text:
family name of white lauan
Not all countries that have shown an interest in the name are listed in the bar graph. A submission from Canada says the name Luan means "Circle. Luan Peters is an actress and singer. Submit the origin and/or meaning of Luan to us below, luan was also found in the following language(s): Albanian, Mandarin, and Tetum. Luan was born on March 13th, 1991 in Campo Grande. Bo was born on April 16th, 1965 in Harbin. Longer bars in the bar graph indicate that people in the country are more interested in the name. This Dipterocarpaceae article is a stub. She plays Guitar, Piano, and Harmonica. Luan is a soccer player for the Grêmio Foot-Ball Porto Alegrense, Tanabi Esporte Clube, América Futebol Clube (SP), Grêmio Catanduvense de Futebol, Brazil national under-20 football team, and Brazil national under-23 football team. From 1880 to 2018, the Social Security Administration has recorded 1,041 babies born with the first name Luan in the United States. It excludes Near Threatened (NT), Data Deficient (DD), and prehistoric species. Luan was born on May 10th, 1971 in Junik. See other popular names in California, New York, or Massachusetts. Version 2011.1. Please select from the list of all origins below: Survey: Which of the following lists would you find most interesting? It is commonly called (along with some other species in the genus Shorea) Philippine Mahogany or White Lauan.It is endemic to the Philippines.It is threatened by habitat loss which stems from recent human involvement in its natural habitat.. References. U.S. Census Bureau: Frequently Occurring Surnames from the Census 2000 (public domain). A user from Missouri, U.S. says the name Luan is of American origin and means "The joy". Nguyen was born in 1733. Being captive bred. See other popular names in Texas, California, or New Jersey. People with the last name Luan are most frequently Asian, Hawaiian, and Pacific Islander, What is the most accurate origin of the name. What are some names that would belong on a list titled ", the most popular surnames and their ethnicities and origins, According to a user from Australia, the name Luan is of. Currently being captive bred. Luan was born on March 27th, 1993 in São José do Rio Preto, Brazil. Luán was born in Yue (state). Endemic to, Philippine spotted deer, Visayan spotted deer, Negros striped-babbler, Negros tree-babbler, Philippine mouse-deer, Balabac mouse deer, Spotted greenshank, Nordmann's greenshank, Rufous-headed hornbill, Visayan wrinkled hornbill, Walden's hornbill, writhed-billed hornbill, Knifetooth sawfish, narrow sawfish, pointed sawfish, 263 individuals in 2008. Interest is based how many people viewed this name from each country and is scaled based on the total views by each country so that large countries do not always show the most interest. Luan Parle is a pop rock, alternative rock, roots rock, folk rock, country rock, and acoustic music musician. He breathed his last breath in 566. White seraya Family Name Dipterocarpaceae Synonyms Parashorea malaanonan ... Lauan Lauan item Lauan-barik Lauan-danlog Lauan-pula Lauan-puti ... Murut Pendan Philippine mahogany Seraya putah Takuban Takulau Tavoy wood Thingadu Tiaong Urat mata White lauan White seraya Yauaan Regions of Distribution Oceania and S.E. Shorea palosapis is a flowering plant in the Dipterocarpaceae family. White lauan, white seraya Apnit, Bagtikan Native to Brunei Darussalam, Malaysia, and the Philippines Phalaenopsis micholitzii: Endemic to Camarines Sur and Zamboanga: Podocarpus palawanensis: Endemic to Pagdanan Range of Palawan: Rafflesia magnifica: Endemic to Mt. Surviving subspecies endemic to, Agarwood, Aloewood, Eaglewood, Lign-aloes, Native to the Philippines, possibly present in, Possibly extinct. Sunburst turtle, spiny terrapin, spiny turtle, Endemic to Catanduanes Island and some parts of Bicol region, Frog-faced softshell turtle, Cantor's giant softshell, Visayan hornbill, tarictic hornbill, Ticao tarictic hornbill, Less than 1800 individuals remain.
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Of course, you shall have it if you enter six "literary" comments.
Old English isn't too confusing for me considering I am in theatre and Shakespeare is sort of a legend there. I am constantly comparing the new and old English in order to make sense of the words we have today, just as we do when we find out the meaning of words, we compare the modern words to the Latin suffixes and prefixes. Not only did language transfer over, so did the actual story. We have Lion King which was derived from Hamlet.


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text:
Гуанчжоу Спорт Арена — крытый дворец спорта в Китае, расположенный в г. Гуанчжоу. Домашняя арена чемпионата мира по мужскому баскетболу 2019 и Азиатских игр 2010. В основном используется для спортивных мероприятий и концертов.

Общая информация
Разработанная MANICA Architecture в сотрудничестве с Институтом дизайна Гуанчжоу, эта арена не только принимала у себя Азиатские игры в 2010 году, но и стала домом для различных мероприятий мирового уровня, таких как баскетбольные турниры, международные мероприятия и музыкальные концерты.

Проект здания также имеет потенциал, в частности соответствует стандартам НБА, что позволяет ей стать домашней ареной для баскетбольных команд ведущей лиги мира.

Дворец, расположен в быстро развивающемся районе Луоганг, доступен как пешеходам так и автотранспорту.

Арена выделяется своей архитектурной особенностью. Естественные изгибы, которые переводятся во внешнюю часть гармоничного здания и плавные линии из стекла и яркого металла поднимаются и опускаются от фундамента здания до высшей точки. Экстерьер здания является культовым и узнаваемым.

Каждое место было разработано таким образом, чтобы максимально создать комфорт и лучший обзор для всех гостей. Дополнительные помещения включают в себя широкий выбор магазинов, сувенирных лавок и кафе.

Концерты и музыкальные шоу
3 сентября 2011 года ирландская вокальная поп-группа Westlife провела концерт в поддержку альбома Gravity Tour.

18 апреля 2015 года американская певица и автор песен Кэти Перри впервые показала свою музыкальную постановку на китайской земле. Это было её первое шоу в Китае.

3 июня 2017 года Джокер Сюэ впервые выступила на этой арене в рамках тура.

30 августа 2017 года Ариана Гранде впервые выступила на этой арене в рамках тура Dangerous Woman Tour.

Крупные спортивные мероприятия
 2010 — Азиатские игры 2010 года;
 2018 — Чемпионат мира по баскетболу 2019 года.

Примечания

Ссылки
 Сайт арены

Здания и сооружения, построенные в 2010 году в Китае
Спортивные сооружения Гуанчжоу

redpajama_set_name: RedPajamaWikipedia


The fourth lecture from the cycle "Tales from the Plate" will take place on April 29, 2010 at 7:00 P.M. at the Cultural Information Centre, Preradoviceva ulica 5, Zagreb.
Please, ask yourself this question: what is it that you know about the world around you and what is it that you want to believe about it? And then ask yourself: if I learn something new about the world around me, will that shake my present belief?
But don't ask yourself these questions in solitude – accept this invitation and come to the place where you will be able to test them on the spot.
All cow mothers and their children – calves – have been given an opportunity to tell you their story within an hour on April 29 at 7:00 P.M. – the "Tale of a Happy Cow and a Carefree Calf" – and they are asking you to listen.
They are not even aware of how difficult it is to hear their voice... through that huge myth about milk that our civilization has created and that our childhoods have been permeated with, through all that we were taught by people who loved us and wished us good.
They are not aware of how difficult it is to hear their voice through the extraordinarily powerful and incessant messages of the milk industry, which we are stuffed with from all sides – the television, our medical doctors, our schools, and the shelves of our stores.
They are not aware of the fact that their voice is unable to penetrate the walls of our cynicism, sarcasm, and cold rationalizing.
This time they have been given another chance.
Their only hope is that you might want to listen to their story.
So come with your heart open, for them and for yourself.


There's something so satisfying about pulling out a $15 word—the kind that you hardly ever get to use, but fits the situation perfectly. On the other hand, that feeling when you can't quite find the right word for what you're trying to express is incredibly aggravating. Well, we're here to help. Here are 25 weird, obscure, and downright cool words hidden in the English language.
6 Comments on Have You Ever Heard These 25 Obscure English Words?
I'm in love with this list of weird, fun words! It brightened my day.
This is one of your best columns ever! I like this so much I might use the list for Word of the Day, although it will be challenging to work in a few of them. I'm thinking the pogonotrophic pilgarlic archimime in my life won't know their meanings anyway, so I can always use them on him…probably a better idea than the tyrotoxism or defenestration I've been contemplating for his demise.
Great stuff! Thank you. That was very enjoyable and entertaining. I shall certainly memorize and use some of those words in appropriate circumstances.
Don't forget octothorpe, the pound sign (#).
At least I know what they're not.
PS – and vice versa.


text:
NRA Convention Begins, Trump to Speak Friday
May 26, 2022 by NationalWatch
NRA Convention Begins, Trump to Speak Friday (Mark Humphrey/AP File)
By Eric Mack | Thursday, 26 May 2022 08:40 PM
Despite calls for a delay, a move or an outright cancellation of the National Rifle Association convention this weekend in Houston, the show will go on, starting Thursday — with former President Donald Trump vowing to keep his speaking date Friday.
The call for a move of the convention come after Tuesday's mass shooting at a Uvalde, Texas, elementary school that left 19 children and two fourth grade teachers dead.
''Our deepest sympathies are with the families and victims involved in this horrific and evil crime,'' the NRA wrote in a statement in advance of the meetings Friday through Sunday. ''On behalf of our members, we salute the courage of school officials, first responders, and others who offered their support and services.
''Although an investigation is underway and facts are still emerging, we recognize this was the act of a lone, deranged criminal.''
The NRA vowed to use its platform to advocate for securing the safety of children in schools nationwide.
''As we gather in Houston, we will reflect on these events, pray for the victims, recognize our patriotic members, and pledge to redouble our commitment to making our schools secure,'' the statement concluded.
Trump has vowed to deliver his planned speech Friday at the George R. Brown Convention Center in Houston during the leadership forum between 2 and 4 p.m. local time (3 to 5 p.m. Eastern time).
''America needs real solutions and real leadership in this moment, not politicians and partisanship,'' Trump wrote Wednesday from his Truth Social account. ''That's why I will keep my longtime commitment to speak in Texas at the NRA Convention and deliver an important address to America.
''In the meantime, we all continue to pray for the victims, their families, and for our entire nation — we are all in this together!''
Among the other leadership forum speakers are Republicans including Texas Gov. Greg Abbott, South Dakota Gov. Kristi Noem, Sen. Ted Cruz of Texas, Rep. Dan Crenshaw of Texas and North Carolina Lt. Gov. Mark Robinson, according to the NRA.
Teachers Union Leaders to Protest at NRA Convention Friday
Rep. Cheney Starts Reelection Bid; Primary With Trump-Backed Hopeful Looms

