Affine fractional $L^p$ Sobolev inequalities

Abstract

Sharp affine fractional $L^p$ Sobolev inequalities for functions on ${\mathbb {R}}^n$ are established. The new inequalities are stronger than (and directly imply) the sharp fractional $L^p$ Sobolev inequalities. They are fractional versions of the affine $L^p$ Sobolev inequalities of Lutwak, Yang, and Zhang. In addition, affine fractional asymmetric $L^p$ Sobolev inequalities are established.

1 Introduction

Sharp fractional $L^2$ Sobolev inequalities are receiving increasing attention in the last decades. They are central in the study of solutions of equations involving the fractional Laplace operator $(-\Delta )^{1/2}$ which arises naturally in many non-local problems such as the stationary form of reaction-diffusion equations [9], the Signorini problem (and its equivalent formulation as the thin obstacle problem) [2], and the Dirichlet-to-Neumann operator of harmonic functions in the half-space [29]. Also, the general operators $(-\Delta )^s$ for $s \in (0,1)$ arise in stochastic theory, associated with symmetric Levy processes (see [29] and the references therein).

Let $0< s<1$ and $1\le p < n/s$. The fractional $L^p$ Sobolev inequalities state that

$$\begin{aligned} \Vert f\Vert _{{\frac{n p}{n-p s}}}^p \le \sigma _{n,p,s}\int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{\vert f(x)-f(y)\vert ^p}{\vert x-y \vert ^{n+p s}}\,\textrm{d}x\,\textrm{d}y \end{aligned}$$

(1)

for $f\in W^{s,p}({\mathbb {R}}^n)$, the fractional $L^p$ Sobolev space of functions $f\in L^p({\mathbb {R}}^n)$ with finite right side in (1) (see, for example, [27]). In general, the optimal constants $\sigma _{n,p,s}$ and extremal functions are not known (see [7] for a conjecture). Equality is always attained in (1). For $p=1$, the extremal functions of (1) are multiples of indicator functions of balls, and the constants are explicitly known. The only further known case is $p=2$, where the constants $\sigma _{n,2,s}$ can be obtained by duality from Lieb’s sharp Hardy–Littlewood–Sobolev inequalities [18] (see, for example, [10]). The asymptotic behavior of $\sigma _{n,p,s}$ as $s \rightarrow 1^-$ was studied in [5]. Almgren and Lieb [1] and Frank and Seiringer [12] showed that the extremal functions of (1) are radially symmetric and of constant sign.

By a result of Bourgain, Brezis, and Mironescu [4],

$$\begin{aligned} \lim _{s\rightarrow 1^-} p (1-s) \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{\vert f(x)-f(y)\vert ^p}{\vert x-y \vert ^{n+p s}}\,\textrm{d}x\,\textrm{d}y = \alpha _{n,p} \int _{{\mathbb {R}}^n} \vert \nabla f(x)\vert ^p \,\textrm{d}x \end{aligned}$$

for $f\in W^{1,p}({\mathbb {R}}^n)$, the Sobolev space of $L^p$ functions f with weak $L^p$ gradient $\nabla f$, where

$$\begin{aligned} \alpha _{n,p}= \int _{{\mathbb {S}}^{n-1}} \vert \langle \xi ,\eta \rangle \vert ^p \,\textrm{d}\xi \end{aligned}$$

(2)

for any $\eta \in {\mathbb {S}}^{n-1}$. Here, integration on the unit sphere ${\mathbb {S}}^{n-1}$ is with respect to the $(n-1)$-dimensional Hausdorff measure, $\omega _n$ is the volume of the n-dimensional unit ball and $\langle \cdot ,\cdot \rangle$ is the inner product on ${\mathbb {R}}^n$. For $p=1$ and $p=2$, this allows to deduce the sharp $L^p$ Sobolev inequalities from (1) by calculating the limit of $\sigma _{n,p,s}/(1-s)$ as $s\rightarrow 1^-$.

Zhang [32] and Lutwak, Yang, and Zhang [24] obtained the following sharp affine $L^p$ Sobolev inequality that is significantly stronger than the classical $L^p$ Sobolev inequality:

$$\begin{aligned} \Vert f\Vert _{{\frac{n p}{n-p}}}^p \le \sigma _{n,p} \frac{n \omega _n^{\frac{n+p}{n}}}{\alpha _{n,p}} \vert {\Pi ^{*}_{p}\, f}\vert ^{-\frac{p }{n}} \le \sigma _{n,p}\int _{{\mathbb {R}}^n} \vert \nabla f(x) \vert ^p \,\textrm{d}x \end{aligned}$$

(3)

for $f\in W^{1,p}({\mathbb {R}}^n)$ and $1<p<n$, where the inequality between the first and third terms is the classical $L^p$ Sobolev inequality and the optimal constants $\sigma _{n,p}$ were determined by Aubin [3] and Talenti [30]. We have rewritten the explicit constant for the first inequality from [24] using (2). Here $\Pi ^{*}_{p}\, f$ is the $L^p$ polar projection body of f, a convex body associated to f that was introduced with different notation in [24] (see Sect. 2.5), and $\vert {\cdot }\vert$ is the n-dimensional Lebesgue measure.

The main aim of this paper is to establish affine fractional $L^p$ Sobolev inequalities that are stronger than the Euclidean fractional $L^p$ Sobolev inequalities from (1) and are fractional counterparts of (3). The case $p=1$ was studied in [16], so from now on, let $p>1$.

Theorem 1

Let $0<s<1$ and $1<p<n/s$. For $f\in W^{s,p}({\mathbb {R}}^n)$,

$$\begin{aligned}&\Vert f\Vert _{{\frac{n p}{n-p s}}}^p \\&\quad \le \sigma _{n,p,s} n\omega _n^{\frac{n+p s}{n}} \left( \frac{1}{n} \int _{{\mathbb {S}}^{n-1}} \left( \int _0^\infty t^{p s-1} \int _{{\mathbb {R}}^n} \vert f(x+ t\xi )-f(x) \vert ^p\,\textrm{d}x \,\textrm{d}t\right) ^{-\frac{n}{p s}}\,\textrm{d}\xi \right) ^{-\frac{p s}{n}}\\&\quad \le \sigma _{n,p,s}\int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{\vert f(x)-f(y) \vert ^p}{\vert x-y \vert ^{n+p s}}\,\textrm{d}x\,\textrm{d}y. \end{aligned}$$

There is equality in the first inequality if and only if $f=h_{s,p}\circ \phi$ for some $\phi \in \textrm{GL}(n)$, where $h_{s,p}$ is an extremal function of (1). There is equality in the second inequality if f is radially symmetric.

In order to prove Theorem 1, we introduce the s-fractional $L^p$ polar projection body $\Pi ^{*,{ s}}_{{p}} f$ associated to f, defined as the star-shaped set whose gauge function for $\xi \in {\mathbb {S}}^{n-1}$ is

$$\begin{aligned} \Vert {\xi }\Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s}= \int _0^\infty t^{-p s-1}\, \int _{{\mathbb {R}}^n}\vert f(x +t\xi )-f(x)\vert ^p \,\textrm{d}x\,\textrm{d}t \end{aligned}$$

(see Section 3 for details). The affine fractional Sobolev inequality can now be written as

$$\begin{aligned} \Vert f\Vert _{{\frac{n p}{n-p s}}}^p \le \sigma _{n,p,s} n\omega _n^{\frac{n+p s}{n}} \vert {\Pi ^{*,{ s}}_{{p}}f}\vert ^{- \frac{p s}{n}}. \end{aligned}$$

(4)

Since both sides of (4) are invariant under translations of f, and for volume-preserving linear transformations $\phi : {\mathbb {R}}^n\rightarrow {\mathbb {R}}^n$,

$$\begin{aligned} \Pi ^{*,{ s}}_{{p}} \,(f\circ \phi ^{-1}) = \phi \Pi ^{*,{ s}}_{{p}} f, \end{aligned}$$

it follows that (4) is an affine inequality. In Theorem 10, we will show that

$$\begin{aligned} \lim _{s\rightarrow 1^-} p(1-s) \vert {\Pi ^{*,{ s}}_{{p}} f}\vert ^{-\frac{p s}{n}} = \vert {\Pi _{p}^{*}f}\vert ^{-\frac{p}{n}}, \end{aligned}$$

which establishes the connection to the $L^p$ polar projection bodies introduced by Lutwak, Yang and Zhang [24].

In Sect. 4, we introduce fractional asymmetric $L^p$ polar projection bodies as fractional counterparts of the asymmetric $L^p$ polar projection bodies of Haberl and Schuster [14], which in turn are functional versions of the asymmetric $L^p$ polar projection bodies of convex bodies introduced in [19]. We obtain affine fractional asymmetric $L^p$ Sobolev inequalities for non-negative functions that are stronger than the inequalities for the symmetric fractional $L^p$ polar projection bodies.

In the proofs of the main results, we use anisotropic fractional Sobolev norms, which were introduced in [20, 21] and depend on a star-shaped set $K\subset {\mathbb {R}}^n$. In Sect. 10, we discuss which choice of K (with given volume) gives the minimal fractional Sobolev norm and connect it to the corresponding quest for an optimal $L^p$ Sobolev norm solved by Lutwak, Yang, and Zhang [25].

2 Preliminaries

We collect results on function spaces, Schwarz symmetrization, star-shaped sets, anisotropic Sobolev norms, and $L^p$ polar projection bodies that will be used in the following.

2.1 Function spaces

For $p\ge 1$ and measurable $f:{\mathbb {R}}^n\rightarrow {\mathbb {R}}$, let

$$\begin{aligned} \Vert f\Vert _p =\Big (\int _{{\mathbb {R}}^n} \vert f(x)\vert ^p\,\textrm{d}x \Big )^{1/p}. \end{aligned}$$

We set $\{f\ge t\}=\{ x\in {\mathbb {R}}^n: f(x)\ge t\}$ for $t\in {\mathbb {R}}$ and use similar notation for level sets, etc. We say that f is non-zero, if $\{f\ne 0\}$ has positive measure, and we identify functions that are equal up to a set of measure zero. For $p\ge 1$, let

$$\begin{aligned} L^{p}({\mathbb {R}}^n) = \Big \{f: {\mathbb {R}}^n\rightarrow {\mathbb {R}}: f \text { is measurable}, \Vert f\Vert _p < \infty \Big \}. \end{aligned}$$

Here and below, when we use measurability and related notions, we refer to the n-dimensional Lebesgue measure on ${\mathbb {R}}^n$.

