Let $X$ be a compact Hausdorff space and $H$ a Hilbert Space. A spectral measure $E$ relative to $(X,H)$ is a map from the $\sigma$-algebra of all Borel sets of $X$ to the set of projections in $B(H)$ such that

1. $E(\emptyset)=0$$E(X)=1$;
2. $E(S_1\cap S_2)=E(S_1)E(S_2)$ for all Borel sets $S_1,\,S_2$ of $X$;
3. for all $x,y\in H$, the function $E_{x,y}:S\mapsto \langle E(S)x,y\rangle$, is a regular Borel complex measure on $X$.

A Borel measure $\mu$ is a measure defined on Borel sets. If every Borel set in $X$ is both outer and inner regular, then $\mu$ is called regular. A measurable $A\subset X$ is inner and outer regular if

$\mu(A)=\sup\left\{\mu(F):\text{ closed }F\subset A\right\}$, and

$\mu(A)=\inf\left\{\mu(G):A\subset G\text{ open }\right\}$

Denote by $M(X)$ the Banach space of all regular Borel complex measures on $X$, and by $B_\infty(X)$ the C*-algebra of all bounded Borel-measurable complex-valued functions on $X$ (I assume with respect to the Borel $\sigma$-algebra on $\mathbb{C}$).

# Example 2.5.1

Let $X$ be a compact Hausdorff space and let $\mu$ be a positive regular Borel measure on $X$. Define $M_\varphi\in B(L^2(X,\mu))$ by (I’m pretty sure we need to say that $\varphi\in L^\infty(X,\mu)$ here. Also $M_\varphi$ is a product rather than a composition)

$M_\varphi(f)=\varphi f$

That $M_\varphi$ is bounded is given by

$\|M_\varphi\|_2^2=\int |\varphi(x)f(x)|^2\,d\mu(x)\leq \|\varphi\|_\infty \int |f(x)|^2\,d\mu(x)$

which implies that $\|M_\varphi\|\leq \|\varphi\|_\infty$. The operator $M_\varphi$ is called a multiplicative operator. The map

$L^\infty(X,\mu)\rightarrow B(L^2(X,\mu))$$\varphi\mapsto M_\varphi$,

is a *-homomorphism of C*-algebras. In particular, the adjoint of $M_\varphi$ is $M_{\bar{\varphi}}$, and $M_\varphi$ is normal (routine).

If $S$ is a Borel set of $X$, then $\mathbf{1}_{S}$ is a projection in $L^\infty(X,\mu)$, so $E(S)=M_{\mathbf{1}_S}$ is a projection in $B(L^2(X,\mu))$. The map $E:S\mapsto E(S)$ is a spectral measure relative to the pair $(X,L^2(X,\mu))$.

Since the multiplication operators are a very important class we linger with this example a little longer to show that if $\varphi\in L^\infty(X,\mu)$, then $\|M_\varphi\|=\|\varphi\|_\infty$. Suppose this is false. Then $\exists\,\varepsilon>0$ such that $\|\varphi\|_\infty-\varepsilon>\|M_\varphi\|$ and therefore, there is a Borel set $S\subset X$ such that $\mu(S)>0$ and $|\varphi(x)|>\|M_\varphi\|+\varepsilon$ for all $x\in X$. Since $\mu$ is regular,

$\mu(S)=\sup\{\mu(K):K\text{ is compact and }K\subset S\}$,

so we may suppose that $S$ is compact. Then $\mu(S)<\infty$, again by regularity of $\mu$ (?). However

$\|M_\varphi\|^2\mu(S)\geq \|M_{\varphi}(\mathbf{1}_S)\|_2^2$

$=\int |\varphi(x)\mathbf{1}_S(x)|^2\,d\mu(x)$

$\geq \int (\|M_\varphi\|+\varepsilon)^2\mathbf{1}_S(x)\,d\mu(x)$

$=(\|M_\varphi\|+\varepsilon)^2\mu(S)$

and therefore we get $\|M_\varphi\|\geq \|M_\varphi\|+\varepsilon$, a contradiction. Hence $\|M_\varphi\|=\|\varphi\|_\infty$ as claimed.

