SMA 5878 Functional Analysis II - icmc.usp.br · Spectral AnalysisofLinearOperators SMA 5878 Functional Analysis II Alexandre Nolasco de Carvalho Departamento de Matematica Instituto

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

April 01, 2019

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


Dissipative operators and numerical range

Operational calculus

Operational calculus for closed operators


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 each

x ∈ 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 is

dissipative 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), from Lemma 1 we have that

‖(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 A

and 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 .

SinceA 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 and

W (A) be the numerical range of A.

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

satisfies

‖(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 a

self-adjoint operator. It follows that A is closed and densely

defined. If A is bounded above; that is, 〈Au, u〉 ≤ a〈u, u〉 for some

a ∈ R, then C\(−∞, a] ⊂ ρ(A), and

‖(λ− A)−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 uniformly convex and

A : D(A) ⊂ X → X be a closed, densely defined and dissipative

linear operator. If R(I − A) = X, show that

ρ(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 uniformly 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 > M

then,〈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 1






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 operator

with σ(A) = {0}, then A = 0.






Operational calculus in L(X )

Let X be a Banach space over C and A ∈ L(X ). We have alreadyseen that σ(A) is closed, non-empty and bounded.

In fact,

σ(A)⊂{λ ∈ C : |λ|≤ rσ(A)}

and

rσ= infn≥1

‖An‖1n

L(X )≤‖A‖L(X ).






Let γ : [0, 2π] → C be the curve given by γ(t) = re it , t ∈ [0, 2π],with r > rσ(A). We know that, for |λ| > rσ(A),

(λ− A)−1 =∞∑

n=0

λ−n−1An,

and, for j ∈ N,

Aj =1

2πi

∫

γλj (λ− A)−1dλ. (9)

Observe that, the curve γ can be chosen to be any closedrectifiable curve which is homotopic to the above curve in ρ(A).

Emphasise the case j = 0.






So, if p : C → C is a polynomial,

p(A) =1

2πi

∫

γp(λ)(λ− A)−1dλ.

ExerciseLet X be a complex Banach space and A ∈ L(X ). Show that, if

r > ‖A‖L(X ) and γr (t) = re2πit , t ∈ [0, 1] then,

∞∑

n=0

An

n!=

1

2πi

∫

γr

eλ(λ− A)−1dλ.






DefinitionIf X is a Banach space over C and A ∈ L(X ). The class of analytic

functions f : D(f ) ⊂ C → C such that D(f ) is a Cauchy domain

that contains σ(A) is denoted by U(A). For f ∈ U(A) we define

f (A) =1

2πi

∫

+∂Df (λ)(λ− A)−1dλ (10)

where D is a bounded Cauchy domain such that σ(A) ⊂ D and

D ⊂ D(f ).






ExerciseLet X be a Banach space over C and A ∈ L(X ). Show that, if

f , g ∈ U(A) and f , g coincide in an open set that contains σ(A)then, f (A) = g(A).






It is clear that, if f , g ∈ U(A) and α ∈ C, we have that f + g , fgand αf are in em U(A). Besides that, it is easy to see that

f (A) + g(A) = (f + g)(A) and αf (A) = (αf )(A).






Let us show that f (A) ◦ g(A)=(fg)(A). Let D1 and D2 be Cauchydomains such that σ(A) ⊂ D1 ⊂ D1 ⊂ D2 ⊂ D(f ) ∩ D(g). Withthis notation we have that

f (A) =1

2πi

∫

+∂D1

f (λ)(λ− A)−1dλ,

g(A) =1

2πi

∫

+∂D2

g(µ)(µ − A)−1dµ.






Hence

f (A) ◦ g(A) =1

(2πi)2

∫

+∂D1

∫

+∂D2

f (λ)g(µ) (λ−A)−1(µ−A)−1 dµ dλ

=1

(2πi)2

∫

+∂D1

∫

+∂D2

f (λ)g(µ)1

µ− λ[(λ− A)−1−(µ− A)−1] dµ dλ

=1

2πi

∫

+∂D1

f (λ)g(λ)(λ − A)−1 dλ = (fg)(A).






ExerciseLet X be a Banach space over C, B ∈ L(X ) with ‖B‖L(X ) < 1and A = I + B. Show that, if 1 > r > ‖B‖L(X ), α > 0 and

γr (t) = 1 + re2πit , t ∈ [0, 1], then

A−α =∞∑

n=0

(α+ n − 1

n

)

(−1)nBn =1

2πi

∫

γr

λ−α(λ− A)−1dλ.

where

(α+ n− 1

n

)

:=Γ(α+ n)

n! Γ(α)=

α(α + 1) · · · (α+ n − 1)

n!.






Show that A−α−β = A−αA−β for all α, β ∈ (0,∞). In particular,

A−1 =∞∑

n=0

(−1)nBn =1

2πi

∫

γr

λ−1(λ− A)−1dλ and

A−2 =

∞∑

n=0

(n + 1)(−1)nBn =1

2πi

∫

γr

λ−2(λ− A)−1dλ.

Study the positive powers of A.






TheoremLet X be a Banach space over C and A ∈ L(X ). If f ∈ U(A) issuch that f (λ) 6= 0 for all λ ∈ σ(A), then f (A) one-to-one and

onto X with inverse g(A) where g is any function in U(A) thatcoincides with 1

fin an open set that contains σ(A).






Proof: If g = 1fin an open set that contains σ(A) then, g ∈ U(A)

and f (λ)g(λ) = 1 in an open set that contains σ(A). Hence

f (A)g(A) = g(A)f (A) = (fg)(A) = I .







Let X be a Banach space over C and A : D(A) ⊂ X → X a closedlinear operator with non-empty resolvent ρ(A).

