TITLE: minimal even partition among the ones $\geq$ a given partition
QUESTION [2 upvotes]: Let $N$ be an even integer. A partition of $N$ is a sequence $(a_1,a_2,\cdots, a_N)$ of integers such that $a_1 \geq a_2 \geq \cdots \geq a_N \geq 0$, and $\sum_{j=1}^Na_j =N$.
Let $\mathscr P_N$ be the set of partitions of $N$. A partition $\nu \in \mathscr P_N$ is said to be even if $\nu = (\nu_1,\nu_2,\cdots \nu_N)$ satisfy $\nu_i$ is even for $i=1,2,\cdots, N$.
The order "$\leq$" is defined as follows. Let $\nu^1 =(\nu^1_1,\nu^1_2,\cdots,\nu^1_N)$, $\nu^2=(\nu^2_1,\nu^2_2,\cdots,\nu^2_N) \in \mathscr P_N$. We say that $\nu^1 \leq \nu^2$ if $\sum_{j=1}^i \nu^1_j \leq \sum_{j=1}^i \nu^2_j$ for $i =1,2,\cdots,N$. (It is easy to notice that $\mathscr P_N$ has a unique maximal element, which is $(N,0,0,\cdots,0)$, and a unique minimal element, which is $(1,1,\cdots,1)$.)
Let $\lambda \in \mathscr P_N$. Let $$\Gamma =\{\nu \in \mathscr P_N | \lambda \leq \nu \text{ and } \nu \text{ is even}\}.$$
Then does $\Gamma$ always have a unique minimal element (under "$\leq$")? Is there any standard process for me to find a minimal element of $\Gamma$ from $\lambda$?
Thanks to everyone.

REPLY [1 votes]: To find a minimal element:
For convenience, let $\lambda=(\lambda_1, \lambda_2,\cdots, \lambda_N).$
If $\lambda $ is itself even then it is a minimal element. Otherwise, at least one $\lambda_i $  is odd.  Since N is even, and $\sum_{i=1}^{N}{\lambda_i}=N$, number  of $\lambda_i $ which are odd, is even. Now to find the minimal element $m=(m_1, m_2,\cdots, m_N)$ from $\lambda $:
Step-1: Keep/Retain all the even $\lambda_i$ as such in $m .$
Step-2: Add 1 to the first odd $\lambda_j$ and subtract 1 from the immediate next odd $\lambda_k .$ Do this till all $\lambda_i' s $ become even.
Step-3: Rearrange the resulting values so that $m_1\geq m_2\geq....\geq m_N.$
Does it help you?


TITLE: Simplifying a Boolean algebra expression
QUESTION [1 upvotes]: ~ABC + ~BCD + ~A~BC~D + ABCD = ?
The solution book gives this answer:
~AC + CD
I arrived this one:
ACD
My question is if the method used to simplify can affect the result or if should it be equal?
Note: ~ stands for not ( ~A = Not A)
My steps:
~ABC + ~BCD + ~A~BC~D + ABCD =
=CD(~B+AB) + ~AC(B+~B~D)=
=CD(~B+A) + ~AC(B+~D)=
=~BCD + ACD + ~ABC + ~AC~D=
=C(~BD + AD + ~AB + ~A~D)=
=C[D(~B + A + ~AB + ~A~D)]=
=C[D(~B + A + B + ~A~D)]=
=C[D(A + ~A~D)]=
=C[D(A + ~D)]=
=CDA + CD~D=
=ACD

REPLY [1 votes]: It's possible to get an answer that looks different, but it should always be logically equivalent, no matter what route you take.
Unfortunately, your answer is not equivalent to the one provided by the book: If $A$ is False, and $C$ is True, then your expression is False, but the one from the book is True.
Try this: take the second and third term: $B'CD$ and $A'B'CD'$. Notice that they both have $B'C$ in common, so that definitely has to be true in order for one of those terms to be true. However, for the first term to be true, you also need $D$, while for the second you need $A'D'$, and so for one of those to hold you either need $D$ or $A'$ (because you already have $D'$ if not $D$). So, these two terms can be simplified to $B'CD$ and $A'B'C$
Formally:
$B'CD+A'B'CD'= B'C(D + A'D')= B'C(D+A')(D+D')=B'C(D+A')1=B'C(D+A')=B'CD+B'CA'$


TITLE: What creates a unique quaternion?
QUESTION [0 upvotes]: You can create a unique point N from point M and vector V.  You can create a unique vector B from vector A and quaternion Q.  Can you create a unique quaternion Y from a another quaternion X, given some other type of transformation like that of a vector or a quaternion?
I don't know if there would be any practical value to this kind of transformation, but would be interested to know if something like this exists.
Edit:
Hopefully I can clarify,
If I have Point N and translate it by vector V I get a new point M.  Meaning if point N is a point at (0, 0) and vector V has a magnitude of 1 and direction of 1 (1, 1), I get the a new point M at (1, 1).
Same for vectors and quaternions, except I get a new vector from applying a quaternion to a vector. (scalar and versor)
I'm wondering if there is some set 'like a quaternion' that can transform one quaternion into a new quaternion.
Vectors transform points.
Quaternions transform vectors.
"Blank" transforms quaternions?

REPLY [3 votes]: I think a lot of the confusion around your question comes from the fact that you don't really have a 'chain' of the sort you're talking about, because different senses of the word 'transform' are being used in each case.
Very broadly, there's little distinction between 'points' and 'vectors' in the sense that you're specifying; each point can be identified with the vector from the origin to that point.  Viewed from this perspective, the transformation process you're talking about that takes the form of a map (or, if you prefer, a function) from $\mathbb{R}^n\rightarrow\mathbb{R}^n$ (here $\mathbb{R}^n$ just means the space of $n$-dimensional vectors, and $\mathbb{R}$ means we're talking about vectors over the real numbers).  Each vector $\vec{v}$ gives rise to the transformation $T_{\vec{v}}: \mathbb{R}^n\rightarrow\mathbb{R}^n$ given by $T_{\vec{v}}(\vec{w}) = \vec{v}+\vec{w}$.  Note that 'applying' two different vectors $\vec{u}$ and $\vec{v}$ to the same point $\vec{w}$ is the same as applying their sum: $T_\vec{u}(T_\vec{v}(\vec{w})) = T_\vec{u}(\vec{v}+\vec{w}) = \vec{u}+(\vec{v}+\vec{w}) = (\vec{u}+\vec{v})+\vec{w} = T_{\vec{u}+\vec{v}}(w)$.
Now, it's worth noting here that I used '$n$' above — vectors aren't limited to just 3-dimensional space.  The example you gave, for instance, is a two-dimensional vector, and vectors make sense in five dimensions, ten dimensions, and even infinitely many dimensions (some caveats apply!).  On the other hand, quaternions only${}^* $ work in 3 dimensions (${}^* $: and 4, sort of, but that's a more complicated story) and they don't produce all transformations of vectors in 3 dimensions; they only represent rotations of 3-dimensional space.  Each quaternion $\mathbf{q}$ gives rise to a transformation $T_{\mathbf{q}}: \mathbb{R}^3\rightarrow\mathbb{R}^3$ given by $T_{\mathbf{q}}(\vec{v}) = \mathbf{q}\vec{v}\mathbf{q}^{-1}$ , where all of the multiplications are quaternion multiplications and we're representing the vector as a 'purely imaginary' quaternion (one with zero scalar part); even though you're multiplying quaternions here, the result will still be a vector (that is, the scalar part will be zero).  And much like with the vector 'transformation' above, composing transformations — that is, applying multiple quaternions to a vector, one after the other — corresponds to applying their product: $T_{\mathbf{p}}(T_{\mathbf{q}}(\vec{v})) = T_{\mathbf{p}}(\mathbf{q}\vec{v}\mathbf{q}^{-1}) = \mathbf{p}(\mathbf{q}\vec{v}\mathbf{q}^{-1})\mathbf{p}^{-1} = (\mathbf{p}\mathbf{q})\vec{v}(\mathbf{q}^{-1}\mathbf{p}^{-1}) = (\mathbf{p}\mathbf{q})\vec{v}(\mathbf{p}\mathbf{q})^{-1} = T_{\mathbf{p}\mathbf{q}}(\vec{v})$
Now, you may have noticed that even though these two examples are different, they do share a bit of flavor: in both cases, we can say that an object of the given type (a vector or a quaternion) gives rise to a transformation (or a map) from a space to itself, such that composing the transforms corresponds to adding (in the first case) or multiplying (in the second) the objects.  This concept comes up often enough that mathematicians have come up with a special name for it, the group action; you can find more about it under that name.  A group basically just means 'a bunch of abstract things with some special operator for turning two things in the group into a third, satisfying certain conditions'; here, our groups are the group of all vectors (that is, the things in the group are vectors and the operator is vector addition) and the group of all quaternions (where the things in the group are quaternions and the operator is vector multiplication), and the spaces they're acting on are either the space $\mathbb{R}^n$ of $n$-dimensional vectors in the first case or the specific space $\mathbb{R}^3$ of $3$-dimensional vectors in the second.  What you're asking is basically whether there's some other group that acts on the space of all quaternions $\mathbb{H}$ in this same way.  (Note that we call the quaternions '$\mathbb{H}$' in honor of their discoverer Hamilton).
Unfortunately, it turns out that there's an almost-trivial example, and it's closely related to your first example: any group always acts on itself, with the 'transformation' given by the group operation itself.  In the case of vectors, this is exactly your first example; every vector $\vec{v}$ gives rise to a transformation from $\mathbb{R}^n\rightarrow\mathbb{R}^n$ given by the 'group operation' vector addition.  In the case of the quaternions, it means that the quaternions transform themselves; for each quaternion $\mathbf{p}$ there's a transformation $T_{\mathbf{p}}:\mathbb{H}\rightarrow\mathbb{H}$ given by $T_{\mathbf{p}}(\mathbf{q}) = \mathbf{p}\mathbf{q}$.  If you're wondering about other groups with an action on the quaternions - that, unfortunately, I'm not sure of.  Rotations of the quaternions (seen as a general 4-dimensional space) can actually be specified by a pair of unit quaternions $\mathbf{p}$, $\mathbf{q}$ where their action on a given point in that space (that is, a quaternion) $\mathbf{h}$ is given by $T_{\mathbf{p},\mathbf{q}}(\mathbf{h}) = \mathbf{p}\mathbf{h}\mathbf{q}$; Wikipedia's page on 4-dimensional rotations has some more details on this, but I'll warn you in advance that they start to get very boggy very quickly.
Hopefully this gives you, if not quite an answer to your question, some more information on the sort of thing you're looking for!  These concepts (representations and actions) are very core in a lot of areas in mathematics and I highly encourage you to dig more into them.

