Some results about spectral theory over Fréchet spaces
aa r X i v : . [ m a t h . F A ] F e b SOME RESULTS ABOUT SPECTRAL THEORY OVER FR ´ECHET SPACES ´EDER R´ITIS AND LU´IS SALGE
Abstract.
In this study, we present some differences that arise in the spectral analysis ofpseudodifferential operators with constant coefficients, when we use the Fr´echet structureinstead of the Banach structure.Here we show that this change in the topology implies in significant changes on thebehavior of the Laplace’s spectrum, for instance, the resolvent set vanishes, even with abounded domain.The notion of exponential dichotomy for Fr´echet Spaces introduced by the author and itsconnection with the separation of spectrum and dichotomy inspired us to make this work. Introduction
The spectral theory has a key role in the study of qualitative and quantitative of differentialequations.Our work is inpired in the itinerary presented by [1], the constructs the spectral theoryfor closed operators on Banach Spaces, and here we use the definition of pseudodifferentialoperators given by [6].The initial idea to solve this question and the solutions showed here, were studied by[5] and they have given sufficient conditions for a pseudodifferential operator with constantcoefficients defined on S ( R n ) have its spectrum equal to the range of its symbol, hence thespectrum of this kind of this operator is always a connected set.Particularly, if a ( D ) = ∆ when Ω = R n and X = L ( R n ) we have that σ (∆) = σ c (∆) =[0 , ∞ ) and, if Ω is a limited open set with smooth boundary, then σ (∆ D ) = σ p (∆ D ) = { λ j : j ∈ N } , i.e., is the image of a sequence.The following results show that when considering ∆ : H (0 , π ) ⊂ L loc (0 , π ) −→ L loc (0 , π ),which means to change the domain of the operator and its topology, many problems arrise, for instance, the spectrum turned out to be all the complex plane, which is our biggestcontribution for the theory.2. Preliminary Concepts and Results
Definition 2.1.
Fr´echet Space. Let X a topological vector space. Then X is said a Fr´echetspace if it’s Hausdorff, complete and its topology is given by a countable family of seminorms. Supported by Grant: Capes, Brazil.
Theorem 2.2.
Let ( X, ( p j ) j ∈ N ) a Fr´echet space. Given a subspace M ⊂ X such that M = X, then there exists a non-null g ∈ X ′ that satisfies h g, x i = 0 , ∀ x ∈ M. Proof.
Let x ∈ X \ M , so d ( x , M ) > δ for some δ > , where d ( · , · ) is the distance givenby d ( x, y ) = P ∞ j =1 12 j p j ( x − y )1+ p j ( x − y ) for x, y ∈ X. Hence, follows that d ( x , x ) ≥ δ for any x ∈ M, i.e., δ ≤ ∞ X j =1 j p j ( x − x )1 + p j ( x − x ) == n X j =1 j p j ( x − x )1 + p j ( x − x ) + ∞ X j = n +1 j p j ( x − x )1 + p j ( x − x ) ≤≤ n X j =1 j p j ( x − x )1 + p j ( x − x ) + ∞ X n +1 j , (2.1)for every n ∈ N since p j ( x − y )1 + p j ( x − y ) ≤ , ∀ x, y ∈ X, j ∈ N . Now let n ∈ N such that ∞ X j = n +1 j < δ , and, by the inequality 2.1, follows δ ≤ n X j =1 j p j ( x − x )1 + p j ( x − x ) ≤ C n X j =1 p j ( x − x ) , for every x ∈ M. Define g : M M [ x ] −→ C x + λx λ δ , so g is linear and |h g, x + λx i| ≤ (cid:12)(cid:12)(cid:12)D g, − xλ + λx E(cid:12)(cid:12)(cid:12) = | λ | δ ≤≤ | λ | C n X j =1 p j ((1 /λ ) x + x ) = | λ | C n X j =1 p j ( x + λx ) , i.e., g : M L [ x ] → C is linear and continuous. PECTRAL THEORY 3
By the Hahn-Banach Theorem, there exists G ∈ X ′ such that h G, x + λx i = h g, x + λx i for x ∈ M and λ ∈ C , therefore h G, x i = h g, x i = 0 for x ∈ M and h G, x i = h g, x i = δ/ > (cid:3) Definition 2.3.
Let I ⊂ R n a given interval, we define the Sobolev space H ( I ) as follows H ( I ) = n u ∈ L ( I ); ∃ g ∈ L ( I ) com Z I uφ ′ = − Z I gφ, ∀ φ ∈ C ∞ ( I ) o . In this context, the function g is denoted by u ′ and u ′ is said to be the weak derivative of u. Moreover, the usual topology of H ( I ) is determined by the following norm k u k H ( I ) . = k u k L ( I ) + k u ′ k L ( I ) . Given a natural number m ≥ , the Sobolev space H m ( I ) is defined as H m ( I ) = { u ∈ H m − ( I ); u ′ ∈ H m − ( I ) } and its usual topology is determined by the norm given as k u k H m ( I ) . = k u k L ( I ) + m X | α | =1 k D α u k L ( I ) . Definition 2.4.
Let m ∈ N , we denote the closure of C ∞ c ( I ) in H m ( I ) , with the inducedtopology, by H m ( I ) . Now, for any m ∈ R , we can define H m ( R n ) . = n u ∈ L ( I ); ∃ g α ∈ L ( I ) com Z R n u ∂ α φ = ( − | α | Z R n g α φ, ∀ φ ∈ C ∞ ( R n ) o . Theorem 2.5.
Seja m ∈ N temos que H m ( R n ) = { u ∈ S ′ ( R n ); (1 + | ξ | ) m/ F ( u ) ∈ L ( R n ) } . Al´em disso, a norma k u k . = k (1 + | ξ | ) m/ F ( u ) k L ( R n ) (2.2) ´e equivalente a k · k H m ( R n ) . By 2.5 it’s possible to define the Sobolev spaces for any s ∈ R . Let s ∈ R define H s ( R n ) = { u ∈ S ′ ( R n ); (1 + | ξ | ) s/ F ( u ) ∈ L ( R n ) } . Notice that Ω = R n we have that H m ( R n ) = H m ( R n ). [2] Definition 2.6.
