text:
\begin{document}

\title{Extensive escape rate in lattices of weakly coupled expanding maps with holes}
\author{Jean-Baptiste Bardet$^\ast$ and Bastien Fernandez$^\dag$}
\date{}
\maketitle
\begin{center}
$^\ast$ Laboratoire de Math\'ematiques Rapha\" el Salem\footnote{UMR 6085 CNRS - Universit\'e de Rouen} \\ Avenue de l'Universit\'e\\ 76801 Saint \'Etienne du Rouvray, France

$^\dag$ Centre de Physique Th\'{e}orique\footnote{UMR 6207 CNRS - Universit\'e Aix-Marseille II - Universit\'e Aix-Marseille I - Universit\'e Sud Toulon-Var}\\ CNRS Luminy Case 907\\ 13288 Marseille CEDEX 9, France
\end{center}

\begin{abstract}
This paper discusses possible approaches to the escape rate in infinite lattices of weakly coupled maps with uniformly expanding repeller. It is proved that computed-via-volume rates of spatially periodic approximations grow linearly with the period size, suggesting normalized escape rate as the appropriate notion for the infinite system. The proof relies on symbolic dynamics and is based on the control of cumulative effects of perturbations within cylinder sets. A piecewise affine diffusive example is presented that exhibits monotonic decay of the escape rate with coupling intensity.

\smallskip
\noindent \textbf{Mathematics Subject Classification (2010).} Primary 37L60; Secondary 37D50
\end{abstract}

\maketitle

\section{Introduction}
Coupled Map Lattices (CML) have been introduced in the early 1980's as space-time discrete models of reaction-diffusion processes and of other spatially extended systems \cite{CF05,K93}. Originally designed to ascertain the stability of numerical simulation schemes, CML soon appeared to provide a unique opportunity to extend the theory of dynamical systems to realistic infinite dimensional examples with discrete time. In particular, a large effort has been made to define and prove the existence and uniqueness of the (infinite-dimensional analogue of) SRB measure for small perturbations of uniformly expanding maps, see \cite{KL06} and the expanded list of references therein.

Of special interest in dynamical systems with spatial extension is also the definition of dynamical quantifiers such as the entropy \cite{ACFM02,AMU04,CE99} or the escape rate.
The escape rate comes as a natural estimator in presence of 'holes', {\sl i.e.}\ regions in phase space where the dynamics is not defined or where the orbit is lost. Such types of systems often occur in concrete examples; billiards, lattice or hard-spheres gases, etc \cite{GD95}. When the generator of the dynamics is expanding, the existence of holes suggests to consider the repeller, {\sl i.e.}\ the invariant set of points for which the orbit never reaches any hole. A natural problem is then to characterize the exponential rate of escape from a small neighborhood of the repeller.

The goal of this paper is to suggest a suitable definition of the escape rate for infinite lattices of weakly coupled maps with holes. The theory of (finite dimensional) dynamical system contains two distinct approaches to the escape rate, either via the conditionally invariant measure \cite{PY79} or by volume estimates \cite{BR75,ER85}. None of these approaches has an immediate extension to the infinite dimensional setting. On one hand, no proof of existence of an absolutely continuous conditionally invariant measure for CML is available in the literature, even in the simplest setting. On the other hand, the second approach needs to be adapted in order to ensure finiteness of estimated (Lebesgue) volumes at all times.

For simplicity, we deal here with CML with convolution couplings and transitive expanding real maps satisfying the Pianigiani-Yorke conditions \cite{PY79} and we consider the perturbative regime where the symbolic dynamics is well under control (section 2). We first discuss the existence of conditionally invariant measure and some properties of the related escape rate for spatially periodic approximations (section 3.1). A piecewise affine example is analyzed in section 3.2 which shows linear growth of the escape rate with the size of the approximation and also monotonic decay as the coupling intensity increases. This behavior motivates an alternative approach to escape rate that deals with volume estimates of points remaining in the neighborhood of the repeller (section 3.3). In this context, we prove the existence of the escape rate for every spatially periodic approximation of arbitrary systems and its linear growth with the system size. The corresponding asymptotic slope can be regarded as an escape rate per lattice unit for the infinite system. The proof is given in section 4 and essentially relies on the control of cumulative effects of perturbations within cylinder sets.

\section{Weakly coupled lattices of Pianigiani-Yorke maps}
A CML on the lattice $\Z$ is a dynamical system generated by a self-map $F$ in $\ell^\infty(\Z)$ (endowed with uniform topology and associated norm $\|\cdot\|$) whose specificity resides in the following expression. The mapping $F$ is given by the composition $F=C\circ F_0$ of the direct product $F_0$ of identical real maps and of a coupling $C$ \cite{CF05,K93}. In simplest situations, the coupling is the {\bf convolution operator} associated with an arbitrary non-negative and normalized sequence $\{c_n\}_{n\in\Z}$ in $\ell^1(\Z)$, {\sl i.e.}\ we have
\[
(Cx)_s=\sum_{n\in\Z} c_n x_{s-n},\ \forall s\in\Z,
\]
for every configuration $\{x_s\}_{s\in\Z}\in \ell^\infty(\Z)$ (see \cite{AF00} for properties of convolution operators in $\ell^\infty(\Z)$). The individual mapping $F_0$ explicitly writes
\[
(F_0 x)_s=f(x_s),\ \forall s\in\Z.
\]
where $f:\R\mapsto\R$ is a (locally bounded) map  that verifies the following conditions. There exists a finite collection $\{\text{I}_i\}_{i\in {\mathcal A}}$ (where ${\mathcal A}=\{1,\cdots,N\}$, $N\in\N$, $N>1$) of pairwise disjoint bounded and closed intervals with the properties that for each $i\in {\mathcal A}$
\begin{itemize}
\item[(i)] $f$ is of class $C^2$ on $\text{I}_i$ and $\inf\limits_{\text{I}_i}\left|f'\right|>1$,
\item[(ii)] there exists $j\in \mathcal{A}$ such that $\text{I}_j\subset \text{Int}\; f(\text{I}_i)$. Moreover, for every $j$ such that $f(\text{I}_i)\cap \text{I}_j\neq \emptyset$, we have $\text{Int}\; f(\text{I}_i)\supset \text{I}_j$.
\item[(iii)] In addition, we also assume that $f$ is {\bf transitive}, {\sl i.e.}\ for every pair $i,j\in \mathcal{A}$ there is a $t\in\N$ such that $\text{I}_j\subset f^t(\text{I}_i)$.
\end{itemize}
We will denote $\text{I}=\bigcup_i\text{I}_i$. In this context, the region $f(I)\setminus I$ plays the role of holes for the dynamical system $(f,I)$.

\noindent
The conditions on $f$ imply several properties of the dynamics, see e.g.\ \cite{CMS94,CMT98,PY79}, in particular
\begin{itemize}
\item[$\bullet$] the structural stability with respect to $C^1$ perturbations,
\item[$\bullet$] the existence of a repeller $\kappa_f$ on which the dynamics is conjugated to a topological Markov chain ($\kappa_f$ is a Cantor set, forward invariant under $f$ and defined by $\kappa_f=\bigcap\limits_{t\in\N}f^{-t}(\text{I})$),
\item[$\bullet$] the existence of a conditionally invariant probability measure that is absolutely continuous with respect to the Lebesgue measure.
\end{itemize}
The topological Markov chain operates in the set $\Omega\subset {\mathcal A}^\N$  of admissible sequences which are constrained by the condition (ii) ({\sl i.e.}\ the sequence $\{\vartheta^t\}_{t\in\N}$ is {\bf admissible} iff every pair of consecutive symbols $\vartheta^t,\vartheta^{t+1}$ complies with the condition $\text{I}_{\vartheta^{t+1}}\subset \text{Int}\; f(\text{I}_{\vartheta^t})$). Each point in $\kappa_f$ is defined by the intersection
\[
\bigcap_{t\in\N}f^{-t}\left(\text{I}_{\vartheta^t}\right)
\]
for some associated code $\{\vartheta^t\}_{t\in\N}\in \Omega$.

All these properties immediately translate to the uncoupled map lattice $F_0$ (obtained for $C=\text{Id}$, the identity mapping and provided that one only considers periodic lattices when the existence of quasi-invariant measure is concerned). In particular, the symbolic dynamics is given by the action of time translations on sequences of symbol configurations in $\Omega^\Z$. Introducing the notation $\text{\bf I}_\theta=\bigotimes\limits_{s\in\Z}\text{I}_{\theta_s}$ where $\theta=\{\theta_s\}_{s\in\Z}$ is an arbitrary {\bf symbol configuration} in ${\mathcal A}^\Z$, the repeller $K_0=\kappa_f^\Z$ can be regarded as the set of points defined by
\[
\bigcap_{t\in\N}F_0^{-t}\text{\bf I}_{\theta^t}
\]
for some sequence $\{\theta^t\}_{t\in\N}\in \Omega^\Z$. Alternatively, we have $K_0=\bigcap\limits_{t\in\N}F_{0}^{-t}\text{\bf I}$ where $\text{\bf I}=\text{I}^{\Z}$.

\noindent
Intuitively, the dynamical properties listed above are expected to extend to weakly coupled map lattices with symbolic dynamics being unaffected. To formally state this result, we use the norm on $\mathcal{L}(\ell^\infty(\Z))$, also denoted by $\|\cdot\|$, induced by the uniform norm in $\ell^\infty(\Z)$.
\begin{Thm} {\rm \cite{AF00}\footnote{In \cite{AF00}, there are two points in the proof of this theorem that are incorrect. One is the proof of a technical statement (namely Proposition 3.2) and the other is the argument that guarantees that cylinder sets are not empty. Both points are however valid and we provide a corrected proof in Appendix \ref{A-ERRATUM}.}}
Given $f$ with properties {\rm (i)--(iii)}, there exists $\epsilon_f>0$ such that for every $C$ so that $\|\text{Id}-C\|<\epsilon_f$ and every sequence $\{\theta^t\}_{t\in\N}\in\Omega^\Z$, the intersection
\[
\bigcap_{t\in\N}F^{-t}\text{\bf I}_{\theta^t}
\]
is non empty and reduces to a point. All such points form a $F$-invariant, closed, and totally disconnected\footnote{but not compact for the uniform topology} set $K=\bigcap\limits_{t\in\N}F^{-t}\text{\bf I}$ on which $F$ is topologically conjugated to the time translation acting on $\Omega^\Z$.
\label{STABSTRUCT}
\end{Thm}
Notice that, due to the conditions $c_n\geq 0$ and ${\displaystyle\sum_{n\in\Z}}c_n=1$, we have $\|\text{Id}-C\|=2(1-c_0)$ meaning that the condition $\|\text{Id}-C\|<\epsilon_f$ actually involves a single entry of the sequence $\{c_n\}$.

The CML has an obvious {\bf translational symmetry} invariance; namely the map $F$ commutes with translations along the lattice $\Z$. Besides dealing with infinite lattices, this symmetry legitimates focusing on periodic lattices $\Z_L=\Z/L\Z$ (indeed, we have $F\R^{\Z_L}\subset \R^{\Z_L}$ for every $L\in\N$). Restricting to sequences of $L$-periodic symbol configurations in $\Omega^{\Z_L}$  in Theorem \ref{STABSTRUCT} implies that the CML on the periodic lattice of (arbitrary) length $L$ has a limit Cantor set on which the dynamics is conjugated to a topological Markov chain (on a finite alphabet). Following \cite{PY79}, this suggests to investigate the existence of conditionally invariant measures in periodic CML.

\section{Escape rate per lattice unit}
\subsection{Approach via the conditionally invariant measure}
Let $G:X\subset\R^n\to\R^n$ be a mapping acting on some bounded set $X$. A probability measure $\mu$ on $X$ is said to be {\bf conditionally invariant} (or quasi-invariant) if there exists a constant $\alpha\in (0,1)$ such that $\mu\circ G^{-1}=\alpha\mu$ on $X$ \cite{CMS94}. When this measure exists the quantity
\[
\tilde\gamma=-\log\alpha.
\]
is called the {\bf escape rate}. As mentioned before, the conditions (i)--(iii) guarantee the existence for $f$ of a unique absolutely continuous conditionally invariant measure \cite{PY79}.

When $\|\text{Id}-C\|<\epsilon_f$, analogous conditions hold for the restriction of the CML to $\R^{\Z_L}$ ($L\in\N$ arbitrary) with the collection $\{\text{\bf I}_\theta\}_{\theta\in {\mathcal A}^{\Z_L}}$ (see Appendix \ref{A-ERRATUM}). It results that, in the weak coupling regime, a unique absolutely continuous conditionally invariant measure exists for the CML on periodic lattices. An explicit example is provided below.

Furthermore, the same conditions (i)--(iii) for $F\big{|}_{\R^{\Z_L}}$ imply the existence of an invariant measure $\nu$ supported on the limit Cantor set that is a Gibbs state for the potential $\log|F\big{|}_{\R^{\Z_L}}'|$ \cite{CMS94}. Accordingly, the following expression for the escape rate applies \cite{CMT98}
\[
\tilde\gamma=\lambda_\nu-h_\nu
\]
where $\lambda_\nu$ is the sum of (positive) Lyapunov exponents and $h_\nu$ is the Kolmog\-or\-ov-Sinai entropy associated with $\nu$. Thus, in principle, one could have access to the escape rate of periodic lattices of weakly coupled maps based on the knowledge of their Lyapunov exponents and metric entropy (or {\sl vice-versa} as in the example below). However, these quantities are not easily accessible in practice, and their behaviour with the spatial period is mostly unknown. Moreover, the problem of existence of an absolutely continuous conditionally invariant measure for the infinite CML, together with a Gibbs state on the corresponding repeller, remains unsolved.
\begin{figure}[ht]
\centerline{\includegraphics*[height=50mm]{LorenzMapeps}}
\caption{Graph of the Lorenz-type map $f(a)=ax+(1-a)H(x-\frac{1}{2})$ for $a>2$, together with the points $\underline{x}<0<1/a<\overline{x}$.}
\label{LORENZMAP}
\end{figure}

\subsection{Motivating example: CML with Lorenz-type map}
Consider the coupling operator $C$ derived from the one-dimensional discrete Laplacian
\[
(Cx)_s=x_s+\frac{\epsilon}{2}(x_{s-1}-2x_s+x_{s+1}),\quad \forall s\in\Z
\]
where $\epsilon\in [0,1]$. The local map $f$ is piecewise affine and the points 0 and 1 are fixed points, namely
\[
f(x)=ax+(1-a)H(x-\frac{1}{2}),\quad \forall x\in\R
\]
where $a>2$ and $H$ is the Heaviside function, see Figure \ref{LORENZMAP}.
This map $f$ clearly satisfies the assumptions (i)--(iii) with $N=2$ and the corresponding intervals are given by
\[
\text{I}_1=[\underline{x},\overline{x}]\quad\text{and}\quad
\text{I}_2=1-[\underline{x},\overline{x}]
\]
where $\frac{1}{a}<\overline{x}<\frac{1}{2}$ and $\underline{x}<0$ are arbitrary. We have $\text{Int}\; f(\text{I}_i)\supset \text{I}_1\cup \text{I}_2$ for $i=1,2$.

