sma 5878 functional analysis ii€¦ · spectral analysisoflinearoperators(continued) sma 5878...

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 20, 2019

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

Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators

Dual operators

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− Ais 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 − Ais 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.



Compact operators

Let X ,Y be Banach spaces over K. We say that a linear operatorK : X → Y is compact if K (BX

1 (0)) is relatively compact in Y .

We denote by K(X ,Y ) the space of compact linear operatorsK : X → Y .



ExerciseLet X = C ([a, b],C) and k ∈ C ([a, b]× [a, b],C). DefineK ∈ L(X ) by

(Kx)(t) =

∫ b

a

k(t, s)x(s)ds.

Show that K ∈ L(X ) and, using the Arzela-Ascoli Theorem, showthat K ∈ K(X ).



TheoremLet X ,Y be Banach spaces over K. Then K(X ,Y ) is a closedsubspace of L(X ,Y ).



Proof: If K(X ,Y ) ∋ Knn→∞−→ K ∈ L(X ,Y ) in the topology of

L(X ,Y ), given ǫ > 0 there exists nǫ ∈ N such that

K (BX1 (0)) ⊂ Knǫ(B

X1 (0)) + BY

ǫ (0).

From this it easily follows that K (BX1 (0)) is totally bounded

(hence relatively compact) in Y .



ExerciseLet X = ℓ2(C) and A : X → X given by A{xn} =

{xnn+1

}

. We

already know that A is bounded and that 0 ∈ σc(A). Show that Ais compact.



TheoremLet X ,Y ,Z be Banach spaces over K, A ∈ L(X ,Y ) andB ∈ L(Y ,Z ),

(a) If A ∈ K(X ,Y ) or B ∈ K(Y ,Z ), then B ◦ A ∈ K(X ,Z ),

(b) If A ∈ K(X ,Y ), then A∗ ∈ K(Y ∗,X ∗) and

(c) If A ∈ K(X ,Y ) and R(A) is a closed subspace of Y , thatR(A) is finite dimensional.



Proof: The proofs of (a) and (c) are left to the reader asexercises. To prove (b) we show that if {x∗n} is a subsequence inA∗(BY ∗

1 (0)), then it has a convergent subsequence.

Consider the space C (A(BX1 (0)),K). Note that, for y∗ ∈ BY ∗

1 (0)and z ∈ A(BX

1 (0)) there exists x ∈ BX1 (0) such that z = Ax and,

consequently,

|y∗(z)| = |y∗(Ax)| ≤ ‖A‖L(X ,Y ).

Besides that, if z1, z2 ∈ A(BX1 (0))

|y∗(z1)− y∗(z2)| ≤ ‖z1 − z2‖Y .



Hence F = {y∗∣∣A(BX

1(0))

: y∗ ∈ BY ∗

1 (0)} is a uniformly bounded and

equicontinuous family of functions in C (A(BX1 (0)),K).

From the Arzela-Ascoli Theorem, if x∗n = y∗n ◦ A = A∗ ◦ y∗n comy∗n ∈ BY ∗

1 (0), there is a subsequence y∗nk of {y∗n} such that

supx∈BX

1 (0)

|x∗nk (x) − x∗nl (x)| = supx∈BX

1 (0)

|y∗nk ◦ A(x)− y∗nl ◦ A(x)|

= supz∈A(BX

1 (0))

|y∗nk (z)− y∗nl (z)|k,l→∞−→ 0.

Therefore {x∗n} has a convergent subsequence to some x∗ ∈ X ∗

and the proof of (b) is complete.



If X is a Banach space, a projection P : X → X is a boundedlinear operator such that P2 = P . Note that, P ∈ K(X ) if andonly if Z = R(P) is finite dimensional.

In fact, if Z is finite dimensional, then any bounded subset of Z isrelatively compact and consequently P(BX

1 (0)) is relativelycompact. On the other hand, if P(BX

1 (0)) ⊃ BZ1 (0) is relatively

compact, it follows from Theorem 6.5 in [Brezis] that Z is finitedimensional.

