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)=0E(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

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