1 Introduction

1.1 Background

For an open set \(\Omega \subset \mathbb{R}^N\), we indicate by \(W^{1,2}_0(\Omega )\) the closure of \(C^\infty _0(\Omega )\) in the Sobolev space \(W^{1,2}(\Omega )\). We then consider the following quantity

$$\begin{aligned} \lambda _1(\Omega ):=\inf _{u\in W^{1,2}_0(\Omega )\setminus \{0\}}\frac{\int _\Omega |\nabla u|^2\,\mathrm{d}x}{\int _\Omega |u|^2\,\mathrm{d}x}, \end{aligned}$$

which coincides with the bottom of the spectrum of the Dirichlet–Laplacian on \(\Omega\). Observe that for a general open set, such a spectrum may not be discrete and the infimum value \(\lambda _1(\Omega )\) may not be attained. Whenever a minimizer \(u_1\in W^{1,2}_0(\Omega )\) of the problem above exists, we call \(\lambda _1(\Omega )\) the first eigenvalue of the Dirichlet–Laplacian on \(\Omega\).

By definition, such a quantity is different from zero if and only if \(\Omega\) supports the Poincaré inequality

$$\begin{aligned} c\,\int _\Omega |u|^2\,\mathrm{d}x\le \int _\Omega |\nabla u|^2\,\mathrm{d}x,\quad \text{for every}\quad u\in C^\infty _0(\Omega ). \end{aligned}$$

It is well-known that this happens for example if \(\Omega\) is bounded or with finite measure or even bounded in one direction only. However, in general it is quite complicated to give more general geometric conditions, assuring positivity of \(\lambda _1\). In this paper, we will deal with the two-dimensional case \(N=2\).

In this case, there is a by now classical result which asserts that

$$\begin{aligned} \lambda _1(\Omega )\ge \frac{C}{r_\Omega ^2}, \end{aligned}$$
(1.1)

for every simply connected set \(\Omega \subset \mathbb{R}^2\). Here \(C>0\) is a universal constant and the geometric quantity \(r_\Omega\) is the inradius of \(\Omega\), i.e., the radius of a largest disk contained in \(\Omega\). More precisely, this is given by

$$\begin{aligned} r_\Omega :=\sup \left\{ \rho >0\,:\,\exists \, x\in \Omega \,\text{ such that }\,B_\rho (x)\subset \Omega \right\} . \end{aligned}$$

Inequality (1.1) is in scale invariant form, by recalling that \(\lambda _1\) scales like a length to the power \(-2\), under dilations.

Such a result is originally due to Makai (see [25, equation (5)]). It permits in particular to prove that for a simply connected set in the plane, we have the following remarkable equivalence

$$\begin{aligned} \lambda _1(\Omega )>0 \Longleftrightarrow r_\Omega <+\infty . \end{aligned}$$
(1.2)

Indeed, if the inradius is finite, we immediately get from (1.1) that \(\lambda _1(\Omega )\) must be positive. The converse implication is simpler and just based on the easy (though sharp) inequality

$$\begin{aligned} \lambda _1(\Omega )\le \frac{\lambda _1(B_1)}{r_\Omega ^2}. \end{aligned}$$

Here \(B_1\subset \mathbb{R}^2\) is any disk with radius 1 and the estimate simply follows from the monotonicity with respect to set inclusion of \(\lambda _1\), together with its scaling properties.

The proof in [25] runs very similarly to that of the Faber–Krahn inequality, based on symmetrization techniques (see [20, Chapter 3]). It starts by rewriting the Dirichlet integral and the \(L^2\) norm by using the Coarea Formula; then the key ingredient is a clever use of a particular quantitative isoperimetric inequality in \(\mathbb{R}^2\) (a Bonnesen-type inequality), which permits to obtain a lower bound in terms of \(r_\Omega\) only.

It should be noticed that Makai’s result has been overlooked or neglected for some years and then rediscovered independently by Hayman, by means of a completely different proof, see [19, Theorem 1]. For this reason, we will call (1.1) the Makai–Hayman inequality.

It is interesting to remark that the result by Makai is quantitatively better than the one by Hayman: indeed, the former obtains (1.1) with \(C=1/4\), while the latter is only able to get the poorer constant \(C=1/900\) by his method of proof.

This could suggest that the attribution of this result to both authors is maybe too generous. On the contrary, we will show in this paper that, in despite providing a poorer constant, the method of proof by Hayman is elementary, flexible and robust enough to be generalized to other situations, where Makai’s and other approaches become too complicate or do not seem feasible.

In any case, we point out that the exact determination of the sharp constant in (1.1), i.e.,

$$\begin{aligned} C_{MH}:=\inf \left\{ \lambda _1(\Omega )\,r_\Omega ^2\, :\, \Omega \subset \mathbb{R}^2 \text{ simply connected with } r_\Omega <+\infty \right\} , \end{aligned}$$

is still a challenging open problem. The best result at present is that

$$\begin{aligned} 0.6197<C_{MH}<2.13, \end{aligned}$$

obtained by Bañuelos and Carroll (see [2, Corollary 1] for the lower bound and [2, Theorem 2] for the upper bound). The upper bound has then been slightly improved by Brown in [14], by using a refinement of the method by Bañuelos and Carroll.

Inequality (1.1) has also been obtained by Ancona in [1], by using yet another proof. His result comes with the constant \(C=1/16\), much better than Hayman’s one, but still worse than that obtained by Makai. The proof by Ancona is quite elegant: it is based on the use of conformal mappings and the so-called Koebe’s one quarter Theorem (see [22, Chapter 12]), which permits to obtain the following Hardy inequality for a simply connected set in the plane

$$\begin{aligned} \frac{1}{16}\,\int _\Omega \frac{|\varphi |^2}{\mathrm {dist}(x,\partial \Omega )^2}\,\mathrm{d}x\le \int _\Omega |\nabla \varphi |^2\,\mathrm{d}x,\quad \text{for every }\quad \varphi \in C^\infty _0(\Omega ). \end{aligned}$$

From this, inequality (1.1) is easily obtained (with \(C=1/16\)), by observing that

$$\begin{aligned} r_\Omega =\sup _{x\in \Omega } \mathrm {dist}(x,\partial \Omega ), \end{aligned}$$

and then using the definition of \(\lambda _1(\Omega )\). The conformality of the Dirichlet integral plays a central role in this proof. The result by Ancona is quite remarkable, as the Hardy inequality is proved without any regularity assumption on \(\partial \Omega\). A generalization of this result can be found in [23].

1.2 Goal of the paper and main results

Our work is aimed at investigating the validity of a result analogous to (1.1) for fractional Sobolev spaces. In order to be more precise, we need some definitions at first. Let \(0<s<1\) and let us recall the definition of Gagliardo–Slobodeckiĭ seminorm

$$\begin{aligned} {[}u]_{W^{s,2}(\mathbb{R}^N)}= \left( \iint _{\mathbb{R}^N\times \mathbb{R}^N} \frac{|u(x)-u(y)|^2}{|x-y|^{N+2\,s}}\,\mathrm{d}x\,\mathrm{d}y\right) ^\frac{1}{2}. \end{aligned}$$

Accordingly, we consider the fractional Sobolev space

$$\begin{aligned} W^{s,2}(\mathbb{R}^N)=\left\{ u\in L^2(\mathbb{R}^N)\, :\, [u]_{W^{s,2}(\mathbb{R}^N)}<+\infty \right\} , \end{aligned}$$

endowed with the norm

$$\begin{aligned} \Vert u\Vert _{W^{s,2}(\mathbb{R}^N)}=\Vert u\Vert _{L^2(\mathbb{R}^N)}+[u]_{W^{s,2}(\mathbb{R}^N)}. \end{aligned}$$

Finally, we consider the space \({\widetilde{W}}^{s,2}_0(\Omega )\), defined as the closure of \(C^\infty _0(\Omega )\) in \(W^{s,2}(\mathbb{R}^N)\). Observe that by definition the elements of \({\widetilde{W}}^{s,2}_0(\Omega )\) have to be considered on the whole \(\mathbb{R}^N\) and they come with a natural nonlocal homogeneous Dirichlet condition “at infinity,” i.e., they identically vanish on the complement \(\mathbb{R}^N\setminus \Omega\).

We then consider the quantity

$$\begin{aligned} \lambda ^s_1(\Omega ):=\inf _{u\in \widetilde{W}^{s,2}_0(\Omega )\setminus \{0\}}\dfrac{[u]^2_{W^{s,2}(\mathbb{R}^N)}}{\Vert u\Vert ^2_{L^2(\Omega )}}. \end{aligned}$$
(1.3)

Again by definition, this quantity is nonzero if and only if the open set \(\Omega\) supports the fractional Poincaré inequality

$$\begin{aligned} c\,\int _\Omega |u|^2\,\mathrm{d}x\le \iint _{\mathbb{R}^N\times \mathbb{R}^N} \frac{|u(x)-u(y)|^2}{|x-y|^{N+2\,s}}\,\mathrm{d}x\,\mathrm{d}y,\quad \text{ for every }\quad u\in C^\infty _0(\Omega ). \end{aligned}$$

This happens, for example, if \(\Omega\) is an open bounded set (see [9, Lemma 2.4]). As in the local case, whenever the infimum in (1.3) is attained, this quantity will be called first eigenvalue of the fractional Dirichlet–Laplacian of order s. We recall that the latter is the linear operator denoted by the symbol \((-\Delta )^s\) and defined in weak form by

$$\begin{aligned} \langle (-\Delta )^s u,\varphi \rangle =\iint _{\mathbb{R}^N\times \mathbb{R}^N} \frac{(u(x)-u(y))\,(\varphi (x)-\varphi (y))}{|x-y|^{N+2\,s}}\,\mathrm{d}x\,\mathrm{d}y,\quad \text{for every }\quad \varphi \in C^\infty _0(\Omega ). \end{aligned}$$

In this paper, we want to inquire to which extent the Makai–Hayman inequality (1.1) still holds for \(\lambda _1^s\) defined above, still in the case of simply connected sets in the plane. Our main results assert that such an inequality is possible, provided s is “large enough.” More precisely, we have the following

Theorem 1.1

(Fractional Makai–Hayman inequality) Let \(1/2<s<1\) and let \(\Omega \subset \mathbb{R}^2\) be an open simply connected set, with finite inradius \(r_\Omega\). There exists an explicit universal constant \(\mathcal{C}_s>0\) such that

$$\begin{aligned} \lambda _1^s(\Omega )\ge \frac{\mathcal{C}_s}{r_\Omega ^{2\,s}}. \end{aligned}$$
(1.4)

Moreover, \(\mathcal{C}_s\) has the following asymptotic behaviorsFootnote 1

$$\begin{aligned} \mathcal{C}_s\sim \left( s-\frac{1}{2}\right) ,\quad \text{for}\quad s\searrow \frac{1}{2}\quad \text{and}\quad \mathcal{C}_s\sim \frac{1}{1-s},\quad \text{for}\quad s\nearrow 1. \end{aligned}$$

Remark 1.2

We point out that the constant \(\mathcal{C}_s\) appearing in the above estimate exhibits the sharp asymptotic dependence on s, as \(s\nearrow 1\). Indeed, by recalling that for every open set \(\Omega \subset \mathbb{R}^N\) we have (see [8, Lemma A.1])

$$\begin{aligned} \limsup _{s\nearrow 1} (1-s)\,\lambda _1^s(\Omega )\le C_N\,\lambda _1(\Omega ), \end{aligned}$$
(1.5)

from Theorem 1.1 we can obtain the usual Makai–Hayman inequality for the Dirichlet–Laplacian, possibly with a worse constant. We recall that (1.5) is based on the fundamental asymptotic result by Bourgain, Brezis and Mironescu for the Gagliardo–Slobodeckiĭ seminorm, see [6]. We refer to [13] for some interesting refinements of such a result.

