Spectral Theory of Elliptic Differential Operators

Spectral theory of elliptic diﬀerential operators 802628S

Lecture Notes 1st Edition Second printing

Valeriy Serov University 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 the following axioms are satisﬁed:

1) To every pair x, y H there corresponds a vector x + y, called the sum, with the properties: ∈ 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 denote y := x. ∈ ∈ 1 − 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 inner product 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 is called 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 n (x, y)= xjyj, jX=1 where x =(x ,...,x ) Cn and y =(y ,...,y ) Cn. 1 n ∈ 1 n ∈ 1 Example 1.2. On the linear space C[a, b] of continuous complex-valued functions, the formula b (f,g)= f(x)g(x)dx Za 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. ∈ ∈ 1, α = β 2) a system xα α∈A H orthonormal if (xα, xβ) = δα,β = , where A { } ⊂ 0, α = β  6 is some index set.  3) x := (x, x) is called the length of x H. k k ∈ k N Exercise 1. Prove the Theorem of Pythagoras: If xj j=1, k is an orthonormal system in an» inner product space H, then { } ∈

2 k k x 2 = (x, x ) 2 + x (x, x )x j j j k k j=1 | | − j=1 X X

for every x H. ∈ k Exercise 2. Prove Bessel’s inequality: If xj j=1, k is an orthonormal system then { } ≤ ∞ k (x, x ) 2 x 2 , | j | ≤ k k jX=1 for every x H. ∈ Exercise 3. Prove the Cauchy-Schwarz-Bunjakovskii inequality:

(x, y) x y , x, y H. | | ≤ k k k k ∈ Prove also that ( , ) is continuous as a map from H H to C. · · × If H is an inner product space, then

x := (x, x) k k has the following properties: » 1) x 0 for every x H and x = 0 if and only if x = 0. k k ≥ ∈ k k 2) λx = λ x for every x H and λ C. k k | | k k ∈ ∈ 3) x + y x + y for every x, y H. This is the triangle inequality. k k ≤ k k k k ∈ 2 The function = ( , ) is thus a norm on H. It is called the norm induced by the inner product.k·k · · Every inner product» space H is a normed space under the induced norm. The neighborhood of x H is the open ball Br(x)= y H : x y < r . This system of neighborhoods deﬁnes∈ the norm topology on H{such∈ that:k − k }

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 every ε> 0 there exists n N {such} that⊂ x x <ε for k, j n . 0 ∈ k k − jk ≥ 0 ∞ 2) A sequence xj j=1 H is said to be convergent if there exists x H such that for every ε>{ 0} there⊂ exists n N such that x x <ε whenever∈ j n . 0 ∈ k − jk ≥ 0 3) The inner product space H is complete space if every Cauchy sequence in H converges.

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

2) If x ∞ converges to x H then { j}j=1 ∈

lim xj = x . j→∞ k k k k Deﬁnition. (J. von Neumann, 1925) A Hilbert space is an inner product space which is complete (with respect to its norm topology).

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

x + y 2 + x y 2 = 2 x 2 + 2 y 2 . k k k − k k k k k Exercise 5. Prove that if in a normed space H the parallelogram law holds, then there is an inner product on H such that x 2 =(x, x) and that this inner product is deﬁned by the polarization identity k k 1 (x, y) := x + y 2 x y 2 + i x + iy 2 i x iy 2 . 4 k k − k − k k k − k − k

Exercise 6. Prove thatÄ on C[a, b] the norm ä f = max f(x) k k x∈[a,b] | | 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 Cn.

2) The matrix space Mn(C) consisting of n n -matrices whose elements are complex numbers. For A, B M (C) the inner product× is given by ∈ n n ∗ (A, B)= akjbkj = Tr(AB ), k,jX=1 T where B∗ = B .

3) The sequence space l2(C) deﬁned by

∞ l2(C) := x ∞ , x C : x 2 < . { j}j=1 j ∈ | j| ∞  jX=1  The estimates  

x + y 2 2 x 2 + y 2 , λx 2 = λ 2 x 2 | j j| ≤ | j| | j| | j| | | | j| and Ä 1 ä x y x 2 + y 2 | j j| ≤ 2 | j| | j| imply that l2(C) is a linear space. Let us deﬁne the inner product by Ä ä ∞ (x, y) := xjyj jX=1 2 C (k) ∞ 2 C and prove that l ( ) is complete. Suppose that x k=1 l ( ) is a Cauchy sequence. Then for every ε> 0 there exists n N{ such} that∈ 0 ∈ ∞ 2 x(k) x(m) = x(k) x(m) 2 <ε2 − | j − j | j=1 X

for k,m n . It implies that ≥ 0 x(k) x(m) <ε, j = 1, 2,... | j − j | or that x(k) ∞ is a Cauchy sequence in C for every j = 1, 2,.... Since C is { j }k=1 a complete space then x(k) ∞ converges for every ﬁxed j = 1, 2,... i.e. there { j }k=1 exists xj C such that ∈ (k) xj = lim xj . k→∞ This fact and l x(k) x(m) 2 <ε2, l N | j − j | ∈ jX=1 4 imply that l l (k) (m) 2 (k) 2 2 lim x x = x xj ε m→∞ | j − j | | j − | ≤ jX=1 jX=1 for all k n and l N. Therefore the sequence ≥ 0 ∈ l s := x(k) x 2, k n l | j − j| ≥ 0 jX=1 is a monotone increasing sequence which is bounded from above by ε2. Hence this sequence has a limit with the same upper bound i.e.

∞ l (k) 2 (k) 2 2 xj xj = lim xj xj ε . | − | l→∞ | − | ≤ jX=1 jX=1 That’s why we may conclude that

x x(k) + x(k) x x(k) + ε k k ≤ − ≤

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.

f(x) 2dx < . ZΩ | | ∞ It is a linear space with the inner product

(f,g)= f(x)g(x)dx ZΩ and the Riesz-Fisher theorem reads as: L2(Ω) is a Hilbert space.

k 2 5) The Sobolev spaces W2 (Ω) consisting of functions f L (Ω) whose weak or distributional derivatives Dαf also belong to L2(Ω) up∈ to order α k, k = k | | ≤ 1, 2,.... On the space W2 (Ω) the natural inner product is

(f,g)= Dαf(x)Dαg(x)dx. Ω |αX|≤k Z

Deﬁnition. Let H be an inner product space. For any subspace M H the orthogonal complement 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 since 0 M always. ∩ { } ∈ 5 Deﬁnition. A closed subspace of a Hilbert space H is a linear subspace of H which is closed (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 space H. 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 = inf x y := d(x, M). k k y∈M k − k Proof. Let x H. Then ∈ d := d(x, M) inf x y x u ≡ y∈M k − k ≤ k − k ∞ for any u M. The deﬁnition of inﬁmum implies that there exists a sequence uj j=1 M such that∈ { } ⊂ d = lim x uj . j→∞ k − k The parallelogram law implies that

2 2 uj uk = (uj x)+(x uk) k − k k − − k 2 2 2 uj + uk = 2 uj x + 2 x uk 4 x . k − k k − k − − 2

Since (uj + uk)/2 M then ∈ u u 2 2 u x 2 + 2 x u 2 4d2 2d2 + 2d2 4d2 = 0 k j − kk ≤ k j − k k − kk − → − ∞ as j, k . Hence uj j=1 M is a Cauchy sequence in the Hilbert space H. It means that→ ∞ there exists{ u} H⊂such that ∈

u = lim uj. j→∞

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

d = lim x uj = x u . j→∞ k − k k − k Let us denote v := x u and show that v M ⊥. For any y M, y = 0 introduce the number − ∈ ∈ 6 (v, y) α = . − y 2 k k 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 ≤ k − − k k k k k | | k k (v, y)(v, y) (v, y)(y,v) (v, y) 2 (v, y) 2 = d2 + | | = d2 | | . − y 2 − y 2 y 2 − y 2 k k k k k k k k This inequality implies that (y,v) = 0. It means that v M ⊥. In order to prove ∈ ⊥ uniqueness assume that x = u1 + v1 = u2 + v2, where u1, u2 M and v1,v2 M . It follows that ∈ ∈ u u = v v M M ⊥. 1 − 2 2 − 1 ∈ ∩ But M M ⊥ = 0 so that u = u and v = v . ∩ { } 1 2 1 2 Corollary 1 (Riesz-Frechet theorem). If T is a linear continuous functional on the Hilbert space H then there exists a unique h H such that T (x)=(x, h) for all x H. Moreover, T = h . ∈ ∈ k kH→C k k Proof. If T 0 then h = 0 will do. If T = 0 then there exists v H such that ≡ 6 0 ∈ T (v0) = 0. Let 6 M := u H : T (u) = 0 . { ∈ } Since T is linear and continuous then M is a closed subspace. It follows from Thereom 1 that H = M M ⊥ ⊕ i.e. every x H has the unique representation as x = u + v. Since v0 = 0 then ⊥ ∈ 6 v0 M . Therefore, for every x H, we can deﬁne ∈ ∈ e e T (x) u := x v0. − T (v0) Then T (u) = 0 i.e. u M. It follows that ∈

T (x) 2 T (x) 2 (x, v0)=(u, v0)+ v0 = v0 T (v0) k k T (v0) k k or T (v0) T (v0) T (x)= 2 (x, v0)= x, 2 v0 , v v ! k 0k k 0k which is of the desired form. The uniqueness of h can be seen as follows. If T (x) = 2 (x, h)=(x, h) then (x, h h) = 0 for all x H. In particular h h =(h h, h h)= − ∈ − − −

0 i.e. h = h. It remains to prove the statement about the norm T H →C = T . Firstly, e e k ke k ek e e T = sup T (x) = sup (x, h) h . k k kxk≤1 | | kxk≤1 | | ≤ k k

On the other hand T (h/ h ) = h implies that T h . Thus T = h . This ﬁnishes the proof. k k k k k k ≥ k k k k k k

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

⊥ M ⊥⊥ := M ⊥ = M.

Proof. It is not so diﬃcult to check that Ä ä ⊥ M ⊥ = M .

That’s why Ä ä ⊥ ⊥ M ⊥⊥ = M and Theorem 1 implies that Ä ä ⊥ ⊥ H = M M , H = M M ⊥⊥. ⊕ ⊕ Uniqueness of this representation guarantees that M ⊥⊥ = M. Ä ä Ä ä Remark. In the frame of this theorem we have that

x 2 = u 2 + v 2 , v 2 =(x, v). k k k k k k k k Deﬁnition. Let A H be a subset of an inner product space. The subset ⊂ k span A := x H : x = λ x , x A, λ C  ∈ j j j ∈ j ∈   jX=1  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 ⊂ span B = H

N k C i.e. for every x H and every ε > 0 there exist k and cj j=1 such that ∈ ∈ { } ⊂ k x c x <ε, x B. j j j − j=1 ∈ X

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

3) An orthonormal system B = xα α∈A in H which is a basis is called an orthonormal basis. { }

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

8 ∞ Theorem 2 (Characterization of an orthonormal basis). Let B = xj j=1 be an orthonormal system in a separable Hilbert space H. Then the following{ statements} are equivalent: 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, x ) = 0, j = 1, 2,... implies that x = 0. ∈ j 3) Every x H has the Fourier expansion ∈ ∞ x = (x, xj)xj jX=1 i.e. k x (x, x )x 0, k . j j − j=1 → → ∞ X

This means that B is an orthonormal basis.

4) Every pair x, y H satisﬁes the completeness relation ∈ ∞ (x, y)= (x, xj)(y, xj). jX=1

5) Every x H satisﬁes the Parseval equality ∈ ∞ x 2 = (x, x ) 2. k k | j | jX=1

Proof. 1) 2) Suppose that there is z H, z = 0 such that (z, xj) = 0 for all j =⇒ 1, 2,.... Then ∈ 6 z B′ := , x , x ,... z 1 2 k k is an orthonormal system in H. This® fact implies´ that B is not maximal. It contradicts 1) and proves 2).

2) 3) Given x H introduce the sequence ⇒ ∈ k (k) x = (x, xj)xj. jX=1 Theorem of Pythagoras and Bessel’s inequality (Exercises 1 and 2) imply that

k 2 x(k) = (x, x ) 2 x 2 . | j | ≤ k k j=1 X

9 It follows that ∞ (x, x ) 2 | j | jX=1 converges. That’s why, for m < k,

k 2 x(k) x(m) = (x, x ) 2 0 − | j | → j=m+1 X

as k,m . Hence x(k) is a Cauchy sequence in H. Thus there exists y H such that→ ∞ ∈ ∞ (k) y = lim x = (x, xj)xj. k→∞ jX=1 Next, since the inner product is continuous we deduce that

(k) (y, xj) = lim (x , xj)=(x, xj) k→∞

for any j = 1, 2,.... Therefore (y x, xj) = 0 for any j = 1, 2,.... Part 2) implies that y = x and part 3) follows. − 3) 4) Let x, y H. We know from part 3) that ⇒ ∈ ∞ ∞ x = (x, xj)xj, y = (y, xk)xk. jX=1 kX=1 ∞ Continuity of the inner product and orthonormality of xj j=1 allow us to conclude that { } ∞ ∞ ∞ (x, y)= (x, xj)(y, xk)(xj, xk)= (x, xj)(y, xj). jX=1 kX=1 jX=1 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 it ⇒ which is orthogonal to B. Parseval’s equality gives then ∈ ∞ 1= z 2 = (z, x ) 2 = 0. k k | j | jX=1 This contradiction proves the result.

∞ Exercise 8. Let xj j=1 be an orthonormal system in an inner product space H. Let x H, c k C{ and} k N. Prove that ∈ { j}j=1 ⊂ ∈ k k x (x, x )x x c x . j j j j − j=1 ≤ − j=1 X X

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. The space ∈ ∈ N(A) := x D(A) : Ax = 0 { ∈ } is called the nullspace (or the kernel) of A. The space R(A) := y H : y = Ax for some x D(A) { ∈ ∈ } is called the range of A. Both N(A) and R(A) are linear subspaces of H. We say that A is bounded if there exists M > 0 such that Ax M x , x D(A). k k ≤ k k ∈ 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 := inf M : Ax M x , x D(A) k kH→H { k k ≤ k k ∈ } or equivalently

A H→H = sup Ax . k k kxk=1 k k 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 ∈ × b Kf(s)=c K(s,t)f(t)dt, f L2(a, b). Za ∈ Let us prove that K isc bounded. Indeed,

b b b 2 2 c 2 Kf 2 = Kf(s) ds = K(s,t)f(t)dt ds L (a,b) a | | a a Z Z Z b b c c 2 2 2 2 = (K(s, ), f)L ds K(s, ) L 2 f 2 ds a · ≤ a k · k L Z Z b b b = K(s,t) 2 dt f(t) 2dt ds Za Za | | Za | | 2 2 = K 2 f 2 k ÇkL ((a,b)×(a,b)) k kL (a,b) å 11 where we have made use of Cauchy-Schwarz-Bunjakovskii inequality. That’s why we have

K K 2 . L2→L2 ≤ k kL ((a,b)×(a,b))

The norm c K 2 := K k kL ((a,b)×(a,b)) HS

is called the Hilbert-Schmidt norm of K. c

Example 2.2 (Diﬀerential operator incL2). Consider the diﬀerential operator d A := i 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 ¶ © 1 1 1 ′ 1 ′ ′ (Af, g)= if (t)g(t)dt = i fg 0 f(t)g (t)dt = f(t)ig (t)dt =(f, Ag) Z0 | − Z0 Z0 for all f,g D(A). Let us nowñ consider the sequenceô ∈

un(t) := sin(nπt), n = 1, 2,.... Clearly u D(A) and n ∈ 1 2 2 1 un L2 = sin(nπt) dt = . k k Z0 | | 2 But

1 2 1 2 d 2 2 2 1 2 2 Aun L2 = i sin(nπt) dt =(nπ) cos(nπt) dt =(nπ) =(nπ) un L2 . k k 0 dt 0 | | 2 k k Z Z

Therefore A is not bounded. This shows that D(A)= H is an essential assumption in Exercise 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 defined as 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 inner product in H H can be defined as × ×

((x1; y1) , (x2; y2))H×H := (x1, x2)H +(y1, y2)H for any (x ; y ) , (x ; y ) H H. 1 1 2 2 ∈ × 12 Deﬁnition. The operator A is called closed if Γ(A) = Γ(A). We denote this fact by A = A. By deﬁnition, the criterion for closedness is that

x D(A) n ∈ x D(A) xn x ∈  → ⇒ y = Ax. Ax y  n →   The reader is asked to verify that it is also possible to use seemingly weaker, but equivalent, criterion:

xn D(A) ∈w x D(A) xn x ∈  → ⇒ y = Ax, Ax w y  n →   where x w x indicates weak convergence in the sense that n → (x , y) (x, y) n → for all y H. ∈ Deﬁnition. Let A and A1 be two linear operators in a Hilbert space H. We say that A1 is an extension of A (and A is a restriction of A1) if D(A) D(A1) and Ax = A1x for all x D(A). We denote this fact by A A and A = A ⊂ . ∈ ⊂ 1 1|D(A)

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

A = A1. A\⊂A1 A1=A1

Here, by A A , we mean the operator whose domain is D(A A ) := D(A ) D(A ) 1 ∩ 1 1 ∩ 1 1 ∩ 1 and (A1 A1)x := A1x = A1x, x D(A1 A1), › ∩ ∈ ∩ › › whenever A A = A and A A = A . ⊂ 1 1 ⊂ 1 1 If A is closable then Γ ›A = Γ(A). › › Deﬁnition. Consider the subspace› › Ä ä D∗ := v H : there exists h H such that (Ax, v)=(x, h) for all x D(A) . { ∈ ∈ ∈ } The operator A∗ with the domain D(A∗) := D∗ and mapping A∗v = h is called the adjoint 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 sense that 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)= χ (x)g(x),g L2 for some n N , ∈ n ∈ ∈ where ¶ 0, 0 x 1/n © χn(x)= ≤ ≤ 1, 1/n < x 1.  ≤ It is clear that D(A)= L2(0, 1). For v D(A∗) we have ∈ 1 1 −α −α ∗ (Af, v)= x χn(x)g(x)v(x)dx = f(x)x v(x)dx =(f, A v). Z0 Z0 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 ¶ © xα, 1/n < x 1 fn(x)= ≤ 0, 0 x 1/n.  ≤ ≤ Then f D(A) and  n ∈ 1, 1/n < x 1 Afn(x)= ≤ 0, 0 x 1/n.  ≤ ≤ If we assume that A = A then 

fn D(A) α ∈ α x D(A) fn x ∈  → ⇒ 1= Axα. Af 1  n →   But xα / D(A). This contradiction shows us that A is not closed. It is not bounded either 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 then A = A∗.

Proof. 1) Let us deﬁne in H H the linear and bounded operator V as the mapping Ä ä × V :(u; v) (v; u). → − It has the property V 2 = I. The equality (Au, v)=(u, A∗v) for u D(A) and v 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∗)= V Γ(A) . Since the orthogonal complement is always closed then Γ (A∗) is closed. This proves 1).

2) Assume D(A∗) = H. ThenÄ we canä deﬁne A∗∗ := (A∗)∗ and due to part 1) we may conclude that Γ(A∗∗) V Γ(A∗). ⊥ ⊥⊥ Next, since Γ(A) is a closed subspace of H H we have Γ(A)= Γ(A) . Since V 2 = I and V is bounded then × − ⊥⊥ ⊥ ⊥ Ä ä Γ(A)= V 2Γ(A) = V V Γ(A) = (V Γ(A∗))⊥ − − −

by 1). Hence Ä ä Å Ä ä ã Γ(A) V Γ(A∗). ⊥ It follows that Γ(A)=Γ(A∗∗) or Γ(A)=Γ(A∗∗) or A = A∗∗. This proves 2) in one direction. Let us assume now that A is closable but D(A∗) = ∗ 6 H. It is equivalent to the fact that there exists u0 = 0 such that u0 D(A ). In ∗ 6 ⊥ that 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). Since A is closable∈ and⊂V is× bounded then (u ; 0) V Γ(A) or ∈ 0 ∈ V (u ; 0) = (0; u ) Γ(A) − 0 0 ∈

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

15 3) Since A is closable then

1) 2) 2) ∗ A∗ = A∗ = (A∗)∗∗ =(A)∗∗∗ =(A∗∗)∗ = A .

