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

 $\bigvee_{\alpha\in I}W_{\alpha}$

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 $W_{\alpha}$.11 1 Let $\mathfrak{L}$ be the set of all closed subspaces of $H$. Set inclusion $\subseteq$ is a partial order on $\mathfrak{L}$. $\mathfrak{L}$ is a lattice: if $M,N\in\mathfrak{L}$, then $M\cap N$ is the greatest element in $\mathfrak{L}$ that is contained both in $M$ and in $N$, and $M\vee N$ is the least element in $\mathfrak{L}$ that contains each of $M$ and $N$. $\mathfrak{L}$ is a complete lattice: if $M_{\alpha}\in\mathfrak{L}$, then $\bigcup_{\alpha}M_{\alpha}$ is the greatest element in $\mathfrak{L}$ that is contained in each $M_{\alpha}$, and $\bigvee_{\alpha}M_{\alpha}$ is the least element in $\mathfrak{L}$ that contains each $M_{\alpha}$. (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

 $V^{\perp}=\{f\in H:\textrm{\left\langle f,g\right\rangle=0 for all g\in V}\},$

called the orthogonal complement of $V$. If $V$ is a subspace of $H$ then $V^{\perp}$ 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

 $\left\|f-g_{0}\right\|=\inf_{g\in V}\left\|f-g\right\|,$

and that $f-g_{0}\in V^{\perp}$. Then, $f=g_{0}+(f-g_{0})\in V+V^{\perp}$. Therefore, if $V$ is a closed subspace of $H$ then

 $H=V+V^{\perp}.$ (1)

If $V$ and $W$ are subspaces of $H$, we write $V\perp W$ if $\left\langle v,w\right\rangle=0$ for all $v\in V,w\in W$. If $V,W$ are closed subspaces of $H$ and $V\perp W$, we can show that $V+W=V\vee W$, and thus $V+W$ is in particular a closed subspace of $H$.22 2 If $z_{n}\in V+W$ and $z_{n}\to z\in H$, write $z_{n}=v_{n}+w_{n}$, $v_{n}\in V$ and $w_{n}\in W$. Using $V\perp W$ we get $\left\|z_{n}-z_{m}\right\|^{2}=\left\|v_{n}-v_{m}\right\|^{2}+\left\|w_{n}-w_{% m}\right\|^{2}$, 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 $V\perp W$ for which $V+W$ is not itself closed. By induction, we obtain $V_{1}+\cdots+V_{n}=\bigvee_{k=1}^{n}V_{k}$. If $V_{k}$, $k\geq 1$, are mutually orthogonal closed subspaces of $H$, we define

 $\bigoplus_{k=1}^{\infty}V_{k}=\bigvee_{k=1}^{\infty}V_{k},$

called the orthogonal direct sum of the subspaces $V_{k}$. The orthogonal direct sum $\bigoplus_{k=1}^{\infty}V_{k}$ is equal to the closure of the set of all finite sums of the form $\sum v_{k}$, where $v_{k}\in V_{k}$.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

 $H=V\oplus V^{\perp}.$

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 $P_{V}:H\to H$ defined in the following way: if $f\in H=V\oplus V^{\perp}$, let $f=g+h,g\in V,h\in V^{\perp}$, and set $P_{V}(f)=g$. The image of $P_{V}$ is $V$: if $f\in\textrm{im\,}P_{V}$ then there is some $f_{0}\in H$ such that $P_{V}f_{0}=f$ and $f_{0}=g+h$, $g\in V,h\in V^{\perp}$, 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 $f\in\textrm{im\,}P$.