REPLY [2 votes]: First of all, vectors are not the only things that you can apply to points to create points. More generally you can apply affine transformations, which include Euclidean transformations such as rotations and reflections (and therefore include everything you can get from quaternions). One says that the group of affine transformations is the automorphism group of affine space. Following the philosophy of the Erlangen program, this allows us to define affine geometry as the study of things invariant under the affine group in the same way that Euclidean geometry is the study of things invariant under the Euclidean group.
If you want things to apply to quaternions to get other quaternions, you want the automorphism group of the quaternions, in some suitable sense. There are various structures you might ask this group to preserve, and depending on exactly what you care about you'll get different answers. The automorphisms $x \mapsto qxq^{-1}$ (where $x, q$ are quaternions and $q$ is invertible) are the inner automorphisms; they preserve the structure of the quaternions as an algebra over $\mathbb{R}$ and in fact as a *-algebra. They form a group isomorphic to the special orthogonal group $\text{SO}(3)$ of rotations fixing the origin in $\mathbb{R}^3$ and I strongly suspect they give all automorphisms of the quaternions as a real *-algebra.
Keep in mind that quaternions can only be used to represent rotations in $3$ and $4$ dimensions. In higher dimensions, one appropriate generalization is to use Clifford algebras (see geometric algebra).


text:
\section{Analysis}
Firstly, let us review on the following proposition known in free probability theory.
\begin{prop}[\cite{Belinschi2003atoms}]\label{prop:atom}
Let  $\mu$ and $\nu$ be compactly supported probability distributions on $\R$.  Then $\mu \boxtimes \nu$ has an atom at $c \in \R$ if and only if the following three conditions hold : (i) $a \in \R$ (resp.\,$b \in \R$) is an atom of $\mu$ (resp.\,$\nu$), (ii) $c = ab$, and (iii)  $\mu(\{a\}) +  \nu(\{b\}) -1 > 0$.
Furthermore, if $c$ is an atom then $\mu \boxtimes \nu(\{c\}) = \mu(\{a\}) +  \nu(\{b\}) -1$.
\end{prop}

Then we have  the following recurrence equation of the maximum eigenvalue.
\begin{lemma}\label{lemma:atom}
Fix $L \in \N$.
Let  $\beta_1 = 1 $ and $\beta_{\ell}  =   1  - \sum_{k=1}^{\ell-1}(1 - \alpha_k) $  for $\ell \geq2$.
Assume that  $ \beta_L > 0$.
Then for any $\ell \leq  L$,  the value $|| \mu_\ell||_\infty$  is an atom of $\mu_\ell$ with weight  $\beta_{\ell}$.
Furthermore, we have $ || \mu_\ell ||_\infty  =    q_{\ell-1} + \sigma_\ell^2 \gamma_{\ell-1}  || \mu_{\ell-1} ||_\infty$ for $\ell \neq 1$.
\end{lemma}
\begin{proof}
Let us define $\lambda_\ell \in \R$ recursively by $\lambda_{\ell} = q_{\ell-1} + \sigma_\ell^2 \gamma_{\ell-1} \lambda_{\ell-1} (\ell \geq 2)$ and $\lambda_1=q_0$.
Firstly we prove that $\lambda_{\ell}$ is an atom with weight $\beta_\ell$ of $\mu_\ell$ for $\ell \leq L$.
In the case $\ell = 1$, we have $\mu_1 = \delta_1 = \delta_{\lambda_1}$.
Fix $\ell > 1$ and assume that $\lambda_{\ell-1}$ is an atom of $\mu_{\ell-1}$ with weight $\beta_{\ell-1}$.
Now $\beta_{\ell-1} + \alpha_{\ell-1} - 1 = \beta_\ell \geq \beta_L > 0$.
Hence by \cref{prop:atom},  $\nu_{\ell-1} \boxtimes \mu_{\ell-1}$ has an atom $\gamma_{\ell-1}\lambda_{\ell-1}$ with weight $\beta_\ell$. Therefore $\mu_{\ell}$ has the atom $\lambda_\ell$ with weight $\beta_\ell$.
The claim follows from the induction on $\ell$.
To complete the proof, we only need to show that $\lambda_\ell = ||\mu_\ell||_\infty$. Clearly $\lambda_\ell \leq  ||\mu_\ell||_\infty$.
Note that $||\mu \boxtimes \nu ||_\infty \leq ||\mu||_\infty || \nu ||_\infty$.
Then $||\mu_\ell||_\infty \leq  q_{\ell-1}   + \sigma_\ell^2 \gamma_{\ell-1}||\mu_{\ell-1}||_\infty$.
Thus it holds that $ ||\mu_\ell||_\infty  \leq \lambda_\ell$ since $||\mu_1||_\infty = q_0 =  \lambda_1$.
Hence the claim follows.
\end{proof}

Now we have prepared to prove the desired theorem.
\begin{thm}\label{thm:maximum-appendix}

Consider Assumption 4.1 and 4.2.
Then for sufficiently larger $L$,  it holds that $||\mu_L||_\infty$  is an atom of $\mu_L$ with weight $1- (L-1)(1-\alpha_{L-1})$,  and
\begin{align}
     \lim_{L\to\infty}L^{-1}||\mu_L||_\infty  = q \eps_2^{-1} \left[ 1 - \exp\left(-\eps_2\right) \right].
\end{align}
In particular, the limit has the expansion $q(1 - \eps_2/2) + O(\eps_2^2)$ as the further limit $\eps_2 \to 0$.
\end{thm}
\begin{proof}
Since $\eps_1 < 1$, we have $ 1- \alpha_{L-1} < (L-1)^{-1}$ for sufficiently large $L$. Then $\beta_L = 1 -(L-1)(1-\alpha_{L-1}) >0$. Hence by \cref{lemma:atom},  for any $\ell \leq L$,  it holds that $||\mu_\ell||_\infty$ is an atom of $\mu_\ell$ and  $||\mu_L||_\infty = q_L  \sum_{\ell=0}^{L-1}(\sigma_L^2\gamma_{L-1})^\ell$. Then by the same discussion as  Proposition~4.4, the assertion follows.
\end{proof}

\cref{thm:maximum-appendix} shows that the maximum eigenvalue of the conditional FIM $H_L$ is $O(L)$ as $L \to \infty$.
Furthermore, we emphasize that the weight $1 - (L-1)(1-\alpha_{L-1}) \sim 1 - \eps_1$ of the maximal eigenvalue $||\mu_{L}||$ is close to $1$.
Therefore, eigenvalues of the dual conditional FIM $H_L$  concentrates around $qL(1+\eps_2/2)$, and the dual FIM approximates the scaled identity operator.
Clearly the same property holds for  non-zero eigenvalues of the conditional FIM $\mc{I}(\theta|x)$.

config: arxiv


text:
\begin{document}

\title[Boundary regularity for second BVP of Monge-Amp\`ere equations]
{Boundary regularity for the second boundary-value problem of Monge-Amp\`ere equations in dimension two}

\author[S. Chen]
{Shibing Chen}
\address
{Centre for Mathematics and Its Applications,
The Australian National University,
Canberra, ACT 0200, AUSTRALIA}
\email{Shibing.Chen@anu.edu.au}

\author[J. Liu]
{Jiakun Liu}
\address
	{
	School of Mathematics and Applied Statistics,
	University of Wollongong,
	Wollongong, NSW 2522, AUSTRALIA}
\email{jiakunl@uow.edu.au}

