Spectral gap of a class of unbounded, positive-definite operators

Transcription

1 of a class of unbounded, positive-definite operators University of Alberta 2015 COSY, Waterloo - June 15-19

2 Setting The space R n of n-dimensional vectors, x = (x 1,..., x n ) is endowed with a measure: dµ n (x) = e φn(x) dx where φ n (x) = φ(x 1 ) + + φ(x n ) for some positive, bounded φ : R R, with φ bounded away from 0

3 Setting, cont d The space of real-valued functions of n real variables is endowed with: an inner product: < f, g >= f (x)g(x)dµ n (x) R n a quadratic form: q n (f ) = n 1 (( xi 1 2 xi + xi+1 )f ) 2 dµ n (x) R n i=2

4 Setting, cont d The space of real-valued functions of n real variables is endowed with: an inner product: < f, g >= f (x)g(x)dµ n (x) R n a quadratic form: q n (f ) = n 1 (( xi 1 2 xi + xi+1 )f ) 2 dµ n (x) R n i=2

5 Setting, cont d The space of real-valued functions of n real variables is endowed with: an inner product: < f, g >= f (x)g(x)dµ n (x) R n a quadratic form: q n (f ) = n 1 (( xi 1 2 xi + xi+1 )f ) 2 dµ n (x) R n i=2

6 Setting, cont d The quadratic form q n defines an operator Q n on functions of n real values, via the equality q n (f ) =< Q n f, f >. The operator is symmetric positive unbounded with discrete spectrum that includes 0

7 Aim Describe the magnitude of the first gap in the spectrum of the operator Q n as a function of n. This amounts to bound from below the quadratic form q n (f ) by the L 2 norm < f, f >.

8 Properties of the quadratic form Connection with Toeplitz matrices Let T n (p) be the Toeplitz matrix: T n (p) = (1) associated to the polynomial p(x) = 1 2x + x 2 = (x 1) 2 We have that q n (f ) = T n (p)( x f ) 2 dµ n (x) R n where x = ( x1,..., xn ) T and x 1,..., x n 2 = x x 2 n

9 Properties of the quadratic form Connection with Toeplitz matrices Let T n (p) be the Toeplitz matrix: T n (p) = (1) associated to the polynomial p(x) = 1 2x + x 2 = (x 1) 2 We have that q n (f ) = T n (p)( x f ) 2 dµ n (x) R n where x = ( x1,..., xn ) T and x 1,..., x n 2 = x x 2 n

10 Properties of the quadratic form Connection with Toeplitz matrices Let T n (p) be the Toeplitz matrix: T n (p) = (1) associated to the polynomial p(x) = 1 2x + x 2 = (x 1) 2 We have that q n (f ) = T n (p)( x f ) 2 dµ n (x) R n where x = ( x1,..., xn ) T and x 1,..., x n 2 = x x 2 n

11 Properties of the quadratic form, cont d Kernel of the quadratic form Kernel of the Toeplitz matrix T n (p) is the span of the rows of ( ) B = n Kernel of the quadratic form q includes two linear functions l 1 (x) = x 1 + x x n and l 2 (x) = x 1 + 2x x n Kernel of the quadratic form q includes only functions of the form f (x 1 + x x n, x 1 + 2x x n )

12 Properties of the Toeplitz matrix The lowest eigenvalue of the square matrix T n (p)t n (p) T satisfies λ min (T n (p)t n (p) T ) 1 n 2α, i.e. for all sufficiently large n Above d 1 n 2α λ min(t n (p)t n (p) T ) d 2 n 2α d 1 and d 2 are positive constants, independent of n α = 2 is the maximal order of the zeros in the complex unit circle of p(z) = 1 2z + z 2 = (1 z) 2

13 Lower bound for the quadratic form Theorem There exits a constant C > 0 independent of n such that for any function f that integrates to 0 on any 2-dimensional flat parallel to the span of the linear functions l 1 and l 2, we have R n f 2 dµ n (x) C λ min (T n (p)t n (p) T ) R n T n (p)( x f ) 2 dµ n (x) Notation: λ min A stands for the smallest eigenvalue of A

14 Proof Change R n coordinates from x to (α, β), where α R n 2 and β R 2 and x = T n (p) T α + B T β Establish a sequence of inequalities

15 Proof: First Inequality Define ψ β : R n 2 R by ψ β (α) = φ n (T n (p) T α + B T β) The Hessian matrix of ψ β is a positive defined, bounded away from zero matrix for any β (Hessψ β )(α) = T n (p)(hessφ)(x)t n (p) T CT n (p)t n (p) T Cλ min (T n (p)t n (p) T )Identity n 2

16 Proof: Second Inequality Poincare inequality holds for the measure e ψ β(α) dα since Bakry-Emery criterion is satisfied by the Hessian matrix of ψ β R n 2 g 2 (α)e ψ β(α) dα C λ min (T n (p)t n (p) T α g 2 e ψβ(α) dα ) R n 2 for any g with zero average on R n 2 with respect to e ψ β(α) dα

17 Proof: Third Inequality Select g β (α) = f (T n (p) T α + B T β) in the second inequality and integrate with respect to β gβ 2 (α)e ψβ(α) dαdβ R n C λ min (T n (p)t n (p) T α g β 2 e ψβ(α) dαdβ ) R n Apply a change of variable x = T n (p) T α + B T β in both integrals. Note ( α g β )(α) = T n (p)( x f )(x)

18 Generalization The theorem holds for any polynomial p

19 Main result Corollary The operator Q n has a spectral gap of size at least 1/n 4. This is implied by the fact that the lowest eigenvalue of T n (p)t n (p) T decays to 0 as 1/n 4.