Maximization of the second positive Neumann eigenvalue for planar domains
aa r X i v : . [ m a t h . SP ] J a n MAXIMIZATION OF THE SECOND POSITIVE NEUMANNEIGENVALUE FOR PLANAR DOMAINS
ALEXANDRE GIROUARD, NIKOLAI NADIRASHVILI, AND IOSIF POLTEROVICH
Abstract.
We prove that the second positive Neumann eigenvalue ofa bounded simply-connected planar domain of a given area does notexceed the first positive Neumann eigenvalue on a disk of a twice smallerarea. This estimate is sharp and attained by a sequence of domainsdegenerating to a union of two identical disks. In particular, this resultimplies the Polya conjecture for the second Neumann eigenvalue. Theproof is based on a combination of analytic and topological arguments.As a by-product of our method we obtain an upper bound on the secondeigenvalue for conformally round metrics on odd-dimensional spheres. Introduction and main results
Neumann eigenvalues of planar domains.
Let Ω be a boundedplanar domain. The domain Ω is said to be regular if the spectrum of theNeumann boundary value problem on Ω is discrete. This is true, for instance,if Ω satisfies the cone condition, that is there are no outward pointing cusps(see [NS] for more refined conditions and a detailed discussion).Let 0 = µ < µ (Ω) ≤ µ (Ω) ≤ · · · ր ∞ be the Neumann eigenvalues ofa regular domain Ω. According to a classical result of Szeg¨o ([Sz], see also[SY, p. 137], [Hen, section 7.1]), for any regular simply-connected domainΩ(1.1.1) µ (Ω) Area(Ω) ≤ µ ( D ) π ≈ . π, where D is the unit disk, and µ ( D ) is the square of the first zero of thederivative J ′ ( x ) of the first Bessel function of the first type. The proof ofSzeg¨o’s theorem relies on the Riemann mapping theorem and hence worksonly if Ω is simply-connected. However, inequality (1.1.1) holds without thisassumption, as was later shown by Weinberger [We].The P´olya conjecture for Neumann eigenvalues [Po1] (see also [SY, p.139]) states that for any regular bounded domain Ω(1.1.2) µ k (Ω) Area(Ω) ≤ k π for all k ≥
1. This inequality is true for all domains that tile the plane, e.g.,for any triangle and any quadrilateral [Po2]. It follows from the two-termasymptotics for the eigenvalue counting function ([Iv], [Me]) that for anydomain there exists a number K such that (1.1.2) holds for all k > K . Date : November 1, 2018.
Inequality (1.1.1) implies that (1.1.2) is true for µ . The best one couldshow for k ≥ µ k ≤ πk ([Kro]). In the present paper we consider thecase k = 2. Our main result is Theorem 1.1.3.
Let Ω be a regular simply-connected planar domain. Then (1.1.4) µ (Ω) Area(Ω) ≤ µ ( D ) π ≈ . π, with the equality attained in the limit by a family of domains degeneratingto a disjoint union of two identical disks. The second part of the theorem immediately follows from (1.1.4). Indeed,if Ω is a disjoint union of two identical disks then (1.1.4) is an equality.Joining the two disks by a passage of width ǫ we can construct a family ofsimply-connected domains such that the left-hand side in (1.1.4) tends to2 µ ( D ) π as ǫ → k = 2 for anyregular simply-connected planar domain. Remark . It would be interesting to check the bound (1.1.4) for non-simply connected domains. We believe it remains true in this case as well.
Remark . All estimates discussed in this section have analogues in theDirichlet case. For example, (1.1.1) is the Neumann counterpart of thecelebrated Faber-Krahn inequality ([Fa, Kra1], see also [Hen, section 3.2]),which states that among all bounded planar domains of a given area, thefirst Dirichlet eigenvalue is minimal on a disk. Similarly, Theorem 1.1.3 canbe viewed as an analogue of the result due to Krahn and Szeg¨o ([Kra2],[Hen, Theorem 4.1.1]), who proved that among bounded planar domains ofa given area, the second Dirichlet eigenvalue is minimized by the union oftwo identical disks.1.2.
Eigenvalue estimates on spheres.
Let ( S n , g ) be a sphere of dimen-sion n ≥ g . Let0 < λ ( S n , g ) ≤ λ ( S n , g ) ≤ · · · ր ∞ be the eigenvalues of the Laplacian on ( S n , g ). Hersch [Her] adapted theapproach of Szeg¨o to prove that λ ( S , g ) Area( S , g ) ≤ π for any Rie-mannian metric g , with the equality attained on a sphere with the standardround metric g . In order to obtain a similar estimate in higher dimen-sions, one needs to restrict the Riemannian metrics to a fixed conformalclass [EI]. Indeed, in dimension ≥
3, if one only restricts the volume, λ isunbounded [CD]. In particular, it was shown in [EI] (see also [MW]) thatfor any metric g in the class [ g ] of conformally round metrics,(1.2.1) λ ( S n , g ) Vol( S n , g ) n ≤ n ω /nn , AXIMIZATON OF THE SECOND POSITIVE NEUMANN EIGENVALUE 3 where ω n = 2 π n +12 Γ (cid:0) n +12 (cid:1) is the volume of the unit round n -dimensional sphere. This result can beviewed as a generalization of Hersch’s inequality, since all metrics on S areconformally equivalent to the round metric g . A similar problem for higher eigenvalues is much more complicated. Itwas proved in [CE, Corollary 1] that(1.2.2) λ ck ( S n , [ g ]) := sup g ∈ [ g ] λ k ( S n , g ) Vol( S n , g ) n ≥ n ( k ω n ) /n , The number λ ck ( S n , [ g ]) is called the k -th conformal eigenvalue of ( S n , [ g ]).It was shown in [Na] that for k = 2 and n = 2 the inequality in (1.2.2) is anequality, and the supremum is attained by a sequence of surfaces tending toa union of two identical round spheres. We conjecture that the same is truein all dimensions: Conjecture 1.2.3.
The second conformal eigenvalue of ( S n , [ g ]) equals (1.2.4) λ c ( S n , [ g ]) = n (2 ω n ) /n for all n ≥ . As a by-product of the method developed for the proof of Theorem 1.1.3,we prove un upper bound for λ c ( S n , [ g ]) when the dimension n is odd (thiscondition is explained in Remark 4.3.8). Our result is in good agreementwith Conjecture 1.2.3. Theorem 1.2.5.
