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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.

For $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges in $B$. The exterior of this disk is connected and belongs to both algebra resolvent sets. Since each Banach algebra spectrum is compact, each algebra resolvent set has exactly one unbounded component: any unbounded component meets that exterior, and two components meeting a common connected exterior must coincide. Part (ii) says that the component of ${\rho }_A(x)$ meeting the exterior is also a whole component of ${\rho }_B(x)$. Hence the unbounded components are identical.

Use ${\sigma }_A(x)$ and ${\rho }_A(x)$ for Banach algebra spectrum and algebra resolvent set. An inverse in the smaller unital algebra is also an inverse in the larger one, so ${\rho }_B(x)\subseteq {\rho }_A(x)$.

If $\lambda \in {\rho }_B(x)$ and $r=(\lambda 1-x{)}^{-1}\in B$, then for $|\mu -\lambda |\Vert r\Vert <1$ the Neumann series givesThus ${\rho }_B(x)$ is open. To prove relative closedness, suppose ${\lambda }_n\in {\rho }_B(x)$ and ${\lambda }_n\to \lambda \in {\rho }_A(x)$. Continuity of inversion in $A$ gives $({\lambda }_{n}1-x{)}^{-1}\to (\lambda 1-x{)}^{-1}$. Since $B$ is norm closed, the limiting inverse belongs to $B$, so $\lambda \in {\rho }_B(x)$. Consequently ${\rho }_B(x)$ is relatively clopen in ${\rho }_A(x)$.

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

Here “subspace” is understood to mean a closed closed linear subspace. Closedness matters: the finitely supported sequences $c_{00}$ are an infinite-dimensional algebraic subspace of ${\ell }^p$, but contain no infinite-dimensional complete subspace. Indeed, any such subspace would be the countable union of its intersections with the finite-dimensional coordinate spaces, contrary to the Baire category theorem. We prove the assertion with the intended closedness convention, for $1\le p<\infty$.

Write $E={\ell }^p$, the l-p sequence space. Choose positive numbers ${\delta }_n$ with ${\sum }_{n}2{\delta }_n<1$. Starting with $m_0=0$, choose $x_n\in X$ with $\Vert x_n{\Vert }_p=1$ and its first $m_{n-1}$ coordinates zero. This is possible because the kernel of finitely many coordinate linear functionals on an infinite-dimensional vector space is still infinite-dimensional. Choose $m_n>m_{n-1}$ so that the part of $x_n$ after coordinate $m_n$ has norm below ${\delta }_n$. Let $v_n$ be its restriction to $I_n=(m_{n-1},m_n]$, and put $u_n=v_n/\Vert v_n{\Vert }_p$. The triangle inequality givesThe vectors $u_n$ form a normalized block basic sequence. Their supports are disjoint, soConsequently $U=\overline{\mathrm{span}}\{u_n\}$ is isometric to the l-p sequence space.

For each block choose a norm-one linear functional ${\varphi }_n$ supported on $I_n$ with ${\varphi }_n(u_n)=1$. The Hahn-Banach theorem supplies such a functional, including when $p=1$. DefineThe series converges in the l-p sequence space, sinceAlso $Q{u}_n=u_n$, so $Q$ is a bounded projection onto $U$. Now define the bounded linear operatorThis series converges in operator norm and $\Vert A-I\Vert \le {\sum }_{n}2{\delta }_n<1$. The Neumann series therefore makes $A$ invertible. Disjoint support gives $A{u}_n=x_n$. Thus $Y=AU=\overline{\mathrm{span}}\{x_n\}$ lies in the closed subspace $X$, is isomorphic to the l-p sequence space, and is the range of the bounded projection $AQ{A}^{-1}$ on the whole space $E$. This proves the required copy of ${\ell }^p$ is complemented in the ambient ${\ell }^p$, not just in $X$.

Now let $Z$ be an infinite-dimensional complemented subspace of $E$. The construction supplies $Y\subset Z$ with $Y\cong E$ and a bounded projection $R:E\to Y$. Its restriction to $Z$ gives the topological direct sumIf $P:E\to Z$ is a bounded projection, then $(I-R)P$ is a bounded projection onto $W$: its range lies in $W$, and it fixes every element of $W$. Hence $W$ is a complemented subspace of $E$, so $E\cong W\oplus V$ for some closed $V$.

We give the absorption argument in the Pełczyński decomposition method explicitly. Reindexing coordinates by a bijection between ${\mathbb{N}}^2$ and $\mathbb{N}$ yields $E\cong ({⨁}_{n\ge 1}E{)}_p$. Applying the fixed Banach space isomorphism $E\cong W\oplus V$ in every coordinate givesThe extra $W$ is absorbed by shifting the countably many $W$ summands. The maps are bounded with bounded inverses because the same two-space decomposition is used uniformly in each summand; we may equip each finite direct sum with its equivalent $p$-sum norm. Since $Z\cong E\oplus W$, $Z\cong {\ell }^p$.

