The Algebra of Unbounded Continuous Functions on a ... - Project Euclid

The Algebra of Unbounded Continuous Functions on a Stonean Space and Unbounded Operators F o t i o s C . Pa l i og i an n i s

1. Introduction The investigation of commutative operator algebras by means of function space techniques is due to M. H. Stone [7]. The notion of a space such that the closure of every open set G, clos(G), is open (thus, clos(G) is clopen) was introduced by Stone in [8]. Such spaces are called extremely disconnected. Extremely disconnected spaces are also characterized as those topological spaces X for which (i) the interior of a closed subset F of X, int(F ), is clopen, or (ii) disjoint open subsets of X have disjoint closures. A compact Hausdorff extremely disconnected space X is also known as a Stonean space. If A is an abelian von Neumann algebra then A is isomorphic with C(X), where X is a Stonean space (see [5, Thm. 5.2.1]). In [4] (and [5]), Kadison studies a class of unbounded continuous complexvalued (real-valued) functions on an extremely disconnected space X (called normal functions and self-adjoint functions and denoted by N(X) and S(X), respectively), and he proves that N(X) is an algebra [4, Thm. 2.11]. Starting with an abelian von Neumann algebra A, Kadison introduces N(A), the algebra of (normal) operators affiliated with A and S(A), the algebra of self-adjoint operators affiliated with A [4, Thm. 3.3], extending the isomorphism of A with C(X) to a ∗ -isomoprhism of N(A) onto N(X) [4, Thm. 4.1]. In this direction, one is enabled to obtain the spectral theorem for self-adjoint and normal operators (see also [2]). In this article, we present a closely related approach to the study of N(X), S(X), and the spectral theorem for unbounded self-adjoint operators. We begin in Section 2 with a theorem (Theorem 2.1) on continuous extensions from open dense subsets of extremely disconnected spaces (see also [3, p. 96]). Theorem 2.1 leads to a substantial simplification of the proof that N(X) is an algebra, and it plays a key role in our development. We continue, in Section 3, with a discussion on the spectral analysis of a function in S(X), and we give an alternative proof of the fact that S(X) is a boundedly complete lattice. In Section 4 we prove the spectral theorem and characterizations of the spectrum and the spectral projections for unbounded self-adjoint operators. Received September 23, 1997. Revision received January 5, 1998. Michigan Math. J. 46 (1999).

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F o t i o s C . Pa l i og i an n i s