Let n ∈ N be odd and let ( S n , g ) be a n -dimensional spherewith a conformally round metric g ∈ [ g ] . Then (1.2.6) λ ( S n , g ) Vol( S n , g ) n < ( n + 1) π n +12 Γ( n )Γ( n )Γ( n + ) ! /n Remark . Note that the Dirichlet energy is not conformally invariantin dimensions n ≥ , . n . Moreover, the difference between the two constants tends to 0as the dimension n → ∞ , and hence (1.2.6) is “asymptotically sharp” asfollows from (1.2.2). Remark . It was conjectured in [Na] that if n = 2 then (1.2.2) is anequality for all k ≥
1, with the maximizer given by the union of k identicalround spheres. One could view it as an analogue of the P´olya conjecture(1.1.2) for the sphere. Note that a similar “naive” guess about the maximizer ALEXANDRE GIROUARD, NIKOLAI NADIRASHVILI, AND IOSIF POLTEROVICH of the k -th Neumann eigenvalue of a planar domain is false: a union of k equal disks can not maximize µ k for all k ≥
1, because, as one could easilycheck, this would contradict Weyl’s law. For the same reason, (1.2.2) cannot be an equality for all k ≥ n ≥ Plan of the paper.
The paper is organized as follows. In sections 2.1–2.5 we develop the “folding and rearrangement” technique based on the ideasof [Na] and apply it to planar domains. The topological argument used in theproof of Theorem 1.1.3 is presented in section 2.6. In section 2.7 we completethe proof of the main theorem using some facts about the subharmonicfunctions. In sections 3.1 and 3.2 we prove the auxiliary lemmas used inthe proof of Theorem 1.1.3. In section 4.1 we present a somewhat strongerversion of the classical Hersch’s lemma ([Her]). In sections 4.2 and 4.3 weadapt the approach developed in sections 2.1-2.7 for the case of the sphere.In section 4.4 we use the modified Rayleigh quotient to complete the proofof Theorem 1.2.5.
Acknowledgments.
We are very grateful to L. Parnovski for a stimulatingquestion that has lead us to Theorem 1.1.3, and to M. Levitin for manyuseful discussions on this project. We would also like to thank B. Colboisand L. Polterovich for helpful remarks.2.
Proof of Theorem 1.1.3
Standard eigenfunctions for µ on the disk. Let D = (cid:8) z ∈ C (cid:12)(cid:12) | z | < (cid:9) be the open unit disk. Let J be the first Bessel function of the first kind,and let ζ ≈ .
84 be the smallest positive zero of the derivative J ′ . Set f ( r ) = J ( ζr ) . Given R ≥ s = ( R cos α, R sin α ) ∈ R , define X s : D → R by(2.1.1) X s ( z ) = f ( | z | ) z · s | z | = Rf ( r ) cos( θ − α ) , where r = | z | , θ = arg z , and z · s denotes the scalar product in R . Thefunctions X s are the Neumann eigenfunctions corresponding to the doubleeigenvalue µ ( D ) = µ ( D ) = ζ ≈ . . The functions X e and X e form a basis for this space of eigenfunctions(where the vectors { e , e } form the standard basis of R ).2.2. Renormalization of measure.
We say that a conformal transforma-tion T of the disk renormalizes a measure dν if for each s ∈ R , Z D X s ◦ T dν = 0 . (2.2.1) AXIMIZATON OF THE SECOND POSITIVE NEUMANN EIGENVALUE 5
Finite signed measures on D can be seen as elements of the dual of thespace C ( D ) of continuous functions. As such, the norm of a measure dν is(2.2.2) k dν k = sup f ∈ C ( D ) , | f |≤ (cid:12)(cid:12)(cid:12)(cid:12)Z D f dν (cid:12)(cid:12)(cid:12)(cid:12) The following result is an analogue of Hersch’s lemma (see [Her], [SY]).
Lemma 2.2.3.
For any finite measure dν on D there exists a point ξ ∈ D such that dν is renormalized by the automorphism d ξ : D → D defined by d ξ ( z ) = z + ξξz + 1 . Proof.
Set M = R D dν and define the continuous map C : D → D by C ( ξ ) = 1 M f (1) Z D ( X e , X e ) ( d ξ ) ∗ dν = 1 M f (1) Z D ( X e ◦ d ξ , X e ◦ d ξ ) dν Let e iθ ∈ S = ∂ D . For any z ∈ D ,lim ξ → e iθ d ξ ( z ) = e iθ . This means that the map C can be continuously extended to the closure D by C = id on ∂ D . By the same topological argument as in Hersch’s lemma(and as in the proof of the Brouwer fixed point theorem), a continuous map C : D → D such that C ( ξ ) = ξ for ξ ∈ ∂ D must be onto. Hence, there existssome ξ ∈ D such that C ( ξ ) = 0 ∈ D . (cid:3) Lemma 2.2.4.
For any finite measure dν the renormalizing point ξ isunique.Proof. First, let us show that if the measure dν is already renormalized then ξ = 0. Suppose that D ∋ η = 0 renormalizes dν . Without loss of generalityassume that η is real and positive (if not, apply a rotation). Setting s = 1,by Lemma 3.1.1 we get that X s ( d η ( z )) > X s ( z ) for all z ∈ D and hence Z D X s ◦ d η dν > Z D X s dν = 0 , which contradicts the hypothesis that η renormalizes dν .Now let dν be an arbitrary finite measure which is renormalized by ξ ∈ D .Assume η ∈ D also renormalizes dν . Let us show that η = ξ . Taking intoaccount that d − ξ ◦ d ξ = d = id, we can write( d η ) ∗ dν = ( d η ◦ d − ξ ) ∗ ( d ξ ) ∗ dν. A straightforward computation shows that d η ◦ d − ξ = 1 − η ¯ ξ − ¯ ηξ d α , ALEXANDRE GIROUARD, NIKOLAI NADIRASHVILI, AND IOSIF POLTEROVICH where α = d − ξ ( η ) and (cid:12)(cid:12)(cid:12) − η ¯ ξ − ¯ ηξ (cid:12)(cid:12)(cid:12) = 1. This implies that dα renormalizes( d ξ ) ∗ dν which is already renormalized. Hence, as we have shown above, α = d − ξ ( η ) = 0, and therefore ξ = η . (cid:3) Given a finite measure, we write Γ( dν ) ∈ D for its unique renormalizingpoint ξ ∈ D . Corollary 2.2.5.
