text:
\begin{document}

\date{\vspace{-5ex}}

\maketitle

\begin{abstract}
In 1991, Shinichi Kotani proved a theorem giving
a sufficient condition to conclude that a function $f(x)$
on $\Z^d$ decays like $|x|^{-(d-2)}$ for large $x$, assuming that its Fourier transform
$\hat f(k)$
is such that $|k|^{2}\hat f(k)$ is well behaved for $k$ near zero.
The proof was not published.
We prove an extension of Kotani's Theorem, based on Kotani's unpublished proof.
\end{abstract}

\noindent
Keywords: Fourier transform decay.

\medskip \noindent
MSC2010 Classification:  42B05.

\section{Introduction and main result}

Our purpose here is to revive and extend a neglected theorem of Shinichi Kotani that was stated
without its proof as \cite[Theorem~1.6.1]{MS93}.
The theorem  gives
a sufficient condition to conclude that a function $f(x)$
on $\Z^d$ decays like $|x|^{-(d-2)}$ for large $x$, assuming that its Fourier transform
$\hat f(k)$
is such that $|k|^{2}\hat f(k)$ is well behaved for $k$ near zero.
We state Kotani's Theorem and prove an extension.
Our proof is adapted from
Kotani's proof of \cite[Theorem~1.6.1]{MS93}
in his unpublished 1991 handwritten notes in Japanese
\cite{Kota91}, which were translated into English at that time by Takashi Hara.

Let $\mathbb{T}^d = (\R / 2 \pi \Z )^d$ denote the torus.
For a summable function $f:\Z^d \to \C$ we define its Fourier transform by
\begin{equation}
    \hat f(k) = \sum_{x\in\Z^d}f(x)e^{ik\cdot x}
    \qquad
    (k \in \T^d).
\end{equation}
The inverse Fourier transform is
\begin{equation}
    f(x) = \int_{\T^d}\hat{f}(k) e^{-i k \cdot x } \frac{dk}{(2\pi)^{d}}
    \qquad
    (x \in \Z^d).
\end{equation}
For $d>2$, we define the dimension-dependent constants
\begin{equation}
\lbeq{nddef}
    a_d=\frac{\Gamma(\frac{d-2}{2})}{4 \pi^{d/2}},
    \qquad
    n_d= \begin{cases}
    d-2 & (d>4)
    \\
    d-1 & (d=3,4).
    \end{cases}
\end{equation}
For an open set $U\subset \T^d$, $C^n(U)$ denotes the space of functions $\hat f:U \to \C$
with derivatives $\nabla^\alpha \hat f$ continuous for all $|\alpha| \le n$.
The following is Kotani's Theorem in its original form, as stated in \cite[Theorem~1.6.1]{MS93}.
Kotani's proof was not published.

\begin{theorem}
\label{thm:kotani-original}
Let $d >2$.
Given $\hat{f} \in C^{d-2} (\mathbb{T}^d \backslash \{0\})$, let $\hat{h}(k)=|k|^2\hat{f}(k)$.
Suppose that there is a neighbourhood  $0 \in U \subset \mathbb{T}^d$ such that
$\hat{h} \in  	C^{n_d}(U)$.
Then as $|x| \to \infty$,
\begin{equation}
\lbeq{gxasy-original}
	f(x) =
	a_d \hat{h}(0) \frac{1}{|x|^{d-2}} + o\left(\frac{1}{|x|^{d-2}} \right)  .
\end{equation}
\end{theorem}

In \cite[Example~1.6.2]{MS93}, it is shown that the hypothesis of existence of
at least $d-2$
derivatives for $\hat{h}$ cannot be relaxed.  Namely, the
example exhibits a function $\hat{f}$ on $\mathbb{T}^d$, for $d \ge 3$,
with $\hat{h}(k) =|k|^2 \hat{f}(k)$ having $d-3$ but not
$d-2$ derivatives in a neighbourhood of $k=0$, and with $f(x)$
not bounded above by a multiple of $|x|^{-(d-2)}$ for large $x$.

\begin{example}[Random walk Green function]
\label{ex:srw}
Let $d>2$.
Let $D$ be the one-step transition function of a random walk
on $\Z^d$, i.e., $D:\Z^d\to [0,1]$ with $\sum_{x\in\Z^d} D(x)=1$.
Suppose that $D$ obeys $\Z^d$-symmetry and has moments up to and including order
$n_d+2$.
Then $\hat D(k)$ has a Taylor expansion about $k=0$ to order $n_d+2$.
The Green function $C(x)$ is the inverse Fourier transform of $\hat C(k) = \frac{1}{1-\hat D(k)}$.
Let $\sigma^2 = \sum_x |x|^2 D(x)$.
Then $\hat h(k) = |k|^2 \hat C(k)$
obeys $\hat h(0)=2d\sigma^{-2}$.  Moreover, $1/\hat h(k) = |k|^{-2}[1-\hat D(k)]$
has an expansion about $k=0$ to order $n_d$, with constant term $\sigma^2/(2d)$,
so $\hat h\in C^{n_d}(\T^d)$.  It then follows from Theorem~\ref{thm:kotani-original}
 that
\begin{equation}
\lbeq{RWasy}
    C(x) = a_d\frac{2d}{\sigma^2}\frac{1}{|x|^{d-2}}
    + o\left(\frac{1}{|x|^{d-2}} \right).
\end{equation}
This provides an alternate approach to obtaining leading Green function asymptotics,
compared to, e.g., \cite{Uchi98,LL12}.  Fourier analysis is also used in \cite{Uchi98,LL12}, but
the analysis in the proof of Theorem~\ref{thm:kotani-original} is relatively mild.
\end{example}

\begin{rk}
The hypothesis of existence of $n_d+2$ moments in Example~\ref{ex:srw} is
more restrictive than
the hypothesis in \cite[Theorem~2]{Uchi98}  to conclude \refeq{RWasy}, which requires that
$D$ has at least $d-2$ moments for $d>4$,
at least two moments for $d=3$,  while for
$d=4$ the requirement is $\sum_x |x|^2 (\log |x|) D(x)<\infty$.
However, unlike \cite[Theorem~2]{Uchi98}, Theorem~\ref{thm:kotani-original} allows negative
values for $D(x)$.
The hypothesis in Theorem~\ref{thm:kotani-original} differs from the one in \cite[Theorem~1.4]{Hara08}
(see also \cite[Section~7.1]{Uchi98}),
which concerns functions of the form $\hat f(k) = [1-\hat J(k)]^{-1}$ with $J$ subject
to certain hypotheses in $x$-space including the relatively mild estimate
$|J(x)|\le O(|x|^{-(d+2+\epsilon)})$ with $\epsilon \ge 0$, again without
requiring $J(x) \ge 0$.  The proof of \cite[Theorem~1.4]{Hara08} is considerably more
involved than the proof of Theorem~\ref{thm:kotani}
presented here.
\end{rk}