This ﬁnishes the proof. Ä ä

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

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

2 where u0 0, u0 L (R) is ﬁxed and f0 = 0 is an arbitrary but ﬁxed constant. We consider A6≡on the∈ domain 6

2 2 1 D(A)= u L (R) : f0u(x) dx < = L (R) L (R). R ∈ Z | | ∞ ∩ It is known that L2(Rß) L1(R) = L2(R). Thus A is™ densely deﬁned. Let v be an element of D(A∗). Then∩

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

It means that ∗ Ä ä A v =(v, u0)f0. But (v, u )f must belong to L2(R). Since (v, u )f is a constant and f = 0 then 0 0 0 0 0 6 (v, u0) must be equal to 0. Thus u D(A∗) 0⊥ which implies that u D(A∗). 0⊥ Since u = 0 then D(A∗) = H. Thus A∗ exists but is not densely deﬁned. 0 6 6 Exercise 11. Assume that A is closable. Prove that D(A) can be obtained as the closure of D(A) by the norm

1/2 Au 2 + u 2 . k k k k Deﬁnition. Let A : H H with D(A)= H. We say that A is → Ä ä 1. symmetric if A A∗; ⊂ 2. self-adjoint if A = A∗;

∗ 3. essentially self-adjoint if A = A.

Ä ä

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 closed extension of A then ∗ A A A∗ = A ⊂ ⊂ i.e. A is also symmetric. Ä ä 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 is bounded.

2,3) Let us assume that u0 D(A) and Au0 = 0. Then for any v D(A) we obtain that ∈ ∈ 0=(Au0,v)=(u0, Av). −1 It means that u0 H and therefore u0 = 0. It follows that A exists and D(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 (A−1)∗ then ∈ ∈ ∗ (u, v)=(A−1Au, v)=(Au, A−1 v). Ä ä This equality implies that ∗ A−1 v D(A∗)Ä ä ∈ and ∗ ÄA∗ Aä −1 v = v. (2.1)

Ä ä 17 Similarly, if u D(A−1) and v D(A∗) then ∈ ∈ (u, v)=(AA−1u, v)=(A−1u, A∗v)

and therefore ∗ A∗v D A−1 ∈ and ∗ A−1 AÄÄ∗v = äv.ä (2.2) It follows from (2.1) and (2.2) that (A∗)−1 exists and (A∗)−1 =(A−1)∗. Ä ä Exercise 12. Let A and B be injective operators. Prove that if A B then A−1 B−1. ⊂ ⊂ Since A A∗ we have by Exercise 12 that ⊂ ∗ A−1 (A∗)−1 = A−1 ⊂ i.e. A−1 is also symmetric. But D(A−1)= H. That’s why we may conclude that H = D(A−1) D (A−1)∗ H and hence ÄD(A−ä1) = D (A−1)∗ = H. Thus A−1 is self-adjoint⊂ and bounded⊂ (Hellinger-Toeplitz theorem). Finally,

∗ Ä ä A−1 = A−1 =(A∗)−1 Ä ä

if and only if A = A∗. Ä ä

Theorem 3 (Basic criterion of self-adjointness). If A A∗ then the following statements 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. u ⇒D(A∗)= D(A) and Au = iu . Then ∈ − 0 ∈ 0 0 i(u , u )=(iu , u )=(Au , u )=(u , Au )=(u , iu )= i(u , u ). 0 0 0 0 0 0 0 0 0 0 − 0 0 ∗ ∗ This implies that u0 = 0 i.e. N(A iI)= 0 . The proof of N(A + iI)= 0 is left to the reader. − { } { }

18 2) 3) Since A = A and N(A∗ iI)= 0 then, for example, the equation A∗u = iu ⇒ has only the trivial solution± u = 0.{ It} implies that R(A iI)= H. For otherwise− − there exists u0 = 0 such that u0 R(A iI). It means that for any u D(A) we have 6 ⊥ − ∈ ((A iI)u, u ) = 0 − 0 ∗ ∗ ∗ and therefore u0 D(A + iI) and (A + iI)u0 =0or A u0 = iu0, u0 = 0. This contradiction proves∈ that R(A iI) = H. Next, since A is closed− then6 Γ(A) is also 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 k − k − − k k − k k = Au 2 + u 2 , u D(A). k k k k ∈ That’s why if (A iI)un v0 then Aun and un are convergent i.e. Aun ′ ′ − → ′ → v0, un u0 and un D(A). The closedness of A implies that u0 D(A) and ′ →′ ∈ ′ ′ ∈ v0 = Au0 i.e. (A iI)un Au0 iu0 = v0. It means that R(A iI) is a closed set i.e. R(A iI)=− R(A→ iI)=−H. The proof of R(A + iI)=−H is left to the reader. − − 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 there exists v D(A) such∈ that − ∈ 0 ∈ (A iI)v =(A∗ iI)u. − 0 − It is clear that u v D(A∗) (since A A∗) and − 0 ∈ ⊂ (A∗ iI)(u v ) = (A∗ iI)u (A∗ iI)v =(A∗ iI)u (A iI)v − − 0 − − − 0 − − − 0 = (A iI)v (A iI)v = 0. − 0 − − 0 Hence u v N(A∗ iI). − 0 ∈ − Exercise 13. Let A be a linear and densely deﬁned operator in the Hilbert space H. 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∗). − { }

2 d Exercise 14. Let H = L (0, 1),A := i dt and D (A)= f L2(0, 1) : f ′ L2(0, 1), f(0) = f(1)eiγ,γ R . γ ∈ ∈ ∈ 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-adjoint and 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 = 0. k k 6 2) P is a projector if and only if P ⊥ := I P is a projector. − ⊥ 3) H = R(P ) R(P ), P = I and P ⊥ = 0. ⊕ |R(P ) |R(P ) 4) There is one-to-one correspondence between projectors on H and closed linear subspaces of H. More precisely, if M H is a closed linear subspace then there ⊂ exists a projector PM : H M and, conversely, if P : H H is a projector then R(P ) is a closed linear→ subspace. →

5) If e N ,N is an orthonormal system then { j}j=1 ≤ ∞ N P x := (x, e )e , x H N j j ∈ jX=1 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, k k k k k k k k P ∗P P ∗ P P 2 k k ≤ k k k k ≤ k k and

P 2 = sup P x 2 = sup (P x, P x) = sup (P ∗P x, x) sup P ∗P x k k kxk=1 k k kxk=1 kxk=1 ≤ kxk=1 k k = P ∗P . k k Therefore P = P 2 or P = 1 if P = 0. k k k k k k 6 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 ) = R(P ⊥) . First ∈ ∈ ⊥ assume that w R(P ⊥) i.e. (w, (I P )x) = 0 for all x H. This is ∈ − ∈ equivalent to Ä ä (w, x)=(w,Px)=(Pw,x), x H Ä ä ∈ ⊥ or Pw = w. Hence w R(P ) and so we have proved that R(P ⊥) R(P ). ∈ ⊂ For the opposite embedding we let w R(P ). Then there exists xw H such that w = P x . If z R(P ⊥) then z = ∈P ⊥x =(I P )x for some x ∈H. Thus w ∈ z − z Ä zä∈ (w, z)=(P x , (I P )x )=(P x , x ) (P x , P x ) = 0 w − z w z − w z ⊥ since P is a projector. Therefore w R(P ⊥) and we may conclude that ⊥ ∈ R(P ) = R(P ⊥) . This fact allows us to conclude that R(P ) = R(P ) and H = R(P ) R(P ⊥). Moreover, it is easyÄ to checkä by deﬁnition that P = I ⊕ |R(P ) and P ⊥ = 0. |R(PÄ ) ä 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 P : H M ∈ ∈ ∈ M → as PM x = u. 2 2 It is clear that PM x = PM u = u = PM x i.e. PM = PM . Moreover, if y H then y = u + v , u M,v M ⊥ and ∈ 1 1 1 ∈ 1 ∈

(PM x, y)=(u, u1 + v1)=(u, u1)=(u + v, u1)=(u + v, PM y)=(x, PM y)

∗ i.e. PM = 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 = c e , c 2 < .  ∈ j j | j| ∞  jX=1 jX=1    Then M is a closed subspace of H. If we deﬁne a linear operator PM as

∞ P x := (x, e )e , x H M j j ∈ jX=1 then by Bessel’s inequality we obtain that P x M and M ∈ P x x . k M k ≤ k k

21 It means that PM is a bounded linear operator into M. But PM ej = ej and thus P 2 x = P x for all x H. Next, for all x, y H we have M M ∈ ∈ ∞ ∞ ∞ (PM x, y) = (x, ej)ej, y = (x, ej)(ej, y)= (x, (y,ej)ej) jX=1 jX=1 jX=1 Ñ ∞ é = x, (y,ej)ej =(x, PM y) jX=1 Ñ é ∗ i.e. PM = 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 or equal 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 denote A> 0, and say that A is positive≤ , if A c I for some c > 0. ≥ ≥ 0 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) P x Qx for all x H. k k ≤ k k ∈ 3) R(P ) R(Q). ⊂ 4) P = P Q = QP .

Proof. 1) 2) Follows immediately from (P x, x)=(P 2x, x)=(P x, P x)= P x 2 . ⇔ k k 3) 4) Assume R(P ) R(Q). Then QP x = P x or QP = P . Conversely, if QP = P ⇔ then clearly R(P⊂) R(Q). Finally, P = QP = P ∗ =(QP )∗ = P ∗Q∗ = P Q. ⊂ 2) 4) If 4) holds then P x = P Qx and P x = P Qx Qx for all x H. ⇔ Conversely, if P x Qx then P x k= QPk x +kQ⊥P xkimplies ≤ k k that ∈ k k ≤ k k 2 P x 2 = QP x 2 + Q⊥P x QP x 2 . k k k k ≤ k k

Hence 2 Q⊥P x = 0

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

22 Exercise 15. Let P ∞ be a sequence of projectors with P P for each j = { j}j=1 j ≤ j+1 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 k k k k ∈ 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 ∗ = I i.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 a spectral family if the following conditions are satisfied: { } 1) E is a projector for all λ R. λ ∈ 2) E E for all λ<µ. λ ≤ µ 3) E is right continuous with respect to the strong operator topology i.e. { λ}

lim Esx Etx = 0 s→t+0 k − k for all x H. ∈ 4) E is normalized as follows: { λ}

lim Eλx = 0, lim Eλx = x λ→−∞ k k λ→+∞ k k k k for all x H. The latter condition can also be formulated as ∈

lim Eλx x = 0. λ→+∞ k − k

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

EλEµ = Emin{λ,µ}.

Proposition 3. For every ﬁxed x, y H, (Eλx, y) is a function of bounded variation with respect to λ R. ∈ ∈ 23 Proof. Let us deﬁne E(α, β] := E E , α<β. β − α Then E(α, β] is a projector. Indeed,

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

(E(α, β])2 = (E E )(E E )= E2 E E E E + E2 β − α β − α β − α β − β α α = E E E + E = E(α, β]. β − α − α α Another property is that

E(α , β ]x E(α, β]y, x, y H 1 1 ⊥ ∈ if β α or β α . To see this for β α calculate 1 ≤ ≤ 1 1 ≤ (E(α , β ]x, E(α, β]y) = (E x E x, E y E y) 1 1 β1 − α1 β − α = (E x, E y) (E x, E y) (E x, E y)+(E x, E y) β1 β − α1 β − β1 α α1 α = (x, E y) (x, E y) (x, E y)+(x, E y) = 0. β1 − α1 − β1 α1 Let now λ <λ < <λ . 0 1 ··· n Then

n n E x, y E x, y = (E(λ ,λ ]x, y) λj − λj−1 | j−1 j | j=1 j=1 X X n Ä ä Ä ä = (E(λ ,λ ]x, E(λ ,λ ]y) | j−1 j j−1 j | jX=1 n E(λ ,λ ]x E(λ ,λ ]y ≤ k j−1 j k k j−1 j k jX=1 1/2 1/2 n n E(λ ,λ ]x 2 E(λ ,λ ]y 2 ≤ k j−1 j k k j−1 j k jX=1 jX=1 Ñn né Ñ é = E(λ ,λ ]x E(λ ,λ ]y j−1 j j−1 j j=1 j=1 X X

= E(λ0,λn]x E( λ 0,λn]y x y . k k k k ≤ k k k k 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 continuous function f(λ) we may conclude that the limit

n n ∗ ∗ lim f(λj )(E(λj−1,λj]x, y) = lim f(λj )E(λj−1,λj]x, y , ∆→0 ∆→0 jX=1 jX=1 Ñ é ∗ where λj [λj−1,λj], α = λ0 < λ1 < < λn = β and ∆ = max1≤j≤n λj−1 λj exists and∈ by deﬁnition this limit is ··· | − |

β f(λ)d(Eλx, y), x, y H. Zα ∈ It can be shown that this is equivalent to the existence of the limit in H

n ∗ lim f(λj )E(λj−1,λj]x, ∆→0 jX=1 which we denote by β f(λ)dEλx. Zα Thus β β f(λ)d(Eλx, y)= f(λ)dEλx, y , x, y H. Zα Zα ∈ For the spectral representation of self-adjointÇ operatorså one needs not only integrals over ﬁnite intervals but also over whole line which is naturally deﬁned as the limit

∞ β ∞ f(λ)d(Eλx, y) = lim f(λ)d(Eλx, y)= f(λ)dEλx, y −∞ α→−∞ α −∞ Z β→∞ Z Z if it exists. Deriving first some basic properties of the integralÅ just definedã one can check that ∞ β β f(λ)d(EλEβx, y)= f(λ)d(Eλx, y) := lim f(λ)d(Eλx, y), x, y H. α→−∞ Z−∞ Z−∞ Zα ∈ ∞ Theorem 1. Let Eλ λ=−∞ be a spectral family on the Hilbert space H and let f be a real-valued continuous{ } function on the line. Define ∞ 2 D := x H : f(λ) d(Eλx, x) < ∈ Z−∞ | | ∞ (or D := x H : ∞ f(λ)ßdE x exists ). Let us define on this™ domain an operator A ∈ −∞ λ as R ∞ ¶ (Ax, y)= f(λ)d(Eλ©x, y), x D(A) := D, y H Z−∞ ∈ ∈ (or Ax = ∞ f(λ)dE x, x D(A)). Then A is self-adjoint and satisfies −∞ λ ∈ R E(α, β]A AE(α, β], α<β. ⊂

25 Proof. It can be shown that the integral

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

v E(α, β]v = v E v + E v (I E )v + E v < ε. k − k k − β α k ≤ k − β k k α k On the other hand,

∞ β (Ax, y) = f(λ)d(Eλx, y) = lim f(λ)d(Eλx, y) −∞ α→−∞ α Z β→∞ Z β β = lim f(λ)d(x, Eλy) = lim x, f(λ)dEλy α→−∞ α α→−∞ α β→∞ Z β→∞ Z

β Ç å = x, lim f(λ)dEλy =(x, Ay). α→−∞ α β→∞ Z Ñ é In order to prove that A = A∗ it remains to show that D(A∗) D(A). Let u D(A∗). Then ⊂ ∈ β ∗ (E(α, β]z, A u)=(AE(α, β]z, u)= f(λ)d(Eλz, u) Zα for any z H. This equality implies that ∈ β ∞ ∞ ∗ (z, A u) = lim f(λ)d(Eλz, u)= f(λ)d(Eλz, u)= f(λ)d(z,Eλu) α→−∞ α −∞ −∞ β→∞ Z Z Z ∞ = f(λ)d(Eλu, z)= (Au, z)=(z, Au), Z−∞

26 where the integral exists because (z, A∗u) exists. Hence u D(A) and A∗u = Au. For the second claim we ﬁrst calculate ∈ ∞ E(α, β]Ax =(Eβ Eα) Ax =(Eβ Eα) f(λ)dEλx − − Z−∞ ∞ ∞ β α = f(λ)dEλEβx f(λ)dEλEαx = f(λ)dEλx f(λ)dEλx Z−∞ − Z−∞ Z−∞ − Z−∞ β ∞ = f(λ)dEλx = f(λ)dEλ (Eβ Eα) x = A (Eβ Eα) x = AE(α, β]x Zα Z−∞ − − for any x D(A). Since the left hand side is deﬁned on D(A) and the right hand side on all of H∈ then the latter is an extension of the former.

Exercise 17. Let A be as in Theorem 1. Prove that ∞ 2 2 Au = f(λ) d(Eλu, u) k k Z−∞ | | 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 E ∞ . ∈ { λ}λ=−∞ Theorem 2 (J. von Neumann’s spectral theorem). Every self-adjoint operator A on the Hilbert space H has a unique spectral representation i.e. there is a unique spectral family E ∞ such that { λ}λ=−∞ ∞ Ax = λdEλx, x D(A) Z−∞ ∈ (i.e. (Ax, y)= ∞ λd(E x, y), x D(A), y H), where D(A) is defined as −∞ λ ∈ ∈ R ∞ 2 D(A)= x H : λ d(Eλx, x) < . ∈ Z−∞ ∞ Proof. At first we assume thatß this theorem holds when A is™ bounded, that is, there is a unique spectral family F ∞ such that { µ}µ=−∞ ∞ Au = µdFµu, u H Z−∞ ∈ since D(A)= H in this case. But F 0 for µ M, where µ ≡ µ ≡ m = inf (Ax, x), M = sup (Ax, x). kxk=1 kxk=1

That’s why the spectral representation has a view

M Au = µdFµu, u H. Zm ∈ 27 Let us consider now an unbounded operator which is semibounded from below i.e.

(Au, u) m (u, u), u D(A) ≥ 0 ∈ 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.k The normk ≤ estimate follows from 2 (v, A−1v) A−1v , v D(A−1). ≥ ∈

−1 −1 Since A is bounded then D(A ) is a closed subspace in H. But self-adjointness of A 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 ≤ ≤ k k ∈ we may conclude in this case that m 0, M 1 and ≥ ≤ 1 −1 A v = µdFµv, v H, Z0 ∈ −1 where Fµ is the spectral family of A . Let us note that F1 = I and F0 = 0. They follow 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 1 1 Bεu := dFµu, u D(A). Zε µ ∈ For every v H we have ∈ 1 1 1 1 1 −1 1 −1 1 1 BεA v = dFµ(A v)= dFµ λdFλv = d λd(FµFλv) Zε µ Zε µ Z0 Zε µ Z0 1 1 µ 1 1 1 = d λdFλv = µdFÇ µv = ådFµv = F1vÇ Fεv = v Fåεv. Zε µ Zε Zε µ Zε − − Since every spectral familyÅ is rightã continuous then

−1 lim BεA v = v ε→0+0 exists. For every u D(A) we have similarly, ∈ 1 1 µ −1 1 A Bεu = µdFµ(Bεu)= µd dFλu = u Fεu Z0 Zε Zε λ − and hence Ç å −1 lim A Bεu = u ε→0+0

28 exists. These two equalities mean that

−1 −1 lim Bε = A = A ε→0+0 exists and the spectral representation Ä ä 1 1 1 1 A = dFµ = lim dFµ ε→0+ Z0 µ Zε µ holds. If we deﬁne Eλ = I F 1 , 1 λ< then − λ ≤ ∞ 1 1 ∞ A = dE 1 = λdEλ. µ − Z0 µ Z1 Exercise 19. Prove that this E is a spectral family. { λ} Domain D(A) can be characterized as

∞ 1 2 1 D(A)= u H : λ d(Eλu, u) < = u H : 2 d(Fµu, u) < . ∈ Z1 ∞ ∈ Z0 µ ∞ This proves theß 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

p (A) := a I + a A + + a AN , N N, N 0 1 ··· N ∈

where aj R for j = 0, 1,...,N. Then pN (A) is also self-adjoint and bounded with ∈ pN (A) sup pN (t) . k k ≤ |t|≤kAk | |

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

f(A) p (A) < ε. k − N k 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 | | ≤ | | ≤ k k k k k k 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

b L(f)= f(x)dν(x), Za where ν is a measure that satisﬁes the conditions

1) L(f) 0 for f 0 ≥ ≥ 2) L(f) ν(K) f , where K [a, b] is compact and | | ≤ k kK ⊂ f = max f(x) . k kK x∈K | | Step 5. It follows from Step 4 that

M (Au, v)= λdν(λ; u, v). Zm Step 6. It is possible to prove that ν(λ; u, v) is a self-adjoint bilinear form. That’s why we may conclude that there exists a self-adjoint and bounded operator Eλ such that ν(λ; u, v)=(Eλu, v).

