TITLE: Finding the MVUE using Rao-Blackwell Theorem
QUESTION [5 upvotes]: The number of breakdowns Y per day for a certain machine is a Poisson random variable with mean $\lambda$. The daily cost of repairing these break downs is given by $C=3Y^2$ If $Y_1, Y_2, ..., Y_n$ denote the observed number of breakdowns for $n$ independently selected days find an MVUE for $E(C)$.
We can use the Rao-Blackwell Theorem.
We know that $E(C) = E(3Y^2)=3[V(Y) + (E(Y))^2]$ and $E(Y)=\lambda=V(Y)$. With some calculations we see that $E(Y^2)= \lambda + \lambda^2$
$\sum_{i=1}^n Y_i=\bar Y$ is a sufficient statistics for $ \lambda$  So I am assuming we can replace $\lambda$ with $\bar {Y}$
I am unsure where to go from here. Can someone help me pull the strings together?

REPLY [1 votes]: $$E(Y^2)=V(Y) + E(Y)^2$$
It is from the definition of variance.  Just moved around.


TITLE: Where does gravitational waves' energy go?
QUESTION [15 upvotes]: Following the measurement of gravitational waves, many sources described them and explained they carry energy away. What I don't get is how this energy will get transfered back to anything else.
If the fabric of space-time itself is vibrating, it would seem to be in impossible for any physical object to gain this energy.

What am I missing?
How would one hypothetically get energy out of gravitational waves?
If impossible, does the universe end up with nothing but GW?

REPLY [16 votes]: To a first approximation gravitational waves are never dissipated. They just spread out into the universe gradually getting fainter.
Gravitational waves are exceedingly difficult to dissipate for the same reason that they are exceedingly hard to generate in the first place. They couple very weakly to matter. The gravitational wave detected by LIGO stretched than compressed the Earth by a factor of about $10^{-21}$. The Earth is squidgy, meaning that if you stretch it then let it relax you don't get as much energy out as you put in - the rest goes to heating up the Earth. So in principle some of the energy in the gravitational wave was dissipated as it deformed the Earth. However in practice the fraction of its energy that the wave lost is utterly insignificant. Possibly the bits of the wave that hit Jupiter and the Sun lost a bit more energy, but remember that most of the wave passed through the Solar System without hitting any matter at all.
However gravitational waves do get fainter with time for two reasons. Firstly the gravitational wave from a black hole merger propagates roughly in a plane so its intensity falls off as a factor of somewhere between $\frac{1}{r}$ and $\frac{1}{r^2}$, where $r$ is distance away from the source. Secondly the energy in the wave is diluted as the universe expands. Actually the expansion not only dilutes the energy but it also red shifts it, so if $a$ is the scale factor by which the universe has expanded the energy of the wave falls as $\frac{1}{a^4}$.
You ask:

If impossible, does the universe end up with nothing but GW?

but this isn't going to happen simply because it's so hard to produce gravitational waves. The matter currently lying around in the universe is mostly going to remain lying around in the universe for the foreseeable future.
Actually gravitational waves aren't unique in not being dissipated. Light interacts very strongly with matter, but most light emitted by objects in the universe isn't going to hit any matter simply because the universe is mostly empty space. For example most of the photons emitted in the cosmic microwave background (CMB) haven't hit anything in the 13.8 billion years since, which is of course why we can still see the CMB.


TITLE: On the size of residue class
QUESTION [1 upvotes]: Let $n \in \mathbb{N}$ be a odd number. Let $S \subseteq \{1,3,5,7,...,n-2,n\}$ and $|S|$ is even number. Let $R_i^k=\{a \mid a \in S \text{ } \&\text{ } a\equiv i \pmod k\}$ where $0\leq i\leq k-1$.
$\textbf{Question:}$ What is the upper bound $N$ on $k$ such that $\forall S$ such that $|S|>0$ is even $\exists k \leq N \quad \exists i$ such that $|R_i^k|$  is odd number. One can see that $N \leq n+1$. What is best upper bound one can get? Is it possible to get $O(\sqrt{n})$ or $O(n^{1/3})$ or better.

REPLY [1 votes]: Not a full answer, but intuition that the answer is probably $\tilde{O}(n^\frac12)$ and a proof that the answer is $\Omega(n^\frac12)$.
First we show that the answer is at least $\Omega(n^\frac12).$ For some notation, for $1 \le i \le n$ let $x_i = 0$ depending on whether $i \in S$ or not. Now, consider all $k \le \frac{1}{3} n^\frac12$ say, and the values $R_i^k$ for each of these. For each of the $R_i^k$ to be even they would have to satisfy linear equations over $\mathbb{F}_2$ of the form $$x_i + x_{i+k} + \dots + x_{i+jk} \equiv 0 \pmod{2}, $$ where $i+jk$ is the largest integer at most $n$ that is $i \pmod{k}.$ If we only consider $k \le \frac{1}{3}n^\frac12$, we have $< \frac{n}{2}$ total equations, hence they cannot force $x_1 = x_2 = \dots = x_n = 0$.
I don't have a proof of the other direction, but these equations feel mostly independent at least when $k$ is prime; which is why I expect an answer of $\tilde{O}(n^\frac12)$ to be correct. I'll keep thinking; this feels like it'll need analytic NT if it's possible.


text:
\section{Basic absolute  Linear Algebra}

In this section we describe several aspects of absolute Linear Algebra, partially and loosely
following the Kapranov-Smirnov document \cite{KapranovUN}.  We will usually only consider finite or infinitely countable dimensional vector spaces;
in the second chapter of the author, detailed considerations will be made on dimensions of any cardinality.

\subsection{Structural setting and mantra}

As we want to see $\Fun$ as a field which is different from $\F_2$, one often depicts $\Fun$ as the set $\{0,1\}$ for which we only have the following
operations:
\begin{equation}
0\cdot 1 = 0 = 0\cdot 0\ \ \mbox{and}\ \ 1\cdot 1 = 1.
\end{equation}

So in absolute Linear Algebra we are not allowed to have addition of vectors and we have to define everything in terms of scalar multiplication. \\

\subsection{Field extensions of $\Fun$}

Formally, for each $m \in \mathbb{N}^\times$ we define the {\em field extension}\index{field extension} $\Fun^m$\index{$\Fun^m$} of $\Fun$ of degree $m$ as the set
$\{0\} \cup \mu_m$, where $\mu_m$ is the (multiplicatively written) cyclic group of order $m$, and $0$ is an absorbing element for the extended
multiplication to $\{0\} \cup \mu_m$.\\

\subsection{Vector spaces over $\mathbb{F}_1^{(n)}$}

At the level of $\mathbb{F}_1$ we cannot make a distinction between
affine spaces and vector spaces (as a torsor, nothing happens), so
in the vein of the previous section, a {\em vector/affine space}\index{vector space over $\F_1^n$}{affine space over $\F_1^n$} over $\mathbb{F}_1^n$, $n \in \mathbb{N}^\times$, is a triple $V = (\mathbf{0},X,\mu_n)$, where $\mathbf{0}$
is a distinguished point and $X$ a set, and
where $\mu_n$ acts freely on $X$. Each $\mu_n$-orbit corresponds to a direction. If $n = 1$, we get the notion considered in the previous section.
If the dimension is countably infinite, $\mu_n$ may be replaced by $\mathbb{Z}, +$ (the infinite cyclic group). Another definition is needed when the dimension is larger | we will come back to this issue in due course.\\

\subsection{Basis}

A {\em basis}\index{basis} of the $d$-dimensional $\mathbb{F}_1^n$-vector space $V = (\mathbf{0},X,\mu_n)$ is a set of $d$ elements in $X$ which are
two by two contained in different $\mu_n$-orbits (so it is a set of representatives of the $\mu_n$-action); here, formally, $X$ consists of $dn$ elements, and $\mu_n$ is the cyclic group with $n$ elements. (If $d$ is not finite one selects exactly one element in each $\mu_n$-orbit.)
If $n = 1$, we only have $d$ elements in $X$
(which expresses the fact that the $\mathbb{F}_1$-linear group indeed is the symmetric group) - as such we obtain the {\em absolute basis}\index{absolute!basis}.

Once a choice of a basis $\{b_i \vert i \in I\}$ has been made, any element $v$ of $V$ can be uniquely written as $b_j^{\alpha^u}$, for
unique $j \in I$ and $\alpha^u \in \mu_n = \langle \alpha \rangle$. So we can also represent $v$ by a $d$-tuple with exactly one nonzero entry, namely $b_j^{\alpha^u}$ (in
the $j$-th column).

\subsection{Dimension}

In the notation of above, the {\em dimension}\index{dimension of vector space} of $V$ is given by  $\mathrm{card}(V)/n = d$ (the number of $\mu_n$-orbits).

\subsection{Field extension}

Let $V$ be a (not necessarily finite dimensional) $d$-space over $\mathbb{F}_1^n$, $n$ finite, so that $\vert X = X_V \vert = dn$. For any positive integral divisor $m$ of $n$, with $n = mr$,
$V$ can also be seen as a $dr$-space over $\mathbb{F}_1^{m}$. Note that there is a unique cyclic subgroup $\mu_m$ of $\mu_n$ of size $m$, so
there is only one way to do it (since we have  to preserve the structure of $V$ in the process).\\

In terms of affine spaces, interpretation over a subfield can be depicted as follows.\\

\begin{equation}
\begin{array}{ccc}
\AG(d,\mathbb{F}_1^n) &\longrightarrow &\AG(dr,\mathbb{F}_1^{m})\\
&&\\
\downarrow& &\downarrow\\
&&\\
(X,\mu_n)& \longrightarrow& (X,\mu_m)
\end{array}
\end{equation}

\subsection{Projective completion}

By definition, the {\em projective completion}\index{projective!completion} of a combinatorial affine space $\AG(n,\mathbb{K})$, $n \in \mathbb{N}$ and $\mathbb{K}$ a field, is the projective space $\PG(n,\mathbb{K})$ of the same dimension and defined over the same field, which one obtains by adding a ``hyperplane at infinity". The latter is a projective spave of one dimension less of which the subspaces represent parallel classes of subspaces of $\AG(n,\mathbb{K})$. For example, if $n = 2$ and $\mathbb{K} = \mathbb{R}$, we add a line at infinity which consists of parallel classes of affine lines.

Following the aforementioned considerations on projective completion, we immediately have the details on field extension for projective $\mathbb{F}_1^n$-spaces: starting from a projective $\mathbb{F}_1^n$-space $\mathbf{P} = \PG(d,\mathbb{F}_1^n)$, we choose an arbitrary point $x$, construct
the affine space $\mathbf{P} \setminus \{x\}$, blow up as above, and then projectively complete.\\

From the motivic point of view (which will be considered in the second chapter of the author in this volume), projective completion is extremely important: we refer the reader to the
aforementioned chapter for the meaning of the mysterious identity

\begin{equation}
[\mathbf{P}^n(k)] = \mathbf{1} + \mathbb{L} + \mathbb{L}^2 + \cdots + \mathbb{L}^n.
\end{equation}

\subsection{Direct sums}

One defines a {\em direct sum}\index{direct sum} of (not necessarily finite dimensional) vector spaces $V$ and $W$, both defined over $\mathbb{F}_1^n$, as
\begin{equation}
V \oplus W := V \coprod W,
\end{equation}
where the distinguished points $\mathbf{0}_V$ and $\mathbf{0}_W$ are identified.

\begin{theorem}[Dimension Theorem]
We have that
\begin{equation}
\mathrm{dim}(V \oplus W) =  \mathrm{dim}(V) + \mathrm{dim}(W).
\end{equation}
\end{theorem}

\medskip
\subsection{Tensor products}

For defining the {\em tensor product}\index{tensor product} we start with vector spaces $V$ and $W$ defined over $\mathbb{F}_1^n$ and put
\begin{equation}
V \otimes W := V\times W,
\end{equation}
 the vector space corresponding to the Cartesian product of free $\mu_n$-sets. Here, we identify $\mathbf{0}_V \times W$ with
 $V \times \mathbf{0}_W$.