The next theorem is an extension of Theorem~\ref{thm:kotani-original}
with an explicit error estimate and a weaker hypothesis on
$\hat h$.
For $p \in [1,\infty)$, we use the norms
\begin{equation}
    \|\hat f\|_{L^p}^p = \int_{\T^d}|\hat f(k)|^p\frac{dk}{(2\pi)^d}  .
\end{equation}
For $n\in\N$, the Sobolev space $W^{n,p}$ denotes the space of functions on $\T^d$ which have all
derivatives of order less than or equal to $n$ in $L^p(\T^d)$, with norm
\begin{equation}
    \|\hat f\|_{W^{n,p}} = \sum_{|\alpha| \le n} \|\nabla^\alpha \hat f\|_{L^p}.
\end{equation}

\begin{theorem}
\label{thm:kotani}
Let $d > 2$.  Choose $p_d \in (\frac{d}{d-2},2]$ for $d>4$, and set $p_d=2$ for $d=3,4$.
Let $U,V$ be open subsets of $\T^d$ with $0 \in V \subset \overline{V} \subset U$.
Let $\hat f$ be a continuous function on $\T^d \setminus \{0\}$ such that $\hat f \in W^{d-2,1}(\T^d\setminus V)$.
Let $\hat{h}(k)=|k|^2\hat{f}(k)$ be continuous on $\T^d$
and suppose that $\hat{h} \in W^{n_d,p_d}(U)$.
Then
\begin{equation}
\lbeq{gxasy}
	f(x) =
	\frac{a_d}{|x|^{d-2}} \left(  \hat{h}(0) + R_f(x)\right)
\end{equation}
with $\lim_{x \to \infty}R_f(x)=0$.
 Moreover, $R_f$ obeys the upper bound
\begin{equation}
\lbeq{Rbound}
    |R_f(x)| \le
    c
    \left(
    \|\hat f  \|_{W^{d-2,1}(\T^d \setminus V)}
    +
    \|  \hat h   \|_{W^{n_d,p_d}(U)}
    \right)
    ,
\end{equation}
with constant $c$ depending on $U,V,d$.
\end{theorem}

The bound \refeq{Rbound} does not decay in $x$ but it does provide a bound
$|f(x)| \le c_1 |x|^{-(d-2)}$ with explicit dependence of $c_1$ on $\hat f$.
Note that in \refeq{Rbound} the norm of $\hat f$
is on $\T^d \setminus V$, so the possibly singular behaviour of $\hat f$ at the origin
does not contribute.  Also, the hypothesis on $\hat h$ permits singular behaviour of
derivatives of $\hat h$.\footnote{An application of Theorem~\ref{thm:kotani} to the lace expansion
is given in Version~1 of this arXiv
submission; subsequently a more direct proof was discovered
that does not require Theorem~\ref{thm:kotani} for that purpose \cite{Slad20_lace}.}

\section{Proof of Theorem~\ref{thm:kotani}}

Our proof of Theorem~\ref{thm:kotani} is adapted from
Kotani's unpublished proof of Theorem~\ref{thm:kotani-original}.
The proof  uses the Fourier transform
\begin{equation}
    \hat f(k) = \int_{\R^d} f(y) e^{ik\cdot y} dy
    \qquad
    (k \in \R^d)
\end{equation}
of suitable functions $f:\R^d \to \C$, with inverse Fourier transform
\begin{equation}
\lbeq{RdinvFT}
    f(y) = \int_{\R^d} \hat f(k) e^{-ik\cdot y} \frac{dk}{(2\pi)^d}
    \qquad
    (y \in \R^d).
\end{equation}
Two items in the proof are deferred to two lemmas that follow the proof.

\begin{proof}[Proof of Theorem~\ref{thm:kotani}]
Let $\hat\chi \in C^\infty (\T^d)$
be a fixed bump function supported in $U$, with  $\hat\chi(k) =1$
for $k\in V$ and $\hat\chi (k) \in [0,1]$ for all $k$.
We set $\hat{g}(k)=\hat f(k)(1-\hat\chi(k))$, and make the decomposition
\begin{align}
    f(x) &
    = \int_{\T^d} \hat f(k)  \hat\chi(k) e^{-ik\cdot x} \frac{dk}{(2\pi)^d}
    + \int_{\T^d} \hat g(k)  e^{-ik\cdot x} \frac{dk}{(2\pi)^d}
    \nnb &
    =
    I_1(x)+I_2(x).
\end{align}
By hypothesis, $\nabla^{\alpha} \hat g \in L^1(\T^d)$ for $|\alpha| \le d-2$,
so by the
Riemann--Lebesgue Lemma (and integration by parts)
\begin{equation}
    I_2(x) = o(|x|^{-(d-2)}).
\end{equation}
There is also the quantitative estimate, for any component $i$,
\begin{equation}
    |I_2(x)| \le \frac{1}{x_i^{d-2}} \| \nabla^{d-2}_i \hat g\|_{L^1(\T^d)}.
\end{equation}
The $C^\infty$ function $\hat\chi$ depends only on the neighbourhoods $U,V$ and does not depend on $\hat f$.  There is therefore a constant depending only on $d,U,V$ (and the choice of $\hat\chi$) such that
\begin{equation}
    |I_2(x)| \le \frac{c}{|x|^{d-2}} \|  \hat f   \|_{W^{d-2,1}(\T^d \setminus V)}.
\end{equation}
Thus $I_2(x)$ obeys the desired error estimate and
the main term is $I_1(x)$.

