Lower Bounds for Laplacian and Fractional Laplacian Eigenvalues
aa r X i v : . [ m a t h . DG ] F e b Lower Bounds for Laplacian andFractional Laplacian Eigenvalues
Guoxin Wei He-Jun Sun and Lingzhong Zeng
Abstract:
In this paper, we investigate eigenvalues of Laplacian on a bounded domainin an n -dimensional Euclidean space and obtain a sharper lower bound for the sum of itseigenvalues, which gives an improvement of results due to A. D. Melas [15]. On the otherhand, for the case of fractional Laplacian ( − ∆) α/ | D , where α ∈ (0 , Let D ⊂ R n be a bounded domain with piecewise smooth boundary ∂D in an n -dimensional Euclidean space R n . Let λ i be the i -th eigenvalue of the fixed membraneproblem: ( ∆ u + λu = 0 , in D,u = 0 , on ∂D, (1.1)where ∆ is the Laplacian in R n . It is well known that the spectrum of this eigenvalueproblem is real and discrete:0 < λ ≤ λ ≤ λ ≤ · · · → + ∞ , where each λ i has finite multiplicity which is repeated according to its multiplicity.If we use the notations V ol ( D ) and ω n to denote the volume of D and the volumeof the unit ball in R n , respectively, then Weyl’s asymptotic formula asserts that theeigenvalues of the fixed membrane problem (1.1) satisfy the following formula: λ k ∼ π ( ω n V ol ( D )) n k n , k → + ∞ . (1.2)From the above asymptotic formula, it follows directly that1 k k X i =1 λ i ∼ nn + 2 4 π ( ω n V ol ( D )) n k n , k → + ∞ . (1.3) Key words and phrases: eigenvalues, lower bound, Laplacian, fractional Laplacian.The first author and the second author were supported by the National Natural Science Foun-dation of China (Grant Nos.11001087, 11001130). . Wei, H.-J. Sun and L. Zeng P´olya [17] proved that λ k ≥ π ( ω n V ol ( D )) n k n , for k = 1 , , · · · , (1.4)if D is a tiling domain in R n . Furthermore, he put forward the following: Conjecture of P ´o lya. If D is a bounded domain in R n , then the k -th eigenvalue λ k of the fixed membrane problem satisfies λ k ≥ π ( ω n V ol ( D )) n k n , for k = 1 , , · · · . (1.5)On the Conjecture of P´olya, Berezin [2] and Lieb [13] gave a partial solution. Inparticular, Li and Yau [13] proved the Berezin-Li-Yau inequality as follows:1 k k X i =1 λ i ≥ nn + 2 4 π ( ω n V ol ( D )) n k n , for k = 1 , , · · · . (1.6)The formula (1.3) shows that the result of Li and Yau is sharp in the sense of average.From this inequality (1.6), one can derive λ k ≥ nn + 2 4 π ( ω n V ol ( D )) n k n , for k = 1 , , · · · , (1.7)which gives a partial solution for the conjecture of P´olya with a factor nn + 2 . Weprefer to call this inequality (1.6) as Berezin-Li-Yau inequality instead of Li-Yauinequality because (1.6) can be obtained by a Legendre transform of an earlierresult by Berezin [2] as it is mentioned [14]. Recently, improvements to the Berezin-Li-Yau inequality given by (1.6) for the fixed membrane problem have appeared, forexample see [10, 15, 20]. In particular, A.D.Melas [15] has improved the estimate(1.6) to the following:1 k k X i =1 λ i ≥ nn + 2 4 π ( ω n V ol ( D )) n k n + 124( n + 2) V ol ( D ) Ine ( D ) , for k = 1 , , · · · , (1.8)where Ine ( D ) =: min a ∈ R n Z D | x − a | dx is called the moment of inertia of D . After a translation of the origin, we can assumethat the center of mass is the origin and Ine ( D ) = Z D | x | dx. By taking a value nearby the extreme point of the function f ( τ ) (given by ( ?? )),we add one term of lower order of k − n to its right hand side, which means that weobtain a sharper result than (1.8). In fact, we prove the following:2 ower Bounds for Laplacian and Fractional Laplacian Eigenvalues Theorem 1.1.
Let D be a bounded domain in an n-dimensional Euclidean space R n . Assume that λ i , i = 1 , , · · · , is the i -th eigenvalue of the eigenvalue problem(1.1). Then the sum of its eigenvalues satisfies k k X j =1 λ j ≥ nk n n + 2 ω − n n (2 π ) V ol ( D ) − n + 124( n + 2) V ol ( D ) Ine ( D )+ nk − n n + 2) ω n n (2 π ) − V ol ( D ) Ine ( D ) ! V ol ( D ) n . (1.9)Furthermore, we consider the fractional Laplacian operators restricted to D ,and denote them by ( − ∆) α/ | D , where α ∈ (0 , − ∆) α/ u ( x ) =: P.V. Z R n u ( x ) − u ( y ) | x − y | n + α dy, where P.V. denotes the principal value and u : R n → R . Define the characteristicfunction χ D : t χ D ( t ) by χ D ( t ) = ( , x ∈ D, , x ∈ R n \ D, then the special pseudo-differential operator can be represented as the Fourier trans-form of the function u [11, 19], namely( − ∆) α/ | D u := F − [ | ξ | α F [ uχ D ]] , where F [ u ] denotes the Fourier transform of a function u : R n → R : F [ u ]( ξ ) = b u ( ξ ) = 1(2 π ) n Z R n e − ix · ξ u ( x ) dx. It is well known that the fractional Laplacian operator ( − ∆) α/ can be consideredas the infinitesimal generator of the symmetric α -stable process [3–6, 23]. Supposethat a stochastic process X t has stationary independent increments and its transitiondensity (i.e., convolution kernel) p α ( t, x, y ) = p α ( t, x − y ) , t > , x, y ∈ R n isdetermined by the following Fourier transformExp( − t | ξ | α ) = Z R n e iξ · y p α ( t, y ) dy, t > , ξ ∈ R n , then we can say that the process X t is an n -dimensional symmetric α -stable process with order α ∈ (0 ,
2] in R n (also see [4, 5, 23]). Remark 1.1.
