TITLE: Write $1/z$ as a power series
QUESTION [0 upvotes]: Show that the function $f(z)=1/z$ can be represented as a power series in a ball $B(z_0,r)$, where $z_0 \neq 0$. Find the radius of convergence of this power series.
$$f(z)=\frac1z = \frac1{z_0+z-z_0}= \frac1{z_0(1+(z-z_0)/z_0)}$$
Then what do I do?

REPLY [1 votes]: $$f(z)=\dfrac{1}{z_0\left(1+\dfrac{z-z_0}{z_0}\right)}=\dfrac{1}{z_0}\cdot\dfrac{1}{1+\dfrac{z-z_0}{z_0}}=\dfrac{1}{z_0}\cdot\sum_{k=0}^\infty (-1)^k\left(\dfrac{z-z_0}{z_0}\right)^k$$
$$\therefore f(z)=\dfrac{1}{z_0}\left[1-\left(\dfrac{z-z_0}{z_0}\right)+\left(\dfrac{z-z_0}{z_0}\right)^2-\left(\dfrac{z-z_0}{z_0}\right)^3+\ldots\right]$$


TITLE: In a family with two children, what are the chances, if the oldest child is a girl, that both children are girls?
QUESTION [0 upvotes]: If $X$ is the random variable representing the number of girls out of 2 children. The probability to get a child of a given sex rahter than the other is 1/2. X equals 0 with probability 1/4, 1 with probability 1/2, and 2 with probability 1/4.
We can define a new random variable $Z$ the number of girls conditional on the older child being a girl. According to this book (p. 73), when $Z = 1$ the probability is 1/2 and 2 with probability 1/2.
However, when I compute it :
$$\frac{\overbrace{1/2}^{probability \ of \ [the \ number \ of \ girls = 1]}}{\underbrace{1/2}_{probability \ of \ the \ older \ child \ being \ a \ girl}} = 1$$
What do I get wrong ?

REPLY [0 votes]: We can define a new random variable $Z$ the number of girls conditional on the older child being a girl.

No we cannot.  Disregard the book; it is incoherent.  Random variables are not defined as conditional; it is the probability measures that are defined as such.
What we can define is the random variable $Y$ as indicating that the older child is a girl ($0$ if not, $1$ if so).  Then we have the conditional probability measures for $X$ under the condition $Y=1$: $$\mathsf P(X=x\mid Y=1)=\begin{cases}\tfrac 12 &:& x=1 \text{ or }x=2\\[1ex]0&:&\text{otherwise}\end{cases}$$

Thus is defined as $$\mathsf P(X=x\mid Y=1)=\dfrac{\mathsf P(X-Y=x-1\cap Y=1)}{\mathsf P(Y=1)}\\=\mathsf P(X-Y=x-1)$$


TITLE: Localization as dual to quotienting?
QUESTION [5 upvotes]: Why does localization look so close to quotienting? Consider the ismorphism theorem:

For a ring map $\phi: R_1 \rightarrow R_2$, let $\ker\phi \equiv \{ r_1\in R_1 : \phi(r_1) = 0 \}$. Then, $\ker \phi$ is an ideal of $R_1$, and there exists an epi-mono factorization of $\phi$ into $R_1 \twoheadrightarrow R_1 /\ker\phi \hookrightarrow R_2$  where $\hookrightarrow$ is injective and $\twoheadrightarrow$ is surjective.

Now consider the similar theorem for localization, where I denote by $R  \hat \times S \equiv S^{-1}R$ for notational suggestivity. Now I write down:

For a ring map $\psi : R_1 \rightarrow R_2$, Let $\operatorname{rek}\psi \equiv \{ r_1 \in R_1 : \psi(r_1) = 1 \}$. Then $\operatorname{rek}\psi$ is a multiplicative subset of $R_1$ and there exists a mono-epi factorization of $\psi$ into $R_1 \hookrightarrow R_1 \hat \times \psi \twoheadrightarrow R_2$

I can build a table:

$\phi \leftrightarrow \psi$.
$\ker \leftrightarrow\operatorname{rek}$.
injection $\leftrightarrow$ surjection.
$/ \leftrightarrow \hat\times$
ideal $\leftrightarrow$ multiplicative subset.
$0 \leftrightarrow 1$.

to convert from the quotienting into localization. Is there some "deep" going on here for this duality? This $0 \leftrightarrow 1$ business makes me hopeful that there might be something deeper / categorical lurking in the background.
EDIT: I had only commutative rings in mind when I wrote this. Please feel free to take assumptions on $R$ as required (Commutative, Noetherian, for example), if that allows us to explain this "duality".

REPLY [4 votes]: Fabio gives a very nice answer to your question, but doesn't directly address an important point of confusion in your original post/comments, so I'm adding this answer for posterity. In general, a map $\psi:R_1\rightarrow R_2$  will absolutely not induce an epimorphism $\text{rek}(\psi)^{-1}R_1\twoheadrightarrow R_2$, even if we take the stronger definition $\text{rek}(\psi)=\psi^{-1}(R_2^\times)$ given by Fabio. For instance, if every element of $\text{rek}(\psi)$ is already a unit in $R_1$, then we will just have $\text{rek}(\psi)^{-1}R_1=R_1$, and so using this it is easy to come up with examples where the induced map is not epi.
For instance, take $R_1=\mathbb{Q}$, and $R_2$ any field extension of $\mathbb{Q}$ with a non-trivial automorphism $\alpha$ that fixes $\mathbb{Q}$ pointwise, with $\psi:R_1\hookrightarrow R_2$ the inclusion map. Then $\text{rek}(\psi)=\mathbb{Q}^\times$ , so $\text{rek}(\psi)^{-1}\mathbb{Q}=\mathbb{Q}$ and the induced map to $R_2$ is just $\psi$, which is certainly not an epimorphism. (E.g. $\alpha\circ\psi=\text{id}_{R_2}\circ\psi$ but $\alpha\neq\text{id}_{R_2}$).
Indeed, the polynomial ring example you give in the comments of your post does not hold in general either. If we let $R_1=\mathbb{R}[x]$ and $R_2=\mathbb{R}[x,y]$, with $\psi:R_1\hookrightarrow R_2$ again the inclusion map, then once again $\text{rek}(\psi)=R_1^\times$ but $\psi$ is certainly not epi.
The problem in all of these examples is that $R_2$ can be very big compared to the image of $R_1$; hopefully the above examples clarify that point. (Note however, that – provided $R_2\neq\{0\}$ – the map $R_1\hookrightarrow \text{rek}(\psi)^{-1}R_1$ will still be injective, even if we use Fabio's stronger definition of $\text{rek}(\psi)$, because no element of $\text{rek}(\psi)$ can be a zero-divisor in $R_1$.)


TITLE: How to show a simply-connected negatively-curved n-dimensional Riemannian manifold is homeomorphic to $R^n$?
QUESTION [0 upvotes]: I saw this word in a paper , but there is not proof . And a simple hint is given that for fixing a poing $0\in M$ , mapping the $T_0M$ noto $M$ via exponential map. But it is not enough to say it is homeomorphic . How to show it ?

REPLY [1 votes]: The homeomorphism is given by mapping point $rv \in T_0M$, where $ r \in \Bbb R^{+}$ and $\|v\| = 1$, to the point to $\exp_0(rv) \in M$.
You still have to show that this map is injective, but that's what the negative curvature is for: you use it to estimate the distance to the nearest "focal point" of $\exp$, which turns out to be infinitely far away.


text:
\section{The geometric extended \(S_{n+1}\) action on $Y_{\Fc_{A_{n-1}}}$} \label{sec:action1}

We recall that there  is a well know `extended' \(S_{n+1}\) action on the De Concini-Procesi (real or complex)  model \(Y_{\Fc_{A_{n-1}}}\):  it comes from the isomorphism with the moduli space \({\overline M_{0,n+1}}\)
and the character of the resulting representation  on cohomology has been computed in \cite{rains2009} in the real case, and in \cite{getzler} in the complex case.

In order to describe how \(S_{n+1}\) acts on the strata   of \(Y_{\Fc_{A_{n-1}}}\) it is sufficient to show the corresponding action on \(\Fc_{A_{n-1}}\), since these strata are indexed by the elements of \(\Bc(n-1)\).

Let \(\Delta=\{\alpha_0,\alpha_1,...,\alpha_{n-1}\}\) be a basis for the root system \(\Phi_{A_n}\) of type \(A_n\) (we added to a basis of \(A_{n-1}\) the extra root \(\alpha_0\)).

We identify in the standard way  \(S_{n+1}\) with the group which permutes \(\{0,1,...,n\}\) and    \(s_{\alpha_0}\) with the transposition \((0,1)\). Therefore    \(S_{n}\), the subgroup generated by \(\{s_{\alpha_1},...,s_{\alpha_{n-1}}\}\), is identified with the subgroup which permutes \(\{1,...,n\}\).

