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Let $C$ consist of continuous functions on the closed unit disk whose boundary values extend to a member of the disk algebra. The extension is unique by the maximum modulus principle; the extension map $T:C\to A(\overline{\mathbb{D}})$ is a surjective unital algebra homomorphism of norm at most one. Its kernel $I$ is the ideal of functions vanishing on the boundary. A character vanishing on $I$ factors through $T$, so it is $f\mapsto T(f)(\lambda )$ for some closed-disk point $\lambda$. For a character $\chi$ not vanishing on $I$, choose $h\in I$ with $\chi (h)\ne 0$. The rule $F\mapsto \chi (hF)/\chi (h)$ extends $\chi$ to a character of all continuous functions on the disk: multiplicativity follows from $\chi (hF)\chi (hG)=\chi (h)\chi (hFG)$. It is therefore evaluation at an interior point, since its value at $h$ is nonzero. These two closed-disk families agree exactly on their boundaries. Their glued compact space maps continuously and bijectively to the Hausdorff character space of an algebra, giving the stated homeomorphism. The algebra is closed because uniform convergence of boundary data gives uniform convergence of their analytic extensions.

Every commutative unital C-star algebra $A$ is isometrically star-isomorphic to $C({\Phi }_A)$ through its Gelfand transform, where ${\Phi }_A$ is its character space of an algebra. Thus abstract algebra elements can be treated as continuous functions on a compact Hausdorff space.

This is the isometric commutative C-star specialization of the Gelfand representation.

For a unital commutative complex Banach algebra $A$, the Gelfand transform sends $a\in A$ to the continuous function $\hat{a}(\phi )=\phi (a)$ on its character space of an algebra. It is a contractive unital algebra homomorphism into $C({\Phi }_A)$.

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.

For the unheaded representation request, retain the question's standing unital hypothesis. Let $K$ be the character space of an algebra of $A$, with its weak-star or Gelfand topology. The standard commutative Banach-algebra results allowed here give a nonempty compact Hausdorff character space and ${\sigma }_A(a)=\{\phi (a):\phi \in K\}$. In particular, character compactness follows from the Banach-Alaoglu theorem: the equations $\phi (1)=1$ and $\phi (ab)=\phi (a)\phi (b)$ define a weak-star closed subset of the dual unit ball.

The Gelfand transform is the unital homomorphismEvery element of the commutative C-star algebra is normal. Part i and the character spectrum formula therefore giveThus $\Gamma$ is an isometry, hence injective, and its range is closed because $A$ is complete. Part ii gives $\Gamma (a^{\ast })=\overline{\Gamma (a)}$. The range contains constants and separates points of $K$, since two distinct characters differ on some element of $A$. The complex Stone-Weierstrass theorem states that a unital, conjugation-closed subalgebra of $C(K)$ separating points of a compact Hausdorff space is uniformly dense. Its application here, together with closedness of the range, yieldsThis proves the Commutative Gelfand--Naimark theorem from the preceding parts and the stated approximation theorem.

The word “every” in the printed deduction must be read under the initial unital hypothesis. If nonunital commutative C-star algebras are included, the compact-space statement is false: the nonzero algebra $c_0$ has no multiplicative identity, whereas $C(K)$ for nonempty compact $K$ does. The general nonunital version is $C_0(K)$ on a locally compact Hausdorff space, with vanishing at infinity. This qualification preserves the proved unital result.

For the final real spectrum criterion for a normal C-star element, the ambient algebra may be noncommutative. Let $B$ be the norm-closed unital subalgebra generated by $x$ and $x^{\ast }$. Normality makes these two generators commute. The involution preserves their polynomial algebra and is continuous, so $B$ is a commutative C-star subalgebra.

We first show that $x$ has the same spectrum in $B$ as in $A$, without assuming spectral permanence. Put $\Omega =\mathbb{C}\setminus {\sigma }_A(x)$. A compact subset of the real axis has connected complement: upper and lower half-plane paths join around either end of a bounded interval containing it, and each omitted real point can be joined locally to the upper half-plane. DefineThis set is nonempty because the Neumann series puts the inverse in $B$ for $|\lambda |>\Vert x\Vert$. It is relatively open: from an inverse in $B$, the local inverse formulastays in $B$ for small $h$. It is relatively closed by continuity of the ambient resolvent and norm-closedness of $B$. Connectedness gives $V=\Omega$. Conversely an inverse in $B$ is an inverse in $A$, soThis is the connected-resolvent case of spectrum in a closed unital subalgebra.