redpajama_set_name: RedPajamaCommonCrawl


text:
C.H.U.D. (1984)
Written by: Shepard Abbott (story), Parnell Hall (screenplay)
Directed by: Douglas Cheek
Starring: John Heard, Daniel Stern and Christopher Curry
Reviewed by: Brett Gallman
� Is that what you're going to tell the papers? That you're afraid of monsters? "
Toxic waste--remember when everyone was kind of obsessed with this stuff? I guess it was a natural extension of our nuclear paranoia, which got distilled down to a sludge that frequently oozed onto theater screens in the 1980s and gave birth to all manner of deformities and mutants that replaced the oversized, monstrous creatures that populated films earlier in the post-nuclear age. If you needed an excuse for a horror movie, toxic waste was the catch-all plot device to create monsters or zombies--shit, Lloyd Kaufman practically built Tromaville on it, and it was once used to actually dispose of Jason Voorhees. As such, the 80s were often very melty and gooey, and C.H.U.D. is one of the more memorable products of this cinematic bowel movement.
It�s also one of the more overcooked entries, as, for some reason, there�s about a half-dozen characters all investigating some strange occurrences in the subterranean depths of New York City. There�s a police captain (Christopher Curry) whose wife recently disappeared during an evening stroll with her dog (a prologue reveals to us that she was actually swiped into the sewers by some unseen force), a soup kitchen runner (Daniel Stern), an investigative journalist (J.C. Quinn), and a freelance photographer and his girlfriend (John Heard and Kim Griest), all of whom begin to discover that something monstrous is indeed terrorizing the homeless population that dwell in the city�s sewers.
As you can see, this is actually a pretty damn remarkable cast for a movie that�s about �Cannibalistic Humanoid Underground Dwellers,� which is the term coined for the disgusting creatures lurking below the streets. And it�s a good thing the movie is well-cast too since you spend a lot of time with the characters; if you�re expecting this to be 90 minutes of outrageous, gross thrills, go watch Street Trash instead. C.H.U.D. is actually more concerned with the mechanizations behind its creatures. While it�s not exactly a hard-boiled police procedural, most of the plot revolves around these various characters uncovering the conspiracy that created these things, and it�s typical government-phobia stuff that reveals the real monsters be sitting in our elected offices, where they prey upon the disenfranchised (in this case, they even go far as to dump toxic waste on them).
This is perhaps more heady than you�d expect, but it also tends to make C.H.U.D. chug a bit more than it should since everyone involved doesn�t even come together until the last act. Plus, there�s no real sense of mystery since we know what�s going on down in the sewers, and we actually see a chud in full view early on so you might be a little bummed that it takes a while to get back around to them. This unusually great cast makes it work okay, though--Curry does the �hard-wired, cop-on-the-edge� routine quite well, and Stern actually makes for a bit more of a straight-laced assistant. Stern is a guy who probably should have had a better career; he often manages to find an affable quality within his offbeat characters, and he does so here as the kindly �reverend� Shepherd, who is still kind of quirky and offbeat. Likewise, John Heard is one of those infinitely affable good guys, so it�s not like we�re watching total bores the whole time. Even the likes of Sam McMurray and John Goodman show up in bit parts as cops.
Still, most of the appeal lies in that final act, when C.H.U.D. finally gets some energy and turns into more of a horror movie, complete with some slick practical effects and gore gags. Even the score goes full-on Goblin style as the title characters finally begin to emerge and wreak a little bit of havoc (and have some havoc wreaked upon them). There�s even a little bit of intensity as the film cuts back and forth between the streets and the sewers, where Stern and Heard find themselves trapped, unable to escape both the chuds and the mandate that�s ordered all the sewers to be locked down so they can be fumigated. At its best, though, C.H.U.D. can hardly be considered riveting--it�s grungy and looks like it was doused in toxic waste, and it never fully plumbs the depths of its subversiveness to become anything more than a standard monster movie all gussied up in some slimy, 80s clothes.
As an extension of creature features, C.H.U.D. perhaps reveals that we were no longer worried about science so much as our government, as it�s boiled in post-Watergate distrust, with the most slimy characters actually being the cabal of seedy politicians. Unfortunately, it misses a few opportunities to be truly sick and smart, preferring instead to coast on these familiar beats that probably felt rotten even in �84. The 80s saw better films of this type, but C.H.U.D. still has quite a cult following, perhaps because its title actually became a part of our lexicon. Anchor Bay released it on DVD over a decade ago, so it�s out of print but still reasonable on the secondary market. If you go through the effort to track it down, you�ll be greeted with a presentation that still holds up alright--the transfer is anamorphic and appropriately grimy, and the mono track is good enough. Extras include a trailer, a still gallery, and a commentary with director Douglas Cheek, writer Shepard Abbott, Heard, Stern, and Curry. I couldn�t worry about seeking it out since it�s currently streaming on Netflix, though, since C.H.U.D. is kind of a chud itself at times. Rent it!


text:
\section{Introduction}

Let $M$ be a connected, noncompact Riemannian manifold of dimension $N\ge2$. We recall that $M$ is said to be stochastically complete if the lifetime of Brownian paths on $M$ is a.s.~infinite or, in analytical terms, if the free heat semigroup preserves the identity in the sense that
\[
\int_Mp(t,x,y)\,{d}y=1
\]
for all $(x,t)\in M\times(0,+\infty)$, where $p$ is the (minimal) heat kernel on $M$ and ${d}y$ denotes the Riemannian measure on $M$. Note that $M$ need not be geodesically complete.

A number of analytic and/or geometric conditions ensuring stochastic completeness, or incompletess, of $M$, are by now available, and it would be hopeless to give a full list of the related literature, for which we refer to the comprehensive discussions provided by Grigor'yan e.g.~in \cite{GrigHK, Grig, GrigBams}, but we mention at least that either curvature conditions, volume growth properties and function theoretic conditions, possibly originating from different contexts but applicable to the present problem as well, can be successfully used to verify stochastic completeness, see e.g.~\cite{A, D, G, GrigTo, H, I1, I, IM, K, M, M2, PRS, PRS2, S, T, Y}. To give a flavour of the curvature conditions that guarantee stochastic completeness, one can restrict the attention to Cartan-Hadamard manifolds and notice that qualitatively, in such situation, stochastic completeness holds when curvature does not diverge to minus infinity faster than quadratically, and it does not hold otherwise (see e.g.~\cite{M} for a precise statement).

Of particular interest for the present discussion is the well-known fact that stochastic completeness is related, in fact equivalent, to two other analytic properties. In fact, having fixed two positive constants $\lambda, T$ and a \it bounded \rm function $u_0$, the following three properties turn out to be equivalent (see \cite[Theorem 8.18]{GrigHK}):
\begin{itemize}

\item $M$ is stochastically complete;
\item The equation $\Delta v=\lambda v$ does not admit any nonnegative, nontrivial, bounded solutions;
\item The Cauchy problem
\begin{equation}\label{heat}
\begin{cases}
u_t = \Delta u & \text{in} \ M \times[0,T] \, , \\
u(\cdot,0)=u_0 & \text{in} \ M \, ,
\end{cases}
\end{equation}
has a unique nonnegative solution in $L^\infty(M\times(0,T))$ for $u_0\in L^\infty(M)$ with $u_0\ge 0$.
\end{itemize}

Uniqueness, for bounded data, of solutions to the heat equation on Riemannian manifolds is therefore another form of stochastic completeness, and is in particular known to hold under some clear geometric assumptions, see e.g.~\cite{IM, M} or Corollary \ref{compl} below.

In the recent years, the issue of uniqueness for \it nonlinear \rm evolution equations posed on Riemannian manifolds has been the object of some investigation. In particular, the \it porous medium equation \rm (see \cite{V} for a thorough discussion of this equation in the Euclidean case)
\begin{equation}\label{pme}
\begin{cases}
u_t=\Delta u^m & \text{in } M\times[0,T] \, , \\
u(\cdot,0) = u_0  & \text{in } M \, ,
\end{cases}
\end{equation}
where one has to assume that $m>1$ so that the diffusion coefficient $mu^{m-1}$ vanishes when $u$ does, has been investigated e.g.~in \cite{GMP, GMP2, GMV}. Among other issues, uniqueness results for suitable classes of data, possibly requiring appropriate curvature bounds, have been proved in these papers.

Much less is known so far on \eqref{pme} in the \it singular \rm case $m<1$, which goes under the name of \it fast diffusion equation \rm since the diffusion coefficient $mu^{m-1}$ diverges  when $u$ vanishes, thus forcing infinite speed of propagation and, in fact, even the possibility that solutions vanish in finite time. This has been indeed proved to happen, for suitable data, when $M$ admits a spectral gap (i.e.~$\min \sigma(-\Delta)>0$, where $\sigma(-\Delta)$ is the $L^2$ spectrum of $-\Delta$), which is the case for instance on Cartan-Hadamard manifolds when Sec$\,\le-k$ for a suitable $k>0$ (see \cite{Mc}), and in particular on the hyperbolic space (see \cite{BGV}). It has to be noted that, in the Euclidean case $M=\mathbb{R}^N$, \emph{existence and uniqueness} of solutions to \eqref{pme} hold instead even when $u_0$ is merely required to be \it locally integrable\rm, see \cite{HP}. Little seems to be known in this connection on noncompact manifolds other than $\mathbb{R}^N$. Uniqueness for \it strong \rm solutions is presently known only when curvature is allowed to be negative, but decaying to zero at least quadratically, see \cite{BS}.

We shall establish here that equivalence between stochastic completeness, a concept a priori related to a \it linear \rm evolution on $M$, and uniqueness of nonnegative, bounded solutions to \eqref{pme} is still valid in the fast diffusion range $m<1$. We devote Section \ref{case-fde} to a deeper discussion of this quite significant case. In fact we shall prove a more general result, namely that the above mentioned equivalence holds if \eqref{pme} is replaced by
\begin{equation}\label{pmephi}\begin{cases}
u_t=\Delta \phi(u) & \text{in } M\times(0,T) \, , \\
u(\cdot, 0)=u_0 & \text{in } M \, ,
\end{cases}
\end{equation}
where $\phi$ is an \it arbitrary \rm concave, increasing positive \rm function, regular outside the origin and satisfying $\phi(0)=0$.  This provides, thanks to \cite[Theorem 8.18]{GrigHK}), what is in our view a rather unexpected connection between properties of \it linear \rm and \it nonlinear \rm evolutions on manifolds. As a byproduct, one gets immediately that uniqueness of nonnegative bounded solutions to the nonlinear evolution \eqref{pmephi} for a given concave, increasing, positive and regular outside the origin structure function $\phi$ (including the linear case $\phi(x)=x$) is \it equivalent \rm to the same property for the same problem associated with \it any \rm such structure function $\phi$, a fact that we find remarkable. We stress that, to our knowledge, these are the first results available in such a general framework.

Problem \eqref{pmephi} is deeply related to the semilinear elliptic equation
\begin{equation}\label{elliptic-phi}
\Delta W=\phi^{-1}(W) \qquad \text{in } M \, .
\end{equation}
Indeed, the existence of a nonnegative, nontrivial, bounded solution to \eqref{elliptic-phi} is also equivalent to stochastic incompleteness, or analogously, stochastic completeness is equivalent to the fact that $ W \equiv 0 $ is the only nonnegative, nontrivial, bounded solution to \eqref{elliptic-phi}. This elliptic result is a byproduct of our methods of proof (which in particular do not exploit any kind of continuity of solutions), even though in the specific case of the Laplace-Beltrami operator was basically already known: see Remark \ref{rem-ell} below.

