#  A new approach to Steiner symmetrization of coercive convex functions

Youjiang Lin  School of Mathematical Sciences, Peking University, Beijing,
100871, China; Department of Mathematics, Department of Mathematics, Shanghai
University, Shanghai, 200444, China<EMAIL_ADDRESS>,  Gangsong Leng
Department of Mathematics, Shanghai University, Shanghai, 200444, China
<EMAIL_ADDRESS>and  Lujun Guo  Department of Mathematics, Shanghai
University, Shanghai, People’s Republic of China 200444<EMAIL_ADDRESS>
###### Abstract.

In this paper, a new approach of defining Steiner symmetrization of coercive
convex functions is proposed and some fundamental properties of the new
Steiner symmetrization are proved. Further, using the new Steiner
symmetrization, we give a different approach to prove a functional version of
the Blaschke-Santaló inequality due to Ball [2].

###### Key words and phrases:

Rearrangements of functions; Steiner symmetrizations; Coercive convex
functions; Blaschke-Santaló inequality

2010 Mathematics Subject Classification.46E30, 52A40.

The authors would like to acknowledge the support from China Postdoctoral
Science Foundation Grant 2013M540806, National Natural Science Foundation of
China under grant 11271244 and National Natural Science Foundation of China
under grant 11271282 and the 973 Program 2013CB834201.

##  1\. Introduction

The purpose of this paper is to introduce a new way of defining Steiner
symmetrization for coercive convex functions, and to explore its applications.
Our new definition is motivated by and can be regarded as an improvement of a
functional Steiner symmetrization of [1]. In particular, our new definition
has a key property: the invariance of integral, which is not true for the
definition of [1]. Moreover, our definition provides a new approach to the
familiar functional Steiner symmetrization (see [7, 8]), but we do not use
geometric Steiner symmetrization and our approach is more suitable for certain
functional problems.

Steiner symmetrization was invented by Steiner [32] to prove the isoperimetric
inequality. For over 160 years Steiner symmetrization has been a fundamental
tool for attacking problems regarding isoperimetry and related geometric
inequalities [17, 18, 32, 33]. Steiner symmetrization appears in the titles of
dozens of papers (see e.g. [4, 5, 6, 8, 10, 13, 16, 20, 21, 24, 26, 27, 31])
and plays a key role in recent work such as [19, 25, 34, 35].

Steiner symmetrization is a type of rearrangement. In the 1970s, interest in
rearrangements was renewed, as mathematicians began to look for geometric
proofs of functional inequalities. Rearrangements were generalized from smooth
or convex bodies to measurable sets and to functions in Sobolev spaces.
Functional Steiner symmetrization, as a kind of important rearrangement of
functions, has been studied in [1, 7, 8, 9, 11, 12, 14]. In [7], Brascamp,
Lieb, and Luttinger established that the spherical symmetrization of a
nonnegative function can be approximated in $L^{p}(\mathbb{R}^{n+1})$ by a
sequence of Steiner symmetrizations and rotations. In [8], Burchard proved
that Steiner symmetrization is continuous in $W^{1,p}(\mathbb{R}^{n+1})$,
$1\leq p<\infty$, for every dimension $n\geq 1$, in the sense that
$f_{k}\rightarrow f$ in $W^{1,p}$ implies $Sf_{k}\rightarrow Sf$ in $W^{1,p}$.
In [14], Fortier gave a thorough review and exposition of results regarding
approximating the symmetric decreasing rearrangement by polarizations and
Steiner symmetrizations.

For a nonnegative measurable function $f$, the familiar definition of its
Steiner symmetrization (see [7, 8, 9, 14]) is defined as following:

######  Definition 1.

For a measurable function $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{+}$, let $m$
denote the Lebesgue measure, if $m([f>t])<+\infty$ for all $t>0$, then its
Steiner symmetrization is defined as

$\displaystyle\bar{S}_{u}f(x)=\int_{0}^{\infty}\mathcal{X}_{S_{u}E(t)}(x)dt,$
(1.1)

where $S_{u}E(t)$ is the Steiner symmetrization of the level set
$E(t):=\\{x\in\mathbb{R}^{n}:f(x)>t\\}$ about $u^{\perp}$ and
$\mathcal{X}_{A}$ denotes the characteristic function of set $A$.

During the study of the analogy between convex bodies and log-concave
functions, Artstein-Klartag-Milman in [1] defined another functional Steiner
transformation as follows:

######  Definition 2.

For a coercive convex function
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ and a hyperplane
$H=u^{\bot}$ ($u\in S^{n-1}$) in $\mathbb{R}^{n}$, for any $x=x^{\prime}+tu$,
where $x^{\prime}\in H$ and $t\in\mathbb{R}$, we define the Steiner
symmetrization $\widetilde{S}_{u}f$ of $f$ about $H$ by

$\displaystyle(\widetilde{S}_{u}f)(x)=\inf_{t_{1}+t_{2}=t}[\frac{1}{2}f(x^{\prime}+2t_{1}u)+\frac{1}{2}f(x^{\prime}-2t_{2}u)].$
(1.2)

In this paper, we introduce a new way of defining the functional Steiner
symmetrization for coercive convex functions. A function
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$, not identically
$+\infty$, is called convex if

$f(\alpha x+(1-\alpha)y)\leq\alpha f(x)+(1-\alpha)f(y)$

for all $x$, $y\in\mathbb{R}^{n}$ and for $0\leq\lambda\leq 1$. A convex
function $f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ is called
coercive if $\lim_{|x|\rightarrow+\infty}f(x)=+\infty$.

######  Definition 3.

For a coercive convex function
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ and a hyperplane
$H=u^{\bot}$ ($u\in S^{n-1}$) in $\mathbb{R}^{n}$, for any $x=x^{\prime}+tu$,
where $x^{\prime}\in H$ and $t\in\mathbb{R}$, we define the Steiner
symmetrization $S_{u}f$ of $f$ about $H$ by

$\displaystyle(S_{u}f)(x)=\sup_{\lambda\in[0,1]}\inf_{t_{1}+t_{2}=t}[\lambda
f(x^{\prime}+2t_{1}u)+(1-\lambda)f(x^{\prime}-2t_{2}u)].$ (1.3)

Our definition $S_{u}f$ is motivated by and can be regarded as an improvement
of $\widetilde{S}_{u}f$ in Definition 2. When compared with $\bar{S}_{u}f$ in
Definition 1, our definition symmetrizes a parabola-like (one-dimension) cure
once at a time instead of symmetrizing the level set as in $\bar{S}_{u}f$.

The rest of the paper is organized as follows. In Section 2, we explore the
analogy between convex bodies and coercive convex functions using our new
definition (see Table 1).

Table 1. A contrast between convex bodies and coercive convex functions on
Steiner symmetrization

| Convex bodies | Coercive convex functions
---|---|---
1 |  For a convex body $K$, $S_{u}K$ is still a convex body and symmetric about $u^{\perp}$. |  For a coercive convex function $f$, $S_{u}f$ is still a coercive convex function and symmetric about $u^{\perp}$.
2 |  $Vol_{n}(S_{u}K)=Vol_{n}(K)$. |  $\int_{\mathbb{R}^{n}}\exp(-S_{u}f)=\int_{\mathbb{R}^{n}}\exp(-f)$.
3 |  $K$ can be transformed into an unconditional body using $n$ Steiner symmetrizations. |  $f$ can be transformed into an unconditional function using $n$ Steiner symmetrizations.
4 |  For any convex bodies $K_{1}\subset K_{2}$, then $S_{u}K_{1}\subset S_{u}K_{2}$. |  For any coercive convex functions $f_{1}\leq f_{2}$, then $S_{u}f_{1}\leq S_{u}f_{2}$.
5 |  If $K$ is a symmetric about $z$, then $S_{u}K$ is symmetric about $z|u^{\perp}$. |  If $f$ is even about $z$, then $S_{u}f$ is even about $z|u^{\perp}$.
6 |  If the sequence $\\{K_{i}\\}$ converges in the Hausdorff metric to $K$, then the sequence $\\{S_{u}K_{i}\\}$ will converge to $S_{u}K$. |  If the sequence $\\{\exp(-f_{i})\\}$ converges in the $L^{p}$ distance to $\exp(-f)$, then the sequence $\\{\exp(-S_{u}f_{i})\\}$ will converge to $\exp(-S_{u}f)$.
7 |  There is a sequence of directions $\\{u_{i}\\}$ so that the sequence of convex bodies $K_{i}=S_{u_{i}}\dots S_{u_{1}}K$ converges to the ball with the same volume as $K$. |  There is a sequence of directions $\\{u_{i}\\}$ so that the sequence of log-concave functions $\exp(-f_{i})$, where $f_{i}=S_{u_{i}}\dots S_{u_{1}}f$, converges to a radial function with the same integral as $\exp(-f)$.

In Section 3, we will elaborate on the relations between Definition 3 and
Definitions 1, 2. In Section 4, we give a completely different approach to
prove a functional version of the Blaschke-Santaló inequality due to Ball [2].

##  2\.  The functional Steiner symmetrization

We first study the one-dimensional case. In Definition 3, when $n=1$,
$S^{0}=\\{-1,1\\}$ and $H=\\{0\\}$, it is clear that
$(S_{1}f)(x)=(S_{-1}f)(x)$ for any $x\in\mathbb{R}$. Let $Sf$ denote Steiner
symmetrization of one-dimensional function, then

$\displaystyle Sf(x)=\sup_{\lambda\in[0,1]}\inf_{x_{1}+x_{2}=x}[\lambda
f(2x_{1})+(1-\lambda)f(-2x_{2})].$ (2.1)

######  Theorem 1.

If $f:\mathbb{R}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ is a coercive convex
function, then $Sf(x)$ is a coercive even convex function and for any
$s\in\mathbb{R}$,

$\displaystyle Vol_{1}([f\leq s])=Vol_{1}([Sf\leq s]),$ (2.2)

where $[f\leq s]=\\{x\in\mathbb{R}:f(x)\leq s\\}$ denotes the sublevel set of
$f$.

The following lemma is straightforward, and we omit its proof.

######  Lemma 1.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a coercive convex function, then we
have

(i) If $a=\inf f(t)$, then $a\in(-\infty,+\infty)$ and
$f^{-1}(a)=\\{x\in\mathbb{R}:f(x)=a\\}$ is a nonempty finite closed interval
$[\mu,\nu]$, where $\mu$ may equal to $\nu$.

(ii) $f(t)$ is strictly decreasing on the interval $(-\infty,\mu]$ and
strictly increasing on the interval $[\nu,+\infty)$.

(iii) If $f(c)=f(d)$ and $c<d$, then $\mu<d$ and $c<\nu$.

(iv) For $c$ and $d$ given in (iii), we have the right derivative
$f^{\prime}_{r}(d)\geq 0$ for $f$ is increasing on $[\mu,+\infty)$, we also
have $f^{\prime}_{r}(c)\leq 0$ for $f$ is decreasing on $(-\infty,\nu]$.

(v) For two intervals $[a,a+t_{0}]$ and $[b,b+t_{0}]$ with the same length
$t_{0}>0$, if $f(a)=f(a+t_{0})$, then either $f(b)\geq f(a)$ or
$f(b+t_{0})\geq f(a+t_{0})$.

Proof of Theroem 1. First, we show that $Sf$ is even. For any
$x\in\mathbb{R}$, by (2.1), we have

$\displaystyle Sf(-x)$ $\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{x_{2}\in\mathbb{R}}[\lambda
f(-2x_{2}-2x)+(1-\lambda)f(-2x_{2})]$ (2.3) $\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{x_{2}\in\mathbb{R}}[\lambda
f(2x_{2}-2x)+(1-\lambda)f(2x_{2})]$ $\displaystyle=$
$\displaystyle\sup_{\lambda^{\prime}\in[0,1]}\inf_{x_{2}\in\mathbb{R}}[\lambda^{\prime}f(2x_{2})+(1-\lambda^{\prime})f(2x_{2}-2x)]$
$\displaystyle=$ $\displaystyle Sf(x),$

which implies that $Sf$ is even.

