Some Remarks on Petty Projection of Log-concave Functions

Abstract

In this note we study the Petty projection of a log-concave function, which has been recently introduced in [9]. The aim of this note is to report a mistake in Theorem 5.2 of [9] and to give correct new inequalities involving this new notion.

1 Introduction

Let $K\subset \mathbb R^n$ be a convex body, i.e., a convex and compact set with non-empty interior, whose boundary is denoted by $\partial K$. Moreover, let ${\mathcal {K}}^n$ be the set of all convex bodies in $\mathbb R^n$. For these and most of the forthcoming definitions and ideas on Convex Geometry, we recommend the books [14] and [3].

If the origin is an interior point of K, the polar body $K^\circ$ of K is

$$\begin{aligned} K^\circ = \{x\in \mathbb {R}^n: \langle x,y\rangle \le 1\text { for any }y\in K\}, \end{aligned}$$

which is also a convex body with the origin in its interior.

A convex body $K\in {\mathcal {K}}^n$ is uniquely defined by its support function, defined by

$$\begin{aligned} h_K(x)=\sup \{\langle x,y\rangle :y\in K\}. \end{aligned}$$

(1)

For $C\subset \mathbb R^n$ of affine dimension $k\in \{1,\dots ,n\}$, we denote by $\textrm{vol}_k(C)$ its volume measured inside the affine hull of C, and moreover we also write $\textrm{vol}(C)=\textrm{vol}_n(C)$.

The mixed volume of two convex bodies K ($n-1$ times) and L can be defined by

$$\begin{aligned} V_1(K,L)= \frac{1}{n} \lim _{\varepsilon \rightarrow 0^+}\frac{{{\,\textrm{vol}\,}}(K+\varepsilon L)-{{\,\textrm{vol}\,}}(K)}{\varepsilon }. \end{aligned}$$

There is a unique finite measure $S(K,\cdot )$ on the unit euclidean sphere $\mathbb {S}^{n-1}$, called the surface area of K, so that

$$\begin{aligned} V_1(K,L)=\frac{1}{n}\int _{\mathbb {S}^{n-1}}h_L(v)\,dS(K,v), \end{aligned}$$

(cf. [11]). When K has a $\mathcal {C}^2$ boundary $\partial K$ with positive curvature, the density of $S(K,\cdot )$ with respect to the Lebesgue measure on $\mathbb {S}^{n-1}$ is the reciprocal of the Gauss curvature of $\partial K$.

For any $x\in \mathbb {R}^n$ we let $x^\bot$ be the $(n-1)$-subspace orthogonal to x, and let $P_{x^\bot }C$ be the orthogonal projection of C onto $x^\bot$. Then the projection body $\Pi K$ of $K\in {\mathcal {K}}^n$ is the centrally symmetric convex body given by its support function

$$\begin{aligned} h_{\Pi K}(u)={{\,\textrm{vol}\,}}_{n-1}(P_{u^\bot }K), \end{aligned}$$

for every $u\in \mathbb {S}^{n-1}$. Using standard properties of the mixed volume $V_1(\cdot ,\cdot )$ (cf. [14]) it is easy to see that

$$\begin{aligned} h_{\Pi K}(u)=\frac{1}{2}\int _{\mathbb S^{n-1}}|\langle u,v\rangle |dS(K,v). \end{aligned}$$

(2)

In fact,

$$\begin{aligned} \int _{\mathbb S^{n-1}}|\langle u,v\rangle |dS(K,v)= \int _{\mathbb {S}^{n-1}}h_{L_u}(v)\,dS(K,v), \end{aligned}$$

where $L_u=[-u,u]=\{tu:t\in [-1,1]\}$. Using Fubini’s formula,

$$\begin{aligned} {{{\,\textrm{vol}\,}}(K+\varepsilon L_u)-{{\,\textrm{vol}\,}}(K)}=2\varepsilon \int _{P_{u^\bot }K}dx'=2\varepsilon {{\,\textrm{vol}\,}}_{n-1}({P_{u^\bot }K}). \end{aligned}$$

Then

$$\begin{aligned} n V_{1}(K,L_u) =\lim _{\varepsilon \rightarrow 0^+}\frac{{{\,\textrm{vol}\,}}(K+\varepsilon L_u)-{{\,\textrm{vol}\,}}(K)}{\varepsilon }= 2{{\,\textrm{vol}\,}}_{n-1}({P_{u^\bot }K}) =2h_{\Pi K}(u). \end{aligned}$$

Finally, the polar projection body $\Pi ^\circ K$ is the polar body of $\Pi K$.

A function $f:\mathbb {R}^n\rightarrow [0,\infty )$ is log-concave if $\log f$ is concave, i.e., if

$$\begin{aligned} f((1-\lambda )x+\lambda y)\ge f(x)^{1-\lambda }f(y)^\lambda , \end{aligned}$$

for every $x,y\in \mathbb R^n$, $\lambda \in (0,1)$. Then $f=e^{-\varphi }$ for a convex function $\varphi :\mathbb {R}^n\rightarrow [-\infty ,\infty )$. Moreover, let $\mathcal F(\mathbb R^n)=\{f\text { log-concave with }f\in L^1(\mathbb R^n)\}$.