The renormalizing point Γ( dν ) ∈ D depends continuouslyon the measure dν .Proof. Let ( dν n ) be a sequence of measures converging to the measure dν inthe norm (2.2.2). Without loss of generality suppose that dν is renormalized.Let ξ n ∈ D ⊂ D be the unique element such that d ξ n renormalizes dν n . Let( ξ n k ) be a convergent subsequence, say to ξ ∈ D . Now, by definition of ξ n there holds 0 = lim k →∞ | Z D X s ( d ξ nk ) ∗ dν n k | = | Z D X s ( d ξ ) ∗ dν | , and hence d ξ renormalizes dν . Since we assumed that dν is normalized, byuniqueness we get ξ = 0. Therefore, 0 is the unique accumulation point ofthe set ξ n ∈ D and hence by compactness we get ξ n →
0. This completesthe proof of the lemma. (cid:3)
Corollary 2.2.5 will be used in the proof of Lemma 2.5.3, see section 3.2.2.3.
Variational characterization of µ . It follows from the Riemannmapping theorem and Lemma 2.2.3 that for any simply-connected domainΩ there exists a conformal equivalence φ : D → Ω, such that the pullbackmeasure dµ ( z ) = φ ∗ ( dz ) = | φ ′ ( z ) | dz satisfies for any s ∈ S Z D X s ( z ) dµ ( z ) = 0 . (2.3.1)Using a rotation if necessary, we may also assume that Z D X e ( z ) dµ ( z ) ≥ Z D X s ( z ) dµ ( z ) . (2.3.2)for any s ∈ S . The proof of Theorem 1.1.3 is based on the followingvariational characterization of µ (Ω): µ (Ω) = inf E sup = u ∈ E R D |∇ u | dz R D u dµ (2.3.3)where E varies among all two-dimensional subspaces of the Sobolev space H ( D ) that are orthogonal to constants, that is for each f ∈ E , R D f dµ = 0.Note that the Dirichlet energy is conformally invariant in two dimensions,and hence the numerator in (2.3.3) can be written using the standard Eu-clidean gradient and the Lebesgue measure. AXIMIZATON OF THE SECOND POSITIVE NEUMANN EIGENVALUE 7
Folding of hyperbolic caps.
It is well-known that the group of au-tomorphisms of the disk coincides with the isometry group of the Poincar´edisk model of the hyperbolic plane [Bea, section 7.4]. Therefore, for any ξ ∈ D , the automorphism d ξ ( z ) = z + ξξz + 1is an isometry. Note that we have d = id and d ξ (0) = ξ for any ξ .Let γ be a geodesic in the Poincar´e disk model, that is a diameter orthe intersection of the disk with a circle which is orthogonal to ∂ D . Eachconnected component of D \ γ is called a hyperbolic cap on D . The space ofhyperbolic caps is parametrized as follows. Given ( r, p ) ∈ ( − , × S let a r,p = d rp ( a ,p ) , where a ,p = { x ∈ D : x · p > } is the half-disk such that p is the center of its boundary half-circle. Thelimit r → ∂ D (that is, a → p ), while the limit r → − D (that is, a → D ). Given p ∈ D , we define the automorphism PSfrag replacements r ppa ,p a r,pd rp −→ R p ( z ) = − p ¯ z . It is the reflection with respect to the line going through0 and orthogonal to the segment joining 0 and p . For each cap a r,p , let usdefine a conformal automorphism τ a = d rp ◦ R p ◦ d − rp . One can check that this is the reflection with respect to the hyperbolicgeodesic ∂a r,p . In particular, τ a ( a ) = D \ a and τ a is the identity on ∂a .2.5. Folding and rearrangement of measure.
Given a measure dµ on D and a hyperbolic cap a ⊂ D , the folded measure dµ a is defined by dµ a = ( dµ + τ ∗ a dµ on a, D \ a. Clearly, the measure dµ a depends continuously in the norm (2.2.2) on thecap a ⊂ D . For each cap a ∈ D let us construct the following conformalequivalence ψ a : D → a . First, observe that it follows from the proof of the ALEXANDRE GIROUARD, NIKOLAI NADIRASHVILI, AND IOSIF POLTEROVICH
PSfrag replacements T a −→ φ b −→ T ′ a −→ ψ a a b DD Riemann mapping theorem [Ta, p.342] that there exists a family φ a : a → D of conformal equivalences depending continuously on the cap a such thatlim a → D φ a = id pointwise. Let ξ ( a ) = Γ( dµ a ) be the normalizing point for themeasure dµ a and set T a = d ξ ( a ) . The measure ( T a ) ∗ dµ a is supported in thecap b = T a ( a ). Pushing this measure to the full disk using φ b : b → D leadsto the measure ( φ b ◦ T a ) ∗ dµ a . Let η ( a ) = Γ(( φ b ◦ T ) ∗ dµ a ) and set T ′ a := d η ( a ) : D → D The conformal equivalence ψ a : D → a is defined by ψ a = (cid:0) T ′ a ◦ φ b ◦ T a (cid:1) − . The pull-back by ψ a of the folded measure is(2.5.1) dν a = ψ ∗ a dµ a It is clear from the above construction that dν a is a normalized measure onthe whole disk. We call dν a the rearranged measure . It also follows fromthe construction that the conformal transformations ψ a : D → a dependcontinuously on a and(2.5.2) lim a → D ψ a = id : D → D in the sense of the pointwise convergence. We will make use of the followingimportant property of the rearranged measure. Lemma 2.5.3.
If a sequence of hyperbolic caps a ∈ D degenerates to apoint p ∈ ∂ D , the limiting rearranged measure is a “flip-flop” of the originalmeasure dµ : lim a → p dν a = R ∗ p dµ. (F)We call (F) the flip-flop property. The proof of Lemma 2.5.3 will bepresented at the end of the paper.2.6. Maximizing directions.
Given a finite measure dν on D , considerthe function V : R → R defined by V ( s ) = Z D X s dν. AXIMIZATON OF THE SECOND POSITIVE NEUMANN EIGENVALUE 9
This function is a quadratic form since the mapping R × R → R definedby ( s, t ) Z D X s X t dν is symmetric and bilinear (the latter easily follows from (2.1.1)). In partic-ular, V ( s ) = V ( − s ) for any s .Let R P = S / Z be the projective line. We denote by [ s ] ∈ R P theelement of the projective line corresponding to the pair of points ± s ∈ S .We say that [ s ] ∈ R P is a maximizing direction for the measure dν if V ( s ) ≥ V ( t ) for any t ∈ S . The measure dν is called simple if there isa unique maximizing direction. Otherwise, the measure dν is said to be multiple . We have the following Lemma 2.6.1.
