text:
Mario Andretti Michael Andretti Sam Hornish Jr. Marco Andretti Tony Stewart Sports Automobile racing Formula One IndyCar Indianapolis 500
Sam Hornish Jr. Tony Stewart
Andretti at ease at Indy after stepping back from racing
By JENNA FRYER - May. 20, 2021 05:03 PM EDT
Marco Andretti climbs into his car during practice for the Indianapolis 500 auto race at Indianapolis Motor Speedway, Thursday, May 20, 2021, in Indianapolis. (AP Photo/Darron Cummings)
INDIANAPOLIS (AP) — His car so far is just OK and he could use a lot more speed ahead of Indianapolis 500 qualifying. That might normally send Marco Andretti into a tailspin, but his mindset is a whole lot different these days.
Andretti is at Indianapolis Motor Speedway as a part-time racer for the first time in his career. He decided in January he no longer had the heart for a full-time IndyCar season and made the Indy 500 his only race on the schedule.
As he readied for his third day of practice Thursday, the 34-year-old Andretti found himself "super happy, relieved and relaxed" for perhaps the first time in 16 trips to the "Greatest Spectacle in Racing."
"I think the thing I love about it is now I can form it into more of my own baby and not just what everybody else wants me to do," Andretti said. "I'm able to say yea or nay on different opportunities and do it for fun and for the purity of driving, which is why I love this sport."
Andretti is a third-generation racer in the most famous family in motorsports. His grandfather Mario Andretti is considered one of the greatest ever and a winner in IndyCar, Formula One and NASCAR. His 1969 Indianapolis 500 victory is the only race win for seven different Andretti racers and he ranks second on IndyCar's all-time wins list.
Michael Andretti also made it to Formula One — just one underwhelming season — but ranked third on IndyCar's all-time wins list when he retired. He turned his focus to the business of racing with Andretti Autosport, which fields 15 cars across seven different series around the globe.
The IndyCar team is the Andretti baby and Marco's decision to step away took an Andretti off the track for the first time since his 2006 debut. He said Thursday choosing a limited schedule wasn't difficult and although his father thinks his son struggled to inform him he wanted out, Marco thinks he made it easier on an Andretti organization struggling to find funding for his car.
"Honestly, it was more of a relief to him because we would have had to find more money," Marco Andretti said. "You know, even my close friends, they'll be like, 'Are you OK with everything?' And I'm like, 'Guys, yes. I really am."
His career has never come easy despite his dazzling debut as a 19-year-old in his first Indy 500. He passed his father to take the lead with two laps remaining and the win was within reach until Sam Hornish Jr. passed Andretti 100 yards from the finish line.
It was the closest Marco Andretti ever came to Indy success and he won just two races in a 15-year career. The pressure became so intense that any joy in the job was gone.
His new approach is perfect.
"I can still win the crown jewel, I know we can, and it's the one that matters," he said of the Indy 500. "And I'm free to do other things and I'm able to push myself, make myself by really uncomfortable to see if I can adapt to different types of cars and do it with friends and hang out with my cousin and drive with whoever I want. It's been really nice."
He's slated next month to compete in Tony Stewart's six-race Superstar All-Star Racing Experience at dirt tracks across the country and later this year team with cousin Jarett Andretti on the Andretti Autosport sports car team.
So for this week he is relaxed at Indy for the first time he can remember. He doesn't think he's got a shot at the pole — Andretti won the pole last year but was then the first driver since 2001 not to lead a single lap — so he's already thinking about strategy that could help him get to the front in the May 30 race.
And when that's over, his schedule is his to do what he chooses.
For now, that includes returning to Indianapolis every year.
"I want to just keep coming back here as long as I know I can be competitive and I think as long as I can keep coming back and just being annoying, hopefully one day I can win this thing," he said. "That's my bucket list. Everything else is just pure fun. Not that this isn't, this is the real business, but this is my main focus."


Chargers star wide receiver Keenan Allen suffered a kidney injury during his team's loss to the Ravens, and has been in the hospital since Sunday night. Now it looks like he might miss the rest of the season.
It's been a very bad week for NFL injuries.


text:
ASWC
Anderson and Fox to Fill Vacant ASWC Senate Seats
Eleanor Matson, Staff Reporter
Sophomore Paris Fox and senior Jack Anderson will fill the two vacant seats in the Associated Students of Whitman College (ASWC) senate. Last week a special election was held to determine who would become the new sophomore and senior senators, and the results of the special election were released this past Saturday.
Looking forward to the upcoming year, Fox discussed some of her plans for her time as an ASWC senator. One of her plans is to build more connections between Whitman and the larger Walla Walla community, so students will feel a closer bond with the surrounding community.
Sophmore senator Paris Fox. Photo by Caroline Ashford Arya
"I would like to foster more community interaction between Walla Walla and Whitman and have professionals come in from the community and talk to us about different things. I think Walla Walla is a great place and is often underrated," Fox said.
She also stressed the importance of addressing sexual assault on campus, as well as the divide between Greek and non-Greek students and the relationship between the two.
"I think that as far as sexual assault prevention and resources [goes]…[it] feels like there is a divide between the Greek community and the non-Greek community because Greeks are organized…and it is easy to get information out to them through liaisons and through [their] SVP (Sexual Violence Prevention) representatives. But for our non-Greek students it might be a less obvious pathway to resources," Fox said. "I think fostering more community between Greek and non-Greek affiliated students and somehow creat[ing] very clear pathways to resources that students need is important."
She also discussed plans related to sexual assault in regards to Title IX and her hopes to continue to refine the reporting system.
"I would like to get involved and help make…Title IX [and] Sexual Assault Prevention programs the best that [they] can be," she said.
Anderson also touched on these issues in his platform for the senate race. He hopes that during his tenure as a ASWC senator he can continue conversations about sexual assault and unify Greek and non-Greek students into a whole community. He did not develop his stance much further, however.
"By the end of the year, I'm hopeful that we will continue in our efforts to make the school more sustainable, free from sexual violence, while opening more outlets for more expression on campus," said Anderson in an email to The Whitman Wire.
Senior senator Jack Anderson. Photo by Caroline Ashford Arya
In his campaign video, Anderson elaborated more on his ideas to create more outlets for creative expression.
"It will be my mission to have [important] conversations and connect the campus into an interconnected whole. A good place to start for improving student engagement within the campus is to provide additional avenues for artistic performance and creative expression to play a greater role in how students experience student life," Anderson said in his video.
This past Sunday was the first senate meeting of the new school year, and was the first chance for both of these new senators to interact with ASWC. After this meeting, Fox mentioned that she looks forward to the chance to interact with more people on campus and to be more knowledgeable about events happening around campus.
"I would like to see the inner workings of ASWC and things that are happening at Whitman [during] my time here," Fox said. "Also, I am excited to get to know everyone…and to talk to the sophomore class and fellow students about change happening at Whitman and what they would like to be changed."
Associated Students of Whitman College
Paris Fox
Whitman College
Virtual events, emergency funding and academic communities: ASWC leaders reflect on semester goals
Bye bye, Betsy DeVos
Upcoming event: "Strategies for Responding to Election-Related Stress"
Upcoming event: "Becoming a Voting Rights Activist"
Under new leadership, ASWC adjusts to virtual learning

redpajama_set_name: RedPajamaCommonCrawl


Here are a few maps you might find useful.
Humboldt County Map, by Humboldt County.
Humboldt Bay Water Trails Map, by Explore North Coast, 2004 and updated by Redwood Community Action Agency, 2017.
Map of Humboldt Bay Beaches, Dunes and Wetlands, by Friends of the Dunes, 2013.
Nautical Chart 18622 of Humboldt Bay, by NOAA, Office of Coast Survey.
Nautical Chart 18605 Trinidad Harbor, by NOAA, Office of Coast Survey.
Nautical Chart 18600 Trinidad to Cape Blanco, by NOAA, Office of Coast Survey.
North Coast Refuges, an index map by the Humboldt Bay National Wildlife Refuge.
And, don't forget to check out some of our Favorite Places.


text:
Missing Article Test
by Angie Badillo | Tue., Jan. 29, 2019 1:24 PM
Halsey is addressing her split from G-Eazy, sharing what she learned from their breakup in a candid new interview.
The "Him & I" duo first called it quits in July after one year together, before getting back together about a month later. But by the end of October, a source told E! News that the couple had broken up once again. Though Halsey never made a public statement regarding the split, she did feel the pressure to speak out about the end of the relationship.
TheImageDirect.com
"The biggest lesson I learned was to make art, not headlines," she tells Glamour. "Because it can become quite easy, in the social media generation, to go from being a musician to becoming a personality."
Halsey goes on to talk about her song "Without Me," saying that it's about caring for a person so much that you end up losing sight of yourself.
"I call myself a collector; I collect things from people and use them to widen my artistic repertoire, so that I am writing from a culmination of experiences from the world," she shares. "But I'm an imitator as well, because I'm so passionately putting myself in other people's shoes all the time."


text:
\section{Introduction}\label{sec:intro}

This paper is concerned with the study of Hamiltonian system which are deeply studied because they model various physical systems. More precisely, we consider the following  ODEs on the $d=2d'$ dimensional torus $\T^d \simeq \mathbb R^d/ \Z^d$

\begin{align}\label{ode}
\begin{cases}
\dot{\gamma} (t) = J \nabla H(t, \gamma(t))
\\
\gamma (0) = x_0,
\end{cases}
\end{align}
where $H : \T^d \rightarrow \mathbb R$ is the autonomous Hamiltonian function, $x_0 \in \mathbb R^d$, $\gamma : [0,1] \rightarrow \T^d$ is a curve and $J $ is the $d \times d$ matrix
 $$J=\begin{pmatrix}
 0_{d'}  & I_{d'} \\
 -I_{d'} & 0_{d'}
 \end{pmatrix}.$$

  Such a $\gamma$ is called integral curve of the {\em Hamiltonian  vector field} $J \nabla H$ starting from $x_0$, more precisely we give the following definition.
\begin{definition}\label{defn:int-curve}
Let  $u: (0,1) \times \T^d \to \mathbb R^d$ be a Borel map. We say that $\gamma \in AC([0,1]; \T^d)$ is an integral curve of $u$ starting at $x$ if $\gamma(0)=x$ and $\gamma'(t) = u(t, \gamma(t))$ for a.e. $t\in [0,1]$.
\end{definition}
With such definition we recall that if the vector field is Lipschitz, which in our case means that $H \in W^{2, \infty}(\T^d, \mathbb R)$, then by Cauchy-Lipschitz theorem we have uniqueness of integral curves of \eqref{ode} for every starting point $x_0 \in \T^d$.

In the regular setting, i.e. when the Hamiltonian is in $W^{2, \infty}(\T^d)$, we have that the integral curves characterize the unique solution of the following continuity equation
\begin{align} \label{d_continuity}
\begin{cases}
\partial_t \rho + \diver_x (\rho J \nabla H) = 0,
\\
\rho(\cdot, 0)= \rho_0(\cdot),
\end{cases}
\end{align}
where the unknown is the density $\rho: [0,1] \times \T^d \rightarrow \mathbb R$, while the initial datum $\rho_0 \in L^\infty (\T^d)$ and the vector field $J \nabla H \in W^{1, \infty}(\T^d, \mathbb R^d)$ are given. More precisely, in this setting the unique solution (uniqueness holds thanks to a simple application of Gr\"onwall lemma) of \eqref{d_continuity}  is given by\footnote{We denote with $X(t, \cdot)_\# (\mu) $ the pushforward of the measure $\mu$ through the map $X$ at the fixed time $t$. }

\begin{equation}\label{RLF_continuity}
\rho(t, \cdot ) \mathscr{L}^d=X(t,\cdot )_\# ( \rho(0, \cdot ) \mathscr{L}^d),
\end{equation}

where $X : [0,1] \times \T^d \to \mathbb R^d$ is  called the flow related to $J \nabla H$ and is defined collecting all the integral curves, namely $X(t, x_0)= \gamma_{x_0}(t)$, where $\gamma_{x_0}$ is the unique integral curve of \eqref{ode} starting from $x_0$.  The last result is classical and is known in the literature as Liouville's theorem.

 A lot of efforts were made in the last decades to understand what happens in the non regular setting.
The first classical failing is the nonuniqueness of integral curves with vector fields $ u \in C^{0, \alpha} (\T^d, \mathbb R^d)$ for $\alpha \in [0,1)$, that means that the previous results cannot be extended easily to less regular spaces.

From now on we will refer to solutions of \eqref{d_continuity} in the sense of distribution, so the reasonable assumption to talk about that is that $H \in W^{1,q}(\T^d, \mathbb R^d)$ and $\rho \in L^1 ((0,1), L^p(\T^d))$.

 We now summarize the main results of the last decades that are important to understand our contribution. To do so we will present all the results in the particular case of autonomous hamiltonian vector fields, even if they do not require neither the autonomous hypothesis nor the hamiltonian.  The interested reader can find the general results in the references we provide.

DiPerna and Lions proved in \cite{DPL89} a result that implies the uniqueness of \eqref{d_continuity} solutions in the class of densities $\rho \in L^{\infty}((0,1), L^p(\T^d))$ once given $\rho_0 \in L^{p}(\T^d)$ and the autonomous hamiltonian vector field $H \in W^{2, q}(\T^d)$, with $p$ and $q$ such that

\begin{equation}
 \frac{1}{p} + \frac{1}{q} \leq 1.
\end{equation}

The authors in \cite{DPL89} consider also a selection of integral curves, which defines  the so-called ``Regular Lagrangian flow''. In turn, this characterizes the unique solution $\rho$ by \eqref{RLF_continuity} using as flow the ``Regular Lagrangian flow''.
To be precise here we give the definition of the ``Regular  Lagrangian flow'' with the compressibility condition following Ambrosio's definition (see in \cite[Section 6]{A04}).

\begin{definition}[Regular Lagrangian flow]\label{RLF}
Let $u:(0,1)\times\T^d\to \mathbb R^d$ be Borel. We say that a Borel map $X:[0,1]\times \mathbb R^d\to \mathbb R^d$ is a regular Lagrangian flow of $u$ if
	\begin{enumerate}[(i)]
		\item for $\mathscr{L}^d$-a.e. $x\in \T^d$, $t\mapsto X(t,x)$ is integral curve of $u$ with $X(0,x)=x$,
		\item there is a constant $C>0$ such that for every $t\in[0,1]$, $X(t,.)_\#\mathscr{L}^d\leq C\mathscr{L}^d$.
	\end{enumerate}
\end{definition}

In \cite{A08} Ambrosio proved the superposition principle, which implies the following: every {\em non negative } solution $\rho \in L^1$ to \eqref{d_continuity} with an autonomous hamiltonian vector field such that  $ \rho J \nabla H \in L^{1}$ are transported by a ``generalized flow'', which is roughly speaking a measure supported on integral curves of the associated vector field. See \cite[Theorem 3.2]{A08} for the precise result, known in the literature as ``Ambrosio's superposition principle''. This result connects the ODE \eqref{ode} with the PDE \eqref{d_continuity} in a larger class of regularity where uniqueness of solutions of \eqref{d_continuity}  fails and hence it is crucial for us (see in \cite{MoSz2019AnnPDE,MS20,MoSz2019Calc,BDLC20}).

