text:
\section{Unfoldings in $\DR$.}
In this section, we first give the proof of  Theorem \ref{t:unfoldings} which describes some non-trivial equisingular unfoldings for homogeneous foliations in $\DR$. Notice that, in this theorem, if the initial $1-$form is polynomial then the unfolding is also polynomial. In this way, we provide non-trivial deformations of global foliations in $\mathbb{CP}^2$ of constant topological type, a situation that is generically impossible (see \cite{IL}). In fact, each of these unfoldings compactifies to a codimension one foliation in a projective space of bigger dimension.

Then, we establish that these unfoldings are really \emph{new} examples by constructing some having no special property of integrability whereas all known examples until now presented a Louvillian first integral.

Finally, we prove Corollary \ref{c:universal unfolding} providing examples of universal equisingular unfoldings

\subsection{Proof of Theorem \ref{t:unfoldings}.}
\label{s:unfoldings}
Let us consider the rational map
\[
\Lambda\left(x,y,c\right)=\left(1+\frac{\left\langle c,y\right\rangle }{x}\right)\cdot\left(x,y\right)
\]
where $\left\langle c,y\right\rangle =\sum_{i=1}^{n-1}c_{i}y^{i}.$ By construction, it has an indeterminacy set at $x=0$ and fixes every other lines passing through the origin.

The $1$-form $\Omega$ in $\mathbb{C}^2\times \mathbb{C}^{n-1}$ as defined in the statement of Theorem \ref{t:unfoldings} is the pull-back of $\left.\Omega\right|_{c=0}=\omega $ by $\Lambda$ up to some multiplication by a meromorphic unit $u$. In
particular, $\Omega$ is an integrable $1-$form since we have
\[
u^{2}\Omega\wedge \dd\Omega=\Lambda^{*}\left(\left.\Omega\right|_{c=0}\wedge\left.\mbox{d}\Omega\right|_{c=0}\right)=0.
\]
After the blow-up $E$ of the singular locus $\left\{ x=0,y=0\right\} $, $\Omega$ is written in the coordinates of the blow-up $y=xt$
\[
E^{*}\Omega=\left(R\left(1,t\right)+\sum_{i=1}^{n-1}a_{i}\left(x+\left\langle c,tx\right\rangle \right)t^{i}\right)\mbox{d}t+Q\left(1,t\right)\dd\left(x+\left\langle c,tx\right\rangle \right).
\]
The induced foliation restricted to a fibre of $(x,y,c)\mapsto c$ over a point $c=(c_1,\ldots,c_{n-1})$ such that $1/c_1$ is not a root of $R(1,t)=0$ lies in $\mathcal{D}(n)$.
The tangency locus with the exceptional divisor $x=0$ is equal $\left\{ x=0,t=t_{i}\right\} $
where $t_{i}$ is a solution to of $R\left(1,t\right)=0.$ Since the curves ${\{x=0,t=t_{i}\}}=0$ are contained in a invariant hypersurface of $E^{*}\Omega$, the $1-$form $\Omega$ defines an equireducible unfolding of $\left.\Omega\right|_{c=c_0}$ for any $c_0$ lying in the Zariski open set $U=\{c\in\mathbb{C}^{n-1}: c_1t_i\neq 1, \forall t_i\}$

Suppose now that $\Omega$ is trivial along a certain smooth submanifold
of the space of parameter. Then, there exists a germ of application
\[
c:t\in\mathbb{C}\to\left(c_{1}\left(t\right),c_{2}\left(t\right),\cdots,c_{p}\left(t\right)\right)
\]
with $c\left(0\right)=0$ and $c^{\prime}\left(0\right)\neq\left(0,\cdots,0\right)$
such that $\left.\Omega\right|_{c\left(t\right)}$ is a trivial unfolding
of one variable. Now $\left.\Omega\right|_{c\left(t\right)}$ is written
\begin{eqnarray*}
\left(R\left(x,y\right)+\sum_{i=1}^{n-1}a_{i}\left(x+\left\langle c_{i}\left(t\right),y\right\rangle \right)y^{i}x^{n-i}\right)\omega_{R}+Q\left(x,y\right) \dd\left(x+\left\langle c_{i}\left(t\right),y\right\rangle \right)=\\
\qquad\left(\cdots\right)\omega_{R}+Q\left(x,y\right)\dd x+Q\left(x,y\right)\sum_{i=1}^{n-1}c_{i}\left(t\right)iy^{i-1}\dd y+\left(Q\left(x,y\right)\sum_{i=1}^{n-1}c_{i}^{\prime}\left(t\right)y^{i}\right)\dd t
\end{eqnarray*}
Following \cite{M}, the triviality of $\left.\Omega\right|_{c\left(t\right)}$ implies that the coefficient of $\dd t$ belongs
to the ideal generated by the coefficients of $\dd x$ and $\dd y$. If there
exists such a relation, we can evaluate it for $t=0$ and find polynomial functions $A$ and $B$ such that
\begin{equation}\label{superbieneq}
PQ=A\left(Q-yW\right)+xBW.
\end{equation}
where $P\left(y\right)=\sum_{i=1}^{n-1}c_{i}^{\prime}\left(0\right)y^{i}$. Since, $Q$ and $W$ are relatively prime, then there exists $\Delta$ such that $xB-yA=\Delta Q$ and thus $A=P-W\Delta$. Therefore, rewriting (\ref{superbieneq}) yields
\begin{equation}\label{eq:jetnunfolding}
-yP-Q\Delta+yW\Delta+xB=0
\end{equation}
Since the orders of $Q$ and $W$ at $0$ are $n+2$ and $n$, evaluating the jet of order $n$ of the above equality gives
\[-yP+x\textup{Jet}_n(B)=0\]
and thus $P=0$ which is impossible. This proves that the unfolding is non-trivial for $c_0=0$. To prove it for $c_0\in U$, it suffices to remark that the jet $n$ of equation (\ref{eq:jetnunfolding}) remains exactly the same if we impose $c(0)=c_0$ instead of $c_0(0)=0$.

\subsection{An example with no special integrability property}
As already explained, the  Theorem \ref{t:unfoldings} provides the first \emph{non trivial} examples of unfolding of singularities of foliations in $\mathbb{C}^2$. Actually, the only examples known until now were the unfoldings of singularities admitting a Louvillian first integral. Let us be more specific.: in this situation, the foliation is given by a closed form \[\omega=f_1f_2\cdots f_p\sum_{i=1}^p \lambda_i\frac{\dd f_i}{f_i}\]
where the $f_i$´s are holomorphic functions and the $\lambda_i$'s are complex numbers. In such situation, it is enough to unfold the holomorphic function $f_1f_2\cdots f_p$ to deduce an unfolding of $\omega_0$. Indeed, consider any analytical topologically trivial deformation $f_{i,\epsilon}$ of $f_i,\ f_{i,0}=f_i$ with $\epsilon\in \left(\mathbb{C},0\right).$ The one form of $\mathbb{C}^{2+1}$ defined by
 \[\Omega=f_{1,\epsilon}f_{2,\epsilon}\cdots f_{p,\epsilon}\sum_{i=1}^p \lambda_i\frac{\dd f_{i,\epsilon}}{f_{i,\epsilon}}\]is naturally closed and thus integrable. Therefore, it defines an unfolding of $\omega$.

To ensure that our theorem produces new examples, we are going to exhibit a foliation in $\DR$ that does not admit any Louvillian first integral.

Suppose  $\mathcal{F}$ lies in $\mathcal{D}(1)$ and admitting a Louvillian first integral. After the blow-up and the restriction of this first integral to $E$, we can see that $h=H(\F)$ admits a Louvillian first integral on $E$, that is there exists a non-constant holomorphic germ $f$ on $E$ such that $f\circ h=h$ and $df$ extends to a meromorphic closed one-form on $E$.  We know that such a Liouvillian function on a rational curve admits at most a countable number of singularities, that is, homotopy classes of paths along which the analytical extension of the germ of function is impossible. Below, we produce an example of periodic map $h$ such that any first integral has an uncountable number of singularities. Any foliation admitting this germ $h$ as holonomy will not admit a Liouvillian first integral.

\begin{lem}\label{l:singularity}
Let $D\subset\mathbb{C}$ be a region containing $0$ and $h:D\rightarrow D$ a holomorphic mapping, $h(0)=0$ and the germ at $0$ of $h$ satisfies $h^{\circ n}=\text{Id}$ for $n\in\mathbb{N}$. Suppose that the set of singularities of $h$ in $\partial D$ has an accumulation point $p$ and there exists a continuous extension of $h$ to a neighbourhood of $p$ in $\partial D$ satisfying $h(p)\in D$. Then any non-constant holomorphic first integral $f$ of $h$ that is defined and holomorphic on $D$ has a singularity at $p$.
\end{lem}
\begin{pf}
Before starting the proof,  remark that there is an example of an integral having the said properties, namely  $f=Id+h+h\circ h+\ldots+h\circ\ldots^{n-1}\circ h$.
Let $\gamma:[0,1]\rightarrow\overline{D}$ be a path that satisfies $\gamma^{-1}(D)=[0,1)$, $\gamma(0)=0$ and $\gamma(1)$ is a singularity of $h$.  If a first integral $f$ admits analytic extension, we denote its extension by $f_{\gamma}$ and $\lim_{t\rightarrow 1}h(\gamma(t))\in D$, then part of the graph of $h$ is contained in the set $$\{(x,y)\in\text{dom}(f_{\gamma})\times D: f(y)=f_{\gamma}(x)\}.$$
If $f'(h(\gamma(1))\neq 0$, the implicit function theorem tells us that $h$ extends analytically to $\gamma(1)$, which is not possible, so the only possibility is that for every singularity $q\in\partial D$ of $h$ where $h$ extends continuously with value in $D$ and $f$ extends analytically , $f'(h(q))=0$. Suppose for a contradiction that $f$ extends analytically to $p$. Since there exists continuous extension of $h$ to a neighbourhood of $p$ in $\partial D$, for a sequence $p_n$ of singularities of $h$ that accumulate on $p$ we have $f'(h(p_n))=0$ for all $n$. The convergence of $h(p_n)$ to $h(p)\in D$ then tells us that $f$ is constant, contrary to assumption. Hence $p$ is a singularity for $f$.
\end{pf}

To finish let us provide an example of such an $h$ with curves of singularities. Let $D$ be a simply connected plane region bounded by a Jordan curve of class $\mathcal{C}^1$ that is not analytic such that for the rotation $\theta$ of angle $2\pi/n$ around a point $p\in D$, the intersection points of $\partial D$ and $\theta(\partial D)$ are points of transversality between the curves. Let $D_p$ be the connected component of $D\cap\theta(D)$ containing $p$ and $A=\partial D_p\cap h^{-1}(D_p)$. Now, if $\varphi:\overline{\mathbb{D}}\rightarrow \overline{D}$ is the homeomorphic extension of Riemann's mapping Theorem between $\mathbb{D},0$ and $D,p$ to the boundaries given by Carath\' eodory's Theorem, then the continuous map $h:\overline{D_0}\rightarrow \overline{D_0}$ defined by $h=\varphi^{-1}\circ\theta\circ\varphi$ on $D_0=\varphi^{-1}(D_p)$, has order $n$ at $0$ and singularities at each of the points of $\partial D_0$ (see \cite{PM}, p. 628). By construction the values of the extension on $\varphi^{-1}(A)$ belong to $D_0$, so Lemma \ref{l:singularity} guarantees that the points of $\varphi^{-1}(A)$ are singular for any germ of first integral of $h$ around $0$ that is holomorphic on $D_0$.

In fact we have proven that any non-constant first integral of $h$ will not extend to any point in a curve, which is not a countable union of complex analytic sets. Hence no foliation admitting $h$ as holonomy germ will not admit a first integral that is holomorphic (and possibly multivalued) outside a countable union of  analytic sets.

