text:
\section{Deep Neural Network Approximation} \label{sec:method}
The ODE \eqref{eq:ode} defines a map from the initial condition to the solution at $t=T$ as below.
\begin{equation*}
\x(T)=\PPhi_\f(\x_0, \aalpha(\cdot, \omega)|_{[0, T]}, T)
\end{equation*}
Suppose $\alpha(t, \omega)=\alpha_n$ for $t\in [t_n, t_{n+1}]$, then by the mean value theorem we have
\begin{align*}
\x_{n+1}&=\x_n+\int_{t_n}^{t_{n+1}} \f(\x(s), \aalpha(s, \omega))\, ds\\
&=\x_n+\f(\x(\tau), \aalpha(\tau, \omega)), \quad \tau\in [t_n, t_{n+1}]\\
&=\x_n+\f(\x(\tau), \aalpha_n)\\
&=\x_n+\f(\PPhi_\f(\x_n, \aalpha(\cdot, \omega)|_{[t_n, \tau]}, \Delta_n), \aalpha_n)\\
&=\x_n+\f(\PPhi_\f(\x_n, \aalpha_n, \Delta_n), \aalpha_n).
\end{align*}

Then if we define the effective increment $\pphi_\f(\x_n, \aalpha_n, \Delta_n)=\f(\PPhi_\f(\x_n, \aalpha_n, \Delta_n), \aalpha_n)$, we have the following discrete flow map.
\begin{equation*}
\x_{n+1}=\x_n+\pphi_\f(\x_n, \aalpha_n, \Delta_n).
\end{equation*}

We want to use the neural network to approxiamte the effective increment $\pphi_\f(\x_n, \aalpha_n, \Delta_n)$.

\subsection{Single Parameters}

\subsection{Multiple Parameters}

\section{Uncertainty Quantification}

Epistemic uncertainty quantification...

config: arxiv


TITLE: Internalising the functor action on morphisms (e.g. to exponential objects)
QUESTION [7 upvotes]: Part of what it means to be a functor between two categories is to have a map of morphisms e.g. $F$ sends $f: A \to B$ to $Ff: FA \to FB$.
Suppose $F$ is a functor from a category to itself, and that category has (some or all) exponential objects. $B^A$ intuitively corresponds to an object of morphisms $A \to B$, so I'd hope to be able to apply the morphism-mapping of the functor $F$ to get $FB^{FA}$. I can do this in $\mathbf{Set}$ in an obvious way, but I can't see how to frame that idea category-theoretically so that I might generalise it. So my questions are:

Under what conditions can I "internalise" the morphism map of $F$ to get a morphism $B^A \to FB^{FA}$?
If that doesn't work, what if I replace $B^A$ with the hom-functor of a category enriched over itself, or some other internal notion of function?

REPLY [5 votes]: This difficulty is exactly the reason why enriched category theory isn't vacuous. There is a result of Kock which spells out necessary and sufficient conditions for a functor to have an enrichment – without ever mentioning internal homs! Here it is.
Definition. Let $\mathcal{C}$ be a monoidal closed category (such as a cartesian closed category). A strength for an endofunctor $T : \mathcal{C} \to \mathcal{C}$ is a natural transformation $\sigma_{X, Y} : X \otimes T Y \to T (X \otimes Y)$ satisfying these axioms:
$\newcommand{\id}{\textrm{id}}$

$T \lambda_X \circ \sigma_{I, X} = \lambda_{T X}$, where $I$ is the monoidal unit and $\lambda_X : I \otimes X \to X$ is the unitor.
$\sigma_{X, Y \otimes Z} \circ (\id_X \otimes \sigma_{Y,Z}) \circ \alpha_{X,Y,T Z} = T \alpha_{X,Y,Z} \circ \sigma_{X \otimes Y, Z}$, where $\alpha_{X,Y,Z} : (X \otimes Y) \otimes Z \to X \otimes (Y \otimes Z)$ is the associator.

Theorem. There is a bijection between natural transformations of type
$$\sigma_{X,Y} : X \otimes T Y \to T (X \otimes Y)$$
and natural transformations of type
$$t_{X,Y} : \mathcal{H}om (X, Y) \to \mathcal{H}om (T X, T Y)$$
such that $\sigma_{X,Y}$ is a strength for $T$ if and only if $t$ is an enrichment for $T$.
Proof. I briefly indicate where the bijection comes from; the details can be found in Theorem 3.2.17 of my notes. Suppose $\sigma$ is given. To construct $t$, it is enough to construct morphisms $\mathcal{H}om (X, Y) \otimes T X \to T Y$ satisfying an appropriate extranaturality condition. But there is a fairly obvious one we can use, namely $T (\textrm{ev}_{X, Y}) \circ \sigma_{\mathcal{H}om (X, Y), X}$. Conversely, given $t$, we construct $\sigma$ as the tensor–hom transpose of $t_{Y, X \otimes Y} \circ D_{X, Y}$, where $D_{X, Y} : X \to \mathcal{H}om (Y, X \otimes Y)$ is the unit of the tensor–hom adjunction.
Ultimately, however, the idea comes from the following observation: ordinary functors between cocomplete categories have a canonical comparison map between copowers (by the universal property thereof), and the same is true for enriched functors between cocomplete enriched categories, where "copower" is replaced by "tensor product". This extra structure tells us how to handle "non-discrete" families of morphisms in enriched categories.


TITLE: Cutting convex sets
QUESTION [19 upvotes]: Any bounded convex set of the Euclidean plane can be cut into two convex pieces of equal area and circumference.
Can one cut every bounded convex set of the Euclidean plane into an arbitrary number $n$
of convex pieces having equal area and circumference?
The solution of this problem for $n=2$ is generically unique. Are there other values of $n$
(assuming that the problem is possible) where this happens?
More generally, given a $d-$dimensional bounded convex set $C$ in the Euclidean space $\mathbf E^d$ of dimension $d$. for which values of $n$ can one cut $C$ into $n$ convex pieces of equal area with boundaries of equal $(d-1)-$dimensional area? ($n=2$ is again easy, but the solution is no longer generically unique if $d>2$). Are there values for $(n,d)$ for which the solution always exists in a generically unique way (or for which the number of solutions is generically finite)?

REPLY [2 votes]: There is a generalization of this question, where "area" and "circumference" are replaced by arbitrary "nice" measures (for the purpose of this answer, say absolutely continuous measures) $\mu$ and $\nu$ on $\mathbb{R}^2$. Bárány and Matoušek have a nice paper on the subject.
Even more generally, fix nice probability measures $\mu_1,\ldots,\mu_i$ on $\mathbb{R}^2$.
A $k$-fan in $\mathbb{R}^2$ consists of $k$ rays (semi-infinite lines) $r_1,\ldots,r_k$ emanating from a point, listed in some clockwise order. (In fact $k$-fans are also allowed to emanate from the point at infinity, i.e., a set of $k$ parallel lines is considered to be a $k$-fan.) Write $C_k$ for the region proceeding $r_k$ in the clockwise order.
Given a vector $\alpha=(\alpha_1,\ldots,\alpha_k)$ with non-negative entries summing to one,
say that $\mu_1,\ldots,\mu_r$ can be simultaneously $\alpha$-partitioned if
there exists a $k$-fan such that $\mu_i(C_j)=\alpha_j$ for each $i=1,\ldots,r$ and $j=1,\ldots,k$. (If $\alpha_1=\ldots=\alpha_k=1/k$ say that the measures can be simultaneously equipartitioned. This case, with $k=2$, is closest to the original )
Bárány and Matoušek have a whole host of results about when such partitions exist and do not exist. Here are just a couple:

For any $k \geq 5$ and any $\alpha$, there are two measures that can not be simultaneously $\alpha$-partitioned.
For any $\alpha=(\alpha_1,\alpha_2)$, any two measures can be simultaneously $\alpha$-partitioned, even if the center of the fan is specified in advance.

No one knows, for example, if any two measures can be simultaneously equipartitioned into four parts. Karasev seems to have a paper where he proves that any two measures can be simultaneously equipartitioned into $q$ convex parts, whenever the number of parts is a prime power. (This was first  achieved for three parts -- this is the result by Bárány et al that Joseph O'Rourke mentioned.) I am unclear on the relation between this and the result of Hubard and Aronov, mentioned by Joseph O'Rourke in his answer.
Higher-dimensional versions have also been considered but much is open. For example, for any three measures in $\mathbb{R}^3$ can one always find a convex $3$-partition of space so that each measure has measure $1/3$ on each part? (I heard Bárány say in a seminar that the version with $3$ replaced by a power of $2$, is known to be true; but I didn't note down a reference.)


TITLE: Prescribing the Lie derivative of the metric?
QUESTION [5 upvotes]: This is a question that arises from my research problem. Suppose $(M,g)$ is a compact Riemannian manifold with boundary and $g$ is smooth up to the boundary (if you like, take $M$ to be diffeomorphic to a closed ball). Let $h$ be a symmetric traceless 2-tensor on $M$ (in my case this is actually the traceless Ricci $\stackrel\circ {Ric}$).
Now, I have shown that $\int_M \langle h, \mathcal L_Xg\rangle dV=0$ for any vector field $X$ on $M$. I wonder does it follow that $h=0$? (Well, may be not...)
Naturally, one would try to see if there is $X$ such that $\mathcal L_X g=h$. While this seems plausible, I have no idea when it will hold.
In case this is not always possible, instead one ask: Let $\epsilon>0$ and $p\ge 1$, does there exists functions $\lambda, \mu$ on $ M$ with $\lambda\ge 1$, and a vector field $X$ such that
  $\|\mathcal L_X g -(\lambda h+ \mu g)\|_{L^p(M)}<\epsilon$?
Well, if they don't hold in general, can we impose conditions on $(M,g)$ so that one of them hold?

REPLY [9 votes]: If $h\in S^2(TM^*)$ is a symmetric $(0,2)$-tensor such that $\int_M \langle h,\mathcal L_X g\rangle =0$ for all vector fields $X$ on $M$, then $h$ need not be zero. In fact, the above is precisely equivalent to $\delta h=0$, where $$\delta=\nabla^*\big|_{S^2(TM^*)} \colon S^2(TM^*)\to \Omega^1(M)$$ is the divergence operator, i.e., the formal $L^2$-adjoint of the covariant derivative $\nabla \colon \Omega^1(M)$ $\to S^2(TM^*)$.
The above claim follows from an infinitesimal version of Ebin's slice theorem for the pull-back action of the diffeomorphism group $\mathcal D:=Diff(M)$ on $S^2(TM^*)$, see this paper of Berger and Ebin, JDG '69. The statement I am referring to is the $L^2$-orthogonal decomposition: $$S^2(TM^*)=\ker \delta\oplus Im \ \delta^*.$$
Given any element $g\in S^2(TM^*)$, e.g., a Riemannian metric, the space $\ker\delta$ is the tangent space to the slice at $g$ to the $\mathcal D$-action and $Im\ \delta^*=T_g \mathcal D(g)$ is the tangent space to the orbit of $g\in S^2(TM^*)$. The latter is precisely formed by tensors of the form $\mathcal L_X g$, where $X$ is some field on $M$. In fact, consider $\phi_g\colon\mathcal D\to S^2(TM^*)$, $\phi_g(\eta)=\eta^*(g)$, and let $\eta_t\in\mathcal D$ be a curve of diffeomorphisms, with $\eta_0=id$ and $\dot\eta_0=X$. Then $$\mathrm d\phi_g(id)X=\frac{\mathrm d}{\mathrm dt}\phi_g(\eta_t)\big|_{t=0}=\frac{\mathrm d}{\mathrm dt}\eta_t^*(g)\big|_{t=0}=\mathcal L_X g=2\delta^*(X^b),$$ where $X^b=g(X,\cdot)$ is the $1$-form dual to the field $X$. The last equality follows from $\delta^*(\omega)=\tfrac12\mathcal L_{X_\omega}g$, where $X_\omega$ is the vector field dual to the $1$-form $\omega$. Note the above line also proves that $Im \ \delta^*=T_g\mathcal D(g)$.
Thus, going back to your original question, the above result of Ebin and Berger tells you that your symmetric traceless tensor $h$ is geometric, in the sense that it is tangent to the slice of the $\mathcal D$-action on $S^2(TM^*)$, or, equivalently, $L^2$-orthogonal to the tangent space to the $\mathcal D$-orbit. In some sense, this means that it ``descends'' to an object in the quotient space of tensors modulo diffeomorphisms (where, e.g., the moduli space of Riemannian metrics should live).
The above observations also answer your question regarding what symmetric $(0,2)$-tensors are of the form $\mathcal L_Xg$; namely, they are precisely the ones in $Im \ \delta^*$. Hope this helps...


TITLE: Sequence Function?
QUESTION [1 upvotes]: Problem: What is the close-form/ recursive equation/ generating fucntion  of the following sequence, which has following 256 entries:
1 7 7 7 7 9 9 9 7 9 9 9 7 9 9 9 7 11 11 11 11 13 13 13 11 13 13 13 11 13 13 13 7 11 11 11 11 13 13 13 11 13 13 13 11 13 13 13 7 11 11 11 11 13 13 13 11 13 13 13 11 13 13 13 7 11 11 11 11 13 13 13 11 13 13 13 11 13 13 13 9 13 13 13 13 15 15 15 13 15 15 15 13 15 15 15 9 13 13 13 13 15 15 15 13 15 15 15 13 15 15 15 9 13 13 13 13 15 15 15 13 15 15 15 13 15 15 15 7 11 11 11 11 13 13 13 11 13 13 13 11 13 13 13 9 13 13 13 13 15 15 15 13 15 15 15 13 15 15 15 9 13 13 13 13 15 15 15 13 15 15 15 13 15 15 15 9 13 13 13 13 15 15 15 13 15 15 15 13 15 15 15 7 11 11 11 11 13 13 13 11 13 13 13 11 13 13 13 9 13 13 13 13 15 15 15 13 15 15 15 13 15 15 15 9 13 13 13 13 15 15 15 13 15 15 15 13 15 15 15 9 13 13 13 13 15 15 15 13 15 15 15 13 15 15 15
Origin/background of Problem: These number are generated from a divide and conquer algorithm. For example if we have a 8 bit binary-string and we are asked to find the location of 1's in minimal number of questions then, the above sequence is the # of questions asked for each possible binary 8-bit word i.e from 00000000, 00000001 ....11111111 the corresponding # of questions are elements of above sequence i.e 1,7,..15, respectively.
Algorithm:
1) First question: Is is equal to ZERO (0000 0000) ? The answer is No, for our case.
2) Then divide the original 8 bit word into two 4 bit segments, and again ask the same question for each of the two 4 bit words. So for our case it would be YES for the first segment (0000) and NO for the other segment (0001)?
3) Now, this time I will be questioning only the segment where I got NO. In our case it was 0001. Then, I will now again divide this 4-bit segment into two segments and pose the same question. Hence, is 00 equal to ZERO? answer : YES. For the other segment , 01, the answer is NO.
4) This is the final step. I will again divide the 2-bit word into two 1-bits, i-e 0 and 1. So, my first question is is 0 equal to 0? answer is YES. And for the other remaining bit, is 1 equal to 0? Answer is NO.
So, I asked a total of 7 questions to find the location of 1 in a binary word of 0000 0001. Similarly, we will go through other binary words.

