37.4 The Spectral Theorem

Proof. (1): Let $x, y \in H$, $\seqf{B_j}\subset \cb_{X}$ be disjoint Borel sets, and $B = \bigsqcup_{j = 1}^{n} B_{j}$, then for each $1 \le i < j \le n$, $E(B_{i})(H) \perp E(B_{j})(H)$, so by the Cauchy-Schwarz inequality and the Pythagorean Theorem,

As the above holds for all finite sequences of disjoint Borel sets, $\norm{E_{x, y}}_{\text{var}}\le \norm{x}_{H} \norm{y}_{H}$.

(2): For each $B \in \cb_{X}$, $E(B)$ is a projection, so $E_{x, x}(B) = \dpn{E(B)x, x}{H}\ge 0$.

(3): For each $x, y \in H$ and $B, C \in \cb_{X}$,

By linearity, $fdE_{x, y}\in \mathscr{E}$ for all $f \in \Sigma(X; \complex)$. For each $f \in \Sigma(X; \complex)$, the mapping $\mu \mapsto f d\mu$ is continuous in the total variation norm, so $fd\mu \in \mathscr{E}$ for all $\mu \in \mathscr{E}$ and $f \in \Sigma(X; \complex)$. By Proposition 16.1.10, $\Sigma(X; \complex)$ is dense in $L^{1}(\mu; \complex)$ for all $\mu \in \mathscr{E}$. Therefore $fd\mu \in \mathscr{E}$ for all $f \in L^{1}(\mu; \complex)$ and $\mu \in \mathscr{E}$.

Finally, let $\mu \in \mathscr{E}$ and $\nu \in M_{R}(X; \complex)$ with $\nu \ll \mu$, then by the Radon-Nikodym Theorem, there exists $f \in L^{1}(\mu; \complex)$ such that $d\nu = f d\mu \in \mathscr{E}$.

(4): By (3), for any $\mu \in \mathscr{E}$ and $f \in L^{1}(\mu; \complex)$, $fd\mu \in \mathscr{E}$ as well. By Proposition 38.8.1, there exists a unique weak*-continuous involution and separately weak*-continuous product on $\mathscr{E}^{*}$ making $\mathscr{E}^{*}$ a commutative unital $C^{*}$-algebra, and $J|_{C(X; \complex)}$ a unital *-homomorphism. Since

1. (i)

   $J$ is $\sigma(B^{\infty}(X; \complex), M_{R}(X; \complex))$-$\sigma(\mathscr{E}^{*}, \mathscr{E})$ continuous.

2. (ii)

   Conjugation on $B^{\infty}(X; \complex)$ is $\sigma(B^{\infty}(X; \complex), M_{R}(X; \complex))$-continuous.

3. (iii)

   Multiplication on $B^{\infty}(X; \complex)$ is separately $\sigma(B^{\infty}(X; \complex), M_{R}(X; \complex))$-continuous.

the mapping $J$ is a unital *-homomorphism.$\square$

Proof. (1): Let $\phi \in \mathscr{E}^{*}$ and $x, y \in H$, then by Lemma 37.4.2,

Since the above holds for all $x, y \in H$, $I_{E}(\phi) \in B(H)$ with $\norm{I_E(\phi)}_{B(H)}\le \norm{\phi}_{\mathscr{E}^{*}}$.

(2): For each $x, y \in H$, $E_{x, y}\in \mathscr{E}$. Since $\angles{\int \phi dE \cdot x, y}_{H}= \dpn{E_{x, y}, \phi}{\mathscr{E}}$ for every $\phi \in \mathscr{E}^{*}$, $I_{E}$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $B(H)$.

(3): By Lemma 26.6.4, the simple functions $\Sigma(X; \complex)$ are uniformly dense in the bounded Borel functions $B^{\infty}(X; \complex)$. Since

1. (i)

   $I_{E}$ restricted to $J(\Sigma(X; \complex))$ is a *-homomorphism.

2. (ii)

   Multiplication and conjugation are continuous in the uniform norm on $B^{\infty}(X; \complex)$

3. (iii)

   Composition and adjunction are continuous in the operator norm on $B(H)$

the map $I_{E}$ restricted to $J(B^{\infty}(X; \complex))$ is a *-homomorphism by continuity.

By Goldstine’s Theorem, $C(X; \complex) \subset B^{\infty}(X; \complex)$ is weak*-dense in $C(X; \complex)^{**}$, so $J(C(X; \complex))$ is weak*-dense in $\mathscr{E}^{*}$. As

1. (i)

   $I_{E}$ restricted to $J(B^{\infty}(X; \complex))$ is a *-homomorphism.