\subsection{Global universal equisingular unfoldings in $\DR(2)$}
In this subsection we prove Corollary \ref{c:universal unfolding}. The form $\Omega(x,y,c)=[y^2+a(x)y(x+cy)+b(x)](x\dd y- y\dd x)+(x+cy)^4\dd x$ defined in the statement can be pulled back by the biholomorphism $(x,y,c)\mapsto (x+cy,y,c)$ to obtain $$[y^2+a(x-cy)yx+b(x)](x\dd y- y\dd x)+x^4\dd (x-cy). $$
This form corresponds precisely to the one appearing in Theorem \ref{t:unfoldings} up to a sign on $c$. Hence it is a non-trivial equisingular unfolding around each fixed parameter $c\in \mathbb{C}\setminus\{t: t^2+a(0)t+b(0)=0\}$. Since the dimension of the base space of the obtained unfolding coincides with the dimension of the universal equisingular unfolding of $\mathcal{F}_c$ , they are equivalent. It is worth mentioning that the topological type of the parameters that were excluded is not contained in $\mathcal{D}$. It is actually singular after blow-up. However these allow to compactify the foliation along the parameter space.

Let us analyze the analytic invariants of $\mathcal{F}_c$ along the parameter space. By construction $H(\mathcal{F}_c)=H(\mathcal{F}_0)$ for all admissible $c$'s. The position of the divisors $q(\mathcal{F}_c)$ defined in the proof of Theorem \ref{t:modulispacesDR} depend on $c$ by a holomorphic (possibly multivalued) non-constant function that assumes any value in $\mathbb{C}\setminus |T(\mathcal{F}_0)|$. By Theorems \ref{t:invariantsinDR} and \ref{t:modulispacesDR} we cover all analytic classes of foliations in $\mathcal{D}(2)$ having the same holonomy $H(\mathcal{F}_0)$.

It remains to see that any choice of rotations around points in $E$ is realized as the holonomy of a normal form.   Once we have a holomorphic germ in $\mathcal{D}(2)$ with given holonomy generators we can consider the coefficients $a$ and $b$ of one of its normal forms in Corollary \ref{c:normalformssmalln} as $\mathcal{F}_0$. The realization part can be found in section \ref{s:holonomy}.

config: arxiv


TITLE: Probability USC Problem Error?
QUESTION [3 upvotes]: A person rolls four fair six-sided die. What is the probability that the person
rolls exactly one 1 and exactly one 2?
This is rather a simple probability problem, but I scored incorrectly for some reason. I first discerned how many total possibilities there were which was $\frac{1}{6^4}$. I then figured out how many checked out. I did $\frac{(1*1*4*4)(4!)}{6^4}$; one way to choose 1, one way to choose 2, and since there can only be exactly one of 1 and 2, the other two die must be 4. I then multiplied by 4! for 24 different arrangements. I resulted in an answer of $\frac{8}{27}$ but the answer is $\frac{4}{27}$.
Help is appreciated.

REPLY [1 votes]: Multinomial probability. Consider three kinds of outcomes in each of $n=4$ trials:  $1$'s, $2$'s, and
'something else', with probabilities $(1/6), (1.6),$ and $(4/6),$ respectively.
$$P(\text{One 1 and One 2}) = {4 \choose {1,1,2}}(1/6)^1(1/6)^1(4/6)^2
= \frac{4!}{1!\cdot 1!\cdot 2!}\frac{4^2}{6^4} = \frac{4}{27} = 0.1481481.$$
Simulation. In R statistical software we simulate the 4-roll experiment
a million times. At the end, the vector event has a million TRUEs and FALSEs; the mean of such a logical vector is its proportion of TRUEs.
With a million iterations, the result 0.148 should be accurate to at least two decimal places (in this particular case, three places).
set.seed(625)  # for exact same simulation, retain this; omit for fresh simulation
m = 10^6; event = logical(m)
for(i in 1:m){
  s = sample(1:6, 4, repl=T)
  event[i] = (sum(s==1)==1) & (sum(s==2)==1)
  }
mean(event)
## 0.147672           # aprx probability 'One 1 and One 2'
2*sd(event)/sqrt(m)
## 0.0007095495       # 95% margin of simulation error

Note: Second, more efficient, logically equivalent simulation. It uses
pseudorandom numbers in exactly the same way as did the first simulation, so it gets
exactly the same result starting with the same seed.
set.seed(625)
m = 10^6;  die = c(1, 10, 0, 0, 0, 0)  # 1=1, 10=2, 0=Else
s = replicate( m, sum(sample(die, 4, repl=T)) )
mean(s==11)
## 0.147672


TITLE: Understanding of proof Ostrik Theorem 1
QUESTION [1 upvotes]: I am currently reading the proof of Theorem 1 on page 1o in Victor Ostrik's Module categories (https://arxiv.org/abs/math/0111139).
I don't understand why the functor he defines $N \mapsto \mathrm{Hom}(N, M)$ goes to $\mathcal{C}$ - why is $\mathrm{Hom}(N, M)$ part of $\mathcal{C}$?
Sorry if I am just blind... I'm thankful for any hint :D

REPLY [2 votes]: In the paper, the functor is not defined by $N\mapsto \operatorname{Hom}(N,M)$, it is actually defined by $N\mapsto \underline{\operatorname{Hom}}(N,M)$. This bifunctor $\underline{\operatorname{Hom}}$ is an internal hom-functor, so it is an object of $\mathcal{C}$ (as in Definition 11).


TITLE: One of characterizations of projective modules over noetherian ring of finite global dimension
QUESTION [3 upvotes]: Let $A$ be noetherian ring of finite global dimension, $M$ be finitely generated module. Then i want to prove that $\mathop{Ext^i}(M,A) = 0, i>0 \implies M -$ projective. Since in this case projectivity is the local property and localization is interchanble with $\mathop{Ext}$ then it's enough to prove this in the case when $A$ is local. But i still have no idea how to prove it..

REPLY [3 votes]: Hint: Use induction to reduce to $\text{pdim}_A(M)\leq 1$. In this case pick a presentation $0\to P\to Q\to M\to 0$ with $P,Q$ finitely generated and projective and consider the associated extension class.
Remark: Dropping the finiteness of the global dimension, but assuming instead that the injective dimension of $_AA$ is finite, one can still deduce that $M$ has an infinite projective resolution $0\to M\to P^0\to P^1\to ...$ to the right. This generalizes the claim above since if $\text{gldim}(A)<\infty$, such an infinite syzyzygy is necessarily projective. In general, the modules which admit an infinite projective resolution to the right are called Gorenstein projective (or (maximal) Cohen-Macaulay); the quotient $\text{G-Proj}(A)/\text{Proj}(A)$ can be taken as a 'measure' for the singularity of $A$ and is therefore called Singularity Category of $A$.


text:
\begin{document}

\twocolumn[
\icmltitle{Submission and Formatting Instructions for \\
           International Conference on Machine Learning (ICML 2020)}

\icmlsetsymbol{equal}{*}

\begin{icmlauthorlist}
\icmlauthor{Aeiau Zzzz}{equal,to}
\icmlauthor{Bauiu C.~Yyyy}{equal,to,goo}
\icmlauthor{Cieua Vvvvv}{goo}
\icmlauthor{Iaesut Saoeu}{ed}
\icmlauthor{Fiuea Rrrr}{to}
\icmlauthor{Tateu H.~Yasehe}{ed,to,goo}
\icmlauthor{Aaoeu Iasoh}{goo}
\icmlauthor{Buiui Eueu}{ed}
\icmlauthor{Aeuia Zzzz}{ed}
\icmlauthor{Bieea C.~Yyyy}{to,goo}
\icmlauthor{Teoau Xxxx}{ed}
\icmlauthor{Eee Pppp}{ed}
\end{icmlauthorlist}

\icmlaffiliation{to}{Department of Computation, University of Torontoland, Torontoland, Canada}
\icmlaffiliation{goo}{Googol ShallowMind, New London, Michigan, USA}
\icmlaffiliation{ed}{School of Computation, University of Edenborrow, Edenborrow, United Kingdom}

\icmlcorrespondingauthor{Cieua Vvvvv}{c.vvvvv@googol.com}
\icmlcorrespondingauthor{Eee Pppp}{ep@eden.co.uk}

\icmlkeywords{Machine Learning, ICML}

\vskip 0.3in
]

\printAffiliationsAndNotice{\icmlEqualContribution}

\begin{abstract}
This document provides a basic paper template and submission guidelines.
Abstracts must be a single paragraph, ideally between 4--6 sentences long.
Gross violations will trigger corrections at the camera-ready phase.
\end{abstract}

\section{Electronic Submission}
\label{submission}

Submission to ICML 2020 will be entirely electronic, via a web site
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\begin{center}
\textbf{\texttt{http://icml.cc/}}
\end{center}

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\end{itemize}

\subsection{Submitting Papers}

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\medskip

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\subsubsection{Camera-Ready Author Information}
\label{final author}

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   \IF{$x_i > x_{i+1}$}
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\end{document}


text:
(***********************************************************************************
 * Copyright (c) 2016-2018 The University of Sheffield, UK
 *
 * All rights reserved.
 *
 * Redistribution and use in source and binary forms, with or without
 * modification, are permitted provided that the following conditions are met:
 *
 * * Redistributions of source code must retain the above copyright notice, this
 *   list of conditions and the following disclaimer.
 *
 * * Redistributions in binary form must reproduce the above copyright notice,
 *   this list of conditions and the following disclaimer in the documentation
 *   and/or other materials provided with the distribution.
 *
 * THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
 * AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE
 * DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE LIABLE
 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR
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 * CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY,
 * OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
 * OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
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 * SPDX-License-Identifier: BSD-2-Clause
 ***********************************************************************************)

section\<open>The Core DOM\<close>
text\<open>This theory is the main entry point of our formalization of the core DOM.\<close>

theory Core_DOM
imports
  "Core_DOM_Heap_WF"
begin

end

subset_name: curated


TITLE: Sufficient condition for a locally ringed space to be an affine scheme
QUESTION [2 upvotes]: Let $X$ be a locally ringed space.
Let $\mathcal{O}$ be its structure sheaf.
Let $A = \Gamma(X, \mathcal{O})$.
Let $x \in X$.
Let $m_x$ be the maximal ideal of $\mathcal{O}_x$.
Let $f \in A$.
We denote by $f_x$ the image of $f$ by the canonical homomorphism $A \rightarrow \mathcal{O}_x$.
We denote by $f(x)$ the image of $f_x$ by the canonical homomorphism $\mathcal{O}_x \rightarrow \mathcal{O}_x/m_x.$
We write $X_f = \{x \in X|\ f(x) \neq 0\}$.
Is the following proposition true?
Proposition
(1) $X_f$ is open for every $f \in A$.
(2) Let $f_1,\dots,f_n$ be elements of $A$.
Suppose $A = (f_1,\dots,f_n)$ and $X_{f_i}$ is an affine scheme for every $i$.
Then $X$ is an affine scheme.
EDIT
As Zhen Lin and QiL pointed out, (2) is probably false.
However, if $X$ is a Noetherian topological space, (2) is true.
EDIT2
As QiL's edited answer shows, (2) is true after all.
I leave the previous EDIT as it is for the sake of honesty.
EDIT3
This is just a remark.
Suppose $A = (f_1,\dots,f_n)$.
I will show that $X = \bigcup X_{f_i}$.
There exist $a_1,\dots,a_n \in A$ such that $1 = a_1f_1 + \cdots + a_nf_n$.
Let $x \in X$.
Then $1 = a_1(x)f_1(x) + \cdots + a_n(x)f_n(x)$.
Hence $f_i(x) \neq 0$ for some $i$.
Hence $x \in X_{f_i}$ as desired.