2. N(X ) and S(X ) Theorem 2.1. Let X be an extremely disconnected space, and let Y be a compact Hausdorff space. Suppose that U is an open dense subset of X. If f : U → Y is a continuous function, then f has a unique continuous extension f˜ on X. Proof. The uniqueness is clear, since if two continuous functions agree on a dense subset then they agree everywhere. T We prove the existence. For each y ∈ Y, let Ay = G∈Ny f −1(G) (closure denotes the closure in X), where Ny = { G : G is open in Y, y ∈ G }. Clearly, Ay is a closed (possibly empty) subset of X for all y ∈ Y. Now, if x ∈ X then there is a net {x d } in U such that x d → x. Since Y is compact, the net {f (x d )} has a cluster point, say y, in Y. We claim that x ∈ Ay . / f −1(G). If x ∈ / Ay , then there is an open set G in Y such that y ∈ G and x ∈ Hence there exists a d 0 such that, for d ≥ d 0 , x d ∈ / f −1(G). In particular, for d ≥ / G. Since G is a neighborhood of y and y is a cluster point for d 0 we have f (x d ) ∈ {f (x d )}, this is a contradiction. Thus, x ∈ Ay . We define f˜(x) = y for x ∈ Ay . This is well-defined; for if y1 6= y 2 then Ay1 ∩ Ay 2 = ∅. To see this, suppose y1 6= y 2 . Then there exist open disjoint sets G1, G2 in Y with y1 ∈ G1 and y 2 ∈ G2 . Hence, f −1(G1 ) and f −1(G2 ) are disjoint and open in U. Therefore, they are disjoint and open in X. Since X is extremely disconnected, their closures are disjoint as well. Thus, Ay1 ∩ Ay 2 = ∅. To see that f˜(x) = f (x) for all x ∈ U, let D be any directed set and take x d = x for all d ∈ D. Then f (x d ) = f (x) and f (x d ) → f (x), so x ∈ Af (x) . Hence, f˜(x) = f (x). It remains to show the continuity of f˜. Let F be any closed subset of Y and T NF = { G : G is open in Y and F ⊆ G }. We claim that f˜−1(F ) = G∈NF f −1(G) (which immediately gives the continuity of f˜ ). In fact, if x ∈ f˜−1(F ) then f˜(x) = T y ∈ F. Hence x ∈ Ay ⊆ G∈NF f −1(G). / F. Choose G1, G2 disjoint open Conversely, if x ∈ / f˜−1(F ) then f˜(x) = y ∈ sets in Y such that y ∈ G1 and F ⊆ G2 . The same argument as before gives that T f −1(G1 ) and f −1(G2 ) are disjoint. Therefore, Ay ∩ G∈NF f −1(G) = ∅. Since T / G∈NF f −1(G). x ∈ Ay , we have x ∈ ˙ = C ∪ {∞} denote the one-point compactification of the complex plane C Let C ¨ = [−∞, +∞] the two-point compactification of the real line R. and R ˙ such Definition. Let X be a Stonean space. A continuous function f : X → C, that Uf = { x : f (x) 6= ∞ } is (open) dense in X, is called a normal function on X. We denote by N(X) the set of normal functions on X. ¨ such that Uf = { x : −∞ < f (x) < +∞ } A continuous function f : X → R, is (open) dense in X, is called a self-adjoint function on X. We denote by S(X) the set of self-adjoint functions on X.

Unbounded Continuous Functions on a Stonean Space

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Let f ∈ N(X). We define f ∗ to be the unique element of N(X) that extends f¯ defined on Uf . Proposition 2.2. N(X) is a ∗ -algebra containing C(X), and S(X) is the subalgebra of self-adjoint elements of N(X). Proof. By definition of f ∗, f ∗ ∈ N(X) whenever f ∈ N(X). To see that N(X) is an algebra, suppose that f and g are in N(X). Write Uf = { x : f (x) 6= ∞ } and Ug = { x : g(x) 6= ∞ }. Then Uf ∩ Ug is open and dense in X, and both f + g and fg are defined and continuous on Uf ∩ Ug . By Theorem 2.1, ˙ and f ·g, respecf +g and fg both have unique continuous extensions on X, f +g ˙ and ·), N(X) becomes tively. Now it is easy to see that, with these operations (+ an algebra with the constant function 1 as unit and C(X) as a subalgebra. It is also easy to see that, for f ∈ N(X), f = f ∗ iff there is a unique g ∈ S(X) ˙ is defined by ¨ →C such that f = θ B g, where θ : R λ if λ ∈ R, θ (λ) = ∞ if λ = ±∞. Thus S(X) is the subalgebra of self-adjoint elements (f ∗ = f ) of N(X). Note that f is invertible in N(X) precisely when 1/f makes sense on a dense open set of X, iff int{ x : f (x) = 0 } = ∅. Note also that, if f ∈ N(X) and e is a projection in C(X)—that is, e = X G (the characteristic function of G) with G a clopen set in X—then f (x) if x ∈ G, f · e(x) = 0 if x ∈ / G.