\author[X.-J. Wang]
{Xu-Jia Wang}
\address
{Centre for Mathematics and Its Applications,
The Australian National University,
Canberra, ACT 0200, AUSTRALIA}
\email{Xu-Jia.Wang@anu.edu.au}

\thanks{This work was supported by ARC FL130100118 and ARC DP170100929.}

\subjclass[2000]{35J96, 35J25, 35B65.}

\keywords{Monge-Amp\`ere equation, global regularity.}

\begin{abstract}
In this paper, we introduce an iteration argument to prove that a convex solution to the Monge-Amp\`ere equation
$\det D^2 u =f $ in dimension two subject to the natural boundary condition $Du(\Omega) = \Omega^*$
is $C^{2,\alpha}$ smooth up to the boundary.
We establish the estimate under the sharp conditions that the inhomogeneous term $f\in C^\alpha$
and the domains are convex and $C^{1,\alpha}$ smooth. When $f\in C^0$ (resp.
$1/C<f<C$ for some positive constant $C$), we also obtain the global $W^{2,p}$ (resp. $W^{2,1+\epsilon}$) regularity.
\end{abstract}

\maketitle

\baselineskip=16.4pt
\parskip=3pt

\section{Introduction}
In this paper we study the second boundary-value problem for the Monge-Amp\`ere equation,
\beq\label{MAE1}
\left\{\begin{array}{ll}
\det D^2u = f &\quad\mbox{in }\Om,\\
Du(\Om) = \Om^*, &
\end{array}
\right.
\eeq
where $\Om, \Om^*$ are two bounded domains in $\R^2$ and $f$ is a given positive function.
Problem \eqref{MAE1} received intensive attention in the last two decades,
due to its wide range of applications, such as in optimal transportation, minimal Lagrangian graphs,
seismology, image processing, machine learning,
see for instances \cite{Bre, BW, EFY, HZTA, Vi} and references therein.

The regularity of solutions to \eqref{MAE1} is a focus of attention in optimal transportation \cite {Vi}.
In a landmark paper \cite{C96}, Caffarelli established the global $C^{2,\alpha'}$ regularity for the problem \eqref{MAE1},
assuming that $\Om$ and $\Om^*$ are uniformly convex with $C^2$ boundary and $f\in C^\alpha(\overline\Om)$.
Under stronger assumptions on domains and $f$,
the global smooth solution was first obtained by Delano\"e \cite{De} in dimension two
and later on by Urbas \cite{Ur} in higher dimensions.
The uniform convexity and smoothness of domains played a critical role
in the above mentioned papers.

Very recently in \cite{CLW} we discovered a different proof for the global $C^{2,\alpha}$ regularity,
assuming the domains $\Om, \Om^*$ are convex only (instead of uniformly convex) with merely $C^{1,1}$ boundaries,
and $f\in C^\alpha(\overline\Om)$ for the same $\alpha\in(0,1)$.
Moreover, we obtained the global $W^{2,p}$ regularity for all $p\geq1$, when $f\in C^0(\overline\Om)$.
We introduced a completely different method to obtain the uniform obliqueness,
which is the key estimate in proving the boundary regularity.

In particular in dimension two, it was proved that the uniform obliqueness holds under much weaker conditions,
including the case when $\Om$ and $\Om^*$ are bounded convex domains with $C^{1,\alpha}$ boundaries.
See  \cite[\S4.1]{CLW}.
On the other hand, it was proved by Caffarelli \cite {C96} that
the uniform density  holds for general convex domains in dimensions two.
Combining the above properties, in this paper
we are able to prove the following global regularity results for problem \eqref{MAE1}
in dimension two.
The key point is that by using an iteration argument, we can improve the regularity from $u\in C^{1,\alpha}$ to $u\in C^{1,\alpha'}$ for a greater $\alpha'>\alpha$. By estimating the increment $|\alpha'-\alpha|\geq \delta$ for a universal constant $\delta>0$, we then obtain that $u\in C^{1,\alpha}$ for all $\alpha<1$.

\vspace{5pt}

The first one is the global $C^{2,\alpha}$ regularity under sharp conditions.

\begin{theorem}\label{thm1}
Assume that $\Omega, \Omega^*$ are two bounded convex domains in $\mathbb{R}^2$.
Assume that $\partial\Omega, \partial\Omega^* \in C^{1,\alpha}$, and $f\in C^\alpha(\overline\Omega)$ is positive,
for some $\alpha\in(0,1)$.
Let $u$ be a convex solution to \eqref{MAE1}.
Then we have the estimate
	\begin{equation}\label{main}
		\|u\|_{C^{2,\alpha}(\overline\Omega)} \leq C
	\end{equation}
for the same $\alpha$, where $C>0$ is a constant depending on $f$ and the domains.
\end{theorem}

\vspace{5pt}
The second one is the global $W^{2,p}$ regularity.

\begin{theorem}\label{main01}
Assume that $\Omega, \Omega^*$ are two bounded convex domains in $\mathbb{R}^2$.
Assume that $\partial\Omega, \partial\Omega^* \in C^{1,\varepsilon}$ for some $\varepsilon>0$, and $f\in C^0(\bom)$ is positive.
Let $u$ be a convex solution to \eqref{MAE1}.
Then we have the estimate
\beq\label{esti01}
\|u\|_{W^{2,p}(\bom)}\le C
\eeq
for all $p\ge 1$,
where $C$ is a constant depending only $n, p, f, \Om$, and $\Om^*$.
\end{theorem}

When $f\equiv1$, very recently Savin and Yu \cite{SY} obtained the global $W^{2,p}$ estimate for
arbitrary bounded convex domains $\Om, \Om^*\subset \R^2$.
Note that the interior $W^{2,p}$ estimate for the Monge-Amp\`ere equation was proved by Caffarelli \cite {C90}
and the global $W^{2,p}$ estimate for the Dirichlet problem was proved by Savin \cite {Sa}.

\vspace{5pt}
Our third regularity result is the global $W^{2,1+\epsilon}$ regularity.

\begin{theorem}\label{epsilon}
Assume that $\Omega, \Omega^*$ are two bounded convex domains in $\mathbb{R}^2$.
Assume that $\lambda^{-1}<f<\lambda$, for a constant $\lambda>0$.
Let $u$ be a convex solution to \eqref{MAE1}.
Then we have the estimate
\beq\label{epsest}
\|u\|_{W^{2,1+\epsilon}(\bom)}\le C,
\eeq
where $\epsilon, C>0$ are universal constants depending only $\lambda$ and the domains $\Om, \Om^*$.
\end{theorem}

The interior $W^{2,1+\epsilon}$ regularity was obtained by De Philippis, Figalli and Savin in \cite{DFS},
which extends the $W^{2, 1}$ estimate of De Philippis and Figalli \cite {DF}.
Our proof is based on the technique in \cite {DFS} and the uniform density in \cite{C96, CLW}.
By the technique in \cite {DFS} and the uniform density in \cite{CLW},
one can also obtain the global $W^{2,1+\epsilon}$ estimate in high dimensions. That is

\begin{thma}
Assume that $\Omega, \Omega^*$ are two bounded convex domains in $\mathbb{R}^n$, $n>2$.
Assume that $\partial\Omega, \partial\Omega^* \in C^{1,1}$, and $\lambda^{-1}<f<\lambda$, for a constant $\lambda>0$.
Let $u$ be a convex solution to \eqref{MAE1}.
Then we have the estimate
\beq\label{epsestnd}
\|u\|_{W^{2,1+\epsilon}(\bom)}\le C,
\eeq
where $\epsilon, C>0$ are universal constants depending only $n, \lambda$ and the domains $\Om, \Om^*$.
\end{thma}

\vspace{5pt}

This paper is organised as follows:
In \S2, we give some preliminary results on the uniform density and uniform obliqueness properties.
In \S3, we give the proof of Theorems \ref{thm1} and \ref{main01}, in which the key ingredient is to improve the geometric shape of boundary sub-level sets by an iteration argument.
In \S4, we prove Theorems \ref{epsilon} and 1.3\textprime.

