Spectral theory of normal C*-algebras

In functional analysis, every C*-algebra is isomorphic to a subalgebra of the C*-algebra $$\mathcal{B}(H)$$ of bounded linear operators on some Hilbert space $$H.$$ This article describes the spectral theory of closed normal subalgebras of $$\mathcal{B}(H)$$. A subalgebra $$A$$ of $$\mathcal{B}(H)$$ is called normal if it is commutative and closed under the $$\ast$$ operation: for all $$x,y\in A$$, we have $$x^\ast\in A$$ and that $$xy = yx$$.

Resolution of identity
Throughout, $$H$$ is a fixed Hilbert space.

A projection-valued measure on a measurable space $$(X, \Omega),$$ where $$\Omega$$ is a σ-algebra of subsets of $$X,$$ is a mapping $$\pi : \Omega \to \mathcal{B}(H)$$ such that for all $$\omega \in \Omega,$$ $$\pi(\omega)$$ is a self-adjoint projection on $$H$$ (that is, $$\pi(\omega)$$ is a bounded linear operator $$\pi(\omega) : H \to H$$ that satisfies $$\pi(\omega) = \pi(\omega)^*$$ and $$\pi(\omega) \circ \pi(\omega) = \pi(\omega)$$) such that $$\pi(X) = \operatorname{Id}_H \quad$$ (where $$\operatorname{Id}_H$$ is the identity operator of $$H$$) and for every $$x, y \in H,$$ the function $$\Omega \to \Complex$$ defined by $$\omega \mapsto \langle \pi(\omega)x, y \rangle$$ is a complex measure on $$M$$ (that is, a complex-valued countably additive function).

A resolution of identity on a measurable space $$(X, \Omega)$$ is a function $$\pi : \Omega \to \mathcal{B}(H)$$ such that for every $$\omega_1, \omega_2 \in \Omega$$:  $$\pi(\varnothing) = 0$$; $$\pi(X) = \operatorname{Id}_H$$; for every $$\omega \in \Omega,$$ $$\pi(\omega)$$ is a self-adjoint projection on $$H$$; for every $$x, y \in H,$$ the map $$\pi_{x, y} : \Omega \to \Complex$$ defined by $$\pi_{x,y}(\omega) = \langle \pi(\omega) x, y \rangle$$ is a complex measure on $$\Omega$$; $$\pi\left(\omega_1 \cap \omega_2\right) = \pi\left(\omega_1\right) \circ \pi\left(\omega_2\right)$$; if $$\omega_1 \cap \omega_2 = \varnothing$$ then $$\pi \left(\omega_1 \cup \omega_2\right) = \pi\left(\omega_1\right) + \pi\left(\omega_2\right)$$; 

If $$\Omega$$ is the $$\sigma$$-algebra of all Borels sets on a Hausdorff locally compact (or compact) space, then the following additional requirement is added: for every $$x, y \in H,$$ the map $$\pi_{x, y} : \Omega \to \Complex$$ is a regular Borel measure (this is automatically satisfied on compact metric spaces). </ol> Conditions 2, 3, and 4 imply that $$\pi$$ is a projection-valued measure.

Properties
Throughout, let $$\pi$$ be a resolution of identity. For all $$x \in H,$$ $$\pi_{x, x} : \Omega \to \Complex$$ is a positive measure on $$\Omega$$ with total variation $$\left\|\pi_{x, x}\right\| = \pi_{x, x}(X) = \|x\|^2$$ and that satisfies $$\pi_{x, x}(\omega) = \langle \pi(\omega) x, x \rangle = \|\pi(\omega) x\|^2$$ for all $$\omega \in \Omega.$$