We can extend the integral defining $I_1(x)$ to all of $\R^d$.
We define a function $\hat s$ on $\R^d$ by
\begin{equation}
    \hat s(k) =
    \begin{cases}
        \hat h(k) \hat\chi(k) & (k\in U)
        \\
        0 & (k \in \R^d\setminus U).
    \end{cases}
\end{equation}
By Lemma~\ref{lem:s},
the inverse Fourier transform $s$ of $\hat s$, defined by \refeq{RdinvFT}, satisfies
$s \in L^t(\R^d)$ for $t \in [1,\infty)$ and
$\|s\|_{L^1(\R^d)} \le c \|\hat h\|_{W^{n_d,p_d}(U)}$.
By definition, and by the fact\footnote{The
identity \refeq{I1} is well known when $\hat s$ is
a Schwartz function, e.g., \cite[Theorem~2.4.6]{Graf10}.
Our $\hat s$, although continuous and of compact support, need not be a Schwartz function.
The following argument shows that \refeq{I1}
holds nevertheless.  Fix a $C^\infty$ function
$\hat\psi:\R^d\to [0,\infty)$ with compact support,
such that $\hat\psi_\epsilon(k)=\epsilon^{-d}\hat\psi(k/\epsilon)$
is an approximate identity (as in \cite[Definition~1.2.15]{Graf10}).  In particular,
$\|\hat\psi_\epsilon\|_{L^1}=1$.
Let $\hat s_\epsilon(k)=(\hat s*\hat\psi_\epsilon)(k)$; then $\hat s_\epsilon$ is a Schwartz function so
\refeq{I1} holds with $s$ replaced by $s_\epsilon$.
Also, $\|\hat{s}_\epsilon\|_{L^\infty} \le \|\hat s\|_{L^\infty} < \infty$,
and $|s_\epsilon| = |s\psi_\epsilon| \le |s|$ since $\psi_\epsilon(y) \le \|\hat\psi_\epsilon\|_{L^1}=1$.
Then we obtain \refeq{I1} by letting
$\epsilon \to  0$, using $\hat s_\epsilon(k) \to \hat s(k)$,
$\psi_\epsilon (y) \to 1$, and dominated convergence.}
that $|k|^{-2}$ is the Fourier transform
of $a_d|y|^{-(d-2)}$,
\begin{equation}
\lbeq{I1}
    I_1(x) = \int_{\R^d} \frac{\hat s(k)}{|k|^2} e^{-ik\cdot x} \frac{dk}{(2\pi)^d}
    =
    a_d \int_{\R^d} \frac{1}{|x-y|^{d-2}}s(y)dy.
\end{equation}
Given $\epsilon \in (0,1)$, we further decompose
\begin{align}
    I_1(x)
    &=
    a_d \int_{|y|<\epsilon |x|} \frac{1}{|x-y|^{d-2}}s(y)dy
    +
    a_d \int_{|y|\ge \epsilon |x|} \frac{1}{|x-y|^{d-2}}s(y)dy
    \nnb & =
    J_1(x) + J_2(x).
\end{align}

Since $s\in L^1(\R^d)$, the definition of $\hat h$ gives
\begin{equation}
    \hat h(0) = \hat s(0) = \int_{\R^d} s(y) dy.
\end{equation}
Therefore,
\begin{align}
    |x|^{d-2}I_1(x) -  a_d \hat h(0)
    &=
    |x|^{d-2}J_1(x) + |x|^{d-2}J_2(x) -  a_d \int_{\R^d}s(y)dy
    \nnb & =
    |x|^{d-2}J_2(x) -
    a_d \int_{|y|<\epsilon |x|} \left(1- \frac{|x|^{d-2}}{|x-y|^{d-2}}  \right) s(y)dy
    - a_d \int_{|y|\ge \epsilon |x|}s(y)dy.
\end{align}
By Lemma~\ref{lem:J2},  $|x|^{d-2}J_2(x)=o(1)$ and also
$J_2(x) = O(\|\hat h\|_{W^{n_d,p_d}(U)})$
as in \refeq{Rbound}.
By Lemma~\ref{lem:s}, the final term is also $o(1)$ and at most
$O(\|\hat h\|_{W^{n_d,p_d}(U)})$.
For the middle term, we write $\hat x = x/|x|$ and
use the fact that when $|y|<\epsilon |x|$ (now we take $\epsilon$ small) we have
\begin{equation}
    \left| 1 - \frac{|x|^{d-2}}{|x-y|^{d-2}} \right|
    =
    \left| 1 - \frac{1}{|\hat x- y/|x||^{d-2}} \right|
    \in
    \left( 1-\frac{1}{(1-\epsilon)^{d-2}},1 - \frac{1}{(1+\epsilon)^{d-2}} \right).
\end{equation}
Therefore, since $s \in L^1(\R^d)$, the second term is $o(1)$ by dominated convergence,
and it is $O(\|\hat h\|_{W^{n_d,p_d}(U)})$ since $\|s\|_{L^1(\R^d)}$
is by Lemma~\ref{lem:s}.

This completes the proof, subject to the following two lemmas.
\end{proof}

We write $n$ in place of $n_d$ (recall \refeq{nddef}),
$p$ in place of $p_d$ (specified in Theorem~\ref{thm:kotani}), and
let $q$ be the dual index defined by
$\frac 1q + \frac 1p =1$.
Both lemmas use the function
\begin{equation}
    u(y)=|y|^{n }s(y) \qquad (y \in \R^d).
\end{equation}

\begin{lemma}
\label{lem:s}
Under the assumptions of Theorem~\ref{thm:kotani},
$\|u\|_{L^{q}(\R^d)} \le c\|\hat h\|_{W^{n,p}(U)}$,
$\lim_{y\to\infty}u(y)=0$,
$|u(y)| \le c \|\hat h\|_{W^{n,p}(U)}$,
$\|s\|_{L^1(\R^d)} \le c \|\hat h\|_{W^{n,p}(U)}$,
and $s\in L^t(\R^d)$ for all $t \in [1,\infty)$.
\end{lemma}

\begin{proof}
By hypothesis, $\nabla_i^{n}\hat s \in L^p(\R^d)$.  Its inverse Fourier transform
is essentially $y_i^{n}s(y)$.
By H\"older's inequality, $|y|^n \le d^{n-2} \sum_{i=1}^d |y_i|^n$.
Since $p \in (1,2]$ by hypothesis, we can apply
the Hausdorff--Young inequality to see that
(the constant $c$ can change from one occurrence to the next)
\begin{equation}
\lbeq{ubd1}
    \|u\|_{L^q(\R^d)} \le c\max_i \|y_i^{n}s(y)\|_{L^q(\R^d)}
    \le c\max_i \|\nabla_i^{n}\hat s\|_{L^p(U)}
    \le c  \| \hat h\|_{W^{n,p}(U)}
    .
\end{equation}
Also,
\begin{equation}
    |u(y)| \le c\max_i |y_i^{n} s(y)|
    = c\max_i \Big|\int_{\R^d} \nabla_i^{n}\hat s(k) e^{-ik\cdot y} \frac{dk}{(2\pi)^d}\Big|.
\end{equation}
Because $\nabla_i^{n}\hat s$ has compact support and is in $L^p$, it is also in $L^1$,
so the right-hand side goes to zero as $|y|\to\infty$, by the Riemann--Lebesgue Lemma.
It is also bounded above by a multiple of the $L^p$ norm controlled in \refeq{ubd1}.
This proves the three statements for $u$.