\vskip5pt

\section{Preliminary}

\subsection{Uniform density}

Let $u$ be a convex solution to \eqref{MAE1}.
We extend $u$ to a global convex function as follows,
	\begin{equation*}
		\tilde u(x) := \sup\{\ell(x) : \ell \mbox{ is affine, } \ell \leq u \mbox{ in }\Omega, \nabla\ell\in\Omega^*\}, \quad\mbox{for }x\in\mathbb{R}^2.
	\end{equation*}
For simplicity, we still write the extended function $\tilde u$ as $u$.
Given a point $x_0\in\overline\Omega$, we denote the ``centred" sub-level set of $u$ at $x_0$ of height $h$ by
	\begin{equation*}
		S_h^c[u](x_0) = \left\{x\in\mathbb{R}^2 : u(x) < \hat\ell(x) + h \right\},
	\end{equation*}
where $\hat\ell$ is an affine function such that $\hat\ell(x_0)=u(x_0)$, and $x_0$ is the mass centre of $S_h^c[u](x_0)$.
We also write $S_h^c[u](x_0)$ as $S_h^c[u]$ or $S_h^c(x_0)$ when no confusion arises.

\begin{lemma}\label{LeUni}
Assume that $\Omega, \Omega^*\subset\R^2$ are bounded convex domains and $0\in\partial\Omega$.
Then there is a positive constant $\delta_0$, independent of $u$ and $h$, such that
	\begin{equation}\label{unid}
		\frac{\Vol\left(\Omega\cap S_h^c(0)\right)}{\Vol\left(S_h^c(0)\right)} \geq \delta_0.
	\end{equation}
\end{lemma}

The uniform density \eqref{unid} was obtained by Caffarelli in \cite{C96} assuming that $\partial\Omega$ is polynomially convex at $0$.
As remarked in \cite{C96} that in dimension two, any bounded convex domain is polynomially convex, therefore the uniform density \eqref{unid} is a free gift from \cite{C96}.
Alternatively, in \cite{CLW} we provide a different proof in any dimension, relaxing the polynomial convexity to the convexity of domains $\Omega, \Omega^*$ with $C^{1,\gamma}$ boundaries for some $\gamma\leq1$.

As a corollary of Lemma \ref{LeUni}, we have the following duality property that
\begin{corollary}\label{dcoro}
Let $v$ be the dual potential function on $\Om^*$, then
\begin{itemize}
\item[$(i)$] $S^c_h[u]$ is conjugate to $S^c_h[v]$, namely if $A(S^c_h[u])\approx B_{\sqrt{h}}$ by an affine transform $A$,
then $A'^{-1}(S_h^c[v])\approx B_{\sqrt{h}},$ where $A'$ is the transpose of $A$.

\item[$(ii)$] $\forall$ $x\in S^c_h[u],$ $y\in S^c_h[v]$, we have
$$|x\cdot y| \lesssim h.$$
\end{itemize}
\end{corollary}

\subsection{Obliqueness}

\begin{lemma}\label{lem obli}
Assume that $\Omega, \Omega^*\subset\R^2$ are two convex domains {with $C^{1,\alpha}$ boundaries for some $\alpha>0$,}
and $f$ is positive and continuous.
Let $0\in\partial\Omega$ and the image $Du(0)=0\in\partial\Omega^*$.
Then there exists a positive constant $\mu$ such that
	\begin{equation}\label{sobli}
		\langle \nu(0), \nu^*(Du(0)) \rangle \geq\mu>0,
	\end{equation}
where $\nu$ and $\nu^*$ are the  unit inner normal of $\Om$ and $\Om^*$, respectively.
\end{lemma}

The uniform obliqueness  \eqref{sobli} in dimension two was proved in \cite[\S4.1]{CLW}.
More generally, in \cite[\S4.1]{CLW} it merely assumes that locally $\pom=\{x_2=\rho(x_1)\}$ for a convex function $\rho\geq0$, and there exists an even, continuous function $\sigma$ satisfying $\sigma(t)=o(|t|)$ as $t\to0$ and monotone increasing for $t>0$, such that
\beqs
\rho(t) \leq \sigma(t)\qquad\mbox{for } t\leq0.
\eeqs
Typically by choosing $\rho(t)=|t|^{1+\alpha}$ for some $\alpha>0$, we then obtain \eqref{sobli}.

\begin{remark}
The uniform obliqueness was proved by Caffarelli \cite{C96} for uniformly convex domains with $C^2$ boundary,
and by Delano\"e \cite{De} and Urbas \cite {Ur} for uniformly convex domains with $C^{3,1}$ boundary.
\end{remark}

\vspace{5pt}

\section{Proof of Theorems \ref{thm1} and \ref{main01}}

By the uniform obliqueness \eqref{sobli}, we may assume that locally near the origin,
$$\Omega=\{x_2\geq \rho(x_1)\},\qquad \Omega^*=\{x_2\geq \rho^*(x_1)\},$$
where $\rho, \rho^*\in C^{1,\alpha}$, for some $\alpha>0$, are convex functions satisfying $\rho(0)=\rho^*(0)=0$ and $\rho'(0)=(\rho^*)'(0)=0$.

By John's lemma, we write
$$S^c_h[u]\approx E_1:=\{x\in\R^2 : \frac{(x_1-k_1x_2)^2}{a_1^2}+\frac{x_2^2}{b_1^2}=1\},$$
and
$$S^c_h[v] \approx E_2:=\{y\in\R^2 : \frac{(y_1-k_2y_2)^2}{a_2^2}+\frac{y_2^2}{b_2^2}=1\}.$$

By global $C^{1,\delta}$ regularity of $u, v$, we have
	\begin{equation}
		b_1, b_2\gtrsim h^{\frac{1}{1+\delta}},\quad\mbox{ for some } \delta\in(0,1).
	\end{equation}
Denote $\beta:=\frac{1}{1+\delta},$ we have $\beta\in (\frac{1}{2}, 1).$
Letting $p$ (resp. $q$) be the intersection of the positive $e_2$-axis and $\p E_1$ (resp. $\p E_2$),
we actually have $p_2, q_2\gtrsim h^\beta.$

Note that by the uniform density property, we have
	\[ |S^c_h[u]|\approx |S^c_h[v]| \approx h, \]
 where $|E|=\Vol(E)$ denotes the Lebesgue measure of the set $E$.

\begin{proof}[Proof of Theorems]
We shall prove Theorems \ref{thm1} and \ref{main01} simultaneously.
The proof is divided into three steps.

\noindent{\bf Step 1:}
We show that $u$ is $C^{1,\gamma}$ tangentially for some $\gamma$ that can be computed explicitly.
Recall that we already knew $u\in C^{1,\delta}$ for some $\delta\in(0,1)$, thus tangentially
	\[ a_1 \gtrsim h^{\frac{1}{1+\delta}} = h^\beta. \]
We shall improve $\delta$ to a bigger $\gamma>\delta$.

Indeed, suppose to the contrary that $a_1 \lesssim h^{\frac{1}{1+\gamma}}$, namely
	\begin{equation}
		a_1\lesssim h^{\frac{1}{2}+\epsilon},\quad\mbox{ for  $\epsilon$ satisfying }\gamma=\frac{1-2\epsilon}{1+2\epsilon}.
	\end{equation}
Then, since $|S^c_h[u]|\approx h\approx a_1b_1,$ we have
	\begin{equation}
		b_1\gtrsim h^{\frac{1}{2}-\epsilon.}
	\end{equation}

Take a point $x\in\partial\Omega$ such that $x_1=h^{\frac{1}{2}+\frac{1}{1+\eta}\epsilon}$, where $\eta>0$ is a small constant to be determined later.
 Note that since $\rho\in C^{1,\alpha}$, we have
$$x_2=\rho(x_1)\lesssim h^{(\frac{1}{2}+\frac{1}{1+\eta}\epsilon)(1+\alpha)}.$$

Denote $\tilde{x}=Tx,$ where $T$ is the affine transformation normalising $S^c_h[u],$ namely
\[  \tilde{x}_1=\frac{x_1-k_1x_2}{a_1},\qquad \tilde{x}_2=\frac{x_2}{b_1}\]
such that $TE_1=B_1.$
We \emph{claim} that $x_1-k_1x_2\approx x_1$ (that implies $\tilde{x}_1\rightarrow \infty$ as $h\rightarrow 0$).
Indeed, $k_1\leq \frac{a_1}{p_2}\lesssim h^{\frac{1}{2}+\epsilon-\beta}.$  Hence, $$k_1x_2\lesssim h^{\frac{1}{2}+\epsilon-\beta+(\frac{1}{2}+\frac{1}{1+\eta}\epsilon)(1+\alpha)}.$$
In order for $k_1x_2$ to be a lower order term comparing to $x_1,$ we only need
$$\frac{1}{2}+\epsilon-\beta+(\frac{1}{2}+\frac{1}{1+\eta}\epsilon)(1+\alpha)>\frac{1}{2}+\frac{1}{1+\eta}\epsilon,$$
or equivalently
\[ \frac{1+\eta+\alpha}{1+\eta}\epsilon > \beta - \frac12(1+\alpha).  \]
It suffices to have $\frac{1}{1+\eta}\epsilon-\beta+\frac{1}{2}(1+\alpha)\geq 0.$ Therefore, we can take
$$\epsilon=\epsilon_\beta=(\eta+1)(\beta-\frac{1}{2}-\frac{\alpha}{2}).$$
Note that if $\beta\leq \frac{1}{2}+\frac{\alpha}{2},$ we can take $\epsilon$ as any positive small constant.
On the other hand it is straightforward to compute that
$$\tilde{x}_2=\frac{x_2}{b_1}\lesssim h^{(\frac{1}{2}+\frac{1}{1+\eta}\epsilon)(1+\alpha)-\frac{1}{2}+\epsilon}\rightarrow 0$$
as $h\rightarrow 0.$
Hence, the above estimates and convexity imply that as $h\to0$, the limit domain is independent of $e_1$ direction. Therefore, as long as $\epsilon_\beta>0,$ we obtain that $u$ is
tangentially $C^{1,\gamma}$ with $\gamma=\frac{1-2\epsilon_\beta}{1+2\epsilon_\beta}$, by using the argument in \cite{C96, CLW}.

