Spectral Theory for BoundedSelf-adjoint Operators

Roland Strömberg

U.U.D.M. Project Report 2006:5

Examensarbete i matematik, 20 poängHandledare och examinator: Andreas Strömbergsson

Juni 2006

Department of Mathematics

Uppsala University

1 Introduction

Spectral theory for a self-adjoint operator is a quite complicated topic. Ifthe operator at hand is compact the theory becomes, if not trivial, lesscomplicated. Consider first the case of a self-adjoint operator A : V → Vwith V finite dimensional. The complete spectral decomposition of A canbe stated in a quite elementary fashion: We know that we can represent Awith a matrix [A] and compute its eigenvalues and corresponding eigenvec-tors. The spectral theorem says that there exists a total orthonormal setof eigenvectors ψ1 · · · ψn such that [A] can be diagonalized, or equivalentlythat there exists a unitary matrix [U ] and eigenvalues λ1, · · · , λn so thatfor each ψ ∈ H, ([U ][A][U−1]ψi) = λiψi. This result carries over to thecase where V is a infinite dimensional separable Hilbert space if we assumethat V is a compact. In my thesis I have focused on bounded self-adjointoperators in full generality.

For these operators I present the spectral theorem in three differentforms; functional calculus form, multiplication operator form and projections-valued measure form. The proofs of these theorems that I have used areclassical results in functional analysis, but there are many more in the math-ematical literature. For example, a general version of the spectral theoremfor self-adjoint operators by Stone and von Neumann (1929-1932), and morerecently a version concerning unbounded operators in Hilbert spaces wasstated 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 concernedwith 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 thesense that it is reminiscent of the spectral theorem for compact self-adjointmatrices stated above. We can leisurely say that the second form saysthat every bounded self-adjoint operator is a multiplication operator on asuitable measure space. Now, it is quite evident that we need a rather heavymathematical machinery even to state the theorem. We clearly need somemeasure theory and for that we introduce the so called spectral measures.But the first step in reaching the second formulation of the spectral theoremis to define the functional calculus. We formulate this calculus first forthe continuous case then for Borel sets. The first theorem (Theorem 1)concerning continuous functional calculus states that there exists a map fromthe set of continuous functions on the spectrum of a bounded self-adjointoperator A, σ(A) to the set of bounded linear operators on a Hilbert spaceH, 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 firstversion of the spectral theorem is basically Theorem 1 extended to Borel

1

functions on R. The fact that we are dealing with a separable Hilbert spaceH enables us to decompose H into a direct sum of invariant subspaces. Thenon each subspace we can find a unitary operator so that we get a relationanalogous to the matrix case above. This enables us to state the spectraltheorem 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. Ifwe have a bounded self-adjoint operator A and a Borel set Ω at hand wecall PΩ ≡ χΩ(A) a spectral projection of A. We further call a family of suchprojections PΩ | Ω is an arbitrary Borel set a projection valued measureif it obeys a number of certain conditions. The third form of the spectraltheorem says that there is a one-one correspondence between a boundedself-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 sinceone of the most recurring operator in mathematical analysis, the differentialoperator, is unbounded. This section is merely a brief introduction and doesnot exhaust the topic of unbounded operators.

2 The Functional calculus

Let A be a given fixed operator and f a continuous function. How can onedefine f(A) to make sense? If f is given by a finite polynomial

∑nk=0 akx

k

we want f(A) to be∑n

k=0 akAk. We also want f to be well defined in the

sense of convergence of the sum. If f is given by∑∞

k=0 akxk with radius

of convergence R and ||A|| < R then∑∞

k=0 akAk converges in L(H) (the

set of bounded linear operators on H). So if f is given by an infinite seriesthen f(A) is given by

∑∞k=0 ak ·Ak. The theorem below states the existence

and uniqueness of an operator between the family of all continuous complexfunctions defined on the spectrum of A, and L(H).

Theorem 1 (Continuous functional calculus.) Let A be a boundedself-adjoint operator on a Hilbert space H. Then there exists a unique mapΦ : 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 ||∞.

2

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 ifP (x) = xn then Φ(P ) = An. Now the additive homomorphism propertyasserts that if P (x) =

∑nk=0 akx

k then Φ(P ) =∑n

k=0 akAk.

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 LinearTransform theorem. To prove the equality between the norms above a specialcase of (e) that holds for arbitrary bounded operators is proved.

Lemma 1 Let P (x) =∑n

k=0 akxk and P (A) =

∑nk=0 akA

k. Then σ[P (A)] =P (λ)|λ ∈ σ(A).