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 $e\ne 0,1$.

A C-star algebra is a complex Banach algebra with an antilinear algebra involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a^{\ast }{)}^{\ast }=a$ and the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. First take a nonzero unital algebra. Its identity has norm one, since $\Vert 1{\Vert }^2=\Vert 1\Vert$. Alsoand exchanging $a,a^{\ast }$ proves $\Vert a^{\ast }\Vert =\Vert a\Vert$. Thus the algebra involution is continuous. An element $h=h^{\ast }$ is a Hermitian C-star element, and an element satisfying $x{x}^{\ast }=x^{\ast }x$ is a normal C-star element.

We establish the spectral facts needed for the representation, to avoid assuming the conclusion. For a Hermitian C-star element $h$, the Banach algebra exponential $u_t=e^{ith}$ is unitary: taking adjoints of its absolutely convergent series gives $u_t^{\ast }=e^{-ith}$, and multiplication gives $u_t{u}_t^{\ast }=1$. Hence $\Vert u_t\Vert =1$. If $\lambda \in \sigma (h)$, then $e^{it\lambda }\in \sigma (u_t)$. Indeed, the entire factorizationcan be evaluated by its convergent power series at $h$. If the resulting product were invertible, its commuting factors would make $h-\lambda 1$ invertible, a contradiction. The spectral bound therefore gives $|e^{it\lambda }|\le 1$ for every real $t$. Both signs of $t$ force $\mathrm{Im}\lambda =0$. Thus Hermitian elements have real spectrum.

For completeness, the spectral radius formula needed below follows from the same resolvent theory. Submultiplicativity givesFor the existence of the limit, write $n=qj+r$ with $0\le r<j$ and bound $\Vert a^n\Vert \le \Vert a^j{\Vert }^q\Vert a^r\Vert$; this bounds the upper limit by every $j$th root, while the infimum bounds the lower limit. If a power vanishes, the conclusion follows directly. For $|\lambda |>\gamma$, the resolvent series converges by the root test, so $r(a)\le \gamma$. Conversely, for any $R>r(a)$, contour integration of the resolvent givesAt a larger circle this follows by termwise integration of the Neumann series; deformation to radius $R$ is justified by norm analyticity and scalar Cauchy theory as in Question 4. Therefore $\Vert a^n\Vert \le R^{n+1}{\sup }_{|\lambda |=R}\Vert R_a(\lambda )\Vert$, so $\gamma \le R$. Letting $R\downarrow r(a)$ proves $r(a)=\gamma$.

For a normal C-star element $a$, its adjoint commutes with it. The C-star identity giveswhere the middle norm equality uses the identity for the Hermitian element $a^{\ast }a$. Every power is again normal, so $\Vert a^{2^j}\Vert =\Vert a{\Vert }^{2^j}$. The spectral-radius formula yieldsThis derives the spectral radius norm equality for normal elements from the axioms.

Now let $A$ be commutative and unital. Its character space of an algebra $\Delta (A)$ is compact Hausdorff in the weak-star topology. Every algebra character has norm at most one by the argument in Question 4, and this space is the weak-star closed subset of the dual unit ball defined by $\varphi (1)=1$ and $\varphi (ab)=\varphi (a)\varphi (b)$. Compactness follows from the Banach-Alaoglu theorem; weak-star separation makes it Hausdorff.

We also need the full character-spectrum correspondence, not just one inclusion. Every proper ideal is contained in a maximal ideal. Such a maximal ideal $M$ is closed: if its closure contained one, it would contain an element within norm distance less than one of the identity, which is invertible by the Neumann series, contradicting properness. Its closure is therefore proper and maximality gives equality. The quotient $A/M$ is a complex Banach division algebra. By nonemptiness of its spectrum, every quotient element $b$ has some $\lambda$ for which $b-\lambda 1$ is noninvertible; in a division algebra this means $b=\lambda 1$. Thus the quotient is the scalars, proving the needed Gelfand-Mazur theorem conclusion directly and producing a algebra character with kernel $M$.