\subsection{Statement of the main result}

We denote by $ \mathfrak{C} $ the class of all functions $ \phi: \mathbb{R}^+ \to \mathbb{R}^+ $ satisfying the following assumptions:
\begin{gather*} \label{g1}
\phi \ \text{is \emph{concave}} \, , \\ \label{g2}
\phi \ \text{is \emph{strictly increasing} with} \ \phi(0)=0 \, , \\ \label{g3}
\phi \in C(\mathbb{R}^+) \cap C^1(\mathbb{R}^+ \setminus \{ 0 \}) \, .
\end{gather*}

Let $ M $ be a connected, noncompact Riemannian manifold. For any $ T \in (0,+\infty] $, any nonnegative $ u_0 \in L^\infty(M) $ and any $ \phi \in \mathfrak{C} $, we can consider the nonlinear parabolic problem
\begin{equation}\label{gen-zero-bdd}
\begin{cases}
u_t = \Delta \phi(u) & \text{in } M \times (0,T) \, , \\
u(\cdot,0)=u_0  & \text{in } M  \, ,
\end{cases}
\end{equation}
which always has an $ L^\infty(M\times(0,T)) $ nonnegative solution, by standard results. As mentioned above, we shall also be concerned with the problem of existence of nonnegative, nontrivial, bounded solutions to the semilinear elliptic equation \eqref{elliptic-phi}. For the precise meaning of solution to \eqref{elliptic-phi} and \eqref{gen-zero-bdd}, we refer to Section \ref{preliminaries}.

\medskip

We are now ready to state the main result of this paper.

\begin{thm}\label{main-teo}
Let $ M $ be a connected, noncompact Riemannian manifold. Let $ \phi \in \mathfrak{C} $, $ T \in (0,+\infty] $ and $ u_0 \in L^\infty(M)$, with $ u_0 \ge 0 $. Then the following assertions are equivalent:
\begin{enumerate}[(a)]
\item $ M $ is stochastically incomplete. \label{a}
\item The Cauchy problem \eqref{gen-zero-bdd} admits at least two nonnegative solutions in $ L^\infty(M\times(0,T)) $. \label{b}
\item The semilinear elliptic equation \eqref{elliptic-phi} admits a nonnegative, nontrivial, bounded solution. \label{c}
\end{enumerate}
\end{thm}

Theorem~\ref{main-teo} is of course equivalent to the following one, that nevertheless we state separately for the reader's convenience.
\begin{thm}\label{main-teob}
Let $ M $ be a connected, noncompact Riemannian manifold. Let $ \phi \in \mathfrak{C} $, $ T \in (0,+\infty] $ and $ u_0 \in L^\infty(M)$, with $ u_0 \ge 0 $. Then the following assertions are equivalent:
\begin{enumerate}[(a)]
\item[(a')] $ M $ is stochastically complete.
\item[(b')] The Cauchy problem \eqref{gen-zero-bdd} has a unique nonnegative solution in $ L^\infty(M\times(0,T)) $.
\item[(c')] The semilinear elliptic equation \eqref{elliptic-phi} does not admit any nonnegative, nontrivial, bounded solutions. \label{c'}
\end{enumerate}
\end{thm}

As an immediate Corollary of Theorem~\ref{main-teo},
we have the following result, which follows by simply noting that $\phi$, $T$ and $u_0$ are arbitrary in the above statement.

\begin{cor}\label{cor1} Let $ M $ be a connected, noncompact Riemannian manifold. The following statements are equivalent, and each of them is equivalent to stochastic incompleteness (resp.~completeness):
\begin{itemize}
\item For some bounded datum $u_0\ge0$, some function $ \phi \in \mathfrak{C} $ and some $T\in(0,+\infty]$ the Cauchy problem \eqref{gen-zero-bdd} admits at least two nonnegative solutions (resp.~a unique nonnegative solution) in $ L^\infty(M\times(0,T)) $;
\item For all bounded data $u_0\ge0$, all function $ \phi \in \mathfrak{C} $ and all $T\in(0,+\infty]$ the Cauchy problem \eqref{gen-zero-bdd} admits at least two nonnegative solutions (resp.~a unique nonnegative solution) in $ L^\infty(M\times(0,T)) $;
\item For some function $ \phi \in \mathfrak{C} $ the semilinear elliptic equation \eqref{elliptic-phi} admits (resp.~does not admit) a nonnegative, nontrivial, bounded solution;
\item For all functions $ \phi \in \mathfrak{C} $ the semilinear elliptic equation \eqref{elliptic-phi} admits (resp.~does not admit) a nonnegative, nontrivial, bounded solution.
\end{itemize}

\end{cor}

\medskip
\begin{oss}\label{rem-ell}
\rm Notice that $\phi(u)=u$ belongs to $ \mathfrak{C} $. Hence Corollary \ref{cor1} implies that uniqueness for solutions to the generalized fast diffusion equation \eqref{gen-zero-bdd} holds, for (nonnegative) bounded data and $\phi\in \mathfrak{C}$, if and only if uniqueness holds for solutions to the heat equation \eqref{heat} for bounded data. Similarly, existence of nonnegative, nontrivial, bounded solutions to the semilinear elliptic problem \eqref{elliptic-phi} with $\phi\in \mathfrak{C}$ holds if and only if the linear equation $\Delta v=\lambda v$ admits a nonnegative, nontrivial, bounded solution for some, hence all $\lambda>0$. Let us focus on the more general semilinear equation
\begin{equation}\label{elliptic-general}
\Delta W=f(W) \qquad \text{in } M \, .
\end{equation}
Actually, as a consequence of previous results, the elliptic equivalence theorem is even stronger: from \cite[Theorem 3.1 and condition (vii) at page 43]{PRS}, which in fact goes back to \cite{PRS2}, one deduces that $M$ is stochastically complete if and only if for every strictly increasing $ f \in C^0(\mathbb{R}^+)  $, with $ f(0)=0 $, equation \eqref{elliptic-general} does not admit any nonnegative, nontrivial, bounded solution. In other words, one can drop the convexity assumption on $ \phi^{-1} $. See also \cite{MV} for generalizations to suitable nonlinear diffusion operators. However, for the methods of \cite{PRS,PRS2} to work the continuity of $W$ is essential, which for the Laplace-Beltrami operator is for free due to elliptic regularity solutions are at least $C^{1,\alpha}$), but may not hold for more general operators: we refer to Remark \ref{rrr} below. Vice versa, if $ M $ is stochastically incomplete then a nonnegative, nontrivial solution to $ \Delta v = \lambda v $ with $ \| v \|_\infty=1 $ is always a subsolution to \eqref{elliptic-general}, provided $ \lambda $ is chosen to be the maximum of $ f^\prime $ in $ [0,1] $. This does not require convexity either, only $ f \in C^1(\mathbb{R}^+) $. Once a nonnegative, nontrivial, bounded subsolution exists then a bounded solution lying above the subsolution can be constructed.
\end{oss}

\smallskip
\begin{oss}\label{reg}\rm
The problem of optimal regularity of merely \emph{distributional} solutions to \eqref{gen-zero-bdd} is, to our knowledge, presently open, even in the Euclidean setting, although it is known under some additional assumptions (for instance local finiteness of the energy of solutions, see e.g.~\cite{DUV}).
\end{oss}

\smallskip

\begin{oss}\label{rrr}\rm
The methods used here are in principle applicable to more general second order differential operators, not only the Laplace-Beltrami one, being based essentially only on suitable comparison principles, see Appendix \ref{app}. We stress that for our arguments to work, we do not need any kind of continuity of solutions to \eqref{elliptic-phi} or \eqref{gen-zero-bdd}.
\end{oss}

\medskip

The paper is organized as follows. In Section \ref{case-fde} we explain part of our strategy by focusing on the very important model case $ \phi(u)=u^m $ with $ m \in (0,1) $. In Section \ref{preliminaries} we briefly recall the concept of distributional solution for the differential equations considered. In Section \ref{proofmain} we prove our main result in the form of Theorem \ref{main-teo}. Appendix \ref{app} contains concise proofs of the comparison principles used in our argument. Appendix \ref{cor} states, for the reader's convenience, a list of known geometric and analytic conditions that ensure that a manifold is stochastically complete or incomplete, that in particular give explicit criteria on $M$ ensuring uniqueness, or nonuniqueness, of solutions to \eqref{gen-zero-bdd}.

\section{The fast diffusion case}\label{case-fde}

As it has already been mentioned in the Introduction, Theorems \ref{main-teo}--\ref{main-teob} include as a particular case $ L^\infty $ uniqueness and nonuniqueness sharp results for the \emph{fast diffusion equation} \eqref{pme} on manifolds, which corresponds to $ \phi(u)=u^m $ for any $ m \in (0,1) $. Such equation has widely been studied on Euclidean space: we quote, without any claim to completeness, the references \cite{HP,PZ0,PZ,V,BBDGV,BDGV}. More recently, it has also been investigated on manifolds, mainly with negative curvature: see \cite{BGV,GM} and \cite[Section 4]{BS}.

More specifically, as concerns existence and uniqueness issues, in the celebrated paper \cite{HP} M.A.~Herrero and M.~Pierre set up a whole theory for data merely in $ L^1_{\mathrm{loc}} $. Indeed, \cite[Theorem 2.1]{HP} establishes that for any $ m \in (0,1) $ and $ u_0 \in L^1_{\mathrm{loc}}(\mathbb{R}^N) $ there exists a (distributional) solution $ u $ of \eqref{pme} (at least up to some finite $T>0$), which is moreover locally bounded for positive times if $ m $ is larger than the critical value $ (N-2)^+/N $ \cite[Theorem 2.2]{HP}. This is by itself a remarkable result, since it means that the behaviour at spatial infinity of $ u_0(x) $ does not count at all, in sharp contrast with the linear case, namely the heat equation (see also the Introduction to \cite{GMP} for a detailed discussion on these topics). Furthermore, uniqueness holds in such a wide class \cite[Theorem 2.3]{HP}, under the additional assumption that solutions are \emph{strong}, i.e.~that $u_t$ is an $ L^1_{\mathrm{loc}}(\mathbb{R}^N \times (0,T)) $ function.

Existence for data in $ L^1_{\mathrm{loc}} $ also holds on manifolds regardless of curvature or volume-growth assumptions, since the proof of \cite[Theorem 2.1]{HP} is purely local. On the other hand, an analogue of the Herrero-Pierre result was proved in \cite{BS} (see Theorem 4.9 there), provided the Ricci curvature is bounded from below by an inverse quadratic function of the distance from a reference point, i.e.~negative curvatures must vanish at least with a quadratic rate at infinity.

