TITLE: Isomorphisms between subspaces
QUESTION [2 upvotes]: Let $U$ and $W$ two subspaces of a vector space $V$.  Let $T: U \times W \to V$ be a function, defined by $T(u, w) = u + w$.
Show that

i) $T$ is a linear transformation,
ii) Image of $T$ is the subspace $U+W$,
iii) $\ker T = \{(u,-u) \mid u \in U \cap W\}$ is isomorphic to $U \cap W$.

Well, i) is rather easy.
Any thoughts about ii) and iii)?
Thanks!

REPLY [1 votes]: For (ii) note that
$$
U+W=\{u+w:u\in U, w\in W\}
$$
But then
\begin{align*}
\DeclareMathOperator{Image}{Image}\Image(T)
&= \{T(u,w):u\in U,w\in W\} \\
&= \{u+w:u\in U,w\in W\} \\
&= U+W
\end{align*}

REPLY [0 votes]: For ii), there isn't much to say: it's the definition of the sum of two subspaces.
As for iii), just use the definition of the kernel:
$$(u,w)\in\ker T\iff w=-u$$
so $u\in W$ (hence in $U\cap W$ , as well as $w$), and a vector in $\ker T$ can be written as: $$(u,w)=(u,-u)\in (U\cap W)\times(U\cap W)$$
Conversely, it is trivial that such a vector is in $\ker T$.


TITLE: Why is $\int d^3r_1 \, d^3 r_2 \, \frac{F(\vec{r}_1) F^*(\vec{r}_2)}{|\vec{r}_1 - \vec{r}_2|}$ positive?
QUESTION [3 upvotes]: I have the following quantity:
$$
I = \int_{\mathbb{R}^3 \times \mathbb{R}^3} d^3r_1 \, d^3 r_2 \, \frac{F(\vec{r}_1) F^*(\vec{r}_2)}{|\vec{r}_1 - \vec{r}_2|}
$$
$I$ is obviously real by symmetry. How can I show that it is also positive semidefinite?

REPLY [3 votes]: Recall that the solution to Poisson's equation $\nabla^2 J = F$ is given by $J(r_2) = -\frac{1}{4\pi}\int \frac{F(r_1){\rm d}r_1}{|r_1-r_2|}$. Using this the integral can be written
$$I = -4\pi\int J(r_2)\nabla^2 J^*(r_2){\rm d}r_2 = 4\pi\int |\nabla J(r_2)|^2 {\rm d}r_2 \geq 0$$
where we used integration by parts (as usual assuming boundary terms vanish) to arrive at the last equality.


TITLE: Utility of displacements potentials in geophysics
QUESTION [2 upvotes]: In the elasticity theory, you can derive a wave equation from the fundamental equation of motion for an elastic linear homogeneous isotropic medium:
$\rho \partial^2_t \overline{u} = \mu \nabla^2 \overline{u} + (\mu+\lambda) \nabla(\nabla \cdot \overline{u})$
But in the seismology tradition, you introduce scalar and vector potentials for the P and S components of the displacements, derive wave equations for them and use them.
Now, in electrodynamics you can derive from Maxwell's equations the wave equations for Fields and for Potentials; but there you use the potentials because they compose a quadrivector. In geophysics what's the convenience of it?

REPLY [1 votes]: Also, adding to what j.c. states:
If you have a complicated constitutive relation between the electric field and the electric displacement field, e.g. nonlinear and in terms of a convolution (no instantanious reaction of the medium), it might be very complicated to find a representation in terms of potentials obaying a wave equation using a non-perturbative ansatz.


text:
\section{Introduction}
\begin{figure}[!t]
\centering
\includegraphics[width=0.4\textwidth]{figbdry_and_mobius.png}
\caption{A visualization of two weighting vectors.  The set in the top figure is supported within four disjoint components, and they live in $\R^2$. The set in the bottom figure is supported on an embedding of M\"{o}bius strip, and it lives in $\R^3$. In both images, the weight of each point is represented using color and point size.}
\label{fig:numerical_example}
\end{figure}
{\em Magnitude} is a scalar quantity that has meaning for many different kinds of data, and as with other scalar quantities such as rank, diameter, and measure, it has wide applicability, an intuitive interpretation and a solid theoretical foundation. Magnitude has been discovered, and rediscovered multiple times in both practical and theoretical contexts. In this paper, our goal is to apply recent developments drawn from magnitude theory to machine learning, and to empirically demonstrate characteristics of magnitude that, while implicitly described by abstract theoretical results, have not, to our knowledge, been explicitly stated before, nor have they been leveraged for practical purpose.

Informally, magnitude aims to quantify the effective number of points in a space. Our aim is more subtle: we wish to identify \textit{which points} are considered ``effective'' and ``important.''  We do this using the {\em weighting vector}. The weighting vector appears naturally in the definition of magnitude, and we find that the weighting vector, under appropriate conditions, serves as an effective \textit{boundary detector}.   It is this behavior that makes the weighting vector especially well suited for machine learning tasks.

\subsection{Background, notation and examples}
We now define magnitude and the weighting vector, we present several canonical examples, and we state several central theorems of the field. While our focus is largely on subsets of $\R^n$, we note that the concept {\em magnitude} and {\em weighting vector} can be defined for far more general types of sets.
\begin{definition}
Let $X$ be a finite metric space with metric $d$. Denote the number of points in $X$ by $\abs{X}$. The \textit{similarity matrix} of $X$ is defined to be $\z_X(i, j) := \exp(-d(x_i, x_j))$ for $1 \leq i,  j \leq \abs{X}$. Whenever the inverse of $\z_X$ exists, we define the \textit{weighting vector} of $X$ to be
    \begin{align*}
        w_X \coloneqq \z_X^{-1}\one,
    \end{align*}
where $\one$ is the $\abs{X} \times 1$ column vector of all ones. The \textit{magnitude} of $X$ is defined to be the quantity
    \begin{align*}
        \Mag{X} \coloneqq \one^Tw_X = \one^T \z_X^{-1} \one.
    \end{align*}
That is, $\Mag{X}$ is the sum of all the entries of the weighting vector $w_X$.
\end{definition}

\begin{example}
When $X$ is a finite subset of Euclidean space, $\z_X$ is a symmetric positive definite matrix [Theorem 2.5.3, \cite{Leinster2013TheMO}]. In particular, $\z_X^{-1}$ is guaranteed to exist. Hence, the weighting vector and magnitude exist for finite subsets of $\R^n$.
\end{example}

\begin{example}
Given an undirected, unweighted graph $G$, one can define a metric space whose points are given by the vertices of $G$, and whose metric is taken to be the length of the shortest path between two vertices. The weighting vector of this metric space is not guaranteed to exist.
\end{example}

\begin{definition}\label{def:infinitesets}
For an arbitrary subset $X \subseteq \R^n$, the magnitude of $X$ is defined as
\begin{align*}
    \Mag{X} = \sup\{ \Mag{Y} \mid Y \text{ is a finite subset of } X \}.
\end{align*}
\end{definition}
\begin{example}
In 1 dimension, and for $t>0$, one has that $\Mag{[0,t]}=1+t/2$. This was shown by Leinster in  \cite{Leinster2013TheMO}. The magnitude of the ball with radius $r$ in $\R^{2n+1}$ is a rational function of $r$, and this was recently demonstrated by Barcel\'{o} and Carbery~\cite{Barcelo18CptEuclid}.
\end{example}

For a finite metric space $(X, d)$, and any $t \in [0, \infty]$, we can define a new metric space $(tX, td)$ in the following way. The points of $tX$ are the same as those of $X$, and the metric $td$ is $d$ scaled by $t$: $td(x, y) \coloneqq t\cdot d(x, y)$. The \textit{magnitude function of $X$} is the map $t \mapsto \Mag{tX}$, and it is well-defined whenever $\z_{tX}$ is invertible. Although the inverse of $\z_{tX}$ may not be defined in general, it has been shown in [Proposition 2.2.6 \cite{Leinster2013TheMO}] that for finite subsets of $\R^n$, the magnitude function is analytic on $(0, \infty)$. We also have the following:
\begin{theorem}[Proposition 2.2.6 \cite{Leinster2013TheMO}]\label{theorem_t_inf_finite_set}
For $X \subset \R^n$ finite, $\lim_{t \rightarrow \infty} \Mag{tX} = \abs{X}$.
\end{theorem}

The above proposition is one of the reasons underlying the informal interpretation of magnitude as quantifying the effective number of points in a space. The following very recent theorem gives a connection between the magnitude of $X \subset \R^n$ and the $n$-volume of $X$.
\begin{theorem}[Theorem 1 \cite{Barcelo18CptEuclid}]\label{theorem_vol}
For $X \subset \R^n$ nonempty and compact, we have
\begin{align*}
    \lim_{t \rightarrow 0^+} \Mag{tX}
    &=
    1, \text{\hspace{3pt} and}\\
    \lim_{t \rightarrow \infty}\frac{\Mag{tX}}{t^n}
    &=
    \frac{\Vol{X}}{n!\Vol{B_n}},
\end{align*}
\noindent where $B_n \subset \R^n$ is the unit ball.
\end{theorem}

\subsection{Properties of the weighting vector}
The weighting vector plays a central role in the applications that are discussed below, but it is not obvious by inspection of its definition what useful information the individual entries of the weighting vector carries. To provide some intuition about this vector, we now highlight its key features. Our present aim is to convey a qualitative sense of things, so our focus is on numerical examples and basic facts. Note that the ability of the weighting vector to perform boundary detection is more than conjecture: it may be completely explained using harmonic analysis~\cite{folland1999real,meckes2015magnitude}. But for reasons of space and  scope, we limit our focus.

Let $X\subset \R^n$ be a finite set and recall that the weighting vector of $X\subset \R^n$ is defined as $w_X\coloneqq \zeta_{X}^{-1}\one$. This vector is related to the magnitude of $X$ through $\Mag{X}=\one'\zeta_{X}^{-1}\one=\one'w_{X}$. Note that while $X\subset\R^n$, the vector $w_X\subset \R^{\abs{X}}$, i.e., the dimension of the weighting vector is a function of the
size of $X$, and not of the dimension~$n$. Also note that the entries of $w_X$ may be indexed in a canonical way by $x\in X$. We call $w_X(x)$, the weight of $x$.

Since all operations involved in evaluating $w_X$ are continuous, the weighting vector of a small perturbation of $X$ will approximate the weighting vector of $X$ itself. More precisely, let $X_\epsilon\coloneqq\left\{x+\epsilon \eta_x :x\in X, \norm{\eta_x}\leq 1\right\}$. Then for all $x\in X$, one has $\lim_{\epsilon\rightarrow 0}w_{X_{\epsilon}}(x+\epsilon\eta_x)=w_X(x)$. Since the similarity matrix $\zeta_{X}$ is positive definite, its inverse exists and is also positive definite. Thus, $\one'\zeta_{X}^{-1}\one = \one'w_{X}>0$. Although the average value of the entries of $w_X$ is guaranteed to be positive, it may happen that $w_X(x)<0$ holds for some $x\in X$.  It is currently unknown what, if any, significance to ascribe to the sign of $w_X(x)$.

Let $f:\R^{n}\rightarrow\R^{m}$ be an affine isometry, i.e., $f(x)=Qx+q$ for some $Q\in \R^{m\times n}$ and $q\in \R^m$, and $f$ satisfies $\norm{z-w} = \norm{f(z)-f(w)}$ for all $z,w\in\R^{n}$.
Since $f$ is an isometry, the similarity matrices of $X$ and $f(X)$ agree (i.e., $\zeta_{X}=\zeta_{f(X)}$), and as a result, $w_X(x)=w_{f(X)}(f(x))$. By the Mazur-Ulam theorem, all surjective isometries between normed spaces are necessarily affine. Thus, we have that the weighting vector is invariant under transformation by {any} surjective isometry. More concretely, in Figure~\ref{fig:numerical_example}, the weightings that are displayed are independent of the location and the orientation of the sets.

When $t>0$ is large, the scaled space $tX$ has the property that all points in it are far from each other.  Thus, one has $\lim_{t\rightarrow\infty}\zeta_{tX}=\lim_{t\rightarrow\infty}\zeta_{tX}^{-1} = I$. By Theorem~\ref{theorem_t_inf_finite_set}, the magnitude function satisfies $\lim_{t\rightarrow\infty}\Mag{tX}=\abs{X}$. Combining these observations, we find that for all $x\in X$,  $\lim_{t\rightarrow\infty} w_{tX}(tx)=1$. The closely-related concept of a {\em scattered space} appears in prior work (Definition 2.1.2, \cite{Leinster2013TheMO}), where under conditions far more general than considered here, it is shown that scattered spaces have well-defined magnitudes, and hence, weighting vectors.

Conversely, when $t>0$ is small, each entry of the similarity matrix $\zeta_{tX}$ is close to 1. In particular, the limiting matrix is the rank-1 matrix $\one\one'$, which does not have an inverse.  However, by Theorem~\ref{theorem_vol}, one has $\lim_{t\rightarrow 0^+}\one'w_{tX}(x)=1$.  Empirically, when $t>0$ is very small, we find that the weights of ``interior points'' of the global space of $X$ are small, while the ``extreme'' points of $X$---especially points that live nearest the convex hull of $X$---are significantly larger.