This result means that the map $\varphi\mapsto M_\varphi$ is in fact an isometric *-isomorphism of $L^\infty(X,\mu)$ onto a C*-subalgebra of $B(L^2(X,\mu))$. We therefore have $\sigma(M_\varphi)=\sigma(\varphi)$.

# Lemma 2.5.1

Let $X$ be a compact Hausdorff space, let $H$ be a Hilbert space, and suppose that $\mu_{x,y}\in M(X)$ for all $x,y\in H$. Suppose also that for each Borel set $S\subset X$ the function

$\sigma_S:H^2\rightarrow \mathbb{C}$$(x,y)\mapsto \mu_{x,y}(S)$,

is a sesquilinear form. Then for each $f\in B_\infty(X)$ the function

$\sigma_f:H^2\rightarrow \mathbb{C}$$(x,y)\mapsto\int f\,d\mu_{x,y}$

is a sesquilinear form.

## Proof

Suppose $f$ is simple, so we can write (finite sum) $f=\sum_j \lambda_j\mathbf{1}_{S_j}$, where $\{S_j\}\subset\mathcal{B}(X)$ are pairwise disjoint Borel Sets and $\{\lambda_j\}\subset \mathbb{C}$. Then

$\int f\,d\mu_{x,y}=\sum_{j=1}^n\lambda_j\int \mathbf{1}_{S_j}\,d\mu_{x,y}=\sum_{j=1}^n\lambda_j\mu_{x,y}(S_j)$.

The set of sesquilinear forms on $H$ is a vector space with the point-wise defined operations, and we have just shown that $\sigma_f$ is a linear combination of the sesquilinear forms $\sigma_{S_j}$.

Now let $f\in B_\infty(X)$. Then $f$ is the uniform limit of a sequence $\{f_n\}\subset \mathcal{E}(X)$. Hence

$\int |f_n-f|\,d|\mu_{x,y}|\leq \|f_n-f\|_\infty |\mu_{x,y}|(X)$, so

$\int f\,d\mu_{x,y}=\lim_{n\rightarrow \infty}\int f_n\,d\mu_{x,y}$ for each $x,y\in H$.

It immediately follows that $\sigma_f$ is a sesquilinear form on $H$ (if we prove that the vector space of bounded sesquilinear forms is complete – this isn’t too difficult using Th. 2.3.6 – to all bounded sesquilinear forms there exists a bounded operator $T$ such that $\sigma(x,y)=\langle Tx,y\rangle$. Using the completeness of $B(H)$ finishes the proof$\bullet$

# Theorem 2.5.2

Let $X$ be a compact Hausdorff space, $H$ a Hilbert space, and $E$ a spectral measure relative to $(X,H)$. Then for each $f\in B_\infty(X)$ the function

$\sigma_f:H^2\rightarrow \mathbb{C}$$(x,y)\mapsto \int f\,dE_{x,y}$

is a bounded sesquilinear form on $H$, and $\|\sigma_f\|\leq \|f\|_\infty$.

## Proof

That $\sigma_f$ is a sesquilinear form follows from the preceding lemma, we need only show $\|\sigma_f\|\leq \|f\|_\infty$. Suppose that $X=S_1\cup\cdots\cup S_n$ disjointly, where $\{S_i\}\subset \mathcal{B}(X)$. Then (here we use the fact that the $E(S_j)$ are projections)

$\sum_{j=1}^n\left|\langle E(S_j)x,y\rangle\right|=\sum_{j=1}^n\left|\langle E(S_j)x,E(S_j)x\rangle\right|$

$\leq \left(\sum_{j=1}^n\|E(S_j)x\|^2\right)^{1/2}\left(\sum_{j=1}^n\|E(S_j)y\|^2\right)^{1/2}$