REPLY [1 votes]: Write the number in base 4.  Count the number of nonzero digits.
If there none, it will take 1 question.
If there is one, it will take 7 questions.
If there are three, it will take 13 questions.
If there are four, it will take 15 questions.
If there are two in the same half, it will take 9 questions.
If there are two in opposite halves, it will take 11 questions.


text:
\begin{document}
\title[A contractible Levi-flat hypersurface]
{A contractible Levi-flat hypersurface
which is a determining set for pluriharmonic functions}
\author{Franc Forstneri\v c}
\address{Institute of Mathematics, Physics and Mechanics,
University of Ljubljana, Jadranska 19, 1000 Ljubljana, Slovenia}
\email{franc.forstneric@fmf.uni-lj.si}
\thanks{Supported by grants P1-0291 and J1-6173, Republic of Slovenia.}

\subjclass [2000]{Primary 32V05, 32V25; secondary 57R30}
\date{February 22, 2005}
\keywords{Levi-flat hypersurfaces, pluriharmonic functions}

\begin{abstract}
We find a real analytic Levi-flat hypersurface in $\C^2$
containing a bounded contractible domain which is a determining
set for pluriharmonic functions.
\end{abstract}

\maketitle

\section{The main result}
A real hypersurface $M$ in an $n$-dimensional complex manifold
is {\em Levi-flat} if it is foliated by complex manifolds of
dimension $n-1$; this {\em Levi foliation} is as smooth as
$M$ itself according to Barrett and Forn\ae ss \cite{BF}.
If $M$ is real analytic, it is locally near every point
defined by a {\em pluriharmonic function}
$v$: $dd^c v=2i\partial\overline \partial v =0$.
One might expect that an oriented, real analytic, Levi flat hypersurface
admits a pluriharmonic defining function on any topologically simple
relatively compact domain, perhaps under an additional
analytic asumption such as the existence of a
fundamental system of Stein neighborhoods (see e.g.\  Theorem 2 in
\cite{T}, p.\ 298). Here we show that, on the contrary,
even a most simple domain in a real analytic Levi flat
hypersurface may be a {\em determining set for pluriharmonic functions}.

\begin{theorem}
\label{Maintheorem}
There exist an ellipsoid $B\subset \C^2$ and a
real analytic, Levi-flat hypersurface $M\subset\C^2$
intersecting the boundary $bB$ transversely such that
the Levi foliation of $M$ has trivial holonomy
and $A=M\cap B$ satisfies the following:
\begin{itemize}
\item[(i)] $\overline {A}$ is diffeomorphic to the three-ball and
admits a Stein neighborhood basis.
\item[(ii)] Any real analytic function on $A$ which is
constant on Levi leaves is constant.
\item[(iii)] Any pluriharmonic function in a connected open
neighborhood of $A$ in $\C^2$ which vanishes on $A$ is identically zero.
\end{itemize}
\end{theorem}

The Levi foliation of $M$ in our proof is a {\em simple foliation}
(\cite{God1}, p.\ 79) whose leaves are complex discs.
Likely one can also obtain a similar example in the ball of $\C^2$.
On the other hand, for any compact subset $A$ in a
real analytic, simply connected Levi-flat hypersurface $M$
there is a {\em smooth} defining function $v$ for $M$
whose pluricomplex Laplacian $dd^c v$ is {\em flat} on $A$;
this suffices for the construction of Stein neighborhood basis of
certain compact subsets of $M$ \cite {FL}.

We mention that D.\ Barrett gave an example of a
{\em compact} real analytic Levi-flat hypersurface
with trivial holonomy and without a global pluriharmonic
defining function (Theorem 3 in \cite{Ba}).
His example is the quotient of $S^1\times\C^*$
by $(\theta,z)\to (\phi(\theta),2z)$ where $\phi$  is a real analytic
diffeomorphism of the circle $S^1$ which is topologically
but not diffeomorphically conjugate to a rotation.

\section{A real analytic foliation of $\R^2$ without analytic
first integrals}
Our construction of the hypersurface $M$ in theorem \ref{Maintheorem}
is based on the following.

\begin{proposition}
\label{proposition}
Let $D$ be the open unit disc in $\R^2$.
There exists a real analytic foliation $\cF$ of\ $\R^2$ by closed
lines such that any real analytic function on $D$ which is
constant on every leaf of the restricted foliation $\cF|_D$ is constant.
\end{proposition}

\begin{remark}
While we cannot exclude the possibility
that an example of this kind is contained in the vast
literature on the subject, we could not find a precise
reference in some of the standard sources concerning foliations
of the plane (\cite{HR}, \cite{Ha}, \cite{God1}, \cite{God2}, \cite{CC}).
It is known that every smooth foliation  of $\R^2$ by lines has a
global continuous first integral but in general not one of
class $\cC^1$, not even in the analytic case (Wazewsky \cite{W});
however, there exists a smooth first integral without critical
points on any relatively compact subset (Kamke \cite{K}).
\end{remark}

\begin{proof}
Let $(x,y)$ be  coordinates on $\R^2$. Define subsets
$E_1,E_2\subset \R^2$ by
$$
	E_1 = \{(x,y)\in \R^2\colon x<-1 \ {\rm or}\ y>0\}, \quad
	E_2 = \{(x,y)\in \R^2\colon x>1 \  {\rm or}\ y>0\}.
$$
Let $\cF_j$ denote the restriction of the foliation $\{y=c\}_{c\in\R}$
to $E_j$ $(j=1,2)$. Let $\psi$ be a real analytic orientation
preserving diffeomorphism of the half line $(0,+\infty)$,
so $\lim_{t \downarrow 0}\psi(t)=0$. (We do not require
that $\psi$ extends analytically to a neighborhood of $0$.) Then
$\phi(x,y)=(x,\psi(y))$ is a real analytic diffeomorphism
of the upper half plane $E_{1,2}= E_1\cap E_2=\{(x,y)\in \R^2\colon y>0\}$
onto itself which maps every leaf of $\cF_1|_{E_{1,2}}$ to
a leaf of $\cF_2|_{E_{1,2}}$. Let $E$ be the quotient of the
topological (disjoint) sum $E_1\sqcup E_2$ obtained by identifying a point
$(x,y)\in E_1$ belonging to $E_{1,2}$ with the point
$\phi(x,y)\in E_2$. The foliations $\cF_j$ $(j=1,2)$ amalgamate
into a real analytic foliation $\cF$ on $E$.

By construction
$E$ is a real analytic manifold homeomorphic to $\R^2$,
and hence there exists a real analytic diffeomorphism
of $E$ onto $\R^2$. (This follows in particular from
the classification theorem for simply connected Riemann
surfaces.) We identify $E$ with $\R^2$ and denote the resulting
real analytic foliation of $\R^2$ by $\cF=\cF_\psi$.
Let $\pi\colon \R^2 \to Q=\R^2/\cF$
denote the projection onto the space of leaves.
$Q$ admits the structure of a non-Hausdorff real analytic manifold
such that $\pi$ is a real analytic submersion. (The real analytic
structure on $Q$ is obtained by declaring the restriction
of $\pi$ to any local analytic transversal $\ell$ to $\cF$
to be a diffeomorphism of $\ell$ onto the open set $\pi(\ell)\subset Q$.
For the details see \cite{Ha}, \cite{HR}.)
In our case $Q$ is the quotient of the topological
sum $\R_1 \sqcup \R_2$ of two copies of
the real axis obtained by identifying a point $t>0$ in $\R_1$
with the point $\psi(t)\in \R_2$ (no identifications are made
for points $t\le 0$). The only pair of branch points
in $Q$ (i.e., points without a pair of disjoint neighborhoods)
are those corresponding to $0\in \R_1$ and $0\in \R_2$.

\begin{lemma}
\label{lemma1}
If $\psi$ is flat at origin
$(\lim_{t\downarrow 0} \psi^{(k)}(t)=0$ for $k\in \N)$
then every real analytic function on $\R^2$ which is
constant on every leaf of $\cF_\psi$ is constant.
\end{lemma}

\begin{proof}
A real analytic function $f$ on $\R^2$ which is constant on the
leaves of the foliation $\cF_\psi$ is of the form $f=h\circ \pi$ for some
real analytic function $h\colon Q\to \R$, where $Q$ is the space of leaves.
From our construction of the foliation it follows that $h$ is given by
a pair of real analytic functions $h_j \colon \R\to \R$
$(j=1,2)$ satisfying $h_1(t)=h_2(\psi(t))$ for $t>0$.
As $t\downarrow 0$, the flatness of $\psi$ at $0$ implies
that the derivative $h_1'$ is flat at $0$. Hence $h_1$, and
therefore also $h_2$, are constant.
\end{proof}

Fix $\psi$ and consider the following pair of subsets
of  $E_1$ resp.\ $E_2$:
\begin{eqnarray*}
	D_1 &=& \{(x,y)\in \R^2 \colon -3<x<-2,\ -1< y <+2\}, \\
	D_2 &=& \{(x,y)\in\R^2  \colon 2<x<3,\  -1<y< \psi(2) \} \cup\\
	    &&  \cup\, \{(x,y)\in\R^2 \colon  -3<x<3,\ \psi(1)<y<\psi(2)\}.
\end{eqnarray*}
Let $D$ be the quotient of the disjoint sum $D_1 \sqcup D_2$
obtained by identifying any point $(x,y) \in D_1$ such that $1<y<2$
with the point $\phi(x,y)=(x,\psi(y))\in D_2$. Clearly $D$ is a
simply connected domain with compact closure in $E \simeq \R^2$,
and the space of leaves $Q_D = D/\cF$ is a  non-Haudorff manifold
with a simple branch at $t=1\in\R_1$ resp.\ $\psi(1)\in \R_2$.

\begin{lemma}
\label{lemma2}
If $\psi$ is flat at the origin then every real analytic function
$f$ on $D$ which is constant on every leaf of $\cF_\psi|_D$ is constant.
\end{lemma}

\begin{proof}
As in lemma \ref{lemma1} such $f$ is of the form
$f=h\circ \pi$ for some real analytic function
$h$ on $Q_D=D/\cF_\psi$.
Such $h$ is given by a pair of real analytic functions
$h_1\colon (-1,2)\to\R$, $h_2\colon (-1,\psi(2))\to\R$
satisfying $h_1(t)=h_2(\psi(t))$ for $1< t< 2$.
By analyticity this  relation persists on the largest
interval on which both sides are defined, which is $(0,2)$.
By flatness of $\psi$ at $0$ we conclude as in lemma \ref{lemma1} that
$h_1$ and $h_2$ must be constant.
\end{proof}

Let $\cF=\cF_\psi$ be the foliation of $\R^2$ constructed
above with the diffeomorphism $\psi(t)=t e^{-1/t}$ of $(0,+\infty)$
(which is flat at $0$). Let $D\subset\subset \R^2$ satisfy the
conclusion of lemma \ref{lemma2}. Choose a disc containing $D$;
clearly lemma \ref{lemma2} still holds for this disc, and by
an affine change of coordinates on $\R^2$ we may assume this
to be the unit disc. This completes the proof
of proposition \ref{proposition}.
\end{proof}

\begin{remark}
Proposition \ref{proposition} holds for any foliation
$\cF_\psi$ constructed above for which the diffeomorphism
$\psi$ of $(0,+\infty)$ is such that $h\circ \psi$ does not
extend as a real analytic function to a neighborhood of $0$ for
any real analytic function $h$ near $0$. An example is $t^\alpha$ for
an irrational $\alpha>0$.
The foliation of $\R^2$ determined by the algebraic 1-form
$\omega= (\alpha-x)(1+x)dy - xdx$ has the space of leaves
$\cC^1$-diffeomorphic to the `simple branch' $Q$ determined
by $\psi(t)=t^\alpha$ (\cite{God2}, p.\ 120); hence it might be possible
to find a disc $D\subset \R^2$ satisfying the proposition
\ref{proposition} for this foliation. These examples indicate that
a real analytic foliation of $\R^2$ only rarely admits
real analytic first integrals on large compact subsets.
\end{remark}

\section{Proof of theorem 1.1}
Let $\cF$ be a real analytic foliation of $\R^2$ furnished
by the proposition \ref{proposition}
such that any real analytic function on $D= \{x_1^2+x_2^2<1\}\subset \R^2$
which is constant on the leaves of $\cF|_D$ is constant.
Denote by $(x_1+iy_1,x_2+iy_2)$ the coordinates on $\C^2$
and identify $\R^2$ with the plane $\{y_1=0, y_2=0\} \subset \C^2$.
Complexifying the leaves of $\cF$ we obtain the Levi foliation
of a closed, real analytic, Levi-flat hypersurface $M$
in an open tubular neighborhood $\Omega\subset \C^2$ of $\R^2$.
Set $B =\{x_1^2+x_2^2+ c(y_1^2+y_2^2)<1\}$ where $c>0$
is chosen sufficiently large such that $\overline B\subset \Omega$.
Note that $B\cap \R^2=D$. A generic choice of $c$
insures that $M$ intersects $bB$ transversely
(since transversality holds along $bD \cap M$).
Set $A=M\cap B\subset\subset M$. If $B$ is sufficiently thin
(which is the case if $c$ is sufficiently large)
then clearly $\overline A$ is diffeomorphic to the closed ball
in $\R^3$. If a real analytic function $u\in \cC^\omega(A)$
is constant on every Levi leaf of $A$ then $u|_D$ is constant on
every leaf of $\cF|_D$ and hence is constant. Thus $A$ satisfies
property (ii) in theorem \ref{Maintheorem}.

The foliation $\cF$ of $\R^2$ is transversely orientable
and hence admits a transverse real analytic vector field $\nu$.
Its complexification is a holomorphic vector field $w$
in a neighborhood of $\R^2$ in $\C^2$ such that
$iw$ is transverse to $M$ in a neighborhood of
$\overline B$, provided that $B$ is chosen sufficiently thin.
Moving $M$ off itself to either side by a short time
flow of $iw$ in a neighborhood of $\overline B$
we obtain thin neighborhoods of
$\overline A$ with two Levi-flat boundary components;
intersecting these with $rB$ for $r>1$ close to $1$
gives a fundamental system of Stein neighborhoods of $\overline A$.

Suppose that $v$ is a real pluriharmonic function
in a connected open neighborhood of $A$ such that $v|_A=0$.
For every point $x\in A$ there is an open connected
neighborhood $U_x\subset B$ and a pluriharmonic function $u_x$ on $U_x$,
determined up to a real constant, such that
$u_x+iv$ is holomorphic on $U_x$.
Since $A$ is contractible, $H^1(A,\R)=0$ and hence the
collection $\{u_x\}_{x\in A}$ can be assembled into a pluriharmonic
function $u$ in a neighborhood of $A$ such that $u+iv$ is holomorphic.
Since $v|_A=0$, $u$ is constant on every Levi leaf on $A$
and hence constant by property (ii) of $A$. Thus $v$ is constant
and hence identically zero. This proves theorem \ref{Maintheorem}.

\smallskip
\it Acknowledgements. \rm
I wish to thank C.\ Laurent--Thi\'ebaut for having bro\-ught
the question to my attention and  Nils \O vrelid for suggesting
to consider complexified foliations.

\bibliographystyle{amsplain}


text:
Given empirical data $\{\xi_i\}_{i=1}^N$, we define the empirical distribution embedding as $\mu_{\hat P}=\kme{\frac1N}{\xi_i}$.
We give the representation theorem for kernel DRO problem~\eqref{eq:kdro}.