The image of a projection is closed.55 5 We call $P:H\to H$ 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

 $H=\textrm{im\,}P_{V}\oplus(\textrm{im\,}P_{V})^{\perp}=\textrm{im\,}P_{V}% \oplus\ker P_{V}.$

If $P_{V}$ is a projection onto a closed subspace $V$ of $H$, we check that $P_{V}\in B(H)$. If $g\in V,h\in V^{\perp}$, then, because $\left\langle g,h\right\rangle=0$,

 $\left\|P_{V}(g+h)\right\|^{2}=\left\|g\right\|^{2}\leq\left\|g\right\|^{2}+% \left\|h\right\|^{2}=\left\langle g,g\right\rangle+\left\langle h,h\right% \rangle=\left\langle g+h,g+h\right\rangle=\left\|g+h\right\|^{2},$

so $\left\|P_{V}\right\|\leq 1$. If $V\neq\{0\}$, let $g\in V$, $g\neq 0$. Then

 $\left\|\frac{g}{\left\|g\right\|}\right\|=1,\qquad\left\|P_{V}\left(\frac{g}{% \left\|g\right\|}\right)\right\|=\left\|\frac{g}{\left\|g\right\|}\right\|=1.$

Hence if $P_{V}$ is a projection and $P_{V}\neq 0$, then $\left\|P_{V}\right\|=1$. Of course, if $P_{V}=0$ then $\left\|P_{V}\right\|=0$. Using $H=V\oplus V^{\perp}$, we check that if $f_{1},f_{2}\in H$ then

 $\left\langle P_{V}f_{1},f_{2}\right\rangle=\left\langle f_{1},P_{V}f_{2}\right\rangle,$

hence $P_{V}$ is self-adjoint.

We call $P\in B(H)$ idempotent if $P^{2}=P$.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

• $\ker P=(\textrm{im\,}P)^{\perp}$

• $\left\|P\right\|=1$

4 The lattice of projections

Two important projections: $\textrm{id}_{H}$ 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 $v\in H$,

 $\left\langle Tv,v\right\rangle\geq 0.$

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

 $\left\langle Sv,v\right\rangle\geq\left\langle Tv,v\right\rangle.$

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

Let $P\in B(H)$ be a projection. As $H=\textrm{im\,}P\oplus(\textrm{im\,}P)^{\perp}=\textrm{im\,}P\oplus\ker P$, if $v\in H$ then $v=v_{1}+v_{2}$, $v_{1}\in\textrm{im\,}P,v_{2}\in(\textrm{im\,}P)^{\perp}$,

 $\left\langle Pv,v\right\rangle=\left\langle v_{1},v_{1}+v_{2}\right\rangle=% \left\langle v_{1},v_{1}\right\rangle+\left\langle v_{1},v_{2}\right\rangle=% \left\langle v_{1},v_{1}\right\rangle\geq 0.$

Hence $P\geq 0$.

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

• $P\leq Q$

• $\textrm{im\,}P\subseteq\textrm{im\,}Q$

• $QP=P$

• $PQ=P$

• $\left\|Pv\right\|\leq\left\|Qv\right\|$ for all $v\in H$

The set of projections is a complete lattice.77 7 If $P,Q\in B(H)$ are projections, then the infimum of $P$ and $Q$ is the projection onto $\textrm{im\,}P\cap\textrm{im\,}Q$, and the supremum of $P$ and $Q$ is the projection onto $\textrm{im\,}P\vee\textrm{im\,}Q$. See Problem 96 of Paul Halmos’s Hilbert Space Problem Book.

5 Sums of projections

If $P,Q\in B(H)$ are projections, we write $P\perp Q$ if $\textrm{im\,}P\perp\textrm{im\,}Q$, 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 $P\perp Q$, in which case $P+Q$ is a projection onto $\textrm{im\,}P\oplus\textrm{im\,}Q$. The product $PQ$ is a projection if and only if $PQ=QP$, in which case it is a projection onto $\textrm{im\,}P\cap\textrm{im\,}Q$.

