Mittag-Leffler and Runge's Theorem

By [L1], choose a compact Jordan set $J$ such that $K\subseteq \mathrm{int}J\subseteq J\subseteq U$, and write $J$ as a finite union of closed cells from one square grid. Give every cell boundary its positive orientation. Each edge internal to $J$ then occurs twice with opposite orientations; cancel those pairs and let $\mathrm{\Gamma }$ be the finite chain of the remaining oriented frontier edges. At every grid vertex the incoming and outgoing coefficients balance, so $\mathrm{\Gamma }$ is a cycle. Its trace is the frontier of $J$, hence ${\mathrm{\Gamma }}^{\ast }\subseteq U\setminus K$.

Let $p\in \mathrm{int}J$ lie on no grid line. Summing the positively oriented boundaries of all cells gives the same integral and index as $\mathrm{\Gamma }$, because the two orientations of every internal edge cancel by [L3]. For one grid cell $Q$, the four-edge continuous argument of $\zeta -p$ makes one positive turn when $p\in \mathrm{int}Q$ and returns with zero net turn when $p\mathrm{\notin }Q$; hence [L2] gives $n({\mathrm{\partial }}^{+}Q,p)=1$ in the first case and $0$ in the second. Exactly one cell containing $p$ contributes $1$, so additivity in [L3] gives $n(\mathrm{\Gamma },p)=1$.

Fix $z\in K$. Choose a disc $D(z,r)\subseteq \mathrm{int}J$ and a point $p\in D(z,r)$ on no grid line. The disc misses ${\mathrm{\Gamma }}^{\ast }$ and is connected, so local constancy in [L4] and step 2.1 give $n(\mathrm{\Gamma },z)=n(\mathrm{\Gamma },p)=1$.

Choose $\mathrm{\Gamma }$ as in [L1]. Because $n(\mathrm{\Gamma },z)=1$ on $K$ and ${\mathrm{\Gamma }}^{\ast }\subseteq U$, the cycle is null-homologous in $U$ and [L2] gives $f(z)=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{f(\zeta )}{\zeta -z}d\zeta (z\in K)\mathrm{.}$

Decompose $\mathrm{\Gamma }$ into finitely many oriented line segments ${\gamma }_j:[a_j,b_j]\to \mathbb{C}$. For each $j$, the function ${\varphi }_j(t,z)=f({\gamma }_j(t)){\gamma }_j^{\mathrm{\prime }}(t)\mathrm{/}({\gamma }_j(t)-z)$ is continuous on the compact set $[a_j,b_j]\times K$, because ${\mathrm{\Gamma }}^{\ast }\cap K=\mathrm{\varnothing }$. By [L3], each ${\varphi }_j$ is uniformly continuous there, so a fine enough Riemann sum approximates ${\int }_{a_j}^{b_j}{\varphi }_j(t,z)dt$ uniformly in $z\in K$.

Summing those edgewise Riemann sums gives a rational function of the form $r(z)={\sum }_{\nu }{c}_{\nu }\mathrm{/}({\xi }_{\nu }-z)$ with sample points ${\xi }_{\nu }\in {\mathrm{\Gamma }}^{\ast }$. Choosing the mesh so that the total edgewise error is below $\epsilon$ and using step 1.1 yields ${\sup }_{z\in K}\mathrm{\mid }f(z)-r(z)\mathrm{\mid }<\epsilon$.

Fix one disc step of the chain, say a closed disc $\bar{D}(c,\rho )$ disjoint from $K$ and two points $a,b\in D(c,\rho )$. [given, L1] Define $S\subseteq D(c,\rho )$ to be the set of points $u$ such that $(z-a{)}^{-1}$ can be approximated uniformly on $K$ by rational functions with only pole $u$. Certainly $a\in S$.

