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Use a nonzero complex unital Banach algebra $B$ with submultiplicative norm, and write its unit as $e$. The complex scalar field is the usual convention in this spectral statement; for a real algebra one uses its complexification. We derive the needed analytic facts from norm-convergent series and scalar complex analysis.

If $\Vert a\Vert <1$, completeness makes the Neumann series ${\sum }_{j=0}^{\infty }{a}^j$ converge. Multiplying its finite partial sums by $e-a$ gives $e-a^{m+1}$, which tends to $e$. Hence its sum is the inverse of $e-a$. If $b$ is invertible, then $b+h=b(e+b^{-1}h)$ is invertible whenever $\Vert b^{-1}h\Vert <1$. This proves that the invertible elements form an open set and that inversion is locally norm-continuous.

Define the Banach algebra spectrum and the resolvent of an element byFor $|\lambda |>\Vert x\Vert$, the Neumann series givesThus $\sigma (x)$ is bounded by $\Vert x\Vert$ and is closed, because invertibility is open. Also $R(\lambda )\to 0$ in norm at infinity; for example its norm is at most $\Vert e\Vert /(|\lambda |-\Vert x\Vert )$, since $\Vert e\Vert \ge 1$ in a nonzero submultiplicative unital algebra. The standard normalization $\Vert e\Vert =1$ is not needed.

For points where the Banach algebra resolvent exists $\lambda ,\mu$, multiply out the inverses to get the resolvent identityFor ${\lambda }_0$ outside the Banach algebra spectrum, put $R_0=R({\lambda }_0)$. Locally,This series and its differentiated series converge uniformly on every smaller disk. It therefore proves $R$ is a Banach-space-valued holomorphic function, establishing complex differentiability in the Banach-space norm and gives $R^{\prime }=-R^2$, without assuming a Banach-algebra-valued analytic theorem. In particular every continuous complex linear functional $\varphi$ makes $\varphi (R(\lambda ))$ a scalar holomorphic function.

The Banach algebra spectrum is nonempty. Otherwise $R$ would be defined and analytic everywhere. Its local norm continuity and decay at infinity make it globally bounded. For every continuous linear functional $\varphi$, the scalar entire function $\varphi (R(\lambda ))$ is bounded and tends to zero at infinity, so it is identically zero: the scalar Cauchy estimate for its derivative on circles of radius $r$ is $M/r$, which tends to zero, and the limiting value then fixes its constant as zero. Such functionals separate points by the Hahn-Banach theorem. Explicitly a bounded real functional can be extended from the real span of any nonzero element using its norm, and complexifying it by $\varphi (v)=\ell (v)-i\ell (iv)$ gives a bounded complex linear functional nonzero on that element. Therefore $R$ itself would be zero, contradicting $(\lambda e-x)R=e$. We have proved the Banach algebra spectrum is a nonempty compact subset of the complex plane.

We next establish the spectral radius formula. The power norms are submultiplicative. If some $x^m=0$, their root norms are eventually zero. Otherwise, for any fixed $m$, write $n=qm+r$ with $0\le r<m$ to obtainTaking roots gives ${\mathrm{limsup}}_n\Vert x^n{\Vert }^{1/n}\le \Vert x^m{\Vert }^{1/m}$. Taking the infimum in $m$, and using that every root norm is at least that infimum, proves existence ofFor $|\lambda |>{\rho }_0$, the root test makes $\sum x^j/{\lambda }^{j+1}$ converge, even when $|\lambda |\le \Vert x\Vert$. The same telescoping multiplication proves it is a two-sided inverse. HenceFor the reverse bound fix $R>r_{\sigma }$. If $R_1>\max (R,\Vert x\Vert )$, integrate the uniformly convergent exterior series term by term on $|\lambda |=R_1$, obtainingThe same integral equals the one on $|\lambda |=R$: applying an arbitrary continuous linear functional reduces contour deformation through the Banach algebra spectrum-free annulus to the scalar Cauchy theorem, and separation of points restores equality in $B$. This proves the resolvent Cauchy coefficient formula rather than assuming vector-valued analytic calculus. Using the norm bound for the allowed vector integral givesTaking $n$th roots yields ${\rho }_0\le R$, and letting $R\downarrow r_{\sigma }$ finishes the proof:

