$M$ has maximal group image isomorphic to $G\ast\mathbb{Z}$ which is a one-
relator group and thus has decidable word problem by Magnus’ Theorem. It then
follows from [22, Theorem 3.3] that $M$ has decidable word problem. But in
Theorem 7.4 it is proved that decidability of the word problem for $M$ implies
decidability of the membership problem for $Q$ in $G$, hence the membership
problem for $Q$ in $G$ is decidable, as claimed.

In the proof of Theorem 7.4 it is shown that the identity map on
$B\cup\\{x,t\\}$ induces an isomorphism between the inverse monoids

$M=\operatorname{Inv}\bigl{\langle}B,x,t\>|\>r=1,\quad
tz_{1}st\overline{z_{1}}s\ldots stz_{k}st\overline{z_{k}}s=1\bigr{\rangle}$

and

$T=\operatorname{Inv}\bigl{\langle}B,x,t\>|\>r=1,\quad(tz_{1}t^{-1})(tz_{1}^{-1}t^{-1})=1,\quad\ldots,\quad(tz_{k}t^{-1})(tz_{k}^{-1}t^{-1})=1\bigr{\rangle}.$

It follows that the identity map induces an isomorphism between the maximal
group images of these inverse monoids

$\operatorname{Gp}\bigl{\langle}B,x,t\>|\>r=1,\quad
tz_{1}st\overline{z_{1}}s\ldots
stz_{k}st\overline{z_{k}}s=1\bigr{\rangle}\rightarrow\operatorname{Gp}\bigl{\langle}B,x,t\>|\>r=1\bigr{\rangle}.$

(In general if the identity map induces an isomorphism
$\operatorname{Inv}\bigl{\langle}A\>|\>R\bigr{\rangle}\rightarrow\operatorname{Inv}\bigl{\langle}A\>|\>S\bigr{\rangle}$
then for any additional set of relations $T$ the identity map will also induce
an isomorphism between $\operatorname{Inv}\bigl{\langle}A\>|\>R\cup
T\bigr{\rangle}\rightarrow\operatorname{Inv}\bigl{\langle}A\>|\>S\cup
T\bigr{\rangle}$ since $S$ and $R$ are consequence of each other, and hence
the same is true of $S\cup T$ and $R\cup T$. And so in particular it is true
when $T$ is the set of relations $aa^{-1}=1=aa^{-1}$ added to get the maximal
group image.) Now set $u\equiv r$ and $v\equiv tz_{1}st\overline{z_{1}}s\ldots
stz_{k}st\overline{z_{k}}s$ and $X=B\cup\\{x,t\\}$. Then the identity map on
$X$ induces isomorphisms

$\operatorname{Gp}\bigl{\langle}X\>|\>u=1,v=1\bigr{\rangle}\rightarrow\operatorname{Gp}\bigl{\langle}X\>|\>u=1\bigr{\rangle}.$

It follows from Theorem 7.4 that the maximal group image of $M$ is isomorphic
to $G\ast\mathbb{Z}$. But the group
$\operatorname{Gp}\bigl{\langle}X\>|\>vuv^{-1}=1\bigr{\rangle}$ we have
constructed is isomorphic to the maximal group image
$\operatorname{Gp}\bigl{\langle}X\>|\>u=1,v=1\bigr{\rangle}$ of $M$, so this
completes the proof of the theorem. ∎

######  Corollary 7.6.

There is a positive two-relator group
$\operatorname{Gp}\bigl{\langle}A\>|\>u=1,v=1\bigr{\rangle}$, where $u,v\in
A^{+}$, with an undecidable prefix membership problem.

###### Proof.

This follows from Theorem 7.5 together with Theorem 3.8 that shows there are
positive one-relator groups containing finitely generated submonoids in which
membership is undecidable. ∎

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# An affine isoperimetric inequality for log-concave functions∗

Zengle Zhang1 and Jiazu Zhou2,3,∗∗<EMAIL_ADDRESS><EMAIL_ADDRESS>1
Key Laboratory of Group and Graph Theories and Applications, Chongqing
University of Arts and Sciences, Chongqing 402160, China. 2 School of
Mathematics and Statistics, Southwest University, Chongqing 400715, China. 3
College of Science, Wuhan University of Science and Technology, Wuhan 430081,
China.

###### Abstract.

The authors gave an affine isoperimetric inequality [38] that gives a lower
bound for the volume of a polar body and the equality holds if and only if the
body is a simplex. In this paper, we give a functional isoperimetric
inequality for log-concave functions that contains the affine isoperimetric
inequality of Lutwak, Yang and Zhang in [38].

###### Key words and phrases:

LYZ ellipsoid; LYZ polar inequality; log-concave function; affine
isoperimetric inequality; reverse affine isoperitmetric inequality.

###### 2010 Mathematics Subject Classification:

52A40, 52A41

*Supported in part by NSFC (No. 12071378, No. 12301071) and the Science and Technology Research Program of Chongqing Municipal Education Commission (No. KJQN202201339);

**The corresponding author

##  1\. Introductions

The isoperimetric problem was known in Ancient Greece. However, the first
mathematically rigorous proof was obtained only in the 19th century by
Weierstrass based on works of Bernoulli, Euler, Lagrange and others. The
isoperimetric problem is equivalent to the isoperimetric inequality that is
bounding by the surface area and volume of the geometric domain $K$ in the
Euclidian space $\mathbb{R}^{n}$ with equality if and only if $K$ is a
standard ball. The natural generalizations of the isoperimetric inequality are
Alexandrov-Fenchel inequalities that are bounding by mixed volumes of convex
bodies in integral and convex geometry analysis. Blascheke-Santaló inequality,
differential affine isoperimetric inequality, Busemann-Petty centroid
inequality, Petty projection inequality and more affine isoperimetric
inequalities are found with equalities for ellipsoids, polar bodies or
simplices [8, 14, 15, 27, 34, 35, 36, 37, 39, 41, 46].

During past decades, the reverse affine isoperimetric inequalities, with
simplices, cubes or polar bodies as extremal, received more attentions.
Usually, reverse affine isoperimetric inequalities are much harder to
establish. One of the known open problem about the reverse affine
isoperimetric inequalities is the Mahler conjecture.

Let $\mathcal{K}_{o}^{n}$ denote the set of compact convex sets that contain
the origin in their interiors in $\mathbb{R}^{n}$. Let
$K\in\mathcal{K}_{o}^{n}$, the polar body $K^{\circ}$ of $K$, is defined by

$\displaystyle K^{\circ}=\\{x\in\mathbb{R}^{n}:x\cdot y\leq 1,\quad\text{for
all}\quad y\in K\\},$ (1.1)

and $x\cdot y$ denotes the inner product of $x$ and $y$.

If $K\in\mathcal{K}_{o}^{n}$ is symmetric, Mahler[40] conjectured that

$\displaystyle|K||K^{\circ}|\geq\frac{4^{n}}{n!},$ (1.2)

where $|K|$ denotes the volume of $K$, with equality if and only if either $K$
or $K^{\circ}$ is a parallelepiped.

