# An Orlicz-Besov Poincaré Inequality via John Domains

Authors: Hongyan Sun
Journal: Journal of Function Spaces
 (2019)

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## Abstract

Denote byB˙⁎α,ϕ(Ω) the intrinsic Orlicz-Besov space, where α∈R, ϕ is a Young function, and Ω⊂Rn is a domain. For α∈(-n,0) and optimal ϕ, via John domains, we establish criteria for bounded domains Ω⊂Rn supporting an Orlicz-Besov Poincaré inequality. ‖u-uΩ‖Ln/|α|(Ω)≤C‖u‖B˙⁎α,ϕ(Ω)∀u∈B˙⁎α,ϕ(Ω). This extends the known criteria for bounded domains supporting Sobolev-Poincaré inequality and its fractional analogue.


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## Body

## 1. Introduction

Letn≥2 in this paper. Recall that a bounded domain Ω⊂Rn is called aJohn domain with respect to some x0∈Ω and C>0 if, for each x∈Ω, there is a rectifiable curve γ:[0,l]→Ω parameterized by arc-length (called aJohn curve for x) such that γ(0)=x, γ(l)=x0, and d(γ(t),Ω∁)>Ct for all t∈[0,l].It is well known that John domains essentially characterize bounded domains ofRn supporting Poincaré-Sobolev inequality and some fractional analogue in the literature. Precisely, Bojarski [1] proved that, for any 1≤p<n, a John domain Ω⊂Rn always supports the following Sobolev-Poincaré inequality: there exists a constant C≥1 such that (1)u-uΩLnp/n-pΩ≤C∇uLpΩwhenever the distributional derivative ∇u∈Lp(Ω). In this paper, uE is the average of u in set E with 0<|E|<∞. When 0<s<1 and 1≤p<n/s, as proved by Hurri-Syrjänen and Vähäkangas [2] (see also Dyda, Ihnatsyeva, and Vähäkangas [3]), a John domain Ω⊂Rn always supports the following fractional Sobolev-Poincaré inequality: there exists a constant C≥1 such that (2)u-uΩLnp/n-psΩ≤C∫Ω∫x-y<1/2distx,∂Ωux-uypx-yn+spdxdy1/pwhenever u∈L1(Ω). Moreover, as proved by Zhou [4], when 0<s<1 and p∈[1,n/s), a John domain Ω⊂Rn always supports the following Hajłasz-Sobolev Poincaré inequality: there exists a constant C≥1 such that(3)infc∈R⁡u-cLnp/n-psΩ≤CgLpΩ,whenever g is an intrinsic s-order Hajłasz upper gradient of u; that is,(4)ux-uy≤x-ysgx+gyfor almost allx∈Ωand  almost ally∈Bx,12distx,∂Ω.Conversely, suppose that Ω is a bounded simply connected planar domain and a bounded domain that is quasiconformally equivalent to some uniform domain when n≥3. Buckley-Koskela [5] proved that if Ω satisfies (1), then Ω is a John domain. If either (3) or (2) holds, then Ω is also proven to be a John domain; see [2–4].The main purpose is to prove that John domains essentially characterize bounded domains ofRn supporting an Orlicz-Besov Poincaré inequality as below. Let ϕ be a Young function in [0,∞); that is, ϕ∈C([0,∞)) is convex and satisfies ϕ(0)=0, ϕ(t)>0 for t>0 and limt→∞⁡ϕ(t)=∞. Given any α∈R and domain Ω⊂Rn, as motivated by the Orlicz-Besov spaces considered in [6] and also semi-norms that appeared in the right hand side of (2) and (3), we consider the intrinsic Orlicz-Besov space B˙∗α,ϕ(Ω), which is defined as the collection of all measurable functions u in Ω whose (semi-)norms (5)uB˙∗α,ϕΩ≔infλ>0:∫Ω∫x-y<1/2distx,∂Ωϕux-uyλx-yαdxdyx-y2n≤1are finite. Modulo constant function B˙∗α,ϕ(Ω) is a Banach space. In the spirit of (1), (2), and (3), we say that a bounded domain Ω supports an Orlicz-Besov Poincaré inequality with respect to α∈(-n,0) and ϕ if there exists a constant C≥1 depending on n, α, ϕ, and Ω such that (6)u-uΩLn/αΩ≤CuB˙∗α,ϕΩ∀u∈B˙∗α,ϕΩ.Theorem 1.