3. The Spectral Analysis of a Self-Adjoint Function For real-valued functions f, g in C(X), we will write f ≤ g if f (x) ≤ g(x) for all x ∈ X. W functions in C(X). We deW Let {fα }α∈ be a collection of real-valued note by α∈ fα the l.u.b.{ fα : α ∈ }, that is, α∈ fα = f is such that V fa ≤ f for all α ∈ , and if fα ≤ g for all α ∈ then f ≤ g. Similarly, α∈ fα denotes the g.l.b.{ fα : α ∈ }. Definition. Let {eλ }λ∈R be a collection of projections in C(X) and Gλ = { x ∈ X : eλ(x) = 1 }. The family {eλ }λ∈R is called a resolution of the identity in C(X) if ¡S W (i) λ∈R eλ = 1 ⇐⇒ clos¡ λ∈R Gλ = X, T V e = 0 ⇐⇒ int ¡ λ∈R Gλ = ∅, (ii) T V λ∈R λ (iii) µ>λ eµ = eλ ⇐⇒ int µ>λ Gµ = Gλ for all λ ∈ R. Clearly, condition (iii) implies that {eλ }λ∈R is monotonic in λ. Proposition 3.1. There is a bijective correspondence between S(X) and the collection of all resolutions of the identity in C(X).


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Proof. Given any ϕ ∈ S(X), if Gλ = int{ x : ϕ(x) ≤ λ } for λ ∈ R, then eλ = X Gλ defines a resolution of the identity in C(X). Moreover, Gλ = int{ x : ϕ(x) ≤ λ } is equivalent to { x : ϕ(x) < λ } ⊆ Gλ ⊆ { x : ϕ(x) ≤ λ }, which in turn is equivalent to ϕ · (1− eλ ) ≥ λ(1 − eλ ) and ϕ · eλ ≤ λeλ for all λ ∈ R. Conversely, let {eλ }λ∈R be any resolution of the identity in C(X), and let Gλ = ¨ by { x : eλ(x) = 1 }. Define the function ϕ : X → R c ¡S  on λGλ ,  +∞ T ϕ(x) = −∞ on λGλ ,  S T sup{ λ : x ∈ / Gλ } = inf{ λ : x ∈ Gλ } on λGλ ∼ λGλ . It is easy to see that ϕ as defined satisfies the condition { x : ϕ(x) < λ } ⊆ Gλ ⊆ { x : ϕ(x) ≤ λ } for all λ ∈ R, and this condition determines ϕ uniquely. We now show that ϕ is continous. Suppose ϕ(x 0 ) = +∞. Let R be any positive real number. Choose any λ > R. Then x 0 ∈ (Gλ )c and for all x ∈ (Gλ )c we have ϕ(x) ≥ λ > R. Thus, U = (Gλ )c is an open neighborhood of x 0 , and ϕ(U ) ⊆ (R, +∞]. Similarly, suppose ϕ(x 0 ) = −∞. Given any R > 0, choose λ < −R. Then U = Gλ is an open neighborhood of x 0 and ϕ(U ) ⊆ [−∞, R). Now suppose that ϕ(x 0 ) is finite. Let α, β ∈ R such that α < ϕ(x 0 ) < β. Choose λ, µ ∈ R such that α < λ < ϕ(x 0 ) < µ < β. Then U = Gµ ∩ (Gλ )c is µ]T ⊆ (α, β). an open neighborhood of x 0 , and ¡ϕ(U S ) ⊆ [λ, c and λ Gλ both have empty interior. To see that ϕS∈ S(X),T note that λ Gλ Hence, Uϕ = λ Gλ ∼ λ Gλ is dense in X. Theorem 3.2. Let ϕ ∈ S(X) and with {eλ }λ∈R be the resolution of the identity in C(X) defined by ϕ. For J = (α, β] with −∞ < α < β < +∞, let fJ = eβ − eα . If 5 = {λ 0 , λ1, . . . , λ n } with α = λ 0 < λ1 < · · · < λ n = β is any partition of [α, β], if ξ j ∈ [λj −1, λj ] for j = 1, 2, . . . , n, and if k5k = max j =1,2,. . . ,n (λj − λj −1), then ° ° n X ° ° °ϕ · fJ − ° ξ (e − e ) ≤ k5k; J λj λj −1 ° ° that is, ϕ · fJ =

Rβ

j =1

C(X)