Let ${\rm dom}f:=\\{x\in\mathbb{R}^{n}:f(x)<+\infty\\}$ denote the effective
domain of $f$. To prove the remaining part of the theorem, we shall consider
two cases: ${\rm dom}f=\mathbb{R}$ and ${\rm dom}f\neq\mathbb{R}$.

Case (1) ${\rm dom}f=\mathbb{R}$. There are two steps.

First Step. We shall prove that $Sf(0)=\inf f$ and for any $x>0$, there exists
some $x^{\prime}\in\mathbb{R}$ such that

$\displaystyle Sf(x)=f(x^{\prime})=f(x^{\prime}-2x).$ (2.4)

Let $x=0$, by (2.1), we have

$\displaystyle Sf(0)$ $\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{x_{1}+x_{2}=0}[\lambda
f(2x_{1})+(1-\lambda)f(-2x_{2})]$ (2.5) $\displaystyle=$
$\displaystyle\inf_{x_{1}\in\mathbb{R}}f(2x_{1})=\inf_{x\in\mathbb{R}}f(x).$

For $x>0$, since $f$ is coercive and convex, there exists some
$x^{\prime}\in\mathbb{R}$ satisfying

$\displaystyle f(x^{\prime})=f(x^{\prime}-2x).$ (2.6)

Indeed, let $f_{x}(x_{1}):=f(x_{1})-f(x_{1}-2x)$, $a=\inf f$ and
$f^{-1}(a)=[\mu,\nu]$, by Lemma 1(ii), $f_{x}(x_{1})<0$ if $x_{1}<\mu$ and
$f_{x}(x_{1})>0$ if $x_{1}>\nu$. Since $f(x_{1})$ and $f(x_{1}-2x)$ are convex
functions about $x_{1}\in\mathbb{R}$ and any convex function is continuous on
the interior of its effective domain, thus $f_{x}(x_{1})$ is continuous in
$\mathbb{R}$. Therefore, there exists some $x^{\prime}$ such that
$f_{x}(x^{\prime})=0$.

Now we prove $Sf(x)=f(x^{\prime})$, where $x>0$ and $x^{\prime}$ satisfies
equality (2.6). Let $G_{x}(\lambda)$ be a function about $\lambda\in[0,1]$
defined as

$\displaystyle G_{x}(\lambda):=\inf_{x_{1}\in\mathbb{R}}[\lambda
f(2x_{1})+(1-\lambda)f(2x_{1}-2x)].$ (2.7)

For any $\lambda\in[0,1]$, choose $x_{1}=\frac{x^{\prime}}{2}$, we have

$\displaystyle G_{x}(\lambda)\leq\lambda
f(x^{\prime})+(1-\lambda)f(x^{\prime}-2x)=f(x^{\prime}).$ (2.8)

Thus, $Sf(x)=\sup_{\lambda\in[0,1]}G_{x}(\lambda)\leq f(x^{\prime})$.

On the other hand, we prove that there exists some $\lambda_{0}\in[0,1]$ such
that $G_{x}(\lambda_{0})=f(x^{\prime})$. Since $f$ is a convex function
defined in $\mathbb{R}$ and by Theorem 1.5.2 in [30], both the right
derivative $f^{\prime}_{r}$ and the left derivative $f^{\prime}_{l}$ exist and
$f^{\prime}_{l}\leq f^{\prime}_{r}$.

######  Claim 1.

There exists some $\lambda_{0}\in[0,1]$ satisfying

$\displaystyle\lambda_{0}f^{\prime}_{r}(x^{\prime})+(1-\lambda_{0})f^{\prime}_{r}(x^{\prime}-2x)=0.$
(2.9)

Proof of Claim 1. Since $f(x^{\prime})=f(x^{\prime}-2x)$ and $x>0$, by Lemma
1(iv), we have $f_{r}^{\prime}(x^{\prime})\geq 0$ and
$f_{r}^{\prime}(x^{\prime}-2x)\leq 0$, thus
$f_{r}^{\prime}(x^{\prime})-f_{r}^{\prime}(x^{\prime}-2x)\geq 0$.

(i) If $f^{\prime}_{r}(x^{\prime})-f^{\prime}_{r}(x^{\prime}-2x)>0$, then
(2.9) can be obtained by choosing

$\displaystyle\lambda_{0}=\frac{-f^{\prime}_{r}(x^{\prime}-2x)}{f^{\prime}_{r}(x^{\prime})-f^{\prime}_{r}(x^{\prime}-2x)}.$
(2.10)

(ii) If $f^{\prime}_{r}(x^{\prime})-f^{\prime}_{r}(x^{\prime}-2x)=0$, then
$f_{r}^{\prime}(x^{\prime})=f_{r}^{\prime}(x^{\prime}-2x)=0$, thus, for any
$\lambda_{0}\in[0,1]$, we can get (2.9). $\Box$

Choose a $\lambda_{0}$ satisfying (2.9), we define

$\displaystyle\Phi_{\lambda_{0}}(x_{1})=\lambda_{0}f(2x_{1})+(1-\lambda_{0})f(2x_{1}-2x).$
(2.11)

Since $f$ is a convex function, then $\Phi_{\lambda_{0}}$ is a convex function
about $x_{1}$. By (2.9), we have that the right derivative and the left
derivative of $\Phi_{\lambda_{0}}$ at $x_{1}=\frac{x^{\prime}}{2}$ satisfy

$\displaystyle\Phi_{\lambda_{0}r}^{\prime}(x_{1})|_{x_{1}=\frac{x^{\prime}}{2}}=2\lambda_{0}f_{r}^{\prime}(x^{\prime})+2(1-\lambda_{0})f_{r}^{\prime}(x^{\prime}-2x)=0,$
(2.12)

and
$\Phi_{\lambda_{0}l}^{\prime}(x_{1})|_{x_{1}=\frac{x^{\prime}}{2}}\leq\Phi_{\lambda_{0}r}^{\prime}(x_{1})|_{x_{1}=\frac{x^{\prime}}{2}}=0$.

By (2.6), (2.11) and the fact that if a convex function
$f:\mathbb{R}\rightarrow\mathbb{R}$ satisfies $f^{\prime}_{r}(x_{0})\geq 0$
and $f^{\prime}_{l}(x_{0})\leq 0$ then
$f(x_{0})=\min\\{f(x):x\in\mathbb{R}\\}$, we have

$\displaystyle\inf_{x_{1}\in\mathbb{R}}\Phi_{\lambda_{0}}(x_{1})=\Phi_{\lambda_{0}}(\frac{x^{\prime}}{2})=f(x^{\prime}).$
(2.13)

By (2.11) and (2.13), we have

$\displaystyle Sf(x)=\sup_{\lambda\in[0,1]}G_{x}(\lambda)\geq
G_{x}(\lambda_{0})=\inf_{x_{1}\in\mathbb{R}}\Phi_{\lambda_{0}}(x_{1})=f(x^{\prime}).$
(2.14)

Thus, we have $Sf(x)=f(x^{\prime})=f(x^{\prime}-2x)$.

Second Step. We shall prove that $Sf$ is coercive and convex, and for any
$s\in\mathbb{R}$, $Vol_{1}([Sf\leq s])=Vol_{1}([f\leq s])$.

First, we prove that $Sf$ is coercive. Suppose that there exists $M_{0}>0$ and
a sequence $\\{x_{n}\\}$ satisfying $|x_{n}|>n$ and $Sf(x_{n})<M_{0}$ for any
positive integer $n$, then by (2.4), there exists $x_{n}^{\prime}$ such that

$\displaystyle Sf(x_{n})=f(x_{n}^{\prime})=f(x_{n}^{\prime}-2x_{n})<M_{0}.$
(2.15)

Since
$2\max\\{|x_{n}^{\prime}|,|x_{n}^{\prime}-2x_{n}|\\}\geq|x_{n}^{\prime}|+|x_{n}^{\prime}-2x_{n}|\geq
2|x_{n}|>2n$, there is a sequence $\\{y_{n}\\}$, where $y_{n}=x_{n}^{\prime}$
if $|x_{n}^{\prime}|\geq|x_{n}^{\prime}-2x_{n}|$ and
$y_{n}=x_{n}^{\prime}-2x_{n}$ if
$|x_{n}^{\prime}|\leq|x_{n}^{\prime}-2x_{n}|$, satisfying
$\lim_{n\rightarrow+\infty}|y_{n}|=+\infty$ and $f(y_{n})<M_{0}$, which is
contradictory with $f$ is coercive.

Next, we prove that $Sf$ is a convex function on $\mathbb{R}$. First, we prove
that $Sf(x)$ is increasing on $[0,+\infty)$. In fact, by (2.4), for any
$0<x_{1}<x_{2}$, there exist $x_{1}^{\prime}$ and $x_{2}^{\prime}$ such that
$Sf(x_{i})=f(x_{i}^{\prime})=f(x_{i}^{\prime}-2x_{i})$ $(i=1,2)$. By Lemma
1(iii), for $\mu$ and $\nu$ given in Lemma 1, we have $x_{i}^{\prime}>\mu$
$(i=1,2)$ and $x_{i}^{\prime}-2x_{i}<\nu$ $(i=1,2)$. If
$f(x_{1}^{\prime})>f(x_{2}^{\prime})$, since $f$ is increasing on the interval
$[\mu,+\infty)$, then $x_{1}^{\prime}>x_{2}^{\prime}$. By $0<x_{1}<x_{2}$, we
have $x_{1}^{\prime}-2x_{1}>x_{2}^{\prime}-2x_{2}$. Since $f$ is decreasing on
the interval $(-\infty,\nu]$, we have $f(x_{1}^{\prime}-2x_{1})\leq
f(x_{2}^{\prime}-2x_{2})$, which is a contradiction. The contradiction means
that $f(x_{1}^{\prime})\leq f(x_{2}^{\prime})$, thus $Sf$ is increasing on
$[0,+\infty)$. Since $Sf$ is even, to prove $Sf$ is convex on $\mathbb{R}$, it
suffices to prove that $Sf$ is convex on $[0,+\infty)$.

For any $0\leq x_{1}<x_{2}$ and $0<\alpha<1$, by (2.4), let $x_{1}^{\prime}$,
$x_{2}^{\prime}$ and $x_{0}\triangleq(\alpha x_{1}+(1-\alpha)x_{2})^{\prime}$
be three real numbers satisfying

$\displaystyle Sf(x_{1})=f(x^{\prime}_{1})=f(x_{1}^{\prime}-2x_{1}),$ (2.16)
$\displaystyle Sf(x_{2})=f(x^{\prime}_{2})=f(x_{2}^{\prime}-2x_{2}),$ (2.17)
$\displaystyle Sf(\alpha x_{1}+(1-\alpha)x_{2})=f(x_{0})=f(x_{0}-2(\alpha
x_{1}+(1-\alpha)x_{2})).$ (2.18)

Since $f$ is a convex function, we have

$\displaystyle\alpha f(x_{1}^{\prime})+(1-\alpha)f(x_{2}^{\prime})\geq
f(\alpha x_{1}^{\prime}+(1-\alpha)x_{2}^{\prime}),$ (2.19)
$\displaystyle\alpha
f(x_{1}^{\prime}-2x_{1})+(1-\alpha)f(x_{2}^{\prime}-2x_{2})$ (2.20)
$\displaystyle\geq$ $\displaystyle f(\alpha
x_{1}^{\prime}+(1-\alpha)x_{2}^{\prime}-2(\alpha x_{1}+(1-\alpha)x_{2})).$

Since $f(x_{0})=f(x_{0}-2(\alpha x_{1}+(1-\alpha)x_{2}))$ and both
$[x_{0}-2(\alpha x_{1}+(1-\alpha)x_{2}),x_{0}]$ and $[\alpha
x^{\prime}_{1}+(1-\alpha)x^{\prime}_{2}-2(\alpha x_{1}+(1-\alpha)x_{2}),\alpha
x^{\prime}_{1}+(1-\alpha)x^{\prime}_{2}]$ have the same length $2(\alpha
x_{1}+(1-\alpha)x_{2})>0$, by Lemma 1(v), either

$\displaystyle f(\alpha x_{1}^{\prime}+(1-\alpha)x_{2}^{\prime})\geq f(x_{0})$
(2.21)

or

$\displaystyle f(\alpha x_{1}^{\prime}+(1-\alpha)x_{2}^{\prime}-2(\alpha
x_{1}+(1-\alpha)x_{2}))$ (2.22) $\displaystyle\geq$ $\displaystyle
f(x_{0}-2(\alpha x_{1}+(1-\alpha)x_{2})).$

If (2.21) holds, then we use (2.19) and if (2.22) holds, then we use (2.20),
by (2.16)-(2.18), $Sf$ is a convex function.