Letα∈(-n,0) and ϕ be a Young function satisfying (7)Λ_ϕα≔supx>0⁡∫01ϕt1-αxϕxdttn+1<∞,(8)Λ¯ϕα≔supx>0⁡∫1∞ϕt-αxϕxdttn+1<∞.
(i) IfΩ⊂Rn is a John domain, then Ω supports the Orlicz-Besov Poincaré inequality (6).
(ii) Assume thatΩ⊂Rn is a bounded simply connected planar domain, or a bounded domain that is quasiconformally equivalent to some uniform domain when n≥3. If Ω supports the Orlicz-Besov Poincaré inequality (6), then Ω is a John domain.The assumption (7) above guarantees Cc1(Ω)⊂B˙∗α,ϕ(Ω) and hence the nontriviality of B˙∗α,ϕ(Ω); see Lemma 4. For the optimality of (7), see Remark 5. The assumption (8) always fails if α≤-n. Indeed, since ϕ∈C([0,∞)) is convex, ϕ(0)=0, and ϕ(t)>0 for all t>0, there exists c>0 such that ϕ(t)-ϕ(1)>c(t-1) for all t∈[0,∞). Then we always have (9)Λ¯ϕα≥supx>0⁡∫1∞ϕ1+ct-αx-1ϕxdttn+1=supx>0⁡ϕ1-cnϕx+cxϕx∫1∞dttn+α+1=∞.For α∈(-n+1,0), Young functions satisfying (7) and (8) include tp[ln(1+t)]γ with p∈(n/(1-α),n/|α|) and γ>1 or γ=0, tn/(1-α)[ln(1+t)]γ with γ>1 and also their convex combinations. For α∈(-n,-n+1], Young functions satisfying (7) and (8) include tp[ln(1+t)]γ with p∈(1,n/|α|) and γ≥1 or γ=0 and also their convex combinations.Moreover, for anys∈(0,1) and 1≤p<n/s, let α=s-n/p and ϕ(t)=tp. Then α∈(-n,0), np/(n-ps)=n/|α| and ϕ satisfies (7) and (8). In this case, we always have (10)uB˙∗α,ϕΩ=∫Ω∫x-y<1/2distx,∂Ωux-uypx-yn+spdxdy1/p.We see that (6) coincides with (2). Thus, Theorem 1 extends above criteria for bounded domains supporting fractional Poincaré inequality given in [2, 3] (see also [4]).We prove Theorem1 in Section 3. To prove Theorem 1 (i), we use Boman’s chain rule for John domains and an Orlicz-Besov imbedding in cubes under (7) and (8) as proved by [7] (see Lemma 8) and also borrow some ideas from [1, 4]. To prove Theorem 1 (ii), we use (7) and (8) to calculate the B˙∗α,ϕ(Ω)-norms of some cut-off functions in a precise way; see Lemmas 6 and 7. This allows us to prove the LLC(2) property of domains supporting inequality (6); see Proposition 9. By borrowing some ideas from [4, 5, 8], we then obtain Theorem 1 (ii).Here we make some conventions on the notations or notion used in this paper. Throughout the paper,C denotes positive constants, which depends only on n,α,ϕ,Ω but whose value might be changed from line to line. We write A≲(≳)B if there exists a constant C>0 such that A≤(≥)CB. For any x∈Rn and A⊂Rn, dist(x,A) denotes the distance from x to A and diamA denotes the diameter of the set A.


## 2. Some Basic Properties

The following properties of Young functions are given in [7, Lemma2.1].Lemma 2.
Letα∈(-n,0) and ϕ be a Young function.
(i) Ifϕ satisfies (7), then (11)ϕxs≤22nΛ_ϕαϕ21-αxsn/1-α∀0<s≤1,x>0.
(ii) Ifϕ satisfies (8), then ϕ(xs-α)s-n→0 as s→∞ for any x>0 and (12)ϕxs≤23nΛ¯ϕαϕxs-n/α∀s≥1,x>0.Next we show that intrinsic Orlicz-Besov functions are locally integrable.Lemma 3.
Letα∈(-n,0) and ϕ be a Young function. For any bounded domain Ω⊂Rn, we have B˙∗α,ϕ(Ω)⊂Lloc1(Ω) as sets.Proof.
