Projection-valued measures and spectral integrals

Jordan Bell
April 16, 2014

1 Introduction

The purpose of these notes is to precisely define all the objects one needs to talk about projection-valued measures and the spectral theorem.

2 Orthogonal complements

Let H be a separable complex Hilbert space.

If W is a subset of H, the span of W is the set of all finite linear combinations of elements of W. If Wα, αI, are subsets of H, let

αIWα

denote the closure of the span of αIWα. It is straightforward to check that this is the intersection of all closed subspaces of H that contain each of the Wα.11 1 Let 𝔏 be the set of all closed subspaces of H. Set inclusion is a partial order on 𝔏. 𝔏 is a lattice: if M,N𝔏, then MN is the greatest element in 𝔏 that is contained both in M and in N, and MN is the least element in 𝔏 that contains each of M and N. 𝔏 is a complete lattice: if Mα𝔏, then αMα is the greatest element in 𝔏 that is contained in each Mα, and αMα is the least element in 𝔏 that contains each Mα. (For comparison, the set of finite dimensional subspaces of an infinite dimensional Hilbert space is a lattice but is not a complete lattice.)

If V is a subspace of H, define

V={fH:f,g=0 for all gV},

called the orthogonal complement of V. If V is a subspace of H then V is a closed subspace of H.

Let V be a closed subspace of H and let fH. It is a fact that there is a unique g0V such that

f-g0=infgVf-g,

and that f-g0V. Then, f=g0+(f-g0)V+V. Therefore, if V is a closed subspace of H then

H=V+V. (1)

If V and W are subspaces of H, we write VW if v,w=0 for all vV,wW. If V,W are closed subspaces of H and VW, we can show that V+W=VW, and thus V+W is in particular a closed subspace of H.22 2 If znV+W and znzH, write zn=vn+wn, vnV and wnW. Using VW we get zn-zm2=vn-vm2+wn-wm2, which we use to prove the claim. On the other hand, there are examples of closed subspaces V,W of a Hilbert space that do not satisfy VW for which V+W is not itself closed. By induction, we obtain V1++Vn=k=1nVk. If Vk, k1, are mutually orthogonal closed subspaces of H, we define

k=1Vk=k=1Vk,

called the orthogonal direct sum of the subspaces Vk. The orthogonal direct sum k=1Vk is equal to the closure of the set of all finite sums of the form vk, where vkVk.33 3 See Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 26, §13, Theorem 2.

If V is a closed subspace of H, then (1) is

H=VV.

The subobjects of a Hilbert space that we care about are mostly closed subspaces, because they are themselves Hilbert spaces (a closed subset of a complete metric space is itself a complete metric space). Generally, when talking about one type of object, we care mostly about those subsets of it that are the same type of object, which is why, for example, that we are so glad to know when closed subspaces are mutually orthogonal, because then their sum is itself a closed subspace.44 4 Paul Halmos in his Introduction to Hilbert Space and the Theory of Spectral Multiplicity uses the term linear manifold to refer to what I call a subspace and subspace to refer to what I call a closed subspace. I think even better terms would be linear subspace and Hilbert subspace, respectively; this would emphasize the category of objects with which one is working.

3 Projections

Let V be a closed subspace of H. The projection onto V is the map PV:HH defined in the following way: if fH=VV, let f=g+h,gV,hV, and set PV(f)=g. The image of PV is V: if fim PV then there is some f0H such that PVf0=f and f0=g+h, gV,hV, so by the definition of PV we get PVf0=g, giving f=gV; on the other hand, if fV then PVf=f, so fim P.

The image of a projection is closed.55 5 We call P:HH a projection if there is some closed subspace V of H such that P is the projection onto V, and this V will be the image of P. We often talk about a projection in H without specifying what it is a projection onto. Hence if P is a projection, we have

H=im PV(im PV)=im PVkerPV.

If PV is a projection onto a closed subspace V of H, we check that PVB(H). If gV,hV, then, because g,h=0,

PV(g+h)2=g2g2+h2=g,g+h,h=g+h,g+h=g+h2,

so PV1. If V{0}, let gV, g0. Then

gg=1,PV(gg)=gg=1.

Hence if PV is a projection and PV0, then PV=1. Of course, if PV=0 then PV=0. Using H=VV, we check that if f1,f2H then

PVf1,f2=f1,PVf2,

hence PV is self-adjoint.

