text:
\begin{document}
\title{The C*-algebra of Boidol's group}

\author[Y.-F. Lin]{Ying-Fen Lin}
\address{Mathematical Sciences Research Centre,
  Queen's University Belfast, Belfast BT7 1NN, United Kingdom}
\email{y.lin@qub.ac.uk}

\author[J. Ludwig]{Jean Ludwig}
\address{Universit\'e de Lorraine, Institut Elie Cartan de Lorraine,  UMR 7502, Metz, F-57045, France}
\email{jean.ludwig@univ-lorraine.fr}

\keywords{C*-algebra, 4-dimensional exponential Lie groups, algebra of
operator fields, unitary spectrum}

\begin{abstract}
We consider Boidol's group,
whose Lie algebra is an extension of the Heisenberg Lie algebra by the reals
with the roots $1$ and $-1 $. It is the only connected
solvable Lie group of dimension less than or equal to four
whose group C*-algebra had not yet been determined.
In this paper we describe its C*-algebra as an algebra of
operator fields defined over the spectrum of the group.
\end{abstract}
\maketitle

\section{Introduction and notations}\label{intro}
Let $\A$ be a C*-algebra and $\widehat \A$ be its unitary spectrum. The
C*-algebra $ l^\iy(\wh\A) $  of all bounded operator fields defined over
$\widehat \A$ is given by
\begin{eqnarray}\label{}
 \nn  l^\iy(\widehat \A):=\{A=(A(\pi)\in\B(\H_\pi))_{\pi\in\widehat \A}; \no
A_\iy
:=\sup_{\pi}\noop {A(\pi)}<\iy\},
\end{eqnarray}
where $ \H_\pi $ is the Hilbert space of $\pi$. Let $\F$ be the Fourier
transform of $ \A $, i.e.,
\begin{eqnarray}\label{}
\nn \F(a):=\hat a:=(\pi(a))_{\pi\in\widehat \A} \quad \text{for} \quad a\in \A.
\end{eqnarray}
It is an injective, hence isometric, homomorphism from $ \A $ into $
l^\iy(\widehat
\A)$. {{Therefore}} one can analyse the C*-algebra $ \A $ by
recognising the elements of
$\F(\A)$ inside the (big) C*-algebra $l^\iy(\widehat \A)$.

We know that the unitary spectrum  $ \widehat{C^*(G)} $ of the C*-algebra $
C^*(G) $ of a locally compact group $G$ can
be identified with the unitary dual $ \widehat G $ of $ G $. Furthermore, if $G$
is an
\textit{exponential} Lie group, i.e., it is a connected, simply connected
solvable Lie group for which
the exponential mapping $\text{exp}:\g
\to G$ from the Lie algebra $ \g $ to its
Lie group $ G $ is a diffeomorphism, then the
Kirillov-Bernat-Vergne-Pukanszky-Ludwig-Leptin theory shows that there is a
canonical
homeomorphism $ K:\g^*/G \to \widehat G $ from the space of coadjoint orbits of
$ G$ in the linear dual space $ \g^* $ onto the unitary dual space $ \widehat G
$ of $ G $
(see \cite{Lep-Lud} for details and references). In this case, one can
identify the unitary spectrum $ \widehat{C^*(G)} $ of the C*-algebra
of an exponential Lie group with the space $ \g^*/G $ of coadjoint orbits of
the group $ G $. On the other hand, since connected Lie groups are second
countable, the algebra $
C^*(G) $ and its dual space $ \widehat G $ are then separable topological
spaces. This allows us
to work in $ \widehat G $ with sequences instead of nets.

The method of describing group
C*-algebras as algebras of
operator fields defined on the dual spaces  was first introduced in
\cite{Fe} and \cite{Lee1}.
{ In serial}, the C*-algebra of $ax+b$-like groups \cite{Lin-Lud},
the C*-algebras of the Heisenberg groups and of the threadlike groups
\cite{Lu-Tu},
the C*-algebras of the { affine automorphism} groups $G_{n,\mu} $
in
\cite{ILL1}, \cite{ILL2}, and the C*-algebra of the group $\T\ltimes H_1 $
 \cite{Lu-Re}  were all characterised as algebras of operator fields defined on
the
 corresponding spectrum of the groups.
In this way, the C*-algebra of every exponential Lie group
of dimension less than or equal to $4 $ has been explicitly determined with one
exception,
namely \emph{Boidol's group} $G $, which is an extension of the Heisenberg
group by
the reals with the roots $1 $ and $-1 $.
This group can be considered to be {\em wild} with respect to its group algebra
$\l1G $, since it is not star-regular (see~\cite{Boi, BLS}), is not hermitian
\cite{Boi1} and { it} does not have the
Wiener property (see~\cite{Mo}) as all the other simply connected solvable Lie
groups of dimension
less than or equal to $4$ do.

{ In this paper, we consider {{Boidol's group $G=\exp
\g$,  which is of dimension 4}}.{{ We can
write down { precisely} the dual space $\widehat G$ using the
structure of the space $
\g^*/G$ of coadjoint orbits}}  and determine its topology. {{We
decompose this orbit space}} into the union of a finite sequence
$(\Ga_i=S_i\setminus
S_{i-1})_{i= 0}^d$ {{of relatively closed }}subsets. On each of the
sets $\Gamma_i$, the orbit space topology is Hausdorff and the main question is
{{to understand the operator fields $ \hat a, a\in C^*(G)$}}, in
particular the behaviour of the operators $\hat{a}(\gamma)$, $\gamma\in
\Gamma_i$,  when $\gamma$ approaches elements in $S_{i-1}$. For each of these
sets $\Gamma_i$, we obtain different conditions for the C*-algebra.}

{ We first recall the following definitions which were given in
\cite{Be-Be-Lu}.}

\begin{definition}\label{lcd}\rm
Let $d$ be a natural number.
\begin{itemize}\label{}
\item  Let $S$ be  a topological space.
We say that $S $ is \textit{locally compact of step $\leq d$}, if
there exists a
finite increasing family $ \emptyset \ne  S_0\subset S_{1}\subset\cdots\subset
S_d=S $ of closed
subsets of $S $,  such that   the subsets $\Gamma_0=S_0$ and
$ \Gamma_i:=S_{i}\setminus S_{i-1}$,  $i=1,\dots, d$,
are  locally compact and Hausdorff in  their relative topologies.

\item   Let $S$ be    locally compact of step $\leq d$, and let $\{\H_i\}_{i=1,
\dots, d}$ be Hilbert spaces.
For a closed subset $ M \subset S $, denote by $ CB(M,\H_i) $ the unital
$C^*$-algebra of all uniformly
bounded operator fields  $ (\psi(\gamma)\in \B(\H_i))_{\gamma\in M\cap
\Gamma_i, i=1,\dots, d}$,
which are
operator norm continuous on the subsets $ \Gamma_i \cap M$ for every $
i\in\{0,\dots, d\} $ with  $ \Gamma_i\cap M\ne\emptyset $ and such that
$\ga\mapsto \psi(\ga) $ goes to 0 in operator norm if $\ga $ goes to infinity
on $M$.
We provide the algebra $ CB(M,\H_i) $ with the infinity-norm
\begin{align*}
\no{\varphi}_{M}=\sup\left\{\no{\varphi(\gamma)}_{\B(\H_i)}; M \cap \Gamma_i\ne
\emptyset, \, \gamma\in M \cap \Gamma_i\right\}.
\end{align*}
\item  Let $S $ be a set. For every $s\in S $, choose a Hilbert space $\H_s $.
We define the C*-algebra $l^\iy(S) $ of uniformly bounded operator fields
defined over $S $ by
\begin{eqnarray*}
 l^\iy(S):=\{(\ph(s))_{s\in S}; \ \ph(s)\in  \B(\H_s), s\in S, \ \sup_{s\in
S}\noop{\ph(s)}<\iy\}.
 \end{eqnarray*}
Here $\B(\H) $ denotes  the algebra of bounded linear operators on the Hilbert
space $\H $.
 \end{itemize}
\end{definition}

{ Let $\A$ be a C*-algebra, if $(\pi,\H) $ and $(\pi',\H') $ are two
equivalent irreducible
unitary representations of $\A $, we denote
by $U_{\pi,\pi'}:H_{\pi}\to \H_{\pi'} $ the corresponding  intertwining
operator (which is unique up to multiplication with some $\zeta \in\C  $ of
modulus 1). If $\va(\pi) $ is a bounded linear operator on $\H $ then
$\va(\pi') $ is the operator
\begin{eqnarray*}\label{}
 \nn \va(\pi'):=U_{\pi,\pi'}\circ  \va(\pi)\circ
U_{\pi,\pi'}^*{{\in \B(\H_{\pi'})}}.
 \end{eqnarray*}}

\begin{definition}\label{norcontspec}\rm
{Let $\A $ be a separable liminary C*-algebra such that
  the spectrum $\widehat{\A} $ of $\A$ is a locally compact space of step $\leq
d$,
  { with closed subsets $S_i$ of $\widehat{\A}$ and}
 \begin{align*}
  \emptyset= S_{-1}\subset  S_0\subset S_{1}\subset \cdots \subset
S_d=\widehat{\A}.
\end{align*}

 Denote by $\F:\A\to l^\iy(\wh \A) $ the Fourier transform of $\A $, i.e., for
$a\in \A $, let
\begin{eqnarray*}
 \F(a)(\ga)=\hat a(\ga):=\pi_\ga(a), \quad \ga\in \Ga_i, \, i= 0,\cdots, d.
 \end{eqnarray*}
We say that $\F(\A )$ is \textit{continuous of step} $\leq d $, if the set $
S_0 $ is the collection of all characters of $\A$ and if
for every $ \gamma\in \Gamma_i $  there is a Hilbert space $\H_i$ such that a
concrete realisation $ (\pi_\gamma,\H_i) $ of
$\gamma $ on the Hilbert space $ \H_i $ and
 $\F(\A)\res{\GA_i} $ is contained in $CB(\wh \A,\H_i) $, for every $0\leq
i\leq d $}.
 \end{definition}

\section{Boidol's group }\label{boidol group}

\subsection{}\label{Boidol def}
Let $\g $ be the real Lie algebra of dimension 4 with a
basis $\{ T,X,Y,Z\} $
and the non-trivial brackets
\begin{eqnarray*}
\nn  [ T,X]=  -X, \, [T,Y]= Y, \, [ X,Y]= Z.
 \end{eqnarray*}
The simply connected connected group
$G $ with Lie algebra $\g $, which we call Boidol's group,  can be realized on
$\R^4 $ with the multiplication
\begin{eqnarray*}
 (t,x,y,z)\cdot (t', x' , y' ,{z'})=(t+t', e^{t'}x+x',e^{-t'}y+y',
z+z'+\frac{1}{2}(e^{t'}xy'-e^{-t'}x'y)).
 \end{eqnarray*}
The inverse of
$(t,x,y,z) $ is given by
\begin{eqnarray*}
 (t,x,y,z)\inv=(-t,-e^{-t}x, -e^{t}y,-z).
 \end{eqnarray*}
 We see that the subgroup $\ZZ:=\{ (0,0,0,z); z\in\R\} $ is the centre of $G $.
{Furthermore
\begin{eqnarray*}
 G=\R\ltimes H,
 \end{eqnarray*}
is the semi-direct product of $\R $ acting on the Heisenberg group $H
=\{0\}\times \R^3$.}

{{\subsection{The coadjoint orbit space}
We have three different kinds of coadjoint orbits in $\g^* $.

1) The orbits in general position: For $\rh, \la\in\R\times \R^*$, let
$\ell=\ell_{\rh,\la}=(\rh,0,0,\la)\in\g^* $.
Then the stabilizer $G(\ell)  $ of $\ell $ in $G $ is given by
$G(\ell)=\exp{\R T}\cdot \exp{\R Z} $ and  the coadjoint orbit $O_{\rh,\la} $
of $\ell_{\rh,\la} $ is the subset

 \begin{eqnarray*}\label{}
 \nn O_{\rh,\la}=
 \{ (\frac{\rh\la +xy}{\la})T^*+xX^*+y Y^*+\la Z^*; x,y\in\R\}
 \end{eqnarray*}
of $\g^* $.

2) The orbits of dimension 2 vanishing on $Z $: For $(\al,\be)\in\R^2,
\al^2+\be^2\ne 0 $, the orbit
$O_{\al,\be,0}={{ \Ad^*(G)\ell_{\al,\be,0}}}$, of the
element $\ell_{\al,\be,0}= \al X^*+\be Y^* $ is given by
 \begin{eqnarray*}\label{}
 \nn O_{\al,\be,0} &= &
 \{ u T^*+(e^t\al )X^*+(e^{-t} \be) Y^*; t,u\in\R\}\\
 \nn  &= &
\{ u T^*+( \vert{\be }\vert {\al}e^t )X^*+sign(\be) (e^{-t}) Y^*; t,u\in\R\}\\
\nn  &= &
\{ u T^*+(sign({\al})e^t )X^*+( \vert{\al }\vert \be e^{-t}) Y^*; t,u\in\R\}.
 \end{eqnarray*}

3) The real characters: For every $\ta\in\R $ we have the real  character
$\ell_\ta:=\ta T^* $ of $\g
$.}}

This gives us the following partition of $\g^* $,
\begin{align*}
\g^*=\GA_3\bigcup\GA_2\bigcup\GA_1\bigcup\GA_0,
\end{align*}
where
\begin{enumerate}\label{}
\item[(i)]
$ \GA_3=\{O_{\rh,\la}; \rh\in\R,\la\in\R^*\}.$

 The subset
 \begin{eqnarray*}\label{}
 \nn S_3 :=  \{ \ell_{\rh,\la}\vert  (\rh,\la)\in \R\times \R^*\}
 \end{eqnarray*}
is a section for the $G $-orbits in $\GA_3 $.
\item[(ii)] Let
\begin{align*}
 \GA_{2,\ve,\si}:= \{O_{\rh \ve,\si,0};  \rh\in\R_{>0},\ve,\si\in\{+,-\}\},
 \end{align*}
and
\begin{align*}\label{}
 \nn \GA_{2}:= \bigcup_{\ve,\si\in\{ +,-\}}\GA_{2,\ve,\si}.
 \end{align*}
For $\om\in\R^*$ and $\si \in\{ +,-\} $, let
\begin{eqnarray*}
 \ell_{\om,\si,0}:=(0, \om,\si ,0)\in O_{\om,\si,0}.
 \end{eqnarray*}
 The subset $S_2$ of $\GA_2 $, { defined by}
 \begin{align*}\label{}
 \nn S_2:= \{ \ell_{\om,\si,0}; \si\in\{ +,-\},\om\in \R^*\},
 \end{align*}
is  a section for the the $G $-orbits in $\GA_2 $.
 \item[(iii)]

$ \GA_{1}:= \{O_{1,0,0},O_{-1,0,0}, O_{0,1,0},O_{0,-1,0} \}.$

The subset $S_1$ of $\GA_1$, { given by}
 \begin{align*}\label{}
 S_1:= \{ \ell_{\si,0,0},\ell_{0,\si,0}; \si\in\{ +,-\}\},
 \end{align*}
  is a section for the the $G $-orbits in $\GA_1 $.
 \item[(iv)] Finally let
\begin{align*}
 \GA_0:= \{\ell_\ta; \ta\in\R\},
 \end{align*}
where $\ell_\ta:=(\ta,0,0,0)$.
 \end{enumerate}

{\begin{remark}\label{closed}
\rm   It follows from the description of the coadjoint orbits that
 the orbits in $ \Ga_3$, $ \Ga_2$ and $ \Ga_0$ are closed in $ \g^*$, but
the 4 orbits in $ \Ga_1$  are not.
Since the canonical mapping $ K: \g^*/G\to \widehat{G} $  is a homeomorphism,
it follows that for every closed coadjoint orbit $ \Om\subset \g^*/G$  the
irreducible representations $ \pi_\Om=K(\Om)$  associated to $ \Om$  sends the C*-algebra
of $ G$  onto the algebra of compact operators $ \K(\H_{\pi_\Om})
$ on the Hilbert space $\H_{\pi_\Om} $  of $
\pi_\Om$.
\end{remark}}

\begin{proposition}\label{toporbitspace}
The orbit space topology is Hausdorff on $\GA_3 $. {{In}} $\GA_2 $
the subsets
$\GA_{2,\ve,\si}$, $\ve, \si\in \{+, -\}$, are open and
Hausdorff, $\GA_1 $ is discrete and $\GA_0 $
is homeomorphic to $\R $.
\end{proposition}

\begin{proof}The proof follows easily from the fact that a
sequence of coadjoint orbits $(O_k)_{k\in\N}\subset \g* $ converges to an orbit
$O $ if and only if for every $\ell\in O $ and every $k\in\N $ there exists an
$\ell_k\in O_k $ such that $\limk \ell_k=\ell $
(see \cite{Lep-Lud}).
\end{proof}

\begin{proposition}\label{limitGamma3toGamma2}
Let $\ol O:= (O_{{\rh_k,\la_k}})_k$ be a properly converging sequence in $\GA_3$
such that $\limk \la_k= 0$. Then
\begin{enumerate}\label{}
\item the sequence $(\om_k)_k:=(\rh_k\la_k)_k $ converges to some
$\om\in\R $;
\item  if $\om\ne 0 $, then the limit set $L $ of the sequence $\ol O $ is the
two point set $\{O_{\om,- 1,0}, O_{-\om,1,0}\} $;
\item  if $\om=0 $ then the  limit set $L $ is the subset $\GA_1\bigcup \GA_0 $
of the orbit space.

 \end{enumerate}

\end{proposition}
\begin{proof}
(1) Suppose that the sequence $(O_{\rh_k,\la_k})\subset \GA_3  $ converges to
some $O\in \GA_2\cup\GA_1\cup\GA_0 $.
Take { $\ell=t T^*+xX^*+yY^*\in 0 $}. Then we have for any $k\in\N $
an element
$\ell_k= (\frac{\rh_k\la_k +x_ky_k}{\la_k})T^*+x_kX^*+y_k Y^*+\la_k Z^*\in O_k$
which converges to $\ell $. In particular,
$\limk y_k=y, \limk x_k=x $ and $\limk \frac{\rh_k\la_k +x_ky_k}{\la_k}=t $.
Let
$\om:=-xy $.
Since $\lim_{k\to \infty} \la_k=0 $ it follows that $\limk \rh_k\la_k=\om $.

(2) Suppose that $\om\ne 0$. Let $O
$ be a limit point  of the sequence $(O_{\rh_k,\la_k})_k $. Then $O$ is
contained in $\GA_2\cup\GA_1\cup\GA_0. $ Let $\ell=tT^*+xX^*+yY^*\in O $.
Since $0\ne\om=-xy $ we have that $O\in\GA_2 $. We can assume that $t=0, \val
x= \vert{\om }\vert  $ and $\val y=1$. If $y=1 $, then $x=-\om $
and conversely if $y=-1 $ then $x=\om$.