\jz{We will show that contrary to commonly assumed, the bound is not necessarily tight.}

\subsection{Derivation 1}
\begin{equation}
  \begin{aligned}
  &\mnz_x&  &\bigg\{F(x):=\underset{P,\mu}{\sup}  \int l(x,z) \ dP(z) \bigg\}\\
  &\sjt& & \|\mu - \mu_{\hat P}\|_\rkhs\leq \epsilon\\

  &&&     \int{\phi(z)}{\ dP(z)} = \mu\\
  &&& P\in \mathcal M, \mu\in \rkhs,
  \end{aligned}
  \label{eq:kdro_signed_meas}
\end{equation}
We note there is a key difference between \eqref{eq:kdro_signed_meas} and \eqref{eq:kdro} is that the worst case measure in \eqref{eq:kdro_signed_meas} may not be a probability measure.
Nonetheless, we first show the \eqref{eq:kdro_signed_meas} admits the following RKHS representer.

\begin{lemma}
Let $(l^*, \mu_{p^*})$ be the optimal solution pair to \eqref{eq:kdro_signed_meas}.
Then $l^*$ and  $p^*$ both admit finite-term KME expansion at the given data points., i.e.
  $$
  l^* = \kme{\beta_i}{\xi_i},  \mu_{p^*} = \kme{\alpha_i}{\xi_i}\\
  $$
  for some $\alpha_i, \beta_i, i=1...N$.
\end{lemma}

**Proof** We rewrite the objective as
$$
\min _ {l\in \mathcal L}   \pip{l}{\mu_{\hat P}}_\rkhs +  \max_{\mu_P} \hip{l'}{\mu_P - \mu_{\hat P}}
$$
By virtue of Cauchy-Schwarz inequality in RKHS, we can analytically solve the inner maximization problem.
$$
\min _ {l\in \mathcal L}   \pip{l}{\mu_{\hat P}}_\rkhs +  \epsilon\hnorm{l}.
$$
The above bound is tight and is attained at $\mu_P = \mu_{\hat P} + \epsilon \frac{l}{\hnorm{l}}$. By the RKHS representer theorem of Schoelkopf et al., we have it that the optimal solution to the above problem can be represented as
$$
l^* = \kme{\beta_i}{\xi_i}.
$$

Then we naturally obtain
$$
\mu_P = \mu_{\hat P} + \epsilon \frac{l^*}{\hnorm{l^*}}.
$$
Since both of the summands are finitely represented, we have
$$
\mu_{p^*} = \kme{\alpha_i}{\xi_i}
$$
for some $\alpha_i, i=1...N$.

\subsection{Extending to all continuous functions on compact domain}
We now use the fact that some RKHS (e.g. Gaussian) is dense on C[compact] to derive the existence of extremal distributions represented by finitely many kernel expansion.

Given empirical data $\{\xi_i\}_{i=1}^N$, we define the empirical distribution embedding as $\mu_{\hat P}=\kme{\frac1N}{\xi_i}$.
\begin{theorem}
(RKHS representer of extreme distribution)

Let $l$ be a continuous function on a compact domain $\mathcal X$, and $P^*$ be the worst-case distribution of the DRO problem $\ref{eq:dro2}$. For any $\epsilon>0$,  there exists an distribution $P_\epsilon$ such that
$$
|\pip{l}{P_\epsilon} -\pip{l}{P^*}| \leq\epsilon,
$$
where the RKHS embedding of $P^*$ admits the finite-term representation
$$
\mu_{p_\epsilon} = \kme{\alpha_i}{\xi_i}.
$$
for some $\alpha_i,, i=1...N$.

Further more, if we let $\epsilon\to 0$, we **have**
$$
\mu_{p^*} = \kme{\alpha_i}{\xi_i}.
$$
for some $\alpha_i,, i=1...N$.
\end{theorem}
[**remark**] this theorem tells us that, for solving the DRO problem, it surfices to consider the representation on the empirical expansion. DRO constraint $\| \kme{\alpha^i_t}{\hat x^i} -\kme{\frac1N}{x(u,\xi^i)}\|_\rkhs\leq\epsilon.$ is a robust bound, rather than a sample-based approximation to the $\|P- \hat P\|_{MMD}\leq\epsilon$.

In addition, we don't need to find new expansion points $\xi_i$, which amounts to nonlinear optimization problem such as in [bach herding, chen, welling]

\textbf{proof.}

outline: get $ l'$ such that $|l-l'|_\infty\leq\epsilon_2$, then $p_\epsilon$ is arbitrarily close to true extremal distribution. Then finally, we use the fact that the convergence sequence is Cauchy [need verify].

cf. Apple notes

\subsection{Extremal representation of conv. comb. of Diracs}

If we constrain $\sum\alpha_i = 1$, we imply the extreme distribution is the convex combination of Diracs. This **would be a strong result**!

At least, this is a **robust lower bound**


text:
\begin{document}

\title{Bounds for solid angles of lattices of rank three}
\author{Lenny Fukshansky \and Sinai Robins}

\address{Department of Mathematics, Claremont McKenna College, 850 Columbia Avenue, Claremont, CA 91711-6420}
\email{lenny@cmc.edu}
\address{Division of Mathematical Sciences, Nanyang Technological University, SPMS-MAS-03-01, 21 Nanyang Link, Singapore 637371 }
\email{rsinai@ntu.edu.sg}
\subjclass[2000]{Primary: 52C07, 05B30; Secondary: 11H06, 51M25}
\keywords{lattices, solid angles, spherical configurations}

\begin{abstract} We find sharp absolute constants $C_1$ and $C_2$ with the following property: every well-rounded lattice of rank 3 in a Euclidean space has a minimal basis so that the solid angle spanned by these basis vectors lies in the interval $[C_1,C_2]$. In fact, we show that these absolute bounds hold for a larger class of lattices than just well-rounded, and the upper bound holds for all. We state a technical condition on the lattice that may prevent it from satisfying the absolute lower bound on the solid angle, in which case we derive a lower bound in terms of the ratios of successive minima of the lattice. We use this result to show that among all spherical triangles on the unit sphere in $\mathbb R^N$ with  vertices on the minimal vectors of a lattice, the smallest possible area is achieved by a configuration of minimal vectors of the (normalized) face centered cubic lattice in $\mathbb R^3$. Such spherical configurations come up in connection with the kissing number problem.
\end{abstract}

\maketitle

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\def\B{{\mathcal B}}
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\def\E{{\mathcal E}}
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\def\K{{\mathcal K}}
\def\L{{\mathcal L}}
\def\Ll{{\mathfrak L}}
\def\M{{\mathcal M}}
\def\Mm{{\mathfrak M}}
\def\Pp{{\mathfrak P}}
\def\Aa{{\mathfrak A}}
\def\Ss{{\mathfrak S}}
\def\N{{\mathcal N}}
\def\PP{{\mathcal P}}
\def\R{{\mathcal R}}
\def\s{{\mathcal S}}
\def\V{{\mathcal V}}
\def\W{{\mathcal W}}
\def\X{{\mathcal X}}
\def\Y{{\mathcal Y}}
\def\H{{\mathcal H}}
\def\OO{{\mathcal O}}
\def\aaa{{\mathbb A}}
\def\cee{{\mathbb C}}
\def\Nn{{\mathbb N}}
\def\pee{{\mathbb P}}
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\def\gmn{{\mathbb G_m^N}}
\def\qbar{{\overline{\mathbb Q}}}
\def\DL{{\underline{\Delta}}}
\def\DU{{\overline{\Delta}}}
\def\eps{{\varepsilon}}
\def\vek{{\varepsilon_k}}
\def\ahat{{\hat \alpha}}
\def\bhat{{\hat \beta}}
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\def\rank{\operatorname{rank}}
\def\aut{\operatorname{Aut}}
\def\lcm{\operatorname{lcm}}
\def\sgn{\operatorname{sgn}}
\def\spn{\operatorname{span}}
\def\md{\operatorname{mod}}
\def\Norm{\operatorname{Norm}}
\def\dim{\operatorname{dim}}
\def\det{\operatorname{det}}
\def\Vol{\operatorname{Vol}}
\def\rk{\operatorname{rk}}
\def\md{\operatorname{mod}}
\def\sqp{\operatorname{sqp}}
\def\Aut{\operatorname{Aut}}
\def\GL{\operatorname{GL}}
\def\Sim{\operatorname{Sim}}

\section{Introduction}
\label{intro}

Given an integer $N \geq 2$, let $\real^N$ be the Euclidean space with the usual norm $\|\ \|$ on it. The kissing number problem in dimension $N$ asks for the maximal number of non-overlapping unit balls in $\real^N$ that touch another unit ball. The answer to this problem is currently only known in dimensions $N=2,3,4,8,24$ (see \cite{bachoc}, \cite{conway}, \cite{musin}, and \cite{odlyzko}). In fact, in dimension 3 this was the subject of a famous argument between Isaac Newton and David Gregory, where Newton claimed that the kissing number is 12 while Gregory believed it was 13; two different proofs that this number is 12 finally appeared in the 1950-s, by Sch\"{u}tte and van der Waerden \cite{waerden} and by Leech \cite{leech} (see also \cite{conway}, \cite{hsiang} for details, including an extensive bibliography), although there were previous unsuccessful and incomplete attempts. The kissing number problem can be reformulated as follows: find the maximal configuration of points on the unit sphere in $\real^N$ such that the {\it angular separation} between any pair of these points is at least $\pi/3$; by angular separation between two points we mean the smallest angle between the vectors connecting these two points to the origin in the plane spanned by these vectors. Such configurations are often expected to come from sets of {\it minimal vectors} (i.e. vectors of smallest nonzero norm) of lattices, at least this is the case in all dimensions where the kissing number is known. For example, in dimension three an optimal configuration of 12 points is given by the set of minimal vectors of the (normalized) face centered cubic lattice $A_3$. Define a {\it spherical lattice-minimal triangle} to be a non-degenerate spherical triangle with vertices at the endpoints of minimal vectors of a lattice; as we will discuss below, the minimality condition forces the angular separation between every pair of these points to be at least $\pi/3$. The above consideration raises the following naturally related question.

\begin{quest} \label{quest1} Given a spherical lattice-minimal triangle on a unit sphere in $\real^N$, what is the minimal possible two-dimensional spherical area it can have?
\end{quest}

\noindent
In other words, although minimal vectors of the normalized lattice $A_3$ form the largest (with respect to cardinality) configuration of points on the unit sphere with angular separation at least $\pi/3$, and hence produce spherical triangles of small area (at least on the average), could it be that minimal vectors of some other lattice form some spherical triangle of even smaller area? Our first result answers this question.

\begin{thm} \label{sp_area} Any spherical lattice-minimal triangle on a unit sphere in $\real^N$ has area at least
\begin{equation}
\label{sp1}
0.551285598\hdots = 4\pi \times 0.043869914\hdots,
\end{equation}
which is precisely the area of the spherical triangle formed by the vectors
\begin{equation}
\label{sp_sharp}
\left( \begin{matrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{matrix} \right),\ \left( \begin{matrix} 0 \\ 1/\sqrt{2} \\ 1/\sqrt{2} \end{matrix} \right),\ \left( \begin{matrix} 1/\sqrt{2} \\ 0 \\ 1/\sqrt{2} \end{matrix} \right)
\end{equation}
on the unit sphere in $\real^3$. These are precisely minimal basis vectors of the face centered cubic lattice $A_3$, normalized to lie on the unit sphere.
\end{thm}

To prove Theorem \ref{sp_area}, we use a somewhat more general and technical result, which we present next. We start with some basic notation. Let $B_N$ be a unit ball in $\real^N$. Given a lattice $\Lambda \subset \real^N$ of rank~$r$, we define its successive minima
$$0 < \lambda_1 \leq \dots \leq \lambda_r$$
by
$$\lambda_i = \inf \{ \lambda \in \real_{>0} : \Lambda \cap \lambda B_N \text{ contains } i \text{ linearly independent vectors} \}.$$
There exists a collection of linearly independent vectors $\bx_1,\dots,\bx_r$ in $\Lambda$ such that $\|\bx_i\| = \lambda_i$ for each $1 \leq i \leq r$; we will refer to them as vectors {\it corresponding} to successive minima. When $r \leq 4$, these vectors form a basis for $\Lambda$, which is precisely a minimal basis; this is not necessarily true for $r \geq 5$ (see for instance \cite{pohst}). Let us also write
$$S(\Lambda) = \left\{ \bx \in \Lambda : \|\bx\| = \lambda_1 \right\}$$
for the set of all minimal vectors of $\Lambda$. In the special case when $S(\Lambda)$ contains $r$ linearly independent vectors, i.e. when
$$\lambda_1 = \dots = \lambda_r,$$
$\Lambda$ is called a {\it well-rounded} lattice, abbreviated WR. WR lattices and configurations of their minimal vectors play an important role in discrete optimization problems (see \cite{martinet}, \cite{conway}, \cite{hsiang}). In particular, spherical configurations which give good kissing numbers always come from WR lattices.

We will say that a basis $\bx_1, \dots, \bx_r$ for a lattice $\Lambda \subset \real^N$ of rank $r$ is {\it minimal} (often referred to as {\it Minkowski reduced} in the literature - see for instance \cite{lek}) if $\bx_1 \in S(\Lambda)$ and for each $2 \leq i \leq r$, $\bx_i \in \Lambda$ is a shortest vector such that the collection of vectors $\bx_1,\dots,\bx_i$ is extendable to a basis in~$\Lambda$. Minimal bases for lattices are extensively studied for their importance in number theory and discrete geometry, as well as in computer science and engineering applications (see \cite{pohst}, \cite{near:ort}, \cite{esm}). In particular, if a lattice happens to have an orthogonal basis, it is easy to prove that this basis must be minimal. In fact, it turns out that even suitably defined ``near-orthogonality" of the basis vectors is sufficient for a basis to be minimal (or ``almost minimal" - here we mean reduced vs Minkowski reduced). This idea has been successfully exploited by the famous LLL algorithm \cite{LLL}. It should be mentioned that in addition to the notion of Minkowski reduction that we are using here other basis reduction procedures exist, most notably the Hermite-Korkine-Zolotareff reduction (see for instance \cite{martinet} for details); the common general principle behind various reduction procedures is the minimization of the orthogonality defect.

In the reverse direction, one can ask how ``close" to orthogonal does a minimal basis have to be? When $r=2$, the answer is classical: if $\bx_1,\bx_2$ is a minimal basis for a lattice $\Lambda \subset \real^N$ of rank two, then the (normalized) angle between these vectors has to lie in the interval~$[1/6, 1/3]$; moreover, for every such lattice there exists a minimal basis with the (normalized) angle between vectors lying in the interval~$[1/6, 1/4]$ (since we will use this result, we include its proof in section~\ref{2D} to make our presentation self-contained). It is natural to ask for analogues of this statement for lattices of higher rank, starting with~$r=3$ as follows.

\begin{quest} \label{quest2} Do there exist absolute constants $C_1$ and $C_2$ such that every lattice of rank three has a minimal basis that spans a solid angle in the interval $[C_1,C_2]$?
\end{quest}

Notice that when $\Lambda$ is a WR lattice of rank three in $\real^N$, then $S(\Lambda)$ contains a minimal basis. In fact, by scaling if necessary, we can assume that $\lambda_1 = \lambda_2 = \lambda_3 = 1$, and then any three linearly independent vectors in $S(\Lambda)$ form a minimal basis and span a spherical lattice-minimal triangle on the unit sphere centered at the origin in $\real^N$. On the other hand, vertices of a spherical lattice-minimal triangle on the unit sphere centered at the origin in $\real^N$ form a minimal basis for a WR lattice of rank three with $\lambda_1 = \lambda_2 = \lambda_3 = 1$. Since the solid angle in question is equal to the normalized spherical area of the corresponding spherical triangle, there is a direct connection between Question \ref{quest2} restricted to just WR lattices and Question \ref{quest1}. In this note, we answer Question \ref{quest2} completely for a wide class of lattices including all WR lattices and provide a partial answer (with the constant $C_1$ depending on the lattice) for all the remaining lattices. We then use this result, along with our method, to answer Question \ref{quest1}.