For the first example take the commutative algebra of dual numbers $B=\mathbb{C}[\epsilon ]/({\epsilon }^2)$ with norm $\Vert a+b\epsilon \Vert =|a|+|b|$. Completeness follows from finite dimension, and the product estimate makes the norm submultiplicative. For $x=\epsilon$, $x\ne 0$ but $x^2=0$, so $\rho (x)=0$. Directly, $\lambda e-\epsilon$ has inverse ${\lambda }^{-1}e+{\lambda }^{-2}\epsilon$ for $\lambda \ne 0$, and its Banach algebra spectrum is $\{0\}$. For the second example take $B=\mathbb{C}$ with its absolute-value norm and $x=1$. Its Banach algebra spectrum is $\{1\}$, giving $\rho (x)=\Vert x\Vert =1$.

Suppose $b$ were a left inverse of an algebra element $x$, so $bx=1$. Then $b{x}_n\to 1$, and $b{x}_n$ is invertible for all sufficiently large $n$ by the Neumann series. For such $n$, $b=(b{x}_n)x_n^{-1}$ is a product of invertibles, hence invertible. The equation $bx=1$ then gives $x=b^{-1}$, a contradiction.

If instead $xb=1$, then $x_{n}b\to 1$, so $x_{n}b$ is eventually invertible. Now $b=x_n^{-1}(x_{n}b)$ is invertible, and again $x=b^{-1}$, a contradiction. This proves that a noninvertible limit of invertibles has neither type of one-sided inverse. No finite-dimensional assumption is used.

If $\Vert x_n^{-1}\Vert \Vert x-x_n\Vert <1$, thenis invertible by the Neumann series, a contradiction. Thus every $n$ satisfiesHere $x\ne x_n$ because $x_n$ is invertible and $x$ is not. Since $x_n\to x$, this proves the inverse norm divergence at noninvertible boundary points:

In a unital Banach algebra, the Neumann series gives $(1-u{)}^{-1}={\sum }_{k=0}^{\infty }{u}^k$ whenever $\Vert u\Vert <1$: the series converges in norm by submultiplicativity and completeness, and multiplication of its partial sums on either side gives $1-u^{N+1}\to 1$. Consequently, if $g$ is invertible and $\Vert g^{-1}\Vert \Vert h-g\Vert <1$, writeThe second factor has a Neumann series inverse, so $h$ is invertible. This exhibits a ball around each $g$ in the group of invertible elements of a Banach algebra, proving that group is an open set.

The same series proves continuity of inversion in a Banach algebra, with the useful quantitative estimateThe right side tends to zero as $h\to g$. Since applying inversion twice gives the original element, it is a bijection whose inverse is the same continuous map. Thus inversion is a homeomorphism of the open group of invertibles onto itself.

For the final unheaded request, defineIt is open by the preceding argument and contains $0$. If ${\lambda }_0$ were a boundary point, a sequence ${\lambda }_n\in E$ would converge to ${\lambda }_0$, while openness gives ${\lambda }_0\notin E$. Then $1-{\lambda }_{n}a\to 1-{\lambda }_{0}a$ is a noninvertible limit of invertibles. The one-sided-inverse argument in part (ii) shows that this limit has neither a left inverse of an algebra element nor a right inverse of an algebra element, contradicting the hypothesis. Therefore $E$ has empty boundary; it is both open and closed in the connected complex plane, and $E=\mathbb{C}$.

