SMA 5878 Functional Analysis II › pessoas › ... › Aulas › Aula05-english.pdf · Spectral AnalysisofLinearOperators SMA 5878 Functional Analysis II Alexandre Nolasco de Carvalho

Spectral Analysis of Linear Operators

SMA 5878 Functional Analysis II

Alexandre Nolasco de Carvalho

Departamento de MatematicaInstituto de Ciencias Matematicas and de Computacao

Universidade de Sao Paulo

March 25, 2019

Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II

Spectral Analysis of Linear OperatorsSymmetric and self-adjoint operators

Min-Max Characterisation of Eigenvalues

Symmetric and self-adjoint operators

Let H be a Hilbert space with inner product 〈·, ·〉H : H × H → K

and A : D(A) ⊂ H → H be a densely defined operator. Theadjoint operator A• of A is defined by

D(A•) = {u ∈ H : v 7→ 〈Av , u〉H : D(A) → K is bounded}

and if u ∈ D(A•), A•u is the only element of H such that

〈v ,A•u〉H = 〈Av , u〉H ,∀v ∈ D(A).




RemarkIf H is a Hilbert space over C, E : H → H∗ defined byEu(v) = 〈v , u〉H , is a conjugated isometry between H and H∗.The identification between H and H∗ consists in identifying u withEu. If A∗ : D(A∗) ⊂ X ∗ → X ∗ is the dual of A, thenA• = E−1 ◦ A∗ ◦ E.




RemarkAlso note that, even though E and E−1 be linear conjugatedoperators, E−1 ◦ A∗ ◦ E is a linear operator by double conjugation.We cal both A• and A∗ ajoints A and we denote both by A∗ but itis important to note that, if A = αB then A• = αB• while thatA∗ = αB∗. From this, (λI−A)•= λI−A• while (λI−A)∗=λI ∗−A∗.




If no confusion may arise we will use the notation A∗ to denote thedual and the adjoint operators, indistinctively. In that case we mayalso refer to both as the adjoint operator.




DefinitionLet H be a Hilbert space over K with inner product 〈·, ·〉. We saythat an operator A : D(A) ⊂ H → H is symmetric (also calledHermitian when K = C) if D(A) = H and A ⊂ A•; that is,〈Ax , y〉 = 〈x ,Ay〉 for all x , y ∈ D(A). We say that A is self-adjointif A = A•.




ExerciseLet H is a Hilbert space. If A : D(A) ⊂ H → H is a denselydefined operator, then A• : D(A•) ⊂ H → H is closed. Besidesthat, if A is closed, then A• is densely defined.

ExerciseLet H be a Hilbert space over K. Show that, ifA : D(A) ⊂ H → H is symmetric and λ ∈ K is an eigenvalue of A,then λ ∈ R. Besides that,

inf‖x‖H=1

〈Ax , x〉 ≤ λ ≤ sup‖x‖H=1

〈Ax , x〉.




ExerciseLet H = C

n with the usual inner product. If A = (ai , j)ni , j=1 is a

matrix with complex coefficients that represents a linear operatorA ∈ L(H), find A• and A∗.

ExerciseLet H be a Hilbert space over K with inner product 〈·, ·〉 andA : D(A) ⊂ H → H is a densely defined operator. Show thatG (A•) = {(−Ax , x) : x ∈ D(A)}⊥ (here M⊥ represents theortogonal M).




Proposition

Let H be a Hilbert space over K with inner product 〈·, ·〉. IfA : D(A) ⊂ H → H is a self-adjoint operator, which is injectiveand with dense image, then A−1 is self-ajoint.




Proof: Since A is self-adjoint, it is easy to see that

{(x ,−Ax) : x ∈ D(A)}⊥ = {(Ax , x) : x ∈ D(A)} = G (A−1).

Since A is injective and has dense image, it follows from theprevious exercise,

G ((A−1)•) = {(−A−1x , x) : x ∈ R(A)}⊥ = G (A−1).

Hence A−1 = (A−1)•.




TheoremLet H be a Hilbert space over K with inner product 〈·, ·〉. IfA : D(A) ⊂ H → H is a symmetric and surjective linear operator,then A is self-adjoint.