This operator is idempotent and we may deﬁne Eλ 0 for λ

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

∞ Au = λdEλu, u D(A). Z−∞ ∈ For every continuous function f we can deﬁne

∞ 2 Df := u H : f(λ) d(Eλu, u) < . ∈ Z−∞ | | ∞ This set is a linear subspaceß of H. For every u D and v H™let us deﬁne the linear ∈ f ∈ functional ∞ ∞ L(v) := f(λ)d(Eλu, v)= f(λ)dEλu, v . Z−∞ Z−∞ This functional is continuous because it is bounded. Indeed, Å ã ∞ 2 ∞ 2 2 2 2 2 L(v) f(λ)dEλu v = f(λ) d(Eλu, u) v = c(u) v . | | ≤ Z−∞ k k Z−∞ | | k k k k

30 By the Riesz-Frechet theorem this functional can be expressed in the form of an inner product i.e. there exists z H such that ∈ ∞ f(λ)d(Eλu, v)=(z,v), v H. Z−∞ ∈ We set z := f(A)u, u D ∈ f i.e. ∞ (f(A)u, v)= f(λ)d(Eλu, v). Z−∞ Remark. Since in general f is not real-valued then f(A) is not a self-adjoint operator in general. Example 3.1. Consider λ i f(λ)= − , λ R λ + i ∈ Denote ∞ λ i UA := f(A)= − dEλ. Z−∞ λ + i The operator UA is called the Cayley transform. Since f(λ) = 1 then Df = D(UA)= H and | |

∞ β 2 2 UAu = f(λ) d(Eλu, u) = lim d(Eλu, u) = lim ((Eβu, u) (Eαu, u)) k k −∞ | | α→−∞ α α→−∞ − Z β→∞ Z β→∞ 2 2 2 = lim Eβu Eαu = u α→−∞ k k − k k k k β→∞ by normalization ofÄ Eλ . Hence UA isä an isometry. There is one-to-one correspondence between self-adjoint{ operators} and their Cayley transforms. Indeed,

U =(A iI)(A + iI)−1 A − is equivalent to I U = 2i(A + iI)−1 − A I + U = 2A(A + iI)−1  A or  A = i(I + U )(I U )−1. A − A Example 3.2. Consider 1 f(λ)= , λ R, z C, Im z = 0. λ z ∈ ∈ 6 − Denote ∞ −1 1 Rz := (A zI) = dEλ. − −∞ λ z Z − 31 The operator Rz is called the resolvent of A. Since 1 1

λ z ≤ Im z − | | for all λ R then R is bounded and deﬁned on whole H. ∈ z Exercise 20. Let A = A∗ with spectral family E . Let u D(f(A)) and v D(g(A)). λ ∈ ∈ Prove that ∞ (f(A)u, g(A)v)= f(λ)g(λ)d(Eλu, v). Z−∞ ∗ Exercise 21. Let A = A with spectral family Eλ. Let u D(f(A)). Prove that f(A)u D(g(A)) if and only if u D((gf)(A)) and that ∈ ∈ ∈ ∞ (gf)(A)u = g(λ)f(λ)dEλu. Z−∞ 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 from H to D(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 operator −1 Rz := (A zI) is an analytic function from ρ(A) to B(H; H), the set of all linear operators− in H. Furthermore, the resolvent identity

R R =(z ξ)R R , z,ξ ρ(A) z − ξ − z ξ ∈ ′ 2 holds and Rz =(Rz) . 2) If A = A∗ then z ρ(A) if and only if there exists C > 0 such that ∈ z (A zI)u C u k − k ≥ z k k for all u D(A). ∈

Proof. 1) Assume that z0 ρ(A). Then Rz0 is a bounded linear operator from H to D(A) and thus r := R −∈1 > 0. Let us deﬁne for z z < r the operator k z0 k | − 0| G := (z z )R . z0 − 0 z0 Then G is bounded with G < 1. Hence it deﬁnes the operator z0 k z0 k ∞ (I G )−1 = (G )j − z0 z0 jX=0 because this Neumann series converges. But for z z < r we have | − 0| A zI =(A z I)(I G ) − − 0 − z0 or (A zI)−1 =(I G )−1R . − − z0 z0 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 G )−1R x = (z z )j (R )j+1 x − − z0 z0 − 0 z0 jX=0 n j j+1 = lim (z z0) (Rz ) x. n→∞ − 0 jX=0 −1 It follows from this representation that Rz = (A zI) is an analytic function from ρ(A) to B(H; H). Next we denote −

n s x := (z z )j (R )j+1 x. n − 0 z0 jX=0 It is clear that s x D(A) and that lim s x = y. Moreover, n ∈ n→∞ n

lim (A zI)snx = x. n→∞ −

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

y D(A) n ∈ y D(A) yn y ∈  → ⇒ x =(A zI)y. (A zI)y x  − − n →   Hence y =(A zI)−1x D(A) and therefore ρ(A) is open. The resolvent identity is proved by straightforward− ∈ calculation

R R = R (A ξI)R R (A zI)R = R [(A ξI) (A zI)] R z − ξ z − ξ − z − ξ z − − − ξ = (z ξ)R R . − z ξ Finally, the limit Rz Rξ 2 lim − = lim RzRξ =(Rz) z→ξ z ξ z→ξ − ′ 2 exists and hence Rz =(Rz) exists. It proves this part. 2) Assume that A = A∗. If z ρ(A) then by deﬁnition R maps from H to D(A). ∈ z Hence there exists Mz > 0 such that R v M v , v H. k z k ≤ z k k ∈ Since u = R (A zI)u for any u D(A) then we get z − ∈ u M (A zI)u , u D(A). k k ≤ z k − k ∈ This is equivalent to 1 (A zI)u u , u D(A). k − k ≥ Mz k k ∈

34 Conversely, if there exists Cz > 0 such that (A zI)u C u , u D(A). k − k ≥ z k k ∈ then (A zI)−1 is bounded. Since A is self-adjoint then (A zI)−1 is deﬁned over − − whole H. Indeed, if R(A zI) = H then there exists v0 = 0 such that v0 R(A zI). This means that − 6 6 ⊥ − (v , (A zI)u) = 0, u D(A) 0 − ∈ or (Au, v0)=(zu,v0) or ∗ (u, A v0)=(u, zv0). Thus v D(A∗) and A∗v = zv . Since A = A∗ then v D(A) and Av = zv or 0 ∈ 0 0 0 ∈ 0 0 (A zI)v = 0. − 0 It is easy to check that (A zI)u 2 = (A zI)u 2 for any u D(A). Therefore k − k k − k ∈ (A zI)v = (A zI)v C v . k − 0k k − 0k ≥ z k 0k Hence v = 0 and D ((A zI)−1)= R(A zI)= H. It means that z ρ(A). 0 − − ∈

Corollary 1. If A = A∗ then σ(A) = , σ(A)= σ(A) and σ(A) R. 6 ∅ ⊂ Proof. If z = α + iβ C with Im z = β = 0 then ∈ 6 (A zI)x 2 = (A αI)x iβx 2 = (A αI)x 2 + β 2 x 2 β 2 x 2 . k − k k − − k k − k | | k k ≥ | | k k It implies (see part 2) of Theorem 1) that z ρ(A). It means that σ(A) R. Since A = A∗ and therefore closed then the spectrum∈ σ(A) is closed as a complement⊂ of an open set (see part 1) of Theorem 1). It remains to prove that σ(A) = . Assume on the contrary that σ(A)= . Then 6 ∅ ∅ the resolvent Rz is an entire analytic function. Let us prove that Rz is uniformly bounded with respect to z C. Introduce the functional k k ∈ T (y):=(R x, y), x = 1, y H. z z k k ∈

Then Tz(y) is a linear functional on the Hilbert space H. Moreover, since Rz is bounded for any (ﬁxed) z C then ∈ T (y) R x y R y = C y . | z | ≤ k z k k k ≤ k zk k k z k k

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

sup Tz = c0 < . z∈C k k ∞

35 That’s why we have

T (y) = (R x, y) c y , x = 1, z C. | z | | z | ≤ 0 k k k k ∈

It implies that Rzx c0 i.e. Rz c0. By Liouville theorem we may conclude now that R constant.k k But ≤ by J.k vonk Neumann’s ≤ spectral theorem z ≡ ∞ 1 Rz = dEλ, −∞ λ z Z − where E is a spectral family of A = A∗. Due to the estimate { λ} 1 R k zk ≤ Im z | | we may conclude that Rz 0 as Im z . Hence Rz 0. This contradiction ﬁnishes the proof. k k → | | → ∞ ≡ Exercise 22. [Weyl’s criterion] Let A = A∗. Prove that λ σ(A) if and only if there exists x D(A), x = 1 such that ∈ n ∈ k nk

lim (A λI)xn = 0. n→∞ k − k

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

σ (A)= λ σ(A) : N(A λI) = 0 . p { ∈ − 6 { }} It means that (A λI)−1 does not exist i.e. there exists a non-trivial u D(A) such − ∈ that Au = λu. The complement σ(A) σp(A) is the continuous spectrum σc(A). The discrete spectrum is the set \

σ (A)= λ σ (A) : dim N(A λI) < and λ is isolated in σ(A) . d { ∈ p − ∞ } The set σ (A) := σ(A) σ (A) is called the essential spectrum of A. ess \ d In the frame of this deﬁnition, the complex plane can be divided into regions ac- cording to C = ρ(A) σ(A), ∪ σ(A)= σ (A) σ (A) p ∪ c and σ(A)= σ (A) σ (A), d ∪ ess with all the unions being disjoint. Remark. If A = A∗ then 1) λ σ (A) means that (A λI)−1 exists but is not bounded. ∈ c − 36 2) σess (A)= σc(A) eigenvalues of inﬁnite multiplicity and their accumulation points ∪{ accumulation points of σ (A) . } ∪ { d } Exercise 23. Let A = A∗ and λ ,λ σ (A). Prove that if λ = λ then 1 2 ∈ p 1 6 2 N(A λ I) N(A λ I). − 1 ⊥ − 2 Exercise 24. Let e ∞ be an orthonormal basis in H and let s ∞ C be some { j}j=1 { j}j=1 ⊂ sequence. Introduce the set

∞ D = x H : s 2 (x, e ) 2 < .  ∈ | j| | j | ∞  jX=1  Deﬁne   ∞ Ax = s (x, e )e , x D. j j j ∈ jX=1 Prove that A = A and that σ(A)= s : j = 1, 2,... . Prove also that { j } ∞ −1 1 (A zI) x = (x, ej)ej − sj z jX=1 − for any z ρ(A) and x D. ∈ ∈ Exercise 25. Prove that the spectrum σ(U) of a unitary operator U lies on the unit circle in C.

∗ Theorem 2. Let A = A and let E R be its spectral family. Then { λ}λ∈ 1) µ σ(A) if and only if E E = 0 for every ε> 0. ∈ µ+ε − µ−ε 6 2) µ σ (A) if and only if E E = 0. Here E := lim E in the sense ∈ p µ − µ−0 6 µ−0 ε→0+ µ−ε of strong operator topology.

Proof. 1) Suppose that µ σ(A) but there exists ε> 0 such that E E = 0. ∈ µ+ε − µ−ε Then by spectral theorem we obtain for any x D(A) that ∈ ∞ 2 2 2 (A µI)x = (λ µ) d(Eλx, x) (λ µ) d(Eλx, x) k − k Z−∞ − ≥ Z|λ−µ|≥ε − µ−ε ∞ 2 2 ε d(Eλx, x)= ε + d(Eλx, x) ≥ Z|λ−µ|≥ε Z−∞ Zµ+ε = ε2 (E x, x)+ x 2 (E x, x) = ε2 x 2 . µ−ε k k − ñ µ+ε ô k k This inequality means (see part 2) of Theorem 1) that µ σ(A) but µ ρ(A). This contradiction proves 1) in oneî direction. Conversely, if 6∈ ó ∈

Pn := Eµ+ 1 Eµ− 1 = 0 n − n 6 37 N ∞ for all n then there is a sequence xn n=1 such that xn R(Pn) i.e. xn = Pnxn i.e. x ∈D(A) and x = 1. For this sequence{ } it is true that∈ n ∈ k nk ∞ 2 2 2 (A µI)xn = (λ µ) d(EλPnxn, Pnxn)= (λ µ) d(Eλxn, xn) k − k Z−∞ − Z|λ−µ|≤1/n − ∞ 1 1 2 1 2 d(Eλxn, xn)= 2 xn = 2 0 ≤ n Z−∞ n k k n → as n . Hence, this sequence satisﬁes Weyl’s criterion (see Exercise 22) and therefore µ→ ∞σ(A). ∈ 2) Suppose µ R is an eigenvalue of A. Then there is x D(A), x = 0 such that ∈ 0 ∈ 0 6 ∞ 2 2 0= (A µI)x0 = (λ µ) d(Eλx0, x0). k − k Z−∞ − In particular, for all n N and large enough ε> 0 we have that ∈ n n 2 2 2 0 = (λ µ) d(Eλx0, x0) ε d(Eλx0, x0)= ε ((En Eµ+ε)x0, x0) Zµ+ε − ≥ Zµ+ε − = ε2 (E E )x 2 . k n − µ+ε 0k Thus we may conclude that 0= E x E x . n 0 − µ+ε 0 Similarly we can get that 0= E x E x . −n 0 − µ−ε 0 Letting n and ε 0 we obtain → ∞ →

x0 = Eµx0, 0= Eµ−0x0.

Hence x =(E E )x 0 µ − µ−0 0 and therefore E E = 0. µ − µ−0 6 Conversely, deﬁne the projector

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

E y = E P y = E E y E E y = P y = y. λ λ λ µ − λ µ−0 For λ<µ we have that

E y = E E y E E y = E y E y = 0. λ λ µ − λ µ−0 λ − λ 38 Hence ∞ ∞ 2 2 2 (A µI)y = (λ µ) d(Eλy, y)= (λ µ) dλ(y, y) = 0. k − k Z−∞ − Zµ − That’s why Ay = µy and y D(A), y = 0 i.e. µ is an eigenvalue of A or µ σ (A). ∈ 6 ∈ p

Remark. The statements of Theorem 2 can be reformulated as

1) µ σ (A) if and only if E E = 0. ∈ p µ − µ−0 6 2) µ σ (A) if and only if E E = 0. ∈ c µ − µ−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 H → ∞ into precompact sets in H1 i.e. for every bounded sequence xn n=1 H the sequence Kx ∞ H contains a convergent subsequence. { } ⊂ { n}n=1 ⊂ 1 If K : H H is compact then the following statements hold. → 1 1) K maps every weakly convergent sequence in H into a norm convergent sequence in H1.

2) If H = H1 is separable then every compact operator is a norm limit of a sequence of 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 an orthonormal basis e ∞ . Consider for any n = 1, 2,... the projector { j}j=1 n P x := (x, e )e , x H. n j j ∈ jX=1 Then P P and (I P )x 0 as n . Deﬁne n ≤ n+1 k − n k → → ∞

dn := sup K(I Pn)x K(I Pn) . kxk=1 k − k ≡ k − k

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

lim dn := d 0 n→∞ ≥ exists. Let us choose y R(I P ), y = 1 such that n ∈ − n k nk d K(I P )y = Ky . k − n nk k nk ≥ 2 39 Then

(y , x) = ((I P )y , x) = (y , (I P )x) y (I P )x 0, n | n | | − n n | | n − n | ≤ k nk k − n k → → ∞ w for any x H. It means that yn 0. Compactness of K implies that Kyn 0. Thus d = 0. That’s∈ why → → d = K KP 0. n k − nk → 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 is an eigenvalue of A. ± k k

Proof. Since A = sup (Ax, x) k k kxk=1 | | then there exists a sequence x with x = 1 such that n k nk

A = lim (Axn, xn) . k k n→∞ | |

Actually, we can assume that limn→∞(Axn, xn) exists and equals, say, a. Otherwise ∗ we would take a subsequence of xn . Since A = A then a is real and A = a . Due to the fact that any bounded set{ of} the Hilbert space is weakly compactk k (unit| ball| in our case) we can choose a subsequence of xn , say, xkn which converges weakly i.e. w { } { } x x. Compactness of A implies that Ax y. Next we observe that kn → kn → Ax ax 2 = Ax 2 2a(Ax , x )+ a2 A 2 2a(Ax , x )+ a2 k kn − kn k k kn k − kn kn ≤ k k − kn kn = 2a2 2a(Ax , x ) 2a2 2a2 = 0, − kn kn → − as n . Hence → ∞ Axkn axkn 0 − → xkn x Axk y →  n → ⇒ Ax = ax. x w x  kn →   Since x = 1 then x = 1 also. Hence x = 0 and a is an eigenvalue of A. k kn k k k 6 Theorem 3 (Riesz-Schauder). Suppose A = A∗ is compact. Then

1) A has a sequence of real eigenvalues λj = 0 which can be enumerated in such a way that 6 λ λ λ . | 1| ≥ | 2| ≥ · · · ≥ | j| ≥ · · ·

2) If there are inﬁnitely many eigenvalues then limj→∞ λj = 0 and 0 is the only accumulation point of λ . { j}

3) The multiplicity of λj is ﬁnite.

40 ∞ 4) If ej is the normalized eigenvector for λj then ej j=1 is an orthonormal system and { } ∞ ∞ Ax = λ (x, e )e = (Ax, e )e , x H. j j j j j ∈ jX=1 jX=1 It means that e ∞ is an orthonormal basis on R(A). { j}j=1 5) σ(A)= 0,λ ,λ ,...,λ ,... while 0 is not necessarily an eigenvalue of A. { 1 2 j }

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

(Ax, e1)=(x, Ae1)=(x, λ1e1)= λ1(x, e1) = 0 for any x H . The restriction of the inner product of H to H makes H a Hilbert ∈ 1 1 1 space (since H1 is closed) and the restriction of A to H1, denoted by A1 = A , is |H1 again a self-adjoint compact operator which is mapping in H1. Clearly, its norm is bounded by the norm of A i.e. A A . Applying Lemma to A on H we get an k 1k ≤ k k 1 1 eigenvalue λ2 with λ2 = A1 and a normalized eigenvector e2 with e2 e1. It is clear | | k k ⊥ ⊥ that λ2 λ1 . Next introduce the closed subspace H2 = (span e1,e2 ) . Again, A leaves| H| ≤invariant | | and thus A := A = A is a self-adjoint{ compact} operator in 2 2 1 H2 H2 H . Applying Lemma to A on H we| obtain|λ with λ = A and a normalized 2 2 2 3 | 3| k 2k eigenvector e3 with e3 e2 and e3 e1. This process in the inﬁnite dimensional Hilbert ⊥ ⊥ ∞ space leads us to the sequence λj j=1 such that λj+1 λj and corresponding normalized eigenvectors. Since λ{ >}0 and monotone| decreasing| ≤ | | then there is a limit | j|

lim λj = r. j→∞ | |

Clearly r 0. Let us prove that r = 0. If r> 0 then λj r> 0 for each j = 1, 2,... or ≥ | | ≥ 1 1 < . λ ≤ r ∞ | j| Hence the sequence of vectors ej yj := λj w is bounded and therefore there is a weakly convergent subsequence yjk y. Com- → √ pactness of A implies the strong convergence of Ayjk ejk . But ejk ejm = 2 for k = m. This contradiction proves 1) and 2). ≡ k − k 6 Exercise 26. Prove that if H is an inﬁnite dimensional Hilbert space then the identical operator 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

n P x := (x, e )e , x H. n j j ∈ jX=1 Then I P is a projector onto (span e ,...,e )⊥ H and hence − n { 1 n} ≡ n

A(I Pn)x A (I Pn)x λn+1 x 0 k − k ≤ k kHn k − k ≤ | | k k → as n . Since → ∞ n n APnx = (x, ej)Aej = λj(x, ej)ej jX=1 jX=1 and A(I P )x = Ax AP x 0, n k − n k k − n k → → ∞ then ∞ Ax = λj(x, ej)ej jX=1 and part 4) follows. Finally, Exercise 24 gives immediately that

σ(A)= 0,λ ,λ ,...,λ ,... . { 1 2 j } This ﬁnishes the proof.

