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A uniform algebra is natural exactly when functions with no common zero generate the unit ideal. If a generated ideal is proper, it is contained in a maximal ideal. That maximal ideal is closed: the closure of a proper ideal stays proper, since an element within distance less than one of the identity is invertible by the Neumann series. The Gelfand-Mazur theorem identifies the maximal-ideal quotient with the complex numbers, yielding a character of an algebra. Naturality would make that character a common-zero evaluation, a contradiction. Conversely, the asserted unit-ideal property makes the zero sets of every finite collection in a character's kernel intersect. Compactness gives a point common to all these zero sets; applying this to $f-\phi (f)1$ proves $\phi (f)=f(x)$. For $C(K)$, one can directly take $g_j=\overline{f_j}/{\sum }_k|f_k{|}^2$.

A uniform algebra on $K$ is natural if every character of an algebra is evaluation at a point of $K$. The evaluation embedding then identifies its character space homeomorphically with $K$. A natural uniform algebra is characterized by the finite-generator criterion for a natural uniform algebra. In particular, $C(K)$ is natural; the disk algebra is natural when its underlying space is the closed disk.

The prescribed-pole form of Runge theorem is as follows. Let $K\subset \mathbb{C}$ be compact, let $f$ be holomorphic on a neighborhood of $K$, and let $S\subset \hat{\mathbb{C}}\setminus K$ meet every connected component of that complement. Then, for every $ϵ>0$, there is a rational function $R$, all of whose poles belong to $S$, with ${\sup }_K|f-R|<ϵ$. A pole at infinity means a polynomial part is allowed. In particular, if the complement is connected, choose $S=\{\infty \}$ and obtain the polynomial Runge theorem.

First approximate using poles anywhere off $K$. Choose a bounded polygonal neighborhood $V$ of $K$ whose closure lies in the neighborhood on which $f$ is holomorphic, and whose boundary is disjoint from $K$. Such a $V$ comes from a sufficiently fine finite grid covering a small closed neighborhood of $K$. Orient the outer boundary positively and boundaries of holes negatively. The Cauchy integral formula givesBecause $\partial V$ and $K$ have positive distance, the integrand is uniformly continuous on their product, separately along each boundary edge. Its Riemann sums converge uniformly for $z\in K$. Those sums are rational functions ${\sum }_j{c}_j/({\zeta }_j-z)$ with poles ${\zeta }_j\notin K$. Thus arbitrary exterior poles suffice.

To move them to the prescribed set, let $A$ be the uniform closure on $K$ of rational functions with poles only in $S$. It is a closed unital algebra; it contains the coordinate function when $\infty \in S$. DefineThis set is open relative to $\mathbb{C}\setminus K$. Indeed, if $a\in E$ and $|b-a|<\mathrm{dist}(a,K)$, the uniformly convergent expansionplaces the new reciprocal in $A$, using its algebra and closedness properties. It is also relatively closed: if $a_m\in E$ tends to $a\notin K$, then $(z-a_m{)}^{-1}$ tends uniformly on $K$ to $(z-a{)}^{-1}$. Each bounded component contains a finite selected pole and hence meets $E$. The unbounded component also meets $E$ if it contains a finite selected pole; if its selected pole is infinity, take $|a|>{\sup }_K|z|$ and usea uniform polynomial expansion. Being both open and closed and meeting every component, $E$ is all of $\mathbb{C}\setminus K$. Every initial Cauchy-sum reciprocal therefore belongs to $A$, and so does $f$. This proves Runge theorem, including its prescribed-pole assertion, rather than merely polynomial approximation on a special compact set.

For the requested pointwise polynomial approximation of a half-plane sign, take $n\ge 2$ andThese three compact sets are disjoint, and their union $K_n$ has connected complement: there are horizontal gaps above and below the segment, and every complementary point can be joined through these gaps or around a rectangle to the exterior of $[-n,n{]}^2$. Define $h_n$ to be $1,0,-1$ on disjoint open neighborhoods of these three sets. It is holomorphic on their union. By the polynomial Runge theorem, choose $P_n$ with ${\sup }_{K_n}|P_n-h_n|<1/n$.

