

Similarly, for $s > \frac{3}{2}$, we have

$$ L^1(\Phi_{\frac{r}{2}} dx) \subset H^{-s}(\Phi_\gamma dx). $$

We state two further estimates:

**Lemma 1.** Let $u \in H^s(\Phi_\gamma dx)$. Then

$$ \|\nabla u\|_{H^{s-1}(\Phi_\gamma dx)} \le C \|u\|_{H^s(\Phi_\gamma dx)}. $$

If $s \ge 0$,

$$ \|u\|_{H^s(\Phi_\gamma dx)} \approx \|u\|_{L^2(\Phi_\gamma dx)} + \|\nabla u\|_{H^{s-1}(\Phi_\gamma dx)}. $$

*Proof.* We have

$$ \|\nabla u\|_{H^{s-1}(\Phi_\gamma dx)} = \|\Phi^{\frac{\gamma}{2}} \nabla u\|_{H^{s-1}} \le \|\nabla(\Phi^{\frac{\gamma}{2}} u)\|_{H^{s-1}} + \frac{|\gamma|}{2} \|\Phi^{\frac{\gamma}{2}} u \frac{\nabla \Phi}{\Phi}\|_{H^{s-1}}. $$

Since $\frac{\nabla \Phi}{\Phi}$ is bounded with all its derivatives, we find that

$$ \|\Phi^{\frac{\gamma}{2}} u \frac{\nabla \Phi}{\Phi}\|_{H^{s-1}} \le C \|\Phi^{\frac{\gamma}{2}} u\|_{H^{s-1}} \le C \|\Phi^{\frac{\gamma}{2}} u\|_{H^s} $$

and we easily conclude since $\|\nabla(\Phi^{\frac{\gamma}{2}} u)\|_{H^{s-1}} \le \|\Phi^{\frac{\gamma}{2}} u\|_{H^s}$.

If $s \ge 0$, we have

$$ \|u\|_{H^s(\Phi_\gamma dx)} = \|\Phi^{\frac{\gamma}{2}} u\|_{H^s} \approx \|\Phi^{\frac{\gamma}{2}} u\|_{L^2} + \|\nabla(\Phi^{\frac{\gamma}{2}} u)\|_{H^{s-1}}. $$
We have

$$ \|\nabla(\Phi_{\frac{\gamma}{2}}^{\frac{\gamma}{2}}u)\|_{H^{s-1}} \le \|\nabla u\|_{H^{s-1}(\Phi_{\gamma} dx)} + C \|\Phi_{\frac{\gamma}{2}}^{\frac{\gamma}{2}}u\|_{H^{s-1}} $$

with

$$ \|\Phi_{\frac{\gamma}{2}}^{\frac{\gamma}{2}}u\|_{H^{s-1}} \le \|\Phi_{\frac{\gamma}{2}}^{\frac{\gamma}{2}}u\|_{2} $$

if $0 \le s \le 1$ and

$$ \|\Phi_{\frac{\gamma}{2}}^{\frac{\gamma}{2}}u\|_{H^{s-1}} \le \|\Phi_{\frac{\gamma}{2}}^{\frac{\gamma}{2}}u\|_{2}^{\frac{1}{s}} \|\Phi_{\frac{\gamma}{2}}^{\frac{\gamma}{2}}u\|_{H^{s}}^{1-\frac{1}{s}} \le \frac{1}{s\epsilon^s} \|\Phi_{\frac{\gamma}{2}}^{\frac{\gamma}{2}}u\|_{2} + \epsilon^{\frac{s}{s-1}} (1-\frac{1}{s}) \|\Phi_{\frac{\gamma}{2}}^{\frac{\gamma}{2}}u\|_{H^{s}} $$

if $s > 1$ and $\epsilon > 0$.

□

**Lemma 2.** For $1 \le p \le +\infty$ and $0 < t \le 1$ we have

$$ \| \int \frac{1}{(\sqrt{t} + |x - y|)^4} f(y) \, dy \|_{L^p(\Phi_4 dx)} \le C t^{-\frac{1}{2}} \|f\|_{L^p(\Phi_4 dx)}. $$

*Proof.* This is obvious for $p = +\infty$ since $\int \frac{1}{(\sqrt{t} + |x-y|)^4} dy = C t^{-\frac{1}{2}}$.
For $p = 1$, we have to prove that

$$ \| \frac{1}{(1 + |x|)^4} \int \frac{1}{(\sqrt{t} + |x - y|)^4} (1 + |y|)^4 g(y) \, dy \|_1 \le C \|g\|_1. $$

As $(1 + |y|)^4 \le 16((1 + |x|)^4 + |x - y|^4)$, we have

$$ \int \frac{(1 + |y|^4)}{(1 + |x|)^4 (\sqrt{t} + |x - y|)^4} dx \le 16 \left( \int \frac{dx}{(1 + |x|)^4} + \int \frac{dx}{(\sqrt{t} + |x - y|)^4} \right) \le C t^{-\frac{1}{2}} $$

and we conclude by Fubini.

For $1 < p < +\infty$, we conclude by interpolation.

□

Remark: we have as well

$$ \| \int \frac{1}{(\sqrt{t} + |x - y|)^4} f(y) \, dy \|_{L^p(\Phi_\gamma dx)} \le C t^{-\frac{1}{2}} \|f\|_{L^p(\Phi_\gamma dx)} $$

for $0 \le \gamma \le 4$, $1 \le p \le +\infty$ and $0 < t \le 1$ (by interpolation between $L^p(dx)$ and $L^p(\Phi_4 dx)$).

Our next result deals with the Leray projection operator acting on the divergence of a tensor in $L^1(\Phi_4 dx)$:

**Proposition 1** (Leray projection).

Let $\mathbb{F} \in L^1(\mathbb{R}^3, \Phi_4 dx)$. Then there exists a unique pair $(\mathbf{b}_1, \mathbf{b}_2)$ such that

$$ \mathrm{div}(\mathbb{F}) = \mathbf{b}_1 + \mathbf{b}_2 $$

with
• $\mathbf{b}_1 \in H^\sigma(\Phi_\gamma dx)$ and $\mathrm{div} \, \mathbf{b}_1 = 0$

• $\mathbf{b}_2 \in H^\sigma(\Phi_\gamma dx)$ and $\nabla \wedge \mathbf{b}_2 = 0$

• $\lim_{\tau \to +\infty} e^{\tau \Delta} \mathbf{b}_2 = 0$ in $S'(\mathbb{R}^3)$.

for $\gamma > 7$ and $\sigma < -\frac{5}{2}$.

$\mathbf{b}_1$ is called the Leray projection of $\mathrm{div}(\mathbb{F})$ and we write

$$ \mathbf{b}_1 = \mathbb{P} \mathrm{div}(\mathbb{F}). $$

*Proof.* Uniqueness is obvious: if $\mathrm{div} \, \mathbb{F} = \mathbf{b}_1 + \mathbf{b}_2 = \mathbf{b}'_1 + \mathbf{b}'_2$ and if $\mathbf{b} = \mathbf{b}_2 - \mathbf{b}'_2$, then

$$ \Delta \mathbf{b} = \nabla \wedge (\nabla \wedge (\mathbf{b}_2 - \mathbf{b}'_2)) - \nabla(\mathrm{div}(\mathbf{b}_1 - \mathbf{b}'_1)) = 0 $$

while

$$ \mathbf{b} = \lim_{\tau \to +\infty} - \int_0^\tau e^{s\Delta} \Delta \mathbf{b} \, ds = 0. $$

We now construct $\mathbf{b}_2$. We want to have

$$ \Delta \mathbf{b}_2 = \nabla(\mathrm{div} \, \mathbf{b}_2) = \nabla(\mathrm{div}(\mathrm{div} \, \mathbb{F})) = \nabla\left(\sum_{i=1}^3 \sum_{j=1}^3 \partial_i \partial_j F_{i,j}\right) $$

and

$$ \mathbf{b}_2 = - \int_0^{+\infty} e^{s\Delta} \nabla(\mathrm{div}(\mathrm{div} \, \mathbb{F})) \, ds. $$

Let $\theta \in \mathcal{D}$ such that $0 \le \theta \le 1$, $\theta$ is supported in the ball $B(0, 2)$ and $\sum_{k \in \mathbb{Z}^3} \theta(x-k) = 1$. For $\sigma \in \mathbb{R}$ and $\gamma \in \mathbb{R}$, we write

$$ \| \mathbf{b}_2 \|_{H^\sigma(\Phi_\gamma dx)} \le \sum_{k \in \mathbb{Z}^3} \sum_{j \in \mathbb{Z}^3} \left\| \frac{1}{(1 + |x|^2)^{\frac{\gamma}{2}}} \theta(x-k) \int_0^{+\infty} e^{s\Delta} \nabla \mathrm{div}(\mathrm{div}(\theta(x-j)\mathbb{F})) \, ds \right\|_{H^\sigma}. $$

We take $\gamma > 7$ and $\sigma < -\frac{5}{2}$. When $|j-k| \ge 8$, we write

$$ \begin{aligned} & \left\| \frac{1}{(1 + |x|^2)^{\frac{\gamma}{2}}} \theta(x-k) \int_0^{+\infty} e^{s\Delta} \nabla \mathrm{div}(\mathrm{div}(\theta(\cdot - j)\mathbb{F})) \, ds \right\|_{H^\sigma} \\ & \le C \left\| \frac{1}{(1 + |x|^2)^{\frac{\gamma}{2}}} \theta(x-k) \int_0^{+\infty} e^{s\Delta} \nabla \mathrm{div}(\mathrm{div}(\theta(x-j)\mathbb{F})) \, ds \right\|_1 \\ & \le C' \left\| \frac{1}{(1 + |x|^2)^2} \theta(x-k) \int (\int_0^{+\infty} |(\nabla \otimes \nabla)\nabla(W_s(x-y))| \, ds) \theta(y-j) |\mathbb{F}(y)| \, dy \right. \\ & \qquad = C'' \left\| \frac{1}{(1 + |x|^2)^2} \theta(x-k) \int \frac{1}{|x-y|^4} \theta(y-j) |\mathbb{F}(y)| \, ds \right\|_1 \\ & \qquad \le C''' \left\| \frac{1}{(1 + |x|^2)^2} \theta(x-k) \int \frac{1}{(1 + |x-y|^2)^2} \theta(y-j) |\mathbb{F}(y)| \, ds \right\|_1 \end{aligned} $$
with

$$
\begin{align*}
& \sum_{k \in \mathbb{Z}^3} \sum_{j \in \mathbb{Z}^3} \left\| \frac{1}{(1 + |x|^2)^2} \theta(x - k) \int \frac{1}{(1 + |x - y|^2)^2} \theta(y - j) |\mathbb{F}(y)| \, ds \right\|_1 \\
&= \left\| \frac{1}{(1 + |x|^2)^2} \int \frac{1}{(1 + |x - y|^2)^2} |\mathbb{F}(y)| \, ds \right\|_1 \\
&\le C' \left\| \mathbb{F} \right\|_{L^1(\Phi_4 \, dx)}
\end{align*}
$$

(by Lemma 2).

For $|j-k| < 8$, we remark that $\frac{(1+|k|)^\gamma}{(1+|x|^2)^{\frac{\gamma}{2}}}$ $\theta(x-k)$ is smooth and bounded
(with all its derivatives) independently from $k$, so that

$$
\begin{align*}
& \cdot \left\| \frac{1}{(1 + |x|^2)^{\frac{\gamma}{2}}} \theta(x-k) \int_0^{+\infty} e^{s\Delta} \nabla \operatorname{div}(\operatorname{div}(\theta(\cdot - j)\mathbb{F})) ds \right\|_{H^\sigma} \\
&\le C \frac{1}{(1 + |k|)^\gamma} \left\| \mathbb{P} \operatorname{div}(\theta(\cdot - j)\mathbb{F}) \right\|_{H^\sigma}. \\
&\le C' \frac{1}{(1 + |k|)^\gamma} \left\| \theta(\cdot - j)\mathbb{F} \right\|_{H^{\sigma+1}} \\
&\le C'' \frac{1}{(1 + |k|)^\gamma} \left\| \theta(\cdot - j)\mathbb{F} \right\|_1 \\
&\le C''' \frac{(1 + |j|)^4}{(1 + |k|)^\gamma} \left\| \mathbb{F} \right\|_{L^1(\Phi_4 \, dx)}.
\end{align*}
$$

We may conclude, as

$$
\sum_{k \in \mathbb{Z}^3} \sum_{j \in \mathbb{Z}^3, |j-k|<8} \frac{(1+|j|)^4}{(1+|k|)^\gamma} < +\infty.
$$

Finally, we study $e^{\tau\Delta}\mathbf{b}_2$ when $\tau \to +\infty$. We write

$$
\begin{align*}
\| e^{\tau \Delta} \mathbf{b}_2 \|_{L^1(\Phi_4 \, dx)} &\le C \left\| \frac{1}{(1 + |x|^2)^2} \int \left( \int_0^{+\infty} |(\nabla \otimes \nabla) \nabla(W_{s+\tau}(x-y))| \, ds \right) |\mathbb{F}(y)| \, dy \right\| \\
&\le C'' \left\| \frac{1}{(1 + |x|^2)^2} \int \frac{1}{(\sqrt{\tau} + |x-y|^2)^2} |\mathbb{F}(y)| \, ds \right\|_1
\end{align*}
$$

with, for $\tau > 1$,

$$
\frac{1}{(1 + |x|^2)^2} \frac{1}{(\sqrt{\tau} + |x - y|^2)^2} |\mathbb{F}(y)| \le \frac{1}{(1 + |x|^2)^2 (1 + |x - y|^2)^2} |\mathbb{F}(y)| \in L^1(\mathbb{R}^2).
$$

By dominated convergence, we find that $\lim_{\tau \to +\infty} \|e^{\tau\Delta}\mathbf{b}_2\|_{L^1(\Phi_4 \, dx)} = 0$. $\square$
## 2 Weak solutions for the Stokes equations in

$$L^1((0, T), L^1(\mathbb{R}^3, \frac{dx}{(1+|x|)^4}))$$

In this section, we consider the Stokes equations

$$\left\{
\begin{array}{l}
\partial_t \mathbf{u} = \Delta \mathbf{u} - \mathbb{P} \operatorname{div}(\mathbb{F}) \\
\operatorname{div} \mathbf{u} = 0 \\
\lim_{t \to 0} \mathbf{u}(t, \cdot) = \mathbf{u}_0
\end{array}
\right.
\qquad (6)$$

where the tensor $\mathbb{F} = (F_{i,j})_{1 \le i,j \le 3}$ belongs to $L^1((0, T), L^1(\mathbb{R}^3, \frac{dx}{(1+|x|)^4}))$, $\operatorname{div}(\mathbb{F}) = \mathbf{b}$ with $b_j = \sum_{i=1}^3 \partial_i F_{i,j}$ and where the solution $\mathbf{u}$ belongs to $L^1((0, T), L^1(\mathbb{R}^3, \frac{dx}{(1+|x|)^4}))$. [Remark: we don't study the existence of such a solution, we assume in this section that it exists.]

