\documentclass [12pt] {amsart}

\headheight=6.15pt
\textheight=574pt
\textwidth=432pt
\oddsidemargin=18.88pt
\evensidemargin=18.88pt
\topmargin=14.21pt

\title{Covering numbers and ``low $M^{*}$-estimate'' for quasi-convex bodies}

\author{A. E. Litvak}
\thanks{Litvak's and Milman's research was partially supported by
BSF. Research at MSRI  is supported in part by NSF grant DMS-9022140.}

\address{\hskip-\parindent A. E. Litvak,
 Department of Mathematics, Tel Aviv University, Ramat Aviv, Israel}
\email{alexandr@math.tau.ac.il}

\author{V. D. Milman}
\address{\hskip-\parindent V. D. Milman,
 Department of Mathematics, Tel Aviv University, Ramat Aviv, Israel}
\email{vitali@math.tau.ac.il}

\author{A. Pajor}
\address{\hskip-\parindent A. Pajor,
Universite de Marne-la-Vall\'ee, Equipe de Mathematiques,
                2 rue de la Butte Verte, 93166 Noisy-le-Grand Cedex, France}
\email{pajor@math.univ-mlv.fr}

\def\kkk{\qed}
\def\ee{ \eps _0/C^p }
\def\l{ \left}
\def\r{ \right}
\def\nek{,\ldots,}
\def\eps{\varepsilon}
\def\alp{\alpha}
\def\ome{\Omega}
\def\lam{\lambda}
\def\Ker{{\rm Ker}}
\def\hb{\hfill\break}
\def\nor{\parallel}
\def\str{\longrightarrow}
\def\tt{\theta}
\def\P{C(\alpha ,p)  }
\def\Q{C(\gamma ,q)  }
\def\we{we }
\def\We{We }
\def\bb{ J.Bastero, J.Bernu\'es, and A.Pe\~na }
\newcommand{\RR}{\mbox{\rm $~\vrule height6.5pt width0.5pt
depth0.3pt\!\!$R}}
\newcommand{\R}{\RR ^n}
\newcommand{\Rk}{\RR ^k}
\newcommand{\Rm}{\RR ^m}
\newcommand{\RP}{\RR ^{+}}
%\def\R{ {\rm R}^n  }
%\def\Rm{ {\rm R}^m  }
%\def\Rk{ {\rm R}^k  }
\newcommand{\Z}{\mbox{\rm $~\vrule height6.5pt width0.5pt
depth0.3pt\!\!$Z}}
\newcommand{\N}{\mbox{\rm $~\vrule height6.5pt width0.5pt
depth0.3pt\!\!$N}}
\def\fk{ finite-dimensional version of Krivine's Theorem}
\def\bm{ Brunn-Minkowski inequality }
\def\pc{ $ p-$convex   }
\def\qs{ quasi-norm}
\def\qn{ $ q-$norm}
\def\no{ \nor . \nor }
\def\nx{ \nor x \nor }
\def\ny{ \nor y \nor }
\def\nxy{ \nor x+y \nor }
\def\na{ \nor a \nor }
\def\nao{ \nor a \nor _0 }
\def\nap{ \nor a \nor _p }
\def\naoq{ \nor a \nor _0^q }
\def\nua{ \nu _a(\eps ) }
\def\fa{ f_a }
\def\nuf{ \nu _{\fa } (\eps ) }
\def\tp{ (t,\pi  ) }
\def\me{ \mu  \{(t,\pi ) \ : \ | \fa ^q  ( \tp ) }
\def\bl{  balls}
\def\ex { \exp (-\eps ^2 n /64  ) }
\def\o{ \tt }
\def\tp{ (t,\pi  ) }
\def\tp{ (t,\pi  ) }
\renewcommand{\thefootnote}{\fnsymbol{ \ }}
\begin{document}

%\date{}



\begin{abstract}
%
%%%***  We give estimates on covering numbers and
  This article gives  estimates on covering numbers and
diameters of random proportional
sections and projections of  symmetric quasi-convex
bodies in $\R$.
%%%*** These results were known in convex case and played
   These results were known for the convex case and played
an essential role in development
of the theory. Because duality relations can not be applied in
the quasi-convex setting, new ingredients were
introduced that give new understanding
%%%*** also in convex case.
 for the convex case as well.
%
%%
%
\end{abstract}

\maketitle




%00000000000000000000000000000000000000000000000000000000000000000000
%\begin{description}
%00000000000000000000000000000000000000000000000000000000000000000000
  % \medskip
   \section{Introduction  and notation}
%\item[{ 1. Introduction and notation. }]
%\end{description}


