TITLE: How many quadratic non-residues are there in the image of $f(a) = a^2 - x^2$?
QUESTION [2 upvotes]: When reading about Cipolla's algorithm on Wikipedia, I found that the number of quadratic non-residues in the image of $f:\Bbb F \longrightarrow \Bbb F, f(a) = a^2 - x^2$ where $\Bbb F$ is a finite field of prime cardinality $p$ and $x \in \Bbb F$, is $\frac{p-1} 2$. Why?

I was thinking that there might be some sort of symmetry between the nonzero quadratic residues and the quadratic nonresidues, but I don't see it.
[edit]
Wherever I previously said quadratic residue, it nows says quadratic nonresidue.

REPLY [0 votes]: $$f(a) = a^2 - x^2 = a^2\left(1 - \left( \frac x a \right)^2 \right)$$
Now $x/a$ can be anything, so we've got the $1 - $ an arbitrary quadratic residue. The problem becomes counting how many quadratic residues there are in the image of $f(x) = 1 - x^2$ (I found it distracting to make the constant in the function $x$).
So let's say we've got $$1 - x^2 = y^2 \iff 1 = x^2 + y^2$$ This looks like a circle. So we have to count the number of points on the unit circle, and then get rid of repeated values of $x$.
Maybe we can do that considering lines from $(-1,0)$ to the circle. Every point on the circle has such a line, and so we just have to show that for any slope we get two points only one of which is $(-1,0)$ (therefore making it one-to-one). The non-$(-1,0)$ solution to this is well known: $$(\frac{2t}{1 + t^2}, \frac{1-t^2} {1+t^2})$$ where $t$ is the slope of the line. So we have that every point yields a unique line and every line yields a point, so one-to-one as claimed.
Now divide the number of points on the circle (which is $p+1$, when counting every slope value and the vertical line) by $2$ and then add $1$ to get $\frac{p+1}{2} + 1 = \frac{p + 3} 2$


TITLE: Matrix-vector multiplication/cross product problem
QUESTION [3 upvotes]: How can I generally solve equations of the form $\mathbf{A} \mathbf{w} =
\begin{pmatrix} x \\ y \\ z \end{pmatrix}
\times \mathbf{w}$ for the matrix $\mathbf{A},$ where $\mathbf{w}$ can be any vector? I recognize that you could just set $\mathbf{w}$ to a vector with simple values, such as $\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$, but doing so still isn't helpful. Also, $x,$ $y,$ and $z$ are entirely independent variables.

REPLY [2 votes]: OK, let's put it other way as $\mathbf{w}\times \mathbf{v}=-\mathbf{A}\mathbf{w}$.
We can write the the cross product as vector-matrix multiplication:
$$\mathbf{w}\times\mathbf{v} =[\mathbf{w}]_\times \mathbf{v}=\begin{bmatrix}\,0&\!-w_{3}&\,\,w_{2}\\\,\,w_{3}&0&\!-w_{1}\\-w_{2}&\,\,w_{1}&\,0\end{bmatrix}\mathbf{v}.$$
So you can write your equation as a system of linear equations
$$[\mathbf{w}]_\times \mathbf{v}=-\mathbf{A}\mathbf{w}.$$
Matrix $[\mathbf{w}]_\times$ has rank $2$ and its nullspace is spanned by $[w_1,\,w_2,\,w_3]^\top$.
Now depending on whether you assume $w_2\neq 0$ or $w_3\neq 0$, you can transform this system and find a particular solution. However, this solution can be found only if $\langle\mathbf{w},\mathbf{Aw}\rangle=0$. In particular, this implies that $\mathbf{A}^\top=-\mathbf{A}$.


TITLE: Abstract algebra book suggestion
QUESTION [2 upvotes]: I have been suggested Artin or Herstein's books. What do you think is more rigorous but at the same time clear and good to read?

REPLY [2 votes]: Lang's Algebra is very nice, and van der Waerden (vol1, vol2) is the best I know of, together with Mac Lane & Birkhoff. I find the latter significantly harder to follow. There is also Dummit & Foote, at a level equivalent to Lang.
I don't own Artin's Algebra, but I'd say it's also at the level of Lang, and it has a very good reputation.
Edit: the other books I have are of Emil Artin. The Artin whose lectures the books by van der Waerden were originally based on (see here on Wikipedia). Michael Artin is actually his son, and is probably the author you are refering to.


TITLE: Formula for $U(N)$ integration wanted
QUESTION [11 upvotes]: Before you jump on the "duplicate" buttom, let me say that I do not want to hear about Weingarten calculus and I do not want to see a character of the symmetric group.
What I would like is a formula for the (normalized) Haar measure integral
$$
\int_{U(N)} g_{i_1 j_1}\cdots g_{i_n j_n} {\bar{g}}_{k_1 l_1}\cdots {\bar{g}}_{k_n l_n}\ d\mu(g)
$$
of the form
$$
\left.\mathcal{D}\ g_{i_1 j_1}\cdots g_{i_n j_n} {\bar{g}}_{k_1 l_1}\cdots {\bar{g}}_{k_n l_n}\right|_{g=\bar{g}=0}
$$
where $\mathcal{D}$ is an explicit constant coefficient differential operator of infinite order.
Of course, in this formula the $g$'s and $\bar{g}$'s are treated are $2N^2$ completely unrelated formal variables.
As an example of what I would like, in the case of $SU(N)$ and $\bar{g}$-free monomials
$$
\mathcal{D}=\sum_{n=0}^{\infty} \frac{0!1!\cdots (N-1)!}{n!(n+1)!\cdots(n+N-1)!}\ ({\rm det}(\partial g))^n
$$
works.
As per the "Additional remark" in my second answer to this MO question, I had a vague recollection of seeing a math-physics paper with
such a formula, but maybe my memory is faulty. So I think it's better to ask the experts.

REPLY [6 votes]: To expand on my comments, this paper https://arxiv.org/pdf/hep-th/9209083v2.pdf by Shatashvili deals
with ``correlation functions'' of Haar unitary matrices of the form
$$
\int_{U(N)}^{} d\mu(U) e_{}^{tr(UAU_{}^{-1}B)}
U_{i_1j_1}^{}\bar U_{k_1\ell_1}^{}\ldots
U_{i_mj_m}^{}\bar U_{k_m\ell_m},
$$
and provides a certain combinatorial formula for these. Then setting $A=0$ would probably
recover what you're asking about.
The same correlation functions (and an alternative formula for them) are also discussed in
https://arxiv.org/pdf/hep-th/0502041.pdf


text:
\section{Monotone Neural Network}
\label{sec:monotone}

In this work, we introduce a window-based method to utilize FNNs to learn the dynamics of the system~\eqref{eq:system}.

We first show that this method can reduce the learning error in general (\cref{sub:window}), as well as how to impose the monotonicity and stability constraints
(\cref{sub:monotonicity constraint,sub:stablity constraint}, respectively).

\subsection{Window-Based Learning Method}
\label{sub:window}

Conventionally (e.g.,~\cite{richards_LyapunovNeuralNetwork_2018,chang_NeuralLyapunovControl_2019,kolter_LearningStableDeep_2019}), to learn the dynamics $f$ of the system \eqref{eq:system},
an FNN $\hat{f}_\gamma$ with the weights $\gamma$ may be trained to predict the next state $x(t+1)$ from the current state $x(t)$.

Thus, to predict a future $x(t+T)$ from the state $x(t)$, the $T$-fold composition $\hat{f}^{(T)}_\gamma(x(t))= \underbrace{  \hat f_\gamma( \hat f_\gamma(\cdots \hat f_\gamma}_\text{$T$ in total} (\cdot)))$
of $\hat{f}_\gamma$ can be used. In that case, the prediction error is captured by the following lemma.

\begin{lemma} \label{lem:err_propagation}
Let us assume that (i)~$\exists a,b\in\mathbb{R}$
s.t. $a \Vert x_1-x_2 \rVert \leq \Vert f(x_1)-f(x_2) \Vert \leq b\Vert x_1-x_2\Vert$~and $a\Vert x_1-x_2 \Vert \leq \Vert \hat f_\gamma(x_1)-\hat f_\gamma(x_2) \Vert \leq b \Vert x_1-x_2\Vert$
for all $x_1,x_2 \in \real^n$; and
(ii) each element in the one step estimation error vector
$\delta_{\gamma} (x(t+i-1)) = f(x(t+i-1)) - \hat f_{\gamma}(x(t+i-1))$

distributes uniformly in $[-\epsilon,\epsilon]$ for all $i \in [1,T]$. Then, an approximate lower bound of variance of each element in $x(t+T) - \hat{f}^{(T)}_\gamma(x(t))$ when iteratively predicting $x(t+T)$ from $x(t)$ is $(\sum_{i=1}^{T} a^{2(i-1)}) \epsilon^2/3$.
\end{lemma}
\vspace{-0.1 in}
Assumption $(i)$ captures that $f$ is a  Lipschitz function and $\hat f_\gamma$ in general captures the gradient properties of $f$. Assumption $(ii)$ refers to a similar case analyzed in~\cite{van2016deep}, which implies if an $\hat f_\gamma$ is trained with stochastic gradient-based algorithms on same training data sufficiently large number of times (each time is with a unique initialization), then all predictions are assumed to be centered at the corresponding ground truth value with some disturbance bounded by $\epsilon$.

The proof of Lemma \ref{lem:err_propagation}
can be found in Appendix~\ref{sec:l2}.

Note that, from the lemma, for $a>1$ (i.e., when the system is unstable), the accumulated variance increases exponentially with respect to~$T$.

To reduce the prediction error, we propose to train a set of FNNs $\mathcal{F}_\theta=\{\hat{f}_{\theta_i} \mid i = 1, \ldots, q \}$
using a  $q$-window of previous states $x(t:t-q+1) = \big( x(t), \ldots, x(t-q+1) \big)$, i.e., $\mathcal{F}_\theta$ predicts $x(t+1)$ from $x(t:t-q+1)$, where $\theta_i$ refers to the weights of the $i^\text{th}$ estimator in $\mathcal{F}_\theta$ and $\theta$ represents all the parameters in $\mathcal{F}_\theta$ in general.
Specifically, we can rewrite \eqref{eq:system} as
\begin{align}\label{eq:rewrite_system}
    x(t+1) = p_1 f(x(t)) + p_2 f^{(2)}(x(t-1)) + \dots + p_q f^{(q)}(x(t-q+1)),
\end{align}
where $q$ is the window length and $\sum_{i=1}^q p_i = 1$. Then $x(t+1)$ can be approximated by $\mathcal{F}_\theta$ as
\begin{align}\label{eq:sys_est_constrained}
\hat x(t+1) = p_1 \hat{f}_{\theta_1}(x(t)) + p_2 \hat{f}_{\theta_2}(x(t-1)) + \dots + p_q \hat{f}_{\theta_q}(x(t-q+1)),
\end{align}
where $\sum_{i=1}^q p_i = 1$, $p_i \geq 0$, and ideally after training each $\hat f_{\theta_i^*}$ predicts $f^{(i)}(x)$ from $x$.

In this case, all the $\theta_i$'s can be obtained by solving the following
optimization problem with a dual objective~--~i.e.,
\vspace{-12pt}

\begin{align}\label{eq:constrained_opt}
    &\min_{\theta_i,p_i}  \mathbb{E}_{x(\cdot)\sim \rho } [\Vert x(t+1)-\sum_{i=1}^q p_i \hat{f}_{\theta_i}(x(t-i+1) \Vert ^2 ]
    \quad  s.t. ~~ \sum_{i=1}^q p_i = 1, \text{ and } p_i>0;
\end{align}
here, $\rho$ is the state visitation distribution from a random initial state $x(0)$, $t\in\{0,...,H\}$ and $H$ is the system time horizon. In general,~\eqref{eq:constrained_opt} can be achieved by either solving the corresponding primal-dual problem, or a two-step method which starts with training $\hat f_{\theta_i}$'s separately followed by solving the linear programming problem w.r.t $p_i$'s. Note that given $x(t:t-q+1)$, we can estimate $x(t+T)$ using recursion as shown in Appendix~\ref{app:algs} Alg.~\ref{alg:recursion_F}.
In the following theorem we show that if a set of FNNs $\mathcal{F}_\theta$ is trained using~\eqref{eq:constrained_opt}, then the variance of approximation error is upper bounded.

\begin{theorem}\label{the:error}
Let us assume that: (i) there exists $0<\eta\ll 1$ such that $\forall i\in\{1,...q\}$, $| p_i-1/q| \leq \eta$;
(ii) $\exists a,b$ s.t. $a\lVert x_1-x_2\rVert \leq \lVert f(x_1)-f(x_2)\rVert \leq b\lVert x_1-x_2 \rVert$ and $a\lVert x_1-x_2 \rVert \leq \lVert \hat f_{\theta_i}(x_1)-\hat f_{\theta_i}(x_2)\rVert\leq b\lVert x_1-x_2\rVert$  $\forall i\in\{1,...,q\}$;
(iii) $b^k/q^k \approx 0$ for all $k\geq 4$; and (iv) each element in the error vector $\delta_{\theta_i} (x(t-i+1)) = \hat f_{\theta_i}(x(t-i+1)) - f^{(i)}(x(t-i+1))$
distributes uniformly in $[-\epsilon,\epsilon]$ for all $i\in\{1,...,q\}$. If $\mathcal{F}_\theta$ is trained using

\eqref{eq:constrained_opt}, the variance of each element in $x(t+T)-\hat x(t+T)$ when iteratively predicting $x(t+T)$ using $q$-window is approximately upper bounded by $(1+ \frac{b^2+\epsilon^2/3}{q})\frac{\epsilon^2}{3q}$.
\end{theorem}
\vspace{-0.1 in}
Theorem~\ref{the:error} proof and rationale of assumptions are provided in Appendix~\ref{sec:proofs}, whereas the relatively strong condition $(i)$ is, for the analyzed systems (from Section~\ref{sec:case}), numerically verified to hold  in Appendix~\ref{app:verify}. Note that the error bound from \cref{the:error} does not increase with $T$;
this is also numerically validated in \cref{sec:case} for $(q=100)$-window learning.
From Lemma~\ref{lem:err_propagation} and Theorem~\ref{the:error}, the variance of estimating $x(t+T)$ using the $q$-window approach \eqref{eq:sys_est_constrained} is upper bounded by the recursive prediction $\hat f_\gamma^{(T)}(x(t))$~when it holds that
\begin{equation}\label{eq:q_a_b_relation}
    (q+b^2+\epsilon^2/3)/q^2 \leq (\sum_{i=1}^{T} a^{2(i-1)}).
\end{equation}
Note that~\eqref{eq:q_a_b_relation} may not always hold in general; and we analyze two nonlinear systems in Section~\ref{sec:case},
where one~\eqref{eq:q_a_b_relation} satisfies and the other does not.
However, in Section~\ref{sec:case}, we show numerically that even when~\eqref{eq:q_a_b_relation} is not satisfied, the prediction error using $q$-window learning does not increase significantly compared to the error of $\hat f_\gamma^{(T)} (x(t))$.

Practically, when approximating $x(t+1)$, we can introduce a bias term $b$ into~\eqref{eq:sys_est_constrained}~--~i.e.,
\begin{align}\label{eq:sys_est_bias}
\hat x(t+1) = p_1 \hat{f}_{\theta_1}(x(t)) + p_2 \hat{f}_{\theta_2}(x(t-1)) + \dots + p_q \hat{f}_{\theta_q}(x(t-q+1)) + z.
\end{align}
Since all the estimators in $\mathcal{F}$ are FNNs, the $p_i$'s and $z$ in the right side of \eqref{eq:sys_est_bias} can be interpreted as the weights of a fully connected layer (with bias term added) that takes as input the output of all the estimators in $\mathcal{F}$. Thus, all the parameters can be encapsulated into a meta-NN $\hat{f}_\theta$ with weights $\theta$,~as\footnote{Without loss of generality, we slightly abuse $\theta$ here, as well as in the rest of the paper, to represent the weights within the meta-NN instead of referring to all the parameters in $\mathcal{F}$ in general as in \eqref{eq:sys_est_constrained}.}
\begin{equation}\label{eq:sys_est_final}
    \hat x(t+1) = \hat{f}_\theta(x(t),x(t-1),\dots,x(t-q+1)),
\end{equation}
which can be trained by minimizing the loss
\begin{equation}\label{eq:loss}
    \mathbb{E}_{x(\cdot)\sim \rho}[\Vert x(t+1)-\hat{f}_\theta(x(t),x(t-1),\dots,x(t-q+1)) \Vert ^2].
\end{equation}

In practice, the expectation in loss \eqref{eq:loss}
is substituted by the empirical training loss
for a batch of $N$ given sample paths
of the time horizon $H$,~i.e.,~
\begin{equation} \label{eq:empirical loss}
    J(\hat{f}_\theta) = \frac{1}{N} \sum_{i=1}^{N}
    \frac{1}{H - q - 1} \sum_{t = q-1}^{H - 1}
    \big\Vert
    x_i(t+1) - \hat{f}_\theta(x_i(t),x_i(t-1),\dots,x_i(t-q+1))
    \big\Vert^2,
\end{equation}
where $N$ is the batch size and $x_i(\cdot)$ represents the state obtained from the $i^\text{th}$ sample path for all $i\in\{1,...,N\}$.

