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An algebra involution on a complex associative algebra is a conjugate-linear map satisfying $(xy{)}^{\ast }=y^{\ast }{x}^{\ast }$ and $(x^{\ast }{)}^{\ast }=x$. It fixes the identity when one exists. A star algebra is an algebra equipped with such a map. These algebraic identities do not themselves impose norm continuity or the C-star identity.

A Banach star-algebra is a complex Banach algebra equipped with a continuous algebra involution. Thus the involution is conjugate-linear, reverses products, and squares to the identity. Some conventions additionally require $\Vert a^{\ast }\Vert =\Vert a\Vert$; this is the isometric convention. In the unital case the involution fixes the unit. The C-star identity is a stronger additional condition. The algebra of continuous complex-valued functions on a compact Hausdorff space, with pointwise operations, complex conjugation and the supremum norm, satisfies both the isometric condition and the C-star identity.

A C-star algebra is a complex Banach algebra with an algebra involution $a\mapsto a^{\ast }$ satisfying the C-star identityThe algebra involution is conjugate-linear, reverses products and obeys $(a^{\ast }{)}^{\ast }=a$. An identity need not be present.^[ref]

Examples include complex matrices with the operator norm and conjugate transpose, bounded linear operators on a Hilbert space with the operator adjoint, and $C_0(X)$ on a Hausdorff space that is a locally compact space with pointwise complex conjugation and the supremum norm. The identity ties the norm to the algebraic product and involution: it implies spectral radius norm equality for normal elements and makes an injective C-star homomorphism an isometry. This rigidity permits continuous functional calculus for normal C-star elements.

A C-star isomorphism is a bijective C-star homomorphism: a complex-linear multiplicative map preserving the involution. Between unital C-star algebras it automatically preserves the unit. It preserves the Banach algebra spectrum, since it preserves invertibility in both directions. Applying the C-star identity and the spectral norm formula to $a^{\ast }a$ gives $\Vert \Phi (a){\Vert }^2=\Vert \Phi (a^{\ast }a)\Vert =\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$, so it is automatically isometric.

A C-star subalgebra is a norm-closed subalgebra closed under the involution, with the inherited C-star identity. A unital inclusion in this context uses the same identity as the ambient C-star algebra. The spectral permanence for C-star algebras says the inclusion preserves the spectrum of an element.

The left multiplication representation of a C-star algebra is isometric: test $L_a$ on $a^{\ast }/\Vert a\Vert$ to get $\Vert L_a\Vert =\Vert a\Vert$. For a nonunital algebra, its closed image plus the identity operator is the algebraic unitization. For a formal $z=a+\lambda 1$ and $b\in A$, $\Vert zb{\Vert }^2=\Vert b^{\ast }{z}^{\ast }zb\Vert \le \Vert b\Vert \Vert L_{z^{\ast }z}b\Vert$. Thus $\Vert L_z{\Vert }^2\le \Vert L_{z^{\ast }z}\Vert \le \Vert L_{z^{\ast }}\Vert \Vert L_z\Vert$; applying the same estimate to $z^{\ast }$ proves the algebra involution is isometric and the C-star identity holds. This gives a unitization without first assuming an operator representation on a Hilbert space.

An injective C-star homomorphism preserves norms. For unital maps, reduce to the commutative algebra generated by $1$ and $a^{\ast }a$. Its induced map on character spaces has dense, hence full, image; otherwise a nonzero continuous function vanishing on that image would lie in the kernel. The C-star identity then gives norm preservation for $a$.

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 unital Banach star-algebra is a complex unital Banach algebra with a continuous conjugate-linear algebra involution satisfyingThe algebra norm is complete and submultiplicative, with the unit conventionally normalized to norm one. Many definitions additionally require the involution to be isometric, $\Vert a^{\ast }\Vert =\Vert a\Vert$; the function algebra here satisfies this stronger convention as well. A Banach star-algebra does not by definition require the extra C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$.