Let $\Lambda$ be a lattice of rank 3 in $\real^N$ and let $\bx_1,\bx_2,\bx_3$ be a fixed minimal basis for $\Lambda$, corresponding to successive minima $\lambda_1,\lambda_2,\lambda_3$, respectively. We will write $\theta_{ij}$ for the angle between the vectors $\bx_i$ and $\bx_j$, $1 \leq i < j \leq 3$, and will refer to $\theta_{12},\theta_{13},\theta_{23}$ as the {\it vertex angles} corresponding to this minimal basis. We also define the ratios of successive minima
\begin{equation}
\label{KKK}
K_{12} = \frac{\lambda_1}{\lambda_2},\ K_{23} =  \frac{\lambda_2}{\lambda_3},\ K_{13} = \frac{\lambda_1}{\lambda_3} = K_{12} K_{23}.
\end{equation}
Then clearly $\max \{ K_{12}, K_{23} \} \leq 1$, and $\min \{ K_{12}, K_{23} \} < 1$ if and only if $\Lambda$ is not WR. Let us also define
\begin{equation}
\label{nu}
\nu := \frac{1}{4} \cos^{-1} \left( \frac{1}{2} \left( 2K_{12} K_{23} \cos \theta_{12} + 2K_{12} \cos \theta_{13} - K_{12}^2 K_{23} - K_{23}  \right) \right) < \frac{\pi}{6}.
\end{equation}

\begin{rem} \label{solid_measure} Our measure of three-dimensional solid angles is always normalized to be between 0 and 1, i.e. we divide by $4\pi$, the maximal possible solid angle which corresponds to the full sphere in $\real^3$.
\end{rem}

\noindent
We can now state our next result.

\begin{thm} \label{main_3} Let $\Lambda \subset \real^N$ be a lattice of rank 3. Then there exists a minimal basis $\bx_1,\bx_2,\bx_3$ for $\Lambda$ such that
\begin{equation}
\label{mn3.1}
\pi/3 \leq \theta_{12}, \theta_{13} \leq \pi/2,\ \pi/3 \leq \theta_{23} \leq 2\pi/3.
\end{equation}
Fix any minimal basis $\bx_1,\bx_2,\bx_3$ for $\Lambda$ satisfying (\ref{mn3.1}), and let $\Omega$ be the solid angle  formed by these vectors. Then $\Omega \leq 0.125$. If in addition
\begin{equation}
\label{mn3.2}
\theta_{23} \leq  \cos^{-1} \left( \cos \theta_{12} + \cos \theta_{13} - 1 \right),
\end{equation}
then $\Omega \geq 0.043869914\hdots$ The above condition is satisfied in particular by every WR lattice, and the bounds on $\Omega$ are sharp as demonstrated in Remark \ref{sharp} below. On the other hand, if
$$\cos^{-1} \left( \cos \theta_{12} + \cos \theta_{13} - 1 \right) < \theta_{23} \leq 2\pi/3,$$
then in fact
$$\theta_{23} \leq \cos^{-1} \left( \frac{1}{2} \left( 2K_{12} K_{23} \cos \theta_{12} + 2K_{12} \cos \theta_{13} - K_{12}^2 K_{23} - K_{23}  \right) \right) < \frac{2\pi}{3},$$
and
$$\Omega \geq \frac{1}{\pi} \tan^{-1} \left( \tan \nu\ \sqrt{\frac{1 - 3 \tan^2 \nu}{3 - \tan^2 \nu}} \right) > 0,$$
where $\nu$ is as in \eqref{nu}.
\end{thm}

In section~\ref{2D} we discuss some basic two-dimensional lemmas, which will be later used in our main argument. Theorem \ref{main_3} is proved in section~\ref{3D}: it follows immediately by combining Lemmas \ref{min_bas}, \ref{solid_3}, \ref{solid_other}, Corollary \ref{solid_wr}, and Remark \ref{sharp}. In  section~\ref{sphere}, we use the techniques developed in section~\ref{3D} to prove Theorem~\ref{sp_area}. We are now ready to proceed.
\bigskip

\section{Preliminary two-dimensional lemmas}
\label{2D}

We start with some basic lemmas about the angles between minimal basis vectors for lattices of rank~2, which we will use in our argument for lattices of rank~3.

\begin{lem} \label{vec_angles} Let $\bx_1$ and $\bx_2$ be nonzero vectors in $\real^N$ so that the angle $\theta$ between them satisfies either $0 < \theta < \frac{\pi}{3}$ or $\frac{2\pi}{3} < \theta < \pi$. Then
$$\min \left\{ \| \bx_1 - \bx_2 \|,  \| \bx_1 + \bx_2 \| \right\} < \max \{ \| \bx_1 \|, \| \bx_2 \| \}.$$
\end{lem}

\proof
Notice that either
$$\frac{1}{2} < \cos \theta = \frac{\bx_1^t \bx_2}{\|\bx_1\| \|\bx_2\|} < 1,\text{ or } -1 < \cos \theta = \frac{\bx_1^t \bx_2}{\|\bx_1\| \|\bx_2\|} < -\frac{1}{2}.$$
In the first case:
\begin{eqnarray*}
\|\bx_1 - \bx_2\|^2 & = & (\bx_1-\bx_2)^t (\bx_1-\bx_2) = \|\bx_1\|^2 + \|\bx_2\|^2 - 2 \bx_1^t \bx_2 \\
& < & \|\bx_1\|^2 + \|\bx_2\|^2 - \|\bx_1\| \|\bx_2\| < \max \{ \| \bx_1 \|, \| \bx_2 \| \}^2.
\end{eqnarray*}
In the second case:
\begin{eqnarray*}
\|\bx_1 + \bx_2\|^2 & = & (\bx_1+\bx_2)^t (\bx_1+\bx_2) = \|\bx_1\|^2 + \|\bx_2\|^2 + 2 \bx_1^t \bx_2 \\
& < & \|\bx_1\|^2 + \|\bx_2\|^2 - \|\bx_1\| \|\bx_2\| < \max \{ \| \bx_1 \|, \| \bx_2 \| \}^2.
\end{eqnarray*}
\endproof

\begin{lem} \label{angle} Let $\Lambda \subset \real^N$ be a lattice of rank 2 with successive minima $\lambda_1 \leq \lambda_2$, and let $\bx_1,\bx_2$ be vectors in $\Lambda$ corresponding to $\lambda_1,\lambda_2$, respectively. Let $\theta$ be the angle between $\bx_1$ and $\bx_2$. Then
$$\pi/3 \leq \theta \leq 2\pi/3.$$
\end{lem}

\proof
Clearly $\theta \in (0, \pi)$. Assume that either $0 < \theta < \pi/3$ or $\frac{2\pi}{3} < \theta < \pi$, then Lemma \ref{vec_angles} implies that
$$\min \left\{ \| \bx_1 - \bx_2 \|,  \| \bx_1 + \bx_2 \| \right\} < \| \bx_2 \| = \lambda_2,$$
which contradicts the definition of $\lambda_2$ since the vectors $\bx_1$ and $\bx_1\pm \bx_2$ are linearly independent.
\endproof

\begin{lem} \label{plane_angles} Let $\Lambda \subset \real^N$ be a lattice of full rank with successive minima
$$0 < \lambda_1 \leq \dots \leq \lambda_N,$$
and vectors $\bx_1,\dots,\bx_N$ corresponding to these successive minima, respectively, chosen in such a way that all of them lie in the half-space $x_N \geq 0$. For every pair of indices $1 \leq i < j \leq N$, let $\theta_{ij}$ be the angle between the vectors $x_i$ and $x_j$. Then
$$\pi/3 \leq \theta_{ij} \leq 2 \pi/3.$$
\end{lem}

\proof
Let $\Lambda_{ij} = \spn_{\zed} \{ \bx_i, \bx_j \}$, then the successive minima $\mu_1, \mu_2$ of $\Lambda_{ij}$ are:
$$\mu_1 = \lambda_i \leq \mu_2 = \lambda_j.$$
Indeed, it is clear that $\mu_1 \leq \mu_2$ and $\mu_1 \leq \lambda_i$, $\mu_2 \leq \lambda_j$, so suppose for instance that $\mu_1 < \lambda_i$. Then there exists a vector $\bwy = a_i \bx_i + a_j \bx_j$, such that $0 \neq a_i, a_j \in \zed$ and
\begin{equation}
\label{a1}
\|\bwy\| = \mu_1 < \lambda_i.
\end{equation}
Then the collection of vectors
\begin{equation}
\label{a2}
\bx_1,\dots,\bx_{i-1},\bwy,\bx_{i+1},\dots,\bx_N
\end{equation}
must be linearly independent in $\Lambda$, and so by definition of successive minima, $\mu_1 \geq \lambda_i$, which is a contradiction. If, on the other hand, we assume that $\mu_2 < \lambda_j$, then we can apply the same argument as above replacing $\mu_1$ with $\mu_2$ in (\ref{a1}) and inserting $\bwy$ for $\bx_j$ instead of $\bx_i$ in (\ref{a2}) to reach the same contradiction. Therefore $\lambda_i,\lambda_j$ are the successive minima of $\Lambda_{ij}$, and hence $\bx_i,\bx_j$ are vectors corresponding to successive minima. Now the conclusion follows by Lemma \ref{angle} above.
\endproof

\bigskip

\section{Bounds for solid angles: proof of Theorem \ref{main_3}}
\label{3D}

In this section we prove a collection of lemmas, which together comprise the result of Theorem \ref{main_3}. First we need to fix our choice of a minimal basis for a lattice of rank~3, which is accomplished by the following lemma.

\begin{lem} \label{min_bas}  Let $\Lambda \subset \real^N$ be a lattice of rank 3. Then there exists a minimal basis $\bx_1,\bx_2,\bx_3$ for $\Lambda$ such that
\begin{equation}
\label{min_angle}
\pi/3 \leq \theta_{12}, \theta_{13} \leq \pi/2,\ \pi/3 \leq \theta_{23} \leq 2\pi/3,
\end{equation}
where $\theta_{ij}$ is the angle between vectors $\bx_i$ and $\bx_j$ for all $1 \leq i < j \leq 3$.
\end{lem}

\proof
Let
$$0 < \lambda_1 \leq \lambda_2 \leq \lambda_3$$
be successive minima of $\Lambda$, and let $\bx_1,\bx_2,\bx_3$ be the vectors corresponding to $\lambda_1, \lambda_2, \lambda_3$ respectively. Then Lemma \ref{plane_angles} implies that $\pi/3 \leq \theta_{ij} \leq 2\pi/3$. In fact, if $\theta_{12} > \pi/2$, replace $\bx_2$ with $- \bx_2$, and if $\theta_{13} > \pi/2$, replace $\bx_3$ with $- \bx_3$. Then $\bx_1,\bx_2,\bx_3$ is a minimal basis satisfying (\ref{min_angle}), as required.
\endproof

In what follows, the measure of three-dimensional solid angles is normalized as specified in Remark~\ref{solid_measure}, however we assume it is converted back to steradians when we compute values of trigonometric functions of $\Omega$.

\begin{lem} \label{solid_3} Let $\Lambda \subset \real^N$ be a lattice of rank 3, and let $\bx_1,\bx_2,\bx_3$ be a minimal basis for $\Lambda$ guaranteed by Lemma \ref{min_bas}. Let $\Omega$ be the solid angle  formed by these vectors. Then $\Omega \leq 0.125$. If in addition
\begin{equation}
\label{cos-1}
\theta_{23} \leq  \cos^{-1} \left( \cos \theta_{12} + \cos \theta_{13} - 1 \right),
\end{equation}
then $\Omega \geq 0.043869914\hdots$
\end{lem}

\proof
With our choice of the basis $\bx_1,\bx_2,\bx_3$, we have
\begin{equation}
\label{tij.0}
\pi/3 \leq \theta_{12}, \theta_{13} \leq \pi/2,\ \pi/3 \leq \theta_{23} \leq 2\pi/3,
\end{equation}
Now make the choice of $\pm \bx_3$ that makes $\theta_{23}$ as small as possible for (\ref{tij.0}) to hold; notice in particular that if $\theta_{13} = \pi/2$, then $\pm \bx_3$ can be chosen to ensure that $\pi/3 \leq \theta_{23} \leq \pi/2$. From now on we will always assume this choice of $\bx_1, \bx_2, \bx_3$. Let
\begin{equation}
\label{ts1}
\alpha = \frac{\theta_{12} + \theta_{13}}{4},\ \beta = \frac{\theta_{12}-\theta_{13}}{4},\ c = \frac{\theta_{23}}{4},
\end{equation}
and so
\begin{equation}
\label{ts2.0}
\frac{\pi}{6} \leq \alpha \leq \frac{\pi}{4},\ -\frac{\pi}{24} \leq \beta \leq \frac{\pi}{24},\ \frac{\pi}{12} \leq c \leq \frac{\pi}{6}.
\end{equation}
Let $\Omega$ be the solid angle spanned by the vectors $\bx_1,\bx_2,\bx_3$, then L'Huilier's theorem (see, for instance, \cite{sphere_trig}) implies that
\begin{equation}
\label{tan1.1}
\tan \left( \frac{\Omega}{4} \right)^2 = \tan \left( \alpha + c \right) \tan \left( \alpha - c \right) \tan \left( c+ \beta \right) \tan \left( c - \beta \right),
\end{equation}
when $\Omega$ is measured in steradians. Notice that for each $c \in (-\pi/4,\pi/4)$,
\begin{equation}
\label{beta}
\tan \left( c + \beta \right) \tan \left( c - \beta \right) = \frac{\tan^2 c - \tan^2 \beta}{1 - \tan^2 c \tan^2 \beta}
\end{equation}
is a decreasing function of $|\beta|$, and so (\ref{beta}) is minimized when $\beta$ is as large as possible, and maximized when $\beta = 0$.
\smallskip

