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Write $K={\sigma }_A(x)$ and $R_x(\zeta )=(\zeta 1-x{)}^{-1}$. Suppose first that a unital complex-algebra homomorphism sends $Z$ to $x$. If $\lambda \notin U$, the holomorphic function $1/(Z-\lambda )$ belongs to $\mathcal{O}(U)$ and is the multiplicative inverse of $Z-\lambda$. Its image is an inverse for $x-\lambda 1$. Thus existence forces $K\subset U$, even before continuity is used.

For the converse, assume $K\subset U$. Choose a bounded finite union of polygonal regions $V$ with $K\subset V$, $\overline{V}\subset U$, and boundary avoiding $K$. Orient its boundary cycle $\Gamma$ positively around $V$, including negative orientations on holes. Its winding number is one near $K$ and zero outside $U$. DefineThe integral exists in $A$ because the integrand is continuous on finitely many compact contour pieces. The Cauchy theorem shows it is unchanged by replacing $\Gamma$ with another such spectral cycle: the difference has zero winding around the possible singularities in $K$. Banach-valued Cauchy statements follow by applying bounded linear functionals. We use the scalar Cauchy integral formula in its cycle form: a cycle with index one at a point and surrounding only a holomorphy region integrates $f(\zeta )/(\zeta -z)$ to $2\pi if(z)$.

Linearity is immediate and, for a fixed cycle,This is a bound by one compact-uniform seminorm, proving continuity for local uniform convergence. To obtain ${\Theta }_x(1)=1$, deform the resolvent integral, which is holomorphic on $\mathbb{C}\setminus K$, to a large circle and integrate its Neumann series. Similarly $\zeta R_x(\zeta )=1+x{R}_x(\zeta )$ gives ${\Theta }_x(Z)=x$.

For multiplicativity choose an outer spectral cycle for $f$ and an inner one for $g$, nested in $U$ so the inner cycle lies inside the outer region. The resolvent identity isSubstitute it into the double integral for ${\Theta }_x(f){\Theta }_x(g)$. In the term containing $R_x(\eta )$, first integrate $f(\zeta )/(\zeta -\eta )$ around the outer cycle, obtaining $f(\eta )$. In the term containing $R_x(\zeta )$, first integrate $g(\eta )/(\zeta -\eta )$ around the inner cycle, obtaining zero because $\zeta$ lies outside it. Interchange is valid for the continuous integrands on disjoint compact cycles. HenceThis proves the required continuous unital homomorphism, the holomorphic functional calculus.

For uniqueness use the following version of the allowed Runge theorem: on any open $U\subseteq \mathbb{C}$, rational functions with all finite poles outside $U$ approximate each holomorphic function locally uniformly; poles at infinity are allowed, giving polynomials. A unital homomorphism sending $Z$ to $x$ is forced on polynomials and on every inverse $(Z-\lambda {)}^{-1}$, $\lambda \notin U$, and hence on these rational functions. Continuity then forces its value on their limits. Thus ${\Theta }_x$ is unique. This argument works for disconnected $U$ as well and does not assume polynomials alone are dense. It proves continuity and uniqueness of holomorphic functional calculus.

We will use the character description of spectra, with a short justification. A character of an algebra $\phi$ is a nonzero complex multiplicative linear functional; it satisfies $\phi (1)=1$. Applying it to an inverse shows $\phi (a)\in {\sigma }_A(a)$, so $|\phi (a)|\le \Vert a\Vert$ and the character is automatically continuous. Conversely, for noninvertible $a-\lambda 1$, its principal ideal is proper because $A$ is commutative. Put it in a maximal ideal $M$. The closure of $M$ is proper: an ideal containing an element sufficiently close to one contains an invertible element and thus the identity. Maximality makes $M$ closed. The quotient is a complex Banach division algebra. By the nonempty-spectrum result of Question 3, each quotient element differs from some scalar by a noninvertible element, which in a division algebra must be zero. Thus the quotient is $\mathbb{C}$ and its quotient map is a character taking $a$ to $\lambda$. This provesIt is the spectrum equals character values in a commutative Banach algebra statement and does not assume semisimplicity.

