[proofplan] We first replace $T$ by its restriction to the cyclic reducing subspace generated by $h$, so that polynomial vectors $p(T,T^*)h$ are dense. The continuous functional calculus for the restricted normal operator produces a positive linear functional on continuous functions, and the [Riesz representation theorem](/theorems/218) represents this functional by a finite positive Borel measure. The map sending a polynomial vector $p(T,T^*)h$ to the function $p$ is then an isometry, because the same measure computes both norms. Density and Stone-Weierstrass extend this isometry to a unitary, and the intertwining relation follows by checking multiplication by $z$ on polynomial vectors. [/proofplan]

[step:Restrict the operator to the cyclic reducing subspace generated by $h$]Let \begin{align*} H_h := \overline{\{p(T,T^*)h : p \in \mathbb{C}[z,\bar z]\}}. \end{align*} For every $p \in \mathbb{C}[z,\bar z]$, both $z p$ and $\bar z p$ belong to $\mathbb{C}[z,\bar z]$. Therefore \begin{align*} T p(T,T^*)h = (z p)(T,T^*)h \end{align*} and \begin{align*} T^* p(T,T^*)h = (\bar z p)(T,T^*)h. \end{align*} It follows that $H_h$ is invariant under both $T$ and $T^*$, hence $H_h$ reduces $T$. Define the restricted operator \begin{align*} A := T|_{H_h} \in \mathcal{L}(H_h). \end{align*} This agrees with the operator $A$ named in the theorem statement. Since $T$ is normal and $H_h$ reduces $T$, the operator $A$ is normal on the [Hilbert space](/page/Hilbert%20Space) $H_h$. Let \begin{align*} K := \sigma(A) \subset \mathbb{C}. \end{align*} The spectrum of a bounded operator is compact, so $K$ is compact.[/step]

[guided]The theorem is cyclic, so the first task is to isolate the part of the Hilbert space actually generated by the vector $h$. We define \begin{align*} H_h := \overline{\{p(T,T^*)h : p \in \mathbb{C}[z,\bar z]\}}. \end{align*} This is the closed linear span of all vectors obtained by applying mixed polynomials in $T$ and $T^*$ to $h$. We need $H_h$ to reduce $T$, not merely to be invariant under $T$. The reason is that the restriction of a normal operator remains normal on a reducing subspace, and normality is needed for the continuous functional calculus. Let $p \in \mathbb{C}[z,\bar z]$. Since multiplication by $z$ and multiplication by $\bar z$ keep us inside $\mathbb{C}[z,\bar z]$, we have \begin{align*} T p(T,T^*)h = (z p)(T,T^*)h \end{align*} and \begin{align*} T^* p(T,T^*)h = (\bar z p)(T,T^*)h. \end{align*} Thus the dense subspace of polynomial vectors is invariant under both $T$ and $T^*$. Because $T$ and $T^*$ are bounded, this invariance passes to the closure $H_h$. Hence $H_h$ reduces $T$. Now define \begin{align*} A := T|_{H_h} \in \mathcal{L}(H_h). \end{align*} Since $H_h$ reduces $T$, the adjoint of $A$ is $A^*=T^*|_{H_h}$. Therefore \begin{align*} AA^* = T T^*|_{H_h} \end{align*} and \begin{align*} A^*A = T^* T|_{H_h}. \end{align*} Normality of $T$ gives $TT^*=T^*T$, so $AA^*=A^*A$. Hence $A$ is normal. Finally, set \begin{align*} K := \sigma(A). \end{align*} Because $A$ is bounded, its spectrum is a nonempty compact subset of $\mathbb{C}$.[/guided]