Inequality (1.2) was proved by Mahler [40] for $n=2$, the $3$-dimensional case
was solved by Iriyeh and Shibata [28], and the case when $n\geq 4$ is still
open. One can see [8, 9, 10, 14, 15, 25, 26, 27, 34, 35, 36, 37, 38, 39, 41,
46, 47, 48, 49] for more references on Mahler’s conjecture, affine
isoperimetric inequalities and reverse affine isoperimetric inequalities.

Let $K\in\mathcal{K}_{o}^{n}$. The radial function of $\Gamma_{-2}K$, usually
called the LYZ ellipsoid, is defined by

$\displaystyle\rho_{\Gamma_{-2}K}(u)^{-2}=\frac{1}{|K|}\int_{\partial
K}|u\cdot n_{K}(x)|^{2}h_{K}(n_{K}(x))^{-1}d\mathcal{H}^{n-1}(x),\quad u\in
S^{n-1},$ (1.3)

where $n_{K}(x)$ denotes the outer unit normal vector of at $x\in\partial K$,
$h_{K}$ is the support function of $K$ and $\mathcal{H}^{n-1}$ is the
$(n-1)$-dimensional Hausdorff measure. Another known reverse affine
isoperimetric inequality is the following LYZ polar inequality [38].

LYZ polar inequality. Let $K\in\mathcal{K}_{o}^{n}$, then

$\displaystyle|K^{\circ}||\Gamma_{-2}K|\geq\frac{(n+1)^{\frac{n+1}{2}}\omega_{n}}{n!n^{\frac{n}{2}}}$
(1.4)

with equality if and only if $K$ is a simplex whose centroid is at the origin.

Recently, Fang and Zhou [20] introduced the LYZ ellipsoid of log-concave
functions by solving an extremum problem. A function
$f:\mathbb{R}^{n}\to\mathbb{R}$ is called a log-concave function if $\log f$
is concave. Log-concave functions would be considered as extension of convex
bodies. Usually, the results of convex bodies can be covered by taking log-
concave functions, such as the Gaussian function or the characteristic
function of convex bodies. The researches on log-concave functions originated
from Ball’s Phd thesis [7], in which the Blaschke-Santaló inequality of even
log-concave function was established. In recent years, some considerable
progress has been made in establishing the geometric inequalities of log-
concave function. Colesanti and Fragalà[17] established the Minkowski
inequality of log-concave function via the Prékopa-Leindler inequality. Fang,
Xing and Ye[19] further gave $L_{p}$ Minkowski inequalities and solved the
existence of the even $L_{p}$ Minkowski problem of log-concave functions for
$p>1$. John and Löwner ellipsoids for log-concave functions and their related
inequalities were given in [3, 32]. Some affine isoperimetric inequalities of
the affine surface areas of log-concave functions were obtained in [12, 13].
Other inequalities, such as Pólya-Szegö inequality, Rogers-Shephard inequality
and its reverse and some stability results for log-concave functional
inequalites can be found in [1, 2, 3, 4, 5, 6, 10, 11, 16, 22, 23, 29, 30, 31,
32, 33].

In this paper, for a convex function
$\varphi:\mathbb{R}^{n}\to\mathbb{R}\cup\\{\infty\\}$, we use $\varphi^{\ast}$
to denote the Legendre transform of $\varphi$, and
$f^{\circ}=e^{-\varphi^{\ast}}$ to denote the dual of a log-concave function
$f$. The integral of $f$ on $\mathbb{R}^{n}$ is denoted by $J(f)$ and the
domain of $\varphi$ is denoted by ${\rm
dom}(\varphi)=\\{x:\varphi(x)<\infty\\}$. For any function
$f:\mathbb{R}^{n}\to\mathbb{R}$ and $t\in\mathbb{R}$, the upper level set of
$f$ is defined by

$K_{t}(f)=\\{x\in\mathbb{R}^{n}:f(x)\geq t\\}.$

Let ${\rm LC}_{n}$ denote the class of all upper semi-continuous log-concave
functions $f=e^{-\varphi}$ with ${\rm dom}(\varphi)=\mathbb{R}^{n}$ and
$0<J(f)<\infty$.

Let $f\in\ {\rm LC}_{n}$, for the LYZ ellipsoid $\Gamma_{-2}f$ of the log-
concave function $f$, defined in Definition 4.1, we have the following affine
isoperimetric inequality, the functional LYZ polar inequality of (1.4).

Theorem. Let $f=e^{-\varphi}\in{\rm LC}_{n}$ and $\mu_{f}$ be the surface area
measure of $f$. Suppose that $\varphi^{\ast}(x)>0$ on
$\mathbb{R}^{n}\backslash\\{o\\}$ and $-\log
t=\sup\\{\varphi^{\ast}(x/\varphi^{\ast}(x)):x\in{\rm supp}(\mu_{f})\\}$, then

$\displaystyle|K_{t}(f^{\circ})|J(\Gamma_{-2}f)\geq\frac{8^{\frac{n}{2}}(n+1)^{\tfrac{n+1}{2}}\Gamma(\frac{n}{2}+1)\omega_{n}}{n!n^{n}}.$
(1.5)

Moreover, if $\varphi(x)>\varphi(o)$ for all
$x\in\mathbb{R}^{n}\backslash\\{o\\}$, then equality holds if and only if
$f(x)=e^{-{\|x\|_{K}}+1}$ and $K$ is a simplex whose centroid is at the
origin.

Note that the inequality (1.5) is affine invariant, that is, the inequality
(1.5) does not change under compositions of $f$ with affine transformations of
$\mathbb{R}^{n}$. Inequality (1.5) includes the LYZ polar inequality as a
special case. In particular, when $f(x)=e^{-{\|x\|_{K}}+1}$ and $K$ is a
convex body contains origin in its interior, the inequality (1.5) becomes the
LYZ polar inequality (1.4). A detailed derivation can be found in Section 5.
The supremum $-\log t$ may achieve infinity and in this case the inequality
(1.5) is strict. However, $-\log t$ is often bounded and some examples of the
function $f$ will be given in Section 5.

This paper is organized as follows. In Section 2, we provide some basics of
convex bodies and log-concave functions. The Ball-Barth inequality for
isotropic embeddings, which is a key tool to establish our main theorem, is
introduced in Section 3. In Section 4, the definition of the LYZ ellipsoid of
the log-concave function $\Gamma_{-2}f$ is given, and its properties is
listed. In Section 5, we give the proof of the main theorem and show that
LYZ’s polar inequality is a direct result of inequality (1.5) ((Theorem 5.1).

##  2\. Preliminaries and notations

In this section, we list some preliminaries and notations about convex bodies
and log-concave functions. Good general references for the theory of convex
bodies and log-concave functions are provided by the books of Gardner [24],
Schneider[45] and Rockfellar [43].