It means that initially given $u\in C^{1,\delta}$, or equivalently, given a $\beta = \frac{1}{1+\delta}$, we can obtain that $u\in C^{1,\gamma}$ tangentially with $\gamma=\frac{1-2\epsilon_\beta}{1+2\epsilon_\beta}$, or equivalently, improve $\beta$ to a better (smaller) $\beta_1$ with
\[ \beta_1 = \frac{1}{1+\gamma} = \frac12+\epsilon_\beta. \]

\noindent{\bf Step 2.} From the previous step, $h^{\frac{1}{2}+\epsilon_\beta}e_1\in S^c_h[u].$
By Corollary \ref{dcoro}, $S^c_h[v]$ is conjugate to $S^c_h[u]$ and
$$\left|y\cdot (h^{\frac{1}{2}+\epsilon_\beta}e_1) \right|\lesssim h,\qquad \forall\, y\in S^c_h[v].$$
 It implies that the length of the projection of  $S^c_h[v]$ to $e_1$-axis is bounded by
$h^{\frac{1}{2}-\epsilon_\beta}.$ Since $|E_2|\approx h,$ we have that $q_2\geq h^{\beta_1},$ where $\beta_1=\frac{1}{2}+\epsilon_\beta.$

Recall that originally we merely
have $q_2\gtrsim h^\beta.$
Now we \emph{claim} that $\beta-\beta_1\geq c_\alpha$ for some constant $c_\alpha$ depending only on $\alpha.$
Indeed $$\beta-\beta_1=\beta-\frac{1}{2}-(\eta+1)(\beta-\frac{1}{2}-\frac{\alpha}{2})=\frac{\alpha}{2}+\frac{\eta}{2}+\frac{\alpha}{2}\eta-\eta\beta>\frac{\alpha}{4}$$
provided $\eta$ is sufficiently small.

\noindent{\bf Step 3.}
Now, we can repeat {\bf Step 1} for $v$ by replacing $\beta$ by $\beta_1,$ namely using $q_2\geq h^{\beta_1}$,
Then by {\bf Step 2}, $\beta_1$ is improved to $\beta_2,$ namely in turn, we obtain $p_2\geq h^{\beta_2}$ and
$\beta_1-\beta_2>\frac{\alpha}{4}.$
After finite steps, we have $\epsilon_{\beta_m}<0$ for some finite number $m\in\ZZ$.
This implies that $u, v$ are tangentially $C^{1,\gamma}$ for any $\gamma\in (0,1).$

Once having the tangential $C^{1,\gamma}$ regularity for any $\gamma\in (0,1)$ and the uniform obliqueness \eqref{sobli}, we can use the same argument in \cite[\S5]{CLW} to obtain the global $C^{2,\alpha}$ and $W^{2,p}$ regularities, respectively.
\end{proof}

\vspace{5pt}

\section{Proof of Theorems \ref{epsilon} and 1.3\textprime}
In this section, we prove the global $W^{2,1+\epsilon}$ regularities, by using the geometric properties of boundary sub-level sets established in \cite{CLW}, and the argument of De Philippis-Figalli \cite{DF}, De Philippis-Figalli-Savin \cite{DFS} for the interior $W^{2,1}$, $W^{2,1+\epsilon}$ estimates for Monge-Amp\`ere equations.

For any $x\in \overline{\Omega}\subset\R^n,$ denote the sub-level set
$$S_h(x)=\{z\in \overline{\Omega} : \ u(z)<u(x)+Du(x)\cdot (z-x)+h\}.$$
Recall the following properties established in \cite[\S2]{CLW}:

\begin{itemize}
\item[$i$)] $S_{h/C}^c(x)\cap \overline{\Omega}\subset S_h(x)\subset S_{Ch}^c(x)\cap \overline{\Omega}$ for a universal constant $C>0$;

\item[$ii$)] $Du(S^c_h(x))$ is conjugate to $S^c_h(x),$ namely, let $A$ be an affine transformation such that
$A\left(S^c_h(x)\right)\approx B_{h^{\frac{1}{2}}}$, then $A'^{-1} \left(Du(S^c_h(x))\right)\approx B_{h^{\frac{1}{2}}}$, where $A'$ is the transpose of $A$;

\item[$iii$)] $|S_h(x)|\approx |S_h^c(x)|\approx h^{\frac{n}{2}}.$
\end{itemize}

\begin{remark}
Note that the properties $i$)-$iii$) hold in $\R^2$ for any convex domain, see Corollary \ref{dcoro}.
In $\R^n$, $n>2$, we assume the domains are $C^{1,1}$ and convex.
\end{remark}

Caffarelli and Gutierrez \cite{CG} showed that the sub-level sets $S_h(x)\Subset \Omega$ have engulfing properties similar to those of balls.
In the following lemma, we show that those engulfing properties also hold for the sub-level sets
not necessarily compactly contained in $\Omega$.

\begin{lemma}[Engulfing property]\label{eng}
There exist universal constants $r_0>0$ and $C>1$ such that
for any $x_1\in \overline{\Omega}$, $x_2\in S_h(x_1)$ and $h\leq r_0$, we have $S_h(x_1)\subset S_{Ch}(x_2).$
\end{lemma}

\begin{proof}
Without loss of generality we may assume $x_1=0, u(0)=0$ and $Du(0)=0$. By properties $i$)--$iii$), up to an affine transform $A$, we have
$S_h(0)\approx B_{h^{\frac{1}{2}}}$, $Du(S_h(0))\approx B_{h^{\frac{1}{2}}}.$
Hence $|Du(x_2)|\lesssim h^{\frac{1}{2}}.$
Therefore, for any $z\in S_h(0)$ we have
\begin{equation}
u(z)-\big(u(x_2)+Du(x_2)\cdot (z-x_2)\big)\leq h+|Du(x_2)||z-x_2|\leq Ch,
\end{equation}
which implies that $z\in S_{Ch}(x_2).$
\end{proof}

Using the above lemma it is easy to deduce that
\begin{lemma}\label{engulfing} There is a universal constant $\delta>0$ such that for any $x_1, x_2\in\overline{\Omega}$ we have
that if $h_1\leq h_2$ and $S_{\delta h_1}(x_1)\cap S_{\delta h_2}(x_2)\neq \emptyset$, then $S_{\delta h_1}(x_1)\subset S_{h_2}(x_2)$.
\end{lemma}
\begin{proof}
Let $z\in S_{\delta h_1}(x_1)\cap S_{\delta h_2}(x_2)\neq \emptyset,$ by Lemma \ref{eng} we can find some universal constant $C$ so that
$S_{\delta h_i}(x_i)\subset S_{C\delta h_2}(z)$ for $i=1, 2.$ Then, since $x_2\in S_{C\delta h_2}(z),$ by Lemma \ref{eng} again, we have that
 $S_{C\delta h_2}(z)\subset  S_{C^2\delta h_2}(x_2).$ The conclusion of the lemma follows by taking $\delta=\frac{1}{C^2}.$
\end{proof}

It is well known that the property of sub-level sets
stated in Lemma \ref{engulfing} implies the following Vitali covering lemma:
\begin{lemma}[Vitali covering]\label{vital}
 Let $D$ be a compact subset of $\overline{\Omega}$
and let $\{S_{h_x}(x)\}_{x\in D}$ be a family of sub-level sets with $h_x\leq r_0.$ Then, there exist a finite number of
sub-level sets $\{S_{h_{x_i}}(x_i)\}_{i=1,\ldots, m}$ such that
$$D\subset \bigcup_{i=1}^mS_{h_{x_i}}(x_i)$$
with $\{S_{\delta h_{x_i}}(x_i)\}_{i=1,\ldots, m}$ disjoint, where $\delta>0$ is a universal constant.
\end{lemma}

For any $x_0\in \overline{\Omega},$  as in \cite{DFS} we define ${\bf a}(S_h(x_0)):=\|A\|^2$ as the normalised size of $S_h(x_0),$ where $A$ is the affine transformation normalising $S_h(x_0)$ such that
$A\left(S_h(x_0)\right)\approx B_{h^{\frac{1}{2}}}.$

\begin{lemma}\label{decay}
Assume that $\lambda^{-1}<f<\lambda$, for a constant $\lambda>0$.
There exists a universal constant $K>0$ such that
for any $x_0\in \overline{\Omega},$ we have
\begin{equation}\label{decay1}
\int_{S_h(x_0)}\|D^2u\|\leq K{\bf a} \big|\{K^{-1}{\bf a}\leq \|D^2u\|\leq K{\bf a}\}\cap S_{\delta h}(x_0)\big|,
\end{equation}
where ${\bf a}:={\bf a}(S_h(x_0)).$
\end{lemma}

\begin{proof}
By subtracting a linear function we may assume that $u(x_0)=0$, $Du(x_0)=0.$
Let $A$ be the affine transformation such that $A\left(S_h(x_0)\right)\approx B_{h^{\frac{1}{2}}}$ with $\det\, A=1.$
Let $\bar{u}(x):=u(A^{-1}x).$ Then, $\det D^2\bar{u}=\tilde{f},$ where $\lambda^{-1}<\tilde{f}(x):=f(A^{-1}x)<\lambda.$
Since $D\bar{u}(A(S_h(x_0)))\approx B_{h^{\frac{1}{2}}},$ we have $|D\bar{u}(z)|\lesssim h^{\frac{1}{2}}$ for any $z\in  A(S_h(x_0))$ and
$|\partial (AS_h(x_0))|\approx h^{\frac{n-1}{2}}.$
Hence,
\begin{equation}\label{es11}
\int_{S_h(x_0)}\|D^2u\|\leq {\bf a}\int_{A(S_h(x_0))}\|D^2\bar{u}\|={\bf a}\int_{\partial (A(S_h(x_0)))}\bar{u}_\nu\approx {\bf a} h^{\frac{n}{2}}\leq C_1{\bf a}|S_{\delta h}(x_0)|,
\end{equation}
where $C_1$ is a universal constant.

