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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.

The Riesz-Markov-Kakutani representation theorem identifies the dual of the complex space of continuous functions on a compact space $C(K)$ with finite regular complex Borel measures on $K$: each bounded linear functional has a unique representationHere $|\mu |$ is the variation measure; positive functionals correspond exactly to positive regular measures. This is the measure representation theorem, rather than the Hilbert-space representation of a functional by a vector.

For the spectral construction use the convention that the Hilbert inner product is linear in its first argument. A unital subalgebra of $\mathcal{B}(H)$ is understood to contain $I_H$: an algebra whose abstract identity is a smaller projection could not give a resolution normalized by $P(K)=I_H$. On a compact Hausdorff space, a resolution of the identity is understood to be a normalized regular projection-valued measure: its scalar measures are regular, its values are orthogonal projections, and it is countably additive in the strong operator topology. Regularity is part of the usual convention needed for the uniqueness assertion; the statement is interpreted in this sense.

The Commutative Gelfand--Naimark theorem makes the Gelfand transform an isometric unital star-isomorphism $A\to C(K)$, where $K={\Phi }_A$ is compact Hausdorff. Denote its inverse by $\pi :C(K)\to A\subset \mathcal{B}(H)$. For $x,y\in H$, the bounded linear functional $f\mapsto ⟨\pi (f)x,y⟩$ has norm at most $\Vert x\Vert \Vert y\Vert$. Riesz-Markov-Kakutani gives a regular complex measure ${\mu }_{x,y}$ such thatUniqueness makes these measures sesquilinear in $x,y$. For $f\ge 0$, the identity $\pi (f)=\pi (\sqrt{f})\pi (\sqrt{f}{)}^{\ast }$ shows positivity, so ${\mu }_{x,x}$ is positive with total mass $\Vert x{\Vert }^2$.

For every Borel set $B$, the bounded sesquilinear form $(x,y)\mapsto {\mu }_{x,y}(B)$ determines an operator $P(B)$ by Hilbert-space Riesz representation theorem, withIn particular $0\le P(B)\le I$ and $P(B)$ is self-adjoint. We now verify the projection identity rather than assume it.

For continuous $f$, uniqueness of the representing measure applied to continuous test functions givesThese identities imply $P(B)\pi (f)=\pi (f)P(B)$. Testing once more against a continuous $g$ givesThe restriction $1_B{\mu }_{x,y}$ of a finite regular Borel measure is still regular, so uniqueness in Riesz-Markov-Kakutani yields ${\mu }_{P(B)x,y}=1_B{\mu }_{x,y}$. Consequently, for any Borel $C$,Taking $C=B$ proves that $P(B)$ is an orthogonal projection. Also $P(K)=I$ and $P(\varnothing )=0$. For disjoint $B_j$, these projections are orthogonal, and scalar countable additivity gives weak countable additivity. If $B={\bigcup }_j{B}_j$, then $R_N=P(B)-{\sum }_{j\le N}P(B_j)$ is the projection of the remaining union andThus countable additivity holds in the strong operator topology. The scalar measures are the regular measures already constructed. This completes the scalar-measure construction of a projection-valued measure.

By the integral theorem permitted in the question, continuous $f$ satisfies $⟨(\int fdP)x,y⟩=\int fd{\mu }_{x,y}=⟨\pi (f)x,y⟩$, and hence $\int fdP=\pi (f)$. For $T\in A$, this proves the spectral theorem for a commutative operator algebra:Any other regular resolution giving these integrals has the same scalar integrals on all of $C(K)$; Riesz-Markov-Kakutani uniqueness forces the same scalar measures and therefore the same projections on every Borel set.

For nonempty open $U\subset K$, compact Hausdorff normality supplies a nonzero continuous function $f$ supported in $U$. If $P(U)=0$, the stated squared-norm identity for spectral integrals gives $\pi (f)=\int fdP=0$, contradicting the isometry of $\pi$. This proves the full support of a faithful spectral measure propertyFaithfulness of the representation is essential here.

The spectral theorem for normal operators says that a bounded normal operator $T$ on a nonzero complex Hilbert space has a unique regular projection-valued measure $E$ on $\sigma (T)$ such that $T=\int zdE(z)$. Its support is all of $\sigma (T)$, and bounded Borel functions have the associated Borel functional calculus for a normal operator.