2. (ii)

   The involution $\phi \mapsto \ol \phi$ is weak*-continuous on $\mathscr{E}^{*}$.

3. (iii)

   The adjunction $T \mapsto T^{*}$ is weak-operator continuous on $B(H)$.

4. (iv)

   The product $(\phi, \psi) \mapsto \phi \psi$ is separately weak*-continuous on $\mathscr{E}^{*}$.

5. (v)

   The composition $(S, T) \mapsto ST$ is separately weak-operator continuous on $B(H)$.

the map $I_{E}$ is a *-homomorphism by the weak* to weak-operator continuity established in (2). Since $E(X) = I_{B(H)}$, $I_{E}$ is a unital *-homomorphism.

Finally, let $\phi \in \mathscr{E}^{*}$ with $I_{E}(\phi) = 0$, then $\dpn{I_E(\phi)x, y}{H}= \dpn{E_{x, y}, \phi}{\mathscr{E}}= 0$ for all $x, y \in H$. As $\mathscr{E}$ is the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, $\phi = 0$. Therefore $I_{E}$ is an injective unital *-homomorphism.$\square$

Proof, [Theorem 20.2, Zhu93]. (1): By the Gelfand-Naimark Theorem, $\Gamma_{A}: A \to C(\Omega(A); \complex)$ is a *-isomorphism. For each $x, y \in H$, $\Gamma_{A}^{-1}$ induces a mapping

which, by the Riesz Representation Theorem, takes the form of a complex Radon measure on $\Omega(A)$. Thus by the uniqueness part of the Riesz Representation Theorem, such a spectral measure must be unique if it exists.

Since $\norm{E_{x, y}}_{C(\Omega(A); \complex)^*}\le \norm{x}_{H}\norm{y}_{H}$ for all $x, y \in H$, $\bracsn{E_{x, y}|x, y \in H}$ induces a bounded linear map

with $J_{E}(f) = \Gamma_{A}^{-1}(f)$ for all $f \in C(\Omega(A); \complex)$.

For any $C \in \cb_{\Omega(A)}$, $\one_{C}$ is a projection in $B^{\infty}(\Omega(A); \complex)$. So to see that

defines a spectral measure, it is sufficient to show that $J_{E}$ is a *-homomorphism.

Let $x, y \in H$, then as $\Gamma_{A}$ is a *-isomorphism, for any $f, g \in C(\Omega(A); \complex)$,

As the above holds for all $g \in C(\Omega(A); \complex)$, $fE_{x, y}= E_{x, J_E(f)^*y}$. Now, fix $\phi \in B^{\infty}(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$,

so $\phi E_{x, y}= E_{J_E(\phi)x, y}$ for all $\phi \in B^{\infty}(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^{\infty}(\Omega(A); \complex)$,

and $J_{E}$ is a homomorphism.

Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_{A}$ is a *-isomorphism, $J_{E}(f) = \Gamma_{A}^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)}= \dpn{J_E(f)x, x}{H}\in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^{\infty}(\Omega(A); \real)$ and $x \in H$, $\dpn{J_E(\phi)x, x}{H}= \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*}\in \real$ as well. Therefore $J_{E}(\phi)$ is self-adjoint, and $J_{E}$ is a *-homomorphism.

(2): By Definition 37.4.3, $I_{E}$ is an injective unital *-homomorphism, so it is sufficient to show that $I_{E}(\mathscr{E}^{*}) = B$.

Let $J: C(\Omega(A); \complex) \to \mathscr{E}^{*}$ be defined by $\dpn{\mu, J(f)}{\mathscr{E}}= \int_{\Omega(A)}f d\mu$ for each $\mu \in \mathscr{E}$ and $f \in C(\Omega(A); \complex)$. By Goldstine’s Theorem, $C(\Omega(A); \complex)$ is weak*-dense in $C(\Omega(A); \complex)^{**}$, so $J(C(\Omega(A); \complex))$ is weak*-dense in $\mathscr{E}^{*}$. Since $I_{E}$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $B(H)$, $I_{E}(\mathscr{E}^{*}) \subset B$ by Proposition 5.5.3.

On the other hand, by the Banach-Alaoglu Theorem, $\ol{B_{\mathscr{E}^*}(0, 1)}$ is weak*-compact, so $I_{E}(\ol{B_{\mathscr{E}^*}(0, 1)})$ is weak-operator compact. As $\Gamma_{A}: A \to C(\Omega(A); \complex)$ is an isometric isomorphism, $I_{E}(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_A(0, 1)}$. By the Kaplansky Density Theorem, $I_{E}(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_B(0, 1)}$, and $I_{E}(\mathscr{E}^{*}) = B$.$\square$

Proof, [Theorem 1.47, Fol16]. (1): Fix $i \in I$ and let $\mu_{i} = E_{\xi_i, \xi_i}$, then for any $C \in \cb_{\Omega(A)}$ with $E_{x, y}(C) = 0$ for all $x, y \in \ol{A\xi_i}$, $\mu_{i}(C) = 0$. By (1) and (2) of Lemma 37.4.2, $\mu_{i}$ is a finite positive Radon measure.