λ deλ (in the Riemann–Stieltjes sense). P Proof. Set ϕ 5,ξ = jn=1 ξ j (eλj − eλj −1 ). For x ∈ Gβ ∼ Gα we have eλ 0 (x) = eα (x) = 0 and eλ n (x) = eβ (x) = 1. Hence there exists a unique j = 1, 2, . . . , n such that eλj −1 (x) = 0 and eλj (x) = 1. Then ϕ 5,ξ (x) = ξ j and ϕ ·fJ (x) = ϕ(x) ∈ [λj −1, λj ], so α

|ϕ · fJ (x) − ϕ 5,ξ (x)| = |ϕ(x) − ξ j | < (λj − λj −1) ≤ k5k. For x ∈ Gα we have ϕ · fJ (x) = 0 and eλ 0 (x) = eα (x) = 1. Hence, ϕ 5,ξ (x) = 0 and the estimate trivially holds. For x ∈ / Gβ we have ϕ · fJ (x) = 0 and eλ n (x) = eβ (x) = 0, and again the estimate trivially holds. Remark 3.3. Note that ϕ, the element of S(X) R β associated with {eλ }λ∈R is the unique element of S(X) satisfying ϕ · fJ = α λ deλ . In fact, if ψ ∈ S(X) with


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S S ψ ·fJ = ϕ ·fJ for all J = (α, β], then ψ = ϕ on J { x : fJ (x) 6= 0 } = λ Gλ ∼ T λ Gλ , which is dense in X. Therefore, ψ = ϕ everywhere. Definition. Let {eλ }λ∈R be a resolution of the identity in C(X). We define eλ− = W µλ Gµ ⊆ α∈ { x : ϕ α (x) ≤ λ } and so Gλ = ¡T int µ>λ Gµ . Now let ϕ 0 be the (unique) function in S(X) that corresponds to the resolution of the identity {eλ }λ∈R . Note now that Gλ = int{ x : ϕ 0 (x) ≤ λ }. From Lemma 3.5, we have ϕ α ≤ ϕ 0 for all α ∈ . Moreover, if ψ ∈ S(X) is such that ϕ α ≤ ψ for all α ∈ and if Hλ = int{ x : ψ(x) ≤ λ }, then Hλ ⊆ int{ x : ϕ α (x) ≤ λ } for all α ∈ . It follows that Hλ ⊆ Gλ and, from Lemma 3.5 again, we get ϕ 0 ≤ ψ. This completes the proof.

4. The Spectral Theorem Let B(H ) be the algebra of bounded linear operators on a Hilbert space H, and let Op(H ) be the set of unbounded densely defined linear operators on H. We recall that, for A, B ∈ Op(H ), B is called an extension of A, denoted by A ⊂ B, if D(A) ⊆ D(B) and Ax = Bx for all x ∈ D(A). Let A ∈ Op(H ) be a closed operator, and let T ∈ B(H ). We say that T commutes with A if TA ⊂ AT ; that is, if x ∈ D(A) then Tx ∈ D(A) and TAx = ATx. We denote by {A}0 the set of all operators in B(H ) that commute with the operator A in the foregoing sense: {A}0 = { T ∈ B(H ) : TA ⊂ AT }. It is easy to see that {A}0 is a subalgebra of B(H ) that is closed in the strong operator topology (s.o.t.). Note also that T ∈ {A}0 iff T ∗ ∈ {A∗ }0 . Thus, {A}0 ∩ {A∗ }0 is a von Neumann algebra. We write {A}00 = {{A}0 }0 for the commutant of {A}0 . Definition. Let A be a von Neumann algebra of operators on H, and let A ∈ Op(H ) be a closed operator. We say that A is affiliated with A, denoted AηA, when A0 ⊂ {A}0 . We denote by S(A) the family of self-adjoint operators affiliated with the algebra A. Note that AηA iff A0 ⊂ {A}0 ∩ {A∗ }0 iff {{A}0 ∩ {A∗ }0 }0 ⊂ A. Note also that W ∗ (A) = {{A}0 ∩ {A∗ }0 }0 is the smallest von Neumann algebra with which A is affiliated, and is referred to as the von Neumann algebra generated by A. Clearly, if A is self-adjoint (A = A∗ ), then AηA

iff W ∗ (A) = {A}00 ⊂ A.

