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In a commutative complex unital Banach algebra, a maximal ideal is closed: its closure remains proper, since an element within distance less than one of the identity is invertible by the Neumann series. The quotient is a complex division Banach algebra, hence is the complex field by the Gelfand-Mazur theorem. Its quotient map is a character of an algebra. Conversely a nonzero multiplicative complex linear functional has value one at the identity, is onto the complex field, and has maximal kernel. It is automatically continuous because its value at each element belongs to that element's spectrum. The kernel determines the algebra character uniquely since the quotient is one-dimensional.

In a commutative complex unital Banach algebra, a maximal ideal is closed: its closure remains proper, since an element within distance less than one of the identity is invertible by the Neumann series. The quotient is a complex division Banach algebra, hence is the complex field by the Gelfand-Mazur theorem. Its quotient map is a character of an algebra. Conversely a nonzero multiplicative complex linear functional has value one at the identity, is onto the complex field, and has maximal kernel. It is automatically continuous because its value at each element belongs to that element's spectrum. The kernel determines the character uniquely since the quotient is one-dimensional.

For a commutative complex unital Banach algebra, maximal ideals correspond bijectively to characters of an algebra by taking kernels. Give the maximal ideal space the Gelfand topology through this correspondence. It is a compact Hausdorff space, since the characters form a weak-star closed subset of the dual unit ball, which is compact by the Banach-Alaoglu theorem.

A maximal left ideal is a proper left ideal contained in no larger proper left ideal. In a unital Banach algebra it is norm closed: its closure is a left ideal, and if that closure were the entire algebra it would contain elements of $L$ within distance less than one of $1$, hence invertible by the Neumann series. A proper left ideal contains no invertible element. The quotient $A/L$ is consequently a quotient Banach space and a simple module.

If $a$ belongs to a unital Banach algebra and $\Vert a\Vert <1$, thenAbsolute convergence and multiplication of partial sums prove the identity.

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.

Suppose $a_n\to a$ in a unital Banach algebra, with each $a_n$ invertible. If $ba=1$, then $b{a}_n\to 1$, so $b{a}_n$ is invertible eventually by the Neumann series. Then $b=(b{a}_n)a_n^{-1}$ is invertible, and $a=b^{-1}$. The analogous argument with $a_{n}b\to 1$ treats a right inverse of an algebra element. Thus a noninvertible such limit admits neither type of one-sided inverse.

In a unital Banach algebra, for $\lambda \ne 0$ the existence of $R=(\lambda 1-ab{)}^{-1}$ gives $(\lambda 1-ba{)}^{-1}={\lambda }^{-1}(1+bRa)$, as verified by multiplying on either side. Interchanging $a,b$ proves the converse. Zero may differ: for the unilateral shift operator $S$ on ${\ell }^2({\mathbb{N}}_0)$, $S^{\ast }S=1$ has spectrum $\{1\}$ and $S{S}^{\ast }=1-P_0$ has spectrum $\{0,1\}$.

A normed algebra is an algebra over a field with a norm satisfying $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$. Its completion is a Banach algebra, since multiplication extends continuously. Completeness is additional structure, rather than part of the definition of a normed algebra.

A normed division algebra is a nonzero unital normed algebra in which every nonzero element has a two-sided multiplicative inverse. The complex Gelfand-Mazur theorem forces such an algebra to be $\mathbb{C}$, even when completeness is not initially assumed: embed it in its Banach algebra completion and use nonemptiness of the Banach-algebra spectrum.

A algebra character $\phi$ is nonzero, linear and multiplicative, with $\phi (1)=1$. Since $\phi (a)\in {\sigma }_A(a)$, it satisfies $|\phi (a)|\le \Vert a\Vert$ and is automatically continuous. If $h=h^{\ast }$ and $t\in \mathbb{R}$, then $(e^{ith}{)}^{\ast }=e^{-ith}$, so $e^{ith}$ is a Unitary element of a C-star algebra and has norm one. HenceUsing both signs of $t$ forces $\mathrm{Im}\phi (h)=0$. Write $x=h+ik$ with $h=(x+x^{\ast })/2$ and $k=(x-x^{\ast })/(2i)$ both self-adjoint C-star elements. Their algebra character values are real, and thereforeThis is the conjugation in the original PDF; the converted TeX loses the bar.

