# Affine Hardy–Littlewood–Sobolev inequalities

Julián Haddad  Departamento de Análisis Matemático, Facultad de Matemáticas,
Universidad de Sevilla, Sevilla, Spain<EMAIL_ADDRESS>and  Monika Ludwig
Institut für Diskrete Mathematik und Geometrie, Technische Universität Wien,
Wiedner Hauptstraße 8-10/1046, 1040 Wien, Austria<EMAIL_ADDRESS>
###### Abstract.

Sharp affine Hardy–Littlewood–Sobolev inequalities for functions on
$\mathbb{R}^{n}$ are established, which are significantly stronger than (and
directly imply) the sharp Hardy–Littlewood–Sobolev inequalities by Lieb and by
Beckner, Dou, and Zhu. In addition, sharp reverse inequalities for the new
inequalities and the affine fractional $L^{2}$ Sobolev inequalities are
obtained for log-concave functions on $\mathbb{R}^{n}$.

2000 AMS subject classification: 26D15 (26B25, 26D10, 46E35, 52A40)

##  1\. Introduction

Lieb [19] established the following sharp Hardy–Littlewood–Sobolev
inequalities (HLS inequalities):

(1)
$\gamma_{n,\alpha}\|f\|_{\frac{2n}{n+\alpha}}^{2}\geq\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{|x-y|^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y$

for $0<\alpha<n$ and non-negative $f\in L^{p}(\mathbb{R}^{n})$ with
$p={2n}/({n+\alpha})$. There is equality if and only if
$f(x)=a(1+\lambda\,|x-x_{0}|^{2})^{-(n+\alpha)/2}$ for $x\in\mathbb{R}^{n}$
with $a\geq 0$, $\lambda>0$ and $x_{0}\in\mathbb{R}^{n}$. Here $|\cdot|$
denotes the Euclidean norm in $\mathbb{R}^{n}$ and
$\|f\|_{p}^{p}=\int_{\mathbb{R}^{n}}|f(x)|^{p}\,\mathrm{d}x$ while
$L^{p}(\mathbb{R}^{n})$ is the space of measurable functions
$f:\mathbb{R}^{n}\to\mathbb{R}$ with $\|f\|_{p}<\infty$. The constant is given
by

(2)
$\gamma_{n,\alpha}=\pi^{\frac{n-\alpha}{2}}\frac{\Gamma(\frac{\alpha}{2})}{\Gamma(\frac{n+\alpha}{2})}\Big{(}\frac{\Gamma(n)}{\Gamma(\frac{n}{2})}\Big{)}^{\frac{\alpha}{n}},$

where $\Gamma$ is the gamma function. The HLS inequalities (1) can be
considered as weak Young inequalities and are equivalent by duality to sharp
fractional $L^{2}$ Sobolev inequalities (for more information, see [20, 5, 4,
9]).

The HLS inequalities are invariant under translations, rotations, and
inversions, but not under volume-preserving linear transformations. For
geometric questions, affine inequalities, that is, inequalities that are
unchanged under translations and volume-preserving linear transformations
turned out to be very powerful. The best-known example is the Petty projection
inequality for convex bodies (that is, compact convex sets) in
$\mathbb{R}^{n}$, which is stronger than the Euclidean isoperimetric
inequality and directly implies it (see, for example, [31, 10], and see [25,
14, 34, 32] for related results in the $L^{p}$ Brunn–Minkowski theory and for
more general sets). Very recently, Milman and Yehudayoff [28] established
isoperimetric inequalities for affine quermassintegrals, which are
significantly stronger than the isoperimetric inequality for the classical
quermassintegrals (thereby confirming a conjecture by Lutwak [24]). Gaoyong
Zhang’s affine Sobolev inequality [34] is an affine version of the classical
$L^{1}$ Sobolev inequality. It was extended to functions of bounded variation
by Tuo Wang [32], and corresponding results for $L^{p}$ Sobolev inequalities
were established by Lutwak, Yang, and Zhang [26] and Haberl and Schuster [15].
Fractional Petty projection inequalities were recently obtained by the authors
in [16] and affine fractional $L^{p}$ Sobolev inequalities in [16, 17]. In all
cases, the affine inequalities are significantly stronger than (and imply)
their Euclidean counterparts.

The main aim of this paper is to establish sharp affine HLS inequalities that
are stronger than Lieb’s sharp HLS inequalities (1).

######  Theorem 1.

For $\,0<\alpha<n$ and non-negative $f\in
L^{{2n}/({n+\alpha})}(\mathbb{R}^{n})$,

(3) $\gamma_{n,\alpha}\|f\|_{\frac{2n}{n+\alpha}}^{2}\geq
n\omega_{n}^{\frac{n-\alpha}{n}}\Big{(}\frac{1}{n}\int_{{\mathbb{S}}^{n-1}}\big{(}\int_{0}^{\infty}t^{\alpha-1}\int_{\mathbb{R}^{n}}f(x)f(x+t\xi)\,\mathrm{d}x\,\mathrm{d}t\big{)}^{\frac{n}{\alpha}}\,\mathrm{d}\xi\Big{)}^{\frac{\alpha}{n}}\\\
\geq\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{|x-y|^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y.$

There is equality in the first inequality precisely if
$f(x)=a(1+|\phi(x-x_{0})|^{2})^{-(n+\alpha)/2}$ for $x\in\mathbb{R}^{n}$ with
$a\geq 0$, $\phi\in\operatorname{GL}(n)$ and $x_{0}\in\mathbb{R}^{n}$. There
is equality in the second inequality if $f$ is radially symmetric.

Here, ${\mathbb{S}}^{n-1}$ is the unit sphere in $\mathbb{R}^{n}$ and
integration on ${\mathbb{S}}^{n-1}$ is with respect to the $(n-1)$-dimensional
Hausdorff measure while $\omega_{n}$ is the $n$-dimensional volume of the
$n$-dimensional unit ball.

To prove Theorem 1, for given $0<\alpha<n$, we introduce the star-shaped set
$\operatorname{S}_{\alpha}\\!f$ associated to $f$, defined by its radial
function for $\xi\in{\mathbb{S}}^{n-1}$ as

(4)
$\rho_{\operatorname{S}_{\alpha}\\!f}(\xi)^{\alpha}=\int_{0}^{\infty}t^{\alpha-1}\int_{\mathbb{R}^{n}}f(x)f(x+t\xi)\,\mathrm{d}x\,\mathrm{d}t$

(see Section 3 for details). The first inequality from Theorem 1 now can be
written as

(5) $\gamma_{n,\alpha}\|f\|_{\frac{2n}{n+\alpha}}^{2}\geq
n\omega_{n}^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f|^{\frac{\alpha}{n}},$

where $|\cdot|$ denotes $n$-dimensional Lebesgue measure. Since both sides of
(5) are invariant under translations of $f$ and

$\operatorname{S}_{\alpha}\\!\,(f\circ\phi^{-1})=\phi\operatorname{S}_{\alpha}\\!f$

for volume-preserving linear transformations
$\phi:\mathbb{R}^{n}\to\mathbb{R}^{n}$, it follows that inequality (5) is
indeed an affine inequality.

As a critical tool in the proof of Theorem 1, we introduce an anisotropic
version of the right side of (1). In addition, we use the Riesz rearrangement
inequality, its equality case due to Burchard [3], and the original HLS
inequalities for radially symmetric functions.

Recently, Dou and Zhu [7] and Beckner [2] obtained sharp HLS inequalities also
for $\alpha>n$. In Section 5, we will establish sharp affine HLS inequalities
for $\alpha>n$, which are stronger than (and directly imply) their results.

In Section 6, for $E\subset\mathbb{R}^{n}$ measurable, we consider
$\operatorname{S}_{\alpha}\\!\operatorname{1}_{E}$, where
$\operatorname{1}_{E}$ is the indicator function of $E$. For an
$n$-dimensional convex body $E\subset\mathbb{R}^{n}$, we show that
$\operatorname{S}_{\alpha}\\!\operatorname{1}_{E}$ is proportional to the
radial $\alpha$-mean body of $E$, an important notion that was introduced by
Gardner and Zhang [12]. Sharp isoperimetric inequalities for radial
$\alpha$-mean bodies were recently obtained in [16]. Sharp reverse
inequalities were already established by Gardner and Zhang [12]. They
generalize Zhang’s reverse Petty projection inequality [33], and equality is
attained precisely for $n$-dimensional simplices.

In Section 7, we establish reverse inequalities also in the functional
setting.

######  Theorem 2.

For $\,0<\alpha<n$ and log-concave $f\in L^{2}(\mathbb{R}^{n})$,

$\frac{\Gamma(n+1)^{\frac{\alpha}{n}}}{\Gamma(\alpha)}\,|\operatorname{S}_{\alpha}\\!f|^{\frac{\alpha}{n}}\geq\|f\|_{2}^{2-\frac{2\alpha}{n}}\|f\|_{1}^{\frac{2\alpha}{n}}\geq\|f\|_{\frac{2n}{n+\alpha}}^{2}.$

There is equality in the first inequality if
$f(x)=a\,e^{-\|x-x_{0}\|_{\Delta}}$ for $x\in\mathbb{R}^{n}$ with $a\geq 0$,
$x_{0}\in\mathbb{R}^{n}$, and $\Delta$ an $n$-dimensional simplex having a
vertex at the origin.

Here, $\|\cdot\|_{\Delta}$ is the gauge function of
$\Delta\subset\mathbb{R}^{n}$. The second inequality follows from Hölder’s
inequality; for the proof of the first inequality, see Section 7.

We remark that it is easy to see that for general, non-negative $f\in
L^{2}(\mathbb{R}^{n})$, no non-trivial reverse inequality can hold. In Section
7, we establish reverse inequalities also for $\alpha>n$ (see Theorem 15) and
obtain results for $s$-concave functions for $s>0$. Moreover, we will
establish reverse affine fractional $L^{2}$ Sobolev inequalities.

##  2\. Preliminaries

We collect results on symmetrization, star-shaped sets, log-concave and
$s$-concave functions, and fractional polar projection bodies.

###  2.1. Symmetrization

Let $E\subseteq\mathbb{R}^{n}$ be a Borel set of finite measure. The Schwarz
symmetral of $E$, denoted by $E^{\star}$, is the closed, centered Euclidean
ball with the same volume as $E$.