Proof: First we show that A and A∗ are both injective. Ifx ∈ D(A) and Ax = 0, we have that 〈Ax , y〉 = 〈x ,Ay〉 for ally ∈ D(A). Consequently, from the fact that R(A) = X , we havethat x = 0. To see that A∗ is injective we proceed the same way.




Now we show that A is closed. In fact, ifD(A∗) ⊃ D(A) ∋ xn → x ∈ X and Axn = A∗xn → y , thenx ∈ D(A∗) and A∗x = y . Since A is surjective, there existsw ∈ D(A) such that Aw = A∗w = A∗x and from the injectivity ofA∗ we have that w = x . With this x ∈ D(A) and Ax = y , showingthat A is closed.

It follows from the Closed Graph Theorem that a A has inverseA−1 ∈ L(X ). Clearly A−1 is self-adjoint and, from a previousresult, it follows that A is self-adjoint.




Our next theorem and the previous theorem are the main tools toshow that a linear operator is symmetric.

Theorem (Friedrichs)

Let X be a Hilbert space over K and A : D(A) ⊂ X → X asymmetric operator for which there exists a α ∈ R such that

〈Ax , x〉 ≤ α‖x‖2 or 〈Ax , x〉 ≥ α‖x‖2 (1)

for all x ∈ D(A). Then A admits a self-adjoint extension thatpreserves the bound (1).




Proof: We prove only the case 〈Ax , x〉 ≥ α‖x‖2 for all x ∈ D(A)and for some α ∈ R. The remaining case can be deduced from thisconsidering the operator −A.

Also, we only consider the case α = 1 for the general case can bededuced from this considering the operator A+ (1− α)I .




In D(A) consider the inner productD(A)× D(A) ∋ (x , y) 7→ 〈Ax , y〉 ∈ K. Clearly the norm

D(A) ∋ x 7→ ‖x‖ 12= 〈Ax , x〉

12 ∈ R

+ from this inner product

satisfies ‖x‖ 12≥ ‖x‖. Denote by X

12 the completion of D(A)

relatively to the norm ‖ · ‖ 12.




Let us show that X12 , as a set, is in a bijective correspondance with

a subset of the completion of D(A) relatively to the norm ‖ · ‖. Itis clear that all sequences {xn} in D(A) that are Cauchy relativelyto the norm ‖ · ‖ 1

2is also Cauchy relatively to the norm ‖ · ‖.




To conclude the injectivity we show, by reduction to absurd, that if{xn} is a Cauchy sequence relatively to the norm ‖ · ‖ 1

2and for

which limn→∞ ‖xn‖ 12= a > 0, we cannot have that

limn→∞ ‖xn‖ = 0.

If the thesis is false, we have that

2Re〈Axn, xm〉 = 〈Axn, xn〉+ 〈Axm, xm〉 − 〈A(xn − xm), (xn − xm)〉m,n→∞−→ 2a2

which is an absurd since 〈Axn, xm〉m→∞−→ 0.




Since X is complete, X12 can be identified with a subset of X .

Let D = D(A∗) ∩ X12 . Since D(A) ⊂ D(A∗), we must have that

D(A) ⊂ D ⊂ D(A∗). We define A taking the restriction of A∗ to Dand showing that A is self-adjoint.




Let us first show that A is symmetric. If x , y ∈ D there aresequences {xn} and {yn} in D(A) such that ‖xn − x‖ 1

2

n→∞−→ 0

‖yn − y‖ 12

n→∞−→ 0.

It follows that

limm→∞

limn→∞

〈Axn, ym〉 = limn→∞

limm→∞

〈Axn, ym〉 = 〈x , y〉 12

exists and coincides with

limn→∞

limm→∞

〈Axn, ym〉 = limn→∞

〈Axn, y〉 = limn→∞

〈xn, Ay〉 = 〈x , Ay〉 and with

limm→∞

limn→∞

〈Axn, ym〉 = limm→∞

〈x ,Aym〉 = limm→∞

〈Ax , ym〉 = 〈Ax , y〉.

Hence A is symmetric.




To conclude the proof it is enough to show that A is surjective andthis is done in the following way.

Let y ∈ X and consider the linear functional f : D(A) → K givenby f (x) = 〈x , y〉.