Proof. Take arbitrary λ ∈ σ(A). Then A− λ does not have an inversein 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 lowerdegree than P . Now P (A) − P (λ) = (A − λ) Q(A), so P (A) − P (λ) failsto have an inverse in L(H). Hence P (λ) ∈ σ(P (A)). For µ ∈ σ(P (A)) letP (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) =

∑nk=0 anA

n then P (A)∗ =(∑n

k=0 anAn)∗ =

∑nk=0 an An = P (A). Further (PP )(A) is self-adjoint:

(PP )(A)∗ = P (A)∗(P (A))∗ = P (A)P (A)∗∗ = P (A)P (A) = (PP )(A). Sosupλ∈σ(PP (A)) |λ| = r(PP (A)) = ||PP (A)||, where r is the spectral radius of

PP (A). And then by Lemma 1 : supλ∈σ(PP (A)) |λ| = supλ∈σ(A) |PP (λ)|. So

||P (A)||2 = ||P (A)∗P (A)|| = ||PP (A)|| = supλ∈ σ(PP (A))

|λ|

= supλ∈σ(A)

|PP (λ)| = supλ∈σ(A)

|P (λ)||P (λ)|

= supλ∈σ(A)

|P (λ)|2 = ( supλ∈σ(A)

|P (λ)|)2.

3

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. Thecomplex Stone-Weierstrass theorem tell us that the closure of the above setis C(σ(A)). Starting with (a): Let P and Q be polynomials. Then

Φ(PQ) = PQ(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 Φ obeys (a),(b) and (c) then Φ agrees withΦ on polynomials and thus by continuity on C(σ(A)).

(d) If Aψ = λψ, then Φ(P )ψ = P (A)ψ = (∑n

k=0 an ·An)ψ = (c0 + c1A+c2A

2 + .......c2An)ψ = c0ψ+c1Aψ+c2A

2ψ+ .......c2Anψ = (c0 +c1λ+c2λ

2 +.......c2λ

n)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 handλ ∈ Range(f) then there is a µ ∈ σ(A) such that f(µ) = λ. Let ε > 0 begiven. According to Weierstrauss approximation theorem is a polynomialP (t) such that ||P − f ||∞ < ε. Let Q(t) = P (t+ µ) and make the approachQ(t) = cnt

n + 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 Hand 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

4

and ||P − f ||∞ < ε. Next we turn to the choice of the vector ψ. LetK = ||(cn(A − µ)n−1 + cn−1(A − µ)n−2 + ... + c1)||. Since we assumedthat µ ∈ σ(A) Weyls criterion says that there is a unit vector ψ so that||(A− µ)v|| < ve

K+1 . 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)].

5

3 The Spectral measures

Let A be a fixed bounded self-adjoint operator on the Hilbert space H. Thatis 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-Markovstheorem there exists a unique measure µψ on the compact set σ(A) with(ψ, f(A)ψ) =

∫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 functionson 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) =∫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 lemmashows 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) =∫σ(A) g(λ)dµφ, as desired. Where the last equality

is due to (φ, g(A)φ) = (1/4)[∫σ(A) g(λ)dµ2φ−

∫σ(A) g(λ)dµ0−i

∫σ(A) g(λdµ(1+i)φ+

i∫σ(A) g(λ)dµ(1−i)φ] = (1/4)[4

∫σ(A) g(λ)dµφ−0

∫σ(A) g(λ)dµ−2i

∫σ(A) g(λ)dµφ+

2i∫σ(A) g(λ)dµφ = (1/4)[4 − 0 − 2i + 2i]

∫σ(A) g(λ)dµφ =

∫σ(A) g(λ)dµφ by

usage of the polarization identity above, the integral formula and from thefact 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 measuredµµ.

The theorem stated below is basically Theorem 1 with the domain of Φextended 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 Φ : B(R) −→ L(H) such that:(a) Φ is an algebraic *-homomorphism.(b) Φ is norm continuous, that is ||Φ(f)||L(H) ≤ ||f ||∞.

(c) If f(x) = x then Φ(f) = A.(d) If fn(x) −→ f(x) for each x and ||fn||∞ ≤ M for some real M thenΦ(fn) → Φ(f) strongly.Moreover Φ has the properties:

(e) If Aψ = λψ then Φ(f)ψ = f(λ)ψ.(f) If f ≥ 0 then Φ(f) ≥ 0.(g) If BA = AB then Φ(f)B = BΦ(f).

Proof. Only item (d) in the theorem is proved. This is because theother statements follow from Theorem 1.

6

The proof of (d) is left as an exercise in [2]. Fix an arbitrary vectorψ ∈ H and let µψ be the associated spectral measure . First is noted thatfor every n: |fn − f |2 ∈ L1(R, dµψ) since with M as in Theorem 2 (d)∫|fn(x)− f(x)|2dµψ ≤ (M +M)2

∫dµψ = (M +M)2µ(R) <∞ because µψ

is a Baire measure. And further for every n: |fn − f |2 is dominated by gµψ − a.e. where g(x) = (M +M)2 and

∫|g(x)|µψ <∞. Thus, by Lebesgue

dominated convergence theorem∫σ(A) |fn − f |2 dµψ → 0. So |fn − f |2 → 0

µψ a.e.. Hence we get ||Φ(fn)(ψ) − Φ(f)(ψ)||2 = ||Φ(fn − f)ψ||2 = (Φ(fn −f)ψ, Φ(fn−f)ψ) = (ψ, Φ(fn−f)∗Φ(fn−f)ψ) = (ψ, Φ(fn − f)Φ(fn−f)ψ) =(ψ, Φ(|fn−f |2)ψ) =

∫σ(A) |fn−f |2dµψ → 0. Where the first, third and fourth

equality are due to the homomorphism property and the last by definition.