For every nonzero $\mu$, put $\lambda ={\mu }^{-1}$. The identity $\mu 1-a=\mu (1-\lambda a)$ proves invertibility, so the spectrum of an element $a$ is contained in $\{0\}$. It is nonempty: if the resolvent of an element existed for all $\mu \in \mathbb{C}$, it would be an entire algebra-valued function tending to zero at infinity by the Neumann series. Applying any bounded linear functional and then the Liouville theorem would make it zero; the Hahn-Banach theorem separates points and would force the resolvent itself to be zero, contradicting its inverse equation. Hence

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.

First, a maximal left ideal $L$ is closed. Its norm closure is again a left ideal, since multiplication is continuous. If that closure were all of $A$, some $\ell \in L$ would satisfy $\Vert 1-\ell \Vert <1$ and would be invertible by the Neumann series. But a proper left ideal cannot contain an invertible element: ${\ell }^{-1}\ell =1$ would put $1$ in $L$. Maximality therefore forces $\overline{L}=L$.

Consequently $X=A/L$ is a nonzero quotient Banach space. It is a simple module for left multiplication by $A$, because the inverse image of any submodule under $A\to A/L$ is a left ideal containing $L$, and hence is either $L$ or $A$. Since $La\subseteq L$, right multiplication defines a well-defined bounded module endomorphismIndeed, changing $u$ by an element of $L$ changes $ua$ by an element of $L$, and taking the infimum over representatives proves the quotient norm bound. Right multiplication commutes with the left action by associativity.

Choose $\lambda$ in the nonempty spectrum of an element of $T$ in the Banach algebra $\mathcal{B}(X)$. Both the kernel and range of $T-\lambda I$ are submodules. If this module endomorphism were nonzero, simplicity would force its kernel to be zero and its range to be all of $X$. It would be a bounded bijection of Banach spaces, with bounded inverse by the bounded inverse theorem, contradicting the choice of $\lambda$. Thus $T=\lambda I$. Applying this identity to $1+L$ yields $a-\lambda 1\in L$. If also $a-\mu 1\in L$, then $(\lambda -\mu )1\in L$; since $1\notin L$, $\lambda =\mu$. ThereforeThis is the scalar right action on a maximal-left-ideal quotient; it does not assume that $L$ is a two-sided ideal or that $A/L$ is an algebra.

Now consider the center of an associative algebra $Z$. For each $u\in A$, the map $z\mapsto zu-uz$ is a continuous linear map, so its kernel is closed. Their intersection $Z$ is therefore closed. The identity belongs to $Z$, as do sums and scalar multiples of its elements. If $z,w\in Z$, then for every $u\in A$,Thus $zw\in Z$. Also $zw=wz$, since $z$ commutes with every element of $A$, including $w$. Hence the center is a closed commutative unital subalgebra, and is a Banach algebra in the inherited norm.

For every $z\in Z$, the inclusion $Lz=zL\subseteq L$ allows the first argument to apply. Let $\chi (z)$ be the unique scalar for which $z-\chi (z)1\in L$. Equivalently, the induced right multiplication on $A/L$ is $T_z=\chi (z)I$. Sincethe map $\chi :Z\to \mathbb{C}$ is a complex-linear unital character of an algebra. It is surjective because $\chi (\alpha 1)=\alpha$, and its kernel is precisely $L\cap Z$. ConsequentlyIndeed any ideal properly containing the kernel would have a nonzero image in the field $\mathbb{C}$ and thus would be all of $Z$. The algebra character is also continuous: $|\chi (z)|=\Vert \chi (z)I\Vert =\Vert T_z\Vert \le \Vert z\Vert$, although continuity is not needed for maximality.

An algebra involution is a map $x\mapsto x^{\ast }$ satisfyingThus it is a conjugate-linear map, reverses multiplication, and has square equal to the identity map. These are algebraic axioms; norm continuity, when included in a convention for a Banach star algebra, is an additional requirement and is not used here. A Hermitian element of a star algebra satisfies $h^{\ast }=h$.