**Theorem 2** (Stokes equations in weighted Lebesgue space).

Let $0 < T < +\infty$ and $\mathbb{F} \in L^1((0, T), L^1(\mathbb{R}^3, \frac{dx}{(1+|x|)^4}))$. Let $\mathbf{u}$ be a solution of the Stokes equation

$$\partial_t \mathbf{u} = \Delta \mathbf{u} - \mathbb{P} \operatorname{div}(\mathbb{F}) \qquad (7)$$

with $\mathbf{u} \in L^1((0, T), L^1(\mathbb{R}^3, \frac{dx}{(1+|x|)^4}))$. Then we have

$$\partial_t \mathbf{u} \in L^1((0, T), H^{-4}(\Phi_8 dx)) \quad \text{and} \quad \mathbf{u} \in C([0, T], H^{-4}(\Phi_8 dx)).$$

In particular, if $\varphi \in D(\mathbb{R})$ with $\varphi(0) = 1$ and $\varphi(T) = 0$, we have

$$\lim_{t \to 0} \mathbf{u}(t, \cdot) = - \int_{0<t<T} \partial_t (\varphi \mathbf{u}) \, dt. \qquad (8)$$

Moreover, we have

$$\operatorname{div}(\mathbf{u}(t, \cdot)) = e^{t\Delta} \operatorname{div}(\mathbf{u}(0, \cdot)).$$

*Proof.* As $L^1(\Phi_4 dx) \subset H^{-2}(\Phi_8 dx)$, we have $\mathbf{u} \in L^1((0, T), H^{-2}(\Phi_8 dx))$ and $\Delta \mathbf{u} \in L^1((0, T), H^{-4}(\Phi_8 dx))$. From Proposition 1, we know that, for $\gamma > 7$ and $\sigma < -5/2$,

$$\|\mathbb{P}(\operatorname{div} \mathbb{F})(t, \cdot)\|_{H^{\sigma}(\Phi_{\gamma} dx)} \le C_{\gamma, \sigma} \|\mathbb{F}(t, \cdot)\|_{L^1(\Phi_4 dx)},$$

hence $\mathbb{P}(\operatorname{div} \mathbb{F}) \in L^1((0, T), H^{-4}(\Phi_8 dx))$.

From $\partial_t \mathbf{u} \in L^1((0, T), H^{-4}(\Phi_8 dx))$ and $\mathbf{u} \in L^1((0, T), H^{-4}(\Phi_8 dx))$, we conclude that $\mathbf{u} \in C([0, T], H^{-4}(\Phi_8 dx))$. Finally, we write

$$\partial_t \operatorname{div} \mathbf{u} = \operatorname{div} \partial_t \mathbf{u} = \operatorname{div} \Delta \mathbf{u} = \Delta \operatorname{div} \mathbf{u}$$

so that $\operatorname{div} \mathbf{u}(t, \cdot) = e^{t\Delta}(\operatorname{div} \mathbf{u}(0, \cdot))$. $\square$

<figure><img src="image_2.png" /></figure>
**Theorem 3** (Solutions bounded in weighted Lebesgue space).

Let $0 < T < +\infty$, $1 < p < +\infty$ and $\mathbb{F} \in L^\infty((0, T), L^p(\mathbb{R}^3, \frac{dx}{(1+|x|)^4}))$. Let $\mathbf{u}$ be a distribution defined on $(0, T) \times \mathbb{R}^3$. Then the following assertions are equivalent:

* assertion A1: $\mathbf{u}$ is a solution of the Stokes equation

$$ \left\{ \begin{array}{l} \partial_t \mathbf{u} = \Delta \mathbf{u} - \mathbb{P} \operatorname{div}(\mathbb{F}) \\ \operatorname{div} \mathbf{u} = 0 \end{array} \right. \qquad (9) $$

with $\mathbf{u} \in L^\infty((0, T), L^p(\mathbb{R}^3, \frac{dx}{(1+|x|)^4}))$.

* assertion A2: there exists a tempered distribution $\mathbf{u}_0$ on $\mathbb{R}^3$ such that $\operatorname{div} \mathbf{u}_0 = 0$, $\mathbf{u}_0 \in L^p(\mathbb{R}^3, \frac{dx}{(1+|x|)^4})$ and

$$ \mathbf{u} = e^{t\Delta}\mathbf{u}_0 - \int_0^t e^{(t-s)\Delta}\mathbb{P}\operatorname{div}(\mathbb{F})ds. $$

We then have $\lim_{t \to 0} \mathbf{u}(t, \cdot) = \mathbf{u}_0$.

*Proof.* (A1) $\implies$ (A2): Let $\mathbf{u}$ be a solution of the Stokes equation

$$ \partial_t \mathbf{u} = \Delta \mathbf{u} - \mathbb{P} \operatorname{div}(\mathbb{F}) \qquad (10) $$

with $\mathbf{u} \in L^\infty((0, T), L^p(\mathbb{R}^3, \frac{dx}{(1+|x|)^4}))$. As $\int \frac{dx}{(1+|x|)^4} < +\infty$, we have $\mathbf{u} \in L^1((0, T), L^1(\mathbb{R}^3, \frac{dx}{(1+|x|)^4}))$ and $\mathbb{F} \in L^1((0, T), L^1(\mathbb{R}^3, \frac{dx}{(1+|x|)^4}))$. By Theorem 2, we know that $\mathbf{u} \in C([0, T], H^{-4}(\Phi_8 dx))$. We may then write $\lim_{t \to 0} \mathbf{u}(t, \cdot) = \mathbf{u}_0$ as a strong limit in $H^{-4}(\Phi_8 dx)$ but as well, since $\mathbf{u}(t, \cdot)$ is bounded in $L^p(\Phi_4 dx)$, as a weak-* limit in $L^p(\Phi_4 dx)$.

(A2) $\implies$ (A1): From classical estimates on Oseen's tensor (see for instance section 4.5 in [8]), we have

$$ |e^{(t-s)\Delta}\mathbb{P}\operatorname{div}(\mathbb{F})| \le C \int \frac{1}{(\sqrt{t-s} + |x-y|)^4} |\mathbb{F}(s, y)| dy $$

and, by Lemma 2, for $0 < t < T$,

$$ \begin{aligned} \| \int_0^t e^{(t-s)\Delta} \mathbb{P} \operatorname{div}(\mathbb{F}) ds \|_{L^p(\Phi_4 dx)} &\le C \int_0^t \max(\frac{1}{\sqrt{t-s}}, 1) \| \mathbb{F}(s, \cdot) \|_{L^p(\Phi_4 dx)} ds \\ &\le C' \sqrt{t} (1 + \sqrt{t}) \| \mathbb{F} \|_{L^\infty((0,T), L^p(\Phi_4 dx))}. \end{aligned} $$

Similarly, we write

$$ |e^{t\Delta}\mathbf{u}_0(x)| \le C \int \frac{\sqrt{t}}{(\sqrt{t} + |x - y|)^4} |\mathbf{u}_0(x)| dy $$

and

$$ \|e^{t\Delta}\mathbf{u}_0\|_{L^p(\Phi_4 dx)} \le C \max(1, \sqrt{t}) \|\mathbf{u}_0\|_{L^p(\Phi_4 dx)} \quad \square $$
## 3 Mollified equations

We want to find a weak solution to the Navier-Stokes equations

$$ egin{cases} 
\partial_t \mathbf{u} = \Delta \mathbf{u} - \mathbb{P} \operatorname{div}(\mathbf{u} \otimes \mathbf{u}) \\ 
\operatorname{div} \mathbf{u} = 0 \\ 
\lim_{t \to 0} \mathbf{u}(t, \cdot) = \mathbf{u}_0 
\end{cases} 
\quad (11) $$

globally in time when $\mathbf{u}_0$ is divergence free and $\mathbf{u}_0 \in L^p(\frac{dx}{(1+|x|)^\gamma})$ with $2 < p < +\infty$ and $0 < \gamma < 2$.

Following Leray [9], we replace the Navier-Stokes equations (11) with the mollified equations

$$ egin{cases} 
\partial_t \mathbf{u}_{\epsilon, \alpha} = \Delta \mathbf{u}_{\epsilon, \alpha} - \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{u}_{\epsilon, \alpha})) \otimes \mathbf{u}_{\epsilon, \alpha}) \\ 
\operatorname{div} \mathbf{u}_{\epsilon, \alpha} = 0 \\ 
\lim_{t \to 0} \mathbf{u}_{\epsilon, \alpha}(t, \cdot) = \mathbf{u}_0 
\end{cases} 
\quad (12) $$

where $\varphi \in \mathcal{D}(\mathbb{R}^3)$ with $\varphi \ge 0$, $\int \varphi(x) dx = 1$, $\varphi(x) = 0$ if $|x| > 1$ and $\varphi_{\epsilon}(x) = \frac{1}{\epsilon^3} \varphi(\frac{x}{\epsilon})$ and where $\theta \in \mathcal{D}(\mathbb{R}^3)$ with $0 \le \theta \le 1$, $\theta(x) = 1$ if $|x| \le 1$, $\theta(x) = 0$ if $|x| \ge 2$ and $\theta_{\alpha}(x) = \theta(\alpha x)$.

We shall prove that equations (12) have a solution

$$ \mathbf{u}_{\epsilon, \alpha} \in \cap_{0<T<+\infty} L^{\infty}((0,T), L^p(\Phi_{\gamma} dx)). $$

However, the control of $\mathbf{u}_{\epsilon, \alpha}$ with respect to $\epsilon$ and $\alpha$ is not good enough when $(\epsilon, \alpha)$ goes to $(0, 0)$: we find that

$$ \sup_{0<t<T} \|\mathbf{u}_{\epsilon, \alpha}(t, \cdot)\|_{L^p(\Phi_{\gamma} dx)} \le C_{\epsilon, \alpha, T, \mathbf{u}_0} \quad \text{with} \quad \lim_{\epsilon, \alpha \to (0,0)} C_{\epsilon, \alpha, T, \mathbf{u}_0} = +\infty. $$

Instead of $L^p(\Phi_{\gamma} dx)$, we follow Calderón [2] and we shall work in $L^2(\Phi_2 dx) + L^r$ (with $3 < r < \infty$) and prove that, for every $T > 0$, we have

$$ \sup_{0<t<T} \|\mathbf{u}_{\epsilon, \alpha}(t, \cdot)\|_{L^2(\Phi_2 dx) + L^r} \le C_{T, \mathbf{u}_0} < +\infty \quad (13) $$

and

$$ \| \mathbf{u}_{\epsilon, \alpha}(t, \cdot) \|_{L^2((0,T), H^{\frac{1}{2}}(\Phi_4 dx))} \le C_{T, \mathbf{u}_0} < +\infty. \quad (14) $$

We shall see that estimates (13) and (14) are sufficient to grant some sequence $\mathbf{u}_{\epsilon_n, \alpha_n}$ is weakly convergent in $L^2((0, T), \Phi_4 dx)$ to a solution $\mathbf{u}$ of the Navier-Stokes equations.
### 3.1 First estimates in the norm of $L^p(\Phi_\gamma dx)$.

In this section, $2 \le p < +\infty$ and $2 \le \gamma \le 4$.

**Lemma 3.** If $f \in L^p(\Phi_\gamma dx)$ then

$$ \|\varphi_\epsilon * (\theta_\alpha f)\|_{\infty} \le \|\varphi\|_{\frac{p}{p-1}} \frac{1}{\epsilon^\frac{\gamma}{p}} (1 + \frac{2}{\alpha})^\frac{\gamma}{p} \|f\|_{L^p(\Phi_\gamma dx)}. $$

*Proof.* Just write

$$ \|\varphi_\epsilon * (\theta_\alpha f)\|_{\infty} \le \|\varphi_\epsilon\|_{\frac{p}{p-1}} \|\theta_\alpha \Phi^{-\frac{\gamma}{p}}\|_{\infty} \|\Phi^\frac{\gamma}{p} f\|_p. \quad \square $$

**Lemma 4.** If $\mathbf{u}_0 \in L^p(\Phi_\gamma dx)$, then $e^{t\Delta}\mathbf{u}_0 \in \mathcal{C}([0, +\infty[, L^p(\Phi_\gamma dx))$ and

$$ \|e^{t\Delta}\mathbf{u}_0\|_{L^p(\Phi_\gamma dx)} \le C \max(1, \sqrt{t}) \|\mathbf{u}_0\|_{L^p(\Phi_\gamma dx)}. $$

*Proof.* We write again

$$ |e^{t\Delta}\mathbf{u}_0(x)| \le C \int \frac{\sqrt{t}}{(\sqrt{t} + |x - y|)^4} |\mathbf{u}_0(y)| dy $$

and

$$ \|e^{t\Delta}\mathbf{u}_0\|_{L^p(\Phi_\gamma dx)} \le C \max(1, \sqrt{t}) \|\mathbf{u}_0\|_{L^p(\Phi_\gamma dx)}. $$

Thus, for $0 < T < +\infty$, convolution with the heat kernel is a bounded map from $L^p(\Phi_\gamma dx)$ to $L^\infty((0, T), L^p(\Phi_\gamma dx))$.