We now turn to explain our contribution in this context. Large part of this paper is dedicated to prove a nonuniqueness result for positive solutions of the continuity equation \eqref{d_continuity} for Hamiltonian autonomous vector fields, more precisely we will prove the following theorem.

\begin{theorem}\label{t_ce}
Let
${{d\geq 4}}$ be an even integer, $p \in (1,\infty), r \in [1,\infty]$ be such that
\begin{equation}
\label{hp:exp}
\frac 1p + \frac 1r > 1+ \frac 1{d-1}
\end{equation}
and denote by $p'$ the dual exponent of $p$, i.e. $\frac{1}{p}+\frac{1}{p'}=1$.
Then for every $T>0$ there exists an autonomous Hamiltonian $H \in W^{2,r}(\T^d; \mathbb R)$ and a  nonconstant $\rho \in C ([0,T], L^1 (\T^d))$ such that $\rho J \nabla H \in C ([0,T], L^1 (\T^d))$ and \eqref{d_continuity} holds with initial data $\rho(0, \cdot) =1$ and for which $\rho \geq c_0$ for some positive constant $c_0$. Moreover, if $p' \geq d-1$, we have the Hamiltonian $H \in C(\T^d; \mathbb R)$.
\end{theorem}

The theorem above is proved using the ``convex integration type'' techniques borrowed from a groundbreaking work of Modena and Sz\'ekelyhidi \cites{MoSz2019AnnPDE,MoSz2019Calc} and subsequently improved  by Modena and Sattig \cite{MS20}.
 We refer to \cite{DLSZ09,DLSZ13,DLSZ17,Is18,BV} and the references therein for the birth of this and related lines of research. See also some  recent results related to convex integration in \cite{CDRS21,CL20,CL20NS,CL21,SP21}.

As a consequence of Ambrosio's superposition principle and Theorem \ref{t_ce} we prove a non uniqueness result of integral curves for autonomous Hamiltonian vector fields. This strategy has already been used in the work of Bru\'e, Colombo and De Lellis \cite[Theorem 1.3]{BDLC20}.

\begin{theorem} \label{t_main}
For any even integer $d \geq 4$ and any real number $r< d-1$  there is an autonomous  Hamiltonian  $H \in C(\T^d; \mathbb R) \cap W^{2,r}(\T^d; \mathbb R)$ such that the following holds for every Borel map $v$ with $v= J \nabla H \ \mathcal{L}^{d}-a.e.:$

(NU)
There is a measurable set $A \subset \T^d$ with positive Lebesgue measure such that for every $x \in A$ there

\hspace{0.8cm}  are at least two integral curves of $v$ starting at $x$.
\end{theorem}

The case $d=2$ is not included in our theorem: indeed in the 2 dimensional setting our theorem statement would tell us that $\frac{1}{p} + \frac{1}{r} > 2$ which is impossible for $p,r \geq 1$. Moreover, in \cite[Theorem 5.2]{ABC09}, the authors proved a uniqueness result  in dimension $2$ which implies the uniqueness for the continuity equation \eqref{d_continuity} in the class $\rho \in L^1((0,1) \times \mathbb R^2)$ for autonomous bounded Hamiltonian vector fields such that $H \in W^{2,1} \cap W^{1, \infty}$ (see \cite[Section 2.15 (iii)]{ABC09} for this implication).
They also explain in \cite[Section 6.2]{ABC09} the additional assumption that is needed in dimensions $d>2$, which in turn is necessary and in general not satisfied by $W^{2,r}(\T^d)$ with $r<d-1$, because of our result Theorem \ref{t_main}.
 We highlight that the Sard property, which is a key ingredient in \cite[Theorem 5.2]{ABC09},  is true in Sobolev spaces $W^{d,1}$ without any Lipschitz assumption thanks to \cite{BKK15}, where $d$ is the dimension of the space.

Finally we discuss the ``conservation of the Hamiltonian'' along integral curves. It is well known that for a smooth Hamiltonian vector field, the Hamiltonian is constant along integral curves of the flow.  Indeed, in physics  it represents the total energy of the system. For Sobolev fields, the same conclusion is true for the integral curves of the associated Regular Lagrangian Flow. This can be proved by an approximation argument. However, the Hamiltonian is not necessarily constant along the non-unique integral curves constructed in Theorem \ref{t_main}; this is the content of the next theorem.

\begin{theorem}\label{thm:hamilnotcons}
For any integer $d \geq 4$, and any real number $r < d-1$, there is an autonomous Hamiltonian  $H \in C(\T^d; \mathbb R) \cap W^{2,r}(\T^d; \mathbb R)$ such that the following holds for every $v = J \nabla H$ $\mathcal{L}^d- $a.e.. There is a measurable set $A \subset \T^d$ with positive Lebesgue measure such that for every $x \in A$ there exists $\gamma_x$ integral curve of $v$ starting from $x$ and
 $$H (\gamma_x(0)) > H (\gamma_x(1)),$$
in particular $H$ is not conserved along some integral curves starting from a non negligible set.
\end{theorem}

We now highlight the main new technical ideas in the convex integration scheme used to prove Theorem \ref{t_ce}:
\begin{itemize}
    \item We perturb the autonomous vector field in a ``universal way'', i.e. independently from the previous error (that in the literature is called ``Reynolds error''). This allows us to preserve the autonomous property of the Hamiltonian.

    \item We notice that it is not necessary to have $\rho \in L^p$ and $J \nabla H \in L^{p'}$, but only need that $\rho, \rho J \nabla H  \in L^1$ in order to make sense of a distributional solution to \eqref{d_continuity} and to apply Ambrosio superposition principle.
    \item We construct autonomous Hamiltonian vector fields (used in the scheme to perturb the previous Hamiltonian vector field)\footnote{This functions used in the perturbation step are called in the literature ``building blocks''.}.

\end{itemize}

\section{Preliminary Lemmas}

\subsection{Geometric lemma} We start with an elementary geometric fact, namely that every vector in $\mathbb R^d$ can be written as a positive linear combination of elements in a suitably chosen finite subset $\Lambda$ of $\mathbb{Q}^d\cap \partial B_1$. This is reminiscent of the geometric lemma in \cite{DLS13} and it is proved in \cite[Lemma 3.1]{BDLC20}.

\begin{lemma}\label{lemma:geom}
	There exists a finite set $\{ \xi\}_{\xi \in \Lambda} \subseteq \partial B_1\cap \mathbb{Q}^d$ and smooth non-negative coefficients $a_\xi(R)$ such that for every $R \in \partial B_1$
	\[
	R= \sum_{\xi \in \Lambda} a_\xi(R) \xi\, .
	\]
	Moreover, for each $\xi \in \Lambda$ there exists $\xi^\perp, \xi_1, .., \xi_{d-2}$ such that $\{ \xi, \xi^\perp, \xi_1, ..,\xi_{d-2} \} \subset \partial B_1 \cap \Q^d $ form an orthonormal  basis of $\mathbb R^d$ and $J \xi^\perp = \xi$.
	Finally, since we will periodize functions, let $n_\ast \in \N$  be
$$ \max_{\xi \in \Lambda} l_\xi, $$
where $l_{\xi}$ is
	  the l.c.m. of the denominators of the rational numbers $\xi, \xi^\perp, \xi_1,..,\xi_{d-2}$.
\end{lemma}

And we also recall the following result from \cite[Lemma 4.2]{BDLC20} to get the property of building blocks disjoint supports.

\begin{lemma}\label{lemma:disjointsupports}
Let $d \geq 3$, $\frac14 >\rho>0$ and $\Lambda \subseteq \mathbb S^{d-1} \cap \mathbb Q^d$ be a finite number of vectors. Then there exists $\mu_0:= \mu_0 (d , \Lambda) >0$ and a family of vectors $\{v_\xi\}_{\xi \in \Lambda} \subseteq \mathbb R^d$ such that the periodized cylinders
$v_\xi+ B_{2\rho \mu^{-1}} + \mathbb R \xi + \mathbb Z^d
$ are disjoint as $\xi$ varies in $\Lambda$, provided $\mu \geq \mu_0$.
\end{lemma}

\subsection{Antidivergences}
We recall that the operator $\nabla \Delta^{-1}$ is an anti-divergence when applied to smooth vector fields of $0$ mean. As shown in \cite[Lemma 2.3]{MoSz2019AnnPDE} and \cite[Lemma 3.5]{MS20}, however, the following lemma introduces an improved anti-divergence operator, for functions with a particular structure.

\begin{lemma}\label{lemma23}(Cp. with \cite[Lemma 3.5]{MS20})
	Let $\lambda \in \N$ and $f, g : \T^d \to \mathbb R$ be smooth functions, and $g_\lambda= g(\lambda x)$.
	Assume that $\int g = 0$. Then if we set $\ensuremath{\mathcal{R}} (f g_\lambda) = f \nabla \Delta^{-1} g_\lambda -\nabla \Delta^{-1} (\nabla f \cdot \nabla \Delta^{-1}g_\lambda+\int fg_\lambda)$, we have that $\diver  \ensuremath{\mathcal{R}} (f g_\lambda) = f g_\lambda-\int fg_\lambda$ and for some $C:=C({k,p})$
	\begin{equation}
	\label{ts:antidiv}
	\|D^k \ensuremath{\mathcal{R}} (f g_\lambda)\|_{L^p} \leq C \lambda^{k-1} \|f\|_{C^{k+1}} \| g\|_{W^{k,p}} \qquad \mbox{for every } k\in \N, p\in [1,\infty].
	\end{equation}
\end{lemma}
\begin{proof}
	It is enough to combine \cite[Lemma 3.5]{MS20} and the remark in \cite[page 12]{MS20}.
\end{proof}

\subsection{Slow and fast variables}
Finally we recall the following improved H\"older inequality, stated as in \cite[Lemma 2.6]{MoSz2019AnnPDE} (see also \cite[Lemma 3.7]{BV}).
If $\lambda \in \N$ and $f,g:\T^d \to \mathbb R$ are smooth functions, then we have
\begin{equation}
\label{eqn:impr-holder}
\| f(x) g(\lambda x) \|_{L^p} \leq \| f \|_{L^p} \| g \|_{L^p} + \frac{C(p)\sqrt d \|f \|_{C^1} \|g\|_{L^p}}{\lambda^{1/p}}
\end{equation}
and
\begin{equation}
\label{eqn:l26}
\Big| \int f(x) g(\lambda x) \, dx \Big| \leq\Big| \int f(x) \Big(g(\lambda x) - \int g \Big) \, dx \Big| + \Big| \int f \Big |\cdot \Big| \int g \Big| \leq \frac{\sqrt d \|f \|_{C^1} \|g\|_{L^1}}{\lambda} + \Big| \int f \Big| \cdot \Big | \int g \Big| .
\end{equation}

\section{Building blocks}
Let $0<\rho<\frac 14$ be a constan
. We consider $\varphi\in C^\infty_c (\mathbb R^{d-1})$ and $\psi\in C^\infty_c (\mathbb R^{d-1})$ which satisfy
	$$\varphi \in C^{\infty}_c(B_\rho) \qquad \int \varphi =1, \qquad  \varphi \geq 0, \qquad
	$$
	and
	$$\psi \in C^\infty_c(B_{2\rho}), \qquad \int \psi=0, \qquad \psi (x_1,x_2,..,x_{d-1}) = x_1  \mbox{ on }B_\rho.$$
Given $\mu \gg 1$ we define
\begin{align*}
\overline{\varphi}_\mu(x):= \mu^{(d-1)/p} \varphi(\mu x),
\\
\overline{\psi}_\mu(x):= \mu^{(d-1)/p'} \psi(\mu x).
\end{align*}

By an abuse of notation, we periodize $ \overline{\varphi} , \overline{\psi}_\mu $ so that the functions are treated as periodic functions defined on $\T^{d-1}$. These periodic functions will allow us to define our building blocks, defined on $\T^d$.

Given $\Lambda$ and $n_\ast \in \N$ as in Lemma \ref{lemma:geom}, for any $\xi \in \Lambda$ (we recall the notation of $\xi^\perp, \xi_1,..,\xi_{d-2}$ as in Lemma \ref{lemma:geom}) and for any $\sigma >0$, we define $\tilde{\Theta}_{\xi, \mu, \sigma}, \tilde{H}_{\xi,\mu} : \T^d \to \mathbb R$ and the autonomous Hamiltonian vector field $X_{\tilde{H}_{\xi,\mu}} : \T^d \to \mathbb R^d$

\begin{align*}
{\Theta}_{\xi,\mu, \sigma} (x) := \sigma n_\ast^{d-1} \overline{\varphi}_\mu (n_\ast \xi^\perp \cdot (x - v_{\xi}), n_\ast \xi_1 \cdot (x - v_{\xi}), n_\ast \xi_2 \cdot (x - v_{\xi}), .., n_\ast \xi_{d-2}  \cdot (x - v_{\xi})),
\\
{H}_{\xi,\mu}(x) := \frac{1}{\mu}   \overline{\psi}_\mu (n_\ast \xi^\perp \cdot (x - v_{\xi}), n_\ast \xi_1 \cdot (x - v_{\xi}), n_\ast \xi_2 \cdot (x - v_{\xi}), .., n_\ast \xi_{d-2}  \cdot (x - v_{\xi})),
\\
X_{{H}_{\xi,\mu}} := J \nabla {H}_{\xi,\mu}(x).
\end{align*}
where  $\mu \geq \mu_0(d, \Lambda)$ and  $\{v_{\xi} \}_{\xi \in \Lambda}$ are given by Lemma \ref{lemma:disjointsupports} in order to get the following property on the supports of these family of functions
$$\text{supp} X_{H_{\xi, \mu}}   \cap \text{supp} \Theta_{\xi',\mu,\sigma} = \text{supp} X_{H_{\xi, \mu}} \cap \text{supp} X_{H_{\xi', \mu}} = \text{supp} \Theta_{\xi',\mu,\sigma} \cap \text{supp} \Theta_{\xi,\mu,\sigma} = \emptyset,$$
for any $\xi \neq \xi' \in \Lambda$.
\begin{remark}
In the previous definition we just multiply $\varphi$ by $\sigma$, because we will need autonomous vector fields, but the natural choice is to split $\sigma>0$ as $\sigma^{1/p}$ and $\sigma^{1/p'}$ as in \cite[Section 4]{BDLC20}.
\end{remark}

Finally, by standard computations, we have proved the following fundamental lemma for our building blocks.

\begin{lemma} \label{lemma:buildingblocks}
Let $d \geq 4$, $\Lambda \subset \partial B_1 \cap \Q^d$ be a finite set. Then there exists $\mu_0 >0$ such that the following holds.