Finally, we prove that $Vol_{1}([f\leq s])=Vol_{1}([Sf\leq s])$ for any
$s\in\mathbb{R}$. Since $Sf(x)$ is an even convex function, $Sf(0)=\inf Sf$.
Since $Sf(0)=\inf f$ by (2.5), thus $\inf Sf=\inf f$. Let $a=\inf Sf=\inf f$,
$(Sf)^{-1}(a)=[-\delta,\delta]$, and $f^{-1}(a)=[\mu,\nu]$.

If $s=a$, then $Vol_{1}([f\leq s])=\nu-\mu$ and $Vol_{1}([Sf\leq s])=2\delta$.
Next, we prove $\nu-\mu=2\delta$. By Lemma 1, $Sf$ is strictly decreasing on
$(-\infty,-\delta)$ and strictly increasing on $(\delta,+\infty)$, and $f$ is
strictly decreasing on $(-\infty,\mu)$ and strictly increasing on
$(\nu,+\infty)$. For $\delta\geq 0$, if $\nu-\mu>2\delta$, then
$x_{0}:=\delta+\frac{\nu-\mu-2\delta}{2}>\delta$, thus $Sf(x_{0})>Sf(\delta)$,
which is contradictory with

$\displaystyle Sf(x_{0})$ $\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{x_{1}\in\mathbb{R}}[\lambda
f(2x_{1})+(1-\lambda)f(2x_{1}-2x_{0})]$ (2.23) $\displaystyle\leq$
$\displaystyle\sup_{\lambda\in[0,1]}[\lambda
f(\nu)+(1-\lambda)f(\nu-2x_{0})]=a,$

where inequality is by choosing $x_{1}=\frac{\nu}{2}$ and last equality is by
$\nu-2x_{0}=\mu$. Thus, $\nu-\mu\leq 2\delta$. Thus if $\delta=0$, then
$\mu=\nu$. For $\delta>0$, by (2.4), there exists $\delta^{\prime}$ such that
$Sf(\delta)=f(\delta^{\prime})=f(\delta^{\prime}-2\delta)=a$, which implies
that $\nu-\mu\geq 2\delta$. Thus, $\nu-\mu=2\delta$.

If $s>a$, by Lemma 1, equality (2.4), and $Sf$ is even, there is a unique
$x>0$ and a unique $x^{\prime}\in\mathbb{R}$ such that
$Sf(-x)=Sf(x)=s=f(x^{\prime})=f(x^{\prime}-2x)$, thus we have $Vol_{1}([f\leq
s])=Vol_{1}([Sf\leq s])=2x$.

If $s<a$, then $[Sf\leq s]=[f\leq s]=\emptyset$, thus $Vol_{1}([f\leq
s])=Vol_{1}([Sf\leq s])=0$.

Case (2) ${\rm dom}f\neq\mathbb{R}$. There exist eight cases for ${\rm dom}f$:
1) $[\alpha,\beta]$; 2) $(\alpha,\beta)$; 3) $(\alpha,\beta]$; 4)
$[\alpha,\beta)$; 5) $(-\infty,\beta]$; 6) $(-\infty,\beta)$; 7)
$[\alpha,+\infty)$; 8) $(\alpha,+\infty)$. We need only prove our conclusion
for ${\rm dom}f=(\alpha,\beta)$. By the same method we can prove our
conclusion for other cases. For ${\rm dom}f=(\alpha,\beta)$, there exist three
cases: (i) $f$ is decreasing on $(\alpha,\beta)$; (ii) $f$ is increasing on
$(\alpha,\beta)$; (iii) $f$ is decreasing on $(\alpha,\gamma]$ and increasing
on $[\gamma,\beta)$ for some $\gamma\in(\alpha,\beta)$. Cases (i) and (ii) are
corresponding to the cases of
$\lim_{\gamma\rightarrow\beta,\gamma<\beta}\gamma$ and
$\lim_{\gamma\rightarrow\alpha,\gamma>\alpha}\gamma$ in case (iii),
respectively, thus we need only prove our conclusion for case (iii).

If
$\lim_{x\rightarrow\alpha,x>\alpha}f(x)=\lim_{x\rightarrow\beta,x<\beta}f(x)$,
following the proof of Case (1) (i.e., ${\rm dom}f=\mathbb{R}$), we have that
$Sf$ is convex on $(-\frac{\beta-\alpha}{2},\frac{\beta-\alpha}{2})$ and
$Vol_{1}([Sf\leq s])=Vol_{1}([f\leq s])$ for any
$s<\lim_{x\rightarrow\alpha,x>\alpha}f(x)$.

If
$\lim_{x\rightarrow\alpha,x>\alpha}f(x)\neq\lim_{x\rightarrow\beta,x<\beta}f(x)$,
we may assume that

$\displaystyle\lim_{x\rightarrow\alpha,x>\alpha}f(x)=b>\lim_{x\rightarrow\beta,x<\beta}f(x)=c.$
(2.24)

If $c=a=\inf f$, then $f$ is decreasing on $(\alpha,\beta)$. Thus we may
suppose that $c>a$. Let $\gamma\in(\alpha,\beta)$ satisfy $f(\gamma)=c$. If
$|x|<\frac{\beta-\gamma}{2}$, by the proof of Case (1), there exists
$x^{\prime}\in(\gamma,\beta)$ such that
$Sf(x)=f(x^{\prime})=f(x^{\prime}-2x)$.

Step 1. We shall prove that for $|x|\geq\frac{\beta-\gamma}{2}$ and
$|x|<\frac{\beta-\alpha}{2}$,

$\displaystyle Sf(x)=f(\beta-2|x|).$ (2.25)

Since $Sf$ is even, we may assume $\frac{\beta-\gamma}{2}\leq
x<\frac{\beta-\alpha}{2}$. For any $\lambda\in[0,1]$,

$\displaystyle\inf_{x_{1}\in\mathbb{R}}[\lambda
f(2x_{1})+(1-\lambda)f(2x_{1}-2x)]$ (2.26) $\displaystyle\leq$
$\displaystyle\lambda\lim_{t\rightarrow\beta,t<\beta}f(t)+(1-\lambda)f(\beta-2x)$
$\displaystyle=$ $\displaystyle\lambda c+(1-\lambda)f(\beta-2x).$

Since $\frac{\beta-\gamma}{2}\leq x<\frac{\beta-\alpha}{2}$, then
$\alpha<\beta-2x\leq\gamma$. Since $f$ is decreasing on $(\alpha,\gamma]$,
thus $f(\beta-2x)\geq f(\gamma)=c$. Thus, by (2.26), we have

$\displaystyle Sf(x)$ $\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{x_{1}\in\mathbb{R}^{n}}[\lambda
f(2x_{1})+(1-\lambda)f(2x_{1}-2x)]$ (2.27) $\displaystyle\leq$
$\displaystyle\sup_{\lambda\in[0,1]}[\lambda
c+(1-\lambda)f(\beta-2x)]=f(\beta-2x).$

On the other hand, we prove that $Sf(x)\geq f(\beta-2x)$. Since ${\rm
dom}f=(\alpha,\beta)$ and $\inf_{x_{1}\in\mathbb{R}}[\lambda
f(x_{1})+(1-\lambda)f(x_{1}-2x)]=\inf f$ for $\lambda=0$ or $\lambda=1$, we
have

$\displaystyle
Sf(x)=\sup_{\lambda\in(0,1)}\inf_{x_{1}\in(\alpha+2x,\beta)}[\lambda
f(x_{1})+(1-\lambda)f(x_{1}-2x)].$ (2.28)

By $b>c>a$, if $f^{-1}(a)=[\mu,\nu]$, then $\alpha<\mu\leq\nu<\beta$, thus $f$
is strictly decreasing on $(\alpha,\mu]$ and strictly increasing on
$[\nu,\beta)$.

######  Claim 2.

For a fixed $\beta^{\prime}\in(\nu,\beta)\cap(\alpha+2x,\beta)$, there exists
$\delta>0$ such that function

$\displaystyle G_{x}(x_{1}):=\lambda f(x_{1})+(1-\lambda)f(x_{1}-2x)$ (2.29)

is decreasing on $(\alpha+2x,\beta^{\prime}]$ for any $0<\lambda<\delta$.

Proof of Claim 2. For $x_{1}\in(\alpha+2x,\beta^{\prime}]$, the right
derivative of $G_{x}(x_{1})$

$\displaystyle G_{xr}^{\prime}(x_{1})$ $\displaystyle=$ $\displaystyle\lambda
f_{r}^{\prime}(x_{1})+(1-\lambda)f_{r}^{\prime}(x_{1}-2x)$ (2.30)
$\displaystyle\leq$ $\displaystyle\lambda
f_{r}^{\prime}(\beta^{\prime})+(1-\lambda)f_{r}^{\prime}(\beta^{\prime}-2x),$

where the inequality is by the right derivative of a convex function is
increasing on the interior of its effective domain. Since
$\beta^{\prime}\in(\nu,\beta)\cap(\alpha+2x,\beta)$ and
$x\in[\frac{\beta-\gamma}{2},\frac{\beta-\alpha}{2})$, then
$\beta^{\prime}-2x\in(\alpha,\gamma+\beta^{\prime}-\beta)$, thus
$f_{r}^{\prime}(\beta^{\prime})>0$ and $f_{r}^{\prime}(\beta^{\prime}-2x)<0$
for $f$ is strictly increasing on $(\nu,\beta)$ and strictly decreasing on
$(\alpha,\gamma]$. Thus, by (2.30), we choose

$\displaystyle\delta=\frac{-f_{r}^{\prime}(\beta^{\prime}-2x)}{f_{r}^{\prime}(\beta^{\prime})-f_{r}^{\prime}(\beta^{\prime}-2x)},$
(2.31)

then $G_{xr}^{\prime}(x_{1})<0$ on $(\alpha+2x,\beta^{\prime}]$ for any
$\lambda\in(0,\delta)$. Therefore, $G_{x}(x_{1})$ is decreasing on
$(\alpha+2x,\beta^{\prime}]$ for any $\lambda\in(0,\delta)$. $\Box$
By (2.28) and Claim 2, we have that

$\displaystyle Sf(x)$ $\displaystyle=$
$\displaystyle\sup_{\lambda\in(0,1)}\inf_{x_{1}\in(\alpha+2x,\beta)}[\lambda
f(x_{1})+(1-\lambda)f(x_{1}-2x)]$ (2.32) $\displaystyle\geq$
$\displaystyle\sup_{\lambda\in(0,\delta)}\inf_{x_{1}\in(\alpha+2x,\beta)}[\lambda
f(x_{1})+(1-\lambda)f(x_{1}-2x)]$ $\displaystyle=$
$\displaystyle\sup_{\lambda\in(0,\delta)}\inf_{x_{1}\in[\beta^{\prime},\beta)}[\lambda
f(x_{1})+(1-\lambda)f(x_{1}-2x)]$ $\displaystyle\geq$
$\displaystyle\sup_{\lambda\in(0,\delta)}[\lambda
f(\beta^{\prime})+(1-\lambda)f(\beta-2x)]$ $\displaystyle=$ $\displaystyle
f(\beta-2x),$

where the second inequality is by
$x_{1}\in[\beta^{\prime},\beta)\subset(\nu,\beta)$ and $\beta^{\prime}-2x\leq
x_{1}-2x<\beta-2x\leq\gamma$ and $f$ is strictly increasing on $(\nu,\beta)$
and strictly decreasing on $(\alpha,\gamma]$, and the last equality is by
$f(\beta-2x)\geq f(\beta^{\prime})$.

