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A algebra character $\phi$ is nonzero, linear and multiplicative, with $\phi (1)=1$. Since $\phi (a)\in {\sigma }_A(a)$, it satisfies $|\phi (a)|\le \Vert a\Vert$ and is automatically continuous. If $h=h^{\ast }$ and $t\in \mathbb{R}$, then $(e^{ith}{)}^{\ast }=e^{-ith}$, so $e^{ith}$ is a Unitary element of a C-star algebra and has norm one. HenceUsing both signs of $t$ forces $\mathrm{Im}\phi (h)=0$. Write $x=h+ik$ with $h=(x+x^{\ast })/2$ and $k=(x-x^{\ast })/(2i)$ both self-adjoint C-star elements. Their algebra character values are real, and thereforeThis is the conjugation in the original PDF; the converted TeX loses the bar.

For the final unheaded request, let $B=C^{\ast }(1,x)\subseteq A$. Normality makes this a commutative unital C-star subalgebra. We first justify spectral permanence, so that we do not silently use the possibly larger algebra spectrum in a mere non-star subalgebra. A self-adjoint C-star element $h$ in a C-star algebra has real algebra spectrum: $e^{ith}$ has norm one, and $\lambda \in \sigma (h)$ implies $e^{it\lambda }\in \sigma (e^{ith})$ for each real $t$. The latter assertion follows from the factorization $e^{ith}-e^{it\lambda }1=(h-\lambda 1)g_t(h)$ with commuting factors. An invertible product of commuting factors would make $h-\lambda 1$ invertible. Both signs of $t$ then force $\lambda \in \mathbb{R}$.

A compact subset of the real line has connected planar complement. Apply Question 1 to the self-adjoint C-star element $h\in B$: all of ${\rho }_A(h)$ is its unbounded component and is retained in $B$. Thus ${\sigma }_A(h)={\sigma }_B(h)$. If $a\in B$ is invertible in $A$, both $a^{\ast }a$ and $a{a}^{\ast }$ are self-adjoint C-star elements and invertible in $A$, hence invertible in $B$. The elements $(a^{\ast }a{)}^{-1}{a}^{\ast }$ and $a^{\ast }(a{a}^{\ast }{)}^{-1}$ are respectively a left and a right inverse in $B$, so coincide and give its inverse. This proves spectral permanence for C-star algebras.

In a commutative unital complex Banach algebra, algebra spectra are the ranges of the characters. Indeed a noninvertible $b-\lambda 1$ lies in a maximal ideal. Such an ideal is closed, since its proper closure cannot contain an element sufficiently near $1$. The quotient is a complex Banach division algebra; nonemptiness of the algebra spectrum in that quotient makes each element scalar, producing an algebra character with value $\lambda$ at $b$. Conversely, an algebra character value cannot correspond to an invertible difference.

Apply this to $B$. The preceding involution result gives $\phi (x^{\ast })=\overline{\phi (x)}$, and spectral permanence applies also to $p(x,x^{\ast })$. ConsequentlyThus the second argument is conjugated as in the original PDF, not a second copy of $\lambda$. This is the spectral mapping rule for a star polynomial in one normal element.

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

Put $K=\sigma (T)$ and use the continuous functional calculus $\pi :C(K)\to \mathcal{B}(H)$. It is a unital star-homomorphism taking the coordinate function to $T$. This follows from the commutative C-star theorem applied to $C^{\ast }(I,T,T^{\ast })$, together with spectral permanence for C-star subalgebras, which is among the permitted C-star results.

Take the Hilbert space inner product linear in its first variable. For a vector $\xi$, let $H_{\xi }=\overline{\{\pi (f)\xi :f\in C(K)\}}$. This cyclic subspace is invariant under every $\pi (f)$ and its adjoint, so it is reducing. The functional $f\mapsto ⟨\pi (f)\xi ,\xi ⟩$ is positive because $f\ge 0$ implies $\pi (f)=\pi (\sqrt{f}{)}^{\ast }\pi (\sqrt{f})$. It has, by the Riesz-Markov-Kakutani representation theorem, a finite regular positive Borel measure ${\mu }_{\xi }$ on $K$. The identitymakes $f\mapsto \pi (f)\xi$ an isometry on its quotient by the functions of zero $L^2$ norm. Density of continuous functions in $L^2({\mu }_{\xi })$ extends it to a unitary $U_{\xi }:L^2({\mu }_{\xi })\to H_{\xi }$. Multiplication by a continuous function intertwines with $\pi$ on this subspace.

