[proofplan] By the [Commutative Gelfand-Naimark Theorem](/theorems/2689), the Gelfand transform $\Gamma: A \to C(K)$, $T \mapsto \hat{T}$, is an isometric, unital $*$-isomorphism. Inverting it gives an isometric, unital $*$-homomorphism $\gamma := \Gamma^{-1}: C(K) \to A \subseteq \mathcal{L}(H)$, sending continuous functions on $K$ to operators on $H$. The strategy is to extend $\gamma$ from $C(K)$ to $L^\infty(K, \mathcal{B}(K))$, producing a $*$-homomorphism $\Psi$ that contains the resolution of the identity in disguise: $P(E) := \Psi(\mathbb{1}_E)$ defines projections, and the multiplicativity $\mathbb{1}_{E \cap F} = \mathbb{1}_E \mathbb{1}_F$ together with multiplicativity of $\Psi$ produces the multiplicativity axiom for $P$. The extension uses the [Riesz Representation Theorem](/theorems/221) on $C(K)$: each pair $x, y \in H$ produces a bounded functional $\hat{T} \mapsto (Tx, y)_H$ on $C(K)$ which is represented by a regular complex Borel measure $\mu_{x,y}$, and the family $\{\mu_{x,y}\}_{x,y \in H}$ is the data of a sesquilinear form that defines $\Psi(f)$ for any bounded Borel $f$. [/proofplan] [step:Apply the Commutative Gelfand-Naimark Theorem to obtain the inverse $\gamma: C(K) \to A$] By hypothesis, $A$ is a commutative, unital $C^*$-subalgebra of $\mathcal{L}(H)$. The [Commutative Gelfand-Naimark Theorem](/theorems/2689) applies and produces an isometric, unital $*$-isomorphism \begin{align*} \Gamma: A &\to C(K) \\ T &\mapsto \hat{T}, \end{align*} where $K = \Phi_A$ is the character space of $A$ with the Gelfand topology and $\hat{T}(\chi) := \chi(T)$ for $\chi \in K$. The hypothesis "commutative, unital $C^*$" of theorem 2689 is exactly our hypothesis on $A$. The conclusion of the theorem provides that $K$ is compact Hausdorff and that $\Gamma$ is a bijective isometry. Inverting, set \begin{align*} \gamma := \Gamma^{-1}: C(K) \to A \subseteq \mathcal{L}(H). \end{align*} Then $\gamma$ is an isometric, unital $*$-homomorphism, and $\gamma(\hat{T}) = T$ for every $T \in A$. [/step] [step:Construct regular complex Borel measures $\mu_{x,y}$ on $K$ via the Riesz Representation Theorem] Fix $x, y \in H$. Define the linear functional \begin{align*} \Lambda_{x,y}: C(K) &\to \mathbb{C} \\ f &\mapsto (\gamma(f)x, y)_H. \end{align*} *Boundedness.* For $f \in C(K)$, \begin{align*} |\Lambda_{x,y}(f)| = |(\gamma(f)x, y)_H| \le \|\gamma(f)\|_{\mathcal{L}(H)} \|x\|_H \|y\|_H = \|f\|_{C(K)} \|x\|_H \|y\|_H, \end{align*} using Cauchy-Schwarz in $H$ and the isometry $\|\gamma(f)\|_{\mathcal{L}(H)} = \|f\|_{C(K)}$ from Step 1. Hence $\Lambda_{x,y}$ is a bounded linear functional on $C(K)$ with norm at most $\|x\|_H \|y\|_H$. By the [Riesz Representation Theorem](/theorems/221) for $C(K)$ (with $K$ compact Hausdorff): every bounded linear functional $\Lambda$ on $C(K)$ is represented uniquely by a regular complex Borel measure $\mu$ on $K$ via $\Lambda(f) = \int_K f \, d\mu$, with $\|\Lambda\| = \|\mu\|_1$ (total variation norm). Applying this to $\Lambda_{x,y}$, there exists a unique regular complex Borel measure $\mu_{x,y}$ on $K$ with \begin{align*} (\gamma(f)x, y)_H = \int_K f \, d\mu_{x,y} \quad \text{for all } f \in C(K), \end{align*} and $\|\mu_{x,y}\|_1 \le \|x\|_H \|y\|_H$. *Sesquilinearity.