Spectral theory of elliptic differential operators

802628S

Lecture Notes1st Edition

Second printing

Valeriy SerovUniversity of Oulu

2009

Edited by Markus Harju

Contents

1 Inner product spaces and Hilbert spaces 1

2 Symmetric operators in the Hilbert space 11

3 J. von Neumann’s spectral theorem 20

4 Spectrum of self-adjoint operators 33

5 Quadratic forms. Friedrichs extension. 48

6 Elliptic differential operators 52

7 Spectral function 61

8 Fundamental solution 64

9 Fractional powers of self-adjoint operators 85

Index 106

i

1 Inner product spaces and Hilbert spaces

A collection of elements is called a complex (real) vector space (linear space) H if thefollowing axioms are satisfied:

1) To every pair x, y ∈ H there corresponds a vector x+y, called the sum, with theproperties:

a) x+ y = y + x

b) x+ (y + z) = (x+ y) + z ≡ x+ y + z

c) there exists unique 0 ∈ H such that x+ 0 = x

d) for every x ∈ H there exists unique y1 ∈ H such that x+y1 = 0. We denotey1 := −x.

2) For every x ∈ H and every λ, µ ∈ C there corresponds a vector λ · x such that

a) λ(µx) = (λµ)x ≡ λµx

b) (λ+ µ)x = λx+ µx

c) λ(x+ y) = λx+ λy

d) 1 · x = x.

Definition. For a linear space H every mapping (·, ·) : H ×H → C is called an innerproduct or a scalar product if

1) (x, x) ≥ 0 and (x, x) = 0 if and only if x = 0

2) (x, y + z) = (x, y) + (x, z)

3) (λx, y) = λ(x, y)

4) (x, y) = (y, x)

for every x, y, z ∈ H and λ ∈ C. A linear space equipped with an inner product iscalled an inner product space.

An immediate consequence of this definition is that

(λx+ µy, z) = λ(x, z) + µ(y, z),

(x, λy) = λ(x, y)

for every x, y, z ∈ H and λ, µ ∈ C.

Example 1.1. On the complex Euclidean space H = Cn the standard inner product

is

(x, y) =n∑

j=1

xjyj,

where x = (x1, . . . , xn) ∈ Cn and y = (y1, . . . , yn) ∈ C

n.

1

Example 1.2. On the linear space C[a, b] of continuous complex-valued functions, theformula

(f, g) =∫ b

af(x)g(x)dx

defines an inner product.

Definition. Suppose H is an inner product space. One calls

1) x ∈ H orthogonal to y ∈ H if (x, y) = 0.

2) a system xαα∈A ⊂ H orthonormal if (xα, xβ) = δα,β =

1, α = β

0, α 6= β, where A

is some index set.

3) ‖x‖ :=»

(x, x) is called the length of x ∈ H.

Exercise 1. Prove the Theorem of Pythagoras : If xjkj=1, k ∈ N is an orthonormal

system in an inner product space H, then

‖x‖2 =k∑

j=1

|(x, xj)|2 +

∥∥∥∥∥∥x−

k∑

j=1

(x, xj)xj

∥∥∥∥∥∥

2

for every x ∈ H.

Exercise 2. Prove Bessel’s inequality : If xjkj=1, k ≤ ∞ is an orthonormal system

thenk∑

j=1

|(x, xj)|2 ≤ ‖x‖2 ,

for every x ∈ H.

Exercise 3. Prove the Cauchy-Schwarz-Bunjakovskii inequality :

|(x, y)| ≤ ‖x‖ ‖y‖ , x, y ∈ H.

Prove also that (·, ·) is continuous as a map from H ×H to C.

If H is an inner product space, then

‖x‖ :=»

(x, x)

has the following properties:

1) ‖x‖ ≥ 0 for every x ∈ H and ‖x‖ = 0 if and only if x = 0.

2) ‖λx‖ = |λ| ‖x‖ for every x ∈ H and λ ∈ C.

3) ‖x+ y‖ ≤ ‖x‖ + ‖y‖ for every x, y ∈ H. This is the triangle inequality .

2

The function ‖·‖ =»

(·, ·) is thus a norm on H. It is called the norm induced by theinner product .

Every inner product space H is a normed space under the induced norm. Theneighborhood of x ∈ H is the open ball Br(x) = y ∈ H : ‖x− y‖ < r. This systemof neighborhoods defines the norm topology on H such that:

1) The addition x+ y is a continuous map H ×H → H.

2) The scalar multiplication λ · x is a continuous map C ×H → H.

Definition. 1) A sequence xj∞j=1 ⊂ H is called a Cauchy sequence if for everyε > 0 there exists n0 ∈ N such that ‖xk − xj‖ < ε for k, j ≥ n0.

2) A sequence xj∞j=1 ⊂ H is said to be convergent if there exists x ∈ H such thatfor every ε > 0 there exists n0 ∈ N such that ‖x− xj‖ < ε whenever j ≥ n0.

3) The inner product space H is complete space if every Cauchy sequence in Hconverges.

Corollary. 1) Every convergent sequence is a Cauchy sequence.

2) If xj∞j=1 converges to x ∈ H then

limj→∞

‖xj‖ = ‖x‖ .

Definition. (J. von Neumann, 1925) A Hilbert space is an inner product space whichis complete (with respect to its norm topology).

Exercise 4. Prove that in an inner product space the norm induced by this innerproduct satisfies the parallelogram law

‖x+ y‖2 + ‖x− y‖2 = 2 ‖x‖2 + 2 ‖y‖2 .

Exercise 5. Prove that if in a normed space H the parallelogram law holds, thenthere is an inner product on H such that ‖x‖2 = (x, x) and that this inner product isdefined by the polarization identity

(x, y) :=1

4

Ä‖x+ y‖2 − ‖x− y‖2 + i ‖x+ iy‖2 − i ‖x− iy‖2

ä.

Exercise 6. Prove that on C[a, b] the norm

‖f‖ = maxx∈[a,b]

|f(x)|

is not induced by an inner product.

Exercise 7. Give an example of an inner product space which is not complete.

3

Next we list some examples of Hilbert spaces.

1) The Euclidean spaces Rn and C

n.

2) The matrix spaceMn(C) consisting of n×n -matrices whose elements are complexnumbers. For A,B ∈Mn(C) the inner product is given by

(A,B) =n∑

k,j=1

akjbkj = Tr (AB∗),

where B∗ = BT.

3) The sequence space l2(C) defined by

l2(C) :=

xj∞j=1, xj ∈ C :

∞∑

j=1

|xj|2 <∞ .

The estimates

|xj + yj|2 ≤ 2Ä|xj|2 + |yj|2

ä, |λxj|2 = |λ|2|xj|2

and

|xjyj| ≤1

2

Ä|xj|2 + |yj|2

ä

imply that l2(C) is a linear space. Let us define the inner product by

(x, y) :=∞∑

j=1

xjyj

and prove that l2(C) is complete. Suppose that x(k)∞k=1 ∈ l2(C) is a Cauchysequence. Then for every ε > 0 there exists n0 ∈ N such that

∥∥∥x(k) − x(m)∥∥∥2

=∞∑

j=1

|x(k)j − x

(m)j |2 < ε2

for k,m ≥ n0. It implies that

|x(k)j − x

(m)j | < ε, j = 1, 2, . . .

or that x(k)j ∞k=1 is a Cauchy sequence in C for every j = 1, 2, . . .. Since C is

a complete space then x(k)j ∞k=1 converges for every fixed j = 1, 2, . . . i.e. there

exists xj ∈ C such that

xj = limk→∞

x(k)j .

This fact andl∑

j=1

|x(k)j − x

(m)j |2 < ε2, l ∈ N

4

imply that

limm→∞

l∑

j=1

|x(k)j − x

(m)j |2 =

l∑

j=1

|x(k)j − xj|2 ≤ ε2

for all k ≥ n0 and l ∈ N. Therefore the sequence

sl :=l∑

j=1

|x(k)j − xj|2, k ≥ n0

is a monotone increasing sequence which is bounded from above by ε2. Hencethis sequence has a limit with the same upper bound i.e.

∞∑

j=1

|x(k)j − xj|2 = lim

l→∞

l∑

j=1

|x(k)j − xj|2 ≤ ε2.

That’s why we may conclude that

‖x‖ ≤∥∥∥x(k)

∥∥∥+∥∥∥x(k) − x

∥∥∥ ≤∥∥∥x(k)

∥∥∥+ ε

and x ∈ l2(C).

4) The Lebesgue space L2(Ω), where Ω ⊂ Rn is an open set. The space L2(Ω)

consists of all Lebesgue measurable functions f which are square integrable i.e.∫

Ω|f(x)|2dx <∞.

It is a linear space with the inner product

(f, g) =∫

Ωf(x)g(x)dx

and the Riesz-Fisher theorem reads as: L2(Ω) is a Hilbert space.

5) The Sobolev spaces W k2 (Ω) consisting of functions f ∈ L2(Ω) whose weak or

distributional derivatives Dαf also belong to L2(Ω) up to order |α| ≤ k, k =1, 2, . . .. On the space W k

2 (Ω) the natural inner product is

(f, g) =∑

|α|≤k

∫

ΩDαf(x)Dαg(x)dx.

Definition. LetH be an inner product space. For any subspaceM ⊂ H the orthogonalcomplement of M is defined as

M⊥ := y ∈ H : (y, x) = 0, x ∈M .

Remark. It is clear that M⊥ is a linear subspace of H. Moreover, M ∩M⊥ = 0 since0 ∈M always.

5

Definition. A closed subspace of a Hilbert space H is a linear subspace of H which isclosed (i.e. M = M) with respect to the induced norm.

Remark. The subspace M⊥ is closed if M is any subset of a Hilbert space.

Theorem 1 (Projection theorem). Suppose M is a closed subspace of a Hilbert spaceH. Then every x ∈ H has the unique representation as

x = u+ v,

where u ∈M and v ∈M⊥, or equivalently,

H = M ⊕M⊥.

Moreover, one has that

‖v‖ = infy∈M

‖x− y‖ := d(x,M).

Proof. Let x ∈ H. Then

d := d(x,M) ≡ infy∈M

‖x− y‖ ≤ ‖x− u‖

for any u ∈M . The definition of infimum implies that there exists a sequence uj∞j=1 ⊂M such that

d = limj→∞

‖x− uj‖ .

The parallelogram law implies that

‖uj − uk‖2 = ‖(uj − x) + (x− uk)‖2

= 2 ‖uj − x‖2 + 2 ‖x− uk‖2 − 4∥∥∥∥x−

uj + uk

2

∥∥∥∥2

.

Since (uj + uk)/2 ∈M then

‖uj − uk‖2 ≤ 2 ‖uj − x‖2 + 2 ‖x− uk‖2 − 4d2 → 2d2 + 2d2 − 4d2 = 0

as j, k → ∞. Hence uj∞j=1 ⊂ M is a Cauchy sequence in the Hilbert space H. Itmeans that there exists u ∈ H such that

u = limj→∞

uj.

But M = M implies that u ∈M . By construction one has that

d = limj→∞

‖x− uj‖ = ‖x− u‖ .

Let us denote v := x− u and show that v ∈M⊥. For any y ∈M, y 6= 0 introduce thenumber

α = −(v, y)

‖y‖2 .

6

Since u− αy ∈M we have

d2 ≤ ‖x− (u− αy)‖2 = ‖v + αy‖2 = ‖v‖2 + (v, αy) + (αy, v) + |α|2 ‖y‖2

= d2 − (v, y)(v, y)

‖y‖2 − (v, y)(y, v)

‖y‖2 +|(v, y)|2‖y‖2 = d2 − |(v, y)|2

‖y‖2 .

This inequality implies that (y, v) = 0. It means that v ∈ M⊥. In order to proveuniqueness assume that x = u1 + v1 = u2 + v2, where u1, u2 ∈ M and v1, v2 ∈ M⊥. Itfollows that

u1 − u2 = v2 − v1 ∈M ∩M⊥.

But M ∩M⊥ = 0 so that u1 = u2 and v1 = v2.

Corollary 1 (Riesz-Frechet theorem). If T is a linear continuous functional on theHilbert space H then there exists a unique h ∈ H such that T (x) = (x, h) for all x ∈ H.Moreover, ‖T‖H→C

= ‖h‖ .Proof. If T ≡ 0 then h = 0 will do. If T 6= 0 then there exists v0 ∈ H such thatT (v0) 6= 0. Let

M := u ∈ H : T (u) = 0 .Since T is linear and continuous then M is a closed subspace. It follows from Thereom1 that

H = M ⊕M⊥

i.e. every x ∈ H has the unique representation as x = u + v. Since v0 6= 0 thenv0 ∈M⊥. Therefore, for every x ∈ H, we can define

u := x− T (x)

T (v0)v0.

Then T (u) = 0 i.e. u ∈M . It follows that

(x, v0) = (u, v0) +T (x)

T (v0)‖v0‖2 =

T (x)

T (v0)‖v0‖2

or

T (x) =T (v0)

‖v0‖2 (x, v0) =

(x,T (v0)

‖v0‖2 v0

),

which is of the desired form. The uniqueness of h can be seen as follows. If T (x) =

(x, h) = (x, h) then (x, h−h) = 0 for all x ∈ H. In particular∥∥∥h− h

∥∥∥2

= (h−h, h−h) =

0 i.e. h = h. It remains to prove the statement about the norm ‖T‖H→C= ‖T‖. Firstly,

‖T‖ = sup‖x‖≤1

|T (x)| = sup‖x‖≤1

|(x, h)| ≤ ‖h‖ .

On the other hand T (h/ ‖h‖) = ‖h‖ implies that ‖T‖ ≥ ‖h‖. Thus ‖T‖ = ‖h‖. Thisfinishes the proof.

7

Corollary 2. If M is a linear subspace of a Hilbert space H then

M⊥⊥ :=ÄM⊥

ä⊥= M.

Proof. It is not so difficult to check that

M⊥ =ÄM

ä⊥.

That’s why

M⊥⊥ =(ÄM

ä⊥)⊥

and Theorem 1 implies that

H = M ⊕ÄM

ä⊥, H =

ÄM

ä⊥ ⊕M⊥⊥.

Uniqueness of this representation guarantees that M⊥⊥ = M .

Remark. In the frame of this theorem we have that

‖x‖2 = ‖u‖2 + ‖v‖2 , ‖v‖2 = (x, v).

Definition. Let A ⊂ H be a subset of an inner product space. The subset

spanA :=

x ∈ H : x =

k∑

j=1

λjxj, xj ∈ A, λj ∈ C

is called the linear span of A.

Definition. Let H be a Hilbert space.

1) A subset B ⊂ H is called a basis of H if B is linearly independent in H and

spanB = H

i.e. for every x ∈ H and every ε > 0 there exist k ∈ N and cjkj=1 ⊂ C such

that ∥∥∥∥∥∥x−

k∑

j=1

cjxj

∥∥∥∥∥∥< ε, xj ∈ B.

2) The Hilbert space is called separable if it has a countable or finite basis.

3) An orthonormal system B = xαα∈A in H which is a basis is called an orthonor-mal basis .

By the Gram-Schmidt orthonormalization we may conclude that every separableHilbert space has an orthonormal basis.

8

Theorem 2 (Characterization of an orthonormal basis). Let B = xj∞j=1 be an or-thonormal system in a separable Hilbert space H. Then the following statements areequivalent:

1) B is maximal i.e. it is not a proper subset of any other orthonormal system.

2) For every x ∈ H the condition (x, xj) = 0, j = 1, 2, . . . implies that x = 0.

3) Every x ∈ H has the Fourier expansion

x =∞∑

j=1

(x, xj)xj

i.e. ∥∥∥∥∥∥x−

k∑

j=1

(x, xj)xj

∥∥∥∥∥∥→ 0, k → ∞.

This means that B is an orthonormal basis.

4) Every pair x, y ∈ H satisfies the completeness relation

(x, y) =∞∑

j=1

(x, xj)(y, xj).

5) Every x ∈ H satisfies the Parseval equality

‖x‖2 =∞∑

j=1

|(x, xj)|2.

Proof. 1)⇒ 2) Suppose that there is z ∈ H, z 6= 0 such that (z, xj) = 0 for allj = 1, 2, . . . . Then

B′ :=

®z

‖z‖ , x1, x2, . . .

´

is an orthonormal system in H. This fact implies that B is not maximal. Itcontradicts 1) and proves 2).

2)⇒3) Given x ∈ H introduce the sequence

x(k) =k∑

j=1

(x, xj)xj.

Theorem of Pythagoras and Bessel’s inequality (Exercises 1 and 2) imply that

∥∥∥x(k)∥∥∥2

=k∑

j=1

|(x, xj)|2 ≤ ‖x‖2 .

9

It follows that∞∑

j=1

|(x, xj)|2

converges. That’s why, for m < k,

∥∥∥x(k) − x(m)∥∥∥2

=k∑

j=m+1

|(x, xj)|2 → 0

as k,m → ∞. Hence x(k) is a Cauchy sequence in H. Thus there exists y ∈ Hsuch that

y = limk→∞

x(k) =∞∑

j=1

(x, xj)xj.

Next, since the inner product is continuous we deduce that

(y, xj) = limk→∞

(x(k), xj) = (x, xj)

for any j = 1, 2, . . .. Therefore (y−x, xj) = 0 for any j = 1, 2, . . .. Part 2) impliesthat y = x and part 3) follows.

3)⇒4) Let x, y ∈ H. We know from part 3) that

x =∞∑

j=1

(x, xj)xj, y =∞∑

k=1

(y, xk)xk.

Continuity of the inner product and orthonormality of xj∞j=1 allow us to con-clude that

(x, y) =∞∑

j=1

∞∑

k=1

(x, xj)(y, xk)(xj, xk) =∞∑

j=1

(x, xj)(y, xj).

4)⇒5) Take y = x in part 4).

5)⇒1) Suppose that B is not maximal. Then we can add a unit vector z ∈ H to itwhich is orthogonal to B. Parseval’s equality gives then

1 = ‖z‖2 =∞∑

j=1

|(z, xj)|2 = 0.

This contradiction proves the result.

Exercise 8. Let xj∞j=1 be an orthonormal system in an inner product space H. Letx ∈ H, cjk

j=1 ⊂ C and k ∈ N. Prove that∥∥∥∥∥∥x−

k∑

j=1

(x, xj)xj

∥∥∥∥∥∥≤∥∥∥∥∥∥x−

k∑

j=1

cjxj

∥∥∥∥∥∥.

10

2 Symmetric operators in the Hilbert space

Assume that H is a Hilbert space. A linear operator from H to H is a mapping

A : D(A) ⊂ H → H,

where D(A) is a linear subspace of H and A satisfies the condition

A(λx+ µy) = λAx+ µAy

for all λ, µ ∈ C and x, y ∈ D(A). The space D(A) is called the domain of A. Thespace

N(A) := x ∈ D(A) : Ax = 0is called the nullspace (or the kernel) of A. The space

R(A) := y ∈ H : y = Ax for somex ∈ D(A)is called the range of A. Both N(A) and R(A) are linear subspaces of H. We say thatA is bounded if there exists M > 0 such that

‖Ax‖ ≤M ‖x‖ , x ∈ D(A).

We say that A is densely defined if D(A) = H. In such case A can be extended to Aex

which will be defined on the whole H with the same norm estimate and we may define

‖A‖H→H := infM : ‖Ax‖ ≤M ‖x‖ , x ∈ D(A)or equivalently

‖A‖H→H = sup‖x‖=1

‖Ax‖ .

Exercise 9 (Hellinger-Toeplitz). Suppose that D(A) = H and

(Ax, y) = (x,Ay), x, y ∈ H.

Prove that A is bounded.

Example 2.1 (Integral operator in L2). Suppose that K(s, t) ∈ L2((a, b) × (a, b)).Define the integral operator K as

Kf(s) =∫ b

aK(s, t)f(t)dt, f ∈ L2(a, b).

Let us prove that K is bounded. Indeed,

∥∥∥Kf∥∥∥2

L2(a,b)=

∫ b

a|Kf(s)|2ds =

∫ b

a

∣∣∣∣∣

∫ b

aK(s, t)f(t)dt

∣∣∣∣∣

2

ds

=∫ b

a

∣∣∣(K(s, ·), f)L2

∣∣∣2ds ≤

∫ b

a‖K(s, ·)‖2

L2

∥∥∥f∥∥∥2

L2ds

=∫ b

a

Ç∫ b

a|K(s, t)|2dt

∫ b

a|f(t)|2dt

åds

= ‖K‖2L2((a,b)×(a,b)) ‖f‖

2L2(a,b)

11

where we have made use of Cauchy-Schwarz-Bunjakovskii inequality. That’s why wehave ∥∥∥K

∥∥∥L2→L2

≤ ‖K‖L2((a,b)×(a,b)) .

The norm‖K‖L2((a,b)×(a,b)) :=

∥∥∥K∥∥∥

HS

is called the Hilbert-Schmidt norm of K.

Example 2.2 (Differential operator in L2). Consider the differential operator

A := id

dt

in L2(0, 1) with the domain

D(A) =¶f ∈ C1[0, 1] : f(0) = f(1) = 0

©.

First of all we have that D(A) = L2. Moreover, integration by parts gives

(Af, g) =∫ 1

0if ′(t)g(t)dt = i

ñfg|10 −

∫ 1

0f(t)g′(t)dt

ô=∫ 1

0f(t)ig′(t)dt = (f,Ag)

for all f, g ∈ D(A). Let us now consider the sequence

un(t) := sin(nπt), n = 1, 2, . . . .

Clearly un ∈ D(A) and

‖un‖2L2 =

∫ 1

0| sin(nπt)|2dt =

1

2.

But

‖Aun‖2L2 =

∫ 1

0

∣∣∣∣∣id

dtsin(nπt)

∣∣∣∣∣

2

dt = (nπ)2∫ 1

0| cos(nπt)|2dt = (nπ)2 1

2= (nπ)2 ‖un‖2

L2 .

Therefore A is not bounded. This shows that D(A) = H is an essential assumption inExercise 9.

We assume later on that D(A) = H i.e. that A is densely defined in any case.

Definition. The graph Γ(A) of a linear operator A in the Hilbert space H is definedas

Γ(A) := (x; y) ∈ H ×H : x ∈ D(A) and y = Ax .Remark. The graph Γ(A) is a linear subspace of the Hilbert space H ×H. The innerproduct in H ×H can be defined as

((x1; y1) , (x2; y2))H×H := (x1, x2)H + (y1, y2)H

for any (x1; y1) , (x2; y2) ∈ H ×H.

12

Definition. The operator A is called closed if Γ(A) = Γ(A). We denote this fact byA = A.

By definition, the criterion for closedness is that

xn ∈ D(A)

xn → x

Axn → y

⇒x ∈ D(A)

y = Ax.

The reader is asked to verify that it is also possible to use seemingly weaker, butequivalent, criterion:

xn ∈ D(A)

xnw→ x

Axnw→ y

⇒x ∈ D(A)

y = Ax,

where xnw→ x indicates weak convergence in the sense that

(xn, y) → (x, y)

for all y ∈ H.

Definition. Let A and A1 be two linear operators in a Hilbert space H. We say thatA1 is an extension of A (and A is a restriction of A1) if D(A) ⊂ D(A1) and Ax = A1xfor all x ∈ D(A). We denote this fact by A ⊂ A1 and A = A1|D(A).

Definition. We say that A is closable if A has an extension A1 and A1 = A1. Theclosure of A, denoted by A, is the smallest closed extension of A if it exists, i.e.

A =⋂

A⊂A1

A1=A1

A1.

Here, by A1∩›A1, we mean the operator whose domain is D(A1∩›A1) := D(A1)∩D(›A1)and

(A1 ∩ ›A1)x := A1x = ›A1x, x ∈ D(A1 ∩ ›A1),

whenever A ⊂ A1 = A1 and A ⊂ ›A1 = ›A1.

If A is closable then ΓÄAä

= Γ(A).

Definition. Consider the subspace

D∗ := v ∈ H : there existsh ∈ H such that (Ax, v) = (x, h) for allx ∈ D(A) .

The operator A∗ with the domain D(A∗) := D∗ and mapping A∗v = h is called theadjoint operator of A.

Exercise 10. Prove that A∗ exists as unique linear operator.

13

Remark. The adjoint operator is maximal among all linear operators B (in the sensethat B ⊂ A∗) which satisfy

(Ax, y) = (x,By)

for all x ∈ D(A) and y ∈ D(B).

Example 2.3. Consider the operator

Af(x) := x−αf(x), α > 1/2

in the Hilbert space H = L2(0, 1). Let us define

D(A) :=¶f ∈ L2(0, 1) : f(x) = χn(x)g(x), g ∈ L2 for somen ∈ N

©,

where

χn(x) =

0, 0 ≤ x ≤ 1/n

1, 1/n < x ≤ 1.

It is clear that D(A) = L2(0, 1). For v ∈ D(A∗) we have

(Af, v) =∫ 1

0x−αχn(x)g(x)v(x)dx =

∫ 1

0f(x)x−αv(x)dx = (f,A∗v).

That’s why we may conclude that

D(A∗) =¶v ∈ L2 : x−αv ∈ L2

©.

Let us show that A is not closed. To see this take the sequence

fn(x) =

xα, 1/n < x ≤ 1

0, 0 ≤ x ≤ 1/n.

Then fn ∈ D(A) and

Afn(x) =

1, 1/n < x ≤ 1

0, 0 ≤ x ≤ 1/n.

If we assume that A = A then

fn ∈ D(A)

fn → xα

Afn → 1

⇒xα ∈ D(A)

1 = Axα.

But xα /∈ D(A). This contradiction shows us that A is not closed. It is not boundedeither since α > 1/2.

