In a series of ground-breaking work by Lutwak, Yang and Zhang [7, 8], the classical Brunn-Minkowski theory emerged at the turn of the 19th into the 20th century and then the $L_{p}$ Brunn-Minkowski theory originated from Lutwak's seminal work [4, 5], were remarkably generalized to the more broad framework, which is the so-called Orlicz-Brunn-Minkowski theory.

Within the Orlicz-Brunn-Minkowski theory, Orlicz projection body is spontaneously the important object. In retrospect, the two classical inequalities which connect the volume of a convex body with that of its polar projection body are the Petty and Zhang projection inequalities. The Petty projection inequality led to the affine Sobolev inequality [9] that is stronger than the classical Sobolev inequality and yet is independent of any underlying Euclidean structure. The $L_{p}$ analogue of projection bodies and the celebrated Petty projection inequality was established in [1] by Lutwak, Yang and Zhang, and independently derived by Campi and Gronchi [2] using an alternate approach. Recently, Lutwak, Yang and Zhang [7] established the corresponding Orlicz version.

We consider convex $\phi: \Bbb R \rightarrow [0,\infty)$ such that $\phi(0)=0$. This means that $\phi$ must be decreasing on $(-\infty,0]$ and increasing on $[0,\infty)$. We will assume throughout that one of these is happening strictly so; i.e., $\phi$ is either strictly decreasing on $(-\infty,0]$ or strictly increasing on $[0,\infty)$. The class of such $\phi$ will be denoted by $\mathcal{C}$.

Let $K$ be a convex body in ${\Bbb R}^{n}$ that contains the origin in its interior and has volume $|K|$. For $\phi\in\mathcal{C}$, the Orlicz projection body $\Pi_{\phi}K$ of $K$ is defined as the body whose support function is given by

$h_{\Pi_{\phi}K}(x)=\inf \Bigg\{\lambda >0: \int_{\partial K} \phi \Bigg(\frac{x \cdot v(y)}{\lambda y \cdot v(y)} \Bigg )y \cdot v(y) d \mathcal{H} ^{n-1} (y) \leq n |K| \Bigg\},$

where $v(y)$ is the outer unit normal of $\partial K$ at $y \in \partial K$, where $x \cdot v(y)$ denotes the inner product of $x$ and $v(y)$, and $\mathcal{H}^{n-1}$ is $(n-1)$-dimensional Hausdorff measure.

With $\phi_{1}(t)= | t |$, it turns out that for $u \in S^{n-1}$, $h_{\Pi_{\phi _{1}}K}(u)=\frac{c_{n}}{| K |} |K_{u}|,$ where $| K_{u} |$ denotes the $(n-1)$-dimensional volume of $K_{u}$, the image of the orthogonal projection of $K$ onto the subspace $u^{\perp}$. Thus $\Pi_{\phi_{1}}K=\frac{c_{n}}{| K |}\Pi K,$ where $\Pi K$ is the classical projection body of $K$ introduced by Minkowski.

With $\phi_{p}(t)=| t |^{p}$, and $p \geq 1$, $\Pi_{\phi_{p}}K=\frac{c_{n,p}}{| K |^{\frac{1}{p}}}\Pi_{p} K,$ where $\Pi _{p} K$ is the $L_{p}$ projection body of $K$, defined as the convex body whose support function is given by

$h_{\Pi_{p}K}(x)=\Bigg\{ \int_{\partial K} |x \cdot v(y) | ^{p} |y \cdot v(y)| ^{1-p} d\mathcal{H} ^{n-1} (y) \Bigg \}^{1/p}.$

In this paper, we demonstrate the fact that the Orlicz projection bodies of ellipsoids are still ellipsoids in $\mathbb{R}^{n}$. As examples, we compute two concrete support functions of Orlicz projection bodies of the unit ball for two specific convex functions.

