# Measure comparison problems for dilations of convex bodies

Malak Lafi and Artem Zvavitch

###### Abstract.

We study a version of the Busemann-Petty problem for $\log$-concave measures
with additional information about dilates. One of our main tools is a an
analog of of the classical large deviation principle to the case of
$\log$-concave measures depending on a norm of a convex body, which we hope
will be of independent interest.

###### Key words and phrases:

Busemann-Petty problem, log-concavity, large deviation.

###### 2020 Mathematics Subject Classification:

Primary: 52A20, 52A21; Secondary: 46T12, 60F10

Both authors are supported in part by the U.S. National Science Foundation
Grant DMS-1101636 and the United States - Israel Binational Science Foundation
(BSF) Grant 2018115

##  1\. Introduction

We denote by ${\mathbb{R}}^{n}$ the $n-$dimensinal Euclidean space equipped
with the inner product $\langle\cdot,\cdot\rangle$ and the standard
orthonormal basis $\\{e_{1},\dots,e_{n}\\}$, and denote by $|\cdot|$ the
$\ell^{2}$ Euclidean norm on ${\mathbb{R}}^{n}$. For measurable set
$A\subset{\mathbb{R}}^{n}$ we refer to its volume (the Lebesgue measure) by
$|A|,$ and its boundary by $\partial A.$ The notation $B_{2}^{n}$ stands for
the closed unit ball in ${\mathbb{R}}^{n},$ and ${\mathbb{S}}^{n-1}$ for the
unit $n-$dimensional sphere. A convex body is a convex, compact set with
nonempty interior. Furthermore, a convex body $K$ is symmetric if $K=-K$. A
measure $\mu$ is $\log$-concave on ${\mathbb{R}}^{n}$ if for every pair of
Borel sets $A$ and $B$ in $\mathbb{R}^{n}$ and $0<\lambda<1$ we have

$\mu(\lambda A+(1-\lambda)B)\geq\mu(A)^{\lambda}\mu(B)^{1-\lambda},$

where the addition here is the Minkowski sum which is defined as the set
$A+B=\\{a+b:a\in A,b\in B\\},$ and the constant multiple (dilation) of a set
$A$ is defined as $\alpha A=\\{\alpha a:a\in A\\}$ where
$\alpha\in\mathbb{R}$. It follows from the Prékopa-Leindler inequality [16,
17], that if a measure $\mu$ that is defined on the measurable subsets of
$\mathbb{R}^{n}$, is generated by a $\log$-concave density function, then the
measure $\mu$ is also $\log$-concave. Furthermore, Borell provides
characterization for $\log$-concave measures [3], precisely a locally finite
and regular Borel measure $\mu$ is $log$-concave, if and only if, its density
is $log$-concave function, that is the density of the form $e^{-\psi(x)}$,
where $\psi:\mathbb{R}^{n}\to\ [0,\infty)$ is convex.

In 1956, Busemann and Petty [5] posed the following volume comparison problem:
Let $K$ and $L$ be symmetric convex bodies in $\mathbb{R}^{n}$ so that the
$(n-1)$-dimensional volume of every central hyperplane section of $K$ is
smaller than the same section for $L$. Does it follow that the $n$-dimensional
volume of $K$ is smaller than the $n$-dimensional volume of $L$? In the late
90’s, the Busemann-Petty problem was solved as a result of many works [8, 9,
12, 13, 20]. The answer is affirmative when $n\leq 4,$ and negative whenever
$n\geq 5$. It is natural to consider an analog of the Busemann-Petty problem
for a more general class of measures. The first result in this direction was a
solution of the Gaussian analog of the Busemann-Petty problem [21]. It turns
out that the answer for the Busemann-Petty problem will be the same if we
replace the volume with the Gaussian measure. Moreover it was proved in [22]
that the answer is the same if we replace the volume with any measure with
continuous, positive and even density.

V. Milman asked whether the answer to the Gaussian Busemann-Petty problem
would change in a positive direction if we compared not only the Gaussian
measure of sections of the bodies but also the Gaussian measure of sections of
their dilates, that is, consider two convex symmetric bodies
$K,L\subset\mathbb{R}^{n}$, such that

(1) $\gamma_{n-1}(rK\cap\xi^{\perp})\leq\gamma_{n-1}(rL\cap\xi^{\perp}),\hskip
14.22636pt\forall\xi\in\mathbb{S}^{n-1},\hskip 14.22636pt\forall r>0,$

where $\xi^{\perp}$ denotes the central hyperplane perpendicular to $\xi.$
Does it follow that

$\gamma_{n}(K)\leq\gamma_{n}(L)?$

Here $\gamma_{n}$ denotes the standard Gaussian measure on $\mathbb{R}^{n}$,
and

$\gamma_{n-1}(K\cap\xi^{\perp})=\frac{1}{(\sqrt{2\pi})^{n-1}}\int_{K\cap\xi^{\perp}}e^{-\frac{|x|^{2}}{2}}dx.$

The addition of dilation to the Busemann-Petty problem, clearly, would not
change anything in case of the volume measure. Still in the case of of the
$\log$-concave measures the behaviour of the measure of a dilation of a convex
body is very interesting, we refer to [1, 6, 11, 14] for just a few examples
of such results.

Even though the dilation adds some strength to the condition of the bodies,
the answer to the dilation problem is positive for $n\leq 4$ and negative for
$n\geq 7$. That leaves the problem open for $n=5,6.$

To show the strength of the condition of the dilates, it was proved in [23]
that the dilation problem has an aﬃrmative answer when $K$ is a dilate of a
centered Euclidean ball: Consider a star bodies $L\subset\mathbb{R}^{n}$ and
assume there exists $R>0,$ such that

$\gamma_{n-1}(rRB_{2}^{n}\cap\xi^{\perp})\leq\gamma_{n-1}(rL\cap\xi^{\perp}),\hskip
14.22636pt\forall\xi\in\mathbb{S}^{n-1},\hskip 14.22636pt\forall r>0$

then it follows that $RB_{2}^{n}\subseteq L.$

In this paper we would like to study generalizations of the above fact. In
particular, we study measures $\mu$ for which we have an affirmative answer
for the following problem:

######  Question 1.

Consider a convex, symmetric body $K\subset{\mathbb{R}}^{n}$ such that for
every $t$ large enough and for some $R>0$

$\mu(tRB_{2}^{n})\leq\mu(tK),$

does it follow $RB_{2}^{n}\subseteq K?$

In Section 2 we will present a solution for Question 1 for the case of
$\log$-concave, rotation invariant probability measure $\mu.$

In Section 3 we consider a more general case, instead of comparing $K$ with
$B_{2}^{n}$, we will compare $K$ with another convex, symmetric body. Let us
denote by $\|x\|_{L}$ the Minkowski functional of $L$ which is defined to be
$\|x\|_{L}=\min\\{\lambda\geq 0:x\in\lambda L\\}.$

######  Question 2.

Let $K,L\subset\mathbb{R}^{n}$ be convex, symmetric bodies, and let $\mu$ be a
$\log$-concave probability measure, with density $e^{-\phi(\|x\|_{L})}$, where
$\phi:[0,\infty)\to[0,\infty)$ is an increasing convex function. If for every
$t$ large enough and some $R>0$

$\mu(tRL)\leq\mu(tK),$

does it follow that $RL\subseteq K$?