Theorem 1. Let A be linear and densely defined operator. Then

1) A∗ = A∗.

14

2) A is closable if and only if D(A∗) = H. In this case A∗∗ := (A∗)∗ = A.

3) If A is closable thenÄAä∗

= A∗.

Proof. 1) Let us define in H×H the linear and bounded operator V as the mapping

V : (u; v) → (v;−u).

It has the property V 2 = −I. The equality (Au, v) = (u,A∗v) for u ∈ D(A) andv ∈ D(A∗) can be rewritten as

(V (u;Au), (v;A∗v))H×H = 0.

It implies that Γ (A∗)⊥V Γ(A) and Γ (A∗)⊥V Γ(A). It means (see Theorem 1 in

Section 1) that Γ (A∗) =ÄV Γ(A)

ä⊥. Since the orthogonal complement is always

closed then Γ (A∗) is closed. This proves 1).

2) Assume D(A∗) = H. Then we can define A∗∗ := (A∗)∗ and due to part 1) wemay conclude that

Γ (A∗∗)⊥V Γ (A∗).

Next, since Γ(A) is a closed subspace of H×H we have Γ(A) =ÄΓ(A)

ä⊥⊥. Since

V 2 = −I and V is bounded then

Γ(A) = −ÄV 2Γ(A)

ä⊥⊥= −

ÅV

ÄV Γ(A)

ä⊥ã⊥

= − (V Γ(A∗))⊥

by 1). HenceΓ(A)⊥V Γ (A∗).

It follows thatΓ(A) = Γ(A∗∗)

orΓ(A) = Γ(A∗∗)

orA = A∗∗.

This proves 2) in one direction. Let us assume now that A is closable butD(A∗) 6=H. It is equivalent to the fact that there exists u0 6= 0 such that u0⊥D(A∗). Inthat case for any v ∈ D(A∗) the element (u0; 0) ∈ H × H is orthogonal to(v;A∗v) ∈ Γ(A∗) ⊂ H ×H. This is equivalent to (see 1)) (u0; 0) ∈ V Γ(A). SinceA is closable and V is bounded then (u0; 0) ∈ V Γ(A) or

−V (u0; 0) = (0;u0) ∈ Γ(A)

or A(0) = u0. Linearity of A implies u0 = 0. This contradiction proves 2).

15

3) Since A is closable then

A∗ 1)= A∗

2)= (A∗)∗∗ = (A)∗∗∗ = (A∗∗)∗

2)=

ÄAä∗.

This finishes the proof.

Example 2.4. Consider the Hilbert space H = L2(R) and the operator

Au(x) = (u, f0)u0(x),

where u0 6≡ 0, u0 ∈ L2(R) is fixed and f0 6= 0 is an arbitrary but fixed constant. Weconsider A on the domain

D(A) =ßu ∈ L2(R) :

∫

R

|f0u(x)|dx <∞™

= L2(R) ∩ L1(R).

It is known that L2(R) ∩ L1(R) = L2(R). Thus A is densely defined. Let v be anelement of D(A∗). Then

(Au, v) = ((u, f0)u0, v) = (u, f0)(u0, v) =Äu, (u0, v)f0

ä= (u, (v, u0)f0) .

It means thatA∗v = (v, u0)f0.

But (v, u0)f0 must belong to L2(R). Since (v, u0)f0 is a constant and f0 6= 0 then(v, u0) must be equal to 0. Thus

u0⊥D(A∗)

which implies thatu0⊥D(A∗).

Since u0 6= 0 then D(A∗) 6= H. Thus A∗ exists but is not densely defined.

Exercise 11. Assume that A is closable. Prove that D(A) can be obtained as theclosure of D(A) by the norm

Ä‖Au‖2 + ‖u‖2

ä1/2.

Definition. Let A : H → H with D(A) = H. We say that A is

1. symmetric if A ⊂ A∗;

2. self-adjoint if A = A∗;

3. essentially self-adjoint ifÄAä∗

= A.

16

Remark. A symmetric operator is always closable and its closure is also symmetric.Indeed, if A ⊂ A∗ then D(A) ⊂ D(A∗). Hence

H = D(A) ⊂ D(A∗) ⊂ H

implies that D(A∗) = H. That’s why A is closable. Since A is the smallest closedextension of A then

A ⊂ A ⊂ A∗ =ÄAä∗

i.e. A is also symmetric.

Some properties of symmetric operator A are:

1) A ⊂ A = A∗∗ ⊂ A∗

2) A = A = A∗∗ ⊂ A∗ if A is closed.

3) A = A = A∗∗ = A∗ if A is self-adjoint.

4) A ⊂ A = A∗∗ = A∗ if A is essentially self-adjoint.

Theorem 2 (J. von Neumann). Assume that A ⊂ A∗.

1) If D(A) = H then A = A∗ and bounded.

2) If R(A) = H then A = A∗ and A−1 exists and is bounded.

3) If A−1 exists then A = A∗ if and only if A−1 = (A−1)∗.

Proof. 1) Since A ⊂ A∗ then H = D(A) ⊂ D(A∗) ⊂ H and hence D(A) = D(A∗) =H. Thus A = A∗ and the Hellinger-Toeplitz theorem (Exercise 9) says that A isbounded.

2,3) Let us assume that u0 ∈ D(A) and Au0 = 0. Then for any v ∈ D(A) we obtainthat

0 = (Au0, v) = (u0, Av).

It means that u0⊥H and therefore u0 = 0. It follows that A−1 exists andD(A−1) = R(A) = H. Hence (A−1)

∗exists. Let us prove that (A∗)−1 exists

too and (A∗)−1 = (A−1)∗. Indeed, if u ∈ D(A) and v ∈ D

Ä(A−1)

∗äthen

(u, v) = (A−1Au, v) = (Au,ÄA−1

ä∗v).

This equality implies that ÄA−1

ä∗v ∈ D(A∗)

andA∗

ÄA−1

ä∗v = v. (2.1)

17

Similarly, if u ∈ D(A−1) and v ∈ D(A∗) then

(u, v) = (AA−1u, v) = (A−1u,A∗v)

and thereforeA∗v ∈ D

ÄÄA−1

ä∗ä

and ÄA−1

ä∗A∗v = v. (2.2)

It follows from (2.1) and (2.2) that (A∗)−1 exists and (A∗)−1 = (A−1)∗.

Exercise 12. Let A and B be injective operators. Prove that if A ⊂ B thenA−1 ⊂ B−1.

Since A ⊂ A∗ we have by Exercise 12 that

A−1 ⊂ (A∗)−1 =ÄA−1

ä∗

i.e. A−1 is also symmetric. But D(A−1) = H. That’s why we may conclude thatH = D(A−1) ⊂ D

Ä(A−1)

∗ä ⊂ H and hence D(A−1) = DÄ(A−1)

∗ä= H. Thus

A−1 is self-adjoint and bounded (Hellinger-Toeplitz theorem). Finally,

A−1 =ÄA−1

ä∗= (A∗)−1

if and only if A = A∗.

Theorem 3 (Basic criterion of self-adjointness). If A ⊂ A∗ then the following state-ments are equivalent:

1) A = A∗.

2) A = A and N(A∗ ± iI) = 0.

3) R(A± iI) = H.

Proof. 1)⇒2) Since A = A∗ then A is closed. Suppose that u0 ∈ N(A∗ − iI) i.e.u0 ∈ D(A∗) = D(A) and Au0 = iu0. Then

i(u0, u0) = (iu0, u0) = (Au0, u0) = (u0, Au0) = (u0, iu0) = −i(u0, u0).

This implies that u0 = 0 i.e. N(A∗ − iI) = 0. The proof of N(A∗ + iI) = 0is left to the reader.

18

2)⇒3) Since A = A and N(A∗±iI) = 0 then, for example, the equation A∗u = −iuhas only the trivial solution u = 0. It implies that R(A− iI) = H. For otherwisethere exists u0 6= 0 such that u0⊥R(A− iI). It means that for any u ∈ D(A) wehave

((A− iI)u, u0) = 0

and therefore u0 ∈ D(A∗ + iI) and (A∗ + iI)u0 = 0 or A∗u0 = −iu0, u0 6= 0. Thiscontradiction proves that R(A− iI) = H. Next, since A is closed then Γ(A) isalso closed and due to the fact that A is symmetric we have

‖(A− iI)u‖2 = ((A− iI)u, (A− iI)u) = ‖Au‖2 − i(u,Au) + i(Au, u) + ‖u‖2

= ‖Au‖2 + ‖u‖2 , u ∈ D(A).

That’s why if (A − iI)un → v0 then Aun and un are convergent i.e. Aun →v′0, un → u′0 and un ∈ D(A). The closedness of A implies that u′0 ∈ D(A) andv′0 = Au′0 i.e. (A− iI)un → Au′0 − iu′0 = v0. It means that R(A− iI) is a closedset i.e. R(A− iI) = R(A− iI) = H. The proof of R(A + iI) = H is left to thereader.

3)⇒1) Assume that R(A± iI) = H. Since A ⊂ A∗ it suffices to show that D(A∗) ⊂D(A). For every u ∈ D(A∗) we have (A∗ − iI)u ∈ H. Part 3) implies that thereexists v0 ∈ D(A) such that

(A− iI)v0 = (A∗ − iI)u.

It is clear that u− v0 ∈ D(A∗) (since A ⊂ A∗) and

(A∗ − iI)(u− v0) = (A∗ − iI)u− (A∗ − iI)v0 = (A∗ − iI)u− (A− iI)v0

= (A− iI)v0 − (A− iI)v0 = 0.

Hence u− v0 ∈ N(A∗ − iI).

Exercise 13. Let A be a linear and densely defined operator in the Hilbert spaceH. Prove that

H = N(A∗) ⊕R(A).

By this exercise we know that

H = N(A∗ − iI) ⊕R(A+ iI).

But in our case R(A+ iI) = H. Hence N(A∗ − iI) = 0 and therefore u = v0.Thus D(A) = D(A∗).

Exercise 14. Let H = L2(0, 1), A := i ddt

and

Dγ(A) =¶f ∈ L2(0, 1) : f ′ ∈ L2(0, 1), f(0) = f(1)eiγ, γ ∈ R

©.

Prove that A is self-adjoint on Dγ(A).

19

3 J. von Neumann’s spectral theorem

Definition. A bounded linear operator P on a Hilbert space H which is self-adjointand idempotent i.e. P 2 = P is called an orthogonal projection operator or a projector .

Proposition 1. Let P be a projector.

1) ‖P‖ = 1 if P 6= 0.

2) P is a projector if and only if P⊥ := I − P is a projector.

3) H = R(P ) ⊕R(P⊥), P |R(P ) = I and P |R(P⊥) = 0.

4) There is one-to-one correspondence between projectors on H and closed linearsubspaces of H. More precisely, if M ⊂ H is a closed linear subspace then thereexists a projector PM : H → M and, conversely, if P : H → H is a projectorthen R(P ) is a closed linear subspace.

5) If ejNj=1, N ≤ ∞ is an orthonormal system then

PNx :=N∑

j=1

(x, ej)ej, x ∈ H

is a projector.

Proof. 1) Since P = P ∗ and P = P 2 then P = P ∗P . Hence ‖P‖ = ‖P ∗P‖. But‖P ∗P‖ = ‖P‖2. Indeed,

‖P ∗P‖ ≤ ‖P ∗‖ ‖P‖ ≤ ‖P‖2

and

‖P‖2 = sup‖x‖=1

‖Px‖2 = sup‖x‖=1

(Px, Px) = sup‖x‖=1

(P ∗Px, x) ≤ sup‖x‖=1

‖P ∗Px‖

= ‖P ∗P‖ .

Therefore ‖P‖ = ‖P‖2 or ‖P‖ = 1 if P 6= 0.

2) Since P is linear and bounded then the same is true about I − P . Moreover,

(I − P )∗ = I − P ∗ = I − P

and(I − P )2 = (I − P )(I − P ) = I − 2P + P 2 = I − P.

20

3) It follows immediately from I = P + P⊥ that every x ∈ H is of the form u + v,

where u ∈ R(P ) and v ∈ R(P⊥). Let us prove that R(P ) =ÄR(P⊥)

ä⊥. First

assume that w ∈ÄR(P⊥)

ä⊥i.e. (w, (I − P )x) = 0 for all x ∈ H. This is

equivalent to(w, x) = (w,Px) = (Pw, x), x ∈ H

or Pw = w. Hence w ∈ R(P ) and so we have proved thatÄR(P⊥)

ä⊥ ⊂ R(P ).For the opposite embedding we let w ∈ R(P ). Then there exists xw ∈ H suchthat w = Pxw. If z ∈ R(P⊥) then z = P⊥xz = (I−P )xz for some xz ∈ H. Thus

(w, z) = (Pxw, (I − P )xz) = (Pxw, xz) − (Pxw, Pxz) = 0

since P is a projector. Therefore w ∈ÄR(P⊥)

ä⊥and we may conclude that

R(P ) =ÄR(P⊥)

ä⊥. This fact allows us to conclude that R(P ) = R(P ) and

H = R(P ) ⊕R(P⊥). Moreover, it is easy to check by definition that P |R(P ) = Iand P |R(P⊥) = 0.

4) If M ⊂ H is a closed subspace then Theorem 1 in Section 1 implies that x =u + v ∈ H, where u ∈ M and v ∈ M⊥. In that case let us define PM : H → Mas

PMx = u.

It is clear that P 2Mx = PMu = u = PMx i.e. P 2

M = PM . Moreover, if y ∈ H theny = u1 + v1, u1 ∈M, v1 ∈M⊥ and

(PMx, y) = (u, u1 + v1) = (u, u1) = (u+ v, u1) = (u+ v, PMy) = (x, PMy)

i.e. P ∗M = PM . Hence PM is a projector. If P is a projector then we know from

part 3) that M := R(P ) is closed subspace of H.

5) Let us assume that N = ∞. Define M as

M :=

x ∈ H : x =

∞∑

j=1

cjej,∞∑

j=1

|cj|2 <∞ .

Then M is a closed subspace of H. If we define a linear operator PM as

PMx :=∞∑

j=1

(x, ej)ej, x ∈ H

then by Bessel’s inequality we obtain that PMx ∈M and

‖PMx‖ ≤ ‖x‖ .

21

It means that PM is a bounded linear operator into M . But PMej = ej and thusP 2

Mx = PMx for all x ∈ H. Next, for all x, y ∈ H we have

(PMx, y) =

Ñ∞∑

j=1

(x, ej)ej, y

é=

∞∑

j=1

(x, ej) (ej, y) =∞∑

j=1

(x, (y, ej)ej)

=

Ñx,

∞∑

j=1

(y, ej)ej

é= (x, PMy)

i.e. P ∗M = PM . The case of finite N requires no convergence questions and is left

to the reader.

Definition. A bounded linear operator A on the Hilbert space H is smaller than orequal to a bounded operator B on H if

(Ax, x) ≤ (Bx, x), x ∈ H.

We denote this fact by A ≤ B. We say that A is non-negative if A ≥ 0. We denoteA > 0, and say that A is positive, if A ≥ c0I for some c0 > 0.

Remark. In the frame of this definition (Ax, x) and (Bx, x) must be real for all x ∈ H.

Proposition 2. For two projectors P and Q the following statements are equivalent:

1) P ≤ Q.

2) ‖Px‖ ≤ ‖Qx‖ for all x ∈ H.

3) R(P ) ⊂ R(Q).

4) P = PQ = QP .

Proof. 1)⇔2) Follows immediately from (Px, x) = (P 2x, x) = (Px, Px) = ‖Px‖2 .

3)⇔4) Assume R(P ) ⊂ R(Q). Then QPx = Px or QP = P . Conversely, if QP = Pthen clearly R(P ) ⊂ R(Q). Finally, P = QP = P ∗ = (QP )∗ = P ∗Q∗ = PQ.

2)⇔4) If 4) holds then Px = PQx and ‖Px‖ = ‖PQx‖ ≤ ‖Qx‖ for all x ∈ H.Conversely, if ‖Px‖ ≤ ‖Qx‖ then Px = QPx+Q⊥Px implies that

‖Px‖2 = ‖QPx‖2 +∥∥∥Q⊥Px

∥∥∥2 ≤ ‖QPx‖2 .

Hence ∥∥∥Q⊥Px∥∥∥2

= 0

i.e. Q⊥Px = 0 for all x ∈ H. Hence P = QP = PQ.

22

Exercise 15. Let Pj∞j=1 be a sequence of projectors with Pj ≤ Pj+1 for each j =1, 2, . . .. Prove that limj→∞ Pj := P exists and that P is a projector.

Definition. Any linear map A : H → H with the property

‖Ax‖ = ‖x‖ , x ∈ H

is called an isometry .

Exercise 16. Prove that

1) A is an isometry if and only if A∗A = I.

2) Every isometry A has an inverse A−1 : R(A) → H and A−1 = A∗|R(A).

3) If A is an isometry then AA∗ is a projector on R(A).

Definition. A surjective isometry U : H → H is called a unitary operator .

Remark. It follows that U is unitary if and only if it is surjective and U∗U = UU∗ = Ii.e. (Ux, Uy) = (x, y) for all x, y ∈ H.

Definition. Let H be a Hilbert space. The family of operators Eλ∞λ=−∞ is called aspectral family if the following conditions are satisfied:

1) Eλ is a projector for all λ ∈ R.

2) Eλ ≤ Eµ for all λ < µ.

3) Eλ is right continuous with respect to the strong operator topology i.e.

lims→t+0

‖Esx− Etx‖ = 0

for all x ∈ H.

4) Eλ is normalized as follows:

limλ→−∞

‖Eλx‖ = 0, limλ→+∞

‖Eλx‖ = ‖x‖

for all x ∈ H. The latter condition can also be formulated as

limλ→+∞

‖Eλx− x‖ = 0.

Remark. It follows from the previous definition and Proposition 2 that

EλEµ = Eminλ,µ.

Proposition 3. For every fixed x, y ∈ H, (Eλx, y) is a function of bounded variationwith respect to λ ∈ R.

23

Proof. Let us defineE(α, β] := Eβ − Eα, α < β.

Then E(α, β] is a projector. Indeed,

E(α, β]∗ = E∗β − E∗

α = Eβ − Eα = E(α, β]

i.e. E(α, β] is self-adjoint. It is also idempotent due to

(E(α, β])2 = (Eβ − Eα)(Eβ − Eα) = E2β − EαEβ − EβEα + E2

α

= Eβ − Eα − Eα + Eα = E(α, β].

Another property is that

E(α1, β1]x⊥E(α, β]y, x, y ∈ H

if β1 ≤ α or β ≤ α1. To see this for β1 ≤ α calculate

(E(α1, β1]x,E(α, β]y) = (Eβ1x− Eα1x,Eβy − Eαy)

= (Eβ1x,Eβy) − (Eα1x,Eβy) − (Eβ1x,Eαy) + (Eα1x,Eαy)

= (x,Eβ1y) − (x,Eα1y) − (x,Eβ1y) + (x,Eα1y) = 0.

Let nowλ0 < λ1 < · · · < λn.

Then

n∑

j=1

∣∣∣ÄEλj

x, yä−

ÄEλj−1

x, yä∣∣∣ =

n∑

j=1

|(E(λj−1, λj]x, y)|

=n∑

j=1

|(E(λj−1, λj]x,E(λj−1, λj]y)|

≤n∑

j=1

‖E(λj−1, λj]x‖ ‖E(λj−1, λj]y‖

≤Ñ

n∑

j=1

‖E(λj−1, λj]x‖2

é1/2 Ñn∑

j=1

‖E(λj−1, λj]y‖2

é1/2

=

∥∥∥∥∥∥

n∑

j=1

E(λj−1, λj]x

∥∥∥∥∥∥

∥∥∥∥∥∥

n∑

j=1

E(λj−1, λj]y

∥∥∥∥∥∥= ‖E(λ0, λn]x‖ ‖E(λ0, λn]y‖ ≤ ‖x‖ ‖y‖ .

Here we have made use of orthogonality, normalization and the Cauchy-Schwarz-Bunjakovskii inequality.

24

Due to Proposition 3 we can define a Stieltjes integral. Moreover, for any continuousfunction f(λ) we may conclude that the limit

lim∆→0

n∑

j=1

f(λ∗j) (E(λj−1, λj]x, y) = lim∆→0

Ñn∑

j=1

f(λ∗j)E(λj−1, λj]x, y

é,

where λ∗j ∈ [λj−1, λj], α = λ0 < λ1 < · · · < λn = β and ∆ = max1≤j≤n |λj−1 − λj|exists and by definition this limit is

∫ β

αf(λ)d(Eλx, y), x, y ∈ H.

It can be shown that this is equivalent to the existence of the limit in H

lim∆→0

n∑

j=1

f(λ∗j)E(λj−1, λj]x,

which we denote by ∫ β

αf(λ)dEλx.

Thus ∫ β

αf(λ)d(Eλx, y) =

Ç∫ β

αf(λ)dEλx, y

å, x, y ∈ H.

For the spectral representation of self-adjoint operators one needs not only integralsover finite intervals but also over whole line which is naturally defined as the limit

∫ ∞

−∞f(λ)d(Eλx, y) = lim

α→−∞

β→∞

∫ β

αf(λ)d(Eλx, y) =

Å∫ ∞

−∞f(λ)dEλx, y

ã

if it exists. Deriving first some basic properties of the integral just defined one cancheck that∫ ∞

−∞f(λ)d(EλEβx, y) =

∫ β

−∞f(λ)d(Eλx, y) := lim

α→−∞

∫ β

αf(λ)d(Eλx, y), x, y ∈ H.

Theorem 1. Let Eλ∞λ=−∞ be a spectral family on the Hilbert space H and let f be areal-valued continuous function on the line. Define

D :=ßx ∈ H :

∫ ∞

−∞|f(λ)|2d(Eλx, x) <∞

™

(or D :=¶x ∈ H :

∫∞−∞ f(λ)dEλx exists

©). Let us define on this domain an operator A

as

(Ax, y) =∫ ∞

−∞f(λ)d(Eλx, y), x ∈ D(A) := D, y ∈ H

(or Ax =∫∞−∞ f(λ)dEλx, x ∈ D(A)). Then A is self-adjoint and satisfies

E(α, β]A ⊂ AE(α, β], α < β.

25

Proof. It can be shown that the integral

∫ ∞

−∞f(λ)d (Eλx, y)

exists for x ∈ D and y ∈ H. Thus (Ax, y) is well-defined. Let v be any element of Hand let ε > 0. Then, by normalization, there exists α < −R and β > R with R largeenough such that

‖v − E(α, β]v‖ = ‖v − Eβv + Eαv‖ ≤ ‖(I − Eβ)v‖ + ‖Eαv‖ < ε.

On the other hand,

∫ ∞

−∞|f(λ)|2d(EλE(α, β]v, E(α, β]v) =

∫ ∞

−∞|f(λ)|2d(EλE(α, β]v, v)

=∫ ∞

−∞|f(λ)|2d(EλEβv, v) −

∫ ∞

−∞|f(λ)|2d(EλEαv, v)

=∫ β

−∞|f(λ)|2d(Eλv, v) −

∫ α

−∞|f(λ)|2d(Eλv, v)

=∫ β

α|f(λ)|2d(Eλv, v) <∞.

These two facts mean that E(α, β]v ∈ D and D = H. Since f(λ) = f(λ) then A issymmetric. Indeed,

(Ax, y) =∫ ∞

−∞f(λ)d(Eλx, y) = lim

α→−∞

β→∞

∫ β

αf(λ)d(Eλx, y)

= limα→−∞

β→∞

∫ β

αf(λ)d(x,Eλy) = lim

α→−∞

β→∞

Çx,∫ β

αf(λ)dEλy

å

=

Ñx, lim

α→−∞

β→∞

∫ β

αf(λ)dEλy

é= (x,Ay).

In order to prove that A = A∗ it remains to show that D(A∗) ⊂ D(A). Let u ∈ D(A∗).Then

(E(α, β]z, A∗u) = (AE(α, β]z, u) =∫ β

αf(λ)d(Eλz, u)

for any z ∈ H. This equality implies that

(z, A∗u) = limα→−∞

β→∞

∫ β

αf(λ)d(Eλz, u) =

∫ ∞

−∞f(λ)d(Eλz, u) =

∫ ∞

−∞f(λ)d(z, Eλu)

=∫ ∞

−∞f(λ)d(Eλu, z) = (Au, z) = (z, Au),

26

where the integral exists because (z, A∗u) exists. Hence u ∈ D(A) and A∗u = Au. Forthe second claim we first calculate

E(α, β]Ax = (Eβ − Eα)Ax = (Eβ − Eα)∫ ∞

−∞f(λ)dEλx

=∫ ∞

−∞f(λ)dEλEβx−

∫ ∞

−∞f(λ)dEλEαx =

∫ β

−∞f(λ)dEλx−

∫ α

−∞f(λ)dEλx

=∫ β

αf(λ)dEλx =

∫ ∞

−∞f(λ)dEλ (Eβ − Eα) x = A (Eβ − Eα) x = AE(α, β]x

for any x ∈ D(A). Since the left hand side is defined on D(A) and the right hand sideon all of H then the latter is an extension of the former.

Exercise 17. Let A be as in Theorem 1. Prove that

‖Au‖2 =∫ ∞

−∞|f(λ)|2d(Eλu, u)

if u ∈ D(A).

Exercise 18. Let H = L2(R) and Au(t) = tu(t), t ∈ R. Define D(A) on which A = A∗

and evaluate the spectral family Eλ∞λ=−∞.