\noindent
Simple algebra shows that the condition $\|\text{Id}-C\|<\epsilon_f$ in Theorem \ref{STABSTRUCT} becomes $a(1-2\epsilon)>2$ in this case. Indeed, when $a(1-2\epsilon)>2$, for every $\theta\in\{1,2\}^\Z$, the image $F\text{\bf I}_\theta$ covers the product $(\text{I}_1\cup\text{I}_2)^\Z$ (and the expansiveness condition $\|(C\circ F_0)'\|\geq a(1-\epsilon)>1$ holds). This condition is optimal because when $a(1-2\epsilon)\leq 2$, some sequences of symbol configurations in $\{1,2\}^{\Z_2\times\N}$ are not admissible \cite{FG04}.

Let $L>1$ be fixed in $\N$ and assume that $a(1-2\epsilon)>2$. Then the CML on the $L$-periodic lattice is a piecewise expanding map of constant derivative for which every point in $\left(\text{I}_1\cup\text{I}_2\right)^{\Z_L}$ has $2^L$ pre-images.
As a consequence, the uniform measure on this set is conditionally invariant.

\noindent
Computing the determinant of the derivative $(C\circ F_0)'$ yields the following expression for the corresponding escape rate
\begin{equation}
\tilde\gamma_L(\epsilon)=L\log\left(\frac{a}{2}\right)+\sum_{k=1}^L\log\hat{c}\left(\frac{k}{L},\epsilon\right)
\label{TAUXFUITE}
\end{equation}
where $\hat{c}(\omega,\epsilon)=1-\epsilon(1-\cos2\pi\omega)>0$. In particular, since the Lyapunov exponents are in this case trivially given by $\log a\; \hat{c}\left(\frac{k}{L},\epsilon\right)$ ($k=1,\cdots,\lfloor\frac{L}{2}\rfloor$) this expression shows that the Kolmogorov-Sinai entropy of the measure supported on the Cantor set is given by $L\log 2$, {\sl i.e.}\ it corresponds to the topological entropy.

\noindent
The expression (\ref{TAUXFUITE}) suggests the following additional comments.
\begin{itemize}
\item[$\bullet$] As a sum of monotonically decreasing functions of $\epsilon$, the escape rate $\tilde\gamma_L(\epsilon)$ decreases when $\epsilon$ increases. In other words, increasing the interaction between sites slows down escape from the repeller, as expected.
\item[$\bullet$] The rate $\tilde\gamma_L(\epsilon)$ is an extensive quantity that linearly diverges with the period of configurations.
\end{itemize}
As for the definition of the entropy in spatially extended systems \cite{ACFM02,CE99}, the linear divergence suggests to normalize the escape rate by the period length, {\sl i.e.}\ to consider the ratio
\[
\frac{\tilde\gamma_L(\epsilon)}{L}=\log\left(\frac{a}{2}\right)+\frac{1}{L}\sum\limits_{k=1}^L\log\hat{c}\left(\frac{k}{L},\epsilon\right).
\]
The quantity $\frac{1}{L}\sum\limits_{k=1}^L\log\hat{c}\left(\frac{k}{L},\epsilon\right)$ represents a Riemann sum over $[0,1]$ for the fonction $\omega\mapsto\log\hat{c}(\omega,\epsilon)$. Thus an {\bf escape rate per lattice unit} can be defined in the limit of large periods
\[
\lim\limits_{L\to\infty}\frac{\tilde\gamma_L(\epsilon)}{L}=\log\left(\frac{a}{2}\right)+\int_0^1\log\left(1-\epsilon(1-\cos2\pi\omega)\right)d\omega
\]
This normalized rate also decreases monotonically when $\epsilon$ increases.

It is interesting to note that all the arguments above extend to CML with any convolution coupling. The expression for the escape rate per lattice unit remains unchanged
\[
\lim\limits_{L\to\infty}\frac{\tilde\gamma_L}{L}=\log\left(\frac{a}{2}\right)+\int_0^1\log|\hat{c}\left(\omega\right)|d\omega\quad\text{if}\quad a(2c_0-1)>2
\]
where $\hat{c}\left(\omega\right)=\sum\limits_{n\in\Z}c_ne^{2i\pi n\omega}$ is the (discrete) Fourier transform of the sequence $\{c_n\}$.

The explicit computations here crucially depend on the assumption that the derivative $(C\circ F_0)'$ is constant in phase space. In order to prove the existence of the escape rate per lattice unit for a broader class of CML, we use a more direct approach that relies on estimating the volume of the sets of points that remain located in the neighborhood of the repeller.

\subsection{Approach via volumes}
A direct definition of the escape rate can be formulated as follows \cite{BR75,ER85}. Given a mapping $G:X\subset\R^n\to\R^n$ ($X$ bounded) with an invariant set $J$, the {\bf escape rate} from an $\eta$-neighborhood $U$ of $J$ ($\eta>0$) is the limit (if it exists)
\[
\gamma(U)=-\lim\limits_{T\to\infty}\frac{1}{T}\log\text{Vol}(U_T)
\]
where $\text{Vol}(U_T)$ is the Lebesgue volume of the set $U_T$ of points $x$ such that $G^tx\in U$ for all $0\leq t<T$. In general, the limit does not depend on $U$ provided that $\eta$ is small enough. Moreover the escape rate of Axiom A attractors of diffeomorphisms on compact manifolds  satisfies the following variational principle \cite{BR75}
\[
\gamma(U)=\sup\left\{\lambda^+_\mu-h_\mu\ :\ \mu\ \text{ergodic invariant probability, supp}(\mu)\subset J\right\}
\]
Eckmann and Ruelle \cite{ER85} conjectured that this relation should be valid in a more general setting. This has been recently proved for repellers of some one-dimensional maps \cite{BDM07} and a counter-example has also been constructed \cite{BBS99}.

Back to CML on the lattice $\Z$, an escape rate can be introduced directly by limiting the dynamics to periodic configurations in $\Z_L$. Given an $\eta$-neighborhood $U\subset\ell^\infty(\Z)$ of the invariant set $K$ for the infinite CML, we consider the following limit
\[
\gamma_L(U)=-\lim\limits_{T\to\infty}\frac{1}{T}\log\text{Vol}(U_T^L)
\]
where $U_T^L$ is the set of periodic configurations $x\in\R^{\Z_L}$ such that $F^tx\in U$ for all $0\leq t\leq T$. (Of note, all configurations $F^tx$ must then also be $L$-periodic and the Lebesgue volume is computed in $\R^{\Z_L}$.)

\noindent
Following \cite{BS88,PS91}, we say that a convolution operator $C$ is of {\bf finite range} iff there exists $\zeta>1$ such that we have
\[
\sum\limits_{n\in\Z}\zeta^{|n|}c_n<+\infty
\]
for the corresponding sequence $\{c_n\}_{n\in\Z}$. As announced before, the existence of the limit $\frac{\gamma_L(U)}{L}$ can be proved for any CML with short range couplings and local map satisfying the conditions (i)--(iii).
\begin{Thm}
Given $f$ with properties {\rm (i)--(iii)} and a short range $C$ such that $\|\text{Id}-C\|<\epsilon_{f}$, for any sufficiently small $\eta$-neighborhood $U$ of $K$, the escape rate $\gamma_L(U)$ is well-defined for every $L>1$ and the following limit exists and is independent of $U$
\[
\gamma_\infty=\lim_{L\to\infty}\frac{\gamma_L(U)}{L}
\]
\label{THMTAUXFUIT}
\end{Thm}
The quantity $\gamma_\infty$ provides a definition of the (normalized) escape rate at the 'thermodynamics' limit in terms of estimates for the finite dimensional approximations. Of note, this definition does not rely on the existence of a conditionally invariant measure for the infinite system, nor on the direct computation of volumes in this limit. We conclude by a list of open questions
\begin{itemize}
\item[$\bullet$] Prove or disprove that
\[
\gamma_\infty=\lim_{T\to\infty}\frac{1}{T}\lim_{L\to\infty}\frac{1}{L}\log\text{Vol}(U_T^L)
\]
(see comment after Lemma \ref{MAJOSUM} below).
\item[$\bullet$] For the piecewise affine example of section 3.2, the quantities $\tilde\gamma_L$ and $\gamma_L(U)$ coincide (see Lemma \ref{BOUNDS} below). Prove that this relation holds for CML with arbitrary Pianigiani-Yorke maps.
\item[$\bullet$] Prove the Eckmann-Ruelle conjecture for periodic approximations of the CML.
\item[$\bullet$] Following \cite{B02}, prove that $\gamma_\infty$ corresponds to the escape rate of the $\Z^2$-action generated by $F$ and the spatial translations.
\end{itemize}

\section{Proof of Theorem \ref{THMTAUXFUIT}}
\paragraph{First step: Reducing neighborhood to union of basic sets.}
We claim that w.l.o.g.\ one can rely on the set $\text{\bf I}\cap\R^{\Z_L}=\bigcup\limits_{\theta\in {\mathcal A}^{\Z_L}}(\text{\bf I}_\theta\cap\R^{\Z_L})$ \footnote{The equality here follows from the fact that the original intervals $\text{I}_i$ are disjoints; hence to be a periodic configuration in a basic set $\text{\bf I}_\theta$ imposes that the symbol configuration $\theta\in {\mathcal A}^{\Z_L}$ must be periodic.} to compute the escape rate of the periodic CML.

\noindent
First, the set $K$ is included in the disjoint union $\text{\bf I}={\displaystyle\bigcup_{\theta\in {\mathcal
A}^\Z}}\text{\bf I}_\theta$ and each component $K\cap \text{\bf I}_\theta$ is at positive distance
from the complement $\ell^\infty(\Z)\setminus \text{\bf I}_\theta$ (see proof of condition (ii) in
Appendix \ref{A-ERRATUM}). Therefore, there exists $\eta_0>0$ such that for every $\eta<\eta_0$, the
$\eta$-neighborhood $U$ of $K$
is contained in $\text{\bf I}$. By passing to intersections
with $\R^{\Z_L}$, it follows that the escape rate computed based on the $\eta$-neighborhood $U$ cannot exceed the rate computed with the union of basic sets.

\noindent
To prove the converse inequality, consider an $\eta$-neighborhood $U$ contained in $\text{\bf I}$. The assumptions that the sets $\text{\bf I}_\theta$ are pairwise disjoints and that the intersection $F\text{\bf I}_\theta\cap \text{\bf I}_{\theta'}$ is non empty iff $\text{Int}\; F\text{\bf I}_\theta\supset \text{\bf I}_{\theta'}$ imply that
\[
\bigcap_{t=0}^{T} F^{-t}\text{\bf I}=\bigcup\limits_{\{\theta^t\}_{t=0}^{T}\in\Omega_{T+1}^{\Z}}\bigcap_{t=0}^{T}F^{-t}\text{\bf I}_{\theta^t}
\]
(where $\Omega_T$ represents the set of admissible words of length $T$ in ${\mathcal A}^{T}$). For every $T\in\N$, the latter set is contained in an $\eta_T$-neighborhood of $K$ for some $\eta_T>0$ such that$\lim\limits_{T\to\infty}\eta_T=0$. Therefore, for $T^*$ large enough such that $\eta_T<\eta$, we obtain that
\[
\bigcap_{t=0}^{T^*-1} F^{-t}\text{\bf I}\subset U
\]
which implies that the escape rate obtained with the union of basic sets cannot exceed the rate computed based on neighborhoods.

\noindent
Given a periodic symbol configuration $\theta\in {\mathcal A}^{\Z_L}$, let $\text{\bf I}^L_\theta=\text{\bf I}_\theta\cap\R^{\Z_L}$ be the corresponding set of periodic configurations. Analogous arguments to as before and the CML translation symmetry show that in order to estimate the escape rate of CML on periodic lattices, we may consider the volume of
\[
U_T^L=\bigcup\limits_{\{\theta^t\}_{t=0}^{T}\in\Omega_{T+1}^{\Z_L}}\bigcap\limits_{t=0}^{T}F^{-t}\text{\bf I}_{\theta^t}^L.
\]
\bigskip

\paragraph{Second step: Volume computation and partition function.}
The sets $\text{\bf I}_{\{\theta^t\}_{t=0}^{T}}^L=\bigcap\limits_{t=0}^{T}F^{-t}\text{\bf I}_{\theta^t}^L$ are pairwise disjoints; hence
\[
\text{Vol}\left(U_T^L\right)=\sum\limits_{\{\theta^t\}_{t=0}^{T}\in\Omega_{T+1}^{\Z_L}}\text{Vol}(\text{\bf I}_{\{\theta^t\}_{t=0}^{T}}^L).
\]
The volumes in the RHS trivially read
\[
\text{Vol}(\text{\bf I}_{\{\theta^t\}_{t=0}^{T}}^L)=\int_{\text{\bf I}_{\{\theta^t\}_{t=0}^{T}}^L}dx
\]
The map $F^{T}$ is one-to-one on each $\text{\bf I}_{\{\theta^t\}_{t=0}^{T}}^L$ and we have $F\text{\bf I}^L_{\theta^t}\supset \text{\bf I}^L_{\theta^{t+1}}$ for all $0\leq t\leq T-1$ when $\{\theta^t\}_{t=0}^{T}$ is admissible. It results that $F^{T}\text{\bf I}_{\{\theta^t\}_{t=0}^{T}}^L= \text{\bf I}^L_{\theta^{T}}$. Introducing $y=F^Tx\in\R^{\Z_L}$, we obtain
\begin{align*}
\text{Vol}\left(\text{\bf I}_{\{\theta^t\}_{t=0}^{T}}^L\right)=&\int_{\text{\bf I}_{\theta^{T}}^L}\prod_{t=0}^{T-1}\left|F'\circ F^{t-T}|_{\text{\bf I}_{\{\theta^k\}_{k=t}^T}}y\right|^{-1}dy\\
=&\left|C\right|_L^{-T}\int_{\text{\bf I}_{\theta^{T}}^L}\prod_{0\leq t<T\atop s\in\Z_L}\left|f'(F^{t-T}|_{\text{\bf I}_{\{\theta^k\}_{k=t}^T}}y)_s\right|^{-1}dy
\end{align*}
where $\left|C\right|_L$ is the determinant of the restriction $C\big{|}_{\R^{\Z_L}}$.