To produce an upper bound for $\Omega$, let us write
\begin{equation}
\label{xy}
x_1 = \pi/2 - \theta_{12},\ x_2 = \pi/2 - \theta_{13},\ y = \theta_{23} - \pi/2, \text{ and } x = x_1+x_2,
\end{equation}
so that
\begin{equation}
\label{xy1}
\alpha = \frac{\pi}{4} - \frac{x}{4},\ c = \frac{\pi}{8} + \frac{y}{4}, \text{ where } 0 \leq x \leq \frac{\pi}{3},\ -\frac{\pi}{6} \leq y \leq \frac{\pi}{6},
\end{equation}
and $y$ is as small as possible for (\ref{xy1}) to hold. Then (\ref{tan1.1}), along with the fact that (\ref{beta}) is maximized when $\beta=0$, implies that
\begin{equation}
\label{tan1.3.2}
\tan \left( \frac{\Omega}{4} \right)^2 \leq \tan \left( \frac{3\pi}{8} - \frac{x}{4} + \frac{y}{4} \right) \tan \left( \frac{\pi}{8} - \frac{x}{4} - \frac{y}{4}  \right) \tan \left( \frac{\pi}{8} + \frac{y}{4} \right)^2
\end{equation}
It is not difficult to observe that with the constraints of (\ref{xy1}) satisfied, the right hand side of (\ref{tan1.3.2}) is a decreasing function of $x$, so to maximize it we can assume that $x=0$, meaning that $\theta_{12} = \theta_{13} = \pi/2$. In this case $\pi/3 \leq \theta_{23} \leq \pi/2$, meaning that $-\pi/6  \leq y \leq 0$, and
\begin{equation}
\label{tan1.3.3}
\tan \left( \frac{\Omega}{4} \right)^2 \leq \tan \left( \frac{3\pi}{8} + \frac{y}{4} \right) \tan \left( \frac{\pi}{8} - \frac{y}{4}  \right) \tan \left( \frac{\pi}{8} + \frac{y}{4} \right)^2,
\end{equation}
where the right hand side of (\ref{tan1.3.3}) is an increasing function of $y$. Therefore
$$\tan \left( \frac{\Omega}{4} \right)^2 \leq \tan \left( \frac{3\pi}{8} \right) \tan \left( \frac{\pi}{8} \right)^3,$$
and the upper bound for $\Omega$ follows.
\smallskip

Next we produce a lower bound for $\Omega$. Assume that in addition to (\ref{tij.0}), the inequality (\ref{cos-1}) is also satisfied. Then
\begin{equation}
\label{tij}
\pi/3 \leq \theta_{12}, \theta_{13} \leq \pi/2,\ \pi/3 \leq \theta_{23} \leq  \min \left\{ \cos^{-1} \left( \cos \theta_{12} + \cos \theta_{13} - 1 \right), 2\pi/3 \right\},
\end{equation}
and $\bx_1,\bx_2,\bx_3$ is chosen so that $\theta_{23}$ is as small as possible for (\ref{tij}) to hold. Then
\begin{equation}
\label{ts2}
\frac{\pi}{6} \leq \alpha \leq \frac{\pi}{4},\ -\frac{\pi}{24} \leq \beta \leq \frac{\pi}{24},\ \frac{\pi}{12} \leq c \leq \min \left\{ \frac{1}{4} \cos^{-1} \left( 2 \cos 2\alpha \cos 2 \beta - 1 \right), \frac{\pi}{6} \right\}.
\end{equation}
The right hand side of (\ref{tan1.1}) is easily checked to be an increasing function of $c$ when the constraints of (\ref{ts2}) are satisfied (we use Sage mathematical software package \cite{sage} - see illustration in section~\ref{sage_code}). Therefore for each fixed pair of values of $\alpha$ and $\beta$,
\begin{equation}
\label{tan1.2}
\tan \left( \alpha + \frac{\pi}{12} \right) \tan \left( \alpha - \frac{\pi}{12} \right) \tan \left( \frac{\pi}{12}+ \beta \right) \tan \left( \frac{\pi}{12} - \beta \right) \leq \tan \left( \frac{\Omega}{4} \right)^2,
\end{equation}
when (\ref{cos-1}) holds. Since $\theta_{12}, \theta_{13} \geq \pi/3$, it follows that
$$4 |\beta| = |\theta_{12}-\theta_{13}| \leq 4 \alpha - 2\pi/3,$$
and so (\ref{beta}) is minimized when $\beta = \min \{ \alpha - \pi/6, \pi/24\}$. We can also assume without loss of generality that $0 < \Omega < 1/2$ (i.e. the measure of $\Omega$ in steradians is between 0 and $2\pi$), and so $\tan \left( \frac{\Omega}{4} \right)^2$ is an increasing function of $\Omega$.

First assume that $\alpha - \pi/6 \leq \pi/24$, then $\alpha \leq 5\pi/24$ and the above observations combined with (\ref{tan1.2}) imply that
\begin{equation}
\label{tan1.3}
\tan \left( \frac{\Omega}{4} \right)^2 \geq \tan \left( \alpha + \frac{\pi}{12} \right) \tan \left( \alpha - \frac{\pi}{12} \right)^2 \tan \left( \frac{\pi}{4} - \alpha \right) \geq \tan \left( \frac{\pi}{12} \right)^3.
\end{equation}

Next assume that $\alpha - \pi/6 \geq \pi/24$, then $\alpha \geq 5\pi/24$ and the above observations combined with (\ref{tan1.2}) imply that
\begin{eqnarray}
\label{tan1.3.1}
\tan \left( \frac{\Omega}{4} \right)^2 & \geq & \tan \left( \alpha + \frac{\pi}{12} \right) \tan \left( \alpha - \frac{\pi}{12} \right) \tan \left( \frac{\pi}{8} \right) \tan \left( \frac{\pi}{24} \right) \nonumber \\
& \geq & \tan \left( \frac{7\pi}{24} \right) \tan \left( \frac{\pi}{8} \right)^2 \tan \left( \frac{\pi}{24} \right) > \tan \left( \frac{\pi}{12} \right)^3.
\end{eqnarray}
The lower bound for $\Omega$ now follows in the case when (\ref{cos-1}) is satisfied.
\endproof
\smallskip

\begin{cor} \label{solid_wr} Let $\Lambda \subset \real^N$ be a well-rounded lattice of rank 3, and let $\bx_1,\bx_2,\bx_3$ be a minimal basis for $\Lambda$ guaranteed by Lemma \ref{min_bas}. Then (\ref{cos-1}) is satisfied, and so
$$0.043869914\hdots \leq \Omega \leq 0.125,$$
by Lemma \ref{solid_3}.
\end{cor}

\proof
Let $\lambda$ be the common value of successive minima of $\Lambda$, and notice that
$$\| \bx_1-\bx_2-\bx_3 \|^2 = \lambda^2(3 + 2(-\cos \theta_{12} - \cos \theta_{13} + \cos \theta_{23} )) \geq \lambda^2,$$
which means that $-\cos \theta_{12} - \cos \theta_{13} + \cos \theta_{23}  \geq -1$.
\endproof

\begin{rem} \label{sharp} The bounds of Lemma \ref{solid_3} are sharp. The lower bound is achieved by the fcc (face centered cubic) lattice $A_3$ (same as $D_3$) in $\real^3$ with the choice of a minimal basis
$$\bx_1 = \left( \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right),\ \bx_2 = \left( \begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right),\ \bx_3 = \left( \begin{matrix} 1 \\ 0 \\ 1 \end{matrix} \right),$$
so that the angles between these vectors are
$$\theta_{12} = \theta_{13} = \theta_{23} = \pi/3.$$
The upper bound is achieved by the integer lattice $\zed^3$ in $\real^3$ with the standard choice of a minimal basis
$$\bx_1 = \left( \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right),\ \bx_2 = \left( \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right),\ \bx_3 = \left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right),$$
so that the angles between these vectors are
$$\theta_{12} = \theta_{13} = \theta_{23} = \pi/2.$$
Notice that both of these lattices are well-rounded, and so in fact the bounds of Corollary \ref{solid_wr} are sharp.
\end{rem}
\bigskip

Next we consider the situation when (\ref{cos-1}) is not satisfied, in which case we can derive a lower bound for the solid angle formed by the minimal basis vectors of a lattice depending on the ratios of successive minima.

\begin{lem} \label{solid_other} Let $\Lambda \subset \real^N$ be a lattice of rank 3, let $\bx_1,\bx_2,\bx_3$ be a minimal basis for $\Lambda$, corresponding to successive minima $\lambda_1, \lambda_2, \lambda_3$, as guaranteed by Lemma \ref{min_bas}, and let $\Omega$ be the solid angle  formed by these vectors. By (\ref{tij.0}), we can assume that
$$\pi/3 \leq \theta_{12}, \theta_{13} \leq \pi/2,\ \pi/3 \leq \theta_{23} \leq 2\pi/3,$$
and $\theta_{23}$ is as small as possible for these inequalities to hold. Suppose in addition that
\begin{equation}
\label{cos-1-2}
\cos^{-1} \left( \cos \theta_{12} + \cos \theta_{13} - 1 \right) < \theta_{23} \leq 2\pi/3.
\end{equation}
Let $K_{12}$, $K_{13}$, and $K_{23}$ be as in \eqref{KKK}, then $\min \{ K_{12}, K_{23} \} < 1$ and
\begin{equation}
\label{O-nu}
\Omega \geq \frac{1}{\pi} \tan^{-1} \left( \tan \nu\ \sqrt{\frac{1 - 3 \tan^2 \nu}{3 - \tan^2 \nu}} \right) > 0,
\end{equation}
where
\begin{equation}
\label{nu_def}
\nu = \frac{1}{4} \cos^{-1} \left( \frac{1}{2} \left( 2K_{12} K_{23} \cos \theta_{12} + 2K_{12} \cos \theta_{13} - K_{12}^2 K_{23} - K_{23}  \right) \right) < \frac{\pi}{6}.
\end{equation}
\end{lem}

\proof
We use the notation from the proof of Lemma \ref{solid_3}. The first implication of (\ref{cos-1-2}) is that $-\frac{1}{2} \leq \cos \theta_{23}  < -1 + \cos \theta_{12} + \cos \theta_{13}$, so in particular $\cos \theta_{12} + \cos \theta_{13} \geq 1/2$. Combining this observation with (\ref{tij.0}), we obtain
\begin{equation}
\label{tij_2}
\frac{\pi}{3} \leq \theta_{12} \leq \frac{\pi}{2},\ \frac{\pi}{3} \leq \theta_{13} \leq \cos^{-1} \left( \frac{1}{2} - \cos \theta_{12} \right),
\end{equation}
in addition to (\ref{cos-1-2}).

Another implication of (\ref{cos-1-2}) is that $\Lambda$ cannot be well-rounded, by Corollary \ref{solid_wr}. Therefore $\min \{ K_{12}, K_{23} \} < 1$, and thus $K_{13} < 1$. Notice that the vectors $\bx_1,\bx_2,\bx_1-\bx_2-\bx_3$ are linearly independent, and hence
\begin{eqnarray*}
\lambda_3^2 & \leq & \|\bx_1-\bx_2-\bx_3\|^2 \\
& = & \lambda_3^2 \left( 1 + K_{13}^2 + K_{23}^2 - 2K_{13} K_{23} \cos \theta_{12} - 2K_{13} \cos \theta_{13} + 2K_{23} \cos \theta_{23} \right).
\end{eqnarray*}
Therefore, since $\cos \theta_{13} \geq 1/2 - \cos \theta_{12}$, $\cos \theta_{12} \leq 1/2$, and $K_{13} = K_{12} K_{23}$,
\begin{eqnarray*}
\cos \theta_{23} & \geq & -\frac{1}{2K_{23}} \left( K_{13}^2 + K_{23}^2 - 2K_{13} K_{23} \cos \theta_{12} - 2K_{13} \cos \theta_{13} \right) \\
& = & -\frac{1}{2} \left( K_{12}^2 K_{23} + K_{23} - 2K_{12} K_{23} \cos \theta_{12} - 2K_{12} \cos \theta_{13} \right) \\
& \geq & -\frac{1}{2} \left( K_{12}^2 K_{23} + K_{23} - K_{12} + 2K_{12} (1 - K_{23}) \cos \theta_{12}  \right) \\
& \geq & -\frac{K_{23}}{2} \left( K_{12}^2  -  K_{12} + 1  \right) > - \frac{1}{2},
\end{eqnarray*}
since $\min \{ K_{12}, K_{23} \} < 1$ and $\max \{ K_{12}, K_{23} \} \leq 1$. Therefore
\begin{equation}
\label{cos_K}
\theta_{23} \leq \cos^{-1} \left( \frac{1}{2} \left( 2K_{12} K_{23} \cos \theta_{12} + 2K_{12} \cos \theta_{13} - K_{12}^2 K_{23} - K_{23}  \right) \right) < \frac{2\pi}{3}.
\end{equation}
\smallskip

Let now $\nu$ be as in (\ref{nu_def}). It can now be easily checked that the right hand side of (\ref{tan1.1}) is a decreasing function of $c$ when the constraints of (\ref{cos-1-2}) and (\ref{tij_2}) are satisfied (we use Sage mathematical software package \cite{sage} - see illustration in section~\ref{sage_code}), and $c \leq \nu < \pi/6$ by (\ref{cos_K}). Therefore for each fixed pair of values of $\alpha$ and $\beta$,
\begin{equation}
\label{tan1.4}
\tan \left( \alpha + \nu \right) \tan \left( \alpha - \nu \right) \tan \left( \nu + \beta \right) \tan \left( \nu - \beta \right) \leq \tan \left( \frac{\Omega}{4} \right)^2,
\end{equation}
when (\ref{cos-1-2}) holds. The left hand side of (\ref{tan1.4}) is now minimized when $\alpha$ is as small as possible. Notice that
$$4\alpha \geq 2\pi/3 + 4|\beta|,$$
and hence
\begin{equation}
\label{tan1.5}
 \tan \left( \frac{\Omega}{4} \right)^2 \geq \tan \left( \frac{\pi}{6} + |\beta| + \nu \right) \tan \left( \frac{\pi}{6} + |\beta| - \nu \right) \tan \left( \nu + \beta \right) \tan \left( \nu - \beta \right),
\end{equation}
where $0 \leq |\beta| \leq \pi/24$. The right hand side of (\ref{tan1.5}) is easily checked to be an increasing function of $|\beta|$, so to minimize take $|\beta| = 0$, therefore
\begin{equation}
\label{tan1.6}
\tan \left( \frac{\Omega}{4} \right)^2 \geq \left( \frac{1 - 3 \tan^2 \nu}{3 - \tan^2 \nu} \right) \tan^2 \nu,
\end{equation}
and (\ref{O-nu}) now follows from (\ref{tan1.6}), since  the left hand side of (\ref{tan1.6}) is an increasing function of $\Omega$, as was indicated in the proof of Lemma \ref{solid_3}.
\endproof
\bigskip

\section{Area of spherical triangles: proof of Theorem~\ref{sp_area}}
\label{sphere}

In this section we prove Theorem~\ref{sp_area}. Consider a spherical lattice-minimal triangle $T$ on the unit sphere centered at the origin in $\real^N$, and let $\bx_1,\bx_2,\bx_3$ be minimal vectors of some lattice $L$ corresponding to the vertices of this triangle. Let $\Omega$ be the three-dimensional solid angle spanned by $\bx_1,\bx_2,\bx_3$, measured in steradians (i.e. not normalized as in Remark~\ref{solid_measure}), then the measure of $\Omega$ is precisely the spherical two-dimensional area of $T$. Hence we want to show that $\Omega$ is greater or equal than the number in (\ref{sp1}).

Let $\Lambda = \spn_{\zed} \{ \bx_1,\bx_2,\bx_3 \}$. Then
$\bx_1,\bx_2,\bx_3$ are minimal vectors in $\Lambda$, since they are minimal vectors in $L$ and $\Lambda \subseteq L$, and
$$\|\bx_1\| = \|\bx_2\| = \|\bx_3\| = 1.$$
Since these vectors form a spherical triangle, they must be linearly independent, and so they form a minimal basis for $\Lambda$ with solid angle $\Omega$, in particular $\Lambda$ is WR with all the successive minima equal to 1. If the vertex angles $\theta_{12},\theta_{13},\theta_{23}$ corresponding to these vectors satisfy~(\ref{mn3.1}), then Theorem~\ref{main_3} implies the result with equality precisely in the case described in (\ref{sp_sharp}), which is a minimal basis for the normalized fcc lattice (see Remark~\ref{sharp}).