Denote by U∞(A) the set of all analytic functions f with domainscontaining the union of σ(A) with the exterior of a compact setand that satisfying limλ→∞ f (λ) = f (∞).






ExerciseLet R > 0, A(0,R ,∞) = (B

C

R(0))c and f : A(0,R ,∞) → C a

bounded analytic function. Show that the limit below exists 1

limλ→∞

f (λ).

1Suggestion: Show that 0 is a removable singularity of the analytic functiong : B 1

R(0)\{0} → C defined by g(λ) = f ( 1

λ).






We define in U∞(A) the equivalence relation R by (f , g) ∈ R if fand g are equal in an open set that contains σ(A) and the exteriorof a ball. We write f ∼ g to denote that (f , g) ∈ R.

ExerciseShow that the relation R ⊂ U∞ × U∞ is an equivalence relation.






Observe that, if D is an unbounded Cauchy domain withD ⊃ A(0, r ,∞) and f : D(f ) ⊂ C → C is a function in U∞(A)with D(f ) ⊃ D, entao

f (ξ) =1

2πi

∫

γr

f (λ)

λ− ξdλ+

1

2πi

∫

∂D+

f (λ)

λ− ξdλ (11)

where r > 0 is such that Br (0) ⊃ Dc , ξ is a point of D with|ξ| < r and γr (t) = re2πit , t ∈ [0, 1].






Hence, making r → ∞ in (11) and using that limλ→∞

f (λ) = f (∞),

we obtain

f (ξ) = f (∞) +1

2πi

∫

∂D+

f (λ)

λ− ξdλ (12)

for all ξ in D.

Using the same reasoning, if ξ is exterior to D, then

0 = f (∞) +1

2πi

∫

∂D+

f (λ)

λ− ξdλ (13)






When f ∈ U∞(A), we define

f (A) = f (∞)I +1

2πi

∫

+∂Df (λ)(λ− A)−1dλ, (14)

where D is an unbounded Cauchy domain such thatσ(A) ⊂ D ⊂ D ⊂ D(f ). Note that f (A) ∈ L(X ) even when A isnot a bounded operator.






ExerciseLet X be a Banach space over C and A : D(A) ⊂ X → X a closed

operator with non-empty resolvent.

a) Show that, if f , g ∈ U∞(A) and f ∼ g then, f (A) = g(A).

b) Show that, if f (λ) = 1 for all λ ∈ C then, f (A) = I .






Let X be a Banach space over C and A : D(A) ⊂ X → X a closedoperator with non-empty resolvent. Se f , g ∈ U∞(A), show thatf (A) ◦ g(A) = (fg)(A).






As before, let D1 and D2 Cauchy domains such thatσ(A) ⊂ D1 ⊂ D1 ⊂ D2 ⊂ D(f ) ∩ D(g). With this notation wehave that

f (A) = f (∞)I +1

2πi

∫

+∂D1

f (λ)(λ − A)−1dλ

and

g(A) = g(∞)I +1

2πi

∫

+∂D2

g(µ)(µ − A)−1dµ.






Using (12) and (11), if λ ∈ ∂D1 and µ ∈ ∂D2, we have that

g(λ) = g(∞) +1

2πi

∫

+∂D2

g(µ)

µ− λdµ

and

0 = f (∞) +1

2πi

∫

+∂D1

f (λ)

λ− µdλ.






Consequently,

f (A) ◦ g(A) = f (∞)g(∞)I

+1

(2πi)2

∫

+∂D1

∫

+∂D2

f (λ)g(µ) (λ − A)−1(µ− A)−1 dµ dλ

+g(∞)

2πi

∫

+∂D1

f (λ) (λ − A)−1dλ+f (∞)

2πi

∫

+∂D1

g(µ) (µ − A)−1dµ






= f (∞)g(∞)I

+1

(2πi)2

∫

+∂D1

∫

+∂D2

f (λ)g(µ)(λ−A)−1−(µ−A)−1

µ− λdµ dλ

+g(∞)

2πi

∫

+∂D1

f (λ) (λ − A)−1dλ+f (∞)

2πi

∫

+∂D1

g(µ) (µ − A)−1dµ






= f (∞)g(∞)I +1

2πi

∫

+∂D1

f (λ)(λ − A)−1

(1

2πi

∫

+∂D2

g(µ)

µ− λdµ

)

dλ

+1

2πi

∫

+∂D2

g(µ)(µ − A)−1

(1

2πi

∫

+∂D1

f (λ)

λ− µdλ

)

dµ

+g(∞)

2πi

∫

+∂D1

f (λ) (λ − A)−1dλ+f (∞)

2πi

∫

+∂D1

g(µ) (µ − A)−1dµ

= f (∞)g(∞)I +1

2πi

∫

+∂D1

f (λ)g(λ)(λ − A)−1 dλ = (fg)(A).






Proceeding exactly as in Theorem 3 we have that the followingresult holds.

TheoremLet X be a Banach space over C and A : D(A) ⊂ X → X be a

closed operator with non-empty resolvent. If f ∈ U∞(A) is suchthat f (λ) 6= 0 for all λ ∈ σ(A) ∪ {∞} then, f (A) is one-to-one and

onto X with inverse g(A) where g is any function of U∞(A) withg ∼ 1

f.






ExerciseLet X be a Banach space over C and A ∈ L(X ). Show that, if

f ∈ U(A) ∩ U∞(A) then, (10) and (14) give rise to the same

operator f (A).


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SMA 5878 Functional Analysis II - icmc.usp.br · Spectral AnalysisofLinearOperators SMA 5878 Functional Analysis II Alexandre Nolasco de Carvalho Departamento de Matematica Instituto