Hardy-type Inequalities - CAPDE Hardy-type Inequalities - CAPDE

Hardy-type InequalitiesJ. Dávila and L. DupaigneMay 31, 20071 IntroductionThe well-known Hardy-Sobolev inequality states that for any given domain Ω ⊂ R n , n ≥ 3 and any u ∈ Cc∞ (Ω),∫K 2 u 2Ω |x|∫Ω2 ≤ |∇u| 2 , (1)RΩ |∇u|2 RΩ u2 /|x| 2 ,where K = (n − 2)/2. Though the constant K 2 is optimal, in the sense that K 2 = inf u∈C ∞ c (Ω)\{0}equality in (1) is never achieved (by any u ∈ H0 1 (Ω)). This fact has lead to the improvement of the inequality invarious ways: Brezis and Vázquez [BV] first showed that if Ω is bounded then for some γ > 0,(∫ ) 2/p ∫γ |u| p + K 2ΩΩu 2|x|∫Ω2 ≤ |∇u| 2 , (2)with p < 2n/(n − 2). Vázquez and Zuazua [VZ] were then able to replace the L p norm in the right-hand sideof (2) by a W 1,q norm for q < 2. Various improvements (involving e.g. weighted L p or W 1,p norms) were alsoobtained and we point the interested reader to [Da, ACR, BFT] and the references therein.One of the consequences of inequality (2) is that the operator L 0 := −∆−µ/|x| 2 has a positive first eigenvalue,in the sense that)inf(|∇u|‖u‖ L 2 (Ω)=1∫Ω2 − µ u2|x| 2 > 0,whenever µ ≤ K 2 .In the first part of this work, given a compact smooth boundaryless manifold Σ ⊂ Ω of codimension k ≠ 2,we look at operators of the formL = −∆ − µd(x) 2 ,where d(x) = dist(x, Σ) and µ ∈ R, and wonder whether an inequality similar to (2) holds.The first results in this direction are due to Marcus, Mizel and Pinchover [MMP] and Matskevich andSobolevskii [MSo]. They showed that if Ω is a convex domain and Σ = ∂Ω then14∫Ωu 2d(x)∫Ω2 ≤ |∇u| 2 . (3)The same authors showed that (3) did not hold in a general domain Ω and provided examples of smooth domainsΩ such that∫Ωinf ∫|∇u|2 < 1u≢0 u 24 .Ω d 2Alternatively, Brezis and Marcus showed in [BM] that the following inequality remains true on a general (smoothbounded) domain Ω :∫1 u 2∫4 Ω d∫Ω2 ≤ |∇u| 2 + C u 2 ,Ω1

where C is some positive constant.Finally, among many other results, Barbatis, Filippas and Tertikas [BFT] extended (3) to the case whereΣ ⊂ Ω is a smooth compact manifold of codimension k, satisfying some geometric condition: they showed thatif ∆d k−2 ≤ 0 in D ′ (Ω \ Σ) then(∫ ) 2/p ∫γ |u| p + H 2 u 2ΩΩ d∫Ω2 ≤ |∇u| 2 ,where H = (k − 2)/2 and 1 ≤ p < 2nn−2 .Our goal here is to drop the assumption ∆d k−2 ≤ 0. Our results are summarized in the following twotheorems :Theorem 1 Let Ω ⊂ R n be an open bounded set and Σ ⊂ Ω be a compact smooth manifold without boundary ofcodimension k ≠ 2. Let H = k−22. Then there exists p > 2 and C > 0, γ > 0 independent of u such that for anyu ∈ Cc ∞ (Ω \ Σ)( ∫γ |u| p) ∫2/p+ H2 u 2∫ΩΩ d∫Ω2 ≤ |∇u| 2 + C u 2 , (4)Ωwhere d(x) = dist(x, Σ), 1 ≤ p < p k and p k is given by1= 1 p k 2 − 2k(n − k + 2)for k > 2 and1= 1 p 1 2 − 1n + 1if k = 1.Theorem 2 Make the same assumptions as those of Theorem 1. Then there exist β > 0 and a neighbourhoodΩ β := {x ∈ Ω : d(x, Σ) < β} of Σ in Ω such that for any u ∈ C ∞ c (Ω β \ Σ)Remark 1 .H 2 ∫Ωu 2d∫Ω2 ≤ |∇u| 2 . (5)• If k ≥ 3 it follows by density that (4) and (5) hold for all u ∈ C ∞ c(Ω).• The exponent p k appearing in Theorem 1 is probably not optimal and we expect that (4) holds for all1 ≤ p < 2n/(n − 2). In fact Maz’ja [Ma] (Corollary 3, Section 2.1.6) proved this result when Σ = {x ∈ R n :x 1 = x 2 = .. = x k = 0}.As a direct consequence of Theorem 1, we see that the first eigenvalue of the operator L = −∆ − µ d −2 is finite,i.e.)λ 1 := inf(|∇u|‖u‖ L 2 (Ω)=1∫Ω2 − µ u2d 2 > −∞,whenever µ ≤ H 2 . We proved in [DD] that in such circumstances there exists an eigenfunction ϕ 1 associated toλ 1 i.e. a solution (in a sense which we shall make precise soon) of⎧⎨ −∆ϕ 1 − µ d 2 ϕ 1 = λ 1 ϕ 1 in Ω⎩ϕ 1 = 0on ∂Ω.Normalizing ϕ 1 by ‖ϕ 1 ‖ L 2 (Ω)=1 and ϕ 1 > 0, we then investigate the behaviour of ϕ 1 near Σ and show that forsome constants C 1 , C 2 > 0,C 1 d(x) −α(µ) ≤ ϕ 1 ≤ C 2 d(x) −α(µ) , (6)where α(µ) = H − √ H 2 − µ.This result enables us to treat two model applications. First we consider the quantity∫ΩJ λ := inf|∇u|2 − λ ∫ ∫Ω u2u≢0Ω u2 /d 22

and extend a result of Brezis and Marcus [BM] stating that J λ is achieved if and only if J λ < H 2 .Our second application is a nonexistence result for positive solutions of the equation−∆u − µ d 2 u = up + λ,completing a study started in [DN]. See Section 4.2 for details.The last purpose of this article is to extend some results in [DD]. This generalization is necessary to includethe case of potentials a(x) = µ dist(x, Σ) −2 . More precisely we shall derive estimates for solutions of the linearequation {−∆u − a(x)u = f in Ωu = 0on ∂Ω,(7)under the assumptions that a ∈ L 1 loc(Ω), a is bounded below, i.e.essinf Ω a > −∞,and( ∫γ |u| r) ∫∫∫2/r+ a(x)u 2 ≤ |∇u| 2 + M u 2 , (8)ΩΩΩΩfor some r > 2, γ > 0, M > 0.Let us now clarify what we mean by a solution of (7).We first define the Hilbert space H as the completion of Cc∞ (Ω) with respect to the norm∫‖u‖ 2 H = (u|u) H := |∇u| 2 − a(x)u 2 + Mu 2Ωwhere M is the same constant that appears in (8). Observe that the definition of H does not change if we replaceM by any larger constant.Given f ∈ H ∗ , we then say that u ∈ H is a solution of (7) if(u|v) H =< f, v > H ∗ ,H +M(u|v) L2 (Ω) ∀v ∈ H.It is convenient at this point to recall some facts that were proved in [DD]. We start by mentioning that Hembeds compactly in L 2 (Ω). In particularL = −∆ − a(x)has a first eigenvalue λ 1 , which is simple. λ 1 is not necessarily positive (Theorem 1 provides examples of potentialsa(x) = H 2 /d(x) 2 for which in general λ 1 can be nonpositive), but when it is, then for f ∈ H ∗ problem (7) has aunique solution u ∈ H.We note here that uniqueness fails if one considers other classes of solutions (see an example in [DD]).The first eigenvalue λ 1 has an associated positive eigenfunction ϕ 1 (it is not only positive a.e. but it alsosatisfies ϕ 1 ≥ c dist(x, ∂Ω) for some c > 0).Solutions in H of an equation like (7) are typically unbounded (see examples in [D, DD, DN]). In [DD] weshowed that when λ 1 > 0 and f ≥ 0, f ≢ 0 then the solution u ∈ H of (7) is bounded below by a positiveconstant times ϕ 1 . We also proved that when λ 1 > 0 and f = 1, the solution u of (7) satisfies u ≤ Cϕ 1 for someC > 0.Our main result is the following :Theorem 3 Let 0 < m < r and suppose thatp >2rm(r − 2)andp ≥rr − m .3

