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Every element of a nonzero complex unital Banach algebra has a nonempty compact spectrum of an element. If the resolvent of an element existed everywhere, composing it with any bounded linear functional would give a bounded entire function tending to zero at infinity. The Liouville theorem and point separation by the Hahn-Banach theorem would force the resolvent to be zero, contradicting the inverse equation. Nonunital algebras are treated by unitization.

For a nonzero complex unital Banach algebra, defineThis is the spectrum of an element; its complement carries the resolvent of an element $R_x(\lambda )=(\lambda 1-x{)}^{-1}$.

If $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges in $A$. Multiplication of finite partial sums followed by a norm limit gives both inverse identities. Thus the spectrum is bounded by $\Vert x\Vert$. The invertible set is open by the allowed assumption, and $\lambda \mapsto \lambda 1-x$ is continuous, so the spectrum is closed. It is therefore compact if nonempty.

To prove nonemptiness, suppose the resolvent exists for every $\lambda$. At a fixed ${\lambda }_0$, the same series applied locally givesfor $|h|\Vert R_x({\lambda }_0)\Vert <1$, proving norm analyticity. Every bounded complex linear functional $\psi \in A^{\ast }$ therefore gives an entire scalar function $\psi (R_x(\lambda ))$. On compact disks it is bounded by continuity of inversion, and the large-$\lambda$ series givesHere the factor $\Vert 1\Vert \ge 1$ handles an algebra norm whose identity was not normalized to one. The Liouville theorem forces each scalar function to be identically zero. The Hahn-Banach theorem separates points of $A$, so $R_x(\lambda )=0$, contradicting $(\lambda 1-x)R_x(\lambda )=1$. HenceThis includes an explicit proof of the nonemptiness of the Banach-algebra spectrum rather than using it as a named result.

For the algebra of entire functions, choose the compact-disk algebra norm on entire functionsIt is finite by continuity on a compact disk, satisfies the norm axioms, and is submultiplicative. If the norm is zero, $f$ vanishes on the open disk, so the identity theorem gives $f=0$ on the plane. Thus it is an actual algebra norm, not merely a seminorm.

No complete algebra norm is possible. For the coordinate function $Z(z)=z$, the function $Z-\lambda 1$ has a zero at $\lambda$ for every complex $\lambda$ and hence cannot have a multiplicative inverse among entire functions. Thus its algebraic spectrum is all of $\mathbb{C}$, independently of any chosen norm. Were an algebra norm complete, the spectral theorem just proved would make that spectrum bounded, a contradiction. Therefore entire functions admit an algebra norm but no complete algebra norm. This distinguishes failure of every complete algebra norm from failure of just the displayed disk norm.

For continuous functions on the plane, construct disjoint plateaux explicitly. LetThe supports of ${\chi }_n$ lie in the disks $|z-n|\le 1/4$, so they are pairwise disjoint and form a locally finite family. Thus $f$ is continuous. On the support where $g_n$ is nonzero, ${\chi }_n=1$ and all other bumps vanish, giving $f{g}_n=n{g}_n$; also $g_n(n)=1$, so no $g_n$ is zero. If an algebra norm existed, submultiplicativity would implyso $n\le \Vert f\Vert$ for every positive integer, impossible. Consequently the algebra of all continuous functions on the plane admits no algebra norm, with no completeness hypothesis needed.

The sequence of norms is bounded. Choose $M>{\sup }_n\Vert x_n\Vert$ and $M>\Vert x\Vert$. By the spectral radius formula, every relevant spectrum lies in $|\lambda |<M$. The compact set $K=\{\lambda :|\lambda |\le M\}\setminus U$ is disjoint from $\sigma (x)$. If nonempty, continuity of the resolvent of an element gives $C={\sup }_K\Vert (\lambda 1-x{)}^{-1}\Vert <\infty$. For large $n$, $C\Vert x_n-x\Vert <1$, andis invertible for every $\lambda \in K$ by the Neumann series. If $K$ is empty there is nothing to exclude. No spectral points can lie outside the radius-$M$ disc either. Consequently $\sigma (x_n)\subset U$ for all sufficiently large $n$. Taking $U$ to be the disc of radius $r(x)+\epsilon$ also proves upper semicontinuity of the spectral radius.

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

A complex Banach algebra is an associative complex algebra with a complete norm satisfying submultiplicativity, $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$. A field has a nonzero identity $e$, and every nonzero element is invertible. We derive the Gelfand-Mazur theorem from these facts.