Let $u\in S$. Choose $r>0$ so that $\bar{D}(u,r)\subseteq D(c,\rho )$ and $r<\mathrm{dist}(K,u)\mathrm{/}2$. [step 1.1, choose, algebra] If $v\in D(u,r)$ and $m\ge 1$, then for $z\in K$ one has $\mathrm{\mid }u-v\mathrm{\mid }<\mathrm{\mid }z-v\mathrm{\mid }$, so $\frac{1}{(z-u{)}^m}=\frac{1}{(z-v{)}^m}{(1-\frac{u-v}{z-v})}^{-m}={\sum }_{\nu =0}^{\mathrm{\infty }}\left(\genfrac{}{}{0px}{}{m+\nu -1}{\nu }\right)\frac{(u-v{)}^{\nu }}{(z-v{)}^{m+\nu }},$ with uniform convergence on $K$. Therefore every rational function with only pole $u$ can be approximated uniformly on $K$ by one with only pole $v$. Since $u\in S$, this shows $v\in S$, so $S$ is open. The same expansion with $u$ and $v$ exchanged shows that whenever $v\in S$ is sufficiently close to $u$, then $u\in S$ as well. Hence $S$ is also closed in $D(c,\rho )$. Because the disc is connected and $S$ is nonempty, $S=D(c,\rho )$, so in particular $b\in S$.

If $m=0$, then $a_m=a_0$, and $r(z)=(z-a_0{)}^{-1}$ proves clause 1 with zero error. Assume $m\ge 1$. Apply step 2.1 successively to the discs of the chain, choosing the $j$-th local error below $\epsilon \mathrm{/}m$. [step 2.1, choose, construct, cases, algebra] The triangle inequality then produces a rational function with only pole $a_m$ and total error below $\epsilon$ on $K$. This proves clause 1.

For clause 2, clause 1 gives a rational function $r$ with only pole $a_m$ and ${\sup }_K\mathrm{\mid }r(z)-(z-a_0{)}^{-1}\mathrm{\mid }<\epsilon \mathrm{/}2$. [step 3.1, algebra, discharge-construct] Write the principal part of $r$ at $a_m$ as ${\sum }_{\mathrm{\ell }=1}^L{d}_{\mathrm{\ell }}(z-a_m{)}^{-\mathrm{\ell }}$. Because $\mathrm{\mid }z\mathrm{\mid }<R<\mathrm{\mid }{a}_m\mathrm{\mid }$ on $K$, each factor $(z-a_m{)}^{-\mathrm{\ell }}=(-a_m{)}^{-\mathrm{\ell }}(1-z\mathrm{/}{a}_m{)}^{-\mathrm{\ell }}$ has a power series in $z\mathrm{/}{a}_m$ that converges uniformly on $K$. Truncating those finitely many series gives a polynomial $p$ with ${\sup }_K\mathrm{\mid }p-r\mathrm{\mid }<\epsilon \mathrm{/}2$. Then ${\sup }_K\mathrm{\mid }p(z)-(z-a_0{)}^{-1}\mathrm{\mid }<\epsilon$, proving the polynomial approximation.

By [L1], choose a rational function $r_0(z)={\sum }_{\nu =1}^N{c}_{\nu }\mathrm{/}(z-a_{\nu })$ whose poles $a_{\nu }$ lie in $\widehat{\mathbb{C}}\setminus K$ and such that ${\sup }_{z\in K}\mathrm{\mid }f(z)-r_0(z)\mathrm{\mid }<\epsilon \mathrm{/}2$.

If $N=0$, then $r_0$ already has no finite poles, so its poles are contained in $P$ and step 1.1 already proves the theorem. Assume from now on that $N\ge 1$.

For each pole $a_{\nu }$, let $C_{\nu }$ be the connected component of $\widehat{\mathbb{C}}\setminus K$ containing it. By [L2], choose $p_{\nu }\in P\cap C_{\nu }$. Applying [L3] to the function $c_{\nu }\mathrm{/}(z-a_{\nu })$ inside $C_{\nu }$, choose a rational function $s_{\nu }$ with poles only at $p_{\nu }$ and ${\sup }_{z\in K}\mid s_{\nu }(z)-\frac{c_{\nu }}{z-a_{\nu }}\mid <\frac{\epsilon }{2N}\mathrm{.}$

Put $r={\sum }_{\nu =1}^N{s}_{\nu }$. Then every pole of $r$ lies in $P$, and the triangle inequality together with steps 1.1 and 3.1 gives ${\sup }_{z\in K}\mathrm{\mid }f(z)-r(z)\mathrm{\mid }\le {\sup }_{z\in K}\mathrm{\mid }f(z)-r_0(z)\mathrm{\mid }+{\sum }_{\nu =1}^N{\sup }_{z\in K}\mid s_{\nu }(z)-\frac{c_{\nu }}{z-a_{\nu }}\mid <\epsilon \mathrm{.}$

Since $\widehat{\mathbb{C}}\setminus K$ is connected, the singleton $P=\{\mathrm{\infty }\}$ meets its unique complementary component. Applying [L1] with that pole set gives rational approximants whose only possible pole is at $\mathrm{\infty }$.

