Definition and basic properties of heat kernels I, An introduction

Transcription

1 Definition and basic properties of heat kernels I, An introduction Zhiqin Lu, Department of Mathematics, UC Irvine, Irvine CA April 23, 2010

2 In this lecture, we will answer the following questions:

3 In this lecture, we will answer the following questions: 1 What is the heat kernel?

4 In this lecture, we will answer the following questions: 1 What is the heat kernel? 2 Why it is so difficult to understand the heat kernel?

5 In this lecture, we will answer the following questions: 1 What is the heat kernel? 2 Why it is so difficult to understand the heat kernel? 3 The basic properties of the heat kernel.

6 In this lecture, we will answer the following questions: 1 What is the heat kernel? 2 Why it is so difficult to understand the heat kernel? 3 The basic properties of the heat kernel.

7 S-T. Yau, R. Schoen Differential Geometry Academic Press, 2006 N. Berline, E. Getzler, M. Vergne Heat kernels and Dirac operators Springer 1992 E.B. Davies, One-parameter semi-groups Academic Press 1980 E.B. Davies, Heat kernels and spectral theory Cambridge University Press, 1990

8 The basic settings: Let M be a Riemannian manifold with the Riemannian metric ds 2 = g ij dx i dx j.

9 The basic settings: Let M be a Riemannian manifold with the Riemannian metric ds 2 = g ij dx i dx j. The Laplace operator is defined as = 1 g x i where (g ij ) = (g ij ) 1, g = det(g ij ). ( g ij g x j ),

10 At least three questions have to be addressed: 1 The Laplace operator as a densely defined self-adjoint operator;

11 At least three questions have to be addressed: 1 The Laplace operator as a densely defined self-adjoint operator; 2 The semi-groupof operators;

12 At least three questions have to be addressed: 1 The Laplace operator as a densely defined self-adjoint operator; 2 The semi-groupof operators; 3 The existence of heat kernel.

13 Finite dimensional case Let A be a positive definite matrix. Then A can be diagonalized. That is, up to a similar transformation by an orthogonal matrix, A is similar to the matrix λ 1... Let E i be the eigenspace with respect to the eigenvalue λ i and let P i : R n E i be the orthogonal projection. Then we can write n A = λ i P i i=1 λ n

14 Infinite dimensional case, an example Let B be the space of L 2 periodic functions on [ π, π]. Let f B. Then we have the Fourier expansion f(x) a (a k cos kx + b k sin kx). k=1 If f is smooth, then the above expansion is convergent to the function.

15 The Laplace operator on one dimensional is = 2 x 2

16 The Laplace operator on one dimensional is = 2 x 2 If f is smooth, then f = ( k 2 a k cos kx k 2 b k sin kx) k=1

17 The Laplace operator on one dimensional is = 2 x 2 If f is smooth, then f = ( k 2 a k cos kx k 2 b k sin kx) k=1 Define P k f = a k cos kx Q k f = b k sin kx

18 The Laplace operator on one dimensional is = 2 x 2 If f is smooth, then f = ( k 2 a k cos kx k 2 b k sin kx) k=1 Define P k f = a k cos kx Q k f = b k sin kx Then we can write = 0 P 0 (k 2 P k + k 2 Q k ) k=1

19 In general, we can write = 0 λ de, where E is the so-called spectral measure.

20 In general, we can write = 0 λ de, where E is the so-called spectral measure. Let L 2 (M) be the space of L 2 functions. The space C 0 (M) is the space of smooth functions on M with compact support.

21 The Laplacian is defined on the space of smooth functions with compact support.

22 The Laplacian is defined on the space of smooth functions with compact support.. It is symmetric.

23 The Laplacian is defined on the space of smooth functions with compact support.. It is symmetric.that is f g = f g

24 The Laplacian is defined on the space of smooth functions with compact support.. It is symmetric.that is f g = f g However, C0 (M) is not a Banach space.

25 The Laplacian is defined on the space of smooth functions with compact support.. It is symmetric.that is f g = f g However, C 0 (M) is not a Banach space. We would like to pick L 2 (M), the space of L 2 functions.

26 The Laplacian is defined on the space of smooth functions with compact support.. It is symmetric.that is f g = f g However, C 0 (M) is not a Banach space. We would like to pick L 2 (M), the space of L 2 functions. Question Can we extend the Laplacian onto L 2 (M)?

27 Answer: No!

28 1 The Laplacian is an unbounded operator; Answer: No!

29 Answer: No! 1 The Laplacian is an unbounded operator; 2 Like most differential operator, it is a closed graph operator;

30 Answer: No! 1 The Laplacian is an unbounded operator; 2 Like most differential operator, it is a closed graph operator; 3 By the Closed Graph Theorem, if can be extended, then it must be a bounded operator, a contradiction.

31 Self-adjoint densely defined operator Definition Let H be a Hilbert space, Given a densely defined linear operator A on H, its adjoint A is defined as follows: 1 The domain of A consists of vectors x in H such that y x, Ay is a bounded linear functional, where y Dom( ); 2 If x is in the domain of A, there is a unique vector z in H such that x, Ay = z, y for any y Dom( ). This vector z is defined to be A x. It can be shown that the dependence of z on x is linear. If A = A (which implies that Dom (A ) = Dom (A)), then A is called self-adjoint.