  Let $ |\cdot | $ be   % \hfill  $ \blacksquare $    \\
a euclidean   norm
%\renewcommand{\thefootnote}{\fnsymbol{ \ }}
%%%%%%%**********************************%%%%%%%%%%\footnote{ Supported in part by BSF.}
on $\R $. Let $D$ be an ellipsoid
associated with this norm.
 Denote  $ A= \sqrt{\frac{n}{k}} \int \limits_{S^{n-1}}
\sqrt{    \sum _{i=1}^{k} x_i^2          }
%(  \sum _{i=1}^{k} x_i^2 )^{1/2}
\ d\sigma (x) $, where $\sigma $
is the normalized rotation invariant measure on the euclidean sphere
$ S^{n-1} $. Then $ A=A(n,k)<1$
 and $ A \str 1 $ as $ n,k \str \infty $.
 For any star-body $K$ in $\R$ define
$ M_K=  \int \limits_{S^{n-1}}
\nx d\sigma (x) $, where  $\nx$ is the gauge of $K$.
Let $M^{*}_K$ be $M_{K^0}$, where $K^0$ is the polar of $K$.
For any subsets $K_1, K_2 $ of $\R$ denote by
$N(K_1,K_2)$
 the smallest number $N$ such that there are
%*********************************************************
 N points $y_1,...,y_N$  in $K_1$ such that
%%$ x_i \in B \ \ (1 \leq i \leq 2^{k-1} ) $ such that
$$ K_1 \subset \bigcup _{i=1}^{N} (y_i+ K_2  ) .$$

 Recall that a body $K$ is
called quasi-convex
if there is a constant $c$
such that $K+K\subset cK$, and given a $ p \in (0,1] $ a
body $K$ is called $p$-convex if for any
$\lam ,\mu >0$ satisfying
$\lam^p + \mu^p =1$ and any points $x,y\in K$
the point $\lam  x + \mu y$ belongs to $K$.
Note that for the gauge $\nor \cdot  \nor \, = \,
\nor \cdot  \nor _K  $  associated with the
the quasi-convex  ($p$-convex) body $K$ the following
inequality holds for any $x, y \in \R $
$$ \nxy \leq C \max\{ \nx  ,\ny \}
\ \ \  % \l(  \nxy \leq 2^{1/p}( \nx  +\ny )\   \r) . $$
 \l(  \nxy ^p \, \leq  \, %2^{1/p}
   \nx ^p   +\ny ^p  \,   \r) . $$
 % Of course
 In particular every $p$-convex
body $K$  is also quasi-convex one and
$K+K\subset 2^{1/p}K$.
%
%**********************************************
%
%
   A
%\len
%\footnoterule{ Supported in part by BSF.}  \newpage
 more delicate result is that for every quasi-convex body
$K$ ( $K+K\subset cK$)
there
 exists a  $q$-convex body $K_0$ such that
$K\subset  K_0 \subset 2c K$,
where $2^{1/q}=2 c$.
 This is Aoki-Rolewicz theorem
([KPR], [R], see also [K], p.47).
 In this note by a body we always mean
 a  centrally-symmetric
 compact
star-body, i.e. a body $K$ satisfying
   $tK \subset  K$ for any $t\in
%(0,1)
  [-1, 1]$,
 however not all results use symmetry. Lemma~2,
 some analog of Lemma~4 for $p$-convex body $K$, %of A.~Pajor
and Theorem~3 hold in non-symmetric case also.

%***********************************************************



% DEFINITIONS



%********************************************************************************


 Let us remind of the so-called
``low $M^{*}$-estimate" result. %  of the second named
%author.
%  ([M], see also [PT]).
\newtheorem{theorem}{Theorem}
\begin{theorem}
 Let $\lam > 0 $ and $n$ be large enough.
%($n> c/(1-\lam )^2 $).
    Let $K$ be a convex body  in $\R$ and
$\nor \cdot  \nor$ be the gauge
of $K$. Then   there exists a subspace $E$ of
 $( \R  , \nor \cdot  \nor  ) $ such that
$ \dim E = [\lam n ] $ and for any
$x \in E $ the following inequality holds
$$ \nx \geq  \frac
{ f(\lam ) }
%{\sqrt{1-\lam }}
{ %c \
M^{*}_K}
 | x | \ \ $$
%where $c$ is an absolute constant.
 for some function $ f(\lam ) $, $0<\lam <1 $.