Finally, we consider weighting vectors of $X\subset\R^n$ that is neither too scattered nor especially concentrated about the origin. As one example, let $X\subset\R^n$ be a regular convex polytope. By a symmetry argument, for all vertices  $x,y\in X$ one has $w_X(x)=w_X(y)$. Thus, modulo a normalizing constant, the weighting vector is completely specified.   Next, consider Figure~\ref{fig:numerical_example}, which displays weightings of two sets that do not have any special symmetry: $X_0\subset \R^2$ which consists of points supported in the union of four disjoint sets, and $X_1\subset\R^3$ which lives on an embedding of the M\"{o}bius strip. Both sets were generated using a uniform random sampling process. In these renderings, every $x\in X_{i}$ has its weight, $w_{X_{i}}(x)$, conveyed both by the marker size and by color, where $i=0,1$. It is clear from these figures that points within the relative interior of some component have low weight, while points in close proximity to some  boundary tend to have larger weight.  It is this empirical observation that leads to the utility of weightings in applications.

We close this section by observing that magnitude, and by extension, weighting vectors, are well-defined on a very general class of sets, including sets that are not necessarily subsets of $\R^n$. It is therefore possible to extend the notion that connects a point's weight and its proximity to a boundary to {\em any} space that has a weighting vector.

\subsection{Related work, paper structure}

An early reference to the concept of magnitude occurs in~\cite{Solow1994}, where it was introduced as a way to measure biological diversity. However, the mathematical motivations were not divulged in this paper. Two decades later, Leinster~\cite{Leinster2013TheMO} placed the magnitude of a metric space within a formal mathematical framework using category theory.  This highly abstract perspective lead to the current era, where it is being explored through many different lenses, including functional analysis \cite{MeckesPosDef, Barcelo18CptEuclid}, harmonic analysis~\cite{meckes2015magnitude} and homology theory \cite{leinster2017magnitude}, where it has been shown to be equivalent to an Euler characteristic.   Much of the prior emphasis has been on a set's  magnitude, and this focus has overshadowed the potential utility of the weighting vector.

Recently, {topological data analysis} has emerged as an approach to the problem of describing the shape of high-dimensional data~\cite{Edelsbrunner2000TopologicalPA, Scopigno04persistencebarcodes, computingPH}. One particularly popular topic within this field is persistent homology \cite{Edelsbrunner2000TopologicalPA}.  Recent efforts have realized magnitude as the Euler characteristic of a homology theory, called magnitude homology \cite{leinster2017magnitude}. It has also been shown that there is a direct relationship between magnitude homology and persistent homology \cite{ otter2018magnitude}; however, the current paper is the first known application of magnitude directly to machine learning.

We now describe the remaining sections of this paper. Section~\ref{sec:properties} presents practical techniques for working with, and computing, the magnitude and weighting vectors of a discrete set. Section~\ref{sec:algorithms} introduces three algorithms that leverage the weighting vector in some essential way. The algorithms perform classification, active learning, and outlier detection.  Section~\ref{sec:results} presents results. We end with concluding remarks in Section~\ref{sec:conclusions}.

config: arxiv


TITLE: Find generating function
QUESTION [2 upvotes]: Define  a  transformation $T_s$ of integer sequence $\{ a_n \}$  by
$$
b_n=T_s(a_n)={n \choose s} \sum_{i=s}^{n-1} \frac{a_i}{{i \choose s}},
$$
for a fixed $s \in \mathbb{N}.$
For instance, if we  aplly  the transformation $T_2$  to the sequence $a_n=1$  then we get the sequence $b_n=n(n-2).$
Maple code for  the sample
    T:=(a,s)->factor(simplify(product(n-i,i=0..s-1)*sum(a(k)/product(k-i,i=0..s-1),k=s..n-1)));
a:=n->1:T(a,2);
                                 n(n-2)

Question. Suppose that $a_n$ has a  generating function (ordinary or  exponential or another one) $a(x).$ What is the generating function of transformed sequence?

REPLY [3 votes]: This can be done step by step.
First note that $\binom{n}{s}/\binom{i}{s}$ can be written as $n(n-1)\cdots(n-s+1)/i(i-1)\cdots(i-s+1)$
Since we have the generating function (with assuming $a_i=0$ for $i< s$)
$$a(x)x^{-s}=\sum_{i=s}^{\infty} a_ix^{i-s}$$
We obtain the following by integrating $s$ times.
Let $A_0(x)=a(x)x^{-s}$, and $A_{k+1}(x)= \int_0^x A_k(t)dt$.
Then
$$A_s(x)=\sum_{i=s}^{\infty} a_i\frac{x^i}{i(i-1)\cdots (i-s+1)}$$
The generating function for $b_n/ n(n-1)\cdots (n-s+1)$ can be obtained from the product
$$\left(\sum_{i=s}^{\infty} a_i\frac{x^i}{i(i-1)\cdots (i-s+1)}\right)\left(\sum_{j=1}^{\infty} x^j\right)$$
Now the generating function for $b_n$ follows from differentiating s times
Again assuming $b_i =0$ for $i< s$, we have
$$B_0(x)= A_s(x) \frac{x}{1-x}$$
$$B_{k+1}(x)=\frac{d}{dx} B_k(x)$$
$$\sum_{n=s}^{\infty} b_n x^n = B_s(x) x^s$$


TITLE: How many points can I prescribe for a diffeomorphism of the plane?
QUESTION [8 upvotes]: I was trying to find out how to construct a $\mathcal C^\infty$ curve that joins two arbitrary line segments. My idea was to use bump functions and the likes, but for that I had to make the line segments lie at $y = 0$ and $y = 1$. Of course, the transformation that transported the segments would have to be a diffeomorphism (at least of some set containing the segments). This I readily found to be equivalent to prescribing the image of four points under some diffeomorphism. A couple ways of doing this occurred to me, but I encountered this question in the process.
What restrictions are there on the cardinal of a set A\subset\Bbb R^2 such that there exists a (auto) diffeomorphism f with f(A) = B for some fixed $B\subset \Bbb R^2? (Obviously #B = # A)
In light of developments, I'll ask for which homeomorphic $A,B\subset\Bbb R^2$ is there a diffeomorphism of $\Bbb R^2$ (or an open set containing $A$,$B$) sending $A$ to $B$?
If $\#A<\infty$ there always exists such an $f$, constructed by appropriate polynomials. (e.g. $f(x,y) = (ax^2+bxy+cy^2,a'x^2+b'xy+c'y^2)$ allows for six points, and so on.
My intuitive guess is that we can do the same for a countable set, but I couldn't justify it myself. For uncountable it seems obvious that the answer is negative in general. Along with an answer for the countable case, I'd like to know if possible of some other examples, possibly more elegant, for the finite case.
Also optional: how does the answer to this question (in a manifold) relate to the geometry of the manifold?

REPLY [6 votes]: The questions you ask have a very wide scope, so I will just give some examples. If diffeomorphism is replaced with homeomorphism, problems like this have a long history with the most famous being the Jordan-Schoenflies problem. Daverman's book Decompostions of Manifolds is a good source for this material, though really this question probably spans all of geometric topology.
The density of $A$ and $B$ in $\Bbb R^2$ and the homeomorphism type of $A$ and $B$ give some obvious invariants of finding a homeomorphism of $\Bbb R^2$ sending $A$ to $B$ (i.e. Hagen's answer). Sometimes these are enough. In fact, it is true that for every subset of the plane $C$ homeomorphic to the Cantor set, there is a homeomorphism of $\Bbb R^2$ which sends $C$ to the standard Cantor set. In $\Bbb R^3$, things become much more complicated. There are sets homeomorphic to Cantor sets (see Antoine's necklace) where there exists no such homeomorphism.
Also in $\Bbb R^3$, we can ask when can whether or not there exists such a diffeomorphism for a pair of smoothly-embedded circles $K_1, K_2$. Gordon and Luecke showed this is true whenever $\Bbb R^3-K_1$ and $\Bbb R^3-K_2$ are homeomorphic, and in one dimension higher there are some wide open questions i.e.

Open Question: Let $S_1,S_2\subset \Bbb R^4$ be smoothly embedded 2-spheres one of which is the standard $S^2\subset \Bbb R^3 \times 0$. If there is a homeomorphism of $\Bbb R^4$ sending $S_1$ to $S_2$ is there a diffeomorphism of $\Bbb R^4$ sending $S_1$ to $S_2$?


TITLE: How to calculate inverse function of a "weird" function like $y=x+\cos x$
QUESTION [3 upvotes]: As the title said, how to calculate the inverse function of $y = f(x) = x + \cos x$ ?
I found a super function $x = g(y)$ like this:
$x = y -\cos(y-\cos(y-\cos(y-cos(...(y-\cos y)...))))$, with infinite nested loop.
And I use matlab to test it, find that the more nests, the lower approximation error.
So I'm conjecturing if $g(y)$ is the true inverse function of $f(x)$.
If yes, why? And can we reduce the form into some "simpler" finite style?
If not, what should it be? And also why?
###############
Note of changes: Sorry for my poor math knowledge, I cannot find a proper symbol describe what "..." exatly means. how about I use programming language?
y = a; % a is a constant
x = y - cos(y);
for i = 1:10000
   x = y - cos(x);
end
disp(x)

REPLY [2 votes]: The sequence in your code is defined by $x_0 = a$ and $x_{n+1} = a - \cos(x_n)$, and you want to set $g(a) = \lim_{n \to \infty} x_n$. Assume for the moment that the limit exists.  Then of course the limit of $x_{n+1}$ is also $g(a)$, so taking the limit of both sides of $x_{n+1} = a - \cos(x_n)$ gives $g(a) = a - \cos(g(a))$, or $g(a) + \cos(g(a)) = a$, or $f(g(a)) = a$.  That is what we would expect for the inverse of $f$, so this is promising.
To complete the proof you must check 2 more things: (1) that the limit exists, and (2) $g(f(a)) = a$ as well.


text:
\begin{document}

\title[Approximation property]{The approximation property for weighted spaces of differentiable functions}
\author[K.~Kruse]{Karsten Kruse}
\address{TU Hamburg \\ Institut f\"ur Mathematik \\
Am Schwarzenberg-Campus~3 \\
Geb\"aude E \\
21073 Hamburg \\
Germany}
\email{karsten.kruse@tuhh.de}

\subjclass[2010]{Primary 46E40, 46A32 Secondary 46E10}

\keywords{approximation property, tensor product, differentiable, weight, vector-valued}

\date{\today}
\begin{abstract}
 We study spaces $\mathcal{CV}^{k}(\Omega,E)$ of $k$-times continuously partially differentiable functions
 on an open set $\Omega\subset\mathbb{R}^{d}$ with values in a locally convex Hausdorff space $E$.
 The space $\mathcal{CV}^{k}(\Omega,E)$ is given a weighted topology generated by a family of weights $\mathcal{V}^{k}$.
 For the space $\mathcal{CV}^{k}(\Omega,E)$ and its subspace $\mathcal{CV}^{k}_{0}(\Omega,E)$ of functions that
 vanish at infinity in the weighted topology we try to answer the question whether their elements can be approximated
 by functions with values in a finite dimensional subspace. We derive sufficient conditions for an affirmative answer to
 this question using the theory of tensor products.
\end{abstract}
\maketitle
\section{Introduction}
\input{Intro}

\section{Notation and Preliminaries}
\input{Notation1}

\input{approx}

\bibliography{biblio}
\bibliographystyle{plain}
\end{document}


text:
\begin{document}

\title[Coarse graining and critical slope]{Coarse graining,
  fractional moments and the critical slope of random copolymers}

 \author{Fabio~Lucio~Toninelli}
 \address{
CNRS and
 ENS Lyon, Laboratoire de Physique,
 46 All\'ee d'Italie, 69364 Lyon, France\\ \\
\textit{E-mail address:} fabio-lucio.toninelli@ens-lyon.fr}

 \begin{abstract} For a much-studied model of random copolymer at a
   selective interface we prove that the slope of the critical curve
   in the weak-disorder limit is strictly smaller than $1$, which is
   the value given by the annealed inequality. The proof is based on a
   coarse-graining procedure, combined with upper bounds on the
   fractional moments of the partition function.  \\ \\ 2000
   \textit{Mathematics
     Subject Classification:  60K35, 82B44, 60K37 } \\ \\
   \textit{Keywords: Copolymers at Selective Interfaces, Fractional
     Moment Estimates, Coarse-graining}\\ \\
\textit{Submitted to EJP on June 10, 2008,
final version accepted on January 30, 2009}
 \end{abstract}

\maketitle

\newpage
\section{Introduction}

We consider a model of {\sl copolymer at a selective interface}
introduced in \cite{cf:GHLO}, which has  attracted much
attention among both theoretical physicists and probabilists (we refer to
\cite{cf:GB} for general references and motivations).

