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For compact $K$ in a plane domain $\Omega$, its holomorphic convex hull isIt is $K$ together with the components of $\hat{\mathbb{C}}\setminus K$ lying entirely in $\Omega$. Inclusion of those components follows from the maximum modulus principle. For exclusion, suppose $w$ lies in a complementary component meeting $\hat{\mathbb{C}}\setminus \Omega$. Choose an allowed pole there. The open-and-closed resolvent argument in the proof of Runge theorem approximates $(z-w{)}^{-1}$ uniformly on $K$ by functions $h$ holomorphic on $\Omega$. Then $1-(z-w)h(z)$ has value one at $w$ and arbitrarily small modulus on $K$, separating $w$ from the hull. For $\Omega =\mathbb{C}$, this is the polynomial hull in one complex variable.

The zero-prescription claim requires a closed discrete subset, meaning no accumulation in $\Omega$. If “discrete” means only that each listed point is isolated, it is false: $\{1/n:n\ge 1\}\subset \mathbb{C}$ is a discrete subset, but any holomorphic function vanishing there is identically zero by the identity theorem for holomorphic functions, and then has additional zeros. In the usual zero-set interpretation, discreteness includes local finiteness. We prove the claim with this necessary interpretation.

First choose an increasing compact exhaustion $K_j$ of $\Omega$ by relatively filled sets: every component of $\hat{\mathbb{C}}\setminus K_j$ meets $\hat{\mathbb{C}}\setminus \Omega$. Such an exhaustion is obtained by exhausting with finite unions of small closed squares lying in $\Omega$, then adding all complementary components relatively compact in $\Omega$. The filled sets remain compact in $\Omega$: a complementary component containing a point within distance $\mathrm{dist}(K_j,\mathbb{C}\setminus \Omega )$ of the complement could join that complement through a ball disjoint from $K_j$, and hence would not be filled. Their boundedness follows by using a disk containing the initial compact. Nesting can be ensured by enlarging each stage first. This is also the geometric description of the holomorphic convex hull proved below.

A finite prescribed set is realized by a finite product of linear factors. For an infinite closed discrete subset, enumerate the distinct points as $a_n$. Every compact contains only finitely many of them, so there are indices $j(n)\to \infty$ with $a_n\notin K_{j(n)}$. Finitely many initial factors can be handled without an approximation requirement. For the remaining factors, let $U_n$ be the complementary component containing $a_n$. It also contains a point $c_n\notin \Omega$ or infinity. Join $a_n$ to $c_n$ by an arc in that component, or to infinity by a ray-like arc. DefineEach $p_n$ is a holomorphic function on $\Omega$, has exactly one simple zero, at $a_n$, and no other zeros there.

The arc is disjoint from $K_{j(n)}$, and $p_n$ has a holomorphic logarithm $L_n$ near that compact. To justify the logarithm, any closed curve avoiding the arc has equal winding numbers about its finite endpoints, so the integral of $p_n^{\prime }/p_n$ around it is zero. For a ray to infinity the winding number about $a_n$ is zero. A primitive of $p_n^{\prime }/p_n$, adjusted by a constant on each component, supplies the logarithm.

The Runge theorem proved in Question 4 approximates $L_n$ uniformly on $K_{j(n)}$ by $h_n\in \mathcal{O}(\Omega )$: all needed finite poles can be chosen outside $\Omega$ because every complementary component meets its complement. Choose the error small enough thatEach $F_n$ still has exactly the simple zero $a_n$. On every fixed compact, all sufficiently late factors satisfy the summable bound above. Hence their infinite product converges locally uniformly. It has a nonzero tail because, once $|F_n-1|<1/2$, its logarithmic tails satisfy $|\log F_n|\le 2|F_n-1|$ and converge absolutely. Thus multiplying by the finitely many earlier factors gives a holomorphic function with precisely the prescribed zeros, and no additional zeros. Repeating a factor to any prescribed finite positive multiplicity gives the corresponding multiplicity version. This is prescribed zeros in a plane domain via Runge approximation, not an invocation of the existence theorem being asked for.

For the specified infinite product, put $a_k=q^{-2k-1}$. On any compact annulus $0<r\le |z|\le R$,The logarithmic-tail estimate therefore proves locally uniform convergence on ${\mathbb{C}}^{\ast }$ and nonvanishing wherever no individual factor vanishes. At a zero of one factor, the product is zero and the remaining product converges to a nonzero value. ConsequentlyAll these zeros are simple and distinct: their moduli distinguish the two strings and their indices. At $z=0$ the reciprocal factors are undefined, so the given expression has no convergence value there. The infinite sequence of zeros tending to zero also rules out continuation across that point as a holomorphic function.

Set $P(z)={\prod }_{k\ge 1}(1-q^{-2k}z)(1-q^{-2k}{z}^{-1})$. Absolute local convergence allows reindexing:Since $1-z=-z(1-z^{-1})$, $\sigma (qz)=-z\sigma (q^{-1}z)$ for every $z\ne 0$. No division by a vanishing factor is involved, so the identity also holds at all the zeros.

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.

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 $K$.