# Show the two versions of Mittag-Leffler Theorem are equivalent (Hormander 1.4.3')

The proof of equivalence is left as an exercise by Hormander (Page 12, An Introduction to Complex Analysis in Several Variables). I had trouble showing First implies the Second proposition (how to set up the open sets).

First:

Theorem 1.4.3. (Mittag-Leffler): Let $$z_j, j =1,2,...$$, be a discrete sequence of different points in the open set $$\Omega$$ and let $$f_j$$ be meromorphic in a neighbourhood of $$z_j$$. Then there exists a meromorphic function $$f$$ in $$\Omega$$ such that f is analytic outside the points $$z_j$$ and $$f-f_j$$ is analytic in a neighbourhood of $$z_j$$ for every $$j$$.

Second:

Theorem 1.4.3'. Let $$\Omega = \bigcup_{j}\Omega_j$$ where $$\Omega_j \subset \mathbb{C}$$ open, $$f_j:\Omega_j \to \mathbb{C}$$ meromorphic and $$f_j-f_k$$ holomorphic in $$\Omega_j \cap\Omega_k$$ for all $$j$$ and $$k$$, one can find $$f$$ meromorphic in $$\Omega$$ such that $$f-f_j$$ is holomorphic in $$\Omega_j \space\forall j$$.

• Ah, no wonder I was so confused. I tracked down the book and found you incorrectly wrote your question... Your latter theorem should say "one can find $f$ meromorphic in $\Omega$ such that...". I was wondering how on Earth the result you were claiming could be equivalent. I have edited the question to fix this mistake... if for some reason you intended the former version, feel free to revert. – Brevan Ellefsen Apr 12 '19 at 5:18
• Thanks you're right, should have double checked that! – Christopher Abe Apr 12 '19 at 6:02
• I found some time tonight to think up a proof. Let me know if you have any questions. – Brevan Ellefsen Apr 13 '19 at 7:23

Consider the following condition in the equivalent statement of Mittag-Leffler:

$$f_j-f_k \text{ is holomorphic in }\Omega_j \cap\Omega_k \text{ for all } j \text{ and } k. \tag{1}$$

Note this statement essentially says that wherever our domains overlap, our functions must have poles at the same locations with the same orders (else the subtraction could not be holomorphic), or in more formal terminology, $$f_j$$ and $$f_k$$ have the same singular behavior on $$\Omega_j \cap \Omega_k.$$

Importantly, this implies the set of all poles coming from all $$f_i$$ must be a discrete set; no pole $$p_j$$ can be an accumulation point, since if this were so then there would be some sequence of poles $$\{p_i\} \to p_j$$ but then there must exist $$k$$ such that $$p_i \in \Omega_j$$ for all $$i \ge k.$$ Since our meromorphic functions must have the same singular behavior on overlaps, we conclude $$\{p_i\}_{i \ge k}$$ are poles of $$f_j,$$ violating $$f_j$$ being meromorphic (since its poles have an accumulation point).

With this out of the way, we are essentially done. Let us recall the classical statement of the Mittag-Leffler Theorem:

Theorem $$\mathbf{1.4.3}$$. (Mittag-Leffler): Let $$z_j, j =1,2,...$$, be a discrete sequence of different points in the open set $$\Omega$$ and let $$f_j$$ be meromorphic in a neighbourhood of $$z_j$$. There exists a meromorphic function $$f$$ in $$\Omega$$ such that $$f$$ is analytic outside the points $$z_j$$ and $$f-f_j$$ is analytic in a neighbourhood of $$z_j$$ for every $$j$$.

We now apply this classical version to all the $$f_i$$ in our equivalent statement. What we do is essentially break up each $$f_i$$ into meromorphic functions with a single pole each to apply Mittag-Leffler, and then recombine the smaller functions into the $$f_i.$$ We then show $$f - f_i$$ is holomorphic on all of $$\Omega_i,$$ and not just in a neighborhood of each pole.

More formally, let $$P_i$$ be the set of poles of $$f_i$$ and for each pole $$p \in P_i,$$ define $$g^i_p$$ to be the restriction of $$f_i$$ to $$\Omega_i \setminus (P_i \setminus p),$$ i.e. $$g^i_p$$ is the restriction of $$f_i$$ having a pole only at $$p.$$ We apply Mittag-Leffler to the set of all $$g^i_p$$ for all $$i$$ and for all $$p \in P_i$$ to find an $$f$$ on $$\Omega$$ such that $$f$$ is holomorphic away from each pole $$p$$ and $$f - g^i_p$$ is holomorphic in a neighborhood of $$p.$$ Now, in such a neighborhood we simply have $$g^i_p = f_i,$$ so we have in fact found $$f$$ on $$\Omega$$ such that $$f$$ is holomorphic away from each pole $$p$$ and $$f - f_i$$ is holomorphic in a neighborhood of pole of $$f_i.$$ Since both $$f$$ and $$f_i$$ are holomorphic on $$\Omega_i \setminus P_i,$$ we conclude $$f - f_i$$ is in fact holomorphic on all of $$\Omega_i.$$ This completes the proof of the equivalent statement. $$\quad\square$$

• Thank you! Really great explanation. – Christopher Abe Apr 14 '19 at 5:06