 There exist two families of functions
$\{ \Theta_{\xi, \mu,\sigma} \}_{\xi, \mu, \sigma} \subset C^\infty(\T^d)$, $\{ H_{\xi, \mu,} \}_{\xi, \mu} \subset C^\infty(\T^d;\mathbb R^d)$, where $\xi \in \Lambda$ and $ \sigma, \mu \in \mathbb R$ such that
 for any  $\mu \geq \mu_0$, $\sigma >0$  we have

 \begin{equation}\label{eqn:itsolves}
	\diver ( X_{H_{\xi, \mu}} \Theta_{\xi, \mu, \sigma})=0,
	\end{equation}
 $$\diver X_{H_{\xi, \mu}}=0,$$
 \begin{equation}
 \label{eqn:avW}
 \int  X_{H_{\xi, \mu}}=0,
 \end{equation}
	\begin{equation}\label{eqn:rightaverage}
	\int  X_{H_{\xi, \mu}} \Theta_{\xi, \mu, \sigma}= \sigma \xi.
	\end{equation}
For any $k\in \N$ and any $s\in [1,\infty]$ one has
\begin{equation}\label{eqn:Thetanorms}
\| D^k \Theta_{\xi, \mu, \sigma} \|_{L^s}\le C(d,k,s, n_\ast) \sigma \mu^{k + (d-1) ( 1/p - 1/s )},
\end{equation}
\begin{equation}\label{eqn:Hamiltonian_norms}
\| D^k  X_{H_{\xi, \mu}} \|_{L^s} \le C(d,k,s, n_\ast) \mu^{ k+ (d-1)(1/p'-1/s)},
\qquad \| D^k  H_{\xi, \mu} \|_{L^s} \le C(d,k,s, n_\ast) \mu^{ k-1+ (d-1)(1/p'-1/s)}.
\end{equation}
Finally, they have pairwise compact disjoint supports for any $\xi \neq \xi'$, namely
\begin{align} \label{eqn:disjointsupp}
\text{supp} X_{H_{\xi, \mu}}   \cap \text{supp} \Theta_{\xi',\mu,\sigma} = \text{supp} X_{H_{\xi, \mu}} \cap \text{supp} X_{H_{\xi', \mu}} = \text{supp} \Theta_{\xi',\mu,\sigma} \cap \text{supp} \Theta_{\xi,\mu,\sigma} = \emptyset,
\end{align}
for any $\xi \neq \xi'$.
\end{lemma}

\section{Iteration scheme}
As in \cite{MoSz2019AnnPDE} we consider the following system of equations in $[0,T] \times \T^d$
\begin{equation}\label{eqn:CE-R}
\partial_t \rho_{q} + \diver (  \rho_{q} J \nabla H_{q})=-\diver R_{q},
\end{equation}
where we observe that  $\diver J \nabla H =0. $
We then fix three parameters $a_0$, $b>0$ and $\beta>0$, to be chosen later only in terms of $d$, $p$, $r$, and for any choice of $a>a_0$ we define
\[
\lambda_0 =a, \quad \lambda_{q+1} = \lambda_q^b \quad \mbox{and}\quad
\delta_{q} = \lambda_{q}^{-2\beta}\, .
\]
The following proposition builds a converging sequence of functions with the inductive estimates
\begin{equation}
\label{eqn:ie-1}
\max_t \|R_q (t, \cdot)\|_{L^1} \leq \delta_{q+1}
\end{equation}
\begin{equation}
\label{eqn:ie-2}
\max_t \left(\| \rho_q (t, \cdot)\|_{C^1} +\|\partial_t \rho_q (t, \cdot)\|_{C^0} + \|H_q \|_{C^1} + \|H_q \|_{W^{2,p'}} + \| H_q \|_{W^{3,r}} \right)
\leq \lambda_{q}^\alpha\, ,
\end{equation}
where $\alpha$ is yet another positive parameter which will be specified later.

\begin{prop}\label{prop:inductive}
	There exist $\alpha,b, a_0, M>5$, $0<\beta<(2b)^{-1}$ such that the following holds. For every $a\geq a_0$, if
	$(\rho_q, H_q, R_q)$ solves \eqref{eqn:CE-R} and enjoys the estimates \eqref{eqn:ie-1}, \eqref{eqn:ie-2}, then there exist $(\rho_{q+1}, H_{q+1}, R_{q+1})$ which solves \eqref{eqn:CE-R}, enjoys the estimates \eqref{eqn:ie-1}, \eqref{eqn:ie-2} with $q$ replaced by $q+1$ and also the following properties:
	\begin{itemize}
		\item[(a)] $\max_t \|( \rho_{q+1}- \rho_q) (t, \cdot)\|_{L^1}  \leq  \delta_{q+1}$
		\item [(b)] $\|H_{q+1}- H_q \|_{W^{2,r}}  + \|H_{q+1}- H_q\|_{W^{1,p'}} \leq  \frac{M}{2^q}$,
		\item [(b+)] If $p' \geq  d-1$, then $\|H_{q+1}- H_q \|_{L^\infty} \leq \frac{M}{ \lambda_{q}} $
		\item [(c)] $\max_t \| \rho_{q+1} J \nabla H_{q+1} - \rho_q J \nabla H_q \|_{L^1} \leq M \delta_{q+1}$
		\item[(d)] $\inf (\rho_{q+1} - \rho_q) \geq - \delta_{q+1}$
		\item[(e)] if for some $t_0>0$ we have that $\rho_q(t, \cdot) = 1$ and $R_q(t, \cdot)=0$ for every $t\in [0,t_0]$, then $\rho_{q+1}(t, \cdot) = 1$ and $R_{q+1}(t, \cdot)=0$  for every $t\in [0,t_0- \lambda_q^{-1-\alpha}]$.
	\end{itemize}
\end{prop}
Our iterative proposition is quite similar to the one proposed in \cite{BDLC20}, but in order to get an autonomous vector field, we put all the `bad' terms form the previous Reynolds error onto the definition of the new density. This gives much worse estimates on the new density: indeed we are unable to control its $L^p$ norm. But the concentration parameter $\mu$ allows us to control the density in the $L^1$ norm. Point (c) in the above proposition allows us to show convergence of $\rho_q J\nabla H_q$ in $L^1$ which was previously done in \cite{BDLC20} using H\"older's inequality.

Also, in order to get an autonomous vector field, we are unable to use the non-autonomous building blocks of \cite{BDLC20} but instead use the autonomous `Mikado flows' that were used as building blocks in \cite{MoSz2019AnnPDE}. The disadvantage of this approach that was already apparent in \cite{MoSz2019AnnPDE}, is that we are unable to get the full dimensional concentration $d$ but are only able to get a $d-1$ in eqn. \ref{hp:exp}.
\subsection{Choice of the parameters}\label{sec:param}
The choice of parameter is very similar to those in \cite[Section 5.1]{BDLC20}.
We define first the constant
\[
\gamma:= \Big(1+\frac 1p\Big) \left(\min \Big\{ \frac{d-1}{p}, \frac{d-1}{p'}, -1-(d-1)  \Big(\frac1{p'}-\frac1r\Big)\Big\}\right)^{-1}>0,
\]
Notice that, up to enlarging $r$, we can assume that the quantity in the previous line is less than $1/2$, namely that $\gamma>2$.
Hence we set $\alpha:=4 + \gamma (d+1)$,
\begin{equation}
\label{eqn:choice-b}
b := \max\{ p, p'\} (3(1+\alpha)(d+2)+2),
\end{equation}
and
\begin{equation}
\label{eqn:choice-beta}
\beta:= \frac 1{2 b} \min\Big\{ p, p', r, \frac{1}{b+1}\Big\}= \frac{1}{2 b(b+1)} .
\end{equation}
Finally, we choose $a_0$ and $M$ sufficiently large (possibly depending on all previously fixed parameters) to absorb numerical constants in the inequalities.
We set
\begin{equation}
\label{eqn:choice-ell}
\ell: = \lambda_{q}^{-1-\alpha},
\end{equation}
\begin{equation}
\label{eqn:choice-mu}\mu_{q+1} := \lambda_{q+1}^{\gamma}.
\end{equation}

\subsection{Convolution}
The convolution step is the same of \cite[Section 5.2]{BDLC20}.
We first perform a convolution of $\rho_q$ and $u_q$ to have estimates on more than one derivative of these objects and of the corresponding error. Let $\phi \in C^\infty_c(B_1)$ be a standard convolution kernel in space-time, $\ell$ as in \eqref{eqn:choice-ell} and define
$$\rho_{\ell} := \rho_{q} \ast \phi_\ell, \qquad H_\ell := H_q \ast \phi_\ell, \qquad u_\ell := u_q \ast \phi_\ell, \qquad R_{\ell} := R_{q} \ast \phi_\ell. $$
We observe that $(\rho_{\ell}, u_\ell, R_{\ell}+ (\rho_{q} u_q)_\ell - \rho_{\ell} u_\ell)$ solves system \eqref{eqn:CE-R}
and  by \eqref{eqn:ie-1}, \eqref{eqn:choice-beta} enjoys the following estimates

\begin{equation}
\label{eqn:r-l-l1}
\|R_\ell \|_{L^1} \leq \delta_{q+1},
\end{equation}
\begin{equation} \label{eqn:rho-ell-conv}
\| \rho_\ell - \rho_q\|_{L^p} \leq  \ell \|\rho_q\|_{C^1} \leq  \ell \lambda_q^\alpha  \leq  \delta_{q+1},
\end{equation}
$$\| u_\ell - u_q\|_{L^{p'}} \leq C \ell \lambda_q^\alpha \leq  \delta_{q+1},$$
$$\| u_\ell - u_q\|_{W^{1,r}} \leq C \ell \lambda_q^\alpha \leq  \delta_{q+1}\, .$$
Indeed note that by \eqref{eqn:choice-beta}
\[
\ell \lambda_q^\alpha = \lambda_q^{-1} = \delta_{q+1}^{\frac{1}{2b\beta}} \ll
\delta_{q+1}
\]
Next observe that
\[
\|\partial_t^N \rho_\ell\|_{C^0} + \| \rho_\ell\|_{C^N} +\| u_\ell\|_{W^{1+N,r}} + \|\partial_t^N u_\ell\|_{W^{1,r}}
\leq  C(N) \ell^{-N+1}( \| \rho_q\|_{C^1} +\| u_q\|_{W^{2,r}}
)\leq  C(N) \ell^{-N+1}\lambda_{q}^\alpha
\]
for every $N\in \N\setminus\{0\}$.
Using the Sobolev embedding $W^{d,r}\subset {W^{d,1}\subset }C^0$ we then conclude
\[
\|\partial_t^N u_\ell\|_{C^0} + \|u_\ell\|_{C^N} \leq C (N) \ell^{-N-d+2} \lambda_q^\alpha\, .
\]
By Young's inequality we estimate the higher derivatives of $R_\ell$ in terms of $\|R_q \|_{L^1}$ to get
\begin{equation}\label{e:D_tR}
\| R_\ell\|_{C^N} +\|\partial_t^N R_\ell\|_{C^0}
\leq \|D^N \rho_\ell \|_{L^\infty} \|R_q \|_{L^1} \leq  C(N) \ell^{-N-d} \leq  C(N) \lambda_{q}^{(1+\alpha)(d+N)}\,
\end{equation}
for every $N\in \N$.
Finally, thanks to \cite[Lemma 5.1]{BDLC20} for the last part of the error we have
\begin{equation}\label{e:commutatore}
\|(\rho_q u_q)_\ell - \rho_\ell u_\ell\|_{L^1} \leq C \ell^2 \lambda_q^{2\alpha} \leq \frac{1}{4} \delta_{q+2} ,
\end{equation}
where we have assumed that $a$ is sufficiently large.

\subsection{Definition of the perturbation} \label{sec:perturbation}

Let $\mu_{q+1}>0$ be as in \eqref{eqn:choice-mu} and let $\chi\in C^\infty_c (-\frac{3}{4}, \frac{3}{4})$ such that $\sum_{n \in \mathbb Z} \chi (\tau - n) =1$ for every $\tau\in \mathbb R$.