\end{theorem}
{\it Remark.} Inequality of this type  was first proved in
[M1] with very poor dependence on
$\lam $ and then improved in [M2]
to $  f(\lam )=C (1-\lam ) $. It
was later shown ([PT]), that one can take $ f(\lam )=C  \sqrt{1-\lam }$
(for different proofs see [M3] and [G]).


 By duality this theorem is equivalent
 to the following
 \medskip
 theorem.
%\theoremstyle{change}
%\newtheorem{claim}{Proposition }
%\begin{claim}
\newline
\bf Theorem 1'    \it
Let $\lam > 0 $ and $n$ be large enough.
%($n> c/(1-\lam )^2 $).
 For any convex body $K$ in $\R$
there exists an orthogonal projection
P of
%XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
%the
 rank $[\lam n ]$ such that
$$ PD \subset
\frac{
%c\
 M_K}
%{\sqrt{1-\lam }}
{ f(\lam )}
 PK \ , $$
where $c$ is an absolute
   \medskip
 constant.




%\end{claim}
\rm
 In this note we will extend
%in this note
 both
theorems to quasi-convex bodies. Because
duality  arguments can not be
applied to a non-convex body these two theorems
become different statements.
 Also ``"$M^*_K$" should be substituted
by an appropriate quantity not involving duality.
%depending
% on $K^0$.
  Note  that by avoiding  the  use of convexity assumption
 \medskip
 we in fact simplified proof also for a convex case.
%Since polar of body $K$ is equal to polar of $convK$,
%  duality argument can not be used
%in a non-convex case.
%So, this two theorems become non-equivalent and, to receive
%such kind of results, we have to use some other technique.
%\newline
%
%00000000000000000000000000000000000000000000000000000000000000
   \section{Main results}
%00000000000000000000000000000000000000000000000000000000000000
%\item[{ 2. Main results. }]



 The following theorem is an extension of Theorem 1'.
\begin{theorem}
 Let $\lam > 0 $ and $n$ be large enough ($n> c/(1-\lam )^2 $).
For any $p$-convex body $K$ in $\R$
there exists an orthogonal projection
P of the rank $[\lam n ]$ such that
$$ PD \subset \frac{A_p M_K}{(1-\lam )^{1+1/p}}  PK \ , $$
%where $A_p$ depends on $p$ only.
where
 $ A_p = const ^{\frac{\ln (2/p)}{p}}$.


\end{theorem}

The proof of this theorem is based on the next three lemmas.
The first one was proved by
W.B.Johnson and J.Lindenstrauss
in [JL].  The second one was proved in [PT]
for convex bodies
%([PTJ])
 and is the dual form of
  Sudakov minoration theorem.
%can easily check that the
%same methods work for quasi-convex bodies.
\newtheorem{lemma}{Lemma}
\begin{lemma}  There is an absolute constant $c$ such that
  if  $\eps > \sqrt{c/k} $ and $N< e^{\eps^2 k/c } $,
 then for any set of points
$y_1, ... , y_N \in \R $ and any orthogonal
projection P of  rank $k$
$$\mu \l( \{ U\in O_n \ | \ \forall j :
 A(1-\eps ) \sqrt{k/n} \    |y_j | \leq
| P U y_j | \leq A(1+\eps ) \sqrt{k/n}  \   | y_j | \} \r)
\geq $$
$$ \geq 1 - \sqrt{\pi /2} \cdot e^{-\eps ^2 k /c} $$
 % Here $c$ is an absolute constant.




\end{lemma}
\begin{lemma}
 Let $K$ be a body such that $K-K \subset aK $. Then
 $N(D, tK) \leq 2 e^{2 n (a M_K / t)^2 }$.
%where $ c_p \leq c^{1/p  } $
%   for some
%absolute
% constant $c$






\end{lemma}
 M.~Talagrand gave a direct simple proof
of this lemma for a convex case ([T]).
   Below, using his idea, we prove more
 general lemma for $p$-convex bodies, so we do not prove
 Lemma 2 now.
%%!!!!!!!!!!                   We will not prove Lemma~2, but will prove more general
%
%
%%!!!!!!!!!!                   lemma for quasi-convex bodies below.
% Below we present a more general lemma of A.~Pajor ([P]) for $p$-smooth
%bodies which is trivially applyed for $p$-convex bodies. So we do not prove
%Lemma 2 now.
\begin{lemma}
 Let $B$ be a star-body, $K$ be a $p$-convex set,
$r \in (0,1) $, $\{x_i\} \subset rB$
 and $B \subset  \bigcup
%_{i=1}^{2^{k-1}}
(x_i+K  ) $.
 Then $ B\subset t_r K$, where $t_r=
\frac{1}{(1-r^p)^{1/p}} $.