We call PB(H) idempotent if P2=P.66 6 Often the term projection is used to refer to a thing we would call an idempotent, and the term orthogonal projection is used to refer to a thing we would call a projection. The following conditions are each equivalent to a nonzero idempotent P being a projection:

  • P is positive

  • P is self-adjoint

  • P is normal

  • kerP=(im P)

  • P=1

4 The lattice of projections

Two important projections: idH is the projection onto H, and 0 is the projection onto {0}.

If TB(H) is self-adjoint, we say that T is positive if, for all vH,

Tv,v0.

If S,TB(H) are self-adjoint, we say that ST if, for all vH,

Sv,vTv,v.

This is equivalent to S-T being a positive operator. Thus, TB(H) is a positive operator if and only if T0. One checks that is a partial order on the set of bounded self-adjoint operators on H.

Let PB(H) be a projection. As H=im P(im P)=im PkerP, if vH then v=v1+v2, v1im P,v2(im P),

Pv,v=v1,v1+v2=v1,v1+v1,v2=v1,v10.

Hence P0.

Let P,QB(H) be projections. It is a fact that the following statements are equivalent:

  • PQ

  • im Pim Q

  • QP=P

  • PQ=P

  • PvQv for all vH

The set of projections is a complete lattice.77 7 If P,QB(H) are projections, then the infimum of P and Q is the projection onto im Pim Q, and the supremum of P and Q is the projection onto im Pim Q. See Problem 96 of Paul Halmos’s Hilbert Space Problem Book.

5 Sums of projections

If P,QB(H) are projections, we write PQ if im Pim Q, which is equivalent to PQ=0 and QP=0; neither one of those by itself is sufficient. It is a fact88 8 See Steven Roman, Advanced Linear Algebra, third ed., p. 78 that P+Q is a projection if and only if PQ, in which case P+Q is a projection onto im Pim Q. The product PQ is a projection if and only if PQ=QP, in which case it is a projection onto im Pim Q.

Paul Halmos shows the following in Question 94 of his Hilbert Space Problem Book: If TnB(H) are self-adjoint such that TnTn+1 for all n and if there is some self-adjoint TB(H) such that TnT for all n, then there is some self-adjoint TB(H) such that TnT in the strong operator topology.99 9 Recall that if TnB(H) and TB(H), we say that TnT in the strong operator topology if for all vH we have TnvTv. (If TnT in B(H) then TnT in the strong operator topology.) This is analogous to how a nondecreasing sequence of real numbers that has an upper bound converges to a real number.

Let Mn, n1, be closed subspaces of H such that MnMn+1 and let Pn be the projection onto Mn. As MnMn+1 we have PnPn+1 for all n. Also, PnidH for all n, as idH is the projection onto H and MnH. Hence by the result from Halmos stated above, there is some self-adjoint PB(H) such that PnP in the strong operator topology. Let

M=n=1Mn.

I claim that P is the projection onto M.1010 10 cf. Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 46, §28, Theorem 1.

If PnB(H), n1, are projections and PiPj for ij, let

Mn=k=1nim Pk.

Define Tnv=k=1nPkv. As Mn is a closed subspace of H, we have

H=MnMn=im P1PnMn.

If v=v1++vn+v, v1im P1,,vnim Pn,vMn, then

Tnv=k=1n(Pk(v1++vn+v))=k=1nvk.

Thus Tn is the projection onto Mn. Therefore, there is some self-adjoint PB(H) such that TnP in the strong operator topology, and with

M=n=1k=1nim Pk=n=1im Pk=n=1im Pn,

P is the projection onto M. k=1nPkP in the strong operator topology, and we denote P by k=1Pk.

6 Definition of projection-valued measures

Let 𝒫(H) be the set of projections in B(H). Let () be the Borel σ-algebra of . A projection-valued measure on is a map E:()𝒫(H) such that

  • E()=0 and E()=idH

  • If Bn(), n1, are pairwise disjoint, then

    E(n1Bn)=n1E(Bn),

    where n1E(Bn) is the limit of 1nNE(Bn) in the strong operator topology.

7 Finite additivity

If E:()𝒫(H) is any function that satisfies E(B1B2)=E(B1)+E(B2) for disjoint B1,B2(), then it satisfies the following four properties. In particular, if E is a projection-valued measure it satisfies them.

  1. 1.

    E()=E()=E()+E(), so E()=0.

  2. 2.

