text:
\begin{document}
\title{Regularity results for the spherically symmetric Einstein-Vlasov system}
\author{H\aa kan Andr\'{e}asson\\
Mathematical Sciences\\ University of Gothenburg\\
        Mathematical Sciences\\ Chalmers University of Technology\\
        S-41296 G\"oteborg, Sweden\\
        email: hand@chalmers.se}
\maketitle

\begin{abstract}
The spherically symmetric Einstein-Vlasov system is considered in Schwarzschild coordinates and in
maximal-isotropic coordinates. An open problem is the issue of global existence for initial data
without size restrictions. The main purpose of the present work is to propose a method of approach for general initial data, which improves the regularity of the terms that need to be estimated compared to previous methods. We prove that global existence holds outside the centre in both these coordinate systems. In the Schwarzschild case we improve the bound on the momentum support obtained in \cite{RRS} for compact initial data. The improvement implies that we can admit non-compact data with both ingoing and outgoing matter. This extends one of the results in \cite{AR1}. In particular our method avoids the difficult task of treating the pointwise matter terms.
Furthermore, we show that singularities never form in Schwarzschild time for ingoing matter as long as $3m\leq r.$ This removes an additional assumption made in \cite{A1}. Our result in maximal-isotropic coordinates is analogous to the result in \cite{R1}, but our method is different and it improves the regularity of the terms that need to be estimated for proving global existence in general.
\end{abstract}

\section{Introduction}
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In the present work we investigate the issue of global existence for the spherically symmetric Einstein-Vlasov system when the initial data is unrestricted in size. The system is analyzed in Schwarzschild coordinates, i.e. in a polar time gauge, and in maximal-isotropic coordinates where a maximal time gauge is imposed. These coordinate systems are often, in the literature, conjectured to be singularity avoiding. However, there is to our knowledge no proof of this statement for any matter model and it would be very satisfying to provide an answer to this conjecture for the Einstein-Vlasov system. Moreover, a proof of global existence would be of great importance due to its relation to the weak cosmic censorship conjecture, cf. \cite{Cu1,Ds1,DR,A1,AKR2,AKR3}. A third motivation for our interest in this problem is the fact that global existence for general data have been obtained for the Vlasov-Poisson system, i.e. Newtonian gravity coupled to Vlasov matter. Batt \cite{B} showed global existence in the spherically symmetric case 1977, and the general case was settled independently by Pfaffelmoser \cite{P}, and Lions and Perthame \cite{LP} in 1991. It is thus natural to ask if similar results can be obtained when Newtonian gravity is replaced by general relativity. One should of course bear in mind that the situations are fundamentally different since in the latter case there exist data which lead to spacetime singularities, cf. \cite{AKR2,AR2,R2}. Nevertheless, as mentioned above, global existence may hold for polar or maximal time slicing.

The issue of global existence for the spherically symmetric Einstein-Vlasov system has previously been investigated in several papers, cf. \cite{RR,RRS,R1,A1,Ds2,AKR1,AKR2,AR1}. Global existence for small initial data has been proved in \cite{RR} and \cite{Ds2} for massive and massless particles respectively. In \cite{AKR2} initial data which guarantee formation of black holes are constructed, and it is proved that for a particular class of such initial data, with a steady state in the interior of the surrounding matter, global existence holds in Schwarzschild time. In \cite{AKR1} global existence is shown to hold in a maximal time gauge for rapidly outgoing matter. The methods of proofs in these cases are all tailored to treat special initial data and they will likely not apply in more general situations. The investigations \cite{RRS,R1,A1,AR1} are conditional in the sense that assumptions are made on the solutions, and not only on the initial data. However, the methods are general and cover large classes of initial data and can thus be thought of as possible approaches for treating the general case. The main purpose of the present work is to propose a method of approach for general initial data, which improves the regularity of the terms that need to be estimated compared to the methods in \cite{RRS,R1,A1,AR1}. Furthermore, the method improves and simplifies some of the previous results. To be more precise, let us discuss the relation between these studies.

An important quantity in the study of the Einstein-Vlasov system and the Vlasov-Poisson system is
\begin{equation}\label{Qcompact}
Q(t):=\sup \{1+|v|:\exists (s,x)\in [0,t]\times \mathbb{R}^3\mbox{ such that
    }f(s,x,v)\not= 0\}.
\end{equation}
Here $f$ is the density function on phase, and $v\in\mathbb{R}^3$ is the momentum. $Q(t)$ measures the support of the momenta, and the content of the continuation criterion for these systems, cf. \cite{RR,R1,B}, is that solutions can be extended as long as $Q(t)$ remains bounded. The definition of $Q$ only applies in the case when the initial data have compact support in the momentum variables. For non-compact initial data, it was shown in \cite{AR1} that solutions can be extended as long as $\tilde{Q}(t)$ is bounded where

\begin{equation}\label{Qsupp}
\tilde{Q}(t)
:=\sup\left\{\frac{1+|V(s,0,r,v)|}{1+|v|} \mid 0\leq s\leq t,\
(r,v)\in \supp \fn \,\right\},
\end{equation}
Here $V$ is a solution of the characteristic system, cf. (\ref{char2}).
The analysis in \cite{RRS,A1,AR1} is carried out in Schwarzschild coordinates. The main result in \cite{RRS} shows that as long as there is no matter in the ball
$$\{x\in\mathbb{R}^3:|x|\leq\epsilon\},$$ the estimate
\begin{equation}\label{Qee}
Q(t)\leq e^{\log{Q(0)}e^{C(\epsilon)t}},
\end{equation}
holds. Here $C(\epsilon)$ is a constant which depends on $\epsilon.$ In view of the continuation criterion this can thus be viewed as a global existence result outside the centre of symmetry for initial data with compact support. The bound is obtained by estimating each term individually in the characteristic equation (\ref{char2}) for the radial momentum. This involves a particular difficulty. Let us consider the term involving $\mu_r$ in (\ref{char2}). The Einstein equations imply that
$$\mu_r=\frac{m}{r^2}e^{2\lambda}+4\pi rpe^{2\lambda}=:T_1+T_2,$$
where $m$ is the quasi local mass and $p$ is the pressure. There is a distinct difference between the terms $T_1$ and $T_2$ due to the fact that $m$ can be regarded as an average, since it is given as a space integral of the energy density $\rho,$ whereas $p$ is a pointwise term. Also the term involving $\lambda_t$ in (\ref{char2}) is a pointwise term in this sense. The method in \cite{RRS} is able to estimate the pointwise terms outside the centre but generally it seems very unpleasant to have to treat these terms. In the present work we give an alternative and simplified proof of the result in \cite{RRS}. In particular our method avoids the pointwise terms by using the fact that the characteristic system can be written in a form such that Green's formula in the plane can be applied. This results in a combination of terms involving second order derivatives which can be substituted for by one of the Einstein equations. This method was first introduced in \cite{A1} but here the set up is different and the application of Green's formula becomes very natural. In addition the bound of $Q$ is improved compared to (\ref{Qee}) and reads $$Q(t)\leq (Q(0)+\frac{Ct}{\epsilon^2})e^{C(1+t)/\epsilon}.$$ This bound implies that also $\tilde{Q}(t)$ is bounded and therefore a consequence of the method is that global existence outside the centre also holds for non-compact initial data. This improves the result in \cite{AR1} where only non-compact data for which the matter is ingoing, and such that the matter keeps on going inwards for all times, are admitted. Another consequence of our method is that global existence holds for ingoing matter as long as $3m(t,r)\leq r.$ Note that in Schwarzschild coordinates $2m(t,r)\leq r$ always, and that there are closed null geodesics if $3m=r$ in the Schwarzschild spacetime. This result was already proved in \cite{A1} but an additional assumption was imposed which now has been dropped.

We now turn to the case of maximal-isotropic coordinates. Rendall shows in \cite{R1} global existence outside the centre in maximal-isotropic coordinates. The bound on $Q(t)$ is again obtained by estimating each term in the characteristic equation, cf. equation (\ref{charW}). In this case there are no pointwise terms in contrast to the case with Schwarzschild coordinates. The terms are however, in analogy with the Schwarzschild case, strongly singular at the centre. The method that we use in Schwarzschild coordinates also applies in this case. The improvement lies in the fact that there is some gain in regularity, i.e. the terms that need to be estimated are less singular compared to the corresponding terms in \cite{R1}. Indeed, these terms can schematically, in both the Schwarzschild and maximal-isotropic case, be written as
\begin{equation}\label{g}
\int_0^{\infty}g(t,\eta)d\eta,
\end{equation}
and the known a priori bounds read
\begin{equation}\label{rg}
\int_0^r\eta g(t,\eta)d\eta.
\end{equation}
Hence, the degree of the singularity is of order one in the radial variable. Roughly, for the methods in \cite{RRS} and \cite{R1} the singularity is of second order. An advantage of our method is clearly that it applies in both cases, and it also turns out that the principal term to be estimated in the two cases is the spacetime integral of the Gauss curvature of the two dimensional quotient manifold ${\cal{M}}/SO(3),$ where ${\cal{M}}$ is the four dimensional spacetime manifold. This is interesting since the Gauss curvature is a coordinate independent quantity. Moreover, the principal term in \cite{Cu2}, p. 1172, has the same form. Finally we mention that our method also applies in the case of maximal-areal coordinates which were used in the study \cite{AKR1}. The analysis in this case is completely analogous and is left out.

The outline of the paper is as follows. The Einstein-Vlasov system in
Schwarzschild coordinates is treated in sections 2-5. In section 2 the system is formulated and
the a priori bounds are given in section 3. Section 4 is devoted to the proof of global existence outside the centre and in section 5 global existence is shown for ingoing matter which satisfies $m/r\leq 1/3.$ Sections 6-8 concern the system in maximal-isotropic coordinates. The system is given in section 6, and in section 7 the necessary a priori bounds are derived. Section 8 is devoted to the proof of the global existence theorem outside the centre.

\section{The Einstein-Vlasov system}
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For an introduction to the Einstein-Vlasov system and kinetic theory we refer to \cite{A2,R1}, and for a careful derivation of the system given below we refer to~\cite{R}.
In Schwarzschild coordinates the spherically symmetric metric takes the form
\begin{equation}
ds^{2}=-e^{2\mu(t,r)}dt^{2}+e^{2\lambda(t,r)}dr^{2}
+r^{2}(d\theta^{2}+\sin^{2}{\theta}d\varphi^{2}).
\end{equation}
The Einstein equations read
\begin{eqnarray}
&\displaystyle e^{-2\lambda}(2r\lambda_{r}-1)+1=8\pi r^2\rho,&\label{ee1}\\
&\displaystyle e^{-2\lambda}(2r\mu_{r}+1)-1=8\pi r^2 p,&\label{ee2}\\
&\displaystyle\lambda_{t}=-4\pi re^{\lambda+\mu}j,&\label{ee3}\\
&\displaystyle e^{-2\lambda}(\mu_{rr}+(\mu_{r}-\lambda_{r})(\mu_{r}+
\frac{1}{r}))-e^{-2\mu}(\lambda_{tt}+\lambda_{t}(\lambda_{t}-\mu_{t}))=
8\pi p_T.&\label{ee4}
\end{eqnarray}
The indicies $t$ and $r$ denote derivatives.
The Vlasov equation for the density distribution function
$f=f(t,r,w,L)$ is given by
\begin{equation}
\partial_{t}f+e^{\mu-\lambda}\frac{w}{E}\partial_{r}f
-(\lambda_{t}w+e^{\mu-\lambda}\mu_{r}E-
e^{\mu-\lambda}\frac{L}{r^3E})\partial_{w}f=0,\label{vlasov}
\end{equation}
where
\begin{equation}
E=E(r,w,L)=\sqrt{1+w^{2}+L/r^{2}}.\label{E}
\end{equation}
Here $w\in (-\infty,\infty)$ can be thought of as the radial component
of the momentum variables, and $L\in [0,\infty)$ is the square of
the angular momentum.
The matter quantities are defined by
\begin{eqnarray}
\rho(t,r)&=&\frac{\pi}{r^{2}}
\int_{-\infty}^{\infty}\int_{0}^{\infty}Ef(t,r,w,L)\;dwdL,\label{rho}\\
p(t,r)&=&\frac{\pi}{r^{2}}\int_{-\infty}^{\infty}\int_{0}^{\infty}
\frac{w^{2}}{E}f(t,r,w,L)\;d
wdL,\label{p}\\
j(t,r)&=&\frac{\pi}{r^{2}}
\int_{-\infty}^{\infty}\int_{0}^{\infty}wf(t,r,w,L),\;dwdL,\label{j}\\
p_T(t,r)&=&\frac{\pi}{2r^{4}}\int_{-\infty}^{\infty}\int_{0}^{\infty}\frac{L}{E}f(t,r,w,L)\;
dwdL.\label{q}
\end{eqnarray}
Here $\rho,p,j$ and $p_T$ are the energy density, the radial pressure, the current and the tangential pressure respectively.
The following boundary conditions are imposed to ensure asymptotic flatness
\begin{equation}
\lim_{r\rightarrow\infty}\lambda(t,r)=\lim_{r\rightarrow\infty}\mu(t,r)=0,\label{bdryaf}
\end{equation}
and a regular centre requires
\begin{equation}\label{bdryrc}
\lambda(t,0)=0,\;t\geq 0.
\end{equation}
We point out that the Einstein equations are not independent and that
e.g. the equations (\ref{ee3}) and (\ref{ee4}) follow by
(\ref{ee1}), (\ref{ee2}) and (\ref{vlasov}).