Given α = 1 , X t is the Cauchy process in R n whose transition den-sities are given by the Cauchy distribution (Poisson kernel) p ( t, x, y ) = c n t ( t + | x − y | ) n +12 , t > , x, y ∈ R n , . Wei, H.-J. Sun and L. Zengwhere c n = Γ( n + 12 ) /π n +12 = 1 √ πω n , is the semiclassical constant that appears in the Weyl estimate for the eigenvaluesof the Laplacian. Remark 1.2.
Given α = 2 , X t is just the usual n -dimensional Brownian motion B t but running at twice the speed, which is equivalent to say that, when α = 2 , we have X t = B t and p ( t, x, y ) = 1(4 πt ) n/ Exp " −| x − y | t , t > , x, y ∈ R n . Let Λ αj and u αj denote the j -th eigenvalue and the corresponding normalizedeigenvector of ( − ∆) α/ | Ω , respectively. Eigenvalues Λ αj (including multiplicities)satisfy 0 < Λ ( α )1 ≤ Λ ( α )2 ≤ Λ ( α )3 ≤ · · · → + ∞ . For the case of α = 1, E. Harrell and S. Y. Yolcu gave an analogue of theBerezin-Li-Yau type inequality for the eigenvalues of the Klein-Gordon operators H ,D := √− ∆ restricted to D in [9]:1 k k X j =1 Λ ( α ) j ≥ nn + 1 π ( ω n V ol ( D )) n ! k n . (1.10)Very recently, S.Y.Yolcu [22] has improved the estimate (1.10) to the following:1 k k X j =1 Λ ( α ) j ≥ n e C n n + 1 V ol ( D ) − n k n + f M n V ol ( D ) n Ine ( D ) k − /n , (1.11)where e C n = π ( ω n ) n and the constant f M n depends only on the dimension n . Moreover,for any α ∈ (0 , k k X j =1 Λ ( α ) j ≥ nn + α π ( ω n V ol ( D )) n ! α k αn . (1.12)Furthermore, S.Y.Yolcu and T.Yolcu [23] refined the Berezin-Li-Yau inequality inthe case of fractional Laplacian ( − ∆) α | D restricted to D :1 k k X j Λ ( α ) j ≥ nn + α π ( ω n V ol ( D )) n ! α k αn + ℓ n + α ) (2 π ) α − ( ω n V ol ( D )) α − n V ol ( D ) Ine ( D ) k α − n , (1.13)where ℓ is given by ℓ = min ( α , αnπ (2 n + 2 − α ) ω n n ) . ower Bounds for Laplacian and Fractional Laplacian Eigenvalues Remark 1.3.
In fact, by a direct calculation, one can check the following inequality: α ≤ αnπ (2 n + 2 − α ) ω n n , which implies ℓ n + α ) (2 π ) α − ( ω n V ol ( D )) α − n V ol ( D ) Ine ( D ) k α − n = α n + α ) (2 π ) α − ( ω n V ol ( D )) α − n V ol ( D ) Ine ( D ) k α − n . The another main purpose of this paper is to provide a refinement of the Berezin-Li-Yau type estimate. In other word, we have proved the following:
Theorem 1.2.
Let D be a bounded domain in an n-dimensional Euclidean space R n . Assume that Λ ( α ) i , i = 1 , , · · · , is the i -th eigenvalue of the fractional Laplacian ( − ∆) α/ | D . Then, the sum of its eigenvalues satisfies k k X j =1 Λ ( α ) j ≥ nn + α (2 π ) α ( ω n V ol ( D )) αn k αn + α n + α ) (2 π ) α − ( ω n V ol ( D )) α − n V ol ( D ) Ine ( D ) k α − n + α ( n + α − C ( n ) n ( n + α ) (2 π ) α − ( ω n V ol ( D )) α − n V ol ( D ) Ine ( D ) ! k α − n , (1.14) where C ( n ) = ( , when n ≥ , , when n = 2 or n = 3 . In particular, the sum of its eigenvalues satisfies k k X j =1 Λ (2) j ≥ nk n n + 2 ω − n n (2 π ) V ol ( D ) − n + 124( n + 2) V ol ( D ) Ine ( D )+ nk − n n + 2) ω n n (2 π ) − V ol ( D ) Ine ( D ) ! V ol ( D ) n , (1.15) when α = 2 . Remark 1.4.
Observing Theorem 1.2, it is not difficult to see that the coefficients(with respect to k α − n ) of the second terms in (1.14) are equal to that of (1.13). Inother word, we can claim that the inequalities (1.14) are sharper than (1.13) sincethe coefficients (with respect to k α − n ) of the third terms in (1.14) are positive. By using Theorem 1.2, we can give an analogue of the Berezin-Li-Yau typeinequality for the eigenvalues of the Klein-Gordon operators H ,D restricted to thebounded domain D : 5 . Wei, H.-J. Sun and L. Zeng Corollary 1.1.
Let D be a bounded domain in an n-dimensional Euclidean space R n .Assume that Λ i , i = 1 , , · · · , is the i -th eigenvalue of the Klein-Gordon operators H ,D . Then, the sum of its eigenvalues satisfies k k X j =1 Λ j ≥ nn + 1 2 π ( ω n V ol ( D )) n k n + 148( n + 1) (2 π ) − ( ω n V ol ( D )) − n V ol ( D ) Ine ( D ) k − n + ( n − C ( n ) n ( n + 1) (2 π ) − ( ω n V ol ( D )) − n V ol ( D ) Ine ( D ) ! k − n , (1.16) where C ( n ) = ( , when n ≥ , , when n = 2 or n = 3 . In order to prove the following Lemma 2.3 , we need the following lemmas given byS.Y.Yolcu and T.Yolcu in [23]:
Lemma 2.1.