The last expression allows one to substitute the analysis of the asymptotic behaviour of $\text{Vol}\left(U_T^L\right)$ by the study of a kind of {\bf partition function} associated with the CML and defined by
\[
Z_{L,T}=\sum\limits_{\{\theta^t\}_{t=0}^{T-1}\in\Omega_T^{\Z_L}}\sup\limits_{x\in \text{\bf I}_{\{\theta^t\}_{t=0}^{T-1}}^L}\prod_{0\leq t<T\atop s\in\Z_L}\left|f'(x_s^t)\right|^{-1}
\]
where we use the notation $x_s^t=(F^tx)_s$. The substitution is justified by the following result.
\begin{Lem}
There exist two numbers $0\leq\underline{c}<\overline{c}$ such that for all $L,T\geq 1$, we have
\[
\underline{c}^L\left|C\right|_L^{-T}Z_{L,T}\leq \text{\rm Vol}\left(U_T^L\right) \leq \overline{c}^L\left|C\right|_L^{-T}Z_{L,T}
\]
\label{BOUNDS}
\end{Lem}
This statement leads to consider the quantity $K_L=\lim\limits_{T\to\infty}\frac{1}{T}\log Z_{L,T}$. If this limit exists, we have $\gamma_L(U)=K_L-\log\left|C\right|_L$ and, since
\[
\lim\limits_{L\to\infty}\frac{\log\left|C\right|_L}{L}=\int_0^1\log|\hat{c}\left(\omega\right)|d\omega
\]
(see section 3.2), to obtain the theorem it only remains to show that the limit $\lim\limits_{L\to\infty}\frac{K_L}{L}$ exists.

\begin{proof}
The proof essentially consists in obtaining a bounded distorsion estimate (see for instance \cite{C96}), that is to say in proving the existence of a number $c_1\geq 1$ such that for any $T\geq1$ and every admissible word $\{\theta^t\}_{t=0}^{T-1}$, we have
\begin{equation}
c_1^{-1}\leq\frac{\prod_{t=0}^{T-1}\left|f'(x_s^t)\right|^{-1}}{\prod_{t=0}^{T-1}\left|f'(y_s^t)\right|^{-1}}\leq c_1,\quad\forall s\in\Z_L,\ x,y\in \text{\bf I}_{\{\theta^t\}_{t=0}^{T-1}}^L
\label{DISTORSION}
\end{equation}
To begin, notice that the definition of $F$ implies that when $C$ is invertible, the following inequality holds for every $L$-periodic configurations $x,y$
\[
\left|f(x_s)-f(y_s)\right|\leq \left\|C^{-1}\right\| \left\|Fx-Fy\right\|\quad \forall s\in\Z_L.
\]
If, in addition, both configurations $x$ and $y$ belong to the same basic set $\text{\bf I}_\theta^L$, then we have
\[
\left|x_s-y_s\right|\leq \alpha\left\|Fx-Fy\right\|\quad \forall s\in\Z_L
\]
where
\[
\alpha=\frac{\left\|C^{-1}\right\|}{\inf\limits_{\text{I}}\left|f'\right|}\leq \frac{1}{(1-\epsilon_f)\inf\limits_{\text{I}}\left|f'\right|}<1
\]
due to the assumption $\|\text{Id}-C\|<\epsilon_{f}$ (see Appendix \ref{A-ERRATUM}). If the configurations $x$ and $y$ belong to the same cylinder set $\text{\bf I}_{\{\theta^t\}_{t=0}^{T-1}}^L$, then the argument can be repeated to obtain
\[
\left|x^t_s-y^t_s\right|\leq \alpha^{T-t}\left\|x^T-y^T\right\|\quad \forall 0\leq t<T,\ s\in\Z_L
\]
In particular, we have $\left|x^t_s-y^t_s\right|\leq \alpha^{T-t}M$ where $M=2\max\limits_{\text{I}}|f|$.
Furthermore, $f$ is supposed to be of class $C^2$ on each interval $\text{I}_i$. It follows that
\begin{equation}
\left|\log\left|f'(x)\right|-\log\left|f'(y)\right|\right|\leq\max\limits_{z\in \text{I}}\left|\frac{f''(z)}{f'(z)}\right| |x-y|\quad \forall x,y\in \text{I}_i
\label{ACCROIFIN}
\end{equation}
By combining the last two inequalities, we get
\[
\left|\log\left|\frac{f'(x_s^t)}{f'(y_s^t)}\right|\right|\leq \beta M \alpha^{T-t}\quad \forall 0\leq t<T,\ s\in\Z_L,\ x,y\in \text{\bf I}_{\{\theta^t\}_{t=0}^{T-1}}^L
\]
where $\beta=\max\limits_{z\in \text{I}}\left|\frac{f''(z)}{f'(z)}\right|$. By summing over $t$ from 0 to $T-1$, the inequality (\ref{DISTORSION}) then easily follows with $c_1=\exp\frac{\alpha\beta M}{1-\alpha}$.
\bigskip

Now, the inequality (\ref{DISTORSION}) implies that for all $x\in \text{\bf I}_{\{\theta^t\}_{t=0}^{T}}^L$, we have
\[
c_1^{-L}\sup_{x\in \text{\bf I}_{\{\theta^t\}_{t=0}^{T-1}}^L}\prod_{0\leq t<T\atop s\in\Z_L}\left|f'(x_s^t)\right|^{-1}\leq \prod_{0\leq t<T\atop s\in\Z_L}\left|f'(x_s^t)\right|^{-1},
\]
which yields
\[
c_1^{-L}\left|C\right|_L^{-T}\text{Vol}(\text{\bf I}_{\theta^{T}}^L)\sup_{x\in \text{\bf I}_{\{\theta^t\}_{t=0}^{T-1}}^L}\prod_{0\leq t<T\atop s\in\Z_L}\left|f'(x_s^t)\right|^{-1}\leq \text{Vol}(\text{\bf I}_{\{\theta^t\}_{t=0}^{T}}^L).
\]
On the other hand, we obviously have
\[
\prod_{0\leq t<T\atop s\in\Z_L}\left|f'(x_s^t)\right|^{-1}\leq \sup_{x\in \text{\bf I}_{\{\theta^t\}_{t=0}^{T-1}}^L}\prod_{0\leq t<T\atop s\in\Z_L}\left|f'(x_s^t)\right|^{-1},
\]
which implies
\[
\text{Vol}(\text{\bf I}_{\{\theta^t\}_{t=0}^{T}}^L)\leq \left|C\right|_L^{-T}\text{Vol}(\text{\bf I}_{\theta^{T}}^L)\sup_{x\in \text{\bf I}_{\{\theta^t\}_{t=0}^{T-1}}^L}\prod_{0\leq t<T\atop s\in\Z_L}\left|f'(x_s^t)\right|^{-1}
\]
The Lemma then immediately follows with $\underline{c}=\frac{1}{c_1}\min\limits_i\left|\text{I}_i\right|$ and $\overline{c}=\max\limits_i\left|\text{I}_i\right|$.
\end{proof}
\bigskip

\paragraph{Third step: Existence of the escape rate $\gamma_L(U)$.}
To begin, notice that the condition $x\in\text{\bf I}_{\{\theta^t\}_{t=0}^{T_1+T_2-1}}^L$ implies $x^{T_1}\in \text{\bf I}_{\{\theta^{t+T_1}\}_{t=0}^{T_2-1}}^L$. Therefore, for every triple $L,T_1,T_2\geq 1$, we have
\begin{align*}
Z_{L,T_1+T_2}\leq&\sum\limits_{\{\theta^t\}_{t=0}^{T_1+T_2-1}\in\Omega_{T_1+T_2}^{\Z_L}}\left(\sup\limits_{x\in \text{\bf I}_{\{\theta^t\}_{t=0}^{T_1-1}}^L}\prod_{0\leq t<T_1\atop s\in\Z_L}\left|f'(x_s^t)\right|^{-1}\right.\\
&\left.\times\sup\limits_{x\in \text{\bf I}_{\{\theta^{t+T_1}\}_{t=0}^{T_2-1}}^L}\prod_{0\leq t<T_2\atop s\in\Z_L}\left|f'(x_s^{t+T_1})\right|^{-1}\right)\\
\leq&\sum\limits_{\{\theta^t\}_{t=0}^{T_1-1}\in\Omega_{T_1}^{\Z_L}}
\sup\limits_{x\in \text{\bf I}_{\{\theta^t\}_{t=0}^{T_1-1}}^L}
\prod_{0\leq t<T_1\atop s\in\Z_L}\left|f'(x_s^t)\right|^{-1}\\
&\times\sum\limits_{\{\theta^t\}_{t=0}^{T_2-1}\in\Omega_{T_2}^{\Z_L}}
\sup\limits_{x\in \text{\bf I}_{\{\theta^t\}_{t=0}^{T_2-1}}^L}
\prod_{0\leq t<T_2\atop s\in\Z_L}\left|f'(x_s^t)\right|^{-1}\\
=&Z_{L,T_1}Z_{L,T_2}
\end{align*}
This property implies that the sequence $\{\log Z_{L,T}\}_{T\geq 1}$ is sub-additive; thus the following exists
\[
\lim\limits_{T\to\infty}\frac{1}{T}\log Z_{L,T}=\inf\limits_{T\geq 1}\frac{1}{T}\log Z_{L,T}
\]
which, together with Lemma \ref{BOUNDS}, proves the existence of $\lim\limits_{T\to\infty}\frac{1}{T}\log\text{Vol}(U_L^T)$.
\bigskip

\paragraph{Fourth step: Analysis of the asymptotic behaviour of $\gamma_L(U)$.}
Since for $\|\text{Id}-C\|<\epsilon_f$, the symbolic dynamics is given by the direct product $\Omega^\Z$, the spatial product in the definition of $Z_{L,T}$ can be decomposed into two independent spatial products, namely
\begin{align*}
Z_{L_1+L_2,T}\leq&\sum\limits_{\{\theta^t_1\}_{t=0}^{T-1}\in\Omega_{T}^{\Z_{L_1}}}\sup\limits_{x\in \text{\bf I}_{\{(\theta_1\theta_2)^t\}_{t=0}^{T-1}}^{L_1+L_2}}\prod_{0\leq t<T\atop 1\leq s\leq L_1}\left|f'(x_s^t)\right|^{-1}\\
&\times\sum\limits_{\{\theta^t_2\}_{t=0}^{T-1}\in\Omega_{T}^{\Z_{L_2}}}\sup\limits_{x\in \text{\bf I}_{\{(\theta_1\theta_2)^t\}_{t=0}^{T-1}}^{L_1+L_2}}\prod_{0\leq t<T\atop L_1+1\leq s\leq L_1+L-2}\left|f'(x_s^t)\right|^{-1}
\end{align*}
where $(\theta_1\theta_2)^t$ denotes the spatial concatenation of the words $\theta_1^t$ et $\theta_2^t$ (which have spatial length $L_1$ and $L_2$ respectively). Accordingly, a relationship has to be established between the supremum involved in the first term of the RHS and the supremum in the definition of $Z_{L_1,T}$, and similarly for the second term with $Z_{L_2,T}$. To that goal we are going to use the following inequalities (which are consequences of (\ref{ACCROIFIN}))
\[
\exp-\beta\sum_{0\leq t<T\atop 1\leq s\leq L}\left|x_s^t-y_s^t\right|\leq
\prod_{0\leq t<T\atop 1\leq s\leq L}\frac{\left|f'(x_s^t)\right|}{\left|f'(y_s^t)\right|}\leq
\exp\beta\sum_{0\leq t<T\atop 1\leq s\leq L}\left|x_s^t-y_s^t\right|
\]
where $x_s^t,y_s^t\in \text{I}_{\theta_s^t}$. It remains to control the double sum involved in these inequalities. This is the scope of the following statement whose proof is given in Appendix \ref{A-MAJOSUM}.
\begin{Lem}
For every $L,T\geq 1$ there exists $\Sigma_{L,T}>0$ with the following property
\[
\sup\limits_{L\geq 1}\lim\limits_{T\to\infty}\frac{1}{T}\Sigma_{L,T}<\infty
\]
such that for every admissible word $\{\theta^t\}_{t=0}^{T-1}\in\Omega_{T}^{\Z_L}$ and every configuration $x,y\in {\displaystyle\bigcap_{t=0}^{T-1}}F^{-t}\text{\bf I}$ with $x_s^t,y_s^t\in \text{I}_{\theta_s^t}$ for all $0\leq t<T$ and $1\leq s\leq L$, we have
\[
\sum_{0\leq t<T\atop 1\leq s\leq L}\left|x_s^t-y_s^t\right|\leq \Sigma_{L,T}
\]
\label{MAJOSUM}
\end{Lem}
When combined with the upper bound above for $Z_{L_1+L_2,T}$, this Lemma implies the following inequality
\[
Z_{L_1+L_2,T}\leq Z_{L_1,T}Z_{L_2,T}e^{\beta(\Sigma_{L_1,T}+\Sigma_{L_2,T})}
\]
Consequently, the limit $\lim\limits_{T\to\infty}\frac{1}{T}\log Z_{L,T}$ is a sub-additive sequence of $L$ up to a constant. Proposition 9.6.4 in \cite{KH95} implies the existence of the double limit $\lim\limits\limits_{L\to\infty}\frac{1}{L}\lim\limits_{T\to\infty}\frac{1}{T}\log Z_{L,T}$, which concludes the proof of Theorem \ref{THMTAUXFUIT}.

\noindent
Of note, the method of proof in Appendix \ref{A-MAJOSUM} does not allow to guarantee that
\[
\lim\limits_{T\to\infty}\frac{1}{T}\lim\limits_{L\to\infty}\frac{1}{L}\Sigma_{L,T}=0
\]
from which commutation of space and time limits would follow in the definition of $\gamma_\infty$.