Since $\hat s$ is continuous and has compact support,
 $\hat s \in L^t(\R^d)$ for all $t \in [1,\infty)$.  Therefore,
by the Hausdorff--Young inequality, $s \in L^r(\R^d)$ for all $r \ge 2$.
It follows that
\begin{equation}
    \int_{|y|\le 1} |s(y)|dy \le c \left( \int_{|y|\le 1} |s(y)|^q dy \right)^{1/q}
    \le c \|s\|_{L^q(\R^d)} \le c \|\hat s\|_{L^p(\R^d)}
    \le
    c\|\hat h\|_{W^{d-2,p}(U)}.
\end{equation}
Also, by H\"older's inequality,
\begin{align}
    \int_{|y|> 1}|s(y)|dy & \le
    \left( \int_{|y|> 1}\frac{1}{|y|^{pn}} dy \right)^{1/p}
    \left( \int_{|y|> 1}|u(y)|^q dy \right)^{1/q}  .
\end{align}
For $d>4$ we have $pn =p(d-2) > d$ by hypothesis, and for $d=3,4$ we have $pn = 2(d-1)>d$.
Therefore the first integral is bounded.  The second is bounded by
$\|u\|_{L^q(\R^d)}$, whose upper bound we have already discussed.  This proves that
$s \in L^1(\R^d)$, with the desired estimate.  Since also $s \in L^t(\R^d)$ for all $t \ge 2$,
in particular for $t=2$, it follows that $s \in L^t(\R^d)$ for all $t \ge 1$.
\end{proof}

\begin{lemma}
\label{lem:J2}
Under the assumptions of Theorem~\ref{thm:kotani},
$|x|^{d-2}J_2(x)$ is both $o(1)$ and $O(\|\hat h\|_{W^{n,p}(U)})$.
\end{lemma}

\begin{proof}
With
$A_{21}= \{|y|\ge \epsilon |x|\} \cap \{|x-y|\ge 1\}$ and
$A_{22}= \{|y|\ge \epsilon |x|\} \cap \{|x-y|< 1\}$,
we make the decomposition
\begin{align}
    J_2(x) & =
    a_d \int_{A_{21}} \frac{1}{|x-y|^{d-2}}s(y)dy
    +
    a_d \int_{A_{22}} \frac{1}{|x-y|^{d-2}}s(y)dy
    \nnb & =
    J_{21}(x)+J_{22}(x).
\end{align}

By H\"older's inequality,
\begin{align}
\lbeq{J21}
    |J_{21}(x)| & \le
    a_d \left( \int_{A_{21}}\frac{1}{|x-y|^{p(d-2)}}\frac{1}{|y|^{pn}} dy \right)^{1/p}
    \left( \int_{A_{21}}|u(y)|^q dy \right)^{1/q} .
\end{align}
For $d>4$, since
$pn=p(d-2)>d$ by hypothesis,
\begin{equation}
    \left( \int_{A_{21}}\frac{1}{|x-y|^{p(d-2)}}\frac{1}{|y|^{p(d-2)}} dy \right)^{1/p}
    \le \frac{1}{(\epsilon |x|)^{d-2}}
    \left( \int_{|x-y| \ge 1}\frac{1}{|x-y|^{p(d-2)}}  dy \right)^{1/p}
    \le \frac{c}{ (\epsilon |x|)^{d-2}}.
\end{equation}
For $d=3,4$, since $pn=2(d-1)$, we use instead
\begin{align}
    \left(  \int_{A_{21}}\frac{1}{|x-y|^{2(d-2)}|y|^{2(d-1)}} dy \right)^{1/2}
    & \le
    \left( \int_{|x-y|\ge 1} \frac{1}{|x-y|^{4(d-2)}} dy\right)^{1/4}
    \left( \int_{|y|\ge \epsilon |x|} \frac{1}{|y|^{4(d-1)}}dy \right)^{1/4},
\end{align}
and observe that
the first factor is finite since $4(d-2)>d$ for $d=3,4$,
and that since $4(d-1)>d$ the
second is $O(|x|^{-\frac 14 (3d-4)}) = O(|x|^{-(d-2)})$.
For any $d>2$, the integral in the second factor of \refeq{J21} obeys
\begin{equation}
\lbeq{A21u}
    \int_{A_{21}}|u(y)|^q dy
    \le
    \int_{|y|\ge \epsilon |x|}|u(y)|^q dy .
\end{equation}
Since $u \in L^q(\R^d)$ by Lemma~\ref{lem:s},
the right-hand side of \refeq{A21u} is $o(1)$ as $x\to\infty$, and
also its $q^{\rm th}$ root is bounded by $\|u\|_{L^q(\R^d)}$ which
at most $O(\|\hat h\|_{W^{n,p}(U)})$ by Lemma~\ref{lem:s}.

Finally,
\begin{equation}
    J_{22}(x)
    \le
    a_d \frac{1}{(\epsilon |x|)^{n}}
    \left(\sup_{|y| \ge \epsilon |x|}|u(y)| \right)
    \int_{|x-y| < 1} \frac{1}{|x-y|^{d-2}}dy
    \le
    c_\epsilon \frac{1}{ |x|^{n}}
    \left(\sup_{|y| \ge \epsilon |x|}|u(y)| \right),
\end{equation}
since the $y$-integral is finite.
For $d>4$ the power on the right-hand side is $|x|^{-(d-2)}$, and for $d=3,4$ it
is $|x|^{-(d-1)} = o(|x|^{-(d-2)})$.
By Lemma~\ref{lem:s}, the supremum factor  is $o(1)$
since $u(y) \to 0$ as $y \to \infty$, and also is at most $O(\|\hat h\|_{W^{n,p}(U)})$.
This completes the proof.
\end{proof}

\section*{Acknowledgements}

This work was supported in part by NSERC of Canada.
I am grateful to Shinichi Kotani for providing his unpublished notes in 1991
with the proof of Theorem~\ref{thm:kotani-original}
that forms the basis for the proof of Theorem~\ref{thm:kotani}
and for recent correspondence
concerning this work,
and to Takashi Hara for his translation of Professor Kotani's notes in  1991.


TITLE: What is the terminology for this type of order-preserving function?
QUESTION [0 upvotes]: As I roughly know, a function $f$ in $\mathbb{R}$ is called order-preserving if $f(x)>f(y)$ for $x>y$, $x,y \in \mathbb{R}$.
May I ask, if anybody knows the terminology for two functions $f$ and $g$ (or what $g$ is to $f$) if $g(x)>g(y)$ for $f(x)>f(y)$ ?
Also, the terminology when $g(x)>g(y)$ for $f(x)<f(y)$?
Many thanks :)

REPLY [1 votes]: I don't know of any standard terminology, but one option would be to define that whenever $P$ is a poset and $f : X \rightarrow P$ is a function, we have $$A(f) = \{(x,y) \in X^2 \mid f(x) < f(y)\}.$$
Then the relation of interest between $f$ and $g$ can be denoted $A(f) \subseteq A(g),$ which can be abbreviated $f \subseteq g : A$ (read: $f$ is a subset of $g$ when viewed under $A$).
By the way, a function that preserves $\leq$ is typically called order-preserving. A function that preserves $<$ might be called "strict-order preserving."


TITLE: Do the following functions exist?
QUESTION [0 upvotes]: While practicing for my topology exam, I stumbled upon the following question from a previous exam:
Give a proof of your answer:
(a) Is there a continuous surjective map from $\mathbb{C}$ to $\mathbb{C} - \{0 \} $?
(b) Is there a continuous surjective map from $\mathbb{R}$ to $\mathbb{R} - \{0 \} $?
(c) Is there a continuous surjective map from $\mathbb{R} - \{ 0 \}$ to $\mathbb{R} $?
(d) Is there a continuous surjective map from $\mathbb{S^1}$ to $\mathbb{S^1} - \{1 \} $?
I think the answer to (a) and (b) is "Yes, there is. Those are the identitiy functions $id_{\mathbb{C} }$ and $id_{\mathbb{R} }$". Is this true? And what about (c) and (d)? And how do you prove it?
Edit: I now know that my proposed functions are no good (they aren't even functions!). So I guess the questions remain: do the continuous surjective maps exist?

