1004/1004.4828.md

A Fractional Hardy-Sobolev-Maz'ya Inequality on the Upper Halfspace

Craig A. Sloane

Georgia Institute of Technology, Atlanta, Georgia 30332-0160

csloane@math.gatech.edu

October 31, 2018

Abstract

We prove several Sobolev inequalities, which are then used to establish a fractional Hardy-Sobolev-Maz'ya inequality on the upper halfspace.

1 Introduction

The present work answers a question by Frank and Seiringer [9] concerning fractional Hardy-Sobolev-Maz'ya inequalities for the upper halfspace in the case $p = 2$ . Let

$${\mathbb{R}}_{+}^n=\{x=(x^{\mathrm{\prime }},x_n)\in {\mathbb{R}}^n:x^{\mathrm{\prime }}\in {\mathbb{R}}^{n-1},x_n>0\}\backslash mathbb\{R\}\_+^n\; =\; \backslash \{x\; =\; (x\text{'},\; x\_n)\; \backslash in\; \backslash mathbb\{R\}^n\; :\; x\text{'}\; \backslash in\; \backslash mathbb\{R\}^\{n-1\},\; x\_n\; >\; 0\backslash \}$$

be the upper halfspace, and let $\Omega$ be a domain in $\mathbb{R}^n$ with nonempty boundary. Then, there exists a fractional Hardy inequality on $\mathbb{R}+^n$ which states that there exists $D{n,p,\alpha} > 0$ so that for all $f \in C_c(\mathbb{R}_+^n)$ ,

$${\int }_{{\mathbb{R}}_{+}^n\times {\mathbb{R}}_{+}^n}\frac{\mathrm{\mid }f(x)-f(y){\mathrm{\mid }}^p}{\mathrm{\mid }x-y{\mathrm{\mid }}^{n+\alpha }}dxdy\ge D_{n,p,\alpha }{\int }_{{\mathbb{R}}_{+}^n}\frac{\mathrm{\mid }f(x){\mathrm{\mid }}^p}{x_n^{\alpha }}dx,(1)\backslash int\_\{\backslash mathbb\{R\}\_+^n\; \backslash times\; \backslash mathbb\{R\}\_+^n\}\; \backslash frac\{|f(x)\; -\; f(y)|^p\}\{|x\; -\; y|^\{n+\backslash alpha\}\}\; dx\; dy\; \backslash geq\; D\_\{n,p,\backslash alpha\}\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; \backslash frac\{|f(x)|^p\}\{x\_n^\backslash alpha\}\; dx,\; \backslash quad\; (1)$$

where $1 \leq p < \infty$ , $0 < \alpha < p$ , and $\alpha \neq 1$ . See, e.g., [7]. Bogdan and Dyda found, in [3], the sharp constant $D_{n,2,\alpha}$ for the $p = 2$ case. Later, the sharp constant, $D_{n,p,\alpha}$ , for general $p$ was found in [9]. Therein, the authors posed the question whether there existed a lower bound to the remainder for the inequality in (1) that is a positive multiple of the $L^{p^*}$ -norm of $f$ , where $p^* = np/(n - \alpha)$ is the critical Sobolev exponent. Such an inequality would be a fractional analogue to the Hardy-Sobolev-Maz'ya inequality on the halfspace. Maz'ya [13] was the first to show the general result

$${\int }_{{\mathbb{R}}_{+}^n}\mathrm{\mid }\mathrm{\nabla }f(x){\mathrm{\mid }}^{2}dx-\frac{1}{4}{\int }_{{\mathbb{R}}_{+}^n}\frac{\mathrm{\mid }f(x){\mathrm{\mid }}^2}{x_n^2}dx\ge C_n{({\int }_{{\mathbb{R}}_{+}^n}\mathrm{\mid }f(x){\mathrm{\mid }}^{\frac{2n}{n-2}}dx)}^{\frac{n-2}{n}}\mathrm{.}\backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; |\backslash nabla\; f(x)|^2\; dx\; -\; \backslash frac\{1\}\{4\}\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; \backslash frac\{|f(x)|^2\}\{x\_n^2\}\; dx\; \backslash geq\; C\_n\; \backslash left(\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; |f(x)|^\{\backslash frac\{2n\}\{n-2\}\}\; dx\; \backslash right)^\{\backslash frac\{n-2\}\{n\}\}.$$

More recently the existence of minimizers for dimensions greater than or equal to 4 [14] and the sharp constant for dimension 3 [4] has been established. Further improvements in the general case have been shown in [10].The following Theorem for the sharp fractional Hardy inequality with remainder was proven by Frank and Seiringer in [9].

THEOREM 1.1. Let $n \geq 1$ , $2 \leq p \leq \infty$ , and $0 < \alpha < p$ with $\alpha \neq 1$ . Then for all $f \in C_c^\infty(\mathbb{R}_+^n)$ ,

$$\begin{array}{r}{\displaystyle {\int }_{\mathrm{\Omega }\times \mathrm{\Omega }}\frac{\mathrm{\mid }f(x)-f(y){\mathrm{\mid }}^p}{\mathrm{\mid }x-y{\mathrm{\mid }}^{n+\alpha }}dxdy-D_{n,p,\alpha }{\int }_{{\mathbb{R}}_{+}^n}\frac{\mathrm{\mid }f(x){\mathrm{\mid }}^p}{x_n^{\alpha }}dx}\\ {\displaystyle \ge c_p{\int }_{\mathrm{\Omega }\times \mathrm{\Omega }}\frac{\mathrm{\mid }{x}_n^{(1-\alpha )\mathrm{/}p}f(x)-y_n^{(1-\alpha )\mathrm{/}p}f(y){\mathrm{\mid }}^p}{\mathrm{\mid }x-y{\mathrm{\mid }}^{n+\alpha }}\frac{dx}{x_n^{(1-\alpha )\mathrm{/}2}}\frac{dy}{y_n^{(1-\alpha )\mathrm{/}2}}}\end{array}(2)\backslash begin\{aligned\}\; \backslash int\_\{\backslash Omega\; \backslash times\; \backslash Omega\}\; \backslash frac\{|f(x)\; -\; f(y)|^p\}\{|x\; -\; y|^\{n+\backslash alpha\}\}\; dx\; dy\; -\; D\_\{n,p,\backslash alpha\}\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; \backslash frac\{|f(x)|^p\}\{x\_n^\backslash alpha\}\; dx\; \backslash \backslash \; \backslash geq\; c\_p\; \backslash int\_\{\backslash Omega\; \backslash times\; \backslash Omega\}\; \backslash frac\{|x\_n^\{(1-\backslash alpha)/p\}\; f(x)\; -\; y\_n^\{(1-\backslash alpha)/p\}\; f(y)|^p\}\{|x\; -\; y|^\{n+\backslash alpha\}\}\; \backslash frac\{dx\}\{x\_n^\{(1-\backslash alpha)/2\}\}\; \backslash frac\{dy\}\{y\_n^\{(1-\backslash alpha)/2\}\}\; \backslash end\{aligned\}\; \backslash quad\; (2)$$

where $0 < c_p \leq 1$ is given by

If $p = 2$ then this is an equality with $c_p = 1$ .