∞ Corollary (Hilbert-Schmidt theorem). The orthonormal system ej j=1 of eigenvectors of a compact self-adjoint operator A in a Hilbert space H is an{ orthonormal} basis if 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 there { } ∈ exists yε R(A) such that ∈ x y < ε/2. k − εk But by Riesz-Schauder theorem

∞ yε = Axε = λj(xε,ej)ej. jX=1 Hence

∞ x y = x λ (x ,e )e < ε/2. ε j ε j j k − k − j=1 X

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

n n ∞ ∞ x (x, e )e x λ (x ,e )e = x λ (x ,e )e + λ (x ,e )e j j j ε j j j ε j j j ε j j − j=1 ≤ − j=1 − j=1 j=n+1 X X X X

∞ < ε/2+ λ (x ,e )e j ε j j j=n+1 X 1/2 ∞

ε/2+ λ 2 (x ,e ) 2 ≤ | j| | ε j | j=Xn+1 Ñ é 1/2 ∞ ε/2+ λ (x ,e ) 2 ≤ | n+1| | ε j | j=Xn+1 ε/2+ λ Ñx <ε é ≤ | n+1| k εk ∞ for n large enough. It means that ej j=1 is a basis in H, and moreover, it is an orthonormal basis. { } ∞ Conversely, if ej j=1 is complete in H then R(A) = H (Riesz-Schauder) and therefore N(A)= {0 }. { } Remark. The condition N(A)= 0 means that A−1 exists and H must be separable in this case. { }

Theorem 4 (Lemma of Riesz). If A is a compact operator on H and µ C then ∈ R(I µA) is closed in H. − Proof. If µ = 0 then R(I µA)= H. If µ = 0 then we assume without loss of generality − 6 that µ = 1. Let f R(I A), f = 0. Then there exists a sequence gn H such that ∈ − 6 { } ⊂ f = lim (I A)gn. n→∞ − We will prove that f R(I A) i.e. there exists g H such that f =(I A)g. Since f = 0 then by the decomposition∈ − H = N(I A) ∈ N(I A)⊥ we can− assume that 6 ⊥ − ⊕ − gn N(I A) and gn = 0 for all n N. ∈Suppose− that g is bounded.6 Then∈ there is a subsequence g such that n { kn } g w g. kn → Compactness of A implies that

Ag h = Ag. kn → Next, g =(I A)g + Ag f + h. kn − kn kn → Hence g = f + Ag i.e. f =(I A)g. −

43 Suppose that gn is not bounded. Then we can assume without loss of generality that g . Introduce a new sequence k nk → ∞ g u := n . n g k nk w Since un = 1 then there exists a subsequence ukn u. Compactness of A gives Au k Auk . Since (I A)g f then → kn → − n → 1 (I A)u = (I A)g 0. − kn g − kn → k kn k It means again that u =(I A)u + Au Au kn − kn kn → ⊥ ⊥ and u = Au i.e. u N(I A). But gn N(I A) . Hence ukn N(I A) and ∈ ⊥ − ∈⊥ − ∈ − further u N(I A) because N(I A) is closed. Since ukn = 1 then u = 1. Therefore∈u = 0 while− − k k k k 6 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 H either 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 the corresponding homogeneous equation) and∈ the set− of all solutions is a ﬁnite dimensional aﬃne 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 then− any solution is of the form ∈ −

f = f + u, u N(I µA) 0 ∈ − 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 and 0 σ (A). \ { } ∈ ess Exercise 29. Consider the Hilbert space H = l2(C) and x x A(x , x ,...,x ,...)=(0, x , 2 ,..., n ,...) 1 2 n 1 2 n 2 for (x1, x2,...,xn,...) l (C). Show that A is compact and has no eigenvalues (even more, 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 does not exist. It means that the Fredholm alternative does not hold for non-compact− but self-adjoint operator. Exercise 31. Let H = L2(Rn) and let

Af(x)= K(x, y)f(y)dy, Rn Z 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 orthonormal system x ∞ such that ∈ { n}n=1 (A λI)x 0 k − nk → as n . → ∞ Proof. We will provide only a partial proof. Suppose that λ σess (A). If λ is an eigenvalue 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 it is clear{ that} − ∞ − ≡ (A λI)x 0. − n → Next, suppose that λ is an accumulation point of σ(A). It means that λ σ(A) and ∈

λ = lim λn, n→∞ where λ = λ , n = m and λ σ(A). Hence for each n = 1, 2,... we have that n 6 m 6 n ∈ E E = 0 λn+ε − λn−ε 6 for all ε> 0. Therefore there exists a sequence r 0 such that n → E E = 0. λn+rn − λn−rn 6 45 That’s why we can ﬁnd a normalized vector xn R(Eλn+rn Eλn−rn ). Since λn = λm ∞ ∈ − 6 for n = m we can ﬁnd xn n=1 as an orthonormal system. By spectral theorem we have 6 { } ∞ 2 2 (A λI)xn = (λ µ) d(Eµxn, xn) k − k −∞ − Z ∞ 2 = (λ µ) d(Eµ(Eλn+rn Eλn−rn )xn, xn) Z−∞ − − λn+rn 2 = (λ µ) d(Eµxn, xn) λn−rn − Z ∞ 2 max (λ µ) d(Eµxn, xn) λ −r ≤µ≤λ +r ≤ n n n n − Z−∞ = max (λ µ)2 0, n . λn−rn≤µ≤λn+rn − → → ∞

Theorem 7 (Weyl). Let A and B be two self-adjoint operators in a Hilbert space. If there 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 an orthonormal system x ∞ such⊂ that ∈ { n}n=1 (A λI)x 0, n . k − nk → → ∞

Deﬁne the sequence yn as

y := (A zI)x (A λI)x +(λ z)x . n − n ≡ − n − n Due to Bessel’s inequality any orthonormal system in the Hilbert space converges weakly to 0. Hence y w 0. We also have n → λ z y λ z x (A λI)x = λ z (A λI)x > | − | > 0 k nk ≥ | − | k nk − k − nk | − | − k − nk 2 for all n n >> 1. Next we take the identity ≥ 0 (B zI)−1 (λ z)−1 y = T y (λ z)−1(A λI)x . − − − n − n − − − n w Since T is compact and yn 0 we deduce that î → ó (B zI)−1 (λ z)−1 y 0. − − − n → Introduce î z := (B zI)−1yó . n − n 46 Then z (λ z)−1y 0 n − − n → or y +(z λ)z 0. n − n → This fact and y > |λ−z| imply that z |λ−z| for all n n >> 1. But k nk 2 k nk ≥ 3 ≥ 0 (B λI)z (B zI)z +(z λ)z = y +(z λ)z 0. − n ≡ − n − n n − n → Due to z |λ−z| > 0 the sequence z ∞ can be chosen as an orthonormal system. k nk ≥ 3 { n}n=1 Thus λ σess (B). This proves that σess (A) σess (B). Finally, since T is compact too 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 the domain 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 for all x D(Q). ∈ ≥ − k k ∈ d) closed (and semibounded) if D(Q) is complete with respect to the norm

2 x Q := Q(x, x)+(λ + 1) x . k k q k k e) bounded (continuous) if there exists M > 0 such that

Q(x, y) M x y | | ≤ k k k k for all x, y D(Q). ∈ Exercise 32. Prove that is a norm and that k·kQ

(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 quadratic form in a Hilbert space H such that

Q(x, x) λ x 2 , x D(Q). ≥ − k k ∈ Then there exists a unique self-adjoint operator A which is semi-bounded from below i.e. (Ax, x) λ x 2 , x D(A). ≥ − k k ∈ 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(x, y)+(λ + 1)(x, y), x, y D(Q) Q ∈ (see Exercise 32). Since Q is closed then D(Q)= D(Q) is a closed subspace of H with respect to the norm . It means that D(Q) with this inner product defines a new k·kQ Hilbert space HQ. It is clear also that x x k kQ ≥ k k for all x H . Thus, for fixed x H, ∈ Q ∈ L(y):=(y, x), y H ∈ Q defines a continuous (bounded) linear functional on the Hilbert space HQ. Applying ∗ ∗ the Riesz-Frechet theorem to HQ we obtain an element x HQ (x D(Q)) such that ∈ ∈ (y, x) L(y)=(y, x∗) . ≡ Q It is clear that the map H x x∗ H ∋ 7→ ∈ Q defines a linear operator J such that

J : H H , Jx = x∗. → Q Hence (y, x)=(y,Jx) , x H, y H . Q ∈ ∈ Q Next we prove that J is self-adjoint and that it has an inverse operator J −1. For any x, 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). Suppose that 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 and therefore 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 now x D(A) and y D(Q) H then ∈ ∈ ≡ Q Q(x, y)=(x, y) (λ + 1)(x, y)=(J −1x, y) (λ + 1)(x, y)=(Ax, y). Q − − 49 The semi-boundedness of A from below follows from that of Q. It remains to prove that this representation for A is unique. Assume that we have two such representations, A1 and A . Then for every x D(A ) D(A ) and y D(Q) we have that 2 ∈ 1 ∩ 2 ∈

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

It follows that ((A A )x, y) = 0. 1 − 2 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)= H . Moreover, ≡ Q 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 family ∞ Eµ µ=0 such that { } ∞ A + λI = µdEµ. Z0 That’s why we can deﬁne the operator

∞ √A + λI := √µdEµ Z0 which is also self-adjoint and non-negative. Then for any x D(A) and y D(Q) we ∈ ∈ have that

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

∗ This means that x D(√A + λI) and y D( √A + λI ). But √A + λI is self- Ä ä adjoint and, therefore,∈ ∈

∗ D(√A + λI)= D( √A + λI Ä)= D(Q)ä H . ≡ Q

Ä ä Theorem 2 (Friedrichs extension). Let A be a non-negative, symmetric linear operator in a Hilbert space H. Then there exists a self-adjoint extension AF of A which is the smallest among all non-negative self-adjoint extensions of A in the sense that its corresponding quadratic form has the smallest domain. This extension AF is called the Friedrichs 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(x, y)+(x, y), x, y D(Q). Q ∈ Then D(Q) becomes an inner product space. This inner product space has a completion HQ with respect to the norm

2 x Q := Q(x, x)+ x . k k q k k Moreover, the quadratic form Q(x, y) has an extension Q1(x, y) to this Hilbert space HQ deﬁned by Q1(x, y) = lim Q(xn, yn) n→∞ H H whenever x =Q lim x , y =Q lim y , x , y D(Q) and these limits exist. The n→∞ n n→∞ n n n ∈ quadratic form Q1 is densely deﬁned, closed, non-negative and symmetric. That’s why Theorem 1, applied to Q1, gives a unique and non-negative, self-adjoint operator AF such that

Q (x, y)=(A x, y), x D(A ) H , y D(Q ) H . 1 F ∈ F ⊂ Q ∈ 1 ≡ Q Since for x, y D(A) one has ∈

(Ax, y)= Q(x, y)= Q1(x, y)=(AF x, 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 form Q (x, y):=(Bx, y) is≥ an extension of Q Q . Hence⊂ B ≡ A Q Q = Q . B ⊃ 1 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 =(x ,...,x ) Ω 1 n ∈ 2) x = x2 + + x2 | | 1 ··· n

3) α =(α1,...,αn) is a multi-index i.e. αj N0 N 0 . » ∈ ≡ ∪ { } a) α = α + + α | | 1 ··· n b) α β if α β for all j = 1, 2,...,n. ≥ j ≥ j c) α + β =(α1 + β1,...,αn + βn) d) α β =(α β ,...,α β ) if α β − 1 − 1 n − n ≥ e) xα = xα1 xαn with 00 = 1 1 ··· n f) α!= α ! α ! with 0! = 1 1 ··· n 1 1 ∂ α α1 αn |α| α 4) Dj = ∂j = = i∂j and D = D1 D ( i) ∂ i i ∂xj − ··· n ≡ − Deﬁnition. An elliptic partial diﬀerential operator A(x, D) of order m on Ω is an operator of the form α A(x, D)= aα(x)D , |αX|≤m where a (x) C∞(Ω) and whose principal symbol α ∈ a(x, ξ)= a (x)ξα, ξ Rn α ∈ |αX|=m is invertible for all x Ω and ξ Rn 0 , that is, a(x, ξ) = 0 for all x Ω and ξ Rn 0 . ∈ ∈ \ { } 6 ∈ ∈ \ { } Assumption 1. We assume that a (x) are real for α = m. α | | 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 that m is even and for any compact set K Ω there exists C > 0 such that ⊂ K a(x, ξ) C ξ m, x Ω, ξ Rn. ≥ K | | ∈ ∈ Assumption 2. We assume that A(x, D) is formally self-adjoint i.e.

A(x, D)= A∗(x, D) := ( 1)|α|Dα(a (x) ). − α · |αX|≤m

52 Exercise 33. Prove that A(x, D)= A∗(x, D) if and only if |β| α β−α aα(x)= ( 1) Cβ D aβ(x), α≤β − |βX|≤m where β! Cα = . β α!(β α)! − Hint: Make use of the generalized Leibniz formula α β α−β β D (fg)= Cα D fD g. βX≤α Assumption 3. We assume that A(x, D) has a divergence form A(x, D) ( 1)|α|Dα(a (x)Dβ), ≡ − αβ |α|=|Xβ|≤m/2 where aαβ = aβα and real for all α and β. We assume also the ellipticity condition a (x)ξαξβ ν ξα 2 = ν ξ2α, αβ ≥ | | |α|=|Xβ|=m/2 |α|X=m/2 |α|X=m/2 where ν > 0 is called the constant of ellipticity. Such operator is called uniformly elliptic. Exercise 34. Prove that ξ2α ξ m ≍ | | |α|X=m/2 i.e. c ξ m ξ2α C ξ m, | | ≤ ≤ | | |α|X=m/2 where c and C are some constants. Example 6.1. Let us consider n A(x, D)= D2 = ∆, x Ω Rn j − ∈ ⊂ jX=1 in H = L2(Ω) and prove that A A∗ with ⊂ D(A)= C∞(Ω) = f C∞(Ω) : supp f = x : f(x) = 0 is compact in Ω . 0 ∈ { 6 } ∞ Let u, v C0 (Ω). Then ∈ ¶ © n n 2 2 (Au, v)L2 = Dj u vdx = ∂j u vdx Ω − Ω Z jX=1 jX=1 Z Ñn é n = ∂j ((∂ju) v) dx + Ä (∂juä) ∂jv dx − Ω Ω jX=1 Z jX=1 Z

= (v u, nx)dx +( u, v)L2 =( Äu, äv)L2 , − Z∂Ω ∇ ∇ ∇ ∇ ∇ 53 where ∂Ω is the boundary of Ω and nx is the unit outward vector at x ∂Ω. Here we have made use of the divergence theorem. In a similar fashion we obtain∈

n 2 ( u, v)L2 = u∂j vdx =(u, ∆v)L2 =(u, Av)L2 . ∇ ∇ − Ω − jX=1 Z Hence A A∗ and A is closable. ⊂ Deﬁnition. Let s 0. The L2-based Sobolev space of order s is deﬁned by ≥ s Rn s Rn 2 Rn 2 s 2 H ( ) W2 ( ) := f L ( ) : (1 + ξ ) f(ξ) dξ < , Rn ≡ ∈ Z | | | | ∞ b where f(ξ) is the Fourier transformß of f(x) given by ™

b f(ξ)=(2π)−n/2 e−i(x,ξ)f(x)dx. Rn Z s Rn b ∞ Rn The space H ( ) is the closure of C0 ( ) with respect to the norm

1/2 f := (1 + ξ 2)s f(ξ) 2dξ . s Rn k k Z | | | | ◦ b s Å ∞ ã By the symbol H (Ω) we denote the closure of C0 (Ω) with respect to the same norm. If s N, say s = k, then the norm is equivalent to the norm ∈ k·ks

2 α 2 f W k(Rn) := D f dx, k k 2 Rn | | Z |αX|≤k ◦ α 2 Rn k where D f are the generalized derivatives in L ( ). We also see the space W2 (Ω) ∞ 2 as the closure of C (Ω) with respect to the norm k Rn . Moreover, we can also 0 W2 ( ) k k·k consider W2 (Ω). Example 6.2. Recall from Example 6.1 that

∞ ( ∆u, v) 2 =( u, v) 2 , u,v C (Ω). − L ∇ ∇ L ∈ 0 Hence 2 ∞ ( ∆u, u) 2 = u 2 u 2 ∆u 2 , u C (Ω). − L k∇ kL ≤ k kL k kL ∈ 0 Therefore,

2 2 2 2 u W 2 = u L2 + 2 u L2 + ∆u L2 k k 2 k k k∇ k k k 2 2 u L2 + 2 u L2 ∆u L2 + ∆u L2 ≤ k k 2 k k k2 k k2 k 2 u 2 + 2 ∆u 2 2 u , ≤ k kL k kL ≡ k kA

54 where is a norm which corresponds to the operator A = ∆ as follows: k·kA − 2 2 2 u := u 2 + ∆u 2 . k kA k kL k− kL

It is also clear that u A u W 2 . Combining these inequalities gives k k ≤ k k 2 1 u W 2 u A u W 2 √2 k k 2 ≤ k k ≤ k k 2

∞ ∞ for all u C0 (Ω). A completion of C0 (Ω) with respect to these norms leads us to the statement:∈ ◦ 2 D(A)= W2 (Ω). ◦ 2 ∗ Thus A = ∆ on D(A) = W2 (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 all u C∞(Ω) . − ∈ 0 } ¶ If we assume that v W 2(Ω) then it is equivalent to ∈ 2 (u, ( ∆)∗v)=(u, v∗) − ∗ ∗ ∗ 2 Rn Rn i.e. ( ∆) v = v and D(( ∆) ) = W2 (Ω). Finally, for Ω with Ω = we obtain− that − ⊂ 6 A A A∗ (A)∗ ⊂ ⊂ ≡ and A = A and A = (A)∗, that is, the closure of A does not lead us to a self-adjoint operator.6 6

◦ Remark. If Ω= Rn then W 2(Rn) W 2(Rn) and therefore 2 ≡ 2 A = A∗ =(A)∗.