Fix a parameter $\kappa=\frac{2
}{\delta_{q+2}}$ and consider a finite set $\Lambda \subset \partial B_1 \cap \Q^d $ as in Lemma \ref{lemma:geom} and consider the related building blocks implicitly defined in Lemma \ref{lemma:buildingblocks}.
We define the new density, Hamiltonian ,and vector field by adding to $\rho_\ell$, $H_\ell$ and $u_\ell$ a principal term and a smaller corrector, namely we set
\begin{align*}
\rho_{q+1} &:= \rho_{\ell} + \theta_{q+1}^{(p)}+ \theta_{q+1}^{(c)}\,,\\
H_{q+1} &:= H_\ell + h_{q+1}\,,\\
u_{q+1}  &:= u_\ell+  w_{q+1} \  .
\end{align*}
The principal perturbations are given, respectively, by
\begin{align}
\label{eqn:h}
h_{q+1} (x) &= \frac{1}{2^q\lambda_{q+1}} \sum_{\xi \in \Lambda}  H_{\xi , \mu_{q+1}} ( \lambda_{q+1} x),  \\
\label{eqn:w}
w_{q+1} (x) &= \frac{1}{2^q} \sum_{\xi \in \Lambda}  X_{H_{\xi , \mu_{q+1}}} ( \lambda_{q+1} x),  \\
\label{eqn:theta}
\theta^{(p)}_{q+1} (t,x) &= 2^q \sum_{n \ge 12} \sum_{\xi\in \Lambda } \chi (\kappa |R_\ell (t,x)| - n)   a_{\xi}\left(\frac{R_\ell(t,x)}{|R_\ell(t,x)|}\right) \Theta_{\xi, \mu_{q+1}, n /\kappa} ( \lambda_{q+1} x)\, ,
\end{align}

where we understand that $a_\xi \left(\frac{R_\ell(t,x)}{|R_\ell(t,x)|}\right)$ is well defined because the term $\chi(\kappa |R_\ell(t,x)| -n)$ vanishes at points where $R_\ell$ vanishes.
	Furthermore, in the definition of $\theta^{(p)}_{q+1}$ the first sum runs for $n$ in the range
	\begin{equation}
	\label{remark:sum is finite}
		12\leq n \leq  C\ell^{-d}\delta_{q+2}^{-1} \le C \lambda_{q}^{d(1+\alpha) + 2\beta b^2}
	\leq C \lambda_{q}^{d(1+\alpha)+1}.
	\end{equation}
	Indeed $\chi(\kappa |R_\ell(t,x)|-n)=0$ if $n\ge 2
	\delta_{q+2}^{-1}\|R_\ell\|_{C^0}+1$ and by \eqref{e:D_tR} we obtain an upper bound for $n$.
\\
Notice that $u_{q+1}$ is an autonomous Hamiltonian vector field. Indeed, we have that $w_{q+1} = J\nabla h_{q+1}$ and consequently $u_{q+1} = J\nabla H_{q+1}$.\\
The aim of the corrector term for the density is to ensure that the overall addition has zero average:
\[
\theta^{(c)}_{q+1} (t):=- \int \theta_{q+1}^{(p)}(t,x)\, dx.
\]

\section{Proof of the Proposition \ref{prop:inductive}}
In this section, we prove the main iterative proposition \ref{prop:inductive}. The proof is very similar to the one found in \cite{BDLC20} as we have very similar estimates on the building blocks. We provide details for the convenience of the reader.
\begin{lemma}\label{l:uglylemma}
	For $m\in \mathbb N$, $N\in \mathbb N\setminus \{0\}$ and $n\ge 2$ we have
	\begin{align}
	&\|\partial^m_t \chi (\kappa|R_\ell|-n)\|_{C^N} \leq C (m, N) \delta_{q+2}^{-2(N+m)} \ell^{-(N+m)(1+d)}
	\leq
	C(m,N) \lambda_q^{(N+m) (d+2) (1+\alpha)}\\
	&\|\partial^m_t (a_\xi ({\textstyle{\frac{R_\ell}{|R_\ell|}}}))\|_{C^N}\leq C (m, N) \delta_{q+2}^{-N-m} \ell^{-(N+m)(1+d)}
	\leq C(m,N)\lambda_q^{(N+m) (d+2) (1+\alpha)} \qquad\mbox{on $\{\chi (\kappa|R_\ell|-n) >0\}$}.
	\end{align}
\end{lemma}
In the following estimates are crucial the choices of parameters $\ell, b, \beta, \gamma$ and $ \kappa$ fixed in Subsection \ref{sec:param} and the inductive parameters $\lambda_{q+1}, \delta_{q+1}$ and $\mu_{q+1}$.
\subsection{Estimates on the perturbation} Here we show the inductive estimates are satisfied for the perturbed quantities.
\subsubsection{Estimates on the velocity field and Hamiltonian}
\begin{equation}\label{eqn:w}
    \begin{split}
   \|h_{q+1} \|_{W^{1,p'}} \lesssim \|w_{q+1}\|_{L^{p'}} &\leq 2^{-q} |\Lambda| \|X_{H_{\xi, \mu_{q+1}}} (\lambda_{q+1} \cdot)\|_{L^{p'}} \leq 2^{-q} |\Lambda| C \mu_{q+1}^{(d-1)(1/p'-1/p')} \\
    &\leq C |\Lambda| 2^{-q}
    \end{split}
\end{equation}
where we have used \eqref{eqn:Hamiltonian_norms} in the third inequality.
\begin{equation}\label{eqn:w-sobolev}
    \begin{split}
   \|h_{q+1} \|_{W^{2,r}} \lesssim \|w_{q+1}\|_{W^{1,r}} &\leq \|w_{q+1}\|_{L^r} + \|Dw_{q+1}\|_{L^r} \\
    &\leq 2^{-q} |\Lambda| \|X_{H_{\xi, \mu_{q+1}}} (\lambda_{q+1} \cdot)\|_{L^{r}} + 2^{-q} |\Lambda|  \lambda_{q+1} \|D X_{H_{\xi, \mu_{q+1}}} (\lambda_{q+1} \cdot)\|_{L^{r}} \\
    &\leq 2^{-q} |\Lambda| C \mu_{q+1}^{(d-1)(1/p'-1/r)} + 2^{-q} |\Lambda| C \lambda_{q+1} \mu_{q+1}^{1+(d-1)(1/p'-1/r)}\\
    &\leq 2^{-q} |\Lambda| C \lambda_{q+1}^{\gamma(d-1)(1/p'-1/r)} + 2^{-q} |\Lambda| C \lambda_{q+1}^{1 + \gamma(1+(d-1)(1/p'-1/r))}\\
    &\leq 2^{-q} |\Lambda| C \lambda_{q+1}^{-\gamma} + 2^{-q} |\Lambda| C \lambda_{q+1}^{-1/p}\\
    &\leq \delta_{q+1}
    \end{split}
\end{equation}
Here we have used \eqref{eqn:Hamiltonian_norms} in the fourth inequality and \eqref{eqn:choice-mu} in the fifth inequality.
\subsubsection{Estimates on the density}
Using standard estimates, \eqref{eqn:Thetanorms} and \eqref{remark:sum is finite}
\begin{equation}\label{eqn:theta-pert-p}
\begin{split}
\|\theta^{(p)}_{q+1} (t,\cdot)\|_{L^1} &\leq \sum_{n\ge 12} \sum_{\xi\in \Lambda} 2^q \| \chi (\kappa |R_\ell (t,x)| - n)  \textstyle{a_{\xi}\left(\frac{R_\ell(t,x)}{|R_\ell(t,x)|}\right)} \|_{C^0} \|\Theta_{\xi, \mu_{q+1}, n /\kappa} (\lambda_{q+1} x) \|_{L^1}\\
&\leq C |\Lambda| 2^q \sum_{n\ge 12} \frac{n}{\kappa} \mu_{q+1}^{(d-1)(1/p-1)} \leq C |\Lambda| 2^q \kappa^{-1} \lambda_q^{2d(1+\alpha)+2} \mu_{q+1}^{(d-1)(1/p-1)}\\
&\leq C |\Lambda| 2^q \kappa^{-1} \lambda_q^{2d(1+\alpha)+2} \mu_{q+1}^{-(d-1)/p'} \leq \delta_{q+1}
\end{split}
\end{equation}
Now note that by the definition of $\theta^{(c)}_{q+1} (t)$ we have that
\begin{equation}\label{eqn:theta-pert-c}
    |\theta^{(c)}_{q+1} (t)| \leq \|\theta^{(p)}_{q+1} (t,\cdot)\|_{L^1}
\end{equation}
From the above two estimates point $(a)$ of Theorem \ref{prop:inductive} is now proved. Now notice that since $\theta^{(p)}_{q+1}$ is non-negative, we have
\begin{align}
    \inf \theta^{(p)}_{q+1} (t,\cdot) + \theta^{(c)}_{q+1} (t) \geq \theta^{(c)}_{q+1} (t) \geq -\delta_{q+1}
\end{align}
From these computations point $(d)$ of Theorem \ref{prop:inductive} is now proved.
\subsubsection{Estimates for Part (c) of Theorem \ref{prop:inductive}}
Using standard estimates, the improved H\"older inequality \ref{eqn:impr-holder}, \eqref{eqn:Thetanorms} and \eqref{remark:sum is finite}
\begin{equation}
\label{eqn:tw}
    \begin{split}
        \| \theta_{q+1}^{(p)} w_{q+1} \|_{L^1} &\leq \|\sum_{n \ge 12} \sum_{\xi\in \Lambda } \chi (\kappa |R_\ell (t,x)| - n)   a_{\xi}\left(\frac{R_\ell(t,x)}{|R_\ell(t,x)|}\right) \Theta_{\xi, \mu_{q+1}, n /\kappa} ( \lambda_{q+1} x) X_{H_{\xi,\mu_{q+1}}} (\lambda_{q+1} x) \|_{L^1}\\
        & \le C\lambda_{q+1}^{-1} \sum_{n\ge 12}\sum_{\xi\in \Lambda} \|  \chi( \kappa|R_\ell| - n ) \textstyle{a_{\xi}\left( \frac{ R_\ell }{ |R_\ell| } \right)} \|_{C^1}
\| \Theta_{\xi, \mu_{q+1}, n/\kappa}X_{H_{\xi,\mu_{q+1}}}\|_{L^1}
\\&\qquad + \sum_{n\ge 12}\sum_{\xi\in \Lambda} \| \chi( \kappa|R_\ell| - n ) \textstyle{a_{\xi}\left( \frac{ R_\ell }{ |R_\ell| } \right)} \|_{L^1} \| \Theta_{\xi, \mu_{q+1}, n/\kappa}X_{H_{\xi,\mu_{q+1}}}\|_{L^1}
\\& \le C\lambda_{q+1}^{-1} \lambda_{q}^{3(1+\alpha)(d+2)}
+ C \sum_{n\ge 12} \frac{n}{\kappa} \| \chi( \kappa|R_\ell| - n ) \textstyle{a_{\xi}\left( \frac{ R_\ell }{ |R_\ell| } \right)} \|_{L^1}
\\& \le \frac{1}{2}\delta_{q+1} + C \|R_\ell\|_{L^1}\le M \delta_{q+1}.
\end{split}
\end{equation}
We could also prove the above indirectly as we know the product cancels with $R_\ell$ up to a very small error.
\subsection{Estimates on higher derivatives}\label{sec:highderiv} Here we show that the perturbed quantities satisfy the estimate \eqref{eqn:ie-2}. By the choice of $\alpha$, since in particular $\alpha\ge 2+\gamma(d+1)$, we have that
\begin{equation}
\label{eqn:est-rhoC1}
\begin{split}
\| \rho_{q+1}\|_{C^1} &\leq \| \rho_{\ell}\|_{C^1}+ \| \theta_{q+1}\|_{C^1} \leq \| \rho_{q}\|_{C^1}
+ 2^q \sum_{n\ge 12} \sum_{\xi\in \Lambda^{[n]}} \| \chi(\kappa|R_\ell| - n ) \textstyle{ a_{\xi} \left( \frac{ R_\ell }{|R_\ell|}\right)}\|_{C^1} \|\Theta_{\xi, \mu_{q+1},n/\kappa}(\lambda_{q+1}x) \|_{C^1}
\\&\leq C \lambda_q^\alpha +
C 2^q \lambda_q^{3(1+\alpha)(d+2)} \lambda_{q+1}\mu_{q+1}^{1+(d-1)/p}
\leq \lambda_{q+1}^{\alpha}.
\end{split}
\end{equation}
where in the above calculation, we have used \eqref{eqn:Thetanorms}, \eqref{eqn:choice-mu}, \eqref{remark:sum is finite} and Lemma \ref{l:uglylemma}. An entirely similar estimate is valid for $\|\partial_t \rho_{q+1}\|_{C^0}$. Now we estimate the velocity field using \eqref{eqn:Hamiltonian_norms} and \eqref{eqn:choice-mu}:
\begin{equation}
\label{eqn:est-HamC1}
\begin{split}
\| u_{q+1}\|_{C^0} &\leq \| u_{\ell}\|_{C^0}+ \| w_{q+1}\|_{C^0} \leq \| u_{q}\|_{C^0}
+ 2^{-q} \sum_{\xi\in \Lambda}  \|X_{H_{\xi, \mu_{q+1}}}(\lambda_{q+1}x) \|_{C^0}
\\&\leq \lambda_q^\alpha +
C 2^{-q} \mu_{q+1}^{1+(d-1)/p'}
\leq \lambda_{q+1}^{\alpha}.
\end{split}
\end{equation}
Now we estimate the Sobolev norms of $u_{q+1}$ which give us the estimates on $H_{q+1}$.
\begin{equation}
    \begin{split}
        \|u_\ell+ w_{q+1}\|_{W^{1,p'}} &\leq \|u_\ell\|_{W^{1,p'}} + \|w_{q+1}\|_{L^{p'}} + \|Dw_{q+1} (t,\cdot)\|_{L^{p'}} \\
        &\leq C\lambda_q^\alpha + 2^{-q}|\Lambda|C + 2^{-q}|\Lambda|C \|D X_{H_{\xi,\mu_{q+1}}}(\lambda_{q+1} \cdot)\|_{L^{p'}}\\
        &\leq C\lambda_q^\alpha + 2^{-q}|\Lambda|C + 2^{-q}|\Lambda|C \lambda_{q+1} \mu_{q+1}^{1+(d-1)(1/p'-1/p')}\\
        &\leq C\lambda_q^\alpha + 2^{-q}|\Lambda|C + 2^{-q}|\Lambda|C \lambda_{q+1}^{1+\gamma} \leq \lambda_{q+1}^\alpha\\
    \end{split}
\end{equation}
where we have used \eqref{eqn:Hamiltonian_norms} and \eqref{eqn:choice-mu}. Similarly we get
\begin{equation}
    \begin{split}
        \|u_\ell+ w_{q+1}\|_{W^{2,r}} &\leq \|u_\ell\|_{W^{2,r}} + \|w_{q+1}\|_{W^{1,r}} + \|D^2w^{(p)}_{q+1} (t,\cdot)\|_{L^r} \\
    &\leq \|u_\ell\|_{W^{2,r}} + 2^{-q} |\Lambda| C + 2^{-q} |\Lambda| C \lambda_{q+1}^2 \|D^2 X_{H_{\xi, \mu_{q+1}}}\|_{L^{r}} \\
    &\leq \|u_\ell\|_{W^{2,r}} + 2^{-q} |\Lambda| C + 2^{-q}|\Lambda| C \lambda_{q+1}^2 \mu_{q+1}^{2+(d-1)(1/p'-1/r)}\\
    &\leq C\lambda_q^\alpha + 2^{-q} |\Lambda| C + 2^{-q} |\Lambda|C \lambda_{q+1}^{2 + \gamma(2+(d-1)(1/p'-1/r))}\\
    &\leq C\lambda_q^\alpha + 2^{-q} |\Lambda| C + 2^{-q} |\Lambda|C \lambda_{q+1}^{\gamma + 1/p'} \leq \lambda_{q+1}^\alpha
    \end{split}
\end{equation}
where we have used \eqref{eqn:w-sobolev} in the second inequality.

