[proofplan] The four properties are established in order. Linearity is immediate from the linearity of the integrand in $f$; continuity follows from a uniform contour-length estimate giving $\|\Theta_x(f)\|_A \le C\, \|f\|_{[\Gamma]}$ with $C$ independent of $f$, after which the seminorm criterion for continuity into a Banach space yields continuity of $\Theta_x$ from the lcs $\mathcal{O}(U)$ to $A$. For rational functions, we first show $\Theta_x(\mathbb{1}_U) = 1_A$ by deforming $\Gamma$ to a large circle $|z| = R$ (justified by the [Vector-Valued Cauchy Theorem](/theorems/2683)) and computing the expansion of the resolvent as a Neumann series; the case of a general rational function then follows from a polynomial identity for $r(z) - r(x)$ and one further application of theorem 2683. Property 3 reduces to the scalar Cauchy integral formula by pulling $\varphi \in \Phi_A$ inside the integral via [Functionals Commute with Vector-Valued Integrals](/theorems/2682) and using $\varphi((z \cdot 1 - x)^{-1}) = (z - \varphi(x))^{-1}$ together with $\varphi(x) \in \sigma_A(x)$ from the [Spectrum via Characters](/theorems/2677) theorem. Property 4 is then immediate by applying [Spectrum via Characters](/theorems/2677) to $\Theta_x(f)$ and using Property 3. [/proofplan] [step:Establish linearity and continuity of $\Theta_x$] **Linearity.** For $f, g \in \mathcal{O}(U)$ and scalars $\lambda, \mu \in \mathbb{C}$, the integrand of $\Theta_x(\lambda f + \mu g)$ is \begin{align*} (\lambda f(z) + \mu g(z))\, (z \cdot 1 - x)^{-1} = \lambda\, f(z)(z \cdot 1 - x)^{-1} + \mu\, g(z)(z \cdot 1 - x)^{-1}, \end{align*} and the contour integral $\int_\Gamma \cdot\, dz$ is linear in the integrand (it is built from vector-valued Riemann integrals, which are linear). Hence $\Theta_x(\lambda f + \mu g) = \lambda \Theta_x(f) + \mu \Theta_x(g)$. **Continuity.** Let $K_0 := [\Gamma]$ denote the (compact) image of the cycle $\Gamma$ in $\mathbb{C}$. The map $z \mapsto (z \cdot 1 - x)^{-1}$ is continuous on the open set of invertible elements $G(A)$ (continuity of inversion in a Banach algebra), and on $K_0 \subset U \setminus \sigma_A(x)$ each $z \cdot 1 - x$ is invertible (by the definition of the spectrum), so $z \mapsto (z \cdot 1 - x)^{-1}$ is continuous on $K_0$, hence bounded: \begin{align*} M_x := \sup_{z \in K_0} \|(z \cdot 1 - x)^{-1}\|_A < \infty. \end{align*} For any $f \in \mathcal{O}(U)$, the norm bound from [Functionals Commute with Vector-Valued Integrals](/theorems/2682) (specifically the inequality $\|\int_a^b h(t)\, d\mathcal{L}^1(t)\| \le \int_a^b \|h(t)\|\, d\mathcal{L}^1(t)$ applied to each smooth piece of $\Gamma$ parametrised by $\gamma : [a, b] \to U$) gives \begin{align*} \left\|\Theta_x(f)\right\|_A &= \frac{1}{2\pi}\left\|\int_\Gamma f(z)\, (z \cdot 1 - x)^{-1}\, dz\right\|_A \\ &\le \frac{1}{2\pi}\int_\Gamma \|f(z)\, (z \cdot 1 - x)^{-1}\|_A\, |dz| \\ &\le \frac{1}{2\pi} \cdot \ell(\Gamma) \cdot \sup_{z \in K_0} |f(z)| \cdot M_x \\ &= C\, \|f\|_{K_0}, \end{align*} where $C := \frac{\ell(\Gamma) M_x}{2\pi}$, $\ell(\Gamma)$ is the total arc length of $\Gamma$, and $\|f\|_{K_0} = \sup_{z \in K_0}|f(z)|$ is one of the seminorms defining the topology of $\mathcal{O}(U)$ (since $K_0 \subset U$ is compact). Hence $\Theta_x$ is bounded with respect to the seminorm $\|\cdot\|_{K_0}$ on $\mathcal{O}(U)$ and the norm $\|\cdot\|_A$ on $A$. By [Continuity of Linear Maps Between Locally Convex Spaces](/theorems/2636) — applied with source the lcs $\mathcal{O}(U)$ (with seminorm system $\{\|\cdot\|_K : K \subset U \text{ compact}\}$) and target the Banach space $A$ — the bound $\|\Theta_x(f)\|_A \le C\|f\|_{K_0}$ implies continuity of $\Theta_x$. Theorem 2636 requires a single seminorm bound dominating the target norm; we have produced exactly such a bound. [guided] We split the verification into two pieces. **(i) Linearity is bookkeeping.** The integrand is linear in $f$: \begin{align*} (\lambda f + \mu g)(z)\, (z \cdot 1 - x)^{-1} = \lambda\, f(z)(z\cdot 1 - x)^{-1} + \mu\, g(z)(z \cdot 1 - x)^{-1}, \end{align*} and the cycle integral is a finite sum of vector-valued Riemann integrals on smooth pieces, each of which is linear in its integrand by definition. So linearity descends from the integrand to $\Theta_x$. **(ii) Continuity reduces to a single seminorm bound.** The target $A$ is a Banach space, and the source $\mathcal{O}(U)$ is the locally convex space of holomorphic functions on $U$ with the topology of locally uniform convergence — its topology is generated by the seminorm system $\{\|\cdot\|_K : K \subset U\text{ compact}\}$. By [Continuity of Linear Maps Between Locally Convex Spaces](/theorems/2636), a linear map $T : \mathcal{O}(U) \to A$ is continuous iff there exist a compact $K \subset U$ and a constant $C > 0$ with $\|T(f)\|_A \le C\|f\|_K$ for all $f \in \mathcal{O}(U)$. We produce such a $K$ and $C$. Take $K := K_0 = [\Gamma] \subset U$ (compact, as the image of the parametrised cycle). The two ingredients to bound are: (a) $\sup_{z \in K_0} |f(z)| = \|f\|_{K_0}$, and (b) $\sup_{z \in K_0}\|(z \cdot 1 - x)^{-1}\|_A$. The latter is finite — call it $M_x$ — because the resolvent $z \mapsto (z\cdot 1 - x)^{-1}$ is continuous on the resolvent set $\rho_A(x) \supset K_0$ (continuity of the inversion map on the open set of invertible elements is a standard fact in Banach algebras), and a continuous function on a compact set is bounded. Why does $K_0 \subset \rho_A(x)$? Because the cycle $\Gamma$ was chosen with $K_0 = [\Gamma] \subset U \setminus \sigma_A(x)$, i.e.