As initial data it is sufficient to prescribe a distribution function
$\open{f}=\open{f}(r,w,L)\geq 0$ such that
\begin{equation}\label{notsinit}
   \int_0^r 4\pi\eta^2\open{\rho}\,(\eta)\,d\eta < \frac{r}{2}.

\end{equation}
Here we denote by $\open{\rho}$ the energy density induced by the initial
distribution function $\open{f}$. This condition ensures that no trapped surfaces are present
initially. Given $\open{f},$ equations (\ref{ee1}) and (\ref{ee2}) can be solved to give $\lambda$ and $\mu$ at $t=0.$ We will only consider initial data
such that $\fn=0$ if $L>L_2,$ for some $L_2>0,$
and such that $\fn=0$ if $r>R_2,$ for some $R_2>0.$
If in addition the initial data is $C^1$ we say that it is regular.
In most of the previous investigations the condition of compact support on the momentum
variable $w$ has
been included in the definition of regular data. The exception is \cite{AR1} where non-compact initial
data is studied and
a decay condition replaces the assumption of compact support. In this study we also
include data with non-compact support and we impose the decay condition from \cite{AR1}
\begin{equation}\label{iddecay}
\sup_{(r,w,L)\in\R^3} |w|^5 \fn (r,w,L) < \infty.
\end{equation}
We distinguish between the two cases; regular initial data with compact support
and regular initial data which satisfy the decay condition (\ref{iddecay}). We denote these
classes of initial data by $\cal{I}^C$ and $\cal{I}^D$ respectively.

The main results below concern subclasses of $\cal{I}^C$ and $\cal{I}^D.$ Given $R_{1}>0,$ we define
the subclass ${\cal{I}}^C(R_1)$ with radial cut-off by
\[
{\cal{I}}^C(R_1)=\{\fn\in{\cal{I}}^C: \fn=0\mbox{ for }r\leq R_{1}\}.
\]
The subclass ${\cal{I}}^D(R_1)$ is defined analogously.

Let us now write down a couple of
facts about the system (\ref{ee1})-(\ref{bdryrc}).
A solution to the Vlasov
equation can be written
\begin{equation}
f(t,r,w,L)=f_{0}(R(0,t,r,w,L),W(0,t,r,w,L),L),
\label{repre}
\end{equation}
where $R$ and $W$ are solutions of the characteristic system
\begin{eqnarray}
\frac{dR}{ds}&=&e^{(\mu-\lambda)(s,R)}\frac{W}{E(R,W,L)},\label{char1}\\
\frac{dW}{ds}&=&-\lambda_{t}(s,R)W-e^{(\mu-\lambda)(s,R)}\mu_{r}(s,R)E(R,W,L)
\nonumber\\
& &+e^{(\mu-\lambda)(s,R)}\frac{L}{R^3E(R,W,L)},\label{char2}
\end{eqnarray}
such that $(R(s,t,r,w,L),W(s,t,r,w,L),L)=(r,w,L)$ when $s=t$.
This representation shows that $f$ is nonnegative for all $t\geq 0,$ $\|f\|_{\infty}=\|f_{0}\|_{\infty},$ and that $f(t,r,w,L)=0$ if $L>L_{2}.$
The quasi local mass $m$ is defined by
\begin{equation}
m(t,r)=4\pi\int_{0}^{r}\eta^{2}\rho(t,\eta)d\eta,\label{m}
\end{equation}
and by integrating (\ref{ee1}) we find
\begin{equation}
e^{-2\lambda(t,r)}=1-\frac{2m(t,r)}{r}.\label{e2-lambda}
\end{equation}
A fact that we will need is that $$\mu+\lambda\leq 0.$$
This is easily seen by adding the equations (\ref{ee1}) and
(\ref{ee2}), which gives $$\lambda_{r}+\mu_{r}\geq 0,$$ and then using
the boundary conditions on $\lambda$ and $\mu.$ Furthermore, from (\ref{e2-lambda})
we get that $\lambda\geq 0,$ and it follows that $\mu\leq 0.$ We also introduce the notations
$\hat{\mu}$ and $\check{\mu}.$ From equation (\ref{ee2}) we have
\begin{equation}\label{muhatcheck}
\mu(t,r)=-\int_r^{\infty}\frac{m(t,\eta)}{\eta^2}e^{2\lambda}-\int_r^{\infty}4\pi \eta pe^{2\lambda}\, d\eta=:\hat{\mu}+\check{\mu}.
\end{equation}
Finally, we note that in~\cite{RR} and \cite{AR1} local existence theorems are proved for compact
and non-compact initial data respectively,
and it will be used below that solutions exist on some time interval $[0,T[.$

\section{A priori bounds in Schwarzschild coordinates}
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In this section we collect the a priori bounds that we need in the proofs below.
There are two known conserved quantities for the Einstein-Vlasov
system, the number of particles and the ADM mass $M$. Here, we will only need the latter
which is given by
\begin{equation}
M=4\pi\int_{0}^{\infty}r^{2}\rho(t,r)dr.\label{adm}
\end{equation}
The conservation of the ADM mass follows from general arguments but it can easily be
obtained by simply taking the time derivative of the integral expression and
use of the Vlasov equation and the Einstein equations.
The following results are given in \cite{A1} but since the proofs are very short we
have included them here for completeness. By a regular solution we mean a solution which
is launched by regular initial data.
\begin{lemma}
Let $(f,\mu,\lambda)$ be a regular solution to the Einstein-Vlasov system.
Then
\begin{eqnarray}
&\displaystyle\int_{0}^{\infty}4\pi r(\rho+p)e^{2\lambda}e^{\mu+\lambda}dr\leq 1,&\label{expr}\\
&\displaystyle\int_{0}^{\infty}(\frac{m}{r^2}+4\pi rp)e^{2\lambda}e^{\mu}dr\leq
1.&\label{expr2}
\end{eqnarray}
\end{lemma}
\textbf{Proof. }Using the boundary condition (\ref{bdryaf}) we get
\begin{eqnarray*}
1\geq
1-e^{\mu+\lambda}(t,0)&=&\int_{0}^{\infty}\frac{d}{dr}e^{\mu+\lambda}dr\\
&=&\int_{0}^{\infty}(\mu_{r}+\lambda_{r})e^{\mu+\lambda}dr.
\end{eqnarray*}
The right hand side equals (\ref{expr}) by equations (\ref{ee1})
and (\ref{ee2}) which completes the first part of the lemma. The
second part follows by studying $e^{\mu}$ instead of
$e^{\mu+\lambda}.$
\begin{flushright}
$\Box$
\end{flushright}
Next we show that not only $\rho(t,\cdot)\in L^{1},$ which follows from
the conservation of the ADM mass, but that also $e^{2\lambda}\rho(t,\cdot)\in
L^{1}.$
\begin{lemma}
Let $(f,\mu,\lambda)$ be a regular solution to the Einstein-Vlasov
system.
Then
\begin{equation}
\int_{0}^{\infty}r^{2}e^{2\lambda}\rho(t,r) dr\leq
\int_{0}^{\infty}r^{2}e^{2\lambda}\rho(0,r) dr+\frac{t}{8\pi}.\label{r2rhoe2lambda}
\end{equation}
\end{lemma}
\textbf{Proof. }Using the Vlasov equation we obtain
\begin{eqnarray*}
\partial_{t}\left(r^{2}e^{2\lambda}\rho(t,r)\right)&=&
-\partial_{r}\left(r^{2}e^{\mu+\lambda}j\right)-re^{\mu+\lambda}2je^{2\lambda}
\frac{m}{r}\\
&\leq&-\partial_{r}\left(r^{2}e^{\mu+\lambda}j\right)+
\frac{1}{2}re^{\mu+\lambda}(\rho+p)e^{2\lambda}.
\end{eqnarray*}
Here we used that $m/r\leq 1/2$ together with the elementary inequality
$2|j|\leq\rho+p,$ which follows from the expressions
(\ref{rho})-(\ref{j}). In view of (\ref{repre}) and (\ref{char1}) we see that
$\lim_{r\to\infty}r^2j(t,r)=0,$ since the initial data has compact
support in $r$. Since the solution is regular and hence bounded the boundary
term at $r=0$ also vanishes (as a matter of fact spherical symmetry
even implies that $j(t,0)=0$). Thus, by lemma 1 we get
\begin{displaymath}
\frac{d}{dt}\int_{0}^{\infty}r^{2}e^{2\lambda}\rho(t,r) dr\leq
\frac{1}{8\pi},
\end{displaymath}
which completes the proof of the lemma.
\begin{flushright}
$\Box$
\end{flushright}

\section{A regularity result in Schwarzschild coordinates}
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\begin{theorem}\label{theorem1} Let $0<\epsilon<R_1.$ Consider a solution of
the spherically symmetric
Einstein-Vlasov system, launched by initial data in ${\cal{I}}^D(R_1),$
on its maximal time interval $[0,T[$ of existence. If $f(s,r,\cdot,\cdot)=0,$ for
$(s,r)\in [0,t[\times [0,\epsilon],$ then
\begin{equation}\label{Qestimate}
\tilde{Q}(t)\leq (1+\frac{Ct}{\epsilon^2})e^{\frac{C(1+t)}{\epsilon}}.
\end{equation}
In particular, if $f(t,r,\cdot,\cdot)=0$ for $(t,r)\in [0,T[\times [0,\epsilon],$ then $T=\infty.$
\end{theorem}
The last statement in the theorem is a consequence of the continuation criterion derived in \cite{AR1}. The theorem thus improves the result in \cite{AR1} which was restricted to a special class of initial data where all the matter is ingoing. The result holds in particular for compactly supported data, and the bound (\ref{Qestimate}) improves the bound in \cite{RRS}, cf. inequality (\ref{Qee}).

