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

Every proper maximal ideal $M$ of $A$ is closed. Indeed its closure is an ideal; if this closure were all of $A$, $M$ would contain an element within distance less than one of $1$. Such an element is invertible by the Neumann series, forcing $1\in M$. Thus the closure is proper and maximality makes it equal to $M$. The quotient Banach space $A/M$, with its quotient Banach algebra structure, is a complex normed division algebra. By the Gelfand-Mazur theorem, it is $\mathbb{C}$, so the quotient map gives a character of an algebra with kernel $M$. Conversely, the kernel of every character of an algebra is a maximal ideal, since the character is onto $\mathbb{C}$. The Zorn lemma supplies a maximal ideal containing every proper ideal, so the character space of an algebra $\Delta (A)$ is nonempty.

These facts give the exact relation between the character space and the spectrum of an element:One inclusion was proved above. For the other, if $a-\lambda 1$ is noninvertible, the principal ideal it generates is proper because $A$ is commutative. Contain it in a maximal ideal and use its corresponding character of an algebra to obtain $\chi (a)=\lambda$.

Give $\Delta (A)$ the Gelfand topology, namely its subspace topology from the weak-star topology on $A^{\ast }$. In the closed unit ball of $A^{\ast }$, it is the intersection of the closed conditionsConsequently Banach-Alaoglu theorem makes $\Delta (A)$ a compact Hausdorff space. For every $a\in A$, define the Gelfand transform $\hat{a}(\chi )=\chi (a)$. This is a continuous function on $\Delta (A)$ by definition of the Gelfand topology. The Gelfand representation theorem gives a contractive unital algebra homomorphism over a fieldMultiplicativity and linearity follow by evaluating at each character of an algebra; the supremum norm equality follows from the preceding spectrum of an element identity. Its kernel isthe Jacobson radical. Equivalently, its elements have spectrum of an element $\{0\}$. Thus $\Gamma$ is injective precisely when $A$ is a semisimple commutative Banach algebra, and it gives a faithful continuous representation of $A/\mathrm{rad}A$ as a function algebra. Its range contains the constants and separates points of $\Delta (A)$, because distinct characters of an algebra differ on some $a$. An arbitrary Banach algebra need not have an isometric or surjective Gelfand transform, nor a uniformly dense range: those conclusions require further hypotheses.

For the Banach algebra $C(K)$ on a nonempty compact Hausdorff space $K$, all characters of an algebra are evaluation characters. To see this, let $M$ be a maximal ideal. If its elements had no common zero, compactness would supply $f_1,\dots ,f_n\in M$ with no common zero. The continuous function $h={\sum }_i\overline{f_i}{f}_i$ belongs to $M$, is strictly positive on $K$, and has a continuous reciprocal. It is therefore invertible, a contradiction. Hence all elements of $M$ vanish at some $x\in K$, so $M\subseteq \ker {\delta }_x$ and maximality gives equality. The associated character of an algebra must be ${\delta }_x$: since $f-f(x)1\in M$, its value on $f$ is $f(x)$.

The map $x\mapsto {\delta }_x$ is a continuous bijection $K\to \Delta (C(K))$, using separation of points by continuous functions. Compactness and the Hausdorff property make it a homeomorphism. Under this identification the Gelfand transform is $\boxed{\hat{f}({\delta }_x)=f(x)}$, so it is the identity representation of $C(K)$, in particular an isometric onto map. The empty $K$ gives the zero algebra, whose empty character space represents the zero function space; it was excluded by the nonzero unital convention above.

Now let $A$ be a commutative unital C-star algebra. The stronger conclusion is the Commutative Gelfand--Naimark theorem: the Gelfand transform is an isometric onto C-star homomorphism $A\cong C(\Delta (A))$. We prove the additional assertions without assuming this conclusion.

