A disc maximizes Laplace eigenvalues among isoperimetric surfaces of revolution
aa r X i v : . [ m a t h . A P ] O c t A DISC MAXIMIZES LAPLACE EIGENVALUES AMONGISOPERIMETRIC SURFACES OF REVOLUTION
SINAN ARITURK
Abstract.
The Dirichlet eigenvalues of the Laplace-Beltrami operatorare larger on a flat disc than on any other surface of revolution immersedin Euclidean space with the same boundary. Introduction
Let Σ be a compact connected immersed surface of revolution in R withone smooth boundary component. The Euclidean metric on R induces aRiemannian metric on Σ. Let ∆ Σ be the corresponding Laplace-Beltramioperator on Σ. Denote the Dirichlet eigenvalues of − ∆ Σ by0 < λ (Σ) < λ (Σ) ≤ λ (Σ) ≤ . . . Let R be the radius of the boundary of Σ, and let D be a disc in R ofradius R . Let ∆ be the Laplace operator on R , and denote the Dirichleteigenvalues of − ∆ on D by0 < λ ( D ) < λ ( D ) ≤ λ ( D ) ≤ . . . Theorem. If Σ is not equal to D , then for j = 1 , , , . . . , λ j (Σ) < λ j ( D )We remark that there are compact connected surfaces, which are notsurfaces of revolution, embedded in R whose boundary is a circle of radius R and have first Dirichlet eigenvalue larger than λ ( D ). This can be provenwith Berger’s variational formulas [Be].This problem resonates with the Rayleigh-Faber-Krahn inequality, whichstates that the flat disc has smaller first Dirichlet eigenvalue than any otherdomain in R with the same area [F] [K]. Hersch proved that the canonicalmetric on S maximizes the first non-zero eigenvalue among metrics withthe same area [H]. Li and Yau showed the canonical metric on RP maxi-mizes the first non-zero eigenvalue among metrics with the same area [LY].Nadirashvili proved the same is true for the flat equilateral torus, whosefundamental parallelogram is comprised of two equilateral triangles [N1]. Itis not known if there is such a maximal metric on the Klein bottle, butJakobson, Nadirashvili, and Polterovich showed there is a critical metric [JNP]. El Soufi, Giacomini, and Jazar proved this is the only critical metricon the Klein bottle [EGJ].As for the second eigenvalue, the Krahn-Szeg¨o inequality states thatthe union of two discs with the same radius has smaller second Dirichleteigenvalue than any other domain in R with the same area [K]. Nadirashviliproved that the union of two round spheres of the same radius has largersecond non-zero eigenvalue than any metric on S with the same area [N2].It is conjectured that a disc has smaller third Dirichlet eigenvalue than anyother planar domain with the same area. Bucur and Henrot established theexistence of a quasi-open set in R which minimizes for the third eigenvalueamong sets of prescribed Lebesgue measure [BH]. This was extended tohigher eigenvalues by Bucur [Bu].On a compact orientable surface, Yang and Yau obtained upper bounds,depending on the genus, for the first non-zero eigenvalue among metrics ofthe same area [YY]. Li and Yau extended these bounds to compact non-orientable surfaces [LY]. However, Urakawa showed that there are metricson S with volume one and arbitrarily large first non-zero eigenvalue [U].Colbois and Dodziuk extended this to any manifold of dimension three orhigher [CD].For a closed compact hypersurface in R n +1 , Chavel and Reilly obtainedupper bounds for the first non-zero eigenvalue in terms of the surface areaand the volume of the enclosed domain [C, R]. This was extended to highereigenvalues by Colbois, El Soufi, and Girouard [CEG]. Abreu and Freitasproved that for a metric on S which can be isometrically embedded in R asa surface of revolution, the first S -invariant eigenvalue is less than the firstDirichlet eigenvalue on a flat disc with half the area [AF]. Colbois, Dryden,and El Soufi extended this to O ( n )-invariant metrics on S n which can beisometrically embedded in R n +1 as hypersurfaces of revolution [CDE].We conclude this section by reformulating the theorem. Fix a planein R containing the axis of symmetry of Σ. Identify R with this planeisometrically in such a way that the axis of symmetry is identified with { ( x, y ) ∈ R : x = 0 } Define R = { ( x, y ) ∈ R : x ≥ } We may assume ∂ Σ intersects R at the point ( R, L be the lengthof the meridian Σ ∩ R . Let α : [0 , L ] → R be a regular, arc-lengthparametrization of Σ ∩ R with α (0) = ( R, α = ( F α , G α ). Notethat F α ( L ) = 0 and F α is positive over [0 , L ).Let C (0 , L ) be the set of functions w : [0 , L ] → R which are continuouslydifferentiable and vanish at zero. For a non-negative integer k and a positive DISC MAXIMIZES EIGENVALUES AMONG ISOPERIMETRIC SURFACES 3 integer n , define λ k,n ( α ) = min W max w ∈ W R L | w ′ | F α + k w F α dt R L w F α dt Here the minimum is taken over all n -dimensional subspaces W of C (0 , L ).We remark that (cid:26) λ j (Σ) (cid:27) = (cid:26) λ k,n ( α ) (cid:27) Moreover, if we count λ k,n ( α ) twice for k = 0, then the values occur withthe same multiplicity. Define ω : [0 , R ] → R by ω ( t ) = ( R − t, λ k,n ( ω ) similarly to λ k,n ( α ). Then (cid:26) λ j ( D ) (cid:27) = (cid:26) λ k,n ( ω ) (cid:27) Again, if we count λ k,n ( ω ) twice for k = 0, then the values occur withthe same multiplicity. Now to prove the theorem, it suffices to prove thefollowing lemma. Lemma 1. If α does not equal ω , then for any non-negative integer k andany positive integer n , λ k,n ( α ) < λ k,n ( ω )To prove this, we define a neighborhood of the boundary ∂ R and treatthe segments of the curve outside and inside of this neighborhood seperately.For the exterior segment, we simply project α orthogonally onto ω andobserve that this increases the eigenvalue. For the interior segment, weunroll the curve to ω and see that this increases the eigenvalue as well.2. Proof
We first extend the definition of the functionals λ k,n to Lipschitz curves.Let [ a, b ] be a finite, closed interval and let ψ : [ a, b ] → R be a Lipschitzcurve. Write ψ = ( F ψ , G ψ ). Assume that F ψ is positive over [ a, b ). LetLip ( a, b ) be the set of continuous functions w : [ a, b ) → R which vanish at a and are Lipschitz over [ a, c ] for every c in ( a, b ). For a non-negative integer k and a positive integer n , define λ k,n ( ψ ) = inf W max w ∈ W R ba | w ′ | F ψ | ψ ′ | + k w | ψ ′ | F ψ dt R ba w F ψ | ψ ′ | dt SINAN ARITURK
Here the infimum is taken over all n -dimensional subspaces W of Lip ( a, b ).Let H ( ψ, k ) be the set of continuous functions w : [ a, b ) → R which vanishat a and have a weak derivative such that Z ba | w ′ | F ψ | ψ ′ | + k w | ψ ′ | F ψ dt < ∞ In the following lemma, we note that if ψ is a regular piecewise con-tinuously differentiable curve which meets the axis transversally, then theinfimum in the defintion of the functionals λ k,n is attained. Lemma 2.
Let ψ : [ a, b ] → R be a piecewise continuously differentiablecurve. Assume there is a positive constant c such that for all t in [ a, b ] , | ψ ′ ( t ) | ≥ c Write ψ = ( F ψ , G ψ ) . Assume that F ψ is positive over [ a, b ) . Assume that F ψ ( b ) = 0 and F ′ ψ ( b ) < . Let k be a non-negative integer. Then there arefunctions ϕ k, , ϕ k, , ϕ k, , . . . which form an orthonormal basis of H ( ψ, k ) such that, for any positiveinteger n , λ k,n ( ψ ) = R ba | ϕ ′ k,n | F ψ | ψ ′ | + k ϕ k,n | ψ ′ | F ψ dt R ba ϕ k,n F ψ | ψ ′ | dt Each function ϕ k,n has exactly n − roots in ( a, b ) and satisfies the followingequation weakly: (cid:18) F ψ ϕ ′ k,n | ψ ′ | (cid:19) ′ = k | ψ ′ | ϕ k,n F ψ − λ k,n ( ψ ) F ψ | ψ ′ | ϕ k,n Also, λ k, ( ψ ) < λ k, ( ψ ) < λ k, ( ψ ) < . . . We omit the proof which is standard and refer to Gilbarg and Trudinger[GT] and Zettl [Z].Now fix a non-negative integer K and a positive integer N , for theremainder of the article. Let µ = K p λ K,N ( ω )The inequality µ < R is a basic fact about Bessel functions [W]. Let α beas defined in the introduction, and let A = min (cid:26) t ∈ [0 , L ] : F α ( t ) = µ (cid:27) DISC MAXIMIZES EIGENVALUES AMONG ISOPERIMETRIC SURFACES 5
Define β : [0 , L ] → R to be a piecewise continuously differentiable functionsuch that β (0) = ( R,
0) and β ′ ( t ) = ( ( F ′ α ( t ) , t ∈ [0 , A )( F ′ α ( t ) , G ′ α ( t )) t ∈ ( A, L ] Lemma 3.