For notational convenience, we write

$$I_{\alpha ,p}^{\mathrm{\Omega }}(f)={\int }_{\mathrm{\Omega }\times \mathrm{\Omega }}\frac{\mathrm{\mid }f(x)-f(y){\mathrm{\mid }}^p}{\mathrm{\mid }x-y{\mathrm{\mid }}^{n+\alpha }}dxdy,I\_\{\backslash alpha,p\}^\backslash Omega(f)\; =\; \backslash int\_\{\backslash Omega\; \backslash times\; \backslash Omega\}\; \backslash frac\{|f(x)\; -\; f(y)|^p\}\{|x\; -\; y|^\{n+\backslash alpha\}\}\; dx\; dy,$$

and

$$J_{\alpha ,p}^{\mathrm{\Omega }}(f)={\int }_{\mathrm{\Omega }\times \mathrm{\Omega }}\frac{\mathrm{\mid }{x}_n^{(1-\alpha )\mathrm{/}p}f(x)-y_n^{(1-\alpha )\mathrm{/}p}f(y){\mathrm{\mid }}^p}{\mathrm{\mid }x-y{\mathrm{\mid }}^{n+\alpha }}\frac{dx}{x_n^{(1-\alpha )\mathrm{/}2}}\frac{dy}{y_n^{(1-\alpha )\mathrm{/}2}}\mathrm{.}J\_\{\backslash alpha,p\}^\backslash Omega(f)\; =\; \backslash int\_\{\backslash Omega\; \backslash times\; \backslash Omega\}\; \backslash frac\{|x\_n^\{(1-\backslash alpha)/p\}\; f(x)\; -\; y\_n^\{(1-\backslash alpha)/p\}\; f(y)|^p\}\{|x\; -\; y|^\{n+\backslash alpha\}\}\; \backslash frac\{dx\}\{x\_n^\{(1-\backslash alpha)/2\}\}\; \backslash frac\{dy\}\{y\_n^\{(1-\backslash alpha)/2\}\}.$$

Since we are primarily concerned with the case $p = 2$ , we further denote $I_\alpha^\Omega = I_{\alpha,2}^\Omega$ and $J_\alpha^\Omega = J_{\alpha,2}^\Omega$ . Thus, for $p = 2$ , we can rewrite (2) as

$$I_{\alpha }^{{\mathbb{R}}_{+}^n}(f)-D_{n,2,\alpha }{\int }_{{\mathbb{R}}_{+}^n}\mathrm{\mid }f(x){\mathrm{\mid }}^2{x}_n^{-\alpha }dx=J_{\alpha }^{{\mathbb{R}}_{+}^n}(f),(3)I\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(f)\; -\; D\_\{n,2,\backslash alpha\}\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; |f(x)|^2\; x\_n^\{-\backslash alpha\}\; dx\; =\; J\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(f),\; \backslash quad\; (3)$$

The main result of this paper is the following fractional Hardy-Sobolev-Maz'ya inequality for $p = 2$ .

THEOREM 1.2. Let $n \geq 2$ , $1 < \alpha < 2$ . There exists $a_{n,\alpha} > 0$ so that

$${\int }_{{\mathbb{R}}_{+}^n\times {\mathbb{R}}_{+}^n}\frac{\mathrm{\mid }f(x)-f(y){\mathrm{\mid }}^2}{\mathrm{\mid }x-y{\mathrm{\mid }}^{n+\alpha }}dxdy-D_{n,2,\alpha }{\int }_{{\mathbb{R}}_{+}^n}\frac{\mathrm{\mid }f(x){\mathrm{\mid }}^2}{x_n^{\alpha }}dx\ge a_{n,\alpha }{({\int }_{{\mathbb{R}}_{+}^n}\mathrm{\mid }f(x){\mathrm{\mid }}^{2^{\ast }}dx)}^{2\mathrm{/}{2}^{\ast }},(4)\backslash int\_\{\backslash mathbb\{R\}\_+^n\; \backslash times\; \backslash mathbb\{R\}\_+^n\}\; \backslash frac\{|f(x)\; -\; f(y)|^2\}\{|x\; -\; y|^\{n+\backslash alpha\}\}\; dx\; dy\; -\; D\_\{n,2,\backslash alpha\}\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; \backslash frac\{|f(x)|^2\}\{x\_n^\backslash alpha\}\; dx\; \backslash geq\; a\_\{n,\backslash alpha\}\; \backslash left(\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; |f(x)|^\{2^*\}\; dx\; \backslash right)^\{2/2^*\},\; \backslash quad\; (4)$$

for all $f \in C_c^\infty(\mathbb{R}_+^n)$ .

Alternatively, we write (4) as $J_{\alpha}^{\mathbb{R}+^n}(f) \geq a{n,\alpha} |f|_{2^*}^2$ , where $|\cdot|_p$ refers to the $L^p$ -norm with usual Lebesgue measure.

2 Sobolev Inequalities

Herein, we establish two Sobolev-type inequalities that we'll need in the proof of Theorem 1.2. We prove each for the more general $p$ -case. The first inequality we prove is for $I_{\alpha,p}^\Omega$ with respect to convex sets $\Omega$ .THEOREM 2.1. Let $p \geq 2$ , $1 < \alpha < \min{n, p}$ , and let $\Omega \subseteq \mathbb{R}^n$ be convex. Then, for all $f \in C_c^\infty(\Omega)$ , there exists $c_{n,p,\alpha} > 0$ so that

$$I_{\alpha ,p}^{\mathrm{\Omega }}(f)\ge c_{n,p,\alpha }\mathrm{\parallel }f{\mathrm{\parallel }}_{p^{\ast }}^p\mathrm{.}I\_\{\backslash alpha,p\}^\backslash Omega(f)\; \backslash geq\; c\_\{n,p,\backslash alpha\}\; \backslash |f\backslash |\_\{p^*\}^p.$$

Proof. In [12], it is shown

$$I_{\alpha ,p}^{\mathrm{\Omega }}(f)\ge D_{n,p,\alpha }{\int }_{\mathrm{\Omega }}\mathrm{\mid }f(x){\mathrm{\mid }}^p{d}_{\mathrm{\Omega }}(x{)}^{-\alpha }dx,I\_\{\backslash alpha,p\}^\backslash Omega(f)\; \backslash geq\; D\_\{n,p,\backslash alpha\}\; \backslash int\_\{\backslash Omega\}\; |f(x)|^p\; d\_\{\backslash Omega\}(x)^\{-\backslash alpha\}\; dx,$$

for all $f \in C_c(\Omega)^\infty$ , where $d_{\Omega}(x) = \text{dist}(x, \partial\Omega)$ . Further, there exists $S_{n,p,\alpha}$ , see, e.g., [1], Theorems 7.34, 7.47, so that

$$I_{\alpha ,p}^{{\mathbb{R}}^n}(f)\ge S_{n,p,\alpha }\mathrm{\parallel }f{\mathrm{\parallel }}_{p^{\ast }}^p,I\_\{\backslash alpha,p\}^\{\backslash mathbb\{R\}^n\}(f)\; \backslash geq\; S\_\{n,p,\backslash alpha\}\; \backslash |f\backslash |\_\{p^*\}^p,$$

for all $f \in C_c^\infty(\mathbb{R}^n)$ . Thus, if $B(x, r)$ is the ball of radius $r$ centered at $x$ , then since

$${\int }_{{\mathrm{\Omega }}^c}\mathrm{\mid }x-y{\mathrm{\mid }}^{-n-\alpha }dy\le {\int }_{B(x,d_{\mathrm{\Omega }}(x){)}^c}\mathrm{\mid }x-y{\mathrm{\mid }}^{-n-\alpha }dy=\frac{1}{\alpha }\mathrm{\mid }{\mathbb{S}}^{n-1}\mathrm{\mid }{d}_{\mathrm{\Omega }}(x{)}^{-\alpha },\backslash int\_\{\backslash Omega^c\}\; |x\; -\; y|^\{-n-\backslash alpha\}\; dy\; \backslash leq\; \backslash int\_\{B(x,\; d\_\{\backslash Omega\}(x))^c\}\; |x\; -\; y|^\{-n-\backslash alpha\}\; dy\; =\; \backslash frac\{1\}\{\backslash alpha\}\; |\backslash mathbb\{S\}^\{n-1\}|\; d\_\{\backslash Omega\}(x)^\{-\backslash alpha\},$$

we have

where $|\mathbb{S}^{n-1}| = \frac{2\pi^{n/2}}{\Gamma(n/2)}$ is the surface area of the sphere of radius 1 in $\mathbb{R}^n$ . $\square$

The next inequality is a weighted inequality for the term $J_{\alpha,p}^{\mathbb{R}_+^n}(f)$ . We leave the proof to the appendix.