Let $f$ be a non-negative measurable function with superlevel sets of finite
measure. Let $\\{f\geq t\\}=\\{x\in\mathbb{R}^{n}:f(x)\geq t\\}$ for
$t\in\mathbb{R}$. We say that $f$ is non-zero if $\\{f\neq 0\\}$ has positive
measure, and we identify functions that are equal up to a set of measure zero.
The layer cake formula states that

(6) $f(x)=\int_{0}^{\infty}\operatorname{1}_{\\{f\geq t\\}}(x)\,\mathrm{d}t$

for almost every $x\in\mathbb{R}^{n}$ and allows us to recover the function
from its superlevel sets. Here, for $E\subset\mathbb{R}^{n}$, the indicator
function $\operatorname{1}_{E}$ is defined by $\operatorname{1}_{E}(x)=1$ for
$x\in E$ and $\operatorname{1}_{E}(x)=0$ otherwise.

The Schwarz symmetral of $f$, denoted by $f^{\star}$, is defined as

$f^{\star}(x)=\int_{0}^{\infty}\operatorname{1}_{\\{f\geq
t\\}^{\star}}(x)\,\mathrm{d}t$

for $x\in\mathbb{R}^{n}$. Hence $f^{\star}$ is determined a.e. by the
properties of being radially symmetric and having superlevel sets of the same
measure as those of $f$. Note that $f^{\star}$ is also called the symmetric
decreasing rearrangement of $f$. We say that $f^{\star}$ is strictly symmetric
decreasing, if $f^{\star}(x)>f^{\star}(y)$ whenever $|x|<|y|$.

The proofs of our results make use of the Riesz rearrangement inequality (see,
for example, [20, Theorem 3.7]).

######  Theorem 3 (Riesz’s rearrangement inequality).

For $f,g,k:\mathbb{R}^{n}\to\mathbb{R}$ non-negative, measurable functions
with superlevel sets of finite measure,

$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}f(x)k(x-y)g(y)\,\mathrm{d}x\,\mathrm{d}y\leq\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}f^{\star}(x)k^{\star}(x-y)g^{\star}(y)\,\mathrm{d}x\,\mathrm{d}y.$

We will use the characterization of equality cases of the Riesz rearrangement
inequality due to Burchard [3].

######  Theorem 4 (Burchard).

Let $A,B$ and $C$ be sets of finite positive measure in $\,\mathbb{R}^{n}$ and
denote by $\alpha,\beta$ and $\gamma$ the radii of their Schwarz symmetrals
$A^{\star},B^{\star}$ and $C^{\star}$. For
$\,|\alpha-\beta|<\gamma<\alpha+\beta$, there is equality in

$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\operatorname{1}_{A}(y)\operatorname{1}_{B}(x-y)\operatorname{1}_{C}(x)\,\mathrm{d}x\,\mathrm{d}y\leq\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\operatorname{1}_{A^{\star}}(y)\operatorname{1}_{B^{\star}}(x-y)\operatorname{1}_{C^{\star}}(x)\,\mathrm{d}x\,\mathrm{d}y$

if and only if, up to sets of measure zero,

$A=a+\alpha D,\,B=b+\beta D,\,C=c+\gamma D,$

where $D$ is a centered ellipsoid, and $a,b$ and $c=a+b$ are vectors in
$\mathbb{R}^{n}$.

######  Theorem 5 (Burchard).

Let $f,g,k:\mathbb{R}^{n}\to\mathbb{R}$ non-negative, non-zero, measurable
functions with superlevel sets of finite measure such that

(7)
$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}f(x)k(x-y)g(y)\,\mathrm{d}x\,\mathrm{d}y<\infty.$

If at least two of the Schwarz symmetrals $f^{\star},g^{\star},k^{\star}$ are
strictly symmetric decreasing, then there is equality in

(8)
$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}f(x)k(x-y)g(y)\,\mathrm{d}x\,\mathrm{d}y\leq\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}f^{\star}(x)k^{\star}(x-y)g^{\star}(y)\,\mathrm{d}x\,\mathrm{d}y$

if and only if there is a volume-preserving $\phi\in\operatorname{GL}(n)$ and
$a,b,c\in\mathbb{R}^{n}$ with $c=a+b$ such that

$f(x)=f^{\star}(\phi^{-1}x-a),\,k(x)=k^{\star}(\phi^{-1}x-b),g(x)=g^{\star}(\phi^{-1}x-c)$

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

###  2.2. Star-shaped sets and dual mixed volumes

A closed set $K\subseteq\mathbb{R}^{n}$ is star-shaped (with respect to the
origin) if the interval $[0,x]\subset K$ for every $x\in K$. The gauge
function $\|\cdot\|_{K}:\mathbb{R}^{n}\to[0,\infty]$ of a star-shaped set is
defined as

$\|x\|_{K}=\inf\\{\lambda>0:x\in\lambda K\\}$

and the radial function $\rho_{K}:\mathbb{R}^{n}\setminus\\{0\\}\to[0,\infty]$
as

$\rho_{K}(x)=\|x\|_{K}^{-1}=\sup\\{\lambda\geq 0:\lambda x\in K\\}.$

For the $n$-dimensional unit ball ${B^{n}}$, we have
$\|\cdot\|_{{B^{n}}}=|\cdot|$. The $n$-dimensional Lebesgue measure, or volume
of a star-shaped set $K\subset\mathbb{R}^{n}$ with measurable radial function
is given by

$|K|=\frac{1}{n}\int_{{\mathbb{S}}^{n-1}}\rho_{K}(\xi)^{n}\,\mathrm{d}\xi.$

We call $K$ a star body if its radial function is strictly positive and
continuous in $\mathbb{R}^{n}\setminus\\{0\\}$.

Let $\alpha\in\mathbb{R}\backslash\\{0,n\\}$. For star-shaped sets
$K,L\subseteq\mathbb{R}^{n}$ with measurable radial functions, the dual mixed
volume is defined as

$\tilde{V}_{\alpha}(K,L)=\frac{1}{n}\int_{{\mathbb{S}}^{n-1}}\rho_{K}(\xi)^{n-\alpha}\rho_{L}(\xi)^{\alpha}\,\mathrm{d}\xi.$

Note that

$\tilde{V}_{\alpha}(K,K)=|K|.$

For $0<\alpha<n$ and star-shaped sets $K,L\subseteq\mathbb{R}^{n}$ of finite
volume, the dual mixed volume inequality states that

(9) $\tilde{V}_{\alpha}(K,L)\leq|K|^{({n-\alpha})/n}|L|^{\alpha/n}.$

Equality holds if and only if $K$ and $L$ are dilates, where we say that star-
shaped sets $K$ and $L$ are dilates if $\rho_{K}=c\,\rho_{L}$ almost
everywhere on ${\mathbb{S}}^{n-1}$ for some $c\geq 0$. The definition of dual
mixed volume for star bodies is due to Lutwak [23], where also the dual mixed
volume inequality (9) is derived from Hölder’s inequality. For $0<\alpha<n$
and star-shaped sets of finite volume, it follows from (9) that the dual mixed
volume is finite. For star-shaped sets $K,L\subseteq\mathbb{R}^{n}$ and
$\alpha>n$, the dual mixed volume inequality states that

(10) $\tilde{V}_{\alpha}(K,L)\geq|K|^{({n-\alpha})/n}|L|^{\alpha/n}.$

It follows from the equality case of Hölder’s inequality that equality holds
for finite $\tilde{V}_{\alpha}(K,L)$ if and only if $K$ and $L$ are dilates.
See [31, 10] for more information on dual mixed volumes.

###  2.3. Log-concave and $s$-concave functions

A function $f:\mathbb{R}^{n}\to[0,\infty)$ is log-concave, if $x\mapsto\log
f(x)$ is a concave function on $\mathbb{R}^{n}$ with values in
$[-\infty,\infty)$. For $s>0$, a function $f:\mathbb{R}^{n}\to[0,\infty)$ is
$s$-concave, if $f^{s}$ is concave on its support.

The following result is a consequence of the Prékopa–Leindler inequality. It
can be found as Theorem 11.3 in [11]. Log-concave functions are included as
the case $s=0$.

######  Lemma 6.

Let $s\geq 0$. If $f,g\in L^{1}(\mathbb{R}^{n})$ are $s$-concave, then their
convolution is $s/(ns+2)$-concave.

Here, the convolution of $f,g\in L^{1}(\mathbb{R}^{n})$ is defined by

$f*g(x)=\int_{\mathbb{R}^{n}}f(x-y)\,g(y)\,\mathrm{d}y$

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

We also consider the limiting case $s=\infty$. We say that a function
$f:\mathbb{R}^{n}\to[0,\infty)$ is $s$-concave with $s=\infty$ if it is a
multiple of the indicator function of a convex body.

###  2.4. Fractional $L^{2}$ polar projection bodies

For measurable $f:\mathbb{R}^{n}\to\mathbb{R}$ and $0<\alpha<1$, the
$\alpha$-fractional $L^{2}$ polar projection body of $f$, denoted by
$\Pi_{2}^{\ast,\alpha}f$, was defined in [17] by its radial function for
$\xi\in{\mathbb{S}}^{n-1}$ as

(11)
$\rho_{\Pi_{2}^{\ast,\alpha}f}(\xi)^{-2\alpha}=\int_{0}^{\infty}t^{-2\alpha-1}\,\int_{\mathbb{R}^{n}}|f(x+t\xi)-f(x)|^{2}\,\mathrm{d}x\,\mathrm{d}t.$

The fractional Sobolev space $W^{\alpha,2}(\mathbb{R}^{n})$ is the set of all
$f\in L^{2}(\mathbb{R}^{n})$ such that

$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{|f(x)-f(x)|^{2}}{|x-y|^{n+2\alpha}}\,\mathrm{d}x\,\mathrm{d}y<\infty.$

The affine fractional $L^{2}$ Sobolev inequality [17, Theorem 1] states that

(12)
$\|f\|_{{\frac{2n}{n-2\alpha}}}^{2}\leq\sigma_{n,2,\alpha}\,n\omega_{n}^{\frac{n+2\alpha}{n}}|\Pi_{2}^{\ast,\alpha}f|^{-\frac{2\alpha}{n}}$

for $f\in W^{\alpha,2}(\mathbb{R}^{n})$ and $0<\alpha<1$, where
$\sigma_{n,2,\alpha}$ is an explicitly known constant.