At this point, we recall some facts from the basic theory of self-adjoint operators. Let σ(A) denote the spectrum of a self-adjoint operator A. Then σ(A) ⊆ R and V = (iI − A)−1 is a bounded operator with adjoint V ∗ = (−iI − A)−1 (see [1, p. 318]). It is easy to see that V is a normal operator in B(H ). In fact,


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V ∗ V = V V ∗ = (V ∗ − V )/2i. Moreover, {A}0 = {V }0 . Thus, W ∗ (A) = {A}00 = {V }00 , where {V }00 is the abelian von Neumann algebra generated by V (i.e., the s.o.t.-closure of the set of polynomials in V ). Let A be an abelian von Neumann algebra, and let X = XA be the Gelfand space (or maximal ideal space) of A. From the Gelfand–Naimark representation theorem for abelian C ∗ -algebras, the Gelfand map 0 : A → C(X) (where 0(A) = Aˆ is the Gelfand transform of A for A ∈ A) is an isometric ∗ -isomorphism from A onto C(X). As we noted in the introduction, N(A) is a (commutative) ∗ -algebra and the isomorphism 0 extends to a ∗ -isomorphism of N(A) with N(X). Although the algebraic properties of N(A) and the extension of 0 will not be used in the sequel, we shall also extend 0 to a bijection of S(A) with S(X). Theorem 4.1. Let A be a self-adjoint operator, and let A be any abelian von Neumann algebra such that AηA. Let X = XA . Then there exists a unique ϕ ∈ S(X) ˙ ˆ whenever B ∈ A and AB ∈ A. We write 0(A) ˙ such that (AB)ˆ = ϕ · B, = Aˆ = ϕ. Proof. Let V = (iI − A)−1. Since AηA and {V }00 = {A}00 , we have that V ∈ A. Let v = Vˆ ∈ C(X) be the Gelfand transform of V. Note that AV = −(iI − A)V + iV = −I + iV ∈ A. Hence (AV )ˆ = −1 + iv. If F is the projection onto Ker(V ), then F is the largest projection in A such that VF = 0. Therefore, Fˆ = X G , where G is the largest clopen set contained in { x : v(x) = 0 }, that is, G = int{ x : v(x) = 0 }. Since V is one-to-one, F = 0 and so G = ∅. Thus, 1/v exists in N(X). ¯ Furthermore, Define ϕ = −1/v + i. Note that ϕ ∗ − ϕ = −2i + (v¯ − v)/vv. since V ∗ V = V V ∗ = (V ∗ − V )/2i, it follows that ϕ ∗ = ϕ. Hence, ϕ ∈ S(X). Note also that, on { x : v(x) 6= 0 }, an open dense subset of X, ϕv = −1+ iv = (AV )ˆ. Thus, by Theorem 2.1, ϕv has a unique continuous extension, ϕ · v = (AV )ˆ, on X. Now, if C = AB with B, C ∈ A, then VC = VAB = −V (iI − A)B + iVB ⊂ −B + iVB. Since VC ∈ B(H ), it follows that VC = −B + iVB. ˆ Then vc = (−1 + iv)b. This Let b, c ∈ C(X) be such that b = Bˆ and c = C. implies c = (−1/v +i)b on { x : v(x) 6= 0 }. Therefore, c = ϕ ·b; that is, (AB)ˆ = ˆ ϕ · B. To see that the restriction of 0˙ in A is 0, let A be a bounded self-adjoint opera˙ tor in A whose Gelfand transform Aˆ = a. Then v = 1/(i − a) and 0(A) =ϕ= α = 0(A). Lemma 4.2. Let A be a self-adjoint operator, and let A be any abelian von Neu˙ˆ ˆ ⊆ Uϕ = mann algebra such that AηA. Let ϕ = A. Suppose B ∈ A and supp(B) { x : −∞ < ϕ(x) < +∞ }. Then AB ∈ A. ˆ and b = B. ˆ Set G = supp(b) = clos{ x : Proof. Let V = (iI − A)−1, v = V, b(x) 6= 0 } (the support of b). Then G is a clopen set and G ⊆ { x : v(x) 6= 0 }. Now, if e = X G and E ∈ A with Eˆ = e, then EB = B.