Two typical embeddings of all convex bodies onto the set $\mathcal F(\mathbb R^n)$ are given by the mappings that identify K either with the characteristic function $\chi _K(x)=e^{-I^\infty _K(x)}$ or the exponential gauge $e^{-\Vert x\Vert _K}$ of K, where

$$\begin{aligned} I^{\infty }_{K}(x) = \left\{ \begin{array}{lr} 0 &{} \text {if } \quad x \in K\\ \infty &{} \text {otherwise.} \end{array} \right. \quad \text {and}\quad \Vert x\Vert _K=\inf \{t>0:x\in tK\}. \end{aligned}$$

Considering the definition of $h_K$ given by (1), we can write

$$\begin{aligned} h_K(x)=\sup \{\langle x,y\rangle -I_K^\infty (y):y\in \mathbb {R}^n\}=(I_K^\infty )^*(x), \end{aligned}$$

(3)

where, for a convex function $\varphi$,

$$\begin{aligned} \varphi ^*(x)=\sup \{\langle x,y\rangle -\varphi (y):y\in \mathbb {R}^n\} \end{aligned}$$

is the so called Legendre transform of $\varphi$ (cf. [10]). As

$$\begin{aligned} \chi _K=e^{-I_K^\infty }\text { and }h_K=(I_K^\infty )^*, \end{aligned}$$

it is natural to define the support function of a log-concave function $f=e^{-\varphi }$ as

$$\begin{aligned} h_f=\varphi ^*, \end{aligned}$$

(cf. [13]). Note that $h_K^*=I^\infty _K$ for every $K\in \mathcal K^n$ (cf. [10]).

In order to define the polar function of a function $f=e^{-\varphi }$ as a log-concave function, it is natural to search for a transformation T between convex functions so that $f^\circ =e^{-T\varphi }$. Since $(K^\circ )^\circ =K$ and if $K_1\subset K_2$ then $K_1^\circ \supset K_2^\circ$, we have to ask for T to verify $T^2$ to be the identity, and if $\varphi _1\le \varphi _2$, then $T\varphi _1\ge T\varphi _2$. From [5], these properties characterize the Legendre transform, so $T\varphi =\varphi ^*$.

As a consequence, for any log-concave $f:\mathbb {R}^n\rightarrow [0,+\infty )$ with $f=e^{-\varphi }$, its polar function $f^\circ$ is defined by $f^\circ =e^{-\varphi ^*}$ (cf. [5]). With this definition, if $f\in \mathcal F(\mathbb R^n)$ with $0\in \textrm{int}(\textrm{supp}f)$, and $f^\circ (x_M)=\Vert f\Vert _\infty$ for some $x_M\in \mathbb {R}^n$, then $f^\circ \in \mathcal F(\mathbb R^n)$ too (see Theorem 4.3 below). Note that $f^\circ =e^{-h_f}$.

To define the analogue definition of $\Pi f$ for a log-concave f, firstly defined in [9], we take into account the equality for a convex body K

$$\begin{aligned} \int _{\mathbb {S}^{n-1}}|\langle u,v\rangle |\,dS(K,v) = \int _{\mathbb {R}^n}|\langle \nabla \chi _K(x),u\rangle |\,dx, \end{aligned}$$

for $u\in \mathbb {S}^{n-1}$ [see [16], or Proposition 2.2 iii]. We may now generalize (2) to define the Petty projection function $\Pi f$ of f given its support function

$$\begin{aligned} h_{\Pi f}(u)=\frac{1}{2}\int _{\mathbb {R}^n}|\langle \nabla f(x),u\rangle |\,dx, \end{aligned}$$

(see [9]). Note that, by the chain rule, if $f=e^{-\varphi }$, then $\nabla f=-f\nabla \varphi$, and the previous definition admits the form

$$\begin{aligned} h_{\Pi f}(u)=\frac{1}{2}\int _{\mathop {\textrm{supp}}f}|\langle \nabla \varphi (x),u\rangle |f(x)\,dx. \end{aligned}$$

In particular, for any $f\in \mathcal F(\mathbb R^n)$, the polar projection function is given by $\Pi ^\circ f=(\Pi f)^\circ$.

2 Properties and Main Result

The main result here serves as a correction to [9, Theorem 5.2] and introduces a lower bound for the integral of $\Pi ^\circ f$. Let us denote by $B^n_2$ the n-dimensional Euclidean unit ball, $\mathbb S^{n-1}$ its boundary, and let $\omega _n=\textrm{vol}(B^n_2)$ be its volume. Moreover, let $|x|=\sqrt{x_1^2+\cdots +x_n^n}$ be the Euclidean norm for every $x=(x_1,\dots ,x_n)\in \mathbb R^n$.

Theorem 2.1

Let $f\in \mathcal F(\mathbb R^n)$. Then

$$\begin{aligned} \left( \int _{\mathbb R^n}|\nabla f(x)|dx\right) ^n\int _{\mathbb R^n}\Pi ^\circ f(z)dz\ge \omega _n n!\left( \frac{n\omega _n}{\omega _{n-1}}\right) ^n. \end{aligned}$$

Moreover, equality holds if there exists $g:[0,\infty )\rightarrow [0,\infty )$, $g\in \mathcal F(\mathbb R^1)$, such that $f(x)=g(|x|)$ for every $x\in \mathbb R^n$.