One of the core steps in answering Questions 1 and 2 is a generalization of
the classical large deviation principle, which is provided in Lemma 2 and
equation (12) below: Consider two symmetric, convex bodies
$K,L\subset{\mathbb{R}}^{n}$, let $r(K,L)=\max\\{R>0:RL\subset K\\}$, then

$\limsup_{t\to\infty}\frac{\ln{\mu((tK)^{c})}}{\phi(r(K,L)t)}=-1,$

where $\mu$ $\log$-concave probability measure, with density
$e^{-\phi(\|x\|_{L})}$.

Finally in Section 4 we will discuss the generalization of the Dilation
problem for Gaussian measures:

######  Question 3.

Consider a measure $\mu$ with continuous positive density $f$. Let
$\mu_{n-1}(K\cap\xi^{\perp})=\int_{K\cap\xi^{\perp}}f(x)dx.$ Consider two
convex symmetric bodies $K,L\subset\mathbb{R}^{n}$, such that

$\mu_{n-1}(rK\cap\xi^{\perp})\leq\mu_{n-1}(rL\cap\xi^{\perp}),\hskip
14.22636pt\forall\xi\in\mathbb{S}^{n-1},\hskip 14.22636pt\forall r>0,$

does it follow that

$\mu(K)\leq\mu(L)?$

We show that in general, the answer is still negative in dimension $n\geq 5$,
even under the assumption that $f$ is not a constant $\log$-concave function.
We also prove that if we add the requirement for the measure to be rotation
invariant, the answer will be negative in dimension $n\geq 7$, which leaves
the case of rotation invariant $\log$-concave measures open in dimension
$n=5,6.$

Acknowledgments. We are grateful to Mathieu Fradelizi, Dylan Langharst, Fedor
Nazarov and Mokshay Madiman for a number of valuable discussions and
suggestions.

##  2\. The case of rotation invariant measures

We remind that the measure $\mu$ is $\log$-concave probability measure if

$\mu(A)=\int_{A}e^{-\psi(x)}dx,$

where $\psi$ is a convex function and $\mu({\mathbb{R}}^{n})=1.$

######  Theorem 1.

Let $\mu$ be a $\log$-concave probability measure, with rotation invariant
non-constant density, i.e. $\psi(x)=\phi(|x|)$, where
$\phi:[0,\infty)\to[0,\infty)$ increasing convex function. Consider a convex,
symmetric body $K\subset{\mathbb{R}}^{n}$ such that for every $t$ large enough
and some $R>0$

$\mu(tRB_{2}^{n})\leq\mu(tK),$

then $RB_{2}^{n}\subseteq K.$

In order to prove the above theorem, we will need two lemmas.

######  Lemma 1.

Let $B^{n}_{2}$ be the Euclidean ball, and let $R>0,$ then

(2) $\limsup_{t\to\infty}\frac{\ln{\mu((tRB_{2}^{n})^{c})}}{\phi(tR)}=-1.$

###### Proof.

We will first show that the left-hand side inequality (2) is less or equal to
$-1$. Writing the integral in polar coordinates, we get

$\limsup_{t\to\infty}\frac{\ln{\int_{S^{n-1}}\int_{t}^{\infty}e^{-\phi(r)}r^{n-1}drd\theta}}{\phi(t)}=\limsup_{t\to\infty}\frac{\ln{\int_{t}^{\infty}e^{-\phi(r)}r^{n-1}dr}}{\phi(t)},$

we remind that $\lim_{t\to\infty}\phi(t)=\infty.$ Let
$\eta(t)=-(n-1)\ln{t}+\phi(t)$, thus $\eta(t)$ is a convex function. Using
that $\phi$ is convex function, we get that there exists $t_{\circ}\geq 0$
such that $\phi^{\prime}(t)\geq\phi^{\prime}(t_{0})>0$ for all $t>t_{0}$
(where we denote by $\phi^{\prime}(t)$ the left derivative, in the case convex
function $\phi(t)$ is not differentiable at $t$). Thus there exists constant
$a>0$ such that for for all $t$ large enough $\eta^{\prime}(t)>a$. Moreover,
using convexity of $\eta$ we get $\eta(r)\geq\eta(t)+a(r-t)$, for $r>t$. Thus

$\displaystyle\limsup_{t\to\infty}\frac{\ln{\int_{t}^{\infty}e^{-\phi(r)}r^{n-1}dr}}{\phi(t)}$
$\displaystyle\leq\limsup_{t\to\infty}\frac{\ln{\int_{t}^{\infty}e^{-\eta(t)-a(r-t)}dr}}{\phi(t)}$
$\displaystyle=\limsup_{t\to\infty}\frac{\ln{e^{-\eta(t)}}+\ln{\int_{t}^{\infty}e^{-a(r-t)}dr}}{\phi(t)}$
$\displaystyle=\limsup_{t\to\infty}\frac{\ln(t^{n-1}e^{-\phi(t)})+\ln{\frac{1}{a}}}{\phi(t)}$
$\displaystyle=-1+\limsup_{t\to\infty}\frac{\ln\frac{1}{a}}{\phi(t)}$
$\displaystyle=-1.$

Next, we will show that the right-hand side is greater or equal to $-1$. Since
$r>t$ we have

$\displaystyle\limsup_{t\to\infty}\frac{\ln{\int_{t}^{\infty}e^{-\phi(r)}r^{n-1}dr}}{\phi(t)}$
$\displaystyle\geq\limsup_{t\to\infty}\frac{\ln{t^{n-1}\int_{t}^{\infty}e^{-\phi(r)}dr}}{\phi(t)}$
$\displaystyle=\limsup_{t\to\infty}\frac{\ln{\int_{t}^{\infty}e^{-\phi(r)}dr}}{\phi(t)}.$

To finish proving the lemma, we prove the following claim.