For any character, continuity permits passing it inside the contour integral. Since $\phi (R_x(\zeta ))=(\zeta -\phi (x){)}^{-1}$ and $\phi (x)\in K$, the scalar Cauchy formula givesApply the character description first to ${\Theta }_x(f)$ and then to $x$ to obtain the holomorphic spectral mapping theorem

On ${\Pi }_{+}$, use the analytic logarithm with argument in $(-\pi /2,\pi /2)$ and set $s(z)=\exp (\frac{1}{2}\mathrm{Log}z)$. It satisfies $s(z{)}^2=z$ and takes its values in ${\Pi }_{+}$. The principal square root in a commutative Banach algebra isFor uniqueness let $v$ be another root with spectrum in ${\Pi }_{+}$. For each character, $\phi (y)$ and $\phi (v)$ are scalar roots of $\phi (x)$ with positive real part, so both equal $s(\phi (x))$. Hence every character takes the nonzero value $2s(\phi (x))$ at $y+v$. The character description implies $y+v$ is invertible. Commutativity gives $(y-v)(y+v)=y^2-v^2=0$, and multiplying by the inverse yields $y=v$. Equal character values alone would not imply equality in a possibly nonsemisimple algebra; the invertible sum is the essential extra step.

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.

The first assertion needs the C-star identity; a Banach-algebra involution alone is insufficient. For an explicit counterexample, take the dual-number algebraThis is a commutative unital Banach algebra, since it is finite-dimensional andIts involution is conjugate-linear, multiplicative, involutive and isometric. Every character of an algebra kills $\epsilon$, because $\varphi (\epsilon {)}^2=0$. Thus there is just one character, $a+b\epsilon \mapsto a$, and the Gelfand transform sends the nonzero norm-one element $\epsilon$ to zero. It is neither injective nor norm preserving. In particular $\Vert {\epsilon }^{\ast }\epsilon \Vert =0\ne \Vert \epsilon {\Vert }^2$, identifying the missing norm axiom.

Even the claimed compatibility with the involution is not automatic. On ${\mathbb{C}}^2$ with the maximum norm and pointwise multiplication, define $(a,b{)}^{\ast }=(\bar{b},\bar{a})$. This is another isometric involution, but evaluation at the first coordinate takes $(1,0{)}^{\ast }$ to zero rather than to the conjugate of its value $1$ at $(1,0)$. These examples establish involution alone does not imply an isometric Gelfand transform.

We now prove the intended assertion for a commutative complex unital C-star algebra. Let $\mathcal{M}$ be its maximal ideal space, identified with the compact character space of an algebra as in Question 3. The C-star identity and submultiplicativity giveand applying the same inequality to $a^{\ast }$ proves $\Vert a^{\ast }\Vert =\Vert a\Vert$. Thus the involution is continuous. For a self-adjoint element $h$, the exponential series defines $u_t=\exp (ith)$ in norm. Applying the involution term by term gives $u_t^{\ast }=\exp (-ith)$, and multiplying the absolutely convergent series shows $u_t^{\ast }{u}_t=1$. Hence $\Vert u_t\Vert =1$. Every character of an algebra is contractive by Question 3, and soTaking both signs of $t$ forces $\varphi (h)$ to be real. Writingwith $h,k$ self-adjoint, provesThis proves that characters of a C-star algebra respect the involution.

To prove the norm assertion, we also justify the spectral radius formula used here. Put $r(a)={\max }_{\lambda \in \sigma (a)}|\lambda |$. By spectrum equals character values in a commutative Banach algebra,For $R>\Vert a\Vert$, integrating the uniformly convergent Neumann series on $|z|=R$ givesOnly the coefficient of $z^{-1}$ survives. The resolvent is holomorphic outside the spectrum by Question 3, so the Cauchy integral theorem deforms this contour to any circle of radius $R>r(a)$. For algebra-valued integrals the same identity follows by applying every bounded linear functional and then using Hahn-Banach separation. Taking norms givesThe maximum is finite by resolvent continuity on the circle. Taking $n$th roots and then $R\downarrow r(a)$ proves ${\lim }_n\Vert a^n{\Vert }^{1/n}=r(a)$.

