TITLE: Normal curvature visualization
QUESTION [1 upvotes]: I am currently studying Meusnier's theorem that tells that the normal curvature of a curve $\gamma$ drawn on a surface $S$ depends only on the direction of $\dot{\gamma}$. I completely agree with the proof but I have no visualization/intuition of what it represents geometrically.
And I have this problem not only for the normal curvature, geodesic curvature and for curvature in general, I know it represents the "tendency to curve" but I still have a weak visualization of what it means

REPLY [1 votes]: Signed Curvature. Think of a particle traveling along a curve parameterized at unit-speed. Its velocity at time $t$ is normal to its acceleration. Let $L$ be the line passing through the particle and parallel to its acceleration. Let the time elapse by $\delta t$ and repeat the same construction as before getting a new line $L'$. The intersection $L\cap L'$ depends on $t$ and $\delta t$ and converges to a point $\mathbf c$ on $L$ as $\delta t\to0$. The curvature $\kappa$ is the reciprocal of the distance between $p$ and $\mathbf c$. It is positive if $\mathbf c$ is in the direction of the normal and negative otherwise. The circle with center $\mathbf c$ and radius $1/|\kappa|$ has the same tangent at the particle than the curve.
Normal Curvature. Consider a plane $\Pi$ in $\mathbb R^3$ including the normal vector $\mathbf n(p)$ at a given point $p\in S$. The intersection $\Pi+p\cap S$ can be visualized as a curve on $S$. The intersection $\Pi\cap T_p(S)$ is a one-dimensional subspace, i.e., a line. Let $\mathbf v$ be a unit vector on $\Pi\cap T_p(S)$. Note that $\{\mathbf v,\mathbf n(p)\}$ is an orthonormal basis of $\Pi$. In this regard, having $\Pi$ can be considered equivalent to having $\mathbf v$. In addition, $\mathbf v$ is an element of the tangent to the curve $\Pi+p\cap S$. If $\mathbf v_1$ and $\mathbf v_2$ are the principal directions of $S$ at $p$ and we are assuming that $\|\mathbf v\|=1$, we have
$$
    \mathbf v = (\cos\theta)\mathbf v_1 + (\sin\theta)\mathbf v_2
$$
for some convenient value of $\theta$. Since we have some freedom to fix the origin of $\theta$ we may consider it to be the angle measured from $\mathbf v_1$ to $\mathbf v_2$. In other words, $\theta=0$ corresponds to $\mathbf v_1$ and $\theta=\pi/2$ to $\mathbf v_2$. By definition the normal curvature at $\mathbf v$ is
\begin{equation}\label{eq119-1}
    \kappa_{\mathbf n} = \mathbb{II}_p(\mathbf v) = k_1\cos^2\theta + k_2\sin^2\theta.
\end{equation}
Using this information it can be shown that $\kappa_{\mathbf n}$ is the signed curvature of the (plane) curve $\Pi+p\cap S$ at $p$. As $\Pi$ rotates from $0$ to $\pi$, the intersecting curves run over all the possible values of $k_{\mathbf n}$ (which get repeated between $\pi$ and $2\pi$).
In sum, the normal curvature of $\gamma$ at $t_0$ is the signed curvature of the plane curve defined by the intersection of $S$ with the plane spawn by $\mathbf n$ and $\gamma'(t_0)$.
Osculating Plane. A particle traveling along a spatial curve would experience acceleration on any turn of said curve. This acceleration would be caused by the force impelled by the curve on the particle, aimed at annihilating the particle's natural inertia and deviate the direction of its velocity vector so as to constrain it to the trajectory dictated by the curve. Since (in the unit-speed case) the curve can only push in directions that are normal to it, the acceleration of the particle must be in a normal direction too. In this regard, we can see that the motion of the particle is instantaneously confined to the plane defined by the velocity and the acceleration vectors. The osculating plane of the (spatial) curve at a given point is the plane spawn by its tangent and normal vectors.
Geodesic Curvature. In the case where the spatial curve lies on a regular surface, we have two planes: the osculating one and the tangent to the surface. Even though both planes include the vector tangent to the curve, they are in general different because the normal to the curve is not the normal to the surface: the former points to the center of the curvature circle of the curve, the latter is influenced by all the other curves laying on the surface and passing through the point. In particular, we can project the derivative of the tangent vector (i.e., the acceleration of the particle traveling along the curve) onto the tangent plane. The geodesic curvature is the component (a.k.a. shadow) of the acceleration vector $T'$ onto the tangent plane.

For lots of wonderful geometrical insights, explanations and proofs refer to Visual Differential Geometry and Forms by Tristan Needham.


TITLE: history-based strategy versus position-based strategy
QUESTION [3 upvotes]: SUMMARY: suppose a game is played that is a walk on an infinite directed graph without blind points, where the two players each make one step alternatingly.
Suppose that the pay-off of a play is determined by any confinal subset of the play.
In this case, can a winning history-based strategy always be replaced by a winning position-based strategy?
In the comments, Bof answered no. Gabriel Debs in Paris had a counterexample.
Game theorists probably know the answer to this question. Thank you for your interest.
Let $M$ be a move set and $A \subset {^\omega}M$ a pay-off set. Let $\rho: M\cup\{0\} \rightarrow \mathcal{P}M\setminus\{\emptyset\}$ be a rule.
A play is a sequence $x\in{^\omega}M$, where the even moves were made by player I and the odd moves were made by player II.
The first move (made by I) should be in $\rho(0)$.
After one player plays a move $m$, the other player is required to play a move in $\rho(m)$.
If any player makes an illegal move in this sense, the first player to do so loses the game.
If, on the other hand, in a play $x$ all moves were legal, then player I is the winner iff $x\in A$.
A history-based strategy is a mapping $\sigma:{^{<\omega}}M\rightarrow M$. If a player plays $\sigma$, this means that if $s\in{^{<\omega}}M$ is the history of all moves that have been made up to a certain point, then the player plays $\sigma(s)$.
A position-based strategy is a mapping $\tau:M\sqcup\{0\}\rightarrow M$. If a player plays $\tau$, this means that if $s\in{^{<\omega}}M$ is the history of all moves up to a given point, then the player plays
$$
\begin{cases}
\tau(0)\quad&\text{if $s=\emptyset$,}\\
\tau\big(s(len(s)-1)\big)\quad&\text{otherwise.}
\end{cases}
$$
PROBLEM. Suppose that for any legal plays (obeying $\rho$) $x,y \in {^\omega}M$ and $c_1<c_2<\dots<\omega,d_1<d_2<\dots<\omega$ such that $\forall i:x(c_i)=y(d_i)$, we have that $x \in A$ iff $y \in A$.
Also assume that player I has a winning history-based strategy. Can we conclude that I has a winning position-based strategy?
AFFIRMATION IF $|M|<\omega$.
Observe that $A$ is determined by the moves that occur cofinally often in some element of $A$.
Let $G=\big\{m\in M:\exists x\in A,C\in[\omega]^\omega:x(C)=\{m\}\big\}$.
Let $\sigma$ be a winning history-based strategy for I.
This means that for every play $x$ following $\sigma$ we have $|x^{-1}G|=\omega$.
Let
$$
U=\big\{u\in{^\omega}M:\forall i:u(2i)=\sigma\big(u|(2i)\big)\big\}
$$
be the set of won plays according to $\sigma$.
Let
$$
N=\bigcup_{u\in U}u[2\mathbb{N}+1]\subseteq M
$$
be the set of won positions where it is I to move.
Consider $u\in U$ and $i\in\omega$ odd.
There is an $n(u,i)>0$ such that for all $y\in U$ with $y|(i+1)=u|(i+1)$ we have
$$
y[i,n(u,i)]:=\{y(i+1),y(i+2),\dots,y(i+2n(u,i))\}\cap G\neq\emptyset.
$$
(If this were not true, the set
$$
\big\{y|n:y\in U\land y|i=u|i\land n\in\omega\land y[i,n]\cap G\neq\emptyset\big\}
$$
would be an infinite tree of finite width.
It contains an infinite branch by König's lemma, contradicting the fact that $\sigma$ is winning.)
Assume that $n(u,i)$ was chosen minimal with its property.
For $m\in N$, let
$$
n(m)=\min_{u\in U,u(i)=m}n(u,i)
$$
and $u^m,i^m$ the corresponding arguments. Thus $u^m(i^m)=m$ and $n(m)=n(u^m,i^m)$.
Let
$$
\tau(m)=\sigma\big(u^m|(i^m+1)\big).
$$
(Of course, $\tau(0)=\sigma(\emptyset)$.)
Then $\tau$ is a winning position-based strategy for I: let $x$ be a play according to $\tau$ and $i\in\omega$ odd.
Then $m:=x(i)\in N$. For all $y\in U$ with $y|i^m=u^m|i^m$ it holds that $y[i^m,n(m)]\cap G\neq\emptyset$.
We need to find $j>i$ such that $x(j)\in G$. Pick $v\in U$ such that $v|(i^m+1)=u^m|(i^m+1)$, $v(i^m+1)=\tau(m)=x(i+1)$ and $v(i^m+2)=x(i+2)$.
Assume w.l.o.g. that $x(i+1),x(i+2)\notin G$.
We have $\{x(i+1),x(i+2)\}\cup y[i^m+2,n(m)-1]=y[i^m,n(m)]$ for all $y\in U$ with $y|(i^m+3)=v|(i^m+3)$.
This implies that $n(v,i^m+2)\leq n(m)-1$. So $n(x(i+2))\leq n(m)-1$.
We see that $n(x(i))>n(x(i+2))>\dots$ so eventually we find a $j>i$ such that $x(j)\in G$.
Q.E.D.
(Our game is in addition determined in the case $|M|<\omega$, which I saw when applying König's lemma.)
I have a slight feeling that the answer to the problem will be no in general.
If the answer is yes anyways, we naturally extend the question to game lengths larger than $\omega$.

REPLY [3 votes]: The Banach–Mazur (or Banach–Mazur–Oxtoby) game $BM(X)$ on a topological space $X$ is played as follows. A play of the game is an infinite nested sequence $B_1\supseteq W_1\supseteq B_2\supseteq W_2\supseteq\cdots$ of nonempty open sets, where $B_n$ is chosen by Black and $W_n$ by White; Black wins if the intersection of the chosen sets is empty, White wins if it's nonempty. This is a game of the kind you're asking about; $M$ is the set of all nonempty open subsets of $X$. (If you want a smaller set $M$, you could require the moves to be chosen from a given $\pi$-base without, I think, affecting anything important.)
It is known that Black has a winning (historical or perfect-information) strategy if and only if $X$ is not a Baire space, and in this case Black even has a winning positional strategy, also called a stationary strategy or a tactic. On the other hand,  Gabriel Debs, Stratégies gagnantes dans certains jeux topologiques, Fund. Math. 126 (1985), 93–105 (pdf), has constructed an example of a topological space $X$ (a refinement of the usual topology on the real line) such that White has a winning historical strategy in $BM(X)$ but has no winning positional strategy.


text:
\begin{document}
\title[Sampling Theory for Sturm-Liouville Problem ]{Sampling Theory for
Sturm-Liouville Problem with Boundary and Transmission Conditions Dependent
on Eigenparameter\ \ \ }
\author{Fatma H\i ra}
\address[F. H\i ra and N. Alt\i n\i \c{s}\i k]{Ondokuz May\i s University,
Arts and Science Faculty, 55139, Samsun, Turkey }
\email[Corresponding author: F. H\i ra]{fatma.hira@omu.edu.tr}
\author{Nihat Alt\i n\i \c{s}\i k}
\email[N. Alt\i n\i \c{s}\i k]{anihat@omu.edu.tr}
\subjclass[2000]{34L10; 34B24; 41A05; 94A20}
\keywords{Whittaker-Shannon's sampling theory, Kramer's sampling theory,
discontinuous Sturm-Liouville problems}

\begin{abstract}
In this paper, we investigate the sampling analysis associated with
discontinuous Sturm-Liouville problem which has transmission conditions at
two points of discontinuity also contains an eigenparameter in a boundary
condition and two transmission conditions. We establish briefly spectral
properties of the problem and then we prove the sampling theorem associated
with the problem.
\end{abstract}

\maketitle

\section{Introduction}

Let $\sigma >0,$ and denote by $B_{\sigma }^{2},$ the Paley-Wiener space of
all entire functions $f$ of exponential type with band width at most $\sigma
$ which are $L^{2}\left(

\mathbb{R}

\right) $ functions when restricted to $

\mathbb{R}

.$ This space is characterized by the following relation which is given by
Paley and Wiener $\left[ 1\right] ,$ known as the Paley-Wiener theorem:
\begin{equation}
f\left( t\right) \in B_{\sigma }^{2}\Leftrightarrow f\left( t\right) =\frac{1
}{\sqrt{2\pi }}\int\limits_{-\sigma }^{\sigma }e^{ixt}g\left( x\right) dx,
\text{ \ \ \ \ \ }g\in L^{2}\left( -\sigma ,\sigma \right) .  \tag{1.1}
\end{equation}
\qquad \qquad \qquad

The Whittaker-Kotel'nikov-Shannon (WKS) sampling theorem states that if $
f\left( t\right) \in B_{\sigma }^{2},$ then it is completely determined by
its values at the points $t_{k}=\frac{k\pi }{\sigma },k\in

\mathbb{Z}

$ and can be reconstructed by means of the formula
\begin{equation}
f\left( t\right) =\sum\limits_{k=-\infty }^{\infty }f\left( t_{k}\right)
\sin c\left( \sigma t-k\pi \right) ,\text{ \ \ }t\in

\mathbb{C}

,  \tag{1.2}
\end{equation}
where
\begin{equation}
\sin c\left( \sigma t-k\pi \right) =\left\{ \QATOP{\frac{\sin \left( \sigma
t-k\pi \right) }{\sigma t-k\pi },\text{ \ \ \ \ }t\neq \frac{k\pi }{\sigma },
}{1,\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }t=\frac{k\pi }{\sigma }.}\right.
\tag{1.3}
\end{equation}
The series (1.2) is absolutely and uniformly convergent on compact subsets
of $

\mathbb{C}

,$ uniformly convergent on $

\mathbb{R}

,($see $\left[ 2,3\right] ).$

One of the important generalizations of the WKS sampling theorem is the
Paley-Wiener-Levinson theorem which can be stated as follows: Let $\left\{
t_{k}\right\} _{k\in

\mathbb{Z}

}$ be a sequence of real numbers satisfying
\begin{equation}
D:=\sup_{k\in

\mathbb{Z}

}\left\vert t_{k}-\frac{k\pi }{\sigma }\right\vert <\frac{\pi }{4\sigma },
\tag{1.4}
\end{equation}
and let
\begin{equation}
G\left( t\right) :=\left( t-t_{0}\right) \prod\limits_{k=1}^{\infty }\left(
1-\frac{t}{t_{k}}\right) \left( 1-\frac{t}{t_{-k}}\right) .  \tag{1.5}
\end{equation}
Then for any $f\left( t\right) \in B_{\sigma }^{2},$
\begin{equation}
f\left( t\right) =\sum\limits_{k=-\infty }^{\infty }f\left( t_{k}\right)
\frac{G\left( t\right) }{G^{\prime }\left( t_{k}\right) \left(
t-t_{k}\right) },\text{ \ \ \ }t\in

\mathbb{C}

.  \tag{1.6}
\end{equation}
The series (1.6) converges uniformly an compact subsets of $

\mathbb{C}

,($see $\left[ 2,4\right] ).$

Series of the form (1.6) is called a Lagrange-type interpolation.