Apply the just-proved commutative C-star representation to $B$. For every character $\psi$ of $B$, $\psi (x)$ lies in this real spectrum. Part ii makes $\psi (x^{\ast })=\overline{\psi (x)}=\psi (x)$. The Gelfand transform of $B$ is injective by its isometry, soFor a nonunital ambient C-star algebra, perform the same argument in its standard C-star unitization, using the usual unitization spectrum.

Normality cannot be dropped. In $M_2(\mathbb{C})$, takeSince $N^2=0$, its spectrum is $\{0\}$: for $\lambda \ne 0$, $(\lambda I-N{)}^{-1}={\lambda }^{-1}I+{\lambda }^{-2}N$, while $N$ is not invertible. Yet $N\ne N^{\ast }$ andThus a real spectrum alone does not force self-adjointness; normality is essential.

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.

Fix Haar measure $m_G$ and use $\hat{f}(\chi )={\int }_{G}f(x)\overline{\chi (x)}d{m}_G(x)$. A positive-definite function $p$ satisfies ${\sum }_{i,j}{z}_i\overline{z_j}p(x_i-x_j)\ge 0$ for every finite choice of points and coefficients. The Bochner theorem states that a continuous positive-definite function on a locally compact Hausdorff Abelian group has a unique representationwhere ${\mu }_p$ is a Radon measure. Conversely, any finite positive Radon measure on the Pontryagin dual group has a continuous positive-definite inverse transform. The sign convention here is the one compatible with the displayed Fourier transform.

First note that the dual is a locally compact Hausdorff topological group. Its group operations are continuous in the compact-open topology proved in question 3. For local compactness, adjoin the zero functional to the character space of an algebra of $L^1(G)$. All its elements lie in the closed unit ball of $L^1(G{)}^{\ast }$, and the equations $\phi (a\ast b)=\phi (a)\phi (b)$ define a closed subset in the weak-star topology. The Banach-Alaoglu theorem makes that subset compact Hausdorff. Removing the zero functional gives an open, hence locally compact, subspace. This provides the setting for constructing the measure on the dual, without assuming a Fourier inversion theorem.

Let $u^{\ast }(x)=\overline{u(-x)}$ and let $\mathcal{P}$ be the finite sums of convolution squares $u\ast u^{\ast }$, $u\in C_c(G)$. Each such square is continuous with compact support, and is positive-definite becauseIts transform is $|\hat{u}{|}^2$, so every $p\in \mathcal{P}$ has $\hat{p}\ge 0$. Also $p\ast q\in \mathcal{P}$: convolution of two squares is $(u\ast v)\ast (u\ast v{)}^{\ast }$, and finite sums distribute. Insert the Bochner representation of $q$ into $p\ast q$ and use the Fubini theorem to obtainThe representing measure on the right is finite and positive. Uniqueness in the Bochner theorem and commutativity now give the Bochner consistency identity for convolution squares

For every compact $K\subseteq \hat{G}$ there is $p\in \mathcal{P}$ with $\hat{p}>0$ throughout $K$. Indeed, at any ${\chi }_0$ choose $u={\chi }_{0}a$, where $a\ge 0$ is a nonzero member of $C_c(G)$. Then $\hat{u}({\chi }_0)=\int a>0$, and continuity supplies a neighbourhood where it stays nonzero. A finite subcover of $K$ and the sum of the corresponding squares give such a $p$.

For $\psi \in C_c(\hat{G})$, choose $p$ positive in transform on its support and defineThe quotient is set to zero away from the support; its denominator is bounded away from zero there. If $q$ is another choice, the consistency identity, multiplied by $\psi /(\hat{p}\hat{q})$ on that compact support, shows that the integral is unchanged. Choosing one $p$ for a union of supports proves linearity; positivity follows from positivity of ${\mu }_p$. On each fixed compact support it is bounded by ${\mu }_p(K)/{\min }_K\hat{p}$ times the uniform norm. Moreover, for every $p\in \mathcal{P}$ and $\psi \in C_c(\hat{G})$, choosing $q$ positive on the support of $\psi$ givesA nonzero square has ${\mu }_p(\hat{G})=p(0)=\Vert u{\Vert }_2^2>0$; a suitable nonnegative compactly supported $\psi$ consequently makes this last expression positive. Thus $J$ is nonzero.