Suppose that ς : [0 , ∞ ) → [0 , such that ≤ ς ( s ) ≤ Z ∞ ς ( s ) ds = 1 . Then, there exists ǫ ≥ such that Z ǫ +1 ǫ s d ds = Z ∞ s d ς ( s ) ds. Moreover, we have Z ǫ +1 ǫ s d + α ds ≤ Z ∞ s d + α ς ( s ) ds. Lemma 2.2.
For s > , τ > , ≤ b ∈ N , < α ≤ , we have the followinginequality: s b + α ≥ b + αb s b τ α − αb τ b + α + αb τ b + α − ( s − τ ) . In light of Lemma 2.1 and Lemma 2.2, we obtain the following result which willplay important roles in the proof of Theorem 1.1 and Theorem 1.2.
Lemma 2.3.
Let b ( ≥ be a positive real number and µ ( > be defined by (2.13).If ψ : [0 , + ∞ ) → [0 , + ∞ ) is a decreasing function such that − µ ≤ ψ ′ ( s ) ≤ ower Bounds for Laplacian and Fractional Laplacian Eigenvaluesand A := Z ∞ s b − ψ ( s ) ds > , then, we have Z ∞ s b + α − ψ ( s ) ds ≥ b + α ( bA ) b + αb ψ (0) − αb + α b ( b + α ) µ ( bA ) b + α − b ψ (0) b − α +2 b + α ( b + α − b ( b + α ) µ ( bA ) b + α − b ψ (0) b − α +4 b , (2.1) when b ≥ ; we have Z ∞ s b + α − ψ ( s ) ds ≥ b + α ( bA ) b + αb ψ (0) − αb + α b ( b + α ) µ ( bA ) b + α − b ψ (0) b − α +2 b + α ( b + α − b ( b + α ) µ ( bA ) b + α − b ψ (0) b − α +4 b , (2.2) when ≤ b < . In particular, the inequality (2.1) holds when α = 2 and b ≥ .Proof. If we consider the following function ̺ ( t ) = ψ (cid:0) ψ (0) µ t (cid:1) ψ (0) , then it is not difficult to see that ̺ (0) = 1 and − ≤ ̺ ′ ( t ) ≤ . Without loss ofgenerality, we can assume ψ (0) = 1 and µ = 1 . Define E α := Z ∞ s b + α − ψ ( s ) ds. One can assume that E α < ∞ , otherwise there is nothing to prove. By the assump-tion, we can conclude that lim s →∞ s b + α − ψ ( s ) = 0 . Putting h ( s ) = − ψ ′ ( s ) for any s ≥
0, we get0 ≤ h ( s ) ≤ Z ∞ h ( s ) ds = ψ (0) = 1 . By making use of integration by parts, one can get Z ∞ s b h ( s ) ds = b Z ∞ s b − ψ ( s ) ds = bA, . Wei, H.-J. Sun and L. Zeng and Z ∞ s b + α h ( s ) ds ≤ ( b + α ) E α , since ψ ( s ) > . By Lemma 2.1, one can infer that there exists an ǫ ≥ Z ǫ +1 ǫ s b ds = Z ∞ s b h ( s ) ds = bA, (2.3)and Z ǫ +1 ǫ s b + α ds ≤ Z ∞ s b + α h ( s ) ds ≤ ( b + α ) E α . (2.4)Let Θ( s ) = bs b + α − ( b + α ) τ α s b + ατ b + α − ατ b + α − ( s − τ ) , then, by Lemma 2.2, we have Θ( s ) ≥ . Integrating the function Θ( s ) from ǫ to ǫ + 1, we deduce from (2.3) and (2.4), for any τ > ,b ( b + α ) E α − ( b + α ) τ α bA + ατ b + α ≥ α τ b + α − . (2.5)Define f ( τ ) := ( b + α ) τ α bA − ατ b + α + α τ b + α − , (2.6)then we can obtain from (2.5) that, for any τ > ,E α = Z ∞ s b + α − ψ ( s ) ds ≥ f ( τ ) b ( b + α ) . Taking τ = ( bA ) b b + α − b + α ) ( bA ) − b ! b , and substituting it into (2.6), we obtain f ( τ ) = ( bA ) b + αb b − α ( b + α − b + α ) ( bA ) − b ! b + α − b + α ) ( bA ) − b ! αb + α
12 ( bA ) b + α − b b + α − b + α ) ( bA ) − b ! b + α − b . (2.7)By using the Taylor formula, one has for t > t ) αb ≥ αb t + α ( α − b )2 b t + α ( α − b )( α − b )6 b t + α ( α − b )( α − b )( α − b )24 b t , ower Bounds for Laplacian and Fractional Laplacian Eigenvalues and (1 + t ) b + α − b ≥ b + α − b t + ( b + α − α − b t + ( b + α − α − α − − b )6 b t + ( b + α − α − α − − b )( α − − b )24 b t . Putting t = b + α − b + α ) ( bA ) − b > , one has b − αt > τ = ( bA ) b (1 + t ) b , b − α ( b + α − b + α ) ( bA ) − b ! b + α − b + α ) ( bA ) − b ! αb = ( b − αt )(1 + t ) αb ≥ ( b − αt ) " αb t + α ( α − b )2 b t + α ( α − b )( α − b )6 b t + α ( α − b )( α − b )( α − b )24 b t = b − α ( α + b )2 b t − α ( α − b )( α + b )3 b t − α ( α − b )( α − b )( α + b )8 b t − α ( α − b )( α − b )( α − b )24 b t = b − α ( α + b )2 b b + α − b + α ) ( bA ) − b ! − α ( α − b )( α + b )3 b b + α − b + α ) ( bA ) − b ! − α ( α − b )( α − b )( α + b )8 b b + α − b + α ) ( bA ) − b ! − α ( α − b )( α − b )( α − b )24 b b + α − b + α ) ( bA ) − b ! , (2.8)9 . Wei, H.