Operator Theory and Integral Equations 802660S Lecture Notes

Home , Integral equation, Operator (mathematics)

Operator theory and integral equations 802660S

Lecture Notes Second printing

Valery Serov University of Oulu 2012

Edited by Markus Harju Contents

1 Inner product spaces and Hilbert spaces 1

2 Symmetric operators in the Hilbert space 12

3 J. von Neumann’s spectral theorem 25

4 Spectrum of self-adjoint operators 38

5 Quadratic forms. Friedrichs extension. 54

6 Elliptic diﬀerential operators 58

7 Spectral function 67

8 Integral operators with weak singularities. Integral equations of the ﬁrst and second kind. 71

9 Volterra and singular integral equations 81

10 Approximate methods 88

Index 98

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 a 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, j=1 X 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. We say that 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 { } ⊂ 0, α = β A ( 6 is some index set.

3) x := (x,x) is called the length of x H. k k ∈ k Exercise 1. pProve the Theorem of Pythagoras: If xj j=1, k N is an orthonormal system in an inner product space H, then { } ∈

k k 2 2 2 x = (x,xj) + x (x,xj)xj 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 j=1 X 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: p 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 · · p 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. · × → 3) The inner product (x,y): H H C is continuous. × → ∞ 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 that on C[a,b] the norm

f = max f(x) k k x∈[a,b] | | is not induced by an inner product.

3 Exercise 7. Give an example of an inner product space which is not complete. 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| ∞ ( j=1 ) X The estimates

x + y 2 2 x 2 + y 2 , λx 2 = λ 2 x 2 | j j| ≤ | j| | j| | j| | | | j| and 1 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

∞

(x,y) := xjyj j=1 X 2 (k) ∞ 2 and prove that l (C) is complete. Suppose that x k=1 l (C) 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→∞

4 This fact and l x(k) x(m) 2 <ε2, l N | j − j | ∈ j=1 X imply that l l (k) (m) 2 (k) 2 2 lim xj xj = xj xj ε m→∞ | − | | − | ≤ j=1 j=1 X X for all k n and l N. Therefore the sequence ≥ 0 ∈ l s := x(k) x 2, k n l | j − j| ≥ 0 j=1 X 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→∞ | − | ≤ j=1 j=1 X X 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 |Xα|≤k Z

5 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, for all x M . { ∈ ∈ } Remark. It is clear that M ⊥ is a linear subspace of H. Moreover, M M ⊥ = 0 since 0 M always. ∩ { } ∈ 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

u u 2 = (u x)+(x u ) 2 k j − kk k j − − k k u + u 2 = 2 u x 2 +2 x u 2 4 x j k . k j − k k − kk − − 2

Since (u + u )/2 M then j k ∈ 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→∞

6 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 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 = u + v = u + v , where u ,u ∈ M and v ,v M ⊥. It 1 1 2 2 1 2 ∈ 1 2 ∈ 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 . { ∈ } ⊥ Then v0 M ,v0 = 0 and T (v0) = 0. Since T is linear and continuous then M is a closed subspace.∈ It6 follows from Thereom6 1 that

H = M M ⊥ ⊕ i.e. every x H has the unique representation as x = u + v. Therefore, for every x H, we can∈ deﬁne ∈ T (x) u := x v0. e e − 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 , v0 v0 ! k k k k 7 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 e k k ek k 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 k k k k k k ≥ k k k k k k ﬁnishes the proof. 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

⊥ M ⊥ = M .

That’s why ⊥ ⊥ M ⊥⊥ = M and Theorem 1 implies that

⊥ ⊥ H = M M , H = M M ⊥⊥. ⊕ ⊕ Uniqueness of this representation guarantees that M⊥⊥ = M. Remark. In the frame of this theorem we have that

x 2 = u 2 + v 2 , v 2 =(x,v), u 2 =(x,u). k k 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 ∈ ( j=1 ) X 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

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

x cjxj <ε, xj B. − ∈ j=1 X

8 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.

∞ 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 j=1 X i.e. k

x (x,xj)xj 0, k . − → → ∞ 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). j=1 X

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

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).

9 2) 3) Given x H introduce the sequence ⇒ ∈ k (k) x = (x,xj)xj. j=1 X 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

It follows that ∞ (x,x ) 2 | j | j=1 X 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→∞ j=1 X 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) = 0forany 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. j=1 X Xk=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). j=1 j=1 X Xk=1 X 4) 5) Take y = x in part 4). ⇒

10 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 | j=1 X 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,xj)xj x cjxj . − ≤ − j=1 j=1 X X

11 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(Ω Ω), Ω Rn. ∈ × ⊂ Deﬁne the integral operator K as

Kf(s)= K(s,t)f(t)dt, f L2(Ω). b ∈ ZΩ Let us prove that K is bounded.b Indeed,

2 2 Kf b = Kf(s) 2ds = K(s,t)f(t)dt ds 2 L (Ω) Ω | | Ω Ω Z Z Z 2 2 2 b b 2 = (K(s, ), f)L ds K(s, ) 2 f 2 ds · ≤ k · kL L ZΩ ZΩ

= K(s,t) 2 dt f(t) 2dt ds Ω Ω | | Ω | | Z 2Z 2 Z = K 2 f 2 k kL (Ω×Ω) k kL (Ω) 12 where we have made use of Cauchy-Schwarz-Bunjakovskii inequality. That’s why we have

K K L2(Ω×Ω) . L2→L2 ≤ k k

The norm b K L2(Ω×Ω) := K k k HS is called the Hilbert-Schmidt norm of K. b Example 2.2 (Schur test). Assume that p and q are positive measurable functions on b Ω Rn and α and β are positive numbers such that ⊂ K(x,y) p(y)dy αq(x), a.e. in Ω | | ≤ ZΩ and K(x,y) q(x)dx βp(y), a.e. in Ω. | | ≤ ZΩ Then K is bounded and K αβ. L2→L2 ≤ b p Proof. For any f L2(Ω) we have b ∈ 2 2 K(x,y) K(x,y) f(y) dy dx = K(x,y) p(y) | | f(y) dy dx | |·| | | | p(y) | | s ! Z Z Z Z p p K(x,y) K(x,y) p(y)dy | | f(y) 2dy dx ≤ | | p(y) | | Z Z Z f(y) 2 α K(x,y) q(x)dx | | dy ≤ | | p(y) Z Z αβ f(y) 2dy ≤ | | Z by the Cauchy-Schwarz-Bunjakovskii inequalityand Fubini’s theorem. Example 2.3 (Diﬀerential operator in L2). Consider the diﬀerential operator d A := i dt of order 1 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 (Af, g)= if ′(t)g(t)dt = i fg 1 f(t)g′(t)dt = f(t)ig′(t)dt =(f, Ag) |0 − Z0 Z0 Z0 13 for all f,g D(A). Let us now consider the sequence ∈

un(t) := sin(nπt), n =1, 2,....

Clearly u D(A) and n ∈ 1 2 2 1 u 2 = sin(nπt) dt = . k nkL | | 2 Z0 But

1 2 1 2 d 2 2 2 1 2 2 Au 2 = i sin(nπt) dt =(nπ) cos(nπt) dt =(nπ) =(nπ) u 2 . k nkL dt | | 2 k nkL Z0 Z0

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

Example 2.4 (Diﬀerential operator in L2). Consider the diﬀerential operator

d2 d A := p + ip + p 0 dt2 1 dt 2 of order 2 in L2(0, 1) with the domain

D(A)= f C2[0, 1] : f(0) = f(1) = 0 { ∈ } 2 and with real constant coeﬃcients p0,p1 and p2. The fact D(A)= L and integration by parts gives

1 1 1 ′′ ′ (Af, g)= p0 f gdt + ip1 f gdt + p2 f gdt 0 · 0 · 0 · Z 1 Z Z 1 ′ 1 ′ ′ 1 ′ = p0 f g 0 f g dt + ip1 fg 0 f g dt + p2(f,g)L2 | − 0 · | − 0 · 1 Z 1 Z ′ ′ ′ = p0 f g dt ip1 f g dt +(f,p2g)L2 − 0 · − 0 · Z 1 Z ′ 1 ′′ ′ = p fg f g dt +(f,ip g ) 2 +(f,p g) 2 − 0 0 − · 1 L 2 L Z0 1 ′′ ′ = p f g dt +(f,ip g ) 2 +(f,p g) 2 =(f, Ag) 2 0 · 1 L 2 L L Z0 for all f,g D(A). Moreover, for the sequence u (t) = sin(nπt) we have (for suﬃ- ∈ n

14 ciently large n) that

1 2 ′′ ′ 2 Aun L2 = p0(sin(nπt)) + ip1(sin(nπt)) + p2 sin(nπt) dt k k 0 | | Z 1 = (p (nπ)2 p )2 sin2(nπt)+(nπ)2p2 cos2(nπt) dt 0 − 2 1 Z0 1 4 (nπ) 2 2 2 2 2 p0 sin (nπt)+(nπ) p1 cos (nπt) dt ≥ 0 2 Z 1 (nπ)2p2 sin2(nπt)+cos2(nπt) dt ≥ 1 Z0 2 2 1 2 2 2 = 2(nπ) p = 2(nπ) p u 2 . 1 2 1 k nkL So A is unbounded, since 2 2 A 2 2 2(nπ) p k kL →L ≥ 1 for n . → ∞ 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 ∈ × 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. Axn y → 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 ∈ →w ⇒ (y = Ax, Axn y →   15 where x w x indicates weak convergence in the sense that n → (x ,y) (x,y) n → for all y H. ∈ Remark. It is important from the point of view of applications (in particular, for numerical procedures) that the closedness of an operator guarantees the convergence of some process to the ”correct” (right) result.

Deﬁnition. Let A and A1 be two linear operators in a Hilbert space H. We say that A1 is an extension of A (or 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 f (A A )x := A x = A x, x D(A A f), f 1 ∩ 1 1 1 ∈ 1 ∩ 1 whenever A A = A and A A = A . ⊂ 1 1 f ⊂ 1 1 f f If A is closable then Γ A = Γ(A). f f Deﬁnition. Consider the subspace

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. 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 2.5. Consider the operator

Af(x) := x−αf(x), α> 0 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 ∈ ∈ 16 where 0, 0 x 1/n χn(x)= ≤ ≤ 1, 1/n

D(A∗)= v L2 : x−αv L2 . ∈ ∈ Let us show that A is not closed. To see this take the sequence

xα, 1/n

fn D(A) α ∈ α x D(A) fn x ∈ α → ⇒ (1= Ax . Afn 1 → But xα / D(A). This contradiction shows us that A is not closed. It is not bounded either since∈ α> 0.

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

1) A∗ = A∗.

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 × 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.

17 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) . Let us check that the criterion for ⊂ closedness holds i.e. v D(A∗) n ∈ v D(A∗) v v ∈  n ⇒ ∗ ∗ → (y = A v. A vn y → Indeed, for any u D(A) we have ∈ (Au, v ) (Au, v). n → On the other hand, (Au, v )=(u, A∗v ) (u,y). n n → Hence (Au, v)=(u,y). Thus v D(A∗) and y = A∗v. This proves 1). ∈ 2) Assume D(A∗) = H. Then we can 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 − ⊥⊥ ⊥ ⊥ Γ(A)= V 2Γ(A) = V V Γ(A) = (V Γ(A∗))⊥ − − − by 1). Hence 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).

