Minimizers for the fractional Sobolev inequality on domains

Abstract

We consider a version of the fractional Sobolev inequality in domains and study whether the best constant in this inequality is attained. For the half-space and a large class of bounded domains we show that a minimizer exists, which is in contrast to the classical Sobolev inequalities in domains.

Appendix A. Some facts about the Sobolev spaces \(\mathring{H}^\sigma (\mathbb {R}^n)\)

We begin with an improvement of the fractional Sobolev inequality due to Gérard, Meyer and Oru [27]. More general ones can be found in [28]. For the sake of completeness we include a simple proof following the lines of [2, Theorem 1.43]. We use the notation

$$\begin{aligned} \left( e^{t\Delta }u \right) (x):=\int _{\mathbb {R}^n} \frac{1}{(4\pi t)^{n/2}}e^{-|x-y|^2/4t}u(y)\,d y. \end{aligned}$$

Lemma A.1

Let $0<\sigma <1/2$ if $n=1$ and $0<\sigma <1$ if $n\ge 2$. Then there is a constant $C_{n,\sigma }$ such that for all $u\in \mathring{H}^\sigma (\mathbb {R}^n)$,

$$\begin{aligned} \Vert u\Vert _{L^\frac{2n}{n-2\sigma }(\mathbb {R}^n)}\le C_{n,\sigma } \left( \int _{\mathbb {R}^n}\int _{\mathbb {R}^n} \frac{(u(x)-u(y))^2}{|x-y|^{n+2\sigma }}\,d x\,d y\right) ^{\frac{n-2\sigma }{2n}} \left( \sup _{t>0}t^{\frac{n-2\sigma }{4}} \Vert e^{t\Delta }u\Vert _{L^\infty (\mathbb {R}^n)}\right) ^{\frac{2\sigma }{n}} \,. \end{aligned}$$

Proof

We abbreviate $q=\frac{2n}{n-2\sigma }$ and $\alpha =-\frac{n-2\sigma }{4}$ and assume, without loss of generality, that

$$\begin{aligned} \sup _{t>0}t^{-\alpha } \Vert e^{t\Delta }u\Vert _{L^\infty (\mathbb {R}^n)}=1. \end{aligned}$$

Note that

$$\begin{aligned} \int _{\mathbb {R}^n}|u|^q=q \int _{0}^\infty |\{|u|>\lambda \}|\lambda ^{q-1}\,d\lambda . \end{aligned}$$

We bound

$$\begin{aligned} |\{|u|>\lambda \}|\le |\{|e^{t\Delta }u|>\lambda /2\}|+ |\{|e^{t\Delta }u-u|>\lambda /2\}| \end{aligned}$$

and choose $t=(\lambda /2)^{1/\alpha }$, so that the first term on the right hand side is zero. Thus, by Plancherel’s theorem

$$\begin{aligned} |\{|u|>\lambda \}|&\le |\{|e^{(\lambda /2)^{1/\alpha }\Delta }u-u|>\lambda /2\}|\\&\le (2/\lambda )^2\int _{\mathbb {R}^n }|e^{(\lambda /2)^{1/\alpha }\Delta }u-u|^2\,\,d x\\&= (2/\lambda )^2\int _{\mathbb {R}^n }(e^{-(\lambda /2)^{1/\alpha }|\xi |^2}-1)^2|\hat{u}(\xi )|^2\,\,d \xi \,. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{\mathbb {R}^n}|u|^q&\le q \int _{\mathbb {R}^n}|\hat{u}(\xi )|^2\,\,d \xi \int _{0}^\infty (2/\lambda )^2(e^{-(\lambda /2)^{1/\alpha }|\xi |^2}-1)^2\lambda ^{q-1}\,d\lambda \\&= C\int _{\mathbb {R}^n}|\hat{u}(\xi )|^2|\xi |^{2\sigma }\,\,d \xi \end{aligned}$$

with

$$\begin{aligned} C = q \int _{0}^\infty (2/\lambda )^2(e^{-(\lambda /2)^{1/\alpha }}-1)^2\lambda ^{q-1}\,\,d\lambda = \alpha q 2^q \int _0^\infty (e^{-\mu } -1)^2 \mu ^{\alpha (q-2)-1}\,d\mu <\infty \,. \end{aligned}$$

Another application of Plancherel’s theorem concludes the proof of the inequality. $\square$

The following lemma shows that on domains of finite measure with sufficiently regular boundary there is no Sobolev inequality for $\sigma <1/2$. The proof uses ideas from [15].