Step 2. We shall prove that $Sf$ is convex in $\mathbb{R}$. Since $Sf$ is
increasing on $[0,\frac{\beta-\alpha}{2})$ and $Sf$ is even on
$(-\frac{\beta-\alpha}{2},\frac{\beta-\alpha}{2})$. Thus, it suffices to prove
$Sf$ is convex in $[0,\frac{\beta-\alpha}{2})$. For any
$x_{1},x_{2}\in[\frac{\beta-\gamma}{2},\frac{\beta-\alpha}{2})$ and
$\lambda\in(0,1)$, by (2.25) and $f$ is convex function, then

$\displaystyle\lambda Sf(x_{1})+(1-\lambda)Sf(x_{2})$ $\displaystyle=$
$\displaystyle\lambda f(\beta-2x_{1})+(1-\lambda)f(\beta-2x_{2})$ (2.33)
$\displaystyle\geq$ $\displaystyle f(\beta-2(\lambda x_{1}+(1-\lambda)x_{2}))$
$\displaystyle=$ $\displaystyle Sf(\lambda x_{1}+(1-\lambda)x_{2}),$

where the last equality is by $\lambda
x_{1}+(1-\lambda)x_{2}\in[\frac{\beta-\gamma}{2},\frac{\beta-\alpha}{2})$. By
(2.33), $Sf$ is convex on $[\frac{\beta-\gamma}{2},\frac{\beta-\alpha}{2})$.
Because that $Sf$ is convex in $[0,\frac{\beta-\gamma}{2}]$ by the proof in
Case (1), it suffices to prove that the left derivative of $Sf$ at
$x=\frac{\beta-\gamma}{2}$ is less than its right derivative at
$x=\frac{\beta-\gamma}{2}$.

By (2.25), we have

$\displaystyle Sf_{r}^{\prime}(\frac{\beta-\gamma}{2})$ $\displaystyle=$
$\displaystyle\lim_{t\rightarrow
0,t>0}\frac{Sf(\frac{\beta-\gamma}{2}+t)-Sf(\frac{\beta-\gamma}{2})}{t}$
(2.34) $\displaystyle=$ $\displaystyle\lim_{t\rightarrow
0,t>0}\frac{f(\gamma-2t)-f(\gamma)}{t}=-2f_{l}^{\prime}(\gamma).$

For any $t\in(-\frac{\beta-\gamma}{2},0)$, we have
$\frac{\beta-\gamma}{2}+t\in(0,\frac{\beta-\gamma}{2})$. Thus there exist
$x^{\prime},\;x^{\prime\prime}\in(\gamma,\beta)$ such that
$x^{\prime\prime}-x^{\prime}=2(\frac{\beta-\gamma}{2}+t)$ and
$Sf(\frac{\beta-\gamma}{2}+t)=f(x^{\prime})=f(x^{\prime\prime})$. Since

$\displaystyle(x^{\prime}-\gamma)+2(\frac{\beta-\gamma}{2}+t)=(x^{\prime}-\gamma)+(x^{\prime\prime}-x^{\prime})=x^{\prime\prime}-\gamma<\beta-\gamma,$

$x^{\prime}<\gamma-2t$. Let $|t|$ be sufficiently small such that
$\gamma+2|t|<\mu$, where $\mu$ satisfies $f^{-1}(a)=[\mu,\nu]$, then
$f(x^{\prime})>f(\gamma-2t)$ for $f$ is strictly decreasing on $(\gamma,\mu)$.
Then

$\displaystyle Sf_{l}^{\prime}(\frac{\beta-\gamma}{2})$ $\displaystyle=$
$\displaystyle\lim_{t\rightarrow
0,t<0}\frac{Sf(\frac{\beta-\gamma}{2}+t)-Sf(\frac{\beta-\gamma}{2})}{t}=\lim_{t\rightarrow
0,t<0}\frac{f(x^{\prime})-f(\gamma)}{t}$ (2.35) $\displaystyle\leq$
$\displaystyle\lim_{t\rightarrow
0,t<0}\frac{f(\gamma-2t)-f(\gamma)}{t}=-2f_{r}^{\prime}(\gamma).$

Since $f$ is a convex function, then $f_{l}^{\prime}(\gamma)\leq
f_{r}^{\prime}(\gamma)$, by (2.34) and (2.35), we have
$Sf_{l}^{\prime}(\frac{\beta-\gamma}{2})\leq
Sf_{r}^{\prime}(\frac{\beta-\gamma}{2})$.

Step 3. Proof of $Vol_{1}([Sf\leq a])=Vol([f\leq s])$ for any
$s\in\mathbb{R}$.

If $s<c$, the proof is the same as in Case (1).

If $c\leq s<b$, since $f$ is strictly decreasing on $(\alpha,\gamma)$, there
is a unique $x^{\prime}\in(\alpha,\gamma)$ such that $f(x^{\prime})=s$, thus
$[f<s]=[x^{\prime},\beta)$. By (2.25), we have
$Sf(\frac{\beta-x^{\prime}}{2})=f(x^{\prime})=s$, thus
$[Sf<s]=[-\frac{\beta-x^{\prime}}{2},\frac{\beta-x^{\prime}}{2}]$. Therefore,
$Vol_{1}([Sf<s])=Vol_{1}([f<s])=\beta-x^{\prime}$.

If $s\geq b$, then $b<+\infty$ for $s\in\mathbb{R}$, we have
$Vol_{1}([Sf<s])=Vol_{1}([f<s])=\beta-\alpha$. $\Box$

######  Remark.

1) By Theorem 1, for any $x\in\mathbb{R}$, if $x=0$, then $Sf(0)=\inf f$; if
$x\neq 0$, then there exist three cases:

  * i)

$Sf(x)=f(x^{\prime})=f(x^{\prime}-2|x|)$ for some $x^{\prime}\in\mathbb{R}$;

  * ii)

$Sf(x)=f(x_{0}-2|x|)$ for some $x_{0}\in\mathbb{R}$;

  * iii)

$S_{u}f(x)=f(x_{0}+2|x|)$ for some $x_{0}\in\mathbb{R}$.

2) In Theorem 1, there exist three cases for ${\rm dom}Sf$: i) ${\rm
dom}Sf=(-\delta,\delta)$; ii) ${\rm dom}Sf=[-\delta,\delta]$; iii) ${\rm
dom}Sf=\mathbb{R}$. ${\rm dom}Sf=(-\delta,\delta)$ is corresponding to ${\rm
dom}f=(\alpha,\beta)$, ${\rm dom}f=(\alpha,\beta]$ or ${\rm
dom}f=[\alpha,\beta)$, where $\delta=\frac{\beta-\alpha}{2}$. ${\rm
dom}Sf=[-\delta,\delta]$ is corresponding to ${\rm dom}f=[\alpha,\beta]$.
${\rm dom}Sf=\mathbb{R}$ is corresponding to ${\rm dom}f=(-\infty,\beta)$,
${\rm dom}f=(-\infty,\beta]$, ${\rm dom}f=(\alpha,+\infty)$, ${\rm
dom}f=[\alpha,+\infty)$ or ${\rm dom}f=\mathbb{R}$. For a non-empty convex set
$K\subset\mathbb{R}^{n}$ and a hyperplane $H=u^{\perp}$, where $u\in S^{n-1}$,
the Steiner symmetrization $S_{H}K$ of $K$ about $H$ is defined as

$S_{H}K=\\{x^{\prime}+\frac{1}{2}(t_{1}-t_{2})u:\;x^{\prime}\in
P_{H}(K),\;t_{i}\in I_{K}(x^{\prime})\;{\rm for}\;i=1,2\\},$

where $P_{H}(K)=\\{x^{\prime}\in H:\;x^{\prime}+tu\in K\;{\rm
for\;some}\;t\in\mathbb{R}\\}$ is the projection of $K$ onto $H$ and
$I_{K}(x^{\prime})=\\{t\in\mathbb{R}:\;x^{\prime}+tu\in K\\}$. By the above
definition and Definition 3, for coercive convex function
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ and its Steiner
symmetrization $S_{u}f$, we have

$\displaystyle{\rm dom}(S_{u^{\perp}}f)=S_{u^{\perp}}({\rm dom}f).$ (2.36)

We know that ${\rm dom}f$ is convex if $f$ is convex and the Steiner
symmetrization of a non-empty convex set is still a convex set, thus by
(2.36), ${\rm dom}(S_{u^{\perp}}f)$ is a convex set.

3) For a convex function
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$, the epigraph of $f$
is defined as ${\rm epi}f:=\\{(x,y)\in\mathbb{R}^{n+1}:x\in{\rm dom}f,\;y\geq
f(x)\\}$. By the definition of epigraph and Theorem 1, for one-dimensional
coercive convex function $f:\mathbb{R}\rightarrow\mathbb{R}\cup\\{+\infty\\}$,
we have ${\rm cl}({\rm epi}Sf)=S_{e^{\perp}}({\rm cl}({\rm epi}f))$, where $e$
is a unit vector along the $x$-axis and ${\rm cl}A$ denotes the closure of a
subset $A\subset\mathbb{R}^{n}$. Let
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ be a coercive and
convex function and $u\in S^{n-1}$. For any $x^{\prime}\in u^{\perp}$ and
$t\in\mathbb{R}$, if $\tilde{f}(t)=f(x^{\prime}+tu)$ is considered as a one-
dimensional function about $t$, then $S\tilde{f}(t)=S_{u}f(x^{\prime}+tu)$. By
Theorem 1, ${\rm cl}({\rm epi}(S\tilde{f}))=S_{e^{\perp}}({\rm cl}({\rm
epi}\tilde{f}))$. Since $x^{\prime}\in u^{\perp}$ is arbitrary, thus we have

$\displaystyle{\rm cl}({\rm epi}(S_{u}f))=S_{\tilde{u}^{\perp}}({\rm cl}({\rm
epi}f)),$ (2.37)

where $\tilde{u}^{\perp}\subset\mathbb{R}^{n+1}$ denotes the hyperplane
through the origin and orthogonal to the unit vector
$\tilde{u}=(u,0)\in\mathbb{R}^{n+1}$.

Next, by Definition 3 and Theorem 1, we shall prove five propositions which
are corresponding to properties 1-5 in Table 1.

The following lemma is an obvious fact, and we omit its proof.

######  Lemma 2.

For $f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$, let $u\in
S^{n-1}$ and $H=u^{\perp}$, if

  * i)

$f$ is symmetric with respect to hyperplane $H$, i.e., for any $x^{\prime}\in
H$ and $t\in\mathbb{R}$, $f(x^{\prime}+tu)=f(x^{\prime}-tu)$;

  * ii)

for any $x^{\prime}\in H$ and $t_{1}$, $t_{2}\in\mathbb{R}$, if
$|t_{1}|\leq|t_{2}|$, then $f(x^{\prime}+t_{1}u)\leq f(x^{\prime}+t_{2}u)$;

  * iii)

$f$ is convex on half-space $H^{+}:=\\{x^{\prime}+tu:x^{\prime}\in
u^{\perp},t\geq 0\\}$.

Then $f$ is a convex function on $\mathbb{R}^{n}$.