Theorem 2 (J. von Neumann’s spectral theorem). Every self-adjoint operator A onthe Hilbert space H has a unique spectral representation i.e. there is a unique spectralfamily Eλ∞λ=−∞ such that

Ax =∫ ∞

−∞λdEλx, x ∈ D(A)

(i.e. (Ax, y) =∫∞−∞ λd(Eλx, y), x ∈ D(A), y ∈ H), where D(A) is defined as

D(A) =ßx ∈ H :

∫ ∞

−∞λ2d(Eλx, x) <∞

™.

Proof. At first we assume that this theorem holds when A is bounded, that is, thereis a unique spectral family Fµ∞µ=−∞ such that

Au =∫ ∞

−∞µdFµu, u ∈ H

since D(A) = H in this case. But Fµ ≡ 0 for µ < m and Fµ ≡ I for µ > M , where

m = inf‖x‖=1

(Ax, x), M = sup‖x‖=1

(Ax, x).

That’s why the spectral representation has a view

Au =∫ M

mµdFµu, u ∈ H.

27

Let us consider now an unbounded operator which is semibounded from below i.e.

(Au, u) ≥ m0(u, u), u ∈ D(A)

with some constant m0. We assume without loss of generality that (Au, u) ≥ (u, u).This condition implies that A−1 exists, is defined over whole H and ‖A−1‖ ≤ 1. Indeed,A−1 exists and is bounded because Au = 0 if and only if u = 0. The norm estimatefollows from

(v, A−1v) ≥∥∥∥A−1v

∥∥∥2, v ∈ D(A−1).

Since A−1 is bounded then D(A−1) is a closed subspace in H. But self-adjointness ofA means that A−1 = (A−1)

∗. That’s why A−1 is closed and D(A−1) = H i.e. A−1 is

densely defined. Therefore D(A−1) = H and R(A) = H. Since

0 ≤ (A−1v, v) ≤ ‖v‖2 , v ∈ H

we may conclude in this case that m ≥ 0,M ≤ 1 and

A−1v =∫ 1

0µdFµv, v ∈ H,

where Fµ is the spectral family of A−1. Let us note that F1 = I and F0 = 0. Theyfollow from the spectral theorem and from the fact that A−1v = 0 if and only if v = 0.Next, let us define the operator Bε, ε > 0 as

Bεu :=∫ 1

ε

1

µdFµu, u ∈ D(A).

For every v ∈ H we have

BεA−1v =

∫ 1

ε

1

µdFµ(A−1v) =

∫ 1

ε

1

µdFµ

Ç∫ 1

0λdFλv

å=∫ 1

ε

1

µd

Ç∫ 1

0λd(FµFλv)

å

=∫ 1

ε

1

µdÅ∫ µ

ελdFλv

ã=∫ 1

ε

1

µµdFµv =

∫ 1

εdFµv = F1v − Fεv = v − Fεv.

Since every spectral family is right continuous then

limε→0+0

BεA−1v = v

exists. For every u ∈ D(A) we have similarly,

A−1Bεu =∫ 1

0µdFµ(Bεu) =

∫ 1

εµd

Ç∫ µ

ε

1

λdFλu

å= u− Fεu

and hencelim

ε→0+0A−1Bεu = u

28

exists. These two equalities mean that

limε→0+0

Bε =ÄA−1

ä−1= A

exists and the spectral representation

A =∫ 1

0

1

µdFµ = lim

ε→0+

∫ 1

ε

1

µdFµ

holds. If we define Eλ = I − F 1λ, 1 ≤ λ <∞ then

A = −∫ 1

0

1

µdE 1

µ=∫ ∞

1λdEλ.

Exercise 19. Prove that this Eλ is a spectral family.

Domain D(A) can be characterized as

D(A) =ßu ∈ H :

∫ ∞

1λ2d(Eλu, u) <∞

™=

®u ∈ H :

∫ 1

0

1

µ2d(Fµu, u) <∞

´.

This proves the theorem for self-adjoint operators that are semibounded from below.For bounded operators we will only sketch the proof.

Step 1. If A = A∗ and bounded then we can define

pN(A) := a0I + a1A+ · · · + aNAN , N ∈ N,

where aj ∈ R for j = 0, 1, . . . , N . Then pN(A) is also self-adjoint and boundedwith

‖pN(A)‖ ≤ sup|t|≤‖A‖

|pN(t)|.

Step 2. For every continuous function f on [m,M ], where m and M are as above wecan define f(A) as an approximation by pN(A) i.e. we can prove that for anyε > 0 there exists pN(A) such that

‖f(A) − pN(A)‖ < ε.

Step 3. For every u, v ∈ H let us define the functional L as

L(f) := (f(A)u, v).

Then|L(f)| ≤ |(f(A)u, v)| ≤ ‖f(A)‖ ‖u‖ ‖v‖

that is, L(f) is a bounded linear functional on C[m,M ].

29

Step 4. (Riesz’s theorem) Every positive linear continuous functional L on C0[a, b]can be represented in the form

L(f) =∫ b

af(x)dν(x),

where ν is a measure that satisfies the conditions

1) L(f) ≥ 0 for f ≥ 0

2) |L(f)| ≤ ν(K) ‖f‖K , where K ⊂ [a, b] is compact and

‖f‖K = maxx∈K

|f(x)|.

Step 5. It follows from Step 4 that

(Au, v) =∫ M

mλdν(λ;u, v).

Step 6. It is possible to prove that ν(λ;u, v) is a self-adjoint bilinear form. That’swhy we may conclude that there exists a self-adjoint and bounded operator Eλ

such thatν(λ;u, v) = (Eλu, v).

This operator is idempotent and we may define Eλ ≡ 0 for λ < m and Eλ ≡ Ifor λ ≥ M . Thus Eλ∞λ=−∞ is the required spectral family and this theorem isproved.

Let A : H → H be a self-adjoint operator in the Hilbert space H. Then by J. vonNeumann’s spectral theorem we can write

Au =∫ ∞

−∞λdEλu, u ∈ D(A).

For every continuous function f we can define

Df :=ßu ∈ H :

∫ ∞

−∞|f(λ)|2d(Eλu, u) <∞

™.

This set is a linear subspace of H. For every u ∈ Df and v ∈ H let us define the linearfunctional

L(v) :=∫ ∞

−∞f(λ)d(Eλu, v) =

Å∫ ∞

−∞f(λ)dEλu, v

ã.

This functional is continuous because it is bounded. Indeed,

|L(v)|2 ≤∥∥∥∥∫ ∞

−∞f(λ)dEλu

∥∥∥∥2

‖v‖2 =∫ ∞

−∞|f(λ)|2d(Eλu, u) ‖v‖2 = c(u) ‖v‖2 .

30

By the Riesz-Frechet theorem this functional can be expressed in the form of an innerproduct i.e. there exists z ∈ H such that

∫ ∞

−∞f(λ)d(Eλu, v) = (z, v), v ∈ H.

We setz := f(A)u, u ∈ Df

i.e.

(f(A)u, v) =∫ ∞

−∞f(λ)d(Eλu, v).

Remark. Since in general f is not real-valued then f(A) is not a self-adjoint operatorin general.

Example 3.1. Consider

f(λ) =λ− i

λ+ i, λ ∈ R

Denote

UA := f(A) =∫ ∞

−∞

λ− i

λ+ idEλ.

The operator UA is called the Cayley transform. Since |f(λ)| = 1 then Df = D(UA) =H and

‖UAu‖2 =∫ ∞

−∞|f(λ)|2d(Eλu, u) = lim

α→−∞

β→∞

∫ β

αd(Eλu, u) = lim

α→−∞

β→∞

((Eβu, u) − (Eαu, u))

= limα→−∞

β→∞

Ä‖Eβu‖2 − ‖Eαu‖2

ä= ‖u‖2

by normalization of Eλ. Hence UA is an isometry. There is one-to-one correspondencebetween self-adjoint operators and their Cayley transforms. Indeed,

UA = (A− iI)(A+ iI)−1

is equivalent to I − UA = 2i(A+ iI)−1

I + UA = 2A(A+ iI)−1

orA = i(I + UA)(I − UA)−1.

Example 3.2. Consider

f(λ) =1

λ− z, λ ∈ R, z ∈ C, Im z 6= 0.

Denote

Rz := (A− zI)−1 =∫ ∞

−∞

1

λ− zdEλ.

31

The operator Rz is called the resolvent of A. Since

∣∣∣∣1

λ− z

∣∣∣∣ ≤1

|Im z|

for all λ ∈ R then Rz is bounded and defined on whole H.

Exercise 20. Let A = A∗ with spectral family Eλ. Let u ∈ D(f(A)) and v ∈ D(g(A)).Prove that

(f(A)u, g(A)v) =∫ ∞

−∞f(λ)g(λ)d(Eλu, v).

Exercise 21. Let A = A∗ with spectral family Eλ. Let u ∈ D(f(A)). Prove thatf(A)u ∈ D(g(A)) if and only if u ∈ D((gf)(A)) and that

(gf)(A)u =∫ ∞

−∞g(λ)f(λ)dEλu.

Remark. It follows from Exercise 21 that

(gf)(A) = (fg)(A)

on the domain D ((fg)(A)) ∩D ((gf)(A)).

32

4 Spectrum of self-adjoint operators

Definition. Given a linear operator A in the Hilbert space H with domain D(A),D(A) = H, the set

ρ(A) =¶z ∈ C : (A− zI)−1 exists as a bounded operator fromH toD(A)

©

is called the resolvent set of A. Its complement

σ(A) = C \ ρ(A)

is called the spectrum of A.

Theorem 1. 1) If A = A then the resolvent set is open and the resolvent operatorRz := (A − zI)−1 is an analytic function from ρ(A) to B(H;H), the set of alllinear operators in H. Furthermore, the resolvent identity

Rz −Rξ = (z − ξ)RzRξ, z, ξ ∈ ρ(A)

holds and R′z = (Rz)

2.

2) If A = A∗ then z ∈ ρ(A) if and only if there exists Cz > 0 such that

‖(A− zI)u‖ ≥ Cz ‖u‖

for all u ∈ D(A).

Proof. 1) Assume that z0 ∈ ρ(A). Then Rz0 is a bounded linear operator from Hto D(A) and thus r := ‖Rz0‖−1 > 0. Let us define for |z − z0| < r the operator

Gz0 := (z − z0)Rz0 .

Then Gz0 is bounded with ‖Gz0‖ < 1. Hence it defines the operator

(I −Gz0)−1 =

∞∑

j=0

(Gz0)j

because this Neumann series converges. But for |z − z0| < r we have

A− zI = (A− z0I)(I −Gz0)

or(A− zI)−1 = (I −Gz0)

−1Rz0 .

Hence Rz exists with D(Rz) = H and is bounded. It remains to show that R(Rz) ⊂D(A). For x ∈ H we know that

y := (A− zI)−1x ∈ H.

33

We claim that y ∈ D(A). Indeed,

y = (A− zI)−1x = (I −Gz0)−1Rz0x =

∞∑

j=0

(z − z0)j (Rz0)

j+1 x

= limn→∞

n∑

j=0

(z − z0)j (Rz0)

j+1 x.

It follows from this representation that Rz = (A − zI)−1 is an analytic function fromρ(A) to B(H;H). Next we denote

snx :=n∑

j=0

(z − z0)j (Rz0)

j+1 x.

It is clear that snx ∈ D(A) and that limn→∞ snx = y. Moreover,

limn→∞

(A− zI)snx = x.

Denoting yn := snx we may conclude from the criterion for closedness that

yn ∈ D(A)

yn → y

(A− zI)yn → x

⇒y ∈ D(A)

x = (A− zI)y.

Hence y = (A − zI)−1x ∈ D(A) and therefore ρ(A) is open. The resolvent identity isproved by straightforward calculation

Rz −Rξ = Rz(A− ξI)Rξ −Rz(A− zI)Rξ = Rz [(A− ξI) − (A− zI)]Rξ

= (z − ξ)RzRξ.

Finally, the limit

limz→ξ

Rz −Rξ

z − ξ= lim

z→ξRzRξ = (Rz)

2

exists and hence R′z = (Rz)

2 exists. It proves this part.

2) Assume that A = A∗. If z ∈ ρ(A) then by definition Rz maps from H to D(A).Hence there exists Mz > 0 such that

‖Rzv‖ ≤Mz ‖v‖ , v ∈ H.

Since u = Rz(A− zI)u for any u ∈ D(A) then we get

‖u‖ ≤Mz ‖(A− zI)u‖ , u ∈ D(A).

This is equivalent to

‖(A− zI)u‖ ≥ 1

Mz

‖u‖ , u ∈ D(A).

34

Conversely, if there exists Cz > 0 such that

‖(A− zI)u‖ ≥ Cz ‖u‖ , u ∈ D(A).

then (A − zI)−1 is bounded. Since A is self-adjoint then (A − zI)−1 is defined overwhole H. Indeed, if R(A− zI) 6= H then there exists v0 6= 0 such that v0⊥R(A− zI).This means that

(v0, (A− zI)u) = 0, u ∈ D(A)

or(Au, v0) = (zu, v0)

or(u,A∗v0) = (u, zv0).

Thus v0 ∈ D(A∗) and A∗v0 = zv0. Since A = A∗ then v0 ∈ D(A) and Av0 = zv0 or

(A− zI)v0 = 0.

It is easy to check that ‖(A− zI)u‖2 = ‖(A− zI)u‖2 for any u ∈ D(A). Therefore

‖(A− zI)v0‖ = ‖(A− zI)v0‖ ≥ Cz ‖v0‖ .

Hence v0 = 0 and D ((A− zI)−1) = R(A− zI) = H. It means that z ∈ ρ(A).

Corollary 1. If A = A∗ then σ(A) 6= ∅, σ(A) = σ(A) and σ(A) ⊂ R.

Proof. If z = α+ iβ ∈ C with Im z = β 6= 0 then

‖(A− zI)x‖2 = ‖(A− αI)x− iβx‖2 = ‖(A− αI)x‖2 + |β|2 ‖x‖2 ≥ |β|2 ‖x‖2 .

It implies (see part 2) of Theorem 1) that z ∈ ρ(A). It means that σ(A) ⊂ R. SinceA = A∗ and therefore closed then the spectrum σ(A) is closed as a complement of anopen set (see part 1) of Theorem 1).

It remains to prove that σ(A) 6= ∅. Assume on the contrary that σ(A) = ∅. Thenthe resolvent Rz is an entire analytic function. Let us prove that ‖Rz‖ is uniformlybounded with respect to z ∈ C. Introduce the functional

Tz(y) := (Rzx, y), ‖x‖ = 1, y ∈ H.

Then Tz(y) is a linear functional on the Hilbert spaceH. Moreover, since Rz is boundedfor any (fixed) z ∈ C then

|Tz(y)| ≤ ‖Rzx‖ ‖y‖ ≤ ‖Rz‖ ‖y‖ = Cz ‖y‖ .

Therefore Tz(y) is continuous i.e. Tz, z ∈ C is a pointwise bounded family of contin-uous linear functionals. By Banach-Steinhaus theorem we may conclude that

supz∈C

‖Tz‖ = c0 <∞.

35

That’s why we have

|Tz(y)| = |(Rzx, y)| ≤ c0 ‖y‖ , ‖x‖ = 1, z ∈ C.

It implies that ‖Rzx‖ ≤ c0 i.e. ‖Rz‖ ≤ c0. By Liouville theorem we may conclude nowthat Rz ≡ constant. But by J. von Neumann’s spectral theorem

Rz =∫ ∞

−∞

1

λ− zdEλ,

where Eλ is a spectral family of A = A∗. Due to the estimate

‖Rz‖ ≤ 1

|Im z|

we may conclude that ‖Rz‖ → 0 as |Im z| → ∞. Hence Rz ≡ 0. This contradictionfinishes the proof.

Exercise 22. [Weyl’s criterion] Let A = A∗. Prove that λ ∈ σ(A) if and only if thereexists xn ∈ D(A), ‖xn‖ = 1 such that

limn→∞

‖(A− λI)xn‖ = 0.

Definition. Let us assume that A = A. The point spectrum σp(A) of A is the set ofeigenvalues of A i.e.

σp(A) = λ ∈ σ(A) : N(A− λI) 6= 0 .

It means that (A− λI)−1 does not exist i.e. there exists a non-trivial u ∈ D(A) suchthat Au = λu. The complement σ(A) \ σp(A) is the continuous spectrum σc(A). Thediscrete spectrum is the set

σd(A) = λ ∈ σp(A) : dimN(A− λI) <∞ andλ is isolated inσ(A) .

The set σess (A) := σ(A) \ σd(A) is called the essential spectrum of A.

In the frame of this definition, the complex plane can be divided into regions ac-cording to

C = ρ(A) ∪ σ(A),

σ(A) = σp(A) ∪ σc(A)

andσ(A) = σd(A) ∪ σess (A),

with all the unions being disjoint.

Remark. If A = A∗ then

1) λ ∈ σc(A) means that (A− λI)−1 exists but is not bounded.

36

2) σess (A) = σc(A)∪eigenvalues of infinite multiplicity and their accumulation points∪ accumulation points ofσd(A).

Exercise 23. Let A = A∗ and λ1, λ2 ∈ σp(A). Prove that if λ1 6= λ2 then

N(A− λ1I)⊥N(A− λ2I).

Exercise 24. Let ej∞j=1 be an orthonormal basis in H and let sj∞j=1 ⊂ C be somesequence. Introduce the set

D =

x ∈ H :

∞∑

j=1

|sj|2|(x, ej)|2 <∞ .

Define

Ax =∞∑

j=1

sj(x, ej)ej, x ∈ D.

Prove that A = A and that σ(A) = sj : j = 1, 2, . . .. Prove also that

(A− zI)−1x =∞∑

j=1

1

sj − z(x, ej)ej

for any z ∈ ρ(A) and x ∈ D.

Exercise 25. Prove that the spectrum σ(U) of a unitary operator U lies on the unitcircle in C.

Theorem 2. Let A = A∗ and let Eλλ∈R be its spectral family. Then

1) µ ∈ σ(A) if and only if Eµ+ε − Eµ−ε 6= 0 for every ε > 0.

2) µ ∈ σp(A) if and only if Eµ −Eµ−0 6= 0. Here Eµ−0 := limε→0+Eµ−ε in the senseof strong operator topology.

Proof. 1) Suppose that µ ∈ σ(A) but there exists ε > 0 such that Eµ+ε −Eµ−ε = 0.Then by spectral theorem we obtain for any x ∈ D(A) that

‖(A− µI)x‖2 =∫ ∞

−∞(λ− µ)2d(Eλx, x) ≥

∫

|λ−µ|≥ε(λ− µ)2d(Eλx, x)

≥ ε2∫

|λ−µ|≥εd(Eλx, x) = ε2

ñ∫ µ−ε

−∞+∫ ∞

µ+ε

ôd(Eλx, x)

= ε2î(Eµ−εx, x) + ‖x‖2 − (Eµ+εx, x)

ó= ε2 ‖x‖2 .

This inequality means (see part 2) of Theorem 1) that µ 6∈ σ(A) but µ ∈ ρ(A). Thiscontradiction proves 1) in one direction. Conversely, if

Pn := Eµ+ 1n− Eµ− 1

n6= 0

37

for all n ∈ N then there is a sequence xn∞n=1 such that xn ∈ R(Pn) i.e. xn = Pnxn

i.e. xn ∈ D(A) and ‖xn‖ = 1. For this sequence it is true that

‖(A− µI)xn‖2 =∫ ∞

−∞(λ− µ)2d(EλPnxn, Pnxn) =

∫

|λ−µ|≤1/n(λ− µ)2d(Eλxn, xn)

≤ 1

n2

∫ ∞

−∞d(Eλxn, xn) =

1

n2‖xn‖2 =

1

n2→ 0

as n→ ∞. Hence, this sequence satisfies Weyl’s criterion (see Exercise 22) and there-fore µ ∈ σ(A).

2) Suppose µ ∈ R is an eigenvalue of A. Then there is x0 ∈ D(A), x0 6= 0 such that

0 = ‖(A− µI)x0‖2 =∫ ∞

−∞(λ− µ)2d(Eλx0, x0).

In particular, for all n ∈ N and large enough ε > 0 we have that

0 =∫ n

µ+ε(λ− µ)2d(Eλx0, x0) ≥ ε2

∫ n

µ+εd(Eλx0, x0) = ε2((En − Eµ+ε)x0, x0)

= ε2 ‖(En − Eµ+ε)x0‖2 .

Thus we may conclude that0 = Enx0 − Eµ+εx0.

Similarly we can get that0 = E−nx0 − Eµ−εx0.

Letting n→ ∞ and ε→ 0 we obtain

x0 = Eµx0, 0 = Eµ−0x0.

Hencex0 = (Eµ − Eµ−0)x0

and thereforeEµ − Eµ−0 6= 0.

Conversely, define the projector

P := Eµ − Eµ−0.

If P 6= 0 then there exists y ∈ H, y 6= 0 such that y = Py (e.g. any y ∈ R(P ) 6= 0will do). For λ > µ it follows that

Eλy = EλPy = EλEµy − EλEµ−0y = Py = y.

For λ < µ we have that

Eλy = EλEµy − EλEµ−0y = Eλy − Eλy = 0.

38

Hence

‖(A− µI)y‖2 =∫ ∞

−∞(λ− µ)2d(Eλy, y) =

∫ ∞

µ(λ− µ)2dλ(y, y) = 0.

That’s why Ay = µy and y ∈ D(A), y 6= 0 i.e. µ is an eigenvalue of A or µ ∈ σp(A).

Remark. The statements of Theorem 2 can be reformulated as

1) µ ∈ σp(A) if and only if Eµ − Eµ−0 6= 0.

2) µ ∈ σc(A) if and only if Eµ − Eµ−0 = 0.

Definition. Let H and H1 be two Hilbert spaces. A bounded linear operator K :H → H1 is called compact or completely continuous if it maps bounded sets in Hinto precompact sets in H1 i.e. for every bounded sequence xn∞n=1 ⊂ H the sequenceKxn∞n=1 ⊂ H1 contains a convergent subsequence.

If K : H → H1 is compact then the following statements hold.

1) K maps every weakly convergent sequence in H into a norm convergent sequencein H1.

2) If H = H1 is separable then every compact operator is a norm limit of a sequenceof operators of finite rank (i.e. operators with finite dimensional ranges).

3) The norm limit of a sequence of compact operators is compact.

Let us prove 2). Let K be a compact operator. Since H is separable it has anorthonormal basis ej∞j=1. Consider for any n = 1, 2, . . . the projector

Pnx :=n∑

j=1

(x, ej)ej, x ∈ H.

Then Pn ≤ Pn+1 and ‖(I − Pn)x‖ → 0 as n→ ∞. Define

dn := sup‖x‖=1

‖K(I − Pn)x‖ ≡ ‖K(I − Pn)‖ .

Since R(I − Pn) ⊃ R(I − Pn+1) (see Proposition 2 in Section 3) then dn∞n=1 is amonotone decreasing sequence of positive numbers. Hence the limit

limn→∞

dn := d ≥ 0

exists. Let us choose yn ∈ R(I − Pn), ‖yn‖ = 1 such that

‖K(I − Pn)yn‖ = ‖Kyn‖ ≥ d

2.

39

Then

|(yn, x)| = |((I − Pn)yn, x)| = |(yn, (I − Pn)x)| ≤ ‖yn‖ ‖(I − Pn)x‖ → 0, n→ ∞

for any x ∈ H. It means that ynw→ 0. Compactness of K implies that Kyn → 0. Thus

d = 0. That’s whydn = ‖K −KPn‖ → 0.

Since Pn is of finite rank then so is KPn i.e. K is a norm limit of finite rank operators.

Lemma. Suppose A = A∗ is compact. Then at least one of the two numbers ±‖A‖ isan eigenvalue of A.

Proof. Since‖A‖ = sup

‖x‖=1|(Ax, x)|

then there exists a sequence xn with ‖xn‖ = 1 such that

‖A‖ = limn→∞

|(Axn, xn)|.

Actually, we can assume that limn→∞(Axn, xn) exists and equals, say, a. Otherwisewe would take a subsequence of xn. Since A = A∗ then a is real and ‖A‖ = |a|. Dueto the fact that any bounded set of the Hilbert space is weakly compact (unit ball inour case) we can choose a subsequence of xn, say, xkn which converges weakly i.e.xkn

w→ x. Compactness of A implies that Axkn → y. Next we observe that

‖Axkn − axkn‖2 = ‖Axkn‖2 − 2a(Axkn , xkn) + a2 ≤ ‖A‖2 − 2a(Axkn , xkn) + a2

= 2a2 − 2a(Axkn , xkn) → 2a2 − 2a2 = 0,

as n→ ∞. Hence

Axkn − axkn → 0

Axkn → y

xkn

w→ x

⇒xkn → x

Ax = ax.

Since ‖xkn‖ = 1 then ‖x‖ = 1 also. Hence x 6= 0 and a is an eigenvalue of A.

Theorem 3 (Riesz-Schauder). Suppose A = A∗ is compact. Then

1) A has a sequence of real eigenvalues λj 6= 0 which can be enumerated in such away that

|λ1| ≥ |λ2| ≥ · · · ≥ |λj| ≥ · · · .

2) If there are infinitely many eigenvalues then limj→∞ λj = 0 and 0 is the onlyaccumulation point of λj.

3) The multiplicity of λj is finite.

40

4) If ej is the normalized eigenvector for λj then ej∞j=1 is an orthonormal systemand

Ax =∞∑

j=1

λj(x, ej)ej =∞∑

j=1

(Ax, ej)ej, x ∈ H.

It means that ej∞j=1 is an orthonormal basis on R(A).

5) σ(A) = 0, λ1, λ2, . . . , λj, . . . while 0 is not necessarily an eigenvalue of A.