\textit{Proof: }
We consider the quantities $G=E+W$ and $H=E-W,$ which satisfy $G>0,\; H>0.$
Along a characteristic $(R(s),W(s),L)$ we have by (\ref{char1}) and (\ref{char2}).
\begin{equation}
\frac{dG}{ds}=-\left[\lambda_t\frac{W}{E}+\mu_r e^{\mu-\lambda}\right]G
+\frac{Le^{\mu-\lambda}}{R^3E},\label{charG}
\end{equation}
and
\begin{equation}
\frac{dH}{ds}=\left[\lambda_t\frac{W}{E}+\mu_r e^{\mu-\lambda}\right]H
-\frac{Le^{\mu-\lambda}}{R^3E}.\label{charH}
\end{equation}
Let us first consider the quantity $H.$ From (\ref{charH}) we have
\begin{equation}
\frac{dH}{ds}\leq\left[\lambda_t\frac{W}{E}+\mu_r e^{\mu-\lambda}\right]H.\label{2charH}
\end{equation}
It follows that
\begin{equation}
H(t)\leq H(0)e^{\int_{0}^{t}\left[\lambda_t(s,R(s))\frac{W(s)}{E(s)}+\mu_r(s,R(s))e^{(\mu-\lambda)(s,R(s))}
\right]\, ds}.\label{HT}
\end{equation}
Let us denote the curve $(s,R(s)),\; 0\leq s\leq t,$ by $\gamma.$
By using that $$\frac{dR}{dt}=e^{\mu-\lambda}\frac{W}{E},$$
the integral above can be written as the curve integral
\begin{equation}
\int_{\gamma}e^{(-\mu+\lambda)(t,r)}\lambda_{t}(t,r)dr+
e^{(\mu-\lambda)(t,r)}\mu_{r}(t,r)dt.\label{curveH}
\end{equation}
Let $\Gamma$ denote the closed curve
$\Gamma:=\gamma+C_{t}+C_{\infty}+C_{0},$ oriented clockwise, where
\begin{eqnarray}
C_{t}&=&\{(t,r):R(t)\leq r\leq R_{\infty}\},\\
C_{\infty}&=&\{(s,R_{\infty}):t\geq s\geq 0\},\\
C_{0}&=&\{(0,r):R_{\infty}\geq r\geq R(0)\}.\label{Ccurves}
\end{eqnarray}
Here $R_{\infty}\geq R_{2}+t,$ so that $f=0$ when $r\geq R_{\infty}.$
We now apply Green's formula in the plane and use equation (\ref{ee4}) to obtain
\begin{eqnarray}
&\displaystyle\oint_{\Gamma} e^{-\mu+\lambda}\lambda_{t}dr+
e^{\mu-\lambda}\mu_{r}dt&\nonumber\\
&\displaystyle =\int\int_{\Omega}\partial_{t}\left(e^{-\mu+\lambda}
\lambda_{t}\right)-\partial_{r}
\left( e^{\mu-\lambda}\mu_{r}\right)dtdr&\nonumber\\
&\displaystyle=\int\int_{\Omega}\left(\lambda_{tt}+(\lambda_{t}-\mu_{t}\right)
\lambda_{t}-(\mu_{rr}+(\mu_{r}-\lambda_{r})\mu_{r})e^{\mu-\lambda}\,dtdr&
\nonumber\\
&\displaystyle =\int\int_{\Omega}(\mu_{r}-\lambda_{r})\frac{e^{\mu-\lambda}}{r}-8\pi
p_T e^{\mu+\lambda}\,dtdr&
\nonumber\\
&\displaystyle =\int\int_{\Omega}e^{\mu+\lambda}\left(\frac{2m}{r^{3}}
-4\pi(\rho-p)-8\pi p_T\right)\,dtdr.&\label{id}
\end{eqnarray}
Using the hypothesis that matter stays away from the region $r\leq\epsilon,$ we have $R(s)\geq\epsilon,\, 0\leq s\leq t,$ and we get in view of (\ref{expr2}), together with the facts that $\lambda\geq 0,\;\rho-p\geq 0$
and $q\geq 0$,
\begin{equation}\label{Omegaest}
\int\int_{\Omega}e^{\mu+\lambda}\left(\frac{2m}{r^{3}}
-4\pi(\rho-p)-8\pi p_T\right)\,dtdr\leq \frac{2t}{\epsilon}.
\end{equation}
\textbf{Remark 1: }The integrand above has a geometrical meaning. It is the scalar curvature of the quotient manifold ${\cal{M}}/SO(3)$ with metric $ds^2=-e^{2\mu}dt^2+e^{2\lambda}dr^2.$ It is interesting to note that the principal term in \cite{Cu2}, p. 1172, has the same form. \\
Now we wish to estimate the curve integral (\ref{curveH}). We have
\begin{equation}
\int_{\gamma}...=\oint_{\Gamma}...-\int_{C_t}4\pi rje^{2\lambda}dr-\int_{C_{\infty}}
\frac{M}{R_{\infty}^2}dt-\int_{C_0}4\pi rje^{2\lambda}dr.
\end{equation}
Here we used that $\lambda_t=4\pi rje^{\mu+\lambda}$
and that $\mu_r(t,R_{\infty})=\frac{M}{R_{\infty}^2}.$
The integral along $C_0$ is given in terms of the initial data,
\begin{equation}
\big|\int_{C_0}4\pi rje^{2\lambda}dr\big|\leq C.
\end{equation}
By letting $R_{\infty}\to\infty$ the integral along $C_{\infty}$ vanishes and it remains to
estimate the contribution from the integral along $C_t.$ In view of (\ref{r2rhoe2lambda})
we get
\begin{equation}\label{intrj}
\big|\int_{C_t}4\pi rje^{2\lambda}dr\big|\leq \frac{1}{\epsilon}\int_0^{\infty}
4\pi r^2\rho(t,r)e^{2\lambda(t,r)}dr\leq \frac{C}{\epsilon}(1+t).
\end{equation}
The esimate of $G$ is very similar. Indeed, by the hypothesis of the theorem we
have $R(t)\geq \epsilon,$ thus
\begin{equation}
\frac{Le^{\mu-\lambda}}{R^3E}\leq \frac{\sqrt{L}}{\epsilon^2}
\leq\frac{\sqrt{L_2}}{\epsilon^2}.
\end{equation}
We derive in view of (\ref{charG})
\begin{eqnarray}
\displaystyle G(t)&\leq& G(0)e^{\int_{0}^{t}\left[-\lambda_t(s,R(s))
\frac{W(s)}{E(s)}-\mu_r(s,R(s))e^{(\mu-\lambda)(s,R(s))}\right]ds}\nonumber\\
&+&\int_0^t\frac{\sqrt{L_2}}{\epsilon^2}e^{\int_{\tau}^{t}\left[-\lambda_t(s,R(s))\frac{W(s)}{E(s)}-\mu_r(s,R(s))e^{(\mu-\lambda)(s,R(s))}\right]ds}d\tau.\nonumber
\end{eqnarray}
Note that the integrals in the exponent above are identical to the integral in (\ref{HT}) except for the sign.
We can accordingly use the same arguments as above, the only difference is that we use
(\ref{intrj}) twice, for $C_t$ and $C_{\tau}$, and that the integral in (\ref{id}) has
opposite sign. Since $2p_T\leq \rho-p$ it is sufficient to estimate
the integral
\begin{equation}
\int\int_{\Omega} 8\pi(\rho-p)e^{\mu+\lambda}\,dtdr.
\end{equation}
We use the bound (\ref{expr}) to obtain the estimate (\ref{Omegaest}) also in this case.
Hence, we have the following bounds
\begin{equation}
G(t)\leq (G(0)+\frac{Ct}{\epsilon^2})e^{\frac{C(1+t)}{\epsilon}},
\end{equation}
and
\begin{equation}
H(t)\leq H(0)e^{\frac{C(1+t)}{\epsilon}}.
\end{equation}
Since $E=G+H$ this implies that
\begin{equation}
\tilde{Q}(t)\leq (1+\frac{Ct}{\epsilon^2})e^{\frac{C(1+t)}{\epsilon}}.
\end{equation}
This completes the proof of Theorem \ref{theorem1}.
\begin{flushright}
$\Box$
\end{flushright}

\section{Global existence for ingoing matter with $3m\leq r$}
\setcounter{equation}{0}
In this section we consider compactly supported initial data $\open{f}\in{\cal{I}}^C(R_1)$ such that for given $0<L_1<L_2$ and $P>0,$ it holds that $L\geq L_1,$ and  $w<-P$ for all $(w,L)\in\supp\open{f}.$ Furthermore, the initial data have the property that $3m<r$ everywhere. Note that in Schwarzschild spacetime there are closed null geodesics when $3m=r.$ Let us denote this class of initial data by ${\cal{I}}^C(R_1,L_1,P,3)\subset{\cal{I}}^C(R_1).$ By continuity there is a $T_1>0$ such that $w\leq 0$ for $w\in\supp f(t),$ and $3m(t,r)\leq r$ everywhere, for $t\leq T_1.$ We will show that on the time interval $[0,T_1]$ singularities do not form in the evolution. This can be phrased as a global existence result for ingoing matter satisfying $3m\leq r$. A similar result is proved in
\cite{A1} but an additional assumption on the solution is imposed and the present result is thus an improvement.
\begin{theorem}\label{theorem2}
Consider a solution to the spherically symmetric Einstein-Vlasov system launched by initial data
$\open{f}\in {\cal{I}}^C(R_1,L_1,P,3)$. Let $T>0$ be the maximal time interval on
which the solution exists and let $T_1$ be as above. If $T<\infty$ then $T_1<T.$
\end{theorem}
\textit{Proof: }
Let us consider the quantity $He^{\hat{\mu}}$ along a given characteristic $(t,R(t),W(t),L),$
with $W\leq 0.$
In view of (\ref{charH}) and (\ref{muhatcheck}) we have
\begin{equation}
\frac{d}{ds}(He^{\hat{\mu}})=\left[\lambda_t\frac{W}{E}+\mu_r e^{\mu-\lambda}+\hat{\mu}_t+\hat{\mu_r}\frac{W}{E}e^{\mu-\lambda}\right]He^{\hat{\mu}}
-\frac{Le^{\mu-\lambda}}{R^3E}e^{\hat{\mu}}.\label{charHexpmuhat}
\end{equation}
Since $\mu=\hat{\mu}+\check{\mu}$ we have
\begin{equation}
\mu_r e^{\mu-\lambda}+\hat{\mu_r}\frac{W}{E}e^{\mu-\lambda}=\check{\mu}_r e^{\mu-\lambda}
+\hat{\mu}_r e^{\mu-\lambda}(1+\frac{W}{E})=\check{\mu}_r e^{\mu-\lambda}
+\hat{\mu}_r e^{\mu-\lambda}\frac{1+L/R^2}{E(E-W)}.
\end{equation}
Since $H=E-W$ and $$\hat{\mu}_r=\frac{me^{2\lambda}}{r^2},$$ we obtain
\begin{equation}
\frac{d}{ds}(He^{\hat{\mu}})=\left[\lambda_t\frac{W}{E}+\check{\mu}_r e^{\mu-\lambda}+\hat{\mu}_t
\right]He^{\hat{\mu}}
+e^{\mu-\lambda}
\left(\frac{me^{2\lambda}}{R}\frac{(R^2+L)}{R^3E}-\frac{L}{R^3E}\right)e^{\hat{\mu}}.
\label{charHexpmuhatweak}
\end{equation}
We note that if
\begin{equation}\label{weak}
\frac{m(t,R(t))}{R(t)}\leq\frac{1}{3},
\end{equation}
we have
\begin{equation}
\frac{m(t,R(t))}{R(t)}e^{2\lambda(t,R(t))}=\frac{m(t,R(t))}{R(t)(1-\frac{2m(t,R(t))}{R(t)})}\leq 1.
\end{equation}
Hence, as long as (\ref{weak}) holds true we get in view of (\ref{charHexpmuhatweak})
the inequality
\begin{equation}
\frac{d}{ds}(He^{\hat{\mu}})\leq
\left[\lambda_t\frac{W}{E}+\check{\mu}_r e^{\mu-\lambda}+\hat{\mu}_t
\right]He^{\hat{\mu}}
+e^{\mu-\lambda}
\frac{1}{RE}e^{\hat{\mu}}.
\label{charHexpmuhatweakineq}
\end{equation}
Since $ER\geq\sqrt{L_1}$ the last term is bounded. We can without loss of generality
assume that $He^{\hat{\mu}}\geq 1$ on $[0,T_1]$, and we get for $t\leq T_1,$
\begin{eqnarray}
\displaystyle H(t)e^{\hat{\mu}(t,R(t))}&\leq &H(0)e^{\hat{\mu}(0,R(0))}
e^{\frac{t}{\sqrt{L_1}}+\int_0^t \hat{\mu}_t(s,R(s))ds}\nonumber\\
\displaystyle & &\times e^{\int_{0}^{t}
\left[\lambda_t(s,R(s))\frac{W}{E}
+\check{\mu}_r(s,R(s))e^{(\mu-\lambda)(s,R(s))}\right]ds}.\label{Hexphatmuweak}
\end{eqnarray}
Let us denote by $\gamma$ the curve $(t,R(t)),\, 0\leq t\leq T_1.$
The last integral in (\ref{Hexphatmuweak}) can be written as
\begin{equation}
\int_{\gamma}e^{(-\mu+\lambda)(t,r)}\lambda_{t}(t,r)dr+
e^{(\mu-\lambda)(t,r)}\check{\mu}_{r}(t,r)dt.\label{Hcurveweak}
\end{equation}
We will apply the Green formula to this curve integral and we introduce as above
the closed curve $\Gamma=\gamma+C_{T_1}+C_{\infty}+C_0$ where $C_{T_1},\,C_{\infty}$ and
$C_0$ are defined as in (\ref{Ccurves}).
We have
\begin{eqnarray}
&\displaystyle\oint_{\Gamma} e^{-\mu+\lambda}\lambda_{t}dr+
e^{\mu-\lambda}\check{\mu}_{r}ds&\nonumber\\
&\displaystyle =\int\int_{\Omega}\partial_{t}\left(e^{-\mu+\lambda}
\lambda_{t}\right)-\partial_{r}
\left( e^{\mu-\lambda}\check{\mu}_{r}\right)dsdr&\nonumber\\
&\displaystyle =\int\int_{\Omega}\partial_{t}\left(e^{-\mu+\lambda}
\lambda_{t}\right)-\partial_{r}
\left( e^{\mu-\lambda}\mu_{r}\right)dsdr
+\int\int_{\Omega}\partial_{r}
\left(e^{\mu+\lambda}\frac{m}{r^{2}}\right)dsdr.&\nonumber\\
\end{eqnarray}