 Let \(A\) be a subspace in \(\Fc_{A_{n-1}}\) different from \(V\),  let \(\sigma \in S_{n+1}\) and let us consider the subspace \(\sigma A\) according to the natural action of \(S_{n+1}\) on  \(\Fc_{A_{n}}\). Morover  we   denote  by \(\overline A\) the subspace generated by all the roots of \(\Phi_{A_n}\)   that are orthogonal to \(A\). We notice that if

some of the roots contained in \(\sigma A\) have  \(\alpha_0\) in their support
 then \(\sigma \overline A\) belongs to \(\Fc_{A_{n-1}}\).
 Therefore we define the action of $S_{n+1}$ on  \(\Fc_{A_{n-1}}\) as follows. Let $\sigma \in S_{n+1}$ and
 \(A \in \Fc_{A_{n-1}}\).
 We set \(\sigma \cdot V:=V\) and for $A \neq V$ we define $$\sigma \cdot A := \left\{
 \begin{array}{ll}
 \sigma A & \mbox{ if }\sigma A \in  \Fc_{A_{n-1}} \\
  \sigma \overline A & \mbox{ otherwise.}
 \end{array}
 \right.
 $$

Let us now write explicitly    how the above described  action extends to  \(\Bc(n-1)\).  Let \[\Sc=\{V, A_1,A_2,...,A_k, B_1,B_2,...,B_s\}\] be an element of  \(\Bc(n-1)\), i.e. a nested set  in \({\Fc_{A_{n-1}}}\)  that contains \(V\) and let \(\sigma \in S_{n+1}\).
Moreover, let us suppose that, for every subspace \( A_j\),  the subspace \(\sigma A_j\) doesn't belong to \(\Fc_{A_{n-1}}\), while  the subspaces
\(\sigma B_t\) belong to \(\Fc_{A_{n-1}}\). Then \(\sigma \Sc= \{V, ..,\sigma\overline{A_j},.., \sigma B_s,..\}\).
As one can quickly check, \(\sigma \Sc \in \Bc(n-1)\).
\begin{remark}This  action can also be lifted to the minimal spherical model of type \(A_{n-1}\)  (see for instance the exposition in \cite[Section 3]{callegarogaiffi3}).
\end{remark}

config: arxiv


TITLE: Invertibility of a square matrix with zero diagonal elements and positive non-diagonal elements
QUESTION [5 upvotes]: $M$ is square and
$$M(i,j)=0, i=j$$
$$M(i,j)>0, i\ne j$$
Is $M$ full-rank or invertible?
Actually the $M$ I am studying has much stronger properties but I guess the simple conditions above might be enough to make $M$ non-singular. The stronger properties of $M$ are:

All elements in $M$ are non-negative integers between $0$ and $N$;
The sum of each row is equal to $N$.

Edit
$N$ is not the dimension of $M$. It's just a constant positive integer.
I did search for this problem, but it seemed there was no much work on such matrices. It looks easy, but I don't know how to prove it and I couldn't find a counter-example either.

REPLY [7 votes]: The matrix $M$ doesn't have to invertible. I give a simple counterexample here.
$$M=\begin{pmatrix}
0&2&1&1\\
2&0&1&1\\
1&1&0&2\\
1&1&2&0
\end{pmatrix}$$
You will see $r_1+r_2-r_3-r_4=0$ which indicates they are linear dependent.


TITLE: Finite Morphisms of Affine Curves are Surjective
QUESTION [2 upvotes]: I have a question about a step used in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 109):

The statement is that a finite morphism of integral affine curves is surjective.
Indeed I'm familar with some proofs to show this statement but the point I'm really interested is concretely the argument given in this source by the author.
He claims that the Case 2 of the proof of 4.3.5 cannot occure in the case of a finite morphism.
My question is why?
Here the proof of 4.3.5:

REPLY [2 votes]: The argument in the book is too complicated:
A finite morphism of schemes is a closed map, hence the image of Y is closed and irreducible in $X$ and thus that image is equal to $X$.
By the way, my argument does not use that $X$ or $Y$ is affine, nor that $Y$ has dimension $1$!


text:
\begin{document}

\begin{abstract}
 We consider a Dirichlet problem for the mean curvature operator in  the Minkowski spacetime,   obtaining  a necessary and sufficient condition for the   existence of a spacelike solution,  with prescribed mean curvature,
which is the   graph of a function   defined
on a   domain  equal to  the complement in $\R^n$ of the union of a finite number of bounded Lipschitz domains. The mean curvature $H=H(x,t)$ is assumed to have   absolute value  controlled from above by a locally bounded, $L^p$-function, $p\in [1,2n/(n+2)]$,  $n\geq 3$.
\end{abstract}
\maketitle

\section{Introduction and statement of the results}
Due to their importance in general relativity, spacelike hypersurfaces with constant mean curvature or, more generally, with prescribed mean curvature $H$, have been extensively studied since Lichnerowicz's paper \cite{Li44} (cf.  \cite{MaTi80}). Maximal spacelike hypersurfaces ($H=0$) have also attracted the interest of researchers because of  their similarity with minimal hypersurfaces in the Riemannian setting.
In fact, they are  critical points of the area functional and,  under some curvature assumptions on the ambient spacetime, locally maximize it among all nearby spacelike hypersurfaces having the same boundary (cf. \cite{BrFl76}).  Two  very relevant examples  in this perspective  are   the so-called Calabi-Bernstein problem in the Minkowski spacetime \cite{Ca70, ChYa76} and
its counterpart with constant mean curvature hypersurfaces \cite{Treibe82}.
 Since then a lot of related papers have been  appearing; just to recall a few, we refer here to  the  new proofs of the Calabi-Bernstein theorem for surfaces given  in \cite{Ro96, AlPa01},  as well as  to   related Calabi-Bernstein type results in different ambient spaces \cite{EckHui91, AlRoSa95, AlbAl09, CaRoRu11, CaCaLi11} and  to  some existence results in Minkowski spacetime for entire or radial spacelike graphs  under different growth conditions on the mean curvature \cite{BDD,A,A2,P,MP}.

The Dirichlet problem on a bounded open set  for spacelike hypersurfaces, described  as the graph of a function,  with prescribed mean curvature  was studied in the Minkowski spacetime \cite{BarSim83}, in spacetimes conformal to an orthogonal  splitting \cite{Gerhar83} and  for some cosmological spacetimes \cite{Ba84}.  A fairly general case  was considered in \cite{Bartni88}.
Recently, non-smooth critical point theory \cite{Sz86} has been used in \cite{BeJeMa14} to obtain existence and multiplicity of  spacelike solutions in the Minkowski spacetime for the Dirichlet problem  with homogeneous boundary data on a $C^2$ domain,  when the mean curvature is a  function depending on a parameter.
We would like to emphasize that the result in \cite{BarSim83}  does not require assumptions on the regularity of the boundary. Namely, boundary values are considered according to the definition given in \cite[p. 133]{BarSim83} which allows  one   dealing with quite general open bounded subsets of $\R^n$ (for instance bounded domains  with just continuous boundaries are admitted).

Among other results, in \cite{BarSim83} it is also proved  that if a variational solution, with mean curvature not depending on the time coordinate $t$, contains a segment of light ray, then it contains the ray extended to the boundary or to infinity.
This property will be fundamental in the proof of our main result,   Theorem~\ref{maint},  dealing with solutions  vanishing  at infinity of the Dirichlet problem for spacelike  hypersurfaces on an unbounded open subset of $\R^n$.  Theorem~\ref{maint}  can  be considered as an extension of \cite[Theorem 4.1]{BarSim83} to  exterior  domains. A solution is indeed obtained  by a standard minimization argument  (Proposition~\ref{pr:min})  applied to  functional $\mathcal I$ in \eqref{I} which, differently from the area functional, is well-defined  on functions with  square integrable gradient on the exterior domain (see Section~\ref{s3}).
The boundary conditions are as usual encoded in the functional setting adopted (see Section~\ref{s2}). As we need    an extendibility property  (Lemma~\ref{Lext}),   we  reinforce  a bit the regularity of the domain  w.r.t.  \cite{BarSim83}, by considering a Lipschitz boundary. Functional  $\mathcal I$  is then defined on a convex, closed subset (Proposition~\ref{weakly})  of an affine subspace of the homogeneous Sobolev space of locally integrable functions with square integrable partial derivatives which share the same trace on the boundary of the exterior domain. The tangent space of this affine manifold  is the Hilbert space obtained as the completion,   w.r.t.  the $L^2$-norm  of the gradient, of the space of test functions on the exterior domain. This fact allows us to recover  some   embedding  properties  (Lemma~\ref{prop})  in a similar variational setting exploited in  \cite{BoPo16}   to  find   solutions of a  Born-Infeld equation in $\R^n$  (see the end of this introduction for more details).

Let $(\LL^{n+1}, \langle\cdot,\cdot\rangle)$ be the $(n+1)$-dimensional Minkowski spacetime with the following sign convention:  $\langle(\tau,v),(\tau,v)\rangle =-\tau^2+|v|^2$ for  $\tau\in \R$,  $v\in \R^n$, where $|\cdot |$ is the Euclidean norm in $\R^n$. A smooth immersion $\phi\colon\Sigma\rightarrow \LL^{n+1}$ of an $n$-dimensional connected manifold $\Sigma$ is a {\em spacelike hypersurface} if the metric induced by $\phi$ is a Riemannian metric on $\Sigma$.
Let $A$ be the shape operator of $\Sigma$ and  $H:= -\frac{1}{n}{\rm tr}A$ its  {\em mean curvature}.
Assume that $\Sigma$ is an open subset of $\R^n$ and $\phi(x)=\big(x,u(x)\big)$, $x\in\Sigma$; when $u$ is at least  of class  $C^2$, the mean curvature  of this  hypersurface is then equal to
\[
{\rm div}\left(\frac{\nabla u}{\sqrt{1-|\nabla u|^2}}\right)\ =\ nH(x,u),
\]
where ${\rm div}(\cdot)$ is the divergence operator in $\R^n$.