\end{lemma}
{\it Proof: } Obviously $t_r=\max \{ \nx _K \  | \  x\in B \} $.
Since $B \subset  \bigcup
%_{i=1}^{2^{k-1}}
(x_i+K  ) $,
for any point $x$ in $B$ there are  points  $x_0$ in $rB$ and
$y$ in $K$ such that $x=x_0+y$. Then by maximality
of $t_r$ and $p$-convexity of $K$ we have
$t_r^p \leq r^p t_r^p + 1$. That proves the lemma. \kkk % $\Box $
\newline
{\it Remark. } Somewhat similar argument was %often
 used by N.~Kalton in dealing
 with $p$-convex sets.
  % See, e.g. [GK].
 \newline \newline
%
%
%************************************************
%
%
{\it Proof of Theorem 2: } \newline
 Any $p$-convex body
$K$ satisfies $K-K \subset aK$ with $a=2^{1/p}$.
   By Lemma~1 and Lemma~2,
we obtain
 for $c_p \approx  2^{1/p} $
 that if
 %for $c_p \approx  2^{1/p} $
$$ c_p n \l( \frac{M_K }{t}
\r) ^{2} \leq \frac{\eps ^2 k}{c} $$
%($c_p$ depend on $p$ only)
 and $\eps > \sqrt{c/k} $,  then
there exist points $x_1,...,x_N$ in $D$ and an orthogonal
projection $P$
of
%XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
 %the
 rank $k$ such that
$$ PD \subset \bigcup (Px_i + tPK) \ \mbox{ and } \ \
|Px_i| \leq (1+\eps ) \sqrt{ \frac{k}{n}  } |x_i| \ \ .$$
 Let $\lam = k/n$. Denote $r=(1+\eps ) \sqrt{ \lam } $. Lemma 3 gives us
$$
PD \subset t t_r PK \ \mbox{ for }  \
t=\frac{\sqrt{c c_p } M_K}{\eps \sqrt{\lam }}
\ \mbox{ and }
\ \eps  ^2  >  \frac{c}{\lam n} \ , \  r<1        \  \  .$$
Choose $$
\eps  = \frac{1-\sqrt{\lam }}{2 \sqrt{\lam }} \ .$$
Then for $n$ large enough
 %$n$
 we get
$$ PD \subset \frac{A_p M_K}{(1-\lam )^{1+1/p}}  PK \ , $$
for  $ A_p = const ^{\frac{\ln (2/p)}{p}}$ .
%where $A_p$ depends on $p$ only
%(in fact $ A_p = const ^{\frac{\ln (2/p)}{p}}$ ).
This completes the proof. \kkk

 Theorem 2 can be formulated in the global  \medskip  form.
\newline
  \bf
Theorem 2'   \it
 Let $K$ be a $p$-convex
body in $\R$. Then there is an orthogonal
operator $U$ such that \medskip
$$ D \subset   A^{'}_p M_K ( K+UK ) \, , \ \mbox{ where }
A^{'}_p
  = const ^{\frac{\ln (2/p)}{p}}
% \mbox{ depends on } p
%\mbox{  as   } A_p  \mbox{ in Theorem 2. }
\  .$$


\rm
 This theorem can be proved independently, but we show
how it follows from Theorem~2.  \newline
{\it Proof of Theorem 2': }
 It follows from the proof of Theorem~2 that actually the
measure of such projections is large. So we can choose
two orthogonal subspaces $E_1, E_2$ of $\R$ such that $\dim E_1 = [n/2], \
\dim E_2 = [(n+1)/2] $ and
$$ P_i D \subset
%\frac{
A^{''}_p M_K
%}
%{(1-\lam )^{1+1/p}}
P_i K \, , $$
where $P_i$ is the projection on the space $E_i$ ($i=1,2$).
 Denote $I=id_{\R }=P_1+P_2$ and $U=P_1-P_2$. So $P_1=1/2 (I+U)$ and $P_2=1/2(I-U)$.
 Then $U$ is an orthogonal operator and for any $x\in D$ we have
$$x=P_1 x+ P_2 x \subset 1/2 \ A^{''}_p M_K (I+U) K +
1/2 \ A^{''}_p M_K (I-U) K = $$
$$
  =
 A^{''}_p M_K \frac{K+K}{2} +
  A^{''}_p M_K \frac{UK-UK}{2} = A^{'}_p M_K (K+UK)  \ .
$$
That proves Theorem~2'.   \kkk