REPLY [6 votes]: For (a) and (c), we can do it, with (for example) the maps $z\mapsto e^z$ and $x\mapsto x-\frac1x.$ For (b), it is impossible, since the continuous image of any real-valued function on $\Bbb R$ is a singleton, interval, ray, or all of $\Bbb R$--more generally, continuous images of (non-empty) connected sets are again (non-empty) connected. For (d) it is impossible, since a continuous image of a compact set is compact, but $\Bbb S^1-\{1\}$ is not compact.


TITLE: Eisenstein Criterion with Polynomials of Rational Coefficients?
QUESTION [2 upvotes]: I'm wondering if it's possible to apply the Eisenstein criterion to polynomials that have rational, non-integer coefficients? I am thinking about using a shift to go to integers and apply the Eisenstein criterion there. Then, since f(x) is irreducible iff f(x+c) is irreducible, that should show that the original polynomial is irreducible. For instance, with the polynomial $x^5-(3/5)x^4+9x-3$, I wonder if there's some way to shift this so it gets integer coefficients and we can do the procedure I said?

REPLY [2 votes]: Just multiply  the polynomial by the common denominator of the coefficients. Eisenstein criterion works not only for monic polynomials, but also polynomials with leading coefficient not divisible by $p$.


TITLE: Relative $0$-homology of $(\Delta^n, \partial \Delta^n)$
QUESTION [1 upvotes]: Let $\Delta^n$ be the standard $n$-simplex, with $n>0$. Denote with $H_0$ the (simplicial) $0$-homology.
In my book it is written that $H_0(\Delta^n, \partial \Delta^n)=\mathbb{Z}$. But $\Delta^n$ and $\partial \Delta^n$ have the same number of vertices, so $H_0(\Delta^n, \partial \Delta^n)=0$. Is it just a typo of the book? Am I wrong?

REPLY [0 votes]: You're right, and this is presumably a typo.  I would guess that the error arose from the fact that for non-relative homology, $H_0$ being "trivial" means it is $\mathbb{Z}$, rather than $0$.


TITLE: Parallel and Anti-parallel Protons in NMR
QUESTION [2 upvotes]: I'm an Alevel Student trying to understand the concept of NMRI, I understand that when an external magnetic field is applied, the protons either line up with (parallel) or line up against (antiparallel) the external magnetic field. Sources state that protons parallel to the field have lower energy than antiparallel protons and that there are more parallel protons than antiparallel protons.
I have difficulties understanding the concept of:

Difference in energy between anti-parallel and parallel protons. There's an answer in quora https://www.quora.com/Why-does-a-spin-up-electron-have-slightly-more-energy-than-a-spin-down-electron explaining in terms of protons that

If the spin is already aligned with the magnetic field (up), it means that no energy is required to align it with B, which means that the spin up particle has the least potential energy.
If the spin is aligned opposite to the magnetic field (down), it means that the energy required to align it is maximum, which means that spin down particles has the most potential energy.

If the parallel protons have less potential energy, doesn't that mean that it has more kinetic energy than anti-parallel protons? Why are we only so concerned about potential energy?

The fact that there are more parallel than antiparallel protons due to lower energy of parallel protons. Why is this so? In this link https://www.reddit.com/r/askscience/comments/2axmfz/why_do_more_protons_align_themselves_parallel/, the only reason that I could somewhat grasp was based on part of one of the answers:

Always think about Boltzmann's population distribution when you are considering problems like this.

Using my Alevel knowledge (from Boltzmann distribution curve), I understand that in a system, majority of the particles will have lower/ intermediate energy compared to higher energy. Doesn't Boltzmann Distribution curve only apply to kinetic energy and not potential energy?
There are other reasons based on Zeeman Effect which I can't seem to understand due to my limit in knowledge at this moment, but will try to read on it. At this moment, I'm looking for simpler answers that an Alevel student could understand. Thank You.

REPLY [2 votes]: If the parallel protons have less potential energy, doesn't that mean that it has more kinetic energy than anti-parallel protons? Why are we only so concerned about potential energy?

This implies conservation of energy (less potential energy => more kinetic), which is true for closed systems, but here the system is not closed! We bomb this protons with external magnetic field, which pumps a lot of energy into the system. In fact, this external energy is much larger than the energy that characterize the protons movement (which is confined by their interaction with the environment). Therefore, when this magnetic field is applied, we can just omit altogether all the kinetic energy terms - they are tiny in comparison. So we are just interested in how the protons interact and respond to the external field, and here they have two options - to be in parallel to it (lower energy) or anti-parallel to it (higher energy).

Using my Alevel knowledge (from Boltzmann distribution curve), I understand that in a system, majority of the particles will have lower/ intermediate energy compared to higher energy. Doesn't Boltzmann Distribution curve only apply to kinetic energy and not potential energy?

No, the Boltzmann distribution is true for every system with some temperature. It states that the probability of the system to be in a state with some energy $\epsilon$ drops as $\epsilon$ is larger, and the "measure" against we examine this energy is the temperature. So $\epsilon \ll k_B T$ is favorable to $\epsilon \gg k_b T$. So, the lower the temperature, more and more of the protons will tend to go for the lower-energy state, which is the parallel spin configuration. The ratio between them will be determined by the temperature and the energy, as can be read from the Boltzmann distribution.


TITLE: Let R be a PID, B a torsion R module and p a prime in R. Prove that if $pb=0$ for some non zero b in B, then $\text{Ann}(B)$ is a subset of (p)
QUESTION [0 upvotes]: I have tried to show it but to no avail. Its an exercise in chapter 12.1 of Dummit and Foote which I am learning by myself. Thanks in advance for any help.

REPLY [4 votes]: The assumption says that the ideal $\operatorname{Ann}(b):=\{r\in R\mid rb=0\}$ contains $(p)$. Since in a PID, every prime generates a maximal ideal and $\operatorname{Ann}(b)$ is clearly not the whole ring ($1\not\in \operatorname{Ann}(b)$) we must thus have equality $(p)=\operatorname{Ann}(b)$. Clearly $\operatorname{Ann}(B)$ is contained in $\operatorname{Ann}(b)$.