Another important generalization of the WKS sampling theorem is the theorem
of Kramer $\left[ 5\right] ,$ can be stated as follows: Let $I$ be a finite
closed interval, $K\left( x,t\right) $ a function continuous in $t$~such
that $K\left( x,t\right) \in L^{2}\left( I\right) ~$for all $t\in

\mathbb{C}

,$ and let $\left\{ t_{k}\right\} _{k\in

\mathbb{Z}

}$ be a sequence of real numbers such that $\left\{ K\left( x,t_{k}\right)
\right\} _{k\in

\mathbb{Z}

}$ is a complete ortogonal set in $L^{2}\left( I\right) .$ Suppose that
\begin{equation}
f\left( t\right) =\int\limits_{I}K\left( x,t\right) g\left( x\right) dx,
\text{ \ \ \ \ }g\in L^{2}\left( I\right) .  \tag{1.7}
\end{equation}
Then
\begin{equation}
f\left( t\right) =\sum\limits_{k=-\infty }^{\infty }f\left( t_{k}\right)
\frac{\int\limits_{I}K\left( x,t\right) \overline{K\left( x,t_{k}\right) }dx
}{\left\Vert K\left( x,t_{k}\right) \right\Vert _{L^{2}\left( I\right) }^{2}}
.  \tag{1.8}
\end{equation}

Generalization of WKS sampling theorem has been investigated extensively,
see $\left[ 6-9\right] .$ Sampling theorems associated with Sturm-Liouville
problems were investigated in $\left[ 10-15\right] $. Also $\left[ 16,17
\right] $ and $\left[ 18,19\right] $ are the example works with an
eigenparameter in the boundary conditions in direction of sampling analysis
associated with continuous and discontinuous eigenproblems, respectively. In
$\left[ 20\right] ,$ the authors discussed the situation of deriving a
sampling theorem of Kramer type when the kernels are discontinuous. They
introduced the discontinuous Sturm-Liouville problems studied by Kobayashi $
\left[ 21\right] $ to Sturm-Liouville problems with eigenfunctions having
two symmetrically located discontinuities and satisfying symmetric jump (or
transmission) conditions. Their sampling result states that the
transformation
\begin{equation*}
F\left( \lambda \right) =\int\limits_{0}^{\pi }f\left( x\right) u\left(
x,\lambda \right) dx,\text{ \ \ \ \ }f\in L^{2}\left( 0,\pi \right) ,
\end{equation*}
can be reconstructed via the sampling form
\begin{equation*}
F\left( \lambda \right) =\sum\limits_{n=0}^{\infty }F\left( \lambda
_{n}\right) \frac{\omega \left( \lambda \right) }{\left( \lambda -\lambda
_{n}\right) \omega ^{\prime }\left( \lambda _{n}\right) },
\end{equation*}
where $\left\{ \lambda _{n}\right\} _{n=0}^{\infty }$ are the zeros of the
function $\omega \left( \lambda \right) ,$ and they are exactly the
eigenvalues of the discontinuous Sturm-Liouville problem. In this problem,
neither boundary conditions nor transmission conditions contain an
eigenparameter. In $\left[ 22\right] ,$ the author investigated the sampling
analysis associated with discontinuous Sturm-Liouville problems which has
one point of discontinuity and contains an eigenparameter in all boundary
conditions and derived sampling representations for transforms whose kernels
are either solutions or Green's functions.

We consider the boundary value problem:
\begin{equation}
\tau \left( u\right) :=-u^{\shortparallel }\left( x\right) +q\left( x\right)
u\left( x\right) =\lambda u\left( x\right) ,~\ \ \ \ x\in I  \tag{1.9}
\end{equation}
with boundary conditions;
\begin{equation}
B_{1}\left( u\right) :=\beta _{1}u\left( a\right) +\beta _{2}u^{\prime
}\left( a\right) =0,  \tag{1.10}
\end{equation}
\begin{equation}
B_{2}\left( u\right) :=\lambda \left( \alpha _{1}^{\prime }u\left( b\right)
-\alpha _{2}^{\prime }u^{\prime }\left( b\right) \right) -\left( \alpha
_{1}u\left( b\right) -\alpha _{2}u^{\prime }\left( b\right) \right) =0,
\tag{1.11}
\end{equation}
and transmission conditions at two points of discontinuity $c_{1}$and $
c_{2}; $
\begin{equation}
T_{1}\left( u\right) :=u\left( c_{1}^{-}\right) -\delta u\left(
c_{1}^{+}\right) =0,  \tag{1.12}
\end{equation}
\begin{equation}
T_{2}\left( u\right) :=u^{\prime }\left( c_{1}^{-}\right) -\delta u^{\prime
}\left( c_{1}^{+}\right) +\lambda u\left( c_{1}^{-}\right) =0,  \tag{1.13}
\end{equation}
\begin{equation}
T_{3}\left( u\right) :=\delta u\left( c_{2}^{-}\right) -\gamma u\left(
c_{2}^{+}\right) =0,  \tag{1.14}
\end{equation}
\begin{equation}
T_{4}\left( u\right) :=\delta u^{\prime }\left( c_{2}^{-}\right) -\gamma
u^{\prime }\left( c_{2}^{+}\right) +\lambda u\left( c_{2}^{-}\right) =0,
\tag{1.15}
\end{equation}
where $I=[a,c_{1})\cup \left( c_{1},c_{2}\right) \cup
(c_{2},b],~a<c_{1}<c_{2}<b;$ $\lambda $ is a complex spectral parameter$;$ $
q\left( x\right) $ is a given real valued function which is continuous in $
[a,c_{1}),~\left( c_{1},c_{2}\right) $ and $(c_{2},b]$ and has finite limits
$q(c_{1}^{\pm })=\underset{x\rightarrow c_{1}^{\pm }}{\lim }q\left( x\right)
$ and $q(c_{2}^{\pm })=\underset{x\rightarrow c_{2}^{\pm }}{\lim }q\left(
x\right) ;$ $\alpha _{i},\alpha _{i}^{\prime },\beta _{i},$ $\delta ,\gamma
\in

\mathbb{R}

$ $\left( i=1,2\right) ,$ $\left\vert \beta _{1}\right\vert +\left\vert
\beta _{2}\right\vert \neq 0,$ $\delta >0,$ $\gamma \neq 0;$ $c_{i}^{\pm
}:=c_{i}\pm 0,$ $\left( i=1,2\right) $ and $\rho :=\left( \alpha
_{1}^{\prime }\alpha _{2}-\alpha _{1}\alpha _{2}^{\prime }\right) >0.$

In the present work, we investigate the sampling analysis associated with
discontinuous Sturm-Liouville problem (1.9)-(1.15) which has transmission
conditions at two points of discontinuity and contains an eigenparameter in
a boundary condition and two transmission conditions. This is the difference
between our problem and sampling theories associated with discontinuous
eigenproblems studied in the literature. The problem of deriving a sampling
theorem of Kramer type when the problem contains an eigenparameter in two
transmission conditions besides in a boundary conditions does not exist as
far as we know. To derive sampling theorem associated with the problem
(1.9)-(1.15), we study briefly the spectral properties of the problem
(1.9)-(1.15) and then we prove that integral transforms associated with the
problem (1.9)-(1.15) can also be reconstructed in a sampling form of
Lagrange interpolation type.

\section{Spectral Properties}

To formulate a theoretic approach to the problem (1.9)-(1.15) let $
L:=L^{2}(a,c_{1})\oplus L^{2}(c_{1},c_{2})\oplus L^{2}\left( c_{2},b\right) $
and $

\mathbb{C}

^{3}:=

\mathbb{C}

\oplus

\mathbb{C}

\oplus

\mathbb{C}

$ and we define the Hilbert space $H=L\oplus

\mathbb{C}

^{3}$ with an inner product
\begin{equation}
\begin{array}{l}
\left\langle F(.),G(.)\right\rangle _{H}:=\dint\limits_{a}^{c_{1}}f\left(
x\right) \overline{g}\left( x\right) dx+\delta
^{2}\dint\limits_{c_{1}}^{c_{2}}f\left( x\right) \overline{g}\left( x\right)
dx+\gamma ^{2}\dint\limits_{c_{2}}^{b}f\left( x\right) \overline{g}\left(
x\right) dx+ \\
\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\dfrac{\gamma ^{2}}{\rho
}h_{1}\overline{k_{1}}+h_{2}\overline{k_{2}}+\delta h_{3}\overline{k_{3}},
\end{array}
\tag{2.1}
\end{equation}
where

$F(x)=\left(
\begin{array}{c}
f\left( x\right) \\
h_{1} \\
h_{2} \\
h_{3}
\end{array}
\right) ,G(x)=\left(
\begin{array}{c}
g(x) \\
k_{1} \\
k_{2} \\
k_{3}
\end{array}
\right) \in H,$ $f\left( .\right) ,$ $g\left( .\right) \in L^{2}(a,b)$ and $
h_{i},k_{i}\in

\mathbb{C}

$ $\left( i=1,2,3\right) .$

For convenience we put
\begin{equation}
R_{b}\left( u\right) :=\alpha _{1}u\left( b\right) -\alpha _{2}u^{\prime
}\left( b\right) ,\text{ \ \ }R_{b}^{\prime }\left( u\right) :=\alpha
_{1}^{\prime }u\left( b\right) -\alpha _{2}^{\prime }u^{\prime }\left(
b\right) ,  \tag{2.2}
\end{equation}
\begin{equation}
R_{c_{1}}\left( u\right) :=u\left( c_{1}^{-}\right) ,\text{ \ \ \ \ }
R_{c_{1}}^{\prime }\left( u\right) :=u^{\prime }\left( c_{1}^{-}\right)
-\delta u^{\prime }\left( c_{1}^{+}\right) ,  \tag{2.3}
\end{equation}
\begin{equation}
R_{c_{2}}\left( u\right) :=u\left( c_{2}^{-}\right) ,\text{ \ \ \ \ }
R_{c_{2}}^{\prime }\left( u\right) :=\delta u^{\prime }\left(
c_{2}^{-}\right) -\gamma u^{\prime }\left( c_{2}^{+}\right) .  \tag{2.4}
\end{equation}

For functions $f(x),$ which is defined on $I$ and has finite limit $
f(c_{1}^{\pm })=\underset{x\rightarrow c_{1}^{\pm }}{\lim }$ $f(x)$, $
f(c_{2}^{\pm })=\underset{x\rightarrow c_{2}^{\pm }}{\lim }$ $f(x),$ by $
f_{(i)}(x)$ $\left( i=\overline{1,4}\right) $ we denote the functions
\begin{equation}
\begin{array}{l}
f_{(1)}(x):=\left\{
\begin{array}{l}
f(x),\text{ \ }x\in \lbrack a,c_{1}), \\
f(c_{1}^{-}),\text{ \ }x=c_{1},
\end{array}
\right. \ \ \ f_{(2)}(x):=\left\{
\begin{array}{l}
f(x),\text{ \ }x\in \left( c_{1},c_{2}\right) , \\
f(c_{1}^{+}),\text{ \ }x=c_{1},
\end{array}
\right. \\
\\
f_{(3)}(x):=\left\{
\begin{array}{l}
f(x),\text{ \ }x\in \left( c_{1},c_{2}\right) , \\
f(c_{2}^{-}),\text{ \ }x=c_{2},
\end{array}
\right. \text{ \ }f_{(4)}(x):=\left\{
\begin{array}{l}
f(x),\text{ \ }x\in \left( c_{2},b\right] , \\
f(c_{2}^{+}),\text{ \ }x=c_{2},
\end{array}
\right.
\end{array}
\tag{2.5}
\end{equation}
which are defined on $I_{1}:=[a,c_{1}],$ $I_{2}:=\left[ c_{1},c_{2}\right] $
and $I_{3}:=[c_{2},b],$ respectively.

In this space, we define a linear operator $A$ by the domain of definition
\begin{equation}
D\left( A\right) :=\left\{ F\left( x\right) =\left(
\begin{array}{c}
f\left( x\right) \\
R_{b}^{\prime }\left( f\right) \\
R_{c_{1}}\left( f\right) \\
R_{c_{2}}\left( f\right)
\end{array}
\right) \in H\left\vert
\begin{array}{l}
f_{\left( i\right) }\left( .\right) ,\text{ }f_{\left( i\right) }^{\prime
}\left( .\right) \text{ are absolutely continuous } \\
\text{ in }I_{i}\text{ \ }\left( i=1,2,3\right) ;\text{ }\tau \left(
f\right) \in L\text{; } \\
B_{1}\left( f\right) =0,\text{ }T_{1}\left( f\right) =T_{3}\left( f\right)
=0, \\
h_{1}=R_{b}^{\prime }\left( f\right) ,h_{2}=R_{c_{1}}\left( f\right)
,h_{3}=R_{c_{2}}\left( f\right)
\end{array}
\right. \right\}  \tag{2.6}
\end{equation}
and
\begin{equation}
A\left(
\begin{array}{c}
f\left( x\right) \\
R_{b}^{\prime }\left( f\right) \\
R_{c_{1}}\left( f\right) \\
R_{c_{2}}\left( f\right)
\end{array}
\right) =\left(
\begin{array}{c}
\tau \left( f\right) \\
R_{b}\left( f\right) \\
-R_{c_{1}}^{\prime }\left( f\right) \\
-R_{c_{2}}^{\prime }\left( f\right)
\end{array}
\right) ,\text{ \ \ \ \ \ }\left(
\begin{array}{c}
f\left( x\right) \\
R_{b}^{\prime }\left( f\right) \\
R_{c_{1}}\left( f\right) \\
R_{c_{2}}\left( f\right)
\end{array}
\right) \in D\left( A\right) .\text{\ \ }  \tag{2.7}
\end{equation}
Consequently, the problem (1.9)-(1.15) can be rewritten in operator form as $
AF=\lambda F,$ i.e., the problem (1.9)-(1.15) can be considered as the
eigenvalue problem for the operator $A.$