If $\Vert z\Vert <1$, the series $e+z+z^2+\cdots$ converges in the complete algebra. Multiplying its partial sums by $e-z$ and passing to the limit proves the Neumann series inverse. In particular, for any $x\in B$ and $|\lambda |>\Vert x\Vert$,This resolvent of an element tends to zero in norm at infinity; the estimate follows from the geometric bound on $\Vert x^n\Vert$.

We next prove that the spectrum of an element cannot be empty. Suppose every $\lambda e-x$ were invertible. At each ${\lambda }_0$, writeThe Neumann series on a small disk about ${\lambda }_0$ proves that $R$ is analytic there. Consequently, for each continuous linear functional $\ell :B\to \mathbb{C}$, the function $\lambda \mapsto \ell (R(\lambda ))$ is entire. It is bounded outside a large disk by the estimate above and bounded on that disk by continuity. The Liouville theorem makes it constant; its zero limit at infinity makes the constant zero.

Continuous complex linear functionals separate points of $B$. Here is the needed consequence of the Hahn-Banach theorem: for $z\ne 0$, define $g(tz)=t\Vert z\Vert$ on the real span of $z$ and extend it as a bounded real linear functional of norm one. Then $\ell (w)=g(w)-ig(iw)$ is complex linear and continuous, and $\mathrm{Re}\ell (z)=\Vert z\Vert \ne 0$. The real extension theorem is proved explicitly in Question 4(i) below. Thus $\ell (R(\lambda ))=0$ for every $\ell$ forces $R(\lambda )=0$, contradicting $(\lambda e-x)R(\lambda )=e\ne 0$. This proves the nonemptiness of the Banach-algebra spectrum.

Choose $\lambda \in {\sigma }_B(x)$. Since $B$ is a field, the noninvertible element $x-\lambda e$ must be zero. Hence every $x\in B$ is a scalar multiple of $e$. The mapis therefore a bijective unital complex-algebra homomorphism. It and its inverse are bounded, since $\Vert \lambda e\Vert =|\lambda |\Vert e\Vert$. Thus it is an isomorphism of Banach algebras. If the usual normalization $\Vert e\Vert =1$ is included in the definition, it is an isometry; without that normalization, it is still the required continuous algebra isomorphism.

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 $\rho (x)=\Vert x\Vert =1$.

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

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.

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.

The Gelfand-Mazur theorem says that a nonzero complex unital Banach algebra in which every nonzero element is invertible is the complex field itself: $a\mapsto \lambda$ where $a=\lambda 1$. This is an isometric isomorphism under the usual normalization $\Vert 1\Vert =1$; without that normalization its norm is simply $\Vert \lambda 1\Vert =|\lambda |\Vert 1\Vert$. We first prove the required spectral and resolvent facts rather than assume them.

For any element $a$ of a complex unital Banach algebra $A$, define its spectrum of an element byIf $\Vert b\Vert <1$, the partial sums of ${\sum }_{n\ge 0}{b}^n$ converge in the complete algebra. Multiplying a partial sum by $1-b$ on either side gives $1-b^{N+1}\to 1$, proving the Neumann series inverse. For $|\lambda |>\Vert a\Vert$ this givesThus the spectrum is bounded. Under $\Vert 1\Vert =1$, $\Vert R(\lambda ,a)\Vert \le (|\lambda |-\Vert a\Vert {)}^{-1}$; for an unnormalized identity the same estimate with numerator $\Vert 1\Vert$ suffices. Also $\lambda R(\lambda ,a)\to 1$ as $|\lambda |\to \infty$.

If ${\lambda }_0$ is outside the spectrum and $R_0=R({\lambda }_0,a)$, then for $|h|\Vert R_0\Vert <1$,This follows by factoring $({\lambda }_0+h)1-a=({\lambda }_{0}1-a)(1+h{R}_0)$ and applying the just-proved series. The series converges locally in norm, proving that the resolvent set is open, the spectrum is closed, and the resolvent of an element is holomorphic. Subtracting inverse equations also gives the resolvent identityconsistent with the derivative $R^{\prime }(\lambda ,a)=-R(\lambda ,a{)}^2$ from the local series.

The spectrum is nonempty. If it were empty, the resolvent would be an entire algebra-valued function tending to zero at infinity. Choose a bounded linear functional $\ell :A\to \mathbb{C}$ with $\ell (1)=1$, extending the functional on $\mathbb{C}1$ by the complex Hahn-Banach theorem. Then $\ell (R(\lambda ,a))$ is an entire function, bounded outside a disk by the Neumann estimate and bounded inside by continuity and compactness. The Liouville theorem makes it constant, and its decay makes that constant zero. But $\lambda \ell (R(\lambda ,a))\to \ell (1)=1$, a contradiction. We have proved that every spectrum is a nonempty compact subset of the disk $|\lambda |\le \Vert a\Vert$.