Lemma A.2

Let $n\ge 1$, $0<\sigma <1/2$ and let $\Omega \subset \mathbb {R}^n$ be an open set of finite measure such that

$$\begin{aligned} \left| \left\{ x\in \Omega :\ {{\mathrm{dist}}}(x,\Omega ^c)<\delta \right\} \right| = o(\delta ^{2\sigma }) \qquad \text {as}\ \delta \rightarrow 0 \,. \end{aligned}$$

Then

$$\begin{aligned} \inf _{0\not \equiv u \in C^1_c(\Omega )} \frac{I_{n,\sigma ,\Omega }[u]}{\left( \int _{\Omega } |u|^\frac{2n}{n-2\sigma }\,\mathrm {d}x \right) ^\frac{n-2\sigma }{n}} = 0 \,. \end{aligned}$$

Note that, if $\Omega$ is bounded Lipschitz, then $\left| \left\{ x\in \Omega :\ {{\mathrm{dist}}}(x,\Omega ^c)<\delta \right\} \right| \lesssim \delta$ for $\delta$ sufficiently small and therefore $S_{n,\sigma }(\Omega )=0$ for $\sigma <1/2$.

Proof

Let $u_\delta \in C^1_c(\Omega )$ such that $u_\delta (x)=1$ if ${{\mathrm{dist}}}(x,\Omega ^c)\ge \delta$, $0\le u_\delta \le 1$ and $|\nabla u_\delta |\lesssim \delta ^{-1}$. Then

$$\begin{aligned} I_{n,\sigma ,\Omega }[u_\delta ]&\le 2 \int _{\{x\in \Omega :\ {{\mathrm{dist}}}(x,\Omega ^c)<\delta \}} \int _\Omega \frac{(u_\delta (x)-u_\delta (y))^2}{|x-y|^{n+2\sigma }}\,\mathrm {d}y\,\mathrm {d}x\\&= 2\int _{\{x\in \Omega :\ {{\mathrm{dist}}}(x,\Omega ^c)<\delta \}} \left( I(x)+ II(x) \right) \,\mathrm {d}x \end{aligned}$$

where

$$\begin{aligned} I(x)&:= \int _{\{y\in \Omega :\ |x-y|<\delta \}} \frac{(u_\delta (x)-u_\delta (y))^2}{|x-y|^{n+2\sigma }}\,\mathrm {d}y \\&\lesssim \frac{1}{\delta ^2} \int _{\{y\in \Omega :\ |x-y|<\delta \}} \frac{\mathrm {d}y}{|x-y|^{n+2\sigma -2 }} \\&\lesssim \frac{1}{\delta ^{2\sigma }} \end{aligned}$$

and

$$\begin{aligned} II(x)&:= \int _{\{y\in \Omega :\ |x-y|\ge \delta \}} \frac{(u_\delta (x)-u_\delta (y))^2}{|x-y|^{n+2\sigma }}\,\mathrm {d}y \\&\le \int _{\{y\in \Omega :\ |x-y|<\delta \}} \frac{\mathrm {d}y}{|x-y|^{n+2\sigma }} \\&\lesssim \frac{1}{\delta ^{2\sigma }} \,. \end{aligned}$$

Thus,

$$\begin{aligned} I_{n,\sigma ,\Omega }[u_\delta ] \lesssim \delta ^{-2\sigma } \left| \left\{ x\in \Omega :\ {{\mathrm{dist}}}(x,\Omega ^c)<\delta \right\} \right| \rightarrow 0 \qquad \text {as}\ \delta \rightarrow 0 \,. \end{aligned}$$

Since $\int _\Omega u_\delta ^\frac{2n}{n-2\sigma }\,\mathrm {d}x \rightarrow |\Omega |$, we obtain the lemma. $\square$