REPLY [5 votes]: (2) is true. if $X$ is a scheme quasi-compact and quasi-separated (e.g. $X$ has a finite covering by affine open subsets $U_i$ and $U_i\cap U_j$ are also covered by finitely many affine open subsets. This is an exercise in Hartshorne. Note first that under the hypothesis of (2), $X$ is a quasi-compact scheme because it is the union of the affine open subsets $X_{f_i}$. Second, as Keenan pointed out in the comment, $X$ is quasi-separated because $X_{f_i}\cap X_{f_j}$ is actually affine being a standard open subset in $X_{f_i}$ defined by $f_j|_{X_{f_i}}\in O(X_{f_i})$. This implies that for all $f\in A$, $\Gamma(X_f, O_X)=A_f$.
Now consider the canonical map $f : X\to \mathrm{Spec}A=:Y$. For any $i$, $X_{f_i}$ is then isomorphic to $Y_{f_i}$. As the $f_i$'s generate $A$, $X$ (resp. $Y$) is the union of the $X_{f_i}$'s (of the $Y_{f_i}$'s). So $f$ is isomorphism.
If $X$ is not quasi-separated, there is probably a counterexample.


text:
\begin{document}
\maketitle

\footnotetext[1]{Partially supported by Israeli Ministry of Absorption}
\footnotetext[2]{Partially supported by the NSERC Research Grant}

\newcommand{\qed}{\hbox to 0pt{}\hfill$\rlap{$\sqcap$}\sqcup$
\vspace{2mm}}

\begin{abstract}
New explicit conditions of asymptotic and exponential stability
are obtained for the scalar nonautonomous linear delay differential
equation
$$
\dot{x}(t)+\sum_{k=1}^m  a_k(t)x(h_k(t))=0
$$
with measurable delays and coefficients.
These results are compared to known stability tests.
\end{abstract}

{\bf Keywords:} Delay equations,  stability,
explicit conditions

{\bf Running title:} New Stability Conditions

\thispagestyle{empty}

\section{Introduction}
\label{introduct}

In this paper we  continue the study of stability  properties
for the linear differential equation with several delays and an  arbitrary
number of positive and negative coefficients
\begin{equation}
\label{1}
\dot{x}(t)+\sum_{k=1}^m a_k(t)x(h_k(t))=0, ~~t \geq t_0,
\end{equation}
which was begun in \cite{BB1}-\cite{BB3}.

Equation  (\ref{1}) and its special
cases were intensively studied, for example, in
\cite{YS1}-\cite{Mal2}.
In \cite{BB2} we gave a review of stability tests obtained in these
papers.

In almost all papers on stability of delay differential equations
coefficients and delays are assumed to be continuous, which is essentially
used in the proofs of main results.
In real world problems, for example, in biological and ecological models
with seasonal fluctuations of parameters and in economical models with
investments, parameters of differential equations are not necessarily
continuous.

There are also some mathematical reasons to consider differential equations
without the assumption that parameters are continuous functions.
One of the main methods to investigate impulsive differential equations
is their reduction to a non-impulsive differential equation with
discontinuous coefficients.
Similarly, difference equations can sometimes be reduced to the similar
problems for delay differential equations with discontinuous
piecewise constant  delays.

In paper \cite{BB1}  some problems for differential equations
with several delays were reduced to
similar problems for equations with one delay which
generally is not continuous.

One of the purposes of this paper is to extend and partially  improve most popular
stability results for linear delay equations with continuous coefficients and delays
to equations with measurable parameters.

Another purpose is to generalize some results of \cite{BB1,BB2,
BB3}. In these papers, the sum of coefficients was supposed to be
separated from zero and delays were
assumed to be bounded. So the results of these papers
are not applicable, for
example, to the following   equations
$$
\dot{x}(t)+|\sin t| x(t-\tau)=0,
$$$$
\dot{x}(t)+(|\sin t|-\sin t) x(t-\tau)=0,
$$$$
\dot{x}(t)+\frac{1}{t}x(t)+\frac{\alpha}{t}
x \left( \frac{t}{2} \right) =0.
$$
In most results of the present paper
these restrictions are omitted, so we can consider all the equations
mentioned above.
Besides, necessary stability conditions (probably for the first time)
are obtained for equation (\ref{1}) with nonnegative coefficients and
bounded delays.
In particular, if this equation is exponentially stable then the ordinary
differential equation
$$
\dot{x}(t)+\sum_{k=1}^m a_k(t)x(t)=0
$$
is also exponentially stable.

\section{Preliminaries}
\label{prelim}

We consider the scalar linear equation with several
delays (\ref{1}) for $t\geq t_0$ with the initial conditions
(for any $t_0\geq 0$)
\begin{equation}
\label{2}
x(t)=\varphi(t), ~ t<t_0,~ x(t_0)=x_0,
\end{equation}
and under the following assumptions:

(a1) $a_k(t)$ are Lebesgue measurable
 essentially bounded  on $[0,\infty)$ functions;

(a2) $h_k(t)$ are Lebesgue measurable functions,
$$
h_k(t)\leq t, ~\limsup_{t\rightarrow\infty} h_k(t)=\infty;
$$

(a3) $\varphi :(-\infty,t_0)\rightarrow R $ is a  Borel
measurable bounded  function.

We assume conditions
(a1)-(a3) hold for all equations throughout the paper.

\noindent
{\bf Definition.} A locally absolutely continuous
function $x:R\rightarrow R$ is called {\em a solution of the problem}
(\ref{1}), (\ref{2}) if it satisfies the equation (\ref{1})  for almost
all $t\in [t_0,\infty)$ and the equalities (\ref{2}) for $t\leq t_0$.

Below we present a solution representation formula for the nonhomogeneous
equation with locally Lebesgue integrable right-hand side $f(t)$:
\begin{equation}
\label{3}
\dot{x}(t)+\sum_{k=1}^m a_k(t)x(h_k(t))=f(t), ~~t \geq t_0.
\end{equation}

\noindent
{\bf Definition}. A solution  $X(t,s)$ of the problem
$$
\dot{x}(t)+ \sum_{k=1}^m a_k(t)x(h_k(t))=0, ~~t\geq s\geq 0,
$$
$$
x(t)=0,~~t< s,~ x(s)=1,
$$
is called {\em the fundamental function} of (\ref{1}).

\newtheorem{uess}{Lemma}
\begin{uess} $\cite{HL,ABR}$
Suppose conditions (a1)-(a3) hold.
Then the solution of (\ref{3}), (\ref{2})
has the following form
\begin{equation}
\label{4}
x(t)=X(t,t_0)x_0-\int_{t_0}^t X(t,s)\sum_{k=1}^m a_k(s)\varphi(h_k(s))ds
+\int_{t_0}^t X(t,s) f(s)ds,
\end{equation}
where $\varphi(t)=0,~~t\geq t_0$.
\end{uess}

\noindent
{\bf Definition} \cite{HL}.
Eq.~(\ref{1}) is {\em stable}
if for any initial point $t_0$ and number $\varepsilon>0$ there exists
$\delta>0$ such that the inequality
$\sup_{t<t_0}|\varphi(t)|+|x(t_0)|<\delta$
implies $|x(t)|<\varepsilon, ~t\geq t_0,$ for the solution
(\ref{1})-(\ref{2}).

Eq.~(\ref{1}) is {\em asymptotically stable}
if it stable
and all solutions of
(\ref{1})-(\ref{2})  for any initial point $t_0$
tend to zero as $t\rightarrow\infty$.
\vspace{2mm}

In particular, Eq.~(\ref{1}) is asymptotically stable if the fundamental
function is
uniformly bounded: $|X(t,s)|\leq K, ~t\geq s\geq 0$
and all solutions tend to zero as
$t\rightarrow\infty$.

We apply in this paper only these two conditions of
asymptotic stability.

\vspace{2mm}

\noindent
{\bf Definition}. Eq.~(\ref{1}) is
{\em (uniformly) exponentially stable}, if there exist
$M>0, \mu>0$ such that  the solution of problem (\ref{1})-(\ref{2})
has the estimate
$$
|x(t)|\leq M~e^{-\mu (t-t_0)}\left( |x(t_0)|+
\sup_{t<t_0}|\varphi(t)| \right),~~t\geq t_ 0,
$$
where $M$ and $\mu$ do not depend on $t_ 0$.
\vspace{2mm}

\noindent
{\bf Definition}.
The fundamental function  $X(t,s)$ of (\ref{1}) {\em has
an exponential estimation}
if there exist
$K>0, \lambda>0$ such that
\begin{equation}
\label{5}
|X(t,s)|\leq K~e^{-\lambda (t-s)},~~t\geq s\geq 0.
\end{equation}

For the linear equation (\ref{1})  with bounded delays the last  two
definitions are equivalent. For unbounded delays estimation (\ref{5})
implies asymptotic stability of
(\ref{1}).

Under our assumptions the exponential stability does not depend on
values of equation parameters on any finite interval.

\begin{uess}$\cite{GL,BB4}$
Suppose $a_k(t)\geq 0$.
If
\begin{equation}
\label{6}
\int_{\max \{h(t), t_0\}}^t \sum_{i=1}^m a_i(s)ds\leq \frac{1}{e}, ~~
h(t)=\min_k \{h_k(t)\}, ~~t\geq t_0,
\end{equation}
or there exists $\lambda >0$, such that
\begin{equation}
\label{7}
\lambda\geq \sum_{k=1}^m A_ke^{\lambda \sigma_k},
\end{equation}
where
$$
0\leq a_k(t)\leq A_k,~~ t-h_k(t)\leq \sigma_k,~~t\geq t_0,
$$
then $X(t,s)>0,~~t\geq s\geq t_0$, where $X(t,s)$ is the fundamental
function of equation (\ref{1}).
\end{uess}
\begin{uess}$\cite{BB3}$
Suppose $a_k(t)\geq 0$,
\begin{equation}
\label{8}
\liminf_{t\rightarrow\infty}\sum_{k=1}^m a_k(t)>0,
\end{equation}
\begin{equation}
\label{9}
\limsup_{t\rightarrow\infty}(t-h_k(t))<\infty, ~~k=1, \dots , m,
\end{equation}
and there exists $r(t)\leq t$ such that for sufficiently large $t$
$$
\int_{r(t)}^t \sum_{k=1}^m a_k(s)ds\leq\frac{1}{e}.
$$

If
\begin{equation}
\label{10}
\limsup_{t\rightarrow\infty}
\sum_{k=1}^m \frac{ a_k(t)}{\sum_{i=1}^m a_i(t)}
\left|\int_{h_k(t)}^{r(t)} \sum_{i=1}^m a_i(s)ds\right|<1,
\end{equation}
then equation (\ref{1}) is exponentially stable.
\end{uess}
\begin{uess}$\cite{BB3}$
Suppose (\ref{9}) holds and there exists a set of indices $I\subset \{1,\dots,m\}$, such
that  $a_k(t)\geq 0, k\in I$,
\begin{equation}
\label{10a}
\liminf_{t\rightarrow\infty}\sum_{k\in I}a_k(t)>0,
\end{equation}
and the fundamental function of the equation
\begin{equation}
\label{11}
\dot{x}(t)+\sum_{k\in I}a_k(t)x(h_k(t))=0
\end{equation}
is eventually positive. If
\begin{equation}
\label{12}
\limsup_{t\rightarrow\infty} \frac{\sum_{k\not\in I}
|a_k(t)|}{\sum_{k\in I}a_k(t)}<1,
\end{equation}
then equation (\ref{1}) is exponentially stable.
\end{uess}
The following lemma is a consequence of Corollary 2 \cite{BB5}
obtained for impulsive delay differential equations.

\begin{uess}
Suppose for equation (\ref{1}) condition (\ref{9}) holds and this equation is exponentially
stable. If
$$
\int_0^{\infty}\sum_{k=1}^n |b_k(s)|ds<\infty, ~\limsup_{t\rightarrow\infty}
(t-g_k(t))<\infty,~ g_k(t)\leq t,
$$
then the equation
$$
\dot{x}(t)+\sum_{k=1}^m a_k(t)x(h_k(t))+\sum_{k=1}^n b_k(t)x(g_k(t))=0
$$
is exponentially stable.
\end{uess}

The following elementary result will be used in the paper.

\begin{uess}
The ordinary differential equation
\begin{equation}
\label{12b}
\dot{x}(t)+a(t)x(t)=0
\end{equation}
is exponentially stable if and only if there exists $R>0$ such that
\begin{equation}
\label{12a}
\liminf_{t\rightarrow\infty}
\int_t^{t+R} a(s)ds >0.
\end{equation}
\end{uess}

The following example illustrates that a stronger than (\ref{12a})
sufficient condition
\begin{equation}
\label{ode2}
\liminf_{t,s \to \infty} \frac{1}{t-s} \int_s^t
a(\tau)~d\tau >0
\end{equation}
is not necessary for the exponential stability of the ordinary
differential equation (\ref{12b}).
\vspace{2mm}

\noindent
{\bf Example 1.}
Consider the equation
$$
\dot{x}(t)+a(t)x(t)=0, \mbox{~~where~~~}
a(t)= \left\{ \begin{array}{ll} 1, & t \in [2n,2n+1), \\
0, & t \in [2n+1,2n+2),   \end{array}
\right. ~~n=0,1,2, \dots
$$
Then $\liminf$ in (\ref{ode2}) equals zero, but
${\displaystyle |X(t,s)|< e~ e^{-0.5(t-s)} }$, so the equation is
exponentially stable.
Moreover, if we consider $\liminf$ in (\ref{ode2}) under the
condition $t-s \geq R$, then it is still zero for any $R \leq 1$.