######  Claim 1.

$\limsup_{t\to\infty}\frac{\ln{\int_{t}^{\infty}e^{-\phi(r)}dr}}{\phi(t)}\geq{-1}.$

Proof of Claim 1. Assume the result is not true, then there exists $\alpha>1$
such that

$\limsup_{t\to\infty}\frac{\ln{\int_{t}^{\infty}e^{-\phi(r)}dr}}{\phi(t)}\leq{-\alpha}.$

Thus, there exists $t_{0}>0$ such that for all $t>t_{0}$ we have,

(3) $\int_{t}^{\infty}e^{-\phi(r)}dr\leq e^{-\alpha\phi(t)}.$

Let $F(t)=\int_{t}^{\infty}e^{-\phi(r)}dr$, note that
$F^{\prime}(t)=-e^{-\phi(t)}$ and thus (3) is equivalent to
$F(t)^{\frac{1}{\alpha}}<-F^{\prime}(t).$ Therefore for $t>t_{0}$ we have

$1<-\frac{F^{\prime}(t)}{F(t)^{\frac{1}{\alpha}}}.$

Integrating both sides of above inequality from $t>t_{0}$ to infinity we get
that $\frac{1}{1-\frac{1}{\alpha}}F(t)^{1-\frac{1}{\alpha}}$ is unbounded,
which gives a contradiction and the claim is proved. This finishes the proof
of Lemma 1. ∎

######  Remark 1.

We note that in Claim 1 we have proved a stronger statement. Indeed, fix
$\alpha>1$ and let

$E=\\{t:\frac{\ln{\int_{t}^{\infty}e^{-\phi(r)}dr}}{\phi(t)}\leq-\alpha\\}$

then $|E|<\infty$. This follows from the fact that, again, for all $t\in E$ we
have that

$1<-\frac{F^{\prime}(t)}{F(t)^{\frac{1}{\alpha}}},$

thus

(4)
$|E|\leq\int_{E}-\frac{F^{\prime}(t)}{F(t)^{\frac{1}{\alpha}}}dt<\int_{t_{0}}^{\infty}-\frac{F^{\prime}(t)}{F(t)^{\frac{1}{\alpha}}}dt<\infty.$

######  Remark 2.

It is tempting to replace limit superior by the actual limit in the statement
of Lemma 1. This may be done in many particular cases of measure $\mu$, but it
is not true in general. Indeed, if we assume that

$\lim_{t\to\infty}\frac{\ln{\int_{t}^{\infty}e^{-\phi(r)}dr}}{\phi(t)}={-1}.$

then there exits $T_{1}>0$ such that for all $t>T_{1}$ we have

$\left|\frac{\ln{\int_{t}^{\infty}e^{-\phi(r)}dr}}{\phi(t)}+1\right|\leq 1.$

In particular

(5) $\int_{t}^{\infty}e^{-(\phi(r)-\phi(t))}dr\geq e^{-\phi(t)}.$

Using convexity of $\phi$ we get that
$\phi(r)-\phi(t)\geq\phi^{\prime}(t)(r-t)$ and thus, combining this with (5)
we get that $\phi^{\prime}(t)\leq e^{\phi(t)},$ for all $t>T_{1}$. Now one can
inductively define an increasing, positive, convex function $\phi$ which has
sufficiently large derivative at a sequence of points $t_{i}\to\infty,$ such
$\phi$ would contradict (5). Indeed, given a function $\phi$ on interval
$[0,t_{0}]$ we will extend $\phi$ to $[t_{0},\infty)$ such that there is
$t>t_{0}$ with $\phi^{\prime}(t)>e^{\phi(t)}.$ Without loss of generality, we
may assume that $t_{0}=0$ and $\phi(0)=a,\phi^{\prime}(0)=b$, for $t\geq 0.$
Let $\phi^{\prime}(t)=\alpha t+b,$ $\alpha>0$ and thus $\phi(t)=\alpha
t^{2}/2+bt$. Our goal is to find $t$ and $\alpha$ such that $\alpha
t+b>e^{\alpha t^{2}/2+bt}.$ Take $t=1/\sqrt{\alpha}$ the previous inequality
becomes $\sqrt{\alpha}+b>e^{1/2+b/\sqrt{\alpha}}$ which is true for $\alpha$
large enough.

We remind that for two convex, symmetric bodies $K,L\subset{\mathbb{R}}^{n}$,
we define $r(K,L)=\max\\{R>0:RL\subset K\\}$. The next lemma may be seen as a
generalization of the classical large deviation principle (see Corollary 4.9.3
in [2]).

######  Lemma 2.

Consider a symmetric body $K\subset{\mathbb{R}}^{n}$, then

$\limsup_{t\to\infty}\frac{\ln{\mu((tK)^{c})}}{\phi(r(K,B_{2}^{n})t)}=-1.$

###### Proof.

Let $R=r(K,B_{2}^{n})$, then $(tK)^{c}\subset(tRB_{2}^{n})^{c}$. Thus using
Lemma 1 we get

$\limsup_{t\to\infty}\frac{\ln{\mu((tK)^{c})}}{\phi(tR)}\leq\limsup_{t\to\infty}\frac{\ln{\mu((tRB_{2}^{n})^{c})}}{\phi(tR)}=-1.$

To obtain the reverse inequality, we denote by $P$ a plank of width $2R$ which
contains $K$. More precisely, using the maximality of $R$ there are tangent
points $y,-y\in R{\mathbb{S}}^{n-1}\cap\partial K$ and we can take
$P=\\{x\in{\mathbb{R}}^{n}:|\langle x,y\rangle|\leq R\\}$. Next,

$\limsup_{t\to\infty}\frac{\ln{\mu((tK)^{c})}}{\phi(tR)}\geq\limsup_{t\to\infty}\frac{\ln{\mu((tP)^{c})}}{\phi(tR)}.$

By the rotation invariant of $\mu$, we may assume that $y=Re_{n}$, and so

$\mu((tP)^{c})=2\int_{tR}^{\infty}\int_{\mathbb{R}^{n-1}}e^{-\phi(|ze_{n}+x|)}dxdz.$

Using the triangle inequality and the polar coordinates we get

$\displaystyle\mu((tP)^{c})$ $\displaystyle\geq
2\int_{tR}^{\infty}\int_{\mathbb{R}^{n-1}}e^{-\phi(z+|x|)}dxdz=2\int_{tR}^{\infty}\int_{\mathbb{S}^{n-2}}\int_{0}^{\infty}e^{-\phi(z+r)}r^{n-2}drd\theta
dz$
$\displaystyle=2|{\mathbb{S}^{n-2}}|\int_{0}^{\infty}r^{n-2}\int_{tR}^{\infty}e^{-\phi(z+r)}dzdr=2|{\mathbb{S}^{n-2}}|\int_{0}^{\infty}r^{n-2}\int_{tR+r}^{\infty}e^{-\phi(z)}dzdr$
$\displaystyle=2|\mathbb{S}^{n-2}|\int_{tR}^{\infty}e^{-\phi(z)}\int_{0}^{z-tR}r^{n-2}drdz=2\frac{|\mathbb{S}^{n-2}|}{n-1}\int_{tR}^{\infty}(z-tR)^{n-1}e^{-\phi(z)}dz.$

Now to finish the proof of lemma 2 we need to prove the following claim.