Note that $\hat f_\theta$ can be used to predict $x(t+T)$ given $x(t:t-q+1)$ by slightly modifying the recursion function in Appendix~\ref{app:algs} Alg.~\ref{alg:recursion_F}, which in turn results in Appendix~\ref{app:algs} Alg.~\ref{alg:recursion_f}.

config: arxiv


TITLE: Complicated Indefinite Integration
QUESTION [3 upvotes]: $\displaystyle\int{\dfrac{\left(\sqrt{x}+2\right)^2}{3\sqrt{x}}\;\mathrm{d}x} =\frac{2}{3}\big(\sqrt{x}+2\big) $
I do not know how to get $\dfrac{2}{3}$. I assume $\sqrt{x}+2 = u$, then $\mathrm{d}u= \dfrac{1}{2}\,x^{1/2}$
But it is wrong answer....
So help me
Thank you

REPLY [2 votes]: Assuming you mean $\displaystyle \int \frac{\left(\sqrt{x}+2\right)^2}{3\sqrt{x}}\,dx$:
Let $\,u=\sqrt{x}+2\,$ and $\,du=\dfrac{dx}{2\sqrt{x}}$
Thus, we now have $$\dfrac{2}{3} \int u^2 \,du=\dfrac{2u^3}{9}+C=\dfrac{2\left(\sqrt{x}+2\right)^3}{9}+C$$


TITLE: Given a series $b_0,b_1,...$, where $b_0=1,b_1 = 2,b_2 = 3$ and $b_n=b_{n-3}+b_{n-2}+b_{n-1}$, show $b_n \leq 3^n$ for $n>2$.
QUESTION [1 upvotes]: I have a series $b_0, b_1...$, where $b_0 = 1$, $b_1 = 2$, $b_2 = 3$ and $b_n = b_{n-3} + b_{n-2}+b_{n-1}$. I have to show, that when $n > 2$ and $n \in \Bbb Z$, then $b_n \leq 3^n$.
I'm using mathematical induction.
Base case $n = 3$
$b_3= b_0 + b_1 + b_2 = 1 + 2 +3 = 6\\  b_3 \leq 3^3 \rightarrow 6 \leq 27$
So I presume that $b_k \leq 3^k$ is true and I should show that $b_{k+1} \leq 3^{k+1}$ is also true, but this is where I get stuck and don't know how to do it.
How should I show this?

REPLY [1 votes]: Take $b_{k+1}$ with the assumption that the claim holds for all $b_j$ when $j\le k$. Now, by definition, $b_{k+1}=b_k+b_{k-1}+b_{k-2}$. The induction hypothesis tells us that $b_n\le 3^n$ for $n\le k$, so $b_{k+1} \le 3^k+3^{k-1}+3^{k-2}<3^k+3^k+3^k=3\cdot3^k=3^{k+1}$.

REPLY [1 votes]: You have to assume it is true
for the 3 previous cases.
If $b_k \le 3^k$
for $k = n-1, n-2, n-3$
then
$\begin{array}\\
b_n
&=b_{n-1}+b_{n-2}+b_{n-3}\\
&\le 3^{n-1}+3^{n-2}+3^{n-3}\\
&=3^n(3^{-1}+3^{-2}+3^{-3})\\
&=3^n\dfrac{9+3+1}{27}\\
&=3^n\dfrac{13}{27}\\
&\lt 3^n\\
\end{array}
$


text:
\begin{document}

\begin{abstract}
We consider the trace map associated with the Fibonacci
Hamiltonian as a diffeomorphism on the invariant surface
associated with a given coupling constant and prove that the
non-wandering set of this map is hyperbolic if the coupling is
sufficiently small. As a consequence, for these values of the
coupling constant, the local and global Hausdorff dimension and
the local and global box counting dimension of the spectrum of the
Fibonacci Hamiltonian all coincide and are smooth functions of the
coupling constant.
\end{abstract}

\maketitle

MSC 2000: 82B44, 37D20, 37D50, 37D30, 81Q10.

\section{Introduction}

The Fibonacci Hamiltonian is the most prominent model in the study
of electronic properties of quasicrystals. It is given by the
discrete one-dimensional Schr\"odinger operator
$$
[H_{V,\omega} u](n) = u(n+1) + u(n-1) + V \chi_{[1-\alpha,1)}(n
\alpha + \omega \!\!\!\! \mod 1) u(n),
$$
where $V > 0$ is the coupling constant, $\alpha =
\frac{\sqrt{5}-1}{2}$ is the frequency, and $\omega \in [0,1)$ is
the phase.

This operator family displays a number of interesting phenomena,
such as Cantor spectrum of zero Lebesgue measure \cite{S89} and
purely singular continuous spectral measure for all phases
\cite{DL}. Moreover, it was recently shown that is also gives rise
to anomalous transport \cite{DT}. We refer the reader to the
survey articles \cite{D00,D07,S95} for further information and
references.

Already the earliest papers on this model, \cite{KKT,OPRSS},
realized the importance of a certain renormalization procedure in
its study. This led in particular to a consideration of the
following dynamical system, the so-called trace map,
$$
T:\Bbb{R}^3\to \Bbb{R}^3, \; T(x,y,z)=(2xy-z,x,y),
$$
whose properties are closely related to all the spectral
properties mentioned above. The existence of the trace map and its
connection to spectral properties of the operators is a
consequence of the invariance of the potential under a
substitution rule. This works in great generality; see the surveys
mentioned above and references therein. In the Fibonacci
case, the existence of a first integral is an additional useful property,
which allows one to restrict $T$ to invariant surfaces.

Fix some coupling constant $V$. For a complete spectral study of
the operator family $\{ H_{V,\omega} \}_{\omega \in [0,1)}$, it
suffices to study $T$ on a single invariant surface $S_V$. This
phenomenon is a peculiarity of the choice of the model (a discrete
Schr\"odinger operator in math terminology or an on-site lattice
model in physics terminology) and does not follow solely from the
symmetries coming from the invariance under the Fibonacci
substitution. For example, in continuum analogs of the Fibonacci
Hamiltonian, the invariant surface will in general be energy-dependent.

Let us denote the restriction of $T$ to the invariant surface
$S_V$ by $T_V$. As we will discuss in more detail below, it is of
interest to study the non-wandering set of this surface
diffeomorphism because it is closely connected to the spectrum of
$H_{V,\omega}$.\footnote{By a strong convergence argument, it
follows that the spectrum of $H_{V,\omega}$ does not depend on
$\omega$. It does, however, depend on $V$.} This correspondence in
turn allows one to show that the spectrum has zero Lebesgue
measure. It is then natural to investigate its fractal dimension.
A number of papers have studied this problem; for example,
\cite{DEGT,LW,Ra}. As pointed out in \cite{DEGT}, the work of
Casdagli, \cite{Cas}, has very important consequences for the
fractal dimension of the spectrum as a function of $V$. Casdagli
studied the map $T_V$ and proved, for $V \ge 16$, that the
non-wandering set is hyperbolic. Combining this with results in
hyperbolic dynamics, it follows that the local and global
Hasudorff and box counting dimensions of the spectrum all coincide
and are smooth functions of $V$. This result was crucial for the
work \cite{DEGT}, which determined the exact asymptotic behavior
of this function of $V$ as $V$ tends to infinity. It was shown
that
$$
\lim_{V \to \infty} \dim \sigma(H_{V,\omega}) \cdot \log V = \log
(1 + \sqrt{2}).
$$

Of course, the asymptotic behavior of the dimension of the
spectrum as $V$ approaches zero is of interest as well. Given the
discussion above, the natural first step is to prove the analogue
of Casdagli's result at small coupling. This is exactly what we do
in this paper. We will show that, for $V$ sufficiently small, the
non-wandering set of $T_V$ is hyperbolic and hence we obtain the
same consequences for the dimension of the spectrum as those
mentioned above in this coupling regime.

The structure of the paper is as follows.
Section~\ref{s.background} gives a more explicit description of
the previous results on the Fibonacci trace map, recalls some
useful general results from hyperbolic dynamics, and states the
main result of the paper --- the hyperbolicity of the
non-wandering set of the trace map for sufficiently small coupling
$V$. Sections~\ref{s.vequzero}--\ref{s.ppe} contain the proof of
this result. More precisely, Section~\ref{s.vequzero} contains a
discussion of the case $V=0$, Section~\ref{s.s04} studies the
dynamics of the trace map near a singular point and formulates the
crucial Proposition~\ref{p.expanding}. Section~\ref{s.s05} shows
how the main result follows from it, and finally,
Section~\ref{s.ppe} contains a proof of
Proposition~\ref{p.expanding}.

After this paper was finished we learned that Serge Cantat
provided a proof of uniform hyperbolicity of the trace map for all
non-zero values of the coupling constant \cite{Can}. Our results
were obtained independently and we use completely different
methods.

\bigskip

\noindent\textit{Acknowledgments.} We wish to express our
gratitude to Helge Kr\"uger, John Roberts, and Amie Wilkinson for
useful discussions.

\section{Background and Main Result}\label{s.background}

In this section we expand on the introduction and state
definitions and previous results more carefully. This will
eventually lead us to the statement of our main result in
Theorem~\ref{t.main} below.

\subsection{Description of the Trace Map and Previous Results}

The main tool that we are using here is the so called
\textit{trace map}. It was originally introduced in \cite{K,KKT};
further useful references include \cite{BGJ,BR,HM,Ro}. Let us quickly
recall how it arises from the substitution invariance of the
Fibonacci potential; see \cite{S87} for detailed proofs of some of
the statements below.

The one step transfer matrices associated with the difference
equation $H_{V,\omega} u = E u$ are given by
$$
T_{V,\omega}(m,E) = \begin{pmatrix} E - V \chi_{[1-\alpha,1)}(m
\alpha + \omega \!\!\!\! \mod 1) & -1 \\
1 & 0 \end{pmatrix}.
$$
Denote the Fibonacci numbers by $\{F_k\}$, that is, $F_0 = F_1 =
1$ and $F_{k+1} = F_k + F_{k_1}$ for $k \ge 1$. Then, one can show
that the matrices
$$
M_{-1}(E) = \begin{pmatrix} 1 & -V \\ 0 & 1
\end{pmatrix} , \quad M_0(E) =
\begin{pmatrix} E & -1 \\ 1 & 0 \end{pmatrix}
$$
and
$$
M_k(E) = T_{V,0}(F_k,E) \times \cdots \times T_{V,0}(1,E) \quad
\text{ for } k \ge 1
$$
obey the recursive relations
$$
M_{k+1}(E) = M_{k-1}(E) M_k(E)
$$
for $k \ge 0$. Passing to the variables
$$
x_k(E) = \frac12 \mathrm{Tr} M_k(E),
$$
this in turn implies
$$
x_{k+1}(E) = 2 x_k(E) x_{k-1}(E) - x_{k-2}(E).
$$
These recursion relations exhibit a conserved quantity; namely, we
have
$$
x_{k+1}(E)^2+x_k(E)^2+x_{k-1}(E)^2 - 2 x_{k+1}(E) x_k(E)
x_{k-1}(E) -1 = \frac{V^2}{4}
$$
for every $k \ge 0$.

Given these observations, it is then convenient to introduce the
\textit{trace map}
$$
T : \Bbb{R}^3 \to \Bbb{R}^3, \; T(x,y,z)=(2xy-z,x,y).
$$
The following function
$$
G(x,y,z) = x^2+y^2+z^2-2xyz-1
$$
is invariant under the action of $T$,\footnote{The function
$G(x,y,z)$ is called the \textit{Fricke character}.} and hence $T$
preserves the family of cubic surfaces\footnote{The surface $S_0$
is called the \textit{Cayley cubic}.}
$$
S_V = \left\{(x,y,z)\in \Bbb{R}^3 : x^2+y^2+z^2-2xyz=1+
\frac{V^2}{4} \right\}.
$$
Plots of the surfaces $S_{0.01}$, $S_{0.1}$, $S_{0.2}$, and
$S_{0.5}$ are given in Figures~\ref{fig:s0.01}--\ref{fig:s0.5},
respectively.

\begin{figure}\begin{minipage}[t]{5cm} \includegraphics[width=1.2\textwidth]{Sonehundredthb.jpg} \par
\caption{The surface $S_{0.01}$.} \label{fig:s0.01}
\end{minipage} \hfill \begin{minipage}[t]{5cm} \includegraphics[width=1.2\textwidth]{Sonetenthb.jpg} \par
\caption{The surface $S_{0.1}$.} \label{fig:s0.1}
\end{minipage} \end{figure}

\begin{figure} \begin{minipage}[t]{5cm} \includegraphics[width=1.2\textwidth]{Sonefifthb.jpg} \par
\caption{The surface $S_{0.2}$.} \label{fig:s0.2}
\end{minipage} \hfill \begin{minipage}[t]{5cm} \includegraphics[width=1.2\textwidth]{Sonehalfb.jpg}
\par \caption{The surface $S_{0.5}$.} \label{fig:s0.5} \end{minipage}
\end{figure}

Denote by $\ell_V$ the line
$$
\ell_V = \left\{ \left(\frac{E-V}{2}, \frac{E}{2}, 1 \right) : E
\in \Bbb{R} \right\}.
$$
It is easy to check that $\ell_V \subset S_V$.

S\"ut\H{o} proved the following central result in \cite{S87}.

\bthm[S\"ut\H{o} 1987]\label{spectrum}  An energy $E$ belongs to
the spectrum of $H_{V,\omega}$ if and only if the positive
semiorbit of the point $(\frac{E-V}{2}, \frac{E}{2}, 1)$ under
iterates of the trace map $T$ is bounded. \ethm

It is of course natural to consider the restriction $T_{V}$ of the
trace map $T$ to the invariant surface $S_V$. That is, $T_{V}:S_V
\to S_V$, $T_{V}=T|_{S_V}$. Denote by $\Omega_{V}$ the set of
points in $S_V$ whose full orbits under $T_{V}$ are bounded. {\it
A priori} the set of bounded orbits of $T_V$ could be different
from the non-wandering set\footnote{A point $p\in M$ of a
diffeomorphism $f:M\to M$ is wandering if there exists a
neighborhood $O(p)\subset M$ such that $f^k(O)\cap O = \emptyset$
for any $k\in \mathbb{Z}\backslash 0$. The non-wandering set of
$f$ is the set of points that are not wandering.} of $T_V$, but
our construction of the Markov partition and analysis of the
behavior of $T_V$ near singularities show that in our case these
two sets do coincide. Notice that this is parallel to the
construction of the symbolic coding in \cite{Cas}.

Let us recall that an invariant closed set $\Lambda$ of a
diffeomorphism $f : M \to M$ is \textit{hyperbolic} if there
exists a splitting of a tangent space $T_xM=E^u_x\oplus E^u_x$ at
every point $x\in \Lambda$ such that this splitting is invariant
under $Df$ and the differential $Df$ exponentially contracts
vectors from stable subspaces $\{E^s_x\}$ and exponentially
expands vectors from unstable subspaces $\{E^u_x\}$. A hyperbolic
set $\Lambda$ of a diffeomorphism $f : M \to M$ is \textit{locally
maximal} if there exists a neighborhood $U(\Lambda)$ such that
$$
\Lambda=\bigcap_{n\in\Bbb{Z}}f^n(U).
$$

The second central result about the trace map we wish to recall is
due to Casdagli; see \cite{Cas}.

\bthm[Casdagli 1986]\label{Casdagli} For $V\ge 16$, the set
$\Omega_{V}$ is a locally maximal hyperbolic set of
$T_{V}:S_V \to S_V$. It is homeomorphic to a Cantor set. \ethm

\subsection{Some Properties of Locally Maximal Hyperbolic Invariant
Sets of Surface Diffeomorphisms}

Given Theorem~\ref{Casdagli}, several general results apply to the
trace map of the strongly coupled Fibonacci Hamiltonian. Let us
recall some of these results that yield interesting spectral
consequences, which are discussed below.

Consider a locally maximal invariant transitive hyperbolic set
$\Lambda\subset M$, $\dim M=2$, of a diffeomorphism $f\in \text{\,
\rm Diff\,}^r(M), r\ge 1$. We have
$\Lambda=\bigcap_{n\in\Bbb{Z}}f^n(U(\Lambda))$ for some neighborhood
$U(\Lambda)$. Assume also that $\dim E^u=\dim E^s=1$. Then, the
following properties hold.