$=\|E(X)x\|\|E(X)y\|=\|x\|\|y\|$

Hence, $\|E_{x,y}\|\leq \|x\|\|y\|$. Therefore,

$\left|\int d\,dE_{x,y}\right|\leq \|f\|_\infty\|E_{x,y}\|\leq \|f\|_\infty\|x\|\|y\|$,

so $\|\sigma_f\|\leq \|f\|_\infty$ $\bullet$

# Theorem 2.5.3

Let $X$ be a compact Hausdorff space, $H$ a Hilbert space, and $E$ a spectral measure relative to $(X,H)$. Then for each $f\in B_\infty(X)$ there is a unique bounded operator $T\in B(H)$ such that

$\int f\,dE_{x,y}= \langle Tx,y\rangle$$\forall\,x\,y\in H$ $\bullet$

We write $\int f\,dE$ for $T$ and call it the integral of $f$ with respect to $E$. Note that $\int \mathbf{1}_S\,dE=E(S)$ for each Borel set $S$.

# Theorem 2.5.4

With the same assumptions on $X,H$ and $E$ as in the last theorem, the map

$\varphi:B_\infty(X)\rightarrow B(H)$$f\mapsto \int f\,dE$

is a unital *-homomorphism.

## Proof

Linearity is routine and boundedness follows from Theorems 2.5.2 and 2.3.6. To show that $\varphi(fg)=\varphi(f)\varphi(g)$ and $\varphi(\bar{f})=\overline{\varphi(f)}$, we need only show these results when $f,g\in\mathcal{E}(X)$, because $\overline{\mathcal{E}(X)}\subset B_\infty (X)$. Hence, by way of linearity, we may suppose that $f=\mathbf{1}_{S_1}$ and $g=\mathbf{1}_{S_2}$. Then

$\varphi(fg)=\int \mathbf{1}_{S_1}\mathbf{1}_{S_2}\,dE=E(S_1\cap S_2)=E(S_1)E(S_2)$

$=\int \mathbf{1}_{S_1}\,dE\times\int \mathbf{1}_{S_2}\,dE=\varphi(f)\varphi(g)$

Also

$\varphi(\bar{f})=\varphi(f)=E(S)=\varphi(f)^*$ $\bullet$

# Theorem 2.5.5

Let $X$ be a compact Hausdorff space and $H$ a Hilbert space, and suppose that $\varphi :C(X)\rightarrow B(H)$ is a unital *-homomorphism. Then there is a unique spectral measure $E$ relative to $(X,H)$ such that

$\varphi(f)=\int f\,dE$, $\forall\,f\in C(X)$

Moreover, if $T\in B(H)$, then $T$ commutes with $\varphi(f)$ for all $f\in C(X)$ iff $T$ commutes with $E(S)$ for all Borel sets $S\subset X$.

## Proof

If $x,y\in H$, then the function

$\tau_{x,y}:C(X)\rightarrow \mathbb{C}$$f\mapsto \langle \varphi(f)x,y\rangle$,

is linear and $\|\tau_{x,y}\|\leq \|x\|\|y\|$. By the Riesz-Kakutani Theorem (to each continuous linear functional on $C(X)$ we may isometrically associate a measure on $X$), there is a unique measure $\mu_{x,y}\in M(X)$ such that $\tau_{x,y}(f)=\int f\,d\mu_{x,y}$ for all $f\in C(X)$. Also $\|\mu_{x,y}\|=\|\tau_{x,y}\|$ (here the “norm” on $M(X)$ is total variation). Since the function

$H^2\rightarrow\mathbb{C}$$(x,y)\mapsto \langle \varphi(f)x,y\rangle$

is sesquilinear, the maps from $H\rightarrow M(X)$ given by

$x\mapsto \mu_{x,y}$ and $y\mapsto \mu_{x,y}$

are, respectively, linear and conjugate-linear. Hence for each $f\in B_\infty(X)$ the function