 If the dimensions of $V$ and $W$ are respectively $d$ and $e$, then $V \otimes W$ consists of $den^2$ elements, so is of dimension $den$ over $\mathbb{F}_1^n$. In order to have a sensible notion of tensor product we have to eliminate the $n$-factor. We do this by identifying  $(x,y)$ with $(x^\nu,y^{\nu^{-1}})$ for any $\nu$ in $\mu_n$ and call the corresponding vector space $V \otimes W$ (so the latter is the quotient of $V \times W$ by the anti-diagonal action of $\mu_n$). If we denote the image of  $(x,y)$ in $V \otimes W$ by $x \otimes y$, then the identification merely says we can pull the $\mu_n$-action through the tensor sign:

 \begin{equation}
(x \otimes y)^{\nu} = x^{\nu} \otimes y = x \otimes y^{\nu},
 \end{equation}
 with $\nu \in \mu_n$ arbitrary. The set $V \otimes W$ is equipped with this $\mu_n$-action.

 \begin{theorem}[Dimension for tensor product]
 We have that
 \begin{equation}
 \mathrm{dim}(V \otimes W) = \mathrm{dim}(V)\mathrm{dim}(W).
 \end{equation}
\end{theorem}

\subsection{Linear automorphisms}

A {\em linear automorphism}\index{linear!automorphism} $\alpha$ of an $\mathbb{F}_1^n$-vectorspace $V$ with basis $\{b_i\}$  is of the form
\begin{equation}
\alpha(b_i) = b_{\sigma(i)}^{\beta_i}
\end{equation}
for some power $\beta_i$ of the primitive $n$-th root of unity $\alpha$, and some permutation $\sigma \in \mathbf{S}_d$.
Then we have that

\begin{equation}
\GL_d(\mathbb{F}_1^n) \cong \mathbf{S}_d \wr (\mu_n)^d.
\end{equation}

Elements of $\GL_d(\mathbb{F}_1^n)$ can be written as $d \times d$-matrices with precisely one element of $\mu_n$ in each row or column (and
conversely, any such element determines an element of $\GL_d(\mathbb{F}_1^n)$).

(Note that the reason that rows and columns have only one nonzero element is that we do not have addition in our vector space.)

(In this setting, $\mathbf{S}_d$ is represented by $d\times d$-matrices with in each row and column exactly one $1$ | permutation matrices.)

\subsection{Determinants}

Using the setting of the precious paragraph, we define the {\em determinant}\index{determinant}
\begin{equation}
\mathrm{det}(A) = \prod \beta_i \in \mu_n.
\end{equation}

One verifies that the determinant is multiplicative and independent of the choice of basis.\\

\subsubsection{Examples}

Scalar multiplication by $\nu \in \mu_n$ gives an automorphism on any $d$-dimensional $\mathbb{F}_1^n$-vectorspace $V$ and the corresponding determinant clearly $\nu^d$. That is, the det-functor ``remembers'' the dimension modulo $n$. These mod $n$ features are a recurrent theme in absolute Linear Algebra.

Another example, which will become relevant when we come to reciprocity laws, is the following.

Take $n=2$. Then, an $\mathbb{F}_1^2$-vector space $V$ of dimension $d$ is a set $V$ consisting of $2d$ elements equipped with a free involution. Any linear automorphism $\alpha$ is represented by a $d \times d$-matrix $A$ having one nonzero entry in every row and column being equal to $+1$ or $-1$. Hence, the determinant $\mathrm{det}(A)$ is in $\{+1,-1\}$ (having put $\alpha = -1$).

On the other hand, by definition, the linear automorphism $\alpha$ determines a permutation on the $2d$ non-zero elements of $V$ (the elements of $X_V$).
In fact, it is a permutation on the $d$ $\mu_2$-orbits.
The connection between these two interpretations is that $\mathrm{det}(A) = \mathrm{sgn}(A)$; the determinant gives the sign of the permutation.

\subsection{Power residue symbol}

For a prime power $q = p^k$ with $q \equiv 1 \mod{n}$, the roots of unity $\mu_n$ are contained in $\mathbb{F}_q^\times$, so that $\mathbb{F}_q$ is a vectorspace over $\mathbb{F}_1^n$. For any field unit $a \in \mathbb{F}_q^\times$ we have the power residue symbol

\begin{equation}
\left(\begin{array}{c}
a\\
\mathbb{F}_q
\end{array}\right)_n
= a^{\frac{q - 1}{n}} \in \mu_n.
\end{equation}

On the other hand, multiplication by $a$ is a linear automorphism $A$ on the $\mathbb{F}_1^n$-vectorspace $\mathbb{F}_q$ and hence we can look at its determinant $\mathrm{det}(A)$. The characteristic one interpretation of a classical lemma by Gauss now asserts:

\begin{theorem}
The power residue symbol equals $\mathrm{det}(A)$.
\end{theorem}

config: arxiv


text:
\begin{document}

\title{\bf A new contraction family\\ for porous medium\\ and fast diffusion equations}
\author{G. Chmaycem$^{b,*}$, M. Jazar\footnote{LaMA-Liban, Lebanese University, P.O. Box 37 Tripoli,
Lebanon. E-mail: ghada.chmaycem@gmail.com (G. Chmaycem),
mjazar@laser-lb.org (M. Jazar).
\newline \indent $\,\,{}^{b}$ Universit\'{e} Paris-Est,
CERMICS, Ecole des Ponts ParisTech, 6 et 8 avenue Blaise Pascal,
Cit\'e Descartes Champs-sur-Marne, 77455 Marne-la-Vall\'ee Cedex 2,
France. E-mail: chemaycg@cermics.enpc.fr (G. Chmaycem),
monneau@cermics.enpc.fr (R. Monneau)} , R. Monneau$^{b}$}

\date{June 14, 2014}
\maketitle

\centerline{\small{\bf{Abstract}}}
{\small{In this paper, we present a surprising  two-dimensional contraction family for porous medium and fast diffusion equations.
This approach provides new a priori estimates on the solutions, even for the standard heat equation.

}}\hfill\break

 \noindent{\small{\bf{AMS Classification:}}} {\small{{35K55, 35K65}.}}\hfill\break
 \noindent{\small{\bf{Keywords:}}} {\small{contraction, porous medium equation, fast diffusion.

 }}\hfill\break

\section{Introduction}

In this paper, we answer a long standing open question about the existence of new contractions for porous medium type equations.
For $m>0$ and $d\ge 1$, we consider
nonnegative solutions $U(t,x)$ of  the following normalized equation
\begin{equation}\label{eq::1}
m U_t = \Delta U^m \quad \mbox{on}\quad (0,+\infty)\times \Omega= Q \quad \mbox{with}\quad \Omega=\R^d.
\end{equation}
For $m>1$, this equation is called the porous medium equation (PME), while for $m<1$, it is called the fast diffusion equation (FDE).
We choose the following standard definition of solution:
\begin{defi}\label{defi::1}{\bf (Notion of solution)}\\
Let $m>0$.  We say that $U\ge 0$ is a solution of (\ref{eq::1}) with intial data $0\le U_0\in L^1(\Omega)$,
if $U, U^m\in L^1_{loc}(Q)$ and for every function $\eta\in C_c^\infty(\overline{Q})$ with compact support, we have
$$\int\hspace{-0,3cm}\int_Q \left\{\frac{U^m}{m}\Delta \eta + U\eta_t\right\}\ dxdt+ \int_\Omega U_0(x)\eta(0,x) \ dx= 0.$$
\end{defi}
We recall that for $m>0$ and any initial data satisfying
\begin{equation}\label{eq::r1}
U_0\in L^1(\Omega)\cap L^{\infty}(\Omega)\quad \mbox{with}\quad U_0\ge 0,
\end{equation}
there is exactly one solution $U\in C([0,+\infty); L^1(\Omega))\cap L^{\infty}(Q)$ in the sense of Definition \ref{defi::1} (see \cite{V1,V2}).
In the special case where $m>m_c(d):=\max\left(0, \frac{d-2}{d}\right)$, it is well known that it is possible to remove the condition $U_0\in L^\infty(\Omega)$ to get existence and uniqueness in some weaker space. To present our results in a concise and  unified way for $m>0$, we will stick under assumptions (\ref{eq::r1}).

In any dimensions, it is known (see \cite{V} and the references therein), that this equation has at least three types of contractions:
in $L^1$, in $H^{-1}$ and finally for the $2$-Wasserstein distance if $m>m_c(d)$. This last contraction property has been discovered by Otto in \cite{O2} and published in \cite{O1} (see also later \cite{CMV}).
It is also known for the PME, that for the $p$-Wasserstein distance, there is no contraction for $d\ge 2$ and $p>p_1(m,d)$, while there is contraction for any $p\in [1,+\infty]$ in dimension $d=1$ (see \cite{V}).
Note also that in \cite{V2} (Theorem A.5 p. 583), the author proves that the PME is not contractive with respect to $L^p$ norm for $m\ge 2$ if $p>p_2(m,d)$.

In this paper, for $0<m<2$, we present a new family of contractions for this equation in any dimensions, which extends the $L^1$ contraction properties. Our contraction can be seen as the fourth known contraction for this equation.
Even for the case $m=1$, our approach leads to new results for the standard heat equation.\\
More precisely, for $U$ and $V$ two nonnegative solutions of (\ref{eq::1}), we show that the following quantity
\begin{equation}\label{eq::r2}
\int_{\Omega}|U^\alpha-V^\alpha|^p
\end{equation}
is a Lyapunov functional which is nonincreasing in time for all $(\alpha,p)$ in some admissible set.\\
For convenience, we will work in the whole paper with
$$n=m-1.$$
For $0<|n|<1$, we define the admissible (convex) set (which is skeched on Figure \ref{fig-Kn})

\begin{equation}\label{eq::2}
K_{|n|}=\left\{(\alpha,p)\in\left[|n|,1\right]\times {\mathbb{R}}; \quad
P_-(\alpha)\leq p\leq P_+(\alpha),\quad \mbox{with}\quad P_\pm(\alpha)=1+\frac{2}{n^2}(1-\alpha)(\alpha\pm \sqrt{\alpha^2-n^2})\right\}.
\end{equation}
For $n=0$, we also set $K_0=\left\{(\alpha,p)\in\left(0,1\right]\times {\mathbb{R}}; \quad  \alpha p \ge 1\right\}$.
One may wonder if contraction (\ref{eq::r2}) is related or not to some gradient flow structure of the equation.
Indeed, for a given positive solution $U$ of (\ref{eq::1}), we can set $u=U^\alpha$
which solves
\begin{equation}\label{eq::1-bis}
u_t = u^{\gamma-1}\left( u\Delta u +\bar \gamma\vert \nabla u\vert^2\right) \quad \mbox{with} \quad \gamma = \frac{n}{\alpha}\end{equation}
and $\bar \gamma = \gamma-1+\alpha^{-1}$.
In the special case where   $\bar \gamma =\gamma/2$,
it can be seen that equation (\ref{eq::1-bis}) is the negative $L^2$-gradient flow of some energy, i.e. it solves
$$u_t = -\nabla_{L^2} E(u) \quad\quad  \mbox{with}\quad \quad E(u)=\int_\Omega u^\gamma \frac{|\nabla u|^2}{2}.$$
Moreover, it is easy to check (computing the hessian of $E$) that $E$ is convex if $\gamma\in [-1,0]$. This corresponds exactly to the points $(\alpha,p)=(1+\frac{n}{2},2)\in K_{|n|}$ that we capture for $n\in [-\frac23,0]$.
This interpretation is similar to the derivation of Yamabe flow (see for instance \cite{Y}) with the difference here that the exponent of the fast diffusion is not directly related to the space dimension.
Nevertheless, except this very exceptional case, contraction (\ref{eq::r2}) does not seem to be related to any gradient flow structure.
\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.8\textwidth]{Kn.png}
\caption{The set $K_{|n|}$. }\label{fig-Kn}
\end{center}
\end{figure}

The paper is organized as follows.
In Section 2, we present our main results. In Section 3, we prove the new contraction property.
Finally in Section 4, we prove several gradient decay estimates.