Assume α is not equal to β and λ K,N ( α ) ≥ λ K,N ( ω ) . Then λ K,N ( α ) < λ K,N ( β ) Proof.
Fix a number p in (0 , α p : [0 , L ] → R to be a regularpiecewise continuously differentiable curve such that α p (0) = ( R,
0) and α ′ p ( t ) = ( ( F ′ α ( t ) , pG ′ α ( t )) t ∈ [0 , A )( F ′ α ( t ) , G ′ α ( t )) t ∈ ( A, L ]We first show that λ K,N ( α ) < λ K,N ( α p )By Lemma 2, there is a N -dimensional subspace Φ of H ( α p , K ) such that λ K,N ( α p ) = max w ∈ Φ R L | w ′ | F α | α ′ p | + K w | α ′ p | F α dt R L w F α | α ′ p | dt Moreover Φ is contained in Lip (0 , L ) and the maximum over Φ is onlyattained by scalar multiples of a function ϕ K,N which has exactly N − , L ). Let v be a function in Φ such that R L | v ′ | F α | α ′ | + K v | α ′ | F α dt R L v F α | α ′ | dt = max w ∈ Φ R L | w ′ | F α | α ′ | + K w | α ′ | F α dt R L w F α | α ′ | dt Note this quantity is at least λ K,N ( α ), which is at least λ K,N ( ω ). It followsthat R L | v ′ | F α | α ′ | + K v | α ′ | F α dt R L v F α | α ′ | dt ≤ R L | v ′ | F α | α ′ p | + K v | α ′ p | F α dt R L v F α | α ′ p | dt If equality holds, then v must vanish on a set of positive measure. In eithercase, we obtain λ K,N ( α ) ≤ R L | v ′ | F α | α ′ | + K v | α ′ | F α dt R L v F α | α ′ | dt < λ K,N ( α p )Now we repeat the argument to obtain λ K,N ( α p ) ≤ λ K,N ( β ) SINAN ARITURK
Let ε >
0. There is an N -dimensional subspace W of Lip (0 , L ) such thatmax w ∈ W R L | w ′ | F α | β ′ | + K w | β ′ | F α dt R w F α | β ′ | dt < λ K,N ( β ) + ε Let u be a function in W such that R L | u ′ | F α | α ′ p | + K u | α ′ p | F α dt R L u F α | α ′ p | dt = max w ∈ W R L | w ′ | F α | α ′ p | + K w | α ′ p | F α dt R L w F α | α ′ p | dt Note this quantity is at least λ K,N ( α p ), which is at least λ K,N ( ω ). It followsthat R L | u ′ | F α | α ′ p | + K u | α ′ p | F α dt R L u F α | α ′ p | dt ≤ R L | u ′ | F α | β ′ | + K u | β ′ | F α dt R L u F α | β ′ | dt Now we obtain λ K,N ( α p ) ≤ λ K,N ( β ) + ε Therefore, λ K,N ( α ) < λ K,N ( β ) (cid:3) Write β = ( F β , G β ). Define F γ : [0 , L ] → R by F γ ( t ) = ( min { F β ( s ) : s ∈ [0 , t ] } t ∈ [0 , A ] F β t ∈ [ A, L ]Let G γ = G β . Let γ = ( F γ , G γ ). Note that γ : [0 , L ] → R is Lipschitz. Lemma 4.
Assume λ K,N ( β ) ≥ λ K,N ( ω ) . Then λ K,N ( β ) ≤ λ K,N ( γ ) Proof.