Assume that f ∈ H ∗ satisfies ‖ϕ 1−m1 f‖ L p (Ω) < ∞ and that u ∈ H is a solution of (7). Then|u(x)| ≤ C(‖ϕ 1−m1 f‖ L p (Ω) + ‖u‖ L 2)ϕ 1 (x), a.e. x ∈ Ω.A simple corollary of this result, obtained by choosing m = 2 is the following:Corollary 1 Suppose f ∈ H ∗ satisfies f/ϕ 1 ∈ L p (Ω) for some p > and that u ∈ H is a solution of (7).Then|u(x)| ≤ C(‖f/ϕ 1 ‖ p + ‖u‖ L 2)ϕ 1 (x), a.e. x ∈ Ω.The paper is organized as follows : Section 2 is devoted to the proof of Theorems 1 and 2. In Section 3 wederive (6) whereas Section 4 is dedicated to the afore mentioned applications. Finally we prove Theorem 3 inSection 5.2 Hardy Inequalities2.1 Proof of Theorem 1The object of this subsection is to prove Theorem 1. Our arguments are based on improvements of the onedimensionalHardy inequality inspired by [BM] and a decomposition of L 2 functions in spherical harmonics takenfrom [VZ].We start out with a series of three lemmas, which yield a refined version of the classical Hardy inequality inR k (see (20)). The first lemma deals with radial functions :rr−2Lemma 1 Let k ≠ 2 and H = (k − 2)/2. There exists a constant C > 0 depending only on k such that∫ [1 (du ) ]22 ∫ 1− H 2 u20 dr r 2 r k−1 2dr + C u 2 r k−1 dr ≥0∫ [1 (du ) ]22+ H 2 u2dr r 2 r k dr + 1 ∫ 1 (2 d( ) ) 2r r H u(r) dr4 dr00(9)for all u ∈ C ∞ c (0, 1 2 ).Proof. Let u ∈ C ∞ c(0, 1 2 ) and v(r) = rH u(r). A standard computation yields[ (du ) ]2 ( ) 2− H 2 u2dvdr r 2 r k−1 = r − H d ( v 2). (10)dr drIntegrating, it follows thatA :=∫ 120[ (du ) ]2− H 2 u2dr r 2 r k−1 dr =∫ 120r( ) 2 dvdr. (11)drSimilarly, using (10) and an integration by parts,∫ [1 (du ) ]22B :=+ H 2 u20 dr r 2 r k dr∫ 1 ( )22 ∫ 1dv= r 2 dr + 2H 2 2 u 2 ∫ [1 ( ) ]2 d v20 dr0 r 2 rk dr − H r dr0 dr∫ 1 ( )22 dv= r 2 dr + ( 2H 2 + H ) ∫ 1 2v 2 dr. (12)dr004

Using integration by parts again, it follows that for given ε > 0, there exists C > 0 such that∫ 1∫ 12v 2 2dr = −2 rv dv ∫ 1 ∫ 1 ( )00 dr dr ≤ C 2v 2 2 2 dvr dr + εr dr00 dr∫ 1∫ 1 ( )2= C u 2 r k−1 2 2 dvdr + εr dr. (13)drCollecting (11),(12) and (13), we obtain for ε small enough∫ 1( )22 ∫ 1dv2A − B ≥ r(1 − r − Cε) dr − C u 2 r k−1 dr ≥ 1 dr4000∫ 120r0( ) 2 ∫ 1dv2dr − C u 2 r k−1 dr.dr0The next lemma will help us deal with the nonradial part of a given function u : R k → R.Lemma 2 Let k ≠ 2, H = (k − 2)/2 and c > ¯c > 0. There exists constants C, τ > 0 depending only on k and ¯csuch that∫ [1 (du ) ]22 ∫ 1− (H 2 − c) u20 drr 2 r k−1 2dr + C u 2 r k−1 dr ≥0∫ [1 (du ) ]22 ∫ [1 (du ) ]+ (H 2 + c) u2drr 2 r k 22dr + τ+ c u2dr r 2 r k−1 dr (14)for all u ∈ C ∞ c (0, 1 2 ).Proof. It follows from (9) that ifandthenWe can also rewrite (10) asso that if τ = min( ¯c4H, 1 2 4 ),τD :=∫ 120∫ 120[ (du ) ]2− (H 2 − c) u2drr 2E :=∫ 120∫ 12D − E ≥ c≥ c 2It then follows from (17) and (18) that00∫ 1200∫ 1r k−1 dr + C2u 2 r k−1 dr (15)0[ (du ) ]2+ (H 2 + c) u2drr 2 r k dr (16)u 2r 2 (1 − r)rk−1 dr + 1 4u 2r 2 rk−1 dr + 1 4∫ 120r∫ 120( dvdr( ) 2 ( ) 2 dur k−1 = H 2 u2 dvdr r 2 rk−1 + r − H d ( v 2)dr dr( ) 2 dur k−1 dr ≤ ¯c ∫ 12dr 4 0D − E ≥ τ∫ 120u 2r 2 rk−1 dr + 1 ∫ 12r4 0( ) 2 dur k−1 dr + c ∫ 12dr 4 0( ) 2 dvr drdr) 2dr. (17)( ) 2 dvdr. (18)dru 2r 2 rk−1 dr. (19)Hence (14) holds.□Finally, the following lemma yields the improved Hardy inequality (in R k ) that we will be using in the sequel.□5