$H^2\rightarrow \mathbb{C}$$(x,y)\mapsto \int f\,d\mu_{x,y}$

is a sesquilinear form (Th.2.5.1: if $(x,y)\mapsto \mu_{x,y}$ is sesquilinear so is $(x,y)\mapsto \int f\,d\mu_{x,y}$ where $f\in B_\infty(X)$). Also

$\left|\int f\,d\mu_{x,y}\right|\leq \|f\|_\infty \|\mu_{x,y}\|\leq \|f\|_\infty \|x\|\|y\|$

so this sesquilinear form is bounded with norm bounded by $\|f\|_\infty$. By Theorem 2.3.6 there is a unique operator, $\psi(f)\in B(H)$ say, such that

$\langle\psi(f)x,y\rangle=\int f\,d\mu_{x,y}$$\forall\,x,y\in H$.

Moreover, $\|\psi(f)\|\leq \|f\|_\infty$.

Now suppose that $f\in C(X)$. Then

$\langle\psi(f)x,y\rangle=\int f\,d\mu_{x,y}=\tau_{x,y}(f)=\langle \varphi(f)x,y\rangle$, $\forall\,x,y\in H$,

so $\psi(f)=\varphi(f)$.

It is straightforward to check that the map

$\psi:B_\infty(X)\rightarrow B(H)$, $f\mapsto \psi(f)$

is linear and we already know that it is norm-decreasing. We now show that $\psi$ is a *-homomorphism.

If $f\in C(X)$ and $\bar{f}=f$, then $\varphi(f)$ is hermitian (it is a *-homomorphism), so $\int f\,d\mu_{x,x}=\langle \varphi(f)x,x\rangle$ is a real number. Thus the measure $\mu_{x,x}$ is real (if we can use continuous functions to approximate to $\mathbf{1}_S$ for any Borel set $S\subset X$ – perhaps via Urysohn’s Lemma – then for all Borel sets $\int \mathbf{1}_S\,\mu_{x,x}=\mu_{x,x}=\mu_{x,x}(S)$ which is real and we are done), that is $\overline{\mu_{x,x}}=\mu_{x,x}$, and therefore if $f\in B_\infty(X)_{\text{SA}}$, then $\langle \psi(f)(x),x\rangle=\int f\,d\mu_{x,x}$ is real. Since $x$ is arbitrary this shows that $\psi(f)$ is hermitian. Therefore, $\psi$ preserves involutions.

Let $f\in B_\infty(X)$ and $x\in H$.

### Claim:

If the equation

$\langle \psi(fg)x,x\rangle=\langle \psi(f)\psi(g)x,x\rangle$  (*)

holds for all $g\in C(X)$, then it also holds for all $g\in B_\infty(X)$.

#### Proof

Observe that (*) is equivalent to

$\int gf\,d\mu_{x,x}=\int g\,d\mu_{x,\psi{\bar{f}}x}$  (**)

To prove the assertion, note that if (*) holds for all $g\in C(X)$, then the regular measures $f\,d\mu_{x,x}$ and $\mu_{x,\psi(\bar{f})x}$ are equal because (**) holds for all $g\in C(X)$. Hence (**) holds for all $g\in B_\infty(X)$; that is (*)  holds for all such $g$, as claimed $\bullet$

Since $\varphi$ is a *-homomorphism, (*) holds for all $f,\,g\in C(X)$. Hence, by the assertion, (1_ holds if $f\in C(X)$ and $g\in B_\infty(X)$. Replacing $f,g$ with their conjugates, we get

$\langle \psi(\bar{f}\bar{g})x,x\rangle=\langle \psi(\bar{f})\psi(\bar{g})x,x\rangle$.