From \eqref{es11} we can deduce that
\begin{equation}\label{es12}
\big|\{\|D^2\bar{u}\|\leq 2C_1\}\cap A(S_{\delta h}(x_0))\big|\geq \frac{1}{2}|S_{\delta h}|.
\end{equation}
On the other hand, since $\lambda^{-1}<\det D^2\bar{u}<\lambda,$ we have
\begin{equation}\label{es13}
K^{-1}I\leq D^2\bar{u}<KI\qquad \text{ in}\quad \{\|D^2\bar{u}\|\leq 2C_1\}.
\end{equation}
By the definition of $\bar{u}$  it is straightforward to check that
\begin{equation}\label{es14}
\{K^{-1}I\leq D^2\bar{u}<KI\}\subset A\{K^{-1}{\bf a}\leq \|D^2u\|\leq K{\bf a}\}.
\end{equation}
Now, \eqref{decay1} follows from \eqref{es11}--\eqref{es14}.
\end{proof}

Let $D_k:=\{x\in \overline{\Omega}: \|D^2u(x)\|\geq N^k\},$ where $N$ is some large constant to be determined later. In the following lemma we shall establish a geometric
decay of $\int_{D_k} \|D^2u\|.$

\begin{lemma}\label{gdecay} There exists a universal constant $N$ such that
\begin{equation}\label{geom}
\int_{D_{k+1}} \|D^2u\|\leq (1-\tau)\int_{D_k} \|D^2u\|,
\end{equation}
for some universal constant $\tau\in (0,1).$
\end{lemma}

\begin{proof} Let $N\gg K,$ and to be fixed later.
For any $x\in D_{k+1},$ since ${\bf a}(S_h(x))\rightarrow \|D^2u(x)\|$ as $h\rightarrow 0$ and ${\bf a}(S_\delta(x))$ is bounded by some universal constant, we can find a height $h_x$ such that
${\bf a}(S_{h_x}(x))=KN^k.$

Let $\{S_{h_i}(x_i)\}_{i=1}^m$ be a Vitali cover of $D_{k+1}.$ Then, by Lemma \ref{decay} we have
\begin{equation}\label{decay2}
\int_{S_{h_i}(x_i)}\|D^2u\|\leq K^2N^k \big|\{N^k\leq \|D^2u\|\leq K^2N^k\}\cap S_{\delta h_i}(x_i)\big|,
\end{equation}

Summing up these inequalities and using the facts that $D_{k+1}\subset \bigcup S_{h_i}(x_i)$ and $S_{\delta h_i}(x_i)$ are disjoint, we obtain
\begin{eqnarray}\label{decay3}
\int_{D_{k+1}}\|D^2u\|&\leq& K^2N^k \big|\{N^k\leq \|D^2u\|\leq K^2N^k\}\cap \overline{\Omega}\big|\\
&\leq&C\int_{D_k\backslash D_{k+1}}\|D^2u\|
\end{eqnarray}
provided $N>K^2.$
Estimate \eqref{geom} follows by adding $C\int_{D_{k+1}}\|D^2u\|$ on both sides of \eqref{decay3} and taking $\tau=1/(1+C).$
\end{proof}

Theorems \ref{epsilon} and {1.3\textprime}  easily follows from Lemma \ref{gdecay} by an elementary computation.


TITLE: Computing $\lim_{n\to\infty} [\frac{n-1}{n}]$
QUESTION [1 upvotes]: I have a slight problem computing $\lim_{n\to\infty} [\frac{n-1}{n}]$. ( $[.]$ denotes the floor fonction). I know that the limit cannot "go inside" as the fonction is not continuous in the integers. I have attempted to use the inequality: $[x] \leq x < [x]+1$ but that was also unsuccessful.

REPLY [3 votes]: For $n>1$, $[n-\frac{1}{n}]=n-1$. Thus the limit is $\infty$.

Now that you have changed the question, one should note that
$$
\biggr[\frac{n-1}{n}\biggr]=0
$$
for all $n>1$, and the limit is $0$.

REPLY [1 votes]: $$\left [ n - \frac{1}{n} \right ] > n - \frac{1}{n} - 1 \longrightarrow \infty$$

Now since you've changed the question : $$\left [ \frac{n-1}{n}\right ] = \left [1-\frac{1}{n}\right ] = 0$$
Thus the limit is 0.


TITLE: About the definition of extremely disconnected space
QUESTION [0 upvotes]: a space $X$ is said to be extremely disconnected if every open set has an open closure.
I have a question: the closure is a closed set, but in the above definition, it requires the closure is open, which puzzled me a lot.

REPLY [2 votes]: The correct term is extremally disconnected. Every topological space $X$ has two closed set that are open: $\emptyset$ and $X$. If there are more that those, the space is said to be disconnected. Extremally disconnected spaces have a lot of closed sets which are open, specifically all the closures of open subsets must be.


TITLE: Creating a custom exponential function
QUESTION [1 upvotes]: I'm trying to come up with an exponential function that starts at zero, rises quickly (in $y$) between $x = 0$ and $x = 100$ and then slowly levels off as $x$ continues into infinity.
Something that looks like....

REPLY [4 votes]: Try
$$ y=\mu \frac{1}{e^{-\alpha x} + \epsilon}+\beta$$
Adjust $\alpha$, $\beta$ $\mu$ and $\epsilon$ to meet your needs.
Note:

$\mu$ scales the function
$\beta$ shifts it up
$\alpha$ controls the rate of growth for small $x$
$\epsilon$ controls when the fuction starts to level off

REPLY [2 votes]: If you want a function going from 0 to 100, may be you could use
y = 100 (1 - Exp[-a x])
parameter "a" defines the slope at the origin.


TITLE: Doubt about a paragraph in the book "Algebraic Number Theory by Neukirch".
QUESTION [0 upvotes]: Please refer to the question in the given link.
The question already has an answer here
An element is integral iff its minimal polynomial has integral coefficients.
My question is ---
Here, $A$ is integrally closed in $K$. And $\beta_i$'s $\in \bar{K}$. Don't we have to show that $\beta_i$'s $\in K$ to show they belong to $A$? Because $A$ is integrally closed in $K$ not in $\bar{K}$.
Thank you.

REPLY [1 votes]: Here, $\beta_i$'s belong to $\bar{K}$ which are integral over $A$. So, coefficients of $p(x)$ are integral over $A$. But the coefficients of $p(x)$ belong to $K$. So, coefficients of $p(x)$ belong to $A$ as $A$ is integrally closed in $K$. Hence, $p(x)\in A[x]$.


TITLE: If sequence of derivatives converges uniformly is function sequence increasing?
QUESTION [0 upvotes]: Let $g_n : \mathbb{R} → \mathbb{R}$ be a sequence of functions where $g_n'(x)$ converges to $1+x^2$ uniformly does $g_n$ necessarily converge to an increasing function?
I claimed the answer was no, since we do not know about the convergence of $g_n$ at any point in $\mathbb{R}$ what would be the correct answer?

REPLY [0 votes]: Let $g_n(x) = a_n+ x+ \frac{1}{3}x^3 $. You can choose $a_n$ to be any sequence of numbers.


TITLE: Regular distribution function
QUESTION [0 upvotes]: For a regular distribution F we know it can be identified by a locally integrable function s.t
$$F(\phi) =  \int_{ℝ^N}f(x)\phi(x)dx \space,\space x \in ℝ^N \space,\space \phi(x) \in D(ℝ^N) $$
I've assumed the function f is unique for a given distribution but I can't find anything to confirm or explain why this is, I know this is kind of a simple question but I'm not getting any help from my book or google.

