SMA 5878 Functional Analysis II - icmc.usp.br › ... › andcarva › AnaliseFuncional-II › Aulas › Aula06 … · Spectral AnalysisofLinearOperators SMA 5878 Functional Analysis

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 27, 2019

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

Spectral Analysis of Linear OperatorsMin-Max Characterisation of Eigenvalues

Dissipative operators and numerical range

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 Friedrichs Theorem, A0 has a self-adjoint extension A thatsatisfies 〈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 Friedrichs Theorem we know that (−∞, 1π ) ⊂ 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.




Min-Max Characterisation of Eigenvalues

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. subspace of dimension n of 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.





DefinitionLet X be a Banach space over K. The duality map J : X → 2X

∗is

a multivalued function defined by

J(x) = {x∗ ∈ X ∗ : Re〈x , x∗〉 = ‖x‖2, ‖x∗‖ = ‖x‖}.

From the Hanh-Banach Theorem we have that J(x) 6= ∅.

A linear operator A : D(A) ⊂ X → X is dissipative if for eachx ∈ D(A) there exists x∗ ∈ J(x) such that Re 〈Ax , x∗〉 ≤ 0.




ExerciseShow that, if X ∗ is uniformly convex and x ∈ X, then J(x) is aunitary subset of X ∗.




LemmaThe linear operator A is dissipative if and only if

‖(λ− A)x‖ ≥ λ‖x‖ (1)

for all x ∈ D(A) and λ > 0.




Proof: If A is dissipative, λ > 0, x ∈ D(A), x∗ ∈ J(x) andRe〈Ax , x∗〉 ≤ 0,

‖λx − Ax‖‖x‖ ≥ |〈λx − Ax , x∗〉| ≥ Re〈λx − Ax , x∗〉 ≥ λ‖x‖2

and (1) follows. Conversely, given x ∈ D(A) suppose that (1)holds for all λ > 0.




If y∗λ ∈ J((λ− A)x) and g∗λ = y∗λ/‖y

∗λ‖ we have that

λ‖x‖≤‖λx − Ax‖=〈λx−Ax , g∗λ〉=λRe〈x , g∗

λ〉−Re〈Ax , g∗λ〉

≤ λ‖x‖ − Re〈Ax , g∗λ〉

(2)




Since the unit ball of X ∗ is compact in the weak∗-topology wehave that there exists g∗ ∈ X ∗ with ‖g∗‖ ≤ 1 such that g∗ is alimit point of the sequence {g∗

n} [there is a sub-net (see Apendix)of {g∗

n} that converges to g∗].

From (2) it follows that Re〈Ax , g∗〉 ≤ 0 and Re〈x , g∗〉 ≥ ‖x‖. ButRe〈x , g∗〉 ≤ |〈x , g∗〉| ≤ ‖x‖ and therefore Re〈x , g∗〉 = ‖x‖.

Taking x∗ = ‖x‖g∗ we have that x∗ ∈ J(x) and Re〈Ax , x∗〉 ≤ 0.Thus, for all x ∈ D(A) there exists x∗ ∈ J(x) such thatRe〈Ax , x∗〉 ≤ 0 and A e dissipative.




Theorem (G. Lumer)

Suppose that A is a linear operator in a Banach space X . If A isdissipative and R(λ0 − A) = X for some λ0 > 0, then A is closed,ρ(A) ⊃ (0,∞) and

‖λ(λ− A)−1‖L(X ) ≤ 1,∀λ > 0.




Proof: If λ > 0 and x ∈ D(A), do Lemma 2 temos que

‖(λ− A)x‖ ≥ λ‖x‖.

Now R(λ0 − A) = X , ‖(λ0 − A)x‖ ≥ λ0‖x‖ for x ∈ D(A), so λ0 isin ρ(A) and A is closed. Let Λ = ρ(A) ∩ (0,∞).




Λ is an open subset of (0,∞) for ρ(A) is open, let us prove that Λis a closed subset of (0,∞) to conclude that Λ = (0,∞).

