Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 10 / 4

The holomorphic convex hull is usually defined analytically byWith that convention the requested equivalence is the definition. Its substantive planar content is the geometric descriptionOne can instead define the hull by the right side and prove the analytic equivalence; we do so after proving Runge theorem.

The prescribed-pole form of Runge theorem is as follows. If $K\subset \mathbb{C}$ is compact and $A\subset \mathbb{C}\setminus K$ meets each bounded complementary component, every holomorphic function on a neighborhood of $K$ is uniformly approximable on $K$ by rational functions with finite poles in $A$. Polynomial terms are allowed, accounting for the unbounded component. In particular, if every complementary component meets the complement of a domain $\Omega$, the approximants can be chosen as holomorphic functions on $\Omega$. Connected complement permits polynomial approximants.

For a proof, surround $K$ by the interior of a finite union of small grid squares whose closure lies in the neighborhood where $f$ is a holomorphic function. Orient their boundaries positively and cancel shared edges, obtaining finitely many piecewise linear boundary curves $\Gamma$ a positive distance from $K$. The Cauchy integral formula, applied to the union, givesUniform continuity of this integrand for $z\in K$ makes its Riemann sums converge uniformly on $K$. Each Riemann sum is a finite linear combination of reciprocals $(z-b{)}^{-1}$ with $b\in \Gamma \subset \mathbb{C}\setminus K$.

It remains to move these poles into the allowed components. Let $\mathcal{A}$ be the uniform closure on $K$ of rational functions with finite poles in $A$ and polynomial terms. This is a closed algebra. DefineFor $b\in E$ and $|c-b|<\mathrm{dist}(b,K)$, the uniformly convergent geometric serieslies in $\mathcal{A}$, so $E$ is relatively open. If $b_n\in E$ tends to $b\notin K$, the corresponding reciprocals converge uniformly on $K$, so $E$ is also relatively closed. Thus containing one point of a complementary component means containing the entire component. Each bounded component has its prescribed point in $A\subset E$. The unbounded component meets $E$ because, for $|b|>{\sup }_K|z|$,is uniformly approximated by polynomials. Therefore $E=\mathbb{C}\setminus K$. Approximate the finitely many reciprocal terms in a Riemann sum closely enough to control their combined error. This completes the proof of Runge theorem without assuming its conclusion during the pole-moving argument.

Now let $H$ denote the geometric filled set displayed above. For $w\in K$ the required inequality is automatic. If $w$ lies in a filled component $U$, then $f$ is a holomorphic function near its compact closure and $\partial U\subset K$. The maximum modulus principle gives $|f(w)|\le {\sup }_K|f|$. Hence $H\subseteq {\hat{K}}_{\Omega }$.

Conversely, suppose $w\in \Omega \setminus H$. Its component $V$ of $\mathbb{C}\setminus K$ meets $\mathbb{C}\setminus \Omega$, or is unbounded. To see the alternative, a bounded component contained in $\Omega$ has boundary in $K$, so its closure is compact in $\Omega$ and would have been filled. Apply the same open-and-closed reciprocal argument to the algebra generated by polynomials and reciprocals with poles outside $\Omega$. The component $V$ has a starting reciprocal either at a point outside $\Omega$ or near infinity. Thus there is $h\in \mathcal{O}(\Omega )$ approximating $(z-w{)}^{-1}$ as closely as desired on $K$. Let $M={\sup }_K|z-w|>0$ and choose the error less than $1/(2M)$. ThenThis holomorphic function separates $w$ from the analytic hull. Thus $w\notin {\hat{K}}_{\Omega }$, proving the equivalence with the geometric definition as well. For empty $K$, the filled hull is empty and the constant function one excludes every point when the supremum of the nonnegative modulus over the empty set is taken as zero; the nonempty case is the one used above.