Spectral Theory for Bounded
Self-adjoint Operators
Roland Strömberg
U.U.D.M. Project Report 2006:5
Examensarbete i matematik, 20 poäng
Handledare och examinator: Andreas Strömbergsson Juni 2006
Department of Mathematics
1 Introduction
Spectral theory for a self-adjoint operator is a quite complicated topic. If the operator at hand is compact the theory becomes, if not trivial, less complicated. Consider first the case of a self-adjoint operator A : V → V with V finite dimensional. The complete spectral decomposition of A can be stated in a quite elementary fashion: We know that we can represent A with a matrix [A] and compute its eigenvalues and corresponding eigenvec- tors. The spectral theorem says that there exists a total orthonormal set of eigenvectors ψ1· · · ψn such that [A] can be diagonalized, or equivalently that there exists a unitary matrix [U ] and eigenvalues λ1, · · · , λn so that for each ψ ∈ H, ([U][A][U−1]ψi) = λiψi. This result carries over to the case where V is a infinite dimensional separable Hilbert space if we assume that V is a compact. In my thesis I have focused on bounded self-adjoint operators in full generality.
For these operators I present the spectral theorem in three different forms; functional calculus form, multiplication operator form and projections- valued measure form. The proofs of these theorems that I have used are classical results in functional analysis, but there are many more in the math- ematical literature. For example, a general version of the spectral theorem for self-adjoint operators by Stone and von Neumann (1929-1932), and more recently a version concerning unbounded operators in Hilbert spaces was stated by E.B. Davies in 1992. This version is discussed in a thesis titled
”Functional calculus” by C. Illand at Uppsala university in 2004. In my the- sis I have chosen to follow the the proofs in the book ”Functional analysis, Book 1” by Reed and Simon, [2], and my main work has been concerned with filling in several if not all of the details which are left to the reader in [2].
Of these three forms the second one is the most transparent, in the sense that it is reminiscent of the spectral theorem for compact self-adjoint matrices stated above. We can leisurely say that the second form says that every bounded self-adjoint operator is a multiplication operator on a suitable measure space. Now, it is quite evident that we need a rather heavy mathematical machinery even to state the theorem. We clearly need some measure theory and for that we introduce the so called spectral measures.
But the first step in reaching the second formulation of the spectral theorem is to define the functional calculus. We formulate this calculus first for the continuous case then for Borel sets. The first theorem (Theorem 1) concerning continuous functional calculus states that there exists a map from the set of continuous functions on the spectrum of a bounded self-adjoint operator A, σ(A) to the set of bounded linear operators on a Hilbert space H, L(H), i.e. f 7→ f(A). This map has a number of desirable properties, for example it is a ”conjugate-homomorphism” and continuous. The first version of the spectral theorem is basically Theorem 1 extended to Borel
functions on R. The fact that we are dealing with a separable Hilbert space H enables us to decompose H into a direct sum of invariant subspaces. Then on each subspace we can find a unitary operator so that we get a relation analogous to the matrix case above. This enables us to state the spectral theorem in its second form.
To state the spectral theorem in its third form, projection-valued mea- sure form, we need to introduce some projections as the name indicates. If we have a bounded self-adjoint operator A and a Borel set Ω at hand we call PΩ ≡ χΩ(A) a spectral projection of A. We further call a family of such projections {PΩ| Ω is an arbitrary Borel set } a projection valued measure if it obeys a number of certain conditions. The third form of the spectral theorem says that there is a one-one correspondence between a bounded self-adjoint operator A and a bounded projection valued measure PΩ.
The contents of my thesis also involve a section about unbounded op- erators. The theory of such operators are of uttermost importance since one of the most recurring operator in mathematical analysis, the differential operator, is unbounded. This section is merely a brief introduction and does not exhaust the topic of unbounded operators.
2 The Functional calculus
Let A be a given fixed operator and f a continuous function. How can one define f (A) to make sense? If f is given by a finite polynomial Pn
k=0akxk we want f (A) to be Pn
k=0akAk. We also want f to be well defined in the sense of convergence of the sum. If f is given by P∞
k=0akxk with radius of convergence R and ||A|| < R then P∞
k=0akAk converges in L(H) (the set of bounded linear operators on H). So if f is given by an infinite series then f (A) is given byP∞
k=0ak· Ak. The theorem below states the existence and uniqueness of an operator between the family of all continuous complex functions defined on the spectrum of A, and L(H).
Theorem 1 (Continuous functional calculus.) Let A be a bounded self-adjoint operator on a Hilbert space H. Then there exists a unique map Φ : C(σ(A)) → L(H) with the following properties:
(a) Φ is an algebraic ∗ − homomorphism, that is Φ(fg) = Φ(f)Φ(g);
Φ(λf ) = λ Φ(f ); Φ(1) = I; Φ(f + g) = Φ(f ) + Φ(g); Φ(f) = Φ(f )∗. (b) Φ is continuous, that is ||Φ(f)||L(H) ≤ C ||f||∞.
(c) If f (x) = x then Φ(f ) = A.
Moreover Φ has the additional properties.
(d) IfAψ = λψ then Φ(f )ψ = f (λ)ψ.
(e) σ[Φ(f )] = {f(λ) | λ ∈ σ(A)}.
(f) If f ≥ 0, then Φ(f) ≥ 0.
(g) ||Φ(f)|| = ||f||∞.
We first prove that (a) and (c) uniquely determine Φ(P ) for any poly- nomial P (x). First let P be identity, then Φ(P ) = P (A) = A. If P (x) = x2 then Φ(P ) = A2 (by property (a)). By induction one obtains that if P (x) = xn then Φ(P ) = An. Now the additive homomorphism property asserts that if P (x) =Pn
k=0akxk then Φ(P ) =Pn
k=0akAk.
Weierstrass theorem states that the set of polynomials is dense in C(σ(A)) so the core of the proof is to prove that ||P (A)||L(H) = ||P (x)||C(σ(A)) ≡ supλ∈σ(A)|(P (λ))|.
Existence and uniqueness of Φ then follow from the Bounded Linear Transform theorem. To prove the equality between the norms above a special case of (e) that holds for arbitrary bounded operators is proved.
Lemma 1 Let P (x) =Pn
k=0akxkand P (A) =Pn
k=0akAk. Then σ[P (A)] = {P (λ)|λ ∈ σ(A)}.
Proof. Take arbitrary λ ∈ σ(A). Then A − λ does not have an inverse in L(H). The fact that x = λ is a root of P (x) − P (λ) enables the factor- ization of P (x) − P (λ) as (x − λ)Q(x), where Q is a polynomial with lower degree than P . Now P (A) − P (λ) = (A − λ) Q(A), so P (A) − P (λ) fails to have an inverse in L(H). Hence P (λ) ∈ σ(P (A)). For µ ∈ σ(P (A)) let P (x) − µ = a(x − λ1) · · · (x − λn).