-J. Sun and L. Zeng and b + α − b + α ) ( bA ) − b ! b + α − b = (1 + t ) b + α − b ≥ b + α − b b + α − b + α ) ( bA ) − b ! + ( b + α − α − b b + α − b + α ) ( bA ) − b ! + ( b + α − α − α − − b )6 b b + α − b + α ) ( bA ) − b ! + ( b + α − α − α − − b )( α − − b )24 b b + α − b + α ) ( bA ) − b ! . (2.9)Therefore, we obtain from (2.8) and (2.9) f ( τ ) = ( b + α ) τ α bA − ατ b + α + α τ b + α − ≥ ( bA ) b + αb " b − α ( α + b )2 b b + α − b + α ) ( bA ) − b ! − α ( α − b )( α + b )3 b b + α − b + α ) ( bA ) − b ! − α ( α − b )( α − b )( α + b )8 b b + α − b + α ) ( bA ) − b ! − α ( α − b )( α − b )( α − b )24 b b + α − b + α ) ( bA ) − b ! + α
12 ( bA ) b + α − b " b + α − b b + α − b + α ) ( bA ) − b ! + ( b + α − α − b b + α − b + α ) ( bA ) − b ! + ( b + α − α − α − − b )6 b b + α − b + α ) ( bA ) − b ! + ( b + α − α − α − − b )( α − − b )24 b b + α − b + α ) ( bA ) − b ! = b ( bA ) b + αb + α
12 ( bA ) b + α − b + I + I + I , (2.10)10 ower Bounds for Laplacian and Fractional Laplacian Eigenvalues where I = α ( b + α − b ( b + α ) ( bA ) b + α − b , (2.11) I = α ( b + α − α + 2 b − b b + α − b + α ) ! ( bA ) b + α − b + α ( b + α − α + (5 b − α + ( − b + 8 b + 16)]288 b × b + α − b + α ) ! ( bA ) b + α − b , (2.12) I = αγ b b + α − b + α ) ! ( bA ) b + α − b , and γ = ( α − α − − b )( α − − b )( b + α ) − α ( α − b )( α − b )( α − b ) . Noticing that − α ( α − b )( α − b )( α − b ) ≥ , we have γ ≥ ( α − α − − b )( α − − b )( b + α ) . Define β := ( α − α − − b )( α − − b )( b + α ) , then we have β ≤ γ ≥ β . Therefore, we have I ≥ αβ b b + α − b + α ) ! ( bA ) b + α − b , (2.13)Next, we consider two cases: Case 1: b ≥ When b ≥
4, for any α ∈ (0 , α + (5 b − α + ( − b + 8 b + 16) ≤ (5 b − α + ( − b + 8 b + 20)= − b + (8 + 5 α ) b + 20 − α ≤ − b + (8 + 10) b + 20 ≤ . (2.14)11 . Wei, H.-J. Sun and L. Zeng Since ( bA ) b ≥ b +1) b ≥ (see [8]), one can deduce from (2.12) and (2.14) I ≥ α ( b + α − α + 2 b − b b + α − b + α ) ! ( bA ) b + α − b + α ( b + α − α + (5 b − α + ( − b + 8 b + 16)]1152 b × b + α − b + α ) ! ( bA ) b + α − b = α ( b + α − (cid:2) b ( α + 2 b −
6) + α + (5 b − α + ( − b + 8 b + 16) (cid:3) b × b + α − b + α ) ! ( bA ) b + α − b = α ( b + α − (cid:2) b + ( −
88 + 21 α ) b + ( α − α + 16) (cid:3) b × b + α − b + α ) ! ( bA ) b + α − b . (2.15)On the other hand, we have I ≥ αβ b b + α − b + α ) ! ( bA ) b + α − b , since β ≤ bA ) b ≥ b +1) b ≥ . Therefore, the estimate of the lower bound of I + I can be given by I + I ≥ ( α ( b + α − (cid:2) b + ( −
88 + 21 α ) b + ( α − α + 16) (cid:3) b + αβ b ) × b + α − b + α ) ! ( bA ) b + α − b = α { b ( b + α − b + ( −
88 + 21 α ) b + ( α − α + 16)] + β } b × b + α − b + α ) ! ( bA ) b + α − b . Next, we will verify the following inequality4 b ( b + α − b + ( −
88 + 21 α ) b + ( α − α + 16)] + β ≥ . (2.16)12 ower Bounds for Laplacian and Fractional Laplacian Eigenvalues Indeed, since 0 < α ≤ b ≥
4, we have4 b ( b + α − b + ( −
88 + 21 α ) b + ( α − α + 16)] + β = 4 b ( b + α − b + ( −
88 + 21 α ) b + ( α − α + 16)]+ ( α − α − − b )( α − − b )( b + α ) ≥ b [26 b + ( −
88 + 21 α ) b + ( α − α + 16)] − | ( α − α − − b )( α − − b )( b + α ) |≥ b [26 b + ( −
88 + 21 α ) b + ( α − α + 16)] − | ( b + 2)(2 b + 2)( b + 2) |≥ b [26 b − b + ( α − α )] − b + 2)(2 b + 2)( b + 2) ≥ b (26 b − b ) − b + 2)(2 b + 2)( b + 2)= 204 b − b − b − ≥ b − b − ≥ b − ≥ . (2.17)Thus, it is not difficult to see that the inequality (2.16) follows from (2.17), whichimplies I + I ≥ . Therefore, when b ≥ , we have f ( τ ) ≥ b ( bA ) b + αb + α