We then remark that $L^p(dx)$ is dense in $L^p(\Phi_\gamma dx)$, and that $W^{2,p}$ is dense in $L^p$. If $\mathbf{u}_0 \in W^{2,p}$, then, for $0 \le t \le \tau$, we have

$$ \|e^{t\Delta}\mathbf{u}_0 - e^{\tau\Delta}\mathbf{u}_0\|_{L^p(\Phi_\gamma dx)} \le \|e^{t\Delta}\mathbf{u}_0 - e^{\tau\Delta}\mathbf{u}_0\|_p \le \int_t^\tau \|e^{s\Delta}\Delta\mathbf{u}_0\|_p dx \le (\tau - t)\|\Delta\mathbf{u}_0\|_p. $$

Thus, convolution with the heat kernel is a bounded map from $W^{2,p}$ to $\mathcal{C}([0, T], L^p(\Phi_\gamma dx))$ for the $W^{2,p}$ norm and from $W^{2,p}$ to $L^\infty([0, T], L^p(\Phi_\gamma dx))$ for the $L^p(\Phi_\gamma dx)$ norm. Thus, it is a bounded map from $L^p(\Phi_\gamma dx)$ to $\mathcal{C}([0, T], L^p(\Phi_\gamma dx))$. $\square$

**Lemma 5.** If $\mathbb{F} \in L^\infty((0, T), L^p(\Phi_\gamma dx)$ (where $0 < T < +\infty$), then

$$ \int_0^t e^{(t-s)\Delta} \mathbb{P} \operatorname{div} \mathbb{F} ds \in \mathcal{C}([0, T], L^p(\Phi_\gamma dx)) $$

and, for $0 \le t \le T$,

$$ \| \int_0^t e^{(t-s)\Delta} \mathbb{P} \operatorname{div} \mathbb{F} ds \|_{L^p(\Phi_\gamma dx)} \le C \max(t, \sqrt{t}) \|\mathbb{F}\|_{L^\infty((0,T), L^p(\Phi_\gamma dx))}. $$
Proof. We write

$$ \left| \int_0^t e^{(t-s)\Delta} \mathbb{P} \operatorname{div} \mathbb{F} ds \right| \le C \int_0^t \int \frac{1}{\sqrt{t-s} + |x-y|^4} |\mathbb{F}(s,y)| dy $$

and

$$ \begin{aligned} \| \int_0^t e^{(t-s)\Delta} \mathbb{P} \operatorname{div} \mathbb{F} ds \|_{L^p(\Phi_\gamma dx)} &\le C \int_0^t \max(1, \frac{1}{\sqrt{t-s}}) \|\mathbb{F}(s, \cdot)\|_{L^p(\Phi_\gamma dx)} \\ &\le C'(t + \sqrt{t}) \|\mathbb{F}\|_{L^\infty((0,T), L^p(\Phi_\gamma dx))}. \end{aligned} $$

Thus, for $0 < T < +\infty$, the operator $\mathcal{L}$ defined by

$$ \mathbb{F} \mapsto \mathcal{L}(\mathbb{F}) = \int_0^t e^{(t-s)\Delta} \mathbb{P} \operatorname{div} \mathbb{F} ds $$

is a bounded map from $L^\infty((0, T), L^p(\Phi_\gamma dx))$ to $L^\infty((0, T), L^p(\Phi_\gamma dx))$.

We then remark that $L^\infty((0, T), L^p(\Phi_\gamma dx))$ is embedded in $L^4((0, T), L^p(\Phi_\gamma dx))$ and that $L^\infty((0, T), W^{2,p})$ is dense in $L^4((0, T), L^p(\Phi_\gamma dx))$. The operator $\mathcal{L}$ is a bounded map from $L^4((0, T), L^p(\Phi_\gamma dx))$ to $L^\infty((0, T), L^p(\Phi_\gamma dx))$:

$$ \begin{aligned} \| \int_0^t e^{(t-s)\Delta} \mathbb{P} \operatorname{div} \mathbb{F} ds \|_{L^p(\Phi_\gamma dx)} &\le C \int_0^t \max(1, \frac{1}{\sqrt{t-s}}) \|\mathbb{F}(s, \cdot)\|_{L^p(\Phi_\gamma dx)} ds \\ &\le C \left( \int_0^t \max(1, \frac{1}{\sqrt{t-s}})^{4/3} ds \right)^{3/4} \|\mathbb{F}\|_{L^4((0,T), L^p(\Phi_\gamma dx))} \\ &\le C' \max(t^{3/4}, t^{1/4}) \|\mathbb{F}\|_{L^4((0,T), L^p(\Phi_\gamma dx))}. \end{aligned} $$

If $\mathbb{F} \in L^\infty((0, T), W^{2,p})$, then, for $0 \le t \le T$, we have

$$ \begin{aligned} \| \partial_t \int_0^t e^{(t-s)\Delta} \mathbb{P} \operatorname{div} \mathbb{F} ds \|_{L^p(\Phi_\gamma dx)} &\le \| \partial_t \int_0^t e^{(t-s)\Delta} \mathbb{P} \operatorname{div} \mathbb{F} ds \|_p \\ &= \| \mathbb{P} \operatorname{div} \mathbb{F}(t, \cdot) + \int_0^t e^{(t-s)\Delta} \mathbb{P} \operatorname{div} \Delta \mathbb{F} ds \|_p \\ &\le C (\| \mathbb{F}(t, \cdot) \|_{W^{2,p}} + \int_0^t \frac{1}{\sqrt{t-s}} \| \Delta \mathbb{F}(s, \cdot) \|_p ds) \\ &\le C'(1 + \sqrt{t}) \| \mathbb{F} \|_{L^\infty((0,T), W^{2,p})}. \end{aligned} $$

Thus, $\mathcal{L}$ is a bounded map from $L^\infty((0, T), W^{2,p})$ to $\mathcal{C}([0, T], L^p(\Phi_\gamma dx))$ for the $L^\infty((0, T), W^{2,p})$ norm and from $L^\infty((0, T), W^{2,p})$ to $L^\infty([0, T], L^p(\Phi_\gamma dx))$ for the $L^4((0, T), L^p(\Phi_\gamma dx))$ norm; we find that $\mathcal{L}$ is a bounded map from $L^\infty((0, T), L^p(\Phi_\gamma dx))$ to $\mathcal{C}([0, T], L^p(\Phi_\gamma dx))$. $\square$
**Proposition 2.** Let $2 \le p < +\infty$, $2 \le \gamma \le 4$ and $\mathbf{u}_0 \in L^p(\Phi_\gamma dx)$. The mollified equations

$$ \left\{ \begin{array}{l} \partial_t \mathbf{u}_{\epsilon, \alpha} = \Delta \mathbf{u}_{\epsilon, \alpha} - \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{u}_{\epsilon, \alpha})) \otimes \mathbf{u}_{\epsilon, \alpha}) \\ \lim_{t \to 0} \mathbf{u}_{\epsilon, \alpha}(t, \cdot) = \mathbf{u}_0 \end{array} \right. \qquad (15) $$

have a unique (maximal) solution in $C([0, T, L^p(\Phi_\gamma dx))$.

If the maximal time of existence $T_{\epsilon, \alpha}$ is finite, then $\lim_{t \to T_{\epsilon, \alpha}} \|\mathbf{u}_{\epsilon}(t, \cdot)\|_{L^2(\Phi_4 dx)} = +\infty$.

There exists a constant $C_0 > 0$ such that

$$ T_{\epsilon, \alpha} > \min(1, \frac{\epsilon^3 \alpha^4}{C_0 \|\mathbf{u}_0\|_{L^p(\Phi_\gamma dx)}^2 (2+\alpha)^4}). $$

*Proof.* We consider the fixed point problem in $C([0, T], L^p(\Phi_\gamma dx))$

$$ \mathbf{u} = e^{t\Delta} \mathbf{u}_0 - B_{\epsilon, \alpha}(\mathbf{u}, \mathbf{u}), $$

where

$$ B_{\epsilon, \alpha}(\mathbf{v}, \mathbf{w}) = \int_{0}^{t} e^{(t-s)\Delta} \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v})) \otimes \mathbf{w}) \, ds. $$

We define

$$ R = \sup_{t \in [0, T]} \|e^{t\Delta} \mathbf{u}_0\|_{L^p(\Phi_\gamma dx)} $$

and we want to prove that the map $\mathbf{u} \mapsto e^{t\Delta} \mathbf{u}_0 - B_{\epsilon, \alpha}(\mathbf{u}, \mathbf{u})$ is a contraction in $B_R = \{\mathbf{u} \in C([0, T], L^p(\Phi_\gamma dx)) / \sup_{t \in [0, T]} \|\mathbf{u}(t, \cdot)\|_{L^p(\Phi_\gamma dx)} \le 2R\}$.

Fist, we use Lemma 4 and get that $R \le C_1 \max(1, \sqrt{T}) \|\mathbf{u}_0\|_{L^p(\Phi_\gamma dx)}$.

As $L^p(\Phi_\gamma dx) \subset L^4(\Phi_2 dx)$, we can use Lemmas 3 and 5 and find, for $\mathbf{v}$, $\mathbf{w}$ in $C([0, T], L^p(\Phi_\gamma dx))$,

$$ \|\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}(t, \cdot))\|_{\infty} \le \|\varphi\|_2 \frac{1}{\epsilon^2} \left(\frac{2+\alpha}{\alpha}\right)^2 \|\mathbf{v}(t, \cdot)\|_{L^p(\Phi_\gamma dx)} $$

and, for $0 < t < T$,

$$ \begin{aligned} & \|B_{\epsilon}(\mathbf{v}, \mathbf{w})(t, \cdot)\|_{L^p(\Phi_\gamma dx)} \\ & \le C_2(T + \sqrt{T}) \frac{1}{\epsilon^2} (2+\epsilon)^2 \sup_{t \in [0, T]} \|\mathbf{v}(t, \cdot)\|_{L^p(\Phi_\gamma dx)} \sup_{t \in [0, T]} \|\mathbf{w}(t, \cdot)\|_{L^p(\Phi_\gamma dx)}. \end{aligned} $$

Thus, we will have a contraction in $B_R$ if

$$ 4C_2(T + \sqrt{T}) \frac{1}{\epsilon^2} \left(\frac{2+\alpha}{\alpha}\right)^2 R < 1, $$
in particular if

$$4C_1C_2\sqrt{T}(1+\sqrt{T})^2\frac{1}{\epsilon^2}(\frac{2+\alpha}{\alpha})^2\|\mathbf{u}_0\|_{L^p(\Phi_\gamma dx)} < 1.$$

This proves that that

$$T_{\epsilon, \alpha} > \min(1, \frac{\epsilon^3 \alpha^4}{C_0 \|\mathbf{u}_0\|_{L^p(\Phi_\gamma dx)}^2 (2+\alpha)^4}).$$

If $T_{\epsilon, \alpha}$ is finite and $T < T_{\epsilon, \alpha}$, considering the initial value problem at initial time $T$, we find

$$T_{\epsilon, \alpha} - T > \min(1, \frac{\epsilon^3 \alpha^4}{C_0 \|\mathbf{u}_{\epsilon, \alpha}(T, \cdot)\|_{L^p(\Phi_\gamma dx)}^2 (2+\alpha)^4}).$$

Thus, $\lim_{T \to T_{\epsilon, \alpha}} \|\mathbf{u}_{\epsilon, \alpha}(T, \cdot)\|_{L^p(\Phi_\gamma dx)} = +\infty$. We prove as well that

$$\lim_{T \to T_{\epsilon, \alpha}} \|\mathbf{u}_{\epsilon, \alpha}(T, \cdot)\|_{L^4(\Phi_2 dx)} = +\infty.$$

If this was not the case, then the maximal existence time $\tilde{T}_{\epsilon, \alpha}$ for $\mathbf{u}_{\epsilon, \alpha}$ in $C([0, \tilde{T}_{\epsilon, \alpha}), L^2(\Phi_4 dx)$ would satisfy $\tilde{T}_{\epsilon, \alpha} > T_{\epsilon, \alpha}$ and thus $\mathbf{u}_{\epsilon, \alpha}$ would be bounded in $L^2(\Phi_4 dx)$ on $[0, T_{\epsilon, \alpha})$. For $0 < T < t < T_{\epsilon, \alpha}$, we would have

$$\begin{aligned} & \|\mathbf{u}_{\epsilon, \alpha}(t, \cdot)\|_{L^p(\Phi_\gamma dx)} \\ & \le \|e^{(t-T)\Delta} \mathbf{u}_{\epsilon, \alpha}(T, \cdot)\|_{L^p(\Phi_\gamma dx)} + \|\int_T^t e^{(t-s)\Delta} \mathbf{P} \operatorname{div}((\varphi_\epsilon * (\theta_\alpha \mathbf{u}_{\epsilon, \alpha})) \otimes \mathbf{u}_{\epsilon, \alpha}) ds\|_{L^p(\Phi_\gamma dx)} \\ & \le C \max(1, \sqrt{T_{\epsilon, \alpha} - T}) \|\mathbf{u}_{\epsilon, \alpha}(T, \cdot)\|_{L^p(\Phi_\gamma dx)} \\ & + C_2(T_{\epsilon, \alpha} - T + \sqrt{T_{\epsilon, \alpha} - T}) \frac{1}{\epsilon^2} (\frac{2+\alpha}{\alpha})^2 \sup_{s \in [0, T_{\epsilon, \alpha}]} \|\mathbf{u}_{\epsilon, \alpha}(s, \cdot)\|_{L^2(\Phi_4 dx)} \sup_{s \in [T, t]} \|\mathbf{u}_{\epsilon, \alpha}(s, \cdot)\|_{L^p(\Phi_\gamma dx)}. \end{aligned}$$

If $T$ is close enough to $T_{\epsilon, \alpha}$, so that

$$C_2(T_{\epsilon, \alpha} - T + \sqrt{T_{\epsilon, \alpha} - T}) \frac{1}{\epsilon^2} (\frac{2+\alpha}{\alpha})^2 \sup_{s \in [0, T_{\epsilon, \alpha}]} \|\mathbf{u}_{\epsilon, \alpha}(s, \cdot)\|_{L^2(\Phi_4 dx)} < \frac{1}{2},$$

we would get

$$\sup_{T \le t < T_{\epsilon, \alpha}} \|\mathbf{u}_{\epsilon, \alpha}(t, \cdot)\|_{L^p(\Phi_\gamma dx)} \le 2C \max(1, \sqrt{T_{\epsilon, \alpha} - T}) \|\mathbf{u}_{\epsilon, \alpha}(T, \cdot)\|_{L^p(\Phi_\gamma dx)}$$

in contradiction with $\lim_{t \to T_{\epsilon, \alpha}} \|\mathbf{u}_{\epsilon, \alpha}(t, \cdot)\|_{L^p(\Phi_\gamma dx)} = +\infty$. $\square$

<figure><img src="image_3.png" /></figure>
## 3.2 Calderón's splitting.

Now, for $2 < p < +\infty$ and $0 < \gamma < 2$, we want to study the mollified equations when the initial data $u_0$ belongs to $L^p(\Phi_\gamma dx) \subset L^2(\Phi_4 dx)$ with $\text{div } u_0 = 0$. Following Calderón [2], we will split the solution $u_{\epsilon,\alpha}$ as a sum $u_{\epsilon,\alpha} = v_{\eta,\epsilon,\alpha} + b_{\eta,\epsilon,\alpha} \in L^\infty((0, T_\eta), L^2(\Phi_2 dx)) + L^\infty((0, T_\eta), L^r)$, $r \in (3, +\infty)$. The aim is to get a minoration of the existence time $T_\eta$ independent of $\epsilon$ and $\alpha$.