For the proof sketch, $A=C^{\ast }(I,T)$ is commutative because $T$ commutes with $T^{\ast }$; polynomials in these two operators commute, as do their norm limits. The map ${\Phi }_A\to \sigma (T)$, $\phi \mapsto \phi (T)$, is onto by the character of an algebra formula and spectral permanence for C-star algebras. It is one-to-one because a character of an algebra preserves the star operation and its values on $T,T^{\ast }$ determine it on their dense polynomial algebra. It is therefore a homeomorphism from compact ${\Phi }_A$ to the Hausdorff spectrum. Transport the resolution just constructed through this homeomorphism. It gives the formula for $T$ and full support; conversely a regular resolution for $T$ gives the same integrals for polynomials in $T,T^{\ast }$, hence by density the same continuous functional calculus and the same resolution. This argument works without separability of $H$.

Finally choose disjoint nonempty relatively open sets $U,V\subset \sigma (T)$ around two distinct spectral points, and put $Q=E(U)$. Full support gives $Q\ne 0$ and $E(V)\ne 0$, while $QE(V)=0$, so $Q\ne I$. The spectral integral, or multiplicativity of its Borel calculus with $1_U$, gives $QT=TQ$ and also $Q{T}^{\ast }=T^{\ast }Q$. Thus $Y=QH$ is closed, nonzero and proper, and $T(Qx)=Q(Tx)\in Y$. The spectral projection gives a reducing subspace conclusion isIn fact it is a reducing subspace, since it is also invariant under $T^{\ast }$.

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):

A character of an algebra is a nonzero algebra homomorphism $\phi :A\to \mathbb{C}$. The character space isFor a unital algebra, $\phi (1)=1$. Moreover $\phi (a)\in {\sigma }_A(a)$, since applying $\phi$ shows that $a-\phi (a)1$ cannot be invertible. In a Banach algebra,Thus $\Vert \phi \Vert \le 1$, while $\phi (1)=1$ gives equality.

If $A$ is commutative and $\lambda \in {\sigma }_A(a)$, the proper ideal generated by $a-\lambda 1$ lies in a maximal ideal. The quotient by that maximal ideal is $\mathbb{C}$, and its quotient map is a character taking $a$ to $\lambda$. ThereforeSince every Banach-algebra element has nonempty spectrum, ${\Phi }_A$ is nonempty.

For $X=Y\oplus Z$, let $P,Q$ be the complementary coordinate projections. A character must send each idempotent to zero or one, and $P+Q=I$ forces their values to be different. Choose mutually inverse isomorphisms between $Y$ and $Z$ and regard them as off-diagonal operators $S,T$ on $X$. ThenMultiplicativity would give $\phi (P)=\phi (S)\phi (T)=\phi (T)\phi (S)=\phi (Q)$, a contradiction. Hence ${\Phi }_{\mathcal{B}(X)}$ is empty. This is the absence of characters on an operator algebra with isomorphic complementary summands.

If $a=a^{\ast }$ in a unital C*-algebra, then $e^{ita}$ is unitary for real $t$. If $\lambda \in \sigma (a)$, the spectral mapping theorem gives $e^{it\lambda }\in \sigma (e^{ita})$, whose modulus is one. Varying positive and negative $t$ forces $\mathrm{Im}\lambda =0$, so $\sigma (a)\subset \mathbb{R}$.

If $B\subseteq A$ is a unital C*-subalgebra and $x\in B$ is normal, spectral permanence holds. Indeed, when $x-\lambda$ is invertible in $A$, the positive normal elementhas spectrum bounded away from zero. Continuous functional calculus uniformly approximates its reciprocal by polynomials, placing the reciprocal in $B$. It follows that $(x-\lambda {)}^{-1}\in B$. Thus ${\sigma }_B(x)={\sigma }_A(x)$.

