TITLE: Showing a set is immune (has no infinte r.e. subset)
QUESTION [0 upvotes]: Consider the set $SD = \{ e : \phi_e(0) \downarrow \wedge \forall j < e (\phi_j(0) \neq \phi_e(0) )\}$
Show that $SD$ has no infinite recursively enumerable subset.
I think the idea is to use the recursion theorem somehow to get a contradiction.  We want to be able to set $\phi_e(0)$ so that we can get $\phi_e(0) = \phi_i(0)$ for some $i$ greater than $e$.

REPLY [1 votes]: It is sufficient to show that $SD$ contains no infinite recursive subset $R \subseteq SD$. Assume there is such a subset $R$. Using recursion theorem define the following function
$$
\varphi_e(x) = \begin{cases}
\uparrow,& \text{if } e \in R,\\
\varphi_n(x),& \text{where } n = \mu t(t \in R \wedge t > e),
\text{ if } e \notin R.
\end{cases}
$$
If $e \in R \subseteq SD$, then $\varphi_e(0)\!\uparrow$ and hence $e \notin SD$, a contradiction. It means $e \notin R$, but in this case $\varphi_e(0) = \varphi_t(0)\!\downarrow$ for some $t \in R \subseteq SD$ and $t > e$, contradicting $t \in SD$.


TITLE: Applying a constant force on pendulum and integrating for velocity
QUESTION [1 upvotes]: We apply a constant horizontal force $ F $ to a weight that's connected to a solid rod. I'd like to find out the velocity of the weight when it's at its horizontal position. See image below.

My attempt at solving it:
$$ {dv\over dt} = {F(\theta) \over m} $$
$$ {dv\over ds}{ds\over dt} = {F(\theta) \over m} $$
$$ {dv}{ds\over dt} = {F(\theta) \over m} ds $$
$$ {dv}\cdot v = {r \over m} F(\theta) d\theta $$
$$ {dv}\cdot v = {rF \over m} \cos(\theta) d\theta $$
Solving the integral I get
$$ {v^2 \over 2} = {rF \over m} \sin(\pi/2) = {rF \over m} $$
I'm pretty sure I messed up somewhere because this give's the wrong answer.
Any help is appreciated. Thanks!
Edit: I just realized I didn't take into account $ g $ that's working against us!

REPLY [1 votes]: You forgot the $mgL$ term in your setup. It should be
$$\frac{dv}{dt}=\frac{F\cos\theta}{m}-g\sin\theta$$
Then do the same thing you will get the same answer as the one mentioned by grdgfgr.


TITLE: Determining transformation matrix of $f\colon V \rightarrow V, p \mapsto f(p) := p'+cp$
QUESTION [0 upvotes]: Let $V$ be the vector-space of the polynomial-functions $\in \mathbb{R}$ with $\deg \le n$ and $f \in \text{End}(V)$ with $c \in \mathbb{R}$:
$f\colon V \rightarrow V, p \mapsto f(p) := p'+cp$
How can I determine the transformation matrix of $f$ with respect to the monome-basis $B=(x^0,x,\dots,x^n)$?

REPLY [3 votes]: In order to determine the matrix associated to $f$ with respect to $B$, you have to evaluate $f(x^i)$ for any element $x^i$ of $B$. More precisely, $$f(x^i)=ix^{i-1}+cx^i=q_i(x)$$
For any $i$, the polynomial $q_i(x)$ has coordinates $(0,\dots,\underbrace{i}_{i^{th}\text{ entry}},\underbrace{c}_{(i+1)^{th}\text{ entry}},\dots,0)=v_i$.
By definition, the $i^{th}$ column of the matrix associated to $f$ coincide with $v_i$. Hence the matrix associated to $f$ with respect to $B$ is
$$
\begin{pmatrix}
c & 1 & \dots & \dots & \dots & 0\\
0 & c & 2 & \dots & \dots &0\\
0 & 0 & c & \dots & \dots & 0\\
\dots &\dots & \dots &\dots & \dots &\dots\\
\dots &\dots & \dots &\dots & c &n\\
\dots &\dots & \dots &\dots & \dots &c
\end{pmatrix}
$$


TITLE: Free Group Proofs
QUESTION [1 upvotes]: I am completely new to free groups and I have no idea how to solve this, so if anyone could help that would be amazing. Thanks! The problem goes as follows:
Considering the diagram:

Let $S=\{x,y,z\}$ and let $G$ be the free group $G=F\{a,b\}$. Let $\phi$ be the map $x\mapsto a^2, y\mapsto b^2, z\mapsto ab$. Then the universal property says $\Phi$ is a unique homomorphism $F\to G$ with $\Phi \circ i=\phi$. Prove:
$a)$ $\Phi:F\to G$ is injective
$b)$ the image $\Phi(F)$ is a normal subgroup of $G$
$c)$ $G/\Phi(F) \cong \Bbb Z/2$
Thanks a lot!

REPLY [1 votes]: Though there may be slick proofs involving just the universal properties themselves, it's often helpful to see what these constructions concretely are.

$F$ is the free group on 3 elements; it consists of all possible words made of $x$, $y$, $z$, their inverses, and a unit. For example, $xy^{-1}zyz^{-1}z^{-1}$ is a word in $F$.
$i : 3 \hookrightarrow F$ sends the letters $x, y, z$ to their corresponding "one-letter words" in $F$.
$G$ is the free group on 2 elements; it consists of all possible words made of $a$ and $b$, their inverses, and a unit.
$\varphi : 3\rightarrow G$ sends the letters $x,y,z$ to the words $aa$, $bb$, and $ab$ respectively.
The unique function $\Phi$ must exist because of $F$'s universal property. Hence $\Phi: F\xrightarrow{!}G$ sends words made out of $x$, $y$, $z$ (etc.) into words made out of $a$ and $b$ (etc.) by systematically replacing all instances of $x$ with $aa$, $y$ with $bb$, $z$ with $ab$, and similarly for their inverses.
(In general, the free group property means that if you have a function $\varphi$ sending letters in $L$ to a group $G$, there's a unique function $\Phi$ which extends $\varphi$ to words made of letters in $L$ (i.e. the free group on $L$). That unique function $\Phi$ applies $\varphi$ to each of the letters in the word, then multiplies the results together.)

Proving that $\Phi$ is injective:

Intuitively, you can tell that the translation process mentioned above is uniquely reversible: given a $\Phi$-made word like $aab^{-1}b^{-1}ab$ in $G$, you can find the unique string $xy^{-1}z$ in $F$ that produced it.
You could formalize this process to show that $\Phi$ is injective. Alternatively, another approach is to show that $\Phi$ sends nothing to the identity of $G$ except for the identity of $F$ itself. ("$\Phi$ has a trivial kernel").

Proving that $\Phi(F)$ is a normal subgroup of $G$:

A normal subgroup is one that is closed under conjugation by members of $G$.
Because $G$ is a free group, it is made up of words of $a$ and $b$ and their inverses and unit. Hence we only need to show that it's closed under conjugation by these one-letter words and their inverses.

Proving that $G/\Phi(F)$ is the cyclic group of order 2:

You know that $\Phi(F)$ contains, for example, $aa$, $bb$, and $ab$. When you quotient by a normal subgroup, you cause every member of that subgroup to become equal to the identity.
Note that $\Phi(F)$ also contains $ba$, because $\Phi(yz^{-1}x)=\Phi(y)\Phi(z^{-1})\Phi(x) = bb(ab)^{-1}aa = bbb^{-1}a^{-1}aa = ba$.
Hence in $G/\Phi(F)$, we have that $aa = 1$, $bb = 1$, $ab = 1$, $ba=1$, among other things. Hence all higher powers of $a$ and $b$ are equal to the identity, and $a$ and $b$ are inverses of each other. The group consists of just $\{1, a\}$, and has the same multiplication structure as $\mathbb{Z}/2$.


TITLE: Expected value of distance between min and max of independent events with uniform distribution
QUESTION [1 upvotes]: Let $U_1, \cdots , U_5$ be independent, each with uniform distribution on (0, 1).  Let R be the distance between the minimum and maximum of the $U_i's$.  Find E(R).
I think R might have a Beta distribution.  I really don't know how to solve this.
Thanks!

REPLY [1 votes]: Let $A$ be the minimum, $B$ be the maximum.
Then $R=B-A$.
Thus $E(R)=E(B)-E(A)$.
It is easy to find $E(A)=\int_0^1 5y(1-y)^4 dy=1/6$,  and $E(B)=\int_0^1 5y^5 dy=5/6$
Hence $E(R)=2/3$.


TITLE: Doubt on a simple proof exercise about operators between normed spaces.
QUESTION [1 upvotes]: Exercise. Let $T$ be a bounded linear operator from a normed space $X$ onto a normed space $Y$. Show that if there is a constant $c>0$ such that
$$ \|Tx\| \geqslant c\|x\|, \quad \forall x \in X, $$
then $T$ admits inverse, $T^{-1}$, and it is bounded.
My attempt. To see that $T$ admits inverse, it suffices to show that $\mathcal{N}(T) = \{0_X\},$ where $\mathcal N(T)$ stands for the null space of $T$ (or kernell, depending on the literature). The inclusion $\{0_X\} \subset \mathcal N(T)$ is obvious since $T(0_X) = 0_Y,$ in every scenario. On the other hand, let $x \in X$ be an arbitrary element such that $\mathcal x \in \mathcal N(T),$ i.e., $T(x) = 0_Y.$ Replacing this in the inequality from the exercise, it follows that
$$ \|T(x)\| = 0 \geqslant c\|x\| \Leftrightarrow \|x\| \leqslant 0. $$
By properties of the norm, one knows that $\|x\| \geqslant 0,$ and so, it comes that $\|x\| = 0.$ Again by properties of the norm, it follows that $x = 0_X$, proving that $\mathcal N(T) \subset \{0_X\}.$ Thus, $T$ admits inverse.
Now, we must prove that $T^{-1}$ is bounded. Fix an arbitrary $y \in Y.$ Since $T$ is surjective, we can guarantee that there is some $x \in X$ such that $T(x) = y$ (or equivalently, $T^{-1}(y) = x.$ In particular, this element $x \in X$ satisfies the inequality provided by hypothesis in the exercise. Thus,
$$ \|Tx\| \geqslant c\|x\| \Leftrightarrow \|y\| \geqslant c\|T^{-1}y\| \Leftrightarrow \frac{1}{c}\|y\| \geqslant \|T^{-1}y\|. $$
Since $\frac{1}{c} > 0$ and our choice of $y$ was arbitrary, we might conclude that
$$ \| T^{-1}y \|  \leqslant \frac{1}{c}\|y\|, \quad \forall y \in Y,$$
i.e., $T^{-1}$ is bounded.
My concerns. I can't say where my proof is wrong (if it is) but at the same time I have a (REALLY) weird feeling: I never used the fact that $T$ is bounded, and I also enver used the fact that $T$ is linear during the whole proof. Can someone explain me what is the necessity of this conditions in the exercise. ?

REPLY [1 votes]: You did not use boundedness. In fact, your proof works for linear $T$ without boundedness.
As an example,
consider the following operator $T$, as a linear map from $c_{00}$ to $c_{00}$, where $c_{00}$ is the space of finite sequences (supplied with any $l^p$-norm):
$$
T(x) = (x_1, 2x_2, 3x_3, \dots).
$$
This $T$ is linear from $c_{00}$ to $c_{00}$, bijective, but unbounded. It satisfies your inequality with $c=1$,
and $T^{-1}$ is linear and bounded, it is given by
$$
T^{-1}(x) = (x_1, x_2/2, x_3/3, \dots).
$$


TITLE: Uniform Approximation to a Function
QUESTION [0 upvotes]: Corollary 6.17 in Lee's ISM states that certain smooth maps from manifolds to $\mathbb{R}^n$ can be "uniformly approximated by embeddings." Problem is, I don't know what uniform approximation means. My initial guess at a definition was the following:

Let $f : X \to Y$ be a map where $X$ is a set and $Y$ is a metric space, and let $\mathcal{C} = \{\phi_\alpha\}$ be a collection of functions $\phi_\alpha : X \to Y$. Then $f$ can be uniformly approximated by $\mathcal{C}$ if for all $\delta > 0$, there exists an $\alpha$ such that $|f(x) - \phi_\alpha(x)| < \delta$ for all $x \in X$.

I don't think this definition is right anymore, since Lee claims that projections in the directions of two distinct vectors can be arbitrarily close. This doesn't work with my tentative definition because if $v,w$ are two vectors in $\mathbb{R}^n$ that are not in $\mathbb{R}^{n-1}$ with the standard embedding of $\mathbb{R}^{n-1}$ into $\mathbb{R}^n$), then for all $\delta > 0$, we can find a point $x$ such that $|\pi_v(x) - \pi_w(x)| > \delta$ (where $\pi_v,\pi_w : \mathbb{R}^n \to \mathbb{R}^{n-1}$ are the projections in the directions of $v$ and $w$).
I see two possibilities here. Either (i) my definition actually is correct and the contradiction explained above can be recitifed by noting that the domains considered in the theorem are compact and thus bounded, or (ii) I guessed the wrong definition. How do we rigorously treat the idea of uniform approximation?

REPLY [2 votes]: Your possibility (i) is correct. Notice that Corollary 6.17 applies only to compact manifolds. If $M$ is a compact manifold embedded in $\mathbb R^n$, then $\sup_{x\in M} |\pi_v(x) - \pi_w(x)|$ can be made as small as desired by choosing $v$ sufficiently close to $w$.


TITLE: Proof of $n\ge 6\implies 2n−8\le n^2−8n+ 16$ using induction.
QUESTION [2 upvotes]: I have to do a proof by induction for this theorem:
For each $n\in \mathbb N$ such that $n\ge 6$ we have $$2n−8\le n^2−8n+ 16$$

Is this possible, or should I do a different type of proof. I am confused how to prove that $1\in S$  when $n\ge 6$, so I am thinking maybe I need to first prove $6\in S$ and then handle $n>6 $?
Thank you!

REPLY [3 votes]: The answer given by Shaun is excellent and works in general but notice here that you can factor both sides of $$2n-8\le n^2-8n+16$$
$$2(n-4)\le (n-4)(n-4)$$
and since $$n\ge 6$$ you could simplify this inequality. Do you see how to proceed ?


TITLE: Is there any square matrix with orthonormal rows but not orthonormal columns?
QUESTION [1 upvotes]: An orthogonal matrix necessarily has orthonormal columns, and orthonormal columns necessarily give an orthogonal matrix. Also, orthonormal columns imply orthonormal rows.
But how about the converse of the last statement? Meaning, do orthonormal rows necessarily imply orthonormal columns?
Thanks

REPLY [6 votes]: do orthonormal rows necessarily imply orthonormal columns?

For square matrices, yes. To see this, note that if $M$ has orthonormal rows but not orthonormal columns then the transpose $M^T$ has orthonormal columns but not orthonormal rows, which you know to be impossible.
For non-square matrices we can have orthonormal rows but not orthonormal columns or vice-versa. For example, consider the matrix
$$\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
\end{pmatrix}$$


TITLE: Radial component of velocity in a Rankine-vortex
QUESTION [1 upvotes]: A Rankine vortex is a model of a vortex that assumes an inner region ($r<R$) with constant vorticity, $\omega_0$, which rotates as a rigid solid, and an outer region ($r>R$) with zero vorticity, $\omega = 0$, which behaves as an irrotational flow.
I'm trying to derive the velocity field of this model in two dimensions through the expression of the vorticity in plane polar coordinates,
$$
\omega=\frac{1}{r} \frac{\partial}{\partial r}\left(r v_{\theta}\right)-\frac{1}{r} \frac{\partial v_{r}}{\partial \theta}
$$
I have seen that the radial component of the velocity field in this model is zero: $v_r=0$. Is this part of the definition of the model or a consequence of its postulates about the vorticity?

REPLY [2 votes]: Generally I'd say it is just defined by the velocity field $\boldsymbol{v}$, but even so you can derive it if you are willing to accept that it is cylindrically symmetric.
Cylindrical symmetry means that $\boldsymbol{v}$ must be independent of $\theta$ which gives
$$\omega=\frac{1}{r} \frac{\partial}{\partial r}\left(r v_{\theta}\right).$$
In the inner region, $\omega=\omega_0$ and so we need
$$v_{\theta}=\frac{1}{2}\omega_0r+\frac{C}{r}$$
where $C$ is a constant. However, the velocity must be finite as $r\rightarrow 0$ and so $C=0$. In the outer region, $\omega=0$ which gives $v_{\theta}\propto{1/r}$.
Now we just need to argue that $v_r$ is zero. We can do this by mass conservation. The mass flux through a cylinder of unit length and radius $r$ is $F=2\pi r v_r$ where we have used that $v_r$ is independent of $\theta$. $F$ must be the same for all $r$, so by considering $r\rightarrow0$ with $v_r$ finite, we see that $F=0$ and so $v_r=0$ for all $r$.