(3) If $\om=0 $ then $\limk x_ky_k=0 $. Take $\ell=\si X^*\in \GA_1 $. We can
use  $\ell_k= \si X^*-\si \om_k Y^*+\la_k
Z^*=(\frac{\rh_k\la_k +\si (-\si \om_k)}{\la_k})T^*+\si
X^*-\si \om_k Y^*+\la_k Z^*\in O_k$ and $\om_k:=\rh_k\la_k$ for $k\in\N. $ Then $\limk
\ell_k=\ell $. Similarly for $\ell=\si Y^* $. For $\ell=\al T^*\in \GA_0 $, we
take $x_k:=- \sqrt{\val{\al \la_k-\om_k}}$ and $y_k:=\sqrt{\val{\al
\la_k-\om_k}}sign(\al \la_k-\om_k)$, provided  $\al \la_k-\om_k\ne 0 $.
Otherwise we use $x_k:= 0=:y_k$. Then $\ell_k={ \al T^*}+\la_k
Z^*\in O_k $ and
$\limk \ell_k=\ell $.
\end{proof}

The following proposition can be found in Theorem~2.3 \cite{Lin-Lud}.

\begin{proposition}\label{omiszero}
 Let $\ol O=(O_{\ve\rh_k,\si,0})_k, \rh_k>0$ and $\ve,\si\in\{ +,-\} $, be  a
properly converging sequence in $\GA_2 $ with limits in $\GA_0\cup\GA_1 $.

 \begin{enumerate}\label{}
\item Then $\limk \rh_k=0 $ and the limit set of $\ol O $ is $
\{O_{\ve,0,0},O_{0,\si,0}\}\cup \GA_0$.
\item  The closure of $O_{\si,0,0}$ (resp. $O_{0,\si,0}$) is $O_{\si,0,0}\cup
\GA_0 $ (resp. $O_{0,\si,0}\cup \GA_0 $) for $\si=+$ or $\si=- $.
 \end{enumerate}

 \end{proposition}

\section{The Fourier transform }\label{Fourier}
We let once and for all
\begin{eqnarray}\label{si ep defined}
 \si,\ve\in\{+,-\}.
 \end{eqnarray}
 \begin{definition}\label{pi a defined}
\rm
Let ${{\ga}}:G\to G$ be the automorphism
 \begin{align*}\label{}
 \nn {{\ga}}(t,x,y,z):=  (t,-x,-y,z).
 \end{align*}
For a representation $(\pi,\H) $ of $G$ and an automorphism
${{\al}}:
G\to G $, we define the new representation $(\pi^{{\al}} ,\H)$ of $G
$ by
\begin{eqnarray*}
 \pi^{{\al}}(g):=\pi({{\al}}\inv (g)), \quad g\in G.
 \end{eqnarray*}

 \end{definition}

\subsection{The irreducible representations}

\begin{enumerate}\label{}
\item[(i)]  For $(\rh,\la)\in \GA_3 $, let
\begin{eqnarray*}
 \pi_{\rh,\la}:=\ind P G\ch_{\rh,\la},
 \end{eqnarray*}
{where $P=\exp \p $, $\p:=\text{span}\{T,Y,Z\}$ and
$\ch_{\rh,\la}(\exp tT)\exp(y Y)\exp(z Z):=e^{-i (\rh t+ \la z )}$, $t, y,
z\in\R $.}
The Hilbert space $  L^2(G/P,
\ch_{\rh,\la}) $ of $\pi_{\rh,\la} $ can be realized as $\l2\R $.
Let $(t,x,y,z)\in G, \xi\in {L^2(\R)}, u\in\R $. Then
\begin{align*}
 \pi_{\rh,\la}(t,x,y,z)\xi(u)= e^{t/2}e^{-i\rh t}e^{-i\la z}e^{-i{\la
xy/2}{}}e^{i\la e^t yu}\xi(e^t u-x).
 \end{align*}
Hence, for $F\in\l1G $, $\xi\in L^2(\R),u\in\R, $ we have that
\begin{eqnarray}\label{kernel pirhola}
 \pi_{\rh,\la}(F)\xi(u)\nn  & = &
\int_G F(g)\pi_{\rh,\la}(g)\xi(u)dg\\
&=&
{\int_\R \big(\int_\R e^{t/2} e^{-i\rh t}\hat
F^{3,4}(t,x,\la(\frac{x}{2}-e^tu),\la)dt\big)\xi(e^tu-x)dxdt}\\
 \nn & = &
\int_\R \big(\int_\R e^{t/2} e^{-i\rh t}\hat
F^{3,4}(t,e^tu-x,-\frac{\la}{2}(x+e^tu),\la)dt\big)\xi(x)dx.
 \end{eqnarray}
 {Here $\hat F^{3,4} $ is defined by}
 {\begin{align*}\label{}
 \nn \hat F^{3,4}(t,x,a,b):= \int_{\R^2}F(t,x,y,z)e^{-i(ay+bz)}dydz, \quad
t,x,a,b\in\R.
 \end{align*}
}
Therefore  $\pi_{\rh,\la} $ is  a kernel operator with kernel function
\begin{eqnarray}\label{kerfunc rh la}
 F_{\rh,\la}(u,x)=
 \int_\R e^{t/2} e^{-i\rh t}\hat
F^{3,4}(t,e^tu-x,-\frac{\la}{2}(x+e^tu),\la)dt, \quad u, x\in\R.
 \end{eqnarray}
 {Furthermore, {for $s\in\R $,} we obtain the identity
\begin{align*}\label{}
 \nn \pi_{\rh,\la}^{{\ga}}(t,x,y,z){ \xi}(u)=
e^{t/2}e^{-i\rh t}e^{-i\la
z}e^{-i\la
xy/2}e^{-i\la e^t yu}\xi(e^t u+x),
 \end{align*}
which shows that }
\begin{eqnarray}\label{kernel pirhola s}
 \pi_{\rh,\la}^{{\ga}} (F)\xi(u)\nn  & = &
 \int_{R^4} e^{t/2}e^{-i\rh t}e^{-i\la z}e^{-i{\la xy/2}{}}e^{-i\la e^t
yu}F(t,x,y,z)\xi(e^t u+x)dzdydxdt\\
 & = &
\int_\R \big(\int_\R e^{t/2} e^{-i\rh t}\hat
F^{3,4}(t,-e^tu+x,\frac{\la}{2}(x+e^tu),\la)dt\big)\xi(x)dx.
 \end{eqnarray}
 Let $S:\l2\R\to{\l2\R} $ be the unitary operator defined by
 \begin{align*}\label{}
 \nn S(\xi)(u):= \xi(-u), \quad u\in\R.
 \end{align*}
Then
\begin{eqnarray}\label{pilarh s}
 S\circ  \pi_{\rh,\la}(t,x,y,z)\circ  S(\xi)(u)  &= &
e^{t/2}e^{-i\rh t}e^{-i\la z}e^{-i\la xy/2}{{e^{-i\la e^t yu}\xi(e^t
u+x)}}\\
  \nn&= &
\pi_{\rh,\la}^{{\ga}} (t,x,y,z)(\xi)(u), \quad (t,x,y,z)\in G.
 \end{eqnarray}

We shall need later in section \ref{NCDL} an  equivalent version
of the representations $\pi_{\rh,\la} $.

\begin{definition}\label{Vk pl defined}
\rm
 Let  $\et\in\l2{\R, \frac{dx}{\val x}}$ and $k\in \N$, the operator $V_{k}:\l2{\R, \frac{dx}{\vert x\vert }}  \to \l2{\R,dx}$ defined by
\begin{eqnarray*}
 V_{k}(\et)(s)=\xi_k(s):=\frac{1}{\vert s\vert ^{1/2}}\et(\vert \la_k \vert
s) e^{ i\va_k({ \la_k s )}}, \quad s\in \R,
\end{eqnarray*}
where $\va_k:\R\to \R $ is a measurable  function, is a unitary operator.
We have that
\begin{align*}
 V_{k}^*(\xi)(u)= \frac{\vert u\vert ^{1/2}}{\sqrt{\val
{\la_k}}}e^{- i\va_k(u)}\xi(\frac{u}{\vert \la_k\vert }), \quad u\in\R, \xi\in
\l2 {\R}.
\end{align*}
We can also view $V_k $ as an operator $V_k: \l2{\R_+, \frac{dx}{\val x}}\oplus
\l2{\R_-, \frac{dx}{\val x}}\to \l2{\R_+, {dx}}\oplus  \l2{\R_-, {dx}}$ for
$k\in\N $.
It is easy to see that
\begin{align}\label{ScommVk}
  S\circ  V_k= V_k\circ  S,
\end{align}
where $S:\l2{\R_\si, \frac{du}{ \vert{u }\vert }}\to \l2{\R_{-\si},
\frac{du}{ \vert{u }\vert }} $ is defined as before by
\begin{align*}\label{}
 \nn  S(\et)(u)=  \et(-u), \quad u\in\R_\si.
 \end{align*}
\end{definition}

Let us take now
\begin{eqnarray}\label{vak def}
 \nn \va_k(s):= e^{i\rh_k\ln( \vert{s }\vert )}, \, s\in\R.
\end{eqnarray}
For $ \et\in
C_c^\iy(\R_{>0}), s\in\R$,  and $ \hat
F^{3,4}\in C_c^\iy (\R^4) \subset \l1G$,  we see by {(\ref{kernel pirhola})} that
{\begin{eqnarray}\label{rh al circ Vk}
\pi_{\rh_k,\la_k}(F)(V_{k }(\et))(s)
\nn  &= &
\int_\R \int_\R e^{t/2} e^{-i\rh_k t}\hat
F^{3,4}(t,x,-\frac{\la_k}{2}(-x+2e^ts),\la_k)\\
 \nn & &
\frac{e^{i\rh_k\ln(\val{-\la_kx+\la_k e^ts})}\val{\la_k}^{1/2}}{\val{-\la_k
x+e^t \la_k s }^{1/2}}\et(-\vert \la_k\vert  x+\vert \la_k\vert  e^t s
)dxdt.
 \end{eqnarray}}
Hence, writing $\la_k=\ve  \vert{\la_k }\vert  $, we obtain:
 {\begin{eqnarray}\label{vkstar circ Vk}
 \nn V_{k}^*(\pi_{\rh_k,\la_k}(F)(V_{k }(\et)))(s)
 &= &
{e^{-i\rh_k\ln(\val s)}}\int_\R \int_\R  e^{-i\rh_k t}\hat
F^{3,4}(t,x,\frac{\la_k}{2}x-\ve {e^ts}{},\la_k)\\
 \nn & &
\frac{e^{i\rh_k\ln(\val{-\la_kx+  \ve e^ts})}\val{e^{t}s}^{1/2}}{\val{-\ve\la_k
x+  e^ts }^{1/2}}\et(- \ve\la_k x+  e^t s )dxdt\\
 &= &
\int_\R \int_\R  \hat F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{e^ts}{},\la_k)\\
 \nn & &
\frac{e^{i\rh_k(\ln(\val{-\ve\la_kx+
e^ts}-\ln(\val{e^{t}s}))}\val{e^{t}s}^{1/2}}{\val{-\ve\la_k x+
e^ts}^{1/2}}\et(- \ve\la_k x+ e^t s )dxdt.
 \end{eqnarray}}
 Furthermore, for the automorphism $ \ga$ and $s\in\R $, we have that
 {
 \begin{eqnarray}\label{vkstar s circ Vk}
 V_{k}^*(\pi_{\rh_k,\la_k}^{{\ga}} (F)(V_{k }(\et)))(s)
 \nn  &= &
{e^{-i\rh_k\ln(\val s)}}\int_\R \int_\R  e^{-i\rh_k t}\hat
F^{3,4}(t,x,\frac{\la_k}{2}x+\ve {e^ts}{},\la_k)\\
 \nn & &
\frac{e^{i\rh_k\ln(\val{\la_kx+  \ve e^ts})}\val{e^{t}s}^{1/2}}{\val{\ve\la_k
x+  e^ts }^{1/2}}\et(\ve\la_k x+  e^t s )dxdt\\
\nn
 &= &
\int_\R \int_\R  \hat F^{3,4}(t,x,\frac{\la_k}{2}x+\ve{e^ts}{},\la_k)\\
 \nn & &
\frac{e^{i\rh_k(\ln(\val{\la_kx+  \ve
e^ts}-\ln(\val{e^{t}s}))}\val{e^{t}s}^{1/2}}{\val{\la_k x+  \ve
e^ts}^{1/2}}\et(\ve\la_k x+ e^t s )dxdt\\
 &= &
\int_\R \int_\R  \hat F^{3,4}(t,-x,-\frac{\la_k}{2}x+\ve{e^ts}{},\la_k)\\
 \nn & &
\frac{e^{i\rh_k(\ln(\val{-\la_kx+  \ve
e^ts}-\ln(\val{e^{t}s}))}\val{e^{t}s}^{1/2}}{\val{-\la_k x+  \ve
e^ts}^{1/2}}\et(-\ve\la_k x+ e^t s )dxdt.
 \end{eqnarray}
 }
 \item[{(ii)}]  For $\ell\in\GA_2 \bigcup \GA_1$, let
\begin{eqnarray*}
 \pi_{\ell}:=\ind H G\ch_{\ell},
 \end{eqnarray*}
where $H:=\exp\h $ and $\h:=\text{span}\{X,Y,Z\} $.

Take $\ell=\ell_{\mu,\nu}:=(0,\mu,\nu,0) $ for some $\mu,\nu\in\R $ with
$\mu^2+\nu^2\ne 0 $ and write
\begin{eqnarray*}
 \pi_{\mu,\nu}=\pi_{\ell_{\mu,\nu}}.
 \end{eqnarray*}
We have that $L^2(G/H,\chi_\ell)\simeq\l2\R $ and for $(t,x,y,z)\in G, \va\in
L^2(\R) $ and $v\in\R $:
\begin{align*}
 \pi_{\ell}(t,x,y,z)\va(v) = e^{-i\mu e^{v-t}x}e^{-i \nu e^{t-v}y}\va(v-t).
 \end{align*}
Hence for $F\in\l1G $,
\begin{align*}
 \pi_{\ell}(F)\va(v) = \int_\R \hat F^{2,3,4}(v-t,\mu {e^{t}},\nu
{e^{-t}},0)\va(t)dt.
 \end{align*}
This means that  $\pi_\ell(F) $ is a kernel operator with kernel function
\begin{eqnarray}\label{kerfunc mu nu}
 F_{\ell}(u,t)= \hat F^{2,3,4}(v-t,\mu {e^t},\nu
{e^{-t}},0), \quad v,t\in\R.
 \end{eqnarray}
Let us write an equivalent representation :
We use the multiplication invariant measure $\frac{du}{\val u} $  on $\R_{\si}$.
Let
\begin{eqnarray*}
 U_\si:\l2{\R}\to\l2{\R_\si,\frac{du}{\vert u\vert }}; \, U_\si\xi(u): =
\xi(-\ln (\si u)),u\in\R_\si.
 \end{eqnarray*}
 Then $U_\si  $ is  a unitary operator and
\begin{eqnarray*}
 (U_\si)^*(\et)(s)=\et(\si e^{- s}) \, \text{ for } \, s\in\R,\et\in
\l2{\R_\si,
\frac{du}{\val
u}}.
 \end{eqnarray*}
Let
\begin{eqnarray}\label{ta mu nu def }
 \ta^{\si}_{\ell}:=U_\si\circ \pi_{\ell}\circ U^*_\si.
\end{eqnarray}
Then, identifying $\ell_{\mu,\nu}=(0,\mu,\nu, 0) $ with $(\mu,\nu) $ we obtain
the relation:

\begin{eqnarray*}
 \ta_{\mu,\nu}^{\si}(t,x,y,z)\et(u)\nn
 &= &
(\pi_\ell(t,x,y,z)(U_\si^*(\et)))(-\ln(\si u))\\
 \nn  &= &
e^{-i\mu e^{-\ln(\si u)-t}x}e^{-i \nu e^{t+\ln(\si u)}y}U_\si^*(\et)(-\ln(\si
u)-t)\\
\nn  &= &
e^{-i(\si\mu) u\inv e^{-t}x}e^{-i (\si\nu )u e^{t}y}\et(e^tu),
\quad \et\in\l2{\R_\si, \frac{du}{\val u}}.
 \end{eqnarray*}
 We see that
 \begin{eqnarray}\label{equi - + s}
 (\ta^{\si}_{\mu,\nu}){{^\ga}} (t,x,y,z)\et(u)&=&
 e^{i(\si\mu) u\inv e^{-t}x}e^{i (\si\nu )u e^{t}y}\et(e^tu)\\
 \nn  &= &
(\ta^{\si}_{-\mu,-\nu})(t,x,y,z)\et(u),
\quad \et\in\l2{\R_\si, \frac{du}{\val u}}.
\end{eqnarray}

 On the other hand,
\begin{eqnarray}\label{tamunu circ S}
  S\circ  \ta^\si_{\mu,\nu}(t,x,y,z)\circ  S(\et)(u) &= &
 e^{-i(-\si\mu) u\inv e^{-t}x}e^{-i (-\si\nu )u e^{t}y}\et(e^tu)\\
 \nn&=&
 \ta^{-\si}_{\mu,\nu}(\et)(u),
\quad \et\in\l2{\R_{-\si}, \frac{du}{\val u}}, u\in \R_{-\si}.
\end{eqnarray}
In particular the representations
$ \ta^\si_{\mu,\nu}$ and $ \ta^{-\si}_{\mu,\nu}$ are equivalent.
Furthermore, for $ \et\in L^2(\R_\si) $  and $ F\in L^1(G) $
$(\ell=(0,\mu,\nu,0)) $:
 \begin{eqnarray}\label{ta mu nu }
 \ta_{\mu,\nu}^{\si}(F)\et(u) &= &
\int_G F(t,x,y,z)e^{-i\si\mu u\inv e^{-t}x}e^{-i \si\nu u
e^{t}y}\et(e^tu)dtdxdydz\\
\nn  &= &
\int_\R \hat F^{2,3,4}(t,\si\mu (e^{t}u)\inv ,\si\nu (
e^tu),0)\et(e^tu)dt\\
\nn  &= &
\int_\R \hat F^{2,3,4}(t-\ln(\si u),{ \mu e^{-t} ,\nu
e^t},0)\et(\si
e^t)dt.
 \end{eqnarray}

 \item[(iii)] For $\ell\in \GA_0 $, let
\begin{eqnarray*}
 \pi_{\ell}:=\ch_{\ell}.
 \end{eqnarray*}
Hence for $F\in\l1G $:
\begin{eqnarray*}
 \pi_{\ell}(F)=\int_\R\int_{\R^3}e^{-it\ell}F(t, h)dhdt.
 \end{eqnarray*}

 \end{enumerate}

In order to define the Fourier transform $F\mapsto \hat F, F\in
C^*(G)$,  for every $ \pi\in\widehat{G}$ we choose a concrete representative, namely:
 \begin{eqnarray}\label{fourrier def}
  \hat F(\rh,\la) &:= & \pi_{\rh,\la}(F)\in\K(L^2(\R)),\, \, \rh\in \R,\la\in\R^*,\\
 \nn  \hat F(\mu,\nu)&:= & \ta^+_{\mu,\nu}(F)\in\K(L^2(\R_+)), \, \, \ell_{\mu,\nu}\in{{\Gamma_2}},\\
 \nn  \hat F(\mu,\nu)&:= & \ta^+_{\mu,\nu}(F)\in\B(L^2(\R_+)), \, \, \ell_{\mu, \nu}\in {{\Ga_1}},\\
\nn  \hat F(\ell)&\in & \C,\, \, \ell\in \GA_0.
 \end{eqnarray}
\begin{remark}\label{Grelaud}
\rm   We observe that the topological space $\Ga^0:=\Ga_2\cup\Ga_1\cup\Ga_0 $ is
homeomorphic to the spectrum of the quotient group $G/\ZZ\simeq \R\ltimes \R^2 $
 which is an extension of $\R^2 $ by $\R $ with the roots $+1,-1 $. Note that the C*-algebra of $G/\ZZ $ has
been described in the paper \cite{Lin-Lud}.
 \end{remark}

\section{Norm control of dual limits. }\label{NCDL}

 In this section, we will describe the conditions that our C*-algebra as the image of the Fourier transform must fulfill
and characterise the C*-algebra of Biodol's group (Theorem~\ref{Cster G identified}).