The previous result is complemented by the next one, asserting that for \(0<s\le 1/2\) a fractional Makai–Hayman inequality is not possible. In this way, we see that even for \(s\searrow 1/2\) the asymptotic behavior of \(\mathcal{C}_s\) is optimal, in a sense.

Theorem 1.3

(Counter-example for \(0<s\le 1/2\)) There exists a sequence \(\{\widetilde Q_n\}_{n\in \mathbb{N}}\subset \mathbb{R}^2\) of open bounded simply connected sets such that

$$\begin{aligned} 0<r_{\widetilde{Q}_n}\le C,\quad \text{for every}\quad n\in \mathbb{N}, \end{aligned}$$

and

$$\begin{aligned} \lim _{n\rightarrow \infty }\lambda _1^s(\widetilde{Q}_n)=0,\quad \text{ for every }\quad 0<s\le \frac{1}{2}. \end{aligned}$$

Remark 1.4

In [16, Theorem 1.1], a different counter-example for \(0<s<1/2\) is given. Apart from the fact that in [16] the borderline case \(s=1/2\) is not considered, one could observe that strictly speaking the counter-example in [16] is not a simply connected set, since it is made of countably many connected components.

As it will be apparent to the experienced reader, our example will clearly display the role of fractional \(s-\)capacity in the failure of the Makai–Hayman inequality for \(0<s\le 1/2\) (see for example [26, Chapter 10, Section 4] for fractional capacities). Indeed, the range \(0<s\le 1/2\) is precisely the one for which lines have zero fractional \(s-\) capacity. This implies that, by removing a finite number of segments from an open set, the first eigenvalue \(\lambda _1^s\) remains unchanged, while this operation heavily affects the inradius. However, even if this is the ultimate reason for such a failure, our proof will be elementary and will not explicitly appeal to the properties of capacities.

We point out that, for practical reasons, our sequence \(\{\widetilde Q_n\}_{n\in \mathbb{N}}\) is given by a square with side length \(2\,n\), from which a periodical array of segments is removed. If we scale this sequence by a factor 1/n, we could produce another sequence contradicting the fractional Makai–Hayman, with the additional property of being equi-bounded.

Remark 1.5

Geometric estimates for eigenvalues of \((-\Delta )^s\) aroused great interest in the last years, also in the field of stochastic processes. Indeed, it is well-known that this operator is the infinitesimal generator of a symmetric \((2\,s)\)-stable Lévy process. We recall that the nonlocal homogeneous Dirichlet boundary condition considered above (i.e., \(u\equiv 0\) on \(\mathbb{R}^N\setminus \Omega\)) corresponds to a process where particles are “killed” upon reaching the complement of the set \(\Omega\). The Gagliardo–Slobodeckiĭ seminorm corresponds to the so-called Dirichlet form associated with this process. For more details, we refer for example to [5, Section 2] and the references therein.

In this context, we wish to mention the papers [3, 4, 28], where some geometric estimates for \(\lambda _1^s\) are obtained, by exploiting this probabilistic approach. In particular, the paper [4] is very much related to ours, since in [4, Corollary 1] it is proved the lower bound

$$\begin{aligned} \lambda _1^s(\Omega )\ge \frac{C}{r_\Omega ^{2\,s}}, \end{aligned}$$

in the restricted class of open convex subsets of the plane, with the sharp constant C. This result can be seen as the fractional counterpart of a well-known result for the Laplacian, which goes under the name of Hersch–Protter inequality, see [21, 30].

1.3 Method of proof

As already announced at the beginning, we will achieve the result of Theorem 1.1 by adapting to our setting Hayman’s proof. It is then useful to recall the key ingredients of such a proof. These are essentially two:

  1. 1.

    A covering lemma, asserting that it is possible to cover an open subset \(\Omega \subset \mathbb{R}^2\) with \(r_\Omega <+\infty\) by means of boundary disks, whose radius is universally comparable to \(r_\Omega\) and which do not overlap “too much” with each other. Here by boundary disk we simply mean a disk centered at the boundary \(\partial \Omega\);

  2. 2.

    A Poincaré inequality for boundary disks in a simply connected set.

Point 1. is purely geometrical and thus it can still be used in the fractional setting.

On the contrary, the proof of point 2. is very much local. Indeed, an essential feature of the proof in [19] is the fact that

$$\begin{aligned} |\nabla u|^2\ge \frac{1}{\varrho ^2}\,|\partial _\theta u|^2, \end{aligned}$$
(1.6)

where \((\varrho ,\theta )\) denote the usual polar coordinates. Then one observes that a boundary circle always meets the complement of \(\Omega\), when the latter is simply connected. Thus, taken a function \(u\in C^\infty _0(\Omega )\), the periodic function \(\theta \mapsto u(\varrho ,\theta )\) vanishes somewhere in \([0,2\,\pi ]\). Consequently, it satisfies the following one-dimensional Poincaré inequality on the interval

$$\begin{aligned} \int _0^{2\,\pi } |u(\varrho ,\theta )|^2\,\mathrm{d}\theta \le C\,\int _0^{2\,\pi } |\partial _\theta u(\varrho ,\theta )|^2\,\mathrm{d}\theta . \end{aligned}$$
(1.7)

In a nutshell, this permits to prove point 2. by “foliating” the boundary disk with concentric boundary circles, using (1.7) on each of these circles, then integrating with respect to the radius of the circle and finally appealing to (1.6).

In the fractional case, property (1.6) has no counterpart, because of the nonlocality of the Gagliardo–Slobodeckiĭ seminorm. Consequently, adapting this method to prove a fractional Poincaré inequality for boundary disks is a bit involved. We will achieve this through a lengthy though elementary method, which we believe to be of independent interest.

Remark 1.6

(Other proofs?) We conclude the introduction, by observing that it does not seem easy to prove (1.4) by adapting Makai’s proof, because of the lack of a genuine Coarea Formula for Gagliardo–Slobodeckiĭ seminorms. The proof by Ancona seems to be even more prohibitive to be adapted, because of the rigid machinery of conformal mappings on which is based. In passing, we mention that it would be interesting to know whether his Hardy inequality for simply connected sets in the plane could be extended to fractional Sobolev spaces. For completeness, we refer to [17] for some fractional Hardy inequalities under minimal regularity assumptions.

1.4 Plan of the paper

In Sect. 2, we set the main notations and present some technical tools, needed throughout the paper. In particular, we recall Hayman’s covering lemma from [19] and present a couple of technical results on fractional Sobolev spaces.

In Sect. 3, we prove a Poincaré inequality for boundary disks. This is the main ingredient for the proof of the fractional Makai–Hayman inequality.

Section 4 is then devoted to the proof of Theorem 1.1, while the construction of the counter-example of Theorem 1.3 is contained in Sect. 5.

Finally, in Sect. 6 we highlight some consequences of our main result. Among these, we record a Cheeger-type inequality, a comparison result for \(\lambda _1^s\) and \(\lambda _1\) and the fractional analogue of characterization (1.2).

The paper concludes with “Appendix A,” containing a one-dimensional fractional Poincaré inequality for periodic functions vanishing at one point (see Proposition A.2). This is the cornerstone on which the result in Sect. 3 is built.

2 Preliminaries

2.1 Notations

Given \(x_0\in \mathbb{R}^N\) and \(R>0\), we will denote by \(B_R(x_0)\) the \(N-\)dimensional open ball with radius R and center \(x_0\). When the center coincides with the origin, we will simply write \(B_R\). We indicate by \(\omega _N\) the \(N-\)dimensional Lebesgue measure of \(B_1\), so that by scaling

$$\begin{aligned} |B_R(x_0)|=\omega _N\,R^N. \end{aligned}$$

If \(E\subset \mathbb{R}^N\) is a measurable set with positive measure and \(u\in L^1(E)\), we will use the notation

figure a

For \(0<s<1\) and for a measurable set \(E\subset \mathbb{R}^N\), we will indicate by

$$\begin{aligned} W^{s,2}(E)=\left\{ u\in L^2(E)\, :\, [u]_{W^{s,2}(E)}<+\infty \right\} , \end{aligned}$$

where

$$\begin{aligned} {[}u]_{W^{s,2}(E)}=\left( \iint _{E\times E}\frac{|u(x)-u(y)|^2}{|x-y|^{N+2\,s}}\,dx\,dy\right) ^\frac{1}{2}. \end{aligned}$$

This space will be endowed with the norm

$$\begin{aligned} \Vert u\Vert _{W^{s,2}(E)}=\Vert u\Vert _{L^2(E)}+[u]_{W^{s,2}(E)}. \end{aligned}$$

We observe that the following Leibniz-type rule holds

$$\begin{aligned} {[}u\,v]_{W^{s,2}(E)}\le [u]_{W^{s,2}(E)}\,\Vert v\Vert _{L^\infty (E)}+[v]_{W^{s,2}(E)}\,\Vert u\Vert _{L^\infty (E)},\quad \text{for every }\quad u,v\in W^{s,2}(E)\cap L^\infty (E). \end{aligned}$$
(2.1)

This will be useful somewhere in the paper.

Finally, by \(W^{s,2}_{\mathrm{loc}}(\mathbb{R}^N)\) we mean the collection of functions which are in \(W^{s,2}(B_R)\), for every \(R>0\).

2.2 Technical tools

In order to prove Theorem 1.1, we will need the following covering Lemma, whose proof can be found in [19, Lemma 2]. The result in [19] is stated for bounded sets and, accordingly, the relevant covering is made of a finite number of balls. However, a closer inspection of the proof in [19] easily shows that the same result still holds by removing the boundedness assumption. In this case, the covering could be made of countably infinitely many balls: this is still enough for our purposes. We omit the proof, since it is exactly the same as in [19].

Lemma 2.1

Let \(\Omega \subset \mathbb{R}^2\) be an open set, with finite inradius \(r_\Omega\). Then there exist at most countably many distinct points \(\{z_{n}\}_{n\in \mathbb{N}}\subset \partial \Omega\) such that the family of disks

$$\begin{aligned} \mathfrak {B}=\left\{ B_r(z_n)\right\} _{n\in \mathbb{N}},\quad \text{with} \quad r=r_\Omega \,\left( 1+\sqrt{2}\right) , \end{aligned}$$

is a covering of \(\Omega\). Moreover, \(\mathfrak {B}\) can be split in at most 36 subfamilies \(\mathfrak {B}_1,\ldots ,\mathfrak {B}_{36}\) such that

$$\begin{aligned} B_r(z_n)\cap B_r(z_m)=\emptyset \quad \text{if}\quad B_r(z_{n}),B_r(z_{m})\in \mathfrak {B}_k,\quad \text{with}\ m\not =n, \end{aligned}$$

for every \(k=1,\ldots ,36\).

In the following technical result, we explicitly construct a continuous extension operator for fractional Sobolev spaces defined on a ball. The result is certainly well-known (see for example [18, Theorem 5.4]), but here we pay particular attention to the constant appearing in the continuity estimate (2.2): indeed, this can be taken to be independent of the differentiability index s.