TITLE: How to calculate minimum constant speed to catch up with constantly accelerated moving point
QUESTION [1 upvotes]: My homework wants me to calculate the minimal velocity needed to catch up with a train moving past me.
The train is accelerating with $a_z = const.$. After it drives past me with $v_{0}$, I take 2 seconds to react, after which I sprint after it with $v_p$.
What is the minimum $v_p$ with which I can manage to catch the train?
My attempt so far is this:
I set $x_z = x_p$.
$x_z = \int_{0}^{t} \int_{0}^{t} a_z dt dt$
$    = \int_{0}^{t} a_z t + v_0 dt $
$    = \frac{1}{2} a_z t^2 + v_0 t + x_0 $
$    = \frac{1}{2} a_z t^2 + v_0 t $
and
$x_p = v_p t$
and so I solve the following for t
$x_z = x_p$.
$\frac{1}{2}a_z t^2 + v_0 t = v_p t$
$\frac{1}{2}a_z t + v_0 = v_p$
$\frac{1}{2}a_z t = v_p - v_0$
$t = \frac{2(v_p - v_0)}{a_z}$
And so I am looking for $\min_{v_p} t = \frac{2(v_p - v_0)}{a_z} $
As I have two variables I dont know what to do or minimize. Clearly there is a space of solutions, but I dont know how to find the minimum.
Is my approach correct until here? Can you give me any tips?

REPLY [0 votes]: You’ve correctly determined the equation of motion $s=\frac12a_zt^2+v_0t$ for the train, but as noted in a comment you’ve neglected to take into account the two second delay before you instantly accelerate to the interception speed $v_p$, so the correct equation of motion for your running is $s=v_p(t-2)$.
You want this line to intersect the parabola at $t\ge2$, so equate the two expressions for $s$ to get the equation $$\frac12a_zt^2+v_0t=v_p(t-2).\tag{*}$$ The line’s slope corresponds to your running speed, so for minimal speed this line should be tangent to the parabola. Therefore, choose a positive $v_p$ so that (*) has exactly one real root (hint: examine the discriminant).


TITLE: How often are irrational numbers well-approximated by rationals?
QUESTION [27 upvotes]: Suppose $x\in \mathbb{R}$ is irrational, with irrationality measure $\mu=\mu(x)$; this means that the inequality $|x-\frac{p}{q}|< q^{-\lambda}$ has infinitely many solutions in integers $p,q$ if and only if $\lambda < \mu$. A beautiful theorem of Roth asserts that algebraic numbers have irrationality measure $2$. For $\lambda<\mu$, let $\mathcal{Q}(x,\lambda) \subset \mathbb{N}$ be the (infinite) set of all $q$ occuring in solutions to the aforementioned inequality.
Question: For which pairs $(x,\lambda)$ does $\mathcal{Q}(x,\lambda)$ have positive relative density in the positive integers? For which pairs $(x,\lambda)$ does the cardinality of $\mathcal{Q}(x,\lambda) \cap [1,N]$ grow like a positive power of $N$?

REPLY [8 votes]: An answer to your second question may be found in an old paper of Erdos: Some results on diophantine approximation. Acta Arith. 5 1959 359–369 (1959). (MathSciNet)
He proves that for $1 < \lambda < 2$, for almost every $x$, the set $\mathcal{Q}(x,\lambda)$ grows like $n^\frac{1}{2-\lambda}$.  More precisely, Erdos proves that a one-sided version of your set $$\mathcal{Q}'(x,\lambda) = \big\{ q \geq 1 \ \big| \ \text{there exists } p, \ (q,p) = 1, \text{ such that } 0 < x-\frac pq < q^{-\lambda} \big\}$$ satisfies $$\lim_{n \to \infty} \frac{\big |\mathcal{Q}'(x,\lambda) \cap [1,n] \big| }{\sum_{q=1}^n q^{1-\lambda}} = \frac{12}{\pi^2}.$$ That the set $\mathcal{Q}(x,\lambda)$ grows like $n^\frac{1}{2-\lambda}$ is consistent with the fact that $\mathcal{Q}(x,1)$ is all of $\mathbb{N}$ while the numbers in $\mathcal{Q}(x,2)$ coming from continued fraction convergent denominators grow exponentially.


TITLE: Complexity of $r$-sum as a function of integer sizes
QUESTION [0 upvotes]: Given $n$ distinct integers whose absolute value is of size $c_{n,k}=\lceil n^{1/k}\rceil$ bits ($c_{n,k}$th bit position is always $1$ for absolute value) we know that using dynamic programming we can find if there is a subset that sums to $0$ in $2^{O(n^{1/k})}$ time.
Given $r\in\Bbb N$ what is best known algorithm for complexity of $r$-sum on such situations for case (1) fixed $k>1$ and for case (2) $k=\theta((\log n)^\alpha)$ for some fixed $\alpha\in(0,1)$?
What happens at limit $\alpha\rightarrow1-$?

REPLY [2 votes]: There is a straightforward dynamic programming algorithm with complexity $n \times 2^{O(r \cdot n^{1/k})}$.  This is pretty slow.
For $r=2$, you can use algorithms for the birthday paradox to find a solution in time $O(n)$, if a solution exists.  A solution will exist with high probability.  In fact, for $r=2$, with high probability a solution will exist within the first $n^{1/2k}$ numbers, and we can find a solution in expected running time $O(n^{1/(2k)})$.
There is an algorithm with complexity $O(r \cdot n^{1/(k(1+\lg r))})$.  You start by finding pairs (2-sums) whose sum has many of their high bits equal to zero (by combining pairs of elements, using algorithms for the birthday paradox); then look for 4-sums whose sum has a (slightly smaller) number of high bits equal to zero (by combining pairs of 2-sums); then look for 8-sums (by combining pairs of 4-sums); and so on.  The running time comes from optimizing the exact number of high bits that you ask to be zero at each stage.
For instance, for the case $r=4$, there is an algorithm with complexity $O(n^{1/(3k)})$ for finding at least one such solution.
As far as I know, it is an open problem whether one can do better than this.
See also the following research papers:

A. Blum, A. Kalai, H. Wasserman. "Noise-tolerant learning, the parity problem, and the statistical query model".  JACM, 50(4):506-519, 2003.
P. Camion, J. Patarin.  "The Knapsack hash function proposed at Crypto'89 can be broken".  EUROCRYPT '91, pp.39-53.
R. Schroeppel, A. Shamir.  "A T = O(2^n/2), S = O(2^n/4) algorithm for certain NP-Complete problems".  SIAM J. Computing,  10(3):456--464.
P. Chose, A. Joux, M. Mitton.  "Fast Correlation Attacks: an algorithmic point of view".  EUROCRYPT 2002.


TITLE: Monotonicity of a function in two variables
QUESTION [1 upvotes]: Consider the function $f(x,y) = \dfrac{ax+by}{x+y}$ , where $x,y>0$.
I'd like to find out over what region $f(x,y)$ is monotonically decreasing
(i.e. for $x'>x$ and $y'>y$, then $f(x',y')<f(x,y)~)$. Computing the individual partial derivatives, we get -
$\begin{eqnarray}
\dfrac{\partial f}{\partial x} &=& \dfrac{(a-b)y}{(x+y)^{2}} \\
\dfrac{\partial f}{\partial y} &=& \dfrac{(b-a)x}{(x+y)^{2}}
\end{eqnarray}$
Hence, if $a>b$, $f$ is monotonically increasing over $x$ and decreasing over $y$ (and vice-versa). I'm curious whether the double partial derivative has any significance?
$\begin{eqnarray}
\dfrac{\partial^{2} f}{\partial x \partial y} &=& \dfrac{(a-b)(x-y)}{(x+y)^{3}}
\end{eqnarray}$
Would it be right to say that if $\dfrac{\partial^{2} f}{\partial x \partial y} < 0$ for some region (such as when $x>y$ and $a<b$), then $f$ is monotonically decreasing over this region?
It would be helpful if someone could point me to a reference paper/book for determining the monotonicity of such functions.

