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For a bounded operator on a complex Banach space, partition a disconnected spectrum into two nonempty compact pieces. The holomorphic function equal to one near the first piece and zero near the second defines a Riesz projection. The holomorphic spectral mapping theorem makes it neither zero nor identity. Its closed range is a proper nonzero invariant vector subspace. This need not be an orthogonal projection, and the argument does not require a normal operator.

Write $K={\sigma }_A(x)$ and equip $\mathcal{O}(U)$ with the compact-open topology. All algebra homomorphisms here are complex linear. Choose a finite positively oriented boundary cycle $\Gamma$ in $U\setminus K$ with winding number one on a neighborhood of $K$ and zero outside $U$. Such a cycle can be obtained as the boundary of a finite union of sufficiently small squares whose interior contains $K$ and whose closure is contained in $U$; holes are included with negative boundary orientation. Define the holomorphic functional calculus by the norm-convergent contour integralThe resolvent identity proves that $R$ is a Banach-space-valued holomorphic function off $K$. The Cauchy integral theorem, applied after each bounded linear functional, makes the integral independent of the chosen boundary cycle. The Hahn-Banach theorem then gives equality of the algebra-valued integrals. Moreover,which proves complex linearity and continuity for the compact-open topology.

For the constant function $1$ and the coordinate function $Z$, the contour can be deformed to a large circle because their integrands are holomorphic everywhere outside $K$. There the Neumann series gives $R(z)={\sum }_{n\ge 0}{x}^n{z}^{-n-1}$. Integrating term by term showsTo prove multiplicativity, take nested cycles ${\Gamma }_o,{\Gamma }_i$ around $K$, with the outer cycle winding once on the inner cycle and the inner winding zero on the outer cycle. Use the outer cycle for $f$ and the inner for $g$. The resolvent identity readsIn the resulting double integral, the term containing $R(w)$ gives $f(w)g(w)R(w)$ after the $z$ integral, by the Cauchy integral formula. The term containing $R(z)$ vanishes after the $w$ integral, because $w\mapsto g(w)/(z-w)$ is holomorphic on and inside the inner cycle's region. ThereforeThe integrals are over finite compact curves and have continuous integrands, so exchanging their order is legitimate. This proves the existence of the required continuous unital algebra homomorphism.

For uniqueness, the Runge approximation theorem gives density in $\mathcal{O}(U)$, for the compact-open topology, of rational functions with poles outside $U$. More explicitly, exhaust $U$ by compact sets that are holomorphically convex relative to $U$, and approximate on those sets by rational functions whose poles lie in the complement of $U$; this gives convergence on every compact subset. Any unital algebra homomorphism taking $Z$ to $x$ must take $(Z-\alpha {)}^{-1}$, for $\alpha \notin U$, to $(x-\alpha 1{)}^{-1}$, since it preserves inverse equations. Partial fractions and polynomial algebra therefore fix its value on every such rational function. Continuity fixes its value on their limits. Thus the continuous unital homomorphism with the prescribed coordinate value is unique. Restricting a function to a smaller neighborhood of $K$ gives the same value, either by the contour definition or this uniqueness argument.

To prove the holomorphic spectral mapping theorem, first let $\mu \notin f(K)$. There is an open neighborhood $V\subseteq U$ of $K$ on which $f-\mu$ has no zeros. The holomorphic function $h=1/(f-\mu )$ on $V$ supplies a two-sided inverse throughThus ${\sigma }_A({\Theta }_x(f))\subseteq f(K)$. Conversely, if $\lambda \in K$ and $\mu =f(\lambda )$, the divided differenceis holomorphic on $U$, and multiplicativity givesThe two factors commute. A commuting product $uv$ can be invertible only if both factors are invertible: its inverse commutes with each factor, and $v(uv{)}^{-1}$ is a two-sided inverse for $u$. Invertibility of the displayed product would therefore contradict $\lambda \in K$. This proves

For the root, let $D=\mathbb{C}\setminus (-\infty ,0]$ and $q(z)=\exp (\mathrm{Log}z/3)$, using the principal complex logarithm. Define $y={\Theta }_x(q)$. Since $q(z{)}^3=z$, multiplicativity and the holomorphic spectral mapping theorem giveThis is the principal cube root of a Banach-algebra element.

For uniqueness, suppose $v^3=x$ and ${\sigma }_A(v)\subset S$. Cubing maps the whole open sector $S$ into $D$, sois a continuous unital algebra homomorphism: pullback by cubing is continuous for the compact-open topology, since images of compact sets are compact. It takes the coordinate function to $v^3=x$, and uniqueness already proved gives $F={\Theta }_x$. For $z\in S$, the scalar principal cube root satisfies $q(z^3)=z$, because $3\arg z\in (-\pi ,\pi )$. HenceThis proves uniqueness with the stated spectral sector, without assuming the ambient Banach algebra is commutative or that algebra characters separate its elements.