For every $$\omega_1, \omega_2 \in \Omega$$: <ul> <li>$$\pi\left(\omega_1\right) \pi\left(\omega_2\right) = \pi\left(\omega_2\right) \pi\left(\omega_1\right)$$ (since both are equal to $$\pi\left(\omega_1 \cap \omega_2\right)$$).</li> <li>If $$\omega_1 \cap \omega_2 = \varnothing$$ then the ranges of the maps $$\pi\left(\omega_1\right)$$ and $$\pi\left(\omega_2\right)$$ are orthogonal to each other and $$\pi\left(\omega_1\right) \pi\left(\omega_2\right) = 0 = \pi\left(\omega_2\right) \pi\left(\omega_1\right).$$</li> <li>$$\pi : \Omega \to \mathcal{B}(H)$$ is finitely additive. <li>If $$\omega_1, \omega_2, \ldots$$ are pairwise disjoint elements of $$\Omega$$ whose union is $$\omega$$ and if $$\pi\left(\omega_i\right) = 0$$ for all $$i$$ then $$\pi(\omega) = 0.$$</li> <li>For any fixed $$x \in H,$$ the map $$\pi_x : \Omega \to H$$ defined by $$\pi_{x}(\omega) := \pi(\omega) x$$ is a countably additive $$H$$-valued measure on $$\Omega.$$ </ul>
 * However, $$\pi : \Omega \to \mathcal{B}(H)$$ is additive only in trivial situations as is now described: suppose that $$\omega_1, \omega_2, \ldots$$ are pairwise disjoint elements of $$\Omega$$ whose union is $$\omega$$ and that the partial sums $$\sum_{i=1}^n \pi\left(\omega_i\right)$$ converge to $$\pi(\omega)$$ in $$\mathcal{B}(H)$$ (with its norm topology) as $$n \to \infty$$; then since the norm of any projection is either $$0$$ or $$\geq 1,$$ the partial sums cannot form a Cauchy sequence unless all but finitely many of the $$\pi\left(\omega_i\right)$$ are $$0.$$
 * Here countably additive means that whenever $$\omega_1, \omega_2, \ldots$$ are pairwise disjoint elements of $$\Omega$$ whose union is $$\omega,$$ then the partial sums $$\sum_{i=1}^n \pi\left(\omega_i\right) x$$ converge to $$\pi(\omega)x$$ in $$H.$$ Said more succinctly, $$\sum_{i=1}^{\infty} \pi\left(\omega_i\right) x = \pi(\omega) x.$$
 * In other words, for every pairwise disjoint family of elements $$\left(\omega_i\right)_{i = 1}^{\infty}\subseteq \Omega$$ whose union is $$\omega_{\infty}\in\Omega$$, then $$\sum_{i = 1}^{n} \pi\left(\omega_i\right) = \pi\left(\bigcup_{i=1}^{n} \omega_i\right)$$ (by finite additivity of $$\pi$$) converges to $$\pi\left(\omega_{\infty}\right)$$ in the strong operator topology on $$\mathcal{B}(H)$$: for every $$x\in H$$, the sequence of elements $$\sum_{i=1}^{n}\pi\left(\omega_i\right)x$$ converges to $$\pi\left(\omega_{\infty}\right)x$$ in $$H$$ (with respect to the norm topology).</li>

L∞(π) - space of essentially bounded function
The $$\pi : \Omega \to \mathcal{B}(H)$$ be a resolution of identity on $$(X, \Omega).$$

Essentially bounded functions
Suppose $$f : X \to \Complex$$ is a complex-valued $$\Omega$$-measurable function. There exists a unique largest open subset $$V_f$$ of $$\Complex$$ (ordered under subset inclusion) such that $$\pi\left(f^{-1}\left(V_f\right)\right) = 0.$$ To see why, let $$D_1, D_2, \ldots$$ be a basis for $$\Complex$$'s topology consisting of open disks and suppose that $$D_{i_1}, D_{i_2}, \ldots$$ is the subsequence (possibly finite) consisting of those sets such that $$\pi\left(f^{-1}\left(D_{i_k}\right)\right) = 0$$; then $$D_{i_1} \cup D_{i_2} \cup \cdots = V_f.$$ Note that, in particular, if $$D$$ is an open subset of $$\Complex$$ such that $$D \cap \operatorname{Im} f = \varnothing$$ then $$\pi\left(f^{-1}(D)\right) = \pi (\varnothing) = 0$$ so that $$D \subseteq V_f$$ (although there are other ways in which $$\pi\left(f^{-1}(D)\right)$$ may equal $0$). Indeed, $$\Complex \setminus \operatorname{cl}(\operatorname{Im} f) \subseteq V_f.$$