For every fixed point with positive imaginary part, both coordinate bounds and $1/n\le \mathrm{Im}z$ eventually hold, so it lies in $K_n^{+}$ for all large $n$; the reflected assertion holds for negative imaginary part. Every real point eventually lies in $K_n^0$. ConsequentlyThe discontinuity of the limit does not contradict holomorphicity of the approximating functions: this convergence is pointwise, and is not locally uniform near the real axis.

Finally, identify $K^{\ast }$ as the holomorphic convex hull relative to $\Omega$. If $U$ is a component of $\hat{\mathbb{C}}\setminus K$ lying entirely in $\Omega$, it cannot contain infinity and is bounded. Its boundary lies in $K$, and its closure is compact in $\Omega$. For any holomorphic $f$ on $\Omega$, the maximum modulus principle on $U$ gives $|f(w)|\le {\sup }_{\partial U}|f|\le {\sup }_K|f|$. Thus $U\subseteq K^{\ast }$, as is $K$ itself.

Conversely let $w\in \Omega \setminus K$ lie in a complementary component $U$ that is not entirely in $\Omega$. If it contains a finite point $a\notin \Omega$, form the uniform algebra on $K$ generated by constants and $(z-a{)}^{-1}$. The same open-and-closed reciprocal argument above, now only on the connected component $U$, shows that $(z-w{)}^{-1}$ is in its uniform closure. Its approximants have poles only at $a$ and are holomorphic on $\Omega$. If the available point outside $\Omega$ is infinity, use the polynomial algebra and its large-pole expansion instead. Hence in either case choose $h\in \mathcal{O}(\Omega )$ withThe function $F(z)=1-(z-w)h(z)$ is holomorphic on $\Omega$, has $F(w)=1$, and satisfies ${\sup }_K|F|<1/2$. It separates $w$ from the hull, so $w\notin K^{\ast }$. Combining the two inclusions provesHere $\mathbb{P}$ denotes the Riemann sphere; using it ensures that the unbounded component, which contains infinity, is not accidentally filled. The empty compact set has empty hull by the usual convention; the separation argument above concerns nonempty $K$.

Assume the finite-generator identity holds whenever there is no common zero. Let $\phi \in {\Phi }_A$. For every finite family $f_1,\dots ,f_n\in \ker \phi$, their zero sets must have nonempty intersection: otherwise an identity $\sum f_j{g}_j=1$ would give $0=\phi (1)=1$. These zero sets are closed in compact $K$. The finite intersection property therefore supplies $x\in K$ at which every member of $\ker \phi$ vanishes. For any $f\in A$, the element $f-\phi (f)1$ is in that kernel, so $f(x)=\phi (f)$. Thus $\phi ={\delta }_x$, proving

For $C(K)$, the identity is explicit. If the $f_j$ have no common zero, the continuous function $q={\sum }_j|f_j{|}^2$ is strictly positive on compact $K$. SetThen $\sum f_j{g}_j=1$. The equivalence just proved shows that $C(K)$ is natural.

Next consider the disc algebra $A(\overline{\mathbb{D}})$, the continuous functions on the closed unit disk that are holomorphic in its interior. We show directly that each algebra character is evaluation. Polynomials are uniformly dense in this algebra: for $0<r<1$, $f_r(z)=f(rz)$ tends uniformly to $f$ as $r\uparrow 1$, by uniform continuity. The function $f_r$ is holomorphic on $|z|<1/r$, so its Taylor polynomials converge uniformly on $|z|\le 1$.