##  3\. The Star-shaped Set $\operatorname{S}_{\alpha}\\!f$

Let $f:\mathbb{R}^{n}\to[0,\infty)$ be measurable, $K\subset\mathbb{R}^{n}$
star-shaped with measurable radial function and $\alpha>0$. Anisotropic
fractional Sobolev norms were introduced in [21, 22] and used in [16]. Here,
we introduce

$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{\|x-y\|_{K}^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y,$

an anisotropic version of the functional from (1). Using Fubini’s theorem,
polar coordinates, and (4), we obtain that

$\displaystyle\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{\|x-y\|_{K}^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y$
$\displaystyle=\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(y)f(y+z)}{\|z\|_{K}^{n-\alpha}}\,\mathrm{d}y\,\mathrm{d}z$
$\displaystyle=\int_{{\mathbb{S}}^{n-1}}\int_{0}^{\infty}\rho_{K}(t\xi)^{n-\alpha}\,t^{n-1}\int_{\mathbb{R}^{n}}f(y)f(y+t\xi)\,\mathrm{d}y\,\mathrm{d}t\,\mathrm{d}\xi$
$\displaystyle=\int_{{\mathbb{S}}^{n-1}}\rho_{K}(\xi)^{n-\alpha}\int_{0}^{\infty}t^{\alpha-1}\int_{\mathbb{R}^{n}}f(y)f(y+t\xi)\,\mathrm{d}y\,\mathrm{d}t\,\mathrm{d}\xi$
$\displaystyle=\int_{{\mathbb{S}}^{n-1}}\rho_{K}(\xi)^{n-\alpha}\rho_{\operatorname{S}_{\alpha}\\!f}(\xi)^{\alpha}d\xi.$

Hence,

(13)
$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{\|x-y\|_{K}^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y=n\tilde{V}_{\alpha}(K,\operatorname{S}_{\alpha}\\!f)$

for measurable $f:\mathbb{R}^{n}\to[0,\infty)$ and star-shaped
$K\subset\mathbb{R}^{n}$ with measurable radial function.

Note that, using polar coordinates and Fubini’s theorem, we obtain that

(14)
$\displaystyle\begin{split}|\operatorname{S}_{n}\\!f|&=\frac{1}{n}\int_{{\mathbb{S}}^{n-1}}\rho_{\operatorname{S}_{n}\\!f}(\xi)^{n}\,\mathrm{d}\xi\\\
&=\frac{1}{n}\int_{{\mathbb{S}}^{n-1}}\int_{0}^{\infty}\int_{\mathbb{R}^{n}}t^{n-1}f(x)f(x+t\xi)\,\mathrm{d}x\,\mathrm{d}t\,\mathrm{d}\xi\\\
&=\frac{1}{n}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}f(x)f(x+y)\,\mathrm{d}y\,\mathrm{d}x\\\
&=\frac{1}{n}\|f\|_{1}^{2}\end{split}$

for measurable $f:\mathbb{R}^{n}\to[0,\infty)$.

We remark that for $0<\alpha<n$ and given measurable
$f:\mathbb{R}^{n}\to[0,\infty)$, the dual mixed volume inequality (9) and (13)
imply that

(15)
$\sup\big{\\{}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{\|x-y\|_{K}^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y:K\subset\mathbb{R}\text{
star-
shaped},|K|=\omega_{n}\big{\\}}=n\omega_{n}^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f|^{\frac{\alpha}{n}}$

is attained precisely for a suitable dilate of $\operatorname{S}_{\alpha}\\!f$
if $|\operatorname{S}_{\alpha}\\!f|$ is finite. In this sense,
$\operatorname{S}_{\alpha}\\!f$ is the optimal choice of the star-shaped set
$K$ for given $f$. For $\alpha>n$ and given measurable
$f:\mathbb{R}^{n}\to[0,\infty)$, the dual mixed volume inequality (10) and
(13) imply that

(16)
$\inf\big{\\{}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{\|x-y\|_{K}^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y:K\subset\mathbb{R}\text{
star-
shaped},|K|=\omega_{n}\big{\\}}=n\omega_{n}^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f|^{\frac{\alpha}{n}}$

is attained precisely for a suitable dilate of $\operatorname{S}_{\alpha}\\!f$
if $|\operatorname{S}_{\alpha}\\!f|$ is finite. Again
$\operatorname{S}_{\alpha}\\!f$ is the optimal choice in this sense.

We mention the following property of $\operatorname{S}_{\alpha}\\!f$ for log-
concave $f$.

######  Proposition 7.

If $f:\mathbb{R}^{n}\to[0,\infty)$ is log-concave and in
$L^{1}(\mathbb{R}^{n})$, then $\operatorname{S}_{\alpha}\\!f$ is a convex body
for every $\alpha>0$.

###### Proof.

Since $f$ is log-concave and in $L^{1}(\mathbb{R}^{n})$, so is

$y\mapsto\int_{\mathbb{R}^{n}}f(x)f(x+y)\,\mathrm{d}x.$

This follows from the Young inequality (cf. [20, Theorem 4.2]) and Lemma 6.
Hence, using a result of Ball [1, Theorem 5] (or see [12, Corollary 4.2]), we
obtain the result. ∎

##  4\. Affine HLS Inequalities for $0<\alpha<n$

The following result is an immediate consequence of the Riesz rearrangement
inequality and its equality case from Theorem 5.

######  Lemma 8.

Let $q>0$ and $K\subset\mathbb{R}^{n}$ a star-shaped set with measurable
radial function and $|K|>0$. For a non-zero, measurable function
$f:\mathbb{R}^{n}\to[0,\infty)$ such that

(17)
$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{\|x-y\|_{K}^{q}}\,\mathrm{d}x\,\mathrm{d}y<\infty$

and strictly symmetric decreasing $f^{\star}$, there is equality in

(18)
$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{\|x-y\|_{K}^{q}}\,\mathrm{d}x\,\mathrm{d}y\leq\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f^{\star}(x)f^{\star}(y)}{\|x-y\|_{K^{\star}}^{q}}\,\mathrm{d}x\,\mathrm{d}y$

if and only if $K$ is a centered ellipsoid and $f$ is a translate of
$f^{\star}$.

We require the following lemmas for the proof of Theorem 1.

######  Lemma 9.

Let $\,0<\alpha<n$ and $K\subset\mathbb{R}^{n}$ be star-shaped with measurable
radial function. If $f:\mathbb{R}^{n}\to[0,\infty)$ is non-zero and measurable
and

(19)
$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{\|x-y\|_{K}^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y<\infty,$

then

(20)
$\tilde{V}_{\alpha}(K,\operatorname{S}_{\alpha}\\!f)\leq\tilde{V}_{\alpha}(K^{\star},\operatorname{S}_{\alpha}\\!f^{\star}).$

For $|K|>0$ and $f^{\star}$ strictly symmetric decreasing, there is equality
if and only if $K$ is a centered ellipsoid and $f$ is a translate of
$f^{\star}$.

###### Proof.

By (13) and the Riesz rearrangement inequality, Theorem 3, we have

$\tilde{V}_{\alpha}(K,\operatorname{S}_{\alpha}\\!f)\leq\tilde{V}_{\alpha}(K^{\star},\operatorname{S}_{\alpha}\\!f^{\star}).$

By Lemma 8, there is equality if and only if $K$ is a centered ellipsoid and
$f$ is a translate of $f^{\star}$. ∎

######  Lemma 10.

Let $0<\alpha<n$ and $p=2\alpha/(n+\alpha)$. For non-negative $f\in
L^{p}(\mathbb{R}^{n})$,

(21)
$|\operatorname{S}_{\alpha}\\!f|\leq|\operatorname{S}_{\alpha}\\!f^{\star}|.$

For $f^{\star}$ strictly symmetric decreasing with
$|\operatorname{S}_{\alpha}\\!f^{\star}|<\infty$, there is equality if and
only if $f$ is a translate of $f^{\star}$.

###### Proof.

First, assume that $\,|\operatorname{S}_{\alpha}\\!f|<\infty$. By Lemma 9 with
$K=\operatorname{S}_{\alpha}\\!f$ and the dual mixed volume inequality (9) for
$0<\alpha<n$, we have

(22) $\displaystyle|\operatorname{S}_{\alpha}\\!f|$
$\displaystyle=\tilde{V}_{\alpha}(\operatorname{S}_{\alpha}\\!f,\operatorname{S}_{\alpha}\\!f)$
(23)
$\displaystyle\leq\tilde{V}_{\alpha}((\operatorname{S}_{\alpha}\\!f)^{\star},\operatorname{S}_{\alpha}\\!f^{\star})$
(24)
$\displaystyle\leq|(\operatorname{S}_{\alpha}\\!f)^{\star}|^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f^{\star}|^{\frac{\alpha}{n}}$
(25)
$\displaystyle=|\operatorname{S}_{\alpha}\\!f|^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f^{\star}|^{\frac{\alpha}{n}}.$

The equality case follows from Lemma 9.