To prove translation invariance, fix $\eta \in \hat{G}$ and modulate $p$ by $p_{\eta }(x)=\eta (x)p(x)$. This again belongs to $\mathcal{P}$, since modulation takes $u\ast u^{\ast }$ to $(\eta u)\ast (\eta u{)}^{\ast }$. Direct calculation and Bochner uniqueness giveHence, for ${\psi }_{\eta }(\chi )=\psi ({\eta }^{-1}\chi )$, using $p_{\eta }$ as the denominator function and changing variables gives $J({\psi }_{\eta })=J(\psi )$. The Riesz representation on compactly supported continuous functions now produces a nonzero translation-invariant positive Radon measure $\nu$ on the dual. It is finite on compact sets, so it is Haar measure. This is the compatible dual Haar measure, denoted $m_{\hat{G}}=\nu$.

The identity for $J(\psi \hat{p})$ says, by uniqueness of Radon measures,Thus $\hat{p}\in L^1(\hat{G},\nu )$ and the Bochner representation immediately provesIt also proves the formula for all finite complex linear combinations of these functions. The measure scale is fixed by this identity: at $x=0$, a nonzero square gives $\int |\hat{u}{|}^{2}d\nu =\Vert u{\Vert }_2^2$.

A wider inversion class is $f\in L^1(G)$ with $\hat{f}\in L^1(\hat{G},\nu )$. To prove this extension, choose nonnegative $a_U\in C_c(G)$ with integral one and support in a small identity neighbourhood $U$, and put $p_U=a_U\ast a_U^{\ast }$. These functions form an approximate identity: they are nonnegative, have integral one and support in $U-U$. Also $0\le {\hat{p}}_U=|{\hat{a}}_U{|}^2\le 1$, and ${\hat{p}}_U\to 1$ uniformly on compact subsets of the dual. For this uniform convergence use equicontinuity of compact families of characters: evaluation is jointly continuous, as follows from the translation quotient in question 3, and a finite cover of a compact family gives uniform control near the identity.

Convolving the already established inversion formula for $p_U$ with $f$, the Fubini theorem givesThis is justified by $\Vert f{\Vert }_1{\mu }_{p_U}(\hat{G})<\infty$. Compact-set uniform convergence and an integrable-tail split implyExplicitly, first choose a compact set outside which $\int |\hat{f}|$ is small, use the bound $|{\hat{p}}_U-1|\le 2$ on that tail, and use uniform convergence on the compact remainder. This argument works for neighbourhood-indexed nets, without assuming metrizability. Consequently $f\ast p_U$ converges uniformly to the continuous inverse integral. On the other hand, translation continuity in Lp on a locally compact group gives $f\ast p_U\to f$ in $L^1$; testing these two limits against $C_c(G)$ shows they agree almost everywhere. At every continuity point of the chosen representative of $f$, the shrinking-support approximate identity also converges pointwise to $f(x)$. We obtain Fourier inversion on a locally compact abelian group:valid everywhere for continuous $f\in L^1(G)$ with integrable transform, and in general as equality with a continuous representative almost everywhere. This class includes all convolution squares above, and their linear span.

Finally, for arbitrary $h\in C_c(G)$, apply the inversion formula at zero to $h\ast h^{\ast }$:Thus the integral Fourier transform is linear and norm-preserving on $C_c(G)$. That space is dense in $L^2(G)$, so for $f\in L^2(G)$ choose $h_n\in C_c(G)$ with $h_n\to f$ in $L^2$ and define $\mathcal{F}f$ as the $L^2(\hat{G},\nu )$ limit of ${\hat{h}}_n$. The norm identity makes this limit exist and independent of the approximating sequence, and preserves linearity. This proves the Plancherel theorem for locally compact abelian groups in the requested form:On $L^1\cap L^2$, choose the approximating sequence to converge in both norms; uniform convergence of the integral transforms and their $L^2$ convergence show that this extension agrees almost everywhere with the original transform. For general $L^2$ functions the extension is defined by norm limits, not by presuming that the pointwise integral exists.

