Definition and basic properties of heat

kernels I, An introduction

Zhiqin Lu,Department of Mathematics,UC Irvine, Irvine CA 92697

April 23, 2010

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.

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.

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.

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.

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.

S-T. Yau, R. SchoenDifferential GeometryAcademic Press, 2006

N. Berline, E. Getzler, M. VergneHeat kernels and Dirac operatorsSpringer 1992

E.B. Davies,One-parameter semi-groupsAcademic Press 1980

E.B. Davies,Heat kernels and spectral theoryCambridge University Press, 1990

The basic settings:Let M be a Riemannian manifold with the Riemannian metric

ds2 = gijdxi dxj.

The Laplace operator is defined as

∆ =1√g

∑ ∂

∂xi

(gij√g∂

∂xj

),

where (gij) = (gij)−1, g = det(gij).

The basic settings:Let M be a Riemannian manifold with the Riemannian metric

ds2 = gijdxi dxj.

The Laplace operator is defined as

∆ =1√g

∑ ∂

∂xi

(gij√g∂

∂xj

),

where (gij) = (gij)−1, g = det(gij).

At least three questions have to be addressed:

1 The Laplace operator as a densely defined self-adjointoperator;

2 The semi-groupof operators;

3 The existence of heat kernel.

At least three questions have to be addressed:

1 The Laplace operator as a densely defined self-adjointoperator;

2 The semi-groupof operators;

3 The existence of heat kernel.

At least three questions have to be addressed:

1 The Laplace operator as a densely defined self-adjointoperator;

2 The semi-groupof operators;

3 The existence of heat kernel.

Finite dimensional case

Let A be a positive definite matrix. Then A can bediagonalized. That is, up to a similar transformation by anorthogonal matrix, A is similar to the matrixλ1

. . .

λn

Let Ei be the eigenspace with respect to the eigenvalue λi andlet Pi : Rn → Ei be the orthogonal projection. Then we canwrite

A =n∑i=1

λiPi

Infinite dimensional case, an example

Let B be the space of L2 periodic functions on [−π, π]. Letf ∈ B. Then we have the Fourier expansion

f(x) ∼ a0

2+∞∑k=1

(ak cos kx+ bk sin kx).

If f is smooth, then the above expansion is convergent to thefunction.

The Laplace operator on one dimensional is

∆ =∂2

∂x2

If f is smooth, then

∆f =∞∑k=1

(−k2ak cos kx− k2bk sin kx)

Define

Pkf = ak cos kx

Qkf = bk sin kx

Then we can write

∆ = 0 · P0 −∞∑k=1

(k2Pk + k2Qk)

The Laplace operator on one dimensional is

∆ =∂2

∂x2

If f is smooth, then

∆f =∞∑k=1

(−k2ak cos kx− k2bk sin kx)

Define

Pkf = ak cos kx

Qkf = bk sin kx

Then we can write

∆ = 0 · P0 −∞∑k=1

(k2Pk + k2Qk)

The Laplace operator on one dimensional is

∆ =∂2

∂x2

If f is smooth, then

∆f =∞∑k=1

(−k2ak cos kx− k2bk sin kx)

Define

Pkf = ak cos kx

Qkf = bk sin kx

Then we can write

∆ = 0 · P0 −∞∑k=1

(k2Pk + k2Qk)

The Laplace operator on one dimensional is

∆ =∂2

∂x2

If f is smooth, then

∆f =∞∑k=1

(−k2ak cos kx− k2bk sin kx)

Define

Pkf = ak cos kx

Qkf = bk sin kx

Then we can write

∆ = 0 · P0 −∞∑k=1

(k2Pk + k2Qk)

In general, we can write

∆ = −∫ ∞

0

λ dE,

where E is the so-called spectral measure.

Let L2(M) be the space of L2 functions. The space C∞0 (M)is the space of smooth functions on M with compact support.

In general, we can write

∆ = −∫ ∞

0

λ dE,

where E is the so-called spectral measure.Let L2(M) be the space of L2 functions. The space C∞0 (M)is the space of smooth functions on M with compact support.

The Laplacian ∆ is defined on the space of smooth functionswith compact support.

. It is symmetric.That is∫∆f · g =

∫f ·∆g

However, C∞0 (M) is not a Banach space. We would like topick L2(M), the space of L2 functions.

Question

Can we extend the Laplacian onto L2(M)?

The Laplacian ∆ is defined on the space of smooth functionswith compact support.. It is symmetric.

That is∫∆f · g =

∫f ·∆g

However, C∞0 (M) is not a Banach space. We would like topick L2(M), the space of L2 functions.