Second, assume that $|\operatorname{S}_{\alpha}\\!f|=\infty$. For $k\geq 1$,
define

$f_{(k)}(x)=f(x)\operatorname{1}_{k{B^{n}}}(x).$

Note that $f_{(k)}$ is non-decreasing with respect to $k$ and converges to $f$
pointwise. By the monotone convergence theorem, we obtain

(26)
$\lim_{k\to\infty}\int_{0}^{\infty}t^{\alpha-1}\int_{\mathbb{R}^{n}}f_{(k)}(x)f_{(k)}(x+t\xi)\,\mathrm{d}x\,\mathrm{d}t=\int_{0}^{\infty}t^{\alpha-1}\int_{\mathbb{R}^{n}}f(x)f(x+t\xi)\,\mathrm{d}x\,\mathrm{d}t$

and the convergence is monotone. A second application of the monotone
convergence theorem shows that

(27)
$\lim_{k\to\infty}\int_{{\mathbb{S}}^{n-1}}\left(\int_{0}^{\infty}t^{\alpha-1}\int_{\mathbb{R}^{n}}f_{(k)}(x)f_{(k)}(x+t\xi)\,\mathrm{d}x\,\mathrm{d}t\right)^{\frac{n}{\alpha}}\,\mathrm{d}\xi\\\
=\int_{{\mathbb{S}}^{n-1}}\left(\int_{0}^{\infty}t^{\alpha-1}\int_{\mathbb{R}^{n}}f(x)f(x+t\xi)\,\mathrm{d}x\,\mathrm{d}t\right)^{\frac{n}{\alpha}}\,\mathrm{d}\xi.$

Hence,

(28)
$\lim_{k\to\infty}|\operatorname{S}_{\alpha}\\!f_{(k)}|=|\operatorname{S}_{\alpha}\\!f|=\infty.$

Since $f\in L^{p}(\mathbb{R}^{n})$, the function $f_{(k)}$ has compact support
and $|\operatorname{S}_{\alpha}\\!f_{(k)}|<\infty$ for $k\geq 1$. It is easy
to verify that $(f_{(k)})^{\star}\leq f^{\star}$ a.e., so the first part of
the lemma implies that

(29)
$|\operatorname{S}_{\alpha}\\!f_{(k)}|\leq|\operatorname{S}_{\alpha}\\!(f_{(k)})^{\star}|\leq|\operatorname{S}_{\alpha}\\!f^{\star}|$

for $k\geq 1$. It follows from (28) that
$|\operatorname{S}_{\alpha}\\!f_{(k)}|\to\infty$, which shows that

(30) $|\operatorname{S}_{\alpha}\\!f^{\star}|=\infty,$

which is what we wanted to show. ∎

###  4.1. Proof of Theorem 1

Since $\|f\|_{p}=\|f^{\star}\|_{p}$, we obtain by the classical HLS inequality
(1), by (13) and by Lemma 10 that

(31) $\displaystyle\gamma_{n,\alpha}\|f\|_{p}^{2}$
$\displaystyle\geq\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f^{\star}(x)f^{\star}(y)}{|x-y|^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y$
(32)
$\displaystyle=n\tilde{V}_{\alpha}({B^{n}},\operatorname{S}_{\alpha}\\!f^{\star})$
(33)
$\displaystyle=n\omega_{n}^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f^{\star}|^{\frac{\alpha}{n}}$
(34) $\displaystyle\geq
n\omega_{n}^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f|^{\frac{\alpha}{n}}.$

If there is equality throughout, then $f^{\star}$ realizes equality in the HLS
inequality (1). Hence $f^{\star}(x)=a(1+\lambda\,|x|^{2})^{-n/p}$ for some
$a\geq 0$ and $\lambda>0$. Consequently, $f^{\star}$ is strictly symmetric
decreasing, and we may apply Lemma 10 to obtain the equality case in the first
inequality in Theorem 1.

For the second inequality, we set $K={B^{n}}$ in (13) and apply the dual mixed
volume inequality (9) to obtain

(35)
$\displaystyle\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{|x-y|^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y$
$\displaystyle=n\tilde{V}_{\alpha}({B^{n}},\operatorname{S}_{\alpha}\\!f)\leq
n\omega_{n}^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f|^{\frac{\alpha}{n}}.$

There is equality precisely if $\operatorname{S}_{\alpha}\\!f$ is a ball,
which is the case for radially symmetric functions.

##  5\. Affine HLS Inequalities for $\alpha>n$

Jingbo Dou and Meijun Zhu [7] and William Beckner [2] established sharp HLS
inequalities for $\alpha>n$ (also see [30, 6]):

(36)
$\gamma_{n,\alpha}\|f\|_{\frac{2n}{n+\alpha}}^{2}\leq\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{|x-y|^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y$

for non-negative $f\in L^{p}(\mathbb{R}^{n})$ with $p={2n}/({n+\alpha})$,
where $\gamma_{n,\alpha}$ is defined in (2). There is equality if
$f(x)=a(1+\lambda\,|x-x_{0}|^{2})^{-(n-\alpha)/2}$ for $x\in\mathbb{R}^{n}$
with $a\geq 0$, $\lambda>0$ and $x_{0}\in\mathbb{R}^{n}$.

We will establish sharp HLS inequalities for $\alpha>n$ that strengthen and
imply (36). We require the following lemmas.

The following result is a consequence of the Riesz rearrangement inequality
and Theorem 4. Note that the middle function in (37) has superlevel sets of
infinite measure.

######  Lemma 11.

Let $q>0$ and $K\subset\mathbb{R}^{n}$ a star-shaped set with $0<|K|<\infty$.
For non-negative, non-zero $f\in L^{p}(\mathbb{R}^{n})$ such that

(37)
$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}{f(x){\|x-y\|_{K}^{q}}\,f(y)}\,\mathrm{d}x\,\mathrm{d}y<\infty,$

there is equality in

(38)
$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}{f(x){\|x-y\|_{K}^{q}}\,f(y)}\,\mathrm{d}x\,\mathrm{d}y\geq\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}{f^{\star}(x){\|x-y\|_{K^{\star}}^{q}}\,f^{\star}(y)}\,\mathrm{d}x\,\mathrm{d}y$

if and only if $K$ is a centered ellipsoid and $f$ is a translate of
$f^{\star}$.

###### Proof.

Writing

$\|z\|_{K}^{q}=\int_{0}^{\infty}k_{s}(z)\,\mathrm{d}s$

where $k_{t}(z)=\operatorname{1}_{s^{1/q}(\mathbb{R}^{n}\backslash K)}(z)$,
and using the layer-cake formula (6) for $f$ and $g$, we obtain

(39)
$\displaystyle\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}{f(x){\|x-y\|_{K}^{q}}f(y)}\,\mathrm{d}x\,\mathrm{d}y$
(40)
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\operatorname{1}_{\\{f\geq
r\\}}(x)k_{s}(x-y)\operatorname{1}_{\\{f\geq
t\\}}(x)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}r\,\mathrm{d}s\,\mathrm{d}t.$

The Riesz rearrangement inequality, Theorem 3, implies that

(41)
$\displaystyle\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\operatorname{1}_{\\{f\geq
r\\}}(x)$ $\displaystyle k_{s}(x-y)\operatorname{1}_{\\{f\geq
t\\}}(y)\,\mathrm{d}x\,\mathrm{d}y$ (42)
$\displaystyle=\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\operatorname{1}_{\\{f\geq
r\\}}(x)(1-\operatorname{1}_{s^{1/q}K}(x-y))\operatorname{1}_{\\{f\geq
t\\}}(y)\,\mathrm{d}x\,\mathrm{d}y$ (43)
$\displaystyle=\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\operatorname{1}_{\\{f\geq
r\\}}(x)\operatorname{1}_{\\{f\geq t\\}}(y)\,\mathrm{d}x\,\mathrm{d}y$ (44)
$\displaystyle\phantom{=\,}-\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\operatorname{1}_{\\{f\geq
r\\}}(x)\operatorname{1}_{s^{1/q}K}(x-y)\operatorname{1}_{\\{f\geq
t\\}}(y)\,\mathrm{d}x\,\mathrm{d}y$ (45)
$\displaystyle\geq\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\operatorname{1}_{\\{f\geq
r\\}^{\star}}(x)k_{s}^{\star}(x-y)\operatorname{1}_{\\{f\geq
t\\}^{\star}}(y)\,\mathrm{d}x\,\mathrm{d}y$

for $r,s,t>0$. Note that $\int_{\mathbb{R}^{n}}\operatorname{1}_{\\{f\geq
r\\}}(x)\,\mathrm{d}x<\infty$ for $r>0$, as $f\in L^{p}(\mathbb{R}^{n})$. If
there is equality in (38), then there is a null set $M\subset(0,\infty)^{3}$
such that

(46) $\displaystyle\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}$
$\displaystyle\operatorname{1}_{\\{f\geq
r\\}}(x)\operatorname{1}_{s^{1/q}K}(x-y)\operatorname{1}_{\\{f\geq
t\\}}(y)\,\mathrm{d}x\,\mathrm{d}y$
$\displaystyle=\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\operatorname{1}_{\\{f\geq
r\\}^{\star}}(x)\operatorname{1}_{s^{1/q}K^{\star}}(x-y)\operatorname{1}_{\\{f\geq
t\\}^{\star}}(y)\,\mathrm{d}x\,\mathrm{d}y$

for $(r,s,t)\in(0,\infty)^{3}\backslash M$.

For almost every $(r,t)\in(0,\infty)^{2}$, we have
$(r,s,t)\in(0,\infty)^{3}\backslash M$ for almost every $s>0$. For such
$(r,t)$ with $r\geq t$ and $s>0$ sufficiently small, the assumptions of
Theorem 4 are fulfilled and therefore there are a centered ellipsoid $D$ and
$a,b\in\mathbb{R}^{n}$ (depending on $(r,s,t)$) such that

(47) $\displaystyle\\{f\geq r\\}=a+\alpha D,\quad s^{1/q}K=b+\beta
D,\quad\\{f\geq t\\}=c+\gamma D$

where $c=a+b$. Since $K=s^{-1/q}b+(|K|/|D|)^{1/n}D$, the centered ellipsoid
$D$ does not depend on $(r,s,t)$ and hence the vectors $a$ and $c$ do not
depend on $s$. It follows that $b=0$ and that $K$ is a multiple of $D$. Hence
$a=c$ is a constant vector, which concludes the proof. ∎

######  Lemma 12.

Let $\alpha>n$ and $K\subset\mathbb{R}^{n}$ be star-shaped with measurable
radial function. If $f:\mathbb{R}^{n}\to[0,\infty)$ is non-zero and measurable
and

(48)
$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{\|x-y\|_{K}^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y<\infty,$

then

(49)
$\tilde{V}_{\alpha}(K,\operatorname{S}_{\alpha}\\!f)\geq\tilde{V}_{\alpha}(K^{\star},\operatorname{S}_{\alpha}\\!f^{\star}).$

For $|K|>0$, there is equality if and only if $K$ is a centered ellipsoid and
$f$ is a translate of $f^{\star}$.