Clearly the identity operator I : X → X is compact if and only if Xis finite dimensional and, consequently, if A ∈ K(X ) and X isinfinite dimensional then 0 ∈ σ(A) (if not, I = A ◦ A−1 is compactand dim(X ) < ∞).



TheoremLet X be a Banach space over K and A ∈ K(X ). If λ ∈ K\{0},then N((λ− A)n) is finite dimensional, n = 1, 2, 3, · · · .



Proof: Let us consider first the case n = 1. Clearly N((λ− A)) isclosed and if x ∈ N((λ − A)), x = λ−1Ax . Hence the identityoperator in N((λ − A)) is compact and N((λ− A)) is finitedimensional.

The general case flows from the previous case noting that

(λ− A)n =

n∑

k=0

λn−k

(nk

)

(−1)kAk = λnI − Aλ

where Aλ = −n∑

k=1

λn−k

(nk

)

(−1)kAk ∈ K(X ).



ExerciseLet X be a Banach space over K and T ∈ L(X ). Prove that1 ifN(T n0) = N(T n0+1) then N(T n) = N(T n+1) for all n ≥ n0.

1Sugestion: Show that N(T n+1) = {x ∈ X : Tx ∈ N(T n)}.



TheoremLet X be a Banach space over K, A ∈ K(X ) and λ ∈ K\{0}.There exists n0 ∈ N such that N((λ− A)n+1) = N((λ− A)n) forall n ≥ n0.



Proof: It is enough to prove that there exists n0 ∈ N such thatN((λ − A)n0+1) = N((λ− A)n0).

Clearly N((λ− A)n) is closed and N((λ − A)n) ⊂ N((λ − A)n+1)for all n ∈ N.

Suppose that N((λ− A)n) ( N((λ− A)n+1) for all n ∈ N.

From Riesz Lemma (Lemma 6.1 in [Brezis]), for each n ∈ N, thereexists xn ∈ N((λ − A)n+1) such that ‖xn‖X = 1 and‖xn − x‖X ≥ 1

2 , for all x ∈ N((λ − A)n).



Hence, if 1 ≤ m < n,

Axn − Axm = λxn + (−λxm + (λ− A)xm − (λ− A)xn) = λxn − z ,

where z = −λxm + (λ− A)xm − (λ− A)xn ∈ N((λ − A)n). So

‖Axn − Axm‖X = |λ|‖xn − λ−1z‖x ≥|λ|

2

and {Axn} does not have a convergent subsequence and A is notcompact. This contradiction proves the theorem.



If N(λ− A) 6= {0} we have that λ is an eigenvalue of A; that is,λ ∈ σp(A).

In this case, the geometric multiplicity of λ is the dimension ofN(λ − A) and there is a least positive integer n0 such thatN((λ − A)n0) = N((λ− A)n0+1), we say that N((λ− A)n0) is thegeneralised eigenspace associated to the eigenvalue λ and thatdim(N((λ − A)n0)) is the algebraic multiplicity of λ.



RemarkObserve that, if X is a Banach space over K, λ ∈ K\{0} andA ∈ K(X ), from Theorem 6.6 (c) in [Brezis], R(λ− A) = X if andonly if N(λ − A) = {0}. Hence λ ∈ ρ(A) if and only ifN(λ − A) = {0}. It follows that, all points in σ(A)\{0} areeigenvalues.



LemmaLet X be an infinite dimensional Banach space over K andA ∈ K(X ). If {λn} is a sequence of distinct numbers such that

λn → λ

λn ∈ σ(A)\{0}, ∀n ∈ N.

Then λ = 0; that is, each point in σ(A)\{0} is isolated.



