Projection-valued measures and spectral integrals

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 $W_{\alpha }$, $\alpha \in I$, are subsets of $H$, let

$$\underset{\alpha \in I}{\bigvee }{W}_{\alpha }$$

denote the closure of the span of ${\bigcup }_{\alpha \in I}{W}_{\alpha }$. It is straightforward to check that this is the intersection of all closed subspaces of $H$ that contain each of the $W_{\alpha }$.

If $V$ is a subspace of $H$, define

$$V^{⟂}=\{f\in H:⟨f,g⟩=0\text{ for all }g\in V\},$$

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 $f\in H$. It is a fact that there is a unique $g_0\in V$ such that

$$\parallel f-g_0\parallel =\underset{g\in V}{inf}\parallel f-g\parallel ,$$

and that $f-g_0\in V^{⟂}$. Then, $f=g_0+(f-g_0)\in V+V^{⟂}$. Therefore, if $V$ is a closed subspace of $H$ then

$$H=V+V^{⟂}.$$ (1)

If $V$ and $W$ are subspaces of $H$, we write $V⟂W$ if $⟨v,w⟩=0$ for all $v\in V,w\in W$. If $V,W$ are closed subspaces of $H$ and $V⟂W$, we can show that $V+W=V\vee W$, and thus $V+W$ is in particular a closed subspace of $H$. By induction, we obtain $V_1+\mathrm{\cdots }+V_n={\bigvee }_{k=1}^n{V}_k$. If $V_k$, $k\ge 1$, are mutually orthogonal closed subspaces of $H$, we define

$$\underset{k=1}{\overset{\mathrm{\infty }}{\oplus }}{V}_k=\underset{k=1}{\overset{\mathrm{\infty }}{\bigvee }}{V}_k,$$

called the orthogonal direct sum of the subspaces $V_k$. The orthogonal direct sum ${\oplus }_{k=1}^{\mathrm{\infty }}{V}_k$ is equal to the closure of the set of all finite sums of the form $\sum v_k$, where $v_k\in V_k$.

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

$$H=V\oplus V^{⟂}.$$

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.

3 Projections

Let $V$ be a closed subspace of $H$. The projection onto $V$ is the map $P_V:H\to H$ defined in the following way: if $f\in H=V\oplus V^{⟂}$, let $f=g+h,g\in V,h\in V^{⟂}$, and set $P_V(f)=g$. The image of $P_V$ is $V$: if $f\in \text{im }{P}_V$ then there is some $f_0\in H$ such that $P_V{f}_0=f$ and $f_0=g+h$, $g\in V,h\in V^{⟂}$, so by the definition of $P_V$ we get $P_V{f}_0=g$, giving $f=g\in V$; on the other hand, if $f\in V$ then $P_{V}f=f$, so $f\in \text{im }P$.

The image of a projection is closed. Hence if $P$ is a projection, we have

$$H=\text{im }{P}_V\oplus {(\text{im }{P}_V)}^{⟂}=\text{im }{P}_V\oplus \ker P_V.$$

If $P_V$ is a projection onto a closed subspace $V$ of $H$, we check that $P_V\in B(H)$. If $g\in V,h\in V^{⟂}$, then, because $⟨g,h⟩=0$,

$${\parallel P_V(g+h)\parallel }^2={\parallel g\parallel }^2\le {\parallel g\parallel }^2+{\parallel h\parallel }^2=⟨g,g⟩+⟨h,h⟩=⟨g+h,g+h⟩={\parallel g+h\parallel }^2,$$

so $\parallel P_V\parallel \le 1$. If $V\ne \{0\}$, let $g\in V$, $g\ne 0$. Then

$$\parallel \frac{g}{\parallel g\parallel }\parallel =1,\parallel P_V\left(\frac{g}{\parallel g\parallel }\right)\parallel =\parallel \frac{g}{\parallel g\parallel }\parallel =1.$$

Hence if $P_V$ is a projection and $P_V\ne 0$, then $\parallel P_V\parallel =1$. Of course, if $P_V=0$ then $\parallel P_V\parallel =0$. Using $H=V\oplus V^{⟂}$, we check that if $f_1,f_2\in H$ then

$$⟨P_V{f}_1,f_2⟩=⟨f_1,P_V{f}_2⟩,$$

hence $P_V$ is self-adjoint.