\smallskip

The results entailed by Theorem \ref{main-teo} show that the picture is totally different on a general manifold $M$: if the latter is stochastically incomplete then uniqueness to \eqref{pme} (for $ m \in (0,1) $) \emph{fails} even in the class of bounded solutions. In order to give a flavor of the methods of proof we shall exploit in Section \ref{proofmain}, let us construct an explicit solution to \eqref{pme} with $ u_0 \equiv 0 $ which is not identically zero, on any stochastically incomplete Riemannian manifold $M$. Indeed, thanks to \cite[Theorem 8.18]{GrigHK}, we know that there exists a nonnegative, nontrivial, bounded solution $v$ to
$$
\Delta v = v \quad \text{in } M \, , \qquad \| v \|_{L^\infty(M)}=1 \, .
$$
In particular,
$$
\Delta v \ge v^{\frac{1}{m}} \quad \text{in } M \, ,
$$
so that by a standard monotone procedure (see e.g.~the proof of $(\ref{b}) \Rightarrow (\ref{c})$ in Section \ref{proofmain}) one can find a nonnegative, nontrivial, bounded function $ W $ solving
$$
\Delta W = W^{\frac{1}{m}} \quad \text{in } M \, ,
$$
which is in fact larger than $v$. Hence a routine computation shows that the separable function
$$
u := \left[ (1-m) \,t \right]^{\frac{1}{1-m}} W^{\frac{1}{m}}
$$
is a solution to $ u_t=\Delta u^m $, bounded in $ M \times [0,T] $ for every $ T>0 $. Since $ u(\cdot,0) \equiv 0 $, this yields nonuniqueness for the Cauchy problem \eqref{pme} with $ u_0 \equiv 0 $. Then one could actually exploit the existence of such a solution to construct multiple solutions for \emph{any} initial datum: we refer to the techniques adopted in the proof of $(\ref{a}) \Rightarrow (\ref{b})$.

\section{On the concepts of solution}\label{preliminaries}

We collect here the basic definitions of (distributional or very weak) solutions to \eqref{gen-zero-bdd} and \eqref{elliptic-phi}, as well as sub/supersolutions. We shall also briefly discuss existence of solutions to \eqref{gen-zero-bdd} for bounded initial data, which is rather standard. At the end of the paper, for the reader's convenience, we provide a short Appendix where related comparison principles (on balls) are established, since the latter constitute a key tool of our analysis.

\begin{den}\label{ds}
Let $ \phi \in \mathfrak{C} $, $ T \in (0,+\infty] $ and $ u_0 \in L^\infty(M)$, with $ u_0 \ge 0 $. We say that a nonnegative function $ u \in L^\infty(M \times (0,T)) $ is a distributional solution to problem \eqref{gen-zero-bdd} if it satisfies
\begin{equation}\label{def-sol}
\int_0^T \int_M u \, \xi_t  \, dx dt + \int_0^T \int_M \phi(u) \, \Delta \xi \, dx dt + \int_M u_0(x) \, \xi(0,x) \, dx = 0
\end{equation}
for all $ \xi \in C^2_c( M \times [0,T) ) $. Similarly, we say that a nonnegative function $ u \in L^\infty(M \times (0,T) ) $ is a distributional supersolution [subsolution] to \eqref{gen-zero-bdd} if \eqref{def-sol} is satisfied with ``='' replaced by ``$\le$'' [``$ \ge $''], for all nonnegative $ \xi \in C^2_c( M \times [0,T) ) $.
\end{den}

\begin{pro}\label{minimal}
Let $ \phi \in \mathfrak{C} $, $ T = +\infty $ and $ u_0 \in L^\infty(M)$, with $ u_0 \ge 0 $. There exists a (minimal) solution to \eqref{gen-zero-bdd}, in the sense of Definition \ref{ds}, which in particular satisfies $ \| u(t) \|_\infty \le \| u_0 \|_\infty $ for a.e.\ $ t>0 $.
\end{pro}
\begin{proof}[Sketch of proof]
Let $ B_R $ be the ball of radius $ R>0 $ centered at any point $ o \in M $. One considers the approximate problems
\begin{equation*}\label{sol-approx-zero}
\begin{cases}
\left( u_R \right)_t = \Delta \phi\!\left( u_R \right)   & \text{in } B_R \times \mathbb{R}^+  \, , \\
u_R(\cdot,t) = 0 &  \text{on } \partial B_R \times  \mathbb{R}^+ \, , \\
u_R(\cdot,0) = u_0  & \text{in } B_R  \, ,
\end{cases}
\end{equation*}
which admit a unique (nonnegative) strong energy solution for each $ R>0 $, namely a bounded function $ u_R $ such that $ \nabla \phi(u_R) \in L^2(B_R \times (0,T)) $ and $ (u_R)_t \in L^1(B_R \times (0,T)) $. See \cite{V} and references therein for analogous results in the Euclidean setting, that can easily be extended to the present one. Such a solution is in fact the \emph{standard} one (see Appendix \ref{app}). In this class of solutions and sub/supersolutions the comparison principle holds, so that the sequence $ \{u_R\} $  is pointwise increasing with respect to $R$ and bounded above by $ \| u_0 \|_\infty $. Hence $ u:=\lim_{R\to\infty} u_R $ is indeed a solution to \eqref{gen-zero-bdd}, which is minimal again by comparison on balls. For all of the comparison results we have exploited we refer to Proposition \ref{CP} in the Appendix (when comparing $ u_{R} $ and $ u_{R+1} $ note that the fact that $ u_{R+1} $ is defined in $ B_{R+1} $ rather than the whole $M$ is irrelevant).
\end{proof}

As concerns the elliptic equation \eqref{elliptic-phi}, we give the following standard definition.

\begin{den}\label{vw}
Let $ \phi \in \mathfrak{C} $. We say that a nonnegative function $ W \in L^\infty(M) $ is a distributional solution to equation \eqref{elliptic-phi} if it satisfies
\begin{equation*}\label{def-sol-ell}
 \int_M W \, \Delta \eta  \, dx = \int_M \phi^{-1}(W) \, \eta  \, dx
\end{equation*}
for all $ \eta \in C^2_c( M ) $. Similarly, we say that a nonnegative function $ W \in L^\infty(M ) $ is a distributional supersolution [subsolution] to \eqref{elliptic-phi} if \eqref{vw} is satisfied with ``='' replaced by ``$\le$'' [``$ \ge $''], for all nonnegative $ \eta \in C^2_c( M  ) $.
\end{den}

\section{Proof of the main result}\label{proofmain}

\begin{proof}[Proof of $(\ref{a}) \Rightarrow (\ref{b})$]
First of all, it is convenient to rewrite the differential equation in \eqref{gen-zero-bdd} as follows:
\begin{equation}\label{eq:inv}
\psi^\prime(w) \, w_t = \Delta w \, .
\end{equation}
By virtue of the assumptions on $ \phi $, we know that $ \psi(w)=\phi^{-1}(w) $ is a \emph{convex} function having the same properties as $ \phi $, except that it is defined in the interval $ [0,a) $, with $ a:= \lim_{u \to +\infty} \phi(u) \in (0,+\infty] $. Thanks to \cite[Theorem 8.18]{GrigHK}, stochastic incompleteness ensures the existence of a nonnegative, nontrivial, bounded solution $ v $ to
\begin{equation}\label{v-SI}
\Delta v = \alpha v \qquad \text{in } M \, ,
\end{equation}
for every $ \alpha>0 $; we shall choose later a precise value of $ \alpha $. For simplicity, we can and shall assume that $ \| v \|_\infty = 1 $. The starting point is to look for a subsolution to \eqref{eq:inv} of the form
\begin{equation}\label{eq:ee}
w(x,t) = f(t) \, v(x) \, ,
\end{equation}
where $ f $ is a suitable positive, increasing function that we shall specify below. An immediate computation shows that $ w $ is actually a subsolution to \eqref{eq:inv} if and only if
\begin{equation*}\label{eq:inv-2}
\psi^\prime(f v) \, f_t \le \alpha f \, ,
\end{equation*}
which is trivially implied by
\begin{equation*}\label{eq:inv-3}
\psi^\prime(f) \, f_t \le \alpha f
\end{equation*}
given the monotonicity of $ \psi^\prime $ (recall that $ \psi $ is convex), the fact that $ f $ is increasing and the bound $ \| v \|_\infty = 1 $. So, let us pick $f$ as the solution to the Cauchy problem
\begin{equation}\label{choice-f}
\begin{cases}
f_t = \frac{\alpha f}{\psi^\prime(f)} \, , \\
f(0) = \varepsilon \in (0,a) \, ,
\end{cases}
\end{equation}
where $ \varepsilon$ for the moment is a free parameter.
Clearly, such a function is increasing and, the r.h.s.~of the differential equation being sublinear, it exists for all times $ t \in \mathbb{R}^+ $ (at this stage one could also use the variable $ g=\psi(f) $ and study the differential equation $ g_t=\phi(g) $). Let us set
$$
F(x):= \frac{1}{\alpha} \int_\varepsilon^x \frac{\psi^\prime(y)}{y} \, dy \qquad \forall x \in [\varepsilon,+\infty) \, .
$$
By integrating \eqref{choice-f}, we deduce that
\begin{equation*}\label{choice-f-1}
F(f(t)) = t \qquad \forall t \in \mathbb{R}^+ \, .
\end{equation*}
Because $F$ is locally regular in $ (0,a) $ and  $ \lim_{x \to a^-} F(x) = +\infty $, we infer that necessarily
\begin{equation}\label{choice-f-2}
\lim_{t\to+\infty} f(t) = a \qquad \Longrightarrow \qquad \lim_{t\to+\infty} \psi\! \left( f(t) \right) = + \infty  \, .
\end{equation}
Our next aim is to exhibit a subsolution to \eqref{gen-zero-bdd} which, at a certain time (smaller than $T$), exceeds the $ L^\infty $ norm of the initial datum. To this end, given $ w $ as in \eqref{eq:ee} with the above choices, let us consider the function
$$
\underline{w}:= \left( w-\varepsilon \right) \vee 0 \, .
$$
Clearly, $ \underline{w}(0) = \left( \varepsilon v -\varepsilon \right) \vee 0 = 0 $ and $ \underline{w} $ is still a subsolution to the differential equation \eqref{eq:inv}, since
$$
\Delta \underline{w} = \Delta \left( \left( w-\varepsilon \right) \vee 0 \right) \ge \chi_{w \ge \varepsilon} \, \Delta w \ge  \chi_{w \ge \varepsilon} \, w_t \, \psi^\prime(w) = \underline{w}_t \, \psi^\prime(w) \ge \underline{w}_t \, \psi^\prime(\underline{w}) \, ,
$$
where we have used Kato's inequality along with the fact that $ w_t \ge 0 $ and $ \psi^\prime $ is a nondecreasing function. Hence, going back to the original problem, we have constructed the following subsolution to \eqref{gen-zero-bdd}:
$$
\underline{u}(x,t) := \psi(\underline{w}(x,t)) = \psi\!\left( \left( f(t)\,v(x)-\varepsilon \right) \vee 0 \right) ;
$$
in particular,
\begin{equation}\label{eq-infty}
\left\| \underline{u}(t) \right\|_\infty = \psi\!\left( \left( f(t)-\varepsilon \right) \vee 0 \right) .
\end{equation}
Since $ \lim_{w \to a^-} \psi(w) = +\infty $, we can take $ \varepsilon $ so small that
\begin{equation*}\label{eq-infty-bis}
\psi\!\left( b - 2\varepsilon  \right) > \left\| u_0 \right\|_\infty ,
\end{equation*}
where $ b \in (0,a) $ is any number such that $ \psi\!\left( b \right) > \left\| u_0 \right\|_\infty $. By \eqref{choice-f-2} and \eqref{eq-infty}, we deduce that
\begin{equation}\label{eq-infty-ter}
\left\| \underline{u}(S) \right\|_\infty > \left\| u_0 \right\|_\infty
\end{equation}
provided $ S>0 $ is so large that $ f(S) \ge b-\varepsilon $. In order to make sure that $ S < T $, we can exploit a simple scaling argument: just note that $ f(t)=f_0(\alpha t) $, the function $f_0$ being the solution to \eqref{choice-f} corresponding to $ \alpha=1 $. Hence, if $ S_0 $ is the time at which $ f_0(S_0) = b-\varepsilon $, it is enough to take $ \alpha $ so large that $ S_0 / \alpha < T $.\\
Finally, we are able to construct a \emph{solution} to \eqref{gen-zero-bdd} which stays above $ \underline{u} $ in $ M \times (0,S) $, by means of the following approximate Cauchy-Dirichlet problems (let $ R>0 $):
\begin{equation}\label{sol-approx-dir-bb}
\begin{cases}
\left( u_R \right)_t = \Delta \phi\!\left( u_R \right)   & \text{in } B_R \times \mathbb{R}^+  \, , \\
u_R(\cdot,t) = \psi\!\left (f(t) \vee \phi(\| u_0 \|_\infty)\right) &  \text{on } \partial B_R \times  \mathbb{R}^+ \, , \\
u_R(\cdot,0) = u_0  & \text{in } B_R  \, .
\end{cases}
\end{equation}
It is apparent that $ \underline{u} $ is a subsolution to \eqref{sol-approx-dir-bb} for all $R>0$. As a consequence of the comparison principle on balls (we refer to Proposition \ref{CP}), along with the fact that the boundary condition is given by an increasing function of $t$, we can easily infer that
\begin{equation*}\label{std-comp-bb}
\underline{u} \le u_{R+1} \le u_{R} \le \psi\!\left(f(t) \vee \phi(\| u_0 \|_\infty)\right) \qquad \text{in } B_R \times \mathbb{R}^+
\end{equation*}
for all $ R>0 $. In particular, $ \{ u_R \} $ (e.g.~set to zero outside $B_R$) is a monotone-decreasing sequence of solutions to \eqref{sol-approx-dir-bb} which stays bounded in $ L^\infty(M \times (0,\tau)) $ for all $ \tau>0 $. Hence, by passing to the limit as $ R \to \infty $, we find that $ u_1 := \lim_{R \to \infty} u_R $ is a solution to \eqref{gen-zero-bdd} satisfying the additional estimate
\begin{equation}\label{std-comp-1-bb}
\underline{u} \le u_1 \quad \text{in } M \times \mathbb{R}^+ \qquad \Longrightarrow \qquad \left\| u_0 \right\|_\infty < \left\| u_1(S) \right\|_\infty ,
\end{equation}
where we have exploited \eqref{eq-infty-ter}. On the other hand, one can always construct the \emph{minimal} solution to \eqref{gen-zero-bdd}, namely the one obtained by the same approximation scheme as above, upon replacing the boundary condition in \eqref{sol-approx-dir-bb} with $ 0 $ on $ \partial B_R \times \mathbb{R}^+ $ (see Proposition \ref{minimal}). By proceeding in this way, we end up with another solution $ u_2 $ to \eqref{gen-zero-bdd} (also existing in $ M \times \mathbb{R}^+ $), which in particular satisfies
\begin{equation}\label{std-comp-2-bb}
u_2 \le \left\| u_0 \right\|_{\infty} \qquad \text{in } M \times \mathbb{R}^+ \, .
\end{equation}
Because \eqref{std-comp-1-bb} and \eqref{std-comp-2-bb} are clearly incompatible and $ S < T $, we necessarily deduce that $ u_1 \not \equiv u_2 $ in $ M \times (0,T) $. Actually we have to point out that, by construction, $ u_1 $ is in $ L^\infty(M \times (0,\tau)) $ for all $ \tau>0 $ but does not belong to $ L^\infty(M \times \mathbb{R}^+) $. In order to make $ u_1 $ globally bounded (which is required in the case $ T=+\infty $ only), but still different from $ u_2 $, it is enough to stop it at time $S$ and then restart it e.g.~with a minimal construction.
\end{proof}