\section{Main Results}
\label{main}
\begin{uess}
Suppose $a_k(t)\geq 0$, (\ref{8}),  (\ref{9}) hold and
\begin{equation}
\label{13}
\limsup_{t\rightarrow\infty}
\sum_{k=1}^m \frac{ a_k(t)}{\sum_{i=1}^m a_i(t)}
\int_{h_k(t)}^t \sum_{i=1}^m a_i(s)ds<1+\frac{1}{e}.
\end{equation}
Then equation (\ref{1}) is exponentially stable.
\end{uess}
{\bf Proof.}
By (\ref{8}) there exists function
$r(t)\leq t$ such that for sufficiently large $t$
$$
\int_{r(t)}^t \sum_{k=1}^m a_k(s)ds=\frac{1}{e}.
$$
For this function condition (\ref{10}) has the form
$$
\limsup_{t\rightarrow\infty}
\sum_{k=1}^m \frac{ a_k(t)}{\sum_{i=1}^m a_i(t)}
\left|\int_{h_k(t)}^t \sum_{i=1}^m a_i(s)ds-
\int_{r(t)}^t \sum_{i=1}^m a_i(s)ds\right|
$$$$
=\limsup_{t\rightarrow\infty}
\sum_{k=1}^m \frac{ a_k(t)}{\sum_{i=1}^m a_i(t)}
\left|\int_{h_k(t)}^t \sum_{i=1}^m a_i(s)ds-\frac{1}{e}\right|<1.
$$
The latter inequality follows from (\ref{13}).
The reference to Lemma 3 completes the proof.
\qed

\newtheorem{corollary}{Corollary}
\begin{corollary}
Suppose $a_k(t)\geq 0$, (\ref{8}), (\ref{9}) hold and
\begin{equation}
\label{14}
\limsup_{t\rightarrow\infty}
\int_{\min_k\{h_k(t)\}}^t \sum_{i=1}^m a_i(s)ds<1+\frac{1}{e}.
\end{equation}
Then equation (\ref{1}) is exponentially stable.
\end{corollary}

The following theorem contains stability conditions for equations
with unbounded delays.
We also omit condition (\ref{8}) in Lemma 7.
\newtheorem{guess}{Theorem}
\begin{guess}
Suppose $a_k(t)\geq 0$, condition (\ref{13}) holds,
${\displaystyle \sum_{k=1}^m a_k(t)\not= 0}$ a.e. and
\begin{equation}
\label{21}
\int_0^{\infty}\sum_{k=1}^m a_k(t)dt =\infty,~
\limsup_{t\rightarrow\infty} \int_{h_k(t)}^t \sum_{i=1}^m a_i(s)ds <\infty.
\end{equation}
Then equation  (\ref{1}) is asymptotically stable.

If in addition there exists $R>0$ such that
\begin{equation}
\label{22a}
\liminf_{t,\rightarrow\infty}\int_t^{t+R} \sum_{k=1}^ma_k(\tau)d\tau>0
\end{equation}
then the fundamental function of equation  (\ref{1}) has an exponential estimation.

If condition (\ref{9}) also holds then  (\ref{1}) is exponentially stable.
\end{guess}
{\bf Proof}.
Let ${\displaystyle s=p(t):=\int_0^t \sum_{k=1}^m a_k(\tau)d\tau,~ y(s)=x(t)}$,
where $p(t)$ is a strictly increasing function.
Then $x(h_k(t))=y(l_k(s)),~ l_k(s)\leq s$, ${\displaystyle l_k(s)=\int_0^{h_k(t)}
\sum_{k=1}^m a_k(\tau)d\tau}$
and (\ref{1}) can be rewritten in the form
\begin{equation}
\label{22}
\dot{y}(s)+\sum_{k=1}^m b_k(s)y(l_k(s))=0,
\end{equation}
where
${\displaystyle b_k(s)=\frac{a_k(t)}{\sum_{i=1}^m a_i(t)},~~~
s-l_k(s)=\int_{h_k(t)}^t \sum_{k=1}^m
a_k(\tau)d\tau.}
$
~~Since ${\displaystyle \sum_{k=1}^m b_k(s)=1}$ and \\
${\displaystyle \limsup_{s\rightarrow\infty}(s-l_k(s))<\infty}$,
then Lemma 7  can be applied to equation (\ref{22}).
We have
$$
\limsup_{s\rightarrow\infty}
\sum_{k=1}^m \frac{ b_k(s)}{\sum_{i=1}^m b_i(s)}
\int_{l_k(s)}^s \sum_{i=1}^m b_i(\tau)d\tau
=\limsup_{s\rightarrow\infty}\sum_{k=1}^m  b_k(s)(s-l_k(s))
$$$$
=\limsup_{t\rightarrow\infty}
\sum_{k=1}^m \frac{ a_k(t)}{\sum_{i=1}^m a_i(t)}
\int_{h_k(t)}^t \sum_{i=1}^m a_i(s)ds<1+\frac{1}{e}.
$$
By Lemma 7  equation (\ref{22}) is exponentially
stable.
Due to the first equality in (\ref{21})
$t\rightarrow\infty$ implies $s\rightarrow\infty$.
Hence
${\displaystyle \lim_{t\rightarrow\infty}x(t)=\lim_{s\rightarrow\infty}y(s)=0}$.

Equation (\ref{22}) is exponentially
stable, thus the fundamental function $Y(u,v)$ of equation (\ref{22}) has
an exponential estimation
\begin{equation}
\label{22b}
|Y(u,v)|\leq Ke^{-\lambda (u-v)},~u\geq v\geq 0,
\end{equation}
with $K>0,~ \lambda>0$.
Since ${\displaystyle X(t,s)=Y \left( \int_0^t \sum_{k=1}^m
a_k(\tau)d\tau,\int_0^s \sum_{k=1}^m a_k(\tau)d\tau \right) }$, where
$X(t,s)$ is the fundamental function of (\ref{1}), then (\ref{22b}) yields
$$
|X(t,s)|\leq K\exp\left\{ -\lambda \int_s^t \sum_{k=1}^m a_k(\tau)d\tau
\right\}.
$$
Hence $|X(t,s)|\leq K, ~t\geq s\geq 0$, which together with
${\displaystyle \lim_{t\rightarrow\infty}x(t)=0}$
yields that equation (\ref{1}) is asymptotically stable.

Suppose now that (\ref{22a}) holds.
Without loss of generality we can assume that for some $R>0, \alpha>0$
we have
$$\int_t^{t+R} \sum_{k=1}^m a_k(\tau)d\tau\geq \alpha>0,~ t\geq s\geq 0.
$$
Hence
$$
\exp\left\{ -\lambda \int_s^t \sum_{k=1}^m a_k(\tau)d\tau
\right\}\leq \exp\left\{ \lambda R\sup_{t\geq 0}\sum_{k=1}^m
a_k(t)\right\} e^{-\lambda \alpha (t-s)/R}.
$$

Thus, condition (\ref{22a}) implies the exponential estimate for $X(t,s)$.

The last statement of the theorem is evident.
\qed

\noindent
{\bf Remark.}
The substitution
${\displaystyle s=p(t):=\int_0^t \sum_{k=1}^m a_k(\tau)d\tau,~ y(s)=x(t)}$
was first used in \cite{LSS}.

\begin{corollary}
Suppose ${\displaystyle a_k(t)\geq 0,~ \sum_{k=1}^m
a_k(t)\equiv\alpha>0}$, condition (\ref{9}) holds
 and
\begin{equation}
\label{15}
\limsup_{t\rightarrow\infty}
\sum_{k=1}^m  a_k(t)(t-h_k(t))<1+\frac{1}{e}.
\end{equation}
Then equation (\ref{1}) is exponentially stable.
\end{corollary}
\begin{corollary}
Suppose $a_k(t)=\alpha_k p(t), \alpha_k>0, p(t)>0$ a.e.,
${\displaystyle \int_0^{\infty} p(t)dt =\infty}$ and
\begin{equation}
\label{16}
\limsup_{t\rightarrow\infty}
\sum_{k=1}^m \alpha_k \int_{h_k(t)}^t p(s)ds<1+\frac{1}{e}.
\end{equation}
Then  equation (\ref{1}) is asymptotically stable.

If in addition there exists $R>0$ such that
\begin{equation}
\label{23a}
\liminf_{t\rightarrow\infty}\int_t^{t+R} p(\tau)d\tau>0,
\end{equation}
then the fundamental function of equation  (\ref{1}) has an exponential estimation.

If also (\ref{9}) holds then equation  (\ref{1}) is exponentially stable.
\end{corollary}
\begin{corollary}
Suppose $a(t)\geq 0, b(t)\geq 0, a(t)+b(t)\not= 0$ a.e.,
$$
\int_0^{\infty} (a(t)+b(t))dt =\infty,~~~
\limsup_{t\rightarrow\infty}\int_{h(t)}^t (a(s)+b(s))ds<\infty,
\mbox{~~~and}
$$
\begin{equation}
\label{19}
\limsup_{t\rightarrow\infty} \frac{b(t)}{a(t)+b(t)}\int_{h(t)}^t
(a(s)+b(s))ds<1+\frac{1}{e}.
\end{equation}
Then the following equation is asymptotically stable
\begin{equation}
\label{20}
\dot{x}(t)+a(t)x(t)+b(t)x(h(t))=0.
\end{equation}

If in addition there exists $R>0$ such that
${\displaystyle \liminf_{t\rightarrow\infty}\int_t^{t+R}
(a(\tau)+b(\tau))d\tau>0}$
then the fundamental function of (\ref{20}) has an exponential estimation.

If also ${\displaystyle \limsup_{t\rightarrow\infty}(t-h(t))<\infty}$
then equation (\ref{20}) is exponentially stable.
\end{corollary}

In the following theorem we will omit condition
${\displaystyle \sum_{k=1}^m a_k(t)\not= 0}$ a.e.
of Theorem 1.

\begin{guess}
Suppose $a_k(t)\geq 0$, condition (\ref{14}) and the first inequality
in (\ref{21}) hold.
Then  equation  (\ref{1}) is asymptotically stable.

If in addition (\ref{22a}) holds then the fundamental function of
equation  (\ref{1}) has an exponential estimation.

If also (\ref{9}) holds then equation  (\ref{1}) is exponentially stable.