\subsubsection{Stability}

There is a neighborhood $\mathcal{U}\subset \text{\, \rm
Diff\,}^1(M)$ of the map $f$ such that for every $g\in
\mathcal{U}$, the set $\Lambda_g = \bigcap_{n\in
\Bbb{Z}}g(U(\Lambda))$ is a locally maximal invariant hyperbolic
set set of $g$. Moreover, there is a homeomorphism $h:\Lambda\to
\Lambda_g$ that conjugates $f|_{\Lambda}$ and $g|_{\Lambda_g}$,
that is, the following diagram commutes:
$$
\CD \Lambda @>{f|_{\Lambda}}>>\Lambda \\
@V{h}VV@VV{h}V\\
\Lambda_g @>g|_{\Lambda_g}>>\Lambda_g\endCD
$$

\subsubsection{Invariant Manifolds}

For $x\in \Lambda$ and small $\varepsilon>0$, consider the local
stable and unstable sets
$$
W^s_{\varepsilon}(x)=\{w\in M : d(f^n(x), f^n(w))\le \varepsilon \
\ \text{\rm for all}\ \  n\ge 0\},
$$
$$
W^u_{\varepsilon}(x)=\{w\in M : d(f^n(x), f^n(w))\le \varepsilon \
\ \text{\rm for all}\ \  n\le 0\}.
$$
If $\varepsilon>0$ is small enough, these sets are embedded
$C^r$-disks with $T_xW^s_{\varepsilon}(x)=E^s_x$ and
$T_xW^u_{\varepsilon}(x)=E^u_x$. Define the (global) stable and
unstable sets as
$$
W^s(x)=\bigcup_{n\in \Bbb{N}}f^{-n}(W^s_{\varepsilon}(x)), \ \ \
W^u(x)=\bigcup_{n\in \Bbb{N}}f^{n}(W^u_{\varepsilon}(x)).
$$
Define also
$$
W^s(\Lambda)=\bigcup_{x\in \Lambda}W^s(x)\ \ \ \text{\rm and}\ \ \
W^u(\Lambda)=\bigcup_{x\in \Lambda}W^u(x).
$$

\subsubsection{Invariant Foliations}

A stable foliation for $\Lambda$ is a foliation $\mathcal{F}^s$ of a neighborhood of $\Lambda$ such that
\begin{itemize}
\item[(a)] for each $x\in \Lambda$, $\mathcal{F}(x)$, the leaf
containing $x$, is tangent to $E^s_x$,

\item[(b)] for each $x$ sufficiently close to $\Lambda$,
$f(\mathcal{F}^s(x))\subset \mathcal{F}^s(f(x))$.
\end{itemize}
An unstable foliation $\mathcal{F}^u$ can be defined in a similar way.

For a locally maximal hyperbolic set $\Lambda\subset M$ of a
$C^1$-diffeomorphism $f:M\to M$, $\dim M=2$, stable and unstable
$C^0$ foliations with $C^1$-leaves can be constructed \cite{M}. In
the case of $C^2$-diffeomorphism,  $C^1$ invariant foliations
exist (see, for example, \cite{PT}, Theorem 8 in Appendix 1).

\subsubsection{Local Hausdorff Dimension and Box Counting
Dimension}

Consider, for $x\in \Lambda$ and small $\varepsilon>0$, the set
$W^u_{\varepsilon}(x)\cap \Lambda$. Its Hausdorff dimension does
not depend on $x\in \Lambda$ and $\varepsilon>0$, and coincides
with its box counting dimension (see \cite{MM,T}):
$$
\dim_H W^u_{\varepsilon}(x)\cap \Lambda = \dim_B
W^u_{\varepsilon}(x)\cap \Lambda.
$$
In a similar way,
$$
\dim_H W^s_{\varepsilon}(x)\cap \Lambda= \dim_B
W^s_{\varepsilon}(x)\cap \Lambda.
$$
Denote $h^s = \dim_H W^s_{\varepsilon}(x)\cap \Lambda$ and $h^u =
\dim_H W^u_{\varepsilon}(x)\cap \Lambda$. We will say that $h^s$
and $h^u$ are the \textit{local stable and unstable Hausdorff
dimensions} of $\Lambda$.

For properly chosen small $\varepsilon > 0$, the sets
$W^u_{\varepsilon}(x)\cap \Lambda$ and $W^s_{\varepsilon}(x)\cap
\Lambda$ are dynamically defined Cantor sets (see \cite{PT1} for
definitions and proof), and this implies, in particular, that
$$
h^s<1 \ \ \  \text{\rm and} \ \ \ h^u<1,
$$
see, for example, Theorem 14.5 in \cite{P}.

\subsubsection{Global Hausdorff Dimension}

The Hausdorff dimension of $\Lambda$ is equal to its box
counting dimension, and
$$
\dim_H \Lambda=\dim_B \Lambda = h^s + h^u;
$$
see \cite{MM,PV}.

\subsubsection{Continuity of the Hausdorff Dimension}

The local Hausdorff dimensions $h^s(\Lambda)$ and $h^u(\Lambda)$
depend continuously on $f:M\to M$ in the $C^1$-topology; see
\cite{MM,PV}. Therefore, $\dim_H \Lambda_f = \dim_B \Lambda_f =
h^s(\Lambda_f) + h^u(\Lambda_f)$ also depends continuously on $f$
in the $C^1$-topology. Moreover, for a $C^r$ diffeomorphism
$f:M\to M$, $r\ge2$, the Hausdorff dimension of a hyperbolic set
$\Lambda_f$ is a $C^{r-1}$ function of $f$; see \cite{Ma}.

\brm For hyperbolic sets in dimension greater than two, many of
these properties do not hold in general; see \cite{P} for more
details. \erm

\subsection{Implications for the Trace Map and the Spectrum}

Due to Theorem~\ref{Casdagli}, for every $V\ge 16$, all the
properties from the previous subsection can be applied to the
hyperbolic set $\Omega_{\lambda}$ of the trace map $T_{V}:S_V \to
S_V$.

Moreover, the results in \cite[Section 2]{Cas} imply the following
statement. \blm\label{l.translargecoupling} For $V\ge 16$ and
every $x\in \Omega_{V}$, the stable manifold $W^s(x)$ intersects
the line $\ell_V$ transversally. \elm

The existence of a $C^1$-foliation $\mathcal{F}^s$ allows one to
locally consider the set $W^s(\Omega_{V})\cap \ell_V$ as a
$C^1$-image of the set $W^u_{\varepsilon}(x)\cap\Omega_{V}$. Due
to Theorem \ref{spectrum}, this implies the following properties
of the spectrum $\sigma(H_{V,\omega})$ for $V\ge 16$:

\begin{Cor}\label{c.specprop}
For $V\ge 16$, the following statements hold:

\begin{itemize}

\item[{\rm (i)}] The spectrum $\sigma(H_{V,\omega})$ depends
continuously on $V$ in the Hausdorff metric.

\item[{\rm (ii)}] For every small $\varepsilon>0$ and every $x\in
\sigma(H_{V,\omega})$, we have
\begin{align*}
\dim_H \left( (x-\varepsilon, x+\varepsilon) \cap
\sigma(H_{V,\omega}) \right) & = \dim_B \left( (x-\varepsilon,
x+\varepsilon) \cap \sigma(H_{V,\omega}) \right) \\ & = \dim_H
\sigma(H_{V,\omega}) \\ & =\dim_B \sigma(H_{V,\omega}).
\end{align*}

\item[{\rm (iii)}] The Hausdorff dimension $\dim_H
\sigma(H_{V,\omega})$ is a $C^{\infty}$-function of $V$, and is
strictly smaller than one.

\end{itemize}
\end{Cor}

\subsection{Hyperbolicity of the Trace Map for
Small Coupling}

We are now in a position to state the main result of this paper.

\bthm\label{t.main} There exists $V_0>0$ such that for every $V\in
(0,V_0)$, the following properties hold.

\begin{itemize}

\item[{\rm (i)}] The non-wandering set $\Omega_{V}\subset S_{V}$
of the map $T_{V}:S_{V}\to S_{V}$ is hyperbolic.

\item[{\rm (ii)}] The non-wandering set $\Omega_{V}\subset S_{V}$
is homeomorphic to a Cantor set, and $T_{V}|_{\Omega_{V}}$ is
conjugated to a topological Markov chain $\sigma_{A}:\Sigma^6_A\to
\Sigma^6_A$ with the matrix
$$
A=\begin{pmatrix}
    0 & 0 & 0 & 1 & 1 & 1 \\
    0 & 0 & 1 & 0 & 1 & 1 \\
    0 & 0 & 0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 0 & 0 & 1 \\
    1 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 & 0 & 0 \\
  \end{pmatrix}.
$$

\item[{\rm (iii)}] For every $x\in \Omega_{V}$, the stable
manifold $W^s(x)$ intersects the line $\ell_V$ transversally.

\end{itemize}
\ethm

As before, we obtain the following consequences:

\begin{Cor}\label{c.specpropsl}
With $V_0 > 0$ from Theorem~\ref{t.main}, the following statements
hold for $V \in (0,V_0)$:

\begin{itemize}

\item[{\rm (i)}] The spectrum $\sigma(H_{V,\omega})$ depends
continuously on $V$ in the Hausdorff metric.

\item[{\rm (ii)}] For every small $\varepsilon>0$ and every $x\in
\sigma(H_{V,\omega})$, we have
\begin{align*}
\dim_H \left( (x-\varepsilon, x+\varepsilon) \cap
\sigma(H_{V,\omega}) \right) & = \dim_B \left( (x-\varepsilon,
x+\varepsilon) \cap \sigma(H_{V,\omega}) \right) \\ & = \dim_H
\sigma(H_{V,\omega}) \\ & =\dim_B \sigma(H_{V,\omega}).
\end{align*}

\item[{\rm (iii)}] The Hausdorff dimension $\dim_H
\sigma(H_{V,\omega})$ is a $C^{\infty}$-function of $V$, and is strictly smaller than one.

\end{itemize}
\end{Cor}

We expect these properties to be of similar importance in a study
of the asymptotic behavior of the fractal dimension of the
spectrum as $V \to 0$ as was the case in the large coupling
regime.

\section{Properties of the Trace Map for $V=0$}\label{s.vequzero}

Up to this point, we only considered the case $V > 0$. Since we
will regard the case of small positive $V$ as a small perturbation
of the case $V = 0$, we will also include the latter case in our
considerations. In fact, this section is devoted to the study of
this ``unperturbed case.''

Denote by $\mathbb{S}$ the part of the surface $S_0$ inside of the
cube $\{ |x|\le 1, |y|\le 1, |z|\le 1\}$. The surface $\mathbb{S}$
is homeomorphic to $S^2$, invariant, smooth everywhere except at the four
points $P_1=(1,1,1)$, $P_1=(1,-1,-1)$, $P_1=(-1,1,-1)$, and
$P_1=(-1,-1,1)$, where $\mathbb{S}$ has conic singularities,  and
the trace map $T$ restricted to $\mathbb{S}$ is a factor of a
hyperbolic automorphism of a two torus:
$$
\mathcal{A}(\theta, \varphi)=(\theta + \varphi, \theta)\ (\text{\rm
mod}\ 1).
$$
The semiconjugacy is given by the map
$$
F: (\theta, \varphi) \mapsto (\cos 2\pi(\theta + \varphi), \cos 2\pi
\theta, \cos 2\pi \varphi).
$$
The map $\mathcal{A}$ is hyperbolic, and is given by a matrix $A=\begin{pmatrix}
                                                                   1 & 1 \\
                                                                   1 & 0 \\
                                                                 \end{pmatrix}$
with eigenvalues
$$
\mu=\frac{1+\sqrt{5}}{2}\ \ \text{\rm and} \ \ \
-\mu^{-1}=\frac{1-\sqrt{5}}{2}.
$$
Let us denote by $\mathbf{v}^u, \mathbf{v}^u \in \mathbb{R}^2 $ the
unstable and stable eigenvectors of $A$:
$$
A\mathbf{v}^u=\mu\mathbf{v}^u, \
A\mathbf{v}^s=-\mu^{-1}\mathbf{v}^s, \
\|\mathbf{v}^u\|=\|\mathbf{v}^s\|=1.
$$
Fix some small $\zeta > 0$ and define the stable (resp., unstable)
cone fields on $\mathbb{R}^2$ in the following way:
\begin{align}\label{e.cones}
    K^s_p & =\{\mathbf{v}\in T_p\mathbb{R}^2 :
\mathbf{v}=v^u\mathbf{v}^u+v^s\mathbf{v}^s, \ |v^s| > \zeta^{-1}
|v^u|\},\\
\nonumber    K^u_p & =\{\mathbf{v}\in T_p\mathbb{R}^2 :
\mathbf{v}=v^u\mathbf{v}^u+v^s\mathbf{v}^s, \ |v^u|
> \zeta^{-1} |v^s|\}.
\end{align}
These cone fields are invariant:
$$
\forall\ \mathbf{v}\in K^u_p\ \ \ \ A\mathbf{v}\in K^u_{A(p)},
$$
$$
\forall\ \mathbf{v}\in K^s_p\ \ \ \ A^{-1}\mathbf{v}\in
K^s_{A^{-1}(p)}.
$$
Also, the iterates of the map $A$ expand vectors from the unstable
cones, and the iterates of the map $A^{-1}$ expand vectors from the
stable cones:
$$
\forall\ \mathbf{v}\in K^u_p\ \ \ \ \forall\ n\in \mathbb{N}\ \ \
\ |A^n\mathbf{v}| > \frac{1}{\sqrt{1 + \zeta^2}}
\lambda^n|\mathbf{v}|,
$$
$$
\forall\ \mathbf{v}\in K^s_p\ \ \ \ \forall\ n\in \mathbb{N}\ \ \
\ |A^{-n}\mathbf{v}| > \frac{1}{\sqrt{1 + \zeta^2}}
\lambda^n|\mathbf{v}|.
$$
The families of cones $\{K^s\}$ and $\{K^u\}$ invariant under
$\mathcal{A}$ can be also considered on $\mathbb{T}^2$.

The differential of the semiconjugacy $F$ sends these cone families
to stable and unstable cone families on $\mathbb{S}\backslash \{P_1,
P_2, P_3, P_4\}$. Let us denote these images by $\{\mathcal{K}^s\}$
and $\{\mathcal{K}^u\}$.

\blm\label{l.differential} The differential of the semiconjugacy $DF$ induces a map of
the unit bundle of $\mathbb{T}^2$ to the unit bundle of
$\mathbb{S}\backslash \{P_1, P_2, P_3, P_4\}$. The derivatives of
the restrictions of this map to a fiber are uniformly bounded. In
particular, the sizes of cones in families $\{\mathcal{K}^s\}$ and
$\{\mathcal{K}^u\}$ are uniformly bounded away from zero.  \elm

\begin{proof}
Choose small enough neighborhoods $U_1(P_1), U_2(P_2), U_3(P_3)$,
and $U_4(P_4)$ in $\mathbb{S}$. The complement
$$
\hat{\mathbb{S}} = \mathbb{S} \setminus \Big( \bigcup_{j=1}^4
U_j(P_j) \Big)
$$
is compact, so $F^{-1}(\hat{\mathbb{S}})$ is also compact, and the
action of $DF$ on a fiber of the unit bundle over points of
$F^{-1}(\hat{\mathbb{S}})$ has uniformly bounded derivatives.

Due to the symmetries of the trace map (see, for example, \cite{K}
for the detailed description of the symmetries of the trace map)
and the semiconjugacy $F$, it is enough to consider a neighborhood
$U_1(P_1)$. Hence, it is enough to consider the map $F$ in a
neighborhood of the point $(0,0)$.

The differential of $F$ has the form
$$
DF=-2\pi\begin{pmatrix}
          \sin 2\pi(\theta+\varphi) & \sin 2\pi(\theta+\varphi) \\
          \sin 2\pi\theta & 0 \\
          0 & \sin 2\pi\varphi \\
        \end{pmatrix}.
$$
If $(\theta, \varphi)$ is in a small neighborhood of $(0,0)$, then
$$
DF\sim -4\pi^2\begin{pmatrix}
          \theta+\varphi & \theta+\varphi \\
        \theta & 0 \\
          0 & \varphi \\
        \end{pmatrix}.
$$

Therefore, up to a multiplicative constant and higher order terms,
the image of the vector $(1,0)$ under $DF$ is $(\theta+\varphi,
\theta, 0)$, and the image of the vector $(0,1)$ is
$(\theta+\varphi, 0, \varphi)$. In order to estimate the
derivative of the projective action of $DF$ on a fiber over a
point $(\theta, \varphi)$, it is enough to estimate the angle
between images of basis vectors, and the ratio of the lengths of
the images of these vectors.