\section{Main results}

We state precisely our main results.
With the previous notation, and for $u,v>0$, we define the symmetric matrix for $p>1$ and $\alpha\in (0,1]$:
\begin{equation}\label{Q-alpha-p}
Q_{\alpha,p}(v,u)=(p-1)\begin{pmatrix}
              \displaystyle{v^{\gamma}\left(1+\Gamma\left(\frac{u}{v}-1\right)\right)} & \displaystyle{-\frac{1}{2}(v^{\gamma}+u^{\gamma})}   \\
              \displaystyle{-\frac{1}{2}(v^\gamma+u^\gamma)} & \displaystyle{u^\gamma\left(1+\Gamma\left(\frac{v}{u}-1\right)\right)}
        \end{pmatrix} \quad \quad \mbox{with}\quad \Gamma = \frac{1-\alpha}{\alpha(p-1)}.
\end{equation}
and set $Q_{1,1}(v,u)=0$.
We can check that this matrix is nonnegative for all $u,v>0$ if and only if $(\alpha,p)\in K_{|n|}$ (see Lemma \ref{lem-sign-matrix-Q}).
For $y\in \R$, we set $y_+=\max(0,y)$. Then our main result is the following

\begin{theo}\label{theo-positive-negative-part}{\bf (The new contraction family)}\\
We recall that $\Omega=\R^d$ with $d\geq 1$. Let $|n|<1$ and $U$ (resp. $V$) be the solution of (\ref{eq::1}) in the sense of Definition \ref{defi::1}, with initial data $0\leq U_0\in L^1(\Omega)\cap L^\infty(\Omega)$ (resp. $0\leq V_0\in L^1(\Omega)\cap L^\infty(\Omega)$).
Let $(\alpha,p)\in K_{|n|}$ where $K_{|n|}$ is defined in (\ref{eq::2}). We set
$$u=U^\alpha \quad \mbox{and} \quad v=V^\alpha.$$
Then for $w=u$ and $v$, we have $w\in C(Q)\cap L^\infty(Q)\cap C([0,+\infty);L^p(\Omega))$ and $w\in C^\infty(Q\cap \left\{w>0\right\})$.
Moreover, with the notation  $v(\tau)=v(\tau,\cdot)$, $u(\tau)=u(\tau,\cdot)$, we have for all $t\ge 0$
\begin{equation}\label{eq::4}
\int_{\Omega}\left(v(t)-u(t)\right)_{+}^p~dx + p \int_0^td\tau  \int_{\{v-u>0\}\cap \left\{u>0\right\}} e[v,u]\ dx \le \int_{\Omega}\left(v(0)-u(0)\right)_{+}^p~dx
\end{equation}
with
\begin{equation}\label{eq::r11}
e[v,u]=|v-u|^{p-2}
\begin{pmatrix}
\nabla v  \\
\nabla u
\end{pmatrix}^T
Q_{\alpha,p}(v,u)
\begin{pmatrix}
\nabla v \\
\nabla u
\end{pmatrix}\ge 0, \quad \mbox{for} \quad v>u>0.
\end{equation}
\end{theo}
Note that it is possible to see that there is equality in (\ref{eq::4}), for instance if $u$ and $v$ are smooth and positive on the torus  $\T^d$
instead of the whole space.

\noindent We now define a second symmetric matrix $M_{1,1}=0$ and for $(\alpha,p)\in K_{|n|}\backslash \left\{(1,1)\right\}$
\begin{equation}\label{K-alpha-p}
M_{\alpha,p}=(p-1)\begin{pmatrix}
              1 & \displaystyle{-\Gamma+\frac{\gamma}{2}}   \\
                  &      \\
              \displaystyle{-\Gamma+\frac{\gamma}{2}} & \displaystyle{\Gamma (1-\gamma)}
        \end{pmatrix},
\end{equation}
which turns out to be also nonnegative (see Lemma \ref{lem::r55}).\\

\noindent Then, we have also the following result
\begin{theo}\label{th::r3}{\bf (Decay of the gradient)}\\
Let $U$ be the solution of equation (\ref{eq::1}) with initial data $U_0$, under the assumptions of Theorem \ref{theo-positive-negative-part}.
Assume that $\nabla U_0^\alpha \in L^p(\Omega)$.  Setting $u=U^\alpha$, we have $\nabla u \in L^\infty([0,+\infty);L^p(\Omega))$.
Moreover, with the notation $u(\tau)=u(\tau,\cdot)$, we have for almost every $t\ge 0$:
\begin{equation}\label{eq::r50}
\begin{array}{l}
\displaystyle \int_{\Omega}\left|\nabla u(t)\right|^p~dx
+ p\int_0^t d\tau \int_{\{\nabla u\not= 0\}\cap \left\{u>0\right\}} {\bar e}[u, \nabla u,D^2 u]
dx\quad \le  \int_{\Omega}\left|\nabla u(0)\right|^p~dx.
\end{array}
\end{equation}
with
\begin{equation}\label{eq::r51}
\bar e[u, w, A]=
|w|^{p-2}u^{\gamma-2}
\begin{pmatrix}
u A_0 \\
w^2
\end{pmatrix}^T
M_{\alpha,p}
\begin{pmatrix}
uA_0\\
w^2
\end{pmatrix}+ |w|^{p-2} u^{\gamma}\left(|A_2|^2+p|A_1|^2\right)\ge 0,
\end{equation}
where   we write a general symmetric matrix $A$  as
$$A=\begin{pmatrix}
A_0 & A_1 \\
A_1^T & A_2
\end{pmatrix} \quad \mbox{by blocks in the vector space}\quad \R b \oplus b^\perp\quad \mbox{with}\quad b=\frac{w}{|w|}.$$
\end{theo}

\noindent Again, note that it is possible to see that there is equality in (\ref{eq::r50}), if $U_0$ is smooth and positive with $\Omega=\T^d$.\\

Let us mention that our results will be used in a future work to get information on the anomalous exponents associated to self-similar solutions like the one of Aronson-Graveleau (see \cite{AG}), and also the one of King-Peletier-Zhang (see \cite{K, PZ}).

\section{The new contraction property}\label{sec-proof-theo-wsol}

This section is devoted to the proof of Theorem \ref{theo-positive-negative-part}.
To this end, we first need the following result whose proof is elementary.
\begin{lem}{\bf(Properties of the matrix $Q_{\alpha,p}$)}\label{lem-sign-matrix-Q}\\
For $|n|<1$, let $\displaystyle (\alpha, p)\in \mbox{Int}\ K_{|n|}$ where $K_{|n|}$ is defined in (\ref{eq::2}).\\
\noindent {\bf i) Positive definite matrix}\\
Let $\eta >0$. Then for $v>u>0$ with $1+\eta^{-1}> v/u \ge 1+\eta$, we have
\begin{equation}\label{eq::r21}
Q_{\alpha,p}(v,u)\ge v^\gamma \nu I_2,
\end{equation}
where $I_2$ is the $2\times 2$ identity matrix, $Q_{\alpha,p}$ is defined in (\ref{Q-alpha-p}), and $\nu=\nu(n,\alpha,p,\eta)>0$.\\
\noindent {\bf ii) Bound from below for the top diagonal term}\\
Let $v\ge u>0$, then the top diagonal term of the matrix satisfies
\begin{equation}\label{eq::r27}
(Q_{\alpha,p}(v,u))_{11} \ge v^\gamma \nu_0,
\end{equation}
for some $\nu_0=\nu_0(n,\alpha,p)>0$.
\end{lem}

\noindent \textbf{Proof of Lemma \ref{lem-sign-matrix-Q}}\\
We set $\displaystyle{w=u/v \in (0,1]}$ and start with $\alpha\in (0,1)$ and $p>1$.\\

\textbf{Step 1: on the determinant}\\
A direct computation shows that $\mbox{det}\  Q_{\alpha,p}(v,u)> 0$ if and only if
\begin{equation}\label{eq::r25}
G_{\gamma}(w):=\Gamma\left(1-\Gamma\right)f_1(w)-\frac{1}{4}f_{\gamma}(w)> 0 \quad \mbox{with}\quad f_{\gamma}(w):=\displaystyle{w^{\gamma}}-2+\frac{1}{w^{\gamma}}.
\end{equation}
Note that $G_\gamma(1)=G'_\gamma(1)=0$. We now show that $G_\gamma$ is strictly convex for $w\le 1$, which will imply (\ref{eq::r25}) for $w\not=1$.
Indeed, we have
$$G_{\gamma}''(w)=-w^{-3}F_{\gamma}(w) \quad \mbox{with} \quad F_{\gamma}(w):=\frac{1}{4}\gamma\left((\gamma-1)w^{\gamma+1}+(\gamma+1)w^{1-\gamma}\right)-2\Gamma\left(1-\Gamma\right)\le  F_\gamma(1),$$
where we have used $|\gamma|< 1$ and $w\le 1$ to get the inequality. We see that
$F_{\gamma}(1)< 0$ if and only if $\Gamma$ is in between the two distinct roots
$\displaystyle{\Gamma_{\pm}:=\frac{1\pm \sqrt{1-\gamma^2}}{2}}$,
which is equivalent to $(\alpha, p)\in \mbox{Int}\ K_{|n|}$. This implies the strict convexity of $G_\gamma$.\\

\textbf{Step 2: conclusion}\\
Recall that for $(\alpha,p)\in \mbox{Int}\ K_{|n|}$, we have $\displaystyle p> P_-(\alpha)\ge \frac{1}{\alpha}$.
We deduce that each diagonal term of $Q_{\alpha, p}$ is positive (which shows ii)) and then $\mbox{tr}\ Q_{\alpha, p}(v,u)> 0$.
With Step 1, this implies the result i).
\hfill $\Box$\\