In the next proposition we collect some useful computations needed in this paper, refereed to characteristic functions and exponential gauges of convex bodies.

Proposition 2.2

Let $K\in \mathcal K^n$. Then we have that

• i) $h_{\chi _K}=h_K$.

• ii) $h_{e^{-\Vert \cdot \Vert _K}}=I^\infty _{K^\circ }$.

• iii) $h_{\Pi \chi _K}=h_{\Pi K}$.

• iv) $\Pi \chi _K=e^{-I^\infty _{\Pi K}}=\chi _{\Pi K}$.

• v) $\Pi ^\circ \chi _K=e^{-h_{\Pi K}}$.

• vi) $h_{\Pi e^{-\Vert \cdot \Vert _K}}=(n-1)!h_{\Pi K}$.

• vii) $\int _{\mathbb R^n}e^{-\Vert x\Vert _K}dx=n!\textrm{vol}(K)=n!\int _{\mathbb R^n}\chi _K(x)dx$.

• viii) $\int _{\mathbb R^n}\Pi ^\circ (e^{-\Vert \cdot \Vert _K})(x)dx=n!\Gamma (n)^{-n}\textrm{vol}(\Pi ^\circ K)=\Gamma (n)^{-n}\int _{\mathbb R^n}\Pi ^\circ (\chi _K)(x)dx$.

Let us observe that Theorem 5.2 in [9] is not correct. Indeed, using Proposition 2.2, one can verify that if $f=\chi _{tB^n_2}$, for some $t>0$, then the Theorem 5.2 in [9] becomes $(1-\log t)^{n-1}\le C(n)$, for some constant $C(n)>0$ only depending on the dimension n. Later on, we will discuss how to correct those bounds for the integral of the Petty projection function.

One can also bound from above the term $\int |\nabla f|$ by means of an entropic function under some extra assumptions; indeed, if $a\chi _{B^n_2}\le f$ for some $f\in \mathcal F(\mathbb R^n)$ and $a>0$, then

$$\begin{aligned} \int _{\mathbb R^n}|\nabla f(x)|dx\le n\int _{\mathbb R^n}f(y)dy+\int _{\mathbb R^n}f(z)\log \frac{f(z)}{a\Vert f\Vert _\infty }dz, \end{aligned}$$

(4)

with equality if $f=a\chi _{B^n_2}$ (cf. [1], see also [6]).

The quantity $\textrm{vol}(K)^{n-1}\textrm{vol}(\Pi ^\circ K)$ is an affine invariant, its maximum value is provided by Petty’s projection inequality [12], with equality if and only if K is an ellipsoid, and its minimum value is given by Zhang’s inequality [15], with equality if and only if K is a simplex:

$$\begin{aligned} \frac{{2n\atopwithdelims ()n}}{n^n}\le \textrm{vol}(K)^{n-1}\textrm{vol}(\Pi ^\circ K)\le \left( \frac{\omega _n}{\omega _{n-1}}\right) ^n. \end{aligned}$$

(5)

For any $f\in \mathcal F(\mathbb R^n)$, $f=e^{-\varphi }$, let $\Pi _bf$ be the Petty projection body of f, which is the convex body whose support function is given by

$$\begin{aligned} h_{\Pi _bf}(y)=\int _{supp\,f}|\langle \nabla \varphi (x),y\rangle |f(x)dx,\quad \text {i.e.}\quad h_{\Pi _bf}=2h_{\Pi f}. \end{aligned}$$

In order to avoid future confusion, here we have changed the original name also given by Fang and Zhou [9] (they used the name $\Pi f$, and we insert the subindex b to stress that it is a body). Its polar $\Pi ^\circ _bf$ was firstly introduced in [1], and here once more we change the old naming $\Pi ^\circ f$ by $\Pi ^\circ _bf$, and it is the unit ball of the norm given by

$$\begin{aligned} \Vert y\Vert _{\Pi _b^\circ f}=\int _{\mathbb R^n}|\langle \nabla f(x),y\rangle |dx. \end{aligned}$$

Since $f=e^{-\varphi }$, due to $\nabla f(x)=-e^{-\varphi (x)}\nabla \varphi (x)$, we have that $\Pi ^\circ _bf=(\Pi _bf)^\circ$, as one may expect. The right-hand side of (5) was extended to functional settings by Zhang [16] and it is also known as the affine Sobolev inequality, whereas the left-hand side of (5) was recently extended to log-concave functions in [2],

$$\begin{aligned} \frac{2^{-n}}{n!}\Vert f\Vert _1^{-n-1}\int _{\mathbb R^n}\int _{\mathbb R^n}\min \{f(x),f(y)\}dxdy\le \textrm{vol}(\Pi ^\circ _bf)\le \left( \frac{\omega _n}{2\omega _{n-1}}\right) ^n\Vert f\Vert _{\frac{n}{n-1}}^{-n}.\nonumber \\ \end{aligned}$$

(6)

Moreover, equality holds on the right-hand side if and only if $\frac{f}{\Vert f\Vert _\infty }=\chi _{AB^n_2}$, for any regular $A\in \mathbb R^{n\times n}$, and on the left hand side if and only if $\frac{f}{\Vert f\Vert _\infty }=e^{-\Vert \cdot \Vert _S}$, for any simplex $S\in \mathcal K^n$, with $0 \in S$. One can immediately verify that

$$\begin{aligned} \Pi ^\circ f(y)=e^{-h_{\Pi f}(y)}=e^{-\frac{1}{2}h_{\Pi _bf}(y)}=e^{-\frac{1}{2}\Vert y\Vert _{\Pi ^\circ _bf}}, \end{aligned}$$

(7)

and thus, (6) can be used to give optimal bounds of the integral of $\Pi ^\circ f$ for any $f\in \mathcal F(\mathbb R^n)$.