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.

Give $\mathcal{O}(U)$ the compact-open topology, that is, uniform convergence on compact subsets. Put $R(z)=(z1-x{)}^{-1}$. Choose a finite oriented contour $\Gamma$ in $U\setminus \sigma (x)$ with winding number one on $\sigma (x)$ and zero outside $U$; boundaries of a finite union of sufficiently small rectangles around the compact spectrum provide such a contour. Define the holomorphic functional calculus byThe integral is a Bochner integral along the piecewise smooth contour. It is independent of the admissible contour by the Cauchy integral theorem, since the resolvent is holomorphic off the spectrum. In particular, restricting $f$ to a smaller open spectral neighborhood leaves its value unchanged.

Linearity is immediate, andproves continuity for the compact-open topology. For the constant and coordinate functions, move the corresponding resolvent integrals to a large circle. The norm-convergent expansion $R(z)={\sum }_{n\ge 0}{x}^n{z}^{-n-1}$ there gives ${\Theta }_x(1)=1$ and ${\Theta }_x(u)=x$.

To check multiplication, take two nested admissible contours, with $z$ on the outer one and $w$ on the inner one. The resolvent identity givesIn the double integral for ${\Theta }_x(f){\Theta }_x(g)$, the term with $R(w)$ integrates in $z$ to $f(w)R(w)g(w)$ by the Cauchy integral formula. The term with $R(z)$ integrates in $w$ to zero, because $z$ lies outside the inner region. Thus ${\Theta }_x(fg)={\Theta }_x(f){\Theta }_x(g)$. Nested regions can be chosen with closures inside $U$, so the argument also handles disconnected $U$ and contours with several boundary components.

For uniqueness, use the following form of the Runge theorem: rational functions whose poles lie outside an open set $U$ are dense in $\mathcal{O}(U)$ for uniform convergence on compact subsets; polynomials are included, with a pole at infinity allowed. This follows by a Runge exhaustion of $U$; no connectedness of $U$ is required. Any unital algebra homomorphism sending $u$ to $x$ must send $(u-a{)}^{-1}$ to $(x-a1{)}^{-1}$ for $a\notin U$. Consequently it agrees with the constructed map on every such rational function. Continuity and density give uniqueness. This is continuity and uniqueness of holomorphic functional calculus.

For the holomorphic spectral mapping theorem, first suppose $\mu \notin f(\sigma (x))$. The reciprocal $g=1/(f-\mu )$ is holomorphic on some smaller open neighborhood $V$ of $\sigma (x)$, even if it is not holomorphic on all of $U$. Restriction consistency and multiplicativity on $V$ show that ${\Theta }_x(g)$ is an inverse of ${\Theta }_x(f)-\mu 1$. Thus $\sigma ({\Theta }_x(f))\subseteq f(\sigma (x))$.

Conversely, if $\lambda \in \sigma (x)$, thenis holomorphic on $U$, with the singularity at $\lambda$ removed. HenceIn a commutative algebra, an invertible product has invertible factors. The displayed product cannot be invertible because $x-\lambda 1$ is not. Therefore

For the operator conclusion, work over complex scalars and let $\sigma (T)=K_1\cup K_2$ be a partition into disjoint nonempty compact sets. To apply the preceding commutative-algebra result precisely, take the closed unital subalgebra of $\mathcal{B}(X)$ generated by $T$ and all its resolvents $(zI-T{)}^{-1}$, $z\notin {\sigma }_{\mathcal{B}(X)}(T)$. The generators commute, so this is a resolvent-generated commutative algebra. Its spectrum of $T$ is exactly the operator spectrum: every resolvent outside the latter was included, and an inverse inside the subalgebra would also be an inverse in $\mathcal{B}(X)$.

Choose disjoint open neighborhoods $U_1,U_2$ of $K_1,K_2$, and let $f$ equal one on $U_1$ and zero on $U_2$. The corresponding Riesz projection $P=f(T)$ satisfiesThe last spectrum is computed in the chosen commutative algebra and already ensures $P\ne 0,I$. Its range $Y=PX=\ker (I-P)$ is closed, nonzero and proper. The commutation identity makes it invariant under $T$. This proves that a disconnected spectrum yields a nontrivial invariant subspace. Using all resolvents avoids inadvertently replacing the operator spectrum by the possibly larger spectrum in a polynomial-generated subalgebra.