\subsection{Norm convergence }\label{norm conv}$ $
The following proposition is easy to verify from the expressions (\ref{kerfunc rh la})
and (\ref{kerfunc mu nu}).

\begin{proposition}\label{normcontinuity}
For any $a\in C^*(G)$, the mappings $O\mapsto \pi_O (a)$ are norm continuous on
the
different  sets $\GA_i, i=0,1,\cdots,3  $.
For any sequence $(O_k)_k $ tending to infinity, we have that $\limk \noop
{\pi_{O_k}(a) }=0$.
 \end{proposition}
\subsection{An approximation of $\pi_{\rh_k,\la_k}(F) $}\label{approx}$ $

Let $(\pi_{\rh_k,\la_k} )_k$ be  a properly converging sequence in $\wh G $
such
that $\limk \la_k=0 $ and that $\limk \la_k\rh_k=\om\in\R $. We can assume
(by passing to a subsequence) that $\la_k=\ve \val{\la_k} $ for every $k\in\N $.

 \begin{definition}\label{R_k Mde def}
\rm   Let $c\leq d\in\lbrack -\infty,\infty \rbrack $, $\de\geq 0 $ and
$\si\in\{+,-\} $. We define a family of multiplication operators
  on $\l2\R $ by: for $ s\in\R$ and $\xi\in\l2\R$,
\begin{eqnarray*}
M_{\{c\leq d\}}(\xi)(s)\nn
&:= &
\ch_{ [ c, d]}(s)\xi(s),\\
M_{\{\leq d\}}(\xi)(s)\nn
&:= &
\ch_{ [ -\iy, d]}(s)\xi(s),\\
M_{\{ d\leq \}  }(\xi)(s)\nn
&:= &
\ch_{\lbrack d,\iy \rbrack}\xi(s),\\
 M_{\{\val{}\leq \de\}}(\xi)(s)
 &:=&
 \ch_{{[ -\de,\de]}}(s)\xi(s), \\
 \nn \text{resp. }M_{{\{\val{}\geq\de }\}}(\xi)(s)
 & :=&
\ch_{\lbrack -\iy,-\de \rbrack\cup \lbrack \de,\iy \rbrack}(s)\xi(s).\\

 \end{eqnarray*}

 \end{definition}

\begin{lemma}\label{limk de=0rk to iy}
Suppose that $\limk \la_k=0 $. Take a sequence  $(R_k)_k $  in $\R_+ $ such
that

\begin{eqnarray}\label{Rk conditions}
\limk R_k=+\iy.
 \end{eqnarray}
Then we have that
  \begin{eqnarray}\label{small to 0}
 \limk \noop{V_k^*\circ \pi_{\rh_k,\la_k}(F)\circ V_{k }\circ M_{\{   R_k \leq
\vert{ }\vert\} }}=0,\ F\in C^*(G).
 \end{eqnarray}
 \end{lemma}
 \begin{proof}
By (\ref{kernel pirhola}), it suffices to show that for $F\in \l1{G} $ with
$\hat F^{3,4}\in C_c(\R^4) $ we have that
\begin{eqnarray}\label{hat F 34 is 0}
 \hat F^{3,4}(t,\ve\frac{e^ts}{\la_k}-\frac{\ve x}{\la_k},-\frac{1}{2}(\ve
x+\ve{e^ts}{}), \la_k)=0
 \end{eqnarray}
for $\vert x\vert\geq R_k $ and $k $ large enough.
Now if $s, x$ have the same sign, we have that
$\vert \ve x+\ve{e^ts} \vert\geq R_k $ and thus (\ref{hat F 34 is 0}) is
satisfied. Similarly, if $s $ and $x $ have different signs, it follows that
$\vert \frac{e^ts}{\la_k}-\frac{x}{\la_k}\vert\geq \frac{R_k}{\val{\la_k}} $.

\end{proof}

\begin{lemma}\label{limk de=0}
Suppose that $\limk \la_k=0 $ and $\limk{ {\rh_k\la_k}}=\om\in\R
^*$. Take
a sequence $(R_k)_k $  in $\R_+ $ such that
\begin{eqnarray}\label{Rk conditions}
\limk R_k=+\iy, \ \limk R_k\vert \la_k\vert =0.
 \end{eqnarray}
 Then we have that
  \begin{eqnarray}\label{small to 0}
 \limk \noop{\pi_{\rh_k,\la_k}(F)\circ V_{k }\circ M_{\{ \vert{ }\vert\leq
\vert R_k\la_k\vert\} }}=0, \quad F\in C^*(G).
 \end{eqnarray}
 \end{lemma}

 \begin{proof}
It
suffices to prove the lemma for $F\in\l1G $ such that $\hat F^{3,4}\in
C_c^{\iy}(\R^4) $.  Take  $\et\in \l2{\R, \frac{dx}{\val x}} $ such that
$\et(x)=0 $ for $\val x \geq \val{R_k\la_k}$. Since $\hat F^{3,4}\in C_c(\R^4)
$,  by (\ref{vkstar circ Vk}) there exists a $C>0 $ such that
 \begin{eqnarray*}
  V_{k}^*(\pi_{\rh_k,\la_k}(F)(V_{k }(\et)))(s)=0 \text{ for }\val s\leq
C\val{R_k\la_k}.
 \end{eqnarray*}
 Hence
  \begin{eqnarray*} V_{k}^*(\pi_{\rh_k,\la_k}(F)(V_{k }(\et)))=M_{\{
C\val{R_k\la_k}\leq\}}V_{k}^*(\pi_{\rh_k,\la_k}(F)(V_{k}(\et)).
 \end{eqnarray*}
 Furthermore
 \begin{eqnarray*}
 \nn  & &
V_{k}^*(\pi_{\rh_k,\la_k}(F)(V_{k}(\et)))(s)\\
 \nn  &= &
 {e^{-i\rh_k\ln(\val s)}\val s^{1/2}}\int_\R \big(\int_\R e^{t/2} e^{-i\rh_k
t}\hat
F^{3,4}(t,\ve\frac{e^ts}{\la_k}-x,-\frac{\la_k}{2}(x+\ve\frac{e^ts}{\la_k}),
\la_k)dt\big)\\
 \nn  & &
\frac{e^{i\rh_k\ln(\val{\la_kx})}}{\val{\la_k x}^{1/2}}\et(\ve\la_k x)dx\\
\nn  &\overset{\text{integration by part}}= &
\frac{e^{-i\rh_k\ln(\val s)}\val s^{1/2}} {-i\rh_k }
 \int_\R \big(\int_\R e^{-i\rh_k t}\partial_t\Big(e^{t/2} \hat
F^{3,4}(t,\ve\frac{e^ts}{\la_k}-x,-\frac{\la_k}{2}(x+\ve\frac{e^ts}{\la_k}),
\la_k)\Big)dt\big)\\
 \nn  & &
\frac{e^{i\rh_k\ln(\val{\la_kx})}}{\val{\la_k x}^{1/2}}\et(\ve\la_k x)dx\\
\nn  &= &
\frac{e^{-i\rh_k\ln(\val s)}\val s^{1/2}} {-i \rh_k}
 \int_\R \big(\int_\R e^{-i\rh_k t}
\Big(\frac{1}{2}e^{t/2} \hat
F^{3,4}(t,\ve\frac{e^ts}{\la_k}-x,-\frac{\la_k}{2}(x+\ve\frac{e^ts}{\la_k}),
\la_k)\\
 \nn  & &+ e^{t/2} (\ve\frac{e^ts}{\la_k})\partial_2\hat
F^{3,4}(t,\ve\frac{e^ts}{\la_k}-x,-\frac{\la_k}{2}(x+\ve\frac{e^ts}{\la_k}),
\la_k)
\\
\nn  & &- \frac{\ve}{2}
e^{t/2} {e^ts}\partial_3\hat
F^{3,4}(t,\ve\frac{e^ts}{\la_k}-x,-\frac{\la_k}{2}(x+\ve\frac{e^ts}{\la_k}),
\la_k)\Big)dt\\
 \nn  & &
\frac{e^{i\rh_k\ln(\val{\la_kx})}}{\val{\la_k x}^{1/2}}\et(\ve\la_k x)dx\\
\nn  &= &\frac{1} {\rh_k }
 \frac{e^{-i\rh_k\ln(\val s)}\val s^{1/2}}{\val{\la_k}^{1/2}}\int_\R
\Big(\int_\R e^{-i\rh_k
t}H_k(t,x,s)dt\Big)\frac{e^{i\rh_k\ln(\val{\la_kx})}}{\val{x}^{1/2}}
\et(\ve\la_k x)dx,
 \end{eqnarray*}
where, for some non-negative function $\va\in C_c^{\iy}(\R) $, we
have that
\begin{eqnarray*}
 \val{H_k(t,x,s)}\leq {(1+\vert \frac{e^t s}{\la_k}\vert )}\va(t)\va(\frac{\ve
e^ts}{\la_k}-x)\va({\la_k}x+\ve{e^ts}).
 \end{eqnarray*}

Therefore
\begin{eqnarray*}
 \nn  & &
\no{ V_{k,\si}^*(\pi_{\rh_k,\la_k}(F)(V_{k}\circ M_{\{ \vert{ }\vert\leq
\vert R_k\la_k\vert\} }(\et)))}_2^2\\
\nn  &= &
\int_{\leq\val{C R_k\la_k}}\Big \vert \frac{e^{-i\rh_k\ln(\val s)}\val
s^{1/2}}{\val{\la_k}^{1/2}}\int_{\val x\leq R_k} \int_\R e^{t/2} e^{-i\rh_k
t}\hat
F^{3,4}(t,\ve\frac{e^ts}{\la_k}-x,-\frac{\la_k}{2}(x+\ve\frac{e^ts}{\la_k}),
\la_k)dt\\
 \nn  & &
\frac{e^{i\ve\rh_k\ln(\val{\la_kx})}}{\val{x}^{1/2}}\et(\ve\la_k x)dx\Big\vert
^2\frac{ds}{\val s}\\
\nn  &\leq&
\frac{1}{\rh_k^2}\int_{\leq\val{C R_k\la_k}} \frac{\val
s}{\val{\la_k}}\Big(\int_\R \int_{\val x\leq R_k} \val{H_k(t,x,s)}dt
\frac{1}{\val{x}^{1/2}}\vert \et(\ve\la_k x)\vert dx\Big) ^2\frac{ds}{\val s}\\
\nn  &\leq &
\frac{1}{\rh_k^2 }\int_{\val { x}\leq R_k}
\int_{\val s \leq\val{C R_k\la_k}}\int_\R \frac{1}{\val{\la_k}} \vert
H_k(t,x,s)\vert  \frac{\val{\et(\ve\la_k x)}^2}{\val x}dt  ds dx
\int_{\R}\int_{\val u\leq R_k}\sup_{\val s\leq C
R_k\val{\la_k}}\val{H_k(t,u,s)}dt du\\
\nn  &\leq&
\frac{1}{\rh_k^2 }\int_{\val { x}\leq R_k}
\int_{\val s \leq\val{C R_k\la_k}}\int_\R
\frac{1}{\val{\la_k}}(1+\frac{\val{e^ts}}{\val{\la_k}} )\va(t)\va(\frac{\ve
e^ts}{\la_k}-x)\va(\la_k x+\ve e^ts)\frac{\val{\et(\ve\la_k x)}^2}{\val x}dt
ds
dx\\
\nn  & &
\int_{\R}\int_{\val u\leq R_k}\sup_{\val s\leq C R_k\val{\la_k}}(1+\vert
\frac{e^ts}{\la_k}\vert ) \va(t)\va(\frac{\ve e^ts}{\la_k}-u)\va(\la_k u+\ve
e^t
s)dt du\\
\nn  &\leq&
\frac{CR_k \no\va_1^2\no\va_\iy}{\rh_k^2 }\int_{\val { x}\leq R_k}
\int_{\val s \leq\val{C R_k\la_k}}\int_\R
\frac{1}{\val{\la_k}}(1+\frac{\val{e^ts}}{\val{\la_k}} )\va(t)\va(\ve
\frac{e^ts}{\la_k}-x)\va(\la_k x+\ve e^ts)\frac{\val{\et(\ve\la_k x)}^2}{\val
x}dt  ds dx\\
\nn  &\overset{s=\la_k s'}\leq&
\frac{CR_k \no\va_1^2\no\va_\iy^2}{\rh_k^2 }\int_{\val { x}\leq R_k}
\int_{\val s' \leq\val{C R_k}}\int_\R (1+{\val{e^ts'}} )\va(t)\va({\ve
e^ts'}-x)\frac{\val{\et(\ve\la_k x)}^2}{\val x}dt  ds' dx\\

 \nn &\leq & C\frac{(R_k\la_k)^2}{\om_k^2}\no\et_2^2,
\end{eqnarray*}
for some (new) constants $C>0 $.
 {Then it follows from the condition $\limk \frac{R_k\la_k}{\om_k}=0
$ that }
  \begin{eqnarray*}\label{smaller to 0}
 \limk \noop{V_{k}^*\circ\pi_{\rh_k,\la_k}(F)\circ V_{k }\circ M_{\{
\vert{ }\vert\leq \vert R_k\la_k\vert \}}}=0.
 \end{eqnarray*}

\end{proof}

 \begin{lemma}\label{difference operator lak rhk ta}
 Let $(\pi_{\rh_k,\la_k} )_k$ be  a properly converging sequence in $\wh G $
such that $\la_k=\ve  \vert{\la_k }\vert, k\in\N$, $\limk \la_k=0 $ and that
 $\limk \om_k:= \limk \la_k\rh_k=\om\in\R^* $.
 Take a sequence $(R_k)_k $ in $\R_+ $ such that  $\limk R_k=\iy $ and
\begin{eqnarray}\label{Rk conditions}
 \nn \limk R_k\vert \la_k\vert =0,\ \limk ({R_k^2}{\vert \la_k}\vert
)=\iy\Leftrightarrow \limk \frac{\rh_k}{R_k^2}=0.
 \end{eqnarray}
 Then for $F\in \l1G $ with $\hat F^{3,4}\in C_c(\R^4) $, we have  that
\begin{eqnarray*}\label{limit -om+}
 (a)& &
 \noop{\pi_{\rh_k,\la_k}(F)\circ V_{k}\circ M_{\{\geq 0\}}-V_k\circ
\ta^{+}_{\ve\om_k,-\ve}(F) \circ M_{\{\geq \val{\la_k}R_k\}} }\leq C\big(\frac{
\vert
\om_k \vert }{\val{R_k^2\la_k}}+\frac{1}{R_k}\big),
\\
 (b) & &
  \noop{\pi_{\rh_k,\la_k}(F)\circ V_{k}\circ M_{\{\leq
0\}}-V_k\circ \ta^{-}_{-\ve\om_k,\ve}(F) \circ M_{\{\leq
-\val{\la_k}R_k\}} }\leq C\big(\frac{ \vert \om_k \vert
}{\val{R_k^2\la_k}}+\frac{1}{R_k}\big),
\end{eqnarray*}
 for some constant $C$ depending on $F $.
 \end{lemma}
\begin{proof}

We can assume that $\hat F^{3,4}\in C_c^\iy(\R^4) $, since these functions are
dense in $C^*(G)$.

 Let $ s,t,x\in\R$ and $0<M<\iy, C>0 $ such that $\val t\leq M, \val x\leq M$ and $\val
s\geq C\val{ R_k\la_k}$.  We have that
 \begin{eqnarray}\label{ln(s+small)}
 \ln(\vert  e^ts-\la_k x\vert )-\ln(\vert e^t s\vert )+{\la_k x}( e^ts)\inv
&=&-\la_k x \int_{0}^{1}(\frac{1}{ e^ts-\la_k x v }-\frac{1}{ e^ts })dv\\
\nn  &= &
\frac{-(\la_k x)^2}{e^t s} \int_{0}^{1}\frac{v dv}{ e^ts-\la_k x v } \\
 \nn  &=: &
   \la_k{c_k(t,s, x)}.
 \end{eqnarray}
Here
\begin{eqnarray*}
 c_k(t,s,x)=\frac{-\la_k x^2}{e^t s} \int_{0}^{1}\frac{ vdv}{ e^ts-\la_k x v } .
 \end{eqnarray*}
Hence for $k $ large enough and for some new constant $C>0 $, for all our $s, x,
t $ we have:
\begin{eqnarray*}
 \vert \ln(\vert e^ts-\la_k x\vert )-\ln(\vert e^t s\vert )+{\la_k x}(\vert
e^ts\vert )\inv  \vert &=&
 \vert  \la_k c_k(t,s,x)  \vert \\
 \nn  &\leq &
C\frac{\la_k^2}{s^2}\\
\nn  &\leq &
C\frac{1}{  R_k^2}.
 \end{eqnarray*}
Whence
\begin{eqnarray}\label{one minus e ick}
 \vert e^{i\rh_k\la_kc_k(t,s,x)}-1\vert &\leq &\val{\rh_k\la_kc_k(t,s,x)}\\
\nn  &\leq &
\frac{C\val{\rh_k\la_k}}{\val{\la_k R_k^2}}.
\end{eqnarray}
By (\ref{vkstar circ Vk})  and (\ref{ln(s+small)}) we have that
\begin{eqnarray*}
 V_{k}^*(\pi_{\rh_k,\la_k}(F)(V_{k}(\et)))(s)
 \nn  &= &
\int_\R \int_\R  \hat F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{e^ts}{},\la_k)\\
 \nn & &
\frac{e^{i\rh_k(\ln(\val{-\ve\la_kx+
e^ts}-\ln(\val{e^{t}s}))}\val{e^{t}s}^{1/2}}{\val{-\ve\la_k x+
e^ts}^{1/2}}\et(- \ve\la_k x+ e^t s )dxdt\\
 \nn  &= &
\int_\R \int_\R  \hat F^{3,4}(t,x,\frac{\la_k}{2}x-{ {\ve
e^ts}},\la_k)\\
 \nn & &
\frac{\val{e^{t}s}^{1/2}e^{-i\ve\rh_k{\la_k x} (e^ts)\inv
+i\ve\rh_k\la_k{c_k(t,s,\ve x)}}}{\val{-\ve\la_k x+e^t  s
}^{1/2}}\et(- \ve\la_k x+
e^t s )dxdt,
 \end{eqnarray*}
 for $\et\in L^2(\R, \frac{dx}{\val x}) $.