TITLE: Why can the hyperplanes in $\mathbb{R}^{n+1}$ be represented by means of affine functions $\mathbb{R}^{n+1} \to \mathbb{R}$?
QUESTION [0 upvotes]: Why can the hyperplanes in $\mathbb{R}^{n+1}$ can be represented by affine functions $\mathbb{R}^{n} \to \mathbb{R}$?
Intuitively, it is true. But still, it is not clear for me.
An $n$-dimensional hyperplane in $\mathbb{R}^{n+1}$ defined by $(b,\beta)\in \mathbb{R}^{n+1}$ can be represented as
$$\{x\in \mathbb{R}^n | \langle x, b \rangle = \beta \}.$$
I also know that the hyperplane can also be represented as an affine function
$$\mathbb{R}^n \ni x \mapsto f(x)= \langle b, x\rangle - \beta.$$
Can someone help me deduce the first expression from the second?
Thank you in advance

REPLY [1 votes]: We can represent a hyperplane $\Pi \in \mathbb{R}^{n + 1}$ via a pair $(b, \beta) \in (\mathbb{R}^{n + 1} - \{0\}) \times \mathbb{R}$. Explicitly, $\Pi$ is the set
$$\{ x \in \mathbb{R}^{n + 1} : \langle b, x \rangle = \beta\}.$$
On the other hand, this is precisely the zero set of the affine function $f: \mathbb{R}^{n + 1} \to \mathbb{R}$ defined by
$$x \mapsto \langle b, x \rangle - \beta.$$
Conversely, any affine function $f: \mathbb{R}^{n + 1} \to \mathbb{R}$ can be written as
$$x \mapsto b_1 x_1 + \cdots b_{n + 1} x_{n + 1} - \beta$$ for (unique) coefficients $b_1, \ldots, b_{n + 1}, \beta$. If $f$ is nonconstant, then $b := (b_1, \ldots b_n)$ is nonzero, and
$$x \mapsto \langle b, x\rangle - \beta$$
defines a hyperplane.


text:
\begin{document}

\title{   Leray's backward self-similar solutions to
		   the 3D Navier-Stokes equations in Morrey spaces}
		\author{Quansen Jiu\footnote{
School of Mathematical Sciences, Capital Normal University, Beijing 100048, P. R. China Email: jiuqs@cnu.edu.cn},\,\,\,\, Yanqing Wang\footnote{ Department of Mathematics and Information Science, Zhengzhou University of Light Industry, Zhengzhou, Henan  450002,  P. R. China Email: wangyanqing20056@gmail.com}\;~ and\, \, Wei Wei\footnote{Center for Nonlinear Studies, School of Mathematics, Northwest University, Xi'an, Shaanxi 710127, P. R. China Email: ww5998198@126.com } }
\date{}
\maketitle
\begin{abstract}
In this paper,
it is shown that there
  does not exist a non-trivial Leray's backward self-similar solution to the 3D Navier-Stokes equations with profiles in Morrey spaces $\dot{\mathcal{M}}^{q,1}(\mathbb{R}^{3})$ provided $3/2<q<6$, or in $\dot{\mathcal{M}}^{q,l}(\mathbb{R}^{3})$ provided $6\leq q<\infty$ and $2<l\leq q$.
  This  generalizes the   corresponding
results obtained by
  Ne\v{c}as-R\r{a}u\v{z}i\v{c}ka-\v{S}ver\'{a}k
   \cite[Acta. Math. 176 (1996)]{[NRS]} in $L^{3}(\mathbb{R}^{3})$, Tsai \cite[Arch. Ration. Mech. Anal. 143 (1998)]{[Tsai]} in $L^{p}(\mathbb{R}^{3})$ with $p\geq3$,
  Chae-Wolf \cite[Arch. Ration. Mech. Anal. 225 (2017)]{[CW]} in Lorentz spaces $L^{p,\infty}(\mathbb{R}^{3})$ with $p>3/2$, and Guevara-Phuc
\cite[SIAM J. Math. Anal. 12 (2018)]{[GP2]} in $\dot{\mathcal{M}}^{q,\f{12-2q}{3}}(\mathbb{R}^{3})$ with $12/5\leq q<3$ and in $L^{q, \infty}(\R^3)$ with $12/5\leq q<6$.

  \end{abstract}
	\noindent {\bf MSC(2000):}\quad 35B65, 35D30, 76D05 \\\noindent
	{\bf Keywords:} Navier-Stokes equations;   self-similar solutions; Morrey spaces

	\section{Introduction}
	\label{intro}
	\setcounter{section}{1}\setcounter{equation}{0}
		We study  the following   incompressible Navier-Stokes equations in   three-dimensional space
	\be\left\{\ba\label{NS}
	&u_{t} -\Delta  u+ u\cdot\ti
	u+\nabla \pi=0, ~~\Div u=0,\\
	&u|_{t=0}=u_0,
	\ea\right.\ee
	where $u $ stands for the flow  velocity field, the scalar function $\pi$ represents the   pressure.   The
	initial  velocity $u_0$ satisfies   $\text{div}\,u_0=0$.

In a seminal work \cite{[Leray]}, Leray introduced the backward self-similar solutions to construct  singular solutions to the 3D Navier-Stokes equations \eqref{NS}. The so-called backward self-similar solution is a weak solution $(u,\pi)$  of \eqref{NS} satisfying, for $a>0, T\in \mathbb{R},$
\be\ba
&u(x,t)=\wred{\f{1}{\sqrt{2a(T-t)}}U\bigg(\f{x}{\sqrt{2a(T-t)}}\bigg)}, \label{learysb}\\
&\pi(x,t)=\f{1}{ 2a(T-t)}\Pi\bigg(\f{x}{\sqrt{2a(T-t)}}\bigg),
\ea\ee
where $U=(U_{1},U_{2},U_{3})$ and $\Pi$ are defined in $\mathbb{R}^{3}$,
and the pair $(u(x,t),\pi(x,t))$ is defined in $\mathbb{R}^{3}\times(-\infty,T)$.
We obtain a singular solution at $t=T$ if $U\neq0$ and
\be \label{SNS}
	-\Delta  U+a U+a(y\cdot \nabla)U+U\cdot\nabla U+ \nabla \Pi=0  , ~~\Div U=0, ~~y\in \mathbb{R}^{3}.
	 \ee
The first breakthrough of backward self-similar solutions
was due to  Ne{\v{c}}as-R{\r u}{\v{z}}i{\v{c}}ka-{\v S}ver\'ak \cite{[NRS]}. They ruled out the existence of Leray's backward self-similar solutions to the 3D Navier-Stokes system if a weak solution $ U$ of equations \eqref{SNS}  is in $   L^{3}(\mathbb{R}^{3})$ (see Section 2 for the definition of weak solutions of \eqref{SNS}).
Subsequently, various  results involving non-existence of Leray's backward self-similar non-trivial solutions  were  obtained by
 Tsai in \cite{[Tsai]}. Precisely,  he
proved that Leray's backward self-similar solution is trivial under the condition that $U\in L^{p}(\mathbb{R}^{3})$ with $3<p<\infty$, and the solution $U\in L^{\infty}(\mathbb{R}^{3})$ in system \eqref{SNS} is a constant. In addition, Tsai's second result is that
 there does not exist a non-trivial solution of the form \eqref{learysb} if $u$ satisfy
the  local  energy inequality
\begin{equation}\label{local-energy}
\sup_{t<s<T} \int_{B_{x_0}( r)}\dfrac12|u(x, s)|^2dx+\int_{t}^{T}\int_{B_{x_0}( r)} |\nabla u(x,s)|^2dxds<\infty,
\end{equation}
for some ball $B_{x_0}( r)$ and some $t<T$. Here, $B_{x_0}( r)$ denotes the ball of
center $x_0$ and radius $r$.

Very recently, Chae-Wolf \cite{[CW]} and Guevara-Phuc \cite{[GP2]} independently made progress
in this direction. On one hand,
 Chae-Wolf \cite{[CW]} proved that
if $U$ belong to the Lorentz spaces $ L^{q, \infty}(\R^3)$ with $q>\f32$ or
 \be\label{CW1}
\|  U\|_{L^{q}(B_{y_{0}}(1))}+\|\nabla U\|_{L^{2}(B_{y_{0}}(1))} =o(|y_{0}|^{1/2}), ~~\text{as} ~~|y_{0}|\rightarrow\infty,
\ee
then $U$ must be identically zero. Roughly speaking, the proof in \cite{[CW]} relied on  $\varepsilon$-regularity criteria
without pressure at one scale, the decay at infinity in  Lorentz spaces  and Tsai's first result mentioned above.
On the other hand, Guevara-Phuc \cite{[GP2]} showed that there does not  exist a non-trivial solution under the condition that $U$ is in Morrey spaces $ \dot{\mathcal{M}}^{q,\f{12-2q}{3}}(\mathbb{R}^{3})$ with $12/5\leq q<3$, or $U\in L^{q, \infty}(\R^3)$ with $12/5\leq q<6$, or $U\in L^{6}(\mathbb{R}^{3})$.
Their arguments were based on  Riesz potentials in Morrey spaces, Sobolev spaces with negative indices and Tsai's second aforementioned result.

The definitions of relevant function spaces can be found in Section
2.
Notice that there holds the following embedding relation
\be\label{inclusion} L^{q}(\mathbb{R}^{3})\hookrightarrow L^{q,\infty}(\mathbb{R}^{3})\hookrightarrow \dot{\mathcal{M}}^{q,l}( \mathbb{R}^{3})\hookrightarrow\dot{\mathcal{M}}^{q,1}( \mathbb{R}^{3}),~ 1\leq l<q.\ee
This fact can be found in \cite{[CX]}.

The well-posedness of the 3D Navier-Stokes equations in Morrey spaces was studied in \cite{[MY],[Kato],[Taylor],[CX],[L]}.
Compared with the Lebesgue spaces $L^{q}(\mathbb{R}^{3})$ and  Lorentz spaces $L^{q,\infty}(\mathbb{R}^{3})$, $C_{0}^{\infty}(\mathbb{R}^{3})$ is not dense in
Morrey spaces. It is worth pointing out that Sawano \cite{[Sawano]} showed that $ \dot{\mathcal{M}}^{q,l_{2}}(\mathbb{R}^{n})$  is not dense in $\dot{\mathcal{M}}^{q,l_{1}} (\mathbb{R}^{n})$ if
 $1<l_{1}<l_{2}\leq q$. Based on this, the first objective of this paper is to generalize Guevara and Phuc's result in \cite{[GP2]} to general Morrey spaces. Our result is reformulated as
\begin{theorem}\label{the1.1}
Let $U\in W^{1,2}_{\rm loc}(\R^3)$ be a weak solution  of \eqref{SNS}. If
\be\label{cond1}
U\in \dot{\mathcal{M}}^{q,l}(\mathbb{R}^{3}) ~~ \text{with }~~ 2<l\leq q<\infty,
\ee
  then $U\equiv0$.
\end{theorem}
\begin{remark}
This theorem extends all the known results involving non-existence
of Leray's backward self-similar non-trivial solutions to the 3D
Navier-Stokes equations in \cite{[Tsai],[NRS],[GP2],[CW]}.
\end{remark}

 We follow the path of \cite{[GP2],[Tsai],[NRS]} to prove Theorem \ref{the1.1}. First, in contrast to  \cite{[GP2]}, we construct pressure $\Pi$ in the Morrey space directly under condition \eqref{cond1}
 without the application of
 Riesz potentials in Morrey spaces and dual spaces of Sobolev spaces.
 Second, as \cite{[GP2]}, our main target is   to  deduce  \eqref{local-energy} by deriving the  energy bound \eqref{enebound} in terms of  \eqref{cond1} from the local energy inequality \eqref{loc}.  The key tool in \cite{[GP2]} is to utilize the  duality between $W^{-1,2}(B)$ and the Sobolev space $W^{1,2}_0(B)$ to bound energy flux in local energy inequality \eqref{loc}, namely, for $12/5\leq q<3,$
  \be\ba\label{gpei}
  \left|\int_{T-R^{2}}^{T}\int_{B}(|u|^{2}+2\pi)u\cdot\nabla \phi dxdt\right|
 =&\left|\int_{T-R^{2}}^{T}\langle |u|^{2}+2\pi ,u\cdot\nabla \phi \rangle_{W^{-1,2}(B),W_{0}^{1,2}(B)} dt\right| \\
\leq&C\int_{T-R^{2}}^{T}\lambda^{2-\f6q}(t)\|U\|_{
\mathcal{\dot{M}}^{q,\f{12-2q}{3}}(\mathbb{R}^{3})}^{2} \|\nabla u\|_{L^{2}(B)}dt,
 \ea\ee
 where the Hardy-Littlewood-Sobolev inequality for Riesz potentials in Morrey spaces was used and $\lambda(t)=[2a(T-t)]^{-1/2}$. It is worth remarking that the  upper bound $q<3$  in \cite{[GP2]} comes from the  employment  of Hardy-Littlewood-Sobolev inequality, but this strategy breaks down in the case \eqref{cond1}. Here, we make full use of the Meyer-Gerard-Oru interpolation
inequality \eqref{mgoi} and the fact that Morrey spaces $\dot{\mathcal{M}}^{q,1}(\mathbb{R}^{3})$ can be embedded in Besov spaces $\dot{B}^{-\f{3}{q}}_{\infty,\infty}(\mathbb{R}^{3})$ for $1<q<\infty$. This helps us  control energy flux and the first term in the left hand side of \eqref{gpei} as follows
  $$\ba
&\int_{T-\f{R^{2}}{4}}^{T}\|u\|^{3}_{L^{3}(B(R/2))}dt
\\\leq& C \B(\sup_{T-R^{2}\leq t\leq T}\| u\|^{2}_{L^{2}(B(R))}+\int_{T-R^{2}}^{T}\|\nabla u\|^{2}_{L^{2}(B(R))}dt\B)^{\f32-\f{1}{\alpha}}
\B(\int_{T-\f{R^{2}}{4}}^{T}\lambda^{p-\f{3p}{q}}(t)\|U\|^{p}_{\dot{\mathcal{M}}^{q,1}(\mathbb{R}^{3})}dt\B)^{\f{2}{p\alpha}},
\ea$$
where $\alpha=2/p+3/q<2$.
Additionally, we apply the pressure decomposition  as \cite{[HWZ]}  to bound  the second term in the left hand side of \eqref{gpei}. To sum up, we get the following energy bound
\be\ba \label{enebound}
&\|u\|^2_{L^{\infty}(T-\f{R^{2}}{4},T;L^{2}(B(R/2)))}+\|\nabla u\|^2_{L^{2}(T-\f{R^{2}}{4},T;L^{2}(B(R/2)))} \\ \leq&    CR^{3-2\alpha}\B( \int_{T-R^{2}}^{T}\lambda^{p-\f{3p}q}(t)\|U\|^{p}
_{\mathcal{\dot{M}}^{q,1}(\mathbb{R}^{3})} dt\B) ^{\f2p} +  CR^{\f{6-5\alpha}{2-\alpha}} \B(\int_{T-R^{2}}^{T}\lambda^{p-\f{3p}q}(t)\|U\|^{p}
_{\mathcal{\dot{M}}^{q,1}(\mathbb{R}^{3})} dt\B)^{\f{4}{p(2-\alpha )}}\\& + CR^{1-\f6m} \B(\int_{T-R^{2}}^{T}\lambda^{2-\f{3}m}(t) \|\Pi\|
_{\mathcal{\dot{M}}^{m,1}(\mathbb{R}^{3})} dt \B)^{2}.
\ea\ee
This together with \eqref{local-energy} means Theorem \ref{the1.1}.

It should be mentioned that the extra restriction that $q>2$ and $l>2$ in   \eqref{cond1} resulted from the construction process of pressure $\Pi=\mathcal{R}_i \mathcal{R}_j(U_{i}U_{j})$ in the proof of Theorem \ref{the1.1}. Partially motivated by Chae-Wolf in \cite{[CW]}, our next target is to utilize local suitable weak solutions (see Definition \ref{defilsw}) to remove this restriction.
\begin{theorem}\label{the1.4}
Let $U\in W^{1,2}_{\rm loc}(\R^3)$ be a weak solution  of \eqref{SNS}. If
\be\label{cond1.4}
U\in \dot{\mathcal{M}}^{q,1}(\mathbb{R}^{3}) ~~ \text{with }~~ \f32<q<6,
\ee
  then $U\equiv0$.
\end{theorem}
\begin{remark}
According to \eqref{inclusion}, this theorem is an improvement of corresponding results in \cite{[Tsai],[NRS],[GP2],[CW]}.
\end{remark}
The proof of Theorem \ref{the1.4} is based on a combination of techniques from \cite{[Wolf1],[Tsai],[CW]}. Our starting point is to set up a new Caccioppoli type inequality
 below
\be\ba\label{Caccioppoli inequ}
&\|u\|^{2}_{L^{3}(T-\f{R^{2}}{4},T;L^{\f{18}{5}}(B(\f{R}{2})))}+ \|\nabla u\|^{2}_{L^{2}(T-\f{R^{2}}{4},T;L^{2}(B(\f{R}{2})))}\\
\\
\leq &
  CR^{\f{5q-12}{3q}}\| u\|^{2}_{L^{\f{6q}{2q-3}}(T-R^{2},T;\dot{\mathcal{M}}^{q,1}(B(R)))}
  \\&+CR^{\f{4q-15}{2q-3}}\| u\|^{\f{6q}{2q-3}}_{L^{\f{6q}{2q-3}}(T-R^{2},T;\dot{\mathcal{M}}^{q,1}(B(R)))}
  +CR^{\f{2q-6}{q}}\| u\|^{3}_{L^{\f{6q}{2q-3}}(T-R^{2},T;\dot{\mathcal{M}}^{q,1}(B(R)))}.
\ea\ee
This inequality is derived by local suitable weak solutions to the 3D Navier-Stokes equations \eqref{NS} and
  the aforementioned Meyer-Gerard-Oru interpolation
inequality. Various  Caccioppoli type inequalities were recently established in
 \cite{[JWZ],[CW],[HWZ],[Wolf1],[Wolf2]}.
 All the proofs  rest on
  local suitable weak solutions originated in
  \cite {[Wolf1],[Wolf2]}
  by Wolf.
It is known that any usual suitable weak solution to the Navier-Stokes system enjoys the
local energy inequality \eqref{wloc1} (see \cite[Appendix A, p.1372]{[CW2]}).
The novelty of local suitable weak solutions is that, as stated in \cite{[Wolf1],[Wolf2],[CW]}, the  relevant  local energy
inequality \eqref{wloc1} removed   the non-local effect of the pressure term.
 Based on this, Caccioppoli type inequality \eqref{Caccioppoli inequ} and \eqref{cond1.4}  allow us to derive that
$$
\|\nabla U\|_{L^{2}(B_{y_{0}}(1))}=o(|y_{0}|^{1/2}) ~~~\text{and}~~~
\|  U\|_{L^{3}(B_{y_{0}}(1))}=o(|y_{0}|^{2/3}), ~~\text{as} ~~|y_{0}|\rightarrow\infty.
$$
  However, in light of \eqref{CW1}, this is not enough to show that $U\equiv0$. Notice that \eqref{Caccioppoli inequ} implies $\int_{\mathbb{R}^{3}}|U|^{3}|y|^{-2}dy<\infty$, our approach to overcome this difficulty is to construct the pressure $\Pi$ via $A_p$ weighted inequalities for singular integrals, which enables us to obtain $\pi\in L^{3/2}_{\rm loc}$ in terms of \eqref{learysb}. A similar argument has been used by Tsai in \cite{[Tsai],[Tsai2]} for the proof of his second result \eqref{local-energy}.  Adopting this approach together with the local energy inequality \eqref{wloc1}, we deduce \eqref{local-energy} and finally prove Theorem \ref{the1.4}.