Lemma 3 Let k ≠ 2, H = (k − 2)/2 and β > 0. Let Bβ k denote the ball of Rk centered at the origin and ofradius β. There exist positive constants C = C(β, k), τ = τ(k) and α = α(β, k) such that∫) ∫(|∇u| 2 − H 2 u2Bβk |y| 2 dy + C u 2 dy ≥Bβk∫) ∫∫1β( ) 2|y|(|∇u| 2 + H 2 u22β Bβk |y| 2 dy + τ |∇(u − u 0 )| 2 dv0dy + α r dr (20)Bβk 0 drfor all u ∈ Cc∞ (Bβ k \ {0}) and where u ∫0(r) = u 0 (|y|) = − udσ and v∂Brk 0 (r) = r H u 0 (r).Proof. Let {f i } ∞ i=0 be an orthonormal basis of L2 (S k−1 ), composed of eigenvectors of the Laplace-Beltramioperator ∆| S k−1. The corresponding eigenvalues are given by c i = i(k + i − 2), i = 0, 1, 2.. (see e.g. ??).Any u ∈ Cc ∞ (B k 1 \ {0}) can then be written as2where 1 2 > r > 0, θ ∈ Sk−1 and x = rθ.Furthermore, for g ∈ C(R + , R),∫B k 1/2|∇u| 2 g(|y|)dy ==∫ 120u(x) =∞∑i=0∞∑u i (r)f i (θ)i=0[ (∂u ) ]2+ 1S ∂r r 2 |∇ θu| 2 dθk−1[ (dui ) ]2r k−1 g(r) + c idr r 2 u2 i dr. (21)∫r k−1 g(r) dr∫ 12For i = 0, it follows from (9) that if v 0 (r) = r H u 0 (r),[ (du0∫ 120drwhile (14) implies that for i ≥ 1,[ (dui∫ 120dr) ]2 ∫ 1− H 2 u2 0r 2 r k−1 2dr + C0[ (du0∫ 1200dr) ]2 ∫ 1− (H 2 − c i ) u2 ir 2 r k−1 2dr + C0[ (dui∫ 120dru 2 0r k−1 dr ≥) 2+ H 2 u2 0r 2 ]r k dr + 1 4u 2 i r k−1 dr ≥) ]2+ (H 2 + c i ) u2 ir 2 r k dr + τ∫ 120∫ 120r[ (dui( ) 2 dv0dr, (22)drdr) ]2+ c u2 ir 2 r k−1 dr. (23)Using (22),(23) and (21) with g(r) ≡ 1 for terms involving r k−1 and g(r) = r for terms in r k , (20) follows forβ = 1 2. The general case is obtained by scaling.□Next, we introduce some geometric notation that will be needed in the proof of Theorem 1. DefineΩ β = { x ∈ Ω | dist(x, Σ) < β }.We will work only with β small enough so that the projection π : Ω β → Σ given by |π(x) − x| = dist(x, Σ) is welldefined and smooth.6

Let {V i } i=1,...,m be a family of open disjoint subsets of Σ such thatΣ =m⋃V i , and |V i ∩ V j | = 0 ∀i ≠ j.i=1We can also assume that:a) ∀i = 1, . . . , m there exists a smooth diffeomorphismp i : B n−k1 → U iwhere U i ⊂ Σ is open and V i ⊂ U i ;b) p −1i (V i ), which is an open set in R n−k , has a Lipschitz boundary; andc) there is a smooth choice of unit vectors N1(σ), i . . . , Nk i(σ) ∀σ ∈ U i which form an orthonormal frame for Σon U i ⊂ R n , i.e. ∀σ ∈ U iN i j(σ) ∈ R n , N i j(σ) · N i k(σ) = δ jk , and N i j(σ) · v = 0 ∀v ∈ T σ Σ.Let W i = p −1i (V i ). For z ∈ W i we will also write (abusing the notation) Nj i(z) = N j i(p i(z)). LetF i (y, z) = p i (z) +k∑y j Nj(z),iwhere y = (y 1 , .., y k ) ∈ B k β and z ∈ W i, so that F i is a smooth diffeomorphism between B k β × W i and T i β , wherej=1T i β = π −1 (V i ) ∩ Ω β . (24)ΩΣβT i βV iFigure 1.It follows from the condition |V i ∩ V j | = 0 ∀i ≠ j that |T i β ∩ T j β| = 0 ∀i ≠ j, and hence, for any f ∈ L 1 (Ω β ) wehave:∫m∑∫f = fΩ β i=1 Tβim∑∫(25)=f ◦ F i (y, z) JF i (y, z) dy dz,i=1 W i×Bβk7

where JF i (y, z) stands for the Jacobian of F i at (y, z). We claim thatJF i (y, z) = G i (z)(1 + O(|y|)), (26)where O(|y|) denotes a quantity bounded by |y| (uniformly for z ∈ W i ) and G i (z) is a smooth function, which isbounded away from zero. More preciselyG i (z) = Jp i (z) =√ (Dpi(z) ) ∗Dpi (z).To prove (26) it suffices to observe that JF i (y, z) is smooth and to compute it at y = 0:JF i (0, z) 2 = det (DF i (0, z) ∗ DF i (0, z))( [Dz= det p i |N1, i . . . , Nk] i ∗ [Dz p i |N1, i . . . , Nki ] )⎡⎢= det ⎣ (D zp i ) ∗ D z p i 0⎤⎥⎦0 IProof of Theorem 1. First, observe that it is sufficient to prove the theorem for u with support near Σ. Indeed,following an idea of Vázquez and Zuazua [VZ], let η ∈ Cc∞ (R n ) so that η ≡ 1 in Ω β/2 and supp(η) ⊂ Ω β . Letu ∈ Cc∞ (Ω \ Σ) and write u = u 1 + u 2 where u 1 = ηu, u 2 = (1 − η)u. Suppose that the conclusion of the theoremholds for u 1 . Then∫) ∫ () ∫ ()(|∇u| 2 − H 2 u2Ωd 2 = |∇u 1 | 2 − H 2 u2 1Ωd 2 + |∇u 2 | 2 − H 2 u2 2Ωd∫2 (+ 2 ∇u 1 · ∇u 2 − H 2 u )1u 2d 2 . (27)Since 1 dAlso note thatis bounded away from Σ we have∫∫Ω∫∇u 1 · ∇u 2 =∫=ΩΩ∫≥ −CIt follows from (27), (28) that∫] ∫[|∇u| 2 − H 2 u2d 2 ≥ΩΩΩu 2 2d 2 + u ∫1u 2d 2 ≤ C u 2 .Ω[η(1 − η)|∇u| 2 − |∇η| 2 u 2 + u∇u · ∇η(1 − 2η) ][η(1 − η)|∇u| 2 − |∇η| 2 u 2] − 1 2Using (4) with u 1 we conclude that∫] ∫[|∇u| 2 − H 2 u2d 2 + CΩΩ∫Ω β \Ω β/2u 2 ∇ · (∇η(1− 2η) )u 2 . (28)Ω[] ∫∫|∇u 1 | 2 − H 2 u2 1d 2 + |∇u 2 | 2 − C u 2 .ΩΩΩ( ∫u 2 ≥ γ |u 1 | p) ∫2/p+ |∇u 2 | 2 ,ΩΩfor some γ > 0 independent of u. From here the conclusion of the theorem for u follows easily.8

Let I i denote the quantity∫I i =T i β[|∇u| 2 − H 2 u2d 2 + u2 ], (29)where Tβ i was defined in (24). In what follows we will fix i and show that there is p > 2 and C > 0 independentof u such that( ∫ |u| p) 2/p≤ CIi .TβiFor simplicity, and since i is fixed, we will drop the index i from all the notation that follows.Let us introduce some additional notation:Let us writeũ(y, z) = u(F (y, z))∫(30)ũ 0 (r, z) = − ũ(y, z) ds(y)∂B r(31)v 0 (r, z) = r H ũ 0 (r, z). (32)∇u = ∇ N u + ∇ T uwhere ∇ N u is the gradient of u is the normal direction and ∇ T u is orthogonal to ∇ N u. More precisely, for a pointx = F (y, z)k∑∇ N u(x) = ∇u(x) · N j (z) N j (z).j=1Step 1. There exists C > 0 independent of u such that∫∫CI ≥ |∇ y ũ| 2 |y| dy dz +W ×B k β∫+W∫ β0( ∂v0∂rW ×B k β) 2rdr dz +∫∣ ∇y(ũ(y,z) − ũ0 (y, z) )∣ ∣ 2 dy dzW ×B k β|(∇ T u) ◦ F | 2 dy dz.First note that by (25), there is a constant C > 0 such that∫)I ≥(|∇ N u(F (y, z))| 2 − H 2 ũ2W ×Bβk |y| 2 G(z) dy dz∫− C(|∇ N u(F (y, z))| 2 + H 2 ũ2W ×B k β∫+W ×Bβk(|∇ T u(F (y, z))| 2 + ũ 2) (1 − C|y|)G(z) dy dz.For fixed z we can apply Lemma 3 to the function ũ(·, z). Observe thatand thusLemma 3 then yields∫)(|∇ N u(F (y, z))| 2 − H 2 u2|y| 2B k β∂ũ(y, z)∂y j= ∇u(F (y, z)) · N j (z)|∇ y ũ(y, z)| 2 = |∇ N u(F (y, z))| 2 .(33)|y| 2 )G(z)|y| dy dz (34)∫dy + C ũ 2 dy ≥ 1 ∫|y|(|∇Bβk 2βN u(F (y, z))| 2 + H 2 ũ2Bβk∫+ τB k β∫ β|∇ y (ũ − ũ 0 )| 2 dy + α r0)|y| 2 dy( ) 2 dv0dr.dr(35)9