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 $T_{n}\leq T_{n+1}$ for all $n$ and if there is some self-adjoint $T^{\prime}\in B(H)$ such that $T_{n}\leq T^{\prime}$ for all $n$, then there is some self-adjoint $T\in B(H)$ such that $T_{n}\to T$ in the strong operator topology.99 9 Recall that if $T_{n}\in B(H)$ and $T\in B(H)$, we say that $T_{n}\to T$ in the strong operator topology if for all $v\in H$ we have $T_{n}v\to Tv$. (If $T_{n}\to T$ in $B(H)$ then $T_{n}\to T$ 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 $M_{n}$, $n\geq 1$, 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 $P_{n}\leq P_{n+1}$ for all $n$. Also, $P_{n}\leq\textrm{id}_{H}$ for all $n$, as $\textrm{id}_{H}$ 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

 $M=\bigvee_{n=1}^{\infty}M_{n}.$

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 $P_{n}\in B(H)$, $n\geq 1$, are projections and $P_{i}\perp P_{j}$ for $i\neq j$, let

 $M_{n}=\bigoplus_{k=1}^{n}\textrm{im\,}P_{k}.$

Define $T_{n}v=\sum_{k=1}^{n}P_{k}v$. As $M_{n}$ is a closed subspace of $H$, we have

 $H=M_{n}\oplus M_{n}^{\perp}=\textrm{im\,}P_{1}\oplus\cdots P_{n}\oplus M_{n}^{% \perp}.$

If $v=v_{1}+\cdots+v_{n}+v^{\prime}$, $v_{1}\in\textrm{im\,}P_{1},\ldots,v_{n}\in\textrm{im\,}P_{n},v^{\prime}\in M_{% n}^{\perp}$, then

 $T_{n}v=\sum_{k=1}^{n}\left(P_{k}(v_{1}+\cdots+v_{n}+v^{\prime})\right)=\sum_{k% =1}^{n}v_{k}.$

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

 $M=\bigvee_{n=1}^{\infty}\bigvee_{k=1}^{n}\textrm{im\,}P_{k}=\bigvee_{n=1}^{% \infty}\textrm{im\,}P_{k}=\bigoplus_{n=1}^{\infty}\textrm{im\,}P_{n},$

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

6 Definition of projection-valued measures

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

• $E(\emptyset)=0$ and $E(\mathbb{C})=\textrm{id}_{H}$

• If $B_{n}\in\mathscr{B}(\mathbb{C})$, $n\geq 1$, are pairwise disjoint, then

 $E\Big{(}\bigcup_{n\geq 1}B_{n}\Big{)}=\sum_{n\geq 1}E(B_{n}),$

where $\sum_{n\geq 1}E(B_{n})$ is the limit of $\sum_{1\leq n\leq N}E(B_{n})$ in the strong operator topology.

If $E:\mathscr{B}(\mathbb{C})\to\mathscr{P}(H)$ is any function that satisfies $E(B_{1}\cup B_{2})=E(B_{1})+E(B_{2})$ for disjoint $B_{1},B_{2}\in\mathscr{B}(\mathbb{C})$, then it satisfies the following four properties. In particular, if $E$ is a projection-valued measure it satisfies them.

1. 1.

$E(\emptyset)=E(\emptyset\cup\emptyset)=E(\emptyset)+E(\emptyset)$, so $E(\emptyset)=0$.

2. 2.

If $B_{1},B_{2}\in\mathscr{B}(\mathbb{C})$ and $B_{1}\subseteq B_{2}$, then

 $E(B_{2})=E(B_{2}\setminus B_{1}\cup B_{1})=E(B_{2}\setminus B_{1})+E(B_{1})% \geq E(B_{1}),$

since $E(B_{2}\setminus B_{1})$ is a projection and hence is a positive operator. Therefore, if $B_{1}\subseteq B_{2}$ then $E(B_{1})\leq E(B_{2})$.

3. 3.