18 3) Since A is closable then

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

This ﬁnishes the proof.

Example 2.6. 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

D(A)= u L2(R): f u(x) dx < = L2(R) L1(R). ∈ | 0 | ∞ ∩ ZR It is known that L2(R) 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 ∗ 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 → 1) symmetric if A A∗; ⊂ 2) self-adjoint if A = A∗;

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

19 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. 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. ⊂ Example 2.7. Consider the operator d2 A := dx2 in the Hilbert space H = L2(0, 1) with the domain

D(A)= f C2[0, 1] : f(0) = f(1) = f ′(0) = f ′(1) = 0 . { ∈ } It is clear that D(A) = L2(0, 1) and A is not closed. Moreover, integration by parts gives (Af, g)L2 =(f, Ag)L2 2 ∗ for any f D(A) and g W2 (0, 1). That is, A is symmetric such that A A and ∗ ∈ 2 ∈ ∗ ∗ ⊂ D(A ) = W2 (0, 1). As we know, A = A always. Now we will show that A is the ◦ ◦ 2 2 same diﬀerential operator of order 2 with D(A) = W2 (0, 1), where W2 (0, 1) denotes 2 the closure of D(A) with respect to the norm of Sobolev space W2 (0, 1). Indeed, for any f D(A) we have ∈ 2 2 2 Af 2 + f 2 f 2 k kL k kL ≤ k kW2 and 1 2 2 2 ′ 2 f 2 = Af 2 + f 2 + f dx W2 L L k k k k k k 0 | | Z 1 2 2 ′′ 3 2 3 2 = Af 2 + f 2 ff dx Af 2 + f 2 . k kL k kL − ≤ 2 k kL 2 k kL Z0 It means that 2 2 2 Af 2 + f 2 f 2 . k kL k kL ≍ k kW2 20 Exercise 11 gives now that ◦ 2 D(A)= W2 (0, 1). So we have ﬁnally

◦ 2 ∗∗ 2 D(A) ( D(A)= W2 (0, 1) = D(A ) ( W2 (0, 1). The closure A is symmetric, but not self-adjoint since

◦ ∗ W 2(0, 1) = D(A) = D(A )= D(A∗)= W 2(0, 1). 2 6 2 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).

This equality implies that ∗ A−1 v D(A∗) ∈ and ∗ A∗ A−1 v = v. (2.1) 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 A∗v = v. (2.2) ∗ It follows from (2.1) and (2.2) that (A ∗)−1 exists and (A∗)−1 =(A−1) .

21 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 D(A−1) = D (A−1)∗ = H. Thus A−1 is self-adjoint⊂ and bounded⊂ (Hellinger-Toeplitz theorem). Finally, ∗ A−1 = A−1 =(A∗)−1 if and only if A = A∗.

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. − { } { } 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. − −

22 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∗). − { }

Example 2.8. Assume that some operator A = A is closed and symmetric in the Hilbert space H. Consider the operator A∗A on the domain

D(A∗A)= f D(A): Af D(A∗) . { ∈ ∈ } This operator is self-adjoint. Indeed,

(A∗A)∗ = A∗A∗∗ = A∗A = A∗A.

So A∗A is symmetric. At the same time for any f D(A) it holds that ∈ (A∗Af, f)=(Af, A∗∗f)=(Af, Af)= Af 2 . k kH It means that A∗A is positive. This fact leads to R(A∗A iI)= H, since A∗A iI is invertible in this case. Thus, Theorem 3 gives us that A∗A±is self-adjoint. The same± is true for the operator AA∗ on the domain

D(AA∗)= f D(A∗): A∗f D(A) . { ∈ ∈ } It is clear that in general AA∗ = A∗A. 6 In case the equality holds here the operator A is called normal.

23 d Exercise 14. Let H = L2(0, 1) and A := i . dx 1) Prove that A is closed and symmetric on the domain

◦ D(A)= f L2(0, 1) : f ′ L2(0, 1),f(0) = f(1) = 0 W 1(0, 1). { ∈ ∈ }≡ 2 2) Prove that A is self-adjoint on the domain

D (A)= f L2(0, 1) : f ′ L2(0, 1),f(0) = f(1)eiγ,γ R . γ ∈ ∈ ∈

24 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. Then

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 ∈ j=1 X 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 Px 2 = sup (Px,Px)= sup (P ∗Px,x) sup P ∗Px 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. − − − − −

25 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 ∈ − ∈ (w,x)=(w,Px)=(Pw,x), x H ∈ ⊥ 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 x ⊂H such w that w = Px . If z R(P ⊥) then z = ∈P ⊥x =(I P )x for some x ∈H. Thus w ∈ z − z z ∈ (w,z)=(Px , (I P )x )=(Px ,x ) (Px ,Px )=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 easy to check by deﬁnition that P = I ⊕ |R(P ) and P ⊥ = 0. |R(P ) 4) If M H is a closed subspace then Theorem 1 in Section 1 implies that x = ⊂ ⊥ u + v H, where u M and v M . In that case let us deﬁne PM : H 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| ∞ ( j=1 j=1 ) X X

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 ∈ j=1 X then by Bessel’s inequality we obtain that P x M and M ∈ P x x . k M k ≤ k k

26 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) j=1 ! j=1 j=1 X X X ∞

= x, (y,ej)ej =(x, PM y) j=1 ! X ∗ 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) Px Qx for all x H. k k ≤ k k ∈ 3) R(P ) R(Q). ⊂ 4) P = PQ = QP .

Proof. 1) 2) Follows immediately from (Px,x)=(P 2x,x)=(Px,Px)= Px 2 . ⇔ k k 3) 4) Assume R(P ) R(Q). Then QPx = Px or QP = P . Conversely, if QP = P ⇔ then clearly R(P⊂) R(Q). Finally, P = QP = P ∗ =(QP )∗ = P ∗Q∗ = PQ. ⊂ 2) 4) If 4) holds then Px = PQx and Px = PQx Qx for all x H. ⇔ Conversely, if Px Qx then Px k= QPxk +kQ⊥Pxkimplies ≤ k k that ∈ k k ≤ k k 2 Px 2 = QPx 2 + Q⊥Px QPx 2 . k k k k ≤ k k Hence 2 Q⊥Px =0 i.e. Q⊥Px = 0 for all x H. Hence P = QP = PQ. ∈

27 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. ∈ ∈ 28 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 = (E(λ ,λ ]x,E(λ ,λ ]y) | j−1 j j−1 j | j=1 Xn E(λ ,λ ]x E(λ ,λ ]y ≤ k j−1 j k k j−1 j k j=1 X n 1/2 n 1/2 E(λ ,λ ]x 2 E(λ ,λ ]y 2 ≤ k j−1 j k k j−1 j k j=1 ! j=1 ! X X n n

= E(λj−1,λj]x E(λj−1,λj]y

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.

29 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 j=1 j=1 ! X X ∗ 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(E x,y), x,y H. λ ∈ 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 j=1 X which we denote by β f(λ)dEλx. Zα Thus β β f(λ)d(E x,y)= f(λ)dE x,y , x,y H. λ λ ∈ Zα Zα For the spectral representation of self-adjoint operators 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→−∞ Z−∞ β→∞ Zα Z−∞ if it exists. Deriving first some basic properties of the integral just defined 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 ∞ D := x H : f(λ) 2d(E x,x) < ∈ | | λ ∞ Z−∞ (or D := x H : ∞ f(λ)dE x exists ). Let us define on this domain an operator ∈ −∞ λ A as n R ∞ o (Ax, y)= f(λ)d(E x,y), x D(A) := D,y H λ ∈ ∈ 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. ⊂

30 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→−∞ Z−∞ β→∞ Zα β β = lim f(λ)d(x,Eλy)= lim x, f(λ)dEλy α→−∞ α→−∞ β→∞ Zα β→∞ Zα β = x, lim f(λ)dEλy =(x, Ay). α→−∞ β→∞ Zα ! 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→−∞ β→∞ Zα Z−∞ Z−∞ ∞ = f(λ)d(Eλu,z)= (Au, z)=(z, Au), Z−∞ 31 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 =(E E ) Ax =(E E ) f(λ)dE x β − α β − α λ Z−∞ ∞ ∞ β α = f(λ)dE E x f(λ)dE E x = f(λ)dE x f(λ)dE x λ β − λ α λ − λ Z−∞ Z−∞ Z−∞ Z−∞ β ∞ = f(λ)dE x = f(λ)dE (E E ) x = A (E E ) x = AE(α, β]x λ λ β − α β − α 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 ∞ Au 2 = f(λ) 2d(E u,u) k k | | λ 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 = λdE x, x D(A) λ ∈ Z−∞ (i.e. (Ax, y)= ∞ λd(E x,y),x D(A),y H), where D(A) is defined as −∞ λ ∈ ∈ R ∞ D(A)= x H : λ2d(E x,x) < . ∈ λ ∞ Z−∞ Proof. At first we assume that this theorem holds when A is bounded, that is, there is a unique spectral family F ∞ such that { µ}µ=−∞ ∞ Au = µdF u, u H µ ∈ 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 = µdF u, u H. µ ∈ Zm 32 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). ≥ ∈ Since A−1 is bounded then D(A−1) 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 A−1v = µdF v, v H, µ ∈ 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 B u := dF u, u D(A). ε µ µ ∈ Zε For every v H we have ∈ 1 1 1 1 1 1 1 1 B A−1v = dF (A−1v)= dF λdF v = d λd(F F v) ε µ µ µ µ λ µ µ λ Zε Zε Z0 Zε Z0 1 1 µ 1 1 1 = d λdF v = µdF v = dF v = F v F v = v F v. µ λ µ µ µ 1 − ε − ε Zε Zε Zε Zε Since every spectral family is right continuous then

−1 lim BεA v = v ε→0+0 exists. For every u D(A) we have similarly, ∈ 1 1 µ 1 A−1B u = µdF (B u)= µd dF u = u F u ε µ ε λ λ − ε Z0 Zε Zε and hence −1 lim A Bεu = u ε→0+0

33 exists. These two equalities mean that

−1 −1 lim Bε = A = A ε→0+0 exists and the spectral representation

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

∞ 1 1 D(A)= u H : λ2d(E u,u) < = u H : d(F u,u) < . ∈ λ ∞ ∈ µ2 µ ∞ Z1 Z0 This proves the theorem for self-adjoint operators that are semibounded from below. For bounded operators we will only sketch the proof.

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

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 real-valued 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].

