SMA 5878 Functional Analysis II€¦ · Spectral AnalysisofLinearOperators(Continued) SMA 5878...
Transcript of 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)
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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)∗).
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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∗).
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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)∗).
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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 .
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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 .
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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 ).
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
TheoremLet X ,Y be Banach spaces over K. Then K(X ,Y ) is a closedsubspace of L(X ,Y ).
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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 .
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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 .
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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 ) < ∞).
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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, · · · .
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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 ).
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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)}.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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).
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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 .
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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).
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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,
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
∥∥∥∥
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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}.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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).
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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).
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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∗.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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•.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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〉.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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).
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II
Spectral Analysis of Linear Operators (Continued)Dual operatorsCompact operatorsSymmetric and self-adjoint operators
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)•.
Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II