5The equality statement between the norms in Theorem 1 carries over toTheorem 2 with the aid of the following alteration. Let ||f ||′∞ be the L∞-norm on the space B(R) with the notion of almost everywhere altered asfollows: Choose an orthonormal basis of vectors ψn and define a property tobe true a.e. if it is true a.e. with respect to each µψn

. Then ||Φ(f)||L(H) =||(f)||′∞.

Definition 3 We say that a vector ψ is cyclic for an operator A if finitelinear 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 vectorψ. Then there exists a unitary operator U : H −→ L2(σ(A), dµψ) with(UAU−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 UΦ(f)ψ ≡ f . We now need to show that Uis 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. Wantto 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) =

∫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 cyclicvector so Φ(f)ψ|f ∈ C(σ(A)) = H. The Bounded Linear TransformationTheorem now enables the extension of U to an isometric map of H intoL2(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 theoremis checked. Take f ∈ C(σ(A)). Then (UAU−1f)(λ) = (UAΦ(f)ψ)(λ) =(Ux(A)Φ(f)ψ)(λ) = (UΦ(x)Φ(f)ψ)(λ) = (UΦ(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 (UAU−1fn)(λ) = λfn(λ). Thus (UAU−1f)(λ) = limn→∞(UAU−1fn)(λ) =

7

limn→∞(λfn(λ)) = λf(λ). Where the limits are to be interpreted as limitsin L2(σ(A), dµψ).

5

The next lemma asserts the extension of the previous lemma to arbitraryA, in sense of making sure that A has family of invariant subspaces spanningH so that A is cyclic on each subspace.

Lemma 4 Let A be s self-adjoint operator on a separable Hilbert spaceH. Then there is a direct sum decomposition

⊕Nn=1Hn with N=1,2....., or

∞ so that:(a) If ψ ∈ Hn then Aψ ∈ Hn.

(b) For each n there is a φn which 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 readyto introduce the spectral theorem in multiplication operator form.

Theorem 4 Let A be a bounded self-adjoint operator on a separableHilbert space H. Then there exist measures µnNn , where N=1,2,.... or∞, on σ(A) and a unitary operator U : H → ⊕N

n=1 L2(R, dµn) so that

(UAU−1ψ)n(λ) = λψn where an element ψ ∈ ⊕Nn=1 L

2(R, dµn) is writtenas a N-tuple 〈ψ1(λ), · · · , ψN (λ)〉. We call this realization of A a spectralrepresentation.

Proof. First we decompose H according to Lemma 4. For each n wehave a vector φn cyclic for A |Hn . So by Lemma 3 we have a unitary operatorU : H → L2(σ(A), dµn) where µn is the spectral measure associated withφn , with (UAU−1)ψn(λ) = λψn(λ).

5

The previous theorem indicates that every bounded self-adjoint operatoris a multiplication operator on a suitable measure space with the underlyingmeasures changing as the operator changes. This is made out more explicitlyin the following corollary.

Corollary 5 . Let A be bounded self-adjoint operator on a separableHilbert space H. Then there exists a finite measure space 〈M,µ〉, a boundedfunction F on M , and a unitary map, U : H → L2(M,dµ) such that(UAU−1f)(m) = F (m)f(m).

Proof. Choose cyclic vectors φn so that ||φn|| = 2−n and define µ byµ|Rn = µn(R), where µn is the spectral measure associated with φn, as in theproof of Theorem 4. We further define the set M to be the union of N copiesof R, i.e. M =

⋃Nn=1 R. We denote the n : th copy of R by Rn and define µ by

µ|Rn = µn(R). In the proof of 3 we defined U by UΦ(f)ψ ≡ f with the cyclic

8

vector ψ. Now, if we let f ≡ 1 and use the above definition of U we get theequality Uψ = UIψ = UΦ(1)ψ = 1 in L2(σ(A), dµψ) . Since U is a unitaryoperator we have ||Uψ|| = ||ψ||. So 1 = ||Uψ|| = ||Uψ||2 = ||ψ||2 = (ψ,ψ) =(ψ, 1(A)ψ) =

∫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(

⋃Nn=1 Rn) =

∑Nn=1 µn(R) =

∑Nn=1 ||φn|| =

∑Nn=1 2−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 completeorthonormal set of eigenvectors ψ1, · · · , ψn with Aψi = λiψi. First makethe assumption that the eigenvalues are distinct. Then consider the Diracmeasures δ(x − λi) and its sum µ =