Suppose that {λn}∞n=1 ⊂ Λ, λn → λ > 0, if n is sufficiently large

we have that |λn − λ| ≤ λ/3 then, for all n sufficiently large,‖(λ−λn)(λn−A)−1‖≤|λn−λ|λ−1

n ≤1/2 and I+(λ−λn)(λn−A)−1

is in isomorphism of X .

Thenλ− A =

{I + (λ− λn)(λn − A)−1

}(λn − A) (3)

takes D(A) over X and λ ∈ ρ(A), as desired.




Corollary

Let A be a closed and densely defined linear operator. If both Aand A∗ are dissipative, then ρ(A) ⊃ (0,∞) and

‖λ(λ− A)−1‖ ≤ 1,∀λ > 0.




Proof: From Theorem (G. Lummer) it is enough to prove thatR(I − A) = X .

Since A is dissipative and closed, R(I − A) is a closed subspace ofX .

Let x∗ ∈ X ∗ be such that 〈(I − A)x , x∗〉 = 0 for all x ∈ D(A).This implies that x∗ ∈ D(A∗) and (I ∗ − A∗)x∗ = 0.

Since A∗ is also dissipative it follows from the previous lemma thatx∗ = 0. Consequently R(I −A) is dense in X and since R(I −A) isclosed, R(I − A) = X .




In several examples, the technique used to obtain estimates for theresolvent of a given operator and the localisation of its spectrum isthe localisation of the numerical range (defined next).




If A is a linear operator in a complex Banach space X its numericalrange W (A) is the set

W (A) :={〈Ax , x∗〉 :x ∈D(A), x∗∈X ∗, ‖x‖=‖x∗‖= 〈x , x∗〉=1}. (4)

When X is a Hilbert space

W (A) = {〈Ax , x〉 : x ∈ D(A), ‖x‖ = 1}.




Theorem (Numerical Range)

Let A : D(A) ⊂ X → X be a closed densely defined operator andW (A) be the numerical range of A.

1. If λ /∈ W (A) then λ− A is injective, has closed image andsatisfies

‖(λ− A)x‖ ≥ d(λ,W (A))‖x‖. (5)

where d(λ,W (A)) is the distance of λ to W (A). Besidesthat, if λ ∈ ρ(A),

‖(λ− A)−1‖L(X ) ≤1

d(λ,W (A)). (6)

2. If Σ is open and connected in C\W (A) and ρ(A) ∩Σ 6= ∅,then ρ(A) ⊃ Σ and (6) is satisfied for all λ ∈ Σ.




Proof: Let λ /∈ W (A). If x ∈ D(A), ‖x‖ = 1, x∗ ∈ X ∗, ‖x∗‖ = 1and 〈x , x∗〉 = 1 then,

0<d(λ,W (A))≤|λ−〈Ax , x∗〉|= |〈λx −Ax , x∗〉|≤‖λx −Ax‖ (7)

and therefore λ− A is one-to-one, has closed image and satisfies(5). If, besides that, λ ∈ ρ(A) then, (7) implies (6).




It remains to show that, if Σ intersects ρ(A) then, ρ(A) ⊃ Σ. Tothat end consider the nonempty set ρ(A) ∩Σ.

This set is clearly open in Σ.

But it is also closed since, if λn ∈ ρ(A) ∩ Σ and λn → λ ∈ Σ then,for sufficiently large n, |λ− λn| < d(λn,W (A)).

From this and (6) it follows that |λ− λn| ‖(λn − A)−1‖ < 1, for nsufficiently large. Consequently,λ∈ρ(A) and ρ(A)∩Σ is closed in Σ.

It follows that ρ(A) ∩Σ = Σ that is ρ(A) ⊃ Σ, as desired.




Example

Let H be a Hilbert space over K and A : D(A) ⊂ H → H be aself-adjoint operator. It follows that A is closed and denselydefined. If A is bounded above; that is, 〈Au, u〉 ≤ a〈u, u〉 for somea ∈ R, then C\(−∞, a] ⊂ ρ(A), and

‖(A − λ)−1‖L(X ) ≤M

|λ− a|,

for some constant M ≥ 1, depending only on ϕ, for allλ ∈ Σa,ϕ = {λ ∈ C : |arg(λ− a)| < ϕ}, ϕ < π.




