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The Weierstrass function is defined as: $$\wp(z) = \frac{1}{z^2}+ {\sum\limits_{\omega\in\Gamma'}} \left (\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\right).$$

Fix $r>0$. For $\left|z\right| \leq r,$ we split the sum in two parts: \begin{align} \wp(z) = \frac{1}{z^2}+{\sum_{\substack{\omega\in\Gamma', \\ \left|\omega \right|\leq 2r}}} \left (\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\right) + {\sum_{\substack{\omega\in\Gamma', \\ \left|\omega \right|>2r}}} \left (\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\right). \end{align} Clearly $z=0$ is a pole of $\wp$. Also, the first part of the summation meromorphic function. Now we will show that the second part of the summation is bounded above and therefore cannot have any pole. Note that $$ \frac{1}{(z-\omega)^2}-\frac{1}{\omega^2} = \frac{z(2\omega - z)}{\omega^2 \left(z-w\right)^2}. $$ Assuming $\left|\omega\right| > 2\left|z\right|$, $$\left| 2\omega - z \right| \leq \left| 2\omega \right| + \left| z \right| < \frac{5}{2}\left| \omega \right|,$$ $$ \left| z-\omega \right| \geq \left| \omega \right| - \left| z \right| > \frac{\left| \omega \right|}{2}.$$ This implies $$ \left| \dfrac{1}{(z-\omega)^2}-\dfrac{1}{\omega^2} \right| \leq \dfrac{\left| z \right| \times \dfrac{5}{2} \left| \omega \right|}{\left| \omega \right|^2 \times \dfrac{\left| \omega \right|^2}{4}} = \dfrac{10\left|z\right|}{\left|\omega\right|^3} \leq \dfrac{10r}{\left|\omega\right|^3}$$

We know that $\sum \frac{1}{\omega^3} < \infty$. Hence $\wp$ is a meromorphic function in $|z|\leq r$.

Now how to show that $\wp$ is meromorphic in $\mathbb{C}$?

I have seen some proofs and I am not able to understand how do they proceed after this step.

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    $\begingroup$ You can pick any $r$, so fix $z_0 \in \mathbb {C} $ and pick $r = |z_0|+1$. Then you have that the weierstrass function is meromorphic in $|z| \leq r$ and $|z_0| < r $ by definition. $\endgroup$ – Federico Sep 27 '16 at 8:18
  • $\begingroup$ So how do we conclude that $\wp$ is meromorphic in $\mathbb{C}$? $\endgroup$ – glip-glop Sep 27 '16 at 8:20
  • $\begingroup$ This is a great example of how to both write a good question, and provide solid details and efforts in the OP's answer. I'm just sorry Complex Function Theory is not my thing! +1 $\endgroup$ – Kevin Sep 27 '16 at 8:23
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    $\begingroup$ Being meromorphic is a local property, so if you show that the function is meromorphic in every ball contained in $\mathbb{C}$ and hence in every point of $\mathbb {C} $ (simply choosing the ball big enough) you have showed the function is meromorphic on $\mathbb {C} $. $\endgroup$ – Federico Sep 27 '16 at 8:24
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    $\begingroup$ Show $\sum\limits_{\omega\in\Gamma'} \frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}$ and its derivative converges around $z=0$, so that $\wp(z) - \frac{1}{z^2}$ is holomorphic (and hence analytic) there, then write the series for $\wp'(z)$ and note it is (obviously) doubly periodic $\endgroup$ – reuns Sep 27 '16 at 9:18

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