We choose (and fix once and for all) β > 0 small enough so that 1/(2β) ≥ C + 1. Then multiplying (35) by G(z),integrating over W and combining the result with (34) we conclude that (33) holds.Step 2.‖∇v 0 ‖ 2 L 2 (W ×Bβ 2 )≤ CI. (36)By (33) the partial derivative ∂v0∂ri = 1, . . . , n − k. ButandBut note that ∂p∂z iis a tangent vector, henceIntegrating over W × Bβ k we have∫is bounded in L2 (W × Bβ 2 ∂v0) by CI. We just have to control the derivatives∫∂v 0(r, z) = r H ∂ũ− (y, z) ds(y)∂z i ∂z i∂B r⎡⎤∂ũ(y, z) = ∇u(F (y, z)) · ⎣ ∂p k∑ ∂N j+ y j⎦ .∂z i ∂z i ∂z ij=1|∇ z ũ(y, z)| ≤ C|∇ T u(F (y, z))| + C|y||∇ N u(F (y, z))|.W ×B k βfor some C independent of u by (33). It follows that∫W ×B 2 βby (37).Step 3. There is p > 2 such that∫|∇ z v 0 | 2 dy dz =∫≤W∫≤ C≤ CI∂z i,|∇ z ũ(y, z)| 2 dy dz ≤ CI, (37)∫ βW 0∫ β0W ×B k β∣∫∣∣∣2r 2H+1 − ∇ z ũ(y, z) ds(y)∣∂B r∫r k−1 − |∇ z ũ(y, z)| 2 ds(y) dr dz∂B r|∇ z ũ(y, z)| 2 dy dzMore precisely, for k ≥ 3 one can take any 2 < p < p k where p k is given bydr dz(38)‖ũ 0 ‖ 2 L p (W ×Bβ k )≤ CI. (39)1= 1 p k 2 − 2k(n − k + 2) ,and for k = 1 one can take 2 < p ≤ p 1 where p 1 is given by1= 1 p 1 2 − 1n + 1 .Using Sobolev’s inequality (on W × Bβ 2 ) combined with (36) we obtain∫W∫ β0|v 0 | q r dr dz ≤ CI q/2 ,with q given by 1 q = 1 2 − 1n−k+2 . That is, in terms of ũ 0 we have∫W∫ β0|ũ 0 | q r qH+1 dr dz ≤ CI q/2 . (40)10

We want an estimate for ∫ |ũ 0 | p r k−1 dr dz for some suitable 2 < p < q and for this we use Hölder’s inequality,distinguishing two cases:Case : k ≥ 3. We have∫ ∫ β|ũ 0 | p r k−1 dr dz =WWe then choose α so that0≤∫∫ βW( 0∫ ∫ βW 0C|ũ 0 | p r α r k−2−α r dr dz|ũ 0 | q r αq/p+1 dr, dzαp = H = k − 2 .2) p/q ( ∫ βIn order to have the second factor on the right hand side of (41) finite we need to imposewhich is equivalent to the conditionThus we need p = α/H < p k where p k is given byi.e.p k =k − 2 − α1 − p/qα −2k1 + 4 .q(k−2)2k(k − 2) ( )1 + 4q(k−2)1= 1 p k 2 − 2k(n − k + 2) .Observe that p k > 2. Combining then (40) and (41) finishes this case.Case: k = 1. In this case q is given by 1 q = 1 2 − 1n+1, and we can choose p = q:∫ ∫ β∫ ∫ β|ũ 0 | q r k−1 dr dz = |ũ 0 | q dr dzW0∫≤∫=W 0∫ βW 0∫ βW00|ũ 0 | q r −q/2+1 dr dz|ũ 0 | q r Hq+1 dr dzr k−2−α1−p/q +1 dr) 1−p/q.(41)because −q/2 + 1 < 0.Step 4.‖ũ − ũ 0 ‖ 2 L 2∗ (W ×Bβ k )≤ CI. (42)This is a consequence of Sobolev’s inequality applied to the function ũ − ũ 0 on the domain W × Bβ k . (33) alreadyprovides a bound in L 2 (W × Bβ k) for ∇ y(ũ − ũ 0 ). Hence we only need to obtain a bound for the derivative ofũ − ũ 0 with respect to z. In the case of the function ũ we have it already in (37). For ũ 0 it is derived by acomputation very similar to that at the end of Step 2. Indeed∫∫ ∫ β∣∫∣∣∣2|∇ z ũ 0 | 2 dy dz = r k−1 − ∇ z ũ(y, z) ds(y)∣ dr dzW 0∂B rW ×B k β≤ CI11

which we obtain as in (38).Conclusion. By (39) and (42) we see that‖ũ‖ 2 L p (W ×B k β ) ≤ CIfor some C independent of u. Changing variables and reintroducing the index i we have∫‖u‖ 2 L p (Tβ i ) ≤ C |∇u| 2 − H 2 u2d 2 + u2 .T i βAdding these inequalities over i proves the statement of the theorem.□2.2 A local version of the Hardy inequalityIn this section, we show how to adapt the proof of Theorem 1 to obtain Theorem 2. We first derive variants ofLemmas 1,2,3.Lemma 4 Let k ≠ 2 and H = (k − 2)/2. There exist constants C, β 0 > 0 such that for 0 < β ≤ β 0∫ [β(du ) ]2 ∫ [β(du ) ]2− H 2 u2dr r 2 r k−1 dr ≥+ H 2 u2dr r 2 r k dr (43)00for all u ∈ C ∞ c(0, β).Proof. Given v ∈ C ∞ c(0, 1 2), we have∫ 12Using this and (12) we obtain0∫ 1v 2 2dr = −2 rv dv0∫ 120dr dr ≤ C ∫ 12[ (du ) ]2+ H 2 u2dr r 20( ) 2 dvr 2 dr + 1 ∫ 12v 2 dr.dr 2 0∫ 1 ( )r k 22 dvdr ≤ C r 2 dr.0 drChanging variables, it then follows that for u ∈ Cc ∞ (0, β)∫ [β(du ) ]2 ∫ β+ H 2 u2dr r 2 r k dr ≤ Cβ k−2while (11) becomes0∫ βPicking β small, (43) follows from (44) and (45).A straightforward corollary of the above lemma is :000( ) 2 dvr 2 dr, (44)dr[ (du ) ]2 ∫ β( ) 2− H 2 u2dvdr r 2 r k−1 dr = β k−2 r dr. (45)drLemma 5 Let k ≠ 2, H = (k − 2)/2 and c > 0. There exist constants C, β 0 > 0 such that for 0 < β ≤ β 0∫ [β(du ) ]2 ∫ [β(du ) ]2− (H 2 − c) u2drr 2 r k−1 dr ≥+ (H 2 + c) u2drr 2 r k dr (46)00□for all u ∈ C ∞ c(0, β).Combining these two lemmas, we then obtain :12

