Spectral Theory for Bounded Self-Adjoint Operators

Home , Multiplication operator, Spectral radius, Spectral theorem, Spectral theory, Unbounded operator

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 Uppsala University

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 eigenvectors. The spectral theorem says that there exists a total orthonormal set of eigenvectors ψ ψ such that [A] can be diagonalized, or equivalently 1 · · · n that there exists a unitary matrix [U] and eigenvalues λ , , λ so that 1 · · · n for each ψ H, ([U][A][U −1]ψ ) = λ ψ . This result carries over to the ∈ i i i 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 mathematical 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 thesis 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 f(A). This map has a number of desirable properties, 7→ for example it is a ”conjugate-homomorphism” and continuous. The first version 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 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 measure 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 P χ (A) a spectral projection of A. We further call a family of such Ω ≡ Ω projections P Ω is an arbitrary Borel set a projection valued measure { Ω | } 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 operators. 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 n k define f(A) to make sense? If f is given by a finite polynomial k=0 akx n k P we want f(A) to be k=0 akA . We also want f to be well defined in the P ∞ k sense of convergence of the sum. If f is given by k=0 akx with radius of convergence R and A < R then ∞ a Ak conP verges in L(H) (the || || k=0 k set of bounded linear operators on H). PSo if f is given by an infinite series then f(A) is given by ∞ a Ak. The theorem below states the existence k=0 k · and uniqueness of an opPerator 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) C f . || ||L(H) ≤ || ||∞ (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 polynomial 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 n k n k asserts that if P (x) = k=0 akx then Φ(P ) = k=0 akA . Weierstrass theoremPstates that the set of polynomialsP is dense in C(σ(A)) so the core of the proof is to prove that P (A) = P (x) || ||L(H) || ||C(σ(A)) ≡ sup (P (λ)) . λ∈σ(A) | | 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.

n k n k Lemma 1 Let P (x) = k=0 akx and P (A) = k=0 akA . Then σ[P (A)] = P (λ) λ σ(A) . P P { | ∈ } 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 λ ) (x λ ). − − 1 · · · − n If no λ σ(A) then (P (A) µ)−1 = a−1(A λ )−1 (A λ )−1 exists. i ∈ − − 1 · · · · − n But this contradicts that µ σ(A). ∈ Thus λ σ(A) for some i and hence µ = P (λ) for some λ σ(A). i ∈ ∈ 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) = || || sup P (λ) . λ∈σ(A) | | 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 = 2 n n || ||∗ T (cf. [1, Theorem 3.9-4(e)]). Also, if P (A) = k=0 anA then P (A) = || ||n n ∗ n n P ( k=0 anA ) = k=0 an A = P (A). Further (P P )(A) is self-adjoint: (P P )(A)∗ = P (A)P∗(P (A))∗ = P (A)P (A)∗∗ = P (A)P (A) = (P P )(A). So sup λ = r(P P (A)) = P P (A) , where r is the spectral radius of λ∈σ(P P (A)) | | || || P P (A). And then by Lemma 1 : sup λ = sup P P (λ) . So λ∈σ(P P (A)) | | λ∈σ(A) | | P (A) 2 = P (A)∗P (A) = P P (A) = sup λ || || || || || || λ∈ σ(P P (A)) | | = sup P P (λ) = sup P (λ) P (λ) λ∈σ(A) | | λ∈σ(A) | || | = sup P (λ) 2 = ( sup P (λ) )2. λ∈σ(A) | | λ∈σ(A) | |