\subsection{Estimates on the new Reynold's Stress}
\begin{equation}\label{eqn:new-error}
\begin{split}
-\diver R_{q+1} = & \partial_t \rho_{q+1} + \diver(  \rho_{q+1} u_{q+1}) = \diver ( \theta_{q+1}^{(p)} w_{q+1} - R_\ell) + \partial_t \theta_{q+1}^{(p)}+\partial_t \theta_{q+1}^{(c)}
\\&+ \diver( \theta_{q+1}^{(p)} u_\ell+ \rho_\ell w_{q+1}  +\theta_{q+1}^{(c)}w_{q+1}
) + \diver ( (\rho_q u_q)_\ell - \rho_\ell u_\ell).
\end{split}
\end{equation}
We firstly observe that point (e) is a consequence of standard mollification properties and the definition of the perturbations in Subsection \ref{sec:perturbation}.

Using Lemma \ref{lemma:geom}, the property $\sum_{n \in \mathbb Z} \chi (\tau - n) =1$ we get
\begin{align}
    \theta_{q+1}^{(p)} w_{q+1} - R_\ell &= \sum_{n \ge 12} \sum_{\xi\in \Lambda } \chi (\kappa |R_\ell (t,x)| - n)   a_{\xi}\left(\frac{R_\ell(t,x)}{|R_\ell(t,x)|}\right) \Theta_{\xi, \mu_{q+1}, n /\kappa} ( \lambda_{q+1} x) X_{H_{\xi,\mu_{q+1}}} (\lambda_{q+1} x) - R_\ell \\
    &= \sum_{n \ge 12} \sum_{\xi\in \Lambda } \chi (\kappa |R_\ell (t,x)| - n)   a_{\xi}\left(\frac{R_\ell(t,x)}{|R_\ell(t,x)|}\right) \left(\Theta_{\xi, \mu_{q+1}, n /\kappa} ( \lambda_{q+1} x) X_{H_{\xi,\mu_{q+1}}} (\lambda_{q+1} x) - \frac{n}{\kappa} \xi \right)\\
    &\quad + \sum_{n \ge 12} \sum_{\xi\in \Lambda } \chi (\kappa |R_\ell (t,x)| - n)   \frac{n}{\kappa} a_\xi \left (\frac{R_\ell(t,x)}{|R_\ell(t,x)|} \right) \xi  -R_\ell
\end{align}
and the last sum is ready to ``cancel'' the error $R_\ell$.
Thus, on taking the divergence, we get
\begin{equation}
    \begin{split}
        \diver(\theta_{q+1}^{(p)} w_{q+1} &- R_\ell)\\
        &= \diver \left(\sum_{n \ge 12} \sum_{\xi\in \Lambda } \chi (\kappa |R_\ell (t,x)| - n)   a_{\xi}\left(\frac{R_\ell(t,x)}{|R_\ell(t,x)|}\right) \Theta_{\xi, \mu_{q+1}, n /\kappa} ( \lambda_{q+1} x) X_{H_{\xi,\mu_{q+1}}} (\lambda_{q+1} x) - R_\ell \right) \\
    &= \sum_{n \ge 12} \sum_{\xi\in \Lambda } \nabla \left(\chi (\kappa |R_\ell (t,x)| - n)   a_{\xi}\left(\frac{R_\ell(t,x)}{|R_\ell(t,x)|}\right)\right) \cdot  \left(\Theta_{\xi, \mu_{q+1}, n /\kappa} ( \lambda_{q+1} x) X_{H_{\xi,\mu_{q+1}}} (\lambda_{q+1} x) - \frac{n}{\kappa} \xi \right) \\
    &\quad + \sum_{n \ge 12} \sum_{\xi\in \Lambda } \diver \left(\chi (\kappa |R_\ell (t,x)| - n)   \frac{n}{\kappa} a_\xi \left ( \frac{R_\ell(t,x)}{|R_\ell(t,x)|} \right) \xi \right) - \diver R_\ell
    \end{split}
\end{equation}
For the first term, we apply the convex integration.
The second term
where
\[
\tilde R_\ell:=
\sum_{n \ge 12} \sum_{\xi\in \Lambda } \left(\chi (\kappa |R_\ell (t,x)| - n)   \frac{n}{\kappa} a_\xi \left ( \frac{R_\ell(t,x)}{|R_\ell(t,x)|} \right) \xi \right) = \sum_{n \ge 12} \chi(\kappa |R_\ell|- n)	\frac{R_\ell}{|R_\ell|} \frac{n}{k}.
\]
We have
\begin{align}\label{eqn:tildeRestimate}
\notag
|R_\ell-\tilde R_\ell| & \le \Big|\sum_{n=-1}^{1
} \chi(\kappa |R_\ell|- n) R_\ell \Big|
+\Big| \sum_{n\ge 12} \chi(\kappa |R_\ell|- n)\left( \frac{R_\ell}{|R_\ell|}\frac{n}{k}-R_\ell\right)	\Big|
\\& \le \frac{1
}{\kappa} + \sum_{n\ge 12} \chi(\kappa |R_\ell|- n)\left| |R_\ell|-\frac{n}{\kappa}\right|
\\& \le \frac{1
}{2
}\delta_{q+2}+\frac{3}{4
}\delta_{q+2}
\le \frac{1
}{2
} \delta_{q+2}.
\end{align}
We can now define $R_{q+1}$ which satisfies \eqref{eqn:new-error} as
\begin{equation}
\begin{split}
-R_{q+1} :=& R^{quadr}+ (\tilde R_\ell- R_\ell) + R^{time}  +\theta_{q+1}^{(p)} u_\ell+ \rho_\ell w_{q+1} +[(\rho_q u_q)_\ell - \rho_\ell u_\ell],
\end{split}
\end{equation}
where
\begin{equation}
\label{defn:Rquadr}
R^{quadr}: = \sum_{n\ge 12} \sum_{\xi\in \Lambda} \mathcal{R} \left[  \nabla \left(\chi(\kappa |R_\ell|- n)
\textstyle{a_{\xi}\left(\frac{R_\ell}{|R_\ell|}\right)}\right)\cdot  \left((\Theta_{\xi, \mu_{q+1}, n /\kappa} X_{H_{\xi, \mu_{q+1}}})(\lambda_{q+1}x) - \frac{n}{\kappa} \xi \right)\right],
\end{equation}
\begin{equation}
\label{eqn:Rc}
R^{time}:=\nabla \Delta^{-1} (\partial_t \theta_{q+1}^{(p)} +\partial_t \theta_{q+1}^{(c)}),
\end{equation}

Notice that $R^{quadr}$ is well defined and by \eqref{eqn:rightaverage} the function $(\Theta_{\xi, \mu_{q+1}, n /\kappa} X_{H_{\xi, \mu_{q+1}}})(\lambda_{q+1}x)- \frac{n}{\kappa} \xi$ has $0$ mean.
We have that $ \partial_t \theta_{q+1}^{(p)} +\partial_t \theta_{q+1}^{(c)}$ has $0$ mean, so that $R^{time}$ is well defined.\\
We now estimate in $L^1$ each term in the definition of $R_{q+1}$.
Recall that  the estimate on $\|(\rho_q u_q)_\ell - \rho_\ell u_\ell\|_{L^1}$ has been already established in \eqref{e:commutatore}.

By the property \eqref{ts:antidiv} of the anti-divergence operator $\ensuremath{\mathcal{R}}$, Lemma \ref{l:uglylemma} and \eqref{remark:sum is finite} we have
\begin{align*}
\|R^{quadr}\|_{L^1}  & \leq \frac C {\lambda_{q+1}} \sum_{n\ge 12}\sum_{\xi\in \Lambda}
\| \chi(\kappa |R_\ell|- n)
\textstyle{a_{\xi}\left(\frac{R_\ell}{|R_\ell|}\right)}\|_{C^2}\|\Theta_{\xi, \mu_{q+1}, n/\kappa} X_{H_{\xi, \mu_{q+1}}}(\lambda_{q+1}x)\|_{L^1} \\&\leq C \delta_{q+2} \frac{\lambda_q^{4(1+\alpha)(d+2)+2}}{\lambda_{q+1}} \leq \frac{\delta_{q+2}}{20}.
\end{align*}
To estimate the terms which are linear with respect to the fast variables, we take advantage of the concentration parameter $\mu_{q+1}$. First of all, by Calderon-Zygmund estimates we get
\begin{align*}
\| R^{time} \|_{L^1} \le  C \| \partial_t \theta_{q+1}^{(p)}+\partial_t \theta_{q+1}^{(c)}\|_{L^1}
\le  \| \partial_t \theta_{q+1}^{(p)} \|_{L^1}+ | \partial_t \theta_{q+1}^{(c)} |	.
\end{align*}
Next, notice that
\begin{align}\label{eqn:tpl1}
\| \partial_t \theta_{q+1}^{(p)} \|_{L^1}
& \leq
C 2^q \sum_{n\ge 12}\sum_{\xi\in \Lambda}\| \partial_t\big[ \chi(\kappa |R_\ell|- n) \textstyle{a_{\xi}\left( \frac{R_\ell}{|R_\ell|} \right)}\big]\|_{C^0}
\|\Theta_{\xi, \mu_{q+1}, n /\kappa} \|_{L^1}
\\& \leq C 2^q \delta_{q+2} \lambda_{q}^{3(1+\alpha)(d+2)}\mu_{q+1}^{-d/p'} \leq \frac{\delta_{q+2}}{20}.
\end{align}
From \eqref{eqn:tpl1}, \eqref{eqn:itsolves} and \eqref{eqn:l26} we get
\begin{align*}
| \partial_t \theta_{q+1}^{(c)} |
\le \| \partial_t \theta_{q+1}^{(p)} \|_{L^1}
\le \frac{\delta_{q+2}}{20}
\end{align*}
Similarly, we have that
\begin{equation} \label{eqn:slow_interaction}
\begin{split}
\| & \theta_{q+1}^{(p)} u_\ell + \rho_\ell w_{q+1}\|_{L^1} \leq \| \theta_{q+1}^{(p)}\|_{L^1} \| u_\ell\|_{L^\infty} + \|\rho_\ell\|_{L^\infty} \|w_{q+1}\|_{L^1}\\
&\leq  \sum_{n\ge 12} \sum_{\xi \in \Lambda^{[n]}} 2^q \| \chi(\kappa|R_\ell| - n ) \textstyle{ a_{\xi} \left( \frac{ R_\ell }{|R_\ell|} \right) }\|_{L^\infty} \| \Theta_{\xi, \mu_{q+1},n/\kappa}\|_{L^1} \| u_\ell\|_{L^\infty} + \|\rho_\ell\|_{L^\infty} \|X_{H_{\xi, \mu_{q+1}}}\|_{L^1}\\
&\leq C 2^q \delta_{q+2}^{1/p} \lambda_{q}^{2(1+\alpha)(d+2)} \mu^{-d/p'}_{q+1} + C\delta_{q+2}^{1/p'}
\lambda_{q}^{2(1+\alpha)(d+2)}\mu^{-d/p}_{q+1} \leq \frac{\delta_{q+2}}{20}
\end{split}
\end{equation}
In the last inequality we used  $2\beta b^2\le 1$, the definition of $\gamma$, and  $b(1+1/p)\ge 2(1+\alpha)(d+2)+1$.
\subsection{Proof of items (b), (b+) and (c)}
Firstly we prove point (b). By definitions \eqref{eqn:choice-ell}, \eqref{eqn:h} and by the estimates \eqref{eqn:ie-2}, \eqref{eqn:w-sobolev} we have
\begin{align*}
\| H_{q+1} - H_{q} \|_{W^{2,r}} \leq \| H_{q} - H_{\ell} \|_{W^{2,r}} + \| h_{q+1} \|_{W^{2,r}} \leq
\ell \lambda_q^\alpha + \frac{C|\Lambda|}{\lambda_{q+1}^{1/p}}
\leq  \frac{1}{\lambda_q} + \frac{C|\Lambda|}{\lambda_{q+1}^{1/p}} \leq \frac{M}{\lambda_q},
\end{align*}
and similarly we can estimate $\| H_{q+1} - H_{q} \|_{W^{1,p'}}$ using \eqref{eqn:w} instead of \eqref{eqn:w-sobolev}.
If $p' \geq d-1$, by Lemma \ref{lemma:buildingblocks}, the definition of $\ell$ and \eqref{eqn:ie-2} we get
\begin{align*}
\| H_{q+1} - H_{q} \|_{L^\infty} \leq \| H_{q} - H_{\ell} \|_{L^\infty} + \| h_{q+1} \|_{L^\infty} \leq \frac{1}{\lambda_q} + \frac{C|\Lambda|}{2^q \lambda_{q+1}} \leq \frac{M}{\lambda_q},
\end{align*}
which leads to point (b+).
For point (c), using \eqref{eqn:tw}, \eqref{eqn:slow_interaction}, \eqref{eqn:choice-ell}, $\diver u_\ell =0$ and standard mollification estimates (see for instance \cite[Lemma 5.1]{BDLC20}) we have  the following
\begin{align*}
    \| \rho_{q+1} J \nabla H_{q+1} - \rho_q J \nabla H_q \|_{L^1} &\leq \| (\rho_{\ell} + \theta_{q+1})(u_\ell + w_{q+1}) - \rho_q J \nabla H_q \|_{L^1}\\
    &\leq \| \theta_{q+1}w_{q+1} \|_{L^1} + \| \theta_{q+1} u_\ell + \rho_\ell w_{q+1}\|_{L^1} + \|\rho_{\ell} u_\ell - \rho_q J \nabla H_q \|_{L^1}\\
    &\leq 2\delta_{q+1} + \frac{\delta_{q+2}}{20} + \|\rho_{\ell} u_\ell - (\rho_q u_q)_\ell \|_{L^1} + \|(\rho_q u_q)_\ell - \rho_q J \nabla H_q \|_{L^1}\\
    &\leq M \delta_{q+1}
\end{align*}
Note that the above estimate would imply that $\rho_q J \nabla H_q \to f$ in $L^1$ for some $f \in L^1([0,T] \times \T^d ; \mathbb R^d)$. But as $\rho_q \to \rho$ and $J \nabla H_q \to J \nabla H$ in $L^1$, up to subsequences they converge pointwise a.e. Thus we get that $f = \rho J \nabla H$ a.e. on $[0,T] \times \T^d$.
\section{Proof of the main results}
The proof of Theorem \ref{t_ce} and Theorem \ref{t_main} are quite similar to \cite[Theorem 1.4, Theorem 1.3]{BDLC20} respectively, but we write them here for the convenience of the reader.