[step:Represent the cyclic vector state by a finite Borel measure]We use the continuous functional calculus for bounded normal operators. Since $A$ is normal with spectrum $K$, there is a unital $*$-homomorphism \begin{align*} \Phi : C(K) \to \mathcal{L}(H_h) \end{align*} such that $\Phi(\operatorname{id}_K)=A$, where \begin{align*} \operatorname{id}_K : K \to \mathbb{C}, \qquad \zeta \mapsto \zeta. \end{align*} For $f \in C(K)$, write $f(A):=\Phi(f)$. Define a linear functional \begin{align*} \Lambda_h : C(K) \to \mathbb{C}, \qquad f \mapsto (f(A)h,h)_H. \end{align*} If $f \in C(K)$ satisfies $f(\zeta)\ge 0$ for all $\zeta \in K$, then the positive square root $g:=f^{1/2}$ belongs to $C(K)$ and the $*$-homomorphism property gives \begin{align*} \Lambda_h(f) = (g(A)^*g(A)h,h)_H. \end{align*} Thus \begin{align*} \Lambda_h(f)=\|g(A)h\|_H^2 \ge 0. \end{align*} So $\Lambda_h$ is a positive linear functional. By the [Riesz representation theorem](/theorems/221) for positive linear functionals on $C(K)$, there exists a finite positive regular Borel measure $\mu_h$ on $K$ such that \begin{align*} \Lambda_h(f)=\int_K f(\zeta)\,d\mu_h(\zeta) \end{align*} for every $f \in C(K)$. In particular, \begin{align*} \mu_h(K)=\int_K 1\,d\mu_h(\zeta)=\Lambda_h(1)=\|h\|_H^2 < \infty. \end{align*}[/step]

[guided]The purpose of this step is to turn the cyclic vector $h$ into a measure. Since $A$ is a bounded normal operator on the complex Hilbert space $H_h$ and $K=\sigma(A)$, the continuous functional calculus gives a unital $*$-homomorphism \begin{align*} \Phi : C(K) \to \mathcal{L}(H_h) \end{align*} with $\Phi(\operatorname{id}_K)=A$, where $\operatorname{id}_K:K\to\mathbb{C}$ is the coordinate map $\zeta\mapsto \zeta$. For $f\in C(K)$, write $f(A):=\Phi(f)$. Define \begin{align*} \Lambda_h : C(K) \to \mathbb{C}, \qquad f \mapsto (f(A)h,h)_H. \end{align*} This is linear because $\Phi$ is linear and the Hilbert-space [inner product](/page/Inner%20Product) is linear in the first argument. We now verify positivity. If $f\in C(K)$ satisfies $f(\zeta)\ge 0$ for every $\zeta\in K$, then the pointwise square root $g:=f^{1/2}$ is again a [continuous function](/page/Continuous%20Function) on $K$. Since $\Phi$ is a $*$-homomorphism, $f(A)=g(A)^*g(A)$. Therefore \begin{align*} \Lambda_h(f)=(g(A)^*g(A)h,h)_H. \end{align*} By the defining property of the adjoint, \begin{align*} (g(A)^*g(A)h,h)_H=\|g(A)h\|_H^2\ge 0. \end{align*} Thus $\Lambda_h$ is a positive linear functional on $C(K)$. The [Riesz representation](/theorems/67) theorem for positive linear functionals on $C(K)$ applies because $K$ is compact Hausdorff. It gives a finite positive regular Borel measure $\mu_h$ on $K$ satisfying \begin{align*} \Lambda_h(f)=\int_K f(\zeta)\,d\mu_h(\zeta) \end{align*} for every $f\in C(K)$. The regularity of $\mu_h$ is part of the conclusion and will be used when continuous functions are identified as dense in the corresponding $L^2$ space. Applying the representation to the constant function $1_K:K\to\mathbb{C}$ gives \begin{align*} \mu_h(K)=\int_K 1\,d\mu_h(\zeta)=\Lambda_h(1_K)=\|h\|_H^2<\infty. \end{align*}[/guided]