Hence the closure of A is self-adjoint in that case. Example 6.3. Consider again A = ∆ on D(A)= C∞(Ω) with Ω = Rn. Since − 0 6 2 ( ∆u, u) 2 = u 2 0 − L k∇ kL ≥ then ∆ is non-negative with lower bound λ = 0. That’s why −

Q(u, v):=( u, v) 2 ∇ ∇ L ∞ is a densely deﬁned and non-negative quadratic form with D(Q) D(A) = C0 (Ω). A new inner product is deﬁned as ≡

(u, v) := ( u, v) 2 +(u, v) 2 Q ∇ ∇ L L 55 and 2 2 u Q u W 1(Ω) . k k ≡ k k 2 If we apply now the procedure from Theorem 2 from Section 5 then we obtain the existence of Q1 = Q with respect to the norm Q which will also be non-negative ◦ k·k and closed with D(Q ) W 1(Ω). Next step is to get the Friedrichs extension A as 1 ≡ 2 F A = J −1 I F − ◦ 1 with D(AF ) R(J) W2 (Ω). More careful examination of Theorem 1 of Section 5 leads us to the≡ fact ⊂

◦ ◦ D(A )= W 1(Ω) D(A∗)= W 1(Ω) W 2(Ω). F 2 ∩ 2 ∩ 2 Remark. In general, for symmetric operator, we have

D(A )= u H : Au H F { ∈ Q ∈ } which is equivalent to

D(A )= u H : u D(A∗) . F { ∈ Q ∈ } Exercise 35. Let H = L2(Ω) and A(x, D) = ∆+ q(x), where q(x) = q(x) and q(x) L∞(Ω). Deﬁne A, A∗ and A . − ∈ F Exercise 36. Let f C∞(Ω). Prove that ∈ 0 1 f 2 ε f 2 + f 2 , k ks ≤ k ks1 4ε k ks2 where ε> 0 and s >s>s with s + s 2s. 1 2 1 2 ≥ Consider now bounded Ω Rn and an elliptic operator A(x, D) in Ω of the form ⊂ α β A(x, D)= D (aαβ(x)D ), |α|=|Xβ|≤m/2 where aαβ(x)= aβα(x) are real. Assume that there exists C0 > 0 such that m a (x) C , α , β < | αβ | ≤ 0 | | | | 2 for all x Ω. Assume also that A(x, D) is elliptic, that is, ∈ ( 1)m/2 a (x)ξαξβ ν ξα 2, ν> 0. − αβ ≥ | | |α|=|Xβ|=m/2 |α|X=m/2

56 Theorem 1 (G˚arding’s inequality). Suppose that A(x, D) is as above. Then for any ε> 0 there is Cε > 0 such that

2 2 2 (Af, f)L (Ω) (ν ε) f W m/2(Ω) Cε f L2(Ω) ≥ − k k 2 − k k for any f C∞(Ω). ∈ 0 Proof. Let f C∞(Ω). Then integration by parts yields ∈ 0 α β (Af, f)L2(Ω) = D (aαβ(x)D f)fdx Ω |α|=|Xβ|≤m/2 Z |α| α β = ( 1) aαβ(x)D fD fdx Ω − |α|=|Xβ|=m/2 Z |α| α β + ( 1) aαβ(x)D fD fdx Ω − |α|=|Xβ| 0 and 0 < δ m/2 there is C (δ) > 0 such that ≤ ε (1 + ξ 2)m/2−δ ε(1 + ξ 2)m/2 + C (δ) | | ≤ | | ε for any ξ Rn. ∈ Proof. Let ε> 0 and 0 < δ m/2. If (1+ ξ 2)δ 1 then ≤ | | ≥ ε (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 C (δ). For (1+ ξ 2)δ < 1 we can get ε | | ε m/2−δ 1 δ (1 + ξ 2)m/2−δ < C (δ). | | ε ≡ ε

Ç å

57 s Applying this lemma with δ = 1 to the norm of W2 -spaces we may conclude that

2 2 2 f W m/2−1(Ω) ε1 f W m/2(Ω) + Cε1 f L2(Ω) k k 2 ≤ k k 2 k k for any ε1 > 0. Hence

2 2 2 (Af, f)L (Ω) ν f W m/2(Ω) (C0 + ν) f W m/2−1(Ω) ≥ k k 2 − k k 2 2 2 2 ν f W m/2(Ω) (C0 + ν)ε1 f W m/2(Ω) (C0 + ν)Cε1 f L2(Ω) ≥ k k 2 − k k 2 − k k 2 2 = (ν ε) f W m/2(Ω) Cε f L2(Ω) . − k k 2 − k k This proves the theorem.

Corollary 1. There exists a self-adjoint Friedrichs extension AF of A with domain ◦ D(A )= W m/2(Ω) W m(Ω). F 2 ∩ 2 Proof. It follows from G˚arding’s inequality that

2 (Af, f) 2 C f 2 , f D(A). L (Ω) ≥ − ε k kL (Ω) ∈

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

(A ) (A ) = A + µI µ F ≡ F µ F with domain ◦ D(A )= D((A ) )= W m/2(Ω) D(A∗), F µ F 2 ∩ ◦ m/2 where W2 (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

∗ m D(A )= W2 (Ω).

G˚arding’s inequality has two more consequences. Firstly,

(A ) f C f 2 , C > 0 k F µ kL2 ≥ 0 k kL 0 so that (A )−1 : L2(Ω) L2(Ω). F µ → Secondly, ′ ′ (AF )µf W −m/2(Ω) C0 f W m/2(Ω) , C0 > 0 k k 2 ≥ k k 2 so that ◦ (A )−1 : L2(Ω) W m/2(Ω). F µ → 2

58 Corollary 2. The spectrum σ(A ) = λ ∞ is the sequence of eigenvalues of ﬁnite F { j}j=1 multiplicity with only one accumulation point at + . In short, σ(AF )= σd(AF ). The ∞ ∞ corresponding orthonormal system ψj j=1 of eigenfunctions forms an orthonormal basis and { } ∞ L2 AF f = λj(f,ψj)ψj jX=1 for any f D(A ). ∈ F Proof. We begin with a lemma. Lemma. The embedding ◦ (W m/2(Ω) ֒ L2(Ω 2 → is compact. ◦ ∞ m/2 Proof. It is enough to show that for any ϕk k=1 W2 (Ω) with ϕk W m/2 1 there { } ⊂ k k 2 ≤ exists ϕ ∞ which is a Cauchy sequence in L2(Ω). Since Ω is bounded we have { jk }k=1 1/2 ϕ (ξ) ϕ 2 Ω | k | ≤ k kkL | | i.e. ϕk(ξ) is uniformly bounded. That’s why there exists ϕjk (ξ) which converges Rn pointwise in . Next, ” d 2 2 ϕ”jk ϕjm 2 = ϕjk (ξ) ϕjm (ξ) dξ L Rn k − k Z | − | 2 2 = dϕjk (ξ) ϕjm (ξ) dξ + ϕjk (ξ) ϕjm (ξ) dξ |ξ|r | − | Z ‘ Z d 2 d ϕjk (ξ) ϕjm (ξ) dξ ≤ |ξ| 0. The→ second ∞ term converges to 0 as r because → ∞ ϕjk ϕjm W m/2 2 k − k 2 ≤ Lemma gives us that (A )−1 : L2(Ω) L2(Ω) µ F → is a compact operator. Applying Riesz-Schauder and Hilbert-Schmidt theorems we get

1) σ((A )−1)= 0,µ ,µ ,... with µ µ > 0 and µ 0 as j . µ F { 1 2 } j ≥ j+1 j → → ∞

59 2) µj is of ﬁnite multiplicity 3) (A )−1 ψ = µ ψ , where ψ ∞ is an orthonormal system µ F j j j { j}j=1 4) ψ ∞ forms an orthonormal basis in L2(Ω). { j}j=1 1 Since AF ψj = λjψj with λj = µ then we may conclude that µj − σ(A )= λ ∞ , λ λ ,λ . F { j}j=1 j ≤ j+1 j → ∞

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

∞ (A )−1 f = µ (f,ψ )ψ , f L2(Ω). µ F j j j ∈ jX=1 Exercise 37. Prove that ∞ AF f = λj(f,ψj)ψj jX=1 for any f D(A ). ∈ F 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 form ⊂ α β A(x, D)= D (aαβ(x)D ), |α|=|Xβ|≤m/2 where a = a are real, a C∞(Ω) and bounded for all α and β. We assume that αβ βα αβ ∈ ( 1)m/2 a (x)ξαξβ ν ξ m, ν> 0. − αβ ≥ | | |α|=|Xβ|=m/2 As it was proved above there exists at least one self-adjoint extension of A with D(A)= ∞ C0 (Ω), namely, the Friedrichs extension AF with

◦ D(A )= W m/2(Ω) W m(Ω). F 2 ∩ 2 Let us consider an arbitrary self-adjoint extension A of A. Without loss of generality we assume that A 0. That’s why A has the spectral representation ≥ ∞ A = λdEλ “ “ “ Z0 with domain ∞ 2“ 2 D A = f L (Ω) : λ d(Eλf, f) < . ∈ Z0 ∞ In general case we have no such formula for D A as for the Friedrichs extension AF . Ä ä ß ™ But we can say that “ ◦ m W (Ω) DÄ Aä . 2 ⊂ “ Indeed, since a (x) C∞(Ω) and bounded then A(x, D) can be rewritten in the usual αβ ∈ form Ä ä “ γ A(x, D)= aγ(x)D |γX|≤m with bounded coeﬃcients. Hence f

γ Af L2(Ω) c D f L2(Ω) c f W m(Ω) . k k ≤ k k ≡ k k 2 |γX|≤m This proves the embedding.

∗ 2 Theorem 1 (G˚arding). If A = A then Eλ is an integral operator in L (Ω) such that

Eλf(x)= θ(x,y,λ)f(y)dy, “ “ ZΩ 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)= θ(x,z,λ)θ(z,y,λ)dz ZΩ and θ(x, x, λ)= θ(x,z,λ) 2dz 0 ZΩ | | ≥

3) k sup θ(x, ,λ) L2(Ω) c1λ , x∈Ω1 k · k ≤ where Ω = Ω Ω, k N with k > n and c = c(Ω ). 1 1 ⊂ ∈ 2m 1 1 Remark. It was proved by L. H¨ormander that actually

θ(x, x, λ) c λn/m. ≤ 1 Corollary. Let z ρ A . 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 Ä ä 1) “ “ ∞ d θ(x,y,λ) G(x, y, z)= λ “ 0 λ z Z −

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

Proof. Since z ρ A then J. von Neumann’s spectral theorem gives us ∈ ∞ −1 −1 Ä ä (A zI) f = (λ z) dEλf. “ − Z0 − Next, by Theorem 1 we get “ ∞ −1 −1 (A zI) f = (λ z) dλ θ(x,y,λ)f(y)dy − 0 − Ω Z ∞ Z −1 = (λ z) Ådλθ(x,y,λ) f(y)dyã “ ZΩ Z0 − = ÅG(x, y, z)f(y)dy, ã ZΩ where G(x, y, z) is as in 1). Since

∞ dθ(x,y,λ) ∞ dθ(y,x,λ) G(x, y, z)= = = G(y, x, z) 0 λ z 0 λ z Z − Z − then 2) is also proved.

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

1) θ(x,y,λ) 2 θ(x, x, λ)θ(y,y,λ) | | ≤ 1/2 2) E f(x) θ(x, x, λ) f 2 . | λ | ≤ k kL (Ω) Exercise 39. Prove that

E f(x) E f(x) E f E f θ(x, x, λ) θ(x, x, µ) 1/2 | λ − µ | ≤ k λ − µ kL2(Ω) | − | for any λ> 0 and µ> 0.

Exercise 40. Let us assume that n

∞ θ(x,y,λ)dλ G(x, y, z)= 0 (λ z)2 Z − 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

α A(x, D)= aα(x)D , |αX|≤m where a (x) C∞(Ω) and bounded, a (x) real for α = m and α ∈ α | | α a(x, ξ)= aα(x)ξ > 0 |αX|=m for any x Ω and ξ Rn 0 . Let us define the set ∈ ∈ \ { } Z := z C : arg z < θ θ { ∈ | | } for some fixed 0 <θ<π/2. Lemma 1. There is a constant c > 0 such that, for any z / Z , 0 ∈ θ a(x, ξ)+1+ z a(x, ξ) + 1 z c (a(x, ξ)+1+ z ) , | | ≥ | − | ≥ 0 | | where x Ω and ξ Rn. ∈ ∈ Proof. Let z = λ + iµ / Z . Abbreviate a := a(x, ξ) 0. The first (leftmost) ∈ θ ≥ inequality follows immediately from triangle inequality. Hence it remains to prove the latter (rightmost) inequality. To that end, we start from

Consider now λ > 0. Since z / Zθ then µ h = λ tan θ, see Figure 1. It means that µ2 γλ2, where γ = tan2 ∈θ> 0. Hence| | ≥ | | ≥ 1 (a + 1)2 2λ(a +1)+ λ2 + µ2 (a + 1)2 ε(a + 1)2 λ2 + λ2 + µ2 − ≥ − − ε 1 = (1 ε)(a + 1)2 + 1 λ2 + δµ2 + (1 δ)µ2 − − ε − 1 (1 ε)(a + 1)2 + Ç1 å λ2 + γδλ2 + (1 δ)µ2 ≥ − − ε − 1 = (1 ε)(a + 1)2 + Ç1 å+ γδ λ2 + (1 δ)µ2 − − ε −

64 Ç å z µ b b

h b b

Zθ θ b θ − λ

Figure 1: The geometry of Lemma 1. for any ε> 0 and δ R. Let us choose δ = 1 and ε such that ∈ 2 2 <ε< 1. 2+ γ Then for 1 1 γ 2c2 := min , 1 ε, 1 + > 0 0 2 − − ε 2 we have that Ç å

a + 1 z 2 2c2 (a + 1)2 + λ2 + µ2 = 2c2 (a + 1)2 + z 2 c2 (a +1+ z )2 . | − | ≥ 0 0 | | ≥ 0 | | This proves the lemma. Ä ä Ä ä

Let ϕk(x, ξ, z), k = 0, 1,... be the sequence deﬁned by 1 ϕ (x, ξ, z) = , 0 a(x, ξ) + 1 z − 1 ϕ (x, ξ, z) = [a(x, ξ) + 1 A(x, D + ξ)] ϕ , k = 1, 2,..., k a(x, ξ) + 1 z − k−1 − where α A(x, D + ξ)= aα(x)(D + ξ) |αX|≤m with x Ω, ξ Rn and z / Z . Due to Lemma 1 this sequence is well-deﬁned. ∈ ∈ ∈ θ Exercise 41. Prove that

α β α−β β (D + ξ) f = Cα ξ D f. βX≤α

65 Exercise 42. Prove that Dαf A(x, D)(fg)= A(α)(x, D)g, α! αX≥0 (α) (α) α where A has the symbol A (x, ξ)= ∂ξ 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) , ν = 0, 1,..., (a + 1 z)ν+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 1 ϕ (x, ξ, z)= 0 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.≥ ≤ ≤ b(x, ξ) ϕk = ν+1 ν (a + 1 z) X − and mν l k. Concerning ϕ we have that − ≥ j+1 1 ϕ = [a + 1 A(x, D + ξ)] ϕ j+1 a + 1 z − j (a + 1)−ϕ 1 = j a (x) Cβξα−βDβϕ a + 1 z − a + 1 z α α j − − |αX|≤m βX≤α (a + 1)ϕ 1 = j a (x)Cβ ξγDβϕ a + 1 z − a + 1 z β+γ β+γ j − − |βX|≤m |β+Xγ|≤m := I1 + I2, where I = (a+1)ϕj . The term in I that corresponds to β = 0 and γ = m is 1 a+1−z 2 | | 1 aϕ a (x)ξγϕ = j . −a + 1 z γ j −a + 1 z − |γX|=m −

Hence I1 + I2 can be rewritten as

(a + 1)ϕj aϕj 1 β γ β I1 + I2 = aβ+γ(x)Cβ+γξ D ϕj a + 1 z − a + 1 z − a + 1 z |β|≤m |β+γ|≤m − − − β>X0 X ′ ′ = I1 + I2,

66 where ϕ I′ = j . 1 a + 1 z − Since b(x, ξ) ϕj = ν+1 ν (a + 1 z) X − then ′ b(x, ξ) I1 = ν+2 ν (a + 1 z) X − and mν l j. This implies that mν + m l j + m > j + 1, since m 2. In other words, − ≥ − ≥ ≥ m(ν + 1) l j + 1 − ≥ ′ β ′ and the lemma is proved for I1. Let us consider now D ϕj from I2 i.e.

β β b(x, ξ) β b(x, ξ) D ϕj = D ν+1 = Dx ν+1 , mν l j. ν (a + 1 z) ν (a + 1 z) − ≥ X − X − We prove that diﬀerentiation of b(x, ξ) (a + 1 z)ν+1 − with respect to x leads to b , (a + 1 z)ν+1 Xν e− where mν l = mν l j i.e. the value of mνe l does not change. Indeed, − − ≥ e − e b b′ (ν + 1)ba′ e ∂ = j j , j (a + 1 z)ν+1 (a + 1 z)ν+1 − (a + 1 z)ν+2 − − − ′ ′ where bj and baj are homogeneous polynomials with respect to ξ of orders l and l = l + m, respectively. That’s why e mν l = m(ν + 1) (m + l)= mν l. − − − β This fact holds for any derivativee e D ,β > 0. Now we may conclude that

′ 1 γ b I = Cβ+γ(x)ξ , 2 −a + 1 z (a + 1 z)ν+1 ν |β|≤m |β+γ|≤m e − X β>X0 X − e where γ m 1 and mν l =e mν l j. But multiplication by ξγ, γ m 1 leads us| | finally ≤ − to the sum of− the terms− ≥ | | ≤ − e e 1 b bξγ b ξγ = = 1 , (a + 1 z) (a + 1 z)ν+1 (a + 1 z)ν+2 (a + 1 z)ν+2 − e− e− − e 67 e ‹ e where l = l + γ and 1 | | m(ν +e 1) l = mν + m l γ = mν l + m γ mν l + 1 j + 1. − 1 − − | | − − | | ≥ − ≥ ‹ This finishese the proof.e e Definition. A locally‹ integrable function F ( ,y,z) with parameters y Ω and z / Z · ∈ ∈ θ 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) h − · · i for all ϕ C∞(Ω) or ∈ 0 F (x, y, z)(A∗ zI)ϕ(x)dx = ϕ(y), ZΩ − where A∗(x, D) is formally adjoint to A(x, D) in L2(Ω). Let us deﬁne a new function

k F (x, y, z) := F −1 ϕ (x, , z) (x y), k = 0, 1, 2,.... k j · − jX=0 Ñ é Here F −1 is the inverse Fourier transform of tempered distributions i.e.

F −1u, ϕ := u, F −1ϕ , h i h i where ϕ C∞(Rn) is such that all of its derivatives decay faster than the reciprocal ∈ of any polynomial at inﬁnity and

(F −1f)(x)=(2π)−n ei(x,y)f(y)dy Rn Z 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)F (x, y, z)= δ(x y) H (x, y, z), − k − − k where H (x, y, z)= F −1((a(x, ) + 1 z)ϕ ). k · − k+1 Proof. Since

α i(x−y,ξ) β β α−β i(x−y,ξ) Dx (ϕj(x, ξ, z)e ) = Cα Dx ϕj Dx e βX≤α

Äβ α−β ä Äβ äi(x−y,ξ) = Cα ξ Dx ϕj(x, ξ, z) e βX≤α Ñ é 68 then

k (A(x, D) zI)F (x, y, z) = F −1 ((A(x, D + ) zI)ϕ (x, , z)) − k · − j · jX=0 k = F −1 ((a + 1 z)ϕ (a + 1 A(x, D + ))ϕ ) − j − − · j jX=0 k = F −1 ((a + 1 z)ϕ (a + 1 z)ϕ ) − j − − j+1 jX=0 = F −1 ((a + 1 z)ϕ (a + 1 z)ϕ ) − 0 − − k+1 = F −1(1) F −1 ((a + 1 z)ϕ ) − − k+1 = δ(x y) H (x, y, z). − − k This proves the lemma. Remark. The following representation holds:

−1 bν(x, ) Hk(x, y, z)= F · ν , ν (a(x, ) + 1 z) X · − where ν = 1, 2,... and mν l k + 1. Ç å − ν ≥ 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)+ Fk(x, u, z)hk(u, y, z)du, ZΩ1 where x, y Ω1 = Ω1 Ω, z / Zθ and hk(u, y, z) is an unknown function which is to be determined.∈ Since ⊂ ∈ (A zI)F = δ H − k − k then

(A zI)F = δ Hk + (δ Hk)hkdu = δ Hk + hk Hkhkdu. − − ZΩ1 − − − ZΩ1

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

hk(x, y, z)= Hk(x, y, z)+ Hk(x, u, z)hk(u, y, z)du. ZΩ1 We plan to solve this equation by iterations. In order to do it and obtain estimates of F (x, y, z) it will be necessary to estimate

k −1 −1 bνj Fk = F ϕj = F , νj +1 ν (a + 1 z) jX=0 Xj Xj − Ç å

69 where ν = 0, 1,... and mν l j and j j − j ≥

−1 bν Hk = F ν , ν (a + 1 z) X − where ν = 1, 2,... and mν l k + 1. LetÇ us introduceå the function − ≥ b(x, ) ψ(x, u) := F −1 · (u), (a(x, )+1)ν · where mν l> 0. This deﬁnition is understoodÇ in the senseå of tempered distributions i.e. − b b(x, ξ) ψ(x, ), ϕ = ,F −1ϕ = F −1ϕ(ξ)dξ. ν Rn ν h · i (a + 1) Z (a(x, ξ)+1) Æ ∏ Lemma 4 (Main lemma). Suppose that u = 1. For every Ω1 = Ω1 Ω there is δ > 0 such that for any λ> 0 we have | | ⊂

1) mν l < n, − ψ(x, λu) c λmν−l−ne−δλ; | | ≤ 1

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

3) mν l > n, − ψ(x, λu) c e−δλ, | | ≤ 1 where x Ω and c = c(Ω ). ∈ 1 1 1

Proof. Let us denote by c1 any constant that depends on Ω1. Since a(x, ξ) > 0 for all x Ω and ξ Rn 0 then for any Ω = Ω Ω there is ∈ ∈ \ { } 1 1 ⊂ a constant c1 > 0 such that

a(x, ξ) c ξ m, x Ω , ξ Rn. ≥ 1| | ∈ 1 ∈ This way we can consider the function

b(x, ) ψ (x, u) := F −1 · (u), 0 (a(x, ))ν · where x Ω and u Rn. It is very easy toÇ check thatå ∈ 1 ∈ mν−l−n ψ0(x, λu)= λ ψ0(x, u).