Question

Can we extend the Laplacian onto L2(M)?

The Laplacian ∆ is defined on the space of smooth functionswith compact support.. It is symmetric.That is∫

∆f · g =

∫f ·∆g

However, C∞0 (M) is not a Banach space. We would like topick L2(M), the space of L2 functions.

Question

Can we extend the Laplacian onto L2(M)?

The Laplacian ∆ is defined on the space of smooth functionswith compact support.. It is symmetric.That is∫

∆f · g =

∫f ·∆g

However, C∞0 (M) is not a Banach space.

We would like topick L2(M), the space of L2 functions.

Question

Can we extend the Laplacian onto L2(M)?

The Laplacian ∆ is defined on the space of smooth functionswith compact support.. It is symmetric.That is∫

∆f · g =

∫f ·∆g

However, C∞0 (M) is not a Banach space. We would like topick L2(M), the space of L2 functions.

Question

Can we extend the Laplacian onto L2(M)?

The Laplacian ∆ is defined on the space of smooth functionswith compact support.. It is symmetric.That is∫

∆f · g =

∫f ·∆g

However, C∞0 (M) is not a Banach space. We would like topick L2(M), the space of L2 functions.

Question

Can we extend the Laplacian onto L2(M)?

Answer: No!

1 The Laplacian is an unbounded operator;

2 Like most differential operator, it is a closed graphoperator;

3 By the Closed Graph Theorem, if ∆ can be extended,then it must be a bounded operator, a contradiction.

Answer: No!

1 The Laplacian is an unbounded operator;

2 Like most differential operator, it is a closed graphoperator;

3 By the Closed Graph Theorem, if ∆ can be extended,then it must be a bounded operator, a contradiction.

Answer: No!

1 The Laplacian is an unbounded operator;

2 Like most differential operator, it is a closed graphoperator;

3 By the Closed Graph Theorem, if ∆ can be extended,then it must be a bounded operator, a contradiction.

Answer: No!

1 The Laplacian is an unbounded operator;

2 Like most differential operator, it is a closed graphoperator;

3 By the Closed Graph Theorem, if ∆ can be extended,then it must be a bounded operator, a contradiction.

Self-adjoint densely defined operatorDefinitionLet H be a Hilbert space, Given a densely defined linearoperator A on H, its adjoint A∗ is defined as follows:

1 The domain of A∗ consists of vectors x in H such that

y 7→ 〈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 inH 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 Ais called self-adjoint.

Define H10 (M) be the Sobolev space of the completion of the

vector space Λp(M) under the norm

||η||1 =

√∫M

|η|2dVM +

√∫M

|∇η|2 dVM .

We define the quadratic form Q on H10 (M) by

Q(ω, η) =

∫M

(〈dω, dη〉) dVM

for any ω, η ∈ H10 (M). Then we define

Dom(∆) =φ ∈ H1

0 (M) |∀ψ ∈ C∞(M),∃f ∈ L2(M), s.t. Q(φ, ψ) = −(f, ψ)

.

Proof.We first observe that for any ψ ∈ C∞(M) and φ ∈ H1

0 (M),we have

Q(φ, ψ) = −(φ,∆ψ).

Using this result, the proof goes as follows: for anyφ ∈ Dom(∆), the functionalψ 7→ (∆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 7→ (∆ψ, φ) is bounded. By the Rieszrepresentation theorem, there is a unique f ∈ L2(M) suchthat (∆ψ, φ) = (f, ψ). Thus we must haveQ(φ, ψ) = −(∆ψ, φ) = −(f, ψ).

From the above discussion, we have proved that

TheoremThe Laplace operator has a self-adjoint extension.

DefinitionA one-parameter semigroup of operators on a complex Banachspace B is a family Tt of bounded linear operators, whereTt : B → B parameterized by real numbers t ≥ 0 and satisfiesthe following relations:

1 T0 = 1;

2 If 0 ≤ s, t <∞, then

TsTt = Ts+t.

3 The mapt, f → Ttf

from [0,∞)× B to B is jointly continuous.

The family of bounded operators

e∆t

forms a semi-group.

Even though e−∆t are all bounded operator, the kernel doesn’texist in general.

Definition of operator kernel

Let A be an operator on L2(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 theoperator.

By the above definition, the kernel of an operator doesn’t existin general. For example, let B be a Banach space, and let Ibe the identity map. Then the kernel of I doesn’t exist.

The family of bounded operators

e∆t

forms a semi-group.Even though e−∆t are all bounded operator, the kernel doesn’texist in general.

Definition of operator kernel

Let A be an operator on L2(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 theoperator.