Proposition 2.3

Let $f\in \mathcal F(\mathbb R^n)$. Then

$$\begin{aligned} \int _{\mathbb R^n}\Pi ^\circ f(x)dx=2^nn!\textrm{vol}(\Pi ^\circ _bf). \end{aligned}$$

After writing this note, Fang and Zhou have told us in personal communication that they also noticed their mistake; however, it seems that they have amended it replacing it by the right-hand side of (6) and using Proposition 2.3.

3 Proofs

We start this section by proving Proposition 2.2.

Proof of of Proposition 2.2

• (i) See (3).

• (ii) It is a direct consequence of $\Vert \cdot \Vert _K^*=I^\infty _{K^\circ }$.

• (iii) Here we use a similar argument to the one exhibited in [16, §4]. Let us denote by d(x, A) the Euclidean distance from a point $x \in \mathbb {R}^n$ to a set $A \subset \mathbb {R}^n$. Let $\varepsilon > 0$ and define

  $$\begin{aligned} f_{\varepsilon }(x) = \left\{ \begin{array}{lr} 0 &{} \text {if } d(x,K) \ge \varepsilon \\ 1 - \frac{d(x, K)}{\varepsilon } &{} \text {if } d(x,K) < \varepsilon \end{array} \right. , \end{aligned}$$

  If $d(x, K)>0$ for some $x\in \mathbb R^n$, then there exists a unique $x' \in \partial K$ such that $d(x, K) = |x - x'|$. Let $\nu (x')=\frac{x-x'}{|x - x'|}$ be the outer normal of K at $x'$ and let $D_{\varepsilon } = \{x \in \mathbb {R}^n: 0< d(x, K) < \varepsilon \}$. Then

  $$\begin{aligned} \nabla f_{\varepsilon }(x) = \left\{ \begin{array}{lr} -\varepsilon ^{-1}\nu (x') &{} \text {if } x \in D_{\varepsilon }\\ 0 &{} \text {otherwise}\\ \end{array} \right. , \end{aligned}$$

  from which

  $$\begin{aligned} \frac{1}{2}\int _{\mathbb {R}^n}|\langle \nabla f_{\varepsilon }(x), y \rangle |dx = \frac{1}{2}\int _{D_{\varepsilon }}|\langle \varepsilon ^{-1}\nu (x'), y\rangle |dx = \frac{\varepsilon ^{-1}}{2}\int _{D_{\varepsilon }}|\langle \nu (x'), y\rangle |dx. \end{aligned}$$

  When $\varepsilon \rightarrow 0$, we have that

  $$\begin{aligned} \frac{\varepsilon ^{-1}}{2}\int _{D_{\varepsilon }}|\langle \nu (x'), y\rangle |dx \rightarrow \frac{1}{2}\int _{\partial K}|\langle \nu (x'), y \rangle |d\sigma (\partial K,x'), \end{aligned}$$

  where $d\sigma (\partial K,\cdot )$ is the surface area element of $\partial K$. Since $\lim _{\varepsilon \rightarrow 0}f_{\varepsilon }=\chi _K$ and by (2) we can conclude that

  $$\begin{aligned} h_{\Pi \chi _K}(y)&= \lim _{\varepsilon \rightarrow 0}\frac{1}{2}\int _{\mathbb {R}^n}|\langle \nabla f_{\varepsilon }(x), y \rangle |dx \\&= \frac{1}{2}\int _{\partial K}|\langle \nu (x'), y \rangle |d\sigma (\partial K,x') \\&= \frac{1}{2}\int _{\mathbb {S}^{n-1}}|\langle u, y \rangle |dS(K, u) \\&= h_{\Pi K}(y). \end{aligned}$$

• (iv) Since by definition $\Pi \chi _K(x)=e^{-h_{\Pi \chi _K}^*(x)}$, by iii we can conclude that

  $$\begin{aligned} \Pi \chi _K(x)=e^{-h^*_{\Pi K}(x)}=e^{-I^\infty _{\Pi K}(x)}=\chi _{\Pi K}(x). \end{aligned}$$

• (v) Using iii we have that $\Pi ^{\circ }\chi _K(x) = e^{-h_{\Pi \chi _K}(x)} = e^{-h_{\Pi K}(x)}$, as desired.

• (vi) See [9, Prop. 5.1].