If no λi∈ σ(A) then (P (A) − µ)−1 = a−1(A − λ1)−1· · · ·(A − λn)−1 exists.
But this contradicts that µ ∈ σ(A).
Thus λi∈ σ(A) for some i and hence µ = P (λ) for some λ ∈ σ(A). 5 In order to prove Theorem 1 one more Lemma is needed.
Lemma 2 Let A be a bounded self-adjoint operator. Then ||P (A)|| = supλ∈σ(A)|P (λ)|.
Proof. Before writing out the proof a few preliminaries are needed.
First, for a bounded operator T on a Hilbert space it holds that ||T∗T || =
||T ||2 (cf. [1, Theorem 3.9-4(e)]). Also, if P (A) =Pn
k=0anAnthen P (A)∗ = (Pn
k=0anAn)∗ = Pn
k=0an An = P (A). Further (P P )(A) is self-adjoint:
(P P )(A)∗ = P (A)∗(P (A))∗ = P (A)P (A)∗∗ = P (A)P (A) = (P P )(A). So supλ∈σ(P P (A))|λ| = r(P P (A)) = ||P P (A)||, where r is the spectral radius of P P (A). And then by Lemma 1 : supλ∈σ(P P (A))|λ| = supλ∈σ(A)|P P (λ)|. So
||P (A)||2 = ||P (A)∗P (A)|| = ||P P (A)|| = sup
λ∈ σ(P P (A))
|λ|
= sup
λ∈σ(A)|P P (λ)| = sup
λ∈σ(A)|P (λ)||P (λ)|
= sup
λ∈σ(A)|P (λ)|2= ( sup
λ∈σ(A)|P (λ)|)2.
5
Proof of Theorem 1. Let Φ(P ) = P (A). Then ||Φ(P )||L(H) =
||P (A)||L(H) = ||P ||C(σ(A)) so Φ can be extended to the closure of poly- nomials in C(σ(A)). Due to continuity it is enough to verify (a),(b),(d),(g) for polynomials f and g. The set of polynomials over C constitues a commu- tative algebra containing 1, complex conjugate and separates points. The complex Stone-Weierstrass theorem tell us that the closure of the above set is C(σ(A)). Starting with (a): Let P and Q be polynomials. Then
Φ(P Q) = P Q(A) = P (A)Q(A) = Φ(P )Φ(Q);
Φ(P + Q) = (P + Q)(A) = P (A) + Q(A) = Φ(P ) + Φ(Q);
Φ(λP ) = λP (A) = λΦ(P );
Φ(1) = 1(A) = I;
Φ(P ) = P (A) = P (A)∗.
(b) and (g): Using Lemma 1 yields ||Φ(P )||L(H) = ||P ||C(σ(A)) ≡ supλ∈σ(A)|P (λ)| =
||P ||∞.
(c) P (x) = x =⇒ ΦP = P (A) = 1A = A.
To prove uniqueness, suppose eΦ obeys (a),(b) and (c) then Φ agrees with Φ on polynomials and thus by continuity on C(σ(A)).
(d) If Aψ = λψ, then Φ(P )ψ = P (A)ψ = (Pn
k=0an· An)ψ = (c0+ c1A + c2A2+ ...c2An)ψ = c0ψ + c1Aψ + c2A2ψ + ...c2Anψ = (c0+ c1λ + c2λ2+ ...c2λn)ψ = P (λ)ψ. So applying continuity yields the desired result.
(f) If f ≥ 0 then there is a real function g such that f = g2 and g ∈ C(σ(A)). So Φ(f ) = Φ(g2) = Φ(g)Φ(g) = Φ(g)2, where Φ(g) is self-adjoint.
So Φ(f ) ≥ 0.
(e) The proof of this fact is left as an exercise in [2]. First let λ /∈ Range(f ) and g = (f − λ)−1. Then Φ(g) = Φ((f − λ)−1) = (Φ(f − λ))−1 = ((f − λ)(A))−1 = (f (A) − λ)−1 = (Φ(f ) − λ)−1. If on the other hand λ ∈ Range(f) then there is a µ ∈ σ(A) such that f(µ) = λ. Let ε > 0 be given. According to Weierstrauss approximation theorem is a polynomial P (t) such that ||P − f||∞< ε. Let Q(t) = P (t + µ) and make the approach Q(t) = cntn+ cn−1tn−1+ · · · + c0. Then P (A) = Q(A − µ), and P (µ) = Q(0) = c0. We now choose a suitable unit vector ψ in the Hilbert space H and consider the following computations. ||(f(A) − f(µ))ψ|| = ||(P (A) − P (µ) + ((f − P )(A) − (f − P )(µ))ψ|| = ||Q(A − µ)ψ − c0ψ + (f − P )(A)ψ − (f − P )(µ)ψ|| ≤ ||Q(A − µ)ψ − c0ψ|| + ||(f − P )(A)ψ − (f − P )(µ)ψ|| ≤
||cn(A − µ)nψ + cn−1(A − µ)n−1ψ + ... + c0ψ − c0ψ|| + ||(f − P )(A)ψ|| + ||(f − P )(µ)ψ|| = ||(cn(A − µ)n−1+ cn−1(A − µ)n−2+ ... + c1)(A − µ)ψ|| + ||P − f ||∞+ ||P − f||∞< ||(cn(A − µ)n−1+ cn−1(A − µ)n−2+ ... + c1)(A − µ)ψ|| + ε + ε ≤ ||(cn(A − µ)n−1+ cn−1(A − µ)n−2+ ... + c1)|| · ||(A − µ)ψ|| + 2ε.
Where we have used item (d) in Theorem 1, and the assumptions ||ψ|| = 1
and ||P − f||∞ < ε. Next we turn to the choice of the vector ψ. Let K = ||(cn(A − µ)n−1 + cn−1(A − µ)n−2 + ... + c1)||. Since we assumed that µ ∈ σ(A) Weyls criterion says that there is a unit vector ψ so that
||(A − µ)v|| < K+1ve . If we insert this in the computations above we get that
||(f(A)−f(µ))v|| ≤ K||(A−µ)v||+2ε < 3ε. This holds true for every ε > 0.
Thus by Weyls criterion λ ∈ σ[Φ(f)].
3 The Spectral measures
Let A be a fixed bounded self-adjoint operator on the Hilbert space H. That is A∗ = A and ||A|| < ∞. Let ψ be a vector in H. Then f 7−→ (ψ, f(A)ψ) is a positive linear functional on C(σ(A)). According to Riesz-Markovs theorem there exists a unique measure µψ on the compact set σ(A) with (ψ, f (A)ψ) = R
σ(A)f (λ)dµψ. We call µψ the spectral measure associated with the vector ψ. Let B(R) denote the set of all bounded Borel functions on R. Let g be an arbitrary B(R) function.