∑ni=1 δ(x − λi). What determines

f ∈ L2(R, dµ) is its values at the points λ1 · · · λn. This is because of thenature of the measure chosen. Thus each f ∈ L2(R, dµ) can be written as an-dimensional vector with complex components, that is each f correspondsto 〈f(λ1), · · · , f(λ)n〉. So L2(R, dµ) is really C

n. With this representa-tion at hand it is clear that the function λf corresponds to the vector orn-tuple 〈λ1f(λ1), · · · , λnf(λ)n〉. So the operator A is actually multiplica-tion by λ on L2(R, dµ). However, we can do just as well if we take themeasure µ =

∑ni=1 ciδ(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 eigenvaluesare 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 theadditional assumption that A is compact. By Hilbert-Schmidt theorem weknow that there is a complete orthonormal set of eigenvectors ψn∞n=1 withAψn = λnλn, if H is separable. We can modify the measure used abovewith n replaced with ∞ and multiply each summand with a factor 2−n inorder to keep the measure finite.

Next, we will relate the spectral measure to the spectrum. In order todo 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 havemeasures µ1, · · · , µN for N ∈ N or N = ∞ and µi(Bi) = 0 for Bi ⊂ Mthen it is clear that µi(

⋂Ni 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 bethe complement of an open set. So we remedy this by defining the support

9

to be the closure of the complement of⋂Nn=1Bn. This reasoning is made

precise in the following definition.

Definition 6 . Let µnNn=1 be a family of measures, where n = 1, 2, · · ·or ∞. Then the support of µnNn=1 is the complement of the largest open

set B with µn(B) = 0 for all n; so suppµn =⋃Nn=1 suppµn.

After the definition we state a proposition that proclaims the equalitybetween the spectrum of a self-adjoint operator A and a family of spectralmeasures µnNn=1.

Proposition 6 . Let A be a self-adjoint operator and µnNn=1 a fam-ily of spectral measures in a spectral representation of A. Then σ(A) =suppµnNn=1.

We are now turning to the definition of the essential range of a real-valued function F on a measure space 〈M,µ〉.

Definition 7 . Let F be a real valued function on a measure space〈M,µ〉. We say that λ is in the essential range of F denoted by Essrange(F )if and only if µm | 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 measurespace 〈M,µ〉. 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 leftas an exercise in [2]. First suppose that λ ∈ Essrange(F ). Let ε > 0and 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 thatλ ∈ Essrange(F ) we know that µ(D) > 0. So we can rescale f into a unitvector, call this unit vector f1. Now we look at the norm ||(TF − λ)f1||2 =∫M |F (m)f1(m) − λf1(m)|2dµ(m) =

∫D |F (m)f1(m) − λf1(m)|2dµ(m) ≤(

supm∈D |F (m) − λ|2)·∫D |f1(m)|2 dµ(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 firstequality and that f1 is a unit vector in last equality. Using Weyls criterionwe conclude that λ ∈ σ(TF ).

Conversely suppose that λ ∈ σ(TF ). Weyls criterion tells us that thereexists fn∞n=1 such that ||fn|| = 1 and ||(TF − λ)fn|| → 0. Now letus assume that λ /∈ Essrange(F ). Then there exists an ε0 such thatµ(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 =

10

∫M |(F (m) − λ)f(m)|2dµ(m) ≥

∫M ε20|f(m)|2dµ(m) = ε20 · 1 = ε20. But this

contradicts the existence of a sequence fn∞n=1 with the properties ||fn|| = 1and ||(TF −λ)fn|| → ∞ asserted by Weyls criterion. Thus we conclude thatλ ∈ Essrange(TF ).

5

If a self-adjoint operator A has spectrum σ(A), then representing theoperator as UAU−1, with a unitary U does not change the spectrum. Sothe spectrum is a unitary invariant. A problem with this invariant is thatwe 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 theelements 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 bethe same. Thus, our next task will be to find better invariants that aresimpler than measures. In order to do this we start by decomposing themeasure µ into µ = µpp + µac + µsing, where µpp is the pure point mea-sure, µac is absolutely continuous with respect to Lebesgue measure, andµsing is continuous and singular with respect to Lebesgue measure. Firstwe note that µpp is singular with respect to µac because Lebesgue mea-sure assigns the value zero to point sets and µac is absolutely continuouswith respect to Lebesgue measure. Next, we note that if µac(A) = 0 fora set A then the Lebesgue measure λ(A) = 0 which in turn implies thatµsing(A

c) = 0. At last we se that µpp is singular with respect to µsing becauseµsing is a continuous measure and thus zero on point sets. Knowing thatthese three pieces are mutually singular enables the following decomposition:L2(R, dµ) = L2(R, dµpp)

⊕L2(R, dµac)

⊕L2(R, dµsing). If we are given a

family of spectral measures µn=1Nn=1 we can sum⊕N

n=1 L2(R, dµn;ac) by

defining:

Definition 8 Let A be a bounded self-adjoint operator on H. Let furtherHpp = 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 measuresare mutually singular we have assured the validity of the following theorem.