Given an open set Ω ⊂ R n and s ∈ R we define the local Sobolev space oforder s as H sloc (Ω) = { u ∈ D ′ (Ω); φu ∈ H s (Ω) , ∀ φ ∈ C ∞ c (Ω) } . E.R. ARAG ˜AO-COSTA AND L.M. SALGE
Observation 2.7.
The H sloc (Ω) are Fr´echet spaces. Definition 2.8 (Symbol of a Pseudodifferential Operator of Order m ) . Let a ∈ C ∞ (Ω × R m ) and m ∈ R such that for every compact K ⊂ Ω and multi-index α, β ∈ Z n there exists C K,α,β > with the property | ∂ βx ∂ βξ a ( x, ξ ) | ≤ C K,α,β (1 + | ξ | ) m −| β | , ∀ ξ ∈ R m , x ∈ K. The function a is said to be a symbol of a pseudodifferential operator of order m and theclass of all symbols of pseudodifferential operators of order m is denoted by S m (Ω) . Definition 2.9 (Pseudodifferencial Operator) . Let a ∈ S m (Ω) the pseudodifferential operator a ( x, D ) : S ( R n ) → S ( R n ) is defined as a ( x, D ) ψ ( x ) . = Z R n e πxξ ( a ( x, ξ ) F ψ )( ξ ) dξ, ψ ∈ S ( R n ) , where F denotes the Fourier transform. Observation 2.10.
Let a ∈ C ∞ ( R n ) such that exist m ∈ R and for each multi-index α ∈ Z n we can find a C α > that satisfies | ∂ α a ( ξ ) | ≤ C α (1 + | ξ | ) m −| α | , ∀ ξ ∈ R m . Therefore, if we denote a ( x, ξ ) ˙= a ( ξ ) for each x and ξ ∈ R n , then a ∈ S m ( R n ) and a ( D ) . = a ( x, D ) is said to be a pseudodifferential operator with constant coefficients. Example 2.11.
Let Ω ⊂ R n be an open set. Consider m ∈ N and define p ( x, ξ ) = X | α |≤ m a α ( x ) ξ α in which a α ∈ C ∞ (Ω) . It’s not difficult to see that p ∈ S m (Ω) . Definition 2.12.
Given a m ∈ R , a linear operator A : C ∞ c (Ω) → C ∞ (Ω) is said to be a operator of order m if, for every s ∈ R , we can extend A to an operator A s : H s + m (Ω) ⊂ H s + mloc (Ω) → H sloc (Ω) . Theorem 2.13.
Let p ∈ S m (Ω) , hence p ( x, D ) is an operator of order m. Proof.
Given p ∈ S m (Ω) , let us prove that p is of order m. First of all, observe that if a ∈ C ∞ c (Ω)then the following map u au PECTRAL THEORY 5 if of order 0 . Therefore, to complete the proof of the result we need to show that ap ( x, D ) isof order m, i.e., for every s ∈ R ap ( x, D ) : H s + m (Ω) → H sloc (Ω) u ap ( x, D )( u ) , or equally we can suppose that p ( x, ξ ) = 0 for x / ∈ K in which K ⊂ Ω is compact.Define v . = p ( x, D ) u, as ˆ p ( η, ξ ) = Z R n e − πiηx p ( x, ξ ) dx, it implies thatˆ v ( η ) = Z R n e − πiηy [ p ( x, D ) u ]( y ) dy = Z R n (cid:16) Z R n e − πiηy e πiyξ p ( y, ξ )ˆ u ( ξ ) dξ (cid:17) dy == Z R n ˆ p ( η − ξ, ξ )ˆ u ( ξ ) dξ. Notice that its allowed to apply the Fubini theorem, since p ( · , ξ ) has compact support foreach ξ ∈ R n . Furthermore, for each N ∈ N and s ∈ R we have that | ˆ p ( η, ξ ) | ≤ C N (1 + | η | ) − N (1 + | ξ | ) m for η, ξ ∈ R n . Then(1 + | η | ) s | ˆ v ( η ) | ≤ C N Z R n | η − ξ | ) N (1 + | ξ | ) m + s | ˆ u ( ξ ) | dξ ≤≤ C N (cid:16) Z R n | η − ξ | ) N dξ (cid:17) / (cid:16) Z R n (1 + | ξ | ) m + s ) | ˆ u ( ξ ) | dξ (cid:17) / . Moreover, for para
N > n, it follows that Z R n (1 + | η | ) s | ˆ v ( η ) | dη ≤ C k u k H s + m ( R n ) and hence v ∈ H s ( R n ) . Definition 2.14.
Consider a Fr´echet space X um espa¸co de Fr´echet and a linear operator A : D ( A ) ⊂ X → X. The graph of A is given by the set G ( A ) = { ( u, Au ); u ∈ D ( A ) } ⊂ X × X. The operator A is said to be a closed operator if its graph G ( A ) ⊂ X × X is a closed set. E.R. ARAG ˜AO-COSTA AND L.M. SALGE
Definition 2.15.
Considere a Fr´echet space X and a linear operator A : D ( A ) ⊂ X → X. If there exists a linear closed operator A : D ( A ) ⊂ X → X, in which D ( A ) ⊂ D ( A ) and Au = Au for u ∈ D ( A ) , so A is said to closable. Definition 2.16.
Let X a complex Fr´echet space and a linear operator A : D ( A ) ⊂ X −→ X definide in a subspace D ( A ) of X , which is called A domain.The resolvent set of A , denoted by ρ ( A ) , is the set composed by all numbers λ ∈ C suchthat satisfies the following conditions:(a) The operator λ − A : D ( A ) ⊂ X −→ X is injective.(b) The range of λ − A : D ( A ) ⊂ X −→ X is dense in X .(c) The inverse ( λ − A ) − : R ( λ − A ) ⊂ X −→ X is continuous.If λ ∈ ρ ( A ) , the operator ( λ − A ) − : R ( λ − A ) ⊂ X −→ X is called the resolvent of A on λ . Finally, we define the spectrum of A , which is indicated by σ ( A ) , as the set of numbers λ ∈ C the are not in the resolvent set of A , i.e., σ ( A ) = C \ ρ ( A ) . Next we define respectively the point spectrum, residual spectrum and continuous spectrumas following σ p ( A ) . = { λ ∈ C ; λ − A is not injective } ,σ r ( A ) . = { λ ∈ C ; λ − A is injective R ( λ − A ) = X } ,σ c ( A ) . = { λ ∈ C ; λ − A is injective, R ( λ − A ) = X e ( λ − A ) − : R ( λ − A ) → X is continuous } . Henceforth, we have that σ ( A ) = σ p ( A ) ∪ σ r ( A ) ∪ σ r ( A ) . The subsequent result is a consequence of the Closed Graph Theorem:
Lemma 2.17.
