> [!theoremb] Spectral Theorem (I) > > Let $\alg$ be a [[Commutative Banach Algebra|commutative]] [[C-Star Algebra|C*-subalgebra]] of $L(\ch, \ch)$ containing $I$, $\Sigma = \sigma(\alg)$ be its [[Multiplicative Functional|spectrum]], $\Gamma: \alg \to C(\Sigma)$ be the [[Gelfand Transform|Gelfand transform]], and $\Gamma^{-1}: B(\Sigma) \to L(\ch, \ch)$ be its [[Extended Inverse Gelfand Transform|extended inverse]] on the [[Space of Bounded Borel Functions|bounded Borel functions]]. Then there exists a unique regular [[Projection-Valued Measure|projection-valued measure]] $P$ on $\Sigma$ such that > 1. $T = \angles{\Gamma T, P}$ for all $T \in \alg$. > 2. $\Gamma^{-1}T = \angles{f, P}$ for all $f \in B(\Sigma)$. > > In other words, the inverse Gelfand transform has representation as a projection-valued measure. > > *Proof*. Let $E \in \cb(\alg)$ and define $P(E) = \Gamma^{-1}\one_E = T_{\one_E}$. Since $\one_E^2 = \one_E = \ol{\one}_E$, $P(E)^2 = P(E) = P(E)^*$. As $E$ is a self-adjoint projection, it is an [[Orthogonal Projection|orthogonal projection]]. Since $\Gamma^{-1}$ is a $*$-homomorphism, $P(\Sigma) = 1$ and > $ > P(E \cap F) = \Gamma^{-1}\one_{E \cap F} = \Gamma^{-1}\one_E \cdot \Gamma^{-1}\one_F = P(E)P(F) > $ > Let $\seq{E_j} \subset \cm$ be a disjoint family with $\bigcup_{j \in \nat}E_j = E$, and let $F_n = \bigcup_{j \le n}E_j$, then since $\one_{F_n} \upto \one_E$ under the conditions of the [[Dominated Convergence Theorem]], $\one_{F_n} \to \one_E$ in $B(\Sigma)$, and > $ > \sum_{j \le n}P(E_j) = P(F_n) \to P(E) > $ > in the [[Operator Topologies|weak operator topology]]. On the other hand, since $\one_{E \setminus F_n}$ is also an orthogonal projection, > $ > \begin{align*} > \norm{(P(E) - P(F_n))x}^2 &= \norm{P(E \setminus F_n)} = \angles{P(E \setminus F_n)x, P(E \setminus F_n)x} \\ > &= \angles{P(E \setminus F_n)x, x} = \angles{(P(E) - P(F_n))x, x} > \end{align*} > $ > so the series converges in the strong operator topology. > > $P$ is uniquely determined by $P_{x, y}$, and by assumption $(2)$, > $ > P_{x, y}(E) = \angles{P(E)x, y} = \angles{T_{\one_E}x, y} = \angles{\one_E, \mu_{x, y}} > $ > we have $P_{x, y}$ being uniquely determined by the [[Spectral Measure|spectral measure]] via the [[Riesz Representation Theorem (Radon Measure)|Riesz representation theorem]], which makes $P$ regular. > > Now let $f \in B(\Sigma)$, and $x, y \in \ch$. Since $P_{x, y} = \mu_{x, y}$, $f$ defines a unique linear map $T$ such that > $ > \angles{f, P_{x, y}} = \angles{Tx, y} = \angles{f, \mu_{x, y}} > $ > By properties of the spectral measure, $T = \Gamma^{-1}f$, and $(2)$ is proven, which implies $(1)$. > [!theorem] > > Let $S \in L(\ch, \ch)$, then the following are equivalent: > 1. $S$ commutes with elements of $\alg$. > 2. $S$ commutes with elements of $\bracs{P(E): E \in \cb(\Sigma)}$. > 3. $S$ commutes with $\Gamma^{-1}f = \angles{f, P}$ for all $f \in B(\Sigma)$. > [!theoremb] Spectral Theorem (II) > > $ > \begin{CD} > \ch @>U>> L^2(\mu) \\ > @A{T}AA @AA\phi_T A \\ > \ch @<<U^{-1}< L^2(\mu) > \end{CD} > $ > > Let $\alg$ be a commutative $C^*$-subalgebra of $L(\ch, \ch)$ containing $I$. Then there exists > 1. A [[Semifinite Measure|semifinite]] measure space $(\Omega, \cm, \mu)$. > 2. A unitary map $U: \ch \to L^2(\mu)$. > 3. A $*$-homomorphism $\alg \to L^\infty(\mu)$ with $T \mapsto \phi_T$, where $\phi_T$ acts by pointwise multiplication. > > such that > > 4. $UTU^{-1} = \phi_T$ for all $T \in \alg$. > 5. The homomorphism $T \mapsto \phi_T$ is isometric. > 6. $\Omega$ is a disjoint union of copies of $\sigma(\alg)$, where $\mu$ is finite on each copy. > > *Proof*. Let $v \in \ch$ and $\alg v = \bracs{Tv: T \in \alg}$. Let $\mu_{v, v}$ be the corresponding spectral measure, which is finite and regular $(1)$. For any $T \in \alg$, > $ > \norm{Tv}^2 = \angles{T^*Tv, v} = \angles{\abs{\Gamma T}^2, \mu_{u, v}} > $ > In particular, if $Tv = Sv$, then $\wh T - \wh S = 0$ $\mu$-a.e. Thus $U_v: \alg v \to L^2$ with $Tv \mapsto \Gamma T$ is a well-defined linear isometry, and extends uniquely to a linear isometry on $\ol{\alg v}$. By the [[Method of Successive Approximations]] and the [[Gelfand-Naimark Theorem]], $U(\ol{\alg v})$ is closed and includes $C(\sigma(\alg))$. Since $\mu_{v, v}$ is regular, $C(\sigma(\alg))$ is dense in $L^2$. Thus $U$ is a [[Unitary Map|unitary map]] $(2)$. > > Let $f \in C(\sigma(\alg))$ and $T \in \alg$, define $\phi_T = \Gamma T$, then > $ > UTU^{-1}f = UTT_fv = \Gamma(TT_f) = \Gamma T f > $ > and $(4)$ holds. Since $\Gamma$ is a $*$-homomorphism, for any $R, S, T \in \alg$, > $ > \Gamma(RS)U(Tv) = \Gamma(RS)\Gamma(T) = \Gamma(R)\Gamma(S)\Gamma(T) = \Gamma(R)\Gamma(S)U(Tv) > $ > and $\Gamma(R^*)U(Tv) = \ol{\Gamma(R)}U(Tv)$, so $(3)$ holds. As the Gelfand transform is an isometry, $\norm{\phi_T}_\infty = \norm{UTU^{-1}} = ||{T|_\ol{\alg v}}||$ and $(5)$ holds. > > Let $\seqi{v}$ be a maximal collection of non-zero vectors in $\ch$ such that the subspaces $\ch_i = \ol{\alg v_i}$ are mutually orthogonal. Let $\ch' = \bigoplus_{i \in I}\ch_i$. If $\ch' \subsetneq \ch$, then for any $u \in \ch'$, $v \in \ch'^\perp$, and $T \in \alg$, $\angles{Tv, u} = \angles{v, T^*u} = 0$ because $T^*u \in \ch'$. So $\ol{\alg v}$ can be added to the sum, contradicting the maximality of $\seqi{v}$, and $\ch' = \ch$. > > For each $i \in I$, let $\Sigma_i$ be a copy of $\sigma(\alg)$, inheriting the same topology and Borel $\sigma$-algebra, $\mu_i = \mu_{v, i}$, and $\Omega = \bigsqcup_{i \in I}\Sigma_i$. Define > $ > \cm = \bracs{E \subset \Omega: E \cap \Sigma_i \in \cb(\Sigma_i)} > $ > and > $ > \mu(E) = \sum_{i \in I}\mu_i(E \cap \Sigma_i) > $ > then $\cm$ is a $\sigma$-algebra with $\mu$ being a semi-finite measure on it (because each $\mu$ is regular and each $\Sigma_i$ is compact). This allows identifying $L^2(\mu) = \bigoplus_{i \in I}L^2(\mu_i)$. > > Let $U_i: \ch_i \to L^2(\mu_i)$ as before, and $U = \bigoplus U_i$, then $U$ preserves all existing properties and is defined on $\ch$. As $\norm{T} = \norm{UTU^{-1}} = \norm{\phi_T}_\infty$, the correspondence of operators is an isometry as well.