We want to define g(A)ψ for all vectors ψ so that (ψ, g(A)ψ) =R
σ(A)g(λ)dµψ. Consider the inner product (ψ, g(A)φ) for all pairs of vectors ψ, φ. The fol- lowing guise of the polarization identity gives us (ψ, g(A)φ) = (1/4)
(ψ + φ, g(A)(ψ + φ)) − (ψ − φ, g(A)(ψ − φ))
− i
(ψ + iφ, g(A)(ψ + iφ)) − (ψ − iφ), g(A)(ψ − iφ)
. So define the bounded linear functional T as T (ψ) = (g(A)φ, ψ). One can prove from the definitions that T is linear and bounded.
Under the assumption that T is a bonded linear functional Riesz lemma shows that there is a vector ξ ∈ H such that T (ψ) = (ξ, ψ) for all ψ ∈ H. Now define g(A)φ := ξ. In particular taking ψ = φ we then obtain (g(A)φ, φ) = (φ, g(A)φ) =R
σ(A)g(λ)dµφ, as desired. Where the last equality is due to (φ, g(A)φ) = (1/4)[R
σ(A)g(λ)dµ2φ−R
σ(A)g(λ)dµ0−iR
σ(A)g(λdµ(1+i)φ+ iR
σ(A)g(λ)dµ(1−i)φ] = (1/4)[4R
σ(A)g(λ)dµφ−0R
σ(A)g(λ)dµ−2iR
σ(A)g(λ)dµφ+ 2iR
σ(A)g(λ)dµφ = (1/4)[4 − 0 − 2i + 2i]R
σ(A)g(λ)dµφ = R
σ(A)g(λ)dµφ by usage of the polarization identity above, the integral formula and from the fact that for all vectors v and all complex numbers it holds that µcv= |c|2µv. The last statement can be proved from the definition of the spectral measure dµµ.
The theorem stated below is basically Theorem 1 with the domain of Φ extended to B(R) .
Theorem 2 (Spectral theorem-functional calculus form.)
Let A be a bounded self-adjoint operator on a Hilbert space H. Then there is a unique map ˆΦ : B(R) −→ L(H) such that:
(a) ˆΦ is an algebraic *-homomorphism.
(b) ˆΦ is norm continuous, that is || ˆΦ(f )||L(H)≤ ||f||∞. (c) If f (x) = x then ˆΦ(f ) = A.
(d) If fn(x) −→ f(x) for each x and ||fn||∞ ≤ M for some real M then Φ(fˆ n) → ˆΦ(f ) strongly.
Moreover ˆΦ has the properties:
(e) If Aψ = λψ then ˆΦ(f )ψ = f (λ)ψ.
(f) If f ≥ 0 then ˆΦ(f ) ≥ 0.
(g) If BA = AB then ˆΦ(f )B = B ˆΦ(f ).
Proof. Only item (d) in the theorem is proved. This is because the other statements follow from Theorem 1.
The proof of (d) is left as an exercise in [2]. Fix an arbitrary vector ψ ∈ H and let µψ be the associated spectral measure . First is noted that for every n: |fn− f|2 ∈ L1(R, dµψ) since with M as in Theorem 2 (d) R|fn(x) − f(x)|2dµψ ≤ (M + M)2R
dµψ = (M + M )2µ(R) < ∞ because µψ
is a Baire measure. And further for every n: |fn− f|2 is dominated by g µψ− a.e. where g(x) = (M + M)2 andR
|g(x)|µψ < ∞. Thus, by Lebesgue dominated convergence theorem R
σ(A)|fn− f|2dµψ → 0. So |fn− f|2 → 0 µψ a.e.. Hence we get || ˆΦ(fn)(ψ) − ˆΦ(f )(ψ)||2 = ||ˆΦ(fn− f)ψ||2 = ( ˆΦ(fn− f )ψ, ˆΦ(fn−f)ψ) = (ψ, ˆΦ(fn−f)∗Φ(fˆ n−f)ψ) = (ψ, ˆΦ(fn− f)ˆΦ(fn−f)ψ) = (ψ, ˆΦ(|fn−f|2)ψ) =R
σ(A)|fn−f|2dµψ → 0. Where the first, third and fourth equality are due to the homomorphism property and the last by definition.
5
The equality statement between the norms in Theorem 1 carries over to Theorem 2 with the aid of the following alteration. Let ||f||0∞ be the L∞- norm on the space B(R) with the notion of almost everywhere altered as follows: Choose an orthonormal basis of vectors ψn and define a property to be true a.e. if it is true a.e. with respect to each µψn. Then ||bΦ(f )||L(H)=
||(f)||0∞.
Definition 3 We say that a vector ψ is cyclic for an operator A if finite linear combinations of the elements {Anψ}∞n=0 are dense in H.
The notion of cyclic vectors leads to the lemma below.
Lemma 3 Let A be a bounded self-adjoint operator with cyclic vector ψ. Then there exists a unitary operator U : H −→ L2(σ(A), dµψ) with (U AU−1f )(λ) = λf (λ). Here the equality sign means equality in L2 sense.
Proof. For every continuous function f we define U : {Φ(f)ψ|f ∈ C(σ(A))} → L2(σ(A), dµψ) by U Φ(f )ψ ≡ f. We now need to show that U is well defined and not dependent of choices of representatives. So let f, g ∈ L2(σ(A), dµψ) be two continuous functions such that Φ(f )ψ = Φ(g)ψ. Want to show that f = g. ||Φ(f − g)ψ||2 = (Φ(f − g)ψ, Φ(f − g)ψ) = (ψ, Φ(f − g)∗Φ(f − g)ψ) = (ψ, Φ((f − g)(f − g))ψ) = (ψ, (f − g)(f − g)(A)ψ) = (ψ, |(f −g)(A))|2ψ) =R
σ(A)|f(λ)−g(λ)|2dµψ = 0. Thus f −g = 0, so f = g.
Hence U is well-defined and norm-preserving. By assumption ψ is a cyclic vector so {Φ(f)ψ|f ∈ C(σ(A))} = H. The Bounded Linear Transformation Theorem now enables the extension of U to an isometric map of H into L2(σ(A), dµψ). Since C(σ(A)) = L2(σ(A), dµψ) we know that Range(U ) = L2(σ(A), dµψ). Finally the validity of the formula stated in the theorem is checked. Take f ∈ C(σ(A)). Then (UAU−1f )(λ) = (U AΦ(f )ψ)(λ) = (U x(A)Φ(f )ψ)(λ) = (U Φ(x)Φ(f )ψ)(λ) = (U Φ(xf )ψ)(λ) = (xf )(λ) = x(λ)f (λ) = λf (λ). So if f ∈ L2 then there is a sequence {fn} ∈ C(σ(A)) such that fn→ f and (U AU−1fn)(λ) = λfn(λ). Thus (U AU−1f )(λ) = limn→∞(U AU−1fn)(λ) =
limn→∞(λfn(λ)) = λf (λ). Where the limits are to be interpreted as limits in L2(σ(A), dµψ).