For $0<s<1$ and $p\ge 1$, we define the fractional Sobolev space $W^{s,p}({\mathbb {R}}^n)$ as

$$\begin{aligned} W^{s,p}({\mathbb {R}}^n) = \left\{ f \in L^p({\mathbb {R}}^n) : \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{|f(x)-f(y)|^p}{|x-y|^{n+p s}} \,\textrm{d}x \,\textrm{d}y < \infty \right\} . \end{aligned}$$

For $p\ge 1$, we set

$$\begin{aligned} W^{1,p}({\mathbb {R}}^n) = \big \{f \in L^p({\mathbb {R}}^n) : \vert \nabla f \vert \in L^p({\mathbb {R}}^n)\big \}, \end{aligned}$$

where $\nabla f$ is the weak gradient of f.

2.2 Symmetrization

For a set $E\subset {\mathbb {R}}^n$, the indicator function $1_E$ is defined by $1_E(x)=1$ for $x\in E$ and $1_E(x)=0$ otherwise. 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 $f:{\mathbb {R}}^n\rightarrow {\mathbb {R}}$ be a non-negative measurable function with super-level sets $\{f \ge t\}$ of finite measure. The layer cake formula states that

$$\begin{aligned} f(x) = \int _0^\infty 1_{\{f\ge t\}}(x) \,\textrm{d}t \end{aligned}$$

for almost every $x \in {\mathbb {R}}^n$ and allows us to recover the function from its super-level sets. The Schwarz symmetral of f, denoted by $f^\star$, is defined by

$$\begin{aligned} f^\star (x) = \int _0^\infty 1_{\{f\ge t\}^\star }(x) \,\textrm{d}t \end{aligned}$$

for $x\in {\mathbb {R}}^n$. Hence, $f^\star$ is determined by the properties of being radially symmetric, decreasing, and having super-level sets of the same measure as those of f. Note that $f^\star$ is also called the symmetric decreasing rearrangement of f.

The proofs of our results make use of the Riesz rearrangement inequality, which is stated in full generality, for example, in [6].

Theorem 2

(Riesz’s rearrangement inequality) For $f,g,k:{\mathbb {R}}^n \rightarrow {\mathbb {R}}$ non-negative, measurable functions with super-level sets of finite measure,

$$\begin{aligned} \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} f(x) k(x-y) g(y) \,\textrm{d}x \,\textrm{d}y \le \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} f^\star (x) k^\star (x-y) g^\star (y) \,\textrm{d}x \,\textrm{d}y. \end{aligned}$$

We will use the characterization of equality cases of the Riesz rearrangement inequality due to Burchard [8].

Theorem 3

(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 $\,\vert \alpha - \beta \vert< \gamma < \alpha +\beta$, there is equality in

$$\begin{aligned} \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} 1_A(y) 1_B(x-y) 1_C(x) \,\textrm{d}x \,\textrm{d}y \le \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} 1_{A^\star }(y) 1_{B^\star }(x-y) 1_{C^\star }(x) \,\textrm{d}x \,\textrm{d}y \end{aligned}$$

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

$$\begin{aligned} A=a+\alpha D,\, B = b+\beta D,\, C = c+\gamma D, \end{aligned}$$

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

2.3 Star-shaped sets and star bodies

A 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 $\Vert \cdot \Vert _K : {\mathbb {R}}^n \rightarrow [0,\infty ]$ of a star-shaped set is defined as

$$\begin{aligned} \Vert x\Vert _K = \inf \{ \lambda > 0 : x \in \lambda K\}, \end{aligned}$$

and the radial function $\rho _K:{\mathbb {R}}^n \setminus \{0\} \rightarrow [0,\infty ]$ as

$$\begin{aligned} \rho _K(x) = \Vert x\Vert _K^{-1} = \sup \{\lambda \ge 0:\lambda x \in K\}. \end{aligned}$$

The n-dimensional Lebesgue measure or volume of a star-shaped set K in ${\mathbb {R}}^n$ with measurable radial function is given by

$$\begin{aligned} \vert {K}\vert = \frac{1}{n} \int _{{\mathbb {S}}^{n-1}} \rho _K(\xi )^n\,\textrm{d}\xi . \end{aligned}$$

We call a star-shaped set $K\subset {\mathbb {R}}^n$ a star body if its radial function is strictly positive and continuous in ${\mathbb {R}}^n \setminus \{0\}$. On the set of star bodies, the q-radial sum for $q\ne 0$ of $K,L\subset {\mathbb {R}}^n$ is defined by

$$\begin{aligned} \rho ^q(K{\tilde{+}}_{ q} L,\xi )= \rho ^q(K,\xi )+\rho ^q(L,\xi ) \end{aligned}$$

for $\xi \in {\mathbb {S}}^{n-1}$ (cf. [28, Section 9.3]). The dual Brunn–Minkowski inequality (cf. [28, (9.41)]) states that for star bodies $K,L\subset {\mathbb {R}}^n$ and $q>0$,

$$\begin{aligned} \vert {K{\tilde{+}}_{ -q} L}\vert ^{-q/n} \ge \vert {K}\vert ^{-q/n} +\vert {L}\vert ^{-q/n}, \end{aligned}$$

(5)

with equality precisely if K and L are dilates, that is, there is $\lambda >0$ such that $K=\lambda L$.

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

$$\begin{aligned} \tilde{V}_\alpha (K,L) = \frac{1}{n} \int _{{\mathbb {S}}^{n-1}} \rho _K(\xi )^{n-\alpha } \rho _L(\xi )^{\alpha } \,\textrm{d}\xi . \end{aligned}$$

Note that

$$\begin{aligned} \tilde{V}_\alpha (K,K) = \vert {K}\vert \end{aligned}$$

and that

$$\begin{aligned} \tilde{V}_\alpha (K,L_1{\tilde{+}}_{ \alpha } L_2)=\tilde{V}_\alpha (K, L_1) +\tilde{V}_\alpha (K,L_2) \end{aligned}$$

for star-shaped sets $K, L_1, L_2\subseteq {\mathbb {R}}^n$ with measurable radial functions.

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

$$\begin{aligned} \tilde{V}_{\alpha }(K,L) \le \vert {K}\vert ^{({n-\alpha })/n} \vert {L}\vert ^{ \alpha /n}. \end{aligned}$$

(6)

Equality holds if and only if K and L are dilates, where we say that star-shaped sets K and L are dilates if $\rho _K=\lambda \,\rho _L$ almost everywhere on ${\mathbb {S}}^{n-1}$ for some $\lambda > 0$. The definition of dual mixed volume for star bodies is due to Lutwak [22], where also the dual mixed volume inequality is derived from Hölder’s inequality (also see [28, Section 9.3] or [13, B.29]).

2.4 Anisotropic fractional Sobolev norms

Let $0<s<1$ and $p\ge 1$. For $K\subset {\mathbb {R}}^n$ a star body and $f \in W^{s,p}({\mathbb {R}}^n)$, the anisotropic fractional $L^p$ Sobolev norm of f with respect to K is

$$\begin{aligned} \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{\vert f(x)-f(y)\vert ^p}{\Vert x-y\Vert _K^{n+p s}} \,\textrm{d}x\,\textrm{d}y. \end{aligned}$$

(7)

It was introduced in [21] for K a convex body (also, see [20]). For $K={B^n}$, the Euclidean unit ball, we obtain the classical s-fractional $L^p$ Sobolev norm of f. The limit as $s\rightarrow 1^-$ was determined in [4] in the Euclidean case and in [21] in the anisotropic case. We will also consider the following asymmetric versions of (7),

$$\begin{aligned} \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{( f(x)-f(y))_{+}^{p}}{\Vert x-y\Vert _K^{n+p s}} \,\textrm{d}x\,\textrm{d}y, \quad \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{( f(x)-f(y))_{-}^{ p}}{\Vert x-y\Vert _K^{n+p s}} \,\textrm{d}x\,\textrm{d}y, \end{aligned}$$

where $a_{+} = \max \{a,0\}$ and $a_{-} = \max \{-a, 0\}$ for $a\in {\mathbb {R}}$. The limits as $s\rightarrow 1^-$ were determined in [26].

2.5 $L^p$ polar projection bodies

For $p\ge 1$ and $f \in W^{1,p}({\mathbb {R}}^n)$, the $L^p$ polar projection body is defined as the star body with gauge function given by

$$\begin{aligned} \Vert \xi \Vert _{\Pi _{p}^{*} f}^p = \int _{{\mathbb {R}}^n} \vert \langle \nabla f(x),\xi \rangle \vert ^p \,\textrm{d}x \end{aligned}$$

for $\xi \in {\mathbb {S}}^{n-1}$, were $\langle \cdot ,\cdot \rangle$ denotes the inner product. It is the polar body of a convex body. The definition is due to Lutwak, Yang, and Zhang [24]. For a convex body $K\subset {\mathbb {R}}^n$, they defined the $L^p$ polar projection body (with a different normalization) in [23] by

$$\begin{aligned} \Vert \xi \Vert _{\Pi _{p}^{*} K}^p = \int _{{\mathbb {S}}^{n-1}} \vert \langle \xi , \eta \rangle \vert ^p \,\textrm{d}S_p(K,\eta ), \end{aligned}$$

(8)

where $S_p(K,\cdot )$ is the $L^p$ surface area measure of K (for the definition of $L^p$ surface area measures, see, for example, [28, Section 9.1]).

Asymmetric $L^p$ polar projection bodies of convex bodies were introduced in [19]. For $f \in W^{1,p}({\mathbb {R}}^n)$, the asymmetric $L^p$ polar projection bodies of f are defined as the star bodies with gauge function given by

$$\begin{aligned} \Vert \xi \Vert _{\Pi ^{*}_{p, \pm } f}^p = \int _{{\mathbb {R}}^n} \langle \nabla f(x),\xi \rangle _{\pm }^p \,\textrm{d}x \end{aligned}$$

for $\xi \in {\mathbb {S}}^{n-1}$.