REPLY [1 votes]: $f$ will be unique up to sets of measure zero.  Remember, $f$ really isn't a function but rather an equivalence class of functions who are equal a.e. i.e. a member of $L^1_{loc}$. So if we define the distribution as the integration of a smooth function of compact support again a member of $L^1_{loc}$ then the uniqueness is really a result of integration.   Try working it out, suppose you have two different $f_1,f_2 \in L^1_{loc}$ what can you say about the distributions they define in the above manner?


text:
\section{Conic Reformulation for the Computational Study}\label{sec:conic_relaxation}

Recall that in instances \etaleasy and \general~we consider suppliers with a Uniform choice model, whose demand function is $Q_j(A) = |A|/(|A|+u_{j0})$. For this case, Problem \eqref{eq:aux_problem} corresponds to
\begin{align}\label{eq:conic}
\max_{x} &\quad \sum_{j\in V} \frac{x_j}{x_j +u_{j0}}  \notag \\
s.t. &\quad  \sum_{j \in V} x_j = n, \notag\\
&\quad x_j \in \mathbb{Z}_+, \forall j \in V. \notag
\end{align}
This problem is an integer concave program. Nevertheless, we cannot give it to CVXPY\footnote{\url{https://www.cvxpy.org}} because the objective is not in disciplined convex programming (dcp). In what follows, we see how to write the problem in dcp.
First, make $y_j =x _j +u_{j0}$. Note that $\frac{x_j}{x_j+u_{j0}} = \frac{y_j-u_{j0}}{y_j}=1-\frac{u_{j0}}{y_j}$.
Second, we write $-\frac{u_{j0}}{y_j}$ as a rotated second order cone. To this end, make $t_j \leq -\frac{u_{j0}}{y_j}$. Then, we maximize $\sum_{j \in V} 1+t_j$ subject to $t_j \leq -\frac{u_{j0}}{y_j},y_j=x_j+u_{j0}, \sum_{j \in V} x_j = m, x_j \in \mathbb{Z}_+$.

Take the inequality above: $t_j \leq -\frac{u_{j0}}{y_j} \Leftrightarrow -y_jt_j \geq u_{j0}\Leftrightarrow (-t_j,y_j,\sqrt{\frac{u_{j0}}{2}}) \in \Q^2_r$, where $\Q^2_r$ is the rotated second order cone.
Third, we get the following integer conic program:
\begin{align}
\max_{x} &\quad \sum_{j\in V} 1+t_j  \notag \\
s.t. &\quad  \sum_{j \in V} x_j = n, \notag\\
&\quad x_j \in \mathbb{Z}_+, \forall j \in V.\notag\\
&\quad (-t_j,x_j+u_{j0},\sqrt{\frac{u_{j0}}{2}}) \in \Q^2_r.
\notag
\end{align}
Forth, since CVXPY does not have implemented the rotated quadratic cone, we use the following hack:
\[
(-t_j,,x_j+u_{j0},\sqrt{\frac{u_{j0}}{2}}) \in \Q^2_r \Leftrightarrow \frac{\sqrt{u_{j0}}^2}{x_j+u_{j0}}\leq - t_j \Leftrightarrow \text{\texttt{cvxpy.quad\_over\_lin}}(\sqrt{u_{j0}},x_j+u_{j0})\leq -t_j.
\]
We take an analogous approach when suppliers follow the MNL model in instances \easyinst.


TITLE: What is the algebraic structure of functions with fixed points?
QUESTION [49 upvotes]: So I just noticed that the set of functions with a fixed point
$$f(x_0)=x_0,$$
are closed under composition
$$(f\circ g)(x):=g(f(x)),$$
and with $e(x)=x$, the inverible functions even seem to form a (non-commutative) group.
Then if one chooses another point $x_1$ and restricts the set to the functions which also have $x_1$ as a fixed point, then it is again closed and so on.
If I have one parameterized point (i.e. a curve, or even a couple of those), then solving $f_t(x(t))=x(t)$ for the families $f_t$ should give me morphisms between the functions for different values of $t$.

Are there general considerations regarding this?

And is this somehow related to the characterization of points of a manifold via the ideal of functions which evaluate to $0$ that point?
Edit 1: Might be just a general property of homeomorphisms or something, although I don't associating picking out isolated fixed points with these kind of things.
Edit 2: I now see that this might relate a translation/transformation of points in the manifold to a transformation of the function algebra over that manifold. This has some features: If you take two points $y_1$ and $y_2$ and transformations along the curves $Y_1(t),Y_2(t)$ with $Y_1(0)=y_1, Y_1(1)=y_2$ and $Y_2(0): =y_2, Y_2(1)=y_1$ (they move into each other), then the fuction set with both fixed points $Y_1(t),Y_2(t)$ makes a loop as $\{Y_1(0),Y_2(0)\}=\{Y_1(1),Y_2(1)\}=\{y_1,y_2\}$. The particular form of the curves have an impact on how the function set looks in between.

REPLY [3 votes]: I see that this question hasn't received a full answer and that may be due to that vast number of fields which study fixed points and the lack of specificity of which objects you're interested in (although the points you have made in fact were quite general and apply to many objects). However, I'll try to answer in the context of manifolds.
We begin with a manifold $M$ and the space of all orientation preserving homeomorphisms from $M$ to itself which we will denote by $Homeo(M)$ (if $M$ is smooth we can in fact restrict to diffeomorphisms of $M$). We may also restrict to homeomorphisms which fix a particular set $N\subset M$ and we will denote this subgroup of $Homeo(M)$ by $Homeo(M,N)$. It is a subgroup for the very reasons you mentioned in your post. Now, $Homeo(M)$ is intractable. Therefore, it is often useful to examine this group up to some sort of equivalence. In this case, we may choose to use isotopy equivalence (i.e. two homeomorphisms are equivalent if there exists an arc of homeomorphisms between them). We denote this group up to equivalence as $\pi_0(Homeo(M,N))$.
The reason why certain subgroups which fix a subset of $M$ are interesting can be demonstrated in the following example: consider a closed annulus $A$. Denote the boundary of $A$ as $\partial A\cong S^0\times S^1$. Also consider the subgroup of $Homeo(A)$ which fix $\partial A$, $Homeo(A,\partial A)$. It is easy to visualize why $\pi_0(Homeo(A))\cong 0$ whereas $\pi_0(Homeo(A,\partial A)\cong\mathbb{Z}$. Intuitively, this means that homeomorphisms of the annulus can be undone if we allow isotopies which do not fix the boundary (we can "untwist"); however, working with a fixed boundary gives us a non-trivial mapping class group ($\pi_0(Homeo(M,\partial M)$). This notion is discussed in detail in "A Primer on Mapping Class Groups" by Benson Farb and Dan Margalit. Hope this demonstrates one use of this idea that you discovered. There are many many others.


TITLE: Is there a way to find the specific variable coefficient in a binomial expansion?
QUESTION [3 upvotes]: If a problem asks to find the coefficient of a variable, say, $x^2$, in a large binomial expansion, is there a way to solve without doing the whole expansion (I do it with Pascal's Triangle / Binomial Theorem).
For example, in this problem

The coefficient of $x^2$ in the expansion of $(\frac{1}{x} + 5x)^8$ is equal to the coefficient of $x^4$ in the expansion of $(a+5x)^7$, $a$ is a real number. Find the value of $a$.

I expand it out and get different answers on different tries. Not sure what's the best method to proceed. If anyone could help I would appreciate it so much!

REPLY [1 votes]: This is exactly what the binomial formula is for!
$$(a+b)^n = \sum_{i=0}^n {n \choose i} a^ib^{n-i}$$
I'll show you how to do the first one. Start by plugging in $a= 1/x$ and $b = 5x$ and $n=8$ and simplify:
$$(1/x+5x)^8 = \sum_{i=0}^8 {8 \choose i} (1/x)^i(5x)^{8-i} =   \sum_{i=0}^8 {8 \choose i} 5^{8-i} \frac{x^{8-i}}{x^i} =   \sum_{i=0}^8 {8 \choose i} 5^{8-i}  x^{8-2i}.$$
This is the sum of nine terms, one for each $i=0,1,\ldots, 8$. Notice each term has a different power of $x$. So the coefficient of $x^2$ happens when $n-i = 2$ which is when $i=3$. That means the coefficient is
$${8 \choose 3} 5^{8-3} = {8 \choose 3} 5^{5} $$
which you can simplify.
Do the same for the other binomial and equate the two answers and then solve for $a$.


TITLE: Exterior Derivative of One-Form
QUESTION [0 upvotes]: I'm trying to follow this example from my class lecture notes, but am unable too.
It says we look at plane polar coordinates: $ds^2 = dr^2 + r^2 d\theta^2$.
Then, we introduce an orthonormal basis: $\omega^r = dr$ and $\omega^\theta = r d\theta$.
I get the first exterior derivative $dw^r = 0$ because this is just $d(dr) = d^2 r=0$.
But, I don't understand how they got: $d \omega^\theta = \frac{1}{r}w^r \wedge w^\theta$.
That is, I don't see how the factor of $1/r$ appears.
Thank you.

REPLY [4 votes]: By direct computation, you have $$d \omega^{\theta}=d(r d\theta)=dr\wedge d\theta=\frac{1}{r}dr\wedge rd\theta=\frac{1}{r}\omega^{r}\wedge\omega^{\theta},$$
so the factor $1/r$ appears when comparing $d\theta$ to $\omega^{\theta}$.