Letu∈B˙∗α,ϕ(Ω). For λ>uB˙∗α,ϕΩ, we have (13)∫Ω∫x-y<1/2distx,∂Ωϕux-uyλx-yαdydxx-y2n≤1.By Fubini’s theorem, for almost all x∈Ω we have (14)∫x-y<1/2distx,∂Ωϕux-uyλx-yαdyx-y2n<∞.For any ball B with 8B⋐Ω, one may choose such x satisfying above inequality so that (15)B⊂Bx,12distx,∂Ω∖Bx,116distx,∂Ω.Thus this implies that (16)∫Bϕux-uy16-αλdistx,∂Ωαdy≤2-2ndistx,∂Ω2n∫x-y<1/2distx,∂Ωϕux-uyλx-yαdyx-y2n<∞.By Jensen’s inequality, we have (17)ϕ1B∫Bux-uy16-αλdistx,∂Ωαdy<∞.Notice that ϕ(t)→∞ as t→∞. Then (18)∫Bux-uy16-αλdistx,∂Ωαdy<∞,which implies that (19)⨏Buydy≤ux+⨏Bux-uydy<∞.That is, u∈L1(B) as desired. The proof of Lemma 3 is complete.The assumption (7) guarantees the nontriviality of intrinsic Orlicz-Besov spaces in bounded domains.Lemma 4.
Letα∈(-n,0) and ϕ be a Young function satisfying (7). For any bounded domain Ω⊂Rn, we have Cc1(Ω)⊂B˙∗α,ϕ(Ω) as sets.Proof.
Letu∈Cc1(Ω) and L=DuL∞(Ω). By (7) we have (20)∫Ω∫w-z<1/2distw,∂Ωϕuz-uwλz-wαdzdwz-w2n≤ωn∫Ω∫01/2diamΩϕt1-αλ/Ldttn+1dw≤ωnΩ12diamΩ-n∫01ϕ1/2diamΩ1-αt1-αλ/Ldttn+1≤Λ_ϕαωnΩ12diamΩ-nϕ1/2diamΩ1-αLλ.Since the right hand side is less than 1 when λ>0 is sufficiently large enough, we know that ‖u‖B˙∗α,ϕ(Ω)<∞; that is, u∈B˙∗α,ϕ(Ω) as desired. The proof of Lemma 4 is complete.Remark 5.
We remark that the assumption (7) is optimal to get Cc1(Ω)⊂B˙α,ϕ(Ω) or the nontrivial of B˙α,ϕ(Ω) in the following sense. For α≥1-n and p≥1, let ϕ(t)=tp. By a direct calculation, the Young function ϕ(t) satisfies (7) if and only if p>n/(1-α). On the other hand, let s=n/p+α. If p≤n/(1-α), then s≥1, and observe that (10) also holds for all measurable functions u in Ω. For any ball B, write B˙pps(B) as the space of all functions satifying (21)uB˙ppsB≔∫B∫Bux-uypx-yn+spdxdy1/p<∞.By [9, Theorem4.1&4.2] and s≥1, we know that B˙pps(B) only contains constant functions. If u∈B˙∗α,ϕ(Ω), then u∈B˙pps(B) for any ball B with 2B⊂Ω. Thus u is a constant on every ball B with 2B⊂Ω, and hence by the connectedness of Ω, u is a constant function in Ω.
To end this section, we calculateB˙∗α,ϕ(Ω)-norms of some special functions, which will be used in Sections 2 and 3.
Letz∈Ω and dist(z,∂Ω)≤s<diamΩ. If Ω∖B(z,s)¯ is connected, we write Ωz,s=Ω∖B(z,s)¯; if Ω∖B(z,s)¯ is disconnected, we let Ωz,s be any component of Ω∖B(z,s)¯. For t>r≥s with Ωz,s∖B(z,t)¯≠∅, define a function vz,s,r,t in Ω by (22)vz,s,r,ty=0,y∈Ω∖Ωz,s∖Bz,r;y-z-rt-r,y∈Ωz,s∩Bz,t∖Bz,r;1,y∈Ωz,s∖Bz,t.Lemma 6.
Letα∈(0,n) and ϕ be a Young function satisfying (7) and (8). For any bounded domain Ω⊂Rn and z∈Ω with dist(z,∂Ω)≤s, if Ωz,s is a component Ω∖B(z,s)¯, t>r≥s with Ωz,s∖B(z,t)¯≠∅, we have (23)vz,s,r,tB˙∗α,ϕΩ≤Ct-r-αϕ-1t-rnΩz,s∖Bz,r-1.Here C≥1 is a constant depending only on n, ϕ, and α.Proof.