The following theorem can be proven by the same methods in similar studies $
\left[ 21,23-25\right] .$

\begin{theorem}
\bigskip $\mathit{i)}$ $A$\textit{\ is a symmetric operator in }$H;$\textit{
\ the eigenvalue problem for the operator }$A$\textit{\ and the problem
(1.9)-(1.15) coincide. }$\mathit{ii)}$\textit{\ All eigenvalues and
eigenfunctions of the operator }$A$ \textit{(or the problem (1.9)-(1.15))
are real. }$\mathit{iii)}$\textit{\ Two eigenfunctions }$u\left( x,\lambda
\right) $\textit{\ and }$v\left( x,\mu \right) $\textit{\ corresponding to
different eigenvalues }$\lambda $\textit{\ and }$\mu $\textit{, are
orthogonal, i.e.,}
\begin{equation}
\begin{array}{l}
\dint\limits_{a}^{c_{1}}\mathit{\ }u\left( x,\lambda \right) v\left( x,\mu
\right) dx+\delta ^{2}\dint\limits_{c_{1}}^{c_{2}}u\left( x,\lambda \right)
v\left( x,\mu \right) dx+\gamma ^{2}\dint\limits_{c_{2}}^{b}u\left(
x,\lambda \right) v\left( x,\mu \right) dx+ \\
\text{\ \ \ \ \ }\dfrac{\gamma ^{2}}{\rho }R_{b}^{\prime }\left( u\right)
R_{b}^{\prime }\left( v\right) +R_{c_{1}}\left( u\right) R_{c_{1}}\left(
v\right) +\delta R_{c_{2}}\left( u\right) R_{c_{2}}\left( v\right) =0.
\end{array}
\tag{2.8}
\end{equation}
\end{theorem}

Now, we will construct a special fundamental system of solutions of the
equation (1.9). By virtue of theorem 1.5 in $[26],$ we will define two
solutions of the equation (1.9) as follows:
\begin{equation}
\phi _{\lambda }\left( x\right) =\left\{
\begin{array}{l}
\phi _{1\lambda }\left( x\right) ,\text{ \ \ }x\in \left[ a,c_{1}\right) ,
\\
\phi _{2\lambda }\left( x\right) ,\text{ \ }x\in \left( c_{1},c_{2}\right) ,
\\
\phi _{3\lambda }\left( x\right) ,\text{ \ \ }x\in \left( c_{2},b\right] ,
\end{array}
\right. ,\text{\ }\chi _{\lambda }\left( x\right) =\left\{
\begin{array}{l}
\chi _{1\lambda }\left( x\right) ,\text{ \ \ }x\in \left[ a,c_{1}\right) ,
\\
\chi _{2\lambda }\left( x\right) ,\text{ \ }x\in \left( c_{1},c_{2}\right) ,
\\
\chi _{3\lambda }\left( x\right) ,\text{ \ \ }x\in \left( c_{2},b\right] .
\end{array}
\right.  \tag{2.9}
\end{equation}

Let $\phi _{1\lambda }\left( x\right) =\phi _{1}\left( x,\lambda \right) $
be the solution of the equation (1.9) on $\left[ a,c_{1}\right] ,$ which
satisfies the initial conditions
\begin{equation}
u(a)=\beta _{2},\text{ \ \ \ \ \ \ }u^{\prime }(a)=-\beta _{1}.  \tag{2.10}
\end{equation}

By virtue of theorem 1.5 in $[26],$ after defining this solution we may
define the solution $\phi _{2\lambda }\left( x\right) =\phi _{2}\left(
x,\lambda \right) $ of the equation (1.9) on $\left[ c_{1},c_{2}\right] $ by
means of the solution $\phi _{1\lambda }\left( x\right) $ by the nonstandard
initial conditions
\begin{equation}
u(c_{1})=\frac{1}{\delta }\phi _{1\lambda }\left( c_{1}^{-}\right) ,\text{ \
\ \ \ \ }u^{\prime }(c_{1})=\frac{1}{\delta }\left\{ \phi _{1\lambda
}^{\prime }\left( c_{1}^{-}\right) +\lambda \phi _{1\lambda }\left(
c_{1}^{-}\right) \right\} .  \tag{2.11}
\end{equation}
After defining this solution, we may define the solution $\phi _{3\lambda
}\left( x\right) =\phi _{3}\left( x,\lambda \right) $ of the equation (1.9)
on $\left[ c_{2},b\right] $ by means of the solution $\phi _{2\lambda
}\left( x\right) $ by the nonstandard initial conditions
\begin{equation}
u(c_{2})=\frac{\delta }{\gamma }\phi _{2\lambda }\left( c_{2}^{-}\right) ,
\text{ \ \ \ \ \ }u^{\prime }(c_{2})=\frac{1}{\gamma }\left\{ \delta \phi
_{2\lambda }^{\prime }\left( c_{2}^{-}\right) +\lambda \phi _{2\lambda
}\left( c_{2}^{-}\right) \right\} .  \tag{2.12}
\end{equation}
Thus, $\phi _{\lambda }\left( x\right) =\phi \left( x,\lambda \right) $
satisfies the equation (1.9) on $I,$ yhe boundary condition (1.10) and the
transmission conditions (1.12)-(1.15).

Analogically, first we define the solution $\chi _{3\lambda }\left( x\right)
=\chi _{3}\left( x,\lambda \right) $ of the equation (1.9) on $\left[ c_{2},b
\right] $ by the initial conditions
\begin{equation}
u\left( b\right) =\lambda \alpha _{2}^{\prime }-\alpha _{2},\text{ \ \ \ \ \
\ \ }u^{\prime }\left( b\right) =\lambda \alpha _{1}^{\prime }-\alpha _{1}.
\tag{2.13}
\end{equation}
Again, after defining this solution, we define the solution $\chi _{2\lambda
}\left( x\right) =\chi _{2}\left( x,\lambda \right) $ of the equation (1.9)
on $\left[ c_{1},c_{2}\right] $ by the initial conditions
\begin{equation}
u(c_{2})=\frac{\gamma }{\delta }\chi _{3\lambda }\left( c_{2}^{+}\right) ,
\text{ \ \ \ \ \ }u^{\prime }(c_{2})=\frac{\gamma }{\delta }\left\{ \chi
_{3\lambda }^{\prime }\left( c_{2}^{+}\right) -\frac{\lambda }{\delta }\chi
_{3\lambda }\left( c_{2}^{+}\right) \right\} .  \tag{2.14}
\end{equation}
After defining this solution, we define the solution $\chi _{1\lambda
}\left( x\right) =\chi _{1}\left( x,\lambda \right) $ of the equation (1.9)
on $\left[ a,c_{1}\right] $ by the initial conditions
\begin{equation}
u(c_{1})=\delta \chi _{2\lambda }\left( c_{1}^{+}\right) ,\text{ \ \ \ \ \ }
u^{\prime }(c_{1})=\delta \left\{ \chi _{2\lambda }^{\prime }\left(
c_{1}^{+}\right) -\lambda \chi _{2\lambda }\left( c_{1}^{+}\right) \right\} .
\tag{2.15}
\end{equation}
Thus, $\chi _{\lambda }\left( x\right) =\chi \left( x,\lambda \right) $
satisfies the equation (1.9) on $I,$ the boundary condition (1.11) and the
transmission conditions (1.12)-(1.15).

Since the Wronskian $\ W\left( \phi _{i\lambda },\chi _{i\lambda };x\right) $
are independent on variable $x\in I_{i}$ $\ \left( i=1,2,3\right) $ and $
\phi _{i\lambda }\left( x\right) $ and $\chi _{i\lambda }\left( x\right) $
are the entire functions of the parameter $\lambda $ for each $x\in I_{i},$
then the functions
\begin{equation}
\omega _{i}\left( \lambda \right) :=W\left( \phi _{i\lambda },\chi
_{i\lambda };x\right) =\phi _{i\lambda }\left( x\right) \chi _{i\lambda
}^{\prime }\left( x\right) -\phi _{i\lambda }^{\prime }\left( x\right) \chi
_{i\lambda }\left( x\right) ,\text{ \ }\left( i=1,2,3\right)  \tag{2.16}
\end{equation}
are the entire functions of parameter $\lambda .$ Taking into account
(2.11), (2.12), (2.14) and (2.15), after a short calculation, we get
\begin{equation*}
\omega _{1}\left( \lambda \right) =\delta ^{2}\omega _{2}\left( \lambda
\right) =\gamma ^{2}\omega _{3}\left( \lambda \right)
\end{equation*}
for each $\lambda \in

\mathbb{C}

.$

\begin{corollary}
The zeros of the functions $\omega _{1}\left( \lambda \right) ,$ $\omega
_{2}\left( \lambda \right) $ and $\omega _{3}\left( \lambda \right) $
coincide.
\end{corollary}

Then,\ we may introduce to the consideration the characteristic function $
\omega \left( \lambda \right) $ as
\begin{equation}
\omega \left( \lambda \right) :=\omega _{1}\left( \lambda \right) =\delta
^{2}\omega _{2}\left( \lambda \right) =\gamma ^{2}\omega _{3}\left( \lambda
\right) .  \tag{2.17}
\end{equation}

\begin{theorem}
The eigenvalues of the problem (1.9)-(1.15) are coincided zeros of the
function $\omega \left( \lambda \right) .$
\end{theorem}

\begin{proof}
...
\end{proof}

\begin{lemma}
All eigenvalues $\lambda _{n}$ of the problem (1.9)-(1.15) are simple zeros
of $\omega \left( \lambda \right) .$
\end{lemma}

\begin{proof}
...
\end{proof}

If $\lambda _{n}$ $\left( n=0,1,2,...\right) $ denote the zeros of $\omega
\left( \lambda \right) ,$ then
\begin{equation}
\Phi _{\lambda _{n}}\left( x\right) :=\left(
\begin{array}{c}
\phi _{\lambda _{n}}\left( x\right) \\
R_{b}^{\prime }\left( \phi _{\lambda _{n}}\right) \\
R_{c_{1}}\left( \phi _{\lambda _{n}}\right) \\
R_{c_{2}}\left( \phi _{\lambda _{n}}\right)
\end{array}
\right)  \tag{2.26}
\end{equation}
are the corresponding eigenvectors of the operator $A,$ satisfying the
orthogonality relation
\begin{equation}
\left\langle \Phi _{\lambda _{n}}\left( .\right) ,\Phi _{\lambda _{m}}\left(
.\right) \right\rangle _{H}=0\text{ \ for \ \ }n\neq m.  \tag{2.27}
\end{equation}

Here $\left\{ \phi _{\lambda _{n}}\left( .\right) \right\} _{n=0}^{\infty }$
is a sequence of eigenfunctions of the problem (1.9)-(1.15) corresponding to
the eigenvectors of $A,$ i.e.,
\begin{equation}
\Psi _{\lambda _{n}}\left( x\right) :=\frac{\Phi _{\lambda _{n}}\left(
x\right) }{\left\Vert \Phi _{\lambda _{n}}\left( x\right) \right\Vert _{H}}
=\left(
\begin{array}{c}
\Psi _{\lambda _{n}}\left( x\right) \\
R_{b}^{\prime }\left( \Psi _{\lambda _{n}}\right) \\
R_{c_{1}}\left( \Psi _{\lambda _{n}}\right) \\
R_{c_{2}}\left( \Psi _{\lambda _{n}}\right)
\end{array}
\right) .  \tag{2.28}
\end{equation}

Therefore $\left\{ \chi _{\lambda _{n}}\left( .\right) \right\}
_{n=0}^{\infty }$ is another set of eigenfunctions which is related by $
\left\{ \phi _{\lambda _{n}}\left( .\right) \right\} _{n=0}^{\infty }$ with
\begin{equation}
\chi _{\lambda _{n}}\left( x\right) =k_{n}\phi _{\lambda _{n}}\left(
x\right) ,\text{ \ }x\in I,\text{ }n\in

\mathbb{Z}

\tag{2.29}
\end{equation}
where $k_{n}\neq 0$ are non-zero constants, since all eigenvalues are simple.

\section{Asymptotic Formulas for Eigenvalues and Eigenfunctions}

The asymptotics formulas for eigenvalues and eigenfunctions can be derived
similar to the classical techniques of $\left[ 26\right] ,$ see also $\left[
23-25\right] .$ We state the results briefly.

Let $\phi _{\lambda }\left( .\right) $ be the solutions of the equation
(1.9) defined in section 2, and let $\lambda =s^{2}.$ Then the following
integral equations hold for $k=0$ and $k=1:$
\begin{equation}
\begin{array}{l}
\dfrac{d^{k}}{dx^{k}}\phi _{1\lambda }\left( x\right) =\beta _{2}\dfrac{d^{k}
}{dx^{k}}\left( \cos s\left( x-a\right) \right) -\dfrac{\beta _{1}}{s}\dfrac{
d^{k}}{dx^{k}}(\sin s\left( x-a\right) )+ \\
\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\dfrac{1}{s}\underset{a}{
\overset{x}{\int }}\dfrac{d^{k}}{dx^{k}}\left( \sin s\left( x-y\right)
\right) q\left( y\right) \phi _{1\lambda }\left( y\right) dy,
\end{array}
\tag{3.1}
\end{equation}
\begin{equation}
\begin{array}{l}
\dfrac{d^{k}}{dx^{k}}\phi _{2\lambda }\left( x\right) =\dfrac{1}{\delta }
\phi _{1\lambda }\left( c_{1}^{-}\right) \dfrac{d^{k}}{dx^{k}}\left( \cos
s\left( x-c_{1}\right) \right) +\dfrac{1}{s\delta }\left\{ \phi _{1\lambda
}^{\prime }\left( c_{1}^{-}\right) +s^{2}\phi _{1\lambda }\left(
c_{1}^{-}\right) \right\} \times \\
\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\dfrac{d^{k}}{dx^{k}}\left( \sin
s\left( x-c_{1}\right) \right) +\dfrac{1}{s}\underset{c_{1}}{\overset{x}{
\int }}\dfrac{d^{k}}{dx^{k}}\left( \sin s\left( x-y\right) \right) q\left(
y\right) \phi _{2\lambda }\left( y\right) dy,
\end{array}
\tag{3.2}
\end{equation}
\begin{equation}
\begin{array}{l}
\dfrac{d^{k}}{dx^{k}}\phi _{3\lambda }\left( x\right) =\dfrac{\delta }{
\gamma }\phi _{2\lambda }\left( c_{2}^{-}\right) \dfrac{d^{k}}{dx^{k}}\left(
\cos s\left( x-c_{2}\right) \right) +\dfrac{1}{s\gamma }\left\{ \delta \phi
_{2\lambda }^{\prime }\left( c_{2}^{-}\right) +s^{2}\phi _{2\lambda }\left(
c_{2}^{-}\right) \right\} \times \\
\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\dfrac{d^{k}}{dx^{k}}\left( \sin
s\left( x-c_{2}\right) \right) +\dfrac{1}{s}\underset{c_{2}}{\overset{x}{
\int }}\dfrac{d^{k}}{dx^{k}}\left( \sin s\left( x-y\right) \right) q\left(
y\right) \phi _{3\lambda }\left( y\right) dy.
\end{array}
\tag{3.3}
\end{equation}