First every character of an algebra respects the C-star algebra involution. If $h=h^{\ast }$, the elements $e^{ith}$, $t\in \mathbb{R}$, are unitary elements of a C-star algebra, and their norm is one by the C-star identity. Continuity and multiplicativity give $\chi (e^{ith})=e^{it\chi (h)}$. Thus $|e^{it\chi (h)}|\le 1$ for every real $t$, forcing $\chi (h)$ to be real. Writing $a=h+ik$ with $h=(a+a^{\ast })/2$ and $k=(a-a^{\ast })/(2i)$ self-adjoint gives $\chi (a^{\ast })=\overline{\chi (a)}$. Therefore the range of $\Gamma$ is closed under complex conjugation.

Every element of commutative $A$ is a Normal element of a C-star algebra. For a normal $b$, use the C-star identity, and then the same identity for the self-adjoint element $b^{\ast }b$, to obtainIts powers are also normal, so $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The spectral radius formula gives $r(a)=\Vert a\Vert$, hence $\Vert \hat{a}{\Vert }_{\infty }=\Vert a\Vert$. The Gelfand transform is therefore an isometry, and its range is complete and closed in the supremum norm. It contains constants, separates points and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that range dense, hence all of $C(\Delta (A))$.

The approximation step in Stone-Weierstrass theorem can also be seen directly here. For a unital conjugation-closed point-separating subalgebra $B\subseteq C(K)$, the real-valued part of its uniform closure is closed under absolute values, by polynomial approximation to $|t|$ on bounded intervals, hence under pointwise maxima and minima. Its real-valued functions separate points. Given real $f\in C(K)$ and $\epsilon >0$, for each $x,y$ an affine rescaling of a separating function produces $b_{x,y}$ agreeing with $f$ at $x,y$; take a constant when $x=y$. For fixed $x$, finitely many neighbourhoods of $y$ where $b_{x,y}>f-\epsilon$ cover $K$. Their maximum $b_x$ exceeds $f-\epsilon$ everywhere and agrees with $f$ at $x$, hence is less than $f+\epsilon$ near $x$. Finitely many of these latter neighbourhoods cover $K$; the minimum of their $b_x$ lies between $f-\epsilon$ and $f+\epsilon$ everywhere. Approximate real and imaginary parts separately. This proves the density used above and completes the C-star algebra conclusion.

Invertibility is stable under sufficiently small perturbations, and inversion is continuous. Let $1$ be the identity of the complex unital Banach algebra $A$, with its submultiplicative algebra norm. For $\Vert h\Vert <1$, completeness makes the Neumann seriesconverge in $A$. Multiplying either side of the partial sums by $1-h$ gives $1-h^{N+1}$, which tends to $1$, proving that the sum is a two-sided inverse.

Fix $a\in G(A)$, the group of invertible elements of a Banach algebra. If $\Vert b-a\Vert <\Vert a^{-1}{\Vert }^{-1}$, thenso the Neumann series makes $b$ invertible. This proves that $G(A)$ is open. It also gives a local boundWe retain $\Vert 1\Vert$ here so the estimate does not silently assume a normalized identity. The inverse identitythen impliesThis proves continuity of inversion in a Banach algebra.

For a unital Banach algebra, the spectrum of an element isFor a nonunital Banach algebra, use its unitization of an algebra $A^{+}=A\oplus \mathbb{C}$ with multiplication and normIt is a unital Banach algebra, with identity $(0,1)$, and define ${\sigma }_A(x)={\sigma }_{A^{+}}((x,0))$. In this nonunital convention, $0$ belongs to the spectrum: the scalar coordinate of $(x,0)$ is zero, so it cannot be invertible in $A^{+}$.

The spectrum is nonempty and compact. It suffices to work in a nonzero unital Banach algebra, since unitization reduces the other case to this one. The group of invertible elements of a Banach algebra is open, so the complement of the spectrum of an element is open. For $|\lambda |>\Vert x\Vert$, the Neumann series givesThus ${\sigma }_A(x)$ is closed and lies in $\{\lambda :|\lambda |\le \Vert x\Vert \}$, hence is compact.