The Commutative Gelfand--Naimark theorem says that the Gelfand transform is an isometric unital star-isomorphismfor every commutative unital C*-algebra. If $a$ is positive, continuous functional calculus for the function $t\mapsto \sqrt{t}$ on $\sigma (a)\subset [0,\infty )$ gives a positive $b$ with $b^2=a$. Pointwise uniqueness in $C^{\ast }(1,a)\cong C(\sigma (a))$ gives the unique positive square root. This is the positive square root in a C*-algebra.

In a unital C-star algebra $A$, an element is Hermitian when $x=x^{\ast }$, unitary when $x^{\ast }x=x{x}^{\ast }=1$, and normal when $x^{\ast }x=x{x}^{\ast }$. It is a Positive element of a C-star algebra when $x=y^{\ast }y$ for some $y$, equivalently when it is Hermitian and its spectrum lies in $[0,\infty )$.

Suppose first that $\Vert \tau \Vert =\tau (1)$, and normalize so that both equal one. If $x=x^{\ast }$, the C-star identity givesWriting $\tau (x)=a+ib$, we obtainfor every real $t$, which forces $b=0$. Decomposing an arbitrary $x$ into its Hermitian real and imaginary parts now gives

Let $x=x^{\ast }$. Every character of an algebra on the commutative unital Banach algebra generated by $x$ has norm and value at one equal to one, so the preceding argument makes its value on $x$ real. The character description of the spectrum of an element therefore gives $\sigma (x)\subset \mathbb{R}$. Iterating the C-star identity,so $\Vert x\Vert =r(x)$. If $c=\Vert x\Vert$, then $x+c1$ has nonnegative spectrum and is positive, as is $c1$; henceis a difference of positive elements.

If $0\le x$ and $\Vert x\Vert \le 1$, then spectral translation givesThus $1-x$ is positive and, by the norm--spectral-radius equality for Hermitian elements, $\Vert 1-x\Vert \le 1$.

Return to a norm-one $\tau$ with $\tau (1)=1$. For $0\le x\le 1$, the preceding paragraph and reality on Hermitian elements giveScaling proves that $\tau$ is a positive functional on a C-star algebra. Every character has norm and value at one equal to one, so every character is positive. On $C[0,1]$, the functionalis positive but is not multiplicative, and hence is not a character.

Conversely, let $\tau$ be positive. Writing a Hermitian element as a difference of positive elements shows that $\tau$ is real on Hermitian elements. Positivity offor every $\lambda \in \mathbb{C}$ says that the associated quadratic polynomial is nonnegative. Minimizing it in $\lambda$ gives the Cauchy--Schwarz inequalityTaking $y=1$ yields $|\tau (x){|}^2\le \tau (x^{\ast }x)\tau (1)$. Since $0\le x^{\ast }x\le \Vert x{\Vert }^{2}1$, positivity givesTogether with $\tau (1)\le \Vert \tau \Vert$, this proves $\Vert \tau \Vert =\tau (1)$.

Convex combinations preserve positivity and value one at the identity, so the state space $S(A)$ is convex. If $x$ is normal, the unital C-star subalgebra $C^{\ast }(1,x)$ is commutative and its Gelfand transform identifies it with $C(\sigma (x))$. Choose $\lambda \in \sigma (x)$ with $|\lambda |=\Vert x\Vert$. Evaluation at $\lambda$ is a state taking $x$ to $\lambda$. Its norm-preserving Hahn--Banach extension to $A$ still takes $1$ to one, so the norm criterion makes the extension a state $\tau$ satisfying $|\tau (x)|=\Vert x\Vert$.

The state space is nonempty, convex, and weak-star compact by the Banach-Alaoglu theorem. The Krein-Milman theorem gives an extreme point, so a pure state on a C-star algebra exists. For positive $x$, the preceding norm-attainment result makesa nonempty weak-star compact convex set. It is a face: if a convex combination has the maximal possible value on $x$, each summand does. An extreme point of $F$ exists by Krein--Milman and, because $F$ is a face, is extreme in $S(A)$. It is the required pure state.

A character of an algebra is a nonzero multiplicative complex-linear functional $\phi :A\to \mathbb{C}$, and the character space of an algebra is the set ${\Phi }_A$ of all such characters. Since $A$ is unital, $\phi (1)=1$. Moreover $\phi (a)\in {\sigma }_A(a)$: otherwise $a-\phi (a)1$ would be invertible, while applying $\phi$ to its inverse identity would give $0=1$. The spectral radius estimate therefore yieldsThus every character is continuous and has norm one.