######  Claim 2.

$\limsup_{t\to\infty}\frac{\ln{\int_{tR}^{\infty}(z-tR)^{m}e^{-\phi(z)}dz}}{\phi(Rt)}\geq-1,$

for any $m>0$.

Proof of Claim 2. Making the change of variables we get

$\limsup_{t\to\infty}\frac{\ln{\int_{tR}^{\infty}(z-tR)^{m}e^{-\phi(z)}dz}}{\phi(Rt)}=\limsup_{t\to\infty}\frac{\ln{\int_{t}^{\infty}(r-t)^{m}e^{-\phi(r)}dr}}{\phi(t)}.$

We will first prove the following inductive step: fix $m\geq 1$, then

(6)
$\liminf_{t\to\infty}\frac{\ln{\int_{t}^{\infty}(r-t)^{m}e^{-\phi(r)}dr}}{\ln{\int_{t}^{\infty}(r-t)^{m-1}e^{-\phi(r)}dr}}=1.$

We note that the quantity under the natural logarithm in the above equality is
less than $1$, for $t$ large enough, and thus the denominator and numerator
are negative. It is a bit easier to work with the following fraction when both
the denominator and numerator are non-negative. So we will prove that

$\liminf_{t\to\infty}\frac{-\ln{\int_{t}^{\infty}(r-t)^{m}e^{-\phi(r)}dr}}{-\ln{\int_{t}^{\infty}(r-t)^{m-1}e^{-\phi(r)}dr}}=1.$

We first note that

$\int_{t}^{\infty}(r-t)^{m-1}e^{-\phi(r)}dr=\frac{1}{m}\int_{t}^{\infty}e^{-\phi(r)}d(r-t)^{m}$
$=\frac{1}{m}\int_{t}^{\infty}(r-t)^{m}\phi^{\prime}(r)e^{-\phi(r)}dr\geq\frac{1}{m}\phi^{\prime}(t)\int_{t}^{\infty}(r-t)^{m}e^{-\phi(r)}dr,$

where, again, we denote by $\phi^{\prime}(t)$ the left derivative of $\phi$.
Thus

$-\ln\int_{t}^{\infty}(r-t)^{m-1}e^{-\phi(r)}dr\leq-\ln(\phi^{\prime}(t)/m)-\ln\int_{t}^{\infty}(r-t)^{m}e^{-\phi(r)}dr,$

and

(7)
$\liminf_{t\to\infty}\frac{-\ln{\int_{t}^{\infty}(r-t)^{m}e^{-\phi(r)}dr}}{-\ln{\int_{t}^{\infty}(r-t)^{m-1}e^{-\phi(r)}dr}}\geq$
$\liminf_{t\to\infty}\frac{\ln(\phi^{\prime}(t)/m)-\ln{\int_{t}^{\infty}(r-t)^{m-1}e^{-\phi(r)}dr}}{-\ln{\int_{t}^{\infty}(r-t)^{m-1}e^{-\phi(r)}dr}}\geq
1,$

where we used that $\phi^{\prime}(t)$ is increasing and thus
$\phi^{\prime}(t)>c>0$ for $t>t_{0}$, where $t_{0}>0$. To prove the reverse
inequality let

$F(t)=\int_{t}^{\infty}(r-t)^{m}e^{-\phi(r)}dr\mbox{ and thus
}F^{\prime}(t)=-m\int_{t}^{\infty}(r-t)^{m-1}e^{-\phi(r)}dr.$

Assume that

$\liminf_{t\to\infty}\frac{-\ln{F(t)}}{-\ln(-\frac{1}{m}F^{\prime}(t))}>\alpha>1,$

but then, again there exists $t_{0}>0$ such that for all $t>t_{0},$ we have

$-\ln{F(t)}>-\alpha\ln(-\frac{1}{m}F^{\prime}(t)),$

and thus

$m<\frac{-F^{\prime}(t)}{F(t)^{\frac{1}{\alpha}}}.$

Take an integral over $t\in[x,\infty]$ from both sides to get
$F^{1-\frac{1}{\alpha}}(x)=\infty,$ which is a contradiction. This finishes
the proof of inductive step, but we actually need a bit stronger statement,
similar to Remark 1. Indeed, consider any $\alpha>1$ and let

$E_{m,\alpha}=\\{t:\frac{-\ln{\int_{t}^{\infty}(r-t)^{m}e^{-\phi(r)}dr}}{-\ln{\int_{t}^{\infty}(r-t)^{m-1}e^{-\phi(r)}dr}}>\alpha\\},$

then, using the same ideas as in (4) we get $|E_{m,\alpha}|<\infty$.