Proof. Lemma gives the existence of an eigenvalue λ1 ∈ R with |λ1| = ‖A‖ and anormalized eigenvector e1. Introduce H1 = e⊥1 . Then H1 is a closed subspace of H andA maps H1 into itself. Indeed,

(Ax, e1) = (x,Ae1) = (x, λ1e1) = λ1(x, e1) = 0

for any x ∈ H1. The restriction of the inner product of H to H1 makes H1 a Hilbertspace (since H1 is closed) and the restriction of A to H1, denoted by A1 = A|H1

, isagain a self-adjoint compact operator which is mapping in H1. Clearly, its norm isbounded by the norm of A i.e. ‖A1‖ ≤ ‖A‖. Applying Lemma to A1 on H1 we get aneigenvalue λ2 with |λ2| = ‖A1‖ and a normalized eigenvector e2 with e2⊥e1. It is clearthat |λ2| ≤ |λ1|. Next introduce the closed subspace H2 = (span e1, e2)⊥. Again, Aleaves H2 invariant and thus A2 := A1|H2

= A|H2is a self-adjoint compact operator in

H2. Applying Lemma to A2 on H2 we obtain λ3 with |λ3| = ‖A2‖ and a normalizedeigenvector e3 with e3⊥e2 and e3⊥e1. This process in the infinite dimensional Hilbertspace leads us to the sequence λj∞j=1 such that |λj+1| ≤ |λj| and correspondingnormalized eigenvectors. Since |λj| > 0 and monotone decreasing then there is a limit

limj→∞

|λj| = r.

Clearly r ≥ 0. Let us prove that r = 0. If r > 0 then |λj| ≥ r > 0 for each j = 1, 2, . . .or

1

|λj|≤ 1

r<∞.

Hence the sequence of vectors

yj :=ej

λj

is bounded and therefore there is a weakly convergent subsequence yjk

w→ y. Com-pactness of A implies the strong convergence of Ayjk

≡ ejk. But ‖ejk

− ejm‖ =√

2 fork 6= m. This contradiction proves 1) and 2).

Exercise 26. Prove that ifH is an infinite dimensional Hilbert space then the identicaloperator I is not compact and inverse of a compact operator (if it exists) is not bounded.

Exercise 27. Prove part 3) of Theorem 3.

41

Consider now the projector

Pnx :=n∑

j=1

(x, ej)ej, x ∈ H.

Then I − Pn is a projector onto (span e1, . . . , en)⊥ ≡ Hn and hence

‖A(I − Pn)x‖ ≤ ‖A‖Hn‖(I − Pn)x‖ ≤ |λn+1| ‖x‖ → 0

as n→ ∞. Since

APnx =n∑

j=1

(x, ej)Aej =n∑

j=1

λj(x, ej)ej

and‖A(I − Pn)x‖ = ‖Ax− APnx‖ → 0, n→ ∞

then

Ax =∞∑

j=1

λj(x, ej)ej

and part 4) follows. Finally, Exercise 24 gives immediately that

σ(A) = 0, λ1, λ2, . . . , λj, . . ..

This finishes the proof.

Corollary (Hilbert-Schmidt theorem). The orthonormal system ej∞j=1 of eigenvec-tors of a compact self-adjoint operator A in a Hilbert space H is an orthonormal basisif and only if N(A) = 0.

Proof. Recall from Exercise 13 that

H = N(A∗) ⊕R(A) = N(A) ⊕R(A).

If N(A) = 0 then H = R(A). It means that for any x ∈ H and any ε > 0 thereexists yε ∈ R(A) such that

‖x− yε‖ < ε/2.

But by Riesz-Schauder theorem

yε = Axε =∞∑

j=1

λj(xε, ej)ej.

Hence

‖x− yε‖ =

∥∥∥∥∥∥x−

∞∑

j=1

λj(xε, ej)ej

∥∥∥∥∥∥< ε/2.

42

Making use of the Theorem of Pythagoras, Bessel’s inequality and Exercise 8 yields

∥∥∥∥∥∥x−

n∑

j=1

(x, ej)ej

∥∥∥∥∥∥≤

∥∥∥∥∥∥x−

n∑

j=1

λj(xε, ej)ej

∥∥∥∥∥∥=

∥∥∥∥∥∥x−

∞∑

j=1

λj(xε, ej)ej +∞∑

j=n+1

λj(xε, ej)ej

∥∥∥∥∥∥

< ε/2 +

∥∥∥∥∥∥

∞∑

j=n+1

λj(xε, ej)ej

∥∥∥∥∥∥

≤ ε/2 +

Ñ∞∑

j=n+1

|λj|2|(xε, ej)|2é1/2

≤ ε/2 + |λn+1|Ñ

∞∑

j=n+1

|(xε, ej)|2é1/2

≤ ε/2 + |λn+1| ‖xε‖ < ε

for n large enough. It means that ej∞j=1 is a basis in H, and moreover, it is anorthonormal basis.

Conversely, if ej∞j=1 is complete in H then R(A) = H (Riesz-Schauder) andtherefore N(A) = 0.

Remark. The condition N(A) = 0 means that A−1 exists and H must be separablein this case.

Theorem 4 (Lemma of Riesz). If A is a compact operator on H and µ ∈ C thenR(I − µA) is closed in H.

Proof. If µ = 0 then R(I−µA) = H. If µ 6= 0 then we assume without loss of generalitythat µ = 1. Let f ∈ R(I − A), f 6= 0. Then there exists a sequence gn ⊂ H suchthat

f = limn→∞

(I − A)gn.

We will prove that f ∈ R(I −A) i.e. there exists g ∈ H such that f = (I −A)g. Sincef 6= 0 then by the decomposition H = N(I − A) ⊕ N(I − A)⊥ we can assume thatgn ∈ N(I − A)⊥ and gn 6= 0 for all n ∈ N.

Suppose that gn is bounded. Then there is a subsequence gkn such that

gkn

w→ g.

Compactness of A implies that

Agkn → h = Ag.

Next,gkn = (I − A)gkn + Agkn → f + h.

Hence g = f + Ag i.e. f = (I − A)g.

43

Suppose that gn is not bounded. Then we can assume without loss of generalitythat ‖gn‖ → ∞. Introduce a new sequence

un :=gn

‖gn‖.

Since ‖un‖ = 1 then there exists a subsequence ukn

w→ u. Compactness of A givesAukn → Au. Since (I − A)gn → f then

(I − A)ukn =1

‖gkn‖(I − A)gkn → 0.

It means again thatukn = (I − A)ukn + Aukn → Au

and u = Au i.e. u ∈ N(I − A). But gn ∈ N(I − A)⊥. Hence ukn ∈ N(I − A)⊥ andfurther u ∈ N(I − A)⊥ because N(I − A)⊥ is closed. Since ‖ukn‖ = 1 then ‖u‖ = 1.Therefore u 6= 0 while

u ∈ N(I − A) ∩N(I − A)⊥.

This contradiction shows that unbounded gn cannot occur.

Theorem 5 (Fredholm alternative). Suppose A = A∗ is compact. For given g ∈ Heither the equation

(I − µA)f = g

has the unique solution (µ−1 /∈ σ(A)) and in this case f = (I − µA)−1g or µ−1 ∈σ(A) and this equation has a solution if and only if g ∈ R(I − µA) i.e. g⊥N(I −µA). In this case the general solution of the equation is of the form f = f0 + u,where f0 is a particular solution and u ∈ N(I − µA) (u is the general solution of thecorresponding homogeneous equation) and the set of all solutions is a finite dimensionalaffine subspace of H.

Proof. Lemma of Riesz (Theorem 4) gives

R(I − µA) = N(I − µA)⊥.

If µ−1 /∈ σ(A) then (µ)−1 /∈ σ(A) also. Thus

R(I − µA) = N(I − µA)⊥ = 0⊥ = H.

Since A = A∗ this means that (I − µA)−1 exists and the unique solution is f =(I − µA)−1g.

If µ−1 ∈ σ(A) then R(I − µA) is a proper subspace of H and the equation (I −µA)f = g has a solution if and only if g ∈ R(I−µA). Since the equation is linear thenany solution is of the form

f = f0 + u, u ∈ N(I − µA)

and the dimension of N(I − µA) is finite.

44

Exercise 28. Let A = A∗ be compact. Prove that σp(A) = σd(A) = σ(A) \ 0 and0 ∈ σess (A).

Exercise 29. Consider the Hilbert space H = l2(C) and

A(x1, x2, . . . , xn, . . .) = (0, x1,x2

2, . . . ,

xn

n, . . .)

for (x1, x2, . . . , xn, . . .) ∈ l2(C). Show that A is compact and has no eigenvalues (evenmore, σ(A) = ∅) and is not self-adjoint.

Exercise 30. Consider the Hilbert space H = L2(R) and

(Af)(t) = tf(t).

Show that the equation Af = f has no non-trivial solutions and that (I − A)−1 doesnot exist. It means that the Fredholm alternative does not hold for non-compact butself-adjoint operator.

Exercise 31. Let H = L2(Rn) and let

Af(x) =∫

RnK(x, y)f(y)dy,

where K(x, y) ∈ L2(Rn × Rn) is such that K(x, y) = K(y, x). Prove that A = A∗ and

that A is compact.

Theorem 6 (Weyl). If A = A∗ then λ ∈ σess (A) if and only if there exists an or-thonormal system xn∞n=1 such that

‖(A− λI)xn‖ → 0

as n→ ∞.

Proof. We will provide only a partial proof. Suppose that λ ∈ σess (A). If λ is aneigenvalue of infinite multiplicity then there is an infinite orthonormal system of eigen-vectors xn∞n=1 because dim(Eλ −Eλ−0)H = ∞ in this case. Since (A− λI)xn ≡ 0 itis clear that

(A− λI)xn → 0.

Next, suppose that λ is an accumulation point of σ(A). It means that λ ∈ σ(A) and

λ = limn→∞

λn,

where λn 6= λm, n 6= m and λn ∈ σ(A). Hence for each n = 1, 2, . . . we have that

Eλn+ε − Eλn−ε 6= 0

for all ε > 0. Therefore there exists a sequence rn → 0 such that

Eλn+rn − Eλn−rn 6= 0.

45

That’s why we can find a normalized vector xn ∈ R(Eλn+rn −Eλn−rn). Since λn 6= λm

for n 6= m we can find xn∞n=1 as an orthonormal system. By spectral theorem wehave

‖(A− λI)xn‖2 =∫ ∞

−∞(λ− µ)2d(Eµxn, xn)

=∫ ∞

−∞(λ− µ)2d(Eµ(Eλn+rn − Eλn−rn)xn, xn)

=∫ λn+rn

λn−rn

(λ− µ)2d(Eµxn, xn)

≤ maxλn−rn≤µ≤λn+rn

(λ− µ)2∫ ∞

−∞d(Eµxn, xn)

= maxλn−rn≤µ≤λn+rn

(λ− µ)2 → 0, n→ ∞.

Theorem 7 (Weyl). Let A and B be two self-adjoint operators in a Hilbert space. Ifthere is z ∈ ρ(A) ∩ ρ(B) such that

T := (A− zI)−1 − (B − zI)−1

is a compact operator then σess (A) = σess (B).

Proof. We show first that σess (A) ⊂ σess (B). Take any λ ∈ σess (A). Then there is anorthonormal system xn∞n=1 such that

‖(A− λI)xn‖ → 0, n→ ∞.

Define the sequence yn as

yn := (A− zI)xn ≡ (A− λI)xn + (λ− z)xn.

Due to Bessel’s inequality any orthonormal system in the Hilbert space convergesweakly to 0. Hence yn

w→ 0. We also have

‖yn‖ ≥ |λ− z| ‖xn‖ − ‖(A− λI)xn‖ = |λ− z| − ‖(A− λI)xn‖ >|λ− z|

2> 0

for all n ≥ n0 >> 1. Next we take the identity

î(B − zI)−1 − (λ− z)−1

óyn = −Tyn − (λ− z)−1(A− λI)xn.

Since T is compact and ynw→ 0 we deduce that

î(B − zI)−1 − (λ− z)−1

óyn → 0.

Introducezn := (B − zI)−1yn.

46

Thenzn − (λ− z)−1yn → 0

oryn + (z − λ)zn → 0.

This fact and ‖yn‖ > |λ−z|2

imply that ‖zn‖ ≥ |λ−z|3

for all n ≥ n0 >> 1. But

(B − λI)zn ≡ (B − zI)zn + (z − λ)zn = yn + (z − λ)zn → 0.

Due to ‖zn‖ ≥ |λ−z|3

> 0 the sequence zn∞n=1 can be chosen as an orthonormal system.Thus λ ∈ σess (B). This proves that σess (A) ⊂ σess (B). Finally, since −T is compacttoo we can interchange the roles of A and B and obtain the opposite embedding.

47

5 Quadratic forms. Friedrichs extension.

Definition. Let D be a linear subspace of a Hilbert space H. A function Q : D×D →C is called a quadratic form if

1) Q(α1x1 + α2x2, y) = α1Q(x1, y) + α2Q(x2, y)

2) Q(x, β1y1 + β2y2) = β1Q(x, y1) + β2Q(x, y2)

for all α1, α2, β1, β2 ∈ C and x1, x2, x, y1, y2, y ∈ D. The space D(Q) := D is called thedomain of Q. We say that Q is

a) densely defined if D(Q) = H.

b) symmetric if Q(x, y) = Q(y, x).

c) semibounded from below if there exists λ ∈ R such that Q(x, x) ≥ −λ ‖x‖2 forall x ∈ D(Q).

d) closed (and semibounded) if D(Q) is complete with respect to the norm

‖x‖Q :=√Q(x, x) + (λ+ 1) ‖x‖2.

e) bounded (continuous) if there exists M > 0 such that

|Q(x, y)| ≤M ‖x‖ ‖y‖

for all x, y ∈ D(Q).

Exercise 32. Prove that ‖·‖Q is a norm and that

(x, y)Q := Q(x, y) + (λ+ 1)(x, y)

is an inner product.

Theorem 1. Let Q be a densely defined, closed, semibounded and symmetric quadraticform in a Hilbert space H such that

Q(x, x) ≥ −λ ‖x‖2 , x ∈ D(Q).

Then there exists a unique self-adjoint operator A which is semi-bounded from belowi.e.

(Ax, x) ≥ −λ ‖x‖2 , x ∈ D(A).

Moreover, this operator A defines the quadratic form Q as

Q(x, y) = (Ax, y), x ∈ D(A), y ∈ D(Q)

and D(A) ⊂ D(Q).

48

Proof. Let us introduce an inner product on D(Q) by

(x, y)Q := Q(x, y) + (λ+ 1)(x, y), x, y ∈ D(Q)

(see Exercise 32). Since Q is closed then D(Q) = D(Q) is a closed subspace of H withrespect to the norm ‖·‖Q. It means that D(Q) with this inner product defines a newHilbert space HQ. It is clear also that

‖x‖Q ≥ ‖x‖

for all x ∈ HQ. Thus, for fixed x ∈ H,

L(y) := (y, x), y ∈ HQ

defines a continuous (bounded) linear functional on the Hilbert space HQ. Applyingthe Riesz-Frechet theorem to HQ we obtain an element x∗ ∈ HQ (x∗ ∈ D(Q)) suchthat

(y, x) ≡ L(y) = (y, x∗)Q.

It is clear that the mapH ∋ x 7→ x∗ ∈ HQ

defines a linear operator J such that

J : H → HQ, Jx = x∗.

Hence(y, x) = (y, Jx)Q, x ∈ H, y ∈ HQ.

Next we prove that J is self-adjoint and that it has an inverse operator J−1. For anyx, y ∈ H we have

(Jy, x) = (Jy, Jx)Q = (Jx, Jy)Q = (Jx, y) = (y, Jx).

Hence J = J∗. It is bounded due to Hellinger-Toeplitz theorem (Exercise 9). Supposethat Jx = 0. Then

(y, x) = (y, Jx)Q = 0

for any y ∈ D(Q). Since D(Q) = H then the last equality implies that x = 0 andtherefore N(J) = 0 and J−1 exists. Moreover,

H = N(J) ⊕R(J∗) = R(J)

and R(J) ⊂ HQ. Now we can define a linear operator A on the domain D(A) ≡ R(J)as

Ax := J−1x− (λ+ 1)x, λ ∈ R.

It is clear that A is densely defined and A = A∗ (J−1 is self-adjoint since J is). If nowx ∈ D(A) and y ∈ D(Q) ≡ HQ then

Q(x, y) = (x, y)Q − (λ+ 1)(x, y) = (J−1x, y) − (λ+ 1)(x, y) = (Ax, y).

49

The semi-boundedness of A from below follows from that of Q. It remains to prove thatthis representation for A is unique. Assume that we have two such representations, A1

and A2. Then for every x ∈ D(A1) ∩D(A2) and y ∈ D(Q) we have that

Q(x, y) = (A1x, y) = (A2x, y).

It follows that((A1 − A2)x, y) = 0.

Since D(Q) = H then we must have A1x = A2x. This finishes the proof.

Corollary. Under the same assumptions as in Theorem 1, there exists√A+ λI which

is self-adjoint on D(√A+ λI) ≡ D(Q) = HQ. Moreover,

Q(x, y) + λ(x, y) = (√A+ λIx,

√A+ λIy)

for all x, y ∈ D(Q).

Proof. Since A + λI is self-adjoint and non-negative there exists a spectral familyEµ∞µ=0 such that

A+ λI =∫ ∞

0µdEµ.

That’s why we can define the operator

√A+ λI :=

∫ ∞

0

√µdEµ

which is also self-adjoint and non-negative. Then for any x ∈ D(A) and y ∈ D(Q) wehave that

Q(x, y) + λ(x, y) = ((A+ λI)x, y) = (√A+ λIx,

Ä√A+ λI

ä∗y).

This means that x ∈ D(√A+ λI) and y ∈ D(

Ä√A+ λI

ä∗). But

√A+ λI is self-

adjoint and, therefore,

D(√A+ λI) = D(

Ä√A+ λI

ä∗) = D(Q) ≡ HQ.

Theorem 2 (Friedrichs extension). Let A be a non-negative, symmetric linear operatorin a Hilbert space H. Then there exists a self-adjoint extension AF of A which isthe smallest among all non-negative self-adjoint extensions of A in the sense that itscorresponding quadratic form has the smallest domain. This extension AF is called theFriedrichs extension of A.

50

Proof. Let A be a non-negative, symmetric operator with domain D(A) dense in H,D(A) = H. Its associated quadratic form

Q(x, y) := (Ax, y), x, y ∈ D(Q) ≡ D(A)

is densely defined, non-negative and symmetric. Let us define a new inner product

(x, y)Q = Q(x, y) + (x, y), x, y ∈ D(Q).

ThenD(Q) becomes an inner product space. This inner product space has a completionHQ with respect to the norm

‖x‖Q :=√Q(x, x) + ‖x‖2.

Moreover, the quadratic form Q(x, y) has an extension Q1(x, y) to this Hilbert spaceHQ defined by

Q1(x, y) = limn→∞

Q(xn, yn)

whenever xHQ= limn→∞ xn, y

HQ= limn→∞ yn, xn, yn ∈ D(Q) and these limits exist. The

quadratic form Q1 is densely defined, closed, non-negative and symmetric. That’s whyTheorem 1, applied to Q1, gives a unique and non-negative, self-adjoint operator AF

such that

Q1(x, y) = (AFx, y), x ∈ D(AF ) ⊂ HQ, y ∈ D(Q1) ≡ HQ.

Since for x, y ∈ D(A) one has

(Ax, y) = Q(x, y) = Q1(x, y) = (AFx, y)

then AF is a self-adjoint extension of A.It remains to prove that AF is the smallest non-negative self-adjoint extension of

A. Suppose that B ≥ 0, B = B∗ is such that A ⊂ B. The associated quadratic formQB(x, y) := (Bx, y) is an extension of Q ≡ QA. Hence

QB ⊃ Q = Q1.

This finishes the proof.

51

6 Elliptic differential operators

Let Ω be a domain in Rn i.e. an open and connected set. Introduce the following

notation:

1) x = (x1, . . . , xn) ∈ Ω

2) |x| =»x2

1 + · · · + x2n

3) α = (α1, . . . , αn) is a multi-index i.e. αj ∈ N0 ≡ N ∪ 0.

a) |α| = α1 + · · · + αn

b) α ≥ β if αj ≥ βj for all j = 1, 2, . . . , n.

c) α+ β = (α1 + β1, . . . , αn + βn)

d) α− β = (α1 − β1, . . . , αn − βn) if α ≥ β

e) xα = xα11 · · ·xαn

n with 00 = 1

f) α! = α1! · · ·αn! with 0! = 1

4) Dj = 1i∂j = 1

i∂

∂xj= −i∂j and Dα = Dα1

1 · · ·Dαnn ≡ (−i)|α|∂α

Definition. An elliptic partial differential operator A(x,D) of order m on Ω is anoperator of the form

A(x,D) =∑

|α|≤m

aα(x)Dα,

where aα(x) ∈ C∞(Ω) and whose principal symbol

a(x, ξ) =∑

|α|=m

aα(x)ξα, ξ ∈ Rn

is invertible for all x ∈ Ω and ξ ∈ Rn \ 0, that is, a(x, ξ) 6= 0 for all x ∈ Ω and

ξ ∈ Rn \ 0.

Assumption 1. We assume that aα(x) are real for |α| = m.

Under Assumption 1 either a(x, ξ) > 0 or a(x, ξ) < 0 for all x ∈ Ω and ξ ∈ Rn\0.

Without loss of generality we assume that a(x, ξ) > 0. Assumption 1 implies also thatm is even and for any compact set K ⊂ Ω there exists CK > 0 such that

a(x, ξ) ≥ CK |ξ|m, x ∈ Ω, ξ ∈ Rn.

Assumption 2. We assume that A(x,D) is formally self-adjoint i.e.

A(x,D) = A∗(x,D) :=∑

|α|≤m

(−1)|α|Dα(aα(x)·).

52

Exercise 33. Prove that A(x,D) = A∗(x,D) if and only if

aα(x) =∑

α≤β

|β|≤m

(−1)|β|CαβD

β−αaβ(x),

where

Cαβ =

β!

α!(β − α)!.

Hint: Make use of the generalized Leibniz formula

Dα(fg) =∑

β≤α

CβαD

α−βfDβg.

Assumption 3. We assume that A(x,D) has a divergence form

A(x,D) ≡∑

|α|=|β|≤m/2

(−1)|α|Dα(aαβ(x)Dβ),

where aαβ = aβα and real for all α and β. We assume also the ellipticity condition∑

|α|=|β|=m/2

aαβ(x)ξαξβ ≥ ν∑

|α|=m/2

|ξα|2 = ν∑

|α|=m/2

ξ2α,

where ν > 0 is called the constant of ellipticity . Such operator is called uniformlyelliptic.

Exercise 34. Prove that ∑

|α|=m/2

ξ2α ≍ |ξ|m

i.e.c|ξ|m ≤

∑

|α|=m/2

ξ2α ≤ C|ξ|m,

where c and C are some constants.

Example 6.1. Let us consider

A(x,D) =n∑

j=1

D2j = −∆, x ∈ Ω ⊂ R

n

in H = L2(Ω) and prove that A ⊂ A∗ with

D(A) = C∞0 (Ω) =

¶f ∈ C∞(Ω) : supp f = x : f(x) 6= 0 is compact in Ω

©.

Let u, v ∈ C∞0 (Ω). Then

(Au, v)L2 =∫

Ω

Ñn∑

j=1

D2ju

évdx = −

n∑

j=1

∫

Ω

Ä∂2

juävdx

= −n∑

j=1

∫

Ω∂j ((∂ju) v) dx+

n∑

j=1

∫

Ω(∂ju)

Ä∂jv

ädx

= −∫

∂Ω(v∇u, nx)dx+ (∇u,∇v)L2 = (∇u,∇v)L2 ,

53

where ∂Ω is the boundary of Ω and nx is the unit outward vector at x ∈ ∂Ω. Here wehave made use of the divergence theorem. In a similar fashion we obtain

(∇u,∇v)L2 = −n∑

j=1

∫

Ωu∂2

j vdx = (u,−∆v)L2 = (u,Av)L2 .

Hence A ⊂ A∗ and A is closable.

Definition. Let s ≥ 0. The L2-based Sobolev space of order s is defined by

Hs(Rn) ≡ W s2 (Rn) :=

ßf ∈ L2(Rn) :

∫

Rn(1 + |ξ|2)s|f(ξ)|2dξ <∞

™,

where f(ξ) is the Fourier transform of f(x) given by

f(ξ) = (2π)−n/2∫

Rne−i(x,ξ)f(x)dx.

The space Hs(Rn) is the closure of C∞0 (Rn) with respect to the norm

‖f‖s :=Å∫

Rn(1 + |ξ|2)s|f(ξ)|2dξ

ã1/2

.

By the symbol

Hs(Ω) we denote the closure of C∞0 (Ω) with respect to the same norm.

If s ∈ N, say s = k, then the norm ‖·‖s is equivalent to the norm

‖f‖2W k

2 (Rn) :=∫

Rn

∑

|α|≤k

|Dαf |2dx,

where Dαf are the generalized derivatives in L2(Rn). We also see the space

W k2 (Ω)

as the closure of C∞0 (Ω) with respect to the norm ‖·‖2

W k2 (Rn). Moreover, we can also

consider W k2 (Ω).

Example 6.2. Recall from Example 6.1 that

(−∆u, v)L2 = (∇u,∇v)L2 , u, v ∈ C∞0 (Ω).

Hence(−∆u, u)L2 = ‖∇u‖2

L2 ≤ ‖u‖L2 ‖∆u‖L2 , u ∈ C∞0 (Ω).