3 Fractional $L^p$ polar projection bodies

Let $0<s<1$ and $1< p <n/s$. For a measurable function $f:{\mathbb {R}}^n\rightarrow {\mathbb {R}}$, define the s-fractional $L^p$ polar projection body $\Pi ^{*,{ s}}_{{p}} f$ as the star-shaped set given by the gauge function

$$\begin{aligned} \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} = \int _0^\infty t^{-p s-1} \int _{{\mathbb {R}}^n} \left| f(x+t\xi )-f(x)\right| ^p \,\textrm{d}x \,\textrm{d}t \end{aligned}$$

(9)

for $\xi \in {\mathbb {R}}^n$. Note that $\Vert \cdot \Vert _{\Pi ^{*,{ s}}_{{p}} f}$ is a one-homogeneous function on ${\mathbb {R}}^n$.

Let $K\subset {\mathbb {R}}^n$ be a star body. The following simple calculation turns out to be useful. For $f\in W^{s,p}({\mathbb {R}}^n)$,

$$\begin{aligned}&\int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{|f(x)-f(y)|^p}{\Vert x-y\Vert _K^{n+p s}} \,\textrm{d}x \,\textrm{d}y\\&\quad = \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{|f(y+z)-f(y)|^p}{\Vert z\Vert _K^{n+p s}} \,\textrm{d}z \,\textrm{d}y\\&\quad = \int _{{\mathbb {S}}^{n-1}} \int _0^\infty {\Vert t \xi \Vert _K^{-n-p s}} \int _{{\mathbb {R}}^n} {|f(y+t\xi )-f(y)|^p}\,t^{n-1} \,\textrm{d}y \,\textrm{d}t \,\textrm{d}\xi \\&\quad = \int _{{\mathbb {S}}^{n-1}} \int _0^\infty {\Vert \xi \Vert _K^{-n-p s}} t^{-p s-n} \left\| f(\cdot +t\xi ) - f\right\| _{p}^p t^{n-1} \,\textrm{d}t \,\textrm{d}\xi \\&\quad = \int _{{\mathbb {S}}^{n-1}} \rho _K(\xi )^{n+p s} \int _0^\infty t^{-p s-1} \left\| f(\cdot +t\xi ) - f\right\| _{p}^p \,\textrm{d}t \,\textrm{d}\xi \\&\quad = \int _{{\mathbb {S}}^{n-1}} \rho _K(\xi )^{n+p s} \rho _{\Pi ^{*,{ s}}_{{p}} f}(\xi )^{-p s} d\xi . \end{aligned}$$

Hence,

$$\begin{aligned} \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{|f(x)-f(y)|^p}{\Vert x-y\Vert _K^{n+p s}} \,\textrm{d}x \,\textrm{d}y = n\, \tilde{V}_{-p s}(K, \Pi ^{*,{ s}}_{{p}} f) \end{aligned}$$

(10)

in this case.

Next, we establish basic properties of fractional $L^p$ polar projection bodies.

Proposition 4

For non-zero $f \in W^{s,p}({\mathbb {R}}^n)$, the set $\,\Pi ^{*,{ s}}_{{p}} f$ is an origin-symmetric star body with the origin in its interior. Moreover, there is $c>0$ depending only on f and p such that $\Pi ^{*,{ s}}_{{p}} f \subseteq c\, {B^n}$ for every $s \in (0,1)$.

Proof

First, note that since for $\xi \in {\mathbb {R}}^n$ and $t>0$,

$$\begin{aligned} \int _{{\mathbb {R}}^n}\vert {f(x -t\xi )-f(x)}\vert ^p \,\textrm{d}x=\int _{{\mathbb {R}}^n}\vert f(x)-f(x+t\xi )\vert ^p \,\textrm{d}x, \end{aligned}$$

the set $\Pi ^{*,{ s}}_{{p}} f$ is origin-symmetric.

Next, we show that $\Pi ^{*,{ s}}_{{p}} f$ is bounded. We take $r>1$ large enough so that $\Vert f\Vert _{L^p(r{B^n})} \ge \frac{2}{3} \Vert f\Vert _p$ and easily see that for $t>2r$,

$$\begin{aligned} \Vert f(\cdot +t\xi ) - f(\cdot )\Vert _p&\ge \Vert f(\cdot +t\xi ) - f(\cdot )\Vert _{L^p(r{B^n}-t\xi )}\\&= \Vert f(\cdot ) - f(\cdot -t\xi )\Vert _{L^p(r{B^n})}\\&\ge \Vert f\Vert _{L^p(r{B^n})} - \Vert f(\cdot - t\xi )\Vert _{L^p(r{B^n})}\\&\ge \frac{2}{3}\Vert f\Vert _p - \frac{1}{3} \Vert f\Vert _p. \end{aligned}$$

Hence,

$$\begin{aligned} \int _0^\infty t^{-p s-1} \int _{{\mathbb {R}}^n} \big | {f(x+t\xi )-f(x)} \big |^p \,\textrm{d}x \,\textrm{d}t&\ge \frac{\Vert f\Vert _p^p}{3^p}\, \, \int _r^\infty t^{-p s-1} \,\textrm{d}t\\&\ge \frac{\Vert f\Vert _p^p}{3^p} \, \frac{r^{-p s}}{p s} \ge c, \end{aligned}$$

which implies that $\Pi ^{*,{ s}}_{{p}} f \subseteq c\, {B^n}$ for $c>0$ independent of s.

Now, we show that $\Pi ^{*,{ s}}_{{p}} f$ has the origin in its interior. First observe that for $\xi , \eta \in {\mathbb {R}}^n$, by the triangle inequality and a change of variables,

$$\begin{aligned}&\Vert \xi +\eta \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s}\nonumber \\&\quad = \int _0^\infty t^{-p s-1} \Vert f(\cdot +t\xi +t\eta )-f(\cdot )\Vert _p^p \,\textrm{d}t \nonumber \\&\quad \le \int _0^\infty t^{-p s-1} \left( \Vert f(\cdot +t\xi +t\eta )-f(\cdot +t\xi )\Vert _p +\Vert f(\cdot +t\xi )-f(\cdot )\Vert _p\right) ^p \,\textrm{d}t \nonumber \\&\quad \le \int _0^\infty t^{- p s-1} 2^{p-1} (\Vert f(\cdot +t\eta )-f(\cdot )\Vert _p^p + \Vert f(\cdot +t\xi ) -f(\cdot )\Vert _p^p) \,\textrm{d}t\nonumber \\&\quad = 2^{p-1} \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} + 2^{p-1} \Vert \eta \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s}. \end{aligned}$$

(11)

Using the relation (10) with $K={B^n}$, we get

$$\begin{aligned} \int _{{\mathbb {S}}^{n-1}} \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{ p s} \,\textrm{d}\xi = \frac{1}{n} \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{|f(x) -f(y)|^p}{|x-y|^{n+p s}} \,\textrm{d}x \,\textrm{d}y, \end{aligned}$$

which is finite since $f\in W^{s,p}({\mathbb {R}}^n)$. We choose $r>0$ large enough so that the set $A = \{\xi \in {\mathbb {S}}^{n-1}: \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{s} < r\}$ has positive $(n-1)$-dimensional Hausdorff measure and contains a basis $\{\xi _1, \ldots , \xi _n\} \subseteq A$ of ${\mathbb {R}}^n$. Applying (if necessary) a linear transformation to $\Pi ^{*,{ s}}_{{p}} f$, we may assume without loss of generality that $\xi _i = e_i$ are the canonical basis vectors. For every $x \in {\mathbb {R}}^n$, writing $x = \sum x_i e_i$ and using (11), we get

$$\begin{aligned} \Vert x\Vert _{\Pi ^{*,{ s}}_{{p}} f} \le \Big ( 2^{n(p-1)} \sum _{i=1}^n |x_i|^{p s} \Vert e_i\Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} \Big )^{\frac{1}{p s}} \le d\, |x|, \end{aligned}$$

(12)

where $d>0$ is independent of x. This shows that $\Pi ^{*,{ s}}_{{p}} f$ has the origin as interior point.

Finally, we show that $\Vert \cdot \Vert _{\Pi ^{*,{ s}}_{{p}} f}$ is continuous. For $\xi , \eta \in {\mathbb {R}}^n$, by the triangle inequality and (12), we have

$$\begin{aligned}&\Vert \xi + \eta \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s}\\&\quad = \int _0^\infty t^{-1- p s} \Vert f(\cdot + t \xi + t \eta ) - f(\cdot )\Vert _p^p\,\textrm{d}t\\&\quad \le \int _0^\infty t^{-1- p s} \big ( \Vert f(\cdot + t \eta ) - f(\cdot )\Vert _p + \Vert f(\cdot + t \xi ) - f(\cdot )\Vert _p \big )^p \,\textrm{d}t \\&\quad \le \big (1+|\eta |^{\frac{s}{2}\frac{p}{p-1}}\big )^{p-1}\int _0^\infty {t^{-1- p s}}\left( \frac{\Vert f(\cdot + t \eta ) - f(\cdot )\Vert _p^p}{|\eta |^{\frac{p s}{2}}} + \Vert f(\cdot + t \xi ) - f(\cdot )\Vert _p^p \right) \,\textrm{d}t \\&\quad = \big (1+|\eta |^{\frac{s}{2}\frac{p}{p-1}}\big )^{p-1} \big ( |\eta |^{-\frac{p s}{2}}\Vert \eta \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} + \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} \big )\\&\quad \le \big (1+|\eta |^{\frac{s}{2}\frac{p}{p-1}}\big )^{p-1} \big ( d\, |\eta |^{\frac{p s}{2}} + \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} \big ), \end{aligned}$$

where we used the inequality $a+b \le (1+r^{p/(p-1)})^{(p-1)/p}((r^{-1} a)^p + b^p)^{1/p}$ for $a,b,r > 0$, which is a consequence of Hölder’s inequality.