70 1) Let mν l < n. Let us prove ﬁrst that − ψ (x, u) c | 0 | ≤ 1 ∞ R for u = 1 and x Ω1. Take a function χ(ξ) C0 ( ) such that χ(ξ) 1 for ξ 1. Then| | ∈ ∈ ≡ | | ≤ b b ψ (x, u)= F −1 χ + F −1 (1 χ) := ψ + ψ . 0 aν aν − 0 0 b 1 Rn Since mν l < n then aν LlocÇ( ).å That’sÇ why å − ∈ › › −1 b −n b(x, ξ) i(ξ,u) ψ0(x, u)= F ν χ = (2π) ν χ(ξ)e dξ, a Z|ξ|

For ψ0(x, u) we have the identity›

α −1 α b u ψ0(x, u)= F Dξ ν (1 χ) . › a − Next, we prove that Ç ñ ôå › α b c1 Dξ (1 χ(ξ)) . aν − ≤ (1 + ξ )mν−l+|α|

| |

Indeed, ñ ô b b′ νba′ b D (1 χ(ξ)) = ξj (1 χ) ξj (1 χ) χ′ := I + I + I . ξj aν − aν − − aν+1 − − aν 1 2 3

b c1 Dξj (1 χ(ξ)) . aν − ≤ (1 + ξ )mν−l+1

| |

The required estimate follows ñ now by induction.ô If we choose now α such that α > n + l mν i.e. mν l + α > n then | | − − | | α dξ ′ u ψ0(x, u) c1 = c1. Rn mν−l+|α| ≤ Z (1 + ξ ) | |

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) c , x Ω , u = 1. | | ≤ 1 ∈ 1 | | Let us consider now the main symbol a(x, ζ) for complex ζ. Let ζ = ξ + iη. Then

α β β α−β a(x, ζ) = aα(x)(ξ + iη) = aα(x) Cα ξ (iη) |αX|=m |αX|=m βX≤α α β β α−β = aα(x)ξ + aα(x) Cα ξ (iη) . |αX|=m |αX|=m β<αX For η δ it follows that | | ≤ a(x, ζ) + 1 a(x, ξ) + 1 a (x) Cβ ξ |β| η |α−β| | | ≥ − α α | | | | |αX|=m β<αX m−1 a(x, ξ) + 1 c ξ j η m−j ≥ − 1 | | | | jX=0 m−1 1 a(x, ξ) + 1 c ε ξ m + η m ≥ − 1 | | εj/(m−j) | | jX=0 m m = a(x, ξ) + 1 c1εm ξñ cε η ô − | | − | | c ξ m c εm ξ m + 1 c η m ≥ 1| | − 1 | | − ε| | = (c εc m) ξ m + 1 c η m c ( ξ m + 1), 1 − 1 | | − ε| | ≥ 0 | | where ‹ c = min c εc m, 1 c δm > 0 0 ‹ { 1 − 1 − ε } and the parameters are chosen as

c εc m>‹ 0, 1 c δm > 0 1 − 1 − ε i.e. 1 c 1 m ‹ ε< 1 , δ< . c1m cε We proved that there exists δ > 0 such that Ç å ‹ a(x, ξ + iη) + 1 c ( ξ m + 1), c > 0 | | ≥ 0 | | 0 for all ξ,η Rn, η δ and x Ω . This inequality allows us to extend the function ∈ | | ≤ ∈ 1 b(x, ξ) Dα ei(λu,ξ) ξ (a(x, ξ)+1)ν

ñ 72 ô as an analytic function with respect to ζ = ξ + iη for ξ Rn and η δ. Moreover, this inequality leads to the estimate ∈ | | ≤

α b(x, ξ) c1 Dξ (a(x, ξ)+1)ν ≤ (1 + ξ )mν−l+|α|

| | n ñ ô n for all x Ω1, ξ R and η δ. Let us consider ﬁxed η R with η δ. Cauchy theorem∈ for α >∈ l mν +| n| ≤shows us that ∈ | | ≤ | | −

α −n α b(x, ξ) i(λu,ξ) (λu) ψ(x, λu) = (2π) Dξ e dξ Rn ν Z (a(x, ξ)+1) −n α b(x, ξ + iη) i(λu,ξ+iη) = (2π) Dξ e dξ. Rn ν Z (a(x, ξ + iη)+1) Hence α 1 −λ(u,η) ′ −λ(u,η) (λu) ψ(x, λu) c1 e dξ c1e . | | ≤ Rn (1 + ξ )mν−l+|α| ≤ Z | | α α1 αn α α α Since u = u1 un > 0 for u = 1 then u min|u|=1 u = u0 := c0, where | u | = 1| and| ·c · ·> | 0.| Therefore the| | latter inequality| | ≥ can be transformed| | | | to | 0| 0 ψ(x, λu) c′′λ−|α|e−λ(u,η). | | ≤ 1 If we choose η = δu then we obtain

ψ(x, λu) c λ−|α|e−δλ, λ> 0, u = 1, x Ω , | | ≤ 1 | | ∈ 1 where α > l mν + n. For α = l mν + n + 1 > 0 and for λ 1 we can obtain | | − | | − ≥ ψ(x, λu) c λmν−l−n−1e−δλ c λmν−l−ne−δλ. | | ≤ 1 ≤ 1 Consider now 0 <λ< 1. It is easily seen that b b (λu)α(ψ(x, λu) ψ (x, λu)) = F −1 Dα ν ν (λu) − 0 ξ (a + 1)ν − aν (a + 1)ν aν = F −Ç1 Dñα − ôåb (λu) − ξ (a2 + a)ν ν ν−1 j j −1 Ç α ñ j=0 cνa bν ôå = F Dξ 2 ν (λu) − "P(a + a) #! ν−1 −1 α j=0 bj = F Dξ 2 ν (λu), − (Pa + a)  Ñ  é ‹ where bj is a homogeneous polynomial of order lj = l + mj. Let us denote

ν−1 b I(x, ξ) := j=0 j . ‹ 2 ‹ ν (Pa + a)

73 ‹ It is clear that I(x, ξ) behaves as

I ξ l−mν, ξ 0 ≍ | | | | → and I ξ l−mν−m, ξ . ≍ | | | | → ∞ Combining these two asymptotics yields 1 1 I ≍ ξ mν−l (1 + ξ )m | | | | for all ξ Rn 0 . By the same arguments we have ∈ \ { } 1 1 DαI . | | ≍ ξ mν−l+|α| (1 + ξ )m | | | | Let us choose now α = l mν + n 1 0. Then | | − − ≥ 1 1 DαI . | | ≍ ξ n−1 (1 + ξ )m | | | | This allows us to get DαI L1(Rn) and, therefore, ∈ α 1 ′ (λu) (ψ(x, λu) ψ0(x, λu)) c1 c1. | − | ≤ Rn ξ n−1(1 + ξ )m ≤ Z | | | | Hence ψ(x, λu) ψ (x, λu) c λmν−l−n+1 | − 0 | ≤ 1 for u = 1. It implies that | | ψ(x, λu) c λmν−l−n+1 + ψ (x, λu) c λmν−l−n + λmν−l−n ψ (x, u) | | ≤ 1 | 0 | ≤ 1 | 0 | c λmν−l−n c′ λmν−l−ne−δλ, 0 <λ< 1. ≤ 1 ≤ 1 Thus, 1) is proved for all λ> 0. 2) Let mν l = n. Taking the derivative with respect to λ we obtain − ∂ i(u, ξ)b(x, ξ) ψ(x, λu)= F −1 (λu). ∂λ (a(x, ξ)+1)ν It is equivalent to Ç å

∂ b(x, u, ξ) ψ(x, λu)= F −1 (λu) := ψ(x, λu), ∂λ (a(x, ξ)+1)ν ! e where mν l = mν l 1= n 1 < n. That’s why we can apply part 1) to ψ(x, λu) ‹ and obtain− − − − e ∂ −1 −δλ ψ(x, λu) c1λ e . ∂λ ≤ ‹

74 Next, consider two cases: 0 <λ< 1 and λ 1. In the ﬁrst case ≥ 1 ∂ ψ(x, λu)= ψ(x, τu)dτ + ψ(x, u). − Zλ ∂τ Since ψ(x, u) is bounded for u = 1 then | | 1 1 1 −δτ 1 ψ(x, λu) c1 e dτ + c1 c1 dτ + 1 = c1(1 + log λ ). | | ≤ Zλ τ ≤ Zλ τ | | In the second case we begin with Ç å ∞ ∂ ψ(x, λu)= ψ(x, τu)dτ − Zλ ∂τ because ψ(x, λu) 0 as λ . Hence → → ∞ ∞ ∞ 1 −δτ −δτ ′ −δλ ψ(x, λu) c1 e dτ c1 e dτ = c1e | | ≤ Zλ λ ≤ Zλ for λ 1. That’s why ≥ ψ(x, λu) c (1 + log λ )e−δλ, λ> 0. | | ≤ 1 | | This proves 2).

3) Since mν l > n then the claim follows from above considerations immediately without diﬀerentiation.−

Lemma 5. Let z / Z . Deﬁne the function ∈ θ b(x, ξ) I (x, y, z)= F −1 (x y), ν (a(x, ξ) + 1 z)ν − − Ç å where b(x, ξ) and a(x, ξ) are as in lemma 4, x Ω1 and y Ω. Then there exists δ > 0 such that ∈ ∈

1) mν l < n, − 1 I (x, y, z) c x y mν−l−ne−δ|x−y|(1+|z|) m ; | ν | ≤ 1| − |

2) mν l = n, − 1 1 −δ|x−y|(1+|z|) m I (x, y, z) c 1+ log x y (1 + z ) m e ; | ν | ≤ 1 | − | | |

75 3) mν l > n, − 1 − mν−l−n −δ|x−y|(1+|z|) m I (x, y, z) c (1 + z ) m e . | ν | ≤ 1 | |

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 b(x, ξ) I (x, y, z)= F −1 (x y). ν (a(x, ξ)+1+ z )ν − | | f But it is not so diﬃcult to check thatÇ å

n−mν+l I (x, y, z)=(1+ z ) m ψ(x, λu), ν | | 1 x−y where λ = x y (1 + z f) m and u = . Applying Lemma 4 implies now 1)-3). | − | | | |x−y| Corollary. Let α> 0, z / Z , x Ω and y Ω. Then ∈ θ ∈ 1 ∈ 1) mν l < n + α , − | | 1 DαI (x, y, z) c x y mν−l−|α|−ne−δ|x−y|(1+|z|) m ; | x ν | ≤ 1| − | 2) mν l = n + α , − | | 1 α 1 −δ|x−y|(1+|z|) m D I (x, y, z) c 1+ log x y (1 + z ) m e ; | x ν | ≤ 1 | − | | |

3) mν l > n + α , − | | n+l+|α|−mν 1 α −δ|x−y|(1+|z|) m D I (x, y, z) c (1 + z ) m e . | x ν | ≤ 1 | | Exercise 43. Prove Corollary. Now we are in the position to prove the main theorem. Theorem 1. For any Ω = Ω Ω there is R > 0 such that for all k n, n N and 1 1 ⊂ 0 ≥ ∈ for all z > R0, z / Zθ there exists a fundamantal solution F (x, y, z), x, y Ω1 of an operator| |A(x, D) ∈zI. This fundamental solution has the form ∈ −

F (x, y, z)= Fk(x, y, z)+ Fk(x, u, z)hk(u, y, z)du, ZΩ1 k −1 where Fk = j=0 F ϕj,hk(x, y, z) exists and satisﬁes the estimate

P 1 n−k−1 −δ|x−y|(1+|z|) m h (x, y, z) c (1 + z ) m e . | k | ≤ 1 | |

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

hk(x, y, z)= Hk(x, y, z)+ Hk(x, u, z)hk(u, y, z)du, ZΩ1 where H (x, y, z)= F −1 ((a + 1 z)ϕ )(x y) k − k+1 − with bν(x, ξ) (a(x, ξ) + 1 z)ϕk+1(x, ξ, z)= ν − ν (a(x, ξ) + 1 z) X − and mν l k + 1. − ν ≥ Since k n then mν lν n + 1 > n and, therefore, we can apply part 3) of Lemma 5 and obtain≥ − ≥

n−(mν−l ) 1 1 ν −δ|x−y|(1+|z|) m n−k−1 −δ|x−y|(1+|z|) m Hk(x, y, z) c1 (1 + z ) m e c1(1 + z ) m e . | | ≤ ν | | ≤ | | X

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

∞ (j) hk(x, y, z) := Hk (x, y, z), jX=1 (1) where Hk = Hk and

(j) (j−1) Hk (x, y, z)= Hk(x, u, z)Hk (u, y, z)du, j 2. ZΩ1 ≥

It is clear that hk(x, y, z) of such form is a (formal) solution of the corresponding integral equation for h . It remains merely to prove that for z >> 1, z / Z this series k | | ∈ θ converges uniformly with respect to x and y from Ω1. To this end, let us consider the estimates for iterations of Hk. Indeed,

(2) (1) Hk (x, y, z) Hk(x, u, z) Hk (u, y, z) du | | ≤ ZΩ1 | || | 2 1 n−k−1 m 2 m −δ(|x−u|+|u−y|)(1+|z|) c1 (1 + z ) e du ≤ | | ZΩ1 2 1 2 n−k−1 −δ|x−y|(1+|z|) m c (1 + z ) m Ω e ≤ 1 | | | 1| since x u + u y x y . Here and later Ω denotes the Lebesgue measure of | − | | − | ≥ | − | | 1| Ω1. By induction we get

j 1 (j) n−k−1 j−1 −δ|x−y|(1+|z|) m H (x, y, z) c (1 + z ) m Ω e . | k | ≤ 1 | | | 1|

77 Hence

∞ 1 ∞ j−1 (j) n−k−1 −δ|x−y|(1+|z|) m n−k−1 H c (1 + z ) m e c Ω (1 + z ) m . | k | ≤ 1 | | 1| 1| | | jX=1 jX=1 If we choose z so that n−k−1 c Ω (1 + z ) m < 1 1| 1| | | or m z > R := (c Ω ) k+1−n 1 | | 0 1| 1| − then the series converges uniformly with respect to x, y Ω1. Therefore hk(x, y, z) is well-deﬁned and F (x, y, z) exists. ∈ Corollary 1. The fundamental solution F (x, y, z) satisﬁes the following estimates: 1) m < n, 1 F (x, y, z) c x y m−ne−δ|x−y|(1+|z|) m ; | | ≤ 1| − | 2) m = n,

1 1 −δ|x−y|(1+|z|) m F (x, y, z) c 1+ log x y (1 + z ) m e ; | | ≤ 1 | − | | |

3) m > n, 1 n−m −δ|x−y|(1+|z|) m F (x, y, z) c (1 + z ) m e , | | ≤ 1 | | where x, y Ω and z > R , z / Z . ∈ 1 | | 0 ∈ θ Proof. Since

k k −1 −1 bνj Fk(x, y, z)= F ϕj(x y)= F (x y), νj +1 − ν (a + 1 z) − jX=0 jX=0 Xj − Ç å where mνj lj j then we can apply Lemma 5 with ν = νj + 1 and l = lj for j = 0, 1,...,k− . That’s≥ why we can obtain

mνj +m−lj −n x y , mνj + m lj < n 1 k m | − | 1 − −δ|x−y|(1+|z|)  m Fk c1e 1+ log x y (1 + z ) , mνj + m lj = n | | ≤  l |+n−m−|mν | | − j=0 νj  j j X X (1 + z ) m , mν + m l > n. | | j − j   Let us remember that ν0 = l0 = 0. That’s why the ﬁrst term of this sum can be estimated as

x y m−n, m n. | |    78 But mν l j means that the value mν l grows with respect to j. That’s why j − j ≥ j − j the next terms in this sum have better estimates than I0. This remark allows us to conclude that x y m−n, mn | |  for k = 0, 1,.... In order to obtain estimates for F (x, y, z) it remains to investigate its second term, namely

Fk(x, u, z)hk(u, y, z)du. ZΩ1 Applying the estimates for Fk and hk we obtain

1 n−k−1 m 2 m −δ(|x−u|+|u−y|)(1+|z|) Fk(x, u, z)hk(u, y, z)du c1(1 + z ) e du ZΩ1 ≤ | | ZΩ1

m−n x u , mn | |  1 2 n−k−1 −δ|x−y|(1+|z|) m c (1 + z ) m e ≤ 1 | | m−n Ω1 x u du, m < n | − | 1  RΩ 1+ log x u (1 + z ) m du, m = n ×  1 | − | | |  n −m (1R + z ) m Ω1 , m > n. | | | |  It is clear that  m−n x u du c0 ZΩ1 | − | ≤ if m < n and n−m ′ (1 + z ) m Ω c | | | 1| ≤ 0 if m > n. If m = n then

1 1+ log x u (1 + z ) m du du + log x u du Ω1 | − | | | ≤ Ω1 Ω1 | | − || Z Z Z 1 + log(1+ z ) m du | | ZΩ1 1 c 1+log(1+ z ) m ≤ 1 | | ε c (1 + z ) m , ε> 0. ≤ 1,ε | | Since ε> 0 and small enough then ﬁnally

1 −δ|x−y|(1+|z|) m Fk(x, u, z)hk(u, y, z)du c1e . ZΩ1 ≤

This ﬁnishes the proof.

79 Corollary 2. Let 0 < α R and x, y Ω . Then | | ∈ θ | | 0 ∈ 1 1) m < n + α , | | 1 DαF (x, y, z) c x y m−n−|α|e−δ|x−y|(1+|z|) m ; | x | ≤ 1| − | 2) m = n + α , | | 1 α 1 −δ|x−y|(1+|z|) m D F (x, y, z) c 1+ log x y (1 + z ) m e ; | x | ≤ 1 | − | | |

3) m > n + α , | | n+|α|−m 1 α −δ|x−y|(1+|z|) m D F (x, y, z) c (1 + z ) m e . | x | ≤ 1 | | Exercise 44. Prove Corollary 2. Exercise 45. Let α = m and x y ε. Prove that | | | − | ≥ 1 DαF (x, y, z) c e−δε(1+|z|) m . | x | ≤ ε Let A be an arbitrary self-adjoint extension of an elliptic diﬀerential operator A(x, D) of even order m in L2(Ω), where Ω Rn is a bounded domain with smooth ⊂ boundary. 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)= G(x, y, z)f(y)dy. ZΩ ” Thus (”A zI)G = G (A zI)= I − z z − 2 2 and Gz : L (Ω) L (Ω) is a bounded operator. We know also that for any Ω1 = Ω1 → Z ⊂ Ω there is a fundamental solution“ F”(x, y, z”) for“ z / θ and z > R0. Moreover, we α ∈ | | have the estimates for Dx F (x, y, z) for α m 1, see Corollary for Theorem 1. Our main” 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 if the integral operator ∈ ∈ − Qz := Gz Ez, − “ where Ez is the integral operator with kernel E(x, y, z), has the mapping property ” ” ∞” 1 j Qz : L (Ω1) D(A ) ” → j\=1 for any Ω = Ω Ω. 1 1 ⊂ ” “ 80 Remark. It is easy to obtain the representation

(A zI)E = I + P , − z z where P = (A zI)Q or Q = G P . z − − z z − z z If Ω = Ω Ω then denote “ ” ” 1 1 ⊂ ” “ ” ” ”” ε0 = dist(Ω1,∂Ω) > 0.