• (vii) By definition, $\int _{\mathbb R^n}\chi _K(x)dx=\textrm{vol}(K)$. Second,

  $$\begin{aligned} \begin{aligned} \int _{\mathbb R^n}e^{-\Vert x\Vert _K}dx&= \int _0^1\textrm{vol}(\{x\in \mathbb R^n:e^{-\Vert x\Vert _K}\ge t\})dt\\&=\int _0^1\textrm{vol}(\{x\in \mathbb R^n:\Vert x\Vert _K\le -\log t\})dt\\&=\textrm{vol}(K)\int _0^1\left( \log \frac{1}{t}\right) ^ndt\\&=\textrm{vol}(K)\int _0^\infty s^ne^{-s}ds\\&=\Gamma (n+1)\textrm{vol}(K). \end{aligned} \end{aligned}$$

• (viii) On the one hand, using v and vii we immediately get that

  $$\begin{aligned} \int _{\mathbb R^n}\Pi ^\circ \chi _K(x)dx=\int _{\mathbb R^n}e^{-h_{\Pi K}(x)}dx=\int _{\mathbb R^n}e^{-\Vert x\Vert _{\Pi ^\circ K}(x)}dx =n!\textrm{vol}(\Pi ^\circ K). \end{aligned}$$

  On the other hand, using vi and vii and the 1-homogeneity of the support function we can conclude that

  $$\begin{aligned} \begin{aligned} \int _{\mathbb R^n}\Pi ^\circ (e^{-\Vert \cdot \Vert _K})(x)dx&= \int _{\mathbb R^n}e^{-h_{\Pi e^{-\Vert \cdot \Vert _K}}(x)}dx\\&=\int _{\mathbb R^n}e^{-\Gamma (n)h_{\Pi K}(x)}dx\\&=\Gamma (n)^{-n}\int _{\mathbb R^n}e^{-h_{\Pi K}(x)}dx\\&=\Gamma (n)^{-n}n!\textrm{vol}(\Pi ^\circ K). \end{aligned} \end{aligned}$$

  $\square$

We now prove Theorem 2.1. The main ingredients of it are the integration by polar coordinates and the Jensen inequality [3], which states that if $(X, \Sigma , \mu )$ is a probability space, then for any convex function $\varphi :\mathbb R\rightarrow \mathbb R$ and any $\mu$-integrable function $f:X \rightarrow \mathbb R$, we have that

$$\begin{aligned} \varphi \left( \int _{X}f(x)d\mu (x)\right) \le \int _{X}\varphi \circ f(x)d\mu (x), \end{aligned}$$

and moreover, equality holds if and only if either $\varphi$ is affine or f is independent of x. One can compare the proof below to the one in [9, Thm. 5.2], where we have detected mistakes in (5.17) at the change of variables and at the application of Jensen inequality.

Proof of Theorem 2.1

Let $f=e^{-\varphi }$. Since $\nabla f(x)=-f(x)\nabla \varphi (x)$ and using polar coordinates, we can write

$$\begin{aligned} \begin{aligned} \int _{\mathbb R^n}\Pi ^\circ f(z)dz&=\int _{\mathbb R^n}e^{-\frac{1}{2}\int _{\mathbb R^n}|\langle \nabla f(x),z\rangle |dx}dz \\&=n\omega _n\int _0^\infty \int _{\mathbb S^{n-1}}e^{-\frac{r}{2}\int _{\mathbb R^n}|\langle \nabla f(x),u\rangle |dx}r^{n-1}d\mu (u)dr, \end{aligned} \end{aligned}$$

where $\mu$ is the uniform probability measure in $\mathbb S^{n-1}$. Since $e^x$ is convex, Jensen inequality implies that

$$\begin{aligned}{} & {} \exp \left( \int _{\mathbb S^{n-1}}-\frac{r}{2} \int _{\mathbb R^n}|\langle \nabla f(x),u\rangle |dxd\mu (u)\right) \\{} & {} \quad \le \int _{\mathbb S^{n-1}}\exp \left( -\frac{r}{2}\int _{\mathbb R^n}|\langle \nabla f(x),u \rangle |dx\right) d\mu (u). \end{aligned}$$

Using Fubini and using the fact that

$$\begin{aligned} \int _{\mathbb S^{n-1}}|\langle v,u\rangle |d\mu (u)=\frac{2\omega _{n-1}}{n\omega _n}|v|, \end{aligned}$$

for any $v\in \mathbb R^n$, then

$$\begin{aligned} \begin{aligned} \int _{\mathbb R^n}\Pi ^\circ f(z)dz&\ge n\omega _n\int _0^\infty e^{-\frac{r}{2} \int _{\mathbb R^n}\int _{\mathbb S^{n-1}}|\langle \nabla f(x),u\rangle |d\mu (u)dx}r^{n-1}dr \\&= n\omega _n\int _0^\infty e^{-\frac{r}{2} \int _{\mathbb R^n}\frac{2\omega _{n-1}}{n\omega _n}|\nabla f(x)|dx}r^{n-1}dr\\&=n\omega _n\int _0^\infty e^{-\frac{\omega _{n-1}}{n\omega _n}\Vert \nabla f\Vert _1 r}r^{n-1}dr. \end{aligned} \end{aligned}$$

Letting $t=\frac{\omega _{n-1}}{n\omega _n}\Vert \nabla f\Vert _1r=ar$, then $dt=adr$ and

$$\begin{aligned} \int _0^\infty e^{-ar}r^{n-1}dr=\int _0^\infty e^{-t}t^{n-1}a^{1-n}a^{-1}dt=a^{-n}\Gamma (n). \end{aligned}$$

