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If a compact spectrum of an element avoids zero and zero belongs to its unbounded complementary component, choose a simple arc from zero to infinity avoiding the Banach algebra spectrum. Its complement is a simply connected neighborhood of the Banach algebra spectrum, admitting a holomorphic logarithm. Apply holomorphic functional calculus to that logarithm and the composition rule for holomorphic functional calculus to the exponential to obtain the displayed identity. An invertible matrix over the complex numbers always satisfies the hypothesis because its finite Banach algebra spectrum cannot separate zero from infinity.

The displayed series converges absolutely in a Banach algebra. It is the holomorphic functional calculus value of the analytic branch $z\mapsto \log (1-z)$ on the unit disk. The composition rule for holomorphic functional calculus and $\exp (\log (1-z))=1-z$ give $\exp (\log (1-x))=1-x$. Thus every sufficiently small norm perturbation of the identity is an exponential.

The exponential series is absolutely norm-convergent, since ${\sum }_{n\ge 1}\Vert x^n\Vert /n!\le e^{\Vert x\Vert }-1$. Completeness defines the Banach algebra exponential $e^x\in A$. If $xy=yx$, absolute convergence permits rearrangement of the product series, and the binomial identity givesTaking $y=-x$ yields $(e^x{)}^{-1}=e^{-x}$.

For the spectral hypothesis, put $K={\sigma }_A(u)$. Its complement has an open, hence path-connected, unbounded component containing $0$. Join $0$ to an exterior point by a simple polygonal arc in that component, and continue it to infinity, avoiding $K$. Removing this polygonal slit gives a simply connected open neighborhood $\Omega$ of $K$ on which $z$ has a branch of the complex logarithm $L(z)$. The holomorphic functional calculus therefore defines $x=L(u)$ and its composition rule givesThe relevant slit is chosen through the resolvent, not assumed to be a fixed negative-real-axis cut. This is the logarithm from a spectral slit in a Banach algebra.

Let $H$ be the subgroup of finite products of exponentials. It is a subgroup because $e^a{e}^{-a}=1$ and reversing a product gives its inverse. Each product is joined to $1$ by $t\mapsto e^{t{a}_1}\cdots e^{t{a}_m}$, so $H\subseteq G_0$. Conversely, if $\Vert v-1\Vert <1$, the convergent logarithm seriessatisfies $e^{\log v}=v$, by the scalar analytic identity and functional calculus. Thus $H$ contains a neighborhood of $1$. Its translates make $H$ open, and all other cosets are open too, making $H$ closed. Connectedness forces $G_0\subseteq H$. Finally $g{e}^a{g}^{-1}=e^{ga{g}^{-1}}$, so $H$ is normal. Thereforean open-and-closed normal subgroup. This is the identity component of Banach-algebra invertibles; finite products are essential and are not asserted to be single exponentials.

Give $\mathcal{O}(U)$ the compact-open topology. Choose a finite polygonal cycle $\Gamma$ in $U\setminus {\sigma }_A(x)$ whose winding number is one on the Banach algebra spectrum and zero outside $U$. One can take the oriented boundary of a finite union of small squares surrounding the compact Banach algebra spectrum and contained in $U$. DefineThe integral is an algebra-norm integral of a continuous function on finitely many segments. Cauchy's theorem, applied after any bounded linear functional on $A$, makes it independent of the cycle. Moreoverso it is continuous for the compact-open topology.

Here is a direct verification of the algebra properties and uniqueness using the allowed Runge theorem. The version required is that every holomorphic function on an open plane set is a locally uniform limit of rational functions with poles outside that set, allowing a polynomial part. A rational function $r=p/q$ of this kind has the unambiguous value $r(x)=p(x)q(x{)}^{-1}$, since the roots of $q$ avoid the Banach algebra spectrum. These rational values form a unital algebra homomorphism.

They agree with the contour integral. First $\frac{1}{2\pi i}{\int }_{\Gamma }(z1-x{)}^{-1}dz=1$, by replacing the cycle with a large circle and integrating its Neumann expansion. The identity $z^{j}1-x^j=(z1-x){\sum }_{k=0}^{j-1}{z}^{j-1-k}{x}^k$ then proves the assertion for polynomials. For $a\notin U$, the resolvent identity givesThe cycle has winding number zero at $a$, so integration gives $(x-a1{)}^{-1}$. This identity holds on a neighborhood of $a$ disjoint from the cycle and the Banach algebra spectrum. Differentiating with respect to $a$ gives the analogous result for every power $(z-a{)}^{-k}$, and partial fractions give it for all the rational functions under consideration.