Let $\lambda =s^{2}$ and $\left\vert \func{Im}s\right\vert =t.$ Then the
functions $\phi _{i\lambda }\left( x\right) $ have the following asymptotic
representations for $\left\vert \lambda \right\vert \rightarrow \infty ,$
which hold uniformly for $x\in I_{i}$ $\left( i=1,2,3\right) :$
\begin{equation}
\dfrac{d^{k}}{dx^{k}}\phi _{1\lambda }\left( x\right) =\beta _{2}\dfrac{d^{k}
}{dx^{k}}\left( \cos s\left( x-a\right) \right) +O\left( \frac{1}{s}
e^{t\left( x-a\right) }\right) ,  \tag{3.4}
\end{equation}
\begin{equation}
\dfrac{d^{k}}{dx^{k}}\phi _{2\lambda }\left( x\right) =\dfrac{s\beta _{2}}{
\delta }\cos \left( s\left( c_{1}-a\right) \right) \dfrac{d^{k}}{dx^{k}}
\left( \sin s\left( x-c_{1}\right) \right) +O\left( e^{t\left( x-a\right)
}\right) ,  \tag{3.5}
\end{equation}
\begin{equation}
\dfrac{d^{k}}{dx^{k}}\phi _{3\lambda }\left( x\right) =\dfrac{s^{2}\beta _{2}
}{\delta \gamma }\cos \left( s\left( c_{1}-a\right) \right) \sin \left(
s\left( c_{2}-c_{1}\right) \right) \dfrac{d^{k}}{dx^{k}}\left( \sin s\left(
x-c_{2}\right) \right) +O\left( se^{t\left( x-a\right) }\right) ,  \tag{3.6}
\end{equation}

By substituting (3.6) into the representation
\begin{equation}
\omega \left( \lambda \right) =\gamma ^{2}\left\{ \left( s^{2}\alpha
_{1}^{\prime }-\alpha _{1}\right) \phi _{3\lambda }\left( b\right) -\left(
s^{2}\alpha _{2}^{\prime }-\alpha _{2}\right) \phi _{3\lambda }^{\prime
}\left( b\right) \right\} ,  \tag{3.7}
\end{equation}
then the characteristic function $\omega \left( \lambda \right) $ has the
following asymptotic representation:
\begin{equation}
\omega \left( \lambda \right) =\frac{s^{5}\gamma \beta _{2}\alpha
_{2}^{\prime }}{\delta }\cos \left( s\left( c_{1}-a\right) \right) \sin
\left( s\left( c_{2}-c_{1}\right) \right) \cos \left( s\left( b-c_{2}\right)
\right) +O\left( s^{4}e^{t\left( b-a\right) }\right) .  \tag{3.8}
\end{equation}

\begin{corollary}
The eigenvalues of the problem (1.9)-(1.15) are bounded below.
\end{corollary}

Now we can obtain the asymptotic approximation formula for the eigenvalues
of the problem (1.9)-(1.15). Since the eigenvalues coincide with the zeros
of the entire function $\omega \left( \lambda \right) ,$ it follows that
they have no finite limit. Moreover, we know that all eigenvalues are real
and bounded below. Therefore, we may renumber them as $\lambda _{0}\leq
\lambda _{1}\leq ...,$ listed according to their multiplicity.

\begin{theorem}
The eigenvalues $\lambda _{n}=s_{n}^{2},\left( n=0,1,...\right) $ of the
problem (1.9)-(1.15) have the following asymptotic representation for $
n\rightarrow \infty :$
\begin{equation}
\widetilde{s}_{n}=\frac{\left( n+1/2\right) \pi }{\left( c_{1}-a\right) }
+O\left( \frac{1}{n}\right) ,  \tag{3.9}
\end{equation}
\begin{equation}
\widetilde{\widetilde{s}}_{n}=\frac{n\pi }{\left( c_{2}-c_{1}\right) }
+O\left( \frac{1}{n}\right) ,  \tag{3.10}
\end{equation}
\begin{equation}
\widetilde{\widetilde{\widetilde{s}}}_{n}=\frac{\left( n+1/2\right) \pi }{
\left( b-c_{2}\right) }+O\left( \frac{1}{n}\right) .  \tag{3.11}
\end{equation}
\end{theorem}

\begin{proof}
...
\end{proof}

Then from (3.4)-(3.6) (for $k=0$) and the above theorem, the asymptotic
behaviour of the eigenfunctions:
\begin{equation}
\phi _{\lambda _{n}}\left( x\right) =\left\{
\begin{array}{l}
\phi _{1\lambda _{n}}\left( x\right) ,\text{ \ \ }x\in \left[ a,c_{1}\right)
, \\
\phi _{2\lambda _{n}}\left( x\right) ,\text{ \ }x\in \left(
c_{1},c_{2}\right) , \\
\phi _{3\lambda _{n}}\left( x\right) ,\text{ \ \ }x\in \left( c_{2},b\right]
,
\end{array}
\right.  \tag{3.13}
\end{equation}
of the problem (1.9)-(1.15) is given by
\begin{equation*}
\phi _{\widetilde{\lambda }_{n}}\left( x\right) =\left\{
\begin{array}{l}
\beta _{2}\cos \left( \dfrac{\left( n+1/2\right) \pi }{\left( c_{1}-a\right)
}\left( x-a\right) \right) +O\left( \dfrac{1}{n}\right) ,\text{ \ \ }x\in
\left[ a,c_{1}\right) , \\
O\left( 1\right) ,\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x\in \left( c_{1},c_{2}\right) , \\
O\left( 1\right) ,\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x\in \left( c_{2},b\right] ,
\end{array}
\right.
\end{equation*}
\begin{equation*}
\phi _{\widetilde{\widetilde{\lambda }}_{n}}\left( x\right) =\left\{
\begin{array}{l}
\beta _{2}\cos \left( \dfrac{n\pi }{\left( c_{2}-c_{1}\right) }\left(
x-a\right) \right) +O\left( \dfrac{1}{n}\right) ,\text{ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x\in \left[ a,c_{1}\right) , \\
\dfrac{n\pi }{\left( c_{2}-c_{1}\right) }\dfrac{\beta _{2}}{\delta }\cos
\left( \dfrac{n\pi \left( c_{1}-a\right) }{\left( c_{2}-c_{1}\right) }
\right) \sin \left( \dfrac{n\pi }{\left( c_{2}-c_{1}\right) }\left(
x-c_{1}\right) \right) +O\left( 1\right) ,\text{\ }x\in \left(
c_{1},c_{2}\right) , \\
O\left( 1\right) ,\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ }x\in \left( c_{2},b\right] ,
\end{array}
\right.
\end{equation*}
\begin{equation*}
\phi _{\widetilde{\widetilde{\widetilde{\lambda }}}_{n}}\left( x\right)
=\left\{
\begin{array}{l}
\beta _{2}\cos \left( \dfrac{\left( n+1/2\right) \pi }{\left( b-c_{2}\right)
}\left( x-a\right) \right) +O\left( \dfrac{1}{n}\right) ,\text{ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x\in \left[ a,c_{1}\right) , \\
\dfrac{\left( n+1/2\right) \pi }{\left( b-c_{2}\right) }\dfrac{\beta _{2}}{
\delta }\cos \left( \dfrac{\left( n+1/2\right) \pi \left( c_{1}-a\right) }{
\left( b-c_{2}\right) }\right) \sin \left( \dfrac{\left( n+1/2\right) \pi
\left( c_{2}-c_{1}\right) }{\left( b-c_{2}\right) }\right) \times \\
\sin \left( \dfrac{\left( n+1/2\right) \pi }{\left( b-c_{2}\right) }\left(
x-c_{2}\right) \right) +O\left( 1\right) ,\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x\in \left( c_{1},c_{2}\right) , \\
O\left( 1\right) ,\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ }x\in \left( c_{2},b\right] ,
\end{array}
\right.
\end{equation*}
all these asymptotic formulas hold uniformly for $x.$

\section{The Sampling Theorem}

Now we can derive the sampling theorem associated with the problem
(1.9)-(1.15).

\begin{theorem}
Let
\begin{equation}
\phi _{\lambda }\left( x\right) =\left\{
\begin{array}{l}
\phi _{1\lambda }\left( x\right) ,\text{ \ \ }x\in \left[ a,c_{1}\right) ,
\\
\phi _{2\lambda }\left( x\right) ,\text{ \ }x\in \left( c_{1},c_{2}\right) ,
\\
\phi _{3\lambda }\left( x\right) ,\text{ \ \ }x\in \left( c_{2},b\right] ,
\end{array}
\right.  \tag{4.1}
\end{equation}
be the solution of the equation (1.9) together with the conditions
(2.10)-(2.12). Let $g\left( .\right) \in L^{2}\left( a,b\right) $ and
\begin{equation}
\digamma \left( \lambda \right) =\dint\limits_{a}^{c_{1}}\mathit{\ g}\left(
x\right) \phi _{1\lambda }\left( x\right) dx+\delta
^{2}\dint\limits_{c_{1}}^{c_{2}}\mathit{g}\left( x\right) \phi _{2\lambda
}\left( x\right) dx+\gamma ^{2}\dint\limits_{c_{2}}^{b}\mathit{g}\left(
x\right) \phi _{3\lambda }\left( x\right) dx.  \tag{4.2}
\end{equation}
Then $\digamma \left( \lambda \right) $ is an entire function of exponential
type that can be reconstructed from its values at the points $\left\{
\lambda _{n}\right\} _{n=0}^{\infty }$ via the sampling formula
\begin{equation}
\digamma \left( \lambda \right) =\sum_{n=0}^{\infty }\digamma \left( \lambda
_{n}\right) \frac{\omega \left( \lambda \right) }{\left( \lambda -\lambda
_{n}\right) \omega ^{\prime }\left( \lambda _{n}\right) }.  \tag{4.3}
\end{equation}
The series (4.3) converges absolutely on $

\mathbb{C}

$ and uniformly on compact subset of $

\mathbb{C}

.$ Here $\omega \left( \lambda \right) $ is the entire function defined in
(2.17).
\end{theorem}

\begin{proof}
...
\end{proof}

\begin{remark}
To see that expansion (4.3) is a Lagrange type interpolation, we may replace
$\omega \left( \lambda \right) $ by the canonical product
\begin{equation}
\varpi \left( \lambda \right) =\left\{
\begin{array}{c}
\overset{\infty }{\underset{n=0}{\Pi }}\left( 1-\dfrac{\lambda }{\lambda _{n}
}\right) ,\text{ \ if zero is not an eigenvalue,} \\
\\
\lambda \overset{\infty }{\underset{n=1}{\Pi }}\left( 1-\dfrac{\lambda }{
\lambda _{n}}\right) ,\text{ \ if }\lambda _{0}=0\text{ is an eigenvalue.}
\end{array}
\right.  \tag{4.26}
\end{equation}
\end{remark}

From Hadamard's factorization theorem, see $\left[ 6\right] ,\omega \left(
\lambda \right) =h\left( \lambda \right) \varpi \left( \lambda \right) ,$
where $h\left( \lambda \right) $ is an entire function with no zeros. Thus,
\begin{equation}
\frac{\omega \left( \lambda \right) }{\omega ^{\prime }\left( \lambda
_{n}\right) }=\frac{h\left( \lambda \right) \varpi \left( \lambda \right) }{
h\left( \lambda _{n}\right) \varpi ^{\prime }\left( \lambda _{n}\right) }
\tag{4.27}
\end{equation}
and (4.2), (4.3) remain valid for the function $\digamma \left( \lambda
\right) /h\left( \lambda \right) .$ Hence
\begin{equation}
\digamma \left( \lambda \right) =\overset{\infty }{\underset{n=0}{\sum }}
\digamma \left( \lambda _{n}\right) \frac{h\left( \lambda \right) \varpi
\left( \lambda \right) }{\left( \lambda -\lambda _{n}\right) h\left( \lambda
_{n}\right) \varpi ^{\prime }\left( \lambda _{n}\right) }.  \tag{4.28}
\end{equation}
We may redefine (4.2) by taking kernel $\phi _{\lambda }\left( .\right)
/h\left( \lambda \right) =\overline{\phi }_{\lambda }\left( .\right) $ to get
\begin{equation}
\overline{\digamma }\left( \lambda \right) =\frac{\digamma \left( \lambda
\right) }{h\left( \lambda \right) }=\overset{\infty }{\underset{n=0}{\sum }}
\overline{\digamma }\left( \lambda _{n}\right) \frac{\varpi \left( \lambda
\right) }{\left( \lambda -\lambda _{n}\right) \varpi ^{\prime }\left(
\lambda _{n}\right) }.  \tag{4.29}
\end{equation}

config: arxiv


TITLE: Solving 5th order equation
QUESTION [0 upvotes]: How do you solve $x^5-3x^4+3x^3-x^2=0$?
I want to solve this equation in the shortest and quickest way possible. How can I go about doing so?

REPLY [1 votes]: .\begin{split}
x^5-3x^4+3x^3-x^2 & = x^2(x^3-3x^2+3x-1)
\end{split}
Once you are here, either  you can identify it if you know the cube formulas for $a+b/a-b$, or use the rational root theorem , followed by further division and factorization. This will tell you that:
$$
x^3-3x^2+3x-1 = (x-1)^3
$$
Hence, the answer is $x^2(x-1)^3$. Equating this to $0$ would give you two roots, $0$ (of multiplicity $2$) and $1$ (of multiplicity $3$).


TITLE: Proof that f = {(x,n) e R x Z | n =< x < n + 1} is a function.
QUESTION [1 upvotes]: I have to prove that f = {(x,n) e R x Z | n =< x < n + 1} a function is.
I started with saying (x,n1) and (x,n2) are both elements of the function. This means n1 =< x < n1 + 1 (1) and n2 =< x < n2 + 1 (2). I substracted n1 from (1) and n2 from (2) this gives 0 =< x - n1 < 1 (3) and 0 =< x - n2 < 1 (4) when I substract (3) from (4), this gives 0 =< n1 - n2 < 0 (5). This is impossible since a number can't be both greater than or equal to 0 and smaller than 0. I wanted to work towards n1 - n2 = 0. Can someone help me find the mistake I made?
Thanks!

REPLY [2 votes]: We cannot subtract inequalities!
Consider for example $1<2,\ 0<2$, if you subtract them, you would receive $1<0$ which is false.
It's not working because addition of inequalities does work, and subtraction is $a-b=a+(-b)$, while multiplication by $-1$ reverses the inequality to be subtracted.
If we carefully add appropriate forms of the inequalities instead:
$$0\le (x-n_1)<1\\ -1<(n_2-x)\le 0$$
then we do get $-1<(n_2-n_1)<1$, implying $n_2-n_1=0$ as these are integers.