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\rm
Let us complement Lemma 2 by
 mentioning how
 %estimating
 covering number
$N(K, tD)$
 can be estimated.  In
convex case this estimate is given by Sudakov inequality,
using quantity
$M^{*}$. More precisely, if $K$ is
  a  convex body, then
$$N(K, tD) \leq 2 e^{c n ( M^{*}_K / t)^2 }\ .$$
 Of course,
%we mentioned above
  using  duality for a
non-convex setting leads to a weak
result, and we suggest below
a substitution for quantity $M^{*}$.












%Let us discuss Theorem 1. Using technics of [MP1]
%we can obtain this theorem
%(with factor $(1- \lam )^{-1}$ istead of  $(1- \lam )^{-1/2}$)
% from Lemma 1 and
%well known estimate of
%covering number $N(K, tD) \leq e^{cn(M^{*}/t)^2 }$.
%In quasi-normed case this estimate not so good,
%because we can cover
%$convK$ by such number of translations of $t D$.





% Define the following
%characteristic of $K$,   that allows us estimate
%covering number body $N(K, tD)$  without
%using of $M^{*}_K$.
%Let $$ \tilde M_K = \frac{1}{|K|}
%\int \limits_{K} |x|   dx \   ,$$
%where $|K|$ is volume of $K$.



  For two quasi-convex
%$K, B $
bodies $K, B $  define the following number
$$  M(K, B) = \frac{1}{|K|}
\int \limits_{K} \nx _B    dx \   ,$$
where $|K|$ is volume of $K$, and $\nx _B$ is the gauge of $B$.
Such numbers are considered in [MP1], [MP2] and [BMMP].
 %  We use the following lemma of A.~Pajor ([P]).





% This number does not depend on polar of $K$ and,
%it seems to me, less then $M^{*}_K$. For example,
% if $K=B(l^n_1)$ we
%have $  \tilde M_K \leq const \ n^{-1/2}$ and
%$ M^{*}_K \geq
%const \  n^{-1/2} (\log n )^{1/2}$. Unfortunately,
% there are  examples, when $ \tilde M_K $ has the
%same order as  $ M^{*}_K$
%  ($ K=B(l^n_{\infty })$).
%
% %
%
%
%\begin{lemma}
% Let $K$ be a $p$-convex body and $B$ be a
%star-body such that $B + B \subset aB$.
% Let $\o >0$ and  $c_{\o } = 2^{1/p-1/\o } $
%if  $\o >p$ and  $c_{\o } =1$
%otherwise.
%Then
% $$N(K , tB) \leq 2 c_{\o }^n
%e^{ (c n /\o  )  (a  M(K, B) / t)^{\o} }\ ,\
%\mbox{ where  }  c \mbox{ is the
%absolute
% constant. }$$
%
%\end{lemma}
%%
%
%
\begin{lemma}
 Let $K$ and $B$ be
%  $p$-convex bodies.
%and $B$ be a
 star-bodies.
 %such that
 Assume  $B + B \subset aB$.  For $\o >0$ denote by
 $c_{\o }=  c_{\o } (K) $
 %denote
 the best possible
 constant such that
$$
 \nor x+y \nor _K^{\o } + \nor x-y \nor _K^{\o }
  \leq 2  \cdot c_{\o }^{\o } \l( \nx _K^{\o }  +
\ny _K^{\o }  \r)
\
\mbox{ for any  } \  x, y \in \R \ .
$$
% Let $\o >0$ and  $c_{\o } = 2^{1/p-1/\o } $
%if  $\o >p$ and  $c_{\o } =1$
%otherwise.
  Then
 $$N(K , tB) \leq 2 c_{\o }^n
e^{ (c n /\o  )  (a  M(K, B) / t)^{\o} }\ ,\
\mbox{ where  }  c \mbox{ is an
absolute
 constant. }$$