**Lemma 6.** Let $2 < p < +\infty$ and $0 < \gamma < 2$. Let $r_0 = 2^{\frac{p-\gamma}{2-\gamma}}$ and $\max(r_0, 3) < r < +\infty$. Let $u_0 \in L^p(\mathbb{R}^3, \frac{dx}{(1+|x|^2)^2})$ with $\text{div } u_0 = 0$. Then, for every $\eta > 0$ there exists $v_{0,\eta}$ and $b_{0,\eta}$ such that

$$u_0 = v_{0,\eta} + b_{0,\eta}$$

with

$$v_{0,\eta} \in L^2(\mathbb{R}^3, \frac{dx}{(1+|x|^2)}), \text{div } v_{0,\eta} = 0$$

and

$$b_{0,\eta} \in L^r(\mathbb{R}^3), \text{div } b_{0,\eta} = 0, \|b_{0,\eta}\|_r < \eta.$$

*Proof.* We have the interpolation result $L^p(\Phi_\gamma dx) = [L^2(\Phi_2 dx), L^{r_0}]_{[1-\frac{\gamma}{p}],}$ so that $L^p(\Phi_\gamma dx) \subset L^2(\Phi_2 dx) + L^{r_0}$. Moreover,

$$L^{r_0} = [L^2, L^r]_{[\frac{1}{2}-\frac{1}{r_0}, \frac{1}{2}-\frac{1}{r}]} \subset [L^2, L^r]_{[\frac{1}{2}-\frac{1}{r_0}, \infty)}$$

Let $\delta = \frac{\frac{1}{2}-\frac{1}{r_0}}{\frac{1}{2}-\frac{1}{r}}$. We may split $u_0$ in $u_0 = u_1 + u_2$ with $u_1 \in L^2(\Phi_2 dx)$ and $u_2 \in L^{r_0}$ and, for every $A > 0$, we may split $u_2$ into $v_A + b_A$ with

$$\|v_A\|_{L^2(\Phi_2 dx)} \leq C A^\delta \|u_2\|_{r_0} \text{ and } \|b_A\|_r \leq C A^{\delta-1} \|u_2\|_{r_0}.$$

Moreover, as $\Phi_\gamma$ is a Muckenhoupt weight in the class $A_p$ and $\Phi_2$ is a Muckenhoupt weight in the class $A_2$, we may apply the Leray projection operator and write

$$u_0 = \mathbb{P}(u_0) = \mathbb{P}(u_1 + v_A) + \mathbb{P}(b_A)$$

with

$$\mathbb{P}(u_1 + v_A) \in L^2(\mathbb{R}^3, \frac{dx}{(1+|x|^2)}), \text{div } \mathbb{P}(u_1 + v_A) = 0$$

and

$$\mathbb{P}(b_A) \in L^r(\mathbb{R}^3), \text{div } \mathbb{P}(b_A) = 0, \|\mathbb{P}(b_A)\|_r < C A^{\delta-1} \|u_2\|_{r_0}.$$

We conclude by taking $A$ large enough. □

<figure><img src="image_4.png" /></figure>
For some $T$, we then split the solution $\mathbf{u}_{\epsilon,\alpha}$ (defined in $C([0, T_{\epsilon,\alpha}), L^2(\Phi_4 dx))$) in $\mathbf{u}_{\epsilon,\alpha} = \mathbf{v}_{\eta,\epsilon,\alpha} + \mathbf{b}_{\eta,\epsilon,\alpha}$ where $\mathbf{b}_{\eta,\epsilon,\alpha}$ is a solution in $L^\infty((0, T), L^r)$ of

$$ \mathbf{b}_{\eta, \epsilon, \alpha} = e^{t\Delta} \mathbf{b}_{0, \eta} - B_{\epsilon, \alpha}(\mathbf{b}_{\eta, \epsilon, \alpha}, \mathbf{b}_{\eta, \epsilon, \alpha}) $$

and $\mathbf{v}_{\eta,\epsilon,\alpha}$ is a solution in $\bigcap_{S<T_{\epsilon,\alpha}} L^\infty((0, \min(S, T), L^2(\Phi_4 dx)))$ of

$$ \mathbf{v}_{\eta, \epsilon, \alpha} = e^{t\Delta} \mathbf{v}_{0, \eta} - B_{\epsilon, \alpha}(\mathbf{b}_{\eta, \epsilon, \alpha}, \mathbf{v}_{\eta, \epsilon, \alpha}) - B_{\epsilon, \alpha}(\mathbf{v}_{\eta, \epsilon, \alpha}, \mathbf{b}_{\eta, \epsilon, \alpha}) - B_{\epsilon, \alpha}(\mathbf{v}_{\eta, \epsilon, \alpha}, \mathbf{v}_{\eta, \epsilon, \alpha}). $$

### 3.3 Local estimates in the norm of $L^r$.

We write $W_t$ for the heat kernel $W_t(x) = \frac{1}{(4\pi t)^{\frac{3}{2}}}e^{-\frac{|x|^2}{4t}}$.

**Proposition 3.** Let $r > 3$. There exists a constant $C_1 > 0$ such that, for every $\epsilon > 0$, $\alpha > 0$, $\eta > 0$ and every $\mathbf{b}_{0,\eta} \in L^r$ with $\|\mathbf{b}_{0,\eta}\|_r \le \eta$, the mollified equations

$$ \left\{ \begin{array}{l} \partial_t \mathbf{b}_{\eta, \epsilon, \alpha} = \Delta \mathbf{b}_{\eta, \epsilon, \alpha} - \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{b}_{\eta, \epsilon, \alpha})) \otimes \mathbf{b}_{\eta, \epsilon, \alpha}) \\ \lim_{t \to 0} \mathbf{b}_{\eta, \epsilon, \alpha}(t, \cdot) = \mathbf{b}_{0, \eta} \end{array} \right. \qquad (16) $$

have a unique solution on $(0, T_{[\eta]}) \times \mathbb{R}^3$ (with $T_{[\eta]}^{\frac{1}{2}-\frac{3}{2r}} = \frac{1}{C_1\eta}$) such that

* $\mathbf{b}_{\eta, \epsilon, \alpha} \in C([0, T_{[\eta]}], L^2(\Phi_4 dx))$,

* $\sup_{0 \le t \le T_{[\eta]}} \|\mathbf{b}_{\eta, \epsilon, \alpha}(t, \cdot)\|_r \le 2\eta$,

* $\sup_{0 \le t \le T_{[\eta]}} t^{\frac{1}{2}} \|\nabla \otimes \mathbf{b}_{\eta, \epsilon, \alpha}(t, \cdot)\|_r \le 2 \|\nabla W_1\|_1 \eta$,

* $\sup_{0 < t < T_{[\eta]}} t^{\frac{3}{2r}} \|\mathbf{b}_{\eta, \epsilon, \alpha}(t, \cdot)\|_{\infty} \le 2 \|W_1\|_{\frac{r}{r-1}} \eta$.

*Proof.* First, we remark that $L^r(\mathbb{R}^3) \subset L^2(\Phi_4 dx)$. We have the obvious results for the heat kernel operating on $L^\infty$:

$$ \|e^{t\Delta} f\|_{L^r(dx)} = \|W_t * f\|_r \le \|W_t\|_1 \|f\|_r = \|f\|_r, $$

$$ \|\nabla e^{t\Delta} f\|_{L^r(dx)} = \|\nabla W_t * f\|_r \le \|\nabla W_t\|_1 \|f\|_r = \frac{1}{\sqrt{t}} \|\nabla W_1\|_1 \|f\|_r $$

and

$$ \|e^{t\Delta} f\|_{L^\infty(dx)} = \|W_t * f\|_{\infty} \le \|W_t\|_{\frac{r}{r-1}} \|f\|_r = \frac{1}{t^{\frac{3}{2r}}} \|W_1\|_{\frac{r}{r-1}} \|f\|_r. $$
We consider the fixed point problem

$$ \mathbf{b} = e^{t\Delta}\mathbf{b}_{0,\eta} - B_{\epsilon,\alpha}(\mathbf{b}, \mathbf{b}). $$

We want to prove that the map $\mathbf{b} \mapsto e^{t\Delta}\mathbf{b}_{0,\eta} - B_{\epsilon,\alpha}(\mathbf{b}, \mathbf{b})$ is a contraction in

$$ B_{\eta} = \{\mathbf{b} \in C([0, T], L^2(\Phi_4 dx)) / \begin{array}{l} \sup_{t \in [0,T]} \|\mathbf{b}(t,.)\|_r \le 2\eta, \\ \sup_{0<t \le T} \sqrt{t} \|\nabla \otimes \mathbf{b}(t,.)\|_r \le 2\|W_1\|_1\eta, \\ \sup_{0<t \le T} t^{\frac{3}{2r}} \|\mathbf{b}(t,.)\|_{\infty} \le 2\|W_1\|_{\frac{r-1}{r}}\eta \end{array} \} $$

Let $\mathbf{v}, \mathbf{w} \in C([0, T], L^2(\Phi_4 dx))$ with

$$ \bullet \sup_{0<t<T} \|\mathbf{v}(t,.)\|_r < +\infty, $$

$$ \bullet \sup_{0<t<T} \|\mathbf{w}(t,.)\|_r < +\infty, $$

$$ \bullet \sup_{0<t<T} t^{\frac{3}{2r}} \|\mathbf{v}(t,.)\|_{\infty} < +\infty, $$

$$ \bullet \sup_{0<t<T} t^{\frac{3}{2r}} \|\mathbf{w}(t,.)\|_{\infty} < +\infty, $$

$$ \bullet \sup_{0<t<T} \sqrt{t} \|\nabla \otimes \mathbf{v}(t,.)\|_r < +\infty, $$

$$ \bullet \sup_{0<t<T} \sqrt{t} \|\nabla \otimes \mathbf{w}(t,.)\|_r < +\infty. $$

We have the inequalities

$$ \begin{aligned} \|B_{\epsilon,\alpha}(\mathbf{v}, \mathbf{w})\|_r &\le C \int_0^t \frac{1}{\sqrt{t-s}} \|\varphi_{\epsilon} * (\theta_{\alpha}\mathbf{v}(s, .))\|_r \|\mathbf{w}(s, .)\|_{\infty} ds \\ &\le C \int_0^t \frac{1}{\sqrt{t-s}} \frac{1}{s^{\frac{3}{2r}}} ds \sup_{0<t<T} \|\mathbf{v}(s, .)\|_r \sup_{0<s<T} s^{\frac{3}{2r}} \|\mathbf{w}(s, .)\|_{\infty} \\ &\le C' t^{\frac{1}{2}(1-\frac{3}{r})} \sup_{0<t<T} \|\mathbf{v}(s, .)\|_r \sup_{0<s<T} s^{\frac{3}{2r}} \|\mathbf{w}(s, .)\|_{\infty}, \end{aligned} $$

$$ \begin{aligned} \|B_{\epsilon,\alpha}(\mathbf{v}, \mathbf{w})\|_{\infty} &\le C \int_0^t \frac{1}{\sqrt{t-s}} \|\varphi_{\epsilon} * (\theta_{\alpha}\mathbf{v}(s, .))\|_{\infty} \|\mathbf{w}(s, .)\|_{\infty} ds \\ &\le C \int_0^t \frac{1}{\sqrt{t-s}} \frac{1}{s^{\frac{3}{r}}} ds \sup_{0<t<T} s^{\frac{3}{2r}} \|\mathbf{v}(s, .)\|_{\infty} \sup_{0<s<T} s^{\frac{3}{2r}} \|\mathbf{w}(s, .)\|_{\infty} \\ &\le C' t^{-\frac{3}{2r}} t^{\frac{1}{2}(1-\frac{3}{r})} \sup_{0<t<T} s^{\frac{3}{2r}} \|\mathbf{v}(s, .)\|_{\infty} \sup_{0<s<T} s^{\frac{3}{2r}} \|\mathbf{w}(s, .)\|_{\infty}, \end{aligned} $$
and, for $1 \le j \le 3$, remarking that $\|\partial_j\theta_\alpha\|_3 = \|\partial_j\theta\|_3$,

$$
\begin{align*}
\|\partial_j B_{\epsilon, \alpha}(\mathbf{v}, \mathbf{w})\|_r
&\le C \int_0^t \frac{1}{\sqrt{t-s}} \|\varphi_\epsilon * (\theta_\alpha \mathbf{v}(s, .))\|_{\infty} \|\partial_j \mathbf{w}(s, .)\|_r ds \\
&\quad + C \int_0^t \frac{1}{\sqrt{t-s}} \|\varphi_\epsilon * (\theta_\alpha \partial_j \mathbf{v}(s, .))\|_r \|\mathbf{w}(s, .)\|_{\infty} ds \\
&\quad + C \int_0^t \frac{1}{(t-s)^{1-\frac{3}{2r}}} \|\varphi_\epsilon * (\partial_j \theta_\alpha \mathbf{v}(s, .))\|_3 \|\mathbf{w}(s, .)\|_{\infty} ds \\
&\le C' t^{-\frac{3}{2r}} \left( \sup_{0<t<T} s^{\frac{3}{2r}} \|\mathbf{v}(s, .)\|_{\infty} + \sup_{0<s<T} \sqrt{s} \|\nabla \otimes \mathbf{v}(s, .)\|_r \right) \\
&\quad \times \left( \sup_{0<s<T} s^{\frac{3}{2r}} \|\mathbf{w}(s, .)\|_{\infty} + \sup_{0<s<T} \sqrt{s} \|\nabla \otimes \mathbf{w}(s, .)\|_r \right).
\end{align*}
$$

For the norm

$$
\|\mathbf{v}\|_* = \max\left( \sup_{0<s<T} \|\mathbf{v}(s, .)\|_r, \sup_{0<s<T} \sqrt{s} \|\nabla \otimes \mathbf{v}(s, .)\|_r, \sup_{0<s<T} s^{\frac{3}{2r}} \|\mathbf{v}(s, .)\|_{\infty} \right),
$$

we find

$$
\| B_{\epsilon, \alpha}(\mathbf{v}, \mathbf{w}) \|_* \le C T^{\frac{1}{2} - \frac{3}{2r}} \|\mathbf{v}\|_* \|\mathbf{w}\|_*
$$

(where $C$ doesn't depend on $\epsilon$ nor on $\alpha$), which proves that $B_{\epsilon,\alpha}$ is a contraction on $B_{\eta}$ for $T^{\frac{1}{2}-\frac{3}{2r}}\eta$ small enough. $\square$