TITLE: A line in a proof regarding nth power residues
QUESTION [4 upvotes]: I would appreciate help understanding this highlighted line in a proof in Ireland & Rosen (p. 45). I don't know much group theory although I know the residue classes $\pmod m$ form a multiplicative group whose order is $\phi(m)$.
Prop 4.2.1: If $m\in \mathbb{Z}^{+}$ and $(a,m)=1$ then $a$ is an $n$th power residue iff $a^{\phi(m)/d} \equiv 1\pmod m$ where $d= (n,\phi(m))$:
In the proof it says: If $g$ is a primitive root $\pmod m$ and $a=g^b$ and $x=g^y$ then $x^n\equiv a\pmod m$ is equivalent to $g^{ny} \equiv g^b\pmod m$ (so far so good) which is equivalent to:

If $g^{ny}\equiv g^b \pmod m$ then $ny\equiv b\pmod {\phi(m)}$

Thanks
EDIT Due to my carelessness, I left out a crucial part of the statement of the Proposition. It should be "if $m$ possesses primitive roots, etc" I left my misstatement above as it stands since it resulted in the illuminating answer from @Daniel Fischer below.

REPLY [3 votes]: As written, the condition is only correct when there exists a primitive root modulo $m$, i.e. when the group of units of $\mathbb{Z}/(m)$ is cyclic. In that case, the highlighted line holds because we have $g^{ny} \equiv g^b \pmod{m} \iff g^{ny-b} \equiv 1 \pmod{m}$, and $g^k \equiv 1 \pmod{m}$ if and only if $k$ is a multiple of the order of $g$. By definition, a primitive root modulo $m$ has order $\phi(m)$, so $g^k \equiv 1 \pmod{m} \iff \phi(m) \mid k$, and $\phi(m) \mid k$ is the definition of $k \equiv 0 \pmod{\phi(m)}$. So
\begin{align}
g^{ny} \equiv g^b \pmod{m} &\iff g^{ny-b} \equiv 1 \pmod{m}\\
&\iff ny -b \equiv 0 \pmod{\phi(m)}\\
&\iff ny \equiv b \pmod{\phi(m)}.
\end{align}
As an example that the condition as written does not hold for all $m$, take $m = 15$ and $n = 4$. Then $\phi(m) = 8$, and $d = (n,\phi(m)) = 4$, so the condition is $a^{8/4} = a^2 \equiv 1 \pmod{15}$, but this congruence is satisfied by $1,4,11,14 \pmod{15}$, while $b^4 \equiv 1 \pmod{15}$ for all $b$ coprime to $15$, so the only biquadratic residue modulo $15$ is $1$.
For general $m$, we must replace Euler's totient function $\phi$ with the Carmichael function $\lambda$. For $a$ coprime to $m$, and $n \in \mathbb{N}\setminus \{0\}$, $a$ is an $n^{\text{th}}$-power residue if and only if $a^{\lambda(m)/d} \equiv 1 \pmod{m}$, where $d = \gcd(n,\lambda(m))$. The proof is similar to the proof when the group of units of $\mathbb{Z}/(m)$ is cyclic, one writes the unit group as a product of cyclic groups and argues in each factor in the given way.


TITLE: Measuring Variance of 2D Data Points
QUESTION [1 upvotes]: I have a list of weighted data points on a 2D plane in the form $(x, y)$.  I believe the mean can calculated as $\left(\frac{\sum{x_i\times w_i}}{\sum w_i}, \frac{\sum{y_i\times w_i}}{\sum w_i}\right)$.
What is the best way to calculate a single value which will accurately describe the spread of the data points?

REPLY [1 votes]: You have computed the weighted mean $$\mathbf{\bar x}_* = \sum w_i\mathbf{x}_i$$ (assume $\sum w_i = 1$ for ease of notation). Then why not use weighted variance
$$\sum w_i|\mathbf{x}_i -\mathbf{\bar x}_*|^2$$
or, for an unbiased version, use the unbiased weighted variance
$$\frac{1}{1-\sum w_i^2}\sum w_i|\mathbf{x}_i -\mathbf{\bar x}_*|^2$$
Here, $|\cdot|$ denotes the ordinary Euclidean norm $|\mathbf u| = \sqrt{u_1^2+u_2^2}$, so $|\mathbf{ u }|^2 = u_1^2+u_2^2$.


TITLE: Can A function grow quicker than another but never catch up to it
QUESTION [6 upvotes]: I know this is a bit of a strange question, but it is one that has been on my mind for a little bit now. If we have two real valued functions f(x) and g(x) and they are both everywhere continuous and differentiable, could we choose f(x) and g(x) such that three conditions are satisfied

f(x) > g(x) for all x in R
g'(x) > f'(x) for all x in R
f(x) and g(x) both diverge as x tends to infinity

Essentially, I'm asking if g(x) can grow infinitely but never catch up to f(x)? I know this problem would be simple if f(x) and g(x) converged, but I am curious about if these criterion could be met while maintaining divergence. My initial thought is that no such functions exist, but I am not sure how I could prove that.  If such functions did exist, I suspect g(x) would need to be ln(x) or something of the like, but I am not convinced it is possible for these criterion to all be met. If anyone has insight either way, that would be extremely helpful.

REPLY [8 votes]: Hint: Look at the difference $h:= f-g$. Can you quickly think of a function $h$ which is always positive but whose derivative is always negative?
Now, to make the the third condition true, it seems one only has to add another function $g$ which goes to $\infty$ for $x \to \infty$.
Then with $f = g+h$ you're done.


text:
\begin{document}
\title[Simple Overrings]{Integral Domains whose Simple Overrings are
Intersections of Localizations}

\author{Marco Fontana}
\address{Dipartimento di Matematica\\ Universit\`{a} degli Studi Roma Tre\\
    Largo San L. Murialdo, 1\\
    00146 Roma, Italy}
\email{fontana@mat.uniroma3.it}
\author{Evan Houston}
\address[Evan Houston and Thomas Lucas]{Department of Mathematics\\
    University of North Carolina at Charlotte\\
    Charlotte, NC 28223 U.S.A.}
\email[Evan Houston]{eghousto@email.uncc.edu}
\author{Thomas Lucas}
\email[Thomas Lucas]{tglucas@email.uncc.edu} \subjclass{13A15,
13F05} \keywords{Pr\"ufer domain}
\thanks{All three authors acknowledge support from the Cultural
Co-operation Agreement between Universit\`{a} degli Studi Roma Tre
and The University of North Carolina at Charlotte; the first-named
author was also partially supported by research grant MIUR
2001/2002 (Cofin 2000-MM01192794).}

\date{}
\begin{abstract} Call a domain $R$ an sQQR-domain if each
simple overring of $R$, i.e., each ring of the form $R[u]$ with
$u$ in the quotient field of $R$, is an intersection of
localizations of $R$. We characterize Pr\"ufer domains as
integrally closed sQQR-domains. In the presence of certain
finiteness conditions, we show that the sQQR-property is very
strong; for instance, a Mori sQQR-domain must be a Dedekind
domain.  We also show how to construct sQQR-domains which have
(non-simple) overrings which are not intersections of
localizations.
\end{abstract}
\maketitle

\section*{Introduction} \label{s:intro}

Throughout this work, $R$ will denote an integral domain with
quotient field $K$.  The Kaplansky transform of an ideal $I$ of
$R$ is denoted by $\vo(I)$ and is defined by $$\vo(I)=\{x \in K
\mid \text{for each } a \in I \text{ there is an integer } n(a)
\ge 1 \text{ such that }a^{n(a)}x \in R\}.$$ (The notation
$\vo_R(I)$ will be used when the context involves more than one
ring.)  In \cite{fh} the first- and second-named authors studied
domains $R$ each of whose overrings is a Kaplansky transform of an
ideal $I$ of $R$. This work is in part a sequel to that paper.  It
turns out, however, that our investigations depend more heavily on
the notions of unique minimal overrings and QQR-domains developed
by Gilmer and Heinzer in \cite{gh}.  Recall that a domain $R$ is a
\emph{QQR-domain} if each overring of $R$ is an intersection of
localizations of $R$. Davis \cite{da} showed that a Pr\"ufer
domain must have the QQR-property and asked whether the converse
is true. In their paper, Gilmer and Heinzer showed that the
converse does not hold, and they explored in depth the relation
between the Pr\"ufer and QQR-properties.

We review basic definitions and prove basic facts in
Section~\ref{s:basics}. In particular, we dub the domains of the
title \emph{sQQR-domains}, and we characterize Pr\"ufer domains as
being integrally closed sQQR-domains. The remainder of the paper
is devoted to studying the non-integrally closed case. Since
QQR-domains were characterized in \cite{gh}, we are interested in
sQQR-domains which are not QQR-domains.  Much of
Section~\ref{s:results} is concerned with showing that, in the
presence of many frequently studied finiteness conditions, there
are no such examples.  For instance, we show that a Mori
sQQR-domain is a Dedekind domain, that an sQQR-domain with
Noetherian spectrum is a Pr\"ufer domain, and that a semilocal
sQQR-domain with treed spectrum is a QQR-domain. In
Section~\ref{s:examples} we do provide many (non-local) examples
of sQQR-domains which are not QQR-domains. Finally, in the last
section, we briefly discuss connections with seminormality.

\section{Definitions and basic facts} \label{s:basics}

We begin by collecting some of the results we shall need. Recall
that for an ideal $I$ of $R$, the $v$-closure of $I$ is defined by
$I_v=(I^{-1})^{-1}$ and the $t$-closure by $I_t=\bigcup J_v$,
where the union is taken over all finitely generated subideals $J$
of $I$.  We assume familiarity with these star operations.

\begin{proposition} \label{p:basics} Let $I$ be an ideal of $R$.
Then:
\begin{itemize}
\item[(1)] $\vo(I)=\bigcap \{R_P \mid P \in \textup{Spec}(R), I
\nsubseteq P\}$.
\item[(2)] If $I$ is finitely generated and $S$ is a
multiplicatively closed subset of $R$, then $\vo_R(I)R_S =
\vo_{R_S}(IR_S)$.
\item[(3)] $\vo(I)=\vo(\rad I)$.
\item[(4)] If $I$ has the property that for each $P \in
\textup{Spec}(R)$ with $I \sub P$, we have $\vo(I) \nsubseteq
R_P$, then $\rad I$ is largest ideal $J$ for which
$\vo(I)=\vo(J)$.
\item[(5)] $\vo(I)=\vo(I_t)$.
\end{itemize}
\end{proposition}
\begin{proof} Statement (1) is proved in \cite{h}.  For statement
(2), one sees easily that the Kaplansky transform coincides with
the familiar Nagata transform for finitely generated ideals, and
the Nagata transform is known to localize well.  That
$\vo(I)=\vo(\rad I)$ is \cite[Lemma 3.1(c)]{f}.  For (4) suppose
that $I \sub P$ and that $\vo(J) =\vo(I) \nsubseteq R_P$.  By (1),
$J \sub P$.  Hence $J \sub \rad I$.  Finally, (5) appears in
\cite[Proposition 3.4]{f}.
\end{proof}

\begin{notation} \label{n:colon} Let $R$ be a domain with quotient field $K$.  If $U$
is a subset of $K$, we shall write $(R:U)$ for the fractional
ideal $\{z \in K \mid zU \sub R\}$ and $(R:_R U)$ for the ideal
$\{r \in R \mid rU \sub R\}$.
\end{notation}

\begin{lemma} \label{l:finiteext}
    Let $R$ be a domain, and let $U$ be a subset of $K$.  Then
\begin{itemize}
\item[(1)] $R[U] \subseteq \varOmega(R:_R U)$.
\item[(2)] If $U$ is finite, then $R[U]$ is the Kaplansky
transform of an ideal of $R$ $\lra$ $R[U]=\vo(R:_R U)$.
\end{itemize}
\end{lemma}
\begin{proof} Statement (1) follows easily from the definitions (with the convention
that $\vo(0)=K$).  Suppose that $U$ is finite and that
$R[U]=\vo(I)=\bigcap_{I \nsubseteq P} R_P$ for some ideal $I$ of
$R$. Let $P$ be a prime ideal of $R$ with $I \nsubseteq P$. Then
$U \subseteq R_{P}$, whence $(R:_R U) \nsubseteq P$.  Thus
$\varOmega(R:_R U) = \bigcap_{(R:_R U) \nsubseteq Q} R_{Q}
\subseteq R_{P}$. Hence $\varOmega(R:_R U) \subseteq \bigcap_{I
\nsubseteq P} R_{P} = R[U]$.  This proves (2).
\end{proof}

\begin{lemma} \label{l:same} Let $R$ be a domain with quotient field
$K$.  Then a finitely generated overring of $R$ is a Kaplansky
transform of an ideal of $R$ if and only if it is an intersection
of localizations of $R$.
\end{lemma}
\begin{proof} Let $U$ be a finite subset of $K$, and suppose that
there is a set of primes $\mc P$ in $R$ with $R[U]=\bigcap_{P \in
\mc P} R_P$.  For each $P \in \mc P$, we then have $(R:_R U)
\nsubseteq P$.  Hence $$R[U]=\bigcap_{P \in \mc P} R_P \supseteq
\bigcap_{(R:_R U) \nsubseteq Q} R_Q = \vo(R:_R U) \supseteq
R[U],$$ the last inclusion following from Lemma~\ref{l:finiteext}.
Thus $R[U] = \vo(R:_R U)$. The converse holds in general (without
the finiteness assumption) by Proposition~\ref{p:basics}(1).
\end{proof}

\begin{definition} \label{d:sqqr}  We say that a domain $R$ is an
\emph{sQQR-domain} (respectively, \emph{fQQR-domain}) if each
simple overring (respectively, each finitely generated overring)
of $R$ is an intersection of localizations of $R$.
\end{definition}

Recall from \cite{gh} that a domain is a \emph{QQR-domain} if each
overring of $R$ is an intersection of localizations of $R$.  In
\cite{fh} the authors defined an \emph{$\vo$-domain} to be one for
which each overring is a Kaplansky transform; hence a $\vo$-domain
is a QQR-domain. It was shown in \cite{fh} that the two properties
are not the same. However, if one restricts to finitely generated
overrings, then the properties are the same, as is shown by
Lemma~\ref{l:same}. Since \cite{gh} is a much older paper than
\cite{fh}, we have chosen to use the terminology ``sQQR''
(respectively, ``fQQR'') instead of ``s$\vo$'' (respectively,
``f$\vo$'').