Lemma 2.2

Let \(0<s<1\), there exists a linear extension operator

$$\begin{aligned} \mathcal{E}:W^{s,2}(B_1(x_0))\rightarrow W^{s,2}_\mathrm{loc}(\mathbb{R}^N), \end{aligned}$$

such that for every \(u\in W^{s,2}(B_1(x_0))\) and every \(R>1\) we have

$$\begin{aligned} \Big[ \mathcal{E}[u]\Big] _{W^{s,2}(B_R(x_0))}\le 4\,R^{4\,N}\,[u]_{W^{s,2}(B_1(x_0))},\quad \Vert \mathcal{E}[u]\Vert _{L^2(B_R(x_0))}\le 2\,R^{2\,N}\,\Vert u\Vert _{L^2(B_1(x_0))}. \end{aligned}$$
(2.2)

Proof

Without loss of generality, we can suppose that \(x_0\) coincides with the origin. Then, let us recall the definition of inversion with respect to \(\mathbb{S}^{N-1}\): this is the bijection \(\mathcal{K}:\mathbb{R}^N\setminus \{0\}\rightarrow \mathbb{R}^N\setminus \{0\}\), given by

$$\begin{aligned} \mathcal{K}(x)=\frac{x}{|x|^2},\quad \text{for every }\quad x\in \mathbb{R}^N\setminus \{0\}. \end{aligned}$$

It is easily seen that if \(x\in B_R\setminus B_1\), then \(\mathcal{K}(x)\in B_1\setminus B_{1/R}\). Moreover, we have

$$\begin{aligned} \mathcal{K}^{-1}(x)=\mathcal{K}(x)\quad \text{and}\quad |\mathrm {det}(D\mathcal{K}(x))|=\frac{1}{|x|^{2\,N}},\quad \text{for every}\quad x\in \mathbb{R}^N\setminus \{0\}. \end{aligned}$$

For every \(u\in W^{s,2}(B_1)\), we define the extended function \(\mathcal{E}[u]\) given by

$$\begin{aligned} \mathcal{E}[u](x)={\left\{ \begin{array}{ll} u(x),&{}\text{if}\quad x\in B_1,\\ u(\mathcal{K}(x))&{}\text{if}\quad x\in \mathbb{R}^N\setminus B_1. \end{array}\right. } \end{aligned}$$
(2.3)

It is easily seen that the operator \(u\mapsto \mathcal{E}[u]\) is linear. In order to prove that \(\mathcal{E}[u]\in W^{s,2}_\mathrm{loc}(\mathbb{R}^N)\), together with the claimed estimate (2.2), we take \(R>1\) and we split the seminorm of \(\mathcal{E}[u]\) as follows

$$\begin{aligned} \begin{aligned} \Big [\mathcal{E}[u]\Big ]_{W^{s,2}(B_R)}^2&=[u]^2_{W^{s,2}(B_1)} \\&\quad +\iint _{(B_R\setminus B_1)\times (B_R\setminus B_1)}\frac{|{u}(\mathcal{K}(x))-{u}(\mathcal{K}(y))|^2}{|x-y|^{N+2\,s}}\,\mathrm{d}x\,\mathrm{d}y\\&\quad +2\,\iint _{B_1\times (B_R\setminus B_1)}\frac{|{u}(x)-{u}(\mathcal{K}(y))|^2}{|x-y|^{N+2\,s}}\,\mathrm{d}x\,\mathrm{d}y.\\ \end{aligned} \end{aligned}$$

By performing the change of variable \(z=\mathcal{K}(x)\) in the second term on the right-hand side and the change of variable \(w=\mathcal{K}(y)\) in the second and third terms, we get

$$\begin{aligned} \begin{aligned}&\Big [\mathcal{E}[u]\Big ]_{W^{s,2}(B_R)}^2\\&\quad =[u]^2_{W^{s,2}(B_1)} \\&\qquad +\iint _{(B_1\setminus B_\frac{1}{R})\times (B_1\setminus B_\frac{1}{R})}\frac{|u(z)-u(w)|^2}{|\mathcal{K}^{-1}(z)-\mathcal{K}^{-1}(w)|^{N+2\,s}}\,|\det D \mathcal{K}^{-1}(z)|\,|\det D \mathcal{K}^{-1}(w)|\,\mathrm{d}z\,\mathrm{d}w \\&\qquad +2\,\int _{B_1\times (B_1\setminus B_\frac{1}{R})}\frac{|u(x)-u(w)|^2}{|x-\mathcal{K}^{-1}(w)|^{N+2\,s}}\,|\det D \mathcal{K}^{-1}(w)|\,\mathrm{d}x\,\mathrm{d}w. \end{aligned} \end{aligned}$$

By using the expression for the Jacobian determinant, we then obtain

$$\begin{aligned} \begin{aligned} \Big [\mathcal{E}[u]\Big ]_{W^{s,2}(B_R)}^2&\le [u]^2_{W^{s,2}(B_1)} \\&\quad +R^{4\,N}\,\iint _{(B_1\setminus B_\frac{1}{R})\times (B_1\setminus B_\frac{1}{R})} \frac{|u(z)-u(w)|^2}{|\mathcal{K}^{-1}(z)-\mathcal{K}^{-1}(w)|^{N+2\,s}}\,\mathrm{d}z\,\mathrm{d}w\\&\quad +2\,R^{2\,N}\,\iint _{B_1\times (B_1\setminus B_\frac{1}{R})}\frac{|u(x)-u(w)|^2}{|x-\mathcal{K}^{-1}(w)|^{N+2\,s}}\,\mathrm{d}x\,\mathrm{d}w. \end{aligned} \end{aligned}$$
(2.4)

In order to estimate the last two integrals, it is sufficient to use that

$$\begin{aligned} |\mathcal{K}^{-1}(z)-\mathcal{K}^{-1}(w)|=\left| \frac{1}{|z|^2}\,z-\frac{1}{|w|^2}\,w\right| \ge |z-w|,\quad \text{for every}\quad z,w\in B_1\setminus \{0\}, \end{aligned}$$
(2.5)

and

$$\begin{aligned} |x-\mathcal{K}^{-1}(w)|=\left| x-\frac{1}{|w|^2}\,w\right| \ge |{x}-w|,\quad \text{for every}\quad x,w\in B_1\setminus \{0\}. \end{aligned}$$
(2.6)

Indeed, by taking the square, we see that (2.5) is equivalent to

$$\begin{aligned} \left( \frac{1}{|z|^2}-|z|^2\right) +\left( \frac{1}{|w|^2}-|w|^2\right) \ge 2\,\left( \frac{1}{|z|^2\,|w|^2}-1\right) \,\langle z,w\rangle . \end{aligned}$$

This in turn follows from Young’s inequality

$$\begin{aligned} 2\,\langle z,w\rangle \le |z|^2+|w|^2, \end{aligned}$$

once we multiply both sides by the positive quantity

$$\begin{aligned} \left( \frac{1}{|z|^2\,|w|^2}-1\right) . \end{aligned}$$

As for inequality (2.6), by taking again the square we see that the latter is equivalent to

$$\begin{aligned} \frac{1}{|w|^2}-|w|^2\ge 2\,\left( \frac{1}{|w|^2}-1\right) \,\langle x,w\rangle . \end{aligned}$$
(2.7)

This in turn follows again from Young’s inequality: more precisely, by using that \(|x|<1\), we have

$$\begin{aligned} 2\,\langle x,w\rangle \le |x|^2+|w|^2\le 1+|w|^2, \end{aligned}$$

and if we now multiply both sides by the positive quantity (here we use that \(|w|<1\))

$$\begin{aligned} \left( \frac{1}{|w|^2}-1\right) , \end{aligned}$$

we get (2.7), with some simple algebraic manipulations.

By applying estimates (2.5) and (2.6) in (2.4), we finally get

$$\begin{aligned} \begin{aligned} \Big [\mathcal{E}[u]\Big ]_{W^{s,2}(B_R)}^2&\le [u]^2_{W^{s,2}(B_1)} \\&\quad +R^{4\,N}\,\iint _{(B_1\setminus B_\frac{1}{R})\times (B_1\setminus B_\frac{1}{R})}\frac{|u(z)-u(w)|^2}{|z-w|^{N+2\,s}}\,\mathrm{d}z\,\mathrm{d}w\\&\quad +2\,R^{2\,N}\,\iint _{B_1\times (B_1\setminus B_\frac{1}{R})}\frac{|u(x)-u(w)|^2}{|x-w|^{N+2\,s}}\,\mathrm{d}x\,\mathrm{d}w\\&\le \left( 1+R^{4\,N}+2\,R^{2\,N}\right) \,[u]^2_{W^{s,2}(B_1)}, \end{aligned} \end{aligned}$$

which proves the first estimate in (2.2).

We are left with estimating the \(L^2\) norm of \(\mathcal{E}[u]\). This is simpler and can be done as follows

$$\begin{aligned} \begin{aligned} \int _{B_R}|\mathcal{E}[u](x)|^2\,\mathrm{d}x&=\int _{B_1}|u(x)|^2\,\mathrm{d}x+\int _{B_R\setminus B_1}|u(\mathcal{K}(x))|^2\,\mathrm{d}x\\&\le \int _{B_1}|u(x)|^2\,\mathrm{d}x+R^{2\,N}\,\int _{B_1\setminus B_\frac{1}{R}}|u(z)|^2\,\mathrm{d}z\le \left( 1+R^{2\,N}\right) \,\int _{B_1}|u(x)|^2\,\mathrm{d}x. \end{aligned} \end{aligned}$$

This concludes the proof. \(\square\)

Remark 2.3

Another important feature of the previous result is that, rather than the usual continuity estimate

$$\begin{aligned} \big \Vert \mathcal{E}[u]\big \Vert _{W^{s,2}(B_R)}\le C\,\Vert u\Vert _{W^{s,2}(B_1)}, \end{aligned}$$

for the extension operator, we obtained the more precise estimate (2.2). This will be useful in the next result.

Proposition 2.4

Let \(0<s<1\) and let \(E\subseteq B_R(x_0)\subset \mathbb{R}^N\) be a measurable set, with positive measure. There exists a constant \(\mathcal{M}=\mathcal{M}(N)>0\) such that for every \(u\in W^{s,2}(B_R(x_0))\) we have

$$\begin{aligned} \Vert u-\overline{u}_E\Vert ^2_{L^2(B_R(x_0))}\le \mathcal{M}\,(1-s)\frac{R^{N}}{|E|}\,R^{2\,s}\,[u]^2_{W^{s,2}(B_R(x_0))}. \end{aligned}$$

Proof

By a standard scaling argument, it is sufficient to prove the result for \(R=1\) and \(x_0=0\). For every \(t>0\), we denote by \(Q_t=(-t/2,t/2)^N\) the \(N-\)dimensional open cube centered at the origin, with side length t.

We consider the extension \(\mathcal{E}[u]\) of u to the whole \(\mathbb{R}^N\), as in (2.3). For ease of notation, we will simply write \(\widetilde{u}:=\mathcal{E}[u]\). By using the triangle inequality and the fact that \(B_1\subset Q_2\), we have

$$\begin{aligned} \begin{aligned} \Vert u-\overline{u}_E\Vert ^2_{L^2(B_1)}&\le \Vert {\widetilde{u}}-\overline{u}_E\Vert ^2_{L^2(Q_2)}\\&\le 2\,\Vert \widetilde{u}-\overline{\widetilde{u}}_{Q_2}\Vert ^2_{L^2(Q_2)}+2\,\Vert \overline{\widetilde{u}}_{Q_2}-\overline{u}_E\Vert ^2_{L^2(Q_2)}. \end{aligned} \end{aligned}$$
(2.8)

By using Jensen’s inequality and the fact that \(|Q_2|=2^N\), we can estimate the second term as follows

figure b

Thus from (2.8) we get

$$\begin{aligned} \Vert u-\overline{u}_E\Vert ^2_{L^2(B_1)}\le 2\,\left( 1+\frac{2^N}{|E|}\right) \,\Vert \widetilde{u}-\overline{\widetilde{u}}_{Q_2}\Vert _{L^2(Q_2)}^2. \end{aligned}$$

We can now apply the following fractional Poincaré inequalityFootnote 2 proved by Maz’ya and Shaposnikova (see [27, p. 300])

$$\begin{aligned} \Vert \varphi -{\overline{\varphi }}_{Q_2}\Vert ^2_{L^2(Q_2)}\le C_N\, (1-s)\,[\varphi ]_{W^{s,2}(Q_2)}^2,\quad \text{for every}\quad \varphi \in W^{s,2}(Q_2). \end{aligned}$$
(2.9)

Here \(C_N\) is an explicit dimensional constant. This yields

$$\begin{aligned} \begin{aligned} \Vert u-\overline{u}_E\Vert ^2_{L^2(B_1)}&\le 2\,\left( 1+\frac{2^N}{|E|}\right) \,C_N\, (1-s)\,[\widetilde{u}]_{W^{s,2}(Q_2)}^2\\&\le 2\,\frac{\omega _N+2^N}{|E|}\,C_N\,(1-s)\,[\widetilde{u}]_{W^{s,2}(B_{\sqrt{2}})}^2, \end{aligned} \end{aligned}$$

where we used that \(Q_2\subset B_{\sqrt{2}}\). It is now sufficient to apply Lemma 2.2 with \(R=\sqrt{2}\), to get the claimed conclusion. \(\square\)

3 An expedient Poincaré inequality

The following result is a nonlocal counterpart of [19, Lemma 1] in Hayman’s paper. In the proof we pay due attention to the dependence of the constant on the fractional parameter s, as always.