The setting for this paper is the $n$-dimensional Euclidean space $\mathbb{R}^{n}$. We write $e_{1},\cdots ,e_{n}$ for the standard orthonormal basis of $\mathbb{R}^{n}$. Throughout this paper, $B^{n}=\{ x \in \mathbb{R}^{n}: |x| \leq 1\}$ denotes the unit ball centered at the origin, and $\omega_{n}=|B^{n}|$ denotes its $n$-dimensional volume.

A convex body is a compact convex subset of $\mathbb{R}^{n}$ with nonempty interior. All the convex bodies of $\mathbb{R}^{n}$ will be denoted by $\mathcal{K}_{0}^{n}$. Associated with a convex body $K$ is its support function $h_{K}$ defined on $\mathbb{R}^{n}$ by $h_{K}(x)=\max \{ x \cdot y: y \in K\}.$ Thus, if $y \in \partial K$, then $h_{K}(v_{K}(y))=v_{K}(y) \cdot y,$ where $v_{K}(y)$ denotes an outer unit normal to $\partial K$ at $y$.

For more detailed facts on convex bodies, you can refer the excellent books authored by Schneider [3] and Gardner [6].

In [7], it is proved that the definition of $h_{\Pi_{\phi}K}(x)$ is equivalent to the following definition:

$h_{\Pi_{\phi}K}(x)=\inf \Bigg \{\lambda >0: \int_{S^{n-1}} \phi \Bigg (\frac{x \cdot u}{\lambda h_{K}(u)} \Bigg )h_{K}(u) d S_{K}(u) \leq n |K| \Bigg \},$

or equivalently,

$h_{\Pi_{\phi}K}(x)=\inf \Bigg \{\lambda >0: \int_{S^{n-1}} \phi \Bigg (\frac{1}{\lambda}(x \cdot u)\rho_{K^{\ast}}(u)\Bigg ) d V_{K}(u) \leq 1 \Bigg \}.$

The polar body of $\Pi_{\phi}K$ will be denoted by $\Pi_{\phi}^{\ast}K$.

Since the area measure $S_{K}$ cannot be concentrated on a closed hemisphere of $S^{n-1}$, and since we assume that $\phi$ is strictly increasing on $[0,\infty)$ or strictly decreasing on $(-\infty,0]$, it follows that the function

$\lambda \longmapsto \int_{S^{n-1}} \phi \Bigg (\frac{1}{\lambda}(x \cdot u)\rho_{K^{\ast}}(u) \Bigg )d V_{K}(u)$

is strictly decreasing in $(0,\infty)$. Thus we have

Lemma 1 Suppose $\phi\in\mathcal {C}$, and $K\in \mathcal {K}_{0}^{n}$. If $x_{0}\in \mathbb{R}^{n}\setminus\{0\}$, then

$\int_{S^{n-1}}\phi\Bigg(\frac{x_{0}\cdot v}{\lambda_{0}h_{K}(v)}\Bigg)dV_{K}(v) >1,$

$=1$, or $<1$, respectively, if and only if $h_{\Pi_{\phi}K}(x_{0})>\lambda_{0},$ $=\lambda_{0}$, or $<\lambda_{0}$, respectively.

From Lemma 1, we can show immediately the inclusion relation, which is a monotonicity of Orlicz projection body in some sense.

Theorem 1 If $K\in\mathcal {K}$ and $\phi_{1}, \phi_{2}\in\mathcal {C}$, $ \phi_{1}\leq\phi_{2}$, then $\Pi_{\phi_{1}}K\subseteq\Pi_{\phi_{2}}K$.

Proof $\forall u\in S^{n-1}$, let $h_{\Pi_{\phi_{1}}K}(u)=\lambda.$ In terms of Lemma 1, it has

$\frac{1}{n|K|}\int_{S^{n-1}}\phi_{1}\bigg(\frac{u\cdot \upsilon}{\lambda h_{K}(\upsilon)}\bigg)h_{K}(\upsilon)dS_{K}(\upsilon)=1.$

Since $\phi_{1}\leq\phi_{2}$, we have

$\frac{1}{n|K|}\int_{S^{n-1}}\phi_{2}\bigg(\frac{u\cdot \upsilon}{\lambda h_{K}(\upsilon)}\bigg)h_{K}(\upsilon)dS_{K}(\upsilon)\geq 1$

from Lemma 1 again, we have $h_{\Pi_{\phi_{2}}K}(u)\geq\lambda.$ Therefore $h_{\Pi_{\phi_{1}}K}(u)\leq h_{\Pi_{\phi_{2}}K}(u),$ that is $\Pi_{\phi_{1}}K\subseteq\Pi_{\phi_{2}}K.$ This completes the proof.