Define V = (cid:26) t ∈ [0 , A ] : F β ( t ) = F γ ( t ) (cid:27) By the Riesz sunrise lemma, there are disjoint open intervals ( a i , b i ) suchthat V = [ i ( a i , b i )and F γ is constant over each interval. Suppose λ K,N ( β ) > λ K,N ( γ ). Thenthere is a N -dimensional subspace W of Lip (0 , L ) such thatmax w ∈ W R L | w ′ | F γ | γ ′ | + K w | γ ′ | F γ dt R L | w | F γ | γ ′ | dt < λ K,N ( β )Note that over each interval ( a i , b i ), the function | γ ′ | is zero, so each w in W is constant. Let J = [0 , L ] \ V . The isolated points of J are countable, DISC MAXIMIZES EIGENVALUES AMONG ISOPERIMETRIC SURFACES 7 so at almost every point in J , the curve γ is differentiable with γ ′ = β ′ . If w is a non-zero function in W , then w cannot vanish identically on J , and R J | w ′ | F β | β ′ | + K w | β ′ | F β dt R J | w | F β | β ′ | dt = R L | w ′ | F γ | γ ′ | + K w | γ ′ | F γ dt R L | w | F γ | γ ′ | dt < λ K,N ( β )Also for every w in W , Z V | w ′ | F β | β ′ | + K w | β ′ | F β dt = Z V K w | β ′ | F β dt ≤ λ K,N ( ω ) Z V | w | F β | β ′ | dt Here the inequality is strict unless w is identically zero over V . It followsthat max w ∈ W R L | w ′ | F β | β ′ | + K w | β ′ | F β dt R L | w | F β | β ′ | dt < λ K,N ( β )This is a contradiction. (cid:3) Let L ∗ be the length of γ . Define ℓ : [0 , L ] → [0 , L ∗ ] by ℓ ( t ) = Z t | γ ′ ( u ) | du Define ρ : [0 , L ∗ ] → [0 , L ] by ρ ( s ) = min n t ∈ [0 , L ] : ℓ ( t ) = s o This function ρ need not be continuous, but ζ = γ ◦ ρ is piecewise continu-ously differentiable, and for all t in [0 , L ], ζ ( ℓ ( t )) = γ ( t )Morover ζ is parametrized by arc length. Lemma 5.
This reparametrization satisfies λ K,N ( γ ) ≤ λ K,N ( ζ ) Proof.
Write γ = ( F γ , G γ ) and ζ = ( F ζ , G ζ ). Let w be a function inLip (0 , L ∗ ) such that R L ∗ | w ′ | F ζ | ζ ′ | + K w | ζ ′ | F ζ dt R L ∗ | w | F ζ | ζ ′ | dt < ∞ Define v = w ◦ ℓ . Then v is in Lip (0 , L ), and changing variables yields R L | v ′ | F γ | γ ′ | + K v | γ ′ | F γ dt R L | v | F γ | γ ′ | dt = R L ∗ | w ′ | F ζ | ζ ′ | + K w | ζ ′ | F ζ dt R L ∗ | w | F ζ | ζ ′ | dt It follows that λ K,N ( γ ) ≤ λ K,N ( ζ ). (cid:3) We can now prove Lemma 1 for the case K = 0. SINAN ARITURK
Proof of Lemma 1 for the case K = 0 . Suppose α is not equal to ω and λ K,N ( α ) ≥ λ K,N ( ω )Then α is not equal to β , so by Lemmas 3, 4, and 5 λ K,N ( α ) < λ K,N ( β ) ≤ λ K,N ( γ ) ≤ λ K,N ( ζ )But in this case, ζ = ω , so the proof is complete. (cid:3) For the remainder of the article, we assume that K is positive. Write ζ = ( F ζ , G ζ ). Let P = R − µ . Let χ : [0 , L ∗ ] → R be a piecewisecontinuously differentiable function such that χ (0) = ( R,
0) and for t in[0 , L ∗ ] with t = P , χ ′ ( t ) = (cid:16) F ′ ζ ( t ) , | G ′ ζ ( t ) | (cid:17) Then λ K,N ( ζ ) = λ K,N ( χ ), trivially. Write χ = ( F χ , G χ ). Note that, for t in[0 , P ], χ ( t ) = R − t Also, for every t in [0 , L ∗ ] with t = P , | χ ′ | = 1Let Φ K, , Φ K, , . . . be the functions given by Lemma 2 associated to ω . Let z be the largest root of Φ K,N in (0 , R ). It follows from basic facts aboutBessel functions [W] that z < P and that Φ K,N has no critical pointsin [
P, R ). There is a unique number Λ such that there exists a function u : [ z , P ] → R which is non-vanishing over ( z , P ) and satisfies ( ωu ′ ) ′ + (Λ ω − K ω ) u = 0 u ( z ) = 0 u ′ ( P ) = 0Moreover, Λ < λ K,N ( ω )To compare λ K,N ( χ ) and λ K,N ( ω ), we need the following lemma. Lemma 6.