For nonemptiness of the Banach-algebra spectrum, suppose that $R$ were defined on all of $\mathbb{C}$. At any ${\lambda }_0$, factoring $\lambda 1-x=({\lambda }_{0}1-x)(1+(\lambda -{\lambda }_0)R({\lambda }_0))$ gives a locally convergent power seriesFor every bounded linear functional $\ell \in A^{\ast }$, the scalar function $\ell (R(\lambda ))$ is therefore entire. The bound at infinity makes it bounded outside a disk, while continuity makes it bounded on the disk. By the Liouville theorem, it is constant; since it tends to zero at infinity, it is identically zero. The version of the Hahn-Banach theorem used here says that bounded linear functionals separate points of a normed vector space: if $z\ne 0$, there is $\ell$ with $\ell (z)\ne 0$. Consequently $R(\lambda )=0$ for every $\lambda$, contradicting $(\lambda 1-x)R(\lambda )=1$. This proves the assertion.

Every nonzero complex unital normed division algebra is algebraically and topologically isomorphic to $\mathbb{C}$. This is the Gelfand-Mazur theorem; completeness is not needed in its statement. Let $D$ be a normed division algebra, and take its completion of a normed space using the algebra norm to obtain a unital Banach algebra $\hat{D}$. Submultiplicativity extends multiplication continuously to the completion, and the original identity remains its identity.

For $x\in D$, the spectrum of an element of $x$ in $\hat{D}$ contains some $\lambda$. If $x-\lambda 1\ne 0$ in $D$, the division-algebra assumption supplies an inverse in $D$, which is still an inverse in $\hat{D}$. This contradicts $\lambda \in {\sigma }_{\hat{D}}(x)$. Therefore $x=\lambda 1$. The map $\lambda \mapsto \lambda 1$ is a bijective complex algebra homomorphism, andIt and its inverse are continuous; when $\Vert 1\Vert =1$, it is an isometry.

A complete algebra norm on a function algebra dominates the supremum norm, even before continuity of point evaluations is known. Let $A$ be the given algebra of functions. Form the same artificial unitization $A^{+}$ even if $A$ already has an identity. For each $t\in K$, the algebraic mapis a unital multiplicative complex linear functional. No continuity has been assumed. Since ${\chi }_t(f-f(t)1)=0$, the element $f-f(t)1$ cannot be invertible: applying ${\chi }_t$ to an inverse equation would give $0=1$. Hence $f(t)\in {\sigma }_{A^{+}}(f)$, and the spectrum bound already proved yieldsTaking the supremum proves the supremum bound for a complete function-algebra norm:In particular every $f$ is bounded, and every point evaluation is a bounded linear functional of norm at most $1$. This is an instance of automatic continuity of characters: a character of an algebra on a Banach algebra is bounded because its value at any element belongs to that element's spectrum.

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.

A character of an algebra is a nonzero complex-linear multiplicative map $\phi :A\to \mathbb{C}$. The character space of an algebra ${\Phi }_A$ carries the Gelfand topology, namely pointwise convergence on $A$, equivalently the subspace weak-star topology in $A^{\ast }$.

For a unital complex Banach algebra, $\phi (1)=1$: multiplicativity and nonzeroness force this. If $a-\phi (a)1$ were invertible, applying $\phi$ to its product with its inverse would give $0=1$. Thus $\phi (a)\in {\sigma }_A(a)$. The Neumann series shows ${\sigma }_A(a)\subset \{|z|\le \Vert a\Vert \}$, so $|\phi (a)|\le \Vert a\Vert$ and the character of an algebra is automatically bounded. The same conclusion for a nonunital algebra follows by extending $\phi$ to its unitization of an algebra via $\tilde{\phi }(a,t)=\phi (a)+t$ and using the standard norm $\Vert (a,t)\Vert =\Vert a\Vert +|t|$.

In the algebra $R(K)$ let $u(z)=z$ and fix $\phi \in {\Phi }_{R(K)}$. Write $\lambda =\phi (u)$. If $\lambda \notin K$, the rational function $(u-\lambda {)}^{-1}$ belongs to $R(K)$, which is impossible because a character of an algebra cannot vanish on an invertible element. Thus $\lambda \in K$. For a rational function $r=p/q$ without poles in $K$, choose a reduced denominator having no zeros in $K$. Multiplicativity gives $\phi (r)=p(\lambda )/q(\lambda )=r(\lambda )$. Density of these rational functions and boundedness of $\phi$ then yield $\phi (h)=h(\lambda )$ for every $h\in R(K)$.