Therefore,

‖u‖2W 2

2= ‖u‖2

L2 + 2 ‖∇u‖2L2 + ‖∆u‖2

L2

≤ ‖u‖2L2 + 2 ‖u‖L2 ‖∆u‖L2 + ‖∆u‖2

L2

≤ 2 ‖u‖2L2 + 2 ‖∆u‖2

L2 ≡ 2 ‖u‖2A ,

54

where ‖·‖A is a norm which corresponds to the operator A = −∆ as follows:

‖u‖2A := ‖u‖2

L2 + ‖−∆u‖2L2 .

It is also clear that ‖u‖A ≤ ‖u‖W 22. Combining these inequalities gives

1√2‖u‖W 2

2≤ ‖u‖A ≤ ‖u‖W 2

2

for all u ∈ C∞0 (Ω). A completion of C∞

0 (Ω) with respect to these norms leads us tothe statement:

D(A) =

W 22 (Ω).

Thus A = −∆ on D(A) =

W 22 (Ω). Let us determine D(A∗) in this case. By the

definition of D(A∗) we have

D((−∆)∗) =¶v ∈ L2(Ω) : there exists v∗ ∈ L2(Ω) such that

(−∆u, v) = (u, v∗) for allu ∈ C∞0 (Ω) .

If we assume that v ∈W 22 (Ω) then it is equivalent to

(u, (−∆)∗v) = (u, v∗)

i.e. (−∆)∗v = v∗ and D((−∆)∗) = W 22 (Ω). Finally, for Ω ⊂ R

n with Ω 6= Rn we

obtain thatA ⊂ A ⊂ A∗ ≡ (A)∗

and A 6= A and A 6= (A)∗, that is, the closure of A does not lead us to a self-adjointoperator.

Remark. If Ω = Rn then

W 22 (Rn) ≡ W 2

2 (Rn) and therefore

A = A∗ = (A)∗.

Hence the closure of A is self-adjoint in that case.

Example 6.3. Consider again A = −∆ on D(A) = C∞0 (Ω) with Ω 6= R

n. Since

(−∆u, u)L2 = ‖∇u‖2L2 ≥ 0

then −∆ is non-negative with lower bound λ = 0. That’s why

Q(u, v) := (∇u,∇v)L2

is a densely defined and non-negative quadratic form with D(Q) ≡ D(A) = C∞0 (Ω).

A new inner product is defined as

(u, v)Q := (∇u,∇v)L2 + (u, v)L2

55

and‖u‖2

Q ≡ ‖u‖2W 1

2 (Ω) .

If we apply now the procedure from Theorem 2 from Section 5 then we obtain theexistence of Q1 = Q with respect to the norm ‖·‖Q which will also be non-negative

and closed with D(Q1) ≡

W 12 (Ω). Next step is to get the Friedrichs extension AF as

AF = J−1 − I

with D(AF ) ≡ R(J) ⊂

W 12 (Ω). More careful examination of Theorem 1 of Section 5

leads us to the fact

D(AF ) =

W 12 (Ω) ∩D(A∗) =

W 12 (Ω) ∩W 2

2 (Ω).

Remark. In general, for symmetric operator, we have

D(AF ) = u ∈ HQ : Au ∈ H

which is equivalent to

D(AF ) = u ∈ HQ : u ∈ D(A∗) .

Exercise 35. Let H = L2(Ω) and A(x,D) = −∆ + q(x), where q(x) = q(x) andq(x) ∈ L∞(Ω). Define A,A∗ and AF .

Exercise 36. Let f ∈ C∞0 (Ω). Prove that

‖f‖2s ≤ ε ‖f‖2

s1+

1

4ε‖f‖2

s2,

where ε > 0 and s1 > s > s2 with s1 + s2 ≥ 2s.

Consider now bounded Ω ⊂ Rn and an elliptic operator A(x,D) in Ω of the form

A(x,D) =∑

|α|=|β|≤m/2

Dα(aαβ(x)Dβ),

where aαβ(x) = aβα(x) are real. Assume that there exists C0 > 0 such that

|aαβ(x)| ≤ C0, |α|, |β| < m

2

for all x ∈ Ω. Assume also that A(x,D) is elliptic, that is,

(−1)m/2∑

|α|=|β|=m/2

aαβ(x)ξαξβ ≥ ν∑

|α|=m/2

|ξα|2, ν > 0.

56

Theorem 1 (Garding’s inequality). Suppose that A(x,D) is as above. Then for anyε > 0 there is Cε > 0 such that

(Af, f)L2(Ω) ≥ (ν − ε) ‖f‖2

Wm/22 (Ω)

− Cε ‖f‖2L2(Ω)

for any f ∈ C∞0 (Ω).

Proof. Let f ∈ C∞0 (Ω). Then integration by parts yields

(Af, f)L2(Ω) =∑

|α|=|β|≤m/2

∫

ΩDα(aαβ(x)Dβf)fdx

=∑

|α|=|β|=m/2

∫

Ω(−1)|α|aαβ(x)DαfDβfdx

+∑

|α|=|β|<m/2

∫

Ω(−1)|α|aαβ(x)DαfDβfdx

≥ ν∑

|α|=m/2

∫

Ω|Dαf |2dx− C0

∑

|α|=|β|<m/2

∫

Ω|Dαf ||Dβf |dx

≥ ν∑

|α|≤m/2

∫

Ω|Dαf |2dx− (C0 + ν)

∑

|α|<m/2

∫

Ω|Dαf |2dx

= ν ‖f‖2

Wm/22 (Ω)

− (C0 + ν) ‖f‖2

Wm/2−12 (Ω)

.

Next we make use of the following

Lemma. For any ε > 0 and 0 < δ ≤ m/2 there is Cε(δ) > 0 such that

(1 + |ξ|2)m/2−δ ≤ ε(1 + |ξ|2)m/2 + Cε(δ)

for any ξ ∈ Rn.

Proof. Let ε > 0 and 0 < δ ≤ m/2. If (1 + |ξ|2)δ ≥ 1ε

then

(1 + |ξ|2)−δ ≤ ε.

Hence(1 + |ξ|2)m/2−δ ≤ ε(1 + |ξ|2)m/2

i.e. the claim holds for any positive constant Cε(δ). For (1 + |ξ|2)δ < 1ε

we can get

(1 + |ξ|2)m/2−δ <

Ç1

ε

åm/2−δδ

≡ Cε(δ).

57

Applying this lemma with δ = 1 to the norm of W s2 -spaces we may conclude that

‖f‖2

Wm/2−12 (Ω)

≤ ε1 ‖f‖2

Wm/22 (Ω)

+ Cε1 ‖f‖2L2(Ω)

for any ε1 > 0. Hence

(Af, f)L2(Ω) ≥ ν ‖f‖2

Wm/22 (Ω)

− (C0 + ν) ‖f‖2

Wm/2−12 (Ω)

≥ ν ‖f‖2

Wm/22 (Ω)

− (C0 + ν)ε1 ‖f‖2

Wm/22 (Ω)

− (C0 + ν)Cε1 ‖f‖2L2(Ω)

= (ν − ε) ‖f‖2

Wm/22 (Ω)

− Cε ‖f‖2L2(Ω) .

This proves the theorem.

Corollary 1. There exists a self-adjoint Friedrichs extension AF of A with domain

D(AF ) =

Wm/22 (Ω) ∩Wm

2 (Ω).

Proof. It follows from Garding’s inequality that

(Af, f)L2(Ω) ≥ −Cε ‖f‖2L2(Ω) , f ∈ D(A).

This means that Aµ := A + µI is positive for µ > Cε and therefore Theorem 2 ofSection 5 gives us the existence of

(Aµ)F ≡ (AF )µ = AF + µI

with domain

D(AF ) = D((Aµ)F ) =

Wm/22 (Ω) ∩D(A∗),

where

Wm/22 (Ω) is the domain of the corresponding closed quadratic form (see Theorem

2). If Ω is bounded with smooth boundary ∂Ω then it can be proved that

D(A∗) = Wm2 (Ω).

Garding’s inequality has two more consequences. Firstly,

‖(AF )µf‖L2 ≥ C0 ‖f‖L2 , C0 > 0

so that(AF )−1

µ : L2(Ω) → L2(Ω).

Secondly,‖(AF )µf‖W

−m/22 (Ω)

≥ C ′0 ‖f‖W

m/22 (Ω)

, C ′0 > 0

so that

(AF )−1µ : L2(Ω) →

Wm/22 (Ω).

58

Corollary 2. The spectrum σ(AF ) = λj∞j=1 is the sequence of eigenvalues of finitemultiplicity with only one accumulation point at +∞. In short, σ(AF ) = σd(AF ). Thecorresponding orthonormal system ψj∞j=1 of eigenfunctions forms an orthonormalbasis and

AFfL2

=∞∑

j=1

λj(f, ψj)ψj

for any f ∈ D(AF ).

Proof. We begin with a lemma.

Lemma. The embedding

Wm/22 (Ω) → L2(Ω)

is compact.

Proof. It is enough to show that for any ϕk∞k=1 ⊂

Wm/22 (Ω) with ‖ϕk‖W

m/22

≤ 1 there

exists ϕjk∞k=1 which is a Cauchy sequence in L2(Ω). Since Ω is bounded we have

|”ϕk(ξ)| ≤ ‖ϕk‖L2 |Ω|1/2

i.e. ”ϕk(ξ) is uniformly bounded. That’s why there exists ϕjk(ξ) which converges

pointwise in Rn. Next,

‖ϕjk− ϕjm‖2

L2 =∫

Rn|ϕjk

(ξ) − ‘ϕjm(ξ)|2dξ

=∫

|ξ|<r|ϕjk

(ξ) − ‘ϕjm(ξ)|2dξ +∫

|ξ|>r|ϕjk

(ξ) − ‘ϕjm(ξ)|2dξ

≤∫

|ξ|<r|ϕjk

(ξ) − ‘ϕjm(ξ)|2dξ

+1

(1 + r2)m/2

∫

Rn(1 + |ξ|2)m/2|ϕjk

(ξ) − ‘ϕjm(ξ)|2dξ

=∫

|ξ|<r|ϕjk

(ξ) − ‘ϕjm(ξ)|2dξ + (1 + r2)−m/2 ‖ϕjk− ϕjm‖2

Wm/22

:= I1 + I2.

The first term I1 tends to 0 as k,m → ∞ by the dominated convergence theorem ofLebesgue for any fixed r > 0. The second term converges to 0 as r → ∞ because‖ϕjk

− ϕjm‖Wm/22

≤ 2

Lemma gives us that(Aµ)−1

F : L2(Ω) → L2(Ω)

is a compact operator. Applying Riesz-Schauder and Hilbert-Schmidt theorems we get

1) σ((Aµ)−1F ) = 0, µ1, µ2, . . . with µj ≥ µj+1 > 0 and µj → 0 as j → ∞.

59

2) µj is of finite multiplicity

3) (Aµ)−1F ψj = µjψj, where ψj∞j=1 is an orthonormal system

4) ψj∞j=1 forms an orthonormal basis in L2(Ω).

Since AFψj = λjψj with λj = 1µj

− µ then we may conclude that

σ(AF ) = λj∞j=1, λj ≤ λj+1, λj → ∞.

Moreover, λj has finite multiplicity and ψj are the corresponding eigenfunctions. Wehave also the following representation

(Aµ)−1F f =

∞∑

j=1

µj(f, ψj)ψj, f ∈ L2(Ω).

Exercise 37. Prove that

AFf =∞∑

j=1

λj(f, ψj)ψj

for any f ∈ D(AF ).

Now we may conclude that the corollary is proved.

60

7 Spectral function

Let us consider a bounded domain Ω ⊂ Rn and an elliptic differential operator A(x,D)

in Ω of the formA(x,D) =

∑

|α|=|β|≤m/2

Dα(aαβ(x)Dβ),

where aαβ = aβα are real, aαβ ∈ C∞(Ω) and bounded for all α and β. We assume that

(−1)m/2∑

|α|=|β|=m/2

aαβ(x)ξαξβ ≥ ν|ξ|m, ν > 0.

As it was proved above there exists at least one self-adjoint extension of A with D(A) =C∞

0 (Ω), namely, the Friedrichs extension AF with

D(AF ) =

Wm/22 (Ω) ∩Wm

2 (Ω).

Let us consider an arbitrary self-adjoint extension “A of A. Without loss of generalitywe assume that “A ≥ 0. That’s why “A has the spectral representation

“A =∫ ∞

0λdEλ

with domain

DÄ “A

ä=

ßf ∈ L2(Ω) :

∫ ∞

0λ2d(Eλf, f) <∞

™.

In general case we have no such formula for DÄ “A

äas for the Friedrichs extension AF .

But we can say that

Wm2 (Ω) ⊂ D

Ä “Aä.

Indeed, since aαβ(x) ∈ C∞(Ω) and bounded then A(x,D) can be rewritten in the usualform

A(x,D) =∑

|γ|≤m

aγ(x)Dγ

with bounded coefficients. Hence

‖Af‖L2(Ω) ≤ c∑

|γ|≤m

‖Dγf‖L2(Ω) ≡ c ‖f‖W m2 (Ω) .

This proves the embedding.

Theorem 1 (Garding). If “A = “A∗then Eλ is an integral operator in L2(Ω) such that

Eλf(x) =∫

Ωθ(x, y, λ)f(y)dy,

where θ(x, y, λ) is called the spectral function and has the properties

1) θ(x, y, λ) = θ(y, x, λ)

61

2)

θ(x, y, λ) =∫

Ωθ(x, z, λ)θ(z, y, λ)dz

andθ(x, x, λ) =

∫

Ω|θ(x, z, λ)|2dz ≥ 0

3)supx∈Ω1

‖θ(x, ·, λ)‖L2(Ω) ≤ c1λk,

where Ω1 = Ω1 ⊂ Ω, k ∈ N with k > n2m

and c1 = c(Ω1).

Remark. It was proved by L. Hormander that actually

θ(x, x, λ) ≤ c1λn/m.

Corollary. Let z ∈ ρÄ “A

ä. Then (“A − zI)−1 is an integral operator whose kernel

G(x, y, z) is called the Green’s function corresponding to “A and which has the properties

1)

G(x, y, z) =∫ ∞

0

dλθ(x, y, λ)

λ− z

2) G(x, y, z) = G(y, x, z).

Proof. Since z ∈ ρÄ “A

äthen J. von Neumann’s spectral theorem gives us

(“A− zI)−1f =∫ ∞

0(λ− z)−1dEλf.

Next, by Theorem 1 we get

(“A− zI)−1f =∫ ∞

0(λ− z)−1dλ

Å∫Ωθ(x, y, λ)f(y)dy

ã

=∫

Ω

Å∫ ∞

0(λ− z)−1dλθ(x, y, λ)

ãf(y)dy

=∫

ΩG(x, y, z)f(y)dy,

where G(x, y, z) is as in 1). Since

G(x, y, z) =∫ ∞

0

dθ(x, y, λ)

λ− z=∫ ∞

0

dθ(y, x, λ)

λ− z= G(y, x, z)

then 2) is also proved.

62

Exercise 38. Prove that θ(x, x, λ) is a monotone increasing function with respect toλ and

1) |θ(x, y, λ)|2 ≤ θ(x, x, λ)θ(y, y, λ)

2) |Eλf(x)| ≤ θ(x, x, λ)1/2 ‖f‖L2(Ω).

Exercise 39. Prove that

|Eλf(x) − Eµf(x)| ≤ ‖Eλf − Eµf‖L2(Ω) |θ(x, x, λ) − θ(x, x, µ)|1/2

for any λ > 0 and µ > 0.

Exercise 40. Let us assume that n < m. Prove that

G(x, y, z) =∫ ∞

0

θ(x, y, λ)dλ

(λ− z)2

and that G(·, y, z) ∈ L2(Ω).

63

8 Fundamental solution

Let us consider an elliptic differential operator of even order m in general form

A(x,D) =∑

|α|≤m

aα(x)Dα,

where aα(x) ∈ C∞(Ω) and bounded, aα(x) real for |α| = m and

a(x, ξ) =∑

|α|=m

aα(x)ξα > 0

for any x ∈ Ω and ξ ∈ Rn \ 0. Let us define the set

Zθ := z ∈ C : | arg z| < θ

for some fixed 0 < θ < π/2.

Lemma 1. There is a constant c0 > 0 such that, for any z /∈ Zθ,

a(x, ξ) + 1 + |z| ≥ |a(x, ξ) + 1 − z| ≥ c0 (a(x, ξ) + 1 + |z|) ,

where x ∈ Ω and ξ ∈ Rn.

Proof. Let z = λ + iµ /∈ Zθ. Abbreviate a := a(x, ξ) ≥ 0. The first (leftmost)inequality follows immediately from triangle inequality. Hence it remains to prove thelatter (rightmost) inequality. To that end, we start from

|a+ 1 − z|2 = (a+ 1)2 − 2λ(a+ 1) + λ2 + µ2.

If λ ≤ 0 then

(a+ 1)2 − 2λ(a+ 1) + λ2 + µ2 = (a+ 1)2 + 2|λ|(a+ 1) + λ2 + µ2

≥ (a+ 1)2 + λ2 + µ2 = (a+ 1)2 + |z|2

≥ 1

2(a+ 1 + |z|)2 .

Consider now λ > 0. Since z /∈ Zθ then |µ| ≥ |h| = λ tan θ, see Figure 1. It meansthat µ2 ≥ γλ2, where γ = tan2 θ > 0. Hence

(a+ 1)2 − 2λ(a+ 1) + λ2 + µ2 ≥ (a+ 1)2 − ε(a+ 1)2 − 1

ελ2 + λ2 + µ2

= (1 − ε)(a+ 1)2 +

Ç1 − 1

ε

åλ2 + δµ2 + (1 − δ)µ2

≥ (1 − ε)(a+ 1)2 +

Ç1 − 1

ε

åλ2 + γδλ2 + (1 − δ)µ2

= (1 − ε)(a+ 1)2 +

Ç1 − 1

ε+ γδ

åλ2 + (1 − δ)µ2

64

θ

−θ

bz

bµ

b

λ

bh b

Zθ

Figure 1: The geometry of Lemma 1.

for any ε > 0 and δ ∈ R. Let us choose δ = 12

and ε such that

2

2 + γ< ε < 1.

Then for

2c20 := min

Ç1

2, 1 − ε, 1 − 1

ε+γ

2

å> 0

we have that

|a+ 1 − z|2 ≥ 2c20Ä(a+ 1)2 + λ2 + µ2

ä= 2c20

Ä(a+ 1)2 + |z|2

ä≥ c20 (a+ 1 + |z|)2 .

This proves the lemma.

Let ϕk(x, ξ, z), k = 0, 1, . . . be the sequence defined by

ϕ0(x, ξ, z) =1

a(x, ξ) + 1 − z,

ϕk(x, ξ, z) =1

a(x, ξ) + 1 − z[a(x, ξ) + 1 − A(x,D + ξ)]ϕk−1, k = 1, 2, . . . ,

whereA(x,D + ξ) =

∑

|α|≤m

aα(x)(D + ξ)α

with x ∈ Ω, ξ ∈ Rn and z /∈ Zθ. Due to Lemma 1 this sequence is well-defined.

Exercise 41. Prove that

(D + ξ)αf =∑

β≤α

Cβαξ

α−βDβf.

65

Exercise 42. Prove that

A(x,D)(fg) =∑

α≥0

Dαf

α!A(α)(x,D)g,

where A(α) has the symbol A(α)(x, ξ) = ∂αξ A(x, ξ).

Hint. Check these first for f = ei(x,η) and g = ei(x,ξ).

Lemma 2. Every ϕk, k = 0, 1, . . . is a finite sum of terms of the form

b(x, ξ)

(a+ 1 − z)ν+1, ν = 0, 1, . . . ,

where b(x, ξ) is a homogeneous polynomial of order l with respect to ξ and mν− l ≥ k.

Proof. We prove by induction with respect to k.For k = 0 the claim is immediate because

ϕ0(x, ξ, z) =1

a(x, ξ) + 1 − z

allows us to take ν = 0, l = 0 so that m · 0 − 0 ≥ 0.Assume that the claim holds for 0 ≤ k ≤ j i.e.

ϕk =∑

ν

b(x, ξ)

(a+ 1 − z)ν+1

and mν − l ≥ k. Concerning ϕj+1 we have that

ϕj+1 =1

a+ 1 − z[a+ 1 − A(x,D + ξ)]ϕj

=(a+ 1)ϕj

a+ 1 − z− 1

a+ 1 − z

∑

|α|≤m

aα(x)∑

β≤α

Cβαξ

α−βDβϕj

=(a+ 1)ϕj

a+ 1 − z− 1

a+ 1 − z

∑

|β|≤m

∑

|β+γ|≤m

aβ+γ(x)Cββ+γξ

γDβϕj

:= I1 + I2,

where I1 = (a+1)ϕj

a+1−z. The term in I2 that corresponds to β = 0 and |γ| = m is

− 1

a+ 1 − z

∑

|γ|=m

aγ(x)ξγϕj = − aϕj

a+ 1 − z.

Hence I1 + I2 can be rewritten as

I1 + I2 =(a+ 1)ϕj

a+ 1 − z− aϕj

a+ 1 − z− 1

a+ 1 − z

∑

|β|≤m

β>0

∑

|β+γ|≤m

aβ+γ(x)Cββ+γξ

γDβϕj

= I ′1 + I ′2,

66

whereI ′1 =

ϕj

a+ 1 − z.

Since

ϕj =∑

ν

b(x, ξ)

(a+ 1 − z)ν+1

then

I ′1 =∑

ν

b(x, ξ)

(a+ 1 − z)ν+2

and mν − l ≥ j. This implies that mν +m− l ≥ j +m > j + 1, since m ≥ 2. In otherwords,

m(ν + 1) − l ≥ j + 1

and the lemma is proved for I ′1. Let us consider now Dβϕj from I ′2 i.e.

Dβϕj = Dβ∑

ν

b(x, ξ)

(a+ 1 − z)ν+1=∑

ν

Dβx

b(x, ξ)

(a+ 1 − z)ν+1, mν − l ≥ j.

We prove that differentiation of

b(x, ξ)

(a+ 1 − z)ν+1

with respect to x leads to∑

ν

b

(a+ 1 − z)ν+1,

where mν − l = mν − l ≥ j i.e. the value of mν − l does not change. Indeed,

∂jb

(a+ 1 − z)ν+1=

b′j(a+ 1 − z)ν+1

− (ν + 1)ba′j(a+ 1 − z)ν+2

,

where b′j and ba′j are homogeneous polynomials with respect to ξ of orders l and

l = l +m, respectively. That’s why

mν − l = m(ν + 1) − (m+ l) = mν − l.

This fact holds for any derivative Dβ, β > 0. Now we may conclude that

I ′2 = − 1

a+ 1 − z

∑

ν

∑

|β|≤m

β>0

∑

|β+γ|≤m

Cβ+γ(x)ξγ b

(a+ 1 − z)ν+1,

where |γ| ≤ m − 1 and mν − l = mν − l ≥ j. But multiplication by ξγ, |γ| ≤ m − 1leads us finally to the sum of the terms

1

(a+ 1 − z)

b

(a+ 1 − z)ν+1ξγ =

bξγ

(a+ 1 − z)ν+2=

‹b1(a+ 1 − z)ν+2

,

67

where ‹l1 = l + |γ| and

m(ν + 1) − ‹l1 = mν +m− l − |γ| = mν − l +m− |γ| ≥ mν − l + 1 ≥ j + 1.

This finishes the proof.

Definition. A locally integrable function F (·, y, z) with parameters y ∈ Ω and z /∈ Zθ

is called a fundamental solution of the operator A(·, D) − zI in Ω if

(A(·, D) − zI)F (·, y, z) = δ(· − y)

in the sense of distributions i.e.

〈(A− zI)F (·, y, z), ϕ(·)〉 = ϕ(y)

for all ϕ ∈ C∞0 (Ω) or

∫

ΩF (x, y, z)(A∗ − zI)ϕ(x)dx = ϕ(y),

where A∗(x,D) is formally adjoint to A(x,D) in L2(Ω).

Let us define a new function

Fk(x, y, z) := F−1

Ñk∑

j=0

ϕj(x, ·, z)é

(x− y), k = 0, 1, 2, . . . .

Here F−1 is the inverse Fourier transform of tempered distributions i.e.

〈F−1u, ϕ〉 := 〈u, F−1ϕ〉,

where ϕ ∈ C∞(Rn) is such that all of its derivatives decay faster than the reciprocalof any polynomial at infinity and

(F−1f)(x) = (2π)−n∫

Rnei(x,y)f(y)dy

for f ∈ L1(Rn). Then the following lemma holds.

Lemma 3. In the sense of distributions, the function Fk(x, y, z) satisfies

(A(x,D) − zI)Fk(x, y, z) = δ(x− y) −Hk(x, y, z),

where Hk(x, y, z) = F−1((a(x, ·) + 1 − z)ϕk+1).

Proof. Since

Dαx (ϕj(x, ξ, z)e

i(x−y,ξ)) =∑

β≤α

Cβα

ÄDβ

xϕj

ä ÄDα−β

x ei(x−y,ξ)ä

=

Ñ∑

β≤α

Cβαξ

α−βDβxϕj(x, ξ, z)

éei(x−y,ξ)

68

then

(A(x,D) − zI)Fk(x, y, z) =k∑

j=0

F−1 ((A(x,D + ·) − zI)ϕj(x, ·, z))

=k∑

j=0

F−1 ((a+ 1 − z)ϕj − (a+ 1 − A(x,D + ·))ϕj)

=k∑

j=0

F−1 ((a+ 1 − z)ϕj − (a+ 1 − z)ϕj+1)

= F−1 ((a+ 1 − z)ϕ0 − (a+ 1 − z)ϕk+1)

= F−1(1) − F−1 ((a+ 1 − z)ϕk+1)

= δ(x− y) −Hk(x, y, z).