For any $\et\in L^2(\R,\frac{dx}{ \vert{x }\vert}) $ with $\et(s)=0 $
for $\val s\leq C R_k
\vert{\la_k }\vert $ (where $1>C>0 $ and $k $ is large enough), we have that
$V_{k}^*(\pi_{\rh_k,\la_k}(F)(V_{k}(\et)))(s)=0 $ for $\val s\leq
CR_k\val{\la_k} $.
 For $\val s\geq CR_k\val{\la_k} $ we have that
 \begin{eqnarray*}
 \nn  & &
\big| V_{k}^*(\pi_{\rh_k,\la_k}(F)(V_{k }(\et)))(s)\\
 \nn  & &-
\int_\R   \hat F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{e^ts}{},\la_k)
\frac{\val{e^{t}s}^{1/2}e^{-i\ve\rh_k{\la_k x}( e^ts)\inv }}{\val{-\ve\la_k x+
e^t  s }^{1/2}}\et(- \ve\la_k x+ e^t s )dxdt \big|\\
\nn  &= &
\big|
\int_\R \int_\R  \hat
F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{e^ts}{},\la_k)\val{e^{t}s}^{1/2}\\
 \nn & &
(\frac{e^{-i\ve\rh_k{\la_k x}(\vert e^ts\vert )\inv
+i\ve\rh_k\la_k{c_k(t,s,\ve x)}}-e^{-i\ve\rh_k{\la_k x}(\vert
e^ts\vert )\inv
}}{\val{-\ve\la_k x+  e^t  s }^{1/2}}) \et(-\ve\la_k x+e^t s ) dxdt\big| \\
\nn  &\leq &
\int_\R \int_\R  \vert \hat
F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{e^ts},\la_k)\vert
\frac{\vert e^{i\ve\rh_k\la_k{c_k(t,s,\ve x)}}-1\vert
}{\val{-\ve\la_k x+e^t
s}^{1/2}}\val{e^{t}s}^{1/2}\vert \et(-\ve\la_k x+e^t s )\vert dxdt\\
\nn  &\overset{(\ref{one minus e ick})}{\leq}&
\frac{C \vert \rh_k \la_k\vert }{\val{\la_k R_k^2}}\int_\R \int_\R  \vert \hat
 F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{e^ts},\la_k)\vert
\frac{1 }{\val{-\ve\la_k x+  e^t  s }^{1/2}}\val{e^{t}s}^{1/2}{\vert
\et(-\ve\la_k
x+e^t s )\vert} dxdt.
 \end{eqnarray*}
Therefore, for any $\et\in\l2\R $ with $\et(s)=0 $ for $\val s\leq C R_k
\vert{\la_k }\vert $ (where $1>C>0 $ and $k $ is large enough) we have that
\begin{eqnarray*}
 \nn  & &
\int_{\{\val s\geq C R_k\val{\la_k}\}}\val{ V_{k}^*(\pi_{\rh_k,\la_k}(F)(V_{k
}(\et)))(s)\\
 \nn  & &-
\int_\R \int_\R  \hat F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{e^ts},\la_k)
\frac{\val{e^{t}s}^{1/2}e^{-i\ve\rh_k{\la_k x}( e^ts)\inv }}{\val{-\ve\la_k x+
e^t  s }^{1/2}}\et(-\ve\la_k x+e^t s )dxdt}^2\frac{ds}{\vert s\vert }\\
\nn  &\leq &
\int_{\{\val s\geq C R_k\val{\la_k}\}}{\Big(\frac{C  \vert{\rh_k \la_k
}\vert}{\val{\la_k R_k^2}}\int_\R \int_\R  \vert \hat
F^{3,4}(t,x,\frac{\la_k}{2}x-{{\ve e^ts}}),\la_k)\vert
\frac{1 }{\val{-{\ve\la_k x}+  e^t  s
}^{1/2}}}\val{e^{t}s}^{1/2}\vert {\et(-\ve\la_k x+e^t s )} \vert
dxdt\Big)^2\frac{ds}{\val s}\\
\nn  &\leq &
(\frac{C \vert \rh_k \la_k\vert }{\val{\la_k R_k^2}})^2\int_{\{\val s\geq C
R_k\val{\la_k}\}}{\int_\R \int_\R  \vert e^{t/2}\hat
F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{e^ts}),\la_k)\vert
\frac{1 }{\val{-\ve\la_k x+  e^t  s }}}\vert \et(-\ve\la_k x+e^t s )\vert^2
dxdtds\\
\nn  & &
\cdot \sup_{\val s\geq C R_k\val{\la_k}}\int_\R\int_\R e^{r/2}\vert \hat
F^{3,4}(r,u,\la_k u-\ve e^rs,\la_k)\vert dudr
 \end{eqnarray*}
Choose a function $\va\in C_c^{\iy}(\R,\R_+) $ such that
\begin{eqnarray*}
 \val{e^{t/2}\hat F^{3,4}(t,x,u,\la_k)}\leq \va(t)\va(x)\va(u),\,  k\in\N, u, t, x\in \R.
 \end{eqnarray*}
Then we see that
\begin{eqnarray}\label{difference one}
  & &
\int_{\{\val s\geq C R_k\val{\la_k}\}}\val{
V_{k,\si}^*(\pi_{\rh_k,\la_k}(F)(V_{k
}(\et)))(s)\\
 \nn  & &-
 \int_\R\int_\R  \hat F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{e^ts},\la_k)
\frac{\val{e^{t}s}^{1/2}e^{-i\ve\rh_k{\la_k x}( e^ts )\inv }}{\val{-\ve\la_k x+
 e^t
s }^{1/2}}\et(-\ve\la_k x+e^t s )dxdt}^2\frac{ds}{\vert s\vert }\\
\nn  &\leq &
(\frac{C \vert \rh_k \la_k\vert }{\val{\la_k R_k^2}})^2\int_{\si \R_+}\int_\R
\int_\R e^{-t} \va(t)\va(x)
\frac{1 }{\val{  s }}\vert \et(e^t s )\vert^2 dxdtds\\
\nn  &\leq &
(\frac{C \vert \rh_k \la_k\vert }{\val{\la_k R_k^2}})^2\no\et^2_2
 \end{eqnarray}
 for some new constants $C>0 $.

 Now
 \begin{eqnarray*}
  e^{t} s-\ve\la_k x= e^{t} s\al_k(t,s,x)=e^{t+\ln(\al_k(t,s,x))}s,
 \end{eqnarray*}
where
\begin{eqnarray*}
 \al_k(t,s,x)=1-\frac{\ve\la_k x}{ e^{t}s}.
 \end{eqnarray*}
If $\val {s}\geq C \val{\la_k}R_k $ for some constant $C>0 $, $k
$ large enough and for all $\val t\leq M, \val x\leq M$, it follows that
\begin{eqnarray}\label{al k estimated}
 \vert \ln(\al_k(t,s,x))\vert \leq C\frac{1}{R_k},\\
\nn \val{\partial_t \ln(\al_k(t,s,x))}\leq C\frac{1}{R_k},
 \end{eqnarray}
for some new constant $C>0 $.

Let\begin{eqnarray*}
 r=\ta_k(t) &:= &t+\ln(\al_k(t,s,x)), \\
 \nn \ta\inv_k(r)=\mu_k(r) & =&
t, \quad \val t\leq M, \val x\leq M, \val{s}\geq  C\val{\la_k R_k},
\end{eqnarray*}
we have that $\val{\mu_k(r)-r}=\val{\ln(\al_k(t,s,x))}\leq C\frac{1}{R_k}$.
Then
\begin{eqnarray*}
  & &\int_{\R} \int_\R  \hat F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{ e^ts}{},\la_k)
\frac{\val{e^{t}s}^{1/2}e^{-i\ve\rh_k{\la_k x}( e^ts )\inv }}{\val{-\ve\la_k x+
 e^t  s }^{1/2}}\et(-\ve\la_k x+e^t s )dxdt\\
&= &
\int_{\R}  \int_\R  \hat F^{3,4}(\mu_k(r),x, e^{r}s,\la_k)
\frac{\val{e^{\mu_k(r)}s}^{1/2}e^{-i\ve\rh_k{\la_k x}( e^{\mu_k(r)}s)\inv
}}{\val{e^{r}  s }^{1/2}}\et(e^{r} s )dx\mu'_k(r)dr.
\end{eqnarray*}
Now by (\ref{al k estimated}), for $k $ large enough, we have
\begin{eqnarray*}
   & &\vert \hat F^{3,4}(\mu_k(r),x,{ e^rs},\la_k)
{\val{e^{\mu_k(r)}s}^{1/2}e^{-i\ve\rh_k{\la_k x}( e^{\mu_k(r)}s )\inv
}}\mu'_k(r)\\
 & & \quad -
 \hat F^{3,4}(r,x,{e^rs}{},0)
{\val{e^{r}s}^{1/2}e^{-i\ve\rh_k{\la_k x}( e^rs)\inv }}\vert \\
  &\leq &
\frac{C}{R_k}\va(r)\va(x)\val s^{1/2}.
\end{eqnarray*}

Therefore for $k $ large enough, for any $\et\in \l2{ \R, \frac{ds}{\val s}} $
with $\|\et\|_2\leq 1 $ and $\et(u)=0 $ for $\vert  u\vert \leq R_k \vert{\la_k
}\vert $ we have that
\begin{eqnarray}\label{first approx}
  &  &
\int_{\R}\big| \int_{\R} \int_\R  \hat F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{
e^ts}{},\la_k)
\frac{\val{e^{t}s}^{1/2}e^{-i\ve\rh_k{\la_k x}( e^ts )\inv }}{\val{-\ve\la_k x+
 e^t  s }^{1/2}}\et(-\ve\la_k x+e^t s )dxdt\\
\nn  & &-
\int_{\R}  \int_\R  \hat F^{3,4}(r,x,-{ e^rs}{},0)
{e^{-i\ve\rh_k{\la_k x}( e^{r}s )\inv }}{}\et( e^r s )dxdr{\big|}^{2}\frac{ds}{
\val s }\\
  \nn&= &
  \int_\R\big| \int_{\R}  \int_\R  \hat F^{3,4}(\mu_k(r),x,-e^r s,\la_k)
\frac{\val{e^{\mu_k(r)}s}^{1/2}e^{-i\ve\rh_k{\la_k x}( e^{\mu_k(r)}s )\inv
}}{\val{e^r  s }^{1/2}}\et( e^r s )dx\mu'_k(r)dr\\
\nn  & &-
\int_{\R}  \int_\R  \hat F^{3,4}(r,x,{ -e^rs}{},0)
{e^{-i\ve\rh_k{\la_k x}( e^{r}s )\inv }}{}\et( e^r s
)dxdr\big|^{2}\frac{ds}{\val s}
\\
\nn  &  \leq&
\frac{C}{R_k}.
 \end{eqnarray}

 Let us recall that for $\et\in L^2(\R_\si, \frac{dx}{\val x}) $,
 \begin{eqnarray}\label{ta mu nu }
 \ta_{\mu,\nu}^{\si}(F)\et(u) &= &
\int_G F(t,x,y,z)e^{-i\si\mu u\inv e^{-t}x}e^{-i \si\nu u
e^{t}y}\et(e^tu)dtdxdydz\\
\nn  &= &
\int_\R \hat F^{2,3,4}(t,\mu \si(e^{t}u)\inv ,\si\nu ( e^tu),0)\et(e^tu)dt.
 \end{eqnarray}

Hence  for any $\et\in \l2{\R_+, \frac{ds}{\val s}} $, we have that
\begin{eqnarray*}
  & &\Vert{V_{k}^*\circ\pi_{\rh_k,\la_k}(F)\circ V_{k}\circ M_{\{\geq\val
{R_k\la_k}\}}(\et)-\ta^{+}_{(\ve\om_k,- \ve)}(F)\circ M_{\{\geq
\val{\la_k}R_k\}}(\et)}\Vert _2\\
  \nn  &= &
  \Big(\int_\R
\vert V_{k}^*\circ\pi_{\rh_k,\la_k}(F)\circ V_{k}\circ M_{\{\geq\val
{R_k\la_k}\}}(\et)(s)\\
\nn  & &-
\int_\R \hat F^{2,3,4}(t,\ve\om_k s\inv e^{-t}, -s e^t,0)M_{\{\geq
\val{\la_k}R_k\}}(\et)( e^ts)dt
\vert^{2}\frac{ds}{\val s}\Big)^{1/2}\\
  \nn  &= &
  \Big(\int_\R
\vert V_{k}^*\circ\pi_{\rh_k,\la_k}(F)\circ V_{k}\circ M_{\{\geq\val
{R_k\la_k}\}}(\et)(s)\\
\nn  & &-
\int_\R \int_\R\hat F^{3,4}(t,x,-{ e^ts},0)
{\val{e^{t}s}^{1/2}e^{-i \ve\rh_k{\la_k x}( e^ts )\inv }}M_{\{\geq
\val{\la_k}R_k\}}(\et)( e^ts)dtdx
\vert ^{2}\frac{ds}{\val s}\Big)^{1/2}\\
 \nn  &\overset{(\ref{first approx})}\leq &
  (\frac{C }{R_k})^{1/2} \no\et_2+
\Big(\int_\R
\vert V_{\si}^*\circ\pi_{\rh_k,\la_k}(F)\circ V_{\si}\circ M_{\{\geq\val
{R_k\la_k}\}}(\et)(s)\\
\nn  & &-
\int_\R \int_\R  \hat F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{ e^ts}{}),0)
\frac{\val{e^{t}s}^{1/2}e^{-i\ve\rh_k{\la_k x}( e^ts )\inv }}{\val{-\ve\la_k x+
e^t  s }^{1/2}}M_{\{\geq R_k\val{\la_k}\}}(\et)(-\ve\la_k x+ e^t s
)dxdt\vert
^2\frac{ds}{\vert s\vert }\Big)^{1/2}\\
\nn  &\leq &
  (\frac{C }{R_k})^{1/2} \no\et_2+
\Big(\int_\R
\vert \int_\R \int_\R  \hat F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{ e^ts}{}),0)\\
 \nn & &
\frac{\val{e^{t}s}^{1/2}e^{-i\ve\rh_k{\la_k x} (e^ts)\inv
+i\rh_k\la_k{c_k(t,s,\ve x)}}}{\val{-\ve\la_k x+ e^t  s
}^{1/2}}M_{\{\geq\val
{R_k\la_k}\}}\et(- \ve\la_k x+  e^t s )dxdt\\
\nn  & &-
\int_\R \int_\R  \hat F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{ e^ts}{}),0)
\frac{\val{e^{t}s}^{1/2}e^{-i\ve\rh_k{\la_k x}( e^ts )\inv }}{\val{-\ve\la_k x+
e^t  s }^{1/2}}M_{\{\geq R_k\val{\la_k}\}}(\et)(-\ve\la_k x+ e^t s
)dxdt\vert
^2\frac{ds}{\vert s\vert }\Big)^{1/2}\\
\nn  &\leq &
  (\frac{C }{R_k})^{1/2} \no\et_2+
\Big(\int_\R
\vert \int_\R \int_\R  \hat F^{3,4}(t,x,\frac{\la_k}{2}x-\ve{ e^ts}{}),0)\\
 \nn & &
 \Big\vert
 e^{i\ve\rh_k\la_k{c_k(t,s,\ve x)}}-1\Big\vert
\frac{\val{e^{t}s}^{1/2}e^{-i\ve\rh_k{\la_k x} (e^ts)\inv
}}{\val{-\ve\la_k x+ e^t  s }^{1/2}}
M_{\{\geq\val{R_k\la_k}\}}\et(-\ve \la_k x+  e^t s )dxdt
\vert
^2\frac{ds}{\vert s\vert }\Big)^{1/2}\\
\nn  &\overset{(\ref{one minus e ick})}\leq &
C(\frac{ \vert \rh_k \la_k\vert }{\val{\la_k R_k^2}}+\frac{1
}{R_k})^{1/2}\no\et_2
\end{eqnarray*}
for a constant $C>0 $.  Then for any  $s\in\R $, it follows that
\begin{eqnarray*}
\nn  & &
C(\frac{ \vert \om_k \vert }{\val{R_k^2\la_k}}+\frac{1}{R_k})^{1/2}\\
\nn  &{\geq} &  \noop{ \pi^{\ga}_{\rh_k,\la_k}(F)\circ V_{k}\circ
M_{\{\geq
0\}}-V_k\circ
 (\ta^{+}_{\ve\om_k,-\ve})^{\ga}(F) \circ M_{\{\geq \val{\la_k}R_k\}}
}\\
\nn  &\overset{(\ref{pilarh s}), (\ref{equi - + s})}{=} &
\noop{S\circ  \pi_{\rh_k,\la_k}(F)\circ  S\circ V_{k}\circ M_{\{\geq
0\}}-V_k\circ
\ta^{+}_{-\ve\om_k,\ve}(F) \circ M_{\{\geq \val{\la_k}R_k\}} }  \\
\nn  &\overset{(\ref{ScommVk}), (\ref{tamunu circ S})}=&
\noop{  \pi_{\rh_k,\la_k}(F)\circ V_{k}\circ M_{\{\leq
0\}}-V_k\circ
\ta^{-}_{-\ve\om_k,\ve}(F) \circ M_{\{\leq \val{\la_k}R_k\}} }, \quad \hat
F^{3,4}  \in
C_c^\iy(\R^4).
\end{eqnarray*}

\end{proof}

 \subsection{Convergence in operator norm: $\om\ne 0$}$
 $

Let $(\pi_{\rh_k,\la_k} )_k$ be  a properly converging sequence in $\wh G $
such
that $\limk \la_k=0 $ and $\limk \la_k\rh_k= \om \in \R^*$.  We recall
(Proposition \ref{limitGamma3toGamma2}) that the limit set $L $ of the sequence
$(O_{\rh_k,\la_k})_k $ is the two points set
\begin{eqnarray*}
 L=\{O_{-\om,1,0}, O_{\om,-1,0}\}.
 \end{eqnarray*}
For $k\in \N$, we define $ \si_{k}^\om$  by
\begin{align}\label{control om}
\nn\si_{k}^\om(\va\res L):= V_{k}\circ \Big(\va(\ta^{+}_{\ve\om,-\ve})\circ
M_{\{\geq
\val{\la_k}R_k\}}\oplus
{\va(\ta^{-}_{-\ve\om,\ve})}\circ M_{\{\leq-\val{\la_k}R_k\}}\Big)
\circ V_k^*, \quad \va\in CB(\wh\A).
 \end{align}

 \begin{theorem}\label{lim rhklak is om}
 Suppose that $ \limk \rh_k\la_k =\om\in\R^*$ and $\la_k=\ve \val{\la_k}, k\in\N$.
Then
\begin{eqnarray*}
 \limk \noop{\pi_{\rh_k,\la_k}(F)-\si_{k}^\om(\hat F\res L)}=0
 \end{eqnarray*}
for every $ F\in C^*(G)$.
 \end{theorem}
\begin{proof}
Apply Proposition~\ref{norm conv} and  Lemma~\ref{difference operator lak rhk
ta}.
\end{proof}