We obtain

$$\begin{aligned} \Vert \xi + \eta \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} \le \big (1+|\eta |^{\frac{s}{2} \frac{p}{p-1}}\big )^{p-1} \big ( d\, |\eta |^{\frac{p s}{2}} + \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} \big ). \end{aligned}$$

(13)

Applying inequality (13) to the vectors $\xi + \eta$ and $-\eta$, we get

$$\begin{aligned} \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} = \Vert \xi + \eta - \eta \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} \le \big (1+|-\eta |^{\frac{s}{2}\frac{p}{p-1}}\big )^{p-1} \big ( d\, |-\eta |^{\frac{p s}{2}} + \Vert \xi + \eta \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} \big ), \end{aligned}$$

which implies

$$\begin{aligned} \Vert \xi + \eta \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} \ge \big (1+|\eta |^{\frac{s}{2}\frac{p}{p-1}}\big )^{p-1} \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} - d\, |\eta |^{\frac{p s}{2}}. \end{aligned}$$

(14)

The continuity of $\Vert \cdot \Vert _{\Pi ^{*,{ s}}_{{p}} f}$ now follows from (13) and (14). $\square$

4 Fractional asymmetric $L^p$ polar projection bodies

Let $0<s<1$ and $1< p <n/s$. For a measurable function $f:{\mathbb {R}}^n\rightarrow {\mathbb {R}}$, define the asymmetric s-fractional $L^p$ polar projection bodies $\Pi ^{*,{ s}}_{{p}, +} f$ and $\Pi ^{*,{ s}}_{{p}, -} f$ as the star-shaped sets given by the gauge functions

$$\begin{aligned} \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}, \pm } f}^{p s} = \int _0^\infty t^{-p s-1} \int _{{\mathbb {R}}^n} (f(x+t\xi )-f(x))_{\pm }^{ p} \,\textrm{d}x \,\textrm{d}t \end{aligned}$$

for $\xi \in {\mathbb {R}}^n$. We have $\Pi ^{*,{ s}}_{{p}, -} f= \Pi ^{*,{ s}}_{{p}, +} {(-f)}=-\Pi ^{*,{ s}}_{{p}, +} f$ and state our results just for $\Pi ^{*,{ s}}_{{p}, +} f$. Note that, as in the symmetric case, $\Vert \cdot \Vert _{\Pi ^{*,{ s}}_{{p}, +} f}^{p s}$ is a one-homogeneous function on ${\mathbb {R}}^n$. Also, note that

$$\begin{aligned} \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}} f}^{p s} =\Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}, +} f}^{p s}+ \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}, -} f}^{p s} \end{aligned}$$

(15)

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

Let $K\subset {\mathbb {R}}^n$ be a star body and $f \in W^{s,p}({\mathbb {R}}^n)$. As in (10), we obtain that

$$\begin{aligned} \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{(f(x)-f(y))_{+}^{p}}{\Vert x-y\Vert _K^{n+p s}} \,\textrm{d}x \,\textrm{d}y = n\, \tilde{V}_{-p s}(K, \Pi ^{*,{ s}}_{{p}, +} f). \end{aligned}$$

(16)

In the following proposition, we derive the basic properties of fractional asymmetric $L^p$ polar projection bodies.

Proposition 5

For non-zero $f \in W^{s,p}({\mathbb {R}}^n)$, the set $\,\Pi ^{*,{ s}}_{{p}, +} f$ is a star body with the origin in its interior. Moreover, there is $c>0$ depending only on f and p such that $\Pi ^{*,{ s}}_{{p}, +} f \subseteq c\, {B^n}$ for every $s \in (0,1)$.

Proof

Since the functions $(a)_{+}^{p}$ and $(a)_{-}^{ p}$ are convex, the inequalities $(a+b)_{+}^{p}\ge (a)_{+}^{p} + p (a)_{+}^{p-1} b$ and $(a+b)_{-}^{ p} \ge (a)_{-}^{ p} + p (a)_{-}^{ p-1} b$ hold for $a,b\in {\mathbb {R}}$.

If $\int _{{\mathbb {R}}^n} (f(x))_{+}^{p} \,\textrm{d}x>0$, take $\varepsilon >0$ so small that $\varepsilon +p \varepsilon ^{1/p} \left\| f\right\| _p^{p-1}\le \frac{1}{2} \int _{{\mathbb {R}}^n} (f(x))_{+}^{p} \,\textrm{d}x$, and take $r>0$ so large that $\int _{{\mathbb {R}}^n \setminus r{B^n}} |f(x)|^p \,\textrm{d}x < \varepsilon$. For $z \in {\mathbb {R}}^n \setminus 2r {B^n}$, we obtain by Hölder’s inequality that

$$\begin{aligned}&\int _{r{B^n}}(f(x)-f(x+z))_{+}^{p} \,\textrm{d}x \\&\quad \ge \int _{r{B^n}} (f(x))_{+}^{p}-p\, (f(x))_{+}^{p-1} f(x+z) \,\textrm{d}x \\&\quad \ge \int _{r{B^n}} (f(x))_{+}^{p} \,\textrm{d}x - p \Big (\int _{r{B^n}} (f(x))_{+}^{p} \,\textrm{d}x \Big )^{\frac{p-1}{p}} \Big (\int _{r{B^n}}\left| f(x+z)\right| ^p \,\textrm{d}x\Big )^{\frac{1}{p}}\\&\quad \ge \int _{r{B^n}} (f(x))_{+}^{p} \,\textrm{d}x - p \Big (\int _{{\mathbb {R}}^n} |f(x)|^p \,\textrm{d}x \Big )^{\frac{p-1}{p}} \Big (\int _{{\mathbb {R}}^n\setminus r{B^n}}\left| f(x)\right| ^p \,\textrm{d}x\Big )^{\frac{1}{p}} \\&\quad \ge \int _{{\mathbb {R}}^n} (f(x))_{+}^{p} \,\textrm{d}x - \varepsilon - p \,\Vert f\Vert _p^{p-1} \varepsilon ^{\frac{1}{p}}\\&\quad \ge \frac{1}{2} \int _{{\mathbb {R}}^n} (f(x))_{+}^{p} \,\textrm{d}x. \end{aligned}$$

In case $\int _{{\mathbb {R}}^n} (f(x))_{+}^{p} \,\textrm{d}x=0$ the previous inequality holds trivially for any $r>0$.

By an analogous calculation and eventually increasing the value of r, we obtain that

$$\begin{aligned} \int _{r{B^n}-z }(f(x)-f(x+z))_{+}^{p} \,\textrm{d}x&= \int _{r{B^n}}(f(x)-f(x-z))_{-}^{ p} \,\textrm{d}x\\&\ge \frac{1}{2} \int _{{\mathbb {R}}^n} (f(x))_{-}^{ p} \,\textrm{d}x. \end{aligned}$$

It follows that $\int _{{\mathbb {R}}^n} (f(x)-f(x+z))_{+}^{p} \,\textrm{d}x \ge \frac{1}{2} \Vert f\Vert _p^p$ for every $z \in {\mathbb {R}}^n \setminus 2r {B^n}$ with $r>0$ depending only on f. Finally,

$$\begin{aligned} \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}, +} f} ^{p s}&\ge \int _{2r}^\infty t^{-1-p s} \int _{{\mathbb {R}}^n} (f(x)-f(x+z))_{+}^{p} \,\textrm{d}x \,\textrm{d}t\\&\ge \int _{2r}^\infty t^{-1-p s} \,\textrm{d}t\, \frac{1}{2} \int _{{\mathbb {R}}^n} |f(x)|^p \,\textrm{d}x \\&\ge \frac{(2r)^{-p s}}{p s} \frac{1}{2} \int _{{\mathbb {R}}^n} |f(x)|^p \,\textrm{d}x \\&\ge \frac{(2r)^{-p}}{2 p} \Vert f\Vert _p^p. \end{aligned}$$

Note that $\Pi ^{*,{ s}}_{{p}} f\subset \Pi ^{*,{ s}}_{{p}, +} f$. Hence, it follows from Proposition 4 that $\Pi ^{*,{ s}}_{{p}, +} f$ contains the origin in its interior, that is, there is $d>0$ such that

$$\begin{aligned} \Vert x\Vert _{\Pi ^{*,{ s}}_{{p}, +} f} \le d\, |x| \end{aligned}$$

(17)

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

Finally, we show that $\Vert \cdot \Vert _{\Pi ^{*,{ s}}_{{p}, +} f}$ is continuous. Observe that the inequality $(a+b)_+^p \le (a_+ + b_+)^p$ holds for any $a,b \in {\mathbb {R}}$. Hence, for $\xi , \eta \in {\mathbb {R}}^n$, we obtain that

$$\begin{aligned}&\int _{{\mathbb {R}}^n} (f(x + t \xi + t \eta ) - f(x) )_{+}^{p} \,\textrm{d}x \\&\quad =\int _{{\mathbb {R}}^n} (f(x + t \xi + t \eta ) - f(x + t \xi ) + f(x + t \xi ) - f(x) )_{+}^{p} \,\textrm{d}x \\&\quad \le \int _{{\mathbb {R}}^n} \big ( (f(x + t \xi + t \eta ) - f(x + t \xi ))_{+} + (f(x + t \xi ) - f(x))_{+} \big )^p \,\textrm{d}x \\&\quad \le \int _{{\mathbb {R}}^n} (1+|\eta |^{\frac{s}{2}\frac{p }{(p-1)} })^{p-1} \Big ( \frac{ (f(x + t \xi + t \eta ) - f(x + t \xi ))_{+}^{p}}{|\eta |^{\frac{p s}{2}}} + (f(x + t \xi ) - f(x))_{+}^{p}\Big ) \,\textrm{d}x \\&\quad \le (1+|\eta |^{\frac{s}{2}\frac{p }{(p-1)} })^{p-1} \Big (\frac{\Vert (f(\cdot + t \eta ) - f(\cdot ))_{+}\Vert _p^p}{|\eta |^{\frac{p s}{2}}} + \Vert (f(\cdot + t \xi ) - f(\cdot ))_{+}\Vert _p^p \Big ), \end{aligned}$$

where we used the inequality $a+b \le (1+r^{ p/(p-1)})^{(p-1)/p}((r^{-1} a)^p + b^p)^{1/p}$ for $a,b,r > 0$, which is a consequence of Hölder’s inequality. Thus, integrating and using (17), we obtain

$$\begin{aligned} \Vert \xi + \eta \Vert _{\Pi ^{*,{ s}}_{{p}, +} f}^{p s} \le (1+|\eta |^{\frac{s}{2}\frac{p}{p-1} })^{p-1} ( d\, |\eta |^{\frac{p s}{2}} + \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}, +} f}^{p s} ). \end{aligned}$$

(18)

Applying inequality (18) to the vectors $\xi + \eta$ and $-\eta$, we get

$$\begin{aligned} \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}, +} f}^{p s} = \Vert \xi + \eta - \eta \Vert _{\Pi ^{*,{ s}}_{{p}, +} f}^{ p s} \le (1+|-\eta |^{\frac{s}{2}\frac{p}{p-1}})^{p-1} ( d\, |-\eta |^{\frac{p s}{2}} + \Vert \xi + \eta \Vert _{\Pi ^{*,{ s}}_{{p}, +} f}^{p s} ), \end{aligned}$$

which implies

$$\begin{aligned} \Vert \xi + \eta \Vert _{\Pi ^{*,{ s}}_{{p}, +} f}^{p s} \ge (1+|\eta |^{\frac{s}{2}\frac{p}{p-1}})^{-(p-1)} \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}, +} f}^{ p s} - d\, |\eta |^{\frac{p s}{2}}. \end{aligned}$$

(19)

The continuity of $\Vert \cdot \Vert _{\Pi ^{*,{ s}}_{{p}, +} f}$ now follows from (18) and (19). $\square$

5 The limit of fractional $L^p$ polar projection bodies

We establish the limiting behavior of s-fractional $L^p$ polar projection bodies for $1< p <n/s$ as $s\rightarrow 1^-$ in the symmetric and asymmetric case. For $p=1$, a corresponding result was proved in [16].

