SMA 5878 Functional Analysis II · Spectral Analysis of Linear Operators (Continued) Bounded Linear...

Spectral Analysis of Linear Operators (Continued)

SMA 5878 Functional Analysis II

Alexandre Nolasco de Carvalho

Departamento de MatematicaInstituto de Ciencias Matematicas and de Computacao

Universidade de Sao Paulo

March 18, 2019

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

Spectral Analysis of Linear Operators (Continued)Bounded Linear OperatorsDual operators

Note that, for ξ0 ∈ ρ(A), if |ξ − ξ0| < ‖(ξ0 − A)−1‖−1L(X ) we have

that ξ ∈ ρ(A) and

(ξ − A)−1 =

∞∑

n=0

(ξ0 − ξ)n(ξ0 − A)−n−1. (1)

On the other hand, if |ξ − ξ0| > ‖(ξ0 − A)‖L(X ) we have thatξ ∈ ρ(A) and

(ξ − A)−1 = −

∞∑

n=0

(ξ0 − ξ)−n−1(ξ0 − A)n (2)



Hence, the radius of convergence of the Taylor’s series in (1) is thereciprocal of the spectral radius of the operator (ξ0 − A)−1 whilethe radius of convergence of the Laurent’s series in (2) is thespectral radius of (ξ0 − A). Thus, in the circles{λ ∈ C : |λ− ξ0| = (rσ(ξ0 − A)−1)−1} and{λ ∈ C : |λ− ξ0| = rσ((ξ0 − A))} there are points of σ(A).



O primeiro teorema da aplicacao espectral

Next we show a version of the Spectral Mapping Theorem forpolynomials.

Seja p(λ) = anλn + an−1λ

n−1 + · · ·+ a1λ+ a0, ai ∈ C, 0 ≤ i ≤ n.

If A ∈ L(X ), we define

p(A) = anAn + an−1A

n−1 + · · ·+ a1A+ a0I

and, if B ⊂ C, we define p(B) := {p(b) : b ∈ B}.



TheoremIf A ∈ L(X ) and p : C → C is a polynomial, then

i) σp(p(A)) = p(σp(A)),

ii) σr (p(A)) = p(σr (A))\p(σp(A)),

iii) σc(p(A)) = p(σc(A))\(p(σp(A)) ∪ p(σr (A))) and

iv) σ(p(A)) = p(σ(A)).



Proof: Let p(λ) = anλn + an−1λ

n−1 + · · · + a1λ+ a0, ai ∈ C,0 ≤ i ≤ n be a polinomial.

It is easy to se that for all scalar α ∈ C, and T ∈ L(X ),σ(αT ) = {αλ ∈ C : λ ∈ σ(T )} =: ασ(T ).

Hence, without loss of generality, we may assume that an = (−1)n.

If β1, · · · , βn are the roots of the polynomial q(λ) = µ− p(λ), then

µ− p(A) = q(A) = (β1 − A) · · · (βn − A). (3)



The proof of iv) follows immediately from i), ii) and iii).

i) If µ− p(A) is not injective, it follows from (3) that there is i0with 1 ≤ i0 ≤ n such that (βi0 − A) is not injective.

Reciprocally, if for some i0 with 1 ≤ i0 ≤ n, (βi0 − A) is notinjective, it follows from (3) that µ− p(A) is not injective. Thisshows that σp(p(A)) = p(σp(A)).



ii) If µ ∈ σr (p(A)), µ− p(A) is injective, from (3), (βi − A) isinjective for all 1 ≤ i ≤ n.

Besides that, R(µ− p(A)) is not dense and consequently, for some1 ≤ i0 ≤ n we must have that R(βi0 − A) is not dense. It followsthat βi0 ∈ σr (A) and p(βi0) = µ.

This shows that σr (p(A)) ⊂ p(σr (A))\p(σp(A)).