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

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

∞ 2 ∞ L(v) 2 f(λ)dE u v 2 = f(λ) 2d(E u,u) v 2 = c(u) v 2 . | | ≤ λ k k | | λ k k k k Z−∞ Z−∞

35 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(E u,v)=(z,v), v H. λ ∈ 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 U := f(A)= − dE . A λ + i λ Z−∞ 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→−∞ − Z−∞ β→∞ Zα β→∞ 2 2 2 = lim Eβu Eαu = u α→−∞ k k − k k k k β→∞ by normalization of Eλ . Hence UA is 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 −1 I UA =2i(A + iI) − −1 (I + UA =2A(A + iI) 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 −

36 Denote ∞ 1 R := (A zI)−1 = dE . z − λ z λ Z−∞ − 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 Example 3.3. Suppose that K(x,y) L2(Ω Ω). Deﬁne the integral operator on L2(Ω) as ∈ × Af(x)= K(x,y)f(y)dy. ZΩ Then A∗f(x)= K(y,x)f(y)dy ZΩ and therefore A∗Af(x)= K(y,z)K(y,x)dy f(z)dz. ZΩ ZΩ As we know from Example 2.8, A∗A is self-adjoint on L2(Ω). This fact can also be checked directly, since

K(y,z)K(y,x)dy = K(y,x)K(y,z)dy. ZΩ ZΩ And J. von Neumann’s spectral theorem gives us for this operator and for any s 0 that ≥ kAk2 L2→L2 ∗ s s (A A) = λ dEλ, Z0 ∗ 2 since A A is positive and bounded by A 2 2 . k kL →L 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)). ∩

37 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

n s x := (z z )j (R )j+1 x. n − 0 z0 j=0 X 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)yn x − − → 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 ∈

39 This is equivalent to 1 (A zI)u u , u D(A). k − k≥ Mz k k ∈

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 40 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 ∞ 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 R x c i.e. R c . By Liouville theorem we may conclude now k z k≤ 0 k zk≤ 0 that Rz is constant with respect to z. But by J. von Neumann’s spectral theorem ∞ 1 R = dE , z λ z λ 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) σ (A) is the continuous spectrum∈σ (A). The \ p c 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 according 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.

41 Remark. If A = A∗ then

1) λ σ (A) means that (A λI)−1 exists but is not bounded. ∈ c − 2) σ (A)= σ (A) eigenvalues of inﬁnite multiplicity and their accumulation points ess c ∪{ } 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 ej j=1 be an orthonormal basis in H and let sj j=1 C be some sequence. Introduce{ the} set { } ⊂

∞ D = x H : s 2 (x,e ) 2 < . ∈ | j| | j | ∞ ( j=1 ) X Deﬁne ∞ Ax = s (x,e )e , x D. j j j ∈ j=1 X Prove that A = A and that σ(A)= s : j =1, 2,... . Prove also that { j } ∞ 1 (A zI)−1x = (x,e )e − s z j j j=1 j X − 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 be its spectral family. Then { λ}λ∈R 1) µ σ(A) if and only if E E =0 for every ε> 0. ∈ µ+ε − µ−ε 6

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

Proof. 1) Suppose that µ σ(A) but there exists ε> 0 such that Eµ+ε Eµ−ε = 0. Then by spectral theorem we∈ obtain for any x D(A) that − ∈ ∞ (A µI)x 2 = (λ µ)2d(E x,x) (λ µ)2d(E x,x) k − k − λ ≥ − λ Z−∞ Z|λ−µ|≥ε µ−ε ∞ ε2 d(E x,x)= ε2 + d(E x,x) ≥ λ λ Z|λ−µ|≥ε Z−∞ Zµ+ε = ε2 (E x,x)+ x 2 (E x,x) = ε2 x 2 . µ−ε k k − µ+ε k k 42 This inequality means (see part 2) of Theorem 1) that µ σ(A) but µ ρ(A). This contradiction proves 1) in one direction. Conversely, if 6∈ ∈

Pn := Eµ+ 1 Eµ− 1 =0 n − n 6 ∞ for all n 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 ∞ (A µI)x 2 = (λ µ)2d(E P x , P x )= (λ µ)2d(E x ,x ) k − nk − λ n n n n − λ n n Z−∞ Z|λ−µ|≤1/n 1 ∞ 1 1 d(E x ,x )= x 2 = 0 ≤ n2 λ n n n2 k nk n2 → Z−∞ 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 ∞ 0= (A µI)x 2 = (λ µ)2d(E x ,x ). k − 0k − λ 0 0 Z−∞ In particular, for all n N and large enough ε> 0 we have that ∈ n n 0 = (λ µ)2d(E x ,x ) ε2 d(E x ,x )= ε2((E E )x ,x ) − λ 0 0 ≥ λ 0 0 n − µ+ε 0 0 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

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

E y = E Py = E E y E E y = Py = 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 Hence ∞ ∞ (A µI)y 2 = (λ µ)2d(E y,y)= (λ µ)2d (y,y)=0. k − k − λ − λ 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 ∈ j=1 X 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

44 ∞ 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 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. x w 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 Axkn y → w→ ⇒ (Ax = ax. xk x n → Since x = 1 then x = 1 also. Hence x = 0 and a is an eigenvalue of A. k kn k k k 6

45 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.

∞ 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 ∈ j=1 j=1 X X 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 H1. The restriction of the inner product of H to H1 makes H1 a Hilbert space (since∈ H is closed) and the restriction of A to H , denoted by A = A , is 1 1 1 |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

46 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. Consider now the projector

n P x := (x,e )e , x H. n j j ∈ j=1 X 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 j=1 j=1 X X and A(I P )x = Ax AP x 0, n k − n k k − n k → → ∞ then ∞

Ax = λj(x,ej)ej j=1 X 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

47 But by Riesz-Schauder theorem

∞

yε = Axε = λj(xε,ej)ej. j=1 X Hence

∞

x yε = x λj(xε,ej)ej < ε/2. k − k − j=1 X

Making use of the Theorem of Pythagoras, Bessel’s inequalit y 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+ λj(xε,e j)ej j=n+1 X 1/2 ∞ ε/2+ λ 2 (x ,e ) 2 ≤ | j| | ε j | j=n+1 ! X ∞ 1/2 ε/2+ λ (x ,e ) 2 ≤ | n+1| | ε j | j=n+1 ! X ε/2+ λ x <ε ≤ | 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. { } Proposition 1 (Riesz). If A is a compact operator on H and µ C then the nullspace ∈ of I µA is a ﬁnite-dimensional subspace. − Proof. The nullspace N(I µA) is a closed subspace of H since I µA is bounded. Indeed, for each sequence f− f and f µAf = 0 we have that f− µAf = 0 since n → n − n − A is continuous. The operator A is compact on H and therefore also compact from N(I µA) onto N(I µA), since N(I µA) is closed. Hence, for any f N(I µA) we have− − − ∈ − If =(I µA)f + µAf = µAf − and I is compact on N(I µA). Thus N(I µA) is ﬁnite-dimensional. − −

48 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. − 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. We are now ready to derive the following fundamental result of the Riesz theory.

49 Theorem 5 (Riesz). Let A : H H be a compact linear operator on the Hilbert space H. Then for any µ C the operator→ I µA is injective (i.e. (I µA)−1 exists) if and only if it is surjective∈ (i.e. R(I µA−) = H). Moreover, in this− case the inverse operator (I µA)−1 : H H is bounded.− − → Proof. If (I µA)−1 exists then (I µA∗)−1 exists too and therefore N(I µA∗)=0. Then Lemma− of Riesz (Theorem 4)− and Exercise 13 imply H = R(I µA),− i.e. I µA is surjective. − − Conversely, if I µA is surjective then N(I µA∗) = 0 i.e. I µA∗ injective and so is I µA. − − − It remains− to show that (I µA)−1 is bounded on H if I µA is injective. Assume −1 − − that (I µA) is not bounded. Then there exists a sequence fn H with fn =1 such that− ∈ k k (I µA)−1f n. − n ≥ Define f (I µA)−1f g := n , ϕ := − n . n (I µA)−1f n (I µA)−1f k − nk k − nk Then gn 0 as n and ϕn = 1. Since A is compact we can select a subsequence ϕ such→ that Aϕ → ∞ϕ as k k . But kn kn → n → ∞ ϕ µAϕ = g n − n n and we observe ϕkn µϕ and ϕ N(I µA). Hence ϕ = 0 and this contradicts ϕ = 1. → ∈ − k nk Theorem 6 (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 f is a particular solution and u N(I µA) (u is the general solution of the 0 ∈ − 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. − − 50 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. − ∗ 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, n ZR 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 7 (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 →

51 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

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 ∞ = (λ µ)2d(E (E E )x ,x ) − µ λn+rn − λn−rn n n Z−∞ λn+rn 2 = (λ µ) d(Eµxn,xn) λn−rn − Z ∞ 2 max (λ µ) d(Eµxn,xn) ≤ λn−rn≤µ≤λn+rn − Z−∞ = max (λ µ)2 0, n . λn−rn≤µ≤λn+rn − → → ∞

Theorem 8 (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 σ (A) σ (B). Take any λ σ (A). Then there is an ess ⊂ ess ∈ ess 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 52 for all n n >> 1. Next we take the identity ≥ 0 (B zI)−1 (λ z)−1 y = Ty (λ z)−1(A λI)x . − − − n − n − − − n Since T is compact and y w 0 we deduce that n → (B zI)−1 (λ z)−1 y 0. − − − n → Introduce z := (B zI)−1y . n − n 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.−

53 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

x := Q(x,x)+(λ + 1) x 2. k kQ k k q 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 deﬁned by the quadratic form Q as Q(x,y)=(Ax, y), x D(A),y D(Q), ∈ ∈ which is semi-bounded from below i.e.

(Ax, x) λ x 2 , x D(A) ≥− k k ∈ and D(A) D(Q). ⊂

54 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 − − 55 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- adjoint and, therefore,∈ ∈ ∗ D(√A + λI)= D( √A + λI )= D(Q) H . ≡ Q

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.

56 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

x := Q(x,x)+ x 2. k kQ k k q 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.

57 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 =(αp,...,α ) is a multi-index i.e. α N N 0 . 1 n j ∈ 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 = D D ( i) ∂ i i ∂xj − 1 ··· 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

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

a (x)= ( 1)|β|CαDβ−αa (x), α − β β 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. Xβ≤α Assumption 3. We assume that A(x, D) has a divergence form

A(x, D) Dα(a (x)Dβ), ≡ αβ |α|=X|β|≤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|=X|β|=m/2 |αX|=m/2 |αX|=m/2 where ν > 0 is called the constant of ellipticity. Such operator is called uniformly elliptic.

Exercise 34. Prove that ξ2α ξ m ≍| | |αX|=m/2 i.e. c ξ m ξ2α C ξ m, | | ≤ ≤ | | |αX|=m/2 where c and C are some constants.

Example 6.1. Let us consider

n A(x, D)= D2 = ∆, x Ω Rn j − ∈ ⊂ j=1 X 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 inΩ . 0 ∈ { 6 } n o

59 Let u,v C∞(Ω). Then ∈ 0 n n 2 2 (Au, v) 2 = D u vdx = ∂ u vdx L j − j ZΩ j=1 ! j=1 ZΩ n X Xn = ∂ ((∂ u) v) dx + (∂ u) ∂ v dx − j j j j j=1 ZΩ j=1 ZΩ X X = (v u,n )dx +( u, v) 2 =( u, v) 2 , − ∇ x ∇ ∇ L ∇ ∇ L Z∂Ω 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) 2 = u∂ vdx =(u, ∆v) 2 =(u, Av) 2 . ∇ ∇ L − j − L L j=1 Ω X Z Hence A A∗ and A is closable. ⊂ 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 2 = u 2 + u 2 + ∆u 2 k kW2 k kL k∇ kL k kL 2 2 u 2 + u 2 ∆u 2 + ∆u 2 ≤ k kL k kL k kL k kL 3 2 3 2 3 2 u 2 + ∆u 2 u , ≤ 2 k kL 2 k kL ≡ 2 k kA 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 u 2 . Combining these inequalities gives k kA ≤ k kW2 2 u 2 u u 2 W2 A W2 r3 k k ≤ k k ≤ k k for all u C∞(Ω). A completion of C∞(Ω) with respect to these norms leads us to ∈ 0 0 the statement: ◦ 2 D(A)= W2 (Ω).