Proposition 3.1

(Poincaré for boundary disks) Let \(1/2<s<1\) and let \(\Omega \subset \mathbb{R}^2\) be an open simply connected set, with \(\partial \Omega \not =\emptyset\). There exists a constant \(\mathcal{T}_s>0\) depending on s only, such that for every \(r>0\) and every \(x_0\in \partial \Omega\), we have

$$\begin{aligned} \frac{\mathcal{T}_s}{r^{2\,s}}\,\int _{B_r(x_0)}|u(x)|^2\,\mathrm{d}x\le \iint _{B_r(x_0)\times \mathbb{R}^2}\frac{|u(x)-u(y)|^2}{|x-y|^{2+2\,s}}\,\mathrm{d}x\,\mathrm{d}y,\quad {for\,every }\quad u\in C^\infty _0(\Omega ). \end{aligned}$$

Moreover, \(\mathcal{T}_s\) has the following asymptotic behaviors

$$\begin{aligned} \mathcal{T}_s\sim \left( s-\frac{1}{2}\right) ,\quad \text{for}\quad s\searrow \frac{1}{2}\quad \hbox{and}\quad \mathcal{T}_s\sim \frac{1}{1-s},\quad \text{for}\quad s\nearrow 1. \end{aligned}$$

Proof

Up to scaling and translating, we can assume without loss of generality that \(r=1\) and that \(x_0\) coincides with the origin. We split the proof in three main steps: we first show that it is sufficient to prove the claimed estimate for the boundary ring \(B_1\setminus B_{1/2}\). Then we prove such an estimate and at last we discuss the asymptotic behavior of the constant obtained.

Step 1 Reduction to a ring. Let \(u\in C^\infty _0(\Omega )\), we then estimate the \(L^2\) norm on \(B_1\) as follows

$$\begin{aligned} \begin{aligned} \int _{B_1}|u(x)|^2\,\mathrm{d}x&=\int _{B_1\setminus B_{1/2}}|u(x)|^2\,\mathrm{d}x+\int _{B_{1/2}}|u(x)|^2\,\mathrm{d}x\\&\le \int _{B_1\setminus B_{1/2}}|u(x)|^2\,\mathrm{d}x+2\,\int _{B_{1/2}}|u(x)-\overline{u}_{B_1\setminus B_{1/2}}|^2\,\mathrm{d}x+2\,\int _{B_{1/2}}|\overline{u}_{B_1\setminus B_{1/2}}|^2\,\mathrm{d}x\\&\le \int _{B_1\setminus B_{1/2}}|u(x)|^2\,\mathrm{d}x+2\,\int _{B_1}|u(x)-\overline{u}_{B_1\setminus B_{1/2}}|^2\,\mathrm{d}x+2\,\int _{B_{1/2}}|\overline{u}_{B_1\setminus B_{1/2}}|^2\,\mathrm{d}x\\&\le \int _{B_1\setminus B_{1/2}}|u(x)|^2\,\mathrm{d}x+2\,\int _{B_1}|u(x)-\overline{u}_{B_1\setminus B_{1/2}}|^2\,\mathrm{d}x+\frac{2}{3}\,\int _{B_1\setminus B_{1/2}}|u(x)|^2\,\mathrm{d}x, \end{aligned} \end{aligned}$$

where we used the elementary inequality \((a+b)^2\le 2\,a^2+2\,b^2\) and Jensen’s inequality. If we now apply Proposition 2.4 with \(R=1\) and \(E=B_1\setminus B_{1/2}\), we get

$$\begin{aligned} \int _{B_1}|u(x)|^2\,\mathrm{d}x\le \frac{5}{3}\,\int _{B_1\setminus B_{1/2}}|u(x)|^2\,\mathrm{d}x+\frac{8}{3\,\pi }\,\mathcal{M}\,(1-s)\,[u]^2_{W^{s,2}(B_1)}. \end{aligned}$$

Thus, in order to conclude, it is sufficient to prove that there exists a constant \(C=C(s)>0\) such that

$$\begin{aligned} \int _{B_{1}\setminus B_{1/2}}|u(x)|^2\,\mathrm{d}x\le C\,\iint _{B_1\times \mathbb{R}^2}\frac{|u(x)-u(y)|^2}{|x-y|^{2+2\,s}}\,\mathrm{d}x\,\mathrm{d}y,\quad \text{for every}\quad u\in C^\infty _0(\Omega ). \end{aligned}$$
(3.1)

Step 2 Estimate on the ring. We start with a topological observation. Since we are assuming that \(0\in \partial \Omega\) and that \(\Omega\) is simply connected, we have the following crucial property

$$\begin{aligned} \partial B_\varrho \cap (\mathbb{R}^2\setminus \Omega )\not =\emptyset ,\quad \text{for every }\quad \varrho >0. \end{aligned}$$
(3.2)

Indeed, if this were not true, we would have existence of a circle entirely contained in \(\Omega\) and centered on the boundary of \(\partial \Omega\). Such a circle could not be null-homotopic in \(\Omega\), thus contradicting our topological assumption.

In the rest of the proof, we will use polar coordinates \((\varrho ,\theta )\) and we will make the slight abuse of notation of writing \(u(\varrho ,\theta )\). Then, in light of property (3.2), for each \(\varrho \in (1/2,1)\) there exists \(\theta _\varrho \in [0,2\pi )\) such that \(\theta \mapsto u(\varrho ,\theta )\) must vanish at \(\theta _\varrho\). Hence, for every \(1/2<\varrho <1\) we can apply Proposition A.2 to the function \(\theta \mapsto u(\varrho ,\theta )\) and get

$$\begin{aligned} \int _0^{2\,\pi }|u(\varrho ,\theta )|^2\,\mathrm{d}\theta \le \frac{1}{\mu _s}\,\int _0^{2\pi }\int _0^{2\pi } \frac{|u(\varrho ,\theta )-u(\varrho ,\varphi )|^2}{|\theta -\varphi |^{1+2\,s}_{\mathbb{S}^1}}\,\mathrm{d}\theta \,\mathrm{d}\varphi . \end{aligned}$$

The constant \(\mu _s\) is the same as in Proposition A.2 and

$$\begin{aligned} |\theta -\varphi |_{\mathbb{S}^1}:=\min _{k\in \mathbb{Z}} |\theta -\varphi +2\,k\,\pi |,\quad \text{for every}\quad \theta ,\varphi \in \mathbb{R}. \end{aligned}$$

If we now multiply both sides by \(\varrho\), integrate over the interval (1/2, 1) and write the \(L^2\) norm in polar coordinates, we get

$$\begin{aligned} \begin{aligned} \int _{B_1\setminus B_{1/2}}|u(x)|^2\,\mathrm{d}x&\le \frac{1}{\mu _s}\,\int _\frac{1}{2}^1\int _0^{2\pi }\int _0^{2\,\pi } \frac{|u(\varrho ,\theta )-u(\varrho ,\varphi )|^2}{\varrho ^{1+2\,s}\,{|\theta -\varphi |^{1+2\,s}_{\mathbb{S}^1}}} \,\varrho \,\mathrm{d}\varrho \,\mathrm{d}\theta \,\mathrm{d}\varphi . \end{aligned} \end{aligned}$$
(3.3)

Observe that we further used the fact that \(\varrho \le 1\), to let the term \(\varrho ^{-1-2\,s}\) appear. In order to achieve (3.1), we need to show that the term on the right-hand side can be estimated by a two-dimensional Gagliardo–Slobodeckiĭ seminorm. To this aim, we follow an argument similar to that of [7, Lemma B.2]. At first, it is easily seen that

$$\begin{aligned} \begin{aligned} \left( \varrho \,|\theta -\varphi |_{\mathbb{S}^1}\right) ^{-1-2\,s}&=(1+2\,s)\,\int _0^{+\infty }\left( t+\varrho \,|\theta -\varphi |_{\mathbb{S}^1}\right) ^{-2-2\,s}\,\mathrm{d}t. \end{aligned} \end{aligned}$$

By inserting this in (3.3), we end up with

$$\begin{aligned} \begin{aligned}&\int _{B_1\setminus B_{1/2}} |u(x)|^2\,\mathrm{d}x\\&\quad \le \frac{1+2\,s}{\mu _s}\,\int _0^{2\pi }\int _0^{2\pi }\int _\frac{1}{2}^1\int _0^{+\infty }\frac{|u(\varrho ,\theta )-u(\varrho ,\varphi )|^2}{(t+\varrho \,|\theta -\varphi |_{\mathbb{S}^1})^{2+2\,s}}\,\varrho \,\mathrm{d}\theta \,\mathrm{d}\varphi \,\mathrm{d}\varrho \,\mathrm{d}t\\&\quad \le \frac{2\,(1+2\,s)}{\mu _s}\,\int _0^{2\pi }\int _0^{2\,\pi }\int _\frac{1}{2}^1\int _0^{+\infty }\frac{|u(\varrho ,\theta )-u(\varrho ,\varphi )|^2}{(t+\varrho \,|\theta -\varphi |_{\mathbb{S}^1})^{2+2\,s}}\,\varrho \,(\varrho +t)\,\mathrm{d}\theta \,\mathrm{d}\varphi \,\mathrm{d}\varrho \,\mathrm{d}t, \end{aligned} \end{aligned}$$
(3.4)

where we have used that \(1/2\le \varrho +t\). We now split the set \([0,2\,\pi ]\times [0,2\,\pi ]=J_{-1}\cup J_0\cup J_1\), where

$$\begin{aligned} J_{-1}=\left\{ (\theta ,\varphi )\,:\,\theta \in [0,\pi ], \,\theta +\pi<\varphi \le 2\,\pi \right\} ,\quad J_{1}=\left\{ (\theta ,\varphi )\,:\,\theta \in [\pi ,2\,\pi ],\,0\le \varphi <\theta -\pi \right\} , \end{aligned}$$

and

$$\begin{aligned} J_0=\left\{ (\theta ,\varphi )\,:\,\theta \in [0,2\,\pi ],\, \max \{0,\theta -\pi \}\le \varphi \le \min \{2\,\pi ,\theta +\pi \}\right\} , \end{aligned}$$

see Fig. 1.