We can thus conclude that

$$\begin{aligned} \begin{aligned} \int _{\mathbb R^n}\Pi ^\circ f(z)dz&\ge n!\omega _n\left( \frac{\omega _{n-1}}{n\omega _n}\Vert \nabla f\Vert _1\right) ^{-n}\\&=n!\omega _n\left( \frac{\omega _{n-1}}{n\omega _n}\right) ^{-n}\left( \int _{\mathbb R^n}|\nabla f(x)|dx\right) ^{-n}. \end{aligned} \end{aligned}$$

In the equality case, there must be equality in the inequality above. Hence, by Jensen’s equality case, we must have that $\int _{\mathbb R^n}|\langle \nabla f(x),u\rangle |dx$ is independent of $u\in \mathbb S^{n-1}$. In particular, if $f(x)=g(|x|)$ for some $g:[0,\infty )\rightarrow [0,\infty )$ log-concave and every $x\in \mathbb R^n$, as desired. $\square$

A geometrical consequence of Theorem 2.1 is the following result (cf. [16]), which relates the surface area measure S(K) of a $K\in \mathcal K^n$ with the volume $\textrm{vol}(\Pi ^\circ K)$, and can be also obtained by Hölder inequality in (5).

Corollary 3.1

Let $K\in \mathcal K^n$. Then

$$\begin{aligned} S(K)^n\textrm{vol}(\Pi ^\circ K)\ge \omega _n\left( \frac{n\omega _n}{\omega _{n-1}}\right) ^n. \end{aligned}$$

Moreover, equality holds if $K=B^n_2$.

Proof

Let us particularize Theorem 2.1 taking $f(x)=e^{-\Vert x\Vert _K}$. If we denote by $d\sigma (\partial K,\cdot )$ the surface area element of K, then

$$\begin{aligned} \int _{\mathbb {R}^n}|\nabla f(x)|dx&= \int _{\mathbb {R}^n}e^{-||x||_K}|\nabla ||x||_K|dx \\&= \int _{0}^{\infty }\int _{t\partial K}e^{-||x||_K} |\nabla ||x||_K|\cdot |\nabla ||x||_K|^{-1}d\sigma (t\partial K, x)dt \\&= \int _{0}^{\infty }\int _{t\partial K}e^{-||x||_K}d\sigma (t\partial K, x)dt \\&= \int _{0}^{\infty }\int _{\partial K}e^{-t} t^{n-1}d\sigma (\partial K, y)dt \\&= \int _{0}^{\infty }e^{-t}t^{n-1}dt\int _{\partial K}d\sigma (\partial K, y) \\&= \int _{0}^{\infty }e^{-t}t^{n-1}dt S(K)\\&= \Gamma (n)S(K). \end{aligned}$$

This, together with viii in Proposition 2.2, imply that

$$\begin{aligned} \begin{aligned} \frac{n!}{(n-1)!^n}\textrm{vol}(\Pi ^\circ K)&= \int _{\mathbb R^n}\Pi ^\circ (e^{-\Vert \cdot \Vert _K})(x)dx\\&\ge n!\omega _n\left( \frac{\omega _{n-1}}{n\omega _n}\right) ^{-n}\left( \int _{\mathbb R^n}|\nabla (e^{-\Vert \cdot \Vert _K})(x)|dx\right) ^{-n}\\&=n!\omega _n\left( \frac{\omega _{n-1}}{n\omega _n}\right) ^{-n}(n-1)!^{-n}S(K)^{-n}, \end{aligned} \end{aligned}$$

as desired.

Equality holds if $e^{-\Vert x\Vert _K}$ is independent of $x\in \mathbb S^{n-1}$, for instance, if $K=B^n_2$. $\square$

Now we show Proposition 2.3.

Proof of Proposition 2.3

As a consequence of (7) and vii) in Proposition 2.2, we obtain that

$$\begin{aligned}{} & {} \int _{\mathbb R^n}\Pi ^\circ f(x)dx =\int _{\mathbb R^n}e^{-h_{\Pi f}(x)}dx =\int _{\mathbb R^n}e^{-\frac{1}{2}\Vert x\Vert _{\Pi ^\circ _bf}}dx \\{} & {} \quad = 2^n\int _{\mathbb R^n}e^{-\Vert x\Vert _{\Pi ^\circ _bf}}dx =2^nn!\textrm{vol}(\Pi ^\circ _bf). \end{aligned}$$

$\square$

4 Integrability of Log-concave Functions

In this section we characterize the integrability of log-concave functions in terms of the value of f over all possible rays. Other characterizations of the integrability of log-concave functions were given in [8]. Before stating the next result, we would like to remember that for any log-concave function $f:\mathbb R^n\rightarrow [0,\infty )$, the function

$$\begin{aligned} g(x)=\left\{ \begin{array}{cc}f(x) &{} \text {if }x\in \textrm{int}(\textrm{supp }f),\\ \limsup _{y\rightarrow x,y\in \textrm{int}(\textrm{supp }f)}f(y) &{} \text {if }x\in \partial \,\textrm{supp }f,\\ 0 &{} \text {otherwise} \end{array}\right. \end{aligned}$$

is log-concave, continuous on its support $\mathop {\textrm{supp}}g=\mathop {\textrm{supp}}f=\overline{\{x\in \mathbb R^n:f(x)>0\}}$, and has $\int g=\int f$ (see [7, Lem. 2.1]). Thus, we can always replace f by g and hence we can always extend continuously f to its support.