60 ◦ 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 n n i.e. ( ∆) v = v and D(( ∆) ) = W2 (Ω). Finally, for Ω R with Ω = R 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 and 2 2 u u 1 . k kQ ≡ k kW2 (Ω) 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

61 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 H = L2(Ω) and

A(x, D)= ( + i−→W (x))2 + q(x), − ∇ 1 where −→W is an n-dimensional real-valued vector from W∞(Ω) and q is a real-valued ∞ ∗ function from L (Ω). Deﬁne A, A and AF . 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|=X|β|≤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, ∈ a (x)ξαξβ ν ξα 2, ν> 0. αβ ≥ | | |α|=X|β|=m/2 |αX|=m/2 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 (Af, f)L2(Ω) (ν ε) f m/2 Cε f L2(Ω) ≥ − k kW2 (Ω) − k k for any f C∞(Ω). ∈ 0

62 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|=X|β|≤m/2 Z |α| α β = ( 1) aαβ(x)D fD fdx Ω − |α|=X|β|=m/2 Z |α| α β + ( 1) aαβ(x)D fD fdx Ω − |α|=X|β| 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

k Applying this lemma with δ = 1 to the norm of Sobolev spaces W2 we may conclude that 2 2 2 f m/2−1 ε1 f m/2 + Cε1 f L2(Ω) k kW2 (Ω) ≤ k kW2 (Ω) k k for any ε1 > 0. Hence 2 2 (Af, f)L2(Ω) ν f m/2 (C0 + ν) f m/2−1 ≥ k kW2 (Ω) − k kW2 (Ω) 2 2 2 ν f m/2 (C0 + ν)ε1 f m/2 (C0 + ν)Cε1 f L2(Ω) ≥ k kW2 (Ω) − k kW2 (Ω) − k k 2 2 = (ν ε) f m/2 Cε f L2(Ω) . − k kW2 (Ω) − k k This proves the theorem.

63 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 2 C f 2 , C > 0 k F µ kL ≥ 0 k kL 0 so that (A )−1 : L2(Ω) L2(Ω). F µ → Secondly, ′ ′ (AF )µf −m/2 C0 f m/2 , C0 > 0 k kW2 (Ω) ≥ k kW2 (Ω) so that ◦ (A )−1 : L2(Ω) W m/2(Ω). F µ → 2 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 j=1 X for any f D(A ). ∈ F Proof. We begin with a lemma.

64 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 m/2 1 there { } ⊂ k kW2 ≤ 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. the Fourier transform ϕ (ξ) is uniformly bounded. That’s why there exists ϕ (ξ) k c jk which converges pointwise in Rn. Next, c c 2 2 ϕjk ϕjm L2 = ϕjk (ξ) ϕjm (ξ) dξ k − k n | − | ZR 2 2 = cϕj (ξ) dϕjm (ξ) dξ + ϕj (ξ) ϕjm (ξ) dξ | k − | | k − | Z|ξ|r ϕc (ξ) ϕd(ξ) 2dξ c d ≤ | jk − jm | Z|ξ|

The ﬁrst term I1 tends to 0 as k,m by the dominated convergence theorem of Lebesgue for any ﬁxed r > 0. The→ second ∞ term converges to 0 as r because → ∞ 2 ϕjk ϕjm m/ 2 k − kW2 ≤ 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 → → ∞

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 → ∞

65 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 ∈ j=1 X Exercise 37. Prove that ∞

AF f = λj(f,ψj)ψj j=1 X for any f D(A ). ∈ F Now we may conclude that the corollary is proved.

66 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|=X|β|≤m/2 where a = a are real, a C∞(Ω) and bounded for all α and β. We assume that αβ βα αβ ∈ a (x)ξαξβ ν ξ m, ν> 0. αβ ≥ | | |α|=X|β|=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 ≥ ∞ b b Ab= λdEλ Z0 with domain b ∞ 2 2 D A = f L (Ω) : λ d(Eλf,f) < . ∈ 0 ∞ Z In general case we haveb no such formula for D A as for the Friedrichs extension AF . But we can say that ◦ b W m(Ω) D A . 2 ⊂ Indeed, since a (x) C∞(Ω) and bounded then A(x, D) can be rewritten in the usual αβ b form ∈ γ A(x, D)= aγ(x)D |γX|≤m with bounded coeﬃcients. Hence e

γ Af 2 c D f 2 c f m . k kL (Ω) ≤ k kL (Ω) ≡ k kW2 (Ω) |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

bEλf(bx)= θ(x,y,λ)f(y)dy, ZΩ where θ(x,y,λ) is called the spectral function and has the properties

67 1) θ(x,y,λ)= θ(y,x,λ) 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 b b 1) ∞ d θ(x,y,λb ) G(x,y,z)= λ λ z Z0 −

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

Proof. Since z ρ A then J. von Neumann’s spectral theorem gives us ∈ ∞ b (A zI)−1f = (λ z)−1dE f. − − λ Z0 Next, by Theorem 1 we getb ∞ −1 −1 (A zI) f = (λ z) dλ θ(x,y,λ)f(y)dy − 0 − Ω Z ∞ Z b = (λ z)−1d θ(x,y,λ) f(y)dy − λ ZΩ Z0 = G(x,y,z)f(y)dy, 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) λ z λ z Z0 − Z0 − then 2) is also proved.

68 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

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

Exercise 40. Let us assume that n

Eλf = (f,ψj)ψj = f(y)ψj(y)dyψj(x) Ω λXj <λ λXj <λ Z

= ψj(x)ψj(y) f(y)dy = θ(x,y,λ)f(y)dy Ω   Ω Z λXj <λ Z   i.e. the spectral function θ(x,y,λ) has the following form

θ(x,y,λ)= ψj(x)ψj(y). λXj <λ Hence (see Corollary of Theorem 1 of this chapter) the Green’s function has the form

∞ ψ (x)ψ (y) G(x,y,z)= j j in L2. λ z j=1 j X − 69 If we assume now that n

∞ ψ (x) ψ (y) ∞ ψ (x) ψ (y) G(x,y,z) | j || j | = | j || j | | |≤ 2 2 2 2 j=1 λ + z k k+1 λ + z X j 2 Xk=0 2 ≤λXj <2 j 2 q q1/2 1/2 ∞ 1 ψ (x) 2 ψ (y) 2 (22k + z2)1/2 j j ≤ 2  k k+1 | |   k k+1 | |  Xk=0 2 ≤λXj <2 2 ≤λXj <2 ∞ (2k+1)n/m     2k 2 1/2 . ≤ (2 + z2) Xk=0

Since n

70 8 Integral operators with weak singularities. Inte- gral equations of the ﬁrst and second kind.

Let Ω be a bounded domain in Rn. Then

Af(x)= K(x,y)f(y)dy ZΩ is an integral operator in L2(Ω) with kernel K. Deﬁnition. Integral operator A is said to be operator with weak singularity if its kernel K(x,y) is continuous for all x,y Ω,x = y and there are positive constants M and α (0,n] such that ∈ 6 ∈ K(x,y) M x y α−n, x = y. | |≤ | − | 6 Remark. If K(x,y) is continuous for all x,y Ω and bounded then this integral operator is considered also as an operator with∈ weak singularity.

If we have two integral operators A1 and A2 with kernels K1 and K2, respectively, then we can consider their composition as follows:

(A A )f(x)= K (x,y)A f(y)dy = K (x,y) K (y,z)f(z)dz dy 1 ◦ 2 1 2 1 2 ZΩ ZΩ ZΩ

= K1(x,y)K2(y,z)dy f(z)dz ZΩ ZΩ and analogously

(A A )f(x)= K (x,y)K (y,z)dy f(z)dz 2 ◦ 1 2 1 ZΩ ZΩ assuming that the conditions of Fubini theorem are fulﬁlled. So, we may conclude that the compositions A1 A2 and A2 A1 are again integral operators with the kernels ◦ ◦

K(x,y)= K1(x,z)K2(z,y)dz, ZΩ (8.1) K(x,y)= K2(x,z)K1(z,y)dz, ZΩ e respectively. In general, K(x,y) = K(x,y), that is A1 A2 = A2 A1. Returning to the integral operators6 with weak singularitie◦ 6 s we◦ obtain their very important property. e

Lemma 1. If A1 and A2 are integral operators with weak singularities then A1 A2 as well as A A are also integral operators with weak singularities. Moreover, if◦ 2 ◦ 1 K (x,y) M x y α1−n and K (x,y) M x y α2−n, (8.2) | 1 |≤ 1| − | | 2 |≤ 2| − | 71 then there is M > 0 such that

x y α1+α2−n, α + α n.  The same estimates hold for the kernel K(x,y).

Proof. Using (8.1) and (8.2) we obtain e

K(x,y) M M x z α1−n z y α2−ndz. | |≤ 1 2 | − | | − | ZΩ If α + α

u 1/2, 1/2 u 3/2, u 3/2. | |≤ ≤| |≤ | |≥ In the ﬁrst case u e e u 1 1/2=1/2 | − 0|≥| 0|−| |≥ − and therefore

u e α1−n u α2−ndu 2n−α1 u α2−ndu | − 0| | | ≤ | | Z|u|≤1/2 Z|u|≤1/2 1/2 2n−α1−α2 =2n−α1 rα2−1dr dθ = Sn−1 , (8.5) n−1 α | | Z0 ZS 2 where Sn−1 denotes the area of the unit sphere in Rn. In the| third| case 2 u u e u e u 1 u u = | | | − 0|≥| |−| 0|≥| |− ≥| |− 3| | 3 and we have analogously

∞ 2n−α1−α2 3α2 u e α1−n u α2−ndu 3n−α1 Sn−1 rα1+α2−n−1dr = Sn−1 . | − 0| | | ≤ | | n α α | | Z|u|≥3/2 Z3/2 − 1 − 2 (8.6)

72 In the case 1/2 u 3/2 we have that u e 5/2 and so ≤| |≤ | − 0|≤ 2n−α1−α2 5α1 u e α1−n u α2−ndu 2n−α2 u e α1−ndu = Sn−1 . | − 0| | | ≤ | − 0| α | | Z1/2≤|u|≤3/2 Z|u−e0|≤5/2 1 (8.7) Combining (8.4)-(8.7) we obtain (8.3) for the case α1 + α2 < n. It can be mentioned here that the estimate (8.3) in this case holds also in the case of any (not necessarily bounded) domain Ω. If now α1 + α2 = n then the proof of (8.3) will be a little bit diﬀerent and holds only for bounded domain Ω. Indeed, for any z Ω and ∈ x y x y x z | − | or z y | − | | − |≤ 2 | − |≤ 2 we have in both cases that

n−α2 α2−n α1−n K(x,y) M1M22 x y x z dz | |≤ | − | Ω′ | − | Z |x−y|/2 M M 2n−α2 Sn−1 x y α2−n rα1−1dr ≤ 1 2 | || − | Z0 M M M M = 1 2 Sn−1 or 1 2 Sn−1 . (8.8) α1 | | α2 | | If z Ω does not belong to these balls with radius x y /2 then we consider two cases:∈ z x z y and z x z y . In both| cases− we| have | − |≥| − | | − |≤| − | d dz n−1 dr n−1 2d K(x,y) M1M2 n M1M2 S = M1M2 S log , | |≤ ′ x z ≤ | | r | | x y ZΩ\Ω | − | Z|x−y|/2 | − | (8.9) where d = diamΩ. The estimates (8.8) and (8.9) give us (8.3) in the case α1 + α2 = n. If ﬁnally α1 + α2 > n then since Ω is bounded we can analogously obtain (8.3) in this case. This ﬁnishes the proof.