Taking conjugates of both sides and using the fact that $\psi$ preserves involutions, we get

$\langle\psi(gf)x,x\rangle=\langle\psi(g)\psi(f)x,x\rangle$ (***)

for all $g\in B_\infty(X)$ and $f\in C(X)$. Using the claim again, with roles of $f$ and $g$ reversed, we get that (***) holds for all $f,g\in B_\infty(X)$. Hence $\psi(gf)=\psi(g)\psi(f)$, so $\psi$ is a homomorphism.

If $S\in\mathcal{B}(X)$, define $E(S)=\psi(\mathbf{1}_S)$$E(S)$ is a projection as required, and the map $E:S\mapsto E(S)$ from the $\sigma$-algebra $\mathcal{B}(X)$ to $B(H)$ is a spectral measure relative to $(X,H)$ (just a few calculations –$E(X)=\psi(\mathbf{1})$ Let $g\in B_\infty(X)$. $\psi(g)=\psi(\mathbf{1}_xg)=\psi(\mathbf{1}_X)\psi(g)=E(S)\psi(g)$. Hence, on $\psi(B_\infty(X))$ at least, $E(X)=1$) .We have that $E_{x,y}=\mu_{x,y}\in M(X)$, since $E_{x,y}(S)=\langle E(S)x,y\rangle=\langle \psi(\mathbf{1}_S)x,y\rangle=\int\mathbf{1}_S\,d\mu_{x,y}$.

If $f\in B_\infty(X)$, then

$\left\langle \left(\int f\,dE\right)x,y\right\rangle=\int f\,dE_{x,y}=\int f\,d\mu_{x,y}=\langle \psi(f)x,y\rangle$,

so $\psi(f)=\int f\,dE$. In particular, $\varphi(f)=\int f\,dE$ for all $f\in C(X)$.

To see uniqueness of $E$, suppose that $F$ is another spectral measure relative to $(X,H)$ such that $\varphi(f)=\int f\,dF$ for all $f\in C(X)$. Then

$\int f\,dF_{x,y}=\langle \varphi(f)x,y\rangle=\int f\,dE_{x,y}$.

Hence $F_{x,y}=E_{x,y}$ and therefore $\langle F(S)x,y\rangle=\langle E(S)x,y\rangle$ so $E=F$.

Now suppose $T$ is an operator on $H$ commuting with all of the elements of the range of $\varphi$. Then if $f\in C(X)$,

$\int f\,d\mu_{Tx,y}=\langle \psi(f)Tx,y\rangle=\langle T\psi(f)x,y\rangle=\langle \psi(f)x,T^*y\rangle=\int f\,d\mu_{x,T^*y}$.

Hence $E_{Tx,y}=E_{x,T^*y}$, so $E(S)T=TE(S)$ for all $S\in\mathcal{B}(X)$ ($\langle E(S)Tx,y\rangle=\langle E(S)x,T^*y\rangle$ and take the adjoints on the right).

Conversely now suppose that $T$ commutes with all of the projections $E(S)$. Then

$\langle E(S)Tx,y\rangle=\langle TE(S)x,y\rangle=\langle E(S)x,T^*y\rangle$

so $E_{Tx,y}=E_{x,T*y}$. Hence for every $f\in C(X)$,

$\int f\,dE_{Tx,y}=\int f\,dE_{x,T^*y}$;

that is $\langle \varphi(f)Tx,y\rangle=\langle\varphi(f)x, T^*y\rangle$ so $\varphi(f)T=T\varphi(f)$ $\bullet$

# Spectral Theorem

Let $T$ be a normal operator on a Hilbert space $H$. Then there is a unique spectral measure $E$ relative to $(\sigma(T),H)$ such that $T=\int T\,dE$, where $z$ is the inclusion map of $\sigma(T)$ in $\mathbb{C}$.