Conjugate linearity gives $0^{\ast }=0$. The map $\ast$ is surjective, so every $u$ is $v^{\ast }$ for some $v$. The identities $(v1{)}^{\ast }=1^{\ast }{v}^{\ast }=v^{\ast }$ and $(1v{)}^{\ast }=v^{\ast }{1}^{\ast }=v^{\ast }$ show that $1^{\ast }$ is a two-sided identity; uniqueness of the identity gives $1^{\ast }=1$. Therefore both zero and the identity are Hermitian algebra elements.

For the final unheaded request, first prove equality of the two spectra for a Hermitian element of a star algebra $h\in B$. Write $D=\mathbb{C}\setminus {\sigma }_A(h)$. The spectrum is a compact subset of the real axis by the assumption that $A$ is a Hermitian Banach algebra, so $D$ is connected: points above and below the real axis can be joined by paths passing beyond the endpoints of an interval containing the spectrum; points of the real axis outside the spectrum can first move a short distance vertically. DefineFor large $|\lambda |$, the Neumann series ${\lambda }^{-1}{\sum }_{n\ge 0}(h/\lambda {)}^n$ converges in the closed unital star-subalgebra $B$, so $E$ is nonempty. It is relatively open in $D$ by the same series applied to a perturbation of an already invertible element of $B$. It is relatively closed in $D$, since the resolvent of an element is continuous in $A$ and $B$ is closed. Connectedness gives $E=D$. Conversely an inverse in $B$ is an inverse in $A$, soThis proves the needed spectrum in a closed unital subalgebra equality for Hermitian algebra elements without presupposing that their spectra in $B$ are real.

Now suppose $b\in B$ is invertible in $A$. Both $b{b}^{\ast }$ and $b^{\ast }b$ are Hermitian algebra elements, since reversing products and applying $\ast$ twice fixes them. They are invertible in $A$, by part (ii), so the preceding equality of spectra of Hermitian algebra elements makes them invertible in $B$. Part (ii), applied within $B$, then makes $b$ invertible in $B$. Applying this to $b-\lambda 1$ for every scalar $\lambda$, and using the automatic converse for an inverse already in $B$, proves spectral permanence for Hermitian Banach algebras:Only closedness, a common identity, closure under the algebra involution, and real ambient spectra of Hermitian algebra elements were used; the C-star identity was not assumed.

Put $Z={\sum }_i{Y}_i(\infty )$ and $1=(1,\dots ,1{)}^T$. Sum the bounds in part (a) over the vertices, using linearity of expectation and the Tonelli theorem for the nonnegative series:The bound remains valid if the series diverges, although it then gives no finite estimate. When the spectral radius of $\beta A$ is less than one, the series is a Neumann series, giving the resolvent bound for a discrete SIR epidemic.

Because $A$ is a real symmetric matrix, the spectral theorem gives $\Vert A{\Vert }_2=\rho (A)$. For $\beta \rho <1$, the Neumann series satisfiesLet $m={\sum }_i{X}_i(0)$ be the initial infected count. Since its entries are indicator random variables, $\Vert X(0){\Vert }_2=\sqrt{m}$, while $\Vert 1{\Vert }_2=\sqrt{n}$. The Cauchy-Schwarz inequality and the operator norm bound yieldThus a sufficient asymptotic condition for a small outbreak isFor example, it holds when $m=o(n)$ and $\beta \rho \le 1-\eta$ for a fixed $\eta >0$. For any fixed $\delta >0$, the Markov inequality then givesThis is the spectral condition for a small discrete SIR outbreak. A uniform gap and few initial infections are needed for this conclusion; the finite-population condition $\beta \rho <1$ alone does not control an asymptotically vanishing gap or an initially macroscopic infected set.

On complexified Minkowski spacetime, use the twistor incidence relationThe harmless factor $i$ is a conventional choice; omitting it corresponds to rescaling the complexified coordinates. For fixed $x$, projectivizing the nonzero two-component spinors $\pi$ gives a twistor line $L_x\simeq {\mathbb{CP}}^1$. The two stated patches cover all these incidence lines; they describe the affine-space-time twistor domain with the line $\pi =0$ removed from full projective twistor space.