\subsection{Proof of Theorem \ref{t_ce}}

Without loss of generality we assume $T=1$.
Let $\alpha,b, a_0, M>5$, $\beta>0$ be fixed as in Proposition~\ref{prop:inductive}. Let $a \geq a_0$ be chosen such that
$$ \sum_{q=0}^\infty \delta_{q+1} \leq \frac 1 {32M}.$$
Let $\chi_0$ be a smooth time cut-off which equals $1$ in $[0,1/3]$ and $0$ in $[2/3,1]$,

We set $ \lambda = 20a$ and define the starting triple $(\rho_0,  H_0, R_0)$ of the iteration as follows:
\[
\rho_0 = \chi_0(t) + \Big(1+\frac{\sin( \lambda x_1)}{4}\Big) (1- \chi_0(t)), \qquad H_0 = 0, \qquad R_0 =- \partial_t \chi_0\frac{ \cos (\lambda x_1)}{4 \lambda} e_1\, .
\]
Simple computations show that the tripe enjoys \eqref{eqn:CE-R} with $q=0$. Moreover $\|R_0\|_{L^1} \leq C \lambda^{-1} = (1/20)C \lambda_0^{-1}$ and thus \eqref{eqn:ie-1} is satisfied because $2\beta <1$ (again we need to assume $a_0$ sufficiently large to absorb the constant). Next $\|\partial_t \rho_0\|_{C^0}+\|\rho_0\|_{C^1}\leq C \lambda = 20C \lambda_0$. Since $H_0\equiv 0$ and $\alpha >1$, we conclude that \eqref{eqn:ie-2} is satisfied as well.

Next use Proposition~\ref{prop:inductive} to build inductively $(\rho_q, H_q, R_q)$ for every $q\geq 1$. The sequence $\{\rho_q\}_{q\in \N}$ is Cauchy in $C (L^{1})$ and we denote by $\rho \in  C ([0,1], L^{1})$ its limit.
Similarly the sequence of autonomous Hamiltonians $\{ H_q\}_{q\in \N}$ is Cauchy in $W^{1,p'}$ and $ W^{2,r}$; hence, we define $H \in   W^{1,p'} \cap W^{2,r}$ as its limit. Moreover, thanks to the property (c) and the fact that the sequences $\rho_q$ and $J \nabla H_q$ (up to subsequences) converge pointwise a.e. we get that $\rho_q J \nabla H_q$ converges in $C(L^1)$ to $\rho J \nabla H$.

Clearly $\rho$ and $J \nabla H$ solve the continuity equation and $\rho$ is non-negative on $\T^d$ by
\[
\inf_{\T^d} \rho \geq \inf \rho_0 + \sum_{q=0}^\infty  \inf (\rho_{q+1} - \rho_q) \geq \frac 34 - \sum_{q=0}^\infty \delta_{q+1} \geq \frac 14\, .
\]
Moreover, $\rho$ does not coincide with the solution which is constantly $1$, because
\[
\|\rho - 1\|_{L^p} \geq \|1-\rho_0\|_{L^p} - \sum_{q=0}^\infty  \|\rho_{q+1}-\rho_q\|_{L^p}  \geq  \frac 1 {16}- M \sum_{q=0}^\infty \delta_{q+1} >0.
\]
Finally, since $\rho_0(t, \cdot) \equiv 1 $ for $t\in [0,1/3]$, point (c) in Proposition~\ref{prop:inductive} ensures that  $\rho (t, \cdot) \equiv 1$ for every $t$ sufficiently close to $0$.

\subsection{Proof of Theorem~\ref{t_main}}
We first recall a general fact: if $u$ is an everywhere defined Borel vector field in $L^1([0,T] \times \T^d; \mathbb R^d)$ such that, for a.e. $x \in \T^d$, the integral curve starting from $x$ is unique, then the corresponding continuity equation is well posed in the class of non-negative, $L^1([0,T] \times \T^d)$ solutions for any $L^1$ initial datum such that $\rho u \in L^1([0,T] \times \T^d)$.

Indeed, Ambrosio's superposition principle (see e.g. \cite[Theorem 3.2]{A08}) guarantees that each non-negative, $L^1([0,T] \times \T^d)$ solution such that $\rho u \in L^1([0,T] \times \T^d)$ is transported by integral curves of the vector field, namely (no matter how the Borel representative is chosen) there is a probability measures $\eta$ on the space of absolutely continuous curves, supported on the integral curves of the vector field in the sense of Definition~\ref{defn:int-curve},  such that $\rho(t,x)\, \mathscr{L}^d = (e_t)_\# \eta$ for a.e. $t\in [0,T]$ (where $e_t$ is the evaluation map at time $t$).
Let us consider the disintegration $\{\eta_x\}_{x \in \T^d}$ of $\eta$ with respect to the map $e_0$, which is $\rho_0$-a.e. well defined; since by assumption for a.e. $x \in \T^d$, the integral curve starting from $x$ is unique (and hence coincides with the regular Lagrangian flow),
 we deduce that $\eta_x$ is a Dirac delta on the curve $t \to X(t,x)$ and consequently $\rho(t,\cdot ) \mathscr{L}^d = X(t,\cdot)_\# (\rho_0 \mathscr{L}^d)$. This concludes the proof of the claim.

Let $u= J \nabla H$ be the autonomous Hamiltonian vector field given by Theorem \ref{t_ce} and observe that the Cauchy problem for the continuity equation \eqref{d_continuity} from the initial datum $\rho_0 \equiv 1$ has two different non-negative solutions in $[0,T]$: $\rho^{(1)} \equiv 1$ and the non-constant solution $\rho^{(2)}$ given by Theorem \ref{t_ce}. Hence, by the previous observation we conclude that there exists a set of initial data of positive measure such that the corresponding integral curves are non-unique. Since the fact that the two functions are distinct solutions of the continuity equation is independent of the pointwise representative chosen for the vector field, this completes the proof of Theorem \ref{t_main}.

\section{Non conservation of the Hamiltonian along the trajectories}

\subsection{Proof of Theorem \ref{thm:hamilnotcons}}
We first prove a Theorem in the spirit of Theorem \ref{t_ce}. We fix $p' \geq d-1$.
Let $\alpha,b, a_0, M>5$, $\beta>0$ be fixed as in Proposition~\ref{prop:inductive} and $a \geq a_0$ be chosen such that
$$ \Delta: = \sum_{q=0}^\infty \delta_{q+1} \leq \frac 1 {16}.$$
We fix $ \overline{\psi}_0 \in C^\infty_c ((0,1))$ such that  $\overline{\psi}_0 \geq \Delta$,
$\overline{\psi}_0(x) = \Delta$ for any $x \in [0,1/2]$, $\| \overline{\psi}_0 \|_{L^\infty} \leq 4$, $\int_0^1 \overline{\psi}_0 =1$ (just use the convolution of a proper function and the standard convolution properties).
Then we fix $\overline{H}_0 \in C^\infty_c((0,1/2))$ $\overline{H}_0 \geq 0$ $\int \overline{H}_0 =1,$ $\| \overline{H}_0 \|_{L^\infty} \leq 4 $. We extend these two functions as periodic functions on $\T^d$, imposing that they are independent on the last $d-1$ variables (we call them with the same name with a slight abuse of notation).

Let $\chi_0$ be a smooth time cut-off which equals $1$ in $[0,1/3]$ and $0$ in $[2/3,1]$,

We set $ \lambda = 20a$ and define the starting triple $(\rho_0,  H_0, R_0)$ of the iteration as follows:
\[
{\rho}_0(t,x) = \chi_0(t) + \overline{\psi}(\lambda x) (1- \chi_0(t)), \qquad {H}_0(x) = \overline{H}_0(\lambda x_1), \qquad R_0 =- \partial_t \chi_0 \ensuremath{\mathcal{R}} (\overline{\psi} (\lambda \cdot ) -1) ,
\]
where we have periodized the functions $\overline{\psi}$ and $\overline{H}_0$ with the parameter $\lambda_0$.
Simple computations show that the tripe enjoys \eqref{eqn:CE-R} with $q=0$. Moreover, thanks to Lemma \ref{lemma23}, $\|R_0\|_{L^1} \leq C \lambda^{-1} = (1/20)C \lambda_0^{-1}$ and thus \eqref{eqn:ie-1} is satisfied because $2\beta <1$ (again we need to assume $a_0$ sufficiently large to absorb the constant). Next $\|\partial_t \rho_0\|_{C^0}+\|\rho_0\|_{C^1}\leq C \lambda = 20C \lambda_0$. Since  $\alpha >1$, we conclude that \eqref{eqn:ie-2} is satisfied as well.

Next use Proposition~\ref{prop:inductive} to build inductively $(\rho_q, H_q, R_q)$ for every $q\geq 1$. The sequence $\{\rho_q\}_{q\in \N}$ is Cauchy in $C (L^{1})$ and we denote by $\rho \in  C ([0,1], L^{1})$ its limit.
Similarly the sequence of autonomous Hamiltonians $\{ H_q\}_{q\in \N}$ is Cauchy in $W^{1,p'}$, $ W^{2,r}$ and $L^\infty$ (for the last property we used property (b+) of Proposition \ref{prop:inductive} since $p' >d-1$); hence, we define $H \in  W^{1,p'} \cap W^{2,r} \cap  L^\infty$ as its limit, that is also continuous.
Moreover, thanks to the property (c) and the fact that the sequences $\rho_q$ and $J \nabla H_q$ converge a.e. we get that $\rho_q J \nabla H_q$ converges in $C(L^1)$ to $\rho J \nabla H$.

Clearly $\rho$ and $J \nabla H$ solve the continuity equation and $\rho$ is non-negative on $\T^d$ by
\[
\inf_{\T^d} \rho \geq \inf \rho_0 + \sum_{q=0}^\infty  \inf (\rho_{q+1} - \rho_q) \geq \Delta - \Delta = 0.
\]

Now we apply the Ambrosio's  superposition principle (see e.g. \cite[Theorem 3.2]{A08}) to the non negative solution $\rho$ (note that $\rho J \nabla H \in L^1$), which guarantees that $\rho$
is transported by integral curves of the vector field $J \nabla H$, namely (no matter how the Borel representative is chosen) there is a probability measures $\eta$ on the space of absolutely continuous curves, supported on the integral curves of the vector field in the sense of Definition~\ref{defn:int-curve},  such that $\rho(t,x)\, \mathscr{L}^d = (e_t)_\# \eta$ for a.e. $t\in [0,T]$ (where $e_t$ is the evaluation map at time $t$).
We use the notation $\{\eta_x\}_{x \in \T^d}$, that are probability measure defined by the disintegration of $\eta$ with respect to the map $e_0$, which is $\mathcal{L}^d$-a.e. well defined.
 To conclude the proof of our theorem is sufficient to prove that
$$ \int_{\T^d} \int_{AC} H(\gamma (0)) d \eta_x (\gamma) dx > \int_{\T^d} \int_{AC} H(\gamma (1)) d \eta_x (\gamma) dx,$$
because $\eta_x$ is concentrated, for $\mathcal{L}^d$ a.e. $x \in \T^d$, on the family of absolutely continuous integral curves of $J \nabla H$. Thanks to the superposition principle it is equivalent to prove that
$$ \int_{\T^d}  H(x) \rho(0,x) dx >  \int_{\T^d}  H(x) \rho(1,x) dx,$$
notice that the solution $\rho$ of the continuity equation with respect to the vector field $J \nabla H$ is independent to the pointwise representative of $J \nabla H$, this would conclude the proof.

 By properties (b+), (d) of Proposition \ref{prop:inductive} and the definition of $\rho_0$,
we have the following estimates
\begin{align*}
\int_{\T^d}  H(x) \rho(0,x) dx = \int_{\T^d}  H(x) dx \geq  \int_{\T^d}  H_0(x) dx - \| H - H_0 \|_{L^\infty} \geq 1 - \frac{1}{\lambda_0}.
\end{align*}
and
\begin{align*}
\int_{\T^d} &  H(x) \rho(1,x) dx
\\
&= \int_{\T^d}  (H(x) - H_0(x)) \rho(1,x) dx + \int_{\T^d}  H_0(x)( \rho(1,x) - \rho_0(1,x)) dx + \int_{\T^d}  H_0(x) \rho_0(1,x) dx.
\end{align*}
By property (b+), the definition of $\lambda_{q+1}= \lambda_q^b$ and the definition of $H_0$ and $\rho_0$ we estimate every summand and we get
\begin{align*}
 \int_{\T^d}  (H(x) - H_0(x)) \rho(1,x) dx  \leq \| H - H_0 \|_{L^\infty} \| \rho\|_{L^\infty_tL^1_x} \leq \frac{1}{\lambda_0},
 \\
 \int_{\T^d}  H_0(x)( \rho(1,x) - \rho_0(1,x)) dx \leq \| H_0 \|_{L^\infty} \| \rho - \rho_0 \|_{L^\infty_tL^1_x} \leq 4 \Delta,
 \\
  \int_{\T^d}  H_0(x) \rho_0(1,x) dx \leq 4 \Delta.
\end{align*}
The thesis follows observing that $1 > \frac{2}{\lambda_0} + 8 \Delta$.