Writev=vz,r,t for simplicity. For any x∈Ω and y∈B(x,1/2dist(x,∂Ω)), |v(x)-v(y)|≠0 implies that at least one of x,y lies in Ωz,s∖B(z,r). Thus(24)H≔∫Ω∫Bx,1/2distx,∂Ωϕvx-vyλx-yαdyx-y2ndx≤2∫Ωz,s∖Bz,r∫Ωϕvx-vyλx-yαdyx-y2ndx.Since |v(x)-v(y)|≤|x-y|/(t-r) for all x,y∈Ω with |x-y|≤t-r, and |v(x)-v(y)|≤1 for all x,y∈Ω with |x-y|≥t-r, we further write (25)H≤2∫Ωz,s∖Bz,r∫Bx,t-rϕx-y1-αλt-rdyx-y2ndx+2∫Ωz,s∖Bz,r∫Rn∖Bx,t-rϕx-y-αλdyx-y2ndx≕H1+H2.
By a change of variables and applying (7), we obtain (26)H1≤2∫Ωz,s∖Bz,r∫Bx,t-rϕx-y1-αλt-rdyx-y2ndx=2ωnΩz,s∖Bz,r∫0t-rϕs1-αλt-rdssn+1=2ωnΩz,s∖Bz,rt-rn∫01ϕs1-αλt-rαdssn+1≤2Λ_ϕαωnΩz,s∖Bz,rt-rnϕ1λt-rα.Similarly, by a change of variables and applying (8), we obtain (27)H2≤2∫Ωz,s∖Bz,r∫Rn∖Bx,t-rϕx-y-αλdyx-y2ndx=2ωnΩz,s∖Bz,r∫t-r∞ϕs-αλdssn+1=2ωnΩz,s∖Bz,rt-rn∫1∞ϕs-αλt-rαdssn+1≤2Λ¯ϕαωnΩz,s∖Bz,rt-rnϕ1λt-rα.Therefore,(28)H≤H1+H2≤2Λ_ϕα+Λ¯ϕαωnΩz,s∖Bz,rt-rnϕ1λt-rα.If (29)λ≥λ0≔2Λ_ϕα+Λ¯ϕαωn+1t-r-αϕ-1t-rnΩz,s∖Bz,r-1>0,by the convexity of ϕ, we have H≤1. Thus ‖v‖B˙α,ϕ(Ω)≤λ0 as desired. The proof of Lemma 6 is complete.Letx0∈Ω, z∈Ω, and r>0 such that dist(z,∂Ω)<r≤|x0-z|. Define (30)wx0,z,ry=1rinfγx0,ylγ∩Bz,r,∀y∈Ω,where the infimum is taken over all rectifiable curves γ joining x0 and y.Lemma 7.
Letα∈(-n,0) and ϕ be a Young function satisfying (7) and (8). There exists a constant C such that, for all bounded domains Ω⊂Rn, x0∈Ω, and z∈Ω satisfying dist(z,∂Ω)<r<|x0-z|, we have (31)wx0,z,rB˙∗α,ϕΩ≤Cr-αwheneverΩ∖Bz,r¯isdisconnected.Proof.
Assume thatΩ∖B(z,r)¯ is disconnected. One has dist(z,∂Ω)≤r. Note that if x∈Ω∖B(z,10r) and y∈B(x,1/2dist(x,∂Ω)), we have wx0,z,r(x)=wx0,z,r(y). Indeed, for such x,y, we have (32)distx,∂Ω≤x-z+distz,∂Ω≤x-z+rand hence (33)y-z>x-z-y-x>x-z-12x-z-r2>12x-z-r2>4r.Therefore, B(x,1/2dist(x,∂Ω))∩B(z,2r)=∅. Letting γx,y be the line segment joining x,y contained in B(x,1/2dist(x,∂Ω)), we have γx,y⊂Ω∖B(z,r). For any γx0,x, we know that γx0,x∪γx,y gives a curve joining x0 and y and (34)lγx0,x∪γx,y∩Bz,r=lγx0,x∩Bz,r.Thus we have wx0,z,r(y)≤wx0,z,r(x). Similarly, wx0,z,r(x)≤wx0,z,t(y).
