An asymptotic lower bound for the norm of the Laplace operator on a - PowerPoint PPT Presentation

An asymptotic lower bound for the norm of the Laplace operator on a space of polynomials Christian Rebs Technische Universit¨ at Chemnitz, 17 . 8 . 2017

Introduction A. B¨ ottcher, C.Rebs: On the constants in Markov inequalities for the Laplace operator on polynomials with the Laguerre norm , Asymptotic Analysis 101 (2017), 227-239. For n, N ∈ N we define P N n as the finite dimensional linear space of all complex polynomials f of the form � f i 1 ,...,i N t i 1 1 . . . t i N f ( t 1 , . . . , t N ) = N , n ,..., iN ( i 1 n ) ∈ [0 , 1] N f i 1 ,...,i N ∈ C . We equip these space with the Laguerre norm � � f � 2 := (0 , ∞ ) N | f ( t 1 , . . . , t N ) | 2 e − t 1 . . . e − t N dt 1 . . . dt N . 1

The Laplace operator ∆ = ∂ 2 + . . . + ∂ 2 : P N n → P N n ∂t 2 ∂t 2 1 N is a bounded linear operator on P N n . There exists a constant C = C ( n, N ), such that � ∆ f � ≤ C � f � holds for all f ∈ P N n . The best constant with this property is C = � ∆ � . Aim: Calculate (asymptotic) bounds for � ∆ � 2

Main result: Theorem 1. Let n, N ∈ N and let ω 0 be the positive solution of the equation 2 + 2 cosh( ω ) − ω sinh( ω ) = 0 . Then we have for the operator norm of the Laplace operators on P N with respect to the Laguerre n norm � ( n + 1) 2 � � ∆ � ≥ N ( n + 1) 2 + o ω 2 0 for n → ∞ . 2 With = 0 , 34741 . . . this implies ω 2 0 � ∆ � lim inf Nn 2 ≥ 0 . 1737 . n →∞ 3

The matrix representation of ∆ The norm � ∆ � is the largest singular value of the matrix representation [∆] of ∆ in an orthonormal basis of P N n . We set P n := P 1 n . For k ∈ N 0 we set � t � t 2 � t k � k � k 2! − · · · + ( − 1) k � k L k ( t ) := 1 − 1! + 1 2 k ! k and get an orthonormal basis { L 0 , L 1 , . . . , L n } is P n . 4

The ordinary differential operator D 2 is defined by D 2 : P n → P n , f �→ f ′′ . [1] L. F. Shampine: Some L 2 Markoff inequalities , J. Res. Nat. Bur. Standards 69B (1965), 155-158: The matrix representation of D 2 is the ( n + 1) × ( n + 1) matrix   0 0 1 2 n − 2 n − 1 . . .   0 0 1 n − 3 n − 2 . . .     . ... ... ... ... . .   � D 2 �   . ... ... ... .   = . .   ...   0 1     0 0   0 5

with � N We identify P N j =1 P n . The polynomials L j 1 ⊗ · · · ⊗ L j N with n j 1 , . . . , j N ∈ { 0 , . . . , n } form an orthonormal basis in P N n . The matrix representation of the Laplace operator in this basis is � D 2 ⊗ I ⊗ . . . ⊗ I + I ⊗ D 2 ⊗ I ⊗ . . . ⊗ I + . . . + I ⊗ . . . ⊗ I ⊗ D 2 � [∆] = � D 2 � � D 2 � = ⊗ I n +1 ⊗ . . . ⊗ I n +1 + I n +1 ⊗ ⊗ I n +1 ⊗ . . . ⊗ I n +1 � D 2 � + . . . + I n +1 ⊗ . . . ⊗ I n +1 ⊗ N � D 2 � � = I ( n +1) k − 1 ⊗ ⊗ I ( n +1) N − k . k =1 6