3.4 Local estimates in the norm of L²(Φ₂ dx).

We now want to estimate $\mathbf{v}_{\eta, \epsilon, \alpha}(t, \cdot)$ in $L^2(\Phi_2 dx)$ for $\le t \le T < \min(T_{\epsilon, \alpha}, T_{[\eta]})$.
We have

$$
\mathbf{v}_{\eta, \epsilon, \alpha}(t, \cdot) = e^{t\Delta} \mathbf{v}_{0, \eta} - \mathcal{L}(\mathbb{F}_{\eta, \epsilon, \alpha})
$$

where we defined the operator $\mathcal{L}$ by

$$
\mathbb{F} \mapsto \mathcal{L}(\mathbb{F}) = \int_0^t e^{(t-s)\Delta} \mathbb{P} \, \mathrm{div} \, \mathbb{F} \, ds
$$

and where

$$
\mathbb{F}_{\eta, \epsilon, \alpha} = (\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{u}_{\epsilon, \alpha})) \otimes \mathbf{u}_{\epsilon, \alpha} - (\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{b}_{\eta, \epsilon, \alpha})) \otimes \mathbf{b}_{\eta, \epsilon, \alpha}
$$

**Lemma 7.** $\mathbf{v}_{\eta, \epsilon, \alpha} \in \mathcal{C}([0, T], L^2(\Phi_2 dx)) \cap L^2((0, T), H^1(\Phi_2 dx))$ and $\partial_t \mathbf{v}_{\eta, \epsilon, \alpha} \in L^2((0, T), H^{-1}(\Phi_2 dx))$. In particular, we have

$$
\begin{align*}
\int |\mathbf{v}_{\eta, \epsilon, \alpha}(t, x)|^2 \Phi_2(x) \, dx - \int |\mathbf{v}_{0, \eta}(x)|^2 \Phi_2(x) \, dx \\
&= 2 \int_0^t \int \mathbf{v}_{\eta, \epsilon, \alpha}(s, x) \cdot \partial_t \mathbf{v}_{\eta, \epsilon, \alpha}(s, x) \, \Phi_2(x) \, ds \, dx \\
&= 2 \int_0^t \int \mathbf{v}_{\eta, \epsilon, \alpha}(s, x) \cdot (\Delta \mathbf{v}_{\eta, \epsilon, \alpha}(s, x) - \mathbb{P} \operatorname{div} \mathbb{F}_{\eta, \epsilon, \alpha}(s, x)) \, \Phi_2(x) \, ds \, dx.
\end{align*}
$$
Proof. First, we recall that $\|f\|_{H^s(\Phi_2 dx)} = \|\Phi f\|_{H^s}$. As $\Phi$ is bounded with all its derivatives, we have $\|\Phi f\|_{H^s} \le C_s \|f\|_{H^s}$ for every $s \in \mathbb{R}$. As $\Phi_2$ is a weight in the Muckenhoupt class $A_2$ we have

$$ \|e^{t\Delta} f\|_{L^2(\Phi_2 dx)} \le \|\mathcal{M}_f\|_{L^2(\Phi_2 dx)} \le C \|f\|_{L^2(\Phi_2 dx)} $$

where $\mathcal{M}_f$ is the Hardy-Littlewood maximal function of $f$.

If $\mathbf{w}_0 \in L^2$, we have $e^{t\Delta}\mathbf{w}_0 \in \mathcal{C}([0, T], L^2) \cap L^2((0, T), H^1)$ and $\partial_t(e^{t\Delta}\mathbf{w}_0) \in L^2((0, T), H^{-1})$; in particular, $\Phi e^{t\Delta}\mathbf{w}_0 \in L^2((0, T), H^1)$ and $\partial_t(\Phi e^{t\Delta}\mathbf{w}_0) \in L^2((0, T), H^{-1})$. We thus have, for $0 < t < T$,

$$ \begin{aligned} \|\Phi e^{t\Delta} \mathbf{w}_0\|_2^2 - \|\Phi \mathbf{w}_0\|_2^2 &= 2 \int_0^t \int (\Phi e^{s\Delta} \mathbf{w}_0)(s, x) \cdot \partial_t (\Phi e^{s\Delta} \mathbf{w}_0)(s, x) \, ds \, dx \\ &= 2 \int_0^t \int \Phi(x)^2 e^{s\Delta} \mathbf{w}_0(s, x) \cdot \Delta e^{s\Delta} \mathbf{w}_0(s, x) \, ds \, dx \\ &= \int_0^t \int \Phi(x)^2 (\Delta |e^{s\Delta} \mathbf{w}_0(s, x)|^2 - 2|\nabla e^{s\Delta} \mathbf{w}_0(s, x)|^2) \, ds \, dx. \end{aligned} $$

so that, as $|\Delta\Phi_2| \le C\Phi_2$,

$$ \begin{aligned} 2 \int_0^t \|\nabla e^{s\Delta} \mathbf{w}_0(s, \cdot)\|_{L^2(\Phi_2 dx)}^2 ds \\ = -\|\Phi e^{t\Delta} \mathbf{w}_0\|_2^2 + \|\Phi \mathbf{w}_0\|_2^2 + \int_0^t \int |e^{s\Delta} \mathbf{w}_0(s, x)|^2 \Delta\Phi_2(x) \, ds \, dx \\ \le C \|\mathbf{w}_0\|_{L^2(\Phi_2 dx)}^2 (1+t). \end{aligned} $$

By density of $L^2$ in $L^2(\Phi_2 dx)$, we find that $\mathbf{w}_0 \mapsto e^{t\Delta}\mathbf{w}_0$ is a bounded map from $L^2(\Phi_2 dx)$ to $\mathcal{C}([0, T], L^2(\Phi_2 dx)) \cap L^2((0, T), H^1(\Phi_2 dx))$. Moreover, since $\partial_t e^{t\Delta}\mathbf{w}_0 = \Delta e^{t\Delta}\mathbf{w}_0$, we have that $\partial_t e^{t\Delta}\mathbf{w}_0 \in L^2((0, T), H^{-1}(\Phi_2 dx))$.

A classical result on the heat kernel states that, if $\mathbb{F} \in L^2((0, T), L^2)$, then $\int_0^t e^{(t-s)\Delta} \mathbb{P} \, \mathrm{div} \, \mathbb{F} \in \mathcal{C}([0, T], L^2) \cap L^2((0, T), H^1)$ and thus $\int_0^t e^{(t-s)\Delta} \mathbb{P} \, \mathrm{div} \, \mathbb{F} \in \mathcal{C}([0, T], L^2(\Phi_2 dx)) \cap L^2((0, T), H^1(\Phi_2 dx))$.

As we have

$$ \|\mathbb{F}_{\eta, \epsilon, \alpha}(t, \cdot)\|_2 \le \|\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{u}_{\epsilon, \alpha})\|_{\infty} \left( \int_{|x| \le \frac{2}{\alpha} + \epsilon} |\mathbf{u}_{\epsilon, \alpha}(t, x)|^2 dx \right)^{\frac{1}{2}} \\ + \|\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{b}_{\epsilon, \alpha})\|_{\infty} \left( \int_{|x| \le \frac{2}{\alpha} + \epsilon} |\mathbf{b}_{\epsilon, \alpha}(t, x)|^2 dx \right)^{\frac{1}{2}}, $$

we find that

$$ \mathcal{L}(\mathbb{F}_{\eta, \epsilon, \alpha}) \in \mathcal{C}([0, T], L^2) \cap L^2((0, T), H^1) \subset \mathcal{C}([0, T], L^2(\Phi_2 dx)) \cap L^2((0, T), H^1(\Phi_2 dx)). $$
Moreover, $\partial_t \mathcal{L}(\mathbb{F}_{\eta, \epsilon, \alpha}) = \Delta \mathcal{L}(\mathbb{F}_{\eta, \epsilon, \alpha}) - \mathbb{P} \text{ div } \mathbb{F}_{\eta, \epsilon, \alpha}$ with

$$ \Delta \mathcal{L}(\mathbb{F}_{\eta, \epsilon, \alpha}) \in L^2((0, T), H^{-1}(\Phi_2 \, dx)) $$

and

$$ \mathbb{P} \text{ div}(\mathbb{F}_{\eta, \epsilon, \alpha}) \in L^2((0, T), H^{-1}) \subset L^2((0, T), H^{-1}(\Phi_2 \, dx)). $$

Finally, as $\Phi \mathbf{v}_{\eta, \epsilon, \alpha} \in L^2((0, T), H^1)$ and $\partial_t(\Phi \mathbf{v}_{\eta, \epsilon, \alpha}) \in L^2((0, T), H^{-1})$, one has

$$ \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}(t, \cdot)\|_2^2 - \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}(0, \cdot)\|_2^2 = 2 \int_0^t \int (\Phi \mathbf{v}_{\eta, \epsilon, \alpha})(s, x) \cdot \partial_t (\Phi \mathbf{v}_{\eta, \epsilon, \alpha})(s, x) \, ds \, dx. \quad \square $$

<figure><img src="image_5.png" /></figure>

Lemma 7 will be a key ingredient for controlling the norm of $\mathbf{v}_{\eta, \epsilon, \alpha}$ in $C([0, T], L^2(\Phi_2 \, dx)) \cap L^2((0, T), H^1(\Phi_2 \, dx))$ and the existence time of the solution $\mathbf{v}_{\eta, \epsilon, \alpha}$ independently from $\epsilon$ and $\alpha$.

**Proposition 4.** Let $2 < p < +\infty$ and $0 < \gamma < 2$. There exists a constant $C_2 > 1$ such that, for every $\epsilon > 0$, every $\alpha > 0$, every $\eta > 0$ and every $\mathbf{u}_0 \in L^p(\Phi_\gamma \, dx)$ with $\text{div} \, \mathbf{u}_0 = 0$, writing $\mathbf{u}_0 = \mathbf{b}_{0,\eta} + \mathbf{v}_{0,\eta}$ with $\mathbf{v}_{0,\eta} \in L^2(\Phi_2 \, dx)$, $\text{div} \, \mathbf{v}_{0,\eta} = 0$ and $\|\mathbf{b}_{0,\eta}\|_r < \eta$, $\text{div} \, \mathbf{b}_{0,\eta} = 0$, and writing $\mathbf{u}_{\epsilon,\alpha} = \mathbf{v}_{\eta,\epsilon,\alpha} + \mathbf{b}_{\eta,\epsilon,\alpha}$ with $\mathbf{u}_{\epsilon,\alpha} \in C([0, T_{\epsilon,\alpha}), L^p(\Phi_\gamma \, dx)$ (as described in Proposition 2) and $\mathbf{b}_{\eta,\epsilon,\alpha} \in L^\infty((0, T_{[\eta]}), L^r)$ (as described in Proposition 3), the mollified equation

$$ \mathbf{v}_{\eta, \epsilon, \alpha} = e^{t\Delta} \mathbf{v}_{0, \eta} - B_{\epsilon, \alpha}(\mathbf{b}_{\eta, \epsilon, \alpha}, \mathbf{v}_{\eta, \epsilon, \alpha}) - B_{\epsilon, \alpha}(\mathbf{v}_{\eta, \epsilon, \alpha}, \mathbf{b}_{\eta, \epsilon, \alpha}) - B_{\epsilon, \alpha}(\mathbf{v}_{\eta, \epsilon, \alpha}, \mathbf{v}_{\eta, \epsilon, \alpha}) \quad (17) $$

has a unique solution on $(0, T_{\eta, \epsilon, \alpha}) \times \mathbb{R}^3$ such that

$$ \bullet \ T_{\eta, \epsilon, \alpha} = \min(T_{\epsilon, \alpha}, \frac{1}{C_2} T_{[\eta]}, \frac{1}{C_2} \frac{1}{1+\max(1, \alpha)^6 \|\mathbf{v}_{0, \eta}\|_{L^2(\Phi_2 \, dx)}^4}) $$

$$ \bullet \ \mathbf{v}_{\eta, \epsilon, \alpha} \in C([0, T_{\eta, \epsilon, \alpha}], L^2(\Phi_2 \, dx)) \text{ with} $$

$$ \sup_{0 \le t \le T_{\eta, \epsilon, \alpha}} \|\mathbf{v}_{\eta, \epsilon, \alpha}(t, \cdot)\|_{L^2(\Phi_2 \, dx)} \le 2 \|\mathbf{v}_{0, \eta}\|_{L^2(\Phi_2 \, dx)}, $$

$$ \bullet \ \mathbf{v}_{\eta, \epsilon, \alpha} \in L^2((0, T_{\eta, \epsilon, \alpha}), H^1(\Phi_2 \, dx)) \text{ with} $$

$$ \|\mathbf{v}_{\eta, \epsilon, \alpha}(t, \cdot)\|_{L^2((0, T_{\eta, \epsilon, \alpha}), H^1(L^2(\Phi_2 \, dx)))} \le C_2 \|\mathbf{v}_{0, \eta}\|_{L^2(\Phi_2 \, dx)}. $$

In particular, we have

$$ T_{\epsilon, \alpha} \ge \min \left( \frac{1}{C_2} \left( \frac{1}{C_1 \|\mathbf{b}_{0, \eta}\|_r} \right)^{\frac{2r}{r-3}}, \frac{1}{C_2} \frac{1}{1 + \max(1, \alpha)^6 \|\mathbf{v}_{0, \eta}\|_{L^2(\Phi_2 \, dx)}^4} \right). $$
Proof. We write

$$
\begin{align*}
\partial_t \mathbf{v}_{\eta, \epsilon, \alpha} &= \Delta \mathbf{v}_{\eta, \epsilon, \alpha} - \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})) \otimes \mathbf{b}_{\eta, \epsilon, \alpha}) - \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{b}_{\eta, \epsilon, \alpha})) \otimes \mathbf{v}_{\eta, \epsilon, \alpha}) \\
&\quad - \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})) \otimes \mathbf{v}_{\eta, \epsilon, \alpha}) \\
&= \Delta \mathbf{v}_{\eta, \epsilon, \alpha} - \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})) \otimes \mathbf{b}_{\eta, \epsilon, \alpha}) - \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{b}_{\eta, \epsilon, \alpha})) \otimes \mathbf{v}_{\eta, \epsilon, \alpha}) \\
&\quad - \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha}) \otimes \mathbf{v}_{\eta, \epsilon, \alpha})) - \nabla q_{\eta, \epsilon, \alpha}
\end{align*}
$$

with

$$
q_{\eta, \epsilon, \alpha} = \sum_{1 \le i \le 3} \sum_{1 \le j \le 3} R_i R_j ((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha, i})) \mathbf{v}_{\eta, \epsilon, \alpha, j}).
$$