Now assume that the vertex angles do not satisfy~(\ref{mn3.1}). This means that at least two of the angles $\theta_{12}, \theta_{13}, \theta_{23}$ must lie in the interval $(\pi/2,2\pi/3]$. Since $\Lambda$ is WR we can reindex the three vectors as necessary, and so we can assume without loss of generality that
$$\pi/2 < \theta_{12} \leq \theta_{23} \leq 2\pi/3,\ \pi/3 \leq \theta_{13} \leq \theta_{12}.$$
Let $\alpha$, $\beta$, and $c$ be as in (\ref{ts1}), then
\begin{equation}
\label{sp_bounds}
\frac{5\pi}{24} < \alpha \leq \frac{\pi}{3},\ 0 \leq \beta \leq \frac{\pi}{12},\ \frac{\pi}{8} \leq c \leq \frac{\pi}{6}.
\end{equation}
Now $\Omega$ is given by (\ref{tan1.1}), and since it is proportional to the spherical area of our triangle, we want to understand how small can it be. The right hand side of (\ref{tan1.1}) is easily checked to be an increasing function of $c$ when the constraints of (\ref{sp_bounds}) are satisfied (we use Sage mathematical software package \cite{sage} - see illustration in section~\ref{sage_code}); also, as discussed in the proof of Lemma~\ref{solid_3}, to minimize $\Omega$ we need to maximize $|\beta|$, so we can take $\beta=\pi/12$. Therefore for each fixed value of $\alpha$ and $\beta$,
\begin{eqnarray}
\label{sp_tan1}
\tan \left( \frac{\Omega}{4} \right)^2 & \geq & \tan \left( \alpha + \frac{\pi}{8} \right) \tan \left( \alpha - \frac{\pi}{8} \right) \tan \left( \frac{\pi}{8}+ \beta \right) \tan \left( \frac{\pi}{8} - \beta \right) \nonumber \\
& \geq &  \tan \left( \alpha + \frac{\pi}{8} \right) \tan \left( \alpha - \frac{\pi}{8} \right) \tan \left( \frac{5\pi}{24} \right) \tan \left( \frac{\pi}{24} \right) \nonumber \\
& > & \tan \left( \frac{\pi}{3} \right) \tan \left( \frac{5\pi}{24} \right) \tan \left( \frac{\pi}{12} \right) \tan \left( \frac{\pi}{24} \right) > \tan \left( \frac{\pi}{12} \right)^3,
\end{eqnarray}
which means that $\Omega$ is greater than the number in (\ref{sp1}), and hence finishes the proof of the theorem.
\bigskip

\section{Appendix: Sage code}
\label{sage_code}

Here we present the Sage code illustrating the increasing/decreasing behavior of functions used in the arguments above. Here is the code corresponding to the proof of Lemma~\ref{solid_3}:

\begin{sageblock}
#auto
@interact
def plotter(x = slider(pi/3,pi/2), y = slider(pi/3,pi/2)):
        z = var('z')
        a, b, c = (x+y)/4, (x-y)/4, z/4
        expression(z) = tan(a+c)*tan(a-c)*tan(c+b)*tan(c-b)
        zmin, zmax = pi/3, min(2*pi/3,arccos(cos(x)+cos(y)-1))
        color = (1,.25,0)
        plot(expression, z, zmin, zmax, rgbcolor=color)
            .show(xmin=pi/2,xmax=2*pi/3,ymin=0,ymax=0.3)
\end{sageblock}

\noindent
Here is the code corresponding to the proof of Lemma~\ref{solid_other}:

\begin{sageblock}
#auto
@interact
def plotter(x = slider(pi/3,pi/2), y = slider(pi/3,pi/2)):
    if arccos(-1+cos(x)+cos(y)) <= 2*pi/3:
        z = var('z')
        a, b, c = (x+y)/4, (x-y)/4, z/4
        expression(z) = tan(a+c)*tan(a-c)*tan(c+b)*tan(c-b)
        zmin, zmax = arccos(-1+cos(x)+cos(y)), 2*pi/3
        color = (1,.25,0)
        plot(expression, z, zmin, zmax, rgbcolor=color)
            .show(xmin=pi/3,xmax=2*pi/3,ymin=0,ymax=.1)
    else:
       print 'Bad Domain'
\end{sageblock}

\noindent
Here is the code corresponding to the proof of Theorem~\ref{sp_area}:

\begin{sageblock}
#auto
@interact
def plotter(x = slider(pi/2,2*pi/3), y = slider(pi/3,2*pi/3)):
    if y <= x:
        z = var('z')
        a, b, c = (x+y)/4, (x-y)/4, z/4
        expression(z) = tan(a+c)*tan(a-c)*tan(c+b)*tan(c-b)
        zmin, zmax = x, 2*pi/3
        color = (1,.25,0)
        plot(expression, z, zmin, zmax, rgbcolor=color)
            .show(xmin=pi/3,xmax=2*pi/3,ymin=0,ymax=.5)
    else:
       print 'Bad Domain'
\end{sageblock}
\bigskip

{\bf Acknowledgment.} We would like to thank Achill Sch\"{u}rmann and the anonymous referees for their helpful comments on the subject of this paper, and Andrew Ohana for helping us with some computations in Sage.
\bigskip

\bibliographystyle{plain}
\bibliography{fukshansky_robins}

\end{document}


text:
\section{Discussion}

We have developed a natural and substantial extension of traditional label
spreading for semi-supervised learning that can incorporate a broad range of
nonlinearities into the spreading process. These nonlinearities come from mixing
functions that operate on higher-order information associated with the data.
Given the challenges in developing optimization results involving nonlinear
functions, it is remarkable that we can achieve a sound theoretical framework
for our expressive spreading process. We provided guarantees on convergence of
the iterations to a unique solution, and we showed that the process optimizes a
meaningful objective function.
Finally, the convergence result in Theorem~\ref{thm:general-convergence}
is more general than what we considered in our experiments. This provides a
new way to analyze nonlinearities in graph-based methods that we expect will be
useful for future research.

\subsection*{Acknowledgments} We thank Matthias Hein for supplying the code that implements the HTV algorithm in \cite{hein2013total}.
We thank David Gleich and Antoine Gautier for helpful discussions.
ARB is supported by NSF Award DMS-1830274,
ARO Award W911NF19-1-0057,
ARO MURI,
and JPMorgan Chase \& Co.
FT is partially supported by INdAM-GNCS.

This research was funded in part by JPMorgan Chase \& Co. Any views or opinions expressed herein are
solely those of the authors listed, and may differ from the views and opinions expressed by JPMorgan Chase
\& Co. or its affiliates. This material is not a product of the Research Department of J.P. Morgan Securities
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TITLE: Prove that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}} = 0$ without substituting
QUESTION [3 upvotes]: I found this possible solution:
Let $r^2=x^2 + y^2, x = r \cos(\theta)$ and $y=r\sin(\theta)$. Then:
$$
\begin{split}
\lim_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}}
 &= \lim_{r \to 0} \frac{(r \cos(\theta))^2}{\sqrt{r^2}} \\
 &= \lim_{r \to 0} \frac{r^2 \cos^2(\theta)}{r} \\
 &= \lim_{r \to 0} r \cos^2(\theta)
\end{split}$$
Since $\lvert \cos^2(\theta)\rvert \leq 1$ and $\lim\limits_{r \to 0} r = 0$, we have:
$$
\lim_{r \to 0} r \cos^2(\theta)
 = 0
 = \lim_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}}
$$
I learned this substitution in this class.
First of all I would like to know if my solution is possible, but I would also would like to see a prove that does not use this kind of substitution.

REPLY [1 votes]: You can consider that the following generalized limit:
$$\lim_{(x,y) \to (0,r)} \frac{x^2}{\sqrt{x^2 +y^2}}=0.$$
Observe that,
$$\begin{align}\frac{x^2}{\sqrt{x^2}}&=\frac{x^2}{|x|}=|x|\\
&≥\frac{x^2}{\sqrt{x^2 +y^2}}\\
&≥0\end{align}$$
This implies
$$\begin{align}0&=\lim_{(x,y) \to (0,0)}|x|\\
&=\lim_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}}
\end{align}$$
This also shows that
$$\begin{align}\lim_{(x,y) \to (0,0)}|x|&=\lim_{(x,y) \to (0,r)} \frac{x^2}{\sqrt{x^2 +y^2}}&\\
&=0.
\end{align}$$
where, $r\in\mathbb R$.


TITLE: Show that the set of all vectors in $S$ with non-negative components is a generating set.
QUESTION [0 upvotes]: $S = \{(x_1, ... ,x_n ) : x_1 = ... = x_m = 0\}$ for any fixed $m,\ 1 \leq m < n$
Show that the set of all vectors in $S$ with non-negative components is a generating set.
According to my proof set of all vectors is in generating set. How can I restrict my proof to only the set of non-negative components?

REPLY [1 votes]: Observe that every vector $(x_1, \dots, x_n)$ with $x_1 = \cdots = x_m = 0$ for some integer $1 \leq m \leq n - 1$ can be written as $x_1 \mathbf e_1 + \cdots + x_n \mathbf e_n,$ where the vectors $\mathbf e_i$ have a $1$ in the $i$th place and $0$s elsewhere. (One other way of saying this is that the $j$th entry of $\mathbf e_i$ is the Kronecker delta $\delta_{ij}.$) Consequently, the set $E = \{\mathbf e_1, \dots, \mathbf e_n \}$ generates $S.$ Consider the set $N$ of all vectors in $S$ with non-negative components. Certainly, the set $E \cup N$ generates $S,$ as $E$ alone generates $S.$ But in view of the fact that $E \subseteq N,$ we have that $E \cup N = N,$ hence $N$ generates $S,$ as desired. QED.


text:
\begin{document}

\begin{abstract}
We prove that  the flag kernel singular integral operators of  Nagel-Ricci-Stein on a homogeneous group are bounded on  $L^p$, $1<p<\infty$. The gradation associated with the kernels is the natural gradation of the underlying Lie algebra. Our main tools are the Littlewood-Paley theory and a symbolic calculus combined in the spirit of Duoandikoetxea and Rubio de Francia.
\end{abstract}

\maketitle
\section{Introduction}
Flag kernels on homogeneous groups have been introduced by Nagel-Ricci-Stein \cite{nagel} in their study of quadratic $CR$-manifolds. They can be regarded as a generalization of Calder\'on-Zygmund singular kernels with singularities extending over the whole of the hyperspace $x_1=0$, where $x_1$ is the top level variable. The definition is complex (see below), as it involves cancellation conditions for each variable separately. However, the descritption of flag kernels in terms of their Fourier transforms is much simpler and bears a striking resemblance to that of the symbols of convolution operators considered independently by  the author (in, e.g. \cite{arkiv2007}).

In Nagel-Ricci-Stein \cite{nagel} we find an $L^p$-boundedness theorem for the very special flag kernels where the associated gradation consists of commuting subalgebras of the underlying Lie algebra of the homogeneous group. The natural question of what happens if the gradation is the natural gradation of the homogeneous Lie algebra is left open. The aim of this paper is to answer the question in the affirmative. We prove that such flag kernels give rise to bounded operators.

The smooth  symbolic calculus mentioned above has been adapted to an extended class of flag kernels of small (positive and negative) orders and combined with a variant of the Littlewood-Paley theory built on a stable semigroup of measures with smooth densities very similar to the Poisson kernel on the Euclidean space. The strong maximal function of Christ \cite{christ} is also instrumental. The  approach has been inspired by the well-known paper by Duoandicoetxea and Rubio de Francia \cite{duoandi}. The dependence of the present paper on Duoandicoetxea and Rubio de Francia \cite{duoandi} is evident throughout.

The class of flag kernels dealt with here is in fact an algebra. For this the reader is referred to  \cite{colloquium2010} where also the $L^2$-boundedness of flag kernels is proved solely by means of the symbolic calculus.

After this paper had been completed,  a preprint of Nagel-Ricci-Stein-Wainger \textit{Singular integrals with flag kernels on homogeneous groups I,} has been made available, where the $L^p$-boundedness theorem for flag kernels is proved. This comprehensive treatment of flag kernels on homogeneous groups has been announced for some time. Professor Stein has lectured a couple of times on the subject, see, e.g. \cite{stein}. The authors also use a version of Littlewood-Paley theory but otherwise the approach differs from the one presented here in many respects, the most important being our use of the symbolic calculus and partitions of unity related to a stable semigroup of measures. That is why we believe that what  is presented here has an independent  value and may count as a contribution to the theory.

\section{Preliminaries}
Let $\Ge$ be a nilpotent Lie algebra with a fixed Euclidean structure and $\Ges$ its dual.  Let $\delta_tx=tx$, $t>0$ be a family of  dilations on $\Ge$ and let
\[
\Ge_j=\{x\in\Ge: \delta_tx=t^{p_j}\cdot x\},
\hspace{2em}
1\le j\le d,
\]
where $1=p_1< p_2<\dots <p_d$. Denote by
\[
Q_j=p_j\cdot\dim\Ge_j
\]
the homogenous dimension of $\Ge_j$. The homogeneous dimension of $\Ge$ is
\[
Q=\sum_{j=1}^dQ_j.
\]

We have
\begin{equation}\label{grad}
\Ge=\bigoplus_{j=1}^d\Ge_j,
\qquad
\Ges=\bigoplus_{j=1}^d\Ges_j
\end{equation}
and
\[
[\Ge_i,\Ge_j]\subset \left\{
\begin{array}{ll}
\Ge_k, & {\rm if \ }  p_i+p_j=p_k,\cr \{0\}, & {\rm if \ } p_i+p_j\notin{\mathcal{P},}
\end{array}
\right.
\]
where $\mathcal{P}=\{p_j:1\le j\le d\}$.

Let
\[
x\to|x|\approx\sum_{j=1}^d\|x_j\|^{1/p_j}
\]
be a homogeneous norm on $\Ge$ smooth away from the origin. Let also
\[
|x|_j=|(x_1,x_2,\dots,x_j,0,\dots,0)|,
\qquad
1\le j\le d.
\]
In particular, $|x|_1=|x_1|$, and $|x|_d=|x|$. Another notation will be applied to $\Ges$. For $\xi\in\Ges$,
\[
|\xi|_j=|(0,\dots,0,\xi_j,\xi_{j+1},\dots,\xi_d)|,
\qquad
1\le j\le d.
\]
In particular, $|\xi|_1=|\xi|$, and $|\xi|_d=|\xi_d|$.

We shall also regard $\Ge$ as a Lie group with the
Campbell-Hausdorff multiplication
\[
xy=x+y+r(x,y),
\]
where $r(x,y)$ is the (finite) sum of terms of order at least $2$ in the Campbell-Hausdorff
series for $\Ge$. Under this identification the homogeneous ideals
\[
 \Ge^{(k)}=\bigoplus_{j=k}^d\Ge_j
\]
are normal subgroups.