If $r=P\mathrm{/}Q$ is such a rational function in lowest terms and $Q$ has positive degree, then a zero of $Q$ would give a finite pole of $r$. Therefore $Q$ is constant, so $r$ is a polynomial. Hence the approximants from step 1.1 are polynomials.

Choose an increasing exhaustion $K_1\subseteq K_2\subseteq \cdots$ of compact subsets of $\mathrm{\Omega }$ with ${\bigcup }_n{K}_n=\mathrm{\Omega }$ and $K_n\subseteq \mathrm{int}{K}_{n+1}$. By recursively applying [L1] and filling every complementary component of the chosen Jordan neighbourhood that lies entirely in $\mathrm{\Omega }$, we may also require that every connected component of $\widehat{\mathbb{C}}\setminus K_n$ meets $P$.

Apply [L2] to each $K_n$ with tolerance $2^{-n}$. This gives a rational function $r_n$ with poles in $P$ and ${\sup }_{z\in K_n}\mathrm{\mid }f(z)-r_n(z)\mathrm{\mid }<2^{-n}\mathrm{.}$

Fix a compact set $K\subseteq \mathrm{\Omega }$. Choose $N$ with $K\subseteq K_N$. Then for every $n\ge N$ one has $K\subseteq K_n$, so step 2.1 gives ${\sup }_{z\in K}\mathrm{\mid }f(z)-r_n(z)\mathrm{\mid }\le 2^{-n}$. Hence $r_n\to f$ uniformly on $K$. Since $K$ was arbitrary, [L3] gives local-uniform convergence on $\mathrm{\Omega }$.

Choose radii $R_n\uparrow \mathrm{\infty }$ so that $A\cap \mathrm{\partial }D(0,R_n)=\mathrm{\varnothing }$ and $A\cap \bar{D}(0,R_n)$ is finite for every $n$. [given, L1, choose] Put $S_1:=A\cap \bar{D}(0,R_1),S_n:=A\cap (\bar{D}(0,R_n)\setminus \bar{D}(0,R_{n-1}))(n\ge 2)\mathrm{.}$ Then each $S_n$ is finite and $A={⨆}_{n\ge 1}{S}_n$. For $n\ge 2$, the finite sum $f_n(z):={\sum }_{a_k\in S_n}{p}_k(z)$ is holomorphic on a neighbourhood of $\bar{D}(0,R_{n-1})$, because every pole it carries lies outside that disc. Set likewise $f_1(z):={\sum }_{a_k\in S_1}{p}_k(z)$.

By [L2], for each $n\ge 2$ choose a polynomial $q_n$ such that [L2, step 1.1, construct] ${\sup }_{\mathrm{\mid }z\mathrm{\mid }\le R_{n-1}}\mathrm{\mid }{f}_n(z)-q_n(z)\mathrm{\mid }<2^{-n}\mathrm{.}$ Put $g_1:=f_1$ and $g_n:=f_n-q_n$ for $n\ge 2$. Then every $g_n$ is meromorphic on $\mathbb{C}$, has the same principal parts as the finitely many $p_k$ with $a_k\in S_n$, and is holomorphic on $\bar{D}(0,R_{n-1})$.

Fix a compact set $K\subseteq \mathbb{C}\setminus A$. Choose $\mathrm{\ell }$ with [step 2.1, choose, algebra, discharge-construct] $K\subseteq \bar{D}(0,R_{\mathrm{\ell }})$. For every $n\ge \mathrm{\ell }+1$, one has $K\subseteq \bar{D}(0,R_{n-1})$, so step 2.1 gives ${\sup }_K\mathrm{\mid }{g}_n\mathrm{\mid }\le 2^{-n}$. Hence ${\sum }_{n\ge 1}{g}_n$ converges uniformly on $K$. Near any point $a_k\in S_{\mathrm{\ell }}$, all terms except $g_{\mathrm{\ell }}$ are holomorphic, so the sum $f:={\sum }_{n\ge 1}{g}_n$ is meromorphic on $\mathbb{C}$ and has principal part $p_k$ at $a_k$.