Here is the lattice proof of the Stone-Weierstrass theorem in its complex self-adjoint form. Let $B$ be the uniform closure of the given unital star-subalgebra in $C(\Omega )$, and let $B_{\mathbb{R}}$ be its real-valued part. Closure preserves multiplication and complex conjugation, so $B$ is still a unital star-algebra. Real and imaginary parts lie in $B_{\mathbb{R}}$, and separation by the complex algebra implies separation by its real-valued part.

For $h\in B_{\mathbb{R}}$, polynomial approximation of absolute value on a bounded interval gives $|h|\in B_{\mathbb{R}}$. For example, this polynomial approximation follows from the binomial series for the square root, whose coefficients give uniform convergence on the relevant closed interval. Thusalso lie in $B_{\mathbb{R}}$. It is therefore a lattice of functions.

To see how the lattice gives density, fix a real continuous target $f$ and a tolerance $ϵ>0$. For any two distinct points $x,y$, separation and constants provide $h_{xy}\in B_{\mathbb{R}}$ with $h_{xy}(x)=f(x)$ and $h_{xy}(y)=f(y)$, by an affine rescaling of a separating function. For $x=y$, use the constant $f(x)$.

Fix $x$. The sets where $h_{xy}>f-ϵ$ cover $\Omega$ as $y$ varies. A finite subcover and a pointwise maximum give $g_x\in B_{\mathbb{R}}$ with $g_x>f-ϵ$ everywhere and $g_x(x)=f(x)$. The neighborhoods where $g_x<f+ϵ$ now cover $\Omega$ as $x$ varies. A second finite subcover and a pointwise minimum give $h\in B_{\mathbb{R}}$ satisfyingThis is lattice approximation from two-point approximation. Using half the desired tolerance gives strict uniform error below the desired bound. Since $B_{\mathbb{R}}$ is closed, it contains every real continuous function. Splitting a complex function into real and imaginary parts proves $\boxed{B=C(\Omega )}$.

The annulus test. For the closed annulus, the family holomorphic in the interior is not dense in the full continuous-function algebra. A uniform limit on the annulus is locally uniform in its interior, so locally uniform convergence of holomorphic functions makes that limit holomorphic there. Thus the proposed family is closed in $C(\Omega )$. It is proper because $z\mapsto \overline{z}$ is continuous but not holomorphic on the annulus interior.

There is also an explicit continuous functional detecting the obstruction. Choose a small closed disk of radius $r$ contained in the annulus interior and let $L(h)$ be the integral of $h(z)dz$ around its positively oriented boundary. The Cauchy integral theorem gives $L(h)=0$ for every member of the holomorphic family, while parametrizing the circle gives $L(\overline{z})=2\pi i{r}^2\ne 0$. Since $|L(h)|\le 2\pi r\Vert h{\Vert }_{\infty }$, no uniform approximation of $\overline{z}$ by that family is possible.

A C-star algebra is a complex Banach algebra with a conjugate-linear involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a^{\ast }{)}^{\ast }=a$ and $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. It is unital when it has an identity for multiplication. For $B(H)$ use composition and the Hilbert adjoint. Completeness in the operator norm is given; composition is submultiplicative and the adjoint has the required algebraic properties. For a unit vector $\xi$,Taking the supremum and using the opposite bound $\Vert T^{\ast }T\Vert \le \Vert T^{\ast }\Vert \Vert T\Vert =\Vert T{\Vert }^2$ proves the C-star identity. Its identity is $I_H$, proving $B(H)$ is a unital C-star algebra.

Compact operators form a norm-closed linear space and are stable under composition with bounded operators. They are also adjoint stable. To see this directly, choose a finite $\epsilon$-net of the compact image of the unit ball under $T$, and let $P$ project orthogonally onto the span of its centers. Then $\Vert (I-P)T\Vert \le \epsilon$. The finite-rank operator $T^{\ast }P$ approximates $T^{\ast }$ with the same error, so $T^{\ast }$ is compact by norm closure. Thus $\mathcal{K}(H)$ is a C-star algebra, with the inherited norm and operations.