A measure dν is multiple if and only if V ( s ) does not dependon s ∈ S .Proof. Since V ( s ) is a symmetric quadratic form, it can be diagonalized.This means that there exists an orthonormal basis ( v , v ) of R , such thatfor any s = αv + βv ∈ D we have V ( s ) = M α + mβ . for some numbers0 < m ≤ M. It is clear now that the measure dν is multiple if and only if M = m , and therefore V ( s ) takes the same value for all s ∈ S . (cid:3) Note that by (2.3.2), [ e ] is a maximizing direction for the measure dµ . Proposition 2.6.2.
If the measure dµ is simple, then there exists cap a ⊂ D such that the rearranged measure dν a is multiple. The proof of this proposition is based on a topological argument, some-what more subtle than the one used in the proof of Lemma 2.2.3. This isa proof by contradiction. We assume the measure dµ as well as the mea-sures dν a to be simple. Given a cap a ⊂ D , let [ s ( a )] ∈ R P be the uniquemaximizing direction for dν a . Since dν a depends continuously on a and X s depends continuously on s , it follows that the map a [ s ( a )] is continuous.Let us understand the behavior of the maximizing directions as the cap a degenerates to the full disk and to a point. Lemma 2.6.3.
Assume the measures dµ as well as each dν a to be simple.Then lim a → D [ s ( a )] = [ e ](2.6.4) lim a → e iθ [ s ( a )] = [ e iθ ] . (2.6.5) Proof.
First, note that formula (2.6.4) immediately follows from (2.5.2) and(2.3.2). Let us prove (2.6.5). Set p = e iθ . Lemma 2.5.3 implies(2.6.6) lim a → p Z D X s dν a = Z D X s R ∗ p dµ = Z D X s ◦ R p dµ = Z D X R p s dµ. Since [ e ] is the unique maximizing direction for D , the right hand sideof (2.6.6) is maximal for R p s = ± e . Applying R p on both sides we get s = ± e iθ and hence [ s ] = [ e iθ ]. (cid:3) Proof of Proposition 2.6.2.
Suppose that for each cap a ⊂ D the measure dν a is simple. Recall that the space of caps is identified with ( − , × S .Define h : ( − , × S → R P by by h ( r, p ) = [ s ( a r,p )] . It follows fromLemma 2.6.3) that h extends to a continuous map on [ − , × S such that h ( − , e iθ ) = [ e ] , h (1 , e iθ ) = [ e iθ ] . This means that h is a homotopy between a trivial loop and a non-contractibleloop on R P . This is a contradiction. (cid:3) Test functions.
Assume that dµ is simple. By Proposition 2.6.2 andLemma 2.6.1 there exists a cap a ⊂ D such that Z D X s dν a ( z )does not depend on the choice of s ∈ S . Let a ∗ = D \ a . Definition 2.7.1.
Given a function u : a → R , the lift of u , ˜ u : D → R isgiven by ˜ u ( z ) = ( u ( z ) if z ∈ a,u ( τ a z ) if z ∈ a ∗ . Given u : a → R we have Z a u dµ a = Z a u dµ + Z a ∗ u ◦ τ a dµ = Z D ˜ u dµ, For every s ∈ R , set u sa = X s ◦ ψ − a : a → R . We will use the two-dimensional space E = (cid:8) ˜ u sa (cid:12)(cid:12) s ∈ R (cid:9) of test functions in the variational characterization (2.3.3) of µ . Proposition 2.7.2.
For each s ∈ R (2.7.3) R D |∇ ˜ u sa | dz R D (˜ u sa ) dµ ≤ µ ( D ) . We split the proof of Proposition 2.7.2 in two parts.
Lemma 2.7.4.
For any hyperbolic cap a ⊂ D , Z D |∇ ˜ u sa | dz = (cid:18) π Z r =0 f ( r ) r dr (cid:19) µ ( D ) . AXIMIZATON OF THE SECOND POSITIVE NEUMANN EIGENVALUE 11
Lemma 2.7.5. Z D (˜ u sa ) dµ ≥ π (cid:18)Z r =0 f ( r ) r dr (cid:19) . (2.7.6) Proof of Lemma 2.7.4.
It follows from the definition of the lift that Z D |∇ ˜ u sa | dz = Z a |∇ u sa | dz + Z a ∗ |∇ ( u sa ◦ τ a ) | dz. By conformal invariance of the Dirichlet energy, the two terms on the righthand side are equal, so that Z D |∇ ˜ u sa | dz = 2 Z a |∇ u sa | dz = 2 Z a |∇ ( X s ◦ ψ − a ) | dz = 2 Z D |∇ X s | dz ←− (by conformal invariance)= 2 µ ( D ) Z D X s dz ←− (since X s is the first eigenfunction on a disk)(2.7.7)It follows from (2.1.1) that given two orthogonal directions s, t ∈ S we have Z D ( X s + X t ) dz = Z D f ( | z | ) dz. Therefore, by symmetry we get Z D X s dz = 12 Z D f ( | z | ) dz = π Z r =0 f ( r ) r dr. This completes the proof of the lemma. (cid:3)
To prove Lemma 2.7.5 we use the following result.
Lemma 2.7.8.
The rearranged measure dν a on D can be represented as dν a = δ ( z ) dz , where δ : D → R is a subharmonic function.Proof. Indeed, dν a = ψ ∗ a dµ a , where the measure dµ a on the cap a is obtainedas the sum of measures dµ and τ ∗ a dµ . Both measures dµ and τ ∗ a dµ corre-spond to flat Riemannian metrics on a , because dµ is the pullback of theEuclidean measure dz on the domain Ω by the conformal map φ : D → Ω(see section 2.3). Since the maps ψ a and τ a are also conformal, one has ψ ∗ a dµ = α ( z ) dz and ψ ∗ a ( τ ∗ a dµ ) = β ( z ) dz for some subharmonic functions α ( z ) , β ( z ). Indeed, the metrics corresponding to these measures are flat(they are pullbacks by ψ a of flat metrics on a that we mentioned above),and the well-known formula for the Gaussian curvature in isothermal coor-dinates yields ∆ log α ( z ) = ∆ log β ( z ) = 0 (cf. [BR, p. 663]). Therefore, α ( z ) and β ( z ) are subharmonic as exponentials of harmonic functions [Le,p. 45]. Finally, dν a = δ ( z ) dz , where δ ( z ) = α ( z ) + β ( z ) is subharmonic as asum of subharmonic functions. This completes the proof of the lemma. (cid:3) Proof of Lemma 2.7.5.