Let $X$ be the character space of an algebra, consisting of nonzero multiplicative complex linear functionals on $A$. We construct the required Gelfand transform and prove its properties.

Characters exist. A maximal proper ideal exists by the axiom of choice and is closed: if it were dense it would contain an element within distance one of the identity, hence an invertible element, contradicting properness. The quotient is a complex unital Banach division algebra, and the Gelfand-Mazur theorem identifies it with $\mathbb{C}$, producing a character. The same argument applied to the proper ideal generated by $a-\lambda$ provesEvery character has norm one, since its value belongs to the spectrum and is bounded by $\Vert a\Vert$, while $\chi (1)=1$. The character equations are weak-star closed in the dual unit ball. Thus Banach-Alaoglu theorem makes $X$ compact Hausdorff; separability of $A$ makes it metrizable.

The algebra is inverse-closed in $B(H)$. If $a\in A$ is invertible there, $a^{\ast }a\ge cI$ for some $c>0$. For $M=\Vert a^{\ast }a\Vert$, the Neumann series for $(a^{\ast }a{)}^{-1}=M^{-1}[I-(I-a^{\ast }a/M){]}^{-1}$ converges in $A$, since the positive contraction in brackets has norm at most $1-c/M<1$. Therefore $a^{-1}=(a^{\ast }a{)}^{-1}{a}^{\ast }\in A$. This proves spectral permanence for C-star algebras in the present concrete setting. In particular a self-adjoint element has real spectrum in $A$, so characters take real values on it. Decomposing $a$ into self-adjoint real and imaginary parts gives $\chi (a^{\ast })=\overline{\chi (a)}$.

Define $\hat{a}(\chi )=\chi (a)$. This is a unital star-homomorphism to $C(X)$. Every element of the abelian star-algebra is normal, and the C-star identity gives $\Vert a^{2^j}\Vert =\Vert a{\Vert }^{2^j}$. The spectral radius formula therefore gives $r_A(a)=\Vert a\Vert$. The displayed spectrum identity now yields $\Vert \hat{a}{\Vert }_{\infty }=\Vert a\Vert$, proving isometry and injectivity. Its image is closed because $A$ is complete, contains the constants, is closed under conjugation, and separates distinct characters. The Stone-Weierstrass theorem makes that image dense in $C(X)$, hence equal to it. $\boxed{A\cong C(X)}$ isometrically as a star-algebra.

A uniform algebra on a compact Hausdorff space $K$ is a closed complex subalgebra $A\subseteq C(K)$ containing the constants and separating points of $K$, with the supremum norm. It is therefore a commutative unital Banach algebra. Its character space of an algebra ${\Phi }_A$, also called its carrier space, consists of nonzero multiplicative linear functionals, equipped with the Gelfand topology.

For $x\in K$, evaluation gives the character of an algebra ${\delta }_x(f)=f(x)$. The map $\delta :K\to {\Phi }_A$ is continuous: composing it with the coordinate $\phi \mapsto \phi (f)$ gives the continuous function $f$ for every $f\in A$. It is injective because $A$ separates points. The Gelfand topology is Hausdorff because distinct algebra characters differ on some element of $A$. A continuous injection from compact $K$ into a Hausdorff space is a homeomorphism onto its image: closed subsets of $K$ have compact, hence closed, images. Its image is also compact and closed. Thusis the asserted homeomorphism. A natural uniform algebra is one for which this image is all of ${\Phi }_A$.

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 $C(\varnothing )=\{0\}$.

Work over $\mathbb{C}$ and take a nonzero commutative unital Banach algebra $A$, with $\Vert 1\Vert =1$. A character of an algebra is a nonzero multiplicative complex linear functional $\chi :A\to \mathbb{C}$. It satisfies $\chi (1)=1$. Moreover $\chi (a)\in {\sigma }_A(a)$: otherwise $a-\chi (a)1$ would be invertible, although its image under $\chi$ is zero. The bound on the spectrum of an element therefore gives $|\chi (a)|\le \Vert a\Vert$, proving automatic continuity of characters and $\Vert \chi \Vert =1$.