For any ﬁxed ε such that 0 <ε<ε deﬁne the function χ(x) C∞(Ω) as 0 ∈ 0 1, x Ωε/2 χ(x)= ∈ 1 0, x Ω Ωε,  ∈ \ 1

ε/2 ε/2 ε ε/2 ε where dist(Ω1,∂Ω1 ) = dist(Ω1 ,∂Ω1) = ε/2 with Ω1 Ω1 Ω1 Ω and ε ⊂ ⊂ ⊂ dist(Ω1,∂Ω) > 0, see Figure 2.

Ω

ε Ω1

ε/2 Ω1

Ω1

χ = 1

χ = 0

Figure 2:

ε Let us consider a fundamental solution F (x, y, z) for x, y Ω1. Define the function E by ∈ E(x, y, z) := χ(x)F (x, y, z) ε which is well-defined for all x Ω and y Ω1. Let us assume that E(x, y, z) 0 for ε ∈ ∈ ≡ x Ω and y Ω Ω1. Thus, E(x, y, z) is defined for all x, y Ω. It is also clear ∈ ∈ \ ε/2 ∈ that E(x, y, z)= F (x, y, z) for x, y Ω1 . 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 is more, − 1 k ε m k −δ 2 (1+|z|) A Q f c (1 + z ) e f 1 z k,ε L (Ω1) L2(Ω) ≤ | | k k “ for any k N. ∈ Proof. Since E(x, y,“ z)” C∞(Ω) for x = y then ∈ 0 6 (A zI)E(x, y, z) = (A(x, D) zI)E(x, y, z)= χ(x)(A(x, D) zI)F (x, y, z) − − − 1 + DαχA(α)(x, D)F (x, y, z)= δ(x y)+ P (x, y, z), α>0 α! − “ X where x Ω, y Ω1 and P (x, y, z) is defined by the second term in the sum. It is ∈ ∈ α ε ε/2 also easy to see that D χ = 0 only for x Ω1 Ω1 . Since y Ω1 then we may 6 ε ∈ ε/2 \ ∈ conclude that P (x, y, z) = 0 for all x Ω1 Ω1 and y Ω1. This fact implies that x y ε/2. Next, since6 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 − 1 −δ ε (1+|z|) m P (x, y, z) c e 2 . | | ≤ ε Moreover, we prove that

1 k k −δ ε (1+|z|) m A P (x, y, z) c (1 + z ) e 2 | | ≤ ε,k | | 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 a (x)DαF = zF a (x)DαF. α x − α x |αX|=m |α|≤Xm−1 It follows from ellipticity that a (x)DαF c DαF . α x ≥ 1 | | |αX|=m |αX|=m Hence 1 1 α α −δ ε (1+|z|) m −δ ε (1+|z|) m D F c z F + c D F c z e 2 + e 2 | | ≤ | || | | | ≤ | | |αX|=m |α|≤Xm−1 1 −δ ε (1+|z|) m = c(1 + z )e 2 Å ã | | by corollaries 1 and 2. Next, since x y ε/2 then | − | ≥ Dαχ AP A(x, D)P (x, y, z)= A(x, D) A(α)F ≡ α! α>X0 1 1 α+β (β) (α) γ = D χA (x, D) A (x, D)F = cγD χBγ(x, D)F, “ α! β! α>X0 βX≥0 γ>X0 82Ä ä where Bγ(x, D) is an operator of order 2m γ 2m 1. If γ < m then the corresponding terms in this sum can be estimated− | | from ≤ above− as | |

1 −δ ε (1+|z|) m cεe 2 .

If γ m then by ellipticity we obtain again that | | ≥ B (x, D)F DβA(x, D)F γ ≍ or B (x, D)F Dβ(zF ), γ ≍ where β = γ m m 1. Hence, we obtain in that case | | | | − ≤ − 1 −δ ε (1+|z|) m B (x, D)F c z e 2 . | γ | ≤ ε| | If we combine these two estimates then we obtain

1 −δ ε (1+|z|) m AP (x, y, z) c (1 + z )e 2 | | ≤ ε | | and by induction 1 k k −δ ε (1+|z|) m A “P (x, y, z) c (1 + z ) e 2 . | | ≤ ε,k | | Next we write

AQ = AG P “= A zI + zI G P = P zG P = P + zQ . z − z z − − z z − z − z z − z z This implies that î ó “” “”” AQz“f Pzf””+ z ”Qzf ”. ” ” ” L2 ≤ L2 | | L2

But Q = G P and G is a bounded operator in L 2(Ω). Hence z − z z z “” ” ” AQzf Pzf + c z Pzf . L2(Ω) ≤ L2(Ω) | | L2(Ω) ” ”” ”

1 So it remains to estimate Pzf . If f L (Ω1) then “” L2(Ω) ” ∈ ”

P f(x) = P (x, y, z)f(y)dy P (x, y, z) f(y) dy z ” | | ZΩ1 ≤ ZΩ1 | || | 1 ε m −δ 2 (1+|z|) c e f 1 . ε L (Ω 1) ” ≤ k k It follows that

1 1/2 1 ε m ε m −δ 2 (1+|z|) 1/2 −δ 2 (1+|z|) Pzf cεe f 1 dx = cε Ω1 e f 1 . L2(Ω) L (Ω1) L (Ω1) ≤ k k Ω1 | | k k Z

Applying the identity Ç å ” k k−1 k−1 A Q = A P + z A Q z − z z 83 “ ” “ ” “ ” we obtain

84 9 Fractional powers of self-adjoint operators

∗ If A = A and A> 0, say A c I,c > 0, then for z / Z we know that (A zI)−1 ≥ 0 0 ∈ θ − ≡ 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 x y m−|α|−n, mn + α | | | |  for α m 1 and x, y Ω, and Q(x, y, z) is given by | | ≤ − ∈ Q(x, y, z)= G(x, u, z)P (u, y, z)du. − ZΩ

The integral operator Qz with this kernel has the mapping property ∞ j Q : L1(Ω ) ∆(A) D(A ), z 1 → ≡ ” j\=1 and for each k N ∈ ” “ 1“ k ε m k −δ 2 (1+|z|) A Q f c(1 + z ) e f 1 . z L (Ω1) L2(Ω) ≤ | | k k

Concerning the estimates for Q(x, y, z) we have “ ” x y m−n, mn | |  only for x Ω1 and y Ω because G(x, y, z) is simply equal to the fundamental ∈ ∈ s solution in that case. Next, for any s> 0 we can deﬁne A by the spectral theorem as ∞ s s A = λ dEλ 0 Z “ with ∞ s 2 2s D(A )= f L“(Ω) : λ d(Eλf, f) < . ∈ Z0 ∞ Example 9.1. If A = AF > ß0 then ™ “ ∞ s s AF f = λjfjej(x), “ jX=1 ∞ where f =(f,e ) 2 , e is an orthonormal basis of eigenvectors and j j L (Ω) { j}j=1 ∞ D(As )= f L2(Ω) : λ2s f 2 < . F  ∈ j | j| ∞  jX=1    85 Since we consider A c I,c > 0 then we can also deﬁne ≥ 0 0 ∞ −τ −τ A = λ dEλ, τ> 0, “ Zc0 with ∞ −τ “ 2 −2τ D(A )= f L (Ω) : λ d(Eλf, f) < . ∈ Zc0 ∞ Suppose that 0 <τ < 1. Then® ´ “ −τ −τ −τ ∞ t dt y=t/λ ∞ λ y ∞ = dy = λ−τ y−τ (1 + y)−1dy Z0 λ + t Z0 1+ y Z0 r=(1+y)−1 1 = λ−τ (1 r)−τ rτ−1dr = λ−τ B(τ, 1 τ) Z0 − − Γ(τ)Γ(1 τ) π = λ−τ − = λ−τ Γ(1) sin πτ

−τ for any λ > 0. That’s why we can rewrite the spectral representation for A , 0 < τ < 1 in the form

∞ ∞ ∞ −τ −τ −τ sin τπ t dt “ A = λ dEλ = dEλ Zc0 π Zc0 Z0 λ + t ∞ ∞ ∞ sin τπ −τ −1 sin τπ −τ = t (λ + t)® dEλ dt ´= t G−tdt, “ π Z0 Zc0 π Z0 because z = t / Z and, therefore,® G is well-deﬁned.´ − ∈ θ −t ‘ −τ Theorem 1. If 0 <τ < 1 then A is the sum of the integral operators E(τ) and Q(τ) whose kernels are ‘

∞ “sin τπ −τ “ Eτ (x, y)= t E(x, y, t)dt “ π Z0 − and ∞ sin τπ −τ Qτ (x, y)= t Q(x, y, t)dt, π Z0 − respectively. Moreover,

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

G−t = E−t + Q−t since G(x, y, t)=‘E(x,‘ y, t)+‘Q(x, y, t). − − − Let us obtain now the estimates for DαE (x, y), α m 1. Since x τ | | ≤ − ∞ α sin τπ −τ α Dx Eτ (x, y) t Dx E(x, y, t) dt | | ≤ π Z0 | − | we make use of the estimates for DαE . | x | 1) Let m < n + α . Then mτ

2) Let m = n + α . Then again mτ

3) In the case m > n + α our task is to estimate | | ∞ n+|α| 1 m α −τ −1+ m −δ|x−y|t Dx Eτ (x, y) c t (1 + t) e dt | | ≤ Z0 in three subcases.

87 n+|α| a) Let mτ < n + α . Then τ + 1 m < 1. Let us change the variables 1 | | − u := x y t m . Then the integral in question equals | − | m n+|α|−m ∞ u m c x y mτ−m um(1−τ)−1 1+ e−δudu | − | 0 x y Z | − | or ∞ Ç Ç å ån+|α|−m c x y mτ−|α|−n um(1−τ)−1 ( x y m + um) m e−δudu. | − | Z0 | − | It can be estimated from above by ∞ c x y mτ−|α|−n um(1−τ)−1un+|α|−me−δudu | − | Z0 which equals ∞ c x y mτ−|α|−n un+|α|−mτ−1e−δudu | − | Z0 or c 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| = 1 and we consider the integral | | − m ∞ n+|α|−m c x y mτ−|α|−n um(1−τ)−1 ( x y m + um) m e−δudu | − | Z0 | − | in two parts: |x−y| ∞ du + du := I1 + I2. Z0 Z|x−y| We have

|x−y| m(1−τ)−1 n+|α|−m n+|α|−m m(1−τ) I1 c u x y du = c x y x y ≤ Z0 | − | | − | | − | = c x y n+|α|−mτ = c | − | and ∞ ∞ m(1−τ)−1 n+|α|−m −δu −1 −δu I2 c u u e du = c u e du ≤ Z|x−y| Z|x−y| c (1 + log x y ) . ≤ | | − || c) Let mτ >n + α . Then τ + 1 n+|α| > 1 and | | − m ∞ n+|α| 1 m α −τ −1+ m −δ|x−y|t Dx Eτ (x, y) c t (1 + t) e dt | | ≤ 0 Z ∞ −τ −1+ n+|α| c t (1 + t) m dt = c ≤ Z0 because the last integral converges.

88 Hence the second part of this theorem is proved. Finally, using Theorem 2 of section 8 we can obtain ∞ k −τ k A Q(τ)f t A Q−tf dt 2 2 L (Ω) ≤ Z0 L (Ω) ∞ 1 ε m −τ k − δ 2 (1+t) c t (1 + t) e f 1 dt c f 1 , L (Ω1) L (Ω1) “ “ ≤ Z0 “ ‘ k k ≤ k k for any k = 0, 1,.... This ﬁnishes the proof.

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

K (x, y) E (x, y)+ Q (x, y) τ ≡ τ τ “ such that x y mτ−n, mτ n,  where x, y Ω. But Eτ (x, y) 0 forx Ω Ω1, y Ω and for x Ω, y Ω Ω1. The ∈ ≡ ∈ \ ∈ 1 ∈ ∈ \ integral operator Q(τ) with the kernel Qτ (x, y) maps from L (Ω1) to ∆(A). But if we take into account Qτ (x, y) we can get only the estimate

“ x y mτ−n, mτ n,  where x Ω , y Ω and c = c(Ω ). If now τ 1 then we use the fact that ∈ 1 ∈ 1 1 ≥ −τ −τ −τ −τ A = A 1 A 2 A l , ◦ ◦···◦ where τ1 + τ2 + + τl = τ with 0 < τj < 1, j = 1, 2,...,l. Due to this fact we ··· −τ may conclude that A is an“ integral“ operator“ with the“ kernel Kτ (x, y) which can be calculated as

Kτ (x, y)= Kτ1 (x,“ u1)du1 Kτ2 (u1, u2)du2 Kτl−1 (ul−2, ul−1)Kτl (ul−1, y)dul−1. ZΩ ZΩ ··· ZΩ

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

x y m(τ1+τ2)−n, mτ + mτ < n | − | 1 2 Kτ1+τ2 (x, y) c1 1+ log x y , mτ1 + mτ2 = n | | ≤  | | − || 1, mτ1 + mτ2 > n,  where x, y Ω and c = c(Ω ).  ∈ 1 1 1

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

−α1 −α2 −α1 −α2 n−(α1+α2) x u u y du x u u y du c0 x y . Rn ZΩ | − | | − | ≤ Z | − | | − | ≤ | − |

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

x u −α1 u y −α2 du c(Ω)(1 + log x y ) . ZΩ | − | | − | ≤ | | − ||

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

x u −α1 u y −α2 du c(Ω). ZΩ | − | | − | ≤ Proof of lemma. Since

Kτ1+τ2 (x, y) Kτ1 (x, u) Kτ2 (u, y) du | | ≤ ZΩ1 | || | then we can apply the estimates for Kτ1 and Kτ2 with 0 < τ1 < 1, 0 < τ2 < 1, α = n mτ and α = n mτ . 1 − 1 2 − 2

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

mτ1−n mτ2−n Kτ1+τ2 (x, y) c1 x u u y du | | ≤ ZΩ1 | − | | − | c x y n−(n−mτ1)−(n−mτ2) = c x y m(τ1+τ2)−n. ≤ 1| − | 1| − |

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

K (x, y) c (1 + log x y ) . | τ1+τ2 | ≤ 1 | | − ||

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

K (x, y) c . | τ1+τ2 | ≤ 1

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

mτ1−n Kτ1+τ2 (x, y) c1 x u (1 + log u y ) du | | ≤ ZΩ1 | − | | | − || mτ1−n −ε cε x u u y du cε ≤ ZΩ1 | − | | − | ≤

if we choose 0 <ε

90 Remark. It follows by induction that

x y mτ−n, mτ n,   where x, y Ω1,c1 = c(Ω1) and τ > 0. These estimates can be extended to hold for x Ω and∈y Ω. ∈ 1 ∈ n Theorem 2. Suppose that τ > 2m . Then

−τ A f c1 f 2 , ∞ L (Ω) L (Ω1) ≤ k k

where Ω = Ω Ω and c = c(Ω ). 1 1 ⊂ 1 “ 1 Proof. For f L2(Ω) and τ > 0 it follows from lemma above and Cauchy-Schwarz- Bunjakovskii inequality∈ that

−τ A f sup Kτ (x, y) f(y) dy ∞ L (Ω1) ≤ x∈Ω1 ZΩ | || |

mτ−n x y , mτ n,  1/2  x y 2mτ−2n, mτ n, ê  c f 2  ≤ 1 k kL (Ω)  because τ > n if and only if 2mτ 2n> n which makes the latter integral converge. 2m − −

Now let us assume for simplicity that A = AF > 0 is the Friedrichs extension. In that case ∞ AF f = λjfjej(x), “ jX=1 2 where λj > 0,λj and fj are the Fourier coeﬃcients of f L (Ω) with respect to e (x) ∞ . The orthonormal→ ∞ system of eigenvectors e (x) ∞∈ forms an orthonormal { j }j=1 { j }j=1 basis in L2(Ω). By the spectral theorem we know also that

∞ D(As )= f L2(Ω) : λ2s f 2 < . F  ∈ j | j| ∞  jX=1    91 Corollary. Let Ω Ω be a compact set and let τ > n . Then 1 ⊂ 2m ∞ λ−2τ e (x) 2 c j | j | ≤ 1 jX=1 holds uniformly with respect to x Ω with c = c(Ω ). ∈ 1 1 1 Proof. Since ∞ −τ −τ AF f = λj fjej(x) jX=1 then Theorem 2 implies that

1/2 ∞ ∞ λ−τ f e (x) c f 2 , x Ω . j j j 1 j 1 j=1 ≤ j=1 | | ∈ X X Ñ é

The left hand side of this inequality can be considered as the inner product of

−τ −τ ~e =(λ1 e1(x),...,λj ej(x),...) and f~ =(f1,..., fj,...) in l2(C). Hence, ~ ~ (~e, f)l2 c1 f . ≤ l2

By duality we may conclude that

~e 2 c k kl ≤ 1 or ∞ λ−2τ e (x) 2 c2. j | j | ≤ 1 jX=1

Theorem 3. The Fourier series ∞ fjej(x) jX=1 converges absolutely and uniformly on each compact set Ω1 Ω for each function f τ n ⊂ from D(AF ) with τ > 2m . Proof. Using the corollary above and Cauchy-Schwarz-Bunjakovskii inequality we obtain

1/2 1/2 1/2 ∞ ∞ ∞ ∞ f e (x) λ2τ f 2 λ−2τ e (x) 2 c λ2τ f 2 , | j|| j | ≤ j | j| j | j | ≤ 1 j | j| jX=1 jX=1 jX=1 jX=1 Ñ é Ñ é Ñ é where τ > n . But the latter number series converges since f D(Aτ ). 2m ∈ F

92 Remark. We know that ◦ W m(Ω) D(A ). 2 ⊂ F Then for τ N we may conclude that ∈ ◦ W mτ (Ω) D(Aτ ). 2 ⊂ F n This embedding implies that if τ > 2m or mτ > n/2 then the Fourier series corre- ◦ mτ sponding to f W2 (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 spectral theorem we can characterize D(A) as follows: f D(A) if and only if ∈ ∞ 2 (1 + λ )d(Eλf, f) < Z0 ∞ and deﬁne a new norm f := f + Af . k kD(A) k kH k kH Deﬁnition. Let G(t) t>0 be a family of bounded linear operators from H to H. This family 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 depend onk t orkf. ≤ k k ∈ 3) lim G(t)f f = 0 for f H. t→+0 k − kH ∈ 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 by the formula { } G(t) I lim − f Af = 0 t→0 t − H with domain D(A) consisting of all f H such that ∈ G(t) I lim − f t→0 t 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 positive function on Rn 0 , which is positively homogeneous of order m > 0 i.e. ω(tξ) = t mω(ξ). Let us\ define { } the family G(t) by the formula | | { }t>0 G(t)f := F −1 e−tω(ξ)F f , f L2(Rn). ∈ It is clear that G(t) : L2(Rn) L2(Rn). Moreover, → Ä ä 93 1)

G(t+s)f = F −1 e−(t+s)ω(ξ)F f = F −1 e−tω(ξ)FF −1 e−sω(ξ)F f = G(t)(G(s)f).