Remark. In the case α1 + α2 = n, since for any 0

log t C t−ε, ε> 0 | |≤ ε instead of logarithmic singularity in (8.3) we may consider weak singularity for the kernel K(x,y) as K(x,y) M x y −ε, | |≤ ε| − | where ε> 0 can be chosen appropriately. Let A be integral operator in L2(Ω) with weak singularity. Then since 0 < α n we have ≤ x y α−ndy β and x y α−ndx β, | − | ≤ | − | ≤ ZΩ ZΩ 73 where β = sup x y α−ndy < . | − | ∞ x∈Ω ZΩ Schur test (see Example 2.2) shows that A is bounded in L2(Ω) and

A 2 2 Mβ. k kL (Ω)→L (Ω) ≤ We can prove even more. Theorem 1. The integral operator with weak singularity is compact in L2(Ω). Proof. Let us introduce the function

1, 0 t σ χσ(t)= ≤ ≤ (0, t>σ. Then for any σ > 0 we may write

K(x,y)= χ ( x y )K(x,y)+(1 χ ( x y ))K(x,y) := K (x,y)+ K (x,y). σ | − | − σ | − | 1 2

The integral operator with kernel K2(x,y) is Hilbert-Schmidt operator for any σ > 0 since K (x,y) 2dxdy M 2 x y 2α−2ndxdy | 2 | ≤ | − | ZΩ ZΩ ZZσ≤|x−y|≤d is ﬁnite. That’s why it is compact in L2(Ω) (see Exercise 31). For the integral operator A1 with kernel K1(x,y) we proceed as follows:

2 ∗ A f 2 =(A f, A f) 2 =(f, A A f) 2 , (8.10) k 1 kL (Ω) 1 1 L (Ω) 1 ◦ 1 L (Ω) ∗ where A1 is adjoint operator with kernel K∗(x,y)= χ ( x y )K(y,x) 1 σ | − | which is also with weak singularity. Using Lemma 1 we can estimate the right hand side of (8.10) from above as

1 K (x,y) f(x) f(y) dxdy K (x,y) f(x) 2dxdy | σ || || | ≤ 2 | σ || | ZΩ ZΩ ZΩ ZΩ 1 + K (x,y) f(y) 2dxdy, (8.11) 2 | σ || | ZΩ ZΩ where Kσ(x,y) is the kernel of weak singularity i.e.

x y 2α−n, αn/2   74 where ε> 0 can be chosen as small as we want. Let us note also that the deﬁnition of χσ(t) implies that Kσ(x,y)=0 for x y > 2σ. Thus (see (8.10) and (8.11)) we have (α< 2n) | − |

2 2α−n 2 A f 2 M x y f(x) dxdy k 1 kL (Ω) ≤ | − | | | ZZ|x−y|≤2σ 2α 2 2α−n 2 (2σ) n−1 M f(x) x y dydx = M f 2 S 0, ≤ | | | − | k kL (Ω) 2α | | → ZΩ Z|x−y|≤2σ as σ 0. It means that →

A 2 2 0, σ 0. k 1kL (Ω)→L (Ω) → → The same fact is valid for the cases α n/2. Thus, ≥

A A 2 2 A 2 2 0 k − 2kL (Ω)→L (Ω) ≤ k 1kL (Ω)→L (Ω) → as σ 0. But A2 is compact for any σ > 0, therefore A is also compact as the limit of compact→ operators. This completes the proof. We want now to expand the analysis of the integral operators with weak singularity defined on domains in Rn to the integral operators with weak singularity defined on surfaces of dimension n 1. Assume that ∂Ω is− the boundary of a bounded domain of class C1. It means, roughly speaking, that at any point x ∂Ω there is a tangent plane with normal vector ν(x) which is continuous function∈ on ∂Ω and surface differential dσ(y) in the neighborhood of each point x ∂Ω satisfies the inequality (see Kress book) ∈ dσ(y) c ρn−2dρdθ, ≤ 0 where (ρ,θ) are the polar coordinates in the tangent plane with origin x and c0 is independent of x. According to the dimension n 1 of the surface ∂Ω, an integral operator in L2(∂Ω) i.e. − Af(x)= K(x,y)f(y)dσ(y) Z∂Ω is said to be with weak singularity if its kernel K(x,y) is continuous for all x,y ∂Ω,x = y and there are constants M > 0 and α (0,n 1] such that ∈ 6 ∈ − K(x,y) M x y α−(n−1), x = y. | |≤ | − | 6 If K(x,y) is continuous everywhere we require that K is bounded on ∂Ω ∂Ω. We can provide now the following theorem. ×

Theorem 2. The integral operator with weak singularity is compact in L2(∂Ω).

Proof. The same as for Theorem 1.

75 There is very useful and quite general result for the integral operators with weak singularity both for domains and surfaces in Rn.

Theorem 3. The integral operator with weak singularity transforms bounded functions into continuous functions.

Proof. We give the proof for the domains in Rn. The proof for surfaces in Rn is the same. Let x,y Ω and x y <δ. Then ∈ | − | Af(x) Af(y) ( K(x,z) + K(y,z) ) f(z) dz | − |≤ | | | | | | Z|x−z|<2δ + K(x,z) K(y,z) f(z) dz | − || | ZΩ\{|x−z|<2δ} α−n α−n M f ∞ ( x z + y z )dz ≤ k kL (Ω) | − | | − | Z|x−z|<2δ

+ f ∞ K(x,z) K(y,z) dz := I + I . k kL (Ω) | − | 1 2 ZΩ\{|x−z|<2δ} Since z y x z + x y then | − |≤| − | | − | 3δ α n−1 α−1 n−1 (3δ) I 2M f ∞ S r dr =2M S f ∞ 0, δ 0. 1 ≤ k kL (Ω) | | | | k kL (Ω) α → → Z0 On the other hand for x y <δ and x z 2δ we have that | − | | − |≥ y z x z x y > 2δ δ = δ. | − |≥| − |−| − | − So the continuity of the kernel K outside of the diagonal implies that

K(x,z) K(y,z) 0, δ 0, − → → uniformly in z Ω x z < 2δ . Since Ω is bounded we obtain that I2 0 as δ 0. This ﬁnishes∈ \ the {| proof.− | } → → Exercise 41. Prove that if A is as in Theorem 3 then f(x) + Af(x) C(Ω) for f L2(Ω) implies f C(Ω). ∈ ∈ ∈ We are now in the position to extend the solvability conditions (Fredholm alternative, see Theorem 6 of Chapter 4) to the equations in the Hilbert space with compact but not necessarily self-adjoint operators.

Theorem 4 (Fredholm alternative II). Suppose A : H H is compact. For any µ C either the equations → ∈ (I µA)f = g, (I µA∗)f ′ = g′ (8.12) − −

76 have the unique solutions f and f ′ for any given g and g′ from H or the corresponding homogeneous equations

(I µA)f =0, (I µA∗)f ′ =0 (8.13) − − have nontrivial solutions such that

dim N(I µA) = dim N(I µA∗) < − − ∞ and in this case equations (8.12) have the solutions if and only if

g N(I µA∗) g R(I µA) ⊥ − ⇔ ∈ − g′ N(I µA) g′ R(I µA∗) ⊥ − ⇔ ∈ − respectively. Proof. Lemma of Riesz (see Theorem 4 of Chapter 4) and Exercise 13 give

R(I µA)= N(I µA∗)⊥ − − R(I µA∗)= N(I µA)⊥. − − Let us ﬁrst prove that always

dim N(I µA) = dim N(I µA∗). − − These two dimensions are finite due to Riesz (see Proposition 1 of Chapter 4). Since every compact operator is a norm limit of a sequence of operators of finite rank (see Chapter 4 for details), for any µ C,µ = 0 we have ∈ 6 I µA = µA +(I µA ), − − 0 − 1 where A is of finite rank and µA < 1. Then (I µA )−1 exists and 0 k 1k − 1 (I µA )−1(I µA)= I µ(I µA )−1A := I A , − 1 − − − 1 0 − 2 where A is of finite rank too. Analogously, since (I µA∗)−1 exists then 2 − 1 (I µA∗)(I µA∗)−1 = I µA∗(I µA∗)−1 := I A∗, − − 1 − 0 − 1 − 2 ∗ where A2 is adjoint to A2 and is of finite rank too. These representations allow us to conclude that

g N(I µA) g N(I A ) ∈ − ⇔ ∈ − 2 g′ N(I µA∗) (I µA∗)−1g′ N(I µA∗) ∈ − 2 ⇔ − 1 ∈ − Thus, it suﬃces to show that the number of independent solutions of the equations

′ ∗ ′ g = A2g, g = A2g

77 are equal. ∗ Since we know that the ranks of A2 and A2 are ﬁnite we may represent the mappings ∗ ∗ of the operators I A2 and I A2 as the mappings of matrices I M2 and I M2 − −∗ − − with adjoint matrices M2 and M2 . But the ranks of the adjoint matrices are equal and ′ ∗ ′ therefore the number of independent solutions of the equations g = A2g and g = A2g are equal. The next step is: if R(I µA) = H then N(I µA∗) = 0 and consequently N(I µA) = 0 and R(I − µA∗) = H (see Exercise− 13). This{ } means that both (I −µA)−1 and{ (I} µA∗)−1 −exist and the unique solutions of (8.12) are given by − − f =(I µA)−1g, f ′ =(I µA∗)−1g′. − − If N(I µA) and N(I µA∗) are not zero then R(I µA) and R(I µA∗) are proper subspaces− of H and equations− (8.12) have the solutions− if and only if− g R(I µA), g′ R(I µA∗). ∈ − ∈ − It is equivalent (see Exercise 13) to g N(I µA∗), g′ N(I µA). ⊥ − ⊥ −

We will demonstrate now this Fredholm alternative for the integral operators. Let Ω Rn be any domain and let ⊂ Af(x)= K(x,y)f(y)dy ZΩ be a compact integral operator in L2(Ω). Then its adjoint is deﬁned as

A∗f(x)= K(y,x)f(y)dy. ZΩ Hence, the Fredholm alternative for these operators reads as: either the equations

f(x) µ K(x,y)f(y)dy = g(x) − ZΩ (8.14) f ′(x) µ K(y,x)f ′(y)dy = g′(x) − ZΩ are uniquely solvable for any g and g′ from L2(Ω) or the equations

f(x)= µ K(x,y)f(y)dy ZΩ (8.15) f ′(x)= µ K(y,x)f ′(y)dy ZΩ have the same (ﬁnite) number of linearly independent solutions. And in this case equations (8.14) are solvable if and only if g and g′ are orthogonal to any solution f and f ′ of the equations (8.15), respectively.