This proves the lemma.

Remark. The following representation holds:

Hk(x, y, z) =∑

ν

F−1

Çbν(x, ·)

(a(x, ·) + 1 − z)ν

å,

where ν = 1, 2, . . . and mν − lν ≥ k + 1.

Next, we look for the fundamental solution F (x, y, z) of A(x,D) − zI in the form

F (x, y, z) = Fk(x, y, z) +∫

Ω1

Fk(x, u, z)hk(u, y, z)du,

where x, y ∈ Ω1 = Ω1 ⊂ Ω, z /∈ Zθ and hk(u, y, z) is an unknown function which is tobe determined. Since

(A− zI)Fk = δ −Hk

then

(A− zI)F = δ −Hk +∫

Ω1

(δ −Hk)hkdu = δ −Hk + hk −∫

Ω1

Hkhkdu.

Hence F is the fundamental solution if and only if hk satisfies the equation

hk(x, y, z) = Hk(x, y, z) +∫

Ω1

Hk(x, u, z)hk(u, y, z)du.

We plan to solve this equation by iterations. In order to do it and obtain estimates ofF (x, y, z) it will be necessary to estimate

Fk =k∑

j=0

F−1ϕj =∑

j

∑

νj

F−1

Çbνj

(a+ 1 − z)νj+1

å,

69

where νj = 0, 1, . . . and mνj − lj ≥ j and

Hk =∑

ν

F−1

Çbν

(a+ 1 − z)ν

å,

where ν = 1, 2, . . . and mν − l ≥ k + 1. Let us introduce the function

ψ(x, u) := F−1

Çb(x, ·)

(a(x, ·) + 1)ν

å(u),

where mν− l > 0. This definition is understood in the sense of tempered distributionsi.e.

〈ψ(x, ·), ϕ〉 =

Æb

(a+ 1)ν, F−1ϕ

∏=∫

Rn

b(x, ξ)

(a(x, ξ) + 1)νF−1ϕ(ξ)dξ.

Lemma 4 (Main lemma). Suppose that |u| = 1. For every Ω1 = Ω1 ⊂ Ω there is δ > 0such that for any λ > 0 we have

1) mν − l < n,|ψ(x, λu)| ≤ c1λ

mν−l−ne−δλ;

2) mν − l = n,|ψ(x, λu)| ≤ c1(1 + | log λ|)e−δλ;

3) mν − l > n,|ψ(x, λu)| ≤ c1e

−δλ,

where x ∈ Ω1 and c1 = c(Ω1).

Proof. Let us denote by c1 any constant that depends on Ω1.Since a(x, ξ) > 0 for all x ∈ Ω and ξ ∈ R

n \ 0 then for any Ω1 = Ω1 ⊂ Ω there isa constant c1 > 0 such that

a(x, ξ) ≥ c1|ξ|m, x ∈ Ω1, ξ ∈ Rn.

This way we can consider the function

ψ0(x, u) := F−1

Çb(x, ·)

(a(x, ·))ν

å(u),

where x ∈ Ω1 and u ∈ Rn. It is very easy to check that

ψ0(x, λu) = λmν−l−nψ0(x, u).

70

1) Let mν − l < n. Let us prove first that

|ψ0(x, u)| ≤ c1

for |u| = 1 and x ∈ Ω1. Take a function χ(ξ) ∈ C∞0 (R) such that χ(ξ) ≡ 1 for |ξ| ≤ 1.

Then

ψ0(x, u) = F−1

Çb

aνχ

å+ F−1

Çb

aν(1 − χ)

å:= ›ψ0 +

››ψ0.

Since mν − l < n then baν ∈ L1

loc(Rn). That’s why

›ψ0(x, u) = F−1

Çb

aνχ

å= (2π)−n

∫

|ξ|<L

b(x, ξ)

a(x, ξ)νχ(ξ)ei(ξ,u)dξ,

where suppχ ⊂ |ξ| < L. Hence

|›ψ0(x, u)| ≤ c1

∫

|ξ|<L|ξ|l−mνdξ = c′1.

For››ψ0(x, u) we have the identity

uα››ψ0(x, u) = F−1

ÇDα

ξ

ñb

aν(1 − χ)

ôå.

Next, we prove that∣∣∣∣∣D

αξ

ñb

aν(1 − χ(ξ))

ô∣∣∣∣∣ ≤c1

(1 + |ξ|)mν−l+|α|.

Indeed,

Dξj

ñb

aν(1 − χ(ξ))

ô=b′ξj

aν(1 − χ) −

νba′ξj

aν+1(1 − χ) − b

aνχ′ := I1 + I2 + I3.

It is clear that I3 ≡ 0 for |ξ| < 1 and |ξ| > L. Hence |I3| ≤ c1 and has compactsupport. For |ξ| > 1 the term I1 satisfies

|I1| ≤ c1|ξ|l−1

|ξ|mν= c1|ξ|l−mν−1.

The same estimate holds for I2 too. Since I1 ≡ 0 and I2 ≡ 0 for |ξ| < 1 then combiningthese estimates we obtain

∣∣∣∣∣Dξj

ñb

aν(1 − χ(ξ))

ô∣∣∣∣∣ ≤c1

(1 + |ξ|)mν−l+1.

The required estimate follows now by induction. If we choose now α such that |α| >n+ l −mν i.e. mν − l + |α| > n then

∣∣∣∣uα››ψ0(x, u)

∣∣∣∣ ≤ c1

∫

Rn

dξ

(1 + |ξ|)mν−l+|α|= c′1.

71

Since we can choose coordinates so that u = (|u|, 0, . . . , 0) we obtain

∣∣∣∣››ψ0(x, u)

∣∣∣∣ ≤ c1, x ∈ Ω1, |u| = 1.

The same arguments as above lead us also to

|ψ(x, u)| ≤ c1, x ∈ Ω1, |u| = 1.

Let us consider now the main symbol a(x, ζ) for complex ζ. Let ζ = ξ + iη. Then

a(x, ζ) =∑

|α|=m

aα(x)(ξ + iη)α =∑

|α|=m

aα(x)∑

β≤α

Cβαξ

β(iη)α−β

=∑

|α|=m

aα(x)ξα +∑

|α|=m

aα(x)∑

β<α

Cβαξ

β(iη)α−β.

For |η| ≤ δ it follows that

|a(x, ζ) + 1| ≥ a(x, ξ) + 1 −∑

|α|=m

aα(x)∑

β<α

Cβα |ξ||β||η||α−β|

≥ a(x, ξ) + 1 − c1m−1∑

j=0

|ξ|j|η|m−j

≥ a(x, ξ) + 1 − c1m−1∑

j=0

ñε|ξ|m +

1

εj/(m−j)|η|m

ô

= a(x, ξ) + 1 − c1εm|ξ|m − cε|η|m≥ ‹c1|ξ|m − c1εm|ξ|m + 1 − cε|η|m= (‹c1 − εc1m)|ξ|m + 1 − cε|η|m ≥ c0(|ξ|m + 1),

wherec0 = min ‹c1 − εc1m, 1 − cεδ

m > 0

and the parameters are chosen as

‹c1 − εc1m > 0, 1 − cεδm > 0

i.e.

ε <‹c1c1m

, δ <

Ç1

cε

å 1m

.

We proved that there exists δ > 0 such that

|a(x, ξ + iη) + 1| ≥ c0(|ξ|m + 1), c0 > 0

for all ξ, η ∈ Rn, |η| ≤ δ and x ∈ Ω1. This inequality allows us to extend the function

Dαξ

ñb(x, ξ)

(a(x, ξ) + 1)ν

ôei(λu,ξ)

72

as an analytic function with respect to ζ = ξ + iη for ξ ∈ Rn and |η| ≤ δ. Moreover,

this inequality leads to the estimate∣∣∣∣∣D

αξ

ñb(x, ξ)

(a(x, ξ) + 1)ν

ô∣∣∣∣∣ ≤c1

(1 + |ξ|)mν−l+|α|

for all x ∈ Ω1, ξ ∈ Rn and |η| ≤ δ. Let us consider fixed η ∈ R

n with |η| ≤ δ. Cauchytheorem for |α| > l −mν + n shows us that

(λu)αψ(x, λu) = (2π)−n∫

RnDα

ξ

b(x, ξ)

(a(x, ξ) + 1)νei(λu,ξ)dξ

= (2π)−n∫

RnDα

ξ

b(x, ξ + iη)

(a(x, ξ + iη) + 1)νei(λu,ξ+iη)dξ.

Hence

|(λu)αψ(x, λu)| ≤ c1

∫

Rn

1

(1 + |ξ|)mν−l+|α|e−λ(u,η)dξ ≤ c′1e

−λ(u,η).

Since |uα| = |u1|α1 · · · |un|αn > 0 for |u| = 1 then |uα| ≥ min|u|=1 |uα| = |uα0 | := c0,

where |u0| = 1 and c0 > 0. Therefore the latter inequality can be transformed to

|ψ(x, λu)| ≤ c′′1λ−|α|e−λ(u,η).

If we choose η = δu then we obtain

|ψ(x, λu)| ≤ c1λ−|α|e−δλ, λ > 0, |u| = 1, x ∈ Ω1,

where |α| > l −mν + n. For |α| = l −mν + n+ 1 > 0 and for λ ≥ 1 we can obtain

|ψ(x, λu)| ≤ c1λmν−l−n−1e−δλ ≤ c1λ

mν−l−ne−δλ.

Consider now 0 < λ < 1. It is easily seen that

(λu)α(ψ(x, λu) − ψ0(x, λu)) = F−1

ÇDα

ξ

ñbν

(a+ 1)ν− bνaν

ôå(λu)

= −F−1

ÇDα

ξ

ñ(a+ 1)ν − aν

(a2 + a)νbν

ôå(λu)

= −F−1

(Dα

ξ

[∑ν−1j=0 c

jνa

jbν

(a2 + a)ν

])(λu)

= −F−1

ÑDα

ξ

∑ν−1

j=0‹bj

(a2 + a)ν

é

(λu),

where ‹bj is a homogeneous polynomial of order ‹lj = l +mj. Let us denote

I(x, ξ) :=

∑ν−1j=0

‹bj(a2 + a)ν

.

73

It is clear that I(x, ξ) behaves as

I ≍ |ξ|l−mν , |ξ| → 0

andI ≍ |ξ|l−mν−m, |ξ| → ∞.

Combining these two asymptotics yields

I ≍ 1

|ξ|mν−l

1

(1 + |ξ|)m

for all ξ ∈ Rn \ 0. By the same arguments we have

|DαI| ≍ 1

|ξ|mν−l+|α|

1

(1 + |ξ|)m.

Let us choose now |α| = l −mν + n− 1 ≥ 0. Then

|DαI| ≍ 1

|ξ|n−1

1

(1 + |ξ|)m.

This allows us to get DαI ∈ L1(Rn) and, therefore,

|(λu)α(ψ(x, λu) − ψ0(x, λu))| ≤ c1

∫

Rn

1

|ξ|n−1(1 + |ξ|)m≤ c′1.

Hence|ψ(x, λu) − ψ0(x, λu)| ≤ c1λ

mν−l−n+1

for |u| = 1. It implies that

|ψ(x, λu)| ≤ c1λmν−l−n+1 + |ψ0(x, λu)| ≤ c1λ

mν−l−n + λmν−l−n|ψ0(x, u)|≤ c1λ

mν−l−n ≤ c′1λmν−l−ne−δλ, 0 < λ < 1.

Thus, 1) is proved for all λ > 0.

2) Let mν − l = n. Taking the derivative with respect to λ we obtain

∂

∂λψ(x, λu) = F−1

Çi(u, ξ)b(x, ξ)

(a(x, ξ) + 1)ν

å(λu).

It is equivalent to

∂

∂λψ(x, λu) = F−1

(b(x, u, ξ)

(a(x, ξ) + 1)ν

)(λu) := ‹ψ(x, λu),

where mν − l = mν − l− 1 = n− 1 < n. That’s why we can apply part 1) to ‹ψ(x, λu)and obtain ∣∣∣∣∣

∂

∂λψ(x, λu)

∣∣∣∣∣ ≤ c1λ−1e−δλ.

74

Next, consider two cases: 0 < λ < 1 and λ ≥ 1. In the first case

ψ(x, λu) = −∫ 1

λ

∂

∂τψ(x, τu)dτ + ψ(x, u).

Since ψ(x, u) is bounded for |u| = 1 then

|ψ(x, λu)| ≤ c1

∫ 1

λ

1

τe−δτdτ + c1 ≤ c1

Ç∫ 1

λ

1

τdτ + 1

å= c1(1 + | log λ|).

In the second case we begin with

ψ(x, λu) = −∫ ∞

λ

∂

∂τψ(x, τu)dτ

because ψ(x, λu) → 0 as λ→ ∞. Hence

|ψ(x, λu)| ≤ c1

∫ ∞

λ

1

λe−δτdτ ≤ c1

∫ ∞

λe−δτdτ = c′1e

−δλ

for λ ≥ 1. That’s why

|ψ(x, λu)| ≤ c1(1 + | log λ|)e−δλ, λ > 0.

This proves 2).

3) Since mν − l > n then the claim follows from above considerations immediatelywithout differentiation.

Lemma 5. Let z /∈ Zθ. Define the function

Iν(x, y, z) = F−1

Çb(x, ξ)

(a(x, ξ) + 1 − z)ν

å(x− y),

where b(x, ξ) and a(x, ξ) are as in lemma 4, x ∈ Ω1 and y ∈ Ω. Then there existsδ > 0 such that

1) mν − l < n,

|Iν(x, y, z)| ≤ c1|x− y|mν−l−ne−δ|x−y|(1+|z|)1m ;

2) mν − l = n,

|Iν(x, y, z)| ≤ c1(1 +

∣∣∣log(|x− y|(1 + |z|) 1

m

)∣∣∣)e−δ|x−y|(1+|z|)

1m ;

75

3) mν − l > n,

|Iν(x, y, z)| ≤ c1(1 + |z|)−mν−l−nm e−δ|x−y|(1+|z|)

1m .

Proof. We know from Lemma 1 that

a(x, ξ) + 1 − z ≍ a(x, ξ) + 1 + |z|.

This fact allows us to consider, without loss of generality, the function

Iν(x, y, z) = F−1

Çb(x, ξ)

(a(x, ξ) + 1 + |z|)ν

å(x− y).

But it is not so difficult to check that

Iν(x, y, z) = (1 + |z|)n−mν+lm ψ(x, λu),

where λ = |x− y|(1 + |z|) 1m and u = x−y

|x−y|. Applying Lemma 4 implies now 1)-3).

Corollary. Let α > 0, z /∈ Zθ, x ∈ Ω1 and y ∈ Ω. Then

1) mν − l < n+ |α|,

|DαxIν(x, y, z)| ≤ c1|x− y|mν−l−|α|−ne−δ|x−y|(1+|z|)

1m ;

2) mν − l = n+ |α|,

|DαxIν(x, y, z)| ≤ c1

(1 +

∣∣∣log(|x− y|(1 + |z|) 1

m

)∣∣∣)e−δ|x−y|(1+|z|)

1m ;

3) mν − l > n+ |α|,

|DαxIν(x, y, z)| ≤ c1(1 + |z|)n+l+|α|−mν

m e−δ|x−y|(1+|z|)1m .

Exercise 43. Prove Corollary.

Now we are in the position to prove the main theorem.

Theorem 1. For any Ω1 = Ω1 ⊂ Ω there is R0 > 0 such that for all k ≥ n, n ∈ N andfor all |z| > R0, z /∈ Zθ there exists a fundamantal solution F (x, y, z), x, y ∈ Ω1 of anoperator A(x,D) − zI. This fundamental solution has the form

F (x, y, z) = Fk(x, y, z) +∫

Ω1

Fk(x, u, z)hk(u, y, z)du,

where Fk =∑k

j=0 F−1ϕj, hk(x, y, z) exists and satisfies the estimate

|hk(x, y, z)| ≤ c1(1 + |z|)n−k−1m e−δ|x−y|(1+|z|)

1m .

76

Proof. As we already know F (x, y, z) of such form is a fundamental solution of theoperator A(x,D) − zI if and only if the function hk(x, y, z) solves the equation

hk(x, y, z) = Hk(x, y, z) +∫

Ω1

Hk(x, u, z)hk(u, y, z)du,

whereHk(x, y, z) = F−1 ((a+ 1 − z)ϕk+1) (x− y)

with

(a(x, ξ) + 1 − z)ϕk+1(x, ξ, z) =∑

ν

bν(x, ξ)

(a(x, ξ) + 1 − z)ν

andmν − lν ≥ k + 1.

Since k ≥ n then mν − lν ≥ n+ 1 > n and, therefore, we can apply part 3) of Lemma5 and obtain

|Hk(x, y, z)| ≤ c1∑

ν

(1 + |z|)n−(mν−lν )m e−δ|x−y|(1+|z|)

1m ≤ c1(1 + |z|)n−k−1

m e−δ|x−y|(1+|z|)1m .

Next, we look for hk(x, y, z) as the series of iterations of Hk i.e.

hk(x, y, z) :=∞∑

j=1

H(j)k (x, y, z),

where H(1)k = Hk and

H(j)k (x, y, z) =

∫

Ω1

Hk(x, u, z)H(j−1)k (u, y, z)du, j ≥ 2.

It is clear that hk(x, y, z) of such form is a (formal) solution of the correspondingintegral equation for hk. It remains merely to prove that for |z| >> 1, z /∈ Zθ this seriesconverges uniformly with respect to x and y from Ω1. To this end, let us consider theestimates for iterations of Hk. Indeed,

|H(2)k (x, y, z)| ≤

∫

Ω1

|Hk(x, u, z)||H(1)k (u, y, z)|du

≤ c21((1 + |z|)n−k−1

m

)2∫

Ω1

e−δ(|x−u|+|u−y|)(1+|z|)1m du

≤ c21((1 + |z|)n−k−1

m

)2|Ω1|e−δ|x−y|(1+|z|)

1m

since |x− u| + |u− y| ≥ |x− y|. Here and later |Ω1| denotes the Lebesgue measure ofΩ1. By induction we get

|H(j)k (x, y, z)| ≤

(c1(1 + |z|)n−k−1

m

)j|Ω1|j−1e−δ|x−y|(1+|z|)

1m .

77

Hence

∞∑

j=1

|H(j)k | ≤ c1(1 + |z|)n−k−1

m e−δ|x−y|(1+|z|)1m

∞∑

j=1

(c1|Ω1|(1 + |z|)n−k−1

m

)j−1.

If we choose z so thatc1|Ω1|(1 + |z|)n−k−1

m < 1

or|z| > R0 := (c1|Ω1|)

mk+1−n − 1

then the series converges uniformly with respect to x, y ∈ Ω1. Therefore hk(x, y, z) iswell-defined and F (x, y, z) exists.

Corollary 1. The fundamental solution F (x, y, z) satisfies the following estimates:

1) m < n,

|F (x, y, z)| ≤ c1|x− y|m−ne−δ|x−y|(1+|z|)1m ;

2) m = n,

|F (x, y, z)| ≤ c1(1 +

∣∣∣log(|x− y|(1 + |z|) 1

m

)∣∣∣)e−δ|x−y|(1+|z|)

1m ;

3) m > n,

|F (x, y, z)| ≤ c1(1 + |z|)n−mm e−δ|x−y|(1+|z|)

1m ,

where x, y ∈ Ω1 and |z| > R0, z /∈ Zθ.

Proof. Since

Fk(x, y, z) =k∑

j=0

F−1ϕj(x− y) =k∑

j=0

∑

νj

F−1

Çbνj

(a+ 1 − z)νj+1

å(x− y),

where mνj − lj ≥ j then we can apply Lemma 5 with ν = νj + 1 and l = lj forj = 0, 1, . . . , k. That’s why we can obtain

|Fk| ≤ c1e−δ|x−y|(1+|z|)

1m

k∑

j=0

∑

νj

|x− y|mνj+m−lj−n, mνj +m− lj < n

1 +∣∣∣log

(|x− y|(1 + |z|) 1

m

)∣∣∣ , mνj +m− lj = n

(1 + |z|)lj+n−m−mνj

m , mνj +m− lj > n.

Let us remember that ν0 = l0 = 0. That’s why the first term of this sum can beestimated as

|I0| ≤ c1e−δ|x−y|(1+|z|)

1m

|x− y|m−n, m < n

1 +∣∣∣log

(|x− y|(1 + |z|) 1

m

)∣∣∣ , m = n

(1 + |z|)n−mm , m > n.

78

But mνj − lj ≥ j means that the value mνj − lj grows with respect to j. That’s whythe next terms in this sum have better estimates than I0. This remark allows us toconclude that

|Fk(x, y, z)| ≤ c1e−δ|x−y|(1+|z|)

1m

|x− y|m−n, m < n

1 +∣∣∣log

(|x− y|(1 + |z|) 1

m

)∣∣∣ , m = n

(1 + |z|)n−mm , m > n

for k = 0, 1, . . .. In order to obtain estimates for F (x, y, z) it remains to investigate itssecond term, namely ∫

Ω1

Fk(x, u, z)hk(u, y, z)du.

Applying the estimates for Fk and hk we obtain∣∣∣∣∫

Ω1

Fk(x, u, z)hk(u, y, z)du∣∣∣∣ ≤ c21(1 + |z|)n−k−1

m

∫

Ω1

e−δ(|x−u|+|u−y|)(1+|z|)1m du

×

|x− u|m−n, m < n

1 +∣∣∣log

(|x− u|(1 + |z|) 1

m

)∣∣∣ , m = n

(1 + |z|)n−mm , m > n

≤ c21(1 + |z|)n−k−1m e−δ|x−y|(1+|z|)

1m

×

∫Ω1

|x− u|m−ndu, m < n∫Ω1

(1 +

∣∣∣log(|x− u|(1 + |z|) 1

m

)∣∣∣)du, m = n

(1 + |z|)n−mm |Ω1|, m > n.

It is clear that ∫

Ω1

|x− u|m−ndu ≤ c0

if m < n and(1 + |z|)n−m

m |Ω1| ≤ c′0

if m > n. If m = n then∫

Ω1

(1 +

∣∣∣log(|x− u|(1 + |z|) 1

m

)∣∣∣)du ≤

∫

Ω1

du+∫

Ω1

|log |x− u|| du

+ log(1 + |z|) 1m

∫

Ω1

du

≤ c1(1 + log(1 + |z|) 1

m

)

≤ c1,ε(1 + |z|) εm , ε > 0.

Since ε > 0 and small enough then finally∣∣∣∣∫

Ω1

Fk(x, u, z)hk(u, y, z)du∣∣∣∣ ≤ c1e

−δ|x−y|(1+|z|)1m .

This finishes the proof.

79

Corollary 2. Let 0 < |α| < m, z /∈ Zθ, |z| > R0 and x, y ∈ Ω1. Then

1) m < n+ |α|,

|DαxF (x, y, z)| ≤ c1|x− y|m−n−|α|e−δ|x−y|(1+|z|)

1m ;

2) m = n+ |α|,

|DαxF (x, y, z)| ≤ c1

(1 +

∣∣∣log(|x− y|(1 + |z|) 1

m

)∣∣∣)e−δ|x−y|(1+|z|)

1m ;

3) m > n+ |α|,

|DαxF (x, y, z)| ≤ c1(1 + |z|)n+|α|−m

m e−δ|x−y|(1+|z|)1m .

Exercise 44. Prove Corollary 2.

Exercise 45. Let |α| = m and |x− y| ≥ ε. Prove that

|DαxF (x, y, z)| ≤ cεe

−δε(1+|z|)1m .

Let “A be an arbitrary self-adjoint extension of an elliptic differential operatorA(x,D) of even order m in L2(Ω), where Ω ⊂ R

n is a bounded domain with smoothboundary. Without loss of generality we assume that “A ≥ 0. For z /∈ Zθ we know that(“A− zI)−1 exists and is an integral operator with kernel G(x, y, z) which is called the

Green’s function. We denote this inverse by ”Gz i.e.

”Gzf(x) =∫

ΩG(x, y, z)f(y)dy.

Thus(“A− zI)”Gz = ”Gz(“A− zI) = I

and ”Gz : L2(Ω) → L2(Ω) is a bounded operator. We know also that for any Ω1 = Ω1 ⊂Ω there is a fundamental solution F (x, y, z) for z /∈ Zθ and |z| > R0. Moreover, wehave the estimates for Dα

xF (x, y, z) for |α| ≤ m− 1, see Corollary for Theorem 1. Ourmain task now is to obtain estimates for G(x, y, z).

Definition. A function E(x, y, z), x, y ∈ Ω, z /∈ Zθ is called a parametrix for “A− zI ifthe integral operator

”Qz := ”Gz − ”Ez,

where ”Ez is the integral operator with kernel E(x, y, z), has the mapping property

”Qz : L1(Ω1) →∞⋂

j=1

D( “A j)

for any Ω1 = Ω1 ⊂ Ω.

80

Remark. It is easy to obtain the representation

(“A− zI)”Ez = I + ”Pz,

where ”Pz = −(“A− zI)”Qz or ”Qz = −”Gz”Pz.

If Ω1 = Ω1 ⊂ Ω then denote

ε0 = dist(Ω1, ∂Ω) > 0.

For any fixed ε such that 0 < ε < ε0 define the function χ(x) ∈ C∞0 (Ω) as

χ(x) =

1, x ∈ Ωε/21

0, x ∈ Ω \ Ωε1,

where dist(Ω1, ∂Ωε/21 ) = dist(Ω

ε/21 , ∂Ωε

1) = ε/2 with Ω1 ⊂ Ωε/21 ⊂ Ωε

1 ⊂ Ω anddist(Ωε

1, ∂Ω) > 0, see Figure 2.