Since $E{|}_{L_x}$ is holomorphically trivial, choose a global bundle frame on that line. Comparing its columns with the bundle frames in the two patch trivializations gives invertible matrices $H(x,\pi )$ and $\tilde{H}(x,\pi )$, holomorphic on their respective patches, withThis is the holomorphic splitting of a twistor patching matrix. It follows directly from existence of the global bundle frame, rather than from an arbitrary factorization of an arbitrary matrix: nontrivial line bundles would obstruct such a nonsingular splitting.

The factors can be chosen holomorphically in $x$ near any given $x_0$. An explicit local argument is useful. Normalize the known splitting at $x_0$ and set $Q(x,\lambda )=\tilde{H}(x_0,\lambda {)}^{-1}F(x,\lambda )H(x_0,\lambda )$ on a fixed circular contour in the overlap. It is close to the identity when $x$ is close to $x_0$. Let $P_{+}$ select strictly positive Laurent powers, and seek $q=1+$ positive powers satisfyingIn a holomorphic contour-function norm the operator has norm below one for sufficiently small variation, so its Neumann series converges and is holomorphic in $x$. Then $Qq=\tilde{q}$ has only nonpositive powers and is holomorphic on the exterior patch including infinity. Both factors are close to the identity and invertible. They extend over their full patches through $Q$ and its inverse. Taking $H=H(x_0)q$ and $\tilde{H}=\tilde{H}(x_0)\tilde{q}$ gives the required local family. This is also the elementary Laurent series reason that $H^1({\mathbb{CP}}^1,\mathcal{O})=0$ leaves no infinitesimal obstruction.

Raise primed two-component spinor indices with the antisymmetric two-component spinor form and define $V_A={\pi }^{A^{\prime }}{\partial }_{A{A}^{\prime }}$, holding $\pi$ fixed. Incidence givesThus $V_{A}F=0$ for a pulled-back function depending only on twistor coordinates. Differentiating the splitting and multiplying by the inverse factors givesThe left expression is holomorphic on one two-component spinor patch, the right on the other, and each is homogeneous of degree one in $\pi$. They therefore glue to a global matrix-valued section of $\mathcal{O}(1)$ on the line.

Such a section is linear in its homogeneous coordinates. To see this explicitly, set $\lambda ={\pi }_{0^{\prime }}/{\pi }_{1^{\prime }}$ on $U$. The normalized expression is an entire matrix function of $\lambda$; on the other patch its quotient by $\lambda$ is holomorphic at infinity. Hence it has at most linear growth. Cauchy estimates force all its derivatives of order at least two to vanish. Entry by entry it is $a_A(x)+\lambda b_A(x)$, or invariantlyThe coefficients are holomorphic in $x$. This proves existence of the desired space-time one-form, with no remaining dependence on $\pi$ in its coefficients.

A second splitting must differ by right multiplication $H\mapsto Hg(x)$ and $\tilde{H}\mapsto \tilde{H}g(x)$. Indeed the comparison matrix is holomorphic on both patches of a compact complex projective line and therefore constant along the line. The resulting transformation isthe gauge connection transformation law. Local coefficient matrices consequently glue to a holomorphic gauge connection on the bundle whose fiber at $x$ is the space of holomorphic sections of $E{|}_{L_x}$. If a single global matrix-valued one-form is wanted on all $M_{\mathbb{C}}\simeq {\mathbb{C}}^4$, choose a basis at the origin and parallel-transport it along the straight radial paths to each $x$. Linear transport exists on every finite path and varies holomorphically with $x$; this gives a global holomorphic bundle frame. The assertion does not require flat connection in all four directions, only a fixed choice of paths for a trivialization. On more general space-time domains the intrinsic object is a gauge connection with these gauge-related local one-forms.