If $\alpha$ is the angle between $DF_{(\theta, \varphi)}(1,0)$ and
$DF_{(\theta, \varphi)}(0,1)$, then
$$
\cos \alpha \sim
\frac{(\theta+\varphi)^2}{\sqrt{(\theta+\varphi)^2+\theta^2}\cdot
\sqrt{(\theta+\varphi)^2+\varphi^2}}=\frac{1}{\sqrt{1+\frac{\theta^2+\varphi^2}{(\theta+\varphi)^2}
+\frac{\theta^2\varphi^2}{(\theta+\varphi)^4}}}.
$$
We have
$$
\frac{\theta^2+\varphi^2}{(\theta+\varphi)^2}\ge \frac{1}{2},\ \ \
\text{\rm and}\ \ \
\frac{\theta^2\varphi^2}{(\theta+\varphi)^4}\ge 0,
$$
so $\cos \alpha \le \sqrt{\frac{2}{3}}+0.001<1$ if $(\theta,
\varphi)$ is close enough to $(0,0)$.

Now let us estimate the ratio of the lengths of $DF_{(\theta,
\varphi)}(1,0)$ and $DF_{(\theta, \varphi)}(0,1)$. Up to higher
order terms it is equal to
$$
\frac{\sqrt{\varphi^2+2\theta^2+2\theta\varphi}}{\sqrt{2\varphi^2+\theta^2+2\theta\varphi}}=\sqrt{\frac{1+2t^2+2t}{2+t^2+2t}}=
\sqrt{\frac{2(t+\frac{1}{2})^2+\frac{1}{2}}{(t+1)^2+1}},
$$
where $t=\frac{\theta}{\varphi}\in \mathbb{R} \cup{\{\infty\}}$,
and this function is bounded from above and bounded away from
zero. Lemma \ref{l.differential} is proved.
\end{proof}

\section{The Structure of the Trace Map in a Neighborhood of a Singular
Point}\label{s.s04}

Due to the symmetries of the trace map it is enough to consider
the dynamics of $T$ in a neighborhood of $P_1=(1,1,1)$. Let
$U\subset \mathbb{R}^3$ be a small neighborhood of $P_1$ in
$\mathbb{R}^3$. Let us consider the set $Per_2(T)$ of periodic
points of $T$ of period 2.

\begin{figure} \begin{minipage}[t]{5cm} \includegraphics[width=1.2\textwidth]{line1.jpg} \par
\caption{$S_{0.1}$ and $Per_2(T)$ near $(1,1,1)$.}
\label{fig:sing0.0} \end{minipage} \hfill
\begin{minipage}[t]{5cm}
\includegraphics[width=1.2\textwidth]{line2.jpg}
\par \caption{$S_{0.2}$ and $Per_2(T)$ near $(1,1,1)$.} \label{fig:sing0.2}
\end{minipage} \end{figure}

\blm\label{l.periodtwo} We have
$$
Per_2(T)=\left\{(x,y,z) : x \in (-\infty, \tfrac{1}{2})\cup
(\tfrac{1}{2}, \infty), \ y=\frac{x}{2x-1}, \ z=x\right\}.
$$
\elm
\begin{proof}
Direct calculation.
\end{proof}
Notice that in a neighborhood $U$ the intersection $I\equiv
Per_2(T)\cap U$ is a smooth curve that is a normally hyperbolic
with respect to $T$ (see, for example, \cite{PT}, Appendix 1, for
the formal definition of normal hyperbolicity). Therefore, the
local center-stable manifold  $W_{loc}^{cs}(I)$ and the local
center-unstable manifold $W^{cu}_{loc}(I)$ defined by
$$
W_{loc}^{cs}(I)=\left\{p\in U : T^n(p) \in U \ \text{\rm for all}\
n\in \mathbb{N}\right\},
$$
$$
W_{loc}^{cu}(I)=\left\{p\in U : T^{-n}(p) \in U \ \text{\rm for
all}\ n\in \mathbb{N}\right\}
$$
are smooth two-dimensional surfaces. Also, the local strong stable
manifold $W^{ss}_{loc}(P_1)$ and the local strong unstable
manifold $W^{uu}_{loc}(P_1)$ of the fixed point $P_1$, defined by
$$
W^{ss}_{loc}(P_1)=\left\{p\in W^{cs}_{loc}(I) : T^n(p)\to P_1 \
\text{\rm as}\ n\to +\infty\right\},
$$
$$
W^{uu}_{loc}(P_1)=\left\{p\in W^{cu}_{loc}(I) : T^{-n}(p)\to P_1 \
\text{\rm as}\ n\to +\infty\right\},
$$
are smooth curves.

Let $\Phi:U\to \mathbb{R}^3$ be a smooth change of coordinates such that $\Phi(P_1)=(0,0,0)$ and
\begin{itemize}
\item
$\Phi(I)$ is a part of the line $\{x=0, z=0\}$;
\item
$\Phi(W^{cs}_{loc}(I))$ is a part of the plane $\{z=0\}$;
\item
$\Phi(W^{cu}_{loc}(I))$ is a part of the plane $\{x=0\}$;
\item
$\Phi(W^{ss}_{loc}(P_1))$ is a part of the line $\{y=0, z=0\}$;
\item
$\Phi(W^{uu}_{loc}(P_1))$ is a part of the line $\{x=0, y=0\}$.
\end{itemize}
Denote $f=\Phi\circ T\circ \Phi^{-1}$.

In this case,
$$
Df(0,0,0)=D(\Phi\circ T\circ \Phi^{-1})(0,0,0)=\begin{pmatrix}
                                       \lambda^{-1} & 0 & 0 \\
                                       0 & -1 & 0 \\
                                       0 & 0 & \lambda \\
                                     \end{pmatrix},
$$
where $\lambda$ is a largest eigenvalue of the differential
$DT(P_1):T_{P_1}\mathbb{R}^3\to T_{P_1}\mathbb{R}^3$,
$$
DT(P_1)=\begin{pmatrix}
                                       2 & 2 & -1 \\
                                       1 & 0 & 0 \\
                                       0 & 1 & 0 \\
                                     \end{pmatrix}, \ \ \ \lambda=\frac{3+\sqrt{5}}{2}=\mu^2.
$$
Let us denote $\frak{S}_V=\Phi(S_V)$. Then, away from $(0,0,0)$,
the family $\{\frak{S}_V\}$ is a smooth family of surfaces,
$\frak{S}_0$ is diffeomorphic to a cone, contains lines the
$\{y=0, z=0\}$ and $\{x=0, y=0\}$, and at each non-zero point on
these lines, it has a quadratic tangency with a horizontal or
vertical plane; compare Figures~\ref{fig:sing0.0} and
\ref{fig:sing0.2}.

We will use the variables $(x,y,z)$ for coordinates in
$\mathbb{R}^3$. For a point $p\in \mathbb{R}^3$, we will denote
its coordinates by $(x_p, y_p, z_p)$.

In order to study the properties of the map $f$ (i.e., of the map
$T$ in a small neighborhood of singularities), we need the
following statement.

\bprop\label{p.expanding} Given $C_1>0, C_2>0, \lambda>1$,
$\varepsilon \in (0,\frac{1}{4})$, and $\eta>0$, there exist
$\delta_0=\delta_0(C_1, C_2, \lambda, \varepsilon)$, $N_0\in
\mathbb{N}$, $N_0=N_0(C_1, C_2, \lambda, \varepsilon, \delta_0)$,
and $C=C(\eta)>0$ such that for
any $\delta\in (0, \delta_0)$,
the following
holds.

Let $f:\mathbb{R}^3\to \mathbb{R}^3$ be a $C^2$-diffeomorphism such that
\begin{itemize}
\item[{\rm (i)}] $\|f\|_{C^2}\le C_1$; \item[{\rm (ii)}] The plane
$\{z=0\}$ is invariant under iterates of $f$; \item[{\rm (iii)}]
$\|Df(p)-A\|<\delta$ for every $p\in \mathbb{R}^3$, where
$$
A=\begin{pmatrix}
    \lambda^{-1} & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & \lambda \\
  \end{pmatrix}
$$
is a constant matrix.
\end{itemize}
Introduce the following cone field in $\mathbb{R}^3$:
\beq\label{e.coneformula} K_p=\{v\in T_p\mathbb{R}^3, \
\mathbf{v}=\mathbf{v}_{xy}+\mathbf{v}_z : |\mathbf{v}_z|\ge C_2
\sqrt{|z_p|}|\mathbf{v}_{xy}|\}. \eneq Given a  point $p=(x_p,
y_p, z_p)$ such that $0<z_p<1$, denote by $N=N(p)$ the smallest
integer $N\in \mathbb{N}$ such that $f^N(p)$ has $z$-coordinate
larger that 1.

If $N(p)\ge N_0$ {\rm (}i.e., if $z_p$ is small enough{\rm )} then
\beq\label{e.expansionforN} |Df^N(\mathbf{v})|\ge
\lambda^{\frac{N}{2}(1-4\varepsilon)}|\mathbf{v}| \ \ \ \text{for
any } \ \  \mathbf{v}\in K_p, \eneq and if
$Df^N(\mathbf{v})=\mathbf{u}=\mathbf{u}_{xy}+\mathbf{u}_z$, then
\beq\label{e.u} |\mathbf{u}_{xy}|<2\delta^{1/2}|\mathbf{u}_{z}|.
\eneq

Moreover, if $|\mathbf{v}_z|\ge \eta|\mathbf{v}_{xy}|$ then
\beq\label{e.expansionfork} |Df^k(\mathbf{v})|\ge
C\lambda^{\frac{k}{2}(1-4\varepsilon)}|\mathbf{v}| \ \ \ \text{for
each } \ \  k=1, 2, \ldots, N. \eneq \enprop

Returning to our specific situation at hand, if, for a given
$\delta$, the neighborhood $U$ is small enough, then at every
point $P\in U$ the differential $D(\Phi\circ T\circ \Phi^{-1})(P)$
satisfies the condition (iii) of Proposition \ref{p.expanding}.
Also, since the tangency of $\frak{S}_0$ with the horizontal plane
is quadratic, there exists $C_2>0$ such that every vector tangent
to $\frak{S}_0$ from the cone $D\Phi(\mathcal{K}^u)$ also belongs
to the cone \eqref{e.coneformula}. The same holds for vectors
tangent to $\frak{S}_V$ from continuations of cones
$D\Phi(\mathcal{K}^u)$ if $V$ is small enough. Therefore,
Proposition~\ref{p.expanding} can be applied to all those vectors.

We postpone the proof of Proposition~\ref{p.expanding} to
Section~\ref{s.ppe} and first show how to use it to prove Theorem
\ref{t.main}.

\section{Proof of Theorem~\ref{t.main} Assuming
Proposition~\ref{p.expanding}}\label{s.s05}

In order to prove the hyperbolicity of $\Omega_V$ we construct
only the unstable cone field and prove the unstable cone
condition. Due to the symmetry of the trace map, the stable cones
can be constructed in the same way.

Let $U'$ be a neighborhood of $\{P_1, P_2, P_3, P_4\}$ where the
results of the previous section can be applied. Since
$\mathbb{S}\backslash U'$ is compact, $F^{-1}(\mathbb{S}\backslash
U')$ is also compact. Denote
$$
\widetilde{C}=2\max_{p\in \mathbb{S}\backslash U'} \left\{
\|DF^{-1}(p)\|, \|DF(F^{-1}(p))\| \right\} < \infty.
$$
Take any $p\in \mathbb{S}\backslash U'$ and any $\mathbf{v}\in
T_p\mathbb{S}, \mathbf{v}\in \mathcal{K}^u.$ If $ T^n(p)\in
\mathbb{S}\backslash U'$, then \beq\label{e.ctilda}
\|DT^n(\mathbf{v})\|=\|D(F\circ \mathcal{A}^n\circ
F^{-1})(\mathbf{v})\|\ge \widetilde{C}^{-2} \mu^n \|\mathbf{v}\|.
\eneq Fix a small $\varepsilon>0$ and $n^*\in \mathbb{N}$ such
that $\widetilde{C}^{-2}\mu^{n^*}\ge \mu^{n^*(1-\varepsilon)}.$

\blm There exists a neighborhood $U^*\subset U'$ of the singular
set $\{P_1, P_2, P_3, P_4\}$ such that if $p\not \in U^*$ but
$T^{-1}(p)\in U^*$, and $n_0$ is the smallest positive integer
such that $T^{n_0}(p)\in U^*$, then the finite orbit $\{p, T(p),
\ldots, T^{n_0-1}(p)\}$ contains at least $n^*$ points outside of
$U'$. \elm
\begin{proof}
Take a point $p_1\in W^{uu}_{loc}(P_1)\subset U$, denote
$p_2=T(p_1)$, and consider the closed arc $J\subset W^{uu}(P_1)$
between points $p_1$ and $p_2$. For any point $p\in J$, denote by
$m(p)$ the smallest number $m\in \mathbb{N}$ such that the finite
orbit $\{p, T(p), \ldots, T^{m}(p)\}$ contains $n^*$ points
outside of $\overline{U'}$. Notice that the function $m(p)$ is
upper semi-continuous, and therefore (due to compactness of $J$)
bounded. Let $M\in \mathbb{N}$ be an upper bound. The set
$\bigcup_{i=0}^MT^i(J)$ is compact and does not contain the
singularities $P_1, P_2, P_3$ and $P_4$. Therefore, there exists
$\xi>0$ so small that if $\text{\rm dist}(q, J)<\xi$, then the
distance between any of the first $M$ iterates of $q$ and any of
the points $P_1, P_2, P_3, P_4$ is greater than $\xi$, and the
finite orbit $\{q, T(q), \ldots, T^{M}(q)\}$ contains at least
$n^*$ points outside of $\overline{U'}$. Now take $\xi'>0$ so
small that any point in $\xi'$-neighborhood of $P_1$ whose orbit
follows $W^{uu}(P_1)$ towards $p_1$ hits the $\xi$-neighborhood of
$J$ before leaving $U'$. Now we can take the $\xi'$-neighborhood
of the set $P_1, P_2, P_3, P_4$ as $U^*$.
\end{proof}

For small $V$, denote by $\mathbb{S}_{V, U^*}$ the bounded
component of $S_V\backslash U^*$. The family $\{\mathbb{S}_{V,
U^*}\}_{V\in [0,V_0)}$ of surfaces with boundary depends smoothly
on the parameter and has uniformly bounded curvature. For small
$V$, a projection $\pi_V:\mathbb{S}_{V, U^*} \to \mathbb{S}$ is
defined. The map $\pi_V$ is smooth, and if $p\in \mathbb{S}$,
$q\in \mathbb{S}_{V, U^*}$, and $\pi_V(q)=p$, then $T_p
\mathbb{S}$ and $T_q S_V$ are close. Denote by $\mathcal{K}_V^u$
(resp., $\mathcal{K}_V^s$) the image of the cone $\mathcal{K}^u$
(resp., $\mathcal{K}^s$) under the differential of $\pi_V^{-1}$.

Our choice of $n^*$ guarantees that the following statement holds.
 \blm\label{l.uprimeexpansion} There exists $V_0>0$ such that for
every $V\in [0, V_0)$, the following holds: if $\{q, T(q), T^2(q),
\ldots, T^n(q)\}\subset \mathbb{S}_{V, U^*}$, $\mathbf{v}\in
\mathcal{K}_V^u\subset T_qS_V$, $q, T^n(q)\in \mathbb{S}_{V, U'}$,
and $n\ge n^*$, then
 $$
 \|DT^n(\mathbf{v})\|\ge \mu^{n(1-2\varepsilon)}\|\mathbf{v}\|.
 $$
 \elm
 \blm\label{l.exansionlong}
There exist $V_0>0$ and $C>0$ such that for any $V\in [0, V_0)$,
we have that if $q\in \mathbb{S}_{V, U'}$, $\mathbf{v}\in
\mathcal{K}_V^u\subset T_qS_V$, and $T^n(q)\in \mathbb{S}_{V,
U'}$, then
 $$
 \|DT^n(\mathbf{v})\|\ge C\mu^{n(1-4\varepsilon)}\|\mathbf{v}\|.
 $$
 \elm
 \begin{proof}
 Let us split the orbit $\{q, T(q), T^2(q), \ldots, T^n(q)\}$ into several intervals
 $$
 \{q, T(q), T^2(q), \ldots, T^{k_1-1}(q)\}, \{T^{k_1}(q),  \ldots, T^{k_2-1}(q)\}, \ldots, \{T^{k_s}(q), \ldots, T^n(q)\}
 $$
 in such a way that the following properties hold:
 \begin{enumerate}
 \item
 for each $  i=1, 2, \ldots, s$, the points $T^{k_i-1}(q)$ and $T^{k_i}(q)$ are outside of $U'$;
 \item
 if $\{T^{k_i}(q),  \ldots, T^{k_{i+1}-1}(q)\}\cap U^*\ne \emptyset$, then $\{T^{k_i+1}(q),  \ldots, T^{k_{i+1}-2}(q)\}\subset U'$;
 \item
 for each $  i=1, 2, \ldots, s-1$, we have either $k_{i+1}-k_i\ge n^*$ or $\{T^{k_i}(q),  \ldots, T^{k_{i+1}-1}(q)\}\cap U^*\ne \emptyset$.
 \end{enumerate}
 Such a splitting exists due to the choice of $U^*$ above.

