[proofplan] The construction is the **Cauchy integral formula in the Banach algebra**: choose a contour $\gamma$ in $U$ that winds once around $\sigma_A(x)$ and zero times around $\mathbb{C} \setminus U$, and define \begin{align*} \Theta_x(f) := \frac{1}{2\pi i} \oint_\gamma f(\zeta)(\zeta \cdot 1 - x)^{-1}\, d\zeta, \end{align*} the integral being an $A$-valued contour integral of a continuous function (the integrand is well-defined since $\zeta \in U \setminus \sigma_A(x)$ along $\gamma$, hence $\zeta \cdot 1 - x \in G(A)$). Independence of $\gamma$ comes from the holomorphy of $\zeta \mapsto (\zeta \cdot 1 - x)^{-1}$ on $\mathbb{C} \setminus \sigma_A(x)$ (the resolvent identity) and the fact that any two such contours are homologous in $U \setminus \sigma_A(x)$. Continuity, linearity, and $\Theta_x(\operatorname{id}) = x$ are direct. Multiplicativity is the **resolvent equation calculation**: applying it to a product $fg$ and using $\frac{1}{(\zeta - x)(\eta - x)} = \frac{1}{\eta - \zeta}\left(\frac{1}{\zeta - x} - \frac{1}{\eta - x}\right)$ gives, after a Fubini and a Cauchy integral, $\Theta_x(fg) = \Theta_x(f)\Theta_x(g)$. The character formula $\varphi(\Theta_x(f)) = f(\varphi(x))$ follows by passing $\varphi$ through the integral (functionals commute with vector-valued contour integrals, theorem 2682) and applying the Cauchy integral formula for the scalar holomorphic function $f$ at the point $\varphi(x) \in \sigma_A(x) \subset U$. The spectral mapping identity $\sigma_A(\Theta_x(f)) = f(\sigma_A(x))$ then follows from the description $\sigma_A(z) = \{\varphi(z) : \varphi \in \Phi_A\}$ (theorem 2677). Uniqueness comes from density of polynomials in $\mathcal{O}(U)$ on compact sets (Runge's theorem, the case where $f$ is holomorphic in a fixed neighbourhood of a contour). [/proofplan] [step:Choose an admissible contour winding around $\sigma_A(x)$ inside $U$] The spectrum $\sigma_A(x)$ is a non-empty compact subset of $\mathbb{C}$ (theorem 2669), and $U \subseteq \mathbb{C}$ is open with $\sigma_A(x) \subset U$. We construct a finite collection of piecewise-$C^1$ closed curves $\gamma$ contained in $U \setminus \sigma_A(x)$ such that the winding number satisfies \begin{align*} n(\gamma, \lambda) = \begin{cases} 1 & \lambda \in \sigma_A(x), \\ 0 & \lambda \in \mathbb{C} \setminus U. \end{cases} \end{align*} Such a contour exists by a standard construction: cover $\sigma_A(x)$ by finitely many open squares of side $\delta < \frac{1}{2}\operatorname{dist}(\sigma_A(x), \mathbb{C} \setminus U)$ (positive by compactness of $\sigma_A(x)$ and openness of $U$), each square contained in $U$, and let $\gamma$ be the union of those edges of the squares that bound a square containing a point of $\sigma_A(x)$ on the inside but a point of $\mathbb{C} \setminus \sigma_A(x)$ on the outside, oriented counterclockwise around the squares they bound. Cancellation of internal edges produces a finite collection of closed curves with the required winding numbers. Call any such system of curves an **admissible contour for $(x, U)$** and denote it $\gamma$. [/step] [step:Define $\Theta_x(f)$ as a Banach-algebra-valued contour integral and verify well-definedness] For each $f \in \mathcal{O}(U)$, the function \begin{align*} F_f : U \setminus \sigma_A(x) &\to A \\ \zeta &\mapsto f(\zeta)\,(\zeta \cdot 1 - x)^{-1} \end{align*} is well-defined: for $\zeta \in U \setminus \sigma_A(x)$, $\zeta \notin \sigma_A(x)$ means $\zeta \cdot 1 - x \in G(A)$, so the resolvent $(\zeta \cdot 1 - x)^{-1} \in A$ exists. The map $\zeta \mapsto (\zeta \cdot 1 - x)^{-1}$ is holomorphic on $\rho_A(x) = \mathbb{C} \setminus \sigma_A(x)$ as an $A$-valued function: differentiability follows from the **resolvent equation** \begin{align*} (\zeta \cdot 1 - x)^{-1} - (\eta \cdot 1 - x)^{-1} = (\eta - \zeta)\,(\zeta \cdot 1 - x)^{-1}(\eta \cdot 1 - x)^{-1}, \end{align*} which gives $\frac{d}{d\zeta}(\zeta \cdot 1 - x)^{-1} = -(\zeta \cdot 1 - x)^{-2}$. Multiplying by the scalar holomorphic $f$ keeps $F_f$ holomorphic on $U \setminus \sigma_A(x)$, hence continuous on the image of $\gamma$. Define \begin{align*} \Theta_x(f) := \frac{1}{2\pi i} \oint_\gamma f(\zeta)(\zeta \cdot 1 - x)^{-1}\, d\zeta \in A, \end{align*} where the integral is the Banach-space-valued contour integral (a Bochner integral against arc-length, well-defined since $F_f$ is continuous on the compact image of $\gamma$). Independence from the choice of admissible contour: if $\gamma, \gamma'$ are two admissible contours for $(x, U)$, then their formal difference $\gamma - \gamma'$ is a cycle in $U \setminus \sigma_A(x)$ (closed: each individual contour is closed), and $n(\gamma - \gamma', \lambda) = n(\gamma, \lambda) - n(\gamma', \lambda) = 0$ for $\lambda \in \mathbb{C} \setminus (U \setminus \sigma_A(x))$ — at points of $\sigma_A(x)$ both equal $1$, and at points of $\mathbb{C} \setminus U$ both equal $0$. By the **Cauchy theorem for cycles homologous to zero in an open set** applied to the holomorphic $A$-valued integrand $F_f$ on $U \setminus \sigma_A(x)$, \begin{align*} \oint_\gamma F_f\, d\zeta - \oint_{\gamma'} F_f\, d\zeta = \oint_{\gamma - \gamma'} F_f\, d\zeta = 0. \end{align*} The Banach-valued Cauchy theorem reduces to the scalar case by composing with continuous linear functionals $\ell \in A^*$: for each $\ell$, $\zeta \mapsto \ell(F_f(\zeta))$ is scalar-holomorphic on $U \setminus \sigma_A(x)$, and the scalar Cauchy theorem for cycles homologous to zero gives $\oint_{\gamma - \gamma'} \ell(F_f)\, d\zeta = 0$; since $A^*$ separates points on $A$ (Hahn–Banach), the $A$-valued integral itself vanishes. Hence $\Theta_x(f)$ is independent of $\gamma$. [/step] [step:Verify $\Theta_x$ is linear, continuous, and unital] **Linearity**: For $f, g \in \mathcal{O}(U)$ and $\alpha, \beta \in \mathbb{C}$, the integral is linear in the integrand: \begin{align*} \Theta_x(\alpha f + \beta g) = \frac{1}{2\pi i}\oint_\gamma (\alpha f + \beta g)(\zeta)(\zeta \cdot 1 - x)^{-1}\, d\zeta = \alpha \Theta_x(f) + \beta \Theta_x(g). \end{align*} **Continuity**: Equip $\mathcal{O}(U)$ with the topology of uniform convergence on compact subsets. Fix the admissible contour $\gamma$ and let $K \subseteq U \setminus \sigma_A(x)$ be its image (compact). For $f \in \mathcal{O}(U)$, \begin{align*} \|\Theta_x(f)\|_A \le \frac{1}{2\pi}\,\ell(\gamma)\, \sup_{\zeta \in K} |f(\zeta)|\, \sup_{\zeta \in K}\|(\zeta \cdot 1 - x)^{-1}\|_A, \end{align*} where $\ell(\gamma)$ is the arc length of $\gamma$, and the supremum of the resolvent norm on $K$ is finite by continuity of the resolvent and compactness of $K$. So if $f_n \to f$ uniformly on $K$, then $\Theta_x(f_n) \to \Theta_x(f)$ in $A$. Hence $\Theta_x : \mathcal{O}(U) \to A$ is continuous. **Unital**: We compute $\Theta_x(1) = 1$. The constant function $1$ gives integrand $(\zeta \cdot 1 - x)^{-1}$. We expand using the Neumann series at infinity. Take a contour $\gamma_R$ that is a large circle of radius $R > \|x\|$ centred at $0$, traversed once counterclockwise; for $R$ large, the small admissible contour $\gamma$ and $\gamma_R$ together bound an annular region in $\rho_A(x)$ (since $\sigma_A(x)$ is contained in the interior of $\gamma$), so $\oint_\gamma = \oint_{\gamma_R}$ for the constant integrand $(\zeta \cdot 1 - x)^{-1}$. On $|\zeta| = R > \|x\|$: \begin{align*} (\zeta \cdot 1 - x)^{-1} = \zeta^{-1}\sum_{n=0}^\infty \zeta^{-n} x^n, \end{align*} converging absolutely and uniformly on $|\zeta| = R$. Integrating term by term using $\frac{1}{2\pi i}\oint_{|\zeta|=R} \zeta^{-k}\, d\zeta = \delta_{k,1}$: \begin{align*} \Theta_x(1) = \frac{1}{2\pi i}\oint_{|\zeta|=R}(\zeta \cdot 1 - x)^{-1}\, d\zeta = \sum_{n=0}^\infty \left(\frac{1}{2\pi i}\oint_{|\zeta|=R}\zeta^{-n-1}\, d\zeta\right) x^n = x^0 = 1. \end{align*} [/step] [step:Verify $\Theta_x(\operatorname{id}_U) = x$ by the same Neumann-series computation] Take $f = \operatorname{id}_U$, so $f(\zeta) = \zeta$. Repeat the calculation on the large circle $|\zeta| = R$: \begin{align*} \Theta_x(\operatorname{id}) = \frac{1}{2\pi i}\oint_{|\zeta|=R} \zeta(\zeta \cdot 1 - x)^{-1}\, d\zeta = \sum_{n=0}^\infty \left(\frac{1}{2\pi i}\oint_{|\zeta|=R}\zeta^{-n}\, d\zeta\right) x^n. \end{align*} The integral $\frac{1}{2\pi i}\oint_{|\zeta|=R}\zeta^{-n}\, d\zeta$ equals $1$ if $n = 1$ and $0$ otherwise. Hence $\Theta_x(\operatorname{id}) = x$. [/step] [step:Establish multiplicativity via the resolvent identity and a double-contour argument] Let $f, g \in \mathcal{O}(U)$. Choose two admissible contours $\gamma$ and $\gamma'$ for $(x, U)$ with $\gamma'$ in the interior of $\gamma$ (i.e.\ $\gamma'$ winds once around $\sigma_A(x)$ and lies inside the region bounded by $\gamma$ in $U \setminus \sigma_A(x)$). Both contours represent $\Theta_x$ — we use $\gamma$ in the integral defining $\Theta_x(f)$ and $\gamma'$ in the integral defining $\Theta_x(g)$: \begin{align*} \Theta_x(f)\Theta_x(g) = \frac{1}{(2\pi i)^2}\oint_\gamma \oint_{\gamma'} f(\zeta)g(\eta)(\zeta \cdot 1 - x)^{-1}(\eta \cdot 1 - x)^{-1}\, d\eta\, d\zeta. \end{align*} The product of resolvents factors via the **resolvent equation**: for $\zeta \neq \eta$ both in $\rho_A(x)$, \begin{align*} (\zeta \cdot 1 - x)^{-1}(\eta \cdot 1 - x)^{-1} = \frac{1}{\eta - \zeta}\left[(\zeta \cdot 1 - x)^{-1} - (\eta \cdot 1 - x)^{-1}\right]. \end{align*} (Multiply both sides by $(\zeta \cdot 1 - x)(\eta \cdot 1 - x)$ to verify directly.) Substituting, \begin{align*} \Theta_x(f)\Theta_x(g) &= \frac{1}{(2\pi i)^2}\oint_\gamma \oint_{\gamma'}\frac{f(\zeta)g(\eta)}{\eta - \zeta}\bigl[(\zeta \cdot 1 - x)^{-1} - (\eta \cdot 1 - x)^{-1}\bigr]\, d\eta\, d\zeta\\ &= \underbrace{\frac{1}{2\pi i}\oint_\gamma f(\zeta)(\zeta \cdot 1 - x)^{-1}\left(\frac{1}{2\pi i}\oint_{\gamma'}\frac{g(\eta)}{\eta - \zeta}\, d\eta\right) d\zeta}_{(\mathrm{I})}\\ &\quad - \underbrace{\frac{1}{2\pi i}\oint_{\gamma'}g(\eta)(\eta \cdot 1 - x)^{-1}\left(\frac{1}{2\pi i}\oint_\gamma \frac{f(\zeta)}{\eta - \zeta}\, d\zeta\right) d\eta}_{(\mathrm{II})}. \end{align*} The