For anyx,y∈Ω, we have (35)wx0,z,rx-wx0,z,ry≤1rx-y.Indeed, letting γx,y be the line segment joining x,y, we have l(γx,y∩B(z,s))≤|x-y|. Since γx0,x∪γx,y connects x0 and y for any γx0,x, we have (36)wx0,z,ry≤wx0,z,rx+1rx-y.Similarly, by changing the roles of x,y, we also have (37)wx0,z,rx≤wx0,z,ry+1rx-y.
Therefore, we obtain(38)H≔∫Ω∫x-y≤1/2distx,∂Ωϕwx0,z,rx-wx0,z,ryλx-yαdxdyx-y2n≤∫Ω∩Bz,10r∫x-y≤1/2distx,∂Ωϕx-y1-αλrdyx-y2ndx. Note that, for x∈Ω∩B(z,10r), one has (39)distx,∂Ω≤10r+distz,∂Ω<12r.By (7), we obtain (40)H≤∫Ω∩Bz,10rωn∫06rϕt1-αλrdttn+1dx≤Mϕ61-αr-αλfor some M≥1. Thus for any λ>61-αM[ϕ-1(1)]-1r-α, by Jensen’s inequality, we have H≤1, which implies that vB˙∗α,ϕ(Ω)≤Cr-α as desired.


## 3. Proof of Theorem1

With the aid of the following result given in [7, Theorem1.1], we are able to prove Theorem 1 (i) as below. For any cube Q⊂Rn and any measurable functions u in Q, we write (41)uB˙α,ϕQ≔infλ>0:∫Q∫Qϕux-uyλx-yαdxdyx-y2n≤1.Lemma 8.
Letα∈(-n,0) and ϕ be a Young function satisfying (7) and (8). Then there exists a constant C≥1 depending on n,α,ϕ such that (42)u-uQLn/αQ≤CuB˙α,ϕQ∀cubesQ⊂Rnandu∈B˙α,ϕQ.Proof of Theorem1.
(i) Assume thatΩ is a bounded John domain. By Boman [10] and Buckley et al. [11], Ω enjoys the following chain property: for every integer κ>1, there exist a positive constant C(κ,Ω) and a collection F of cubes such that
(a)Q⊂κQ⊂Ω for all Q∈F, Ω=⋃Q∈FQ and (43)∑Q∈FχκQ≤Cκ,ΩχΩ;
(b)Q0∈F is fixed cube; for any other Q∈F, one can find a sequence {Qj}j=1N from F satisfying that, for all 0≤j≤N-1, we have (44)Q=QN⊂Cκ,ΩQj,Qj+1Cκ,Ω≤Qj≤Cκ,ΩQj+1,Qj∩Qj+1≥min⁡Qj,Qj+1Cκ,Ω.
Setκ=5n. By Ω⊂⋃Q∈FQ given in (a), we write (45)∫Ωuz-uQ0n/αdz≤∑Q∈F∫Quz-uQ0n/αdz≲∑Q∈F∫Quz-uQn/αdz+∑Q∈FQ·uQ-uQ0n/α≕I1+I2.
By Lemma8, we have (46)∫Quz-uQn/αdz≤CuB˙α,ϕQn/α,and hence (47)I1≤C∑Q∈FuB˙α,ϕQn/α.Let λQ=uB˙α,ϕQ, for λ=1/M(∑Q∈FλQn/|α|)|α|/n with M=23nΛ¯ϕ(α)C(κ,Ω)+1. For any x,z∈Q, by 2nQ⊂κQ⊂Ω as in (a), we have |x-z|≤1/2dist(x,∂Ω). Thus by (12) and convexity of ϕ, we get (48)H≔∫Ω∫x-y<1/2distx,∂ΩϕMux-uy∑P∈FλPn/αα/nx-yαdxdyx-y2n≥1Cκ,Ω∑Q∈F∬Q×QϕMux-uyx-yαλQλQ∑P∈FλPn/αα/ndxdyx-y2n≥1Cκ,Ω123nΛ¯ϕα∑Q∈FλQn/α∑P∈FλPn/α∬Q×QϕMux-uyx-yαλQdxdyx-y2n≥MCκ,Ω123nΛ¯ϕα∑Q∈FλQn/α∑P∈FλPn/α∬Q×Qϕux-uyx-yαλQdxdyx-y2n=MCκ,Ω123nΛ¯ϕα>1.We then obtain (49)∑Q∈FuB˙α,ϕQn/α≤MuB˙∗α,ϕΩn/α,and hence I1≤C(uB˙∗α,ϕ(Ω))n/|α|.