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Define c = e · 1/v. If C ∈ A with Cˆ = c, then c ∈ C(X) and VC = E. Thus, AB = AEB = AVCB ∈ A. Definition. A resolution of the identity in B(H ) is a family of projections {Eλ }λ∈R in B(H ) satisfying: W (i) Vλ∈R Eλ = I, (ii) V λ∈R Eλ = 0, and (iii) µ>λ Eµ = Eλ for all λ ∈ R. We are now ready to prove the spectral theorem for unbounded self-adjoint operators. Theorem 4.3 (Spectral Theorem). Let A be an unbounded self-adjoint operator on H. Then there exists a unique resolution of the identity {Eλ }λ∈R in B(H ) such that: (i) for any interval J = (α, β], if FJ = Eβ − Eα then AFJ is a bounded selfRβ adjoint operator on H and AFJ = α λ dEλ ; (ii) x ∈ D(A) iff the net {AFJ x}J ∈J converges and in fact Z Ax = lim AFJ x = lim

α

β

λ dEλ x =

Z

∞

−∞

λ dEλ x.

Moreover, Eλ ∈ {A}00 , and {Eλ }λ∈R is called the spectral family of A. ˙ Proof. (i) Let A = {A}00 and X = XA . If ϕ = Aˆ and eλ = X Gλ , where Gλ = int{ x : ϕ(x) ≤ λ }, then {eλ }λ∈R is a resolution of the identity in C(X). For J = (α, β], let fJ = eβ − eα . From Theorem 3.2, ϕ · fJ ∈ C(X) and ϕ · fJ = Rβ ˆ ˆ α λ deλ . Now take Eλ ∈ A such that Eλ = eλ , and let FJ = fJ . Then {Eλ }λ∈R is a resolution of the identity in B(H ) and FJ = Eβ − Eα . Note that supp(fJ ) = Gβ ∼ Gα ⊆ Uϕ . Hence, by Lemma 4.2, AFJ ∈ A. Moreover, (AFJ )ˆ = ϕ · fJ is real-valued, hence AFJ is self-adjoint. At the same Rβ time, since the Gelfand map is an isometry, AFJ = α λ dEλ . (ii) Note that FJ ∈ A ⊆ A0 = {A}0 and so FJ A ⊂ AFJ for all J. Let J be theS directed set of half-open intervals J = (α, β] in R W ordered by inclusion. { x : f (x) = 6 0 } is dense in X, it follows that Since J J ∈J J ∈J fJ = 1. Hence, W F = I. Therefore, F ↑ I in the strong operator topology. J J ∈J J Now, if x ∈ D(A), then AFJ x = FJ Ax → Ax. Conversely, suppose AFJ x → y. Then, since A is closed and FJ x → x, we have x ∈ D(A) and Ax = y. It remains to prove the uniqueness of the spectral family. Suppose {Eλ0 }λ∈R is another resolution of the identity satisfying (i) and (ii). Let B be the abelian von Neumann algebra generated by {Eλ0 }λ∈R . Let FJ0 = Eβ0 − Eα0 ∈ B. By (i), AFJ0 is the limit in the uniform operator topology (hence, s.o.t.) of a net of operators in B, so AFJ0 ∈ B. If B ∈ B 0 and x ∈ D(A), then by (ii) we have A(BFJ0x) = BAFJ0x → BAx. At the same time BFJ0x → Bx. Since A is closed, we conclude that B ∈ {A}0 . Thus, A ⊆ B.