If it has a multiplicative identity $E$, apply $ER=R$ to every rank-one operator with arbitrary prescribed range vector $\xi$. Evaluating that equation on a vector sent to $\xi$ gives $E\xi =\xi$. Hence $E=I_H$. This identity is compact exactly when $H$ is finite dimensional: finite dimension gives compactness of the unit ball, while in infinite dimension an infinite orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Therefore $\mathcal{K}(H)$ is unital exactly in finite dimension, with the zero-algebra convention understood if $H=0$.

Use the source's notation for the vanishing-at-zero ideal $I=\{f\in C[0,1]:f(0)=0\}$. It is the kernel of bounded evaluation at zero, hence closed. It is stable under pointwise multiplication and complex conjugation, so inherits the C-star identity and completeness. If it had an identity $e$, multiplication against $f(t)=t$ would give $e(t)=1$ for $t>0$, contradicting continuity and $e(0)=0$. Thus $I$ is a non-unital C-star subalgebra. This explicit meaning of the notation matters because the usual notation $C_0$ for a compact space has a different convention.

Define the unitization map by $\Phi (f,\lambda )=f+\lambda 1$. Its inverse is $h\mapsto (h-h(0),h(0))$. The specified multiplication and involution are preserved by direct substitution. The pullback norm $\Vert (f,\lambda )\Vert =\Vert f+\lambda 1{\Vert }_{\infty }$ is a C-star norm, so equals the unique unitization norm supplied in the question. Therefore $\Phi$ is a unital isometric C-star isomorphism onto $C[0,1]$.

For the approximation put $h(s)=a(\sqrt{s})$ on $[0,1]$, so $h$ is continuous and $h(0)=0$. The Stone-Weierstrass theorem supplies a polynomial $p(s)$ with $\Vert h-p{\Vert }_{\infty }<\epsilon /2$. Then $b(t)=p(t^2)-p(0)$ is a polynomial in the positive even powers of $t$, with zero constant term, andThis gives exactly the required finite form.

The spectrum of an element is ${\sigma }_A(a)=\{\lambda \in \mathbb{C}:a-\lambda 1_A\text{ is not invertible in }A\}$. A Normal element of a C-star algebra satisfies $a^{\ast }a=a{a}^{\ast }$.

For a Hermitian C-star element $h$, the C-star identity gives $\Vert h^2\Vert =\Vert h{\Vert }^2$. For normal $a$, commutation givesAll powers remain normal, so induction yields $\Vert a^{2^k}\Vert =\Vert a{\Vert }^{2^k}$. Taking the spectral-radius limit along these powers givesLet $\phi :A\to B$ be a C-star homomorphism, without assuming continuity or preservation of the ambient unit. Put $p=\phi (1_A)$, a projection in a C-star algebra. If $p=0$, then $\phi =0$. Otherwise its image lies in the unital corner $pBp$ and it is unital as a map into that corner. An inverse in $A$ maps to an inverse in the corner, so ${\sigma }_{pBp}(\phi (b))\subseteq {\sigma }_A(b)$. Apply this to the normal element $b=a^{\ast }a$ and use the C-star identity in both algebras:Thus every C-star homomorphism is norm non-increasing, including nonunital maps.

A Unitary element of a C-star algebra satisfies $u^{\ast }u=u{u}^{\ast }=1$. A projection in a C-star algebra satisfies $p=p^{\ast }=p^2$.

If $q=u^{\ast }u$ is a projection in a C-star algebra, the C-star identity gives $\Vert u(1-q){\Vert }^2=\Vert (1-q)q(1-q)\Vert =0$. Hence $u=uq=u{u}^{\ast }u$. Conversely $u{u}^{\ast }u=u$ implies $(u^{\ast }u{)}^2=u^{\ast }u$, and $u^{\ast }u$ is a Hermitian C-star element, so it is a projection in a C-star algebra. Taking adjoints of the triple-product equality gives the corresponding equality for $u^{\ast }$, provingFor nonzero $u$, let $p=u{u}^{\ast }$ and $q=u^{\ast }u$. These are projections, $u=pu=uq$, and $\Vert u\Vert =1$. The rectangular corner $pAq$ is closed because it is the range of the bounded idempotent map $a\mapsto paq$.