If $a-\lambda 1$ is noninvertible, the ideal it generates is proper, since commutativity would otherwise supply its inverse. A containing maximal ideal therefore supplies a algebra character with $\varphi (a)=\lambda$. The converse inclusion was already proved. HenceThe Gelfand transform $a\mapsto \hat{a}$, $\hat{a}(\varphi )=\varphi (a)$, is a unital algebra homomorphism into $C(\Delta (A))$. If $h=h^{\ast }$, the reality of its spectrum makes $\varphi (h)$ real. Writing $a=h+ik$ with Hermitian C-star elements $h,k$ gives $\varphi (a^{\ast })=\overline{\varphi (a)}$. Thus the transform respects the algebra involution. Every element of a commutative C-star algebra is normal, and thereforeIt is a linear isometry, hence injective and with closed image because $A$ is complete. The image contains constants, separates algebra characters by their definition, and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that image dense in $C(\Delta (A))$, and closedness makes it all of that space. We have proved the Commutative Gelfand--Naimark theorem:These steps show exactly how completeness, the algebra involution and the C-star norm identity enter the theorem. A general Banach algebra need not have an injective or norm-preserving Gelfand transform.

The general commutative theorem also has a nonunital form. Here is an elementary construction of the necessary C-star unitization, without assuming the representation theorem. Left multiplication $L_a$ on $A$ is isometric in operator norm: the upper bound is submultiplicativity, and testing on $a^{\ast }/\Vert a\Vert$ proves the reverse bound for nonzero $a$. If $A$ is nonunital, adjoin the identity operator to the closed algebra $L(A)$ inside the bounded operators on the Banach space $A$. The sum $L(A)+\mathbb{C}I$ is closed, since it is a finite-dimensional extension of a closed subspace, and $I\notin L(A)$. A left identity in a C-star algebra would, by taking adjoints, give a right identity; the two coincide, so $I\in L(A)$ would contradict nonunitality.

For the formal element $z=a+\lambda 1$, define its algebra involution by $z^{\ast }=a^{\ast }+\overline{\lambda }1$ and norm by $\Vert L_z\Vert$. For every $b\in A$,Taking suprema over the unit ball and using submultiplicativity gives $\Vert L_z{\Vert }^2\le \Vert L_{z^{\ast }z}\Vert \le \Vert L_{z^{\ast }}\Vert \Vert L_z\Vert$. Apply the same inequalities to $z^{\ast }$ to obtain $\Vert L_z\Vert =\Vert L_{z^{\ast }}\Vert$ and then the C-star identity. This is the C-star unitization by left multiplication. The original algebra is a closed ideal in it.

For commutative nonunital $A$, apply the proved unital theorem to $\tilde{A}$. The scalar quotient algebra character $\epsilon (a+\lambda 1)=\lambda$ identifies $A=\ker \epsilon$ with the continuous functions on $\Delta (\tilde{A})$ that vanish at $\epsilon$. Removing this one algebra character gives a locally compact Hausdorff space; those functions are exactly its continuous functions vanishing at infinity. Restriction and extension of nonzero algebra characters identify this space with $\Delta (A)$: a algebra character extends uniquely by $\tilde{\varphi }(a+\lambda 1)=\varphi (a)+\lambda$, and every algebra character other than $\epsilon$ restricts nontrivially. Thus the full nonunital conclusion is $A\cong C_0(\Delta (A))$.

Return to a normal C-star element $x$ in a unital C-star algebra, possibly noncommutative. Let $B=C^{\ast }(1,x)$, the closed unital star algebra generated by $x$ and $x^{\ast }$. It is commutative because the two generators commute. We first prove the spectral permanence for C-star algebras needed to use the ambient spectrum rather than a potentially larger subalgebra spectrum.

If $h=h^{\ast }\in B$, both its ambient and subalgebra spectra are real. The complement of the compact real set ${\sigma }_A(h)$ is connected. Within this complement, the set of $\lambda$ for which $(\lambda 1-h{)}^{-1}\in B$ is nonempty by the large-parameter Neumann series; it is open by a local inverse series in $B$ and closed by continuous inversion in $A$ and closedness of $B$. It is therefore the entire complement, proving ${\sigma }_B(h)={\sigma }_A(h)$. If a general $b\in B$ is invertible in $A$, then $b^{\ast }b$ and $b{b}^{\ast }$ are invertible Hermitian elements, whose inverses lie in $B$. The elements $(b^{\ast }b{)}^{-1}{b}^{\ast }$ and $b^{\ast }(b{b}^{\ast }{)}^{-1}$ give a left and a right inverse for $b$ inside $B$, and these inverses coincide. Applying this to every $b-\lambda 1$ proves ${\sigma }_B(b)={\sigma }_A(b)$ for all $b\in B$.

By the proved commutative theorem, $B\cong C(\Delta (B))$. The mapis onto by the character-spectrum correspondence and spectral permanence. It is injective because algebra characters respect adjoints and are determined by their values on the star polynomials in one normal element, which are dense in $B$. It is a continuous bijection from a compact space to a Hausdorff space, hence a homeomorphism. Consequently the inverse Gelfand transform, under this identification, suppliesIt is a continuous unital star homomorphism and a linear isometry. Its image is exactly $B$. If another such map sends $Z$ to $x$, it sends $\overline{Z}$ to $x^{\ast }$ and agrees on every polynomial in $Z,\overline{Z}$. These functions contain constants, separate points of the compact spectrum and are closed under conjugation; Stone-Weierstrass gives their uniform density. Continuity proves uniqueness on all of $C({\sigma }_A(x))$. This is the requested continuous functional calculus, including its isometry and image assertion.