By Spectral Theorem I, for each $S, T \in A$ and $C \in \cb_{\Omega(A)}$,

so $\Gamma_{A}S \cdot \ol{\Gamma_AT}dE_{\xi_i, \xi_i}= dE_{S\xi_i, T\xi_i}\ll \mu_{i}$. By (1) of Lemma 37.4.2 and completeness of $L^{1}(\mu_{i}; \complex)$, $\bracsn{E_{x, y}|x, y \in \ol{A\xi_i}}$ is absolutely continuous with respect to $\mu_{i}$. Therefore for any $C \in \cb_{\Omega(A)}$, $\mu_{i}(C) = 0$ if and only if $E_{x, y}(C) = 0$ for all $x, y \in \ol{A \xi_i}$.

(2): Let $i, j \in I$ with $i \ne j$, $x \in \ol{A\xi_i}$, $y \in \ol{A\xi_j}$, and $f \in C(\Omega(A); \complex)$, then since $\ol{A\xi_i}\perp \ol{A\xi_j}$,

As the above holds for all $f \in C(\Omega(A); \complex)$, $E_{x, y}= 0$.

Given that $\seqi{\xi}$ is maximal, $x = \sum_{i \in I}P_{i}x$ for all $x \in H$. Thus for any $x, y \in H$,

(3): Let $T \in A$ with $\Gamma_{A} T = 0$ $\mu_{i}$-almost everywhere for all $i \in I$. By (1), $E_{P_ix, P_iy}\ll \mu_{i}$ for all $i \in I$. Thus for any $x, y \in H$,

Therefore $C(\Omega(A); \complex)$ may be identified as a subspace of $[l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)]$.

(4): By (2), for each $x, y \in H$, $E_{x, y}= \sum_{i \in I}E_{P_ix, P_iy}$. By (1), $E_{P_ix, P_iy}\ll \mu_{i}$ for all $i \in I$, so $\mathscr{E}\subset \mathscr{M}([l^{1}(I); L^{1}(\mu_{i}; \complex)])$.

On the other hand, for each $i \in I$, since $\mu_{i}$ is a Radon measure, $C(\Omega(A); \complex)$ is dense in $L^{1}(\mu_{i}; \complex)$ by Proposition 25.1.7. As

and $\mathscr{E}\subset M_{R}(\Omega(A); \complex)$ is closed, $\mathscr{E}\supset \bracsn{f d\mu_i|f \in L^1(\mu_i; \complex)}$.

Finally, given that the above holds for all $i \in I$, $\mathscr{E}= \mathscr{M}([l^{1}(I); L^{1}(\mu_{i}; \complex)])$. By Theorem 16.5.6 and Theorem 16.3.4, $[l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)] = [l^{1}(I); L^{1}(\mu_{i}; \complex)]^{*}$, so $\mathscr{E}^{*}$ may be identified with its image under $\mathscr{M}^{*}$.

(5): Fix $i \in I$, then for any $S, T \in A$ with $S\xi_{i} = T\xi_{i}$,

so $\Gamma_{A} S = \Gamma_{A} T$ $\mu_{i}$-almost everywhere. Thus the mapping

is well-defined. Moreover, for any $S, T \in A$,

so $U_{i}$ extends into an isometry between $\ol{A\xi_i}$ and $L^{2}(\mu_{i}; \complex)$. Thus the mapping

is an isometry between $H$ and $[l^{2}(I); L^{2}(\mu_{i}; \complex)]$ such that $U(Tx) = \Gamma_{A}T \cdot Ux$ for all $x \in H$ and $T \in A$.

Finally, given that

1. (i)

   By Goldstine’s Theorem, $C(\Omega(A); \complex)$ is weak*-dense in $\mathscr{E}^{*}$.

2. (ii)

   The weak* topology on $[l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)]$ is equal to the weak operator topology of $[l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)]$ acting on $[l^{2}(I); L^{2}(\mu_{i}; \complex)]$.

3. (iii)

   $A$ is weak-operator dense in $B$.

4. (iv)

   By Spectral Theorem I, the isomorphism $\phi \mapsto \int \phi dE$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $B$.

the mapping $U$ is a unitary equivalence between $\mathscr{E}^{*}$ acting on $[l^{2}(I); L^{2}(\mu_{i}; \complex)]$ and $B$ acting on $H$.$\square$