5
The next lemma asserts the extension of the previous lemma to arbitrary A, in sense of making sure that A has family of invariant subspaces spanning H so that A is cyclic on each subspace.
Lemma 4 Let A be s self-adjoint operator on a separable Hilbert space H. Then there is a direct sum decomposition LN
n=1Hn with N =1,2..., or
∞ so that:
(a) If ψ ∈ Hn then Aψ ∈ Hn.
(b) For each n there is a φnwhich is cyclic for A|Hn, i.e. Hn={f(A)ψn| f ∈ C(σ(A))}.
After these two lemmas we have reached the point where we are ready to introduce the spectral theorem in multiplication operator form.
Theorem 4 Let A be a bounded self-adjoint operator on a separable Hilbert space H. Then there exist measures {µn}Nn, where N =1,2,.... or
∞, on σ(A) and a unitary operator U : H → LN
n=1L2(R, dµn) so that (U AU−1ψ)n(λ) = λψn where an element ψ ∈ LN
n=1L2(R, dµn) is written as a N-tuple hψ1(λ), · · · , ψN(λ)i. We call this realization of A a spectral representation.
Proof. First we decompose H according to Lemma 4. For each n we have a vector φncyclic for A |Hn. So by Lemma 3 we have a unitary operator U : H → L2(σ(A), dµn) where µn is the spectral measure associated with φn , with (U AU−1)ψn(λ) = λψn(λ).
5
The previous theorem indicates that every bounded self-adjoint operator is a multiplication operator on a suitable measure space with the underlying measures changing as the operator changes. This is made out more explicitly in the following corollary.
Corollary 5 . Let A be bounded self-adjoint operator on a separable Hilbert space H. Then there exists a finite measure space hM, µi, a bounded function F on M , and a unitary map, U : H → L2(M, dµ) such that (U AU−1f )(m) = F (m)f (m).
Proof. Choose cyclic vectors φn so that ||φn|| = 2−n and define µ by µ|Rn = µn(R), where µnis the spectral measure associated with φn, as in the proof of Theorem 4. We further define the set M to be the union of N copies of R, i.e. M =SN
n=1R. We denote the n : th copy of R by Rnand define µ by µ|Rn = µn(R). In the proof of 3 we defined U by U Φ(f )ψ ≡ f with the cyclic
vector ψ. Now, if we let f ≡ 1 and use the above definition of U we get the equality U ψ = U Iψ = U Φ(1)ψ = 1 in L2(σ(A), dµψ) . Since U is a unitary operator we have ||Uψ|| = ||ψ||. So 1 = ||Uψ|| = ||Uψ||2= ||ψ||2 = (ψ, ψ) = (ψ, 1(A)ψ) =R
σ(A)1dµψ = µψ(σ(A)) = µψ(R). Hence in our this setting we have ||φn|| = 2−n= µn(Rn). It is now clear that µ as defined above is finite, for µ(M ) = µ(SN
n=1Rn) = PN
n=1µn(R) = PN
n=1||φn|| = PN
n=12−n < ∞.
Thus µ is a finite measure.
5
With the following example the corollary above is put into use.
Example 1
Consider a n × n self-adjoint matrix A. This matrix A can be diago- nalized by the finite spectral theorem. Or equivalently, A has a complete orthonormal set of eigenvectors ψ1, · · · , ψn with Aψi = λiψi. First make the assumption that the eigenvalues are distinct. Then consider the Dirac measures δ(x − λi) and its sum µ = Pn
i=1δ(x − λi). What determines f ∈ L2(R, dµ) is its values at the points λ1· · · λn. This is because of the nature of the measure chosen. Thus each f ∈ L2(R, dµ) can be written as a n-dimensional vector with complex components, that is each f corresponds to hf(λ1), · · · , f(λ)ni. So L2(R, dµ) is really Cn. With this representa- tion at hand it is clear that the function λf corresponds to the vector or n-tuple hλ1f (λ1), · · · , λnf (λ)ni. So the operator A is actually multiplica- tion by λ on L2(R, dµ). However, we can do just as well if we take the measure eµ = Pn
i=1ciδ(x − λi), where ci > 0 for every index i. Then A is also represented by as multiplication by λ on L2(R, dµ). If the eigenvalues are not distinct, we can not represent a self-adjoint operator as multipli- cation on L2(R, dµ) with only one measure. These results can be gener- alized to a self-adjoint operator A on infinite dimensional spaces with the additional assumption that A is compact. By Hilbert-Schmidt theorem we know that there is a complete orthonormal set of eigenvectors {ψn}∞n=1with Aψn = λnλn, if H is separable. We can modify the measure used above with n replaced with ∞ and multiply each summand with a factor 2−n in order to keep the measure finite.
Next, we will relate the spectral measure to the spectrum. In order to do so we will first define the notion of support of a measure.
Definition 5 . Let µ be Borel measure on Rn and let B be the largest open subset such that µ(B) = 0. We then define supp(µ) := Bc.
We also need to define the support for a sequence of measures. If we have measures µ1, · · · , µN for N ∈ N or N = ∞ and µi(Bi) = 0 for Bi ⊂ M then it is clear that µi(TN
i Bi) = 0 for all i. So the support of the sequence should be the complement of the above set. Only we want the support to be the complement of an open set. So we remedy this by defining the support
to be the closure of the complement of TN
n=1Bn. This reasoning is made precise in the following definition.
Definition 6 . Let {µn}Nn=1 be a family of measures, where n = 1, 2, · · · or ∞. Then the support of {µn}Nn=1 is the complement of the largest open set B with µn(B) = 0 for all n; so supp{µn} =SN
n=1supp{µn}.
After the definition we state a proposition that proclaims the equality between the spectrum of a self-adjoint operator A and a family of spectral measures {µn}Nn=1.
Proposition 6 . Let A be a self-adjoint operator and {µn}Nn=1 a fam- ily of spectral measures in a spectral representation of A. Then σ(A) = supp{µn}Nn=1.
We are now turning to the definition of the essential range of a real- valued function F on a measure space hM, µi.
Definition 7 . Let F be a real valued function on a measure space hM, µi. We say that λ is in the essential range of F denoted by Essrange(F ) if and only if µ{m | λ − ε < F (m) < λ + ε} > 0 for all ε > 0.
We can now use this definition to state the next proposition.
Proposition 7 . Let F be a bounded real-valued function on a measure space hM, µi. Let TF be the operator on L2(M, dµ) given by TF(g)(m) = F (m)g(m). Then σ(TF) is the essential range of F .