3 5 Proof of Theorem 1. Let Φ(P ) = P (A). Then Φ(P ) = || ||L(H) P (A) = P so Φ can be extended to the closure of poly- || ||L(H) || ||C(σ(A)) 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 ) = P sup P (λ) = || ||L(H) || ||C(σ(A)) ≡ λ∈σ(A) | | P . || ||∞ (c) P (x) = x = ΦP = P (A) = 1A = A. ⇒ To prove uniqueness, suppose Φ obeys (a),(b) and (c) then Φ agrees with Φ on polynomials and thus by conetinuity on C(σ(A)). n n (d) If Aψ = λψ, then Φ(P )ψ = P (A)ψ = ( k=0 an A )ψ = (c0 + c1A + 2 n 2 P n · 2 c2A +...... c2A )ψ = c0ψ +c1Aψ +c2A ψ +...... c2A ψ = (c0 +c1λ+c2λ + n ...... c2λ )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) = c tn + c tn−1 + + c . Then P (A) = Q(A µ), and P (µ) = n n−1 · · · 0 − 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 c ψ + (f P )(A)ψ − − − || || − − 0 − − (f P )(µ)ψ Q(A µ)ψ c ψ + (f P )(A)ψ (f P )(µ)ψ − || ≤ || − − 0 || || − − − || ≤ c (A µ)nψ + c (A µ)n−1ψ + ... + c ψ c ψ + (f P )(A)ψ + (f || n − n−1 − 0 − 0 || || − || || − P )(µ)ψ = (c (A µ)n−1 + c (A µ)n−2 + ... + c )(A µ)ψ + P || || n − n−1 − 1 − || || − f + P f < (c (A µ)n−1 + c (A µ)n−2 + ... + c )(A µ)ψ + ||∞ || − ||∞ || n − n−1 − 1 − || ε + ε (c (A µ)n−1 + c (A µ)n−2 + ... + c ) (A µ)ψ + 2ε. ≤ || n − n−1 − 1 || · || − || 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. Let || − ||∞ K = (c (A µ)n−1 + c (A µ)n−2 + ... + c ) . Since we assumed || n − n−1 − 1 || that µ σ(A) Weyls criterion says that there is a unit vector ψ so that ∈ (A µ)v < ve . If we insert this in the computations above we get that || − || K+1 (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. That is A∗ = A and A < . Let ψ be a vector in H. Then f (ψ, f(A)ψ) || || ∞ 7−→ 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) = f(λ)dµ . We call µ the spectral measure associated Rσ(A) ψ ψ 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) = g(λ)dµ . Rσ(A) ψ Consider the inner product (ψ, g(A)φ) for all pairs of vectors ψ, φ. The following 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) = g(λ)dµ , as desired. Where the last equality Rσ(A) φ is due to (φ, g(A)φ) = (1/4)[ g(λ)dµ g(λ)dµ i g(λdµ + Rσ(A) 2φ−Rσ(A) 0− Rσ(A) (1+i)φ i g(λ)dµ ] = (1/4)[4 g(λ)dµ 0 g(λ)dµ 2i g(λ)dµ + Rσ(A) (1−i)φ Rσ(A) φ− Rσ(A) − Rσ(A) φ 2i g(λ)dµ = (1/4)[4 0 2i + 2i] g(λ)dµ = g(λ)dµ by Rσ(A) φ − − Rσ(A) φ Rσ(A) φ 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 µ = c 2µ . cv | | v 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 Φˆ : B(R) L(H) such that: −→ (a) Φˆ is an algebraic *-homomorphism. (b) Φˆ is norm continuous, that is Φˆ (f) f . || ||L(H) ≤ || ||∞ (c) If f(x) = x then Φ(ˆ f) = A. (d) If f (x) f(x) for each x and f M for some real M then n −→ || n||∞ ≤ Φ(ˆ f ) Φ(ˆ f) strongly. n → Moreover Φˆ has the properties: (e) If Aψ = λψ then Φ(ˆ f)ψ = f(λ)ψ. (f) If f 0 then Φ(ˆ f) 0. ≥ ≥ (g) If BA = AB then Φˆ(f)B = BΦˆ(f).