This gives (as div $\mathbf{v}_{\eta, \epsilon, \alpha} = 0$)

$$
\begin{align*}
2\partial_t \mathbf{v}_{\eta, \epsilon, \alpha} \cdot \mathbf{v}_{\eta, \epsilon, \alpha} &= \Delta(|\mathbf{v}_{\eta, \epsilon, \alpha}|^2) - 2|\nabla \otimes \mathbf{v}_{\eta, \epsilon, \alpha}|^2 \\
&\quad - 2\mathbf{v}_{\eta, \epsilon, \alpha} \cdot (\mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})) \otimes \mathbf{b}_{\eta, \epsilon, \alpha}) + \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{b}_{\eta, \epsilon, \alpha})) \otimes \mathbf{v}_{\eta, \epsilon, \alpha})) \\
&\quad - \operatorname{div}(|\mathbf{v}_{\eta, \epsilon, \alpha}|^2 (\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha}))) + 2|\mathbf{v}_{\eta, \epsilon, \alpha}|^2 \operatorname{div}(\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})) \\
&\quad - 2 \operatorname{div}(q_{\eta, \epsilon, \alpha} \mathbf{v}_{\eta, \epsilon, \alpha})
\end{align*}
$$

with

$$
\operatorname{div}(\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})) = \varphi_{\epsilon} * (\mathbf{v}_{\eta, \epsilon, \alpha} \cdot \nabla \theta_{\alpha}).
$$

Integrating against $\Phi(x)^2 dx$, we obtain

$$
\frac{d}{dt} \| \mathbf{v}_{\eta, \epsilon, \alpha} \|_{L^2(\Phi_2 \, dx)}^2 + 2 \| \nabla \otimes \mathbf{v}_{\eta, \epsilon, \alpha} \|_{L^2(\Phi_2 \, dx)}^2 = \sum_{k=1}^{6} A_k
$$

with:

• $A_1 = \int \Phi^2 \Delta(|\mathbf{v}_{\eta, \epsilon, \alpha}|^2) \, dx = \int |\mathbf{v}_{\eta, \epsilon, \alpha}|^2 \Delta(\Phi^2) \, dx \le C \|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^2$ (since $|\Delta(\Phi)^2| \le C\Phi^2$);

• $A_2 = -2 \int \Phi^2 \mathbf{v}_{\eta, \epsilon, \alpha} \cdot \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})) \otimes \mathbf{b}_{\eta, \epsilon, \alpha}) \, dx$; since the Riesz transforms are bounded on $L^2(\Phi_2 \, dx)$ and since

$$
| \int \Phi^2 f \partial_j g \, dx | \le \| \Phi f \|_{H^1} \| \Phi \partial_j g \|_{H^{-1}} \le C (\| \Phi f \|_2 + \| \Phi \nabla f \|_2) \| \Phi g \|_2,
$$

we have (since $\|\varphi_{\epsilon} * f\|_{L^2(\Phi_2 \, dx)} \le C \|\mathcal{M}_f\|_{L^2(\Phi_2 \, dx)} \le C' \|f\|_{L^2(\Phi_2 \, dx)}$)

$$
\begin{align*}
A_2 &\le C (\|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)} + \|\nabla \otimes \mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}) \|(\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})) \otimes \mathbf{b}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)} \\
&\le C' (\|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)} + \|\nabla \otimes \mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}) \|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)} \|\mathbf{b}_{\eta, \epsilon, \alpha}\|_{\infty} \\
&\le C'' (\|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)} + \|\nabla \otimes \mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}) \|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)} \eta t^{-\frac{3}{2r}} \\
&\le \frac{1}{10} \|\nabla \otimes \mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^2 + C''' \|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^2 (1 + \eta^2 t^{-\frac{3}{r}});
\end{align*}
$$
• $A_3 = -2 \int \Phi^2 \mathbf{v}_{\eta, \epsilon, \alpha} \cdot \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{b}_{\eta, \epsilon, \alpha})) \otimes \mathbf{v}_{\eta, \epsilon, \alpha}) dx$; with similar computations as for $A_2$ we find

$$A_3 \le \frac{1}{10} \| \nabla \otimes \mathbf{v}_{\eta, \epsilon, \alpha} \|_{L^2(\Phi_2 \, dx)}^2 + C''' \| \mathbf{v}_{\eta, \epsilon, \alpha} \|_{L^2(\Phi_2 \, dx)}^2 (1 + \eta^2 t^{-\frac{3}{r}});$$

• $A_4 = - \int \Phi^2 \operatorname{div}(|\mathbf{v}_{\eta, \epsilon, \alpha}|^2 (\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha}))) dx$; since $|\nabla \Phi^2| \le C \Phi^3$, we have

$$A_4 \le C \int \Phi^3 |\mathbf{v}_{\eta, \epsilon, \alpha}|^2 |\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})| dx;$$

as $\Phi \mathbf{v}_{\eta, \epsilon, \alpha} \in L^2$, we have $\Phi(\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})) \in L^2$; on the other hand, we have $\Phi \mathbf{v}_{\eta, \epsilon, \alpha} \in L^2$, we have $\Phi \mathbf{v}_{\eta, \epsilon, \alpha} \in H^1$, so that $\Phi \mathbf{v}_{\eta, \epsilon, \alpha} \in L^3 \cap L^6$; thus, we have

$$\begin{aligned}
A_4 &\le C \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}\|_6 \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}\|_3 \|\Phi(\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha}))\|_2 \\
&\le C' \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}\|_6 \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}\|_3 \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}\|_2 \\
&\le C'' (\|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)} + \|\nabla \otimes \mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^{\frac{3}{2}})^{\frac{3}{2}} \|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^{\frac{3}{2}} \\
&\le \frac{1}{10} \|\nabla \otimes \mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^2 + C''' \|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^2 (1 + \|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^4);
\end{aligned}$$

• $A_5 = -2 \int \Phi^2 \operatorname{div}(q_{\eta, \epsilon, \alpha} \mathbf{v}_{\eta, \epsilon, \alpha}) dx$; similarly, we have

$$A_5 \le C \int \Phi^3 |q_{\eta, \epsilon, \alpha} ||\mathbf{v}_{\eta, \epsilon, \alpha}|| dx \le \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}\|_6 \|\Phi^2 q_{\eta, \epsilon, \alpha}\|_{6/5};$$

we recall that

$$q_{\eta, \epsilon, \alpha} = \sum_{1 \le i \le 3} \sum_{1 \le j \le 3} R_i R_j ((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha, i})) \mathbf{v}_{\eta, \epsilon, \alpha, j})$$

with $\Phi \mathbf{v}_{\eta, \epsilon, \alpha} \in L^3$ and $\Phi(\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})) \in L^2$; thus,

$$\Phi^2((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})) \otimes \mathbf{v}_{\eta, \epsilon, \alpha}) \in L^{\frac{6}{5}}$$

or, equivalently, $(\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha})) \otimes \mathbf{v}_{\eta, \epsilon, \alpha} \in L^{\frac{6}{5}}(\Phi^{\frac{12}{5}} dx)$; as $\frac{12}{5} < 3$, $\Phi^{\frac{12}{5}}$ belongs to the Muckenhoupt class $A_{\frac{6}{5}}$, and the Riesz transforms are bounded on $L^{\frac{6}{5}}(\Phi^{\frac{12}{5}} dx)$; we thus get

$$\begin{aligned}
A_5 &\le C \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}\|_6 \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}\|_3 \|\Phi(\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{v}_{\eta, \epsilon, \alpha}))\|_2 \\
&\le \frac{1}{10} \|\nabla \otimes \mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^2 + C' \|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^2 (1 + \|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^4);
\end{aligned}$$
• $A_6 = 2 \int \Phi^2 |\mathbf{v}_{\eta, \epsilon, \alpha}|^2 (\varphi_{\epsilon} * (\mathbf{v}_{\eta, \epsilon, \alpha} \cdot \nabla \theta_{\alpha})) dx$; we have

$$ 
\begin{aligned}
A_6 &\le C \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}\|_6 \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}\|_3 \|\varphi_{\epsilon} * (\mathbf{v}_{\eta, \epsilon, \alpha} \cdot \nabla \theta_{\alpha})\|_2 \\
&\le C \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}\|_6 \|\Phi \mathbf{v}_{\eta, \epsilon, \alpha}\|_3 \|\mathbf{v}_{\eta, \epsilon, \alpha} \cdot \nabla \theta_{\alpha}\|_2;
\end{aligned}
$$

as $|\nabla \theta_{\alpha}(x)| \le C \max(\alpha, 1) \Phi(x)$), we find

$$A_6 \le \frac{1}{10} \|\nabla \otimes \mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^2 + C' \|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^2 (1 + \max(\alpha, 1)^6 \|\mathbf{v}_{\eta, \epsilon, \alpha}\|_{L^2(\Phi_2 \, dx)}^4).$$

Thus, integrating on $(0, t)$ for $0 < t < T$ with $T < \min(T_{\epsilon, \alpha}, T_{[\eta]})$, we obtain, for $C_T = \sup_{0<t<T} \|\mathbf{v}_{\eta, \epsilon, \alpha}(t, \cdot)\|_{L^2(\Phi_2 \, dx)}^2$,

$$C_T \le \|\mathbf{v}_{0, \eta}\|_{L^2(\Phi_2 \, dx)}^2 + CTC_T + CC_T T^{1-\frac{3}{r}} \eta^2 + CT \max(1, \alpha)^6 C_T^3.$$

We obtain $C_T \le 4 \|\mathbf{v}_{0, \eta}\|_{L^2(\Phi_2 \, dx)}^2$ if

$$CT(1 + 16 \max(1, \alpha)^6 \|\mathbf{v}_{0, \eta}\|_{L^2(\Phi_2 \, dx)}^4) < \frac{1}{4}$$

and

$$CT^{1-\frac{3}{r}} \eta^2 < \frac{1}{2}. \quad \square$$

## 3.5 Rescaling and global estimates

The minoration on $T_{\epsilon, \alpha}$ given in Proposition 4 depends on $\eta$, i.e. on $\|\mathbf{b}_{0, \eta}\|_r$ and on $\|\mathbf{v}_{0, \eta}\|_{L^2(\Phi_2 \, dx)}$. As a matter of fact, following Fernández-Dalgo & Lemarié-Rieusset [3] and Bradshaw, Kucavica & Tsai [1], we can partly get rid of this restriction:

**Theorem 4.** Let $2 < p < +\infty$ and $0 < \gamma < 2$. For every $\mathbf{u}_0 \in L^p(\Phi_\gamma \, dx)$ with $\mathrm{div} \, \mathbf{u}_0 = 0$, let $\mathbf{u}_{\epsilon, \alpha}$ be the (maximal) solution of the mollified equations

$$ \left\{ \begin{array}{l} \partial_t \mathbf{u}_{\epsilon, \alpha} = \Delta \mathbf{u}_{\epsilon, \alpha} - \mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{u}_{\epsilon})) \otimes \mathbf{u}_{\epsilon, \alpha}) \\ \lim_{t \to 0} \mathbf{u}_{\epsilon, \alpha}(t, \cdot) = \mathbf{u}_0 \end{array} \right. \qquad (18) $$

in $C([0, T_{\epsilon}), L^p(\Phi_\gamma \, dx))$. For every $T > 0$, there exists $\alpha_{T, \mathbf{u}_0} > 0$ such that, for every $0 < \epsilon$ and every $0 < \alpha < \alpha_{T, \mathbf{u}_0}$, one has $T_{\epsilon, \alpha} > T$. Moreover, one may split $\mathbf{u}_{\epsilon, \alpha}$ as $\mathbf{u}_{\epsilon, \alpha} = \mathbf{b}_{\epsilon, \alpha, T} + \mathbf{v}_{\epsilon, \alpha, T}$ on $(0, T) \times \mathbb{R}^3$ with

$$\bullet \quad \mathbf{b}_{\epsilon, \alpha, T} \in C([0, T], L^2(\Phi_4 \, dx)),$$

$$\bullet \quad \sup_{0 \le t \le T} \|\mathbf{b}_{\epsilon, \alpha, T}(t, \cdot)\|_r \le C_{3,T,\mathbf{u}_0},$$
• $\sup_{0<t\le T} t^{\frac{1}{2}} \|\nabla \otimes \mathbf{b}_{\epsilon,\alpha,T}(t, \cdot)\|_r \le C_{3,T,\mathbf{u}_0}$,

• $\sup_{0<t\le T} t^{\frac{3}{2r}} \|\mathbf{b}_{\epsilon,\alpha,T}(t, \cdot)\|_{\infty} \le C_{3,T,\mathbf{u}_0}$,

• $\mathbf{v}_{\epsilon,\alpha,T} \in C([0,T], L^2(\Phi_2 dx))$ with

$$ \sup_{0 \le t \le T} \| \mathbf{v}_{\epsilon, \alpha, T}(t, \cdot) \|_{L^2(\Phi_2 dx)} \le C_{3,T,\mathbf{u}_0}, $$

• $\mathbf{v}_{\eta,\epsilon,\alpha} \in L^2((0,T), H^1(\Phi_2 dx))$ with

$$ \| \mathbf{v}_{\epsilon, \alpha, T}(t, \cdot) \|_{L^2((0,T), H^1(L^2(\Phi_2 dx)))} \le C_{3,T,\mathbf{u}_0} $$

where $C_{3,T,\mathbf{u}_0}$ doesn't depend on $\epsilon$ nor on $\alpha$. In particular, we have

• $\sup_{0<\epsilon,0<\alpha<\alpha_T} \sup_{0<t<T} \|\mathbf{u}_{\epsilon,\alpha}(t, \cdot)\|_{L^2(\Phi_7^2 dx)} \le C_{4,T,\mathbf{u}_0}$

• $\sup_{0<\epsilon,0<\alpha<\alpha_T} \|\mathbf{u}_{\epsilon,\alpha}(t, \cdot)\|_{L^2((0,T),H^{1/2}(\Phi_2 dx))} \le C_{4,T,\mathbf{u}_0}$

• $\sup_{0<\epsilon,0<\alpha<\alpha_T} \|\partial_t \mathbf{u}_{\epsilon,\alpha}(t, \cdot)\|_{L^2((0,T),H^{-4}(\Phi_8 dx))} \le C_{4,T,\mathbf{u}_0}$