REPLY [0 votes]: While @Ted has provided a generalized answer, the following procedure yields a specific condition that can lead to monotonically decreasing $f(x,y)$ as a function of the increments i.e. $\Delta x = (x'-x)$ and $\Delta y = (y'-y)$.
$
\begin{eqnarray}
\Delta f &=& f(x',y') - f(x,y) \\[5pt] &=& \dfrac{a(x+\Delta x)+b(y+\Delta y)}{x+\Delta x + y+\Delta y} -\dfrac{ax+by}{x+ y}  \\[5pt] &=&  \dfrac{a\cdot[(x+\Delta x)(x+y)-x(x+\Delta x + y+\Delta y)] + b\cdot[(y+\Delta y)(x+y)-y(x+\Delta x + y+\Delta y)]}{(x+\Delta x + y+\Delta y)(x+y)} \\[5pt] &=& \dfrac{a \cdot (\Delta x~y - \Delta y~x) + b\cdot ( \Delta y~x - y~\Delta x) }{(x+\Delta x + y+\Delta y)(x+y)} \\[5pt] &=& \dfrac{ (a-b) \cdot (\Delta x~y - \Delta y~x)}{(x+\Delta x + y+\Delta y)(x+y)}
\end{eqnarray}
$
Given that $x,y,\Delta x,\Delta y$ >0, $~~~$then $\Delta f < 0 $ when
$
\begin{eqnarray}
\text{1.} ~~a<b ~~\text{and}~~ \dfrac{\Delta x}{x} > \dfrac{\Delta y}{y} \\[5pt]
\text{2.} ~~a>b ~~\text{and}~~ \dfrac{\Delta x}{x} < \dfrac{\Delta y}{y}
\end{eqnarray}
$


TITLE: Consider $x^4 \pmod {pq}$, with $p = q = 3 \pmod4$.
QUESTION [1 upvotes]: Consider $x^4 \pmod {pq}$, with $p = q = 3 \pmod 4$.
Would someone explain to me why exactly one of the four square roots of $x^4 \pmod {pq}$ is also a square?
This result was given without proof and I do not understand.
What I have tried:
Edit: This boils down to proving that some $\exists$ $a: a^2 \cong x^4 \pmod {pq}$, $a$ is a quadratic residue $\pmod {pq}$
Since $p = q = 3 \pmod 4$, I know that the value of the Legendre symbol ($\frac{-1}{pq})$ is $-1$.
So, if the four candidates are $x,-x,y,-y$, one of $x,-x$ and one of $y,-y$ is not a quadratic residue, due to the multiplicity of the Legendre Symbol.
Without loss of generality, let's assume $x$ and $y$ are left as candidates. Without loss of generality, how can we prove $y$ is not a square/quadratic residue $\pmod {pq}$?

REPLY [0 votes]: We have equations $y^4 \equiv x^4 \pmod p $   if p|x then 0 is the soln. which is $0^2$
otherwise $(yx^{-1})^4 \equiv 1 \pmod p$.
As gcd(4,p-1)= 2 as($p  \equiv 3 \pmod 4$),by fermat's little theorem it reduces to solve $(yx^{-1})^2 \equiv 1 \pmod  p$ i.e. $y^2 \equiv x^2 \pmod p $ i.e. $y = (+1 or -1)x \pmod p$
Now it is easy to observe that either x or -x is a square mod p is a square. Because $p \equiv 3 \pmod 4 $
Let g be generator of $(\Bbb Z / p \Bbb Z)^*$ . Then $g^{(p-1)/2} = -1$ and (p-1)/2 is odd. Take i s.t. $g^i = x$, if i is even we are done. If i is odd $-x = (-1)*x = g^{i + (p-1)/2}$ is a square. Thus it is a square mod p.
Similarly either x or -x is a square mod q . And by chinese remainder thm we are done!


TITLE: Force the moon exerts on the earth
QUESTION [3 upvotes]: Determine the force that the moon exerts on the earth. Note that since the
diameter of the earth, 12,742 km, is not insignificant compared to the distance to
the moon, 384,400 km, the gravitational field of the moon is not uniform over the
volume of the earth. Therefore, to get an accurate result you will need to integrate
the gravitational field ($\vec{g} = -GM_{moon}\frac{\hat{R}}{R^{2}}$) times the density of the earth over the
volume of the earth (you may assume the earth is spherical and of uniform
density). It may be expedient to provide arguments why certain components of the
force can be ignored during the integration.
Compare you result to the approximate solution obtained by assuming that both
the moon and earth are point masses.
--
My thoughts: I'm assuming the only component of the force that matters is the force directed radially towards the moon (if that makes any sense).

REPLY [1 votes]: Indeed, by symmetry the only component is directed along the line that joins the centers of moon and earth. The force can be most easily found using the virtual work method: Compute the potential energy of the earth in the gravitational field of the moon. Let $d$ be the distance between their centers, and ${\bf r}$ a vector from the center of the earth to a point inside the eart, let $\theta$ be the angle bewteen the line joining the centers and ${\bf r}$, and $\rho$ the density if the earth. Then the potential energy is
$$
U = - \int_{Earth} \frac{GM_{moon} \rho}{\sqrt{d^2 +r^2 -2rd \cos \theta}} r^2 dr d\Omega
$$
Now expand in Legendre polynomials to obtain
$$
U = - \int_{0}^{R}\int_{S} GM_{moon}{ \rho} (1/d) \sum_{l=0}^{\infty}\left(\frac{r}{d}\right)^l P_l (\cos\theta) r^2  d\Omega dr
$$
$d\Omega$ is the differential solid angle, $S$ is the unit sphere and $R$ the radius of the earth.
In this integral only the term $l=0$ contribtes due to the orthogonality of the Legendre polynomials, the integration over $S$ gives $4\pi$. Overall you obtain
$U= -GM_{moon}\frac{4\pi R^3}{3}\frac{\rho}{d}  = -GM_{moon}M_{earth}/d$
The force is now given by a variation of $d$, i.e. $F= -\frac{\partial U}{\partial d}$ which gives
$
F = -\frac{GM_{moon}M_{earth}}{d^2}
$
This is the same result as if they were point masses, therefore the answer doesn't change, one can see explicitely from the calculation above that it will not change even when $\rho$ depends on $r$ (to see it doesnt change for any $\rho({\bf r})$) we would need to change our origin of ${\bf r}$ to the center of mass of the earth.