Lemma 6 Let k ≠ 2, H = (k − 2)/2 and β > 0. Let Bβ k denote the ball of Rk centered at the origin and ofradius β. There exist positive constants C, β 0 such that for β ≤ β 0∫) (|∇u| 2 − H 2 u2|y| 2 dy ≥ C ∫)|y|(|∇u| 2 + H 2 u2β|y| 2 dy (47)for all u ∈ C ∞ cB k β(Bβ k \ {0}) and where u ∫0(r) = u 0 (|y|) = − udσ and v∂Brk 0 (r) = r H u 0 (r).As in Lemma 3, for a fixed value β = β 0 > 0 the proof is an application of the decomposition of a function inspherical harmonics. A simple scaling then yields the β-dependence of the constant appearing in (47).B k βProof of Theorem 2. Instead of (29), we now consider∫]J i :=[|∇u| 2 − H 2 u2d 2 . (48)Using the notation of (30) we then have by (25), (26),∫)∫J i ≥(|∇ N u(F i (y, z))| 2 − H 2 ũ2|y| 2 G(z) dy dz−CW ×B k βT i βW ×B k β)|y|(|∇ N u(F i (y, z))| 2 + H 2 ũ2|y| 2 G(z) dy dz ≥ 0,where we used Lemma 6 with β > 0 small in the last inequality. Adding the above estimates over i yields thedesired result.□3 Remarks on the potential a(x) = µ dist(x, Σ) −2For 0 < µ ≤ H 2 we consider the potentialand let L denote the operatora(x) = µd(x) 2Lu = −∆u − a(x)u.Note that a(x) and L depend on µ but we will omit this dependence from the notation.Recall that we defined the Hilbert space H as the completion of Cc∞ (Ω) with respect to the norm∫‖u‖ 2 H = |∇u| 2 − a(x)u 2 + Mu 2 . (49)ΩIf µ < H 2 then by Theorem 1, H coincides with H 1 0 (Ω).The main concern in this section is to obtain a precise description of the behavior near Σ of the first eigenfunctionϕ 1 of the operator L. Indeed, we shall prove:Lemma 7 There are positive constants C 1 , C 2 such thatfor x in a neighborhood of Σ, where α(µ) is given byC 1 d(x) −α(µ) ≤ ϕ 1 (x) ≤ C 2 d(x) −α(µ) (50)α(µ) = H − √ H 2 − µ. (51)Note that when µ = H 2 we have −α(µ) = 1 − k/2. Thus ϕ 1 ∉ H 1 0 (Ω) in this case. Before proving the abovelemma it will be necessary to show that if µ = H 2 then d 1−k/2 (appropriately modified so that it is zero on ∂Ω)belongs to H. We prove this and a little more next.13

Lemma 8 Let µ = H 2 and definev s (x) = η(x)d(x) 1−k/2 (− log d(x)) −swhere η ∈ C ∞ c (Ω) is a cut-off function such that η ≡ 1 in a neighborhood of Σ and η(x) = 0 for d(x) ≥ 1/2.Then v s ∈ H if and only if s > −1/2.Remark 2 This lemma was stated in [VZ] in the case where Σ is a point.Proof. Let us recall and also introduce some notation:Ω r = { x ∈ R N | d(x) < r }.Σ r = ∂Ω r = { x ∈ R N | d(x) = r }.By Papus theorems, the N − 1 dimensional area of Σ r is given by|Σ r | n−1 = ω k−1 r k−1 |Σ| n−k ,where ω k−1 is the area of the unit sphere in R k and | · | j denotes the j-dimensional Lebesgue measure.First we prove that v s ∈ H for s > −1/2. For this purpose it is enough to exhibit a sequence f ε ∈ H suchthat ‖f ε ‖ H ≤ C with C independent of ε and such that f ε → v s a.e. as ε → 0; we takef ε = ηd 1−k/2+ε (− log d) −s , ε > 0.Clearly f ε ∈ H 1 0 (Ω) ⊂ H, ∫ Ω f 2 ε ≤ C and f ε is smooth away from Σ. Thus to estimate ‖f i ‖ H it is sufficient toverify that for a fixed R > 0 small ∫Ω R|∇f ε | 2 − a(x)f 2 ε ≤ C (52)with C independent of ε.Near Σ, η ≡ 1 andso that|∇f ε | 2 = d −k+2ε( (1 − k/2 + ε) 2 (− log d) −2s + s(2 − k + 2ε)(− log d) −2s−1 + s 2 (− log d) −2s−2) .∫1ω k−1 |Σ| n−k∫ R|∇f ε | 2 = (1 − k/2 + ε) 2Ω R 0∫ R+ s 2 r 2ε−1 (− log r) −2s−2 dr.0r 2ε−1 (− log r) −2s dr + s(2 − k + 2ε)∫ R0r 2ε−1 (− log r) −2s−1 drNote that the last integral on the right hand side above is bounded independently of ε for s > −1/2, that isTherefore∫1ω k−1 |Σ| n−kIntegrating by partsΩ R(∫ R0r 2ε−1 (− log r) −2s−2 dr = O(1).|∇f ε | 2 − H 2 f ε2 )d 2 = ε(2 − k + ε)∫ R0+ s(2 − k + 2ε)r 2ε−1 (− log r) −2s dr∫ R0r 2ε−1 (− log r) −2s−1 dr + O(1).(53)∫ R0r 2ε−1 (− log r) −2s dr = 1 2ε R2ε (− log R) −2s − s ε∫ R0r 2ε−1 (− log r) −2s−1 dr14

and substituting in (53) yields∫1|∇f ε | 2 − H 2 f 2 )εω k−1 |Σ| n−k d 2 = 2 − k + ε R 2ε (− log R) −2s + εs2Ω R(Integrating by parts again∫ R0= εs∫ Rr 2ε−1 (− log r) −2s−1 dr = 1 2ε R2ε (− log R) −2s−1 − 2s + 12ε0∫ R0r 2ε−1 (− log r) −2s−1 dr + O(1)r 2ε−1 (− log r) −2s−1 dr + O(1). (54)∫ R0( ) 1r 2ε−1 (− log r) −2s−2 dr = O .εAfter substitution in (54) we obtain finally the estimate (52). Hence v s ∈ H for s > − 1 2 .Our argument to show that v s ∉ H for s ≤ −1/2 relies on the intuitive idea that ∫ (−∆v − a(x)v + Mv)v =‖v‖ 2 H . To exploit this idea, let us first compute ∆v s near Σ, where η ≡ 1. Writingy(t) = t 1−k/2 (− log t) −s .Then near Σ, since |∇d| 2 = 1,We recall here the fact (see [DN]) thatwhere g ∈ L ∞ . Hence,∆v s = y ′′ (d) + y ′ (d)∆d.∆d = k − 1d+ g∆v s = −H 2 d −k/2−1 (− log d) −s + s(s + 1)d −k/2−1 (− log d) −s−2 + (1 − k/2)gd −k/2 (− log d) −s+ sgd −k/2 (− log d) −s−1so that∆v s + H 2 v sd 2 = s(s + 1)d−k/2−1 (− log d) −s−2 + (1 − k/2)gd −k/2 (− log d) −s + sgd −k/2 (− log d) −s−1 . (55)Observe that ∆v s , v s /d 2 ∈ L 1 (Ω) and that equation (55) holds in the sense of distributions. Since we also havethat ∇v s ∈ L 1 (Ω), it follows that for any ϕ ∈ Cc∞ (Ω)∫ () ∫(v s |ϕ) H = −∆v s − H 2 v2 sΩd 2 ϕ + M v s ϕΩ∫(− s(s + 1)d −k/2−1 (− log d) −s−2 + (1 − k/2)gd −k/2 (− log d) −s + sgd −k/2 (− log d) −s−1 )ϕ (56)Ω R∫+ (−∆v s − H 2 v sΩ\Ω Rd∫Ω2 )ϕ + M v s ϕ.By density (56) holds also if ϕ is Lipschitz and ϕ = 0 on ∂Ω.Let us consider first the case s ≠ −1, so that s(s + 1) ≠ 0, and suppose that s < −1/2 and v s ∈ H. Thenthere exists v n ∈ Cc∞ (Ω) such that v n → v s in H. Note that since the injection H ⊂ L 2 (Ω) is continuous, bypassing to a subsequence we also have that v n → v s a.e.. Recall from [DN] (inequality (1.4) of Lemma 1.1) thatfor u ∈ H , u + ∈ H and ‖u + ‖ H ≤ ‖u‖ H . As a consequence v n + → v s in H and a.e.. Using v n + in (56) we concludethat∫d −k/2−1 (− log d) −s−2 v n + ≤ CΩ Rwith C independent of n. But then Fatou’s lemma implies that∫d −k (− log d) −2s−2 < ∞Ω R15