By the above definition, the kernel of an operator doesn’t existin general. For example, let B be a Banach space, and let Ibe the identity map. Then the kernel of I doesn’t exist.

The family of bounded operators

e∆t

forms a semi-group.Even though e−∆t are all bounded operator, the kernel doesn’texist in general.

Definition of operator kernel

Let A be an operator on L2(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 theoperator.

By the above definition, the kernel of an operator doesn’t existin general. For example, let B be a Banach space, and let Ibe the identity map. Then the kernel of I doesn’t exist.

The family of bounded operators

e∆t

forms a semi-group.Even though e−∆t are all bounded operator, the kernel doesn’texist in general.

Definition of operator kernel

Let A be an operator on L2(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 theoperator.

By the above definition, the kernel of an operator doesn’t existin general. For example, let B be a Banach space, and let Ibe the identity map. Then the kernel of I doesn’t exist.

Definition of heat kernel

Let M be a complete Riemannian manifold. Then there existsheat kernel H(x, y, t) ∈ C∞(M ×M × R+) such that

(e∆tf)(x) =

∫M

H(x, y, t)f(y)dy

for any L2 function f . The heat kernel satisfies

1 H(x, y, t) = H(y, x, t)

2 limt→0+

H(x, y, t) = δx(y)

3 (∆− ∂∂t

)H = 0

4 H(x, y, t) =∫MH(x, z, t− s)H(x, y, s)dz, t > s > 0

An example

Let M be a compact manifold. Let fj(x) be an orthonormalbasis of eigenfunctions. Let

∆fj = −λjfj

Then we an write the heat kernel as a series

H(x, y, t) =∞∑j=0

e−λjtfj(x)fj(y)

Let Ω be a bounded domain of Rn 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 =∞∑j=1

e−λjtajfj(x),

where aj =∫Mφ(x)fj(x)dx.

Conclusion: If a1 6= 0, then eλ1tf(t, x)→ a1f1(x) 6= 0

Let Ω be a bounded domain of Rn 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 =∞∑j=1

e−λjtajfj(x),

where aj =∫Mφ(x)fj(x)dx.

Conclusion: If a1 6= 0, then eλ1tf(t, x)→ a1f1(x) 6= 0

Let Ω be a bounded domain of Rn 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 =∞∑j=1

e−λjtajfj(x),

where aj =∫Mφ(x)fj(x)dx.

Conclusion: If a1 6= 0, then eλ1tf(t, x)→ a1f1(x) 6= 0

Let Ω be a bounded domain of Rn 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 =∞∑j=1

e−λjtajfj(x),

where aj =∫Mφ(x)fj(x)dx.

Conclusion: If a1 6= 0, then eλ1tf(t, x)→ a1f1(x) 6= 0

A theorem of Brascamp-Lieb

Theorem [Brascamp-Lieb]

Let Ω be a bounded convex domain with smooth boundary inRn. Let f1(x) > 0 be the first eigenfunction. Then log f1(x) isa concave function.

Ideas of the proof

1 Construct a log-concave function φ(x) with a1 6= 0;

2 Prove that along the flow, the log-concavity is preserved.

A theorem of Brascamp-Lieb

Theorem [Brascamp-Lieb]

Let Ω be a bounded convex domain with smooth boundary inRn. Let f1(x) > 0 be the first eigenfunction. Then log f1(x) isa concave function.

Ideas of the proof

1 Construct a log-concave function φ(x) with a1 6= 0;

2 Prove that along the flow, the log-concavity is preserved.

A theorem of Brascamp-Lieb

Theorem [Brascamp-Lieb]

Let Ω be a bounded convex domain with smooth boundary inRn. Let f1(x) > 0 be the first eigenfunction. Then log f1(x) isa concave function.

Ideas of the proof

1 Construct a log-concave function φ(x) with a1 6= 0;

2 Prove that along the flow, the log-concavity is preserved.

Construction of the log-concavity

function

1 Since Ω is a convex domain, the local defining functionϕ(x) is log-concave neat the boundary.

2 Let φ = ϕe−C|x|2. Then for sufficient large C, φ is

log-concave.

Construction of the log-concavity

function

1 Since Ω is a convex domain, the local defining functionϕ(x) is log-concave neat the boundary.

2 Let φ = ϕe−C|x|2. Then for sufficient large C, φ is

log-concave.

Maximum Principle

Let g = log f . Then

∂g

∂t= ∆g + |∇g|2.

For any i, we have

∂(−gii)∂t

= ∆(−gii) + 2gkiigk + 2∑k

g2ki