Let $S:=\{S_n\}_{n\ge0}$ be the symmetric Simple Random Walk on $\Z$ started
at $S_0=0$, with law $\bP^{\rm SRW}$ such that the increments
$\{S_i-S_{i-1}\}_{i\ge0}$ are IID and $\bP^{\rm SRW}(S_1=\pm1)=1/2$.
The partition function of the model of size $N$ is given by
\begin{eqnarray}
  \label{eq:Zsrw}
  Z_{N,\go}:=\bE^{\rm SRW}\left[e^{-\gl\sum_{n=1}^N(\go_n+h)(1-\sign(S_n))}
\ind_{\{S_N=0\}}
\right]
\end{eqnarray}
where $h\ge0$, $\gl\ge0$ and $\{\go_n\}_{n\ge1}$ is a sequence of IID
standard Gaussian random variables (the quenched disorder).  We adopt
the convention that, if $S_n=0$, then $\sign(S_n):=\sign(S_{n-1})$.
One interprets $\gl$ as the inverse temperature (or coupling strength)
and $h$ as an ``asymmetry parameter'': if $h>0$, since the $\go_n$'s
are centered, the random walk overall prefers to be in the upper
half-plane ($S\ge0$).  It is known that the model undergoes a
delocalization transition: if the asymmetry parameter exceeds a
critical value $h_c(\gl)$ then the fraction of ``monomers'' $S_n$,
$n\le N$, which are in the upper half-plane tends to $1$ in the
thermodynamic limit $N\to\infty$ (delocalized phase), while if
$h<h_c(\gl)$ then a non-zero fraction of them is in the lower
half-plane (localized phase). What mostly attracts attention is the
slope, call it $m_c$, of the curve $\gl\mapsto h_c(\gl)$ in the limit
$\gl\searrow 0$: $m_c$  is expected to be a universal quantity, i.e.,
independent of the details of the law $\bP^{\rm SRW}$ and of the
disorder distribution (see next section for a more extended discussion
on this point). Already the fact that the limit slope is well-defined
and positive is highly non-trivial \cite{cf:BdH}.

Until now, all what was known rigorously about $m_c$ is that $2/3\le
m_c\le 1$, but numerically the true value seems to be rather around
$0.83$ \cite{cf:CGG}. The upper bound comes simply from annealing,
i.e., from Jensen's inequality, as explained in next section.  Our
main new result is that $m_c$ is strictly smaller than $1$.  The proof
works through a coarse-graining procedure in which one looks at the
system on the length-scale $k(\gl,h)$, given by the inverse of the
annealed free energy. The other essential ingredient is a
change-of-measure idea to estimate fractional moments of the partition
function (this idea was developed in \cite{cf:GLT} and \cite{cf:DGLT},
and used in the context of copolymers in \cite{cf:BGLT}).
Coarse-graining schemes, implemented in a way very different from
ours, have already played an important role in this and related
polymer models; we mention in particular \cite{cf:BdH}, \cite{cf:BCT}
and \cite{cf:AZy}.

\section{The general copolymer model}

As in \cite{cf:BGLT}, we consider a more general copolymer model which
includes \eqref{eq:Zsrw} as a particular case. Since the critical
slope is not proven to exist in this general setting, Theorem
\ref{th:slope} will involve a $\limsup$ instead of a limit.

Consider a renewal process $\tau:=\{\tau_0,\tau_1,\ldots\}$ of law
$\bP$, where $\tau_0:=0$ and $\{\tau_i-\tau_{i-1}\}_{i\in\N}$ is an IID
sequence of integer-valued random variables. We call
$K(n):=\bP(\tau_1=n)$ and we assume that $\sum_{n\in\N}K(n)=1$ (the
renewal is recurrent) and that $K(\cdot)$ has a power-law tail:
\begin{eqnarray}
\label{eq:K}
  K(n)\stackrel{n\to\infty}\sim \frac {{\mathcal C}_K}{n^{1+\ga}}
\end{eqnarray}
with $\ga>0$ and ${\mathcal C}_K>0$. As usual, the notation
$A(x)\stackrel{x\to x_0}\sim B(x)$ is understood to mean that $\lim_{x\to
  x_0}A(x)/B(x)=1$.

The copolymer model we are going to define depends on two parameters
 $\gl\ge0$ and $h\ge0$, and on a sequence
$\go:=\{\go_1,\go_2,\ldots\}$ of IID standard Gaussian random
variables (the quenched disorder), whose law is denoted by $\bbP$.
For a given system size $N\in\N$ and disorder realization $\go$, we
define the partition function
 $ Z_{N,\go}:=Z_{N,\go}(\gl,h)$ as
\begin{eqnarray}
  \label{eq:Z}
  Z_{N,\go}:=\bE\left[\prod_{j\ge 1:\tau_j\le N}
\frac{1+e^{-2\gl h(\tau_j-\tau_{j-1})-2\gl\sum_{i=(\tau_{j-1}+1)}^{\tau_j}\go_i}}2
\ind_{\{N\in\tau\}}
\right],
\end{eqnarray}
where $\ind_{\{A\}}$ is the indicator function of the event $A$.

To see that the ``standard copolymer model'' \eqref{eq:Zsrw} is a
particular case of \eqref{eq:Z}, let $\tau:=\{n\ge0\in\N:S_n=0\}$ and
as a consequence $K(n):=\bP^{\rm SRW}(\inf\{k>0:S_k=0\}=n)$. It is
known that in this case $K(\cdot)$ satisfies \eqref{eq:K} with
$\ga=1/2$, see \cite[Ch. III]{cf:feller1} (the fact that in this case
\eqref{eq:K} holds only for $n\in2\N$, while $K(n)=0$ for $n\in 2\N+1$
due to the periodicity of the simple random walk, entails only
elementary modifications in the arguments below).  Next, observe that
if $s_i\in\{-1,+1\}$ denotes the sign of the excursion of $S$ between
the successive returns to zero $\tau_{i-1}$ and $\tau_i$, under
$\bP^{SRW}$ the sequence $\{s_i\}_{i\in\N}$ is IID and symmetric (and
independent of the sequence $\tau$).  Therefore, performing the
average on the $\{s_i\}_i$ in \eqref{eq:Zsrw} one immediately gets
\eqref{eq:Z}.

\medskip

The infinite-volume free energy is defined as
\begin{eqnarray}
  \tf(\gl,h):=\lim_{N\to\infty}\frac1N\bbE\log Z_{N,\go}\ge0,
\end{eqnarray}
where existence of the limit is a consequence of superadditivity of the
sequence
$\{\bbE\log Z_{N,\go}\}_N$ and the inequality $\tf\ge0$ is immediate from
\begin{eqnarray}
  Z_{N,\go}\ge \frac{K(N)}2,
\end{eqnarray}
which is easily seen inserting in the expectation in right-hand side of
\eqref{eq:Z} the indicator function $\ind_{\{\tau_1=N\}}$.
One usually defines the critical line in the $(\gl,h)$ plane as
\begin{eqnarray}
  h_c(\gl):=\sup\{h\ge0:\,\tf(\gl,h)>0\}.
\end{eqnarray}
From Jensen's inequality $\bbE\log Z_{N,\go}\le \log \bbE Z_{N,\go}$
(the ``annealed bound'') one obtains the immediate inequality
\begin{eqnarray}
\label{eq:annb}
  h_c(\gl)\le \gl.
\end{eqnarray}
Indeed, one has
\begin{eqnarray}
\label{eq:Zann}
  \bbE Z_{N,\go}=
\bE\left[\prod_{j\ge 1:\tau_j\le N}
\frac{1+e^{2\gl(\gl- h)(\tau_j-\tau_{j-1})}}2
\ind_{\{N\in\tau\}}
\right],
\end{eqnarray}
from which it is not difficult to prove that
\begin{eqnarray}
\label{eq:Fann}
  \lim_{N\to\infty}\frac1N\log \bbE Z_{N,\go}=2\gl(\gl-h)\ind_{\{h<\gl\}},
\end{eqnarray}
and therefore the claim \eqref{eq:annb}.  For $h\ge\gl$,
\eqref{eq:Fann} follows from the fact that the right-hand side of
\eqref{eq:Zann} is bounded above by $1$, while the left-hand side of
\eqref{eq:Fann} is always non-negative. For $h<\gl$, just observe that
\begin{eqnarray}
\label{eq:UBann}
   \bbE Z_{N,\go}\le \bE\left[\prod_{j\ge 1:\tau_j\le N}
e^{2\gl(\gl- h)(\tau_j-\tau_{j-1})}
\ind_{\{N\in\tau\}}
\right]=\bP(N\in\tau)e^{2\gl(\gl-h)N}
\end{eqnarray}
and that
\begin{eqnarray}
  \bbE Z_{N,\go}\ge
 \bE\left[\prod_{j\ge 1:\tau_j\le N}
\frac{1+e^{2\gl(\gl- h)(\tau_j-\tau_{j-1})}}2
\ind_{\{\tau_1=N\}}
\right]=K(N)\frac{1+e^{2\gl(\gl-h)N}}2.
\end{eqnarray}
The limit \eqref{eq:Fann} is called {\sl annealed free energy}.

\subsubsection{What is known about the critical  line and its slope at the origin}

The critical point is known to satisfy the bounds
  \begin{eqnarray}
\label{eq:baundi}
    \frac{\gl}{1+\ga}\le h_c(\gl)<\gl.
  \end{eqnarray}
  The upper bound, proven recently in \cite[Th. 2.10]{cf:BGLT}, says that the
  annealed inequality \eqref{eq:annb} is strict for every $\gl$. The
  lower bound was proven in \cite{cf:BG} for the model \eqref{eq:Zsrw}
  and in the general situation \eqref{eq:Z} in \cite{cf:GB}, and is
  based on an idea by C.  Monthus \cite{cf:Monthus}.  We mention that
  (the analog of) the lower bound in \eqref{eq:baundi} was recently
  proven in \cite{cf:T_fractmom} and \cite{cf:BCT} to become optimal in
  the limit $\gl\to\infty$ for the {\sl ``reduced copolymer model''}
  introduced in \cite[Sec. 4]{cf:BG} (this is a copolymer model where the
  disorder law $\bbP$ depends on the coupling parameter $\gl$).

As already mentioned, much
  attention has been devoted to the slope of the critical curve at the
  origin, in short the ``critical slope'',
  \begin{eqnarray}
\label{eq:limite}
      \lim_{\gl\searrow0}\frac{h_c(\gl)}\gl.
  \end{eqnarray}
  Existence of such limit is not at all obvious (and indeed was proven
  \cite{cf:BdH} only in the case of the ``standard copolymer model''
  \eqref{eq:Zsrw}), but is expected to hold in general. While the
  proof in \cite{cf:BdH} was given in the case $\bbP(\go_1=\pm1)=1/2$,
  it was shown in \cite{cf:GT_ptrf} (by a much softer argument) that
  the results of \cite{cf:BdH} imply (always for the model
  \eqref{eq:Zsrw}) that the slope exists and is the same in the
  Gaussian case we are considering here. Moreover, the critical slope
  is expected to be a function only of $\ga$ and not of the full
  $K(\cdot)$, at least for $0<\ga<1$, and to be independent of the
  choice of the disorder law $\bbP$, as long as the $\go_n$'s are IID,
  with finite exponential moments, centered and of variance $1$. In
  contrast, it is known that the critical curve $\gl\mapsto h_c(\gl)$
  {\sl does} in general depend on the details of $K(\cdot)$ (this
  follows from \cite[Prop.  2.11]{cf:BGLT}) and of course on the
  disorder law $\bbP$.  The belief in the {\sl universality of the
    critical slope} is supported by the result of \cite{cf:BdH} which,
  beyond proving that the limit \eqref{eq:limite} exists, identifies
  it with the critical slope of a continuous copolymer model, where
  the simple random walk $S$ is replaced by a Brownian motion, and the
  $\go_n$'s by a white noise.

Until
  recently, nothing was known about the value of the critical slope,
  except for
\begin{eqnarray}
\label{eq:liminfsup}
  \frac1{1+\ga}\le \liminf_{\gl\searrow0}\frac{h_c(\gl)}\gl\le
 \limsup_{\gl\searrow0}\frac{h_c(\gl)}\gl\le 1,
\end{eqnarray}
which follows from \eqref{eq:annb} and from the lower bound in
\eqref{eq:baundi} (note that the strict upper bound \eqref{eq:baundi}
does not imply a strict upper bound on the slope).
None of these bounds
is believed to be optimal. In particular, as we mentioned in the
introduction, for the standard copolymer model \eqref{eq:Zsrw}
numerical simulations \cite{cf:CGG} suggest a value around $0.83$ for
the slope.  This situation was much improved in \cite{cf:BGLT}: if
$\ga>1$, then \cite[Ths. 2.9 and 2.10]{cf:BGLT}
  \begin{eqnarray}
    \max\left(\frac1{\sqrt{1+\ga}},\frac12\right)\le\liminf_{\gl\searrow0}\frac{h_c(\gl)}\gl\le
\limsup_{\gl\searrow0}\frac{h_c(\gl)}\gl<1.
  \end{eqnarray}
  Note that $\ga>1$ and $\ga\le 1$ are profoundly different
  situations: the inter-arrival times of the renewal process have finite mean
  in the former case and infinite mean in the latter.  Moreover, it
  was proven in \cite[Th. 2.10]{cf:BGLT} that there exists $\ga_0<1$
  (which can be estimated to be around $0.65$), such that
  if $\ga\ge\ga_0$
  \begin{eqnarray}
    \liminf_{\gl\searrow0}\frac{h_c(\gl)}\gl>\frac1{1+\ga}.
  \end{eqnarray}
Note that this does not cover the case of the standard copolymer model
\eqref{eq:Zsrw}, for which $\ga=1/2$.