###### Proof.

By (13) and Lemma 11, we have

$\tilde{V}_{\alpha}(K,\operatorname{S}_{\alpha}\\!f)\geq\tilde{V}_{\alpha}(K^{\star},\operatorname{S}_{\alpha}\\!f^{\star}).$

By Lemma 11, there is equality if and only if $K$ is a centered ellipsoid and
$f$ is a translate of $f^{\star}$. ∎

######  Lemma 13.

Let $\alpha>n$ and $f:\mathbb{R}^{n}\to[0,\infty)$ measurable. If
$\,|\operatorname{S}_{\alpha}\\!f|<\infty$, then

(50)
$|\operatorname{S}_{\alpha}\\!f|\geq|\operatorname{S}_{\alpha}\\!f^{\star}|.$

There is equality if and only if $f$ is a translate of $f^{\star}$.

###### Proof.

By Lemma 12 with $K=\operatorname{S}_{\alpha}\\!f$ and the dual mixed volume
inequality (10) for $\alpha>n$, we have

(51) $\displaystyle|\operatorname{S}_{\alpha}\\!f|$
$\displaystyle=\tilde{V}_{\alpha}(\operatorname{S}_{\alpha}\\!f,\operatorname{S}_{\alpha}\\!f)$
(52)
$\displaystyle\geq\tilde{V}_{\alpha}((\operatorname{S}_{\alpha}\\!f)^{\star},\operatorname{S}_{\alpha}\\!f^{\star})$
(53)
$\displaystyle\geq|(\operatorname{S}_{\alpha}\\!f)^{\star}|^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f^{\star}|^{\frac{\alpha}{n}}$
(54)
$\displaystyle=|\operatorname{S}_{\alpha}\\!f|^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f^{\star}|^{\frac{\alpha}{n}}.$

The equality case follows from Lemma 12. ∎

We are now in the position to prove affine HLS inequalities for $\alpha>n$.

######  Theorem 14.

For $\alpha>n$ and non-negative $f\in L^{{2n}/({n+\alpha})}(\mathbb{R}^{n})$,

(55) $\gamma_{n,\alpha}\|f\|_{\frac{2n}{n+\alpha}}^{2}\leq
n\omega_{n}^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f|^{\frac{\alpha}{n}}\\\
\leq\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{|x-y|^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y.$

There is equality in the first inequality precisely if
$f(x)=a(1+|\phi(x-x_{0})|^{2})^{-(n-\alpha)/2}$ for $x\in\mathbb{R}^{n}$ with
$a\geq 0$, $\phi\in\operatorname{GL}(n)$ and $x_{0}\in\mathbb{R}^{n}$. There
is equality in the second inequality if $f$ is radially symmetric.

###### Proof.

For the first inequality, we may assume that $|\operatorname{S}_{\alpha}\\!f|$
is finite. Since $\|f\|_{p}=\|f^{\star}\|_{p}$, we obtain by the HLS
inequality (36), by (13) and by Lemma 13 that

(56) $\displaystyle\gamma_{n,\alpha}\|f\|_{p}^{2}$
$\displaystyle\leq\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f^{\star}(x)f^{\star}(y)}{|x-y|^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y$
(57)
$\displaystyle=n\tilde{V}_{\alpha}({B^{n}},\operatorname{S}_{\alpha}\\!f^{\star})$
(58)
$\displaystyle=n\omega_{n}^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f^{\star}|^{\frac{\alpha}{n}}$
(59) $\displaystyle\leq
n\omega_{n}^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f|^{\frac{\alpha}{n}}.$

If there is equality throughout, then $f^{\star}$ realizes equality in the HLS
inequality (36). Hence $f^{\star}(x)=a(1+\lambda\,|x|^{2})^{-n/p}$ for some
$a\geq 0$ and $\lambda>0$. By Lemma 13, we obtain the equality case.

For the second inequality, assume that

$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{|x-y|^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y<\infty.$

We set $K={B^{n}}$ in (13) and apply the dual mixed volume inequality (10) to
obtain

(60)
$\displaystyle\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)f(y)}{|x-y|^{n-\alpha}}\,\mathrm{d}x\,\mathrm{d}y$
$\displaystyle=n\tilde{V}_{\alpha}({B^{n}},\operatorname{S}_{\alpha}\\!f)\geq
n\omega_{n}^{\frac{n-\alpha}{n}}|\operatorname{S}_{\alpha}\\!f|^{\frac{\alpha}{n}}.$

There is equality precisely if $\operatorname{S}_{\alpha}\\!f$ is a ball,
which is the case for radially symmetric functions. ∎

Next, we state a sharp reverse of the first inequality from Theorem 14 for
log-concave functions.

######  Theorem 15.

For $\,\alpha>n$ and log-concave $f\in L^{2}(\mathbb{R}^{n})$,

$\frac{\Gamma(n+1)^{\frac{\alpha}{n}}}{\Gamma(\alpha)}\,|\operatorname{S}_{\alpha}\\!f|^{\frac{\alpha}{n}}\leq\|f\|_{2}^{2-\frac{2\alpha}{n}}\|f\|_{1}^{\frac{2\alpha}{n}}\leq\|f\|_{\frac{2n}{n+\alpha}}^{2}.$

There is equality in the first inequality if
$f(x)=a\,e^{-\|x-x_{0}\|_{\Delta}}$ for $x\in\mathbb{R}^{n}$ with $a\geq
0,x_{0}\in\mathbb{R}^{n}$ and $\Delta$ an $n$-dimensional simplex having a
vertex at the origin.

The proof of this result will be given in Section 7.

##  6\. Radial Mean Bodies

Let $E\subset\mathbb{R}^{n}$ be a convex body. For $\alpha>-1$ and $\alpha\neq
0$, Gardner and Zhang [12] defined the radial $\alpha$-th mean body of $E$, by
its radial function for $\xi\in{\mathbb{S}}^{n-1}$, as

(61)
$\rho_{\operatorname{R}_{\alpha}\\!E}(\xi)^{\alpha}=\frac{1}{|E|}\int_{E}\rho_{E-x}(\xi)^{\alpha}\,\mathrm{d}x$

for $\alpha\neq 0$ and as

$\log(\rho_{\operatorname{R}_{0}\\!E}(\xi))=\frac{1}{|E|}\int_{E}\log(\rho_{E-x}(\xi))\,\mathrm{d}x.$

They showed that $\operatorname{R}_{p}\\!E$ is a star body for $\alpha>-1$ and
a convex body for $\alpha\geq 0$. This also follows from Proposition 7 and
equation (68), which we will establish below. Gardner and Zhang [12] also
showed that for $\alpha>-1$ and $\xi\in{\mathbb{S}}^{n-1}$,

(62)
$\rho_{\operatorname{R}_{\alpha}\\!E}(\xi)^{\alpha}=\frac{1}{(\alpha+1)|E|}\int_{E|\xi^{\perp}}|E\cap(\ell_{\xi}+y)|_{1}^{\alpha+1}\,\mathrm{d}y,$

where $\ell_{\xi}=\\{t\xi:t\in\mathbb{R}\\}$ is the line in direction $\xi$
and $|\cdot|_{1}$ denotes one-dimensional volume while $E|\xi^{\perp}$ is the
image of the orthogonal projection of $E$ to the hyperplane orthogonal to
$\xi$.

Let $E\subset\mathbb{R}^{n}$ be convex. If $\alpha>0$, then the definition of
$\operatorname{S}_{\alpha}\\!\operatorname{1}_{E}$ implies that

(63)
$\displaystyle\rho_{\operatorname{S}_{\alpha}\\!\operatorname{1}_{E}}(\xi)^{\alpha}$
$\displaystyle=\int_{0}^{\infty}t^{\alpha-1}|E\cap(E+t\xi)|_{1}\,\mathrm{d}t$
(64)
$\displaystyle=\int_{0}^{\infty}t^{\alpha-1}\int_{E|\xi^{\perp}}(|E\cap(\ell_{\xi}+y)|_{1}-t)_{\scriptscriptstyle+}\,\mathrm{d}y\,\mathrm{d}t$
(65)
$\displaystyle=\int_{E|\xi^{\perp}}\int_{0}^{\infty}t^{\alpha-1}(|E\cap(\ell_{\xi}+y)|_{1}-t)_{\scriptscriptstyle+}\,\mathrm{d}t\,\mathrm{d}y$
(66)
$\displaystyle=\int_{E|\xi^{\perp}}\int_{0}^{|E\cap(\ell_{\xi}+y)|_{1}}t^{\alpha-1}(|E\cap(\ell_{\xi}+y)|_{1}-t)\,\mathrm{d}t\,\mathrm{d}y$
(67)
$\displaystyle=\frac{1}{\alpha(\alpha+1)}\int_{E|\xi^{\perp}}|E\cap(\ell_{\xi}+y)|_{1}^{\alpha+1}\,\mathrm{d}y.$

Hence,

(68)
$\operatorname{S}_{\alpha}\\!\operatorname{1}_{E}=\Big{(}\frac{|E|}{\alpha}\Big{)}^{1/\alpha}\operatorname{R}_{\alpha}\\!E$

for $\alpha>0$. If $-1<\alpha<0$, then, using (11), we obtain that

(69)
$\displaystyle\rho_{\Pi_{2}^{\ast,-\alpha/2}\operatorname{1}_{E}}(\xi)^{\alpha}$
$\displaystyle=\int_{0}^{\infty}t^{\alpha-1}|E\Delta(E+t\xi)|\,\mathrm{d}t$
(70)
$\displaystyle=\int_{0}^{\infty}t^{\alpha-1}\int_{E|\xi^{\perp}}2\min\\{|E\cap(\ell_{\xi}+y)|_{1},t\\}\,\mathrm{d}y\,\mathrm{d}t$
(71)
$\displaystyle=2\int_{E|\xi^{\perp}}\int_{0}^{\infty}t^{\alpha-1}\min\\{|E\cap(\ell_{\xi}+y)|_{1},t\\}\,\mathrm{d}t\,\mathrm{d}y$
(72)
$\displaystyle=2\int_{E|\xi^{\perp}}\int_{0}^{|E\cap(\ell_{\xi}+y)|_{1}}t^{\alpha}\,\mathrm{d}t+\int_{|E\cap(\ell_{\xi}+y)|_{1}}^{\infty}|E\cap(\ell_{\xi}+y)|_{1}\,t^{\alpha-1}\,\mathrm{d}t\,\mathrm{d}y$
(73)
$\displaystyle=-\frac{2}{\alpha(\alpha+1)}\int_{E|\xi^{\perp}}|E\cap(\ell_{\xi}+y)|_{1}^{\alpha+1}\,\mathrm{d}y,$

where $E\Delta F$ is the symmetric difference of $E,F\subset\mathbb{R}^{n}$.
Hence, we obtain that

(74)
$\Pi_{2}^{\ast,-\alpha/2}\operatorname{1}_{E}=\Big{(}\frac{2|E|}{-\alpha}\Big{)}^{1/\alpha}\operatorname{R}_{\alpha}\\!E$

for $-1<\alpha<0$.