For the first integral above we use (\ref{id}) and we obtain
the identity
\begin{eqnarray}
& &\phantom{GH} \displaystyle\oint_{\Gamma} e^{-\mu+\lambda}\lambda_{t}dr+
e^{\mu-\lambda}\check{\mu}_{r}ds\nonumber\\
& &\displaystyle =\int\int_{\Omega}e^{\mu+\lambda}\left(\frac{2m}{r^{3}}
-4\pi(\rho-p)-8\pi p_T\right)\,dtdr\nonumber\\
& &\phantom{G} \displaystyle +\int\int_{\Omega}e^{\mu+\lambda}\left((\mu_r+\lambda_r)\frac{m}{r^2}+4\pi\rho-\frac{2m}{r^3}\right)dtdr\nonumber\\
& &\displaystyle=\int\int_{\Omega}4\pi e^{\mu+\lambda}\left[
(\rho+p)e^{2\lambda}\frac{m}{r}+2p-\rho\right] drds\nonumber\\
& &\phantom{H} \displaystyle+\int\int_{\Omega}\int_{-\infty}^{\infty}\int_{0}^{\infty}
\frac{4\pi^{2} e^{\mu+\lambda}}{r^{2}E}f(t,r,w,F)dFdw\,drds.\label{id2}
\end{eqnarray}
Here we used that $\mu_r+\lambda_r=4\pi r(\rho+p)e^{2\lambda}$ and that
\begin{equation}\label{almostq}
2p_T=\rho-p-\int_{-\infty}^{\infty}\int_{0}^{\infty}
\frac{4\pi^{2} e^{\mu+\lambda}}{r^{2}E}f(t,r,w,F)dFdw.
\end{equation}
Recall from (\ref{Hexphatmuweak}) that the integral involving $\hat{\mu}$ should also
be taken into account. Since
\begin{equation}
\hat{\mu}_{t}(t,r)=\int_r^{\infty}4\pi je^{2\lambda}e^{\mu+\lambda}d\eta,
\end{equation}
we get
\begin{equation}
\int_0^t\hat{\mu}_{t}(s,R(s))ds=\int\int_{\Omega}4\pi je^{2\lambda}e^{\mu+\lambda}dsdr.
\end{equation}
Hence we obtain with $C_B:=C_{T_1}+C_{\infty}+C_0,$
\begin{eqnarray}
&\displaystyle\oint_{\gamma} e^{-\mu+\lambda}\lambda_{t}dr+
e^{\mu-\lambda}\check{\mu}_{r}ds+\int_{0}^{T}\hat{\mu}_t(s,r) ds&\nonumber\\
&\displaystyle=\int\int_{\Omega}4\pi e^{\mu+\lambda}\left[
(\rho+p)e^{2\lambda}\frac{m}{r}+2p-\rho+je^{2\lambda}\right] drds&\nonumber\\
&\displaystyle+\int\int_{\Omega}\int_{-\infty}^{\infty}\int_{0}^{\infty}
\frac{4\pi^{2} e^{\mu+\lambda}}{r^{2}E}f(t,r,w,F)dFdw\,drds&\nonumber\\
&\displaystyle -\int_{C_B}e^{-\mu+\lambda}\lambda_{t}dr+
e^{\mu-\lambda}\check{\mu}_{r}ds.&\label{green3}
\end{eqnarray}
The integral over $C_B$ equals
\begin{eqnarray}
&\displaystyle-\int_{C_B}e^{-\mu+\lambda}\lambda_{t}dr+
e^{\mu-\lambda}\check{\mu}_{r}ds=\int_{R(T_1)}^{\infty} 4\pi rj(T_1,r)e^{2\lambda(T_1,r)}dr&
\nonumber\\
&\displaystyle+\int_{0}^{T_1}4\pi R_{\infty}p(s,R_{\infty})e^{(\mu+\lambda)(s,R_{\infty})}ds-
\int_{R(0)}^{\infty}4\pi rj(0,r)e^{2\lambda(0,r)}dr.&
\end{eqnarray}
The first integral on the right hand side is negative since matter is ingoing,
the second vanishes when $R_{\infty}$ is large since $p$ has compact support
and the third integral depends only on the initial data.
For the second integral in (\ref{green3}) we note that
\begin{equation}
\int_{-\infty}^{\infty}\int_{0}^{\infty}
\frac{4\pi^2 e^{\mu+\lambda}}{r^{2}E}f(t,r,w,F)dFdw\leq\frac{4\pi}{L_{1}}r^2\rho,
\end{equation}
which in view of (\ref{adm}) implies that it is bounded. It remains to show that the first
integral in (\ref{green3}) is bounded.
Since $1+2\frac{m}{r}e^{2\lambda}=e^{2\lambda}$ we find that the integrand is non-positive
\begin{eqnarray*}
&\displaystyle(\rho+p)e^{2\lambda}\frac{m}{r}+2p-\rho+
je^{2\lambda}&\nonumber\\
&\displaystyle =
-(\rho-p)(1-e^{2\lambda}\frac{m}{r})+(p+j)e^{2\lambda}\leq
0.&\nonumber
\end{eqnarray*}
Here we used that $p\leq\rho, \;e^{2\lambda}m/r\leq 1,$ and that
$p+j\leq 0$ in view of (\ref{p}) and (\ref{j}) since matter is ingoing.
Hence $He^{\hat{\mu}}$ is bounded on $[0,T_1].$ Now, from
the characteristic equation (\ref{char1}) we have that
\begin{equation}
\frac{d}{ds}R^{-1}=\frac{-W}{R^2E}e^{\mu-\lambda}\leq \frac{-W}{R\sqrt{L_1}}e^{\hat{\mu}}\leq
\frac{He^{\hat{\mu}}}{R\sqrt{L_1}}\leq C(T_1)R^{-1}.
\end{equation}
It follows that for any characteristic $(t,R(t),W(t)),\;R(T_1)\geq\epsilon,$ for some $\epsilon>0,$ and Theorem \ref{theorem1}
thus applies which shows that the solution can be extended beyond $t=T_1.$ Hence, if $T<\infty,$ then  $T_1<T,$ and the proof of Theorem \ref{theorem2} is complete.
\begin{flushright}
$\Box$
\end{flushright}

\section{The Einstein-Vlasov system in maximal-isotropic coordinates}
\setcounter{equation}{0}
In \cite{R1} the Einstein-Vlasov system is studied in maximal-isotropic coordinates where the metric reads
\[
ds^2=-\alpha(t,R)dt^2+A^2(t,R)[(dR+\beta(t,R)dt)^2+R^2(d\theta^2+\sin^2{\theta}d\phi^2)].
\]
The condition that the hypersurfaces of constant time are maximal implies, cf. \cite{R1}, that the field equations take the following form:

\begin{eqnarray}
& &\partial_R(R^2\frac{A_{R}}{2\sqrt{A}})=
-\frac18A^{5/2}R^2(\frac32K^2+16\pi\rho)\label{A}\\
& &\alpha_{RR}+\frac{2}{R}\alpha_{R}+\frac{1}{A}A_{R}\,\alpha_{R}=\alpha A^2(\frac32 K^2+4\pi(\rho+p+2p_T))\label{alpha}\\
& &K_{R}+3(\frac{1}{A}A_{R}+\frac{1}{R})K=8\pi j\label{K}\\
& &\beta_{R}-\frac{1}{R}\beta=\frac32 \alpha K\label{beta}\\
& &A_t=-\alpha KA+\partial_R(\beta A)\label{At}\\
& &K_t=-\frac{1}{A^2}\alpha_{RR}+\frac{1}{A^3}A_{R}\alpha_{R}
+\beta K_{R}+4\pi\alpha(2p_T-p-\rho)\nonumber\\
& &\phantom{555555}
+\alpha[-\frac{2}{A^3}A_{RR}+\frac{2}{A^4}(A_{R})^2-\frac{2}{RA^3}A_{R}]\label{Kt}
\label{Kt}
\end{eqnarray}
The indicies $t$ and $R$ denote derivatives but sometimes we also use $\partial_t$ and $\partial_R$.
In \cite{R1} the Vlasov equation is written in different coordinates than we use here. The relation is as follows. Let $x^{i}$ be the coordinates $(R\sin{\theta}\cos{\phi},R\sin{\theta}\cos{\phi},R\cos{\theta})$ and define an orthonormal frame by $e_i=A^{-1}\partial/\partial x^{i}.$ In \cite{R1} the mass shell is coordinatized by $(t,x^{a},v^{i}),$ where $v^{i}$ denote the components of a vector in the orthonormal frame, and the Vlasov equation in \cite{R1} is accordingly written in these variables. Now, let $w=(v\cdot x)/R,$ and let $L=A^2R^2(v^2-(x\cdot v)^2/R^2),$ then the Vlasov equation for the density distribution function $f=f(t,r,w,L)$ takes the form
\begin{equation}
\partial_{t}f+(\frac{\alpha}{A}\frac{w}{E}-\beta)\partial_{R}f
+(-\frac{E\alpha_{R}}{A}+\alpha Kw+\frac{\alpha L}{EA^3R^2}(\frac{1}{R}+\frac{A_{R}}{A}))
\partial_{w}f=0,\label{vlasovmi}
\end{equation}
where
\begin{equation}
E=E(R,w,L)=\sqrt{1+w^{2}+\frac{L}{A^2R^{2}}}.\label{E}
\end{equation}
The matter quantities are defined by
\begin{eqnarray}
\rho(t,R)&=&\frac{\pi}{A^2R^{2}}
\int_{-\infty}^{\infty}\int_{0}^{\infty}Ef(t,r,w,L)\;dwdL,\label{rhomi}\\
p(t,R)&=&\frac{\pi}{A^2R^{2}}\int_{-\infty}^{\infty}\int_{0}^{\infty}
\frac{w^{2}}{E}f(t,r,w,L)\;d
wdL,\label{pmi}\\
j(t,R)&=&\frac{\pi}{A^2R^{2}}
\int_{-\infty}^{\infty}\int_{0}^{\infty}wf(t,r,w,L),\;dwdL,\label{jmi}\\
p_T(t,R)&=&\frac{\pi}{2A^4R^{4}}\int_{-\infty}^{\infty}\int_{0}^{\infty}\frac{L}{E}f(t,r,w,L)\;
dwdL.\label{qmi}
\end{eqnarray}
To ensure asymptotical flatness we impose the boundary conditions
\begin{equation}\label{bdrys}
\lim_{R\to\infty}A(t,R)=1,\;\lim_{R\to\infty}\alpha(t,R)=1,\;\lim_{R\to\infty}\beta(t,R)=0,
\end{equation}
and to ensure a regular centre we require that
\begin{equation}\label{regularctr}
A(t,0)=1,\;\;\forall t\geq 0.
\end{equation}
In the case of Schwarzschild coordinates the constraint equations are easily solved for any given $\open{f}\in{\cal{I}}^D.$ This is not the case in maximal-isotropic coordinates. Here we assume that initial data are given, such that the constraint equations are satisfied, and such that $0\leq\open{f}\in C^1$ has compact support, $\open{f(R,w,L)}=0$ whenever $R<R_1, \;R>R_2$ or if $L>L_2,$ for given $0<R_1<R_2<\infty,$ and $L_2>0.$ Hence, we assume  properties which are similar to the initial data class ${\cal{I}}^C(R_1),$ with the notable difference that the condition (\ref{notsinit}) is not imposed since in these coordinates trapped surfaces are admitted. Let us denote this class of initial data by ${\cal{J}}^C(R_1).$

For a given metric the solution of the Vlasov equation is given by
\begin{equation}
f(t,R,w,L)=f_0({\cal{R}}(0,t,R,w,L),{\cal{W}}(0,t,R,w,L)),
\end{equation}
where $\cal{R}$ and $\cal{W}$ are solutions of the characteristic system
\begin{eqnarray}
\frac{d\cal{R}}{ds}&=&(\frac{\alpha}{A}\frac{\cal{W}}{E}-\beta),\label{charGamma}\\
\frac{d\cal{W}}{ds}&=&-\frac{E\alpha_{R}}{A}+\alpha K{\cal{W}}
+\frac{\alpha L}{EA^{3}{\cal{R}}^2}(\frac{1}{\cal{R}}+\frac{A_{R}}{A}).\label{charW}
\end{eqnarray}
Here ${\cal{R}}(t,t,R,w,L)=R,$ and ${\cal{W}}(t,t,R,w,L)=w.$