###  2.1. Basics regarding convex bodies

Let $e_{1},\cdots,e_{n}$ denote the standard Euclidean basis in the
$n-$dimensional Euclidean space $\mathbb{R}^{n}$. For $x,y\in\mathbb{R}^{n}$,
$x\cdot y$ stands for the inner product of $x,y$, and $|x|$ for the Euclidean
norm of $x$. Let $K$ be a convex body (non-empty compact convex set of
$\mathbb{R}^{n}$) contains the origin in its interior. The Minkowski function
of $K$ is defined by

$\displaystyle\|x\|_{K}=\inf\\{\lambda\geq 0:x\in\lambda K\\}.$ (2.1)

Clearly, if $K$ is the unit ball $B^{n}$ in $\mathbb{R}^{n}$, the Minkowski
function $\|\cdot\|_{K}$ becomes the Euclidean norm.

For convex body $K$, its support function $h_{K}:\mathbb{R}^{n}\to\mathbb{R}$
is defined by

$\displaystyle h_{K}(x)=\max\\{x\cdot y:y\in K\\},$

and its radial function
$\rho_{K}:\mathbb{R}^{n}\backslash\\{o\\}\to\mathbb{R}$ is defined by

$\displaystyle\rho_{K}(x)=\max\\{\lambda\geq 0:\lambda x\in K\\}.$

The support function and Minkowski function are homogeneous of degree $1$
while the radial function is homogeneous of degree $-1$. From the definitions
of the support, Minkowski and radial functions and the definition of the polar
body, one can see that

$\displaystyle\|x\|_{K}=h_{K^{\circ}}(x)=\frac{1}{\rho_{K}(x)},$ (2.2)

for any $x\neq o$. Let $GL(n)$ denote the general linear group of
$\mathbb{R}^{n}$. For $T\in GL(n)$,

$\displaystyle(TK)^{\circ}=T^{-t}K^{\circ},$ (2.3)

where $T^{-t}$ is the inverse of transpose of $T$.

###  2.2. Functional setting

Let $\varphi:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$. If for
every $x,y\in\mathbb{R}^{n}$ and $\lambda\in[0,1]$,

$\varphi((1-\lambda)x+\lambda y)\leq(1-\lambda)\varphi(x)+\lambda\varphi(y),$

then $\varphi$ is a convex function. Let
$\text{dom}(\varphi)=\\{x\in\mathbb{R}^{n}:\varphi(x)<\infty\\}.$ Clearly,
$\text{dom}(\varphi)$ is a convex set for any convex function $\varphi$. The
_Legendre trasformation_ of $\varphi$ is the convex function defined by

$\displaystyle\varphi^{*}(y)=\sup_{x\in\mathbb{R}^{n}}\left\\{x\cdot
y-\varphi(x)\right\\}\quad\quad\forall y\in\mathbb{R}^{n}.$ (2.4)

Clearly, $\varphi(x)+\varphi^{*}(y)\geq x\cdot y$ for all
$x,y\in\mathbb{R}^{n}$, there is an equality if and only if
$x\in\text{dom}(\varphi)$ and $y$ is the gradient of $\varphi$ at $x$. Hence,

$\varphi^{*}(\nabla\varphi(x))+\varphi(x)=x\cdot\nabla\varphi(x).$

The convex function
$\varphi:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ is _lower semi-
continuous_ , if the subset $\\{x\in\mathbb{R}^{n}:\varphi(x)\leq t\\}$ is a
closed set for any $t\in(-\infty,+\infty]$. If $\varphi$ is a lower semi-
continuous convex function, then also $\varphi^{*}$ is a lower semi-continuous
convex function, and $\varphi^{**}=\varphi$.

Let $f$ be a log-concave function. The total mass of $f$ is defined by

$\displaystyle J(f)=\int_{\mathbb{R}^{n}}f(x)dx.$

For $f=e^{-\varphi},g=e^{-\psi}$ and $\alpha,\beta>0$, the Asplund sum of $f$
and $g$ is defined by

$\displaystyle\alpha\cdot f\oplus\beta\cdot
g=e^{-(\alpha\varphi^{\ast}+\beta\psi^{\ast})^{\ast}}.$

Let $f=e^{-\varphi},g=e^{-\psi}\in{\rm LC}_{n}$. Rotem [44] gives the
following first variational formula

$\displaystyle\delta J(f,g)=\lim_{t\rightarrow 0^{+}}\frac{J(f\oplus t\cdot
g)-J(f)}{t}=\int_{\mathbb{R}^{n}}\psi^{\ast}(x)d\mu_{f}(x),$ (2.5)

where $\mu_{f}$ is the surface area measure of $f$, which is defined by

$\mu_{f}(A)=\int_{\nabla\varphi(x)\in A}f(x)dx$

for any $A\subset\mathbb{R}^{n}$.

##  3\. The Ball-Barth inequality

Let $\mu$ be a positive Borel measure $\mu$ on $S^{n}$. A positive semi-
definite $n\times n$ matrix $[\mu]$ can be generated by $\mu$, which is
defined by

$\displaystyle[\mu]=\int_{S^{n}}u\otimes ud\mu(u)$ (3.1)

where $u\otimes u$ is the rank $1$ matrix generated by $u\in S^{n}$, or,
equivalently by

$\displaystyle v\cdot[\mu]v=\int_{S^{n}}|v\cdot u|^{2}d\mu(u)$ (3.2)

for all $v\in S^{n}$. A positive Borel measure $\mu$ is said to be an
isotropic measure if

$\displaystyle[\mu]=I_{n},$ (3.3)

where $I_{n}$ is the identity matrix. If $\mu$ is isotropic, summing the
equation (3.2) with $v=e_{i},i=1,\cdots,n+1$, one has

$\displaystyle\mu(S^{n})=n+1.$ (3.4)

The following Ball-Barth inequality was established in [36]. The Ball-Barth
inequality. _Let $\mu$ an isotropic measure on $S^{n}$ and
$l:S^{n}\rightarrow(0,\infty)$ be a continuous function. Then_

$\displaystyle\det\int_{S^{n}}l(u)u\otimes
ud\mu(u)\geq\exp\left\\{\int_{S^{n}}\log l(u)d\mu(u)\right\\},$ (3.5)

_with equality if and only if $l(u_{1})\cdots l(u_{n+1})$ is a constant for
linearly independent $u_{1},\cdots,u_{n+1}$ in ${\rm supp}(\nu)$._

The isotropic embedding from $\mathbb{R}^{n}$ to $S^{n}$ is given as follows.

######  Definition 3.1.

Let $(\mathbb{R}^{n},\mu)$ be a Borel measure space. A continuous function
$h:\mathbb{R}^{n}\rightarrow S^{n}$ is an isotropic embedding of
$(\mathbb{R}^{n},\mu)$ into $S^{n}$ if

$\displaystyle\int_{\mathbb{R}^{n}}|u\cdot h(x)|^{2}d\mu(x)=1$ (3.6)

for all $u\in S^{n}$.

Let $\mu$ and $h$ have been given in Definition 3.1. Summing the equation
(3.6) with $u=e_{1},\cdots,e_{n},e_{n+1}$, one has

$\displaystyle\mu(\mathbb{R}^{n})=n+1.$ (3.7)

Define a new Borel measure $\nu$ on $S^{n}$, which is the push-forward of
$\mu$ by $h$, by

$\displaystyle\int_{S^{n}}g(u)d\nu(u)=\int_{\mathbb{R}^{n}}g(h(x))d\mu(x)$

for any Borel function $g:S^{n}\to\mathbb{R}$. Definition 3.1 shows that $\nu$
is an isotropic measure on $S^{n}$, by applying the Ball-Barthe inequality to
$\nu$, we can obtain the following Ball-Barthe inequality for isotropic
embeddings.