Theorem 9 H = Hpp⊕Hac

⊕Hsing. Each of these subspaces are in-

variant under A. A | Hpp has a complete set of eigenvectors, A | Hac hasonly 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.

11

Definition 10 .

σpp(A)=λ | λ is an eigenvalueσcont(A) = σ(A | Hcont ≡ Hsing

⊕Hac)

σac(A) = σ(A | Hsing)σsing(A) = σ(A | Hsing).

Before moving on to the question on multiplicity free operators we arestating a proposition. It is worth noting that it may be the case thatσac

⋃σsing

⋃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 unitarilyequivalent to multiplication by x on L2(R, dµ). That is when do we needonly one spectral measure? In the preceding example we saw that this wasthe case only when A did not have any repeated eigenvalues. With thisexample in mind we make the following definition.

Definition 11 A bounded self-adjoint operator A is called multiplicityfree 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 withoutproof.

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 thecandidates for the measure were

∑ni=1 aiδ(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:

∫C |F |dµ < ∞. So we can let

12

ν(C) =∫C |F |dµ and thus dν = Fdµ is a Borel measure. Now we define a

map U : L2(R, dν) → L2(R, dµ) where U is given by (Uf)(λ) =√F (λ)f(λ).

It is easy to see that this map is injective; for if (Uf)(λ) = (Ug)(λ) then√F (λ)f(λ) =

√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 everyf ∈ L2(R, dν) it holds true that (Uf/

√F )(λ) = f(λ). Finally (Uf,Ug) =∫ √

F (λ)f(λ)√F (λ)g(λ)dν =

∫F (λ)f(λ)g(λ)dν =

∫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 termsof ν. By the Radon-Nikodym theorem it is clear that if ν and µ have thesets of measure zero then dν = Fdµ with F 6= 0 µ − a.e.. If on the otherhand dν = Fdµ with F µ− a.e. non-zero then µ(A) =

∫A dFdµ. So µ and

ν have the same sets of measure zero. We can now use this reasoning todefine what is meant for two measures to be equivalent.

Definition 13 . Two Borel measures µ and ν are said to be equivalentif and only if they have the same sets of measure zero. An equivalence class〈µ〉 is called a measure class.

We are finally reaching our goal to decide when two operators are unitarilyequivalent.

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) andAν on L2(R, dν) by (Aνf)(λ) = λf(λ). Then Aµ and 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 listthem according to multiplicity, i.e. list all eigenvalues of multiplicity one,multiplicity two and so on. So we need a way of distinguish operators ofdifferent 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⊕m

n=1 L2(R, dµ).

Using the notion of uniform multiplicity we state a proposition concerningequivalent measures.

Proposition 10 . If A is unitarily equivalent to multiplication by λ on⊕mk=1 L

2(R, dµ) and on⊕m

k=1 L2(R, dν) then n = m and µ and ν are equiv-

alent measures.

13

3.4 Disjoint measure classes

In the canonical listening of the eigenvalues in the finite dimensional case werequire the lists to be disjoint. This prevents us from counting an eigenvaluewith a multiplicity higher than one several times. For instance if we have aneigenvalue of multiplicity two we do not want to count it as an eigenvalueof multiplicity one and then as an eigenvalue of multiplicity two. In thedefinition below we are making an analogy for measures.

Definition 15 We call two measure classes disjoint if any µ ∈ 〈µ〉 andν ∈ 〈ν〉 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 iftheir spectral multiplicity measure classes are identical.

Theorem 16 . Let A be a bounded self-adjoint operator on a Hilbertspace H. Then there is a decomposition H = H1

⊕H2

⊕ · · ·⊕H∞ so that(a) Each Hm is invariant under A.(b) A|Hm has uniform multiplicity m.(c) The measure classes 〈µm〉 associated with the spectral representation ofA|Hm are mutually disjoint.Furthermore, (a)-(c) uniquely determine the subspaces H1, · · · ,Hm, · · · ,H∞

and the measure classes 〈µ1〉, · · · , 〈µm〉, · · · , 〈µ∞〉.

14

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 charateristicfunction χΩ of this set, which is a Borel function. In the previous sectionwe extended our continuous functional calculus to Borel functional calculuswhich enables us to consider functions as above. We now turn attention toour first definition.

Definition 17 Let A be a bounded self-adjoint operator and Ω a Borelset 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 alleigenvectors of A whose eigenvalues belong to Ω. We can easily see that PΩ

is an orthogonal projection, that is PΩPΩ = PΩ = P ∗Ω if we consider χΩ.

For if χΩ(x) = 1 then χΩ(x)2 = 1 and χΩ(x) = 1. Next we consider thefamily of projections PΩ | Ω is a Borel set and state its properties in thefollowing proposition.