interchange of order of integration is Fubini for vector-valued continuous integrands on compact contours. Evaluate the inner scalar integrals using the Cauchy integral formula. For $(\mathrm{I})$: $\zeta$ lies on $\gamma$, which is **outside** $\gamma'$ in $U$, so $\zeta$ is in the exterior of the cycle $\gamma'$. Hence $n(\gamma', \zeta) = 0$ and the **Cauchy integral formula for the exterior** \begin{align*} \frac{1}{2\pi i}\oint_{\gamma'}\frac{g(\eta)}{\eta - \zeta}\, d\eta = n(\gamma', \zeta)\, g(\zeta) = 0 \end{align*} gives that the inner integral in $(\mathrm{I})$ is $0$. Hence $(\mathrm{I}) = 0$. For $(\mathrm{II})$: $\eta$ lies on $\gamma'$, which is **inside** $\gamma$, so $n(\gamma, \eta) = 1$. The Cauchy integral formula gives \begin{align*} \frac{1}{2\pi i}\oint_\gamma \frac{f(\zeta)}{\zeta - \eta}\, d\zeta = n(\gamma, \eta)\, f(\eta) = f(\eta). \end{align*} Note the sign: $\frac{1}{\eta - \zeta} = -\frac{1}{\zeta - \eta}$, so the inner integral in $(\mathrm{II})$ becomes $-f(\eta)$. Therefore \begin{align*} (\mathrm{II}) = \frac{1}{2\pi i}\oint_{\gamma'} g(\eta)(\eta \cdot 1 - x)^{-1}\bigl(-f(\eta)\bigr)\, d\eta = -\Theta_x(fg). \end{align*} Combining $(\mathrm{I}) - (\mathrm{II}) = 0 - (-\Theta_x(fg)) = \Theta_x(fg)$: \begin{align*} \Theta_x(f)\Theta_x(g) = \Theta_x(fg). \end{align*} [guided] The multiplicativity calculation is the heart of the construction — let us go through it slowly. **Setting up the double integral.** We start with $\Theta_x(f)\Theta_x(g)$ as the product of two single contour integrals. The clever choice is to use **different** admissible contours $\gamma$ and $\gamma'$, with $\gamma'$ inside $\gamma$. Both wind once around $\sigma_A(x)$, but they live at different "radii" in $U \setminus \sigma_A(x)$. \begin{align*} \Theta_x(f)\Theta_x(g) = \frac{1}{(2\pi i)^2}\oint_\gamma \oint_{\gamma'} f(\zeta)g(\eta)(\zeta \cdot 1 - x)^{-1}(\eta \cdot 1 - x)^{-1}\, d\eta\, d\zeta. \end{align*} **The resolvent trick.** The product of resolvents at distinct points satisfies a partial-fraction-like identity: \begin{align*} (\zeta \cdot 1 - x)^{-1}(\eta \cdot 1 - x)^{-1} = \frac{1}{\eta - \zeta}\bigl[(\zeta \cdot 1 - x)^{-1} - (\eta \cdot 1 - x)^{-1}\bigr]. \end{align*} Why is this true? Multiply both sides on the left by $(\zeta \cdot 1 - x)$ and on the right by $(\eta \cdot 1 - x)$: \begin{align*} 1 &= \frac{1}{\eta - \zeta}\bigl[(\eta \cdot 1 - x) - (\zeta \cdot 1 - x)\bigr] = \frac{1}{\eta - \zeta}(\eta - \zeta)\cdot 1 = 1. \end{align*} Confirmed. This identity uses commutativity of $A$ in the form $(\zeta \cdot 1 - x)^{-1}(\eta \cdot 1 - x) = (\eta \cdot 1 - x)(\zeta \cdot 1 - x)^{-1}$, which holds because $1$ commutes with everything and powers/inverses of $x$ commute among themselves. **Splitting the double integral.** Substituting the resolvent identity gives a sum of two terms (call them $(\mathrm{I})$ and $-(\mathrm{II})$). The key observation is that each is **iterated**: the inner integral is now scalar (since the resolvent occurring is independent of the inner variable), and the outer integral is $A$-valued. **Inner integral evaluation.