If $r_n\to f$ and $s_n\to g$ locally uniformly, the contour estimate makes their evaluated sequences converge. Since $r_n{s}_n\to fg$ locally uniformly and multiplication in $A$ is continuous, their limits satisfy ${\Theta }_x(fg)={\Theta }_x(f){\Theta }_x(g)$. Also ${\Theta }_x(1)=1$ and ${\Theta }_x(Z)=x$. Conversely any continuous unital complex-algebra homomorphism sending $Z$ to $x$ must send $Z-a$ to $x-a1$ and its reciprocal to the inverse. It therefore agrees on rational functions and then, by Runge density and continuity, on all of $\mathcal{O}(U)$. This proves existence and uniqueness of the holomorphic functional calculus.

For its spectral mapping property, first note compatibility on a smaller open spectral neighborhood $W\subseteq U$: composing restriction to $W$ with the calculus there gives the calculus on $U$ by uniqueness. If $\mu \notin g({\sigma }_A(x))$, choose such a $W$ where $\mu -g$ never vanishes. Thenwith the right side evaluated on $W$. Conversely, for $\lambda \in {\sigma }_A(x)$, write $g(z)-g(\lambda )=(z-\lambda )q(z)$ with $q$ holomorphic on $U$. The evaluated factors commute. If their product were invertible, both factors would be invertible, contradicting $\lambda \in {\sigma }_A(x)$. ThusPullback $h\mapsto h\circ g$ is continuous from $\mathcal{O}(V)$ to $\mathcal{O}(U)$ because the image under $g$ of a compact set is compact. The composite $h\mapsto {\Theta }_x(h\circ g)$ is a continuous unital homomorphism taking the coordinate function to $y={\Theta }_x(g)$. Uniqueness of the calculus at $y$ provesFor the exponential conclusion, let $K={\sigma }_A(x)$. There is a simple polygonal arc from zero through its unbounded complementary component to outside a disk containing $K$, continued by a ray to infinity and avoiding $K$. Loops in an initial polygonal path can be removed to make the arc simple. The complement of this slit is a simply connected open neighborhood of $K$ avoiding zero, and hence admits a holomorphic logarithm $\ell$. Set $y={\Theta }_x(\ell )$. The composition rule givesThe calculus of the entire exponential agrees with the usual algebra exponential because its power-series partial sums converge locally uniformly. Finally an invertible matrix over the complex numbers has a finite Banach algebra spectrum avoiding zero. Removing finitely many plane points leaves a connected set, so the hypothesis always holds: every invertible matrix over the complex numbers is an exponential of a matrix over the complex numbers.

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 $e\ne 0,1$.

Fix an admissible contour $\Gamma$ with interior $D$ compactly contained in $U$ and containing $\sigma (x)$. Apply the preceding spectral inclusion to $D$, so that it also contains $\sigma (x_n)$ for large $n$. Uniformly on the compact contour, the Neumann series makes $R(\zeta ,x_n)$ bounded, and the resolvent identity givesThe resolvents therefore converge uniformly on $\Gamma$. Both holomorphic functional calculus values use that same contour, soHence ${\Theta }_{x_n}(f)\to {\Theta }_x(f)$ for each fixed $f\in \mathcal{O}(U)$.

Define the holomorphic function $e$ on the disconnected open set $U$ to be one on $V$ and zero on $W$. Since $e^2=e$, its holomorphic functional calculus value $p={\Theta }_x(e)$ is an idempotent. The spectral mapping theorem and the nonempty spectral intersection with $V$ show $1\in \sigma (p)$; thus $p\ne 0$. These are the Riesz projection values selecting the indicated spectral component.

Part (b) gives $p_n={\Theta }_{x_n}(e)\to p$. If $\sigma (x_n)$ avoided $V$, spectral mapping would give $\sigma (p_n)=\{0\}$. A nonzero idempotent always has $1$ in its spectrum: if $1-p_n$ were invertible, $p_n(1-p_n)=0$ would force $p_n=0$. Thus avoiding $V$ forces $p_n=0$. That is impossible eventually, since $p\ne 0$ and $p_n\to p$. Therefore $\sigma (x_n)\cap V\ne \varnothing$ eventually. This is persistence of a separated spectral component, not a claim that arbitrary individual spectral points vary continuously.