\subsection*{Acknowledgements}
The authors would like to thank Thomas Alazard, Camillo De Lellis and Maria Colombo for introducing them to the problem. The authors also thank Elio Marconi and Riccardo Tione for advice on the introduction. VG has been supported by the National Science Foundation under Grant No. DMS-FRG-1854344 and MS has been supported by the SNSF Grant 182565.

redpajama_set_name: RedPajamaArXiv


Details: Details: Details: Details: Details: Details: Details: Details: Details: Citation Text: (1) Ohio Marriages,1800-1958: Groom's Name: Alfred Bently Groom's Birth Date: Groom's Birthplace: Groom's Age: Bride's Name: Polly Mack Bride's Birth Date: Bride's Birthplace: Bride's Age: Marriage Date: 20 Oct 1816 Marriage Place: Muskingum, Ohio Groom's Father's Name: Groom's Mother's Name: Bride's Father's Name: Bride's Mother's Name: Groom's Race: Groom's Marital Status: Groom's Previous Wife's Name: Bride's Race: Bride's Marital Status: Bride's Previous Husband's Name: Indexing Project (Batch) Number: M51450-1 System Origin: Ohio-ODM Source Film Number: 865142 Reference Number:.
Details: Details: Details: Details: Details: Details: Details: Details: Details: Citation Text: (1) Ohio Marriages,1800-1958: Elizabeth Bently Ohio Marriages, 1800-1958 birth: christening: baptism: death: burial: residence: parents: spouse: Josephas Powel child: marriage: 15 Sep 1836 ??? Muskingum,Ohio.


text:
Posted on April 1, 2016 in Friday's News Quiz
1. According to chief trauma surgeon Dr. Stefaan Nijs at a Brussels hospital, those injured in the ISIS terrorist attack in Brussels have sustained severe and "life-changing" injuries, including:
a) loss of limbs
b) shrapnel injuries
c) severe burns
2. Belgian surgeon Stevaan Nijs described the British victims' reactions to their injuries as:
a) "courageous and strong"
b) "hysterical and inconsolable"
c) "angry and bitter"
d) "indifferent"
3. During an interview this week, presidential candidate ________________________________, when asked if he/she would consider nuclear retaliation after this week's terror attacks in Brussels, said they would never rule anything out and doesn't want to say "because, at a minimum, I want them [ISIS] to think maybe that we would use it." Adding, "The fact is that we need unpredictability. When you ask a question like that…the enemy is watching…. I frankly don't want the enemy to know how I'm thinking. But with that being said, I don't rule out anything."
a) Bernie Sanders
b) Ted Cruz
c) Hillary Clinton
d) Donald Trump
4. The mission of the newly created Marine Corps Cyberspace Warfare Group (MCCWG) is to man, train and equip Marine cyberspace mission teams to perform both defensive and offensive operations in support of U.S. Cyber Command and Marine Forces Cyberspace Command. The ___________________________ are also actively recruiting cyber soldiers and standing up their own cyber units and programs.
a) Navy, Army, Air Force and National Guard
b) Navy, National Guard, Air Force and Coast Guard
c) Navy, Army, Air Force and Coast Guard
d) Navy, Army, Air Force and Merchant Marines
5. U.S. Military communications has become vulnerable to theft because it has been put onto computer networks. Everything from power grids, banking, government operations to defense contractor weapons' plans have shifted online in the past few decades. That information is all a target for both ___________________ and criminal organizations that are becoming increasingly sophisticated at cybertheft.
a) school-sponsored hackers' clubs
b) state-sponsored hackers
6. During a townhall meeting with Marines in November, Marine Corps Commandant Gen. Robert Neller said that _______________ had stolen military secrets from the United States. He said specifically, "While we've been fighting, our adversaries, many of them in this part of the world — pick one: China, North Korea, Iran, Russia — what have they been doing? Making money, buying new gear, stealing all of our secrets." He also said, "Ever look at all the _________________ equipment? What's it look like? It looks like our stuff. How is that? They stole our stuff, fair and square."
a) China/Chinese
b) Iran/Iranian
c) Russia/Russian
d) North Korea/North Korean
7. A superdelegate is a delegate to the Democratic National Convention who is seated automatically and chooses for whom they want to vote, even if the voters in their state chose a different candidate. Superdelegates include distinguished party leaders, and elected officials, including all Democratic members of the House and Senate and sitting Democratic governors. At the 2016 Democratic National Convention the superdelegates will make up approximately ______________ of the total number of delegates.
a) 1%
b) 3%
c) 15%
d) 50%
8. When asked by a NY Daily News reporter this week, the Democratic superdelegates from New York said they will support Hillary Clinton:
a) only if she wins the primary
b) if neither Sanders nor Clinton wins the NY primary
c) even if opponent Bernie Sanders wins the NY primary
9. The Pentagon announced recently that they had launched a cyberoffensive against __________________. This was the first major integration of U.S. Cyber Command into a major battlefield operation since the command was established in 2009.
a) ISIS in Syria
b) Russia in Syria
c) Israel
d) Turkey
10. The disclosure of a government-sanctioned cyberattack represents a shift in America's war-fighting strategy and power projection. __________________ publicly acknowledged launching cyberwar.
a) We are the only country that hasn't, until now
b) All of our allies have
c) Until now, only Russia and China have
d) No other nation has


Since we moved into our new digs at the former-SGI building today I decided to put Ev's old Bloggercam back in service. Behind me you can see, Biz Stone, Eric Case, Steve D, and Graham. Expect for the usual cast of characters to make on screen cameos. Updated every 15 minutes whether you like it or not.


Before we continue, there are two things about the SQL procedure that you need to know.
(1) The order of the clauses matters in the SELECT statement.
In our previous example, listing the FROM clause before the SELECT clause will result in an error.
(2) There is no semi-colon or comma separating the clauses.
Placing a comma or semi-colon between the clauses will also result in an error.


text:
Mountain At My Gates
by Foals
Album: What Went Down (2015)
Foals frontman Yannis Philippakis told NME the story of this song: "I'd recorded the beginning riff on my phone ages ago," he said. "At the beginning it had a baggy feel, but became less so with more work."
"The central image - 'I see a mountain at my gates' was from me getting more interested in seeing what would come out lyrically where there wasn't a pre-conceived idea," Philippakis added. "Normally I write voraciously in books and journals, then harvest a lot of that for the record. This, though, came out instantaneously in the room."
The video sees Foals perform their song live in a sidewalk alley between mountains while interacting with the viewer. The clip was directed by Nabil (Frank Ocean, Kanye West, Alt-J) with whom the band worked with on previous visuals for "Bad Habit" and "Late Night." It was filmed using GoPro cameras, using their spherical Virtual Reality technology. This enables viewers using Chrome or the YouTube app to interact with the visual.
More songs from Foals
Lyrics to Mountain At My Gates
UmbrellaRihanna
When Rihanna's "Umbrella" was a hit in the summer of 2007, it rained constantly in London, prompting their newspaper The Sun to suggest a "Rihanna Curse."
Live ForeverOasis
The Oasis song "Live Forever" was written in response to "I Hate Myself And I Want To Die" by Nirvana. "Kids don't need to hear that nonsense," said Noel Gallagher.
Piano ManBilly Joel
Billy Joel is surprised that "Piano Man" is so successful. He called it "an old, long song about a guy at a depressing piano bar."
King TutSteve Martin
The comedian Steve Martin had a hit in 1978 with "King Tut." The Nitty Gritty Dirt Band, who Martin would open for on tour, were his backing band on the song.
I Missed AgainPhil Collins
Phil Collins' "I Missed Again" was originally "I Miss You, Babe," and was a very somber song about his recent divorce. Collins decided to lighten it up and inject some humor into the song.
Velcro FlyZZ Top
The music video for ZZ Top's "Velcro Fly" was choreographed by Paula Abdul.
Jon Oliva of Trans-Siberian OrchestraSongwriter Interviews
Writing great prog metal isn't easy, especially when it's for 60 musicians.
Elton JohnFact or Fiction
Does he have beef with Gaga? Is he Sean Lennon's godfather? See if you can tell fact from fiction in the Elton John edition.
Steve Morse of Deep PurpleSongwriter Interviews
Deep Purple's guitarist since 1994, Steve talks about writing songs with the band and how he puts his own spin on "Smoke On The Water."
Michael BoltonSongwriter Interviews
Into the vaults for this talk with Bolton from the '80s when he was a focused on writing songs for other artists.
Dick Wagner (Alice Cooper/Lou Reed)Songwriter Interviews
The co-writer/guitarist on many Alice Cooper hits, Dick was also Lou Reed's axeman on the Rock n' Roll Animal album.
Angelo Moore of FishboneSongwriter Interviews
Fishbone has always enjoyed much more acclaim than popularity - Angelo might know why.


text:
Breaking: Fort Hood Terrorist Found Guilty on 13 Counts of Murder
Army Major and Fort Hood terrorist Nodal Hasan has been convicted of 13 counts of murder and 32 counts of attempted murder in 2009 shooting rampage.
It should be noted that 14 lives were taken that day, including that of an unborn child.
Hood shooter Nidal Hassan was found guilty on all counts of premeditated murder and attempted murder Friday, making him eligible for the death penalty.
A military jury deliberated for roughly one day before announcing the verdict.
Hasan sat emotionless, stroking his beard as the verdict was delivered.
He was charged with 13 counts of pre-meditated murder and 32 counts of attempted pre-meditated murder in the Nov. 2009 shooting attack at the sprawling Texas military base.
Hasan, an Army psychiatrist, had acted as his own attorney at the 13-day trial, but questioned only three of nearly 90 witnesses and presented only a single piece of evidence in his defense: an evaluation from his boss that called him a good soldier. He also chose not to make a closing statement.
Sentencing is set for Monday.
Remember that the Pentagon had actually releases a report on the shooting rampage which failed to mention the word "Islam" or "Muslim."
Now perhaps, since the trial clearly demonstrated this was an act of terrorism, the administration could go ahead and do the right thing and stop referring to this as "workplace violence."
Obama's insistence on referring to the attack as "workplace violence" means that the victims are denied medical benefits they would otherwise receive.
The President has turned his back and forgotten the victims of a heinous terrorist attack. Where is the outrage? Or is sympathy solely reserved for the terrorists themselves under this administration?
Chris Gibson to Announce Retirement
Biden Hammers Hillary Clinton For Handling of 2016 Campaign


text:
Never Grow Old
Action Drama Thriller ... Western
IMDb: 5.6 - 400+ votes
$ 12 000 000 Budget
Country: Ireland
Director: Ivan Kavanagh
Writer: Ivan Kavanagh
Watch Never Grow Old 2019 Online Free Movie
An Irish undertaker profits when outlaws take over a peaceful American frontier town, but his family comes under threat as the death toll rises.
Triple Threat Action - 2019 A hit contract is taken out on a billionaires daughter intent on bringing down a major crime syndicate. A down and out team of mercenaries must take on a group of professional assassins and stop them... IMDb 6.0 - 1k votes Damsel Comedy - 2018 As Samuel Alabaster travels across the American Frontier, on a journey to marry the love he has always looked for, Penelope, life becomes more and more dangerous. Accompanied by his miniature horse,... IMDb 5.6 - 1k votes The Hobbit: The Battle of the Five Armies Adventure - 2014 After the Dragon leaves the Lonely Mountain, the people of Lake-town see a threat coming. Orcs, dwarves, elves and people prepare for war. Bilbo sees Thorin going mad and tries to help. Meanwhile,... IMDb 7.4 - 433k votes Dracula Untold Action - 2014 At the turn of the century, the young lord Vlad and his family live a peaceful life ruling over their small kingdom, but when a Turk warlord demands from Vlad a thousand boys and his son to create an... IMDb 6.3 - 162k votes Star Trek V: The Final Frontier Action - 1989 Captain Kirk and his crew must deal with Mr. Spock's long-lost half-brother who hijacks the Enterprise for an obsessive search for God at the center of the galaxy. IMDb 5.5 - 51k votes Triple Frontier Action - 2019 Five former Special Forces operatives reunite to plan a heist in a sparsely populated multi-border zone of South America. For the first time in their prestigious careers these unsung heroes undertake... IMDb 8.1 - 500+ votes A Fistful of Lead Western - 2018 Four of the West's most infamous outlaws carry out a daring bank heist in the gold-rush town, Bath Water. As the posse takes chase, things take a turn for the worse, as the bandits realize they've... IMDb 6.3 - 92 votes Dementia 13 Fantasy - 2017 Follows an old-money family still dealing with the death of its youngest daughter several years later. While honoring the daughter's death, a long con, an axe-wielding serial killer, and a vengeful... IMDb 4.4 - 600+ votes Emma Fielding: More Bitter Than Death Mystery - 2019 While hosting the annual gathering of the Association for the Study of American Archaeology at Kenzer College, Emma Fielding is thrust into the midst of a murder mystery when the President of her... IMDb 6.6 - 77 votes Lizzie Biography - 2018 In 1892, after the Borden family welcomes a new Irish maid called Bridget Sullivan, she and Lizzie become friends. The friendship between these women becomes something more, even as Lizzie's... IMDb 5.7 - 3k votes They Shall Not Grow Old Documentary - 2018 Through ground breaking computer restoration technology, filmmaker Peter Jackson's team creates a moving real-to-life depiction of the WWI, as never seen before in restored, vivid colorizing &... IMDb 8.5 - 10k votes Black '47 Action - 2018 Set in Ireland during the Great Famine, the drama follows an Irish Ranger who has been fighting for the British Army abroad, as he abandons his post to reunite with his family. Despite experiencing... IMDb 6.9 - 5k votes American Dream/American Knightmare Documentary - 2018 Documentary of the life and exploits of gangsta rap producer Suge Knight. IMDb 5.6 - 100+ votes Demonia Undertaker Action - 2017 Alien bounty hunters from another planet are sent to stop Tanash, a possessed demonic serial killer who comes to earth to take over the bodies of humans before killing them. Staleth, played by... IMDb 3.2 - 81 votes The Dark Knight Rises Action - 2012 Despite his tarnished reputation after the events of The Dark Knight, in which he took the rap for Dent's crimes, Batman feels compelled to intervene to assist the city and its police force which is... IMDb 8.4 - 1360k votes Sgt. Stubby: An American Hero Adventure - 2018 The true story of a stray dog who joins his new master on the battlefields of the First World War. For his valorous actions, Sgt. Stubby is still recognized as the most decorated dog in American... IMDb 6.9 - 1k votes


Capital markets may have overcapitalized the art blending industry amid a whirlwind of new IPOs. In the connection of this "hot" IPO advertise every individual organization's valuation may appear to be sensible. Then again, after cautious investigation of every organization's monetary explanation endless supply of experts' gauges of the business' development prospects, it is misty whether the specialty preparing industry is overcapitalized. This could be another "hot" then "crash" IPO industry, similar to biotech or the PC plate drive industry.
3. What is your assessment of the intrinsic value of Boston Beer's stock at the time of the case? What should be its IPO price?
4. Do you think the total market value of Redhook, Pete's and Boston Beer makes sense given the total size and profitability of the beer industry and the craft-brewing segment?