See [12, 16] for information on sharp affine isoperimetric inequalities for
radial mean bodies.

##  7\. Reverse Affine HLS and Fractional $L^{2}$ Sobolev Inequalities

We prove Theorem 2 and Theorem 15 for log-concave functions and derive results
for $s$-concave functions for $s>0$. In addition, we establish reverse affine
fractional $L^{2}$ Sobolev inequalities.

###  7.1. An auxiliary result

We require the following result (see, for example, [29, Lemma 2.6], where it
is said to be a consequence of [27]). A simple computation gives the equality
case.

######  Lemma 16.

Let $\omega:[0,\infty)\to[0,\infty)$ be decreasing with

(75) $0<\int_{0}^{\infty}t^{\alpha-1}\omega(t)\,\mathrm{d}t<\infty$

for $\alpha>0$. If $\varphi:[0,\infty)\to[0,\infty)$ is non-zero, with
$\varphi(0)=0$, and such that $t\mapsto\varphi(t)$ and $t\mapsto\varphi(t)/t$
are increasing on $(0,\infty)$, then

(76)
$\zeta(\alpha)=\left(\frac{\int_{0}^{\infty}t^{\alpha-1}\omega(\varphi(t))\,\mathrm{d}t}{\int_{0}^{\infty}t^{\alpha-1}\omega(t)\,\mathrm{d}t}\right)^{\frac{1}{\alpha}}$

is a decreasing function of $\alpha$ on $(0,\infty)$. Moreover, $\zeta$ is
constant on $(0,\infty)$ if $\varphi(t)=\lambda t$ on $[0,\infty)$ for some
$\lambda>0$.

Note that (75) and the assumptions on $\varphi$ and $\omega$ imply that both
integrals in (76) are finite and that the conditions of [29, Lemma 2.6] are
fulfilled.

We will extend Lemma 16 to $\alpha\in(-1,\infty)$. Let $\omega$ be as in Lemma
16 and $\alpha>0$. For any $t_{0}>0$, we have

(77)
$\int_{0}^{\infty}t^{\alpha-1}\omega(t)\,\mathrm{d}t=\int_{t_{0}}^{\infty}t^{\alpha-1}\omega(t)\,\mathrm{d}t-\int_{0}^{t_{0}}t^{\alpha-1}(\omega(0)-\omega(t))\,\mathrm{d}t+\omega(0)\frac{t_{0}^{\alpha}}{\alpha}.$

If, in addition,

(78) $\int_{0}^{\infty}t^{\alpha-1}(\omega(0)-\omega(t))\,\mathrm{d}t<\infty$

for $\alpha\in(-1,0)$, then the right side of (77) is finite for
$\alpha\in(-1,0)$. Moreover, it is equal to

$\int_{0}^{\infty}t^{\alpha-1}(\omega(t)-\omega(0))\,\mathrm{d}t,$

which is the analytic continuation of
$\alpha\mapsto\int_{0}^{\infty}t^{\alpha-1}\omega(t)\,\mathrm{d}t$ to $(-1,0)$
(see, for example, [13, Section 1.3]).

For $\omega(t)=e^{-t}$ and $\omega(t)=(1-st)_{\scriptscriptstyle+}^{1/s}$ with
$s>0$, we obtain the well-known analytic continuation formulas for the Gamma
and Beta functions,

(79)
$\Gamma(\alpha)=\begin{cases}\int_{0}^{\infty}t^{\alpha-1}e^{-t}\,\mathrm{d}t&\text{for
}0<\alpha,\\\\[4.0pt]
\int_{0}^{\infty}t^{\alpha-1}(e^{-t}-1)\,\mathrm{d}t&\text{for
}-1<\alpha<0,\end{cases}$

and

(80)
$s^{-\alpha}\operatorname{B}(\alpha,1+\frac{1}{s})=\begin{cases}\int_{0}^{\infty}t^{\alpha-1}(1-st)_{\scriptscriptstyle+}^{1/s}\,\mathrm{d}t&\text{for
}0<\alpha,\\\\[4.0pt]
\int_{0}^{\infty}t^{\alpha-1}((1-st)_{\scriptscriptstyle+}^{1/s}-1)\,\mathrm{d}t&\text{for
}-1<\alpha<0.\end{cases}$

We extend Lemma 16 to $\alpha\in(-1,\infty)$ using the analytic continuation
formula from (77). Special cases of the following lemma were obtained by
Koldobsky, Pajor, and Yaskin [18, Proof of Lemma 3.3] and Fradelizi, Li, and
Madiman [8, Theorem 6.1].

######  Lemma 17.

Let $\omega:[0,\infty)\to[0,\infty)$ be decreasing with

$0<\int_{0}^{\infty}t^{\alpha-1}\omega(t)\,\mathrm{d}t<\infty$

for $\alpha>0$ and

$0<\int_{0}^{\infty}t^{\alpha-1}(\omega(0)-\omega(t))\,\mathrm{d}t<\infty$

for $-1<\alpha<0$. If $\varphi:[0,\infty)\to[0,\infty)$ is non-zero, with
$\varphi(0)=0$, and such that $t\mapsto\varphi(t)$ and $t\mapsto\varphi(t)/t$
are increasing on $(0,\infty)$, then

(81)
$\zeta(\alpha)=\begin{cases}\displaystyle\left(\frac{\int_{0}^{\infty}t^{\alpha-1}\omega(\varphi(t))\,\mathrm{d}t}{\int_{0}^{\infty}t^{\alpha-1}\omega(t)\,\mathrm{d}t}\right)^{\frac{1}{\alpha}}&\text{for
}\alpha>0\\\\[12.0pt]
\displaystyle\exp\Big{(}\int_{0}^{\infty}\frac{\omega(\varphi(t))-\omega(t)}{t\,\omega(0)}\,\mathrm{d}t\Big{)}&\text{for
}\alpha=0\\\\[8.0pt]
\displaystyle\left(\frac{\int_{0}^{\infty}t^{\alpha-1}(\omega(\varphi(t))-\omega(0))\,\mathrm{d}t}{\int_{0}^{\infty}t^{\alpha-1}(\omega(t)-\omega(0))\,\mathrm{d}t}\right)^{\frac{1}{\alpha}}&\text{for
}-1<\alpha<0\end{cases}$

is a continuous, decreasing function of $\alpha$ on $(-1,+\infty)$. Moreover,
$\zeta$ is constant on $(-1,\infty)$ if $\varphi(t)=\lambda t$ on $[0,\infty)$
for some $\lambda>0$.

###### Proof.

First, we show that $\zeta$ is well-defined. Take $t_{0}>0$ such that
$\varphi(t_{0})>0$, and set $v_{0}=\varphi(t_{0})/{t_{0}}$. Since
$t\mapsto\varphi(t)/t$ is increasing, we have $\varphi(t)\leq v_{0}t$ for
$0\leq t\leq t_{0}$ and $\varphi(t)\geq v_{0}t$ for $t\geq t_{0}$.

For $\alpha\in(-1,\infty)$, we have

(82)
$\displaystyle\begin{split}v_{0}\int_{t_{0}}^{\infty}t^{\alpha-1}\omega(\varphi(t))\,\mathrm{d}t&\leq
v_{0}\int_{t_{0}}^{\infty}t^{\alpha-1}\omega(v_{0}t)\,\mathrm{d}t\\\
&=\int_{v_{0}}^{\infty}\left(\frac{t}{v_{0}}\right)^{\alpha-1}\omega(t)\,\mathrm{d}t,\end{split}$

where the last integral is finite and increasing with respect to $\alpha$, and

(83)
$\displaystyle\begin{split}v_{0}\int_{0}^{t_{0}}t^{\alpha-1}(\omega(0)-\omega(\varphi(t)))\,\mathrm{d}t&\leq
v_{0}\int_{0}^{t_{0}}t^{\alpha-1}(\omega(0)-\omega(v_{0}t))\,\mathrm{d}t\\\
&=\int_{0}^{v_{0}}\left(\frac{t}{v_{0}}\right)^{\alpha-1}(\omega(0)-\omega(t))\,\mathrm{d}t,\end{split}$

where the last integral is finite and decreasing with respect to $\alpha$. Now
(82), (83), and (77) together with the dominated convergence theorem show that
$\zeta$ is well-defined and continuous on $(-1,0)$ and $(0,\infty)$.

For $\alpha=0$, we have

(84)
$\displaystyle\begin{split}\int_{0}^{t_{0}}&\Big{|}\frac{\omega(\varphi(t))-\omega(t)}{t}\Big{|}\,\mathrm{d}t\\\
&\leq\int_{0}^{t_{0}}t^{-1}(\omega(0)-\omega(\varphi(t)))\,\mathrm{d}t+\int_{0}^{t_{0}}t^{-1}(\omega(0)-\omega(t))\,\mathrm{d}t\\\
&\leq\int_{0}^{v_{0}}t^{-1}(\omega(0)-\omega(t))\,\mathrm{d}t+\int_{0}^{t_{0}}t^{-1}(\omega(0)-\omega(t))\,\mathrm{d}t\end{split}$

and

(85)
$\displaystyle\begin{split}\int_{t_{0}}^{\infty}\left|\frac{\omega(\varphi(t))-\omega(t)}{t}\right|\,\mathrm{d}t&\leq\int_{v_{0}}^{\infty}t^{-1}\omega(t)\,\mathrm{d}t+\int_{t_{0}}^{\infty}t^{-1}\omega(t)\,\mathrm{d}t.\end{split}$

The monotonicity of the last integrals of (82) and (83) shows that the
integrals in (84) and (85) are finite. Hence $\zeta(0)$ is well-defined.