Choose an increasing compact exhaustion $C_1\subseteq C_2\subseteq \cdots$ of $\mathrm{\Omega }$ with $C_n\subseteq C_{n+1}^{\circ }$ and ${\bigcup }_n{C}_n=\mathrm{\Omega }$. [given, L1, L3, construct] Recursively apply [L1] to choose compact Jordan neighbourhoods of the $C_n$ and then fill every complementary component lying entirely in $\mathrm{\Omega }$. This gives an increasing exhaustion $K_1\subseteq K_2\subseteq \cdots$ of compact subsets of $\mathrm{\Omega }$ such that every connected component of $\widehat{\mathbb{C}}\setminus K_n$ meets $\widehat{\mathbb{C}}\setminus \mathrm{\Omega }$. Because $A$ is discrete, each set $A_n:=A\cap (K_n\setminus K_{n-1})(K_0:=\mathrm{\varnothing })$ is finite. Let $f_n(z):={\sum }_{a\in A_n}{p}_a(z)\mathrm{.}$ Then $f_n$ is meromorphic on $\mathrm{\Omega }$ and holomorphic on a neighbourhood of $K_{n-1}$.

Fix a set $P\subseteq \widehat{\mathbb{C}}\setminus \mathrm{\Omega }$ meeting every connected component of $\widehat{\mathbb{C}}\setminus \mathrm{\Omega }$. [L2, step 1.1, choose] Step 1.1 makes $P$ meet every connected component of $\widehat{\mathbb{C}}\setminus K_{n-1}$ as well. Put $r_1:=0$ and $h_1:=f_1$. For each $n\ge 2$, apply [L2] to $f_n$ on a neighbourhood of $K_{n-1}$ and choose a rational function $r_n$ with poles in $P$ such that ${\sup }_{K_{n-1}}\mathrm{\mid }{f}_n-r_n\mathrm{\mid }<2^{-n}\mathrm{.}$ Then $h_n:=f_n-r_n$ is meromorphic on $\mathrm{\Omega }$, has the same principal parts as $f_n$ on $A_n$, and is uniformly small on $K_{n-1}$.

For a compact set $K\subseteq \mathrm{\Omega }\setminus A$, choose $N$ with [step 2.1, algebra, discharge-construct] $K\subseteq K_{N-1}$. Then for every $n\ge N$, step 2.1 gives ${\sup }_K\mathrm{\mid }{h}_n\mathrm{\mid }\le 2^{-n}$. Therefore ${\sum }_n{h}_n$ converges uniformly on $K$. Only finitely many layers $A_n$ meet a given compact set, so $f:={\sum }_{n\ge 1}{h}_n$ is meromorphic on $\mathrm{\Omega }$, and its principal part at each $a\in A_n$ is exactly $p_a$.

Fix $z\in \mathbb{C}\setminus \mathbb{Z}$. For $N$ large enough that $z,-z\in R_N^{\circ }$, let $R_N$ be the positively oriented rectangle with [L2, L3, given, algebra] vertices $\pm (N+1\mathrm{/}2)\pm i(N+1\mathrm{/}2)$ and set $F_N(w):=\frac{\pi \cot (\pi w)}{w^2-z^2}\mathrm{.}$ By [L2], the poles of $F_N$ inside $R_N$ are the integers $n$ with $\mathrm{\mid }n\mathrm{\mid }\le N$ and the points $w=\pm z$. The residue at an integer $n$ is $(n^2-z^2{)}^{-1}$, while the residues at $w=z$ and $w=-z$ sum to $\frac{\pi \cot (\pi z)}{2z}+\frac{\pi \cot (-\pi z)}{-2z}=\frac{\pi \cot (\pi z)}{z}\mathrm{.}$ Therefore [L3] gives ${\int }_{\mathrm{\partial }{R}_N}{F}_N(w)dw=2\pi i(\frac{\pi \cot (\pi z)}{z}+{\sum }_{n=-N}^N\frac{1}{n^2-z^2})\mathrm{.}$