Let $0<s<1$ and $1< p <n/s$. Set $p'=p/(p-1)$. We say that $f_k \rightarrow f$ weakly in $L^p({\mathbb {R}}^n)$ if

$$\begin{aligned} \int _{{\mathbb {R}}^n} f_k(x) g(x) \,\textrm{d}x \rightarrow \int _{{\mathbb {R}}^n} f(x) g(x) \,\textrm{d}x \end{aligned}$$

for every $g \in L^{p'}({\mathbb {R}}^n)$ as $k\rightarrow \infty$. Set $B_{p', +}= \{g \in L^{p'}({\mathbb {R}}^n) : g \ge 0, \Vert g\Vert _{p'} \le 1 \}$.

We require the following lemmas.

Lemma 6

The following statements hold.

• (1) For $f \in L^p({\mathbb {R}}^n)$,

  $$\begin{aligned} \Vert f_{+}\Vert _p = \sup _{g \in B_{p', +}} \int _{{\mathbb {R}}^n} f(x) g(x) \,\textrm{d}x. \end{aligned}$$

• (2) Let $f_k, f\in L^p({\mathbb {R}}^n)$. If $f_k \rightarrow f$ weakly in $L^p({\mathbb {R}}^n)$ as $k\rightarrow \infty$, then

  $$\begin{aligned} \liminf _{k\rightarrow \infty } \Vert (f_k)_{+}\Vert _p \ge \Vert f_{+}\Vert _p. \end{aligned}$$

• (3) Assume $f_k$ is a bounded sequence in $L^p({\mathbb {R}}^n)$. If

  $$\begin{aligned} \lim _{k\rightarrow \infty }\int _{{\mathbb {R}}^n} f_k(x) g(x)\,\textrm{d}x = \int _{{\mathbb {R}}^n} f(x) g(x)\,\textrm{d}x \end{aligned}$$

  for every g in a dense subset $D \subseteq L^{p'}({\mathbb {R}}^n)$, then $f_k \rightarrow f$ weakly in $L^p({\mathbb {R}}^n)$ as $k\rightarrow \infty$.

Proof

First, we prove (1). Let $g \in B_{p', +}$ and write $f = f_{+} - f_{-}$. Since $f_{-}$ and g are non-negative, it follows from Hölder’s inequality that

$$\begin{aligned} \int _{{\mathbb {R}}^n} f(x) g(x)\,\textrm{d}x \le \int _{{\mathbb {R}}^n} f_{+}(x) g(x) \,\textrm{d}x\le \Vert f_{+}\Vert _p. \end{aligned}$$

For the opposite inequality, take $g = \Vert f_{+}\Vert _p^{-{p}/{p'}} f_{+}^{{p}/{p'}}$ and note that $g \in B_{p', +}$ and

$$\begin{aligned} \int _{{\mathbb {R}}^n} f(x) g(x)\,\textrm{d}x= & {} \Vert f_{+}\Vert _p^{-\frac{p}{p'}} \int _{{\mathbb {R}}^n} f(x) f_{+}(x)^{\frac{p}{p'}} \,\textrm{d}x \le \Vert f_{+}\Vert _p^{-\frac{p}{p'}} \int _{{\mathbb {R}}^n} f_{+}(x)^p\,\textrm{d}x \\= & {} \Vert f_{+}\Vert _p. \end{aligned}$$

Next, we prove (2). Fix $k_0$ and $g_0 \in B_{p', +}$. By (1), we have

$$\begin{aligned} \int _{{\mathbb {R}}^n} f_{k_0}(x) g_0(x) \,\textrm{d}x \le \sup _{g \in B_{p', +}} \int _{{\mathbb {R}}^n} f_{k_0}(x) g(x)\,\textrm{d}x = \Vert (f_{{k_0}})_{+}\Vert _p. \end{aligned}$$

Since this inequality holds for every $k_0$,

$$\begin{aligned} \int _{{\mathbb {R}}^n} f(x) g_0(x) \,\textrm{d}x = \lim _{k\rightarrow \infty } \int _{{\mathbb {R}}^n} f_k(x) g_0(x)\,\textrm{d}x \le \liminf _{k\rightarrow \infty } \Vert (f_{k})_{+}\Vert _p. \end{aligned}$$

Thus, by (1),

$$\begin{aligned} \Vert f_{+}\Vert _p = \sup _{g \in B_{p',+}} \int _{{\mathbb {R}}^n} f(x)g(x)\,\textrm{d}x \le \liminf _{k\rightarrow \infty } \Vert (f_{k})_{+}\Vert _p. \end{aligned}$$

Finally, we prove (3). Take $c \ge \max \{\Vert f_k\Vert _p, \Vert f\Vert _p\}$. Let $\varepsilon > 0$ and $g \in L^{p'}({\mathbb {R}}^n)$. Take $h \in D$ such that $\Vert g-h\Vert _{p'} < \varepsilon /(2c)$. Then

$$\begin{aligned}&\Big | \int _{{\mathbb {R}}^n} f_k(x) g(x)\,\textrm{d}x - \int _{{\mathbb {R}}^n} f(x) g(x)\,\textrm{d}x \Big |\\&\quad \le \Big | \int _{{\mathbb {R}}^n} f_k(x) (g(x) - h(x))\,\textrm{d}x \Big | + \Big | \int _{{\mathbb {R}}^n} f_k(x) h(x) \,\textrm{d}x - \int _{{\mathbb {R}}^n} f(x) h(x)\,\textrm{d}x \Big |\\&\qquad + \Big | \int _{{\mathbb {R}}^n} f(x) (g(x) - h(x)) \,\textrm{d}x \Big |\\&\quad \le c \varepsilon /(2c) + \Big | \int _{{\mathbb {R}}^n} f_k(x) h(x)\,\textrm{d}x - \int _{{\mathbb {R}}^n} f(x) h(x) \,\textrm{d}x \Big | + c \varepsilon /(2c) \end{aligned}$$

and the statement follows. $\square$

Lemma 7

For $f \in W^{1,p}({\mathbb {R}}^n)$ and fixed $\xi \in {\mathbb {S}}^{n-1}$,

$$\begin{aligned} \lim _{t\rightarrow 0} \Big \Vert \Big ( \frac{f(\cdot +t\xi )-f(\cdot )}{t} \Big )_{+} \Big \Vert _{p}^p = \int _{{\mathbb {R}}^n} \langle \nabla f(x), \xi \rangle _{+}^p \,\textrm{d}x. \end{aligned}$$

Proof

Let $g:{\mathbb {R}}^n\rightarrow {\mathbb {R}}$ be a smooth function with compact support. Write $\textrm{div}_x$ for the divergence taken with respect to the variable x. Using integration by parts, we obtain for $\xi \in {\mathbb {S}}^{n-1}$ and $t>0$,

$$\begin{aligned} \int _{{\mathbb {R}}^n} g(x) \frac{f(x+t\xi )-f(x)}{t} \,\textrm{d}x&= \int _{{\mathbb {R}}^n} f(x) \frac{g(x-t\xi )-g(x)}{t} \,\textrm{d}x\\&= - \int _{{\mathbb {R}}^n} f(x) \int _0^1 \langle \nabla g(x-r t\xi ), \xi \rangle \,\textrm{d}r \,\textrm{d}x\\&= - \int _{{\mathbb {R}}^n} f(x) \textrm{div}_x\Big (\int _0^1 g(x-r t\xi ) \,\textrm{d}r\, \xi \Big )\,\textrm{d}x \\&= \int _{{\mathbb {R}}^n} \Big (\int _0^1 g(x-r t\xi ) d r \Big ) \langle \nabla f(x),\xi \rangle \,\textrm{d}x. \end{aligned}$$

By Minkowski’s integral inequality $\Vert \int _0^1 g(\cdot -r t\xi ) \,\textrm{d}r \Vert _{p'} \le \Vert g\Vert _{p'}$, and we deduce

$$\begin{aligned} \Big \Vert \frac{f(\cdot +t\xi )-f(\cdot )}{t} \Big \Vert _p \le \Vert \langle \nabla f(\cdot ),\xi \rangle \Vert _p < \infty . \end{aligned}$$

Hence, $\frac{f(\cdot +t\xi )-f(\cdot )}{t}$ is uniformly bounded in $L^p({\mathbb {R}}^n)$ on $(0,\infty )$.

By Lemma 6(3),

$$\begin{aligned} \lim _{t\rightarrow 0}\int _{{\mathbb {R}}^n} g(x) \frac{f(x+t\xi )-f(x)}{t} \,\textrm{d}x = \int _{{\mathbb {R}}^n} g(x) \langle \nabla f(x), \xi \rangle \,\textrm{d}x \end{aligned}$$

for every $g \in L^{p'}({\mathbb {R}}^n)$. Hence, $\frac{f(\cdot +t\xi )-f(\cdot )}{t}$ converges weakly to $\langle \nabla f(\cdot ), \xi \rangle$ as $t\rightarrow 0$.