The following lemma is a consequence of Lemma \ref{l.differential}
and property~\eqref{e.u} from Proposition \ref{p.expanding}.
 \blm\label{l.coneinvariance}
Suppose $q\in \mathbb{S}_{V, U'}$, $\mathbf{v}\in
\mathcal{K}_V^u\subset T_qS_V$, $T(q)\in U'$, and $l\in
\mathbb{N}$ is the smallest number such that $T^l(q)\not \in U'$.
If $l$ is large enough and $V$ is small enough, then
 $$
 DT^l(q)(\mathbf{v})\in \mathcal{K}_V^u\subset T_{T^l(q)}S_V.
 $$
 \elm
Apply Lemma \ref{l.coneinvariance} to those intervals in the
splitting that are contained in $U'$.  Notice that largeness of
$l$ can be provided by the choice of $U^*$. Together with
Proposition \ref{p.expanding} applied to these intervals, and
Lemma \ref{l.uprimeexpansion} applied to intervals that do not
intersect $U^*$, this guarantees uniform expansion of
$\mathbf{v}$. The first and the last interval may have length
greater than $n^*$, and then Lemma \ref{l.uprimeexpansion} can be
applied, or smaller than $n^*$, but then taking a small enough
constant $C$ (say, $C<\widetilde{C}^{-2}$) will compensate for the
lack of uniform expansion on these intervals. Lemma
\ref{l.exansionlong} is proved.
 \end{proof}

For every small $V>0$, there exists $\eta_V>0$ ($\eta_V\to 0$ as
$V\to 0$) such that if $p\not \in U'$, $T(p)\in U'$, and
$\mathbf{v}\in \mathcal{K}_V^u(p)$, then for the vector
$\mathbf{w}\equiv D\Phi^{-1}(p)(\mathbf{v})$,
$\mathbf{w}=\mathbf{w}_z+ \mathbf{w}_{xy}$, we have
$|\mathbf{w}_z|\ge \eta_V|\mathbf{w}_{xy}|$. An application of
Proposition \ref{p.expanding} together with Lemma
\ref{l.exansionlong} proves uniform expansion of vectors from
$\mathcal{K}_V^u$. Due to the symmetries of the trace map, all
vectors from $\mathcal{K}_V^s$ are expanded by iterates of
$T^{-1}$. Thus, hyperbolicity of the set $\Omega_V$ follows and
Theorem~\ref{t.main}~(i) is proved.

The Markov partition for $T|_{\mathbb{S}}$ (i.e., for $V=0$) was
presented explicitly by Casdagli in~\cite[Section~2]{Cas}; compare
Figure~\ref{fig.Casdagli-Markov}.

\begin{figure}
\setlength{\unitlength}{1mm}
\begin{picture}(120,120)

\put(10,10){\framebox(100,100)}

\put(5,5){(0,0)}

\put(110,5){(1,0)}

\put(5,112){(0,1)}

\put(110,112){(1,1)}

\put(10,10){\line(161,100){36.2}}

\put(60,10){\line(-61,100){13.8}}

\put(40,16){$1$}

\put(60,10){\line(161,100){36.2}}

\put(110,10){\line(-61,100){13.8}}

\put(90,16){$1$}

\put(60,110){\line(-161,-100){36.2}}

\put(10,110){\line(61,-100){13.8}}

\put(25,100){$1$}

\put(110,110){\line(-161,-100){36.2}}

\put(60,110){\line(61,-100){13.8}}

\put(75,100){$1$}

\put(10,60){\line(61,-100){22.1}}

\put(18,28){$5$}

\put(10,60){\line(161,100){22.1}}

\put(10,110){\line(61,-100){22.1}}

\put(18,78){$5$}

\put(110,60){\line(-61,100){22.1}}

\put(102,38){$5$}

\put(110,60){\line(-161,-100){22.1}}

\put(110,10){\line(-61,100){22.1}}

\put(102,88){$5$}

\put(60,60){\line(-161,-100){36.3}}

\put(60,60){\line(161,100){36.3}}

\put(60,60){\line(-61,100){22.1}}

\put(60,60){\line(61,-100){22.1}}

\put(88,46.35){\line(-161,-100){14.3}}

\put(32,73.65){\line(161,100){14.3}}

\put(37.88,46.28){\line(61,-100){8.4}}

\put(82.12,73.72){\line(-61,100){8.4}}

\put(34,34){$4$}

\put(84,84){$4$}

\put(34,84){$3$}

\put(84,34){$3$}

\put(34,60){$2$}

\put(84,60){$2$}

\put(60,34){$6$}

\put(60,84){$6$}

\end{picture}
\caption{The Markov partition for
$T|_{\mathbb{S}}$.}\label{fig.Casdagli-Markov}
\end{figure}

Since the Markov partition is formed by finite pieces of  strong
stable and strong unstable manifolds of the periodic points of $T$
and these manifolds depend smoothly on the parameter $V$, there
exists a Markov partition for $\Omega_V$ with the same matrix as
for $V=0$ (see \cite[Appendix 2]{PT} for more details on Markov
partitions for 2-dimensional hyperbolic maps). This establishes
Theorem~\ref{t.main}~(ii).

\blm\label{l.transversality} If $V$ is small enough, then the line
$\ell_V$ is transversal to the cone field $\mathcal{K}_V^s$. \elm

\begin{proof}
For $V=0$ and small enough $\zeta$ in \eqref{e.cones}, this is
true since $F^{-1}(l_0)=\{\theta=-\varphi\}$, and the vector $(1,
-1)$ is not an eigenvector of $A$. Therefore this is also true for
every sufficiently small $V$ by continuity.
\end{proof}

In order to show that $\ell_V$ is also transversal to the stable
manifolds of $\Omega_V$ inside of $U^*$, let us consider the rectifying
coordinates $\Phi:U^*\to \mathbb{R}^3$ again and define the
central-unstable cone field in $\Phi(U^*)$:
$$
K_p^{cu}=\left\{\mathbf{v}\in T_p\mathbb{R}^3,
\mathbf{v}=\mathbf{v}_x+\mathbf{v}_{yz} :
|\mathbf{v}_{yz}|>\zeta^{-1}|\mathbf{v}_x|\right\}.
$$
Since $\Phi(l_0)$ is transversal to the plane $\{x=0\}$, the curve
$\Phi(\ell_V)$ is transversal to this invariant cone field if
$\zeta, V$, and $U^*$ are small enough. Every stable manifold of
$\Omega_V$ in the rectifying coordinates is tangent to this
central-unstable cone field, and together with Lemma
\ref{l.transversality} this implies that $\ell_V$ is transversal
to stable manifolds of $\Omega_V$. This shows
Theorem~\ref{t.main}~(iii) and hence concludes the proof of
Theorem~\ref{t.main}.

\section{Proof of Proposition~\ref{p.expanding}}\label{s.ppe}

\subsection{Properties of the Recurrent Sequences}

In this subsection we formulate and prove several lemmas on
recurrent sequences that will be used in the next subsection to
prove Proposition~\ref{p.expanding}.

\blm\label{l.dD} Given $C_1>0$, $C_2>0$, $\lambda>1$, and
$\varepsilon \in (0,\frac{1}{4})$, there exist
$\delta_0=\delta_0(C_1, C_2, \lambda, \varepsilon)$ and $N_0\in
\mathbb{N}$, $N_0=N_0(C_1, C_2, \lambda, \varepsilon, \delta_0)$,
such that for every $\delta\in (0, \delta_0)$ and every $N\ge
N_0$, the following holds. Suppose that the sequences
$\{d_i\}_{i=0}^N$, $\{D_i\}_{i=0}^N$ are defined by the initial
conditions \beq\label{e.initialDd} \left\{
  \begin{array}{ll}
    d_0=1 \\
    D_0\ge C_2\frac{1}{(\lambda+\delta)^{N/2}},
  \end{array}
\right.
\eneq
and recurrence relations
\beq\label{e.recurrenceDd}
\left\{
  \begin{array}{ll}
    d_{k+1}=(1+2\delta)d_k+\delta D_k, \\
    D_{k+1}=(\lambda-\delta)D_k-C_1b_kd_k,
  \end{array}
\right.
\eneq
where
$$
b_k=(\lambda-\delta)^{-N+k}.
$$
Then
$$
\left\{
  \begin{array}{ll}
    d_N\le 2\delta^{1/2}D_N \\
    D_N\ge
D_0\lambda^{N(1-\varepsilon)}>\lambda^{\frac{N}{2}(1-4\varepsilon)}.
  \end{array}
\right.
$$
 \elm
\brm\label{r.coneDd} Notice that Lemma \ref{l.dD} implies also that $$
\frac{d_N}{D_N}\le 2\delta^{1/2}.
$$
This inequality will allow us to obtain a small cone (of size of
order $\delta^{1/2}$) where the image of a vector after leaving a
neighborhood of a singularity is located. \erm

\begin{proof}[Proof of Lemma~\ref{l.dD}.]
Let us denote
$$
\Lambda_{-}=\lambda^{1-\varepsilon},\ \ \
\Lambda_{+}=\lambda^{1+\varepsilon}.
$$
If $\delta$ is small enough, then
$1<\Lambda_{-}<\lambda-\delta<\lambda+\delta<\Lambda_{+}$. We will
prove by induction that \beq\label{e.inductiveassumptions} \left\{
  \begin{array}{ll}
    d_k\le (1+2\delta+\delta^{1/2})\max\{d_{k-1}, \delta^{1/2} D_{k-1}\} \\
    D_k\ge \Lambda_{-}D_{k-1}\ge  \Lambda_{-}^k D_0.
  \end{array}
\right. \eneq
It is clear that these inequalities for $k=N$ imply Lemma \ref{l.dD}.

Let us first check the base of induction. If $N$ is large, then
$D_0<1$, so $\max\{d_0, \delta^{1/2}D_0\}=d_0=1$. So we have
 $$
 d_1=(1+2\delta)d_0+\delta D_0<1+3\delta<(1+2\delta+\delta^{1/2})d_0.
 $$
 This is the first inequality in (\ref{e.inductiveassumptions}) for $k=1$.

 Also we have
\begin{align*}
D_1 & = (\lambda-\delta)D_0-C_1b_0d_0 \\
& = (\lambda-\delta)D_0\left(1-\frac{C_1b_0d_0}{D_0}\right) \\
& \ge (\lambda-\delta) D_0 \left(1-\frac{C_1(\lambda+\delta)^{N/2}}{C_2(\lambda-\delta)^{N}}\right)\\
& > (\lambda - \delta) D_0 \left( 1 - \frac{C_1\Lambda_{+}^{N/2}}{C_2\Lambda_{-}^N} \right) \\
& =( \lambda - \delta ) D_0 \left( 1 - \frac{C_1}{C_2}
\lambda^{-\frac{N}{2} ( 1 - 3 \varepsilon)} \right),
\end{align*}
and since
$(\lambda-\delta)\left(1-\frac{C_1}{C_2}\lambda^{-\frac{N}{2}(1-3\varepsilon)}\right)>\Lambda_{-}$
if $N$ is large enough, we have $D_1>\Lambda_{-}D_0$. We checked
the base of induction.

We proceed to proving the induction step. Assume that for some
$k$, the inequalities (\ref{e.inductiveassumptions}) hold. Let us
show that these inequalities also hold for $k+1$. We have
$$
d_{k+1}=(1+2\delta)d_k+\delta D_k.
$$
If $\delta^{1/2}D_k\le d_k$, then
\begin{align*}
d_{k+1} & \le ( 1 + 2 \delta ) d_k + \delta^{1/2} d_k \\
& = ( 1 + 2 \delta + \delta^{1/2} ) d_k \\
& = ( 1 + 2 \delta + \delta^{1/2}) \max\{d_{k-1}, \delta^{1/2}
D_{k-1}\}.
\end{align*}
If $\delta^{1/2}D_k> d_k$, then
\begin{align*}
d_{k+1} & \le (1+2\delta)\delta^{1/2}D_k+\delta D_k \\
& = \delta^{1/2} D_k ( 1 + 2 \delta + \delta^{1/2} ) \\
& = ( 1 + 2 \delta + \delta^{1/2})\max\{d_{k-1}, \delta^{1/2}
D_{k-1}\}.
\end{align*}

In order to estimate $D_{k+1}$, we need more information on
$\{d_i\}_{i=0}^k$. \blm\label{l.dsubk} Assume that $k^*\le N$ and
\eqref{e.inductiveassumptions} holds for all $k=1, \ldots, k^*$.
Assume also that for $k=1, 2, \ldots, l-1$, we have
$\delta^{1/2}D_k\le d_k$ and $\delta^{1/2}D_l>d_l$. Then,
$\delta^{1/2}D_k>d_k$ for every $k=l, l+1, \ldots, k^*$. \elm
\begin{proof}
If $\delta$ is small enough, $1+2\delta+\delta^{1/2}<\Lambda_{-}$.
Therefore, if $\delta^{1/2}D_l>d_l$, then
\begin{align*}
d_{l+1} & = (1+2\delta)d_l+\delta D_l \\
& < (1 + 2\delta+\delta^{1/2})\delta^{1/2} D_l \\
& < \Lambda_{-}\delta^{1/2}D_l \\
& \le \delta^{1/2} D_{l+1}.
\end{align*}
In the same way, $d_{l+2}<\delta^{1/2}D_{l+2}$, and so on.
\end{proof}
Notice that Lemma \ref{l.dsubk} immediately implies the following
statement. \blm\label{l.dsubk1} Assume that $k^*\le N$ and
\eqref{e.inductiveassumptions} holds for all $k=1, \ldots, k^*$.
If $d_k\ge \delta^{1/2}D_k$ for some $k\in\{1, 2, \ldots, k^*\}$,
then $d_k\le (1+2\delta +\delta^{1/2})^k$. \elm

Now let us estimate $D_{k+1}$. If $d_k<\delta^{1/2}D_k$, then
\begin{align*}
D_{k+1} & = (\lambda-\delta)D_k
\left(1-\frac{C_1b_kd_k}{(\lambda-\delta)D_k} \right) \\
& \ge (\lambda-\delta)D_k \left(1 -
\left(\frac{C_1}{\lambda-\delta} \right)
\frac{\delta^{1/2}}{(\lambda-\delta)^{N-k}}\right) \\
& > (\lambda-\delta)D_k(1-C_1\delta^{1/2}) \\
& > \Lambda_{-}D_k
\end{align*}
if $\delta$ is small enough.