    If B1,B2() and B1B2, then

    E(B2)=E(B2B1B1)=E(B2B1)+E(B1)E(B1),

    since E(B2B1) is a projection and hence is a positive operator. Therefore, if B1B2 then E(B1)E(B2).

  3. 3.

    Let B1,B2(). We have

    E(B1)=E(B1B2)+E(B1B2),

    and

    E(B2)=E(B1B2)+E(B2B1),

    and

    E(B1B2)=E(B1B2)+E(B1B2)+E(B2B1),

    and combining these gives, for any B1,B2(),

    E(B1)+E(B2) = E(B1B2)+E(B1B2)+E(B1B2)+E(B2B1)
    = E(B1B2)+E(B1B2).
  4. 4.

    Let B1,B2(). Multiplying both sides of the above equation on the left by E(B2) gives

    E(B1)E(B2)+E(B2)E(B2)=E(B1B2)E(B2)+E(B1B2)E(B2).

    As E(B1B2) and E(B2) are projections and E(B1B2)E(B2), we have

    E(B1B2)E(B2)=E(B2),

    which with E(B2)E(B2)=E(B2) gives

    E(B1)E(B2)+E(B2)=E(B1B2)+E(B2).

    Hence, for any B1,B2(),

    E(B1)E(B2)=E(B1B2).

8 Complex measures

Suppose that E:()𝒫(H) is a function such that E()=idH, and that for all v,wH, the function Ev,w:() defined by

Ev,w(B)=E(B)v,w

is a complex measure. I will show that E is a projection-valued measure. Let Bn(), n1, be pairwise disjoint.1111 11 cf. Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 59, §36, Theorem 3. If B1,B2() are disjoint, then for all v,wH,

E(B1B2)v,w = Ev,w(B1B2)
= Ev,w(B1)+Ev,w(B2)
= E(B1)v,w+E(B2)v,w
= (E(B1)+E(B2))v,w,

and since this holds for all v,wH, we obtain E(B1B2)=E(B1)+E(B2). Therefore, from §7, if B1,B2() are disjoint then

E(B1)E(B2)=E(B1B2)=E()=0.

If vH and vn=E(Bn)v, then, for mn,

vn,vm=E(Bn)v,E(Bm)v=E(Bm)E(Bn)v,v=0v,0=0.

For an, for the sequence n=1Nanvn to converge in H it is equivalent to n=1anvn2<;1212 12 e.g. Walter Rudin’s Functional Analysis, p. 295, Theorem 12.6: if xnH are pairwise orthogonal, not necessarily of unit norm, then for n=1xn to converge is equivalent to n=1xn2<. but vnv, so it suffices to show that n=1|an|2<. Using that if T is a projection then T is self-adjoint and T2=T, and that Ev,v is a complex measure,

n=1E(Bn)v2 = n=1NE(Bn)v,E(Bn)v
= n=1E(Bn)E(Bn)v,v
= n=1E(Bn)v,v
= n=1Ev,v(Bn)
= Ev,v(n=1Bn)
= E(n=1Bn)v,v
= E(n=1Bn)v2
v.

Therefore, the sequence n=1NE(Bn)v converges in H; namely, n=1NE(Bn) converges in the strong operator topology. Let P be its limit. By §5, P is the projection onto n=1im E(Bn).

For v,wH,

E(n=1Bn)v,w = Ev,w(n=1Bn)
= n=1Ev,w(Bn)
= n=1E(Bn)v,w
= Pv,w,

and since this is true for all v,wH, we obtain

E(n=1Bn)=P,

where P is the limit of n=1NE(Bn) in the strong operator topology. This completes the proof that E is a projection-valued measure.

On the other hand, if E:()𝒫(H) is a projection-valued measure, we can show that for each v,wH the function Ev,w:() defined by Ev,w(B)=E(B)v,w is a complex measure.