Then f is a bounded linear functional with respect to the norm‖ · ‖ 1

2and can be extended to a bounded linear functional defined

in X12 . From Riesz Representation Theorem, there exists y ′ ∈ X

12

such that

f (x) = 〈x , y〉 = 〈x , y ′〉 12= 〈Ax , y ′〉, ∀x ∈ D(A).




Hence y ′ ∈ D(A∗) ∩ X12 and A∗y ′ = Ay ′ = y showing that A is

surjective. From here we also conclude that (−∞, 1) ⊂ ρ(A).

Since A is symmetric and surjective, it follows that A is self-adjointand is an extension of A. It is easy to see that 〈Ax , x〉 ≥ ‖x‖2 forall x ∈ D(A). This concludes the proof.




Example

Let X = L2(0, π) and D(A0) = C 20 (0, π) the set of functions which

are twice continuously differentiable functions and have compactsupport in (0, π). Define A0 : D(A0) ⊂ X → X by

(A0φ)(x) = −φ′′(x), x ∈ (0, π).




It is easy to see that A0 is symmetric and that 〈A0φ, φ〉 ≥2π2 ‖φ‖

2X

for all φ ∈ D(A0).

From Theorem 2, A0 has a self-adjoint extension A that satisfies〈Aφ, φ〉 ≥ 2

π2 ‖φ‖2X for all φ ∈ D(A).

Note that, the space X12 from Friedrichs theorem is, in this

example the closure of D(A) in the norm H1(0, π) and therefore

X12 = H1

0 (0, π).




On the other hand D(A∗) is characterised by

D(A∗0) = {φ ∈ X : ∃φ∗ ∈ X such that 〈−u′′, φ〉 = 〈u, φ∗〉, ∀u ∈ D(A0)}

and A∗0φ=−φ′′ for all φ∈D(A∗

0).

Hence, D(A)=H2(0, π)∩H10 (0, π) and Au=−u′′ for all u∈D(A).




Also from Theorem 2 we know that (−∞, 2π2 ) ⊂ ρ(A). In

particular 0 ∈ ρ(A) and if φ ∈ D(A), we have that

|φ(x)− φ(y)| ≤ |x − y |12‖φ′‖L2(0,π) = |x − y |

12 〈Aφ, φ〉

12 .

Hence, if B is a bounded subset of D(A) with the graph norm,then supφ∈B ‖φ′‖L2(0,π) < ∞ and the family B of functions isequicontinuous and bounded in C ([0, π],R) with the uniformconvergence topology.




It follows from the Arzela-Ascoli Theorem that B is relativelycompact in C ([0, π],R) and consequently B is relatively compactin L2(0, π).

From a previous exercise it follows that A−1 is a compact operator.

It follows that σ(A) = {λ1, λ2, λ3, · · · } where λn = n2 ∈ σp(A)

with eigenfunctions φn(x) =(

2π

)12 sen(nx), n ∈ N.





In this section we introduce min-max characterisations ofeigenvalues of compact and self-adjoint operators. To presentthese characterisations we will employ the following result




LemmaLet H be a Hilbert space over K and A ∈ L(H) be a self-adjointoperator, then

‖A‖L(H) = sup‖u‖=1‖v‖=1

|〈Au, v〉| = sup‖u‖=1

|〈Au, u〉|.




Proof: It is enough to prove that

‖A‖L(H) = sup‖u‖=1‖v‖=1

|〈Au, v〉| ≤ sup‖u‖=1

|〈Au, u〉| := a.

If u, v ′ ∈ H, ‖u‖ = ‖v ′‖ = 1, |〈Au, v ′〉| e iα = 〈Au, v ′〉 andv = e−iαv ′, we have that

|〈Au, v ′〉| = 〈Au, v〉 =1

4[〈A(u + v), u + v〉 − 〈A(u − v), u − v〉]

≤a

4[‖u + v‖2 + ‖u − v‖2] ≤ a.

This completes the proof.




ExerciseShow that, if 0 6= A ∈ L(H) is self-adjoint, then A is notquasinilpotent.




TheoremLet H be a Banach space over K and A ∈ K(H) be a self-adjointoperator such that 〈Au, u〉 ≥ 0 for all u ∈ H. Then,

1. λ1 :=sup{〈Au, u〉 :‖u‖=1} is an eigenvalue and exists v1∈H,‖v1‖=1 such that λ1=〈Av1, v1〉. Besides that Av1=λ1v1.