THEOREM 2.2. Let $p \geq 2$ , $1 < \alpha < \min{n, p}$ . Then, there exists $d_{n,p,\alpha} > 0$ so that

$$J_{\alpha ,p}^{{\mathbb{R}}_{+}^n}(f)\ge d_{n,p,\alpha }{({\int }_{{\mathbb{R}}_{+}^n}\mathrm{\mid }f(x){\mathrm{\mid }}^q{x}_n^{-n+nq\mathrm{/}{p}^{\ast }}dx)}^{p\mathrm{/}q},J\_\{\backslash alpha,p\}^\{\backslash mathbb\{R\}\_+^n\}(f)\; \backslash geq\; d\_\{n,p,\backslash alpha\}\; \backslash left(\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; |f(x)|^q\; x\_n^\{-n+nq/p^*\}\; dx\; \backslash right)^\{p/q\},$$

where $q = p \left( \frac{n + \frac{\alpha-1}{2}}{n-1} \right)$ .

Since the remainder of the paper deals with the case $p = 2$ , we write $c_{n,\alpha} = c_{n,2,\alpha}$ and $d_{n,\alpha} = d_{n,2,\alpha}$ .

In the next section, we'll show we can minimize over a certain class of functions that are decreasing, albeit not symmetrically. The crux of the proof of Theorem 1.2 in the following section is to decompose this function by truncation and use these two Sobolev inequalities to appropriately bound the $L^{2^*}$ -norms of the resulting "upper" and "lower" functions.### 3 Class of Minimizing Functions

In this section, we determine the properties of those functions that minimize our Rayleigh quotient

$${\mathrm{\Phi }}_{\alpha }(f):=\frac{I_{\alpha }^{{\mathbb{R}}_{+}^n}(f)-D_{n,2,\alpha }{\int }_{{\mathbb{R}}_{+}^n}\mathrm{\mid }f(x){\mathrm{\mid }}^2{x}_n^{-\alpha }dx}{\mathrm{\parallel }f{\mathrm{\parallel }}_{2^{\ast }}^2}=\frac{J_{\alpha }^{{\mathbb{R}}_{+}^n}(f)}{\mathrm{\parallel }f{\mathrm{\parallel }}_{2^{\ast }}^2}\mathrm{.}\backslash Phi\_\backslash alpha(f)\; :=\; \backslash frac\{I\_\backslash alpha^\{\backslash mathbb\{R\}\_+^n\}(f)\; -\; D\_\{n,2,\backslash alpha\}\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; |f(x)|^2\; x\_n^\{-\backslash alpha\}\; dx\}\{\backslash |f\backslash |\_\{2^*\}^2\}\; =\; \backslash frac\{J\_\backslash alpha^\{\backslash mathbb\{R\}\_+^n\}(f)\}\{\backslash |f\backslash |\_\{2^*\}^2\}.$$

To minimize this functional, we'd like to do a rearrangement, but we have the restriction that the function must have support in the upper halfspace. Still, we can rearrange the function along hyperplanes parallel to the boundary of $\mathbb{R}_+^n$ .

In addition, we consider the conformal transformation $T : B(\mathbf{0}, 1) \rightarrow \mathbb{R}_+^n$ , where $\mathbf{0}$ is the origin in $\mathbb{R}^n$ . If we write $\eta(\omega) = 2 / (|\omega'|^2 + (\omega_n + 1)^2)$ , where $\omega = (\omega', \omega_n)$ , $\omega' \in \mathbb{R}^{n-1}$ , $\omega_n \in \mathbb{R}$ , then

$$T\omega =\left(\frac{2{\omega }^{\mathrm{\prime }},1-\mathrm{\mid }\omega {\mathrm{\mid }}^2}{\mathrm{\mid }{\omega }^{\mathrm{\prime }}{\mathrm{\mid }}^2+({\omega }_n+1{)}^2}\right)=\eta (\omega )({\omega }^{\mathrm{\prime }},\frac{1-\mathrm{\mid }\omega {\mathrm{\mid }}^2}{2})\mathrm{.}T\backslash omega\; =\; \backslash left(\; \backslash frac\{2\backslash omega\text{'},\; 1\; -\; |\backslash omega|^2\}\{|\backslash omega\text{'}|^2\; +\; (\backslash omega\_n\; +\; 1)^2\}\; \backslash right)\; =\; \backslash eta(\backslash omega)\; \backslash left(\; \backslash omega\text{'},\; \backslash frac\{1\; -\; |\backslash omega|^2\}\{2\}\; \backslash right).$$

Note that $T$ is an involution, and its Jacobian is $\eta(\omega)^n$ . See, e.g., Appendix, [6]. We define

$$\tilde{f}(\omega ):=\eta (\omega {)}^{n\mathrm{/}{2}^{\ast }}f(T\omega ),\backslash tilde\{f\}(\backslash omega)\; :=\; \backslash eta(\backslash omega)^\{n/2^*\}\; f(T\backslash omega),$$

and since $\eta(Tx) = 1/\eta(x)$ , then

$$f(x)=\eta (x{)}^{n\mathrm{/}{2}^{\ast }}\tilde{f}(Tx)\mathrm{.}f(x)\; =\; \backslash eta(x)^\{n/2^*\}\; \backslash tilde\{f\}(Tx).$$

Thus, if $\text{supp } \tilde{f} \subseteq B(\mathbf{0}, 1)$ , then $\text{supp } f \subseteq \mathbb{R}_+^n$ . Indeed, for any $0 < R < 1$ , if $\text{supp } \tilde{f} \subseteq B(\mathbf{0}, R)$ , then $\text{supp } f \subseteq B^R$ , where we define

$$B^R:=\{(x^{\mathrm{\prime }},x_n)\in {\mathbb{R}}^n:\mathrm{\mid }{x}^{\mathrm{\prime }}{\mathrm{\mid }}^2+{(x_n-\frac{1+R^2}{1-R^2})}^2\le {\left(\frac{2R}{1-R^2}\right)}^2\},(5)B^R\; :=\; \backslash left\backslash \{\; (x\text{'},\; x\_n)\; \backslash in\; \backslash mathbb\{R\}^n\; :\; |x\text{'}|^2\; +\; \backslash left(\; x\_n\; -\; \backslash frac\{1\; +\; R^2\}\{1\; -\; R^2\}\; \backslash right)^2\; \backslash leq\; \backslash left(\; \backslash frac\{2R\}\{1\; -\; R^2\}\; \backslash right)^2\; \backslash right\backslash \},\; \backslash quad\; (5)$$

using $|Tx|^2 = \frac{|x'|^2 + (x_n - 1)^2}{|x'|^2 + (x_n + 1)^2}$ .

We use these results throughout the remainder of this paper. These provide a new “ball” picture in which to consider our inequality and minimization problem. Among other things, we can also perform a rotation of $\tilde{f}$ on the ball. It turns out that repeated application of this rotation, along with the rearrangement mentioned above, result in a limiting function that is radial in the ball picture and whose Rayleigh quotient is always smaller than that of the original.

This first result follows from Carlen and Loss [6].