text:
Q: Implement re.search() functionality in Groovy or Java I need a Groovy/Java function to search for groups in a string based on regular expression
Ex:
function("([\w-]+)-([\w.-]+)\.([\w.-]+)" ,"commons-collections-3.2.2.jar" )
should return a list ["commons-collections" , "3.2.2" , "jar"]
Python can do this by
>> import re

>> re.search("([\w-]+)-([\w.-]+)\.([\w.-]+)" ,"commons-collections-3.2.2.jar" )

>> print(result.groups())

output is ("commons-collections" , "3.2.2" , "jar")

A: It is a simple and basic task in groovy. Any way I hope this answer will help you.
"commons-collections-3.2.2.jar".findAll(/([\w-]+)-([\w.-]+)\.([\w.-]+)/) {
    println it
}

This will produce the output :
[commons-collections-3.2.2.jar, commons-collections, 3.2.2, jar]

Update :
As @tim_yates mentioned in comment,
println "commons-collections-3.2.2.jar".findAll(/([\w-]+)-([\w.-]+)\.([\w.-]+)/) { it.tail() }

This provides better output than above and also more specific to the task.
Output:
[[commons-collections, 3.2.2, jar]]

redpajama_set_name: RedPajamaStackExchange


text:
class WHmac: public node::ObjectWrap {
public:
	static v8::Local<v8::Object> NewInstance() {
		v8::Local<v8::Function> cons = Nan::New(constructor());
		return Nan::NewInstance(cons).ToLocalChecked();
	}
	static v8::Local<v8::Object> NewInstance(int argc, v8::Local<v8::Value> argv[]) {
		v8::Local<v8::Function> cons = Nan::New(constructor());
		return Nan::NewInstance(cons, argc, argv).ToLocalChecked();
	}

	static const char* ClassName;

	static void Init(v8::Local<v8::Object> exports);

	static NAN_METHOD(New);

	static NAN_METHOD(Generate);
	static NAN_METHOD(Export);
	static NAN_METHOD(Import);
	static NAN_METHOD(Sign);
	static NAN_METHOD(Verify);

	Handle<ScopedHMAC> data;

protected:
	static inline Nan::Persistent<v8::Function> & constructor() {
		static Nan::Persistent<v8::Function> my_constructor;
		return my_constructor;
	}

};

#include "common.h"

#endif // OSSL_W_HMAC_H_INCLUDE

redpajama_set_name: RedPajamaGithub


We serve shoppers worldwide with companies ranging from interior design, inside structure to furnishings design. As a substitute of meshing a color scheme with a way of place, designer Irakli Zaria used rich gold and turquoise as an antidote to gloomy London days on this stylish pied-a-terre "In a spot the place there are such cloudy skies, it is not sensible to have a grey inside," he stated.
If the software packages we reviewed are a bit a lot for you, chances are you'll need to take a look at Planner 5D It is free and much easier to learn and use than all the interior design programs we reviewed. Each inside design undertaking begins with a planning section. Massive and tiny corporations typically hire interior designers as employees on common working hours.
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They have numerous DIY's and behind the scenes with the designers they characteristic on their TELEVISION reveals (hi, Joanna Gaines!). It is important to understand that interior design and interior decorating are two separate professions. Clients love the Decorist online inside design expertise as a result of it saves them each money and time.


Baobab House is the perfect beachfront property for a peaceful vacation surrounded by nature. Wake up to the sound of African bird songs; swim in the warm Indian Ocean; walk on miles of sandy beach, where you will rarely encounter people. Relax under the baobabs and watch the sun setting over the dunes.Previous guests have raved about the quality of the staff including Bernad's excellent cooking.
This private house is on an acre of beachfront property with five large ancient baobab trees. The Baobab tree is indigenous, some of them more than 500 years old, with many beautiful stories, and they give a peaceful atmosphere throughout the whole property. Along side the Baobabs, there are also several acacias, another indigenous tree, providing shady areas when you've had enough sun on the beach.
Website by Lamu Software. All content © Baobab House Lamu .


text:
Welbitz Wins ITEXPO's Idea SHOWCASE
By Paula Bernier February 16, 2018
It was a sweep. Both the audience and the judges at ITEXPO's IDEA Showcase Thursday picked Welbitz as the winner.
The company went up against fellow startups Bekindr, chirrp, Observables, NationPay, and Park Capital. Each company had just a few minutes to give its elevator pitch. After each pitch, a panel of judges had the opportunity to ask a few questions. Then, the audience and the judges voted for what they considered the best idea. Welbitz won the audience vote by a wide margin.
Welbitz aims to make it easier to detect breast cancer.
The company has developed a connected device and a mobile app. The device is a thermometer patch that a woman puts on her breast. If one breast is significantly warmer than the another, that can signal breast cancer, Welbitz says.
The patented device has received FDA 510(k) clearance. And the company expects to bring it to pharmacies in the not too distant future.
Welbitz said this device and its mobile app will encourage women to do self-exams more frequently.
As for Bekindr, the presenter talked about how the company encourages people to share their stories. But it was unclear what the business model or product is.
A company called chirrp talked about its artificial intelligence-based platform. It enables businesses to interact with their customers on multiple channels. That way customers don't have to wait on hold to reach businesses by phone. And businesses can deliver personalized conversations for customers based on the data they've collected.
NationPay told its story as well. The company provides cardless banking solutions. In this case, a bank creates a BankNet on the NationPay wallet for 5000 NPT. The bank selects the fees for up to 3 signatories for minting and burning operations. The fee for the use of a token is 1 per or less. Any person can add the BankNet as a token wallet account, but only a client can mint tokens from their bank account at that bank to this token wallet account or burn tokens back to that bank account. And the Client can send the token to unbanked friends or family around the world, or trade the token for tokens of a foreign bank needed by a friend and then send those tokens to the friend.
Observables has a router that does premises management. It can help IT integrators, managed services providers, value-added resellers and others connect, manage, monitor, and control connected devices at customer locations and software-defined local area networks.
And Park Capital talked about a platform to deliver concierge services. Several noteworthy brands will be launching services, based on the white-label version of the platform, in the next couple of months.


Ridgeway on German and Ridgeway on 23rd assisted living communities emphasize independence and privacy to each of their residents, both today and tomorrow. If you or a loved one needs some assistance with the activities of daily living, or if you simply want a comfortable, supportive home to make life more enjoyable, Ridgeway is the place for you. Our combination of high quality services and premier housing offers privacy and independence, along with the personal care and assistance you may need. With diverse services and amenities, there is something to fit any lifestyle. Our highest priority is the well being of our residents.
Residents enjoy the privacy of their own apartment and Ridgeway is designed to include spacious, beautiful common areas of our facilities. Whether you are visiting with friends or family, or looking for a quiet area to read, you'll enjoy the comfortable atmosphere found at Ridgeway.


Fresh! Fresh! Fresh! That's what you can come to expect from Southern HomeBrew when it comes to yeast for your home wine making pleasure. We also stock chemicals, flavorings and oak for your home wine making.
Southern HomeBrew - your one stop shop!


text:
San Francisco Ballet: Liang / Marston / Pita at Sadler's Wells
San Francisco Ballet: Liang / Marston / Pita at Sadler's Wells | Theatre review
Ghazaleh Golpira
Shot by Ambra Vernuccio
Ghazaleh Golpira Shot by Ambra Vernuccio
With Helgi Tomasson at the helm, it is no doubt that San Francisco Ballet was destined for great things. Since his appointment as the company's artistic director in 1985, Tomasson has not only contemporised ballet; he has revolutionised it. The company, one of the most renowned in the world, bridges the gap between dance and gymnastics, combining elegance with the experimental to create an eclectic piece of theatre that takes this delicate art form to a whole new and breath-taking level.
Leading the way with four triple bills, Programme B – comprising the choreography of Edwaard Liang, Cathy Marston and Arthur Pita – sees Sadler's Wells experience a total of three set changes, ranging from the simple to the thrillingly complex, the spiritual to the materialistic. However, it is Pita's Björk Ballet which is the stand-out contender of the night. His sequence, an almost classicised Cirque du Soleil, is the perfect by-product of what one might label a well-rehearsed carnival. Imbued with beauty, particularly during the troupe's interaction with a bunch of magnetic chandeliers which surreptitiously emerge from the sky, is precious. The use of a reflective glass floor aids the narrative, creating a rippling effect which has an almost celestial quality to it, making it seem as though the masked fisherman are truly walking on water. Then, as if this were not stunning enough, out emerges an inclined rectangular board, upon which the dancers perform their mid-air aerobics, whilst their peers funnel around and under the device. It is a hugely explosive and colourful piece – synchronised to the melancholic voice of Icelandic singer Björk herself – and makes for a hauntingly beautiful experience.
Gone are the days when tutus were the thing, for in these shows there is not one in sight. Classical and neoclassical ballet are about to be wholly refined and the San Francisco team are here to show the world how.
Photos: Ambra Vernuccio
San Francisco Ballet is at Sadler's Wells from 29th May until 8th June 2019. For further information or to book visit the theatre's website here.
Related Itemsfeaturedreview
29th May - 8th June 2019
Modern Table Dance Company: Sok-do (Velocity) at The Place | Theatre review
Dido at the Roundhouse | Live review


text:
Q: How can I change "Mount Point" that has been changed to normal? previously I has been trying to install Kali Linux on my usb.
Now here is the problem, I have used the command
sudo dd if=kali.dmg of=/dev/rdisk1 bs=512k

which I think it changed or replaced my OS X main drive.
I still believe that my OS X drive still exists because the "hard drive" icon that I modified still appeared on the screen.

I decided to check all the drives on my laptop and here is the result using
df on terminal and here is the result.

Does any if these is my OS X that has been replaced?
PS. I checked this by install new OS X on external hard drive.

A: If you dd'd it to your actual drive's partition, then what you're seeing on the bootloader is simply an entry for an OS that doesn't exist anymore. dd, unfortunately, does not change or replace anything except the data. It's a low level copy/replace utility with tons of features.
In Mac OS X the drives are denoted as /dev/diskX where X is an integer. The partitions are denoted as /dev/diskXsX where X is also an integer.
Also, the raw files to the hard drives and to the partitions are /dev/rdiskX and /dev/rdiskXsX respectively.
From the man hdiutil on Mac, you get the following.

/dev/rdisk nodes are character-special devices, but are "raw" in the
  BSD sense and force block-aligned I/O.  They are closer to the
  physical disk than the buffer cache.  /dev/disk nodes, on the other
  hand, are buffered block-special devices and are used primarily by the
  kernel's filesystem code.

In simple terms, you just overwritten your hard drive and you did it in the most efficient and fastest way possible by using rdisk instead of disk.


The Hotel Bauer & Sporting is a prestigious four-star superior hotel overlooking the sea in the spellbinding Jesolo pine groves. It is the ideal place for a relaxing stay surrounded by greenery.
The Hotel Bauer & Sporting has a selection of elegant rooms, finely appointed suites and comfortable family suites. Guests can take advantage of the spacious, well-lit lounges, the American bar and the large, modern covered terrace with stunning views of the sea in the day and the Mediterranean garden with soft, romantic lighting in the evening. Free Wi-Fi is available in the bedrooms, lounges, bar and garden.
For sport and fun in the open air, there are two large pools with crystal clear water where adults can take a relaxing dip. One of the pools is always heated to 29°C and it has hydro-massage jets and neck massage showers. There is an extremely well-equipped gym and bicycles are available for both adults and children.
There are areas for our younger guests in both of our pools, one of which is heated to 29°C. In addition there are two play areas: one inside and one in the garden, surrounded by greenery.
With its colourful umbrellas and enticing sun loungers, our private beach is the perfect place to take it easy and get an impressive tan.
It's fast and easy and you will get all relevant news about our offers and more.


text:
Q: Duplicate Blog ID in array I have two tables first one is tbl_blogs that contains all the blog categories such as, men, women etc.
The second table contains the blog's information of related category.
 $query1=mysqli_query($link,"select * from tbl_blog_cat");
    $array1=array();
    $array2=array();
    $rows=array();
    $row=array();
    if(mysqli_affected_rows($link)){
    while($row=mysqli_fetch_array($query1)){

        $catId=$row['id'];
        $catName=$row['title'];

        $q2=mysqli_query($link,"select * from tbl_blogs where
           blog_cat=$catId");
        if(mysqli_affected_rows($link)){

            while($row2=mysqli_fetch_array($q2)){

                    $rows['BlogId']=$row2['BlogId'];
                    $rows['Title']=$row2['Title'];
                    $rows['blog_cat']=$row2['blog_cat'];

                   $rows['Description']=$row2['Description'];

                  $rows['CoverImage']=$row2['CoverImage'];
                    array_push($array1,$rows);

            }

            $array2[$catName]=$array1;

        }

    }
    echo json_encode($array2);
     }

-----------------    here is my output -------------------
The second array of women category has a duplicate id of the previous category.
I'm expecting to remove or it let me know if there is any mistake in my code.
thanks.enter code here
  { "Men": [ { "BlogId": "25", "Title": "layer", "blog_cat": "5", "Description": "a sheet, quantity, or thickness of material, typically one of several, covering a surface or body", "CoverImage": "Screenshot_from_2018-12-28_23-12-51.png" } ], "Women": [ { "BlogId": "25", "Title": "layer", "blog_cat": "5", "Description": "a sheet, quantity, or thickness of material, typically one of several, covering a surface or body", "CoverImage": "Screenshot_from_2018-12-28_23-12-51.png" }, { "BlogId": "19", "Title": "Blog Test Man", "blog_cat": "6", "Description": "no dsc", "CoverImage": "Screenshot_from_2019-01-07_16-54-44.png" }, { "BlogId": "22", "Title": "layer", "blog_cat": "6", "Description": "a sheet, quantity, or thickness of material, typically one of several, covering a surface or body", "CoverImage": "Screenshot_from_2018-12-28_23-12-51.png" }, { "BlogId": "24", "Title": "Blog Test Man", "blog_cat": "6", "Description": "no dsc", "CoverImage": "Screenshot_from_2019-01-07_16-54-44.png" } ] }

A: You keep populating $array1, but you don't clear it when you start a new category.
Try putting $array1=array(); after your first while