\appendix
\section{Proof of Pianigiani-Yorke conditions for weakly coupled map lattices}\label{A-ERRATUM}
We want to prove that, when $\|\text{Id}-C\|$ is sufficiently small (depending only on $f$), the symbolic dynamics remains unaffected and the conditions (i) and (ii) hold for the CML $C\circ F_0$ with infinite collection $\{\text{\bf I}_{\theta}\}_{\theta\in {\mathcal A}^\Z}$. The transitivity condition (iii) then naturally follows from the assumption that $f$ is transitive.

The condition (ii) implies that the cylinder set $\bigcap\limits_{t=0}^{T} F^{-t}\text{\bf I}_{\theta^t}$ is non-empty for every admissible word $\{\theta^t\}_{t=0}^{T}\in\Omega_{T+1}^\Z$. The condition (i) guarantees that its diameter goes to 0 as $T\to\infty$. Hence $\bigcap\limits_{t=0}^{\infty} F^{-t}\text{\bf I}_{\theta^t}$ consists of a single point. The other properties in Theorem \ref{STABSTRUCT} are easily shown and left to the reader.
\bigskip

\paragraph{Proof of the condition (i).} The map $f$ being $C^2$ on $\text{I}$ implies that $F_0$ and thus $C\circ F_0$ are $C^2$ on $\text{\bf I}$. Moreover, if $\|\text{Id}-C\|\leq \epsilon<1$, then the inverse $C^{-1}$ exists and we have
\[
\left\|C^{-1}\right\|\leq \sum_{k\in\N}\|\text{Id}-C\|^k\leq\frac{1}{1-\epsilon},
\]
and thus $\|Cu\|\geq (1-\epsilon)\|u\|$ for all $u\in\ell^\infty(\Z)$. It follows that the norm $\|(C\circ F_0)'\|$ of the derivative is bounded below on $\text{\bf I}$ by $(1-\epsilon)\inf\limits_{\text{I}}\left|f'\right|$, a bound which is larger than 1 when $\epsilon$ is sufficiently small.
\bigskip

\paragraph{Proof of the condition (ii).} As before, we assume that $\|\text{Id}-C\|\leq\epsilon<1$.
We need to prove that, if $\epsilon$ is small enough, the cylinders $\text{\bf I}_\theta$
and their images satisfy the same dichotomy as in the uncoupled case:
for any $\theta,\theta'\in {\mathcal A}^\Z$,

(a) either there exists $s\in \Z$ such that
$\text{I}_{\theta'_s}\cap f\left(\text{I}_{\theta_s}\right)=\emptyset$, in which case
$$
\text{\bf I}_{\theta'}\cap F\text{\bf I}_\theta=\emptyset
$$

(b) or $\text{I}_{\theta'_s}\subset \text{Int}\; f\left(\text{I}_{\theta_s}\right)$ for all $s\in\Z$, in which case
$$
\text{\bf I}_{\theta'}\subset  \text{Int}\; F\text{\bf I}_\theta
$$

Let us first denote, for any $i\in\mathcal{A}$, $f(I_{i})=[\alpha_{i}, \beta_{i}]$.
We note also that, by compactness, we can quantify the properties of the local map $f$, {\it i.e.}
choose $\gamma>0$ and $\delta>0$ such that, for any $i,j\in\mathcal{A}$,
\begin{align*}
I_{j}\cap f(I_{i})=\emptyset \quad
& \Longrightarrow \quad I_{j}\cap [\alpha_{i}-\gamma,\beta_{i}+\gamma]=\emptyset\\
I_{j}\subset \text{Int}\; f\left(\text{I}_{\theta_s}\right) \quad
& \Longrightarrow \quad I_{j}\subset (\alpha_{i}+\delta,\beta_{i}-\delta)
\end{align*}

Let also $M=2\max\limits_{\text{I}}|f|<\infty$. Then, for any $x\in F_0\text{\bf I}$ and any $s\in \Z$, we have
$$
|x_{s}-(C(x))_{s}|\leq \|\text{Id}-C\| \|x\|\leq \epsilon\frac{M}2.
$$
This estimate directly implies the following inclusion as soon as $\epsilon\leq \frac{2\gamma}M$
$$
F\text{\bf I}_{\theta}=C\left(\bigotimes_{s\in\Z}[\alpha_{\theta_s},\beta_{\theta_s}]\right)\subset
\bigotimes_{s\in\Z}[\alpha_{\theta_s}-\gamma,\beta_{\theta_s}+\gamma].
$$
From this, one immediately gets that $\text{\bf I}_{\theta'}\cap F\text{\bf I}_\theta=\emptyset$
in case (a).

For the other case, let us first remark that, by the method of point (i), one has
$\|\text{Id}-C^{-1}\|\leq\frac\epsilon{1-\epsilon}$.
Hence just as before, as soon as
$\epsilon<\frac\delta{\delta+r} \Leftrightarrow\frac\epsilon{1-\varepsilon}<\frac\delta r$
and for any $\theta$ from case (b), one has
$$
C^{-1}\left(\bigotimes_{s\in\Z}[\alpha_{\theta_s}+\delta,\beta_{\theta_s}-\delta]\right)\subset
\bigotimes_{s\in\Z}[\alpha_{\theta_s},\beta_{\theta_s}],
$$
which implies
$$
\text{\bf I}_{\theta'}\subset  \text{Int}\; \bigotimes_{s\in\Z}[\alpha_{\theta_s}
+\delta,\beta_{\theta_s}-\delta] \subset \text{Int}\; C \left(\bigotimes_{s\in\Z}
[\alpha_{\theta_s},\beta_{\theta_s}]\right)=\text{Int}\; F\text{\bf I}_\theta
$$

\section{Proof of Lemma \ref{MAJOSUM}}\label{A-MAJOSUM}
Let $\iota={\big(\inf\limits_{\text{I}}|f'|\big)}^{-1}$, let again $M$ be the diameter of the smallest ball containing $F\text{\bf I}$. We assume that we assume that $\|\text{Id}-C\|< 1$. As a consequence, the inverse $C^{-1}$ exists and is a convolution operator \cite{AF00}. Let $\{c_n^{(-1)}\}_{n\in\Z}$ be the sequence representing $C^{-1}$. If $x,y\in\R^\Z$ are two configurations such that $x_s,y_s\in \text{I}_{\theta_s}$ for all $1\leq s\leq L$, we have
\begin{align*}
|x_s-y_s|\leq &\iota\sum\limits_{n\in\Z}\left|c_n^{(-1)}\right|\left|\left(Fx\right)_{s-n}-\left(Fy\right)_{s-n}\right|\\
\leq &\iota M\sum\limits_{n<s-L}\left|c_n^{(-1)}\right|+\iota\sum\limits_{n=s-L}^{s-1}\left|c_n^{(-1)}\right|\left|\left(Fx\right)_{s-n}-\left(F_x\right)_{s-n}\right|\\
&+\iota M\sum\limits_{n>s-1}\left|c_n^{(-1)}\right|
\end{align*}
Let $P_{[1,L]}$ be defined in $\R^\Z$ as the canonical projection
\[
\left(P_{[1,L]}x\right)_s=\left\{\begin{array}{cl}
x_s&\text{if}\ 1\leq s\leq L\\
0&\text{otherwise}
\end{array}\right.
\]
and consider the configurations $\overline{H}$ et $\underline{H}$ defined by
\[
\underline{H}_s=\left\{\begin{array}{cl}
0&\text{if}\ s\geq 0\\
M&\text{if}\ s< 0
\end{array}\right.
\quad\text{and}\quad
\overline{H}_s=\left\{\begin{array}{cl}
M&\text{if}\ s\geq 0\\
0&\text{if}\ s< 0
\end{array}\right.
\]
Consider also the operator ${\mathcal C}$ defined by
\[
({\mathcal C}x)_s=\iota\sum\limits_{n\in\Z}\left|c_n^{(-1)}\right|x_{s-n},\quad \forall s\in\Z,\ x\in\ell^\infty(\Z)
\]
and the translation operator defined by $(Rx)_s=x_{s-1}$. The inequality above can be rewritten under the following compact form
\[
\left|x_s-y_s\right|\leq \left({\mathcal C}P_{[1,L]}\left(\left|Fx-Fy\right|\right)\right)_s+\left({\mathcal C}R\left(\underline{H}+R^{L}\overline{H}\right)\right)_s
\]

\noindent
By repeating the argument for the iterates, it results that if $x,y\in\R^\Z$ are two configurations such that $x_s^t,y_s^t\in \text{I}_{\theta_s^t}$ for all $1\leq s\leq L$ and $0\leq t<T$, then we have
\[
|x_s^t-y_s^t|\leq \left({\mathcal C}P_{[1,L]})^{T-t}\left(\underline{H}+\overline{H}\right)\right)_s+\left(\sum\limits_{k=0}^{T-t-1}\left({\mathcal C}P_{[1,L]}\right)^k{\mathcal C}R\left(\underline{H}+R^{L}\overline{H}\right)\right)_s
\]
It follows that $\sum\limits_{0\leq t<T \atop 1\leq s\leq L}|x_s^t-y_s^t|\leq \Sigma'_{L,T}+\Sigma''_{L,T}$ where
\[
\Sigma'_{L,T}=\sum\limits_{0\leq t<T \atop 1\leq s\leq L}\left(\left({\mathcal C}P_{[1,L]}\right)^{T-t}\left(\underline{H}+\overline{H}\right)\right)_s
\]
and
\[
\Sigma''_{L,T}=\sum\limits_{1\leq t\leq T \atop 1\leq s\leq L}t\left(\left({\mathcal C}P_{[1,L]}\right)^{T-t}{\mathcal C}R\left(\underline{H}+R^{L}\overline{H}\right)\right)_s
\]
In particular, this implies the existence of an upper bound for the finite sums $\sum\limits_{0\leq t<T \atop 1\leq s\leq L}|x_s^t-y_s^t|$. It remains to prove the asymptotic behaviours.

The condition $\|\text{Id}-C\|<\epsilon_{f}$ implies that $\|{\mathcal C}\|<1$ and thus $\left\|{\mathcal C}P_{[1,L]}\right\|<1$. Writing the Neumann series for this operator yields the following bounds (where the second inequality follows from the fact that the entries of $\left({\mathcal C}P_{[1,L]}\right)^t$ are all positive).
\[
\Sigma'_{L,T}\leq \sum_{t\geq 1 \atop 1\leq s\leq L}\left(\left({\mathcal C}P_{[1,L]}\right)^{t}\left(\underline{H}+\overline{H}\right)\right)_s\leq M L\sum_{t\geq 1}\|{\mathcal C}\|^t\leq\frac{M L\|{\mathcal C}\|}{1-\|{\mathcal C}\|}
\]

Let us now study the limit $\lim\limits_{T\to\infty}\frac{\Sigma''_{L,T}}{T}$. A direct calculation shows that, for every $x\in [0,1)$, the following limit holds
\[
\lim\limits_{T\to\infty}\frac{1}{T}\sum\limits_{t=1}^{T}t x^{T-t}=\frac{1}{1-x}.
\]
This relation extends to operator components when the corresponding norm is smaller than 1. Consequently, using that $\left\|{\mathcal C}P_{[1,L]}\right\|<1$, we have
\begin{align*}
\lim\limits_{T\to\infty}\frac{1}{T}\sum\limits_{t=1}^{T}t\left(({\mathcal C}P_{[1,L]})^{T-t}{\mathcal C}R\left(\underline{H}+R^{L}\overline{H}\right)\right)_s=&\left|\left(\left(\text{Id}-{\mathcal C}P_{[1,L]})^{-1}{\mathcal C}R(\underline{H}\right.\right.\right.\\
&+\left.\left.\left. R^{L}\overline{H}\right)\right)_s\right|
\end{align*}
The assumption that $C$ is of finite range in Theorem \ref{THMTAUXFUIT} implies that the response $(C^{-1}\Delta)_n=c_n^{(-1)}$ (where $\Delta_s=\delta_{s,0}$) is exponentially localized around 0, {\sl i.e.}\ there exist $m_1>0$ and $0<\zeta_1<1$ such that \cite{BM97}
\[
\iota|c_n^{(-1)}|\leq m_1\zeta_1^{|n|},\quad \forall n\in\Z
\]
Using that any convolution operator commutes with pointwise limits of boun\-ded sequences in $\ell^\infty(\Z)$ \cite{AF00}, it follows that
\[
({\mathcal C}R\underline{H})_s\leq Mm_1\sum\limits_{n\geq s}\zeta_1^{|n|}\quad\text{and}\quad
({\mathcal C}R^{L+1}\overline{H})_s\leq Mm_1\sum\limits_{n\leq s-L-1}\zeta_1^{|n|},\quad \forall s\in\Z
\]

\noindent
Furthermore, the operator $\text{Id}-{\mathcal C}P_{[1,L]}$ is accordingly of finite range for every $L>1$ (when the finite range property is extended to bounded linear operators $A$ of the form
\[
(A x)_s=\sum\limits_{n\in\Z}a_{n,s}x_{s-n},\quad \forall s\in\Z
\]
{\sl i.e.}\ $\sup\limits_{s\in\Z}\sum\limits_{n\in\Z}\zeta^{|n|}|a_{n,s}|<\infty$ for some $\zeta>1$.) The associated decay rate $\zeta_{C}$ is identical to that of $C^{-1}$ and is thus independent of $L$. Using again \cite{BM97}, it results that, for every $n\in\Z$, the response $\left(\text{Id}-{\mathcal C}P_{[1,L]}\right)^{-1}R^n\Delta$ is exponentially localized around $n$. Moreover the decay rate is independent of $L$ because it is controlled by $\zeta_{C}$ and $\frac1{1-\|{\mathcal C}\|}$). Hence there exist $m_2>0$ and $0<\zeta_2<1$ such that for all $n\in\Z$, we have
\[
\left|\left(\left(\text{Id}-{\mathcal C}P_{[1,L]}\right)^{-1}R^n\Delta\right)_s\right|\leq m_2\zeta_2^{|n-s|},\quad\forall s\in\Z,L>1.
\]
Furthermore by using Neumann series and the decomposition $x=\sum\limits_{s\in\Z}x_sR^s\Delta$ for a configuration $x=\left\{x_s\right\}_{s\in\Z}\in\ell^\infty(\Z)$, one shows that
\[
\left(\left(\text{Id}-{\mathcal C}P_{[1,L]}\right)^{-1}{\mathcal C}R\underline{H}\right)_s=\sum\limits_{n\in\Z}({\mathcal C}R\underline{H})_{n}\left(\left(\text{Id}-{\mathcal C}P_{[1,L]}\right)^{-1}R^n\Delta\right)_s,\quad \forall s\in\Z
\]

\noindent
Combining the previous estimates leads to
\begin{align*}
\left|\left(\left(\text{Id}-{\mathcal C}P_{[1,L]}\right)^{-1}{\mathcal C}R\underline{H}\right)_s\right|\leq&Mm_1m_2\sum\limits_{n\in\Z}\zeta_2^{|n|}\sum\limits_{m\geq s-n}\zeta_1^{|m|}\\
\leq&\frac{Mm_1m_2}{1-\zeta_1}\left(\frac{\zeta_1^s}{1-\zeta_1\zeta_2}+\zeta_2^{s}(\frac{1+\zeta_1}{1-\zeta_2}+\frac{\zeta_1^2}{1-\zeta_1\zeta_2})\right.\\
&+\left.\zeta_2\frac{\zeta_1^s-\zeta_2^{s}}{\zeta_1-\zeta_2}\right),
\end{align*}
for all $s\geq 1$; and similarly
\begin{align*}
\left|\left(\left(\text{Id}-{\mathcal C}P_{[1,L]}\right)^{-1}{\mathcal C}R^{L+1}\overline{H}\right)_s\right|&\leq\frac{Mm_1m_2}{1-\zeta_1}\left(\frac{\zeta_1^{L+1-s}}{1-\zeta_1\zeta_2}\right.\\
&+\left.\zeta_2^{L+2-s}(\frac{1+\zeta_1}{1-\zeta_2}+\frac{\zeta_1^2}{1-\zeta_1\zeta_2})\right.\\
&+\left.\frac{\zeta_1^{L+2-s}-\zeta_2^{L+2-s}}{\zeta_1-\zeta_2}\right),
\end{align*}
for all $s\leq L$. Together with the upper bound above on $\Sigma'_{L,T}$, by summing over $s$, we finally conclude that
\[
\sup\limits_{L\geq 1}\lim\limits_{T\to\infty}\frac{1}{T}\Sigma_{L,T}<\infty.
\]

\noindent
{\bf Acknowledgments}

\noindent
BF would like to thank the Courant Institute (New York University) for hospitality during the paper preparation. His support in this phase was partly provided by CNRS and by the EU Marie Curie fellowship PIOF-GA-2009-235741.