\section{A priori bounds in maximal-isotropic coordinates}
\setcounter{equation}{0}
In this section we collect the bounds needed in the next section.
The following estimates are derived in \cite{R1} by using the results of \cite{MO},
\begin{equation}\label{os}
A|K|\leq\frac{2}{R},\;\;\frac{|A_{R}|}{A}\leq\frac{2}{R}.
\end{equation}
Moreover, it follows from equation (\ref{beta}), cf. \cite{R1}, that
\begin{equation}\label{betaestimate}
|\beta|\leq 3.
\end{equation}
Another useful bound follows from conservation of the ADM mass $M,$ which is given by, cf. \cite{R1},
\begin{equation}\label{Mint}
M=\frac14\int_0^{\infty}A^{5/2}R^2(\frac32 K^2+16\pi\rho)\,dR.
\end{equation}
The following estimate plays the role of Lemma 1 in Schwarzschild coordinates.
\begin{lemma}\label{lemma4mi}
Let $(f,\alpha,A,\beta)$ be a regular solution of the Einstein-Vlasov system (\ref{A})-(\ref{regularctr}). Then
\begin{equation}
\int_0^{\infty}\alpha A^2\eta(\frac32 K^2+4\pi(\rho+p+2p_T))\,d\eta\leq 1.
\end{equation}
\end{lemma}
\textit{Proof: }Consider the second order equation (\ref{alpha}) for $\alpha$. The left hand side of
(\ref{alpha}) equals
\[
\frac{1}{R^2A}\partial_R(R^2A\alpha_{R}),
\]
which implies that
\[
\alpha_{R}(R)=\frac{1}{R^2A}\int_0^R\alpha A^3\eta^2(\frac32 K^2+4\pi(\rho+p+2p_T))\,d\eta.
\]
By using the boundary condition $\alpha(t,\infty)=1$ we get
\begin{equation}
1\geq 1-\alpha(0)=\int_0^{\infty}\frac{1}{R^2A}\big(\int_0^R \alpha A^3\eta^2(\frac32 K^2+4\pi(\rho+p+2p_T))\,d\eta\big)dR.
\end{equation}
From equation (\ref{A}) we get that $A$ is monotonic and decreasing. Changing the order of integration
and using the monotonicity of $A$ we get
\begin{eqnarray}\label{alphaA}
1&\geq&\int_0^{\infty}\big(\int_{\eta}^{\infty}\frac{1}{R^2A(\eta)}dR\big)\alpha A^3\eta^2(\frac32 K^2+4\pi(\rho+p+2p_T))\,d\eta\nonumber\\
&=&\int_0^{\infty}\alpha A^2\eta(\frac32 K^2+4\pi(\rho+p+2p_T))\,d\eta.
\end{eqnarray}
\begin{flushright}
$\Box$
\end{flushright}
We also need to establish a decay result for large $R.$
\begin{lemma}\label{lemmadecay}
A solution of the Einstein-Vlasov system (\ref{A})-(\ref{regularctr}) satisfies
\begin{equation}
\int_0^t\frac{\alpha_R(s,R)}{A(s,R)}\,ds\to 0\mbox{ as }R\to\infty.
\end{equation}
\end{lemma}
\textit{Proof: }
Using (\ref{Mint}) together with equation (\ref{A}) we obtain
\[
\frac{A_R}{\sqrt{A}}\geq -\frac{M}{R^2},
\]
which in view of the boundary condition $A(\infty)=1$ implies that
\begin{equation}\label{Aestimate}
A(t,R)\leq (1+\frac{M}{2R})^2.
\end{equation}
From (\ref{charGamma}) and (\ref{betaestimate}) we obtain that all characteristics originating
from the support of the matter have $R(t)\leq R_2+4t=:R_3(t).$
From equation (\ref{K}) we have in view of (\ref{os})
\begin{eqnarray}\label{RAK}
|(R^3A^3K)(t,R)|&=&\big|\int_0^R 8\pi\eta^3 A^4 j\,d\eta\big|
=\big|\int_0^{R_3(t)} 8\pi\eta^3 A^4 j\,d\eta\big|\nonumber\\
&=&|(R_{3}^3A^3K)(t,R_3(t))|\leq 2R_3(t)^2A^2(t,R_3(t)).
\end{eqnarray}
From (\ref{Aestimate}) we have
\[
A(t,R_3(t))\leq (1+\frac{M}{2R_2})^2\leq C.
\]
Thus, since $A\geq 1,$
\begin{equation}\label{Kestimate}
K(R)\leq \frac{CR_3(t)^2}{R^3}=C\frac{(R_2+4t)^2}{R^3}.
\end{equation}
Now we use equations (\ref{Kt}) and (\ref{At}) to derive
\begin{equation}
\partial_t(AKR)=-\frac74 RK^2A+RKA\beta+\frac{2\alpha_R}{A}(1+\frac{A_R}{A}+\frac{A_R^2}{2A^2})+\frac{2\alpha A_R}{A^2},
\end{equation}
where we used (\ref{A}) and (\ref{alpha}) to substitute the second order derivatives of $A$ and $\alpha.$
Since
\[
|\frac{A_R}{\sqrt{A}}|\leq\frac{M}{R^2},
\]
we have for sufficiently large $R$ that
\[
0\geq \frac{A_R}{A}+\frac{A_R^2}{2A^2}\geq -\frac12.
\]
Thus we get
\begin{equation}
0\leq\frac{\alpha_R}{A}\leq \partial_t(AKR)+\frac74 RK^2A-RKA\beta-\frac{2\alpha A_R}{A^2}.
\end{equation}
In view of the decay estimates for $A,\,A_{R}$ and $K$ and the bound of $\beta,$ we obtain
\[
\int_0^t\frac{\alpha_R(s,R)}{A(s,R)}\,ds\to 0\mbox{ as }R\to \infty.
\]
This completes the proof of the lemma.
\begin{flushright}
$\Box$
\end{flushright}

\section{A regularity result in maximal-isotropic coordinates}
\setcounter{equation}{0}
\begin{theorem}\label{theorem3} Let $0<\epsilon<R_1.$ Consider a solution of
the spherically symmetric
Einstein-Vlasov system, launched by initial data in ${\cal{J}}^C(R_1),$
on its maximal time interval $[0,T[$ of existence. If $f(s,R,\cdot,\cdot)=0,$ for
$(s,R)\in [0,t[\times [0,\epsilon],$ then
\begin{equation}\label{Qestimatemi}
Q(t)\leq (Q(0)+\frac{Ct}{\epsilon^2})e^{\frac{C(1+t^2)}{\epsilon}}.
\end{equation}
In particular, if $f(t,R,\cdot,\cdot)=0$ for $(t,R)\in [0,T[\times [0,\epsilon],$ then $T=\infty.$
\end{theorem}
The last statement in the theorem is a consequence of the continuation criterion
derived in \cite{R1}. Indeed, the requirements are that $Q(t)$ and $A(t,0)$ are bounded. Now, since there is no matter in the domain $R\leq\epsilon,$ a bound on $A(t,0)$ follows from equation (\ref{A}) and (\ref{Mint}), cf. the bound (\ref{Aestimate}). The result of the theorem is not new, it is included in \cite{R1}, but as was mentioned in the introduction the method is different and the terms that need to be estimated, in order to obtain global existence without assuming a lower bound of $R,$ are more regular in our approach.\\

\textit{Proof: }The method of proof is to a large extent analogous to the proof in Schwarzschild coordinates. Consider the quantities $G=E(t,{\cal{R}},{\cal{W}})+{\cal{W}}>0$ and
$H=E(t,{\cal{R}},{\cal{W}})-{\cal{W}}>0.$
Along a characteristic $({\cal{R}}(t),{\cal{W}}(t),L)$ we have by
(\ref{charGamma}) and (\ref{charW})
\begin{equation}\label{Gmi}
\frac{dG}{ds}=-[\frac{\alpha_{R}}{A}-\frac{\alpha K{\cal{W}}}{E}]G-\frac{L\alpha K}{2EA^2{\cal{R}}^2}
+\frac{\alpha L}{A^3{\cal{R}}^2E}(\frac{1}{{\cal{R}}}+\frac{A_{R}}{A}),
\end{equation}
and
\begin{equation}\label{Hmi}
\frac{dH}{ds}=[\frac{\alpha_{R}}{A}-\frac{\alpha K{\cal{W}}}{E}]H-\frac{L\alpha K}{2EA^2{\cal{R}}^2}
-\frac{\alpha L}{A^3{\cal{R}}^2E}(\frac{1}{{\cal{R}}}+\frac{A_{R}}{A}).
\end{equation}
First we consider the quantity $H.$
Using equation (\ref{os}) we conclude that
\[
\big|-\frac{L\alpha K}{2EA^3{\cal{R}}^2}
-\frac{\alpha L}{A^3{\cal{R}}^2E}(\frac{1}{{\cal{R}}}+\frac{A_{R}}{A})\big|\leq \frac{C\sqrt{L}}{{\cal{R}}^2}\leq \frac{C}{\epsilon^2}.
\]
Thus

\begin{equation}\label{Hest}
H(t)\leq H(0)e^{\int_0^t \frac{\alpha_{R}}{A}-\frac{\alpha K{\cal{W}}}{E} ds}+ \int_0^t\frac{C}{\epsilon^2}e^{\int_{\tau}^t \frac{\alpha_{R}}{A}-\frac{\alpha K{\cal{W}}}{E} ds}d\tau.
\end{equation}
Let us consider the first of the integrals in the exponent, the second is analogous.
Since
\[
\frac{d{\cal{R}}}{ds}=(\frac{\alpha}{A}\frac{{\cal{W}}}{E}-\beta),
\]
we can write the integral as a curve integral
\[
\int_{\gamma}-(KA)(t,R)dR+(\frac{\alpha_{R}(t,R)}{A(t,R)}-(\beta KA)(t,R))dt,
\]
where $\gamma$ is the curve $(s,{\cal{R}}(s)),\; 0\leq s\leq t.$
Let $\Gamma$ denote the closed curve
$\Gamma:=\gamma+C_{t}+C_{\infty}+C_{0},$ oriented clockwise, where
\begin{eqnarray}
C_{t}&=&\{(t,r):R(t)\leq r\leq R_{\infty}\},\\
C_{\infty}&=&\{(s,R_{\infty}):t\geq s\geq 0\},\\
C_{0}&=&\{(0,r):R_{\infty}\geq r\geq R(0)\}.\label{Ccurves}
\end{eqnarray}
Here $R_{\infty}\geq R_{2}+4t,$ so that $f=0$ when $r\geq R_{\infty}.$
By applying Green's formula in the plane we get
\begin{eqnarray}
&\displaystyle\oint_{\Gamma}-(KA)(t,R)dR
+(\frac{\alpha_{R}(t,R)}{A(t,R)}-(\beta KA)(t,R))dt&\nonumber\\
&\displaystyle =\int\int_{\Omega}\partial_{t}\left(-(KA)(t,R)\right)-\partial_{R}
\left(\frac{\alpha_{R}(t,R)}{A(t,R)}-(\beta KA)(t,R)\right)dtdR=:I_{\Omega}.&\nonumber
\end{eqnarray}
Now we use equation (\ref{Kt}) to substitute for $\partial_tK$ above. We obtain
\begin{equation}
\displaystyle I_{\Omega}=\int\int_{\Omega} \alpha A\big(\frac{2A_{RR}}{A^3}
-\frac{2(A_{R})^2}{A^4}+\frac{2A_{R}}{RA^3}+4\pi(\rho+p-2p_T)\big)+\alpha K^2A
\,dRdt.
\end{equation}
From (\ref{A}) we derive
\[
\frac{A_{RR}}{A^3}=\frac{(A_R)^2}{A^4}
-\frac{A_R}{RA^3}-\frac34 K^2-8\pi\rho.
\]
We thus get
\begin{equation}\label{Iomega}
I_{\Omega}=\int\int_{\Omega} \alpha A\big(\frac{K^2}{4}-\frac{2A_R}{RA^3}
-\frac{R^2A_R^2}{A^4}-4\pi(\rho-p+2p_T)\big).
\end{equation}
\textbf{Remark 2: }The quasi local mass $m$ is given by
\[
m=\frac{r}{2}(1-|\nabla r|^2).
\]
Here $r$ is the area radius, and $\nabla r$ is the gradient of $r.$
In our case $r=AR,$ which implies that
\begin{equation}\label{Hawking}
m=\frac{R^3A^3K^2}{8}-R^2A_{R}-\frac{R^3(A_{R})^2}{2A}.
\end{equation}
It thus follows that $I_{\Omega}$ can be written as
\begin{equation}\label{GC}
I_{\Omega}=\int\int_{\Omega} \alpha A\big(\frac{2m}{A^3R^3}-4\pi(\rho-p+2p_T)\big)\,dRdt.
\end{equation}
This is thus identical to the structure of the corresponding term in Schwarzschild coordinates, cf. Remark 1. Here we will however stick to the form (\ref{Iomega}) for the estimates. \\
To summarize, we have in view of (\ref{Hest}) obtained the estimate
\begin{equation}\label{Is}
H(t)\leq H(0)e^{I_{\Omega}-I_{C_t}-I_{C_{\infty}}-I_{0}},
\end{equation}
where
\begin{eqnarray}
I_{C_t}&=&-\int_{{\cal{R}}(t)}^{R_{\infty}}(KA)(t,R)\, dR,\\
I_{C_0}&=&\int_{{\cal{R}}(0)}^{R_{\infty}}(KA)(0,R)\, dR,
\end{eqnarray}
and
\begin{equation}
I_{C_{\infty}}=\int_0^t \frac{\alpha_{R}(s,R_{\infty})}{A(s,R_{\infty})}
-(\beta KA)(s,R_{\infty}))\,ds.
\end{equation}
We now invoke the a priori bounds derived in the previous section. Since ${\cal{R}}(s)\geq
\epsilon,\, 0\leq s\leq t,$ we have in view of Lemma \ref{lemma4mi},
\[
I_{\Omega}\leq \int\int_{\Omega} \alpha A\frac{2m}{A^3R^3}\,dRdt\leq \frac{2t}{\epsilon}.
\]
Using the bounds (\ref{os}), (\ref{Aestimate}) and (\ref{Kestimate}) we obtain
\[
I_{C_t}\leq \frac{C(1+t^2)}{\epsilon}.
\]
The integral $I_{C_{\infty}}$ vanishes in view of Lemma \ref{lemmadecay} and
$I_{C_0}$ depends only on the initial data. Since the second term in (\ref{Hest}) can
be estimated in the same way we obtain
\[
H(t)\leq (H(0)+\frac{Ct}{\epsilon^2})e^{\frac{C(1+t^2)}{\epsilon}}.
\]
The estimate for $G$ is analogous.
Thus
\[
G(t)\leq (G(0)+\frac{Ct}{\epsilon^2})e^{\frac{C(1+t^2)}{\epsilon}},
\]
and we get
\[
Q(t)\leq (Q(0)+\frac{Ct}{\epsilon^2})e^{\frac{C(1+t^2)}{\epsilon}}.
\]
This completes the proof of Theorem \ref{theorem3}.
\begin{flushright}
$\Box$
\end{flushright}


TITLE: Surreal numbers and function growth
QUESTION [3 upvotes]: I have read in several sources (I didn't keep the references) that surreal numbers are well suited for describing function growth, but none of the texts I have seen went into any detail on this nor did they provide a reference. Google searches on "surreal functions function growth" only got me to the same point. What would be a good starting point to read about this?
Update: I would primarily be interested in simple cases (up to generation $ω$ or, say, $ω^ω$) and only then generalizations to more generic cases. I would expect either something along the lines of
$$\begin{aligned}
u&&\cdots&&&x^u&&(u\text{ real}) \\
ω&&\cdots&&&e^x \\
ε&&\cdots&&&\ln x \\
\end{aligned}$$
(but I don't know what multiplication of surreals would correspond to in this model to reflect that $ω.ε = 1$) or perhaps even
$$\begin{aligned}
u\in\mathbb{R}&&\cdots&&&\text{finite limit} \\
ω&&\cdots&&&x \\
ε&&\cdots&&&1/x \\
ω^2-2ω+3+ε&&\cdots&&&x^2-2x+3+1/x
\end{aligned}$$
(which looks somewhat more consistent but simply replacing $ω$ by $x$ would break at e.g. $ω^ε = \ln ω$).
Introductory stuff is welcome.