Hereafter by a {\em domain} in $\R^n$ we mean an open and connected subset. Given a domain $\Omega$ and a $C^1$  function $u$ on it, its graph $t=u(x_1,\ldots,x_n)=u(x)$ defines a $C^1$ spacelike hypersurface in $\LL^{n+1}$ if and only if the Euclidean gradient of $u$ satisfies $|\nabla u(x)|<1$, for all $x\in \Omega$. In this case, we will say that  $u$ is a {\em  spacelike function} and its graph is then called a spacelike hypersurface.

Instead, a  locally Lipschitz function $u$ is said {\em weakly spacelike} if $|\nabla u|\leq 1$, for {a.e.}  $x\in\Omega$.

We also need the following definition.

\begin{dfn}
Let $V\subset \R^n$; a  function $\psi\colon V\to \R$ is said {\em  spacelike displacing} if its graph is an acausal set,   namely no couple of its points  can be joined by a timelike or lightlike segment.   This is equivalent to  $|\psi(x)-\psi(y)|<|x-y|$, for all $x,y\in V$, $x\neq y$. Moreover,
$\psi:V\to\R$ is said {\em spacelike displacing in $W\subset \R^n$} if $|\psi(x)-\psi(y)|<|x-y|$,  for all $x,y\in V$, $x\neq y$,  with the inner points of the line segment $\overline{xy}$ contained in $W$.
\end{dfn}
\bere\label{nomi}
Notice that  in \cite{BarSim83}  $C^1$ spacelike  functions are named  {\em strictly spacelike}, while spacelike displacing functions in an open subset $V$ of $\R^n$ (and defined on the same $V$) are called {\em spacelike}  (although the graph of  a spacelike displacing function can have degenerate tangent spaces). Furthermore, note  that  if $V$ is open, the graph $G_\psi$ of a continuous  spacelike displacing function  in $V$, $\psi:V\to\R$, is spacelike in the usual sense for  $C^0$ hypersurfaces,   i.e., for each $p\in G_\psi$ there exists a neighbourhood $U$ in $\LL^{n+1}$ such that $G_\psi\cap U$ is acausal and edgeless in $U$ (cf., e.g.,  \cite[p. 213]{EscGal92}, \cite[Definition 14.28]{BeErEa96}). Nevertheless, if $V$ is not convex, the graph of a spacelike displacing $\psi:V\to\R$ in $V$ can be not acausal.
\ere

We recall the definition of a Lipschitz domain.

\begin{dfn}\label{lipdom}
An open subset  $U\subset\R^n$ is  said {\em Lipschitz} if for each $p\in \partial U$ there exist an open ball $B(p,r)\subset \R^n$ and a Lipschitz function $f\colon B(p,r) \to \R$ such that $B(p,r)\cap U =f^{-1}((0,+\infty))$.
\end{dfn}
\bere\label{sepr}
If $U$ is a Lipschitz open subset, then $\partial U=\partial\overline U$ (see, e.g., \cite[Remark 9.59]{Leoni17}). For further use, notice that if  $U\subset \R^n$ is Lipschitz, then $\partial U=\partial (\R^n\setminus \overline U)$.
\ere

We  deal  with Lipschitz exterior domains of $\R^n$. More precisely, we require the following assumption.

\begin{ass}\label{omega}
We  consider  an exterior domain in $\R^n$, $n\geq 3$, defined by means of a finite collection of  bounded Lipschitz domains $\Omega_i$, such that $\overline{\Omega}_i\cap \overline{\Omega}_j= \emptyset$ for all $i,j\in \{1,\ldots, m\}$, $i\neq j$, $m\geq 1$.
Let
\[
\Omega :=\bigcup_{i=1}^{m}\Omega_i \quad \text{and} \quad      \Omega_c:=\R^n\setminus\overline{\Omega}. \]
Notice that by Remark \ref{sepr} $\partial\Omega_c=\partial\Omega$.
\end{ass}

Next let us set our problem.  We consider  the Dirichlet problem
\begin{equation}\label{star}
\begin{cases}
\displaystyle {\rm div}\left(\frac{\nabla u}{\sqrt{1-|\nabla u|^2}}\right)\ =\ nH(x, u)
& \mbox{ in $\Omega_c$}\\
 \displaystyle u=\varphi & \mbox{ on $\partial\Omega$}  \\
  \displaystyle \lim_{|x|\rightarrow +\infty}  u(x)=0
 \end{cases}
\end{equation}
where $\varphi: \partial\Omega\rightarrow\R$ and
$H:\Omega_c\times\R\rightarrow\R$ is   a   Carath\'eodory function\footnote{$H(\cdot,t)$ is measurable in $\Omega_c$
 for all $t\in\R$ and $H(x,\cdot)$ is continuous in $\R$ for a.e. $x\in \Omega_c$.}  satisfying:
\begin{enumerate}[label=(H), ref=H]
\item \label{h}
there exists $h\in L^s(\Omega_c)
\cap L^\infty_{\rm loc}({\Omega}_c)$,
$s\in \left[1,\frac{2n}{n+2}\right]$,
such that
$$
n|H(x,t)|\leq h(x) \quad\text{for a.e. $x\in \Omega_c$  and all $t\in\R$}.
$$
\end{enumerate}

In order to introduce our functional framework (see Section \ref{s2} for more details and remarks),  let us recall that   $\stackrel{\bigcdot }{W}\!\!\mbox{}^{1,2}(\Omega_c)$  is   the homogeneous Sobolev space  of  distributions on $\Omega_c$ with partial derivatives in $L^2(\Omega_c)$. Some useful properties of such a space can be found, for example, in \cite[\S 11]{Leoni17}. \\
Let us finally introduce the space $\mathcal X$ of admissible functions in the variational setting  for   problem \eqref{star}:
\begin{equation}\label{storto}
\mathcal X:=\ \W(\Omega_c)\cap L^{2^*}(\Omega_c)\cap \{u\in C^{0,1}_{\mathrm{loc}}(\Omega_c): \|\nabla u\|_\infty\leq 1\};
\end{equation}
here  $C^{0,1}_{\mathrm{loc}}(\Omega_c)$  denotes the space  of  locally Lipschitz functions on $\Omega_c$ and as usual $2^*=2n/(n-2)$ is the Sobolev critical exponent.

Now we give the definition of {\em weak solution} of \eqref{star} (see also Remark~\ref{rem:ws}).

\begin{dfn}\label{def:ws}
A function $u: \Omega_c \to \R$ is called a {\em weak solution} of \eqref{star}  if it belongs to $ \mathcal X$, $\varphi$ is the trace of $u$ and
\begin{equation}\label{ws}
\int_{\Omega_c}\frac{\nabla u \cdot \nabla v}{\sqrt{1-|\nabla u|^2}}\;{\rm d}x
+n\int_{\Omega_c}H(x,u)v\;{\rm d}x=0,
\quad \hbox{ for all }v\in C^\infty_c(\Omega_c).
\end{equation}
\end{dfn}

We state our main result.
\begin{thm}\label{maint}
Let $\Omega_c$ satisfy Assumption~\ref{omega} and
$H:\Omega_c\times\R\rightarrow\R$ be such that \eqref{h} holds.
Then, there exists a spacelike\footnote{ In the terminology of \cite{BarSim83}  such a solution  is strictly spacelike, cf. Remark~\ref{nomi}.}
weak solution    of \eqref{star}  if and only if
$\varphi:\partial\Omega\rightarrow\R$  is  the trace
of a function $w\in\mathcal{X}$ and moreover $\varphi$ is spacelike displacing in $\Omega_c$ when $\Omega$ is not convex.
\end{thm}
\begin{rmk}
Let us emphasize some points about Theorem~\ref{maint}.
\begin{itemize}
	\item[(1)] The boundary condition, $u=\varphi$ on $\partial \Omega$, in \eqref{star} is meant in a trace  sense, but as we will show in the next section a weak solution  $u$  can   indeed be continuously Lipschitz extended to $\partial \Omega$ (Lemma \ref{Lext}). Therefore, $u_{|\partial\Omega_c}=\varphi$, hence $\varphi$ is a posteriori Lipschitz continuous on $\partial \Omega$.
	\item[(2)] The limit at infinity in \eqref{star} is intended in the classical sense.
	\item[(3)]
 As shown in \cite[p. 148]{BarSim83}  (see also \cite[p. 5]{Bon-Iac2}), as $H$ is locally bounded, by elliptic regularity theory, a spacelike weak solution  $u$  belongs to $W^{2,2}_{\mathrm{loc}}(\Omega_c)$ and $\nabla u$  is locally H\"older. Moreover,  if $H\in C^{k,\alpha}(\Omega_c\times\R)$, $k\in\N\cup\{\infty\}$, then $u\in C^{k+2,\alpha}(\Omega_c)$.
	\item[(4)] If $H$ does not depend on $t$, our statement holds just assuming that $H\in L^s(\Omega_c) \cap L^{\infty}_{\rm loc}(\Omega_c)$. In particular, it holds for $H=0$ and, in such a case, gives the existence of a maximal hypersurface on the exterior domain $\Omega_c$.
		Previous existence (and uniqueness, with respect to a given asymptotic profile) results  for the Dirichlet problem of  maximal graphs on an exterior domain of the Minkowski spacetime have been recently obtained in \cite{HonYua19}. Existence and multiplicity results for radial solutions outside a ball, with  homogeneous boundary condition, have been obtained in \cite{YaLeSi20} for a separable-variables $H$ which is also radial in the $x$ variable.
	We are not aware of  other  results for the Dirichlet problem in an exterior domain when $H\neq 0$.
	\end{itemize}
\end{rmk}
 Since when $\varphi$ is $(1-\epsilon)$-Lipschitz continuous on $\partial\Omega$, for a $\epsilon>0$, we  can ensure that there exists $u\in\mathcal{X}$ such that  $u_{|\partial\Omega}=\varphi$,   we have the following:
\begin{cor}\label{mainc}
Under the assumptions on $\Omega_c$ and $H$ in  Theorem~\ref{maint}, let  $\varphi:\partial\Omega\to \R$ be a $(1-\epsilon)$--Lipschitz continuous. Then, there exists a  spacelike weak   solution of \eqref{star}.
\end{cor}

A further relevant physical motivation to the problem under study is given by the  differential operator
\begin{equation*}
\mathcal Q(u)={\rm div}\left(\dfrac{\nabla u}{\sqrt{1-|\nabla u|^2}}\right)
\end{equation*}
which naturally appears in the Born-Infeld theory.
Almost a century ago, Born and Infeld introduced a new electromagnetic theory in a series of papers \cite{Bnat,BInat,B,BI}
as an alternative to
the classical Maxwell theory. Such a theory was proposed  as a nonlinear model of electrodynamics  having the notable feature of being a fine answer to the well-known infinity energy problem   (the electromagnetic field generated by a point charge has finite energy  in Born-Infeld theory).