In the integrally closed case, the sQQR- and fQQR-properties
coincide (Proposition~\ref{p:prufer}), but we have not been able
to determine whether these properties coincide in general. We can
easily see, however, that the sQQR- and $\vo$-properties are
distinct (even in the integrally closed case). This follows from
the fact that a valuation domain is automatically a QQR-domain
(and therefore an sQQR-domain), whereas a valuation domain need
not be an $\vo$-domain \cite[Example 2.16]{fh}.

In Section~\ref{s:examples}, we produce examples of sQQR-domains
which are not QQR domains.  However, we have not been able to
produce local examples of sQQR-domains which are not QQR. Indeed,
we devote much of the next section to showing that such examples
cannot exist in the presence of certain finiteness assumptions.
\bigskip

\section{Finiteness results} \label{s:results}

We have the following result in the integrally closed case.

\begin{proposition} \label{p:prufer}
    Let $R$ be integrally closed.  Then the following statements are equivalent.
\begin{itemize}
\item[(1)]  $R$ is an sQQR-domain.
\item[(2)]  $R$ is an fQQR-domain.
\item[(3)]  $R$ QQR-domain.
\item[(4)]  $R$ is a Pr\"{u}fer domain.
\end{itemize}
\end{proposition}
\begin{proof} Davis \cite[p. 197]{da} proved (4) $\ra$ (3).  Hence
it suffices to prove (1) $\ra$ (4).  This is a familiar argument
(cf. \cite[proof of Theorem 26.2]{g}).  Let $u \in K$, and
consider the simple overring $R[u^2]$. By hypothesis, this ring is
an intersection of localizations of $R$ and is therefore
integrally closed. Hence $u \in R[u^2]$, and so $u$ satisfies a
polynomial $f$ over $R$ such that some coefficient of $f$ is a
unit of $R$.  It now follows from the $u,u^{-1}$-lemma (as in
\cite[Lemma 19.14]{g}) that $R$ is a Pr\"{u}fer domain.
\end{proof}

\begin{proposition} \label{p:nonunits} Let $R$ be an sQQR-domain
which is not integrally closed, let $x \in \ol R \sm R$, and let
$\mc P$ denote the set of primes of $R$ which do not contain
$(R:_R x)$. Then
\begin{itemize}
\item[(1)] $\bigcup_{P \in \mc P} P$ is the set of nonunits of $R$, and
\item[(2)] $\mc P$ is infinite (so that $R$ has infinitely many
primes).
\end{itemize}
\end{proposition}
\begin{proof} If $s \in R$ but $s \notin \bigcup_{P \in \mc P} P$,
then $s$ is a unit of $\bigcap_{P \in \mc P} R_P=R[x]$ (the
equality following from Lemma~\ref{l:finiteext}). Since $x$ is
integral over $R$, this implies that $s$ is a unit of $R$. This
proves (1). For (2), note that (1) implies that $(R:_R x) \sub
\bigcup_{P \in \mc P} P$; prime avoidance then implies that $\mc
P$ must be infinite.
\end{proof}

This leads to our first restriction on (possible) local examples
of sQQR-domains which are not QQR-domains.

\begin{corollary} \label{c:locmaxnotprinc} Let $(R,M)$ be a local
sQQR-domain which is not integrally closed.  Then
\begin{itemize}
\item[(1)] $\dim R \ge 2$, and
\item[(2)] $M$ is not minimal over a principal ideal.
\end{itemize}
\end{corollary}
\begin{proof} Statement (1) is clear from
Proposition~\ref{p:nonunits}(2).  Statement (2) follows from
Proposition~\ref{p:nonunits}(1), upon observing that for a nonzero
element $a \in M$, $(R:_R 1/a)=(a)$.
\end{proof}

\begin{proposition} \label{p:loc} If $R_M$ is an sQQR-domain
(respectively, fQQR-domain) for each maximal ideal $M$ of $R$,
then $R$ is an sQQR-domain (respectively, fQQR-domain).
\end{proposition}
\begin{proof} For $x \in K$, we have \[R[x]=\bigcap_{M \in
\text{Max}(R)} (R[x])_{R \sm M} = \bigcap_{M \in \text{Max}(R)}
R_M[x].\]  For $M \in \text{Max}(R)$, since $R_M$ is an
sQQR-domain, $R_M[x]$ is an intersection of localizations of $R_M$
and therefore also of $R$. This takes care of the sQQR-property;
the proof for the fQQR-property is similar.
\end{proof}

Despite Proposition~\ref{p:loc}, neither the sQQR-property nor the
fQQR-property is a local property.  (We produce a specific example
of this in Section~\ref{s:examples}.) In fact, in
Proposition~\ref{p:locfindim} below, we show that a domain which
is locally a finite dimensional sQQR-domain is actually a Pr\"ufer
domain (and hence a QQR-domain).

We are able to localize the sQQR-property in some cases.  The
first such case is described in the following result; other cases
will be examined in Remark~\ref{r:locsqqr} below.

\begin{proposition} \label{p:treed} Let $R$ be a semilocal sQQR-domain
with $\textup{Spec}(R)$ treed.  Then $R_M$ is an sQQR-domain for
each maximal ideal $M$ of $R$.
\end{proposition}
\begin{proof} Fix a maximal ideal $M$.  Let $x \in K \sm M$, and let
$\mc P$ (respectively, $\mc Q$) denote the set of primes which do
not contain $(R:_R x)$ and which are contained in $M$
(respectively, are not contained in $M$).  Then we have
$R[x]=(\bigcap_{P \in \mc P} R_P) \bigcap (\bigcap_{Q \in \mc Q}
R_Q)$ by Proposition~\ref{p:basics}.  Set $S=R \sm M$.  Then
$$R_M[x]=R[x]_S = \left(\bigcap_{P \in \mc P} R_P\right)_S \bigcap
\left(\bigcap_{Q \in \mc Q} R_Q\right)_S=\left(\bigcap_{P \in \mc
P} R_P\right) \bigcap \left(\bigcap_{Q \in \mc Q} R_Q\right)_S.$$
We claim that $(\bigcap_{Q \in \mc Q} R_Q)_S \supseteq \bigcap_{P
\in \mc P} R_P$.  This will suffice to establish the result.  To
verify this, select a maximal ideal $N$ different from $M$.  Since
$\text{Spec}(R)$ is treed, Zorn's lemma produces a unique prime
ideal $N'$ which is maximal with respect to being contained in $M
\cap N$ (possibly, $N'=0$). If $(R:_R x) \nsubseteq N'$, then
$$\left(\bigcap_{Q \in \mc Q, Q \sub N} R_Q \right)_S \supseteq
(R_N)_S = R_{N'} \supseteq \bigcap_{P \in \mc P} R_P.$$  If $(R:_R
x) \sub N'$, then no $Q \in \mc Q$ satisfies $Q \sub N$, and the
intersection ``degenerates'' to $K$.  The claim now follows
easily.
\end{proof}

We require the concept of unique minimal overring: A proper
overring $T$ of a domain $R$ is said to be \emph{the unique
minimal overring} of $R$ if each proper overring of $R$ is
actually an overring of $T$. This concept was introduced in
\cite{gh} as a tool for studying QQR-domains.

\begin{proposition} \label{p:uniqueminovrg} Let $(R,M)$ be a local sQQR-domain, and suppose
that $R \subsetneqq \vo(M)$.  Then \begin{itemize}
\item[(1)] $\vo(M)$ is the unique minimal overring of $R$,
\item[(2)] $\vo(M)$ is an sQQR-domain, and
\item[(3)] if $\vo(M)=\ol R$, then $\ol R$ is a Pr\"ufer domain.
\end{itemize}
\end{proposition}
\begin{proof} (1)  Let $T$ be a proper overring of $R$, and let $t \in T \sm R$.  Then
$R[t]=\bigcap_{P \in \mc P} R_P$ for some set $\mc P$ of prime of
$R$ with $M \notin \mc P$.  Since $\vo(M)=\bigcap_{Q \ne M} R_Q$,
we have $\vo(M) \sub R[t] \sub T$.

(2) Pick $x \in K$, and write $R[x]=\bigcap_{P \in \mc P} R_P$ for
some set $\mc P$ of primes of $R$.  Since $\vo(M)$ is the unique
minimal overring of $R$, this yields $R[x]=\vo(M)[x]$; hence
$\vo(M)[x]$ is an intersection of localizations of $R$ and
therefore also of $\vo(M)$.

(3)  An integrally closed sQQR-domain is a Pr\"ufer domain by
Proposition~\ref{p:prufer}.
\end{proof}

\begin{proposition} \label{p:cases} Let $(R,M)$ be a local
sQQR-domain which is not integrally closed and for which $R
\subsetneqq (R:M)$. Then $(R:M)=(M:M)=\vo(M)$, $\vo(M)$ is the
unique minimal overring of $R$, and (exactly) one of the following
must occur:
\begin{itemize}
\item[(1)] $(R:M)$ is local with maximal ideal $M$,
\item[(2)] $(R:M)$ is local with maximal ideal different from $M$, or
\item[(3)] $(R:M)$ has two maximal ideals $N_1,N_2$ with $N_1 \cap N_2 = M$.
\end{itemize}
\end{proposition}
\begin{proof} By Corollary~\ref{c:locmaxnotprinc}, $M$ is not principal.
Since $M$ is maximal, we then have $(R:M)=(M:M)$, and this is a
proper overring of $R$.  Since $(R:M) \sub \vo(M)$, this implies
that $(R:M)=\vo(M)$ and that $\vo(M)$ is the unique minimal
overring of $R$ by Proposition~\ref{p:uniqueminovrg}. That this
situation results in just the three cases mentioned follows from
\cite[Section 2]{gh}.
\end{proof}

The next result shows that in cases (1) and (3) above, $R$ is a
QQR-domain.  We recall that a prime ideal of a domain $D$ is
\emph{unbranched} if $P$ is the only $P$-primary ideal of $D$ and
that in a Pr\"ufer domain this is equivalent to having $P$ be the
union of the prime ideals properly contained in $P$ \cite[Theorem
23.3]{g}.

\begin{proposition} \label{p:localsqqr2} Let $(R,M)$ be a local
sQQR-domain which is not integrally closed and for which $R
\subsetneqq (R:M)$. If $(R:M)$ is local with maximal ideal $M$ or
if $(R:M)$ has two maximal ideals, then
\begin{itemize}
\item[(1)] $(R:M)=\ol{R}$,
\item[(2)] $(R:M)$ is a Pr\"ufer domain, and
\item[(3)] $R$ is a QQR-domain.
\end{itemize}
\end{proposition}
\begin{proof}  We first deal with the case where $(R:M)$ has
maximal ideal $M$.  By \cite[Proposition 2.6]{gh}, $\ol R$ is a
valuation domain with maximal ideal $M$. Hence for $u \in \ol R$,
we have $uM \sub M$, and this implies that $(R:M)= \ol R$.  Thus
$\ol R = \vo(M) = \bigcap_{P \ne M} R_P = \bigcap_{Q \in \mc Q}
\ol R_Q$ for some set $\mc Q$ of primes of $\ol R$, with $Q \cap R
\ne M$ for each $Q \in \mc Q$. Since $\ol R$ is a valuation
domain, this implies that $M$ is unbranched in $\ol R$. By
\cite[Theorem 3.3]{gh}, $R$ is a QQR-domain.

Now suppose that $(R:M)$ has two maximal ideals $N_1$ and $N_2$.
In this case, $\ol R$ is a Pr\"ufer domain, $M$ is an ideal of
$\ol R$ and $N_1 \cap N_2=M$ by \cite[Proposition 2.5]{gh}. Hence
for $u \in \ol R$, we have $Mu \sub M \ol R \sub N_1 \cap N_2 =
M$,  and we have $u \in (R:M)$. Hence $\ol R=(R:M)$. We claim that
each $N_i$ is unbranched.  As before we have $\ol R = \bigcap_{P
\ne M} R_P = \bigcap_{Q \in \mc Q} \ol R_Q$ for some set $\mc Q$
of primes in $\ol R$, with $Q \cap R \ne M$ for each $Q$. Hence
for $Q \in \mc Q$, we must have $Q \ne N_i$ for $i=1,2$. If $N_i$
is branched, we can choose $t \in N_i$ but with $t$ in no other
prime of $\ol R$.  However, this yields $1/t \in \bigcap_{Q \in
\mc Q} \ol R_Q = \ol R$, a contradiction. Thus each $N_i$ is
unbranched, and $R$ is a QQR-domain, again by \cite[Theorem
3.3]{gh}.
\end{proof}

Recall that a \emph{pseudo-valuation domain} is a local domain
$(R,M)$ such that $(R:M)$ is a valuation domain with maximal ideal
$M$. By Proposition~\ref{p:localsqqr2}, we have the following:

\begin{corollary} \label{c:pvd}
    Let $(R,M)$ be a PVD.  Then $R$ is sQQR-domain $\Leftrightarrow$
$R$ is a QQR-domain.
\end{corollary}

We are not able to rule out the possibility of an example as in
case (2) of Proposition~\ref{p:cases}.  The following result
places some restrictions on such an example.

\begin{proposition} \label{p:case2} Let $(R,M)$ be a local
sQQR-domain which is not a QQR-domain.  Suppose that $R$ is as in
case (2) of Proposition~\ref{p:cases}. Set $T=(R:M)$, and let $N$
denote the maximal ideal of $T$.  Then $T$ is non-integrally
closed sQQR-domain, and $(T:N)=T=\vo_T(N)$.
\end{proposition}
\begin{proof}  $T$ is an sQQR-domain by
Proposition~\ref{p:uniqueminovrg}.  Suppose, by way of
contradiction, that $T$ is integrally closed.  Then $T$ is a
valuation domain by Proposition~\ref{p:prufer}.  In particular, we
must have $T=\ol R$. The fact that $T=\bigcap_{P \ne M} R_P$ then
implies that $T=\bigcap_{Q \in \mc Q} T_Q$ for some set $\mc Q$ of
primes of $T$ with $N \notin \mc Q$.  This, in turn, implies that
$N$ is unbranched. However, $M$ is an ideal of $T$ and is
therefore an $N$-primary ideal of $T$ different from $N$, a
contradiction. Thus $T$ is not integrally closed.  Now note that
$(T:M)=(R:M^2) \sub \vo(M) = T$, from which it follows immediately
that $(T:N)=T$. Moreover, $\vo(N)=\bigcap_{Q \nsupseteq N}T_Q
=\bigcap_{Q \nsupseteq MT} T_Q=\bigcap_{P \nsupseteq
M}R_P=\vo_R(M)=T$.
\end{proof}

\begin{corollary} \label{c:possiblelocalexample} If there is a local
sQQR domain $(R,M)$ which is not a QQR-domain, then there is such
an example with $(R:M)=R=\vo(M)$.
\end{corollary}

\begin{proposition} \label{p:locfindim} Let $R$ be a domain
such that $R_M$ is a finite dimensional sQQR-domain for each
maximal ideal $M$ of $R$. Then $R$ is a Pr\"ufer domain.
\end{proposition}
\begin{proof} If $R$ is a local sQQR-domain of dimension 1, then $R$ is a
valuation domain by Corollary~\ref{c:locmaxnotprinc} and
Proposition~\ref{p:prufer}. Let $(R,M)$ be a local sQQR-domain of
finite dimension greater than 1.  By induction, we may assume that
$R_P$ is a valuation domain for each nonmaximal ideal $P$. If
$R=\vo(M)=\bigcap_{P \ne M} R_P$, then $R$ is an integrally closed
sQQR-domain and is therefore a valuation domain by
Proposition~\ref{p:prufer}. Otherwise, $\vo(M)$ is the unique
minimal overring of $R$ by Proposition~\ref{p:uniqueminovrg}, and,
again since each $R_P$ is integrally closed, we have $\vo(M)=\ol
R$.  Hence $\ol R$ is a Pr\"ufer domain by
Proposition~\ref{p:uniqueminovrg}. Moreover, $\ol R$ has at most
two maximal ideals \cite[Corollary 2.2]{gh} and therefore $\ol R$,
and hence also $R$, has only finitely many prime ideals. It now
follows from Proposition~\ref{p:nonunits} that $R=\ol R$.  This
takes care of local $R$. The general case follows easily.
\end{proof}