Let X be a Fr´echet space. If A : D ( A ) ⊂ X −→ X is a closed operator, then ρ ( A ) = (cid:8) λ ∈ C : λ − A : D ( A ) −→ X is bijective } Proof.
Consider λ ∈ C such that λ − A is bijective and let us show that( λ − A ) − : X → X is continuous, therefore we will conclude that λ ∈ ρ ( A ) . PECTRAL THEORY 7
For that purpose, it suffices to show that the graph of the operator ( λ − A ) − is closed,henceforth by the Closed Graph Theorem for Fr´echet spaces we will get that the operator iscontinuous.Let ( f, u ) ∈ G (( λ − A ) − ) , then there exists a sequence ( f j , u j ) ∈ G (( λ − A ) − ) with( f j , u j ) → ( f, u ) in the topology of X × X and ( λ − A ) − f j = u j , com u j ∈ D ( A ) . Therefore, u j → u ( λ − A ) u j = f j → f which implies that ( u, f ) ∈ G (( λ − A )) = G (( λ − A )) , i.e., u ∈ D ( A ) and f = ( λ − A ) u andthen ( f, u ) ∈ G (( λ − A ) − ) which shows that the resolvent operator has closed graph.In the end, consider λ ∈ ρ ( A ) and let us show that λ − A is a bijection. Since λ ∈ ρ ( A ) , it remains to prove that R ( λ − A ) = X. Consider f ∈ X = R ( λ − A ) then there exists a sequence ( u j )) j ∈ N ⊂ D ( A ) such that f = X − lim j →∞ ( λ − A ) u j . Denote f j . = ( λ − A ) u j , in other words, u j = ( λ − A ) − f j . Now we will prove that ( u j ) j ∈ N is a Cauchy sequence. Since f j → f, then ( f j ) j ∈ N is a Cauchy sequence, which means that f j − f l → , as j, l → ∞ . Hence, u j − u l = ( λ − A ) − f j − ( λ − A ) − f l = ( λ − A ) − ( f j − f l ) → , as l, j → ∞ because ( λ − A ) − is continuous.Consequently, there exists a u ∈ X with u j → u and ( f, u ) ∈ G (( λ − A ) − ) or equivalently( u, f ) ∈ G ( λ − A ) = G ( λ − A ) and we have shown that u ∈ D ( A ) and f = ( λ − A ) u ∈ R ( λ − A ) . Theorem 2.18.
Consider X a Fr´echet space. If A : D ( A ) ⊂ X −→ X is closable and A : D ( A ) ⊂ X −→ X its closure, then σ ( A ) = σ ( A ) . Proof.
Clearly, to show that σ ( A ) = σ ( A ) is the same as to prove that ρ ( A ) = ρ ( A ).So, let λ ∈ ρ ( A ), since D ( A ) ⊂ D ( A ) then we have that λ − A | D ( A ) = λ − A consequently we conclude that λ − A is injective.Let f ∈ X = R ( λ − A ) then there exists u ∈ D ( A ) with ( λ − A ) u = f. By the definition ofdomain and range of the closure of A we have that there exists a sequence ( u j ) j ∈ N ∈ D ( A )with u j → u and ( λ − A ) u j → f, hence f ∈ R ( λ − A ) . E.R. ARAG ˜AO-COSTA AND L.M. SALGE
It remains to prove that ( λ − A ) − : R (( λ − A )) ⊂ X → X is a continuous operator. Therefore, consider a sequence ( f j ) j ∈ N ⊂ R ( λ − A ) such that f j → f ∈ R ( λ − A ) , then there is u ∈ D ( A ) with f = ( λ − A ) u. Now we just have to showthat u j → u. Notice that f j = ( λ − A ) u j = ( λ − A ) u j , u j ∈ D ( A ) ⊂ D ( A ) , or equivalently, u j =( λ − A ) − f j and since ( λ − A ) − is continuous we have that u = ( λ − A ) − f = lim j →∞ ( λ − A ) − f j = lim j →∞ u j . Henceforth ρ ( A ) ⊂ ρ ( A ) . Reciprocally, consider λ ∈ ρ ( A ) so λ − A is injective, X = R ( λ − A ) and( λ − A ) − : R ( λ − A ) ⊂ X → X is continuous.Let us prove that λ − A is bijective. Indeed, if u ∈ D ( A ) with ( λ − A ) u = 0 then there isa sequence ( u j ) ⊂ D ( A ) with u j → u and f j . = ( λ − A ) u j → . Note that 0 ∈ R ( λ − A ) , therefore from the continuity ofthe operator ( λ − A ) − we havethat u j = ( λ − A ) − f j → ( λ − A ) − , in other words, u = 0 and it follows that λ − A is injective.Now, we will prove that λ − A is surjective. Let f ∈ X = R ( λ − A ) , so there exists( u j ) ⊂ D ( A ) with f j = ( λ − A ) u j and f j → f. Hence ( f j ) j is a Cauchy sequence, i.e., f j − f l = ( λ − A )( u j − u l ) → , and since ( λ − A ) − is continuous then u j − u l = ( λ − A ) − ( f j − f l ) → . Therefore we conclude that ( u j ) j is a Cauchy sequence, in other words there is a u ∈ X such that u j → u. So we have that ( u, f ) ∈ G ( λ − A ) = G ( λ − A ) , and then u ∈ D ( A ) and f = ( λ − A ) u ∈ R ( λ − A ) . PECTRAL THEORY 9 Spectral Theory on the Schwartz Space and Tempered Distributions
Let a ∈ S m ( R n ) be a symbol of a pseudodifferential operator of order m. We can intuitivelydefine the transpose operator of a ( D ) by a ( D ) ′ : S ′ ( R n ) → S ′ ( R n ) u a ( D ) ′ u . = F ( a F − u ) , in other words for each ψ ∈ S ( R n ) h a ( D ) ′ u, ψ i = hF ( a F − u ) , ψ i = h u, F − ( a F ψ ) = h u, a ( D ) ψ i . (3.1) Definition 3.1.