[step:Define the polynomial-vector transform and prove it is isometric]For each $p \in \mathbb{C}[z,\bar z]$, let $p_K:K\to\mathbb C$ denote the continuous function obtained by restricting $p$ to $K$. Define the dense polynomial-vector subspace \begin{align*} \mathcal{P}_h := \{p(A,A^*)h : p \in \mathbb{C}[z,\bar z]\} \subset H_h. \end{align*} We define the polynomial-vector transform \begin{align*} U_0 : \mathcal{P}_h \to L^2(K,\mathcal{B}(K),\mu_h) \end{align*} first on polynomial representatives by \begin{align*} U_0(p(A,A^*)h) := p_K. \end{align*} This is well-defined as an element of $L^2(K,\mu_h)$. Indeed, suppose $p(A,A^*)h=q(A,A^*)h$. Put $r:=p-q \in \mathbb{C}[z,\bar z]$. Then $r(A,A^*)h=0$, and the functional calculus gives \begin{align*} \|r_K\|_{L^2(K,\mu_h)}^2 = \int_K |r_K(\zeta)|^2\,d\mu_h(\zeta). \end{align*} Since $|r_K|^2=(\overline r r)_K$, the representing identity gives \begin{align*} \int_K |r_K(\zeta)|^2\,d\mu_h(\zeta)=((\overline r r)(A,A^*)h,h)_H. \end{align*} Because the functional calculus is a $*$-homomorphism, \begin{align*} ((\overline r r)(A,A^*)h,h)_H=(r(A,A^*)^*r(A,A^*)h,h)_H. \end{align*} Therefore \begin{align*} \|r_K\|_{L^2(K,\mu_h)}^2=\|r(A,A^*)h\|_H^2=0. \end{align*} Hence $p_K=q_K$ in $L^2(K,\mu_h)$. The same computation with $r=p$ gives \begin{align*} \|U_0(p(A,A^*)h)\|_{L^2(K,\mu_h)}^2=\|p(A,A^*)h\|_H^2. \end{align*} Thus $U_0:\mathcal{P}_h\to L^2(K,\mu_h)$ is an isometry.[/step]

[guided]The natural definition is to send the vector produced by a polynomial to the polynomial itself. More precisely, define \begin{align*} U_0 : \mathcal{P}_h \to L^2(K,\mu_h) \end{align*} on polynomial representatives by \begin{align*} U_0(p(A,A^*)h) := p_K, \end{align*} where $p_K:K\to\mathbb C$ is the restriction of $p$ to $K$. The only possible problem is that the same vector in $H_h$ might have two different polynomial descriptions. We must prove that this ambiguity disappears in $L^2(K,\mu_h)$. Assume \begin{align*} p(A,A^*)h=q(A,A^*)h. \end{align*} Define \begin{align*} r:=p-q \in \mathbb{C}[z,\bar z]. \end{align*} Then \begin{align*} r(A,A^*)h=0. \end{align*} To show that $p_K$ and $q_K$ define the same element of $L^2(K,\mu_h)$, it is enough to show that $r_K$ has zero $L^2$ norm. By definition of the $L^2(K,\mu_h)$ norm, \begin{align*} \|r_K\|_{L^2(K,\mu_h)}^2 = \int_K |r_K(\zeta)|^2\,d\mu_h(\zeta). \end{align*} The function $|r_K|^2$ is the restriction to $K$ of the polynomial $\overline r r$ in $z$ and $\bar z$. By the definition of $\mu_h$ through the vector state, \begin{align*} \int_K |r_K(\zeta)|^2\,d\mu_h(\zeta)=((\overline r r)(A,A^*)h,h)_H. \end{align*} The functional calculus is a $*$-homomorphism, so multiplication of functions becomes multiplication of operators and complex conjugation becomes taking the adjoint. Hence \begin{align*} (\overline r r)(A,A^*)=r(A,A^*)^*r(A,A^*). \end{align*} Substituting this identity gives \begin{align*} \int_K |r_K(\zeta)|^2\,d\mu_h(\zeta)=(r(A,A^*)^*r(A,A^*)h,h)_H. \end{align*} Using the defining property of the Hilbert space adjoint, \begin{align*} (r(A,A^*)^*r(A,A^*)h,h)_H=\|r(A,A^*)h\|_H^2. \end{align*} Since $r(A,A^*)h=0$, we obtain \begin{align*} \|r_K\|_{L^2(K,\mu_h)}^2=0. \end{align*} Therefore $p_K=q_K$ in $L^2(K,\mu_h)$, so $U_0$ is well-defined. The same argument with $q=0$ shows that $U_0$ preserves norms: \begin{align*} \|U_0(p(A,A^*)h)\|_{L^2(K,\mu_h)}^2=\|p(A,A^*)h\|_H^2. \end{align*} Thus $U_0$ is an isometry from the polynomial-vector subspace into $L^2(K,\mu_h)$.[/guided]