######  Proposition 2.1.

If $f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ is a coercive
convex function and $u\in S^{n-1}$, then $S_{u}f$ is a coercive convex
function and symmetric about $u^{\perp}$.

###### Proof.

It is clear that $S_{u}f$ is symmetric about $u^{\perp}$. Indeed, for any
$x^{\prime}\in u^{\perp}$ and $t\in\mathbb{R}$, if we consider
$S_{u}f(x^{\prime}+tu)$ as a one-dimensional function about $t$, then by
Theorem 1 and Definition 3, we have
$S_{u}f(x^{\prime}+tu)=S_{u}f(x^{\prime}-tu)$.

Step 1. We shall prove that $S_{u}f$ is coercive.

Suppose that there exist $M_{0}>0$ and a sequence
$\\{x_{n}\\}_{n=1}^{\infty}\subset\mathbb{R}^{n}$ satisfying that $|x_{n}|>n$
and $S_{u}f(x_{n})<M_{0}$. Next, we shall construct a sequence $\\{y_{n}\\}$
satisfying $|y_{n}|>n$ and $f(y_{n})<M_{0}$, which is contradictory with $f$
is coercive.

For any positive integer $n\geq 1$, let $x_{n}=x_{n}^{\prime}+t_{n}u$ and
$x_{n}^{\prime}\in u^{\perp}$. There exist two cases of $t_{n}\neq 0$ and
$t_{n}=0$.

(1) If $t_{n}\neq 0$, then by Theorem 1, there exist three cases:

i)
$S_{u}f(x_{n})=f(x_{n}^{\prime}+t_{n}^{\prime}u)=f(x_{n}^{\prime}+(t_{n}^{\prime}-2t_{n})u)$
for some $t_{n}^{\prime}\in\mathbb{R}$;

ii) $S_{u}f(x_{n})=f(x_{n}^{\prime}+(t_{0}-2t_{n})u)$ for some
$t_{0}\in\mathbb{R}$;

iii) $S_{u}f(x_{n})=f(x_{n}^{\prime}+(t_{0}+2t_{n})u)$ for some
$t_{0}\in\mathbb{R}$.

For case i), since $|t_{n}^{\prime}|+|t_{n}^{\prime}-2t_{n}|\geq 2|t_{n}|$,
then either $|t_{n}^{\prime}|\geq|t_{n}|$ or
$|t_{n}^{\prime}-2t_{n}|\geq|t_{n}|$. If $|t_{n}^{\prime}|\geq|t_{n}|$, let
$y_{n}=x_{n}^{\prime}+t_{n}^{\prime}u$, then $S_{u}f(x_{n})=f(y_{n})$ and
$|y_{n}|=|x_{n}^{\prime}|+|t_{n}^{\prime}|\geq|x_{n}^{\prime}|+|t_{n}|=|x_{n}|$.
If $|t_{n}^{\prime}-2t_{n}|\geq|t_{n}|$, let
$y_{n}=x_{n}^{\prime}+(t_{n}^{\prime}-2t_{n})u$, then $S_{u}f(x_{n})=f(y_{n})$
and
$|y_{n}|=|x_{n}^{\prime}|+|t_{n}^{\prime}-2t_{n}|\geq|x_{n}^{\prime}|+|t_{n}|=|x_{n}|$.
Since $|x_{n}|>n$, we have $|y_{n}|>n$ and $f(y_{n})=S_{u}f(x_{n})<M_{0}$.

For case ii), since $|t_{0}|+|t_{0}-2t_{n}|\geq 2|t_{n}|$, we have either
$|t_{0}|\geq|t_{n}|$ or $|t_{0}-2t_{n}|\geq|t_{n}|$. If
$|t_{0}-2t_{n}|\geq|t_{n}|$, let $y_{n}=x_{n}^{\prime}+(t_{0}-2t_{n})u$, then
$S_{u}f(x_{n})=f(y_{n})$ and $|y_{n}|\geq|x_{n}|$. If $|t_{0}|\geq|t_{n}|$,
let $y_{n}=x_{n}^{\prime}+t_{0}u$ if $x_{n}^{\prime}+t_{0}u\in{\rm dom}f$,
otherwise let $y_{n}=x_{n}^{\prime}+t_{0}^{\prime}u$, where $t_{0}^{\prime}$
satisfies $x_{n}^{\prime}+t_{0}^{\prime}u\in{\rm dom}f$,
$|x_{n}^{\prime}+t_{0}^{\prime}u|>n$ and
$f(x_{n}^{\prime}+t_{0}^{\prime}u)<f(x_{n}^{\prime}+(t_{0}-2t_{n})u)$, which
can be satisfied for $\lim_{t\rightarrow
t_{0},\;t<t_{0}}f(x_{n}^{\prime}+tu)\leq f(x_{n}^{\prime}+(t_{0}-2t_{n})u)$ by
Theorem 1. Thus, we have $|y_{n}|>n$ and $f(y_{n})<M_{0}$.

For case iii), we can construct $\\{y_{n}\\}$ with the same method as in case
(ii).

(2) If $t_{n}=0$, by Definition 3, we have
$Sf(x_{n})=\inf_{t\in\mathbb{R}}f(x_{n}^{\prime}+tu)$. Since
$S_{u}f(x_{n})<M_{0}$, there exists $y_{n}=x_{n}^{\prime}+t^{\prime}u$ such
that $f(y_{n})<M_{0}$. Since
$|y_{n}|=|x_{n}^{\prime}|+|t^{\prime}|\geq|x_{n}^{\prime}|=|x_{n}|$, we have
$|y_{n}|>n$ and $f(y_{n})<M_{0}$.

Step 2. We shall prove that $S_{u}f$ is convex.

######  Claim 3.

$S_{u}f$ is proper, i.e., $[S_{u}f=+\infty]\neq\mathbb{R}^{n}$ and
$[S_{u}f=-\infty]=\emptyset$.

Proof of Claim 3. For any $x\in\mathbb{R}^{n}$, let $x=x^{\prime}+tu$, where
$x^{\prime}\in u^{\perp}$. Since $f$ is a coercive convex function defined on
$\mathbb{R}^{n}$, one dimensional function $f(x^{\prime}+tu)$ about
$t\in\mathbb{R}$ either is a coercive convex function or is identically
$+\infty$. If $f(x^{\prime}+tu)$ is a coercive convex function, then there
exists $s\in\mathbb{R}$ such that
$s=\inf\\{f(x^{\prime}+tu):t\in\mathbb{R}\\}$. Thus, we have

$Sf(x)=\sup_{\lambda\in[0,1]}\inf_{t_{1}+t_{2}=t}[\lambda
f(x^{\prime}+2t_{1}u)+(1-\lambda)f(x^{\prime}-2t_{2}u)]\geq s,$

which implies that $Sf(x)>-\infty$. If $f(x^{\prime}+tu)$ is identically
$+\infty$, then $S_{u}f(x)=+\infty>-\infty$. By the definition of convex
functions, $f$ is not identically $+\infty$, there exists $x\in\mathbb{R}^{n}$
such that $f(x)<+\infty$. Let $x=x_{0}+tu$, where $x_{0}\in u^{\perp}$, then

$S_{u}f(x_{0})=\inf_{t_{1}\in\mathbb{R}}f(x_{0}+t_{1}u)\leq f(x)<+\infty,$

which implies that $S_{u}f$ is not identically $+\infty$. $\Box$

By Definition 3 and Theorem 1, for any $x^{\prime}\in u^{\perp}$, one-
dimensional function $S_{u}f(x^{\prime}+tu)$ is either an even and coercive
convex function about $t\in\mathbb{R}$ or identically $+\infty$. Thus,
$S_{u}f$ satisfies conditions i) and ii) in Lemma 2. Therefore, to prove that
$S_{u}f$ is convex, it suffices to prove that $S_{u}f$ satisfies condition
iii) of Lemma 2. For any $x$, $y\in\\{x^{\prime}+tu:x^{\prime}\in
u^{\perp},t\geq 0\\}$ and $\lambda\in(0,1)$, if $x\notin{\rm dom}(S_{u}f)$ or
$y\notin{\rm dom}(S_{u}f)$, then $S_{u}f(x)=+\infty$ or $S_{u}f(y)=+\infty$,
thus

$\displaystyle S_{u}f(\lambda x+(1-\lambda)y)\leq\lambda
S_{u}f(x)+(1-\lambda)S_{u}f(y).$ (2.38)

By Remark 2), ${\rm dom}(S_{u}f)$ is convex. Therefore, if $x\in{\rm
dom}(S_{u}f)$ and $y\in{\rm dom}(S_{u}f)$, then $\lambda x+(1-\lambda)y\in{\rm
dom}(S_{u}f)$. Let $x=x^{\prime}+tu$ and $y=y^{\prime}+su$, where
$x^{\prime}$, $y^{\prime}\in u^{\perp}$ and $t\geq 0$ and $s\geq 0$, then
$\lambda x+(1-\lambda)y=[\lambda x^{\prime}+(1-\lambda)y^{\prime}]+[\lambda
t+(1-\lambda)s]u$.

Case 3.1. The case of $t=0$ and $s=0$. For the case we have $x$, $y\in
u^{\perp}$, thus $\lambda x+(1-\lambda)y\in u^{\perp}$. By Definition 3 and
$f$ is convex, we have

$\displaystyle\lambda S_{u}f(x)+(1-\lambda)S_{u}f(y)$ (2.39) $\displaystyle=$
$\displaystyle\lambda\inf_{t\in\mathbb{R}}f(x+tu)+(1-\lambda)\inf_{s\in\mathbb{R}}f(y+su)$
$\displaystyle=$ $\displaystyle\inf_{(t,s)\in\mathbb{R}^{2}}[\lambda
f(x+tu)+(1-\lambda)f(y+su)]$ $\displaystyle\geq$
$\displaystyle\inf_{(t,s)\in\mathbb{R}^{2}}f(\lambda x+(1-\lambda)y+(\lambda
t+(1-\lambda)s)u)$ $\displaystyle=$ $\displaystyle S_{u}f(\lambda
x+(1-\lambda)y).$

Case 3.2. The case of $t>0$ and $s>0$.