Define the reflection operator r : S ( R n ) −→ S ( R n ) given by r ( a )( ξ ) . = a ( − ξ ) . It’s easy to verify that the reflection operator is continuous with the usual topology of theSchwartz space.
Observation 3.2.
We will denote r ( a ) by ˜ a. Futhermore, if ψ ∈ S ( R n ) then F (cid:2) r ( ψ ) (cid:3) ( ξ ) = Z R n e − πiηξ r ( ψ )( η ) dη = Z R n e − πiηξ ψ ( − η ) dη = Z R n e πiηξ ψ ( η ) dη, so F (cid:2) r ( ψ ) (cid:3) ( ψ )( ξ ) = F − ψ ( ξ ) = (cid:0) F ψ (cid:1) ( − ξ ) = r (cid:0) F ψ (cid:1) ( ξ ) and consequently (cid:8) a ( D ) (cid:2) r ( ψ ) (cid:3)(cid:9) ( x ) = Z R n e πixξ a ( ξ ) F (cid:2) r ( ψ ) (cid:3) ( ξ ) dξ = Z R n e πixξ a ( ξ ) (cid:0) F ψ (cid:1) ( − ξ ) dξ = Z R n e πi ( − x ) ξ ˜ a ( ξ ) F ψ ( ξ ) dξ = (cid:2) ˜ a ( D ) ψ (cid:3) ( − x ) = (cid:2) r (˜ a ( D ) ψ ) (cid:3) ( x ) . Therefore a ( D ) = r ◦ ˜ a ( D ) ◦ r. (3.2)In the following result our goal is to study the behavior of the resolvent of pseudodifferentialoperators with constant coefficients considering the topology used on S ( R n ) . We begin, considering the set of all complex numbers λ ∈ C for which • λ − a ( D ) : S ( R n ) −→ S ( R n ) is injective, • R ( λ − a ( D )) = S ( R n ), in which R ( λ − a ( D )) is the closure considering the usualtopology of S ( R n ) and • ( λ − a ( D )) − : R ( λ − a ( D )) ⊂ S ( R n ) → S ( R n ) is continuous with the topologyinduce by S ′ ( R n ), Now a ( D ) ′ is the operator obtained by 3.1 and let ρ ∗ be the set of all complex numbers λ ∈ C which satisfies • λ − a ( D ) ′ : S ′ ( R n ) −→ S ′ ( R n ) is injective, • R ( λ − a ( D ) ′ ) ∗ = S ′ ( R n ), in which R ( λ − a ( D )) ∗ is the closure considering the weakstar topology of S ′ ( R n ) and • ( λ − a ( D ) ′ ) − : R ( λ − a ( D ) ′ ) ⊂ S ′ ( R n ) → S ′ ( R n ) such that for u ∈ R ( λ − a ( D ) ′ )and for every sequence φ j ∈ R ( λ − a ( D ) ′ ) ∩ S ( R n ) with φ j ∗ ⇀ u it is true that( λ − a ( D ) ′ ) − u = S ′ ( R n ) − lim j →∞ ( λ − a ( D ) ′ ) − φ j . (3.3) Theorem 3.3. If a ∈ S m ( R n ) then ρ = ρ ∗ . Proof.
We start proving that ρ ⊂ ρ ∗ . (a) Injection
Given a λ ∈ ρ we will prove that que λ − a ( D ) ′ is injector. For that matter, suppose( λ − a ( D ) ′ ) u = 0, so for each ψ ∈ S ( R n ) we have h u, ( λ − a ( D )) ψ i = h ( λ − a ( D ) ′ ) u, ψ i = 0 , i.e., h u, φ i = 0 , ∀ φ ∈ R ( λ − a ( D )) . Since R ( λ − a ( D )) is a dense set in S ( R n ) it implies that for φ ∈ S ( R n ) h u, φ i = lim j →∞ h u, ( λ − a ( D )) ψ j i = 0 , where ( ψ ) j ∈ N ⊂ S ( R n ) such that φ = S − lim j →∞ ( λ − a ( D )) ψ j . Therefore u = 0and it follows the injectivity of λ − a ( D ) ′ . (b) Range Density
Now let us show that the set R ( λ − a ( D ) ′ ) is dense in S ′ ( R n ) . With that intention,we will first prove that( λ − a ( D ) ′ ) | Sc : S ( R n ) → S ( R n )is dense in S ( R n ) and using the equality λ − a ( D ) ′ = r ◦ ( λ − a ( D )) ◦ r we willconclude that the range of ( λ − a ( D ) ′ ) | S ( R n ) is dense in S ( R n ) . Indeed, given a function ψ ∈ S ( R n ) we have that r ( ψ ) ∈ S ( R n ) . Since the set R ( λ − a ( D )) is dense in S ( R n ) , there exists ( φ j ) ⊂ S ( R n ) with r ( ψ ) = S − lim j →∞ ( λ − a ( D )) φ j . PECTRAL THEORY 11
In other words, r ( ψ ) = S − lim j →∞ ( λ − a ( D )) r ( ψ j ) , where r ( ψ j ) = φ j . Once r : S ( R n ) → S ( R n )is continuous considering the usual topology of S ( R n ) it follows that ψ = r (cid:16) S − lim j →∞ ( λ − a ( D )) r ( ψ j ) (cid:17) = S − lim j →∞ r [( λ − a ( D )) r ( ψ j )] == S − lim j →∞ ( λ − a ( D ) ′ ) ψ j . Hence, ( λ − a ( D ) ′ ) ψ j ∗ ⇀ ψ which implies S ′ ( R n ) = S ( R n ) ∗ = R ( λ − a ( D ) ′ ) ∗ , where the above closures are calculated in the weak* topology and the density isproven.(c) Continuity of the Resolvent Operator