Let Q and z be real numbers with z < z and Q > P . Let ψ : [ z, Q ] → R be continuously differentiable over [ P, Q ] . Assume that, for t in [ z, P ] , ψ ( t ) = ( R − t, Write ψ = ( F ψ , G ψ ) . Assume that F ψ ( Q ) = 0 and F ψ is positive over [ z, Q ) .Assume that | ψ ′ | = 1 over ( P, Q ) and that F ′ ψ ( Q ) < . Let ϕ be a functionin Lip ( z, Q ) such that λ K, ( ψ ) = R Qz | ϕ ′ | F ψ + K ϕ F ψ dt R Qz ϕ F ψ dt DISC MAXIMIZES EIGENVALUES AMONG ISOPERIMETRIC SURFACES 9
Assume that λ K, ( ψ ) > Λ . Then lim t → Q ϕ ( t ) = 0 Also ϕ is differentiable over [ z, Q ) , and over [ P, Q ) , | ϕ ′ | − K ϕ | F ψ | ≤ Furthermore ϕ ′ and ϕF ψ are bounded over [ z, Q ) .Proof. Since | ϕ ′ | F ψ and ϕ /F ψ are integrable, the function ϕ is absolutelycontinuous. Moreover ϕ /F ψ is integrable, but 1 /F ψ is not integrable over( c, Q ) for any c in ( z, Q ). It follows thatlim t → Q ϕ ( t ) = 0By Lemma 2, the function ϕ is continuously differentiable over [ z, Q ), andtwice continuously differentiable over [ z, P ) and ( P, Q ), with( F ψ ϕ ′ ) ′ = K ϕF ψ − λ K,N ( ψ ) F ψ ϕ It is also non-vanishing over ( z, Q ). We may assume that ϕ is positive over( z, Q ). Furthermore, the Picone identity (see, e.g. Zettl [Z]) implies that ϕ ′ ( P ) < F ψ | ϕ ′ | − K ϕ is differentiable over ( P, Q ), and its derivative is − λ K,N ( ψ ) F ψ ϕϕ ′ Therefore, we can prove the inequality by showing thatlim t → Q F ψ | ϕ ′ | = 0Note that ( F ψ | ϕ ′ | ) ′ = 2 K ϕϕ ′ − λ K,N ( ψ ) F ψ ϕϕ ′ Since | ϕ ′ | F ψ and ϕ /F ψ are integrable, it follows that F ψ | ϕ ′ | is absolutelycontinuous. Moreover, the limit as t tends to Q must be zero, because F ψ | ϕ ′ | is integrable and 1 /F ψ is not integrable over ( c, Q ) for any c in ( z, Q ).It remains to show that ϕ ′ and ϕF ψ are bounded over [ z, Q ). Let z ∗ be apoint in [ P, Q ) such that over [ z ∗ , Q ), K F ψ − λ K,N ( ψ ) F ψ > Then ϕ ′ cannot vanish in [ z ∗ , Q ). That is ϕ ′ is negative over [ z ∗ , Q ). Wehave seen that over ( z ∗ , Q ), Kϕ ≥ − F ψ ϕ ′ Now over ( z ∗ , Q ), ϕ ′′ ≥ − F ′ ψ ϕ ′ F ψ − Kϕ ′ F ψ − λ K,N ( ψ ) ϕ In particular, since K ≥
1, lim inf t → Q ϕ ′′ ≥ ϕ ′ is bounded. Since F ′ ψ ( Q ) <
0, it follows from Cauchy’s meanvalue theorem that ϕF is bounded. (cid:3) To compare λ K,N ( χ ) and λ K,N ( ω ) we will unroll χ to ω . The followinglemma describes the homotopy more precisely. Lemma 7.