Ω

Ωε1

Ωε/21

Ω1

χ = 0

χ = 1

Figure 2:

Let us consider a fundamental solution F (x, y, z) for x, y ∈ Ωε1. Define the function

E byE(x, y, z) := χ(x)F (x, y, z)

which is well-defined for all x ∈ Ω and y ∈ Ωε1. Let us assume that E(x, y, z) ≡ 0 for

x ∈ Ω and y ∈ Ω \ Ωε1. Thus, E(x, y, z) is defined for all x, y ∈ Ω. It is also clear

that E(x, y, z) = F (x, y, z) for x, y ∈ Ωε/21 . The function E(x, y, z) is called a smoothed

fundamental solution.

81

Theorem 2. The function E(x, y, z) defined above is parametrix for “A− zI. What ismore, ∥∥∥∥ “Ak ”Qzf

∥∥∥∥L2(Ω)

≤ ck,ε(1 + |z|)ke−δ ε2(1+|z|)

1m ‖f‖L1(Ω1)

for any k ∈ N.

Proof. Since E(x, y, z) ∈ C∞0 (Ω) for x 6= y then

(“A− zI)E(x, y, z) = (A(x,D) − zI)E(x, y, z) = χ(x)(A(x,D) − zI)F (x, y, z)

+∑

α>0

1

α!DαχA(α)(x,D)F (x, y, z) = δ(x− y) + P (x, y, z),

where x ∈ Ω, y ∈ Ω1 and P (x, y, z) is defined by the second term in the sum. It is

also easy to see that Dαχ 6= 0 only for x ∈ Ωε1 \ Ω

ε/21 . Since y ∈ Ω1 then we may

conclude that P (x, y, z) 6= 0 for all x ∈ Ωε1 \ Ω

ε/21 and y ∈ Ω1. This fact implies that

|x− y| ≥ ε/2. Next, since α > 0 then the order of the differential operator A(α)(x,D)is at most m− 1. Hence we can apply the corollaries of Theorem 1 and obtain

|P (x, y, z)| ≤ cεe−δ ε

2(1+|z|)

1m .

Moreover, we prove that

| “AkP (x, y, z)| ≤ cε,k(1 + |z|)ke−δ ε

2(1+|z|)

1m

for any k ∈ N. It suffices to prove this for k = 1 and then use induction. Indeed, since|x− y| ≥ ε/2 then F (x, y, z) is a regular solution of the equation

(A(x,D) − zI)F (x, y, z) = 0.

In other words ∑

|α|=m

aα(x)DαxF = zF −

∑

|α|≤m−1

aα(x)DαxF.

It follows from ellipticity that∑

|α|=m

aα(x)DαxF ≥ c1

∑

|α|=m

|DαF |.

Hence∑

|α|=m

|DαF | ≤ c|z||F | + c∑

|α|≤m−1

|DαF | ≤ cÅ|z|e−δ ε

2(1+|z|)

1m + e−δ ε

2(1+|z|)

1m

ã

= c(1 + |z|)e−δ ε2(1+|z|)

1m

by corollaries 1 and 2. Next, since |x− y| ≥ ε/2 then

“AP ≡ A(x,D)P (x, y, z) = A(x,D)∑

α>0

Dαχ

α!A(α)F

=∑

α>0

∑

β≥0

1

α!

1

β!Dα+βχA(β)(x,D)

ÄA(α)(x,D)F

ä=∑

γ>0

cγDγχBγ(x,D)F,

82

where Bγ(x,D) is an operator of order 2m − |γ| ≤ 2m − 1. If |γ| < m then thecorresponding terms in this sum can be estimated from above as

cεe−δ ε

2(1+|z|)

1m .

If |γ| ≥ m then by ellipticity we obtain again that

Bγ(x,D)F ≍ DβA(x,D)F

orBγ(x,D)F ≍ Dβ(zF ),

where |β| = |γ| −m ≤ m− 1. Hence, we obtain in that case

|Bγ(x,D)F | ≤ cε|z|e−δ ε2(1+|z|)

1m .

If we combine these two estimates then we obtain

|“AP (x, y, z)| ≤ cε(1 + |z|)e−δ ε2(1+|z|)

1m

and by induction

| “AkP (x, y, z)| ≤ cε,k(1 + |z|)ke−δ ε

2(1+|z|)

1m .

Next we write

“A”Qz = −“A”Gz”Pz = −

î “A− zI + zIó ”Gz

”Pz = −”Pz − z”Gz”Pz = −”Pz + z”Qz.

This implies that ∥∥∥“A”Qzf∥∥∥

L2≤∥∥∥”Pzf

∥∥∥L2

+ |z|∥∥∥”Qzf

∥∥∥L2.

But ”Qz = −”Gz”Pz and ”Gz is a bounded operator in L2(Ω). Hence

∥∥∥“A”Qzf∥∥∥

L2(Ω)≤∥∥∥”Pzf

∥∥∥L2(Ω)

+ c|z|∥∥∥”Pzf

∥∥∥L2(Ω)

.

So it remains to estimate∥∥∥”Pzf

∥∥∥L2(Ω)

. If f ∈ L1(Ω1) then

|”Pzf(x)| =∣∣∣∣∫

Ω1

P (x, y, z)f(y)dy∣∣∣∣ ≤

∫

Ω1

|P (x, y, z)||f(y)|dy

≤ cεe−δ ε

2(1+|z|)

1m ‖f‖L1(Ω1) .

It follows that

∥∥∥”Pzf∥∥∥

L2(Ω)≤ cεe

−δ ε2(1+|z|)

1m ‖f‖L1(Ω1)

Ç∫Ω1

dx

å1/2

= cε|Ω1|1/2e−δ ε2(1+|z|)

1m ‖f‖L1(Ω1) .

Applying the identity“Ak ”Qz = − “Ak−1 ”Pz + z “Ak−1 ”Qz

83

we obtain∥∥∥∥ “Ak ”Qzf

∥∥∥∥L2(Ω)

≤∥∥∥∥ “Ak−1 ”Pzf

∥∥∥∥L2(Ω)

+ |z|∥∥∥∥ “Ak−1 ”Qzf

∥∥∥∥L2(Ω)

≤ cε,k(1 + |z|)k−1e−δ ε2(1+|z|)

1m ‖f‖L1(Ω1)

+ cε,k|z|(1 + |z|)k−1e−δ ε2(1+|z|)

1m ‖f‖L1(Ω1)

≤ cε,k(1 + |z|)ke−δ ε2(1+|z|)

1m ‖f‖L1(Ω1)

.

This finishes the proof.

84

9 Fractional powers of self-adjoint operators

If “A = “A∗and “A > 0, say “A ≥ c0I, c0 > 0, then for z /∈ Zθ we know that (“A− zI)−1 ≡

”Gz is an integral operator with kernel G(x, y, z) which can be represented in the form

G(x, y, z) = E(x, y, z) +Q(x, y, z),

where E(x, y, z) is a smoothed fundamental solution satisfying the estimates

|DαxE(x, y, z)| ≤ ce−δ|x−y|(1+|z|)

1m

|x− y|m−|α|−n, m < n+ |α|1 +

∣∣∣log(|x− y|(1 + |z|) 1

m

)∣∣∣ , m = n+ |α|(1 + |z|)n+|α|−m

m , m > n+ |α|for |α| ≤ m− 1 and x, y ∈ Ω, and Q(x, y, z) is given by

Q(x, y, z) = −∫

ΩG(x, u, z)P (u, y, z)du.

The integral operator ”Qz with this kernel has the mapping property

”Qz : L1(Ω1) → ∆(“A) ≡∞⋂

j=1

D( “A j),

and for each k ∈ N∥∥∥∥ “Ak ”Qzf

∥∥∥∥L2(Ω)

≤ c(1 + |z|)ke−δ ε2(1+|z|)

1m ‖f‖L1(Ω1) .

Concerning the estimates for Q(x, y, z) we have

|Q(x, y, z)| ≤ c1e−δ|x−y|(1+|z|)

1m

|x− y|m−n, m < n

1 +∣∣∣log

(|x− y|(1 + |z|) 1

m

)∣∣∣ , m = n

(1 + |z|)n−mm , m > n

only for x ∈ Ω1 and y ∈ Ω because G(x, y, z) is simply equal to the fundamental

solution in that case. Next, for any s > 0 we can define “A sby the spectral theorem as

“A s=∫ ∞

0λsdEλ

with

D( “A s) =

ßf ∈ L2(Ω) :

∫ ∞

0λ2sd(Eλf, f) <∞

™.

Example 9.1. If “A = AF > 0 then

AsFf =

∞∑

j=1

λsjfjej(x),

where fj = (f, ej)L2(Ω), ej∞j=1 is an orthonormal basis of eigenvectors and

D(AsF ) =

f ∈ L2(Ω) :

∞∑

j=1

λ2sj |fj|2 <∞

.

85

Since we consider “A ≥ c0I, c0 > 0 then we can also define

“A−τ=∫ ∞

c0λ−τdEλ, τ > 0,

with

D( “A−τ) =

®f ∈ L2(Ω) :

∫ ∞

c0λ−2τd(Eλf, f) <∞

´.

Suppose that 0 < τ < 1. Then

∫ ∞

0

t−τdt

λ+ t

y=t/λ=

∫ ∞

0

λ−τy−τ

1 + ydy = λ−τ

∫ ∞

0y−τ (1 + y)−1dy

r=(1+y)−1

= λ−τ∫ 1

0(1 − r)−τrτ−1dr = λ−τB(τ, 1 − τ)

= λ−τ Γ(τ)Γ(1 − τ)

Γ(1)= λ−τ π

sin πτ

for any λ > 0. That’s why we can rewrite the spectral representation for “A−τ, 0 <

τ < 1 in the form

“A−τ=

∫ ∞

c0λ−τdEλ =

sin τπ

π

∫ ∞

c0

®∫ ∞

0

t−τdt

λ+ t

´dEλ

=sin τπ

π

∫ ∞

0t−τ

®∫ ∞

c0(λ+ t)−1dEλ

´dt =

sin τπ

π

∫ ∞

0t−τ ‘G−tdt,

because z = −t /∈ Zθ and, therefore, ‘G−t is well-defined.

Theorem 1. If 0 < τ < 1 then “A−τis the sum of the integral operators “E(τ) and

“Q(τ) whose kernels are

Eτ (x, y) =sin τπ

π

∫ ∞

0t−τE(x, y,−t)dt

and

Qτ (x, y) =sin τπ

π

∫ ∞

0t−τQ(x, y,−t)dt,

respectively. Moreover,

|DαxEτ (x, y)| ≤ c

|x− y|mτ−|α|−n, mτ < n+ |α|1 + |log |x− y|| , mτ = n+ |α|1, mτ > n+ |α|

for |α| ≤ m− 1 and x, y ∈ Ω and “Q(τ) has the mapping property

“Q(τ) : L1(Ω1) → ∆(“A).

86

Proof. The first part of this theorem follows immediately from the fact that

‘G−t = ‘E−t + ‘Q−t

sinceG(x, y,−t) = E(x, y,−t) +Q(x, y,−t).

Let us obtain now the estimates for DαxEτ (x, y), |α| ≤ m− 1. Since

|DαxEτ (x, y)| ≤

sin τπ

π

∫ ∞

0t−τ |Dα

xE(x, y,−t)|dt

we make use of the estimates for |DαxE|.

1) Let m < n+ |α|. Then mτ < n+ |α| also and thus

|DαxEτ (x, y)| ≤ c|x− y|m−|α|−n

∫ ∞

0t−τe−δ|x−y|(1+t)

1m dt

≤ c|x− y|m−|α|−n∫ ∞

0t−τe−δ|x−y|t

1m dt

( u := |x− y|t 1m ; t =

um

|x− y|m ; dt =mum−1

|x− y|mdu)

= c|x− y|mτ−|α|−n∫ ∞

0um(1−τ)−1e−δudu

= c|x− y|mτ−|α|−n,

because the last integral converges.

2) Let m = n+ |α|. Then again mτ < n+ |α| and

|DαxEτ (x, y)| ≤ c

∫ ∞

0t−τ

(1 +

∣∣∣log(|x− y|(1 + t)

1m

)∣∣∣)e−δ|x−y|(1+t)

1m dt

≤ c∫ ∞

0t−τ

(1 +

∣∣∣log(|x− y|t 1

m

)∣∣∣)e−δ|x−y|t

1m dt

( u := |x− y|t 1m ; t =

um

|x− y|m ; dt =mum−1

|x− y|mdu)

= c|x− y|mτ−m∫ ∞

0um(1−τ)−1 (1 + |log u|) e−δudu

= c|x− y|mτ−m = c|x− y|mτ−|α|−n

because the last integral converges again.

3) In the case m > n+ |α| our task is to estimate

|DαxEτ (x, y)| ≤ c

∫ ∞

0t−τ (1 + t)−1+

n+|α|m e−δ|x−y|t

1m dt

in three subcases.

87

a) Let mτ < n + |α|. Then τ + 1 − n+|α|m

< 1. Let us change the variables

u := |x− y|t 1m . Then the integral in question equals

c|x− y|mτ−m∫ ∞

0um(1−τ)−1

Ç1 +

Çu

|x− y|

åmån+|α|−mm

e−δudu

or

c|x− y|mτ−|α|−n∫ ∞

0um(1−τ)−1 (|x− y|m + um)

n+|α|−mm e−δudu.

It can be estimated from above by

c|x− y|mτ−|α|−n∫ ∞

0um(1−τ)−1un+|α|−me−δudu

which equals

c|x− y|mτ−|α|−n∫ ∞

0un+|α|−mτ−1e−δudu

orc|x− y|mτ−|α|−n

because the last integral converges in this case (n+ |α| −mτ − 1 > −1).

b) Let mτ = n+ |α|. Then τ +(1 − n+|α|

m

)= 1 and we consider the integral

c|x− y|mτ−|α|−n∫ ∞

0um(1−τ)−1 (|x− y|m + um)

n+|α|−mm e−δudu

in two parts: ∫ |x−y|

0du+

∫ ∞

|x−y|du := I1 + I2.

We have

I1 ≤ c∫ |x−y|

0um(1−τ)−1|x− y|n+|α|−mdu = c|x− y|n+|α|−m|x− y|m(1−τ)

= c|x− y|n+|α|−mτ = c

and

I2 ≤ c∫ ∞

|x−y|um(1−τ)−1un+|α|−me−δudu = c

∫ ∞

|x−y|u−1e−δudu

≤ c (1 + |log |x− y||) .

c) Let mτ > n+ |α|. Then τ +(1 − n+|α|

m

)> 1 and

|DαxEτ (x, y)| ≤ c

∫ ∞

0t−τ (1 + t)−1+

n+|α|m e−δ|x−y|t

1m dt

≤ c∫ ∞

0t−τ (1 + t)−1+

n+|α|m dt = c

because the last integral converges.

88

Hence the second part of this theorem is proved. Finally, using Theorem 2 of section8 we can obtain

∥∥∥∥ “Ak “Q(τ)f∥∥∥∥

L2(Ω)≤

∫ ∞

0t−τ

∥∥∥∥ “Ak ‘Q−tf∥∥∥∥

L2(Ω)dt

≤ c∫ ∞

0t−τ (1 + t)ke−δ ε

2(1+t)

1m ‖f‖L1(Ω1) dt ≤ c ‖f‖L1(Ω1)

,

for any k = 0, 1, . . .. This finishes the proof.

As it was proved, “A−τ, 0 < τ < 1 is an integral operator with the kernel

Kτ (x, y) ≡ Eτ (x, y) +Qτ (x, y)

such that

|Eτ (x, y)| ≤ c

|x− y|mτ−n, mτ < n

1 + |log |x− y|| , mτ = n

1, mτ > n,

where x, y ∈ Ω. But Eτ (x, y) ≡ 0 for x ∈ Ω \Ω1, y ∈ Ω and for x ∈ Ω, y ∈ Ω \Ω1. Theintegral operator “Q(τ) with the kernel Qτ (x, y) maps from L1(Ω1) to ∆(“A). But if wetake into account Qτ (x, y) we can get only the estimate

|Kτ (x, y)| ≤ c1

|x− y|mτ−n, mτ < n

1 + |log |x− y|| , mτ = n

1, mτ > n,

where x ∈ Ω1, y ∈ Ω and c1 = c(Ω1). If now τ ≥ 1 then we use the fact that

“A−τ= “A−τ1 “A−τ2 · · · “A−τl

,

where τ1 + τ2 + · · · + τl = τ with 0 < τj < 1, j = 1, 2, . . . , l. Due to this fact we

may conclude that “A−τis an integral operator with the kernel Kτ (x, y) which can be

calculated as

Kτ (x, y) =∫

ΩKτ1(x, u1)du1

∫

ΩKτ2(u1, u2)du2 · · ·

∫

ΩKτl−1

(ul−2, ul−1)Kτl(ul−1, y)dul−1.

Lemma. Let 0 < τ1 < 1 and 0 < τ2 < 1. Then

|Kτ1+τ2(x, y)| ≤ c1

|x− y|m(τ1+τ2)−n, mτ1 +mτ2 < n

1 + |log |x− y|| , mτ1 +mτ2 = n

1, mτ1 +mτ2 > n,

where x, y ∈ Ω1 and c1 = c(Ω1).

89

Exercise 46. If 0 < α1 < n, 0 < α2 < n and α1 + α2 > n then

∫

Ω|x− u|−α1 |u− y|−α2du ≤

∫

Rn|x− u|−α1 |u− y|−α2du ≤ c0|x− y|n−(α1+α2).

Exercise 47. Let Ω be bounded. If 0 < α1 < n, 0 < α2 < n and α1 + α2 = n then

∫

Ω|x− u|−α1 |u− y|−α2du ≤ c(Ω) (1 + |log |x− y||) .

Exercise 48. Let Ω be bounded. If 0 < α1 < n, 0 < α2 < n and α1 + α2 < n then

∫

Ω|x− u|−α1 |u− y|−α2du ≤ c(Ω).

Proof of lemma. Since

|Kτ1+τ2(x, y)| ≤∫

Ω1

|Kτ1(x, u)||Kτ2(u, y)|du

then we can apply the estimates for Kτ1 and Kτ2 with 0 < τ1 < 1, 0 < τ2 < 1,α1 = n−mτ1 and α2 = n−mτ2.

1) Let mτ1 < n,mτ2 < n and m(τ1 + τ2) < n. Then 0 < α1 < n, 0 < α2 < n andα1 + α2 > n. These conditions imply that

|Kτ1+τ2(x, y)| ≤ c1

∫

Ω1

|x− u|mτ1−n|u− y|mτ2−ndu

≤ c1|x− y|n−(n−mτ1)−(n−mτ2) = c1|x− y|m(τ1+τ2)−n.

2) Let mτ1 < n,mτ2 < n and m(τ1 + τ2) = n. Then α1 + α2 = n and thus

|Kτ1+τ2(x, y)| ≤ c1 (1 + |log |x− y||) .

3) Let mτ1 < n,mτ2 < n and m(τ1 + τ2) > n. Then α1 + α2 < n. That’s why

|Kτ1+τ2(x, y)| ≤ c1.

4) Let mτ1 < n,mτ2 = n. Then m(τ1 + τ2) > n and we obtain

|Kτ1+τ2(x, y)| ≤ c1

∫

Ω1

|x− u|mτ1−n (1 + |log |u− y||) du

≤ cε

∫

Ω1

|x− u|mτ1−n|u− y|−εdu ≤ cε

if we choose 0 < ε < mτ1.

The remaining five cases can be considered in a similar manner.

90

Remark. It follows by induction that

|Kτ (x, y)| ≤ c1

|x− y|mτ−n, mτ < n

1 + |log |x− y|| , mτ = n

1, mτ > n,

where x, y ∈ Ω1, c1 = c(Ω1) and τ > 0. These estimates can be extended to hold forx ∈ Ω1 and y ∈ Ω.

Theorem 2. Suppose that τ > n2m

. Then

∥∥∥∥ “A−τf∥∥∥∥

L∞(Ω1)≤ c1 ‖f‖L2(Ω) ,

where Ω1 = Ω1 ⊂ Ω and c1 = c(Ω1).

Proof. For f ∈ L2(Ω) and τ > 0 it follows from lemma above and Cauchy-Schwarz-Bunjakovskii inequality that

∥∥∥∥ “A−τf∥∥∥∥

L∞(Ω1)≤ sup

x∈Ω1

∫

Ω|Kτ (x, y)||f(y)|dy

≤ c1 supx∈Ω1

∫

Ω|f(y)|

|x− y|mτ−n, mτ < n

1 + |log |x− y|| , mτ = n

1, mτ > n,

dy

≤ c1 supx∈Ω1

‖f‖L2(Ω)

Ü∫

Ω

|x− y|2mτ−2n, mτ < n

(1 + |log |x− y||)2, mτ = n

1, mτ > n,

dy

ê1/2

≤ c1 ‖f‖L2(Ω)

because τ > n2m

if and only if 2mτ−2n > −n which makes the latter integral converge.

Now let us assume for simplicity that “A = AF > 0 is the Friedrichs extension. Inthat case

AFf =∞∑

j=1

λjfjej(x),

where λj > 0, λj → ∞ and fj are the Fourier coefficients of f ∈ L2(Ω) with respect toej(x)∞j=1. The orthonormal system of eigenvectors ej(x)∞j=1 forms an orthonormalbasis in L2(Ω). By the spectral theorem we know also that

D(AsF ) =

f ∈ L2(Ω) :

∞∑

j=1

λ2sj |fj|2 <∞

.

91

Corollary. Let Ω1 ⊂ Ω be a compact set and let τ > n2m

. Then

∞∑

j=1

λ−2τj |ej(x)|2 ≤ c1

holds uniformly with respect to x ∈ Ω1 with c1 = c(Ω1).

Proof. Since

A−τF f =

∞∑

j=1

λ−τj fjej(x)

then Theorem 2 implies that

∣∣∣∣∣∣

∞∑

j=1

λ−τj fjej(x)

∣∣∣∣∣∣≤ c1

Ñ∞∑

j=1

|fj|2é1/2

, x ∈ Ω1.

The left hand side of this inequality can be considered as the inner product of

~e = (λ−τ1 e1(x), . . . , λ

−τj ej(x), . . .)

and~f = (f1, . . . , fj, . . .)

in l2(C). Hence, ∣∣∣(~e, ~f)l2

∣∣∣ ≤ c1∥∥∥~f∥∥∥

l2.

By duality we may conclude that‖~e‖l2 ≤ c1

or∞∑

j=1

λ−2τj |ej(x)|2 ≤ c21.

Theorem 3. The Fourier series∞∑

j=1

fjej(x)

converges absolutely and uniformly on each compact set Ω1 ⊂ Ω for each function ffrom D(Aτ

F ) with τ > n2m

.

Proof. Using the corollary above and Cauchy-Schwarz-Bunjakovskii inequality we ob-tain

∞∑

j=1

|fj||ej(x)| ≤Ñ

∞∑

j=1

λ2τj |fj|2

é1/2 Ñ∞∑

j=1

λ−2τj |ej(x)|2

é1/2

≤ c1

Ñ∞∑

j=1

λ2τj |fj|2

é1/2

,

where τ > n2m

. But the latter number series converges since f ∈ D(AτF ).

92

Remark. We know that

Wm2 (Ω) ⊂ D(AF ).

Then for τ ∈ N we may conclude that

Wmτ2 (Ω) ⊂ D(Aτ

F ).

This embedding implies that if τ > n2m

or mτ > n/2 then the Fourier series corre-

sponding to f ∈

Wmτ2 (Ω) converges absolutely. Our aim is to prove this fact for any τ

such that mτ > n/2.

Let A be a non-negative self-adjoint operator in a Hilbert space H. Due to spectraltheorem we can characterize D(A) as follows: f ∈ D(A) if and only if

∫ ∞

0(1 + λ2)d(Eλf, f) <∞

and define a new norm‖f‖D(A) := ‖f‖H + ‖Af‖H .

Definition. Let G(t)t>0 be a family of bounded linear operators from H to H. Thisfamily is called an equi-bounded, strongly continuous semi-group if

1) G(t+ s)f = G(t)(G(s)f) for s, t > 0 and f ∈ H.

2) ‖G(t)f‖H ≤ M ‖f‖H for t > 0 and f ∈ H with M > 0 which does not dependon t or f .

3) limt→+0 ‖G(t)f − f‖H = 0 for f ∈ H.

Remark. We can complete this definition by G(0) := I.

Definition. The infinitesimal generator A of the semi-group G(t)t>0 is defined bythe formula

limt→0

∥∥∥∥∥G(t) − I

tf − Af

∥∥∥∥∥H

= 0

with domain D(A) consisting of all f ∈ H such that

limt→0

G(t) − I

tf

exists in H.

Remark. In the sense of the previous definition we write G′(0) = A.

Example 9.2. Let H = L2(Rn). Let ω(ξ) be an infinitely differentiable positivefunction on R

n \ 0, which is positively homogeneous of order m > 0 i.e. ω(tξ) =|t|mω(ξ). Let us define the family G(t)t>0 by the formula

G(t)f := F−1Äe−tω(ξ)Ff

ä, f ∈ L2(Rn).

It is clear that G(t) : L2(Rn) → L2(Rn). Moreover,

93

1)

G(t+s)f = F−1Äe−(t+s)ω(ξ)Ff

ä= F−1

Äe−tω(ξ)FF−1

Äe−sω(ξ)Ff

ää= G(t)(G(s)f).

2)

‖G(t)f‖L2 =∥∥∥F−1

Äe−tω(ξ)Ff

ä∥∥∥L2

=∥∥∥e−tω(ξ)Ff

∥∥∥L2

≤ ‖Ff‖L2 = ‖f‖L2 .

