Fractional Sobolev-Poincar′e Inequalities in Irregular Domains?

Chang-Yu GUO

1 Introduction

Recall that a bounded domain ??Rnis a John domain if there is a constant C and a point x0∈ ? so that for each x ∈ ?,one can find a recti fiable curve γ :[0,1]→ ? with γ(0)=x,γ(1)=x0and

for each 0

The condition(1.1)is called a “twisted cone condition” in literature.Thus the condition(1.2)should be called a “twisted cusp condition”.

In the last twenty years,s-John domains have been extensively studied in connection with Sobolev-type inequalities(see[2,7–8,12–13,17]).Recall that a bounded domain ? ? Rn,n ≥ 2 is said to be a(q,p)-Poincar′e domain if there exists a constant Cq,p=Cq,p(?)such that

for all u∈C∞(?)∩L1(?).Here u?=??u(x)dx.When q=p, ? is termed a p-Poincar′e domain and when q>p,we say that ? supports a(q,p)-Sobolev-Poincar′e inequality.Buckley and Koskela[2]have shown that a simply connected planar domain which supports a(,p)-Sobolev-Poincar′e inequality is a 1-John domain.Smith and Stegenga shown that an s-John domain ? is a p-Poincar′e domain,provided that 1 ≤ s<+In particular,if 1≤s

Recently,there has been a growing interest in the study of the so-called fractional(q,p)-Sobolev-Poincar′e inequalities(see for instance[3,9]and the references therein).In this paper,we continue the study of the following fractional(q,p)-Sobolev-Poincar′e inequality in a domain??Rnwith finite Lebesgue measure,n≥2,

where 1≤ p≤ q< ∞,δ∈(0,1),τ∈(0,∞)and the constant C does not depend on u ∈ C(?)∩ L1(?).If ? supports the fractional(q,p)-Sobolev-Poincar′e inequality(1.4),q ≥ p,then we say that ? is a fractional(q,p)-Sobolev-Poincar′e domain1Strictly speaking,we should also indicate the parameter δ in the de finition of a fractional(q,p)-Sobolev-Poincar′e domain.But since we did not emphasize it in the de finition of the fractional(q,p)-Sobolev-Poincar′e inequality either,we keep our current terminology..

From now on,unless otherwise speci fied,δ∈ (0,1)and τ∈ (0,∞)will be fixed constants.Given a function u ∈ C(?)∩L1(?),we de fine gu:? → R as

for x∈?.

It is well-known,due to Maz’ya[15–16],that the validity of a(q,p)-Sobolev-Poincar′e inequality in ? is equivalent to certain capacity-type estimates in ?.Thus one would expect that a similar equivalence result holds in the setting of fractional(q,p)-Sobolev-Poincar′e inequalities as well.Our first main result con firms this expectation.

Theorem 1.1Let ??Rn,n≥2 be a domain with finite Lebesgue measure and 1≤p≤q<∞.Then the following statements are equivalent:

(i)? satis fies the fractional(q,p)-Sobolev-Poincar′e inequality.

(ii)For an arbitrary ball B0 ? ?,there exists a constant C=C(?,p,q,B0,δ,τ)such that

for every measurable set A?? such that=?.The in fimum above is taken over all functions u∈ C(?)∩L1(?)that satisfy u|A ≥ 1 and u|B0=0.

Theorem 1.1 can be regarded as a fractional version of[7,Theorem 1]and it allows us to study the fractional(q,p)-Sobolev-Poincar′e inequalities in irregular domains via capacity estimates.On the other hand,as in the usual Sobolev-Poincar′e case,we have standard techniques for doing capacity estimates.

Our second main result can be regarded as an(un-weighted)fractional version of[7,Theorem 9].

Theorem 1.2Let ? ? Rn,n≥ 2,be an s-John domain.If p

The range for q in Theorem 1.2 is essentially sharp as indicated by the following example.

Example 1.1Given τ,δ∈ (0,1),1 ≤ p

Theorem 1.2 holds for the critical caseas well,provided that s=1 or p=1(see Remark 4.2).We conjecture that Theorem 1.2 holds under the same assumptions for the critical case.