Finally put $D_{A{A}^{\prime }}={\partial }_{A{A}^{\prime }}+{\Phi }_{A{A}^{\prime }}$ and ${\mathcal{D}}_A(\pi )={\pi }^{A^{\prime }}{D}_{A{A}^{\prime }}$. From the defining identity,Thus the columns of $H^{-1}$ are an invertible parallel bundle frame in the two alpha-plane directions. Their matrix commutator must annihilate that bundle frame, soUse the spinor decomposition of gauge curvaturewhere both gauge curvature two-component spinors are symmetric. The second term vanishes under contraction with ${\pi }^{A^{\prime }}{\pi }^{B^{\prime }}$. The first is a quadratic polynomial in the two two-component spinor components; vanishing for every two-component spinor forces each of its coefficients to vanish. Thereforewhich is precisely the ASDYM equation in the convention where the primed symmetric two-component spinor is the self-dual part. Equivalently, the gauge connection is flat when restricted to every alpha-plane. This completes the forward Penrose-Ward correspondence, including both the existence of the gauge connection and its anti-self-duality of gauge curvature.

Work with a nonzero complex unital Banach algebra, so $1\ne 0$. The spectrum of an element isFor $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges in norm and is a two-sided inverse, by multiplying its partial sums and letting the remainder tend to zero. Thus the spectrum lies in $|\lambda |\le \Vert x\Vert$. The assumed openness of the invertible group makes its complement closed, so the spectrum is compact.

To prove nonemptiness, suppose every $\lambda$ has an inverse $R(\lambda )=(\lambda 1-x{)}^{-1}$. Locally,so the resolvent is holomorphic. At infinity its Neumann series gives $\Vert R(\lambda )\Vert =O(|\lambda {|}^{-1})$. For any bounded linear functional $\ell \in A^{\ast }$, the scalar entire function $\ell (R(\lambda ))$ is bounded: it is bounded on compact disks by continuity, and tends to zero outside them. The Liouville theorem makes it constant, and its limit makes that constant zero. The Hahn-Banach theorem separates points of the normed space $A$, so $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Hence the spectrum is nonempty and compact.

On the algebra of entire functions, setIt is finite, homogeneous and satisfies the triangle inequality and submultiplicativity. If it vanishes, the identity theorem makes $f$ identically zero, so it is an algebra norm. However, no algebra norm on the entire-function algebra can be complete. The coordinate function $Z(z)=z$ satisfies $\sigma (Z)=\mathbb{C}$ algebraically: $Z-\lambda$ vanishes at $z=\lambda$ and has no entire multiplicative inverse. A complete algebra norm would make this a Banach algebra, contradicting the proved boundedness of its spectrum. This is the entire function algebra admits no complete algebra norm obstruction; it applies to every proposed complete algebra norm, not only the displayed one.

For all continuous functions on $\mathbb{C}$, choose continuous cutoffsTheir supports are disjoint and escape every compact set. Thus $f(z)={\sum }_{n\ge 1}n{h}_n(z)$ is a locally finite sum and is continuous. Let $g_n(z)=\max (0,1-4|z-3n|)$. This nonzero continuous function is supported where $h_n=1$ and all the other cutoffs vanish. Hence $f{g}_n=n{g}_n$. Any algebra norm would implyso $n\le \Vert f\Vert$ for every positive integer, impossible. Therefore the algebra of all continuous functions on $\mathbb{C}$ admits no algebra norm, even an incomplete one. This proves the continuous functions on the complex plane admit no algebra norm obstruction directly.

Set $a_n=\Vert x^n\Vert$ and $D={\inf }_{n\ge 1}{a}_n^{1/n}$. If some $x^m=0$, all later powers vanish, their root norms tend to zero, and $D=0$. Otherwise fix $m\ge 1$ and write $n=qm+s$, $0\le s<m$. Submultiplicativity givesSince $q/n\to 1/m$, it follows that ${\mathrm{limsup}}_n{a}_n^{1/n}\le a_m^{1/m}$. Taking the infimum over $m$ and using $a_n^{1/n}\ge D$ provesThis establishes existence of the limit and its infimum characterization before identifying it with the spectral radius $r(x)={\max }_{\lambda \in \sigma (x)}|\lambda |$.