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

6 The proof of (d) is left as an exercise in [2]. Fix an arbitrary vector ψ H and let µ be the associated spectral measure . First is noted that ∈ ψ for every n: f f 2 L1(R, dµ ) since with M as in Theorem 2 (d) | n − | ∈ ψ f (x) f(x) 2dµ (M + M)2 dµ = (M + M)2µ(R) < because µ R | n − | ψ ≤ R ψ ∞ ψ is a Baire measure. And further for every n: f f 2 is dominated by g | n − | µ a.e. where g(x) = (M + M)2 and g(x) µ < . Thus, by Lebesgue ψ − R | | ψ ∞ dominated convergence theorem f f 2 dµ 0. So f f 2 0 Rσ(A) | n − | ψ → | n − | → µ a.e.. Hence we get Φ(ˆ f )(ψ) Φ(ˆ f)(ψ) 2 = Φˆ (f f)ψ 2 = (Φ(ˆ f ψ || n − || || n − || n − f)ψ, Φ(ˆ f f)ψ) = (ψ, Φˆ(f f)∗Φˆ(f f)ψ) = (ψ, Φˆ(f f)Φ(ˆ f f)ψ) = n − n − n − n − n − (ψ, Φˆ( f f 2)ψ) = f f 2dµ 0. Where the first, third and fourth | n− | Rσ(A) | n− | ψ → 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 µ . Then Φ(f) = ψn || ||L(H) (f) 0 . b || ||∞ Definition 3 We say that a vector ψ is cyclic for an operator A if finite linear combinations of the elements Anψ ∞ are dense in H. { }n=0 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), dµ ) with −→ ψ (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), dµ ) by UΦ(f)ψ f. We now need to show that U } → ψ ≡ is well defined and not dependent of choices of representatives. So let f, g 2 ∈ L (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) = f(λ) g(λ) 2dµ = 0. Thus f g = 0, so f = g. | − | Rσ(A) | − | ψ − 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 2 2 L (σ(A), dµψ ). Since C(σ(A)) = L (σ(A), dµψ) we know that Range(U) = 2 L (σ(A), dµψ ). Finally the validity of the formula stated in the theorem is 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) = 2 λf(λ). So if f L then there is a sequence fn C(σ(A)) such that fn −1∈ {−1 } ∈ −1 → f and (UAU fn)(λ) = λfn(λ). Thus (UAU f)(λ) = limn→∞(UAU fn)(λ) =

7 limn→∞(λfn(λ)) = λf(λ). Where the limits are to be interpreted as limits 2 in L (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 N H. Then there is a direct sum decomposition n=1 Hn with N=1,2....., or so that: L ∞ (a) If ψ H then Aψ H . ∈ n ∈ n (b) For each n there is a φ which is cyclic for A , i.e. H = f(A)ψ f C(σ(A)) . n |Hn n { n | ∈ } 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 N Hilbert space H. Then there exist measures µn n , where N=1,2,.... or { } N 2 , on σ(A) and a unitary operator U : H L (R, dµn) so that ∞ → Ln=1 (UAU −1ψ) (λ) = λψ where an element ψ N L2(R, dµ ) is written n n ∈ n=1 n as a N-tuple ψ (λ), , ψ (λ) . We call thisLrealization of A a spectral h 1 · · · N i representation.

Proof. First we decompose H according to Lemma 4. For each n we have a vector φn cyclic for A Hn . So by Lemma 3 we have a unitary operator 2 | U : H L (σ(A), dµn) where µn is the spectral measure associated with → −1 φn , with (UAU )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 M, µ , a bounded h i function 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 φ so that φ = 2−n and define µ by n || n|| µ R = µ (R), where µ is the spectral measure associated with φ , as in the | n n n n proof of Theorem 4. We further define the set M to be the union of N copies N of R, i.e. M = R. We denote the n : th copy of R by Rn and define µ by Sn=1 µ R = µ (R). In the proof of 3 we defined U by UΦ(f)ψ f with the cyclic | n n ≡