The above s-John condition on a domain ? is very “geometric” and it provides an effective estimate for capacity.There is another well-known “metric” condition on ? that is sufficient for our capacity estimates.The condition is termed the quasihyperbolic boundary condition in literature and it requires that the quasihyperbolic distance between each point x and a fixed point x0in ? is dominated from above by(a logarithmic function of)its distance to the boundary of ?(see Section 2 below for precise de finitions).With these understood,our third main result can be regarded as a fractional version of[13,Theorems 1.4–1.5]and[10,Theorem 1].

Theorem 1.3Let ? ? Rn,n≥ 2,satisfy the quasihyperbolic boundary condition(2.1)for some β ≤ 1.Then ? is a fractional(q,p)-Sobolev-Poincar′e domain provided that p ∈and q∈

Example 1.2For each q>there exists a domain ? ? Rn,n ≥ 2,satisfying(2.1)which is not a fractional(q,p)-Sobolev-Poincar′e domain.For each 1 ≤ p

Recall that we say a domain ??Rnwith a distinguished point x0has a separation property if there exists a constant C0such that the following property holds:For every x∈?,there exists a curve γ :[0,1]→ ? with γ(0)=x,γ(1)=x0,such that for each t,either

or each y ∈ γ([0,t])Btand x0belongs to different components of ??Bt.

Theorem 1.4Assume that ??Rnis a domain of finite Lebesgue measure that satis fies the separation property with a distinguished point x0.Let 1≤pp,then for each x ∈ ?,there is a curve γ :[0,1]→ ? with γ(0)=x, γ(1)=x0such that

where

The assumptions in Theorem 1.4 can be further relaxed.Indeed,Theorem 1.4 holds if we only assume that the fractional(q,p)-Sobolev-Poincar′e inequality(1.4)holds for all locally Lipschitz continuous functions in ?(see Remark 3.1).

Since this paper generalizes the main results of[2–3,7,9,13]to the fractional setting in a natural way,some of the arguments used in this paper are similar to ones in those papers.In particular,we bene fit a lot from[7,9,13].This paper is organized as follows.Section 2 contains the basic de finitions and Section 3 contains some auxiliary results.We prove our main results,namely,Theorems 1.1–1.2 and Example 1.1 in Section 4.In Section 5,we prove Theorem 1.3 and give the construction of Example 1.2.In the final section,i.e.,Section 6,we discuss the proof of Theorem 1.4 and point out an inaccurate statement,namely,Corollary 4.1 in[2].

2 Notations and De finitions

Recall that the quasihyperbolic metric k?in a domain ? (Rnis de fined to be

where the in fimum is taken over all recti fiable curves γ in ? which join x to y.This metric was introduced by Gehring and Palka in[5].A curve γ joining x to y for which k?(x,y)=is called a quasihyperbolic geodesic.Quasihyperbolic geodesics joining any two points of a proper subdomain of Rnalways exists(see[4,Lemma 1]).

Recall that a domain ? ? Rn,n ≥ 2,is said to satisfy a β-quasihyperbolic boundary condition,β∈(0,1],if there exists a point x0∈? and a constant C0such that

holds for all x∈?.

Let ? be a bounded domain in Rn,n ≥ 2.Then W=W(?)denotes a Whitney decomposition of ?,i.e.,a collection of closed cubes Q ? ? with pairwise disjoint interiors and edges parallel to the coordinate axes,such thatand the diameters of Q∈W belong to the setand satisfy the condition

For j∈Z,we de fine

Note that when we write f(x).g(x),we mean that f(x)≤Cg(x)is satis fied for all x with some fixed constant C ≥ 1.Similarly,the expression f(x)&g(x)means that f(x)≥ C?1g(x)is satis fied for all x with some fixed constant C≥1.We write f(x)≈g(x)whenever f(x).g(x)and f(x)&g(x).

3 Auxiliary Results

We need the following “chain lemma” from[7,Proof of Theorem 9].Note that the condition 3 below is not stated there,however,the proof adapts to our setting and we omit the details.

Lemma 3.1Let ??Rnbe an s-John domain and M>1 a fixed constant.Let B0=B(x0,where x0∈? is the John center.There exists a constant c>0,depending only on ?,M and n,such that given x ∈ ?,there exists a finite “chain” of balls Bi=B(xi,ri),i=0,1,···,k(k depends on the choice of x)that joins x0to x with the following properties:

(1)|Bi∪Bi+1|≤c|Bi∩Bi+1|.