\begin{proof}[Proof of $(\ref{b}) \Rightarrow (\ref{c})$]
Since the \emph{minimal} solution to \eqref{gen-zero-bdd} always exists, which we shall simply denote by $ u $, from assumption $(\ref{b})$ we deduce the existence of another solution $ u_\ast \in L^\infty(M \times (0,T)) $, which is necessarily larger than $ u $ (by minimality) but different from $ u $. By multiplying both $ u_\ast $ and $u$ by $ e^{-t} $ and subtracting, we obtain the following identity:
\begin{equation}\label{eq:ww1}
\left[ e^{-t} \left( u_\ast - u \right) \right]_t = e^{-t} \left[ \Delta \phi(u_\ast) - \Delta \phi(u) \right] - e^{-t} \left( u_\ast - u \right) .
\end{equation}
Now let us integrate \eqref{eq:ww1} between $ t=0 $ and $ t = T $: by exploiting the fact that $ u_\ast \ge u $ and $ u_\ast(\cdot,0) = u(\cdot,0) $, we end up with the inequality
\begin{equation}\label{eq:g1}
\Delta \left[ \int_0^{T} \phi(u_\ast(x,s)) \, \hat{e}(s) \, ds - \int_0^{T} \phi(u(x,s)) \, \hat{e}(s) \, ds \right] \ge \int_0^{T} \left[ u_\ast(x,s) - u(x,s) \right] \hat{e}(s) \, ds \, ,
\end{equation}
valid in the whole $M$, where we set $ \hat{e}(t) := e^{-t} / \left(1-e^{-T}\right) $ if $ T<+\infty $ and $ \hat{e}(t)=e^{-t} $ if $ T=+\infty $. Thanks to the concavity of $ \phi $, Jensen's inequality ensures that
$$
\int_0^{T} \phi\!\left( u_\ast(x,s) - u(x,s) \right) \hat{e}(s) \, ds \le \phi \! \left( \int_0^{T} \left[ u_\ast(x,s) - u(x,s) \right] \hat{e}(s) \, ds \right) ,
$$
so that by \eqref{eq:g1} and the monotonicity of $ \phi $ we infer that
\begin{equation}\label{eq:g2}
\Delta \left[ \int_0^{T} \phi(u_\ast(x,s)) \, \hat{e}(s) \, ds - \int_0^{T} \phi(u(x,s)) \, \hat{e}(s)  \, ds \right] \! \ge \phi^{-1} \! \left( \int_0^{T} \phi\!\left( u_\ast(x,s) - u(x,s) \right) \hat{e}(s) \, ds  \right) .
\end{equation}
Still the concavity of $ \phi $ (in the form of decrease of difference quotients), along with the fact that $ \phi(0)=0 $, yields
\begin{equation}\label{eq:g2-bis}
\phi\!\left( u_\ast(x,s) - u(x,s) \right) \ge \phi(u_\ast(x,s)) - \phi(u(x,s)) \, .
\end{equation}
Hence, from \eqref{eq:g2}, \eqref{eq:g2-bis} and again the monotonicity of $ \phi $, we have:
\begin{equation*}\label{eq:g3}
\begin{aligned}
& \Delta \left[ \int_0^{T} \phi(u_\ast(x,s)) \, \hat{e}(s) \, ds - \int_0^{T} \phi(u(x,s)) \, \hat{e}(s)  \, ds \right] \\
\ge & \, \phi^{-1} \! \left( \int_0^{T} \phi(u_\ast(x,s)) \, \hat{e}(s) \, ds - \int_0^{T} \phi(u(x,s)) \, \hat{e}(s) \, ds \right) .
\end{aligned}
\end{equation*}
Summing up, we have proved that the function
$$
\underline{W}(x) := \int_0^{T} \phi(u_\ast(x,s)) \, \hat{e}(s) \, ds - \int_0^{T} \phi(u(x,s)) \, \hat{e}(s) \, ds \, ,
$$
which is nonnegative, nontrivial and bounded by construction, satisfies
$$
\Delta \underline{W} \ge \phi^{-1}\!\left( \underline{W} \right)  \qquad \text{in } M \, ,
$$
namely it is a \emph{subsolution} to \eqref{elliptic-phi}. On the other hand, it is plain that the constant function $ \overline{W} \equiv \| \underline{W} \|_\infty $ is a \emph{supersolution} to the same equation. We can therefore construct the claimed solution $ W $ by the following standard approximation scheme. For all $ R > 0 $, we solve the Dirichlet problems
\begin{equation}\label{pb-balls-kk}
\begin{cases}
\Delta W_R = \phi^{-1} \! \left( W_R \right)  & \text{in } B_R \, , \\
W_R = \overline{W}  &  \text{on } \partial B_R \, . \\
\end{cases}
\end{equation}
Clearly, for each $ R>0 $, $ \underline{W} $ and $ \overline{W} $ are two ordered sub- and supersolutions, respectively, to \eqref{pb-balls-kk}. By comparison in $B_R$ (see Proposition \ref{CE}), we deduce that
\begin{equation*}\label{eq:est-R-kk}
\underline{W} \le W_{R+1} \le W_R \le \overline{W} \qquad \text{in } B_R \, .
\end{equation*}
Hence, upon letting $ R \to \infty $, we find that $ W:=\lim_{R\to\infty} W_R $ is indeed a nonnegative solution to \eqref{elliptic-phi}, which is nontrivial and bounded since $ W \ge \underline{W} $ and $ W \le \overline{W} $ in $M$.
\end{proof}