\end{lemma}

 Note, that in a case $\o =1$ which correspond to the general
convex case this lemma was announced in [MP2].
 % For completeness of proof we prove this lemma below.
\newline
%
% (Note, that in  case $\o =1$ which correspond to the general
%> convex case this lemma was announced in [MP2].)
%
%
%NNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOTTTTTTTTTTT
%
%
{\it Proof: }
 We follow the idea of M.Talagrand of estimating
covering numbers
in case $K=D$ ([T], see also [BLM] Proposition 4.2).
 Denote the gauge
of $K$ by $\nor \cdot \nor $ and the gauge of $B$ by $ | \cdot  | $.
  Define the measure $\mu  $ by following
$$d\mu =\frac{1}{A} e^{-\nx ^{\o }} dx \ , \mbox{ where }
%\nx = \nx _B
% \mbox{ and }  \ A
%
%
\  A \   \mbox{ chosen such that } \
\int \limits_{\R } d\mu =1  \  .
$$
Let $L=
%\frac{1}{A}
 \int \limits_{\R } |x| d\mu $. Then $\mu \{|x| \leq
2L \} \geq 1/2 $.
Let $x_1,x_2,...$ be a maximal set of points in $K$ such that
$ |x_i-x_j| \geq t$. So the sets  $x_i+\frac{t}{a} B
% \bigcap x_j+t/a int B =
%\emptyset
$ have mutually disjoint interiors.
Let $y_i =\frac{ ab}{t} x_i $ for some $b$.
Then,
by symmetry of $B$ and convexity of the function
$e^t$, we have
$$ \mu \{ y_i+bB \}
=\frac{1}{A}  \int  \limits_{bB }  e^{-\nor x+y_i \nor
 ^{\o }} dx
   = \frac{1}{2 A}
  \int  \limits_{bB }  e^{-\nor x+y_i \nor ^{\o } } +
 e^{-\nor x-y_i \nor ^{\o }}  dx
\geq  $$
 $$
 \geq  \frac{1}{A}
 \int  \limits_{bB }
e^{-\frac{1}{2}(\nor x+y_i \nor ^{\o }  +\nor x-y_i \nor ^{\o } )}  dx
%   \frac{1}{A} =
%%%%%%%%%%=
%%%%%%%%%%\int  \limits_{bD }  e^{-(\nor x \nor
%%%%%%%%%% ^{2 } + \nor y_i \nor
%%%%%%%%%% ^{2 })} dx
%%%%%%%%%%\ ,  $$
%
%
%
%
%
%
%
%
%
%
%
%
%$$
%\geq   \frac{1}{A}  \int  \limits_{bB }  e^{-((\nor x \nor
% ^{p } + \nor y_i \nor
% ^{p })^{\o /p}} dx \geq $$
%$$
  \geq \frac{1}{A}  \int  \limits_{bB }
 e^{- c_{\o }^{\o } (\nor x \nor
 ^{\o } + \nor y_i \nor
 ^{\o })} dx =  $$
$$ =
\frac{1}{A}  e^{-\nor c_{\o }y_i \nor ^{\o }}
\int  \limits_{bB }
e^{-\nor c_{\o } x \nor ^{\o }}  dx  \geq
e^{-(c_{\o } ba/t)^{\o } } c_{\o }^{-n} \mu \{ c_{\o }bB \}
$$
 Choose
$b=2L/ c_{\o }  $.
Then $ \mu \{ c_{\o }bB \} \geq 1/2 $ and, hence,
$$ N(K,tB) \leq
2    c_{\o }^n e^{(2 a L/t)^{\o } } \ . $$
%***********************************************************
 Now compute $L$.
First, the normalization constant $A$ is equal
$$A = \int \limits_{\R }
e^{-\nx ^{\o }} dx =  \int \limits_{\R }
 \int \limits_{\nx }^{\infty } (-e^{-t^{\o }})' dt dx  =
\int \limits_{0 }^{\infty } \o t^{\o -1 } e^{-t^{\o }}
\int \limits_{\nx \leq t } dx dt
= $$
$$
=
\int \limits_{\nx \leq 1 } dx
\int \limits_{0 }^{\infty } \o t^{\o +n-1} e^{-t^{\o }}dt =
|K| \cdot  \Gamma \l(1+\frac{n}{\o } \r) \ ,\
%\mbox { where } \Gamma
%\mbox{ is the gamma-function.}
 $$
  where $ \Gamma $  is the gamma-function.
 The remaining integral is
$$
 \int \limits_{\R } |x| e^{-\nx ^{\o }} dx =
  \int \limits_{\R } |x|
 \int \limits_{\nx }^{\infty } (-e^{-t^{\o }})' dt dx
 = \int \limits_{0 }^{\infty } \o t^{\o -1 } e^{-t^{\o }}
\int \limits_{\nx \leq t } |x| dx dt
= $$
$$
=
\int \limits_{\nx \leq 1 } |x| dx
\int \limits_{0 }^{\infty } \o t^{\o +n} e^{-t^{\o }}dt =
|K|
\cdot
 M(K,B) \cdot  \Gamma \l( 1+\frac{n+1}{\o } \r) \
%,\ \mbox { where } \Gamma
%\mbox{ is the gamma-function}
\ .  $$
 Using Stirling's formula we get
$$L\approx \l( \frac{n}{\o } \r) ^{1/\o } M(K, B) \ .$$
That proves the lemma. \kkk