Fig. 1
figure 1

The partition of \([0,2\,\pi ]\times [0,2\,\pi ]\) needed to define the midpoint function

Full size image

Then, we define the midpoint function by

$$\begin{aligned} \overline{\theta \,\varphi }=\frac{\theta +\varphi }{2}+\ell \,\pi ,\quad \text{if}\quad (\theta ,\varphi )\in J_\ell ,\quad \text{with}\quad \ell =-1,0,1. \end{aligned}$$
(3.5)

Thanks to the triangle inequality, we estimate the numerator in the right-hand side of (3.4) as follows

$$\begin{aligned} \begin{aligned} |u(\varrho ,\theta )-u(\varrho ,\varphi )|^2&\le 2 \left| u\left( \varrho ,\theta \right) -u\left( \varrho +t,\overline{\theta \,\varphi }\right) \right| ^2 +2\left| u\left( \varrho ,\varphi \right) -u\left( \varrho +t,\overline{\theta \,\varphi }\right) \right| ^2. \end{aligned} \end{aligned}$$

As for the denominator, we observe that \(|\theta -\overline{\theta \,\varphi }|_{\mathbb{S}^1}=|\varphi -\overline{\theta \,\varphi }|_{\mathbb{S}^1}\), thus we get

$$\begin{aligned} |\theta -\varphi |_{\mathbb{S}^1}=2\,|\theta -\overline{\theta \,\varphi }|_{\mathbb{S}^1}\ge 2\,|e^{i\,\theta }-e^{i\,\overline{\theta \,\varphi }}|\quad \text{and}\quad |\theta -\varphi |_{\mathbb{S}^1}=2\,|\varphi -\overline{\theta \,\varphi }|_{\mathbb{S}^1}\ge 2\,|e^{i\,\varphi }-e^{i\,\overline{\theta \,\varphi }}|, \end{aligned}$$

where the inequalities come from Lemma A.1. By using this fact, the identity \(|e^{i\overline{\theta \,\varphi }}|=1\) and the triangle inequality again, we can estimate the denominator as

$$\begin{aligned} \begin{aligned} t+\varrho \,|\theta -\varphi |_{\mathbb{S}^1}&\ge t+\varrho \,|e^{i\theta }-e^{i\,\overline{\theta \,\varphi }}|\\&\ge \left| \varrho \,(e^{i\,\theta }-e^{i\,\overline{\theta \,\varphi }}) -t\,e^{i\,\overline{\theta \,\varphi }}\right| =\left| \varrho \,e^{i\,\theta } -(\varrho +t)\,e^{i\,\overline{\theta \,\varphi }}\right| , \end{aligned} \end{aligned}$$

and similarly

$$\begin{aligned} t+\varrho \,|\theta -\varphi |_{\mathbb{S}^1}\ge \left| \varrho \,e^{i\,\varphi }-(\varrho +t)\,e^{i\,\overline{\theta \,\varphi }}\right| . \end{aligned}$$

These allow us to estimate the right-hand side in (3.4) in the following way:

$$\begin{aligned} \begin{aligned}&\int _{B_1\setminus B_{1/2}} |u(x)|^2\,\mathrm{d}x\\&\quad \le \frac{4\,(1+2\,s)}{\mu _s}\,\int _0^{2\,\pi }\int _0^{2\,\pi }\int _\frac{1}{2}^1\int _0^{+\infty }\frac{\left| u\left( \varrho ,\theta \right) -u\left( \varrho +t,\overline{\theta \,\varphi }\right) \right| ^2}{\left| \varrho \,e^{i\,\theta }-(\varrho +t)\,e^{i\,\overline{\theta \,\varphi }}\right| ^{2+2\,s}}\,\varrho \,(\varrho +t)\,\mathrm{d}\theta \,\mathrm{d}\varphi \,\mathrm{d}\varrho \,\mathrm{d}t\\&\qquad +\frac{4\,(1+2\,s)}{\mu _s}\,\int _0^{2\,\pi }\int _0^{2\,\pi }\int _\frac{1}{2}^1\int _0^{+\infty }\frac{\left| u\left( \varrho ,\varphi \right) -u\left( \varrho +t,\overline{\theta \,\varphi }\right) \right| ^2}{\left| \varrho \,e^{i\,\varphi }-(\varrho +t)\,e^{i\,\overline{\theta \,\varphi }}\right| ^{2+2\,s}}\,\varrho \,(\varrho +t)\,\mathrm{d}\theta \,\mathrm{d}\varphi \,\mathrm{d}\varrho \,\mathrm{d}t\\&\quad =\frac{8\,(1+2\,s)}{\mu _s}\,\int _0^{2\,\pi }\int _0^{2\,\pi }\int _\frac{1}{2}^1\int _0^{+\infty }\frac{\left| u\left( \varrho ,\theta \right) -u\left( \varrho +t,\overline{\theta \,\varphi }\right) \right| ^2}{\left| \varrho \,e^{i\,\theta }-(\varrho +t)\,e^{i\,\overline{\theta \,\varphi }}\right| ^{2+2\,s}}\,\varrho \,(\varrho +t)\,\mathrm{d}\theta \,\mathrm{d}\varphi \,\mathrm{d}\varrho \,\mathrm{d}t. \end{aligned} \end{aligned}$$

In the last identity we used that both multiple integrals coincide, by symmetry of the integrands. If we now make the change of variable \(\tau =\varrho +t\) and use the decomposition \([0,2\,\pi ]\times [0,2\,\pi ]=J_{-1}\cup J_0\cup J_1\), we obtain

$$\begin{aligned} \begin{aligned}&\int _{B_1\setminus B_{1/2}} |u(x)|^2\,\mathrm{d}x\\&\quad \le \frac{8\,(1+2\,s)}{\mu _s}\,\sum _{\ell =-1}^1 \iint _{J_\ell }\int _\frac{1}{2}^1\int _\varrho ^{+\infty } \frac{\left| u\left( \varrho ,\theta \right) -u(\tau ,\overline{\theta \,\varphi })\right| ^2}{\left| \varrho \,e^{i\,\theta }-\tau \,e^{i\,\overline{\theta \,\varphi }}\right| ^{2+2\,s}}\,\varrho \,\tau \,\mathrm{d}\theta \,\mathrm{d}\varphi \,\mathrm{d}\varrho \,\mathrm{d}\tau . \end{aligned} \end{aligned}$$
(3.6)

If we now denote

$$\begin{aligned} \widetilde{J}_{-1}= & {} \left\{ (\theta ,\varphi )\,:\,\theta \in [0,\pi ],\, \theta -\pi<\varphi \le 0\right\} \quad \text{and}\\ \widetilde{J}_1= & {} \left\{ (\theta ,\varphi )\,:\,\theta \in [\pi ,2\,\pi ],\,2\,\pi \le \varphi <\theta +\pi \right\} , \end{aligned}$$

use the definition of midpoint function (3.5) and make the change of variables

$$\begin{aligned} \begin{array}{ccc} J_{-1} &{} \rightarrow &{}{\widetilde{J}}_{-1}\\ (\theta ,\varphi ) &{} \mapsto &{} (\theta ,\varphi -2\,\pi ) \end{array}\quad \text{and}\quad \begin{array}{ccc} J_{1} &{} \rightarrow &{}{\widetilde{J}}_{1}\\ (\theta ,\varphi ) &{} \mapsto &{} (\theta ,\varphi +2\,\pi ), \end{array} \end{aligned}$$

we obtain from (3.6)

$$\begin{aligned} \begin{aligned} \int _{B_1\setminus B_{1/2}} |u(x)|^2\,\mathrm{d}x\le \frac{8\,(1+2\,s)}{\mu _s}\,\iint _{\widetilde{J}_{-1}\cup J_0\cup \widetilde{J}_1}\int _\frac{1}{2}^1\int _\varrho ^{+\infty }\frac{\left| u\left( \varrho ,\theta \right) -u\left( \tau ,\frac{\theta +\varphi }{2}\right) \right| ^2}{\left| \varrho \,e^{i\,\theta }-\tau \,e^{i\,\frac{\theta +\varphi }{2}}\right| ^{2+2s}}\,\varrho \,\tau \,\mathrm{d}\theta \,\mathrm{d}\varphi \,\mathrm{d}\varrho \,\mathrm{d}\tau . \end{aligned} \end{aligned}$$

For every \(\theta \in [0,2\,\pi ]\), we now make the change of variable \(\gamma =(\theta +\varphi )/2\), thus the above estimate becomes

$$\begin{aligned} \int _{B_1\setminus B_{1/2}} |u(x)|^2\,\mathrm{d}x \le \frac{16\,(1+2\,s)}{\mu _s}\,\sum _{\ell =-1}^1\iint _{\widehat{J}_\ell }\int _\frac{1}{2}^1\int _\varrho ^{+\infty }\frac{\left| u\left( \varrho ,\theta \right) -u\left( \tau ,\gamma \right) \right| ^2}{\left| \varrho \,e^{i\,\theta }-\tau \,e^{i\,\gamma }\right| ^{2+2s}}\,\varrho \,\tau \,\mathrm{d}\theta \,\mathrm{d}\gamma \,\mathrm{d}\varrho \,\mathrm{d}\tau , \end{aligned}$$
(3.7)

where

$$\begin{aligned} \widehat{J}_{-1}= & {} \left\{ (\theta ,\gamma )\,:\theta \in \left[ 0,\frac{\pi }{2}\right] ,\,\theta -\frac{\pi }{2}<\gamma \le 0\right\} ,\\ \widehat{J}_1= & {} \left\{ (\theta ,\gamma )\,:\,\theta \in \left[ \frac{3}{2}\,\pi ,2\,\pi \right] ,\,2\,\pi \le \gamma <\theta +\frac{\pi }{2}\right\} , \end{aligned}$$

and

$$\begin{aligned} \widehat{J}_0=\left\{ (\theta ,\gamma )\,:\,\theta \in [0,2\,\pi ],\, \max \left\{ 0,\,\theta -\frac{\pi }{2}\right\} \le \gamma \le \min \left\{ 2\pi ,\,\theta +\frac{\pi }{2}\right\} \right\} . \end{aligned}$$

If we now exploit the \(2\,\pi -\)periodicity of the integrand, we have

$$\begin{aligned} \begin{aligned}&\iint _{\widehat{J}_{-1}}\int _\frac{1}{2}^1\int _\varrho ^{+\infty }\frac{\left| u\left( \varrho ,\theta \right) -u\left( \tau ,\gamma \right) \right| ^2}{\left| \varrho \,e^{i\,\theta }-\tau \,e^{i\,\gamma }\right| ^{2+2s}}\,\varrho \,\tau \,\mathrm{d}\theta \,\mathrm{d}\gamma \,\mathrm{d}\varrho \,\mathrm{d}\tau \\&\quad =\iint _{\widehat{J}_{-1}}\int _\frac{1}{2}^1\int _\varrho ^{+\infty }\frac{\left| u\left( \varrho ,\theta \right) -u\left( \tau ,\gamma +2\,\pi \right) \right| ^2}{\left| \varrho \,e^{i\,\theta }-\tau \,e^{i\,(\gamma +2\,\pi )}\right| ^{2+2s}}\,\varrho \,\tau \,\mathrm{d}\theta \,\mathrm{d}\gamma \,\mathrm{d}\varrho \,\mathrm{d}\tau \\&\quad =\iint _{I_{-1}}\int _\frac{1}{2}^1\int _\varrho ^{+\infty }\frac{\left| u\left( \varrho ,\theta \right) -u\left( \tau ,\varphi \right) \right| ^2}{\left| \varrho \,e^{i\,\theta }-\tau \,e^{i\,\varphi }\right| ^{2+2s}}\,\varrho \,\tau \,\mathrm{d}\theta \,\mathrm{d}\varphi \,\mathrm{d}\varrho \,\mathrm{d}\tau , \end{aligned} \end{aligned}$$
(3.8)

where we set \(\varphi =\gamma +2\,\pi\) and

$$\begin{aligned} I_{-1}=\left\{ (\theta ,\varphi )\,:\theta \in \left[ 0,\frac{\pi }{2}\right] ,\,\theta +\frac{3}{2}\,\pi <\varphi \le 2\,\pi \right\} . \end{aligned}$$

Similarly, we can obtain

$$\begin{aligned} \begin{aligned}&\iint _{\widehat{J}_{1}}\int _\frac{1}{2}^1\int _\varrho ^{+\infty }\frac{\left| u\left( \varrho ,\theta \right) -u\left( \tau ,\gamma \right) \right| ^2}{\left| \varrho \,e^{i\,\theta }-\tau \,e^{i\,\gamma }\right| ^{2+2\,s}}\,\varrho \,\tau \,\mathrm{d}\theta \,\mathrm{d}\gamma \,\mathrm{d}\varrho \,\mathrm{d}\tau \\&\quad =\iint _{I_1}\int _\frac{1}{2}^1\int _\varrho ^{+\infty }\frac{\left| u\left( \varrho ,\theta \right) -u\left( \tau ,\varphi \right) \right| ^2}{\left| \varrho \,e^{i\,\theta }-\tau \,e^{i\,\varphi }\right| ^{2+2s}}\,\varrho \,\tau \,\mathrm{d}\theta \,\mathrm{d}\varphi \,\mathrm{d}\varrho \,\mathrm{d}\tau , \end{aligned} \end{aligned}$$
(3.9)

with the change of variable \(\varphi =\gamma -2\,\pi\) and

$$\begin{aligned} I_1=\left\{ (\theta ,\varphi )\,:\,\theta \in \left[ \frac{3}{2}\,\pi ,2\,\pi \right] ,\,0\le \varphi <\theta -\frac{3}{2}\,\pi \right\} . \end{aligned}$$