\subsection{Convergence in operator norm: $\om=0 $}\label{om is  0}$ $

Suppose now that
\begin{eqnarray*}
\limk \la_k=0= \limk \la_k\rh_k.
 \end{eqnarray*}
 We can again assume that $\la_k=\ve \val{\la_k}, k\in\N $.  By  Proposition~\ref{toporbitspace},
 the limit set $L $ of the sequence of representations
$(\pi_{\rh_k,\la_k}) $ is
equal to $\GA_1\cup\GA_0 $.

\begin{lemma}\label{limk de=0 and more}
Suppose that $\limk \la_k=0 $ and $\limk \rh_k \la_k=0 $. Take
a sequence  $(R_k)_k $  in $\R_+ $ such that
\begin{eqnarray}\label{Rk conditions}
\limk R_k=+\iy, \ \limk R_k^{2}\vert \la_k\vert =0.
 \end{eqnarray}
 Then we have that
  \begin{eqnarray}\label{small to 0}
 \limk \noop{\pi_{\rh_k,\la_k}(F)\circ V_{k }\circ M_{\{ \vert{ }\vert\leq
\vert R_k\la_k\vert\} }}=0,\quad F\in C^*(G).
 \end{eqnarray}
 \end{lemma}

  \begin{proof} It suffices to consider $F\in \l1 G $ such that $\hat
F^{3,4}\in C_c^\iy(\R^4) $.
  There exists $M>0 $ such that
  \begin{eqnarray*}
 \hat F^{3,4}({ t},x,y,\la)=0
 \end{eqnarray*}
whenever $ \vert{t }\vert+ \vert{x }\vert+ \vert{y }\vert+ \vert{\la }\vert\geq
M $. We have  that
  \begin{eqnarray}\label{vk star pik Vk}
V_{k}^*(\pi_{\rh_k,\la_k}(F)(V_{k }(\et)))(s)
 \nn  &= &
 \frac{e^{-i\rh_k\ln(\val s)}\val s^{1/2}}{\val{\la_k}^{1/2}}\int_\R
\big(\int_\R e^{t/2} e^{-i\rh_k t}\hat F^{3,4}(t,\ve\frac{e^ts}{\la_k
}-x,-\frac{\la_k}{2}(x+\ve\frac{e^ts}{\la_k}),\la_k)dt\big)\\
 \nn  & &
\frac{e^{i\rh_k\ln(\val{\la_kx})}}{\val x^{1/2}}\et(\ve\la_k x)dx\\
 &= &
 \frac{e^{-i\rh_k\ln(\val s)}\val s^{1/2}}{\val{\la_k}^{1/2}}\int_\R
\big(\int_\R e^{t/2} e^{-i\rh_k t}\hat
F^{3,4}(t,\ve\frac{e^ts}{\la_k}-\frac{\ve
x}{\la_k},-\frac{1}{2}(\ve x+\ve{e^ts}{}),\la_k)dt\big)\\
 \nn  & &
{e^{i\rh_k\ln(\val{x})}\val x^{1/2}}{}\et( x)\frac {dx}{\val x}.
 \end{eqnarray}
Hence for  $C>3e^M $, $k $ large enough, $  \vert{t
}\vert\leq  M$, $  \vert{ x }\vert\leq R_k \vert{ \la_k }\vert$ and $ \vert{s
}\vert\geq C R_k  \vert{\la_k }\vert $, we have that $
\vert{\ve\frac{e^ts}{\la_k}-\frac{\ve x}{\la_k} }\vert> M $ and so
$\hat
F^{3,4}(t,\ve\frac{e^ts}{\la_k}-\frac{\ve x}{\la_k},-\frac{1}{2}(\ve
x+\ve{e^ts}{}),
{ \la_k})=0 $.  Thus,
\begin{eqnarray*}
 M_{\{ \vert{ }\vert\geq C R_k \val{\la_k}\}}\circ
V_{k}^*(\pi_{\rh_k,\la_k}(F)(V_{k }\circ  M_{\{ \vert{ }\vert\leq R_k
\vert{\la_k }\vert\}})=0
 \end{eqnarray*}
 for $k $ large enough. Choose an even continuous function $\va:\R\to \R_+ $ in
$C_c(\R) $ with compact support such that
\begin{eqnarray*}
 \val{\hat F^{3,4}(t,x,y,\la)}\leq \va(t)\va(x)\va(y),\quad t, x,y,\la\in\R.
 \end{eqnarray*}

Now, by  Young's inequality,
\begin{eqnarray*}
 \nn  & &
\noop{M_{ \{\vert{ }\vert\leq C R_k \val{\la_k}\}}\circ
V_{k}^*\circ  \pi_{\rh_k,\la_k}  (F)\circ V_{k }\circ  M_{ \{\vert{ }\vert\leq
R_k \vert{\la_k }\vert\} }}\\
\nn  &\leq &
\sup_{\val x\leq R_k \vert{\la_k }\vert }\int_{ \vert{s }\vert\leq C R_k
\vert{\la_k }\vert}\frac{\val {x s}^{1/2}}{\val{\la_k}^{1/2}}\big(\int_\R
e^{t/2} \vert \hat F^{3,4}(t,\ve\frac{e^ts}{\la_k}-\frac{\ve
x}{\la_k},-\frac{1}{2}(\ve x+\ve{e^ts}{}),\la_k)\vert dt\big)
\frac{ds}{\val s}\\
\nn  & &+
\sup_{\val s\leq C R_k \vert{\la_k }\vert }\int_{ \vert{x }\vert\leq  R_k
\vert{\la_k }\vert}\frac{\val {x s}^{1/2}}{\val{\la_k}^{1/2}}\big(\int_\R
e^{t/2} \vert \hat F^{3,4}(t,\ve\frac{e^ts}{\la_k}-\frac{\ve
x}{\la_k},-\frac{1}{2}(\ve x+\ve{e^ts}{}),\la_k)\vert dt\big)
\frac{dx}{\val x}\\
\nn  &\leq &
\sup_{\val x\leq  R_k \vert{\la_k }\vert }\int_{ \vert{s }\vert\leq C R_k
\vert{\la_k }\vert}\frac{\val x^{1/2}}{\val{\la_k s}^{1/2}}\big(\int_\R e^{t/2}
\va(t)\va({\frac{ e^ts}{\la_k}-\frac{ x}{\la_k}})\va(\frac{1}{2}(x+{e^ts}{}))
dt\big)ds\\
\nn  & &+
\sup_{\val s\leq C R_k \vert{\la_k }\vert }\int_{ \vert{x }\vert\leq R_k
\vert{\la_k }\vert}\frac{\val s^{1/2}}{\val{\la_k x}^{1/2}}\big(\int_\R e^{t/2}
\va(t)\va({\frac{ e^ts}{\la_k}-\frac{ x}{\la_k}})\va(\frac{1}{2}( x+{e^ts}{}))
dt\big)dx\\
\nn  &\leq &
R_k^{1/2}\int_{ \vert{s }\vert\leq C R_k \vert{\la_k }\vert}\frac{1}{\val{
s}^{1/2}}\big(\int_\R e^{t/2} \va(t)\va(\frac{e^ts}{\la_k}-\frac{
x}{\la_k})\va(\frac{1}{2}( x+{e^ts}{})) dt\big)ds\\
\nn  & &+ C^{1/2} R_k^{1/2}\int_{ \vert{x }\vert\leq R_k \vert{\la_k
}\vert}\frac{1}{\val{
x}^{1/2}}\big(\int_\R e^{t/2} \va(t)\va(\frac{e^ts}{\la_k}-\frac{
x}{\la_k})\va(\frac{1}{2}( x+{e^ts}{})) dt\big)dx\\
\nn  &\leq &
R_k^{1/2}\no\va_\iy^2\int_{ \vert{s }\vert\leq C R_k \vert{\la_k
}\vert}\frac{1}{\val{ s}^{1/2}}ds\int_\R e^{t/2} \va(t) dt\\
\nn  & &+ C^{1/2} R_k^{1/2}\no\va_\iy^2\int_{ \vert{x }\vert\leq R_k \vert{\la_k
}\vert}\frac{1}{\val{ x}^{1/2}}dx\int_\R e^{t/2} \va(t) dt\\
\nn  &= &
R_k^{1/2}\no\va_\iy^2{2C^{1/2}R_k^{1/2}\val{\la_k}^{1/2}}\int_\R e^{t/2} \va(t)
dt\\
\nn  & &
+ C^{1/2}R_k^{1/2}\no\va_\iy^2{2R_k^{1/2}\val{\la_k}^{1/2}}\int_\R e^{t/2}
\va(t) dt\\
\nn  &\leq & C R_k \vert{\la_k }\vert^{1/2}
 \end{eqnarray*}
 for some constant $C $.
\end{proof}

\begin{lemma}\label{dek muk}
Let $(\rh_k)_k$ be a real sequences with $\limk
\rh_k=0 $.
Then for any real sequence $(\la_k )_k$ we have that
\begin{eqnarray*}\label{}
 \nn\limk \noop{\pi_{\rh_k,\la_k}(F)-{\pi_{0,\la_k}}(F)}= 0, \quad F\in C^*(G).
 \end{eqnarray*}
 \end{lemma}

\begin{proof} For $k\in \N$, the identity
\begin{eqnarray*}\label{}
 \nn \pi_{\rh_k,\la_k}(F)-\pi_{0,\la_k}(F)=
 \int_{G} { (\chi_{\rh_k}(g)-1)}\pi_{0,\la_k}(g)F(g)dg, \, F\in
C_c(G),
 \end{eqnarray*}
where $\ch_{\rh_k}(\exp {t T}\cdot h)=e^{-i\rh_k t}, t\in\R,h\in H$, shows that
\begin{eqnarray*}\label{}
 \nn \noop{\pi_{\rh_k,\la_k}(F)-{\pi_{0,\la_k}}(F)}\leq
  \vert{\rh_k }\vert\no {F_1}_1,
 \end{eqnarray*}
where $F_1(t,x,y,z):= t F(t,x,y,z)$ and $g=(t,x,y,z)\in G$.
\end{proof}

 \begin{lemma}\label{difference operator lak rhk ta om0}
 Let $(\pi_{\rh_k,\la_k} )_k$ be  a properly converging sequence in $\wh G $
such that $\limk \la_k=0 $ and $\om_k:=\la_k\rh_k, \limk \om_k=0 $.
Take a sequence  $(R_k)_k $  in $\R_+ $ such that  $\limk R_k=\iy $ and
\begin{eqnarray}\label{Rk conditions}
 \limk R_k=\iy,\limk R_k^2\la_k=0, \ \limk
\frac{{\om_k}}{R_k^2\val{\la_k}}=0.
 \end{eqnarray}
 For instance, take $R_k=\frac{1}{\val{\la_k}^{1/3}}, k\in\N,$ if the sequence
$(\rh_k)_k $ is bounded. Otherwise, let $R_k^2=\val{\rh_k}M_k, k\in\N $,
where $\limk M_k=\iy$ and $\limk M_k\om_k=0 $. In the second case we have that
\begin{eqnarray*}\label{}
 \nn \limk \frac{\om_k}{R_k^2\la_k}&=&
 \limk \frac{\rh_k}{R_k^2}\\
 \nn  &= &
\limk\frac{1}{M_k}\\
\nn  &= &
0.
 \end{eqnarray*}
Let $F\in \l1G $ such that $\hat F^{3,4}\in C_c(\R^4) $.
For $\ve\in\{ +,-\} $, $\la_k=\ve \vert{\la_k }\vert $, we have that
\begin{enumerate}\label{limit -om+}
\item[(a)] $\nn\noop{\pi_{\rh_k,\la_k}(F)\circ V_{k}\circ M_{\{\geq
0\}}-V_k\circ
\ta^{+}_{\ve\om_k,-\ve}(F) \circ M_{\{\geq \val{\la_k}R_k\}} }\leq
C\big(\frac{ \vert
\om_k \vert }{\val{R_k^2\la_k}}+\frac{1}{R_k}\big), $
\item[(b)]
 \nn
 $\noop{\pi_{\rh_k,\la_k}(F)\circ V_{k}\circ M_{\{\leq
0\}}-V_k\circ \ta^{-}_{-\ve\om_k,\ve}(F) \circ M_{\{\leq
-\val{\la_k}R_k\}} }\leq C\big(\frac{ \vert \om_k \vert
}{\val{R_k^2\la_k}}+\frac{1}{R_k}\big), $
 \end{enumerate}
for some constant $C $ depending on $F $.
\end{lemma}

\begin{proof} The statements (a) and (b)  are proved in the same way as the
corresponding ones in Lemma~\ref{difference operator lak rhk ta}.
\end{proof}

\begin{definition}\label{Tk def}
Suppose that $\limk \la_k=0 $, $\rh_k\ne 0, k\in\N, $  and $\limk \val
{\rh_k\la_k}=0$.
\rm   Choose two sequences $(S_k), (T_k)_k$ in $\R_{>0}$ such that
$\limk S_k= \iy= \limk T_k $, $\limk \frac{S_k}{T_k}=0 $, $\limk \om_k T_k=0 $
and $R_k\leq \vert{\rh_k}\vert S_k $ for all $k\in \N$,  where the sequence
$(R_k)_k$ is given as above.
Define the subsets $I_{k,j}^{\pm}, k\in\N, j\in\{ 1,2,3\}$, by
{\begin{eqnarray*}\label{}
\nn {J_{k,1}^+}&:=& \rbrack R_k\vert \la_k\vert, \vert \om_k\vert
S_k \rbrack,\ J_{k,1}^-:= \lbrack -\vert \om_k\vert  S_k,-R_k\vert \la_k\vert  \lbrack,\\
 \nn I_{k,1}^+&:= & \rbrack 0, \vert \om_k\vert  S_k \rbrack,\ \ \ \
I_{k,1}^-:= \lbrack -\vert \om_k\vert  S_k,0  \lbrack,\\
 \nn  I_{k,2}^+&:= &] \vert \om_k\vert  S_k, \vert \om_k\vert  T_k ],\
\ \ \ I_{k,2}^-:= ] -\vert \om_k\vert  T_k,-\vert  \om_k\vert  S_k  ], \\
I_{k,3}^+&:= & \rbrack \vert \om_k\vert T_k,\iy \lbrack,\ \ \ \
I_{k,3}^-:= \rbrack -\iy,-\vert \om_k\vert  T_k \rbrack.
 \end{eqnarray*}}
 \end{definition}

\begin{lemma}\label{limk rhk lak=0}
Suppose that $\la_k=\ve \vert{\la_k }\vert$, $\rh_k\ne 0$, for $k\in\N$, and $\limk \la_k=0 $,
 $\limk \val {\rh_k\la_k}=0$. Take a sequence
$(R_k)_k $ in $\R_+ $ as in Lemma ~\ref{difference operator lak rhk ta om0} such that

\begin{eqnarray}\label{Rk conditions}
\limk R_k=+\iy, \limk R_k^2 \la_k=0,
\limk \frac{\om_k}{R_k^2\val{\la_k}}=0.
 \end{eqnarray}
Furthermore use sequences $(S_k)_k,(T_k)_k\subset \R_{>0} $ satisfying the
conditions in  Definition~\ref{Tk def}. Then for any $F\in C^*(G) $, we have
that
 \begin{enumerate}\label{}
\item[(a)] \begin{eqnarray*}
 \limk \noop{ \ta^+_{\om_k,-\ve }(F)-\ta^+_{\om_k,0
}(F)\circ
M_{J_{k,1}^+}\oplus \ta^+_{0,0 }(F)\circ M_{I_{k,2}^+}\oplus
\ta^+_{0,-\ve\vert{\om_k  }\vert  }(F)\circ M_{I_{k,3}^+}}=0,
\end{eqnarray*}
\item[(b)]  \begin{eqnarray*}
\limk \noop{ \ta^-_{-\om_k,\ve }(F)-\ta^-_{-\om_k,0
}(F)\circ
M_{J_{k,1}^-}\oplus \ta^-_{0,0 }(F)\circ M_{I_{k,2}^-}\oplus
\ta^-_{0,\ve\vert{\om_k  }\vert  }(F)\circ M_{I_{k,3}^-}}=0.
 \end{eqnarray*}
 \end{enumerate}

 \end{lemma}
 \begin{proof} First let $F\in \l1G $ such that $\hat F^{2,3,4}\in
C_c^\iy(\R^4) $. Then there exists $\va\in C_c^\iy(\R) $ of non-negative values
such that for all $t, t', x, x', y, y'\in \R$,
\begin{eqnarray*}\label{}
 \nn  \vert{F(t,x,y,0) }\vert &\leq &
 \va(t)\va(x)\va(y) \quad \text{and}\\
 \vert{F(t,x,y,0) -F(t',x', y')}\vert \nn  &\leq &
 \vert{t-t' }\vert\va(x)\va(y)+ \vert{x-x' }\vert\va(t)\va(y)+ \vert{y-y'
}\vert\va(t)\va (x).
 \end{eqnarray*}
Then it follows that
\begin{eqnarray*}\label{}
 & &
 \nn \vert
(\ta^+_{\om_k,{-\ve}}(F)-\ta^+_{\om_k,0}(F))(M_{I_{k,1}^+}
(\et))(u)\vert \\
&= &
 \vert \int_{I_{k,1}^+} (\hat F^{2,3,4}(t-\ln u,\om_k e^{-t},{ -\ve
e^t},0)-\hat
F^{2,3,4}(t-\ln u,\om_k e^{-t},0,0))M_{I_{k,1}^+}\et(e^t)dt \vert\\
&\leq &
{\om_k }\vert S_k\vert \int_{I_{k,1}^+}   \va(t-\ln(u))\va(\om_k
e^{-t})\vert \et(e^t)\vert dt.
 \end{eqnarray*}
Hence
\begin{eqnarray*}\label{}
 \nn \no{(\ta^+_{\om_k,-\ve}(F)-\ta^+_{\om_k,0}(F))\circ  M_{I_{k,1}^+}(\et)}^2
&\leq &
 {\om_k }^2 S_k^2\int_{\R_+} \Big(\int_{I_{k,1}^+}
\va(t-\ln(u))\va(\om_k
e^{-t})\vert \et(e^t)\vert dt\Big)^2\frac{du}{u}\\
\nn  &\leq &
{\om_k }^2 S_k^2 \no\et^2 \no\va_1^2\no\va_\iy.
 \end{eqnarray*}
Therefore,
\begin{eqnarray*}\label{}
 \nn \limk\noop{\ta^+_{\om_k,-\ve}(F)\circ
M_{I_{k,1}^+}-\ta^+_{\om_k,0}(F)\circ  M_{I_{k,1}^+}}= 0.
 \end{eqnarray*}
Since for any $e^t\in I_{k,2}^+ $ we have that
\begin{eqnarray*}\label{}
 \nn \frac{1}{T_k} \leq \frac{ \vert{\om_k }\vert}{e^t}\leq \frac{1}{S_k},
 \end{eqnarray*}
similarly
\begin{eqnarray*}\label{}
 & &
 \nn \vert(\ta^+_{\om_k,{
-\ve}}(F)-\ta^+_{0,0}(F))(M_{I_{k,2}^+}(\et))(u)\vert
\\
&= &
 \vert \int_{I_{k,2}^+} (\hat F^{2,3,4}(t-\ln u,\om_k e^{-t},{ -\ve
e^t},0)-\hat
F^{2,3,4}(t-\ln u,0,0,0))\et(e^t) dt\vert\\
&\leq &
(\frac{1}{S_k}+\frac{1}{T_k} )\no\va_\iy\int_{\R_+}
\va(t-\ln(u))\vert \et(e^t)\vert dt.
 \end{eqnarray*}
This relation implies that $\limk
\noop{\ta^+_{\om_k,-\ve}(F)\circ M_{I_{k, 2}^+}- \ta^+_{0,0}(F)\circ M_{I_{k, 2}^+}}=0 $.