\ avoiding the spectrum, which is the resolvent set $\rho_A(x)$. Now apply the norm bound for vector-valued integrals from [Functionals Commute with Vector-Valued Integrals](/theorems/2682): on each smooth piece $\gamma : [a, b] \to U$ of $\Gamma$, \begin{align*} \left\|\int_a^b f(\gamma(t))\, (\gamma(t)\cdot 1 - x)^{-1}\, \gamma'(t)\, d\mathcal{L}^1(t)\right\|_A \le \int_a^b |f(\gamma(t))|\, \|(\gamma(t) \cdot 1 - x)^{-1}\|_A\, |\gamma'(t)|\, d\mathcal{L}^1(t). \end{align*} Bound the integrand by $\|f\|_{K_0} \cdot M_x \cdot |\gamma'(t)|$ and integrate; sum over the smooth pieces with their integer multiplicities; divide by $2\pi$. The result is \begin{align*} \|\Theta_x(f)\|_A \le \frac{\ell(\Gamma) M_x}{2\pi}\, \|f\|_{K_0} = C\, \|f\|_{K_0}, \end{align*} as required. Theorem 2636 then upgrades this single-seminorm bound to continuity of $\Theta_x$. [/guided] [/step] [step:Compute $\Theta_x(\mathbb{1}_U) = 1_A$ by deforming the contour to a large circle] Choose any radius $R > \|x\|_A$ large enough that $[\Gamma] \subset \{z \in \mathbb{C} : |z| < R\}$. Let $\Gamma_R$ denote the circle $|z| = R$ traversed counterclockwise once. We claim \begin{align*} \frac{1}{2\pi i}\int_\Gamma (z \cdot 1 - x)^{-1}\, dz = \frac{1}{2\pi i}\int_{\Gamma_R} (z \cdot 1 - x)^{-1}\, dz. \tag{$\dagger$} \end{align*} To prove ($\dagger$), consider the Banach-space-valued analytic map \begin{align*} F : \mathbb{C} \setminus \sigma_A(x) &\to A \\ z &\mapsto (z \cdot 1 - x)^{-1}. \end{align*} Analyticity of $F$ on the resolvent set $\rho_A(x) = \mathbb{C} \setminus \sigma_A(x)$ is the standard fact that the resolvent map is $A$-valued holomorphic. Set $V := \{z \in \mathbb{C} : |z| < R + 1\} \setminus \sigma_A(x) \supset [\Gamma] \cup [\Gamma_R]$, an open set on which $F$ is analytic. Form the cycle $\Gamma - \Gamma_R$ in $V$. Its winding numbers satisfy: - For $\omega \in \sigma_A(x)$ with $|\omega| < R$: $n(\Gamma, \omega) = 1$ (by hypothesis on $\Gamma$) and $n(\Gamma_R, \omega) = 1$ (since $|\omega| < R$ and $\Gamma_R$ is the standard counterclockwise circle of radius $R$), so $n(\Gamma - \Gamma_R, \omega) = 0$. - For $\omega \in \sigma_A(x)$ with $|\omega| \ge R$: not possible, because $\sigma_A(x) \subset \{|z| \le \|x\|\} \subset \{|z| < R\}$ by choice of $R$. - For $\omega \in \mathbb{C} \setminus V$ — which means $|\omega| \ge R + 1$ — both $n(\Gamma, \omega) = 0$ (since $[\Gamma] \subset \{|z| < R\}$) and $n(\Gamma_R, \omega) = 0$ (since $\Gamma_R$ has winding number $0$ outside its enclosed disk $\{|z| < R\}$). So $n(\Gamma - \Gamma_R, \omega) = 0$. We have verified that $n(\Gamma - \Gamma_R, \omega) = 0$ for every $\omega \notin V$. Apply the [Vector-Valued Cauchy Theorem](/theorems/2683) to $F$ on $V$ with the cycle $\Gamma - \Gamma_R$. Theorem 2683 requires (a) $V \subset \mathbb{C}$ open (verified), (b) $F : V \to A$ analytic (verified), and (c) $n(\Gamma - \Gamma_R, \omega) = 0$ for all $\omega \notin V$ (verified). It gives \begin{align*} \int_{\Gamma - \Gamma_R} F(z)\, dz = 0_A, \end{align*} which is ($\dagger$). Now compute the right-hand side of ($\dagger$). For $|z| = R > \|x\|_A$, the Neumann series \begin{align*} (z \cdot 1 - x)^{-1} = \frac{1}{z}\, \left(1 - \frac{x}{z}\right)^{-1} = \frac{1}{z}\sum_{n=0}^\infty \frac{x^n}{z^n} = \sum_{n=0}^\infty x^n\, z^{-(n+1)} \end{align*} converges absolutely in $A$ uniformly in $z \in \Gamma_R$, since $\|x^n z^{-(n+1)}\|_A \le \|x\|^n R^{-(n+1)}$ and $\|x\|/R < 1$. The uniform convergence in $A$ allows us to interchange the sum and the (vector-valued) integral over $\Gamma_R$: \begin{align*} \frac{1}{2\pi i}\int_{\Gamma_R} (z\cdot 1 - x)^{-1}\, dz = \sum_{n=0}^\infty x^n \cdot \frac{1}{2\pi i}\int_{\Gamma_R} z^{-(n+1)}\, dz = \sum_{n=0}^\infty x^n \cdot \delta_{n,0} = x^0 = 1_A. \end{align*} The scalar identity $\frac{1}{2\pi i}\int_{\Gamma_R} z^{-(n+1)}\, dz = \delta_{n,0}$ is the standard residue computation. Combining with ($\dagger$), \begin{align*} \Theta_x(\mathbb{1}_U) = \frac{1}{2\pi i}\int_\Gamma (z\cdot 1 - x)^{-1}\, dz = 1_A. \end{align*} [/step] [step:Extend to general rational functions $r$ without poles in $U$] Let $r = p/q$ with $p, q \in \mathbb{C}[z]$, $q$ having no zeros in $U$. We must show $\Theta_x(r) = r(x)$, where $r(x) := p(x)\, q(x)^{-1}$. **Defining $r(x)$.** Since $q$ has no zeros in $U$ and $\sigma_A(x) \subset U$, $q(\sigma_A(x))$ does not contain $0$. By the [Spectral Mapping Theorem for Polynomials](/theorems/2671) applied to the polynomial $q$ and the element $x \in A$, $\sigma_A(q(x)) = q(\sigma_A(x))$. Hence $0 \notin \sigma_A(q(x))$, so $q(x) \in A$ is invertible, and $r(x) := p(x)\, q(x)^{-1}$ is a well-defined element of $A$. The hypothesis required by theorem 2671 — that $q$ is a polynomial in a single variable with complex coefficients and $x \in A$ — is satisfied. **Polynomial identity.** Consider the bivariate polynomial \begin{align*} P(z, w) := p(z)\, q(w) - p(w)\, q(z) \in \mathbb{C}[z, w]. \end{align*} For each fixed $w$, the polynomial $z \mapsto P(z, w)$ vanishes at $z = w$ (since $P(w, w) = 0$), so $z - w$ divides $P(z, w)$ as polynomials in $z$ over $\mathbb{C}[w]$. Therefore there exists a polynomial $S \in \mathbb{C}[z, w]$ with \begin{align*} p(z)\, q(w) - p(w)\, q(z) = (z - w)\, S(z, w). \end{align*} Setting $w = x$ (substituting the algebra element into the polynomial $S$, which is well-defined since $A$ is a commutative algebra and $S$ is a polynomial in two commuting variables), we obtain in $A$ \begin{align*} p(z)\, q(x) - p(x)\, q(z) = (z\cdot 1 - x)\, S(z, x). \tag{$\ddagger$} \end{align*} Multiply both sides of ($\ddagger$) on the left by $q(z)^{-1} q(x)^{-1}$ (valid because $q(z) \ne 0$ for $z \in [\Gamma] \subset U$ and $q(x)$ is invertible): \begin{align*} \frac{p(z)}{q(z)} - \frac{p(x)q(x)^{-1}}{1}\cdot \frac{q(z)}{q(z)}\cdot \frac{q(x)^{-1}}{q(x)^{-1}} \cdot q(x) = (z \cdot 1 - x)\, S(z, x)\, q(z)^{-1}\, q(x)^{-1}. \end{align*} Cleaning up (using commutativity of $A$ and $q(x)^{-1}q(x) = 1_A$), this simplifies to \begin{align*} r(z)\cdot 1_A - r(x) = (z \cdot 1 - x)\, S(z, x)\, q(z)^{-1}\, q(x)^{-1}. \tag{$\sharp$} \end{align*} **Compute $\Theta_x(r)$ via ($\sharp$).