THEOREM 3.1. Let $n \geq 2$ , $f \in L^{2^}(\mathbb{R}+^n)$ . Then, there exists $F \in L^{2^*}(\mathbb{R}+^n)$ such that*

1. 1. $|f|{2^*} = |F|{2^*}$ ,
2. 2. $F$ is nonnegative, symmetric decreasing in hyperplanes parallel to the boundary of $\mathbb{R}_+^n$ ,
3. 3. $\tilde{F}$ is rotationally symmetric, and
4. 4. $\Phi_\alpha(f) \geq \Phi_\alpha(F)$ .Proof. Since $|f(x) - f(y)| \geq | |f|(x) - |f|(y) |$ implies $I_{\alpha}^{\mathbb{R}+^n}(f) \geq I{\alpha}^{\mathbb{R}+^n}(|f|)$ , we can assume $f$ is nonnegative. Then, the first three items of this theorem are a direct result of [6], Theorem 2.4. Indeed, let $Uf$ be the transformation of $f$ obtained by a certain fixed rotation of $\tilde{f}$ , as described in [6]. In particular, using the rotation $\mathbf{R} : (x_1, \dots, x{n-1}, x_n) \mapsto (x_1, \dots, x_n, -x_{n-1})$ , $x_i \in \mathbb{R}$ , $i = 1, \dots, n$ , then $U$ maps

$$f(x)\mapsto \tilde{f}(x)\mapsto \tilde{f}(\mathbf{R}x)\mapsto \eta (x{)}^{n\mathrm{/}{2}^{\ast }}\tilde{f}(\mathbf{R}Tx)\mathrm{.}f(x)\; \backslash mapsto\; \backslash tilde\{f\}(x)\; \backslash mapsto\; \backslash tilde\{f\}(\backslash mathbf\{R\}x)\; \backslash mapsto\; \backslash eta(x)^\{n/2^*\}\; \backslash tilde\{f\}(\backslash mathbf\{R\}Tx).$$

Note how the last transformation mimics the map $\tilde{f} \mapsto f$ . Further, let $V$ be the symmetric decreasing rearrangement in hyperplanes parallel to the boundary of $\mathbb{R}+^n$ . Define $F_k := (VU)^k f$ , then, by Theorem 2.4 in [6], there exists $F \in L^{2^*}(\mathbb{R}+^n)$ that is nonnegative and symmetric decreasing in hyperplanes parallel to the boundary of $\mathbb{R}+^n$ and such that $|f|{2^*} = |F|{2^*}$ , $\tilde{F}$ is radial on the unit ball, and $\lim{k \rightarrow \infty} F_k = F$ in $L^{2^*}(\mathbb{R}_+^n)$ . By passing to a subsequence, we can assume, without loss of generality, $F_k \rightarrow F$ almost everywhere.

As was calculated in [3], there exists a constant $c > 0$ so that we can write the remainder term as

$$J_{\alpha }^{{\mathbb{R}}_{+}^n}(f)=I_{\alpha }^{{\mathbb{R}}_{+}^n}(f)-c{\int }_{{\mathbb{R}}_{+}^n}\frac{\mathrm{\mid }f(x){\mathrm{\mid }}^2}{x_n^{\alpha }}dx\mathrm{.}(6)J\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(f)\; =\; I\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(f)\; -\; c\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; \backslash frac\{|f(x)|^2\}\{x\_n^\backslash alpha\}\; dx.\; \backslash quad\; (6)$$

We claim $\Phi_\alpha(F_k)$ is decreasing. As $F_0 = f$ , it is enough to show that $\Phi_\alpha(f) \geq \Phi_\alpha(VUf)$ .

By a modification of Theorem 7.17 of [11], $I_{\alpha}^{\mathbb{R}+^n}(f)$ decreases under the rearrangement $V$ . From Lemma 6.1 in the Appendix, $I{\alpha}^{\mathbb{R}+^n}(f) = I{\alpha}^{\mathbb{R}+^n}(\tilde{f})$ , the latter of which is invariant under rotations. Hence, $I{\alpha}^{\mathbb{R}+^n}(f)$ is invariant under $U$ . Next, as the rearrangement under $V$ is only along hyperplanes parallel to the boundary of $\mathbb{R}+^n$ (i.e., where $x_n$ is fixed), the integral $\int_{\mathbb{R}_+^n} f^2(x)x_n^{-\alpha} dx$ must be invariant under $V$ , and since

$${\int }_{{\mathbb{R}}_{+}^n}{f}^2(x)x_n^{-\alpha }dx={\int }_{B(0,1)}{\left(\frac{2}{1-\mathrm{\mid }\omega {\mathrm{\mid }}^2}\right)}^{\alpha }{\tilde{f}}^2(\omega )d\omega ,\backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; f^2(x)x\_n^\{-\backslash alpha\}\; dx\; =\; \backslash int\_\{B(0,1)\}\; \backslash left(\; \backslash frac\{2\}\{1-|\backslash omega|^2\}\; \backslash right)^\backslash alpha\; \backslash tilde\{f\}^2(\backslash omega)\; d\backslash omega,$$

it is invariant under $U$ as well. Finally, the $L^{2^*}$ -norm is clearly invariant under $U$ and $V$ .

Therefore, applying Fatou's lemma,

$${\mathrm{\Phi }}_{\alpha }(f)=\frac{J_{\alpha }^{{\mathbb{R}}_{+}^n}(f)}{\mathrm{\parallel }f{\mathrm{\parallel }}_{2^{\ast }}^2}=\frac{J_{\alpha }^{{\mathbb{R}}_{+}^n}(F_0)}{\mathrm{\parallel }F{\mathrm{\parallel }}_{2^{\ast }}^2}\ge \underset{k\to \mathrm{\infty }}{\lim }\frac{J_{\alpha }^{{\mathbb{R}}_{+}^n}(F_k)}{\mathrm{\parallel }F{\mathrm{\parallel }}_{2^{\ast }}^2}\ge \frac{J_{\alpha }^{{\mathbb{R}}_{+}^n}(F)}{\mathrm{\parallel }F{\mathrm{\parallel }}_{2^{\ast }}^2}={\mathrm{\Phi }}_{\alpha }(F)\mathrm{.}\mathrm{\square }\backslash Phi\_\backslash alpha(f)\; =\; \backslash frac\{J\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(f)\}\{\backslash |f\backslash |\_\{2^*\}^2\}\; =\; \backslash frac\{J\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(F\_0)\}\{\backslash |F\backslash |\_\{2^*\}^2\}\; \backslash geq\; \backslash lim\_\{k\; \backslash rightarrow\; \backslash infty\}\; \backslash frac\{J\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(F\_k)\}\{\backslash |F\backslash |\_\{2^*\}^2\}\; \backslash geq\; \backslash frac\{J\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(F)\}\{\backslash |F\backslash |\_\{2^*\}^2\}\; =\; \backslash Phi\_\backslash alpha(F).\; \backslash quad\; \backslash square$$

As a result, we can explicitly write the limiting function in Theorem 3.1 as the product of two radial functions in the ball picture, a specific, known symmetrically increasing function and a symmetrically decreasing function.

THEOREM 3.2. Let $n \geq 2$ , $F \in L^{2^}(\mathbb{R}+^n)$ , where $F$ is nonnegative, symmetric decreasing in hyperplanes parallel to the boundary of $\mathbb{R}+^n$ , and $\tilde{F}$ is rotationally symmetric. Then, there exists a decreasing function $h : [0, 1] \rightarrow [0, \infty]$ , where $h(1) = 0$ , so that*

$$\tilde{F}(\omega )={\left(\frac{2}{1-\mathrm{\mid }\omega {\mathrm{\mid }}^2}\right)}^{n\mathrm{/}{2}^{\ast }}h(\mathrm{\mid }\omega \mathrm{\mid })\mathrm{.}\backslash tilde\{F\}(\backslash omega)\; =\; \backslash left(\; \backslash frac\{2\}\{1-|\backslash omega|^2\}\; \backslash right)^\{n/2^*\}\; h(|\backslash omega|).$$