  Roughly, the following four figures summarize known results about non-existence of Leray's
backward self-similar non-trivial solutions in the framework of Morrey spaces $\dot{\mathcal{M}}^{q,p}(\mathbb{R}^{3})$.

\begin{tikzpicture}[scale=.6,>= stealth]
\node[above] at (4.2,4.6) {\tiny{$p=q$}};
\node[below] at (3.5,-1) {\tiny{Figure 1:  Region of Ne{\v{c}}as-R{\r u}{\v{z}}i{\v{c}}ka-{\v S}ver\'ak, \\ Tsai and Chae-Wolf}};
\draw[->] (0,0)--(7,0)node[right]{\tiny{$q$}};
\foreach \x in {1,...,6}
     		\draw (\x,0) -- (\x,-.1)
		node[anchor=north] {\tiny{\x}};
\draw[->](0,0)--(0,7)node[above]{\tiny{$p$}};
\foreach \y in {1,...,6}
     		\draw (0,\y) -- (.1,\y)
		node[anchor=east] {\tiny{\y}};
\draw (1.5,1.5)--(6.5,6.5);
\draw[fill=white] (1.5,1.5) circle(.05);
\node[below] at (3/2,3/2) {\tiny{(1.5,1.5)}};
\node[below] at (0,-0.05) {\tiny{0}};
\end{tikzpicture} \vspace{.5cm}
\begin{tikzpicture}[scale=.6,>= stealth]
\fill[gray!30] (2.4,2.4) -- ( 3,2)--(3,3);
\draw[->] (0,0)--(7,0)node[right]{\tiny{$q$}};

\foreach \x in {1,...,6}
     		\draw (\x,0) -- (\x,-.1)
		node[anchor=north] {\tiny{\x}};
\draw[->](0,0)--(0,7)node[above]{\tiny{$p$}};
\foreach \y in {1,...,6}
     		\draw (0,\y) -- (.1,\y)
		node[anchor=east] {\tiny{\y}};
\draw[style=dashed] (3,2)--(3,3 );
\draw (2.4,2.4)--(3,2);
\draw (2.4,2.4)--(6,6);
\draw[fill=white] (3,2) circle(.05);
\draw[fill=white] (6,6) circle(.05);
\node[right] at (6,6) {\weixiao{(6,6)}};
\node[below] at (3,2) {\weixiao{(3,2)}};
\node[left] at (12/5,12/5) {\weixiao{(2.4,2.4)}};
\node[below] at (3.5,-0.8) {\tiny{Figure 2: Region of Guevara-Phuc}};
\node[below] at (0,-0.05) {\tiny{0}};
\node[above] at (4.2,4.6) {\tiny{$p=q$}};
\end{tikzpicture} \vspace{.5cm}

\begin{tikzpicture}[scale=.6,>= stealth]
\node[below] at (2,2) {\weixiao{(2,2)}};
\node[below] at (3.8,-0.8) {\tiny Figure 3: Region of Theorem 1.1};
\fill[gray!30] (2,2) -- ( 6.5,2)--(6.5,6.5);
\draw[->] (0,0)--(7,0)node[right]{\tiny{$q$}};
\node[below] at (0,-0.05) {\tiny{0}};
\foreach \x in {1,...,6}
     		\draw (\x,0) -- (\x,-.1)
		node[anchor=north] {\tiny{\x}};
\draw[->](0,0)--(0,7)node[above]{\tiny{$p$}};
\foreach \y in {1,...,6}
     		\draw (0,\y) -- (.1,\y)
		node[anchor=east] {\tiny{\y}};
\draw[style=dashed] (2,2)--(6.5,2 );
\node[above] at (4.2,4.6) {\tiny{$p=q$}};
\draw (2,2)--(6.5,6.5);
\draw[fill=white] (2,2) circle(.05);
\end{tikzpicture} \vspace{.5cm}
\begin{tikzpicture}[scale=.6,>= stealth]
\node[left] at (1.6,1.5) {\weixiao{(1.5,1.5)}};
\node[right] at (6,6) {\weixiao{(6,6)}};
\node[below] at (6,1) {\weixiao{(6,1)}};
\node[below] at (1.5,1) {\weixiao{(1.5,1)}};
\node[below] at (3.8,-0.8) {\tiny Figure 4: Region of  Theorem 1.2};
\fill[gray!30] (1.5,1.5) -- ( 6,6)-- ( 6,1)--(1.5,1);
\draw[->] (0,0)--(7,0)node[right]{\tiny{$q$}};
\node[below] at (0,-0.05) {\tiny{0}};
\foreach \x in {1,...,6}
     		\draw (\x,0) -- (\x,-.1)
		node[anchor=north] {\tiny{\x}};
\draw[->](0,0)--(0,7)node[above]{\tiny{$p$}};
\foreach \y in {1,...,6}
     		\draw (0,\y) -- (.1,\y)
		node[anchor=east] {\tiny{\y}};

\node[above] at (4.2,4.6) {\tiny{$p=q$}};
\draw (1.5,1.5)--(6,6);
\draw (1.5,1)--(6,1);
\draw[fill=white] (1.5,1.5) circle(.05);
\draw[fill=white] (6,6) circle(.05);
\draw[fill=white] (6,1) circle(.05);
\draw[fill=white] (1.5,1) circle(.05);
\draw[style=dashed] (6,1)--(6,6 );
\draw[style=dashed] (1.5,1.35)--( 1.5,1.15);
\end{tikzpicture}

Eventually we would like to mention that, as a by-product of the energy bound and the Caccioppoli type inequality obtained in the proof of Theorem 1.1 and Theorem 1.2, one can establish some
  $\varepsilon$-regularity criteria at one scale in Morrey spaces. To the best knowledge of the authors, this is the first $\varepsilon$-regularity criterion involving Morrey spaces, which is of independent interest.  For the details, see Corollary \ref{the1.3} in Section 3 and Corollary  \ref{the1.5} in Section 4.

This  paper is organized as follows. In the  second section,
we recall the definitions of various function spaces and those of suitable weak solutions including local suitable weak solutions. In addition, we present some auxiliary lemmas.
Section 3 is devoted to the proof of Theorem 1.1. To this end, along the line of \cite{[GP2],[Tsai],[NRS]}, we
construct the pressure $\Pi$ in Morrey spaces, and then localize the  Meyer-Gerard-Oru interpolation
inequality. This together with the local energy inequality yields the energy bound, which concludes  the proof of our first theorem.
In Section 4, we deal with the Caccioppoli type inequality by local suitable weak solutions and local Meyer-Gerard-Oru interpolation
inequality obtained in Section 3.  Then $A_p$ weighted inequalities enable us to  recover the pressure. Finally, applying the local energy inequality \eqref{wloc1} again and \eqref{local-energy}, we complete the proof of Theorem 1.2.

 \section{Function spaces and some known facts} \label{sec2}

\subsection{Function spaces}

In this section, we begin with definitions of various function spaces.
Let $\mathcal{S}\left(\mathbb{R}^{n}\right)$ be the set of all Schwartz functions on $\mathbb{R}^{n}$, endowed with the usual topology, and denote by $\mathcal{S}^{\prime}\left(\mathbb{R}^{n}\right)$ its topological dual, namely, the space of all bounded linear functionals on $\mathcal{S}\left(\mathbb{R}^{n}\right)$ endowed with the weak *-topology.
The classical Sobolev space $W^{1,p}(\Omega)$, with $1\leq  p\leq\infty$, is equipped with the norm $\|f\|_{W^{1,p}(\Omega)}= \|f\|_{L^{p}(\Omega)}+ \|\nabla f\|_{L^{p}(\Omega)}$. A function $f\in W^{1,p}_{\rm loc}(\mathbb{R}^{n})$ if and only if $f\in W^{1,p}(B)$ for every ball $B\subset\mathbb{R}^{n}$. The space  $C^{\infty}_{0}(\Omega)$ is  the set of all  the smooth compactly supported functions on $\Omega$.
Let $W_{0}^{1,p}(\Omega)$ be the completion of $C_{0}^{\infty}(\Omega)$ in the norm
of $W^{1,p}(\Omega)$. Let  $W^{-1,p}(\Omega)$ denote the dual of the Sobolev space $W_{0}^{1,p}(\Omega)$.

Next, we present the definitions of Lorentz spaces and Morrey spaces. For $q\in[1,\infty]$, let
$$
L^{q,\infty}(\mathbb{R}^{3})=\big\{f: f~ \text{is  a measurable function  on}~ \mathbb{R}^{3} ~\text{and} ~\|f\|_{L^{q,\infty}(\mathbb{R}^{3})}<\infty\big\}
$$
be the Lorentz space $L^{q,\infty}$ defined by means of the quasinorm
$$
\|f\|_{L^{q,\infty}(\mathbb{R}^{3})}=
\sup_{\alpha>0}\alpha|\{x\in \mathbb{R}^{3}:|f(x)|>\alpha\}|^{\f{1}{q}},
$$
where $|E|$ represents the three-dimensional Lebesgue measure of a set $E\subset \mathbb{R}^{3}$. The Morrey space $\mathcal{\dot{M}}^{p,l}(\Omega)$, with $1\leq l<\infty$, $1\leq  p\leq\infty$  and a domain $\Omega\subset\mathbb{R}^{3}$, is defined as the space of all measurable functions $f$ on $\Omega$ for which the norm
$$
\|f\|_{\mathcal{\dot{M}}^{p,l}(\Omega)}=\sup_{R>0}
\sup_{x\in \Omega}R^{3(\f1p-\f1l)}\B(\int_{B_{x}(R)\cap\Omega}|f(y)|^{l}dy\B)^{\f1l}<\infty.
$$
Here $B_{x}(R)$ represents the open ball centered at $x\in \mathbb{R}^{3}$ with radius $R>0$. In particular, by using the Lebesgue differentiation theorem, one can easily prove that $\mathcal{\dot{M}}^{\infty,l}(\Omega)=L^{\infty}(\Omega)$.

To give the definition of Besov spaces, we denote $P_{t}=e^{t\Delta}$ as the heat semigroup on $\mathbb{R}^{n}$. For $\alpha<0$, a tempered distribution $f$ on $\mathbb{R}^{n}$ belongs to the Besov space $\dot{B}^{\alpha}_{\infty,\infty}(\mathbb{R}^{n})$ if and only if the following norm
$$
\|f\|_{\dot{B}^{\alpha}_{\infty,\infty}(\mathbb{R}^{n})}=\sup_{t>0}
t^{-\f{\alpha}{2}}\|P_{t}f\|_{L^{\infty}(\mathbb{R}^{n})}
$$ is finite.

As usual, given a Schwartz function $f$ on $\mathbb{R}^{n}$, the Fourier transform $\hat{f}$ of $f$ is given by $$\hat{f}(\xi)=\frac{1}{(2\pi)^{n}}\int_{\mathbb{R}^{n}}f (x)e^{-i\xi\cdot x}\,dx, $$
and the inverse Fourier transform $f^{\vee}$ is defined as
$$
f^{\vee}(\xi)=\widehat{f}(-\xi)
$$
for all $\xi \in \mathbb{R}^{n}.$ Furthermore, for $s\geq 0,$ we define $\Lambda^{s}f $ by
$$\widehat{\Lambda^{s} f}(\xi)=|\xi|^{s}\hat{f}(\xi),$$
where the notation $\Lambda$ stands for the square root of the negative Laplacian $(-\Delta)^{1/2}$. Now introduce
the homogenous Sobolev norm $\|\cdot\|_{\dot{H}^{s}(\mathbb{R}^{n})}$ as follows $$\|f\|_{\dot{H}^{s}(\mathbb{R}^{n})}=
\|\Lambda^{s}f\|_{L^{2}(\mathbb{R}^{n})}.$$
Denote $ H^{s}$ as the standard inhomogenous Sobolev space with the norm $$\|f\| _{H^{s}(\mathbb{R}^{n})}= \|\Lambda^{s}f\|_{L^{2}(\mathbb{R}^{n})}+\|f\|_{L^{2}(\mathbb{R}^{n})}.
$$
For $q\in [1,\,\infty]$, the notation $L^{q}(0,\,T;\,X)$ stands for the set of all measurable functions $f(x,t)$ on the interval $(0,\,T)$ with values in $X$ and $\|f(\cdot,t)\|_{X}$ belonging to $L^{q}(0,\,T)$.
Throughout this paper, we denote
\begin{align*}
     & B(r):=\{y\in \mathbb{R}^{3}|\,|y|< r\}, && & Q(r):=B(r)\times (T-r^2,T).
      \end{align*}
The average integral of a function $h$ on the ball $B(r)$ is defined by
$\overline{h}_{r}:=\frac{1}{|B(r)|}\int_{B(r)}h\,$. For simplicity, we write
$$\ba
\|\cdot\| _{L^{p}L^{q}(Q(r))}:=&\|\cdot\| _{L^{p}(T-r^{2},T;L^{q}(B(r)))} ~~~ \text{and}~~~
 \|\cdot\| _{L^{p}(Q(r))}:=\|\cdot\| _{L^{p}L^{p}(Q(r))}.
 \ea$$
  We will use the summation convention on repeated indices.
 $C$ is an absolute constant which may be different from line to line unless otherwise stated in this paper.
\subsection{Suitable weak solutions}

\begin{definition}\label{defweak}
A divergence-free vector field $U = (U_1,U_2,U_3)\in W^{1,2}_{\rm loc}(\R^3)$  is called a weak solution of \eqref{SNS}, if for all divergence-free vector field $\phi = (\phi_1,\phi_2,\phi_3) \in C^\infty_0(\R^3)$ one has
\begin{equation}\label{weakform}
\int_{\R^3}(  \nabla U\cdot\nabla \phi +[aU+a(y\cdot \nabla)U+(U\cdot\nabla) U]\cdot \phi) dy   = 0.
\end{equation}
\end{definition}

Now,    for the convenience of readers, we recall the \wred{classical definition of   suitable weak solutions} to the Navier-Stokes system \eqref{NS}.

	\begin{definition}\label{defi1}
		A  pair   $(u, \, \pi)$  is called a suitable weak solution to the Navier-Stokes equations \eqref{NS} provided the following conditions are satisfied,
		\begin{enumerate}[(1)]
			\item $u \in L^{\infty}(-T,\,0;\,L^{2}(\mathbb{R}^{3}))\cap L^{2}(-T,\,0;\,\dot{H}^{1}(\mathbb{R}^{3})),\,\pi\in
			L^{3/2}(-T,\,0;L^{3/2}(\mathbb{R}^{3}));$
			\item$(u, ~\pi)$~solves (\ref{NS}) in $\mathbb{R}^{3}\times (-T,\,0) $ in the sense of distributions;
			\item$(u, ~\pi)$ satisfies the following inequality, for a.e. $t\in[-T,0]$,
			\begin{align}
				&\int_{\mathbb{R}^{3}} |u(x,t)|^{2} \phi(x,t) dx
				+2\int^{t}_{-T}\int_{\mathbb{R} ^{3 }}
				|\nabla u|^{2}\phi  dxds\nonumber\\ \leq&  \int^{t}_{-T }\int_{\mathbb{R}^{3}} |u|^{2}
				(\partial_{s}\phi+\Delta \phi)dxds
				+ \int^{t}_{-T }
				\int_{\mathbb{R}^{3}}u\cdot\nabla\phi (|u|^{2} +2\pi)dxds, \label{loc}
			\end{align}
			where non-negative function $\phi(x,s)\in C_{0}^{\infty}(\mathbb{R}^{3}\times (-T,0) )$.\label{SWS3}
		\end{enumerate}
	\end{definition}