(2)d(x,Bi)≤

(3)d(Bi,??)≥ Mri.

(4)

(5)|x?xi|≤and Bk=B(x,).

(6)For any r>0,the number of balls Biwith radius ri>r is less thanwhen s>1.

Recall that for a function f,the Riesz potential Iδ,δ∈ (0,n)of f is de fined by

The following estimate for the Riesz potential is well-known(see for instance[1,Theorem 3.1.4 and Corollary 3.1.5]).

Theorem 3.1Let 0<δ0 such that the weak estimate

holds for every f∈L1(Rn).

The following proposition,which can be regarded as a fractional analogy of[2,Theorem 2.1],is proved in[3,Proposition 6.2].

Proposition 3.1Suppose that ? ? Rnis a domain of finite Lebesgue measure.Let 1 ≤p0 and w ∈ ?.Then there exists a constant C>0 such that

and

if T is the union of all components of ?B(w,d)that do not intersect the ball B0.The constant C depends only on|B0|,|?|,n,p,q, δ and the constant associated to the fractional(q,p)-Sobolev-Poincar′e inequality.

Remark 3.1As in[2],one can check that the conclusion holds whenever the fractional(q,p)-Sobolev-Poincar′e inequality(1.4)with τ=1 holds for every locally Lipschitz continuous functions(see[3,Proof of Proposition 6.2]).

Fix a Whitney cube Q0and assume that x0is the center of Q0.For each cube Q∈W,we choose a quasihyperbolic geodesic γ joining x0to the center of Q and we let P(Q)denote the collection of all the Whitney cubes Q′∈ W which intersect γ.Then the shadow S(Q)of the cube Q is de fined to be

The following lemma is proved in[13,Lemma 2.6].

Lemma 3.2Let ? ? Rn,n≥ 2 be a domain that satis fies the quasihyperbolic boundary condition(2.1).Then for each ε>0,there exists a constant C=C(n,diam ?,ε)such that

We also need the following estimate of the size of the shadow of a Whitney cube Q in terms of the size of Q.The proof can be found in[10,Lemma 6].2I would like to thank Renjin Jiang for sharing the manuscript[10]and Aapo Kauranen for pointing out Lemma 3.3 in their work in[10,Lemma 6].

Lemma3.3Let ? ? Rn,n ≥ 2,be a domain that satis fies the quasihyperbolic boundary condition(2.1).Then there exists a constant C=C(n,d(x0,??))such that

for all Q∈W.Consequently,

4 Main Proofs

Proof of Theorem 1.1We first show that the condition(ii)implies the condition(i).Fix a function u∈ C(?)∩L1(?).Pick a real number b such that both|{x ∈ ? :u(x)≥ b}|and|{x ∈ ? :u(x)≤ b}|are at leastIt suffices to show the fractional(q,p)-Sobolev-Poincar′e inequality with|u?u?|replaced by|u?b|,and by replacing u with u?b,we may assume that b=0.Write v+=max{u,0}and v?= ?min{u,0}.In the sequel,v denotes either v+or v?;all the statements below are valid in both cases.Without loss of generality,we may assume that v≥0.

Fix a ball B0such that 2B0?? ?.We may further assume that v|B0=0.In fact,let φ ∈ C∞(?)satisfy 0≤ φ ≤ 1,spt(φ)? 2B0and φ|B0=1.Note that we may write

The first term φv ∈ C∞(2B0B0)and thus the fractional(q,p)-Sobolev-Poincar′e inequality holds for φv in 2B0B0.On the other hand,the second term(1 ? φ)v ∈ C(?)∩ L1(?)and it vanishes on B0.So if one can prove the fractional(q,p)-Sobolev-Poincar′e inequality for(1? φ)v,then a simple computation,after summing up these two estimates,will imply the fractional(q,p)-Sobolev-Poincar′e inequality for v.

For each j∈Z,we de fine vj(x)=min{2j,max{0,v(x)?2j}}.We next prove the following inequality:

To see it,notice that=0 and≥ 1,where Fj={x ∈ ? :v(x)≥ 2j+1}.So by(1.6),we obtain that

Note thatThus we finally arrive at

which is the desired estimate(4.1).