78 Deﬁnition. Equations (8.14) and (8.15) are called the integral equations of second and ﬁrst kind, respectively. Exercise 42. Consider in L2(a,b) the integral equation

b ϕ(x) ex−yϕ(y)dy = f(x), x [a,b], − ∈ Za where f L2(a,b). Solve this equation and formulate the Fredholm alternative for it. ∈ Example 8.1 (Boundary value problems). Consider the ordinary diﬀerential equation of the second order

′′ ′ a0(x)u (x)+ a1(x)u (x)+ a2(x)u(x)= f(x)

2 1 on the interval [0, 1] with coeﬃcients f,a2 L (0, 1),a1 W2 (0, 1) and with smooth a (x) c > 0 subject to the boundary conditions∈ ∈ 0 ≥ 0 u(0) = u0, u(1) = u1.

Dividing this equation by a0(x) we may consider the boundary value problem in the form ′′ ′ u + a1(x)u + a2(x)u = f, u(0) = u0,u(1) = u1. Using Green’s function G(x,y) of the form

y(1 x), 0 y x 1 G(x,y)= − ≤ ≤ ≤ x(1 y), 0 x y 1 ( − ≤ ≤ ≤ we can rewrite this boundary value problem as 1 u(x)= ϕ (x)+ G(x,y)(a (y)u′ + a (y)u f(y))dy, 0 1 2 − Z0 where ϕ (x)= u (1 x)+ u x. Integration by parts implies 0 0 − 1 1 1 u(x)= ϕ0(x) G(x,y)f(y)dy + G(x,y)a1(y)u(y) 0 − 0 | 1 Z 1 [∂ G(x,y)a (y)+ G(x,y)a′ (y)] u(y)dy + G(x,y)a (y)u(y)dy. − y 1 1 2 Z0 Z0 Since G(x, 1) = G(x, 0) = 0 then this equation can be rewritten as 1 u(x)= ϕ (x) K(x,y)u(y)dy, 0 − Z0 where f 1 ϕ (x)= ϕ (x) G(x,y)f(y)dy 0 0 − Z0 and f K(x,y)= ∂ G(x,y)a (y)+ G(x,y)a′ (y) G(x,y)a (y). y 1 1 − 2

79 Exercise 43. 1) Prove that K(x,y) is a Hilbert-Schmidt kernel on [0, 1] [0, 1]. × 2) Prove that the boundary value problem and this integral equation of the second kind are equivalent.

3) Formulate the solvability condition for the boundary value problem using the Fredholm alternative for this integral operator.

80 9 Volterra and singular integral equations

In this chapter we consider integral equations of special types on the finite interval [a,b]. We consider normed space L∞(a,b), the Lebesgue space, and Cα[a,b], the Hölder space (which are not Hilbert spaces) instead of the Hilbert space L2. The norms of the spaces L∞(a,b) and Cα[a,b] are defined as follows:

f ∞ = inf M : f(x) M a.e. on(a,b) (9.1) k kL (a,b) { | |≤ }

f(x) f(y) f α = f ∞ + sup | − |, (9.2) k kC [a,b] k kL (a,b) x y α x,y∈[a,b] | − | where 0 <α 1. The fact that≤ f belongs to the Hölder space Cα[a,b] is equivalent to the fact that f L∞(a,b) and there is constant c > 0 such that for all h (small enough) ∈ 0 f(x + h) f(x) c h α, | − |≤ 0| | where x,x + h [a,b]. ∈ Definition. Integral equations in L∞(a,b) of the form x f(x)= K(x,y)ϕ(y)dy Za and x ϕ(x)= f(x)+ K(x,y)ϕ(y)dy, (9.3) Za where x [a,b] and supx,y∈[a,b] K(x.y) < , are called Volterra integral equations of the first and∈ second kind, respectively.| | ∞ Theorem 1. For each f L∞(a,b) the Volterra integral equation of the second kind has a unique solution ϕ ∈L∞(a,b) such that ∈ M(x−a) ϕ(x) e f ∞ (9.4) | |≤ k kL (a,b) for any x [a,b] and ∈ M(b−a) ϕ ∞ f ∞ e , (9.5) k kL (a,b) ≤ k kL (a,b) where M = sup K(x,y) . x,y∈[a,b] | | Proof. We introduce the iterations of the equation (9.3) by x ϕj+1(x) := K(x,y)ϕj(y)dy, j =0, 1, 2,... Za with ϕ0 = f. Let us prove by induction that (M(x a))j ϕ (x) − f ∞ , j =0, 1,.... (9.6) | j |≤ j! k kL (a,b)

81 Indeed, this estimate clearly holds for j = 0. Assume that (9.6) is proved for some j 0. Then ≥ x x (M(y a))j ϕj+1(x) K(x,y) ϕj(y) dy M − f L∞(a,b) dy | |≤ a | || | ≤ a j! k k Z x jZ j+1 j+1 (y a) j+1 (x a) = M f ∞ − dy = M f ∞ − . k kL (a,b) j! k kL (a,b) (j + 1)! Za This proves (9.6). Let us introduce the function

∞

ϕ(x) := ϕj(x). (9.7) j=0 X Then, from (9.6) we obtain for all x [a,b] that ∈ ∞ j (M(x a)) M(x−a) ϕ(x) f ∞ − = f ∞ e . | | ≤ k kL (a,b) j! k kL (a,b) j=0 X Thus, the function ϕ(x) is well-deﬁned by the series (9.7) since this series is uniformly convergent with respect to x [a,b]. It remains now to show that∈ this ϕ(x) solves (9.3). Since the series (9.7) converges uniformly we may integrate it term by term and obtain

x ∞ x ∞ K(x,y)ϕ(y)dy = K(x,y)ϕj(y)dy = ϕj+1(x) Za j=0 Za j=0 X∞ X = ϕ (x)+ ϕ (x) f(x)= ϕ(x) f(x). j 0 − − j=1 X So (9.3) holds with this ϕ. The estimate (9.5) then follows from (9.4) immediately. Finally, the uniqueness of this solution follows from (9.5) too.

Corollary. The homogeneous equation

x ϕ(x)= K(x,y)ϕ(y)dy Za has only the trivial solution in L∞(a,b).

Proof. Follows from (9.5). In general, integral equations of the ﬁrst kind are more delicate with respect to the solvability than equations of the second kind. However, in some cases Volterra integral

82 equations of the ﬁrst kind can be treated by reducing them to equations of the second kind. Indeed, consider for x [a,b] ∈ x K(x,y)ϕ(y)dy = f(x) (9.8) Za ∂K ′ and assume that the derivatives ∂x (x,y) and f (x) exist and are bounded and that K(x,x) = 0 for all x [a,b]. Then, diﬀerentiating with respect to x reduces (9.8) to 6 ∈ x ∂K ϕ(x)K(x,x)+ (x,y)ϕ(y)dy = f ′(x) ∂x Za or f ′(x) x ∂K (x,y) ϕ(x)= ∂x ϕ(y)dy. (9.9) K(x,x) − K(x,x) Za Exercise 44. Show that (9.8) and (9.9) are equivalent if f(a)=0.

∂K The second possibility occurs if we assume that ∂y (x,y) exists and is bounded and that K(x,x) = 0 for all x [a,b]. In this case, setting 6 ∈ x ψ(x) := ϕ(y)dy, ψ′ = ϕ Za and performing integration by parts in (9.8) yields x x ∂K f(x)= K(x,y)ψ′(y)dy = K(x,y)ψ(y) x (x,y)ψ(y)dy |a − ∂y Za Za x ∂K = K(x,x)ψ(x) (x,y)ψ(y)dy − ∂y Za or f(x) x ∂K (x,y) ψ(x)= + ∂y ψ(y)dy. K(x,x) K(x,x) Za There is an interesting generalization of equation (9.3) when the kernel has weak singularities. More precisely, we consider (9.3) in the space L∞(a,b) and assume that the kernel K(x,y) satisﬁes the estimate

K(x,y) M x y −α, x,y [a,b],x = y | |≤ | − | ∈ 6 with some 0 <α< 1. If we consider again the iterations

x ϕj(x) := K(x,y)ϕj−1(y)dy, j =1, 2,... Za with ϕ = f, then it can be proved by induction that for all x [a,b] we have 0 ∈ M(x a)1−α j ϕ (x) − f ∞ , j =0, 1,.... (9.10) | j |≤ 1 α k kL (a,b) − 83 Indeed, since this clearly holds for j = 0 assume that it is proved for some j 0. Then ≥ x j x M −α 1−α j ϕj+1(x) K(x,y) ϕj(y) dy M j x y ((y a) ) f L∞(a,b) dy | |≤ a | || | ≤ (1 α) a | − | − k k Z j+1 − x Z M 1−α j −α j ((x a) ) f L∞(a,b) (x y) dy ≤ (1 α) − k k a − −j+1 Z 1−α M 1−α j (x a) ((x a) ) f ∞ − ≤ (1 α)j − k kL (a,b) 1 α − − M(x a)1−α j+1 = − f ∞ . 1 α k kL (a,b) − If we assume now that M(b a)1−α − < 1 1 α − then the series ∞

ϕj(x) j=0 X converges uniformly on the interval [a,b] and the function ϕ deﬁned by

∞

ϕ(x) := ϕj(x) j=0 X solves therefore the nonhomogeneous integral equation (9.3). Moreover, the following estimates hold f ∞ ϕ(x) k kL (a,b) , x [a,b] | |≤ M(x a)1−α ∈ 1 − − 1 α − and f L∞(a,b) ϕ ∞ k k . k kL (a,b) ≤ M(b a)1−α 1 − − 1 α − Exercise 45. Show that the Volterra integral equation of the ﬁrst kind

x ϕ(x)= λ e−(x−y)ϕ(y)dy Z0 has, for any λ, only the trivial solution in L∞(a,b).

Definition. Let 0 <α< 1,ϕ Cα[ a,a] and periodic, i.e. ϕ( a) = ϕ(a). Integral equation in this space of the form∈ − − a ϕ(x + y)dy ϕ(x)= f(x)+ λp.v. , λ C (9.11) y ∈ Z−a 84 is understood in the sense that a ϕ(x + y)dy ϕ(x + y)dy p.v. = lim (9.12) y ε→0 y Z−a Z|y|≥ε,y∈[−a,a] and the function ϕ is extended periodically (with period 2a) on the whole line. Due to (9.12) we have that a dy p.v. =0. y Z−a Thus a ϕ(x + y)dy a ϕ(x + y) ϕ(x) a ϕ(x + y) ϕ(x) p.v. = p.v. − dy = − dy y y y Z−a Z−a Z−a and the latter integral can be understood in the usual sense for periodic ϕ Cα[ a,a] since ∈ − a ϕ(x + y) ϕ(x) a y α a aα − dy c | | dy =2c ξα−1dξ =2c . (9.13) y ≤ 0 y 0 0 α Z−a Z−a Z0 | | α The inequality (9.13) shows us that for any ϕ C [ a,a] the integral in (9.11) is uniformly bounded and also periodic with period∈ 2a. But− even more is true. Proposition 1. For any 2a-periodic ϕ Cα[ a,a] with 0 <α < 1 the integral in (9.11) defines a 2a-periodic function of∈x which− belongs to the same Hölder space Cα[ a,a]. − Proof. Let us denote by g(x) the integral in (9.11). For h > 0 small enough we have | | a ϕ(x + h + y) ϕ(x + h) a ϕ(x + y) ϕ(x) g(x + h) g(x)= − dy − dy − y − y Z−a Z−a ϕ(x + h + y) ϕ(x + h) ϕ(x + y) ϕ(x) = − dy − dy y − y Z|y|≤3|h| Z|y|≤3|h| ϕ(x + h + y) ϕ(x) ϕ(x + y) ϕ(x) + − dy − dy y − y Z|y|≥3|h| Z|y|≥3|h| := I1 + I2.