\begin{lemma}\cite{[HWZ]}\label{lem2}
Let $\Phi$ denote  the standard normalized fundamental solution of Laplace equation in $\mathbb{R}^{3}$. For $0<\xi<\eta$, we consider smooth cut-off function $\psi\in C^{\infty}_{0}(B(\f{\xi+3\eta}{4}))$ such that $0\leq\psi\leq1$ in $B(\eta)$, $\psi\equiv1$ in $B(\f{3\xi+5\eta}{8})$  and $|\nabla^{k}\psi |\leq C/(\eta-\xi)^{k}$ with $k=1,2$ in  $B(\eta)$.  Then we may split  pressure $\pi$ in \eqref{NS} as below
			\be\label{decompose pk}
			\pi(x):=\pi_{1}(x)+\pi_{2}(x)+\pi_{3}(x), \quad x\in B(\f{\xi+\eta}{2}),
			\ee
			where
$$\ba			\pi_{1}(x)=&-\partial_{i}\partial_{j} \Phi  \ast (u_{j}u_{i} \psi) ,\\
			\pi_{2}(x)
			=&2\partial_{i} \Phi  \ast (u_{j}u_{i}\partial_{j}\psi)- \Phi  \ast
			(u_{j}u_{i} \partial_{i}\partial_{j}\psi ) , \\
			\pi_{3}(x)
			=&2\partial_{i} \Phi  \ast (\pi \partial_{i}\psi )- \Phi  \ast (\pi \partial_{i}\partial_{i}\psi ).
			\ea
			$$
 Moreover, there holds
			\begin{align}
				& \| \pi_1\|_{L^{3/2}(Q(\f{\xi+\eta}{2}))}
				\leq C\|  u\|^{2}_{L^{3}(Q(\f{\xi+3\eta}{4}))};\label{p1estimate}\\
				& \|  \pi_2\|_{L^{3/2}(Q(\f{\xi+\eta}{2}))}
				\leq  \f{C\eta^{3}}{(\eta-\xi)^{3}}\|  u\|^{2}_{L^{3}(Q(\f{\xi+3\eta}{4}))};\label{p2estimate}\\
				& \|  \pi_3\|_{L^{1}L^{2}(Q(\f{\xi+\eta}{2}))}
				\leq \f{C\eta^{3/2 }}{(\eta-\xi)^{3}}\|\pi\|_{L^{1}(Q(\f{\xi+3\eta}{4}))}.\label{p3estimate}
			\end{align}
		\end{lemma}
\subsection{Local suitable weak solutions}
We begin with \wgr{the Wolf's} local pressure projection $\mathcal{W}_{p,\Omega}:$ $W^{-1,p}(\Omega)\rightarrow W^{-1,p}(\Omega)$ $(1<p<\infty)$.
 More precisely, for any  $f\in W^{-1,p}(\Omega)$, we define \wgr{$\mathcal{W}_{p,\Omega}(f)= \nabla\pi$}, where $\pi$ satisfies \eqref{GMS}.
Let $\Omega$  be a  bounded domain with $\partial\Omega\in C^{1}$.
According to the $L^p$ theorem of Stokes system in \cite[Theorem 2.1, p.149]{[GSS]},
there exists a unique pair $(u,\pi)\in W^{1,p}(\Omega)\times L^{p}(\Omega)$ such that
\be\label{GMS}
-\Delta u+\nabla\pi=f,~~ \text{div}\,u=0, ~~u|_{\partial\Omega}=0,~~ \int_{\Omega}\pi dx=0.
\ee
Moreover, this pair is subject to the inequality
$$
\|u\|_{\wgr{W^{1,p}}(\Omega)}+\|\pi\|_{\wgr{L^p}(\Omega)}\leq C\|f\|_{\wgr{W^{-1,p}}(\Omega)}.
$$
Let $\nabla\pi= \mathcal{W}_{p,\Omega}(f)$ $(f\in L^p(\Omega))$, then $\|  \pi\|_{L^p(\Omega)}\leq C\|f\|_{L^p(\Omega)},$ where we used the fact that $L^{p}(\Omega)\hookrightarrow W^{-1,p}(\Omega)$.  Moreover, from $\Delta \pi=\text{div}\,f$, we see that $\|  \nabla\pi\|_{L^p(\Omega)}\leq C(\|f\|_{L^p(\Omega)}+ \|  \pi\|_{L^p(\Omega)}) \leq C\|f\|_{L^p(\Omega)}.$
Now, we  present the definition of local suitable weak solutions to Navier-Stokes equations \eqref{NS}.
	\begin{definition}\label{defilsw}
		A  pair   $(u, \,\pi)$  is called a local suitable weak solution to the Navier-Stokes equations \eqref{NS} provided the following conditions are satisfied,
		\begin{enumerate}[(1)]
			\item $u \in L^{\infty}(-T,\,0;\,L^{2}(\mathbb{R}^{3}))\cap L^{2}(-T,\,0;\,\dot{H}^{1}(\mathbb{R}^{3})),\,\pi\in
			L^{3/2}(-T,\,0;L^{3/2}(\mathbb{R}^{3}));$\label{SWS1}
			\item$(u, ~\pi)$~solves (\ref{NS}) in $\mathbb{R}^{3}\times (-T,\,0) $ in the sense of distributions;\label{SWS2}
			\item The local energy inequality reads,
for a.e. $t\in[-T,0]$ and non-negative function $\phi(x,s)\in C_{0}^{\infty}(\mathbb{R}^{3}\times (-T,0) )$,
			 \begin{align}
  &\int_{B(R)}|v|^2\phi (x,t)  d  x+ \int^{t}_{-T }\int_{B(R)}\big|\nabla v\big|^2\wgr{\phi (x,s) dx ds}\nonumber\\  \leq&   \int^{t}_{-T }\int_{B(R)} | v |^2(  \Delta \phi +  \partial_{s}\phi )  d  x d s +\int^{t}_{-T }\int_{B(R)}|v|^{2}u\cdot\nabla \phi    dxds\nonumber\\
& +\int^{t}_{-T }\int_{B(R)} \phi ( u\otimes v : \nabla^{2}\pi_{h} )dxds   +\int^{t}_{-T }\int_{B(R)} \phi \pi_{1}v\cdot\nabla \phi   dxds+\int^{t}_{-T }\int_{B(R)} \phi \pi_{2}v\cdot\nabla \phi   dxds.\label{wloc1}
 \end{align}
\wgr{Here, $\nabla\pi_h=-\mathcal{W}_{p,B(R)}(u)$, $\nabla\pi_1=\mathcal{W}_{p,B(R)}(\Delta u)$, $\nabla\pi_2=-\mathcal{W}_{p,B(R)}(u\cdot\nabla u)$, $v=u+\nabla\pi_h$.}
 In addition, $ \nabla\pi_{h}, \nabla\pi_{1}$ and $\nabla\pi_{2}$ meet the following \wgr{facts}
	\begin{align}   &\|\nabla\pi_{h}\|_{L^p(B(R))}\leq  \wgr{C}\|u\|_{L^p(B(R))}, \label{ph}\\
 &\|\wgr{\pi_{1}}\|_{L^2(B(R))}\leq \wgr{C} \|\nabla u\|_{L^2(B(R))},\label{p1}\\
 &\|\wgr{\pi_{2}}\|_{L^{p/2}(B(R))}\leq \wgr{C} \| |u|^{2}\|_{L^{p/2}(B(R))}.\label{p2}
\end{align}	\end{enumerate}
	\end{definition}

We list some
interior estimates
of the harmonic equation $\Delta h=0$, which will be frequently utilized later. Let $1\leq p,q\leq\infty$ and $0<r<\rho$, then there holds
\be\label{h1}\|\nabla^{k}h\|_{L^{q}
(B(r))}\leq \f{Cr^{\f{3}{q}}}{(\rho-r)^{\f{3}{p}+k}}\|h\|_{L^{p}(B(\rho))}\wgr{,}\ee
\be\label{h2}
 \| h-\overline{h}_{r}\|_{L^{q}
(B(r))}\leq \f{Cr^{\f{3}{q}+1}}{(\rho-r)^{\f{3}{q}+1 }}\|h-\overline{h}_{\rho}\|_{L^{q}(B(\rho))}.\ee
 The proof of \eqref{h1} rests on the mean value property of harmonic functions. This together with
  mean value theorem leads to \eqref{h2}. We leave the details  to the reader.

\subsection{Meyer-Gerard-Oru  interpolation inequality}
For the convenience of the readers, we recall
Meyer-Gerard-Oru  interpolation inequality in \cite{[MGO],[PP]} below
\be\label{mgoi}
\|f\|_{L^{m}(\mathbb{R}^{n})}\leq C \|f\|^{\f2m}_{\dot{H}^{s}(\mathbb{R}^{n})}
 \|f\|^{1-\f2m}_{\dot{B}^{-\f{2s}{m-2}}_{\infty,\infty}(\mathbb{R}^{n})} ~~\text{with} ~~s>0~~\text{and}~~2<m<\infty.\ee
In addition,we have the following embedding relation between Morrey spaces and Besov spaces (see \cite[Section 3.2, Lemma 1]{[PP]})
\be\label{BM}
\|f\|_{\dot{B}^{-\f{3}{q}}_{\infty,\infty}(\mathbb{R}^{3})}\leq C\|f
\|_{\mathcal{\dot{M}}^{q,1}(\mathbb{R}^{3})} ~~\text{with} ~~1<q<\infty.
\ee
These two inequalities play  an important role in our proof.

\section{Proof of Theorem \ref{the1.1} }
In this section, we present the proof of Theorem \ref{the1.1}. In Subsection 3.1, we define $\Pi$ by $\mathcal{R}_i \mathcal{R}_j(U_{i}U_{j})$ and examine that it satisfies the equations \eqref{SNS}. In the second subsection we establish the energy bounds. Invoking energy bounds  and \eqref{local-energy}, we get the desired results in the last subsection.
\subsection{The construction of the pressure}
\begin{lemma} \label{cp}
Suppose that  $U$ is a weak solution of \eqref{SNS} and $U\in \mathcal{\dot{M}}^{2p,2q}(\mathbb{R}^{3})$ with $1<q\leq p <\infty$.
Let $\widetilde{\Pi}=\mathcal{R}_i \mathcal{R}_j(U_{i}U_{j})$. Then there holds
\begin{enumerate}[(1)]
			\item
  \be\label{pm}
\|\widetilde{\Pi}\|_{\mathcal{\dot{M}}^{p,q}(\mathbb{R}^{3})}\leq C \|U\|^{2}_{\mathcal{\dot{M}}^{2p,2q}(\mathbb{R}^{3})}.
\ee
 \item  $\widetilde{\Pi}$ meets
$$-\Delta \Pi=\partial_{i}\partial_{j}(U_{i}U_{j}),$$
in the distributional sense.

\item  $(U,\widetilde{\Pi})$ smoothly solves
\begin{equation}\label{prescon}
-\nu \Delta U+aU+a(y\cdot\nabla)U+(U\cdot\nabla)U+\nabla \Pi=0 \quad {\rm in} \,\,\, \mathbb{R}^3.
\end{equation}
Here the solution $\Pi$ is unique up to a constant.
\end{enumerate}
\end{lemma}

\begin{proof}
(1) The boundedness of
 singular integral operators $R_iR_j$  on Morrey spaces $\mathcal{\dot{M}}^{p,q}(\mathbb{R}^{3})$ with $1<q\leq p <\infty$ means \eqref{pm}.

 (2)
To show the second assertion, assume for a while, we have proved that
 \be\ba\label{cut}
\int_{\mathbb{R}^3} \mathcal{R}_i\mathcal{R}_j( \chi_{B(R)} U_{i}U_{j})  \Delta\varphi dx
=&- \int_{\mathbb{R}^3}\chi_{B(R)} U_iU_j  \partial_i\partial_j \varphi dx, ~~~\varphi\in C^{\infty}_{0}(\mathbb{R}^3),
\ea\ee
where  $\chi_{B(R)}$ is  the characteristic function of $B(R)$. The problem reduces to that of passing to the limit as  $R$ tends to infinity. Indeed, note that, for any $f\in \mathcal{\dot{M}}^{p,q}(\mathbb{R}^{3})$ and $b>3-\f{3q}{p}$,
$$
\int_{\mathbb{R}^{3}}|f|^{q}(1+|x|^{2})^{-\f{b}{2}}dx \leq
C \|f\|^{q}_{\mathcal{\dot{M}}^{p,q}(\mathbb{R}^{3})}.
$$
This fact can be found in \cite{[Kato]}.
To make our paper more self-contained and more readable, we present the proof of the above fact as follows
$$\ba
\int_{\mathbb{R}^{3}}|f|^{q}(1+|x|^{2})^{-\f{b}{2}}dx &\leq
  \int_{|x|\leq1}|f|^{q}(1+|x|^{2})^{-\f{b}{2}}dx
  +\int_{|x|>1}|f|^{q}(1+|x|^{2})^{-\f{b}{2}}dx.
\\&\leq    \int_{|x|\leq1}|f|^{q} dx+\sum_{k=0 }^{\infty}\int_{2^{k}<|x|\leq2^{k+1}}|f|^{q}
(1+|x|^{2})^{-\f{b}{2}}dx\\
&\leq  \|f\|^{q}_{\mathcal{\dot{M}}^{p,q}(\mathbb{R}^{3})}+C\sum_{k=0 }^{\infty}\big(2^{k+1}\big)^{-b-3(\f{q}{p}-1)} \|f\|^{q}_{\mathcal{\dot{M}}^{p,q}(\mathbb{R}^{3})}\\
&\leq  C\|f\|^{q}_{\mathcal{\dot{M}}^{p,q}(\mathbb{R}^{3})}.
\ea$$
In addition, note that $w(x)=(1+|x|^{2})^{-\f{b}{2}}$ satisfies the Muckenhoupt $A_{q}$-condition if $0\leq b<3$. Therefore, for $p<\infty$, we choose $b$ such that $3-\f{3 q}{p}<b<3$. Then, the H\"older inequality and the classical Calder\'on-Zygmund Theorem with $A_{q}$ weights yield that
$$\ba
&\left|\int_{\mathbb{R}^3}\B(\mathcal{R}_i\mathcal{R}_j(U_iU_j )-\mathcal{R}_i\mathcal{R}_j( \chi_{B(R)} U_iU_j )\B)  \Delta\varphi dx\right| \\
&\leq  \left(\int_{\mathbb{R}^3}\left|\mathcal{R}_i\mathcal{R}_j(U_iU_j- \chi_{B(R)} U_iU_j )\right|^q w(x) dx\right)^{\frac1q} \left(\int_{\mathbb{R}^3} |  \Delta\varphi|^{\f{q}{q-1}} w(x)^{\f{1}{1-q}} dx\right)^{1-\frac1q}\\
&\leq C \left(\int_{\mathbb{R}^3}\left| U_iU_j- \chi_{B(R)} U_iU_j  \right|^q w(x) dx\right)^{\frac1q} \left(\int_{\mathbb{R}^3} |  \Delta\varphi|^{\f{q}{q-1}} w(x)^{\f{1}{1-q}} dx\right)^{1-\frac1q}\\
&\leq C \|U_iU_j\|_{\mathcal{\dot{M}}^{p,q}(\mathbb{R}^{3})}
,
\ea$$
Likewise,
$$\ba
&\left|\int_{\mathbb{R}^3}\B(  U_iU_j  - \chi_{B(R)} U_iU_j \B)   \partial_{i}\partial_{j}\varphi dx\right| \\
&\leq C \|U_iU_j\|_{\mathcal{\dot{M}}^{p,q}(\mathbb{R}^{3})}.
\ea$$
Thus we can apply
the Lebesgue's dominated convergence theorem to obtain
$$
\lim_{R\rightarrow\infty}\int_{\mathbb{R}^3}\mathcal{R}_i\mathcal{R}_j( \chi_{B(R)} U_iU_j ) \Delta\varphi dx=\int_{\mathbb{R}^3}\mathcal{R}_i\mathcal{R}_j(U_iU_j )  \Delta\varphi dx,
$$
and
\be\ba\lim_{R\rightarrow\infty}
 \int_{\mathbb{R}^3} \chi_{B(R)} U_{i}U_{j}   \partial_i\partial_j\varphi dx = \int_{\mathbb{R}^3} U_iU_j  \partial_i\partial_j \varphi dx. \ea\ee
Combining this and \eqref{cut}, we know that
  \be\ba\label{cut2}
\int_{\mathbb{R}^3} \mathcal{R}_i\mathcal{R}_j(  U_{i}U_{j})  \Delta\varphi dx
=&- \int_{\mathbb{R}^3} U_iU_j  \partial_i\partial_j \varphi dx,
\ea\ee
which implies the second assertion. Subsequently, we need to prove \eqref{cut} we have assumed.
Since the Fourier transform is a topological isomorphism from
 $\mathcal{S}(\mathbb{R}^{n})$  onto itself, we conclude that $(\Delta\varphi)^{\vee}\in\mathcal{S}(\mathbb{R}^{3})$
 from $\Delta\varphi \in C_{0}^{\infty}(\mathbb{R}^{3})\subset\mathcal{S}(\mathbb{R}^{3})$.
Moreover, there holds $\Delta\varphi= ((\Delta\varphi)^{\vee})^{\wedge}$. Let $s=\min\{q,2\}$.
 According to the definition of the Fourier transform of  tempered distributions and
  $\mathcal{R}_i\mathcal{R}_j( \chi_{B(R)} U_{i}U_{j})\in L^{s}(\mathbb{R}^{3})\subset \mathcal{S}'(\mathbb{R}^{3})$, we discover
   \be\ba\label{cut1}
\int_{\mathbb{R}^3} \mathcal{R}_i\mathcal{R}_j( \chi_{B(R)} U_{i}U_{j})  ((\Delta\varphi)^{\vee})^{\wedge} dx=
\int_{\mathbb{R}^3} (\mathcal{R}_i\mathcal{R}_j( \chi_{B(R)} U_{i}U_{j})) ^{\wedge} (\Delta\varphi)^{\vee} dx.
\ea\ee
We recall the following fact (see \cite[p.76]{[Duoandikoetxea]}) that
 for any $f\in L^{m}(\mathbb{R}^{3})$ with $1<m\leq 2$,
\be\label{xfq}
(\mathcal{R}_i\mathcal{R}_jf)^{\wedge}(\xi)=
 -\f{\xi_{i}\xi_{j}}{|\xi|^{2}}\hat{f}(\xi), ~~~ \xi\in\mathbb{R}^{3}.
 \ee
From
\eqref{cut1} and
\eqref{xfq}, we observe that
 \be\ba\label{cut5}
\int_{\mathbb{R}^3} \mathcal{R}_i\mathcal{R}_j( \chi_{B(R)} U_{i}U_{j})  \Delta\varphi dx
=& \int_{\mathbb{R}^3}( \mathcal{R}_i\mathcal{R}_j( \chi_{B(R)} U_{i}U_{j})) ^{\wedge} (\Delta\varphi)^{\vee} dx
\\
=&- \int_{\mathbb{R}^3} \f{\xi_{i}\xi_{j}}{|\xi|^{2}}(  \chi_{B(R)} U_{i}U_{j} ) ^{\wedge}( |\xi|^{2}\varphi^{\vee}) dx
\\
=&- \int_{\mathbb{R}^3}  \chi_{B(R)} U_{i}U_{j} (\xi_{i}\xi_{j}\,\varphi^{\vee} ) ^{\wedge} dx\\
=&- \int_{\mathbb{R}^3}\chi_{B(R)} U_iU_j  \partial_i\partial_j \varphi dx.
\ea\ee
This confirms \eqref{cut}.

