sma 5878 functional analysis ii€¦ · theorems on perturbations of generators teoremas espectrais...

Theorems on Perturbations of GeneratorsTeoremas Espectrais and Dicotomias

SMA 5878 Functional Analysis II

Alexandre Nolasco de Carvalho

Departamento de MatematicaInstituto de Ciencias Matematicas and de Computacao

Universidade de Sao Paulo

June 06, 2018

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


Trotter Approximation Theorems


TheoremIf A,An ∈ G (M, ω), n ∈ N, the following statements are equivalent

(a) For all x ∈ X and λ with Reλ > ω,(λ− An)−1x

n→∞−→ (λ− A)−1x .

(b) For all x ∈ X and t ≥ 0, eAntxn→∞−→ eAtx .

Besides that, the convergence in (b) is uniform for all t in boundedsubsets of R+.




Trotter-Kato Theorem

Theorem (Trotter-Kato)

If An ∈ G (M, ω) and there exists λ0 with Reλ0 > ω such that

(a) for all x ∈ X , (λ0 − An)−1x → R(λ0)x when n→∞ and

(b) the image of R(λ0) is dense in X ,

then there exists a unique operator A ∈ G (M, ω) such thatR(λ0) = (λ0 − A)−1.




A direct consequence of Theorems 1 and 2 is the following theorem

TheoremLet An ∈ G (M, ω),∀n ∈ N. If for some λ0∈C with Reλ0>ω,

(a) limn→∞(λ0 − An)−1x =: R(λ0)x for all x ∈ X and

(b) the image of R(λ0) is dense in X ,

then, there is a unique operator A ∈ G (M, ω) such thatR(λ0) = (λ0 − A)−1. Besides that, eAntx → eAtx for all x ∈ X ,uniformly for t in bounded subsets of R+.




A little different consequence of the previous results is thefollowing theorem.

Theorem (Trotter)

Let An ∈ G (M, ω) and suppose that

(a) there is a dense subset D of X such that {Anx}n∈N isconvergent for all x ∈ D. Define A : D ⊂ X → X by,Ax = limn→∞ Anx for all x ∈ D,

(b) there is a λ0 with Reλ0 > ω for which (λ0 − A)D is dense inX .

Then A is closable and the closure A of A is in G (M, ω). Besides

that eAntx → eAtx for all x ∈ X , uniformly for t in boundedsubsets of R+.




Proof: If f ∈ D, x = (λ0 − A)f and xn = (λ0 − An)f then,Anf

n→∞−→ Af and xnn→∞−→ x . Furthermore, since

‖(λ0 − An)−1‖L(X ) ≤ M(Reλ0 − ω)−1, it follows that

limn→∞

(λ0 − An)−1x = limn→∞

((λ0 − An)−1(x − xn) + f ) = f ; (1)

that is, (λ0 − An)−1 converges in the image of λ0 − A. From (b)the image of λ0 − A is dense in X and, by hypothesis,‖(λ0 − An)−1‖L(X ) is bounded, uniformly for n ∈ N. It follows that(λ0 − An)−1x converges for all x ∈ X .




Letlimn→∞

(λ0 − An)−1x = R(λ0)x . (2)

From (1) it follows that the image of R(λ0) contains D andtherefore is dense in X . Theorem 2 implies the existence of anoperator A′ ∈ G (M, ω) satisfying R(λ0) = (λ0 − A′)−1. Toconclude the proof we show that A = A′. If x ∈ D then,

limn→∞

(λ0 − An)−1(λ0 − A)x = (λ0 − A′)−1(λ0 − A)x . (3)




On the other hand, since ‖(λ0 − An)−1‖L(X ) is uniformly bounded,

(λ0 − An)−1(λ0 − A)x = (λ0 − An)−1(λ0 − An)x

+ (λ0 − An)−1(An − A)x

= x + (λ0 − An)−1(An − A)xn→∞−→ x ,

given that Anxn→∞−→ Ax for all x ∈ D. Hence,

(λ0 − A′)−1(λ0 − A)x = x , x ∈ D. (4)




But (4) implies that A′x = Ax for x ∈ D and therefore A′ ⊃ A.Since A′ is closed, A is closable. Next we show that A ⊃ A′. Letf ′ = A′x ′. Since (λ0 − A)D is dense in X there is a sequencexn ∈ D such that

fn = (λ0 − A′)xn = (λ0 − A)xnn→∞−→ λ0x

′ − f ′ = (λ0 − A′)x ′.