TITLE: Systematic studies of sum of quadratic polynomials squared
QUESTION [10 upvotes]: I'm wondering if there exists systematic studies of sums of quadratic forms squared, similar to the quadratic forms, which is practically reflected in eigenvalue decomposition (that has huge practical implication). Couple of examples related to the importance of the question.

Principal component analyses (PCA).
Given a set of points $x_i \in \mathbb{R^n}, i=1..k$ find the set of axes $u_1$, ... $u_m$, written as matrix $U \in \mathbb{R^n x R^m}$, and projections $\xi_1$, ..., $\xi_k, \xi_{\circ} \in \mathbb{R^m}$ that minimizes unexplained variance, i.e. solve the following quartic optimization problem
$$ \arg\min_{u_1,\dots, u_n,\\ \xi_1,\dots, \xi_k} \sum \limits_{i} \left( U^T \xi_i - x_i \right)^2$$
By the magic of symmetry it has the solution by singular value decomposition

Generalized PCA. Same as PCA, but now there is a precision matrix $A_i \in \mathbb{R^n x R^n} $ associated with each observable $x_i$. The problem becomes more complicated
$$ \arg\min_{u_1,\dots, u_n,\\ \xi_1,\dots, \xi_k} \sum \limits_{i} \left( A_i U^T \xi_i - x_i \right)^2$$
(when all $A_i$ are identity matrix this problem is equivalent to PCA, when $A_i = A_j, \forall i,j$, and diagonal it is weighted PCA ). This problem can also be solved in polynomial time via semi-definite programming (SDP) -- Since the solution has the form of sums of squares, by NZ Shor (1987) it is convex problem , and Parillo thesis (2000) gives a practical way to compute it via SDP $ \square $

In SDP approach the quartic polynomial is written as a sum of quadratic polynomials squared. Therefore, it is of major importance to know what kind of quartic polynomials can be written as a sum of quadratic forms squared (by the analogy to biquadratic function they can be called biquadratic forms). Most of the literature, I've encountered stop at the point where they find that the minimum of quartic polynomial $p= \sum_k^n (x_k^2-1) + (a^T x)^2, a \in \mathbb{Z^n}$ is encoding partition problem, and there are no arguments why $p$ cannot be represented as a sum of squares of quadratic polynomials, beyond that.
I'm wondering if anyone made systematic studies of polynomials representable by the sum of squares of quadratic polynomials.

REPLY [3 votes]: To the best of my knowledge, there is no such study; furthermore, without some nontrivial advances in the technology of sum-of-squares (SOS) problems, it is not currently clear what the immediate benefit of such a study would be.  (I'll focus on the SOS connection since that, as far as I know, is the best way to solve these general quartic problems.) This statement should be taken in a positive light: I believe there is a lot of research depth surrounding these problems.  I'll substantiate my claim in a few ways, hopefully in ways people find useful..
Firstly, for the most basic problems of the type you discuss, the SVD connection gives a much better solver than the SOS black box; in particular, the latter constructs an SDP with $\binom{n+2}{2}$ terms, where $n$ is the total number of variables in the source optimization problem (for instance, the total number of elements in all unknown matrices; to see where I got these numbers, see lecture 10 from Pablo Parrilo's 2006 course:  http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-972-algebraic-techniques-and-semidefinite-optimization-spring-2006/lecture-notes/lecture_10.pdf ).  This is an SDP you would never want to solve (running time depends on $n$ as $n^{6}$ using an interior point solver?), especially when compared with the ridiculous speed of an SVD solver (using consistent notation, SVD will be something like $\mathcal O(n^{1.5})$; you can de-fudge my computations by tracking the number of columns, rows, and target rank, but it's a disaster no matter how you rectify my negligence).  Along this vein, if you designed a specialized algorithm to solve SOS problems where the max degree within any polynomial is two: this would be amazing, and then the sort of survey you seek would have lots of value.
Secondly, since the basic formulation of these problems is out of the window, one may wonder if certain variants of these problems are well handled by SOS solvers.  As an important example, consider the NMF (non-negative matrix factorization) problem, where the matrix unknowns you are optimizing over (in your above formulation) must now have non-negative entries.  Unfortunately, if you take the standard SDP used to solve these problems (see for instance Pablo Parrilo's notes from above), there's no way to introduce those constraints.  (And since some formulations of the resulting problems are NP-hard, you would now be building an approximation scheme; i.e., this can get nasty.)  Furthermore, there is recent work that exploited the polynomial structure of this problem to build solvers with some guarantees: see http://arxiv.org/abs/1111.0952 by Arora, Ge, Kannan, and Moitra.  They build a few algorithms, however when they solve an "exact" NMF problem (where there is an exact factorization, i.e., one giving objective value 0), they do not use a SOS solver: they use a solver checking feasibility of "semi-algebraic sets", a much more difficult optimization problem which allows the kinds of constraints which NMF raises, but now with exponential running time.
Anyway, to summarize and give some further perspective; since SOS is afaik the only solver for the quartic problems you speak of (i.e., I don't think there is a specialized quartic solver), I've discussed how these solvers have better alternatives for the kinds of quartic problems people care about.  To effectively use SOS tools here, you would either have to build some amazing solver for the quartic case (inner polynomials of degree at most 2), or you would have to find some way to add constraints to these problems.  Otherwise, the connection to SOS problems, while fascinating, does not give you much.
You also mention that you are surprised that the literature you have found does not make this connection.  I think that is mostly due to the newness of practical SOS solvers (even though abstract consideration of SOS problems goes back very far), and what I said above.  In fact, when I first found SOS solvers, it was via Parrilo's notes and papers, and I similarly wondered, "why isn't he talking about PCA-type problems"?  Then I checked the above facts and frowned a lot.  I think it's also a bad sign that Parrilo himself has not, as far as I can tell/skim, discussed these problems outside the reference you mention in his thesis (meanwhile, he has papers on various extensions, and I have a lot of respect for his work in these field: he must have thought about these specific quartic problems many times.. okay I just checked and found a related negative result by him and some colleagues, which has some connections in its introduction: http://arxiv.org/abs/1111.1498 ).


TITLE: Notation for a reflexive predicate that can also have distinct variables?
QUESTION [0 upvotes]: How would you define a predicate $P$ that asserts $P(x,y)$ which can also be reflexive?
The way I did it was:
$\quad$ $I(\exists$ $P.C)=\{x,y \quad | \quad \top \sqsubseteq \exists$ $P.Self$ $\land$ $P(x, y) \}$
I'm trying to state that for some $x$ and $y$, the model can assert $P(x,y)$ but also that $P$ is reflexive.
Is this even valid notation?

REPLY [1 votes]: As far as notation goes, you can just use $P(x,y)$ ... just because $x$ and $y$ are different variables does not mean that $P$ cannot be reflexive: you can always pick the same object for both $x$ and $y$ if needed.