8 vector ψ. Now, if we let f 1 and use the above definition of U we get the ≡ 2 equality Uψ = UIψ = UΦ(1)ψ = 1 in L (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) = 1dµ = µ (σ(A)) = µ (R). Hence in our this setting we Rσ(A) ψ ψ ψ have φ = 2−n = µ (R ). It is now clear that µ as defined above is finite, || n|| n n for µ(M) = µ( N R ) = N µ (R) = N φ = N 2−n < . n=1 n n=1 n n=1 || n|| n=1 ∞ Thus µ is a finiteS measure. P P P 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 ψ , , ψ with Aψ = λ ψ . First make 1 · · · n i i i the assumption that the eigenvalues are distinct. Then consider the Dirac measures δ(x λ ) and its sum µ = n δ(x λ ). What determines − i i=1 − i f L2(R, dµ) is its values at the pointsPλ λ . This is because of the ∈ 1 · · · n 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 f(λ ), , f(λ) . So L2(R, dµ) is really Cn. With this representa- h 1 · · · ni tion at hand it is clear that the function λf corresponds to the vector or n-tuple λ f(λ ), , λ f(λ) . So the operator A is actually multiplica- h 1 1 · · · n ni tion by λ on L2(R, dµ). However, we can do just as well if we take the n measure µ = i=1 ciδ(x λi), where ci > 0 for every index i. Then A is P − 2 also represene ted by as multiplication by λ on L (R, dµ). If the eigenvalues are not distinct, we can not represent a self-adjoint operator as multiplication 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 ψ ∞ with { n}n=1 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) := B c.

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 · · · N ∈ ∞ ⊂ then it is clear that µi( i Bi) = 0 for all i. So the support of the sequence should be the complemenTt 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

9 N to be the closure of the complement of n=1 Bn. This reasoning is made precise in the following deﬁnition. T

N Definition 6 . Let µn n=1 be a family of measures, where n = 1, 2, { } N · · · or . Then the support of µn n=1 is the complement of the largest open ∞ { } N set B with µn(B) = 0 for all n; so supp µn = supp µn . { } Sn=1 { } 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 µ N . { n}n=1 Proposition 6 . Let A be a self-adjoint operator and µ N a fam- { n}n=1 ily of spectral measures in a spectral representation of A. Then σ(A) = supp µ N . { n}n=1 We are now turning to the definition of the essential range of a real- valued function F on a measure space M, µ . h i 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 ) h i 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 M, µ . Let T be the operator on L2(M, dµ) given by T (g)(m) = h i F F 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 f . Now we look at the norm (T λ)f 2 = 1 || F − 1|| F (m)f (m) λf (m) 2dµ(m) = F (m)f (m) λf (m) 2dµ(m) RM | 1 − 1 | RD | 1 − 1 | ≤ sup F (m) λ 2 f (m) 2 dµ(m) = sup F (m) λ 2 f 2 = m∈D | − | · RD | 1 | m∈D | − | || 1|| sup F (m) λ 2 ε2. Here we have used that f = χ , in the ﬁrst m∈D | − | ≤ D equality and that f1 is a unit vector in last equality. Using Weyls criterion we conclude that λ σ(T ). ∈ F Conversely suppose that λ σ(T ). Weyls criterion tells us that there ∈ F exists f ∞ such that f = 1 and (T λ)f 0. Now let { n}n=1 || n|| || F − n|| → us assume that λ / Essrange(F ). Then there exists an ε such that ∈ 0 µ( m λ ε < F (m) < λ + ε ) = 0. This means that F (m) λ ε { | − } | − | ≥ 0 for µ a.e.m. Hence for all unit vectors f we get that (T λ)f 2 = − || F − ||

10 (F (m) λ)f(m) 2dµ(m) ε2 f(m) 2dµ(m) = ε2 1 = ε2. But this RM | − | ≥ RM 0| | 0 · 0 contradicts the existence of a sequence f ∞ with the properties f = 1 { n}n=1 || n|| and (T λ)f asserted by Weyls criterion. Thus we conclude that || F − n|| → ∞ λ Essrange(T ). ∈ F 5 If a self-adjoint operator A has spectrum σ(A), then representing the operator as UAU −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 instance, 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 realize this by noting that σ(A) is equal to the support of the spectral measures 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 measure, 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 measure 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 c µsing(A ) = 0. At last we se that µpp is singular with respect to µsing because µsing is a continuous measure and thus zero on point sets. Knowing that these three pieces are mutually singular enables the following decomposition: 2 2 2 2 L (R, dµ) = L (R, dµpp) L (R, dµac) L (R, dµsing). If we are given a L L family of spectral measures µ N we can sum N L2(R, dµ ) by { n=1}n=1 n=1 n;ac defining: L

Deﬁnition 8 Let A be a bounded self-adjoint operator on H. Let further H = ψ µ is pure point , H = ψ µ absolutely continuous , pp { | ψ } ac { | ψ } H = ψ µ is continuous singular . sing { | ψ } 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 = Hpp Hac Hsing. Each of these subspaces are invariant under A. A H Lhas a Lcomplete set of eigenvectors, A H has | pp | ac only absolutely continuous measure and A H has only continuous sin- | sing gular spectral measure.