\end{guess}
{\bf Proof.}
For simplicity suppose that $m=2$ and consider the equation
\begin{equation}
\label{23}
\dot{x}(t)+a(t)x(h(t))+b(t)x(g(t))=0,
\end{equation}
where $a(t)\geq 0, b(t)\geq 0, \int_0^{\infty}(a(s)+b(s))ds=\infty$
and there exist $t_0\geq 0, \varepsilon>0$ such that
\begin{equation}
\label{24}
\int_{\min\{h(t),g(t)\}}^t (a(s)+b(s))ds<1+\frac{1}{e}-\varepsilon,~t\geq
t_0.
\end{equation}
Let us find $t_1\geq t_0$ such that $e^{-h(t)}<\varepsilon/4$,
$e^{-g(t)}<\varepsilon/4$, $t\geq t_1$,
such $t_1$ exists due to (a2). Then $\int_{\min\{h(t),g(t)\}}^t
e^{-s}ds<\varepsilon/2,~t\geq t_1.$
Rewrite equation (\ref{23}) in the form
\begin{equation}
\label{25}
\dot{x}(t)+(a(t)+e^{-t})x(h(t))+b(t)x(g(t))-e^{-t}x(h(t))=0,
\end{equation}
where $a(t)+b(t)+e^{-t}>0$. After the substitution
${\displaystyle s=\int_0^t (a(\tau)+b(\tau)+e^{-\tau})d\tau}$, $y(s)=x(t)$
equation (\ref{25}) has the form
\begin{equation}
\label{26}
\dot{y}(s)+\frac{a(t)+e^{-t}}{a(t)+b(t)+e^{-t}}y(l(s))
+\frac{b(t)}{a(t)+b(t)+e^{-t}}y(p(s))-\frac{e^{-t}}{a(t)+b(t)+e^{-t}}y(l(s))=0,
\end{equation}
where similar to the proof of Theorem 1
\begin{equation}
\label{27}
s-l(s)=\int_{h(t)}^t (a(\tau)+b(\tau)+e^{-\tau})d\tau,~~~
s-p(s)=\int_{g(t)}^t (a(\tau)+b(\tau)+e^{-\tau})d\tau.
\end{equation}
First we will show that by Corollary 1 the equation
\begin{equation}
\label{28}
\dot{y}(s)+\frac{a(t)+e^{-t}}{a(t)+b(t)+e^{-t}}y(l(s))
+\frac{b(t)}{a(t)+b(t)+e^{-t}}y(p(s))=0
\end{equation}
is exponentially stable.
Since ~~
${\displaystyle
\frac{a(t)+e^{-t}}{a(t)+b(t)+e^{-t}}+\frac{b(t)}{a(t)+b(t)+e^{-t}}=1}$,
then (\ref{8}) holds. Condition (\ref{24}) implies (\ref{9}).
So we have to check only condition (\ref{14}) where the sum under the integral is equal to
1.
By (\ref{24}), (\ref{27}) we have
$$
\int_{\min\{l(s),p(s)\}}^s \!\!\!
1 ds=s-\min\{l(s),p(s)\}, ~~~~~s-l(s)=\int_{h(t)}^t
(a(\tau)+b(\tau)+e^{-\tau})d\tau
$$ $$
=\int_{h(t)}^t (a(\tau)+b(\tau))d\tau
+\int_{h(t)}^t e^{-\tau}d\tau<1+\frac{1}{e}
-\varepsilon+\varepsilon/2=1+\frac{1}{e}-\varepsilon/2, ~t\geq t_1.
$$
The same calculations give $s-p(s)<1+\frac{1}{e}-\varepsilon/2$,
thus condition (\ref{14}) holds.

Hence equation (\ref{28}) is exponentially stable.

We return now to equation (\ref{26}). We have $ds=(a(t)+b(t)+e^{-t})dt$,
then
$$
\int_0^{\infty}\frac{e^{-t}}{a(t)+b(t)+e^{-t}}ds=
\int_0^{\infty}\frac{e^{-t}}{a(t)+b(t)+e^{-t}}(a(t)+b(t)+e^{-t})dt<\infty.
$$
By Lemma 5 equation (\ref{26}) is exponentially stable.
Since $t\rightarrow\infty$ implies $s\rightarrow\infty$ then
${\displaystyle
\lim_{t\rightarrow\infty}x(t)=\lim_{s\rightarrow\infty}y(s)=0}$,
which completes the proof of
the first part of the theorem. The rest of the proof is similar to the
proof of Theorem 1.
\qed

\begin{corollary}
Suppose  ${\displaystyle a(t)\geq 0, \int_0^{\infty} a(t)dt =\infty}$ and
\begin{equation}
\label{17}
\limsup_{t\rightarrow\infty} \int_{h(t)}^t a(s)ds<1+\frac{1}{e}.
\end{equation}
Then the equation
\begin{equation}
\label{18}
\dot{x}(t)+a(t) x(h(t))=0
\end{equation}
is asymptotically stable.
If in addition condition (\ref{12a}) holds
then the fundamental function of (\ref{18})
has an exponential estimation.
If also ${\displaystyle \limsup_{t\rightarrow\infty}(t-h(t))<\infty}$
then equation (\ref{18}) is exponentially stable.
\end{corollary}

Now consider equation  (\ref{1}), where
only some of coefficients are nonnegative.

\begin{guess}
Suppose there exists a set of indices $I\subset \{1,\dots,m\}$ such
that  $a_k(t)\geq 0, k\in I$,
\begin{equation}
\label{30a}
\int_0^{\infty} \sum_{k\in I}a_k(t)dt =\infty, ~~~
\limsup_{t\rightarrow\infty}\int_{h_k(t)}^t
\sum_{i\in I}a_i(s)ds<\infty,~~ k=1,\dots,m,
\end{equation}
\begin{equation}
\label{12A}
\sum_{k\not\in I}|a_k(t)|=0, t\in E,~
\limsup_{t\rightarrow\infty, t\not\in E} \frac{\sum_{k\not\in I}
|a_k(t)|}{\sum_{k\in I}a_k(t)}<1, \mbox{~where~}
E=\left\{ t\geq 0, \sum_{k\in I}a_k(t)=0 \right\}.
\end{equation}

If the fundamental function $X_0(t,s)$ of equation  (\ref{11}) is eventually
positive then all
solutions of equation  (\ref{1}) tend to zero as
$t\rightarrow\infty$.

If in addition there exists $R>0$ such that
\begin{equation}
\label{22d}
\liminf_{t\rightarrow\infty}\int_t^{t+R} \sum_{k\in I} a_k(\tau)d\tau>0
\end{equation}
then the fundamental function of equation  (\ref{1}) has an exponential estimation.

If condition (\ref{9}) also holds then  (\ref{1}) is exponentially stable.
\end{guess}
{\bf Proof.}
Without loss of generality we can assume
$X_0(t,s)>0$, $t\geq s\geq 0$. Rewrite equation (\ref{1}) in the form
\begin{equation}
\label{24a}
\dot{x}(t)+\sum_{k\in I}a_k(t)x(h_k(t))+\sum_{k\not\in I}a_k(t)x(h_k(t))=0.
\end{equation}
Suppose first that ${\displaystyle \sum_{k\in I}a_k(t)\not=0}$ a.e.
After the substitution
${\displaystyle
s=p(t):=\int_0^t \sum_{k\in I} a_k(\tau)d\tau}$, $y(s) = x(t)$
we have
${\displaystyle
x(h_k(t))=y(l_k(s))}$,
${\displaystyle l_k(s)\leq s,~
l_k(s)=\int_0^{h_k(t)} \sum_{i\in I} a_i(\tau)d\tau, ~k=1,\dots,m,
}$
and (\ref{1}) can be rewritten in the form
\begin{equation}
\label{24b}
\dot{y}(s)+\sum_{k=1}^m b_k(s)y(l_k(s))=0,
\end{equation}
where
$
{\displaystyle b_k(s)=\frac{a_k(t)}{\sum_{i\in I}  a_i(t)}.}
$
~
Denote by $Y_0(u,v)$ the fundamental function of the equation
$$
\dot{y}(s)+\sum_{k\in I} b_k(s)y(l_k(s))=0.
$$
We have
$$X_0(t,s)=Y_0\left( \int_0^t \sum_{k\in I} a_k(\tau)d\tau,\int_0^s
\sum_{k\in I} a_k(\tau)d\tau \right),$$ $$
Y_0(u,v)=X_0(p^{-1}(u),p^{-1}(v))>0, ~u\geq v\geq 0.
$$

Let us check that other conditions of Lemma 4 hold for equation  (\ref{24b}).
Since ${\displaystyle \sum_{k\in I}b_k(s)=1}$ then condition (\ref{10a}) is satisfied.
In addition,
$$
\limsup_{s\rightarrow\infty, p^{-1}(s) \not\in E} \frac{\sum_{k\not\in I}
|b_k(s)|}{\sum_{k\in I}b_k(s)}=
\limsup_{t\rightarrow\infty, t \not\in E} \frac{\sum_{k\not\in I}
|a_k(t)|}{\sum_{k\in I}a_k(t)}<1.
$$
By Lemma 4 equation (\ref{24b}) is exponentially stable. Hence for any solution
$x(t)$ of (\ref{1}) we have
${\displaystyle \lim_{t\rightarrow\infty}x(t)=\lim_{s\rightarrow\infty}y(s)=0}$.
The end of the proof is similar to the proof of Theorem 2.
In particular, to remove the condition ${\displaystyle \sum_{k\in
I}a_k(t)\not= 0}$ a.e.
we rewrite the equation by adding the term $e^{-t}$ to one of $a_k(t),
k\in I$.
\qed

\noindent
{\bf Remark.} Explicit positiveness conditions for the fundamental function
were presented in Lemma 2.

\begin{corollary}
Suppose
$$
{\displaystyle
a(t)\geq 0, ~
\int_0^{\infty} a(t)dt =\infty,~
\limsup_{t\rightarrow\infty}\int_{g_k(t)}^t a(s)ds<\infty,}
$$$$
\sum_{k=1}^n|b_k(t)|=0, ~t\in E,~
{\displaystyle\limsup_{t\rightarrow\infty, t\not\in E}\frac{\sum_{k=1}^n
|b_k(t)|}{a(t)}<1},
$$
where $E=\{t\geq 0, a(t)=0\}$.
Then the equation
\begin{equation}
\label{1a}
\dot{x}(t)+a(t)x(t)+\sum_{k=1}^n b_k(t)x(g_k(t))=0
\end{equation}
is asymptotically stable.
If in addition (\ref{12a}) holds
then the fundamental function of (\ref{1a}) has an exponential estimation.
If also ${\displaystyle \limsup_{t\rightarrow\infty}(t-g_k(t))<\infty}$
then (\ref{1a})
is exponentially stable.
\end{corollary}

\begin{guess}
Suppose ${\displaystyle \int_0^{\infty}\sum_{k=1}^m |a_k(s)|ds<\infty}$.
Then all solutions of equation (\ref{1}) are bounded
and (\ref{1}) is not asymptotically stable.
\end{guess}
{\bf Proof.}
For the fundamental function of (\ref{1}) we have the following estimation
$$
|X(t,s)|\leq \exp\left\{ \int_s^t \sum_{k=1}^m
|a_k(\tau)|d\tau\right\}.
$$
Then by solution representation formula (\ref{4}) for any solution $x(t)$ of (\ref{1})
we have
$$
|x(t)|\leq  \exp\left\{ \int_{t_0}^t \sum_{k=1}^m |a_k(s)|ds \right\} |x(t_0)|+
\int_{t_0}^t \exp\left\{ \int_{s}^t \sum_{k=1}^m |a_k(\tau)|d\tau \right\}
\sum_{k=1}^m |a_k(s)||\varphi(h_k(s))|ds
$$$$
\leq \exp\left\{ \int_{t_0}^{\infty} \sum_{k=1}^m |a_k(s)|ds \right\}
\left(|x(t_0)|+\int_{t_0}^{\infty}\sum_{k=1}^m |a_k(s)|ds||\varphi||\right),
$$
where $||\varphi||=\max_{t<0}|\varphi(t)|$.
Then $x(t)$ is a bounded function.

Moreover, ${\displaystyle |X(t,s)| \leq A:= \exp \left\{ \int_{0}^{\infty}
\sum_{k=1}^m |a_k(s)|ds \right\} }$, $t\geq s \geq 0$.
Let us choose $t_0 \geq 0$ such that ${\displaystyle \int_{t_0}^{\infty}
\sum_{k=1}^m |a_k(s)|ds < \frac{1}{2A} }$, then
${\displaystyle
X'_t(t,t_0)+\sum_{k=1}^m a_k(t) X(h_k(t),t_0)=0,~ X(t_0,t_0)=1
}$ implies
${\displaystyle
X(t,t_0) \geq 1 - \int_{t_0}^{\infty} \sum_{k=1}^m |a_k(s)|~A~ds
> 1 - A \frac{1}{2A}= \frac{1}{2},}$
thus $X(t,t_0)$ does not tend to zero, so
(\ref{1}) is not asymptotically stable.
\qed

Theorems 3 and 4  imply the following results.

\begin{corollary}
Suppose $a_k(t)\geq 0$, there exists a set of indices
$I\subset \{1,\dots,m\}$ such that
condition (\ref{12A}) and the
second condition in (\ref{30a}) hold. Then all solutions
of (\ref{1}) are bounded.
\end{corollary}
{\bf Proof.}
If ${\displaystyle \int_0^{\infty}\sum_{k\in I} |a_k(t)|dt=\infty}$,
then all solutions of (\ref{1}) are bounded by Theorem 3.
Let ${\displaystyle \int_0^{\infty}\sum_{k\in I} |a_k(t)|dt<\infty}$.
By (\ref{12}) we have
${\displaystyle \int_0^{\infty}\sum_{k\not\in I} |a_k(t)|dt\leq
\int_0^{\infty}\sum_{k\in I} |a_k(t)|dt<\infty}$.
Then
${\displaystyle \int_0^{\infty}\sum_{k=1}^m |a_k(t)|dt<\infty}$.
By Theorem 4 all solutions of (\ref{1}) are bounded.
\qed

\begin{guess}
Suppose $a_k(t) \geq 0$. If (\ref{1}) is asymptotically
stable, then the ordinary differential equation
\begin{equation}
\label{ODE}
\dot{x}(t)+ \left( \sum_{k=1}^ma_k(t) \right) x(t)=0
\end{equation}
is also asymptotically stable.
If in addition (\ref{9}) holds and (\ref{1}) is exponentially stable,
then (\ref{ODE}) is also exponentially stable.
\end{guess}
{\bf Proof.} The solution of (\ref{ODE}), with the initial condition
$x(t_0)=x_0$, can be presented as \\
${\displaystyle x(t)=x_0 \exp \left\{ -\int_{t_0}^t \sum_{k=1}^ma_k(s)~ds
\right\} }$, so (\ref{ODE}) is asymptotically stable, as far as
\begin{equation}
\label{ode1}
\int_0^{\infty} \sum_{k=1}^ma_k(s)~ds = \infty
\end{equation}
and is exponentially stable if  (\ref{22a}) holds (see Lemma 6).

