[proofplan] The strategy is to specialise the previous theorem **Spectral Theorem for a Commutative Unital $C^*$-Algebra** to the smallest commutative unital $C^*$-subalgebra of $\mathcal{L}(H)$ containing $T$ — namely the subalgebra $A := C^*(T, I)$ generated by $T$, $T^*$, and $I$. Normality of $T$ is precisely the hypothesis that makes $A$ commutative. Once that theorem is applied, we obtain a resolution of the identity $P$ on the character space $K_A := \Phi_A$ of $A$. The key identification is that $K_A$ is canonically homeomorphic to the spectrum $\sigma(T) \subseteq \mathbb{C}$ via $\chi \mapsto \chi(T)$. Pushing $P$ forward along this homeomorphism gives a resolution of the identity over $\sigma(T)$, and the integration formula for $T$ becomes $T = \int \lambda \, dP$. Uniqueness and the commutant identity follow by tracking the same arguments through the homeomorphism. Finally, the **Fuglede–Putnam–Rosenblum Lemma** is needed to upgrade "commutes with $T$" to "commutes with both $T$ and $T^*$", which is what is required for the commutant identity (since $A = C^*(T, I)$ involves $T^*$). [/proofplan] [step:Construct the commutative unital $C^*$-subalgebra $A := C^*(T, I) \subseteq \mathcal{L}(H)$] Define $A$ to be the norm-closure in $\mathcal{L}(H)$ of the set of polynomial expressions in $T$, $T^*$, and the identity $I$: \begin{align*} A := \overline{\{p(T, T^*) : p \in \mathbb{C}\langle X, Y \rangle\}}^{\| \cdot \|_{\mathcal{L}(H)}}. \end{align*} *$A$ is a closed unital $*$-subalgebra.* The polynomial expressions form a unital subalgebra (containing $I = $ constant polynomial $1$) closed under products. The closure is closed under multiplication (continuity of operator multiplication on bounded sets) and addition. The set of polynomial expressions is closed under the involution $A \mapsto A^*$: $(p(T, T^*))^* = \bar{p}(T^*, T)$ (the complex conjugate polynomial with arguments swapped); since polynomials in $(X, Y)$ pair with $(T, T^*)$ symmetrically, the resulting expression is again a polynomial in $T, T^*$. The closure under $*$ persists by continuity of the involution (which is an isometry). *$A$ is commutative.* By hypothesis, $T T^* = T^* T$, so $T$ and $T^*$ commute. Polynomials in two commuting elements commute pairwise. The closure of a commutative set in a Banach algebra remains commutative (continuity of multiplication: if $a_n \to a$ and $b_n \to b$ with $a_n b_n = b_n a_n$, then $ab = \lim a_n b_n = \lim b_n a_n = ba$). Hence $A$ is commutative. *$A$ is a $C^*$-algebra.* $\mathcal{L}(H)$ is a $C^*$-algebra (the $C^*$-identity $\|S^* S\| = \|S\|^2$ is the standard property of the operator adjoint on a Hilbert space). A norm-closed $*$-subalgebra of a $C^*$-algebra inherits the $C^*$-structure. Hence $A$ is a commutative unital $C^*$-subalgebra of $\mathcal{L}(H)$. [/step] [step:Apply the Spectral Theorem for Commutative Unital $C^*$-Algebras to $A$] Apply the previous theorem **Spectral Theorem for a Commutative Unital $C^*$-Algebra** to the algebra $A$ from Step 1. The theorem requires "commutative, unital $C^*$-subalgebra of $\mathcal{L}(H)$", which is exactly the conclusion of Step 1. The theorem produces: - a compact Hausdorff space $K_A := \Phi_A$ (the character space of $A$ with the Gelfand topology), - a unique resolution of the identity $P_A$ of $H$ over $K_A$, such that \begin{align*} \int_{K_A} \hat{S} \, dP_A = S \quad \text{for all } S \in A, \end{align*} where $\hat{S}: K_A \to \mathbb{C}$, $\chi \mapsto \chi(S)$, is the Gelfand transform. Furthermore, by part (ii) of that theorem: \begin{align*} S \in \mathcal{L}(H) \text{ commutes with every } S' \in A \iff S \text{ commutes with every } P_A(E), \, E \in \mathcal{B}(K_A). \end{align*} [/step] [step:Identify the character space $K_A$ with $\sigma(T) \subseteq \mathbb{C}$ via the Gelfand transform of $T$] Define \begin{align*} \Theta: K_A &\to \mathbb{C} \\ \chi &\mapsto \chi(T) = \hat{T}(\chi). \end{align*} We claim $\Theta$ is a homeomorphism from $K_A$ onto $\sigma(T) \subseteq \mathbb{C}$. *Image of $\Theta$ equals $\sigma_A(T)$.* By [Spectrum via Characters](/theorems/2677), for a commutative unital Banach algebra $A$ and any element $S \in A$, the spectrum of $S$ within $A$ is $\sigma_A(S) = \{\chi(S) : \chi \in \Phi_A\} = \hat{S}(\Phi_A)$. Theorem 2677 requires "commutative unital Banach algebra"; $A$ is such by Step 1. Applying with $S = T$, \begin{align*} \Theta(K_A) = \hat{T}(K_A) = \sigma_A(T). \end{align*} *$\sigma_A(T) = \sigma(T)$.* By [Spectral Inclusions and Permanence](/theorems/2688) part (3) applied to the unital $C^*$-subalgebra $A \subseteq \mathcal{L}(H)$ and the normal element $T \in A$: for normal elements, the spectrum within a $C^*$-subalgebra equals the spectrum within the larger $C^*$-algebra. Theorem 2688's hypotheses (unital $C^*$-subalgebra, normal element) are met. Hence $\sigma_A(T) = \sigma_{\mathcal{L}(H)}(T) = \sigma(T) =: K$. So $\Theta: K_A \to K = \sigma(T)$ is well-defined and surjective. *$\Theta$ is injective.* Suppose $\chi_1, \chi_2 \in K_A$ with $\Theta(\chi_1) = \Theta(\chi_2)$, i.e.\ $\chi_1(T) = \chi_2(T)$. Characters are $*$-homomorphisms (this is the content of [Characters Are Star-Homomorphisms](/theorems/2687) applied to the $C^*$-algebra $A$), so $\chi_1(T^*) = \overline{\chi_1(T)} = \overline{\chi_2(T)} = \chi_2(T^*)$. Hence $\chi_1$ and $\chi_2$ agree on $\{T, T^*, I\}$; by linearity and multiplicativity they agree on all polynomials in $T, T^*$; by continuity (characters of a Banach algebra are continuous, [Characters are Continuous](/theorems/2675)) they agree on the closure $A$. So $\chi_1 = \chi_2$. *$\Theta$ is continuous.* The Gelfand topology on $K_A$ is by definition the weakest topology making every $\hat{S}: K_A \to \mathbb{C}$ continuous, so $\hat{T} = \Theta$ is continuous. *$\Theta$ is a homeomorphism.* $K_A$ is compact (Gelfand topology), $\mathbb{C}$ is Hausdorff, and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism onto its image. Hence $\Theta: K_A \to \sigma(T)$ is a homeomorphism. [/step] [step:Push the resolution of the identity $P_A$ forward to $\sigma(T)$ via $\Theta$] Define \begin{align*} P: \mathcal{B}(\sigma(T)) &\to \mathcal{L}(H) \\ E &\mapsto P_A(\Theta^{-1}(E)). \end{align*} Since $\Theta$ is a homeomorphism, $\Theta^{-1}: \sigma(T) \to K_A$ is continuous, hence Borel-measurable, so $\Theta^{-1}(E) \in \mathcal{B}(K_A)$ for every $E \in \mathcal{B}(\sigma(T))$. Hence $P$ is well-defined. *$P$ is a resolution of the identity.