*Proof.* We first recall the results on $\mathbf{u}_{\epsilon,\alpha}$ obtained in Propositions 2, 3, 4. Let $T > 0$. Following Proposition 4, we choose $\eta_T$ with $\eta_T \le \frac{1}{C_1(C_2T)^{\frac{r-3}{2r}}}$ and we split $\mathbf{u}_0$ into $\mathbf{u}_0 = \mathbf{b}_{0,\eta_T} + \mathbf{v}_{0,\eta_T}$ with $\|\mathbf{b}_{0,\eta_T}\|_{\infty} < \eta_T$ and $\mathbf{v}_{0,\eta_T} \in L^2(\Phi_2 dx)$. We then split $\mathbf{u}_{\epsilon,\alpha}$ into $\mathbf{u}_{\epsilon,\alpha} = \mathbf{b}_{\eta_T,\epsilon,\alpha} + \mathbf{v}_{\eta_T,\epsilon,\alpha}$. We rescale $\mathbf{u}_{\epsilon,\alpha}$ into

$$ \begin{aligned} \mathbf{u}_{\lambda, \epsilon, \alpha}(t, x) &= \frac{1}{\lambda} \mathbf{u}_{\epsilon, \alpha} \left( \frac{t}{\lambda^2}, \frac{x}{t} \right) = \frac{1}{\lambda} \mathbf{b}_{\eta_T, \epsilon, \alpha} \left( \frac{t}{\lambda^2}, \frac{x}{t} \right) + \frac{1}{\lambda} \mathbf{v}_{\eta_T, \epsilon, \alpha} \left( \frac{t}{\lambda^2}, \frac{x}{t} \right) \\ &= \mathbf{b}_{\eta_T, \lambda, \epsilon, \alpha}(t, x) + \mathbf{v}_{\eta_T, \lambda, \epsilon, \alpha}(t, x). \end{aligned} $$

We write similarly

$$ \mathbf{u}_{\lambda,0}(x) = \frac{1}{\lambda} \mathbf{u}_0(\frac{x}{\lambda}), \quad \mathbf{v}_{\eta_T,\lambda,0}(x) = \frac{1}{\lambda} \mathbf{v}_{0,\eta_T}(\frac{x}{\lambda}), \quad \mathbf{b}_{\eta_T,\lambda,0}(x) = \frac{1}{\lambda} \mathbf{b}_{0,\eta_T}(\frac{x}{\lambda}). $$

We have

$$ \begin{aligned} \partial_t \mathbf{u}_{\lambda, \epsilon, \alpha}(t, x) &= \frac{1}{\lambda^3} \left( \partial_t \mathbf{u}_{\epsilon, \alpha} \right) \left( \frac{t}{\lambda^2}, \frac{x}{\lambda} \right) \\ &= \frac{1}{\lambda^3} \left( \Delta \mathbf{u}_{\epsilon, \alpha} - \mathbb{P} \operatorname{div} \left( (\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{u}_{\epsilon, \alpha})) \otimes \mathbf{u}_{\epsilon, \alpha} \right) \right) \left( \frac{t}{\lambda^2}, \frac{x}{\lambda} \right) \\ &= \Delta \mathbf{u}_{\lambda, \epsilon, \alpha}(t, x) - \mathbb{P} \operatorname{div} \left( \frac{1}{\lambda} (\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{u}_{\epsilon, \alpha}) \left( \frac{t}{\lambda^2}, \frac{x}{\lambda} \right)) \otimes \mathbf{u}_{\lambda, \epsilon, \alpha}(t, x) \right) \\ &= \Delta \mathbf{u}_{\lambda, \epsilon, \alpha}(t, x) - \mathbb{P} \operatorname{div} \left( (\varphi_{\lambda \epsilon} * (\theta_{\lambda} \mathbf{u}_{\lambda, \epsilon, \alpha}) (t, x)) \otimes \mathbf{u}_{\lambda, \epsilon, \alpha}(t, x) \right) \end{aligned} $$
Thus, $\mathbf{u}_{\lambda, \epsilon, \alpha}$ is the solution of the mollified equation

$$ \mathbf{u}_{\lambda, \epsilon, \alpha} = e^{t\Delta} \mathbf{u}_{\lambda, 0} - B_{\lambda, \epsilon, \alpha}(\mathbf{u}_{\lambda, \epsilon, \alpha}, \mathbf{u}_{\lambda, \epsilon, \alpha}). $$

By Proposition 4, we know that the existence time $T_{\lambda, \epsilon, \alpha}$ of $\mathbf{u}_{\lambda, \epsilon, \alpha}$ can be controlled by below as

$$ T_{\lambda, \epsilon, \alpha} \geq \min\left(\frac{1}{C_2}\frac{1}{(C_1\|\mathbf{b}_{\eta_T, \lambda, 0}\|_r)^{\frac{2r}{r-3}}}, \frac{1}{C_2}\frac{1}{1 + \max(1, \frac{\alpha}{\lambda})^6\|\mathbf{v}_{\eta_T, \lambda, 0}\|_{L^2(\Phi_2 dx)}^4}\right). $$

Thus, we have (since $\lambda^2 \|\mathbf{b}_{\eta_T, \lambda, 0}\|_r^{\frac{2r}{r-3}} = \|\mathbf{b}_{0, \eta_T}\|_r^{\frac{2r}{r-3}}$)

$$ \begin{aligned} T_{\epsilon, \alpha} &= \frac{1}{\lambda^2} T_{\lambda, \epsilon, \alpha} \\ &\geq \min\left(\frac{1}{C_2}\frac{1}{(C_1\|\mathbf{b}_{0, \eta_T}\|_r)^{\frac{2r}{r-3}}}, \frac{1}{C_2\lambda^2}\frac{1}{1 + \max(1, \frac{\alpha}{\lambda})^6\|\mathbf{v}_{\eta_T, \lambda, 0}\|_{L^2(\Phi_2 dx)}^4}\right). \end{aligned} $$

We have

$$ \lambda^2(1 + \|\mathbf{v}_{\eta_T, \lambda, 0}\|_{L^2(\Phi_2 dx)}^4) = \lambda^2 + \left(\int |\mathbf{v}_{0, \eta_T}(x)|^2 \frac{1}{1+|x|^2} \frac{\lambda^2 + \lambda^2|x|^2}{1+\lambda^2|x|^2} dx\right)^2. $$

For every $x \in \mathbb{R}^3$, we have

$$ \sup_{0 < \lambda < 1} \frac{\lambda^2 + \lambda^2 |x|^2}{1 + \lambda^2 |x|^2} = 1 \text{ and } \lim_{\lambda \to 0} \frac{\lambda^2 + \lambda^2 |x|^2}{1 + \lambda^2 |x|^2} = 0. $$

Thus, by dominated convergence, we get that

$$ \lim_{\lambda \to 0} \lambda^2 (1 + \|\mathbf{v}_{\eta_T, \lambda, 0}\|_{L^2(\Phi_2 dx)}^4) = 0. $$

We take a $\lambda_T \in (01)$ such that $\lambda_T^2(1 + \|\mathbf{v}_{\eta_T, \lambda_T, 0}\|_{L^2(\Phi_2 dx)}^4) < \frac{1}{C_2T}$ and we get that $T_{\epsilon, \alpha} = \frac{1}{\lambda_T^2} T_{\lambda_T, \epsilon, \alpha} \geq \max(1, \frac{\alpha}{\lambda_T})^6 T$.

In particular, we have $T_{\epsilon, \alpha} > T$ provided that $\alpha < \alpha_{T, \mathbf{u}_0} = \frac{1}{\lambda_T}$.

With those values of $\eta_T$ and $\lambda_T$, writing $\mathbf{b}_{\epsilon, \alpha, T} = \mathbf{b}_{\eta_T, \epsilon, \alpha}$ and $\mathbf{v}_{\epsilon, \alpha, T}(t, x) = \mathbf{v}_{\eta_T, \epsilon, \alpha}(t, x) = \lambda_T \mathbf{v}_{\eta_T, \lambda_T, \epsilon, \alpha}(\lambda_T^2 t, \lambda_T x)$, we use Proposition 3 and Proposition 4 to get the following estimates on the solution $\mathbf{u}_{\epsilon, \alpha}$ on $(0, T)$:

* $\sup_{0 \leq t \leq T} \|\mathbf{b}_{\epsilon, \alpha, T}(t, \cdot)\|_r \leq 2 \|\mathbf{b}_{0, \eta_T}\|_r$,

* $\sup_{0 < t \leq T} t^{\frac{1}{2}} \|\nabla \otimes \mathbf{b}_{\epsilon, \alpha, T}(t, \cdot)\|_r \leq 2 \|\nabla W_1\|_1 \|\mathbf{b}_{0, \eta_T}\|_r$,

* $\sup_{0 < t \leq T} t^{\frac{3}{2r}} \|\mathbf{b}_{\epsilon, \alpha, T}(t, \cdot)\|_{\infty} \leq C_{3,T, \mathbf{u}_0} \|\mathbf{b}_{0, \eta_T}\|_r$,
• $\sup_{0 \le t \le T} \|\mathbf{v}_{\epsilon, \alpha, T}(t, \cdot)\|_{L^2(\Phi_2 \, dx)} \le \frac{2}{\lambda_T} \|\mathbf{v}_{0, \eta_T}\|_{L^2(\Phi_2 \, dx)}$,

• $\|\mathbf{v}_{\epsilon, \alpha, T}(t, \cdot)\|_{L^2((0,T)(H^1(L^2(\Phi_2 \, dx)))} \le \frac{C_2}{\lambda_T} \|\mathbf{v}_{0, \eta_T}\|_{L^2(\Phi_2 \, dx)}$.

(where we use the inequality

$$ \lambda \|f\|_{L^2(\Phi_2 \, dx)}^2 \le \int \frac{1}{\lambda^2} |f(\frac{x}{\lambda})|^2 \Phi_2(x) \, dx = \int |f(x)|^2 \frac{\lambda}{1 + |\lambda x|^2} \, dx \le \frac{1}{\lambda} \|f\|_{L^2(\Phi_2 \, dx)}^2 $$

for $0 < \lambda \le 1$).

In particular, we have

• $\|\mathbf{u}_{\epsilon, \alpha}(t, \cdot)\|_{L^2(\Phi_2 \, dx)} \le \|\mathbf{v}_{\epsilon, \alpha, T}(t, \cdot)\|_{L^2(\Phi_2 \, dx)} + \|\mathbf{b}_{\epsilon, \alpha, T}(t, \cdot)\|_r$, so that

$$ \sup_{0 < \epsilon, 0 < \alpha < \alpha_T} \sup_{0 < t < T} \|\mathbf{u}_{\epsilon, \alpha}(t, \cdot)\|_{L^2(\Phi_2 \, dx)} \le C_{4,T, \mathbf{u}_0} $$

• We have

$$ \begin{aligned} \|\Phi^2 \mathbf{b}_{\epsilon, \alpha, T}\|_{H^{1/2}} &\le \sqrt{\|\Phi^2 \mathbf{b}_{\epsilon, \alpha, T}\|_2 \|\Phi^2 \mathbf{b}_{\epsilon, \alpha, T}\|_{H^1}} \\ &\le \sqrt{\|\mathbf{b}_{\epsilon, \alpha, T}\|_{L^2(\Phi_4 \, dx)} \|\mathbf{b}_{\epsilon, \alpha, T}\|_{L^2(\Phi_4 \, dx)} + \|\nabla \otimes \mathbf{b}_{\epsilon, \alpha, T}\|_{L^2(\Phi_4 \, dx)}} \\ &\le C \sqrt{\|\mathbf{b}_{\epsilon, \alpha, T}\|_r \sqrt{(\|\mathbf{b}_{\epsilon, \alpha, T}\|_r + \|\nabla \otimes \mathbf{b}_{\epsilon, \alpha, T}\|_r} \\ &\le C'(1 + \sqrt{C_{3,T, \mathbf{u}_0}} t^{-\frac{3}{4r}}) \|\mathbf{b}_{0, \eta_T}\|_r. \end{aligned} $$

As

$$ \begin{aligned} &\|\mathbf{u}_{\epsilon, \alpha}(t, \cdot)\|_{L^2((0,T), H^{1/2}(\Phi_4 \, dx))} \\ &\le \|\mathbf{v}_{\epsilon, \alpha}(t, \cdot)\|_{L^2((0,T), H^1(\Phi_2 \, dx))} + \|\mathbf{b}_{\epsilon, \alpha}(t, \cdot)\|_{L^2((0,T), H^{1/2}(\Phi_4 \, dx))}, \end{aligned} $$

we get that

$$ \sup_{0 < \epsilon, 0 < \alpha < \alpha_T} \|\mathbf{u}_{\epsilon, \alpha}(t, \cdot)\|_{L^2((0,T), H^{1/2}(\Phi_2 \, dx))} \le C_{4,T, \mathbf{u}_0} $$

• From the proof of Theorem 2, we see that $\mathbb{P} \operatorname{div}((\varphi_{\epsilon} * (\theta_{\alpha} \mathbf{u}_{\epsilon, \alpha})) \otimes \mathbf{u}_{\epsilon, \alpha})$ and $\Delta \mathbf{u}_{\epsilon, \alpha}$ are bounded in $L^2((0,T), H^{-4}(\Phi_8 \, dx))$. Thus, we have

$$ \sup_{0 < \epsilon, 0 < \alpha < \alpha_T} \|\partial_t \mathbf{u}_{\epsilon, \alpha}(t, \cdot)\|_{L^2((0,T), H^{-4}(\Phi_8 \, dx))} \le C_{4,T, \mathbf{u}_0}. $$

Theorem 4 is proved. □
# 4 Solutions to the Navier-Stokes equations.

In this final section, we prove Theorem 1. The key tool for going from mollified equations to Navier-Stokes equations will be the following lemma (a simpler variant of the Aubin-Lions theorem):

**Lemma 8 (Rellich-Lions lemma).** Let $\sigma < 0 < s$, $0 < T < +\infty$ and $R > 0$. Let $(u_n)_{n \in \mathbb{N}}$ be a sequence of functions on $(0, T) \times \mathbb{R}^3$ such that $u_n$ is bounded in $L^2((0, T), H^s)$ and $\partial_t u_n$ is bounded in $L^2((0, T), H^\sigma)$ and $u_n(t, x) = 0$ for $|x| > R$. Then there exists a subsequence $(u_{n_k})_{k \in \mathbb{N}}$ and $u_\infty \in L^2((0, T) \times \mathbb{R}^3)$ such that $u_{n_k}$ is strongly convergent to $u_\infty$ in $L^2((0, T), L^2)$.