TITLE: Continuity of a function theory
QUESTION [0 upvotes]: If $f: \mathbb{R}^2 \to \mathbb{R}$ and $ \lim_{(x,y)\rightarrow (0,0)}f(x,y)=0  $
then $f$ is continuous at $(0,0)$?
My teacher said this was true but if you take a function by parts this can be false right?

REPLY [0 votes]: The limit $\lim_{(x,y) \to (0,0)} f(x,y) = 0$ means that
$\forall \epsilon > 0: \exists \delta >0 : d((x,y),(0,0)) < \delta \Rightarrow d(f(x,y),0) < \epsilon$. Presumably we have $f(0,0) = 0$, then indeed $f$ is continuous at $(0,0)$. This is simply the definition of being continuous in a point.
Note that this necessarily implies that
$$
\lim_{y \to 0} f(0,y) =  \lim_{x \to 0} f(x,0) =0.
$$
But the converse doesn't hold as you mention (or at least that's how I interpreted it), meaning that a function can satisfy the above condition but still be discontinuous in $(0,0)$. Consider for example the following function
$$
f(x,y) = \left\{ \begin{matrix} \frac{xy}{x^2 +y^2} & \text{ if } (x,y) \neq (0,0) \\
0 & \text{ if } (x,y) = (0,0)\end{matrix}\right.
$$
In this case we obviously have
$$
\lim_{y \to 0} f(0,y) =  \lim_{x \to 0} f(x,0) =0.
$$
But if we approach $(0,0)$ from the straight line $y=x$, so
$$
\lim_{x \to 0} f(x,x) = \lim_{x \to 0} \frac{x^2}{2x^2} = \frac{1}{2}.
$$
Thus the function $f$ isn't continuous at $0$.


text:
\begin{document}
\maketitle
\baselineskip 18pt
\begin{abstract}
We prove the uniqueness in determining both orders of
fractional time derivatives and spatial derivatives in diffusion equations
by pointwise data.  The proof relies on the eigenfunction expansion and
the asymptotics of the Mittag-Leffler function.
\end{abstract}

\baselineskip 18pt

\noindent

\section{Introduction and main result}

Let $\OOO \subset \R^d$ be a bounded domain with smooth boundary, and
let $\alpha \in (0,2) \setminus \{ 1\}$ and $0<\beta < 1$.
For suitably given $a \in L^2(\OOO)$, let $u_{\alpha,\beta}
= u_{\alpha,\beta}(x,t)$ satisfy
$$\left\{ \begin{array}{rl}
& \pppa u = -A^{\beta} u, \quad x \in \OOO, \, t>0,\\
& u\vert_{\ppp\OOO} = 0, \qquad t>0, \\
& \left\{\begin{array}{rl}
  &u(x,0) = a(x), \quad x \in \OOO \quad \mbox{if $0<\alpha<1$},\\
  & u(x,0) = a(x), \quad \ppp_tu(x,0) = 0 \quad \mbox{if $1<\alpha<2$}.
  \end{array}\right.
\end{array}\right.                     \eqno{(1.1)}
$$
Here $\ppp_t^{\alpha}$ denotes the Caputo derivative:
$$
\ppp_t^{\alpha}v(t) = \frac{1}{\Gamma(n-\alpha)}\int^t_0 (t-s)^{n-\alpha-1}
\frac{d^nv}{ds^n}(s) ds \quad \mbox{for $n-1<\alpha<n$ with $n\in \N$}.
$$
Moreover we set
$$
Au(x) = -\sum_{i,j=1}^d \ppp_i(a_{ij}(x)\ppp_ju(x)) - c(x)u(x), \quad
x\in \OOO
$$
where $a_{ij} = a_{ji} \in C^1(\ooo{\OOO})$ for $1\le i,j\le n$ and
$c \in C(\ooo{\OOO})$, $\le 0$ in $\OOO$, and there exists a constant
$\mu>0$ such that
$$
\sum_{i,j=1}^d a_{ij}(x)\xi_i\xi_j \ge \mu\sum_{i=1}^d \xi_i^2,
\quad x\in \ooo{\OOO}, \, \xi_1, ...., \xi_d \in \R.
$$
To $A$, we attach the domain of the operator:
$$
D(A) = H^2(\OOO) \cap H^1_0(\OOO),
$$
where $H^2(\OOO)$ and $H^1_0(\OOO)$ are usual Sobolev spaces.

By the symmetry and positivity of $A$, we can
define the fractional power $A^{\beta}$ of $A$ (e.g., Pazy \cite{Pa},
Tanabe \cite{Ta}).  We can interpret $2\beta$ as the order of
the spatial fractional derivative because the differential oprerator $A$
is of second order.
A special case $\beta=1$ in (1.1) describes a time-fractional diffusion
equation, and is a macroscopic model for the continuous-time random walk,
which is a meaningful model e.g., for anomalous diffusion in heterogeneous
media.  As for phyiscal backgrounds, see Metzler and Klafter \cite{MK} for
example.

There has been the rapidly increasing literature on mathematical
works for initial boundary value problem (1.1) for time fractional diffusion
equations and we refer only to a very limited number of articles:
Gorenflo, Luchko and Yamamoto \cite{GLY}, Kubica, Ryszewska and Yamamoto
\cite{KRY}, Kubica and Yamamoto \cite{KY}, Sakamoto and Yamamoto
\cite{SY}, Zacher \cite{Za}.

For analyzing such an initial boundary value problem (1.1), we need to
discuss inverse problems which are concerned with a quantitative
procedure for the determination of parameters in (1.1) such as
$\alpha$, $\beta$, $a_{ij}(x)$, etc.

In particular, the orders $\alpha$ and $\beta$ are essential for modelling
the phenomena under consideration.
The main purpose of this article is to establish the uniqueness for
\\
{\bf Inverse problem.}\\
{\it Let $x_0 \in \OOO$ be given.  Determine $\alpha \in (0,2) \setminus
\{1\}$ and $\beta \in (0,1)$ by $u_{\alpha,\beta}(x_0,t)$, $0<t<T$.
}
\\

For the statement of our main result, we need to introduce the eigensystem of
the elliptic operator $A$ of the second order.
Let
$$
0<\la_1 < \la_2 < .... \longrightarrow \infty
$$
be the set of all the eigenvalues of $A$, and
let $\va_{kj}$, $1\le j\le m_k$ be an orthonormal system of eigenfunctions of
$A$ for $\la_k$, $k\in \N$: $A\va_{kj} = \la_k\va_{kj}$, $1\le j \le m_k$.

We further define the Mittag-Leffler function by
$$
E_{\alpha,1}(z) = \sum_{k=0}^{\infty} \frac{z^k}{\Gamma(\alpha k + 1)},
\quad z \in \C.
$$
It is known that $E_{\alpha,1}(z)$ is an entire function in $z \in \C$
(e.g., Podlubny \cite{Po}).

Now we are ready to state our main result.
\\
Thus we have established
\\
{\bf Theorem (uniqueness)}.\\
{\it Let $\gamma > \frac{d}{4}$.  We assume that
$$
a\ge 0, \not\equiv 0 \quad \mbox{or} \quad \le 0, \not\equiv 0 \quad
\mbox{in $\OOO$},
\quad a \in H^{2\gamma}_0(\OOO)            \eqno{(1.2)}
$$
and there exists $k_0\in \N$ such that
$$
\sum_{j=1}^{m_{k_0}} (a, \va_{k_0j})\va_{k_0j}(x_0) \ne 0, \quad \la_{k_0} \ne 1.                                                            \eqno{(1.3)}
$$
Then $u_{\alpha,\beta}(x_0,t)$ for $0<t<T$ uniquely determines
$\alpha \in (0,2)\setminus \{ 1\}$ and $\beta \in (0,1)$.
}
\\

Condition (1.3) is essential for the uniqueness of $\beta$.
Indeed, for simplicity, let each $\la_k$ be a simple eigenvalue:
$m_k=1$ for $k \in \N$.  We write $\va_k := \va_{km_k} = \va_{k1}$
for $k \in \N$.  Let $\la_1 = 1$ and $a=\va_1$.  Then $A^{\beta}a
= a$ for any $\beta>0$.
Then we can readily verify that $v(x,t) = E_{\alpha,1}(-t^{\alpha})
\va_1(x)$ satisfies (1.1) with $a = \va_1$: $u_{\alpha,\beta}(x,t)
= E_{\alpha,1}(-t^{\alpha})\va_1(x)$.
Therefore we cannot determine $\beta$ by $u_{\alpha,\beta}(x_0,t)
= E_{\alpha,1}(-t^{\alpha})\va_1(x_0)$, because the solution is
independent of $\beta$.

The condition (1.3) is satisfied if $a(x_0) \ne 0$ and there exists an
eigenvalue not $1$.

Indeed we have
$$
a(x_0) = \sum_{k=1}^{\infty} \sum_{j=1}^{m_k} (a,\va_{kj})\va_{kj}(x_0).
$$
If $\sum_{j=1}^{m_k} (a,\va_{kj})\va_{kj}(x_0) = 0$ for all $k\in \N$, then
$a(x_0) = 0$.

For the uniqueness for $\alpha$, as is seen in the proof, we need not
(1.2) itself and the uniqueness follows from $(A^{-\beta}a)(x_0) \ne 0$.
\\

As for inverse problems of determining order $\alpha$, we
explain existing related works.
Hatano, Nakagawa, Wang and Yamamoto \cite{HNWY} is an early theoretical
work and we further refer to Ashurov and Umarov \cite{AU}, Janno \cite{J},
Janno and Kinash \cite{JK}, Krasnoschok, Pereverzyev, Siryk and Vasylyeva
\cite{KPSV}, Li and Yamamoto \cite{LY},
Yu, Jing and Qi \cite{YJQ}.  Moreover see a survey chapter Li, Liu and
Yamamoto \cite{LiLiuY}.
In the case of $\OOO = (0,1)$ and $A = -\Delta$ in (1.1),
the article Tatar and Ulusoy \cite{TU}
proves the uniqueness in determining $\alpha$ and $\beta$
simultaneously by $u(x_0,t)$, $0<t<T$ with fixed $x_0 \in (0,1)$
under the assumption
$$
(a,\va_k)_{L^2(0,1)} > 0 \quad \mbox{for all $k\in \N$}.
                                    \eqno{(1.4)}
$$
Here we note that in the one-dimensional $\OOO$, all the eigenvalues
are simple: $m_k=1$ and we write $\va_{km_k} = \va_k$.
The condition (1.4) requires the non-vansihing of all the Fourier coefficients
and is essentially different from our conditions (1.2) - (1.3),
\\

In addition to the inverse problems of determining orders, there are
very many works on other types of inverse problems for fractional equations,
and we here refer to very limited related articles:
Cheng, Nakagawa, Yamamoto and Yamazaki \cite{CNYY},
Li, Zhang, Jia and Yamamoto \cite{LZJY}, Li, Luchko and Yamamoto \cite{LLY},
Ruan, Zhang and Wang \cite{RZW},
Rundell and Zhang \cite{RZ},
Sun, Li and Jia \cite{SLJ}, Zhang, Cheng and Wang \cite{ZCW},
two survey papers Li and Yamamoto \cite{LiY}, Liu, Li and Yamamoto
\cite{LiuLiY}.

This article is composed of three sections.  In Section 2, we will prove the
theorem.
Section 3 is devoted to concluding remarks.

\section{Proof of Theorem}

{\bf First Step.}\\
In terms of the eigenvalues and the eigenfunctions, we can represent
$$
u_{\alpha,\beta}(x,t) = \sumk E_{\alpha,1}(-\la_k^{\beta}t^{\alpha})
\sum_{j=1}^{m_k} (a,\va_{kj})\va_{kj}(x), \quad x\in \OOO, \,
t>0
$$
pointwise under the regularity condition of $a$ in the theorem.
More precisely, we have
$$
\Vert A^{\gamma}u_{\alpha,\beta}(\cdot,t)\Vert_{L^2(\OOO)}
\le C\Vert A^{\gamma}a\Vert_{L^2(\OOO)}
$$
and so we see that
$u_{\alpha,\beta}(\cdot,t) \in C(\ooo{\OOO})$ by the
Sobolev embedding and $4\gamma > d$.