Two conventions in the printed statement need to be made explicit. A unital map into $A$ requires $A$ to have an identity. If nonunital C-star algebras are allowed, for example $A=C_0(\mathbb{R})$, the asserted unital map into $A$ does not exist in general; the unital construction instead has target $\tilde{A}$. Also the image above is the closed unital star algebra $C^{\ast }(1,x)$, not necessarily the smallest nonunital algebra generated by $x,x^{\ast }$. For $x=0$ in $A=\mathbb{C}$, the former is $\mathbb{C}1$ and the latter is $\{0\}$. Thus the intended assertion is correct if “generated” includes the identity; without that convention, this is a counterexample. The distinction is the unital versus nonunital generation by a normal element.

For a nonunital $A$, take $\sigma (x)$ in $\tilde{A}$ and construct the unital calculus there. The scalar quotient algebra character sends $x$ to zero, so it sends $f(x)$ to $f(0)$. Functions with $f(0)=0$ therefore give elements of $A$. They are isometrically identified with $C_0(\sigma (x)\setminus \{0\})$ by restriction and zero extension. Approximating them by star polynomials and subtracting the constant value at zero proves that their image is exactly $C^{\ast }(x)$, with no identity adjoined. This proves the nonunital continuous functional calculus as well as specifying the corrected form of the original claim.

We finish with the positive-square-root theorem, giving a proof that even allows $x$ to be nonnormal. Define a positive element spectrally as a Hermitian element with spectrum in $[0,\infty )$. The calculus of a Hermitian element shows its square is positive. Positivity is closed under sums: for positive $p,q$, let $s=\Vert p\Vert +\Vert q\Vert$. The calculus gives $\Vert \Vert p\Vert 1-p\Vert \le \Vert p\Vert$ and the analogous bound for $q$, hence $\Vert s1-p-q\Vert \le s$. Since $p+q$ has real spectrum, the spectral norm bound forces it into $[0,2s]$. The cone is proper: an element positive and negative has spectrum in $\{0\}$ and therefore norm zero.

To establish positivity of an arbitrary adjoint product, put $b=x^{\ast }x=b^{\ast }$. Let $n=\max (-b,0)$ by the continuous calculus of $b$, and let $d=x{n}^{1/2}$, where the square root of the already positive $n$ is supplied by that same scalar calculus. Multiplicativity givesThus $d^{\ast }d$ has nonpositive spectrum. The nonzero spectra of $d^{\ast }d$ and $d{d}^{\ast }$ agree: for general products $uv,vu$ and $\lambda \ne 0$, direct multiplication verifieswhenever the inverse on the right exists, and the converse follows on exchanging $u,v$. Hence $d{d}^{\ast }$ is also nonpositive. Write $d=u+iv$ with Hermitian C-star elements $u,v$. Thenis positive, as a sum of Hermitian squares, and nonpositive, as a sum of the two nonpositive elements. Properness makes it zero. Therefore $d^{\ast }d=-d{d}^{\ast }$ is both positive and nonpositive, hence zero. The C-star identity gives $d=0$, and $n^2=0$ then implies $n=0$. Thus $b=x^{\ast }x$ is positive. This proof does not use positivity of adjoint products in establishing an earlier step.

Apply the continuous calculus of $b$ to $t\mapsto \sqrt{t}$ on its nonnegative compact spectrum. It gives a positive Hermitian C-star elementFor the originally normal $x$, the equivalent direct formula is $h={\theta }_x(z\mapsto |z|)$. In a nonunital algebra, the unitized construction still lands in $A$, because the square-root function is zero at zero.

For uniqueness, suppose $h^{\prime }=h^{\prime \ast }$ has nonnegative spectrum and $(h^{\prime }{)}^2=b$. It commutes with $b$, and hence with $h$, which is a norm limit of polynomials in $b$. The unital C-star algebra generated by $h,h^{\prime }$ is therefore commutative. Its Gelfand transform, together with spectral permanence, gives two nonnegative continuous functions whose squares are equal. Pointwise uniqueness of the nonnegative real square root makes the functions equal, and injectivity of the transform gives $h=h^{\prime }$. This proves the positive square root in a C-star algebra in full, without an unwarranted cancellation by $h+h^{\prime }$ when zero lies in its spectrum. In the notation ${\mathbb{R}}^{+}$, zero must be included here: a noninvertible $x$, in particular $x=0$, need not have a strictly positive square root.