* The map $(x, y) \mapsto \Lambda_{x,y}$ is linear in $x$ and conjugate-linear in $y$, by linearity of $T \mapsto Tx$ in $x$ and conjugate-linearity of $w \mapsto (w, y)_H$ in $y$. By uniqueness of representation in RRT, the same holds for $\mu_{x,y}$: for $\alpha \in \mathbb{C}$, $\mu_{\alpha x_1 + x_2, y} = \alpha \mu_{x_1, y} + \mu_{x_2, y}$ and $\mu_{x, \alpha y_1 + y_2} = \bar\alpha \mu_{x, y_1} + \mu_{x, y_2}$. *Conjugate symmetry.* For $f \in C(K)$ real-valued, $\bar{f} = f$, so $\gamma(f)^* = \gamma(\bar{f}) = \gamma(f)$, i.e.\ $\gamma(f)$ is Hermitian. Hence \begin{align*} (\gamma(f)x, y)_H = \overline{(\gamma(f)y, x)_H} \implies \int_K f \, d\mu_{x,y} = \overline{\int_K f \, d\mu_{y,x}} = \int_K f \, d\overline{\mu_{y,x}} \end{align*} for all real $f \in C(K)$. By RRT uniqueness applied to the real Borel measures $\operatorname{Re}\mu_{x,y}$, $\operatorname{Im}\mu_{x,y}$, $\operatorname{Re}\overline{\mu_{y,x}}$, $\operatorname{Im}\overline{\mu_{y,x}}$ (real-valued continuous functions separate them), we conclude $\mu_{x,y} = \overline{\mu_{y,x}}$. [/step] [step:Define $\Psi: L^\infty(K, \mathcal{B}(K)) \to \mathcal{L}(H)$ via the bounded sesquilinear form representation] Fix $f \in L^\infty(K, \mathcal{B}(K))$ (bounded Borel measurable, with the essential-supremum norm with respect to *every* regular complex measure simultaneously — equivalently, the supremum norm for bounded Borel functions, modulo the universally null sets). Define \begin{align*} \theta_f: H \times H &\to \mathbb{C} \\ (x, y) &\mapsto \int_K f \, d\mu_{x,y}. \end{align*} *Sesquilinearity.* Linearity in $x$ and conjugate-linearity in $y$ follow from the corresponding properties of $\mu_{x,y}$ (Step 2) by linearity of the integral. *Boundedness.* For all $x, y \in H$, \begin{align*} |\theta_f(x, y)| = \left|\int_K f \, d\mu_{x,y}\right| \le \|f\|_{\infty} \|\mu_{x,y}\|_1 \le \|f\|_{\infty} \|x\|_H \|y\|_H, \end{align*} using the standard estimate $|\int f \, d\mu| \le \|f\|_\infty \|\mu\|_1$ for bounded Borel $f$ and complex measure $\mu$, together with $\|\mu_{x,y}\|_1 \le \|x\|_H \|y\|_H$ from Step 2. *Operator from sesquilinear form.* By the [Riesz Representation Theorem](/theorems/221) for bounded sesquilinear forms on a Hilbert space: every bounded sesquilinear form $\theta: H \times H \to \mathbb{C}$ with $|\theta(x,y)| \le M\|x\|\|y\|$ is represented by a unique $T \in \mathcal{L}(H)$ with $\theta(x, y) = (Tx, y)_H$ and $\|T\|_{\mathcal{L}(H)} \le M$. Applying this to $\theta_f$, there exists a unique $\Psi(f) \in \mathcal{L}(H)$ with \begin{align*} (\Psi(f)x, y)_H = \int_K f \, d\mu_{x,y} \quad \text{for all } x, y \in H, \end{align*} and $\|\Psi(f)\|_{\mathcal{L}(H)} \le \|f\|_\infty$. This defines a map \begin{align*} \Psi: L^\infty(K, \mathcal{B}(K)) \to \mathcal{L}(H). \end{align*} *Compatibility with $\gamma$.