If $d_k \ge \delta^{1/2}D_k$, then due to Lemma \ref{l.dsubk1}, we
have $d_k\le (1+2\delta+\delta^{1/2})^k$, and hence
\begin{align*}
D_{k+1} & = (\lambda-\delta)D_k\left(1-\frac{C_1b_kd_k}{(\lambda-\delta)D_k}\right) \\
& \ge ( \lambda - \delta ) D_k \left( 1 - \left(\frac{C_1}{\lambda-\delta}\right)
\frac{(1 + 2 \delta + \delta^{1/2})^k}{\Lambda_{-}^k D_0(\lambda-\delta)^{N-k}}\right)\\
& \ge ( \lambda - \delta ) D_k \left(1 - \left(
\frac{C_1}{C_2(\lambda-\delta)} \right) \frac{(1 + 2 \delta + \delta^{1/2})^k
(\lambda + \delta)^{N/2}}{\Lambda_{-}^k(\lambda - \delta)^{N-k}}\right)\\
& \ge ( \lambda - \delta) D_k \left(1 - \left(\frac{C_1}{C_2(\lambda-\delta)}\right)
\frac{(1 + 2\delta+\delta^{1/2})^k\Lambda_{+}^{N/2}}{\Lambda_{-}^k\Lambda_{-}^{N-k}}\right)\\
& \ge ( \lambda - \delta) D_k \left(1 - \left(\frac{C_1}{C_2(\lambda-\delta)}\right)
\frac{(1+2\delta+\delta^{1/2})^k \lambda^{\frac{N}{2}(1+\varepsilon)}}{\lambda^{N(1-\varepsilon)}}\right) \\
& \ge ( \lambda - \delta ) D_k \left(1 -
\left(\frac{C_1}{C_2(\lambda-\delta)}\right)
(1+2\delta+\delta^{1/2})^k
\lambda^{-\frac{N}{2}(1-3\varepsilon)}\right)\\
& >\Lambda_{-}D_k
\end{align*}
if $N$ is large enough. This concludes the proof of Lemma
\ref{l.dD}.
\end{proof}

\blm\label{l.Aa} Given $C_1>0$, $C_2>0$, $\lambda>1$, and
$\varepsilon \in (0,\frac{1}{4})$, there exist
$\delta_0=\delta_0(C_1, C_2, \lambda, \varepsilon)$ and $N_0\in
\mathbb{N}$, $N_0=N_0(C_1, C_2, \lambda, \varepsilon, \delta_0)$,
such that for every $\delta\in (0, \delta_0)$ and every $N\ge
N_0$, the following holds. Assume that the sequences
$\{a_k\}_{k=0}^N$ and $\{A_k\}_{k=0}^N$ have the following
properties: \beq\label{e.initialAa} \left\{
  \begin{array}{ll}
    a_0=1 \\
    A_0\ge C_2\sqrt{\widetilde{b}_0}>0,
  \end{array}
\right.
\eneq
and for $k=0, 2, \ldots, N-1$
\beq\label{e.recurrenceAa}
\left\{
  \begin{array}{ll}
    a_{k+1}\le (1+2\delta)a_k+\delta A_k, \\
    A_{k+1}\ge (\lambda-\delta)A_k-C_1\widetilde{b}_k a_k,
  \end{array}
\right.
\eneq
where the sequence $\widetilde{b_k}$ has the properties:
$$
0<\widetilde{b}_0<\widetilde{b}_1 <\ldots <
\widetilde{b}_{N-1}<1\le \widetilde{b}_N ,
$$
and for all $k=0, 2, \ldots, N-1$
$$
(\lambda-\delta)\widetilde{b}_k \le \widetilde{b}_{k+1} \le
(\lambda+\delta)\widetilde{b}_k.
$$
Then,
$$
\left\{
  \begin{array}{ll}
    a_N\le 2\delta^{1/2}A_N \\
    A_N\ge \lambda^{N(1-\varepsilon)}A_0> \lambda^{\frac{N}{2}(1-4\varepsilon)}.
  \end{array}
\right.
$$
\elm
\begin{proof}
Consider the sequences $\{d_k\}$ and $\{D_k\}$ defined by
(\ref{e.initialDd}) and (\ref{e.recurrenceDd}), where we take
$D_0=A_0$. If we prove that for all $k=0, 1, \ldots, N$
\beq\label{e.AaDdinduction} \left\{
  \begin{array}{ll}
    A_k\ge D_k \\
    \frac{A_k}{a_k}\ge \frac{D_k}{d_k},
  \end{array}
\right.
\eneq
then the required inequalities follow from Lemma \ref{l.dD}.

We will prove (\ref{e.AaDdinduction}) by induction. The base of
induction ($k=0$) is provided by our choice of $D_0$. Let us turn
to the induction step. If \eqref{e.AaDdinduction} holds for some
$k$, then
\begin{align*}
\frac{A_{k+1}}{a_{k+1}} & \ge
\frac{(\lambda-\delta)A_k-C_1\widetilde{b}_k
a_k}{(1+2\delta)a_k+\delta
A_k} \\
& =(\lambda-\delta)\delta^{-1}-\frac{(1+2\delta)(\lambda-\delta)\delta^{-1}+C_1\widetilde{b}_k}{(1+2\delta)+\delta\frac{A_k}{a_k}}\\
& \ge (\lambda - \delta) \delta^{-1} - \frac{(1 + 2\delta)(\lambda
- \delta)\delta^{-1}+C_1{b_k}}{(1+2\delta)+\delta\frac{D_k}{d_k}} \\
& =  \frac{D_{k+1}}{d_{k+1}}.
\end{align*}
Also,
\begin{align*}
A_{k+1} & \ge (\lambda-\delta)A_k-C_1b_ka_k \\
& = A_k\left((\lambda-\delta)-C_1\widetilde{b}_k \frac{a_k}{A_k}\right) \\
& \ge D_k\left((\lambda-\delta)-C_1{b_k}\frac{d_k}{D_k}\right) \\
& = D_{k+1}.
\end{align*}
 Lemma \ref{l.Aa} is proved.
\end{proof}

\subsection{Expansion of Vectors From Large Cones}

Finally, we come to the

\begin{proof}[Proof of Proposition~\ref{p.expanding}.]
Take $\mathbf{v}\in K_p$. If $|\mathbf{v}_z|>|\mathbf{v}_{xy}|$,
then the required inequalities follow just from the condition
(iii). Therefore, we can assume that $|\mathbf{v}_{xy}|\ge
|\mathbf{v}_z|$, and normalize the vector $\mathbf{v}$ assuming
that $|\mathbf{v}_{xy}|=1$.

Denote
$$
Df(p)=\begin{pmatrix}
        \nu(p) & m_1(p) & t_1(p) \\
        m_2(p) & e(p) & t_2(p) \\
        s_1(p) & s_2(p) & \lambda(p) \\
      \end{pmatrix}.
$$
Given a vector $\mathbf{v}\in K_p$, denote
$\mathbf{v}'=Df(p)(\mathbf{v})$,
$\mathbf{v}'=\mathbf{v}'_{xy}+\mathbf{v}'_z$,
$\mathbf{v}'_{xy}=\mathbf{v}'_x+\mathbf{v}'_y$. We have
$$
Df(p)(\mathbf{v})=\begin{pmatrix}
        \nu(p) & m_1(p) & t_1(p) \\
        m_2(p) & e(p) & t_2(p) \\
        s_1(p) & s_2(p) & \lambda(p) \\
      \end{pmatrix}\begin{pmatrix}
                     \mathbf{v}_x \\
                     \mathbf{v}_y \\
                     \mathbf{v}_z \\
                   \end{pmatrix}=\begin{pmatrix}
                     \nu(p)\mathbf{v}_x+m_1(p)\mathbf{v}_y+t_1(p)\mathbf{v}_z \\
                     m_2(p)\mathbf{v}_x+e(p)\mathbf{v}_y+t_2(p)\mathbf{v}_z \\
                     s_1(p)\mathbf{v}_x+s_2(p)\mathbf{v}_y+\lambda(p)\mathbf{v}_z \\
                   \end{pmatrix}
$$
The conditions (i)--(iii) of Proposition \ref{p.expanding} imply
that $|\nu(p)|\le \lambda^{-1}+\delta$, $|m_1(p)|, |m_2(p)|,
|t_1(p)|, |t_2(p)|\le \delta$, and $|\lambda(p)|\ge
\lambda-\delta$. Also, if $p$ belongs to the plane $\{z=0\}$, then
$s_1(p)=s_2(p)=0$. Since $\|f\|_{C^2}\le C_1$, for arbitrary $p$,
we have $|s_1(p)|, |s_2(p)|\le C_1z_p$.

Let us use the norm $\|\mathbf{v}\|=|\mathbf{v}_x|+|\mathbf{v}_y|+|\mathbf{v}_z|$.
Then we have
$$
\|\mathbf{v}_{xy}'\|\le
(1+2\delta)\|\mathbf{v}_{xy}\|+\delta\|\mathbf{v}_z\|,
$$
$$
\|\mathbf{v}_z'\|\ge
(\lambda-\delta)\|\mathbf{v}_z\|-C_1z_p\|\mathbf{v}_{xy}\|.
$$
Denote by $\widetilde{b}_k$ the $z$-coordinate of $f^k(p)$, by
$A_k$ the $z$-component of the vector $Df^k(p)(\mathbf{v})$, and
by $a_k$ the $xy$-component of $Df^k(p)(\mathbf{v})$. Applying
Lemma \ref{l.Aa} we get \eqref{e.expansionforN}, and \eqref{e.u}
follows from Remark~\ref{r.coneDd}.

Now if $A_0=|\mathbf{v}_z|\ge \eta$, then (applying
Lemma~\ref{l.dD} for $D_0=\eta$) from \eqref{e.AaDdinduction} and
\eqref{e.inductiveassumptions}, we have
$$
|Df^k(p)(\mathbf{v})|\ge A_k\ge D_k\ge \Lambda_{-}^kD_0\ge
\lambda^{\frac{k}{2}(1-\varepsilon)}\eta>\frac{1}{2}\lambda^{\frac{k}{2}(1-\varepsilon)}\eta
|\mathbf{v}|>C\lambda^{\frac{k}{2}(1-4\varepsilon)} |\mathbf{v}|
$$
for every $k=1, 2, \ldots, N$, where $C=\frac{1}{2}\eta$. Proposition \ref{p.expanding} is proved.
\end{proof}


TITLE: How to prove that $\int_{0}^{\infty}\ln^2(x)\sin(x^2)dx=\frac{1}{32}\sqrt{\frac{\pi}{2}}(2\gamma-\pi+\ln16)^2$
QUESTION [12 upvotes]: Wolfram Alpha provides

$$\int_{0}^{\infty}\ln^2(x)\sin(x^2)dx=\frac{1}{32}\sqrt{\frac{\pi}{2}}(2\gamma-\pi+\ln16)^2\tag{1}$$

But I haven't figured out the way to verify this result.
I know Frullani's Integral
$$\ln(x)= \int_{0}^{\infty}\frac{e^{-t}-e^{-xt}}{t}dt$$
I also know
$$\int_{0}^{\infty}\sin(x^2)~dx=\frac{1}{2}\int_{0}^{\infty}x^{-1/2}\sin(x)~dx$$
Then,
$$\begin{align}
\int_{0}^{\infty}\ln^2(x)\sin(x^2)dx&=\int_{0}^{\infty}\left(\int_{0}^{\infty}\frac{e^{-t}-e^{-xt}}{t}dt\right)\left(\int_{0}^{\infty}\frac{e^{-n}-e^{-xn}}{n}dn\right)\sin(x^2)~dx\\
&=\frac{1}{2}\int_{0}^{\infty}\left(\int_{0}^{\infty}\frac{e^{-t}-e^{-xt}}{t}dt\right)\left(\int_{0}^{\infty}\frac{e^{-n}-e^{-xn}}{n}dn\right)\frac{\sin(x)}{\sqrt{x}}dx\\
&=\frac{1}{2}\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-t}-e^{-xt}}{t}\frac{e^{-n}-e^{-xn}}{n}\frac{\sin(x)}{\sqrt{x}}~dx~dn~dt\\
&=\frac{1}{2}\int_{0}^{\infty}\frac{1}{t}\int_{0}^{\infty}\frac{1}{n}\int_{0}^{\infty}(e^{-t}-e^{-xt})(e^{-n}-e^{-xn})\frac{\sin(x)}{\sqrt{x}}~dx~dn~dt\\
&=\frac{1}{2}\int_{0}^{\infty}\frac{1}{t}\int_{0}^{\infty}\frac{1}{n}\int_{0}^{\infty}(e^{-t-n}-e^{-xn-t}-e^{-xt-n}+e^{-xt-xn})\frac{\sin(x)}{\sqrt{x}}~dx~dn~dt
\end{align}$$
What should I do next?
There is also a general case

$$\int_{0}^{\infty}\ln^2(x^a)\sin(x^2)dx=\frac{a^2}{32}\sqrt{\frac{\pi}{2}}(2\gamma-\pi+\ln16)^2\tag{2}$$

But I think $(2)$ becomes easy to prove if we can prove $(1)$.

REPLY [1 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{R > 0}$ and $\ds{\nu \in \pars{0,1}}$:
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{R}x^{\nu}\exp\pars{\ic x^{2}}\dd x}
\\[5mm] = &\
-\int_{0}^{\pi/4}\pars{R\expo{\ic \theta}}^{\nu}
\exp\pars{\ic R^{2}\expo{2\ic\theta}}R\expo{\ic\theta}\ic\,\dd\theta -
\int_{R}^{0}\pars{r\expo{\ic\pi/4}}^{\nu}
\exp\pars{\ic\bracks{r\expo{\ic\pi/4}}^{2}}\expo{\ic\pi/4}\,\dd r
\\[8mm] = &\
-\overbrace{R^{\nu + 1}\,\ic\int_{0}^{\pi/4}
\exp\pars{\ic\bracks{\nu\theta + R^{2}\cos\pars{2\theta} + \theta}}
\exp\pars{-R^{2}\sin\pars{2\theta}}\dd\theta}^{\ds{\equiv\ \mc{I}\pars{R,\nu}}}
\\[2mm] + &\
\expo{\ic\pars{\nu + 1}\pi/4}\int_{0}^{R}r^{\nu}\expo{-r^{2}}\dd r
\end{align}

Since $\ds{\nu \in \pars{0,1}}$, note that

\begin{align}
0 & < \verts{\mc{I}\pars{R,\nu}} <
R^{\nu + 1}\int_{0}^{\pi/4}\expo{-4R^{2}\theta/\pi}\dd\theta =
{\pi \over 4}\,{1 - \exp\pars{-R^{2}} \over R^{1 - \nu}}
\,\,\,\stackrel{\mrm{as}\ R\ \to\ \infty}{\LARGE\to}\,\,\,
\color{red}{\large 0}
\end{align}

Then,
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\infty}x^{\nu}\sin\pars{x^{2}}\dd x} =
\sin\pars{\bracks{\nu + 1}\,{\pi \over 4}}
\int_{0}^{\infty}r^{\nu}\expo{-r^{2}}\dd r
\\[5mm] \stackrel{r^{2}\ \mapsto\ r}{=}\,\,\,&
{1 \over 2}\,\sin\pars{\bracks{\nu + 1}\,{\pi \over 4}}
\int_{0}^{\infty}r^{\nu/2 - 1/2}\expo{-r}\dd r =
{1 \over 2}\,\sin\pars{\bracks{\nu + 1}\,{\pi \over 4}}
\Gamma\pars{{\nu \over 2} + {1 \over 2}}
\end{align}
and
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\infty}\ln^{2}\pars{x}\sin\pars{x^{2}}\dd x} =
\lim_{\nu \to 0^{+}}
\totald[2]{}{\nu}
\braces{{1 \over 2}\,\sin\pars{\bracks{\nu + 1}\,{\pi \over 4}}
\Gamma\pars{{\nu \over 2} + {1 \over 2}}}
\\[5mm] = &\
\bbx{{1 \over 32}\root{\pi \over 2}
\bracks{\vphantom{\Large A}2\gamma - \pi + \ln\pars{16}}^{2}} \approx 0.0242
\end{align}


text:
\begin{document}

\title{A Lower Bound of the First Eigenvalue of a Closed Manifold with Negative Lower Bound of the Ricci Curvature
\thanks{2000 Mathematics Subject Classification Primary 58J50, 35P15; Secondary 53C21}}

\author{Jun LING}
\date{}

\maketitle

\begin{abstract}
Along the line of the Yang Conjecture, we give a new estimate on
the lower bound of the first non-zero eigenvalue of a closed
Riemannian manifold with negative lower bound of Ricci curvature
in terms of the in-diameter and the lower bound of Ricci
curvature.
\end{abstract}

\section{Introduction}\label{sec-intro}
It has been proved by Li and Yau \cite{liy1} that if $M$ is an
n-dimensional closed Riemannian manifold with Ricci curvature
Ric$(M)$ bounded below by $(n-1)\kappa$, with constant $\kappa<0$,
then the first non-zero eigenvalue $\lambda$ of the Laplacian of
$M$ has the lower bound
\[
\lambda \geq
\frac{1}{2(n-1)d^2}{\exp\{-1-\sqrt{1+4(n-1)^2d^2|\kappa|}\}},
\]
where $d$ is the diameter of $M$.