Lemma 4.1

Let $f:\mathbb R^n\rightarrow [0,\infty )$ be log-concave with $f(x_M)=\Vert f\Vert _\infty$ for some $x_M\in \mathbb R^n$. Then the following are equivalent:

• (1) f is integrable.

• (2) Either $\int f=0$ or there exists no ray $R=x_0+\mathbb R_+u$, $x_0,u\in \mathbb R^n$, $u\ne 0$, for which $f|_R=c>0$.

Proof

We first prove (1) implies (2). Let us suppose that $\int f>0$. Hence, the function f is non-zero in an open ball $B(x_0,r)$, for some $x_0\in \mathbb R^n$ and $r>0$. By continuity of f in $B(x_0,r)$, let us suppose that $f(x)\ge \alpha$, for every $x\in B(x_0,r)$ and for some $\alpha >0$. Moreover, for the sake of contradiction, let us suppose that there exists a ray $R=y_0+\mathbb R_+u$, $y_0,u\in \mathbb R^n$, $u\ne 0$, such that $f|_R=c>0$. Let $z_0\in U=B(x_0,r)\cap (x_0+u^\bot )$ and $t>0$. Let us observe that

$$\begin{aligned} \left( 1-\frac{1}{k}\right) z_0+\frac{1}{k}(y_0+tku)\rightarrow z_0+tu\quad \text { if }\quad k\rightarrow \infty . \end{aligned}$$

Furthermore, we have that

$$\begin{aligned} f\left( \left( 1-\frac{1}{k}\right) z_0+\frac{1}{k}(y_0+tku)\right) \ge f(z_0)^{1-\frac{1}{k}}f(y_0+tku)^\frac{1}{k} \ge \alpha ^{1-\frac{1}{k}}c^\frac{1}{k}, \end{aligned}$$

and thus that

$$\begin{aligned} f(z_0+tu)=\lim _{k\rightarrow \infty }f\left( \left( 1-\frac{1}{k}\right) z_0+\frac{1}{k}(y_0+tku)\right) \ge \alpha . \end{aligned}$$

Note that $z_0+tu\in \textrm{int}(\textrm{supp }f)$. Hence

$$\begin{aligned} \int _{\mathbb R^n} f(x)dx\ge \int _{U+\mathbb R_+u}f(x)dx\ge \alpha \textrm{vol}(U+\mathbb R_+u)=\infty , \end{aligned}$$

thus showing that f is not integrable, a contradiction.

We now show (2) implies (1). If $\int f=0$, then f is integrable. Let us suppose that $\int f>0$. After a suitable translation, let us assume that $f(0)=\Vert f\Vert _\infty$. For every $u\in \mathbb S^{n-1}$, since f is not constant on $\mathbb R_+u$, then there exists $s_u>0$ such that $f(s_uu)<\Vert f\Vert _\infty$. Now, using that f is continuous implies that $t_u=\inf \{s>0:f(su)<\Vert f\Vert _\infty \}$ fulfills $f(t_uu)=\Vert f\Vert _\infty$, for every $u\in \mathbb S^{n-1}$. We now show that if $\{t_u:u\in \mathbb S^{n-1}\}$ is unbounded, we arrive at a contradiction. Indeed, in that case let $\{u_k\}\subset \mathbb S^{n-1}$ be such that $t_{u_k}\rightarrow \infty$ as $k\rightarrow \infty$. Since $\mathbb S^{n-1}$ is compact, there exists a subsequence (which we can suppose w.l.o.g. to be the sequence itself) converging $u_k\rightarrow u_0\in \mathbb S^{n-1}$. For every $t>0$, then

$$\begin{aligned} \frac{t}{t_{u_k}}(t_{u_k}u_k)+\left( 1-\frac{t}{t_{u_k}}\right) 0\rightarrow tu_0\quad \text {if}\quad k\rightarrow \infty . \end{aligned}$$

Note that if $t>0$ is fixed, since $t_{u_k}\rightarrow \infty$, there exists $k_t\in \mathbb N$ such that if $k\ge k_t$ then $t_{u_k}\ge t$. Hence

$$\begin{aligned} f(tu_0) = \lim _{k\rightarrow \infty }f\left( \frac{t}{t_{u_k}}(t_{u_k}u_k)+\left( 1-\frac{t}{t_{u_k}}\right) 0\right) \ge \lim _{k\rightarrow \infty }f(t_{u_k}u_k)^{\frac{t}{t_{u_k}}}f(0)^{1-\frac{t}{t_{u_k}}}=\Vert f\Vert _\infty . \end{aligned}$$