Hereby we will show that the operator( λ − a ( D ) ′ ) − : R ( λ − a ( D ) ′ ) ⊂ S ′ ( R n ) → S ′ ( R n )satisfies the condition 3.3.First, notice that the reflexion operator r is the inverse of itself and that que 3.2is also true when the operator is considered on S ′ ( R n ) . Then we have that( λ − a ( D ) ′ ) − = r ◦ ( λ − a ( D )) − ◦ r, so it remains to prove that the operator( λ − a ( D )) − : R ( λ − a ( D )) ⊂ S ′ ( R n ) → S ′ ( R n )satisfies 3.3.In order for us don’t get confused we will denote the next sets ( λ − a ( D ))[ S ( R n )] ⊂ S ′ ( R n ) and ( λ − a ( D ) ′ )[ S ( R n )] ⊂ S ′ ( R n ) , respectively by, R and R ′ . Consider u ∈ R ( λ − a ( D ) ′ ) ⊂ S ′ ( R n ) , so there are w ∈ S ′ ( R n ) and a sequence( ψ j ) j ∈ N ⊂ S ( R n ) such that u = ( λ − a ( D ) ′ ) w and ψ j ∗ ⇀ w as j → ∞ . Define φ j . = ( λ − a ( D ) ′ ) ψ j , j ∈ N , therefore φ j ∗ ⇀ ( λ − a ( D ) ′ ) w = u as j → ∞ . Particularly, it’s true that φ j − φ l ∗ ⇀ , j, l → ∞ . Since ( λ − a ( D )) − | R is continuous with the weak* topology, it follows that ( λ − a ( D ) ′ ) − | R ′ is also continuous when considering the same topology and so( λ − a ( D ) ′ ) − ( φ j − φ l ) ∗ ⇀ , j, l → ∞ . In other words, [( λ − a ( D ) ′ ) − φ j ] j ∈ N is a Cauchy sequence on S ′ ( R n ) with the weak*topology, which implies that there a unique v ∈ S ′ ( R n ) with( λ − a ( D ) ′ ) − φ j ∗ ⇀ v as j → ∞ . Moreover, for any φ ∈ S ( R n ) h ( λ − a ( D ) ′ ) − φ j , ( λ − a ( D )) φ i = h φ j , φ i −→ h u, φ i and in addition h ( λ − a ( D ) ′ ) − φ j , ( λ − a ( D )) φ i −→ h v, ( λ − a ( D )) φ i . It implies that, for φ ∈ S ( R n ) , we have h u, φ i = h v, ( λ − a ( D )) φ i = h ( λ − a ( D ) ′ ) v, φ i , that is to say S ′ ( R n ) − lim j →∞ ( λ − a ( D ) ′ ) − φ j = v = ( λ − a ( D ) ′ ) − u == ( λ − a ( D ) ′ ) − [ S ′ ( R n ) − lim j →∞ φ j ] . In conclusion, given sequences ( ζ j ) j ∈ N , ( ϕ j ) j ∈ N ⊂ R ( λ − a ( D ) ′ ) ∩ S ( R n ) where ζ j ∗ ⇀ u and ϕ j ∗ ⇀ u, or equivalently ζ j − ϕ j ∗ ⇀ S ( R n ) , hence S ′ ( R n ) lim j →∞ ( λ − a ( D ) ′ ) − ( ϕ j − ζ j ) = 0since, by hypotheses( λ − a ( D ) ′ ) − : R ( λ − a ( D ) ′ ) ∩ S ( R n ) → S ( R n )is continuous with topology induce by S ′ ( R n ) . (a) Injectivity
Now, given a number λ ∈ ρ we shall prove that λ ∈ ρ ∗ . We start noticing that for u ∈ R ( λ − a ( D ) ′ ) and ψ ∈ S ( R n ) we have the following h ( λ − a ( D ) ′ ) − u, ( λ − a ( D )) ψ i = h ( λ − a ( D ) ′ )( λ − a ( D ) ′ ) − u, ψ i = h u, ψ i . (3.4)Provided that ψ ∈ S ( R n ) such that ( λ − a ( D )) ψ = 0 , let us show that ψ ≡ . From 3.4 it follows h u, ψ i = h ( λ − a ( D ) ′ ) − u, ( λ − a ( D )) ψ i = 0 , ∀ u ∈ R ( λ − a ( D ) ′ ) . PECTRAL THEORY 13
Given a distribuition v ∈ S ′ ( R n ) there exists a sequence u j ∈ R ( λ − a ( D ) ′ ) thatconverges to v, so h v, ψ i = lim j →∞ h u j , ψ i = 0 , which means that h v, ψ i = 0 for any distribution v ∈ S ′ ( R n ) . If x ∈ R n , take the distribution δ x ∈ S ′ ( R n ) , which is Dirac’s Delta distributioncomposed with a translation by x . Then ψ ( x ) = h δ x , ψ i = 0and we have obtained that ψ ≡ λ − a ( D ) is injector.(b) Density of the Range
Consider a distribution u ∈ S ′ ( R n ) where h u, ( λ − a ( D )) φ i = 0 , φ ∈ S ( R n ) , which means that h ( λ − a ( D ) ′ ) u, φ i = h u, ( λ − a ( D )) φ i = 0 . We have that ( λ − a ( D ) ′ ) u = 0 and by the injectivity of ( λ − a ( D ) ′ ) , we get that u = 0 . So we conclude that R ( λ − a ( D )) = S ( R n ) . (c) Continuity of the Resolvent Operator
At the end, we shall prove that( λ − a ( D )) − : R ( λ − a ( D )) ⊂ S ( R n ) → S ( R n )is continuous with the topology induced by S ′ ( R n ) . For that matter, consider a sequence ( ψ j ) j ∈ N ⊂ R ( λ − a ( D )) ∩ S ( R n ) such that ψ j ∗ ⇀ , which is the same as h ψ j , φ i −→ , ∀ φ ∈ S ( R n ) . Notice that the resolvent( λ − a ( D ) ′ ) − : R ( λ − a ( D ) ′ ) ⊂ S ′ ( R n ) → S ′ ( R n )satisfies 3.3.Now for u ∈ R ( λ − a ( D )) ⊂ S ′ ( R n ) there is a sequence ( φ j ) j ∈ N ⊂ R ( λ − a ( D )) ∩ S ( R n ) with φ j ∗ ⇀ u, and then( λ − a ( D )) − u = r [( λ − a ( D ) ′ ) − r ( S ′ ( R n ) − lim j →∞ φ j )] == S ′ ( R n ) − lim j →∞ [ r ◦ ( λ − a ( D ) ′ ) − ◦ r ] φ j == S ′ ( R n ) − lim j →∞ ( λ − a ( D )) − φ j , which implies that the following resolvent( λ − a ( D )) − : R ( λ − a ( D )) ⊂ S ′ ( R n ) → S ′ ( R n )also satisfies the property 3.3.Since 0 ∈ S ′ ( R n ) and ψ j ∗ ⇀ , we get that S ′ ( R n ) − lim j →∞ ( λ − a ( D )) − ψ j = ( λ − a ( D )) − S ′ ( R n ) − lim j →∞ ψ j == S ′ ( R n ) − lim j →∞ ( λ − a ( D )) − S ′ ( R n ) , of the resolvent( λ − a ( D )) − : R ( λ − a ( D )) ⊂ S ( R n ) → S ( R n )is concluded.4. General Case for a Linear Differential Operator