Give $\mathcal{O}(U)$ the compact-open topology, that is, uniform convergence on every compact subset of $U$. Choose a bounded finite union of polygonal domains $D$ whose closure lies in $U$, whose interior contains $\sigma (x)$ and whose boundary avoids the spectrum. Orient its boundary $\Gamma$ positively, with holes oriented negatively. It has winding number one on the spectrum and zero outside $U$. For the resolvent of an element $R(\zeta ,x)=(\zeta 1-x{)}^{-1}$ define the holomorphic functional calculus byThis is an integral of a continuous Banach space-valued function. Its value is independent of the admissible contour: the integrand is holomorphic away from the spectrum, and the difference of two such cycles has winding number zero there, so the Cauchy integral theorem applies. The vector-valued theorem follows by applying bounded linear functionals and then the Hahn-Banach theorem.

Linearity is immediate. To prove multiplicativity, choose two admissible contours with the inner domain compactly contained in the outer one, and use the resolvent identityIn the double integral for ${\Theta }_x(f){\Theta }_x(g)$, integrate the first term in the outer variable $\eta$. The Cauchy integral formula gives $g(\zeta )$. For the second term, integration of $f(\zeta )/(\eta -\zeta )$ over the inner contour is zero, since $\eta$ lies outside that inner domain. Therefore the double integral equals ${\Theta }_x(fg)$. The continuous integrands on disjoint compact contours justify exchanging the integrals.

For $f=1$ and $f=Z$, deform to a large circle and use the Neumann series $R(\zeta ,x)={\sum }_{m\ge 0}{x}^m{\zeta }^{-m-1}$. Its relevant coefficients give ${\Theta }_x(1)=1$ and ${\Theta }_x(Z)=x$. The estimateproves continuity in the stated topology.

For uniqueness, the Runge approximation theorem in the required form says that rational functions with poles outside $U$, allowing a pole at infinity, are dense in $\mathcal{O}(U)$ for uniform convergence on compact subsets. Every complex-linear unital homomorphism taking $Z$ to $x$ has prescribed values on polynomials and on $(Z-a{)}^{-1}$ for $a\notin U$: the latter value must be $(x-a1{)}^{-1}$. Hence its rational-function values are prescribed. Continuity and Runge approximation force its values on every holomorphic function. This proves continuity and uniqueness of holomorphic functional calculus, including disconnected $U$.

To prove the spectral mapping theorem, first suppose $\mu \notin f(\sigma (x))$. On a smaller open neighborhood $U^{\prime }$ of the spectrum, $f-\mu$ has no zeros, so $h=1/(f-\mu )$ is holomorphic there. The contour definition is compatible with this restriction. Multiplicativity gives ${\Theta }_x(f)-\mu 1$ the inverse ${\Theta }_x(h)$, excluding $\mu$ from its spectrum.

Conversely, if $\mu =f(\lambda )$ for $\lambda \in \sigma (x)$, write $f(z)-\mu =(z-\lambda )g(z)$, defining $g(\lambda )=f^{\prime }(\lambda )$ at its removable singularity. ThenThese two factors commute. An invertible product of commuting elements makes each factor invertible, by multiplying the inverse of the product by the other factor. That would contradict $\lambda \in \sigma (x)$. Thus

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.

Work over the complex scalar field. If $\lambda \in {\sigma }_A(x)$, the ideal generated by $x-\lambda 1$ is proper: otherwise commutativity would make $x-\lambda 1$ invertible. By Zorn lemma extend it to a maximal ideal $M$. This ideal is closed. Indeed, if its closure were all of $A$, it would contain an element of $M$ sufficiently close to $1$ to be invertible, since the invertible group is open; a proper ideal cannot contain such an element. Its closure is therefore proper, and maximality gives $\overline{M}=M$.

The quotient $A/M$ is a complex unital Banach algebra and a division algebra. The Gelfand-Mazur theorem identifies it with $\mathbb{C}$, giving a character of an algebra $\phi$ with $\phi (x)=\lambda$. Conversely, the argument in (a) shows that every $\phi (x)$ is in the spectrum. ThusHere, as usual, a unital Banach algebra has a nonzero identity; the zero algebra is excluded.