REPLY [4 votes]: I think the clearest account of this idea can be found in this article of Aschenbrenner, van den Dries and van der Hoeven. It is a sort of survey article, easy to read.
The basic ideas are:
-A good framework to speak of the growth of function is that of Hardy fields which are ordered field of sufficiently differentiable real valued functions whose elements and their derivatives are monotonous on neighborhoods of $+\infty$.
-Hardy fields are hard to study, but it is possible in certain cases to describe their general properties (from a model-theoretic standpoint) using formal series called transseries whose model theory is known.
-Transseries are formal but can act as functions on fields of formal series, thus giving a direct link between the growth of a function and the series which represents it.
-Finding correspondences between formal series and functions can give much information about the latter, whereas formal series are subject to more computations and algorithmic methods.
-Transseries can be naturally interpreted as surreal numbers.
-Describing certain functions (with transexponential growth for instance) cannot be done using regular transseries but requires additional types of series, called hyperseries.
-It is conjectured that such hyperseries can also be naturally interpreted as surreal numbers. Provided every surreal number can be represented as a hyperseries, it is believed that the resulting structure has good closure properties in a similar way as regular transseries have good closure properties.

In any case, the middle step between surreal numbers and functions is the notion of transseries, or hyperseries.

At a more basic level, your second line of correspondences is correct in that this is how the functions on the right side are represented as surreal numbers in the natural correspondance between transseries and certain surreal numbers.
This correspondance is the only one which fixes real constants, sends $x$ onto $\omega$ and commutes with $\exp$, $\log$ and with transfinite sums.
The surreal numbers below generation $\omega^{\omega}$ are just sums $\sum \limits_{n \in \mathbb{N}} r_n \omega^{d_n}$ where $(d_n)_{n \in \mathbb{N}}$ is a strictly decreasing sequence of dyadic numbers and $(r_n)_{n \in \mathbb{N}}$ is a sequence of real numbers. Those correspond to formal generalized power series $\sum \limits_{n \in \mathbb{N}} r_n x^{d_n}$ which can act as functions on subclasses of $\mathbf{No}$. Notice that this already flows outside of the class of classically convergent series.
If you allow length $\omega^{\omega}$, you get $\omega^{\omega}=\exp \omega$ itself, which corresponds to the exponential function or the transseries $\operatorname{e}^x$; or $\omega^{\frac{1}{\omega}}=\log \omega$ which corresponds to $\log x$. You also obtain $\omega^{-\omega} \equiv \operatorname{e}^{-x}$, the numbers $\omega^r\equiv x^r$ for non-dyadic $r \in \mathbb{R}$ and the products $\omega^{d\pm\frac{1}{\omega}}\equiv x^d (\log x)^{\pm 1}$ where $d$ is dyadic.
[edit: the following is actually false: I believe every number of generation $<\varepsilon_0$ corresponds to a so-called exponential-logarithmic transseries but I don't have a proof right now. Generation $\varepsilon_0$ brings a lot of more complicated behavior which cannot be described by regular transseries, but rather "nested transseries" $\sqrt{\omega}+\operatorname{e}^{\sqrt{\log \omega}+\operatorname{e}^{\sqrt{\log \log \omega}+\operatorname{e}^{\cdot^{\cdot^{\cdot}}}}}$ or transexponential hyperseries such as $\varepsilon_0 \equiv \operatorname{e}^{\operatorname{e}^{\operatorname{e}^{\cdot^{\cdot^{\cdot}}}}}$.]


TITLE: Will Friction on a cylinder embedded into the earth differ if the method of removal is changed
QUESTION [1 upvotes]: Would it make a difference if a cylinder embedded into the ground is
1)pulled vertically or,
2) removed by rotating with the same force

REPLY [0 votes]: I think this is an interesting question. Our intuition with pulling objects out of the ground is that twisting, or rocking back and forth works better than just pulling straight out.
Some things to think about:

Static friction is usually higher than dynamic or kinetic friction.
The friction force is proportional to the area and the normal force. How you work out the normal force would depend on the situation.
Is the cylinder open or closed at the end. If closed there is probably a suction force as a vacuum is formed, depending on how well the seal is formed. Wet or moist earth would have a better seal than say dry sand. If the earth is sticking to the cylinder you also would not be pulling just the cylinder but the mud or earth with it and that would change the problem. If the soil was more granular, or if the soils was more like a liquid and viscous would change the problem.
We can choose to apply a force pulling the rod directly up, or apply the force radially. The torque being force times the distance from the center of the cylinder. So it is relatively easy to get a larger torque by applying the force further from the center.
A long thin cylinder would be different than a short fat cylinder when it comes to the suction forces.  It would also change the moment of inertia.

So practically speaking, since the area in contact with the earth and presumably the normal forces are the same in both cases. The static friction would be the same, but for a long thin cylinder when you try to rotate, you are not creating a suction force at the base of the cylinder. You may also be able to have a higher torque by the way you apply the force. By breaking the static friction, and rotating first it would be easier to start to pull the rod up since the dynamic friction is less.  Even better,  if a path is created that allow the suction forces to be reduced by equalizing the pressure.


text:
\section{Conclusions}

We propose a novel framework of decentralized algorithms for smooth and strongly convex objectives. The framework provides a unified viewpoint of several well-known decentralized algorithms and, when instantiated with AGD,
achieves optimal convergence rates in theory and state-of-the-art performance in practice. We leave further generalization to (non-strongly) convex  and non-smooth objectives to future work.

config: arxiv


TITLE: Definition of Maximal atlas
QUESTION [11 upvotes]: I some how could not find the definition of maximal atlas on a manifold.
What I see is that an atlas is said to be maximal atlas if it is not contained in any other atlas.
What does this containment actually mean?
Let $\mathcal{A}$  be an atlas and $\mathcal{B}$ be another atlas. When do we say that $\mathcal{A}$ is contained in $\mathcal{B}$?
I was not able to find definition of this.
Another confusion is about union of atlases. Let $\mathcal{A}$ and $\mathcal{B}$ be two atlases. What do we mean by union of atlases? Is it just the union $\{(U,\phi)_{\phi\in \mathcal{A}},(V,\psi)_{\psi\in \mathcal{B}}\}$?
It may happen that this union is not an atlas i.e., there can be two charts $\phi_\mathcal{A}$ and $\psi_{\mathcal{B}}$such that $\phi_{\mathcal{A}}$ and $\psi_{\mathcal{B}}$    are not compatible.
By maximal atlas do I mean an atlas $\mathcal{A}$ such that for any other atlas $\mathcal{B}$, the union as above is not an atlas?
Any reference for the definition is welcome.

REPLY [13 votes]: "Contain" and "union" here literally mean just that.  An atlas $\mathcal{A}$ is a set of charts $(U,\phi)$, and $\mathcal{A}$ is contained in $\mathcal{B}$ if $\mathcal{A}\subseteq\mathcal{B}$: that is, if every chart which is an element of $\mathcal{A}$ is also an element of $\mathcal{B}$.  The union of two atlases is just the set $\mathcal{A}\cup\mathcal{B}$, which as you observe may not be an atlas.
An atlas $\mathcal{A}$ is called maximal if there does not exist any atlas $\mathcal{B}$ such that $\mathcal{A}\subset\mathcal{B}$ (with a strict inclusion).  This is equivalent to saying that if $\mathcal{B}$ is an atlas such that $\mathcal{A}\cup\mathcal{B}$ is an atlas, then $\mathcal{B}\subseteq\mathcal{A}$.


TITLE: Why is formally unramified preserved under base change?
QUESTION [1 upvotes]: Le $f:X\to S$ be a formally unramified morphism of schemes and $T\to S$ another morphism of schemes. I'd like to prove that the base change $X_T\to T$ stays formally unramified. So let $Y$ be a $T$-scheme with $Y_0$ a first order thickening,  $f,g:Y\to X_T$ with $f\circ i=g\circ i$ so we have $\alpha\circ f\circ i=\alpha\circ g\circ i$ so using the formal unramification of $X$ one gets that $\alpha\circ f=\alpha\circ g$ and the following diagram:
$\require{AMScd}$
\begin{CD}
X @<\alpha<< X_T     @<<<  Y_0\\
@VV\lambda V   @VV\beta V @VViV\\
S     @<\gamma <<  T @<\mu<< Y
\end{CD}
Question: how can I conclude that $f=g$? The morphism $\alpha$ is not a monomorphism?
My definition of formally unramified: for every affine $S$-scheme $Y$ and for every closed subscheme $Y_0\subseteq Y$ given by a quasi-coherent ideal $\mathcal{I}\subseteq\mathcal{O}_Y$ of square $\mathcal{I}^2=0$, the canonical map $\text{Hom}_S(Y,X)\to\text{Hom}_S(Y_0,X)$ is injective.

REPLY [1 votes]: The key idea here is the universal property of the fiber product: a morphism $Y\to X_T$ is uniquely equivalent to two morphisms $Y\to X$ and $Y\to T$ which agree when composed with the morphisms $X\to S$ and $T\to S$. If we have a morphism $Y\to X_T$ making the diagram commute, it must be the morphism induced by the universal property of the fiber product by the morphisms $Y\to T$ and $Y\to X$. As the morphism $Y\to T$ is specified and there is at most one morphism $Y\to X$ making the diagram commute (as $X\to S$ is formally unramified), we see that if there is a morphism $Y\to X_T$ making the diagram commute, it is unique.


TITLE: Given limit f(x) = L at x=c , does the function f(x) "always" need to be defined in the close neighborhood of c?
QUESTION [0 upvotes]: According to the epsilon-delta definition of a limit,
L is limit of f(x) at c if
For every arbitrary ε>0 we can find δ>0 such that
for "all" x satisfying c-δ<x<c+δ (except c possibly), f(x) lies in (L-ε,L+ε)
does this definition imply that f(x) has to be defined at "all" points in the interval
(c-δ,c+δ)?

REPLY [1 votes]: Yes and no. Any function $f:A\to B$ of course has to be defined on all of $A$, otherwise it is not a function. However, you may defined $f:[0,1] \to \mathbb{R}$ and still pick $c=0$ and, say, $\delta = 2$. This does not mean that $f$ must be defined on the interval $(-2,0)$, however: The definition says more precisely that for $x\in A$ satisfying $c-\delta<x<c+\delta$, $f(x)$ has to be $\epsilon$-close to $f(c)$. With the quantifier $x\in A$, we do not run into any problems.


TITLE: For $x \geq 2$, prove the inequality $(x+1) cos \frac{ \pi}{x+1}-x cos \frac{\pi}{x}>1$
QUESTION [0 upvotes]: For $x \geq 2$, prove the inequality $(x+1) cos \frac{ \pi}{x+1}-x cos \frac{\pi}{x}>1$.

I am not sure where to begin. A hint is helpful.

REPLY [2 votes]: The mean value theorem
$$g(a) - g(b)= g'(\xi)(a - b)$$ where $\xi\in\left[a,b\right]$.
In your case the function $g(x)=x\cos\left(\frac{\pi}{x}\right)$
Put $a=x+1, b=x$, and get the bound on the derivative.
The derivative is
$$g'(\xi)=\cos\left(\frac{\pi}{\xi}\right)+\left(\frac{\pi}{\xi}\right)\sin\left(\frac{\pi}{\xi}\right)$$
We have to show that $g'(\xi)\gt 1$ for $\xi \in [2,\infty)$.
Claim: $g'(\xi)$ is a monotonically decreasing function. To show this, we have to show that $g''(\xi)$ is always negative for $\xi \in [2,\infty)$.
$$g''(\xi)=-\frac{\pi}{\xi^{3}}\cos\left(\frac{\pi}{\xi}\right)$$
$g''(\xi)\leq 0$ for $\xi \in [2,\infty)$, where equality holds only at $\xi=2$. This implies that $g'(\xi)$ is a decreasing function in $\xi \in [2,\infty)$.
Also observe that
$$g'(2)=\frac{\pi}{2}$$ and $$\\\lim_{\xi\to\infty}g'(\xi)\to 1$$
The function $g'(\xi)$ decreases monotonically from $\frac{\pi}{2}$ at $\xi=2$ and approaches 1 as $\xi$ approaches $\infty$.  In other words,
 $g'(\xi)\gt 1$ for $\xi \in [2,\infty)$.


TITLE: proof that area of convergence is bounded by a circle
QUESTION [2 upvotes]: I'm currently looking into the topic of holomorphic functions and their radii of convergence. While I do understand according to the Cauchy's integral formula why a Taylor series converges in a radius r, which is the distance to the nearest singularity from the centre, then I do not understand why the shape of the area has to be precisely a disk, bounded by a circle, contrary to e.g. an ellipse without encompassing the singularities. How to prove that the power series diverges at any point at a distance greater than r?