\noindent {\bf Proof of Theorem \ref{theo-positive-negative-part}}\\
We only deal with the case $(\alpha,p)\in \mbox{Int}\ K_{|n|}$. Note that the border case can easily be recovered,
by a passage to the limit.
The proof is divided into several steps.\\

\noindent {\textbf{Step 1: Estimate for smooth positive solutions}}\\
Given two nonnegative and smooth initial data $U_0, V_0$ with compact support,
we consider modified initial data for $\varepsilon\in (0,1)$
$$U_{0\varepsilon}= U_0+\varepsilon \quad \mbox{and} \quad  V_{0\varepsilon}=V_0+\varepsilon.$$
According to the standard parabolic theory (see \cite{LSU}),
we can consider the smooth positive functions  $U_\varepsilon$, $V_\varepsilon$, which are the classical solutions of (\ref{eq::1}) respectively associated to initial data $U_{0\varepsilon}$ and $V_{0\varepsilon}$. We have moreover the obvious bounds
\begin{equation}\label{eq::r22}
\varepsilon\le U_\varepsilon, V_\varepsilon \le |U_0|_{L^\infty(\Omega)} + \varepsilon.
\end{equation}
For $u_\varepsilon=U_\varepsilon^\alpha$, $v_\varepsilon = V_\varepsilon^\alpha$, for a cut-off function $\varphi\in C_c^\infty(\Omega)$ satisfying $\varphi\ge 0$ and for $p>1$, we will compute $\frac{d}{dt}\int_\Omega \varphi^2 (v_\varepsilon-u_\varepsilon)^p_+$.
We will use the positivity properties of the matrix $Q_{\alpha,p}$, to control the error term created by the cut-off. To this end,  it is useful to introduce for $\delta \in [0,\varepsilon)$, the function $\Psi_{\delta,p}(w)$ which is an approximation of $\Psi_{0,p}(w)=w^p_+$, given by
$$\Psi_{\delta,p}(w)=\left\{\frac{w^p}{p}-\delta^{p-1}w-\delta^p\left(\frac1p -1\right)\right\}\cdot 1_{\left\{w>\delta\right\}}.$$
Then a direct computation gives
\begin{equation}\label{eq::r12}
\begin{array}{l}
\displaystyle \frac{d}{dt}\int_{\Omega} \varphi^2 \Psi_{\delta,p}\left(v_\varepsilon-u_\varepsilon\right)dx\\
\\
\displaystyle =   \int_{\Omega} \varphi^2 \Psi_{\delta,p}'\left(v_\varepsilon-u_\varepsilon\right)
\left\{v_\varepsilon^{\gamma}\Delta v_\varepsilon-u_\varepsilon^{\gamma}\Delta u_\varepsilon\right\} \ dx
+ \int_{\Omega}\varphi^2 \Psi_{\delta,p}'\left(v_\varepsilon-u_\varepsilon\right)\bar \gamma
\left\{v_\varepsilon^{\gamma-1}\vert \nabla v_\varepsilon\vert^2  -u_\varepsilon^{\gamma-1}\vert \nabla u_\varepsilon\vert^2 \right\}\ dx\\
\displaystyle =
 \int_{\{v_\varepsilon-u_\varepsilon>\delta\}} f\ dx
-\int_{\{v_\varepsilon-u_\varepsilon>\delta\}}
\varphi^2 e[v_\varepsilon,u_\varepsilon]\ dx
+ \delta^{p-1}\int_{\left\{v_\varepsilon-u_\varepsilon>\delta\right\}} \varphi^2 g \ dx
\end{array}
\end{equation}
with the notation $e[\cdot,\cdot]$ defined in (\ref{eq::r11}) and
$$f= -\Psi_{\delta,p}'\left(v_\varepsilon-u_\varepsilon\right)2\varphi \nabla \varphi \cdot
\left\{v_\varepsilon^{\gamma}\nabla v_\varepsilon-u_\varepsilon^{\gamma}\nabla u_\varepsilon\right\}
\quad \mbox{and}\quad g = \frac{(\alpha-1)}{\alpha}\left(|\nabla v_\varepsilon|^2 v_\varepsilon^{\gamma-1} - |\nabla u_\varepsilon|^2 u_\varepsilon^{\gamma-1}\right).$$
Here in (\ref{eq::r12}) we have used equation (\ref{eq::1-bis}) to get the second line, and have done an integration by parts on the first term of the second line, using in particular for $\delta>0$ the chain rule
$\nabla \Psi_{\delta,p}'\left(w\right) = \Psi_{\delta,p}''\left(w\right) \nabla w$
with equality almost everywhere with the convention that the right hand side is zero if $\nabla w=0$ irrespective of whether $\Psi_{\delta,p}''$ is defined. The case $\delta=0$ can be recovered, passing to the limit $\delta\to 0$.\\
Using bounds (\ref{eq::r22}), we see that $0<v_\varepsilon^\gamma, u_\varepsilon^\gamma \le M$ for some  constant $M>0$. Moreover for $v_\varepsilon-u_\varepsilon>\delta>0$, we have by Young inequality
\begin{equation}\label{eq::r33}
|f| \le (v_\varepsilon-u_\varepsilon)^{p-2}\left\{\frac{2M}{\varepsilon\nu} (v_\varepsilon-u_\varepsilon)^2 |\nabla \varphi|^2 + \varphi^2 \varepsilon\nu (v_\varepsilon^\gamma |\nabla v_\varepsilon|^2 +u_\varepsilon^\gamma |\nabla u_\varepsilon|^2)\right\}
\end{equation}
where $\nu=\nu(\delta)>0$ is given in (\ref{eq::r21}) for $\delta>0$.\\

\noindent {\textbf{Step 2: First integral estimate on the gradient}}\\
We now choose $u_\varepsilon\equiv \varepsilon$ and $\delta=0$.
Then estimate (\ref{eq::r33}) still holds true, but with $\nu$ replaced by $\nu_0$  given in (\ref{eq::r27}). This implies
$$\displaystyle \frac{d}{dt}\int_{\Omega} \varphi^2 \Psi_{0,p}\left(v_\varepsilon-\varepsilon\right)dx
\le  \int_{\Omega} C_0 |\nabla \varphi|^2 \Psi_{0,p}(v_\varepsilon-\varepsilon)\ dx
-(1-\varepsilon)\int_{\{v_\varepsilon-\varepsilon>0\}}
\varphi^2 e[v_\varepsilon,\varepsilon]\ dx$$
with $\displaystyle C_0=\frac{2pM}{\varepsilon\nu_0}$.
We then take a sequence of functions $\varphi$ converging towards $\varphi_\lambda(x)=e^{-\lambda |x|}$ for $\lambda>0$. Integrating in time, and passing to the limit $\lambda\to 0$, and using again (\ref{eq::r27}), we get
\begin{equation}\label{eq::r28}
\int_{\Omega} \Psi_{0,p}\left(v_\varepsilon(t)-\varepsilon\right)dx+ (1-\varepsilon)\nu_0 \int_0^t d\tau \int_{\Omega}  v_\varepsilon^\gamma |\nabla v_\varepsilon|^2 \le \int_{\Omega} \Psi_{0,p}\left(v_\varepsilon(0)-\varepsilon\right)dx
\end{equation}

\noindent {\textbf{Step 3:  Refined estimate}}\\
We come back to general $v_\varepsilon,u_\varepsilon$ as in Step 1 and now consider $\delta>0$.
Using (\ref{eq::r33}), we get
$$\begin{array}{l}
\displaystyle \frac{d}{dt}\int_{\Omega} \varphi^2 \Psi_{\delta,p}\left(v_\varepsilon-u_\varepsilon\right)dx\\
\\
\displaystyle\le  \int_{\Omega} C |\nabla \varphi|^2 \Psi_{0,p}(v_\varepsilon-u_\varepsilon)\ dx
-(1-\varepsilon)\int_{\{v_\varepsilon-u_\varepsilon>\delta\}}
\varphi^2 e[v_\varepsilon,u_\varepsilon]\ dx
+\delta^{p-1}\int_{\{v_\varepsilon-u_\varepsilon>\delta\}} \varphi^2 g\ dx
\end{array}$$
with $\displaystyle C=\frac{2pM}{\varepsilon\nu}$.
Note that estimate (\ref{eq::r28})  for both $v_\varepsilon$ and $u_\varepsilon$, controls uniformly in time $\int_\Omega \Psi_{0,p}(v_\varepsilon-u_\varepsilon)\ dx$ and the time integral of $\int_\Omega \delta^{p-1}|g|\ dx$.
Therefore, we can again apply the choice of $\varphi=\varphi_\lambda$, integrate in time, and take the limit $\lambda\to +\infty$ as in Step 2, and then conclude  in the limit $\delta\to 0$
\begin{equation}\label{eq::r30}
\displaystyle \int_{\Omega} \Psi_{0,p}\left(v_\varepsilon(t)-u_\varepsilon(t)\right)dx
+ (1-\varepsilon) \int_0^td\tau\ \int_{\{v_\varepsilon-u_\varepsilon>0\}\cap \left\{u_\varepsilon>0\right\}}
e[v_\varepsilon,u_\varepsilon]\ dx \le \int_{\Omega} \Psi_{0,p}\left(v_\varepsilon(0)-u_\varepsilon(0)\right)dx.
\end{equation}

\noindent {\textbf{Step 4: The limit $\varepsilon\to 0$}}\\
For a point $P_0=(t_0,x_0)$,  let us denote the open parabolic cylinder $Q_r(P_0)=(t_0-r^2,t_0)\times B_r(x_0)$,
where $B_r(x_0)$ is the open ball of center $x_0$ and radius $r>0$.
It is known by Theorem 1.1 in Sacks \cite{S83} that for any smooth solution $U_\varepsilon$ of (\ref{eq::1}) on $Q_{2r}(P_0)$, there exists a modulus of continuity $\omega$ of $U_\varepsilon$ on $Q_{r}(P_0)$, depending only on $r,d,n$ and $|U_\varepsilon|_{L^\infty(Q_{2r}(P_0))}$.
This property is automatically transfered to $u_\varepsilon=U_\varepsilon^\alpha$
with the modulus of continuity $\omega^\alpha$.
This implies, by Ascoli-Arzela theorem, that $U_\varepsilon\to U$, where $U$ is still a solution of (\ref{eq::1}) with initial data $U_0$
in the sense of Definition \ref{defi::1}. We also note that (with $\alpha p\ge 1$)
$$\int_\Omega |u(t,\cdot)-u(s, \cdot)|^p \le \int_\Omega |U(t,\cdot)-U(s, \cdot)|^{\alpha p} \le (2|U|_{L^\infty(Q)})^{\alpha p -1} |U(t,\cdot)-U(s, \cdot)|_{L^1(\Omega)}$$
which shows in particular that $U\in C([0,+\infty);L^1(\Omega))$ implies $u\in C([0,+\infty);L^p(\Omega))$.