\subsubsection{A new upper bound on the critical slope}
Our main result  is that the upper bound in
\eqref{eq:liminfsup} is always strict:
\begin{theorem}
\label{th:slope}
For every $\ga>0$ there exists $\rho(\ga)\in(1/(1+\ga),1)$ such that,
whenever $K(\cdot)$ satisfies \eqref{eq:K},
\begin{eqnarray}
  \label{eq:slope}
  \limsup_{\gl\searrow0}\frac{h_c(\gl)}\gl\le
 \rho(\ga).
\end{eqnarray}
\end{theorem}
It is interesting to note that the upper bound \eqref{eq:slope} depends
only on the exponent $\ga$ and not the details of $K(\cdot)$.
This is coherent with the mentioned belief in universality of the
slope.

The new idea which allows to go beyond the results of \cite[Th. 2.10]{cf:BGLT} is
to bound above the fractional moments of $Z_{N,\go}$ in two steps:
\begin{enumerate}
\item
first we chop the system
into blocks of size $k$, the correlation length of the annealed model, and we decompose
$Z_{N,\go}$ according to which of the blocks contain points of $\tau$
\item
 only at that
point we apply the inequality \eqref{eq:ineqgamma}, where each of the $a_i$ corresponds to one of
the pieces into which the partition function has been decomposed.
\end{enumerate}

\begin{rem}\rm
  Theorem \ref{th:slope} holds in the more general situation where
  $\go$ is a sequence of IID random variables with finite exponential
  moments and normalized so that $\bbE\,\go_1=0,\bbE\, \go_1^2=1$. We
  state the result and give the proof only in the Gaussian case simply
  to keep technicalities at a minimum. The extension to the general
  disorder law can be obtained following the lines of \cite[Sec.
  4.4]{cf:BGLT}.
\end{rem}
\section{Proof of Theorem \ref{th:slope}.}

Fix  $\ga>0$, $1/(1+\ga)<\gga<1$ and define
\begin{eqnarray}
  \label{eq:A}
  A_{N,\gamma}:=\bbE(Z_{N,\go}^\gga).
\end{eqnarray}
The reason why we restrict to $\gga>1/(1+\ga)$ will be clear after
\eqref{eq:condgep}.
From now on we take $h=\rho \gl$, where
the value of
\begin{eqnarray}
\label{eq:rho<1}
  \rho:=\rho(
\ga,\gga)\in(\gamma,1)
\end{eqnarray}
will be chosen close
to $1$ later. Let
\begin{eqnarray}
  k:=\left\lfloor \frac1{\gl^2(1-\rho)}\right\rfloor
\end{eqnarray}
and note that, irrespective of how $\rho$ is chosen, $k$ can be made
arbitrarily large choosing $\gl$ small (which is no restriction since
in Theorem \ref{th:slope} we are interested in the limit $\gl\searrow
0$).  One sees from \eqref{eq:Fann} that, apart from an inessential
factor $2$, $k$ is just the inverse of the annealed free energy, i.e.,
\begin{eqnarray}
  k=\left\lfloor\frac2{\lim_{N\to\infty}(1/N)\log \bbE Z_{N,\go}}\right\rfloor.
\end{eqnarray}

We will show that, if $\rho(\ga,\gga)$ is sufficiently close to
$1$, there exists $\gl_0:=\gl_0(\gga,K(\cdot))>0$ such that for
$0<\gl<\gl_0$ there exists $c:=c(\gga,\gl,K(\cdot))<\infty$ such that
\begin{eqnarray}
\label{eq:auxA}
  A_{N,\gamma}\le c\,\left[K(N/k)\right]^\gamma
\end{eqnarray}
for every $N\in k\N$. In particular, by Jensen's inequality
and the fact that the sequence $\{(1/N)\bbE\log
Z_{N,\go}\}_N$ has a non-negative limit,
\begin{eqnarray}
  \tf(\gl,\rho(\ga,\gga)\gl)=\lim_{n\to\infty}
\frac1{nk}\bbE\log Z_{nk,\go}\le \lim_{n\to\infty}
\frac1{nk\gamma}\log A_{nk,\gamma}=0
\end{eqnarray}
for $\gl< \gl_0$.
 This implies \eqref{eq:slope} with
\begin{eqnarray}
   \rho(\ga):=\inf_{1/(1+\ga)<\gga<1}\rho(\ga,\gga).
\end{eqnarray}

From now on we  assume that $(N/k)$ is integer and we divide the interval
$\{1,\ldots,N\}$ into blocks
\begin{eqnarray}
  \label{eq:blocks}
  B_i:=\{(i-1)k+1,(i-1)k+2,\ldots,ik\}\;\mbox{with\;\;}i=1,\ldots,N/k.
\end{eqnarray}
Set $Z_{i,j}:=Z_{(j-i),\theta^i\go}$ (with the convention
$Z_{i,i}=1$), where $\theta$ is the left shift operator:
$(\theta^a\go)_i:=\go_{i+a}$ for $i,a\in\N$.  We have then the
identity (see Fig. \ref{fig:decompos})
\begin{figure}[h]
\begin{center}
\leavevmode
\epsfysize =4 cm

\psfragscanon
\psfrag{0}[c]{$0=i_0=j_0$}
\psfrag{B1}[c]{$B_1$}
\psfrag{B2}[c]{$B_2$}
\psfrag{B3}[c]{$B_3$}
\psfrag{B4}[c]{$B_4$}
\psfrag{Z1}[c]{$Z_{n_1,j_1}$}
\psfrag{Z2}[c]{$Z_{n_\ell,N}$}
\psfrag{n1}[c]{ $n_1$}
\psfrag{j1}[c]{ $j_1$}
\psfrag{nl}[c]{$n_\ell$}
\psfrag{N}[c]{$N$}
\psfrag{t2}[c]{$\tau_2$}
\psfrag{t3}[c]{$\tau_3$}
\psfrag{t4}[c]{$\tau_4$}
\epsfbox{decompos.eps}
\end{center}
\caption{\label{fig:decompos} A typical configuration which
  contributes to $\hat Z_\go^{(i_1,\ldots,i_\ell)}$ defined in
  \eqref{eq:Zhat}.  In this example we have a number $N/k=4$ of
  blocks, $\ell=2$ and $i_1=2$, while $i_\ell=N/k=4$ by definition
  (cf. \eqref{eq:dec1}).  Big black dots denote the $n_i$'s, white
  dots denote the $j_i$'s, while small dots are all the other points
  of $\tau$.  Note that $j_1-n_1<k$, as it should, and that there is
  no point of $\tau$ between a white dot and the next big black dot.
  In this example, the set $M$ of \eqref{eq:M} is $\{2,3,4\}$, and as
  a consequence $W$ defined in \eqref{eq:W} is $B_2\cup B_3\cup B_4$.
  In words: $n_1$ is the first point of $\tau$ after $0$, $j_1$ is the
  last point of $\tau$ which does not exceed $n_1+k-1$, $n_2$ is the
  first point after $j_1$, and so on.  The index of the block
  containing $n_r$ defines $i_r$. }
\end{figure}

\begin{eqnarray}
\label{eq:dec1}
  Z_{N,\go}=\sum_{\ell=1}^{N/k}\sum_{i_0:=0<i_1<\ldots<i_\ell=N/k}
\hat Z_\go^{(i_1,\ldots,i_\ell)}
\end{eqnarray}
where
\begin{eqnarray}
  \label{eq:Zhat}
  \hat Z_\go^{(i_1,\ldots,i_\ell)}&:=&
  \sum_{n_1\in B_{i_1}}\sum_{j_1=n_1}^{n_1+k-1}\sumtwo{n_2\in B_{i_2}:}{n_2\ge
    n_1+k}\sum_{j_2=n_2}^{n_2+k-1}\ldots \sumtwo{n_{\ell-1}\in B_{i_{\ell-1}}:}
{n_{\ell-1}\ge
    n_{\ell-2}+k}\sum_{j_{\ell-1}=n_{\ell-1}}^{n_{\ell-1}+k-1}\sumtwo
{n_\ell\in B_{N/k}:}{n_\ell\ge n_{\ell-1}+k}\\\nonumber
&&\times \varphi\left((0,n_1]\right)K(n_1)Z_{n_1,j_1}\varphi((j_1,n_2])K(n_2-j_1)
Z_{n_2,j_2}\times\ldots \\\nonumber
&&\times \varphi((j_{\ell-1},n_\ell])K(n_\ell-j_{\ell-1})Z_{n_\ell,N}
\end{eqnarray}
and, for $I\subset \N$,
\begin{eqnarray}
  \label{eq:phi}
  \varphi(I):=\frac{1+e^{-2\gl h |I|-2\gl\sum_{i\in I}\go_i}}2.
\end{eqnarray}
We have then, using the inequality
\begin{eqnarray}
  \label{eq:ineqgamma}
\left(  \sum_{i=1}^n a_i\right)^\gga\le \sum_{i=1}^n a_i^\gga
\end{eqnarray}
which holds for $0\le\gga\le1$ and $a_i\ge0$,
\begin{eqnarray}
  A_{N,\gamma}\le \sum_{\ell=1}^{N/k}\sum_{i_0:=0<i_1<\ldots<i_\ell=N/k}
\bbE\left[(\hat Z_\go^{(i_1,\ldots,i_\ell)})^\gamma\right].
\end{eqnarray}
Define
\begin{eqnarray}
  \label{eq:M}
  M:=M(i_1,\ldots,i_\ell):=\{i_1,i_2,\ldots,i_\ell\}\cup \{i_1+1,i_2+1,\ldots,
i_{\ell-1}+1\},
\end{eqnarray}
\begin{eqnarray}
  \label{eq:W}
  W:=W(i_1,\ldots,i_\ell):=\cup_{u\in M}B_u
\end{eqnarray}
and note that $1\le |M|< 2\ell$. With the conventions of Fig.
\ref{fig:decompos}, $W$ is the union of the blocks $B_i$ which either
contain a big black dot or such that $B_{i-1}$ contains a big black dot.
Note also that, for every $r$, the interval $[n_r,j_r]$ is a subset of $W$.

We want first of all to show that the $\varphi$'s can be effectively
replaced by constants.  To this purpose, we use the inequality
\begin{eqnarray}
  \label{eq:phiphi}
  \varphi((j_r,n_{r+1}])\le 2\varphi((j_r,n_{r+1}]\setminus W)
\,\varphi\left((j_r,n_{r+1}]\cap W\right)
\end{eqnarray}
(with the convention that $\varphi(\emptyset)=1$ and $j_0:=0$),
where $W$ was defined in \eqref{eq:W}.
This is simply due to the fact that, if $I_1$ and $I_2$ are two disjoint
subsets of $\N$, one has $\varphi(I_1\cup I_2)\le 2\varphi(I_1)\varphi(I_2)$.