REPLY [1 votes]: This has nothing to do with the Cauchy Integral Formula or any other complex analysis, it's just simple  inequalities.
Let $R$ be the supremum of $|z|$ such that $\sum a_nz^n$ converges. The definition of $R$ shows that

If $|z|>R$ then $\sum a_nz^n$ diverges.

In fact anything can happen for $|z|=R$, but

If $|z|<R$ then $\sum a_nz^n$ converges.

Proof: The definition of $R$ shows that there exists $w$ with $|w|>|z|$ such that $\sum a_na^n$ converges. Hence $a_nw^n\to0$, so there exists $c$ with $|a_nw^n|\le c$ for all $n$. Now $$\sum|a_nz^n|\le c\sum|z/w|^n<\infty,$$since $|z/w|<1$.


TITLE: Existence of categorical quotient $X/\mathbb{G}_{m,A}$.
QUESTION [1 upvotes]: Let $A$ be an $\bar{\mathbb{F}}_p$-Algebra of finite type (one might assume $A$ to be reduced). Let $X \subset \mathbb{A}_A^d\backslash \{0\}$ be a closed $A$-subscheme together with a group action of $\mathbb{G}_{m,A}$ induced from the action on $\mathbb{A}_A^d\backslash \{0\}$. Under those assumptions, does there exist a categorical quotient $X/\mathbb{G}_{m,A}$ which is a finite type scheme over $A$?
I think this should work in the case $A= \bar{\mathbb{F}}_p$, as $\mathbb{G}_{m,A}$ is then linearly reductive. But I know too few (references) for more general $A$-group scheme actions.

REPLY [1 votes]: You can just take this to be the image of $X$ in $\mathbb{P}_A^{d - 1}$.
To express the above comment about "image" in more precise terms, since for any ring $A$ at all the natural map$$\pi: \mathbb{A}_A^d - \{0\} \to \mathbb{P}_A^{d - 1}$$is a $\mathbb{G}_{m, A}$-torsor, a closed subscheme $X$ of $\mathbb{A}_A^d - \{0\}$ is $\mathbb{G}_{m, A}$-stable if and only if the associated ideal sheaf is stable under the descent datum for $\pi$. Hence, descent theory for quasicoherent sheaves provides a unique closed subscheme $Z \subset \mathbb{P}_A^{d - 1}$ over which $X$ is a $\mathbb{G}_{m, A}$-torsor and it is the desired "quotient" of $X$ (over $A$) in all good senses.


TITLE: A counter-example of a set containing a club which is not a club
QUESTION [2 upvotes]: I wanted to prove that if $C$ is a club (in a $\kappa$ cardinal) and $C\subseteq C'$ then $C'$ is also a club (that is if $C$ is of "mesure" $1$ then $C'$ is too). It is easy to see that $C'$ is unbounded. But I have some problems proving the closeness of $C'$. So, I try to find a $C'$ that is not a club!
Could somebody help me? thanks.

REPLY [4 votes]: First observe that it is true for any club $C$ such that $\mathrm{Lim}\cap\kappa\subseteq C$, since $C$ already contains all the limit points possible.
However if $\delta<\kappa$ is a limit ordinal and $\delta\notin C$ we can take $C'=C\cup\delta$, then in $\delta$ we have that $\sup(C'\cap\delta)=\delta\notin C'$, so it is not a club.


TITLE: Image of $A^\frac{1}{2}$ equal to image of $A$
QUESTION [3 upvotes]: Let $A\in B(H)$ be a closed range positive operator, and $A^\frac{1}{2}$ its square root. Clearly image of $A$ is a subset of image of $A^\frac{1}{2}~~$($R(A)\subset R(A^\frac{1}{2})$). How do we prove $R(A)= R(A^\frac{1}{2})$?
I edite the question

REPLY [5 votes]: This is false. Set $H = \ell^2(\mathbb{N})$ and let $A$ be the multiplication operator defined by
$$ A \left( (x_n)_{n=1}^{\infty} \right) = \left( \frac{x_n}{n} \right)_{n=1}^{\infty}. $$
Then
$$ A^{\frac{1}{2}} \left( (x_n)_{n=1}^{\infty} \right) = \left( \frac{x_n}{\sqrt{n}} \right)_{n=1}^{\infty} $$
and $\left( \frac{1}{n^{1.5}} \right)_{n=1}^{\infty} \in R(A^{\frac{1}{2}})$ but not in $R(A)$.

If the range of $A$ is closed then the result is true. To see why, note that if $x \in \ker(A^{\frac{1}{2}})$ then $Ax = A^{\frac{1}{2}} A^{\frac{1}{2}} x = 0$ so $x \in \ker(A)$ while if $x \in \ker(A)$ then
$$ \left< A^{\frac{1}{2}}x, A^{\frac{1}{2}}x \right> = \left< x, Ax \right> = 0 $$
so $x \in \ker(A^{\frac{1}{2}})$. Now,
$$ R(A) \subseteq R \left(A^{\frac{1}{2}} \right) \subseteq \overline{ R \left( A^{\frac{1}{2}} \right)} = \ker \left( A^{\frac{1}{2}} \right)^{\perp} = \ker(A)^{\perp} = \overline{R(A)} = R(A) $$
and so we get $R(A) = R \left(A^{\frac{1}{2}} \right)$.


TITLE: Directional derivative
QUESTION [5 upvotes]: The governor Ralph has trouble on the bright side of Mercury. The temperature in the wall of the vessel, when it is in the position $(x, y, z)$ is given by $T(x, y, z)=e^{-x^2-2y^2-3z^2}$, where $x$, $y$ and $z$ are measured in meters. Now it is located at $(1,1,1)$.

In what direction should it move to lower the temperature as quickly as possible?
If the vessel is traveling with velocity $e^8$ per second, how quickly will temperature be decreased if it moves in that direction?
Unfortunately the metal of the wall will break if it cooled with rate greater than $\sqrt{14}e^2$ degrees per second. Describe the set of possible directions where it can move to reduce the temperature at a rate not greater than the permissible.

$$$$
I have done the following:

The direction to that the vessel should move so that the temperature decreases as fast as possible is given by :
$$-\nabla T(1, 1, 1)$$
Is this correct??
So, we have the following:
$$\nabla T(x, y, z)=(-2x e^{-x^2-2y^2-3z^2}, -4ye^{-x^2-2y^2-3z^2}, -6ze^{-x^2-2y^2-3z^2}) \\ \Rightarrow -\nabla T(1, 1, 1)=(2e^{-6}, 4e^{-6}, 6e^{-6})$$
Is this correct??
Are we looking for the rate of change at the direction we found at the question $1$ ??
So, is it as followed?? $$grad f \cdot \overrightarrow{v}$$ where $\overline{v}$ is the unit vector of the qquestion $1$ ?? But how can we use the fact that the velocity is $e^8$ ??
Could you give me some hints what we could do ??

$$$$
EDIT:
At the question $3$ are we looking for the directions $\overrightarrow{v}$ such that $$\nabla T \cdot \overrightarrow{v}\leq \sqrt{14}e^2$$ ??

REPLY [2 votes]: Hints:

A direction is usually given by a unit vector. The direction to move to get the fastest cooling would be
$$
-\frac{\nabla T(1,1,1)}{\|\nabla T(1,1,1)\|}
$$
which is the direction of $-\nabla T(1,1,1)$; so, mod the magnitude, your thinking is correct. The direction would therefore be
$$
\frac{(1,2,3)}{\sqrt{14}}
$$
The velocity described by a speed of $e^8$ in the direction given in 1. is
$$
e^8\frac{(1,2,3)}{\sqrt{14}}
$$
The rate of change of $T$ moving at that velocity is
$$
\begin{align}
e^8\frac{(1,2,3)}{\sqrt{14}}\cdot\nabla T(1,1,1)
&=-e^8\frac{\nabla T(1,1,1)}{\|\nabla T(1,1,1)\|}\cdot\nabla T(1,1,1)\\[6pt]
&=-e^8\|\nabla T(1,1,1)\|
\end{align}
$$
Let the direction of travel be $d$, where $|d|=1$. The rate of change of $T$ achieved by moving at speed $e^8$ in direction $d$ is
$$
e^8d\cdot\nabla T(1,1,1)
$$
Now, since $u\cdot v=|u|\,|v|\cos(\theta)$, where $\theta$ is the angle between the directions of $u$ and $v$, the rate of change of $T$ is
$$
e^8\|\nabla T(1,1,1)\|\cos(\theta)
$$
so we need to find $\theta$ so that
$$
-\sqrt{14}e^2\le e^8\|\nabla T(1,1,1)\|\cos(\theta)\le\sqrt{14}e^2
$$


TITLE: Chances when throwing a "D4" (tetrahedron).
QUESTION [1 upvotes]: This is a slightly more advanced example of this question, but instead of having 6 tails or heads resulting from six consecutive coin throws we want to hit all the different 4 sides of D4 tetrahedron dice. We can assume each side has equal $p=1/4$ probability of coming up and that result of consecutive throws are independent.
How many times do we need to throw the D4 to say with 50% probability that all 4 sides has come up in sequence after each other (in any order) at least once?
I am interested in both theoretical approaches and practical methods.

Examples of allowed orders:

1,1,2,1,2,3,4,3,2,...
1,2,1,4,3,2,1,3,3,...
4,3,3,3,2,4,1,2,1,...

REPLY [2 votes]: The system is in state $S_k$, $1\leq k\leq4$, if there are $k$ useful last digits. After the first throw we are with probability $1$ in state $S_1$. State $S_4$ is the terminal state. The Markov matrix for this system is
$$A=\left[\matrix{{1\over4}&{1\over4}&{1\over4}&0\cr
{3\over4}&{1\over4}&{1\over4}&0\cr
0&{1\over2}&{1\over4}&0\cr
0&0&{1\over4}&1\cr}\right]\ .$$
This should be read as follows: When we are in state $S_1$, with probability ${1\over4}$ the next throw leaves us in $S_1$, and with probability ${3\over4}$ the next throw brings us to  $S_2$. When we are in state $S_2$, with probability ${1\over4}$ the next throw brings us back to  $S_1$, with probability ${1\over4}$ the next throw leaves us in $S_2$, and with probability ${1\over2}$ the next throw brings us to $S_3$. Etcetera.
We now have to compute the sequence $$x^{(n)}:=A^{n-1}\left[\matrix{1\cr0\cr0\cr0\cr}\right]\qquad(n\geq1)\ ,$$ and find the first $n$ for which $x^{(n)}_4\geq50\%$. The computation shows that this is the case for $n=12$.


TITLE: multiplying two multi-valued quantities
QUESTION [1 upvotes]: We know that $\sqrt {-1}=\{i,-i\}$, then $\sqrt {-1} \sqrt {-1}=\{i,-i\}\{i,-i\}=\{1,-1\}$ or $\sqrt {-1} \sqrt {-1}=\{i^2, (-i)^2\}=\{-1\}$?

REPLY [0 votes]: Following Omar's advice, $\sqrt{-1}=e^{i(2k+1)\pi/2}$ for $k\in\mathbb{Z}$, we get
$\sqrt{-1}\sqrt{-1}=e^{i(2m+1)\pi/2} e^{i(2n+1)\pi/2}$,
and not $= e^{i(2k+1)\pi/2} e^{i(2k+1)\pi/2}$.
And that simplifies to $e^{ik\pi}=\{1,-1\}$.
This has connection to the anomally that
in general, $\log u+\log u\neq 2\log u$ for a complex numbers $u$.


TITLE: How to find the maximum and minimum of the function $f(x) = \frac{3x}{x^2 -2x + 4}$
QUESTION [2 upvotes]: How would one find the maximum and minimum of such a function: $$f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto f(x) = \frac{3x}{x^2 -2x + 4}$$
I have just been introduced to functions in my calculus class (actually I missed that lectures), and I have more or less have the intuition of maximum and minimum of a function: the maximum should be the greatest $f(x)$ value in a certain range and the minimum the smallest $f(x)$ value.
My first question is: does it makes sense to talk about maximum and minimum of a function in general (instead of just in a certain subset of the domain of the function)?
I need to use maxima to find the maximum and minimum, and then I need to show formally that they are really the maximum and minimum.
My second question is: does anybody know how to find the maximum and minimum of a function using wxMaxima?
My third question is: I know the definition of maximum (and minimum), i.e. a number that is greater or equal (smaller or equal) to all other numbers. My problem is that I am not seeing how would I show it formally.
I used Wolfram Alpha to calculate the maximum and the minimum, and it says that the minimum is $-\frac{1}{2}$ at $-2$ and the maximum is $\frac{3}{2}$ at $2$.