Let χ : [ P, L ∗ ] → R be a continuously differentiable curve,parametrized by arc length. Assume χ ( P ) = ( µ, . Write χ = ( F , G ) ,and assume that F ( L ∗ ) = 0 and F ′ ( L ∗ ) = − . Also assume that F ispositive over [ P, L ∗ ) and G ′ is non-negative over [ P, L ∗ ] . Define a curve χ : [ P, L ∗ ] → R by χ ( t ) = (cid:16) R − t, (cid:17) Then there is a C homotopy χ s : [ P, L ∗ ] → R for s in [0 , with thefollowing properties. The homotopy fixes P , that is χ s ( P ) = ( µ, for all s in [0 , . Each curve in the homotopy is parametrized by arc length, so forall t in [ P, L ∗ ] and for all s in [0 , , | χ ′ s ( t ) | = 1 If we write χ s = ( F s , G s ) , then for all t in [ P, L ∗ ] and for all s in [0 , , ˙ F s ( t ) ≤ Finally, if L ∗ s is defined by L ∗ s = min n t ∈ [ P, L ∗ ] : F s ( t ) = 0 o then F ′ s ( L ∗ s ) < , for all s in [0 , .Proof. Let h : [0 , → R be a continuously differentiable function such that h (0) = 0, h ′ (0) = 0, h (1) = 1, h ′ (1) = 0 and h ′ ( s ) > s in (0 , f : [ P, L ∗ ] → R and f : [ P, L ∗ ] → R , with f ≥ f , we define ahomotopy by f s = (1 − h ( s )) f + h ( s ) f We refer to this homotopy as the monotonic homotopy from f to f via h . DISC MAXIMIZES EIGENVALUES AMONG ISOPERIMETRIC SURFACES 11
There is a continuous function θ : [ P, L ∗ ] → [0 , π ] such that, for all t in[ P, L ∗ ] χ ′ ( t ) = (cid:16) − cos θ ( t ) , sin θ ( t ) (cid:17) Let ε > θ : [ P, L ∗ ] → [0 , π ],which has the following three properties. First for all t in [ P, L ∗ ], θ ( t ) − ε ≤ θ ( t ) ≤ θ ( t )Second θ is continuously differentiable over the set (cid:26) t ∈ [ P, L ∗ ] : θ ( t ) ∈ ( π/ , π ] (cid:27) and θ has finitely many critical points in this set. Third π/ θ . We take the monotonic homotopy from θ to θ via h . The set (cid:26) t ∈ [ P, L ∗ ] : θ ( t ) ≥ π/ (cid:27) consists of finitely many closed intervals [ a , b ] , [ a , b ] , . . . , indexed so that a i > b i +1 for all i . Let U be a small neighborhood of [ a , b ]. Let δ > θ : [ P, L ∗ ] → R by θ ( t ) = ( θ ( t ) t / ∈ U min( θ ( t ) , π − δ ) t ∈ U If U is sufficiently small, then for sufficiently small δ , this function iscontinous. Take the monotonic homotopy from θ to θ via h . Repeat thisfor each of the closed intervals, letting U , U , . . . be small neighborhoods ofeach of the intervals, and letting δ , δ , . . . be small positive numbers. Thisyields finitely many homotopies. Finally, take the monotonic homotopy fromthe last function to the constant zero function via h . Let ˜ θ s : [ P, L ∗ ] → [0 , π ],for s in [0 ,
1] be the composition of all of these homotopies. Then define χ s : [ P, L ∗ ] → R for s in [0 ,
1] to be the C homotopy with χ s ( P ) = ( µ, t in [ P, L ∗ ], χ ′ s ( t ) = (cid:16) − cos ˜ θ s ( t ) , sin ˜ θ s ( t ) (cid:17) If the parameters are sufficiently small, then this homotopy satisfies theproperties. (cid:3)
Now we can compare λ K,N ( χ ) and λ K,N ( ω ). Lemma 8. If χ is not equal to ω , then λ K,N ( χ ) < λ K,N ( ω ) Proof.