We call $P\in B(H)$ idempotent if $P^2=P$. The following conditions are each equivalent to a nonzero idempotent $P$ being a projection:

• $P$ is positive

• $P$ is self-adjoint

• $P$ is normal

• $\ker P={(\text{im }P)}^{⟂}$

• $\parallel P\parallel =1$

4 The lattice of projections

Two important projections: ${\text{id}}_H$ is the projection onto $H$, and $0$ is the projection onto $\{0\}$.

If $T\in B(H)$ is self-adjoint, we say that $T$ is positive if, for all $v\in H$,

$$⟨Tv,v⟩\ge 0.$$

If $S,T\in B(H)$ are self-adjoint, we say that $S\ge T$ if, for all $v\in H$,

$$⟨Sv,v⟩\ge ⟨Tv,v⟩.$$

This is equivalent to $S-T$ being a positive operator. Thus, $T\in B(H)$ is a positive operator if and only if $T\ge 0$. One checks that $\le$ is a partial order on the set of bounded self-adjoint operators on $H$.

Let $P\in B(H)$ be a projection. As $H=\text{im }P\oplus {(\text{im }P)}^{⟂}=\text{im }P\oplus \ker P$, if $v\in H$ then $v=v_1+v_2$, $v_1\in \text{im }P,v_2\in {(\text{im }P)}^{⟂}$,

$$⟨Pv,v⟩=⟨v_1,v_1+v_2⟩=⟨v_1,v_1⟩+⟨v_1,v_2⟩=⟨v_1,v_1⟩\ge 0.$$

Hence $P\ge 0$.

Let $P,Q\in B(H)$ be projections. It is a fact that the following statements are equivalent:

• $P\le Q$

• $\text{im }P\subseteq \text{im }Q$

• $QP=P$

• $PQ=P$

• $\parallel Pv\parallel \le \parallel Qv\parallel$ for all $v\in H$

The set of projections is a complete lattice.

5 Sums of projections

If $P,Q\in B(H)$ are projections, we write $P⟂Q$ if $\text{im }P⟂\text{im }Q$, which is equivalent to $PQ=0$ and $QP=0$; neither one of those by itself is sufficient. It is a fact that $P+Q$ is a projection if and only if $P⟂Q$, in which case $P+Q$ is a projection onto $\text{im }P\oplus \text{im }Q$. The product $PQ$ is a projection if and only if $PQ=QP$, in which case it is a projection onto $\text{im }P\cap \text{im }Q$.

Paul Halmos shows the following in Question 94 of his Hilbert Space Problem Book: If $T_n\in B(H)$ are self-adjoint such that $T_n\le T_{n+1}$ for all $n$ and if there is some self-adjoint $T^{\prime }\in B(H)$ such that $T_n\le T^{\prime }$ for all $n$, then there is some self-adjoint $T\in B(H)$ such that $T_n\to T$ 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 $M_n$, $n\ge 1$, be closed subspaces of $H$ such that $M_n\subseteq M_{n+1}$ and let $P_n$ be the projection onto $M_n$. As $M_n\subseteq M_{n+1}$ we have $P_n\le P_{n+1}$ for all $n$. Also, $P_n\le {\text{id}}_H$ for all $n$, as ${\text{id}}_H$ is the projection onto $H$ and $M_n\subseteq H$. Hence by the result from Halmos stated above, there is some self-adjoint $P\in B(H)$ such that $P_n\to P$ in the strong operator topology. Let

$$M=\underset{n=1}{\overset{\mathrm{\infty }}{\bigvee }}{M}_n.$$

I claim that $P$ is the projection onto $M$.

If $P_n\in B(H)$, $n\ge 1$, are projections and $P_i⟂P_j$ for $i\ne j$, let

$$M_n=\underset{k=1}{\overset{n}{\oplus }}\text{im }{P}_k.$$

Define $T_{n}v={\sum }_{k=1}^n{P}_{k}v$. As $M_n$ is a closed subspace of $H$, we have

$$H=M_n\oplus M_n^{⟂}=\text{im }{P}_1\oplus \mathrm{\cdots }{P}_n\oplus M_n^{⟂}.$$

If $v=v_1+\mathrm{\cdots }+v_n+v^{\prime }$, $v_1\in \text{im }{P}_1,\mathrm{\dots },v_n\in \text{im }{P}_n,v^{\prime }\in M_n^{⟂}$, then

$$T_{n}v=\sum _{k=1}^n\left(P_k(v_1+\mathrm{\cdots }+v_n+v^{\prime })\right)=\sum _{k=1}^n{v}_k.$$

Thus $T_n$ is the projection onto $M_n$. Therefore, there is some self-adjoint $P\in B(H)$ such that $T_n\to P$ in the strong operator topology, and with

$$M=\underset{n=1}{\overset{\mathrm{\infty }}{\bigvee }}\underset{k=1}{\overset{n}{\bigvee }}\text{im }{P}_k=\underset{n=1}{\overset{\mathrm{\infty }}{\bigvee }}\text{im }{P}_k=\underset{n=1}{\overset{\mathrm{\infty }}{\oplus }}\text{im }{P}_n,$$

$P$ is the projection onto $M$. ${\sum }_{k=1}^n{P}_k\to P$ in the strong operator topology, and we denote $P$ by ${\sum }_{k=1}^{\mathrm{\infty }}{P}_k$.