We are now able to consider what happens when various finiteness
conditions are imposed on an sQQR-domain.  Recall that a domain
$R$ is said to be \emph{$v$-coherent} if for each finitely
generated ideal $I$ of $R$, we have $I^{-1}=J_v$ for some finitely
generated fractional ideal $J$ of $R$ and that $R$ is a
\emph{finite conductor domain} if each conductor ideal $(R:_R x)$,
where $x \in K$, is finitely generated.

\begin{proposition} \label{p:vcoh} If $R$ is an sQQR-domain which
is locally finite dimensional, and if, furthermore, $R$ is
$v$-coherent or a finite conductor domain, then $R$ is a Pr\"ufer
domain. In particular, a Noetherian sQQR-domain is a Dedekind
domain.
\end{proposition}
\begin{proof} Let $x \in K$.  If $R$ is a finite conductor domain,
then $(R:_R x)$ is finitely generated, and $R[x]$ is the Kaplansky
transform of a finitely generated ideal. If $R$ is $v$-coherent,
then $(R:_R x)=(R:(1,x))=I_t$ for some finitely generated ideal
$I$. Hence by Proposition~\ref{p:basics} and
Lemma~\ref{l:finiteext}, we again have that $R[x]$ is the
Kaplansky transform of a finitely generated ideal. It then follows
from Proposition~\ref{p:basics} that each localization of $R$ is
also an sQQR-domain. The result now follows from
Proposition~\ref{p:locfindim}.
\end{proof}

Recall that a domain $R$ is said to have \emph{Noetherian
spectrum} if it satisfies the ascending chain condition on radical
ideals and to be a \emph{Mori domain} if it satisfies the
ascending chain condition on divisorial ideals.

\begin{remark} \label{r:locsqqr} From Proposition~\ref{p:loc} and the proof of
Proposition~\ref{p:vcoh}, we obtain that if $R$ is a $v$-coherent
or a finite conductor domain, then $R$ is sQQR if and only if $R$
is locally sQQR.  We shall see later (in the proof of
Theorem~\ref{t:noespecsqqr}) that the sQQR-property is also a
local property for domains with Noetherian spectrum and for Mori
domains.
\end{remark}

\begin{lemma} \label{l:accraddiv} Let $(R,M)$ be a non-integrally
closed local sQQR-domain which satisfies the ascending chain
condition on radicals of conductor ideals.  Then $\vo(M)$ is the
unique minimal overring of $R$.
\end{lemma}
\begin{proof} Since $R$ is an sQQR-domain, for each $x \in K$ we
have $R[x]=\vo(R:_R x)=\vo(\rad(R:_R x))$ by
Proposition~\ref{p:basics} and Lemma~\ref{l:finiteext}. Moreover,
the ideal $(R:_R x)$ satisfies the hypothesis of
Proposition~\ref{p:basics}(4); hence for $x,y \in K$ we have $R[x]
\subsetneqq R[y]$ if and only if $\rad(R:_R y) \subsetneqq
\rad(R:_R x)$.  It then follows by our assumption that we may pick
$u \in \ol R \sm R$ such that $R[u]$ is a minimal extension of $R$
(i.e., there are no rings properly between $R$ and $R[u]$). By
\cite[Lemma 2.3]{gh}, $M$ is the conductor of $R$ in $R[u]$,
whence $R \subsetneqq (R:M) \sub \vo(M)$.  The lemma now follows
from Proposition~\ref{p:uniqueminovrg}.
\end{proof}

It is known that in a Mori domain the radical of a divisorial
ideal is again divisorial. Hence domains with Noetherian spectrum
and Mori domains satisfy the ascending chain condition on radicals
of conductor ideals.

\begin{theorem} \label{t:morisqqr} If $R$ is a Mori sQQR-domain, then $R$
is a Dedekind domain.
\end{theorem}
\begin{proof}
We first handle the case where $R$ is actually assumed to be a
Mori QQR-domain.  In this case, since both properties localize, we
may as well assume that $R$ is also local with maximal ideal $M$.
If $R$ is integrally closed, then $R$ is a valuation domain, and
it is well known that a Mori Pr\"ufer domain is a Dedekind domain
(see, e.g., the proof of \cite[Proposition 2.6]{cgh}). Hence we
also assume that $R$ is not integrally closed.  By \cite[Theorem
3.3]{gh}, $\ol R$ is a Pr\"ufer domain, and the (at most two)
maximal ideals of $\ol R$ are unbranched.  However, if $Q$ is a
nonzero, nonmaximal prime ideal of $\ol R$, then by \cite[Lemma
3.4]{fh}, $R_{Q \cap R}=\ol R_Q$, whence $R_{Q \cap R}$ is a Mori
valuation domain.  It is well known that this implies that $\ol
R_Q$ is a rank one discrete valuation domain, whence $Q$ has
height one. Thus the maximal ideals of $\ol R$ have finite height,
which contradicts the fact that they are unbranched.

For the rest of the proof, we explicitly assume that $R$ is not a
QQR-domain.  Since for an ideal $I$ of a Mori domain, we have
$I_t=I_v=J_v=J_t$ for some finitely generated ideal $J$ of $R$, we
have $\vo(I)=\vo(J)$ by Proposition~\ref{p:basics}, so that the
sQQR-property localizes. Hence we may assume that $R$ is a local
Mori sQQR-domain with maximal ideal $M$. Of course, we may also
assume that $R$ is not integrally closed.  By
Lemma~\ref{l:accraddiv}, $\vo(M)=(R:M)$ is the unique minimal
overring of $R$.  Hence $\vo(M)$ is integral over $R$, and by
\cite[Lemma 2.3]{gh}, $M$ is the conductor of $R$ in $\vo(M)$.  In
particular, $M$ is divisorial.  By Propositions~\ref{p:cases},
\ref{p:localsqqr2}, and \ref{p:case2}, $T=\vo(M)$ is a local
sQQR-domain which is not integrally closed, and the maximal ideal
$N$ of $T$ satisfies $(T:N)=T$, so that $N$ is nondivisorial.
Moreover, since $\vo(M)=(R:M)$, $\vo(M)$ is again a Mori domain by
\cite[I, Th\'eor\`eme 2]{r}. However, as we just showed, this
implies that the maximal ideal $N$ of $T$ is divisorial, a
contradiction.
\end{proof}

\begin{theorem} \label{t:noespecsqqr}  Let $R$ be an sQQR-domain
with Noetherian spectrum.  Then $R$ is a Pr\"ufer domain.
\end{theorem}
\begin{proof} A ring with Noetherian spectrum has the property that
each radical ideal is the radical of a finitely generated ideal.
Hence each Kaplansky transform is the transform of a finitely
generated ideal.  It then follows from Proposition~\ref{p:basics}
that the sQQR-property localizes for domains with Noetherian
spectrum. Hence we may assume that $R$ is a local sQQR-domain with
maximal ideal $M$.  We may also assume that $R$ is not integrally
closed. If $R$ is a QQR-domain, then $M$ is unbranched by
\cite[Theorem 3.3]{gh}. However, $M$ is the radical of a finitely
generated ideal, which contradicts that fact that $M$ is
unbranched.  (To see this, let $M = \rad J$, $J$ finitely
generated.  Since $M$ is unbranched, we must have $M=J$, i.e., $M$
is finitely generated.  However, Nakayama's lemma then guarantees
that $M \ne M^2$, and $M^2$ is then an $M$-primary ideal distinct
from $M$, a contradiction.)  Hence we may assume that $R$ is not a
QQR-domain. By Lemma~\ref{l:accraddiv}, $\vo(M)$ is the unique
minimal overring of $R$, and, as in the proof of
Theorem~\ref{t:morisqqr}, $M$ is the conductor of $R$ in $\vo(M)$,
so that $M$ is divisorial. Also as in the proof of
Theorem~\ref{t:morisqqr}, we obtain that $T=\vo(M)$ is a local
sQQR-domain with nondivisorial maximal ideal $N$.  However, it is
easy to see that the contraction map from $\text{Spec}(T)$ to
$\text{Spec}(R)$ is a homeomorphism, from which it follows that
$T$ also has Noetherian spectrum.  But then the argument just
given shows that $N$ is divisorial, a contradiction.
\end{proof}

A one dimensional domain in which each nonzero element is
contained in only finitely many primes has Noetherian spectrum.
Hence we have the following result.

\begin{corollary} \label{c:onedimprufer}
    Let $R$ be a one-dimensional domain, and assume that each nonzero element of $R$
is contained in only finitely many primes. Then $R$ is an
sQQR-domain $\lra$ $R$ is a Pr\"ufer domain.
\end{corollary}

\begin{theorem} \label{t:semilocsqqr} If $R$ is a semilocal
sQQR-domain with treed spectrum, then $R$ is a QQR-domain.
\end{theorem}
\begin{proof}  By \cite[Theorem 1.9]{gh}, it suffices to prove
that each localization of $R$ is a QQR-domain.  Hence by
Proposition~\ref{p:treed}, we may assume that $R$ is local with
maximal ideal $M$.  We may also assume that $R$ is not integrally
closed.  Let $x \in \ol R \sm R$, and write $R[x]=\bigcap_{(R:_R
x) \nsubseteq P} R_P$.  Since $x$ is integral over $R$, this
intersection cannot contain $R_Q$ for any prime $Q \ne M$.  Thus
we must have $\rad(R:_R x) = M$.  Hence $R[x] = \vo(M)$.  By
Proposition~\ref{p:uniqueminovrg}, $\vo(M)$ is the unique minimal
overring of $R$, and, since $x$ was chosen arbitrarily, we have
$\vo(M)=\ol R$, so that $\ol R$ is a Pr\"ufer domain.  Note that
in this situation $\ol R=\bigcap_{P \ne M}R_P$, and so $M$ must be
the union of the chain of primes properly contained in $M$.  We
claim that each maximal ideal of $\ol R$ is unbranched. If not,
let $N$ be a branched maximal ideal of $\ol R$.  Then, since $\ol
R$ is a Pr\"ufer domain, $N=\rad \ol Ru$ for some $u \in \ol R$
\cite[Theorem 23.3]{g}.  By \cite[Section 2]{gh}, $u^2 \in R$,
whence $M = \rad Ru^2$. However, this contradicts the fact that
$M$ is the union of the chain of primes properly contained in $M$.
Hence $R$ is a QQR-domain by \cite[Theorem 3.3]{gh}.
\end{proof}

\section{Examples} \label{s:examples}

In this section, we construct examples of sQQR-domains which are
not QQR-domains.

\begin{lemma} \label{l:omegapullback}
    Consider the following pullback diagram.
\[
    \begin{CD}
        R   @>>>    D   \\
        @VVV        @VVV    \qquad \qquad   \\
        T   @>\varphi>>    k   \\
    \end{CD}
\]
Here, $T$ is a domain, $k=T/M$ for some maximal ideal $M$ of $T$,
$\varphi$ is the canonical homomorphism, and $D$ is a domain
contained in $k$. Let $I$ be an ideal of $R$ which properly
contains $M$. Then $\varOmega_{R}(I) =
\varphi^{-1}(\varOmega_{D}(\varphi(I)))$.
\end{lemma}
\begin{proof}
    Let $x \in \varOmega_{R}(I)$.  Note that since $I \supsetneqq M$
we have $x \in T$.  Let $u=\varphi(a) \in \varphi(I)$, where $a
\in I$.  Then $xa^{n} \in R$ for some positive integer $n$.  It
follows that $\varphi(x)\varphi(a)^{n} \in D$.  Thus $x \in
\varphi^{-1}(\varOmega_{D}(\varphi(I)))$.  Now let $y \in
\varphi^{-1}(\varOmega_{D}(\varphi(I)))$, and let $b \in I$.  We
have $\varphi(y) \in \varOmega_{D}(\varphi(I))$, so that
$\varphi(yb^{m}) = \varphi(y)\varphi(b)^{m} \in D$ for some
positive integer $m$; that is, $yb^{m} \in R$.  It follows that $y
\in \vo_R(I)$.
\end{proof}

\begin{proposition} \label{p:sqqrpullback}
    Consider the pullback diagram of Lemma~\ref{l:omegapullback},
and assume that $T=V$ is a valuation domain and that $k$ is the
quotient field of $D$.  Then $R$ is an sQQR-domain (respectively,
fQQR-domain) $\lra$ $D$ is an sQQR-domain (respectively,
fQQR-domain).
\end{proposition}
\begin{proof}
    We give the details for the sQQR-case, the fQQR-case being
similar.  Assume that $R$ is an sQQR-domain. Let $u \in k$, and
let $I = \varphi^{-1}(D:_D u)$.  Pick $v \in V$ with
$\varphi(v)=u$. Then  $I = (R:_R v)$, and $R[v] =
\varOmega_{R}(I)$.  Hence by Lemma~\ref{l:omegapullback}, $D[u] =
\varphi(R[v]) = \varphi(\varOmega_{R}(I)) =
\varOmega_{D}(\varphi(I))) = \varOmega_{D}(D:_D u)$.  Thus $D$ is
an sQQR-domain.