It follows from Lemma 16 that $\zeta$ is decreasing on $(0,\infty)$. We show
that it is decreasing on $(-1,0)$. By the change of variables $r=t^{-1}$, we
get

$\zeta(\alpha)^{-1}=\left(\frac{\int_{0}^{\infty}r^{-\alpha-1}(\omega(0)-\omega(\psi(r)^{-1}))\,\mathrm{d}r}{\int_{0}^{\infty}r^{-\alpha-1}(\omega(0)-\omega(r^{-1}))\,\mathrm{d}r}\right)^{-\frac{1}{\alpha}}$

where $\psi(r)=\varphi(r^{-1})^{-1}$. Observe that $\omega(0)-\omega(r^{-1})$
is decreasing and non-negative, and that
${\psi(r)}/r=\left({\varphi(r^{-1})}/{r^{-1}}\right)^{-1}$ is increasing.
Since $-\alpha\in(0,1)$, Lemma 16 implies that $\zeta(\alpha)^{-1}$ is an
increasing function of $\alpha$.

Using (77) and the elementary relation,

$\lim_{\alpha\to
0^{\pm}}\left(\frac{a(\alpha)+\frac{c}{\alpha}}{b(\alpha)+\frac{c}{\alpha}}\right)^{\frac{1}{\alpha}}=\exp\left(\frac{a(0)-b(0)}{c}\right),$

which is valid for a constant $c\neq 0$ and continuous functions
$\alpha\mapsto a(\alpha)$ and $\alpha\mapsto b(\alpha)$, we obtain that

$\lim_{\alpha\to
0^{\pm}}\zeta(\alpha)=\exp\left(\int_{0}^{\infty}\frac{\omega(\varphi(t))-\omega(t)}{t\,\omega(0)}\,\mathrm{d}t\right).$

So $\zeta$ is continuous at $\alpha=0$. ∎

###  7.2. Radial mean bodies of functions

Let $f\in L^{2}(\mathbb{R}^{n})$ be non-zero and non-negative. We define

(86)
$\operatorname{R}_{\alpha}\\!f=\left(\frac{\alpha}{\|f\|_{2}^{2}}\right)^{\frac{1}{\alpha}}\operatorname{S}_{\alpha}\\!f$

for $\alpha>0$ and

(87)
$\operatorname{R}_{\alpha}\\!f=\left(\frac{|\alpha|}{2\|f\|_{2}^{2}}\right)^{\frac{1}{\alpha}}\Pi_{2}^{\ast,-\alpha/2}f$

for $-1<\alpha<0$. In addition, we define $\operatorname{R}_{0}\\!f$ by its
radial function for $\xi\in{\mathbb{S}}^{n-1}$ as

(88)
$\log(\rho_{\operatorname{R}_{0}\\!f}(\xi))=-\gamma+\int_{0}^{\infty}\frac{1}{t}\left(\frac{1}{\|f\|_{2}^{2}}\int_{\mathbb{R}^{n}}f(x)f(x+t\xi)\,\mathrm{d}x-e^{-t}\right)\,\mathrm{d}t,$

where $\gamma$ is Euler’s constant. The definitions (86) and (87) are
compatible with (68) and (74) for $f=\operatorname{1}_{E}$ and a convex body
$E\subset\mathbb{R}^{n}$. Note that (14) implies that

(89) $|\operatorname{R}_{n}\\!f|=\frac{\|f\|_{1}^{2}}{\|f\|_{2}^{2}}$

for non-zero $f\in L^{2}(\mathbb{R}^{n})$.

###  7.3. Results for log-concave functions

We start with a simple calculation, for which we need the following notation.
For $f\in L^{2}(\mathbb{R}^{n})$ and $y\in\mathbb{R}^{n}$, we set

(90) $\mathscr{G}\\!f(y)=\int_{\mathbb{R}^{n}}f(x)\,f(x+y)\,\mathrm{d}x.$

We define the simplex $\Delta_{n}\subset\mathbb{R}^{n}$ as the convex hull of
the origin and the standard basis vectors $e_{1},\dots,e_{n}$. In addition,
let $B_{1}^{n}=\\{(x_{1},\dots,x_{n}):|x_{1}|+\dots+|x_{n}|\leq 1\\}$ and
$\mathbb{R}^{n}_{+}=\\{(x_{1},\dots,x_{n})\in\mathbb{R}^{n}:x_{1},\dots,x_{n}\geq
0\\}$.

######  Lemma 18.

If $f(x)=e^{-\|x\|_{\Delta_{n}}}$ for $x\in\mathbb{R}^{n}$, then

$\mathscr{G}\\!f(y)=\frac{1}{2^{n}}e^{-\|y\|_{B_{1}^{n}}}$

for $y\in\mathbb{R}^{n}$.

###### Proof.

We have

$f(x)=\begin{cases}e^{-(x_{1}+\dots+x_{n})}&\text{ for
}x\in\mathbb{R}^{n}_{+},\\\ 0&\text{ for
}x\not\in\mathbb{R}^{n}_{+}.\end{cases}$

Setting $g=\mathscr{G}\\!f$, we obtain that

(91)
$g(y)=e^{-(y_{1}+\dots+y_{n})}\prod_{i=1}^{n}\int_{\mathbb{R}_{+}\cap(\mathbb{R}_{+}-y_{i})}e^{-2x_{i}}\,\mathrm{d}x_{i}.$

Using

(92)
$2\int_{\mathbb{R}_{+}\cap(\mathbb{R}_{+}-y_{i})}e^{-2x_{i}}\,\mathrm{d}x_{i}=\begin{cases}1&\text{
for }y_{i}\geq 0\\\ e^{2y_{i}}&\text{ for }y_{i}<0\end{cases}$

and $I_{\scriptscriptstyle-}(y)=\\{1\leq i\leq n:y_{i}<0\\}$, we obtain that

(93) $\displaystyle g(y)$
$\displaystyle=\frac{1}{2^{n}}\,e^{-(y_{1}+\dots+y_{n})+2\sum_{i\in
I_{\scriptscriptstyle-}(y)}^{n}y_{i}}=\frac{1}{2^{n}}\,e^{-\|y\|_{B_{1}^{n}}}$

for $y\in\mathbb{R}^{n}$. ∎

We establish the following inclusion relation.

######  Theorem 19.

If $f\in L^{2}(\mathbb{R}^{n})$ is non-zero and log-concave, then

$\frac{1}{\Gamma(\beta+1)^{1/\beta}}\operatorname{R}_{\beta}\\!f\subseteq\frac{1}{\Gamma(\alpha+1)^{1/\alpha}}\operatorname{R}_{\alpha}\\!f$

for $-1<\alpha<\beta<\infty$. There is equality if
$f(x)=a\,e^{-\|x-x_{0}\|_{\Delta}}$ for $x\in\mathbb{R}^{n}$ with
$a>0,x_{0}\in\mathbb{R}^{n}$ and $\Delta$ an $n$-dimensional simplex having a
vertex at the origin.

###### Proof.

Note that $g=\mathscr{G}\\!f$ is even, attains a maximum at $y=0$ and that
$g(0)=\|f\|_{2}^{2}$. Moreover, $g$ is log-concave by Lemma 6.

We fix $\xi\in{\mathbb{S}}^{n-1}$ and see that $t\mapsto g(t\xi)$ is a
positive, decreasing, log-concave function. We apply Lemma 17 to
$\omega(t)=g(0)e^{-t},\varphi(t)=-\log(g(t\xi)/g(0))$ and observe that

$\zeta(\alpha)=\left(\frac{\int_{0}^{\infty}t^{\alpha-1}g(t\xi)\,\mathrm{d}t}{\int_{0}^{\infty}t^{\alpha-1}g(0)e^{-t}\,\mathrm{d}t}\right)^{1/\alpha}=\frac{\rho_{\operatorname{R}_{\alpha}\\!f}(\xi)}{\Gamma(\alpha+1)^{1/\alpha}}$

for $\alpha>0$ and

$\zeta(\alpha)=\left(\frac{\int_{0}^{\infty}t^{\alpha-1}(g(0)-g(t\xi))\,\mathrm{d}t}{\int_{0}^{\infty}t^{\alpha-1}(g(0)-g(0)e^{-t})\,\mathrm{d}t}\right)^{1/\alpha}\\\
=\frac{\rho_{\operatorname{R}_{\alpha}\\!f}(\xi)}{\Gamma(\alpha+1)^{1/\alpha}}$

for $-1<\alpha<0$, where we used formula (79). This proves the inclusion.
There is equality in Lemma 17 if $g(t\xi)=e^{-h(\xi)t}$ for some function
$h:{\mathbb{S}}^{n-1}\to(0,\infty)$. Hence, the equality case follows from
Lemma 18 and the $\operatorname{SL}(n)$ and translation invariance and
homogeneity of $|\operatorname{R}_{\alpha}\\!f|$. ∎

Using (86) and (89), we obtain both Theorem 2 and Theorem 15 immediately from
the above result. Using (87), (89), and Hölder’s inequality, we also obtain
the following inequality from Theorem 19.

######  Theorem 20.

For $\,0<\alpha<1/2$ and log-concave $f\in L^{2}(\mathbb{R}^{n})$,

$\frac{\alpha}{\Gamma(n+1)^{\frac{2\alpha}{n}}\Gamma(1-2\alpha)}\,|\Pi_{2}^{\ast,\alpha}f|^{-\frac{2\alpha}{n}}\leq\|f\|_{2}^{2+\frac{4\alpha}{n}}\|f\|_{1}^{-\frac{4\alpha}{n}}\leq\|f\|_{\frac{2n}{n-2\alpha}}^{2}.$

There is equality in the first inequality if
$f(x)=a\,e^{-\|x-x_{0}\|_{\Delta}}$ for $x\in\mathbb{R}^{n}$ with $a\geq
0,x_{0}\in\mathbb{R}^{n}$ and $\Delta$ an $n$-dimensional simplex having a
vertex at the origin.

The inequality in the previous theorem is a reverse inequality to the affine
fractional $L^{2}$ Sobolev inequality (12).

###  7.4. Results for $s$-concave functions

We obtain the following inclusion relation.