Define $\Theta :pAq\to qAq$ by $\Theta (a)=u^{\ast }a$. It lands in the corner, and its inverse is $b\mapsto ub$. The identities $u{u}^{\ast }a=pa=a$ and $u^{\ast }ub=qb=b$ verify both compositions. Both maps are contractive because $\Vert u\Vert =\Vert u^{\ast }\Vert =1$, so $\Theta$ is an isometry. For the proposed multiplication and involution,Thus the operations correspond exactly to the ordinary operations in the closed C-star algebra $qAq$. This proves associativity, closure, all involution identities and the C*-norm identity at once. Moreover $\Theta (u)=q$, so the unit of $A_u$ is $u$. If $u$ is unitary, $p=q=1$, and the map $a\mapsto u^{\ast }a$ is an onto isometric *-isomorphism from $A_u$ to $A$.

First suppose $x$ is a positive linear functional, without assuming boundedness. Every Hermitian C-star element $h$ is a difference of positive elements, since $\Vert h\Vert 1\pm h\ge 0$. Thus $x(h)$ is real. For a Hermitian C-star element $h$ of norm at most one, $-1\le h\le 1$ gives $|x(h)|\le x(1)$.

For arbitrary $a$, choose a scalar $\zeta$ of modulus one so that $\zeta x(a)=|x(a)|$. The real part of $\zeta a$ is a Hermitian C-star element and has norm at most $\Vert a\Vert$. Since $x$ is real on Hermitian C-star elements, it respects real parts, givingThis proves boundedness and $\Vert x\Vert \le x(1)$. Evaluation at the unit gives the reverse inequality, so positivity implies $\Vert x\Vert =x(1)$.

Conversely let $x$ be bounded with $x(1)=\Vert x\Vert =c\ge 0$. If $c=0$, $x=0$ is positive. Otherwise normalize to $c=1$. For a Hermitian C-star element $h$ and real $t$, the C-star identity givesWriting $x(h)=a+ib$ and squaring yields $-2tb+t^2(a^2+b^2)\le t^2\Vert h{\Vert }^2$. Taking $t\to 0$ through positive and negative values forces $b=0$. Hence $x$ is real on Hermitian C-star elements. If $0\le a\le 1$, then $1-a$ has norm at most one, so $x(1-a)\le 1$ and $x(a)\ge 0$. Scaling proves nonnegativity on every positive element, in particular every $b^{\ast }b$. Therefore

A unital C-star algebra is a complex Banach algebra $A$ with a multiplicative identity $1_A$ and a conjugate-linear involution $a\mapsto a^{\ast }$ satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. A C-star isomorphism $\Phi :A\to B$ is a bijective complex-linear map preserving products and the involution. It carries $1_A$ to $1_B$ and is automatically isometric.

For a commutative unital C-star algebra, let $\Omega$ be its maximal ideal space, equivalently the space of Gelfand characters, with the weak-star topology. A Gelfand character has $\chi (1)=1$ and $\Vert \chi \Vert =1$. The equations $\chi (ab)=\chi (a)\chi (b)$ and $\chi (1)=1$ define a closed in the weak-star topology subset of the dual unit ball. The Banach-Alaoglu theorem makes $\Omega$ a compact Hausdorff space. Maximal ideals give characters by the Gelfand-Mazur theorem, and the Banach algebra spectrum satisfies $\sigma (a)=\{\chi (a):\chi \in \Omega \}$.