Next, we turn our attention to demonstrating  \eqref{prescon}. Before going further, we write
\begin{equation*}
F:= -\nu \Delta U+aU+a(y\cdot\nabla)U+(U\cdot\nabla)U+\nabla \widetilde{\Pi}
\end{equation*}
 and show that $F\equiv 0.$

It follows from  $F=\nabla \widetilde{\Pi}-\nabla \Pi$, div $F=0$ and curl $F=0$ that $\Delta F=0$. Since harmonic functions are analytic, to get $F\equiv0$, it suffices to show that $D^{\alpha}F(0)=0$
for any multi-index $\alpha=(\alpha_1,\alpha_2,\alpha_3)$ with $|\alpha|\geq 0$.
Let $\theta\in C^{\infty}_{0}(\mathbb{R}^{3})$ be any radial function which is supported in $\{x\in \mathbb{R}^{3}|\,|x|<1\}$ and has integral 1. Note that $D^{\alpha}F$ is also harmonic, there holds
\begin{equation}\label{MVP}
D^{\alpha}F(0)=\varepsilon^3\int_{\mathbb{R}^3} D^{\alpha}F(y)\,\theta(\varepsilon y)dy
\end{equation}
for any $\varepsilon>0$. (See \cite[p.275]{[Stein1]} for the details.)

  Integrating by parts, one computes
\begin{equation*}
D^\alpha F(0)=\varepsilon^3\int_{\mathbb{R}^3} D^\alpha F(y)\,\theta(\varepsilon y)dy =(-1)^{|\alpha|}\varepsilon^3\int_{\mathbb{R}^3}  F(y) \varepsilon^{|\alpha|} (D^\alpha\theta)(\varepsilon y)dy.
\end{equation*}
According to this, to show that $D^\alpha F(0)=0$, it suffices to prove that for any $\varphi\in C^{\infty}_{0}(B(1))$,
$$\lim_{\varepsilon\to 0^+}\varepsilon^3\int_{\mathbb{R}^3}F(y)\varphi(\varepsilon y)dy=0.
$$
Integration by parts twice, the H\"older inequality and the definition of Morrey spaces guarantee that
$$\ba
\left|\varepsilon^{3}\int_{\mathbb{R}^{3}}\Delta U\varphi(\varepsilon y)dy\right|=&\left|\varepsilon^{5}\int_{\mathbb{R}^{3}} U\Delta\varphi(\varepsilon y)dy\right|\\
\leq& \varepsilon^{5} \B(\int_{|y|<\f{1}{\varepsilon}}|U|^{2q}dy\B)^{\f{1}{2q}}
\B(\int_{|y|<\f{1}{\varepsilon}}|\Delta\varphi(\varepsilon y)|^{\f{2q}{2q-1}}dy\B)^{1-\f{1}{2q}}\\
\leq& C \varepsilon^{2+\f{3}{2p}}
\|U\|_{\mathcal{\dot{M}}^{2p,2q}(\mathbb{R}^{3})}.
\ea$$
A slight modification of the latter argument yields
$$\ba
\left|\varepsilon^{3}\int_{\mathbb{R}^{3}}U\varphi(\varepsilon y)dy\right| \leq& \,\varepsilon^{3} \B(\int_{|y|<\f{1}{\varepsilon}}|U|^{2q}dy\B)^{\f{1}{2q}}
\B(\int_{|y|<\f{1}{\varepsilon}}|\varphi(\varepsilon y)|^{\f{2q}{2q-1}}dy\B)^{1-\f{1}{2q}}\\
\leq& C \varepsilon^{\f{3}{2p}}
\|U\|_{\mathcal{\dot{M}}^{2p,2q}(\mathbb{R}^{3})}.
\ea$$
In virtue of  integration
by parts once again, we see that
\be\ba
\varepsilon^3\int_{\mathbb{R}^{3}}(y\cdot\nabla) U(y)\varphi(\varepsilon y) dy = -\varepsilon^3\int_{\mathbb{R}^{3}} 3U(y)\varphi(\varepsilon y)dy- \varepsilon^3 \int_{\mathbb{R}^{3}} U(y)\{(\varepsilon y)\cdot \nabla\varphi(\varepsilon y)\}dy.
\ea\ee
In the same manner as above, we can also bound $\left|\varepsilon^3\int_{\mathbb{R}^{3}}(y\cdot\nabla) U(y)\varphi(\varepsilon y) dy\right|$.

On the other hand, it follows from the divergence-free condition that
\be\ba
&\varepsilon^3\int_{\mathbb{R}^{3}}(U\cdot\nabla) U(y)\varphi(\varepsilon y) dy\\
 =& \varepsilon^3\int_{\mathbb{R}^{3}} \text{div}\,U\otimes U(y)\varphi(\varepsilon y)dy\\
=& -\varepsilon^{4}\int_{\mathbb{R}^{3}}  U\otimes U(y)\cdot\nabla\varphi(\varepsilon y)dy,
\ea\ee
which in turn implies that
\be\label{UUe}\ba
&\left|\varepsilon^3\int_{\mathbb{R}^{3}}(U\cdot\nabla) U(y)\varphi(\varepsilon y) dy\right|\\
\leq& \varepsilon^{4} \B(\int_{|y|<\f{1}{\varepsilon}}|U|^{2q}dy\B)^{\f1q}
\B(\int_{|y|<\f{1}{\varepsilon}}|\nabla\varphi(\varepsilon y)|^{\f{q}{q-1}}dy\B)^{1-\f1q}\\
\leq& C \varepsilon^{1+\f{3}{p}}
\|U\|^{2}_{\mathcal{\dot{M}}^{2p,2q}(\mathbb{R}^{3})}.
\ea\ee
Eventually, we need to bound the last term involving $\nabla\widetilde{\Pi}$. Integration by parts gives
\be\ba
\varepsilon^3\int_{\mathbb{R}^{3}}\nabla \widetilde{\Pi}(y)\,\varphi(\varepsilon y) dy
  =-\varepsilon^{4}\int_{\mathbb{R}^{3}}\widetilde{\Pi}(y)\,\nabla\varphi(\varepsilon y)dy.
\ea\ee
Furthermore, a variant of (\ref{UUe}) provides the estimate
$$
\left|\varepsilon^{4}\int_{\mathbb{R}^{3}} \widetilde{\Pi}(y)\,\nabla\varphi(\varepsilon y)dy\right|
\leq C \varepsilon^{1+\f{3}{p}}\|\widetilde{\Pi}\|_{\mathcal{\dot{M}}^{p,q}(\mathbb{R}^{3})} \leq C \varepsilon^{1+\f{3}{p}}\|U\|^{2}_{\mathcal{\dot{M}}^{2p,2q}(\mathbb{R}^{3})}.
$$
This verifies $D^\alpha F(0)=0$ and completes the proof of Lemma \ref{cp}.
\end{proof}
\subsection{Energy bounds }\label{sec4}

\begin{prop}\label{the1.2}Assume that  $(u,  \pi)$ is a suitable weak solution to the 3D Navier-Stokes system \eqref{NS}. Let $\alpha=2/p+3/q$ and the pair  $(p,q)$ satisfy
 \be\label{pqcondi}
1\leq2/p+3/q<2, ~~~\text{with}~~~ 1<p<\infty.
 \ee
 Then there holds, for any $m\geq1$,
\be\ba \label{key ineq}
&\|u\|^2_{L^{\infty}(T-\f{R^{2}}{4},T;L^{2}(B(R/2)))}+\|\nabla u\|^2_{L^{2}(T-\f{R^{2}}{4},T;L^{2}(B(R/2)))} \\ \leq&    CR^{3-2\alpha} \|u\|_{L^{p}(T-R^{2},T;\dot{\mathcal{M}}^{q,1} (B(R)))} ^{2}\\& +  CR^{\f{6-5\alpha}{2-\alpha}} \|u\|_{L^{p}(T-R^{2},T;\dot{\mathcal{M}}^{q,1} (B(R)))}  ^{\f{4}{2-\alpha}} + CR^{1-\f{6}{m}} \|\pi\|^{2}_{L^{1}(T-R^{2},T;\dot{\mathcal{M}}^{m,1} (B(R)))}.
\ea\ee
\end{prop}
\begin{remark}
We refer the readers to \cite{[GP]} and \cite{[HWZ]} for recent progress on energy bounds of suitable weak solutions.
\end{remark}
In the spirit of \cite{[GP],[HWZ]}, we conclude
$\varepsilon$-regularity criteria at one scale in Morrey spaces from the above energy bounds.  Previous  related results in  Lorentz spaces  can be found in \cite{[WWY],[Baker]}. In addition, a summary of
$\varepsilon$-regularity criteria at one scale is given in \cite{[HWZ]}.
\begin{coro}\label{the1.3}
		Let  the pair $(u,  \pi)$ be a suitable weak solution to the 3D Navier-Stokes system \eqref{NS} in $Q(1)$.
		There exists an absolute positive constant $\varepsilon_{1}$
		such that if the pair $(u,\pi)$ satisfy	 \be\label{0il1}\ba
&\|u\|_{ L^{p }(-1,0;\mathcal{\dot{M}}^{q,1}(B(1))) }+\|\pi\|_{ L^{1}(-1,0;\mathcal{\dot{M}}^{m,1}(B(1))) }\leq \varepsilon_{1},\ea\ee
where
$$1\leq 2/p+3/q <2, 1< p<\infty ~\text{and}~ m\geq1,$$
	 then  $u\in L^{\infty}(Q(1/2)).$
	\end{coro}
\begin{remark}
This result extends regularity criteria via Lebesgue spaces in \cite{[HWZ]} to Morrey spaces.
We refer the readers to \cite{[GKT]} for various $\varepsilon$-regularity criteria   at all scales in Lebesgue spaces.
\end{remark}
For abbreviation, we set
$$\ba
 E(r)=\sup_{T-r^{2}\leq t\leq T}\|u(\cdot,t)\|^{2}_{L^{2}(B(r))}+\int_{T-r^{2}}^{T}\|\nabla u\|^{2}_{L^{2}(B(r))}dt .
\ea$$
\begin{lemma}\label{ineq}
For $0<\xi<\eta$, let $r=\f{\xi+3\eta}{4}$ and $\alpha=2/p+3/q$ with the pair $(p,q)$ satisfying
\eqref{pqcondi}. Then
there  exits  an absolute constant $C$  independent of  $\xi$ and $\eta$,~ such that
\be\ba\label{keyineq}
 \int_{T-r^{2}}^{T}\|u\|^{3}_{L^{3}(B(r))}dt
 \leq  C \eta^{\f{3(2-\alpha)}{2\alpha} } \B(1+\f{\eta^{2}}{(\eta-\xi)^{2}}\B)^{\f32-\f{1}{\alpha}}E^{\f32-\f{1}{\alpha}}(\eta)
\B(\int_{T-r^{2}}^{T}\|u\|^{p}_{\dot{\mathcal{M}}^{q,1}(B(\eta))}dt\B)^{\f{2}{p\alpha}}.
\ea\ee
\end{lemma}
\begin{proof}
For any  pair $(p,q)$ satisfying \eqref{pqcondi}, we can
select $p'<p$ such that the value of $\f{2}{p'}+\f3q$ is very close to $2$ and smaller than $2$. Due to the H\"older inequality only in time direction, it is enough to consider the case that $\f2p+\f3q$ is close to 2.
To proceed further, we set
$$m=\f{6\alpha}{3\alpha-2}.$$
Taking $\alpha\rightarrow 2^{-}$, we conclude from some elementary computations that
\be
\label{mxianzhi}
3<m\leq \f{2q+6}{3}.\ee
 Invoking interpolation inequality \eqref{mgoi}, we see that
\be\label{2.9}\ba
\|u\|_{L^{m}(\mathbb{R}^{3})}\leq& C \|u\|^{\f2m}_{\dot{H}^{\f{3(m-2)}{2q}}(\mathbb{R}^{3})}\|u\|^{1-\f2m}
_{\dot{B}_{\infty,\infty}^{-\f3q}(\mathbb{R}^{3})}
\\\leq& C \|u\|^{\f2m}_{\dot{H}^{\f{3(m-2)}{2q}}(\mathbb{R}^{3})}\|u\|^{1-\f2m}
_{\mathcal{\dot{M}}^{q,1}(\mathbb{R}^{3})},
\ea\ee
where we also used \eqref{BM}.

Using the H\"older inequality and \eqref{2.9}, we arrive at
\be\label{2.11}\ba
\|u\|^{3}_{L^{3}(B(r))}\leq& C r^{9(\f13-\f{1}{m})}\|u\|^{3}_{L^{m}(B(r))}\\\leq& C r^{9(\f13-\f{1}{m})}\|u\|^{3}_{L^{m}(\mathbb{R}^{3})}\\
\leq& C r^{9(\f13-\f{1}{m})}\|u\|^{\f6m}_{\dot{H}^{\f{3(m-2)}{2q}}(\mathbb{R}^{3})}
\|u\|^{3-\f6m}_{\mathcal{\dot{M}}^{q,1}(\mathbb{R}^{3})}.
\ea\ee
From the Gagliardo-Nirenberg inequality and \eqref{mxianzhi}, we get
$$
\|u\|_{\dot{H}^{\f{3(m-2)}{2q}}(\mathbb{R}^{3})}\leq C\|  u\|_{L^{2}(\mathbb{R}^{3})}^{\f{2q-3m+6}{2 q}}\|\nabla u\|^{\f{3m-6}{2q}}_{L^{2}(\mathbb{R}^{3})}.
$$
Substituting this into \eqref{2.11}, we deduce that
$$\ba
\|u\|^{3}_{L^{3}(B(r))}
\leq& C r^{9(\f13-\f{1}{m})}\| u\|_{L^{2}(\mathbb{R}^{3})}^{\f{3(2q-3m+6)}{m q}}\|\nabla u\|^{\f{3(3m-6)}{mq}}_{L^{2}(\mathbb{R}^{3})}
\|u\|^{3-\f6m}_{\mathcal{\dot{M}}^{q,1}(\mathbb{R}^{3})}.
\ea$$
Integrating this inequality in time and applying the H\"older inequality, we know that
$$\ba
&\int_{T-r^{2}}^{T}\|u\|^{3}_{L^{3}(B(r))}dt\\
\leq& C r^{9(\f13-\f{1}{m})} \sup_{  T-r^{2}\leq t\leq T}\| u\|_{L^{2}(\mathbb{R}^{3})}^{\f{3(2q-3m+6)}{m q}}\int^{T}_{T-r^{2}}\|\nabla u\|^{\f{3(3m-6)}{mq}}_{L^{2}(\mathbb{R}^{3})}
\|u\|^{3-\f6m}_{\mathcal{\dot{M}}^{q,1}(\mathbb{R}^{3})}dt\\
\leq& C r^{9(\f13-\f{1}{m})} \sup_{  T-r^{2}\leq t\leq T}\| u\|_{L^{2}(\mathbb{R}^{3})}^{\f{3(2q-3m+6)}{m q}}\B(\int_{T-r^{2}}^{T}\|\nabla u\|^{2}_{L^{2}(\mathbb{R}^{3})}dt\B)^{\f{3(3m-6)}{2mq}}\\&\times
\B(\int_{T-r^{2}}^{T}\|u\|^{\f{2q(3m-6)}{2qm-3(3m-6)}}
_{\mathcal{\dot{M}}^{q,1}(\mathbb{R}^{3})}dt\B)^{\f{2qm-3(3m-6)}{2qm}},
\ea$$
namely,
\be\label{4.5}\ba
&\int_{T-r^{2}}^{T}\|u\|^{3}_{L^{3}(B(r))}dt\\
\leq& C r^{\f{3(2-\alpha)}{2\alpha} } \B(\sup_{T-r^{2}\leq t\leq T}\| u\|^{2}_{L^{2}(\mathbb{R}^{3})}+\int_{T-r^{2}}^{T}\|\nabla u\|^{2}_{L^{2}(\mathbb{R}^{3})}dt\B)^{\f32-\f{1}{\alpha}}
\B(\int_{T-r^{2}}^{T}\|u\|^{p}_{\mathcal{\dot{M}}^{q,1}
(\mathbb{R}^{3})}dt\B)^{\f{2}{p\alpha}}.
\ea\ee
Choose a cut-off function $\psi\in C_{0}^{\infty}(B(\eta))$ satisfying $0\leq\psi\leq1$, $\psi\equiv1$ in $B(\f{\xi+3\eta}{4})$ and $|\nabla\psi|\leq \f{C}{|\eta-\xi|}$. Replacing $u$ by $\psi u$ in \eqref{4.5} and taking $r=\f{\xi+3\eta}{4}$, we get
$$\ba
&\int_{T-r^{2}}^{T}\|u\|^{3}_{L^{3}(B(r))}dt
\\\leq& C \eta^{\f{3(2-\alpha)}{2\alpha} } \B(1+\f{\eta^{2}}{(\eta-\xi)^{2}}\B)^{\f32-\f{1}{\alpha}}E^{\f32-\f{1}{\alpha}}(\eta)
\B(\int_{T-r^{2}}^{T}\|u\|^{p}_{\dot{\mathcal{M}}^{q,1}(B(\eta))}dt\B)^{\f{2}{p\alpha}}.
\ea$$
Here we used the fact below
$$
\|\psi u\|_{\dot{\mathcal{M}}^{q,1}(\mathbb{R}^{3})}\leq C
\|u\|_{\dot{\mathcal{M}}^{q,1}(B(\eta))},
$$
which can be derived from the definition of Morrey spaces. This completes the proof.
\end{proof}

\begin{proof}[Proof of  Proposition \ref{the1.2}]
Indeed, consider $0<R/2\leq \xi<\f{3\xi+\eta}{4}<\f{\xi+\eta}{2}<\f{\xi+3\eta}{4}<\eta\leq R$. Let $\phi(x,t)$ be a non-negative smooth function supported in $Q(\f{\xi+\eta}{2})$ such that
$\phi(x,t)\equiv1$ on $Q(\f{3\xi+\eta}{4})$,
$|\nabla \phi| \leq  C/(\eta-\xi) $ and $
|\nabla^{2}\phi|+|\partial_{t}\phi|\leq  C/(\eta-\xi)^{2} .$