By observing that \(I_{-1}\cup \widehat{J}_0\cup I_1\subset [0,2\,\pi ]\times [0,2\,\pi ]\) and that the three sets \(I_{-1},\widehat{J}_0\) and \(I_1\) are pairwise disjoint, from (3.7), (3.8) and (3.9) we finally obtain

$$\begin{aligned} \begin{aligned} \int _{B_1\setminus B_{1/2}} |u(x)|^2\,\mathrm{d}x&\le \frac{16\,(1+2\,s)}{\mu _s}\,\iint _{[0,2\,\pi ]\times [0,2\,\pi ]}\int _\frac{1}{2}^1\int _\varrho ^{+\infty }\frac{\left| u\left( \varrho ,\theta \right) -u\left( \tau ,\varphi \right) \right| ^2}{\left| \varrho \,e^{i\,\theta }-\tau \,e^{i\,\varphi }\right| ^{2+2\,s}}\,\varrho \,\tau \,\mathrm{d}\theta \,\mathrm{d}\varphi \,\mathrm{d}\varrho \,\mathrm{d}\tau \\&\le \frac{16\,(1+2\,s)}{\mu _s}\,\iint _{B_1\times \mathbb{R}^2} \frac{|u(x)-u(y)|^2}{|x-y|^{2+2\,s}}\,\mathrm{d}x\,\mathrm{d}y. \end{aligned} \end{aligned}$$

This concludes the proof of (3.1).

Step 3: asymptotics for the constant. From Step 1 and Step 2, we obtained the Poincaré inequality claimed in the statement, with constant

$$\begin{aligned} \mathcal{T}_s=\left( \frac{80\,(1+2\,s)}{3\,\mu _s}+\frac{8}{3\,\pi }\,\mathcal{M}\,(1-s)\right) ^{-1}. \end{aligned}$$

By using the asymptotics for the constant \(\mu _s\) (see Proposition A.2), we get the desired conclusion. \(\square\)

Remark 3.2

The previous result cannot hold for \(0<s\le 1/2\). Indeed, if the result were true for \(0<s\le 1/2\), this would permit to extend the fractional Makai–Hayman inequality to this range, as well (see the next section). However, this would contradict Theorem 1.3.

4 Proof of Theorem 1.1

Without loss of generality, we can consider \(r_\Omega =1\). We take \(\mathfrak {B}\) and \(\mathfrak {B}_1,\ldots ,\mathfrak {B}_{36}\) to be, respectively, the covering of \(\Omega\) and the subclasses given by Lemma 2.1, made of ball with radius \(r=1+\sqrt{2}\).

We take an index \(k\in \{1,\ldots ,36\}\), then we know that \(\mathfrak {B}_k\) is composed of (possibly) countably many disjoint balls with radius r, centered on \(\partial \Omega\). We indicate by \(B^{j,k}\) each of these balls.

Then, for every \(u\in C^\infty _0(\Omega )\setminus \{0\}\) we have

$$\begin{aligned} \iint _{\mathbb{R}^2\times \mathbb{R}^2}\frac{|u(x)-u(y)|^2}{|x-y|^{2+2\,s}}\,\mathrm{d}x\,\mathrm{d}y\ge \sum _{B^{j,k}\in \mathfrak {B}_k}\iint _{B^{j,k}\times \mathbb{R}^2}\frac{|u(x)-u(y)|^2}{|x-y|^{2+2\,s}}\,\mathrm{d}x\,\mathrm{d}y. \end{aligned}$$
(4.1)

For each ball \(B^{j,k}\), we can apply Proposition 3.1 so to obtain that

$$\begin{aligned} \sum _{B^{j,k}\in \mathfrak {B}_k}\iint _{B^{j,k}\times \mathbb{R}^2}\frac{|u(x)-u(y)|^2}{|x-y|^{2+2\,s}}\,\mathrm{d}x\,\mathrm{d}y\ge \frac{\mathcal{T}_s}{(1+\sqrt{2})^{2\,s}}\,\sum _{B^{j,k}\in \mathfrak {B}_k}\int _{B^{j,k}}|u(x)|^2\,\mathrm{d}x. \end{aligned}$$

We insert this estimate in (4.1) and then sum over \(k=1,\ldots ,36\). We get

$$\begin{aligned} \begin{aligned}&36\,\iint _{\mathbb{R}^2\times \mathbb{R}^2}\frac{|u(x)-u(y)|^2}{|x-y|^{2+2\,s}}\,\mathrm{d}x\,\mathrm{d}y\\&\quad \ge \sum _{k=1}^{36}\sum _{B^{j,k}\in \mathfrak {B}_k}\iint _{B^{j,k}\times \mathbb{R}^2}\frac{|u(x)-u(y)|^2}{|x-y|^{2+2\,s}}\,\mathrm{d}x\,\mathrm{d}y\\&\quad \ge \frac{\mathcal{T}_s}{(1+\sqrt{2})^{2\,s}}\,\sum _{k=1}^{36}\sum _{B^{j,k}\in \mathfrak {B}_k}\int _{B^{j,k}}|u(x)|^2\,\mathrm{d}x\\&\quad \ge \frac{\mathcal{T}_s}{(1+\sqrt{2})^{2\,s}} \int _\Omega |u(x)|^2\,\mathrm{d}x. \end{aligned} \end{aligned}$$

In the last inequality we used that \(\mathfrak {B}\) is a covering of \(\Omega\). By recalling the definition of \(\lambda _1^s(\Omega )\), from the previous chain of inequalities we thus get the claimed estimate (1.1), with constant

$$\begin{aligned} \mathcal{C}_s:=\frac{\mathcal{T}_s}{36\,(1+\sqrt{2})^{2\,s}}. \end{aligned}$$

The asymptotic behavior of \(\mathcal{C}_s\) can now be inferred from that of \(\mathcal{T}_s\), which in turn is contained in Proposition 3.1.

Remark 4.1

For suitable classes of open sets in \(\mathbb{R}^N\) and every \(0<s<1\), it is possible to give a Makai–Hayman-type lower bound on \(\lambda _1^s\), by taking advantage of the nonlocality of the Gagliardo–Slobodeckiĭ seminorm. More precisely, this is possible provided \(\Omega\) satisfies the following mild regularity assumption: there existFootnote 3\(\sigma >1\) and \(\alpha >0 \mbox{ such that }\)

$$\begin{aligned} \frac{|B_{\sigma \, r_\Omega }(x)\setminus \Omega |}{|B_{\sigma \,r_\Omega }(x)|}\ge \alpha ,\quad \text{for every}\, x\in \Omega . \end{aligned}$$
(4.2)

Indeed, in this case for every \(u\in C^\infty _0(\Omega )\) we can simply estimate

$$\begin{aligned} \begin{aligned}&\iint _{\mathbb{R}^N\times \mathbb{R}^N}\frac{|u(x)-u(y)|^2}{|x-y|^{N+2\,s}}\,\mathrm{d}x\,\mathrm{d}y\\&\quad \ge \int _{\Omega }\left( \int _{B_{\sigma \, r_\Omega }(x)\setminus \Omega }\frac{|u(x)|^2}{|x-y|^{N+2\,s}}\,\mathrm{d}y\right) \,\mathrm{d}x\\&\quad \ge \frac{1}{(\sigma \, r_\Omega )^{N+2\,s}}\,\int _{\Omega }|B_{\sigma \, r_\Omega }(x)\setminus \Omega |\,|u(x)|^2\,\mathrm{d}x\\&\quad \ge \frac{\alpha \,\omega _N}{{(\sigma \, r_\Omega )}^{2\,s}}\,\int _{\Omega }|u(x)|^2\,\mathrm{d}x, \end{aligned} \end{aligned}$$

where in the last inequality we used the additional condition (4.2). By arbitrariness of u, we get

$$\begin{aligned} \lambda _1^s(\Omega )\ge \frac{\alpha \,\omega _N}{\sigma ^{2\,s}}\,\frac{1}{r_\Omega ^{2\,s}}. \end{aligned}$$

One could observe that the additional condition (4.2) does not always hold for a simply connected set in the plane. Moreover, the constant obtained in this way is quite poor: first of all, it is not universal. It depends on the parameters \(\alpha\) and \(\sigma\) and it deteriorates as \(\sigma \searrow 1\), since in this case we must have \(\alpha \searrow 0\). Secondly, it does not exhibit the correct asymptotic behavior as s goes to 1.

5 Proof of Theorem 1.3

Let \(0<s\le 1/2\) and \(\{Q_k\}_{k\in \mathbb{N}}\subset \mathbb{R}^2\) be the sequence of open squares \(Q_k=(-k,k)^2\), with \(k\in \mathbb{N}\setminus \{0,1\}\). We introduce the one-dimensional set

$$\begin{aligned} {\Sigma }=\bigcup \limits _{i\in \mathbb{Z}}{\Sigma }^{(i)},\quad \text{where}\quad \Sigma ^{(i)}:=\left\{ (x_1,i)\in \mathbb{R}^2\,:\,|x_1|\ge 1\right\} , \end{aligned}$$

and then define, for every fixed \(k\in \mathbb{N}\setminus \{0,1\}\), the “cracked” square \(\widetilde{Q}_k=Q_k\setminus {\Sigma }\) (see Fig. 2). First of all, we observe that

$$\begin{aligned} r_{\widetilde{Q}_k}=\frac{\sqrt{5}}{2},\quad \text{for every}\, k\ge 2. \end{aligned}$$

Thus, if we can show that

$$\begin{aligned} \lim _{k\rightarrow \infty }\lambda _1^s(\widetilde{Q}_k)=0, \end{aligned}$$
(5.1)

we would automatically get the desired counter-example. We will obtain (5.1) by proving that

$$\begin{aligned} \lambda _1^s(\widetilde{Q}_k)=\lambda _1^s(Q_k),\quad \text{for every}\quad k\ge 2. \end{aligned}$$
(5.2)

Indeed, if this were true, we would have

$$\begin{aligned} \lim _{k\rightarrow \infty }\lambda _1^s(\widetilde{Q}_k)=\lim _{k\rightarrow \infty }\lambda _1^s(Q_k)=\lim _{k\rightarrow \infty } k^{-2\,s}\,\lambda ^s_1(Q_1)=0, \end{aligned}$$

by the scale properties of \(\lambda ^1_s\). This would prove (5.1), as claimed.

Fig. 2
figure 2

The set \(\widetilde{Q}_k\) with \(k=2\). In dashed line, a disk of maximal radius

Full size image

We are thus left with proving (5.2). We already know that

$$\begin{aligned} \lambda _1^s(\widetilde{Q}_k)\ge \lambda _1^s(Q_k), \end{aligned}$$

thanks to the fact that \(\lambda _1^s\) is monotone with respect to set inclusion. In the remaining part of the proof, we focus our attention in proving the opposite inequality.