From the standard parabolic theory  \cite{LSU}, we know that  $U\in C^\infty(Q\cap \left\{U>0\right\})$, with corresponding quantitative estimates on solutions locally bounded from above and below.
Similarly, we have $V_\varepsilon\to V$, and we call $u=U^\alpha$, $v=V^\alpha$.
Using the $C^1$ convergence of $(v_\varepsilon,u_\varepsilon)$ to $(v,u)$ on compact sets inside $\left\{v-u >0\right\}\cap \left\{u> 0\right\}$, we can pass to the limit in (\ref{eq::r30}), and get the same inequality for $\varepsilon=0$ with $(v_\varepsilon,u_\varepsilon)$ replaced by $(v,u)$.
This means (\ref{eq::4}).
We finally conclude to the result for general initial data $U_0,V_0$, by a standard approximation argument in $L^1(\Omega)$.
\hfill $\Box$

\section{Decay of the gradient}\label{sec-gradient}

This section is divided into two subsections. In the first subsection, we give the proof of the gradient decay Theorem \ref{th::r3}.
In the second subsection, we give a directional derivative estimate (Theorem \ref{theo-d-derivative}) as a corollary of our contraction estimate.

\subsection{Decay of the gradient}\label{s4.1}

We start with the following simpler analogue of Lemma \ref{lem-sign-matrix-Q}, whose proof follows from an elementary computation.

\begin{lem}{\bf(Properties of the matrix $M_{\alpha,p}$)}\label{lem::r55}\\
For $|n|<1$, let $\displaystyle (\alpha, p)\in \mbox{Int}\ K_{|n|}$ where $K_{|n|}$ is defined in (\ref{eq::2}).\\
Then we have
\begin{equation}\label{eq::r56}
M_{\alpha,p}\ge \nu_1 I_2,
\end{equation}
where $I_2$ is the $2\times 2$ identity matrix, $M_{\alpha,p}$ is defined in (\ref{K-alpha-p}), and $\nu_1=\nu_1(n,\alpha,p)>0$.
\end{lem}

\noindent {\bf Proof of Theorem \ref{th::r3}}\\
We follow the lines of proof of Theorem \ref{theo-positive-negative-part}.
For $\delta>0$, a direct computation gives with $\displaystyle b_\varepsilon=\frac{\nabla u_\varepsilon}{|\nabla u_\varepsilon|}$
\begin{equation}\label{eq::r53}
\begin{array}{l}
\displaystyle \frac{d}{dt}\int_{\Omega} \varphi^2 \Psi_{\delta,p}\left(|\nabla u_\varepsilon|\right)dx\\
\\
\displaystyle =   \int_{\Omega} \varphi^2 \Psi_{\delta,p}'\left(|\nabla u_\varepsilon|\right)
b_\varepsilon \cdot \nabla \left\{u_\varepsilon^{\gamma-1}\left(u_\varepsilon \Delta u_\varepsilon +\bar \gamma |\nabla u_\varepsilon|^2 \right)\right\} \ dx\\
\displaystyle =
 \int_{\{|\nabla u_\varepsilon|>\delta\}} {\bar f}\ dx
-\int_{\{|\nabla u_\varepsilon|>\delta\}}
\varphi^2 {\bar e}[u_\varepsilon,\nabla u_\varepsilon, D^2 u_\varepsilon]\ dx
+\delta^{p-1} \int_{\{|\nabla u_\varepsilon|>\delta\}}
\varphi^2  {\bar g}
\end{array}
\end{equation}
with the notation $\bar{e}[\cdot,\cdot,\cdot]$ defined in (\ref{eq::r51}) and
$${\bar f}  = \sum_{i=1}^d -2\varphi \nabla_i \varphi \Psi_{\delta,p}'\left(|\nabla u_\varepsilon|\right)
b_\varepsilon \cdot \nabla \left\{u_\varepsilon^{\gamma} \nabla_i u_\varepsilon \right\}
\quad \mbox{and}\quad {\bar g} = \frac{(\alpha-1)}{\alpha} b_\varepsilon \cdot \nabla \left\{u_\varepsilon^{\gamma-1}|\nabla u_\varepsilon|^2\right\}.$$
Here we have integrated by parts the term $\Delta u_\varepsilon$ in the second line of (\ref{eq::r53}).
We then pass to the limit $\delta\to 0$. We replace the fine property (\ref{eq::r21}) of the matrix $Q_{\alpha,p}$, by (\ref{eq::r56}). We finally conclude as in the proof of Theorem \ref{theo-positive-negative-part}. \hfill $\Box$

\subsection{Decay of the directional derivative}\label{s4.2}

For $\xi=(\xi^t,\xi^x)\in \R\times \R^d$, we define the directional derivative
\begin{equation}\label{eq::5}
D_\xi u=\xi^t u_t + \xi ^x\cdot \nabla u.
\end{equation}

In this subsection, we prove the following result as a consequence of our contraction estimate.\\

\begin{theo}\label{theo-d-derivative}{\bf (Decay of the directional derivative)}\\
Let $U$ be the solution of equation (\ref{eq::1}) with initial data $U_0$, under the assumptions of Theorem \ref{theo-positive-negative-part}.
We also set $u=U^\alpha$.
Let $\xi \in  \R\times \R^d$, and let us assume that
\begin{equation}\label{eq::r10}
C_\xi := \limsup_{\eta\to 0^+} \int_{\Omega} \left|\frac{u((0,x)+ \eta\xi)-u(0,x)}{\eta}\right|^p\ dx < +\infty
\end{equation}
With the notation $D_\xi u$ defined in (\ref{eq::5}), we have $D_\xi u \in L^\infty([0,+\infty);L^p(\Omega))$.
Moreover, with the notation $u(\tau)=u(\tau,\cdot)$, we have for almost every $t\ge 0$:
\begin{equation}\label{contraction-deriv}
\begin{array}{l}
\displaystyle \int_{\Omega}\left|D_{\xi}u(t)\right|^p~dx
+ p\int_0^t\int_{\{D_{\xi}u\not= 0\}\cap \left\{u>0\right\}} \bar {\bar e}[u,D_\xi u,\nabla u]
dx\quad \le C_\xi.
\end{array}
\end{equation}
with
\begin{equation}\label{eq::r20}
\bar {\bar e}[u,w,\nabla u]=
|w|^{p-2}u^{\gamma-2}
\begin{pmatrix}
u\nabla w  \\
w\nabla u
\end{pmatrix}^T
M_{\alpha,p}
\begin{pmatrix}
u\nabla w  \\
w\nabla u
\end{pmatrix}\ge 0.
\end{equation}

\end{theo}
In particular (\ref{eq::r10}) is satisfied if $\xi^t=0$ and $D_\xi U_0^\alpha \in L^p(\Omega)$.

\noindent We will use the following result.

\begin{lem}\label{lem-lim-Q-alpha-p}{\bf (Limit of the quadratic term)}\\
For $|n|<1$, let $\displaystyle (\alpha, p)\in K_{|n|}$ where $K_{|n|}$ is defined in (\ref{eq::2}).\\
Let $\Omega_1$ be an open set such that $u,w,z\in C^1(\Omega_1)$ with $u>0$ on $\Omega_1$.
For $\eta \in (0,1)$, let $v_\eta$ be a function satisfying
$$|v_\eta -(u+\eta w + \eta^2 z)|_{C^1(\Omega_1)} =o(\eta^2)$$
Then we have on $\Omega_1$
$$J_\eta(v_\eta, u)
= \eta^{-2 }\begin{pmatrix}
\nabla v_\eta \\
\nabla u
\end{pmatrix}^T
Q_{\alpha,p}\left(v_\eta,u\right)
\begin{pmatrix}
\nabla v_\eta\\
\nabla u
\end{pmatrix}
\quad \underset{\eta \to 0}{\longrightarrow}\quad  J_0=u^{\gamma-2}
\begin{pmatrix}
u\nabla w  \\
w\nabla u
\end{pmatrix}^T
M_{\alpha,p}
\begin{pmatrix}
u\nabla w \\
w\nabla u
\end{pmatrix},$$
where we recall that the matrices $Q_{\alpha, p}$ and $M_{\alpha, p}$ are respectively defined  in (\ref{Q-alpha-p}) and (\ref{K-alpha-p}).
\end{lem}

\noindent \textbf{Proof of Lemma \ref{lem-lim-Q-alpha-p}}\\
The proof is done using a simple Taylor expansion argument.
For $Q_{\alpha,p}(v,u)$, we set $Q_i = \partial^i_v Q_{\alpha,p}(u,u)$ for $i=0,1,2$.
For $V_\eta=\begin{pmatrix}
\nabla v_\eta  \\
\nabla u
\end{pmatrix}$, we set $V_\eta = V_0+ \eta V_1 +\eta^2 V_2 + o(\eta^2)$.
Using the fact that $Q_0 V_0=0 = V_0^TQ_1V_0$, we get
$$J_\eta(v_\eta,u) = \frac12 w^2 V_0^T Q_2  V_0 + 2w V_1^T Q_1 V_0 + V_1^T Q_0 V_1 + o(1)$$
which implies the result.
\hfill $\Box$\\

\noindent {\bf Proof of Theorem \ref{theo-d-derivative}}\\
Let $v_\eta(t,x)=u((t,x)+\eta \xi)$.
We simply apply Theorem \ref{theo-positive-negative-part} to $v_\eta-u$ and $u-v_\eta$ and use the symmetry $e[v,u]=e[u,v]$,
in order to control
$$\int_\Omega \left|\frac{v_\eta(t)-u(t)}{\eta}\right|^p$$
We pass to the limit as $\eta\to 0$ using Lemma \ref{lem-lim-Q-alpha-p} which gives (\ref{eq::r10}).
\hfill $\Box$

\noindent {\bf Aknowledgements}

\noindent The authors thank B. Andreianov for helpful indications on the literature.


TITLE: How can we compute the genus of the graph?
QUESTION [0 upvotes]: $V(G)=\{u_1,\cdots,u_7,v_1,\cdots,v_9, w_1,\cdots,w_5\}.$
 Set
 $$E_1=\{u_1u_j |2\leq j\leq 7\}\cup\{u_1v_j |j=1,3,5,6,8,9\}\cup\{u_1w_3,u_1w_4\};$$
 $$E_2=\{u_2u_j | j= 3, 4,6,7\}\cup\{u_2v_j |j=4,5,6,7\}\cup\{u_2w_5\};$$
 $$E_3=\{u_3u_j | j= 4,5,7\}\cup\{u_3v_j |j=1,7,8,9\}\cup\{u_3w_2\};$$
$$E_4=\{u_4u_j | j= 5,6\}\cup\{u_4v_j |j=1,2,3,4\}\cup\{u_4w_1\};$$
$$E_5=\{u_5u_j | j= 6,7\}\cup\{u_5v_j |j=4,5,6,7\}\cup\{u_5w_5\};$$
$$E_6=\{u_6u_7\}\cup\{u_6v_j |j=1,7,8,9\}\cup\{u_6w_2\};$$
$$E_7=\{u_7v_j |j=1,2,3,4\}\cup\{u_7w_1\};\;\;\;E_8=\{v_1v_5,v_1v_6\};$$
$$E_9=\{v_2v_7 \};\;\;E_{10}=\{v_3v_7 \};\;\;\;E_{11}=\{v_4v_8,v_4v_9 \},$$
$E(G)=\bigcup_{i=1}^{11}E_i$

REPLY [0 votes]: The graph has genus greater than 2:
Since all vertices $w_1, \ldots w_5$ are of valency 1 or 2 they play no role in the embedding and we can just consider the underlying homeomorphic graph $H$ with the 16 vertices $\{ u_1, \ldots, u_7, v_1, \ldots v_9 \}$ and there are 57 edges including $u_2w_5u_5$, $u_3w_2u_6$ and $u_4w_1u_7$. Let us take $n=16$ and $m=57$.
If $H$ was embeddable on the double torus then it would have Euler Characteristic $\chi(H) =-2 = n-m+f$ and so $f=57-16-2= 39$, there are 39 faces. For a simple graph all faces must be triangles or larger, so we know that, counting around each face, we have at least $3f = 117$ edges, and each edge will have been counted twice. However, $2m= 114$ and so there are too many edges to fit $H$ on the double torus.