For the final unheaded request, let $B=C^{\ast }(1,x)\subseteq A$. Normality makes this a commutative unital C-star subalgebra. We first justify spectral permanence, so that we do not silently use the possibly larger algebra spectrum in a mere non-star subalgebra. A self-adjoint C-star element $h$ in a C-star algebra has real algebra spectrum: $e^{ith}$ has norm one, and $\lambda \in \sigma (h)$ implies $e^{it\lambda }\in \sigma (e^{ith})$ for each real $t$. The latter assertion follows from the factorization $e^{ith}-e^{it\lambda }1=(h-\lambda 1)g_t(h)$ with commuting factors. An invertible product of commuting factors would make $h-\lambda 1$ invertible. Both signs of $t$ then force $\lambda \in \mathbb{R}$.

A compact subset of the real line has connected planar complement. Apply Question 1 to the self-adjoint C-star element $h\in B$: all of ${\rho }_A(h)$ is its unbounded component and is retained in $B$. Thus ${\sigma }_A(h)={\sigma }_B(h)$. If $a\in B$ is invertible in $A$, both $a^{\ast }a$ and $a{a}^{\ast }$ are self-adjoint C-star elements and invertible in $A$, hence invertible in $B$. The elements $(a^{\ast }a{)}^{-1}{a}^{\ast }$ and $a^{\ast }(a{a}^{\ast }{)}^{-1}$ are respectively a left and a right inverse in $B$, so coincide and give its inverse. This proves spectral permanence for C-star algebras.

In a commutative unital complex Banach algebra, algebra spectra are the ranges of the characters. Indeed a noninvertible $b-\lambda 1$ lies in a maximal ideal. Such an ideal is closed, since its proper closure cannot contain an element sufficiently near $1$. The quotient is a complex Banach division algebra; nonemptiness of the algebra spectrum in that quotient makes each element scalar, producing an algebra character with value $\lambda$ at $b$. Conversely, an algebra character value cannot correspond to an invertible difference.

Apply this to $B$. The preceding involution result gives $\phi (x^{\ast })=\overline{\phi (x)}$, and spectral permanence applies also to $p(x,x^{\ast })$. ConsequentlyThus the second argument is conjugated as in the original PDF, not a second copy of $\lambda$. This is the spectral mapping rule for a star polynomial in one normal element.

An algebra involution is a conjugate-linear map $a\mapsto a^{\ast }$ satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and $(a^{\ast }{)}^{\ast }=a$; on a unital algebra it fixes $1$. In the Banach-star convention it is continuous, although the following spectral identity uses only its algebraic properties. An element $a$ is invertible precisely when $a^{\ast }$ is: its inverse is $(a^{-1}{)}^{\ast }$. Applying this to $a=\lambda 1-x$ provesFor a nonunital algebra, extend the involution to its unitization of an algebra by $(a,\lambda {)}^{\ast }=(a^{\ast },\overline{\lambda })$ and use the spectrum in that unitization; the same argument proves the identity.

A unital C-star algebra is a complex Banach algebra with an involution satisfying the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. A Normal element of a C-star algebra satisfies $x^{\ast }x=x{x}^{\ast }$. The norm identity for the nonzero identity implies $\Vert 1\Vert =1$.

Work in a nonzero complex unital Banach algebra $B$, with identity $1$. Write $R(x)=\{\lambda \in \mathbb{C}:\lambda 1-x\text{ is invertible}\}$ for the resolvent set, $\sigma (x)=\mathbb{C}\setminus R(x)$ for the spectrum of an element, and $Q(\lambda )=(\lambda 1-x{)}^{-1}$ for its Banach algebra resolvent. Distinguishing the set from the inverse-valued function avoids a notational ambiguity.

The printed formula omits an essential modulus. Complex spectral values cannot be ordered to take the indicated supremum. Even for $x=-1$ in $B=\mathbb{C}$, the spectrum is $\{-1\}$ while the spectral radius is one. The correctly interpreted assertion, proved below, isAll required analytic facts about the Banach algebra resolvent will be derived from norm-convergent series and scalar complex analysis.

For $\Vert a\Vert <1$, the Neumann series ${\sum }_{n\ge 0}{a}^n$ converges in $B$ and is a two-sided inverse of $1-a$: multiplying its finite sums gives $1-a^{N+1}$ and passing to the limit gives $1$. Therefore for $|\lambda |>\Vert x\Vert$,The series converges locally uniformly on that exterior region and tends to zero in norm as $|\lambda |\to \infty$. This statement does not require the optional normalization $\Vert 1\Vert =1$; the zeroth term is $1/\lambda$, and the other terms have the usual geometric bound.