We assume that
$$
u_{\alpha,\beta}(x_0,t) = u_{\alpha_1,\beta_1}(x_0,t), \quad
0<t<T,
$$
for $\alpha, \alpha_1 \in (0,2) \setminus \{1\}$ and
$\beta, \beta_1 \in (0,1)$.
Substituting $x=x_0$, we assume
$$
\sumk E_{\alpha,1}(-\la_k^{\beta}t^{\alpha})
\sum_{j=1}^{m_k} (a,\va_{kj})\va_{kj}(x_0)
= \sumk E_{\alpha_1,1}(-\la_k^{\beta_1}t^{\alpha_1})
\sum_{j=1}^{m_k} (a,\va_{kj})\va_{kj}(x_0), \quad t>0.

$$
By the asymptotics of $\EEE(-z)$ with large $z>0$
(Theorem 1.4: pp.33-34 in Podlubny \cite{Po}), we obtain
$$
u_{\alpha,\beta}(x,t)
= \sumk \sum_{j=1}^{m_k} \frac{(a,\va_{kj})\va_{kj}(x)}
{\Gamma(1-\alpha)\la_k^{\beta}t^{\alpha}}
+ O\left( \frac{1}{t^{2\alpha}}\right)
$$
$$
= \frac{1}{\Gamma(1-\alpha)t^{\alpha}}(A^{-\beta}a)(x)
+ O\left( \frac{1}{t^{2\alpha}}\right) \quad \mbox{as $t \to \infty$.}
                                                                \eqno{(2.1)}
$$
We note that the fractional power $A^{-\beta}$ is defined by
$$
A^{-\beta}a = \frac{\sin \pi\beta}{\pi}
\int^{\infty}_0 \eta^{-\beta}(A+\eta)^{-1}a d\eta
\quad \mbox{in $L^2(\OOO)$}, \quad 0<\beta<1                  \eqno{(2.2)}
$$
(e.g., Pazy \cite{Pa}, Tanabe \cite{Ta}).

{\bf Second Step.}\\
We here prove
\\
{\bf Lemma.}\\
{\it Let $a\ge 0$, $\not\equiv 0$ in $\OOO$ and $a \in H^{2\gamma}_0(\OOO)$
with $\gamma > \frac{d}{4}$.
Then
$$
(A^{-\beta}a)(x) > 0, \quad x\in \OOO.
$$
}
\\
{\bf Proof.}\\
By the strong maximum principle for $\Delta - \eta$ with $\eta\ge 0$, we see
$$
(A+\eta)^{-1}a(x) >  0, \quad x \in \OOO, \, \eta \ge 0.  \eqno{(2.3)}
$$
Indeed we set $w = (A+\eta)^{-1}a$, that is,
$$
\left\{ \begin{array}{rl}
& \sum_{i,j=1}^d \ppp_i(a_{ij}(x)\ppp_jw) - \eta w = -a \le 0
\quad \mbox{in $\OOO$}, \\
& w\vert_{\ppp\OOO} = 0.
\end{array}\right.
$$
The maximum principle (Corollary 4.3 in Renardy and Rogers \cite{RR},
p.106) yields
$$
\min_{\ooo{\OOO}} w \ge \min_{\ppp\OOO} \min\{ w, 0\} = 0
$$
and so
$$
w \ge 0 \quad \mbox{on $\ooo{\OOO}$}.             \eqno{(2.4)}
$$
The strong maximum principle (\cite{RR}, p.109) implies that
$$
\mbox{$w$ cannot achieve a non-positive minimum at any point in $\OOO$},
                                               \eqno{(2.5)}
$$
if $w$ is not a constant function.

Since $w\vert_{\ppp\OOO} = 0$, if $w$ is constant, then $w\equiv 0$, which is
impossible by $a \not\equiv 0$.  Therefore $w$ is not constant.
Therefore (2.5) holds.

By (2.4) and $w\vert_{\ppp\OOO} = 0$, we see that $\min_{\ooo{\OOO}} w = 0$.

Assume that there exists $x_0 \in \OOO$ such that $w(x_0) = 0$, that is,
$w$ attains a non-positive minimum (i.e., $0$) at $x_0\in \OOO$.
This is impossible by (2.5).  Thus (2.3) is verified.

By $a \in D(A^{\gamma}) \subset H^{2\gamma}(\OOO) \subset C(\ooo{\OOO})$
with $\gamma > \frac{d}{4}$,  by (2.2) we have
\begin{align*}
& A^{\gamma}\left(A^{-\beta}a
- \frac{\sin \pi\beta}{\pi}
\int^{\infty}_0 \eta^{-\beta}(A+\eta)^{-1}a d\eta\right) \\
=& A^{-\beta}(A^{\gamma}a)
- \frac{\sin \pi\beta}{\pi}
\int^{\infty}_0 \eta^{-\beta}(A+\eta)^{-1}(A^{\gamma}a) d\eta,
\end{align*}
and so (2.2) holds in $C(\ooo{\OOO})$.  We can substitute
$x=x_0$ in (2.2) to conclude that $(A^{-\beta}a)(x)>0$ for
all $x \in \OOO$.  Thus the proof of Lemma is complete.

{\bf Third Step.}\\
Since $u_{\alpha,\beta}(x_0,t) = u_{\alpha_1,\beta_1}(x_0,t)$ for
$0 < t < T$, the $t$-analyticity means
$u_{\alpha,\beta}(x_0,t) = u_{\alpha_1,\beta_1}(x_0,t)$ for $t>0$.
It follows from (2.1) that
$$
\frac{1}{\Gamma(1-\alpha)t^{\alpha}}(A^{-\beta}a)(x_0)
+ O\left( \frac{1}{t^{2\alpha}} \right)
= \frac{1}{\Gamma(1-\alpha_1)t^{\alpha_1}}(A^{-\beta_1}a)(x_0)
+ O\left( \frac{1}{t^{2\alpha_1}}\right)
$$
as $t \to \infty$.  By Lemma, we see that $p := (A^{-\beta}a)(x_0) \ne 0$ and
$p_1 := (A^{-\beta_1}a)(x_0) \ne 0$.
Let $\alpha < \alpha_1$.  Then
$$
\frac{p}{\Gamma(1-\alpha)} + O\left( \frac{1}{t^{\alpha}}\right)
= \frac{p_1}{\Gamma(1-\alpha_1)t^{\alpha_1-\alpha}}
+ O\left( \frac{1}{t^{2\alpha_1-\alpha}}\right)
$$
as $t \to \infty$.  Letting $t \to \infty$, by $\alpha_1>\alpha$, we
reach $\frac{p}{\Gamma(1-\alpha)} = 0$, which is a contradiction
against $p\ne 0$.
Therefore $\alpha_1 \le \alpha$.  Similarly we can prove $\alpha_1
\ge \alpha$.  Thus we can conclude $\alpha_1 = \alpha$.

{\bf Fourth Step.}\\
Finally we will prove $\beta = \beta_1$.
Since
$$
u_{\alpha,\beta}(x_0,t) = \sum_{k=1}^{\infty}\sum_{j=1}^{m_k}
(a,\va_{kj})\EEE(-\la_k^{\beta}t^{\alpha})\va_{kj}(x_0), \quad
0<t<T,
$$
is analytic in $t$ and convergent pointwise, again by the asymptotics
(e.g., Theorem 1.4 (pp.34-35) in \cite{Po}), for any $N\in \N$, we have
\begin{align*}
& \sum_{\ell=1}^N \frac{(-1)^{\ell+1}}{\Gamma(1-\alpha\ell)}
\frac{1}{t^{\alpha\ell}} \sum_{k=1}^{\infty} \frac{1}{\la_k^{\beta\ell}}
\sum_{j=1}^{m_k} (a,\va_{jk})\va_{jk}(x_0)
+ O\left( \frac{1}{t^{\alpha(N+1)}}\right) \\
= & \sum_{\ell=1}^N \frac{(-1)^{\ell+1}}{\Gamma(1-\alpha\ell)}
\frac{1}{t^{\alpha\ell}} \sum_{k=1}^{\infty} \frac{1}{\la_k^{\beta_1\ell}}
\sum_{j=1}^{m_k} (a,\va_{jk})\va_{jk}(x_0)
+ O\left( \frac{1}{t^{\alpha(N+1)}}\right)
\end{align*}
as $t \to \infty$.
For short descriptions, we set $a_k=
\sum_{j=1}^{m_k} (a,\va_{jk})\va_{jk}(x_0)$, $k\in \N$.
Then
\begin{align*}
& \sum_{\ell=1}^N \frac{(-1)^{\ell+1}}{\Gamma(1-\alpha\ell)}
\frac{1}{t^{\alpha\ell}} \sum_{k=1}^{\infty}
\frac{a_k}{\la_k^{\beta\ell}} + O\left( \frac{1}{t^{\alpha(N+1)}}\right)\\
=& \sum_{\ell=1}^N \frac{(-1)^{\ell+1}}{\Gamma(1-\alpha\ell)}
\frac{1}{t^{\alpha\ell}} \sum_{k=1}^{\infty}
\frac{a_k}{\la_k^{\beta_1\ell}} + O\left( \frac{1}{t^{\alpha(N+1)}}\right)
\end{align*}
as $t \to \infty$ for all $N\in \N$.
Since $\frac{1}{\Gamma(1-\alpha\ell)} > 0$ if
$1-\alpha\ell \not\in \{ 0,-1,-2,-3, ....\}$, we obtain
$$
\sum_{k=1}^{\infty} \frac{a_k}{\la_k^{\beta\ell}}
= \sum_{k=1}^{\infty} \frac{a_k}{\la_k^{\beta_1\ell}}, \quad
\ell \in \N \setminus \left\{ \frac{m}{\alpha} \right\}
_{m\in \N}.                       \eqno{(2.6)}
$$

Let $k_0$ be the minimum natural number satisfying (1.3):
$a_{k_0} \ne 0$.
Then
$$
\sum_{k=k_0}^{\infty} \frac{a_k}{\la_k^{\beta\ell}}
= \sum_{k=k_0}^{\infty} \frac{a_k}{\la_k^{\beta_1\ell}}, \quad
\ell \in \N \setminus \left\{ \frac{m}{\alpha} \right\}
_{m\in \N}.
$$
\\
{\bf Case 1: $\la_{k_0} > 1$.}\\
We assume that $\beta_1 > \beta$.
Multiplying (2.6) with $\la_{k_0}^{\ell\beta}$, we have
$$
a_{k_0}( 1 - (\la_{k_0}^{\beta-\beta_1})^{\ell})
+ \sum_{k=k_0+1}^{\infty} a_k \left( \left( \frac{\la_{k_0}^{\beta}}
{\la_k^{\beta}} \right)^{\ell}
- \left( \frac{\la_{k_0}^{\beta}}{\la_k^{\beta_1}} \right)^{\ell}\right)
= 0                                    \eqno{(2.7)}
$$
for $\ell \in \N \setminus \left\{ \frac{m}{\alpha} \right\}
_{m\in \N}$.
By $\beta - \beta_1 < 0$, $\la_{k_0} > 1$ and
$\left\vert \frac{\la_{k_0}}{\la_k}\right\vert < 1$ for $k \ge k_0+1$,
we have
$\vert \la_{k_0}^{\beta-\beta_1} \vert < 1$ for $k\in \N$ and
$\left\vert \left( \frac{\la_{k_0}}{\la_k} \right)^{\beta} \right\vert$,
$\left\vert \frac{\la_{k_0}^{\beta}}{\la_k^{\beta_1}} \right\vert < 1$ for
$k\ge k_0+1$. Then we choose a sequence $\ell_n
\in \N \setminus \left\{ \frac{m}{\alpha} \right\}_{m\in \N}$
such that $\lim_{n\to\infty} \ell_n = \infty$.

Indeed we assume that such a sequence $\ell_n$ does not exist.
Then there exists $N_0 \in \N$ such that
$$
\left\{ \frac{m}{\alpha} \right\}_{m\in \N} \supset \{N_0, N_0+1, N_0+2,
... \}.
$$
Therefore we can choose $m, m' \in \N$ such that $\frac{m}{\alpha}
= N_0$ and $\frac{m'}{\alpha} = N_0+1$.  Hence the subtraction
yields $\frac{m'-m}{\alpha}= 1$, that is, $\alpha=m'-m$.
This is a contradiction because $0<\alpha<2$ and $\alpha \ne 1$.

Then, applying the
Lebesgue convergence theorem for the series
to the second term on the left-hand side
of (2.7), we reach $a_{k_0} = 0$.  This is impossible and so
$\beta_1 \le \beta$.  Similarly we can prove $\beta_1 \ge \beta$.
Therefore we proved $\beta_1 = \beta$ in the case of $\la_{k_0} > 1$.
\\
{\bf Case 2: $\la_{k_0} < 1$.}

We assume that $\beta_1 > \beta$.
we multiplying (2.6) with $\la_{k_0}^{\ell\beta_1}$.
Then
$$
a_{k_0}\left( \left( \frac{\la_{k_0}^{\beta_1}}{\la_{k_0}^{\beta}}
\right)^{\ell}
- 1 \right)
+ \sum_{k=k_0+1}^{\infty} a_k \left( \left( \frac{\la_{k_0}^{\beta_1}}
{\la_k^{\beta}} \right)^{\ell}
- \left( \frac{\la_{k_0}^{\beta_1}}{\la_k^{\beta_1}} \right)^{\ell}\right)
= 0
$$
for $\ell \in \N \setminus \left\{ \frac{m}{\alpha} \right\}
_{m\in \N}$.
Since $\beta_1 - \beta > 0$, $\la_{k_0} < 1$ and $\la_k > \la_{k_0}$ for
$k \ge k_0+1$, we have
$$
\left\vert \frac{\la_{k_0}^{\beta_1}}{\la_{k_0}^{\beta}}\right\vert < 1,
\quad \left\vert \frac{\la_{k_0}^{\beta_1}}{\la_k^{\beta}} \right\vert,\,
\left\vert \frac{\la_{k_0}^{\beta_1}}{\la_k^{\beta_1}} \right\vert < 1
$$
for $k \ge k_0+1$.  Letting $\ell \to \infty$ within
$\N \setminus \left\{ \frac{m}{\alpha} \right\}_{m\in \N}$,
similarly to Case 1, we see that $-a_{k_0} = 0$,
which is impossible.  Therefore $\beta_1 \le \beta$.
Similarly we can prove $\beta_1 \ge \beta$.
Therefore we proved $\beta_1 = \beta$ in both Cases 1 and 2.
Thus the proof of Theorem is complete.