Proof. We begin by remarking that the proof of this theorem is left as an exercise in [2]. First suppose that λ ∈ Essrange(F ). Let ε > 0 and D = {m | λ − ε < F (m) < λ + ε}. Let further f be the char- acteristic function of this set, that is f = χD. Since we assumed that λ ∈ Essrange(F ) we know that µ(D) > 0. So we can rescale f into a unit vector, call this unit vector f1. Now we look at the norm ||(TF − λ)f1||2 = R
M|F (m)f1(m) − λf1(m)|2dµ(m) = R
D|F (m)f1(m) − λf1(m)|2dµ(m) ≤
supm∈D|F (m) − λ|2
·R
D|f1(m)|2dµ(m) = supm∈D|F (m) − λ|2||f1||2 = supm∈D|F (m) − λ|2 ≤ ε2. Here we have used that f = χD, in the first equality and that f1 is a unit vector in last equality. Using Weyls criterion we conclude that λ ∈ σ(TF).
Conversely suppose that λ ∈ σ(TF). Weyls criterion tells us that there exists {fn}∞n=1 such that ||fn|| = 1 and ||(TF − λ)fn|| → 0. Now let us assume that λ /∈ Essrange(F ). Then there exists an ε0 such that µ({m | λ − ε < F (m) < λ + ε}) = 0. This means that |F (m) − λ| ≥ ε0
for µ − a.e.m. Hence for all unit vectors f we get that ||(TF − λ)f||2 =
R
M|(F (m) − λ)f(m)|2dµ(m) ≥R
Mε20|f(m)|2dµ(m) = ε20· 1 = ε20. But this contradicts the existence of a sequence {fn}∞n=1with the properties ||fn|| = 1 and ||(TF− λ)fn|| → ∞ asserted by Weyls criterion. Thus we conclude that λ ∈ Essrange(TF).
5
If a self-adjoint operator A has spectrum σ(A), then representing the operator as U AU−1, with a unitary U does not change the spectrum. So the spectrum is a unitary invariant. A problem with this invariant is that we can not use it to distinguish between two different operators. For in- stance, an operator with a complete set of eigenfunctions having all ratio- nal numbers in [0, 1] as eigenvalues and the operator which multiply the elements in L2([0, 1], dx) by x both have spectrum [0, 1]. We can also re- alize this by noting that σ(A) is equal to the support of the spectral mea- sures and different sorts of measures having the same support need not be the same. Thus, our next task will be to find better invariants that are simpler than measures. In order to do this we start by decomposing the measure µ into µ = µpp + µac + µsing, where µpp is the pure point mea- sure, µac is absolutely continuous with respect to Lebesgue measure, and µsing is continuous and singular with respect to Lebesgue measure. First we note that µpp is singular with respect to µac because Lebesgue mea- sure assigns the value zero to point sets and µac is absolutely continuous with respect to Lebesgue measure. Next, we note that if µac(A) = 0 for a set A then the Lebesgue measure λ(A) = 0 which in turn implies that µsing(Ac) = 0. At last we se that µppis singular with respect to µsingbecause µsing is a continuous measure and thus zero on point sets. Knowing that these three pieces are mutually singular enables the following decomposition:
L2(R, dµ) = L2(R, dµpp)L
L2(R, dµac)L
L2(R, dµsing). If we are given a family of spectral measures {µn=1}Nn=1 we can sum LN
n=1L2(R, dµn;ac) by defining:
Definition 8 Let A be a bounded self-adjoint operator on H. Let further Hpp = {ψ | µψ is pure point }, Hac = {ψ | µψ absolutely continuous }, Hsing= {ψ | µψ is continuous singular }.
With this decomposition at hand and knowing that these three measures are mutually singular we have assured the validity of the following theorem.
Theorem 9 H = HppL HacL
Hsing. Each of these subspaces are in- variant under A. A | Hpp has a complete set of eigenvectors, A | Hac has only absolutely continuous measure and A | Hsing has only continuous sin- gular spectral measure.
Next we turn to the definition concerning the breakup of the spectrum.
Definition 10 .
σpp(A)={λ | λ is an eigenvalue}
σcont(A) = σ(A | Hcont ≡ HsingL Hac) σac(A) = σ(A | Hsing)
σsing(A) = σ(A | Hsing).
Before moving on to the question on multiplicity free operators we are stating a proposition. It is worth noting that it may be the case that σacS
σsingS
σpp 6= σ. This is due to the definition of σpp as the set of eigenvalues of A and not σ(A | Hpp). The below statement is always true.
Proposition 8
σcont(A) = σac(A)[
σsing(A) σ(A) = σpp(A)[
σcont(A) 3.1 Multiplicity free operators
We are now coming back to the question when an operator A is unitarily equivalent to multiplication by x on L2(R, dµ). That is when do we need only one spectral measure? In the preceding example we saw that this was the case only when A did not have any repeated eigenvalues. With this example in mind we make the following definition.
Definition 11 A bounded self-adjoint operator A is called multiplicity free if and only if A is unitarily equivalent to multiplication by λ on L2(R, dµ) for some measure µ.
In connection with this definition we are stating a theorem given without proof.
Theorem 12 The following statements are equivalent.
(a) A is multiplicity free.
(b) A has a cyclic vector.
(c) {B:AB=BA} is an abelian algebra.
3.2 Measure classes
We are now returning to the question about the non-uniqueness of the mea- sure in the multiplicity free cases. In the example was stated that the candidates for the measure were Pn
i=1aiδ(x − λi) with ai 6= 0. In order to generalize this suppose that dµ on R is given and let F be a measur- able function such that F > 0 µ − a.e. and F is locally L2(R, dµ), mean- ing that for each compact set C ⊂ R: R
C|F |dµ < ∞. So we can let
ν(C) =R
C|F |dµ and thus dν = F dµ is a Borel measure. Now we define a map U : L2(R, dν) → L2(R, dµ) where U is given by (U f )(λ) =p
F (λ)f (λ).
It is easy to see that this map is injective; for if (U f )(λ) = (U g)(λ) then pF (λ)f (λ) =p
F (λ)g(λ) which reduces to f (λ) = g(λ) because F 6= 0 and thus f = g as elements in L2(R, dν). This map is also onto because for every f ∈ L2(R, dν) it holds true that (U f /√
F )(λ) = f (λ). Finally (U f, U g) = R pF (λ)f (λ)p
F (λ)g(λ)dν = R
F (λ)f (λ)g(λ)dν = R
f (λ)g(λ)dµ = (f, g).
So we can conclude that U is unitary operator. So an operator A with spec- tral representation in terms of µ could just as well be represented in terms of ν. By the Radon-Nikodym theorem it is clear that if ν and µ have the sets of measure zero then dν = F dµ with F 6= 0 µ − a.e.. If on the other hand dν = F dµ with F µ − a.e. non-zero then µ(A) =R
AdF dµ. So µ and ν have the same sets of measure zero. We can now use this reasoning to define what is meant for two measures to be equivalent.
Definition 13 . Two Borel measures µ and ν are said to be equivalent if and only if they have the same sets of measure zero. An equivalence class hµi is called a measure class.