At any ${\lambda }_0\in R(x)$, factor ${\lambda }_{0}1-x+h1=({\lambda }_{0}1-x)(1+hQ({\lambda }_0))$. If $|h|\Vert Q({\lambda }_0)\Vert <1$, the Neumann series givesThus the resolvent set is open, $Q$ is norm-continuous there, and the series proves norm differentiability with $Q^{\prime }(\lambda )=-Q(\lambda {)}^2$. Repeated differentiation gives $Q^{(j)}(\lambda )=(-1{)}^{j}j!Q(\lambda {)}^{j+1}$. Direct subtraction of the inverse equations also yields the resolvent identityIn particular, every continuous linear functional $\varphi$ makes $\varphi (Q(\lambda ))$ a scalar holomorphic function; this has been proved, not assumed as an algebra-valued analytic theorem.

The spectrum of an element is closed, and is contained in $\{\lambda :|\lambda |\le \Vert x\Vert \}$ by the exterior series, so it is compact. It is nonempty. Otherwise $Q$ would be defined and norm-continuous on the entire plane, bounded on every fixed closed disk and tending to zero at infinity. Each $\varphi (Q)$ would be a bounded entire scalar function, hence identically zero by Liouville's theorem. Continuous complex linear functionals separate points of $B$, so $Q(\lambda )=0$, impossible for an inverse. This separation follows from the real Hahn-Banach theorem: a continuous real functional $h$ not vanishing on a given vector gives the complex linear functional $\varphi (v)=h(v)-ih(iv)$, which also does not vanish there. We have therefore proved thatis well-defined and finite.

Next establish the existence of the power-norm limit. If $x^m=0$ for some $m$, the limit is zero. Otherwise fix $m$ and write $n=qm+r$, $0\le r<m$. Submultiplicativity givesThe finitely many remainder factors have bounded size, and $q/n\to 1/m$. Hence ${\mathrm{limsup}}_n\Vert x^n{\Vert }^{1/n}\le \Vert x^m{\Vert }^{1/m}$ for every $m$. The opposite lower bound by the infimum is automatic, soFor $|\lambda |>r_0$, the root test proves convergence of $\sum x^n/{\lambda }^{n+1}$ even if $|\lambda |\le \Vert x\Vert$. Its finite sums telescope against $\lambda 1-x$, and their remainder tends to zero, so this sum is an inverse. Thus $s\le r_0$.

For the reverse inequality, fix $R>s$ and first choose $T>\max (R,\Vert x\Vert )$. The exterior Neumann series converges uniformly on $|\lambda |=T$, so termwise vector integration givesOnly its $x^n{\lambda }^{-n-1}$ term contributes. The annulus between radii $R$ and $T$ is in the resolvent set. For every continuous complex linear $\varphi$, the scalar Cauchy integral theorem applied to ${\lambda }^n\varphi (Q(\lambda ))$ makes the two contour integrals equal. Continuous linear functionals commute with vector integration and separate points, so the vector integrals themselves agree. This proves the resolvent Cauchy coefficient formula without presupposing any Banach-algebra-valued analytic-function theorem:Let $M_R={\max }_{|\lambda |=R}\Vert Q(\lambda )\Vert$, finite by continuity. The contour-length estimate gives $\Vert x^n\Vert \le M_R{R}^{n+1}$. Taking $n$th roots gives $r_0\le R$, and letting $R\downarrow s$ gives $r_0\le s$, including $s=0$. Combined with the earlier inequality, this proves the spectral radius formula and the corrected printed identity.

Use the following general Banach algebra facts: in a complex commutative unital Banach algebra, maximal ideals are exactly the kernels of algebra characters, meaning nonzero complex multiplicative linear functionals. Such a character satisfies $\varphi (1)=1$ and $|\varphi (f)|\le \Vert f\Vert$, hence is continuous. The maximal ideal space is given the Gelfand topology, that is, pointwise convergence of characters on algebra elements. These facts follow from closedness of maximal ideals, the Gelfand-Mazur theorem for their quotients, and the spectral bound for character values. They do not identify any of the three spaces by themselves.