On the vertical sides of $\mathrm{\partial }{R}_N$, write $w=\pm (N+1\mathrm{/}2)+iy$. [L1, step 1.1, algebra] $\sin (\pi w)=\pm \cosh (\pi y)$ and $\cos (\pi w)=\mp i\sinh (\pi y)$ by [L1], so $\mathrm{\mid }\cot (\pi w)\mathrm{\mid }=\mathrm{\mid }\tanh (\pi y)\mathrm{\mid }\le 1$. On the horizontal sides, $w=x\pm i(N+1\mathrm{/}2)$, and [L1] gives $\mathrm{\mid }\sin (\pi w){\mathrm{\mid }}^2={\sin }^2(\pi x)+{\sinh }^2(\pi (N+1\mathrm{/}2)),\mathrm{\mid }\cos (\pi w){\mathrm{\mid }}^2={\cos }^2(\pi x)+{\sinh }^2(\pi (N+1\mathrm{/}2)),$ so $\mathrm{\mid }\cot (\pi w)\mathrm{\mid }\le 2$. Also $\mathrm{\mid }w\mathrm{\mid }\ge N+1\mathrm{/}2$ on $\mathrm{\partial }{R}_N$, hence $\mathrm{\mid }{w}^2-z^2\mathrm{\mid }\ge \mathrm{\mid }w{\mathrm{\mid }}^2-\mathrm{\mid }z{\mathrm{\mid }}^2\ge (N+1\mathrm{/}2{)}^2-\mathrm{\mid }z{\mathrm{\mid }}^2$. Thus $\mathrm{\mid }{F}_N(w)\mathrm{\mid }\le C_z\mathrm{/}{N}^2$ on $\mathrm{\partial }{R}_N$, and since the boundary length is $8N+4$, one gets ${\int }_{\mathrm{\partial }{R}_N}{F}_N(w)dw\to 0$ as $N\to \mathrm{\infty }$.

Letting $N\to \mathrm{\infty }$ in step 1.1 and using step 2.1 yields [step 1.1, step 2.1, algebra] $\frac{\pi \cot (\pi z)}{z}=-{\sum }_{n\in \mathbb{Z}}\frac{1}{n^2-z^2}=\frac{1}{z^2}+{\sum }_{n\ge 1}\frac{2}{z^2-n^2}\mathrm{.}$ Multiplying by $z$ gives $\pi \cot (\pi z)=\frac{1}{z}+{\sum }_{n\ge 1}\frac{2z}{z^2-n^2}=\frac{1}{z}+{\sum }_{n\ge 1}(\frac{1}{z-n}+\frac{1}{z+n})\mathrm{.}$ On every compact subset of $\mathbb{C}\setminus \mathbb{Z}$, the last series is bounded termwise by $C_K\mathrm{/}{n}^2$ for all large $n$, so it converges locally uniformly there. This is exactly the claimed expansion.

Let $S_N(z):=\frac{1}{z}+{\sum }_{n=1}^N(\frac{1}{z-n}+\frac{1}{z+n})\mathrm{.}$ By [L1], the holomorphic functions $S_N$ converge locally uniformly on $\mathbb{C}\setminus \mathbb{Z}$ to $\pi \cot (\pi z)$. On any compact set $K\subseteq \mathbb{C}\setminus \mathbb{Z}$, the derivatives satisfy $S_N^{\mathrm{\prime }}(z)=-\frac{1}{z^2}-{\sum }_{n=1}^N(\frac{1}{(z-n{)}^2}+\frac{1}{(z+n{)}^2}),$ and this derivative series converges uniformly on $K$ because its terms are $O(n^{-2})$ there. Therefore [L3] identifies the derivative of the limit with the displayed series on $K$.

Differentiating [L1] gives $-{\pi }^2{\csc }^2(\pi z)=-\frac{1}{z^2}-{\sum }_{n\ge 1}(\frac{1}{(z-n{)}^2}+\frac{1}{(z+n{)}^2})$ by [L2]. Multiplying by $-1$ yields the claimed formula.