The holomorphic functional calculus assigns to every function holomorphic on an open neighbourhood $U$ of ${\sigma }_A(x)$ an element $f(x)\in A$. It is a unital algebra homomorphism, sends the coordinate function to $x$, agrees with evaluation of rational functions having no poles on the spectrum, depends only on the germ near the spectrum, and is continuous for uniform convergence on a fixed surrounding contour. Its formula isHere $\Gamma$ is a finite oriented contour cycle in $U\setminus {\sigma }_A(x)$, with winding number one on the spectrum and zero off $U$. A finite union of suitable polygonal boundary cycles suffices; no connectedness of $U$ is required. Such cycles exist by compactness of the spectrum, and the calculus is independent of the choice. The inverse is the resolvent of an element.

For the exponential, use a circle $|z|=R>\Vert x\Vert$. The uniformly convergent Neumann expansion $(z1-x{)}^{-1}={\sum }_{n\ge 0}{x}^n{z}^{-n-1}$ may be integrated term by term. The scalar Cauchy integral formula gives $\frac{1}{2\pi i}\int e^z{z}^{-n-1}dz=1/n!$, provingThis series converges absolutely in the Banach algebra norm.

A character of an algebra is bounded and takes the resolvent to $(z-\phi (x){)}^{-1}$. Applying it inside the contour integral and using the scalar Cauchy formula gives $\phi (f(x))=f(\phi (x))$. The character of an algebra description of the spectrum, now applied to $f(x)$, proves the spectral mapping theorem:In particular this spectrum is contained in $V$ when $V\supset f(U)$.

For the composition rule for holomorphic functional calculus, choose a relatively compact open neighbourhood $D$ of $f({\sigma }_A(x))$ whose closure is contained in $V$, and a surrounding contour cycle $\Delta$ in $V$ with winding number one on $D$. Choose a sufficiently small neighbourhood $\Omega$ of ${\sigma }_A(x)$ with $\overline{\Omega }\subset U$ and $f(\overline{\Omega })\subset D$, and a contour $\Gamma$ in $\Omega$ surrounding the spectrum. The neighbourhoods and cycles may be finite unions, obtained from finite covers of the relevant compact sets. For $\zeta \in \Delta$ the function $h_{\zeta }(z)=(\zeta -f(z){)}^{-1}$ is holomorphic near $\overline{\Omega }$. Locality and multiplicativity of the calculus giveInsert this in the formula for $g(f(x))$ and interchange the two vector-valued integrals. The scalar inner integral is $g(f(z))$, soThis proves equality as algebra elements. Equality of their character of an algebra values alone would be insufficient in a Banach algebra with nonzero radical.

Finally, $\Vert x\Vert <1$ places the spectrum in the open unit disk. On that disk $f(z)=\log (1-z)=-{\sum }_{n\ge 1}{z}^n/n$ is a holomorphic branch. Its calculus value is the norm-convergent series $y=-{\sum }_{n\ge 1}{x}^n/n$: one may integrate the uniformly convergent scalar series on a circle with $\Vert x\Vert <R<1$. Since $\exp (f(z))=1-z$, the composition rule gives the logarithm of an element near the identity formula

The prescribed-pole form of Runge theorem is this: let $K\subset \mathbb{C}$ be compact, and let $E\subset \hat{\mathbb{C}}\setminus K$ meet every connected component of the complement in the Riemann sphere. If $f$ is holomorphic on a neighbourhood of $K$, then for every $ϵ>0$ there is a rational function with all its poles in $E$ whose uniform distance from $f$ on $K$ is less than $ϵ$. A polynomial is regarded as having its only possible pole at infinity. The empty compact set is trivial, so assume $K\ne \varnothing$.

First suppose $\infty \in E$. Let $A\subset C(K)$ be the uniform rational approximation algebra with prescribed poles, the uniform closure of rational functions whose finite poles lie in $E$. It is a commutative unital Banach algebra, and contains $u(z)=z$. Setwhere the inverse in this definition is the pointwise continuous function on $K$. The set $S$ is relatively open: if the inverse at $\lambda$ belongs to $A$, the Neumann series gives inverses at nearby points. It is relatively closed: when ${\lambda }_n\to \lambda \notin K$, the corresponding scalar functions converge uniformly on $K$, and $A$ is closed.