To complete our proof, let $X(t)=\prod_{i=1}^{m}X_{i}(t)$ where

$X_{i}(t)=\frac{\ln{\int_{t}^{\infty}(r-t)^{i}e^{-\phi(r)}dr}}{\ln{\int_{t}^{\infty}(r-t)^{i-1}e^{-\phi(r)}dr}},$

and let

$Y(t)=\frac{\ln{\int_{t}^{\infty}e^{-\phi(r)}dr}}{\phi(t)}.$

Using (6) we get $\liminf\limits_{t\to{\infty}}X_{i}(t)=1$ and using (2) we
get $\limsup\limits_{t\to\infty}Y(t)=-1.$ Our goal is to prove that

$\limsup_{t\to\infty}X(t)Y(t)\geq-1.$

Assume that this is not true, then there exists $\alpha>1$ such that

$\limsup_{t\to\infty}X(t)Y(t)\leq{-\alpha}<-1.$

Therefore, there exists $t_{0}$ such that for all $t>t_{0}$

(8) $X(t)Y(t)\leq-\alpha.$

Using (7) we may also assume that $X_{i}(t)>0$ for all $t>t_{0}.$ Next,
consider a set

$A:=\\{t>t_{0}:X(t)>\frac{\alpha+1}{2}\\},$

we claim that $|A|<\infty$. Note that

$|\\{t:X(t)>\frac{\alpha+1}{2}\\}|\leq|\\{t:\exists
i=1,\dots,m:X_{i}(t)>\left(\frac{\alpha+1}{2}\right)^{\frac{1}{m}}\\}|$
$<\sum_{i=1}^{m}|{\\{t:X_{i}(t)>(\frac{\alpha+1}{2})^{\frac{1}{m}}\\}}|<\infty.$

We also note that $\frac{2\alpha}{\alpha+1}>1$ and thus

$|\\{t:Y(t)<-\frac{2\alpha}{\alpha+1}\\}|<\infty.$

Finally

$|\\{t>t_{0}:X(t)Y(t)<-\alpha\\}|=|\\{t>t_{0}:Y(t)<-\frac{\alpha}{X(t)}\\}|$
$\leq|A|+|\\{t>t_{0}:Y(t)<-\frac{\alpha}{X(t)}\mbox{ and
}X(t)<\frac{\alpha+1}{2}\\}|$
$\leq|A|+|\\{t:Y(t)<-\frac{2\alpha}{\alpha+1}\\}|<\infty,$

which contradicts with (8). The claim is proved and this finishes the proof of
Lemma 2. ∎

We are now ready to prove Theorem 1.

###### Proof.

We will use the contradiction argument. Let $K\subset{\mathbb{R}}^{n}$ be a
convex, symmetric body such that $\mu(tRB_{2}^{n})\leq\mu(tK)$ holds for every
$t$ large enough and for some $R>0$, but $RB_{2}^{n}\not\subset K.$ Thus, one
can conclude that $K$ contains a maximal Euclidean ball $rRB_{2}^{n}$ with
$r<1$. From the assumption it follows that

$\mu((tRB_{2}^{n})^{c})\geq\mu((tK)^{c}),$

which implies that

(9)
$\frac{\ln{\mu((tRB_{2}^{n})^{c})}}{\phi(tR)}\geq\frac{\ln{\mu((tK)^{c})}}{\phi(trR)}\frac{\phi(trR)}{\phi(tR)}.$

From the convexity of $\phi$ with $r<1$ we have the following inequality

$\phi(trR)=\phi(trR+(1-r)0)\leq r\phi(tR)+(1-r)\phi(0).$

Hence, we get $\lim_{t\to\infty}\frac{\phi(trR)}{\phi(tR)}\leq r$. Taking the
limit superior as $t$ goes to infinity from both sides of the inequality (9),
we obtain $-1\geq-r.$ But this contradicts the fact that $r$ is less than $1$.
Therefore, our assumption that $RB_{2}^{n}\not\subset K$ must be false. ∎

######  Remark 3.

The rotation invariant assumption on $\mu$ in Theorem 1 is necessary. Indeed,
one can construct an example of a $\log$-concave probability measure that is
not rotation invariant in $\mathbb{R}^{2}$ which does not satisfy the
statement of Theorem 1. Consider rectangle
$\Omega=\\{(x,y):|x|\leq\frac{\pi}{2},|y|\leq\frac{1}{2}\\}$, and define the
measure $\mu$ as $\mu(K)=\frac{|K\cap\Omega|}{|\Omega|}$. Take $K=\Omega$, we
have

$B_{2}^{2}\not\subset\Omega,\mbox{ but }|B_{2}^{2}|=|\Omega|=\pi\mbox{ and
}|tB_{2}^{2}|=|t\Omega|,\hskip 2.84544pt\forall t>0.$

Note that $\mu(tB_{2}^{2})\leq\mu(t\Omega);\hskip 2.84544pt\forall t>0$,
indeed, this is equivalent to $|tB_{2}^{2}\cap\Omega|\leq|t\Omega\cap\Omega|$.
If $t\leq 1$, then we have $|tB_{2}^{2}\cap\Omega|\leq|tB_{2}^{2}|=|t\Omega|$,
and if $t\geq 1$ we get $|tB_{2}^{2}\cap\Omega|\leq|\Omega|.$ So, we provided
an example where

$\mu(tB_{2}^{2})\leq\mu(tK);\hskip 14.22636pt\forall t>0$

but $B_{2}^{2}\not\subset K.$

##  3\. The case of density depending on a norm

In this section, we would like to give a proof Theorem 1 in a more general
case, which would answer Question 2. The main idea and computation are in the
same spirit as in the proof of Theorem 1.

######  Theorem 2.

Let $K,L\subset\mathbb{R}^{n}$ be convex, symmetric bodies, and let $\mu$ be a
$\log$-concave probability measure, with density $e^{-\phi(\|x\|_{L})}$, where
$\phi:[0,\infty)\to[0,\infty)$ is increasing, convex function. If for every
$t$ large enough and some $R>0$

$\mu(tRL)\leq\mu(tK),$

then $RL\subseteq K.$

###### Proof.

We have to check Lemma 1 and Lemma 2, i.e. to prove yet another generalization
of the classical large deviation principle (see (12) below).

We claim that for any positive $R$

(10) $\limsup_{t\to\infty}\frac{\ln{\mu((tRL)^{c})}}{\phi(tR)}=-1.$

We can assume $R=1$. Next, write $\mu$ as the following:

$\displaystyle\mu((tL)^{c})$
$\displaystyle=\int_{(tL)^{c}}e^{-\phi(\|x\|_{L})}dx=\int_{(tL)^{c}}\int_{\phi(\|x\|_{L})}^{\infty}e^{-u}dudx$
$\displaystyle=\int_{\mathbb{R}^{n}}\int_{\phi(\|x\|_{L})}^{\infty}\\!\\!\\!\chi_{(tL)^{c}}(x)e^{-u}dudx=\int_{0}^{\infty}\int_{\\{x:\phi(\|x\|_{L})<u\\}}\\!\\!\\!\chi_{(tL)^{c}}(x)e^{-u}dxdu$
$\displaystyle=\\!\int_{0}^{\infty}\\!\\!\\!e^{-u}|\\{x\\!\\!:\|x\|_{L}\in[t,\phi^{-1}(u)]\\}|du=|L|\\!\int_{\phi(t)}^{\infty}((\phi^{-1}(u))^{n}-t^{n})e^{-u}du$
$\displaystyle=|L|\int_{t}^{\infty}(v^{n}-t^{n})\phi^{\prime}(v)e^{-\phi(v)}dv=-|L|\int_{t}^{\infty}(v^{n}-t^{n})de^{-\phi(v)}$
(11) $\displaystyle=n|L|\int_{t}^{\infty}v^{n-1}e^{-\phi(v)}dv.$