%  This lemma is an extension of Lemma 2.
%It follows from the proof, that
%in case $K=D$,
%$$N(D , tB) \leq 2  e^{ (c n   )
%( a M(D, B) / t)^{2} } \ ,\ \mbox{ because of }$$
%$$   \int  \limits_{bD }  e^{-\nor x+y_i \nor _D
% ^2} dx  = \frac{1}{2}
% \int  \limits_{bD }  e^{-\nor x+y_i \nor _D^2} +  e^{-\nor x-y_i \nor _D^2}  dx
%\geq  $$
%$$
% \geq
% \int  \limits_{bD }
%e^{-\frac{1}{2}(\nor x+y_i \nor _D^2  +\nor x+y_i \nor _D^2 )}  dx
%%   \frac{1}{A} =
%=
%\int  \limits_{bD }  e^{-(\nor x \nor
% ^{2 } + \nor y_i \nor
% ^{2 })} dx
%\ ,  $$
% and, by direct computation,   %
%%of course,
%   $M(D,B)= \frac{n}{n+1}M_B$.



  This lemma is an extension of Lemma 2. Indeed,
   since Euclidean space
is
a 2-smooth
 %a
 space,
 %of the  type 2,
 then in case $K=D$ being
an ellipsoid, we have
$c_2(D) =1$.
By direct  computation,  $M(D,B)= \frac{n}{n+1}M_B$.
 Thus,
$$N(D , tB) \leq 2  e^{ (c n   ) (  M_B
%(D, B)
  / t)^{2} }  \, .$$






  Define the following characteristic of $K$,
%that allows us estimate
%covering number body $N(K, tD)$  without
%using of $M^{*}_K$.
 $$ \tilde M_K = \frac{1}{|K|}
\int \limits_{K} |x|   dx \,   .$$
%where $|K|$ is volume of $K$.

By definition, if $K$
is a $p$-convex body, then  for any $x, y \in \R $ holds
$$
 \nor x+y \nor _K^{p } + \nor x-y \nor _K^{p }
  \leq 2  \cdot  \l( \nx _K^{p }  +
\ny _K^{p }  \r)
\,   .
%\mbox{ for any  } \  x, y \in \R \ .
$$
   So, the last lemma shows that for  $p$-convex body $K$
$$N(K , tD) \leq
2  e^{ (c n /p  )  (2
%^{1/p}
   \tilde M_K   / t)^{p} } \, .$$





%%% The last lemma
%%% shows that for $p$-convex
%%%body $K$
%%%%%%%%%%%%%%%%%%%%%%%%%allows us estimate covering number
%%%$$N(K , tB) \leq
%%%2  e^{ (c n /p  )  (2^{1/p}   \tilde M_K   / t)^{p} } \ .$$
%%%%%%%%%%%%%%%%%%%%%%%%%%  for $p$-convex
%%%%%%%%%%%%%%%%%%%%%%%%%%%body $K$.


 The Theorem 3 follows from this
estimate by arguments similar of that in [MP].
% As we mention above this lemma gives us Theorem 4.
\begin{theorem}
 Let $\lam > 0 $ and $n$ be large enough.
%($n> c/(1-\lam )^2 $).
 Let $K$ be a $p$-convex body  in $\R$ and
$\nor \cdot  \nor$ be the gauge
of $K$. Then   there exists  subspace $E$ of
 $( \R  , \nor \cdot  \nor  ) $ such that
$ \dim E = [\lam n ] $ and for any
$x \in E $ the following inequality holds
$$ \nx \geq  \frac{(1-\lam )^{1/2+1/p}}{a_p  \tilde M_K}
 | x | \,  ,$$
where $a_p$ depends on $p$ only (more precisely
$ a_p = const ^{\frac{\ln (2/p)}{p}}$).