######  Proposition 3.1.

If $h:\mathbb{R}^{n}\rightarrow S^{n}$ is an isotropic embedding of the Borel
measure space $(\mathbb{R}^{n},\mu)$ into $S^{n}$, then for each continuous
$l:\mathbb{R}^{n}\rightarrow(0,\infty)$

$\displaystyle\det\int_{\mathbb{R}^{n}}l(x)h(x)\otimes
h(x)d\mu(x)\geq\exp\left\\{\int_{\mathbb{R}^{n}}\log l(x)d\mu(x)\right\\},$
(3.8)

with equality if and only if $l(x_{1}),\cdots,l(x_{n+1})$ is constants for
$x_{1},\cdots,x_{n+1}$ in ${\rm supp}(\nu)$ such that
$h(x_{1}),\cdots,h(x_{n+1})$ are linearly independent.

##  4\. The LYZ ellipsoid of log-concave functions

In this section, we will define the LYZ ellipsoid of log-concave functions,
and introduce its properties.

######  Definition 4.1.

Let $f=e^{-\varphi}\in{\rm LC}_{n}$. Suppose that $\varphi^{\ast}(x)>0$ for
any $x\in{\mathbb{R}^{n}\backslash\\{o\\}}$, then the _LYZ ellipsoid_ of $f$,
denoted by $\Gamma_{-2}f$, is defined as

$\displaystyle-\log\Gamma_{-2}f(x)$ $\displaystyle=\frac{n^{2}}{8\delta
J(f,f)}\int_{\mathbb{R}^{n}}|x\cdot y|^{2}\varphi^{*}(y)^{-1}d\mu_{f}(y)$
$\displaystyle=\frac{n^{2}}{8\delta
J(f,f)}\int_{\mathbb{R}^{n}}|x\cdot\nabla\varphi(y)|^{2}\varphi^{*}(\nabla\varphi(y))^{-1}e^{-\varphi(y)}dy.$
(4.1)

Note that the function $\sqrt{-\log\Gamma_{-2}f(x)}$ is a Minkowski functional
of an ellipsoid. Indeed, if $E$ is an ellipsoid generated by $n\times n$ real
symmetric matrix $A$, that is

$\displaystyle E=\\{x\in\mathbb{R}^{n}:x\cdot Ax\leq 1\\},$

one can check that the Minkowski function of $E$ is given by
$\|x\|_{E}^{2}=x\cdot Ax.$ From this, we see that the function
$\sqrt{-\log\Gamma_{-2}f(x)}$ is a Minkowski functional of an ellipsoid, which
is generated by $n\times n$ real symmetric matrix $M=[m_{ij}(f)]$, where

$\displaystyle m_{ij}(f)={\frac{n^{2}}{8\delta
J(f,f)}}\int_{\mathbb{R}^{n}}(e_{i}\cdot y)(e_{j}\cdot
y)\varphi^{*}(y)^{-1}d\mu_{f}(y).$

######  Property 4.1.

Let $f\in{\rm LC}_{n}$ and $T\in{\rm GL}(n)$. Then

$\displaystyle\Gamma_{-2}(f\circ T)=(\Gamma_{-2}f)\circ T.$ (4.2)

###### Proof.

Let $T^{t}$ be the transpose of $T$. From the definition of Legendre transform
(2.4) and the fact that $Tx\cdot y=x\cdot T^{t}y$, one has

$\displaystyle(\varphi\circ T)^{*}(y)$ $\displaystyle=$
$\displaystyle\sup_{x\in\mathbb{R}^{n}}\left\\{x\cdot y-\varphi(Tx)\right\\}$
$\displaystyle=$ $\displaystyle\sup_{x\in\mathbb{R}^{n}}\left\\{x\cdot
T^{-t}y-\varphi(x)\right\\}$ $\displaystyle=$
$\displaystyle\varphi^{*}(T^{-t}y).$ (4.3)

By (4.1) and the fact that $\nabla(\varphi\circ T)(y)=T^{t}\nabla\varphi(Ty)$,
one has

$\displaystyle\int_{\mathbb{R}^{n}}|x\cdot y|^{2}(\varphi\circ
T)^{*}(y)^{-1}d\mu_{f\circ T}(y)$
$\displaystyle\quad\quad=\int_{\mathbb{R}^{n}}|x\cdot\nabla(\varphi\circ
T)(y)|^{2}(\varphi\circ T)^{*}(\nabla(\varphi\circ
T)(y))^{-1}e^{-\varphi(Ty)}dy$
$\displaystyle\quad\quad=\int_{\mathbb{R}^{n}}|x\cdot
T^{t}\nabla\varphi(Ty)|^{2}\varphi^{*}(\nabla\varphi(Ty))^{-1}e^{-\varphi(Ty)}dy$
$\displaystyle\quad\quad=\int_{\mathbb{R}^{n}}|Tx\cdot\nabla\varphi(Ty)|^{2}\varphi^{*}(\nabla\varphi(Ty))^{-1}e^{-\varphi(Ty)}dy$
$\displaystyle\quad\quad=|\det T|^{-1}\int_{\mathbb{R}^{n}}|Tx\cdot
y|^{2}\varphi^{*}(y)^{-1}d\mu_{f}(y).$

This together with $\delta J(f\circ T,f\circ T)=|\det T|^{-1}\delta J(f,f)$
gives (4.2). ∎

The following corollary shows that the quantity
$|K_{t}(f^{\circ})|J(\Gamma_{-2}(f))$ is invariant under the operation of
$GL(n)$.

######  Corollary 4.1.

Let $f\in{\rm LC}_{n}$ and $T\in{\rm GL}(n)$. Then for any $t>0$

$|K_{t}((f\circ T)^{\circ})|J(\Gamma_{-2}(f\circ
T))=|K_{t}(f^{\circ})|J(\Gamma_{-2}(f)).$

###### Proof.

By (4.3), one has for any $T\in GL(n)$,

$\displaystyle K_{t}((f\circ T)^{\circ})$
$\displaystyle=\\{x\in\mathbb{R}^{n}:f^{\circ}(Tx)\geq
t\\}=\\{x\in\mathbb{R}^{n}:e^{-\varphi^{*}{(T^{-t}x)}}\geq t\\}$
$\displaystyle=\\{T^{t}y\in\mathbb{R}^{n}:e^{-\varphi^{*}{(y})}\geq
t\\}=T^{t}(K_{t}(f^{\circ})).$

Property (4.1) implies that

$J(\Gamma_{-2}(f\circ T))=J((\Gamma_{-2}f)\circ
T)=\int_{\mathbb{R}^{n}}(\Gamma_{-2}f)(Tx)dx=(\det T)^{-1}J(\Gamma_{-2}f).$

Hence

$|K_{t}((f\circ T)^{\circ})|J(\Gamma_{-2}(f\circ
T))=|(K_{t}(f^{\circ}))|J(\Gamma_{-2}f).$

This completes the proof. ∎

When $f=e^{-\frac{\|x\|_{K}^{2}}{2}}$ with $K\in\mathcal{K}_{o}^{n}$,
$\Gamma_{-2}f(x)$ can be calculated precisely.