TITLE: A limit problem involving repeated cosines
QUESTION [0 upvotes]: I was playing around on my calculator and I found something interesting:-
Let's say I take some value $x$ in degrees and apply the following operation: $cos(cos(cos(cos....(x)))))...)$. This always seems to converge to the value $0.999847741531...$, regardless of $x$. It doesn't seem to happen for the $sin$ or $tan$ functions though.
How do I prove that the "limit" of this repeated "$cos$ operation" is the above value (it seems to be an irrational number)? (Assuming that it does converge to that value, of course. I'm not sure if it does though.)
I tried the following:-
Let $y = cos(cos(.....(x))))...)$, where the operation is applied an infinite number of times (Sorry, but I do not know how to express this in a better notation), and where $x$ can take any real value in degrees.
So, $y$ can be expressed as $y=cos(y)$, i.e. $y - cos(y) = 0$. How do I solve this equation for $y$, while also proving that it has only a unique solution, and that the solution is the above-mentioned value?

REPLY [2 votes]: The angle is in degrees, yes?
It is easy to show that the answer is unique. cos(y) is always between -1 and 1, so any solutions must be in this range. cos(y) is always positive in this range, so any solutions have to be in the range 0 to 1. And in this range, y=cos(x) decreases while y=x increases, so they only cross at one point.
A good way to compute the solution to high precision is by the Newton Raphson method. I don't think there is a nice formula for it.
If you only want a few digits (15 or so) then just pressing the "cos" button repeatedly is also a good method of computing the solution. The error decreases by a factor of about 0.017 with every iteration.

REPLY [0 votes]: To first prove that there is a solution to the equation $y-\cos(y)=0$ we can use the intermediate value theorem. Let $f(y)=y-\cos(y)$. Then $f$ is continuous, $$f(0)=-1<0$$ and $$f\left(\frac\pi 2\right)=\frac\pi 2>0,$$ so there exists some $y_0\in (0,\frac \pi 2)$ such that $f(y_0)=0$.
To show that there is a unique such solution, first we observe that the only possible solutions must be in $[-1,1]$ since this is the range of $\cos$. The derivative of $f$ is
$$f'(y) = 1 - \sin(y)$$ which is positive on $[-1,1]$, hence $f$ is strictly increasing on $[-1,1]$. In particular, it reaches the value $0$ only once.
As for computing the exact value, I don't think there is a nice formula for $y_0$. You could use Newton's method to approximate it.

REPLY [0 votes]: To prove the solution is unique, you can just take the derivative of $x-\cos (\frac {\pi x}{180})$ and observe that it is always positive.  You can plot it in Alpha and see it isn't close.
To get the value, you need a one-dimensional root finder.  They are discussed in any numerical analysis text.  Again, Alpha will solve it for you, giving as many decimals as you want.


TITLE: Show that $ \lim_{a \to \infty} \int_0^\infty f(x-a) \, dx = \int_{-\infty}^\infty f(x) \, dx $ for integrable functions $f \in L^1(\mathbb{R})$
QUESTION [2 upvotes]: I noticed in a proof that he needed to state that one integral approached another:
$$ \lim_{a \to \infty} \int_0^\infty f(x-a) \, dx = \int_{-\infty}^\infty f(x) \, dx $$
Does this always hold for integrable functions (Riemann or Lebesgue) functions $f \in L^1(\mathbb{R})$ ?  This seems rather obvious.
$$ \int_{-\infty}^\infty f(x) \, dx - \int_0^\infty f(x-a) \, dx
 =\int_{-\infty}^\infty f(x) \, dx - \int_{-a}^\infty f(x) \, dx
 =\int_{-\infty}^{-a} f(x) \, dx  \stackrel{?}{\to} 0  $$
Here I would have used the linearity of the integral.  Doesn't the integral on the right-hand-side tend to zero?

REPLY [1 votes]: Yes, this is correct.  The convergence of $\int_{-\infty}^{-a} f(x) \, dx$ to $0$ follows from the dominated convergence theorem, for instance (we are integrating the functions $f\cdot 1_{(-\infty,-a)}$ which are dominated by the integrable function $|f|$).


TITLE: Find eigenvalue and root subspace of linear operator $A$
QUESTION [0 upvotes]: $A$ is the matrix:
$$A = \begin{pmatrix} 4 & -5 & 2 \\ 5 & -7 & 3 \\ 6 & -9 & 4 \end{pmatrix}$$
As far As I know to find root subspace we have to find eigenvalue of linear operator, so:
$$\mathrm{det}|A-I{\lambda}| = \begin{vmatrix}4-\lambda & -5 & 2 \\ 5 & -7-\lambda & 3 \\ 6 & -9 & 4-\lambda \end{vmatrix} = \lambda^3-\lambda^2 $$
So:
$$\lambda^3-\lambda^2 = 0 \Rightarrow \lambda(\lambda^2-\lambda) = 0$$
Hence:
$\lambda_1 = 0$ and $\lambda_2 = 1$
How should I proceed to find root subspace?

REPLY [0 votes]: Solve the equations $A.\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)=\lambda\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)$ when $\lambda=0$ and $\lambda=1$.


TITLE: Scott's trick without regularity
QUESTION [11 upvotes]: In ZF(C), one can easily get a class partition of $V$, we can even get an $\mathrm{Ord}$-partition using the Cumulative hierarchy: $P=\{V_{α+1}\setminus V_α\mid α∈\mathrm{Ord}\}$, such a partition let us do stuff like Scott's trick: given a class $A$, we can look at $A∩V_β$ where $β$ is the minimal $β$ so that intersection is not empty.
But the fact that $\bigcup P=V$ is equivalent to the axiom of regularity.
I remember somewhere reading that in $ZF\text{-regularity}$ we can't have Scott's trick like trick.
We can formulate Scott's trick as:

There exists an $\mathrm{Ord}$-partition of $V$ (or equivalently - there exists cumulative hierarchy that sums up to the universe).

In a sense this version of Scott's trick is a "global trick".

While the intuition tells me that $\text{regularity}$ does not follow from this version of Scott's trick, is this true?

REPLY [4 votes]: Yes! Extend $\sf ZF - Reg.$ with the existence of a unique Quine atom $\sf Q=\{Q\}$, take $V$ to be the hierarchy over $\sf Q$, that is: $$\begin{align} & V_\emptyset = \sf Q \\ & V_{\alpha+1}= \mathcal P (V_{\alpha}) \\ & V_\lambda= \bigcup_{\alpha < \lambda} V_\alpha, \text {for limit } \lambda  \\ & V= \bigcup _{\alpha \in \mathrm{Ord}} V_\alpha \end {align}$$
clearly this $V$ has an $\mathrm{Ord}$-partition.


TITLE: $\sigma$-algebra and measurability
QUESTION [0 upvotes]: There is an exercise that I faced in the context of a measure theory and integration course for which I might have a solution, yet I am not entirly sure it's the right one.
Let $B := \{ (0,\infty), (-1,1), (-\frac{1}{2}, \frac{1}{2}), (-\frac{1}{3}, \frac{1}{3}), ... \}$ and let $\mathfrak{A}$ be the $\sigma$-algebra generated by the set $B$.
Now, in the first part I have to decide whether the interval $(-\infty, 0)$ belongs to $\mathfrak{A}$. I would affirm that since $(0, \infty) \in \mathfrak{A}$ and thus, by definition of a $\sigma$-algebra, its complement as well. Is that right?
In the second (and more challenging) part I have to explain whether or not the function $$ f: \mathbb{R} \to \mathbb{R},\, f(x) = \left\{\begin{array}{ll} 1, & \text{falls}\, x > 0 \\
         0, & \text{falls}\, x = 0 \\
         -1, & \text{falls}\, x < 0 \end{array}\right. $$
is measurable with respect to $\mathfrak{A}$. The first thing that I don't understand: Is $\mathfrak{A}$ the $\sigma$-algebra for both preimage and image space? Or does it have to be? There is no more information given than the one I gave above.
According to the definition of measurable functions, the preimage $f^{-1}(A) \in \mathfrak{A}$ for every set A in the $\sigma$-algebra of the image space. In case the answer to the first question above is positive, i.e. $\mathfrak{A}$ is also the $\sigma$-algebra of the image space, then I have only to examine whether $f^{-1}(1)$,$f^{-1}(-1)$ and $f^{-1}(0)$ lies in $\mathfrak{A}$. But this is the case for 0 since it is the countably union of a sets in $\mathfrak{A}$ and similarly for -1 and 1. Is that right?

REPLY [1 votes]: Regarding your first question, you have that $(0,+\infty)^c=(-\infty, 0]$ belongs to $\mathfrak{A}$. Now, if you can prove that $\{0\}\in \mathfrak{A}$, then, you'll have that $(-\infty, 0)\in \mathfrak{A}$, because you can write $(-\infty, 0)$ as an intersection of elements in $\mathfrak A$:
$$(-\infty, 0)=(-\infty, 0] \cap \{0\}^c$$
But note that $\bigcap_{n\in \mathbb N} (-\frac1n, \frac1n)=\{0\}$, and that means $\{0\}\in\mathfrak A$.
Now, as you say, there's no more information, so we'll suppose that $\mathfrak A$ is the $\sigma$-algebra in both spaces. Now, pick $A\in\mathfrak A$, and study if $f^{-1}(A)\in \mathfrak A$. It is enough to see for the sets on $B$, because $$f^{-1}(\cup_n A_n)=\cup_n f^{-1}(A_n) \qquad f^{-1}(\cap_n A_n)=\cap_n f^{-1}(A_n)$$
So, there are two different kind of sets in $B$: $(0,+\infty)$, and $(-\frac1n, \frac1n)$ for $n\in \mathbb N$. Let's see:

$f^{-1}(0,+\infty)=\{x\in\mathbb R: f(x)>0\} = \{x\in\mathbb R: f(x)=1\}=(0,+\infty)\in \mathfrak A$.
If $n\in\mathbb N$, $f^{-1}(-\frac1n, \frac1n)=\{x\in\mathbb R: \frac1n<f(x)<\frac1n\}=\{x\in\mathbb R: f(x)=0\}=\{0\}\in \mathfrak A$


TITLE: Binary Quadratic Forms in Characteristic 2
QUESTION [19 upvotes]: One of the reasons why the classical theory of binary quadratic forms
is hardly known anymore is that it is roughly equivalent to the theory
of ideals in quadratic orders. There is a well known correspondence
which sends the $SL_2({\mathbb Z})$-equivalence class of a form
$$ (A,B,C) = Ax^2 + Bxy + Cy^2 $$
with discriminant
$$ \Delta = B^2 - 4AC $$
to the equivalence class of the ideal
$$ \Big(A, \frac{B - \sqrt{\Delta}}2\Big). $$
One then checks that this gives a group isomorphism between the
primitive forms with discriminant $\Delta$ and the class group
of ideals coprime to the conductor of the order with discriminant $\Delta$.
If we go from forms with integral coefficients to forms with
coefficients in a polynomial ring $k[T]$ for some field $k$,
then everything transforms nicely as long as $k$ has
characteristic $\ne 2$. In characteristic $2$, there is a well
known theory of function fields (instead of $Y^2 = f(T)$,
consider equations $Y^2 + Y h(T) = f(T)$), but I have no idea
what the right objects on the forms side are. There should be
decent objects, since, after all, calculations in the Jacobian
of hyperelliptic curves are performed essentially by composition
and reduction of forms. But I don't thinkg that $Ax^2 + Bxy + Cy^2$
will work; for example, these guys all have square discriminant.
Question: which objects (forms or something else?) correspond to
ideals in quadratic function fields with characteristic $2$?
Added: Thank you for the (three at this point) excellent answers.
I am now almost convinced that binary forms over $F_2[T]$ actually
do work; the reason why I thought they would not was not so much the
square discriminant but the fact that the action of SL$_2$ fixed the
middle coefficient $B$ of $(A,B,C)$, so I didn't get much of a
reduction. But now I see that perhaps this is not so bad after all,
and that some pair $(B,?)$ will be the analog of the discriminant in
the characteristic 2 case.
I knew about Kneser's work on composition, but was convinced (by a
lengthy article of Towber in Adv. Math.) that classical Gauss composition
of forms over PID's, which I was interested in, is quite far away from
composition of quadratic spaces over general rings. Special thanks to
Torsten for showing how to go from quadratic spaces back to forms!