In last years many authors have been focusing their attention on problems related to  $\mathcal Q$ in the whole $\R^n$, $n\ge 1$. In particular, some results for
\[
-{\rm div}\left(\dfrac{\nabla u}{\sqrt{1-|\nabla u|^2}}\right)=\rho, \qquad\hbox{ in }\R^n
\]
can be found in \cite{K,BoPo16,K-corr,Bon-Iac2, BCF, BDPR}, under different assumptions on $\rho$. Here $\rho$ can be considered as an assigned charges source. We also refer to \cite{APS}, where the Born-Infeld equation is coupled with the nonlinear Schr\"odinger one.
 Therefore, Theorem \ref{maint} can also be seen  as an existence result for the Born-Infeld problem on an  exterior  domain with assigned boundary conditions.

\section{Functional setting}\label{s2}
Let us denote by  $\de x$  the Lebesgue measure on $\R^n$ and, unless differently specified, by $\|\cdot\|_r$  the $L^r$-norm in the Lebesgue space $L^r(\Omega_c)$, $1\leq r\leq +\infty$, where $\Omega_c$ is defined in Assumption~\ref{omega}. By $\|\nabla\cdot\|_r$ we mean $\||\nabla\cdot|\|_r$.

Let $V\subset\R^n$ be open and unbounded and let us consider the homogeneous Sobolev  space $D^{1,2}(V)$  defined as  the completion of  $C_c^\infty(V)$, the space of smooth functions with compact support in $V$,   with respect to the $L^2$-norm of the gradient. Moreover, let us denote by $\mathscr{D}'(V)$ the space of distribution on $V$.

The following properties  hold for $D^{1,2}(\Omega_c)$:
\begin{prop}\label{tr0}
\mbox{}
\begin{enumerate}[label=(\arabic{*}), ref=\arabic{*}]
		\item \label{d1} $D^{1,2}(\Omega_c)\hookrightarrow L^{2^*}(\Omega_c)$; moreover
		$D^{1,2}(\Omega_c)\hookrightarrow \mathscr{D}'(\Omega_c)$ and the partial distributional derivatives of its elements are represented by functions in $L^2(\Omega_c)$;
		\item \label{d2}all $u\in D^{1,2}(\Omega_c)$ vanish at infinity,  i.e., ${\rm meas}\{x\in \Omega_c:|u(x)|>t\}<+\infty$,  for all $t>0$;
	\item \label{d3}for all $u\in D^{1,2}(\Omega_c)$, the trace ${\rm Tr}(u)$  on $\partial\Omega_c$ is well-defined and equal to $0$.
\end{enumerate}
\end{prop}
\begin{proof}
\eqref{d1} Since $n\geq 3$, by the Sobolev embedding theorem,  $D^{1,2}(\Omega_c)$ is continuously embedded in $L^{2^*}(\Omega_c)$ (see \cite[\S 15.1]{Mazya11}). As a consequence,
every $u\in D^{1,2}(\Omega_c)$ can be identified with a distribution on $\Omega_c$ and, if $(u_k)_k\subset D^{1,2}(\Omega_c)$ is such that  $\|\nabla u_k-\nabla u\|_{2}\to 0$, then it converges to $u$ in distributional sense. Let now   $(\varphi_k)_k\subset C_c^\infty(\Omega_c)$ be a  Cauchy sequence representing $u\in D^{1,2}(\Omega_c)$; then $(\varphi_k)_k$  is Cauchy in $L^{2^*}(\Omega_c)$ and  for all
$\varphi\in C_c^\infty(\Omega_c)$, $i\in\{1,\dots,n\}$,  we have
\bmln   (\partial_i u,\varphi) := -\int_{\Omega_c} u\partial_i\varphi \,\de x=-\lim_k\int_{\Omega_c}  \varphi_k\partial_i\varphi \,\de x   \\   =\lim_k\int_{\Omega_c} \partial_i \varphi_k\varphi \,\de x=\int_{\Omega_c} \psi_i\varphi \,\de x,
\emln where $\psi_i\in L^2(\Omega_c)$ is the element representing the Cauchy sequence $\partial_i \varphi_k$ in $L^2(\Omega_c)$.

\eqref{d2} Notice that $u\not\in L^{2^*}(\Omega_c)$ if, by contradiction, there exists $\bar t>0$ such that 	${\rm meas}\{x\in \Omega_c:|u(x)|>\bar t\}=+\infty$.

\eqref{d3} Take $R>0$ such that $\overline{\Omega} \subset B(0,R)$ and consider $v$, the restriction to $\Omega_c$ of a smooth function assuming values in the interval $[0,1]$ and equal to $1$ on $B(0,R)$, with compact support on $B(0,2R)$ and such that $|\nabla v|<1$.
Let  $(u_k)_k\subset C_c^\infty(\Omega_c)$ be a sequence  converging to $u$ in $D^{1,2}(\Omega_c)$. Then $v u_k\to vu$ both in $D^{1,2}(\Omega_c)$ and $L^{2^*}(\Omega_c)$, thus   $vu_k\to vu$ in the $H^1$-norm on $B(0,2R)\setminus \overline{\Omega}$, which implies that
$vu\in H^1_0(B(0,2R)\setminus \overline{\Omega})$. As $\partial\Omega_c$ is Lipschitz, $\mathrm{Tr}(vu)=0$ (cf., e.g., \cite[Theorem~18.7]{Leoni17}) and since $v=1$ on $\partial \Omega$, we get that $\mathrm{Tr}(u)_{|\partial \Omega}=0$.
\end{proof}

Actually, the above inclusions and properties  cha\-ra\-cte\-ri\-ze  $D^{1,2}(\Omega_c)$.
In the following we give some useful properties of the homogeneous Sobolev space  $\stackrel{\bigcdot }{W}\!\!\mbox{}^{1,2}(\Omega_c)$ and a relation between it and $D^{1,2}(\Omega_c)$ which is crucial to our purposes.

