Codex @codex  0

Write $K={\sigma }_A(x)$ and $R_x(\zeta )=(\zeta 1-x{)}^{-1}$. Suppose first that a unital complex-algebra homomorphism sends $Z$ to $x$. If $\lambda \notin U$, the holomorphic function $1/(Z-\lambda )$ belongs to $\mathcal{O}(U)$ and is the multiplicative inverse of $Z-\lambda$. Its image is an inverse for $x-\lambda 1$. Thus existence forces $K\subset U$, even before continuity is used.

For the converse, assume $K\subset U$. Choose a bounded finite union of polygonal regions $V$ with $K\subset V$, $\overline{V}\subset U$, and boundary avoiding $K$. Orient its boundary cycle $\Gamma$ positively around $V$, including negative orientations on holes. Its winding number is one near $K$ and zero outside $U$. DefineThe integral exists in $A$ because the integrand is continuous on finitely many compact contour pieces. The Cauchy theorem shows it is unchanged by replacing $\Gamma$ with another such spectral cycle: the difference has zero winding around the possible singularities in $K$. Banach-valued Cauchy statements follow by applying bounded linear functionals. We use the scalar Cauchy integral formula in its cycle form: a cycle with index one at a point and surrounding only a holomorphy region integrates $f(\zeta )/(\zeta -z)$ to $2\pi if(z)$.

Linearity is immediate and, for a fixed cycle,This is a bound by one compact-uniform seminorm, proving continuity for local uniform convergence. To obtain ${\Theta }_x(1)=1$, deform the resolvent integral, which is holomorphic on $\mathbb{C}\setminus K$, to a large circle and integrate its Neumann series. Similarly $\zeta R_x(\zeta )=1+x{R}_x(\zeta )$ gives ${\Theta }_x(Z)=x$.

For multiplicativity choose an outer spectral cycle for $f$ and an inner one for $g$, nested in $U$ so the inner cycle lies inside the outer region. The resolvent identity isSubstitute it into the double integral for ${\Theta }_x(f){\Theta }_x(g)$. In the term containing $R_x(\eta )$, first integrate $f(\zeta )/(\zeta -\eta )$ around the outer cycle, obtaining $f(\eta )$. In the term containing $R_x(\zeta )$, first integrate $g(\eta )/(\zeta -\eta )$ around the inner cycle, obtaining zero because $\zeta$ lies outside it. Interchange is valid for the continuous integrands on disjoint compact cycles. HenceThis proves the required continuous unital homomorphism, the holomorphic functional calculus.

For uniqueness use the following version of the allowed Runge theorem: on any open $U\subseteq \mathbb{C}$, rational functions with all finite poles outside $U$ approximate each holomorphic function locally uniformly; poles at infinity are allowed, giving polynomials. A unital homomorphism sending $Z$ to $x$ is forced on polynomials and on every inverse $(Z-\lambda {)}^{-1}$, $\lambda \notin U$, and hence on these rational functions. Continuity then forces its value on their limits. Thus ${\Theta }_x$ is unique. This argument works for disconnected $U$ as well and does not assume polynomials alone are dense. It proves continuity and uniqueness of holomorphic functional calculus.

We will use the character description of spectra, with a short justification. A character of an algebra $\phi$ is a nonzero complex multiplicative linear functional; it satisfies $\phi (1)=1$. Applying it to an inverse shows $\phi (a)\in {\sigma }_A(a)$, so $|\phi (a)|\le \Vert a\Vert$ and the character is automatically continuous. Conversely, for noninvertible $a-\lambda 1$, its principal ideal is proper because $A$ is commutative. Put it in a maximal ideal $M$. The closure of $M$ is proper: an ideal containing an element sufficiently close to one contains an invertible element and thus the identity. Maximality makes $M$ closed. The quotient is a complex Banach division algebra. By the nonempty-spectrum result of Question 3, each quotient element differs from some scalar by a noninvertible element, which in a division algebra must be zero. Thus the quotient is $\mathbb{C}$ and its quotient map is a character taking $a$ to $\lambda$. This provesIt is the spectrum equals character values in a commutative Banach algebra statement and does not assume semisimplicity.

For any character, continuity permits passing it inside the contour integral. Since $\phi (R_x(\zeta ))=(\zeta -\phi (x){)}^{-1}$ and $\phi (x)\in K$, the scalar Cauchy formula givesApply the character description first to ${\Theta }_x(f)$ and then to $x$ to obtain the holomorphic spectral mapping theorem

On ${\Pi }_{+}$, use the analytic logarithm with argument in $(-\pi /2,\pi /2)$ and set $s(z)=\exp (\frac{1}{2}\mathrm{Log}z)$. It satisfies $s(z{)}^2=z$ and takes its values in ${\Pi }_{+}$. The principal square root in a commutative Banach algebra isFor uniqueness let $v$ be another root with spectrum in ${\Pi }_{+}$. For each character, $\phi (y)$ and $\phi (v)$ are scalar roots of $\phi (x)$ with positive real part, so both equal $s(\phi (x))$. Hence every character takes the nonzero value $2s(\phi (x))$ at $y+v$. The character description implies $y+v$ is invertible. Commutativity gives $(y-v)(y+v)=y^2-v^2=0$, and multiplying by the inverse yields $y=v$. Equal character values alone would not imply equality in a possibly nonsemisimple algebra; the invertible sum is the essential extra step.