Proposition 11 Let PΩ be the family of spectral projections of a boundedself-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 =

⋃∞n=1 Ωn with Ωn

⋂Ωm = for all m 6= n, then

PΩ = s-limN→∞(N∑

n=1

PΩn).

(d) PΩ1PΩ2

= PΩ1

T

Ω2

Proof. (a) Is already proved.(b) Since P = χ(A) per definition we only have to note that χ ≡ 0 fromwhich is follows that P = 0. We continue with the proof of the second partof 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 functionχσ(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 abovelet Ω =

⋃Nn=1 Ωn. Then PΩ = PSN

n=1Ωn

= χSNn=1

Ωn(A) = (

∑Nn=1 χΩn)(A) =

∑Nn=1 χΩn(A) =

∑Nn=1 PΩn . 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

15

the functions χN (x) = χ∪Nn=1

Ω. 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) =

∑Nn=1 χΩn(x) → ∑∞

n=1 χΩn(x) = χ∪∞

n=1Ωn(x) = χΩ(x) when N

goes to infinity. The infinite sum is convergent because the sets consideredare disjoint. Now we apply Theorem 2 (d) to get the desired result.

(d) We get that PΩ1PΩ2

= χΩ1(A)χΩ2

(A) = χΩ1χΩ2

(A), by definitionand the homomorphism-property. So we need to check that χΩ1

χΩ2=

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 Ω1 and Ω2 which makes the left handside equal to one. Thus PΩ1

PΩ2= χΩ1

(A)χΩ2(A) = χΩ1∩Ω2

(A) = PΩ1∩Ω2.

5If we recapitulate for a minute and think about the definition of a mea-sure, especially countable additivity, we see a strong connection betweenthis and condition (c) in Proposition 11. Thus, we are motivated to makethe following definition:

Definition 18 A family of projections PΩ satisfying (a)-(c) is calleda bounded projection-valued measure abbreviated (p.v.m).

As the name projection-valued measure indicates we can use this measurefor integration. If we are given a p.v.m PΩ, then for every Borel set Ω we candefine 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 realbecause PΩ is self-adjoint. We can also easily see that m is a Borel measure:m() = (φ, Pφ) = (φ, 0) = 0, if Ω1,Ω2, . . . are any pairwise disjoint Borelsets then m(

⋃∞n=1 Ωn) = (φ, PS

∞

n=1Ωnφ) = (φ, s-limN→∞

∑Nn=1 PΩnφ) =

limN→∞∑N

n=1(φ, PΩnφ) = limN → ∞∑Nn=1m(Ωn) =

∑∞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) =

∫f(λ)d(φ, Pλφ) for all vectors φ ∈ H. So we can now state our

next theorem:

Theorem 19 If PΩ is a projection-valued measure and f a bounded Borelfunction on supp(PΩ), then there is a unique operator B which we denote∫f(λ)dPλ, so that (φ,Bφ) =

∫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 associatedprojection-valued measure, it can be shown that f(A) =

∫f(λ)dPλ. If we

take f(x) = x we get A =∫λdPλ. We will use this fact later in the proof of

Proposition 12. We are now ready to state the spectral theorem for boundedself-adjoint operators in its third form, that is in p.v.m. form.

16

Theorem 20 There is a one-one correspondence between the boundedself-adjoint operators A and the bounded projection valued measures Ωthat is given by:A 7→ PΩ = χΩ(A)PΩ 7→ A =

∫λdPλ.

Now we put our new knowledge about spectral projections into use whenwe are stating a proposition that gives a condition for a real number λ tobe in the spectrum of a self-adjoint bounded operator A. The proof of thenext 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) = 0and let Ω = (λ0 − ε0, λ0 + ε0). We aim to show that λ0 /∈ σ(A), in otherwords λ0 ∈ ρ(A), where ρ(A) denotes the resolvent of A. Now let φ be anarbitrary 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. Sod(φ, Pλφ) = 0 on Ω. Know we consider ||(A − λ0)φ||2 = ((A − λ0)φ, (A −λ0)φ) = (φ, (A − λ0)

2φ) =∫

(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 wecan make the following estimate. ||(A − λ0)φ||2 ≥ ε20

∫d(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. Weknow that there exists ε0 such that

∫(λ − λ0)

2d(φ, Pλφ) ≥ ε20∫d(φ, Pλφ).

Choose a positive number η < ε0. Since P(λ0−η,λ0+η) 6= 0 there is a vectorψ ∈ 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 getthat P(λ0,ω)P(λ0−η,λ0+η)(φ) = 0 because χ(λ0,ω)χ(λ0−η,λ0+η)(x) = 0. Thuswe have that (ψ, P(λ0 ,ω)ψ) = 0 and hence d(ψ, Pλψ) = 0. The case wherewe are on the left side of this interval is similar. So when integrating overthe interval (λ0 − η, λ0 + η) have that (λ − λ0)

2 ≤ η2 < ε20. But this is acontradiction to the integral-inequality above. Thus P(λ0−η,λ0+η) = 0. 5We 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 spectrumin these parts enables a decomposition of the spectrum of A into two disjointsubsets.