Thus,

xn = (λ0 − A′)−1fnn→∞−→ (λ0 − A′)−1(λ0 − A′)x ′ = x ′ and (5)

Axn = λ0xn − fnn→∞−→ f ′. (6)

From (5) and (6) it follows that f ′ = Ax ′ and A ⊃ A′. HenceA = A′. The remaining statements of the theorem follow directlyfrom Theorem 3.



Spectral decomposition of semigroupsSpectral theorems for semigroups

Spectral decomposition of semigroups

When we study the stability of problems where semigroups oflinear operators are involved, one of the most fundamentalproblems is to determine the spectrum of the semigroup.

In general we known the infinitesimal generator and not theassociated semigroup of linear operators.

So, if we can compute some of the spectral properties of theinfinitesimal generator, we would like to use them to deriveproperties of the spectrum of the associated semigroup.




TheoremSuppose that {T (t) : t ≥ 0} is a strongly continuous semigroupand that, for some t0 > 0, σ(T (t0)) ∩ {λ ∈ C : |λ| = eαt0} = ∅for some α ∈ R. Then, there exists P ∈ L(X ), P2 = P,PT (t) = T (t)P ∀t ≥ 0 such that, if X− = R(P) and X+ = N(P),the restrictions of T (t)

∣∣X±

are in L(X±),

σ(T (t)∣∣X−

) = σ(T (t)) ∩ {λ ∈ C : |λ| < eαt} and

σ(T (t)∣∣X+

) = σ(T (t)) ∩ {λ ∈ C : |λ| > eαt}.There are constants M ≥ 1, δ > 0 such that

‖T (t)∣∣X−‖L(X−) ≤ Me(α−δ)t , ∀t ≥ 0,

{T (t)∣∣X+

; t ≥ 0} extents to a group in L(X+) with

T (t)∣∣X+

= (T (−t)∣∣X+

)−1 for t < 0, and

‖T (t)∣∣X+‖L(X+) ≤ Me(α+δ)t , ∀t ≤ 0.




RemarkFor α = 0, the above separation of the space X together with thedecaying properties is a particular case of exponential dichotomy.A even more special case, but clearly useful, is the case whenσ(T (t0)) ⊂ {λ ∈ C : |λ| < eαt0}, that is, P = I and X+ = {0}then,

‖T (t)‖L(X ) ≤ Me(α−δ)t , t ≥ 0.




Proof: If C is a rectifiable closed simple curve with trace{λ ∈ C : |λ| = eαt0}, define

P =1

2πi

∫C

(λ− T (t0))−1dλ ∈ L(X ).

Then, from what we have seen before, P2 = P and P is acontinuous projection.

It is easy to see that T (t)P = PT (t) for all t ≥ 0. Hence, ifX− = R(P) and X+ = N(P) we have that T (t) takes X+ into X+

and X− em X−.




Also note that σ(T (t0)∣∣X−

) is the part of σ(T (t0)) inside C and

σ(T (t0)∣∣X+

) is the part of σ(T (t0)) outside of C and that the

parts of (λ− T (t0))−1 in X+ and X− coincide with((λ− T (t0))

∣∣X+

)−1 and ((λ− T (t0))∣∣X−

)−1 respectivelly.

Now the spectral radius of T (t0)∣∣X−

is strictly smaller than eαt0 ,

that is,r(T (t0)

∣∣X−

) < e(α−δ)t0 ,

for some δ > 0.