Conversely, evaluation at each $\lambda \in K$ is a character of an algebra. The map $\lambda \mapsto {\delta }_{\lambda }$ is continuous in the Gelfand topology because every $h\in R(K)$ is continuous; its inverse is $\phi \mapsto \phi (u)$, also continuous. Hence the character space of R(K) isThe identification includes the topology, not merely a bijection of sets.

Work over the complex scalar field. If $\lambda \in {\sigma }_A(x)$, the ideal generated by $x-\lambda 1$ is proper: otherwise commutativity would make $x-\lambda 1$ invertible. By Zorn lemma extend it to a maximal ideal $M$. This ideal is closed. Indeed, if its closure were all of $A$, it would contain an element of $M$ sufficiently close to $1$ to be invertible, since the invertible group is open; a proper ideal cannot contain such an element. Its closure is therefore proper, and maximality gives $\overline{M}=M$.

The quotient $A/M$ is a complex unital Banach algebra and a division algebra. The Gelfand-Mazur theorem identifies it with $\mathbb{C}$, giving a character of an algebra $\phi$ with $\phi (x)=\lambda$. Conversely, the argument in (a) shows that every $\phi (x)$ is in the spectrum. ThusHere, as usual, a unital Banach algebra has a nonzero identity; the zero algebra is excluded.

The holomorphic functional calculus assigns to every function holomorphic on an open neighbourhood $U$ of ${\sigma }_A(x)$ an element $f(x)\in A$. It is a unital algebra homomorphism, sends the coordinate function to $x$, agrees with evaluation of rational functions having no poles on the spectrum, depends only on the germ near the spectrum, and is continuous for uniform convergence on a fixed surrounding contour. Its formula isHere $\Gamma$ is a finite oriented contour cycle in $U\setminus {\sigma }_A(x)$, with winding number one on the spectrum and zero off $U$. A finite union of suitable polygonal boundary cycles suffices; no connectedness of $U$ is required. Such cycles exist by compactness of the spectrum, and the calculus is independent of the choice. The inverse is the resolvent of an element.

For the exponential, use a circle $|z|=R>\Vert x\Vert$. The uniformly convergent Neumann expansion $(z1-x{)}^{-1}={\sum }_{n\ge 0}{x}^n{z}^{-n-1}$ may be integrated term by term. The scalar Cauchy integral formula gives $\frac{1}{2\pi i}\int e^z{z}^{-n-1}dz=1/n!$, provingThis series converges absolutely in the Banach algebra norm.

A character of an algebra is bounded and takes the resolvent to $(z-\phi (x){)}^{-1}$. Applying it inside the contour integral and using the scalar Cauchy formula gives $\phi (f(x))=f(\phi (x))$. The character of an algebra description of the spectrum, now applied to $f(x)$, proves the spectral mapping theorem:In particular this spectrum is contained in $V$ when $V\supset f(U)$.

For the composition rule for holomorphic functional calculus, choose a relatively compact open neighbourhood $D$ of $f({\sigma }_A(x))$ whose closure is contained in $V$, and a surrounding contour cycle $\Delta$ in $V$ with winding number one on $D$. Choose a sufficiently small neighbourhood $\Omega$ of ${\sigma }_A(x)$ with $\overline{\Omega }\subset U$ and $f(\overline{\Omega })\subset D$, and a contour $\Gamma$ in $\Omega$ surrounding the spectrum. The neighbourhoods and cycles may be finite unions, obtained from finite covers of the relevant compact sets. For $\zeta \in \Delta$ the function $h_{\zeta }(z)=(\zeta -f(z){)}^{-1}$ is holomorphic near $\overline{\Omega }$. Locality and multiplicativity of the calculus giveInsert this in the formula for $g(f(x))$ and interchange the two vector-valued integrals. The scalar inner integral is $g(f(z))$, soThis proves equality as algebra elements. Equality of their character of an algebra values alone would be insufficient in a Banach algebra with nonzero radical.