\begin{rmk}\label{mainrmk}
Since $\Omega_c$ is a Lipschitz domain, the trace operator  is well-defined  on
$\W(\Omega_c)$. Indeed, for any $R>0$ such that $\overline{\Omega}\subset B(0,R)$ we have $\W\big(B(0,R)\setminus\overline\Omega\big)=W^{1,2}\big(B(0,R)\setminus\overline\Omega\big)$
(see \cite[Corollary 1.1.11]{Mazya11})\footnote{Notice that a Lipschitz domain according to Definition~\ref{lipdom} is called $C^{0,1}$ domain in \cite{Mazya11}; $C^{0,1}$ bounded domains satisfy the cone property required in \cite[Corollary 1.1.11]{Mazya11} (see \cite[Remark 1, p. 15]{Mazya11}).} and then   $\mathrm{Tr}(u)\in L^2(\partial\Omega_c)$, for all $u\in\ \W(\Omega_c)$.  However, the trace operator is evidently not bounded in $\W(\Omega_c)$. Anyway, if $\|\nabla u_k-\nabla u\|_2\to 0$ and $\|u_k-u\|_{2^*}\to 0$, then $\|\mathrm{Tr}(u_k)-\mathrm{Tr}(u)\|_{L^2(\partial\Omega_c)}\to 0$. Moreover, if $u\in C^0(\overline \Omega_c)$, then $\mathrm{Tr}(u)=u_{|\partial \Omega}$. \end{rmk}
 \begin{prop}\label{trace}
The  space $D^{1,2}(\Omega_c)$ is  given by
$$
D^{1,2}(\Omega_c)=\{u\in \ \stackrel{\bigcdot}{W}\!\!\mbox{}^{1,2}(\Omega_c):  u\in L^{2^\ast}(\Omega_c), \ \mathrm{Tr}(u)=0\}.
$$
\end{prop}
\begin{proof}
The first inclusion has  already been  shown in Proposition~\ref{tr0}. For the other one,  let
 $u\in \ \W(\Omega_c)\cap L^{2^\ast}(\Omega_c)$ be such that $\mathrm{Tr}(u)=0$.  As $u\in L^{2^*}(\Omega_c)$ and $|\nabla u|\in L^2(\Omega_c)$, for any $\epsilon>0$ there exists $R>0$ such that $\overline\Omega\subset B(0,R)$, $\|u\|_{L^{2^*}(\R^n\setminus B(0,R))}<\epsilon$ and
 $\|\nabla u\|_{L^{2}(\R^n\setminus B(0,R))}<\epsilon$. Let us take $v$ as in the proof of Proposition~\ref{tr0}, with the further requirement that $|\nabla v|\leq 1/R$ on $A_R:=B(0,2R)\setminus B(0,R)$,  so that   $vu\in H^1_0(B(0,2R)\setminus \overline{\Omega})$. Let $(u_k)_k\subset C^{\infty}_c(B(0,2R)\setminus \overline{\Omega})$ converge to $vu$ in the $H^1$-norm. Then,   taking into account that $\|u\|_{L^{2^*}(A_R)}<\epsilon$ and $\|\nabla u_k\|_{L^2(A_R)}\to \|\nabla(vu)\|_{L^2(A_R)}$,  trivially  extending   $u_k$ on $\R^n\setminus B(0,2R)$, we get that   $u_k\to u$ in $D^{1,2}(\Omega_c)$.
\end{proof}
 Let us now introduce the following subset of $D^{1,2}(\Omega_c)$:
$$\mathcal X_0:=D^{1,2}(\Omega_c)\cap  \{u\in C^{0,1}_{\mathrm{loc}}(\Omega_c): \|\nabla u\|_\infty\leq 1\}.$$ Let $j(u):\R^n\to \R$, $u\in \mathcal X_0$, be defined  as follows:
\[j(u)(x)=\begin{cases}
u(x)&\text{if $x\in \Omega_c$}\\
0&\text{otherwise} .
\end{cases}\]
\begin{lem}\label{isometry}
 The map $j$ is an isometry from $\mathcal X_0$ (as a subset of the Hilbert space $D^{1,2}(\Omega_c)$) into the subset   $D^{1,2}(\R^n)\cap  \{u\in C^{0,1}(\R^n): \|\nabla u\|_{L^\infty(\R^n)}\leq 1\}$ of $D^{1,2}(\R^n)$.
\end{lem}
\begin{proof}
Let $u\in \mathcal X_0$; it is clear that $j(u)\in D^{1,2}(\R^n)$. Hence, $j(u)\in L^{2^*}(\R^n)$ and $|\nabla j(u)|\in L^2(\R^n)$; moreover, $|\nabla j(u)(x)|\leq 1$ for a.e. $x\in \R^n$. By Morrey's embedding theorem and we deduce that $j(u)\in L^{\infty}(\R^n)$, hence  $j(u)\in W^{1,\infty}(\R^n)$ and therefore it is Lipschitz.
\end{proof}

As a consequence of Lemma~\ref{isometry}, $\mathcal X_0$ can be identified with a subset  of $D^{1,2}(\R^n)\cap  \{u\in C^{0,1}(\R^n): \|\nabla u\|_{L^\infty(\R^n)}\leq 1\}$ and then by
\cite[Lemma 2.1]{BoPo16}  we immediately get:
\begin{lem}\label{prop}
As a  subset of the Hilbert space $D^{1,2}(\Omega_c)$, $\mathcal X_0$   satisfies the following properties:
	\begin{enumerate}[label=(\arabic{*}), ref=\arabic{*}]
		\item \label{p1}it is continuously embedded in $W^{1,p}(\Omega_c)$, for all $p\in [2^\ast, + \infty)$;
		\item\label{p2}
		it  is continuously embedded in $L^\infty(\Omega_c)$;
		\item\label{p3} all $u\in \mathcal X_0$ satisfy $\lim_{|x|\to \infty}u(x)= 0$;
		\item\label{p4}
		it is a convex and  weakly closed subset of $D^{1,2}(\Omega_c)$;
		\item\label{p5}
		any  bounded sequence $(u_k)_k\subset \mathcal X_0$ admits a subsequence weakly converging  to some $u\in \mathcal X_0$ and uniformly on compact subsets of $\Omega_c$.
	\end{enumerate}
\end{lem}

\bere\label{rem:ws}
Let us observe that if $u\in \mathcal{X}$  (cf. \eqref{storto}) is a weak solution according to Definition \ref{def:ws}, then the identity in \eqref{ws} also holds for any $v\in \mathcal{X}_0$ by means of a convolution argument and \cite[Lemma 2.10]{BoPo16}.
\ere
\begin{lem}\label{Lext}
Let $\Omega_c$ satisfy Assumption~\ref{omega}. Then, every $u\in C^{0,1}_{\mathrm{loc}}(\Omega_c)$ such that $|\nabla u|\in L^{\infty}(\Omega_c)$ can be extended to a  Lipschitz function on $\overline \Omega_c$.
\end{lem}
\begin{proof}
Let us take $R>0$ such that $\overline\Omega\subset B(0,R)$ and consider the bounded open set $V_R:=B(0,R)\setminus  \overline \Omega$. Let us  show that $u$ is bounded on $V_R$. Clearly $V_R$ is a Lipschitz domain and then it is the union of a finite number of Lipschitz domains  $V_j$ starshaped with respect to  balls $B(y_j, R_j)$ contained in $V_R$ (see \cite[Lemma 1, p. 15]{Mazya11}). Now assume by contradiction that $u_{|V_R}$ is not bounded. Then there exists a sequence  $(x_k)_k\subset V_R$ such that $|u(x_k)|\to +\infty$. Let $j_k$ be one of the indices $j$ such that $x_k\in V_{j_k}$. Then the segment $\overline{y_{j_k}x_{k}}$ is contained in $V_{j_k}$. Hence,
\[
|u(x_k)|\leq |u(y_{j_k})|
+\|\nabla u\|_\infty|x_{k}-y_{j_k}|\leq M+2R\|\nabla u\|_\infty,
\]
where $M:=\max_{j}|u(y_{j})|$, a contradiction.

Thus, $u_{|V_R}\in W^{1,\infty}(V_R)$ and, since $V_R$ has the  extendibility property  (see \cite[Th. 13.17]{Leoni17}\footnote{We point out that the theorem can be applied since $	\partial\Omega=\partial\Omega_c$ is bounded, hence by \cite[p. 424]{Leoni17} being Lipschitz is equivalent to be uniformly Lipschitz.}), it admits an extension $\tilde u\in W^{1,\infty}(\R^n)$. Then
\begin{equation}\label{exte}
\bar u(x):=\begin{cases} \tilde u(x)&\text{if $x\in B(0,R)$}\\
u(x)&\text{otherwise}\end{cases}
\end{equation}
is an extension of $u$ to $\R^n$ such that $|\nabla \bar u|\in L^{\infty}(\R^n)$; then it is Lipschitz on $\R^n$ (see, e.g. \cite[Ex. 11.50-(i)]{Leoni17})  and $\bar u_{|\overline \Omega_c}$ is a  Lipschitz extension of $u$.
\end{proof}

Now we are ready to show
that $\mathcal X$ is included in $W^{1,p}(\Omega_c)$, for all $p\in [2^*,+\infty]$.
\begin{lem}\label{sprimo}
Let  $u\in\mathcal X$; then $u\in W^{1,p}(\Omega_c)$, for all $p\in [2^*,+\infty]$ and moreover $\displaystyle\lim_{|x|\to\infty}u(x)=0$.
\end{lem}
\begin{proof}
 As in the proof of Lemma~\ref{Lext},  $u$ admits a Lipschitz  extension $\bar u$ to $\R^n$. Since $\Omega$ is  bounded,  we get that

\[\bar u \in \ \W(\R^n)\cap L^{2^*}(\R^n)\cap C^{0,1}(\R^n)=D^{1,2}(\R^n)\cap  C^{0,1}(\R^n).
\]
 we deduce that $\bar u\in W^{1,p}(\R^n)$ for all $p\in [2^*,+\infty]$ and $\lim_{|x|\to\infty}\bar u(x)=0$;  plainly, analogous properties hold for~$u$.
\end{proof}

Finally, let $\varphi\in L^{2}(\partial \Omega)$ be such that there exists $w\in\mathcal X$ with $\mathrm{Tr}(w)=\varphi$. By Lemma~\ref{Lext} we  have that $\varphi=w_{|\partial\Omega}$ (in particular $\varphi$ must be Lipschitz on $\partial\Omega$).
Let us set
\[\X :=\{u\in\mathcal X:\mathrm{Tr}(u)=\varphi\}.\]
Then by Proposition~\ref{trace} for every $w\in \X$ we get
\[\X=\{w\}+\mathcal X_0.\]

\begin{prop}\label{weakly}
 $\X$ is convex and weakly closed as   a topological subset  of  the semi-normed vector space $\W(\Omega_c)$.  Moreover,   if $(u_k)_k$ is a bounded sequence in $\X$, then up to a subsequence it weakly converges to some $u\in    \X $ and uniformly converges on compact subsets of $\Omega_c$.
\end{prop}
\begin{proof}
 Fix any $w\in \X$;  since  $\X$ is the translation by $w$ of a convex and weakly closed  subset of $D^{1,2}(\Omega_c)\hookrightarrow\ \W(\Omega_c)$ and the trace operator is linear, we get that $\X$ is convex and weakly closed in $\W(\Omega_c)$.