On the other hand, if µ ∈ p(σr (A))\p(σp(A)), it follows that(βi−A) is injective for all 1≤ i≤n (since σp(p(A)) = p(σp(A))and, for some 1 ≤ i0 ≤ n, R(βi0 − A) is not dense.

From that it follows that µ− p(A) is injective but R(µ− p(A)) isnot dense and µ ∈ σr (p(A)).

This shows that p(σr (A))\p(σp(A)) ⊂ σr (p(A)) and the proof ofii) is complete.



iii) If µ ∈ σc(p(A)) then µ /∈ σp(p(A)) = p(σp(A)) and it followsthat (βi − A) is injective, 1 ≤ i ≤ n.

Also,µ /∈σr (p(A))=p(σr (A))\p(σp(A))andR(βi−A)=X , 1≤ i≤n.

Since µ /∈ρ(A) it follows that R(βi0 − A)(X , for some 1 ≤ i0 ≤ n.

If µ ∈ p(σc(A))\p(σp(A)) ∪ p(σr (A)) then µ = p(βi ) withβi /∈ σp(A) ∪ σr (A), 1 ≤ i ≤ n.

Hence (βi − A) is one-to-one and has dense image, 1 ≤ i ≤ n.

Since βi0 ∈ σc(A), for some 1 ≤ i0 ≤ n, we must have thatR(βi0 − A) ( X and (µ− p(A)) is one-to-one, has dense imageand R(µ− p(A)) ( X . This proves that µ ∈ σc(p(A)).



Example

Let X = ℓ2(C), T : ℓ2(C) 7→ ℓ2(C) the linear operator defined by

T ({x1, x2, x3, · · · }) = {x1, 0, x2, x3, . . .} and p(λ) = λ2 − λ. It iseasy to see that 0 ∈ σr (T ) and therefore p(0) = 0 ∈ p(σr (T )). Onthe other hand, we see that p(0) = 0 /∈ σr (p(T )), since

p(T )({x1, x2, x3, · · · }) = ({0, 0,−x2, x2 − x3, x3 − x4, · · · })

and p(T ) is not injective operator.



Next we give an alternative proof of

rσ(A) = limn→∞

‖An‖1/nL(X ) = inf

n≥1‖An‖

1/nL(X )

using the previous theorem.

In fact, σ(An) = {zn : z ∈ σ(A)} and rσ(A)n = rσ(A

n) ≤ ‖An‖L(X )

and rσ(A) ≤ ‖An‖1/nL(X ). Hence,

rσ(A) = lim supn→∞

‖An‖1/nL(X ) ≤ inf

n∈N‖An‖

1/nL(X ) ≤ lim inf

n→∞‖An‖

1/nL(X ).

So, the limit exists and rσ(A) = infn∈N ‖An‖1/nL(X ).



DefinitionLet X be a Banach space over C and A ∈ L(X ). We say that A is

nilpotent if there is a n0 ∈ N such that An0 = 0 and that A is

quasi-nilpotent if ‖An‖1n

L(X )

n→∞−→ 0.



Example

Let T :ℓ1(C)→ℓ1(C) defined by

T (x1, x2, x3, · · · )=(0, x1,x2

2,x3

3,x4

4, · · · ).

It is easy to see that ‖T n‖L(ℓ1) ≤1n! and therefore T is

quasinilpotent and σ(T ) = {0}.

ExerciseLet X be a Banach space over C, A ∈ L(X ) a given nilpotent

operator and 0 6= λ ∈ C, compute (λ− A)−1.



Dual operators

Next we recall the definition of dual operator. Let X and Y beBanach spaces over the field K (R or C) with duals X ∗ and Y ∗.

If Z is either X or Y , z∗ ∈ Z ∗ we denote its values at a vectorz ∈ Z by 〈z , z∗〉Z ,Z∗ .

The map Z ×Z ∗ ∋ (z , z∗) 7→ 〈z , z∗〉Z ,Z∗ ∈ K is called duality map.