So, we have

$\displaystyle\limsup_{t\to\infty}\frac{\ln\mu((tL)^{c})}{\phi(t)}=\limsup_{t\to\infty}\frac{\ln({n|L|\int_{t}^{\infty}e^{-\phi(v)}v^{n-1}dv})}{\phi(t)}$
$\displaystyle=\limsup_{t\to\infty}\frac{\ln{\int_{t}^{\infty}e^{-\phi(v)}v^{n-1}dv}}{\phi(t)}.$

But we showed the above quantity equals $-1$ and thus Lemma 1 is applied for
the general measure case.

To finish the proof, we must check Lemma 2. In particular, we want to show
that

(12) $\limsup_{t\to\infty}\frac{\ln{\mu((tK)^{c})}}{\phi(r(K,L)t)}=-1,$

for symmetric, convex bodies $K,L\subset{\mathbb{R}}^{n}$, convex, increasing
function $\phi:[0,\infty)\to[0,\infty)$ and measure $\mu$ with density
$e^{-\phi(\|x\|_{L})}$.

Let $R=r(K,L)$, then we have $(tK)^{c}\subset(tRL)^{c}$ from the assumption.
Thus using Lemma 1 we get

$\limsup_{t\to\infty}\frac{\ln{\mu((tK)^{c})}}{\phi(tR)}\leq\limsup_{t\to\infty}\frac{\ln{\mu((tRL)^{c})}}{\phi(tR)}=-1.$

Using that $RL$ is the maximal dilate of $L$ inside $K$ we get that there is,
at least one, pair of points $v,-v\in\partial RL\cap\partial K$. Let $P$ be a
plank created by tangent planes to $RL$ and $K$ at $v$ (and $-v$). Let $n_{v}$
be a normal vector to $\partial RL$ at $v$. Then the width of the plank is
$2Rh_{L}(n_{v})=2h_{K}(n_{v})$: $P=\\{x\in{\mathbb{R}}^{n}:|\langle
x,n_{v}\rangle|\leq Rh_{L}(n_{v})\\}$, where $h_{L}$ is the support function
of $L$ which is defined by $h_{L}(x)=sup\\{\langle x,y\rangle:y\in L\\}.$
Next,

$\limsup_{t\to\infty}\frac{\ln{\mu((tK)^{c})}}{\phi(tR)}\geq\limsup_{t\to\infty}\frac{\ln{\mu((tP)^{c})}}{\phi(tR)}.$

Selecting a proper system of coordinates, we assume that $n_{v}=e_{n}$, let
$a=tRh_{L}(e_{n}),$ we have

$\displaystyle\mu((tP)^{c})$
$\displaystyle=2\int_{a}^{\infty}\int_{\mathbb{R}^{n-1}}e^{-\phi(\|ze_{n}+x\|_{L})}dxdz$
$\displaystyle=2\int_{a}^{\infty}\int_{{\mathbb{R}}^{n-1}}\int^{\infty}_{\phi(\|ze_{n}+x\|_{L})}e^{-u}dudxdz$
$\displaystyle=2\int_{a}^{\infty}\int_{0}^{\infty}\int_{\\{x:\phi(\|ze_{n}+x\|_{L})<u\\}}e^{-u}dxdudz$
$\displaystyle=2\int_{a}^{\infty}\int_{0}^{\infty}e^{-u}\left|\left\\{x\in{\mathbb{R}}^{n-1}:\|ze_{n}+x\|_{L}\leq\phi^{-1}(u)\right\\}\right|dudz.$

Now note that

$\left|\left\\{x\in{\mathbb{R}}^{n-1}:\|ze_{n}+x\|_{L}\leq\phi^{-1}(u)\right\\}\right|=\left|\left\\{x\in{\mathbb{R}}^{n-1}:ze_{n}+x\in\phi^{-1}(u)L\right\\}\right|.$

The above volume is zero if $z>\phi^{-1}(u)h_{L}(e_{n})$ (or
$\phi(z/h_{L}(e_{n}))>u$), for $z\in[0,\phi^{-1}(u)h_{L}(e_{n})]$ we note that
$\phi^{-1}(u)L$ is a convex body and thus contains inside a pyramid
$P^{\prime}$ with base $L\cap{\mathbb{R}}^{n-1}$ and the height
$\phi^{-1}(u)h_{L}(e_{n})$ (with apex $\phi^{-1}(u)h_{L}(e_{n})v/R$), then

$\left|\left\\{x\in{\mathbb{R}}^{n-1}:ze_{n}+x\in\phi^{-1}(u)L\right\\}\right|\geq|P^{\prime}\cap({\mathbb{R}^{n-1}}+ze_{n})|$
$=(\phi^{-1}(u)h_{L}(e_{n})-z)^{n-1}|L\cap{\mathbb{R}}^{n-1}|.$

Thus

$\displaystyle\mu((tP)^{c})$ $\displaystyle\geq
2|L\cap{\mathbb{R}}^{n-1}|\int_{a}^{\infty}\int_{\phi(z/h_{L}(e_{n}))}^{\infty}e^{-u}(\phi^{-1}(u)h_{L}(e_{n})-z)^{n-1}dudz$
$\displaystyle=2|L\cap{\mathbb{R}}^{n-1}|\int_{a}^{\infty}\int_{z/h_{L}(e_{n})}^{\infty}e^{-\phi(u)}\phi^{\prime}(u)(uh_{L}(e_{n})-z)^{n-1}dudz$
$\displaystyle=-2|L\cap{\mathbb{R}}^{n-1}|\int_{a}^{\infty}\int_{z/h_{L}(e_{n})}^{\infty}(uh_{L}(e_{n})-z)^{n-1}de^{-\phi(u)}dz$
$\displaystyle=2(n-1)|L\cap{\mathbb{R}}^{n-1}|\int_{a}^{\infty}\int_{z/h_{L}(e_{n})}^{\infty}e^{-\phi(u)}(uh_{L}(e_{n})-z)^{n-2}dudz$
$\displaystyle=2(n-1)|L\cap{\mathbb{R}}^{n-1}|\int_{a/h_{L}(e_{n})}^{\infty}\int_{a}^{h_{l}(e_{n})u}e^{-\phi(u)}(uh_{L}(e_{n})-z)^{n-2}dzdu$
$\displaystyle=2|L\cap{\mathbb{R}}^{n-1}|\int_{a/h_{L}(e_{n})}^{\infty}e^{-\phi(u)}(uh_{L}(e_{n})-a)^{n-1}du$
$\displaystyle=2h_{L}(e_{n})^{n-1}|L\cap{\mathbb{R}}^{n-1}|\int_{tR}^{\infty}e^{-\phi(u)}(u-tR)^{n-1}du.$