To estimateI2, for each Q∈F write (50)uQ-uQ0≤∑j=0N-1uQj-uQj∩Qj+1+∑j=0N-1uQj∩Qj+1-uQj+1.By (b), Hölder inequality, and Lemma 8, we have (51)uQj-uQj∩Qj+1≤⨏Qj∩Qj+1uQj-uzdz≤Cκ,Ω⨏QjuQj-uzdz≤Cκ,Ω⨏QjuQj-uzn/αdzα/n≤CuB˙α,ϕQjQjα/n.Since Q=QN⊂C(κ,Ω)Qj, j=0,1,…,N given in (b), we have (52)uQ-uQ0≤C∑j=0NuB˙α,ϕQjQjα/n≤C∑P∈F,Q⊂Cκ,ΩPPα/nuB˙α,ϕP.Therefore, by (a), (53)I2≤C∑Q∈FQ∑P∈F,Q⊂Cκ,ΩPPα/nuB˙α,ϕPn/α≤C∑Q∈F∫Q∑P∈FPα/nuB˙α,ϕPχCκ,ΩPxn/αdx≤C∫Ω∑P∈FPα/nuB˙α,ϕPχCκ,ΩPxn/αdx.Recall that by Lemma 4.2 in [1] (see also [12]), for any N>1 and p>1, we have (54)∑Q∈FaQχNQLpΩ≤Cp,N∑Q∈FaQχQLpΩ.Thus, (55)I2≤C∫Ω∑P∈FPα/nuB˙α,ϕPχPxn/αdx
Since (a) implies the bounded overlaps of cubes inF, by α<0, we have (56)∑P∈FPα/nuB˙α,ϕPχPxn/α≤C∑P∈FPα/nuB˙α,ϕPχPxn/α=C∑P∈F1PuB˙α,ϕPn/αχPxfor all x∈Ω. From this, it follows that(57)I2≤C∫Ω∑P∈F1PuB˙α,ϕPn/αχPxdx=C∑P∈FuB˙α,ϕPn/αBy an argument same as I1, we have I2≤C‖u‖B˙∗α,ϕ(Ω).
Combining the estimates forI1 and I2 together, we arrive at (58)∫Ωuz-uQ0n/αdz≤CuB∗α,ϕΩn/α.Since this yields u∈L1(Ω) and (59)∫Ωuz-uΩn/αdz≤C∫Ωuz-uQ0n/αdzwe obtain (6) as desired. This completes the proof of Theorem 1 (i).To prove Theorem1 (ii), we need the following result.Proposition 9.
IfΩ is a bounded domain supporting the B˙∗α,ϕ-imbedding, then Ω has the LLC(2) property; that is, there exists a positive constant b∈(0,1) depending only on α,n,ϕ, and Ω such that, for all z∈Rn and r>0, any pair of points in Ω∖B(z,r)¯ can be joined in Ω∖B(z,br)¯.Proof.
Fix a pointx0∈Ω so that (60)r0≔max⁡distx,∂Ω:x∈Ω=distx0,∂Ω.We claim that there exists b0∈(0,1) such that x,x0 are contained in the same component of Ω∖B(z,b0r)¯ whenever x,x0∈Ω∖B(z,r)¯ for some z∈Ω and r∈(0,2diamΩ).
Note that this claim gives LLC(2) properties ofΩ. Indeed, assume that this claim holds for the moment. First, letting b1=min{1/2,b0/4,b0(1-b0)/2}, we show that x,x0 are contained in the same component of Ω∖B(z,b1r)¯ whenever x,x0∈Ω∖B(z,r)¯ for some z∈Rn and r>0. Indeed, if r>2diamΩ, then, for any y∈B(z,r/2)¯, we have (61)y-x0≥x0-z-z-y≥r-r2=r2>diamΩ,and hence, by Ω⊂B(x0,diamΩ), we have Ω∩B(z,r/2)¯=∅; that is, x,x0 are contained in the connected set Ω=Ω∖B(z,r/2)¯. If r≤2diamΩ and dist(z,Ω)>b0r/4, x,x0 are contained in the connected set Ω=Ω∖B(z,b0r/4)¯. If r≤2diamΩ and dist(z,Ω)≤b0r/4, letting w∈B(z,b0r/2)∩Ω, by B(w,(1-b0/2)r)⊂B(z,r), we see that x,x0∈Ω∖B(w,(1-b0/2)r)¯. By the above claim, x,x0 are contained in the same component of Ω∖B(w,b0(1-b0/2)r)¯. Noting that (62)Bz,b01-b0r2=Bz,b01-b02r-b0r2⊂Bw,b01-b02r,we see that x,x0 are contained in the same component of Ω∖B(z,b0(1-b0)r/2)¯ as desired.