The fractional(q,p)-Sobolev-Poincar′e inequality now follows from the weak type estimates via a standard argument.Write By=B(y,τd(y,??))and Ak=Fk?1Fk,

where

and

For y ∈ Aiand z ∈ Ajwith j?1>i,|v(y)?v(z)|≥ |v(z)|?|v(y)|≥ 2j?2.Hence,

Since the estimate

holds for every k∈Z,(4.2)is valid whenever i≤k≤j and(y,z)∈Ai×Aj.It follows from(4.2)that

Sincechanging the order of the summation yields that the right-hand side of the above inequality is bounded by

The estimate ofis similar.Thus,we have proved that

The desired fractional(q,p)-Sobolev-Poincar′e inequality(1.4)follows from the above inequality as we notice that|u|=v++v?and|v±(y)?v±(z)|≤ |u(y)?u(z)|for all y,z∈ ?.

The implication from the condition(ii)to the condition(i)is easier.To see it, fix a measurable set A? ? such that= ?and a function u∈ C(?)∩L1(?)such that u|A≥ 1 and u|B0=0.If u?≤,then by(1.4),we have

If u?≥then by(1.4)we have

Combining the above two estimates,we conclude that

where C=C(?,B0,p,q,δ,τ).Taking the in fimum over all such u gives(1.6).

Remark 4.1It is clear from the proof above that the condition(ii)of Theorem 1.1 is equivalent to the following condition:For an arbitrary cube Q0? ?,there exists a constant C=C(?,Q0,p,q,δ,τ)such that

for every measurable set A?? with=?.The in fimum above is taken over all functions u ∈ C(?)∩L1(?)that satisfy u|A≥ 1 and u|Q0=0.

Proof of Theorem 1.2Let B0=B(x0,).Assume that p<10 such that

It suffices to show,by Theorem 1.1,that there exists a constant C=C(?,B0,p,q,δ,τ)such that for every measurable set A?? with=?,we have

whenever u ∈ C(?)∩L1(?)satis fies u|A≥ 1 and u|B0=0.Since ? is bounded,we may further assume that diam?=1.

For any x ∈ A,we obtain from Lemma 3.1 a finite chain of balls Bi,i=0,1,···,k,satisfying conditions(1)–(6)in Lemma 3.1 with M>For all i=0,1,···,k,we have

To see this, fix y∈Biand let z be any other point in Bi.Then by the condition(3)in Lemma 3.1,

In order to estimate|A|,we divide A into the “bad” and “good” parts.Setting

where Bx=B(x,),we have|A|≤|G|+|B|.We first estimate|G|.

For x∈G,letbe the associated chain of balls as described before.Then Bx=Bk.By the condition(1)in Lemma 3.1,we have

For a ball Bi,

By(4.3)and the condition(2)in Lemma 3.1,

Thus we conclude that

H¨older’s inequality implies

where κ =Using the condition(6)from Lemma 3.1,one can easily conclude

Therefore,

where the constant C depends only on p,n,?and the constant from s-John condition.

By the condition(2)from Lemma 3.1,Cri≥ |x?y|sfor y∈ Bi,and since p(?κ+δ?<0 according to our choice p≤we obtain

for y∈Bi.For y∈Bi∩(2j+1Bk2jBk),we have|x?y|≈2jrkand hence for such y,

Combining(4.4)with(4.5)leads to

On the other hand,

Comparing the above two estimates,we conclude that there exists an l(depending on?)such that

It follows that

In other words,there exists an Rx≥with

Note that according to our choice of?,the above estimate reduces to the following form:

Applying the Vitali covering lemma to the covering{B(x,Rx)}x∈Eof the set B,we can select pairwise disjoint balls B1,···,Bk,such thatLet ridenote the radius of the ball Bi.Then

We next estimate|B|.Note thatWe may use the Besicovitch covering theorem to select a subcovering{Bxi}i∈Nwith bounded overlap.Since u ≥ 1 on A and uBxi≤we obtain that

for y ∈ A∩Bxi.By the fractional(q,p)-Sobolev-Poincar′e inequality for balls(see for instance[9]),we get

Summing over all balls Bxi,we obtain that

The proof of Theorem 1.2 is now complete.