Every point of $\mathrm{\Omega }$ has a closed disc neighbourhood contained in $\mathrm{\Omega }$, so local finiteness makes each point of $S$ isolated and prevents any accumulation point of $S$ in $\mathrm{\Omega }$. By [L6], enumerate all open discs with rational centre coordinates and positive rational radius in a fixed order. For each $a\in S$, let $k(a)$ be the least index of such a disc containing $a$ and meeting $S$ only at $a$; isolation and density of the rationals ensure its existence. Distinct points have distinct indices. Thus $S$ injects into the nonnegative integers. By [L6], encode the pairs $(k(a),r)$ with $1\le r\le m(a)$ by a fixed pairing of integers, and list the resulting codes in increasing order. This gives the points of $S$ repeated exactly $m(a)$ times, without a choice principle. A compact subset contains only finitely many entries of this list. If the list is finite, $F(z)={\prod }_{a\in S}(z-a{)}^{m(a)}$ works, with empty product $1$. Henceforth assume it is infinite.

If $C:=\mathbb{C}\setminus \mathrm{\Omega }$ is empty, put all entries into a list $(c_j)$ and leave the list $(b_j)$ empty. If $C$ is nonempty, set $\delta (z):=\mathrm{dist}(z,C)$ and split the repeated list, preserving its order, according to $D:=\{z\in \mathrm{\Omega }:\delta (z)<1\mathrm{/}(\mathrm{\mid }z\mathrm{\mid }+1)\}$: $(b_j)$ contains the entries in $D$, and $(c_j)$ those outside $D$. Either sublist may be empty, finite, or infinite. The closedness of $C$ and openness of $\mathrm{\Omega }$ imply $\delta (z)>0$ on $\mathrm{\Omega }$, and [L4] makes $\delta$ continuous.

Suppose $(b_j)$ is infinite. For each $\epsilon >0$, an entry $b_j$ with $\delta (b_j)\ge \epsilon$ satisfies $\mathrm{\mid }{b}_j\mathrm{\mid }<1\mathrm{/}\epsilon$. Such entries lie in the closed bounded set $\{z:\mathrm{\mid }z\mathrm{\mid }\le 1\mathrm{/}\epsilon ,\delta (z)\ge \epsilon \}$, contained in $\mathrm{\Omega }$ and compact by [L4]. Step 1.1 makes their number finite. Thus $\delta (b_j)\to 0$. To specify a nearest boundary point without simultaneous choices, consider the compact nonempty set $C\cap \bar{B}(b_j,\delta (b_j)+1)$. The minimum of $\mathrm{\mid }{b}_j-p\mathrm{\mid }$ on it is $\delta (b_j)$: points outside it have distance greater than $\delta (b_j)+1$, and the definition of the infimum supplies points inside it with distance arbitrarily close to $\delta (b_j)$. The nearest-point set is therefore nonempty compact. First minimize $\mathrm{Re}p$ on that set, then minimize $\mathrm{Im}p$ on the minimizers. This specifies a unique $p_j\in C$ with $\mathrm{\mid }{b}_j-p_j\mathrm{\mid }=\delta (b_j)$.

Suppose $(c_j)$ is infinite. If $C=\mathrm{\varnothing }$, every closed disc is compact in $\mathrm{\Omega }=\mathbb{C}$, so step 1.1 shows that only finitely many entries have modulus at most any fixed $R$. If $C\mathrm{\ne }\mathrm{\varnothing }$, entries with $\mathrm{\mid }{c}_j\mathrm{\mid }\le R$ satisfy $\delta (c_j)\ge 1\mathrm{/}(R+1)$; they lie in the compact subset $\{z:\mathrm{\mid }z\mathrm{\mid }\le R,\delta (z)\ge 1\mathrm{/}(R+1)\}$ of $\mathrm{\Omega }$, which again contains only finitely many entries. In either case $\mathrm{\mid }{c}_j\mathrm{\mid }\to \mathrm{\infty }$. Let $m_0$ be the finite number of zero entries and enumerate the nonzero entries as $(d_j)$. Then $\mathrm{\mid }{d}_j\mathrm{\mid }\to \mathrm{\infty }$.