Set G ( r ) = Z B (0 ,r ) δ ( z ) dz = Z r Z π δ ( ρ e iφ ) ρ dρ dφ. By Lemma 2.7.8 the function δ is subharmonic. The function W ( ρ ) = Z π δ ( ρ e iφ ) dφ is 2 π times the average of δ over the circle of radius ρ , hence it is monotonenon-decreasing in ρ ([Le, p. 46]). Therefore, since r ≤
1, we get as in [SY,p.138] that(2.7.9) G ( r ) = Z r W ( ρ ) ρ dρ = r Z W ( r ρ ) ρ dρ ≤ r Z W ( ρ ) ρ dρ = r G (1) = πr . Now, because ˜ u sa is the lift of u sa = X s ◦ ψ a , we have Z D (˜ u sa ) dµ = Z a ( u sa ) dµ a = Z D X s dν a . Moreover since V a ( s ) doesn’t depend on s ∈ S , V a ( s ) = Z D X s dν a = 12 Z D (cid:0) X e + X e (cid:1) dν a = 12 Z D f ( | z | ) δ ( z ) dz = 12 Z r =0 f ( r ) G ′ ( r ) dr (2.7.10)Integrating by parts and taking into account that G ( r ) ≤ πr due to (2.7.9),we get Z r =0 f ( r ) G ′ ( r ) dr = f (1) G (1) − Z ddr (cid:0) f ( r ) (cid:1) G ( r ) dr ≥ f (1) G (1) − π Z ddr (cid:0) f ( r ) (cid:1) r dr = 2 π Z f ( r ) r dr (2.7.11)This completes the proof of Lemma 2.7.5 and Proposition 2.7.2. (cid:3) Remark . The proof of Lemma 2.7.5 is quite similar to the proof of(1.1.1), see [Sz, p. 348] and [SY, p. 138]. Our approach is somewhat moredirect since it explicitly uses the subharmonic properties of the measure.
Proof of Theorem 1.1.3.
Assume that dµ is simple. Then (1.1.4) immedi-ately follows from Proposition 2.7.2 and the variational characterization(2.3.3) of µ . AXIMIZATON OF THE SECOND POSITIVE NEUMANN EIGENVALUE 13
Suppose now that dµ is multiple. In fact, the proof is simpler in thiscase. Indeed, it follows from Lemma 2.6.1, that any direction [ s ] ∈ R P ismaximizing for dµ so that we can use the space E = (cid:8) X s (cid:12)(cid:12) s ∈ R (cid:9) of test functions in the variational characterization (2.3.3) of µ . Inspectingthe proof of Proposition 2.7.2 we notice that the factor 2 disappears in (2.7.7)and hence in (2.7.3) as well. Therefore, in this case we get using (2.3.3) that µ (Ω) ≤ µ ( D ). This completes the proof of the theorem. (cid:3) Remark . When dµ is multiple, we get a stronger estimate µ (Ω) ≤ µ ( D ) . To illustrate this case, consider Ω = D . Then indeed µ ( D ) = µ ( D ).3. Proofs of auxiliary lemmas
Uniqueness of the renormalizing point.
The following lemma isused in the proof Lemma 2.2.4.
Lemma 3.1.1.
Let r ∈ (0 , and s = 1 . Then X s ( d r ( z )) > X s ( z ) for all z ∈ D .Proof. We have X s ( z ) = f ( | z | ) cos θ and X s ( d r ( z )) = f ( | d r ( z ) | ) cos θ ,where θ = arg z and θ = arg d r ( z ). We need to show(3.1.2) f ( | d r ( z ) | ) cos θ > f ( | z | ) cos θ for all z ∈ D . Note that the function f is monotone increasing, positive onthe interval (0 , f (0) = 0. Set z = a + ib . It is easy to check that for | z | = 0 the inequality in question is satisfied and therefore in the sequel weassume that a + b > | z | and | d r ( z ) | . We note that | z | = | ¯ z | . Since | d r ( z ) | = | z + r || rz + 1 | , we need to compare | z + r | and | r | z | + ¯ z | . This boils down to comparing( a + r ) + b and (( r ( a + b ) + a ) + b , or, equivalently, ( a + r ) and(( r ( a + b ) + a ) . Note that a + b < z ∈ D . We have three cases:(i) a ≥
0. Then | d r ( z ) | > | z | .(ii) a < a + r ≤
0. Then | d r ( z ) | < | z | .(iii) a < a + r > θ and θ .We have: d r ( z ) = z + rrz + 1 = ( a + r ) + ib ( ar + 1) + ibr = ( a + r )( ar + 1) + b r + ib (1 − r )( ar + 1) + b r Taking into account that ar + 1 >
0, we obtain from this formula that incase (iii) cos θ >
0. On the other hand, cos θ < f > Consider now case (i). Using the formula above we get thattan θ = b (1 − r )( a + r )( ar + 1) + b r . If a = 0 then (3.1.2) is true since cos θ = 0 and one may easily check thatcos θ >
0. So let us assume that a = 0. Then tan θ = b/a . Note that thetangent is a monotone increasing function. If b = 0 then θ = θ = 0 and(3.1.2) is satisfied since | d r ( z ) | > | z | . If b = 0, dividing by b and taking intoaccount that a > r > a > − r ( a + r )( ar + 1) + b r . Therefore, if b > θ > tan θ implying 0 < θ < θ < π/ b < θ < tan θ implying that 3 π/ < θ < θ < π .At the same time, in the first case the cosine is monotonely decreasing, andin the second case the cosine is monotonely increasing. Therefore, for any b = 0 we get 0 < cos θ < cos θ , which implies (3.1.2).Finally, consider the case (ii). If ( a + r )( ar + 1) + b r ≥ θ ≥ θ <
0. So letus assume ( a + r )( ar + 1) + b r <
0. If b = 0 then θ = θ = π , hencecos θ = cos θ = − | d r ( z ) | < | z | . If b = 0,as in case (ii) we compare tan θ and tan θ . We claim that again1 a > − r ( a + r )( ar + 1) + b r . Since by our hypothesis the denominators in both cases are negative, it isequivalent to a − ar < a r + ar + a + r + b r . After obvious transformationswe see that this reduces to a + 2 ar + 1 + b = ( a + r ) + (1 − r ) + b > b > π/ < θ < θ < π , and if b < π < θ < θ < π/ θ < cos θ <
0. Together with theinequality | d r ( z ) | < | z | this gives (3.1.2) in case (ii). This completes theproof of the lemma. (cid:3) Proof of Lemma 2.5.3.