######  Theorem 21.

Let $s>0$. If $f\in L^{2}(\mathbb{R}^{n})$ is non-zero and $s$-concave, then

$\frac{1}{\big{(}(n+\frac{2}{s})\operatorname{B}(\beta+1,n+\frac{2}{s})\big{)}^{1/\beta}}\operatorname{R}_{\beta}\\!f\subseteq\frac{1}{\big{(}(n+\frac{2}{s})\operatorname{B}(\alpha+1,n+\frac{2}{s})\big{)}^{1/\alpha}}\operatorname{R}_{\alpha}\\!f$

for $-1<\alpha<\beta<\infty$.

###### Proof.

As in the proof of Theorem 19, note that $g=\mathscr{G}\\!f$ is even,
continuous, and attains a maximum at $y=0$. For $\xi\in S^{n-1}$, it follows
that $t\mapsto g(t\xi)$ is positive and decreasing. By Lemma 6, the function
$g$ is $r$-concave with $r={s}/{(ns+2)}$.

We apply Lemma 17 with $\omega(t)=g(0)(1-rt)_{\scriptscriptstyle+}^{1/r}$ and
$\varphi(t)=(1-(g(t\xi)/g(0))^{r})/r$. We obtain that

$\zeta(\alpha)=\left(\frac{\int_{0}^{\infty}t^{\alpha-1}g(t\xi)\,\mathrm{d}t}{\int_{0}^{\infty}g(0)t^{\alpha-1}(1-rt)_{\scriptscriptstyle+}^{1/r}\,\mathrm{d}t}\right)^{1/\alpha}=\frac{r\,\rho_{\operatorname{R}_{\alpha}\\!f}(\xi)}{(\alpha\operatorname{B}(\alpha,1+\frac{1}{r}))^{1/\alpha}}$

for $\alpha>0$ and

$\zeta(\alpha)=\left(\frac{\int_{0}^{\infty}t^{\alpha-1}(g(0)-g(t\xi))\,\mathrm{d}t}{\int_{0}^{\infty}t^{\alpha-1}(g(0)(1-rt)_{\scriptscriptstyle+}^{1/r}-g(0))\,\mathrm{d}t}\right)^{1/\alpha}\\\
=\frac{r\,\rho_{\operatorname{R}_{\alpha}\\!f}(\xi)}{(\alpha\operatorname{B}(\alpha,1+\frac{1}{r}))^{1/\alpha}}$

for $-1<\alpha<0$, where we used formula (80). The result now follows from
Lemma 17. ∎

For $s\to 0$, we recover Theorem 19 from Theorem 21. For $s\to\infty$ and
$f=\operatorname{1}_{E}$ for a convex body $E\subset\mathbb{R}^{n}$, Theorem
21 implies that

(94)
$\frac{1}{\big{(}n\operatorname{B}(\beta+1,n)\big{)}^{1/\beta}}\operatorname{R}_{\beta}\\!E\subseteq\frac{1}{\big{(}n\operatorname{B}(\alpha+1,n)\big{)}^{1/\alpha}}\operatorname{R}_{\alpha}\\!E$

for $-1<\alpha<\beta$. This recovers Theorem 5.5 by Gardner and Zhang [12],
who showed that there is equality in (94) precisely for $n$-dimensional
simplices. The problem to determine the precise equality conditions in Theorem
19 and Theorem 21 is open.

It follows from the proof of Theorem 21 under the assumptions given there that

(95)
$\frac{\rho_{\operatorname{R}_{\beta}\\!f}(\xi)}{\big{(}(n+\frac{2}{s})\operatorname{B}(\beta+1,n+\frac{2}{s})\big{)}^{1/\beta}}\leq\frac{\rho_{\operatorname{R}_{\alpha}\\!f}(\xi)}{\big{(}(n+\frac{2}{s})\\!\operatorname{B}(\alpha+1,n+\frac{2}{s})\big{)}^{1/\alpha}}$

with equality for $\xi\in{\mathbb{S}}^{n-1}$ if
$\mathscr{G}\\!f(t\xi)=a(1-\lambda t)_{\scriptscriptstyle+}^{1/s}$ for $t>0$
with $a\geq 0$ and $\lambda>0$. The following lemma shows that the inequality
in (95) is sharp in some directions and that the constants in Theorem 21 are
optimal.

######  Lemma 22.

Let $s>0$. If $f(x)=(1-\|x\|_{\Delta_{n}})_{\scriptscriptstyle+}^{1/s}$ for
$x\in\mathbb{R}^{n}$, then

$\mathscr{G}\\!f(y)=a\big{(}1-\frac{1}{2}\|y\|_{B^{n}_{1}}\big{)}_{\scriptscriptstyle+}^{n+2/s}$

for every $y\in\mathbb{R}^{n}$ with $y_{1}+\dots+y_{n}=0$, where
$a=\operatorname{B}(n,1+2/s)/(n-1)!$.

###### Proof.

Let $t_{\scriptscriptstyle-}=\max\\{0,-t\\}$ for $t\in\mathbb{R}$ and
$y_{\scriptscriptstyle-}=((y_{1})_{\scriptscriptstyle-},\dots,(y_{n})_{\scriptscriptstyle-})$
for $y\in\mathbb{R}^{n}$. Setting
$\delta(t)=(1-t)_{\scriptscriptstyle+}^{1/s}$, we have

(96) $\displaystyle\mathscr{G}\\!f(y)$
$\displaystyle=\int_{\mathbb{R}^{n}_{+}\cap(\mathbb{R}^{n}_{+}-y)}\delta\big{(}\sum_{i=1}^{n}x_{i}\big{)}\,\delta\big{(}\sum_{i=1}^{n}(x_{i}+y_{i})\big{)}\,\mathrm{d}x$
(97)
$\displaystyle=\int_{\mathbb{R}^{n}_{+}+y_{\scriptscriptstyle-}}\delta\big{(}\sum_{i=1}^{n}x_{i}\big{)}\,\delta\big{(}\sum_{i=1}^{n}(x_{i}+y_{i})\big{)}\,\mathrm{d}x$
(98)
$\displaystyle=\int_{\mathbb{R}^{n}_{+}}\delta\big{(}\sum_{i=1}^{n}(x_{i}+(y_{i})_{\scriptscriptstyle-})\big{)}\,\delta\big{(}\sum_{i=1}^{n}(x_{i}+y_{i}+(y_{i})_{\scriptscriptstyle-}\big{)}\,\mathrm{d}x$
(99)
$\displaystyle=\int_{\mathbb{R}^{n}_{+}}\delta\big{(}\sum_{i=1}^{n}x_{i}+\sum_{i=1}^{n}(y_{i})_{\scriptscriptstyle-}\big{)}\,\delta\big{(}\sum_{i=1}^{n}x_{i}+\sum_{i=1}^{n}(y_{i})_{\scriptscriptstyle+})\,\mathrm{d}x$
(100)
$\displaystyle=\frac{1}{(n-1)!}\int_{0}^{\infty}r^{n-1}\delta(r+\|y_{\scriptscriptstyle-}\|_{B^{n}_{1}})\,\delta(r+\|y_{\scriptscriptstyle+}\|_{B^{n}_{1}})\,\mathrm{d}r.$

Now note that if
$\|y_{\scriptscriptstyle-}\|_{B^{n}_{1}}=\|y_{\scriptscriptstyle+}\|_{B^{n}_{1}}=\frac{1}{2}\|y\|_{B^{n}_{1}}=t$,
then

(101) $\displaystyle\mathscr{G}\\!f(y)$
$\displaystyle=\frac{1}{(n-1)!}\int_{0}^{\infty}r^{n-1}\delta(r+t)^{2}\,\mathrm{d}r$
(102)
$\displaystyle=\frac{1}{(n-1)!}\int_{0}^{\infty}r^{n-1}(1-r-t)_{\scriptscriptstyle+}^{2/s}\,\mathrm{d}r.$

For $t>1$, this quantity is $0$. Otherwise, we get

(103) $\displaystyle\mathscr{G}\\!f(y)$
$\displaystyle=\frac{1}{(n-1)!}\int_{0}^{1-t}r^{n-1}(1-r-t)^{2/s}\,\mathrm{d}r$
(104)
$\displaystyle=\frac{1}{(n-1)!}(1-t)\int_{0}^{1}((1-t)s)^{n-1}((1-t)-(1-t)r)^{2/s}\,\mathrm{d}r$
(105)
$\displaystyle=\frac{1}{(n-1)!}(1-t)^{n+2/s}\int_{0}^{1}r^{n-1}(1-r)^{2/s}\,\mathrm{d}r,$

which completes the proof. ∎

Using (86), (89) and Hölder’s inequality, we obtain the following inequalities
from Theorem 21.

######  Corollary 23.

Let $s>0$. If $f\in L^{2}(\mathbb{R}^{n})$ is $s$-concave on its support, then

$\frac{(n\operatorname{B}(n,n+1+\frac{2}{s}))^{\frac{\alpha}{n}}}{\operatorname{B}(\alpha,n+1+\frac{2}{s})}\,|\operatorname{S}_{\alpha}\\!f|^{\frac{\alpha}{n}}\geq\|f\|_{2}^{2-\frac{2\alpha}{n}}\|f\|_{1}^{\frac{2\alpha}{n}}\geq\|f\|_{\frac{2n}{n+\alpha}}^{2}$

for $\,0<\alpha<n$ and the inequalities are reversed for $\alpha>n$.

Using (87), (89), and Hölder’s inequalities, we obtain the following
inequality from Theorem 21.

######  Corollary 24.

Let $s>0$. If $f\in L^{2}(\mathbb{R}^{n})$ is $s$-concave on its support, then

$\frac{(n\operatorname{B}(n,n+1+\frac{2}{s}))^{-\frac{2\alpha}{n}}}{2|\operatorname{B}(-2\alpha,n+1+\frac{2}{s})|}\,|\Pi_{2}^{\ast,\alpha}f|^{-\frac{2\alpha}{n}}\leq\|f\|_{2}^{2+\frac{4\alpha}{n}}\|f\|_{1}^{-\frac{4\alpha}{n}}\leq\|f\|_{\frac{2n}{n-2\alpha}}^{2}$

for $\,0<\alpha<1/2$.

The following problems remain open: Are the first inequalities in Corollary 23
and Corollary 24 sharp for $f\not\equiv 0$?

### Acknowledgments

J. Haddad was supported by PAIDI 2020 PY20-00664, and M. Ludwig was supported,
in part, by the Austrian Science Fund (FWF): P 34446.

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