Now assume every nonzero element of $A$ is invertible. For any $a$, choose $\lambda \in {\sigma }_A(a)$. The noninvertible element $a-\lambda 1$ must be zero. Thus every element is a scalar multiple of the identity, uniquely because $1\ne 0$, andNo commutativity assumption was needed for this part; it is forced by the conclusion. Complex scalars are essential to the spectral argument.

For the maximal ideal space and its correspondence with characters of an algebra, now take $A$ to be a commutative complex unital Banach algebra. Every proper ideal lies in an algebraic maximal ideal by Zorn's lemma: a chain's union is an ideal not containing $1$. Every proper ideal also has proper closure. Otherwise it contains an element $m$ with $\Vert 1-m\Vert <1$, which is invertible by the Neumann series, forcing the ideal to contain $1$. A maximal ideal $I$ is therefore closed, since its closure is a proper ideal containing it.

The quotient $A/I$, with its quotient norm, is a complex Banach algebra. Completeness follows from closedness of $I$, and submultiplicativity follows by taking infima over representatives. It is a division algebra: if $a\notin I$, maximality gives $I+Aa=A$, so $ba+i=1$ for some $b\in A$, $i\in I$. Hence $a+I$ is invertible. By the proved Gelfand-Mazur theorem, $A/I\cong \mathbb{C}$, with the identity mapping to $1$. Composing this isomorphism with the quotient map produces a nonzero multiplicative complex linear functional ${\varphi }_I$ with kernel $I$.

Conversely let $\varphi :A\to \mathbb{C}$ be a nonzero multiplicative complex linear functional, without assuming continuity. Nonzeroness and multiplicativity give $\varphi (1)=1$, hence $\varphi (\lambda 1)=\lambda$ and surjectivity onto $\mathbb{C}$. Its kernel is a maximal ideal. If $a-\varphi (a)1$ were invertible, applying $\varphi$ to its product with an inverse would give $0=1$. Therefore $\varphi (a)\in {\sigma }_A(a)$, soThis proves automatic continuity of characters; with normalized identity their operator norm is one. Finally a kernel determines its character uniquely: for every $a$, its class in $A/I=\mathbb{C}1$ is a unique scalar multiple of the unit, and every character with kernel $I$ must take that scalar value. We have obtained the bijectionThe commutativity in this correspondence cannot be dropped: ${\mathrm{Mat}}_2(\mathbb{C})$ has the maximal two-sided ideal $0$, but has no scalar-valued character, since such a character would have zero kernel and give an impossible injective linear map into $\mathbb{C}$.

For use in the next question, the spectrum equals character values in a commutative Banach algebra. We have proved one inclusion. For the other, if $a-\lambda 1$ is noninvertible, its generated ideal is proper, hence lies in a maximal ideal. The corresponding character takes $a$ to $\lambda$. ThusGive ${\mathcal{M}}_A$ the Gelfand topology, namely the weak-star topology of pointwise convergence on $A$ through these characters. It is Hausdorff and compact: the characters lie in the dual unit ball and form a weak-star closed subset, described by the closed equations $\varphi (1)=1$ and $\varphi (ab)=\varphi (a)\varphi (b)$. Apply the Banach-Alaoglu theorem. The Gelfand transform $a\mapsto \hat{a}$, where $\hat{a}(\varphi )=\varphi (a)$, is consequently a contractive unital algebra homomorphism into $C({\mathcal{M}}_A)$; it separates distinct characters. This argument proves compactness and continuity, but does not assert injectivity or isometry for an arbitrary Banach algebra.

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

For an element $x$ of a unital complex Banach algebra $A$, its spectrum of an element isIf $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so the spectrum is bounded. If ${\lambda }_0$ is in the resolvent set, thenis invertible for $\lambda$ sufficiently close to ${\lambda }_0$, again by a Neumann series. Thus the resolvent is open and the spectrum is closed, hence compact.

If the spectrum were empty, the resolvent $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire Banach-space-valued function and would tend to zero at infinity. For every $f\in A^{\ast }$, the scalar entire function $f(R(\lambda ))$ would be bounded and therefore constant by Liouville theorem. It would be zero because of the limit at infinity. The Hahn-Banach theorem would then force $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Therefore ${\sigma }_A(x)$ is nonempty.