Let $M$ be a maximal ideal. Its norm closure is again an ideal. It cannot equal $A$, because then some $m\in M$ would satisfy $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series and forcing $M=A$. Hence $M$ is closed. The quotient $A/M$ is a complex unital Banach division algebra, so the Gelfand-Mazur theorem identifies it with $\mathbb{C}$. Composing the quotient map with this isomorphism gives a character with kernel $M$. Conversely, a character kernel is maximal because its quotient is $\mathbb{C}$.

Now $\lambda \in {\sigma }_A(x)$ exactly when $x-\lambda 1$ is not invertible, equivalently when it lies in some maximal ideal. The preceding result turns that ideal into $\ker \phi$, giving $\phi (x)=\lambda$. The reverse implication follows from the first paragraph, so

The Gelfand topology is the weak-star topology on ${\Phi }_A\subseteq A^{\ast }$. The Gelfand transform isIts values are continuous by the definition of the topology, and multiplicativity and linearity of characters show that it is a unital algebra homomorphism. Finallyso it is continuous.

A character of an algebra is a nonzero multiplicative complex-linear functional $\phi :A\to \mathbb{C}$. The character space of an algebra isIf $A$ is unital, multiplicativity and nonzeroness force $\phi (1)=1$.

For a unital Banach algebra, $\phi (a)$ belongs to ${\sigma }_A(a)$: otherwise $a-\phi (a)1$ would be invertible, while applying $\phi$ to its inverse identity would give $0=1$. Thereforeso every character of an algebra is continuous and has norm one. The nonunital case follows by extending the character to the unitization of an algebra.

The Gelfand topology on ${\Phi }_A$ is the weak-star topology inherited from $A^{\ast }$: a net ${\phi }_i$ converges to $\phi$ exactly when ${\phi }_i(a)\to \phi (a)$ for every $a\in A$. If $A$ is unital, ${\Phi }_A$ lies in the weak-star compact dual unit ball by the Banach-Alaoglu theorem. The equationsdefine a weak-star closed subset, so ${\Phi }_A$ is compact.

If the spectrum of an element $x$ lies in the open right half-plane, holomorphic functional calculus gives the displayed square root with spectrum in that half-plane. It is unique there. Two such roots $y,v$ have character values equal to the unique scalar square root with positive real part. Every character of an algebra is therefore nonzero on $y+v$, so the character description of the spectrum makes $y+v$ invertible. From $(y-v)(y+v)=0$ one obtains $y=v$. Equal character values alone do not prove equality in a possibly nonsemisimple algebra.

For a commutative complex unital Banach algebra, a character of an algebra annihilates $a-\varphi (a)1$, so this element cannot be invertible. Conversely a noninvertible $a-\lambda 1$ generates a proper ideal, which lies in a maximal ideal. The corresponding character then has value $\lambda$ at $a$. Thus the Gelfand transform determines every element's spectrum even when the algebra is not semisimple.

For $B=\overline{\mathbb{C}[1,x]}$ inside $A$, the displayed set is the full spectrum of $x$ in $A$. On a bounded complementary component, the maximum modulus principle gives $|p(\lambda )|\le {\max }_{{\sigma }_A(x)}|p|\le \Vert p(x)\Vert$. Evaluation $p(x)\mapsto p(\lambda )$ therefore extends to a character of an algebra on $B$, showing $\lambda \in {\sigma }_B(x)$. Conversely the unbounded resolvent component survives by resolvent components in a closed unital subalgebra. In particular, if zero lies in that component, $x^{-1}$ belongs to the norm closure of polynomials in $x$.

For an algebra of complex-valued functions with a complete algebra norm, every point value $f(t)$ belongs to the spectrum of an element of $f$ in the artificial unitization. Indeed $(h,\lambda )\mapsto h(t)+\lambda$ is an algebraic unital character of an algebra, and it vanishes on $f-f(t)1$. The spectrum bound gives $|f(t)|\le \Vert f\Vert$ without first assuming that evaluation is continuous.