Proof: We start localising the numerical range of A. First notethat

W (A) = {〈Ax , x〉 : x ∈ D(A), ‖x‖ = 1} ⊂ (−∞, a].

Also, A− a = A∗ − a is dissipative and therefore, from a previousresult, ρ(A− a) ⊃ (0,∞).




From Theorem (Numerical Range) we have thatC\(−∞, a] ⊂ ρ(A) and that

‖(λ− A)−1‖ ≤1

d(λ,W (A))≤

1

d(λ, (−∞, a]).

Besides that, if λ ∈ Σa,ϕ, we have that

1

d(λ, (−∞, a])≤

1

sinϕ

1

|λ− a|

and the result follows.




ExerciseLet X be a Banach space such that X ∗ is strictly convex andA : D(A) ⊂ X → X be a closed, densely defined and dissipativelinear operator. Se R(I − A) = X, show thatρ(A) ⊃ {λ ∈ C : Reλ > 0} and that

‖(λ− A)−1‖L(X ) ≤1

Reλ, for all λ ∈ Σ0,π

2.

Is the hypothesis that X ∗ be strictly convex necessary?




Proposition

Let H be a Hilbert space over K with inner product 〈·, ·〉 andA ∈ L(H) be a self-adjoint operator. If

m = infu∈H‖u‖=1

〈Au, u〉, M = supu∈H‖u‖=1

〈Au, u〉,

then, {m,M} ⊂ σ(A) ⊂ [m,M].




Proof: From the definition of M we have that〈Au, u〉 ≤ M‖u‖2, ∀u ∈ H. From this it follows that, if λ > Mthen,

〈λu − Au, u〉 ≥ (λ−M)︸︷︷︸

>0

‖u‖2. (8)

With that, it is easy to see that a(v , u) = 〈v , λu − Au〉 is asymmetric (a(u, v) = a(v , u) for all u, v ∈ H), continous andcoercive sesquilinear form.




It follows from Lax-Milgram theorem that

〈v , λu − Au〉 = 〈v , f 〉, ∀v ∈ H,

has a unique solution uf for each f ∈ H. It is easy to see that thissolution satisfies

(λ− A)uf = f .

From this it follows that (λ− A) is bijective and (M,∞) ⊂ ρ(A).




Let us show that M∈σ(A). Note that a(u, v)=(Mu−Au, v) is acontinuous, symmetric sesquilinear form and a(u, u) ≥ 0, ∀u ∈ H.Hence

|a(u, v)| ≤ a(u, u)1/2a(v , v)1/2, for all u, v ∈ H,

that is, the Cauchy-Schwarz inequality holds.




It follows that

|(Mu − Au, v)| ≤ (Mu − Au, u)1/2(Mv − Av , v)1/2, ∀u, v ∈ H

≤ C (Mu − Au, u)1/2 ‖v‖

and that

‖Mu − Au‖ ≤ C (Mu − Au, u)1/2, ∀u ∈ H.




Let {un} be a sequence of vectors such that ‖un‖ = 1,〈Aun, un〉 → M. It follows that ‖Mun − Aun‖ → 0. If M ∈ ρ(A)

un = (MI − A)−1(Mun − Aun) → 0

which is in contradiction with ‖un‖ = 1, ∀n ∈ N. It follows thatM ∈ σ(A).

From the above result applied to −A we obtain that(−∞,m) ⊂ ρ(A) and m ∈ σ(A). The proof that σ(A) ⊂ R hasbeen given in Example 2




It follows directly from Proposition 1 (if A ∈ L(H) is self-adjoint,‖A‖ = sup{〈Au, u〉 : u ∈ H, ‖u‖H = 1}) that

Corollary

Let H be a Hilbert space and A ∈ L(H) be a self-adjoint operatorwith σ(A) = {0}, then A = 0.


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SMA 5878 Functional Analysis II - icmc.usp.br › ... › andcarva › AnaliseFuncional-II › Aulas › Aula06 … · Spectral AnalysisofLinearOperators SMA 5878 Functional Analysis