For the ﬁrst integral I1 we have ϕ(x + y + h) ϕ(x + h) ϕ(x + y) ϕ(x) I | − |dy + | − |dy | 1|≤ y y Z|y|≤3|h| | | Z|y|≤3|h| | | y α y α c | | dy + c | | dy ≤ 0 y 0 y Z|y|≤3|h| | | Z|y|≤3|h| | | 3|h| (3 h )α 4c 3α 4c ξα−1dξ =4c | | = 0 h α. (9.14) ≤ 0 0 α α | | Z0

85 For the estimation of I2 we ﬁrst rewrite it as (we change variables in the ﬁrst integral) ϕ(z + x) ϕ(x) ϕ(z + x) ϕ(x) I = − dz − dz 2 z h − z Z|z−h|≥3|h| − Z|z|≥3|h| 1 1 = (ϕ(z + x) ϕ(x)) dz − z h − z Z|z|≥3|h| − ϕ(z + x) ϕ(x) − dz. − z h Z{|z−h|≥3|h|}\{|z|≥3|h|} − Then we have h dz ϕ(z + x) ϕ(x) I ϕ(z + x) ϕ(x) | | + | − |dz | 2|≤ | − | z z h z h Z|z|≥3|h|,z∈[−a,a] | |·| − | Z2|h|≤|z|≤3|h| | − | z α z α c h | | dz + c | | dz ≤ 0| | z 2 z /3 0 z /2 Za≥|z|≥3|h| | |· | | Z2|h|≤|z|≤3|h| | | 3c a 3|h| =2 0 h ξα−2dξ +4c ξα−1dξ · 2 | | 0 Z3|h| Z0 ξα−1 a (3 h )α (3 h )α−1 aα−1 4c 3α =3c h +4c | | =3c h | | + 0 h α 0| | α 1 0 α 0| | 1 α − 1 α α | | − 3|h| − − α α 3 c0 α 4c03 α α 1 4 α < h + h = c03 + h , (9.15) 1 α| | α | | 1 α α | | − − since 0 <α< 1. Estimates (9.14)-(9.15) show that this Proposition is completely proved. If we denote by a ϕ(x + y)dy Aϕ(x) := p.v. (9.16) y Z−a a linear operator on periodic Cα[ a,a], 0 <α< 1 then Proposition 1 gives that A is bounded in this space. But this− operator is not compact there. Nevertheless the following holds.

Corollary. There is λ0 > 0 such that for all λ C, λ < λ0 and for any periodic f Cα[ a,a], 0 <α< 1 the integral equation∈ (9.11| )| has a unique solution in Cα[ a,a∈], 0 <α<− 1. − Proof. Since operator A from (9.16) is a bounded linear operator in the space Cα[ a,a] − then

A α α c k kC →C ≤ 0 with some constant c > 0. If we choose now λ = 1/c then for all λ C, λ < λ , 0 0 0 ∈ | | 0 operator I λA will be invertible in the space Cα[ a,a] since − −

λA α α < 1. k kC →C 86 This fact implies that the integral equation (9.11) can be solved uniquely in this space and the unique solution ϕ can be obtained as

ϕ =(I λA)−1f. − It is equivalent to the fact that (9.11) can be solved by iterations.

87 10 Approximate methods

In this chapter we will study approximate solution methods for equations in the Hilbert space H of the form Aϕ = f, (I A)ϕ = f (10.1) − with bounded or compact operator A. The fundamental concept for approximately solving equations (10.1) is to replace them by the equations

A ϕ = f , (I A )ϕ = f , (10.2) n n n − n n n respectively. For practical purposes, the approximating equations (10.2) will be chosen so that they can be reduced to solving a ﬁnite-dimensional linear system. We will start with some general results which are the basis of our considerations. Theorem 1. Let A : H H be a bounded linear operator with bounded inverse A−1. → Assume that the sequence An : H H of bounded linear operators is norm convergent to A i.e. → A A 0, n . k n − k → → ∞ Then for all n such that A−1(A A) < 1 n − −1 the inverse operators An exist and

−1 −1 A An k−1 k . ≤ 1 A (An A) − k − k Moreover, the solutions of ( 10.1) and (10.2) satisfy the error estimate A−1 ϕn ϕ k−1 k (An A)ϕ + fn f . k − k≤ 1 A (An A) k − k k − k − k − k Proof. Since A−1 exists we may write

A−1A = I A−1(A A ). n − − n Since A−1(A A) < 1 n − for n large enough, for these values of n we have that

−1 I A−1(A A ) − − n exists by the Neumann series. Thus,

−1 −1 A−1A = I A−1(A A ) n − − n or −1 A−1A = I A−1(A A ) n − − n 88 or −1 A−1 = I A−1(A A ) A−1. n − − n The error estimate follows immediately from the representa tion ϕ ϕ = A−1(A A )ϕ + A−1(f f). n − n − n n n −

Theorem 2. Assume that A−1 : H H exist for all n n and their norms are n → ≥ 0 uniformly bounded for such n. Let An A 0 as n . Then the inverse operator A−1 exists and k − k → → ∞

−1 −1 An A k−1 k ≤ 1 An (An A) − k − k for all n n with A−1(A A) < 1. ≥ 0 k n n − k Exercise 46. Prove Theorem 2 and obtain the error estimate in this case.

∞ Deﬁnition. A sequence An n=1 of compact operators in the Hilbert space H is called collectively compact if for{ any} bounded set U H the image ⊂ J = A ϕ : ϕ U, n =1, 2,... { n ∈ } is relatively compact i.e. every sequence from J contains a convergent subsequence.

∞ Exercise 47. Assume that a sequence of compact operators An n=1 is collectively compact and converges pointwise to A in H i.e. { }

lim Anϕ = Aϕ, ϕ H. n→∞ ∈ Prove that the limit operator A is compact. Exercise 48. Under the same assumptions for A ∞ as in Exercise 47 prove that { n}n=1 (A A)A 0, (A A)A 0 k n − k → k n − nk → as n . → ∞ Theorem 3. Let A : H H be a compact operator and let I A be injective. Assume → − that the sequence An : H H is collectively compact and pointwise convergent to A. Then for all n such that →

(I A)−1(A A)A < 1 − n − n the inverse operators (I A )−1 exist and the solutions of (10.1) and (10.2) satisfy − n the error estimate −1 1+ (I A) An ϕn ϕ k −−1 k (An A)ϕ + fn f . k − k≤ 1 (I A) (An A)An k − k k − k − k − − k 89 Proof. By the Riesz theorem (see Theorem 5 of Chapter 4) the inverse operator (I A)−1 exists and is bounded. Due to Exercise 48 −

(A A)A 0, n . k n − nk → → ∞ That’s why for n large enough we have

(I A)−1(A A)A < 1. − n − n This fact allows us to conclude (as in Theorem 1) that (I A )−1 exists and − n −1 −1 1+ (I A) An (I An) k −−1 k . − ≤ 1 (I A) (An A)An − k − − k The error esimate follows from this inequality and the representation

ϕ ϕ =(I A )−1 (A A)ϕ + f f . n − − n n − n −

−1 Corollary. Let An be as in Theorem 3. Assume that the inverse operators (I An) exist and are uniformly bounded for all n n . Then the inverse (I A)−1 exists− if ≥ 0 − (I A )−1(A A)A < 1. − n n −

The solutions of (10.1) and (10.2) satisfy the error estimate

−1 1+ (I An) A ϕn ϕ k −−1 k (An A)ϕ + fn f . k − k≤ 1 (I An) (An A)A k − k k − k − k − − k Theorem 4. Let A : H H be a bounded linear operator with A < 1. Then the successive approximations→ k k

ϕn+1 := Aϕn + f, n =0, 1,... (10.3) converge for each f H and each ϕ H to the unique solution of (10.1). ∈ 0 ∈ Proof. The condition A < 1 implies the existence and boundedness of the inverse operator (I A)−1 andk thek existence of the unique solution of (10.1) as − ϕ =(I A)−1f. − It remains only to show that the successive approximations converge to ϕ for any ϕ H. The deﬁnition (10.3) implies 0 ∈ ϕ ϕ A ϕ ϕ A n ϕ ϕ . k n+1 − nk ≤ k k k n − n−1k≤···≤k k k 1 − 0k

90 Hence for each p N we have ∈ ϕ ϕ ϕ ϕ + + ϕ ϕ k n+p − nk ≤ k n+p − n+p−1k ··· k n+1 − nk A n+p−1 + A n+p−2 + + A n ϕ ϕ ≤ k k k k ··· k k k 1 − 0k A n k k ϕ ϕ 0 ≤ 1 A k 1 − 0k → − k k as n uniformly in p N. It means that ϕn is a Cauchy sequence in the Hilbert space→H ∞. Therefore, there∈ exists unique limit{ }

ϕ = lim ϕn. n→∞ Evidently this ϕ solves (10.1) uniquely. We will return to the integral operators

Af(x)= K(x,y)f(y)dy, (10.4) ZΩ where K(x,y) is assumed to be in L2(Ω Ω). In that case, as we know, A is compact in L2(Ω). × Deﬁnition. A function K (x,y) L2(Ω Ω) is said to be a degenerate kernel if n ∈ × n

Kn(x,y)= aj(x)bj(y), (10.5) j=1 X with some functions a ,b L2(Ω). j j ∈ We consider the integral equation of second kind with degenerate kernel Kn(x,y) i.e. n ϕ (x) a (x)b (y)ϕ (y)dy = f(x) (10.6) n − j j n Ω j=1 Z X in the form n ϕ (x) γ a (x)= f(x), n − j j j=1 X where γj = (ϕn, bj)L2(Ω). It means that the solution ϕn of (10.6) is necessarily represented as n

ϕn(x)= f + γjaj (10.7) j=1 X such that the coeﬃcients γj (which are to be determined) satisfy the linear system

n γ γ (a , b ) 2 =(f, b ) 2 = f , j =1, 2,...,n. (10.8) j − k k j L (Ω) j L (Ω) j Xk=1 91 Hence, the solution ϕn of (10.6) (see also (10.7)) can be obtained whenever we can solve the linear system (10.8) uniquely with respect to γj. Let us consider now the integral equation of second kind with compact self-adjoint operator (10.4) i.e. ϕ(x) Aϕ(x)= f(x). (10.9) − The main idea is to approximate the kernel K(x,y) from (10.9) by the degenerate kernel Kn(x,y) from (10.6) such that

K(x,y) K (x,y) 2 0 (10.10) k − n kL (Ω×Ω) → −1 as n and, in addition, the inverse operators (I An) exist and are uniformly bounded→ ∞ in n. − In that case the system (10.8) is uniquely solvable and we obtain an approximate solution ϕn such that ϕ ϕ 2 0, n . k − nkL (Ω) → → ∞ Indeed, equations (10.6) and (10.9) imply

