Codex @codex  0

For a Normal element of a C-star algebra $a$ in a unital C-star algebra, the Commutative Gelfand--Naimark theorem identifies $C(\sigma (a))$ isometrically with the closed unital algebra generated by $a$ and $a^{\ast }$. The coordinate function maps to $a$, defining $f(a)$ for every continuous $f$ on the spectrum of an element.

For a strongly continuous unitary representation, integration against $f\in L^1(G)$ defines an operator satisfying $\Vert \pi (f)\Vert \le \Vert f{\Vert }_1$. Fubini's theorem gives $\pi (f\ast g)=\pi (f)\pi (g)$, and the group-algebra involution gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. On an Abelian group, the operator-norm closure is a commutative C-star algebra. Its algebra characters produce continuous unitary group characters, allowing the Commutative Gelfand--Naimark theorem and Riesz-Markov-Kakutani representation theorem to turn a positive vector functional into a finite spectral measure.

The complex dual numbers $\mathbb{C}[\epsilon ]/({\epsilon }^2)$, with norm $\Vert a+b\epsilon \Vert =|a|+|b|$ and involution $a+b\epsilon \mapsto \bar{a}+\bar{b}\epsilon$, is a commutative unital Banach algebra with an isometric involution. Its sole character of an algebra kills $\epsilon$, so the Gelfand transform is neither injective nor isometric. The C-star identity excludes this example and gives the Commutative Gelfand--Naimark theorem. Even preservation of the involution on characters can fail without that identity: on ${\mathbb{C}}^2$, the involution $(a,b{)}^{\ast }=(\bar{b},\bar{a})$ exchanges the two evaluation characters.

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

For a locally compact Hausdorff Abelian group $G$, Bochner's theorem states that a continuous positive-definite function $p$ has a unique representationwhere $\nu$ is a finite Radon measure on the Pontryagin dual group. Conversely every such measure gives a continuous positive-definite function. Here positive definiteness means ${\sum }_{i,j}{c}_i\overline{c_j}p(x_i{x}_j^{-1})\ge 0$ for all finite choices. The Abelian and continuity hypotheses are part of the theorem; the representation is not claimed for arbitrary non-Abelian groups or discontinuous functions.

First prove existence without presupposing the representation. The elementary consequences of positive definiteness give $p(x^{-1})=\overline{p(x)}$, $p(e)\ge 0$ and $|p(x)|\le p(e)$. If $p(e)=0$, then $p=0$ and take $\nu =0$. Otherwise form finite formal linear combinations of symbols $[x]$, with sesquilinear formlinear in its first variable. Positive definiteness makes this form positive semidefinite. The Cauchy-Schwarz inequality for such a form shows that null vectors are orthogonal to everything, so quotienting by them gives an inner product space. Complete it to a Hilbert space $H$. Since $G$ is Abelian, translation $U_a[x]=[ax]$ preserves the form and extends to a unitary operator. Its inverse is $U_{a^{-1}}$, so $(U_a)$ is a unitary representation. With $v=[e]$,The translated copies of $v$ span a dense subspace. MoreoverFinite sums and density, together with $\Vert U_a\Vert =1$, give strong continuity on all of $H$. This constructs the cyclic unitary representation of a positive-definite function directly.

For $f\in L^1(G)$ define the integrated unitary representation $\pi (f)=\int f(x)U_{x}dx$, with the integral applied to each vector. Its norm is at most $\Vert f{\Vert }_1$. Fubini gives $\pi (f\ast g)=\pi (f)\pi (g)$, and inversion in the Abelian group gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. Thusis a commutative C-star algebra. It is nondegenerate: choose nonnegative compactly supported functions $u_V$ of Haar integral one with supports shrinking to the identity element. Strong continuity impliesfor every $w\in H$. These form an approximate identity in the representation. Neighbourhoods may form a net, since no metrizability is being assumed here.

The Commutative Gelfand--Naimark theorem identifies $\mathcal{A}$ with $C_0(X)$, where $X$ is its character space. The positive vector functional $\omega (B)=⟨Bv,v⟩$ becomes a bounded positive functional on $C_0(X)$, so the Riesz-Markov-Kakutani representation theorem supplies a finite positive Radon measure $\rho$ withThese general representation theorems are applied to the explicitly constructed algebra; they are not an assumption of Bochner's conclusion.

For each $\theta \in X$, the functional $f\mapsto \theta (\pi (f))$ is a nonzero algebra character of $L^1(G)$, since $\pi (L^1(G))$ is dense in $\mathcal{A}$. By Question 3 it has the form $\int f(x){\chi }_{\theta }(x)dx$. The resulting map $\iota :X\to \hat{G}$ is continuous: pointwise convergence of $\theta$ on $\mathcal{A}$ gives convergence on all $\pi (f)$, and the topological character identification in Question 3 gives compact-open convergence of ${\chi }_{\theta }$. Push $\rho$ forward under $\iota$ to obtain $\nu$. This finite measure is Radon: compact subsets approximating the measure of any inverse-image Borel set have compact images, giving inner regularity; finiteness and complements then give outer regularity.

To recover $p(x)$, note that $U_x\pi (u_V)=\pi (T_x{u}_V)\in \mathcal{A}$. HenceThe left side tends to $p(x)$. On the right, $|{\Phi }_{\chi }(u_V)|\le 1$ and ${\Phi }_{\chi }(u_V)\to 1$ uniformly on compact subsets of $\hat{G}$. For completeness, the evaluation $(\chi ,x)\mapsto \chi (x)$ is jointly continuous: near $x_0$ restrict $x$ to a compact neighbourhood, control the character uniformly there, and use continuity of the fixed character. A finite cover of a compact set of characters then makes those characters equicontinuous at $e$, proving the stated uniform convergence. Continuity of $\iota$ gives uniform convergence on compact subsets of $X$, and the finite Radon measure $\rho$ makes the remaining tails arbitrarily small. Thus passage to the limit is valid even for a net, andTaking $x=e$ also gives the required total mass $p(e)$.