Let $B_{1},B_{2}\in\mathscr{B}(\mathbb{C})$. 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 $B_{1},B_{2}\in\mathscr{B}(\mathbb{C})$,

 $\displaystyle E(B_{1})+E(B_{2})$ $\displaystyle=$ $\displaystyle E(B_{1}\cap B_{2})+E(B_{1}\cap B_{2})+E(B_{1}\setminus B_{2})+E(% B_{2}\setminus B_{1})$ $\displaystyle=$ $\displaystyle E(B_{1}\cap B_{2})+E(B_{1}\cup B_{2}).$
4. 4.

Let $B_{1},B_{2}\in\mathscr{B}(\mathbb{C})$. 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 $E(B_{1}\cup B_{2})\geq E(B_{2})$, 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 $B_{1},B_{2}\in\mathscr{B}(\mathbb{C})$,

 $E(B_{1})E(B_{2})=E(B_{1}\cap B_{2}).$

8 Complex measures

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

 $E_{v,w}(B)=\left\langle E(B)v,w\right\rangle$

is a complex measure. I will show that $E$ is a projection-valued measure. Let $B_{n}\in\mathscr{B}(\mathbb{C})$, $n\geq 1$, be pairwise disjoint.1111 11 cf. Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 59, §36, Theorem 3. If $B_{1},B_{2}\in\mathscr{B}(\mathbb{C})$ are disjoint, then for all $v,w\in H$,

 $\displaystyle\left\langle E(B_{1}\cup B_{2})v,w\right\rangle$ $\displaystyle=$ $\displaystyle E_{v,w}(B_{1}\cup B_{2})$ $\displaystyle=$ $\displaystyle E_{v,w}(B_{1})+E_{v,w}(B_{2})$ $\displaystyle=$ $\displaystyle\left\langle E(B_{1})v,w\right\rangle+\left\langle E(B_{2})v,w\right\rangle$ $\displaystyle=$ $\displaystyle\left\langle(E(B_{1})+E(B_{2}))v,w\right\rangle,$

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 $B_{1},B_{2}\in\mathscr{B}(\mathbb{C})$ are disjoint then

 $E(B_{1})E(B_{2})=E(B_{1}\cap B_{2})=E(\emptyset)=0.$

If $v\in H$ and $v_{n}=E(B_{n})v$, then, for $m\neq n$,

 $\left\langle v_{n},v_{m}\right\rangle=\left\langle E(B_{n})v,E(B_{m})v\right% \rangle=\left\langle E(B_{m})E(B_{n})v,v\right\rangle=\left\langle 0v,0\right% \rangle=0.$

For $a_{n}\in\mathbb{C}$, for the sequence $\sum_{n=1}^{N}a_{n}v_{n}$ to converge in $H$ it is equivalent to $\sum_{n=1}^{\infty}\left\|a_{n}v_{n}\right\|^{2}<\infty$;1212 12 e.g. Walter Rudin’s Functional Analysis, p. 295, Theorem 12.6: if $x_{n}\in H$ are pairwise orthogonal, not necessarily of unit norm, then for $\sum_{n=1}^{\infty}x_{n}$ to converge is equivalent to $\sum_{n=1}^{\infty}\left\|x_{n}\right\|^{2}<\infty$. but $\left\|v_{n}\right\|\leq\left\|v\right\|$, so it suffices to show that $\sum_{n=1}^{\infty}|a_{n}|^{2}<\infty$. 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,

 $\displaystyle\sum_{n=1}^{\infty}\left\|E(B_{n})v\right\|^{2}$ $\displaystyle=$ $\displaystyle\sum_{n=1}^{N}\left\langle E(B_{n})v,E(B_{n})v\right\rangle$ $\displaystyle=$ $\displaystyle\sum_{n=1}^{\infty}\left\langle E(B_{n})E(B_{n})v,v\right\rangle$ $\displaystyle=$ $\displaystyle\sum_{n=1}^{\infty}\left\langle E(B_{n})v,v\right\rangle$ $\displaystyle=$ $\displaystyle\sum_{n=1}^{\infty}E_{v,v}(B_{n})$ $\displaystyle=$ $\displaystyle E_{v,v}\left(\bigcup_{n=1}^{\infty}B_{n}\right)$ $\displaystyle=$ $\displaystyle\left\langle E\left(\bigcup_{n=1}^{\infty}B_{n}\right)v,v\right\rangle$ $\displaystyle=$ $\displaystyle\left\|E\left(\bigcup_{n=1}^{\infty}B_{n}\right)v\right\|^{2}$ $\displaystyle\leq$ $\displaystyle\left\|v\right\|.$

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 $\bigoplus_{n=1}^{\infty}\textrm{im\,}E(B_{n})$.