The local energy inequality \eqref{loc}, the decomposition of  pressure  in Lemma \ref{lem2} and the H\"older inequality ensure that
\begin{align}
 &\int_{B(\f{\eta+\xi}{2})} |u(x,t)|^{2} \phi(x,t) dx
 +2\iint_{Q(\f{\eta+\xi}{2})}
  |\nabla u|^{2}\phi  dxds\nonumber\\\leq &  \f{C\eta^{5/3}}{(\eta-\xi)^{2}}\B(\iint_{Q(\f{\xi+3\eta}{4})} |u|^{3}
dxds\Big)^{2/3}+\f{C}{(\eta-\xi)} \iint_{Q(\f{\xi+3\eta}{4})}  |u|^{3} dxds\nonumber\\&+ \f{C\eta^{3}}{(\eta-\xi)^{4}}  \|u\|^{3}_{L^{3}(Q(\f{\xi+3\eta}{4}))}+\f{C\eta^{3/2}}{(\eta-\xi)^{4}} \|  \pi\|_{L^{1} (Q(\eta))}  \|u\|_{L^{\infty}L^2(Q(\eta)))}\nonumber\\
=&:I+II+III+IV.\label{last3}
 \end{align}
Combining
  \eqref{keyineq} and the Young inequality, we obtain
\begin{align}\nonumber
 I\leq &  \f{C\eta^{3+\alpha}}{(\eta-\xi)^{3\alpha}} \B(1+\f{\eta^{2}}{(\eta-\xi)^{2}}\B)^{ \f{3\alpha-2}{2}}
\|u\|_{L^{p}(T-\eta^{2},T;\dot{\mathcal{M}}^{q,1}(B(\eta)))}^{2}
+\f{1}{6}E(\eta),\\
\nonumber II
\leq & \f{C\eta^{3}}{(\eta-\xi)^{\f{2\alpha}{2-\alpha}}}   \B(1+\f{\eta^{2}}{(\eta-\xi)^{2}}\B)^{ \f{3\alpha-2}{2-\alpha}}
\|u\|_{L^{p}(T-\eta^{2},T;\dot{\mathcal{M}}^{q,1}(B(\eta)))}^{\f{4}{2-\alpha}}+\f{1}{6}E(\eta),\\ III\leq& \f{C\eta^{\f{3(\alpha+2)}{2-\alpha} }}{(\eta-\xi)^{\f{8\alpha}{(2-\alpha)}}}
\B(1+\f{\eta^{2}}{(\eta-\xi)^{2}}\B)^{ \f{3\alpha-2}{2-\alpha}}
\|u\|_{L^{p}(T-\eta^{2},T;\dot{\mathcal{M}}^{q,1}(B(\eta)))}^{\f{4}{2-\alpha}} +\f{1}{6}E(\eta).\label{last2}
\end{align}
In light of the definition of Morrey spaces, we see that
$$\|\pi\|_{L^{1}(B(\eta))}\leq C\eta^{3-\f{3}{m}} \|\pi\|_{\dot{\mathcal{M}}^{m,1}(B(\eta))}.$$
Using the Young inequality again, we conclude that
\be\ba IV  & \leq   \f{C\eta^{3 }}{(\eta-\xi)^{8}} \|  \pi\|^{2}_{L^{1} (Q(\eta))}+\f{1}{6}E(\eta)\\
& \leq \f{C\eta^{9-\f6m }}{(\eta-\xi)^{8}} \|  \pi\|^{2}_{L^{1}(T-\eta^{2},T;\dot{\mathcal{M}}^{m,1}(B(\eta)))} +\f{1}{6}E(\eta).
\label{last1}
\ea\ee
After plugging  \eqref{last2}-\eqref{last1} into \eqref{last3}, we apply
  the  classical Iteration Lemma  \cite[Lemma V.3.1,   p.161]{[Giaquinta]} to finish the proof.
\end{proof}

\begin{proof}[Proof of Corollary \ref{the1.3}]
Recall the $\varepsilon$-regularity criteria below shown in \cite{[HWZ]}: $u\in L^{\infty}(Q(1/2)) $ provided that
\be\label{j0il1}\ba
&\|u\|_{ L^{p }(-1,0;L^{q}(B(1))) }+\|\pi\|_{ L^{1}(Q(1)) }\leq \varepsilon_{3},\ea\ee
where
$$1\leq 2/p+3/q <2 ~~ \text{and }~~1< p<\infty.$$
Note that for any $m\geq 1$,
$$
\|\pi\|_{L^{1}(B(r))}\leq Cr^{3-\f{3}{m}} \|\pi\|_{\dot{\mathcal{M}}^{m,1}(B(r))}.
$$
This together with
the energy  bound \eqref{key ineq}   and
\eqref{j0il1} means \eqref{0il1}.
\end{proof}
\subsection{Proof of Theorem \ref{the1.1}  }

\begin{proof}
Let $\lambda=[2a(T-t)]^{-1/2}$. By means of an elementary change of variables and the definition of Morrey spaces, we have
$$\ba
R^{3(\f{1}{q}-\f{1}{l})}\B(\int_{B_{x}(R)\cap B_{x_{0}}(r)}|u(y,t)|^{l}dy\B)^{\f1l}&\leq   R^{3(\f{1}{q}-\f{1}{l})}\B(\int_{B_{x}(R) }|u|^{l}dy\B)^{\f1l}\\
&=   R^{3(\f{1}{q}-\f{1}{l})}\B(\int_{B_{x}(R) }|\lambda U(\lambda y)|^{l}dy\B)^{\f1l}\\
&\leq\lambda^{1-\f3q}\|U\|
_{\mathcal{\dot{M}}^{q,l}(\mathbb{R}^{3})}.
\ea$$
Hence, it follows from \eqref{pm} that for any $r>0$,
 \be\ba\label{3.14}
&\| u\|_{\mathcal{\dot{M}}^{q,l}(B_{x_{0}}(r))}\leq
 \lambda^{1-\f3q}\|U\|
_{\mathcal{\dot{M}}^{q,l}(\mathbb{R}^{3})},
\\
&\| \pi\|_{\mathcal{\dot{M}}^{q/2,l/2}(B_{x_{0}}(r))}\leq
 \lambda^{2-\f{6}q}\|\widetilde{\Pi}\|
_{\mathcal{\dot{M}}^{q/2,l/2}(\mathbb{R}^{3})}\leq C\lambda^{2-\f{6}q}\|U\|^{2}
_{\mathcal{\dot{M}}^{q,l}(\mathbb{R}^{3})}.
\ea \ee
Recall that $\lambda(t)= (2a(T-t))^{-1/2}$ and choose $p>1$ such that $2-p<p-\f{3p}q<2$, then there hold
 \be\ba
&\int_{T-r^{2}}^{T}\| u\|^{p}_{\mathcal{\dot{M}}^{q,l}(B_{x_{0}}(r))}dt\leq
\int_{T-r^{2}}^{T}\lambda^{p-\f{3p}q}\|U\|^{p}
_{\mathcal{\dot{M}}^{q,l}(\mathbb{R}^{3})} dt<\infty,\\
&\int_{T-r^{2}}^{T}\| \pi\|_{\mathcal{\dot{M}}^{q/2,l/2}(B_{x_{0}}(r))}dt\leq
C\int_{T-r^{2}}^{T}\lambda^{2-\f{6}q} \|U\|^{2}
_{\mathcal{\dot{M}}^{q,l}(\mathbb{R}^{3})} dt<\infty.
\ea
\ee
At this stage, the proof of Theorem \ref{the1.1} follows at once from Proposition \ref{the1.2} and  \eqref{local-energy}.
\end{proof}

\section{Proof of Theorem \ref{the1.4} }
We divide the proof of Theorem \ref{the1.4} into three steps. In
Step 1, utilizing the local suitable weak solutions and local Meyer-Gerard-Oru interpolation
inequality \eqref{keyineq}, we establish a new Caccioppoli type inequality. Step 2 is devoted to constructing the pressure $\Pi$ via  $A_p$ weighted inequalities. In the last step, an application of local energy inequalities leads to the proof of
Theorem \ref{the1.4}.

In what follows, we set
$$\|\cdot\| _{L^{p}\dot{\mathcal{M}}^{q,l}(Q(r))}:=\|\cdot\| _{L^{p}(T-r^{2},T;\dot{\mathcal{M}}^{q,l}(B(r)))}.$$
\subsection{Caccioppoli type inequality}
\begin{prop}\label{cip}
Assume that $u$ is a local suitable weak solution to the Navier-Stokes equations \eqref{NS}. Then there holds for any $q>3/2$,
  $$\ba&\|u\|^{2}_{\wred{L^{3}L^{\f{18}{5}}}(Q(\f{R}{2}))}+ \|\nabla u\|^{2}_{L^{2}(Q(\f{R}{2}))}\\
\leq &
  CR^{\f{5q-12}{3q}}\| u\|^{2}_{L^{\f{6q}{2q-3}}\dot{\mathcal{M}}^{q,1}(Q(R))}
  +CR^{\f{4q-15}{2q-3}}\| u\|^{\f{6q}{2q-3}}_{L^{\f{6q}{2q-3}}\dot{\mathcal{M}}^{q,1}(Q(R))}
  +CR^{\f{2q-6}{q}}\| u\|^{3}_{L^{\f{6q}{2q-3}}\dot{\mathcal{M}}^{q,1}(Q(R))}.
\ea$$
 \end{prop}
 As a straightforward consequence of the above proposition, we establish a  regularity criterion at one scale without pressure in Morrey spaces for local suitable weak solutions to \eqref{NS}. The H\"older inequality together with any result of \cite{[CW],[Wolf1],[Wolf2],[JWZ],[WWZ]} and this proposition  yields the desired.
 \begin{coro}\label{the1.5}
		Let  the pair $(u, \pi)$ be a local suitable weak solution to the 3D Navier-Stokes system \eqref{NS} in $Q(1)$.
		There exists an absolute positive constant $\varepsilon_{2}$
		such that if the pair $(u,\pi)$ satisfy
\be\label{0il}\ba
\int_{-1}^{0}\|u\|^{\f{6q}{2q-3}}_{\mathcal{\dot{M}}^{q,1}(B(1))}dt \leq \varepsilon_{2} ~~\text{with} ~~ q>3/2,
\ea\ee
	 then  $u\in L^{\infty}(Q(1/2)).$
	\end{coro}
\begin{proof}[Proof of Proposition \ref{cip}]
 Consider  $0<R/2\leq r<\f{3r+\rho}{4}<\f{r+\rho}{2}<\rho\leq R$. Let $\phi(x,t)$ be a non-negative smooth function supported in $Q(\f{r+\rho}{2})$ such that
$\phi(x,t)\equiv1$ on $Q(\f{3r+\rho}{4})$,
$|\nabla \phi| \leq  C/(\rho-r) $ and $
|\nabla^{2}\phi|+|\partial_{t}\phi|\leq  C/(\rho-r)^{2} .$