Proof. Let $x \in \mathbb{R}_+^n$ such that $x_n = 1$ , and, recalling that $T$ is an involution, let $\omega = Tx$ . Thus, if we restrict $T$ to the hyperplane

$$H=\{(x^{\mathrm{\prime }},x_n)\in {\mathbb{R}}_{+}^n:x^{\mathrm{\prime }}\in {\mathbb{R}}^{n-1},x_n=1\},H\; =\; \backslash \{(x\text{'},\; x\_n)\; \backslash in\; \backslash mathbb\{R\}\_+^n\; :\; x\text{'}\; \backslash in\; \backslash mathbb\{R\}^\{n-1\},\; x\_n\; =\; 1\backslash \},$$then its image, or stereographic projection, is the sphere

$$S=\{\omega =({\omega }^{\mathrm{\prime }},{\omega }_n):{\omega }^{\mathrm{\prime }}\in {\mathbb{R}}^{n-1},{\omega }_n\in \mathbb{R},\mathrm{\mid }{\omega }^{\mathrm{\prime }}{\mathrm{\mid }}^2+{({\omega }_n+\frac{1}{2})}^2=\frac{1}{4}\}S\; =\; \backslash left\backslash \{\; \backslash omega\; =\; (\backslash omega\text{'},\; \backslash omega\_n)\; :\; \backslash omega\text{'}\; \backslash in\; \backslash mathbb\{R\}^\{n-1\},\; \backslash omega\_n\; \backslash in\; \backslash mathbb\{R\},\; |\backslash omega\text{'}|^2\; +\; \backslash left(\backslash omega\_n\; +\; \backslash frac\{1\}\{2\}\backslash right)^2\; =\; \backslash frac\{1\}\{4\}\; \backslash right\backslash \}$$

whose north and south poles pass through the origin and the point $(0, \dots, 0, -1)$ , respectively. Thus, for all $\omega \in S$ , we have that $\omega_n = -|\omega|^2$ , and $\eta(\omega) = 2/(1 - |\omega|^2)$ . Further,

$$\frac{2}{1-\mathrm{\mid }\omega {\mathrm{\mid }}^2}=\eta (\omega )=\eta (Tx)=\frac{1}{\eta (x)}=\frac{\mathrm{\mid }{x}^{\mathrm{\prime }}{\mathrm{\mid }}^2+4}{2},\backslash frac\{2\}\{1\; -\; |\backslash omega|^2\}\; =\; \backslash eta(\backslash omega)\; =\; \backslash eta(Tx)\; =\; \backslash frac\{1\}\{\backslash eta(x)\}\; =\; \backslash frac\{|x\text{'}|^2\; +\; 4\}\{2\},$$

as $x \in H$ . Then, $|x'|^2 = 4|\omega|^2/(1 - |\omega|^2)$ , and, since $F$ is radial on $H$ , then, for all $\omega \in S$ ,

$$F(T\omega )=F(x^{\mathrm{\prime }},1)=F(\mathrm{\mid }{x}^{\mathrm{\prime }}\mathrm{\mid },1)=F(\frac{2\mathrm{\mid }\omega \mathrm{\mid }}{\sqrt{1-\mathrm{\mid }\omega {\mathrm{\mid }}^2}},1)\mathrm{.}F(T\backslash omega)\; =\; F(x\text{'},\; 1)\; =\; F(|x\text{'}|,\; 1)\; =\; F\backslash left(\backslash frac\{2|\backslash omega|\}\{\backslash sqrt\{1\; -\; |\backslash omega|^2\}\},\; 1\backslash right).$$

Further, as $\tilde{F}(\omega) = \eta(\omega)^{n/2^*} F(T\omega)$ , then

$$\tilde{F}(\omega )={\left(\frac{2}{1-\mathrm{\mid }\omega {\mathrm{\mid }}^2}\right)}^{n\mathrm{/}{2}^{\ast }}F(\frac{2\mathrm{\mid }\omega \mathrm{\mid }}{\sqrt{1-\mathrm{\mid }\omega {\mathrm{\mid }}^2}},1)={\left(\frac{2}{1-\mathrm{\mid }\omega {\mathrm{\mid }}^2}\right)}^{n\mathrm{/}{2}^{\ast }}h(\mathrm{\mid }\omega \mathrm{\mid }),\backslash tilde\{F\}(\backslash omega)\; =\; \backslash left(\backslash frac\{2\}\{1\; -\; |\backslash omega|^2\}\backslash right)^\{n/2^*\}\; F\backslash left(\backslash frac\{2|\backslash omega|\}\{\backslash sqrt\{1\; -\; |\backslash omega|^2\}\},\; 1\backslash right)\; =\; \backslash left(\backslash frac\{2\}\{1\; -\; |\backslash omega|^2\}\backslash right)^\{n/2^*\}\; h(|\backslash omega|),$$

where $h(r) = F\left(\frac{2r}{\sqrt{1-r^2}}, 1\right)$ is a decreasing function and $h(1) = 0$ , as $F$ is radially symmetrically decreasing on $H$ , and $F \in L^{2^*}(\mathbb{R}_+^n)$ .

Note that for each particular radius in the unit ball, the corresponding sphere intersects $S$ . This radius then corresponds to a particular radius in $H$ as given by $|x'|^2 = 4|\omega|^2/(1 - |\omega|^2)$ . But, on the ball, $\tilde{F}$ is a radial function, so if $\omega$ is any point in the unit ball, there exists some rotation $R_\omega$ and $\omega_S \in S$ so that $\omega = R_\omega \omega_S$ . Therefore,

$$\tilde{F}(\omega )=\tilde{F}(R_{\omega }{\omega }_S)={\left(\frac{2}{1-\mathrm{\mid }{R}_{\omega }{\omega }_S{\mathrm{\mid }}^2}\right)}^{n\mathrm{/}{2}^{\ast }}h(\mathrm{\mid }{R}_{\omega }{\omega }_S\mathrm{\mid })={\left(\frac{2}{1-\mathrm{\mid }\omega {\mathrm{\mid }}^2}\right)}^{n\mathrm{/}{2}^{\ast }}h(\mathrm{\mid }\omega \mathrm{\mid }),\backslash tilde\{F\}(\backslash omega)\; =\; \backslash tilde\{F\}(R\_\backslash omega\; \backslash omega\_S)\; =\; \backslash left(\backslash frac\{2\}\{1\; -\; |R\_\backslash omega\; \backslash omega\_S|^2\}\backslash right)^\{n/2^*\}\; h(|R\_\backslash omega\; \backslash omega\_S|)\; =\; \backslash left(\backslash frac\{2\}\{1\; -\; |\backslash omega|^2\}\backslash right)^\{n/2^*\}\; h(|\backslash omega|),$$

for all $\omega \in B(\mathbf{0}, 1)$ . □

4 Proof of Main Result

Proof of Theorem 1.2: From Theorems 3.1 and 3.2, we can assume $\tilde{f}(\omega) = \left(\frac{2}{1 - |\omega|^2}\right)^{n/2^*} h(|\omega|)$ , where $h(r)$ is a decreasing function on $[0, 1]$ and $h(1) = 0$ . Then,

$$f(x)=\eta (x{)}^{n\mathrm{/}{2}^{\ast }}\tilde{f}(Tx)=x_n^{-n\mathrm{/}{2}^{\ast }}h(\mathrm{\mid }Tx\mathrm{\mid }),f(x)\; =\; \backslash eta(x)^\{n/2^*\}\; \backslash tilde\{f\}(Tx)\; =\; x\_n^\{-n/2^*\}\; h(|Tx|),$$

and $I_\alpha^{\mathbb{R}^n}(f), |f|{2^*}^2 < \infty$ . However, we note that $f$ is no longer necessarily in $C_c^\infty(\mathbb{R}+^n)$ .