Proof of the main result For a n × n matrix A we set H ( A ) := 1 2 ( A + A ∗ ). [2] R. A. Horn, C. R. Johnson: Topics in Matrix Analysis , Cambridge University Press, Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, 8th printing, 2007: Theorem 2: [Corollary 3 . 1 . 5 in [2]] For a n × n matrix A we denote by σ 1 ( A ) ≥ σ 2 ( A ) ≥ . . . ≥ σ n ( A ) the singular values A and by λ 1 ( H ( A )) ≥ . . . ≥ λ n ( H ( A )) the eigenvalues of H ( A ). Then we have the estimate σ k ( A ) ≥ λ k ( H ( A )) for all k = 1 , . . . , n . 7

For a n × n matrix A and a m × m matrix B we define the Kronecker sum A ⊕ B := ( I m ⊗ A ) + ( B ⊗ I n ) . Theorem 3: [Theorem 4 . 4 . 5 in [2]] If λ is an eigenvalue of A and if µ is an eigenvalue of B , then λ + µ is an eigenvalue of A ⊕ B . If ν is an eigenvalue of A ⊕ B , then there exist eigenvalues λ of A and µ of B such that ν = λ + µ . 8

We need Kronecker sums with more then two summands: For matrices A 1 of order n 1 , . . . , A N of order n N we set A 1 ⊕ . . . ⊕ A N := A 1 ⊕ ( A 2 ⊕ ( A 3 ⊕ ( . . . ⊕ ( A N − 1 ⊕ A N ) . . . ) . For the Kronecker sum with N summands we get the formula N � A 1 ⊕ . . . ⊕ A N = I � N k = j +1 n k ⊗ A j ⊗ I � j − 1 k =1 n k , j =1 where we set � l k = j n k = 1 for l < j . Eigenvalues: For eigenvalues λ 1 of A 1 , λ 2 of A 2 , . . . , λ N of A N , λ 1 + . . . + λ N is an eigenvalue of A 1 ⊕ . . . ⊕ A N . Every eigenvalue λ of A 1 ⊕ . . . ⊕ A N is a sum λ = λ 1 + . . . + λ N of eigenvalues λ k of A k , k = 1 , . . . , N . 9

Now we get an estimate for the norm of the Laplace operator: � 1 [∆] + [∆] T �� � � ∆ � = σ max ([∆]) ≥ λ max 2  N � D 2 � 1 �  = I ( n +1) k − 1 ⊗ ⊗ I ( n +1) N − k 2 λ max k =1  N � D 2 � T ⊗ I ( n +1) N − k �  + I ( n +1) k − 1 ⊗ k =1   �� D 2 � T � N D 2 � � 1 �   = I ( n +1) k − 1 ⊗ + ⊗ I ( n +1) N − k 2 λ max k =1 ��� D 2 � T � �� D 2 � T �� D 2 � � D 2 � � 1 = + ⊕ . . . ⊕ + 2 λ max �� D 2 � T � D 2 � � N = + 2 λ max . 10

We have to calculate the largest eigenvalue of   0 0 1 2 n − 2 n − 1 . . .   0 0 0 1 n − 2  . . .    . . . .   1 0 . .   ... ...   � D 2 � � D 2 � T = 2 1   + .   ... ... 1 2     . . . .   . . 0 1     n − 2 1 0 0 0 . . .   n − 1 n − 2 2 1 0 0 . . . 11

[3] J. M. Bogoya, A. B¨ ottcher, S. M. Grudsky: Eigenvalues of Hermitian Toeplitz matrices with polynomially increasing entries , Journal of Spectral Theory 2 (2012), 267-292: For α > 0 we define the integral operator K α on L 2 (0 , 1) as � 1 0 | x − y | α f ( y ) dy ( K α f )( x ) := for x ∈ (0 , 1). We denote by µ 1 ( K α ) ≥ µ 2 ( K α ) ≥ . . . > 0 the positive eigenvalues of K α and by L + the index set of this eigenvalues. Theorem 4: [Theorem 1 . 2 in [3]] The operator K 1 has only one positive eigenvalue µ 1 ( K 1 ) = 2 = 0 , 34741 . . . . ω 2 0 Here is ω 0 the positive solution of the equation 2 + 2 cosh( ω ) − ω sinh( ω ) = 0. 12