** Multiply ($\sharp$) on the right by $(z \cdot 1 - x)^{-1}$: \begin{align*} r(z)\, (z \cdot 1 - x)^{-1} - r(x)\, (z \cdot 1 - x)^{-1} = S(z, x)\, q(z)^{-1}\, q(x)^{-1}. \end{align*} Integrate around $\Gamma$ and divide by $2\pi i$: \begin{align*} \Theta_x(r) - r(x)\, \Theta_x(\mathbb{1}_U) = \frac{1}{2\pi i}\int_\Gamma S(z, x)\, q(z)^{-1}\, q(x)^{-1}\, dz. \end{align*} The right-hand side is the $A$-valued contour integral of an $A$-valued function $G(z) := S(z, x)\, q(z)^{-1}\, q(x)^{-1}$, which is analytic on $U$: $z \mapsto S(z, x)$ is a polynomial in $z$ with $A$-coefficients (analytic on all of $\mathbb{C}$), $z \mapsto q(z)^{-1}$ is a scalar holomorphic function on $U$ (since $q$ has no zeros in $U$), and $q(x)^{-1} \in A$ is a constant. Hence $G : U \to A$ is analytic. Apply the [Vector-Valued Cauchy Theorem](/theorems/2683) to $G$ on $U$ with the cycle $\Gamma$. Theorem 2683 requires (a) $U$ open (given), (b) $G : U \to A$ analytic (just verified), and (c) $n(\Gamma, \omega) = 0$ for $\omega \notin U$ (given by hypothesis on $\Gamma$). It yields \begin{align*} \int_\Gamma G(z)\, dz = 0_A. \end{align*} Combined with the previous step ($\Theta_x(\mathbb{1}_U) = 1_A$), \begin{align*} \Theta_x(r) - r(x) \cdot 1_A = 0_A, \qquad \text{i.e.\ } \Theta_x(r) = r(x). \end{align*} [/step] [step:Pull characters $\varphi \in \Phi_A$ inside the integral and apply the scalar Cauchy formula] Let $\varphi \in \Phi_A$ and $f \in \mathcal{O}(U)$. By definition of the character space, $\varphi : A \to \mathbb{C}$ is a continuous unital algebra homomorphism, hence in particular $\varphi \in A^*$ with $\|\varphi\|_{A^*} \le 1$ (characters of unital Banach algebras are continuous of norm $\le 1$, as established in [Characters are Continuous](/theorems/2675)). By [Functionals Commute with Vector-Valued Integrals](/theorems/2682), applied to the Banach space $A$, the continuous map $z \mapsto f(z)\, (z \cdot 1 - x)^{-1}$ along each smooth piece of $\Gamma$, and the bounded linear functional $\varphi \in A^*$: \begin{align*} \varphi(\Theta_x(f)) = \varphi\!\left(\frac{1}{2\pi i}\int_\Gamma f(z)\, (z\cdot 1 - x)^{-1}\, dz\right) = \frac{1}{2\pi i}\int_\Gamma f(z)\, \varphi\!\left((z \cdot 1 - x)^{-1}\right)\, dz, \end{align*} the scalar $\frac{1}{2\pi i}$ and the analytic scalar $f(z)$ being pulled out by linearity of $\varphi$. The hypotheses of theorem 2682 are met: $A$ is a Banach space; the integrand on each smooth piece $\gamma : [a, b] \to U$ is the continuous map $t \mapsto f(\gamma(t))\, (\gamma(t)\cdot 1 - x)^{-1}\, \gamma'(t)$ (continuous in $t$ because each factor is); $\varphi \in A^*$ is bounded linear. **Identify $\varphi((z \cdot 1 - x)^{-1})$.