For $|\lambda |>D$, the root test makes ${\sum }_{n\ge 0}{x}^n/{\lambda }^{n+1}$ converge absolutely. Multiplication of partial sums shows that its limit is $(\lambda 1-x{)}^{-1}$. Thus $r(x)\le D$. For the reverse inequality, take any $\rho >r(x)$ and the circle $|\lambda |=\rho$. The resolvent of an element is holomorphic off the spectrum, andTo justify this identity, first use a radius larger than $\Vert x\Vert$ and integrate the uniformly convergent Neumann series term by term. Then deform to radius $\rho$: the intervening annulus has no spectral points. The Cauchy theorem for these Banach-valued integrals follows by applying bounded linear functionals and using their point separation. Consequently, with $M_{\rho }={\max }_{|\lambda |=\rho }\Vert (\lambda 1-x{)}^{-1}\Vert$,Taking root limits gives $D\le \rho$; let $\rho \downarrow r(x)$. Hence the full spectral radius formula is

For the product comparison, if $\lambda \ne 0$ and $R=(\lambda 1-xy{)}^{-1}$, direct multiplication on both sides verifiesIndeed $(\lambda 1-yx)y=y(\lambda 1-xy)$, and the remaining terms cancel against the inverse equation. Interchanging $x,y$ proves the converse. Thus the nonzero spectra of products in opposite orders agree, soThe possible difference at zero has no effect on the maximum modulus, including when that maximum is zero.

Zero really can differ. In $A=\mathcal{B}({\ell }^2({\mathbb{N}}_0))$, let $S$ be the unilateral shift operator, $S{e}_n=e_{n+1}$, and take $x=S^{\ast }$, $y=S$. Then $xy=1$, whereas $yx=1-P_0$, with $P_0$ the projection onto $e_0$. The latter is zero on $e_0$ and identity on its orthogonal complement, and these blocks also explicitly invert it away from zero and one. Hence

Finally, for $t>r(x)$ the root norms of $(x/t{)}^n$ tend to $r(x)/t<1$, so their norms eventually decay geometrically and the powers converge to zero. Conversely, if the powers converge to zero, choose $N$ with $\Vert (x/t{)}^N\Vert <1$. Homogeneity and the infimum formula implyso $t>r(x)$. The admissible positive $t$ are exactly $(r(x),\infty )$, proving the power-decay characterization of the spectral radiusThis includes quasinilpotent elements with radius zero; no positive endpoint must be attained.

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.

For the unheaded representation request, retain the question's standing unital hypothesis. Let $K$ be the character space of an algebra of $A$, with its weak-star or Gelfand topology. The standard commutative Banach-algebra results allowed here give a nonempty compact Hausdorff character space and ${\sigma }_A(a)=\{\phi (a):\phi \in K\}$. In particular, character compactness follows from the Banach-Alaoglu theorem: the equations $\phi (1)=1$ and $\phi (ab)=\phi (a)\phi (b)$ define a weak-star closed subset of the dual unit ball.

The Gelfand transform is the unital homomorphismEvery element of the commutative C-star algebra is normal. Part i and the character spectrum formula therefore giveThus $\Gamma$ is an isometry, hence injective, and its range is closed because $A$ is complete. Part ii gives $\Gamma (a^{\ast })=\overline{\Gamma (a)}$. The range contains constants and separates points of $K$, since two distinct characters differ on some element of $A$. The complex Stone-Weierstrass theorem states that a unital, conjugation-closed subalgebra of $C(K)$ separating points of a compact Hausdorff space is uniformly dense. Its application here, together with closedness of the range, yieldsThis proves the Commutative Gelfand--Naimark theorem from the preceding parts and the stated approximation theorem.