(ϕ ϕ ) A (ϕ ϕ )=(A A )ϕ. − n − n − n − n Since (I A )−1 exist and are uniformly bounded we have − n ϕ ϕ (I A )−1 A A ϕ 0 k − nk≤ − n k − nk k k → as n by (10.10). The unique solvability of (10.8) (or the uniqueness of ϕn) follows from→ the ∞ existence of the inverse operators (I A )−1. − n We may justify this choice of the degenerate kernel Kn(x,y) by the following con- ∞ 2 2 siderations. Let ej j=1 be an orthonormal basis in L (Ω). Then K(x,y) L (Ω Ω) as a function of x{ }Ω (with parameter y Ω) can be represented by ∈ × ∈ ∈ ∞ K(x,y)= (K( ,y),e ) 2 e (x). · j L j j=1 X Then n

K(x,y) (K( ,y),ej)L2 ej 0 − · → j=1 L2(Ω×Ω) X as n , and we may consider the degenerate kernel K (x,y) in the form → ∞ n n

Kn(x,y)= ej(x)bj(y), j=1 X where b (y)=(K( ,y),e ) 2 . The system (10.8) transforms in this case to j · j L n γ γ (e , (e ,K( ,y)) 2 ) 2 = f . j − k k j · L L (Ω) j Xk=1 92 If, for example, ej are the normalized eigenfunctions of the operator A with corresponding eigenvalues λj then the latter system can be rewritten as

γ λ γ = f , j =1, 2,...,n. j − j j j We assume that λ = 1 so that γ can be uniquely determined as j 6 j f γ = j j 1 λ − j and therefore ϕn is equal to

n f ϕ(x)= f(x)+ j e (x). 1 λ j j=1 j X − A diﬀerent method goes back to Nystr¨om. Let us consider instead of integral operator A with kernel K(x,y) the sequence of numerical integration operators

n (n) (n) (n) Anϕ(x)= αj K(x,xj )ϕ(xj ). (10.11) j=1 X (n) (n) We assume that the points xj and the weights αj are chosen so that

n 2 2 1 (n) (n) (n) Aϕ Anϕ L2 = K(x,y)ϕ(y) αj K(x,xj )ϕ(xj )dy dx 0 k − k Ω Ω − Ω → Z Z | | j=1 X (n) (n) as n . The main problem here is to choose the weights αj and the points xj with→ this ∞ approximation property. The original Nystr¨om method was constructed for continuous kernels K(x,y). In Hilbert spaces it is more natural to consider projection methods.

Deﬁnition. Let A : H H be injective bounded linear operator. Let P : H H → n → n be projection operators such that dim H = n. For given f H the projection method n ∈ generated by Hn and Pn approximates the equation Aϕ = f by the projection equation

P Aϕ = P f, ϕ H. (10.12) n n n n ∈

This projection method is called convergent if there is n0 N such that for each f H the approximating equation (10.12) has a unique solution∈ϕ H for all n n ∈and n ∈ n ≥ 0 ϕ ϕ, n , n → → ∞ where ϕ is the unique solution of the equation Aϕ = f.

93 Theorem 5. A projection method converges if and only if there exist n0 N and M > 0 such that for all n n the operators ∈ ≥ 0 P A : H H n → are invertible and the operators (P A)−1P A : H H are uniformly bounded i.e. n n → (P A)−1P A M, n n . n n ≤ ≥ 0 In case of convergence we have the error estimate

ϕn ϕ (1 + M) inf ψ ϕ . k − k≤ ψ∈Hn k − k

Proof. If a projection method converges then, by deﬁnition, PnA are invertible and the uniform boundedness follows from the Banach–Steinhaus theorem. Conversely, under the assumptions of the theorem ϕ ϕ = ((P A)−1P A I)ϕ. n − n n − Since for all ψ H we have trivially (P A)−1P Aψ = ψ then ∈ n n n ϕ ϕ = ((P A)−1P A I)(ϕ ψ) n − n n − − and the error esimate follows. Remark. Projection methods make sense and we can expect convergence only if the subspaces Hn possess the denseness property inf ψ ϕ 0, n . ψ∈Hn k − k → → ∞ Theorem 6. Assume that A : H H is compact, I A is injective and the projection → − operators Pn : H Hn converge pointwise i.e. Pnϕ ϕ,n for each ϕ H. Then the projection→ method for I A converges. → → ∞ ∈ − Proof. By the Riesz theorem (see Theorem 5 of Chapter 4) operator I A has bounded inverse. Since P ϕ ϕ as n we have P Aϕ Aϕ as n too.− At the same n → → ∞ n → → ∞ time the sequence PnA is collectively compact since A is compact and Pn is of ﬁnite rank. Thus, due to Exercise 48 we have (P A A)P A 0, n . (10.13) k n − n k → → ∞ Then the operators (I P A)−1 exist and are uniformly bounded. Indeed, denoting − n B := I +(I A)−1P A n − n we obtain B (I P A)=(I P A)+(I A)−1P A(I P A) n − n − n − n − n = I (I A)−1(P A A)P A − − n − n = I S . − n

94 But it is easy to see from (10.13) that S 0, n . k nk → → ∞ −1 Hence, both I PnA and Bn are injective. Since PnA is compact then (I PnA) is bounded. As a− consequence of this fact we have that − (I P A)−1 =(I S )−1B . − n − n n Deﬁnition of Bn implies B 1+ (I A)−1 A . k nk≤ − k k Therefore (I P A)−1 is uniformly bounded in n . The exact equation ϕ Aϕ = f n and (10.12k) with− operatork I A lead to − − (I P A)(ϕ ϕ)= P Aϕ Aϕ + P f f, − n n − n − n − which implies also the error estimate

ϕ ϕ (I P A)−1 P Aϕ Aϕ + P f f . k n − k≤ − n k n − k k n − k

Corollary. Under the assumptions of Theorem 6 and provided additionally that P A A 0, n k n − k → → ∞ the approximate equation (10.12) with I A is uniquely solvable for each f H and we have the error estimate − ∈ ϕ ϕ M P ϕ ϕ , k n − k≤ k n − k where M is an upper bound for the norm (I P A)−1 . k − n k −1 Proof. The existence of the inverse operators (I PnA) and their uniform boundedness follows from − I P A =(I A) (P A A)=(I A) I (I A)−1(P A A) , − n − − n − − − − n − −1 (I P A)−1 = I (I A)−1(P A A) (I A)−1 − n − − n − − and −1 −1 (I A) (I PnA) k −−1 k . − ≤ 1 (I A) (PnA A) − k − k k − k The error estimate is the consequence of (ϕ ϕ ) P A(ϕ ϕ )= ϕ P ϕ − n − n − n − n and ϕ ϕ (I P A)−1 P ϕ ϕ . k n − k≤ − n k n − k

95 Let us return back to the projection equation (10.12). It can be rewritten equivalently as (Aϕ f,g)=0, g H . (10.14) n − ∈ n Indeed, if g H then g = P g, P ∗ = P and hence ∈ n n n n 0=(Aϕ f,g)=(Aϕ f, P g)=(P (Aϕ f),g) n − n − n n n − or P (Aϕ f)=0, n n − since Hn is considered here as a Hilbert space. This is the basis for the Galerkin projection method. Assume that e ∞ is an orthonormal basis in the Hilbert space H. Considering { j}j=1

Hn = span(e1,...,en) we have for the solution ϕn of the projection equation (10.12) the representation

ϕn(x)= γjej. (10.15) j=1 X

The task here is to ﬁnd (if possible uniquely) the coeﬃcients γj such that ϕn from (10.15) solves (10.12). Since (10.14) is equivalent to (10.12) we have from (10.15) that

(Aϕ ,g)=(f,g), g H n ∈ n or (Aϕn,ek)=(f,ek)= fk, k =1, 2,...,n or n

γj(Aej,ek)= fk j=1 X or M~γ = f,~ (10.16) where ~γ = (γ1,...,γn), f~ = (f1,...,fn) and M = ajk n×n with ajk = (Aej,ek). If operator A is invertible then the matrix M is invertible{ } too and ~γ can be obtained uniquely as ~γ = M −1f.~

As a result of this consideration we obtain ϕn(x) uniquely from (10.15). It remains only to check that this ϕn converges to the solution of the exact equation Aϕ = f. In order to verify this fact it is enough to apply Theorem 5. We apply now this projection method to the equation (10.9) with compact operator A.

96 Theorem 7. Let A : H H be compact and let I A be injective. Then the Galerkin projection method converges.→ −

Proof. By the Riesz theorem, the operator I A has bounded inverse. That’s why ϕ − n from (10.15) is uniquely deﬁned with γj that satisﬁes the equation (10.16) with matrix M = a ,a = ((I A)e ,e ). Since { jk}n×n jk − j k ∞ P ϕ ϕ 2 = (ϕ,e ) 2 0, n k n − kH | j | → → ∞ j=n+1 X then we may apply Theorem 6 and conclude this theorem.

97 Index adjoint operator, 16 graph, 15 Green’s function, 68 basis, 8 Bessel’s inequality, 2 Hilbert space, 3 bounded, 12 Hilbert-Schmidt norm, 13

Cauchy sequence, 3 idempotent, 25 Cauchy-Schwarz-Bunjakovskii inequality, 2 induced by the inner product, 3 Cayley transform, 36 injective operator, 50 closable, 16 inner product, 1 closed, 15 inner product space, 1 closed subspace, 6 integral equations of second and first kind, closure, 16 79 collectively compact, 89 integral operator, 71 compact operator, 44 isometry, 28 complete space, 3 completeness relation, 9 kernel, 12, 71 constant of ellipticity, 59 Lebesgue space, 5 continuous spectrum, 41 length, 2 convergent sequence, 3 linear operator, 12 criterion for closedness, 15 linear space, 1 degenerate kernel, 91 linear span, 8 densely defined, 12 multi-index, 58 denseness property, 94 discrete spectrum, 41 neighborhood, 3 domain, 12 non-negative operator, 27 norm, 3 elliptic partial differential operator, 58 norm topology, 3 ellipticity condition, 59 normal operator, 23 essential spectrum, 41 nullspace, 12 essentially self-adjoint, 19 extension, 16 open ball, 3 operator with weak singularity, 71 finite rank operator, 44 orthogonal, 2 formally self-adjoint, 58 orthogonal complement, 6 Fourier expansion, 9 orthonormal, 2 Fredholm alternative II, 76 orthonormal basis, 9 Friedrichs extension, 56 parallelogram law, 3 G˚arding’s inequality, 62 Parseval equality, 9 generalized Leibniz formula, 59 point spectrum, 41

98 polarization identity, 3 positive operator, 27 precompact set, 44 principal symbol, 58 projection method, 93 Projection theorem, 6 projector, 25 quadratic form, 54 range, 12 relatively compact, 89 resolvent, 37 resolvent identity, 38 resolvent set, 38 restriction, 16 Riesz-Frechet theorem, 7 scalar product, 1 self-adjoint, 19 semibounded from below, 54 separable, 9 sequence space, 4 Sobolev space, 5 spectral family, 28 spectral function, 67 spectrum, 38 successive approximations, 90 surjective operator, 50 symmetric, 19

Theorem of Pythagoras, 2 triangle inequality, 2 uniformly elliptic operator, 59 unitary operator, 28 vector space, 1 Volterra integral equations of the ﬁrst and second kind, 81