If (\ref{ode1}) does not hold, then by Theorem 4 equation (\ref{1}) is
not asymptotically stable.

Further, let us demonstrate that exponential stability of (\ref{1})
really implies (\ref{22a}).

Suppose for the fundamental function of
(\ref{1}) inequality (\ref{5}) holds and condition (\ref{22a})
is not satisfied.
Then there exists a sequence $\{t_n\}$,
$t_n\rightarrow\infty$,
such that
\begin{equation}
\label{1star}
\int_{t_n}^{t_n+n} \sum_{k=1}^m a_k(\tau)~d\tau
< \frac{1}{n}<\frac{1}{e}, ~~~n\geq 3.
\end{equation}
By (\ref{9}) there exists $n_0>3$ such that $t-h_k(t)\leq n_0,~k=1,\dots,m$.
Lemma 2 implies
$X(t,s)>0$, $t_n\leq s \leq t \leq t_n+n,~n\geq n_0$.
Similar to the proof of Theorem 4 and using the inequality ${\displaystyle
1-x \geq e^{-x}}, ~x>0$,
we obtain
$$X(t_n,t_n+n)  \geq 1-\int_{t_n}^{t_n+n} \sum_{k=1}^m a_k(\tau)~d\tau  \geq
\exp\left\{ - \int_{t_n}^{t_n+n}\sum_{k=1}^m a_k(\tau)~d\tau \right\} >
e^{ -\frac{1}{n}}. $$
Inequality (\ref{5}) implies $|X(t_n+n,t_n)|\leq Ke^{-\lambda n}$.
Hence $Ke^{-\lambda n}\geq e^{ -\frac{1}{n}},~ n\geq n_0 $,
or $K>e^{\lambda n-1/3}$ for any $n\geq n_0 $.
The contradiction proves the theorem.
\qed

Theorems 3 and 5 imply the following statement.

\begin{corollary}
Suppose $a_k(t)\geq 0$ and the fundamental function of equation (\ref{1})
is positive. Then (\ref{1}) is asymptotically stable if
and only if the ordinary differential equation
(\ref{ODE}) is asymptotically stable.

If in addition (\ref{9}) holds then (\ref{1}) is exponentially stable
if and only if (\ref{ODE}) is exponentially stable.
\end{corollary}

\section{Discussion and Examples}

In  paper \cite{BB2} we gave a review of known stability tests for
the linear equation (\ref{1}). In this part we will compare the
new results obtained in this paper with known stability conditions.

First let us compare the results of the present
paper with our papers \cite{BB1}-\cite{BB3}.
In all these three papers we apply the same method
based on Bohl-Perron type theorems and
comparison with known exponentially stable equations.

In \cite{BB1}-\cite{BB3} we considered exponential stability only.
Here we also give explicit conditions for asymptotic stability.
For this type of stability, we omit the requirement that the delays are bounded
and the sum of the coefficients is separated from zero.
We also present some new stability tests, based on the results obtained in
\cite{BB3}.

Compare now the results of the paper with some other known results
\cite{Y,Kz,SYC,GH2,GD,HL}.
First of all  we replace the constant $\frac{3}{2}$ in most of these
tests by the constant $1+\frac{1}{e}$. Evidently
${\displaystyle 1+\frac{1}{e}=1.3678\dots<\frac{3}{2}}$,
so we have a worse constant,
but it is an open problem to obtain $\frac{3}{2}$-stability results for
equations with measurable coefficients and delays.

Consider now equation (\ref{18}) with a single delay. This equation is
well studied beginning with the classical stability result by Myshkis \cite{My}.
We present here 3 statements which cover most of known
stability tests for this equation.
\vspace{2mm}

\noindent
{\bf Statement 1} $\cite{Y}$.
{\em Suppose $a(t)\geq 0, h(t)\leq t$
are continuous functions and
\begin{equation}
\label{29a}
\limsup_{t\rightarrow\infty}\int_{h(t)}^t a(s)ds\leq\frac{3}{2}.
\end{equation}
Then all solutions of (\ref{18}) are bounded.

If in addition
$$
\liminf_{t\rightarrow\infty}\int_{h(t)}^t a(s)ds>0,
$$
and the strict inequality in (\ref{29a}) holds
then equation (\ref{18}) is exponentially stable.}
\vspace{2mm}

\noindent
{\bf Statement 2} $\cite{SYC}$.
{\em Suppose $a(t)\geq 0, h(t)\leq t$
are continuous functions, the strict inequality (\ref{29a})
holds and $\int_0^{\infty} a(s)ds=\infty $. Then all solutions of
(\ref{18}) tend to zero as $t\rightarrow\infty$.
}
\vspace{2mm}

\noindent
{\bf Statement 3} $\cite{GH2,GD}$.
{\em Suppose $a(t)\geq 0, h(t)\leq t$
are measurable functions, $\int_0^{\infty}a(s)ds=\infty$,
$A(t)=\int_0^t a(s)ds$ is a strictly monotone increasing function and
$$
\limsup_{t \rightarrow\infty}\int_{h(t)}^t a(s)ds<
\sup_{0<\omega<\pi/2}\left(\omega+\frac{1}{\Phi(\omega)}\right)\approx
1.45\dots,
$$
$\Phi(\omega)=\int_0^{\infty}u(t,\omega)dt$, where $u(t,\omega)$
is a solution of the initial value problem
$$
y(t)+y(t-\omega)=0,~~ y(t)=0, ~~ t<0, ~~ y(0)=1.
$$
Then equation (\ref{18}) is asymptotically stable.
}
\vspace{2mm}

\noindent
{\bf Example 2.}
Consider the equation
\begin{equation}
\label{42}
\dot{x}(t)+\alpha(|\sin t|-\sin t)x(h(t))=0, h(t)\leq t,
\end{equation}
where $h(t)$ is an arbitrary measurable function such that $t-h(t)\leq
\pi$ and $\alpha >0$.

This equation has the form (\ref{18}) where $a(t)=\alpha(|\sin t|-\sin
t)$.
Let us check that the conditions of  Corollary 5 hold.
It is evident that
$\int_0^{\infty}a(s)ds =\infty.$
We have
$$
\limsup_{t\rightarrow\infty}\int_{h(t)}^ta(s)ds\leq
\limsup_{t\rightarrow\infty}\int_{t-\pi}^ta(s)ds\leq
-\alpha\int_{\pi}^{2\pi}2\sin sds=4\alpha.
$$
If $\alpha<0.25\left(1+\frac{1}{e}\right)$,
then condition (\ref{17}) holds, hence all solutions of equation (\ref{42})
tend to zero as $t\rightarrow\infty$.

Statements 1-3 fail for this equation.
In Statements 1,2 the delay should be continuous.
In Statement 3 function $A(t)=\int_0^t a(s)ds$ should be strictly increasing.
\vspace{2mm}

Consider now the general equation (\ref{1}) with several delays.
The following two statements are well known for this equation.
\vspace{2mm}

\noindent
{\bf Statement 4} $ \cite{Kz}$.
{\em Suppose $a_k(t)\geq 0, h_k(t)\leq t$
are continuous functions and
\begin{equation}
\label{42a}
\limsup_{t\rightarrow\infty}a_k(t)
\limsup_{t\rightarrow\infty}(t-h_k(t))\leq 1.
\end{equation}
Then all solutions of (\ref{1}) are bounded
and 1 in the right hand side of (\ref{42a}) is the best
possible constant.

If ${\displaystyle \sum_{k=1}^m a_k(t)>0}$ and the strict inequality in (\ref{42a}) is
valid then all solutions of
 (\ref{1}) tend to zero as $t\rightarrow\infty$.

If $a_k(t)$ are constants then in (\ref{42a})
the number 1 can be replaced by $3/2$.}
\vspace{2mm}

\noindent
{\bf Statement 5} $ \cite{SYC}$.
{\em Suppose $a_k(t)\geq 0, h_k(t)\leq t$
are continuous, $h_1(t)\leq h_2(t)\leq\dots\leq h_m(t)$
and
\begin{equation}
\label{42b}
 \limsup_{t\rightarrow\infty} \int_{h_1(t)}^t \sum_{k=1}^m a_k(s)ds\leq 3/2.
\end{equation}
Then any solution of (\ref{1}) tends to a constant as $t\rightarrow\infty$.

If in addition ${\displaystyle \int_0^{\infty} \sum_{k=1}^m a_k(s)ds=\infty}$,
then all solutions of (\ref{1}) tend to zero as $t\rightarrow\infty$.          }
\vspace{2mm}

\noindent
{\bf Example 3.}
Consider the equation
\begin{equation}
\label{44}
\dot{x}(t)+\frac{\alpha}{t}x\left( \frac{t}{2}-\sin t
\right) +\frac{\beta}{t}x\left( \frac{t}{2} \right) =0,~ t\geq
t_0>0,
\end{equation}
where $\alpha>0, \beta>0$.
Denote
${\displaystyle
p(t)=\frac{1}{t},~~
h(t)=\frac{t}{2}-\sin t, ~~ g(t)=\frac{t}{2}. }$

We apply Corollary 3.
Since ${\displaystyle \lim_{t\to \infty}
\left[ \ln\left( \frac{t}{2} \right)- \ln\left( \frac{t}{2}-\sin t\right)
\right]=0}$, then
$$\limsup_{t\rightarrow\infty}\left(\alpha\int_{h(t)}^t p(s)ds+
\beta\int_{g(t)}^t p(s)ds\right)
\leq (\alpha+\beta)\ln2.$$
Hence if ${\displaystyle \alpha+\beta<\frac{1}{\ln2}\left(
1+\frac{1}{e} \right)}$ then
equation (\ref{44}) is exponentially stable.
Statement 4 fails for this equation since the delays are unbounded.
Statement 5 fails for this equation since neither $h(t)\leq g(t)$ nor $g(t)\leq h(t)$
holds.
\vspace{2mm}

Stability results where the nondelay term dominates over the delayed
terms are well known
beginning with the book of Krasovskii~\cite{Kr}.
The following result is cited from the monograph~\cite{HL}.
\vspace{2mm}

\noindent
{\bf Statement 6} $\cite{HL}$.
{\em Suppose $a(t), b_k(t), t-h_k(t)$ are bounded continuous functions,
there exist $\delta,k$, $\delta>0, 0<k<1,$ such that
$a(t)\geq \delta$ and  ${\displaystyle \sum_{k=1}^m
|b_k(t)|<k\delta}$.
Then the equation
\begin{equation}
\label{45a}
\dot{x}(t)+a(t)x(t)+\sum_{k=1}^m b_k(t)x(h_k(t))=0
\end{equation}
is exponentially stable. }
\vspace{2mm}

In Corollary 6 we obtained a similar result without the assumption that the parameters of
the equation are continuous functions and the delays are bounded.
\vspace{2mm}