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 allε > 0 where the phrase P is infinite dimensional means that Range(P ) isinfinite dimensional. If λ ∈ σ(A) and there exist ε0 such that P(λ−ε0,λ+ε0)

17

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) intotwo disjoint sets. The next theorem states that the essential spectrum of aself-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 thatλn → λ as n→ ∞. Because λ is a point of accumulation we know that anyopen interval I about λ contains an open interval In about some λn. Sinceλn is in the essential spectrum of A we know that P(λn−ε,λn+ε)(A) is infinitedimensional for every ε > 0 and since In = (λ− ε, λ+ ε) is included in I forsufficiently small ε we conclude that PI(A) is infinite dimensional. This istrue for every open interval I about λ; thus λ ∈ σess(A) and hence σess(A)is closed. 5The following theorem gives conditions for λ ∈ σ(A) to belong to the discretespectrum. 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 onlyif 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 finitedimensional.

Proof. Suppose that λ ∈ σdisc(A), that is there exists ε > 0 so thatRange(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, fori, 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 anis 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 existsa N such that aN = an+1 = aN+2 = · · · . Our next step is to showthat there is no λ′ ∈ (λ − ε/N, λ + ε/N) with λ′ 6= λ that belongs toσ(A). Suppose there is such λ′ ∈ σ(A) and suppose further that λ′ < λ.

18

The case where λ′ > λ is treated in a similar fashion. Choose ε1 > 0so small that λ − ε/N < λ′ + ε1 < λ, then we choose n > N largeenough so that λ′ + ε1 < λ − ε/n. In line with our reasoning above weget that Range(P(λ′−ε1,λ′+ε1)) and Range(P(λ−ε/n,λ+ε/n)) are orthogonalsubspaces of Range(P(λ−ε/N,λ+ε/N)), butwe also know that because anis 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(λ′−ε1,λ′+ε1)) =0, that is P(λ′−ε1,λ′+ε1) = 0. By Proposition 12 we conclude that λ′ /∈σ(A).Still assuming that λ ∈ σdisc(A) we know that there exists some ε0 > 0 suchthat 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)). Thusdim(ψ | Aψ = λψ) ≤ dim(Range(P(λ−ε0 ,λ+ε0)(A)) <∞.

Conversely suppose that both (a) and (b) hold true. Then there exists someε1 > 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. Henceλ ∈ σdisc(A).

5The next theorem concerns the essential spectrum and the proof of thistheorem 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 usedfrequently in the previous sections, namely Weyls criterion that states acondition 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 andonly if there exists a sequence ψn∞n=1 such 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 thatλ ∈ 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 theresolvent set of an operator if and only if there exists c > 0 such that for allx ∈ H : ||(T − λI)x|| ≥ c||x||. This shows that if λ ∈ ρ(A) there cannot bea sequence ψn∞n=1 with the property limn→∞ ||(A− λ)ψn|| = 0.

19

5

5 Unbounded operators

This section about unbounded operators is merely a brief introduction tothe topic. We mention a few motivating examples and state some impor-tant definition and theorems. Not every operator of importance in analysisis bounded. For instance consider the differentiation operator. It is nothard to see that this operator is unbounded. We consider the space of allpolynomials P [t] = p(t) | p is a polynomial . For simplicity lets us pickthe interval [0, 1] and equip it with the norm ||P || = max[0,1]|p(t)|. Wedefine D to be the differentiation operator, that is Dp(t) = p′(t). If wetake the polynomial pn(t) = tn where n is a natural number, then certainly||pn|| = 1 and Dpn(t) = p′(t) = ntn−1. Then ||DPn|| = n||tn−1|| = n. Thus||Txn||/||xn|| = n and n is an arbitrary natural number. So D is an un-bounded operator. Since the differentiation operator is such an importantoperator 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 ununbounded linear operator T : H → H that satisfies (Tx, y) = (x, Ty) forall x, y cannot be defined on all of H.

Theorem 26 If a linear operator T is defined on all of H and satisfies(Tx, y) = (x, Ty) for all x, y then T is bounded.

This theorem suggests that an unbounded operator T is only defined ona dense linear subset of the Hilbert space H. We will in the sequel alwaysassume that the domain of the operator T , denoted by D(T ), is a densesubset of H. We continue with an example of an unbounded operator wherethe Hilbert space is L2(R).