We decompose $h = h_1 + h_0$ by truncation, by fixing $R \in (0, 1)$ so $h_0(r) = \min{h(r), h(R)}$ . Then, $f = f_1 + f_0$ , with the definitions $f_1, f_0$ following from the above. We claim there exists $c, d > 0$ , each dependent on $R, n$ and $\alpha$ , such that

$$J_{{\alpha }^{+}}^{{\mathbb{R}}^n}(f)\ge c\mathrm{\parallel }{f}_1{\mathrm{\parallel }}_{2^{\ast }}^2,(7)J\_\{\backslash alpha^+\}^\{\backslash mathbb\{R\}^n\}(f)\; \backslash geq\; c\; \backslash |f\_1\backslash |\_\{2^*\}^2,\; \backslash quad\; (7)$$

and

$$J_{{\alpha }^{+}}^{{\mathbb{R}}^n}(f)\ge d\mathrm{\parallel }{f}_0{\mathrm{\parallel }}_{2^{\ast }}^2\mathrm{.}(8)J\_\{\backslash alpha^+\}^\{\backslash mathbb\{R\}^n\}(f)\; \backslash geq\; d\; \backslash |f\_0\backslash |\_\{2^*\}^2.\; \backslash quad\; (8)$$

Then, using the triangle and arithmetic-geometric mean inequalities, for all $0 < \lambda < 1$ , we obtain

$$J_{{\alpha }^{+}}^{{\mathbb{R}}^n}(f)\ge \lambda c\mathrm{\parallel }{f}_1{\mathrm{\parallel }}_{2^{\ast }}^2+(1-\lambda )d\mathrm{\parallel }{f}_0{\mathrm{\parallel }}_{2^{\ast }}^2\ge \frac{1}{2}\min \{\lambda c,(1-\lambda )d\}\mathrm{\parallel }f{\mathrm{\parallel }}_{2^{\ast }}^2\mathrm{.}J\_\{\backslash alpha^+\}^\{\backslash mathbb\{R\}^n\}(f)\; \backslash geq\; \backslash lambda\; c\; \backslash |f\_1\backslash |\_\{2^*\}^2\; +\; (1\; -\; \backslash lambda)\; d\; \backslash |f\_0\backslash |\_\{2^*\}^2\; \backslash geq\; \backslash frac\{1\}\{2\}\; \backslash min\backslash \{\backslash lambda\; c,\; (1\; -\; \backslash lambda)\; d\backslash \}\; \backslash |f\backslash |\_\{2^*\}^2.$$Clearly, by fixing $\lambda, R$ not equal to zero or one, the constant is greater than zero. So, by taking the supremum over $\lambda$ and $R$ , the result follows.

First, we prove (7). Note that $\text{supp } h_1 \subseteq [0, R]$ , so $\text{supp } f_1 \subseteq B^R$ , where $B^R$ is as in (5) above. Thus, for all $x, y \in B^R$ ,

It is easy to see that for any $0 < R < 1$ ,

where $A_1$ is dependent on $R, n$ and $\alpha$ . Thus, $J_\alpha^{B^R}(f) + A_1 h^2(R) \geq \frac{1}{2} J_\alpha^{B^R}(f_1)$ . We claim we can apply Theorem 2.1 to $x_n^{\frac{1-\alpha}{2}} f_1(x)$ . Hence, if $x \in B^R$ , then $\frac{1-R}{1+R} \leq x_n \leq \frac{1+R}{1-R}$ , and

Using Theorem 2.2,

$$J_{\alpha }^{{\mathbb{R}}_{+}^n}(f)\ge d_{n,\alpha }{({\int }_{{\mathbb{R}}_{+}^n}\mathrm{\mid }h(\mathrm{\mid }Tx\mathrm{\mid }){\mathrm{\mid }}^q{x}_n^{-n}dx)}^{2\mathrm{/}q}\ge d_{n,\alpha }{h}^2(R){\left({\int }_{B^R}{x}_n^{-n}dx\right)}^{2\mathrm{/}q}=A_2{h}^2(R),J\_\backslash alpha^\{\backslash mathbb\{R\}\_+^n\}(f)\; \backslash geq\; d\_\{n,\backslash alpha\}\; \backslash left(\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; |h(|Tx|)|^q\; x\_n^\{-n\}\; dx\; \backslash right)^\{2/q\}\; \backslash geq\; d\_\{n,\backslash alpha\}\; h^2(R)\; \backslash left(\; \backslash int\_\{B^R\}\; x\_n^\{-n\}\; dx\; \backslash right)^\{2/q\}\; =\; A\_2\; h^2(R),$$

where $A_2$ is also dependent on $R, n$ and $\alpha$ . Therefore,

$$(1+\frac{A_1}{A_2})J_{\alpha }^{{\mathbb{R}}_{+}^n}(f)\ge J_{\alpha }^{B^R}(f)+A_1{h}^2(R)\ge \frac{1}{2}{c}_{n,\alpha }{\left(\frac{1-R}{1+R}\right)}^{2\alpha -2}\mathrm{\parallel }{f}_1{\mathrm{\parallel }}_{2^{\ast }}^2,\backslash left(\; 1\; +\; \backslash frac\{A\_1\}\{A\_2\}\; \backslash right)\; J\_\backslash alpha^\{\backslash mathbb\{R\}\_+^n\}(f)\; \backslash geq\; J\_\backslash alpha^\{B^R\}(f)\; +\; A\_1\; h^2(R)\; \backslash geq\; \backslash frac\{1\}\{2\}\; c\_\{n,\backslash alpha\}\; \backslash left(\; \backslash frac\{1-R\}\{1+R\}\; \backslash right)^\{2\backslash alpha-2\}\; \backslash |f\_1\backslash |\_\{2^*\}^2,$$

which proves (7).

In establishing (8), we use the inequality $\frac{1}{2(n-1)} \leq \frac{(1-S^2)^{n-1}}{S^n} \int_0^S \frac{r^{n-1}}{(1-r^2)^n} dr \leq \frac{1}{n-1}$ , $0 < S < 1$ . Note that $h_0$ is constant on $[0, R]$ , while it is decreasing to zero on $[R, 1]$ . The following establishes how fast $h_0$ vanishes at 1. From Theorem 2.2,

$$\geq d_{n,\alpha} h_0(S)^2 \left( \frac{2^{n-1}}{n-1} |\mathbb{S}^{n-1}| \right)^{2/q} \left( \frac{S^n}{(1-S^2)^{n-1}} \right)^{2/q},$$

where $0 < S < 1$ . Thus,

$$h_0(r{)}^{2^{\ast }}\le d_{n,\alpha }^{-2^{\ast }\mathrm{/}2}{\left(\frac{2^{n-1}}{n-1}\mathrm{\mid }{\mathbb{S}}^{n-1}\mathrm{\mid }\right)}^{-2^{\ast }\mathrm{/}q}{\left(\frac{(1-r^2{)}^{n-1}}{r^n}\right)}^{2^{\ast }\mathrm{/}q}{J}_{\alpha }^{{\mathbb{R}}_{+}^n}(f{)}^{2^{\ast }\mathrm{/}2}\mathrm{.}h\_0(r)^\{2^*\}\; \backslash leq\; d\_\{n,\backslash alpha\}^\{-2^*/2\}\; \backslash left(\; \backslash frac\{2^\{n-1\}\}\{n-1\}\; |\backslash mathbb\{S\}^\{n-1\}|\; \backslash right)^\{-2^*/q\}\; \backslash left(\; \backslash frac\{(1-r^2)^\{n-1\}\}\{r^n\}\; \backslash right)^\{2^*/q\}\; J\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(f)^\{2^*/2\}.$$

Then, we calculate

As $2^* > q$ , the claim follows.