12 ( bA ) b + α − b + α ( b + α − b ( b + α ) ( bA ) b + α − b . Case 2: ≤ b < Uniting the equations (2.11), (2.12) and (2.13), we obtainthe following equation I + I + I ≥ α ( b + α − b ( b + α ) ( bA ) b + α − b + α ( b + α − α + 2 b − b b + α − b + α ) ! ( bA ) b + α − b + α ( b + α − α + (5 b − α + ( − b + 8 b + 16)]288 b × b + α − b + α ) ! ( bA ) b + α − b + αβ b b + α − b + α ) ! ( bA ) b + α − b . Wei, H.-J. Sun and L. Zeng = α ( b + α − b ( b + α ) ( bA ) b + α − b + α ( b + α − b ( b + α ) ( bA ) b + α − b + α ( b + α − ν b b + α − b + α ) ! ( bA ) b + α − b + α ( b + α − ν b b + α − b + α ) ! ( bA ) b + α − b + αβ b b + α − b + α ) ! ( bA ) b + α − b , where ν := ( α + 2 b − , and ν := α + (5 b − α + ( − b + 8 b + 16) . Suppose ν ≤ ν ≤
0, then we have I + I + I ≥ α ( b + α − b ( b + α ) ( bA ) b + α − b + α ( b + α − b b + α − b + α ) ! ( bA ) b + α − b + α ( b + α − α + 2 b − b b + α − b + α ) ! ( bA ) b + α − b + α ( b + α − α + (5 b − α + ( − b + 8 b + 16)]4608 b × b + α − b + α ) ! ( bA ) b + α − b + αβ b b + α − b + α ) ! ( bA ) b + α − b = α ( b + α − b ( b + α ) ( bA ) b + α − b + α ( b + α − b I b + α − b + α ) ! ( bA ) b + α − b + αβ b b + α − b + α ) ! ( bA ) b + α − b , (2.18)where I = 1 + α + 2 b − b + α + (5 b − α + ( − b + 8 b + 16)48 b . ower Bounds for Laplacian and Fractional Laplacian Eigenvalues Noticing that 0 < α ≤ ≤ b <
4, we have I = 48 b + 16 b ( α + 2 b −
6) + α + (5 b − α + ( − b + 8 b + 16)48 b = 74 b + (21 α − b + ( α − α + 16)48 b ≥ b + 21 αb + ( α − α )48 b = 60 b + (21 b − α + α b ≥ b b = 54 b . (2.19)Therefore, we derive from (2.18) and (2.19) I + I + I ≥ α ( b + α − b ( b + α ) ( bA ) b + α − b + 5 α ( b + α − b b + α − b + α ) ! ( bA ) b + α − b + αβ b b + α − b + α ) ! ( bA ) b + α − b ≥ α ( b + α − b ( b + α ) ( bA ) b + α − b + 5 α ( b + α − b b + α − b + α ) ! ( bA ) b + α − b + α ( b + α − β b + α ) b b + α − b + α ) ! ( bA ) b + α − b ≥ α ( b + α − b ( b + α ) ( bA ) b + α − b + α ( b + α − b ( b + α ) + β ]18432( b + α ) b b + α − b + α ) ! ( bA ) b + α − b , since ( bA ) b ≥ b +1) b ≥ . We define a function K ( b ) by letting K ( b ) := 240 b ( b + α ) + β = 240 b ( b + α ) + ( α − α − − b )( α − − b )( b + α ) , where b ∈ [2 , K ( b ) ≥ b ( b + α ) − | ( α − α − − b )( α − − b )( b + α ) |≥ b ( b + α ) − | b )(2 + 2 b )( b + 2) |≥ b ( b + α ) − b )(3 b )(2 b ) ≥ b + 240 αb > , which implies I + I + I ≥ α ( b + α − b ( b + α ) ( bA ) b + α − b . . Wei, H.-J. Sun and L. Zeng For the other cases (i.e., ν ≤ ν > ν > ν ≤
0; or ν > ν > I + I + I ≥ α ( b + α − b ( b + α ) ( bA ) b + α − b . Therefore, when 2 ≤ b ≤ , we have f ( τ ) ≥ b ( bA ) b + αb + α
12 ( bA ) b + α − b + α ( b + α − b ( b + α ) ( bA ) b + α − b . In particular, we can consider the case that α = 2. Noticing that β = 0 when α = 2and b ≥
2, we can claim that I ≥
0. Therefore, when α = 2 and b ≥
2, one candeduce I + I ≥ α ( b + α − α + 2 b − b b + α − b + α ) ! ( bA ) b + α − b + α ( b + α − α + (5 b − α + ( − b + 8 b + 16)]288 b × b + α − b + α ) ! ( bA ) b + α − b = b ( b − b b b + 2) ! ( bA ) b − b + − b + 3 b − b b b + 2) ! ( bA ) b − b ≥ b ( b − b b b + 2) ! ( bA ) b − b + − b + 3 b − b b b + 2) ! ( bA ) b − b = 13 b − b − b b b + 2) ! ( bA ) b − b ≥ b − b − b b b + 2) ! ( bA ) b − b ≥ , which implies f ( τ ) ≥ b ( bA ) b + αb + α
12 ( bA ) b + α − b + α ( b + α − b ( b + α ) ( bA ) b + α − b . This completes the proof of the Lemma 2.3. (cid:3)
In this section, we will prove the Theorem 1.1 and Theorem 1.2 by using the keylemma given in section 2 (i.e., Lemma 2.3).16 ower Bounds for Laplacian and Fractional Laplacian Eigenvalues