REPLY [13 votes]: Classes of binary quadratic forms over any commutative ring R (with no conditions on characteristic, PID, etc.) correspond to certain modules over quadratic extensions of R.  This theory is completely general also in the forms it covers, i.e. covers non-primitive and discriminant zero forms.  We can then restrict our attention to the cases of interest.
For example, when R is an integral domain and the binary quadratic form has non-zero discriminant, the modules associated to the forms are exactly the ideal classes of the quadratic extension.
The construction is as follows, simplied for your case of interest.  From a form $ax^2+bxy+cy^2$, we form a quadratic $R$-algebra $Q:=R[\tau]/(\tau^2+b\tau+ac)$ and a module $M =Rx\oplus Ry$ with $\tau x=-cy-bx$ and $\tau y=ax$.  Given a quadratic algebra and ideal, we can pick an $R$ basis $x,y$ of the ideal and then shift a generator $\tau$ of the quadratic algebra as necessary so that $\tau y$ is a multiple of $x$.  Then you can simply read off the inverse map.
What I've given above is the case when the binary quadratic form, quadratic algebra, and module are free over $R$.  For general $R$, they will only be locally free, and the above gives a construction locally on free patches.  There are also global, coordinate-free descriptions of the correspondence between binary quadratic forms and modules over quadratic algebras.  For the details I refer you to my paper
"Gauss composition over an arbitrary base" http://www.sciencedirect.com/science/article/pii/S0001870810003257


TITLE: gradient from unit vector of steepest direction
QUESTION [0 upvotes]: I have a function $f(x,y)$ whose closed-form expression is elusive. However, the steepest direction
$$v=
\begin{bmatrix}
v_1 \\
v_2 \\
\end{bmatrix}
$$
is known. $||v||=1$ holds.
At a particular pooint $f(x_c, y_c)$,
$$
f(x_c+v_1, y_c+v_2) - f(x_c, y_c)
$$
is also known. Then how to find (or to approximate) the gradient of $f(x,y)$ at $(x_c, y_c)$?

REPLY [2 votes]: The gradient of $f$ is given by
$$\nabla f=\begin{bmatrix}\dfrac{\partial f}{\partial x}\\[6pt] \dfrac{\partial f}{\partial y}\end{bmatrix}$$
We also know that $\nabla f$ points in the same direction as $v$, i.e. that $\nabla f=\lambda v$ for some $\lambda > 0$. Thinking of $\nabla f$ as the change in $f$ as we move towards $v$, we get that $\nabla f\cdot v\approx f\left(x_c+v_1,y_c+v_2\right)-f\left(x_c,y_c\right)$. Since $||v||=1$ and $v$ and $\nabla f$ point in the same direction, $\nabla f\cdot v=||\nabla f||\approx f\left(x_c+v_1,y_c+v_2\right)-f\left(x_c,y_c\right)$. So, we conclude that
$$\nabla f\approx \left[f\left(x_c+v_1,y_c+v_2\right)-f\left(x_c,y_c\right)\right]v$$


TITLE: Why is ROC analysis not used in optimization problems?
QUESTION [1 upvotes]: In machine learning and applied fields of statistics, receiver operating characterization (ROC) analysis is commonly used to select optimal algorithms/models. However, at a lecture I once attended on mathematical optimization, I remember the lecturer saying that ROC analysis just wasn't considered a useful approach to optimization. No justification was offered. As a non-mathematician, I want to know (1) is this true that ROC analysis is not used in optimization problems and (2) if so, why?

REPLY [2 votes]: The lecturer was wrong.  It is now an active area of research (though outside my speciality). For instance, in a paper at NIPS in 2003, they discuss using ROC AUC instead of error as an objective function in training neural networks.


TITLE: If $X \times Y$ is an Alexandrov space, both $X$ and $Y$ are Alexandrov spaces
QUESTION [1 upvotes]: Below is a proof (which I've struggled with for some reason) that $X \times Y$ is an Alexandrov space (arbitrary intersections of open sets are open) only if both $X$ and $Y$ are Alexandrov spaces. It uses the equivalent condition that a space $X$ is an Alexandrov space if and only if there exists a smallest open neighborhood about $x$ for every $x \in X$.
Suppose $X \times Y$ is an Alexandrov space. Let $x \in X$, $y \in Y$. We will prove that $x$ and $y$ have smallest open neighborhoods. Let $U$ be any open set in $X$ containing $x$ and let $V$ be any open set in $Y$ containing $y$. So $U \times V$ is an open set in $X \times Y$ containing $(x, y)$. Since $X \times Y$ is Alexandrov, there exists a smallest open neighborhood $M$ about $(x, y)$. Since $M$ is open, $M = \bigcup_\alpha K_\alpha \times L_\alpha$ where each $K_\alpha$ is open in $X$ and each $L_\alpha$ is open in $Y$. We know that $x$ is contained in one of these $K_\alpha$ (call it $A$) and $y$ is contained in one of these $L_\alpha$ (call it $B$). Therefore, $A \times B$ is an open set containing $(x, y)$. Since $A \times B \subseteq M$ and $M$ is the smallest open neighborhood about $(x, y)$, $M = A \times B$. Futhermore $M = A \times B \subseteq U \times V$ (since $M$ is the smallest open neighborhood), and so $A \subseteq U$ and $B \subseteq V$ (since $A$ and $B$ are nonempty). Therefore, $A$ and $B$ are the smallest open neighborhoods of $x$ and $y$ respectively. Hence $X$ and $Y$ are Alexandrov spaces.
Is this proof correct? I have a feeling that there may be quicker way...

REPLY [2 votes]: Your proof is correct, as far as I can see, but there is an easier way.
Hint: Let $\mathcal{U}$ be any collection of open subsets of $X$, and consider the intersection of all subsets of $X\times Y$ of the form $U\times Y$ with $U\in\mathcal{U}$.


TITLE: Principle of measurable choice
QUESTION [5 upvotes]: Consider the following principle of measurable choice:
Let $X,Y$ be complete separable metric spaces and $E$ a closed $\sigma$-compact subset of $X\times Y$. Then $\pi_1(E)$ is a Borel set in $X$ and there exists a Borel function $\varphi:\pi_1(E)\to Y$ whose graph is contained in $E$. (Here $\pi_1$ denotes the projection of $X\times Y$ on $X$).

Proof.
Since $E$ is $\sigma$-compact we can write $E=\cup_{i=1}^\infty C_i$ with each $C_i$ compact in $X\times Y$. Then $\pi_1(E)=\cup_{i=1}^\infty \pi_1(C_i)$ with each $\pi_1(C_i)$ compact in $X$ since $\pi_1$ is continuous. Compact subsets of metric spaces are closed, hence Borel. Therefore $\pi_1(E)$ is a countable union of Borel sets in $X$, hence Borel.
Fix a countable dense subset $\{y_n:n\in\mathbb{N}\}$ in $Y$. Define $\varphi_1:\pi_1(E)\to Y$ by the rule $\varphi_1(x)=y_{n_1(x)}$, where $n_1(x)$ is the smallest $n$ such that $$E\cap \big[\{x\}\times \bar{B}_{1/2}(y_n)\big]\neq \emptyset \quad\quad\quad (1)$$
where $\bar{B}_{1/2}(y_n):=\{y\in Y: d(y,y_n)\}\leq 1/2\}$. This is well-defined, since $x\in \pi_1(E)$ implies $(x,y)\in E$ for some $y\in Y$, and by density there exists $n$ such that $d(y_n,y)<1/2$. Hence the set of $n$ satisfying $(1)$ is not empty.
Suppose inductively that $\varphi_k:\pi_1(E)\to Y$ has been defined for some $k\geq 1$. Define $\varphi_{k+1}:\pi_1(E)\to Y$ by the rule $\varphi_{k+1}(x)=y_{n_{k+1}(x)}$, where $n_{k+1}(x)$ is the smallest $n$ such that $$E\cap \big[\{x\}\times \bar{B}_{1/2^{k+1}}(y_n)\big]\neq \emptyset \quad \text{and} \quad d(y_n,\varphi_k(x))\leq 1/2^{k-1}\quad\quad\quad (2)$$
This is well-defined, since by definition of $\varphi_k(x)$ there exists $y\in Y$ such that $(x,y)\in E\cap \big[\{x\}\times \bar{B}_{1/2^k}(\varphi_k(x))\big]$, and by density there exists $n$ such that $d(y_n,y)<1/2^{k+1}$. Then $(x,y)\in E\cap \big[\{x\}\times \bar{B}_{1/2^{k+1}}(y_n)\big]$ and by the the triangle inequality $$d(y_n,\varphi_k(x))\leq d(y_n,y)+d(y,\varphi_k(x))\leq 1/2^{k+1}+1/2^k<1/2^{k-1}$$
so the set of $n$ satisfying $(2)$ is not empty.

Question.
At this point in the proof the author says that a simple induction argument shows that each $\varphi_k$ is Borel. This is where am having issues.
Since the range of each $\varphi_k$ is contained in $\{y_n:n\in\mathbb{N}\}$, it suffices to show that each preimage $\varphi^{-1}_k(y_n)$ is Borel in $\pi_1(E)$.
Any ideas on how to show it? Thanks for your help.