So, we have

$\displaystyle\limsup_{t\to\infty}\frac{\ln\mu((tP)^{c})}{\phi(tR)}\geq\limsup_{t\to\infty}\frac{\ln(2h_{L}(e_{n})^{n-1}|L\cap{\mathbb{R}}^{n-1}|\int_{tR}^{\infty}e^{-\phi(u)}(u-tR)^{n-1}du)}{\phi(tR)}$
$\displaystyle=\limsup_{t\to\infty}\frac{\ln\int_{tR}^{\infty}e^{-\phi(u)}(u-tR)^{n-1}du}{\phi(tR)}.$

By Claim 2 the above quantity is greater than or equal to $-1$, thus Lemma 2
is applied here which finishes the proof for our main result. ∎

######  Remark 4.

The proofs for Theorem 1 and Theorem 2 apply similarly to an asymmetric convex
body $K$ with the origin as an interior point of it, the only difference is
that instead of dealing with a plank $P$ in Lemma 2 we will deal with a half-
space. Specifically, for Theorem 1, the half-space
$H=\\{x\in{\mathbb{R}}^{n}:\langle x,y\rangle\leq R\\}$ where $y\in
R{\mathbb{S}}^{n-1}\cap\partial K,$ will do. For Theorem 2, we will use the
half-space defined by $H=\\{x\in{\mathbb{R}}^{n}:\langle x,n_{v}\rangle\leq
Rh_{L}(n_{v})\\},$ where $n_{v}$ is the normal vector to $\partial RL$ at the
tangent point $v$.

##  4\. The Busemann - Petty type problems

In this section, we would like to discuss Question 3. We first note that one
must make some additional assumptions on the measure $\mu$ to avoid a trivial
answer. Indeed, if a measure $\mu$ has a homogeneous density (i.e.
$f(rx)=r^{p}f(x),$ for $r>0$ and $p>1-n$), then the answer is identical to the
one given in [22].

Let us first show that in dimension $n\geq 5$ one can always find a pair of
convex, symmetric bodies $K$ and $L$ and measure $\mu$, such that the answer
to Question 3 is negative. The main idea follows from construction in [23]. To
begin with the following fact:

Fact. If $d\mu=e^{-\phi(\|x\|_{L})}dx$ is a $\log$-concave measure and
$K,L\subset\mathbb{R}^{n}$ are convex, symmetric bodies such that
$|K|\leq|RL|$ for some $R>0,$ then

$\mu(K)\leq\mu(RL).$

###### Proof.

Using calculations similar to (3) we get

$\mu(K)=\int_{0}^{\infty}e^{-u}|K\cap\phi^{-1}(u)L|du$

and

$\mu(RL)=\int_{0}^{\infty}e^{-u}|RL\cap\phi^{-1}(u)L|du.$

To get $\mu(K)\leq\mu(RL)$ we have to check that
$|K\cap\phi^{-1}(u)L|\leq|RL\cap\phi^{-1}(u)L|.$ Indeed, if
$R\leq\phi^{-1}(u),$ then we have

$|K\cap\phi^{-1}(u)L|\leq|K|\leq|RL|=|RL\cap\phi^{-1}(u)L|,$

and if $R\geq\phi^{-1}(u),$ then we get

$|K\cap\phi^{-1}(u)L|\leq|\phi^{-1}(u)L|=|RL\cap\phi^{-1}(u)L|.$

Hence, $\mu(K)\leq\mu(RL),$ for any $R>0.$ ∎

Next, we show that Question 3 has a negative answer for $n\geq 5$.

######  Theorem 3.

For $n\geq 5,$ there are convex symmetric bodies $K,L\subset\mathbb{R}^{n}$
and $\log$-concave measure $\mu$ with density $e^{-\phi(\|x\|_{L})}$, such
that

(13) $\mu(rK\cap\xi^{\perp})\leq\mu(rL\cap\xi^{\perp}),\hskip
14.22636pt\forall\xi\in\mathbb{S}^{n-1},\hskip 14.22636pt\forall r>0,$

but $\mu(K)>\mu(L).$

###### Proof.

Let us assume, towards the contradiction, that Question 3 has an affirmative
answer in $\mathbb{R}^{n}$, for some fixed $n.$ So, for any pair of convex
symmetric bodies $K,L$ that satisfy (13), we would get $\mu(K)\leq\mu(L).$ The
condition on sections (13) will be also satisfied for the dilated bodies $tK$
and $tL$, for all $t>0.$ Therefore, we have

(14) $\mu(tK)\leq\mu(tL),\hskip 14.22636pt\forall t>0.$

Which by definition of $\mu$ means

$\int_{tK}{e^{-\phi(||x||_{L})}dx}\leq\int_{tL}{e^{-\phi(||x||_{L})}dx},$

or equivalently, applying the change of variables $x=tx$, we have

$\int_{K}{e^{-\phi(t||x||_{L})}dx}\leq\int_{L}{e^{-\phi(t||x||_{L})}dx}.$

Using the continuity of $\phi$ and compactness of $K$ and $L$, we can take the
limit for the above inequality as $t\to 0^{+}$, to obtain

$|K|\leq|L|.$

Therefore, we have a relation between the dilation problem for a
$\log$-concave probability measure, with the Busemann-Petty problem for volume
measure, which is if

$\mu(rK\cap\xi^{\perp})\leq\mu(rL\cap\xi^{\perp})),\hskip
14.22636pt\forall\xi\in\mathbb{S}^{n-1},\hskip 14.22636pt\forall r>0,$

then $|K|\leq|L|$.