REPLY [0 votes]: If $x\ne0,$ $$\frac{3x}{x^2-2x+4}=\dfrac3{x-2+4/x}$$
If $x>0,x-2+4/x=\left(\sqrt x-2/\sqrt x\right)^2+4-4\ge2$
Consequently, $$\frac{3x}{x^2-2x+4}\le\dfrac32$$
If $x<0,$ let $x=-y^2,x-2+4/x=-y^2-2-4/y^2=-2-\left(y-2/y\right)^2-4\ge-6$
Consequently, $$\frac{3x}{x^2-2x+4}\ge\dfrac3{-6}=-\dfrac12$$


TITLE: Determining if a linear equation system only has a specific solution given a condition.
QUESTION [0 upvotes]: I have this exercise from the linear algebra course I'm currently taking. However I'm somewhat stuck.
Let $\alpha$ and $\beta$ be real numbers. Show that the linear equation system
\begin{equation*}
  \begin{pmatrix}
    1 & 2\\
    2 & 3\\
    3 & 4
  \end{pmatrix}x=
  \begin{pmatrix}
    \alpha+2\beta\\
    \alpha+2\beta\\
    2\alpha+3\beta
  \end{pmatrix},
\end{equation*}
has solutions if and only if $\alpha+\beta=0$.
Okay, so far I've been able to show, that if $\alpha+\beta=0$, then $\alpha=-\beta$, so the RHS can be written as
$$\begin{pmatrix}
    -\alpha\\
    -\alpha\\
    -\alpha
  \end{pmatrix}$$
I can indeed solve the equation when with gaussian elimination, but I'm not sure how I can show, that the system only has a solution, when $\alpha+\beta=0$. I could come with examples where where is no solution, but I'm unsure how to show it in general for all cases.

REPLY [1 votes]: You can proceed with the Gaussian elimination keeping the $\alpha$ and $\beta$ in place as variables. The result of the process gives you
$$\begin{pmatrix}
1 & 2 \\
0 & -1 \\
0 & 0
\end{pmatrix}x =
\begin{pmatrix}
\alpha + 2\beta \\
-\alpha - 2\beta \\
\alpha + \beta
\end{pmatrix}$$
This system clearly has solutions if and only if the last row on the right side is $0$, which is if and only if $\alpha + \beta = 0$.


TITLE: Composition of two Mobius maps on the unit disc in $\Bbb C$?
QUESTION [1 upvotes]: Let $a,b \in \Bbb D$, the unit disc in the complex plane, and let $\varphi_{a}(z)$ and $\varphi_{b}(z)$ be two typical Mobius maps on $\Bbb D$. Is there a well known formula for the composition of these two maps? Is it another Mobius map?

REPLY [2 votes]: Assuming OP is referring to the unit disk automorphisms (special case of Möbius transformations):
$\varphi$ is a unit disk Automorphism, meaning $\varphi_a(z) = \lambda\frac{z-a}{1-\overline{a}z}$ with $|\lambda|=1 $.
The composition of two unit disk automorphisms remains a unit disk automorphism. You can calculate it exactly by using the matrix representation of Möbius maps, and the property that $T_A\circ T_B = T_{AB}$ for $T$ Mobius and $A,B$ matrices.
Another way to look at it is that $\varphi_a(z)=0$ iff $z=a$. Therefore,
$\varphi_a(z) \circ \varphi_b(z)$ = $\varphi_a(\varphi_b(z))=0$ iff $\varphi_b(z)=a$. Now taking $\varphi_b^{-1}$ on both sides [which is equal $\varphi_{-b}$] we get $a=\varphi_{-b}(z)$.
So $\varphi_a(z) \circ \varphi_b(z) = \varphi_{\varphi_{-b}(a)}(z)$.


text:
\section{Models and background}
\label{sec:models}

This paper studies an infinite-horizon MDP with discounted rewards, as represented by a quintuple $\mathcal{M} = (\cS,\cA, P, r,\gamma)$. Here, $\cS$ and $\cA$ denote respectively the (finite) state space and action space, whereas $\gamma\in (0,1)$ indicates the discount factor.
We use $P:\cS\times\cA \rightarrow \Delta(\cS)$ to represent the probability transition kernel of the MDP, where for each state-action pair $(s,a)\in \cS\times \cA$, $P(s' \mid {s,a})$ denotes the probability of transiting to state $s'$ from state $s$ when action $a$ is executed. The reward function is represented by $r:  \cS\times\cA \rightarrow [0,1]$, such that $r(s,a)$ denotes the immediate reward from state $s$ when action $a$ is taken; for simplicity, we assume throughout that all rewards lie within $[0,1]$.
We focus on the tabular setting which, despite its basic form, is not yet well understood.
See \cite{bertsekas2017dynamic} for an in-depth introduction of this model.

\paragraph{Q-function and the Bellman operator.}
An action selection rule is termed a {\em policy} and represented by a mapping $\pi: \cS \rightarrow \Delta(\cA)$, which maps a state to a distribution over the set of actions.
A policy is said to be {\em stationary} if it is time-invariant.
We denote by $\{s_t,a_t,r_t\}_{t=0}^{\infty}$ a sample trajectory, where $s_t$ (resp.~$a_t$) denotes the state (resp.~the action taken), and $r_t= r(s_{t}, a_{t})$ denotes the reward received at time $t$. It is assumed throughout that the rewards are deterministic and depend solely upon the current state-action pair.
We denote by $V^{\pi}: \cS \rightarrow \real$ the value function of a policy $\pi$, namely,

\begin{align*}
 \forall s\in \cS: \qquad V^{\pi}(s) \defn \mathbb{E} \left[ \sum_{t=0}^{\infty} \gamma^t r(s_t,a_t ) \,\big|\, s_0 =s \right],
\end{align*}

which is the expected discounted cumulative reward received when \emph{(i)} the initial state is $s_0=s$, \emph{(ii)} the actions are taken based on the policy $\pi$ (namely, $a_t \sim \pi(s_t)$ for all $t\geq 0$) and the trajectory is generated based on the transition kernel (namely, $s_{t+1}\sim P(\cdot | s_t, a_t)$). It can be easily verified that $0\leq V^{\pi}(s)\leq \frac{1}{1-\gamma} $ for any $\pi$.
The action-value function (also Q-function) $Q^{\pi}: \cS \times \cA \rightarrow \real$ of a policy $\pi$ is defined by

\begin{equation*}
 \forall (s,a)\in \cS \times \cA: \qquad Q^{\pi}(s,a) \defn \mathbb{E} \left[ \sum_{t=0}^{\infty} \gamma^t r(s_t,a_t ) \,\big|\, s_0 =s, a_0 = a \right],
\end{equation*}

where the actions are taken according to the policy $\pi$ except the initial action (i.e.~$a_t \sim \pi(s_t)$ for all $t\geq 1$).
As is well-known, there exists an optimal policy --- denoted by $\pi^{\star}$ --- that simultaneously maximizes $V^{\pi}(s)$ and $Q^{\pi}(s,a)$ uniformly over all state-action pairs $(s,a)\in (\cS\times\cA)$.
Here and throughout, we shall denote by $V^{\star} :=V^{\pi^{\star}}$ and $Q^{\star} := Q^{\pi^{\star}}$ the optimal value function and the optimal Q-function, respectively.

In addition, the Bellman operator $\mathcal{T}$, which is a mapping from  $\mathbb{R}^{|\cS|\times |\cA|}$ to itself, is defined such that the $(s,a)$-th entry of $\mathcal{T}(Q)$ is given by

\begin{equation*}
	\mathcal{T}(Q)(s,a) := r(s,a) + \gamma \mathbb{E}_{s^{\prime} \sim P(\cdot| s,a)}  \Big[ \max_{a^{\prime}\in \cA} Q(s^{\prime}, a^{\prime}) \Big].
\end{equation*}

It is well known that the optimal Q-function $Q^{\star}$ is the unique fixed point of the Bellman operator.

\paragraph{Sample trajectory and behavior policy.}
Imagine we have access to a sample trajectory
 $\{s_t,a_t,r_t\}_{t=0}^{\infty}$ generated by the MDP $\mathcal{M}$ under a given stationary policy $\pi_{\mathsf{b}}$ --- called a {\em behavior policy}. The behavior policy is deployed to help one learn the ``behavior'' of the MDP under consideration, which
often differs from the optimal policy being sought. Given the stationarity of $\pi_{\mathsf{b}}$, the sample trajectory can be viewed as a sample path of a time-homogeneous Markov chain over all state-action pairs.
Throughout this paper, we impose the following assumption regarding uniform ergodicity \citep{paulin2015concentration}.

\begin{assumption}
	\label{assumption:uniform-ergodic}
	The Markov chain induced by the stationary behavior policy $\pi_{\mathsf{b}}$ is uniformly ergodic.
\end{assumption}

\noindent There are several properties concerning the behavior policy and its resulting Markov chain that play a crucial role in learning the optimal Q-function. Specifically,
denote by $\mu_{\pi_{\mathsf{b}}}$ the stationary distribution (over all state-action pairs) of the aforementioned behavior Markov chain, and define

\begin{align}
	\label{defn:mu-min}
	\mumin:= \min_{(s,a)\in \cS\times \cA} \mu_{\pi_{\mathsf{b}}} (s,a).
\end{align}

Intuitively, $\mumin$ reflects an information bottleneck --- the smaller $\mumin$ is, the more samples are needed in order to ensure all state-action pairs are visited sufficiently many times.
In addition, we define the associated mixing time of the chain as

\begin{align}
	\tmix := \min \Big\{ t ~\Big|~   \max_{ (s_0,a_0) \in \mathcal{S}\times\cA} d_{\mathsf{TV}}\big( P^t( \cdot| s_0,a_0), \mu_{\pi_{\mathsf{b}}} \big) \leq \frac{1}{4} \Big\},
	\label{defn:mixing-time-MC}
\end{align}

where $P^t( \cdot| s_0,a_0)$ denotes the distribution of $(s_t,a_t)$ conditional on the initial state-action pair $(s_0,a_0)$, and  $d_{\mathsf{TV}} (\mu, \nu)$ stands for the total variation distance between two distributions $\mu$ and $\nu$ \citep{paulin2015concentration}.  In words, the mixing time $\tmix$ captures how fast the sample trajectory decorrelates from its initial state. Moreover, we define the cover time associated with this Markov chain as follows

\begin{align}
	\tcovertime := \min \Big\{ t \mid  \min_{(s_0,a_0)\in \cS\times \cA} \mathbb{P}\big( \mathcal{B}_{t}|s_0, a_0  \big) \geq \frac{1}{2} \Big\} ,
	\label{eq:defn-cover-time}
\end{align}

where $\mathcal{B}_{t}$ denotes the event such that all $(s,a)\in \cS\times \cA$ have been visited at least once between time 0 and time $t$, and $\mathbb{P}\big( \mathcal{B}_{t}|s_0, a_0  \big) $ denotes the probability of $\mathcal{B}_t$ conditional on the initial state $(s_0,a_0)$.

\paragraph{Goal.} Given {\em a single} sample trajectory $\{s_t, a_t,r_t\}_{t=0}^{\infty}$ generated by the behavior policy $\pi_{\mathsf{b}}$, we aim to compute/approximate the optimal Q-function $Q^{\star}$ in an $\ell_\infty$ sense.
This setting --- in which a state-action pair can be updated only when the Markovian trajectory reaches it~--- is commonly referred to as {\em asynchronous} Q-learning \citep{tsitsiklis1994asynchronous,qu2020finite} in tabular RL. The current paper focuses on characterizing, in a non-asymptotic manner, the sample efficiency of classical Q-learning and its variance-reduced variant.


TITLE: Combinations for 5 fruits distributed to 7 people
QUESTION [1 upvotes]: If I have 5 different fruits to distribute to 7 people, and each person can receive 0-5 fruits, how do I calculate the number of combinations?
It's been a while since I've calculated these but I would first think the answer is 5$^7$, but then aren't there really 6 possibilities if 0 is also a possibility? Does that make the answer 6$^7$?
Secondly, if I distribute the fruits so that one person can not receive more than one fruit, how is this calculated?
Would it be 5! = 120?

REPLY [3 votes]: First Case:
The first fruit can be given to any of the $7$ people.
The second fruit can also be given to any of the $7$ people.
Similarly, the third,fourth and fifth fruits can also be given to any of the $7$ people.
So the number of choices of distributing the fruits is $7\times 7\times 7\times 7\times 7 = 7^5$.
Second Case:
The $5$ fruits can be distributed among the $7$ people in $\binom{7}{5}\times 5!$ ways since the people receiving $1$ fruit each can be chosen in $\binom{7}{5}$ ways and the fruits being different, they can be given to the five people in $5!$ ways.


TITLE: Verifying the integral solution: $I= \int \frac{\sin x}{\sqrt{1+2\cos x}}dx$
QUESTION [2 upvotes]: Evaluate:
  $$
I= \int \frac{\sin x}{\sqrt{1+2\cos x}}dx
$$

I got an answer which does not match the keys section of the book I'm solving and want to verify which one is correct.
Start by recalling:
$$
\cos 2x = \cos^2 x-\sin^2 x
$$
Therefore:
$$
\sqrt{1+2\cos x} =\sqrt{1+\cos^2 x - \sin^2 x} = \sqrt{2\cos^2 x} = \sqrt2|\cos x|
$$
Thus:
$$
I = \int \frac{\sin x}{\sqrt2 |\cos x|}dx
$$
Substitute $t=\cos x$, then $dt = -\sin x dx$, $dx = - {dt\over \sin x}$, hence:
$$
I = -{1\over \sqrt 2}\int \frac{dt}{|t|} = \boxed{-{\text{sgn}(\cos x)\over \sqrt 2}\ln(|\cos x|)+C}
$$
While the answer section suggests:
$$
I = -\sqrt{1+2\cos x} + C
$$
Which one is correct?

REPLY [2 votes]: Take $u=1+2\cos{x}$
$\frac{-1}{2}du=\sin{x}dx$
So the integral becomes $$\frac{-1}{2}\int \frac{1}{\sqrt{u}}du=-\sqrt{u}+c$$