We suppose that D ⊂ R n is a bounded domain in R n , and then its symmetricrearrangement D ∗ is the open ball with the same volume as D , D ∗ = ( x ∈ R n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | x | < V ol ( D ) ω n ! n ) . By using a symmetric rearrangement of D , one can obtain Ine ( D ) = Z D | x | dx ≥ Z D ∗ | x | dx = nn + 2 V ol ( D ) V ol ( D ) ω n ! n . (3.1)For the case of fractional Laplace operator, let u ( α ) j be an orthonormal eigenfunctioncorresponding to the eigenvalue Λ ( α ) j . Namely, u ( α ) j satisfies ( ( − ∆) α/ u ( α ) j = Λ ( α ) u ( α ) j , in D, R D u ( α ) i ( x ) u ( α ) j ( x ) dx = δ ij , for any i, j, where 0 < α ≤
2. On the other hand, for the case of Laplace operator, we let v j be an orthonormal eigenfunction corresponding to the eigenvalue λ j . Namely, v j satisfies ∆ v j + λ j v j = 0 , in D,v = 0 , on ∂D, R D v i ( x ) v j ( x ) dx = δ ij , for any i, j. Thus, both { u ( α ) j } ∞ j =1 and { v j } ∞ j =1 form an orthonormal basis of L ( D ). Definethe functions ϕ ( α ) j and η j by ϕ ( α ) j ( x ) = ( u ( α ) j ( x ) , x ∈ D, , x ∈ R n \ D, and η j ( x ) = ( v j ( x ) , x ∈ D, , x ∈ R n \ D, respectively. Denote by b η j ( ξ ) and d ϕ ( α ) j ( ξ ) the Fourier transforms of η j ( ξ ) and ϕ ( α ) j ( ξ ),then, for any ξ ∈ R n , we have d ϕ ( α ) j ( ξ ) = (2 π ) − n/ Z R n ϕ ( α ) j ( x ) e i h x,ξ i dx = (2 π ) − n/ Z D u ( α ) j ( x ) e i h x,ξ i dx, and b η j ( ξ ) = (2 π ) − n/ Z R n η j ( x ) e i h x,ξ i dx = (2 π ) − n/ Z D v j ( x ) e i h x,ξ i dx. . Wei, H.-J. Sun and L. Zeng From the Plancherel formula, we have Z R n d ϕ ( α ) i ( x ) d ϕ ( α ) j ( x ) dx = Z R n b η i ( x ) b η j ( x ) dx = δ ij , for any i, j . Since { u ( α ) j } ∞ j =1 and { v j } ∞ j =1 are orthonormal basises in L ( D ), the Besselinequality implies that k X j =1 | d ϕ ( α ) j ( ξ ) | ≤ (2 π ) − n/ Z D | e i h x,ξ i | dx = (2 π ) − n/ V ol ( D ) , (3.2)and k X j =1 | b η j ( ξ ) | ≤ (2 π ) − n/ Z D | e i h x,ξ i | dx = (2 π ) − n/ V ol ( D ) . (3.3)For fractional Laplace operator, we observe thatΛ ( α ) j = Z R n u ( α ) j ( ξ ) · ( − ∆) α/ | Ω u ( α ) j ( ξ ) dξ = Z R n u ( α ) j ( ξ ) · F − [ | ξ | α F [ u ( α ) j ( ξ )]] dξ = Z R n | ξ | α | d u ( α ) j ( ξ ) | dξ, (3.4)since the support of u ( α ) j is D (see [23]). On the meanwhile, for the case of Laplaceoperator, we have (see [12, 15]) λ j = Z R n | ξ | | b v j ( ξ ) | dξ. (3.5)Since ∇ d ϕ ( α ) j ( ξ ) = (2 π ) − n/ Z Ω ixu ( α ) j ( x ) e i h x,ξ i dx, and ∇ b η j ( ξ ) = (2 π ) − n/ Z Ω ixv j ( x ) e i h x,ξ i dx, we obtain k X j =1 |∇ d ϕ ( α ) j ( ξ ) | = k X j =1 |∇ b η j ( ξ ) | = (2 π ) − n Z Ω | ixe i h x,ξ i | dx = (2 π ) − n Ine ( D ) . (3.6)Putting f ( α ) ( ξ ) := k X j =1 | d ϕ ( α ) j ( ξ ) | , ower Bounds for Laplacian and Fractional Laplacian Eigenvalues and f ( ξ ) := k X j =1 | b η j ( ξ ) | , one derives from (3.2) and (3.3) that 0 ≤ f ( α ) ( ξ ) ≤ (2 π ) − n V ol ( D ) and 0 ≤ f ( ξ ) ≤ (2 π ) − n V ol ( D ), it follows from (3.6) and the Cauchy-Schwarz inequality that |∇ f ( α ) ( ξ ) | ≤ k X j =1 | d ϕ ( α ) j ( ξ ) | ! / k X j =1 |∇ d ϕ ( α ) j ( ξ ) | ! / ≤ π ) − n p Ine ( D ) V ol ( D ) , and |∇ f ( ξ ) | ≤ k X j =1 | b η j ( ξ ) | ! / k X j =1 |∇ b η j ( ξ ) | ! / ≤ π ) − n p Ine ( D ) V ol ( D ) , for every ξ ∈ R n . Furthermore, by using (3.4) and (3.5), we have k X j =1 Λ ( α ) j = k X j =1 Z R n | ξ | α | d u ( α ) j ( ξ ) | dξ = k X j =1 Z R n | ξ | α | d ϕ ( α ) j ( ξ ) | dξ = Z R n | ξ | α f ( α ) ( ξ ) dξ, (3.7)and k X j =1 λ j = k X j =1 Z R n | ξ | | b v j ( ξ ) | dξ = k X j =1 Z R n | ξ | | b η j ( ξ ) | dξ = Z R n | ξ | f ( ξ ) dξ. (3.8)From the Parseval’s identity, we derive Z R n f ( α ) ( ξ ) dξ = k X j =1 Z R n | d ϕ ( α ) j ( x ) | dx = k X j =1 Z D | d u ( α ) j ( x ) | dx = k X j =1 Z D | u ( α ) j ( x ) | dx = k. (3.9)Similarly, we have [7, 15] Z R n f ( ξ ) dξ = k X j =1 Z R n | b η j ( x ) | dx = k X j =1 Z D | b v j ( x ) | dx = k X j =1 Z D | v j ( x ) | dx = k. (3.10)19 . Wei, H.-J. Sun and L. Zeng Let h be a nonnegative bounded continuous function on D and h ∗ is its symmetricdecreasing rearrangement, then we have (see [1, 7]) Z R n h ( x ) dx = Z R n h ∗ ( x ) dx = nω n Z ∞ s n − g ( s ) ds (3.11)and Z R n | x | α h ( x ) dx ≥ Z R n | x | α h ∗ ( x ) dx = nω n Z ∞ s n + α − g ( s ) ds, (3.12)where α ∈ (0 ,