For simplicity of notation, if no confusion may arise, we will write〈z , z∗〉 instead of 〈z , z∗〉Z ,Z∗ .



Let A : D(A) ⊂ X → Y be a densely defined operator.

The dual operator A∗ : D(A∗) ⊂ Y ∗ → X ∗ of A is the linearoperator defined by: D(A∗) is the set of all y∗ ∈ Y ∗ such thaty∗ ◦ A : D(A) ⊂ X → K is bounded; that is, the set of all y∗ ∈ Y ∗

for which there exists w∗ ∈ X ∗ satisfying

〈Ax , y∗〉 =(y∗◦A)(x) = w∗(x) = 〈x ,w∗〉, ∀ x ∈ D(A). (4)

If y∗ ∈ D(A∗) we define A∗y∗ := w∗ where w∗ is the uniqueelement of X ∗ satisfying (4).



ExerciseIf X is a Banach space and A : D(A) ⊂ X → Y is a densely

defined linear operator, show that A∗ : D(A∗) ⊂ Y ∗ → X ∗ is a

closed linear operator.

We start with some basic results about dual operators.

LemmaLet X and Y be Banach spaces over K and A ∈ L(X ,Y ); then,A∗ ∈ L(Y ∗,X ∗) and ‖A‖L(X ,Y ) = ‖A∗‖L(Y ∗,X∗).



Proof: For all y∗ ∈ Y ∗, y∗◦A is a bounded linear functional andtherefore it determines a unique element x∗ ∈ X ∗ for which〈x , x∗〉 = 〈Ax , y∗〉, for all x ∈ X . It follows that D(A∗) = Y ∗.

Also,

‖A∗‖L(Y ∗,X∗) = sup‖y∗‖Y ∗≤1

‖A∗y∗‖X∗ = sup‖y∗‖Y ∗≤1

sup‖x‖X≤1

|〈x ,A∗y∗〉|

= sup‖x‖X≤1

sup‖y∗‖X∗≤1

|〈Ax , y∗〉| = sup‖x‖X≤1

‖Ax‖Y

= ‖A‖L(X ,Y ).



LemmaLet X be a reflexive Banach space K. If A : D(A) ⊂ X → X is

closed and densely defined then D(A∗) is dense in X ∗.



Proof: If D(A∗) is not dense in X ∗ there exists x0 ∈ X such thatx0 6= 0 and 〈x0, x

∗〉 = 0 for all x∗ ∈ D(A∗).

Since the graph of A is closed and does not contains (0, x0), fromHahn-Banach Theorem, there are x∗1 and x∗2 in X ∗ such that〈x , x∗1 〉 − 〈Ax , x∗2 〉 = 0 for all x ∈ D(A) and 〈0, x∗1 〉 − 〈x0, x

∗2 〉 6= 0.

It follows that x∗2 6= 0, 〈x0, x∗2 〉 6= 0, x∗2 ∈ D(A∗) and A∗x∗2 = x∗1 .

This implies that 〈x0, x∗2 〉 = 0 which is a contradiction. Thus

D(A∗) is dense in X ∗.



ExerciseExhibit an example of a densely defined operator,

A : D(A) ⊂ X → X such that D(A∗) ( X.



ExerciseThe annihilator of a subset M ⊂ X is the set

M⊥ = {x∗ ∈ X ∗ : 〈x , x∗〉 = 0,∀x ∈ M} and the annihilator of

M∗ ⊂ X ∗ is the set (M∗)⊥ = {x ∈ X : 〈x , x∗〉 = 0,∀x∗ ∈ M∗}.We know that if M ⊂ X is a vector space then (M⊥)⊥ = M.

A subset M∗ ⊂ X ∗ is said total if (M∗)⊥ = {0}. Show that, if

A : D(A) ⊂ X → X is a closed densely defined operator then,

D(A∗) is total.