We are finally reaching our goal to decide when two operators are unitarily equivalent.
Proposition 9 Let µ and ν be Borel measures on R with bounded sup- port. Let the operator Aµ on L2(R, dµ) be given by (Aµf )(λ) = λf (λ) and Aν on L2(R, dν) by (Aνf )(λ) = λf (λ). Then Aµand Aν are unitarily equiv- alent if and only if µ and ν are equivalent measures.
3.3 Operators of uniform multiplicity
In the finite dimensional case we can represent an operator by a matrix.
And if we want a canonical listing of its eigenvalues it is natural to list them according to multiplicity, i.e. list all eigenvalues of multiplicity one, multiplicity two and so on. So we need a way of distinguish operators of different multiplicity. In the light of this we make the following definition.
Definition 14 . A bonded self-adjoint operator is said to be of uni- form multiplicity m if A is unitarily equivalent to multiplication by λ on Lm
n=1L2(R, dµ).
Using the notion of uniform multiplicity we state a proposition concerning equivalent measures.
Proposition 10 . If A is unitarily equivalent to multiplication by λ on Lm
k=1L2(R, dµ) and onLm
k=1L2(R, dν) then n = m and µ and ν are equiv- alent measures.
3.4 Disjoint measure classes
In the canonical listening of the eigenvalues in the finite dimensional case we require the lists to be disjoint. This prevents us from counting an eigenvalue with a multiplicity higher than one several times. For instance if we have an eigenvalue of multiplicity two we do not want to count it as an eigenvalue of multiplicity one and then as an eigenvalue of multiplicity two. In the definition below we are making an analogy for measures.
Definition 15 We call two measure classes disjoint if any µ ∈ hµi and ν ∈ hνi are mutually singular.
Before ending this section about spectral measures we are stating the mul- tiplicity theorem. This theorem says that each bounded self-adjoint op- erator A is described by a family of mutually disjoint measure classes on [−||A||, ||A||] and that two operators are unitarily equivalent if and only if their spectral multiplicity measure classes are identical.
Theorem 16 . Let A be a bounded self-adjoint operator on a Hilbert space H. Then there is a decomposition H = H1L
H2L
· · ·L
H∞ so that (a) Each Hm is invariant under A.
(b) A|Hm has uniform multiplicity m.
(c) The measure classes hµmi associated with the spectral representation of A|Hm are mutually disjoint.
Furthermore, (a)-(c) uniquely determine the subspaces H1, · · · , Hm, · · · , H∞
and the measure classes hµ1i, · · · , hµmi, · · · , hµ∞i.
4 Spectral projections
In this section we present the spectral theorem in projection-valued mea- sure form. We start off by considering a Borel set Ω and the charateristic function χΩ of this set, which is a Borel function. In the previous section we extended our continuous functional calculus to Borel functional calculus which enables us to consider functions as above. We now turn attention to our first definition.
Definition 17 Let A be a bounded self-adjoint operator and Ω a Borel set of R. We call PΩ ≡ χΩ(A) a spectral projection of A.
If we consider the case where the spectrum is discrete we can think of χΩ(A) as the projection onto the closure of the subspace in H spanned by all eigenvectors of A whose eigenvalues belong to Ω. We can easily see that PΩ is an orthogonal projection, that is PΩPΩ = PΩ = PΩ∗ if we consider χΩ. For if χΩ(x) = 1 then χΩ(x)2 = 1 and χΩ(x) = 1. Next we consider the family of projections {PΩ | Ω is a Borel set} and state its properties in the following proposition.
Proposition 11 Let {PΩ} be the family of spectral projections of a bounded self-adjoint operator A. Then {PΩ} has the following properties:
(a) Each PΩ is an orthogonal projection.
(b) P = 0; P(−a,a)= I for some a.
(c) If Ω =S∞
n=1Ωn with ΩnT
Ωm = for all m 6= n, then
PΩ= s-limN →∞( XN n=1
PΩn).
(d) PΩ1PΩ2 = PΩ1TΩ2
Proof. (a) Is already proved.
(b) Since P = χ (A) per definition we only have to note that χ ≡ 0 from which is follows that P = 0. We continue with the proof of the second part of the statement. The fact that we have a bounded operator A with norm
||A|| enables us to conclude that σ(A) ⊂ [−||A||, ||A||]. Here we have used (cf. [1, Theorem 7.3-4]) and (cf. [1, Theorem 9.1-3]). Consider the function χσ(A) which is clearly 1 on σ(A). By Theorem 1 (a) we get χσ(A)(A) = I.
Thus, if take a > ||A|| we χσ(A)(A) = P(−a,a)= I.
(c) First consider the case were N is finite. As in the statement above let Ω =SN
n=1Ωn. Then PΩ = PSN
n=1Ωn = χSN
n=1Ωn(A) = (PN
n=1χΩn)(A) = PN
n=1χΩn(A) =PN
n=1PΩn. The third equality is valid because the sets con- sidered are disjoint and the fourth equality is due to the additive homomorphism- property. Now let us see what happens when N goes to infinity. Consider
the functions χN(x) = χ∪N
n=1Ω. We know that ||χN||∞ ≤ 1 for each N be- cause the sets are disjoint. We also know that for each fixed x ∈ R, χN(x) = χ∪N
n=1Ω(x) =PN
n=1χΩn(x) →P∞
n=1χΩn(x) = χ∪∞n=1Ωn(x) = χΩ(x) when N goes to infinity. The infinite sum is convergent because the sets considered are disjoint. Now we apply Theorem 2 (d) to get the desired result.
(d) We get that PΩ1PΩ2 = χΩ1(A)χΩ2(A) = χΩ1χΩ2(A), by definition and the homomorphism-property. So we need to check that χΩ1χΩ2 = χΩ1∩Ω2. But this certainly holds true, for if the right hand side equals one for a point then that point must be in both Ω1and Ω2which makes the left hand side equal to one. Thus PΩ1PΩ2 = χΩ1(A)χΩ2(A) = χΩ1∩Ω2(A) = PΩ1∩Ω2.
5
If we recapitulate for a minute and think about the definition of a mea- sure, especially countable additivity, we see a strong connection between this and condition (c) in Proposition 11. Thus, we are motivated to make the following definition:
Definition 18 A family of projections {PΩ} satisfying (a)-(c) is called a bounded projection-valued measure abbreviated (p.v.m).