\begin{proof}[Proof of $(\ref{c}) \Rightarrow (\ref{a})$]
As a preliminary step, we observe that \eqref{elliptic-phi} yields
\begin{equation}\label{eq-psi-1}
\Delta W \ge \frac{W}{2} \, \psi^\prime\!\left( \frac W 2 \right)  \qquad \text{in } M \, ,
\end{equation}
where $ \psi(w):=\phi^{-1}(w) $ is the same convex function introduced in the proof of implication $(\ref{a}) \Rightarrow (\ref{b})$. Note that in \eqref{eq-psi-1} we have just used the fact that $ \psi(w) $ lies above the tangent line at $ \frac{w}{2} $, along with the positivity of $ \psi(\frac{w}{2}) $. So from \eqref{eq-psi-1} we can deduce the existence of a nonnegative, nontrivial, bounded function satisfying
\begin{equation*}\label{eq-psi-bis}
\Delta W \ge \frac{W}{2} \, \psi^\prime\!\left( W \right)  \qquad \text{in } M \, ,
\end{equation*}
which for simplicity we shall keep denoting by $W$.
We look for a (local-in-time) subsolution to \eqref{eq:inv} of the form
$$
\underline{w}(x,t) = f(t) \, W(x)
$$
for a suitable positive, increasing function $f$. An elementary computation shows that, to this aim, we need to require that
\begin{equation*}\label{eq-subsol-gen}
\psi^\prime\!\left( f \, W \right) f_t \, W \le f \, \frac{W}{2} \, \psi^\prime(W) \, ,
\end{equation*}
which is trivially implied by
$$
f_t \le \frac{f}{2}
$$
as long as $ f \le 1 $, recalling the fact that $ \psi^\prime $ is nondecreasing. As a consequence, we deduce that (for instance) the function
$$
\underline{w}(x,t) := \frac{e^{\frac{t}{2}}}{2} \, W(x)
$$
is indeed a subsolution to \eqref{eq:inv} at least up to $ t = \log 4 $, namely
$$
\underline{u}(x,t) = \psi\!\left( \frac{e^{\frac{t}{2}}}{2} \, W(x) \right) \qquad \forall t \in [0,\log 4]
$$
is a subsolution to $ u_t = \Delta \phi(u) $, going back to the original variables. Note that
\begin{equation}\label{eq:evo-linf}
\left\| \underline{u}(t) \right\|_\infty =  \psi\!\left( \frac{e^{\frac{t}{2}}}{2} \, \| W \|_\infty \right) .
\end{equation}
We are therefore in position to construct a bounded solution to the Cauchy problem
\begin{equation}\label{k1}
\begin{cases}
u_t = \Delta \phi(u) & \text{in } M \times \mathbb{R}^+ \, , \\
u(\cdot,0)= \left\| \underline{u}(0) \right\|_\infty = \psi\!\left( \frac{1}{2} \, \| W \|_\infty \right) =: c > 0 & \text{in } M  \, ,
\end{cases}
\end{equation}
which is \emph{larger} than the positive constant $ c $. Indeed, it is enough to solve the approximate problems (for all $R>0$)
\begin{equation}\label{k2}
\begin{cases}
\left( u_R \right)_t = \Delta \phi\!\left( u_R \right)   & \text{in } B_R \times \mathbb{R}^+  \, , \\
u_R(\cdot,t) =  \psi\!\left( \frac{e^{\frac{t}{2}}}{2} \, \| W \|_\infty \right) \wedge \psi\!\left( \| W \|_\infty \right) \ge c &  \text{on } \partial B_R \times  \mathbb{R}^+ \, , \\
u_R(\cdot,0) = c >0 & \text{in } B_R  \, ,
\end{cases}
\end{equation}
and let $ R \to \infty $. Since $ \underline{u} $ is a subsolution to \eqref{k2}, for each $ R>0 $, up to $ t=\log 4 $, by arguing similarly to the proof of implication $ (\ref{a}) \Rightarrow (\ref{b}) $ we deduce that $ u_\ast := \lim_{R \to \infty} u_R $ (monotone decreasing limit w.r.t.~$R$) is a solution to \eqref{k1} satisfying
$$
c \le u_\ast \le \psi\!\left( \| W \|_\infty \right) \quad \text{in } M \times \mathbb{R}^+ \qquad \text{and} \qquad u_\ast \ge \underline{u} \quad \text{in } M \times [0,\log 4] \, ;
$$
in particular, recalling \eqref{eq:evo-linf}, there holds
\begin{equation*}\label{kkk}
\left\| u_\ast(t) \right\|_\infty \ge \psi\!\left( \frac{e^{\frac{t}{2}}}{2} \, \| W \|_\infty \right) \qquad \forall t \in [0,\log 4] \, ,
\end{equation*}
so that $ u_\ast $ is a bounded solution to \eqref{k1}, everywhere not smaller than $c$ since such a constant is trivially a subsolution to \eqref{k2} for all $R>0$ (recall again Proposition \ref{CP}), but not identically $c$. By performing the same computations as in the first part of the proof of implication $(\ref{b}) \Rightarrow (\ref{c})$, using $ u \equiv c $ and $ T=+\infty $ there, we obtain the following identity:
\begin{equation}\label{eq:g1-ast}
\Delta \left[ \int_0^{+\infty} \phi(u_\ast(x,s)) \, e^{-s}  \, ds \right] = \int_0^{+\infty} \left[ u_\ast(x,s) - c \right] e^{-s}  \, ds \, .
\end{equation}
Now note that, as $ \phi $ is concave,
\begin{equation}\label{eq:g1-conc}
\phi(u_\ast) \le \phi(c) + \phi^\prime(c) \left( u_\ast - c \right) ,
\end{equation}
where $ \phi^\prime(c)>0 $ because $ \phi $ is strictly monotone. Hence, by combining \eqref{eq:g1-ast} and \eqref{eq:g1-conc}, we end up with the inequality
\begin{equation*}\label{eq:g2-ast}
\Delta \left[ \int_0^{+\infty} \phi(u_\ast(x,s)) \, e^{-s}  \, ds \right] \ge \frac{1}{\phi^\prime(c)} \int_0^{+\infty} \left[ \phi\!\left( u_\ast(x,s) \right) - \phi(c) \right] e^{-s}  \, ds \, ,
\end{equation*}
namely the function
$$
V(x) := \int_0^{+\infty} \left[ \phi\!\left( u_\ast(x,s) \right) - \phi(c) \right] e^{-s}  \, ds
$$
is nonnegative, nontrivial, bounded and satisfies
\begin{equation*}\label{eq:fin}
\Delta V \ge \frac{1}{\phi^\prime(c)} \, V  \qquad \text{in } M \, .
\end{equation*}
Starting from $V$, we can then construct a nonnegative, nontrivial, bounded solution to
$$ \Delta v = \frac{1}{\phi^\prime(c)} \, v \qquad \text{in } M $$
just by performing again the same approximation scheme as in the end of the proof of implication $(\ref{b}) \Rightarrow (\ref{c})$. This ensures that the manifold $M$ is stochastically incomplete, due again to \cite[Theorem 8.18]{GrigHK}.
\end{proof}

\begin{oss} \rm In the proof of $(\ref{a}) \Rightarrow (\ref{b})$ we exploited the assumption of stochastic incompleteness only in terms of existence of nontrivial solutions to \eqref{v-SI} for \it any $\alpha>0$. \rm On the other hand, in the proof of $(\ref{c}) \Rightarrow (\ref{a})$ we established stochastic incompleteness in terms of existence of nontrivial solutions to \eqref{v-SI} for a \emph{specific} $\alpha_0>0$. Nonetheless, it is easy to show directly that this yields existence for \it any $\alpha>0$. \rm Indeed, let $\widetilde v$ be a nonnegative, nontrivial, bounded solution to $\Delta \widetilde v=\alpha_0\widetilde v$. Clearly $\widetilde v$ is a subsolution to $\Delta v=\alpha v$ for every $\alpha\in(0,\alpha_0)$. Moreover, for all $\beta>1$ Kato's inequality ensures that $ v_\beta := \widetilde v^\beta$ is a subsolution to $\Delta v=\beta\alpha_0 v$. Hence our method of proof shows that, under assumption $(\ref{c})$, there is a nonnegative, nontrivial, bounded subsolution to $\Delta v=\alpha v$ for any $\alpha>0$. By exploiting $ v_\beta $ and proceeding exactly as in the end of the proof of  $(\ref{b}) \Rightarrow (\ref{c})$, one can then show the existence of a nonnegative, nontrivial, bounded \emph{solution} to $\Delta v=\alpha v$ for any $\alpha>0$. Going back to our proof of
$(\ref{a}) \Rightarrow (\ref{b})$, in this way we are able to provide a \emph{direct} proof of the implication $(\ref{c}) \Rightarrow (\ref{b})$, thus establishing the equivalence between $ (\ref{b}) $ and $ (\ref{c}) $ independently of $ (\ref{a}) $. The equivalence between $(\ref{c})$ and $(\ref{a})$ can then be deduced as explained in Remark \ref{rem-ell}.
\end{oss}

redpajama_set_name: RedPajamaArXiv


Allowing users to double the length of their posts is a major shift for Twitter as it alters the site's defining characteristic. The company had been facing challenges for some years. While remaining one of the leading social media platforms, growth had been stagnant for some time. Even more serious was the lack of advertising revenue, especially compared to Facebook and other faster-growing sites. Making a change was an attempt to encourage greater engagement. Has it been successful?
Early numbers suggest that since the change the new policy has been successful. According to Twitter CEO Jack Dorsey, engagement has indeed increased over the last few months. Another encouraging number that Dorsey announced is that Twitter made a profit for the first time last quarter. Strangely, however, the average length of tweets has not gone up. The average tweet is actually less than 50 characters, which is less than half of the old maximum. What does all this mean for Twitter and its place on the social media landscape?
Twitter is so frequently in the news that it's easy to lose sight of its many challenges. During the election and beyond, Donald's Trump's often controversial tweets were constantly mentioned. Beyond this, we read about the latest tweets from athlete's celebrities and politicians every day. It's more likely that you'll hear about someone's latest tweet than what they posted on Facebook or Instagram. There's little doubt that Twitter is an active part of social discourse today. At the same time, Twitter has been struggling when it comes to hard numbers.
Last summer, a few months before the company increased the character limit, Twitter admitted that it was losing followers in the United States. In addition to failing to attract new users worldwide, the number of American users fell by about two million. At the same time, the company also lost advertising revenue. In the social media world, Twitter, while maintaining its influence, has either remained in place or fell behind in terms of crucial numbers. There are several reasons for this. Many younger social media users, while maintaining their Facebook accounts, flocked to newer sites such as Instagram and Snapchat. Online advertising revenue, meanwhile, is dominated by the two giants, Google and Facebook. So, while Twitter has held on, it has struggled with growth and keeping pace with its competitors.
Twitter's Future: Encouraging Longer Form Content?
As noted, Twitter has achieved better engagement after increasing the character limit. According to Jack Dorsey, people are retweeting more, following each other with greater frequency and abandoning tweets less often. These are all positive signs that people are using the platform in a more social way. It's also important to remember that the new tweet length is still a recent innovation. Twitter began in 2006, so for 11 years people associated it with 140 characters. It will take some time for everyone to get used to the new feature. It's likely that many casual users don't even realize they can type longer messages.
While there's no evidence that people, on average, are creating longer tweets, this format does open up the site for people who want to use Twitter more like a blog. It's probably no accident that shortly after increasing the character limit, Twitter also introduced a threading feature that makes it easier for people to combine tweets. Prior to this, people could always create "tweetstorms," which were simply a series of tweets. However, this required users to reply to themselves and if other tweets get in the way, continuity was lost. Now, however, users can create a connected series of tweets by pushing a plus button and then posting them all at once.
The threading feature combined with the 280 characters is making it practical for people to post longer content and get into more depth about topics. As more users get comfortable with these features, it could signal the beginning of a whole new Twitter. Of course, the company has other problems to deal with as well, such as increasing ad revenue and dealing with trolls and online bullies. While many users will always use Twitter for the familiar brief tweets, along with accompanying photos and links, we may see the formation of a new blogging platform. Only time will tell if users will embrace these possibilities.