EDIT. I made some progress below. Any feedback is very appreciated!
Define the following correspondence :
$$\phi:X\to \mathcal P(Y), \quad \phi(x)=\big\{y\in Y: (x,y)\in E\big\}$$
Note that $\phi(x)\neq\emptyset$ if and only if $x\in\pi_1(E)$. Let $B$ a closed set in $Y$, and define $E(B)$ by
$$E (B):=\big\{x\in X : \phi(x)\cap B\neq \emptyset\big \}=\pi_1\big[(X\times B) \cap E \big]$$
I claim that $E(B)$ is Borel in $\pi_1(E)$. Indeed $(X\times B) \cap E=\cup_{i=1}^\infty (X\times B) \cap C_i$. For each $i$, $(X\times B) \cap C_i$ is a closed subset of the compact set $C_i$ in $X\times Y$, and is thus compact in $X\times Y$. It follows (as above) that $\pi_1\big[(X\times B) \cap E \big]=\cup_{i=1}^\infty \pi_1 \big[(X\times B) \cap C_i\big]$ is a countable union of compact sets in the metric space $X$, hence Borel in $X$. Clearly, $\pi_1\big[(X\times B) \cap E \big]\subset \pi_1( E)$, and so we conclude that  $E(B)$ is Borel in $\pi_1(E)$.
Now let $A_n=E(B_n)$  with $B_n=\bar{B}_{1/2}(y_n)$ for each $n$. Since each $B_n$ is closed in $Y$, each $A_n$ is Borel in $\pi_1(E)$ by the previous claim. Moreover, we see that
$$\varphi_1^{-1}(y_n)=A_n\setminus \cup_{i=1}^{n-1}A_i \quad\quad\quad (3)$$
for each $n$, and so each preimage is Borel in $\pi_1(E)$. It follows that $\varphi_1$ is Borel measurable.
Now assume inductively that $\varphi_k$ is Borel measurable for some $k\geq 1$. Let $A_n=E(B_n)$  with $B_n=\bar{B}_{1/2^{k+1}}(y_n)$ for each $n$. As before each $A_n$ is Borel in $\pi_1(E)$. Let
$$C_n=A_n\cap \varphi_{k}^{-1}\big[\bar{B}_{1/2^{k-1}}(y_n)\big]$$ for each $n$. Since $\bar{B}_{1/2^{k-1}}(y_n)$ is closed in $Y$ and $\varphi_k$ is Borel measurable by assumption, we have that each $C_n$ is Borel in $\pi_1(E)$. Moreover we have
$$\varphi_{k+1}^{-1}(y_n)=C_n\setminus \cup_{i=1}^{n-1}C_i \quad\quad\quad (4)$$
for each $n$, and so each preimage is Borel in $\pi_1(E)$. It follows that $\varphi_{k+1}$ is Borel measurable.
By induction it follows that each $\varphi_k$ is Borel measurable. Note that in $(3)$ and $(4)$ I assumed each $y_n$ to be distinct. The case of a finite dense subset can be handled in the same way.

REPLY [1 votes]: For completeness I will finish the proof of the theorem. We have constructed a sequence of Borel functions $\varphi_k:\pi_1(E)\to Y$. The next step is to show that this sequence is uniformly Cauchy:
Let $m>n$. Then, by the triangle inequality,
$$d(\varphi_m(x),\varphi_n(x))\leq d(\varphi_m(x),\varphi_{m-1}(x))+\dots+d(\varphi_{n+1}(x),\varphi_n(x)) \leq \frac{1}{2^{m-2}}+\dots+\frac{1}{2^{n-1}}=\sum_{i=n}^{m-1}\frac{1}{2^{i-1}}$$
where the second inequality follows from the definition of $\varphi_k$ in $(2)$ above. Since $\sum_{i=1}^{\infty}\frac{1}{2^{i-1}}<\infty$, for any given $\epsilon>0$ there exists $N$ such that $\sum_{i=N}^{\infty}\frac{1}{2^{i-1}}<\epsilon$. Then $m>n\geq N$ implies $d(\varphi_m(x),\varphi_n(x))<\epsilon$. It follows that $(\varphi_k)$ is uniformly Cauchy.
From $(\varphi_k)$ uniformly Cauchy and the fact that $Y$ is a complete metric space, it follows that $(\varphi_k)$ is uniformly convergent to some function $\varphi:\pi_1(E)\to Y$. We now use the fact that the pointwise limit of  a sequence of measurable functions taking values in a metric space (with its Borel $\sigma$-algebra) is again measurable (see here). Therefore $\varphi$ is Borel measurable.
It remains to show that the graph of $\varphi$ is contained in $E$. Consider the distance function on $X\times Y$ defined by $z\mapsto d[E,z]=\inf\{d(w,z) : w\in E\}$. This function is continuous. Since
$$(x,\varphi_k(x))\to (x,\varphi(x)) \quad \text{as} \quad k\to \infty$$
we obtain $$d[E,(x,\varphi_k(x))]\to d[E,(x,\varphi(x))] \quad \text{as} \quad k\to \infty$$ for all $x\in \pi_1(E)$. Moreover, the definition of $\varphi_k$ in $(1),(2)$ above implies that $d[E,(x,\varphi_k(x))]\leq\frac{1}{2^k}\to 0$ as $k\to\infty$, and so we conclude that $d[E,(x,\varphi(x))]=0$ for all $x\in \pi_1(E)$. As $E$ is closed, this implies that $(x,\varphi(x))\in E$ for all $x\in \pi_1(E)$, as desired.
Remark: Only the completeness and separability of $Y$ was needed.


TITLE: Inequalites between a sum of powers and definite integrals
QUESTION [1 upvotes]: Give a convincing argument that the following inequalities are true:
$$\int_0^n \ x^p\mathrm dx \leq  1 +  2^p + 3^p+ ... + n^p \leq \int_0^{n+1}\ x^p \mathrm dx$$
I've been stuck on this for hours, pretty much lost as to how I would go about solving this question. I was thinking maybe I have to convert the first and third terms into limits of Riemann sums and then the solution would be clearer, but I'm not sure how I'd do even that. Any help is appreciated!
Edit: sorry, forgot to include P is any real number greater than 0, and n is a positive integer. So f(x) is definitely increasing.

REPLY [1 votes]: Nevermind, answer found. I graphed f(x), as well as the summation in the middle. By shifting the graph of the summation over by one unit, its clear that the area of the summation is less than that of the integral of f(x) from 0 to n+1. I think that this suffices as a 'convincing argument'. Thanks all for your help, I'd upvote but I lack the rep to do so.


TITLE: In the definition of the strong operator topology on $L(\mathcal{X}, \mathcal{Y})$, why must $\mathcal{X}$ and $\mathcal{Y}$ be Banach spaces?
QUESTION [1 upvotes]: In Folland, the strong operator topology is defined as the topology on $L(\mathcal{X}, \mathcal{Y})$ induced by the evaluation maps $\{T \mapsto Tx\}_{x \in \mathcal{X}}$, where $\mathcal{X}, \mathcal{Y}$ are Banach spaces. In terms of convergence, we have that $T_\alpha \rightarrow T$ in the strong topology iff $T_\alpha x \rightarrow Tx$ in the norm topology of $\mathcal{Y}$ for all $x \in \mathcal{X}$. My question is, why must $\mathcal{X}, \mathcal{Y}$ be Banach spaces? The "net characterization" still holds if they are considered merely normed vector spaces, I'm pretty sure.

REPLY [3 votes]: I don't think there's any reason why you can't make this definition for the bounded linear operators between normed vector spaces.   Of course any bounded linear map from normed vector space $X$ to normed vector space $Y$ extends uniquely to a bounded linear map from the completion of $X$ to the completion of $Y$, so the extra generality probably doesn't get you much.  When your spaces are complete, analysis is much more convenient (e.g. you have the Baire Category Theorem and its consequences).


TITLE: Why is this module reflexive?
QUESTION [6 upvotes]: Let $A\subset B$ be two integrally closed Noetherian domains with $B$ finitely generated as an $A$-module. Then $B$ is reflexive.

Could you explain me why, please?
Reflexive means that the natural map from $B$ to $B^{**}$, where $B^{**}=\mathrm{Hom}_A(\mathrm{Hom}_A(B,A),A)$, is an isomorphism.

REPLY [2 votes]: For finitely generated modules over noetherian integral domains there are several equivalent conditions for reflexivity which can be very useful in order to solve this type of problems. Let me borrow them from this topic:

Let $M$ be a finitely generated module over an integrally closed noetherian domain $R$. TFAE:

$M$ is reflexive;
$M$ is torsion-free and equals the intersection of its localizations at the prime ideals of height $1$;
$M$ satisfies Serre's condition $(S_2)$ and its support is $\mathrm{Spec}(R)$;
$M$ is the dual of a f.g. module.

For our problem we shall use $2.$ Obviously $B$ is a torsion-free $A$-module. It remains to prove that $B$ is the intersection of its localizations at the prime ideals of $A$ of height $1$. One knows that $B$ is the intersection of its localizations at the prime ideals of height $1$. Now take $x\in \bigcap B_{\mathfrak{p}}$, where $\mathfrak{p}\subset A$ is a height $1$ prime. Let $P\subset B$ a prime ideal of height $1$. Since the extension $A\subset B$ is integral, we have that the height of $\mathfrak{p}=P\cap A$ is least $1$. But we also know that $A$ is integrally closed and $B$ is a domain. This shows that the going-down theorem applies and therefore the height of $\mathfrak{p}$ is $1$. Note now that $B_{\mathfrak{p}}\subset B_P$ and therefore $x\in B_P$. We have proved that $\bigcap B_{\mathfrak{p}}\subset \bigcap B_P=B$ and then $B=\bigcap B_{\mathfrak{p}}$.


TITLE: The splitting field of $x^4-2$ is $\mathbb{Q}[i+2^{1/4}]$.
QUESTION [3 upvotes]: I know the splitting field is generated by $2^{1/4}$ and $i$, I could show $\mathbb{Q} [ 2^{1/4}, i] = \mathbb{Q}[i+2^{1/4}]$ using some algebra.
For the non trivial direction $\mathbb{Q} [ 2^{1/4}, i] \subset \mathbb{Q}[i+2^{1/4}]$. Let us call $\alpha = i+2^{1/4}$, then we know
$$(\alpha-i)^4 - 2 = 0$$
expand the 4th power
$$\alpha^4 - 4\alpha^3 i - 6 \alpha^2 + 4\alpha i + 1 - 2 = 0$$
then we can solve for $i$ in terms of $\alpha$, so $i\in \mathbb{Q}[\alpha]=:\mathbb{Q}[i+2^{1/4}]$. Then $2^{1/4} \in  \mathbb{Q}[i+2^{1/4}]$ as well.
However the hint is to show the orbit of $i+2^{1/4}$ has more than 5 elements under the action of the galois group $Gal(\mathbb Q[2^{1/4}, i]/\mathbb{Q})$. I calculated the Galois group which is $D_8$, and the orbit has more than 5 elements, but how would conclude from this?

REPLY [2 votes]: To elaborate on @Eric Wofsey 's answer, or perhaps show a more direct view of the injection, define a $G=Gal(K/\mathbb{Q})$-action on $K=\mathbb{Q}[2^{1/4},i]$ by $g.x=g(x)$.
Let us look at the orbit $o(x)$ of $x=2^{1/4}+i$ and the stabilizer $stab_G(x)$.
By orbit-stabilizer theorem $|o(x)|=\frac{|G|}{|stab_G(x)|}$.
Say you prove that $|o(x)|\geq5$ and $|G|=8$. Then necessarily $|stab_G(x)|=1$ and so $H=Aut(K/\mathbb{Q}[x])=\{{e}\}$.
We know by the fundamental theorem of Galois theory that $\mathbb{Q}[x]=Fix(H)=K$, the latter being $K$ since $H$ is trivial.