######  Property 4.2.

Let $K\in\mathcal{K}_{o}^{n}$. If $f=e^{-\frac{\|x\|_{K}^{2}}{2}}$, then

$\displaystyle\Gamma_{-2}f(x)=e^{-\frac{\|x\|_{\Gamma_{-2}K}^{2}}{2}}.$

###### Proof.

The normalized cone measure $\sigma_{K}$ of $K$ is defined by

$d\sigma_{K}(z)=\frac{z\cdot
n_{K}(z)}{n|K|}d\mathcal{H}^{n-1}(z)\quad\text{for}\quad z\in\partial K.$

For $y\in\mathbb{R}^{n}$, we write $y=rz$ with $z\in\partial K$, then

$\displaystyle dy=n|K|r^{n-1}drd\sigma_{K}(z).$ (4.4)

Together with the fact that
$\left(\frac{\|x\|_{K}^{2}}{2}\right)^{\ast}=\frac{\|x\|_{K^{\circ}}^{2}}{2}$
for any $x\in\mathbb{R}^{n}$, one has

$\displaystyle\int_{\mathbb{R}^{n}}|x\cdot\nabla\varphi(y)|^{2}\varphi^{*}(\nabla\varphi(y))^{-1}e^{-\varphi(y)}dy$
$\displaystyle\quad\quad=\int_{\mathbb{R}^{n}}|x\cdot\|y\|_{K}\nabla\|y\|_{K}|^{2}\left(\tfrac{\|\|y\|_{K}\nabla\|y\|_{K}\|_{K^{\circ}}^{2}}{2}\right)^{-1}e^{-\tfrac{\|y\|_{K}^{2}}{2}}dy$
$\displaystyle\quad\quad=\int_{\mathbb{R}^{n}}|x\cdot\nabla\|y\|_{K}|^{2}\left(\tfrac{\|\nabla\|y\|_{K}\|_{K^{\circ}}^{2}}{2}\right)^{-1}e^{-\tfrac{\|y\|_{K}^{2}}{2}}dy$
$\displaystyle\quad\quad=2n|K|\int_{0}^{\infty}\int_{\partial
K}r^{n-1}|x\cdot\nabla\|z\|_{K}|^{2}\left(\|\nabla\|z\|_{K}\|_{K^{\circ}}^{2}\right)^{-1}e^{-\frac{r^{2}}{2}}drd\sigma_{K}(z)$
$\displaystyle\quad\quad=2^{\frac{n}{2}}\Gamma\left(\tfrac{n}{2}\right)n|K|\int_{\partial
K}|x\cdot\nabla\|z\|_{K}|^{2}\left(\|\nabla\|z\|_{K}\|_{K^{\circ}}^{2}\right)^{-1}d\sigma_{K}(z).$

By the fact that $\nabla\|z\|_{K}=\frac{n_{K}(z)}{\|n_{K}(z)\|_{K^{\circ}}}$
for $z\in\partial K$ (see [45, Remark 1.7.14]), (2.2) and (1.3), we have

$\displaystyle n|K|\int_{\partial
K}|x\cdot\nabla\|z\|_{K}|^{2}\left(\|\nabla\|z\|_{K}\|_{K^{\circ}}^{2}\right)^{-1}d\sigma_{K}(z)$
$\displaystyle\quad\quad=n|K|\int_{\partial
K}|x\cdot\nabla\|z\|_{K}|^{2}d\sigma_{K}(z)$
$\displaystyle\quad\quad=n|K|\int_{\partial
K}\left|x\cdot\tfrac{n_{K}(z)}{\|n_{K}(z)\|_{K^{\circ}}}\right|^{2}d\sigma_{K}(z)$
$\displaystyle\quad\quad=\int_{\partial K}\left|x\cdot
n_{K}(z)\right|^{2}(z\cdot n_{K}(z))^{-1}d\mathcal{H}^{n-1}(z)$
$\displaystyle\quad\quad=|K|\rho_{\Gamma_{-2}K}(x)^{-2}$
$\displaystyle\quad\quad=|K|\|x\|_{\Gamma_{-2}K}^{2}.$

By using (2.5) and (4.4), we have

$\displaystyle\delta J(f,f)$
$\displaystyle=\frac{1}{2}\int_{\mathbb{R}^{n}}{\|\|y\|_{K}\nabla\|y\|_{K}\|_{K^{\circ}}^{2}}e^{-\tfrac{\|y\|_{K}^{2}}{2}}dy$
$\displaystyle=\frac{n|K|}{2}\int_{0}^{\infty}r^{n+1}e^{-\frac{r^{2}}{2}}dr\int_{\partial
K}\|\nabla\|z\|_{K}\|_{K^{\circ}}^{2}d\sigma_{K}(z)$
$\displaystyle=2^{\frac{n}{2}-2}\Gamma(\tfrac{n}{2})n^{2}|K|.$

Hence

$\displaystyle-\log\Gamma_{-2}f(x)=\frac{\|x\|_{\Gamma_{-2}K}^{2}}{2}.$

This completes the proof. ∎

When $f$ is the Gaussian function $e^{-\frac{|x|^{2}}{2}}$, the following
result can be obtained directly from Property 4.2 and the fact that
$\Gamma_{-2}B^{n}=B^{n}$.

######  Corollary 4.1.

If $f=\gamma_{n}=e^{-\frac{|x|^{2}}{2}}$, then

$\displaystyle\Gamma_{-2}\gamma_{n}=\gamma_{n}.$

##  5\. LYZ polar inequality for log-concave functions

In this section, we will establish the LYZ polar inequality for log-concave
functions. The following lemmas are needed.

######  Lemma 5.1.

Let $f=e^{-\varphi}\in{\rm LC}_{n}$ and
$K_{t}(f)=\\{x\in\mathbb{R}^{n}:f(x)\geq t\\}$ be the upper level set of $f$.
If $\nu$ is a finite, positive Borel measure on $\mathbb{R}^{n}$ and
$\varphi^{\ast}>0$ on $\mathbb{R}^{n}\backslash\\{o\\}$, then

$\displaystyle\int_{\mathbb{R}^{n}}x\varphi^{\ast}(x)^{-1}d\nu(x)\in
rK_{t}(f^{\circ}),$

where $-\log t=\sup\\{\varphi^{\ast}(x/\varphi^{\ast}(x)):{x\in{\rm
supp}(\nu)}\\}$ and $r=|\nu|$.