We note that the two factors in the right-hand side of \eqref{eq:phiphi} are
independent random variables. Moreover, since
\begin{eqnarray}
  (j_r,n_{r+1}]\setminus W=\cup_{i:i_r+1<i<i_{r+1}} B_i,
\end{eqnarray}
we observe that the law of
\begin{eqnarray}
\label{eq:prod}
  \prod_{r=0}^{\ell-1}
\varphi\left((j_r,n_{r+1}]\setminus W\right)
\end{eqnarray}
depends only on $(i_1,\ldots,i_\ell)$ and not on the $n_r$'s and
$j_r$'s, and that, once $(i_1,\ldots,i_\ell)$ is fixed,
\eqref{eq:prod} is the product of $\ell$ independent random variables.
As a consequence,
\begin{eqnarray}
  A_{N,\gamma}\le \sum_{\ell=1}^{N/k}\sum_{i_0:=0<i_1<\ldots<i_\ell=N/k} 2^{\gamma\ell}
\prod_{r=0}^{\ell-1}\bbE \left(
\varphi\left((j_r,n_{r+1}]\setminus W\right)^\gamma\right)
\bbE \left[\left(\check Z^{(i_1,\ldots,i_\ell)}_\go\right)^\gamma\right],
\end{eqnarray}
where
\begin{eqnarray}
  \label{eq:Zcheck}
  \check Z^{(i_1,\ldots,i_\ell)}_\go&:=&
 \sum_{n_1\in B_{i_1}}\sum_{j_1=n_1}^{n_1+k-1}\sumtwo{n_2\in B_{i_2}:}{n_2\ge
    n_1+k}\sum_{j_2=n_2}^{n_2+k-1}\ldots \sumtwo{n_{\ell-1}\in B_{i_{\ell-1}}:}
{n_{\ell-1}\ge
    n_{\ell-2}+k}\sum_{j_{\ell-1}=n_{\ell-1}}^{n_{\ell-1}+k-1}\sumtwo
{n_\ell\in B_{N/k}:}{n_\ell\ge n_{\ell-1}+k}\\\nonumber
&&\times\varphi\left((0,n_1]\cap W\right)K(n_1)Z_{n_1,j_1}
\varphi((j_1,n_2]\cap W)K(n_2-j_1)
Z_{n_2,j_2}\times\ldots\\\nonumber
&&
\times
\varphi((j_{\ell-1},n_\ell]\cap W)K(n_\ell-j_{\ell-1})Z_{n_\ell,N}.
\end{eqnarray}
Thanks to \eqref{eq:ineqgamma}
and to the choice $h=\rho\gl$, for
every $I\subset \N$
\begin{eqnarray}
  \bbE (\varphi(I)^\gamma)\le \frac{1+e^{2\gl^2\gamma(\gamma-\rho)|I|}}
{2^\gamma}\le 2^{1-\gamma},
\end{eqnarray}
where the second inequality is implied by our assumption $\gamma<\rho$,
cf. \eqref{eq:rho<1}.
Then,
\begin{eqnarray}
  A_{N,\gamma}\le \sum_{\ell=1}^{N/k}\sum_{i_0:=0<i_1<\ldots<i_\ell=N/k} 2^\ell
\bbE \left[\left(\check Z^{(i_1,\ldots,i_\ell)}_\go\right)^\gamma\right].
\end{eqnarray}
In order to estimate
the remaining average, we use H\"older's inequality with $p=1/\gamma$
and $q=1/(1-\gamma)$:
\begin{eqnarray}
  \label{eq:Hold}
  \bbE \left[\left(\check Z^{(i_1,\ldots,i_\ell)}_\go\right)^\gamma\right]
=\tilde \bbE\left[\left(\check Z^{(i_1,\ldots,i_\ell)}_\go\right)^\gamma
\frac{\dd \bbP}{\dd \tilde \bbP}(\go)\right]
\le \left(
\tilde \bbE \check Z^{(i_1,\ldots,i_\ell)}_\go\right)^\gamma
\left(\tilde \bbE \left(\frac{\dd \bbP}{\dd \tilde \bbP}\right)^{1/(1-\gamma)}
\right)^{1-\gamma}
\end{eqnarray}
where, under the modified law $\tilde \bbE:=\tilde
\bbE^{(i_1,\ldots,i_\ell)}$, the $\{\go_i\}_{i\in\N}$ are still
Gaussian, independent and of variance $1$, but $\go_i$ has average
$1/\sqrt k$ if $i\in W(i_1,\ldots,i_\ell)$, while $\go_i$ has average
$0$, as under $\bbE$, if $i\notin W(i_1,\ldots,i_\ell)$. Since
$\tilde\bbE$ is still a product measure, it is immediate to check that
\begin{eqnarray}
  \left(\tilde \bbE \left(\frac{\dd \bbP}{\dd \tilde \bbP}\right)^{1/(1-\gamma)}
\right)^{1-\gamma}
=\left[\bbE\, e^{\left(\go_1/\sqrt k+1/(2k)\right)
\frac\gga{1-\gga}}
\right]^{(1-\gga)|W|}=e^{\frac{\gamma |W|}{2k(1-\gga)}}
\le e^{\frac\gamma{1-\gamma}\ell},
\end{eqnarray}
where we used the fact that $|W|=k|M|$ and $|M|\le 2\ell$.  Next, we observe that
\begin{eqnarray}
\label{eq:mnk}
  \tilde \bbE\varphi((j_r,n_{r+1}]\cap W )=
\frac{1+\tilde\bbE e^{-2\gl h |I|-2\gl \sum_{i\in I}\go_i}}2
\end{eqnarray}
with $I=(j_r,n_{r+1}]\cap W$. Thanks to the definition of
$k$ and to $h=\rho \gl$, \eqref{eq:mnk} equals
\begin{eqnarray}
  \frac{1+e^{2\gl^2|I|(1-\rho)-2\gl|I|/\sqrt k}}2\le
\frac{1+e^{2\gl^2|I|(1-\rho)-2|I|\gl^2\sqrt{1-\rho}}}2\le 1.
\end{eqnarray}
In conclusion, we proved
\begin{eqnarray}
  A_{N,\gamma}&\le&\sum_{\ell=1}^{N/k}\sum_{i_0:=0<i_1<\ldots<i_\ell=N/k}
\left(2\, e^{\frac\gamma{1-\gamma}}\right)^\ell\\
\label{eq:sqbr}
&\times&
\left[\sum_{n_1\in B_{i_1}}\ldots \sumtwo{n_\ell\in B_{N/k}:}{n_\ell\ge
n_{\ell-1}+k} K(n_1)\ldots K(n_\ell-j_{\ell-1})
U(j_1-n_1)\ldots U(N-n_\ell)\right]^\gamma
\end{eqnarray}
where, since $[n_r,j_r]$ is a subset of $W$ as observed above,
\begin{eqnarray}
  \label{eq:U}
  U(n):= \bbE Z_{n,\go}(\gl,\rho\gl+1/\sqrt k)
\end{eqnarray}
with the convention that $U(0):=1$.  In \eqref{eq:sqbr} we used
independence of $Z_{n_r,j_r}$ for different $r$'s (recall that
$\tilde\bbE$ is a product measure) to factorize the expectation.  The heart
of the proof of Theorem \ref{th:slope} is the following:
\begin{lemma}
\label{th:lemmagep}
There exists $\gl_0(\gga,K(\cdot))>0$ such that the following holds
for $\gl<\gl_0$.  If, for some $\gep>0$,
  \begin{eqnarray}
    \label{eq:cond1}
    \sum_{j=0}^{k-1}U(j)\le \gep\, Y(k,{\mathcal C}_K,\ga):=\gep\times \left\{
      \begin{array}{lll}
        \frac{k^\ga}{{{\mathcal C}_K}}  & {\rm if} & \ga<1 \\
 & & \\
\frac{2k}{{{\mathcal C}_K} \log k} & {\rm if} & \ga=1\\
 & & \\
 k & {\rm if} & \ga>1
      \end{array}
\right.
  \end{eqnarray}
and
  \begin{eqnarray}
    \label{eq:cond2}
    \sum_{j=0}^{k-1}\sum_{n\ge k}U(j)K(n-j)\le \gep
  \end{eqnarray}
then  the quantity in square brackets in \eqref{eq:sqbr} is bounded above
by
\begin{eqnarray}
  \label{eq:concllemma}
  C_1\, \gep^\ell\,3^{(3+2\ga)\ell} \prod_{r=1}^\ell \frac1{(i_r-i_{r-1})^{1+\ga}}
\end{eqnarray}
where $C_1:=C_1(\gep, k,K(\cdot))<\infty$.
\end{lemma}
Here and in the following, the positive and finite constants $C_i, i\ge1$
depend only on the arguments which are explicitly indicated, while
${{\mathcal C}_K}$ is the same constant which appears in \eqref{eq:K}.  \medskip

Assume that Lemma \ref{th:lemmagep} is true and that
\eqref{eq:cond1}-\eqref{eq:cond2} are satisfied. Then,
\begin{eqnarray}
\label{eq:330}
A_{N,\gamma}\le C_1(\gep, k,K(\cdot))^\gga
\sum_{\ell=1}^{N/k}\sum_{i_0:=0<i_1<\ldots<i_\ell=N/k}&&
\left(3^{\gamma(3+2\ga)}\,2\,e^{\frac\gamma{1-\gamma}}\right)^\ell \gep^{\gga \ell}
\\\nonumber
&&\times\prod_{r=1}^\ell \frac1{(i_r-i_{r-1})^{(1+\ga)\gga}}.
\end{eqnarray}

If moreover  $\gep$ satisfies
\begin{eqnarray}
  \label{eq:condgep}
\xi:=3^{\gga(3+2\ga)}\,2\,e^{\frac\gamma{1-\gamma}}\,\gep^{\gamma} \sum_{n\in\N}\frac1{n^{(1+\ga)\gga}}< 1
\end{eqnarray}
then it follows from \cite[Th. A.4]{cf:GB} that
\begin{eqnarray}
\label{eq:335}
  A_{N,\gamma}\le C_2(\gep,\gga,k,K(\cdot)) {(N/k)^{-(1+\ga)\gga}}
\end{eqnarray}
for every $N\in k\N$. Indeed, the sum in the right-hand
side of \eqref{eq:330} is nothing but the partition function of a
homogeneous pinning model \cite[Ch. 2]{cf:GB} of length $N/k$
 with pinning parameter $\xi$
such that the system is in the delocalized phase
(this is encoded in \eqref{eq:condgep}).
More precisely: to obtain \eqref{eq:335} it is sufficient to apply
Proposition \ref{th:A4} below,
with $\ga$ replaced by $(1+\ga)\gga-1>0$ and $K(\cdot)$ replaced
by
\begin{eqnarray}
  \hat K(n)=\frac 1{n^{(1+\ga)\gga}}\left(\sum_{j\in\N}\frac 1{j^{(1+\ga)\gga}}
\right)^{-1}.
\end{eqnarray}
\begin{proposition}\cite[Th. A.4]{cf:GB}
\label{th:A4}
  If $K(\cdot)$ is a probability on $\N$ which satisfies \eqref{eq:K}
  for some $\alpha>0$, then for every $\xi<1$ there exists
$c=c(K(\cdot),\xi)$ such that for every $N\in\N$
\begin{eqnarray}
  \sum_{\ell=1}^N\sum_{0=i_0<\ldots<i_\ell=N}\xi^\ell \prod_{r=1}^\ell
K(i_r-i_{r-1})\le c\,K(N).
\end{eqnarray}
\end{proposition}

Inequality \eqref{eq:auxA} is then proven; note that $C_2$ depends on
$\gl$ through $k$.  The condition $\gga>1/(1+\ga)$ which we required
since the beginning guarantees that the sum in \eqref{eq:condgep}
converges.  Note that \eqref{eq:condgep} depends on $K(\cdot)$ only
through $\ga$; this is important since we want $\rho$ in
\eqref{eq:slope} to depend only on $\ga$ and not on the whole
$K(\cdot)$.

\bigskip

To conclude the proof of Theorem \ref{th:slope}, we still have to
prove Lemma \ref{th:lemmagep} and to show that
\eqref{eq:cond1}-\eqref{eq:cond2} can be satisfied, with $\gep:=
\gep(\ga,\gga)$ satisfying \eqref{eq:condgep}, for all $\gl\le \gl_0$ and
$h= \gl \rho(\ga,\gga)$,
 if $\gl_0(\gga, K(\cdot))$ is sufficiently small and $\rho(\ga,\gga)$ is close to $1$
 (see also Remark \ref{rem:flow} below).

\medskip

{\sl Proof of Lemma \ref{th:lemmagep}.}
First of all, we get rid of $U(N-n_\ell)$ and effectively we replace
$n_\ell$ by $N$: the quantity in square brackets in \eqref{eq:sqbr}
is upper bounded by
\begin{eqnarray}
\label{eq:t1}
C_3(k,K(\cdot))
\sum_{n_1\in B_{i_1}}\ldots \sum_{j_{\ell-1}=n_{\ell-1}}^{n_{\ell-1}+k-1}&&
K(n_1)\ldots K(n_{\ell-1}-j_{\ell-2})K(N-j_{\ell-1})\\\nonumber
&& \times U(j_1-n_1)\ldots U(j_{\ell-1}-n_{\ell-1}).
\end{eqnarray}
Explicitly, one may take
\begin{eqnarray}
\label{eq:C3}
  C_3(k,K(\cdot))=k \;U(N-n_\ell)\supthree{j,n,N:}{0<j<n}{N-k\le n\le N} \frac{K(n-j)}{K(N-j)}:
\end{eqnarray}
one has
$$
U(N-n_\ell)\le \bbE Z_{N-n_\ell,\go}(\gl,\rho\gl)\le e^{2\gl^2 k}
$$
(recall \eqref{eq:UBann} and $N-n_\ell\le k$), while the supremum in
\eqref{eq:C3} is easily seen from \eqref{eq:K} to depend only on $k$
and $K(\cdot)$.