Choose a maximal family of mutually orthogonal nonzero cyclic reducing subspaces. Their orthogonal sum is all of $H$: otherwise its reducing orthogonal complement contains a nonzero vector generating another such subspace. This argument does not require separability. For a bounded Borel measurable function $g$ on $K$, multiplication $M_g$ on every $L^2({\mu }_{\xi })$ is bounded with norm at most $\Vert g{\Vert }_{\infty }$. Define on that orthogonal sumMultiplication operators satisfy $M_{fg}=M_f{M}_g$, $M_{\overline{g}}=M_g^{\ast }$ and $M_1=I$. The same identities hold for their direct sums, and the norm bound is uniform in $\xi$. Thus ${\beta }_T$ is a norm-decreasing unital star-homomorphism on the algebra of actual bounded Borel measurable functions, and ${\beta }_T(Z)=T$ because it extends $\pi$. No quotient by a single unspecified measure is being substituted for that Borel algebra.

Choose the bounded Borel measurable square root $s(0)=0$ and $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ for $z\ne 0$, taking $\mathrm{Arg}\in [0,2\pi )$. Let $S={\beta }_T(s)$. ThenThus every bounded normal operator has a normal square root. The Borel branch is essential in general: on a circle winding about zero a continuous square-root branch need not exist.

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 definitions in a unital C-star algebra areThus a Hermitian element of a C-star algebra is self-adjoint, a Unitary element of a C-star algebra has inverse $x^{\ast }$, and a Normal element of a C-star algebra commutes with its adjoint. We derive the needed C-star algebra facts directly from $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$, as required.

First $\Vert 1\Vert =\Vert 1{\Vert }^2$ gives $\Vert 1\Vert =1$, andgives $\Vert a\Vert \le \Vert a^{\ast }\Vert$; applying the same argument to $a^{\ast }$ gives equality. Thus the involution is an isometry. For a Unitary element of a C-star algebra $u$,The general Banach algebra spectrum bound gives $|\lambda |\le 1$ for $\lambda \in \sigma (u)$. Such a $\lambda$ is nonzero, and the inverse spectral mapping theorem gives ${\lambda }^{-1}\in \sigma (u^{-1})$, so also $|\lambda {|}^{-1}\le 1$. Hence

If $h=h^{\ast }$, the convergent exponential series and the isometric involution give $(e^{ith}{)}^{\ast }=e^{-ith}$. These commuting exponentials multiply to $1$, so $e^{ith}$ is unitary for real $t$. By the exponential spectral mapping theorem, if $\lambda \in \sigma (h)$ then $e^{it\lambda }\in \sigma (e^{ith})\subseteq \mathbb{T}$. Taking $t=1$ gives $e^{-\mathrm{Im}\lambda }=1$, soOnly general Banach algebra spectral mapping theorem and the defining C-star identity were used.

Now prove spectral permanence for C-star algebras. Let $B\subseteq A$ be the norm-closed C-star subalgebra with the same identity. The algebraic inclusion always gives ${\sigma }_A(b)\subseteq {\sigma }_B(b)$. For $h=h^{\ast }\in B$, both spectra lie in $\mathbb{R}$. If $\lambda \notin {\sigma }_A(h)$, choose nonreal ${\lambda }_n\to \lambda$. Since ${\sigma }_B(h)\subseteq \mathbb{R}$, each $({\lambda }_{n}1-h{)}^{-1}$ belongs to $B$. By continuity of inversion in a Banach algebra, these inverses converge in $A$ to $(\lambda 1-h{)}^{-1}$, which belongs to $B$ because $B$ is closed. Thus $\lambda \notin {\sigma }_B(h)$, proving equality for hermitian elements.

For the requested normal $x\in B$, suppose $b=x-\lambda 1$ is invertible in $A$. The element $b^{\ast }b$ is hermitian and invertible in $A$, so the equality just proved places $(b^{\ast }b{)}^{-1}$ in $B$. ConsequentlySince $b$ already has a two-sided inverse in $A$, multiplying $yb=1$ by that inverse gives $y=b^{-1}$. Hence its inverse belongs to $B$, and

We need one more elementary consequence of the C-star identity: the spectral radius norm equality for normal elements. If $h=h^{\ast }$, then $\Vert h^2\Vert =\Vert h{\Vert }^2$. If $a$ is normal, commutativity of $a,a^{\ast }$ givesEvery power of $a$ is a Normal element of a C-star algebra. Induction gives $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The general spectral radius formula therefore implies

The continuous functional calculus is the unique unital $\ast$-homomorphism $C(K)\to \mathcal{B}(H)$ taking the coordinate function to $T$. For bounded linear operators on a Hilbert space, the adjoint identity gives $\Vert S\Vert ={\sup }_{\Vert v\Vert =\Vert w\Vert =1}|⟨Sv,w⟩|=\Vert S^{\ast }\Vert$. ThereforeThis proves the C-star identity for $\mathcal{B}(H)$ directly; completeness and submultiplicativity come from the operator norm. LetBecause $T$ is a normal operator, this is a commutative unital C-star algebra. Every element of $C$ is a Normal element of a C-star algebra, so the preceding norm identity says its Gelfand transform is isometric:Here $\Delta (C)$ is the compact character space from the general Gelfand representation theorem for commutative Banach algebras.