Remark 4.2In Theorem 1.2,q is assumed to be strictly less thanHowever,one can easily adapt the proof of Theorem 1.2 to show that when s=1 or p=1,q can reach the critical value(the case s=1 has already been proved in[3]).Indeed,we only need to use a variant of Lemma 3.1.Namely,for each x∈?,we may join x to x0via an in finite chain of balls{Bi}i∈Nwith all the properties listed in Lemma 3.1 except the condition(5)in Lemma 3.1 replaced by

as i→ ∞.Then following the proof of Theorem 1.2,we easily deduce the following Rieszpotential-type estimate:

Note that

where β =sδ? (s? 1)n.Thus we conclude that

For s=1 and p>1,the claim follows from the strong-type estimate in Theorem 3.1.For p=1,the claim follows from the weak-type estimate(3.2)and the weak-to-strong principle for fractional Sobolev-Poincar′e inequalities(see[9,Theorem 4.1]).

Proof of Example 1.1We will use the mushroom-like domain as used in[7].The mushroom-like domain ??Rnconsists of a cube Q and an attached in finite sequence of mushrooms F1,F2,···growing on the “top” of the cube.By a mushroom F of size r,we mean a cap C,which is a ball of radius r,and an attached cylindrical stem P of height r and radius rs.The mushrooms are disjoint,and the corresponding cylinders are perpendicular to the side of the cube that we have selected as the top of the cube.We can make the mushrooms pairwise disjoint if the number riassociated with Ficonverges to 0 sufficiently fast as i→∞.We further write P=T∪M∪D,where T is the top-part of P,M is the middle-part of P,and D is the bottom-part of P.

Let uibe a piecewise linear function on ? such that ui=1 on the cap Ci∪ Ti,uiis linear on Miand ui=0 elsewhere.Assume that 1≤s

Note that

On the other hand,

Thus we obtain that for all i∈N,

which is impossible if q>

5 Fractional(q,p)-Sobolev-Poincar′e Inequalities in Domains with Quasihyperbolic Boundary Conditions

Lemma 5.1Fix p and q as in Theorem 1.3.Then there exists a constant C=C(n,p,q,β)such that

whenever E??.

ProofFor simplicity,we write p?,δ=κ =and λ =Then=

Thus

where we have used(3.3)–(3.4)with ε=()κ >0.

The proof of Theorem 1.3 is again based on Theorem 1.1.

Proof of Theorem 1.3Fix Q0?? to be the central Whitney cube containing x0.For each measurable set A? ? with= ?,let u ∈ C(?)∩L1(?)satisfy u|A≥ 1 and u|Q0=0.As in the proof of Theorem 1.2,we divide A into “good” and “bad” parts.Set

We have|A|≤|G|+|B|and we first estimate|B|.

For points x ∈ B,the standard fractional(p?,δ,p)-Sobolev-Poincar′e inequality on cubes provides a trivial estimate

on Whitney cube Q containing x,where p?,δ=Since q

and by summing over all such Whitney cubes,we deduce that

We next estimate|G|and our aim is to show that

so then the conclusion follows from Theorem 1.1.

For each x∈G,let Q(x)be the Whitney cube containing x,for which uQ(x)≥.Then the chaining argument used in the proof of Theorem 1.2 gives us the estimate

Recall that P(Q(x))consists of the collection of all the Whitney cubes which intersect the quasi-hyperbolic geodesic joining x0to the center of Q(x).Strictly speaking,on the right-hand side of(5.3),one should replace Q with λQ,where 1< λ

Integrating(5.3)with respect to the Lebesgue measure and interchanging the order of summation and integration yield

Applying H¨older’s inequality leads to

Applying Lemma 5.1,we find that

which proves(5.2).

Proof of Example 1.2The construction here is similar to that used in the proof of Example 1.1 and thus we only point out the difference.The mushroom-like domain ??Rnconsists of a cube Q and an attached in finite sequence of mushrooms F1,F2,···growing on the “top” of the cube as in Example 1.1.Now,by a mushroom F of size r,we mean a cap C,which is a ball of radius r,and an attached cylindrical stem P of height rτand radius rσ.The mushrooms are disjoint,and the corresponding cylinders are perpendicular to the side of the cube that we have selected as the top of the cube.We can make the mushrooms pairwise disjoint if the number riassociated with Ficonverges to 0 sufficiently fast as i→∞.We further write P=T∪M∪D,where T is the top-part of P,M is the middle-part of P,and D is the bottom-part of P.