** For $z \in [\Gamma] \subset \rho_A(x)$, $z \cdot 1 - x$ is invertible in $A$. Since $\varphi$ is a unital algebra homomorphism, \begin{align*} \varphi(z \cdot 1 - x)\, \varphi((z \cdot 1 - x)^{-1}) = \varphi((z \cdot 1 - x)\, (z \cdot 1 - x)^{-1}) = \varphi(1_A) = 1. \end{align*} Also $\varphi(z \cdot 1 - x) = z\, \varphi(1_A) - \varphi(x) = z - \varphi(x)$. So \begin{align*} \varphi((z\cdot 1 - x)^{-1}) = \frac{1}{z - \varphi(x)}, \end{align*} provided $z - \varphi(x) \neq 0$. By the [Spectrum via Characters](/theorems/2677) theorem applied to the unital commutative Banach algebra $A$ and the element $x \in A$ (the hypothesis: $A$ is a unital commutative Banach algebra and $x \in A$, both as in the present setup), $\varphi(x) \in \sigma_A(x)$. In particular $\varphi(x) \notin [\Gamma]$ (since $[\Gamma] \subset U \setminus \sigma_A(x)$), so $z - \varphi(x) \neq 0$ for all $z \in [\Gamma]$. Substituting, \begin{align*} \varphi(\Theta_x(f)) = \frac{1}{2\pi i}\int_\Gamma \frac{f(z)}{z - \varphi(x)}\, dz. \tag{$\flat$} \end{align*} **Apply the scalar Cauchy integral formula.** The right-hand side of ($\flat$) is the scalar contour integral of $\frac{f(z)}{z - \varphi(x)}$ along $\Gamma$. The function $f$ is holomorphic on $U$, $\varphi(x) \in \sigma_A(x) \subset U$, and the cycle $\Gamma$ has winding number $1$ at $\varphi(x)$ and $0$ outside $U$. The classical scalar **Cauchy Integral Formula on cycles** then gives \begin{align*} \frac{1}{2\pi i}\int_\Gamma \frac{f(z)}{z - \varphi(x)}\, dz = n(\Gamma, \varphi(x))\, f(\varphi(x)) = 1 \cdot f(\varphi(x)) = f(\varphi(x)). \end{align*} Combined with ($\flat$), \begin{align*} \varphi(\Theta_x(f)) = f(\varphi(x)), \end{align*} which is Property 3. [guided] The whole step is the reduction of Property 3 to a scalar Cauchy-formula computation via duality. Let us spell out why each of the three ingredients is needed. **Why pull $\varphi$ inside?** The vector-valued integral $\Theta_x(f)$ lives in $A$, but $\varphi$ takes us to $\mathbb{C}$. To turn $\varphi(\Theta_x(f))$ into a scalar integral we must commute $\varphi$ with the integral. This is exactly what [Functionals Commute with Vector-Valued Integrals](/theorems/2682) does. The hypotheses we verify are exactly those of 2682: - The Banach space $A$. - A continuous map $[a, b] \to A$ on each smooth piece of $\Gamma$. The map is $t \mapsto f(\gamma(t))\, (\gamma(t)\cdot 1 - x)^{-1}\, \gamma'(t)$. Continuity in $t$ holds because $f$ and $\gamma$ are continuous, $\gamma'$ is continuous on each smooth piece, the resolvent is continuous on $\rho_A(x)$, and product of continuous maps into a Banach algebra is continuous. - A bounded linear functional $\varphi \in A^*$. Characters are continuous (theorem [Characters are Continuous](/theorems/2675)), hence in $A^*$. After commuting, \begin{align*} \varphi(\Theta_x(f)) = \frac{1}{2\pi i}\int_\Gamma f(z)\, \varphi\!\left((z\cdot 1 - x)^{-1}\right)\, dz. \end{align*} **Why is $\varphi((z\cdot 1 - x)^{-1}) = 1/(z - \varphi(x))$?