By Lemma 6(2),

$$\begin{aligned} \liminf _{t\rightarrow 0} \Big \Vert \Big (\frac{f(\cdot +t\xi )-f(\cdot )}{t} \Big )_{+} \Big \Vert _{p} \ge \Vert \langle \nabla f(\cdot ), \xi \rangle _{+} \Vert _p. \end{aligned}$$

For the opposite inequality, we recall that for any $g \in B_{p', +}$, the function $x\mapsto \int _0^1 g(x-r t\xi ) \,\textrm{d}r$ is in $B_{p', +}$ as well. Hence,

$$\begin{aligned} \int _{{\mathbb {R}}^n} g(x) \frac{f(x+t\xi )-f(x)}{t} \,\textrm{d}x&= \int _{{\mathbb {R}}^n} \Big (\int _0^1 g(x-r t\xi ) \,\textrm{d}r \Big ) \langle \nabla f(x),\xi \rangle \,\textrm{d}x \\&\le \Vert \langle \nabla f(x), \xi \rangle _{+} \Vert _p. \end{aligned}$$

Again by Lemma 6(1),

$$\begin{aligned} \Big \Vert \Big ( \frac{f(\cdot +t\xi )-f(\cdot )}{t} \Big )_{+} \Big \Vert _{p} \le \Vert \langle \nabla f(\cdot ), \xi \rangle _{+} \Vert _p \end{aligned}$$

for each $t>0$. $\square$

The following result is Lemma 4 in [16].

Lemma 8

If $\varphi :[0,\infty ) \rightarrow [0,\infty )$ be a measurable function with $\lim _{t\rightarrow 0^+} \varphi (t) = \varphi (0)$ and such that $\int _0^\infty t^{-s_0} \varphi (t) \,\textrm{d}t < \infty$ for some $s_0 \in (0,1)$, then

$$\begin{aligned} \lim _{s\rightarrow 1^-} (1-s) \int _0^\infty t^{-s} \varphi (t) \,\textrm{d}t = \varphi (0). \end{aligned}$$

We are now able to prove the main result of this section.

Theorem 9

Let $f \in W^{1,p} ({\mathbb {R}}^n)$. For $\xi \in {\mathbb {S}}^{n-1}$,

$$\begin{aligned} \lim _{s\rightarrow 1^-} (p(1-s))^{\frac{1}{p}}\Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}, +} f} =\Vert \xi \Vert _{\Pi ^{*}_{p, +}f}. \end{aligned}$$

Moreover,

$$\begin{aligned} \lim _{s\rightarrow 1^-} p(1-s) \vert {\Pi ^{*,{ s}}_{{p}, +} f}\vert ^{-\frac{p s}{n}} =\vert {\Pi ^{*}_{p, +}f}\vert ^{-\frac{p}{n}}, \end{aligned}$$

and

$$\begin{aligned} \lim _{s\rightarrow 1^-} p(1-s) \tilde{V}_{-p s}(K, \Pi ^{*,{ s}}_{{p}, +} f)= \tilde{V}_{-p}(K, \Pi ^{*}_{p, +}f) \end{aligned}$$

for every star body $K\subset {\mathbb {R}}^n$.

Proof

Define $\varphi : [0,\infty )\rightarrow [0,\infty )$ by

$$\begin{aligned} \varphi (t) = \Big \Vert \Big ( \frac{f(\cdot +t\xi )-f(\cdot )}{t} \Big )_{+} \Big \Vert _p^p, \end{aligned}$$

and note that $\varphi (t) \le \left( \frac{2\Vert f\Vert _p}{t} \right) ^p$ for $t> 0$. By Lemma 8 and Lemma 7,

$$\begin{aligned} \lim _{s\rightarrow 1^-} p(1-s) \int _0^\infty t^{p(1-s)-1 } \Big \Vert \Big (\frac{f(\cdot +t\xi )-f(\cdot )}{t} \Big )_{+} \Big \Vert _p^p \,\textrm{d}t = \int _{{\mathbb {R}}^n} \langle \nabla f(x), \xi \rangle _{+}^p\,\textrm{d}x. \end{aligned}$$

By Proposition 4, we can use the dominated convergence theorem to obtain

$$\begin{aligned}&\lim _{s\rightarrow 1^-}n\, \vert {(p(1-s))^{-\frac{1}{p s}} \Pi ^{*,{ s}}_{{p}, +} f}\vert \\&\quad = \lim _{s\rightarrow 1^-}\int _{{\mathbb {S}}^{n-1}} \Big ( p(1-s) \int _0^\infty t^{p(1-s)-1} \Big \Vert \Big (\frac{f(\cdot +t\xi )-f(\cdot )}{t} \Big )_{+} \Big \Vert _p^p \,\textrm{d}t \Big )^{-\frac{n}{p s}} \,\textrm{d}\xi \\&\quad = \int _{{\mathbb {S}}^{n-1}} \Big ( \int _{{\mathbb {R}}^n} \langle \nabla f(x), \xi \rangle _{+} ^p \,\textrm{d}x \Big )^{-\frac{n}{p}} \,\textrm{d}\xi \\&\quad = n\, \vert { \Pi ^{*}_{p, +} f}\vert , \end{aligned}$$

and

$$\begin{aligned} \lim _{s\rightarrow 1^-} n p (1-s)\tilde{V}_{-p s}(K, \Pi ^{*,{ s}}_{{p}, +} f)&= \lim _{s\rightarrow 1^-}p(1-s) \int _{{\mathbb {S}}^{n-1}} \Vert \xi \Vert _K^{n+p s} \Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}, +} f}^{p s} \,\textrm{d}\xi \\&= \int _{{\mathbb {S}}^{n-1}} \Vert \xi \Vert _K^{n} \,\Vert \xi \Vert _{\Pi ^{*}_{p, +} f}^p \,\textrm{d}\xi \\&=n\,\tilde{V}_{-p}(K, \Pi ^{*}_{p, +} f), \end{aligned}$$

which completes the proof of the theorem. $\square$

The following result is an immediate consequence of Theorem 9 and (15).

Theorem 10

Let $f \in W^{1,p} ({\mathbb {R}}^n)$. For $\xi \in {\mathbb {S}}^{n-1}$,

$$\begin{aligned} \lim _{s\rightarrow 1^-} (p(1-s))^{\frac{1}{p}}\Vert \xi \Vert _{\Pi ^{*,{ s}}_{{p}} f} =\Vert \xi \Vert _{\Pi _{p}^{*} f}. \end{aligned}$$

Moreover,

$$\begin{aligned} \lim _{s\rightarrow 1^-} p(1-s) \vert {\Pi ^{*,{ s}}_{{p}} f}\vert ^{-\frac{p s}{n}} =\vert {\Pi _{p}^{*}f}\vert ^{-\frac{p}{n}}, \end{aligned}$$

and

$$\begin{aligned} \lim _{s\rightarrow 1^-} p(1-s) \tilde{V}_{-p s}(K, \Pi ^{*,{ s}}_{{p}} f)= \tilde{V}_{-p}(K, \Pi _{p}^{*} f) \end{aligned}$$

(20)

for every star body $K\subset {\mathbb {R}}^n$.

6 Anisotropic fractional Pólya–Szegő inequalities

We will establish anisotropic Pólya–Szegő inequalities for fractional $L^p$ Sobolev norms and their asymmetric counterparts.

Theorem 11

If $f\in L^p({\mathbb {R}}^n)$ is non-negative and $K\subset {\mathbb {R}}^n$ a star body, then

$$\begin{aligned} \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{(f(x) - f(y))_{+}^{p}}{\Vert x-y\Vert _K^{n+p s}} \,\textrm{d}x \,\textrm{d}y \ge \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{(f^\star (x) - f^\star (y))_{+}^{p}}{\Vert x-y \Vert _{K^\star }^{n+p s}} \,\textrm{d}x \,\textrm{d}y. \end{aligned}$$

(21)

Equality holds for non-zero $f\in W^{s, p}({\mathbb {R}}^n)$ if and only if K is a centered ellipsoid and f is a translate of $f^\star \circ \phi$ for some $\phi \in \textrm{SL}(n)$.

Proof

Writing

$$\begin{aligned} \Vert z\Vert _K^{-n-p s} = \int _0^\infty k_t(z) \,\textrm{d}t \end{aligned}$$

where $k_t(z) = 1_{t^{- 1/({n+p s})}K}(z)$, we obtain

$$\begin{aligned} \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{(f(x) - f(y))_{+}^{p}}{\Vert x-y\Vert _K^{n+p s}} \,\textrm{d}x \,\textrm{d}y = \int _0^\infty \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} (f(x) - f(y))_{+}^{p} k_t(x-y) \,\textrm{d}x \,\textrm{d}y \,\textrm{d}t. \end{aligned}$$

Note that

$$\begin{aligned} (f(x) - f(y))_{+}^{p} = p \int _{0}^{\infty } (f(x)-r)_{+}^{p-1} 1_{\{f < r\}}(y) \,\textrm{d}r. \end{aligned}$$

Hence, for $t>0$, it follows from Fubini’s theorem that

$$\begin{aligned}&\int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} (f(x) - f(y))_{+}^{p}\, k_t(x-y) \,\textrm{d}x \,\textrm{d}y\\&\quad =p \int _{0}^{\infty } \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} (f(x)-r)_{+}^{p-1} k_t(x-y) 1_{\{f < r\}}(y) \,\textrm{d}x \,\textrm{d}y \,\textrm{d}r\\&\quad = p \int _{0}^{\infty } \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} (f(x)-r)_{+}^{p-1} k_t(x-y) (1-1_{\{f \ge r\}}(y)) \,\textrm{d}x \,\textrm{d}y \,\textrm{d}r. \end{aligned}$$

Let $r,t>0$. Note that $\int _{{\mathbb {R}}^n} (f(x)-r)_{+}^{p-1} \,\textrm{d}x<\infty$ and that

$$\begin{aligned}&\int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} (f(x)-r)_{+}^{p-1} k_t(x-y) (1-1_{\{f \ge r\}}(y)) \,\textrm{d}x \,\textrm{d}y\\&\quad = p \,\Vert k_t\Vert _1 \int _{{\mathbb {R}}^n} (f(x)-r)_{+}^{p-1} \,\textrm{d}x\\&\quad - p \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} (f(x)-r)_{+}^{p-1} k_t(x-y) 1_{\{f \ge r\}}(y) \,\textrm{d}x \,\textrm{d}y. \end{aligned}$$