If $(u_k)_k$ is a bounded sequence in $\mathcal X$, then $(u_k-w)_k$ is bounded in $D^{1,2}(\Omega_c)$, hence there exists a weakly  converging subsequence to a function $v\in D^{1,2}(\Omega_c)$. This implies that $u_k\rightharpoonup w+v\in \X$. Moreover,  by Lemma~\ref{prop}--\eqref{p2},  $(u_k-w)_k$ is bounded in $L^\infty(\Omega_c)$ and then, by Ascoli-Arzel\`a theorem,  for any given compact set $K\subset \Omega_c$  it admits a  subsequence uniformly converging to $v$ on $K$.
\end{proof}

\section{Proofs of the main results}\label{s3}
Let $\varphi\in L^{2}(\partial \Omega)$ be such that there exists $w\in\mathcal X$ with $\mathrm{Tr}(w)=\varphi$ and consider the functional
\beq\label{I}
{\mathcal{I}}(u)=
\int_{\Omega_c}\left(1-\sqrt{1-|\nabla u|^2}\right) {\rm d}x+
\int_{\Omega_c}G(x,u)\,{\rm d}x
\eeq
with $G(x,t):=n\int_0^tH(x,s)\;{\rm d} s$. Recalling that $\frac{1}{2}t\leq 1-\sqrt{1-t}\leq t$, for all  $t\in [0,1]$,     by  assumption \eqref{h} $\mathcal I$ is well-defined on  $\X$ (see also the proof of the first part of  Lemma \ref{deb}).
Notice also that every spacelike critical point $u$ of $\mathcal I$ weakly satisfies (in the sense of Definition \ref{def:ws}) the first equation in \eqref{star}.
Thus, a  spacelike weak  solution of \eqref{star} can be found if  $\mathcal I$ has a spacelike minimizer on $\X$. In order to prove this last statement we need the following lemmas.
Let us set
$$
\mathcal I_0(u):=\int_{\Omega_c}\left(1-\sqrt{1-|\nabla u|^2}\right){\rm d}x\quad\text{and}\quad \mathcal{G}(u):=\int_{\Omega_c}G(x,u)\, {\rm d}x.
$$

\begin{lem}\label{deb}
Under the assumption on $\Omega_c$ and $H$ in Theorem~\ref{maint}, assume also that $\varphi=\mathrm{Tr}(w)$ for some $w\in\mathcal X$. Then
 $\mathcal{G}$ is well-defined on $\X$ and  sequentially  weakly continuous.
\end{lem}
\begin{proof}
Let us  denote  by $s'$ the conjugate exponent of $s$.   As  $u\in \X\subset \mathcal X$,  by Lemma~\ref{sprimo} we get that $u\in W^{1,p}(\Omega_c)$ for all $p\in [2^*,+\infty]$.
Then for all  $s\in [1,\frac{2n}{n+2}]$,  $s'\geq 2^\ast$ and
$$
|\mathcal{G}(u)|\leq \|h\|_s\|u\|_{s'}<+\infty.
$$
Let us now show that $\mathcal G$ is sequentially weakly continuos on $\X$. Assume that  $(u_k)_k\subset \X$ weakly converges to $u\in \ \W(\Omega_c)$. By Proposition~\ref{weakly}, $u\in\X$;
moreover $(u_k-w)_k\subset \mathcal X_0$ and then it is bounded in $D^{1,2}(\Omega_c)$.
Since $h\in L^s(\Omega_c)$, for a given $\epsilon>0$ there exists $R>0$ such that $\overline\Omega\subset B(0,R)$ and
$$
\|h\|_{L^{s}(\R^n\setminus \overline{B(0,R)})}<\epsilon;
$$
therefore by  Lemma \ref{prop}--\eqref{p1}, there exists $C_1>0$ such that
\begin{align}
\int_{\R^n\setminus \overline{B(0,R)}}|G(x,u_k)|\;{\rm d}x
&\leq \|h\|_{L^{s}(\R^n\setminus \overline{B(0,R)})}\|u_k\|_{s'}
\nonumber
\\
&\leq
\|h\|_{L^{s}(\R^n\setminus \overline{B(0,R)})}\left(\|u_k-w\|_{s'} + \|w\|_{s'}\right)

\leq C_1 \epsilon \label{pippo}
\end{align}
and
\begin{equation}\label{pippa}
\int_{\R^n\setminus \overline{B(0,R)}}|G(x,u)|\;{\rm d}x
\leq C_1 \epsilon.
\end{equation}
Notice that, again by assumption \eqref{h}, taking the bounded subset $B(0,R)\setminus \overline\Omega$, it results in particular $h\in L^1(B(0,R)\setminus\overline\Omega)$ and then
by  Lemma \ref{prop}--\eqref{p2}, for a $C_2>0$ we get
$$
|G(x,u_k)|\leq h(x)\|u_k\|_{\infty} \leq h(x)(\|u_k-w\|_{\infty}+\|w\|_{\infty})\leq C_2 h(x),
$$
for all  $k\in\N$,  a.e.  in $B(0,R)\setminus\overline\Omega$.
By Proposition~\ref{weakly} the bounded sequence $(u_k)_k$ uniformly converges on compact subsets in $\Omega_c$, up to a subsequence. Hence, $(G(x,u_k))_k$ converges to $G(x,u)$ a.e. in $B(0,R)\setminus\overline\Omega$ and  by the Lebesgue's dominated convergence theorem
it follows that
$$
\int_{B(0,R)\setminus\overline\Omega}G(x,u_k)\;{\rm d}x \longrightarrow
\int_{B(0,R)\setminus\overline\Omega}G(x,u)\;{\rm d}x, \quad\hbox{as } k\longrightarrow +\infty.
$$
As the last  convergence actually holds for the whole sequence, being $\epsilon$ arbitrary, by \eqref{pippo} and \eqref{pippa}, we are done.
\end{proof}
Let us now show that $\mathcal I$ has a global minimum point on $\X$.

\begin{prop}\label{pr:min}
Under the assumption on $\Omega$ and $H$ in Theorem~\ref{maint}, the functional $\mathcal{I}$ possesses at least a minimizer in $\mathcal{X}_\varphi$.
\end{prop}

\begin{proof}
By assumption \eqref{h},  Lemma~\ref{sprimo} and \eqref{p1}-\eqref{p2} of Lemma~\ref{prop}, denoting by $s'$ the conjugate exponent of $s$, for all $u\in\X$ we have:
\baln
\mathcal{I}(u)&\geq \frac{1}{2}\|\nabla u\|^2_2-\|h\|_s\|u\|_{s'}\\
&\geq\frac{1}{2}\|\nabla u\|^2_2-\|h\|_s\|u-w\|_{s'}-\|h\|_s\|w\|_{s'}\\
& \geq \frac{1}{2}\|\nabla u\|^2_2-C_1\|h\|_s\|\nabla u-\nabla w\|_{2}-\|h\|_s\|w\|_{s'}\\
&\geq \frac{1}{2}\|\nabla u\|^2_2-C_1\|h\|_s\|\nabla u\|_2-C_1\|h\|_s\|\nabla w\|_{2}-\|h\|_s\|w\|_{s'}
\ealn
hence $\mathcal{I}$ is coercive.  By Proposition~\ref{weakly}, in order to get the existence of a minimizer, we just need to prove that $\mathcal{I}$ is sequentially weakly lower semi-continuous. Actually this holds because the first term $\mathcal{I}_0$ of $\mathcal I$ is convex and strongly continuous (cf.  \cite[Lemma 2.2]{BoPo16} for details), thus it is  sequentially  weakly lower semi-continuous in the semi--normed space $\W(\Omega_c) $
 and  $\mathcal{G}$ is  sequentially  weakly continuous by  Lemma~\ref{deb}.
\end{proof}
The lemma below is based on \cite{BM} and shows that a minimizer $u$ of $\mathcal I$ is a  minimizer also for the analogous  functional  corresponding to  $\mathcal I$ with  $H^*(x)=nH(x,u(x))$ replacing $nH$.
\begin{lem}\label{min}
	Under the assumptions of Lemma~\ref{deb}, assume also that $u$ is a minimizer of $\mathcal I$ on $\X$ and set $H^*(x):=nH(x,u(x))$. Then $u$ is  also  a minimizer   of the functional $\mathcal I^*\colon \X\to \R$,  where
	\[\mathcal I^*(v):=\int_{\Omega_c}\left(1-\sqrt{1-|\nabla v|^2}\right){\rm d}x+\int_{\Omega_c}H^*(x)v(x)\,\de x.\]
\end{lem}
\begin{proof}
	Let $v\in \X$; then   $u_\lambda:=u+\lambda(v-u)\in \X$, for all $\lambda\in[0,1]$. Hence for all $\lambda\in (0,1]$
	$$\mathcal I_0(u_\lambda)+\mathcal G(u_\lambda)\geq \mathcal I_0(u)+\mathcal G(u)
	$$
	and    being  $\mathcal I_0$ convex  we obtain
	\beq\label{withlambda}
	\mathcal I_0(v)-\mathcal I_0(u)-\frac 1 \lambda \big(\mathcal G(u_\lambda)-\mathcal G(u)\big)\geq 0.
	\eeq
	Let $\sigma(x)\in [0,1]$ be such that
	\[
	\frac 1 \lambda \big(\mathcal G(u_\lambda)-\mathcal G(u)\big)= n \int_{\Omega_c}H\big(x,u+\sigma(x)\lambda(v-u)\big)(v-u)\ \de x;
	\]
	hence, by assumption \eqref{h} and Lebesgue's dominated convergence theorem (recall that $v-u\in \mathcal X_0\hookrightarrow L^{\infty}(\Omega_c)$), we get
	\[ n \int_{\Omega_c}H\big(x,u+\sigma(x)\lambda(v-u)\big)(v-u)\ \de x\longrightarrow \int_{\Omega_c}H^*(x)(v-u)\ \de x,\]
	 as $\lambda\to 0$, which by \eqref{withlambda} implies
	\[\mathcal I^*(v)-\mathcal I^*(u)\geq 0.\]
\end{proof}

\bere
Let $\bar u\in \mathcal{X}_\varphi$ be a minimizer of $\mathcal{I}$ found in Proposition \ref{pr:min}. By Lemma \ref{min} all the conclusions of \cite[Proposition 2.7]{BoPo16}, suitable modified, hold and, in particular, we have that
\[
{\rm meas}\{x\in \Omega_c: |\nabla \bar u|=1\}=0
\]
and
\[
\frac{|\nabla \bar u|^2}{\sqrt{1-|\nabla \bar u|^2}}\in L^1(\Omega_c).
\]
However, this is still not enough to conclude that $\bar u$ is a weak solution of problem \eqref{star} in the sense of Definition \ref{def:ws}.