TheoremLet A : D(A) ⊂ X → X be a linear densely defined operator. Then

ρ(A) = ρ(A∗) and ((λ− A)−1)∗ = (λ− A∗)−1,∀λ ∈ ρ(A)



Proof: From the definition of dual (λI − A)∗ = λI ∗ − A∗. If λ− A

is injective and has dense image, let us show that

(1) ((λI − A)−1)∗(λI ∗ − A∗)x∗ = x∗, ∀x∗ ∈ D(A∗) and

(2) (λI ∗ − A∗)((λI − A)−1)∗x∗ = x∗, ∀x∗ ∈ D(((λI − A)−1)∗).



Proof of (1): If x ∈ R(λ− A), x∗ ∈ D(A∗), then

〈x , x∗〉 = 〈(λI −A)(λI −A)−1x , x∗〉 = 〈(λI −A)−1x , (λI ∗−A∗)x∗〉.

It follows that (λI ∗ − A∗)x∗ ∈ D(((λI − A)−1)∗)(R(λI ∗ − A∗) ⊂ D(((λI − A)−1)∗)) and, from the fact thatR(λI − A) = X , we have

((λI − A)−1)∗(λI ∗ − A∗)x∗ = x∗, ∀x∗ ∈ D(A∗).



Proof of (2): If x∗ ∈ D(((λI − A)−1)∗) and x ∈ D(A), then

〈x , x∗〉 = 〈(λI−A)−1(λI−A)x , x∗〉 = 〈(λI−A)x , ((λI−A)−1)∗x∗〉.

Hence ((λI − A)−1)∗x∗ ∈ D(λI ∗ − A∗) and, from the fact thatD(A) = X , we have

(λI ∗ − A∗)((λI − A)−1)∗x∗ = x∗, ∀x∗ ∈ D(((λI − A)−1)∗).



Now we can complete the proof of the theorem. If λ ∈ ρ(A),(λI − A)−1 is bounded and we have that ((λI − A)−1)∗ ∈ L(X ∗).

From (1) and (2) it follows that (λI ∗ − A∗)−1 = ((λI − A)−1)∗

and λ ∈ ρ(A∗).

If λ ∈ ρ(A∗), note that A∗ is closed and, consequently,(λI ∗ − A∗)−1 ∈ L(X ∗).

We already know that λI − A is densely defined.

Let us show that λI − A is injective and has dense image.



To see that λI − A is injective note that, if x ∈ D(A) is such that(λ− A)x = 0 and x∗ ∈ D(A∗), then

0 = 〈(λI − A)x , x∗〉 = 〈x , (λI ∗ − A)∗x∗〉.

Since R(λI ∗ − A∗) = X ∗ we have that x = 0 and therefore λI − A

is injective.



Now, to see that λI − A has dense image note that, if x∗ ∈ X ∗ issuch that 0 = 〈(λI − A)x , x∗〉 for all x ∈ D(A), then x∗ ∈ D(A∗)and 0 = 〈x , (λI − A)∗x∗〉 for all x ∈ D(A).

Since D(A) is dense in X , it follows that (λI − A)∗x∗ = 0 and,since λ ∈ ρ(A∗), it follows that x∗ = 0. With this we have provedthat R(λI − A) is dense in X .



To conclude that λ ∈ ρ(A), it remains to prove that (λI − A)−1 isbounded. If x∗ ∈ X ∗ = R(λI ∗ − A∗) ⊂ D(((λI − A)−1)∗) andx ∈ R(λI − A), from (1) and (2), we have that

|〈(λI − A)−1x , x∗〉| = |〈x , ((λI − A)−1)∗x∗〉| = |〈x , (λI ∗ − A∗)−1x∗〉|

≤ ‖(λI ∗ − A∗)−1‖ ‖x∗‖ ‖x‖

From this it follows that (λ− A)−1 is bounded, proving thatλ ∈ ρ(A) and completing the proof.


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SMA 5878 Functional Analysis II · Spectral Analysis of Linear Operators (Continued) Bounded Linear...

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