Let M be the space of signed finite measureson D endowed with the norm (2.2.2). Recall that the map Γ : M → D isdefined by Γ( dν ) = ξ in such a way that d ξ : D → D renormalizes dν . It iscontinuous by Corollary 2.2.5. The key idea of the proof of the “flip-flop”lemma is to replace the folded measure dµ a by d ˆ µ a := ( τ a ) ∗ dµ. It is clear that(3.2.1) || dµ a − d ˆ µ a || → AXIMIZATON OF THE SECOND POSITIVE NEUMANN EIGENVALUE 15 in the norm (2.2.2) as a degenerates to a point p ∈ ∂ D . At the same time,the next lemma shows that the “flip-flop”property is true for each cap whenthe rearranged measure dν a is replaced by ( dζ a ) ∗ d ˆ µ a , where ζ a = Γ( d ˆ µ a ). Lemma 3.2.2.
Let a = a r,p be a hyperbolic cap. Then ( dζ a ) ∗ d ˆ µ a = ( d ζ a ) ∗ ( τ a ) ∗ dµ = R ∗ p dµ. Proof.
Let us show that ζ a = − rr +1 p . Recall that τ a ( z ) = d rp ◦ R p ◦ d − rp . A simple explicit computation then leads to d ζ a ◦ τ a = R p . This implies Z D X s ◦ d ζ a d ˆ µ a = Z D X s ◦ d ζ a ◦ τ a dµ = Z D X s ◦ R p dµ = Z D X R p s dµ = 0which proves the claim. (cid:3) Let η a := Γ(( d ζ a ) ∗ dµ a ) be the renormalizing vector for the measure ( d ζ a ) ∗ dµ a . Lemma 3.2.3.
As the cap a degenerates to a point p ∈ ∂ D , η a → . Proof.
Since d ζ a is a diffeomorphism, ( d ζ a ) ∗ : M → M is an isometry sothat ( d ζ a ) ∗ dµ a = ( d ζ a ) ∗ ( dµ a − d ˆ µ a ) + ( d ζ a ) ∗ d ˆ µ a = ( d ζ a ) ∗ ( dµ a − d ˆ µ a ) | {z } → +( d ζ a ◦ τ a | {z } R p ) ∗ dµ → ( R p ) ∗ dµ. Here we have used (3.2.1). Continuity of Γ leads to0 = Γ(( R p ) ∗ dµ ) = lim a → p Γ(( d ζ a ) ∗ dµ a ) = lim a → p η a . Note that the first equality follows from (2.3.1) and the identity X s ◦ R p = X R p s that we used earlier. (cid:3) Set q ( a ) = ζ a η a + 1 ζ a η a + 1 , ξ ( a ) = d ζ a ( η a ) = (cid:18) η a + ζ a ζ a η a + 1 (cid:19) . (3.2.4)A direct computation (cf. the proof of Lemma 2.2.4) leads to˜ T a ( z ) := d η a ◦ d ζ a = q ( a ) d ξ ( a ) ( z ) . It follows from its definition that ˜ T a renormalizes dµ a . Hence, Γ( dµ a ) = ξ ( a )and d ξ ( a ) = T a , where the transformation T a was defined in section 2.5. We have T a ∗ dµ a = ( 1 q ( a ) d η a ) ∗ ( d ζ a ) ∗ dµ a = ( 1 q ( a ) d η a ) ∗ ( d ζ a ) ∗ ( d ˆ µ a + ( dµ a − d ˆ µ a )) . Now, it follows from Lemma 3.2.3 that lim a → p q ( a ) = 1 and lim a → p d η a = id,because η a →
0. Therefore, taking into account (3.2.1) we getlim a → p T a ∗ dµ a = lim a → p ( d ξ a ) ∗ d ˆ µ a = R ∗ p dµ. To complete the proof of Lemma 2.5.3 it remains to show that as the cap a degenerates to p , || T a ∗ dµ a − dν a || →
0. By definition dν a = ψ ∗ a dµ , where ψ a = ( T ′ a ◦ φ b ◦ T a ) − (see section 2.5). Let us show that b = T a ( a ) → D as a → p . Indeed, T a = d ξ ( a ) = d ζ a ◦ ( d − ζ a ◦ d ξ ( a ) ) = R p ◦ τ a ◦ ( d − ζ a ◦ d ξ ( a ) ) . Since η a → a → p , it follows from (3.2.4) that the composition d − ζ a ◦ d ξ ( a ) tends to identity. Therefore, the cap T a ( a ) gets closer to D \ R p ( a )when a goes to p and thus lim a → p T a ( a ) = D . This implies lim a → p φ T a ( a ) = idand lim a → p T ′ a = id, and hence lim a → p || T a ∗ dµ a − dν a || = 0. (cid:3) Proof of Theorem 1.2.5
Hersch’s lemma and uniqueness of the renormalizing map.
Theproof of Theorem 1.2.5 is quite similar to the proof of Theorem 1.1.3. Weuse the following notation B n +1 = { x ∈ R n +1 , | x | < } S n = ∂ B n +1 . The standard round metric on S n is g . Given a conformally round metric g ∈ [ g ] we write dg for its induced measure. Given s ∈ R n +1 , define X s : S n → R by X s ( x ) = ( x, s ) . Similarly to (2.3.1) and (2.3.2), we assume that for each s ∈ S n : Z S n X s dg = 0 . (4.1.1) Z S n X e dg ≥ Z S n X s dg. (4.1.2)Given p ∈ S n , R p : R n +1 → R n +1 is the reflection with respect to thehyperplane going through 0 and orthogonal to the segment joining 0 and p ,that is R p ( x ) = x − p, x ) p. AXIMIZATON OF THE SECOND POSITIVE NEUMANN EIGENVALUE 17
Given ξ ∈ B n +1 . define d ξ : B n +1 → B n +1 by(4.1.3) d ξ ( x ) = (1 − | ξ | ) x + (1 + 2( ξ, x ) + | x | ) ξ ξ, x ) + | ξ | | x | . Note that d ξ (0) = ξ and d ξ ◦ d − ξ = id. The map d ξ is a conformal (M¨obius)transformation of S n [SY, p. 142]. Indeed, one can check that for ξ = 0, d ξ = γ ξ ◦ R ξ | ξ | where γ ξ is the spherical inversion with center ξ | ξ | and radius −| ξ | | ξ | . Notethat for n = 1, the map d ξ coincides with the one introduced in Lemma 2.2.3,where complex notation was used for convenience.Similarly to the disk case, the transformation d ξ is said to renormalize ameasure dν on the sphere S n if for each s ∈ R n +1 , Z S n X s ◦ d ξ dν = 0 . (4.1.4)This condition is clearly equivalent to Z S n x i ◦ d ξ dν = 0 , i = 1 , , . . . , n + 1 , which means that the center of mass of the measure ( d ξ ) ∗ dν on S n is at theorigin. The following result is a combination of Hersch’s lemma [Her] and auniqueness result announced in [Na]. Proposition 4.1.5.