Theorem 2 The Orlicz projection body $\Pi_{\phi}E$ of the ellipsoid $E$ is still an ellipsoid.

Proof First, we prove that the Orlicz projection body of the unit ball is still a ball centered at the origin. Suppose $A \in SO(n)$ and $u \in S^{n-1}$, let $z=A^{t}v$. Then

$ $\begin{aligned} &\frac{1}{\omega_{n}}\int_{B^{n}}\phi(\frac{1}{\lambda}Au\cdot\upsilon)d\upsilon =\frac{1}{\omega_{n}}\int_{B^{n}}\phi(\frac{1}{\lambda}u\cdot A^{t}\upsilon)dA^{t}\upsilon =\frac{1}{\omega_{n}}\int_{A^{t}B^{n}}\phi(\frac{1}{\lambda}u\cdot z)dz\\ =&\frac{1}{\omega_{n}}\int_{B^{n}}\phi(\frac{1}{\lambda}u\cdot z)dz. \end{aligned}$ $

It yields $h_{\Pi_{\phi}B^{n}}(Au)=h_{\Pi_{\phi}B^{n}}(u),$ which implies that the Orlicz projection body of the unit ball is still a ball.

Second, suppose the ellipsoid $E=AB^{n}$, $A\in GL(n)$. According to Lemma 2.6 in [7], we have

$\Pi_{\phi}(E)= \Pi_{\phi}(AB^{n})=A^{-t} \Pi_{\phi}(B^{n}).$

In view of the just verified fact, it can conclude that $\Pi_{\phi} E$ is still an ellipsoid.

This completes the proof.

In the following, we compute the support functions of Orlicz projection bodies of the unit ball, when $\phi_{1}(x)=e^{|x|}-1$ and $\phi_{2}(x)=e^{x^{2}}-1$, respectively. Obviously, $\phi_{1}$ and $\phi_{2}$ are both belong to the class $\mathcal{C}$.

(1) $\phi_{1}(x)=e^{|x|}-1$. We know that

$\begin{align*} h_{\Pi_{\phi_{1}}B^{n}}(e_{1})&=\inf\{\lambda>0: \int_{S^{n-1}}\phi_{1}(\frac{1}{\lambda}e_{1}\cdot u)dS(u)\leq n\omega_{n}\}\\ &=\inf \{\lambda>0: \int_{S^{n-1}}(e^{\frac{|u_{1}|}{\lambda}}-1)dS(u)\leq n\omega_{n}\}. \end{align*}$

From Lemma 1, we have

$\begin{align*} &\int_{S^{n-1}}(e^{\frac{|u_{1}|}{\lambda_{0}}}-1)dS(u)= n\omega_{n}\Longleftrightarrow h_{\Pi_{\phi_{1}}B^{n}}(e_{1})=\lambda_{0}.\end{align*}$

So, it has

$\begin{align*} &\int_{S^{n-1}}e^{\frac{|u_{1}|}{\lambda_{0}}}dS(u)=2n\omega_{n}\\ \Longrightarrow&\omega_{n-1}\int_{-1}^{1}e^{\frac{|u_{1}|}{\lambda_{0}}}(1-u_{1}^{2})^{\frac{n-3}{2}}du_{1}=2n\omega_{n}\\ \Longrightarrow&\int_{0}^{1}e^{\frac{u_{1}}{\lambda_{0}}}(1-u_{1}^{2})^{\frac{n-3}{2}}du_{1}=\frac{n\omega_{n}}{\omega_{n-1}}\\ \Longrightarrow&\sum\limits_{k=0}^{\infty}\frac{1}{k!\lambda_{0}^{k}}\int_{0}^{1}u_{1}^{k}(1-u_{1}^{2})^{\frac{n-3}{2}}du_{1}=\frac{n\omega_{n}}{\omega_{n-1}}. \end{align*}$