H.~C.~Yang \cite{yang} obtained later the following estimate
\[
\lambda \geq \frac{\pi^2}{d^2}\exp\{-C_n\sqrt{(n-1)|\kappa|d^2}\},
\]
where $C_n=\max\{\sqrt{n-1}, \sqrt{2}\}$. Yang further conjectured
that
\[
\lambda \geq \frac12(n-1)\kappa + \frac{\pi^2}{d^2} ,
\]

Along the line of the Yang Conjecture, we give a new estimate on
the lower bound of the first non-zero eigenvalue of a closed
Riemannian manifold with negative lower bound of Ricci curvature
in terms of the in-diameter and the lower bound of Ricci
curvature. Instead of using the Zhong-Yang's canonical function or
the "midrange" of the normalized eigenfunction of the first
eigenvalue in the proof,  we use a function $\xi$ that the author
constructed in \cite{ling1} for the construction of the suitable
test function and use the structure of the nodal domains of the
eigenfunction. That provides a new way to sharpen the bound. We
have the following result.
\begin{theorem} \label{main-thm}
If $M$ is an $n$-dimensional closed Riemannian manifold and if the
Ricci curvature of $M$ has a lower bound
\begin{equation}\label{ricci-bound}
\textup{Ric}(M)\geq (n-1)\kappa
\end{equation}
for some constant $\kappa<0$, then the first non-zero eigenvalue
$\lambda$ of the Laplacian $\Delta$ of $M$ satisfies the
inequality
\[
\lambda \geq
\frac{1}{1-(n-1)\kappa/(2\lambda)}\,\frac{\pi^2}{\tilde{d}^2}>0
\]
and $\lambda$ has the following lower bound,
\begin{equation}                    \label{main-bound}
\lambda \geq \frac12(n-1)\kappa+\frac{\pi^2}{\tilde{d}^2},
\end{equation}
where $\tilde{d}$ is the diameter of the largest interior ball in
the nodal domains of the first eigenfunction.
\end{theorem}

Note that from the proof of the Theorem \ref{main-thm}, the
in-diameter $\tilde{d}$ can be replaced by the larger of the
diameters of the two nodal domains of an eigenfunction of the
first non-zero eigenvalue.

If Ric$(M)\geq (n-1)\kappa$ with constant $\kappa>0$, it is known
that the first non-zero eigenvalue $\lambda$ has a lower bound as
the above. Therefore the lower bound in (\ref{main-bound}) is
universal for all three cases, no matter constant $\kappa>0, =0$
or $<0$.

We derive some preliminary estimates and conditions for test
functions in the next section and construct the needed test
function and prove the main result in the last section.

\section{Preliminary Estimates}\label{sec-pre-es}
Let $v$ be a normalized eigenfunction of the first non-zero
eigenvalue $\lambda$ of the Laplacian $\Delta$ such that
\begin{equation}                                \label{v-c-con}
\sup_{M}v=1, \quad \inf_{M}v=-k
\end{equation}
with $0<k\leq 1$. The function $v$ satisfies the following
equation
\begin{equation}                                 \label{v-c-eq}
\Delta v=-\lambda v\quad \textrm{in }M,
\end{equation}
where $\Delta$ is the Laplacian of $M$.

We first use gradient estimate in \cite{li}-\cite{liy1} and
 \cite{sy} to derive following estimate.
\begin{lemma}       \label{pre-es-c}
The function $v$ satisfies the following
\begin{equation}                        \label{basic3-c}
\frac{\left |\nabla v\right |^2}{b^2-v^2} \leq\lambda(1+\beta),
\end{equation}
where $\beta = -(n-1)\kappa/\lambda>0$ and $b>1$ is an arbitrary
constant.
\end{lemma}

\begin{proof}\quad
Consider the function
\begin{equation}                \label{p-of-x-def}
P(x)=|\nabla v|^2+Av^2,
\end{equation}
where $A=\lambda -(n-1)\kappa+\epsilon$ for small $\epsilon>0$.
Function $P$ must achieve its maximum at some point $x_0\in M$. We
claim that $\nabla v (x_0)=0$.

If on the contrary, $\nabla v (x_0)\not=0$, then we can rotate the
local orthonormal frame about $x_0$ such that
\[
|v_1(x_0)|=|\nabla v(x_0)|\not=0\qquad \textrm{and}\qquad
v_i(x_0)=0,\quad i\geq2.
\]
Since $P$ achieves its maximum at $x_0$, we have
\[
\nabla P(x_0)=0 \qquad \textrm{and}\qquad \Delta P(x_0)\leq 0.
\]
That is, at $x_0$
\[
0=\frac12\nabla_{i} P=\sum_{j=1}^{n}v_jv_{ji}+Avv_i,
\]
\begin{equation}                \label{d1}
v_{11}=-Av \qquad \textrm{and}\qquad v_{1i}=0\quad i\geq 2,
\end{equation}
and
\begin{eqnarray}
&0 &\geq \Delta P(x_0) =\sum_{i, j=1}^{n}\left(v_{ji}v_{ji}+v_{j}v_{jii}+Av_{i}v_{i}+Avv_{ii}\right)\nonumber\\
&{} &=\sum_{i, j=1}^{n}\left(v_{ji}^2+v_j(v_{ii})_j +\textrm{R}_{ji}v_{j}v_{i}+Av_{ii}^2 +Av v_{ii}\right)\nonumber\\
&{} &=\sum_{i, j=1}^{n}v_{ji}^2+\nabla v\nabla(\Delta v) +\textrm{Ric}(\nabla v,\nabla v)+A|\nabla v|^2 +Av\Delta v\nonumber\\
&{} &\geq v_{11}^2+\nabla v\nabla(\Delta v) + (n-1)\kappa|\nabla v|^2+A|\nabla v|^2 +Av\Delta v\nonumber\\
&{} &=(-Av)^2-\lambda |\nabla v|^2 + (n-1)\kappa|\nabla v|^2+A|\nabla v|^2 -\lambda Av^2\nonumber\\
&{} &=[A-\lambda + (n-1)\kappa]|\nabla
v|^2+A(A-\lambda)v^2,\nonumber
\end{eqnarray}
where we have used (\ref{d1}) and (\ref{ricci-bound}). Therefore
at $x_0$,
\begin{equation}
0\geq [A-\lambda + (n-1)\kappa]|\nabla v|^2+A(A-\lambda)v^2.
\label{d2}
\end{equation}
That is,
\[
\epsilon |\nabla v(x_0)|^2 +
[-(n-1)\kappa+\epsilon][\lambda-(n-1)\kappa+\epsilon]v(x_0)^2\leq
0.
\]
Thus $\nabla v(x_0) = 0$. This contradicts the assumption $\nabla
v(x_0)\not= 0$. So the above claim is right.

Therefore we have $\nabla v(x_0)=0$, and
\[
P(x_0)=|\nabla v(x_0)|^2 +Av(x_0)^2=Av(x_0)^2\leq A.
\]
Now for all $x\in M$ we have
\[
|\nabla v(x)|^2+Av(x)^2=P(x)\leq P(x_0)\leq A
\]
and
\[
|\nabla v(x)|^2\leq A (1-v(x)^2).
 \]
Letting $\epsilon \rightarrow 0$ in the above inequality, the
estimate (\ref{basic3-c}) follows.
\end{proof}

We want to improve the above upper bound in (\ref{basic3-c})
further and proceed in the following way.

Define a function $Z$ on $[-\sin^{-1}(k/b),\sin^{-1}(1/b)]$ by
\[
Z(t)=\max_{x\in M,t=\sin^{-1} \left(v\left(x\right)/b\right)}
\frac{\left |\nabla v\right |^2}{b^2-v^2}/\lambda.
\]
The estimate in (\ref{basic3-c}) becomes
\begin{equation}                                \label{basic5-c}
Z(t)\leq 1+\beta\qquad \textrm{on }
[-\sin^{-1}(k/b),\sin^{-1}(1/b)].
\end{equation}

Throughout this paper let
\begin{equation}       \label{delta-def}
\alpha=\frac12 (n-1)\kappa<0 \qquad \textrm{and} \qquad \delta =
\alpha/\lambda<0.
\end{equation}

We have the following theorem on conditions for the test function.

\begin{theorem}                                           \label{thm-barrier-c}
If the function $z:\ [-\sin^{-1}(k/b),\sin^{-1}(1/b)]\mapsto
\mathbf{R}^1$ satisfies the following
\begin{enumerate}
 \item $z(t)\geq Z(t) \qquad t\in [-\sin^{-1}(k/b),\sin^{-1}(1/b)]$,
 \item there exists some $x_0\in M$
       such that at point $t_0=\sin^{-1} (v(x_0)/b)$ \linebreak
       $z(t_0)=Z(t_0)$,
 \item $z(t_0)\geq 1$,
 \item $z'(t_0)\sin t_0\leq 0$,
\end{enumerate}
then we have the following
\begin{equation}\label{barrier-eq-c}
0\geq-\frac12z''(t_0)\cos^2t_0 + z'(t_0)\cos t_0\sin t_0+z(t_0) -1
+2\delta \cos^2t_0.
\end{equation}
\end{theorem}

\begin{proof}\quad
Define
\[ J(x)=\left\{ \frac{\left |\nabla v\right |^2}{b^2-v^2}
-\lambda z \right\}\cos^2t,
\]
where $t=\sin^{-1}(v(x)/b)$. Then
\[ J(x)\leq 0\quad\textrm{for } x\in M
\qquad \textrm{and} \qquad J(x_0)=0.
\]
If $\nabla v(x_0)=0$, then
\[ 0=J(x_0)=-\lambda z\cos^2 t.
\]
This contradicts Condition 3 in the theorem. Therefore
\[ \nabla v(x_0)\not=0.
\]

The Maximum Principle implies that
\begin{equation}                                                \label{es1}
\nabla J(x_0)=0\qquad \textrm{and}\qquad \Delta J(x_0)\leq 0.
\end{equation}
$J(x)$ can be rewritten as
\[  J(x)=\frac{1}{b^2}|\nabla v|^2-\lambda z\cos^2t.
\]
Take normal coordinates about $x_0$.  (\ref{es1}) is equivalent to
\begin{equation}                                              \label{es2}
\frac{2}{b^2}\sum_{i}v_iv_{ij}\Big|_{x_0}=\lambda\cos t[z' \cos t
-2z\sin t]t_j\Big|_{x_0}
\end{equation}
and
\begin{eqnarray}                                              \label{es3}
0&\geq&\frac{2}{b^2}\sum_{i,j}v_{ij}^2+\frac{2}{b^2}\sum_{i,j}v_iv_{ijj}
 -\lambda (z''|\nabla t|^2+z'\Delta t)\cos^2t \\
 & &+4\lambda z'\cos t\sin t |\nabla t|^2 -
\lambda z\Delta\cos^2t\Big|_{x_0}.\nonumber
\end{eqnarray}
Rotate the frame so that $v_1(x_0)\not=0$ and $v_i(x_0)=0$ for
$i\geq 2$. Then (\ref{es2}) implies
\begin{equation}                                             \label{es4}
v_{11}\Big|_{x_0}=\frac{\lambda b}{2}(z'\cos t-2z\sin t)
\Big|_{x_0}\quad\text{and}\quad v_{1i} \Big|_{x_0}=0\ \text{for }
i\geq2.
\end{equation}
Now we have
\begin{eqnarray}
|\nabla v|^2
\Big|_{x_0}&=&\lambda b^2z\cos^2t\Big|_{x_0},\nonumber\\
 |\nabla t|^2
\Big|_{x_0}&=&\frac{|\nabla v|^2}{b^2-v^2}=\lambda z
\Big|_{x_0},\nonumber\\
\frac{\Delta v}{b}\Big|_{x_0} &=&\Delta \sin t =\cos t\Delta
t-\sin t |\nabla t|^2
\Big|_{x_0},\nonumber\\
\Delta t\Big|_{x_0}&=&\frac{1}{\cos t}(\sin t|\nabla
t|^2+\frac{\Delta v}{b})
\nonumber\\
 &=&\frac{1}{\cos t} [ \lambda z\sin t-\frac{\lambda}{b}v] \Big|_{x_0}, \quad\textrm{and}
\nonumber\\
\Delta\cos^2t\Big|_{x_0}&=&\Delta \left(1-\frac{v^2}{b^2}\right)
 =-\frac{2}{b^2}|\nabla v|^2-\frac{2}{b^2}v\Delta v
\nonumber\\
  &=&-2\lambda z\cos^2t+\frac{2}{b^2}\lambda v^2\Big|_{x_0}. \nonumber
\end{eqnarray}
Therefore,
\begin{eqnarray}
& {}&
\frac{2}{b^2}\sum_{i,j}v_{ij}^2\Big|_{x_0}\geq\frac{2}{b^2}v_{11}^2
\nonumber\\
& {}& =\frac{\lambda ^2}{2}(z')^2\cos^2t-2\lambda ^2zz'\cos t\sin
t
      +2\lambda ^2z^2\sin^2 t\Big|_{x_0}\nonumber,
\end{eqnarray}
\begin{eqnarray}
\frac{2}{b^2}\sum_{i,j}v_iv_{ijj}\Big|_{x_0}
&=&\frac{2}{b^2}\left(\nabla v\,\nabla
       (\Delta v)+\textrm{Ric}(\nabla v,\nabla v)\right)\nonumber\\
&\geq& \frac{2}{b^2}(\nabla v\,\nabla (\Delta v)+(n-1)\kappa|\nabla v|^2)\nonumber\\
 &=&-2\lambda^2z\cos^2t+4\alpha \lambda z\cos^2t\Big|_{x_0},\nonumber
\end{eqnarray}
\begin{eqnarray}
&{}&  -\lambda (z''|\nabla t|^2+ z'\Delta t)\cos^2t\Big|_{x_0}\nonumber\\
&{}&=-\lambda^2 zz''\cos^2t-
\lambda^2zz'\cos t\sin t\nonumber\\
&{}&{ }+\frac{1}{b}\lambda^2z'v\cos t \Big|_{x_0},\nonumber
\end{eqnarray}
and
\begin{eqnarray}
&{}&4\lambda z'\cos t\sin t|\nabla t|^2-\lambda z\Delta
\cos^2t\Big|_{x_0}
\nonumber\\
&{}&=4\lambda^2zz'\cos t\sin
t+2\lambda^2z^2\cos^2t-\frac{2}{b}\lambda^2zv\sin t
\Big|_{x_0}.\nonumber
\end{eqnarray}
Putting these results into (\ref{es3}) we get
\begin{eqnarray}                                                   \label{es5}
0&\geq&-\lambda^2zz''\cos^2t+ \frac{\lambda^2}{2}(z')^2\cos^2t
+\lambda^2z'\cos t\left(z\sin t +\sin t\right)
  \nonumber\\
 & & {}+2\lambda^2z^2-2\lambda^2z +4\alpha \lambda z\cos^2t
 \Big|_{x_0},
\end{eqnarray}
where we used (\ref{es4}). Now
\begin{equation}                                                    \label{es6}
z(t_0)>0,
\end{equation}
by Condition 3 in the theorem. Dividing two sides of (\ref{es5})
by $2\lambda^2z\Big|_{x_0}$, we have
\begin{eqnarray}
0&\geq&-\frac12z''(t_0)\cos^2t_0 +\frac12z'(t_0)\cos t_0\left(\sin t_0+\frac{\sin t_0}{z(t_0)}\right) +z(t_0) \nonumber\\
 & & {}  -1 +2\delta \cos^2t_0+\frac{1}{4z(t_0)}(z'(t_0))^2\cos^2t_0.\nonumber
\end{eqnarray}
Therefore,
\begin{eqnarray}
0&\geq&-\frac12z''(t_0)\cos^2t_0 + z'(t_0)\cos t_0\sin t_0+z(t_0) -1 +2\delta \cos^2t_0\nonumber\\
 & & {}+\frac12z'(t_0)\sin t_0\cos
 t_0[\frac{1}{z(t_0)}-1]+\frac{1}{4z(t_0)}(z'(t_0))^2\cos^2t_0.                               \label{es6.1}
\end{eqnarray}
By the conditions 3 and 4 in the theorem, the last two terms are
nonnegative. Therefore (\ref{barrier-eq-c}) follows.
\end{proof}

\section{Proof of the Main Result}\label{sec-proof}
We now prove our main result.
\begin{proof}[Proof of Theorem \ref{main-thm}]\quad
Let
\begin{equation}                               \label{z-def}
z(t)=1+\delta\xi(t),
\end{equation}
where $\xi$ is the functions defined by (\ref{xi-def}) in Lemma
\ref{xi-lemma} below and $\delta$ is the negative constant in
(\ref{delta-def}). We claim that
\begin{equation}                        \label{4.1-c}
Z(t)\leq z(t)\qquad \textrm{on } [-\sin^{-1}(k/b),\sin^{-1}(1/b)].
\end{equation}
Lemma \ref{xi-lemma} implies that for $t\in
[-\sin^{-1}(k/b),\sin^{-1}(1/b)] $, we have the following
\begin{eqnarray}
& &{}\frac{1}{2}z''\cos ^2t-z'\cos t\sin t-z
    =-1+ 2\delta\cos^2t,               \label{z-eq}\\
& &{}z'(t)\sin t\leq 0,\qquad (\textrm{since }\delta < 0)\quad\textrm{and}\label{z'-geq0}\\
& &{}z(t) \geq z(\frac{\pi}{2})=1. \label{z-min}
\end{eqnarray}