In the same way, we have that
\begin{eqnarray*}\label{}
 & &
 \nn \vert (\ta^+_{\om_k,-\ve}(F)-\ta^+_{0,-\ve\vert
\om_k\vert }(F))(M_{I_{k,3}^+}(\et))(u)\vert \\
&= &
 \vert \int_{\R} (\hat F^{2,3,4}(t-\ln u,\om_k e^{-t},{ -\ve
e^t},0)- \hat
F^{2,3,4}(t-\ln u, 0, { -\ve \vert \om_k \vert
e^t},0))M_{I_{k,3}^+}\et(e^t)dt \vert\\
&= &
 \vert \int_{T_k}^\iy (\hat F^{2,3,4}(t+\ln( \vert{\om_k }\vert ) -\ln
u,\frac{1}{sign (\om_k) e^t}, { -\ve \vert{\om_k }\vert e^t},0)\\
& & -\hat F^{2,3,4}(t+\ln( \vert{\om_k }\vert )-\ln u, 0, {
-\ve\vert{\om_k }\vert
e^t},0)) \et( \vert{\om_k }\vert e^t) dt \vert\\
&\leq &
\frac{1}{T_k} \no\va_\iy\int_{I_{k,3}^+}
\va(t+\ln( \vert{\om_k }\vert )-\ln(u))\vert \et( \vert{\om_k }\vert
e^t)\vert dt.
 \end{eqnarray*}
 Hence
 \begin{eqnarray*}\label{}
& &
\nn \no{(\ta^+_{\om_k,-\ve}(F)-\ta^+_{0,-\ve \vert{\om_k }\vert }(F))\circ
M_{I_{k,3}^+}(\et)}^2 \\
&\leq &
 \frac{\no\va_\iy^2}{T_k^2}\int_{\R_+} \Big(\int_{I_{k,3}^+}
\va(t+\ln( \vert{\om_k }\vert )-\ln(u))M_{I_{k,3}^+}\et( \vert{\om_k }\vert
e^t)dt\Big)^2\frac{du}{u}\\
&\leq &
 \frac{\no\va_\iy^2}{T_k^2}\int_\R\va(t+\ln( \vert{\om_k }\vert
)-\ln(u))\int_{I_{k,3}^+}
\va(t+\ln( \vert{\om_k }\vert )-\ln(u))\vert \et( \vert{\om_k }\vert
e^t)\vert^2 dt\frac{du}{u} \\
&\leq&
\frac{\no\va_\iy^2 \no\va_1}{T_k^2}\int_{\R}
\va(t-s)\vert \et(
e^t)\vert^2 dt\\
\nn  &= &
\frac{\no\va_\iy^2 \no\va_1^2}{T_k^2} \no\et^2 .
 \end{eqnarray*}

 We proceed similarly for the second relation.

\end{proof}

\begin{definition}\label{sik def om=0}
\rm   Let $(\pi_{\rh_k,\la_k})_{k\in\N} $ be a properly converging sequence
with limit set $L=\Ga_1\cup\Ga_0 $
(i.e.  $\limk \la_k=0, \limk \om_k= 0 $ where $\om_k:= \rh_k \la_k=\rh_k\ve
 \vert{\la_k }\vert $) and
$\rh_k\ne0, k\in\N$. We then choose a
sequence $(R_k)_k\subset \R_{>0} $ such that
\begin{eqnarray*}
 \limk R_k=+\iy, \limk R_k^2\la_k=0, \limk
\frac{\om_k}{R^2_k\la_k}=0 .
 \end{eqnarray*}
We take the sequences $(S_k)_k, (T_k)_k $ as in Definition
{\ref{Tk def}}.
For $k\in\N $ and $\va\in l^\iy(\widehat{ G})$, define the bounded linear
operator
{{$\si_k^{0,\om_k,\ve}(\va \res L) $ (resp. $(\si_k^{ {0,
\om_k,\ve}})'(\va\res L)
$) on $\l2\R$ by
\begin{eqnarray*}
 \si_k^{0,\om_k,\ve}(\va\res L)&:=&
  V_k\circ  s_k^{0,\om_k, \ve, +}(\va\res L)\circ  V_k^*\oplus V_k\circ
s_k^{0,\om_k,\ve,-}(\va\res L)\circ  V_k^*,
 \end{eqnarray*}
 where
 \begin{eqnarray*}\label{}
 \nn s_k^{0,\om_k,\ve,+}(\va\res L) &:= &
 \va(\ta^+_{ \om_k,0
})\circ M_{J_{k,1}^+}\oplus \va(\ta^+_{0,0 })\circ M_{I_{k,2}^+}\oplus
\va(\ta^+_{0,-\ve\vert{\om_k  }\vert  })\circ M_{I_{k,3}^+}
 \end{eqnarray*} and
 \begin{eqnarray*}\label{}
 \nn s_k^{0,\om_k,\ve,-}(\va\res L) &:= &
   \va(\ta^-_{-\om_k,0})\circ
M_{J_{k,1}^-}\oplus \va(\ta^-_{0,0 })\circ M_{I_{k,2}^-}\oplus
\va(\ta^-_{0,\ve\vert{\om_k  }\vert  })\circ M_{I_{k,3}^-}.
 \end{eqnarray*}
}}

\end{definition}

\begin{theorem}\label{lim rhklak is om}
 Suppose that $\limk \la_k=0 $, $ \limk \rh_k\la_k =0$ and $\la_k=\ve\vert{\la_k
 }\vert$ for $k\in\N $. Then we have
\begin{eqnarray*}
 & &
 \limk \noop{\pi_{\rh_k,\la_k}(F)- {{\si_{k}^{0,\ve\om_k,\ve}(\hat
F\res L)}}}= 0.
 \end{eqnarray*}
for every $ F\in C^*(G)$.
 \end{theorem}
\begin{proof}
Apply
  Lemma \ref{difference operator lak rhk ta om0} and
Lemma \ref{limk rhk lak=0}.
\end{proof}

\subsection{The final theorem}\label{fth}

The representations $\ta \in \GA_0 $
 do not map $C^*(G) $ into the algebra $\K $ of compact
operators on
$\l2{\R_{+},\frac{dx}{x}} $.  Following \cite{Lin-Lud}, we define
the compact condition which will
give us  the distance of $\ta(F), F\in C^*(G), $ to the algebra $\K $.

Choose $q\in C_c(\R^2,\R_+) $ such that $\int_{\R^2} q(x,y)dxdy=1 $.
For $f\in\l1\R $, define the function $E(f)\in\l1{\R^3}\simeq \l1{G/\ZZ} $ by
\begin{eqnarray*}\label{}
 \nn E(f):= f(t)q(x,y) \, \text{ for all } (t,x,y)\in\R^3.
 \end{eqnarray*}
We see that for any $\ell=(\mu,\nu)\in  \R^2 $:
\begin{eqnarray*}\label{}
 \nn \pi_{\mu,\nu}(E(f))(\et)(u)= \int_\R f(u-t)\hat q(\mu e^{-t},\nu
e^t)\et(t)dt,\quad \et\in\l2\R, u\in\R.
 \end{eqnarray*}
This formula shows that
\begin{eqnarray*}\label{}
 \nn \pi_{\mu,\nu}(E(f))(\et)= l(f)(q_{\mu,\nu}\cdot \et),
 \end{eqnarray*}
where $l(f) $ is the operator $(l(f)\xi)(u):= \int_{\R}f(t)\xi(u-t){
dt}, \xi\in\l2\R,
u\in\R $,  given by the regular representation of the group $(\R,+) $ on the
Hilbert space ${\l2\R} $, and $q_{\mu,\nu} $ is the continuous
bounded
function $q_{\mu,\nu}(t):=\hat q(e^{-t}\mu,e^t\nu), t\in\R $.

Hence we have that
\begin{eqnarray*}\label{}
 \nn \noop{\pi_{\mu,\nu}(E(f))} \leq \noop{l(f)}\no{q_{\mu,\nu}}_\iy.
 \end{eqnarray*}
This shows that the mapping $E $ extends to a bounded linear mapping of
$C^*(\R)
$ into $C^*(G/\ZZ) $.
 We then have the bounded linear mapping
\begin{eqnarray}\label{Edef}
 \si_0: C_0(\Ga_0)\to C^*(G/\ZZ) \ \text{ given by } \ \si_0(\ps):= E(\F\inv(\ps)),
 \end{eqnarray}
where $\F\inv: C_0(\R)\to C^*(\R) $ is the inverse Fourier transform.

\begin{definition}\label{compact condition}
An operator field $\va $ is said to satisfy the \textit{compact condition} if
for $\ve\in\{ +, -\} $, the operators
$\va(\pi_{\ve,0})-\pi_{\ve,0}(\si_0(\va\res{\Ga_0}))$ and
$\va(\pi_{0,\ve})-\pi_{0,\ve}(\si_0(\va\res{\Ga_0})))$  are compact.
\end{definition}

 We define in the following the subset $D^*(G)$ of $ l^\iy(\wh{G}) $ which will
be our desired C*-algebra of Boidol's group.

\begin{definition}\label{def:DstarG}
\rm
Let $ D^*(G) $ be the subset of $ l^\iy(\wh{G}) $
as the set of all the operator fields $ \ph $ defined over $ \wh{G} $ such
that
\begin{enumerate}\label{}
\item  the mapping $\ga\mapsto \ph(\ga) $ vanishes at infinity,

\item
 \begin{enumerate}
 \item the mapping $\ga\mapsto \ph(\pi_\ga)$ is norm continuous on the set
$\GA_3$,
 \item for any $\ga\in\GA_3 $ the operator $\ph(\ga) $ is compact,

 \item  for every properly converging sequence $(\pi_{\rh_k,\la_k})_k$ in
$\Ga_3 $ with limit set $L\in \Ga_2 $ (i.e. $\limk \om_k=\om\ne 0 $,
where $\om_k= \rh_k \la_k$ and $\la_k=\ve  \vert{\la_k
}\vert $), we have that
\begin{eqnarray*}\label{}
 \nn \limk\noop{\ph(\pi_{\rh_k, \la_k}})- \si^\om_k(\ph\res L)=  0,
 \end{eqnarray*}
\item  for  every properly converging sequence $(\pi_{\rh_k,\la_k})_k$ in
$\Ga_3 $ with limit set $L=  \Ga_1\cup\Ga_0 $, (i.e. $\limk \om_k= 0 $),
admitting a sub-sequence, we have
\begin{eqnarray*}\label{}
 \nn \limk\noop{\ph(\pi_{\rh_k,\la_k})- \si_k^{0,\ve\om_k,\ve}(\ph\res L)}= 0,
 \end{eqnarray*}
 \end{enumerate}
\item
\begin{enumerate}\label{}

\item the mappings $\ga\mapsto \ph(\pi_\ga)$ are norm continuous
 on the sets $\GA_j$ for $j=0,1, 2$  (in particular the function
$\ph\res{\GA_0} $ is
contained in $C_\iy(\GA_0)=C_\iy(\R) \simeq C^*(\R)$),

 \item   for any $\ga\in \GA_{2}$, the operator  $\ph(\ga) $ is compact,
 \item   for every properly converging sequence $(\ta_{\om_k,-\ve})_k$ in
$\Ga_2$, $\ve\in\{ +,-\}$, with limit set $L= \Ga_1\cup \Ga_0 $ (i.e. $\limk
\om_k= 0 $), we have that
{\begin{eqnarray*}\label{}
& &
\limk \noop{ \ph(\ta^+_{\om_k,-\ve })-{ s_k^{0,\om_k,\ve,+}(\ph\res
L)}}= 0\\
  \text{and }\\
& &
\limk \noop{ \ph(\ta^-_{-\om_k,\ve })-{ s_k^{0,\om_k,\ve,-}(\ph\res
L)}}= 0.
 \end{eqnarray*}}

\item  the operator ${\ph(\ga),\ga\in \GA_1,} $ satisfies the compact condition,
 \end{enumerate}

\item  the adjoint operator field $\ph^* $ satisfies the same  conditions.
 \end{enumerate}
 \end{definition}

\begin{remark}\label{cstar GZZ}
\rm   It has been shown in \cite{Lin-Lud} that the family of uniformly bounded
operator fields defined over the set $\Ga^0:=\Ga_2\cup\Ga_1\cup\Ga_0 $
satisfying the conditions $(1) $, $(3)$ and  $(4) $ forms a
C*-algebra, which is isomorphic to $C^*(G/\ZZ) $, { where
$\ZZ$ is the centre of $G$}.

On the other hand, we have the natural projection $P_{G/\ZZ}:C^*(G)\to C^*(G/\ZZ) $ defined on
$\l1G $ by
\begin{eqnarray*}\label{}
 \nn P_{G/\ZZ}F(g)= \int_{\ZZ}F(gu)du, \quad g\in G.
 \end{eqnarray*}
This implies that for every operator field $\va\in D^*(G) $, there exists
$F_\va\in C^*(G) $ such that $\va\res {\GA^0}={
\widehat{F_\va}\res{\GA^0} }$. In
particular  for any sequence $(\rh_k,\la_k)_k\subset \R\times \R^*
$ in which $\limk \rh_k\la_k $ exists, we have  a subsequence satisfying
\begin{eqnarray*}\label{}
 \nn \limk\noop{\va(\pi_{\rh_k,\la_k})-\pi_{\rh_k,\la_k}(F_\va)}=  0
 \end{eqnarray*}
for all these $\va $'s.
 \end{remark}

We show that the set defined in Definition~\ref{def:DstarG} is a
C*-subalgebra of $l^\iy(\widehat{G})$.

\begin{lemma}\label{Dstar is cstar}
The subspace $D^*(G) $ of $l^\iy(\widehat{G}) $ is a $C^* $-algebra with
spectrum equal to $\widehat{G} $.
 \end{lemma}
\begin{proof} It is easy to see that $D^*(G) $ is a closed involutive subspace
of $l^\iy(\widehat{ G}) $. Let us show that it is also a subalgebra.
 Let $\va,\va' $ be two elements
of $D^*(G) $. By Remark \ref{cstar GZZ}, there exist $a,a'\in C^*(G/\ZZ) $
such that $\va\res{\Ga^0}= \hat a, \va'\res{\Ga^0}= \hat a'$. Therefore we  only
need to check if the conditions  (2-b) and (2-c) work for the product $\va\circ
\va'
$.
Now, $C^*(G) $ being an algebra, for any $\om\in{ \R^*} $ by Theorem~\ref{lim rhklak is om}
we have that
\begin{eqnarray*}\label{}
 \nn \limk\noop{\si_k^\om((\widehat{a\cdot a'})
\res{L})-\si_k^\om(\hat a
\res{L})\circ  \si_k^\om( \hat a' \res{L})}= 0
 \end{eqnarray*}
 and so
\begin{eqnarray*}\label{}
 \nn \limk \noop{\va\circ
\va'(\pi_{\rh_k,\la_k})-{\si_k^\om({\va\circ
\va'}\res{L})}}
&= &
 \limk\noop{\va\circ  \va'(\pi_{\rh_k,\la_k})-\si_k^\om((\widehat {a\cdot
a'}) \res{L})}\\
\nn  &= &
 \limk\noop{\va(\pi_{\rh_k,\la_k})\circ
\va'(\pi_{\rh_k,\la_k})-\si_k^\om(\hat a
\res{L})\circ \si_k^\om(\hat a'
\res{L}) }\\
\nn  &= &
0.
 \end{eqnarray*}
Hence $D^*(G) $ is  a C*-algebra, which contains the Fourier transform of
$C^*(G) $. It is clear that the spectrum of $D^*(G) $ contains $\widehat{G} $
since point evaluations are irreducible for $\widehat{C^*(G)} $.
Take now $\pi\in\widehat{D^*(G)} $. Let $R^0:D^*(G)\to\D^*(G/\ZZ) $ be the
restriction map and let $K^0 $ be its kernel in $D^*(G) $.
Then $D^*(G)/K^0\simeq D^*(G/\ZZ)\simeq C^*(G/\ZZ) $.
On the other hand,it follows immediately from the conditions (1)
to (3) that the ideal $K^0 $ of $D^*(G) $ is just the algebra
$C_\iy(\GA_3,\K) $ of continuous mappings defined on the locally compact space
 $\GA_3 $ with values in the algebra $\K $ of compact operators on $\l2\R
$.
If $\pi(K^0)=(0) $, then  $\pi $ can be identified with  an irreducible
representation of $C^*(G/\ZZ) $ and so there exists $\pi' \in \GA^0$, such that
{$\pi(\va)=\va(\pi'), \va\in D^*(G) $}. If
$\pi(K^0)\ne (0) $, then $\pi $ defines an irreducible  representation of
the
algebra $C_\iy(\GA_3,\K) $ and hence there exists  $\pi_{\rh,\la}\in \GA_3 $
such that $\pi(\va)=\va(\pi_{\rh,\la}),\va\in K^0 $, and finally
$\pi $ is the evaluation at $\pi_{\rh,\la} $ for every $\va\in D^*(G) $.
\end{proof}

{ Now we have our main theorem in characterising the C*-algebra $C^*(G)$ of
Boidol's group by our description of the Fourier transform of $C^*(G)$ onto
$D^*(G)$.}

\begin{theorem}\label{Cster G identified}
The Fourier transform defined in (\ref{fourrier def}) is an isomorphism
of the C*-algebra of  Boidol's group $G $ onto the C*-algebra $D^*(G) $.
 \end{theorem}
\begin{proof} Using Lemma \ref{Dstar is cstar}, the
Stone-Weierstrass theorem for C*-algebras (see \cite{Di}) tells us  that
the subalgebra $\widehat{C^*(G)} $ of $D^*(G) $ is equal to $D^*(G) $.