Next we turn to the deﬁnition concerning the breakup of the spectrum.

11 Deﬁnition 10 .

σ (A)= λ λ is an eigenvalue pp { | } σ (A) = σ(A H H H ) cont | cont ≡ sing ac σ (A) = σ(A H ) L ac | sing σ (A) = σ(A H ). sing | sing 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 σ σ σ = σ. This is due to the deﬁnition of σ as the set of ac sing pp 6 pp eigenSvaluesSof A and not σ(A H ). The below statement is always true. | pp Proposition 8

σ (A) = σ (A) σ (A) cont ac [ sing σ(A) = σ (A) σ (A) pp [ cont

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 measure in the multiplicity free cases. In the example was stated that the candidates for the measure were n a δ(x λ ) with a = 0. In order i=1 i − i i 6 to generalize this suppose that dµPon 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: F dµ < . So we can let ⊂ RC | | ∞

12 ν(C) = F dµ and thus dν = F dµ is a Borel measure. Now we deﬁne a RC | | map U : L2(R, dν) L2(R, dµ) where U is given by (Uf)(λ) = F (λ)f(λ). → p 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 = 0 and p p 6 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 (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). R p p R R So we can conclude that U is unitary operator. So an operator A with spectral 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 = 0 µ a.e.. If on the other 6 − hand dν = F dµ with F µ a.e. non-zero then µ(A) = dF dµ. So µ and − RA 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 µ is called a measure class. h i 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- 2 port. Let the operator Aµ on L (R, dµ) be given by (Aµf)(λ) = λf(λ) and 2 Aν on L (R, dν) by (Aνf)(λ) = λf(λ). Then Aµ and Aν are unitarily equivalent 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 uniform multiplicity m if A is unitarily equivalent to multiplication by λ on m L2(R, dµ). Ln=1 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 m 2 R m 2 R k=1 L ( , dµ) and on k=1 L ( , dν) then n = m and µ and ν are equiv- Lalent measures. L

13 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 µ and ∈ h i ν ν are mutually singular. ∈ h i Before ending this section about spectral measures we are stating the multiplicity theorem. This theorem says that each bounded self-adjoint operator 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 = H1 H2 H∞ so that L L · · · L (a) Each Hm is invariant under A. (b) A H has uniform multiplicity m. | m (c) The measure classes µ associated with the spectral representation of h mi A H are mutually disjoint. | m Furthermore, (a)-(c) uniquely determine the subspaces H , , H , , H 1 · · · m · · · ∞ and the measure classes µ , , µ , , µ . h 1i · · · h mi · · · h ∞i

14 4 Spectral projections

In this section we present the spectral theorem in projection-valued measure 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 P χ (A) a spectral projection of A. Ω ≡ Ω

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. 2 For if χΩ(x) = 1 then χΩ(x) = 1 and χΩ(x) = 1. Next we consider the family of projections P Ω is a Borel set and state its properties in the { Ω | } following proposition.

Proposition 11 Let P be the family of spectral projections of a bounded { Ω} self-adjoint operator A. Then P has the following properties: { Ω} (a) Each PΩ is an orthogonal projection. (b) P = 0; P(−a,a) = I for some a. ∞ (c) If Ω = Ωn with Ωn Ωm = for all m = n, then Sn=1 T 6 N P = s-lim ( P ). Ω N→∞ X Ωn n=1

(d) PΩ1 PΩ2 = PΩ1 T Ω2

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 χ (A) = P = I. || || σ(A) (−a,a) (c) First consider the case were N is finite. As in the statement above N N let Ω = Ωn. Then PΩ = P N = χ N (A) = ( χΩn )(A) = n=1 Sn=1 Ωn Sn=1 Ωn n=1 N S N P n=1 χΩn (A) = n=1 PΩn . The third equality is valid because the sets con- Psidered are disjoinPt 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 N . We know that χN ∞ 1 for each N be- ∪n=1Ω || || ≤ cause the sets are disjoint. We also know that for each fixed x R, χN (x) = N ∞ ∞ ∈ χ N (x) = χΩn (x) χΩn (x) = χ∪ Ωn (x) = χΩ(x) when N ∪n=1Ω n=1 → n=1 n=1 goes to infinitPy. The infinite sumP is convergent because the sets considered are disjoint. Now we apply Theorem 2 (d) to get the desired result.