For $v,w\in H$,

 $\displaystyle\left\langle E\left(\bigcup_{n=1}^{\infty}B_{n}\right)v,w\right\rangle$ $\displaystyle=$ $\displaystyle E_{v,w}\left(\bigcup_{n=1}^{\infty}B_{n}\right)$ $\displaystyle=$ $\displaystyle\sum_{n=1}^{\infty}E_{v,w}(B_{n})$ $\displaystyle=$ $\displaystyle\sum_{n=1}^{\infty}\left\langle E(B_{n})v,w\right\rangle$ $\displaystyle=$ $\displaystyle\left\langle Pv,w\right\rangle,$

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

 $E\left(\bigcup_{n=1}^{\infty}B_{n}\right)=P,$

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 $E:\mathscr{B}(\mathbb{C})\to\mathscr{P}(H)$ is a projection-valued measure, we can show that for each $v,w\in H$ the function $E_{v,w}:\mathscr{B}(\mathbb{C})\to\mathbb{C}$ defined by $E_{v,w}(B)=\left\langle E(B)v,w\right\rangle$ is a complex measure.

9 Spectral integrals

Let $\mathfrak{B}(\mathbb{C})$ be the set of bounded measurable functions $\mathbb{C}\to\mathbb{C}$.1313 13 This is not $L^{\infty}(\mathbb{C})$, 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 $\mathscr{L}^{\infty}(\mathbb{C})$ of essentially bounded measurable functions. When does one speak about $\mathscr{L}^{\infty}(\mathbb{C})$? $\mathscr{L}^{\infty}(\mathbb{C})$ is a vector space; let $\mathscr{N}(\mathbb{C})$ be the set of those functions that are equal to $0$ almost everywhere in $\mathbb{C}$; $\mathscr{N}(\mathbb{C})$ is a vector space; then $L^{\infty}(\mathbb{C})$ is the vector space quotient $\mathscr{L}^{\infty}(\mathbb{C})/\mathscr{N}(\mathbb{C})$. It is a complex vector space, and we define the norm $\left\|f\right\|=\sup_{z\in\mathbb{C}}|f(z)|$; one checks that $\mathfrak{B}(\mathbb{C})$ 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 $\phi:H\times H\to\mathbb{C}$ is sesquilinear (we take sesquilinear to mean linear in the first entry and conjugate linear in the second entry) and $M=\sup_{\left\|v\right\|=\left\|w\right\|=1}|\phi(v,w)|<\infty,$ then there exists a unique $A\in B(H)$ such that $\phi(v,w)=\left\langle Av,w\right\rangle,\qquad v,w\in H,$ and $\left\|A\right\|=M$. One also has to prove that for each $f\in\mathfrak{B}$, $\phi(v,w)=E_{v,w}(f)$ is sesquilinear. that if $E:\mathscr{B}(\mathbb{C})\to\mathscr{P}(H)$ is a projection-valued measure and $f\in\mathfrak{B}$, then there is a unique $A\in B(H)$ such that, for all $v,w\in H$,

 $\left\langle Av,w\right\rangle=E_{v,w}(f)=\int fdE_{v,w}=\int_{\mathbb{C}}f(% \lambda)dE_{v,w}(\lambda),$

and $\left\|A\right\|\leq 2\left\|f\right\|$. We write

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

We can check that $\mathfrak{B}(\mathbb{C})$ 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 $\alpha\in\mathbb{C}$ and $f,g\in\mathfrak{B}(\mathbb{C})$, then1515 15 See Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 60, §37, Theorem 2.