2) Ä ä Ä Ä ää −1 −tω(ξ) −tω(ξ) G(t)f 2 = F e F f = e F f F f 2 = f 2 . k kL L2 L2 ≤ k kL k kL

3) Ä ä −1 −tω(ξ) −tω(ξ) G(t)f f 2 = F e F f F f = e 1 F f k − kL − L2 − L2

F f 2 = f 2 , t 0 → k kL k kL → Ä ä Ä ä by the Lebesgue theorem. Also by Lebesgue theorem we have

G(t)f f e−tω(ξ) 1 lim − =H lim F −1 − F f = F −1 (ω(ξ)F f) Af. t→0 t t→0 t ! − ≡

The domain of A is

2 D(A)= f L : ω(ξ)F f 2 < . ∈ k kL ∞ For instance, if ω(ξ)= ξ 2 then A ∆ and D(A)= W 2(Rn). | | ¶ ≡ 2 © Example 9.3. Let A = A∗ 0. Deﬁne ≥ ∞ itA itλ G(t) := e e dEλ. ≡ Z0 Then

∞ ∞ ∞ ∞ i(t+s)λ itλ isλ itλ isµ G(t+s)= e dEλ = e e dEλ = e dEλ e dEµ = G(t)G(s). Z0 Z0 Z0 Z0 2) ∞ 2 itλ 2 2 G(t)f = e d(Eλf, f)= f . k k Z0 | | k k 3) ∞ 2 itλ 2 G(t)f f = e 1 d(Eλf, f) 0, t 0. k − k Z0 | − | → → and G(t)f f ∞ eitλ 1 ∞ − = − dEλf i λdEλf iA, t 0. t Z0 t → Z0 ≡ →

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 domain of Ak. We deﬁne the≥ interpolation space Dα,p, where α>∈ 0 and 1 p as follows: Set ≤ ≤ ∞

K(t, f) := inf ( f0 H + t f1 Dk ) , f H, 0 β then Dα,p Dβ,q ⊂ for any p and q. Theorem 4. Let G(t) be an equi-bounded, strongly continuous semi-group on H with inﬁnitesimal generator A. Then 1) K(t, f) h(t, f) + min(1,t) f , ≍ k kH where h(t, f) = sup G(s)f f . s

h(t, f) = sup G(s)f f H sup G(s)f0 f0 H + sup G(s)f1 f1 H s

(M + 1) f0 H + sup G(s)f1 f1 H . ≤ k k s

95 We can thus write t t G(t)f f = G(s)Afds = AG(s)fds. − Z0 Z0 That’s why

t t sup G(s)f1 f1 H sup G(s)Af1 H ds M Af1 H ds s

K(t, f) = inf f0 H + t f1 D(A) t inf f0 H + f1 D(A) = t f H . f=f0+f1 k k k k ≥ f=f0+f1 k k k k k k If t 1 then ≥ K(t, f) inf f0 H + f1 D(A) = f H . ≥ f=f0+f1 k k k k k k Thus h(t, f) + min(1,t) f (M + 1)K(t, f)+ K(t, f)=(M + 2)K(t, f). k kH ≤ For the other half of 1) we argue as follows. By the deﬁnition of inﬁmum we have

K(t, f) inf f0 H + t f1 D(A) f H ≡ f=f0+f1 k k k k ≤ k k under the choice f 0 and f = f . Therefore 1 ≡ 0 K(t, f) f h(t, f)+ f = h(t, f) + min(1,t) f ≤ k kH ≤ k kH k kH for t 1. If 0

t t = t−1 (G(s)f f)ds + G(s)fds 0 − 0 Z H Z D(A)

t t

sup G(s)f f H + G(s)fds + AG(s)fds ≤ s

sup G(s)f f H + Mt f H + G(t)f f H ≤ s 0 Z0 is an equi-bounded, strongly continuous semi-group with the inﬁnitesimal generator k k k k k iA . Since D(iA ) = D(A ) and (H,D(iA ))θ,2 = (H,D(A ))θ,2 then it follows from Theorem 4 that ∞ 1/2 −2θ 2 dt f k f + t h(t, f) (H,D(A ))θ,2 H k k ≍ k k Z0 t ∞ 1/2 Ç −2θ å 2 dt = f H + t G(t)f f k k Z0 k − k t ∞ ∞ 1/2 Ç −2θ itλk 2 å 2 dt = f H + t e 1 d Eλf k k Z0 Z0 | − | k k t 1/2 ∞ ∞ itλk 2 Ç 2 e 1 dt å = f H + d Eλf | 2−θ | k k Z0 k k Z0 t t ! 1/2 k ∞ ∞ iξ 2 ξ=tλ 2θk 2 e 1 = f H + λ d Eλf | 2−θ+1 | dξ k k Z0 k k Z0 ξ ! ∞ 1/2 2θk 2 kθ = f H + cθ λ d Eλf = f H + cθ A f , k k 0 k k k k H Z since the inner integral with respectÅ to ξ converges forã 0 <θ< 1.

97 Remark. The corollary above allows us to conclude that

(H,D(Ak)) D(Akθ) (H,D(Ak)) θ,1 ⊂ ⊂ θ,∞ for any k N and 0 <θ< 1. ∈ Let AF be Friedrichs extension of an elliptic diﬀerential operator of order m with ∞ 2 an orthonormal basis ej j=1 of eigenvectors in L (Ω), where Ω is bounded. Let also ∗ { } AF = AF > 0.

2 k n ,1 Theorem 5 (Peetre). For any f (L (Ω),D(A )) n ,1 D 2m the series ∈ 2mk ≡ ∞ fjej(x), jX=1 where fj =(f,ej)L2(Ω) converges absolutely and uniformly in compact parts of Ω. Proof. Since A has pure discrete spectrum λ ∞ then F { j}j=1

Eλf(x)= fjej(x) λXj <λ and, therefore, θ(x,y,λ)= ej(x)ej(y). λXj <λ We will use G˚arding’s estimate

1/2 1/2 ∞ 2 2 fj fj = f L2 . (9.1) ν ν+1 | | ≤ | | k k 2 ≤λXj <2 jX=1 Ñ é Ñ é

98 Since

−k k −k k −k k −k k fj =(f,ej)= λj (f,λj ej)= λj (f, AF ej)= λj (AF f,ej)= λj (AF f)j then we have also

1/2 1/2 2 −2k k 2 fj = λj (AF f)j ν ν+1 | | ν ν+1 | | 2 ≤λXj <2 2 ≤λXj <2 Ñ é Ñ é 1/2 −kν k 2 −kν 2 (AF f)j 2 f Dk . (9.2) ≤ ν ν+1 | | ≤ k k 2 ≤λXj <2 Ñ é 2 k Applying (9.1) to f0 and (9.2) to f1 we get for f = f0 + f1, f0 L (Ω), f1 D(AF ) that ∈ ∈ 1/2 2 −kν fj f0 L2 + 2 f1 Dk . ν ν+1 | | ≤ k k k k 2 ≤λXj <2 Ñ é It follows that 1/2 2 −νk fj K(2 , f). ν ν+1 | | ≤ 2 ≤λXj <2 Ñ é That’s why we obtain

∞ ∞ ∞ νn −νk νn −νk fjej(x) c1 2 2m K(2 , f) c1 2 2m K(2 , f). | | ≤ ≤ ν=−∞ jX=1 νX=0 X Next we are going to prove that

∞ ∞ νn −νk − n dt 2 2m K(2 , f) c t 2mk K(t, f) ν=−∞ ≤ 0 t X Z if k N and k > n . Note ﬁrst that ∈ 2m K(t, f) max(1,t/s)K(s, f). (9.3) ≤ Indeed, if t/s 1 then ≥ t t K(t, f)=inf( f + t f ) = inf f + s f K(s, f). k 0k k 1k k 0k s k 1k ≤ s

If t/s < 1 then immediately K(t, f) K(s, fÅ). Using (9.3) weã obtain ≤ 2−kν K(2−kν, f) max 1, K(t, f) ≤ t ! or 2kνK(2−kν, f) t−1K(t, f), t< 2−kν. ≤ 99 Next, ∞ ∞ 2−kν − n dt − n dt t 2mk K(t, f) = t 2mk K(t, f) . −k(ν+1) 0 t ν=−∞ 2 t Z X Z Therefore,

ν n −νk νk −νk ν n −k −1 ν n −k −1 − n +1 2 2m K(2 , f) = 2 K(2 , f)2 ( 2m ) t K(t, f)2 ( 2m ) ct K(t, f)t 2mk ≤ ≤ −k(ν+1) −kν n for 2 2m . It implies that

ν n −νk 1 − n 1 2 2m K(2 , f) ct 2mk K(t, f) , t ≤ t

−k(ν+1) −kν n where 2 2m . Integrating this inequality with respect to t over the interval in question yields

2−kν 2−kν ν n −νk dt − n dt 2 2m K(2 , f) c t 2mk K(t, f) −k(ν+1) −k(ν+1) Z2 t ≤ Z2 t or 2−kν ν n −νk ′ − n dt 2 2m K(2 , f) c t 2mk K(t, f) . −k(ν+1) ≤ Z2 t Finally we obtain

∞ ∞ ∞ 2−kν ν n −νk − n dt fjej c1 2 2m K(2 , f) c1 t 2mk K(t, f) −k(ν+1) | | ≤ ν=−∞ ≤ ν=−∞ 2 t jX=1 X X Z ∞ − n dt = c1 t 2mk K(t, f) . Z0 t n ,1 But the latter integral is finite if and only if f D 2m . ∈ ∞ ∞ Rn Let ϕj(ξ) j=0 be a partition of unity i.e. j=0 ϕj(ξ) 1, ξ , ϕj 0, ϕj ∞ Rn { } −j ∞ ≡ ∈ ≥ ∈ C0 ( ), ϕj(ξ) = ϕ(2 ξ), j = 1, 2,..., where ϕ P C0 with supp ϕ 1/4 ξ 1 and supp ϕ ξ < 1 . ∈ ⊂ { ≤ | | ≤ } 0 ⊂ {| | } Definition. Let s R and 1 p . A function f belongs to the Besov space Bs (Rn) for 1 p<∈ if ≤ ≤ ∞ 2,p ≤ ∞ 1/p ∞ p/2 jps 2 f Bs (Rn) := 2 fj(x) dx < , k k 2,p Rn | | ∞ jX=0 Z Ñ é −1 Å s Rn ã where fj(x)= F (ϕj(ξ)F f)(x). The space B2,∞( ) is defined to consist of functions f for which 1/2 js 2 f s Rn := sup 2 fj(x) dx < B2,∞( ) Rn k k 0≤j≤∞ Z | | ∞ with fj as above. Å ã

100 Exercise 49. Prove that Bs (Rn) Bs (Rn) 2,p ⊂ 2,∞ for 1 p< . ≤ ∞ In the case 1 p< the previous deﬁnition is equivalent to ≤ ∞ 1/p ∞ p/2 p/2 j(ps+ np ) j 2 2 2 2 ϕ(η)F f(2 η) dη + ϕ0(ξ)F f(ξ) dξ <  1/4≤|η|≤1 | | |ξ|<1 | |  ∞ jX=1 Z Z   while in the caseÇ p = we might equivalentlyå requireÇ that å ∞ 1/2 ϕ(η)F f(2jη) 2dη C2−js Z1/4≤|η|≤1 | | ≤ and Ç å 1/2 2 ϕ0(ξ)F f(ξ) dξ C Z|ξ|<1 | | ≤ for some constant C. Ç å Exercise 50. Prove that Bs (Rn) L∞(Rn) 2,p ⊂ 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 (Rn),s> 0, 1 p if and only if ∈ 2,p ≤ ≤ ∞ p 1/p ∞ ω(2)(t; Dαf) dt f s f 2 + , B2,p L s−k k k ≍ k k Z0 " t # t ! where k N , α = k,s k > 0 and ∈ 0 | | − (2) ω (t; g) := sup g( + h) 2g + g( h) L2 . |h|

Let ω(ξ) from Example 9.2 be equal to ξ m with m> 0. Then | | m G(t)= F −1(e−t|ξ| F ) is the semi-group with the inﬁnitesimal generator

Af = F −1( ξ mF f)= ( ∆)m/2f. − | | − − It is also true that

D(A)= f L2(Rn) : ξ mF f L2 = W m(Rn). ∈ | | ∈ 2 For this generator A the following theorem holds. ¶ © 101 Theorem 6. If 0 <θ< 1, 1 p and m> 0 then ≤ ≤ ∞ 2 Rn mθ Rn (L ( ),D(A))θ,p = B2,p ( ).

Proof. Let f Bmθ(Rn) and let ϕ ∞ be a partition of unity. Then ∈ 2,p { j}j=0 ∞ F f = ϕjF f jX=0 so that ∞ ∞

f 2 = F f 2 ϕ F f f . k kL k kL ≤ k j kL2 ≤ k jkL2 jX=0 jX=0 Moreover,

Af = AF −1(ϕ F f)= F −1( ξ mF (F −1ϕ F f)) = F −1( ξ mϕ F f). j j − | | j − | | j Hence Af = ξ mϕ F f 2jm f k jkL2 k| | j kL2 ≤ k jkL2 because supp ϕ 2j−2 ξ 2j , j = 1, 2,..., supp ϕ ξ < 1 and, therefore, j ⊂ { ≤ | | ≤ } 0 ⊂ {| | } ξ m 2jm | | ≤ for ξ supp ϕ . Next we prove that ∈ j ∞ h(t, f) c min(1,t2jm) f . (9.4) ≤ k jkL2 jX=0 Indeed, if t 1 then ≥ ∞

h(t, f) = sup G(s)f f (M + 1) f L2 (M + 1) fj L2 . s

∞ ∞ Mt Af Mt 2jm f . ≤ k jkL2 ≤ k jkL2 jX=0 jX=0 This proves (9.4). We also know from Theorem 4 that

∞ 1/p −θ p dt f 2 f 2 + t h(t, f) . (L ,D(A))θ,p L k k ≍ k k Z0 t Ç å Ä ä

102 It follows from (9.4) that

p 1/p ∞ 1/p ∞ ∞ −θ p dt −θ jm dt t h(t, f) c t min(1,t2 ) fj L2 0 t ≤ 0 k k t Z Z jX=0 Ç å Ñ∞ Ñ∞ 1/pé é Ä ä −θ jm p dt c t min(1,t2 ) fj L2 . ≤ 0 t k k jX=0 Z Ç å Let us calculate the integral appearing in theÄ last estimate. Wäe have

1/p −jm 1/p ∞ p dt 2 t−θ min(1,t2jm) = t(1−θ)p−1dt 2jm Z0 t Z0 ∞ 1/p Ç å + Ç t−θp−1dt å Ä ä −jm Z2 = c′2−(1−θ)jm2jm + c′′2jθm = c2jθm. Å ã Thus ∞ jmθ f 2 Rn f 2 + c 2 fj 2 . k k(L ( ),D(A))θ,p ≤ k kL k kL jX=0 Hence

f (L2(Rn),D(A)) c f Bmθ(Rn) . k k θ,p ≤ k k 2,1 This inequality means that

Bmθ(Rn) (L2(Rn),D(A)) . 2,1 ⊂ θ,p It is also possible to prove sharper embedding

Bmθ(Rn) Bmθ(Rn) (L2(Rn),D(A)) . 2,1 ⊂ 2,p ⊂ θ,p It remains to prove the opposite inequality, namely

f Bmθ(Rn) c f (L2(Rn),D(A)) . k k 2,p ≤ k k θ,p To that end, note that

m f (x) = F −1(ϕ F f)= F −1(ϕ (e−t|ξ| 1)−1F (G(t)f f)) j j j − − −jm −j m t=2= F −1(ϕ (e−(|ξ|2 ) 1)−1F (G(2−jm)f f)). j − − Since supp ϕ 2j−2 ξ 2j then j ⊂ { ≤ | | ≤ } 1 ξ 2−j 1 4 ≤ | | ≤ or 1 m ( ξ 2−j)m 1 4 ≤ | | ≤

Ç å 103 for ξ supp ϕ . Hence ∈ j −(|ξ|2−j )m −1 −jm fj 2 = ϕj(e 1) F (G(2 )f f) k kL − − L2

1 −jm −jm −jm F (G(2 )f f) c G(2 )f f ch(2 , f). ≤ 1 e−(1/4)m − L2 ≤ − L2 ≤

− By the deﬁnition of Besov spaces we have

1/p 1/p ∞ ∞ jps p jps −jm p f Bs = 2 fj 2 c 2 h(2 , f) . k k 2,p k kL ≤ jX=0 jX=0 Ñ é Ñ é As in the proof of Theorem 5 it can be checked that the convergence of the latter series is equivalent to the convergence of the integral

1/p ∞ p dt t−θh(t, f) Z0 t with s = mθ, 0 <θ< 1. Ç å Ä ä Exercise 51. Prove this fact. Thus, the required inequality is proved and, therefore, we have the embedding (L2(Rn),D(A)) Bmθ(Rn). θ,p ⊂ 2,p This ﬁnishes the proof. Corollary. If Ω Rn is a bounded domain with smooth boundary then ⊂ ◦ ◦ 2 m mθ (L (Ω), W2 (Ω))θ,p = B2,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 operator of even order m. Since ◦ W mk(Ω) D(Ak) 2 ⊂ for any k N then ∈ ◦ L2(Ω), W mk(Ω) L2(Ω),D(Ak) 2 θ,p θ,p ⊂ or Ç å ◦ Ä ä Bmkθ(Ω) Dθk,p. 2,p ⊂ Peetre’s theorem means that the corresponding Fourier series converges absolutely and ◦ n ,1 n/2 uniformly if f D 2m . This implies that the same is true for f B2,1 (Ω). We proved also that if f ∈D(Aσ) with σ > n then the Fourier series converges∈ absolutely. Since ∈ 2m ◦ L2(Ω), W mk(Ω) L2(Ω),D(Ak) = D(Akθ) 2 θ,2 θ,2 ⊂ Ç å Ä104 ä or ◦ Bmkθ(Ω) D(Akθ) 2,2 ⊂ ◦ then the Fourier series converges absolutely for f Bn/2+ε(Ω),ε> 0. ∈ 2,2

105 Index K-functional, 95 G˚arding’s inequality, 57 generalized Leibniz formula, 53 adjoint operator, 13 graph, 12 basis, 8 Green’s function, 62 Besov space, 100 Hilbert space, 3 Bessel’s inequality, 2 Hilbert-Schmidt norm, 12 bounded, 11 idempotent, 20 Cauchy sequence, 3 induced by the inner product, 3 Cauchy-Schwarz-Bunjakovskii inequality, 2 infinitesimal generator, 93 Cayley transform, 31 inner product, 1 closable, 13 inner product space, 1 closed, 13 isometry, 23 closed subspace, 6 closure, 13 kernel, 11 compact operator, 39 complete space, 3 Lebesgue space, 5 completeness relation, 9 length, 2 constant of ellipticity, 53 linear operator, 11 continuous spectrum, 36 linear space, 1 convergent sequence, 3 linear span, 8 criterion for closedness, 13 multi-index, 52 densely defined, 11 non-negative operator, 22 discrete spectrum, 36 norm, 3 domain, 11 norm topology, 3 elliptic partial differential operator, 52 nullspace, 11 ellipticity condition, 53 orthogonal, 2 equi-bounded, strongly continuous semi- orthogonal complement, 5 group, 93 orthonormal, 2 essential spectrum, 36 orthonormal basis, 8 essentially self-adjoint, 16 extension, 13 parallelogram law, 3 parametrix, 80 finite rank operator, 39 Parseval equality, 9 formally self-adjoint, 52 partition of unity, 100 Fourier expansion, 9 point spectrum, 36 Friedrichs extension, 50 polarization identity, 3 functional of Peetre, 95 positive operator, 22 fundamental solution, 68 precompact set, 39

106 principal symbol, 52 Projection theorem, 6 projector, 20 quadratic form, 48 range, 11 resolvent, 32 resolvent identity, 33 resolvent set, 33 restriction, 13 Riesz-Frechet theorem, 7 scalar product, 1 self-adjoint, 16 semibounded from below, 48 separable, 8 sequence space, 4 smoothed fundamental solution, 81 Sobolev space, 5, 54 spectral family, 23 spectral function, 61 spectrum, 33 symmetric, 16

Theorem of Pythagoras, 2 triangle inequality, 2 uniformly elliptic operator, 53 unitary operator, 23 vector space, 1

107