The first term is finite since $\{f > r\}$ has finite measure, $f \in L^{\frac{n p}{n - p s}}({\mathbb {R}}^n)$ and $\frac{n p}{n - p s} > p-1$. Clearly, the first term is invariant under Schwarz symmetrization. For the second term, by the Riesz rearrangement inequality, Theorem 2, we have

$$\begin{aligned}&\int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} (f(x)-r)_{+}^{p-1} k_t(x-y) 1_{\{f \ge r\}}(y) \,\textrm{d}x \,\textrm{d}y \\&\quad \le \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} (f^\star (x)-r)_{+}^{p-1} k_t^\star (x-y) 1_{\{f^\star \ge r\}}(y) \,\textrm{d}x \,\textrm{d}y \end{aligned}$$

for $r, t>0$. Note that

$$\begin{aligned} (f(x) - r)_{+}^{p-1} = (p-1) \int _{0}^{\infty } (\tilde{r}-r)_{+}^{p-2} 1_{\{f \ge \tilde{r}\}}(x) \,\textrm{d}\tilde{r}, \end{aligned}$$

and that the corresponding equation holds for $f^\star$. Hence, if there is equality in (21), then, for $(\tilde{r},r, t)\in (0,\infty )^3\backslash M$ with $\vert {M}\vert =0$, we have

$$\begin{aligned}&\int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} 1_{\{f \ge \tilde{r}\}}(x) 1_{t^{- 1/({n+p s})}K}(x-y) 1_{\{f \ge r\}}(y) \,\textrm{d}x \,\textrm{d}y \\&\quad = \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} 1_{\{f^\star \ge \tilde{r}\}}(x) 1_{t^{- 1/({n+p s})}K^\star }(x-y) 1_{\{f^\star \ge r\}}(y) \,\textrm{d}x \,\textrm{d}y. \end{aligned}$$

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

$$\begin{aligned} \{f \ge \tilde{r}\}= a + \alpha D,\quad t^{-1/({n+p s})} K = b + \beta D,\quad \{f \ge r\}= c + \gamma D \end{aligned}$$

where $c= a+ b$. Since $K= t^{1/({n+ps})}b + (\vert {K}\vert /\vert {D}\vert )^{1/n} D$, the centered ellipsoid D does not depend on $(\tilde{r},r,t)$ and also a, c do not depend on t. It follows that $b=0$ and that K is a multiple of D. Hence, $a=c$ is a constant vector, which concludes the proof. $\square$

The following result is a variation of [17, Theorem 3.1].

Theorem 12

If $f\in L^p({\mathbb {R}}^n)$ is non-negative and $K\subset {\mathbb {R}}^n$ a star body, then

$$\begin{aligned} \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{|f(x) - f(y)|^p}{\Vert x-y\Vert _K^{n+p s}} \,\textrm{d}x \,\textrm{d}y \ge \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{|f^\star (x) -f^\star (y)|^p}{\Vert x-y\Vert _{K^\star }^{n+p s}} \,\textrm{d}x \,\textrm{d}y. \end{aligned}$$

Equality holds for non-zero $f\in W^{s, p}({\mathbb {R}}^n)$ if and only if K is a centered ellipsoid and f is a translate of $f^\star \circ \phi$ for some $\phi \in \textrm{SL}(n)$.

Proof

Since

$$\begin{aligned} \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{(f(x) - f(y))_{-}^{ p}}{\Vert x-y\Vert _K^{n+p s}} \,\textrm{d}x \,\textrm{d}y =\int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{(f(x) - f(y))_{+}^{p}}{\Vert x-y\Vert _{-K}^{n+p s}} \,\textrm{d}x \,\textrm{d}y, \end{aligned}$$

the result follows from Theorem 11 for K and $-K$. $\square$

7 Affine fractional Pólya–Szegő inequalities

We establish affine Pólya–Szegő inequalities for fractional asymmetric and symmetric $L^p$ polar projection bodies.

Theorem 13

If $f\in W^{s,p}({\mathbb {R}}^n)$ is non-negative, then

$$\begin{aligned} \vert {\Pi ^{*,{ s}}_{{p}, +} f}\vert ^{-p s/n} \ge \vert {\Pi ^{*,{ s}}_{{p}, +} f^\star }\vert ^{- p s/n}. \end{aligned}$$

(22)

Equality holds if and only if f is a translate of $f^\star \circ \phi$ for some $\phi \in \textrm{SL}(n)$.

Proof

By Theorem 11, (16) and the dual mixed volume inequality, we obtain for $K\subset {\mathbb {R}}^n$ a star body that

$$\begin{aligned} \tilde{V}_{-p s} (K, \Pi ^{*,{ s}}_{{p}, +} f)&\ge \tilde{V}_{-p s} (K^\star , \Pi ^{*,{ s}}_{{p}, +} f^\star )\\&\ge \vert {K^\star }\vert ^{({n+p s})/{n}} \vert {\Pi ^{*,{ s}}_{{p}, +} f^\star }\vert ^{- p s/n}\\&= \vert {K}\vert ^{({n+p s})/{n}} \vert {\Pi ^{*,{ s}}_{{p}, +} f^\star }\vert ^{- p s/n}. \end{aligned}$$

Setting $K = \Pi ^{*,{ s}}_{{p}, +} f$, we see that

$$\begin{aligned} \vert {\Pi ^{*,{ s}}_{{p}, +} f}\vert = \tilde{V}_{-p s} (\Pi ^{*,{ s}}_{{p}, +} f, \Pi ^{*,{ s}}_{{p}, +} f) \ge \vert {\Pi ^{*,{ s}}_{{p}, +} f}\vert ^{({n+p s})/{n}} \vert {\Pi ^{*,{ s}}_{{p}, +} f^\star }\vert ^{-p s/n}, \end{aligned}$$

which completes the proof of the inequality. By Theorem 11, there is equality in (7) if and only if f is a translate of $f^\star \circ \phi$ for some $\phi \in \textrm{SL}(n)$. $\square$

The following result is obtained in the same way as Theorem 13 by replacing Theorem 11 with Theorem 12.

Theorem 14

If $f\in L^p({\mathbb {R}}^n)$ is non-negative, then

$$\begin{aligned} \vert {\Pi ^{*,{ s}}_{{p}} f}\vert ^{-p s/n} \ge \vert {\Pi ^{*,{ s}}_{{p}} f^\star }\vert ^{- p s/n}. \end{aligned}$$

Equality holds for $f\in W^{s, p}({\mathbb {R}}^n)$ if and only if f is a translate of $f^\star \circ \phi$ for some $\phi \in \textrm{SL}(n)$.

We remark that by Theorem 10 we obtain from Theorem 14 in the limit as $s\rightarrow 1^-$ that

$$\begin{aligned} \vert {\Pi _{p}^{*} f}\vert ^{-p /n} \ge \vert {\Pi _{p}^{*} f^\star }\vert ^{- p /n}, \end{aligned}$$

which is equivalent to the Pólya–Szegő inequality for $L^p$ projection bodies by Cianchi, Lutwak, Yang, and Zhang [11, Theorem 2.1]. Similarly, by Theorem 9 we obtain from Theorem 13 in the limit as $s\rightarrow 1^-$ that

$$\begin{aligned} \vert {\Pi ^{*}_{p, +} f}\vert ^{-p /n} \ge \vert {\Pi ^{*}_{p, +} f^\star }\vert ^{- p /n}, \end{aligned}$$

which is equivalent to the Pólya–Szegő inequality for asymmetric $L^p$ projection bodies by Haberl, Schuster and Xiao [15, Theorem 1].

8 Affine fractional asymmetric $L^p$ Sobolev inequalities

We establish the following affine fractional asymmetric $L^p$ Sobolev inequalities and show that they are stronger than Theorem 1.

Theorem 15

Let $0<s<1$ and $1<p<n/s$. For non-negative $f\in W^{s,p}({\mathbb {R}}^n)$,

$$\begin{aligned} \Vert f\Vert _{{\frac{n p}{n-p s}}}^p \le 2\,\sigma _{n,p,s} n\omega _n^{\frac{n+p s}{n}}\vert {\Pi ^{*,{ s}}_{{p}, +} f}\vert ^{- \frac{p s}{n}} \le 2 \sigma _{n,p,s}\int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{(f(x) -f(y))_{+}^{p}}{\vert x-y \vert ^{n+p s}}\,\textrm{d}x\,\textrm{d}y. \end{aligned}$$

There is equality in the first inequality if and only if $f=h_{s,p}\circ \phi$ for some $\phi \in \textrm{GL}(n)$ where $h_{s,p}$ is an extremal function of (1). There is equality in the second inequality if f is radially symmetric.

Proof

By Theorem 13,

$$\begin{aligned} \vert {\Pi ^{*,{ s}}_{{p}, +} f}\vert ^{-p s/n} \ge \vert {\Pi ^{*,{ s}}_{{p}, +} f^\star }\vert ^{- p s/n}, \end{aligned}$$

with equality if f is a translate of $f^\star \circ \phi$ for some $\phi \in \textrm{SL}(n)$. Since $f^\star$ is radially symmetric, $\Pi ^{*,{ s}}_{{p}, +} f^\star =\Pi ^{*,{ s}}_{{p}, -} f^\star$ is a ball. Hence, it follows from (16) that

$$\begin{aligned} 2 n\omega _n^{\frac{n+p s}{n}} \vert {\Pi ^{*,{ s}}_{{p}, +} f^\star }\vert ^{-\frac{p s}{n}}&= 2 \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{( f^\star (x) -f^\star (y))_{+}^{p}}{\vert x-y \vert ^{n+p s}}\,\textrm{d}x\,\textrm{d}y\\&= \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{\vert f^\star (x)-f^\star (y)\vert ^p}{\vert x-y \vert ^{n+p s}}\,\textrm{d}x\,\textrm{d}y. \end{aligned}$$

The fractional Sobolev inequality (1) shows that

$$\begin{aligned} \sigma _{n,p,s}\int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{\vert f^\star (x)-f^\star (y)\vert ^p}{\vert x-y \vert ^{n+p s}}\,\textrm{d}x\,\textrm{d}y\ge \Vert f^\star \Vert _{{\frac{n p}{n-p s}}}^p. \end{aligned}$$

Combining these inequalities and their equality cases, we complete the proof of the first inequality of the theorem.