* For $f \in C(K) \subseteq L^\infty(K, \mathcal{B}(K))$, the defining identity for $\mu_{x,y}$ gives \begin{align*} (\Psi(f)x, y)_H = \int_K f \, d\mu_{x,y} = (\gamma(f)x, y)_H, \end{align*} holding for all $x, y \in H$, so $\Psi(f) = \gamma(f)$. In particular $\Psi|_{C(K)} = \gamma$, so $\Psi$ extends the inverse Gelfand transform. As consequences: \begin{align*} \Psi(\hat{T}) = \gamma(\hat{T}) = T \quad \text{for all } T \in A, \qquad \Psi(1) = \gamma(1) = I. \end{align*} [/step] [step:Verify $\Psi$ is a $*$-homomorphism] We check linearity, the involution-preservation $\Psi(\bar{f}) = \Psi(f)^*$, and multiplicativity $\Psi(fg) = \Psi(f) \Psi(g)$ for $f, g \in L^\infty(K, \mathcal{B}(K))$. *Linearity.* The form $\theta_f$ depends linearly on $f$ (linearity of the integral in the integrand), and the operator-from-form map is itself linear, so $\Psi(\alpha f + g) = \alpha \Psi(f) + \Psi(g)$. *Involution.* For $x, y \in H$, \begin{align*} (\Psi(f)^* x, y)_H = \overline{(\Psi(f) y, x)_H} = \overline{\int_K f \, d\mu_{y, x}} = \int_K \bar{f} \, d\overline{\mu_{y, x}} = \int_K \bar{f} \, d\mu_{x,y} = (\Psi(\bar{f}) x, y)_H, \end{align*} using $\overline{\mu_{y,x}} = \mu_{x,y}$ (Step 2). Holding for all $x, y$, this gives $\Psi(f)^* = \Psi(\bar{f})$. *Multiplicativity.* This is the deepest step; we prove it by showing two pushforward identities of measures. [claim:For all $T \in A$ and $x, y \in H$, $\hat{T} \, d\mu_{x,y} = d\mu_{Tx, y}$ as regular complex Borel measures on $K$.] [proof] Both measures are regular complex Borel measures on $K$. By RRT uniqueness it suffices to show \begin{align*} \int_K f \, \hat{T} \, d\mu_{x,y} = \int_K f \, d\mu_{Tx, y} \quad \text{for all } f \in C(K). \end{align*} Both sides involve continuous integrands, and both are linear continuous functionals of $f \in C(K)$. They agree provided they agree on the dense subspace $\{f = \hat{T}'\} = C(K)$ itself — i.e.\ on all of $C(K)$. We compute both sides using the defining identity of $\mu$: *RHS.* $\int_K f \, d\mu_{Tx, y} = (\gamma(f) Tx, y)_H$. *LHS.* For $f = \hat{T}'$ with $T' \in A$, $f \cdot \hat{T} = \hat{T}' \cdot \hat{T} = \widehat{T' T}$ (Gelfand transform is multiplicative on $A$). Hence \begin{align*} \int_K f \cdot \hat{T} \, d\mu_{x,y} = \int_K \widehat{T'T} \, d\mu_{x,y} = ((T'T)x, y)_H = (T'(Tx), y)_H = (\gamma(\hat{T}')(Tx), y)_H = (\gamma(f)(Tx), y)_H. \end{align*} Since $\gamma$ inverts the Gelfand transform on $A$, every $f \in C(K)$ has the form $\hat{T}'$ for a unique $T' \in A$, so the identity holds for all $f \in C(K)$. RRT uniqueness gives $\hat{T} \, d\mu_{x,y} = d\mu_{Tx, y}$. [/proof] [/claim] [claim:For all $f \in L^\infty(K, \mathcal{B}(K))$ and $x, y \in H$, $f \, d\mu_{x,y} = d\mu_{x, \Psi(f)^* y}$.] [proof] Both measures are regular complex Borel measures on $K$. By RRT uniqueness it suffices to test against $g \in C(K)$: \begin{align*} \int_K g \, d\mu_{x, \Psi(f)^* y} = (\gamma(g)x, \Psi(f)^* y)_H = (\Psi(f) \gamma(g) x, y)_H. \end{align*} We need this to equal $\int_K g f \, d\mu_{x,y}$, which by the defining identity for $\Psi$ equals $(\Psi(gf) x, y)_H$. So we must show $\Psi(f) \gamma(g) = \Psi(gf)$ as operators in $\mathcal{L}(H)$, applied to $x$. Equivalently — using $\gamma(g) = \Psi(g)$ from Step 3 — we must show $\Psi(f) \Psi(g) = \Psi(gf)$ when $g \in C(K)$. We prove this by another application of RRT. For $T \in A$ (so $g = \hat{T}$ for some $T$), by the previous claim, for any $w, z \in H$, \begin{align*} (\Psi(f) T w, z)_H = \int_K f \, d\mu_{Tw, z} = \int_K f \cdot \hat{T} \, d\mu_{w, z} = (\Psi(f \hat{T}) w, z)_H, \end{align*} holding for all $w, z$, so $\Psi(f) T = \Psi(f \hat{T})$, i.e.\ $\Psi(f) \gamma(\hat{T}) = \Psi(f \hat{T})$. Since every $g \in C(K)$ has the form $\hat{T}$, this gives $\Psi(f) \gamma(g) = \Psi(fg)$ for all $g \in C(K)$. Since pointwise multiplication on $L^\infty(K, \mathcal{B}(K))$ is commutative, $fg = gf$, so $\Psi(fg) = \Psi(gf)$. Hence $\Psi(f) \gamma(g) = \Psi(fg) = \Psi(gf)$, which gives \begin{align*} (\Psi(f) \gamma(g) x, y)_H = (\Psi(gf) x, y)_H = \int_K gf \, d\mu_{x,y}, \end{align*} proving the claim. [/proof] [/claim] *Multiplicativity, finally.* Combine the two claims. For $f, g \in L^\infty(K, \mathcal{B}(K))$ and $x, y \in H$, applying the second claim with the integrand $g$ gives $g \, d\mu_{x,y} = d\mu_{x, \Psi(g)^* y}$, hence \begin{align*} (\Psi(fg) x, y)_H = \int_K f g \, d\mu_{x,y} = \int_K f \, d\mu_{x, \Psi(g)^*y} = (\Psi(f) x, \Psi(g)^* y)_H = (\Psi(g) \Psi(f) x, y)_H. \end{align*} Holding for all $x, y \in H$, we obtain $\Psi(fg) = \Psi(g) \Psi(f)$. Since $L^\infty(K, \mathcal{B}(K))$ is a commutative algebra under pointwise multiplication, $fg = gf$, so $\Psi(g) \Psi(f) = \Psi(gf) = \Psi(fg)$. Reversing the roles of $f$ and $g$ in the displayed identity gives $\Psi(gf) = \Psi(f) \Psi(g)$ as well, so $\Psi(fg) = \Psi(f) \Psi(g) = \Psi(g) \Psi(f)$. Multiplicativity holds and the image $\Psi(L^\infty(K, \mathcal{B}(K)))$ is commutative. Hence $\Psi$ is a unital $*$-homomorphism. [/step] [step:Define $P(E) := \Psi(\mathbb{1}_E)$ and verify the resolution-of-the-identity axioms] For each $E \in \mathcal{B}(K)$, the indicator $\mathbb{1}_E \in L^\infty(K, \mathcal{B}(K))$. Set \begin{align*} P(E) := \Psi(\mathbb{1}_E) \in \mathcal{L}(H). \end{align*} We check the axioms (R1)-(R5) of a resolution of the identity (Step 1 of proof 2692). (R1) *$P(E)$ is an orthogonal projection.