###### Proof.

Let

$\displaystyle x_{0}=\int_{\mathbb{R}^{n}}x\varphi^{\ast}(x)^{-1}d\nu(x).$

By the convexity of $\varphi^{\ast}$, we have

$\displaystyle\varphi^{\ast}\left(\frac{x_{0}}{r}\right)\leq\int_{\mathbb{R}^{n}}\varphi^{\ast}\left(\frac{x}{\varphi^{\ast}(x)}\right)\frac{d\nu(x)}{r}\leq-\log
t.$

Hence

$\displaystyle x_{0}\in r\left\\{x\in\mathbb{R}^{n}:f^{\circ}(x)\geq
t\right\\}.$

Then $x_{0}\in rK_{t}(f)$. ∎

Note that $-\log t$ may reach infinity, but it is often bounded for many log-
concave functions, such as $f=e^{-\varphi}\in{\rm LC}_{n}$ with
$\varphi^{\ast}(x)>0$ for $x\in\mathbb{R}^{n}\backslash\\{o\\}$ and ${\rm
dom}(\varphi^{\ast})=\mathbb{R}^{n}$. We first need to show
$J(f^{\circ})<\infty$. For the Legendre transformation of $\varphi$,

$\varphi^{\ast}(y)=\sup_{x\in\mathbb{R}^{n}}x\cdot
y-\varphi(x)\geq|y|-\varphi\left(\frac{y}{|y|}\right),$

we have $\varphi^{\ast}(y)\to\infty$ as $|y|\to\infty$ and hence
$J(f^{\circ})<\infty$. Moreover, $f^{\circ}\in{\rm LC}_{n}$. [17, Lemma 2.5]
shows that there exists two constants $a>0$ and $b$ such that

$\displaystyle\varphi^{\ast}(x)\geq a|x|+b$ (5.1)

for any $x\in\mathbb{R}^{n}$. Hence

$\lim_{|x|\to\infty}\frac{|x|}{\varphi^{\ast}(x)}\leq\lim_{|x|\to\infty}\frac{|x|}{a|x|+b}=\frac{1}{a},$

that is, for any $\varepsilon>0$, there exists a constant $M$ such that
${x}/{\varphi^{\ast}(x)}\in\left(\frac{1}{a}+\varepsilon\right)B^{n}$ for
$x\in(MB^{n})^{c}$ and the unit ball $B^{n}$ in $\mathbb{R}^{n}$. Then
$\varphi^{\ast}\left({x}/{\varphi^{\ast}(x)}\right)$ is bounded for any
$x\in(M_{0}B^{n})^{c}$ as the continuity of $\varphi^{\ast}$. Set
$m=\min\\{\varphi^{\ast}(x):x\in MB^{n}\\}$, then
$x/\varphi^{\ast}(x)\in(M/m)B^{n}$ for any $x\in MB^{n}$. Hence
$\varphi^{\ast}\left({x}/{\varphi^{\ast}(x)}\right)$ is bounded on $MB^{n}$
and $-\log t$ is a finite constant.

The following lemma is the key for the equality condition of our main
inequality.

######  Lemma 5.2.

Let $\varphi$ be convex function and dom$(\varphi)=\mathbb{R}^{n}$ and
$\varphi(x)>\varphi(o)$ for all $x\in\mathbb{R}^{n}\backslash\\{o\\}$. Suppose
that $\varphi$ satisfies the equality

$\displaystyle\varphi(x)=x\cdot\nabla\varphi(x)-c$ (5.2)

a.e. on $\mathbb{R}^{n}$, where $c$ is a constant. Then
$\varphi(x)=\|x\|_{K}-c$ for some $K\in\mathcal{K}_{o}^{n}$.

###### Proof.

Firstly, we consider the case when $c=0$, in which case we claim that

$\displaystyle\varphi(\lambda x)=\lambda\varphi(x)$ (5.3)

for all $x\in\mathbb{R}^{n}$ and any $\lambda\geq 0$. Since
$|\nabla\varphi(x)|$ is finite near the origin, hence
$|\varphi(o)|=\lim_{|x|\to 0}|x\cdot\nabla\varphi(x)|=0$, which implies that
(5.3) holds for $\lambda=0$. Otherwise, if $\varphi(\lambda
x)\neq\lambda\varphi(x)$ for some points $x\neq o$, where $\varphi$ is
differentiable, then by the convexity of $\varphi(x)$, one has

$\displaystyle\varphi(y)-\varphi(x)\geq\nabla\varphi(x)\cdot(y-x)$

for all $y\in\mathbb{R}^{n}$. Note that the above inequality is strictly when
$y=o$ due to the assumptions $\varphi(x)>\varphi(o)$ and $\varphi(\lambda
x)\neq\lambda\varphi(x)$. Then by taking $y=o$, we have
$\varphi(x)<\nabla\varphi(x)\cdot x$, which contradicts with (5.2). Thus
$\varphi(\lambda x)=\lambda\varphi(x)$ at every differentiable points.
Moreover, this also holds on whole $\mathbb{R}^{n}$ as $\varphi$ is continuous
and $\varphi$ is differentiable almost everywhere. Combined with the convexity
of $\varphi$, one sees $\varphi$ is sublinear. By the fact that every
sublinear function from $\mathbb{R}^{n}$ to $\mathbb{R}$ is a support function
of a convex body, we have $\varphi(x)=h_{K}(x)$ for some convex body $K$.
Since $\varphi(x)>\varphi(o)=0$ for all $o\neq x\in\mathbb{R}^{n}$, one can
see that $K$ is a convex body that contains the origin in its interior. In
this case, one has $\varphi(x)=h_{K}(x)=\|x\|_{K^{\circ}}$ and (5.2) follows.
∎

For a real function $\tau:(0,\ +\infty)\rightarrow R$, let

$\displaystyle\int_{0}^{\tau(t)}e^{-l}dl=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{t}e^{-l^{2}}dl.$
(5.4)

The transformation $T:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n+1}$ is defined
by

$\displaystyle Ty=\int_{\mathbb{R}^{n}}h(x)\frac{\tau(y\cdot
h(x))}{h_{i}(x)}d\nu(x),$

for $y\in\mathbb{R}^{n+1}$, with $h_{i}=h\cdot e_{i}$ for $i=1,\cdots,n+1$.

By the Ball-Barth inequality, we obtain the following lemma.

######  Lemma 5.3.

Let $(\mathbb{R}^{n},\nu)$ be a Borel measure space and $h:\mathbb{R}^{n}\to
S^{n}$ be an isotropic embedding of the Borel measure space
$(\mathbb{R}^{n},\nu)$ into $S^{n}$. Then

$\displaystyle(n+1)^{\frac{n+1}{2}}\leq\int_{T(\mathbb{R}^{n+1})}e^{-x_{i}}dx,$
(5.5)

with equality if and only if $h_{i}$ is constant on ${\rm supp}(\nu)$, and
there exists a $C>0$ with respect to $y$ such that

$\prod_{j=1}^{n+1}{\tau^{\prime}(y\cdot h(x_{i}))}=C$

for $x_{1},\cdots,x_{n+1}\in{\rm supp}(\nu)$ with $h(x_{1}),...,h(x_{n+1})$
linearly independent.