For $x=x^{\prime}+tu\in{\rm dom}(S_{u}f)$, by Theorem 1, there exist three
cases:

$a_{1}$) There exists some $t^{\prime}\in\mathbb{R}$ such that

$\displaystyle
S_{u}f(x)=f(x^{\prime}+t^{\prime}u)=f(x^{\prime}+(t^{\prime}-2t)u);$ (2.40)

$a_{2}$) There exists some $t_{0}\in\mathbb{R}$ such that

$\displaystyle
S_{u}f(x)=f(x^{\prime}+(t_{0}-2t)u)\geq\lim_{t_{0}^{\prime}\rightarrow
t_{0},t_{0}^{\prime}<t_{0}}f(x^{\prime}+t_{0}^{\prime}u);$ (2.41)

$a_{3}$) There exists some $t_{0}\in\mathbb{R}$ such that

$\displaystyle
S_{u}f(x)=f(x^{\prime}+(t_{0}+2t)u)\geq\lim_{t_{0}^{\prime}\rightarrow
t_{0},t_{0}^{\prime}>t_{0}}f(x^{\prime}+t_{0}^{\prime}u).$ (2.42)

For $y=y^{\prime}+su\in{\rm dom}(S_{u}f)$, by Theorem 1, there exist three
cases:

$b_{1}$) There exists some $s^{\prime}\in\mathbb{R}$ such that

$\displaystyle
S_{u}f(y)=f(y^{\prime}+s^{\prime}u)=f(y^{\prime}+(s^{\prime}-2s)u);$ (2.43)

$b_{2}$) There exists some $s_{0}\in\mathbb{R}$ such that

$\displaystyle
S_{u}f(y)=f(y^{\prime}+(s_{0}-2s)u)\geq\lim_{s_{0}^{\prime}\rightarrow
s_{0},s_{0}^{\prime}<s_{0}}f(y^{\prime}+s_{0}^{\prime}u);$ (2.44)

$b_{3}$) There exists some $s_{0}\in\mathbb{R}$ such that

$\displaystyle
S_{u}f(y)=f(y^{\prime}+(s_{0}+2s)u)\geq\lim_{s_{0}^{\prime}\rightarrow
s_{0},s_{0}^{\prime}>s_{0}}f(y^{\prime}+s_{0}^{\prime}u).$ (2.45)

We may assume that

$\displaystyle f(x^{\prime}+t_{0}u)=\lim_{t_{0}^{\prime}\rightarrow
t_{0},t_{0}^{\prime}<t_{0}}f(x^{\prime}+t_{0}^{\prime}u)\;\;\;{\rm
for\;\;case}\;\;a_{2}),$ $\displaystyle
f(x^{\prime}+t_{0}u)=\lim_{t_{0}^{\prime}\rightarrow
t_{0},t_{0}^{\prime}>t_{0}}f(x^{\prime}+t_{0}^{\prime}u)\;\;\;{\rm
for\;\;case}\;\;a_{3}),$ $\displaystyle
f(y^{\prime}+s_{0}u)=\lim_{s_{0}^{\prime}\rightarrow
s_{0},s_{0}^{\prime}<s_{0}}f(y^{\prime}+s_{0}^{\prime}u)\;\;\;{\rm
for\;\;case}\;\;b_{2}),$ $\displaystyle
f(y^{\prime}+s_{0}u)=\lim_{s_{0}^{\prime}\rightarrow
s_{0},s_{0}^{\prime}>s_{0}}f(y^{\prime}+s_{0}^{\prime}u)\;\;\;{\rm
for\;\;case}\;\;b_{3}).$ (2.46)

Let $(\tilde{t}_{1},\tilde{t}_{2})$ be a pair of real numbers satisfying

$(\tilde{t}_{1},\tilde{t}_{2})=\left\\{\begin{aligned}
(t^{\prime}-2t,t^{\prime})\;\;\;{\rm for\;\;case}\;\;a_{1})\\\
(t_{0}-2t,t_{0})\;\;\;{\rm for\;\;case}\;\;a_{2})\\\
(t_{0},t_{0}+2t)\;\;\;{\rm for\;\;case}\;\;a_{3}).\end{aligned}\right.$ (2.47)

Let $(\tilde{s}_{1},\tilde{s}_{2})$ be a pair of real numbers satisfying

$(\tilde{s}_{1},\tilde{s}_{2})=\left\\{\begin{aligned}
(s^{\prime}-2s,s^{\prime})\;\;\;{\rm for\;\;case}\;\;b_{1})\\\
(s_{0}-2s,s_{0})\;\;\;{\rm for\;\;case}\;\;b_{2})\\\
(s_{0},s_{0}+2s)\;\;\;{\rm for\;\;case}\;\;b_{3}).\end{aligned}\right.$ (2.48)

Since $f$ is convex and by (2.40-2.45), for $i=1,2$, we have

$\displaystyle\lambda S_{u}f(x)+(1-\lambda)S_{u}f(y)$ (2.49)
$\displaystyle\geq$ $\displaystyle\lambda
f(x^{\prime}+\tilde{t}_{i}u)+(1-\lambda)f(y^{\prime}+\tilde{s}_{i}u)$
$\displaystyle\geq$ $\displaystyle f(\lambda
x^{\prime}+(1-\lambda)y^{\prime}+(\lambda\tilde{t}_{i}+(1-\lambda)\tilde{s}_{i})u).$

By (2.47) and (2.48), we have

$\displaystyle[\lambda\tilde{t}_{2}+(1-\lambda)\tilde{s}_{2}]-[\lambda\tilde{t}_{1}+(1-\lambda)\tilde{s}_{1}]$
(2.50) $\displaystyle=$
$\displaystyle\lambda(\tilde{t}_{2}-\tilde{t}_{1})+(1-\lambda)(\tilde{s}_{2}-\tilde{s}_{1})=2[\lambda
t+(1-\lambda)s].$

By $\lambda x+(1-\lambda)y=\lambda x^{\prime}+(1-\lambda)y^{\prime}+(\lambda
t+(1-\lambda)s)u$ and Definition 3, we have

$\displaystyle S_{u}f(\lambda x+(1-\lambda)y)$ (2.51) $\displaystyle=$
$\displaystyle\sup_{\delta\in[0,1]}\inf_{\omega\in\mathbb{R}}\left[\delta
f(\lambda x^{\prime}+(1-\lambda)y^{\prime}+\omega u)\right.$
$\displaystyle\left.+(1-\delta)f(\lambda
x^{\prime}+(1-\lambda)y^{\prime}+(\omega-2(\lambda t+(1-\lambda)s)u)\right]$
$\displaystyle\leq$ $\displaystyle\sup_{\delta\in[0,1]}\left[\delta f(\lambda
x^{\prime}+(1-\lambda)y^{\prime}+(\lambda\tilde{t}_{2}+(1-\lambda)\tilde{s}_{2})u)\right.$
$\displaystyle\left.+(1-\delta)f(\lambda
x^{\prime}+(1-\lambda)y^{\prime}+(\lambda\tilde{t}_{1}+(1-\lambda)\tilde{s}_{1})u)\right]$
$\displaystyle\leq$ $\displaystyle\max_{i=1,2}f(\lambda
x^{\prime}+(1-\lambda)y^{\prime}+(\lambda\tilde{t}_{i}+(1-\lambda)\tilde{s}_{i})u)$
$\displaystyle\leq$ $\displaystyle\lambda S_{u}f(x)+(1-\lambda)S_{u}f(y),$

where the first inequality is by choosing
$\omega=\lambda\tilde{t}_{2}+(1-\lambda)\tilde{s}_{2}$ and (2.50), and the
last inequality is by (2.49).

Case 3.3. The case of $t=0$ and $s>0$ (or $t>0$ and $s=0$). In this case,
there exists $t_{0}$ such that

$\displaystyle S_{u}f(x)=\lim_{t\rightarrow t_{0},\;x+tu\in{\rm
dom}f}f(x+tu).$ (2.52)

We may assume that

$\displaystyle f(x+t_{0}u)=\lim_{t\rightarrow t_{0},\;x+tu\in{\rm
dom}f}f(x+tu).$ (2.53)

In the proof of Case 3.2, let $\tilde{t}_{1}=\tilde{t}_{2}=t_{0}$, we can get
the required inequality. ∎

######  Proposition 2.2.

Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ be a coercive
convex function and $u\in S^{n-1}$, then

$\displaystyle\int_{\mathbb{R}^{n}}e^{-(S_{u}f)(x)}dx=\int_{\mathbb{R}^{n}}e^{-f(x)}dx.$
(2.54)

###### Proof.

By (2.37), for any $t\in\mathbb{R}$, we have ${\rm cl}[S_{u}f<t]=S_{u}({\rm
cl}[f<t])$. Since Steiner symmetrization of convex sets preserves volume,
$Vol([S_{u}f<t])=Vol([f<t])$. By Fubini’s theorem, we have

$\displaystyle\int_{\mathbb{R}^{n}}e^{-(S_{u}f)(x)}dx$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}}Vol([S_{u}f<t])e^{-t}dt$ (2.55)
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}}Vol([f<t])e^{-t}dt=\int_{\mathbb{R}^{n}}e^{-f(x)}dx.$

∎

######  Lemma 3.

Let $u_{1},u_{2}\in S^{n-1}$ and $\langle u_{1},u_{2}\rangle=0$. If
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ is a coercive convex
function and $f$ is symmetric about $u_{1}^{\perp}$, then $S_{u_{2}}f$ is
symmetric about both $u_{1}^{\perp}$ and $u_{2}^{\perp}$.

###### Proof.

By Proposition 2.1, $S_{u_{2}}f$ is symmetric about $u_{2}^{\perp}$. Next, we
prove that $S_{u_{2}}f$ is symmetric about $u_{1}^{\perp}$. Since $\langle
u_{1},u_{2}\rangle=0$, then $u_{1}\in u_{2}^{\perp}$ and $u_{2}\in
u_{1}^{\perp}$. For any $x^{\prime}\in u_{1}^{\perp}$, let
$x^{\prime}=x^{\prime\prime}+t_{x^{\prime}}u_{2}$, where
$x^{\prime\prime}=x^{\prime}|u_{2}^{\perp}$. Then
$x^{\prime\prime}=x^{\prime}-t_{x^{\prime}}u_{2}\in u_{1}^{\perp}$, thus
$x^{\prime\prime}+tu_{2}\in u_{1}^{\perp}$. Because that $x^{\prime\prime}\in
u_{2}^{\perp}$ and $u_{1}\in u_{2}^{\perp}$, thus $x^{\prime\prime}+tu_{1}\in
u_{2}^{\perp}$. Thus, for any $x^{\prime}\in u_{1}^{\perp}$ and
$t\in\mathbb{R}$, we have

$\displaystyle(S_{u_{2}}f)(x^{\prime}+tu_{1})=(S_{u_{2}}f)(x^{\prime\prime}+tu_{1}+t_{x^{\prime}}u_{2})$
(2.56) $\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{t_{1}+t_{2}=t_{x^{\prime}}}[\lambda
f(x^{\prime\prime}+tu_{1}+2t_{1}u_{2})+(1-\lambda)f(x^{\prime\prime}+tu_{1}-2t_{2}u_{2})]$
$\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{t_{1}+t_{2}=t_{x^{\prime}}}[\lambda
f(x^{\prime\prime}-tu_{1}+2t_{1}u_{2})+(1-\lambda)f(x^{\prime\prime}-tu_{1}-2t_{2}u_{2})]$
$\displaystyle=$
$\displaystyle(S_{u_{2}}f)(x^{\prime\prime}-tu_{1}+t_{x^{\prime}}u_{2})$
$\displaystyle=$ $\displaystyle(S_{u_{2}}f)(x^{\prime}-tu_{1}),$

where the second equality is by $f$ is symmetric about $u_{1}^{\perp}$ and
$x^{\prime\prime}+tu_{2}\in u_{1}^{\perp}$. This completes the proof. ∎

We say that a function $f:\mathbb{R}^{n}\mapsto\mathbb{R}\cup\\{+\infty\\}$ is
unconditional if $f(x_{1},\dots,x_{n})=f(|x_{1}|,\dots,|x_{n}|)$ for every
$(x_{1},\dots,x_{n})\in\mathbb{R}^{n}$.

######  Proposition 2.3.

Any coercive convex function
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ can be transformed
into an unconditional function $\bar{f}$ using $n$ Steiner symmetrizations.

###### Proof.

Let $\\{u_{1},\dots,u_{n}\\}$ be an orthonormal basis of $\mathbb{R}^{n}$. By
Proposition 2.1 and Lemma 3, $S_{u_{n}}\cdots S_{u_{1}}f$ is symmetric about
$u_{i}^{\bot}$, $i=1,\cdots,n$, which implies that $f$ can be transformed into
an unconditional function $\bar{f}=S_{u_{n}}\cdots S_{u_{1}}f$ using $n$
Steiner symmetrizations. ∎

######  Proposition 2.4.

Let $f_{1}:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ and
$f_{2}:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ be coercive
convex functions and $u\in S^{n-1}$. If $f_{1}\leq f_{2}$ (which implies that
$f_{1}(x)\leq f_{2}(x)$ for any $x\in\mathbb{R}^{n}$), then $S_{u}f_{1}\leq
S_{u}f_{2}$.