Recall that by convention $i_0=j_0=0$, and let also from now on
$n_\ell:=N$ and $n_0:=0$. We do the following:
\begin{itemize}
\item for every $1\le r\le \ell$ such that $i_r>i_{r-1}+2$
(which guarantees that between $j_{r-1}$ and $n_r$ there is at least
one full block), we use
  \begin{eqnarray}
    K(n_r-j_{r-1})\le 3^{\ga+2} \frac {{\mathcal C}_K}{(i_r-i_{r-1})^{1+\ga}k^{1+\ga}}.
  \end{eqnarray}
This is true under the assumption that $k\ge k_0$
with $k_0(\gga,K(\cdot))$ large enough, i.e., $\gl\le
\gl_0(\gga,K(\cdot))$ with $\gl_0$ small, since
$n_r-j_{r-1}\ge k(i_r-i_{r-1}-2)$ and
$\sup_{n>2}(n/(n-2))^{1+\ga}=3^{1+\ga}$.
\item for every $1\le r\le \ell$ such that $i_r\le i_{r-1}+2$, we
  leave $K(n_r-j_{r-1})$ as it is.
\end{itemize}
Then, \eqref{eq:t1} is bounded above by
\begin{eqnarray}
  \label{eq:t2}
 && C_3(k,K(\cdot))\frac{3^{(\ga+2)\ell}}{k^{(1+\ga)|J|}}\prod_{r\in J}
\frac {{\mathcal C}_K}{(i_r-i_{r-1})^{(1+\ga)}}\\\nonumber
&& \times\sum_{n_1\in B_{i_1}}\ldots \sum_{j_{\ell-1}=n_{\ell-1}}^{n_{\ell-1}+k-1}
\left(\prod_{r\in\{1,\ldots,\ell\}\setminus J}
K(n_r-j_{r-1})\right)U(j_1-n_1)\ldots U(j_{\ell-1}-n_{\ell-1}),
\end{eqnarray}
where
\begin{eqnarray}
  \label{eq:J}
  J:=J(i_1,\ldots,i_\ell):=\{1\le r\le \ell:i_r>i_{r-1}+2\}.
\end{eqnarray}
Now we can sum over $j_r,n_r,$ $1\le r<\ell$, using the two assumptions
\eqref{eq:cond1}-\eqref{eq:cond2}. We do this in three steps:
\begin{itemize}
\item First, for every $r\in J$ we sum over the allowed values of
  $j_{r-1}$ (provided that $r>1$, otherwise there is no
  $j_{r-1}$ to sum over) using \eqref{eq:cond1} and the constraint $0\le
  j_{r-1}-n_{r-1}<k$.  The sum over all such $j_{r-1}$ gives at
  most
\begin{eqnarray}
C_{4}(\gep,k,K(\cdot))
\gep^{|J|}Y(k,{\mathcal C}_K,\ga)^{|J|}
\end{eqnarray}
where $C_{4}$ takes care of the fact that possibly $1\in J$.
At this point we are left with the task of summing
\begin{eqnarray}
  \prod_{r\in\{1,\ldots,\ell\}\setminus J}\left[
K(n_r-j_{r-1})U(j_{r-1}-n_{r-1})\right]
\end{eqnarray}
(observe that $U(j_0-n_0)=1$) over all the allowed values
of $n_r,1\le r<\ell$ and of $j_{r-1}, r\in \{1,\ldots,\ell\}\setminus J$.
\item Secondly, using \eqref{eq:cond2}, we see that if
  $r\in\{1,\ldots,\ell\}\setminus J$ then the sum of
  $K(n_r-j_{r-1})U(j_{r-1}-n_{r-1})$ over the allowed values of
  $n_r$ and $j_{r-1}$ gives at most $ \gep$ (or $1$ if $r=1$).  The
  contribution from all the $n_r,j_{r-1}$ with
  $r\in\{1,\ldots,\ell\}\setminus J$ is therefore at most
\begin{eqnarray}
  C_{5}(\gep,k,K(\cdot))\gep^{\ell-|J|}.
\end{eqnarray}
This is best seen if one starts to sum on  $n_r,j_{r-1}$ for the largest
value of  $r\in\{1,\ldots,\ell\}\setminus J$
and then proceeds to the second largest, and so on.
\item Finally, the sum over all the $n_r$'s with $r\in J$ is trivial
(the summand does not depend on the $n_r$'s) and gives at
  most $k^{|J|}$.
\end{itemize}
In conclusion, we have upper bounded \eqref{eq:t2} by
\begin{eqnarray}
\label{eq:lastfact}
&& C_{6}(\gep,k,K(\cdot))3^{(3+2\ga)\ell} \gep^{\ell}\prod_{r=1
}^\ell\frac1{(i_r-i_{r-1})^{1+\ga}}\left[\frac{{\mathcal C}_K Y(k,{\mathcal C}_K,\ga)}
{k^{\ga}}\right]^{|J|}.
\end{eqnarray}
If $0<\ga<1$ it is clear from the definition of $Y(k,{\mathcal C}_K,\ga)$ that
the last factor equals $1$ and \eqref{eq:concllemma}
is proven. For $\ga\ge1$,
$$
\frac{{\mathcal C}_K Y(k,{\mathcal C}_K,\ga)}{k^\ga}
$$
can be made as small as wished with $k$ large (i.e. choosing $\gl_0$
small), so that we can again assume that the last factor in
\eqref{eq:lastfact} does not exceed $1$. Lemma \ref{th:lemmagep} is proven.
\qed

\medskip

Finally, we have:
\begin{proposition}
\label{th:cond12}
Let $ \gl_0(\gga,K(\cdot))>0$ be sufficiently small.
There exists  $\gep:=\gep(\ga,\gga)>0$ satisfying
  \eqref{eq:condgep} and $\rho:=\rho(\ga,\gga)\in (\gga,1)$ such that
conditions
  \eqref{eq:cond1}-\eqref{eq:cond2} are satisfied for all $0<\gl<\gl_0$
  and $h=\rho\gl$.
\end{proposition}

{\sl Proof of Proposition \ref{th:cond12}.}
We have by direct computation
\begin{eqnarray}
\label{eq:Uj}
  U(j)&=&\bE\left[\prod_{1\le n:\tau_n\le j}
\frac{1+e^{(2\gl^2(1-\rho)-2\gl/\sqrt k)(\tau_n-\tau_{n-1})}}2
\ind_{\{j\in\tau\}}\right]\\\nonumber
&\le&
\bE\left[\prod_{1\le n:\tau_n\le j}
\frac{1+e^{-\frac1{k\sqrt{1-\rho}}(\tau_n-\tau_{n-1})}}2
\ind_{\{j\in\tau\}}\right]\\
\label{eq:condiz}
&=&\bP(j\in\tau)\bE\left[\left.\prod_{1\le n:\tau_n\le j}
\frac{1+e^{-\frac1{k\sqrt{1-\rho}}(\tau_n-\tau_{n-1})}}2
\right|j\in\tau\right]
\end{eqnarray}
where in the inequality we assume that
$2\sqrt{1-\rho}<1$
(it is important that this condition does  not depend on $\gl$).
Of course, from \eqref{eq:Uj} we see that for every $j$
\begin{eqnarray}
U(j)\le \bP(j\in\tau).
\end{eqnarray}
Moreover, we know from
\cite[Th. B]{cf:Doney} that, if $0<\ga<1$,
\begin{eqnarray}
  \label{eq:Doney}
  \bP(j\in\tau)\stackrel{j\to\infty}\sim \frac{\ga\sin(\pi\ga)}
{\pi}\frac1{{{\mathcal C}_K} j^{1-\ga}},
\end{eqnarray}
while for $\ga=1$
\begin{eqnarray}
  \label{eq:a=1}
   \bP(j\in\tau)\stackrel{j\to\infty}\sim\frac{1}{{{\mathcal C}_K}\log j}
\end{eqnarray}
(cf. for instance  \cite[Th. A.6]{cf:GB}).
For every $a\in(0,1)$  one has then for $k$ sufficiently large, i.e., for
$\gl_0$ small,
\begin{eqnarray}
  \label{eq:somma1}
  \sum_{j=0}^{ak}U(j)\le   \sum_{j=0}^{ak}\bP(j\in\tau)\le
Y(ak,{\mathcal C}_K,\ga)
\end{eqnarray}
(we recall that $Y(\cdot,\cdot,\cdot)$ was defined in \eqref{eq:cond1}).
We need also the following fact:
\begin{lemma}
\label{th:lemma43}
For every $\ga>0$ there exists $C_7(\ga)<\infty$ such that the following holds.
If $0<\ga<1$ then, say, for $q>2$
  \begin{eqnarray}
\label{eq:fa}
    \limsup_{N\to\infty}\bE
\left[\left.
\prod_{1\le j:\tau_j\le N}\frac{1+e^{-(q/N)(\tau_j-\tau_{j-1})}}2
\right|N\in\tau
\right]
\le C_7(\ga)\frac{(\log q)^2}{q^\ga}.
  \end{eqnarray}
If  $\ga\ge1$ then, for every $q>0$,
  \begin{eqnarray}
\label{eq:faga1}
    \limsup_{N\to\infty}\bE
\left[\left.
\prod_{1\le j:\tau_j\le N}\frac{1+e^{-(q/N)(\tau_j-\tau_{j-1})}}2
\right|N\in\tau
\right]
\le C_7(\ga) e^{-q/2}.
  \end{eqnarray}
\end{lemma}
Lemma \ref{th:lemma43} was proven in \cite[Lemma 4.3]{cf:BGLT}.  We add a few
side remarks about its proof in Appendix \ref{sec:lemma43}.

The upper bound \eqref{eq:fa} is certainly not optimal, but
it gives us an estimate which vanishes for $q\to\infty$ and
which depends only on $\ga$ and $q$, which is all we need in the following.
Let us mention that in the case of the standard copolymer model,
using the property that for every $N\in2\N$ and every $k\in\{0,2,\ldots,N\}$
\begin{eqnarray}
  \bP^{\rm SRW}\left[\left.|\{1\le n\le N:\sign(S_n)=-1\}|=k
\right|N\in\tau\right]=\frac1{(N/2)+1}
\end{eqnarray}
(see \cite[Ch. III.9]{cf:feller1}) we obtain for every $q>0$
\begin{eqnarray}
  \lim_{N\to\infty}\bE^{\rm SRW}
\left[\left.
\prod_{1\le j:\tau_j\le N}\frac{1+e^{-(q/N)(\tau_j-\tau_{j-1})}}2
\right|N\in\tau
\right]=\frac{1-e^{-q}}q.
\end{eqnarray}

\subsubsection{Proof of \eqref{eq:cond1}}
Fix $\gep:=\gep(\ga,\gga)$ which satisfies \eqref{eq:condgep} and choose
\begin{eqnarray}
  \label{eq:agep}
a:=a(\gep,\ga):=\left\{
\begin{array}{lll}
(\gep/2)^{1/\ga}& {\rm if} &0<\ga<1\\
& & \\
\gep/4  & {\rm if} &\ga=1\\
 & & \\
\gep/2 & {\rm if} &\ga>1.
\end{array}
\right.
\end{eqnarray}
Via \eqref{eq:somma1} one finds (for $k$ sufficiently large)
\begin{eqnarray}
\label{eq:asac2}
  \sum_{j=0}^{ak}U(j)\le \frac\gep2 Y(k,{\mathcal C}_K,\ga).
\end{eqnarray}
Next we observe that, for $ak\le j<k$, choosing $\gl_0$ small and
$\rho$ sufficiently close to $1$ (how close, depending on $K(\cdot)$
only through the exponent $\ga$) we have $U(j)\le
(\gep/2)\bP(j\in\tau)$. This just follows from Lemma \ref{th:lemma43}
above (applied with $N\sim a k$ and $q=a/\sqrt{1-\rho}$) and from
\eqref{eq:condiz}, since $\gl_0$ small implies $k$ large.  As a
consequence,
\begin{eqnarray}
\label{eq:asac}
  \sum_{j=ak}^{k-1}U(j)\le \frac\gep2 Y(k,{\mathcal C}_K,\ga)
\end{eqnarray}
and \eqref{eq:cond1} follows.

\begin{rem}
\label{rem:flow}
  \rm It is probably useful to summarize the logic of the proof of
  \eqref{eq:cond1} (similar observations hold for the proof of
  \eqref{eq:cond2} below). Given $\ga$, one first fixes $1/(1+\ga)<\gga<1$, then
  $\varepsilon(\ga, \gga)$ which satisfies \eqref{eq:condgep}, then
  $a=a(\varepsilon,\ga)$ as in \eqref{eq:agep} and
  $\rho=\rho(\ga,\gga)<1$ such that the right-hand side of Eqs.
  \eqref{eq:fa}-\eqref{eq:faga1} is smaller than $(\gep/2)$ when $q$
  is replaced by $a/\sqrt{1-\rho}$. Once all these parameters are
  fixed, one chooses $\gl_0$ sufficiently small (i.e. $k$ sufficiently
  large) so that for all $\gl<\gl_0,h=\rho\gl$ the estimates
  \eqref{eq:asac2}-\eqref{eq:asac} hold.