(d) We get that PΩ1 PΩ2 = χΩ1 (A)χΩ2 (A) = χΩ1 χΩ2 (A), by deﬁnition and 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 hand side equal to one. Thus PΩ1 PΩ2 = χΩ1 (A)χΩ2 (A) = χΩ1∩Ω2 (A) = PΩ1∩Ω2 . 5 If we recapitulate for a minute and think about the deﬁnition of a measure, 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 P satisfying (a)-(c) is called { Ω} 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 Ω , Ω , . . . are any pairwise disjoint Borel 1 2 sets then m( ∞ Ω ) = (φ, P ∞ φ) = (φ, s-lim N P φ) = n=1 n Sn=1 Ωn N→∞ n=1 Ωn N S N ∞P limN→∞ (φ, PΩ φ) = limN m(Ωn) = m(Ωn). We Pn=1 n → ∞ Pn=1 Pn=1 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, P φ) for all vectors φ H. So we can now state our R λ ∈ next theorem:

Theorem 19 If PΩ is a projection-valued measure and f a bounded Borel function on supp(PΩ), then there is a unique operator B which we denote f(λ)dP , so that (φ, Bφ) = f(λ)d(φ, P φ), for all φ H. R λ R λ ∈ 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) = f(λ)dP . If we R λ take f(x) = x we get A = λdP . We will use this fact later in the proof of R λ 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.

16 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 P = χ (A) 7→ { Ω} { Ω } P A = λdP . { Ω} 7→ R λ 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 (A) = 0 for any ε > 0. ∈ (λ−ε,λ+ε) 6

Proof. Suppose that there exists ε0 > 0 such that P(λ0−ε0,λ0+ε0)(A) = 0 and let Ω = (λ ε , λ + ε ). We aim to show that λ / σ(A), in other 0 − 0 0 0 0 ∈ words λ ρ(A), where ρ(A) denotes the resolvent of A. Now let φ be an 0 ∈ 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, P φ) = 0 on Ω. Know we consider (A λ )φ 2 = ((A λ )φ, (A λ || − 0 || − 0 − λ )φ) = (φ, (A λ )2φ) = (λ λ )2d(φ, P φ). Here we have used the re- 0 − 0 R − 0 λ 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 λ )2 ε2. So we ∈ − 0 ≥ 0 can make the following estimate. (A λ )φ 2 ε2 d(φ, P φ) = ε2 φ 2. || − 0 || ≥ 0 R λ 0|| || Thus we get that (A λ )φ ε φ . So by (cf. [1, Theorem 9.1-2(e)]) || − 0 || ≥ 0|| || λ ρ(A). 0 ∈ Next suppose that λ ρ(A) and P = 0 for any ε > 0. We 0 ∈ (λ0−ε,λ0+ε) 6 know that there exists ε such that (λ λ )2d(φ, P φ) ε2 d(φ, P φ). 0 R − 0 λ ≥ 0 R λ Choose a positive number η < ε . Since P = 0 there is a vector 0 (λ0−η,λ0+η) 6 ψ H such that P (φ) = ψ for some φ H. Know if we con- ∈ (λ0−η,λ0+η) ∈ 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 (λ η, λ + η) have that (λ λ )2 η2 < ε2. But this is a 0 − 0 − 0 ≤ 0 contradiction to the integral-inequality above. Thus P = 0. (λ0−η,λ0+η) 5 We are now deﬁning what is meant by the essential spectrum and the discrete 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 σ (A), if and only if P is infinite dimensional for all ∈ ess (λ−ε,λ+ε) ε > 0 where the phrase P is infinite dimensional means that Range(P ) is infinite dimensional. If λ σ(A) and there exist ε such that P ∈ 0 (λ−ε0,λ+ε0)