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NEW YORK, Sept. 20, 2018 (GLOBE NEWSWIRE) — In a further move towards strengthening its lead in the fragmented and expanding business aviation market, Vista Global, the aviation group founded by Thomas Flohr, announces the acquisition of XOJET's business aviation fleet and commercial operations. XOJET's regulated air carrier will continue to be independently operated in a joint venture under U.S. control and leadership, in accordance with U.S. Department of Transportation criteria.
XOJET is the leading on demand private aviation platform in North America and its innovative model is unrivaled in the industry. XOJET's four distinct membership programs serve the whole spectrum of business aviation needs on its owned fleet, whilst also providing its customers with additional flexible off-fleet alternatives. All of XOJET's products are asset lite, a business model which is fully aligned with prevailing global macro trends and preferred by both corporates and UHNWIs.
True to its innovative culture, XOJET uses machine learning technology to drive its dynamic pricing, thereby offering extraordinary value to its members and on demand clients. With a fleet of 43 branded super-mid size business jets, XOJET has performed strongly in recent years. In the first half of 2018, the company saw its total revenue rise by 12%, while off-fleet business grew 22% in revenue.
The acquisition is a major strategic step forward for Vista Global, allowing the group to build on its existing global offering and to reinforce its already established position in North America, the largest business aviation market in the world.
XOJET's fleet and commercial offerings (other than the U.S. controlled air carrier) will join Vista Global, the world's most comprehensive business aviation group. Vista Global is the first group that has built the scale to offer global private flying solutions for all customers, setting a new benchmark in the industry. Following today's announcement and upon closing of this strategic acquisition, Vista Global will have over 1,500 employees in 20 offices worldwide.
The most cutting edge technology to support aviation companies' customer services.
Completion of the transaction is subject to customary closing conditions, including the expiration or termination of the applicable waiting period under the Hart-Scott-Rodino Antitrust Improvements Act. There is no financing condition to the transaction. The transaction is currently expected to close in the fourth quarter of 2018. XOJET's regulated air carrier will be operated in a joint venture under U.S. control and leadership, in accordance with U.S. Department of Transportation criteria.
Jefferies LLC is serving as exclusive financial advisor and Latham & Watkins LLP as principal legal advisor to Vista Global. Perella Weinberg Partners is acting as exclusive financial advisor to Mubadala Investment Company ("Mubadala") and White & Case LLP is serving as principal legal advisor to Mubadala. Acquisition financing with non-recourse to the other three Vista Global businesses has been provided under the lead of J.P. Morgan, with Jefferies LLC.
Vista Global Holding is the world leader in business flight solutions. A global group headquartered at the DIFC in Dubai, Vista Global integrates a unique portfolio of companies offering asset lite solutions to cover all key aspects of business aviation: guaranteed and on demand global flight coverage; aircraft leasing and finance; and cutting edge aviation technology.
Founded by Thomas Flohr, the group's mission is to further industrialize and consolidate the fragmented business aviation industry, and to lead the change to provide customers with the most advanced flying solutions and the very best value, anywhere, anytime around the globe. The group owns VistaJet, Vista Lease, TechX and at the closing of the transaction described herein, XOJET's fleet and commercial operations. Vista Global's knowledge and understanding of all facets of the industry deliver the best end-to-end offering, service and technology to any business aviation customer or platform.


Discover the rich heritage and culture of India and have a date with the royal Bengal tiger in this 8 days itinerary. Here on this tour you will visit Delhi & New Delhi- consists of seven ancient cities. Agra- the Taj city, Ranthambore National park – world famous for tiger sightings located at 480 Km from Delhi and lastly you will visit Jaipur – worlds first planned city on our Shilpshartra by a Hindu scholar Vidyadhar Bhattacharya.
Today, after breakfast, you will be driven to Agra (225Km/ 4hrs). On arrival, head on for full day sightseeing of Agra, include visit to Agra Fort & Taj Mahal.
the outer wall. There are number of exquisite building like Moti Masjid-a white marvel mosque akin to a perfect pearl, Diwan-e-Am, Diwan-e-Khaas, Jehangir's Palace, Khaas Mahal, Shish Mahal and Musamman or Samman Burj-where Shahjahan was held captive in 1666 A.D.
Taj Mahal, completed in 1653 A.D. built by the Mughal Emperor Shah Jahan as the final resting place for his Queen, Mumtaz. It is India's most fascinating and beautiful monument, perfectly symmetrical monuments took 22 years (1630-1652) and 20,000 workers, Masons and Jewelers to build. Set amidst landscaped gardens Taj Mahal on the bank of the Yamuna River. It can be observed like a Mirage from the Agra Fort from where Emperor Shah Jahan stared at it, for the last eight years his life as a prisoner of his son Aurangzeb.
Sikandra- the mausoleum of emperor Akbar represent his philosophy and secular outlook, combining the best of Hindu and Muslim architectures in a superlative region. Completed in 1613 A.D. by his son Jahangir, it is one of the well preserved monuments. Itmad-ud-Daulah is the tomb of Mirza Ghyas Beg, a Persian who had obtained service in Akbar Court. The Marvel tomb was built by Emperor Jehangir's Queen, Noorjahan, for her father Mirza Ghyas Beg during 1622-1628 A.D.
Today, after breakfast check out from hotel and you will be driven for Ranthambhore National park (295Km/5 ½ hrs approx.). Ranthambore national Park one of the world famous and most renowned national park in Northern India. It is located in the southeastern Rajasthan, which is about 180 km from Jaipur. It covers an area of 1,334 sq km along with its nearby sanctuaries like - the Mansingh Sanctuary and the Kaila Devi Sanctuary. The park is majorly famous for its tigers and is one of the best locations in India to see the majestic predators in its natural habitat. The tigers can be easily spotted even during the day time busy at their ordinary quest- hunting and taking proper care of their young ones.
On arrival at Ranthambhore check in to hotel and after lunch get ready for afternoon safari at 1430 hrs approx.
You day starts early morning wake up call and tea before you head on early morning Game drive. Our local naturalists are extremely knowledgeable about park and the tiger movements since they grew up there and spent many years working as a naturalist. Having a couple days ensures tiger sightings and if you are lucky may catch a glimpse of them playing, hunting, sleeping and drinking. Return late morning to the lodge where the hot breakfast awaits you. Relax in your room or by the pool side a perfect treat for the hot day. Later after lunch, around 3 pm head on for the afternoon game drive. A treat for every wildlife and nature lover, watching tigers roaming around, verdant greenery, a gamut of other species of chirpy birds and animals is priceless and worth enough to be explored at least once in a life.
After breakfast, you will be driven to Jaipur (180 Km / 4Hrs), upon arrival, check in to your hotel.
Jaipur the pink city and its pink color associated with hospitality in Rajasthan, spreading out pink carpet for its visitors. Planned by young architect, Vidyadhar Bhattacharya, Jaipur was built by Maharaja Sawai Jai Singh II in 1727 AD. Laid in a grid system, with straight avenues, roads, streets, lanes cris cross the city with rows of shops on either side of main bazaars arranged in nine rectangular city sectors (Chowkris), based on the principles of "Shilpshastra", an epochal treatise on Hindu architecture.
On arrival at Jaipur, check in at your hotel and proceed to your half day sightseeing tour of Jaipur city tour visit the City Palace, Jantar Mantar and Hawa Mahal. The city palace houses museum with superb collection of Rajasthani costumes and armory of Mughals and Rajputs including swords of different shapes and sizes with chased handles, some of them inlaid with enamel and embellished with Jewels and encased in magnificent Scabbards. The art gallery has excellent collection of miniature paintings, carpets and rare artifacts. Evening is at leisure.
Today after an early breakfast, check out from your hotel and head on for an excursion to Amber fort. The former capital has a beautiful complex of palaces, halls, gardens and temples built by Raja Man Singh, Mirza Raja Jai Singh and Sawai Jai Singh over a period of about two centuries. It emerges from the placid waters of Mootha Lake and is approachable only through steep path. Enjoy short Elephant ride to reach the fort located on the hill.
After sightseeing, enjoy a farewell lunch at city restaurant and depart for New Delhi (285 km/ 5hrs approx.) On arrival at New Delhi, get transfer to International airport for your onward destination with exciting and sweet memories of Culture & wildlife trip to India.
Accommodation in Double Room / Cottages at all places.
Accommodation in Delhi, Agra & Jaipur on breakfast basis.
Accommodation in Ranthambore on fullboard basis.
Game Drives in National Parks in an open 4X4 Safari vehicle on sharing basis.
Entrance fee to Monuments and National Parks.
Services of a professional English-speaking. Guide/naturalist on safari and sightseeings.


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I find out a not so happy behaviour on the FSSO Controller Agent that makes some troubleshooting harder.
For example: a user logs in at time 10:14 and is working on his workstation. When I check the FSSO Logon Users at 10:27 I can see that the user is logged in with status OK. If I check the status one minute later the user is not in the list. But the user is still on the FortiGate. A few seconds later the user is in the list again with the logon time 10:14 as at the begining. After some time the user is missing until the next logon.
Is this a correct behaviour?
I checked it in the lab with 3 users with the same results.
Also when the user disappears from the primary FSSO Collector Agent it is still in the list on the secondary Collector Agent. Some time later the same user is missing on the second Collecotr Agent also.
Why the worksation check failed? DNS registration, port 445 and Remote Registry is enabled - otherwise the status after a short time would be Not Verified.
Another question: Would the user log entry be removed when the user leaves the workstation and locks his screen?
User missing from the list.
- every single collector makes his own idea about logged on users and due to processing delays and timers it might happen that list and total amounts of logged on users slightly differs between the collectors even when they handle the very same DCs/Domain.
- why? could be misconfigured WMI (or even missing WMI support but it should be supported since Win2000), also it could be that user under which Collector runs has not enough access rights to contact WMI.
- it might also appear that 192.168.221.100 is not a Windows machine, but some MacOS with domain conenctor, which generates 4624 logon event ID in DC and FSSO will process it (WinSec/WinSecWMI polling), and MacOS most probably wont respond on WMI.
- check WMI and OS versions or disable WMI on Collector ..
The WMI sesttings helped me, now it is working as expected.
I'm having the same difficulties using the CA (without dc agent).
The service account is member of the event log readers group.
The agent status is showing multiple logon events with correct timestamps.
But, no user logons are listed on the agent.
CA 5.0.0271 is installed on a non DC and polling the DC using regular event log polling method.