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

The first two of these together with $\left\|E(f)\right\|\leq 2\left\|f\right\|$ show that $E:\mathfrak{B}(\mathbb{C})\to B(H)$ is a bounded linear map. If $f,g\in\mathfrak{B}(\mathbb{C})$, 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 $B_{1},B_{2}\in\mathscr{B}(\mathbb{C})$ then $E_{E(B_{1})v,w}(B_{2})=\left\langle E(B_{2})E(B_{1})v,w\right\rangle=\left% \langle E(B_{1}\cap B_{2})v,w\right\rangle=E_{v,w}(B_{1}\cap B_{2}).$

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

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

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

10 The spectrum of a projection-valued measure

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

 $\sigma(E)=\mathbb{C}\setminus\bigcup_{\alpha\in I}U_{\alpha};$

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 $n\geq 1$, let $D_{n}=\{|z|\leq n\}$. $D_{n}$ is compact, so there are $\alpha_{n,1},\ldots,\alpha_{n,N}$ 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 $1\leq k\leq N$,

 $\displaystyle E\left(\bigcup_{k=1}^{N}U_{\alpha_{n,k}}\right)$ $\displaystyle=$ $\displaystyle E(U_{\alpha_{n,1}})+E\left(\bigcup_{k=2}^{N}U_{\alpha_{n,k}}% \right)-E\left(U_{\alpha_{n,1}}\cap\bigcup_{k=2}^{N}U_{\alpha_{n,k}}\right)$ $\displaystyle=$ $\displaystyle E\left(\bigcup_{k=2}^{N}U_{\alpha_{n,k}}\right)-E(U_{\alpha_{n,1% }})E\left(\bigcup_{k=2}^{N}U_{\alpha_{n,k}}\right)$ $\displaystyle=$ $\displaystyle E\left(\bigcup_{k=2}^{N}U_{\alpha_{n,k}}\right)$ $\displaystyle=$ $\displaystyle\cdots$ $\displaystyle=$ $\displaystyle 0,$

therefore $E(D_{n})=0$ (it will be $\leq 0$, 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 $\bigcup_{n=1}^{\infty}B_{n}=\mathbb{C}$, and as the $B_{n}$ are pairwise disjoint we get

 $E(\mathbb{C})=E\left(\bigcup_{n=1}^{\infty}B_{n}\right)=\sum_{n=1}^{\infty}E(B% _{n})=0,$

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

Here are some facts about projection-valued measures. $E(\mathbb{C}\setminus\sigma(E))=0$.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 $\sigma$-algebra of $\mathbb{C}$ is regular, as he shows on the page after that.

Let $\mathfrak{B}_{E}(\mathbb{C})$ be the set of bounded measurable functions $\sigma(E)\to\mathbb{C}$, and let $\left\|f\right\|_{E}=\sup_{z\in\sigma(E)}|f(z)|$. If $E$ is a projection-valued measure with compact spectrum and $f:\mathbb{C}\to\mathbb{C}$ is continuous, then1818 18 Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 62, §38, Theorem 2.

 $\left\|E(\chi_{\sigma(E)}f)\right\|=\left\|f\right\|_{E}.$

(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 $A=E(\chi_{\sigma(E)}\lambda)$, 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 $T\in B(H)$ is invertible if and only if there is some $\alpha>0$ such that $\left\|Tv\right\|\geq\alpha\left\|v\right\|$ for all $v\in H$.

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

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 $E:\mathscr{B}(\mathbb{C})\to\mathscr{P}(H)$, with $\sigma(E)$ compact and $\sigma(E)\subset\mathbb{R}$, such that

 $A=E(\chi_{\sigma(E)}\lambda)=\int_{\sigma(E)}\lambda dE(\lambda).$

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