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Rβ Now, if Y is the Gelfand space of B and eλ0 = (Eλ0 )ˆ, then ϕ · fJ0 = α λ deλ0 and Rβ ϕ · fJ = α λ deλ in Y. Because such a representation is unique, it follows that eλ0 = eλ for all λ. Therefore, Eλ0 = Eλ for all λ ∈ R. Remark 4.4. For any x ∈ H, let µx be the unique Borel measure on R satisfying µx (J ) = (FJ x, x) = kFJ xk2 = (Eβ x, x) − (Eα x, x). Part (ii) of the spectral theorem can be rewritten equivalently as follows: Z ∞ λ2 dµx (λ) < ∞. x ∈ D(A) iff −∞

For this, first note that FJ FK = FJ ∩K for J, K ∈ J (since Eλ Eµ = Emin(λ,µ) ). Furthermore, we have °2 °2 °X °X ° ° ° n ° n 2 ° ° ° ξ j (Eλj − Eλj −1 )x ° = lim ° ξ j FJj x ° kAFJ xk = lim ° ° n→∞ n→∞ j =1

= lim

n→∞

Z =

J

n X j =1

j =1

ξ j2 kFJj xk2 = lim

λ2 dµx (λ) →

n→∞

Z

∞

−∞

n X j =1

ξ j2 µx (Jj )

λ2 dµx (λ).

Now, if x ∈ D(A) then AFJ x → Ax. So kAFJ xk2 → kAxk2 . Therefore, Z ∞ λ2 dµx (λ) = kAxk2 < ∞. −∞

R∞ Conversely, suppose that −∞ λ2 dµx (λ) < ∞. Let ε > 0 and γ < α < β < δ be any real numbers. If J = (α, β], K = (γ , δ], L = (γ , α], and M = (β, δ], then by choosing J large enough we have Z λ2 dµx (λ) < ε. kAFK x − AFJ xk2 = kAFL xk2 + kAFM xk2 ≤ R−J

Hence, the net {AFJ x}J ∈J is Cauchy and so converges in H. R ∞ Note also that, for x ∈ D(A), the representation (Ax, x) = −∞ λ dµx (λ) is R∞ valid. By the polarization identity, (Ax, y) = −∞ λ dµx,y (λ) for x ∈ D(A) and y ∈ H, where µx,y (J ) = (FJ x, y). This is the classical form of the spectral decomposition of a self-adjoint operator (see e.g. [6, Thm. 13.30]). For the proof of the following lemma, see [5, Lemma 5.6.1] (and replace the sequence by a net). Lemma 4.5. W If {Fd } is an increasing net of projections on the Hilbert space H such that d Fd = I, and if A 0 is a linear operator with dense domain S d Fd (H ) = (D 0 ) such that A 0 Fd is a bounded self-adjoint operator on H, then A 0 is closable and its closure is the unique self-adjoint operator satisfying AFd = A 0 Fd for all d.