[step:Extend the isometry to a unitary map onto $L^2(K,\mu_h)$]By definition of $H_h$, the subspace $\mathcal{P}_h$ is dense in $H_h$. Since $U_0$ is an isometry, it extends uniquely by continuity to an isometry \begin{align*} U_h : H_h \to L^2(K,\mathcal{B}(K),\mu_h). \end{align*} It remains to prove that the range is all of $L^2(K,\mathcal{B}(K),\mu_h)$. The image of $U_0$ contains the restrictions to $K$ of all polynomials in $z$ and $\bar z$. This algebra contains the constants, separates points of $K$, and is closed under complex conjugation. By the [Stone-Weierstrass theorem](/theorems/886), its uniform closure is $C(K)$. Since $K$ is compact and $\mu_h$ is a finite regular Borel measure on $K$, the standard density of continuous functions in $L^2$ for finite regular Borel measures gives that $C(K)$ is dense in $L^2(K,\mathcal{B}(K),\mu_h)$. Hence the range of the isometry $U_h$ is dense. The range of an isometry is closed, so the range is all of $L^2(K,\mathcal{B}(K),\mu_h)$. Therefore $U_h$ is unitary. For the constant polynomial $1$, we have $1(A,A^*)h=h$, so \begin{align*} U_h h=1_K. \end{align*}[/step]

[guided]The extension of $U_0$ is automatic once we know it is an isometry on a dense subspace. By definition, $\mathcal{P}_h$ is dense in $H_h$, and therefore $U_0$ has a unique continuous isometric extension \begin{align*} U_h : H_h \to L^2(K,\mathcal{B}(K),\mu_h). \end{align*} The remaining point is surjectivity. The image of $U_0$ contains every restriction to $K$ of a polynomial in $z$ and $\bar z$. These restricted polynomials form a complex subalgebra of $C(K)$, contain the constant functions, separate points of $K$ because the coordinate function $z$ separates distinct points of $K\subset\mathbb{C}$, and are closed under complex conjugation. The Stone-Weierstrass theorem applies to this subalgebra and implies that its uniform closure is all of $C(K)$. Next, because $K$ is compact and $\mu_h$ is a finite regular Borel measure on $K$, continuous functions are dense in $L^2(K,\mathcal{B}(K),\mu_h)$. Hence the range of $U_h$ contains a dense subspace of $L^2(K,\mathcal{B}(K),\mu_h)$. The range of an isometry is closed: if $U_h x_n\to y$ in $L^2(K,\mathcal{B}(K),\mu_h)$, then $(x_n)$ is Cauchy in $H_h$ because $U_h$ preserves norms, so $x_n\to x$ for some $x\in H_h$, and continuity gives $y=U_hx$. Thus the range is both dense and closed, hence it is the whole target space. Therefore $U_h$ is unitary. Finally, the constant polynomial $1$ satisfies $1(A,A^*)h=h$, and the definition of $U_h$ on polynomial vectors gives \begin{align*} U_h h=1_K. \end{align*}[/guided]