Therefore $f|_{\mathbb R_+u_0}=\Vert f\Vert _\infty$, contradicting the hypothesis. Thus $\{t_u:u\in \mathbb S^{n-1}\}$ is bounded. If $t>t_u$, then $f(tu)<\Vert f\Vert _\infty$. Hence let $t_*>t_u$ be such that $f(t_*u)\le c<\Vert f\Vert _\infty$ for every $u\in \mathbb S^{n-1}$ and some $c>0$. Observe that for every $t\ge t_*$ and every $u\in \mathbb S^{n-1}$ we have that

$$\begin{aligned} f(t_*u)=f\left( \frac{t_*}{t}(tu)+\left( 1-\frac{t_*}{t}\right) 0\right) \ge f(tu)^\frac{t_*}{t}f(0)^{1-\frac{t_*}{t}}. \end{aligned}$$

Thus, integrating in polar coordinates, we get that

$$\begin{aligned} \begin{aligned} \int _{\mathbb R^n}f(x)dx&= \int _{\mathbb R^n\setminus B(0,t_*)}f(x)dx+\int _{B(0,t_*)}f(x)dx \\&= n\omega _n\int _{\mathbb S^{n-1}}\int _{t_*}^\infty t^{n-1}f(tu)dtd\mu (u) + \int _{B(0,t_*)}f(x)dx \\&\le n\omega _n\int _{\mathbb S^{n-1}}\int _{t_*}^\infty t^{n-1}f(0)\left( \frac{f(t_*u)}{f(0)}\right) ^{\frac{t}{t_*}}dtd\mu (u) + \int _{B(0,t_*)}f(x)dx \\&\le n\omega _nf(0)\int _{t_*}^\infty t^{n-1}\left( \frac{c}{f(0)}\right) ^\frac{t}{t_*}dt +\int _{B(0,t_*)}f(x)dx, \end{aligned} \end{aligned}$$

where $\mu$ is the uniform probability measure in $\mathbb S^{n-1}$. Since the first integral is finite as $0<c/f(0)<1$ and the second one is finite as f is bounded and $B(0,t_*)$ is bounded too, hence we obtain that f is integrable. $\square$

Remark 4.2

The existence of $x_M\in \mathbb R^n$ in Lemma 4.1 is necessary. Indeed, the function $f:\mathbb R\rightarrow [0,\infty )$ where $f(x):=e^{-e^x}$, which is monotonically decreasing on $\mathbb R$, is log-concave, it fulfills (2) (since it is not constant over any ray) but does not fulfill (1) (simply noticing that $f(x)\ge e^{-1}$ for every $x\le 0$).

Theorem 4.3

Let $f:\mathbb R^n\rightarrow \mathbb R$ be log-concave with $0\in \textrm{int}(\textrm{supp}(f))$ and $f^\circ (x_M)=\Vert f^\circ \Vert _\infty$ for some $x_M\in \mathbb R^n$. Then $f^\circ$ is integrable.

Proof

Let us suppose that $f=e^{-\varphi }$. We show now that $\varphi ^*$ (and thus $f^\circ$) is not constant over any ray $y_0+\mathbb R_+u$, for any $y_0,u\in \mathbb R^n$, $u\ne 0$, thus concluding by Lemma 4.1 that $f^\circ$ is integrable. Since $\varphi$ is continuous and $0\in \textrm{int}(\textrm{supp }f)$, let us suppose that $\varphi (x)\le C$ whenever $|x|\le \delta$ for some $C,\delta >0$. Let us consider $y_0+tu$, $t>0$. Then

$$\begin{aligned}{} & {} \varphi ^*(y_0+tu)=\sup _{z}(\langle y_0+tu,z\rangle -\varphi (z))\\{} & {} \quad \ge \delta |y_0+tu|-\varphi \left( \delta \frac{y_0+tu}{|y_0+tu|}\right) \ge \delta |y_0+tu|-C, \end{aligned}$$

thus showing that $\varphi ^*$ is not constant over any ray $y_0+\mathbb R_+u$, as desired. $\square$

Remark 4.4

The integrability of $f^\circ$ can be easily deduced by using some Blaschke-Santaló functional inequality

$$\begin{aligned} \int f\int f^\circ \le (2\pi )^n \end{aligned}$$

but only for certain particular translations of f (for instance, when the Santaló point of f is the origin, see [4]). However, the comment in [10, Rmk. 2] is not correct (where the authors said that "All of our results hold, with the same proofs, for log-concave functions that reach their maximum at the origin"), since $f^\circ$ is not necessarily integrable if $f(0)=\Vert f\Vert _\infty$. For instance, letting

$$\begin{aligned} f(x)=\left\{ \begin{array}{cc}e^{-\frac{x^2}{2}} &{} \text {if }x\ge 0 \\ 0 &{} \text {otherwise,}\end{array}\right. \end{aligned}$$

if $f=e^{-\varphi }$, then

$$\begin{aligned} \varphi (x)=\left\{ \begin{array}{cc}\frac{x^2}{2} &{} \text {if }x\ge 0 \\ \infty &{} \text {otherwise}\end{array}\right. \quad \text {and} \quad \varphi ^*(x)=\left\{ \begin{array}{cc}\frac{x^2}{2} &{} \text {if }x\ge 0 \\ 0 &{} \text {otherwise,}\end{array}\right. \end{aligned}$$

thus having that $f^\circ (x)=e^0=1$ if $x<0$, and hence $f^\circ$ would not be integrable.