For any finite measure dν on S n , there exists a uniquepoint ξ ∈ B n +1 such that d ξ renormalizes dν . Moreover, the dependence ofthe point ξ ∈ B n +1 on the measure dν is continuous.Proof. The existence of ξ is precisely Hersch’s lemma (see [Her], [SY, p.144], [LY, p. 274]).Let us prove uniqueness. First, let us show that if dν is a renormalizedmeasure then ξ = 0. It follows from (4.1.3) by a straightforward computationthat if B n +1 ∋ ξ = 0 then X ξ ( x ) < X ξ ( d ξ ( x )) for any x ∈ S n . Assume that d ξ renormalizes dν for some ξ = 0. Then0 = Z S n X ξ dν < Z S n X ξ ◦ d ξ dν = 0 , and we get a contradiction.Now, let dν be an arbitrary finite measure and suppose that it is renor-malized by d ξ and d η . Writing d η = d η ◦ d − ξ ◦ d ξ we get(4.1.6) Z S n X s ◦ d η ◦ d − ξ d ˜ σ = 0where the measure d ˜ σ = ( d ξ ) ∗ dσ is renormalized. At the same time, it easyto check that d η ◦ d − ξ = R ◦ d d − ξ ( η ) , where R is an orthogonal transformation.Indeed, since − d − ξ ( η ) = d ξ ( − η ) we have d η ◦ d − ξ ◦ d d ξ ( − η ) (0) = d η ( − η ) = 0 , and it is well known that any M¨obius transformation of the unit ball pre-serving the origin is orthogonal [Bea, Theorem 3.4.1]. Since R preserves thecenter of mass at zero, it follows from (4.1.6) that d d − ξ ( η ) renormalizes themeasure d ˜ σ , which is already renormalized. Therefore, as we have shownabove, d − ξ ( η ) = 0 and hence ξ = η .Similarly to Corollary 2.2.5, uniqueness of the renormalizing point impliesthat its dependence on the measure is continuous. (cid:3) Spherical caps, folding and rearrangement.
The set C of all spher-ical caps is parametrized as follows: given p ∈ S n let a ,p = { x ∈ S n : ( x, p ) > } be the half-sphere centered at p . Given − < r <
1, let a r,p = d rp ( a ,p ) . To every spherical cap a ∈ C we associate a folded measure: dµ a = ( dg + τ ∗ a dg on a, a ∗ , where a ∗ = S n \ a ∈ C is the cap adjacent to a , and τ a is the conformalreflection with respect to the boundary circle of a . That is, for a = a r,p τ a = d rp ◦ R p ◦ d − rp . Let ξ ( a ) ∈ B n +1 be the unique point such that d ξ ( a ) renormalizes dµ a . Weobtain a rearranged folded measure dν a = ( d ξ ( a ) ) ∗ dµ a . (4.2.1)4.3. Maximizing directions.
Given a finite measure dν on S n , define V ( s ) = Z S n X s dν. Let R P n be the projective space and let [ s ] ∈ R P n be the point correspond-ing to ± s ∈ S n . We say that [ s ] ∈ R P n is a maximizing direction for dν if V ( s ) ≥ V ( t ) for all t ∈ S n . We say that the spherical cap is simple if themaximizing direction is unique. Otherwise, similarly to Lemma 2.6.1, thereexists a two-dimensional subspace W ⊂ R n +1 such that any s ∈ W ∩ S n is amaximizing direction for dν . In particular for each s, t ∈ W , V ( s ) = V ( t ) . In this case the measure dν is called multiple . Proposition 4.3.1.
Let g ∈ [ g ] be a conformally round metric on a sphere S n of odd dimension. If the measure dg is simple then there exists a sphericalcap such that the rearranged folded measure dν a is multiple. The proof of Proposition 4.3.1 is similar to the proof of Proposition 2.6.2.We assume the measures dg as well as each dν a to be simple. Given acap a ⊂ S n let [ s ( a )] ∈ R P be the unique maximizing direction for dν a .The map a [ s ( a )] is continuous. The following spherical version of the“flip-flop” property is proved exactly as Lemma 2.5.3. AXIMIZATON OF THE SECOND POSITIVE NEUMANN EIGENVALUE 19
Lemma 4.3.2.
If a sequence of spherical caps a ∈ C degenerates to a point p ∈ S n , the limiting rearranged measure is a “flip-flop” of the original mea-sure dg : lim a → p dν a = R ∗ p dg. (4.3.3)Similarly to Lemma 2.6.3 we study the maximizing directions for degen-erating caps. Lemma 4.3.4.
Suppose the measures dg as well as each dν a are simple.Then lim a → S n [ s ( a )] = [ e ](4.3.5) lim a → p [ s ( a )] = [ R p e ] . (4.3.6) Proof of Proposition 4.3.1.