Here we present the main result of this paper which is a study for the spectrum of a lineardifferential operator with constant coefficients. Given a symbol a ∈ S m ( R ) we studied thespectrum of the following operator a ( D ) : H s + m ( I ) ⊂ H sloc ( I ) → H sloc ( I )and compared with the spectrum of its dual a ( D ) ∗ : D ( a ( D ) ∗ ) ⊂ H − sc ( I ) → H − sc ( I )such that D ( a ( D ) ∗ ) . = { g ∈ H − sc ( I ); g ◦ a ( D ) : H s + mloc ( I ) ⊂ H sloc ( I ) → C ´e cont´ınuo } . Before stating the next theorem we need a definition and a more simplified version of theTheorem 6 .
36 from [4, pg - 216].
Definition 4.1.
Given a ( D ) : S ′ ( R n ) → S ′ ( R n ) a differential operator with constantcoefficients, we say that a ( D ) is hypoelliptic if, for any u ∈ S ′ ( R n ) , it’s true thatsing supp [ a ( D ) u ] = sing supp u. Theorem 4.2 (H¨ormander) . Let a ( ξ ) = P | α |≤ m a α ξ α be a symbol of a differential operator, a ( D ) , of order m > . The following are equivalent statements:(1) a ( D ) is hypoelliptic;(2) There exists δ, C, R > such that | a ( α ) ( ξ ) | ≤ C | ξ | − δ | α | | a ( ξ ) | for every α and every ξ ∈ R n such that | ξ | > R ; (3) There exists δ > such that if f ∈ H sloc (Ω) , in which Ω ⊂ R n is an open set, everysolution of a ( D ) u = f is an element of H s + δmloc (Ω) . PECTRAL THEORY 15
Theorem 4.3.
Let a ( D ) : H s + m ( I ) ⊂ H sloc ( I ) → H sloc ( I ) an hypoelliptic differential operatorof order m. Then there exists < δ ≤ such that H − s + mc ( I ) ⊂ D [ a ( D ) ∗ ] = H − s + δmc ( I ) where a ( D ) ∗ is the dual of a ( D ) . Proof.
First of all, let u ∈ H s + m and ψ, χ K ∈ C ∞ c ( K ) , where K ⊂ I is fixed compact and χ K ≡ K. It follows that |h ψ, a ( D ) u i| = (cid:12)(cid:12)(cid:12)D ψ, X | α |≤ m a α ∂ α u E(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)D X | α |≤ m ( − | α | a α ∂ α ψ, u E(cid:12)(cid:12)(cid:12) == (cid:12)(cid:12)(cid:12)D X | α |≤ m ( − | α | a α ∂ α ψ, χ K u E(cid:12)(cid:12)(cid:12) ≤ C k ψ k H − s + m ( R ) k χ K u k H s ( R ) , since k X | α |≤ m ( − | α | a α ∂ α ψ k H − s ( R ) = k (1 + ξ ) − s/ X | α |≤ m a α ξ α ψ ( ξ ) k L ( R ) ≤≤ C k (1 + ξ ) ( − s + m ) / F ψ k L ( R ) ≤ k ψ k H − s + m ( R ) and supp [ P | α |≤ m ( − | α | a α ∂ α ψ ] ⊂ supp ψ ⊂ K. Hence, if g ∈ H − s + mc ( K ) there exists a sequence ( ψ l ) l ∈ N ⊂ H − s + mc ( K ) that converges for ψ with the norm k · k H − s + m ( R ) so |h g, a ( D ) u i| = lim l →∞ |h ψ l , a ( D ) u i| ≤ lim l →∞ C k ψ l k H − s + m ( R ) k χ K u k H s ( R ) == C k g k H − s + m ( R ) k χ K u k H s ( R ) . Furthermore, since g ∈ S ′ ( R n ) thus X | α |≤ m a α ∂ α g = S ′ ( R n ) − lim l →∞ X | α |≤ m a α ∂ α ψ l and supp a ( D ) ∗ g ⊂ supp g ⊂ K. So we conclude that g ∈ D [ a ( D ∗ )] , i.e., H − s + mc ( I ) ⊂ D [ a ( D ) ∗ ] . It remains to show that there exists 0 < δ ≤ D [ a ( D ) ∗ ] ⊂ H − s + δm ( I ) . Indeed, given g ∈ D [ a ( D ) ∗ ] we have, by definition of the domain of the operator a ( D ) ∗ , that there exists M > |h g, a ( D ) u i| ≤ M p j ( u ) , u ∈ H s + m ( I ) , where p j ( · ) is a seminorm of the space H sloc ( I ) . Note that a ( D ) ∗ g = P | α |≤ m ( − | α | a α ∂ α g as distributions in S ′ ( R n ) and it follows thatsupp a ( D ) ∗ g ⊂ supp g. Besides that, the map T : C ∞ c ( I ) → C φ
7→ h g, a ( D ) φ i is such that |h g, a ( D ) φ i| ≤ M p j ( φ ) ≤ M k φ k H s ( R ) and then a ( D ) ∗ g ∈ [ H s ( R )] ′ = H − s ( R ) . Joining the facts we conclude that a ( D ) ∗ g ∈ H − sc ( I ) and hence a ( D ) ∗ g ∈ H − sloc ( I ) , since a ( D ) is hypoelliptic we have that a ( D ) ∗ is also hypoelliptic so, from 4.2, there exists 0 < δ ≤ g ∈ H − s + δmloc ( I ) . In other words, g ∈ H − s + δmloc ( I ) com supp g ⊂ I, i.e., g ∈ H − s + δmc ( I )and the proof is complete. (cid:3) Observation 4.4.