\end{rem}

\subsubsection{Proof of \eqref{eq:cond2}}
If we choose $b:=b(\gep,\ga)$ small, $j\le bk$ implies $k-j\ge k/2$.
Therefore,
\begin{eqnarray}
  \sum_{j=0}^{bk}\sum_{n\ge k}U(j)K(n-j)\le
C_{8}(\ga)\frac{C_{K}}{ k^\ga} \sum_{j=0}^{bk}\bP(j\in\tau)\le
\frac\gep2
\end{eqnarray}
(if $k$ is sufficiently large and $b(\gep,\ga)$ is suitably
small). As for the rest of the sum: again from Lemma \ref{th:lemma43}
and \eqref{eq:condiz}, one has $U(j)\le (\gep/2) \bP(j\in\tau)$ for
every $bk\le j<k$.  Then,
\begin{eqnarray}
   \sum_{j=bk}^{k-1}\sum_{n\ge k}U(j)K(n-j)\le
\frac\gep2 \sum_{j=0}^{k-1}\sum_{n\ge k}\bP(j\in\tau)K(n-j)=\frac\gep2.
\end{eqnarray}
In the last equality, we used the fact that
\begin{eqnarray}
\label{eq:spiegaz}
  \sum_{j=0}^{k-1}\bP(j\in\tau)K(n-j)
\end{eqnarray}
is the $\bP$-probability that the first point of $\tau$ which does not
precede $k$ equals $n$. The sum over $n\ge k$ of \eqref{eq:spiegaz} then
clearly equals $1$, since $\tau$ is recurrent.
\qed

\appendix

\section{Remarks on the proof of Lemma \ref{th:lemma43}.}
\label{sec:lemma43}

The proof of Lemma \ref{th:lemma43} given in \cite[Lemma 4.3]{cf:BGLT} works as follows.
Let $X_N:=\max\{n=0,\ldots,N:n\in\tau\}$, i.e., the last
point of $\tau$ up to $N$.
First of all one shows that
\begin{eqnarray}
\label{eq:nocondiz}
&&\limsup_{N\to\infty} \bE
\left[\left.
\prod_{1\le j:\tau_j\le N}\frac{1+e^{-(q/N)(\tau_j-\tau_{j-1})}}2
\right|N\in\tau
\right]\\
\label{eq:nocondiz2}
&&
\label{eq:C9}
\le C_{9}(\ga) \limsup_{N\to\infty}
 \bE
\left[
\frac{1+e^{-(q/N)(N-X_N)}}2
\prod_{1\le j:\tau_j\le N}\frac{1+e^{-(q/N)(\tau_j-\tau_{j-1})}}2
\right].
\end{eqnarray}
We detail below the proof of this inequality in order to leave no
doubts on the fact that the constant $C_9$  depends
only on $\ga$. This was not emphasized in the proof of \cite[Lemma
4.3]{cf:BGLT} since it was not needed there.

For $\ga\ge1$, it follows from \cite[Eqs. (4.23) and (4.49)]{cf:BGLT}
that the $\limsup$ in the right-hand side of \eqref{eq:C9} is actually
a limit, and equals $\exp(-q/2)$ (the $q/8$ which appears in \cite[Eq.
(4.49)]{cf:BGLT} can be immediately improved into $q/2$). As a side
remark, the expectation in \eqref{eq:C9}, irrespective of the value of
$\ga$ and $N$, is not smaller than $\exp(-q/2)$; this just follows
from the convexity of the exponential function:
$$
\frac{1+e^{-(q/N)x}}2\ge e^{-q/(2N)\,x}.
$$
For $0<\ga<1$, the $\limsup$ in
\eqref{eq:C9} does not exceed $C_{10}(\ga)(\log q)^2/q^\ga$, as was
proven in \cite[Eq. (4.43)]{cf:BGLT}.

Finally we prove \eqref{eq:nocondiz2}, which is quite standard.
The expectation in \eqref{eq:nocondiz} is bounded above by
\begin{eqnarray}
&& \bE
\left[\left.
\frac{1+e^{-(q/N)(N/2-X_{N/2})}}2
\prod_{1\le j:\tau_j\le N/2}\frac{1+e^{-(q/N)(\tau_j-\tau_{j-1})}}2
\right|N\in\tau
\right]\\
&&=\sum_{i=0}^{N/2} \bE
\left[\left.
\frac{1+e^{-(q/N)(N/2-i)}}2
\prod_{1\le j:\tau_j\le N/2}\frac{1+e^{-(q/N)(\tau_j-\tau_{j-1})}}2
\right|X_{N/2}=i
\right]\\&&\times\bP(X_{N/2}=i|N\in\tau)
\end{eqnarray}
and \eqref{eq:nocondiz2} follows if we can prove that
\begin{eqnarray}
  \limsup_{N\to\infty}\max_{0\le i\le N/2}\frac{\bP(X_{N/2}=i|N\in\tau)}
{\bP(X_{N/2}=i)}\le C_{11}(\ga).
\end{eqnarray}
To show this, we  use repeatedly
\eqref{eq:Doney} and \eqref{eq:K}. We start from the identity
\begin{eqnarray}
  \frac{\bP(X_{N/2}=i|N\in\tau)}
{\bP(X_{N/2}=i)}=\frac{
\sum_{j=(N/2)+1}^N K(j-i)\bP(N-j\in\tau)
}{\bP(N\in\tau)\sum_{j=(N/2)+1}^\infty K(j-i)}.
\end{eqnarray}
The denominator is lower bounded, uniformly in $0\le i\le N/2$,
by
\begin{eqnarray}
  \bP(N\in\tau)\sum_{j=(N/2)+1}^\infty K(j)\ge \frac{C_{12}(\ga)}{N},
\end{eqnarray}
where the last inequality holds for $N$ sufficiently large.
As for the numerator: always for $N$ sufficiently large,
\begin{eqnarray}
  \sum_{j=(N/2)+1}^{(3/4)N} K(j-i)\bP(N-j\in\tau)\le
C_{13}\bP(N\in\tau)\sum_{j=(N/2)+1}^{\infty}K(j-i)
\end{eqnarray}
and, uniformly on $0\le i\le N/2$,
\begin{eqnarray}
   \sum_{j=(3/4)N}^{N} K(j-i)\bP(N-j\in\tau)\le C_{14}\frac{{\mathcal C}_K}{N^{1+\ga}}
\sum_{j=0}^{N}\bP(j\in\tau)\le \frac{C_{15}(\ga)}{N}.
\end{eqnarray}
\qed

\section*{Acknowledgments}
This work was partially supported by ANR, grant POLINTBIO and grant
LHMSHE. I wish to thank the anonymous referees for the careful reading of the
manuscript and for several useful comments.


TITLE: Solving $E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$
QUESTION [4 upvotes]: $$E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$$
I got no idea how to find the solution to this. Can someone put me on the right track?
Thank you very much.

REPLY [1 votes]: $$\text{Let us check the value of }\frac a{\sin\theta}+\frac b{\cos\theta}$$
$$\frac a{\sin\theta}+\frac b{\cos\theta}=\frac{a\cos\theta+b\sin\theta}{\cos\theta\sin\theta}$$
Putting $a=r\sin\alpha,b=r\cos\alpha$ where $r>0$
Squaring & adding we get $r^2=a^2+b^2\implies r=+\sqrt{a^2+b^2}$
$$\implies \frac a{\sin\theta}+\frac b{\cos\theta}=\frac{2\sqrt{a^2+b^2}(\sin\theta\cos\alpha+\cos\theta\sin\alpha)}{\sin2\theta}$$
$$=2\sqrt{a^2+b^2}\cdot\frac{\sin(\theta+\alpha)}{\sin2\theta}\text{ as }\sin2\theta=2\sin\theta\cos\theta$$
Now, the solution of $P\sin x= Q\sin A $ is general intractable unless  $P=0$ or $Q=0$ or $P=\pm Q\ne0$
Here the coefficients of $\sin(\theta+\alpha),\sin2\theta$ can not be $0$
So, either $\sin2\theta=\sin(\theta+\alpha)$ or $\sin2\theta=-\sin(\theta+\alpha)$
$$\begin{array}{|c|c|c|}
 \hline \text{ Case } & \sin2\theta=\sin(\theta+\alpha) & \sin2\theta=-\sin(\theta+\alpha)=\sin(-\theta-\alpha)  \text{ as }\sin(-x)=-\sin x      \\
 \hline \text{General Solution} &  2\theta=n180^\circ+(-1)^n(\theta+\alpha)\text{  where }n\text{ is any integer } & 2\theta=n180^\circ+(-1)^n(-\alpha-\theta)\text{  where }n\text{ is any integer }    \\
 \hline n=2m & \alpha=\theta-m360^\circ\equiv\theta\pmod{360^\circ} & \alpha=m360^\circ-3\theta\equiv-3\theta \\
\hline n=2m+1 & \alpha=(2m+1)180^\circ-3\theta\equiv 180^\circ-3\theta & \alpha=\theta-(2m+1)180^\circ\equiv\theta+180^\circ  \\
 \hline
  \end{array} $$
Here $a=1,b=-\sqrt3$ and $\theta=10^\circ$
Taking $\sin2\theta=\sin(\theta+\alpha), \alpha=\theta=10^\circ$ or $=180^\circ-3\theta=150^\circ$
$\implies r=+\sqrt{a^2+b^2}=2$ and $\cos \alpha=\frac br=-\frac{\sqrt3}2$ and $\sin\alpha=\frac ar=\frac12\implies \alpha$ lies in the 2nd Quadrant, $\implies \alpha=150^\circ$
$$\text{So,} \frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}=2\sqrt{1^2+(-\sqrt3)^2}\cdot\frac{\sin(10^\circ+150^\circ)}{\sin(2\cdot10^\circ)}=2\cdot2=4$$


text:
theory FL_Logical_Equivalence
imports
  FL_Validity
begin

section \<open>(Strong) Logical Equivalence\<close>

text \<open>The definition of formulas is parametric in the index type, but from now on we want to work
with a fixed (sufficiently large) index type.\<close>

locale indexed_effect_nominal_ts = effect_nominal_ts satisfies transition effect_apply
  for satisfies :: "'state::fs \<Rightarrow> 'pred::fs \<Rightarrow> bool" (infix "\<turnstile>" 70)
  and transition :: "'state \<Rightarrow> ('act::bn,'state) residual \<Rightarrow> bool" (infix "\<rightarrow>" 70)
  and effect_apply :: "'effect::fs \<Rightarrow> 'state \<Rightarrow> 'state" ("\<langle>_\<rangle>_" [0,101] 100) +
  assumes card_idx_perm: "|UNIV::perm set| <o |UNIV::'idx set|"
      and card_idx_state: "|UNIV::'state set| <o |UNIV::'idx set|"
begin

  definition FL_logically_equivalent :: "'effect first \<Rightarrow> 'state \<Rightarrow> 'state \<Rightarrow> bool" where
    "FL_logically_equivalent F P Q \<equiv>
       \<forall>x::('idx,'pred,'act,'effect) formula. x \<in> \<A>[F] \<longrightarrow> (P \<Turnstile> x \<longleftrightarrow> Q \<Turnstile> x)"

  text \<open>We could (but didn't need to) prove that this defines an equivariant equivalence relation.\<close>

end

end

subset_name: curated


TITLE: Error in uniform convergence question?
QUESTION [1 upvotes]: Suppose that $(X,\ d)$ and $(Y,\ \rho)$ are metric spaces, that
  $f_n:X\to Y$ is continuous for each $n$, and that $(f_n)$ convergence
  pointwise to $f$ on $X$. If there exists a sequence $(x_n)$ in $X$
  such that $x_n\to x$ in $X$ but $f_n(x_n)\not\to f(x)$, show that
  $(f_n)$ does not converge uniformly to $f$ on $X$.

I've managed to "prove" that the question is self-contradictory, so please find my error.
$\forall n\in\mathbb{N}$, $f_n$ is continuous at $x$. Let $\epsilon>0$. $\forall n\in\mathbb{N}$ $\exists\delta>0$ s.t. $\rho(f_n(y),\ f_n(x))<\epsilon/2$ for all $y\in X$ s.t. $d(y,\ x)<\delta$. ------------ (1)
Since $x_m\to x$, $\exists N_1\in\mathbb{N}$ s.t. $d(x_m,\ x)<\delta$ $\forall m\ge N_1$. ------------- (2)
From (1) and (2),
$\forall n$, $\forall m\ge N_1\implies d(x_m,\ x)<\delta\implies \rho(f_n(x_m),\ f_n(x))<\epsilon/2$ ------------- (3)
Also, $f_n(y)\to f(y)$ for all $y\in X$ due to pointwise convergence. So, $\exists N_2\in\mathbb{N}$ s.t. $\rho(f_n(y),\ f(y))<\epsilon/2$ $\forall n\ge N_2$ and $\forall y\in X$. ----------- (4)
Let $N_3=\max(N_1,\ N_2)$. Suppose $n\ge N_3$. Then $n\ge N_1$ and $n\ge N_2$.
$\begin{aligned}
\implies\rho(f_n(x_n), f(x))&\le\rho(f_n(x_n),\ f_n(x))+\rho(f_n(x),\ f(x)) \\
&<\epsilon/2+\epsilon/2\text{ [Using (3) and (4)]} \\
&=\epsilon
\end{aligned}$
I have thus "proved" that $f_n(x_n)\to f(x)$, contradicting the question. Where have I gone wrong?