If t > 0, for each m ∈ N there are n = n(m) ∈ N and τ ∈ [0, t0)such that mt = nt0 + τ . It is clear that n(m)

m→∞−→ ∞ and

r(T (t)∣∣X−

) = limm→∞

‖T (mt)∣∣X−‖

1m

L(X−)

= limn→∞

‖T (nt0 + τ)∣∣X−‖

tnt0+τ

L(X−)

≤ limn→∞

‖T (nt0)∣∣X−‖

tnt0+τ

L(X−)‖T (τ)∣∣X−‖

tnt0+τ

L(X−)

= r(T (t0)∣∣X−

)t/t0 < e(α−δ)t




Also, there is an integer N ≥ 1 such that Nt0 ≥ t, consequently

T (Nt0 − t)(T (t0)

∣∣X+

)−Nis the inverse of T (t)

∣∣X+

, that is, T (−t)∣∣X+

and arguing as abovewe can show that

r(T (t)∣∣X+

) < e(α+δ)t , t < 0.

It is easy to see that (considering the components in both spaces)

σ(T (t)) = σ(T (t)∣∣X+

) ∪ σ(T (t)∣∣X−

), t > 0,

and the statements about the spectral radius prove the estimateson the spectrum.




The estimates for the norms are simple. For example, sincer(T (t0)

∣∣X−

) < e(α−δ)t0 ,

‖T (nt0)∣∣X−‖1/nL(X−) < e(α−δ)t0

when n is large, hence

‖T (nt0)∣∣X−‖L(X−) ≤ M0e

n(α−δ)t0

for all n ≥ 0 and some M0 ≥ 1.




So, for n = 0, 1, 2, · · · and 0 ≤ τ < t0,

‖T (nt0 + τ)∣∣X−‖L(X−) ≤ M0e

n(α−δ)t0‖T (τ)∣∣X−‖L(X−)

≤ Me(α−δ)(nt0+τ)

where M = M0 sup0≤τ≤t0

e−(α−δ)τ‖T (τ)∣∣X−‖L(X−).




Spectral theorems for semigroups

It follows from the Spectral Mapping Theorem thatσ(f (A)) = f (σe(A)) when A is a closed operator with non-emptyresolvent and f ∈ U∞(A), this does not hold, in general, if A isunbounded and f /∈ U∞(A).

Since C 3 λ 7→ eλt ∈ C does not belong to U∞(A) for Aunbounded, in general, we cannot say that σ(eAt) = eσe(A)t .

Next we study the relations between the spectrum of a semigroupand the spectrum of its generator.




LemmaLet {eAt : t ≥ 0} be a strongly continuous semigroup. If

Bλ(t)x =

∫ t

0eλ(t−s)eAsxds (7)

then,(λ− A)Bλ(t)x = eλtx − eAtx , ∀x ∈ X (8)

andBλ(t)(λ− A)x = eλtx − eAtx , ∀x ∈ D(A). (9)




Proof: For all λ and t fixed, Bλ(t) defined by (7) is an operator inL(X ). Besides that, for all x ∈ X we have

eAh − I

hBλ(t)x =

eλh−1

h

∫ t

heλ(t−s)eAsxds +

eλh

h

∫ t+h

teλ(t−s)eAsxds

− 1

h

∫ h

0eλ(t−s)eAsxds

h→0+

−→ λBλ(t)x + eAtx − eλtx .




Consequently, Bλ(t)x ∈ D(A) andABλ(t)x = λBλ(t)x + eAtx − eλtx , proving (8).

It is clear that, for x ∈ D(A), ABλ(t)x = Bλ(t)Ax and (9)follows.




TheoremLet {eAt : t ≥ 0} be a strongly continuous semigroup. Then,

σ(eAt) ⊃ etσ(A), t ≥ 0. (10)




Proof: Let eλt ∈ ρ(eAt) and Q = (eλt − eAt)−1. From (8) and(9) we deduce that

(λ− A)Bλ(t)Qx = x , ∀x ∈ X

andQBλ(t)(λ− A)x = x , ∀x ∈ D(A).




Since Bλ(t) and Q comute we also have that

Bλ(t)Q(λ− A)x = x , ∀x ∈ D(A).

Thus, λ ∈ ρ(A), Bλ(t)Q = (λ− A)−1 and ρ(eAt)\{0} ⊂ etρ(A).

This same reasoning implies that λ+ 2kπit ∈ ρ(A), for all k ∈ Z,

which implies eλt /∈ etσ(A) and proves (10).


sma 5878 functional analysis ii€¦ · theorems on perturbations of generators teoremas espectrais...

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