Now let $B\subseteq A$ be a closed unital subalgebra containing $x$. Invertibility in $B$ implies invertibility in $A$, soOn each connected component of $\mathbb{C}\setminus {\sigma }_A(x)$, either the resolvent belongs to $B$ everywhere or nowhere. Indeed, membership holds on a neighbourhood of any one such point by its local Neumann expansion, and the same argument makes the set of such points relatively closed by taking limits in the closed subalgebra $B$. The unbounded component belongs to the first case because the geometric resolvent series lies in $B$ for large $|\lambda |$. ConsequentlyThis is the spectrum in a closed unital subalgebra theorem.

For the bilateral shift $T$ on ${\ell }^1(\mathbb{Z})$, both $T$ and $T^{-1}$ are isometries. The geometric-series argument applied to $T$ for $|\lambda |>1$ and to $T^{-1}$ for $|\lambda |<1$ givesFor $|\lambda |=1$, normalize the vector given by $x_n={\lambda }^{-n}$ on $-N\le n\le N$ and zero elsewhere. Only its two boundary coordinates contribute to $(T-\lambda I)x$, so the norm of that image tends to zero. Thus $T-\lambda I$ is not bounded below and cannot be invertible. Hence

The algebra generated by $T$ is the norm closure of polynomials in nonnegative powers of $T$. It does not contain $T^{-1}$: every $p(T)e_0$ is supported in nonnegative coordinates, whereas $T^{-1}{e}_0=e_{-1}$. Therefore $0\in {\sigma }_{\mathcal{A}}(T)$. The preceding hole theorem then fills the unique bounded component of $\mathbb{C}\setminus \mathbb{T}$, giving

Let $M={\max }_{S}u$ and suppose $u(z_0)=M$. The sub-mean inequality on every closed disk contained in $S$ givesEquality holds throughout. If $u$ were strictly below $M$ at one point of the circle, upper semicontinuity would make it uniformly below $M$ on a small arc, contradicting equality of the average. Thus $u=M$ on every sufficiently small circle centered at $z_0$, and hence throughout a neighborhood of $z_0$.

The set $\{u=M\}$ is therefore open. It is also closed because upper semicontinuity makes $\{u<M\}$ open. Since the domain is connected and $\{u=M\}$ is nonempty, it is all of $S$. This proves the maximum principle for subharmonic functions.

LetThe resolvent set is open, and the resolvent of an element is operator-valued holomorphic on each of its components. Fix $z_0$ in the resolvent set. For any $\delta >0$, choose unit vectors $x,y$ such thatThe scalar function $f(z)=⟨R(z)x,y⟩$ is holomorphic. Its modulus is subharmonic, soLetting $\delta \downarrow 0$ proves that the resolvent norm is subharmonic on the resolvent component containing $z_0$.

Use the convention that the reciprocal resolvent norm is zero on the spectrum. Suppose a bounded component $U$ ofcontained no spectral point. A spectral point in its boundary would belong to the same pseudospectral component, so $\overline{U}$ is contained in the resolvent set. On $\partial U$ one has $\Vert R(z)\Vert \le {ϵ}^{-1}$, whereas inside $U$ one has $\Vert R(z)\Vert >{ϵ}^{-1}$. Continuity on the compact set $\overline{U}$ makes the resolvent norm attain a maximum at an interior point. The subharmonic maximum principle would make it constant, contradicting its boundary values. Hence every bounded component of the pseudospectrum contains spectrum:

For $x$ in a unital complex algebra $A$, the spectrum of an element isFor a nonunital algebra one uses its unitization.

Now let $A$ be a Banach algebra. The invertible group is open, so the resolvent set is open and the spectrum is closed. If $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so ${\sigma }_A(x)$ is contained in the closed disc of radius $\Vert x\Vert$ and is therefore compact.

If the spectrum were empty, $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire $A$-valued function. For each $f\in A^{\ast }$, the scalar function $f(R(\lambda ))$ is bounded: it tends to zero at infinity by the Neumann series and is bounded on every compact disc. The Liouville theorem makes it identically zero. Since the Hahn-Banach theorem separates points, this would give $R(\lambda )=0$, contradicting its invertibility. Hence the spectrum is nonempty.

For a closed linear operator $A$, the resolvent is the bounded inverse $(zI-A{)}^{-1}$ when $z$ lies in the resolvent set of an operator. This definition includes unbounded $A$. It specializes to the resolvent of an element when $A$ is an element of a Banach algebra.

The algebra resolvent set consists of those scalars for which $\lambda 1-x$ is invertible in the algebra. It is open by the Neumann series, and the inverse-valued map is the resolvent of an element. A closed subalgebra can have a smaller algebra resolvent set because an ambient inverse need not lie in it.