TITLE: Can a positive-definite diagonal matrix have square roots that are positive-definite but not diagonal?
QUESTION [0 upvotes]: Consider a a diagonal $n\times n$ matrix $M$ with real diagonal elements $m_{11},m_{22},\dots,m_{nn}>0$. I am interested in finding positive-definite $n \times n$ matrices $X$ that solve
\begin{equation}M=X^2\,.\tag*{(1)}\end{equation}
An obvious solution $X=diag(\sqrt{m_{11}},\sqrt{m_{22}},\dots,\sqrt{m_{nn}})$. Is that the only positive-definite solution?
If I do not restrict my search to positive-definite $M$ and $X$, then I can find $2\times 2$ $M$ for which non-diagonal $X$ solve equation (1). There are plenty of examples on wikipedia. All examples listed there are not positive-definite. I haven't been able to come up with a positive-definite example, thus my hypothesis that $X=diag(\sqrt{m_{11}},\sqrt{m_{22}},\dots,\sqrt{m_{nn}})$ is the only positive-definite solution for positive-definite $M$.

REPLY [1 votes]: Among all square roots of a positive semidefinite matrix $M$, exactly one of them is positive semidefinite. This fact is usually summarised as "every positive semidefinite matrix has a unique positive semidefinite square root". This positive semidefinite square root is usually denoted by $M^{1/2}$ or $\sqrt{M}$.
In your case, since $M=\operatorname{diag}(m_{11},\ldots,m_{nn})$ is positive semidefinite and $X=\operatorname{diag}(\sqrt{m_{11}},\ldots,\sqrt{m_{nn}})$ is already a positive semidefinite square root of $M$, the only positive semidefinite square root of $M$ is $X$. So, the answer to the question in the title is no, $M$ has not any non-diagonal positive definite square root.


TITLE: Finding the fixed points of a recurrence relation (and systems of) analytically?
QUESTION [0 upvotes]: How would I go about finding the fixed points of the following recurrence?
$$X_n = 2X_{n-1}(2- 3X_{n-1}) + X_{n-1}$$
And therein, determining their stability analytically? Also, how does one find the fixed points among a system of recurrences, i.e.:
$$\Delta a_n = a_n - a_{n-1} = 3a_{n-1} - 6$$
$$\Delta b_n = b_n - b_{n-1} = -3a_{n-1} + 2b_{n-1}$$
Is there a simple way to do so for both that I am unaware of?

REPLY [0 votes]: Also note that you can use the derivative of $$f(x^*)$$ to analyze
the stability of points.  If $$|f(x^*)|< 1$$ We have a stable point, and if $$|f(x^*)|> 1$$ we have a unstable point, where $$x^*$$ can be found by solving  $$x^*=f(x^*)$$ In your case, both points are unstable. This is a slightly more strait forward approach to your problem, however, the first post is correct as well.


TITLE: Combinatoric - How many ways are?
QUESTION [0 upvotes]: We making package of 9 items.
Package must contain 4 items of set A and 5 items of set B.
$A=\{x,y,z...\}$ - size of set A is 15 items
$B=\{d,e,..,H,G\}$ - size of set B is 17 items
How many ways are to compilation package, if $H$ may be included to both $A$ and $B$ set and every package must contain $G$ ?
So if $H$ may be included to both $A$ and $B$:
$A=\{x,y,z...,H\}$ - size is 16 items
$B=\{d,e,..,H,G\}$ - size is still 17 items
It is 2 possibilities:
a) choose H from A, G from B
b) chose H, G from B
All ways for choosing are
$\binom{15}{3}\binom{1}{1}\binom{15}{4}\binom{1}{1}+\binom{15}{4}\binom{15}{3}\binom{1}{1}\binom{1}{1}=1242150$
Is that correct?

REPLY [0 votes]: Revised answer in view of OP's clarifications
You almost got it right, but the mistake is that you made choice of $H$ compulsory.
We need to correct that, and since $G$ is compulsory, we can just exclude it it, and take that we want $4$ each from $A$ and $B$
Apart from $H$, $15$ elements each are to be considered in the $2$ sets,
and we might not select $H$ at all, or select it from $A\;$ XOR $B$
You did not consider the case when $H$ is not selected at all, so just add $\binom{15}4\binom{15}4$ to your answer
1242150 + 1863225 = 3105375


TITLE: Let $A=[a_{ij}]\in Mat_n(\mathbb{R})$ the matrix defined Calulate $det(A)$
QUESTION [0 upvotes]: Let $A=[a_{ij}]\in Mat_n(\mathbb{R})$ the matrix defined by:
$$a_{ij}=\begin{eqnarray}n+1 &&&&&&&&\text{if $i=j$}\\
1 &&&&&&&&\text{if $i\not=j$}
\end{eqnarray}$$
Calulate $\det(A)$
My work:
By definition $$A=\begin{pmatrix}
n+1&&1&&1&&1&&...&&1\\
1&&n+1&&1&&1&&...&&1\\
1&&1&&n+1&&1&&...&&1\\
.\\
.\\
.\\
1&&1&&1&&1&&...&&n+1
\end{pmatrix}$$
Apply elemental operation by rows, we have:
$$\begin{pmatrix}
n+1&&1&&1&&1&&...&&1\\
1&&n+1&&1&&1&&...&&1\\
1&&1&&n+1&&1&&...&&1\\
.\\
.\\
.\\
1&&1&&1&&1&&...&&n+1
\end{pmatrix}\equiv
\begin{pmatrix}
1&&1&&1&&1&&...&&n+1\\
1&&n+1&&1&&1&&...&&1\\
1&&1&&n+1&&1&&...&&1\\
.\\
.\\
.\\
n+1&&1&&1&&1&&...&&1
\end{pmatrix}\\\\
\equiv
\begin{pmatrix}
1&&1&&1&&1&&...&&n+1\\
0&&n&&0&&0&&...&&-n\\
0&&0&&n&&0&&...&&-n\\
.\\
.\\
.\\
n&&0&&0&&0&&...&&-n
\end{pmatrix}
$$
$$
\equiv
\begin{pmatrix}
1&&1&&1&&1&&...&&n+1\\
0&&n&&0&&0&&...&&-n\\
0&&0&&n&&0&&...&&-n\\
.\\
.\\
.\\
0&&-n&&-n&&-n&&...&&-n^2-2n
\end{pmatrix}
\equiv
\begin{pmatrix}
1&&1&&1&&1&&...&&n+1\\
0&&n&&0&&0&&...&&-n\\
0&&0&&n&&0&&...&&-n\\
.\\
.\\
.\\
0&&0&&-n&&-n&&...&&-n^2-3n
\end{pmatrix}
\equiv
\begin{pmatrix}
1&&1&&1&&1&&...&&n+1\\
0&&n&&0&&0&&...&&-n\\
0&&0&&n&&0&&...&&-n\\
.\\
.\\
.\\
0&&0&&0&&-n&&...&&-n^2-4n
\end{pmatrix}
\equiv
\begin{pmatrix}
1&&1&&1&&1&&...&&n+1\\
0&&n&&0&&0&&...&&-n\\
0&&0&&n&&0&&...&&-n\\
.\\
.\\
.\\
0&&0&&0&&0&&...&&n^2-n^n
\end{pmatrix}=B
$$
Then,
$\det(A)=-\det(B)=-\det\begin{pmatrix}
1&&1&&1&&1&&...&&n+1\\
0&&n&&0&&0&&...&&-n\\
0&&0&&n&&0&&...&&-n\\
.\\
.\\
.\\
0&&0&&0&&0&&...&&n^2-n^n
\end{pmatrix}=-n^{n-1}+n^n$
I have a little doubt in the $n\times n$ element, I think is a little different. Can someone help me?

REPLY [1 votes]: Writing $A=nI_n+U$, where $U$ is the matrix with all elements equal to one, the eigenvalues $\lambda$ are solutions to the characteristic polynomial $\det(U-(\lambda-n)I_n)=\det(U-\lambda'I_n)$, where $\lambda'=\lambda-n$. In this question was shown that $U$ has all eigenvalues $\lambda'$ equal to $0$, except for one that is equal to $n$. Then $A$ has all eigenvalues $\lambda$ equal to $n$, except for one that is equal to $2n$.  As the determinant is the product of eigenvalues, $\det(A)=n^{n-1}\cdot 2n=2n^n$.


TITLE: I'm trying to compute the factor group of $(\Bbb Z_9\times\Bbb Z_{12})/\langle(4,6)\rangle$.
QUESTION [1 upvotes]: I'm trying to compute the factor group of $(\Bbb Z_9\times\Bbb Z_{12})/\langle(4,6)\rangle$.
I know the order is $6$ but I'm not sure how to compute the cosets without writing out the $108$ terms of $\Bbb Z_9\times\Bbb Z_{12}$.
Is there a way to find the cosets without all the tedious work?

REPLY [1 votes]: The factor group is an abelian group of order $6$ and so is cyclic of order $6$.
Indeed, the coset of $([1]_9,[1]_{12})$ has order $6$ in the factor group.
The elements of the factor group are the cosets of $([n]_9,[n]_{12})$ for $n=0,\dots,5$.