###### Proof.

Deriving both sides of (5.4) with respect to $t$, we have

$\displaystyle-\tau(t)+\log\tau^{\prime}(t)=-\log\sqrt{\pi}-t^{2}.$

Taking $t=y\cdot h(x)$, one has

$\displaystyle-|y\cdot h(x)|^{2}=\log\sqrt{\pi}-\tau(y\cdot
h(x))+\log\frac{\tau^{\prime}(y\cdot h(x))}{h_{i}(x)}+\log h_{i}(x).$ (5.6)

Now we integrate (5.6) over all $x\in\mathbb{R}^{n}$ with respect to the
measure $d\nu$. From the definition of isotropic embedding (3.6), the integral
of the left hand side on (5.6) is equal to

$\displaystyle-\int_{\mathbb{R}^{n}}|y\cdot h(x)|^{2}|d\nu(x)=-|y|^{2}.$ (5.7)

Next we deal with the integral of the last term on the right hand side of
(5.6). From (3.7), one can see that the measure $\frac{1}{n+1}d\nu$ is a
probability measure. By the fact that the $L_{0}$-mean of a function is
dominated by its $L_{2}$-mean on a probability space and (3.6), one has

$\displaystyle\exp\left(\frac{1}{n+1}\int_{\mathbb{R}^{n}}\log
h_{i}(x)d\nu(x)\right)\leq\left(\frac{1}{n+1}\int_{\mathbb{R}^{n}}|h_{i}(x)|^{2}d\nu(x)\right)^{\frac{1}{2}}=\left(\frac{1}{n+1}\right)^{\frac{1}{2}},$
(5.8)

with equality if and only if $h_{i}(x)$ is constant on ${\rm supp}(\nu)$.
Hence

$\displaystyle\int_{\mathbb{R}^{n}}\log
h_{i}(x)d\nu(x)\leq\log\left(\frac{1}{n+1}\right)^{\frac{n+1}{2}}.$

Calculating the derivative of $Ty$ with respect to $y$, we have

$\displaystyle dTy=\int_{\mathbb{R}^{n}}h(x)\otimes
h(x)\frac{\tau^{\prime}(y\cdot h(x))}{h_{i}(x)}d\nu(x).$

From Proposition 3.1, we infer that

$\displaystyle\det(dTy)$ $\displaystyle\geq$
$\displaystyle\exp\left\\{\int_{\mathbb{R}^{n}}\log\frac{\tau^{\prime}(y\cdot
h(x))}{h_{i}(x)}d\nu(x)\right\\},$ (5.9)

with equality if and only if

$\displaystyle\prod_{j=1}^{n+1}\frac{\tau^{\prime}(y\cdot
h(x_{j}))}{h_{i}(x_{j})}$

is constant for $x_{1},\cdots,x_{n+1}\in{\rm supp}(\mu)$ such that
$h(x_{1}),\cdots,h(x_{n+1})$ are linearly independent.

Combining (5.6), (5.7), (5.8), (5.9) and the fact that
$\nu(\mathbb{R}^{n})=n+1$, we have

$\displaystyle\exp\\{-|y|^{2}\\}\leq\left(\frac{\pi}{n+1}\right)^{\frac{n+1}{2}}\det(d(Ty))\exp\\{-e_{i}\cdot
Ty\\}.$

Integrating this inequality over all $y\in\mathbb{R}^{n+1}$ gives the desired
inequality (5.5).∎

Now we are ready to prove our functional isoperimetric inequality for log-
concave functions.

######  Theorem 5.1.

Let $f=e^{-\varphi}\in{\rm LC}_{n}$ and $\mu_{f}$ be the surface area measure
of $f$. Suppose that $\varphi^{\ast}(x)>0$ on
$\mathbb{R}^{n}\backslash\\{o\\}$ and $-\log
t=\sup\\{\varphi^{\ast}(x/\varphi^{\ast}(x)):x\in{\rm supp}(\mu_{f})\\}$, then

$\displaystyle|K_{t}(f^{\circ})|J(\Gamma_{-2}f)\geq\frac{8^{\frac{n}{2}}(n+1)^{\tfrac{n+1}{2}}\Gamma(\frac{n}{2}+1)\omega_{n}}{n!n^{n}}.$
(5.10)

Moreover, if $\varphi(x)>\varphi(o)$ for all
$x\in\mathbb{R}^{n}\backslash\\{o\\}$, then equality holds if and only if
$f(x)=e^{-{\|x\|_{K}}+1}$ and $K$ is a simplex whose centroid is at the
origin.

###### Proof.

From Property 4.1, (2.3) and the fact that $\sqrt{-\log\Gamma_{-2}f(x)}$ is a
Minkowski function of an ellipsoid, we only need to prove (5.10) holds for
$\Gamma_{-2}f(x)=e^{-{|x|^{2}}/{2}}$. This means that the measure

$\nu(\cdot)=\frac{n^{2}}{4\delta J(f,f)}(\varphi^{*})^{-1}\mu_{f}(\cdot)$

is isotropic. Let $c_{n}={2/n}$. For any $x\in\mathbb{R}^{n}$, define a
function $h:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n+1}$ by
$h(x)=(x,c_{n}\varphi^{\ast}(x)),$ and $\bar{h}:\mathbb{R}^{n}\rightarrow
S^{n}$ by $\bar{h}={h}/{|h|}.$ Assume that $y=(z,s)\in\mathbb{R}^{n+1}$. By
the fact that the barycenter of $\mu_{f}$ is at the origin, we have

$\displaystyle\int_{\mathbb{R}^{n}}|y\cdot\bar{h}(x)|^{2}|h(x)|^{2}d\nu(x)$
$\displaystyle\quad\quad=\int_{\mathbb{R}^{n}}|(z,s)\cdot(x,c_{n}\varphi^{\ast}(x))|^{2}d\nu(x)$
$\displaystyle\quad\quad=\int_{\mathbb{R}^{n}}|z\cdot
x|^{2}d\nu(x)+2sc_{n}z\cdot\int_{\mathbb{R}^{n}}xd\mu_{f}(x)+\frac{4s^{2}}{n^{2}}\int_{\mathbb{R}^{n}}(\varphi^{\ast})^{2}d\nu(x)$
$\displaystyle\quad\quad=|z|^{2}+s^{2}$ $\displaystyle\quad\quad=|y|^{2}.$

Hence $\bar{h}:\mathbb{R}^{n}\rightarrow S^{n}$ is an isotropic embedding of
the Borel measure space $(\mathbb{R}^{n},|h|^{2}d\nu)$ into $S^{n}$. From
Lemma 5.3, one can see that

$\displaystyle(n+1)^{\frac{n+1}{2}}\leq\int_{\mathbb{R}^{n+1}}e^{-e_{n+1}\cdot
Ty}|dTy|dy=\int_{T(\mathbb{R}^{n+1})}e^{-e_{n+1}\cdot z}dz.$ (5.11)