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text:
Boston Red Sox vs. Los Angeles Angels Preview and Pick – Bucholtz vs. Lackey
Boston Red Sox (57-40) +135, 7 1/2 at Los Angeles Angels (57-38), 10 pm Eastern Friday
by Zman of Predictem.com
First place in the East meets first place in the West as the Boston Red Sox visit the Los Angeles Angels for the opener of a three-game weekend series Friday night in Anaheim.
Baseball bookmakers are listing Los Angeles and starting pitcher John Lackey (6-2, 2.46) as -145 home favorites for Friday's game, with a total of 7 1/2, while Boston and Clay Buchholz (2-4, 5.70) are getting +135 as road underdogs.
As the "second half" of this MLB season begins, the Angels own the best record in the American League, and lead the second-place Oakland A's by six games in the AL's Western Division.
Meanwhile, Boston won five of its last six games going into the All-Star break, and leads the second-place Tampa Bay Rays by a half-game in the AL East.
In the first meeting between these two teams this season, LA took two of three games in Boston back in April. And two of those games went over their totals, as the games averaged 11.7 total runs.
Last year, the Red Sox took six of 10 games from the Angels during the regular season, with the totals going 6-3-1, as the games averaged 10.6 total runs. Then the Sox swept LA out of the first round of the playoffs in three games, by scores of 4-0, 6-3, and 9-1.
Boston is only 21-29 on the road this year, 8-14 as underdogs, and 48-49 vs. the run line.
LA is 26-20 at home this season, 37-26 as favorites, and 49-46 vs. the run line.
Buchholz is making his second ML start since coming back from the DL. Last Friday vs. Baltimore, Buchholz gave up four ER and 10 baserunners (hits + walks) in five IP. Boston is 2-7 in Buchholz' starts this year, the totals 6-2.
Buchholz has started one game in his career vs. the Angels, allowing three ER and 11 baserunners in six innings of an LA win last August.
Over his last three starts, including a brutal outing last Thursday in which he got knocked around for 15 hits in 5 2/3 innings vs. Texas, Lackey has allowed 11 ER and 30 BR in 20 1/3 IP. Los Angeles is 8-3 in Lackey's starts this year, the totals 2-8.
Lackey started thrice against the Red Sox last year, including once in the playoffs, and gave up 13 ER on 32 BR in 15 2/3 IP. The Angels lost all three of those games, and two went over their totals.
Boston ranks 2nd in the majors so far this year in team batting at .282, 2nd in team OBP at .356, 2nd in team slugging at .451, and is averaging 5.1 runs per game.
Los Angeles ranks 19th in batting at .257, 26th in OBP at .318, 22nd in slugging at .388, and is averaging 4.3 RPG.
Over its last five games, the Boston bullpen has given up four ER and 23 BR in 14 2/3 IP.
Over its last five games, the LA pen has allowed 11 ER on 27 BR in 19 IP.
On the personnel front, Sox slugger David Ortiz is rehabbing a wrist injury in the minors.
The totals are 43-46 in Sox games this year, 35-55 in Angels games, and 17-26 in games played at Angel Stadium, which are averaging 8.85 total runs per this season.
Zman's Pick: We like the Angels at -145 tonight.


The typical American diet leaves a great deal lacking in most categories. Americans settle for fast food rather than real food at home, convenience over healthy options. While fast paced lives rarely leave time for a well-balanced meal, there are many options available for adding supplements to your diet in forms of vitamins. Of course, before adding any supplements to your diet, speak with your primary care physician to ensure that you consume the correct amount.
Calcium – The teeth and jaw bone are both primarily made out of calcium. Therefore, a calcium deficiency can lead to jaw bone deterioration and tooth loss. Increasing your consumption of dairy products (if no dietary limitations prohibit this) and adding a calcium supplement to your daily routine can help protect your entire mouth.
Vitamin D – If your body has a Vitamin D deficiency, then your body is inhibited from absorbing the necessary amount of calcium. Because calcium is essential to your teeth and jaw health, a deficiency in Vitamin D will also contribute to a deficiency in calcium. Symptoms of a Vitamin D deficit include dry mouth, a metallic taste, and a burning sensation in the mouth.
Vitamin A – Vitamin A is directly linked to gum health. The mucous membrane that surrounds and protects the gum tissues is maintained by Vitamin A. Therefore, Vitamin A not only keeps the gums healthy, but it also promotes healing when the gums are damaged or infected.
Vitamin B – B vitamins are important for the overall health of the mouth, and they are easily absorbed with the consumption of meat and fish. Those who are vegetarians or who rarely eat meat need to talk with their physician about taking a Vitamin B supplement. Receding gums, oral sensitivity, and regular tooth aches are often linked to a Vitamin B deficiency.
Vitamin C – Vitamin C is well known for its immunity boosting properties. Vitamin C helps to ward off infections that can be damaging to your oral and overall health.
Your body needs a sufficient supply of vitamins and minerals to remain healthy and perform properly each day. If your diet is lacking in Calcium, Vitamin D, Vitamin A, Vitamin B, and/or Vitamin C, speak with your physician about adding supplements to your daily diet in order to promote optimal oral health.


Samsung Galaxy S10+ could come with an Infinity-O Display panel, as seen from a leaked screen protector. The protector shows two cutouts for the dual selfie camera setup. Also, it suggests thin bezels on the model that is likely to sit against the regular Galaxy S10 and Galaxy S10 Lite. The Galaxy S10+ is also speculated to feature a 6.4-inch QHD+ (1440×3040 pixels) display. Samsung is rumoured to offer various camera-focused features on the Galaxy S10 series to take on the competition. The new flagship family could also include a Holographic technology.
Apart from showing the cutouts, the screen protector showed in the leaked images highlights thin bezels of the Galaxy S10+. There are some few images comparing bezels available on the screen protector with what's available on the Huawei Mate 20 and the iPhone X.
Previous reports have claimed that the Samsung Galaxy S10 will come with a 93.4 percent screen-to-body ratio. The South Korean giant also recently confirmed that the new flagship will have an "industry-leading" screen-to-body ratio.
Particularly in case of the Samsung Galaxy S10+, the Infinity-O Display panel is speculated to have QHD+ (1440×3040 pixels) resolution. The handset could also have a 19:9 aspect ratio.


Yirim Seck is in a good place right now. With three songs left to mix and master on his upcoming album, "Hear Me Out II" and a few shows in the works, the Central District-based rapper with Senegalese roots has gotten his momentum back.
This new wind is "a long time coming," says Seck. After years of working 9-to-5 construction jobs to support his family, the art got sidelined. Then last year, one of his employers tanked, forcing him to endure long commutes for weatherization jobs far out of the Seattle area just as his rent was doubling.
Seck now works independent labor contracts, "which is much more lucrative," he says.
It's like living versus surviving, he says. As a descendent of one of the first few black families who have remained in the Central District since the 1940s — maintaining a sense of Black privilege and family wealth to pass onto his daughters is also a priority.
The hope is that his art is taking him there.
Catch Yirim Seck next at Seattle Art Museum during One Vibe Africa's Kijiji Fest this Saturday, Feb. 11 at 7 p.m.
On Friday, March 3 through Sunday, March 5, Yirim will be presenting a collaborative work with filmmaker/conceptual artist DK Pan at City Arts Magazine's annual "Genre Bender." Shows start at 6:3o p.m. on March 3, 8 p.m. March 4 and 2 p.m. on March 5 at Langston Hughes Performing Arts Institute.
Watch Yirim speak on power and privilege on Thursday, March 23, from 7 p.m. to 9 p.m. as Northwest African American Museum continues its "Complex Exchange" panel discussion series at Seattle Art Museum.
Tags: #BlackLivesMatter, Black privilege, Kijiji Fest, Langston Hughes Performing Arts Institute, Northwest African American Museum, One Vibe Africa, Seattle Art Museum, Seattle Central District, Seattle hip-hop, Yirim Seck.


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Devotion: An Epic Story of Heroism, Friendship, and Sacrifice
Devotion tells the inspirational story of the U.S. Navy's most famous aviator duo, Lieutenant Tom Hudner and Ensign Jesse Brown, and the Marines they fought to defend. A white New Englander from the country-club scene, Tom passed up Harvard to fly fighters for his country. An African American sharecropper's son from Mississippi, Jesse became the navy's first black carrier pilot, defending a nation that wouldn't even serve him in a bar.
While much of America remained divided by segregation, Jesse and Tom joined forces as wingmen in Fighter Squadron 32. Adam Makos takes us into the cockpit as these bold young aviators cut their teeth at the world's most dangerous job—landing on the deck of an aircraft carrier—a line of work that Jesse's young wife, Daisy, struggles to accept.
Deployed to the Mediterranean, Tom and Jesse meet the Fleet Marines, boys like PFC "Red" Parkinson, a farm kid from the Catskills. In between war games in the sun, the young men revel on the Riviera, partying with millionaires and even befriending the Hollywood starlet Elizabeth Taylor. Then comes the war no one expected, in faraway Korea.
Devotion takes us soaring overhead with Tom and Jesse, and into the foxholes with Red and the Marines as they battle a North Korean invasion. As the fury of the fighting escalates and the Marines are cornered at the Chosin Reservoir, Tom and Jesse fly, guns blazing, to try and save them. When one of the duo is shot down behind enemy lines and pinned in his burning plane, the other faces an unthinkable choice: watch his friend die or attempt history's most audacious one-man rescue mission.
A tug-at-the-heartstrings tale of bravery and selflessness, Devotion asks: How far would you go to save a friend?
Adam Makos, 464 pgs, Ballantine Books; Reprint edition, March 21, 2017


Whip-its: The high you can buy at Wal-Mart.
While most people use nitrous oxide canisters to make whipped cream, partyers inhale them for a brief, euphoric high.
Nitrous oxide is characterized as an inhalant, used by those who like to "huff" chemicals.
People can buy canisters online or in the kitchen supply aisle of stores. Balloons filled with the gas are often sold in the parking lots of concerts. Some just extract the gas out of cans of Reddi-Wip.
A quick flash is often accompanied by dizziness, lack of coordination and slurred speech. Since the effect is so brief, users tend to inhale again and again, increasing their feelings of giddiness.
Deprivation of oxygen to the brain, called hypoxia, can cause many long-term problems, including memory loss.
Demi Moore was hospitalized this week after a scary episode reportedly involving inhalation of nitrous oxide.


We produce a waste range of battens for One-Design boat. Tapered leech and full batten for main and jib sail, made of layers of glass fiber stratified with vinylester resin.
Quick Batten has itself the goal to provide the client with the widest selection of battens to allow the best setting of the sail. For this reason, we are starting the production of pultruded battens suitable for One-Design boats: leech and full batten for main and jib sail, made of glass fibre and epoxy resin.
Pultruded battens will be available with a width of mm 10 (thickness inch 0,100, inch 0,140 and inch 0,200) and a width of mm 15 (thickness inch 0,140, inch 0,200 and inch 0,265).


Jimmy's mother complained that her son was beginning to exhibit bizarre behaviors in the past two weeks, barking at passing traffic, hailing Satan, and asking permission to burn bibles. Jimmy spent a few months in substance abuse residential program and was seeing me for aftercare. He and I had been working together for about two months before mom complained about the odd behaviors. A day before one of Jimmy's appointments, his mom called to reschedule because Jimmy was suffering from a sinus infection.
A week later I met with Jimmy and his mother. Mom expressed her concerns and Jimmy said he was only talking like that to tease her. He said he got a kick out of seeing her reaction whenever he brought up Satan and bible burning. I challenged Jimmy by asking, "At whose expense are you reaching your goal?" Jimmy responded with chagrin, "my mom's." I asked Jimmy about his motivation to hurt his mother in this way. He complained, "She irritates me," and he went on to give examples. He said mom uses a baby voice when she talks to him, and that irritated him.
I took Jimmy through five rounds of tapping on irritations and clutching pain. During the tapping process I incorporated phrases from the FeelGoood app that is based on the teachings of Abraham Hicks. After tapping, Jimmy was able to test the clutching pain in his neck and said it didn't exist anymore. I tested the irritation by asking his mom to use her baby voice with Jimmy, who only smiled and gave her a hug, something he had been avoiding for about two weeks according to mom.