Finally, $\Vert x\Vert <1$ places the spectrum in the open unit disk. On that disk $f(z)=\log (1-z)=-{\sum }_{n\ge 1}{z}^n/n$ is a holomorphic branch. Its calculus value is the norm-convergent series $y=-{\sum }_{n\ge 1}{x}^n/n$: one may integrate the uniformly convergent scalar series on a circle with $\Vert x\Vert <R<1$. Since $\exp (f(z))=1-z$, the composition rule gives the logarithm of an element near the identity formula

The prescribed-pole form of Runge theorem is this: let $K\subset \mathbb{C}$ be compact, and let $E\subset \hat{\mathbb{C}}\setminus K$ meet every connected component of the complement in the Riemann sphere. If $f$ is holomorphic on a neighbourhood of $K$, then for every $ϵ>0$ there is a rational function with all its poles in $E$ whose uniform distance from $f$ on $K$ is less than $ϵ$. A polynomial is regarded as having its only possible pole at infinity. The empty compact set is trivial, so assume $K\ne \varnothing$.

First suppose $\infty \in E$. Let $A\subset C(K)$ be the uniform rational approximation algebra with prescribed poles, the uniform closure of rational functions whose finite poles lie in $E$. It is a commutative unital Banach algebra, and contains $u(z)=z$. Setwhere the inverse in this definition is the pointwise continuous function on $K$. The set $S$ is relatively open: if the inverse at $\lambda$ belongs to $A$, the Neumann series gives inverses at nearby points. It is relatively closed: when ${\lambda }_n\to \lambda \notin K$, the corresponding scalar functions converge uniformly on $K$, and $A$ is closed.

Each bounded complementary component meets $E$ in a finite point $a$, and $(u-a1{)}^{-1}$ is an allowed rational function. The unbounded component meets $S$ because for $|\lambda |>\Vert u\Vert$ the geometric series for $(u-\lambda 1{)}^{-1}$ belongs to $A$. Being both open and closed in the complement, $S$ therefore contains every component, so it is the entire complement. Conversely, evaluation at any $z\in K$ prevents $u-z1$ from being invertible. HenceApply the holomorphic functional calculus in $A$ to $f$ near $K$. For every $z\in K$, the evaluation character of an algebra and the contour formula from (b) give $f(u)(z)=f(z)$. Thus $f{|}_K\in A$. By the definition of this uniform closure, the required rational approximations exist.

If $\infty \notin E$, choose a finite point $a\in E$ in the component containing infinity. The Möbius transformation $w=(z-a{)}^{-1}$ sends $K$ to a compact subset $K^{\prime }$ of the plane and sends $E$ to a set $E^{\prime }$ containing infinity. It preserves complementary components, so $E^{\prime }$ meets every component of the complement of $K^{\prime }$. The transformed function $F(w)=f(a+1/w)$ is holomorphic near $K^{\prime }$, since $0\notin K^{\prime }$. The case already proved approximates $F$ by rational functions with poles in $E^{\prime }$. Pulling them back gives rational approximations to $f$ on $K$ with poles only in $E$: a pole at infinity in the $w$-plane becomes a pole at $a$, and every other pole has its prescribed preimage. In particular the pullbacks do not acquire a pole at the original infinity, since $0\notin E^{\prime }$.

This proves the full prescribed-pole theorem, including the case where a pole at infinity is forbidden. If $\mathbb{C}\setminus K$ is connected, take $E=\{\infty \}$ to obtain the polynomial Runge theorem as a corollary.

For the first norm-one assertion, use the usual unital Banach algebra convention $\Vert 1\Vert =1$. A character of an algebra is a nonzero complex linear multiplicative functional, initially without any continuity assumption. It satisfies $\phi (1)=1$, since otherwise $\phi (a)=\phi (a1)=0$ for every $a$. Moreover, $a-\phi (a)1$ cannot be invertible: applying $\phi$ to an inverse identity would give $0=1$. ThusThis proves automatic continuity of characters and $\Vert \phi \Vert \le 1$. Evaluation at the normalized identity gives the reverse inequality, soIf a convention permits $\Vert 1\Vert >1$, only $\Vert \phi \Vert \le 1$ follows in general: on $\mathbb{C}$ with norm $2|z|$, the identity character has norm $1/2$. The later completion argument needs only the upper bound.