Let $\nabla\pi_{h}=\mathcal{W}_{3,B(\f{r+3\rho}{4})}(u)$, then,
from \eqref{ph}-\eqref{p2}, we have
\begin{align}
&\|\nabla \pi_{h}\|_{L^{3}(Q(\f{r+3\rho}{4}))}\leq C\|u\|_{L^{3}(Q(\f{r+3\rho}{4}))},\label{wp1}\\
 &\|  \pi_{1}\|_{L^{2}(Q(\f{r+3\rho}{4}))}\leq C\|\nabla u\|_{L^{2}(Q(\f{r+3\rho}{4}))},\label{wp2}\\
 &\|  \pi_{2}\|_{L^{\f{3}{2}}(Q(\f{r+3\rho}{4}))}\leq C\|  |u|^{2}\|_{L^{\f{3}{2}}(Q(\f{r+3\rho}{4}))}.\label{wp3}
 \end{align}
Since $v=u+\nabla\pi_{h}$, the H\"older inequality and \eqref{wp1} allow us to write
\begin{align}
 \iint_{Q(\rho)} \wred{| v|^2}\Big|  \Delta \phi^{4}+  \partial_{t}\phi^{4}\Big|    \leq& \f{C}{(\rho-r)^{2}}\iint_{Q(\f{r+\rho}{2})} \B(|u|^{2}+|\nabla\pi_{h}|^{2}\B)
\nonumber\\\leq& \f{C\rho^{5/3}}{(\rho-r)^{2}}\B(\iint_{Q(\f{r+\rho}{2})} |u|^{3}+|\nabla\pi_{h}|^{3}\B)^{\f{2}{3}}
\nonumber\\\leq& \f{C\rho^{5/3}}{(\rho-r)^{2}}\|u\|_{L^{3}(Q(\f{r+3\rho}{4}))}^{2}.\label{53.2} \end{align}
 The H\"older  inequality, $v=u+\nabla\pi_{h}$ and  \eqref{wp1}  ensure
 \begin{align}
 \left|\iint_{Q(\rho)}|v|^{2}\phi^{3}u\cdot\nabla \phi\right|
\leq \f{C}{(\rho-r)}
 \| u \|^{3}_{L^{3}(Q(\f{r+3\rho}{4}))}.
 \end{align}
 It follows from the interior estimate of harmonic functions \eqref{h1} and \eqref{wp1} that
$$\ba\|\nabla^{2}\pi_{h} \|_{L^{20/7}(Q(\f{r+\rho}{2}))}&\leq
\f{\wred{C} (r+\rho)  }{(\rho-r)^{ 2}}
\|\nabla\pi_{h} \|_{L^{3}(Q( \f{r+3\rho}{4} ))}\\
&\leq
\f{ C\rho  }{(\rho-r)^{ 2}}
\|u \|_{L^{3}(Q( \f{r+3\rho}{4}))},
\ea$$
which in turn implies that
\begin{align}
&\left|\iint_{Q(\rho)} \phi^{4}( u\otimes v :\nabla^{2}\pi_{h} )\right|  \nonumber\\
\leq&
\| v\phi^{2}\|_{L^{3}(Q(\f{r+\rho}{2}))}
\| u \|_{L^{3}(Q(\f{r+\rho}{2}))}\|\nabla^{2}\pi_{h} \|_{L^{3}(Q(\f{r+\rho}{2}))}
\nonumber\\
\leq &\f{ C\rho  }{(\rho-r)^{ 2}}
\|u \|^{3}_{L^{3}(Q( \f{r+3\rho}{4} ))}.
\end{align}
The H\"older inequality, \eqref{wp2} and Young's inequality yield that
\begin{align}
\left|\iint_{Q(\rho)} \phi^{3} \pi_{1}v\cdot\nabla \phi\right|
&\leq \f{C}{(\rho-r)}\| v\|_{L^{2}(Q(\f{r+\rho}{2}))}
\| \pi_{1} \|_{L^{2}(Q(\f{r+\rho}{2}))} \nonumber\\
&\leq \f{C}{(\rho-r)^{2}}\| v\|^{2}_{L^{2}(Q(\f{r+\rho}{2}))}
+\f{1}{16}\| \pi_{1} \|^{2}_{L^{2}(Q(\f{r+3\rho}{4}))} \nonumber\\
&\leq \f{C\rho^{5/3}}{(\rho-r)^{2}}\|u\|_{L^{3}(Q(\f{r+3\rho}{4}))}^{2}+\f{1}{16}\| \nabla u\|^{2}_{L^{2}(Q(\f{r+3\rho}{4}))}.
\end{align}
We deduce by the H\"older inequality  and \eqref{wp3}   that
\begin{align}
\left|\iint_{Q(\rho)} \phi^{3} \pi_{2}v\cdot\nabla \phi\right|
 \leq \f{C}{(\rho-r)}\| v\phi^{2}\|_{L^{3}(Q(\f{r+\rho}{2}))}
\| \pi_{2} \|_{L^{\f{3}{2}}(Q(\f{r+\rho}{2}))} \leq\f{ C }{(\rho-r)}
\|u \|^{3}_{L^{3}(Q(\f{r+3\rho}{4} ))}.\label{locp5}
\end{align}
Inserting \eqref{53.2}-\eqref{locp5} into the local energy inequality \eqref{wloc1}, we arrive at
\begin{align}
\sup_{T-\rho^{2}\leq t\leq T}\int_{B(\rho)}|v\phi^{2}|^2   + \iint_{Q(\rho)}\big|\nabla( v\phi^{2})\big|^2  \leq& \f{C\rho^{5/3}}{(\rho-r)^{2}}\|u\|_{L^{3}(Q(\f{r+3\rho}{4}))}^{2} +\f{ C\rho  }{(\rho-r)^{ 2}}
\|u \|^{3}_{L^{3}(Q( \f{r+3\rho}{4} ))}\nonumber\\&+\f{ C }{(\rho-r)}
\|u \|^{3}_{L^{3}(Q( \f{r+3\rho}{4} ))}+\f{1}{16}\| \nabla u\|^{2}_{L^{2}(Q(\f{r+3\rho}{4}))}.\label{keyl}
  \end{align}
The interior estimate of harmonic functions
  \eqref{h1} and \eqref{wp1} provide the bound
\be\ba\label{6.10}
\|\nabla\pi_{h}\|^{2}_{L^{3}L^{\f{18}{5}}(Q(r))}&\leq \f{Cr^{\f{5}{3}}}{(\rho-r)^{2}}\|\nabla\pi_{h}\|^{2}_{L^{3}(Q(\f{r+3\rho}{4}))}\leq \f{Cr^{\f{5}{3}}}{(\rho-r)^{2}}
\|u\|^{2}_{L^{3}(Q(\f{r+3\rho}{4}))}.
\ea\ee
Combining the triangle inequality, \eqref{keyl} and \eqref{6.10}, we discover that
\be\ba\label{j4.12}
 \|u\|^{2}_{L^{3}L^{\f{18}{5}}(Q(r))} \leq& \|v\|^{2}_{\wred{L^{3}L^{\f{18}{5}}}(Q(r))}+\|\nabla\pi_{h}\|^{2}_{\wred{L^{3}L^{\f{18}{5}}}(Q(r))}\\
\leq& C\B\{\|v\|_{\wred{L^{\infty}L^2}(Q(r))}^{2}+\|\nabla v\|_{L^{2}(Q(r))}^{2}\B\}+\f{Cr^{\f{5}{3}}}{(\rho-r)^{2}}\|u\|^{2}_{L^{3}(Q(\f{r+3\rho}{4}))}\\
\leq & \f{C\rho^{5/3}}{(\rho-r)^{2}}\|u\|_{L^{3}(Q(\f{r+3\rho}{4}))}^{2}+\f{ C\rho  }{(\rho-r)^{ 2}}
\|u \|^{3}_{L^{3}(Q( \f{r+3\rho}{4} ))} \\&+\f{ C }{(\rho-r)}
\|u \|^{3}_{L^{3}(Q( \f{r+3\rho}{4} ))}+\f{1}{16}\| \nabla u\|^{2}_{L^{2}(Q(\f{r+3\rho}{4}))}.
\ea\ee
According to \eqref{h1} and \eqref{wp1} again, we ascertain
$$
 \|\nabla^{2} \pi_{h}\|^{2}_{L^{2}(Q(r))}\leq \f{Cr^{3}}{(\rho-r)^{5}} \|\nabla\pi_{h}\|^{2}_{L^{2}(Q(\f{r+\rho}{2}))}\leq \f{Cr^{3}\rho^{5/3}}{(\rho-r)^{5}} \| u \|^{2}_{L^{3}(Q(\f{r+3\rho}{4}))}.
$$
Owing to  the triangle \wred{inequality and \eqref{keyl}}, we have
\begin{align}
  \|\nabla u\|^{2}_{L^{2}(Q(r))}\leq & \|\nabla v\|^{2}_{L^{2}(Q(r))}+
 \|\nabla^{2} \pi_{h}\|^{2}_{L^{2}(Q(r))}\nonumber\\ \leq &
  \B\{\f{C\rho^{5/3}}{(\rho-r)^{2}}+\f{Cr^{3}\rho^{5/3}}{(\rho-r)^{5}} \B\} \| u \|^{2}_{L^{3}(Q(\f{r+3\rho}{4}))}\nonumber\\&
  +\B\{\f{ C\rho  }{(\rho-r)^{ 2}} +\f{ C }{(\rho-r)}\B\}
\|u \|^{3}_{L^{3}(Q( \f{r+3\rho}{4} ))}+\f{1}{16}\| \nabla u\|^{2}_{L^{2}(Q(\f{r+3\rho}{4}))}. \label{65.11}
\end{align}
Adding \eqref{j4.12} to \eqref{65.11}, we derive that
\begin{align}
   &\|u\|^{2}_{L^{3}L^{\f{18}{5}}(Q(r))}+\|\nabla u\|^{2}_{L^{2}(Q(r))}\\\leq&
  \B\{\f{C\rho^{5/3}}{(\rho-r)^{2}}+\f{Cr^{3}\rho^{5/3}}{(\rho-r)^{5}} \B\} \| u \|^{2}_{L^{3}(Q(\f{r+3\rho}{4}))}\nonumber\\&
  +\B\{\f{ C\rho  }{(\rho-r)^{ 2}} +\f{ C }{(\rho-r)}\B\}
\|u \|^{3}_{L^{3}(Q( \f{r+3\rho}{4} ))}+\f{1}{8}\| \nabla u\|^{2}_{L^{2}(Q(\f{r+3\rho}{4}))}. \label{j65.11}
\end{align}
Thus, the key ingredient is to control $\|u \|^{3}_{L^{3}}$  on the right hand side of the above inequality.
To this end, invoking the H\"older inequality and \eqref{mgoi}, we see that
\be\label{6.12}\ba
\|u\|^{3}_{L^{3}(B(\f{r+3\rho}{4}))}\leq& C \rho^{9(\f13-\f{3}{2q+6})}\|u\|^{3}_{L^{\f{2q+6}{3}}(B(\f{r+3\rho}{4}))}\\
\leq& C \rho^{\f{3(2q-3)}{2q+6}}\|u\|^{\f{18}{2q+6}}_{\dot{H}^{1}(\mathbb{R}^{3})}
\|u\|^{3-\f{18}{2q+6}}_{\dot{\mathcal{M}}^{q,1}(\mathbb{R}^{3})}.
\ea\ee
Choose a cut-off function $\psi\in C_{0}^{\infty}(B(\rho))$ satisfying $0\leq\psi\leq1$, $\psi\equiv1$ in $B(\f{r+3\rho}{4})$ and $|\nabla\psi|\leq \f{C}{|\rho-r|}$.
In view of triangle inequality and the classical Poincar\'e inequality, we deduce that
\be\ba
\|\nabla [(u-\bar{u}_{B(\rho)})\phi]\|_{L^{2}({B(\rho)})}
\leq& \|\phi\nabla u \|_{L^{2}({B(\rho)})}+\|\nabla\phi (u-\bar{u}_{B(\rho)})\|_{L^{2}({B(\rho)})}\\
\leq&   (C+\f{C\rho}{\rho-r}) \|  \nabla u\|_{L^{2}({B(\rho)})} \\
\leq&    \f{C\rho}{\rho-r}  \|  \nabla u\|_{L^{2}({B(\rho)})}.
\ea\ee
We calculate
\be\label{6.142}\ba
\|(u-\bar{u}_{B(\rho)})\phi\|_{\dot{\mathcal{M}}^{q,1}
(\mathbb{R}^{3})}&\leq
\|u\phi\|_{\dot{\mathcal{M}}^{q,1}(B(\rho))}
+\|\phi\bar{u}_{B(\rho)}\|_{\dot{\mathcal{M}}^{q,1}(B(\rho))}\\
&\leq C
\|u\|_{\dot{\mathcal{M}}^{q,1}(B(\rho))}.
\ea\ee
Thanks to the triangle inequality again and \eqref{6.12}-\eqref{6.142}, we infer that$$\ba
\|u\|^{3}_{L^{3}(B(\f{r+3\rho}{4}))}
\leq& C \rho^{\f{3(2q-3)}{2q+6}}
\|u-\bar{u}_{B(\rho)}\|^{3}_{L^{\f{2q+6}{3}}(B(\f{r+3\rho}{4}))}
+\|\bar{u}_{B(\rho)}\|^{3}_{L^{3}(B(\f{r+3\rho}{4}))}\\\leq& C \rho^{\f{3(2q-3)}{2q+6}}
\|\phi(u-\bar{u}_{B(\rho)})\|^{3}_{L^{\f{2q+6}{3}}(B(\rho))}
+\|\bar{u}_{B(\rho)}\|^{3}_{L^{3}(B(\f{r+3\rho}{4}))}\\
\leq& C \rho^{\f{3(2q-3)}{2q+6}}\|(u-\bar{u}_{B(\rho)})\phi\|^{\f{18}{2q+6}}_{\dot{H}^{1}(\mathbb{R}^{3})}
\|(u-\bar{u}_{B(\rho)})\phi\|^{3-\f{18}{2q+6}}_{\dot{\mathcal{M}}^{q,1}(\mathbb{R}^{3})}+C
\rho^{3-\f{9}{q}}\|u\|^{3}_{\dot{\mathcal{M}}^{q,1}(B(\rho))}\\
\leq&    \f{C\rho^{ \f{8q-3 }{2q+6}}}{(\rho-r)}   \|\nabla u\|^{\f{18}{2q+6}}_{L^{2}(B(\rho))}
\|u\|^{3-\f{18}{2q+6}}_{\dot{\mathcal{M}}^{q,1}(B(\rho))}+C
\rho^{3-\f{9}{q}}\|u\|_{\dot{\mathcal{M}}^{q,1}(B(\rho))}^{3}.
\ea$$

From the H\"older inequality, we know that
$$\ba
&\int^{T}_{T-\f{(r+3\rho)^{2}}{16}}\|u\|^{3}_{L^{3}(B(\f{r+3\rho}{4}))}dt\\
\leq& \f{C\rho^{\f{ 8q-3}{2q+6}}}{(\rho-r)} \int^{T}_{T-\f{(r+3\rho)^{2}}{16}}\|\nabla u\|^{\f{18}{2q+6}}_{L^{2}(B(\rho))}
 \|u\|^{\f{6q}{2q+6}}_{{\dot{\mathcal{M}}^{q,1}(B(\rho))}}dt+
C\rho^{3-\f{6}{q}} \int^{T}_{T-\f{(r+3\rho)^{2}}{16}}\|u\|^{3}_{\dot{\mathcal{M}}^{q,1}(B(\rho))}dt\\
\leq& \f{C\rho^{\f{ 8q-3}{2q+6}}}{(\rho-r)}\B(\int^{T}_{T-\f{(r+3\rho)^{2}}{16}}\|\nabla u\|^{2}_{L^{2}(B(\rho))}dt\B)^{\f{9}{2q+6}}
\B(\int^{T}_{T-\f{(r+3\rho)^{2}}{16}}
 \|u\|^{\f{6q}{2q-3}}_{{\dot{\mathcal{M}}^{q,1}(B(\rho))}}dt\B)^{\f{2q-3}{2q+6}}\\&+C
\rho^{3-\f{6}{q}}\B(\int^{T}_{T-\f{(r+3\rho)^{2}}{16}}
\|u\|^{\f{6q}{2q-3}}_{\dot{\mathcal{M}}^{q,1}(B(\rho))}dt\B)^{\f{2q-3}{2q}},
\ea$$
where we have used the fact that $\f{18}{2q+6}<2$ and $3\leq \f{6q}{2q-3}$.

Plugging this inequality into \eqref{j65.11} and using the Young inequality, we obtain
   $$\ba
&\|u\|^{2}_{L^{3}L^{\f{18}{5}}(Q(r))}+ \|\nabla u\|^{2}_{L^{2}(Q(r))}
  \\\leq&
  \B\{\f{C\rho^{5/3}}{(\rho-r)^{2}}+\f{Cr^{3}\rho^{5/3}}{(\rho-r)^{5}} \B\}^{\f{q+3}{q}}\f{\rho^{\f{8q-3}{3q}}}{(\rho-r)^{\f{2q+6}{3q}}} \| u \|^{2}_{L^{\f{6q}{2q-3}}\dot{\mathcal{M}}^{q,1}(Q(\rho))}\\
  &+\B\{\f{C\rho^{5/3}}{(\rho-r)^{2}}+\f{Cr^{3}\rho^{5/3}}{(\rho-r)^{5}} \B\}
  \rho^{\f{2q-4 }{q}}\|u\|^{2}_{L^{\f{6q}{2q-3}}\dot{\mathcal{M}}^{q,1}(Q(\rho))}\\&
  +\B\{\f{ C\rho  }{(\rho-r)^{ 2}} +\f{ C }{(\rho-r)}\B\}^{\f{2q+6}{2q-3}}\f{\rho^{\f{8q-3}{2q-3}}}{(\rho-r)^{\f{2q+6}{2q-3}}}
\|u \|^{\f{6q}{2q-3}}_{L^{\f{6q}{2q-3}}\dot{\mathcal{M}}^{q,1}(Q( \rho))}\nonumber\\
&+\B\{\f{ C\rho  }{(\rho-r)^{ 2}} +\f{ C }{(\rho-r)}\B\}\rho^{\f{3q-6}{q}}\|u \|^{3}_{L^{\f{6q}{2q-3}}\dot{\mathcal{M}}^{q,1}(Q( \rho))}+\f{3}{16}\| \nabla u\|^{2}_{L^{2}(Q(\rho))}.
 \ea $$
Now, we are in a position to apply  the classical \wred{Iteration Lemma} \cite[Lemma V.3.1,   p.161]{[Giaquinta]} to find that
$$\ba
&\|u\|^{2}_{\wred{L^{3}L^{\f{18}{5}}}(Q(\f{R}{2}))}+ \|\nabla u\|^{2}_{L^{2}(Q(\f{R}{2}))}\\
\leq &
  CR^{\f{5q-12}{3q}}\| u\|^{2}_{L^{\f{6q}{2q-3}}\dot{\mathcal{M}}^{q,1}(Q(R))}
  +CR^{\f{4q-15}{2q-3}}\| u\|^{\f{6q}{2q-3}}_{L^{\f{6q}{2q-3}}\dot{\mathcal{M}}^{q,1}(Q(R))}
  +CR^{\f{2q-6}{q}}\| u\|^{3}_{L^{\f{6q}{2q-3}}\dot{\mathcal{M}}^{q,1}(Q(R))}.
\ea$$
  This achieves the proof of \wred{this proposition}.
\end{proof}

\subsection{Proof of Theorem \ref{the1.4}}
\begin{proof}
It follows from \eqref{3.14} and $\f32<q<6$ that for any $r>0$,
$$
\int_{T-R^{2}}^{T}\| u\|^{\f{6q}{2q-3}}_{\mathcal{\dot{M}}^{q,1}(B(r))}dt\leq
\int_{T-R^{2}}^{T}\lambda^{\f{6q}{2q-3}(1-\f{3}q)}\|U\|^{\f{6q}{2q-3}}
_{\mathcal{\dot{M}}^{q,1}(\mathbb{R}^{3})} dt<\infty.
$$
Thus, this together with Proposition 4.1 implies
\be\label{5.15}
\|u\|^{2}_{L^{3}(Q(\f{R}{2}))}+ \|\nabla u\|^{2}_{L^{2}(Q(\f{R}{2}))}<\infty.
\ee
Let $\lambda_{0}=(2aT)^{-1/2}$. Changing the order of integration, we apply \eqref{5.15} to deduce that
\be\ba \label{6.1}
&\iint_{Q(1)}|u|^{3}dxdt=\int_{\mathbb{R}^{3}}|U|^{3}
 \lambda_{0}^{2}\min\big\{|y|^{-2},
 \lambda_{0}^{-2}\big\}dy,\\
 &\iint_{Q(1)}|\nabla u|^{2}dxdt=\int_{\mathbb{R}^{3}}|\nabla U|^{2}
 2\lambda_{0}^{2}\min\big\{|y|^{-1},\lambda_{0}^{-1}\big\}dy,
\ea\ee
which in turn implies that
$$
\|\nabla U\|_{L^{2}(B_{y_{0}}(1))}=o(|y_{0}|^{1/2}) ~~~\text{and}~~~
\|  U\|_{L^{3}(B_{y_{0}}(1))}=o(|y_{0}|^{2/3}), ~~\text{as} ~~|y_{0}|\rightarrow\infty.
$$
Next, to employ the local energy inequality \eqref{loc} and \eqref{local-energy}, we take into account
  recovering  pressure $\Pi$ via \eqref{6.1}.
Observe that \eqref{6.1} yields
\be\label{6.2}
 \int_{\mathbb{R}^{3}}|U|^{3}|y|^{-2}dy<\infty,
\ee
we can define $$\widetilde{\Pi}=\mathcal{R}_{i}\mathcal{R}_{j}(U_{i}U_{j}),$$
where $U = (U_1,U_2,U_3)$ is determined by \eqref{6.2}.
Due to the classical Calder\'on-Zygmund Theorem with $A_{3/2}$ weights, there holds
\be\label{6.14}
\|\widetilde{\Pi}|y|^{-\f{4}{3}}\|_{L^{3/2}(\mathbb{R}^{3})}\leq C\|U_{i}U_{j}|y|^{-\f{4}{3}}\|_{L^{3/2}(\mathbb{R}^{3})}.
\ee
This means the local integrability of $\widetilde{\Pi}$.
Thus, Weyl's lemma guarantees that $\widetilde{\Pi}$ is smooth.

To proceed further, the fact that $U\in W^{1,2}_{\rm loc}(\mathbb{R}^{3})$ and $\int_{\mathbb{R}^{3}}|U|^{3}(1+|y|)^{-2}dy<\infty$ allows us to  revise the proof of Lemma \ref{cp} to show  that
$\widetilde{\Pi}$ satisfies  $-\Delta \Pi=\partial_{i}\partial_{j}(U_{i}U_{j})$  in the distributional sense.
Here, we just prove that $\nabla \Pi=\nabla \widetilde{\Pi}$.

We use the same notations given in the proof of Lemma \ref{cp}.
By integration by parts, we have
$$\ba
\varepsilon^{3}\left|\int_{\mathbb{R}^{3}}\Delta U\varphi(\varepsilon y)dy\right|=&\varepsilon^{5}\left|\int_{\mathbb{R}^{3}} U|y|^{-\f{2}{3}}|y|^{\f{2}{3}}\Delta\varphi(\varepsilon y)dy\right| \\ \leq& \varepsilon^{5}
\B(\int_{|y|<\f{1}{\varepsilon}}|U|^{3}|y|^{-2}dy\B)^{\f13}
\B(\int_{|y|<\f{1}{\varepsilon}}|y||\Delta\varphi(\varepsilon y)|^{\f{3}{2}}dy\B)^{ \f23}\\
\leq& C\varepsilon^{\f{7}{3}} \B(
\int_{\mathbb{R}^{3}}
|U|^{3}|y|^{-2}dy\B)^{\f13}.
\ea$$
Exactly as the above, we get
$$\ba
\varepsilon^{3}\left|\int_{\mathbb{R}^{3}}U\varphi(\varepsilon y)dy\right| &\leq \varepsilon^{3} \B(\int_{|y|<\f{1}{\varepsilon}}|U|^{3}|y|^{-2}dy\B)^{\f13}
\B(\int_{|y|<\f{1}{\varepsilon}}|y||\varphi(\varepsilon y)|^{\f{3}{2}}dy\B)^{ \f23}\\
&\leq C\varepsilon^{\f{1}{3}}\B(
\int_{\mathbb{R}^{3}}
|U|^{3}|y|^{-2}dy\B)^{\f13}.
\ea$$
In view of  integrating
by parts once again, we see that
\be\ba
\varepsilon^3\int_{\mathbb{R}^{3}}(y\cdot\nabla) U(y)\varphi(\varepsilon y) dy = -\varepsilon^3\int_{\mathbb{R}^{3}} 3U(y)\varphi(\varepsilon y)dy- \varepsilon^3 \int_{\mathbb{R}^{3}} U(y)\{(\varepsilon y)\cdot \nabla\varphi(\varepsilon y)\}dy.
\ea\ee
Hence, proceeding as the above, we can also control $\left|\varepsilon^3\int_{\mathbb{R}^{3}}(y\cdot\nabla) U(y)\varphi(\varepsilon y) dy\right|$.