## Proof

Let $\varphi:C(\sigma(T))\rightarrow B(H)$ be the functional calculus at $T$. By the preceding theorem, there exists a unique spectral measure $E$ relative to $(\sigma(T),H)$ such that $\varphi(f)=\int f\,dE$ for all $f\in C(\sigma(T))$. In particular, $T=\varphi(z)=\int z\,dE$. If $F$ is another spectral measure such that $T=\int z\,dF$, then $\int f\,dF=\int f\,dE=\varphi(f)$ for all $f\in C(\sigma(T))$, since $1$ and $z$ generate $C(\sigma(T))$. Therefore $E=F$  $\bullet$

The spectral measure constructed here is called the resolution of the identity for $T$. Since $f(T)=\int f\,dE$ for all $f\in C(\sigma(T))$, we can define $f(T)=\int f\,dE$ for all $f\in B_\infty(\sigma(T))$. The unital *-homomorphism

$B_\infty(\sigma(T))\rightarrow B(H)$$f\mapsto f(T)$

is called the Borel functional calculus at $T$.

If $R\in B(H)$ commutes with both $T$ and $T^*$, then $R$ commutes with $f(T)$ for all $f\in B_\infty(\sigma(T))$. For in this case $R$ commutes with all polynomials in $T$ and $T^*$, and since $1$ and $z$ generate $C(\sigma(T))$ by the Stone-Weierstrass theorem, $R$ commutes with  $f(T)$ for all $f\in C(\sigma(T))$. By Th.2.5.5 ($T\in B(H)$ commutes with $\varphi(f)$ for all $f\in C(X)$ iff $T$ commutes with $E(S)$ for all $S\in\mathcal{B}(X)$), therefore, $R$ commutes with $E(S)$ for all $S\in\mathcal{B}(\sigma(T))$. It follows that $E_{x,R^*y}=E_{Rx,y}$ for all $x,y\in H$. Hence if $f\in B_\infty(\sigma(T))$,

$\langle Rf(T)x,y\rangle=\int f\,dE_{x,R^*y}$

$=\int f\,dE_{Rx,y}$

$=\langle f(T)Rx,y\rangle$.

Therefore $Rf(T)=f(T)=R$.

Incidentally if $S\in\mathcal{B}(\sigma(T))$, then $\mathbf{1}_S(T)=E(S)$.

# Theorem 2.5.7

Let $T$ be a normal operator on a Hilbert space $H$, and suppose that $g:\mathbb{C}\rightarrow \mathbb{C}$ is a continuous function. Then

$(g\circ f)(T)=g(f(T))$

for all $f\in B_\infty(\sigma(T))$.

## Proof

The result is easily seen (have to do this yet!) by first showing it for $g$ a polynomial in $z$ and $\bar{z}$, and then observing that an arbitrary continuous function $g:\mathbb{C}\rightarrow \mathbb{C}$ is a uniform limit of such polynomials on the compact disc $\Delta=\{\lambda\in\mathbb{C}:|\lambda|\leq \|f\|_\infty\}$, using the Stone-Weierstrass theorem applied to $C(\Delta)$ $\bullet$

# Theorem 2.5.8

Let $U$ be a unitary operator in $B(H)$, where $H$ is a Hilbert space. Then there exists a hermitian operator $S\in B(H)$ such that $T=e^{iS}$ and $\|S\|\leq 2\pi$.

## Proof

The function

$f:[0,2\pi)\rightarrow \mathbb{T}$$t\mapsto e^{it}$

is a continuous bijection with Borel measurable inverse $g$. Since $\sigma(U)\subset \mathbb{T}$, we can set $S=g(U)$. The operator $S$ is self-adjoint because $g$ is real-valued. Moreover, $\|S\|\leq \|g\|_\infty\leq 2\pi$. By Th 2.5.7, $(f\circ g)(U)=f(g(U))=f(S)=e^{iS}$. But $(f\circ g)(\lambda)=\lambda$ for all $\lambda\in\mathbb{T}$, so $(f\circ g)(U)=U$. Therefore, $U=e^{iS}$ $\bullet$