2. Inductively,

λn :=sup{〈Au, u〉 :‖u‖=1 and u⊥vj , 1≤ j≤n−1} ∈σp(A) (1)

and exists vn ∈ H, ‖vn‖ = 1, vn ⊥ vj , 1 ≤ j ≤ n − 1, suchthat λn = 〈Avn, vn〉. Besides that Avn = λnvn.

3. If Vn = {F ⊂ H : F is a vec. subesp. of dimension n de H},

λn = infF∈Vn−1

sup{〈Au, u〉 : ‖u‖ = 1, u ⊥ F}, n ≥ 1 and (2)

λn = supF∈Vn

inf{〈Au, u〉 : ‖u‖ = 1, u ∈ F}, n ≥ 1. (3)




Proof: We consider only the case K = C and λ1 > 0 leaving theremaining cases as exercises for the reader.

1.Let {un} be a sequence in H with ‖un‖=1 and 〈Aun, un〉n→∞−→ λ1.

Taking subsequences if necessary, {un} converges weakly to v1 ∈ Hand {Aun} converges strongly to Av1.

Hence 〈Av1, v1〉 = λ1.




Let us show that the sequence {un} converges strongly.

From the previous lemma we know that 0 < λ1 = ‖A‖L(H) andfrom the fact that {un} converges weakly to v1 we have that0 < ‖v1‖ ≤ 1. Hence,

limn→∞

‖Aun − λ1un‖2 = lim

n→∞‖Aun‖

2 − 2λ1 limn→∞

〈Aun, un〉+ λ21

= ‖Av1‖2 − λ2

1 ≤ 0.




Since {Aun} converges strongly to Av1, {Aun − λ1un} convergesstrongly to zero and λ1 > 0, it follows that {un} converges stronglyto v1, ‖v1‖ = 1 and Av1 = λ1v1. This concludes the proof of 1.




2. The proof of this item follows from 1. simply noting that theorthogonal of Hn−1 = span{v1, · · · , vn−1} is invariant by A andrepeating the procedure for the restriction of A to H⊥

n−1, n ≥ 2.This concludes the proof of 2.




3. We first prove expression (2). If G = span{v1, · · · , vn−1} wehave, from (1), that

λn = sup{〈Au, u〉 : ‖u‖ = 1, u ⊥ G}

≥ infF∈Vn−1

sup{〈Au, u〉 : ‖u‖ = 1, u ⊥ F}.




On the other hand, let F ∈ Vn−1 and w1, · · · ,wn−1 anorthornormal set of F . Choose u =

∑ni=1 αivi such that ‖u‖ = 1

and u ⊥ wj , 1 ≤ j ≤ n − 1. Hence∑n

i=1 |αi |2 = 1 and

〈Au, u〉 =

n∑

i=1

|αi |2λi ≥ λn.




This implies

sup{〈Au, u〉 : ‖u‖ = 1, u ⊥ F} ≥ λn, for all F ∈ Vn−1

and completes the proof of (2).




We now prove (3). If G = span{v1, · · · , vn} and u ∈ G , ‖u‖ = 1,we have that u =

∑ni=1 αivi with

∑ni=1 |αi |

2 = 1 e

〈Au, u〉 =

n∑

i=1

|αi |2λi ≥ λn.

This implies that

supF∈Vn

inf{〈Au, u〉 : ‖u‖ = 1, u ∈ F} ≥ λn.




Conversely, given F ∈ Vn choose u ∈ F , ‖u‖ = 1, such thatu ⊥ vj , 1 ≤ j ≤ n − 1. It follows, from 2., that 〈Au, u〉 ≤ λn andconsequently

inf{〈Au, u〉 : ‖u‖ = 1, u ∈ F} ≤ λn, for all F ∈ Vn.

This completes the proof of (3).




ExerciseIf A : D(A) ⊂ H → H is positive, self-adjoint and (〈Au, u〉 > 0 forall u ∈ D(A)) and has compact resolvent, find the min-maxcharacterisation for the eigenvalues of A.


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SMA 5878 Functional Analysis II › pessoas › ... › Aulas › Aula05-english.pdf · Spectral AnalysisofLinearOperators SMA 5878 Functional Analysis II Alexandre Nolasco de Carvalho