Letx,y∈Ω∖B(z,r)¯ for some z∈Rn and r>0. If |x0-z|>r0/8diamΩr, then x0∈Ω∖B(z,r0/8diamΩr)¯. Since x∈Ω∖B(z,r0/8diamΩr)¯, by the above argument, we know that x0,x∈Ωx0 for some component Ωx0 of Ω∖B(z,b1r0/8diamΩr)¯. Similarly, since y∈Ω∖B(z,r0/8diamΩr), x0,y∈Ωx0. Thus x,y∈Ωx0. If |x0-z|≤r0/8diamΩr, then r-r0/8diamΩr≤diamΩ, which implies r≤2diamΩ, and hence r0/4diamΩr≤r0/2. Thus, (63)Bz,r08diamΩr⊂Bx0,r04diamΩr⊂Bx0,r02⊂Bx0,r0⊂Ω,which implies Ω∖B(z,r0/8diamΩr)¯ is a connected set. Note that x,y∈Ω∖B(z,r0/8diamΩr)¯. This gives the desired LLC(2) properties of Ω.
In the following, we prove above claim by findingb0. Assume that x,x0∈Ω∖B(z,r)¯ for some z∈Ω and 0<r<2diamΩ. Set (64)bx,z,r≔sup⁡c∈0,1,x,x0are  contained  in  the  same  component  ofΩ∖Bz,cr¯.Without loss of generality, we assume that bx,z,r≤1/10; otherwise we take b0=1/10. Ωx denotes the component of Ω∖B(z,2bx,z,rr) containing x. To find b0, it suffices to prove (65)1C112-2bx,z,rr≤Ωx1/n≤C12bx,z,rr,where C1≥1 is a constant independent of x,z,r. Indeed, set b0=1/[10(2C12+1)]>0. By (65), we have bx,z,r>1/[4(C12+1)]>b0, and hence x,x0 are contained in the same component of Ω∖B(z,b0r)¯ as desired.
To prove (65), write c0=2bx,z,r<1/5. We also denote by Ωx0 the component of Ω∖B(z,c0r)¯ containing x0. Then (66)Bx0,12r0⊂Ωx0⊂Ω∖Ωx.Indeed, by the definition of bx,z,r, r0≤maxy∈B(z,c0r)∩Ω⁡|x0-y|≤r+c0r+dist(x0,B(z,r))≤6/5r+dist(x0,B(z,r)). Note that d(x0,B(z,c0r))≥d(x0,z)-1/5r=dist(x0,B(z,r))+r-1/5r=dist(x0,B(z,r))+4/5r. So d(x0,B(z,c0r))≥r0/2.
Proof of Ω x 1 / n ≲ c 0 r. Define (67)wy=1c0rinfγx0,ylγ∩Bz,c0r,∀y∈Ω,where the infimum is taken over all rectifiable curves γ joining x0 and y.
SinceΩ∖B(z,c0r) is disconnected, by Lemma 7, we have ‖w‖B˙∗α,ϕ(Ω)≤C(c0r)|α| and hence by the imbedding assumption, we have (68)w-wΩLn/α≤Cc0rα.On the other hand, by (66) one has w=0 in B(x0,1/2r0). Thus (69)wLn/αΩ≤w-wΩLn/αΩ+Ωα/nwΩ-wBx0,1/2r0≤Cw-wΩLn/αΩ.Since w(y)≥1 for all y∈Ωx, we have ‖w‖Ln/|α|(Ω)≥|Ωx||α|/n. By this, (69), and (68), we have (70)Ωxα/n≤Cc0rα,which gives |Ωx|1/n≤C(c0r) as desired.