6 Definition of projection-valued measures

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

• $E(\mathrm{\varnothing })=0$ and $E(ℂ)={\text{id}}_H$

• If $B_n\in ℬ(ℂ)$, $n\ge 1$, are pairwise disjoint, then

  $$E\left(\bigcup _{n\ge 1}{B}_n\right)=\sum _{n\ge 1}E(B_n),$$

  where ${\sum }_{n\ge 1}E(B_n)$ is the limit of ${\sum }_{1\le n\le N}E(B_n)$ in the strong operator topology.

7 Finite additivity

If $E:ℬ(ℂ)\to 𝒫(H)$ is any function that satisfies $E(B_1\cup B_2)=E(B_1)+E(B_2)$ for disjoint $B_1,B_2\in ℬ(ℂ)$, then it satisfies the following four properties. In particular, if $E$ is a projection-valued measure it satisfies them.

1. $E(\mathrm{\varnothing })=E(\mathrm{\varnothing }\cup \mathrm{\varnothing })=E(\mathrm{\varnothing })+E(\mathrm{\varnothing })$, so $E(\mathrm{\varnothing })=0$.

2. If $B_1,B_2\in ℬ(ℂ)$ and $B_1\subseteq B_2$, then

   $$E(B_2)=E(B_2\setminus B_1\cup B_1)=E(B_2\setminus B_1)+E(B_1)\ge E(B_1),$$

   since $E(B_2\setminus B_1)$ is a projection and hence is a positive operator. Therefore, if $B_1\subseteq B_2$ then $E(B_1)\le E(B_2)$.

3. Let $B_1,B_2\in ℬ(ℂ)$. We have

   $$E(B_1)=E(B_1\cap B_2)+E(B_1\setminus B_2),$$

   and

   $$E(B_2)=E(B_1\cap B_2)+E(B_2\setminus B_1),$$

   and

   $$E(B_1\cup B_2)=E(B_1\cap B_2)+E(B_1\setminus B_2)+E(B_2\setminus B_1),$$

   and combining these gives, for any $B_1,B_2\in ℬ(ℂ)$,

   $E(B_1)+E(B_2)$ $=$ $E(B_1\cap B_2)+E(B_1\cap B_2)+E(B_1\setminus B_2)+E(B_2\setminus B_1)$

   $=$ $E(B_1\cap B_2)+E(B_1\cup B_2).$

4. Let $B_1,B_2\in ℬ(ℂ)$. Multiplying both sides of the above equation on the left by $E(B_2)$ gives

   $$E(B_1)E(B_2)+E(B_2)E(B_2)=E(B_1\cap B_2)E(B_2)+E(B_1\cup B_2)E(B_2).$$

   As $E(B_1\cup B_2)$ and $E(B_2)$ are projections and $E(B_1\cup B_2)\ge E(B_2)$, we have

   $$E(B_1\cup B_2)E(B_2)=E(B_2),$$

   which with $E(B_2)E(B_2)=E(B_2)$ gives

   $$E(B_1)E(B_2)+E(B_2)=E(B_1\cap B_2)+E(B_2).$$

   Hence, for any $B_1,B_2\in ℬ(ℂ)$,

   $$E(B_1)E(B_2)=E(B_1\cap B_2).$$

8 Complex measures

Suppose that $E:ℬ(ℂ)\to 𝒫(H)$ is a function such that $E(ℂ)={\text{id}}_H$, and that for all $v,w\in H$, the function $E_{v,w}:ℬ(ℂ)\to ℂ$ defined by

$$E_{v,w}(B)=⟨E(B)v,w⟩$$

is a complex measure. I will show that $E$ is a projection-valued measure. Let $B_n\in ℬ(ℂ)$, $n\ge 1$, be pairwise disjoint. If $B_1,B_2\in ℬ(ℂ)$ are disjoint, then for all $v,w\in H$,