Suppose λ K,N ( χ ) ≥ λ K,N ( ω ). Let ϕ K, , ϕ K, , ϕ K, , . . . be the func-tions given by Lemma 2 associated to the curve χ . Let z be the largest rootof ϕ K,N . Define χ : [ z, L ∗ ] → R by χ = χ (cid:12)(cid:12)(cid:12) [ z,L ∗ ] It follows from Lemma 2 that λ K,N ( χ ) = λ K, ( χ )Define ω : [ z, R ] → R by ω ( t ) = ( R − t, z < z and λ K,N ( ω ) ≥ λ K, ( ω )Let χ s : [ P, L ∗ ] → R be the homotopy discussed in Lemma 7. Extend thedomain of each curve χ s to [ z, L ∗ ], by defining, for all s in [0 ,
1] and for all t in [ z, P ], χ s ( t ) = χ ( t ) = ( R − t, s in [0 , χ s = ( F s , G s ) and define L ∗ s = min (cid:26) t ∈ [ z, L ∗ ] : F s ( t ) = 0 (cid:27) Then define ω s = χ s (cid:12)(cid:12)(cid:12) [ z,L ∗ s ] These functions map into R . Note ω agrees with the previous defintionand ω = χ . We will show that the function s λ K, ( ω s )is monotonically increasing over [0 , σ in(0 ,
1] where λ K, ( ω σ ) > Λ.We first show the function s λ K, ( ω s )is lower semicontinuous. Fix a point σ in [0 ,
1] such thatlim inf s → σ λ K, ( ω s ) < ∞ Let { s k } be a sequence in [0 ,
1] converging to σ such thatlim k →∞ λ K, ( ω s k ) = lim inf s → σ λ K, ( ω s ) DISC MAXIMIZES EIGENVALUES AMONG ISOPERIMETRIC SURFACES 13
By Lemma 2, for each s in [0 , ϕ s in Lip ( z, L ∗ s ) suchthat λ K, ( ω s ) = R L ∗ s z | ϕ ′ s | F s + K ϕ s F s dt R L ∗ s z ϕ s F s dt We may assume that each function ϕ s is normalized so that Z L ∗ s z | ϕ s | F s dt = 1For s in [0 , ℓ s : [ z, L ∗ σ ] → [ z, L ∗ s ] be a linear function with ℓ s ( z ) = z and ℓ s ( L ∗ σ ) = L ∗ s . Define W s = ϕ s ◦ ℓ s , for s in [0 , τ : [0 , → R by τ ( s ) = R L ∗ σ z | W ′ s | F σ + K W s F σ dt R L ∗ σ z W s F σ dt Changing variables yields τ ( s ) = R L ∗ s z | ℓ ′ s | | ϕ ′ s | ( F σ ◦ ℓ − s ) + K ϕ s ( F σ ◦ ℓ − s ) dt R L ∗ s z ϕ s ( F σ ◦ ℓ − s ) dt For s in [0 , s : [0 , L ∗ ] → R byΨ s ( t ) = ( F σ ◦ ℓ − s ( t ) F s ( t ) t ∈ [0 , L ∗ s )1 t ∈ [ L ∗ s , L ∗ ]Note that lim s → σ Ψ s = 1and the convergence is uniform. This follows from the fact that the functions( s, t ) F σ ◦ ℓ − s ( t )and ( s, t ) F s ( t )are both differentiable at the point ( σ, L ∗ σ ) and their derivatives at this pointare equal. Now we see thatlim s → σ Z L ∗ s z ϕ s F s dt − Z L ∗ s z ϕ s ( F σ ◦ ℓ − s ) dt = 0Similarly, lim k →∞ Z L ∗ sk z | ϕ ′ s k | F s k dt − Z L ∗ sk z | ϕ ′ s k | ( F σ ◦ ℓ − s k ) dt = 0Also, lim k →∞ Z L ∗ sk z K ϕ s k F s k dt − Z L ∗ sk z K ϕ s k ( F σ ◦ ℓ − s k ) dt = 0 It follows that lim k →∞ (cid:16) λ K, ( ω s k ) − τ ( s k ) (cid:17) = 0Moreover τ ( s ) ≥ λ K, ( ω σ ) for all s in [ σ, s → σ λ K, ( ω s ) ≥ λ K, ( ω σ )This proves that the function s λ K, ( ω s )is lower semicontinuous.Next we show the function s λ K, ( ω s )is upper semicontinuous. Fix a point σ in [0 , ϕ σ in Lip ( z, L ∗ σ ) such that λ K, ( ω σ ) = R L ∗ σ z | ϕ ′ σ | F σ + K ϕ σ F σ dt R L ∗ σ z ϕ σ F σ dt For s in [0 , ℓ s : [ z, L ∗ σ ] → [ z, L ∗ s ] be a linear function with ℓ s ( z ) = z and ℓ s ( L ∗ σ ) = L ∗ s . Define V s = ϕ σ ◦ ℓ − s , for s in [0 , λ K, ( ω σ ) = R L ∗ s z | ℓ ′ s | | V ′ s | ( F σ ◦ ℓ − s ) + K V s ( F σ ◦ ℓ − s ) dt R L ∗ s z V s ( F σ ◦ ℓ − s ) dt Then define Υ : [0 , → R byΥ( s ) = R L ∗ s z | V ′ s | F s + K V s F s dt R L ∗ s z V s F s dt For s in [0 , s : [0 , L ∗ ] → R byΨ s ( t ) = ( F σ ◦ ℓ − s ( t ) F s ( t ) t ∈ [0 , L ∗ s )1 t ∈ [ L ∗ s , L ∗ ]As before, lim s → σ Ψ s = 1and the convergence is uniform. Now we see thatlim s → σ Z L ∗ s z V s F s dt − Z L ∗ s z V s ( F σ ◦ ℓ − s ) dt = 0Similarly, lim s → σ Z L ∗ s z | V ′ s | F s dt − Z L ∗ s z | V ′ s | ( F σ ◦ ℓ − s ) dt = 0 DISC MAXIMIZES EIGENVALUES AMONG ISOPERIMETRIC SURFACES 15
Also, lim s → σ Z L ∗ s z K V s F s dt − Z L ∗ s z K V s ( F σ ◦ ℓ − s ) dt = 0It follows that lim s → σ Υ( s ) = λ K, ( ω σ )Moreover Υ( s ) ≥ λ K, ( ω s ) for all s in [0 , σ ]. Therefore,lim sup s → σ λ K, ( ω s ) ≤ λ K, ( ω σ )This proves that the function s λ K, ( ω s )is upper semicontinuous, hence continuous. We remark that Cheeger andColding [CC] proved a general theorem regarding continuity of eigenvalues.Now we show the left lower Dini derivative of the function s λ K, ( ω s )is non-negative at every point σ in (0 ,
1] such that λ K, ( ω σ ) > Λ. Fix σ in(0 ,
1] and assume that λ K, ( ω σ ) > ΛBy Lemma 2, there is a function ϕ σ in Lip (0 , L ∗ σ ) such that λ K, ( ω σ ) = R L ∗ σ z | ϕ ′ σ | F σ + K ϕ σ F σ dt R L ∗ σ z ϕ σ F σ dt By Lemma 6, lim t → L ∗ σ ϕ σ ( t ) = 0Also ϕ ′ and ϕF σ are bounded over [ z, L ∗ ). Over [ P, L ∗ σ ], | ϕ ′ σ | − K ϕ σ | F σ | ≤ s in [0 , σ ], L ∗ s ≥ L ∗ σ Define a function ξ : [0 , σ ] → R by ξ ( s ) = R L ∗ σ z | ϕ ′ σ | F s + K ϕ σ F s dt R L ∗ σ z | ϕ σ | F s dt Now λ K, ( ω s ) ≤ ξ ( s ) for s in [0 , σ ], and λ K, ( ω σ ) = ξ ( σ ). Also ξ is leftdifferentiable at σ with ∂ − ξ ( σ ) = R L ∗ σ P ( | ϕ ′ σ | − K ϕ σ | F σ | − λ K, ( ω σ ) ϕ σ ) ˙ F σ dt R L ∗ σ z | ϕ σ | F σ dt The function ˙ F σ is non-positive. That is, ∂ − ξ ( σ ) ≥
0. This implies that thelower left Dini derivative of the function s λ K, ( ω s )is non-negative at σ . That is, the lower left Dini derivative is non-negativeat every point σ in (0 ,
1] such that λ K, ( ω σ ) > Λ. Since the function is alsocontinuous and λ K, ( ω ) > Λ, it follows that the function is monotonicallyincreasing. Moreover, if χ is not equal to ω , then for some σ , the function˙ F σ is not identically zero, which yields ∂ + ξ ( σ ) <
0. This implies that thelower left Dini derivative of the function s λ K, ( ω s )is negative at some point in [0 , λ K, ( χ ) = λ K, ( ω ) < λ K, ( ω )This yields λ K,N ( χ ) < λ K,N ( ω ). (cid:3) Proof of Lemma 1.
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