2] and g ( | x | ) = h ∗ ( x ). Putting δ := sup |∇ h | , then we can obtain − δ ≤ g ′ ( s ) ≤ s . More detail information on symmetric decreasing rearrangementswill be found in [1, 7, 18].To be brief, we will drop the superscript α to denote f ( α ) by f and let f = f .Assume that f ∗ i is the symmetric decreasing rearrangement of f i ( i = 1 , k = Z R n f i ( ξ ) dξ = Z R n f ∗ i ( ξ ) dξ = nω n Z ∞ s n − φ i ( s ) ds, (3.14)where φ i ( x ) = f ∗ i ( | x | ) and i = 1 , f i , and noting that δ i ≤ π ) − n p Ine (Ω)
V ol (Ω) := σ, (3.15)where δ i = sup |∇ f i | , we obtain from (3.13) − σ ≤ − δ i ≤ φ ′ i ( s ) ≤ , where i = 1 ,
2. By(3.1), we have σ ≥ π ) − n ( nn + 2 ) ω − n n V ol ( D ) n +1 n ≥ (2 π ) − n ω − n n V ol ( D ) n +1 n , since n ≥
2. Moreover, by using (3.7), (3.8) and (3.12), we have k X j =1 Λ ( α ) j = Z R n | ξ | α f ( α ) ( ξ ) dξ = Z R n | ξ | α f ( ξ ) dξ ≥ Z R n | ξ | α f ∗ ( ξ ) dξ = nω n Z ∞ s n + α − φ ( ξ ) dξ, (3.16)and k X j =1 λ j = Z R n | ξ | f ( ξ ) dξ = Z R n | ξ | f ( ξ ) dξ ≥ Z R n | ξ | f ∗ ( ξ ) dξ = nω n Z ∞ s n +1 φ ( ξ ) dξ. (3.17)20 ower Bounds for Laplacian and Fractional Laplacian EigenvaluesProof of Theorem 1.1. In order to apply Lemma 2.3, from (3.14), (3.15) and thedefinition of A , we take b = n, ψ ( s ) = φ ( s ) , A = knω n , and µ = σ = 2(2 π ) − n p V ol ( D ) Ine ( D ) . Therefore, we can obtain from Lemma 2.3 and (3.17) that k X j =1 λ j = nω n E ≥ nω n n ( n + 2) " n ( nA ) n +2 n φ (0) − n + nAφ (0) σ + n ( nA ) n − n n + 2) σ φ (0) n +2 n = ω n n + 2 " n kω n ! n +2 n t − n + kω n t σ + n ( kω n ) n − n n + 2) σ t n +2 n = nn + 2 ω − n n k n +2 n t − n + kt n + 2) σ + nω n n k n − n n + 2) σ t n +2 n , where t = φ (0). Let F ( t ) = nn + 2 ω − n n k n +2 n t − n + kt n + 2) σ + nω n n k n − n n + 2) σ t n +2 n , then one can has F ′ ( t ) = − n + 2 ω − n n k n +2 n t − n +2 n + kt n + 2) σ + 4 n + 2144( n + 2) σ ω n n k n − n t n +2 n . Since F ′ ( t ) is increasing on (0 , (2 π ) − n V ol ( D )], then it is easy to see that F ( t ) isdecreasing on (0 , (2 π ) − n V ol ( D )] if F ′ ((2 π ) − n V ol ( D )) <
0. Indeed, F ′ ((2 π ) − n V ol ( D )) ≤ − n + 2 ω − n n k n +2 n ((2 π ) − n V ol ( D )) − n +2 n + k ((2 π ) − n V ol ( D ))3( n + 2) h (2 π ) − n ω − n n V ol ( D ) n +1 n i + (4 n + 2) ω n n k n − n ((2 π ) − n V ol ( D )) n +2 n n + 2) h (2 π ) − n ω − n n V ol ( D ) n +1 n i = − n + 2 (2 π ) n +2 ω − n n k n +2 n V ol ( D ) − n +2 n + 13( n + 2) (2 π ) n ω n n kV ol ( D ) − n +2 n + 4 n + 2144( n + 2) (2 π ) n − ω n n k n − n V ol ( D ) − n +2 n = (2 π ) n kn + 2 ω n n V ol ( D ) − n +2 n J , . Wei, H.-J. Sun and L. Zeng where J = 13 + 4 n + 2144( n + 2) (2 π ) − ω n n k − n − π ) k n ω − n n <
13 + 4( n + 2)144( n + 2) (2 π ) − ω n n − π ) ω − n n = 13 + 136 (2 π ) − ω n n − π ) ω − n n <
13 + 172 − < , which implies that F ′ ((2 π ) − n V ol ( D )) < . Here, we use the inequality ω nn (2 π ) < .We can replace φ (0) by (2 π ) − n V ol ( D ) to obtain1 k k X j =1 λ j ≥ nk n n + 2 ω − n n (2 π ) V ol ( D ) − n + 124( n + 2) V ol ( D ) Ine ( D )+ nk − n n + 2) ω n n (2 π ) − V ol ( D ) Ine ( D ) ! V ol ( D ) n . since σ = 2(2 π ) − n p V ol ( D ) Ine ( D ) . This completes the proof of Theorem 1.1. (cid:3)
Next, we will give the proof of Theorem 1.2.