\section{Concluding remarks.}
\begin{itemize}
\item
We can discuss the inverse problem by data
$$
\int_{\OOO} u_{\alpha,\beta}(x,t) \rho(x) dx, \quad 0<t<T
$$
with suitable weight $\rho(x)$
or
$$
\frac{\ppp u_{\alpha,\beta}}{\ppp \nu_A}(x_0,t), \quad 0<t<T,
$$
where $x_0 \in \ppp\OOO$ is suitably given and
$\frac{\ppp v}{\ppp \nu_A}$ denotes the conormal derivative:
$\frac{\ppp v}{\ppp \nu_A}(x) = \sum_{i,j=1}^d a_{ij}(x)\nu_j(x)$
with the outward unit normal vector $(\nu_1(x), ..., \nu_d(x))$ to
$\ppp\OOO$.
\item
The uniqueness is the primary theoretical issue for the inverse problem.
The stability is also important but it is open so far.
\end{itemize}

\section*{Acknowledgments}
The author thanks Professor Takeshi Saito (The University of Tokyo)
for the improvement of the argument in the proof of Theorem.
The author was supported by Grant-in-Aid for Scientific Research (S)
15H05740 of Japan Society for the Promotion of Science and
by The National Natural Science Foundation of China
(no. 11771270, 91730303).
This work was prepared with the support of the "RUDN University Program 5-100".

config: arxiv


TITLE: Determining the value of the unknown variables
QUESTION [0 upvotes]: Let f(x) = (ax+sin2x)/3x  for  x>0
       a(x+1) for x=0

       ae^bx for x<0

Questions:
a) Determine all the values of a and b for which f is a continuous function
b)Determine whether there are values of a and b for which f is a differentiable function.
So using the properties of continuity by first equation the limit of x->0+ to x->0- based on the fact that the function is continuous , I found that the value of a is 1 then since the function is continuous , I can also conclude limit of x->0- will be equal to f(0) and hence computer the value of b as 0.
Hence based on that, I concluded that the for the corresponding values the given function is continuous. I know that continuity does not imply differentiability. How do I go about computing the answer for the second question? Any form of help or hints are appreciated.

REPLY [0 votes]: You need to calculate right derivative of your function, namely, of $\frac{ax+\sin(2x)}{3x}$ at $0$, and set it equal to the left derivative at $0$. You already know $a=1$. You need to calculate $\lim_{x\rightarrow0^{+}}\frac{\partial}{\partial x}\frac{x+\sin(2x)}{3x}$, for which I am getting $0$. The left derivative at $x=0$ equals $b\exp(b\cdot0)=b$.


TITLE: how to prove that $ \lim x_n^{y_n}=\lim x_n^{\lim y_n}$?
QUESTION [2 upvotes]: Let $(x_n)$ and $(y_n)$ be two sequences of real numbers, such that:
$\lim x_n = a > 0$ and $\lim y_n = b \in \mathbb R$
I need to prove that:
$$\lim \left( x_n ^{y_n} \right) = a^b$$
I tried using Cauchy's criterion:
$|x_n-a|<\epsilon$ and $|y_n-b|<\epsilon$, and then we need to prove that $\left| x_n ^{y_n} -a^b\right|<\epsilon$, but I didn't find a way of proving it.

REPLY [6 votes]: You just need to know how to prove that $x_n y_n \to xy$ if $x_n \to x$ and $y_n \to y$.
Next, $\exp$ and $\ln$ are continuous functions, so we have:
$$\lim x_n ^{y_n} = \lim e^{y_n \ln(x_n)} = e^{\lim y_n \ln(x_n)} = e^{b \ln(\lim x_n)} = e^{b\ln(a)} = a^b $$

REPLY [0 votes]: Hint: Begin trying to prove that $\lim_n\ln x_n=\ln a$. Then try to prove that $\lim_n (y_n \ln x_n)=b \ln a$. Finally prove that $\ln(\lim_n x_n)=\lim_n \ln x_n$.


TITLE: Complex integration of a function around a triangle.
QUESTION [0 upvotes]: Please help me how to solve this problem. I have no idea how to initiate the solution.

REPLY [2 votes]: $\gamma_1: z = t\\
\gamma_2: z = a + ti\\
\gamma_3: z = a - t + (a-t)i$
Take the derivatives to find $dz$ for each contour
$\int_0^a t \ dt + \int_0^a a (i \ dt) + \int_0^a (a-t)(-1-i\ dt)$


TITLE: Prove that $P(X \le n)-P(X+Y \le n)= \alpha P(X+Y=n)$
QUESTION [1 upvotes]: Let $X$ and $Y$ be mutually independent random variables taking non-negative integer values.
Prove that $P(X \le n)-P(X+Y \le n)= \alpha P(X+Y=n)$ holds for $n=0,1,2,...$ and for some $\alpha>0$ iff $P[Y=n]=\frac{1}{1+ \alpha}(\frac{\alpha}{1+ \alpha})^n , n=0,1,2,...$
I have done enough research on this problem and I have come to know that this is Madan L Puri's Characterization of Geometric DIstribution.
Can anyone provide me a proof?

REPLY [0 votes]: Hint Define $x_k = P(X = k)$ and $y_k = P(Y = k)$. Then write
$$P(X+Y\le n) = \sum_{i+j \le n} x_i y_j$$
and
$$P(X+Y = n) = \sum_{i+j = n} x_i y_j, \qquad P(X\le n) = \sum_{i=0}^n x_i$$
Then write the desired equality with $x_i$'s an $y_j$'s. On both sides, collect the terms in factor of a given $x_k$ and see if the geometric law for $Y$ gives the result.
I haven't checked the result but this is the strategy I would try.


TITLE: Maclaurin's series of a complex function
QUESTION [0 upvotes]: I'm attempting to express the following function (let $\xi$ be a parameter)
$$f(z) = \dfrac{1}{(1 - 2\xi z + z^2)^2} $$
So far, I have thought about getting the expression for $g(z) = \dfrac{1}{1 - 2\xi z + z^2} $ first, then multiply the series. However, I don't know how to effectively expand this in a neighbor of $0$, as I can only use the expansion $\dfrac{1}{1-a}$ when $|a| < 1$.
Is there any way to expand the series without taking derivatives? Thank you.

REPLY [2 votes]: Note that it's
$$ \frac{1}{2z} \frac{d}{d\xi} \frac{1}{1-2\xi+z^2}. $$
In fact we do have an expression for the latter fraction, although not really in closed form:
$$ \frac{1}{1-2\xi+z^2} = \sum_{n=0}^{\infty} U_n(\xi) z^n, $$
where $U_n$ is the Chebyshev polynomial of the second kind. So we find that
$$ \frac{1}{(1-2\xi z+z^2)^2} = \sum_{n=0}^{\infty} \frac{1}{2}U_{n+1}'(\xi) z^n. $$
These are also a special case of the ultraspherical/Gegenbauer polynomials, which have generating function $(1-2\xi z +z^2)^{-\alpha}$.


TITLE: Showing the convergence of the product for $ (1-\frac{1}{k})$
QUESTION [1 upvotes]: What I am trying to calculate is the following limit:
$$\lim\limits_{n \to \infty}\prod_{k=2}^{n}\left( 1-\frac{1}{k}\right).$$
I know that it approaches 0 as n approaches infinity, the problem I have is actually showing it. So far, I tried to express it as
$$\lim\limits_{n \to \infty}\prod_{k=2}^{n}\frac{k-1}{k}$$
hoping that this would simplify the problem, but it didn't help too much. Likewise, proving that it is monotonically increasing didn't either, in fact it made me doubt that it converges towards 0 for a bit. I tried around a few more possible solutions, but most of them are garbage that isn't worth the time to write down here, so I'd be thankful for any nudge into the right direction as to how to solve this.

REPLY [2 votes]: Hint:
$$
\prod_{k=2}^n \frac{k-1}{k}=\frac{1\cdot 2 \cdot 3 \cdot \cdots \cdot (n-1)}{2\cdot 3\cdot 4 \cdot \cdots \cdot n}=\frac{1}{n}
$$


TITLE: Diagonalization of Singular Matrix
QUESTION [1 upvotes]: If I have $n\times n$ matrix $M$ such that this matrix is singular i.e $\det(M) = 0$ and $M$ is symmetric i.e $M = \text{transpose}(M)$ but not diagonal. Now what we can say about the its eigenvalues, can we say that such matrix will not have $n$ distinct eigenvalues?
2nd thing if a matrix does not have $n$ distinct eigenvalues and matrix is not diagonal then can we say that such matrix will not have $n$ independent eigenvectors?
If above $2$ conditions holds can we say we cannot diagonalize such matrix?
Regards
Ahsan

REPLY [0 votes]: To your first question: every symmetric matrix (with real entries) is diagonalizable, and symmetric matrices can have real eigenvalues with any multiplicity.  A matrix is singular if and only if $0$ is one of its eigenvalues.
To your second question: consider as a counter-example
$$
\pmatrix{1&1&1\\1&1&1\\1&1&1}
$$
this matrix is diagonalizable, and it has eigenvalues $\{0,0,3\}$.


TITLE: Is it possible that $\sum n a_n^2$ converges but $\sum a_n$ diverges?
QUESTION [6 upvotes]: Yesterday I was thinking about a problem, when an interesting question appeared:
Does there exist a sequence $a_n \geq 0$ of non-negative real numbers such that $\sum_{n \geq 1} n a_n^2 < \infty$ and $\sum_{n \geq 1} a_n = \infty$?
After thinking about it for a while, I couldn't come up with either an example of this or a proof that this is impossible. Is this known (I didn't manage to find it online)?

Comments:
One thing I noticed is that, if $p > 1$ is fixed, we may assume without loss of generality that $$\frac{1}{n^p} <a_n < \frac{1}{n} \forall n.$$ The reason the following:
Assume there exists $a_n$ that satisfies the above two conditions, and let $S = \left\{n \in \mathbb{N}, n \geq 1 | a_n < \frac{1}{n}\right\}$. Then for every $n \geq 1, n \notin S$, we have $a_n \leq na_n^2$, thus $$\sum_{n \notin S} a_n \leq \sum_{n \geq 1} n a_n^2 < \infty $$
Therefore, since $\sum_{n \geq 1} a_n = \infty$, it follows that $\sum_{n \in S} a_n = \infty$. If we take $b_n = \begin{cases} a_n, &n \in S \\ 0, &n \notin S  \end{cases}$, $b_n$ is a non-negative sequence also satisfying the above two properties. Therefore, we may assume wlog that $a_n < \frac{1}{n} \forall n$.
Now, if $p>1$ is fixed we may also assume wlog that $a_n > \frac{1}{n^p}$, because if $a_n$ satisfies the two properties, then $$b_n = \begin{cases} a_n, &a_n > \frac{1}{n^p} \\ \frac{1}{n^{(p+1)/2}}, &a_n \leq \frac{1}{n^p}  \end{cases}$$
also satisfies them.

REPLY [9 votes]: You could probably have success with $a_n=\frac1{(n+1)\log(n+1)}$.
$$\sum_{n\ge 2}\frac1{n\log(n)^2}$$
converges by twice applying the Cauchy condensation test or the integral test.

REPLY [7 votes]: Let
$$a_n=\frac1{n\ln n}$$
then using the  integral test the series
$$\sum_n a_n$$
is divergent but the series
$$\sum_n na_n^2$$
is convergent.

REPLY [6 votes]: An example such that $\sum\limits_nna_n^2$ converges and $\sum\limits_na_n$ diverges: $a_n=\frac1{n\log n}$.


TITLE: Let G be group of order $125$. Will it have a subgroup of order $25$?
QUESTION [6 upvotes]: Let $G$ be a group of order $125$. Does it have a subgroup of order $25$ ? For any abelian group, the answer is yes. But what about non-abelian groups? I cannot understand. Can anyone help me?

REPLY [1 votes]: Here is a concrete variant of @egreg's answer, using the key fact that $Z(G)$ is not trivial.
If there is an element of order $125$ or $25$, then we're done.
Otherwise, all elements have order $1$ or $5$.
Let $a\in Z(G)$ with $a\ne e$ and $b \in G$ with $b\notin \langle a \rangle$.
Then the set of products $a^i b^j$ is a subgroup of order $25$.


text:
\subsection*{Differential Evolution}

Differential evolution~\cite{Storn_96, Shokrollahi_00}.

\subsubsection*{Initialization}

Conditioning values for $\pi_i$ and $\sigma_i$.

\begin{equation}
\sum_{i = 2}^{d_{s_{max}}} \pi_i = 1
\quad
\sum_{j = 2}^{d_{s_{max}}} \sigma_j = 1
\label{eq:pi-sigma-1}
\end{equation}

\begin{equation}
\pi_i + \sigma_i \leq 1
\label{eq:pi-sigma-2}
\end{equation}

\subsubsection*{Mutation}

$NP$ population size.

\begin{equation}
v_{i, G + 1} = p_{i, G} + F * (p_{r_1, G} - p_{r_2, G})
\end{equation}

\noindent where $r_1$ and $r_2$ are chosen randomly from $[1, NP]$.

\subsubsection*{Recombination}

\begin{equation}
u_{i, G + 1} = (u_{0i, G + 1}, u_{1i, G + 1}, ..., u_{Li, G + 1})
\end{equation}


text:
\section{Tables}\label{sec:tables}
In this section, we aggregate our computations of the Chern-Mather classes and local Euler obstructions.
The computations of the Euler obstructions in $\LG(4,8)$ rely on the recurrence relations obtained by Boe and Fu \cite{boe.fu} and have been checked using the results of this note.

\subsection{Schubert Varieties in Cominuscule spaces}

We recall some facts about diagrams indexing the Schubert varieties in cominuscule spaces.
Our main reference is \cite{BCMP:qkchev}.

Let $G/P$ be a cominuscule space
corresponding to the simple root $\alpha_P$.
Recall from \cref{sec:preliminaries} that the set of positive roots $R^+$ is equipped with a partial
order $<$. Let \rp be the subset of those roots $\alpha \in R^+$ such that $\alpha \ge \alpha_P$. This
is a lattice under the partial order $<$.
A {\em lower-order ideal} in \rp\ is a subset satisfying the following condition:
for any pair of elements $i,j\in\rp$ with $i\in I$ and $j\leq i$, we have $j\in I$.
Following \cite{proctor:bruhat}, the Weyl group elements $w \in W^P$
are in bijection with the lower-order ideals in \rp.

Fix a Hasse diagram for \rp, (cf.~e.g.~\cref{rootPoset}). Then the
lower-order ideal $I_w$ of $w \in W^P$ gives the {\em diagram of $w$}.
This generalizes the usual Young diagram associated to a Schubert variety
in the classical Lie types; for an equivalent model using quivers, see \cite{perrin:small-res}.
In particular, the nodes of $I_w$, which are also the boxes of the diagram of $w$, are given by
positive roots in \rp. These are precisely the positive roots in the inversion set of $w$; thus
the length of $w$ equals the number of boxes in the diagram of $w$. As explained in \cite[\S 3]{BCMP:qkchev},
each box in the diagram of $w$ may also be labelled by a simple reflection, and these labels can be used
to obtain a minimal word for $w$. For instance, the Hasse diagram for the Cayley plane $E_6/P_6$, and the
associated diagram, are given in the \cref{rootPoset}.
 The shaded parts give the diagram $(5,2,1)$, corresponding to
the element $w \in W^P$ with a
reduced expression $w=s_5s_4s_2s_1s_3s_4s_5s_6$; see also \cref{ss:examples}.

\begin{center}
\begin{tabular}{c}
\ytableausetup{centertableaux,boxsize=normal}
\begin{ytableau}*(green) 6& *(green){5} &  *(green){4} & *(green) {3} & *(green) {1}
\\ \none &  \none & *(green) {2} & *(green) {4} & 3 \\ \none & \none & \none & *(green) {5} & {4} & 2
\\ \none & \none & \none & 6 & 5 & 4 & {3} & 1 \end{ytableau}
\end{tabular}
\end{center}

\subsection{Type A: Grassmannians}
The Schubert subvarieties of $\Gr(k,n)$ are indexed by Young diagrams (or partitions) $\lambda= (\lambda_1 \ge \ldots \ge \lambda_k)$ such that $\lambda_1 \le n-k$ and $\lambda_k \ge 0$.
The Schubert variety $X_\lambda$ has dimension $|\lambda| = \lambda_1 + \ldots + \lambda_k$.

\Cref{TBL:Gr36} lists the Mather classes of Schubert varieties in $\Gr(3,7)$.
The expansion of the Chern-Mather class of a Schubert variety
in terms of ordinary Schubert classes
is given in the column indexed by the corresponding partition.
Following \cref{sub:stability}, we see that this table also contains the Mather classes of all $\Gr(k,n)$ with $k \le 3$ and $n-k\le 4$.

\subsection{Type C: Lagrangian Grassmannians}
Let $G/P = \LG(n,2n)$, the variety parametrizing the Lagrangian subspaces of a ${2n}$ dimensional symplectic vector space.
The Schubert subvarieties of $G/P$ are indexed by {\em strict partitions}
$\lambda = (\lambda_1 > \lambda_2 > \ldots > \lambda_k)$, where $\lambda_1 \le n$, $\lambda_k >0$, and $0\le k \le n$.
As before, we have $\dim X_\lambda=|\lambda|$.

\Cref{TBL:cMaLG48} lists the Mather classes of Schubert varieties in $\LG(4,8)$;
The expansion of the Chern-Mather class of a Schubert variety
in terms of ordinary Schubert classes
is given in the column indexed by the corresponding partition.
\Cref{TBL:eulerLG4} list the Euler obstructions of the Schubert varieties in $\LG(4,8)$.

\subsection{Type \texorpdfstring{$E_6$}{E6}: the Cayley plane}
\Cref{TBL:cMaE6} lists the Mather classes of some Schubert subvarieties of the cominuscule space $E_6/P_6$;
The expansion of the Chern-Mather class of a Schubert variety
in terms of ordinary Schubert classes
is given in the column indexed by the corresponding partition.

\subsection{Stability}
\label{sub:stability}
Our `homological' indexing conventions for Schubert varieties gives a stability property for the Mather classes in the ordinary
Grassmannians. and for the maximal isotropic Grassmannians in type C and D. We explain this for the
ordinary Grassmannian, following \cite[\S 2.1]{aluffi.mihalcea:csm}.

Fix a partition $\lambda = (\lambda_1, \ldots , \lambda_k)$ included in the $k \times (n-k)$ rectangle.
Fix the standard flag $F_\bullet: F_1 \subset \ldots \subset F_n = \C^n$ where
$F_i = \langle e_1, \ldots , e_i \rangle$. The Schubert
variety $X_\lambda \subset \Gr(k,n)$ is defined by
\[ X_\lambda = \{ V : \dim V \cap F_{\lambda_{k+1- i} +i} \ge i \} \/. \]
If the $k \times (n-k)$ diagram is included in the $k' \times (n'-k')$ diagram then one can define an embedding
$i: \Gr(k,n) \hookrightarrow \Gr(k',n')$ by $i(V) = \langle e_1 , \ldots , e_{k'-k} \rangle \oplus \widetilde{V}$, where
$\widetilde{V}$ is obtained from $V$ by shifting the indices of basis elements according to the rule
$e_j \mapsto e_{j+k'-k}$. With this definition,
$i(X_\lambda) = X_\lambda$, and it follows that
\begin{equation*}

i_* \cMa(X_\lambda) = \cMa(X_\lambda) \in A_*(\Gr(k',n')) \/.
\end{equation*}
For instance, the Schubert variety $X_\Box \subset \Gr(1,3)$ is $\left\{\langle ae_1 + be_2 \rangle \subset \C^3\mid[a:b] \in \bP^1\right\}$.

Under the inclusion $\Gr(1,3) \hookrightarrow \Gr(3,7)$ the image of $X_\Box$ is the Schubert variety parametrizing the dimension $3$ subspaces $\langle e_1, e_2, ae_3+be_4 \rangle \in \Gr(3,7)$.

One may define similar embeddings, $i: \LG(n,2n) \hookrightarrow \LG(n',2n')$, and
$i: \OG(n,2n) \hookrightarrow \OG(n',2n')$ with $n\leq n'$.
In all these cases, we have have $i(X_\lambda)=X_\lambda$, (cf.~\cite[\S6.2,\S7.2]{smt},

and hence $i_* \cMa(X_\lambda) = \cMa(X_\lambda)$.
We leave it to the reader to check the
details in other types.

\endinput


TITLE: Proof of equality of sets
QUESTION [1 upvotes]: For field $F$ whose charateristic is not 2
(Definition of field characteristic: https://mathworld.wolfram.com/FieldCharacteristic.html),
$\{(a+b, a-b)|a,b \in F\} = \{(a, b)|a,b \in F\}$ ?
It seems true intuitively, but how can I prove ?
And is it false when F's characteristic is not 2?

REPLY [1 votes]: $$ a = 2^{-1} (a+b) + 2^{-1} (a-b) $$
$$ b = 2^{-1} (a+b) - 2^{-1} (a-b) $$
Pair $(1,0)$ creates problems in characteristic $2$, as there are no two elements $x$ and $y$ such that:
$$ 1 = x+y \; \; \mathrm{and} \;\; 0 = x-y.$$


TITLE: How to approach the negative root of the series $\sum\limits_{k=1}^{\infty} \sigma_1(k) x^k$
QUESTION [3 upvotes]: Let's define
$$
f(x) = \sum\limits_{k=1}^{\infty} \sigma_1(k) x^k, \quad \text{where } \sigma_1(k) = \sum\limits_{d|k}d
$$
Is there a way to approach the root $x_0$ of this series, where $-1 < x_0 < 0$ ? I tried to squeeze it and using geometric series, but its too far away.

REPLY [3 votes]: With the help of Mathematica I defined
$$f(x)=\sum _{k=1}^{1000} \sigma _1(k) x^k$$
then  I applied the Newton's method starting from $x_0=-0.411248$
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
$$
\begin{array}{r|l}
n   &x_n\\
\hline
0 &-0.411248\\
1 &-0.4112484791789996\\
2 &-0.4112484791779548\\
3 &-0.4112484791779548\\
4 &-0.4112484791779547\\
5 &-0.4112484791779548\\
\end{array}
$$
bonus
Define Pochhammer symbol
$$Poch(q)=\prod _{k=0}^{\infty } \left(1-q^{k+1}\right)$$
The root of $Poch'(q)$ in the interval $(-1,0)$ is the same as the root of $\sum _{k=1}^{\infty} \sigma _1(k) x^k$ in the same interval.


TITLE: Rescaling of a linear program
QUESTION [1 upvotes]: I have a linear program and I need to check if it has a solution.
Some coefficients entering the program are much smaller than others. E.g., some coefficients are of the order $10^{-10}$, while the others are of the order of $10^{-2}$.
When coefficients are very small, many optimization algorithms do not work properly and, to address the issue, some of them arbitrarily drop coefficients with small values.
Therefore, I need to rescale the linear program.
However, if I rescale by multiplying everything, e.g., by $10^7$, the small coefficients become acceptable, but the others become too large.
What can be a correct way of rescaling?

This is the linear program written in a compact way:
$$
\begin{aligned}
&(1)\quad  A  \text{ }x = b\\
&(2)\quad C \text{ }x\leq c\\
\end{aligned}
$$
$A$ and $C$ contain the coefficients with magnitude of different order as mentioned above. $A$ is $m\times n$ with $n>m$. $C$ is $j\times n$ with $n>j$. $A$ has rank $m$. $C$ has rank $h<j$.
If I had only $A$, I think I could use the full reduced form of $A$. But, here, I also have $C$.

This is the linear program in details (if needed):
The unknown of the linear program is an array $\lambda$ of size $\overbrace{(K+1)\times (K+1) \times ...\times (K+1)}^{D \text{ times}}\times L$. I denote its $(k_1,k_2,...,k_D,y)-th$ element by $\lambda(k_1,k_2,...,k_D,y)$.
The coefficients of the linear program are three arrays, $\alpha$, $\beta$, and $\delta$.
$\alpha$ has size $\overbrace{(K+1)\times (K+1) \times ...\times (K+1)}^{D \text{ times}}\times \overbrace{L \times L}^{2 \text{  times}}$. I denote its $(k_1,k_2,...,k_D,y, y')$-th element by $\alpha(k_1,k_2,...,k_D,y, y')$.
$\beta$ has size $\overbrace{(K+1)\times (K+1) \times ...\times (K+1)}^{D \text{ times}}$. I denote its $(k_1,k_2,...,k_D)$-th element by $\beta(k_1,k_2,...,k_D)$.
$\delta$ has size $L\times 1$. I denote its $y$-th element by $\delta(y)$.
This is the linear program:
$$
\begin{aligned}
&(1) \quad  \sum_{k_1=0}^K  \sum_{k_2=0}^K... \sum_{k_D=0}^K \lambda(k_1,..., k_D,y) \alpha(k_1,...,k_D,y, y')\geq 0 \quad \forall y,y' \text{ with } y\neq y'\\
&(2) \quad \lambda(k_1,..., k_D,y) \geq 0 \quad \forall k_1,..., k_D, y\\
& (3) \sum_{y=1}^L \lambda(k_1,..., k_D,y) =1\quad \forall k_1,..., k_D\\
& (4) \sum_{k_1=0}^K  \sum_{k_2=0}^K... \sum_{k_D=0}^K \lambda(k_1,..., k_D,y) \beta(k_1,..., k_D)=\delta(y) \quad \forall y
\end{aligned}
$$
The arrays that some very small numbers when $K$ is large are $\alpha$ and $\beta$.

REPLY [0 votes]: It is unlikely that linearly rescaling will work, as rescaling is approximately orthogonal to the underlying double precision.
This is because IEEE764 doubles store floating point numbers as $(\pm 1) x^y$ where $x$ has up to $53$ bits of precision and $y$ has up to $11$ bits. The significand $x$ carries only about 15 base-10 digits of precision. Scaling each coefficient approximately only effects the exponents $y$, but the precision issues you're seeing are due to insufficient precision in the significands.
If the coefficients of the matrices $A$ and $C$ vary by $8$ orders of magnitude, then double precision is incapable of recognizing cancellation among the larger coefficients. For example, if the $a_i$ are each of size $10^8$ and the $b_i$ are each roughly constant, then $a_1 a_2 + b_1 \approx a_1 a_2$ in double precision arithmetic, as $b_1$ is too small to affect the precision of the significant in $a_1 a_2$.
It might be possible that using higher precision (e.g. 128-bit quad precision, or arbitrary precision) would work. It might also be necessary to try to modify your linear program to be better conditioned --- but this can be hard.


TITLE: Find the conditions on a and b such that the system below has a unique solution, no solutions and infinitely many solutions
QUESTION [0 upvotes]: The system:
$x+ay=1$
$bx+y=2$
Note: I have to find the unique and infinitely many solutions
I set-up the augmented matrix but my answers didn't come from any operations:
What I got: -The system has a unique solution (x=1, y=2) if a=b=0.
-The system has no solution if a=b=1, then we will have a contradiction.
-The system has infinitely many solutions when $a ≠ 0, a ≠ 1, b ≠ 0, b ≠ 1,$ then solutions will be $ay+x=1$ and $bx+y=2$.
What am I doing wrong and is there a way I could get to this other than simply observing since the matrix already seems reduced?

REPLY [0 votes]: You can write your system as $Au=v$, where $A = \begin{bmatrix}1 & a \\ b & 1\end{bmatrix}$, $u=\begin{bmatrix}x \\ y\end{bmatrix}$ and $v=\begin{bmatrix}1 \\ 2\end{bmatrix}$.
Case 1: Unique solution
It is known that if $A$ is invertible, then the system has unique solution. This is because we can write $$u=A^{-1}v.$$ Therefore, we look for a condition that $A$ is invertible. Such condition is that $\det{A} \neq 0$. In your case: $$1\cdot1-a\cdot b \neq 0.$$
Case 2: Multiple solutions
This happens when the two equations describe the same situation. Each of $x+ay=1$ and $bx+y=2$ describe two lines in the $\mathbb{R}^2$ plane. The solution is given by the points $(x,y)$ which describe the intersection of the two lines. If we seek the case where we have infinite solution points $(x,y)$, we mean that the two lines are identical. This happens if the equations are the same. As $x+ay=1$ is equivalent to $2x+2ay=2$ (observe how the right-hand side is the same for both equations now), we can deduce that $b=2$ and $a=\dfrac{1}{2}$ yield the same equation. In conseqence, we have infinitely-many solutions.
Case 3: No solutions
This part does not need to be treated as you already determined the values for $a,b$ which yield solutions (unique or infinitely many). So, to have no solutions, you require
$$ab = 1\text{ and } b\neq2.$$
Observe how the condition above excludes first unique solution and then infinite case.
Hope it helps and clarifies the process of finding $a$ and $b$.