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Proposition 4.6. If {Eλ }λ∈R is a resolution of the identity in B(H ) and A is an abelian von Neumann algebra containing {Eλ }λ∈R , then there is a self-adjoint operator A in S(A) whose spectral family is {Eλ }λ∈R . Moreover, if X = XA then the mapping 0˙ : S(A) → S(X) is a bijection. Proof. Let eλ = Eˆ λ for all λ ∈ R. Then {eλ }λ∈R is a resolution of the identity in C(X). Let ϕ ∈ S(X) be the (unique) self-adjoint function associated to {eλ }λ∈R Rβ (Proposition 3.1). Then ϕ · fJ ∈ C(X) and ϕ · fJ = α λ deλ , with J = (α, β] and fJ = eβ − eα . If, for each J ∈ J, FJ and AJ are the operators in A whose Gelfand transforms (in C(X)) are fJ andWϕ · fJ , respectively, then {FJ }J ∈J is an increasing net of projections such that J ∈J FJ = I, and AJ isSa bounded self-adjoint operator. Define an operator A 0 with domain D 0 = J ∈J FJ (H ) by A 0 x = AJ x if x ∈ FJ (H ). We claim that A 0 is well-defined; for this, note first that since (ϕ ·fJ )fK = ϕ · fJ ∩K , it follows that AJ FK = AJ ∩K . Now, if x is also in FK (H ), then x = FJ x, x = FK x, and AJ x = AJ FK x = AJ ∩K x = AKFJ x = AK x. From Lemma 4.5, A 0 is closable and its closure A (= A¯ 0 ) is self-adjoint. Since A 0 ⊂ A and A 0 FJ = AJ is everywhere defined, we have AFJ = A 0 FJ = AJ . Rβ Therefore, AFJ = α λ dEλ . Next, we show that FJ A ⊂ AFJ for all J. Suppose x ∈ D(A). Then there is a sequence {x n } in D 0 such that x n → x and A 0 x n → Ax (since A = A¯ 0 ). Hence, FJ A 0 x n → FJ Ax. Note that, for each n, there exists a K such that x n = FK x n . Hence, FJ A 0 x n = FJ A 0 FK x n = FJ AK x n = AKFJ x n = A 0 FJ x n . Now, FJ x n → FJ x and AFJ x n = A 0 FJ x n = FJ A 0 x n → FJ Ax. Since A is closed, FJ x ∈ D(A) and AFJ x = FJ Ax. The same argument as in the proof of Theorem 4.3 gives that x ∈ D(A) iff the net {AFJ x}J ∈J converges (and AFJ x → Ax). To see that A is affiliated with A, suppose that T is in A0 and that x ∈ D(A). Then TFJ x → Tx (since FJ x → x) and ATFJ x = AFJ Tx = TAFJ x → TAx. Since A is closed, Tx ∈ D(A) and ATx = TAx. Thus, A0 ⊂ {A}0 . It is now clear that {Eλ }λ∈R is the spectral family of A (by uniqueness, as in ˙ Theorem 4.3). Moreover, since A is in S(A), 0(A) makes sense and (AFJ )ˆ = ˙ ˙ · fJ = ϕ · fJ for all J (since (AFJ )ˆ = ϕ · fJ ). Thus, 0(A) · fJ . Therefore, 0(A) ˙ 0(A) = ϕ. (Invoking Theorem 3.2, this also proves that 0˙ : S(A) → S(X) is a bijection.) Corollary 4.7. If A is a self-adjoint operator and {Eλ }λ∈R is its spectral family, then W ∗ (A) = {A}00 = { Eλ : λ ∈ R }00 . Proof. Take A = { Eλ : λ ∈ R }00 in Proposition 4.6. Then A is affiliated with A. Therefore, {A}00 ⊆ A. On the other hand, since Eλ ∈ {A}00 , we have { Eλ : λ ∈ R }00 ⊆ {A}00 . Proposition 4.8. Let A be a self-adjoint operator, and let A be any abelian ˙ˆ von Neumann algebra such that AηA. Let ϕ = A. Then σ(A) = ϕ(Uϕ ).


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Proof. If λ ∈ / σ(A), then B = (λI − A)−1 ∈ A and I = (λI − A)B. Taking ˆ Thus, λ ∈ Gelfand transforms, this gives 1 = (λ − ϕ) · b with b = B. / ϕ(Uϕ ). Conversely, if λ ∈ / ϕ(Uϕ ) then b ≡ (1/(λ − ϕ)) ∈ C(X). Take B ∈ A such that Bˆ = b. Since supp(b) ⊆ Uϕ , AB ∈ A. Now b · (λ − ϕ) = (λ − ϕ) · b = 1 and so B(λI − A) ⊂ I = (λI − A)B. Thus, λ ∈ / σ(A). The spectral family {Eλ }λ∈R of a self-adjoint operator completely determines the spectrum of the operator. Theorem 4.9. Let A ∈ Op(H ) be a self-adjoint operator, {Eλ }λ∈R its resolution of the identity, and Wα, β ∈ R with α < β. Then σ(A) ∩ (α, β) = ∅ iff Eβ − − Eα = 0, where Eβ − = µ