Finally, we show that we can approximate $x_n^{\frac{1-\alpha}{2}} f_1(x)$ by functions in $C_c^\infty(B^R)$ . Define

$$g_c(x)=\max \{x_n^{\frac{1-\alpha }{2}}{f}_1(x)-c,0\}g\_c(x)\; =\; \backslash max\backslash \{x\_n^\{\backslash frac\{1-\backslash alpha\}\{2\}\}\; f\_1(x)\; -\; c,\; 0\backslash \}$$

almost everywhere. Then, by monotone convergence,

$$I_{\alpha }^{B^R}(g_c)\to I_{\alpha }^{B^R}(x_n^{\frac{1-\alpha }{2}}{f}_1(x))I\_\{\backslash alpha\}^\{B^R\}(g\_c)\; \backslash rightarrow\; I\_\{\backslash alpha\}^\{B^R\}\backslash left(x\_n^\{\backslash frac\{1-\backslash alpha\}\{2\}\}\; f\_1(x)\backslash right)$$

and

$$\mathrm{\parallel }{g}_c{\mathrm{\parallel }}_{2^{\ast }}^2\to {\parallel x_n^{\frac{1-\alpha }{2}}{f}_1(x)\parallel }_{2^{\ast }}^2\backslash |g\_c\backslash |\_\{2^*\}^2\; \backslash rightarrow\; \backslash left\backslash |\; x\_n^\{\backslash frac\{1-\backslash alpha\}\{2\}\}\; f\_1(x)\; \backslash right\backslash |\_\{2^*\}^2$$

as $c \rightarrow 0$ . Now, $\text{supp } g_c \subsetneq B^R$ and $g_c \in L^2(B^R)$ , so $I_{\alpha}^{\mathbb{R}^n}(g_c) < \infty$ from (6). Denote

$$\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{W^{\alpha \mathrm{/}2,2}({\mathbb{R}}^n)}=\sqrt{\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2^2+I_{\alpha }^{{\mathbb{R}}^n}(\cdot )},\backslash |\backslash cdot\backslash |\_\{W^\{\backslash alpha/2,2\}(\backslash mathbb\{R\}^n)\}\; =\; \backslash sqrt\{\backslash |\backslash cdot\backslash |\_2^2\; +\; I\_\{\backslash alpha\}^\{\backslash mathbb\{R\}^n\}(\backslash cdot)\},$$

and let $W_0^{\alpha/2,2}(\mathbb{R}^n)$ be the completion of $C_c^\infty(\mathbb{R}^n)$ with respect to $|\cdot|_{W^{\alpha/2,2}(\mathbb{R}^n)}$ . Then it is known that

see, e.g. [1], [2]. Since $\text{supp } g_c \subsetneq B^R$ , there exists a sequence ${g_c^j} \subset C_c^\infty(B^R)$ so that $|g_c - g_c^j|{W^{\alpha/2,2}(\mathbb{R}^n)} \rightarrow 0$ as $j \rightarrow \infty$ . Hence, $I{\alpha}^{B^R}(g_c^j) \rightarrow I_{\alpha}^{B^R}(g_c)$ and $|g_c^j|_2^2 \rightarrow |g_c|_2^2$ as $j \rightarrow \infty$ . $\square$

5 Conclusion

Consider the general Hardy-Sobolev-Maz'ya inequality

$${\int }_{{\mathbb{R}}_{+}^n\times {\mathbb{R}}_{+}^n}\frac{\mathrm{\mid }f(x)-f(y){\mathrm{\mid }}^p}{\mathrm{\mid }x-y{\mathrm{\mid }}^{n+\alpha }}dxdy-D_{n,p,\alpha }{\int }_{{\mathbb{R}}_{+}^n}\frac{\mathrm{\mid }f(x){\mathrm{\mid }}^p}{x_n^{\alpha }}dx\ge a_{n,p,\alpha }{({\int }_{{\mathbb{R}}_{+}^n}\mathrm{\mid }f(x){\mathrm{\mid }}^{p^{\ast }}dx)}^{p\mathrm{/}{p}^{\ast }},\backslash int\_\{\backslash mathbb\{R\}\_+^n\; \backslash times\; \backslash mathbb\{R\}\_+^n\}\; \backslash frac\{|f(x)\; -\; f(y)|^p\}\{|x\; -\; y|^\{n+\backslash alpha\}\}\; dx\; dy\; -\; D\_\{n,p,\backslash alpha\}\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; \backslash frac\{|f(x)|^p\}\{x\_n^\backslash alpha\}\; dx\; \backslash geq\; a\_\{n,p,\backslash alpha\}\; \backslash left(\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; |f(x)|^\{p^*\}\; dx\; \backslash right)^\{p/p^*\},$$where $p \geq 2$ , $1 < \alpha < \min{n, p}$ . It is still unknown whether $a_{n,p,\alpha} > 0$ for $p > 2$ . Still, other than Theorem 3.1, the elements of this paper have either already been proven for general $p$ or extend quite easily. Indeed, Theorem 3.1 cannot be extended because Lemma 6.1 is not true for $p \neq 2$ . However, for functions that are symmetrically decreasing about a point in the upper halfspace or that can be rearranged about a point in the upper halfspace, while still maintaining support there, and where the Hardy term increases due to the rearrangement, then a fixed $a_{n,p,\alpha} > 0$ can be found.

Acknowledgement. I am very thankful to Michael Loss for countless valuable discussions and especially for collaboration on Theorem 2.2. This work was partially supported by NSF Grant DMS 0901304.

6 Appendix

Proof of Theorem 2.2: This proof uses the idea from Theorem 4.3, [11] to write the integral in terms of its layer cake representation. We can assume that $f \geq 0$ , since

$$J_{\alpha }^{{\mathbb{R}}_{+}^n}(f)=I_{\alpha }^{{\mathbb{R}}_{+}^n}(f)-2{\kappa }_{n,\alpha }{\int }_{{\mathbb{R}}_{+}^n}\frac{\mathrm{\mid }f(x){\mathrm{\mid }}^2}{x_n^{\alpha }}dx\ge I_{\alpha }^{{\mathbb{R}}_{+}^n}(\mathrm{\mid }f\mathrm{\mid })-2{\kappa }_{n,\alpha }{\int }_{{\mathbb{R}}_{+}^n}\frac{\mathrm{\mid }f(x){\mathrm{\mid }}^2}{x_n^{\alpha }}dx=J_{\alpha }^{{\mathbb{R}}_{+}^n}(\mathrm{\mid }f\mathrm{\mid })\mathrm{.}J\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(f)\; =\; I\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(f)\; -\; 2\backslash kappa\_\{n,\backslash alpha\}\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; \backslash frac\{|f(x)|^2\}\{x\_n^\backslash alpha\}\; dx\; \backslash geq\; I\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(|f|)\; -\; 2\backslash kappa\_\{n,\backslash alpha\}\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; \backslash frac\{|f(x)|^2\}\{x\_n^\backslash alpha\}\; dx\; =\; J\_\{\backslash alpha\}^\{\backslash mathbb\{R\}\_+^n\}(|f|).$$

We need a few preliminary results. Let $1_\Omega$ be the indicator function on the set $\Omega$ , then, for any $s \in \mathbb{R}$ ,

These next results follow from the Appendix in [6]. Let $t \geq 0$ , so $t^p = p(p-1) \int_0^\infty (t-a)_+ a^{p-2} da$ . Then, letting $a \geq 0$ ,

where $1_A$ is the indicator function on $A$ . Let $g(x) = x_n^{\frac{1-\alpha}{p}} f(x)$ . Then, where $d_{n,p,\alpha}$ is a generic constant, and using the results above,

We write

$$\lambda (a)={\int }_{{\mathbb{R}}_{+}^n}{1}_{\{g(x)>a\}}\frac{dx}{x_n^{(1-\alpha )\mathrm{/}2}},\backslash lambda(a)\; =\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; 1\_\{\backslash \{g(x)\; >\; a\backslash \}\}\; \backslash frac\{dx\}\{x\_n^\{(1-\backslash alpha)/2\}\},$$

and

so $u(a, c) \geq u(b, c)$ , $\lambda(a) \geq \lambda(b)$ if $b \geq a$ . Further, since $\alpha > 1$ , we obtain

where $D = \int_{\mathbb{R}+^n} 1{{|y| < 1}} \frac{dy}{y_n^{(1-\alpha)/2}}$ . Using Fubini,