Let $u_{1}^{2}=t$, it gives

$\begin{aligned} &\sum\limits_{k=0}^{\infty}\frac{1}{2k!\lambda_{0}^{k}}\beta(\frac{k+1}{2},\frac{n-1}{2})=\frac{n\omega_{n}}{\omega_{n-1}} \\ \Longrightarrow&\sum\limits_{k=0}^{\infty}\frac{1}{2k!\lambda_{0}^{k}}\frac{\Gamma(\frac{k+1}{2})\Gamma(\frac{n-1}{2})}{\Gamma(\frac{k+n}{2})}=\frac{n\omega_{n}}{\omega_{n-1}}\\ \Longrightarrow&\sum\limits_{k=0}^{\infty}\frac{1}{k!\lambda_{0}^{k}}\frac{\Gamma(\frac{k}{2}+\frac{1}{2})}{\Gamma(\frac{k}{2}+\frac{n}{2})}=\frac{n(n-1)\pi^{\frac{1}{2}}}{\Gamma(1+\frac{n}{2})}, \end{aligned}$

(1)

which gives the required formula which $h_{\Pi_{\phi_{1}}B^{n}}$ is satisfied.

(2) $\phi_{2}(x)=e^{x^{2}}-1$. We know that

$\begin{align*} h_{\Pi_{\phi_{2}}B^{n}}(e_{1})&=\inf\{\lambda>0: \int_{S^{n-1}}\phi_{2}(\frac{1}{\lambda}e_{1}\cdot u)dS(u)\leq n\omega_{n}\}\\ &=\inf\{\lambda>0: \int_{S^{n-1}}(e^{\frac{u_{1}^{2}}{\lambda^{2}}}-1)dS(u)\leq n\omega_{n}\}. \end{align*}$

From Lemma 1, we have

$\begin{align*} &\int_{S^{n-1}}(e^{\frac{u_{1}^{2}}{\lambda_{1}^{2}}}-1)d S(u)= n\omega_{n} \Longleftrightarrow h_{\Pi_{\phi_{2}}B^{n}}(e_{1})=\lambda_{1}.\end{align*}$

So, it has

$\begin{align*} &\int_{S^{n-1}}e^{\frac{u_{1}^{2}}{\lambda_{1}^{2}}}d S(u)= 2n\omega_{n}\\ \Longrightarrow&\omega_{n-1}\int_{-1}^{1}e^{\frac{u_{1}^{2}}{\lambda_{1}^{2}}}(1-u_{1}^{2})^{\frac{n-3}{2}}du_{1}=2n\omega_{n}\\ \Longrightarrow&\int_{0}^{1}e^{\frac{u_{1}^{2}}{\lambda_{1}^{2}}}(1-u_{1}^{2})^{\frac{n-3}{2}}du_{1}=\frac{n\omega_{n}}{\omega_{n-1}}\\ \Longrightarrow&\sum\limits_{k=0}^{\infty}\frac{1}{k!\lambda_{1}^{2k}}\int_{0}^{1}u_{1}^{2k}(1-u_{1}^{2})^{\frac{n-3}{2}}du_{1}=\frac{n\omega_{n}}{\omega_{n-1}}. \end{align*}$