In expressions like $D^{\alpha}$ or $x^{\alpha}$ we shall use multiindices
\[
\alpha=(\alpha_1,\alpha_2,\dots,\alpha_d),
\]
where
\[
\alpha_k=(\alpha_{k1},\alpha_{k1},\dots,\alpha_{kn_k}),
\qquad
n_k=\dim\Ge_k=\dim\Ges_k,
\]
are themselves multiindices with positive integer entries corresponding to the spaces $\Ge_k$ or $\Ges_k$. The homogeneous length of $\alpha$ is defined by
\[
|\alpha|=\sum_{k=1}^d|\alpha_k|,
\qquad|\alpha_k|=p_k(\alpha_{k1}+\alpha_{k2}+\dots+\alpha_{kn_k}).
\]

The Schwartz space of smooth functiions which vanish rapidly at infinity along with their derivatives will be denoted by $\Schw(\Ge)$. For a tempered distribution $K$, that is a continuous linear functional on $\Schw(\Ge)$, we shall write
\[
 \langle K,f\rangle=\int_{\Ge}f(x)K(x)\,dx,
\qquad
f\in\Schw(\Ge),
\]
without implying thereby that $K$ is a locally integrable function.

 Even though the flag kernels are our prime concern here we need a broader class of kernels to properly deal with them. In \cite{studia2010}, we proposed a natural generalization of  the flag kernels of Nagel-Ricci-Stein. Let
\[
 \|f\|_{(k)}=\max_{|\alpha|\le Q_k+1}\sup_{x\in\Ge_k}(1+|x|)^{Q_k+1}|D^{\alpha}f(x)|
\]
be a fixed norm in the Schwartz space $\Schw(\Ge_k)$. Let
\[
\mathcal{N}=\{\nu=(\nu_1,\nu_2,\dots,\nu_d): |\nu_k|<Q_k, \ 1\le k\le d\}.
\]
Let $\nu\in\mathcal{N}$. We define the class $\Fe(\nu)$ by induction on the homogeneous step $d$. When $d=0$ the elements of $\Fe(\emptyset)$ are simply constants. If $d\ge1$, we say that a distribution $K\in\Schw^{\star}(\Ge)$ is in $\Fe(\nu)$ if it is smooth away from the hyperspace $x_1=0$ and satisfies the following conditions:

i) For every multiindex $\alpha$,
\begin{equation}\label{size}
|D^{\alpha}K(x)|\le C_{\alpha}|x|_1^{-\nu_1-Q_1-|\alpha_1|}|x|_2^{-\nu_2-Q_2-|\alpha_2|}\dots
|x|_d^{-\nu_d-Q_d-|\alpha_d|}
\end{equation}
for $x_1\neq0$;

ii) For any $1\le k\le d$,
\begin{equation}\label{cancellation}
<K_{R,\phi},f>=R^{-\nu_k}\int_{\Ge}\phi(Rx_k)f(x_1,\dots,x_{k-1},x_{k+1},\dots,x_d)K(x)\,dx
\end{equation}
is in $\Fe(\nu_{(k)})$ on $\oplus_{j\neq k}\Ge_j$, where $\nu_{(k)}=(\nu_1,\dots,\nu_{k-1},\nu_{k+1},\dots,\nu_d)$, and this is uniform in $\phi\in\Schw(\Ge_1)$ with $|\phi\|_{(k)}\le1$ and $R>0$.  (Note that the meaning  of \textit{uniform boundedness} of a family of members of $\Fe(\nu)$ is obvious in the case $d=0$ and, for $d\ge1$, can be defined by induction.)

 For every $N$, we define a norm $\|\cdot\|_{\nu,N}$ in $\Fe(\nu)$ as the maximum of all the bounds occurring in the definition. First, we let
\[
 s_N^{\nu}(P)=\max_{|\alpha|\le N}\sup_{x_1\neq0}\prod_{k=1}^d|x|_k^{Q_k+\nu_k+|\alpha_k|}|D^{\alpha}K(x)|.
\]
and, if $d=1$,
\[
 \|K\|_{\nu_1,N}=s_N^{\nu_1}(K)+\sup_{|\phi\|_{(1)}\le1}\sup_{R>0}R^{-\nu_1}|<K,\phi\circ\delta_R>|.
\]

If $d>1$, we let
\[
 \|K\|_{\nu,N}=s_N^{\nu}(K)+\max_{1\le k\le d}\sup_{\|\phi\|_{(k)}\le1}\sup_{R>0}\|K_{R,\phi}\|_{\nu_{(k)},N}.
\]
Thus, $\Fe(\nu)$ can be regarded as a locally convex topological vector space. Let us remark that  $\Fe(0)=\Fe(0,0,\dots,0)$ is exactly the class of flag kernels of Nagel-Ricci-Stein \cite{nagel} (see Corollary 3.7 of \cite{studia2010}).

For a $K\in\Schw^{\star}(\Ge)$, let
\[
<\widetilde{K},f>=\int_{\Ge}f(x^{-1})K(dx),
\qquad
f\in\Schw(\Ge).
\]
The following three propositions have been proved in \cite{colloquium2010} and \cite{studia2010}.

\begin{proposition}[(Theorem 2.5 of \cite{colloquium2010})]\label{l2bound_flag}
 Let $K\in\Fe(0)$ be a flag kernel on $\Ge$. The convolution operator
$f\to f\star\widetilde{K}$
defined initially on $\Schw(\Ge)$ extends uniquely to a bounded operator on $L^2(\Ge)$. \end{proposition}

\begin{proposition}[(Proposition 1.5 of \cite{studia2010})]\label{gen_flag}
Let $\nu\in\mathcal{N}$. A distribution $K$ is in $\Fe(\nu)$ if and only if its Fourier transform is locally integrable, smooth for $\xi_d\neq0$, and satisfies
\begin{equation}\label{multiplier}
|D^{\alpha}\widehat{K}(\xi)|\le C_{\alpha}|\xi|_1^{\nu_1-|\alpha_1|}\dots|\xi|_d^{\nu_d-|\alpha_d|},
\qquad
\xi_d\neq0.
\end{equation}
\end{proposition}
\noindent
Cf. also the original Theorem 2.3.9 of Nagel-Ricci-Stein \cite{nagel} for kernels $K\in\Fe(0)$.
\begin{proposition}[Theorem 4.8 of \cite{studia2010}]\label{composition_flag}
Let $\nu,\mu,\nu+\mu\in\mathcal{N}$. Let $K\in\Fe(\nu)$, $L\in\Fe(\mu)$. Let $\phi=\otimes_{k=1}^d\phi_k\in C_c^{\infty}(\Ge)$ be equal to $1$ in a neighbourhood of $0$. There exists a $P=P_{K,L}\in\Fe(\nu+\mu)$ such that
\[
P=\lim_{\e\to0}K_{\e}\star L
\]
in the sense of distributions, where
\[
<K_{\e},f>=\int_{\Ge}\phi(\e x)f(x)K(dx),
\qquad
f\in\Schw(\Ge).
\]
Moreover, the mapping $(K,L)\to P_{K,L}$ is continuous.
 \end{proposition}

\section{Semigroups of measures}

Following Folland-Stein \cite{folland}, we say that a function $\phi$ belongs to the class $\mathcal{R}(a)$, where $a>0$, if it is smooth and
\begin{equation}\label{rclass}
|D^{\alpha}\phi(x)|\le C_{\alpha}(1+|x|)^{-Q-a-|\alpha|},
\qquad
{\rm all \ } \alpha.
\end{equation}

\begin{proposition}\label{fourier}
Let $\phi\in\mathcal{R}(a)$ for some $0<a<1$ and let $\int\phi=0$. Then $\phi\in\Fe(a)$.
\end{proposition}
\begin{proof}
The size condition (\ref{size}) follows by (\ref{rclass}). To verify the cancellation condition (\ref{cancellation}) let $f\in\Schw(\Ge)$ and $R>0$. Then
\begin{align*}
\int_{\Ge}&f(Rx)\phi(x)\,dx=\int_{\Ge}\big(f(Rx)-f(0)\big)\phi(x)\,dx \\
&\le\int_{|x|\le R^{-1}}\big(f(Rx)-f(0)\big)\phi(x)\,dx
+\int_{|x|\ge R^{-1}}\big(f(Rx)-f(0)\big)\phi(x)\,dx \\
&\le\|f\|\left(R\int_{|x|\le R^{-1}}|x|^{-Q-a+1}\,dx+2\int_{|x|\ge R^{-1}}|x|^{-Q-a}\,dx\right)\\
&\le CR^a\|f\|,
\end{align*}
where $\|\cdot\|$ is a Schwartz class norm.
\end{proof}

Let
\[
\langle P,f\rangle
=\lim_{\e\to}\int_{|x|\ge\e}\Big(f(0)-f(x)\Big)\frac{dx}{|x|^{Q+1}},
\qquad
f\in\Schw(\Ge).
\]
The distribution $P$ is an infinitesimal generator of a continuous semigroup of probability measures with smooth densities
\[
h_t(x)=t^{-Q}h(t^{-1}x),
\]
where $h\in\eR(1)$ and $P^Nh\in\eR(N)$ for $N=1,2,\dots$. In other words,
\[
h_t\star h_s=h_{t+s},
\qquad
t,s>0.
\]
and
\[
\frac{d}{dt}\big|_{t=0}<h_t,f>=-<P,f>,
\qquad
f\in\Schw(\Ge),
\]
The operator $Pf=f\star P$ is essentially selfadjoint with $\Schw(\Ge)$ for its core domain. The reader is referred to \cite{inventiones1986} for proofs and details.

For $0<{a}<1$
\begin{equation}\label{m}
\langle P^{a},f\rangle=\frac{1}{\Gamma(-{a})}\int_0^{\infty}t^{-1-{a}}\langle
\delta_0-h_t, f\rangle\,dt=\frac{1}{\Gamma(1-{a})}\int_0^{\infty}t^{-{a}}\langle
Ph_t, f\rangle\,dt
\end{equation}
defines a homogeneous distribution smooth away from the origin (cf., e.g. Yosida \cite{yosida}).

\begin{proposition}\label{pm}
For every $0<a<1$,
\[
P^{a}h\in\eR(a)
\qquad
{\rm and}
\qquad
\int_{\Ge}P^{a}h(x)\, dx=0.
\]
\end{proposition}

\begin{proof}
By (\ref{m}),
\[
P^{a}h(x)=\frac{1}{\Gamma(1-{a})}\int_0^{\infty}t^{-{a}}Ph_{t+1}(x)\,dt,
\]
whence
\begin{align*}
|D^{\alpha}P^{a}h(x)|&\le\frac{C_{\alpha}}{\Gamma(1-{a})}\int_0^{\infty}\frac{t^{-{a}}\,dt}{(t+1+|x|)^{Q+1+|\alpha|}}\\
&\le C_{\alpha}'\int_0^{\infty}\frac{t^{-{a}}\,dt}{(\frac{t}{1+|x|}+1)^{Q+1+|\alpha|}}\cdot(1+|x|)^{-Q-1-|\alpha|}\\
&\le C_{\alpha}''\int_0^{\infty}\frac{t^{-{a}}\,dt}{(t+1)^{Q+1+|\alpha|}}\cdot(1+|x|)^{-Q-{a}-|\alpha|},
\end{align*}
as required.

Now, for every $t>0$,
\[
\int h_t\,dx=1.
\]
Therefore,
\[
\int Ph_t\,dx=-\frac{d}{dt}\int h_t\,dx=0,
\qquad
t>0.
\]
which combined with (\ref{m}) gives the second part of the assertion.
\end{proof}

\section{Littlewood-Paley theory}
From now on we fix the function $\phi=P^{1/2}h_{1/2}$.
\begin{remark}\label{fixfi}
By the results of the previous section, $\phi$ is a smooth function satisfying the estimates
\begin{equation}\label{fiestim}
|D^{\alpha}\phi(x)|\le C_{\alpha}(1+|x|)^{-Q-1/2-|\alpha|}.
\end{equation}
Moreover, $\phi\in\Fe(1/2)$.
\end{remark}
\begin{lemma}\label{fi}
We have
\[
f=\int_0^{\infty}f\star\phi_t\star\phi_t\,\frac{dt}{t},
\qquad
f\in\Schw(\Ge).
\]
\end{lemma}
\begin{proof}
By the semigroup properties,
\[
 -\frac{d}{dt}f\star h_t=f\star Ph_t=\frac{1}{t}f\star(\phi_t\star\phi_t),
\]
whence
\[
\int_{\e}^Mf\star\phi_t\star\phi_t\frac{dt}{t}=f\star h_{\e}- f\star h_M.
\]
Now, if $\e\to0$ and  $M\to\infty$, the expression on the right hand side tends to $f$ in the sense of distributions.
\end{proof}

Let $T=(t_1,\dots,t_d)\in\R_+^d$. We shall regard $\R_+^d$ as a product of copies of the multiplicative group $\R^+$. We shall write
\[
T^{a}=(t_1^{a},\dots, t_d^{a}),
\qquad
TS=(t_1s_1,\dots,t_ds_d),
\qquad
\frac{dT}{T}=\frac{dt_1\dots dt_d}{t_1\dots t_d},
\qquad
a\in\R.
\]

 Let $\phi_k$ be the counterpart of  $\phi$  for $\Ge$ replaced by $\Ge^{(k)}$, $1\le k\le d$. Let
\[
\Phi_k=\delta_k\otimes\phi_k,
\]
where $\delta_k$ stands for the Dirac delta at $0\in\oplus_{j=1}^{k-1}\Ge_j$. Let
\[
\Phi=\Phi_1\star\Phi_2\star\dots\star\Phi_d,
\]
and
\[
\Phi_T=(\Phi_1)_{t_1}\star\dots\star(\Phi_d)_{t_{d}},
\qquad
T\in\R_+^d.
\]
\begin{corollary}\label{unity}
We have
\[
\Phi\in|\Fe|(1/2):=\bigcap_{\e\in\{-1,1\}}\Fe(\e_1/2,\dots,\e_d/2).
\]
Furthermore,
\[
f=\int_{\R_+^d}f\star\Phi_T\star\widetilde{\Phi_T}\,\frac{dT}{T},
\qquad
f\in\Schw(\Ge).
\]
\end{corollary}
\begin{proof}
By  Remark \ref{fixfi},
\[
\Phi_k\in\Fe(0,\dots0,1/2,0,\dots,0)\cap\Fe(0,\dots0,-1/2,0,\dots,0),
\]
where the only nonzero term stands on the $k$-th position. Therefore the  first part of our assertion
follows by Proposition \ref{composition_flag}. The second one is a consequence of Lemma\nobreak \ \ref{fi}.
\end{proof}
\begin{proposition}\label{paley}
The Paley-Littlewood square function
\[
G_{\Phi}(f)(x)=\left(\int_{\R_+^d}|f\star\Phi_T(x)|^2\,\frac{dT}{T}\right)^{1/2},
\]
is bounded as an operator on $L^p(\Ge)$. In other words, for every $1<p<\infty$, there is a constant $C_{\phi,p}>0$ such that
\[
\|G_{\Phi}(f)\|_p\le C_{\phi,p}\|f\|_p,
\qquad
f\in\Schw(\Ge).
\]
\end{proposition}
\begin{proof}
The proof is implicitly contained in Folland-Stein \cite{folland} (see Theorem 6.20.b and Theorem 7.7) so we dispense ourselves with presenting all details.

We start with defining some Hilbert spaces and operators. Let $X_0=\C$ and
\[
X_k=L^2(\R^k_+,\frac{dT}{T}),
\qquad
1\le k\le d.
\]
For a given $x\in\Ge$, let $F_k(x):X_{k-1}\to X_k$ be given by
\[
F_k(x)m(t_1,\dots,t_{k-1},t_k)=(\phi_k)_{t_k}(x_k,\dots,x_d)m(t_1,\dots,t_{k-1}).
\qquad
m\in X_{k-1}.
\]
Finally, let $W_k:C_c(\Ge,X_{k-1})\to C_0(\Ge,X_k)$ be the operator
\[
W_kf(x)(T,t_k)=(f\star F_k)(x)(T,t_k)=\int_{\Ge^{(k)}}(\phi_k)_{t_k}(y)f(xy)(T)\,dy,
\]
where $T=(t_1,\dots,t_{k-1})$. Note that $W_k$ acts only on $(x_k,\dots,x_d)$-variable.