As the name projection-valued measure indicates we can use this measure for integration. If we are given a p.v.m PΩ, then for every Borel set Ω we can define a real number m(Ω) = (φ, PΩφ). Here it is worth noting that m de- pends on the choice of the vector φ ∈ H. We remark that m(Ω) is indeed real because PΩ is self-adjoint. We can also easily see that m is a Borel measure:
m( ) = (φ, P φ) = (φ, 0) = 0, if Ω1, Ω2, . . . are any pairwise disjoint Borel sets then m(S∞
n=1Ωn) = (φ, PS∞
n=1Ωnφ) = (φ, s-limN →∞PN
n=1PΩnφ) = limN →∞PN
n=1(φ, PΩnφ) = limN → ∞PN
n=1m(Ωn) = P∞
n=1m(Ωn). We are now going to use the symbol d(φ, Pλφ) for integration with respect to (φ, PΩ). By the usage of Riesz lemma we can find a unique operator B with (φ, Bφ) =R
f (λ)d(φ, Pλφ) for all vectors φ ∈ H. So we can now state our next theorem:
Theorem 19 If PΩis a projection-valued measure and f a bounded Borel function on supp(PΩ), then there is a unique operator B which we denote Rf (λ)dPλ, so that (φ, Bφ) =R
f (λ)d(φ, Pλφ), for all φ ∈ H.
After this theorem it is worth while making a remark.
Remark. If A is a bounded self-adjoint operator and PΩ its associated projection-valued measure, it can be shown that f (A) = R
f (λ)dPλ. If we take f (x) = x we get A =R
λdPλ. We will use this fact later in the proof of Proposition 12. We are now ready to state the spectral theorem for bounded self-adjoint operators in its third form, that is in p.v.m. form.
Theorem 20 There is a one-one correspondence between the bounded self-adjoint operators A and the bounded projection valued measures {Ω}
that is given by:
A 7→ {PΩ} = {χΩ(A)}
{PΩ} 7→ A =R λdPλ.
Now we put our new knowledge about spectral projections into use when we are stating a proposition that gives a condition for a real number λ to be in the spectrum of a self-adjoint bounded operator A. The proof of the next proposition stated is left unproved in [2].
Proposition 12 λ ∈ σ(A) if and only if P(λ−ε,λ+ε)(A) 6= 0 for any ε > 0.
Proof. Suppose that there exists ε0 > 0 such that P(λ0−ε0,λ0+ε0)(A) = 0 and let Ω = (λ0− ε0, λ0+ ε0). We aim to show that λ0 ∈ σ(A), in other/ words λ0 ∈ ρ(A), where ρ(A) denotes the resolvent of A. Now let φ be an arbitrary vector in H and let m be the corresponding Borel measure on R, as above. Due to our assumption we know that m(Ω) = (φ, PΩφ) = 0. So d(φ, Pλφ) = 0 on Ω. Know we consider ||(A − λ0)φ||2 = ((A − λ0)φ, (A − λ0)φ) = (φ, (A − λ0)2φ) =R
(λ − λ0)2d(φ, Pλφ). Here we have used the re- mark stated above. The value of this integral on Ω is zero because d(φ, Pλφ) is zero on Ω. We also know that for λ /∈ Ω we have (λ − λ0)2 ≥ ε20. So we can make the following estimate. ||(A − λ0)φ||2 ≥ ε20
Rd(φ, Pλφ) = ε20||φ||2. Thus we get that ||(A − λ0)φ|| ≥ ε0||φ||. So by (cf. [1, Theorem 9.1-2(e)]) λ0 ∈ ρ(A).
Next suppose that λ0 ∈ ρ(A) and P(λ0−ε,λ0+ε) 6= 0 for any ε > 0. We know that there exists ε0 such that R
(λ − λ0)2d(φ, Pλφ) ≥ ε20
R d(φ, Pλφ).
Choose a positive number η < ε0. Since P(λ0−η,λ0+η) 6= 0 there is a vector ψ ∈ H such that P(λ0−η,λ0+η)(φ) = ψ for some φ ∈ H. Know if we con- sider for instance P(λ0,ω) with ω > λ0 and apply this vector to ψ we get that P(λ0,ω)P(λ0−η,λ0+η)(φ) = 0 because χ(λ0,ω)χ(λ0−η,λ0+η)(x) = 0. Thus we have that (ψ, P(λ0,ω)ψ) = 0 and hence d(ψ, Pλψ) = 0. The case where we are on the left side of this interval is similar. So when integrating over the interval (λ0− η, λ0+ η) have that (λ − λ0)2 ≤ η2 < ε20. But this is a contradiction to the integral-inequality above. Thus P(λ0−η,λ0+η) = 0. 5 We are now defining what is meant by the essential spectrum and the dis- crete spectrum of an operator A. This notion of breaking up the spectrum in these parts enables a decomposition of the spectrum of A into two disjoint subsets.
Definition 21 We say that λ belongs to the essential spectrum of A, denoted λ ∈ σess(A), if and only if P(λ−ε,λ+ε) is infinite dimensional for all ε > 0 where the phrase P is infinite dimensional means that Range(P ) is infinite dimensional. If λ ∈ σ(A) and there exist ε0 such that P(λ−ε0,λ+ε0)
is finite dimensional, we say that λ is in the discrete spectrum of A, denoted λ ∈ σdisc(A).
So, as mentioned before we now have a new decomposition of σ(A) into two disjoint sets. The next theorem states that the essential spectrum of a self-adjoint bounded operator is always closed.
Theorem 22 The essential spectrum of A, σess(A), is always closed.
Proof. We take an arbitrary sequence {λn} ∈ σess(A) and assume that λn→ λ as n → ∞. Because λ is a point of accumulation we know that any open interval I about λ contains an open interval In about some λn. Since λnis in the essential spectrum of A we know that P(λn−ε,λn+ε)(A) is infinite dimensional for every ε > 0 and since In= (λ − ε, λ + ε) is included in I for sufficiently small ε we conclude that PI(A) is infinite dimensional. This is true for every open interval I about λ; thus λ ∈ σess(A) and hence σess(A) is closed. 5
The following theorem gives conditions for λ ∈ σ(A) to belong to the discrete spectrum. The proof of this theorem is left as an exercise in [2].
Theorem 23 λ is in the discrete spectrum of A, λ ∈ σdisc(A) if and only if both of the following conditions hold:
(a) λ is an isolated point of σ(A), meaning that there exists ε0 such that (λ − ε0, λ + ε0) ∩ σ(A) = {λ}.
(b) λ is an eigenvalue of finite multiplicity, that is {ψ | Aψ = λψ} is finite dimensional.