*Proof.* Extend $u_n$ to $\mathbb{R} \times \mathbb{R}^3$ in the following way: let $\omega \in \mathcal{D}(\mathbb{R})$ with $\omega(t) = 1$ for $|t - \frac{T}{2}| \le \frac{3T}{4}$ and $\omega(t) = 0$ for $|t - \frac{T}{2}| \ge \frac{7T}{8}$. Define $v_n$ on $\mathbb{R} \times \mathbb{R}^3$ by $v_n(t, x) = u_n(t, x)$ if $0 \le t \le T$, $\omega(t)u_n(-t, x)$ if $-T \le t \le 0$, $u_n(2T - t)$ if $T \le t \le 2T$ and $= 0$ if $t \notin [-T, 2T]$. Then $v_n$ is bounded in $L^2(\mathbb{R}, H^s)$ and $\partial_t v_n$ is bounded in $L^2(\mathbb{R}, H^\sigma)$. Taking the Fourier transform on $\mathbb{R} \times \mathbb{R}^3$ defined by

$$ \check{F}(\tau, \xi) = \iint F(t, x)e^{-i(t\tau + x \cdot \xi)} dt dx, $$

we find that $(1 + |\xi|^2)^{s/2} \check{v}_n$ is bounded in $L^2(\mathbb{R} \times \mathbb{R}^3)$ and $\tau(1 + |\xi|^2)^{\sigma/2} \check{v}_n$ is bounded in $L^2(\mathbb{R} \times \mathbb{R}^3)$. As $\sigma < s$, $(1 + |\tau|^2)^{1/2}(1 + |\xi|^2)^{\sigma/2} \check{v}_n$ is bounded in $L^2(\mathbb{R} \times \mathbb{R}^3)$. Let $s_0 = \frac{s}{1-\sigma+s}$; we have $0 < s_0 < 1$ and $s_0 = (1 - s_0)s + s_0\sigma$, so that

$$ \begin{aligned} (1 + \tau^2 + |\xi|^2)^{s_0/2} &\le (1 + \tau^2)^{s_0/2} (1 + |\xi|^2)^{s_0/2} \\ &= ((1 + \tau^2)^{1/2} (1 + |\xi|^2)^{\sigma/2})^{s_0} (1 + |\xi|^2)^{s/2} \end{aligned} $$

and thus $(1 + \tau^2 + |\xi|^2)^{s_0/2} \check{v}_n$ is bounded in $L^2(\mathbb{R} \times \mathbb{R}^3)$, or equivalently $v_n$ is bounded in $H^{s_0}(\mathbb{R} \times \mathbb{R}^3)$. As $s_0 > 0$ and as all the $v_n$ are supported in the compact set $[-T/2, 2T] \times \bar{B}(0, R)$, we may apply Rellich's theorem and get that there exists a subsequence $(v_{n_k})_{k \in \mathbb{N}}$ that is strongly convergent in $L^2(\mathbb{R} \times \mathbb{R}^3)$. The subsequence $(u_{n_k})_{k \in \mathbb{N}}$ is then strongly convergent in $L^2((0, T), L^2(\mathbb{R}^3))$. $\square$

Theorem 1 will then be proved in the following way:

**Proposition 5.** There exists a sequence $(\epsilon_n, \alpha_n)_{n \in \mathbb{N}}$ and a vector field $\mathbf{u}$ such that:

* $\lim_{n \to +\infty} \epsilon_n = \lim_{n \to +\infty} \alpha_n = 0$,

* $\mathbf{u}_{\epsilon_n, \alpha_n}$ is strongly convergent to $\mathbf{u}$ in $L^2((0, T), \Phi_4 dx)$ for every $T > 0$.
Moreover, $\mathbf{u}$ is a solution on $(0, +\infty) \times \mathbb{R}^3$ of the Navier-Stokes equations

$$ egin{cases} 
\partial_t \mathbf{u} = \Delta \mathbf{u} - \mathbb{P} \operatorname{div}(\mathbf{u} \otimes \mathbf{u}) \\ 
\operatorname{div} \mathbf{u} = 0 \\ 
\lim_{t \to 0} \mathbf{u}(t, \cdot) = \mathbf{u}_0 
\end{cases} 
\quad (19) $$

and, for every $0 < T < +\infty$, we have $\mathbf{u} \in L^\infty((0,T), L^2(\frac{dx}{1+|x|^2})) + L^\infty((0,T), L^r)$.

*Proof.* Let $\mathbb{N}^2$ be enumerated as $\{(j_n, k_n) / n \ge 1\}$. We start with a sequence $(\epsilon_{0,n}, \alpha_{0,n})_{n \in \mathbb{N}}$ which converges to $(0,0)$. We choose sequences $(\epsilon_{q,n}, \alpha_{q,n})_{n \in \mathbb{N}}$ by induction on $q$, such that the sequence $(\epsilon_{q+1,n}, \alpha_{q+1,n})_{n \in \mathbb{N}}$ will be a subsequence of the sequence $(\epsilon_{q,n}, \alpha_{q,n})_{n \in \mathbb{N}}$.

Assume that we have chosen the sequence $(\epsilon_{q,n}, \alpha_{q,n})_{n \in \mathbb{N}}$ for some $q \ge 0$. We take $\omega_{q+1} \in \mathcal{D}(\mathbb{R}^3)$ such that $\omega_{q+1}(x) = 1$ for $|x| \le R_{q+1} = 2^{k_{q+1}}$ and $= 0$ for $|x| > 2R_{q+1}$. Let $T_{q+1} = 2^{j_{q+1}}$. For $\alpha_{q,n} < \alpha_{T_{q+1}, \mathbf{u}_0}$, $\omega_{q+1}\mathbf{u}_{\epsilon_{q,n}, \alpha_{q,n}}$ is bounded in $L^2((0, T_{q+1}), H^{1/2})$ and supported in $\{x \in \mathbb{R}^3 / |x| \le 2R_{q+1}\}$, while $\partial_t(\omega_{q+1}\mathbf{u}_{\epsilon_{q,n}, \alpha_{q,n}})$ is bounded in $L^2((0, T_{q+1}), H^{-4})$. We may apply Lemma 8 and choose a subsequence $(\epsilon_{q+1,n}, \alpha_{q+1,n})_{n \in \mathbb{N}}$ such that $\omega_{q+1}\mathbf{u}_{\epsilon_{q+1,n}, \alpha_{q+1,n}}$ is strongly convergent in $L^2((0, T_{q+1}), L^2)$ and thus $\mathbf{u}_{\epsilon_{q+1,n}, \alpha_{q+1,n}}$ is strongly convergent in $L^2((0, T_{q+1}) \times B_{R_{q+1}})$.

We then use Cantor's diagonal argument and define $(\epsilon_n, \alpha_n) = (\epsilon_{n,n}, \alpha_{n,n})$. We have

* $\lim_{n \to +\infty} \epsilon_n = \lim_{n \to +\infty} \alpha_n = 0$,

* $\mathbf{u}_{\epsilon_n, \alpha_n}$ is strongly convergent in $L^2((0, T), L^2(B_R))$ for every $T > 0$ and every $R > 0$.

With similar arguments, we can grant that, for any $T_k$, we have that $\mathbf{u}_{\epsilon_n, \alpha_n} = \mathbf{v}_{\epsilon_n, \alpha_n, T_k} + \mathbf{b}_{\epsilon_n, \alpha_n, T_k}$ on $(0, T_k) \times \mathbb{R}^3$ with the strong convergence of $\mathbf{v}_{\epsilon_n, \alpha_n, T_k}$ to $\mathbf{v}_{T_k}$ and of $\mathbf{b}_{\epsilon_n, \alpha_n, T_k}$ to $\mathbf{b}_{T_k}$ in $L^2((0, T_k), L^2(B_R))$ for every $R > 0$.

Let $\mathbf{u}$ be the limit of $\mathbf{u}_{\epsilon_n, \alpha_n}$. We have, for $R > 0$ and $\chi_R$ the characteristic function of the ball $B_R$ (and for $\alpha_n < \alpha_{T_k, \mathbf{u}_0}$),

$$ egin{align*} 
[\bullet] \|\mathbf{b}_{T_k} - \mathbf{b}_{\epsilon_n, \alpha_n, T_k} \|_{L^2((0, T_k), L^2(\Phi_4 dx))} \\ 
&\le \frac{1}{R^{\frac{1}{4}}} \|\mathbf{b}_{T_k} - \mathbf{b}_{\epsilon_n, \alpha_n, T_k} \|_{L^2((0, T_k), L^2(\Phi_{7/2} dx))} + \|\chi_R(\mathbf{b}_{T_k} - \mathbf{b}_{\epsilon_n, \alpha_n, T_k}) \|_{L^2((0, T_k), L^2)} \\ 
&\le \frac{C}{R^{\frac{1}{4}}} \|\mathbf{b}_{T_k} - \mathbf{b}_{\epsilon_n, \alpha_n, T_k} \|_{L^2((0, T_k), L^r)} + \|\mathbf{b}_{T_k} - \mathbf{b}_{\epsilon_n, \alpha_n, T_k} \|_{L^2((0, T_k) \times B_R)} 
\end{align*} $$

As $\|\mathbf{b}_{T_k} - \mathbf{b}_{\epsilon_n, \alpha_n, T_k}\|_{L^\infty((0, T_k), L^r)} \le C_{T_k, \mathbf{u}_0}$, we find that

$$ \limsup_{n \to +\infty} \|\mathbf{b}_{T_k} - \mathbf{b}_{\epsilon_n, \alpha_n, T_k}\|_{L^2((0, T_k), L^2(\Phi_4 dx))} \le C C_{T_k, \mathbf{u}_0} \sqrt{T_k} \frac{1}{R^{\frac{1}{4}}}. $$
Letting $R$ go to $+\infty$, we find $\lim_{n \to +\infty} \|\mathbf{b}_{T_k} - \mathbf{b}_{\epsilon_n, \alpha_n, T_k}\|_{L^2((0,T_k), L^2(\Phi_4 dx))} = 0$.
[●] Similarly, we have

$$ \begin{aligned} & \|\mathbf{v}_{T_k} - \mathbf{v}_{\epsilon_n, \alpha_n, T_k}\|_{L^2((0,T_k), L^2(\Phi_4 dx))} \\ & \leq \frac{1}{R} \|\mathbf{v}_{T_k} - \mathbf{v}_{\epsilon_n, \alpha_n, T_k}\|_{L^2((0,T_k), L^2(\Phi_2 dx))} + \|\chi_R(\mathbf{v}_{T_k} - \mathbf{v}_{\epsilon_n, \alpha_n, T_k})\|_{L^2((0,T_k), L^2)} \end{aligned} $$

As $\|\mathbf{v}_{T_k} - \mathbf{v}_{\epsilon_n, \alpha_n, T_k}\|_{L^\infty((0,T_k), L^2(\Phi_2 dx))} \leq C_{T_k, \mathbf{u}_0}$, we find

$$ \lim_{n \to +\infty} \|\mathbf{v}_{T_k} - \mathbf{v}_{\epsilon_n, \alpha_n, T_k}\|_{L^2((0,T_k), L^2(\Phi_4 dx))} = 0. $$

[●] We remark that we have the inequalities

$$ \|\varphi_{\epsilon_n} * (\theta_{\alpha_n} f)\|_r \leq \|f\|_r, $$

$$ \|\varphi_{\epsilon_n} * (\theta_{\alpha_n} f)\|_{L^2(\Phi_2 dx)} \leq \|\mathcal{M}_f\|_{L^2(\Phi_2 dx)} \leq C \|f\|_{L^2(\Phi_2 dx)} \quad (\text{as } \Phi_2 \text{ is a Muckenhoupt weight}), $$

$$ \|\varphi_{\epsilon_n} * (\theta_{\alpha_n} f)\|_{L^2(B_R)} = \|\varphi_{\epsilon_n} * (\theta_{\alpha_n} \chi_{R+1} f)\|_{L^2(B_R)} \le \|f\|_{L^2(B_{R+1})} \quad (\text{for } \epsilon_n < 1), $$
so that we have

$$ \lim_{n \to +\infty} \|\varphi_{\epsilon_n} * (\theta_{\alpha_n}(\mathbf{b}_{T_k} - \mathbf{b}_{\epsilon_n, \alpha_n, T_k}))\|_{L^2((0,T_k), L^2(\Phi_4 dx))} = 0 $$

and

$$ \lim_{n \to +\infty} \|\varphi_{\epsilon_n} * (\theta_{\alpha_n}(\mathbf{v}_{T_k} - \mathbf{v}_{\epsilon_n, \alpha_n, T_k}))\|_{L^2((0,T_k), L^2(\Phi_4 dx))} = 0. $$

[●] Similarly, we have the inequalities

$$ \|f - \varphi_{\epsilon_n} * (\theta_{\alpha_n} f)\|_r = \|\varphi_{\epsilon_n} * ((1 - \theta_{\alpha_n})f)\|_r \le \|f\|_r, $$

$$ \|f - \varphi_{\epsilon_n} * (\theta_{\alpha_n} f)\|_{L^2(\Phi_2 dx)} \le \|\mathcal{M}_{(1-\theta_{\alpha_n})f}\|_{L^2(\Phi_2 dx)} \le C \|f\|_{L^2(\Phi_2 dx)}, $$

$$ \|f - \varphi_{\epsilon_n} * (\theta_{\alpha_n} f)\|_{L^2(B_R)} = \|\varphi_{\epsilon_n} * ((1 - \theta_{\alpha_n})\chi_{R+1} f)\|_{L^2(B_R)} \le \|(1 - \theta_{\alpha_n})\chi_{R+1} f\|_2 \quad (\text{for } \epsilon_n < 1). $$

As we have $(1 - \theta_{\alpha_n})\chi_{R+1} = 0$ for $\alpha_n < \frac{1}{R+1}$, we find that

$$ \lim_{n \to +\infty} \|\mathbf{b}_{T_k} - \varphi_{\epsilon_n} * (\theta_{\alpha_n} \mathbf{b}_{T_k})\|_{L^2((0,T_k), L^2(\Phi_4 dx))} = 0 $$

and

$$ \lim_{n \to +\infty} \|\mathbf{v}_{T_k} - \varphi_{\epsilon_n} * (\theta_{\alpha_n} \mathbf{v}_{T_k})\|_{L^2((0,T_k), L^2(\Phi_4 dx))} = 0. $$

Combining all those estimates, we find that for every $T > 0$

$$ \lim_{n \to +\infty} \|\mathbf{u} \otimes \mathbf{u} - (\varphi_{\epsilon_n} * (\theta_{\alpha_n} \mathbf{u}_{\epsilon_n, \alpha_n})) \otimes \mathbf{u}_{\epsilon_n, \alpha_n}\|_{L^1((0,T), L^2(\Phi_4 dx))} = 0. $$

From the proof of Theorem 2, we see that

$$ \lim_{n \to +\infty} \Delta \mathbf{u}_{\epsilon_n, \alpha_n} - \mathbb{P} \operatorname{div}((\varphi_{\epsilon_n} * (\theta_{\alpha_n} \mathbf{u}_{\epsilon_n, \alpha_n})) \otimes \mathbf{u}_{\epsilon_n, \alpha_n}) = \Delta \mathbf{u} - \mathbb{P} \operatorname{div}(\mathbf{u} \otimes \mathbf{u}) $$

in $L^1((0, T), H^{-4}(\Phi_8 dx))$. **u** is a solution of the Navier–Stokes equations (19). □
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