Fig. 3
figure 3

The dashed line encloses one of the sets \(\Sigma _{k,n}^{(i)}\)

Full size image

At this aim, for every \(n\in \mathbb{N}\setminus \{0\}\) we introduce the neighborhoods (Fig. 3)

$$\begin{aligned} \Sigma _{k,n}^{(i)}=\left\{ x\in \mathbb{R}^2\, :\, \mathrm {dist}(x,\Sigma ^{(i)}\cap Q_k)\le \frac{1}{n+1}\right\} , \quad \text{for}\quad i\in \{-(k-1),\ldots ,k-1\}, \end{aligned}$$

and consider a sequence of cut-off functions \(\{\varphi _n^{(i)}\}_{n\in \mathbb{N}\setminus \{0\}}\subset C^\infty _0(\Sigma ^{(i)}_{k,2\,n})\) such that

$$\begin{aligned} 0\le \varphi ^{(i)}_n\le 1,\quad \varphi ^{(i)}_n\equiv 1\,\text{on}\,\,\, \Sigma ^{(i)}_{k,4\,n},\quad |\nabla \varphi ^{(i)}_n(x)|\le C\,n, \end{aligned}$$

for some constant \(C>0\), independent of n. Observe that by construction we have

$$\begin{aligned} \mathrm {spt}(\varphi ^{(i)}_n)\cap \mathrm {spt}(\varphi ^{(j)}_n)=\emptyset ,\quad \text{for}\quad i\not =j, \end{aligned}$$

By using an interpolation inequality (see [10, Corollary 2.2]) and the properties of the cut-off functions, we can estimate the energy of each \(\varphi _n^{(i)}\) as follows

$$\begin{aligned} \begin{aligned}{}[\varphi ^{(i)}_n]^2_{W^{s,2}(\mathbb{R}^2)}&\le C\,\left( \int _{\Sigma _{k,2\,n}^{(i)}} |\varphi ^{(i)}_n|^2\,\mathrm{d}x\right) ^{1-s}\, \left( \int _{\Sigma _{k,2\,n}^{(i)}}|\nabla \varphi ^{(i)}_n|^2\,\mathrm{d}x\right) ^s\\&\le C\,|\Sigma _{k,2\,n}^{(i)}|^{1-s}\,|\Sigma _{k,2\,n}^{(i)}|^s\,n^{2\,s}\le C\,n^{2\,s-1}, \end{aligned} \end{aligned}$$

for a constant \(C>0\) independentFootnote 4 of n. In particular, for every \(i\in \{-(k-1),\ldots ,k-1\}\) we have

$$\begin{aligned} \lim _{n\rightarrow +\infty }[\varphi ^{(i)}_n]^2_{W^{s,2}(\mathbb{R}^2)}=0,\quad \text{if}\quad 0<s<\frac{1}{2}, \end{aligned}$$
(5.3)

while

$$\begin{aligned} \sup _{n\ge 1}\,[\varphi ^{(i)}_n]^2_{W^{s,2}(\mathbb{R}^2)}\le C,\quad \text{if}\quad s=\frac{1}{2}. \end{aligned}$$
(5.4)

From now on, for ease of notation, we denote

$$\begin{aligned} \Phi _{k,n}=\sum \limits _{i=-(k-1)}^{k-1}\varphi ^{(i)}_n\in C^\infty _0(Q_{2\,k}). \end{aligned}$$

Due to the different behaviors (5.3) and (5.4), we need to consider the cases \(0<s<1/2\) and \(s=1/2\) separately.

Case \(0<s<1/2\). For every \(u\in C^\infty _0(Q_k)\setminus \{0\}\), we simply take

$$\begin{aligned} u_{n}=\left( 1-\Phi _{k,n}\right) \,u, \end{aligned}$$

and observe that \(u_{n}\in C^\infty _0(\widetilde{Q}_k)\) for every \(n\in \mathbb{N}\setminus \{0\}\). Since each \(u_n\) is admissible for problem (1.3), we get

$$\begin{aligned} \sqrt{\lambda _1^s(\widetilde{Q}_k)}\le \frac{[u_{n}]_{W^{s,2}(\mathbb{R}^2)}}{\Vert u_n\Vert _{L^2(\widetilde{Q}_k)}}\le \frac{[u]_{W^{s,2}(\mathbb{R}^2)}+\Vert u\Vert _{L^\infty (\mathbb{R}^2)}\,\left[ 1-\Phi _{k,n}\right] _{W^{s,2}(\mathbb{R}^2)}}{\Vert u\,(1-\Phi _{k,n})\Vert _{L^2(\widetilde{Q}_k)}}, \end{aligned}$$
(5.5)

where in the last inequality we have used the Leibniz-type rule (2.1) and the fact \(|1-\Phi _{k,n}|\le 1\). We now observe that

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert u\,(1-\Phi _{k,n})\Vert _{L^2(\widetilde{Q}_k)}=\Vert u\Vert _{L^2(Q_k)}, \end{aligned}$$

which follows from a standard application of the Lebesgue Dominated Convergence Theorem, together with the properties of \(\Phi _{k,n}\). Moreover, it holds

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\left[ 1-\Phi _{k,n}\right] _{W^{s,2}(\mathbb{R}^2)}=0. \end{aligned}$$

This simply follows by using the definition of \(\Phi _{k,n}\), the triangle inequality and (5.3). By using these two limits in (5.5), we get

$$\begin{aligned} \sqrt{\lambda _1^s(\widetilde{Q}_k)}\le \lim _{n\rightarrow \infty }\frac{[u]_{W^{s,2}(\mathbb{R}^2)}+\Vert u\Vert _{L^\infty (\mathbb{R}^2)}\,\left[ 1-\Phi _{k,n}\right] _{W^{s,2}(\mathbb{R}^2)}}{\Vert u\,(1-\Phi _{k,n})\Vert _{L^2(\widetilde{Q}_k)}}=\dfrac{[u]_{W^{s,2}(\mathbb{R}^2)}}{\Vert u\Vert _{L^2(Q_k)}}. \end{aligned}$$

By arbitrariness of \(u\in C^\infty _0(Q_k)\setminus \{0\}\), we get

$$\begin{aligned} \lambda ^s_1(\widetilde{Q}_k)\le \lambda _1^s(Q_k). \end{aligned}$$

and thus the desired conclusion (5.2).

Borderline case \(s=1/2\). This is more delicate, we cannot use directly the sequence \(\{\Phi _{k,n}\}_{n\in \mathbb{N}\setminus \{0\}}\) to construct an approximation of \(u\in C^\infty _0(Q_k)\). Indeed, by owing to (5.4), we can now guarantee that \(\{\Phi _{k,n}\}_{n\in \mathbb{N}\setminus \{0\}}\) only converges weakly to 0 in \(W^{1/2,2}(\mathbb{R}^2)\) as n goes to \(\infty\), up to a subsequence.

In order to “boost” such a sequence, we make a suitable application of Mazur’s Lemma (see for example [24, Theorem 2.13]). More precisely, we define the sequence \(\{F_{k,n}\}_{n\in \mathbb{N}\setminus \{0\}}\subset L^2(\mathbb{R}^2\times \mathbb{R}^2)\), given by

$$\begin{aligned} F_{k,n}(x,y)=\frac{\Phi _{k,n}(x)-\Phi _{k,n}(y)}{|x-y|^{1+\frac{1}{2}}}. \end{aligned}$$

By construction, we have that

$$\begin{aligned} \Vert F_{k,n}\Vert _{L^2(\mathbb{R}^2\times \mathbb{R}^2)}=[\Phi _{k,n}]_{W^{\frac{1}{2},2}(\mathbb{R}^2)}\le C, \end{aligned}$$

and \(\{F_{k,n}\}_{n\in \mathbb{N}\setminus \{0\}}\) converges weakly to 0 in \(L^2(\mathbb{R}^2\times \mathbb{R}^2)\), up to a subsequence. Thanks to Mazur’s Lemma, we can enforce this weak convergence to the strong one, by passing to a sequence of convex combinations. More precisely, we know that for every \(n\in \mathbb{N}\setminus \{0\}\) there exists

$$\begin{aligned} \big \{t_\ell (n)\big \}_{\ell =1}^n\subset [0,1],\quad \text{such that} \quad \sum _{\ell =1}^n t_\ell (n)=1, \end{aligned}$$

and such that the new sequence made of convex combinations

$$\begin{aligned} \widetilde{F}_{k,n}(x,y)=\sum _{\ell =1}^n t_\ell (n)\,F_{k,\ell }(x,y), \end{aligned}$$

strongly converges in \(L^2(\mathbb{R}^2\times \mathbb{R}^2)\) to 0, as n goes to \(\infty\). Observe that by construction we have

$$\begin{aligned} \begin{aligned}&\Vert \widetilde{F}_{k,n}\Vert ^2_{L^2(\mathbb{R}^2\times \mathbb{R}^2)}\\&\quad =\left\| \sum _{\ell =1}^n t_\ell (n)\,F_{k,\ell }\right\| ^2_{L^2(\mathbb{R}^2\times \mathbb{R}^2)}\\&\quad =\iint _{\mathbb{R}^2\times \mathbb{R}^2} \left| \sum _{\ell =1}^n t_\ell (n)\frac{\Phi _{k,\ell }(x)-\Phi _{k,\ell }(y)}{|x-y|^{1+\frac{1}{2}}}\right| ^2\,\mathrm{d}x\,\mathrm{d}y\\&\quad =\iint _{\mathbb{R}^2\times \mathbb{R}^2} \frac{\left| \sum _{\ell =1}^n t_\ell (n)\,\Phi _{k,\ell }(x)-\sum _{\ell =1}^n t_\ell (n)\,\Phi _{k,\ell }(y)\right| ^2}{|x-y|^{3}}\,\mathrm{d}x\,\mathrm{d}y. \end{aligned} \end{aligned}$$

Thus, if we set

$$\begin{aligned} \widetilde{\Phi }_{k,n}=\sum _{\ell =1}^n t_\ell (n)\,\Phi _{k,\ell }\in C^\infty _0(Q_{2\,k}), \end{aligned}$$

the previous observations give that

$$\begin{aligned} \lim _{n\rightarrow \infty } [\widetilde{\Phi }_{k,n}]^2_{W^{\frac{1}{2},2}(\mathbb{R}^2)}=\lim _{n\rightarrow \infty }\Vert \widetilde{F}_{k,n}\Vert ^2_{L^2(\mathbb{R}^2\times \mathbb{R}^2)}=0. \end{aligned}$$
(5.6)

Moreover, by using the fractional Poincaré inequality with \(s=1/2\) for the open bounded set \(Q_{2\,k}\), we also have

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert \widetilde{\Phi }_{k,n}\Vert ^2_{L^2(Q_{2\,k})}\le \frac{1}{\lambda _{1}^\frac{1}{2}(Q_{2\,k})}\,\lim _{n\rightarrow \infty } [\widetilde{\Phi }_{k,n}]_{W^{\frac{1}{2},2}(\mathbb{R}^2)}^2=0. \end{aligned}$$
(5.7)

We take as in the previous case \(u\in C^\infty _0(Q_k)\setminus \{0\}\). In order to approximate u with functions compactly supported in \(\widetilde{Q}_k\), we now define

$$\begin{aligned} \widetilde{u}_n=(1-{\widetilde{\Phi }}_{k,n})\,u. \end{aligned}$$

We observe that this function belongs to \(C^\infty _0(\widetilde{Q}_k)\). Indeed, observe that

$$\begin{aligned} \Phi _{k,\ell }(x)=1,\quad \text{for every}\quad x\in \Sigma ^{(i)}_{k,4\,\ell },\, i\in \{-(k-1),\ldots ,k-1\}\quad \text{and}\quad \ell \in \{1,\ldots ,n\}, \end{aligned}$$

thus in particular

$$\begin{aligned} \widetilde{\Phi }_{k,n}(x)=\sum _{\ell =1}^n t_\ell (n)\,\Phi _{k,\ell }(x)=\sum _{\ell =1}^n t_\ell (n)=1,\quad \text{ for every}\quad x\in \Sigma ^{(i)}_{k,4\,n},\, i\in \{-(k-1),\ldots ,k-1\}, \end{aligned}$$

thanks to the fact that

$$\begin{aligned} \Sigma ^{(i)}_{k,4\,n}\subset \Sigma ^{(i)}_{k,4\,\ell },\quad \text{for}\quad \ell \in \{1,\ldots ,n\}. \end{aligned}$$

Clearly, we still have

$$\begin{aligned} |1-{\widetilde{\Phi }}_{k,n}|\le 1\quad \text{and}\quad \lim _{n\rightarrow \infty } \Vert \widetilde{u}_n\Vert _{L^2(\widetilde{Q}_k)}=\Vert u\Vert _{L^2(Q_k)}. \end{aligned}$$
(5.8)

The second fact in (5.8) can be proved by observing that

$$\begin{aligned} \begin{aligned}&\left| \int _{{\widetilde{Q}}_k} |\widetilde{u}_n|^2\,\mathrm{d}x-\int _{Q_k} |u|^2\,\mathrm{d}x\right| \\&\quad =\left| \int _{{\widetilde{Q}}_k} |u|^2\,\Big [|1-{\widetilde{\Phi }}_{k,n}|^2-1\Big ]\,\mathrm{d}x\right| \\&\quad \le \int _{{\widetilde{Q}}_k} |u|^2\,\Big [1-|1-{\widetilde{\Phi }}_{k,n}|^2\Big ]\,\mathrm{d}x\\&\quad \le 2\,\int _{{\widetilde{Q}}_k} |u|^2\,\Big [1-|1-{\widetilde{\Phi }}_{k,n}|\Big ]\,\mathrm{d}x\le 2\,\Vert u\Vert ^2_{L^\infty (Q_k)}\,\int _{{\widetilde{Q}}_k} |\widetilde{\Phi }_{k,n}|\,\mathrm{d}x, \end{aligned} \end{aligned}$$

and then using (5.7).