\end{proof}


TITLE: Which universities teach true infinitesimal calculus?
QUESTION [19 upvotes]: My colleague and I are currently teaching "true infinitesimal calculus" (TIC), in the sense of calculus with infinitesimals, to a class of about 120 freshmen at our university, based on the book by Keisler.  Two of my colleagues in Belgium are similarly teaching TIC at two universities there. I am also aware of such teaching going on in France in the Strasbourg area, based on Edward Nelson's approach, though I don't have any details on that.
Which universities teach true infinitesimal calculus? Anyone with any additional information in this direction is requested to provide it.
This is cross-posted here.
A colleague in Italy has recently told me about a conference on using infinitesimals in teaching in Italian highschools. This NSA (nonstandard analysis) conference was apparently well attended (over 100 teachers showed up). Anybody with more information about this (who to contact, what the current status of the proposal is, etc.) is hereby requested to provide such information here.
Our experiences teaching calculus with infinitesimals were detailed in this 2017 publication in Humanistic Mathematics.

REPLY [0 votes]: In june 2018 I became aware of a program teaching calculus with infinitesimals in Puerto Rico, which was the subject of this 2018 publication in The Mathematics Enthusiast.


TITLE: Object classifiers in 1-toposes
QUESTION [9 upvotes]: In a Grothendieck $\infty$-topos, it is known that, for arbitrarily large regular cardinals $\kappa$, there is a classifier for the class of relatively $\kappa$-compact morphisms. It is also easy to show that this is not the case in 1-toposes, because we might have isomorphic, but not equal, such morphisms classified by the same map. However, we should be able to recover at least part of the definition of an object classifier. Namely, I need to know that in a Grothendieck 1-topos, for arbitrarily large regular cardinals $\kappa$, there is a map $t: U' \to U$ such that for every relatively $\kappa$-compact morphism $f: X \to Y$ there exists a pullback square
$\require{AMScd}$
\begin{CD}
X @>>> U'\\
@VfVV @VVtV\\
Y @>>> U
\end{CD}
(not necessarily unique and such that the map $Y \to U$ doesn't necessarily only classify $f$). I feel that this should definitely be true, but I can't find it anywhere in the literature. It would be very nice to have a reference for it, or a confutation in the unfortunate case I'm wrong. Thanks!

REPLY [2 votes]: If $\mathcal{X}$ is a topos, then there is an adjunction $\mathcal{X} \leftrightarrows \mathcal{P(C)}$ with a category of presheaves of sets, where the right adjoint is fully faithful and accessible and the left is left exact.
The idea is very similar to the one spelled out in https://www2.mathematik.tu-darmstadt.de/~streicher/NOTES/lift.pdf, except that where its authors first find a suitable class of relatively $\kappa$-compact morphisms in $\mathcal{P(C)}$ and then consider the class of all morphisms in $\mathcal{X}$ that become relatively $\kappa$-compact in $\mathcal{P(C)}$, my analysis allows to find some $\kappa$ such that the desired class is that of all relatively $\kappa$-compact morphisms in $\mathcal{X}$. A complete proof can be found in Appendix A of http://www.tac.mta.ca/tac/volumes/37/5/37-05.pdf
I'd like to point out that neither strategy is more general than the other. The reason of this is that a class of relatively $\kappa$-compact morphisms in $\mathcal{X}$ need not be sent precisely to a class of relatively $\lambda$-compct morphisms in $\mathcal{P(C)}$ and, conversely, the preimage of a class of relatively $\lambda$-compact morphisms in $\mathcal{P(C)}$ need not be precisely a class of relatively $\kappa$-compact morphisms in $\mathcal{X}$.
However, all relatively $\kappa$-compact morphisms found in the latter proof are indeed sent to relatively $\kappa$-compact morphisms in $\mathcal{P(C)}$.


TITLE: Credit Given - Geometricly Modeling Infinity with 3 planes and 9 circles - Ratio of Circles
QUESTION [2 upvotes]: Refer to the attached diagram sketch to help visualize the equation.
I am requesting help with an interesting math problem. Basically, I am diagraming infinity using three planes. These planes connect with each other at either side at 120 degree angles.
These connection lines never really connect at the center, as they go on into infinity, but appear to (think repeating 9). Three perfect circles are placed on each of the planes. When viewed 2 dimensionally, the circles appear to be ellipses.
The Problem: What is the width/height ratio of these illusional ellipses? For computational purposes, put the three perfect circles on each plane touching each other. The actual width of the minor axis  for the smallest illusional ellipse can be set to 1 for computational purposes (or really any number you fee like).
The Diagram: I have used the 1:3 ratio to make the circles in this diagram (120 by 360, 360 by 1080, and 1080 by 3240). I suspect that this ratio is not correct; however, the answer should be fairly close to this ratio. I am not entirely sure if the ratio will be different for each of the circles, but it could be.
Thank you in advance for your time working on this problem. If you have seen it before, great! If not, hopefully you will enjoy the challenge. Solving this problem in needed to complete a logo for a non-profit organization. Credit will be given to whoever solves this.

REPLY [1 votes]: As @JTrana suggests in a comment, the answer depends on how much foreshortening you want to do. If you look down on a tripod, you'll see the legs make $120^\circ$ angles, no matter how "open" the tripod is, so that part is unchanging; circles fitted between the legs of the tripod, however, will appear as ellipses of varying eccentricities for different degrees of tripod open-ness.
Since there's no inherent foreshortening to go along with the apparent $120^\circ$ separation of the legs, then, you may use any ellipse you like in your logo, secure in the knowledge that it corresponds to some amount of foreshortening. This gives you the artistic freedom to use ellipses you find aesthetically pleasing (aspect ratio = golden ratio, perhaps?).
If, however, you really-really want the image to represent the shadow of a particular 3D configuration, then we can do that, too. I'll walk you through it.

Let $OABC$ be a translucent triangular pyramid (ie, a tetrahedron) with equilateral base $\triangle ABC$ and identical isosceles lateral faces such as $\triangle OBC$. (The lateral edges of the pyramid are like the legs of the tripod.) A circle painted on $\triangle OBC$ will cast an elliptical shadow onto the base; one dimension of this ellipse (the one parallel to $\overline{BC}$ will match the diameter of the circle; the perpendicular dimension will undergo the same deformation as the shadow cast by face $\triangle OBC$ itself.
Now, the shadow of $\triangle OBC$ is $\triangle PBC$, where $P$, the shadow of $O$, is the center of $\triangle ABC$. Observe that here, too, that one dimension of the shadow (namely $\overline{BC}$) matches that of the face, while the perpendicular dimension has been deformed. Writing $M$ for the midpoint of $\overline{BC}$, we can express the deformation factor as
$$k = \frac{|\overline{PM}|}{|\overline{OM}|}$$
so that the shadow ellipses have aspect ratios $1:k$.
Since $\triangle ABC$ is equilateral, we have that $|\overline{PM}| = \frac{1}{3}|\overline{AM}|$; moreover, $|\overline{AM}| = \frac{\sqrt{3}}{2}|\overline{BC}|$. Thus, we can write
$$k = \frac{1}{2\sqrt{3}}\;\;\frac{|\overline{BC}|}{|\overline{OM}|} = \frac{\sqrt{3}}{6}\;\frac{\text{base of } \triangle OBC}{\text{altitude of } \triangle OBC}$$
If you like, we can express $k$ in terms of other aspects of the tetrahedron.

If the edges of base $\triangle ABC$ have length $b$, while the lateral edges have length $\ell$, then applying the Pythagorean Theorem to $\triangle OMB$ yields $$|\overline{OM}|^2 + \left(\frac{1}{2} b\right)^2 = \ell^2 \qquad \text{so that} \qquad k = \frac{b\;\sqrt{3}}{3\sqrt{4\ell^2 - b^2}}$$
If the lateral edges make an angle of $\theta$ with each other, then $|\overline{OM}| = \ell \cos(\theta/2)$ and $|\overline{BC}| = 2 \ell \sin(\theta/2)$, so that $$k = \frac{\sqrt{3}}{3}\;\frac{\sin(\theta/2)}{\cos(\theta/2)} = \frac{\sqrt{3}}{3}\;\tan\frac{\theta}{2}$$
If the pyramid is in fact a regular tetrahedron ---with all edges equal and making $60^\circ$ angles with each other--- then either of the above reduces to $$k = \frac{1}{3}$$
If the pyramid is a right-corner tetrahedron ---with lateral edges at right angles to each other--- then we have $$k = \frac{\sqrt{3}}{3}$$

... and so forth.


TITLE: Why is change in Gibbs free energy not always zero? ($q = T \Delta S$)
QUESTION [0 upvotes]: We know that ,
$ \Delta G = \Delta H - T \Delta S$
We know that $ \Delta H =  q = T \Delta S$
so,
$\Delta G = 0$
Where is my mistake here?

REPLY [1 votes]: The change in Gibbs free energy is zero for a reversible process (as long as no non-PV work is being done).  The equivalence that you provide, $\Delta H = q = T \Delta S$, is only valid for reversible processes at constant $T$ and $P$.  If the process is irreversible then $T \Delta S \neq q$, and thus $\Delta G \neq 0$.


text:
\section{Establishing WSC/WSS - Supplementary Details}
\label{app:wsc-wss}
Recall that for any $r > 0$ and $M > 0$, $\cS_M(r)$ denotes the set of all diagonal $M$-truncated weight matrices \stir could possibly generate with respect to models residing in the radius $R$ ball centered at $\vwo$ i.e.
\[
\cS_M(r) := \bc{S = \diag(\vs), \vs_i = \min\bc{\frac1{\abs{\ip\vw{\vx_i}-y_i}},M}, \vw \in \cB_2(\vwo,r)},
\]
then we have the following result.
\begin{lemma}
\label{lem:wsc-wss}
Suppose the data covariates $X = \bs{\vx_1,\ldots,\vx_n}$ are generated from an isotropic $R$-sub-Gaussian distribution $\cD$, and $G$ denotes the set of uncorrupted points (as well as the size of that set) then there exists a constant $c$ that depends only on the distribution $\cD$ such that for any fixed value of $M > 0$,
\[
\left.
\begin{array}{r}
\P{\exists S \in \cS_M\br{\frac1M}: \lambda_{\min}(X_GS_GX_G^\top) < 0.99c\cdot GM}\\
\vspace*{-2ex}\\
\P{\exists S \in \cS_M\br{\frac1M}: \lambda_{\max}(X_GS_GX_G^\top) > 1.01\cdot GM}
\end{array}
\right\} \leq \exp\br{-\Om{n - d\log(d + n)}},
\]
where the constants inside $\Om{\cdot}$ are clarified in the proof. In particular, if $\cD$ is the standard Gaussian $\cN(\vzero,I_d)$, then we can take $c = 0.96$.
\end{lemma}
\begin{proof}
The bound for the largest eigenvalue follows directly due to the fact that all weights are upper bounded by $M$ and hence $X_GS_GX_G^\top \preceq M\cdot X_GX_G^\top$ and applying Lemma~\ref{lem:ssc-sss}. For the bound on the smallest eigenvalue, notice that Lemma~\ref{lem:point-conv} shows us that for any fixed $S \in \cS_M(\frac1M)$, i.e. a set of $M$-truncated weights that correspond to some fixed model $\vw \in \cB_2\br{\vwo,\frac1M}$, we have
\[
\P{\lambda_{\min}(X_GS_GX_G^\top) < 0.995 c\cdot GM} \leq 2\cdot9^d\exp\bs{-\frac{mn(0.005c)^2}{8R^4}}
\]
Recall that we let $R_X := \max_{i \in [n]}\ \norm{\vx_i}_2$ denote the maximum Euclidean length of any covariate. However, Lemma~\ref{lem:approx} shows us that if $\vw^1,\vw^2 \in \cB_2\br{\vwo,\frac1M}$ are two models such that $\norm{\vw^1 - \vw^2}_2 \leq \tau$ then, conditioned on the value of $R_X$, the following holds \emph{almost surely}.
\[
\abs{\lambda_{\min}(X_GS^1_GX_G^\top) - \lambda_{\min}(X_GS^2_GX_G^\top)} \leq 2G\tau M^2R_X^3
\]
This prompts us to initiate a uniform convergence argument by setting up a $\tau$-net over $\cB_2\br{\vwo,\frac1M}$ for $\tau = \frac c{400R_X^3M}$. Note that such a net has at most $\br{\frac{800R_X^3}c}^d$ elements by applying standard covering number bounds for the Euclidean ball \cite[Corollary 4.2.13]{Vershynin2018}. Taking a union bound over this net gives us
\begin{align*}
\P{\exists S \in \cS_M\br{\frac1M}: \lambda_{\min}(X_GS_GX_G^\top) < 0.99c\cdot GM} &\leq 2\cdot\br{\frac{7200R_X^3}c}^d\exp\bs{-\frac{mn(0.005c)^2}{8R^4}}\\
&\leq \exp\br{-\Om{n - d\log(d + n)}},
\end{align*}
where in the last step we used Lemma~\ref{lem:rx} to bound $R_X = \bigO{R\sqrt{d + n}}$ with probability at least $1 - \exp(-\Om n)$. For the specific bound on the constant $c$ for various distributions, including the Gaussian distribution, we refer the reader to Section~\ref{app:c-values}.
\end{proof}

The proof of the above result relies on several intermediate results which we prove in succession below. In the first result Lemma~\ref{lem:point-exp}, we establish expected bounds on the extremal singular values of the matrix $X_GS_GX_G^\top$ corresponding to a fixed model $\vw \in \cB_2\br{\vwo,\frac1M}$. In the next result Lemma~\ref{lem:point-conv}, we establish the same result, but this time with high probability instead of in expectation. The next result Lemma~\ref{lem:approx} establishes that extremal singular values corresponding to two models close to each other must be (deterministically) close.

\begin{lemma}[Pointwise Expectation]
\label{lem:point-exp}
With the same preconditions as in Lemma~\ref{lem:wsc-wss}, there must exist a constant $c > 0$ that depends only on $\cD$ such that for any fixed $S \in \cS_M(\frac1M)$, and fixed vector unit $\vv \in S^{d-1}$, we have
\[
c\cdot GM \leq \E{\vv^\top X_GS_GX_G^\top\vv} \leq GM.
\]
In particular, if $\cD$ is the standard Gaussian $\cN(\vzero,I_d)$, then we can take $c = 0.96$.
\end{lemma}
\begin{proof}
Let $\vx \sim \cD$ and let $y = \ip{\vwo}{\vx}$. Then if we let $\vDelta := \frac{\vw - \vwo}{\norm{\vw - \vwo}_2}$ (note that $\norm{\vw - \vwo} \leq \frac1M$), then we have $s = \min\bc{\frac1{\abs{\ip{\vw}{\vx} - y}}, M} \geq M\cdot\min\bc{\frac1{\abs{\ip\vDelta\vx}},1}$ as well as $s \leq M$. Then by linearity of expectation we have
\[
\E{\vv^\top X_GS_GX_G^\top\vv} = \E{\sum_{i\in G}\vs_i\ip{\vx_i}{\vv}^2} = G\cdot\E{s\cdot\ip\vx\vv^2} \leq GM\cdot\E{\ip\vx\vv^2} = GM,
\]
since $\cD$ is isotropic. We also get
\[
\E{\vv^\top X_GS_GX_G^\top\vv} = G\cdot\E{s\cdot\ip\vx\vv^2} \geq GM\cdot\E{\min\bc{\frac1{\abs{\ip\vDelta\vx}},1}\cdot\ip\vx\vv^2} \geq c\cdot GM,
\]
where, for any distribution $\cD$ over $\bR^d$, we define the constant $c$ as
\[
c := \inf_{\vu,\vv \in S^{d-1}}\bc{\Ee{\vx\sim\cD}{\min\bc{\frac1{\abs{\ip\vu\vx}},1}\cdot\ip\vx\vv^2}}.
\]
This concludes the proof. For the specific bound on the constant $c$ for various distributions, including the Gaussian distribution, we refer the reader to Section~\ref{app:c-values}.
\end{proof}

\begin{lemma}[Pointwise Convergence]
\label{lem:point-conv}
With the same preconditions as in Lemma~\ref{lem:wsc-wss}, for any fixed $S \in \cS_M(\frac1M)$,
\[
\left.
\begin{array}{r}
\P{\lambda_{\min}(X_GS_GX_G^\top) < 0.995 c\cdot GM}\\
\vspace*{-2ex}\\
\P{\lambda_{\max}(X_GS_GX_G^\top) > 1.005 \cdot GM}
\end{array}
\right\}
\leq 2\cdot9^d\exp\bs{-\frac{mn(0.005c)^2}{8R^4}}
\]
\end{lemma}
\begin{proof}
Note that for any square symmetric matrix $A \in \bR^{d \times d}$, we have $c - \delta \leq \lambda_{\min}(A) \leq \lambda_{\max}(A) \leq c + \delta$ for some $\delta > 0$ iff $\abs{\vv^\top A\vv - c} \leq \delta$ for all $\vv \in S^{d-1}$ which itself happens iff $\norm{A - c\cdot I}_2 \leq \delta$. Now, if $\cN_\epsilon$ denotes an $\epsilon$-net over $S^{d-1}$, then for any square symmetric matrix $B \in \bR^{d\times d}$, we have $\norm{B}_2 \leq (1-2\epsilon)^{-1}\sup_{\vv \in \cN_\epsilon}\abs{\vv^\top B\vv}$. Thus, setting $B = A - c\cdot I$ and $\epsilon = 1/4$, we have $\norm{A - c\cdot I}_2 \leq 2\sup_{\vv \in \cN_{1/4}}\abs{\vv^\top A\vv - c}$.