If $a$ is invertible, $\phi (a)\phi (a^{-1})=1$, so every character is nonzero on $a$. Conversely, if $a$ is not invertible, its principal ideal is proper and lies in a maximal ideal $M$ by Zorn lemma. The ideal $M$ is closed: if its closure were the whole algebra, it would contain $m$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series, which is impossible for a proper ideal. Maximality therefore gives $\overline{M}=M$. The quotient $A/M$ is a complex Banach division algebra, hence is $\mathbb{C}$ by the Gelfand-Mazur theorem. The quotient map yields a character vanishing at $a$. Consequently

Now let $A=C(K)$. Its evaluation characters give the map $x\mapsto {\delta }_x\in {\Phi }_A$. Every character has this form. Indeed, its kernel is a proper ideal. If the functions in that kernel had no common zero, compactness would give finitely many $f_1,\dots ,f_m$ in the kernel with no common zero. Then $h={\sum }_j\overline{f_j}{f}_j$ belongs to the kernel and is everywhere positive, hence is invertible in $C(K)$, a contradiction. At a common zero $x$, the fact that $f-\phi (f)1$ belongs to the kernel gives $f(x)=\phi (f)$ for every $f$. Continuous functions distinguish points of the compact Hausdorff space $K$, so the map is bijective. It is continuous for the Gelfand topology, and the compact-to-Hausdorff continuous bijection theorem givesFor $K=\varnothing$ the norm comparison is trivial; assume henceforth that $K\ne \varnothing$.

Let $B$ be the completion for the other algebra norm $\Vert \cdot {\Vert }_1$. It remains commutative and unital, with $A$ embedded densely. The restriction of any $\psi \in {\Phi }_B$ is a nonzero character of $A$, since it takes $1$ to $1$, and is automatically continuous for the original supremum norm. Thus $R(\psi )=\psi {|}_A$ maps into ${\Phi }_A$. It is injective because two continuous characters agreeing on the dense subalgebra $A$ agree on $B$, and it is continuous because each coordinate $R(\psi )(f)=\psi (f)$ is weak-star continuous.

The character space ${\Phi }_B$ is weak-star compact even if its identity was not normalized: the spectral bound puts all its characters in $B_{B^{\ast }}$, and the conditions $\psi (1)=1$, $\psi (bc)=\psi (b)\psi (c)$ define a weak-star closed set. Apply Banach-Alaoglu theorem. Consequently $R$ is a homeomorphism onto a closed subset $L\subseteq K$.

Suppose $x\in U=K\setminus L$. The compact Hausdorff space $K$ is normal, so choose an open $V$ with $x\in V\subseteq \overline{V}\subseteq U$. By the Urysohn lemma, choose $g\in C(K)$ with $g(x)=1$ and $g=0$ on $K\setminus V$, and choose $f\in C(K)$ with $f=0$ on $\overline{V}$ and $f=1$ on $L$. Then $fg=0$. Every character $\psi$ of $B$ restricts to evaluation at a point of $L$, so $\psi (f)=1$. The invertibility criterion proved above makes $f$ invertible in $B$. Multiplying $fg=0$ by $f^{-1}$ in $B$ gives $g=0$, contradicting $g(x)=1$. Therefore $U=\varnothing$ and $R$ is onto $K$.

For each $x\in K$, choose a character ${\psi }_x\in {\Phi }_B$ extending ${\delta }_x$. Its spectral bound givesTaking the supremum over $x$ proves the minimality of the supremum norm on C(K):

The Beurling–Gelfand spectral radius formula states that for every element $a$ of a complex Banach algebra,For a Hermitian element of a C-star algebra $a=a^{\ast }$, the C-star identity gives $\Vert a^2\Vert =\Vert a{\Vert }^2$. Iterating,The subsequence $n=2^m$ in the spectral-radius limit therefore proves $\boxed{r(a)=\Vert a\Vert }$.