We can now use \({\widetilde{u}}_n\) as a competitor for the variational problem defining \(\lambda _1^s({\widetilde{Q}}_k)\) and proceed exactly as in the case \(0<s<1/2\), by using (5.6) and (5.8). This finally concludes the proof.

Remark 5.1

With the notation above, we obtain in particular that the infinite complement comb \(\Theta :=\mathbb{R}^2\setminus \Sigma\) is an open simply connected set such that

$$\begin{aligned} r_{\Theta }=\frac{\sqrt{5}}{2}\quad \text{and}\quad \lambda _1^s(\Theta )=0,\quad \text{for}\quad 0<s\le \frac{1}{2}. \end{aligned}$$

Indeed, by domain monotonicity and (5.1), we have

$$\begin{aligned} 0\le \lambda _1^s(\Theta )\le \lim _{k\rightarrow \infty } \lambda _1^s(\widetilde{Q}_k)=0. \end{aligned}$$

6 Some consequences

We highlight in this section some consequences of our main result, by starting with a fractional analogue of property (1.2) seen in the Introduction.

Corollary 6.1

Let \(\Omega \subset \mathbb{R}^2\) be an open simply connected set. Then we have:

  • for \(1/2<s<1\)

    $$\begin{aligned} \lambda _1^s(\Omega )>0 \Longleftrightarrow r_\Omega <+\infty ; \end{aligned}$$

  • for \(0<s\le 1/2\)

    $$\begin{aligned} \lambda _1^s(\Omega )>0 \Longrightarrow r_\Omega <+\infty , \end{aligned}$$

    but

    $$\begin{aligned} r_{\Omega} < + \infty \nRightarrow \lambda_{1}^{s}(\Omega )>0 \end{aligned}$$

Proof

Let \(0<s<1\) and assume that \(\lambda _1^s(\Omega )>0\). Let \(r>0\) be such that there exists \(x_0\in \Omega\) with \(B_r(x_0)\subset \Omega\). By using the monotonicity of \(\lambda _1^s\) with respect to set inclusion, we get

$$\begin{aligned} \lambda _1^s(\Omega )\le \lambda _1^s(B_r(x_0))=\frac{\lambda _1^s(B_1)}{r^{2\,s}}. \end{aligned}$$

The previous estimate gives

$$\begin{aligned} r<\left( \frac{\lambda _1^s(B_1)}{\lambda _1^s(\Omega )}\right) ^\frac{1}{2\,s}. \end{aligned}$$

By taking the supremum over admissible r, we get \(r_\Omega <+\infty\) by definition of inradius.

For the converse implication in the case \(s>1/2\), it is sufficient to apply Theorem 1.1. Finally, by taking \(\Theta\) as in Remark 5.1, we get an open set with finite inradius, but vanishing \(\lambda _1^s\) for \(0<s\le 1/2\). \(\square\)

Our main results permit to compare two different Sobolev spaces, built up of functions “vanishing at the boundary.” More precisely, let us denote by \(\mathcal{D}^{s,2}_0(\Omega )\) the completion of \(C^\infty _0(\Omega )\) with respect to the norm

$$\begin{aligned} u\mapsto [u]_{W^{s,2}(\mathbb{R}^N)},\quad \text{for every}\quad u\in C^\infty _0(\Omega ). \end{aligned}$$

Observe that this is indeed a norm on \(C^\infty _0(\Omega )\). We refer to [11] for more details on this space. We also recall that by \({\widetilde{W}}^{s,2}_0(\Omega )\) we denote the closure of \(C^\infty _0(\Omega )\) in \(W^{s,2}(\mathbb{R}^N)\).

We have the following

Corollary 6.2

Let \(1/2<s<1\) and let \(\Omega \subset \mathbb{R}^2\) be an open simply connected set, with finite inradius. Then

$$\begin{aligned} \mathcal{D}^{s,2}_0(\Omega )={\widetilde{W}}^{s,2}_0(\Omega ). \end{aligned}$$

On the contrary, for \(0<s\le 1/2\) and \(\Theta\) the infinite complement comb of Remark 5.1, the two spaces

$$\begin{aligned} \mathcal{D}^{s,2}_0(\Theta ) \quad \text{and}\quad \widetilde{W}^{s,2}_0(\Theta ), \end{aligned}$$

cannot be identified with each other.

Proof

For \(1/2<s<1\) and an open simply connected set \(\Omega \subset \mathbb{R}^2\), by Theorem 1.1 the two norms

$$\begin{aligned} {[}u]_{W^{s,2}(\mathbb{R}^2)}\quad \text{and}\quad \Vert u\Vert _{W^{s,2}(\mathbb{R}^2)}, \end{aligned}$$

are equivalent on \(C^\infty _0(\Omega )\). This proves the first point.

As for the second statement, it is sufficient to observe that \({\widetilde{W}}^{s,2}_0(\Theta )\) is always continuously embedded in \(L^2(\Theta )\), by its very definition. On the other hand, for \(0<s\le 1/2\) such an embedding does not hold for \(\mathcal{D}^{s,2}_0(\Theta )\), since \(\lambda _1^s(\Theta )=0\) by Remark 5.1. \(\square\)

We now show how Theorem 1.1 implies some fractional versions of the classical Cheeger’s inequality, a fundamental result in Spectral Geometry. At this aim, for an open set \(\Omega \subset \mathbb{R}^N\) we recall the definition of Cheeger constant

$$\begin{aligned} h_1(\Omega )=\inf \left\{ \frac{P(E)}{|E|}\, :\, E\subset \Omega\, \text{bounded and measurable with }\, |E|>0\right\} , \end{aligned}$$

and \(s-\)Cheeger constant (for \(0<s<1\))

$$\begin{aligned} h_s(\Omega )=\inf \left\{ \frac{P_s(E)}{|E|}\, :\, E\subset \Omega \text{ bounded and measurable with } |E|>0\right\} , \end{aligned}$$

see [9] for some properties of this constant. Here P stands for the perimeter of a set in the sense of De Giorgi, while \(P_s\) is the \(s-\)perimeter of a set, defined by

$$\begin{aligned} P_s(E)=[1_E]_{W^{s,1}(\mathbb{R}^N)}=\iint _{\mathbb{R}^N\times \mathbb{R}^N} \frac{|1_E(x)-1_E(y)|}{|x-y|^{N+s}}\,\mathrm{d}x\,\mathrm{d}y, \end{aligned}$$

for any measurable set \(E\subset \mathbb{R}^N\). Then we have the following

Corollary 6.3

(Fractional Cheeger inequality) Let \(1/2<s<1\) and let \(\Omega \subset \mathbb{R}^2\) be an open simply connected set, with finite inradius. Then we have

$$\begin{aligned} \lambda _1^s(\Omega )\ge \mathcal{C}_s\left( \frac{h_1(\Omega )}{2}\right) ^{2\,s}, \end{aligned}$$

and

$$\begin{aligned} \lambda _1^s(\Omega )\ge \mathcal{C}_s\,\left( \frac{\pi }{P_s(B_1)}\,h_s(\Omega )\right) ^2. \end{aligned}$$

where \(\mathcal{C}_s\) is the same constant as in Theorem 1.1.

Proof

Let \(r<r_\Omega\), by definition of inradius there exists a disk \(B_r(x_0)\subset \Omega\). By using this disk as a competitor for the minimization problem defining \(h_1(\Omega )\), we get

$$\begin{aligned} h_1(\Omega )\le \frac{2\,\pi \,r}{\pi \,r^2}=\frac{2}{r}. \end{aligned}$$

By taking the supremum over admissible r, we get

$$\begin{aligned} h_1(\Omega )\le \frac{2}{r_\Omega }. \end{aligned}$$

By raising to the power \(2\,s\) and using Theorem 1.1, we get the first inequality. The second one can be obtained in exactly the same way. \(\square\)

Finally, we have the following result, which permits to compare \(\lambda _1^s(\Omega )\) and \(\lambda _1(\Omega )\), for simply connected sets in the plane. We refer to [12, Theorem 6.1] and [15, Theorem 4.5] for a similar result in general dimension \(N\ge 2\), under stronger regularity assumptions on the sets.

Corollary 6.4

(Comparison of eigenvalues) Let \(1/2<s<1\) and let \(\Omega \subset \mathbb{R}^2\) be an open simply connected set, with finite inradius. Then we have

$$\begin{aligned} \alpha _s\,\Big (\lambda _1(\Omega )\Big )^s\le \lambda _1^s(\Omega )\le \beta _s\,\Big (\lambda _1(\Omega )\Big )^s, \end{aligned}$$
(6.1)

where \(\alpha _s,\beta _s\) are two positive constants depending on s only, such that

$$\begin{aligned}&\alpha _s\sim \left( s-\frac{1}{2}\right) ,\quad \text{for}\quad s\searrow \frac{1}{2},\quad \text{and}\quad \alpha _s\sim \frac{1}{1-s},\quad \text{for}\quad s\nearrow 1,\\&\beta _s\sim \frac{1}{1-s},\quad \text{for}\quad s\nearrow 1. \end{aligned}$$

Proof

The upper bound follows directly from the general result of [12, Theorem 6.1], see Eq. (6.1) there. From this reference, we can also extract a value for the constant \(\beta _s\), which is given by

$$\begin{aligned} \beta _s=\frac{4^{1-s}}{s\,(1-s)}\,\pi . \end{aligned}$$

For the lower bound, the proof is similar to that of Corollary 6.3, it is sufficient to join the estimate

$$\begin{aligned} \lambda _1(\Omega )\le \frac{\lambda _1(B_1)}{r_\Omega ^2}, \end{aligned}$$

with Theorem 1.1. This gives the claimed estimate, with constant

$$\begin{aligned} \alpha _s=\frac{\mathcal{C}_s}{(\lambda _1(B_1))^s}, \end{aligned}$$

and \(\mathcal{C}_s\) is the same as in (1.4). \(\square\)

Remark 6.5

The lower bound in estimate (6.1) degenerates as s approaches 1/2. This behavior is optimal: indeed, observe that for the set \(\Theta\) of Remark 5.1 we have

$$\begin{aligned} \lambda _1(\Theta )>0\quad \text{and}\quad \lambda _1^s(\Theta )=0,\quad \text{for}\quad 0<s\le \frac{1}{2}. \end{aligned}$$

The first fact follows from the classical Makai–Hayman inequality (1.1), for example. Thus the lower bound cannot hold for this range of values.