Proof of Theorem 1.2. : Define the function φ ( x ) by φ ( | x | ) := f ∗ ( x ). Thenwe know that φ : [0 , + ∞ ) → [0 , (2 π ) − n V (Ω)] is a non-increasing function withrespect to | x | . Taking b = n, ψ ( s ) = φ ( s ) , A = knω n , and µ = σ = 2(2 π ) − n p V ol ( D ) Ine ( D ) , we can obtain from Lemma 2.3 and (3.16) that k X j =1 Λ ( α ) j ≥ nω n Z ∞ s n + α − φ ( s ) ds ≥ nω n (cid:16) kω n (cid:17) n + αn n + α φ (0) − αn + αω n (cid:16) kω n (cid:17) n + α − n n + α ) σ φ (0) n − α +2 n + α ( n + α − ω n (cid:16) kω n (cid:17) n + α − n C ( n ) n ( n + α ) σ φ (0) n − α +4 n , (3.18)22 ower Bounds for Laplacian and Fractional Laplacian Eigenvalues where C ( n ) = ( , when n ≥ , , when n = 2 or n = 3 . Moreover, we define a function ξ ( t ) by letting ξ ( t ) = nω n n + α kω n ! n + αn t − αn + αω n n + α ) σ kω n ! n + α − n t n − α +2 n + α ( n + α − ω n C ( n ) n ( n + α ) σ kω n ! n + α − n t n − α +4 n . (3.19)Differentiating (3.19) with respect to the variable t , it is not difficult to see that ξ ′ ( t ) = αω n n + α kω n ! n + αn t − αn − " − n − α + 2)12 nσ kω n ! − n t n +2 n + (4 n − α + 4)( n + α − C ( n ) n ( n + α ) σ kω n ! − n t n +4 n . (3.20)Letting ζ ( t ) = ξ ′ ( t )( n + ααω n )( kω n ) − n + αn t αn +1 , (3.21)and noting that σ ≥ (2 π ) − n ω − n n V ol ( D ) n +1 n , we can obtain from (3.20) and (3.21)that ζ ( t ) = − n − α + 2)12 nσ kω n ! − n t n +2 n + (4 n − α + 4)( n + α − C ( n ) n ( n + α ) σ kω n ! − n t n +4 n ≤ − n − α + 2)12 n (2 π ) − n ω − n n V ol ( D ) n +1) n kω n ! − n t n +2 n + (4 n − α + 4)( n + α − C ( n ) n ( n + α )(2 π ) − n ω − n n V ol ( D ) n +1) n kω n ! − n t n +4 n . (3.22)It is easy to see that the right hand side of (3.22) is an increasing function of t . Therefore, if the right hand side of (3.22) is less than 0 when we take t =23 . Wei, H.-J. Sun and L. Zeng (2 π ) − n V ol ( D ), which is equivalent to say that ζ ( t ) ≤ − n − α + 2)12 n k − n ω n n (2 π ) + (4 n − α + 4)( n + α − C ( n ) n ( n + α ) k − n ω n n (2 π ) ≤ , (3.23)we can claim from (3.23) that ξ ′ ( t ) ≤ , (2 π ) − n V (Ω)] . By a direct calculation,we can obtain ζ ( t ) ≤ − n − α + 2)12 n + (4 n − α + 4)( n + α − C ( n ) n ( n + α ) ≤ − n + n )12 n + (4 n + 2 n )( n + n ) C ( n ) n = −
34 + 24 C ( n ) ≤ , (3.24)since ω nn (2 π ) < . Thus, it is easy to see from (3.21) and (3.24) that ξ ′ ( t ) ≤
0, whichimplies that ξ ( t ) is a decreasing function on (0 , (2 π ) − n V ol ( D )] . On the other hand, we notice that 0 < φ (0) ≤ (2 π ) − n V ol ( D ) and right handside of the formula (3.18) is ξ ( φ (0)), which is a decreasing function of φ (0) on(0 , (2 π ) − n V ol ( D )]. Therefore, φ (0) can be replaced by (2 π ) − n V ol ( D ) in (2.1) whichgives the following inequality:1 k k X j =1 Λ ( α ) j ≥ nn + α (2 π ) α ( ω n V ol ( D )) αn k αn + α n + α ) (2 π ) α − ( ω n V ol ( D )) α − n V ol ( D ) Ine ( D ) k α − n + α ( n + α − C ( n ) n ( n + α ) (2 π ) α − ( ω n V ol ( D )) α − n V ol ( D ) Ine ( D ) ! k α − n , where C ( n ) = ( , when n ≥ , , when n = 2 or n = 3 . In particular, when α = 2, we can get the inequality (1.15) by using the samemethod as the proof of Theorem 1.1.This completes the proof of Theorem 1.2. (cid:3) Acknowledgment.
The authors wish to express their gratitude to Prof. Q.-M.Cheng for continuous encouragement and enthusiastic help.24 ower Bounds for Laplacian and Fractional Laplacian Eigenvalues
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