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Write $K={\sigma }_A(x)$ and $R(\lambda )=(\lambda 1-x{)}^{-1}$. Choose a bounded finite union of small polygonal regions $V$ with $K\subset V$, $\overline{V}\subset U$ and $\partial V\cap K=\varnothing$. Its positively oriented boundary cycle $\Gamma$ has winding number one on $K$ and zero outside $V$; boundaries of holes have negative orientation. Such regions can be obtained from a sufficiently fine grid around the compact set $K$. This accommodates disconnected spectra and neighborhoods with holes.

Define the holomorphic functional calculus by the norm-convergent contour integralThe integrand is continuous on a finite rectifiable cycle, so the integral exists in the Banach space $A$. Its value is independent of the chosen cycle: on $U\setminus K$ the integrand is holomorphic, and cycles with the stated winding numbers are homologous there. Apply the scalar Cauchy integral theorem to every bounded linear functional and then use the Hahn-Banach theorem to recover equality of the algebra-valued integrals. Equivalently, a common refinement of the polygonal regions gives the same contour-deformation argument.

Linearity is immediate. For the entire functions $1,Z$, enlarge the integration cycle to a sufficiently large circle in the resolvent set, using their entire extension. Integrating the uniformly convergent Neumann series term by term givesTo prove multiplicativity, choose two nested admissible cycles, ${\Gamma }_o$ outside ${\Gamma }_i$, both surrounding $K$ and lying in $U$. The resolvent identity givesUse ${\Gamma }_o$ for $f$ and ${\Gamma }_i$ for $g$, multiply the two integrals, and insert this identity. In the term with $R(\mu )$, the outer Cauchy integral formula replaces $f(\lambda )/(\lambda -\mu )$ by $f(\mu )$. In the term with $R(\lambda )$, the inner integral of $g(\mu )/(\lambda -\mu )$ is zero because $\lambda$ is outside the inner region. ThusThe exchanges of integrals are justified by continuous integrands on compact separated cycles. This proves the unital algebra-homomorphism property, not just an identity of algebra character values.

For continuity in local uniform convergence, fix one cycle and boundThe last supremum is a defining compact-set seminorm of $\mathcal{O}(U)$. Hence the map is continuous.

If another continuous unital algebra homomorphism $\Psi$ sends $Z$ to $x$, its values on polynomials are forced. For every $\zeta \notin U$, the function $(Z-\zeta {)}^{-1}$ exists in $\mathcal{O}(U)$, so multiplicativity forcesTherefore $\Psi$ and ${\Theta }_x$ agree on all rational functions with poles outside $U$. The permitted Runge theorem makes these functions dense in $\mathcal{O}(U)$ for local uniform convergence, including when $U$ is disconnected. Continuity makes the maps agree everywhere. ThusThis proves the continuity and uniqueness of holomorphic functional calculus with the exact topology required.

For an algebra character $\varphi$, multiplicativity and nonzeroness give $\varphi (1)=1$. An invertible element cannot have zero algebra character value, so $\varphi (x)\in {\sigma }_A(x)$. The same observation for each algebra element and the spectral bound from Question 3 gives $|\varphi (b)|\le \Vert b\Vert$; thus a algebra character is automatically a bounded linear functional. In particular,Apply $\varphi$ inside the contour integral and use the Cauchy integral formula with winding number one at $\varphi (x)$:No separation of algebra elements by algebra characters is used; such separation can fail for a general commutative Banach algebra.

For the final clause, the two closed spectral pieces $X,Y$ are compact and have positive distance. Choose disjoint open neighborhoods $U_X,U_Y$ and let $f$ be identically one on $U_X$ and zero on $U_Y$. This locally constant function is holomorphic on their union, and $f^2=f$. Set $e={\Theta }_a(f)$. Multiplicativity gives $e^2=e$, and the proved algebra character formula givesThis is the spectral idempotent from a separated spectrum. It is also nontrivial without assuming that algebra characters separate points. If $e=0$, choose $\lambda \in X$ and define $g=0$ on $U_X$, $g(z)=1/(z-\lambda )$ on $U_Y$. Then $(Z-\lambda )g=1-f$, so $(a-\lambda 1){\Theta }_a(g)=1$, with commuting factors, making $a-\lambda 1$ invertible despite $\lambda \in {\sigma }_A(a)$. This is impossible. If $e=1$, exchange the two pieces and use $\lambda \in Y$ for the analogous contradiction. Hence $e\ne 0,1$.