We denote by T n [ a 0 , . . . , a n − 1 ] the hermitian Toeplitz matrix   a 0 a 1 . . . a n − 1   ¯ a 1 a 0 . . . a n − 2   T n = T n [ a 0 , . . . a n − 1 ] =  .   . . . ... . . . . . .  ¯ a n − 1 ¯ a n − 2 . . . a 0 For the eigenvalues � λ 1 ( T n ) ≤ � λ 2 ( T n ) ≤ . . . ≤ � λ n ( T n ) of T n we have: Theorem 5: [Theorem 1 . 1 in [3]] We assume a k = k α + o ( k α ) for k → ∞ , α > 0. Then the eigenvalues of T n = T n [ a 0 , . . . , a n − 1 ] satisfy, as n → ∞ , λ n +1 − l ( T n ) = µ l ( K α ) n α +1 + o ( n α +1 ) � for l ∈ L + . 13

� D 2 � � D 2 � T = T n +1 [0 , 0 , 1 , . . . , n − 1] = T n +1 [ a 0 , . . . , a n ] with We have + a k = k + o ( k ) for k → ∞ . Now we apply Theorem 1 . 1 from [3] and get for the largest eigenvalue � D 2 � � D 2 � T of + �� D 2 � T � D 2 � � � � � + = T n +1 [0 , 0 , 1 , . . . , n − 1] λ max λ n +1 � � � = T n +1 [0 , 0 , 1 , . . . , n − 1] λ n +2 − 1 � ( n + 1) 2 � µ 1 ( K 1 )( n + 1) 2 + o = � ( n + 1) 2 � 2 ( n + 1) 2 + o = ( n → ∞ ) . ω 2 0 This leads for all N ∈ N to the estimate �� D 2 � T � D 2 � � � ( n + 1) 2 � � ∆ � ≥ N = N ( n + 1) 2 + o + 2 λ max ω 2 0 for n → ∞ . 14

An upper bound � � D 2 � � � � D 2 �� � N � � � � We have � ∆ � = � . k =1 I ( n +1) k − 1 ⊗ ⊗ I ( n +1) N − k � ≤ N � � From [1] L. F. Shampine: Some L 2 Markoff inequalities , J. Res. Nat. Bur. Standards 69B (1965), 155-158: we know � � D 2 � � � � → 1 µ 2 = 0 . 28441 . . . , n 2 where µ is the smallest positive solution of 1 + cos µ cosh µ = 0. This leads to � ∆ � � ∆ � 0 . 1737 ≤ lim inf Nn 2 ≤ lim sup Nn 2 ≤ 0 . 2845 . n →∞ n →∞ 15

N = 1 10 50 100 500 1000 n � ∆ � 0 . 2829 0 . 2844 0 . 2844 0 . 2844 0 . 2844 n 2 N = 2 10 50 100 500 1000 n � ∆ � 0 . 2107 0 . 2175 0 . 2183 0 . 2189 0 . 2190 2 n 2 N = 3 10 20 30 50 100 n � ∆ � 0 . 1950 0 . 1993 0 . 2008 0 . 2020 0 . 2029 3 n 2 N = 4 10 20 30 50 100 n � ∆ � 0 . 1855 0 . 1906 0 . 1923 0 . 1937 0 . 1947 4 n 2 N = 5 10 20 30 50 n � ∆ � 0 . 1804 0 . 1857 0 . 1875 0 . 1889 5 n 2 N = 6 10 20 30 n � ∆ � 0 . 1770 0 . 1824 0 . 1843 6 n 2 N = 7 10 n � ∆ � 0 . 1745 7 n 2 16