For the converse, let $x \in K$.  If $x \in V$ we can use the same
techniques to show that $R[x] = \varOmega_{R}(R:_R x)$.  If $x
\notin V$, then $x^{-1} \in M \subseteq R$, and it is well known
(and easy to show) that $R[x] = \varOmega_{R}(x^{-1}R)$.
\end{proof}

\begin{proposition} \label{p:notqqr} Let $T$ be a Pr\"ufer domain with the following
properties: \begin{itemize}
\item[(1)] $T$ is one dimensional with $\textup{Max}(T)=\{M\} \cup \mc
P$, with $\mc P$ infinite,
\item[(2)] Each element of $M$ is in $P$ for almost all $P \in \mc P$ (equivalently,
each finitely generated ideal contained in $M$ is contained in $P$
for almost all $P \in \mc P$),
\item[(3)] Each element $a \in T \sm M$ satisfies $a \notin P$ for infinitely many
$P \in \mc P$, and
\item[(4)] $T/M$ admits a proper subfield $F$ such that there are no fields between
$F$ and $T/M$.
\end{itemize}

Then $R$ is an sQQR-domain but is not a QQR-domain, where $R$ is
defined by the following pullback diagram: \[
    \begin{CD}
        R   @>>>    F   \\
        @VVV        @VVV    \qquad \qquad   \\
        T   @>\varphi>>    T/M.   \\
    \end{CD}
\]
Moreover, $R_M$ is not an sQQR-domain.
\end{proposition}
\begin{proof} Let $K$ denote the common quotient field of $R$
and $T$.  Suppose that $\mc Q$ is an infinite subset of $\mc P$.
Then for $u \in \bigcap_{Q \in \mc Q} T_Q$, we have that the
finitely generated ideal $(T:_T u) \nsubseteq Q$ for each $Q \in
\mc Q$, whence by (2) $(T:_T u) \nsubseteq M$.  Thus $u \in T_M$.
That is, $\bigcap_{Q \in \mc Q} T_Q \sub T_M$ for each infinite
subset $\mc Q$ of $\mc P$.

Now let $x \in K \sm R$.  We consider two cases.

Case 1.  Suppose that $T[x] \sub T_M$.  Then $(T:_T x) \nsubseteq
M$, whence by (3) $(T:_T x) \nsubseteq P$ for infinitely many $P
\in \mc P$. Thus $T[x]=\bigcap_{Q \in \mc Q} T_Q$ for some
infinite subset $\mc Q$ of $\mc P$.  Since $(T:_T x) \nsubseteq
M$, we can choose $t \in T \sm M$ with $tx \in T$. Now $t \notin
M$ implies that $\varphi(t) \ne 0$, and we may find $t' \in T$
with $\varphi(tt')=1$.  Let $r=tt'$.  Then $r \in R \sm M$, and
$rx \in T$.  If $rx \notin R$, then, since $rx \in T$ and $R \sub
T$ is a minimal extension (there are no rings properly between $R$
and $T$), we have $T=R[rx] \sub R[x]$, whence
$R[x]=T[x]=\bigcap_{Q \in \mc Q} T_Q = \bigcap_{Q \in \mc Q} R_{Q
\cap R}$.  Thus we may assume that $rx \in R$, so that $(R:_R x)
\nsubseteq M$, and we have $R[x] \sub R_M$.  We shall show, in
fact, that $R[x] = T[x] \cap R_M$, which will complete the proof
in this case.  Note that $M[x]$ is a maximal ideal of $T[x]$, and
it follows that $M[x]$ is also a maximal ideal of $R[x]$.  Thus we
have the following pullback diagram: \[
    \begin{CD}
        R[x]   @>>>    R[x]/M[x]   \\
        @VVV        @VVV    \qquad \qquad   \\
        T[x]   @>>>    T[x]/M[x].
    \end{CD}
\]  Moreover, $R/M \sub R[x]/M[x] \sub R_M/MR_M$, whence $R[x]/M[x] = F$.  Similarly,
$T[x]/M[x] = T/M$.  Hence $R[x] \sub T[x]$ is a minimal extension.
However, we have $R[x] \sub T[x] \cap R_M \subsetneqq T[x]$,
whence $R[x] = T[x] \cap R_M$, as desired.

Case 2.  Suppose that $T[x] \nsubseteq T_M$.  Then $(T:_T x) \sub
M$, so that $(T:_T x) \sub P$ for almost all $P \in \mc P$ by (2).
Hence $T[x] = T_{P_1} \cap \cdots \cap T_{P_n}$ for some finite
subset $\{P_1, \ldots, P_n\}$ of $\mc P$. It follows that
$MT[x]=T[x]$. However, $MT[x]=M[x]=MR[x]\sub R[x]$, so that
$R[x]=T[x] = \bigcap_{j=1}^n T_{P_j} = \bigcap_{j=1}^n R_{P_j \cap
R}$. This completes the proof that $R$ has the sQQR-property.

Now by \cite[Theorem 1.9]{gh} the QQR-property is a local
property.  Moreover, $M$ is branched ($R$ is one dimensional),
whence by \cite[Theorem 3.3]{gh}, $R_M$ does not have the
QQR-property.  Hence $R$ is not a QQR-domain.  Finally, observe
that, since $R_M$ is not integrally closed,
Corollary~\ref{c:locmaxnotprinc} assures that $R_M$ is not an
sQQR-domain.
\end{proof}

There remains the construction of domains with the properties
described in Proposition~\ref{p:notqqr}.  Begin with an almost
Dedekind domain $T_0$ which is not Dedekind (so that there are
necessarily infinitely many maximal ideals), which has precisely
one non-finitely generated maximal ideal $M$, and which satisfies
property (2) of the proposition; \cite[Example 2, pp. 338-339]{g3}
is one such example. We claim that condition (3) of the
proposition is then automatically satisfied.  To see this pick $a
\in T_0 \sm M$; then $(M,a)=T_0$, and we may choose $c \in M$ so
that $(a,c)=T_0$. Since $c$ lies in almost all $P_i$, $a$ fails to
be in almost all $P_i$. As for condition (4), if it is not already
satisfied, replace $T_0$ by $T=T_0(X)$ ($T_0(X)=T_0[X]_S$, where
$S$ is the multiplicatively closed subset of $T_0[X]$ consisting
of all polynomials having unit content). By \cite[Proposition
36.7]{g}, $T$ is again an almost Dedekind domain.  Moreover, the
maximal ideals of $T$ are just the extensions of the maximal
ideals of $T_0$; hence $T$ has properties (1)-(3) above.  The
pertinent residue field is $T/MT(X)=(T_0/M)(X)$, which admits
appropriate subfields $F$, e.g., $F=(T_0/M)(X^2)$.

Since valuation domains are (s)QQR-domains,
Propositions~\ref{p:sqqrpullback} and \ref{p:notqqr} can be used
to produce examples of arbitrary dimension of sQQR-domains which
are not QQR-domains.

\section{Seminormality} \label{s:seminormality}

Recall that if $R$ is a domain with quotient field $K$, then $R$
is \emph{seminormal} if whenever $x \in K$ with $x^2,x^3 \in R$,
we have $x \in R$.  We show below that a QQR-domain is seminormal,
but we have not been able to determine whether an sQQR-domain (or
even an fQQR-domain) need be seminormal.  The example of an
sQQR-domain which is not a QQR-domain discussed at the end of
Section~\ref{s:examples} is seminormal (it is also an
fQQR-domain), so a seminormal sQQR-domain need not be a
QQR-domain, but we do not know whether a local seminormal
sQQR-domain (or fQQR-domain) must be a QQR-domain.

\begin{proposition} \label{p:qqrseminormal}  A QQR-domain is
seminormal.
\end{proposition}
\begin{proof} It suffices to establish the result locally.
Hence we assume that $(R,M)$ is a local QQR-domain, and we may as
well assume that $R$ is not Pr\"ufer.  By \cite[Section 2 and
Theorem 3.3]{gh} either $\ol R$ is a valuation domain with maximal
ideal $M$ or $\ol R$ is a Pr\"ufer domain with two maximal ideals
$N_1,N_2$ such that $N_1 \cap N_2 = M$.  In particular, $M$ is a
radical ideal of $\ol R$.  Now let $x \in K$ be such that $x^2,x^3
\in R$. Then $x \in \ol R$.  If $x^2 \notin M$, then $x=x^3/x^2
\in R$. If $x^2 \in M$, then since $M$ is a radical ideal of $\ol
R$, we have $x \in M \sub R$, as desired.
\end{proof}

We close with a characterization of seminormal fQQR-domains.

\begin{proposition} \label{p:fqqrseminormal} The following
statements are equivalent for an fQQR-domain $R$.
\begin{itemize}
\item[(1)] Each overring of $R$ is seminormal.
\item[(2)] Each integral finitely generated overring of $R$ is seminormal.
\item[(3)] $R[x]=R[x^2,x^3]$ for each $x \in K$.
\item[(4)] $\rad(R:_R x) = \rad(R:_R x^2) \cap \rad(R:_R x^3)$ for
each $x \in K$.
\item[(5)] $R$ is seminormal.
\end{itemize}
If $R$ satisfies any of these equivalent conditions, then $\ol R$
is a Pr\"ufer domain.
\end{proposition}
\begin{proof} The implications (3) $\lra$ (1) $\ra$ (2)
$\ra$ (5) are straightforward. Let $x \in K$. Then, using
Proposition~\ref{p:basics}, we have $R[x^2,x^3]=R[x]$ $\lra$
$\vo(R:_R (x^2,x^3)) = \vo(R:_R x)$.  As in the proof of
Lemma~\ref{l:accraddiv}, this last equality holds $\lra$
$\rad(R:_R (x^2,x^3) = \rad(R:_R x)$. Since $(R:_R (x^2,x^3)) =
(R:_R x^2) \cap (R:_R x^3)$, this proves (3) $\lra$ (4).  Assume
(5), and let $a \in (R:_R (x^2,x^3))$. Then $ax^2,ax^3 \in R$,
whence $(ax)^2,(ax)^3 \in R$. Since $R$ is seminormal, $ax \in R$.
Thus $\rad(R:_R x^2) \cap \rad(R:_R x^3) \sub \rad(R:_R x)$. The
converse always holds. Hence (5) $\ra$ (4).  Finally, the last
statement holds by \cite[Theorem 2.3]{adh}.
\end{proof}


TITLE: Function of moderate growth: history, motivation, and uses
QUESTION [5 upvotes]: I recently came across functions of moderate growth via Are functions of moderate growth a bornological space? and I was wondering, what are some concrete uses or applications of this space?  Where does it appear and why was it introduced historically?
I'm most interested in the continuous variant, discussed in Section 5 of Garrett - Examples of function spaces.
There is no Wikipedia page so I ask here.

REPLY [1 votes]: In addition to @Abdelmalek Abdesselam's and @Johannes Hahn's good points, I can add a few things from my own context:
One very non-negotiable thing is that Eisenstein series, with a significant role in the spectral theory of automorphic forms, are not in the relevant $L^2$ space, but are only (in an appropriate sense) "of moderate growth".
The simplest explanatory analogue is the case of Fourier transforms on the real line, and the very standard apparatus there (as mentioned by Abdelmalek and Johannes). To make one comparison, we can realize that to apply a tempered distribution $u$ to the Fourier inversion integral $f(x)=\int e^{2\pi ixy}\widehat{f}(y)\,dy$ it is not obvious, and, in fact, not necessarily correct, that $u(f)=\int u(e^{2\pi ixy})\widehat{f}(y)\,dy$. Namely, the exponential is not a Schwartz function.
Nevertheless, the (purely imaginary…) exponentials are bounded, smooth, etc. So, for example, compactly supported distributions can be sensibly applied to them (compatibly with everything else), and can move inside the Fourier inversion integral.
But in many circumstances one wants to have somewhat more general (though still fairly docile) distributions, not just compactly supported ones. Here, already just on the real line (not to mention the automorphic contexts…), it is possible to tell some needless lies. The basis of the possible lies is that (on the real line), continuous, compactly-supported functions are not sup-norm dense in the collection of all bounded continuous functions. (It's not even about measurability….) Thus, the translation action of $\mathbb R$ on bounded continuous functions on it (with sup norm) is not continuous. (E.g., $\sin(x^2)$). This is very bad… and cannot be counted a pathology, but, rather, a misunderstanding of the proper topology.
I think the fundamental point is that the sup-norm closure of continuous, compactly-supported functions is continuous functions going to $0$ at infinity. Thus, the translation action of $\mathbb R$ is continuous on such functions, with sup norm.
That is, rather than looking just at bounded continuous functions on $\mathbb R$, we look at the spaces $V_t$ (with $t$ real…) of continuous functions $f$ such that $\lim_{x\to \infty}\lvert x\rvert^t\cdot \lvert f(x)\rvert=0$. The continuous, compactly-supported functions are dense in each, so the translation action of $\mathbb R$ is (mercifully) continuous on each.
So, the most technically useful, and topologically useful, version of "bounded continuous" is $\bigcap_{t>0} V_t$, which describes a slightly larger space as a (projective) limit….
Hilariously, the colimit over $t$ (an ascending union) does produce the same topology as the naïve version (the latter incorrect on each limitand).


TITLE: Prove $W_{1} \cup W_{2} ... \cup W_{n}$ is simply connected
QUESTION [0 upvotes]: Given a wedge of circles contains $S_{1},S_{2},...S_{n}$ . They have a common point $p$ . For each $i$ , let $q_{i} \in S_{i}$ is a point different from $p$ . Let $W_{i} = S_{i} - q_{i}$ .
Prove : $\bigcup_{i=1}^{n} W_{i}$ is simply connected

REPLY [0 votes]: Let us consider the following homotopy: $p$ is immobile; for any point $a\in S_i$, where $a$ is neither of $p$ and $q_i$, move $a$ continuously towards $p$ with a constant speed such that in the moment $t=1$ it achieve $p$. There are two arcs on $S_i$, along which we can move $a$, but exactly one of them do not contain $q_i$ - choose it. Thus we can contract all the space to $p$, hence it is simply connected.


TITLE: Help with induction question?
QUESTION [0 upvotes]: Given a graph $G = (V, E)$, the complement of G is the graph $G'$ deﬁned on the same vertex set $V$ , however, an edge is present in $G'$ provided that it is not in $G$. Prove by induction that if $G$ is not connected, then $G'$ must be connected.

REPLY [1 votes]: HINT: Suppose that $G$ is a minimal counterexample: neither $G$ nor $G'$ is connected, but the result is true for all graphs with fewer vertices than $G$. If $\deg(v)=0$ for all $v\in V$, $G'$ is connected, and if $G$ is complete, then $G$ is connected, so there is at least one vertex $v\in V$ such that $$0<\deg(v)<|V|-1\;.$$ Let $H$ be the graph that remains when you remove from $G$ the vertex $v$ and all edges incident at $v$. By hypothesis at least one of $H$ and $H'$ is connected. There is at least one vertex $w$ in $H$ such that $vw$ is an edge of $G$, and there is at least one vertex $u$ in $H$ such that $vu$ is an edge in $G'$. (Why?) Use this fact to show that if $H$ is connected, then so is $G$, and if $H'$ is connected, so is $G'$. In either case this contradicts the assumption that $G$ is a counterexample to the theorem.
(Again, you can easily rephrase this as a traditional proof by induction.)