Let $\vx \sim \cD$ and $t = \sqrt{\min\bc{\frac1{\abs{\ip{\vw - \vwo}\vx}},M}} \leq \sqrt M$ and for any fixed $\vv \in S^{d-1}$, let $Z := t\cdot\ip\vx\vv$. Then we have
\[
\norm{Z}_{\psi_2} = \sup_{p \geq 1} p^{-1/2}\br{\E{|Z|^p}}^{1/p} \leq \sqrt M\cdot\sup_{p \geq 1} p^{-1/2}\br{\E{|\ip\vx\vv|^p}}^{1/p} = R\sqrt M,
\]
where the last step follows by observing that since $\cD$ is $R$-sub-Gaussian, $\norm{\ip{\vx_1}{\vv}}_{\Psi_2} \leq R$. Thus, $Z$ is $R\sqrt M$-sub-Gaussian. This implies $Z^2$ is $MR^2$-subexponential (see \cite[Lemma 2.7.6]{Vershynin2018}), as well as $Z^2 - \bE Z^2$ is $2MR^2$-subexponential by centering and applying the triangle inequality. Note that Lemma~\ref{lem:point-exp} implicitly establishes that $\mu := \bE Z^2 \in [cM,M]$. Let $Z_1,Z_2,\ldots,Z_G$ be independent realizations of $Z$ with respect to a fixed vector $\vv$. Then we have
\begin{align*}
\P{\abs{\vv^\top X_GS_GX_G^\top\vv - G\mu} \geq \varepsilon\cdot GM} &= \P{\abs{\sum_{i \in G}(Z^2_i - \mu)} \geq \varepsilon\cdot GM}\\
&\leq 2\exp\bs{-m\cdot\min\bc{\frac{(\varepsilon\cdot GM)^2}{4M^2R^4G},\frac{\varepsilon\cdot GM}{2MR^2}}}\\
&\leq 2\exp\bs{-\frac{mn\varepsilon^2}{8R^4}}
\end{align*}
where $m > 0$ is a universal constant and in the last step we used $G \geq n/2$ and w.l.o.g. we assumed that $\varepsilon \leq 2R^2$. Taking a union bound over all $9^d$ elements of $\cN_{1/4}$, we get
\begin{align*}
\P{\norm{X_GS_GX_G^\top - G\mu\cdot I}_2 \geq \varepsilon\cdot GM} &\leq \P{\max_{\vv\in\cN_{1/4}}\abs{\vv^\top X_GS_GX_G^\top\vv - G\mu} \geq \frac\varepsilon2\cdot GM}\\
&\leq 2\cdot9^d\exp\bs{-\frac{mn\varepsilon^2}{8R^4}}
\end{align*}
Setting $\varepsilon = 0.005c$ and noticing that $\mu \in [cM,M]$ by Lemma~\ref{lem:point-exp} finishes the proof.
\end{proof}

\begin{lemma}[Approximation Bound]
\label{lem:approx}
Consider two models $\vw^1,\vw^2 \in \bR^d$ such that $\norm{\vw^1 - \vw^2}_2 \leq \tau$ and let $\vs^1,\vs^2$ denote the $M$-truncated weight vectors they induce i.e. $s^j_i = \min\bc{M,\frac1{\abs{\ip{\vw^j}{\vx_i} - y_i}}}, j = 1,2$. Also let $S^1 = \diag(\vs^1)$ and $S^2 = \diag(\vs^2)$. Then for any $X = [\vx_1,\ldots,\vx_n] \in \bR^{d\times n}$ such that $\norm{\vx_i}_2 \leq R_X$ for all $i$,
\[
\abs{\lambda_{\min}(XS^1X^\top) - \lambda_{\min}(XS^2X^\top)} \leq 2n\tau M^2R_X^3
\]
\end{lemma}
\begin{proof}
We have the following four cases with respect to the weights $s^j_i = \min\bc{M,\frac1{\abs{\ip{\vw^j}{\vx_i} - y_i}}}, j = 1,2$ these two models generate on any data point $\vx_i \in \cB_2(R_X)$. Note that we do not assume that these data points are generated from $\cD$, just that they are bounded inside the ball $\cB_2(R_X)$. Also note that although $\abs{s^1_i - s^2_i} \leq M$ trivially holds by virtue of truncation, such a result is not sufficient for us since our later analyses would like to be able to show $\abs{s^1_i - s^2_i} \leq \frac M{1000}$ by setting $\tau$ to be really small.
\begin{description}
	\item[Case 1]: $\abs{\ip{\vw^1}{\vx_i} - y_i} \leq \frac1M$ and $\abs{\ip{\vw^2}{\vx_i} - y_i} \leq \frac1M$. Here $s^1_i = s^2_i = M$ i.e. $\abs{s^1_i - s^2_i} = 0$.
	\item[Case 2]: $\abs{\ip{\vw^1}{\vx_i} - y_i} > \frac1M$ but $\abs{\ip{\vw^2}{\vx_i} - y_i} \leq \frac1M$. In this case $s^2_i = M > s^1_i$. Thus,
	\[
	\abs{s^1_i - s^2_i} = M - \frac1{\abs{\ip{\vw^1}{\vx_i} - y_i}} \leq M - \frac1{\abs{\ip{\vw^2}{\vx_i} - y_i} + \tau R_X} \leq M  - \frac M{1 + \tau MR_X} < 2\tau M^2R_X
	\]
	\item[Case 3]: $\abs{\ip{\vw^1}{\vx_i} - y_i} \leq \frac1M$ but $\abs{\ip{\vw^2}{\vx_i} - y_i} > \frac1M$. This is similar to Case 2 above.
	\item[Case 4]: $\abs{\ip{\vw^1}{\vx_i} - y_i} > \frac1M$ and $\abs{\ip{\vw^2}{\vx_i} - y_i} > \frac1M$. In this case we have
	\[
	\abs{\frac1{\abs{\ip{\vw^1}{\vx_i} - y_i}} - \frac1{\abs{\ip{\vw^2}{\vx_i} - y_i}}} \leq \frac{\abs{\ip{\vw^1-\vw^2}{\vx_i}}}{\abs{\ip{\vw^1}{\vx_i} - y_i}\cdot\abs{\ip{\vw^2}{\vx_i} - y_i}} \leq 2\tau M^2R_X
	\]
\end{description}
This tells us that $\norm{\vs^1 - \vs^2}_1 \leq 2n\tau M^2R_X$. Now, if we let $S^1 = \diag(\vs^1)$ and $S^2 = \diag(\vs^2)$, then for any unit vector $\vv \in S^{d-1}$, denoting $R_X := \max_{i \in [n]}\ \norm{\vx_i}_2$ we have
\[
\abs{\vv^\top XS^1X^\top\vv - \vv^\top XS^2X^\top\vv} = \abs{\sum_{i=1}^n\br{\vs^1_i-\vs^2_i}\ip{\vx_i}{\vv}^2} \leq \norm{\vs^1 - \vs^2}_1\cdot\max_{i\in[n]}\ \ip{\vx_i}{\vv}^2 \leq \norm{\vs^1 - \vs^2}_1\cdot R_X^2 \leq 2n\tau M^2R_X^3.
\]
This proves that $\norm{XS^1X^\top - XS^2X^\top}_2 \leq 2n\tau M^2R_X^3$ and concludes the proof.
\end{proof}

\subsection{Calculation of Distribution-specific Constants}
\label{app:c-values}
The WSC/WSS bounds from Lemma~\ref{lem:wsc-wss} are parametrized by a constant $c$ that lower bounds on the singular values of the matrix $X_GS_GX_G^\top$. Recall that for any covariate distribution $\cD$, the constant is defined as
\[
c := \inf_{\vu,\vv \in S^{d-1}}\bc{\Ee{\vx\sim\cD}{\min\bc{\frac1{\abs{\ip\vu\vx}},1}\cdot\ip\vx\vv^2}}.
\]
Below we present some interesting cases where this constant is lower bounded.
\begin{description}
	\item[Centered Isotropic Gaussian] For the special case of $\cD = \cN(\vzero,I_d)$, notice that by rotational symmetry, we can, without loss of generality, take $\vu = (1,0,0,\ldots,0)$ and $\vv = (v_1,v_2,0,0,\ldots,0)$ where $v_1^2 + v_2^2 = 1$. Thus, if we consider $x_1,x_2 \sim \cN(0,1)$ i.i.d. then $c \geq \inf_{(v_1,v_2) \in S^1} f(v_1,v_2)$ where
	\begin{align*}
		f(v_1,v_2) &= \Ee{x_1,x_2\sim\cN(0,1)}{\min\bc{\frac1{\abs{x_1}},1}\cdot(v_1^2x_1^2 + v_2^2x_2^2 + 2v_1v_2x_1x_2)}\\
		&= \Ee{x_1,x_2\sim\cN(0,1)}{\min\bc{\frac1{\abs{x_1}},1}\cdot(v_1^2x_1^2 + v_2^2x_2^2)}\\
		&= \Ee{x_1\sim\cN(0,1)}{\min\bc{\frac1{\abs{x_1}},1}\cdot(v_1^2x_1^2 + v_2^2)}\\
		&= \sqrt\frac2\pi\br{\int_0^1 (v_1^2 t^2 + v_2^2) e^{-t^2/2}dt + \int_1^\infty \br{v_1^2t + \frac{v_2^2}t} e^{-t^2/2}dt}\\
		&\geq 0.6827\cdot v_1^2 + 0.9060\cdot v_2^2
	\end{align*}
	where in the second step we used the independence of $x_1,x_2$ and $\E{x_2} = 0$, in the third step we used independence once more and $\E{x_2^2} = 1$, and in the last step we used standard bounds on the error function and the exponential integral. This gives us $c \geq \inf_{(v_1,v_2) \in S^1}\ \bc{0.6827\cdot v_1^2 + 0.9060\cdot v_2^2} \geq 0.68$.
	\item[Centered Non-isotropic Gaussian] For the case of $\cD = \cN(\vzero,\Sigma)$, we have $\vx \sim \cD \equiv \Sigma^{1/2}\cdot\cN(\vzero,I_d)$. Thus, for any fixed unit vector $\vv$, we have $\ip\vv\vx \sim \ip{\tilde\vv}\vz$ where $\tilde\vv = \Sigma^{-1/2}\vv$ and $\vz\sim\cN(\vzero,I)$. We also have $\norm{\tilde\vv}_2 \in \bs{\frac1{\sqrt\Lambda},\frac1{\sqrt\lambda}}$ where $\lambda = \lambda_{\min}(\Sigma)$ and $\Lambda = \lambda_{\max}(\Sigma)$. Note that we must insist on having $\lambda = \lambda_{\min}(\Sigma) > 0$ failing which, as the calculations show below, there is no hope of expecting $c$ to be bounded away from $0$. Now for any fixed vectors $\vu,\vv$ we first perform rotations so that we have $\tilde\vu = (u,0,0,\ldots,0)$ and $\tilde\vv = (v_1,v_2,0,0,\ldots,0)$ where we can assume w.l.o.g. that $u \geq 0$. Note that since $\bc{\norm{\tilde\vu}_2,\norm{\tilde\vv}_2} \in \bs{\frac1{\sqrt\Lambda},\frac1{\sqrt\lambda}}$, we have $(v_1,v_2) \in S^r$ and $r, u \in \bs{\frac1{\sqrt\Lambda},\frac1{\sqrt\lambda}}$. This gives us $c \geq \inf_{(v_1,v_2) \in S^r} f(v_1,v_2)$ where
	\begin{align*}
		f(v_1,v_2) &= \Ee{x_1,x_2\sim\cN(0,1)}{\min\bc{\frac1{u\cdot\abs{x_1}},1}\cdot(v_1^2x_1^2 + v_2^2x_2^2 + 2v_1v_2x_1x_2)}\\
		&= \Ee{x_1,x_2\sim\cN(0,1)}{\min\bc{\frac1{u\cdot\abs{x_1}},1}\cdot(v_1^2x_1^2 + v_2^2x_2^2)}\\
		&= \Ee{x_1\sim\cN(0,1)}{\min\bc{\frac1{u\cdot\abs{x_1}},1}\cdot(v_1^2x_1^2 + v_2^2)}\\
		&= \frac1u\sqrt\frac2\pi\br{\int_0^\frac1uu(v_1^2t^2 + v_2^2)e^{-t^2/2}dt + \int_\frac1u^\infty\br{v_1^2t + \frac{v_2^2}t}e^{-t^2/2}dt}\\
		&\geq \frac1u\sqrt\frac2\pi\br{\int_0^\frac1uu(v_1^2t^2 + v_2^2)e^{-\frac12\br{\frac1u}^2}dt + v_1^2e^{-\frac12\br{\frac1u}^2} + \frac{v_2^2}2\int_{\frac12\br{\frac1u}^2}^\infty \frac1ze^{-z}dz}\\
		&\geq \frac1u\sqrt\frac2\pi\br{e^{-\frac12\br{\frac1u}^2}\br{\frac{v_1^2}{3u^2} + v_2^2} + v_1^2e^{-\frac12\br{\frac1u}^2} + \frac{v_2^2}4e^{-\frac12\br{\frac1u}^2}\log\br{1+4u^2}}\\
		&\geq \sqrt{\frac{2\lambda}\pi}e^{-\frac\Lambda2}\br{v_1^2\br{1 + \frac\lambda3} + v_2^2\br{1 + \frac14\log\br{1+\frac4\Lambda}}}\\
		&\geq \sqrt{\frac{2\lambda}\pi}e^{-\frac\Lambda2}(v_1^2 + v_2^2)\\
		&= r^2\sqrt{\frac{2\lambda}\pi}e^{-\frac\Lambda2}\\
		&\geq \frac1\Lambda\sqrt{\frac{2\lambda}\pi}e^{-\frac\Lambda2}
	\end{align*}
	where in the second and third steps we used independence of $x_1,x_2$, $\E{x_2} = 0$ and $\E{x_2^2} = 1$ as before, and in the sixth step we used lower bounds on the exponential integral.
	\item[Non-centered Isotropic Gaussian] We discuss two techniques to handle the case of non-centered covariates.
		\begin{itemize}
			\item \textbf{Pairing Trick} This technique requires changes to the data points and relies on the fact that the difference of two i.i.d. non-centered Gaussian random variables is a centered Gaussian random variable with double the variance. Thus, given $n$ covariates $\vx_1,\ldots,\vx_n \sim \cN(\vmu,I_d)$ and corresponding responses $y_1,\ldots,y_n$, create $n/2$ data points (assume without loss of generality that $n$ is even) $\tilde\vx_i = \frac{\vx_i - \vx_{i+n/2}}{\sqrt2}$ and $\tilde y_i = \frac{y_i - y_{i+n/2}}{\sqrt2}$. Clearly $\tilde\vx_i \sim \cN(\vzero, 2\cdot I_d)$. However, this method has drawbacks since it is likely to increase the proportion of corrupted data points. If $\alpha$ fraction of the original points were corrupted, at most $2\alpha$ fraction of the new points would be corrupted.
			\item \textbf{Direct Centering} Suppose we have data from a distribution $\cD = \cN(\vmu, I_d)$. As earlier, by rotational symmetry, we can take $\vu = (1,0,0,\ldots,0), \vv = (v_1,v_2,0,0,\ldots,0)$ and $\vmu = (\mu_1,\mu_2,\mu_3,0,0,\ldots,0)$. Assume $\norm\vmu_2 = \rho$ and, without loss of generality, $\rho \geq 2$. Letting $\ip\vmu\vv =: p \leq \rho$ and $x_1,x_2,x_3 \sim \cN(0,1)$ i.i.d. gives $c \geq \inf_{(v_1,v_2) \in S^1}\ f(v_1,v_2)$ where, as before, independence of $x_1,x_2,x_3$ and the fact that $\E{x_2} = 0$ and $\E{x_2^2} = 1$, gives us
	\[
	f(v_1,v_2) = \Ee{x_1\sim\cN(0,1)}{\min\bc{\frac1{\abs{x_1 + \mu_1}},1}\cdot((p + v_1x_1)^2 + v_2^2)}
	\]
	Now, since $(v_1,v_2) \in S^1$ we get two cases (recall that we have assumed w.l.o.g. $\rho \geq 2$)
	\begin{description}
		\item[Case 1: $v_2^2 \geq \frac12$] In this case $f(v_1,v_2) \geq \frac12\Ee{x_1\sim\cN(0,1)}{\min\bc{\frac1{\abs{x_1 + \mu_1}},1}} \geq \Om{\exp^{-\rho^2/2}\log\br{1+\frac1{\rho^2}}}$.
		\item[Case 2: $v_1^2 \geq \frac12$] In this case, if $x_1 \geq 2\sqrt2\rho$, then $\abs{v_1x_1 + p} \geq \frac{v_1x_1}2$, as well as $\abs{x_1+\mu_1} \leq 2x_1$.
		\begin{align*}
			f(v_1,v_2) &\geq \Ee{x_1\sim\cN(0,1)}{\min\bc{\frac1{\abs{x_1 + \mu_1}},1}(p + v_1x_1)^2\cdot\ind{x_1 \geq 2\sqrt2\rho}}\\
			&\geq \Ee{x_1\sim\cN(0,1)}{\min\bc{\frac1{2x_1},1}\frac{x_1^2}8\cdot\ind{x_1 \geq \max 2\sqrt2\rho}} \geq \frac1{16}e^{-4\rho^2}
		\end{align*}
	\end{description}
	Since the value $\rho$ influences the final bound on $c$ very heavily, it is advisable to avoid a large $\rho$ value. One way to ensure this is to algorithmically center the covariates i.e. use $\tilde\vx_i := \vx_i - \hat\vmu$ where $\hat\vmu := \frac1n\sum_{i=1}^n\vx_i$. This would (approximately) center the covariates and ensure an effective value of $\rho \approx \bigO{\sqrt\frac d n}$
	\end{itemize}
	\item[Bounded Sub-Gaussian] Suppose our covariate distribution has bounded support i.e. $\supp(\cD) \subset \cB_2(\rho)$ for some $\rho > 0$. Assume $\rho \geq 1$ w.l.o.g. Also, using the centering trick above, assume that $\Ee{\vx\sim\cD}\vx = \vzero$. Then we have $\abs{\ip\vu\vx} \leq \rho$ which implies $\min\bc{\frac1{\abs{\ip\vu\vx}},1} \geq \frac1\rho$. Let $\Sigma$ denote the covariance of the distribution $\cD$ and let $\lambda:=\lambda_{\min}(\Sigma)$ denote its smallest eigenvalue. This gives us $c \geq \frac1\rho\Ee{\vx \sim \cD}{\ip\vx\vv^2} \geq \frac\lambda\rho$.
\end{description}

config: arxiv


TITLE: Value of lim $\lim_{n\to\infty}\frac xn \ln\left(\frac{x}{n} \right)$
QUESTION [0 upvotes]: How can I calculate this limit?
$$\lim_{n\to\infty}\frac xn \ln\left(\frac{x}{n} \right)$$
I think it is a $0 \cdot \infty$ case, but I have no other ideas. I looked again at the properties of $\ln$, thinking that it might help me, but I still didn't figure it out.
Thanks.

REPLY [1 votes]: Substitute $t:= \frac{n}{x}$, then the limit becomes
$$\lim_{t\to \infty}\frac{\ln\left(\frac{1}{t}\right)}{t} = -\lim_{t\to\infty}\frac{\ln(t)}{t}$$
Now, since $\ln(x) < x$, we have
$$0 \le \frac{\ln(t)}{t} = \frac{2\color{red}{\ln\left(\sqrt t\right)}}{\sqrt t \color{red}{\sqrt t}}\le \frac{2}{\sqrt t}\cdot\color{red}1$$ Therefore,$$
0 \le \frac{\ln(t)}{t} \le \frac{2}{\sqrt t}$$
Finally, by squeeze theorem, we have
$$\lim_{t\to\infty}\frac{\ln(t)}{t} = 0$$
which also means that
$$\lim_{n\to\infty}\frac xn \ln\left(\frac{x}{n} \right)= 0$$