We claim that
$$
W_k:L^2(\Ge,X_{k-1})\to L^2(\Ge,X_k)
$$
is an isometry. In fact, by definition of $\Phi_k$,
\begin{align*}
\|W_kf\|_{L^2(\Ge,X_k}^2&=\int_{\Ge}\|W_kf(x)\|_{X_k}^2\,dx\\
&=\int_{\Ge}dx\int_0^{\infty}\frac{dt}{t}
\int_{\R^{k-1}_+}\frac{dT}{T}\int_{\Ge^{(k)}}|(\phi_k)_t(y)f(xy)(T)|^2\,dy\\
&=\int_{\R^{k-1}_+}\frac{dT}{T}\int_0^{\infty}\frac{dt}{t}
\int_{\Ge}\int_{\Ge^{(k)}}|(\phi_k)_t(y)f(xy)(T)|^2\,dydx,\\
&=\int_{\R^{k-1}_+}\frac{dT}{T}\int_0^{\infty}\frac{dt}{t}<f_T\star(\Phi_k)_t,f_T\star(\Phi_k)_t>
=\|f\|_{L^2(\Ge,X_{k-1})}^2,
\end{align*}
where $f_T(x)=f(x)(T)$.

Another property of $W_k$ that is needed is the following.  For every $\alpha$
\begin{equation}\label{Kestimate}
\|D^{\alpha}F_k(x)\|_{(X_{k-1},X_k)}\le C_{\alpha}|x|_k^{-Q-|\alpha|}.
\end{equation}
This follows readily from (\ref{fiestim}) specialized to $\phi_k$:
\[
 |D^{\alpha}\phi_k(x)|\le C_{\alpha}(1+|x|_k)^{-Q-1/2-|\alpha|}.
\]
As a bounded operator from $L^2(\Ge,X_{k-1})$ to $L^2(\Ge,X_k)$ satisfying (\ref{Kestimate}) is $W_k$  a vector-valued kernel of type $0$, and, by Theorem 6.20.b of Folland-Stein \cite{folland},  maps $L^p(\Ge,X_{k-1})$ into $L^p(\Ge,X_k)$ boundedly for every $1<p<\infty$.

This implies our assertion. In fact,
\[
G_{\Phi}(f)(x)=\|f\star F_1\star\dots\star F_d(x)\|_{X_d},
\]
and therefore
\[
 \|G_{\Phi}(f)\|_{L^p(\Ge)}=\|T_dT_{d-1}\dots T_1f\|_{L^p(\Ge,X_d)}
\le C\|f\|_{L^p(\Ge,X_0)}=C\|f\|_p.
\]
\end{proof}

A word of comment on the symbol $\Phi_T$ would be appropriate here. The notation may suggest that the functions $\Phi_T$ are  dilates of a single function. They are not, but they have estimates of this form, which is our justification. The same applies to the symbol $K_T$ below. In the next section we are going to use the same notation for the ``real'' dilates of a function. We hope the reader will not get confused.

\section{The strong maximal function}

For a function $F$ on $\Ge$ and a $T\in\R^d_+$, let
\[
F_T(x)=F_{(t_1,t_2,\dots t_d)}(x)=t_1^{-Q_1}t_2^{-Q_2}\dots t_d^{-Q_d}F(t_1x_1,t_2x_2,\dots,t_dx_d).
\]

\textit{The strong maximal function} on $\Ge$ is defined by
\[
{\bf M}f(x)=\sup_{T\in\R_+^d}\int_{|y|\le1}|f(x(Ty)^{-1})\,dy=\sup_{T}|f\star(\chi_B)_T(x)|,
\]
where $\chi_B$ stands for the characteristic function of the unit ball $B=\{x\in\Ge:|x|\le1\}$, and $Ty=(t_1y_1,\dots,t_dy_d)$.  A theorem of Michael Christ asserts that for every $1<p<\infty$ there exists a constant $C>0$ such that
\[
\|{\bf M}f\|_p\le C\|f\|_p,
\qquad
f\in L^p(\Ge),
\]
that is, ${\bf M}$ is of $(p,p)$ type (see Christ \cite{christ}).

We shall need the following corollary to the Christ theorem. Let
\[
\gamma(t)=\min\{t,t^{-1}\},
\qquad
t>0.
\]
\begin{corollary}\label{mchrist}
Let
\[
F(x)=\Pi_{j=1}^d\gamma(|x_j|)^a|x_j|^{-Q_j},
\qquad
x\neq0,
\]
for some $a>0$.Then the maximal fuction
\[
M_Ff(x)=\sup_{T\in\R_+^d}|f\star F_T(x)|
\]
is of $(p,p)$ type for $1<p<\infty$.
\end{corollary}

\begin{proof}
Let $B_j$ be the unit ball in $\Ge_j$ and let $|B_j|$ be the Lebesgue measure of $B_j$. Let $D=B_1\times\dots\times B_d$. Then for every simple positive function $h\le F$ of the form
\[
h(x)=\sum_{R}c_R\chi_{D}(R^{-1}x),
\qquad
R=(r_1,r_2,\dots,r_d)\in\R_+^d,
\]
we have
\[
h_T(x)=\sum_{R}c_Rr_1^{Q_1}r_2^{Q_2}\dots r_d^{Q_d}(\chi_D)_{RT}(x)
=\frac{C\|h\|_1}{|D|}(\chi_D)_{RT}(x),
\]
and therefore
\[
M_Ff(x)\le\frac{C\|F\|_1}{|D|}{\bf M}f(x),
\]
which completes the proof.
\end{proof}

\section{Flag kernels}

We keep the notation established in previous sections.
\begin{lemma}
Let
\[
K_{T,S}=\widetilde{\Phi_{TS}}\star K\star{\Phi}_{T},
\qquad
T,S\in\R^d_+.
\]
Then $K_{T,S}\in\Fe(0)$ uniformly, and satisfy the estimates
\begin{equation}\label{F0}
|D^{\alpha}\widehat{K}_{T,S}(\xi)|\le C_{\alpha}\gamma(S)^{1/2}|\xi|_1^{-|\alpha_1|}\dots|\xi|_d^{-|\alpha_d|},
\end{equation}
where
\[
 \gamma(S)=\gamma(s_1)\gamma(s_2)\cdots\gamma(s_d).
\]
\end{lemma}

\begin{proof}
By the first part of Corollary \ref{unity}, $\Phi_T\in|\Fe|(1/2)$ with bounds uniformly proportional to $\gamma(T)^{1/2}$. Note that
\[
\gamma(TS)\le\gamma(T)\cdot\gamma(S).
\]
Thus,  our assertion follows by  Proposition \ref{composition_flag}.
\end{proof}

We let
$$
K_T=K\star\Phi_T,
\qquad
T\in\R_+^d.
$$

\begin{lemma}\label{maxi}
For every $T$, $K_{T}$ is an integrable function, and the maximal operator
\begin{equation}\label{max}
K_{\Phi}^{\star}f(x)=\sup_{T}|f\star|\widetilde{K}_{T}|(x)|
\end{equation}
is of type $(p,p)$ for all $1<p<\infty$.
\end{lemma}
\begin{proof}
Observe that by Proposition \ref{l2bound_flag}, $K_{T}\in L^2(\Ge)$ so it is a function. Moreover,
by Corollary \ref{unity} and Proposition \ref{composition_flag}, it is a smooth away from $x_1=0$, and satisfies
\[
|K_{T}(x)|\le C\gamma(T)^{1/2}
\gamma(|x|_1)^{1/2}|x|_1^{-Q_1}\dots\gamma(|x|_d)^{1/2}|x|_d^{-Q_d}
\]
uniformly in $T$ so that $K_{T}\le CF_T$, where $F_T$ is a dilate of
\[
F(x)=\gamma(|x|_1)^{1/2}|x|_1^{-Q_1}\dots\gamma(|x|_d)^{1/2}|x|_d^{-Q_d}.
\]
This shows that $K_{T}$ is integrable. The second part of our claim follows by Corollary \ref{mchrist} and the above.
\end{proof}

We turn to the main result of this paper. The reader may wish to compare the proof we give with that of  Theorem B  and the preceding lemma of Duoandicoetxea-Rubio de Francia \cite{duoandi}.

\begin{theorem}\label{main}
Let $K$ be a flag kernel on $\Ge$. Then  the singular integral operator
\[
f\to f\star\widetilde{K},
\qquad
f\in\Schw(\Ge),
\]
extends uniquely to a bounded operator on $L^p(\Ge)$ for all  $1<p<\infty$.
\end{theorem}

\begin{proof}
Let $f,h\in\Schw(\Ge)$. We have
\begin{align*}\label{decomposition}
<f\star\widetilde{K},h>
&=\int_{\R_+^d}\frac{dS}{S}\int_{\R_+^d}\frac{dT}{T}
<f\star\Phi_{T},h\star\Phi_{TS}\star\widetilde{\Phi_{TS}}\star K\star\Phi_T>\\
&=\int_{\R_+^d}\frac{dS}{S}\int_{\R_+^d}\frac{dT}{T}
<f_T,h_{TS}\star K_{T,S}>,
\end{align*}
where
\[
f_{T}=f\star{\Phi}_{T},
\qquad
h_{TS}=h\star\Phi_{TS},
\qquad
K_{T,S}=\widetilde{\Phi_{TS}}\star K\star\Phi_T.
\]

We are going to estimate
\[
<L_Sf,h>=\int_{\R_+^d}\frac{dT}{T}<f_{T},h_{TS}\star K_{T,S}>
\]
for a given $S$. Let us start with $L^2$-estimates. We have
\[
|<L_Sf,h>|
\le\left(\int_{\Ge}\int_{\R_+^d}|f_{T}(x)|^2\,\frac{dT}{T}dx\right)^{1/2}
\cdot\left(\int_{\Ge}\int_{\R_+^d}|h_{TS}\star K_{T,S}(x)|^2\,\frac{dT}{T}dx\right)^{1/2}.
\]

By (\ref{F0}) and Proposition \ref{l2bound_flag}, the operators $f\to f\star K_{T,S}$ are  bounded with norm estimates uniformly proportional to $\gamma(S)^{1/2}$ so that, by  Proposition \ref{paley},
\begin{align*}
|<L_Sf,h>|&\le C\gamma(S)^{1/2}\,\|G_{\Phi}(f)\|_2\,\|G_{\Phi}(h)\|_2\\
&\le C_1\gamma(S)^{1/2}\|f\|_2\|h\|_2,
\end{align*}
that is,
\begin{equation}\label{l2}
\|L_Sf\|_2\le C_1\gamma(S)^{1/2}\|f\|_2,
\qquad
f\in\Schw(\Ge).
\end{equation}
For $1<p<2$ and $f,h\in\Schw(\Ge)$,
\begin{align*}
|<L_Sf,h>|&
\le\int_{\Ge}\left(\int_{\R_+^d}|f_{T}(x)|^2\frac{dT}{T}\right)^{1/2}
\left(\int_{\R_+^d}|h_{TS}\star K_{T,S}(x)|^2\,\frac{dT}{T}\right)^{1/2}\,dx\\
&\le C_1\|G_{\Phi}(f)\|_p\left(\int_{\Ge}\left(\int_{\R_+^d}|h_{TS}\star K_{T,S}(x)|^2\,\frac{dT}{T}\right)^{q/2}dx\right)^{1/q}\\
&=C_2\|f\|_p\cdot\left\|\int_{\R_+^d}|h_{TS}\star K_{T,S}(\cdot)|^2\,\frac{dT}{T}\right\|_{q/2}^{1/2},
\end{align*}
where  $1/p+1/q=1$. Note that $q>2$. Thus, there exists a nonnegative function $u$ with $\|u\|_r=1$, where $2/q+1/r=1$, such that
\[
\left\|\int_{\R_+^d}|h_{TS}\star K_{T,S}(\cdot)|^2\,\frac{dT}{T}\right\|_{q/2}
=\int_{\Ge}\int_{\R_+^d}|h_{TS}\star K_{T,S}(x)|^2\,\frac{dT}{T}\cdot u(x)\, dx.
\]
Now,
\begin{align*}
h_{TS}\star K_{T,S}&=(h\star\Phi_{TS})\star(\widetilde{\Phi_{TS}}\star K\star\Phi_T)\\
&=(h\star\Phi_{TS}\star\widetilde{\Phi_{TS}})\star(K\star\Phi_T)=h'_{TS}\star K_T.
\end{align*}
Recall also that, by Lemma \ref{maxi}, $K_T$ are integrable functions. Therefore, by Lemma\nobreak\  \ref{maxi} again,
 \begin{align}\label{nomax}
\left\|\int_{\R_+^d}|h'_{TS}\star K_{T}(\cdot)|^2\,\frac{dT}{T}\right\|_{q/2}
&\le C_1\int_{\R_+^d}\int_{\Ge}|h'_{TS}|^2\star|K_{T}|(x)\cdot u(x)\, dx\frac{dT}{T}\notag\\
&\le C_2\int_{\Ge}\int_{\R^d}|h'_{TS}(x)|^2\,\frac{dT}{T}\cdot K_{\Phi}^{\star}u(x)\,dx\notag\\
&\le C_3\|G_{\Phi}(h)\|_q^2\cdot\|K^{\star}_{\Phi}u\|_r\le C_4\|h\|_q^2,
\end{align}
where we have used the estimate
 \begin{align*}
|h'_{TS}\star K_{T}(x)|^2&\le\left(\int_{\Ge}|h'_{TS}(xy^{-1})|\cdot|K_{T}(y)|^{1/2}\cdot|K_{T}(y)|^{1/2}\,dy\right)^2\\
&\le\int_{\Ge}|h'_{TS}|^2(xy^{-1})\cdot|K_{T}|(y)\,dy\cdot\int_{\Ge}|K_{T}(y)|\,dy\\
&\le C|h'_{TS}|^2\star|K_{T}|(x),
\end{align*}
the integrals
\[
\int_{\Ge}|K_{T}(x)|\,dx\le C
\]
being uniformly bounded, as can be seen from the proof of Lemma \ref{maxi}. Therefore,
\begin{equation}\label{lp}
\|L_Sf\|_p\le C_1\|f\|_p.
\end{equation}

\medskip
Now, by interpolating between (\ref{l2}) and (\ref{lp}), we get
\[
\|L_Sf\|_p\le C_2\gamma(S)^{\e_p}\|f\|_p,
\]
where $\e_p>0$ depends only on $p$, and, finally,
\[
\|f\star\widetilde{K}\|_p\le C_3\left(\int_{\R_+^d}\gamma(S)^{\e_p}\,\frac{dS}{S}\right)\cdot\|f\|_p
=C_4\|f\|_p,
\]
which proves our case for $1<p\le2$. The result for $2<p<\infty$ follows by duality.
\end{proof}

\section*{Acknowledgements}
I wish to extend thanks to Alexander Nagel, Fulvio Ricci, and Elias M. Stein for their interest in my work and an inspiring conversation. I am also indebted to Fran{\c c}ois Piquard for pointing out an editorial omission in the initial version of the manuscript.

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