17 is ﬁnite dimensional, we say that λ is in the discrete spectrum of A, denoted λ σ (A). ∈ disc 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 λ σ (A) and assume that { n} ∈ ess λ λ as n . Because λ is a point of accumulation we know that any n → → ∞ open interval I about λ contains an open interval In about some λn. Since

λn is in the essential spectrum of A we know that P(λn−ε,λn+ε)(A) is infinite dimensional for every ε > 0 and since I = (λ ε, λ + ε) is included in I for n − sufficiently small ε we conclude that PI (A) is infinite dimensional. This is true for every open interval I about λ; thus λ σ (A) and hence σ (A) ∈ ess ess 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 σ (A) if and only ∈ disc if both of the following conditions hold:

(a) λ is an isolated point of σ(A), meaning that there exists ε0 such that (λ ε , λ + ε ) σ(A) = λ . − 0 0 ∩ { } (b) λ is an eigenvalue of ﬁnite multiplicity, that is ψ Aψ = λψ is ﬁnite { | } dimensional.

Proof. Suppose that λ σ (A), that is there exists ε > 0 so that ∈ disc Range(P ) is finite dimensional. For every n 1 define a := (λ−ε,λ+ε) ≥ n dim(Range(P )). We will show that a is decreasing, i.e. (λ−ε/n,λ+ε/n) { n} a a . Let us decompose (λ ε/n, λ + ε/n) into three disjoint sets; n+1 ≥ n − (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 I , I , I . ∪ 1 2 3 By Proposition 11 (d) it is the case that PIi PIj = 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 . Thus a = dim(Range(P ) (λ+ε/(n+1),λ+ε/n) n+1 (λ−ε/(n+1),λ+ε/(n+1)) ≤ dim(Range(P )) = a . So we can hereby conclude that a (λ−ε/n,λ+ε/n) n { n} is a decreasing sequence of non-negative integers and must as such be constant for some sufficiently large index. This is to say that there exists a N such that a = a = a = . Our next step is to show N n+1 N+2 · · · that there is no λ0 (λ ε/N, λ + ε/N) with λ0 = λ that belongs to ∈ − 6 σ(A). Suppose there is such λ0 σ(A) and suppose further that λ0 < λ. ∈

18 0 The case where λ > λ is treated in a similar fashion. Choose ε1 > 0 so small that λ ε/N < λ0 + ε < λ, then we choose n > N large − 1 enough so that λ0 + ε < λ ε/n. In line with our reasoning above we 1 − 0 0 get that Range(P(λ −ε1,λ +ε1)) and Range(P(λ−ε/n,λ+ε/n)) are orthogonal subspaces of Range(P ), butwe also know that because a (λ−ε/N,λ+ε/N) { n} is constant for some N that dim(Range(P(λ−ε/n,λ+ε/n))) = an = aN = 0 0 dim(Range(P(λ−ε/N,λ+ε/N))). Thus we must have that Range(P(λ −ε1,λ +ε1)) = 0 0 , that is P 0 0 = 0. By Proposition 12 we conclude that λ / { } (λ −ε1,λ +ε1) ∈ σ(A). Still assuming that λ σ (A) we know that there exists some ε > 0 such ∈ disc 0 that Range(P ) is of ﬁnite dimension. Since λ (λ ε , λ + (λ−ε0,λ−ε0) { } ⊂ − 0 ε ) σ(A) we know that ψ Aψ = λψ Range(P (A)). Thus 0 ∩ { | } ⊂ (λ−ε0,λ+ε0) dim( ψ Aψ = λψ ) dim(Range(P (A)) < . { | } ≤ (λ−ε0,λ+ε0) ∞ Conversely suppose that both (a) and (b) hold true. Then there exists some ε > 0 so that (λ ε , λ ε ) σ(A) = λ and dim( ψ Aψ = λψ ) < . 1 − 1 − 1 ∩ { } { | } ∞ Because of property (a) we have that Range(P (A)) = ψ Aψ = (λ−ε1,λ+ε1) { | λψ and thus for ε Range(P (A)) is of ﬁnite dimension. Hence } 1 (λ−ε1,λ+ε1) λ σ (A). ∈ disc 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 σ ss if and only if one or more of the following conditions holds. ∈ e (a) λ σ (A) σ σ (A). (b) λ is a limit point of σ (A) ∈ cont ≡ ac ∪ sing pp (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 operator 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 ψ ∞ such that ψ = 1 and lim (A { n}n=1 || n|| n→∞ || − λ)ψ = 0. λ σ (A) if and only if the ψ can be chosen to be orthog- n|| ∈ ess { n} 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 ψ ∞ with the property lim (A λ)ψ = 0. { n}n=1 n→∞ || − n||