Proof: Since λn ∈ σp(A), let xn 6= 0 such that (λn −A)xn = 0 andXn = [x1, . . . , xn]. We show that Xn ( Xn+1, ∀n ∈ N. It is enoughto show that {x1, . . . , xn} is a linearly independent set of vectors,for each n ∈ N∗. Suppose, by induction, that {x1, . . . , xn} is alinearly independent set of vectors and let us show that

{x1, · · · , xn+1} is also linearly independent. If xn+1 =n∑

i=1

αixi ,

thenn∑

i=1

λn+1αixi = λn+1xn+1 = Axn+1 =

n∑

i=1

αiλixi .



From this it follows that

n∑

i=1

αi (λn+1 − λi )xi = 0 and therefore α1 = · · · = αn = 0.

With that xn+1 = 0, which is a contradiction. Therefore{x1, · · · , xn+1} is a linearly independent set of vectors.

Since x1 6= 0 we obtain that {x1, · · · , xn} is a linearly independentset of vectors for all n ∈ N and Xn ( Xn+1, for all n ∈ N.

Also note that (λn − A)Xn ⊂ Xn−1 (since (λn − A)xn = 0).



Applying Riesz Lemma (Lemma 6.1 in [Brezis]), we construct {yn}

such that yn ∈ Xn, ‖yn‖ = 1 and dist(yn,Xn−1) ≥1

2for n ≥ 2. If

2 ≤ m < n, then

Xm−1 ⊂ Xm ⊂ Xn−1 ⊂ Xn.

and,



∥∥∥∥

Aynλn

−Aymλm

∥∥∥∥=

∥∥∥∥

∈Xn−1︷︸︸︷

(λm − A)ymλm

−(λn − A)yn

λn

− ym + yn

∥∥∥∥

≥ dist(yn,Xn−1) ≥1

2.

If λn → λ 6= 0, then the sequence

{ynλn

}

is bounded and, from the

fact that A is compact,

{Aynλn

}

has a convergent subsequence,

which leads to a contradiction. Hence λ = 0.



Our next theorem synthesises the results obtained above relativelyto the spectrum of a compact operator.

TheoremLet X be a Banach space over a field K and A ∈ K(X ). Then allpoints in σ(A)\{0} are eigenvalues, σ(A) contains at most acountable number of points and the set of accumulation points ofσ(A) is either ∅ or {0}.



Frequently the compact operators arise as inverse of unboundedoperators. These operators are called operators with compactresolvent that we define next.

DefinitionLet X be a Banach space over K and A : D(A) ⊂ X → X a closedoperator with non-empty resolvent. We say that A has compact

resolvent if, for some λ0 ∈ ρ(A), (λ0 − A)−1 ∈ K(X ).

It is a simple consequence of the resolvent identity that if A hascompact resolvent, then (λ− A)−1 is compact for all λ ∈ ρ(A).



Example

Let X = {f ∈ C ([0, 1],K) : f (0) = 0} and A : D(A) ⊂ X → X thelinear operator defined byD(A) = {f ∈ C 1([0, 1],K) : f (0) = f ′(0) = 0} and Af = f ′ forf ∈ D(A). It is easy to see that A is a closed densely definedclosed operator and that 0 ∈ ρ(A). To see that A has compactresolvent, it is enough to apply the Arzela-Ascoli Theorem.

ExerciseLet A : D(A) ⊂ X → X be a closed operator with 0 ∈ ρ(A). InD(A) define the graph norm ‖x‖G(A) = ‖x‖+ ‖Ax‖ and denote byY the space D(A) endowed with the norm ‖ · ‖G(A). Show that Yis a Banach space and that if Y is compactly embedded in X , thenA has compact resolvent.



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〉, is a conjugated isometry between H and H∗. Theidentification between H and H∗ consists in identifying u with Eu.If A∗ : D(A∗) ⊂ X ∗ → X ∗ is the dual of A, then A• = 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 = Cn 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)•.


sma 5878 functional analysis ii€¦ · spectral analysisoflinearoperators(continued) sma 5878...

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