Proof of ( 1 / 2 - c 0 ) r ≤ C Ω x 1 / n For integer j≥1, take cj∈(cj-1,1] such that (71)Ωx∖Bz,cjr=2-1Ωx∖Bz,cj-1r=2-jΩx.Define a function v in Ω by (72)vy=0,y∈Ω∖Ωx;y-z-c0rc1r-c0r,y∈Ωx∩Bz,c1r;1,y∈Ωx∖Bz,c1r.By Lemma 6, we have (73)vB˙∗α,ϕ≤Cc1r-c0r-αϕ-1c1r-c0rnΩx-1.Hence, by the imbedding assumption, we obtain (74)v-vΩLn/αΩ≤Cc1r-c0r-αϕ-1c1r-c0rnΩx-1.By (66), we have v(y)=0 for all y∈B(x0,1/2r0). Similarly to (69), we have vLn/|α|(Ω)≤Cv-vΩLn/|α|(Ω). Thus (75)v-vΩLn/α≥CΩx∖Bx,c1r-α/n=C12Ωx-α/n.Therefore (76)Ωx-α/n≤Cc1r-c0r-αϕ-1c1r-c0rnΩx-1.This further yields that (77)Ωxc1r-c0rnϕCc1r-c0r-αΩxα/n≥1.By Lemma 2, ϕ(s-αC)s-n→0 as s→∞, and hence, there exists a constant Λ(ϕ,C) such that (78)c1-c0r≤Λϕ,CΩx1/n.
Ifc1≥1/2, then (78) implies that (1/2-c0)r≤Λ(ϕ,C)|Ωx|1/n as desired.
Ifc1<1/2, then for j≥1, define (79)ujy=0,y∈Ω∖Ωx∪Ωx∩Bz,cjr;z-y-cjrcj+1r-cjr,y∈Ωx∩Bz,cj+1r∖Bz,cjr;1,y∈Ωx∖Bz,cj+1r.Let j0=inf{j≥1:cj≥1/2}. Applying Lemma 6, (71), and (66), by an argument similar to (78), we get (80)cj+1-cjr≤Λϕ,CΩx∖Bz,cjr1/n=Λϕ,C2-j/nΩx1/n∀1≤j≤j0.This implies that (81)12-c0r≤cj+1-c0r≤∑j=0j0-1cj+1-cjr≤2Λϕ,CΩx1/nas desired.Using Proposition9, we are ready to prove Theorem 1 (ii).Proof of Theorem1.
(ii) Suppose thatΩ is a simply connected planar domain, or a bounded domain that is quasiconformally equivalent to some uniform domain when n≥3. Assume that Ω supports the Orlicz-Besov Poincaré inequality (6).
By [5, 13], we know that Ω has a separation property with respect to x0∈Ω and some constant C0≥1, that is, for every x∈Ω, there exists a curve γ:[0,1]→Ω with γ(0)=x,γ(1)=x0 such that for each t∈[0,1], either γ([0,t])⊂B¯≡B(γ(t),C0d(γ(t),Ω∁))¯ or each y∈γ([0,t])∖B¯ belongs to a different component of Ω∖B¯ than x0.
For anyx∈Ω, let γ be a curve as above. It then suffices to prove (82)distγt,Ω∁≳diamγ0,t∀t∈0,1.Indeed, by the arguments in [14, pp.385-386] and [15, pp.7-8], (82) allows us to modify γ to get a John curve for x. Here we omit the details.
To prove (82), by Proposition 9, we know that Ω enjoys the LLC(2) property. Let N=2+C0/b, where b is the constant in LLC(2). For any t∈(0,1], if dist(γ(t),Ω∁)≥dist(x0,Ω∁)/N, then (83)γ0,t⊂Ω⊂Bγt,diamΩNdistγt,Ω∁distx0,Ω∁,which implies(84)distγt,Ω∁≥distx0,Ω∁2NdiamΩdiamγ0,t.Assume dist(γ(t),Ω∁)<dist(x0,Ω∁)/N. It is enough to prove that (85)γ0,t⊂Bγt,N-1distγt,Ω∁¯.Indeed, this gives (86)distγt,Ω∁≥12N-1diamγ0,t.This and (84) give (82) as desired. We prove (85) by contradiction. If (85) is not true, then there exists y∈γ([0,t])∖B(γ(t),(N-1)d(γ(t),Ω∁))¯. Since (87)x0-γt≥distx0,Ω∁-dγt,Ω∁>N-1dγt,Ω∁,by Proposition 9, x0 and y are contained in the same component of Ω∖B(γ(t),b(N-1)d(γ(t),Ω∁))¯. Noting that b(N-1)≥C0, we have that x0 and y are contained in the same component of Ω∖B(γ(t),C0d(γ(t),Ω∁))¯, which is contradicted with the separation property. Thus, (85) holds. This finishes the proof of Theorem 1 (ii).



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