Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 6 / 4

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 ${\Phi }_A$.

We prove the forward direction of the finite-generator criterion for a natural uniform algebra. Suppose that $A$ is a natural uniform algebra, and let $f_1,\dots ,f_n$ have no common zero on $K$. The idealis all of $A$. Otherwise it is contained in a maximal ideal $M$, whose associated character of an algebra vanishes on all the $f_j$. Since every algebra character is an evaluation character under naturality, this would give a common zero, a contradiction. Therefore $1\in J$, meaning that

Here are the Banach-algebra details behind the algebra character associated with $M$. A proper ideal cannot have dense closure: if it contained an element within distance less than one of $1$, the Neumann series would make that element invertible, forcing the ideal to contain $1$. Thus the closure of a proper ideal is proper. A maximal ideal, obtained from Zorn's lemma, is therefore closed. Its quotient is a complex unital Banach algebra in which every nonzero element is invertible. The Gelfand-Mazur theorem identifies this quotient with $\mathbb{C}$, so the quotient map is the required algebra character. No closedness of the originally generated ideal $J$ is needed.

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.