which is impossible for s < −1/2.For the case s = −1 the argument above does not work. We see that in this case, if Σ is flat in on open setof Ω then actually−∆w + H 2 w d 2 = 0,in that open set, wherew := v (−1) = ηd 1−k/2 (− log d).So we argue as follows: let −1/2 < s < 0. We are going to show that (L + M)v s ≥ (L + M)w near Σ. Ifwe assume that w ∈ H, then we can apply the maximum principle and deduce that v s ≥ εw near Σ, which isimpossible. Indeed ,by formula (55)and(L + M)w = −(1 − k/2)gd −k/2 (− log d) + gd −k/2 + Md 1−k/2 (− log d),(L + M)v s = −s(s + 1)d −k/2−1 (− log d) −s−2 − (1 − k/2)gd −k/2 (− log d) −s − sgd −k/2 (− log d) −s−1+ Md 1−k/2 (− log d) −s .Thus, for any ε > 0 there is a neighborhood Ω R of Σ such that(L + M)(εw − v s ) ≤ 0 in Ω R . (57)Pick ε > 0 such that εw − v s ≤ 0 in ∂Ω R . Under the hypothesis w ∈ H we can use a version of the maximumprinciple to deduce thatεw − v s ≤ 0 in Ω R .Indeed we have (assuming w ∈ H) that (εw − v s ) + ∈ H. Hence the function{(εw − v s ) + in Ω Rz =0 in Ω \ Ω ralso belongs to H. Let z n ∈ Cc∞ (Ω) be such that z n → z in H. Note that (57) holds in the sense of distributionsand hence testing (57) with z n+ we see thatLetting n → ∞ we get(εw − v s |z + n ) H ≤ 0.‖z‖ H = (εw − v s |z) H ≤ 0.Thus z ≡ 0 which implies that εw ≤ v s in Ω R , which concludes the proof of Lemma 8.Remark 3 To show that v s ∈ H for s > −1/2 one may be tempted to use other approximating sequences, anda very natural one isf i = min(v s , i), i = 1, 2, . . .Again it would be sufficient to establish that for a fixed R > 0 small∫|∇f i | 2 − a(x)fi2 ≤ CΩ Rwith C independent of i. For i large let r i > 0 be such thatr 1−k/2i (− log r i ) −s = iso that r i → 0 as i → ∞. A computation (that we omit) shows that∫1|∇f i | 2 − a(x)fi 2 = k − 2 (− log r i ) −2s − k − 2 (− log R) −2sw k−1 Ω R42+ s22s + 1We see that the above quantity remains bounded as i → ∞ only for s ≥ 0 !((− log R) −2s−1 − (− log r i ) −2s−1) .□16

Remark 4 The above example shows that for m ∈ N there exists v m ∈ H with ‖v m ‖ H = 1 andSince v = min(v, m) + (v − m) + we also have‖ min(v m , m)‖ H → ∞ as m → ∞.‖(v m − m) + ‖ H → ∞ as m → ∞.H is thus quite different from H 1 0 (Ω), in the sense that truncation operators like the one above are not uniformlybounded in H whereas it is always true that for any v ∈ H 1 0 (Ω), min(v, m) → v in the H 1 topology.Proof of Lemma 7. We will give a proof using a comparison argument with a suitable function. First let usrecall that in a neighborhood of Σ∆d = k − 1 + g(x),dwhere g is a bounded function. HenceLd −α = −∆d −α − µ d−αd 2 = − ( −α 2 − α(k − 2) + µ − α(k − 1)gd ) d −α−2 (58)Let α = α(µ) as given by (51). This implies that α 2 − α(k − 2) + µ = 0. ThenL(d −α + Cd −α+1 ) = − ( −α(k − 1)g + C((−α + 1) 2 + (−α + 1)(k − 2) + µ) + (−α + 1)(k − 1)gd ) d −α−1 .Instead of working with the operator L = −∆ − a(x) consider L + M where M is large so that (4) holds (with Creplaced by M; this is the same M that we use in the definition of the space H.) Then, since (−α + 1) 2 + (−α +1)(k − 2) + µ) > 0 we conclude that(L + M)(d −α + Cd −α+1 ) = − ( −α(k − 1)g + C((−α + 1) 2 + (−α + 1)(k − 2) + µ) + (−α + 1)(k − 1)gd ) d −α−1≤ 0+ M(d −α + Cd −α+1 )in a fixed neighborhood Ω R , R > 0 of Σ. On the other hand, the first eigenfunction ϕ 1 of L satisfies(59)(L + M)ϕ 1 = (λ 1 + M)ϕ 1 ≥ 0. (60)Now, both functions ϕ 1 and d −α + Cd −α+1 are smooth away from Σ and near Σ so that one can find ε > 0such that ε(d −α + Cd −α+1 ) ≤ ϕ 1 in ∂Ω R . We can use now the same version of the maximum principle as in theprevious lemma to deduce thatε(d −α + Cd −α+1 ) ≤ ϕ 1 in Ω R .For the estimate ϕ 1 ≤ C 2 d −α(µ) we need a result from [DD].Theorem 4 Let Ω be a bounded smooth domain. Assume that ã ∈ L 1 loc (Ω), ã is bounded below (i.e. inf Ω ã > −∞)and that it satisfies( ∫γ |u| r) ∫∫2/r+ ã(x)u 2 ≤ |∇u| 2ΩΩΩfor some γ > 0 and r > 2.Let ϕ 1 > 0 denote the first eigenfunction for the operator L = −∆ − ã(x) with zero Dirichlet boundarycondition, normalized by ‖ϕ 1 ‖ L 2 (Ω) = 1 and ζ 0 denote the solution of{−∆ζ0 − ã(x)ζ 0 = 1 in Ωζ 0 = 0on ∂Ω.Then there exists C = C(Ω, γ(a), r) > 0 such thatC −1 ζ 0 ≤ ϕ 1 ≤ Cζ 0 .17