$⟨E(B_1\cup B_2)v,w⟩$ $=$ $E_{v,w}(B_1\cup B_2)$

$=$ $E_{v,w}(B_1)+E_{v,w}(B_2)$

$=$ $⟨E(B_1)v,w⟩+⟨E(B_2)v,w⟩$

$=$ $⟨(E(B_1)+E(B_2))v,w⟩,$

and since this holds for all $v,w\in H$, we obtain $E(B_1\cup B_2)=E(B_1)+E(B_2)$. Therefore, from §7, if $B_1,B_2\in ℬ(ℂ)$ are disjoint then

$$E(B_1)E(B_2)=E(B_1\cap B_2)=E(\mathrm{\varnothing })=0.$$

If $v\in H$ and $v_n=E(B_n)v$, then, for $m\ne n$,

$$⟨v_n,v_m⟩=⟨E(B_n)v,E(B_m)v⟩=⟨E(B_m)E(B_n)v,v⟩=⟨0v,0⟩=0.$$

For $a_n\in ℂ$, for the sequence ${\sum }_{n=1}^N{a}_n{v}_n$ to converge in $H$ it is equivalent to ; but $\parallel v_n\parallel \le \parallel v\parallel$, so it suffices to show that . Using that if $T$ is a projection then $T$ is self-adjoint and $T^2=T$, and that $E_{v,v}$ is a complex measure,

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\parallel E(B_n)v\parallel }^2$ $=$ ${\displaystyle \sum _{n=1}^N}⟨E(B_n)v,E(B_n)v⟩$

$=$ ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}⟨E(B_n)E(B_n)v,v⟩$

$=$ ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}⟨E(B_n)v,v⟩$

$=$ ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{E}_{v,v}(B_n)$

$=$ $E_{v,v}\left({\displaystyle \bigcup _{n=1}^{\mathrm{\infty }}}{B}_n\right)$

$=$ $⟨E\left({\displaystyle \bigcup _{n=1}^{\mathrm{\infty }}}{B}_n\right)v,v⟩$

$=$ ${\parallel E\left({\displaystyle \bigcup _{n=1}^{\mathrm{\infty }}}{B}_n\right)v\parallel }^2$

$\le$ $\parallel v\parallel .$

Therefore, the sequence ${\sum }_{n=1}^{N}E(B_n)v$ converges in $H$; namely, ${\sum }_{n=1}^{N}E(B_n)$ converges in the strong operator topology. Let $P$ be its limit. By §5, $P$ is the projection onto ${\oplus }_{n=1}^{\mathrm{\infty }}\text{im }E(B_n)$.

For $v,w\in H$,

$⟨E\left({\displaystyle \bigcup _{n=1}^{\mathrm{\infty }}}{B}_n\right)v,w⟩$ $=$ $E_{v,w}\left({\displaystyle \bigcup _{n=1}^{\mathrm{\infty }}}{B}_n\right)$

$=$ ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{E}_{v,w}(B_n)$

$=$ ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}⟨E(B_n)v,w⟩$

$=$ $⟨Pv,w⟩,$

and since this is true for all $v,w\in H$, we obtain

$$E\left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)=P,$$

where $P$ is the limit of ${\sum }_{n=1}^{N}E(B_n)$ in the strong operator topology. This completes the proof that $E$ is a projection-valued measure.

On the other hand, if $E:ℬ(ℂ)\to 𝒫(H)$ is a projection-valued measure, we can show that for each $v,w\in H$ the function $E_{v,w}:ℬ(ℂ)\to ℂ$ defined by $E_{v,w}(B)=⟨E(B)v,w⟩$ is a complex measure.

9 Spectral integrals

Let $𝔅(ℂ)$ be the set of bounded measurable functions $ℂ\to ℂ$. It is a complex vector space, and we define the norm $\parallel f\parallel ={sup}_{z\in ℂ}|f(z)|$; one checks that $𝔅(ℂ)$ is a Banach space. Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 60, §37, proves that if $E:ℬ(ℂ)\to 𝒫(H)$ is a projection-valued measure and $f\in 𝔅$, then there is a unique $A\in B(H)$ such that, for all $v,w\in H$,

$$⟨Av,w⟩=E_{v,w}(f)=\int f𝑑E_{v,w}={\int }_{ℂ}f(\lambda )𝑑E_{v,w}(\lambda ),$$

and $\parallel A\parallel \le 2\parallel f\parallel$. We write

$$A=E(f)=\int f𝑑E={\int }_{ℂ}f(\lambda )𝑑E(\lambda ).$$

We can check that $𝔅(ℂ)$ is a $C^{*}$-algebra, with $f^{*}$ defined by $f^{*}(z)=\overline{f(z)}$. It is a fact that $B(H)$ is a $C^{*}$-algebra. If $\alpha \in ℂ$ and $f,g\in 𝔅(ℂ)$, then

$$E(\alpha f)=\alpha E(f),E(f+g)=E(f)+E(g);E(f^{*})={(E(f))}^{*}.$$ (2)