Example 2 We takeH = L2(R) and letD(T ) = f ∈ L2(R) |∫

Rx2|f(x)|2dx <

∞. For f ∈ D(T ) we define the operator T by (Tf)(x) = xf(x). We nowaim to show that T is unbounded. If we choose a function f ∈ D(T ) suchthat f(x) 6= 0 near infinity, we realize that we can make ||Tf || =

∫Rx2|f(x)|2

arbitrarily big. Thus ||T || = sup||f ||=1||Tf || = ∞. That is T is unbounded.It we want Range(T ) to be L2(R) we must stick to the restriction we madeabove. For if f /∈ D(T ) then

∫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 ithas 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.

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Definition 27 Let H be a Hilbert space and T : D(T ) → H with D(T ) ⊂H. Then we call T a closed linear operator if its graph Γ(T ) = 〈x, Tx〉 |x ∈ D(T ) is a closed subset of H × H. Here H × H is a Hilbert spaceequipped with the inner product (〈x1, y1〉), 〈x2, y2〉) = (x1, y1) + (x2, y2).

Another important notion concerning unbounded operators is that of ex-tending an operator. We can define the extension of an operator in twoways which are equivalent. We list them both for sake of clarity.

Definition 28 Let T1 and T2 be linear operators on the Hilbert space H.We say that T2 is an extension of T1 denoted by T1 ⊂ T2 if and only ifD(T1) ⊂ D(T2) and T1 = T2 |D(T1). We call T2 a proper extension of T1 ifand only if D(T2) −D(T1) 6= .

Definition 29 With the notion from above kept intact we say that T1 ⊂T2 if and only Γ(T1) ⊂ Γ(T2).

If an operator is not closed we can always try to find a closed extension T .If we find such an operator we then say that the operator T is closable andT is the closure of T . We state this precisely in the following definition.

Definition 30 We call an operator T closable if it has a closed extensionT . The minimal of these extension are called the closure of T denoted T .Here minimal means that every extension T1 of T is an extension of T .

For symmetric operators, that is operators such that (Tx, y) = (x, Ty) forall x, y ∈ D(T ), it is the case that they always have unique extension. wewill introduce these operators more properly later. Next, we state and provea theorem that says that the graph of the closure of an operator equals theclosure of the graph of the operator.

Theorem 31 Let T be an operator on a Hilbert space H with D(T ) ⊂ H.It T is closable, then Γ(T ) = Γ(T ).

Proof. First suppose that S is an arbitrary extension of T . Then Γ(T ) ⊂Γ(S) be definition and if we take the closure we get that Γ(T ) ⊂ Γ(S) =Γ(S). So if we have that 〈0, ψ〉 ∈ Γ(T ) then 〈0, ψ〉 ∈ Γ(S) and ψ = S0 = 0.Let us now define the operator R : D(R) → H where D(R) = ψ | 〈ψ, φ〉 ∈Γ(T ) for some φ , by Rψ = φ where φ ∈ H is the unique vector so that〈ψ, φ〉 ∈ Γ(T ). Let us look at the graph of R. We see that Γ(R) = 〈ψ,Rψ〉 |ψ ∈ D(R) = 〈ψ, φ〉 | 〈ψ, φ〉 ∈ Γ(T ) = Γ(T ). 5We now define the adjoint operator in the unbounded case.

Definition 32 Let T be a linear operator on a Hilbert space H with D(T )dense in H. Let D(T ∗) be the set of φ ∈ H for which there is an η ∈ Hwith (Tψ, ϕ) = (ψ, η) for all ψ ∈ D(T ). For ϕ ∈ D(T ∗) we define T ∗ϕ = η.We call T ∗the adjoint operator of T . Further by Riesz lemma ϕ ∈ H if andonly if |(Tψ, ϕ)| ≤ C||ψ||.

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As in the bounded case it holds that S ⊂ T implies T ∗ ⊂ S∗. It is alsoworth to note that the domain of T ∗ need not be dense in H which is thecase for bounded operators. We can even have that D(T ∗) = 0. Next wedefine what it means for an unbounded operator to be symmetric.

Definition 33 A densely defined operator T on a Hilbert space H iscalled symmetric or Hermitian if T ⊂ T ∗, that is, if D(T ) ⊂ D(T ∗)and Tϕ = T ∗ϕ for all ϕ ∈ D(T ). Equivalently, T is symmetric if and onlyif (Tϕ, ψ) = (ϕ, Tψ) for all ϕ,ψ ∈ D(T ).

Now we use this definition to define what it means for an operator to beself-adjoint.

Definition 34 With the notation kept intact from the previous definitionwe say that T is self-adjoint if T = T ∗, that is, if and only if T is symmetricand D(T ) = D(T ∗).

The three main spectral theorems for bounded self-adjoint operators, i.e.Theorem 2, Theorem 4 and Theorem 20, all carry over to the unboundedcase almost without any changes at all.

References

[1] E. Kreyszig, Introductory functional analysis with applications,Wiley Classics Library Edition, John Wiley & Sons, New York,1989.

[2] M. Reed and B. Simon, Methods of modern mathematical physics.I. Functional analysis, Academic Press Inc., New York, 1980.

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