\noindent
{\bf Example 4.}
Consider the equation
\begin{equation}
\label{45}
\dot{x}(t)+\frac{1}{t}x(t)+\frac{\alpha}{t}x\left( \frac{t}{2} \right) =0,
~~~ t\geq t_0>0.
\end{equation}
If $\alpha<1$ then by Corollary 6  all solutions of equation (\ref{45})
tend to zero.
The delay is unbounded, thus Statement 6 fails for this equation.
\vspace{2mm}

In \cite{LP} the authors considered a delay autonomous equation with linear and
nonlinear parts, where  the differential equation with the linear part only has
a positive fundamental function and the linear part dominates over the nonlinear one.
They generalized the early result of Gy\"{o}ri \cite{G} and some results
of \cite{FH}.

In Theorem 4 we obtained a similar result for a linear nonautonomous equation
without the assumption that coefficients and delays are continuous.
\vspace{2mm}

We conclude this paper with some open problems.
\vspace{2mm}

\noindent
{\bf Open Problem 1.}
{\em Prove or disprove that in Corollary 5 the constant
${\displaystyle 1+\frac{1}{e}}$
can be  changed by the  constant ${\displaystyle \frac{3}{2} }$.}

Note that  all known proofs with the constant $\frac{3}{2}$ apply
methods which are not applicable for equations with measurable
parameters.
\vspace{2mm}

\noindent
{\bf Open Problem 2.}
{\em Suppose $a_k(t)\geq 0$, conditions  (\ref{8}), (\ref{9}) hold and equation
(\ref{1}) is exponentially stable.

Prove or disprove that for any $b_k(t)$, $0\leq b_k(t)\leq a_k(t)$,
where ${\displaystyle \liminf_{t\rightarrow\infty} \sum_{k=1}^m b_k(t)>0}$,
the equation
$$
\dot{x}(t)+\sum_{k=1}^m  b_k(t)x(h_k(t))=0
$$
with the same delays as in (\ref{1})
is also exponentially stable.

Obtain similar result for the asymptotic stability.
}
\vspace{2mm}

The solution of the following problems would improve Theorems 1 and 5, respectively.
\vspace{2mm}

\noindent
{\bf Open Problem 3.}
{\em Suppose (\ref{8}), (\ref{9}) hold and
$$
\limsup_{t\rightarrow\infty}
\sum_{k=1}^m \frac{ |a_k(t)|}{\sum_{i=1}^m a_i(t)}
\int_{h_k(t)}^t \sum_{i=1}^m a_i(s)ds<1+\frac{1}{e}.
$$
Prove or disprove that equation (\ref{1}) is exponentially stable.
}
\vspace{2mm}

\noindent
{\bf Open Problem 4.}
{\em Suppose (\ref{1}) is exponentially   stable. Prove or
disprove that the
ordinary differential equation (\ref{ODE})
is also exponentially (asymptotically)  stable, without restrictions on
the signs of coefficients $a_k(t) \geq 0$, as in Theorem 5.
}


TITLE: classical solutions of PDE with mixed boundary conditions
QUESTION [13 upvotes]: Nowadays people usually consider PDEs in weak formulations only, so I have a hard time finding statements about the existence of classical solutions of the Poisson equation with mixed Dirichlet-Neumann boundary conditions.
Maybe someone here can help me and point to a book or article where I can find sufficient conditions on the right hand side that guarantee the existence of a $C^2$ solution.
Note that the unique solvability in $H^1(\Omega)$ follows, e.g., from the paper on discrete maximum principles by Karatson and Korotov in Numer. Math. 99 (2005), 669-698. So (a slightly more precise version of) my question boils down to finding conditions such that this solution is actually C^2 in the interior and continuous differentiable at the boundary.

REPLY [3 votes]: I believe the papers by Gary Lieberman will answer your question: one is Mixed boundary value problem for elliptic and parabolic differential equations of second order. Another one is Optimal holder regularity for mixed boundary value problems. Thanks


TITLE: Algebraic and rational parts of a real number
QUESTION [6 upvotes]: Let $\alpha$ be a positive real number. Does it make sense to define the closest rational to $\alpha$ as the number $R(\alpha)=\frac{p_1}{p_2}$ such that $p_1,p_2$ are positive co-prime integers minimizing $p_2 \cdot |p_2\alpha - p_1|$? Clearly, there are going to be some irrational numbers for which this makes no sense, for instance $\alpha=\frac{1+\sqrt{5}}{2}$. My guess is that such numbers are rare: there are countably infinitely many of them, or their set has Lebesgue measure 0. Do we have $R(\pi) = \frac{355}{113}$? I looked at the fist 12 convergents of $\pi$, and $\frac{355}{113}$ achieves the minimum.
This is related to approximations of irrational numbers, continued fractions, and the irrationality measure of a number. More generally, we could define the closest algebraic number of degree $d$, as the number $R_d(\alpha)$ defined as follows.
$$P^*= \arg\min_P\Big(H(P)\Big)^{\mu(d)}\cdot |P(\alpha)|,\\
R_d(\alpha) = \Big(H(P^*)\Big)^{\mu(d)}\cdot |P^*(\alpha)|
$$
where $P$ is any polynomial of degree $d$ with integer coefficients, with
highest and lowest coefficients not equal to $0$, and $H(P)$ is the
height of $P$, that is, its highest coefficient in absolute value. To make this work for most $\alpha$, how should we choose $\mu(d)$? Does $\mu(d)=d$ work? It seems to work if $d=1$.
As of now, as far as I know, all results involving $d>1$ are conjectures. Related material includes the Wirsing conjecture. See also the "Generalizations" section in the Wikipedia article on Roth's theorem, here.
Update on Feb 8, 2021: It is possible that the best approximation, if $d$ is an even integer, may be a complex number. Also, see my new question here, about approximations by dyadic fractions. The plan is to look at approximations using the first $n$ digits of $\alpha\in [0, 1]$ in base $b$, where (say) $b=\sqrt{2}$, focusing on values of $n$ where a long run of zeros start, leading to approximations of transcendental numbers by quadratic irrationals. This is briefly discussed in the comments in my new question.

REPLY [10 votes]: Let $\alpha$ be an irrational. We shall consider its continued fraction $[a_0;a_1,a_2,\dots]$. Recall some basic results about convergents of continued fractions (see e.g. here): letting $p_n,q_n$ be the sequence of numerators and denominators of convergents, for any $n>1$ we have $q_{n+1}=a_{n+1}q_n+q_{n-1}>a_{n+1}q_n$ and
$$\left|\alpha-\frac{p_n}{q_n}\right|<\frac{1}{q_nq_{n+1}}<\frac{1}{a_{n+1}q_n^2},$$
hence $q_n|q_n\alpha-p_n|<\frac{1}{a_{n+1}}$. Therefore if $a_{n+1}$ are unbounded, then $q_n|q_n\alpha-p_n|$ does not attain a minimum. Therefore $R(\alpha)$ does not exist for those $\alpha$.
Therefore $R(\alpha)$ can only exist for badly approximable numbers. It is known that those numbers form a set of measure zero, and most natural constants besides quadratic irrationalities, including $\pi$ and all higher degree algebraic irrationals, are conjectured to not lie in it. Therefore $R(\pi)$ probably doesn't exist.
On the other hand, if $\alpha$ is badly approximable, then this still doesn't necessarily mean $R(\alpha)$ necessarily exists, as you note with $\alpha=\frac{1+\sqrt{5}}{2}$. In fact I believe it won't exist for any quadratic irrational. However, using the bound
$$\left|\alpha-\frac{p_n}{q_n}\right|>\frac{1}{q_n(q_{n+1}+q_n)}>\frac{1}{(a_{n+1}+2)q_n^2},$$
we at the very least get that for those numbers the quantity $q_n|q_n\alpha-p_n|$ is bounded away from zero (note that $q|q\alpha-p|$ can only be smaller than $1/2$ if $p/q$ is a convergent, so we don't lose much from looking at just looking at convergents).
Last remark I have is that there are uncountably many $\alpha$ for which $R(\alpha)$ exists. Indeed, from the above considerations it follows easily that this is the case if for some $N$ we have that the continued fraction of $\alpha$ contains a partial denominator $N$, but from some point on all denominators are at most $N-2$.


text:
\section{Third Isomorphism Theorem}
Tags: Isomorphisms, Named Theorems, Isomorphism Theorems

\begin{theorem}
=== [[Third Isomorphism Theorem/Groups|Groups]] ===
{{:Third Isomorphism Theorem/Groups}}
=== [[Third Isomorphism Theorem/Rings|Rings]] ===
{{:Third Isomorphism Theorem/Rings}}
\end{theorem}

config: wiki


TITLE: Bounding differences in probabilities $P(f(X)>a)-P(g(x)>a)$ by $P(|f(x)-g(x)| > \epsilon)$
QUESTION [0 upvotes]: Let $X$ be a random variable and $f,g$ be two nice functions. In this context, I view $g$ as a good approximation to $f$.
I am interested in obtaining bounds for the following probability which measures the discrepancy in the distributions of $f(x)$ and $g(x)$. Can I bound $|P(f(x) <a) - P(g(x)<a)|$ by a quantity of the form $P(|f(x)-g(x)|>\epsilon)$? If I'm able to do so, I could then use inequalities that bound the discrepancy between $f$ and $g$. It doesn't really have to be $P(|f(x)-g(x)|>\epsilon)$ but any probability of the form $P(|f(x)-g(x)| \in B)$ for some set $B$ will do.
The closest inequality to this that I've seen comes from Wikipedia: see this link. But it is not quite what I need.

REPLY [0 votes]: You don't specify the quantifiers on $a,\varepsilon$, and not even whether you wish to get upper or lower bounds of the first quantity in terms of the second. So this answer is going to stay very high level.
You most likely will not get any meaningful bound between the two quantities, since they are dealing with qualitatively different things; observe that the quantity
$$
\lvert \mathbb{P}\{ f(X) < a \} - \mathbb{P}\{ g(X) < a \} \rvert \tag{1}
$$
says something about the distributions of $f(X)$, $g(X)$ (for instance, its value would be unchanged when replacing $g(X)$ by $g(Y)$, for $Y$ an independent copy of $X$). The second,
$$
\mathbb{P}\{ \lvert f(X) - g(X)  \rvert > \varepsilon \} \tag{2}
$$
says something about the random variables $f(X),g(X)$ themselves (not just their distribution).

As an example: take $f$ to be the identity $f(x)=x$, $g$ to be the negation $g(x)=-x$, and let $X$ be a Rademacher random variable (i.e., uniform on $\{-1,1\}$). (For a similar example, you could take $X$ to be a standard Gaussian.)
Then, for all $a\in\mathbb{R}$,
$$
\lvert \mathbb{P}\{ f(X) < a \} - \mathbb{P}\{ g(X) < a \} \rvert = 0
$$
while
$$
\mathbb{P}\{ \lvert f(X) - g(X)  \rvert > \varepsilon \} = 1
$$
for every $\varepsilon < 2$.


TITLE: Is there a quotient of $k[X,Y]$ which is an integral domain but not a subring of $k[x_1, \dots, x_n]$
QUESTION [2 upvotes]: Question:

Let $k$ be the field $\Bbb Q$ or $\Bbb R$.
  Can we find a prime ideal $P$ of $k[X,Y]$ such that $k[X,Y]/P$ is not isomorphic to a subring of $k[x_1, \dots, x_n]$ (whatever $n \geq 0$ is)?

My question above comes from the following observation: the ring $\Bbb C[x,y]/(y^2-x^3)$ is isomorphic to the subring $\Bbb C[t^2,t^3] $ of $\Bbb C[t] \subset \Bbb C[t,y]$ (but not isomorphic to $\Bbb C[t]$ itself, however). The quotients of $\Bbb C[x_1, \dots, x_m]$ occur frequently in algebraic geometry, and I was a bit surprised to see that such a quotient could be… a subring of $\Bbb C[x_1, \dots, x_m]$.