* $\mathbb{1}_E^2 = \mathbb{1}_E$, so by multiplicativity $P(E)^2 = P(E)$. $\overline{\mathbb{1}_E} = \mathbb{1}_E$, so by involution-preservation $P(E)^* = P(E)$. An idempotent self-adjoint operator is an orthogonal projection. (R2) *$P(\varnothing) = 0$, $P(K) = I$.* $\mathbb{1}_\varnothing = 0$, $\Psi(0) = 0$. $\mathbb{1}_K = 1$, $\Psi(1) = \gamma(1) = I$. (R3) *Multiplicativity $P(E \cap F) = P(E) P(F)$.* $\mathbb{1}_{E \cap F} = \mathbb{1}_E \mathbb{1}_F$, so multiplicativity of $\Psi$ gives $P(E \cap F) = \Psi(\mathbb{1}_E \mathbb{1}_F) = \Psi(\mathbb{1}_E) \Psi(\mathbb{1}_F) = P(E) P(F)$. (R4) *Disjoint additivity.* If $E \cap F = \varnothing$, $\mathbb{1}_{E \cup F} = \mathbb{1}_E + \mathbb{1}_F$, so by linearity $P(E \cup F) = P(E) + P(F)$. (R5) *$P_{x,y}$ is a regular complex Borel measure with $\|P_{x,y}\|_1 \le \|x\| \|y\|$.* By construction \begin{align*} P_{x,y}(E) = (P(E)x, y)_H = (\Psi(\mathbb{1}_E)x, y)_H = \int_K \mathbb{1}_E \, d\mu_{x,y} = \mu_{x,y}(E). \end{align*} So $P_{x,y} = \mu_{x,y}$ on Borel sets, and $\mu_{x,y}$ is a regular complex Borel measure with $\|\mu_{x,y}\|_1 \le \|x\| \|y\|$ from Step 2. Hence $P$ is a resolution of the identity over $K$. *The integration formula $\int_K \hat{T} \, dP = T$.* By the previous theorem [Integration Against $P$ — Existence and Properties](/theorems/2692), the integration map $\Phi_P: L^\infty(P) \to \mathcal{L}(H)$ is the unique isometric, unital $*$-homomorphism characterised by $(\Phi_P(f)x, y)_H = \int_K f \, dP_{x,y} = \int_K f \, d\mu_{x,y}$. But this is exactly the defining identity of our $\Psi$ in Step 3 (after restriction to $L^\infty(P)$, which equals $L^\infty(K, \mathcal{B}(K))$ modulo $P$-null sets). So $\Psi = \Phi_P$ on $L^\infty(P)$. In particular for $T \in A$, \begin{align*} \int_K \hat{T} \, dP = \Phi_P(\hat{T}) = \Psi(\hat{T}) = T. \end{align*} [/step] [step:Prove uniqueness of $P$] Suppose $P'$ is another resolution of the identity of $H$ over $K$ with $\int_K \hat{T} \, dP' = T$ for all $T \in A$. Let $\Phi_{P'}$ be the corresponding integration map. Then for all $T \in A$ and $x, y \in H$, \begin{align*} \int_K \hat{T} \, dP'_{x,y} = (T x, y)_H = \int_K \hat{T} \, dP_{x,y}. \end{align*} Both $P_{x,y}$ and $P'_{x,y}$ are regular complex Borel measures on $K$ (from axiom (R5) for both $P$ and $P'$), and the previous identity holds for every $\hat{T}$, $T \in A$. The Gelfand transform's image $\hat{A} = \{\hat{T} : T \in A\}$ equals $C(K)$ (Step 1 — $\Gamma$ is bijective). Hence the two regular complex Borel measures $P_{x,y}$ and $P'_{x,y}$ assign the same integral to every $f \in C(K)$. By RRT uniqueness, $P_{x,y} = P'_{x,y}$ for all $x, y \in H$. Hence $(P(E) x, y)_H = (P'(E) x, y)_H$ for all $E \in \mathcal{B}(K)$, $x, y \in H$, and $P(E) = P'(E)$. [/step] [step:Prove (i) $P(U) \ne 0$ for non-empty open $U$] Let $U \subseteq K$ be a non-empty open set. By **Urysohn's lemma** for the compact Hausdorff space $K$, there exists $f \in C(K)$ with $f: K \to [0, 1]$, $f \not\equiv 0$, and $\operatorname{supp}(f) \subseteq U$. We verify the hypotheses: $K$ is compact Hausdorff by Step 1, hence normal. Pick any point $p \in U$ (possible since $U \ne \varnothing$); since $K$ is Hausdorff, $\{p\}$ is closed. The two disjoint closed sets are $\{p\} \subseteq U$ and $K \setminus U$ (closed since $U$ is open). Urysohn produces a continuous $f: K \to [0,1]$ with $f \equiv 1$ on $\{p\}$ and $f \equiv 0$ on $K \setminus U$, in particular $f(p) = 1$ so $f \not\equiv 0$, and $\operatorname{supp}(f) \subseteq \overline{U}$. Refining via a slightly smaller open set inside $U$ (using normality again) yields $\operatorname{supp}(f) \subseteq U$. Pull back through $\gamma$. Set $T_0 := \gamma(\sqrt{f}) \in A$. Note $\sqrt{f} \in C(K)$ since $f \ge 0$. Then $T_0$ is Hermitian (since $\sqrt{f}$ is real-valued), and $T_0^2 = \gamma(\sqrt{f})^2 = \gamma(f)$ by multiplicativity. Suppose for contradiction $P(U) = 0$. Then for all $x \in H$, \begin{align*} 0 \le P_{x,x}(U) = (P(U) x, x)_H = 0. \end{align*} Since $P_{x,x}$ is a positive regular Borel measure (it equals $\mu_{x,x}$, and $\mu_{x,x}(E) = (P(E) x, x)_H = \|P(E) x\|^2 \ge 0$), and $P_{x,x}(U) = 0$ on the open set $U$, the integrand $f$ vanishes outside $U$ (since $\operatorname{supp}(f) \subseteq U$), so \begin{align*} \int_K f \, dP_{x,x} = \int_U f \, dP_{x,x} \le \|f\|_\infty \, P_{x,x}(U) = 0. \end{align*} Hence $\int_K f \, dP_{x,x} = 0$ for every $x \in H$. But for any $x \in H$, \begin{align*} \|T_0 x\|_H^2 = (T_0^* T_0 x, x)_H = (T_0^2 x, x)_H = (\gamma(f) x, x)_H = \int_K f \, d\mu_{x,x} = \int_K f \, dP_{x,x} = 0, \end{align*} so $T_0 x = 0$ for all $x \in H$, i.e.\ $T_0 = 0$. Applying $\Gamma$, $\sqrt{f} = \Gamma(T_0) = 0$. Squaring, $f = 0$. This contradicts $f \not\equiv 0$. Hence $P(U) \ne 0$. [/step] [step:Prove (ii) the commutant identity] Let $S \in \mathcal{L}(H)$. *$(\Rightarrow)$.* Suppose $S$ commutes with every $T \in A$. For $T \in A$ and $x, y \in H$, \begin{align*} (STx, y)_H = (T x, S^* y)_H = \int_K \hat{T} \, d\mu_{x, S^*y}, \\ (TSx, y)_H = \int_K \hat{T} \, d\mu_{Sx, y}. \end{align*} Equating (since $ST = TS$), the regular complex Borel measures $\mu_{x, S^*y}$ and $\mu_{Sx, y}$ assign the same integral to every $\hat{T} = f \in C(K) = \hat{A}$. By RRT uniqueness, $\mu_{x, S^*y} = \mu_{Sx, y}$. Evaluating on $E \in \mathcal{B}(K)$: $(P(E) x, S^* y)_H = (P(E) S x, y)_H$, i.e.\ $(SP(E)x, y)_H = (P(E) S x, y)_H$. Holding for all $x, y$, $SP(E) = P(E)S$. *$(\Leftarrow)$.* Suppose $S$ commutes with every $P(E)$. By property (iii) of [Integration Against $P$](/theorems/2692), $S$ commutes with $\Phi_P(f)$ for every $f \in L^\infty(P)$. In particular, for $T \in A$, $T = \int_K \hat{T} \, dP = \Phi_P(\hat{T})$, so $S$ commutes with $T$. This completes the proof of (ii) and of the theorem. [/step]