9 Spectral integrals

Let 𝔅() be the set of bounded measurable functions .1313 13 This is not L(), the set of equivalence classes of essentially bounded measurable functions, where two functions are equivalent if they are equal almost everywhere. Moreover, it is not even the set () of essentially bounded measurable functions. When does one speak about ()? () is a vector space; let 𝒩() be the set of those functions that are equal to 0 almost everywhere in ; 𝒩() is a vector space; then L() is the vector space quotient ()/𝒩(). It is a complex vector space, and we define the norm f=supz|f(z)|; one checks that 𝔅() is a Banach space. Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 60, §37, proves1414 14 The operator A is the operator obtained from the following statement, which itself follows from the Riesz representation theorem. If ϕ:H×H is sesquilinear (we take sesquilinear to mean linear in the first entry and conjugate linear in the second entry) and M=supv=w=1|ϕ(v,w)|<, then there exists a unique AB(H) such that ϕ(v,w)=Av,w,v,wH, and A=M. One also has to prove that for each f𝔅, ϕ(v,w)=Ev,w(f) is sesquilinear. that if E:()𝒫(H) is a projection-valued measure and f𝔅, then there is a unique AB(H) such that, for all v,wH,

Av,w=Ev,w(f)=f𝑑Ev,w=f(λ)𝑑Ev,w(λ),

and A2f. We write

A=E(f)=f𝑑E=f(λ)𝑑E(λ).

We can check that 𝔅() is a C*-algebra, with f* defined by f*(z)=f(z)¯. It is a fact that B(H) is a C*-algebra. If α and f,g𝔅(), then1515 15 See Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 60, §37, Theorem 2.

E(αf)=αE(f),E(f+g)=E(f)+E(g);E(f*)=(E(f))*. (2)

The first two of these together with E(f)2f show that E:𝔅()B(H) is a bounded linear map. If f,g𝔅(), then1616 16 From Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 61, §37, Theorem 3. To understand the proof by Halmos (which is symbolically convincing because of our familiarity with the permissible moves one can make when integrating functions using complex measures), keep in mind that if B1,B2() then EE(B1)v,w(B2)=E(B2)E(B1)v,w=E(B1B2)v,w=Ev,w(B1B2).

E(f)E(g)=E(fg).

This and the third statement in (2) show that E:𝔅()B(H) is a homomorphism of C*-algebras:

From the fact that E:𝔅()B(H) is a homomorphism of C*-algebras, it follows in particular that E(f) is a normal operator for each f𝔅().

10 The spectrum of a projection-valued measure

If E:()𝒫(H) is a projection-valued measure, let Uα,αI, be those open sets Uα such that E(Uα)=0. The spectrum of E is

σ(E)=αIUα;

this may also be called the support of E, and is analogous to the support of a nonnegative measure. Since σ(E) is the complement of a union of open sets, it is closed.

For each n1, let Dn={|z|n}. Dn is compact, so there are αn,1,,αn,N such that

Dnk=1NUαn,k.

But, using §7 and the fact that E(Uαn,k)=0 for 1kN,

E(k=1NUαn,k) = E(Uαn,1)+E(k=2NUαn,k)-E(Uαn,1k=2NUαn,k)
= E(k=2NUαn,k)-E(Uαn,1)E(k=2NUαn,k)
= E(k=2NUαn,k)
=
= 0,

therefore E(Dn)=0 (it will be 0, and as a projection is a positive operator it must equal 0). Let Bn=Dn+1Dn; as Bn is a subset of Dn+1 we get E(Bn)=0. We have n=1Bn=, and as the Bn are pairwise disjoint we get

E()=E(n=1Bn)=n=1E(Bn)=0,

contradicting that E()=idH. Therefore, if E is a projection-valued measure then its spectrum is not empty.

Here are some facts about projection-valued measures. E(σ(E))=0.1717 17 Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 62, §38, Theorem 1. His statement is about regular measures, but a projection-valued measure on the Borel σ-algebra of is regular, as he shows on the page after that.

Let 𝔅E() be the set of bounded measurable functions σ(E), and let fE=supzσ(E)|f(z)|. If E is a projection-valued measure with compact spectrum and f: is continuous, then1818 18 Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 62, §38, Theorem 2.

E(χσ(E)f)=fE.

(We often talk about projection-valued measures whose spectrum is compact; since their spectrum is closed, to demand that the spectrum of a projection-valued measure is compact is to demand that it is bounded.)

If E is a projection-valued measure with compact spectrum and if A=E(χσ(E)λ), then1919 19 Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 64, §39, Theorem 2. The proof uses the fact that TB(H) is invertible if and only if there is some α>0 such that Tvαv for all vH.

σ(A)=σ(E).

11 Statement of the spectral theorem

The spectral theorem, proved by in Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 69, §43, Theorem 1, states the following:

If AB(H) is self-adjoint, then there exists a unique projection-valued measure E:()𝒫(H), with σ(E) compact and σ(E), such that

A=E(χσ(E)λ)=σ(E)λ𝑑E(λ).

Since σ(E), we can write E:()𝒫(H).