For the second inequality, we set $K={B^n}$ in (16) and apply the dual mixed volume inequality (6) to obtain

$$\begin{aligned} \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{(f(x)-f(y))_{+}^{p}}{\vert x-y \vert ^{n+p s}}\,\textrm{d}x\,\textrm{d}y = n \tilde{V}_{-p s}({B^n}, \Pi ^{*,{ s}}_{{p}, +} f)\ge n \omega _n^{\frac{n+p s}{n}}\vert {\Pi ^{*,{ s}}_{{p}, +} f}\vert ^{-\frac{p s}{n}}. \end{aligned}$$

There is equality precisely if $\Pi ^{*,{ s}}_{{p}, +} f$ is a ball, which is the case for radially symmetric functions. $\square$

Note that it follows from the definition of fractional symmetric and asymmetric $L^p$ polar projection bodies that

$$\begin{aligned} \Pi ^{*,{ s}}_{{p}} f =\Pi ^{*,{ s}}_{{p}, +} f {\tilde{+}}_{ -p s} \Pi ^{*,{ s}}_{{p}, -} f. \end{aligned}$$

We use the dual Brunn–Minkowski inequality (5) and obtain that

$$\begin{aligned} \vert {\Pi ^{*,{ s}}_{{p}} f}\vert ^{-\frac{p s}{n}} \ge \vert {\Pi ^{*,{ s}}_{{p}, +} f }\vert ^{-\frac{p s}{n}} +\vert { \Pi ^{*,{ s}}_{{p}, -} f}\vert ^{-\frac{p s}{n}}, \end{aligned}$$

with equality precisely if the star bodies $\Pi ^{*,{ s}}_{{p}, +} f$ and $\Pi ^{*,{ s}}_{{p}, -} f$ are dilates. Thus, it follows that for non-negative f, Theorem 15 implies Theorem 1, and it is, in general, substantially stronger than Theorem 1. Of course, they coincide for even functions.

9 Affine fractional $L^p$ Sobolev inequalities: Proof of Theorem 1

For non-negative f, the first inequality in Theorem 1 follows from Theorem 15, as mentioned before. For general f and $x,y\in {\mathbb {R}}^n$, we use

$$\begin{aligned} \vert f(x) - f(y) \vert \ge \big \vert \vert f(x) \vert - \vert f(y) \vert \big \vert , \end{aligned}$$

where equality holds if and only if f(x) and f(y) are both non-negative or non-positive. We obtain

$$\begin{aligned} \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{\vert f(x)-f(y)\vert ^p}{\vert x-y\vert ^{n+s p}}\,\textrm{d}x\,\textrm{d}y \ge \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{\big \vert \vert f(x)\vert -\vert f(y)\vert \big \vert ^p}{\vert x-y\vert ^{n+s p}}\,\textrm{d}x\,\textrm{d}y, \end{aligned}$$

with equality if and only if f has constant sign for almost every $x,y \in {\mathbb {R}}^n$. Using the result for $\vert f\vert$, we obtain the first inequality of the theorem and its equality case.

For the second inequality, we set $K={B^n}$ in (10) and apply the dual mixed volume inequality (6) as in the proof of Theorem 15.

10 Optimal fractional $L^p$ Sobolev bodies

The following important question was asked by Lutwak, Yang and Zhang [25] for a given $f\in W^{1,p}({\mathbb {R}}^n)$ and $1\le p< n$: For which origin-symmetric convex bodies $K\subset {\mathbb {R}}^n$ is

$$\begin{aligned} \inf \left\{ \int _{{\mathbb {R}}^n}\Vert \nabla f(x) \Vert ^p_{K^*}\,\textrm{d}x: K \text { origin-symmetric convex body}, \vert {K}\vert =\omega _n\right\} \end{aligned}$$

(23)

attained? An optimal $L^p$ Sobolev body of f is a convex body where the infimum is attained.

Lutwak, Yang and Zhang [25] showed that the infimum in (23) is attained (up to normalization) at the unique origin-symmetric convex body ${\langle {f}\rangle }_{p}$ in ${\mathbb {R}}^n$ such that

$$\begin{aligned} \int _{{\mathbb {S}}^{n-1}} g(\xi )\,\textrm{d}S_p({\langle {f}\rangle }_{p},\xi )=\int _{{\mathbb {R}}^n} g(\nabla f (x))\,\textrm{d}x \end{aligned}$$

(24)

for every even $g\in C({\mathbb {R}}^n)$ that is positively homogeneous of degree p, where $S_p(K,\cdot )$ is the $L_p$ surface area measure of K. Setting $g=\Vert \cdot \Vert _{K^*}$, they obtain from the $L^p$ Minkowski inequality that

$$\begin{aligned} \frac{1}{n}\int _{{\mathbb {R}}^n}\Vert \nabla f(x) \Vert ^p_{K^*}\,\textrm{d}x =V_p({\langle {f}\rangle }_{p}, K)\ge \vert {{\langle {f}\rangle }_{p}}\vert ^{(n-p)/n} \vert {K}\vert ^{p/n}, \end{aligned}$$

(25)

with equality precisely if K and ${\langle {f}\rangle }_{p}$ are homothetic (see [28, Section 9.1] for the definition of the $L_p$ mixed volume $V_p(\cdot ,\cdot )$ and the $L^p$ Minkowski inequality). Hence, they obtain from their solution to their functional version (24) of the $L^p$ Minkowski problem that ${\langle {f}\rangle }_{p}$ is the optimal $L^p$ Sobolev body associated to f. Tuo Wang [31] obtained corresponding results for $f\in BV({\mathbb {R}}^n)$ and $p=1$.

Let $0<s<1$ and $1<p<n/s$. The results by Lutwak, Yang and Zhang [25] suggest the following question for a given $f\in W^{s,p}({\mathbb {R}}^n)$: For which star bodies $L\subset {\mathbb {R}}^n$ is

$$\begin{aligned} \inf \left\{ \int _{{\mathbb {R}}^n} \int _{{\mathbb {R}}^n} \frac{\vert f(x)-f(y) \vert ^p}{\Vert x-y\Vert _L^{n+p s}}\,\textrm{d}x\,\textrm{d}y: L \text { star body}, \vert {L}\vert =\omega _n \right\} \end{aligned}$$

(26)

attained? An optimal s-fractional $L^p$ Sobolev body of f is a star body where the infimum is attained.

By (10) and the dual mixed volume inequality (6),

$$\begin{aligned} \frac{1}{n}\int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{\vert f(x) -f(y)\vert ^p}{\Vert x-y\Vert _L^{n+p s}}\,\textrm{d}x\,\textrm{d}y =\tilde{V}_{-p s}(L, \Pi ^{*,{ s}}_{{p}} f)\ge \vert {L}\vert ^{(n+p s)/n} \vert {\Pi ^{*,{ s}}_{{p}} f}\vert ^{-(p s)/n}, \end{aligned}$$

and there is equality precisely if L is a dilate of $\Pi ^{*,{ s}}_{{p}} f$. Hence, $\Pi ^{*,{ s}}_{{p}} f$ is the unique optimal s-fractional $L^p$ Sobolev body associated to f.

To understand how the solutions to (23) and (26) are related, we use the following result: For $f\in W^{1,p}({\mathbb {R}}^n)$ and $L\subset {\mathbb {R}}^n$ a star body,

$$\begin{aligned} \lim _{s\rightarrow 1^-}p (1-s)\,\int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{\vert f(x)-f(y)\vert ^p}{\Vert {x-y}\Vert ^{n+p s}_{L}} \,\textrm{d}x\,\textrm{d}y = \int _{{\mathbb {R}}^n} \Vert \nabla f(x) \Vert _{\textrm{Z}_p^*L}\,\textrm{d}x, \end{aligned}$$

(27)

where the convex body $\textrm{Z}_pK$, defined for $\xi \in {\mathbb {S}}^{n-1}$ by

$$\begin{aligned} h_{\textrm{Z}_pL}(\xi )^p = \int _{{\mathbb {S}}^{n-1}} \vert \langle \xi , \eta \rangle \vert ^p \rho _L(\eta )^{n+p} \,\textrm{d}\eta , \end{aligned}$$

is a multiple of the $L^p$ centroid body of L. This can be proved as in [21], where the corresponding result was established for a convex body L (with a different normalization of $\textrm{Z}_pL$). It also follows from Theorem 10. Indeed, by (10) and (20),

$$\begin{aligned} \lim _{s\rightarrow 1^-}p(1-s)\,\int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n} \frac{\vert f(x)-f(y)\vert ^p}{\Vert {x-y}\Vert ^{n+p s}_{L}} \,\textrm{d}x\,\textrm{d}y= \tilde{V}_{-p}(L, \Pi _{p}^{*}f). \end{aligned}$$

Using that

$$\begin{aligned} \Pi _{p}^{*}f= \Pi _{p}^{*} {\langle {f}\rangle }_{p} \end{aligned}$$

(28)

for $f\in W^{1,p}({\mathbb {R}}^n)$, which follows from (24) by setting $g= \vert \langle \cdot ,\eta \rangle \vert ^p$ for $\eta \in {\mathbb {S}}^{n-1}$ and using (8) and (9) (cf. [25]), and that

$$\begin{aligned} V_p(K, \textrm{Z}_pL)=\tilde{V}_{-p}(L, \Pi _{p}^{*} K) \end{aligned}$$

(29)

for K a convex body and L a star body, a well-known relation that follows from Fubini’s theorem, we now obtain (27) from the first equation in (25).

Using (27), we obtain from (26) in the limit as $s\rightarrow 1^-$ for a given $f\in W^{1,p}({\mathbb {R}}^n),$ the following question: For which star bodies $L\subset {\mathbb {R}}^n$ is

$$\begin{aligned} \inf \Big \{\int _{{\mathbb {R}}^n}\Vert \nabla f(x) \Vert _{\textrm{Z}_p^*L}\,\textrm{d}x: L \text { star body}, \vert {L}\vert =\omega _n\Big \} \end{aligned}$$

(30)

attained? By (25) and the dual mixed volume inequality (6), we have

$$\begin{aligned} \frac{1}{n}\int _{{\mathbb {R}}^n} \Vert \nabla f(x) \Vert ^p_{\textrm{Z}_p^*L}\,\textrm{d}x = V_p({\langle {f}\rangle }_{p},\textrm{Z}_pL) = \tilde{V}_{-p}(L, \Pi _{p}^{*} f) \ge \vert {L}\vert ^{(n+p)/n} \vert {\Pi _{p}^{*} f}\vert ^{-p/n}, \end{aligned}$$

with equality precisely if L and $\Pi _{p}^{*}f$ are dilates, where we have used (28) and (29). From Theorem 10, we obtain that a suitably scaled sequence of optimal s-fractional Sobolev bodies converges to a multiple of the optimal body for (30) as $s\rightarrow 1^-$.