32 Define H0(M) 1 be the Sobolev space of the completion of the vector space Λ p (M) under the norm η 1 = η 2 dv M + η 2 dv M. M We define the quadratic form Q on H0(M) 1 by Q(ω, η) = ( dω, dη ) dv M for any ω, η H 1 0(M). Then we define M Dom( ) = { φ H 1 0(M) ψ C (M), f L 2 (M), s.t. Q(φ, ψ) = (f, ψ) }. M

33 Proof. We first observe that for any ψ C (M) and φ H 1 0(M), we have Q(φ, ψ) = (φ, ψ). Using this result, the proof goes as follows: for any φ Dom( ), the functional ψ ( ψ, φ) = Q(ψ, φ) = (f, ψ) is a bounded functional. Thus φ Dom( ). On the other hand, if φ Dom( ), then the functional ψ ( ψ, φ) is bounded. By the Riesz representation theorem, there is a unique f L 2 (M) such that ( ψ, φ) = (f, ψ). Thus we must have Q(φ, ψ) = ( ψ, φ) = (f, ψ). From the above discussion, we have proved that Theorem The Laplace operator has a self-adjoint extension.

34 Definition A one-parameter semigroup of operators on a complex Banach space B is a family T t of bounded linear operators, where T t : B B parameterized by real numbers t 0 and satisfies the following relations: 1 T 0 = 1; 2 If 0 s, t <, then T s T t = T s+t. 3 The map t, f T t f from [0, ) B to B is jointly continuous.

35 The family of bounded operators forms a semi-group. e t

36 The family of bounded operators e t forms a semi-group. Even though e t are all bounded operator, the kernel doesn t exist in general.

37 The family of bounded operators e t forms a semi-group. Even though e t are all bounded operator, the kernel doesn t exist in general. Definition of operator kernel Let A be an operator on L 2 (M). If there is a function A(x, y) such that Af(x) = A(x, y)f(y)dy for all functions f, then we call A(x, y) is the kernel of the operator.

38 The family of bounded operators e t forms a semi-group. Even though e t are all bounded operator, the kernel doesn t exist in general. Definition of operator kernel Let A be an operator on L 2 (M). If there is a function A(x, y) such that Af(x) = A(x, y)f(y)dy for all functions f, then we call A(x, y) is the kernel of the operator. By the above definition, the kernel of an operator doesn t exist in general. For example, let B be a Banach space, and let I be the identity map. Then the kernel of I doesn t exist.

39 Definition of heat kernel Let M be a complete Riemannian manifold. Then there exists heat kernel H(x, y, t) C (M M R + ) such that (e t f)(x) = H(x, y, t)f(y)dy for any L 2 function f. The heat kernel satisfies 1 H(x, y, t) = H(y, x, t) 2 lim t 0 + H(x, y, t) = δ x(y) 3 ( t )H = 0 M 4 H(x, y, t) = H(x, z, t s)h(x, y, s)dz, t > s > 0 M

40 An example Let M be a compact manifold. Let f j (x) be an orthonormal basis of eigenfunctions. Let f j = λ j f j Then we an write the heat kernel as a series H(x, y, t) = e λjt f j (x)f j (y) j=0

41 Let Ω be a bounded domain of R n with smooth boundary.

42 Let Ω be a bounded domain of R n with smooth boundary. Consider the equation { f t = f f(0, x) = φ(x), where φ is the initial value function.

43 Let Ω be a bounded domain of R n with smooth boundary. Consider the equation { f t = f f(0, x) = φ(x), where φ is the initial value function.then f(t, x) = M where a j = M φ(x)f j(x)dx. H(x, y, t)φ(y)dy = e λjt a j f j (x), j=1

44 Let Ω be a bounded domain of R n with smooth boundary. Consider the equation { f t = f f(0, x) = φ(x), where φ is the initial value function.then f(t, x) = M H(x, y, t)φ(y)dy = e λjt a j f j (x), j=1 where a j = M φ(x)f j(x)dx. Conclusion: If a 1 0, then e λ 1t f(t, x) a 1 f 1 (x) 0

45 A theorem of Brascamp-Lieb Theorem [Brascamp-Lieb] Let Ω be a bounded convex domain with smooth boundary in R n. Let f 1 (x) > 0 be the first eigenfunction. Then log f 1 (x) is a concave function. Ideas of the proof

46 A theorem of Brascamp-Lieb Theorem [Brascamp-Lieb] Let Ω be a bounded convex domain with smooth boundary in R n. Let f 1 (x) > 0 be the first eigenfunction. Then log f 1 (x) is a concave function. Ideas of the proof 1 Construct a log-concave function φ(x) with a 1 0;

47 A theorem of Brascamp-Lieb Theorem [Brascamp-Lieb] Let Ω be a bounded convex domain with smooth boundary in R n. Let f 1 (x) > 0 be the first eigenfunction. Then log f 1 (x) is a concave function. Ideas of the proof 1 Construct a log-concave function φ(x) with a 1 0; 2 Prove that along the flow, the log-concavity is preserved.

48 Construction of the log-concavity function 1 Since Ω is a convex domain, the local defining function ϕ(x) is log-concave neat the boundary.

49 Construction of the log-concavity function 1 Since Ω is a convex domain, the local defining function ϕ(x) is log-concave neat the boundary. 2 Let φ = ϕe C x 2. Then for sufficient large C, φ is log-concave.

50 Maximum Principle Let g = log f. Then For any i, we have ( g ii ) t g t = g + g 2. = ( g ii ) + 2g kii g k + 2 k g 2 ki