Let $P\in\mathbf{R}^1$ and $t_0\in
[-\sin^{-1}(k/b),\sin^{-1}(1/b)]$ such that
\[ P=\max_{t\in [-\sin^{-1}(k/b),\sin^{-1}(1/b)]}\left(Z(t)-z(t)\right)=Z(t_0)-z(t_0).
\]
Thus
\begin{equation}\label{4.2}
Z(t)\leq z(t)+P\quad \textrm{on }
[-\sin^{-1}\frac{k}{b},\sin^{-1}\frac1b]\quad\textrm{and}\quad
Z(t_0)=z(t_0)+P.
\end{equation}
Suppose that $P>0$. Then $z+P$ satisfies the conditions in Theorem
\ref{thm-barrier-c}. (\ref{barrier-eq-c}) implies that
\begin{eqnarray}
&{}&z(t_0)+P=Z(t_0)\nonumber\\
&{}&\leq  \frac12(z+P)''(t_0)\cos^2 t_0-(z+P)'(t_0)\cos t_0
 \sin t_0+1-2\delta \cos^2 t_0\nonumber\\
&{}&=\frac12z''(t_0)\cos^2t_0-z'(t_0)\cos t_0\sin
t_0+1-2\delta \cos^2 t_0\nonumber\\
&{}&=z(t_0).\nonumber
\end{eqnarray}
This contradicts the assumption $P>0$. Thus $P\leq 0$ and
(\ref{4.1-c}) must hold. That means
\begin{equation}\label{4.3-c}
\sqrt{\lambda}\geq \frac{|\nabla t|}{\sqrt{z(t)}}.
\end{equation}

Note that the eigenfunction $v$ of the first nonzero eigenvalue
has exactly two nodal domains $D^+=\{x: v(x)>0\}$ and $D^-=\{x:
v(x)<0\}$ and the nodal set $v^{-1}(0)$ is compact. Take $q_1$ on
$M$ such that $v(q_1)=1 =\sup_M v$ and and $q_2\in v^{-1}(0)$ such
that distance $d(q_1, q_2) = \textrm{ distance }d(q_1,
v^{-1}(0))$. Let $L$ be the minimum geodesic segment between $q_1$
and $q_2$. We integrate both sides of (\ref{4.3-c}) along $L$ and
change variable and let $b\rightarrow 1$.  Let $d_+$, $d_-$ be the
diameter of the largest interior ball in $D^+$, $D^-$
respectively, then
\[
d_+=2r_+ \qquad\textup{and}\qquad r_+=\max_{x\in D_+}
\textup{dist}(x, v^{-1}(0))
\]
and
\[
d_-=2r_- \qquad\textup{and}\qquad r_-=\max_{x\in D_-}
\textup{dist}(x, v^{-1}(0)).
\]
 Then $\tilde{d}=\max\{d_+, d_-\}$
\begin{equation}\label{4.4-c}
\sqrt{\lambda}\,\frac{d_+}{2}\geq \int_{L}\, \frac{|\nabla
t|}{\sqrt{z(t)}}\, dl=\int_0^{\frac{\pi}{2}}
\frac{1}{\sqrt{z(t)}}\,dt \geq \frac{\left(\int_0^{\pi/2}\
\,dt\right)^\frac32}{(\int_0^{\pi/2}\ z(t)\,dt)^{\frac12}} \geq
\left( \frac{(\frac{\pi}{2})^3}{\int_0^{\pi/2}\  z(t)\,dt}
\right)^{\frac12}.
\end{equation}
Square the two sides, we get
\[
\lambda \geq \frac{\pi^3}{2(d_+)^2\int_0^{\pi/2} \ z(t)\,dt}.
\]
Now
\[
\int_0^{\frac{\pi}{2}}\ z(t)\,dt=\int_0^{\frac{\pi}{2}}\ [1+
\delta \xi(t)]\,dt=\frac{\pi}{2}(1-\delta),
\]
by (\ref{xi-int}) in Lemma \ref{xi-lemma}. Therefore
\[
\lambda \geq \frac{\pi^2}{(1-\delta)(d_+)^2}\quad\textrm{and}\quad
\lambda \geq  \frac12(n-1)\kappa + \frac{\pi^2}{(d_+)^2}.
\]
Since $\tilde{d}\geq d_+$  and $\tilde{d}\geq d_-$, we complete
the proof.
\end{proof}

We now state and prove Lemma \ref{xi-lemma} used in the proof of
Theorem \ref{main-thm}.
\begin{lemma}                           \label{xi-lemma}
Let
\begin{equation}                        \label{xi-def}
\xi(t)=\frac{\cos^2t+2t\sin t\cos t +t^2-\frac{\pi^2}{4}}{\cos^2t}
\qquad \textrm{on}\quad [-\frac{\pi}{2},\frac{\pi}{2}\,].
\end{equation}
Then the function $\xi$  satisfies the following
\begin{eqnarray}
& &{}\frac{1}{2}\xi''\cos ^2t-\xi'\cos t\sin t-\xi
    =2\cos^2t\quad \textrm{in }(-\frac{\pi}{2},\frac{\pi}{2}\,),          \label{xi-eq}\\
& &{}\xi'\cos t -2\xi\sin t =4t\cos t                      \label{xi-eq2}\\
& &{}\int_0^{\frac{\pi}{2}}\xi(t)\, dt= -\frac{\pi}{2}          \label{xi-int}\\
& &{}1-\frac{\pi^2}{4}=\xi(0)\leq\xi(t)\leq\xi(\pm
\frac{\pi}{2})=0\quad
\textrm{on }[-\frac{\pi}{2},\frac{\pi}{2}\,],                                \nonumber\\
& &{} \xi' \textrm{ is increasing on }
[-\frac{\pi}{2},\frac{\pi}{2}\,] \textrm{ and }
\xi'(\pm \frac{\pi}{2}) =\pm \frac{2\pi}{3},                     \nonumber\\
& &{}\xi'(t)< 0 \textrm{ on }(-\frac{\pi}{2},0)\textrm{ and \ }
\xi'(t)>0 \textrm{ on }(0,\frac{\pi}{2}\,), \nonumber \\
& &{}\xi''(\pm\frac{\pi}{2})=2, \ \xi''(0)=2(3-\frac{\pi^2}{4})
\textrm{ and \ } \xi''(t)> 0 \textrm{ on }
[-\frac{\pi}{2},\frac{\pi}{2}\,], \nonumber\\
& &{}(\frac{\xi'(t)}{t})'>0 \textrm{ on } (0,\pi/2\,)\textrm{ and
\ } 2(3-\frac{\pi^2}{4})\leq \frac{\xi'(t)}{t}\leq \frac43
\textrm{ on } [-\frac{\pi}{2},\frac{\pi}{2}\,],\nonumber \\
& &{}\xi'''(\frac{\pi}{2})=\frac{8\pi}{15}, \xi'''(t)< 0 \textrm{
on }(-\frac{\pi}{2},0) \textrm{ and \ }  \xi'''(t)>0 \textrm{ on
}(0,\frac{\pi}{2}\,). \nonumber
\end{eqnarray}
\end{lemma}

\begin{proof}[Proof of Lemma \ref{xi-lemma}]\quad
For convenience, let $q(t)= \xi'(t)$, i.e.,
\begin{equation}                                       \label{q-def}
q(t) = \xi'(t) = \frac{2(2t\cos t +t^2\sin t +\cos^2 t \sin t
-\frac{\pi^2}{4}\sin t)}{\cos^3 t}.
\end{equation}
Equation (\ref{xi-eq}) and the values $\xi(\pm \frac{\pi}{2})=0$,
$\xi(0)=1-\frac{\pi^2}{4}$ and $\xi'(\pm \frac{\pi}{2}) =\pm
\frac{2\pi}{3}$ can be verified directly from (\ref{xi-def}) and
(\ref{q-def}) .  The values of $\xi''$ at $0$ and $\pm
\frac{\pi}{2}$ can be computed via (\ref{xi-eq}). By
(\ref{xi-eq2}), $(\xi(t)\cos^2 t)' =4t\cos^2 t$. Therefore
\newline $\xi(t)\cos^2 t=\int_{\frac{\pi}{2}}^t \ 4s\cos^2 s\,ds$,
and
\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\
\xi(t)\,dt=2\int_0^{\frac{\pi}{2}}\
\xi(t)\,dt=-8\int_0^{\frac{\pi}{2}}\left( \frac{1}{\cos^2(t)}
\int_t^{\frac{\pi}{2}}\ s\cos^2s\,ds\right)\,dt
\]
\[
=-8\int_0^{\frac{\pi}{2}}\left(\int_0^s\
\frac{1}{\cos^2(t)}\,dt\right)\ s\cos^2s\,ds
=-8\int_0^{\frac{\pi}{2}}\ s\cos s\sin s\,ds=-\pi.
\]
It is easy to see that $q$ and $q'$ satisfy the following
equations
\begin{equation}                                         \label{q-eq}
\frac12 q''\cos t -2q'\sin t -2q\cos t = -4 \sin t,
\end{equation}
and
\begin{equation}                                       \label{q'-eq}
\frac{\cos^2 t}{2(1+\cos^2 t)}(q')''-\frac{2\cos t\sin t}{1+\cos^2
t}(q')'-2(q')=-\frac{4}{1+\cos^2 t}.
\end{equation}
The last equation implies $q'=\xi''$ cannot achieve its
non-positive local minimum at a point in $(-\frac{\pi}{2},
\frac{\pi}{2})$. On the other hand, $\xi''(\pm\frac{\pi}{2})=2$,
by equation (\ref{xi-eq}), $\xi(\pm \frac{\pi}{2})=0$ and
$\xi'(\pm \frac{\pi}{2})=\pm \frac{2\pi}{3}$. Therefore
$\xi''(t)>0$ on $[-\frac{\pi}{2},\frac{\pi}{2}]$ and $\xi'$ is
increasing. Since $\xi'(t)=0$, we have $\xi'(t)< 0$ on
$(-\frac{\pi}{2},0)$ and $\xi'(t)>0$ on $(0,\frac{\pi}{2})$.

Similarly, from the equation
\begin{eqnarray}                                           \label{q''-eq}
&\frac{\cos^2 t}{2(1+\cos^2 t)}(q'')'' -\frac{\cos t\sin t
(3+2\cos^2 t)}{(1+\cos^2 t)^2}(q'')' -\frac{2(5\cos^2 t+\cos^4 t)}{(1+\cos^2 t)^2}(q'') \nonumber\\
&=-\frac{8\cos t\sin t}{(1+\cos^2 t)^2}
\end{eqnarray}
we get the results in the last line of the lemma.

Set $h(t)=\xi''(t)t-\xi'(t)$. Then $h(0)=0$  and $h'(t)=
\xi'''(t)t>0$ in $(0,\frac{\pi}{2})$. Therefore
$(\frac{\xi'(t)}{t})'=\frac{h(t)}{t^2}>0$ in $(0,\frac{\pi}{2})$.
Note that $\frac{\xi'(-t)}{-t}= \frac{\xi'(t)}{t}$,
$\frac{\xi'(t)}{t}|_{t=0}=\xi''(0)=2(3-\frac{\pi^2}{4})$ and
$\frac{\xi'(t)}{t}|_{t=\pi/2}=\frac43$. This completes the proof
of the lemma.
\end{proof}


TITLE: Prove that the locus is a parabola
QUESTION [1 upvotes]: The point P(x,y) moves in XY plane such as that its distance from a fixed point (0,-1) is equal to its distance from the line Y=1. Prove that the locus is a parabola. Find it's focus, directrix, vertex, axis of symmetry and focal length.
I really need help and don't have much of an idea of how to do this.

REPLY [1 votes]: Since we have
$$(x-0)^2+\{y-(-1)\}^2=|y-1|^2,$$
we have
$$x^2=4\cdot (-1)\cdot y.$$
The vertex is $(0,0)$, the axis of symmetry is $x=0$, the focus is $(0,-1)$, the directrix is $y=-(-1)$.


TITLE: Z transform of $u(n)n$
QUESTION [0 upvotes]: I know that the z transform of $u(n)$ is $\frac{z}{z-1}$ and z transform of $n$ is $\frac{z}{(z-1)^2}$. However i don't know how i can find the transform of their product

REPLY [1 votes]: It is not correct. $x[n]=n$ has no z-transform since the following sum$$\sum_{n=-\infty}^\infty nz^{-n}$$does not exist for $z\ne 0$.


TITLE: How can you state how abstract interpretation gets around Rice's Theorem succinctly?
QUESTION [2 upvotes]: At this question, abstract interpretation has a nice in-depth look. However, I'd like someone to clearly and very precisely state how abstract interpretation takes the result of Rice's Theorem over the Turing-complete programming languages/Turing machines, and uses abstract interpretation to get similar results for decidability for non-trivial program properties; or, rather, how does abstract interpretation get around Rice's Theorem/the Halting Problem?

REPLY [7 votes]: To be as clear and concise as possible, I'd try saying this

Abstract interpretation may only over-approximate properties of programs: the most precise abstract value of a program $P$ may be $a$, but any algorithm for computing an abstract value will in general compute some value $a'\geq a$ for $P$ (possibly $\top$ in the worst case).

REPLY [1 votes]: This is logically redundant with the answers already given by Cody and Mikolas, but here's an intuition that might help.  The property that any abstract interpretation proves about a program is guaranteed to be correct, but sometimes this is because the answer it brings back is "I don't know".  That answer may be useless, but it isn't wrong.  Basically, if you want a decidable version of a decision problem, you have to add this third option, so that the possible results are not "yes"/"no", but "yes"/"no"/"I give up".


TITLE: Proof by induction that $2^n - 1 > n^2$
QUESTION [2 upvotes]: i want to prove that $\forall n\geq 5$
$$2^{n}-1 > n^{2}$$
so the basis is trivial, and in the induction step (n+1), i stuck. i get :
$(n+1)^{2} =  n^{2} + 2n + 1 < (2^{n} -1)+ 2n+1 = 2^{n} + 2n $
what i need to do from here?
tnx!

REPLY [0 votes]: You know that $2^n$ grows a lot quicker than $n^2$, so you can be quite generous. Since $n ≥ 5$,
$(n+1)^2 = n^2 + 2n + 1 < n^2 + 3n < n^2 + 5n ≤ 2n^2 ≤ 2 (2^n - 1) = 2^{n+1}-2 < 2^{n+1}-1$.


TITLE: Galois groups and prescribed ramification
QUESTION [7 upvotes]: What is known about finite groups $G$ for which there exists a Galois extension $K$ of $\mathbb{Q}$ ramified only at $2$ such that $\text{Gal}(K/\mathbb{Q}) \cong G$ ? More generally, which groups can be realized over $\mathbb{Q}$ with no ramification outside a given (finite) set of primes?
I am thus interested in results of two kinds:
Realization of specific groups.
Examples of groups which are not realizable in this manner, and restrictions on groups which are.

REPLY [8 votes]: Concerning the question of Pablo that follows Joël's answer:
If $k$ is an algebraically closed field, then the situation is completely understood, thanks to work of Grothendieck (in characteristic 0) completed by the proof of Abhyankar's conjecture by Raynaud and Harbater.
Precisely, let $C/k$ be a smooth affine curve, complement of $r\geq1$ points in a projective smooth connected curve of genus $g$, over the alg. closed field $k$.
Let $F=k(C)$ be the field of rational functions on $C$.
Extensions of $C$ that are unramified on $C$ correspond to connected
étale covers of $C$.
If $\mathop{\rm char}(k)=0$, then Grothendieck showed (this is the main result of SGA 1) that a finite group $G$ is the Galois group of a connected étale cover of $C$ if and only if it is generated by $2g+r-1$ elements. More precisely, the Galois group of the maximal extension of $k(C)$ unramified on $C$ is a free profinite group on $2g+r-1$ elements. It is worth observing the fundamental group of a compact connected Riemann surface of genus $g$ deprived from $r\geq1$ points is  a free group on $2g+r-1$ generators. (The case $r=0$ is also treated by Grothendieck, $G$ then needs to be generated by $2g$ elements $a_1,b_1,\dots,a_g,b_g$ satisfying the relation $(a_1,b_1)(a_2,b_2)\dots(a_g,b_g)=1$.)
If $\mathop{\rm char}(k)=p>0$, then $C$ has « more » étale covers. For example,
if $C=\mathbf A^1_k$, then $f\colon C\to C$ given by $f(t)=t-t^p$ is a connected étale cover of degree $p$. Raynaud (when $C=\mathbf A^1_k$) and Harbater (in general) showed that a finite group $G$ is the Galois group of a connected étale cover of $C$ if and only if the quotient of $G$ by the (normal) subgroup generated by its $p$-Sylow subgroups is generated by $2g+r-1$ elements.
Specific examples had been exhibited by many mathematicians, such as Abhyankar itself (he has a long series of papers in this style) and Nori (when $G$ is a finite group of Lie type).