It is easy to show that ? satis fies the β-quasihyperbolic boundary condition(2.1)if σ =≤ τ(see for instance[13,Example 5.5]).We next show that ? is not a fractional(q,p)-Sobolev-Poincar′e domain if

When τ= σ =(5.5)implies that ? is a β-quasihyperbolic boundary condition which does not support a fractional(q,p)-Sobolev-Poincar′e inequality.This veri fies Example 1.2.

Let uibe a piecewise linear function on ? such that ui=1 on the cap Ci∪ Ti,and uibe linear on Miand ui=0 elsewhere.Assume that the fractional(q,p)-Sobolev-Poincar′e inequality holds on ?.

Note that

On the other hand,

Thus we obtain that for all i∈N,

which is impossible if

6 Necessary Conditions for the Fractional(q,p)-Sobolev-Poincar′e Domains

Proof of Theorem 1.4Fix x ∈ ?.Pick a curve γ :[0,1] → ? with γ(0)=x and γ(1)=x0as in the de finition of separation property.

Let 0

where

A bounded domain ? ? Rnwith a distinguished point x0satisfying(1.7)with φ(t)=is termed s-diam John in[6].It was proved in[6]that,for s>1,s-diam John domains are not necessarily s-John.

In[2,Corollary 4.1],it was stated that if a bounded domain ??Rnsatis fies a separation property and supports a(q,p)-Sobolev-Poincar′e inequality(1.3)with q>p,then ? is s-John with s=One could immediately check that the proof given there is only sufficient to deduce that ? is s-diam John with s=In fact,combining[6,Example 5.1]and[2,Section 4],one can produce an s-diam John domain ? ?Rnwith s=such that ? supports a(q,p)-Sobolev-Poincar′e inequality.Moreover,? is not s′-diam John whenever s′

We next brie fly discuss how to construct such an example in the plane(it works in higher dimensions as well).Set

where 0<α<β≤1 will be speci fied later.The idea is very simple:We first use the mushroom-like domain ?′? R2as is constructed in[2](with different choices of parameters)and then modify ?′to be a spiral domain ? as in[6,Example 5.1].

The mushroom-like domain ?′? R2consists of a cube Q and an attached in finite sequence of mushrooms F1,F2,···growing on the “top” of the cube as in Example 1.1.Now,by a mushroom F of size r,we mean a cap C,which is a ball of radius r,and an attached cylindrical stem C(r).The mushrooms are disjoint,and the corresponding cylinders are perpendicular to the side of the cube that we have selected as the top of the cube.We can make the mushrooms pairwise disjoint if the numbers riassociated with Ficonverge to 0 sufficiently fast as i→∞.

Note first that if β =with n=2,then C(r)satis fies the(q,p)-Sobolev-Poincar′e inequality uniformly in r(see[2]).Letμ =sβ =and p?=One can show that ?′is a(q,p)-Sobolev-Poincar′e domain if

holds with n=2(see[2]).Note also that ? is-John.

We next bend each mushroom Fito make it spiral so that the resulting domain ? is an s-diam John domain.According to our choice,s=One can check that if β =then(6.1)reduces to

Since p1.For anysatisfying(6.2)and=it is easy to check that>=s.It is clear that ?′and ? are bi-Lipschitz equivalent,so the(q,p)-Sobolev-Poincar′e inequality holds in ? as well.Moreover, ? satis fies all the required properties.

One could also modify the above example to the fractional(q,p)-Sobolev-Poincar′e case,but the computations will be too complicated,so we omit them in the present paper.

AcknowledgementsThe author wants to express his gratitude to Antti V V¨ah¨akangas.for posing the question in Jyv¨askyl¨a analysis seminar,which is the main motivation of the current paper,for sharing the manuscript(see[3]),and for numerous useful comments on the manuscript.The author also wants to thank academy Professor Pekka Koskela for the helpful discussions.

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