3)

‖G(t)f − f‖L2 =∥∥∥F−1

Äe−tω(ξ)Ff − Ff

ä∥∥∥L2

=∥∥∥Äe−tω(ξ) − 1

äFf

∥∥∥L2

→ ‖Ff‖L2 = ‖f‖L2 , t→ 0

by the Lebesgue theorem. Also by Lebesgue theorem we have

limt→0

G(t)f − f

tH= lim

t→0F−1

(e−tω(ξ) − 1

tFf

)= −F−1 (ω(ξ)Ff) ≡ Af.

The domain of A is

D(A) =¶f ∈ L2 : ‖ω(ξ)Ff‖L2 <∞

©.

For instance, if ω(ξ) = |ξ|2 then A ≡ ∆ and D(A) = W 22 (Rn).

Example 9.3. Let A = A∗ ≥ 0. Define

G(t) := eitA ≡∫ ∞

0eitλdEλ.

Then

1)

G(t+s) =∫ ∞

0ei(t+s)λdEλ =

∫ ∞

0eitλeisλdEλ =

∫ ∞

0eitλdEλ

∫ ∞

0eisµdEµ = G(t)G(s).

2)

‖G(t)f‖2 =∫ ∞

0|eitλ|2d(Eλf, f) = ‖f‖2 .

3)

‖G(t)f − f‖2 =∫ ∞

0|eitλ − 1|2d(Eλf, f) → 0, t→ 0.

andG(t)f − f

t=∫ ∞

0

eitλ − 1

tdEλf → i

∫ ∞

0λdEλf ≡ iA, t→ 0.

94

We will consider now J. Peetre’s method of real interpolation or the K-method.We have A = A∗ ≥ 0 in a Hilbert space H. For any k ∈ N denote by Dk the domainof Ak. We define the interpolation space Dα,p, where α > 0 and 1 ≤ p ≤ ∞ as follows:Set

K(t, f) := inff=f0+f1

(‖f0‖H + t ‖f1‖Dk) , f ∈ H, 0 < t <∞,

where f0 ∈ H and f1 ∈ Dk. This functional is called the functional of Peetre or theK-functional . Then, if 0 < α < k the space Dα,p is the space of all f ∈ H such that

Ç∫ ∞

0

Ät−

αkK(t, f)

äp dt

t

å1/p

<∞

with the norm

‖f‖Dα,p :=

Ç∫ ∞

0

Ät−

αkK(t, f)

äp dt

t

å1/p

<∞.

We shall denote also(H,Dk)α

k,p := Dα,p.

Remark. If p ≤ q thenDα,p ⊂ Dα,q

and if α > β thenDα,p ⊂ Dβ,q

for any p and q.

Theorem 4. Let G(t) be an equi-bounded, strongly continuous semi-group on H withinfinitesimal generator A. Then

1)K(t, f) ≍ h(t, f) + min(1, t) ‖f‖H ,

where h(t, f) = sups<t ‖G(s)f − f‖H .

2)

‖f‖(H,D(A))θ,p≍ ‖f‖H +

Ç∫ ∞

0

Ät−θh(t, f)

äp dt

t

å1/p

,

where 0 < θ < 1 and 1 ≤ p ≤ ∞.

Proof. Let us assume that f = f0 + f1, where f0 ∈ H and f1 ∈ D(A). Then

h(t, f) = sups<t

‖G(s)f − f‖H ≤ sups<t

‖G(s)f0 − f0‖H + sups<t

‖G(s)f1 − f1‖H

≤ (M + 1) ‖f0‖H + sups<t

‖G(s)f1 − f1‖H .

Since G′(0) = A then G′(t) = AG(t) = G(t)A. Indeed,

G′(t) = limτ→0

G(t+ τ) −G(t)

τ= G(t) lim

τ→0

G(τ) − I

τ

= limτ→0

G(τ) − I

τG(t) = G(t)A = AG(t).

95

We can thus write

G(t)f − f =∫ t

0G(s)Afds =

∫ t

0AG(s)fds.

That’s why

sups<t

‖G(s)f1 − f1‖H ≤∫ t

0sups<t

‖G(s)Af1‖H ds ≤M∫ t

0‖Af1‖H ds

= Mt ‖Af1‖H ≤Mt ‖f1‖D(A) .

Therefore

h(t, f) ≤ (M + 1) ‖f0‖H +Mt ‖f1‖D(A) ≤ (M + 1)K(t, f).

Note that also min(1, t) ‖f‖H ≤ K(t, f). Indeed, if 0 < t < 1 then

K(t, f) = inff=f0+f1

(‖f0‖H + t ‖f1‖D(A)

)≥ t inf

f=f0+f1

(‖f0‖H + ‖f1‖D(A)

)= t ‖f‖H .

If t ≥ 1 thenK(t, f) ≥ inf

f=f0+f1

(‖f0‖H + ‖f1‖D(A)

)= ‖f‖H .

Thus

h(t, f) + min(1, t) ‖f‖H ≤ (M + 1)K(t, f) +K(t, f) = (M + 2)K(t, f).

For the other half of 1) we argue as follows. By the definition of infimum we have

K(t, f) ≡ inff=f0+f1

(‖f0‖H + t ‖f1‖D(A)

)≤ ‖f‖H

under the choice f1 ≡ 0 and f = f0. Therefore

K(t, f) ≤ ‖f‖H ≤ h(t, f) + ‖f‖H = h(t, f) + min(1, t) ‖f‖H

for t ≥ 1. If 0 < t < 1 then we put

f1 = t−1∫ t

0G(s)fds, f0 = f − f1

for any f ∈ H. Then we have

K(t, f) ≤ ‖f0‖H + t ‖f1‖D(A) =

∥∥∥∥∥t−1∫ t

0G(s)fds− f

∥∥∥∥∥H

+ t

∥∥∥∥∥t−1∫ t

0G(s)fds

∥∥∥∥∥D(A)

=

∥∥∥∥∥t−1∫ t

0(G(s)f − f)ds

∥∥∥∥∥H

+

∥∥∥∥∥

∫ t

0G(s)fds

∥∥∥∥∥D(A)

≤ sups<t

‖G(s)f − f‖H +

∥∥∥∥∥

∫ t

0G(s)fds

∥∥∥∥∥H

+

∥∥∥∥∥

∫ t

0AG(s)fds

∥∥∥∥∥H

≤ sups<t

‖G(s)f − f‖H +Mt ‖f‖H + ‖G(t)f − f‖H

≤ h(t, f) +Mt ‖f‖H + h(t, f) ≤ max(2,M) (h(t, f) + t ‖f‖H) .

96

This completes the proof of 1).Since

‖f‖(H,D(A))θ,p=

Ç∫ ∞

0

Ät−θK(t, f)

äp dt

t

å1/p

then part 1) implies that

‖f‖(H,D(A))θ,p≍

Ç∫ ∞

0

Ät−θ min(1, t)

äp dt

t

å1/p

‖f‖H +

Ç∫ ∞

0

Ät−θh(t, f)

äp dt

t

å1/p

=

®∫ 1

0t(1−θ)p−1dt+

∫ ∞

1

dt

t1+θp

´1/p

‖f‖H

+

Ç∫ ∞

0

Ät−θh(t, f)

äp dt

t

å1/p

≍ ‖f‖H +

Ç∫ ∞

0

Ät−θh(t, f)

äp dt

t

å1/p

.

This completes the proof.

Corollary. Let A be a self-adjoint and non-negative operator in a Hilbert space H.Then

(H,D(Ak))θ,2 = D(Akθ)

for any k ∈ N and 0 < θ < 1.

Proof. Due to Example 9.3 we may conclude that the family

G(t) := eitAk

=∫ ∞

0eitλk

dEλ, t > 0

is an equi-bounded, strongly continuous semi-group with the infinitesimal generatoriAk. Since D(iAk) = D(Ak) and (H,D(iAk))θ,2 = (H,D(Ak))θ,2 then it follows fromTheorem 4 that

‖f‖(H,D(Ak))θ,2≍ ‖f‖H +

Ç∫ ∞

0t−2θh(t, f)2dt

t

å1/2

= ‖f‖H +

Ç∫ ∞

0t−2θ ‖G(t)f − f‖2 dt

t

å1/2

= ‖f‖H +

Ç∫ ∞

0t−2θ

∫ ∞

0|eitλk − 1|2d ‖Eλf‖2 dt

t

å1/2

= ‖f‖H +

(∫ ∞

0d ‖Eλf‖2

∫ ∞

0

|eitλk − 1|2t2θ

dt

t

)1/2

ξ=tλk

= ‖f‖H +

(∫ ∞

0λ2θkd ‖Eλf‖2

∫ ∞

0

|eiξ − 1|2ξ2θ+1

dξ

)1/2

= ‖f‖H + cθ

Å∫ ∞

0λ2θkd ‖Eλf‖2

ã1/2

= ‖f‖H + cθ∥∥∥Akθf

∥∥∥H,

since the inner integral with respect to ξ converges for 0 < θ < 1.

97

Remark. The corollary above allows us to conclude that

(H,D(Ak))θ,1 ⊂ D(Akθ) ⊂ (H,D(Ak))θ,∞

for any k ∈ N and 0 < θ < 1.

Let AF be Friedrichs extension of an elliptic differential operator of order m withan orthonormal basis ej∞j=1 of eigenvectors in L2(Ω), where Ω is bounded. Let alsoAF = A∗

F > 0.

Theorem 5 (Peetre). For any f ∈ (L2(Ω), D(Ak)) n2mk

,1 ≡ Dn

2m,1 the series

∞∑

j=1

fjej(x),

where fj = (f, ej)L2(Ω) converges absolutely and uniformly in compact parts of Ω.

Proof. Since AF has pure discrete spectrum λj∞j=1 then

Eλf(x) =∑

λj<λ

fjej(x)

and, therefore,θ(x, y, λ) =

∑

λj<λ

ej(x)ej(y).

We will use Garding’s estimate

θ(x, x, λ) =∑

λj<λ

|ej(x)|2 ≤ c1λnm ,

where c1 = c(Ω1) can be chosen independent of x ∈ Ω1,Ω1 = Ω1 ⊂ Ω. We have

∞∑

j=1

|fjej(x)| ≍∞∑

ν=0

∑

2ν≤λj<2ν+1

|fjej(x)|

≤∞∑

ν=0

Ñ∑

2ν≤λj<2ν+1

|fj|2é1/2 Ñ

∑

2ν≤λj<2ν+1

|ej(x)|2é1/2

≤ c∞∑

ν=0

2ν n2m

Ñ∑

2ν≤λj<2ν+1

|fj|2é1/2

.

It is clear that

Ñ∑

2ν≤λj<2ν+1

|fj|2é1/2

≤Ñ

∞∑

j=1

|fj|2é1/2

= ‖f‖L2 . (9.1)

98

Since

fj = (f, ej) = λ−kj (f, λk

j ej) = λ−kj (f,Ak

F ej) = λ−kj (Ak

Ff, ej) = λ−kj (Ak

Ff)j

then we have also

Ñ∑

2ν≤λj<2ν+1

|fj|2é1/2

=

Ñ∑

2ν≤λj<2ν+1

λ−2kj |(Ak

Ff)j|2é1/2

≤ 2−kν

Ñ∑

2ν≤λj<2ν+1

|(AkFf)j|2

é1/2

≤ 2−kν ‖f‖Dk . (9.2)

Applying (9.1) to f0 and (9.2) to f1 we get for f = f0 + f1, f0 ∈ L2(Ω), f1 ∈ D(AkF )

that Ñ∑

2ν≤λj<2ν+1

|fj|2é1/2

≤ ‖f0‖L2 + 2−kν ‖f1‖Dk .

It follows that Ñ∑

2ν≤λj<2ν+1

|fj|2é1/2

≤ K(2−νk, f).

That’s why we obtain

∞∑

j=1

|fjej(x)| ≤ c1∞∑

ν=0

2νn2mK(2−νk, f) ≤ c1

∞∑

ν=−∞

2νn2mK(2−νk, f).

Next we are going to prove that

∞∑

ν=−∞

2νn2mK(2−νk, f) ≤ c

∫ ∞

0t−

n2mkK(t, f)

dt

t

if k ∈ N and k > n2m

. Note first that

K(t, f) ≤ max(1, t/s)K(s, f). (9.3)

Indeed, if t/s ≥ 1 then

K(t, f) = inf (‖f0‖ + t ‖f1‖) = infÅ‖f0‖ +

t

ss ‖f1‖

ã≤ t

sK(s, f).

If t/s < 1 then immediately K(t, f) ≤ K(s, f). Using (9.3) we obtain

K(2−kν , f) ≤ max

(1,

2−kν

t

)K(t, f)

or2kνK(2−kν , f) ≤ t−1K(t, f), t < 2−kν .

99

Next, ∫ ∞

0t−

n2mkK(t, f)

dt

t=

∞∑

ν=−∞

∫ 2−kν

2−k(ν+1)t−

n2mkK(t, f)

dt

t.

Therefore,

2ν n2mK(2−νk, f) = 2νkK(2−νk, f)2ν( n

2m−k) ≤ t−1K(t, f)2ν( n

2m−k) ≤ ct−1K(t, f)t−

n2mk

+1

for 2−k(ν+1) < t < 2−kν and k > n2m

. It implies that

2ν n2mK(2−νk, f)

1

t≤ ct−

n2mkK(t, f)

1

t,

where 2−k(ν+1) < t < 2−kν and k > n2m

. Integrating this inequality with respect to tover the interval in question yields

2ν n2mK(2−νk, f)

∫ 2−kν

2−k(ν+1)

dt

t≤ c

∫ 2−kν

2−k(ν+1)t−

n2mkK(t, f)

dt

t

or

2ν n2mK(2−νk, f) ≤ c′

∫ 2−kν

2−k(ν+1)t−

n2mkK(t, f)

dt

t.

Finally we obtain

∞∑

j=1

|fjej| ≤ c1∞∑

ν=−∞

2ν n2mK(2−νk, f) ≤ c1

∞∑

ν=−∞

∫ 2−kν

2−k(ν+1)t−

n2mkK(t, f)

dt

t

= c1

∫ ∞

0t−

n2mkK(t, f)

dt

t.

But the latter integral is finite if and only if f ∈ Dn

2m,1.

Let ϕj(ξ)∞j=0 be a partition of unity i.e.∑∞

j=0 ϕj(ξ) ≡ 1, ξ ∈ Rn, ϕj ≥ 0, ϕj ∈

C∞0 (Rn), ϕj(ξ) = ϕ(2−jξ), j = 1, 2, . . ., where ϕ ∈ C∞

0 with suppϕ ⊂ 1/4 ≤ |ξ| ≤ 1and suppϕ0 ⊂ |ξ| < 1.

Definition. Let s ∈ R and 1 ≤ p ≤ ∞. A function f belongs to the Besov spaceBs

2,p(Rn) for 1 ≤ p <∞ if

‖f‖Bs2,p(Rn) :=

Ñ∞∑

j=0

2jpsÅ∫

Rn|fj(x)|2dx

ãp/2é1/p

<∞,

where fj(x) = F−1 (ϕj(ξ)Ff) (x). The space Bs2,∞(Rn) is defined to consist of functions

f for which

‖f‖Bs2,∞(Rn) := sup

0≤j≤∞2js

Å∫Rn

|fj(x)|2dxã1/2

<∞

with fj as above.

100

Exercise 49. Prove thatBs

2,p(Rn) ⊂ Bs

2,∞(Rn)

for 1 ≤ p <∞.

In the case 1 ≤ p <∞ the previous definition is equivalent to

∞∑

j=1

2j(ps+np2 )

Ç∫1/4≤|η|≤1

|ϕ(η)Ff(2jη)|2dηåp/2

+

Ç∫|ξ|<1

|ϕ0(ξ)Ff(ξ)|2dξåp/2

1/p

<∞

while in the case p = ∞ we might equivalently require that

Ç∫1/4≤|η|≤1

|ϕ(η)Ff(2jη)|2dηå1/2

≤ C2−js

and Ç∫|ξ|<1

|ϕ0(ξ)Ff(ξ)|2dξå1/2

≤ C

for some constant C.

Exercise 50. Prove thatBs

2,p(Rn) ⊂ L∞(Rn)

if s > n/2 and 1 ≤ p ≤ ∞.

Remark. It was proved by Peetre and O.V. Besov and S.M. Nikol’skii that f ∈Bs

2,p(Rn), s > 0, 1 ≤ p ≤ ∞ if and only if

‖f‖Bs2,p

≍ ‖f‖L2 +

(∫ ∞

0

[ω(2)(t;Dαf)

ts−k

]pdt

t

)1/p

,

where k ∈ N0, |α| = k, s− k > 0 and

ω(2)(t; g) := sup|h|<t

‖g(· + h) − 2g + g(· − h)‖L2 .

Let ω(ξ) from Example 9.2 be equal to |ξ|m with m > 0. Then

G(t) = F−1(e−t|ξ|mF )

is the semi-group with the infinitesimal generator

Af = −F−1(|ξ|mFf) = −(−∆)m/2f.

It is also true that

D(A) =¶f ∈ L2(Rn) : |ξ|mFf ∈ L2

©= Wm

2 (Rn).

For this generator A the following theorem holds.

101

Theorem 6. If 0 < θ < 1, 1 ≤ p ≤ ∞ and m > 0 then

(L2(Rn), D(A))θ,p = Bmθ2,p (Rn).

Proof. Let f ∈ Bmθ2,p (Rn) and let ϕj∞j=0 be a partition of unity. Then

Ff =∞∑

j=0

ϕjFf

so that

‖f‖L2 = ‖Ff‖L2 ≤∞∑

j=0

‖ϕjFf‖L2 ≤∞∑

j=0

‖fj‖L2 .

Moreover,

Afj = AF−1(ϕjFf) = −F−1(|ξ|mF (F−1ϕjFf)) = −F−1(|ξ|mϕjFf).

Hence‖Afj‖L2 = ‖|ξ|mϕjFf‖L2 ≤ 2jm ‖fj‖L2

because suppϕj ⊂ 2j−2 ≤ |ξ| ≤ 2j, j = 1, 2, . . ., suppϕ0 ⊂ |ξ| < 1 and, therefore,

|ξ|m ≤ 2jm

for ξ ∈ suppϕj. Next we prove that

h(t, f) ≤ c∞∑

j=0

min(1, t2jm) ‖fj‖L2 . (9.4)

Indeed, if t ≥ 1 then

h(t, f) = sups<t

‖G(s)f − f‖ ≤ (M + 1) ‖f‖L2 ≤ (M + 1)∞∑

j=0

‖fj‖L2 .

If 0 < t < 1 then

h(t, f) = sups<t

∥∥∥∥∫ s

0G(τ)Afdτ

∥∥∥∥L2

≤M sups<t

∫ s

0‖Af‖L2 dτ ≤Mt ‖Af‖L2

≤ Mt∞∑

j=0

‖Afj‖L2 ≤Mt∞∑

j=0

2jm ‖fj‖L2 .

This proves (9.4). We also know from Theorem 4 that

‖f‖(L2,D(A))θ,p≍ ‖f‖L2 +

Ç∫ ∞

0

Ät−θh(t, f)

äp dt

t

å1/p

.

102

It follows from (9.4) that

Ç∫ ∞

0

Ät−θh(t, f)

äp dt

t

å1/p

≤ c

Ñ∫ ∞

0

Ñt−θ

∞∑

j=0

min(1, t2jm) ‖fj‖L2

épdt

t

é1/p

≤ c∞∑

j=0

Ç∫ ∞

0

Ät−θ min(1, t2jm)

äp dt

t

å1/p

‖fj‖L2 .

Let us calculate the integral appearing in the last estimate. We have

Ç∫ ∞

0

Ät−θ min(1, t2jm)

äp dt

t

å1/p

=

Ç∫ 2−jm

0t(1−θ)p−1dt

å1/p

2jm

+Å∫ ∞

2−jmt−θp−1dt

ã1/p

= c′2−(1−θ)jm2jm + c′′2jθm = c2jθm.

Thus

‖f‖(L2(Rn),D(A))θ,p≤ ‖f‖L2 + c

∞∑

j=0

2jmθ ‖fj‖L2 .

Hence‖f‖(L2(Rn),D(A))θ,p

≤ c ‖f‖Bmθ2,1 (Rn) .

This inequality means that

Bmθ2,1 (Rn) ⊂ (L2(Rn), D(A))θ,p.

It is also possible to prove sharper embedding

Bmθ2,1 (Rn) ⊂ Bmθ

2,p (Rn) ⊂ (L2(Rn), D(A))θ,p.

It remains to prove the opposite inequality, namely

‖f‖Bmθ2,p (Rn) ≤ c ‖f‖(L2(Rn),D(A))θ,p

.

To that end, note that

fj(x) = F−1(ϕjFf) = F−1(ϕj(e−t|ξ|m − 1)−1F (G(t)f − f))

t=2−jm

= F−1(ϕj(e−(|ξ|2−j)m − 1)−1F (G(2−jm)f − f)).

Since suppϕj ⊂ 2j−2 ≤ |ξ| ≤ 2j then

1

4≤ |ξ|2−j ≤ 1

or Ç1

4

åm

≤ (|ξ|2−j)m ≤ 1

103

for ξ ∈ suppϕj. Hence

‖fj‖L2 =∥∥∥ϕj(e

−(|ξ|2−j)m − 1)−1F (G(2−jm)f − f)∥∥∥

L2

≤ 1

1 − e−(1/4)m

∥∥∥F (G(2−jm)f − f)∥∥∥

L2≤ c

∥∥∥G(2−jm)f − f∥∥∥

L2≤ ch(2−jm, f).

By the definition of Besov spaces we have

‖f‖Bs2,p

=

Ñ∞∑

j=0

2jps ‖fj‖pL2

é1/p

≤ c

Ñ∞∑

j=0

2jpsh(2−jm, f)p

é1/p

.

As in the proof of Theorem 5 it can be checked that the convergence of the latter seriesis equivalent to the convergence of the integral

Ç∫ ∞

0

Ät−θh(t, f)

äp dt

t

å1/p

with s = mθ, 0 < θ < 1.

Exercise 51. Prove this fact.

Thus, the required inequality is proved and, therefore, we have the embedding

(L2(Rn), D(A))θ,p ⊂ Bmθ2,p (Rn).

This finishes the proof.

Corollary. If Ω ⊂ Rn is a bounded domain with smooth boundary then

(L2(Ω),

Wm2 (Ω))θ,p =

Bmθ2,p (Ω)

for any m > 0 such that mθ − 1/2 is not an integer.

Let A be a self-adjoint and non-negative extension of an elliptic differential operatorof even order m. Since

Wmk2 (Ω) ⊂ D(Ak)

for any k ∈ N thenÇL2(Ω),

Wmk2 (Ω)

å

θ,p

⊂ÄL2(Ω), D(Ak)

äθ,p

or

Bmkθ2,p (Ω) ⊂ Dθk,p.

Peetre’s theorem means that the corresponding Fourier series converges absolutely and

uniformly if f ∈ Dn

2m,1. This implies that the same is true for f ∈

Bn/22,1 (Ω). We proved

also that if f ∈ D(Aσ) with σ > n2m

then the Fourier series converges absolutely. SinceÇL2(Ω),

Wmk2 (Ω)

å

θ,2

⊂ÄL2(Ω), D(Ak)

äθ,2

= D(Akθ)

104

or

Bmkθ2,2 (Ω) ⊂ D(Akθ)

then the Fourier series converges absolutely for f ∈

Bn/2+ε2,2 (Ω), ε > 0.

105

Index

K-functional, 95

adjoint operator, 13

basis, 8Besov space, 100Bessel’s inequality, 2bounded, 11

Cauchy sequence, 3Cauchy-Schwarz-Bunjakovskii inequality, 2Cayley transform, 31closable, 13closed, 13closed subspace, 6closure, 13compact operator, 39complete space, 3completeness relation, 9constant of ellipticity, 53continuous spectrum, 36convergent sequence, 3criterion for closedness, 13

densely defined, 11discrete spectrum, 36domain, 11

elliptic partial differential operator, 52ellipticity condition, 53equi-bounded, strongly continuous semi-

group, 93essential spectrum, 36essentially self-adjoint, 16extension, 13

finite rank operator, 39formally self-adjoint, 52Fourier expansion, 9Friedrichs extension, 50functional of Peetre, 95fundamental solution, 68

Garding’s inequality, 57generalized Leibniz formula, 53graph, 12Green’s function, 62

Hilbert space, 3Hilbert-Schmidt norm, 12

idempotent, 20induced by the inner product, 3infinitesimal generator, 93inner product, 1inner product space, 1isometry, 23

kernel, 11

Lebesgue space, 5length, 2linear operator, 11linear space, 1linear span, 8

multi-index, 52

non-negative operator, 22norm, 3norm topology, 3nullspace, 11

orthogonal, 2orthogonal complement, 5orthonormal, 2orthonormal basis, 8

parallelogram law, 3parametrix, 80Parseval equality, 9partition of unity, 100point spectrum, 36polarization identity, 3positive operator, 22precompact set, 39

106

principal symbol, 52Projection theorem, 6projector, 20

quadratic form, 48

range, 11resolvent, 32resolvent identity, 33resolvent set, 33restriction, 13Riesz-Frechet theorem, 7

scalar product, 1self-adjoint, 16semibounded from below, 48separable, 8sequence space, 4smoothed fundamental solution, 81Sobolev space, 5, 54spectral family, 23spectral function, 61spectrum, 33symmetric, 16

Theorem of Pythagoras, 2triangle inequality, 2

uniformly elliptic operator, 53unitary operator, 23

vector space, 1

107