###### Proof.

By Definition 3 and $f_{1}\leq f_{2}$, for $x=x^{\prime}+tu$, where
$x^{\prime}\in u^{\perp}$, we have

$\displaystyle S_{u}f_{1}(x)$ $\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{t_{1}+t_{2}=t}[\lambda
f_{1}(x^{\prime}+2t_{1}u)+(1-\lambda)f_{1}(x^{\prime}-2t_{2}u)]$ (2.57)
$\displaystyle\leq$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{t_{1}+t_{2}=t}[\lambda
f_{2}(x^{\prime}+2t_{1}u)+(1-\lambda)f_{2}(x^{\prime}-2t_{2}u)]$
$\displaystyle=$ $\displaystyle S_{u}f_{2}(x).$

∎

We say a function $f$ is even about point $z\in\mathbb{R}^{n}$ if
$f(z+x)=f(z-x)$ for any $x\in\mathbb{R}^{n}$. Let $z|H$ denote the projection
of $z$ onto hyperplane $H$.

######  Proposition 2.5.

Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ be a coercive
convex function and $u\in S^{n-1}$, if $f$ is even about $z$, then $S_{u}f$ is
even about $z|u^{\perp}$.

###### Proof.

For any $x\in\mathbb{R}^{n}$, let $x=x^{\prime}+tu$, where
$x^{\prime}=x|u^{\perp}$. Let $z=z^{\prime}-t_{0}u$, where
$z^{\prime}=z|u^{\perp}$. By Definition 3, we have

$\displaystyle(S_{u}f)(z^{\prime}+x)=(S_{u}f)(z^{\prime}+x^{\prime}+tu)=(S_{u}f)(z^{\prime}+x^{\prime}-tu)$
(2.58) $\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{t_{1}+t_{2}=-t}[\lambda
f(z^{\prime}+x^{\prime}+2t_{1}u)+(1-\lambda)f(z^{\prime}+x^{\prime}-2t_{2}u)]$
$\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{t_{2}\in\mathbb{R}}[\lambda
f(z+t_{0}u+x^{\prime}-2t_{2}u-2tu)+(1-\lambda)f(z+t_{0}u+x^{\prime}-2t_{2}u)]$
$\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{t_{2}\in\mathbb{R}}[\lambda
f(z+x^{\prime}-2t_{2}u-2tu)+(1-\lambda)f(z+x^{\prime}-2t_{2}u)]$
$\displaystyle=$
$\displaystyle\sup_{\lambda^{\prime}\in[0,1]}\inf_{t_{2}\in\mathbb{R}}[\lambda^{\prime}f(z+x^{\prime}-2t_{2}u)+(1-\lambda^{\prime})f(z+x^{\prime}-2t_{2}u-2tu)],$

where the second equality is by $S_{u}f$ is symmetric about $u^{\perp}$ and
the fifth equality is by replacing $t_{0}-2t_{2}$ by $-2t_{2}$.

On the other hand, since $f$ is even about $z$, we have

$\displaystyle(S_{u}f)(z^{\prime}-x)=(S_{u}f)(z^{\prime}-x^{\prime}-tu)$
(2.59) $\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{t_{1}+t_{2}=-t}[\lambda
f(z^{\prime}-x^{\prime}+2t_{1}u)+(1-\lambda)f(z^{\prime}-x^{\prime}-2t_{2}u)]$
$\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{t_{1}\in\mathbb{R}}[\lambda
f(z+t_{0}u-x^{\prime}+2t_{1}u)+(1-\lambda)f(z+t_{0}u-x^{\prime}+2t_{1}u+2tu)]$
$\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{t_{1}\in\mathbb{R}}[\lambda
f(z-x^{\prime}+2t_{1}u)+(1-\lambda)f(z-x^{\prime}+2t_{1}u+2tu)]$
$\displaystyle=$
$\displaystyle\sup_{\lambda\in[0,1]}\inf_{t_{1}\in\mathbb{R}}[\lambda
f(z+x^{\prime}-2t_{1}u)+(1-\lambda)f(z+x^{\prime}-2t_{1}u-2tu)],$

where the last equality is by $f$ is even about $z$. By (2.58) and (2.59), we
have $(S_{u}f)(z^{\prime}+x)=(S_{u}f)(z^{\prime}-x)$ for any
$x\in\mathbb{R}^{n}$. ∎

##  3\. The relation between new definition and former definitions

###  3.1.  The relation between Definition 3 and Definition 2.

The relation can be generalized as follows:

(i) $S_{u}f$ is in general larger than $\widetilde{S}_{u}f$ (look at Example
1).

(ii) For one-dimensional coercive convex function
$f:\mathbb{R}\rightarrow\mathbb{R}\cup\\{+\infty\\}$, if $f$ is symmetric
about an axes $x=x_{0}$, i.e., $f(x_{0}-x)=f(x_{0}+x)$ for any
$x\in\mathbb{R}$, then $Sf=\widetilde{S}f$.

(iii) For $n$-dimensional coercive convex function
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ and $u\in S^{n-1}$,
if for any $x^{\prime}\in u^{\perp}$, one-dimensional function
$f(x^{\prime}+tu)$ about $t\in\mathbb{R}$ is symmetric about an axes
$t=t_{0}$, then $S_{u}f=\widetilde{S}_{u}f$.

######  Example 1.

For one-dimensional coercive convex function

$f(x)=\left\\{\begin{aligned} x^{3}\;\;{\rm if}\;\;x\geq 0,\\\ x^{2}\;\;{\rm
if}\;\;x\leq 0.\end{aligned}\right.$ (3.1)

We compare $Sf$ with $\widetilde{S}f$, where

$Sf(x)=\sup_{\lambda\in[0,1]}\inf_{x_{1}+x_{2}=x}[\lambda
f(2x_{1})+(1-\lambda)f(-2x_{2})]$

and

$\widetilde{S}f(x)=\inf_{x_{1}+x_{2}=x}[\frac{1}{2}f(2x_{1})+\frac{1}{2}f(-2x_{2})].$

By calculation, we can get that

$\widetilde{S}f(x)=\left\\{\begin{aligned}
\frac{(-12x-1)\sqrt{1+12x}+18x+1}{27}+2x^{2}\;\;{\rm if}\;\;x\geq 0,\\\
\frac{(12x-1)\sqrt{1-12x}-18x+1}{27}+2x^{2}\;\;\;{\rm if}\;\;x\leq
0.\end{aligned}\right.$ (3.2)

and

$\displaystyle Sf(x)=g^{-1}(|x|),$ (3.3)

where $g^{-1}$ is the inverse function of

$\displaystyle g(x)=\frac{1}{2}(\sqrt[3]{x}+\sqrt{x}),\;\;x\in[0,\infty).$
(3.4)

By Matlab, we can draw their figures (see Figure 1). In the figure, we can
find that the level sets of $Sf$ and $f$ have the same size and
$Sf>\widetilde{S}f$.

###  3.2.  The relation between Definition 3 and Definition 1.

In this section, we show that the two definitions are same for log-concave
functions (Theorem 3.2).

######  Lemma 3.1.

Let $F=e^{-f}$ be a log-concave function, where
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is a coercive convex function, then
$[\bar{S}_{u}F>t]=S_{u}([F>t])$.

###### Proof.

By Definition 1, if $\bar{S}_{u}F(x)>t$, then $x\in S_{u}([F>t])$. On the
other hand, if $x\in S_{u}([F>t])$, since $S_{u}([F>t])$ is an open set and
$F$ is continuous, then there exists $t^{\prime}>t$ such that $x\in
S_{u}([F>t^{\prime}])$, by (1.1), we have $\bar{S}_{u}F(x)>t$. ∎

######  Theorem 3.2.

Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ be a coercive convex function and
$u\in S^{n-1}$, then $e^{(-S_{u}f)}=\bar{S}_{u}(e^{-f})$, where $S_{u}f$ and
$\bar{S}_{u}(e^{-f})$ are given in (1.3) and (1.1), respectively.

###### Proof.

For $t>0$, we have

$\displaystyle[e^{(-S_{u}f)}>t]=[S_{u}f<-\ln t]=S_{u}([f<-\ln
t])=S_{u}([e^{-f}>t]),$ (3.5)

where the second equality holds by (2.37).

By Lemma 3.1, we have $[\bar{S}_{u}(e^{-f})>t]=S_{u}([e^{-f}>t])$, thus
$[e^{(-S_{u}f)}>t]=[\bar{S}_{u}(e^{-f})>t]$. Using the “layer-cake
representation”, we have

$\displaystyle
e^{(-S_{u}f)}=\int_{0}^{\infty}\mathcal{X}_{[e^{(-S_{u}f)}>t]}(x)dt=\int_{0}^{\infty}\mathcal{X}_{[\bar{S}_{u}(e^{-f})>t]}(x)dt=\bar{S}_{u}(e^{-f}).$
(3.6)

∎

The continuity and convergence of Steiner symmetrization in $L^{p}$ space have
been proved in many papers [7, 8, 9, 14], especially Proposition 3 and Theorem
2 in [14] are corresponding to the properties 6-7 in Table 1.

##  4\. Application to functional Blaschke-Santaló inequality

We can use the new definition to prove some important inequalities, such as
functional Blaschke-Santaló inequality, Prékopa-Leindler inequality for log-
concave functions, Hardy-Littlewood inequality for log-concave functions, etc.
As an illustration, here we only use it to prove the functional Blaschke-
Santaló inequality for even convex functions.

For a convex body $K\subset\mathbb{R}^{n}$, its polar about $z$ is defined by
$K^{z}=\\{x\in\mathbb{R}^{n}:\sup_{y\in K}\langle x-z,y-z\rangle\leq 1\\}$.
For a log-concave function $f:\mathbb{R}^{n}\rightarrow[0,\infty)$, its polar
about $z$ is defined by

$\displaystyle f^{z}(x)=\inf_{y\in\mathbb{R}^{n}}\frac{e^{-\langle
x-z,y-z\rangle}}{f(y)}.$ (4.1)

To better understand this definition recall the classical Legendre transform:
For a function $\phi:\mathbb{R}^{n}\rightarrow\mathbb{R}$, its Legendre
transform about $z$ is defined by
$\mathcal{L}^{z}\phi(x)=\sup_{y\in\mathbb{R}^{n}}[\langle
x-z,y-z\rangle-\phi(y)]$. From above definition of polarity, if
$f(x)=e^{-\phi(x)}$, where $\phi(x)$ is a convex function, then
$f^{z}(x)=e^{-\mathcal{L}^{z}\phi(x)}$. Since
$\mathcal{L}^{z}(\mathcal{L}^{z}\phi)=\phi$ for a convex function $\phi$,
$(f^{z})^{z}=f$. For $z=0$, we denote $\mathcal{L}^{0}\phi=\mathcal{L}\phi$.

For a convex body $K$, its Santaló point $s(K)$ satisfies
$Vol(K^{s(K)})=\min_{z}Vol(K^{z})$. The Blaschke-Santaló inequality [3, 29]
states that $Vol(K)Vol(K^{s(K)})\leq Vol(B_{2}^{n})^{2}$, where
$B_{2}^{n}=\\{x\in\mathbb{R}^{n}:|x|\leq 1\\}$ is the Euclidean ball
($|\cdot|$ denote the Euclidean norm). The functional Blaschke-Santaló
inequality of log-concave functions is the analogue of Blaschke-Santaló
inequality of convex bodies. If $f$ is a nonnegative integrable function on
$\mathbb{R}^{n}$ such that $f^{0}$ has its barycenter at $0$, then

$\int_{\mathbb{R}^{n}}f(x)dx\int_{\mathbb{R}^{n}}f^{0}(y)dy\leq\left(\int_{\mathbb{R}^{n}}e^{-\frac{1}{2}|x|^{2}}dx\right)^{2}=(2\pi)^{n}.$

In the special case where the function $f$ is even, this result follows from
an earlier inequality of Ball [2]; and in [15], Fradelizi and Meyer prove
something more general (see also [22]). Recently, Lehec [23] gave a direct
proof of the functional Blaschke-Santaló inequality.