Finally, we point out that,  by \cite[Proposition~1.1]{BarSim83}, the minimizer is unique if
$H$ is non-decreasing in  $t$.

\ere

We can now conclude the proof of our main result.
\begin{proof}[Proof of Theorem \ref{maint}.]
As a first step we want to prove that if $\varphi$ is the trace of a function in $\mathcal X_\varphi$, then each minimizer $\bar u$ of $\mathcal{I}$ in $\mathcal{X}_\varphi$ is not only weakly spacelike, but  spacelike as well, and so it is a weak solution of \eqref{star}.

By Lemma~\ref{min} $\bar u$ is a minimizer for the functional $\mathcal I^*$ and, since  $H$  is locally bounded by   \cite[Theorem 3.2]{BarSim83}, we infer that any segment  of a lightlike geodesic possibly contained in the graph of $\bar u$ can be extended until it reaches a point on the graph of $\varphi$. This fact and the local estimates in the proof of  \cite[Theorem 4.1]{BarSim83} imply that  the subset  $K\subset \Omega_c$ where a minimizer could be non-regular  is precisely given by the points which are the projections on $\Omega_c$ of light rays  and lightlike segments (i.e., resp.,  lines or half-lines and  segments whose tangent vectors are  lightlike) contained in the graph $G_{\bar u}$, such that, respectively, no point or at least one  point or both endpoints  belong to $\partial\Omega_c$. Thus, if $\Omega$ is convex, $K$ might  contain only lines or half-lines (without, potentially, their points in the boundary) which are projections of light rays in $G_{\bar u}$, but this is incompatible with the fact that $\lim_{|x|\to\infty}\bar u(x)=0$ (recall Lemma~\ref{sprimo}). If $\Omega$ is not convex and there exists a segment $\overline{xy}$ contained in $K$ whose endpoints $x,y$ belong to $\partial \Omega$,  we would have
$|\varphi(x)-\varphi(y)|=|\bar u(x)-\bar u(y)| = |x-y|$, in contradiction with the spacelike displacing assumption on $\varphi$.
Therefore,  $K=\emptyset$ and $\bar u$ is spacelike.

Let us now prove the reverse implication of the theorem.
We observe that, according to Definition~\ref{def:ws}, any spacelike weak  solution $\bar u$ of \eqref{star}
is locally $1$--Lipschitz, then by Lemma~\ref{Lext} $\varphi=\bar u_{|\partial\Omega}$ and if $\Omega$ is not convex and there exists a segment $\overline{xy}$ in $\Omega_c$ connecting two points $x, y\in\partial \Omega$, then $|\varphi(x)-\varphi(y)|=|\bar u(x)-\bar u(y)|<|x-y|$, i.e., $\varphi$ is spacelike displacing in $\Omega_c$.

\end{proof}
If $\varphi$ is $(1-\epsilon)$--Lipschitz, we can show that it is the trace of a function in $\mathcal X_\varphi$.
\begin{proof}[Proof of Corollary~\ref{mainc}]
We can extend $\varphi$ to a bounded $(1-\epsilon)$--Lipschitz function $\psi$ on $\R^n$ such that $\min\psi=\min\varphi$ and $\max\psi=\max\varphi$ (see, e.g., \cite[p. 243]{Leoni17}). Let us take $R>0$ such that $\overline{\Omega}\subset B(0,R)$, $\epsilon R >\|\varphi\|_\infty$ and a smooth function
$v:\R^n\to [0,1]$ with compact support in $B(0,2R)$,  equal to $1$ on $B(0,R)$ and such that $|\nabla v|\leq 1/R$. As $|\nabla(v\psi)|\leq \frac{\|\varphi\|_\infty}{R}+1-\epsilon<1$, the restriction of $v\psi$ to $\Omega_c$ belongs to $\X$. Therefore $\varphi$ is the trace of a function in $\mathcal X_\varphi$ and we can conclude by Theorem \ref{maint}.
\end{proof}


text:
\documentclass[11pt,a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\usepackage{isabelle,isabellesym}

% this should be the last package used
\usepackage{pdfsetup}

% urls in roman style, theory text in math-similar italics
\urlstyle{rm}
\isabellestyle{it}

\begin{document}

\title{Young's Inequality for Increasing Functions}
\author{Lawrence C. Paulson}
\maketitle

\begin{abstract}
Young's inequality states that
$$ ab \leq \int_0^a f(x)dx + \int_0^b f^{-1}(y) dy $$
where $a\geq 0$, $b\geq 0$ and $f$ is strictly increasing and continuous.
Its proof is formalised following the development by Cunningham and Grossman~\cite{cunningham-youngs}.
Their idea is to make the intuitive, geometric folklore proof rigorous by reasoning about step functions.
The lack of the Riemann integral makes the development longer than one would like, but their argument is reproduced faithfully.
\end{abstract}

\newpage
\tableofcontents
\newpage

% include generated text of all theories
\input{session}

\bibliographystyle{abbrv}
\bibliography{root}

\end{document}

subset_name: curated


TITLE: Numerical solution to integro-differential equation (Wigner Weisskopf Problem)
QUESTION [1 upvotes]: The time dynamics of an atom interacting with a reservoir of spectral density $J(\omega)$ are obtained by solving the following integro-differential equation $[1]$:
$$ \dot{c}(t) = \int_0^t dt'\int_0^{\infty}d\omega J(\omega)e^{i(\omega_0-\omega)(t-t')}c(t') $$
This can be solved analytically, for example when $J(\omega)$ is Lorentzian $[1]$. But how could I solve this numerically (in python), given $J(\omega)$ as an array of data? Is it always possible?
References:
$[1]$: $\textit{The Theory of Open Quantum Systems, Petruccione, Breuer p468}$

REPLY [1 votes]: This equation can be solved via Laplace transforms, which is even possible in a semi-analytical fashion without direct numerical schemes such as discretization. See e.g. https://journals.aps.org/pra/abstract/10.1103/PhysRevA.89.033820 .


TITLE: How do you find the maximum integer value of $(3\cos \theta-5)^2$ without derivatives?
QUESTION [1 upvotes]: The problem is as follows:

The owner of a bakery invests $3450\,USD$ in order to prepare $150$
cakes of $1\,kg.$ for his shop. Assuming that he would sell each for
$(B+2)\,USD$, where $B$ is the maximum integer value which it can take
the expression from below:
$$B=(3\cos \theta-5)^2$$
It is known that angle $\theta$ is acute.
Assuming that at the end of the month he is able to sell all his
cakes. Find the profit which it was obtained from the sale of all the
cakes.

The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\textrm{300 USD}\\
2.&\textrm{400 USD}\\
3.&\textrm{350 USD}\\
4.&\textrm{450 USD}\\
\end{array}$
The thing here I'm assuming is that what it is being requested is in short:
Where $P$ is the profit.
$$P=150(B+2)\times 30-3450$$
Thus in the end the question here is how to get the maximum integer value of $B$, as this was specifically stated in the problem.
$B=(3\cos \theta-5)^2$
The only thing which I can remember is that to maximize you could use derivatives. So for that previous equation become into:
$B'=2(3\cos \theta-5)(-3\sin\theta)=-6\sin\theta(3\cos \theta-5)$
From solving this and equating it to zero:
$B=-6\sin\theta(3\cos \theta-5)$
$\sin\theta= 0$
which reduces to
$\theta = 0^{\circ}$
and
$3\cos \theta-5=0$, but this would mean:
$\cos\theta=\frac{5}{3}$
But this solution is not possible in the real set of numbers because $-1<\cos\theta<1$.
In this situation the maximum value attained for that function would be when $\theta = 0^{\circ}$.
This would mean:
$B=(3\cos 0^{\circ}-5)^2=(3-5)^2=4$
But this doesn't seem to be the right approach because the result will at the end be negative?.
Why the approach using derivatives isn't working? Can this be done using another method in precalculus?.
Upon further inspecting the equation. I'm getting the idea that to maximize $B$ I would want $\cos\theta$ to be the less as possible.
Since it mentions that it want an integer value maybe it would mean?
$\cos\theta=\frac{1}{3} \approx 70.53^{\circ}$
Hence:
$B=(1-5)^2=16$
Hence $16\times 150$
Another thing which it came to my mind was:
$0<\cos \theta < 1$
Thus
$-5 < 3\cos \theta-5<-2$
$25 < (3\cos \theta -5)^2< 4$
Although this doesn't seems logical, but would it meant that
the maximum integer value is 24. Hence
$P=150(24+2)-3450=450$
Thus the answer would it be that number?.
Again I'm confused. Can someone help me here?. I'm confused because the sitation doesn't seem to work with derivatives and if using a definition of the boundaries for the cosine function it seems to cause a contradiction. What's wrong here?. Please take the time to read all my effort because it wasn't easy for me and what it would help me the most is an explanation where did my strategy failed?