For the Wiener algebra, write $f={\sum }_{n\in \mathbb{Z}}{a}_n{u}^n$, where $u(e^{it})=e^{it}$ and $\Vert f{\Vert }_A=\sum |a_n|$. The Fourier series converges absolutely in the algebra norm. If $\varphi$ is a character, set $z=\varphi (u)$. Since $u$ is invertible and $\Vert u{\Vert }_A=\Vert u^{-1}{\Vert }_A=1$, multiplicativity and the norm bound give $|z|\le 1$, $|z^{-1}|\le 1$. Thus $|z|=1$. Continuity now givesConversely, for every $z\in \mathbb{T}$, evaluation is nonzero, linear and multiplicative, and $|f(z)|\le \sum |a_n|$ makes it continuous. Distinct $z$ give distinct characters because they have different values on $u$. Hence the character space of the Wiener algebra is the circle.

For the analytic Wiener algebra, the same expression uses only $n\ge 0$. A character is again determined by $z=\varphi (u)$, but $u^{-1}$ is no longer in this algebra, so only $|z|\le 1$ is forced. Continuity gives $\varphi (f)={\sum }_{n\ge 0}{a}_n{z}^n$. Conversely, every $z$ in the closed unit disk defines evaluation on this absolutely convergent power series. It is bounded and multiplicative, since multiplication of the series is absolutely convergent discrete convolution. Distinct $z$ again differ on $u$. Therefore the character space of the analytic Wiener algebra is the closed disk, including its interior and boundary, not just the original boundary circle.

For $C(\mathbb{T})$ with the supremum norm, let $M=\ker \varphi$ for a character. Suppose the functions in $M$ had no common zero. At each point choose some $f\in M$ nonzero there. Continuity supplies a neighborhood where it remains nonzero; compactness gives a finite subcover associated to $f_1,\dots ,f_k\in M$. Thenis continuous and strictly positive everywhere. It belongs to $M$ because $M$ is an ideal, but it is invertible in $C(\mathbb{T})$ with continuous inverse $1/g$. This would imply $1\in M$, a contradiction. There is therefore a common zero $z$ of $M$. The point-evaluation kernel $I_z=\{f:f(z)=0\}$ is maximal, and $M\subseteq I_z$ forces equality. Since $\varphi (1)=1$, writing $f=f(z)1+(f-f(z)1)$ then proves $\varphi (f)=f(z)$. Conversely every evaluation is a character, and $u$ distinguishes the points. Thus the character space of continuous functions on the circle is the circle.

These are topological identifications as well as bijections. The evaluation map $z\mapsto {\varphi }_z$ is continuous for the Gelfand topology, because each fixed algebra element is a continuous function of $z$: absolute convergence gives uniform convergence on the circle or closed disk in the first two cases. The parameter spaces are compact, while the character spaces are Hausdorff as subspaces of the pointwise topology. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Consequently the three maximal ideal spaces are, respectively,In each case the maximal ideal associated to a point is the kernel of the evaluation just described. The spelling “Weiner” in the source refers to the Wiener algebra.

A C-star algebra is a complex Banach algebra with a conjugate-linear involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. The inequality $\Vert a{\Vert }^2\le \Vert a^{\ast }\Vert \Vert a\Vert$, and then the same inequality for $a^{\ast }$, show that the involution is isometric. In a nonzero unital C-star algebra, the identity has norm one.

For a Hermitian algebra element $h$, the exponential $e^{ith}$ is unitary for real $t$ and has norm one. If $\lambda \in \sigma (h)$, then $e^{it\lambda }\in \sigma (e^{ith})$: factoring $e^{itz}-e^{it\lambda }=(z-\lambda )q(z)$ with entire $q$ proves this, since the evaluated factors commute. Hence $|e^{it\lambda }|\le 1$ for all positive and negative $t$, forcing $\mathrm{Im}\lambda =0$. Thus Hermitian algebra elements have real Banach algebra spectrum.

For a normal C-star element $a$, commutation of $a,a^{\ast }$ givesThe middle equality of norms uses the C-star identity on the Hermitian algebra element $a^{\ast }a$. Iterating along powers $2^n$ and applying the spectral radius formula proves $r(a)=\Vert a\Vert$.