Here $Ty$ is given by

$\displaystyle Ty$
$\displaystyle=\int_{\mathbb{R}^{n}}\bar{h}(x)\frac{\tau(y\cdot\bar{h}(x))}{e_{n+1}\cdot\bar{h}(x)}|h(x)|^{2}d\nu(x)$
$\displaystyle=\int_{\mathbb{R}^{n}}h(x)\frac{\tau(y\cdot\bar{h}(x))}{e_{n+1}\cdot
h(x)}|h(x)|^{2}d\nu(x)$
$\displaystyle=\int_{\mathbb{R}^{n}}((c_{n}\varphi^{\ast}(x))^{-1}x,1)\tau(y\cdot\bar{h}(x))|h(x)|^{2}d\nu(x).$

Lemma 5.1 implies that

$\displaystyle
Ty\in\bigcup_{r>0}c_{n}^{-1}rK_{{t}}(f^{\circ})\times\\{r\\}=:D.$

Therefore,

$\displaystyle\int_{T(\mathbb{R}^{n+1})}e^{-e_{n+1}\cdot z}dz$
$\displaystyle\leq\int_{D}e^{-e_{n+1}\cdot
z}dz=\int_{0}^{\infty}\int_{c_{n}^{-1}rK_{t}}e^{-r}dxdr$
$\displaystyle=\frac{n^{n}}{2^{n}}\int_{0}^{\infty}r^{n}e^{-r}dr|K_{t}(f^{\circ})|=\frac{n^{n}}{2^{n}}\Gamma(n+1)|K_{t}(f^{\circ})|.$

This together with the fact that
$J(e^{-\frac{|x|^{2}}{2}})=2^{\frac{n}{2}}\Gamma(\frac{n}{2}+1)\omega_{n}$
gives (5.10).

From the equality condition of inequality (5.5), we have that $\varphi^{\ast}$
is a positive constant on ${\rm supp}(\tau(y\cdot\overline{h})|h|^{2}\nu)$,
that is, $\varphi^{\ast}=c_{0}>0$ on ${\rm supp}(\mu_{f})$ due to $\tau>0$ on
$\mathbb{R}$ and $|h(x)|^{2}>0$ on $\mathbb{R}^{n+1}$. Since ${\rm
dom}(\varphi)=\mathbb{R}^{n}$, one can see that

$\displaystyle\varphi^{\ast}(\nabla\varphi(x))=x\cdot\nabla\varphi(x)-\varphi(x)=c_{0}$
(5.12)

almost everywhere on $\mathbb{R}^{n}$. If $\varphi(x)>\varphi(o)$ for any
$x\in\mathbb{R}^{n}\backslash\\{o\\}$, Lemma 5.2 implies that
$\varphi(x)=\|x\|_{K}-c_{0}$ for some convex body $K\in\mathcal{K}^{n}_{o}$.
In this case, by (4.4), one can calculate that

$\displaystyle\delta J(f,f)=c_{0}e^{c_{0}}n\Gamma(n)|K|$ (5.13)

and

$\displaystyle\int_{\mathbb{R}^{n}}|x\cdot\nabla\varphi(y)|^{2}\varphi^{*}(\nabla\varphi(y))^{-1}e^{-\varphi(y)}dy$
$\displaystyle\quad\quad=c_{0}^{-1}e^{c_{0}}\int_{\mathbb{R}^{n}}|x\cdot\nabla\|y\|_{K}|^{2}e^{-\|y\|_{K}}dy$
$\displaystyle\quad\quad=c_{0}^{-1}e^{c_{0}}n|K|\int_{0}^{\infty}\int_{\partial
K}|x\cdot\nabla\|z\|_{K}|^{2}r^{n-1}e^{-r}drd\sigma_{K}(z)$
$\displaystyle\quad\quad=c_{0}^{-1}e^{c_{0}}\Gamma(n)n|K|\int_{\partial
K}|x\cdot\nabla\|z\|_{K}|^{2}drd\sigma_{K}(z)$
$\displaystyle\quad\quad=c_{0}^{-1}e^{c_{0}}\Gamma(n)|K|\|x\|_{\Gamma_{-2}K}^{2}.$

This together with the definition of $-\log\Gamma_{-2}(f)$ and (5.13) shows
that

$\displaystyle-\log\Gamma_{-2}(e^{-\|x\|_{K}+c_{0}})=\frac{n}{8c_{0}^{2}}\|x\|_{\Gamma_{-2}K}^{2}.$

For any $c>0$ and convex body $K$,

$\displaystyle J(e^{-c\|x\|_{K}^{2}})$
$\displaystyle=\int_{\mathbb{R}^{n}}e^{-c\|x\|_{K}^{2}}dx$
$\displaystyle=n|K|\int_{0}^{\infty}\int_{\partial
K}r^{n-1}e^{-cr^{2}}drd\sigma_{K}(x)$
$\displaystyle=c^{-\tfrac{n}{2}}\Gamma(\tfrac{n}{2}+1)|K|.$

Hence

$\displaystyle
J(\Gamma_{-2}(e^{-\|x\|_{K}+c}))=\frac{8^{\frac{n}{2}}c_{0}^{n}}{n^{\frac{n}{2}}}\Gamma(\tfrac{n}{2}+1)|\Gamma_{-2}K|.$

By the definition of Legendre transform and $\varphi(x)=\|x\|_{K}-c_{0}$, we
have

$\varphi^{\ast}(x)=\left\\{\begin{aligned} &\ c_{0},&x\in K^{\circ};\\\
&+\infty,&x\notin K^{\circ}.\end{aligned}\right.$

For any $x\in{\rm supp}(\mu_{f})\subset K^{\circ}$ and $c_{0}\geq 1$, we
obtain

$-\log
t=\sup\left\\{\varphi^{\ast}\left(\frac{x}{\varphi^{\ast}(x)}\right):x\in{\rm
supp}{(\mu_{f})}\right\\}=c_{0}.$

Hence

$\displaystyle K_{t}(f^{\circ})=\\{x\in\mathbb{R}^{n}:\varphi^{\ast}(x)\leq
c_{0}\\}=K^{\circ}$

and the inequality (5.10) becomes

$\displaystyle|K^{\circ}||\Gamma_{-2}K|\geq\frac{(n+1)^{\tfrac{n+1}{2}}\omega_{n}}{c_{0}^{n}n!n^{\frac{n}{2}}}.$
(5.14)

Let $conv(A)$ denote the convex hull of $A\subset\mathbb{R}^{n}$ and
$\overline{A}$ denote the closure of $A$. For the surface area $\mu_{f}$ of
the log-concave function, [18, (16)] says that

${\rm conv}({\rm supp}(\mu_{f}))=\overline{\\{\varphi^{\ast}<+\infty\\}},$

therefore ${\rm conv}({\rm supp}(\mu_{f}))=K^{\circ}$.

For $x\in\partial({\rm supp}(\mu_{f}))\subset\partial(K^{\circ})$ and
$c_{0}<1$, we have $x/c_{0}\notin K^{\circ}$ and $-\log t=+\infty$. Therefore
$|K_{t}(f^{\circ})|=+\infty$ and the inequality (5.10) is strict. Hence the
best constant of (5.14) is $c_{0}=1$ and the inequality (5.14) becomes the LYZ
polar inequality (1.4). Therefore the equality of (5.14) holds if and only if
$c_{0}=1$ and $K$ is a simplex with centroid at the origin. ∎

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