REPLY [1 votes]: Since we are told that $\theta$ is acute, we have $0 < \theta < \pi/2$.  On this interval, $0 < \cos \theta < 1$, so $3 \cos \theta < 5$.  Hence $$B= (3 \cos \theta - 5)^2 = 9 \left(\frac{5}{3} - \cos \theta\right)^2$$ which shows that as $\cos \theta \to \pi/2$, $B$ increases; specifically, at $\theta = 0$, $B = 25$.  But since $\theta = 0$ is not attainable, the maximum integer value of $B$ must be $B = 24$, for which we can backsolve and get $\theta = \arccos \frac{5 - 2\sqrt{6}}{3}$.  We don't actually need to solve for this specific value of $\theta$, only know that it exists.  The profit is therefore $150(26)-3450 = 450.$

REPLY [0 votes]: Since $\theta$ is acute
$$0 \lt \cos \theta \leq 1$$
$$\implies 0 \lt 3\cos\theta \leq3$$
$$\implies -5 \lt 3\cos\theta -5 \leq 2$$
$$(3\cos\theta-5)^2 \lt 25$$
Therefore $B = 24$
Yes, the final approach that you made is right - there would be some acute theta for which $B=24$, you aren't really concerned in finding the angle, just it's final value in terms of B


TITLE: transitive subgroups in $S_n$ that are isomorphic to $S_k$ for $k\leq n$
QUESTION [5 upvotes]: Assume that there is a transitive subgroup  $H$ of $S_n$ w.r.t. the standard action $S_n\curvearrowright \{1,2,\cdots, n\}:=X$ such that $H\cong S_k$. Is there any sharp estimate on the upper bound of $n$ in terms of $k$ for large $n$, say for all $n\geq 5$?
For example, one can show that $S_n=HK$, where $K$ is the stabilizer subgroup of $S_n$ for any point in $X$ and hence $K\cong S_{n-1}$. Hence we know $n!\leq k!\cdot (n-1)!$, thus, $n\leq k!$. In particular, is this estimate sharp for $n\geq 5$?
Thanks!

REPLY [4 votes]: Here is a summary of the answers in the comments.
As runway44 says, the answer to the question asked is yes, $k!$ is a sharp upper bound, because the regular representation of $S_k$ is transitive on $n=k!$ points.
A possibly more interesting question would be to as for a lower bound for $n$ in terms of $k$ with $n>k$. This is $k(k-1)/2$ for $k>2$ and $k \ne 5$. For $k=5$ we can have $n=6$.


TITLE: Finding the dimension of a polynomial that is a subspace of $V$
QUESTION [0 upvotes]: Let $V$ be the set of all real polynomials $p$ of degree at most $2$, satisfying $p (0) = p(1)$. I have determined that $V$ is a subspace, but now must determine the dimension of $V$.
Say $p(x) = a + bx + cx^2$
I have plugged in numbers to get: $a + b+ c = a$
Standard basis of P2 is $[1,x,x^2]$
Now, I'm not sure how to find the basis and thus dimension, considering that it must satisfy $P(0) = P(1)$.

REPLY [0 votes]: For $p\in V$ let $p(x)=A(x-1/2)^2+B(x-1/2)+C$ with constants $A,B,C.$ Since $p(0)=p(1),$ we have $B=0$. So $p(x)=A(x-1/2)^2+C.$  Clearly $f=(x-1/2)^2$ and $g=1$ are linearly independent, and $p=Af+Cg.$


TITLE: Is there a canonical extension of Euler's totient function to the reals?
QUESTION [5 upvotes]: Assumption: if a set $X\times Y$, exists such that $\forall\textbf{u},\textbf{v}\in X\times Y. \left(u_1=v_1\implies u_2=v_2\right)$, then a function $f$ exists such that $f:X\to Y$.

The set of points $\Phi_\mathbb{N}=\left\{(n,\phi(n))\mid n\in\mathbb{N}\right\}$ describing the graph/image of the totient function is a subset of the reals.
By definition, $\forall \textbf{x},\textbf{y}\in\mathbb{R}^2$, where $x_1< y_1.\exists \textbf{z}\in\mathbb{R}^2:x_1<z_1<y_1$. Thus, there is a continuous set of points between any two $\textbf{a},\textbf{b}\in\Phi_\mathbb{N}$. It follows, that there is a a set $\Phi_\mathbb{R}\supset\Phi_\mathbb{N}$ which is the graph of a continuous function $\phi:\mathbb{R}\to\mathbb{R}$, ($n\in\mathbb{N}\implies \phi(n)\in\mathbb{N}$).
Thus, it is at least possible to extend the totient function to the reals without breaking the original definition. Is there a canonical way to do this? If not, what is the most appropriate way to extend the totient function?

Note: An extension to the complex numbers is perfectly acceptable too, I just assumed it would be best to "keep it real".
Note: tagged analysis because that's where I'm going with this.

Edit: This might help... probably not, though
For coprime $a$ and $n$
$$a^{\phi(n)}\equiv1\mod{n}\qquad\text{(Euler's Theorem)}$$
For $x,y\in\mathbb{R}$
$$x\ \text{mod}\ y^{*}=\frac{|x|}{2\pi}\sum_{k=1}^\infty\frac{1}{k}\Im\left[\exp\left(\frac{2\pi kix}{y}\right)\right]$$
So
$$a^{\phi(n)}\ \text{mod}\ n=\frac{|a^{\phi(n)}|}{2\pi}\sum_{k=1}^\infty\frac{1}{k}\Im\left[\exp\left(\frac{2\pi kia^{\phi(n)}}{n}\right)\right]=1\implies$$
$$|a^{\phi(n)}|=2\pi\left(\sum_{k=1}^\infty\frac{1}{k}\Im\left[\exp\left(\frac{2\pi kia^{\phi(n)}}{n}\right)\right]\right)^{-1}=$$
$$-4\pi i\left(\ln\left(1-\exp\left(\frac{2\pi ia^{\phi(n)}}{n}\right)\right)-\ln\left(\exp\left(-\frac{2\pi ia^{\phi(n)}}{n}\right)\left(-1+\exp\left(\frac{2\pi ia^{\phi(n)}}{n}\right)\right)\right)\right)^{-1}$$
Assuming $a^{\phi(n)}$ is positive, let $a^\phi(n)=u$ (because I'm running out of space)
$$u=-4\pi i\left(\ln\left(1-\exp\left(\frac{2\pi iu}{n}\right)\right)-\ln\left(\exp\left(-\frac{2\pi iu}{n}\right)\left(-1+\exp\left(\frac{2\pi iu}{n}\right)\right)\right)\right)^{-1}$$
Solve for for $u$ (not sure how), then substitute $a^{\phi(n)}$ back in and take the base-$a$ logarithm to get $\phi(n)$
Assuming all the algebra is correct, this might work...
$^*$ If I understand correctly, we can translate $x\equiv z\mod{y}$ to $x\ \text{mod}\ y=z$, using $\text{mod}$ as a binary operation.

REPLY [7 votes]: Let's first try to define $\phi$ on $\Bbb Q$; presumably, any canonical definition on $\Bbb R$ would have a restriction to $\Bbb Q$ that would meet with our focus-on-$\Bbb Q$ approval. It will be convenient to define $f(n):=n^{-1}\phi(n)$.
For $n\in\Bbb N$ we have $f(n)=\prod_{p\in\Bbb P\land p|n}(1-p^{-1})$, and it is natural to define $\phi(0)=0$ (since there are no integers from $1$ to $0$ exclusive to be coprime to $0$) and, if we want to extend to negative integers, presumably $f(-n)=f(n)$ since the same primes divide $-n$. If for coprime $m,\,n\in\Bbb Z$ with $n>0$ we demand $f(m/n)=f(m)f(n)$ [sic], so as to include $1-1/p$ factors for every prime appearing in $m/n$'s prime factorisation, we have a canonical extension to $\Bbb Q$. In particular, this definition implies $\phi(p^k)=p^k(1-p^{-1})$ for $p\in\Bbb P,\,k\in\Bbb Z\backslash\{0\}$; unsurprisingly, as $p^k\to 0$ we also have $\phi\to 0$.
However, I don't think we can extend this to $\Bbb R$. The only obvious definition is $\phi(x):=\lim_{n\to\infty}\phi(x_n)$ for rationals $x_n$ satisfying $\lim_{n\to\infty}x_n=x$. But I would be very surprised if, for all irrational $x$, this definition is independent of the sequence chosen. In fact, $\phi$ as defined above isn't even continuous at $0$. For example, take your favourite distinct primes $p,\,q$, then define $x_n:=p^n/q^{a_n}$, with $a_n$ just large enough that $x_n<1/n$. Then $\phi(x_n)=(1-1/p)(1-1/q)$ for all $n$.


text:
{\bf Problem.} What is the remainder when the sum of the first 102 counting numbers is divided by 5250?  (The first counting number is 1.)
{\bf Level.} Level 3
{\bf Type.} Algebra
{\bf Solution.} For all $n$, $1 + 2 + \dots + n = n(n + 1)/2$, so $1 + 2 + \dots + 102 = 102 \cdot 103/2 = 5253$.  The remainder when 5253 is divided by 5250 is $\boxed{3}$.

set_name: MATH