The essential range of $$f$$ is defined to be the complement of $$V_f.$$ It is the smallest closed subset of $$\Complex$$ that contains $$f(x)$$ for almost all $$x \in X$$ (that is, for all $$x \in X$$ except for those in some set $$\omega \in \Omega$$ such that $$\pi(\omega) = 0$$). The essential range is a closed subset of $$\Complex$$ so that if it is also a bounded subset of $$\Complex$$ then it is compact.

The function $$f$$ is essentially bounded if its essential range is bounded, in which case define its essential supremum, denoted by $$\|f\|^{\infty},$$ to be the supremum of all $$|\lambda|$$ as $$\lambda$$ ranges over the essential range of $$f.$$

Space of essentially bounded functions
Let $$\mathcal{B}(X, \Omega)$$ be the vector space of all bounded complex-valued $$\Omega$$-measurable functions $$f : X \to \Complex,$$ which becomes a Banach algebra when normed by $$\|f\|_{\infty} := \sup_{x \in X}|f(x) |.$$ The function $$\|\,\cdot\,\|^{\infty}$$ is a seminorm on $$\mathcal{B}(X, \Omega),$$ but not necessarily a norm. The kernel of this seminorm, $$N^{\infty} := \left\{ f \in \mathcal{B}(X, \Omega) : \|f\|^{\infty} = 0 \right\},$$ is a vector subspace of $$\mathcal{B}(X, \Omega)$$ that is a closed two-sided ideal of the Banach algebra $$\left(\mathcal{B}(X, \Omega), \| \cdot \|_{\infty}\right).$$ Hence the quotient of $$\mathcal{B}(X, \Omega)$$ by $$N^{\infty}$$ is also a Banach algebra, denoted by $$L^{\infty}(\pi) := \mathcal{B}(X, \Omega) / N^{\infty}$$ where the norm of any element $$f + N^{\infty} \in L^{\infty}(\pi)$$ is equal to $$\|f\|^{\infty}$$ (since if $$f + N^{\infty} = g + N^{\infty}$$ then $$\|f\|^{\infty} = \| g \|^{\infty}$$) and this norm makes $$L^{\infty}(\pi)$$ into a Banach algebra. The spectrum of $$f + N^{\infty}$$ in $$L^{\infty}(\pi)$$ is the essential range of $$f.$$ This article will follow the usual practice of writing $$f$$ rather than $$f + N^{\infty}$$ to represent elements of $$L^{\infty}(\pi).$$

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Spectral theorem
The maximal ideal space of a Banach algebra $$A$$ is the set of all complex homomorphisms $$A \to \Complex,$$ which we'll denote by $$\sigma_A.$$ For every $$T$$ in $$A,$$ the Gelfand transform of $$T$$ is the map $$G(T) : \sigma_A \to \Complex$$ defined by $$G(T)(h) := h(T).$$ $$\sigma_A$$ is given the weakest topology making every $$G(T) : \sigma_A \to \Complex$$ continuous. With this topology, $$\sigma_A$$ is a compact Hausdorff space and every $$T$$ in $$A,$$ $$G(T)$$ belongs to $$C \left(\sigma_A\right),$$ which is the space of continuous complex-valued functions on $$\sigma_A.$$ The range of $$G(T)$$ is the spectrum $$\sigma(T)$$ and that the spectral radius is equal to $$\max \left\{ |G(T)(h)|: h \in \sigma_A \right\},$$ which is $$\leq \|T\|.$$

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The above result can be specialized to a single normal bounded operator.