Every algebra character of a unital Banach algebra is bounded, with $|\phi (f)|\le \Vert f\Vert$: if $f-\phi (f)1$ were invertible, applying $\phi$ would contradict invertibility, so $\phi (f)$ belongs to the spectrum of an element, which is contained in the disk of radius $\Vert f\Vert$ by the Neumann series. For the coordinate function $z$, put $\lambda =\phi (z)$; then $|\lambda |\le 1$. Multiplicativity gives $\phi (p)=p(\lambda )$ for every polynomial. By polynomial density and continuity, $\phi (f)=f(\lambda )$ for every member of the disc algebra. Hence

Finally consider the algebra $C$ in the last request. For each $f\in C$, its holomorphic extension $g$ from the boundary is unique by the maximum modulus principle. Define $T(f)=g\in A(\overline{\mathbb{D}})$. This is a surjective unital algebra homomorphism, since it restricts to the identity on the disc algebra. AlsoThe algebra $C$ is closed: if $f_n\to f$ uniformly, the associated $T(f_n)$ are uniformly Cauchy by this bound, so their limit belongs to the closed disc algebra and agrees with $f$ on the boundary. Constants and the coordinate function belong to $C$, so it is itself a uniform algebra on the disk. Its idealis contained in $C$, and $C/I$ is the disc algebra.

If a algebra character $\chi$ on $C$ vanishes on $I$, it factors through $T$ and the preceding result givesIf $\chi$ does not vanish on $I$, choose $h\in I$ with $\chi (h)\ne 0$. For $F\in C(\overline{\mathbb{D}})$, defineThe product $hF$ is in $I$, so this is well defined. It is unital and linear, andshows multiplicativity. Since $C(\overline{\mathbb{D}})$ is natural, $\tilde{\chi }$ is evaluation at a point $x$ of the closed disk. The equality $h(x)=\tilde{\chi }(h)=\chi (h)\ne 0$ places $x$ in the open disk. On $C$, $\tilde{\chi }(f)=\chi (f)$, so $\chi$ itself is evaluation at that interior point.

We therefore obtain two families: actual evaluations $f\mapsto f(x)$ on one closed disk, and analytic-extension evaluations $f\mapsto T(f)(\lambda )$ on a second closed disk. They agree on the boundary. They are distinct at interior points: functions in $I$ distinguish actual interior evaluations from every analytic-extension evaluation, and the coordinate function distinguishes points within each family. ThusThis is also a topological identification. Each disk family is continuous in the Gelfand topology, and they agree along their boundary, giving a continuous bijection from the glued disks to ${\Phi }_C$. The glued disks form a compact space, explicitly homeomorphic to the sphere by sending $(x,y)$ in the two copies to $(x,y,\pm \sqrt{1-x^2-y^2})$. Since the character space is Hausdorff, the continuous bijection is a homeomorphism. This boundary-analytic disk algebra with doubled character space has more algebra characters than evaluations on its original disk.

A uniform algebra on a compact Hausdorff space $K$ is a closed complex subalgebra $A\subseteq C(K)$ containing the constants and separating points of $K$, with the supremum norm. It is therefore a commutative unital Banach algebra. Its character space of an algebra ${\Phi }_A$, also called its carrier space, consists of nonzero multiplicative linear functionals, equipped with the Gelfand topology.

For $x\in K$, evaluation gives the character of an algebra ${\delta }_x(f)=f(x)$. The map $\delta :K\to {\Phi }_A$ is continuous: composing it with the coordinate $\phi \mapsto \phi (f)$ gives the continuous function $f$ for every $f\in A$. It is injective because $A$ separates points. The Gelfand topology is Hausdorff because distinct algebra characters differ on some element of $A$. A continuous injection from compact $K$ into a Hausdorff space is a homeomorphism onto its image: closed subsets of $K$ have compact, hence closed, images. Its image is also compact and closed. Thusis the asserted homeomorphism. A natural uniform algebra is one for which this image is all of ${\Phi }_A$.

A uniform algebra on a compact Hausdorff space $K$ is a closed complex subalgebra of $C(K)$, with the supremum norm, containing the constants and separating points. Evaluation embeds $K$ continuously and injectively in the character space of an algebra. The embedding is a homeomorphism onto a closed subset because its domain is compact and its target is Hausdorff. Extra characters need not be point evaluations on the specified $K$.