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Let $h=h^{\ast }$. The exponential $u_t=e^{ith}$ is defined by its norm-convergent power series. Continuity and conjugate linearity of the involution give $u_t^{\ast }=e^{-ith}$, and multiplication of the commuting exponential series gives $u_t^{\ast }{u}_t=u_t{u}_t^{\ast }=1$. Thus $u_t$ is a Unitary element of a C-star algebra and, by the C-star identity, $\Vert u_t\Vert =1$.

A character of an algebra is automatically continuous with $|\phi (a)|\le \Vert a\Vert$, as proved in Question 5. Passing it through the exponential series givesThe left side is $e^{-t\Im \phi (h)}$. Considering both signs of $t$ forces $\Im \phi (h)=0$. So characters take real values on self-adjoint C-star elements.

Now write $x=h+ik$ withboth self-adjoint C-star elements. ThenThis proves that characters of a C-star algebra respect the involution. The argument itself does not require commutativity, although a noncommutative algebra need not have any characters.

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.

A character of an algebra is a nonzero complex linear multiplicative map $\varphi :A\to \mathbb{C}$. It necessarily satisfies $\varphi (1)=1$. If $\lambda =\varphi (x)$, then $x-\lambda 1$ cannot be invertible, since applying $\varphi$ to an inverse identity would give $0=1$. Therefore $\varphi (x)\in \sigma (x)$. The Neumann series bound $\sigma (x)\subseteq \{z:|z|\le \Vert x\Vert \}$ givesproving automatic continuity of characters. In particular every algebra character belongs to the closed unit ball of $A^{\ast }$; no prior continuity was needed.

The character space of an algebra is ${\Phi }_A$, the set of all algebra characters. Its Gelfand topology is the weakest topology making every evaluation $\varphi \mapsto \varphi (x)$ continuous, equivalently the weak-star topology inherited from $A^{\ast }$. Within the weak-star compact ball $B_{A^{\ast }}$, it is the closed set specified byThese equations are closed conditions on finitely many evaluations at a time. By the Banach-Alaoglu theorem, ${\Phi }_A$ is compact, and it is Hausdorff because distinct characters differ on an evaluation.

To obtain every spectral value from a algebra character, take $\lambda \in \sigma (x)$. Commutativity makes the ideal $(x-\lambda 1)A$ proper, so it is contained in an algebraic maximal ideal $\mathfrak{m}$. Every such maximal ideal is norm closed: its closure is an ideal, and cannot be all of $A$, because an element of $\mathfrak{m}$ sufficiently close to $1$ would be invertible by the Neumann series. Maximality then makes its closure equal to itself.

The quotient $A/\mathfrak{m}$ is a complex Banach division algebra, hence is $\mathbb{C}$ by the Gelfand-Mazur theorem. Composing the quotient map with this scalar identification gives a algebra character with $\varphi (x)=\lambda$. Together with the first inclusion,The same maximal-ideal argument shows that the algebra character space of a nonzero algebra is nonempty.

Now suppose $A$ is a commutative C-star algebra. All its elements are normal. The C-star identity givessince the self-adjoint element $h=x^{\ast }x$ satisfies $\Vert h^2\Vert =\Vert h{\Vert }^2$. Iterating yields $\Vert x^{2^n}\Vert =\Vert x{\Vert }^{2^n}$. The spectral radius formula along this subsequence provesThis is spectral radius norm equality for normal elements, applied to the commutative case.

For a self-adjoint $h$, the element $v_t=\exp (ith)$ is unitary for every real $t$, so $\Vert v_t\Vert =1$. Algebra character continuity and multiplicativity giveIts modulus is $\exp (-t\mathrm{Im}\varphi (h))$, forcing $\varphi (h)\in \mathbb{R}$. Write $x=h+ik$ with $h,k$ self-adjoint. ThenThis proves that characters of a C-star algebra respect the involution. The conjugation bar is present in the original PDF and must be retained; the TeX aid omits it.

The Commutative Gelfand--Naimark theorem states that a complex commutative unital C-star algebra is isometrically star-isomorphic to $C(K)$ for a compact Hausdorff space, canonically $K={\Phi }_A$. Its Gelfand transform isIt is a unital algebra homomorphism; the conjugation identity just proved makes it preserve the involution. The algebra character description of the spectrum and the spectral-radius equality giveThus it is injective and isometric, and its range is closed because $A$ is complete. The range contains constants, is closed under complex conjugation, and separates points of ${\Phi }_A$: two distinct algebra characters differ on some element of $A$.

The complex Stone-Weierstrass theorem says that a self-adjoint unital subalgebra of $C(K)$ separating points is uniformly dense. For completeness, its familiar lattice argument explains the last step. The real part of its uniform closure is closed under absolute value, by polynomial approximation of $|t|$ on bounded real intervals, and hence under finite maxima and minima. Constants and point separation permit a real function matching any given real continuous $g$ at any chosen pair of points. Fixing the first point, take a maximum of finitely many such functions to obtain a function above $g-\epsilon$ everywhere and equal to $g$ at that first point. It is below $g+\epsilon$ in a neighborhood of that point. A finite cover by these neighborhoods and the minimum of their associated functions then lies between $g-\epsilon$ and $g+\epsilon$ everywhere. Real and imaginary parts give density for complex functions.

Apply this to $\mathcal{G}(A)$. Its range is both dense and closed, so it is all of $C({\Phi }_A)$. ThereforeThis proves the commutative theorem in the setting of the question. For the zero algebra the corresponding compact space is empty and the representation is $C(\varnothing )=\{0\}$.