Proof. Suppose that λ ∈ σdisc(A), that is there exists ε > 0 so that Range(P(λ−ε,λ+ε)) is finite dimensional. For every n ≥ 1 define an :=
dim(Range(P(λ−ε/n,λ+ε/n))). We will show that {an} is decreasing, i.e.
an+1 ≥ an. Let us decompose (λ − ε/n, λ + ε/n) into three disjoint sets;
(λ − ε/n, λ + ε/n) = (λ − ε/n, λ − ε/(n + 1)] ∪ (λ − ε/(n + 1), λ + ε/(n + 1)) ∪ [λ + ε/(n + 1), λ + ε/n). Let us call these three intervals I1, I2, I3. By Proposition 11 (d) it is the case that PIiPIj = PIi∩Ij = P = 0, for i, j ∈ {1, 2, 3}, and then by (cf. [1, Theorem 9.5-3(e)]) the corresponding pro- jections project onto three pairwise orthogonal subspaces. By Proposition 11 (c) we get that P(λ−ε/n,λ+ε/n)= P(λ−ε/n,λ−ε/(n+1))+ P(λ−ε/(n+1),λ+ε/(n+1))+ P(λ+ε/(n+1),λ+ε/n). Thus an+1 = dim(Range(P(λ−ε/(n+1),λ+ε/(n+1))) ≤ dim(Range(P(λ−ε/n,λ+ε/n))) = an. So we can hereby conclude that {an} is a decreasing sequence of non-negative integers and must as such be con- stant for some sufficiently large index. This is to say that there exists a N such that aN = an+1 = aN +2 = · · · . Our next step is to show that there is no λ0 ∈ (λ − ε/N, λ + ε/N) with λ0 6= λ that belongs to σ(A). Suppose there is such λ0 ∈ σ(A) and suppose further that λ0 < λ.
The case where λ0 > λ is treated in a similar fashion. Choose ε1 > 0 so small that λ − ε/N < λ0 + ε1 < λ, then we choose n > N large enough so that λ0 + ε1 < λ − ε/n. In line with our reasoning above we get that Range(P(λ0−ε1,λ0+ε1)) and Range(P(λ−ε/n,λ+ε/n)) are orthogonal subspaces of Range(P(λ−ε/N,λ+ε/N )), butwe also know that because {an} is constant for some N that dim(Range(P(λ−ε/n,λ+ε/n))) = an = aN = dim(Range(P(λ−ε/N,λ+ε/N ))). Thus we must have that Range(P(λ0−ε1,λ0+ε1)) = {0}, that is P(λ0−ε1,λ0+ε1) = 0. By Proposition 12 we conclude that λ0 ∈/ σ(A).
Still assuming that λ ∈ σdisc(A) we know that there exists some ε0> 0 such that Range(P(λ−ε0,λ−ε0)) is of finite dimension. Since {λ} ⊂ (λ − ε0, λ + ε0) ∩ σ(A) we know that {ψ | Aψ = λψ} ⊂ Range(P(λ−ε0,λ+ε0)(A)). Thus dim({ψ | Aψ = λψ}) ≤ dim(Range(P(λ−ε0,λ+ε0)(A)) < ∞.
Conversely suppose that both (a) and (b) hold true. Then there exists some ε1> 0 so that (λ − ε1, λ − ε1) ∩ σ(A) = {λ} and dim({ψ | Aψ = λψ}) < ∞.
Because of property (a) we have that Range(P(λ−ε1,λ+ε1)(A)) = {ψ | Aψ = λψ} and thus for ε1 Range(P(λ−ε1,λ+ε1)(A)) is of finite dimension. Hence λ ∈ σdisc(A).
5
The next theorem concerns the essential spectrum and the proof of this theorem is left unproved.
Theorem 24 Let A be a bounded self-adjoint operator and let λ ∈ σ(A).
Then λ ∈ σess if and only if one or more of the following conditions holds.
(a) λ ∈ σcont(A) ≡ σac∪ σsing(A). (b) λ is a limit point of σpp(A) (c) λ is an eigenvalue of infinite multiplicity.
Before we end this section we present a theorem that has been used frequently in the previous sections, namely Weyls criterion that states a condition for a real number λ to be in the spectrum of an self-adjoint oper- ator A. We only prove the first part of this theorem.
Theorem 25 Let A be a bounded self-adjoint operator. Then σ(A) if and only if there exists a sequence {ψn}∞n=1such that ||ψn|| = 1 and limn→∞||(A−
λ)ψn|| = 0. λ ∈ σess(A) if and only if the {ψn} can be chosen to be orthog- onal.
Proof. First we note that σ(A)c = ρ(A). So we can prove that λ ∈ ρ(A) if and only if Weyls criterion does not hold. However this is equiv- alent to (cf. [1, Theorem 9.1-2]) that says that a number λ belongs to the resolvent set of an operator if and only if there exists c > 0 such that for all x ∈ H : ||(T − λI)x|| ≥ c||x||. This shows that if λ ∈ ρ(A) there cannot be a sequence {ψn}∞n=1 with the property limn→∞||(A − λ)ψn|| = 0.
5
5 Unbounded operators
This section about unbounded operators is merely a brief introduction to the topic. We mention a few motivating examples and state some impor- tant definition and theorems. Not every operator of importance in analysis is bounded. For instance consider the differentiation operator. It is not hard to see that this operator is unbounded. We consider the space of all polynomials P [t] = {p(t) | p is a polynomial }. For simplicity lets us pick the interval [0, 1] and equip it with the norm ||P || = max[0,1]|p(t)|. We define D to be the differentiation operator, that is Dp(t) = p0(t). If we take the polynomial pn(t) = tnwhere n is a natural number, then certainly
||pn|| = 1 and Dpn(t) = p0(t) = ntn−1. Then ||DPn|| = n||tn−1|| = n. Thus
||T xn||/||xn|| = n and n is an arbitrary natural number. So D is an un- bounded operator. Since the differentiation operator is such an important operator we cannot simply do without the theory of unbounded operators.
When dealing with unbounded operators the domain on which they are de- fined becomes of utter importance. The next theorem proclaims that un unbounded linear operator T : H → H that satisfies (T x, y) = (x, T y) for all x, y cannot be defined on all of H.
Theorem 26 If a linear operator T is defined on all of H and satisfies (T x, y) = (x, T y) for all x, y then T is bounded.
This theorem suggests that an unbounded operator T is only defined on a dense linear subset of the Hilbert space H. We will in the sequel always assume that the domain of the operator T , denoted by D(T ), is a dense subset of H. We continue with an example of an unbounded operator where the Hilbert space is L2(R).
Example 2We take H = L2(R) and let D(T ) = {f ∈ L2(R) |R
Rx2|f(x)|2dx <
∞}. For f ∈ D(T ) we define the operator T by (T f)(x) = xf(x). We now aim to show that T is unbounded. If we choose a function f ∈ D(T ) such that f (x) 6= 0 near infinity, we realize that we can make ||T f|| =R
Rx2|f(x)|2 arbitrarily big. Thus ||T || = sup||f ||=1||T f|| = ∞. That is T is unbounded.
It we want Range(T ) to be L2(R) we must stick to the restriction we made above. For if f /∈ D(T ) then R
Rx2|f(x)|2 = ∞ so xf(x) /∈ L2(R). So D(T ) is defined above is the largest domain of T for which Range(T ) is in LR. The differential operator which was mentioned in the beginning of this sec- tion was found to be unbounded. However it is closed, a property which it has in common with many linear operators occuring in practical problems.
The property of being closed is defined in term of the graph of the operator.
Thus, we define.