Proof of Lemma 7 continued. We use the above theorem with ã = a − M. In view of this result it suffices toshow thatζ 0 ≤ Cd −α(µ) .Using (58) and taking α = α(µ) we have(L + M)(d −α − Cd −α+1 ) = − ( −α(k − 1)g − C((−α + 1) 2 + (−α + 1)(k − 2) + µ) + (−α + 1)(k − 1)gd ) d −α−1≥ 1+ M(d −α − Cd −α+1 )in Ω R if we choose R > 0 small and C > 0 large enough. Now take C 1 large so that ζ 0 ≤ C 1 (d −α − Cd −α+1 )in ∂Ω R . Using the maximum principle as before we deduce that ζ 0 ≤ C 1 (d −α − Cd −α+1 ), which finishes theproof.□Remark 5 The fact that d 1−k/2 ∈ H for µ = H 2 was used in the proof above at the point where the maximumprinciple was applied. That argument requires that both functions that one would like to compare are in H. Ingeneral, if one of these functions does not belong to H then the maximum principle cannot be applied; see [DD]for an example.4 Some applications4.1 Minimizers for the Hardy inequalityWe start this section extending a result of Brezis and Marcus [BM] regarding the quantityJ λ =u∈C ∞ cinf(Ω)\{0} ∫Ω |∇u|2 − λu 2∫Ω u2 /d(x) 2 , (61)where as usual d(x) = dist(x, Σ).The case studied in [BM] corresponds to Σ = ∂Ω, and an interesting feature that the authors found in thatwork is the following, which we state in our situation:Theorem 5 Fix λ ∈ R.Then the infimum in (61) is achieved (in H 1 0 (Ω)) if and only ifJ λ < H 2 .Proof. To prove that the condition J λ < H 2 is sufficient for the infimum in (61) to be achieved, one just needsto mimic the arguments in [BM] so we skip this step.We prove the converse, that is, the claim that if J λ = H 2 then the infimum is not achieved, with an argumentsimilar in spirit to that of [BM]. Suppose that the infimum is achieved by a function u ∈ H0 1 (Ω), which we canassume to be non-negative and not identically zero. Assume also that J λ = H 2 . Then u satisfies−∆u − H 2 ud(x) 2 = λu.It follows that λ is the first eigenvalue for the operator −∆ − H 2 /d 2 and that u > 0. Moreover u has to be amultiple of ϕ 1 (for this results see e.g. [DD, Lemma 2.3]). But by (50) we know that ϕ 1 ∼ d 1−k/2 . This showson one hand that ∫ Ω u2 /d 2 = ∞. But Hardy’s inequality (4) implies on the other hand that ∫ Ω u2 /d 2 < ∞. □4.2 Study of a semilinear problemIn this section, we return to the study of a semilinear problem studied in [DN]. For p > 1, 0 < µ ≤ H 2 and λ > 0consider the equation ⎧⎪ ⎨⎪ ⎩−∆u − µd(x) 2 u = u p + λ in Ωu > 0 in Ωu = 0 on ∂Ω,(62)18

where as usual d(x) = dist(x, Σ). We showed in [DN] that (at least for small values of µ > 0) there exists acritical exponentp 0 = 1 + 2 with α(µ) = H − √ Hα(µ)2 − µsuch that (62) admits no solution (in any reasonable sense) for p > p 0 and λ > 0 whereas for some λ ∗ = λ ∗ (p)solutions exist when p < p 0 and 0 < λ ≤ λ ∗ (and again no solution exists when λ > λ ∗ ). However the criticalcase p = p 0 remained open. Using Lemma 7 in combination with Theorem 4, and following Proposition 6.1 of[DN], one can prove the following :Theorem 6 Given any λ > 0, Problem (62) with p = p 0 admits no solution.5 Estimate for solutions of some singular equationsIn what follows we will use the method developed in [DD] to prove Theorem 3. The idea is to work with w = u/ϕ 1 ,which satisfies an elliptic equation to which Moser’s iteration technique can be applied. In the argument it isdesirable to approximate the potential a(x) by bounded ones. In order to get the convergence of the correspondingsolutions, it is convenient to rewrite the equation (7) aswhere−∆u − ã(x)u = C 0 u + fã = a − C 0and C 0 is chosen large enough, larger than C in (7) (although it will be taken even larger at one point below).We observe that now for any h ∈ H ∗ the equation{−∆v − ãv = h in Ω(63)v = 0 on ∂Ωhas a unique solution v ∈ H. Let us also note that the first eigenfunction for the operator −∆ − ã is still ϕ 1 .Let us state a result which is a kind of Sobolev inequality with weight (see a proof in [DD]).Lemma 9 Assume that a satisfies (8). Then for any 2 ≤ q ≤ r there is a constant C depending only Ω, r andγ(a) such that( ∫ ϕ s 1|w| q) ∫2/q≤ C ϕ 2 1(|∇w| 2 + w 2) (64)Ωfor all w ∈ C 1 (Ω), where s is given by the relationLemma 10 Let 0 < m < r and suppose thatandΩsr = q − 2r − 2 .p >p ≥Then for f ∈ H ∗ , the unique solution v to (63) satisfies2rm(r − 2) , (65)rr − m . (66)|v(x)| ≤ C‖ϕ 1−m1 h‖ p ϕ 1 (x), a.e. x ∈ Ω.Remark 6 If m ≥ 1, the assumption h ∈ H ∗ can be dropped since one can prove that ‖h‖ H ∗19≤ C‖ϕ 1−m1 h‖ p .

Proof of Remark 6. If ‖ϕ 1−m1 h‖ p = +∞, there is nothing to prove. Otherwise h is locally integrable and forϕ ∈ Cc ∞ (Ω), ∣∫∣∣∣ hϕ∣ ≤ ‖hϕ1−m 1 ‖ p ‖ϕϕ m−11 ‖ p ′ ≤ ‖hϕ 1−m1 ‖ p ‖ϕ 1 ‖ m/m′mp ‖ϕ‖ ′ mp ′,Ωwhere we used Hölder’s inequality twice. Now (66) implies that mp ′ ≤ r, so we end up with∫∣ hϕ∣ ≤ C‖hϕ1−m 1 ‖ p ‖ϕ‖ H ∀ϕ ∈ Cc ∞ (Ω),Ωwhich is the desired result.□Proof of Lemma 10. First we note that it is sufficient to prove this result for a bounded potential a, as longas the constants that appear in the estimates only depend on the constants r, γ, C appearing in (7) and Ω. Thisis the same argument employed in [DD] and we will just sketch it here. Consider ã k = min(ã, k), and the firsteigenfunction ϕ k 1 and solution v k of (63) with the potential a replaced by a k . Then ϕ k 1 → ϕ 1 in H and v k → v.Furthermore, ã k satisfies( ∫γ |u| r) ∫∫2/r+ ã(x)u 2 ≤ |∇u| 2 , ∀u ∈ Cc ∞ (Ω).ΩΩΩSo it is enough to establish the results for ã k . We will assume then that ã is bounded. Then all functions involvedbelong to C 1,α (Ω).By working with h + and h − we can assume that h ≥ 0 and hence v ≥ 0. SetThen w satisfies the equationw = v ϕ 1.−∇ · (ϕ 2 1∇w) = ϕ 1 h − (C 0 + λ 1 )ϕ 1 vMultiplying the equation by w 2j−1 and integrating in Ω we find∫∫2j − 1j∫Ω2 ϕ 2 1|∇w j | 2 = ϕ 1 w 2j−1 h − (C 0 + λ 1 )Ω∫∫= ϕ 1 w 2j−1 h − (C 0 + λ 1 )ΩΩΩϕ 1 vw 2j−1ϕ 2 1w 2j (67)Using the variant of Sobolev’s inequality (64) applied to w j with s = mp ′ we obtain( ∫ ϕ mp′1 w qj) ∫2/q≤ C ϕ 2 1(|∇w j | 2 + w 2j ) (68)Ωwhere q is given byq = 2 + mp ′ r − 2 .rWe note that by (66) we have mp ′ ≤ r and therefore we can indeed apply Lemma 9. Combining (67) with (68)we get( ∫ ϕ mp′1 w qj) 2/q Cj 2 ∫≤ ϕ 1 w 2j−1 h + C(1 − j2Ω 2j − 1 Ω2j − 1 (C 0 + λ 1 )) ∫ ϕ 2 1w 2jΩWe make C 0 larger if necessary, so that for j ≥ 1 the second term on the right hand side is negative. Therefore( ∫ ϕ mp′1 w qj) ∫2/q≤ Cj ϕ 1 w 2j−1 h.ΩBy Hölder’s inequality∫Ωϕ 1 w 2j−1 h ≤ΩΩ( ∫ Ωϕ mp′1 w (2j−1)p′) 1/p ′ ‖ϕ 1−m1 h‖ p20