[step:Show that the restricted operator becomes multiplication by the coordinate function]Let $p \in \mathbb{C}[z,\bar z]$. On the polynomial vector $p(A,A^*)h \in \mathcal{P}_h$, we have \begin{align*} U_h A p(A,A^*)h = U_h((z p)(A,A^*)h). \end{align*} By the definition of $U_h$ on polynomial vectors, \begin{align*} U_h((z p)(A,A^*)h)=(z p)_K. \end{align*} Since $M_z p_K=(z p)_K$, this gives \begin{align*} U_h A p(A,A^*)h = M_z U_h p(A,A^*)h. \end{align*} Thus $U_h A=M_z U_h$ on the dense subspace $\mathcal{P}_h$. Both $U_h A$ and $M_z U_h$ are bounded maps from $H_h$ to $L^2(K,\mathcal{B}(K),\mu_h)$, so equality extends to all of $H_h$. Therefore \begin{align*} U_h A U_h^{-1}=M_z. \end{align*} Since $A=T|_{H_h}$, this is exactly \begin{align*} U_h \, T|_{H_h} \, U_h^{-1}=M_z. \end{align*}[/step]

[guided]We prove the intertwining relation first on the dense polynomial-vector subspace, where the computation is algebraic. Let $p\in\mathbb{C}[z,\bar z]$. Since $A$ corresponds to multiplication by the coordinate function in the polynomial functional calculus, we have \begin{align*} A p(A,A^*)h=(z p)(A,A^*)h. \end{align*} Applying $U_h$ and using its definition on polynomial vectors gives \begin{align*} U_h A p(A,A^*)h=U_h((z p)(A,A^*)h)=(z p)_K. \end{align*} On the other hand, $U_h p(A,A^*)h=p_K$, and by definition of the multiplication operator $M_z$, \begin{align*} M_z U_h p(A,A^*)h=M_zp_K=(z p)_K. \end{align*} Thus \begin{align*} U_h A p(A,A^*)h=M_z U_h p(A,A^*)h \end{align*} for every polynomial vector $p(A,A^*)h\in\mathcal{P}_h$. The subspace $\mathcal{P}_h$ is dense in $H_h$. The map $U_hA:H_h\to L^2(K,\mathcal{B}(K),\mu_h)$ is bounded because $U_h$ and $A$ are bounded. The map $M_zU_h:H_h\to L^2(K,\mathcal{B}(K),\mu_h)$ is bounded because $U_h$ is bounded and $M_z$ is bounded; boundedness of $M_z$ follows from compactness of $K$, since $|\zeta|$ is bounded on $K$. Therefore equality on the dense subspace extends by continuity to all of $H_h$: \begin{align*} U_h A=M_zU_h. \end{align*} Since $U_h$ is unitary, multiplying on the right by $U_h^{-1}$ gives \begin{align*} U_h A U_h^{-1}=M_z. \end{align*} Finally $A=T|_{H_h}$, so this is precisely \begin{align*} U_h \, T|_{H_h} \, U_h^{-1}=M_z. \end{align*}[/guided]

[step:Identify the compact set with the spectrum of the restricted operator] We chose \begin{align*} K=\sigma(A)=\sigma(T|_{H_h}). \end{align*} The preceding step shows that $A$ is unitarily equivalent to $M_z$ on $L^2(K,\mu_h)$. Therefore the compact set in the construction is already $\sigma(T|_{H_h})$, as claimed. Equivalently, if one starts from a larger compact set carrying the same measure representation, then the essential support of $\mu_h$ may be used instead. The spectrum of multiplication by the coordinate function on $L^2(K,\mu_h)$ is the essential range of the coordinate function, which is precisely the support of $\mu_h$ after replacing $K$ by that support. This replacement leaves $L^2(K,\mu_h)$ unchanged up to the canonical restriction unitary and gives the same multiplication operator. Hence $K$ may be taken to be $\sigma(T|_{H_h})$. This completes the construction of $\mu_h$, $U_h$, and the multiplication model for the cyclic normal operator. [/step]