By convention (4.1.2), [ e ] is the unique maxi-mizing direction for dg . Recall that the space of caps has been identifiedwith ( − , × S n . The continuous map h : [ − , × S n → R P n is defined by h ( r, p ) = [ e ] for r = − , [ s ( a r,p )] for − < r < , [ R p e ] for r = 1 . That is, h is an homotopy between a constant map and the map φ : S n → R P n defined by φ ( p ) = [ R p e ] . We will show that this is impossible when n is oddby computing its degree. The map φ lifts to the map ψ : S n → S n definedby ψ ( p ) = − R p e = 2( e , p ) p − e . (4.3.7)The two solutions of ψ ( p ) = e are e and − e . It is easy to check that sincethe dimension n is odd, both differentials D e ψ : T e S n → T e S n D − e ψ : T − e S n → T e S n preserve the orientation. This implies deg( ψ ) = 2 . Moreover, the quotientmap π : S n → S n has degree 2 for n odd. It follows thatdeg( φ ) = deg( π ◦ ψ )= deg( π )deg( ψ ) = 4 . Since the degree of a map is invariant under homotopy, this is a contradic-tion. (cid:3)
Remark . In even dimensions one of the differentials D ± e preserves theorientation and the other reverses it. Therefore, deg( φ ) = 0 and the proof ofProposition 4.3.1 does not work in this case. In dimension two the existenceof a multiple cap was proved in [Na] using a more sophisticated topologicalargument.4.4. Test functions and the modified Rayleigh quotient.
Let g bethe standard round metric on the sphere S n , so that(4.4.1) ω n := Z S n dg = 2 π n +12 Γ( n +12 ) . Let g ∈ [ g ] be a conformally round Riemannian metric of volume one,that is R S n dg = 1 . The Rayleigh quotient of a non-zero function u ∈ H ( S n )is R ( u ) = R S n |∇ g u | g dg R S n u dg . We use the following variational characterization of λ ( g ): λ ( g ) = inf E sup = u ∈ E R ( u )(4.4.2)where E varies among all two-dimensional subspaces of the Sobolev space H ( S n ) that are orthogonal to constants, in the sense that for each f ∈ E , R S n f dg = 0 . Following [FN], we use a modified Rayleigh quotient : R ′ ( u ) = (cid:0)R S n |∇ g u | ng dg (cid:1) /n R S n u dg . It follows from Holder inequality that R ( u ) ≤ R ′ ( u ) for each 0 = u ∈ H ( S n ). It is easy to check that R S n |∇ g u | ng dg is conformally invariant foreach dimension n so that we can rewrite the modified Rayleigh quotient asfollows: R ′ ( u ) = (cid:0)R S n |∇ u | n dg (cid:1) /n R S n u dg where the gradient and it’s norm are with respect to the round metric g .Assume that dg is simple and let a ⊂ S n be a spherical cap such that dν a ismultiple. Let W ⊂ R n +1 be the corresponding two dimensional subspace ofmaximizing directions. Given a function u : a → R , the lift of u , ˜ u : S n → R is defined exactly as in Definition 2.7.1. Proposition 4.4.3.
Given s ∈ W ⊂ R n +1 , the function u sa = X s ◦ d ξ ( a ) : a → R is such that R ′ (˜ u sa ) < ( n + 1) π n +12 Γ( n )Γ( n )Γ( n + ) ! /n . AXIMIZATON OF THE SECOND POSITIVE NEUMANN EIGENVALUE 21
Proof.
The conformal invariance of the numerator in R ′ ( u ) implies(4.4.4) (cid:18)Z S n |∇ g ˜ u sa | ng dg (cid:19) /n = (cid:18)Z a |∇ g u sa | ng dg (cid:19) /n + (cid:18)Z a ∗ |∇ g ( u sa ◦ τ a ) | ng dg (cid:19) /n = (cid:18) Z a |∇ g u sa | ng dg (cid:19) /n = Z d ξ ( a ) ( a ) |∇ g X s | ng dg ! /n < (cid:18) Z S n |∇ g X s | ng dg (cid:19) /n Here the second equality follows from conformal invariance. To obtain theinequality at the end we again use the conformal invariance as well as the factthat d ξ ( a ) ( a ) ( S n . To estimate the denominator in the modified Rayleighquotient we first note that for any x = ( x , . . . x n +1 ) ∈ S n , n +1 X j =1 ˜ u e j a ( x ) = n +1 X j =1 x j = 1 . Therefore, given that R S n dg = 1 we obtain: n +1 X j =1 Z S n (˜ u e j a ) dg = 1Now, since W is a subspace of maximizing directions for the measure dν a defined by (4.2.1), for each s ∈ W we have(4.4.5) Z S n (˜ u sa ) dg ≥ n + 1 . Set K n := Z S n |∇ g X s | ng dg . Combining (4.4.4) and (4.4.5) we get(4.4.6) R ′ (˜ u sa ) ≤ ( n + 1) (2 K n ) /n . Proposition 4.4.3 then follows from the lemma below.
Lemma 4.4.7.
The constant K n is given by K n = 2 π n +12 Γ( n )Γ( n )Γ( n + ) . Proof.
Recall that g is the standard round metric on the unit sphere S n .If we consider X s ( x ) = ( x, s ) as a function on R n +1 then its gradient is justthe constant vector s : grad R n +1 X s = s. This means that for any point p ∈ S n the gradient of the function X s : S n → R at p is the projection of s on the tangent space T p S n : ∇ X s ( p ) = s − ( s, p ) p. Therefore, taking into account that | s | = | p | = 1, we get |∇ X s ( p ) | n = ( | s − ( s, p ) p | ) n/ = (1 − ( s, p ) ) n/ , and hence K n = Z S n (1 − ( s, p ) ) n/ dg . Let θ be the angle between the vectors p and s . Making a change of variableswe obtain K n = ω n − Z π (1 − cos θ ) n/ (sin θ ) n − dθ = ω n − Z π sin n − θ dθ, where ω n − is the volume of the standard round sphere S n − given by (4.4.1).The calculation of a table integral [GR, 3.621(4)] Z π sin n − θ dθ = √ π Γ( n )Γ( n + )completes the proofs of Lemma 4.4.7 and Proposition 4.4.3. (cid:3) Remark . It follows from H¨older inequality that R ( u ) = R ′ ( u ) if andonly if u is a constant function. Since ∇ g X s = const we get a strict in-equality R ′ (˜ u sa ) > R (˜ u sa ). This is why the estimate (1.2.6) is not sharp. Inthe context of the first eigenvalue, a similar difficulty was encountered in[Ber, Lemma 4.15]) and overcame in [EI]. To apply the approach of [EI]we need to have a spherical cap of multiplicity n + 1; existence of a cap ofmultiplicity two proved in Proposition 4.3.1 is not enough for this purpose. Proof of Theorem 1.2.5.
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E-mail address : [email protected] Laboratoire d’Analyse, Topologie, Probabilit´es UMR 6632, Centre de Math´ematiqueset Informatique, Universit´e de Provence, 39 rue F. Joliot-Curie, 13453 Mar-seille Cedex 13, France;
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