\end{theorem}
{\it Proof: }
   By Lemma 4 there are points $x_1, ... , x_N$ in $K$,
such that $N<
e^{c_p n (\tilde M_K / t )^p}$ and
for any $x \in K$ there exists some $x_i$ such that $|x-x_i|<t$.
 By Lemma 1 there exists an orthogonal projection $P$
on a subspace of dimension
$\delta n$ such that
for $$c_p n \l( \frac{\tilde M_K}{t} \r) ^p <
\frac{\eps ^2
\delta n}{c} \ \mbox{ and } \
\eps > \sqrt{\frac{c}{\delta n}} $$ we have
$$b|x_i|
=(1-\eps )A \sqrt{\delta } |x_i| \leq |Px_i|
\leq (1+\eps )A \sqrt{\delta } |x_i| $$
for every $x_i$.
Let $E=\Ker P$. Then $\dim E=\lam n$, where $\lam = 1-\delta$.
 Take $x$ in $K \bigcap E $.
There is $x_i$ such that $|x-x_i|<t$.
Hence
$$|x|\leq |x-x_i| +|x_i| \leq t+\frac{|Px_i|}{b} =
t+\frac{|P(x-x_i)|}{b}
\leq
$$
$$
\leq  t+\frac{|x-x_i|}{b} \leq t(1+\frac{1}{b})
\leq    \frac{const \cdot t }{(1-\eps) \sqrt{\delta }} $$
 Therefore  for
$n$ large enough and $$t=\l(
\frac{const\cdot c_p }{\eps ^2 \delta }
\r) ^{1/p } \tilde M_K $$ we
 get
 %have
 $$ \nx \geq  \frac{ const \cdot
\eps ^2 (1-\eps ) \delta ^{1/2+1/p}}{ c_p^{1/p}  \tilde M_K}
 | x | \, .$$
To obtain our result take $\eps $, say, equal to $1/2$.
  \kkk










 As was noted in [MP2]
% Note that
 in some cases
 $  \tilde M_K  << M^{*} $ and then
Theorem~3 gives better estimate
than Theorem~1 even for a convex body
(in some range of $\lam $).
% For example
%([MP2]),
 As an example,
 %for
 $K=B(l^n_1)$,
$  \tilde M_K \leq const \cdot  n^{-1/2}$,
but $ M^{*}_K \geq
const \cdot  n^{-1/2}   \medskip   (\log n )^{1/2}$.
%
%
%
%
%
%
%
%
%
%   This number does not depend on polar of $K$ and,
%it seems to me, less then $M^{*}_K$. For example,
% if $K=B(l^n_1)$ we
%have $  \tilde M_K \leq const \ n^{-1/2}$ and
%$ M^{*}_K \geq
%const \  n^{-1/2} (\log n )^{1/2}$. Unfortunately,
% there are  examples, when $ \tilde M_K $ has the
%same order as  $ M^{*}_K$
%  ($ K=B(l^n_{\infty })$).
%
%
%
%
%
%
%
%
%
%00000000000000000000000000000000000000000000000000000000000000
\section{Additional remarks}
%00000000000000000000000000000000000000000000000000000000000000
%\item[{ 3. Additional remarks. }]
%\end{description}
%\smallskip
%\noindent{\bf  3. Additional remarks. }


 In fact, during the proof of Theorem 2
%we prove
% the following fact, which can be useful.
 a more general fact was proved.
\medskip \newline
%\bigskip
%\vspace*{1in}
\bf Fact.
\it Let $D$ be an ellipsoid and $K$ be a $p$-convex body. Let
 $$N(D,  K)\leq e^{\alpha  n} \ . $$   Denote for an integer $1\leq k \leq n$
the ratio $\lam =k/n $.
Then for some absolute constant $c$ and
$$\gamma =c \sqrt{\alpha }  \ ,\  \ k\in (\gamma ^2 n, (1-2 \gamma )^2 n) $$
there exists an orthogonal projection $P$ of rank $k$ such that
 $$
    \l( p(1- \sqrt{\lam })/2 \r) ^{1/p}
%}{\tt }
PD \subset PK  \ .
                                                               $$
\rm




 In terms of  entropy numbers this means
% On the language of entropy numbers this means
$$
   \frac{\l( p(1-\sqrt{k/n} )/2 \r) ^{1/p}}{e_k (D,K) } PD \subset PK  \ ,
                                                                           $$
where $e_k (D, K)= \inf \{ \eps >0 \  | \  N(D, \eps K ) \leq 2^{k-1} \} $ .

%Point out, that Theorem~2 can be obtained from this results.
  It is worth to point out that Theorem~2 can be obtained from this
results.

  We thank E.~Gluskin for his remarks on the first
draft of this note.


\section*{References}
{\footnotesize

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%(errata insert).46A15

\end{description}
}


\end{document}