Conversely, a finite positive Radon measure givesIts transform is continuous. Indeed, choose a compact set of characters with arbitrarily small measure outside it; joint continuity of evaluation gives uniform continuity in $x$ near the specified point on that compact set, while the outside contribution is bounded by twice its measure. This proves continuity without using an unjustified dominated-convergence assertion for arbitrary nets.

Finally suppose two finite Radon measures have the same transform. Multiply by $f\in L^1(G)$ and integrate over $G$. Fubini shows that the measures agree on every function $\chi \mapsto {\Phi }_{\chi }(f)$. Question 3 showed that these functions are uniformly dense in $C_0(\hat{G})$. Boundedness of both measure functionals extends equality to all of $C_0(\hat{G})$; uniqueness in the Riesz representation theorem gives equality of the measures. This proves existence, uniqueness and the converse, completing Bochner's theorem. In particular $p(e)=1$ corresponds exactly to a probability measure on the dual group.

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.

For an element $a$ of a complex unital Banach algebra, its spectrum of an element isThe nonzero spectra of products in opposite orders agree:For $\lambda \ne 0$, if $R=(\lambda 1-ab{)}^{-1}$ exists, direct multiplication on both sides verifiesInterchanging $a,b$ proves the converse. Zero may differ. On the Hilbert space ${\ell }^2({\mathbb{N}}_0)$, let $S$ be the unilateral shift operator $S{e}_n=e_{n+1}$ and take $a=S^{\ast }$, $b=S$. Then $ab=S^{\ast }S=I$, whereas $ba=S{S}^{\ast }=I-P_0$, where $P_0$ is projection onto the first coordinate. HenceBoth values for the second operator occur on its two nonzero invariant summands; every other value has an inverse obtained separately on those summands.

We use the spectral definition of a Positive element of a C-star algebra: $h$ is positive when $h=h^{\ast }$ and $\sigma (h)\subseteq [0,\infty )$. The continuous functional calculus, supplied by the Commutative Gelfand--Naimark theorem for the algebra generated by a Normal element of a C-star algebra and its adjoint, sends $F\in C(\sigma (h))$ to $F(h)$. It preserves products, sums, conjugation and the identity, is isometric, and satisfies $\sigma (F(h))=F(\sigma (h))$. For a self-adjoint element the spectrum is real, so real-valued functions produce self-adjoint elements.

For example, $F(t)=\sqrt{t}$ gives the positive square root in a C-star algebra of a Positive element of a C-star algebra. It satisfies $(h^{1/2}{)}^2=h$. Its uniqueness follows from the composition rule: if $d\ge 0$ and $d^2=h$, then the continuous functional calculus in the algebra generated by $d$ gives $\sqrt{d^2}=d$. For positive invertible $h$, the functions $t^{-1}$ and $t^{-1/2}$ give positive inverses and inverse square roots. If $0\le h$ and $r\ge \Vert h\Vert$, the function $r-t$ gives $r1-h\ge 0$; the same calculus givesThese transfer elementary scalar inequalities for functions of one element to the algebra. They do not justify applying a scalar inequality to arbitrary noncommuting elements.

To prove closure under addition without assuming an order theorem, let $a,b\ge 0$, set $\alpha =\Vert a\Vert$, $\beta =\Vert b\Vert$ and $s=\alpha +\beta$. If $s=0$, both elements are zero. Otherwise,The element $a+b$ is self-adjoint, so any $\lambda \in \sigma (a+b)$ is real and satisfies $|s-\lambda |\le s$. Thus $\lambda \ge 0$, provingWe will also use that the positive cone is proper: if $h\ge 0$ and $-h\ge 0$, then $\sigma (h)\subseteq \{0\}$, and the norm formula for self-adjoint elements gives $h=0$.

We next prove $a^{\ast }a\ge 0$ without presupposing that positivity is defined by such products. Put $h=a^{\ast }a$, which is self-adjoint, and form its negative part by the continuous functional calculus:This is positive. Let $c=a{h}_{-}^{1/2}$. Since $h_{-}$ commutes with $h$,The nonzero product-spectrum identity proved above gives $\sigma (c{c}^{\ast })\setminus \{0\}=\sigma (c^{\ast }c)\setminus \{0\}$. As $c{c}^{\ast }$ is self-adjoint, it too has spectrum in $(-\infty ,0]$, so $c{c}^{\ast }\le 0$.

Write $c=u+iv$ with $u=u^{\ast }$ and $v=v^{\ast }$. The continuous functional calculus makes $u^2$ and $v^2$ positive, and the addition result givesBut both summands on the left are nonpositive, so their sum is also nonpositive. Properness of the cone forces $c^{\ast }c+c{c}^{\ast }=0$. Now $c^{\ast }c=-c{c}^{\ast }$ is positive as well as nonpositive, so $c^{\ast }c=0$. The C-star identity gives $c=0$, whence $h_{-}^2=0$ and $h_{-}=0$. Thus $h$ has no negative spectral part:This proves the positivity of adjoint products from spectral positivity and the equivalence with the usual definition by elements of the form $d^{\ast }d$: the reverse direction is the factorization $h=(h^{1/2}{)}^{\ast }{h}^{1/2}$ for $h\ge 0$.

Finally, if $a\ge 0$ and $c\in A$, writeThe preceding result therefore provesIn particular, congruence preserves positivity in a C-star algebra, and $a\le b$ implies $c^{\ast }ac\le c^{\ast }bc$. These facts will justify the noncommutative order manipulations in the numbered parts.

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.