Let $u_{1}^{2}=t$, it gives

$\begin{aligned} &\sum\limits_{k=0}^{\infty}\frac{1}{2k!\lambda_{1}^{2k}}\beta(k+\frac{1}{2},\frac{n-1}{2})=\frac{n\omega_{n}}{\omega_{n-1}}\\ \Longrightarrow&\sum\limits_{k=0}^{\infty}\frac{1}{2k!\lambda_{1}^{2k}}\frac{\Gamma(k+\frac{1}{2})\Gamma(\frac{n-1}{2})}{\Gamma(k+\frac{n}{2})}=\frac{n\omega_{n}}{\omega_{n-1}}\\ \Longrightarrow&\sum\limits_{k=0}^{\infty}\frac{1}{k!\lambda_{1}^{2k}}\frac{\Gamma(k+\frac{1}{2})}{\Gamma(k+\frac{n}{2})}=\frac{n(n-1)\pi^{\frac{1}{2}}}{\Gamma(1+\frac{n}{2})}, \end{aligned}$

(2)

which gives the required formula which $h_{\Pi_{\phi_{2}}B^{n}}$ is satisfied.

In view of the very similarity between formulas (1) and (2), we set out to compare which number is more larger between $\lambda_{0}$ and $\lambda_{1}$. For this aim, we consider the monotonicity of the following function $f(x)$.

Lemma 2 The function $f(x)=\frac{\Gamma(x+\frac{1}{2})}{\Gamma(x+\frac{n}{2}+1)}$ is strictly decreasing on $[0, \infty)$ with respect to $x$.

Proof Constructing a function $F(x)=\ln\frac{\Gamma(x+\frac{1}{2})}{\Gamma(x+\frac{n}{2}+1)}=\ln\Gamma(x+\frac{1}{2})-\ln\Gamma(x+\frac{n}{2}+1).$ Since function $\Gamma(x)$is infinitely differentiable, we have

$\Gamma(x)=\lim\limits_{m\rightarrow\infty}\frac{m!m^{x}}{(m+x)(m-1+x)\cdot\cdot\cdot(1+x)x},\forall x\in(0, \infty),$

then

$\Gamma(x+\frac{1}{2})=\lim\limits_{m\rightarrow\infty}\frac{m!m^{x+\frac{1}{2}}}{(m+x+\frac{1}{2})(m-1+x+\frac{1}{2})\cdot\cdot\cdot(1+x+\frac{1}{2})(x+\frac{1}{2})}.$

By the above expansion and the continuity of the natural logarithmic function, $\ln\Gamma(x+\frac{1}{2})$ can be written as

$\ln\Gamma(x+\frac{1}{2})=\lim\limits_{m\rightarrow\infty}\Big(\ln m!+(x+\frac{1}{2})\ln m-\sum\limits_{j=0}^{m}\ln(x+\frac{1}{2}+j)\Big).$

Since this sequence is absolutely convergent, we may interchange differentiation and limits

$\frac{d}{dx}\Big(\ln\Gamma(x+\frac{1}{2})\Big)=\lim\limits_{m\rightarrow\infty}\Big(\ln m-\sum\limits_{j=0}^{m}\frac{1}{x+\frac{1}{2}+j}\Big).$

It gives

$F'(x)=\lim\limits_{m\rightarrow\infty}\sum\limits_{j=0}^{m}(\frac{1}{x+\frac{n}{2}+1+j}-\frac{1}{x+\frac{1}{2}+j})<0(n\geq1, n\in Z).$

We obtain that $f(x)=\frac{\Gamma(x+\frac{1}{2})}{\Gamma(x+\frac{n}{2}+1)}$ is strictly decreasing on $[0,\infty)$ in terms with $x$. This completes the proof.

Hence, according to Lemma 2, we obtain the following results $\lambda_{0} > \lambda_{1}^{2},$ or equivalently,

$h_{\Pi_{\phi_{1}}B^{n}}(e_{1})>h_{\Pi_{\phi_{2}}B^{n}}^{2}(e_{1}),$

where $\phi_{1}(x)=e^{|x|}-1, \phi_{2}(x)=e^{x^{2}}-1,$ that is

$\frac{V(\Pi_{\phi_{1}}B^{n})}{V(B^{n})}>\bigg(\frac{V(\Pi_{\phi_{2}}B^{n})}{V(B^{n})}\bigg)^{2}. $