* Each axiom for $P$ follows from the corresponding axiom for $P_A$ together with set-theoretic identities for $\Theta^{-1}$: (R1) $P(E) = P_A(\Theta^{-1}(E))$ is an orthogonal projection (from the same axiom for $P_A$). (R2) $P(\varnothing) = P_A(\varnothing) = 0$. $P(\sigma(T)) = P_A(K_A) = I$. (R3) For $E, F \in \mathcal{B}(\sigma(T))$, $\Theta^{-1}(E \cap F) = \Theta^{-1}(E) \cap \Theta^{-1}(F)$ (preimage under any function commutes with intersection), so $P(E \cap F) = P_A(\Theta^{-1}(E) \cap \Theta^{-1}(F)) = P_A(\Theta^{-1}(E)) P_A(\Theta^{-1}(F)) = P(E) P(F)$. (R4) Disjoint additivity: $E \cap F = \varnothing \implies \Theta^{-1}(E) \cap \Theta^{-1}(F) = \Theta^{-1}(\varnothing) = \varnothing$, so $P(E \cup F) = P_A(\Theta^{-1}(E \cup F)) = P_A(\Theta^{-1}(E) \cup \Theta^{-1}(F)) = P_A(\Theta^{-1}(E)) + P_A(\Theta^{-1}(F)) = P(E) + P(F)$. (R5) For $x, y \in H$, $P_{x,y}(E) := (P(E)x, y)_H = (P_A(\Theta^{-1}(E)) x, y)_H = (P_A)_{x,y}(\Theta^{-1}(E))$. The map $E \mapsto (P_A)_{x,y}(\Theta^{-1}(E))$ is the *pushforward* of the regular complex Borel measure $(P_A)_{x,y}$ along $\Theta$, and the pushforward of a regular complex Borel measure under a homeomorphism is regular. The total variation is preserved: $\|P_{x,y}\|_1 = \|(P_A)_{x,y}\|_1 \le \|x\| \|y\|$. Hence $P$ is a resolution of the identity over $\sigma(T)$. [/step] [step:Verify $T = \int_{\sigma(T)} \lambda \, dP(\lambda)$] By Step 2 applied to the element $T \in A$ (with the integral against $P_A$ giving the operator $T$), \begin{align*} T = \int_{K_A} \hat{T} \, dP_A. \end{align*} We translate this through the homeomorphism $\Theta$. By the change-of-variables formula for integration of an $L^\infty$ function against a pushforward measure: for any Borel $f: \sigma(T) \to \mathbb{C}$, \begin{align*} \int_{K_A} (f \circ \Theta) \, dP_A = \int_{\sigma(T)} f \, dP, \end{align*} where the integrals are interpreted via the integration map $\Phi$ from the previous theorem **Integration Against $P$**. Setting $f(\lambda) := \lambda$ (the identity function on $\sigma(T) \subseteq \mathbb{C}$), the composition $f \circ \Theta: K_A \to \mathbb{C}$, $\chi \mapsto \chi(T) = \hat{T}(\chi)$, equals $\hat{T}$. Hence \begin{align*} \int_{K_A} \hat{T} \, dP_A = \int_{\sigma(T)} \lambda \, dP(\lambda), \end{align*} and consequently \begin{align*} T = \int_{\sigma(T)} \lambda \, dP(\lambda). \end{align*} *Justification of the change-of-variables formula.* The integration map for $P$ is the unique isometric, unital $*$-homomorphism $\Phi_P: L^\infty(P) \to \mathcal{L}(H)$ with $(\Phi_P(f) x, y)_H = \int_{\sigma(T)} f \, dP_{x,y}$ (by the previous theorem **Integration Against $P$**). The map $f \mapsto \Phi_{P_A}(f \circ \Theta)$ is also an isometric, unital $*$-homomorphism $L^\infty(P) \to \mathcal{L}(H)$ (composition with the homeomorphism $\Theta$ is an isometric $*$-isomorphism $L^\infty(P) \to L^\infty(P_A)$). It satisfies the same characterising property: \begin{align*} (\Phi_{P_A}(f \circ \Theta) x, y)_H = \int_{K_A} (f \circ \Theta) \, d(P_A)_{x,y} = \int_{\sigma(T)} f \, dP_{x,y}, \end{align*} where the second equality is the standard change-of-variables formula for pushforward measures. By uniqueness of the integration map, $\Phi_{P_A}(f \circ \Theta) = \Phi_P(f)$ for all $f \in L^\infty(P)$. [/step] [step:Prove uniqueness of $P$] Suppose $P'$ is another resolution of the identity of $H$ over $\sigma(T)$ with $T = \int_{\sigma(T)} \lambda \, dP'(\lambda)$. Pull $P'$ back to $K_A$ via $\Theta$: define $P'_A(F) := P'(\Theta(F))$ for $F \in \mathcal{B}(K_A)$. By the same change-of-variables argument as Step 5 (in reverse), $P'_A$ is a resolution of the identity over $K_A$ with \begin{align*} \int_{K_A} \hat{T} \, dP'_A = \int_{\sigma(T)} \lambda \, dP' = T. \end{align*} We need this identity for *all* $S \in A$, not just $T$, to invoke uniqueness in the previous theorem on commutative unital $C^*$-algebras. Apply the integration map. For any $S \in A$, $\hat{S} \in C(K_A)$ is in the image of the Gelfand transform. The [Commutative Gelfand-Naimark Theorem](/theorems/2689) gives that the Gelfand transform $A \to C(K_A)$ is surjective; explicitly, every $f \in C(K_A)$ has the form $\hat{S}$ for some $S \in A$. Now: \begin{align*} \int_{K_A} \hat{S} \, dP'_A = \Phi_{P'_A}(\hat{S}), \end{align*} and we want to show this equals $S$ for every $S \in A$. The integration map $\Phi_{P'_A}: L^\infty(P'_A) \to \mathcal{L}(H)$ restricted to $C(K_A) \subseteq L^\infty(P'_A)$ is a unital $*$-homomorphism. The element $\hat{T}$ generates $C(K_A)$ as a unital $*$-subalgebra: by **Stone-Weierstrass** applied to the compact Hausdorff $K_A$, the unital $*$-subalgebra generated by $\hat{T}$ in $C(K_A)$ separates points (since $\hat{T} = \Theta$ is injective by Step 3) and contains the constants (the unit), so it equals $C(K_A)$. Hence the unital $*$-homomorphism $\Phi_{P'_A}|_{C(K_A)}$ is determined by its value on $\hat{T}$ — namely $\Phi_{P'_A}(\hat{T}) = T$ by hypothesis. The unique unital $*$-homomorphism $C(K_A) \to A \subseteq \mathcal{L}(H)$ with $\hat{T} \mapsto T$ is the inverse Gelfand transform $\Gamma^{-1}$, by the [Commutative Gelfand-Naimark Theorem](/theorems/2689). Therefore $\Phi_{P'_A}(\hat{S}) = \Gamma^{-1}(\hat{S}) = S$ for all $S \in A$. By uniqueness in the previous theorem on commutative unital $C^*$-algebras, $P'_A = P_A$ on $K_A$. Pushing forward via $\Theta$, $P' = P$ on $\sigma(T)$. [/step] [step:Establish the commutant identity using the Fuglede–Putnam–Rosenblum Lemma] Let $S \in \mathcal{L}(H)$. We show \begin{align*} ST = TS \iff S \circ P(E) = P(E) \circ S \quad \text{for all } E \in \mathcal{B}(\sigma(T)). \end{align*} *$(\Leftarrow)$.* Suppose $S$ commutes with $P(E)$ for all $E \in \mathcal{B}(\sigma(T))$. Equivalently, $S$ commutes with $P_A(F)$ for all $F \in \mathcal{B}(K_A)$ (via $F = \Theta^{-1}(E)$). By the previous theorem on commutative unital $C^*$-algebras part (ii), $S$ commutes with every element of $A$, in particular with $T \in A$. *$(\Rightarrow)$.* Suppose $ST = TS$. We need $S$ to commute with every element of $A$, in particular with $T^*$. Apply the previous theorem **Fuglede–Putnam–Rosenblum Lemma** to the unital $C^*$-algebra $\mathcal{L}(H)$ and the elements $x = T$, $y = T$, $z = S$. The hypotheses require $x, y$ normal: by hypothesis $T$ is normal. The hypothesis $xz = zy$ becomes $TS = ST$, which holds. The lemma concludes $x^* z = z y^*$, i.e.\ $T^* S = S T^*$, which gives $S T^* = T^* S$. Hence $S$ commutes with both $T$ and $T^*$. By linearity, $S$ commutes with every polynomial in $T$ and $T^*$. By continuity of multiplication, $S$ commutes with every element of the closure $A$. By the previous theorem on commutative unital $C^*$-algebras part (ii), $S$ commutes with $P_A(F)$ for all $F \in \mathcal{B}(K_A)$, i.e.\ $S$ commutes with $P(E)$ for all $E \in \mathcal{B}(\sigma(T))$. This proves both directions of the commutant identity and completes the proof of the spectral theorem. [/step]