19 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 important 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 p(t) . We || || [0,1]| | define D to be the differentiation operator, that is Dp(t) = p0(t). If we n take the polynomial pn(t) = t where n is a natural number, then certainly p = 1 and Dp (t) = p0(t) = ntn−1. Then DP = n tn−1 = n. Thus || n|| n || n|| || || T x / x = n and n is an arbitrary natural number. So D is an un- || n|| || n|| 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 defined 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 2 We take H = L2(R) and let D(T ) = f L2(R) x2 f(x) 2dx < { ∈ | RR | | . 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) = 0 near infinity, we realize that we can make T f = x2 f(x) 2 6 || || RR | | arbitrarily big. Thus T = sup T f = . That is T is unbounded. || || ||f||=1|| || ∞ It we want Range(T ) to be L2(R) we must stick to the restriction we made 2 2 2 above. For if f / D(T ) then R x f(x) = so xf(x) / L (R). So D(T ) ∈ R | | ∞ ∈ R is defined above is the largest domain of T for which Range(T ) is in L . The differential operator which was mentioned in the beginning of this section 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.

20 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, T x {h i | x D(T ) is a closed subset of H H. Here H H is a Hilbert space ∈ } × × equipped with the inner product ( x , y ), x , y ) = (x , y ) + (x , y ). h 1 1i h 2 2i 1 1 2 2 Another important notion concerning unbounded operators is that of ex- tending an operator. We can deﬁne the extension of an operator in two ways 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 T is an extension of T denoted by T T if and only if 2 1 1 ⊂ 2 D(T ) D(T ) and T = T . We call T a proper extension of T if 1 ⊂ 2 1 2 |D(T1) 2 1 and only if D(T ) D(T ) = . 2 − 1 6 Definition 29 With the notion from above kept intact we say that T 1 ⊂ T if and only Γ(T ) Γ(T ). 2 1 ⊂ 2 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 and T 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 extension T . The minimal of these extension are called the closure of T denoted T . Here e minimal means that every extension T1 of T is an extension of T . For symmetric operators, that is operators such that (T x, y) = (x, T y) for all x, y D(T ), it is the case that they always have unique extension. we ∈ will introduce these operators more properly later. Next, we state and prove a theorem that says that the graph of the closure of an operator equals the closure 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. h i ∈ h i ∈ Let us now define the operator R : D(R) H where D(R) = ψ ψ, φ → { | h i ∈ 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 h i ∈ {h i | ψ D(R) = ψ, φ ψ, φ Γ(T ) = Γ(T ). ∈ } {h i | h i ∈ } 5 We 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 H ∈ ∈ with (T ψ, 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 and ∈ only if (T ψ, ϕ) C ψ . | | ≤ || ||

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

Definition 33 A densely defined operator T on a Hilbert space H is called symmetric or Hermitian if T T ∗, that is, if D(T ) D(T ∗) ⊂ ⊂ and T ϕ = T ∗ϕ for all ϕ D(T ). Equivalently, T is symmetric if and only ∈ if (T ϕ, ψ) = (ϕ, T ψ) for all ϕ, ψ D(T ). ∈ Now we use this definition to define what it means for an operator to be self-adjoint.

Deﬁnition 34 With the notation kept intact from the previous deﬁnition we say that T is self-adjoint if T = T ∗, that is, if and only if T is symmetric and 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 unbounded case 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.