Every element is normal because $B$ is commutative. For self-adjoint $h$, the C-star identity gives $\Vert h^2\Vert =\Vert h{\Vert }^2$. Therefore, for arbitrary $a$,Iterating yields $\Vert a^{2^n}{\Vert }^{1/2^n}=\Vert a\Vert$, so the proved spectral radius formula givesThe Gelfand transform is thus isometric and injective. Its range is closed, since $B$ is complete and an isometry takes norm-Cauchy sequences back to norm-Cauchy sequences. It contains the constants, separates the characters, and is closed under complex conjugation by the proved involution identity. The Stone-Weierstrass theorem says that such a unital self-adjoint point-separating subalgebra of $C(\mathcal{M})$ is uniformly dense: apply its real version to real and imaginary parts. Closedness then makes the range all of $C(\mathcal{M})$. Hence the Gelfand transform is an isometric star-isomorphism under the C-star hypothesis, establishing the Commutative Gelfand--Naimark theorem in the form required.

For the finite-dimensional operator application, assume first $V\ne 0$ and let $A$ be the unital algebra generated by $\alpha ,\beta ,{\alpha }^{\ast },{\beta }^{\ast }$. Their pairwise commutation makes $A$ commutative; it is closed under the adjoint and norm closed because it is finite-dimensional. With the operator norm it is a C-star algebra. Indeed inner product duality gives $\Vert T^{\ast }\Vert =\Vert T\Vert$, andThus the missing hypothesis in the first paragraph really does hold for this algebra, independently of any spectral theorem.

The proved result identifies $A$ with $C({\mathcal{M}}_A)$. This character space is finite. For if it had $N$ distinct points, point separation would give, for each pair, a continuous function taking values $1$ and $0$ at the chosen two points. Products of these functions give $N$ functions with Kronecker-delta values on the selected points, hence $N$ linearly independent functions. Taking $N>\dim A$ is impossible. Write ${\mathcal{M}}_A=\{{\varphi }_1,\dots ,{\varphi }_m\}$, a finite Hausdorff and therefore discrete space.

Let $p_j$ be the indicator of $\{{\varphi }_j\}$ and let ${\pi }_j$ be its inverse Gelfand transform in $A$. Since $p_j^2=p_j$, ${\bar{p}}_j=p_j$, $p_j{p}_k=0$ for $j\ne k$, and ${\sum }_j{p}_j=1$, the injective star-isomorphism givesThus the ${\pi }_j$ are mutually orthogonal linear projections, and in particular commute. Put ${\lambda }_j={\varphi }_j(\alpha )$ and ${\mu }_j={\varphi }_j(\beta )$. The pointwise identities $\hat{\alpha }={\sum }_j{\lambda }_j{p}_j$ and $\hat{\beta }={\sum }_j{\mu }_j{p}_j$ pull back toThe ranges are simultaneous eigenspaces: on ${\pi }_{j}V$, the two operators act as the scalars ${\lambda }_j,{\mu }_j$. This proves that finite commuting normal operators have joint spectral projections. If $V=0$, the same conclusion holds with empty sums.

For every character of an algebra $\varphi$, the element $a-\varphi (a)1$ cannot be invertible, so $\varphi (a)\in {\sigma }_A(a)$. Conversely, if $\lambda \in {\sigma }_A(a)$, the principal ideal $A(a-\lambda 1)$ is proper: otherwise $b(a-\lambda 1)=1$ for some $b$, which in this commutative algebra gives an inverse. Part (b) puts that ideal inside a maximal ideal $M$, and the corresponding algebra character from part (c) has $\varphi (a)=\lambda$. ThereforeHere $\hat{a}(\varphi )=\varphi (a)$ is the Gelfand transform. The spectrum equals character values in a commutative Banach algebra result does not require the Gelfand transform to be injective: the algebra need not be semisimple.