and therefore( ∫ ϕ mp′1 w qj) 1/(qj) ( ∫ ≤ (Cj)1/(2j)ϕ mp′1 w (2j−1)p′) 1/(2jp ′ )‖ϕ1−m1 h‖ 1/(2j)p . (69)Ω ΩObserve now that condition (65) is equivalent toq > 2p ′and therefore a standard iteration argument yields the result.Let indeed j 0 = 1/2 + m/2 and for k ≥ 1, define j kinductively by(2j k − 1)p ′ = qj k−1 . (70)One can easily show that {j k } is increasing and converges to +∞ as k → +∞, so that if(∫ ) 1/qjk (∫θ k = ϕ mp′1 w qj kΩϕ mp′1Ω) −1/qjk,{θ k } k is increasing and converges to ‖w‖ ∞ as k → ∞. Observe in passing that since ϕ 1 ∈ H and mp ′ ≤ r,(∫ )< ∞.Equation (69) then yieldsθ k ≤ (Cj k ) 12j k(∫Ωϕ mp′1ϕ mp′1Ω) −1/(qjk )+1/(2j k p ′ )θqj k−12j k p ′k−1( ∫ ϕ (1−m)p1 h p) 1/(2j k p). (71)Ω( ∫ ) 1/pNow, either {θ k } k remains bounded byΩ ϕ(1−m)p 1 h p for all k, in which case passing to the limit providesthe desired inequality, either there exists a smallest integer k 0 such that for k ≥ k 0 − 1,θ k ≥Using this inequality in (71), we obtain for k ≥ k 0 ,Applying (72) inductively, it follows that‖w‖ ∞ ≤θ k ≤ (Cj k ) 1/2j k( ∫ ϕ (1−m)p1 h p) 1/p.Ω( ∫ “ ”) 1ϕ mp′ j − 1k q + 12p ′1θ k−1 . (72)Ω[∞∏( ∫ “ ”]) 1(Cj k ) 1/2j kϕ mp′ j − 1k q + 12p ′1θ k0−1. (73)k=kΩ 0Starting from (70), a straightforward computation shows that for some c > 0j k =( q2p ′ ) kj 0 +( ) k q2p − 1 ( ) k ′ qq ∼ c2p− 1 2p ′ as k → ∞.′Since q/2p ′ > 1, we then conclude that the infinite product in the right-hand side of (73) converges to somefinite constant.If k 0 ≥ 2, applying again (71) for k = k 0 − 1, we also haveθ k0−1 ≤ Cθqj k0 −22j k0 −1 p′k 0−2( ∫ ϕ (1−m)p1 h p) 1/(2j k0 −1p) ( ∫≤ C ϕ (1−m)p1 h p) 1/p, (74)ΩΩ21

where we used the minimality of k 0 in the last inequality. Combining (74) and (73) yields the desired result.If k 0 = 1 then by (69),( ∫ ) 1/(qj0) ( ∫ϕ mp′1 w qj0 ≤ C v r) m/(2j 0r)( ∫ ϕ (1−m)p1 h p) 1/(2j 0p), (75)Ω ΩΩwhere we used Hölder’s inequality and the fact that mp ′ ≤ r, which follows from (66). Now,∫γ‖v‖ 2 r ≤ ‖v‖ 2 H = hv ≤ ‖hϕ 1−m1 ‖ p ‖ϕ1 m−1 v‖ p ′ ≤ ‖hϕ 1−m1 ‖ p ‖w‖ ∞ ‖ϕ m 1 ‖ p ′ ≤ C‖hϕ 1−m1 ‖ p ‖w‖ ∞ . (76)Using (73), (75) and (76), we obtainΩ‖w‖ ∞ ≤ C‖hϕ 1−m1 ‖ m2(m+1) + 1m+12(m+1)∞ ,p ‖w‖ mwhich after simplification yields the desired result.□Lemma 11 Let 0 < m < r and suppose thatp

Remark 7 A direct consequence of the above lemma is that if instead of assumingwe assume thatthen the conclusion is that for all 1 ≤ α < ∞(∫where the constant C may depend on α.Ωϕ mp′1p

Indeed, Lemma 11 applied to u 2 with p = p 1 = 2 and m 1 = 1 implieswhere p 2 is given by( ∫ Ωϕ 2 1∣ u ∣2 ∣∣p 2) 1/p2≤ C‖u‖L 2, (82)ϕ 11= 1 (− 1 − 2 )p 2 p 1 qand q = 2 + 2 r−2r . Inequality (80) combined with (82) shows that (81) holds for p 2.We continue this process using Lemma 11 repeatedly with p and m in that lemma given by1= 1 (− 1 − 2 ),p k+1 p k qm k = 2(p k − 1)p kand q = 2 + 2 r−2r. At each step we obtain (inductively)( ∫ ϕ 2 1∣ u ∣2 ∣∣p k+1) 1/pk+1( ∫ ≤ C ϕ 2 1∣ u ∣ ∣∣p k) 1/pkΩ ϕ 1 Ω ϕ 1≤ C(‖ϕ 1−m1 f‖ p + ‖u‖ L 2).This together with (80) proves that (81) holds for p k+1 . We can continue in this way providedor equivalentlyLet ¯k be the first time that we findso that (81) still holds for ¯k. If p¯k >rr−21(− 1 − 2 )> 0,p k qp k

[Da] E.B. Davies, The Hardy constant, Quart. J. Math. Oxford Ser. (2) 46 (1995), 417–431.[DD]J. Dávila and L. Dupaigne, Comparison results for PDEs with a singular potential, Proc. Roy. Soc. EdinburghSect. A 133 (2003), 61–83.[D] L. Dupaigne, A nonlinear elliptic PDE with the inverse square potential, J. Anal. Math. 86 (2002),359–398.[DN]L. Dupaigne and G. Nedev, Semilinear elliptic PDE’s with a singular potential, Adv. Differential Equations7 (2002), 973–1002.[Ma] V. Maz’ja, Sobolev spaces. Springer Series in Soviet Mathematics. Springer-Verlag, Berlin, 1985.[MMP] M. Marcus, V. Mizel, and Y. Pinchover, On the best constant for Hardy’s inequality in R n , Trans. Amer.Math. Soc. 350 (1998), 3237–3255.[MSo][VZ]T. Matskewich, P.E. Sobolevskii, The best possible constant in generalized Hardy’s inequality for convexdomain in R n , Nonlinear Anal. 28 (1997), 1601–1610.J.L. Vázquez and E. Zuazua, The Hardy inequality and the asymptotic behavior of the heat equation withan inverse square potential, J. Funct. Anal. 173, (2001), 103 – 153.25

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