Another example the quotient $\Bbb C[x,y]/(xy-1)$, i.e. the localization $\Bbb C[x]_{x}$, is clearly (isomorphic to) a subring of $\Bbb C(x)$. Thinking about this, I ended up with the following surprising result:

The quotient ring $\Bbb C[x_1, \dots, x_m]/I$ is an integral domain if and only if it is isomorphic to a subring of $\Bbb C[x_1, \dots, x_n]$ for some $n \geq 0$.

Proof:
The direction $\Longleftarrow$ is easy, since a subring of an integral domain is an integral domain.
But the direction $\implies$ is less clear.
Assume $K[x_1, \dots, x_m]/I$ is an integral domain, where $K$ is any field of characteristic $0$ of cardinality $2^{\aleph_0}$ (e.g. $K=\Bbb C$).
Denote by $L$ the algebraic closure of the fraction field of $K[x_1, \dots, x_m]/I$.
Since the algebraically closed field $L$ has characteristic $0$ (because $I \cap \Bbb Z \subset I \cap K = \{0\}$) and its cardinality is $2^{\aleph_0}$, it is isomorphic to $\Bbb C$ (see for instance here).
Therefore, $K[x_1, \dots, x_m]/I \subset L \cong \Bbb C$, and we are done (with $n=0$).
$\hspace{5cm}\blacksquare$
(In particular, $\Bbb R[x]$ is isomorphic to a subring of $\Bbb C$ ! This is less difficult to see for $\Bbb Q[x] \cong \Bbb Q[\pi] \subset \Bbb R \subset \Bbb C$).
However, I should mention that I didn't find an explicit embedding $\Bbb C[x,y](y^2-x(x+1)) \hookrightarrow \Bbb C[x_1, \dots, x_n]$ for instance.
(Notice that $Q(x,y):=y^2-xP(x) \in \Bbb C[x,y]$ is irreducible whenever $P \in \Bbb C[x]$ satisfy $P(0) \neq 0$, i.e. $x \not\mid P$, so that $(Q)$ is a prime ideal of the UFD $\Bbb C[x,y]$. It is indeed sufficient to apply Eisenstein critertion, $x$ being irreducible in the UFD $\Bbb C[x]$).
The argument above strongly relies on uniqueness properties of algebraically closed fields, so it can't be applied to my original question about $k=\Bbb Q,k=\Bbb R$. The quotient $k[x,y]/P$ is a finitely generated $k$-algebra, but I don't see how to relate this to some polynomial algebra.
I tried to work with $k[x,y]/(y^2-x^3-x)$ and $n=1$. In particular, I looked for the existence of polynomials $P,Q \in k[t]$ such that $P(t)^2 = Q(t)^3+Q(t)$. It was a bit long, and anyway I'm not if this is useful. Notice that the group of units of $k[x,y]/(xy-1)$ is $k^* \times \Bbb Z$, which is isomorphic to a subgroup of $k^*$.
Thank you for your help!

REPLY [2 votes]: The field $\mathbf{R}[X]/(X^2+1)$ is  in fact a field but not contained in $\mathbf{R}[X_1,X_2,\ldots,X_n]$. Here is the idea for producing more such examples.  Hilbert's Nullstellenstatz gives bijective correspondence between maximal ideals and points of affine space only for algebraically closed fields. For other fields $K$  we can find maximal ideals which correspond only to a point in an extension field. SO quotients by such maximal ideals will not be a subring of a polynomial ring over $K$.


TITLE: Is every group a nontrivial subgroup of a larger group?
QUESTION [1 upvotes]: Let $G$ be a group and $\kappa$ be a cardinal number strictly bigger than $|G|$.

(1) Is there a group $H$ for which $G$ is a nontrivial subgroup of $H$?
(2) Is there a group $H$ for which $G$ is a nontrivial subgroup of $H$ and $ \ |H|= \kappa \ $?

Let $ \ \mu: H \times H \to H \ $ be the binary operation of the group $H$. That is, $(H, \mu)$ is the group. I say that a group $(J, \nu)$ is a subgroup of $(H, \mu)$ if, and only if,
(i) $ \quad J \subset H \ $;
(ii) $ \quad \mu|_{J \times J} = \nu \ $.
Here I can only use ZFC set theory. That is, equality is "set theory equality" and identifications or isomorphisms are strictly prohibited.

This question appeared because in a semigroup $S$ without identity it is very easy to add an extra element (set) $ \ e \notin S \ $ such that $ \ S \cup \{ e \} \ $ is a semigroup with identity $e$, just defining by hand what the operation do on it. I wonder if this is possible with other elements and other structures as well.

REPLY [5 votes]: Once you know that the answer work "up to isomorphism", then the answer is positive in the set theoretic sense as well.
Suppose that $(H,*_H)$ is isomorphic to a subgroup of $(G',*_{G'})$. Namely, there is a monomorphism $\varphi\colon H\to G'$. Let $G$ be the set $H\cup(\{H\}\times(G'\setminus\operatorname{range}(\varphi)))$. We can replace $\{H\}$ by $\{x\}$ for which $H\cap\{x\}\times G'=\varnothing$, and of course there is a proper class of such $x$'s.
Now define $\Phi\colon G\to G'$ as follows:
$$\Phi(x)=\begin{cases}\varphi(x) & x\in H\\ y & x=(H,y)\end{cases},$$
namely either $\varphi(x)$ or the projection onto $G'$. And define $$x*_Gy=\Phi^{-1}(\Phi(x)*_{G'}\Phi(y)).$$
It is not hard to check that $(G,*)$ is a group and that $H$ is indeed a subgroup of $G$. Of course, there are no restrictions on the cardinality of $G$.


TITLE: A category $I$ is filtered iff any colimits of $I$ in the category of sets commutes with any finite limits
QUESTION [1 upvotes]: A category is said to be filtered if:

it is not empty
for any objects $i, j$, there exists an object $k$ and
maps $i \to k, j \to k $.
for any objects $i, j$ and for any maps $u, v : i \to j $, there exists $j \to k$ such that the two maps $i \to k$ coincides.

The “only if” part is fundamental.
I want to show the “if” part.
The condition 2 is easy: it follows from the commutativity of filtered colimits with direct products.
But I can’t show 1 and 3.
I think that 3 follws from the commutativity with equalizers.
And 1. seems to be wrong:
The empty category satisfies the condition of the title, but does not satisfies the condition 1.

REPLY [4 votes]: Here's a general result, which covers the case of filtered colimits and finite limits.
Proposition. Let $\mathcal{I}$ and $\mathcal{J}$ be small categories. If colimits of shape $\mathcal{I}$ commute with limits of shape $\mathcal{J}$ in $\textbf{Set}$, then  every diagram $\mathcal{J}^\textrm{op} \to \mathcal{I}$ admits a cocone in $\mathcal{I}$.
Proof. Let $F : \mathcal{J}^\textrm{op} \to \mathcal{I}$ be a diagram and consider the diagram $\mathcal{I} (F -, -) : \mathcal{J} \times \mathcal{I} \to \textbf{Set}$.
We have a comparison map:
$$\textstyle \varinjlim_\mathcal{I} \varprojlim_\mathcal{J} \mathcal{I} (F -, -) \longrightarrow \varprojlim_\mathcal{J} \varinjlim_\mathcal{I} \mathcal{I} (F -, -)$$
For each object $i$ in $\mathcal{J}$, $\varinjlim_\mathcal{I} \mathcal{I} (i, -) \cong 1$.
(This is basically a consequence of the Yoneda lemma.)
Hence, the RHS is $\varprojlim_\mathcal{J} 1$, which is just $1$.
On the other hand, the LHS is a quotient of the disjoint union of $\varprojlim_\mathcal{J} \mathcal{I} (F -, j)$ as $j$ varies, so if the comparison map is a bijection  then there exists some object $j$ in $\mathcal{J}$ for which $\varprojlim_\mathcal{J} \mathcal{I} (F -, j)$ is not empty.
But $\varprojlim_\mathcal{J} \mathcal{I} (F -, j)$ is the set of cocones from $F$ to $j$, so we are done. ■
The above argument actually only uses the hypothesis that the comparison map is a surjection. If we use the hypothesis that the comparison map is a bijection, we get a stronger conclusion:
Proposition. Let $\mathcal{I}$ and $\mathcal{J}$ be small categories. If colimits of shape $\mathcal{I}$ commute with limits of shape $\mathcal{J}$ in $\textbf{Set}$, then for every diagram $F : \mathcal{J}^\textrm{op} \to \mathcal{I}$, the category ${}^{F /} \mathcal{I}$ of cocones under $F$ is (inhabited and) connected.
Finally, a word of warning: these are necessary conditions, but not sufficient.


TITLE: Primes in a group: number theory
QUESTION [5 upvotes]: Characterize the primes for which 3 is not a square in $Z_p$. Compute $T_{17}$ where $T_p:=\{a+\sqrt{3}b:a^2-3b^2=1\} \subset Z_p[\sqrt{3}]$. Compute $T_{17}$. What are the orders of each of the elements in $T_{17}$?
I'm stuck on the first part.... I don't know how to really characterize these primes or what that means.
For computing $T_{17}$, I am assuming I need to find the contained elements. Do I need to check for all possible combinations of a and b? or is there a shortcut of some sort?
For example, I can see that $a=7$, $b=4$ works with $7+4\sqrt{3}$ since $7^2-3(16)=1$.

REPLY [2 votes]: Case 1: $p\equiv 1\mod 4$
By quadratic reciprocity, we have $x^2\equiv 3\mod p$ is solvable if and only if $x^2\equiv p\mod 3$ is solvable.  The latter has a solution if and only if $p\equiv 1\mod 3$ (remembering that $p\ne 3$) so that in this case we have $p\equiv 1\mod 12$.
Case 2: $p \equiv 3\mod 4$
By quadratic reciprocity, we have $x^2\equiv 3\mod p$ is solvable if and only if $x^2\equiv p\mod 3$ is not solvable.  Again, the latter has a solution if and only if $p\equiv 1\mod 3$, so that in this case the original congruence has a solution if and only if $p\equiv 11\mod 12$.
The two cases show that the primes such that $3$ is not a square in $\mathbb{Z}_p$ are exactly those congruent to $5$ and $7$ modulo $12$.
To compute $T_{17}$, you could use symmetry to only check values $0,1,\ldots, 8$ because if $(a,b)$ is a solution to $a^2-3b^2=1$, then so is $(\pm a,\pm b)$.  In any case, we obtain the following $18$ elements, recorded as ordered pairs:
$$T_{17}=\{(\pm1,0),(\pm2,\pm1),(\pm 5,\pm5),(\pm7,\pm 4),(\pm 8,\pm 2)\}$$


TITLE: Endomorphism of $\mathcal{M}_n(\mathbb{R})$ such that $ f({}^t M)={}^t f(M)$
QUESTION [2 upvotes]: I and a friend are trying to find all endomorphisms $f$ of $\mathcal{M}_n(\mathbb{R})$ such that $f({}^t M)={}^t f(M)$ for all $M$. We believe they are of the form $M\mapsto\lambda M+\mu {}^t M$ for a fixed $(\lambda,\mu)\in\mathbb{R}^2$. Any help is appreciated, thank you.

REPLY [3 votes]: Let $f$ be a linear endomorphism on $M_n(\mathbb R)$. Then
$$
f(M^T)=f(M)^T\quad\forall M\tag{1}
$$
if and only if
$$
f(M)=\frac12\left(g(M)+g(M^T)^T\right)\quad\forall M\tag{2}
$$
for some endomorphism $g$.
Given $g$, it is straightforward to verify $(1)$ when $f$ is defined by $(2)$. Conversely, given any $f$ that satisfies $(1)$, condition $(2)$ is satisfied by taking $g=f$.
Alternatively, note that $(1)$ is satisfied if and only if $f$ preserves both symmetric and skew-symmetric matrices. Hence such an $f$ takes the form of $f(M)=h\left(\frac{M+M^T}{2}\right)+k\left(\frac{M-M^T}{2}\right)$ where $h$ is an endomorphism defined on the space $\mathcal H_n$ of all symmetric matrices and $k$ is an endomorphism defined on the space $\mathcal K_n$ of all skew-symmetric matrices. Therefore the dimension of the space of all such $f$s is
$$
\dim\operatorname{End}(\mathcal H_n)+\dim\operatorname{End}(\mathcal K_n)
=\left[\frac{n(n+1)}{2}\right]^2+\left[\frac{n(n-1)}{2}\right]^2
=\frac{n^2(n^2+1)}{2}.
$$