The Gelfand transform is $\hat{a}(\chi )=\chi (a)$. It preserves products, constants and the involution; the latter uses the fact that characters of a C-star algebra respect the involution. Every element of a commutative C-star algebra is normal, so the C-star identity and the spectral radius formula giveThus its image is a closed self-adjoint subalgebra of $C(\Omega )$, contains the constants, and separates characters. The complex Stone-Weierstrass theorem makes that image dense and therefore all of $C(\Omega )$. This is the Commutative Gelfand--Naimark theorem.

For the convergent sequence space $c$, coordinatewise multiplication satisfies $\Vert \alpha \beta {\Vert }_{\infty }\le \Vert \alpha {\Vert }_{\infty }\Vert \beta {\Vert }_{\infty }$, and coordinatewise complex conjugation is an involution. The constant sequence $1$ is the identity, andIt is complete: a uniform limit of convergent sequences is convergent, so $c$ is a closed linear subspace of the Banach space ${\ell }_{\infty }$. Hence it is a commutative unital C-star algebra.

TakeEvery finite point is isolated, and the neighbourhoods of $\infty$ contain all but finitely many integers. This is a compact Hausdorff space: an open cover has a set containing $\infty$, leaving only finitely many points to cover. DefineContinuity at $\infty$ is exactly convergence of the sequence. Thus $\Phi$ is onto $C(\Omega )$, and pointwise operations show it is a C-star isomorphism with $\Vert \Phi \alpha {\Vert }_{\infty }=\Vert \alpha {\Vert }_{\infty }$. Explicitly, these are all Gelfand characters: each $\chi (e_n)$ is $0$ or $1$ and at most one is $1$. If $\chi (e_n)=1$, multiplication by $e_n$ gives $\chi (\alpha )={\alpha }_n$. If all vanish, density of finitely supported sequences in l-p in $c_0$ gives $\chi {|}_{c_0}=0$, so $\chi (\alpha )=\lim {\alpha }_n$.

For a self-adjoint C-star element $h$, the C-star identity gives $\Vert h^2\Vert =\Vert h{\Vert }^2$. If $x$ is a normal C-star element, thenAll powers are normal C-star elements, so induction gives $\Vert x^{2^j}\Vert =\Vert x{\Vert }^{2^j}$. The general spectral radius formula now gives $r(x)=\Vert x\Vert$.

One can also see the needed growth estimate directly. For $R>r(x)$ the resolvent Cauchy coefficient formula givesThe formula follows by the Neumann series on a larger circle and contour deformation through the resolvent annulus. Put $m=2^j$, take $m$th roots and let $j\to \infty$ to get $\Vert x\Vert \le R$. Let $R\downarrow r(x)$; the reverse inequality already follows from the norm bound on the algebra spectrum. ThusThis proves spectral radius norm equality for normal elements from the norm identity.

Submultiplicativity and the C-star identity give $\Vert x{\Vert }^2=\Vert x^{\ast }x\Vert \le \Vert x^{\ast }\Vert \Vert x\Vert$. For $x\ne 0$ this implies $\Vert x\Vert \le \Vert x^{\ast }\Vert$. Apply the same argument to $x^{\ast }$ and use $x^{\ast \ast }=x$ for the reverse inequality; $x=0$ is immediate. Hence

An algebra involution is a conjugate-linear map $a\mapsto a^{\ast }$ satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and $(a^{\ast }{)}^{\ast }=a$; on a unital algebra it fixes $1$. In the Banach-star convention it is continuous, although the following spectral identity uses only its algebraic properties. An element $a$ is invertible precisely when $a^{\ast }$ is: its inverse is $(a^{-1}{)}^{\ast }$. Applying this to $a=\lambda 1-x$ provesFor a nonunital algebra, extend the involution to its unitization of an algebra by $(a,\lambda {)}^{\ast }=(a^{\ast },\overline{\lambda })$ and use the spectrum in that unitization; the same argument proves the identity.

A unital C-star algebra is a complex Banach algebra with an involution satisfying the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. A Normal element of a C-star algebra satisfies $x^{\ast }x=x{x}^{\ast }$. The norm identity for the nonzero identity implies $\Vert 1\Vert =1$.

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

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.