From [11], Theorem 1.13, $g^q(x) = \int_0^\infty q a^{q-1} 1_{{g(x) > a}} da$ . Thus, we denote

$$\mathrm{\parallel }g{\mathrm{\parallel }}_{q(\mu )}^q={\int }_{{\mathbb{R}}_{+}^n}\mathrm{\mid }g(x){\mathrm{\mid }}^q\frac{dx}{x_n^{(1-\alpha )\mathrm{/}2}}={\int }_0^{\mathrm{\infty }}q{a}^{q-1}\lambda (a)da\ge \lambda (b)b^q\mathrm{.}\backslash |g\backslash |\_\{q(\backslash mu)\}^q\; =\; \backslash int\_\{\backslash mathbb\{R\}\_+^n\}\; |g(x)|^q\; \backslash frac\{dx\}\{x\_n^\{(1-\backslash alpha)/2\}\}\; =\; \backslash int\_0^\backslash infty\; q\; a^\{q-1\}\; \backslash lambda(a)\; da\; \backslash geq\; \backslash lambda(b)\; b^q.$$

Using the substitution $c = \left(\frac{\lambda(b)}{D}\right)^{\frac{2}{2n+\alpha-1}} t$ , and the identity $1 - \frac{p}{q} = \frac{\alpha+1}{2n+\alpha-1}$ , then

$$= d_{n,p,\alpha} \left( \int_{\mathbb{R}_+^n} |f(x)|^q x_n^{\frac{\alpha-1}{2}-q\frac{\alpha-1}{p}} dx \right)^{p/q}.$$

Since $\frac{\alpha-1}{2} - q\frac{\alpha-1}{p} = -n + \frac{nq}{p^*}$ , we are done. $\square$

LEMMA 6.1. Let $n \geq 2$ , $0 < \alpha < 2$ , and $f \in C_c(\mathbb{R}^n)$ . Then, $I_\alpha^{\mathbb{R}^n}(f) = I_\alpha^{\mathbb{R}^n}(\tilde{f})$ , with the understanding that $I_\alpha^{\mathbb{R}^n}(\tilde{f}) = \infty$ if $I_\alpha^{\mathbb{R}^n}(f) = \infty$ .

Proof. First, let us define the set

Using the transformation $T$ , as discussed in the remarks prior to Theorem 3.1, and results from [6],

We show that the limit on the last line is zero for all $\epsilon > 0$ ; thus, establishing our result. We write $x = (x', x_n)$ , $x' \in \mathbb{R}^{n-1}$ , $x_n \in \mathbb{R}$ . Then,

where the complement in the last integral is with respect to the half line $[0, \infty)$ . This product of integrals is zero as the left integral is finite, while the right integral is zero, for all $\epsilon > 0$ . Indeed, for the right integral, note that there is no singularity so long as $t \neq 1$ . If we split the integral into $t$ above and below 1, then the latter must be finite, so the integral is finite if the sum is. We compute$$\begin{aligned} \int_{[1-\epsilon, \frac{1}{1-\epsilon}]^c} dt (t^{\alpha-1} - t^{n-1}) \int_{-1}^1 ds \frac{(1-s^2)^{(n-3)/2}}{(t^2+1-2st)^{(n+\alpha)/2}} &= \int_0^{1-\epsilon} dt (t^{\alpha-1} - t^{n-1}) \int_{-1}^1 ds \frac{(1-s^2)^{(n-3)/2}}{(t^2+1-2st)^{(n+\alpha)/2}} \ &\quad + \int_{\frac{1}{1-\epsilon}}^{\infty} dt (t^{\alpha-1} - t^{n-1}) \int_{-1}^1 ds \frac{(1-s^2)^{(n-3)/2}}{(t^2+1-2st)^{(n+\alpha)/2}}. \end{aligned}$$

In fact, the sum is zero, as

Lastly, we consider the left integral. From [8], there exists $c > 0$ so that

$${\int }_{{\mathbb{R}}^n}{f}^2(x)\mathrm{\mid }x{\mathrm{\mid }}^{-\alpha }dx\le c{I}_{\alpha }^{{\mathbb{R}}^n}(f)\mathrm{.}\backslash int\_\{\backslash mathbb\{R\}^n\}\; f^2(x)\; |x|^\{-\backslash alpha\}\; dx\; \backslash leq\; c\; I\_\{\backslash alpha\}^\{\backslash mathbb\{R\}^n\}(f).$$

Hence, if we assume that $I_{\alpha}^{\mathbb{R}^n}(f) < \infty$ , then

as desired. If $I_{\alpha}^{\mathbb{R}^n}(f)$ is not finite, then we need to ask whether $\int_{\mathbb{R}^n} \tilde{f}^2(x', x_n - 1) |x|^{-\alpha} dx < \infty$ . If so, then the result still holds. If this is not true, then since $\int_{\mathbb{R}^n} \tilde{f}^2(x', x_n - 1) |x|^{-\alpha} dx \leq c I_{\alpha}^{\mathbb{R}^n}(\tilde{f})$ , then $I_{\alpha}^{\mathbb{R}^n}(\tilde{f})$ is not finite as well. $\square$

References

• [1] R. Adams, J. Fournier, Sobolev Spaces. Second edition. Acad. Press., New York, 2003.
• [2] K. Bogdan, K. Burdzy, and Z.Q. Chen, Censored Stable Processes, Probab. Theory Related Fields 127, 89-152 (2003).
• [3] K. Bogdan, B. Dyda, The Best Constant in a Fractional Hardy Inequality, arXiv:0807.1825v1 [math.AP] (2008).
• [4] R. Benguria, R. Frank, M. Loss, The Sharp Constant in the Hardy-Sobolev-Maz'ya Inequality in the Three Dimensional Upper Half-space, Math. Res. Lett. 15, no. 4, 613-622 (2008).
• [5] E. Carlen, M. Loss, Extremals of Functionals with Competing Symmetries, J. Funct. Anal. 88 no. 2, 437-456 (1990).
• [6] E. Carlen, M. Loss, On the Minimization of Symmetric Functionals. Special Issue Dedicated to Elliott H. Lieb. Rev. Math. Phys. 6, no. 5A, 1011-1032 (1994).
• [7] B. Dyda, A Fractional Order Hardy Inequality, Ill. J. Math. 48, no 2, 575-588 (2004).
• [8] R. Frank, R. Seiringer, Non-linear Ground State Representations and Sharp Hardy Inequalities. J. Funct. Anal. 255, 34073430.(2008).
• [9] R. Frank, R. Seiringer, Sharp Fractional Hardy Inequalities in Half-Spaces, arXiv:0906.1561v1 [math.FA] (2009).- [10] S. Filippas, A. Tertikas, J. Tidblom, On the Structure of Hardy-Sobolev-Maz'ya Inequalities (English summary), J. Eur. Math. Soc. 11, no. 6, 1165–1185 (2009).
• [11] E. Lieb, M. Loss. Analysis. Graduate Studies in Mathematics, vol. 14. Amer. Math. Society, 2^nd edition, 2001.
• [12] M. Loss, C. Sloane, Hardy Inequalities for Fractional Integrals on General Domains, arXiv:0907.3054v1 [math.AP] (2009).
• [13] V. Mazya. Sobolev Spaces. Translated from the Russian by T. O. Shaposhnikova. Springer Series in Soviet Mathematics. Springer-Verlag, Berlin, 1985.
• [14] A. Tertikas, K. Tintarev, On Existence of Minimizers for the Hardy-Sobolev-Maz'ya Inequality, Ann. Mat. Pura Appl. 186, 645662 (2007).