Thanks to the  divergence-free condition, we arrive at
\be\ba
&\varepsilon^3\int_{\mathbb{R}^{3}}(U\cdot\nabla) U(y)\varphi(\varepsilon y) dy\\
=&-\varepsilon^{4}\int_{\mathbb{R}^{3}}  U\otimes U(y)\cdot\nabla\varphi(\varepsilon y)dy,
\ea\ee
which in turn implies that
\be\label{UUe2}\ba
&\varepsilon^3\left|\int_{\mathbb{R}^{3}}(U\cdot\nabla) U(y)\varphi(\varepsilon y) dy\right| \\
\leq& \varepsilon^{4} \B(\int_{|y|<\f{1}{\varepsilon}}|U|^{3}|y|^{-2}dy\B)^{\f23}
\B(\int_{|y|<\f{1}{\varepsilon}}|y|^{4}|\nabla\varphi(\varepsilon y)|^{3}dy\B)^{\f{1}3}\\
\leq& C\varepsilon^{\f{5}{3}}\B(
\int_{\mathbb{R}^{3}}
|U|^{3}|y|^{-2}dy\B)^{\f23}.
\ea\ee
It remains to bound the last term involving pressure $\widetilde{\Pi}$. Integration by parts gives
\be\ba
\varepsilon^3\int_{\mathbb{R}^{3}}\nabla \widetilde{\Pi}(y)\,\varphi(\varepsilon y) dy
  =-\varepsilon^{4}\int_{\mathbb{R}^{3}}  \widetilde{\Pi}(y)\nabla\varphi(\varepsilon y)dy.
\ea\ee
Furthermore, due to (\ref{6.14}), a variant of (\ref{UUe2}) provides the estimate
$$
\varepsilon^{4}\left|\int_{\mathbb{R}^{3}}  \widetilde{\Pi}(y)\nabla\varphi(\varepsilon y)dy\right|
\leq C\varepsilon^{\f{5}{3}}\B(
\int_{\mathbb{R}^{3}}
|U|^{3}|y|^{-2}dy\B)^{\f23}.
$$
Therefore, we get $\nabla\Pi=\nabla \widetilde{\Pi}$.
Then we can define $\pi$ by $\widetilde{\Pi}$ via \eqref{learysb}.
Note that
$$\iint_{Q(1)}|\pi|^{\f32}dxdt=\int_{\mathbb{R}^{3}}|\widetilde{\Pi}|^{\f32}
 \lambda_{0}^{2}\min\big\{|y|^{-2},
 \lambda_{0}^{-2}\big\}dy.$$
From this and \eqref{6.14}, we know that
$\pi\in L^{3/2}(Q(1 ))$. This together with $u\in L^{3}(Q(1 ))$ implies \eqref{local-energy}  via local energy inequality \eqref{loc}, which concludes the proof of Theorem \ref{the1.4}.
\end{proof}

\section*{Acknowledgement}
The authors would like to express their sincere gratitude to Dr. Daoguo Zhou for short discussion
involving  interpolation inequality in Morrey spaces.
Q. Jiu was partially supported by the National Natural Science Foundation of
China (NNSFC) (No. 11671273, No. 11931010), key research project of the Academy for Multidisciplinary Studies of CNU, and Beijing Natural Science Foundation (BNSF) (No. 1192001).
The research of Wang was partially supported by  the National Natural Science Foundation of China under grant (No. 11971446 and  No. 11601492).
The research of Wei was partially supported by the National Natural Science Foundation of China under grant (No. 11601423, No. 11701450, No. 11701451, No. 11771352, No. 11871057) and Scientific Research Program Funded by Shaanxi Provincial Education Department (Program No. 18JK0763).

config: arxiv


TITLE: If $A$ and $B$ are skew-symmetric matrices, then is $A+B$ a symmetric matrix?
QUESTION [0 upvotes]: Based on the title of the question I edited to be clear:
We have: $A^T=-A$, $B^T=-B$. Then, is $A+B$ a symmetric matrix?

REPLY [2 votes]: Suppose $A$ and $B$ are skew-symmetric, i.e., $A^T = -A$ and $B^T = -B$, then the sum is also skew-symmetric as we have
$$(A+B)^T=A^T+B^T=-A-B=-(A+B)$$
and not necessarily a symmetric matrix.


TITLE: Calculate $\int_{\mathbb{R}}e^{ix\omega}(\frac{\sin(\omega)}{\omega})^2 d\omega$ using convolution theorem.
QUESTION [2 upvotes]: I have study Fourier transformation and found this exercise: Calculate $\int_{\mathbb{R}}e^{ix\omega}(\frac{\sin(\omega)}{\omega})^2 d\omega$ using convolution theorem. I have notice that integral looks like inverse Fourier transformation, where $(\frac{\sin(\omega)}{\omega})^2$ is some function's Fourier transformation. Is there any hint how I can use convolution theorem to solve this?

REPLY [2 votes]: Let the Fourier Transform of $f(\omega)$ be given by
$$F(x)=\int_{-\infty}^\infty f(\omega)e^{i\omega x}\,d\omega$$
We have the Fourier Transform pairs
$$\begin{align}
f(\omega) &\leftrightarrow F(x)\\\\
f^2(\omega) &\leftrightarrow \frac{1}{2\pi}F(x)*F(x)\tag1\\\\
\frac{\sin(\omega)}{\omega}&\leftrightarrow \pi\,\text{rect}(x/2)\tag2\\\\
\frac{\sin^2(\omega)}{\omega^2}&\leftrightarrow \frac{1}{2\pi}\left(\pi\,\text{rect}(x/2)\right)*\left(\pi\,\text{rect}(x/2)\right)=\pi\, \text{tri}(x/2)\tag3
\end{align}$$
where $\text{rect}(t)$ and $\text{tri}(t)$ are the Rectangle Function and Triangle Function, respectively.

NOTES:
The pair in $(1)$ expresses the result of the Convolution Theorem.
To arrive at $(1)$, we use the result $\int_{-\infty}^\infty \frac{\sin(ax)}{x}\,dx=\pi \text{sgn}(a)$.  Proceeding we have
$$\begin{align}
\int_{-\infty}^\infty \frac{\sin(\omega)}{\omega}e^{i\omega x}\,d\omega&=\frac1{2i}\int_{-\infty}^\infty \frac{e^{i\omega(x+1)}-e^{i\omega(x-1)}}{\omega}\,d\omega\\\\
&=\frac\pi2 \text{sgn}(x+1)-\frac\pi2 \text{sgn}(x-1)\\\\
&=\pi\,\text{rect}(x/2)
\end{align}$$
And finally, using $(1)$ and $(2)$ we have
$$\begin{align}
\int_{-\infty}^\infty \left(\frac{\sin(\omega)}{\omega}\right)^2 e^{i\omega x}\,d\omega&=\frac1{2\pi}\int_{-\infty}^\infty \left(\pi\,\text{rect}((x-x')/2)\right)\left(\pi\,\text{rect}(x'/2)\right)\,dx'\\\\
&=\pi\,\text{tri}(x/2)
\end{align}$$


TITLE: Sampling algorithms on convex polytopes
QUESTION [2 upvotes]: Let $f=\mathbf{c}\cdot\mathbf{x}$ be the optimization objective function whose parameter vector $\mathbf{x}\in\mathbb{R}^n$ is subject to the following constraints in the very well-known linear-programming problem
$$
\mathbf{A}\mathbf{x}=\mathbf{b}\in\mathbb{R}^m\;,\quad \mathbf{0}\leq\mathbf{x}\leq\mathbf{v}
$$
Suppose that the space of feasible solutions of the above linear system is a convex polytope; my questions are the following:

How are the vertices of said polytope found in linear programming? Kindly reference the theoretical argument for that.
What are the MCMC algorithms to sample points from the polytope s.t. it eventually finds the optimal point for the object function?

REPLY [3 votes]: Here is a link for software and references: https://rdrr.io/cran/volesti/man/sample_points.html


TITLE: local minima of function in two variables
QUESTION [0 upvotes]: We just started to learn multivariable calculus, and we got the next definition:
We say that a point $x_{0} \in \mathbb{R}^{n}$ is a local minimum of a function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ if there exists a $\delta > 0$ for which the following holds:
$\left\| x-x_{0}\right\| _{\infty } \leq \delta \Rightarrow f\left( x\right) \geq f\left( x_{0}\right)$
given the function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ such that $f\left( x_{1}, x_{2}\right) = 2\left(x_{1}\right)^{2} + \left(x_{2}\right)^{2} + 2x_{1}x_{2} - 2$, find it's local minimum, if such a point exists. If there is a local minimum, prove directly from the above definition that it is a local minimum.
So I started by calculate the partial derivatives and got the critic point $x_{0} = \left(0, 0\right)$.
Now I now it's a minimum of $f$, but I got stuck on proving it from the definition.
My try was:
Let $\delta = 1>0$ and denote $x = \left( x_{1}, x_{2}\right)$. We can notice that:

$\left\| x-x_{0}\right\| _{\infty } \leq \delta \Leftrightarrow \left\| x\right\| _{\infty } \leq 1 \Leftrightarrow \left\| \left( x_{1}, x_{2}\right) \right\| _{\infty } \leq 1 \Leftrightarrow max\left\{ \left| x_{1}\right| ,\left| x_{2}\right| \right\} \leq 1 \Leftrightarrow \left| x_{1}\right| \leq 1\wedge \left| x_{2}\right| \leq 1 \Leftrightarrow -1\leq x_{1}\leq 1 \wedge -1\leq x_{2}\leq 1$

So, if $\left\| x-x_{0}\right\| _{\infty } \leq \delta$, then:
$f\left( x_{1}, x_{2}\right) = 2\left(x_{1}\right)^{2} + \left(x_{2}\right)^{2} + 2x_{1}x_{2} - 2 \geq 2\cdot \left( -1\right) ^{2} + \left( -1\right) ^{2} + 2 \cdot \left( -1\right) \cdot \left( -1\right) - 2 = 3 > -2 = f\left( 0, 0\right)$

But then I noticed that if I put the point $\left( \dfrac{1}{2}, \dfrac{1}{2} \right)$ in $f$ I get $f \left( \dfrac{1}{2}, \dfrac{1}{2} \right) = -\dfrac{3}{4} <3$, and I couldn't find my mistake in my proof.

REPLY [1 votes]: You wrote
$f\left( x_{1}, x_{2}\right) = 2\left(x_{1}\right)^{2} + \left(x_{2}\right)^{2} + 2x_{1}x_{2} - 2 \geq 2\cdot \left( -1\right) ^{2} + \left( -1\right) ^{2} + 2 \cdot \left( -1\right) \cdot \left( -1\right) - 2 = 3 > -2 = f\left( 0, 0\right).$
In this inequality you used that $2x_1x_2 \ge 2 \cdot (-1) \cdot (-1)=2$  for $|x_1| \le 1 $ and $|x_2| \le 1 .$
But this is false. Take $x_1=x_2= 1/4.$
My way:
we have $f(x_1,x_2)= (x_1+x_2)^2+x_1^2-2 \ge -2$  for all $(x_1,x_2)$ and $f(0,0) =-2.$
Hence
$$\min \{f(x_1,x_2): (x_1,x_2) \in \mathbb R^2\}=-2=f(0,0).$$


TITLE: How to compute $\zeta'(0)$?
QUESTION [9 upvotes]: I want to compute $\zeta'(0)$. I read that $\zeta'(0) = -\frac{\ln(2\pi)}{2}$. I try to use the fact that
$$
\zeta(1-s) = \frac{1}{\pi} (2 \pi)^{1-s} \cos(\frac{\pi s}{2}) \Gamma(s) \zeta(s)
$$
Take derivative from both sides:
\begin{align*}
-\zeta'(1-s) = (\frac{1}{\pi}(2 \pi)^{1-s} \Gamma(s)\zeta(s))'\cos(\frac{\pi s}{2}) + (\frac{1}{\pi}(2 \pi)^{1-s} \Gamma(s)\zeta(s))(-\sin(\frac{\pi s}{2}) \frac{\pi}{2})
\end{align*}
I try to let $s \to 1$ to get the answer, but $\zeta(s)\sin(\frac{\pi s}{2})$ diverges, and I don't know what will happen to the first term. I am not so familiar with the expansion of $\Gamma(s)$ or $\zeta(s)$. Sorry for asking this but I just don't know where to start.

REPLY [2 votes]: Euler-Maclaurin Sum Formula
In this answer, I use the Euler-Maclaurin Sum Formula to get, for $s\gt-1$,
$$
\hspace{-1cm}\sum_{k=1}^n\log(k)\,k^{-s}=\frac{\log(n)\,n^{1-s}}{1-s}-\frac{n^{1-s}}{(1-s)^2}-\zeta'(s)+\frac12\log(n)\,n^{-s}+O\!\left(\log(n)\,n^{-1-s}\right)\tag1
$$
Setting $s=0$ gives the logarithm of Stirling's Formula:
$$
\sum_{k=1}^n\log(k)=\log(n)\,n-n-\zeta'(0)+\frac12\log(n)+O\!\left(\log(n)\,n^{-1}\right)\tag2
$$
Matching the constant in Stirling's Formula gives
$$
\bbox[5px,border:2px solid #C0A000]{\zeta'(0)=-\frac12\log(2\pi)}\tag3
$$

Dirichlet $\boldsymbol{\eta}$
In this answer, it is shown that
$$
\eta'(0)=\frac12\log\left(\frac\pi2\right)\tag4
$$
Differentiating
$$
\eta(s)=\left(1-2^{1-s}\right)\zeta(s)\tag5
$$
gives
$$
\eta'(s)=\left(1-2^{1-s}\right)\zeta'(s)+\log(2)2^{1-s}\zeta(s)\tag6
$$
Plugging in $s=0$, and using $(4)$ and $\zeta(0)=-\frac12$, as shown in this answer, gives
$$
\bbox[5px,border:2px solid #C0A000]{\zeta'(0)=-\frac12\log(2\pi)}\tag7
$$

Integration by Parts with a Riemann-Stieltjes Integral
$$
\begin{align}
\sum_{k=1}^n\frac1{k^s}
&=\int_{1^-}^{n^+}\frac{d\lfloor x\rfloor}{x^s}\tag{8a}\\
&=\frac1{s-1}-\frac1{(s-1)\,n^{s-1}}-\int_{1^-}^{n^+}\frac{d\left(\{x\}-\frac12\right)}{x^s}\tag{8b}\\
&=\frac1{s-1}-\frac1{(s-1)\,n^{s-1}}+\frac1{2n^s}+\frac12-s\int_1^n\frac{\{x\}-\tfrac12}{x^{s+1}}\,\mathrm{d}x\tag{8c}
\end{align}
$$
For $s\gt-1$, the integral in $\text{(1c)}$ converges as $n\to\infty$.
As discussed in this answer, $\zeta(s)$ is the constant term in $\text{(8c)}$. Therefore, for $\mathrm{Re}(s)\gt-1$,
$$
\zeta(s)=\frac1{s-1}+\frac12-s\int_1^\infty\frac{\{x\}-\tfrac12}{x^{s+1}}\,\mathrm{d}x\tag9
$$
and therefore,
$$
\zeta'(s)=-\frac1{(s-1)^2}-\int_1^\infty\frac{\{x\}-\tfrac12}{x^{s+1}}+s\int_1^\infty\frac{\{x\}-\tfrac12}{x^{s+1}}\,\log(x)\,\mathrm{d}x\tag{10}
$$
Plugging in $s=0$ gives
$$
\begin{align}
\zeta'(0)
&=-1-\sum_{k=1}^\infty\int_0^1\frac{x-\frac12}{k+x}\,\mathrm{d}x\tag{11a}\\
&=-1-\sum_{k=1}^\infty\left(\color{#C00}{1}-\left(k+\frac12\right)\log\left(\frac{\color{#090}{k+1}}{\color{#00F}{k}}\right)\right)\tag{11b}\\
&=-1-\lim_{n\to\infty}\left(\color{#C00}{n}-\color{#090}{\sum_{k=2}^{n+1}\left(k-\frac12\right)\log(k)}+\color{#00F}{\sum_{k=1}^n\left(k+\frac12\right)\log(k)}\right)\tag{11c}\\
&=-1-\lim_{n\to\infty}\left(n-\left(n+\frac12\right)\log(n+1)+\sum_{k=1}^n\log(k)\right)\tag{11d}\\
&=\bbox[5px,border:2px solid #C0A000]{-\frac12\log(2\pi)}\tag{11e}
\end{align}
$$
Explanation:
$\text{(11a)}$: break the domain into unit intervals and sum
$\text{(11b)}$: integrate
$\text{(11c)}$: write the infinite sum as a limit
$\phantom{\text{(11c):}}$ substitute $k\mapsto k-1$ in the green sum
$\text{(11d)}$: pull out the $k=n+1$ term and combine the rest
$\text{(11e)}$: apply Stirling's Formula