A number of very interesting counterexamples to the Busemann Pettry problem
were shown by Papedimitrakis [15]; Gardner [7] Gardner, Koldobsky and
Schlumprecht [9]: there are convex symmetric bodies $K,L$ in $\mathbb{R}^{n}$
for $n\geq 5$ such that

(15) $|K\cap\xi^{\perp}|\leq|L\cap\xi^{\perp}|,\hskip
14.22636pt\forall\xi\in\mathbb{S}^{n-1},$

but

(16) $|K|>|L|.$

Note that because the volume measure is homogeneous the condition on sections
(15) is also true for dilates of $K$ and $L$, so we have

(17) $|rK\cap\xi^{\perp}|\leq|rL\cap\xi^{\perp}|,\hskip
14.22636pt\forall\xi\in\mathbb{S}^{n-1},\hskip 14.22636pt\forall r>0.$

Now, applying the Fact to (17), we get that

$\mu(rK\cap\xi^{\perp})\leq\mu(rL\cap\xi^{\perp}),\hskip
14.22636pt\forall\xi\in\mathbb{S}^{n-1},\hskip 14.22636pt\forall r>0,$

by (14), we have

$\mu(tK)\leq\mu(tL),\hskip 14.22636pt\forall t>0,$

dividing by $t^{n}$ and taking the limit of the above inequality as $t\to
0^{+}$ we get

$|K|\leq|L|,$

and this contradicts (16).

∎

It is interesting to note that the measure $\mu$ constructed above is very
specific. For example, we can not use this construction directly with the
assumption that $\mu$ is rotation invariant.

Still, we can show that the answer to Question 3 is negative in
${\mathbb{R}}^{n}$ for $n\geq 7$ even when $\mu$ is a $\log$-concave measure
with rotation invariant density.

######  Theorem 4.

For dimension $n\geq 7,$ and $d\mu=e^{-\phi(|x|)}dx$ there is a convex
symmetric body $K\subset\mathbb{R}^{n}$ such that

$\mu(rK\cap\xi^{\perp})\leq\mu(rB_{2}^{n}\cap\xi^{\perp}),\hskip
14.22636pt\forall\xi\in\mathbb{S}^{n-1},\hskip 14.22636pt\forall r>0,$

but $\mu(K)>\mu(B_{2}^{n}).$

###### Proof.

Giannopoulos [10] and Bourgain [4] constructed an example in $\mathbb{R}^{n}$
for $n\geq 7$ of convex body $K\subset\mathbb{R}^{n}$, that satisfies

$|K\cap\xi^{\perp}|\leq|B_{2}^{n}\cap\xi^{\perp}|,\hskip
14.22636pt\forall\xi\in\mathbb{S}^{n-1},$

but $|K|>|B_{2}^{n}|$. We will obtain the result by taking the same convex
body $K$ and $B_{2}^{n}$ as provided in [10, 4] and applying the same steps as
in the previous proof. ∎

## References

  * [1] S. G. Bobkov and F. Nazarov, Sharp dilation-type inequalities with fixed parameter of convexity, Zap. Nauchn. Sem. POMI, 351 (2007), 54–78. English translation in J. Math. Sci. (N.Y.), 152 (2008), 826–839.
  * [2] V. Bogachev, Gaussian measures. Mathematical Surveys and Monographs, 62. American Mathematical Society, Providence, RI, 1998. xii+433 pp.
  * [3] C. Borell, Convex set functions in d-space, Periodica Mat. Hungaria 6 (1975), no. 2, 111–136.
  * [4] J. Bourgain, On the Busemann-Petty problem for perturbations of the ball, Geom. Funct. Anal. 1 (1991), 1-13.
  * [5] H. Busemann and C.M. Petty, Problems on convex bodies, Math. Scand. 4 (1956), 88–94.
  * [6] D. Cordero-Erausquin, M. Fradelizi, and B. Maurey, The (B) conjecture for the Gaussian measure of dilates of symmetric convex sets and related problems, J. Funct. Anal. 214 (2004), no. 2, 410–427.
  * [7] R. J. Gardner, Intersection bodies and the Busemann-Petty problem, Trans. Amer. Math. Soc. 342 (1994), 435–445.
  * [8] R.J. Gardner, Geometric Tomography. Second edition, Cambridge University Press, Cambridge, 2006, xxii+492 pp.
  * [9] R.J. Gardner, A. Koldobsky and Th. Schlumprecht, An analytic solution to the Busemann-Petty problem on sections of convex bodies, Annals of Math. 149 (1999), 691–703.
  * [10] A. Giannopoulos, A note on a problem of H. Busemann and C. M. Petty concerning sections of symmetric convex bodies, Mathematika 37 (1990), 239–244.
  * [11] M. Fradelizi Concentration inequalities for s–concave measures of dilations of Borel sets and applications, Electron. J. Probab. 14 (2009), no. 71, 2068–2090.
  * [12] A. Koldobsky, Intersection bodies, positive definite distributions and the Busemann-Petty problem, Amer. J. Math. 120 (1998), 827–840.
  * [13] E. Lutwak, Intersection bodies and dual mixed volumes, Adv. Math. 71 (1988), 232–261.
  * [14] R. Latala and K. Oleszkiewicz, Gaussian measures of dilations of convex symmetric sets, Ann. Probab. 27 (1999) 1922–1938.
  * [15] M. Papadimitrakis, On the Busemann-Petty problem about convex, centrally symmetric bodies in ${\mathbb{R}}^{n}$, Mathematika 39 (1992), 258–266.
  * [16] A. Prékopa, Logarithmic concave measures with application to stochastic programming, Acta Sci. Math. 32 (1971), 301–316
  * [17] A. Prékopa, On logarithmic concave measures and functions, Acta Sci. Math. 34 (1973), 335–343.
  * [18] W. Rudin, Functional Analysis. Second edition. International Series in Pure and Applied Mathematics. McGraw-Hill, Inc., New York, 1991. xviii+424 pp.
  * [19] R. Schneider, Convex Bodies: The Brunn-Minkowski Theory, Second expanded edition. Encyclopedia of Mathematics and Its Applications 151, Cambridge University Press, Cambridge, 2014.
  * [20] G. Zhang, A positive answer to the Busemann-Petty problem in four dimensions , Annals of Math. 149 (1999), 535–543.
  * [21] A. Zvavitch, Gaussian measure of sections of convex bodies, Adv. Math. 188 (2004), no. 1, 124–136.
  * [22] A. Zvavitch, The Busemann-Petty problem for arbitrary measures, Math. Ann. 331 (2005), 867–887.
  * [23] A. Zvavitch, Gaussian measure of sections of dilates and translations of convex bodies, Adv. Math. 41/2 (2008), 247-254

Department of Mathematical Sciences, Kent State University, Kent, OH USA.
E-mail address<EMAIL_ADDRESS><EMAIL_ADDRESS>