Each algebra character preserves the involution. Indeed, write $a=h+ik$ with $h,k$ hermitian; $\chi (h)\in {\sigma }_C(h)\subseteq \mathbb{R}$ and similarly for $k$, so $\chi (a^{\ast })=\overline{\chi (a)}$. Therefore the continuous mapis injective: its value determines $\chi (T^{\ast })$ and hence its value on all the dense polynomials. It is surjective because the general character description of the spectrum gives $\{\chi (T):\chi \in \Delta (C)\}={\sigma }_C(T)$, and spectral permanence for C-star algebras identifies this with ${\sigma }_{\mathcal{B}(H)}(T)=K$. A continuous function that is a bijection from a compact space to a Hausdorff space is a homeomorphism, so we identify $\Delta (C)$ with $K$.

Under this identification, the Gelfand transform sends $T$ to $u(z)=z$ and $T^{\ast }$ to $\overline{u}$. Its image is an isometric, and therefore closed, unital self-adjoint subalgebra of $C(K)$. It separates points because it contains $u$. The complex Stone-Weierstrass theorem makes the image dense, hence equal to all of $C(K)$. Inverting the Gelfand transform and including $C$ into $\mathcal{B}(H)$ givesan isometric unital C-star homomorphism with $u(T)=T$.

To prove uniqueness without assuming automatic continuity of C-star homomorphisms, let $\Psi$ be any other such map. Since every $f\in C(K)$ is normal, $\Psi (f)$ is normal. A unital algebra homomorphism preserves inverses, soUsing the normal-element norm identity yields $\Vert \Psi (f)\Vert =r(\Psi (f))\le \Vert f{\Vert }_{\infty }$. Thus $\Psi$ is contractive. It agrees with $\Phi$ on polynomials in $u,\overline{u}$, since both send them to the same polynomials in $T,T^{\ast }$. These polynomials are dense by the Stone-Weierstrass theorem, so continuity proves $\Psi =\Phi$. This establishes the continuous functional calculus with no unproved theorem specific to C-star algebras.

A disconnected spectrum gives a nontrivial closed invariant subspace. Write $K=K_1\bigsqcup K_2$ with $K_1,K_2$ nonempty and both open and closed in $K$. The indicator function $f=1_{K_1}$ is continuous on $K$, even though no such continuity is needed across the gap outside $K$. Let $P=f(T)$. The continuous functional calculus givesIts isometry gives $\Vert P\Vert =1$ and $\Vert I-P\Vert =1$, so $P\ne 0,I$. The linear projection has closed range $Y=PH=\ker (I-P)$. Its range is nonzero and proper, and $T(Px)=P(Tx)\in Y$ proves invariance. Since it also commutes with $T^{\ast }$, the same subspace is reducing. This proves disconnected spectrum gives a reducing subspace.

The figure can be realized without any eigenvalues: take $H=L^2(K,dA)$, where $dA$ is planar area on the two closed disks, and let $T$ multiply by $z$. Its adjoint multiplies by $\overline{z}$, so it is a normal operator. An eigenvector for $\lambda$ would be supported on the area-zero singleton $\{\lambda \}$, hence would be zero in $L^2$. Outside $K$, multiplication by $1/(z-\lambda )$ is a bounded inverse to $T-\lambda I$. For $\lambda \in K$, normalized indicator functions of $K\cap \{|z-\lambda |<\epsilon \}$ have $\Vert (T-\lambda I)h\Vert \le \epsilon$, excluding a bounded inverse. Thus its spectrum is exactly $K$. The linear projection in the figure multiplies by $1_{K_1}$, and its range consists of functions supported on the left disk.

For normal $x$, the closed unital C-star subalgebra generated by $x,x^{\ast }$ is commutative. When the ambient spectrum is real, its complement is connected. Membership of the resolvent in this subalgebra is both open and closed on that complement, and holds at large modulus by the Neumann series. Thus the subalgebra spectrum is the same real set. Its isometric Gelfand transform takes $x$ to a real-valued function and $x^{\ast }$ to its conjugate, proving equality. Normality is necessary: the nonzero matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ has spectrum $\{0\}$ but is not self-adjoint.