Now suppose $A$ is commutative and unital. Every maximal ideal is closed: if it were dense, it would contain an element within distance less than one of the identity, hence an invertible element by the Neumann series, which is impossible for a proper ideal. Its quotient is a complex Banach division algebra and is therefore $\mathbb{C}$ by the Gelfand-Mazur theorem. Maximal ideals thus give characters. Conversely, every character $\phi$ satisfies $\phi (a)\in {\sigma }_A(a)$, because applying it to an inverse of $a-\phi (a)1$ would give a contradiction. Characters are consequently contractive. The converse spectral inclusion follows by placing the proper ideal generated by $a-\lambda 1$ in a maximal ideal. We have provedThe character space ${\Phi }_A$ is a weak-star closed subset of the dual unit ball, characterized by $\phi (1)=1$ and $\phi (ab)=\phi (a)\phi (b)$. The Banach-Alaoglu theorem makes it compact Hausdorff. Define the Gelfand transform by $\hat{a}(\phi )=\phi (a)$. For Hermitian $h$, its values are real; writing $a=h+ik$ with both terms Hermitian shows $\widehat{a^{\ast }}=\overline{\hat{a}}$. In the commutative algebra every element is normal, soThe transform is therefore an isometric unital star-homomorphism and has closed range in $C({\Phi }_A)$. Its range separates characters by their definition and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that range dense, hence equal to $C({\Phi }_A)$. This proves the Commutative Gelfand--Naimark theorem:For a nonunital C-star algebra, its C-star unitization can be constructed without assuming an operator representation on a Hilbert space. Represent $A$ by left multiplication on itself, which is isometric because testing $L_a$ on $a^{\ast }/\Vert a\Vert$ gives $\Vert L_a\Vert =\Vert a\Vert$. Adjoin the identity operator and set $\Vert z\Vert =\Vert L_z\Vert$. For $z=a+\lambda 1$ and $b\in A$, the C-star identity in $A$ gives $\Vert zb{\Vert }^2=\Vert b^{\ast }{z}^{\ast }zb\Vert \le \Vert b\Vert \Vert L_{z^{\ast }z}b\Vert$. Consequently $\Vert L_z{\Vert }^2\le \Vert L_{z^{\ast }z}\Vert \le \Vert L_{z^{\ast }}\Vert \Vert L_z\Vert$. Applying this also to $z^{\ast }$ proves that the new involution is isometric and that its C-star identity holds. The extension by the identity is complete, since the isometric image of $A$ is closed and one adjoins just one dimension. Apply the unital result to a commutative $A^{+}$; the character $\infty$ given by $a+\lambda 1\mapsto \lambda$ has kernel $A$. Under the isomorphism, this kernel consists of the continuous functions vanishing at $\infty$. Thus $A\cong C_0({\Phi }_A)$ in the nonunital case, with locally compact character space ${\Phi }_A={\Phi }_{A^{+}}\setminus \{\infty \}$.

For a Hermitian $h$ in a possibly noncommutative algebra, apply this result to the commutative unital C-star subalgebra $C^{\ast }(1,h)$ in the unitization. Its Banach algebra spectrum at $h$ equals the ambient Banach algebra spectrum: both Banach algebra spectra are real, the complement of the ambient compact real Banach algebra spectrum is connected, and the resolvent-component result of Question 1 retains that whole complement in the smaller algebra. Characters of this subalgebra are determined by their values at $h$, so its character space is homeomorphic to $\sigma (h)$. We obtain the continuous functional calculus $C(\sigma (h))\to C^{\ast }(1,h)$.

Interpret ${\mathbb{R}}^{+}$ here as $[0,\infty )$. If $\sigma (h)\subseteq [0,\infty )$, apply this calculus to $t\mapsto \sqrt{t}$, obtaining a Hermitian $k$ with $k^2=h$. In a nonunital algebra this element remains in $A$ because the defining scalar function vanishes at zero. Conversely, if $h=k^2$ with $k=k^{\ast }$, spectral mapping theorem and the real Banach algebra spectrum of $k$ give $\sigma (h)=\{t^2:t\in \sigma (k)\}\subseteq [0,\infty )$. Hence nonnegative Banach algebra spectrum is equivalent to being the square of a Hermitian algebra element. A strict-positive interpretation would exclude $h=0$, and would not be equivalent to the assertion about an arbitrary Hermitian square.