** We use the scalar Cauchy integral formula: \begin{align*} \frac{1}{2\pi i}\oint_C \frac{h(w)}{w - w_0}\, dw = n(C, w_0)\, h(w_0) \end{align*} for $h$ holomorphic in a neighbourhood of $C \cup \mathrm{int}(C)$ and $w_0$ inside (winding $1$) or outside (winding $0$). In $(\mathrm{I})$ we integrate $\frac{g(\eta)}{\eta - \zeta}$ over $\gamma'$, where $\zeta$ is on $\gamma$ — **outside** $\gamma'$. So winding number $0$, integral is $0$. In $(\mathrm{II})$ we integrate $\frac{f(\zeta)}{\eta - \zeta} = -\frac{f(\zeta)}{\zeta - \eta}$ over $\gamma$, where $\eta$ is on $\gamma'$ — **inside** $\gamma$. So winding number $1$, integral is $-f(\eta)$. **Closing the calculation.** $(\mathrm{II})$ collects to $\frac{1}{2\pi i}\oint_{\gamma'}g(\eta)(\eta \cdot 1 - x)^{-1}(-f(\eta))\, d\eta = -\Theta_x(fg)$ (using $\gamma'$ as the admissible contour for $\Theta_x(fg)$, valid since $fg \in \mathcal{O}(U)$). The minus signs combine: $(\mathrm{I}) - (\mathrm{II}) = 0 + \Theta_x(fg) = \Theta_x(fg)$. Hence $\Theta_x(f)\Theta_x(g) = \Theta_x(fg)$. **Why this is striking.** The proof converts an algebraic identity in $A$ ($\Theta_x$ is multiplicative) into two scalar Cauchy integral evaluations. The non-commutativity of $A$ never enters because we are in a commutative algebra; commutativity is precisely what makes the resolvent identity work. (For non-commutative $A$, the calculation goes through provided $x$ is fixed and we work with the commutative subalgebra it generates — which is the version used to extend HFC to general unital Banach algebras.) [/guided] [/step] [step:Establish the character formula $\varphi(\Theta_x(f)) = f(\varphi(x))$] Fix $\varphi \in \Phi_A$. By [Characters are Continuous](/theorems/2675), $\varphi : A \to \mathbb{C}$ is a bounded linear functional with $\|\varphi\|_{A^*} \le 1$. Continuous linear functionals commute with vector-valued contour integrals (a standard property of the Bochner integral, used in this database via the **Functionals Commute with Vector-Valued Integrals** result). Applying $\varphi$ to the defining contour integral: \begin{align*} \varphi(\Theta_x(f)) &= \varphi\!\left(\frac{1}{2\pi i}\oint_\gamma f(\zeta)(\zeta \cdot 1 - x)^{-1}\, d\zeta\right)\\ &= \frac{1}{2\pi i}\oint_\gamma f(\zeta)\,\varphi\bigl((\zeta \cdot 1 - x)^{-1}\bigr)\, d\zeta. \end{align*} Apply $\varphi$ multiplicatively to the resolvent: since $\varphi$ is multiplicative and $\varphi(1) = 1$, \begin{align*} \varphi\bigl((\zeta \cdot 1 - x)^{-1}\bigr) = \varphi(\zeta \cdot 1 - x)^{-1} = (\zeta - \varphi(x))^{-1} \end{align*} (valid because $\varphi(\zeta \cdot 1 - x) = \zeta - \varphi(x) \ne 0$ for $\zeta$ on $\gamma \subset \rho_A(x)$, since $\varphi(x) \in \sigma_A(x)$ by [Spectrum via Characters](/theorems/2677) (ii)). Thus \begin{align*} \varphi(\Theta_x(f)) = \frac{1}{2\pi i}\oint_\gamma \frac{f(\zeta)}{\zeta - \varphi(x)}\, d\zeta. \end{align*} By the scalar Cauchy integral formula applied to the holomorphic function $f \in \mathcal{O}(U)$ at the point $\varphi(x) \in \sigma_A(x) \subset U$, and using $n(\gamma, \varphi(x)) = 1$: \begin{align*} \frac{1}{2\pi i}\oint_\gamma \frac{f(\zeta)}{\zeta - \varphi(x)}\, d\zeta = f(\varphi(x)). \end{align*} Hence $\varphi(\Theta_x(f)) = f(\varphi(x))$ for all $\varphi \in \Phi_A$ and $f \in \mathcal{O}(U)$. [/step] [step:Establish the spectral mapping $\sigma_A(\Theta_x(f)) = f(\sigma_A(x))$] By [Spectrum via Characters](/theorems/2677) (ii) applied to $\Theta_x(f) \in A$: \begin{align*} \sigma_A(\Theta_x(f)) = \{\varphi(\Theta_x(f)) : \varphi \in \Phi_A\}. \end{align*} By the character formula from Step 6, this becomes \begin{align*} \sigma_A(\Theta_x(f)) = \{f(\varphi(x)) : \varphi \in \Phi_A\} = f\bigl(\{\varphi(x) : \varphi \in \Phi_A\}\bigr). \end{align*} Applying [Spectrum via Characters](/theorems/2677) (ii) again, this time to $x$: \begin{align*} \{\varphi(x) : \varphi \in \Phi_A\} = \sigma_A(x). \end{align*} Combining, \begin{align*} \sigma_A(\Theta_x(f)) = f(\sigma_A(x)). \end{align*} [/step] [step:Establish uniqueness of $\Theta_x$] Suppose $\Theta'_x : \mathcal{O}(U) \to A$ is another continuous unital algebra homomorphism with $\Theta'_x(\operatorname{id}) = x$. We show $\Theta'_x = \Theta_x$. **Polynomials.** For any polynomial $p(\zeta) = \sum_{k=0}^n c_k \zeta^k$, both $\Theta_x$ and $\Theta'_x$ are unital homomorphisms with $\Theta_x(\operatorname{id}) = \Theta'_x(\operatorname{id}) = x$. Linearity, multiplicativity, and unitality determine \begin{align*} \Theta_x(p) = \sum_{k=0}^n c_k x^k = \Theta'_x(p). \end{align*} **Rational functions with poles outside $U$.** For $\zeta_0 \notin U$, the resolvent $(\operatorname{id} - \zeta_0 \cdot 1)^{-1}$ exists in $\mathcal{O}(U)$ as $\zeta \mapsto \frac{1}{\zeta - \zeta_0}$. Both $\Theta_x$ and $\Theta'_x$ send this to $(x - \zeta_0 \cdot 1)^{-1}$ (which exists in $A$ because $\zeta_0 \notin U \supseteq \sigma_A(x)$, so $\zeta_0 \notin \sigma_A(x)$, so $x - \zeta_0 \cdot 1 \in G(A)$): apply each homomorphism to the equation $(\operatorname{id} - \zeta_0)\cdot \frac{1}{\operatorname{id} - \zeta_0} = 1$ in $\mathcal{O}(U)$ to conclude $(x - \zeta_0 \cdot 1)\cdot \Theta_x\left(\frac{1}{\operatorname{id} - \zeta_0}\right) = 1$, hence $\Theta_x\left(\frac{1}{\operatorname{id} - \zeta_0}\right) = (x - \zeta_0 \cdot 1)^{-1}$, and likewise for $\Theta'_x$. By linearity, $\Theta_x$ and $\Theta'_x$ agree on the algebra $\mathcal{R}_U$ of rational functions with poles in $\mathbb{C} \setminus U$. **Density and continuity.** By [Runge's Approximation Theorem](/theorems/2681), for every $f \in \mathcal{O}(U)$ and every compact $K \subseteq U$, there exists a sequence $(r_n)$ of rational functions with poles in $\mathbb{C} \setminus U$ such that $r_n \to f$ uniformly on $K$. Choose $K$ to be a compact set containing $\gamma$ and a neighbourhood of $\sigma_A(x)$ in $U$ (for instance, $K = \{\zeta \in U : \operatorname{dist}(\zeta, \sigma_A(x)) \le \delta\} \cup \mathrm{image}(\gamma)$ for $\delta < \operatorname{dist}(\sigma_A(x), \mathbb{C} \setminus U)/2$); then $r_n \to f$ uniformly on $K$ implies $r_n \to f$ in the topology of $\mathcal{O}(U)$ on a neighbourhood of $K$, in particular as $n \to \infty$ in any seminorm $\sup_{\zeta \in K'} | \cdot |$ for $K' \subseteq K$ compact. By continuity of $\Theta_x$ and $\Theta'_x$, \begin{align*} \Theta_x(f) = \lim_n \Theta_x(r_n) = \lim_n \Theta'_x(r_n) = \Theta'_x(f). \end{align*} Hence $\Theta_x = \Theta'_x$. [/step] <!-- illustration-needed: the admissible contour for the holomorphic functional calculus — show the open set U as a region of the complex plane, the compact spectrum sigma_A(x) inside it as a connected blob, and the multi-component contour gamma winding once around sigma_A(x) and lying in U \\ sigma_A(x); for the multiplicativity step, also show two nested contours gamma and gamma' with gamma' in the interior of gamma -->