** Because $\varphi$ is a **unital algebra homomorphism**: it preserves products and the identity. The relation $1_A = (z\cdot 1 - x)\,(z\cdot 1 - x)^{-1}$ in $A$ becomes, under $\varphi$, the scalar relation $1 = \varphi(z\cdot 1 - x)\cdot \varphi((z\cdot 1 - x)^{-1}) = (z - \varphi(x))\cdot \varphi((z\cdot 1 - x)^{-1})$, so $\varphi((z\cdot 1 - x)^{-1}) = 1/(z - \varphi(x))$. The denominator does not vanish on $[\Gamma]$ because $\varphi(x) \in \sigma_A(x)$ by [Spectrum via Characters](/theorems/2677) and $[\Gamma] \cap \sigma_A(x) = \emptyset$ by choice of $\Gamma$. **Why does the scalar Cauchy formula apply?** Because we have packaged the result into a scalar integral with the holomorphic function $f$ and the simple pole $z = \varphi(x) \in U$: \begin{align*} \frac{1}{2\pi i}\int_\Gamma \frac{f(z)}{z - \varphi(x)}\, dz = n(\Gamma, \varphi(x))\, f(\varphi(x)) = 1 \cdot f(\varphi(x)). \end{align*} The winding-number value is $1$ at every point of $\sigma_A(x)$ by construction of $\Gamma$, and $\varphi(x) \in \sigma_A(x)$ by theorem 2677. The scalar Cauchy integral formula on cycles (the homological Cauchy formula) then delivers $f(\varphi(x))$. The conceptual remark: Property 3 reduces a Banach-algebra identity $\varphi(\Theta_x(f)) = f(\varphi(x))$ — an identity in $\mathbb{C}$ between two outputs of $A$-valued constructions — to the **scalar** Cauchy integral formula, by converting the vector resolvent $(z\cdot 1 - x)^{-1}$ to the scalar resolvent $(z - \varphi(x))^{-1}$ via the homomorphism property of $\varphi$. This reduction relies essentially on $\varphi$ being multiplicative — generic elements of $A^*$ would not give a useful scalar resolvent. [/guided] [/step] [step:Deduce $\sigma_A(\Theta_x(f)) = f(\sigma_A(x))$ from Property 3 and Spectrum via Characters] By the [Spectrum via Characters](/theorems/2677) theorem applied to the unital commutative Banach algebra $A$ and the element $\Theta_x(f) \in A$ (hypothesis: $A$ unital commutative Banach algebra and $\Theta_x(f) \in A$ — both standing assumptions in this setup), \begin{align*} \sigma_A(\Theta_x(f)) = \{\varphi(\Theta_x(f)) : \varphi \in \Phi_A\}. \end{align*} By Property 3 (Step 4), $\varphi(\Theta_x(f)) = f(\varphi(x))$ for every $\varphi \in \Phi_A$. Substituting, \begin{align*} \sigma_A(\Theta_x(f)) = \{f(\varphi(x)) : \varphi \in \Phi_A\}. \end{align*} By [Spectrum via Characters](/theorems/2677) applied a second time, this time to $x$, \begin{align*} \sigma_A(x) = \{\varphi(x) : \varphi \in \Phi_A\}. \end{align*} The set on the right above is $\sigma_A(x)$, so as $\varphi$ ranges over $\Phi_A$, $\varphi(x)$ ranges over $\sigma_A(x)$, and \begin{align*} \{f(\varphi(x)) : \varphi \in \Phi_A\} = \{f(\lambda) : \lambda \in \sigma_A(x)\} = f(\sigma_A(x)). \end{align*} Therefore $\sigma_A(\Theta_x(f)) = f(\sigma_A(x))$, which is Property 4. This completes the proof of all four properties. [/step]