The first two of these together with $\parallel E(f)\parallel \le 2\parallel f\parallel$ show that $E:𝔅(ℂ)\to B(H)$ is a bounded linear map. If $f,g\in 𝔅(ℂ)$, then

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

This and the third statement in (2) show that $E:𝔅(ℂ)\to B(H)$ is a homomorphism of $C^{*}$-algebras:

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

10 The spectrum of a projection-valued measure

If $E:ℬ(ℂ)\to 𝒫(H)$ is a projection-valued measure, let $U_{\alpha },\alpha \in I$, be those open sets $U_{\alpha }\subseteq ℂ$ such that $E(U_{\alpha })=0$. The spectrum of $E$ is

$$\sigma (E)=ℂ\setminus \bigcup _{\alpha \in I}{U}_{\alpha };$$

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

For each $n\ge 1$, let $D_n=\{|z|\le n\}$. $D_n$ is compact, so there are ${\alpha }_{n,1},\mathrm{\dots },{\alpha }_{n,N}$ such that

$$D_n\subseteq \bigcup _{k=1}^N{U}_{{\alpha }_{n,k}}.$$

But, using §7 and the fact that $E(U_{{\alpha }_{n,k}})=0$ for $1\le k\le N$,

$E\left({\displaystyle \bigcup _{k=1}^N}{U}_{{\alpha }_{n,k}}\right)$ $=$ $E(U_{{\alpha }_{n,1}})+E\left({\displaystyle \bigcup _{k=2}^N}{U}_{{\alpha }_{n,k}}\right)-E\left(U_{{\alpha }_{n,1}}\cap {\displaystyle \bigcup _{k=2}^N}{U}_{{\alpha }_{n,k}}\right)$

$=$ $E\left({\displaystyle \bigcup _{k=2}^N}{U}_{{\alpha }_{n,k}}\right)-E(U_{{\alpha }_{n,1}})E\left({\displaystyle \bigcup _{k=2}^N}{U}_{{\alpha }_{n,k}}\right)$

$=$ $E\left({\displaystyle \bigcup _{k=2}^N}{U}_{{\alpha }_{n,k}}\right)$

$=$ $\mathrm{\cdots }$

$=$ $0,$

therefore $E(D_n)=0$ (it will be $\le 0$, and as a projection is a positive operator it must equal $0$). Let $B_n=D_{n+1}\setminus D_n$; as $B_n$ is a subset of $D_{n+1}$ we get $E(B_n)=0$. We have ${\bigcup }_{n=1}^{\mathrm{\infty }}{B}_n=ℂ$, and as the $B_n$ are pairwise disjoint we get

$$E(ℂ)=E\left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)=\sum _{n=1}^{\mathrm{\infty }}E(B_n)=0,$$

contradicting that $E(ℂ)={\text{id}}_H$. Therefore, if $E$ is a projection-valued measure then its spectrum is not empty.

Here are some facts about projection-valued measures. $E(ℂ\setminus \sigma (E))=0$.

Let ${𝔅}_E(ℂ)$ be the set of bounded measurable functions $\sigma (E)\to ℂ$, and let ${\parallel f\parallel }_E={sup}_{z\in \sigma (E)}|f(z)|$. If $E$ is a projection-valued measure with compact spectrum and $f:ℂ\to ℂ$ is continuous, then

$$\parallel E({\chi }_{\sigma (E)}f)\parallel ={\parallel f\parallel }_E.$$

(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 $A=E({\chi }_{\sigma (E)}\lambda )$, then

$$\sigma (A)=\sigma (E).$$

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 $A\in B(H)$ is self-adjoint, then there exists a unique projection-valued measure $E:ℬ(ℂ)\to 𝒫(H)$, with $\sigma (E)$ compact and $\sigma (E)\subset ℝ$, such that

$$A=E({\chi }_{\sigma (E)}\lambda )={\int }_{\sigma (E)}\lambda 𝑑E(\lambda ).$$

Since $\sigma (E)\subset ℝ$, we can write $E:ℬ(ℝ)\to 𝒫(H)$.