= Solution
The <holomorphic convex hull> is usually defined analytically by
$$
\widehat K_\Omega=\{w\in\Omega:|f(w)|\leq\sup_K|f|\text{ for every }f\in\mathcal O(\Omega)\}.
$$
With that convention the requested equivalence is the definition. Its substantive planar content is the geometric description
$$
\boxed{\widehat K_\Omega
=K\ \cup\!\bigcup\{U:U\text{ is a component of }\Omega\setminus K,
\ \overline U\text{ is compact in }\Omega\}.}
$$
One 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, gives
$$
f(z)=\frac1{2\pi i}\int_\Gamma\frac{f(\zeta)}{\zeta-z}\,d\zeta\qquad(z\in K).
$$
Uniform 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. Define
$$
E=\{b\in\mathbb C\setminus K:(z-b)^{-1}\in\mathcal A\}.
$$
For $b\in E$ and $|c-b|<\operatorname{dist}(b,K)$, the uniformly convergent <geometric series>
$$
\frac1{z-c}=\sum_{j=0}^\infty\frac{(c-b)^j}{(z-b)^{j+1}}
$$
lies 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 $|b|>\sup_K|z|$,
$$
\frac1{z-b}=-\frac1b\sum_{j=0}^\infty\left(\frac zb\right)^j
$$
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 $|f(w)|\leq\sup_K|f|$. Hence $H\subseteq\widehat K_\Omega$.
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 $1/(2M)$. Then
$$
F(z)=1-(z-w)h(z),\qquad F(w)=1,\qquad\sup_K|F|<\tfrac12.
$$
This <holomorphic function> separates $w$ from the analytic hull. Thus $w\notin\widehat K_\Omega$, 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.
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