kuniga.me > NP-Incompleteness > Mittag-Leffler's Theorem
Mittag-Leffler's Theorem
17 Jun 2025
Magnus Gustaf “Gösta” Mittag-Leffler was a Swedish mathematician. After earning a PhD at Uppsala University, Mittag-Leffler attended lectures on elliptic functions from Charles Hermite in Paris and Karl Weierstrass in Berlin, which influenced his works.
Mittag-Leffler was an advocate of women’s rights. He helped Sofia Kovalevskaya become full professor of mathematics in Stockholm, the first woman to do so in Europe. As a member of the Nobel Prize Committee, Mittag-Leffler was responsible for convincing the committee to include Marie Curie in the Nobel prize in physics, instead of just Pierre Curie.
In this post we’d like to study the Mittag-Leffler’s Theorem.
The theorem says that if we have a given set of numbers, there exists a function that is holomorphic everywhere except at the poles corresponding to these numbers. More precisely:
Theorem 1. Let $\Omega$ be an open set in $\mathbb{C}$ and $E$ a subset of $\Omega$ without limit points. Let $Q_a$ be a polynomial of the form:
\[Q_a(z) = \sum_{n = 1}^{N_a} \frac{c_{a, n}}{(z - a)^n}\]Then there exists a meromorphic function $f(z)$ in $U$ defined as:
\[f(z) = \left(\sum_{k = 1}^{\infty} Q_{p_k}(z) - q_k\right) + g(z)\]for $p_k \in E$, polynomials $q_k$ and a holomorphic function $g(z)$. The poles of this function corresponds to $E$. Additionally, $Q_{p_k}(z)$ is the singular part of the Laurent series expansion of $f(z)$ around $p_k$.
Now, if we assume $E$ is infinite, we can't use the same argument as before because $(1.1)$ is an infinite series which might not converge, so we need more care.
To make the discussion easier, we label its elements $p_1, p_2, \cdots$ such that $\abs{p_1} \le \abs{p_1} \le \cdots$. Let's consider a given $p_k$. For $\abs{z} \lt \abs{p_k}$, we have that $Q_{k}$ is holomorphic, because $z \ne p_k$ and hence it has no poles. Thus it has a convergent Taylor series. Let $q_{k}$ be the first $d_k$ terms of the Taylor series expansion around 0.
Define $M_k$ as the upper bound of $Q_{p_k}$ for $\abs{z} \lt \abs{p_k} / 2$, that is: $$M_{k} = \sup_{\abs{z} \lt \abs{p_k} / 2} \abs{Q_{p_k}(z)}$$ According to Lemma 2, If we subtract $Q_{p_k}(z)$ by $q_k$, we get the following upperbound for $\abs{z} \lt \abs{p_k} / 4$: $$\abs{Q_{p_k}(z) - q_k} \le M_k 2^{-d_k}$$ Since we have freedom to choose $d_k$, we can choose one such that $M_k 2^k \lt 2^{d_k}$ since $M_k$ and $k$ are independent from our choice of $d_k$. Thus we have: $$\abs{Q_{p_k}(z) - q_k} \lt M_k 2^{-k}$$ Now define $$(1.2) \quad f(z) = \sum_{k = 1}^{\infty} Q_{p_k}(z) - q_k$$ This is the meromorphic function we're looking for, but we need to prove it exists by showing that the series on the right hand side converges.
Let $K$ be any compact set in $\Omega$ that doesn't contain any points of $E$. We can find $N$ such that the disk $\abs{z} \lt \abs{p_N} / 4$ contains $K$. Here we need to use the hypothesis that $E$ doesn't have a limit point, because we need $\lim_{k \rightarrow \infty} p_k = \infty$ to be able to find such $N$.
For $z \in K$ we have that $$\abs{f_n(z)} \le \sum_{k = n}^{\infty} \abs{Q_{p_k}(z) - q_k} \lt \sum_{k = n}^{\infty} 2^{-k}$$ converges for all $n \ge N$, which passes Weierstrass M-test [2]. Since $N$ is finite, the series $$\sum_{k = 1}^{N} Q_{p_k}(z) - q_k$$ also converges. We conclude that $f(z)$ exists for any compact set in $\Omega$ that doesn't contain any points of $E$. For $p_k \in E$, we don't need to show that the series converges to $f(z)$. In fact $p_k$ is a pole for $Q_{p_k}(z)$, so the value of the series is infinity anyways.
For any $p_k$, we have that the series form a Laurent series around $p_k$. Since $q_k$ are parts of a Taylor series, the only pole is from $Q_{p_k}(z)$ and thus it forms the singular / principal part (i.e. terms with negative powers) of the Laurent series around $p_k$.
We can thus conclude that $f(z)$ is a meromorphic function in $\Omega$ with poles in $E$. If we add any holomorphic function $g(z)$ to $f(z)$, it preserves these properties.
Note that we cannot combine the sum of $q_k$ (which are polynomials and holomorphic on their own) with $g(z)$ because the series of $q_k$ might not converge.