Note that if a ( D ) ellptic then δ = 1 and consequently D ( a ( D ) ∗ ) = H − s + mc ( I ) . Now we will show the results obtained comparing the sets σ ( a ( D )) and σ ( a ( D ) ∗ ) , where a ( D ) : H s + m ( I ) ⊂ H sloc ( I ) → H sloc ( I )is a differential operator with constant coefficients and a ( D ) ∗ : D ( a ( D ∗ ) ⊂ H − sc ( I ) → H − sc ( I )is its dual and hypoelliptic. Theorem 4.5.
Under the above conditions, we have the following σ p ( a ( D )) ∪ σ r ( a ( D )) ⊂ σ p ( a ( D ) ∗ ) ∪ σ r ( a ( D ) ∗ ) ,σ p ( a ( D )) ∗ ⊂ σ p ( a ( D )) ∪ σ r ( a ( D )) e σ r ( a ( D ) ∗ ) ⊂ σ p ( a ( D )) ∪ σ c ( a ( D )) . Proof.
We proceed the proof showing separately that σ p ( a ( D )) ⊂ σ ( a ( D ) ∗ ) , σ r ( a ( D )) ⊂ σ ( a ( D ) ∗ ) and σ c ( a ( D )) ⊂ σ ( a ( D ) ∗ ) . Step I: σ r ( a ( D )) ⊂ σ p ( a ( D ) ∗ )Given λ ∈ σ r ( a ( D )) we have that λ − a ( D ) is injector and R ( λ − a ( D )) = H sloc ( I ) , from2.2 follows that there exists g = 0 a functional belonging to H − sc ( I ) with h g, ( λ − a ( D )) u i = 0 , ∀ u ∈ H s + m ( I ) . From the above identity, it follows that g ∈ D [ a ( D ) ∗ ] = H − s + mc ( I ) such that h ( λ − a ( D ) ∗ ) g, u i = h g, ( λ − a ( D )) u i = 0 , ∀ u ∈ H s + m ( I ) , i.e., ( λ − a ( D ) ∗ ) g = 0 with g = 0 and so λ ∈ σ p ( a ( D ) ∗ ) . Step II: σ p ( a ( D ) ∗ ) ⊂ σ p ( a ( D )) ∪ σ r ( a ( D ))On the other hand, given λ ∈ σ p ( a ( D ) ∗ ) it means that ( λ − a ( D ) ∗ ) g = 0 for some g ∈ H − s + mc ( I ) non-null. Thus h ( λ − a ( D ) ∗ ) g, u i = 0 , ∀ u ∈ H sloc ( I ) . Particularly, it is true that h g, ( λ − a ( D )) u i = h ( λ − a ( D ) ∗ ) g, u i = 0 , ∀ u ∈ H s + m ( I ) . PECTRAL THEORY 17 where g = 0 and, once more by 2.2, we conclude that R ( λ − a ( D )) = H sloc ( I ) and so λ ∈ σ p ( a ( D )) ∪ σ r ( a ( D )) . Step III: σ p ( a ( D )) ⊂ σ p ( a ( D ) ∗ ) ∪ σ r ( a ( D ) ∗ )If λ ∈ σ p ( a ( D )) then, for a non-null u ∈ H s + m ( I ) , ( λ − a ( D )) u = 0 . Thereafter, for any g ∈ H − sc ( I ) holds h g, ( λ − a ( D )) u i = 0 and, particularly, h g, ( λ − a ( D )) u i = 0 , ∀ g ∈ D ( a ( D ) ∗ ) = H − s + mc ( I ) . We only have two possibilities, ( λ − a ( D ) ∗ ) is injective or not. If λ − a ( D ) ∗ is not injectivethen λ ∈ σ p ( a ( D ) ∗ ) ⊂ σ ( a ( D ) ∗ ) . If λ − a ( D ) ∗ is injective, for g ∈ D ( a ( D ) ∗ ) , then h ( λ − a ( D ) ∗ ) g, u i = h g, ( λ − a ( D ) u i = 0 , because λ ∈ σ p ( a ( D )) . Suppose that R ( λ − a ( D ) ∗ ) = H − sc ( I ) , by the previous identity wemust have that u = 0 , but it contradicts the inicial hypothesis, hence R ( λ − a ( D ) ∗ ) = H − sc ( I )and we conclude σ p ( a ( D )) ⊂ σ p ( a ( D ) ∗ ) ∪ σ r ( a ( D ) ∗ ) . Briefly, we have concluded that σ p ( a ( D )) ∪ σ r ( a ( D )) ⊂ σ p ( a ( D ) ∗ ) ∪ σ r ( a ( D ) ∗ ) . Step IV: σ r ( a ( D ) ∗ ) ⊂ σ p ( a ( D )) ∪ σ c ( a ( D ))Consider λ ∈ C with λ − a ( D ) injective and R ( λ − a ( D )) = H sloc ( I ) , i.e., λ / ∈ σ p ( a ( D )) ∪ σ r ( a ( D )) so, by the previous steps, we must have λ / ∈ σ p ( a ( D )) ∗ , that is to say that λ − a ( D ) ∗ is injective. Let us show that if ( λ − a ( D )) − : R ( λ − a ( D )) ⊂ H sloc ( I ) → H sloc ( I ) is continuous,then R ( λ − a ( D ) ∗ ) = H − sc ( I ) , i.e., if λ ∈ ρ ( a ( D )) it follows λ / ∈ σ r ( a ( D ) ∗ ) or, equivalently, σ r ( a ( D ) ∗ ) ⊂ σ ( a ( D )) . Firstly, we must prove the following identities (cid:2) ( λ − a ( D )) − (cid:3) *( λ − a ( D ) ∗ ) g | R ( λ − a ( D )) = g | R ( λ − a ( D )) , g ∈ D ( a ( D ) ∗ ) . (4.1)( λ − a ( D ) ∗ )[( λ − a ( D )) − ]* g = g, g ∈ D [( λ − a ( D )) − ]* . (4.2)Indeed, if g ∈ D ( a ( D )*) it follows that h ( λ − a ( D )*) g, ( λ − a ( D )) − f i = h g, f i , f ∈ R ( λ − a ( D ))and so ( λ − a ( D ) ∗ ) ◦ g is continuous, proving that R ( λ − a ( D )*) ⊂ D (cid:2) ( λ − a ( D )) − (cid:3) * and4.1 is true.Now let g ∈ D (cid:2) ( λ − a ( D )) − (cid:3) * then h [( λ − a ( D )) − ]* g, ( λ − a ( D )) u i = h g, ( λ − a ( D )) − ( λ − a ( D )) u i = h g, u i , u ∈ H s + m ( I )hence [( λ − a ( D )) − ]* g ◦ ( λ − a ( D )) : H s + m ( I ) → C is continuous with the induced topologyby H sloc ( I ) , thus [( λ − a ( D )) − ]* g ∈ D ( a ( D )*) the identity 4.2 is true, i.e., g = ( λ − a ( D )*)[( λ − a ( D )) − ]* g, g ∈ D [( λ − a ( D )) − ]*and so D [( λ − a ( D )) − ]* ⊂ R ( λ − a ( D ) ∗ ) , from which we conclude the identity D [( λ − a ( D )) − ]* = R ( λ − a ( D ) ∗ ) . By the continuity of ( λ − a ( D )) − then R ( λ − a ( D ) ∗ ) = D [( λ − a ( D )) − ] ∗ = H − sc ( I ) . Therefore, if λ ∈ ρ ( a ( D )) it follows that λ / ∈ σ r ( a ( D ) ∗ ) , i.e., σ r ( a ( D ) ∗ ) ⊂ σ ( a ( D )) . By the first step, we have that σ r ( a ( D ) ∗ ) ⊂ σ p ( a ( D )) ∪ σ c ( a ( D )) , since the first stepimplied that σ r ( a ( D ) ∗ ) ∩ σ r ( a ( D )) = ∅ . (cid:3) Theorem 4.6.
Under the above conditions we have that σ r ( a ( D ) ∗ ) = σ c ( a ( D )) = C so σ ( a ( D )) = σ ( a ( D ) ∗ ) = C . Proof.
Notice that λ ∈ σ r ( a ( D ) ∗ ) iff λ − a ( D ) ∗ is injective and R ( λ − a ( D ) ∗ ) = H − sc ( I ) . Firstly, let’s show that σ p ( a ( D ) ∗ ) = ∅ . Indeed, if ( λ − a ( D ) ∗ ) g = 0 for some g ∈ D ( a ( D ) ∗ )it follows that g ( t ) = P j ≤ m C j e β j t for C ∈ C and β , . . . , β m ∈ C roots of λ − a ∗ ( ξ ) = λ − X | α |≤ m ( − | α | a α ξ α . but since D ( a ( D ) ∗ ) ⊂ H − s + δmc ( I ) , g has compact support it implies that g ≡ σ p ( a ( D ) ∗ ) = ∅ . We claim that there exists a non-null u ∈ H sloc ( I ) such that h u, ( λ − a ( D ) ∗ ) g i = 0 , ∀ g ∈ D ( a ( D ) ∗ ) (4.3)and so it follows that R ( λ − a ( D ) ∗ ) = H − sc ( I ) . Indeed, notice that if u ∈ C ∞ ( I ) ⊂ H sloc ( I ) then h u, ( λ − a ( D ) ∗ ) g i = h ( λ − a ( D )) u, g i , g ∈ D ( a ( D ) ∗ ) . So take u ( t ) = e ξ t , where ξ ∈ C is a root of the polynomial( λ − a ( ξ )) = λ − X | α |≤ m a α ξ α then u ∈ C ∞ ( I ) is non-null and satisfies ( λ − a ( D )) u = 0 , henceforth h u, ( λ − a ( D ) ∗ ) g i = h ( λ − a ( D )) u, g i = 0 , ∀ g ∈ D ( a ( D ) ∗ )and follows that σ r ( a ( D ) ∗ ) = C . It remains to show that σ p ( a ( D )) = ∅ . First notice that ( λ − a ( D )) u = 0 implies u ( t ) = P j ≤ m D j e β j t for D j ∈ C , β , . . . , β m ∈ C are roots of 4.3. However, in order to u ∈ For m = 2 and λ − a ∗ ( ξ ) has only the root β , proceed as in ODE, i.e., the solution is given by g ( t ) = C e β t + C te β t . For the general case, proceed the same way for every root with multiplicity biggerthan one. If m = 2 and λ − a ( ξ ) has only the root β , proceed as in ODE, i.e., a solution is given by u ( t ) = D e β t + D te β t . For the general case, proceed the same way for every root with multiplicity bigger thanone.
PECTRAL THEORY 19 H s + m ( I ) we must have that u to be zero on ∂I and then u ≡ , i.e., σ p ( a ( D )) = ∅ . Finally, C = σ r ( a ( D ) ∗ ) ⊂ σ p ( a ( D )) ∪ σ c ( a ( D )) = σ c ( a ( D )) and so σ c ( a ( D )) = C . (cid:3) References [1] A. E. Taylor, Introduction to functional analysis, John Wiley, 1958.[2] H. Brezis,
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