Each bounded complementary component meets $E$ in a finite point $a$, and $(u-a1{)}^{-1}$ is an allowed rational function. The unbounded component meets $S$ because for $|\lambda |>\Vert u\Vert$ the geometric series for $(u-\lambda 1{)}^{-1}$ belongs to $A$. Being both open and closed in the complement, $S$ therefore contains every component, so it is the entire complement. Conversely, evaluation at any $z\in K$ prevents $u-z1$ from being invertible. HenceApply the holomorphic functional calculus in $A$ to $f$ near $K$. For every $z\in K$, the evaluation character of an algebra and the contour formula from (b) give $f(u)(z)=f(z)$. Thus $f{|}_K\in A$. By the definition of this uniform closure, the required rational approximations exist.

If $\infty \notin E$, choose a finite point $a\in E$ in the component containing infinity. The Möbius transformation $w=(z-a{)}^{-1}$ sends $K$ to a compact subset $K^{\prime }$ of the plane and sends $E$ to a set $E^{\prime }$ containing infinity. It preserves complementary components, so $E^{\prime }$ meets every component of the complement of $K^{\prime }$. The transformed function $F(w)=f(a+1/w)$ is holomorphic near $K^{\prime }$, since $0\notin K^{\prime }$. The case already proved approximates $F$ by rational functions with poles in $E^{\prime }$. Pulling them back gives rational approximations to $f$ on $K$ with poles only in $E$: a pole at infinity in the $w$-plane becomes a pole at $a$, and every other pole has its prescribed preimage. In particular the pullbacks do not acquire a pole at the original infinity, since $0\notin E^{\prime }$.

This proves the full prescribed-pole theorem, including the case where a pole at infinity is forbidden. If $\mathbb{C}\setminus K$ is connected, take $E=\{\infty \}$ to obtain the polynomial Runge theorem as a corollary.

The holomorphic functional calculus assigns to $a$ in a unital complex Banach algebra and every function holomorphic near $\sigma (a)$ the elementwhere $\Gamma$ winds once around the spectrum. It is a unital algebra homomorphism, extends polynomial evaluation, is independent of the admissible contour, and obeys the spectral mapping theorem

Let $z_K\in \mathcal{R}(K)$ be the coordinate function. For a character $\phi$, put $\lambda =\phi (z_K)$. If $\lambda \notin K$, then $(z_K-\lambda {)}^{-1}$ is one of the admitted rational functions, contradicting the fact that $z_K-\lambda$ lies in the kernel of $\phi$. Thus $\lambda \in K$. For every rational function $r$ without poles on $K$,Continuity of characters and uniform density extend this identity to every member of $\mathcal{R}(K)$. Conversely evaluation at every $\lambda \in K$ is a character. Therefore the character space of R(K) is naturally $K$.

Runge approximation theorem says that if $K\subset \mathbb{C}$ is compact, $f$ is holomorphic on a neighbourhood of $K$, and one chooses one point in every bounded component of $\mathbb{C}\setminus K$, then $f$ can be approximated uniformly on $K$ by rational functions whose finite poles belong only to the chosen points.

To prove it, surround $K$ by finitely many small rectangles contained in the domain of $f$ and apply the Cauchy integral formula on their oriented boundaries:Riemann sums approximate this integral uniformly on $K$ by rational functions with poles on $\Gamma$. If a pole $b$ lies in a component containing the selected point $a$, choose a polygonal path from $b$ to $a$ inside that component and subdivide it finely. The resolvent identityallows the pole to be moved step by step along the path, with arbitrarily small uniform error on $K$. Moving every pole proves the theorem.

For an open set $U$, choose a compact exhaustion $K_1\subset K_2\subset \cdots$ such that every component of $\mathbb{C}\setminus K_n$ meets $\mathbb{C}\setminus U$. Runge's theorem approximates a holomorphic function on $K_n$ by a rational function with all poles outside $U$. A diagonal choice gives convergence uniformly on every compact subset, proving that such rational functions are dense in the space of holomorphic functions $\mathcal{O}(U)$ with its compact-open topology.

Finally, equality of the Gelfand transforms givesLet this common compact spectrum be $K$. The divided differenceis holomorphic near $K\times K$. The two-variable holomorphic functional calculus for the commuting pair $(x_1,x_2)$ gives an element $u=g(x_1,x_2)$ satisfyingFor every character,An element of a commutative unital Banach algebra is invertible exactly when its Gelfand transform has no zero. Thus $u$ is invertible, and $f(x_1)=f(x_2)$ implies $x_1=x_2$. This is injectivity through a holomorphic functional calculus with nonvanishing derivative.