REPLY [1 votes]: For a correct proof (by contradiction), show that $f(x_n) \not\to f(x)$ is impossible if convergence is uniform using
$$\rho(f_n(x_n), f(x)) \leqslant \rho(f_n(x_n), f(x_n)) + \rho(f(x_n), f(x))$$
Note that $f$ must be continuous when the sequence of continuous functions $f_n$ is uniformly convergent. Thus each term on the RHS is smaller than $\epsilon/2$ for sufficiently large $n$  -- the first by the uniform convergence $f_n \to f$ and the second by the convergence $x_n \to x$ and the continuity of $f$.


text:
\begin{definition}[Definition:Jacobi's Equation of Functional/Dependent on N Functions]
Let:
:$\ds \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
be a [[Definition:Real Functional|(real) functional]], where $\map {\mathbf y} a = A$ and $\map {\mathbf y} b = B$.
Let:
:$\ds \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h} \rd x$
be a [[Definition:Quadratic Functional|quadratic functional]], where:
:$P_{ij} = \dfrac 1 2 F_{y_i'y_j'}$
:$Q_{ij} = \dfrac 1 2 \paren {F_{y_i y_j} - \dfrac \d {\d x} F_{y_i y_j'} }$
Then the [[Definition:Euler's Equation for Vanishing Variation|Euler's equation]] of the latter [[Definition:Real Functional|functional]]:
:$-\map {\dfrac \d {\d x} } {\mathbf P \mathbf h'} + \mathbf Q \mathbf h = \mathbf 0$
is called '''Jacobi's Equation''' of the former [[Definition:Real Functional|functional]].
{{NamedforDef|Carl Gustav Jacob Jacobi|cat = Jacobi}}
\end{definition}

config: wiki


TITLE: Show that the angular momentum is conserved (Noether)
QUESTION [4 upvotes]: It is the first time I get in touch with classical mechanics and this stuff...
Suppose a mechanical system has $n$ degrees of freedom described by coordinated $q\in V\subset\mathbb{R}^n$, set $v=\dot{q}$. In the situation of particles under the influence of some forces we have the Lagrange function
$$
L(v,q)=\frac{1}{2}vMv-U(q).~~~(*)
$$
where $M$ is a positive diagonal matrix with the masses of the particles as entries and $U$ is the potential corresponding to the forces. Let
$$
p(v,q)=\frac{\partial L}{\partial v}(v,q)
$$
denote the momentum.

Theorem of Noether Let $\Phi(t,q)$ be the flow generated by $f(q)$. If $\Phi$ leaves the Lagrangian invariant, then
    $$
I(v,q)=p(v,q)\cdot f(q)
$$
    is a constant of motion.

Now there is the following task:

Consider $L(v,q)$ from $(*)$ in $\mathbb{R}^3$ with $M=m\mathbb{I}_3$ and suppose $U(q)=U(\lvert q\rvert)$ is rotation invariant. Show that the angular momentum $l= x\wedge p$ is conserved in this case. Here $\wedge$ denotes the cross product in $\mathbb{R}^3$.

I think I have to apply the theorem of Noether, but I do not know how to do so. The first problem for me already is to calculate $p(v,q)$.

REPLY [2 votes]: If we have a Lagrangain $L(\vec{v},\vec{q})$ and the (infinitesimal) transformation $\vec{q}\to \vec{q} + \epsilon \vec{\sigma}(t)$ leaves it invariant then Noether's theorem states that
$$I = \frac{dL}{d\vec{v}}\cdot \vec{\sigma}$$
is a conserved quantity. For rotations about a direction $\vec{n}$ we have $\vec{\sigma} = \vec{n}\times \vec{q}$ so
$$I = \vec{p} \cdot (\vec{n}\times \vec{q}) = \vec{n}\cdot (\vec{q}\times \vec{p}) = \vec{n}\cdot \vec{l}$$
Since $\vec{n}$ is arbitrary we get that $\vec{l}$ is conserved. For more information see for example this note.

One can also solve this problem explicitly. For the case $M = m I_n$ and $U(\vec{q})  = U(q)$ where $q=|\vec{q}|$ we find
$$\frac{dL}{d\vec{v}} = m\vec{v} \equiv \vec{p},~~~\frac{dL}{d\vec{q}} = -\frac{dU(q)}{dq}\frac{\vec{q}}{q}$$
so from the Euler-Lagrange equations the equation of motion is
$$\frac{d\vec{p}}{dt} = \frac{dU}{dq}\frac{\vec{q}}{q}$$
Using the equation of motion we can calculate the time derivative of $\vec{l} = \vec{q}\times \vec{p}$ to find
$$\frac{d\vec{l}}{dt} = \frac{d\vec{q}}{dt}\times \vec{p} +  \vec{q}\times \frac{d\vec{p}}{dt} = \frac{\vec{p}\times \vec{p}}{m} + \frac{dU}{dq}\frac{\vec{q}\times\vec{q}}{q} = 0$$
and angular momentum is conserved.


TITLE: Gibbs free energy + maximum work
QUESTION [0 upvotes]: The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a closed system; this maximum can be attained only in a completely reversible process. This maximum work is equal to $H-TS$. What kind of energy is this $TS$ energy? From where does it come if it is not from the internal energy of the system?

REPLY [2 votes]: So the internal energy $U$ of a system is subject to a thermodynamic relation which looks like $$dU = T~dS-P~dV+\sum_i\mu_i~dN_i$$where $T$ is the temperature, $S$ is the entropy, $P$ is the pressure, $V$ is the volume, $N_i$ are the number of atoms of various species $i$, and $\mu_i$ are their chemical potentials. You can view this as a definition of temperature, pressure, and chemical potential. The symbol $d$, if you have never seen it before, comes from the word "difference" and means "a really tiny change in," so the equation reads, "a really tiny change in internal energy is equal to the temperature times a really tiny change in entropy minus..."
Now this is great if entropy, volume, and the numbers of particles are fixed: it forces the internal energy to stay fixed as well. But what if, say, heat energy can flow into or out of the system? If the system is in good thermal contact with an "environment" that maintains temperature $T$, then it will "borrow energy" from that environment until the temperatures equal each other. It's the same if volume is not fixed: it will "borrow volume" from the environment until the pressures equal each other.
The mathematical trick that we play here is called a "Legendre transform" and amounts to observing that $d(TS) = S~dT + T~dS:$ by subtracting an appropriate quantity from the internal energy $U$ we can produce a relation which again does not vary. For the Gibbs free energy $G = U + PV - TS,$ the relation becomes:$$dG = V~dP - S~dT + \sum_i \mu_i~dN_i,$$and we use it when the system is in contact with an environment with which it doesn't share particles, but does share volume and heat flows. If the pressure, temperature, and particles remain constant, then the Gibbs free energy remains constant even if the internal energy changes. Maybe, say, you've got a constant pressure on a gas condensing into a liquid at its boiling temperature: the internal energy is probably changing, but the Gibbs free energy is not.
If there is a Gibbs-energy change which doesn't come from or go into the internal energy U, then it is coming from or going into the environment that the system is connected with, either in the form of volume expansion or in the form of added entropy.


TITLE: The basic facts about $\mathbb{R}[x,y].$
QUESTION [4 upvotes]: There's all this basic stuff about $\mathbb{R}[x,y]$ that I don't know, and this impacts negatively on my ability to teach e.g. Year 10 and 11 mathematics effectively (for example).
So let $R$ denote a ring. A well-known result says that if $R$ is a UFD, then $R[x]$ is, too. This tells us that $\mathbb{R}[x,y]$ is a UFD. So prime elements are the same as irreducible elements, and every element of $\mathbb{R}[x,y]$ factors uniquely as a product of such elements. That's about all I know. So, here's a few questions designed to help me patch up my knowledge.

Are the irreducible elements in $\mathbb{R}[x,y]$ well-known and easy to describe, and if so, what are they? When one variable is concerned, the discriminant tells us everything we need to know, of course.
It would seem that both the "circle polynomial" $x^2+y^2-1$ and the "hyperbola polynomial" $x^2-y^2-1$ are both irreducible in $\mathbb{R}[x,y]$. Is that correct? If so, how do we know?
At the risk of asking a meaningless question:) assuming they're both irreducible, why is zero-locus of $x^2+y^2-1$ connected, but the zero-locus of $x^2-y^2-1$ is not?
More generally, can we say anything about the zero-loci of irreducible polynomials that, say, a high school student can understand?
Does homogenization somehow help us? If so, how?

REPLY [1 votes]: What you are asking about is essentially the basics of classical algebraic geometry.  Instead of answering your question directly, I would like to discuss two things that your question made me think of.
First, there are a lot of strange phenomena that happen because $\mathbb R$ is not algebraically closed.  Things are substantially simpler when you work over the complex numbers.  Suddenly, the zero locus of $y^2-x^2-1$ isn't two branches, but one giant sheet.  It just turned out that you had a slice of the sheet that contained two separate parts.  Moreover, the map $(a,b)\mapsto (ia,b)$ gives a bijection between the points in your hyperbola and circle examples.  Two objects that seemed to be different are actually the same if you change your perspective, which has interesting consequences.
Second, homogenization is very useful.  I don't know what your background is, but likely you have studied compactness and seen the wonderful things that you can do with it.  If you have seen how you can add a point at infinity to the euclidean plane to make a sphere (an instance of the one point compactification), you know that we can sometimes better understand things by replacing them with compact versions, where things are slightly better behaved.
It turns out that homogenization gives an algebraic analogue of compactification.  Given a field $k$, define $n$ dimensional projective space over $k$ \mathbb P^n(k) to be the quotient of $k^{n+1}-0$ by the equivalence relation $(a_0,a_1,\ldots, a_n)\sim (\lambda a_0, \lambda a_1,\ldots, \lambda a_n)$ for $\lambda \in k-\{0\}$.  When $k=\mathbb R$ this is the same as taking the $n$-dimensional sphere and identifying antipodal points.
Every point of projective space falls into two sets, those where $a_0=0$, and those where $a_0\neq 0$.  For points of the latter type, every point is equivalent to a unique point of the form $(1,a_1,a_2,\ldots, a_n)$.  Thus, projective space may be thought of as taking $k^n$ adjoining the points of the form $[0:a_1:a_2:\ldots:a_n]$ "at infinity" (where the notation stands for the equivalence class of the point).  This is an algebraic way of compactifying $k^n$.
What does this have to do with homogenization?  Suppose that $f$ is a homogenous polynomial of degree $d$ in $n+1$ variables.  Then $f(\lambda x)=\lambda^d f(x)$, and so if $x$ is a root of $f$, then so is $\lambda x$.  This means the zero locus of $f$ can be viewed as living in $\mathbb P^n(k)$ instead of just living in $k^{n+1}$.
Moreover, given a polynomial $g$ in $n$ variables, we can form its homogenizations $\widetilde{g}$ in $n+1$ variables, and (letting $x_0$ be the new variable), $g(x_1,x_2,\ldots,x_n)=0 \Leftrightarrow \widetilde{g}(x_0,x_1,\ldots, x_n)=0$.  Therefore, the zero locus of $\widetilde{g}$ in projective space is the zero locus of $g$ with added points at infinity.
So why do we care?  One reason is that adding back these points at infinity can simplify matters, just as adding in complex points can.  Things that looked different can be seen to be different views of the same object.  But other things can happen too.  For an example, I give Bezout's theorem.
Theorem: Let $f,g \in \mathbb C[x,y]$ of degrees $m$ and $n$, and suppose that $f$ and $g$ have no nontrivial factors in common.  Then, considering the intersections in projective space and properly counting the multiplicities of the intersection (e.g., the curve $y=x^3$ intersects the curve $y=0$ with multiplicity $3$ at $(0,0)$), the zero loci of $f$ and $g$ intersect in exactly $mn$ points.
Compare this to the result given by Hagen.  You would like the number of points of intersection to be more than just bounded by the product of the degrees.  Unfortunately, some points of intersection are complex, but even taking care of these, sometimes two curves can intersect at infinity (e.g., in the projective plane every pair of lines has a unique point of intersection, but if you take two parallel lines, they intersect at infinity).  Thus, by adding in complex points and points at infinity, we can strengthen the link between the algebraic side and the geometric side.


TITLE: Complete graph edge colouring in two colours: lower bound for number of monochromatic triangles
QUESTION [0 upvotes]: Say $K_n$ is a complete graph. Show that any coloring of edges of $K_n$ with $n \ge 6$ in two colors contains at least $$\frac1{20}\binom{n}3$$ monochromatic triangles.  Any ideas on how to use Ramsey theory to solve this?
Any help is much appreciated!

REPLY [0 votes]: Gerry Meyerson's answer will actually get you the better result that the number of monochromatic numbers is at least $\frac1{10}\binom n3.$ Here is an easier way to prove the weaker result stated in the question using Ramsey's theorem for $2$-colorings of $K_6.$
Let $M$ be the number of monochromatic triangles. Let $P$ be the number of ordered pairs $(X,Y)$ where $X$ is a monochromatic triangle and $Y$ is a $K_6$ containing $a.$
On the one hand, $P=M\binom{n-3}3,$ since each monochromatic triangle is contained in $\binom{n-3}3\ $ $K_6$'s.
On the other hand, $P\ge\binom n6,$ since each $K_6 contains at least one monochromatic triangle.
It follows that $M\binom{n-3}3\ge\binom n6,$ whence (assuming $n\ge6$)
$$M\ge\frac{\binom n6}{\binom{n-3}3}=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{6!}\cdot\frac{3!}{(n-3)(n-4)(n-5)}=\frac{n(n-1)(n-2)\cdot3!}{6!}=\frac{n(n-1)(n-2)}{3!}\cdot\frac{3!}{4\cdot5\cdot6}=\frac1{20}\binom n3.$$
P.S. Actually, if we know that every $2$-edge-colored $K_6$ contains at least two monochromatic triangles, then we know that $P\ge2\binom n6,$ so the final result is improved to $M\ge\frac1{10}\binom n3.$