In this paper, inspired by the proof of K. Ball [2] for Santaló inequality for
centrally symmetric convex bodies, we prove functional Blaschke-Santaló
inequality for even convex functions. For the non-even case, we can prove the
inequality by the similar method, but we don’t prove it here.

######  Theorem 4.1.

(K. Ball, [2]) Let $f:\mathbb{R}^{n}\rightarrow[0,\infty)$ be an even convex
function. Assume that $0<\int e^{-f}<\infty$. Then

$\displaystyle\int e^{-f}\int e^{-\mathcal{L}f}\leq(2\pi)^{n}.$ (4.2)

First, we give the following lemmas.

######  Lemma 4.2.

Let $f:\mathbb{R}^{n}\rightarrow[0,\infty)$ be an even convex function and
$u\in S^{n-1}$. Assume that $0<\int e^{-f}<\infty$. Then

$\displaystyle\int e^{-\mathcal{L}f}\leq\int e^{-\mathcal{L}(S_{u}f)}.$ (4.3)

###### Proof.

After a linear transformation, it may be supposed that
$H=u^{\perp}=\\{(x_{i})_{i=1}^{i=n}:x_{n}=0\\}$. For $f$ and $t\in\mathbb{R}$,
we define a new function $f_{(t)}(x^{\prime}):=f(x^{\prime}+tu)$, where
$x^{\prime}\in H$.

By the definition of Steiner symmetrization, for
$x^{\prime}=x_{1}^{\prime}+x_{2}^{\prime}$, where $x^{\prime}$,
$x_{1}^{\prime}$ and $x_{2}^{\prime}\in H$, let $(x^{\prime},t)$ denote
$x^{\prime}+tu$, we have

$\displaystyle(\mathcal{L}(S_{u}f))_{(t)}(x^{\prime})=(\mathcal{L}(S_{u}f))(x^{\prime}+tu)$
(4.4) $\displaystyle=$ $\displaystyle\sup_{(y^{\prime},s)\in
H\times\mathbb{R}}\left[\langle(x^{\prime},t),(y^{\prime},s)\rangle-(S_{u}f)(y^{\prime}+su)\right]$
$\displaystyle=$ $\displaystyle\sup_{(y^{\prime},s)\in
H\times\mathbb{R}}[\langle(x^{\prime},t),(y^{\prime},s)\rangle-\sup_{\lambda\in[0,1]}\inf_{s_{1}+s_{2}=s}(\lambda
f(y^{\prime}+2s_{1}u)+(1-\lambda)f(y^{\prime}-2s_{2}u))]$ $\displaystyle=$
$\displaystyle\sup_{(y^{\prime},s)\in
H\times\mathbb{R}}\inf_{\lambda\in[0,1]}\sup_{s_{1}+s_{2}=s}[\langle(x^{\prime},t),(y^{\prime},s)\rangle-(\lambda
f(y^{\prime}+2s_{1}u)+(1-\lambda)f(y^{\prime}-2s_{2}u))]$ $\displaystyle\leq$
$\displaystyle\sup_{(y^{\prime},s)\in
H\times\mathbb{R}}\sup_{s_{1}+s_{2}=s}[\langle(x^{\prime},t),(y^{\prime},s)\rangle-(\frac{1}{2}f(y^{\prime}+2s_{1}u)+\frac{1}{2}f(y^{\prime}-2s_{2}u))]$
$\displaystyle=$ $\displaystyle\sup_{(y^{\prime},s)\in
H\times\mathbb{R}}\sup_{s_{1}\in\mathbb{R}}[\langle(x^{\prime},t),(y^{\prime},s)\rangle-(\frac{1}{2}f(y^{\prime}+2s_{1}u)+\frac{1}{2}f(y^{\prime}+2(s_{1}-s)u))]$
$\displaystyle\leq$ $\displaystyle\frac{1}{2}\sup_{(y^{\prime},s)\in
H\times\mathbb{R}}\sup_{s_{1}\in\mathbb{R}}[\langle(2x_{1}^{\prime},t),(y^{\prime},2s_{1})\rangle-f(y^{\prime}+2s_{1}u)]$
$\displaystyle+\frac{1}{2}\sup_{(y^{\prime},s)\in
H\times\mathbb{R}}\sup_{s_{1}\in\mathbb{R}}[\langle(2x_{2}^{\prime},-t),(y^{\prime},2s_{1}-2s)\rangle-f(y^{\prime}+2(s_{1}-s)u)]$
$\displaystyle=$
$\displaystyle\frac{1}{2}[(\mathcal{L}f)(2x_{1}^{\prime}+tu)+(\mathcal{L}f)(2x_{2}^{\prime}-tu)],$

where the first inequality is by choosing $\lambda=\frac{1}{2}$ and the second
inequality is by $\sup\sup(A+B)\leq\sup\sup A+\sup\sup B$.

Since $x_{1}^{\prime}$ and $x_{2}^{\prime}$ are arbitrary, by (4.4), we can
get

$\displaystyle\left(e^{-(\mathcal{L}(S_{u}f))_{(t)}}\right)(x^{\prime})\geq\sup_{x_{1}^{\prime}+x_{2}^{\prime}=x^{\prime}}\left(e^{-\frac{1}{2}(\mathcal{L}f)_{(t)}(2x_{1}^{\prime})}\times
e^{-\frac{1}{2}(\mathcal{L}f)_{(-t)}(2x_{2}^{\prime})}\right).$ (4.5)

By (4.5) and Prékopa-Leindler inequality, we have

$\displaystyle\int_{H}e^{-(\mathcal{L}(S_{u}f))_{(t)}}(x^{\prime})dx^{\prime}$
$\displaystyle\geq$
$\displaystyle\left(\int_{H}e^{-(\mathcal{L}f)_{(t)}(x^{\prime})}dx^{\prime}\right)^{\frac{1}{2}}\left(\int_{H}e^{-(\mathcal{L}f)_{(-t)}(x^{\prime})}dx^{\prime}\right)^{\frac{1}{2}}$
(4.6) $\displaystyle=$
$\displaystyle\int_{H}e^{-(\mathcal{L}f)_{(t)}(x^{\prime})}dx^{\prime},$

where the last equality is by $\mathcal{L}f$ is even (since $f$ is even).
Thus, by Fubini’s theorem, we can get the desired inequality. ∎

######  Lemma 4.3.

Let $h(t)$ be an increasing convex function defined on $[0,+\infty)$ and
$\int_{0}^{+\infty}e^{-h(t)}dt<\infty$. Let $\mathcal{L}(h(|\cdot|))$ denote
the Legendre transform of function $h(|x|)$ defined on $\mathbb{R}^{n}$. Then

$\displaystyle\int_{\mathbb{R}^{n}}e^{-h(|x|)}dx\int_{\mathbb{R}^{n}}e^{-(\mathcal{L}(h(|\cdot|)))(x)}dx\leq(2\pi)^{n}.$
(4.7)

###### Proof.

By spherical coordinate transformation, we have

$\displaystyle\int_{\mathbb{R}^{n}}e^{-h(|x|)}dx=\int_{S^{n-1}}\left[\int_{0}^{+\infty}e^{-h(r)}r^{n-1}dr\right]d\omega.$
(4.8)

For any $x\in\mathbb{R}^{n}$, let $x=t_{x}\theta_{x}$, where
$\theta_{x}=\frac{x}{|x|}\in S^{n-1}$ for $|x|\neq 0$ and $\theta_{x}$ is any
unit vector for $|x|=0$, and $t_{x}=|x|$. Then, we have

$\displaystyle\mathcal{L}(h(|\cdot|))(x)$ $\displaystyle=$
$\displaystyle\sup_{y\in\mathbb{R}^{n}}(\langle x,y\rangle-h(|y|))$
$\displaystyle=$ $\displaystyle\sup_{\theta_{y}\in S^{n-1},t_{y}\geq
0}(\langle t_{x}\theta_{x},t_{y}\theta_{y}\rangle-h(t_{y}))=\sup_{t_{y}\geq
0}(t_{x}t_{y}-h(t_{y})).$

Thus, we have

$\displaystyle\int_{\mathbb{R}^{n}}e^{-(\mathcal{L}(h(|\cdot|)))(x)}dx=\int_{S^{n-1}}\left[\int_{0}^{+\infty}\left(e^{-\sup_{t\geq
0}(rt-h(t))}\right)r^{n-1}dr\right]d\omega.$ (4.9)

For $r\in[0,+\infty)$, let $f_{1}(r)=\left(e^{-h(r)}\right)r^{n-1}$,
$f_{2}(r)=\left(e^{-\sup_{t\geq 0}(rt-h(t))}\right)r^{n-1}$ and
$f_{3}(r)=\left(e^{-\frac{r^{2}}{2}}\right)r^{n-1}$. Next, we shall prove that

$\displaystyle\int_{0}^{+\infty}f_{1}(r)dr\int_{0}^{+\infty}f_{2}(r)dr\leq\left(\int_{0}^{+\infty}f_{3}(r)dr\right)^{2}.$
(4.10)

Let $g_{i}(t)=f_{i}(e^{t})e^{t}$ for $i=1,2,3,$ then
$\int_{0}^{+\infty}f_{i}(r)dr=\int_{\mathbb{R}}g_{i}(t)dt$ and for every
$s,t\in\mathbb{R}$,
$g_{1}(s)g_{2}(t)\leq\left(g_{3}(\frac{s+t}{2})\right)^{2}$. Hence inequality
(4.10) follows from Prékopa-Leindler inequality.

By (4.8), (4.9) and (4.10), we have

$\displaystyle\int_{\mathbb{R}^{n}}e^{-h(|x|)}dx\int_{\mathbb{R}^{n}}e^{-(\mathcal{L}(h(|\cdot|)))(x)}dx$
$\displaystyle=$
$\displaystyle\omega_{n}^{2}\int_{0}^{+\infty}f_{1}(r)dr\int_{0}^{+\infty}f_{2}(r)dr\leq\omega_{n}^{2}\left(\int_{0}^{+\infty}e^{-\frac{r^{2}}{2}}r^{n-1}dr\right)^{2}=(2\pi)^{n},$

where $\omega_{n}=n\pi^{n/2}/\Gamma(1+\frac{n}{2})$ is the surface area of
Euclidean unit ball. ∎

Proof of Theorem 4.1. By the integral invariance under Steiner symmetrization
(Proposition 2.2), for any $u\in S^{n-1}$, we have

$\displaystyle\int_{\mathbb{R}^{n}}e^{-(S_{u}f)(x)}dx=\int_{\mathbb{R}^{n}}e^{-f(x)}dx.$
(4.11)

By (4.11) and Lemma 4.2, we have

$\displaystyle\int e^{-f}\int e^{-\mathcal{L}f}\leq\int e^{-S_{u}f}\int
e^{-\mathcal{L}(S_{u}f)}.$ (4.12)

By property 7 in Table 1, for log-concave function $e^{-f}\in
L^{1}(\mathbb{R}^{n})$, there exists a sequence of directions
$\\{u_{i}\\}_{i=1}^{\infty}\subset S^{n-1}$ such that
$e^{-S_{u_{1},\dots,u_{i}}f}$ converges to a radial function
$e^{-h(|\cdot|)}$, where $h(t)$ is a one-dimensional increasing convex
function defined on $[0,+\infty)$. By (4.12) and Lemma 4.3 and the continuity
of integral in $L^{1}(\mathbb{R}^{n})$, we have

$\displaystyle\int e^{-f}\int e^{-\mathcal{L}f}$ $\displaystyle\leq$
$\displaystyle\lim_{i\rightarrow+\infty}\int e^{-S_{u_{1},\dots,u_{i}}f}\int
e^{-\mathcal{L}(S_{u_{1},\dots,u_{i}}f)}$ (4.13) $\displaystyle=$
$\displaystyle\int e^{-h(|\cdot|)}\int
e^{-\mathcal{L}(h(|\cdot|))}\leq(2\pi)^{n}.$

This completes the proof. $\Box$

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