Finally, use the two continuous real functions $t\mapsto \sqrt{\max (t,0)}$ and $t\mapsto \sqrt{\max (-t,0)}$ on $\sigma (x)$. Their calculus values $u,v$ are Hermitian, belong to $A$ also in the nonunital case, and satisfy

Let $G$ be locally compact, Abelian and Hausdorff, and fix a Haar measure. Write $T_{x}f(y)=f(x^{-1}y)$ and use convolutionThe L1 group algebra is a commutative Banach algebra. Here a group character means a continuous homomorphism $\chi :G\to \mathbb{T}$, not the trace of an arbitrary group representation. A Gelfand character is a nonzero multiplicative complex-linear functional on the convolution algebra. We use the conventionWith the alternative standard Fourier convention involving $\overline{\chi }$, one simply conjugates every group character in this identification.

For such $\chi$, the functional is bounded by $\Vert f{\Vert }_1$, and Fubini's theorem, the substitution $y=xz$, and the homomorphism law giveIt is nonzero: if $u\ge 0$ is nonzero in $C_c(G)$, then $f=\overline{\chi }u$ has ${\Phi }_{\chi }(f)=\int u>0$.

Conversely let $\Phi$ be an algebra character. It is automatically bounded, with $|\Phi (f)|\le \Vert f{\Vert }_1$. Indeed, extend it to the unitization by $\tilde{\Phi }(f+\lambda 1)=\Phi (f)+\lambda$. If $|\Phi (f)|>\Vert f{\Vert }_1$, the Neumann series makes $f-\Phi (f)1$ invertible, whereas its character value is zero, contradicting multiplicativity of an invertible element and its inverse.

Choose $g$ with $\Phi (g)\ne 0$ and put $\chi (x)=\Phi (T_{x}g)/\Phi (g)$. Commutativity of the group gives $(T_{x}f)\ast g=f\ast (T_{x}g)$; applying $\Phi$ yieldsThus the ratio is independent of the chosen $g$. Applying the relation successively to $T_x{T}_{y}g=T_{xy}g$ gives $\chi (xy)=\chi (x)\chi (y)$, and $\chi (e)=1$. In particular it never vanishes.

The translation map $x\mapsto T_{x}g$ is continuous in $L^1$. For $g\in C_c(G)$, translated supports near a fixed $x$ lie in one compact set; continuity on that set and a finite cover give uniform convergence of the translated values, hence $L^1$ convergence because the compact set has finite Haar measure. For general $g$, approximate by $C_c(G)$ and use $\Vert T_{x}g{\Vert }_1=\Vert g{\Vert }_1$. Therefore $\chi$ is continuous. For all integers $n$,Positive powers exclude modulus greater than one; negative powers exclude modulus less than one. Hence $|\chi (x)|=1$.

Finally, the $L^1$-valued convolution formula $f\ast g=\int f(x)T_{x}gdx$ and boundedness of $\Phi$ giveCancel $\Phi (g)$ to obtain the displayed formula for every $f$. Different characters give different functionals: if their continuous difference is nonzero at a point, integration against a suitably phased compactly supported test function in a small neighbourhood detects that difference. Thus the correspondence is bijective.

It also identifies the character space of an Abelian L1 group algebra topologically with the Pontryagin dual group. For a net converging uniformly on compact subsets in $\hat{G}$, convergence of the functionals follows first for compactly supported $f$, then for all $L^1$ functions using the common norm bound one. Conversely suppose the functionals converge in the Gelfand topology. Choose $g$ whose limiting character value is nonzero; its values are eventually bounded away from zero. For compact $K\subset G$, the set $\{T_{x}g:x\in K\}$ is compact in $L^1$. Convergence of uniformly bounded linear functionals is uniform on this compact set: a finite norm-net reduces it to convergence at finitely many functions, with the common bound controlling the approximation error. The ratio formula then gives uniform convergence of the group characters on $K$. This proves equality with the compact-open topology, without assuming first countability of $G$.

For later use, these transforms form a self-adjoint subalgebra dense in $C_0(\hat{G})$. The set of algebra characters together with the zero functional is weak-star compact by Banach-Alaoglu theorem and the closed multiplicativity equations. Removing zero gives a locally compact Hausdorff character space; each evaluation $\Phi \mapsto \Phi (f)$ vanishes at infinity, since its modulus-at-least-$ϵ$ set is compact and avoids zero. Transforms separate characters, and none vanishes identically at a given character. They are closed under conjugation because $f^{\ast }(x)=\overline{f(x^{-1})}$ satisfies ${\Phi }_{\chi }(f^{\ast })=\overline{{\Phi }_{\chi }(f)}$. The locally compact Stone-Weierstrass theorem therefore gives the claimed density.

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.

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 $e\ne 0,1$.