Let $\theta :A\to B$ be a unital C-star homomorphism. If $a-\lambda 1$ is invertible in $A$, its inverse is sent to an inverse of $\theta (a)-\lambda 1$ in $B$. Thus ${\sigma }_B(\theta (a))\subseteq {\sigma }_A(a)$ without assuming continuity of $\theta$. For Hermitian $a$, $\theta (a)$ is also Hermitian, soFor arbitrary $x\in A$, apply this inequality to $x^{\ast }x$ and the C-star identity in both algebras:Hence $\boxed{\Vert \theta (x)\Vert \le \Vert x\Vert }$ for every $x$.

Now assume $\theta$ is injective. For a fixed $x$, it is enough to prove norm preservation on $C=C^{\ast }(1,x^{\ast }x)$, a commutative unital C-star algebra: the preceding square-norm identity will then recover the norm of $x$. By the Commutative Gelfand--Naimark theorem, identify $C$ with $C(K)$. Let $D=\overline{\theta (C)}\subseteq B$. It is also a commutative unital C-star algebra, and its character space ${\Phi }_D$ is compact. Each $\chi \in {\Phi }_D$ restricts along $\theta$ to a character of $C(K)$, hence to evaluation at a point $q(\chi )\in K$. The map $q:{\Phi }_D\to K$ is continuous, so $L=q({\Phi }_D)$ is closed.

If $L\ne K$, the Urysohn lemma provides a nonzero $f\in C(K)$ vanishing on $L$. Then $\chi (\theta (f))=f(q(\chi ))=0$ for every $\chi \in {\Phi }_D$. Faithfulness of the Gelfand transform in $D$ gives $\theta (f)=0$, contradicting injectivity. Thus $q$ is onto. The isometric Gelfand transform of $D$ now givesFinally,This proves that an injective C-star homomorphism is isometric without assuming its image is closed before norm preservation has been established.

For an element $x$ of a unital complex Banach algebra $A$, its spectrum of an element isIf $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so the spectrum is bounded. If ${\lambda }_0$ is in the resolvent set, thenis invertible for $\lambda$ sufficiently close to ${\lambda }_0$, again by a Neumann series. Thus the resolvent is open and the spectrum is closed, hence compact.

If the spectrum were empty, the resolvent $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire Banach-space-valued function and would tend to zero at infinity. For every $f\in A^{\ast }$, the scalar entire function $f(R(\lambda ))$ would be bounded and therefore constant by Liouville theorem. It would be zero because of the limit at infinity. The Hahn-Banach theorem would then force $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Therefore ${\sigma }_A(x)$ is nonempty.

Now let $B\subseteq A$ be a closed unital subalgebra containing $x$. Invertibility in $B$ implies invertibility in $A$, soOn each connected component of $\mathbb{C}\setminus {\sigma }_A(x)$, either the resolvent belongs to $B$ everywhere or nowhere. Indeed, membership holds on a neighbourhood of any one such point by its local Neumann expansion, and the same argument makes the set of such points relatively closed by taking limits in the closed subalgebra $B$. The unbounded component belongs to the first case because the geometric resolvent series lies in $B$ for large $|\lambda |$. ConsequentlyThis is the spectrum in a closed unital subalgebra theorem.

For the bilateral shift $T$ on ${\ell }^1(\mathbb{Z})$, both $T$ and $T^{-1}$ are isometries. The geometric-series argument applied to $T$ for $|\lambda |>1$ and to $T^{-1}$ for $|\lambda |<1$ givesFor $|\lambda |=1$, normalize the vector given by $x_n={\lambda }^{-n}$ on $-N\le n\le N$ and zero elsewhere. Only its two boundary coordinates contribute to $(T-\lambda I)x$, so the norm of that image tends to zero. Thus $T-\lambda I$ is not bounded below and cannot be invertible. Hence

The algebra generated by $T$ is the norm closure of polynomials in nonnegative powers of $T$. It does not contain $T^{-1}$: every $p(T)e_0$ is supported in nonnegative coordinates, whereas $T^{-1}{e}_0=e_{-1}$. Therefore $0\in {\sigma }_{\mathcal{A}}(T)$. The preceding hole theorem then fills the unique bounded component of $\mathbb{C}\setminus \mathbb{T}$, giving