Intuitively the terms $q_k$ are used to keep the growth of $Q_{p_k}$ under control and make sure it converges. In the special case in which
\[\sum_{k = 1}^{\infty} Q_{p_k}(z)\]is convergent in $\Omega \setminus E$, we can assume $f(z)$ takes the simpler form:
\[f(z) = \left(\sum_{k = 1}^{\infty} Q_{p_k}(z) \right) + g(z)\]Example
Suppose we want to find the series expansion of $f(z) = \pi^2 / \sin^2(\pi z)$. We can first find the poles of this expression and use the Mittag-Leffler theorem to find a series that converges to it given those poles.
The poles for this function are the points for which $\sin(\pi z) = 0$, which is any integer $n \in \mathbb{Z}$. So the set of poles is $\mathbb{Z}$, which does not have a limit point, a condition for us to use the Mittag-Leffler theorem.
From Lemma 3, the singular part of the Laurent expansion of $\pi^2 / \sin^2(\pi z)$ around a pole $n \in \mathbb{Z}$ is $1 / (z - n)^2$, so we have $Q_n(z) = 1 / (z - n)^2$. And we have that:
\[\sum_{n \in \mathbb{Z}} Q_n(z) = \sum_{n \in \mathbb{Z}} \frac{1}{(z - n)^2}\]is uniformily convergent, which allows us to use the special case of the Mittag-Leffler’s Theorem to conclude that:
\[\frac{\pi^2}{\sin^2(\pi z)} = \left(\sum_{n \in \mathbb{Z}} \frac{1}{(z - n)^2}\right) + g(z)\]We now wish to show $g(z) = 0$. It’s possible to show that if $z = x + iy$, then for $\abs{y} \rightarrow \infty$, both $\pi^2 / \sin^2(\pi z)$ and $\sum_{n \in \mathbb{Z}} 1 / (z - n)^2$ tend uniformily to 0 and thus if $x$ is bounded, say $0 \le x \le 1$, then $\abs{g(z)}$ is also bounded.
Since both $\pi^2 / \sin^2(\pi z)$ and $\sum_{n \in \mathbb{Z}} 1 / (z - n)^2$ are periodic with period 1, $\abs{g(z)}$ is bounded for any “strip” $n \le x \le n + 1$, and thus bounded in the whole plane. By Liouville’s Theorem [1], this means $g(z)$ is a constant. And since we know $g(z)$ assumes the value $0$ for $\abs{y} \rightarrow \infty$, that constant must be 0.
We can thus conclude that:
\[\frac{\pi^2}{\sin^2(\pi z)} = \sum_{n \in \mathbb{Z}} \frac{1}{(z - n)^2}\]Appendix
Lemma 2. Let $Q_{p_k}$, $q_k$, $d_k$ and $M_k$ be as defined in the proof of Theorem 1. Then if $\abs{z} \lt \abs{p_k} / 4$:
\[\abs{Q_{p_k}(z) - q_k} \le M_k 2^{-d_k}\]We already established that if $\abs{z} \lt \abs{p_k}$ then $Q_{p_k}(z)$ is holomorphic, so it has a Taylor series expansion and that $q_k$ is the first $d_k$ terms of it, so if we subtract, we're left with the remaining terms: $$g_{p_k}(z) = \sum_{j = d_k + 1}^\infty c_j z^{j}$$ By Cauchy’s Estimate [1], we have that since $g_{p_k} \lt M_k$ in its domain and thus: $$\abs{c_j} \le \frac{M_k}{r^j}$$ Where $r$ is the radius of a circle around the origin on which $g_{p_k}$ is holomorphic, so we can pick $r = \abs{p_k} / 2$, yielding: $$\abs{c_j} \le \frac{2^j M_k}{p_k^j}$$ Since $\abs{z} \lt \abs{p_k} / 4$ we have: $$\abs{c_j z^j} \le \frac{2^j M_k \abs{p_k}^j}{\abs{p_k}^j 4^j} = M_k 2^{-j}$$ We can sum the terms from $d_k + 1$ to infinity, which forms a geometric series: $$S = 2^{-(d_k + 1)} + 2^{-(d_k + 2)} + 2^{-(d_k + 3)} + \cdots$$ and $$2S = 2^{-d_k} + 2^{-(d_k + 1)} + 2^{-(d_k + 2)} + 2^{-(d_k + 3)} + \cdots$$ So $S = 2^{-d_k}$ and $$\abs{g_{p_k}(z)} \le \sum_{j = d_k + 1}^\infty \abs{c_j z^{j}} \le M^k \sum_{j = d_k + 1}^\infty 2^{-j} = M^k 2^{-d_k}$$
Lemma 3. The singular part of the Laurent expansion of $\pi^2 / \sin^2(\pi z)$ around $n \in \mathbb{Z}$ is $1 / (z - n)^2$ for $n \in \mathbb{Z}$.
