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Let $$\wp(z)=\frac{1}{z^2}+\sum_{w \in \Lambda^*} \left[\frac{1}{(z+w)^2}-\frac{1}{w^2}\right] $$

be the Weierstrass elliptic function with $\Lambda=\Bbb{Z}+\Bbb{Z}\tau$, $\Lambda^*=\Lambda-0$. I want to show that $\wp(z+w)=\wp(z)$ whenever $w \in \Lambda$, without using differentiation. [Stein, Complex Analysis p.279]

Suggested hint: For large $R$, $\wp(z)=\wp^R(z)+O(1/R)$ where $$\wp^R(z)=\frac{1}{z^2}+\sum_{0<|w|<R} \left[\frac{1}{(z+w)^2}-\frac{1}{w^2}\right]$$ Also observe that $\wp^R(z+1)-\wp^R(z)$ and $\wp^R(z+\tau)-\wp^R(z)$ are $O(\sum_{R-c<|w|<R+c} |w|^{-2})=O(1/R)$

But I can't understand the hint. I know that for $|z|<R$, $\sum_{|w|>2R} \left[\frac{1}{(z+w)^2}-\frac{1}{w^2}\right]$ is $O(1/|w|^3)$ (uniformly) so defines a holomorphic function in $|z|<R$. But is the above hint also correct? $O(1/R)$ is ambiguous and restriction of $z$ seems to needed. So if I understood $\wp^R(z)$ as $$\wp^R(z)=\frac{1}{z^2}+\sum_{0<|w|<2R} \left[\frac{1}{(z+w)^2}-\frac{1}{w^2}\right] \quad (|z|<R)$$ But still don't know how to compare $\wp^R(z+1)-\wp^R(z)$. What should I do?

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3 Answers 3

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Let us for a moment pretend we could split the series defining $\wp$ in the following way:

$$\begin{align} \wp(z) &= \frac{1}{z^2} + \sum_{\omega \in \Lambda^\ast} \left(\frac{1}{(z+\omega)^2} - \frac{1}{\omega^2}\right)\\ &= \frac{1}{z^2} + \sum_{\omega\in \Lambda^\ast} \frac{1}{(z+\omega)^2} - \sum_{\omega\in\Lambda^\ast} \frac{1}{\omega^2}\\ &= \sum_{\omega\in \Lambda} \frac{1}{(z+\omega)^2} - \sum_{\omega\in\Lambda^\ast} \frac{1}{\omega^2}. \end{align}$$

Then it would be easy to see the periodicity of $\wp$ with respect to $\Lambda$, as adding some $\omega_0 \in \Lambda$ to $z$ would not influence the second (constant) sum, and in the first sum, it would only be a re-indexing,

$$\sum_{\omega\in\Lambda} \frac{1}{(z+\omega_0+\omega)^2} = \sum_{\omega_0+\omega\in\Lambda} \frac{1}{\bigl(z+(\omega_0+\omega)\bigr)^2}= \sum_{\omega\in \Lambda} \frac{1}{(z+\omega)^2}.$$

Alas, although the "result" is correct, every $\omega_0\in\Lambda$ is a period of $\wp$, the way to obtain it isn't valid. We cannot split the series in such a way since the parts don't converge absolutely (that is, after all, the reason for the correction terms $\frac{1}{\omega^2}$).

So back to reality. We can freely split and rearrange finite sums, and if we take the sum over a suitable large finite subset of the lattice, the terms in both corresponding sums will mostly be the same, and only few terms will appear in one sum but not the other, so that the difference of the two sums is small. The proof makes this hand-waving precise.

Assuming for the moment the given estimates that for large $R$, we have

$$\begin{align} \wp(z) - \wp^R(z) &\in O\left(\frac1R\right),\\ \wp^R(z+1) - \wp^R(z) &\in O\left(\frac1R\right),\tag{1}\\ \wp^R(z+\tau) - \wp^R(z) &\in O\left(\frac1R\right), \end{align}$$

for all $z$ with $\lvert z\rvert \leqslant 1$ (or $\lvert z\rvert \leqslant K$ for some arbitrary positive $K$), we can simply argue

$$\begin{align} \lvert \wp(z+1) - \wp(z)\rvert &= \lvert \wp(z+1)-\wp^R(z+1) + \wp^R(z+1) - \wp^R(z) + \wp^R(z) - \wp(z)\rvert\\ &\leqslant \lvert \wp(z+1)-\wp^R(z+1)\rvert + \lvert\wp^R(z+1) - \wp^R(z)\rvert + \lvert\wp^R(z) - \wp(z)\rvert\\ &\leqslant \frac{C}{R} \end{align}$$

for all $\lvert z\rvert \leqslant 1$, some constant $C$ independent of $R$, and all $R \geqslant R_{\min}$. But $\lvert \wp(z+1)-\wp(z)\rvert \leqslant \frac{C}{R}$ for all large enough $R$ can only hold if $\wp(z+1) = \wp(z)$. Since that identity holds in a non-empty open set, and $\mathbb{C}\setminus \Lambda$ is connected, it follows that $1$ is a period of $\wp$. The argument for $\tau$ is the same.

We must now verify that we have the estimates $(1)$.

$$\left\lvert\frac{1}{(z+\omega)^2}-\frac{1}{\omega^2}\right\rvert = \left\lvert\frac{z^2+2z\omega}{\omega^2(z+\omega)^2}\right\rvert.$$

For $\lvert\omega\rvert \geqslant 2(2+\lvert\tau\rvert)$ we have $\lvert z+a\rvert \leqslant \lvert\omega\rvert/2$ for $\lvert z\rvert \leqslant 1$ and $a \in \{0,1,\tau\}$, so we have

$$\left\lvert\frac{1}{(z+\omega)^2}-\frac{1}{\omega^2}\right\rvert \leqslant \frac{C_1}{\lvert \omega\rvert^3}.$$

Hence

$$\lvert \wp(z+a) - \wp^R(z+a)\rvert = \left\lvert\sum_{\lvert\omega\rvert\geqslant R} \left(\frac{1}{(z+a+\omega)^2} - \frac{1}{\omega^2}\right) \right\rvert \leqslant C_1 \sum_{\lvert\omega\rvert\geqslant R} \frac{1}{\lvert\omega\rvert^3}.$$

Since $\operatorname{Im}\tau \neq 0$, there is a $c > 0$ with $\lvert x+y\tau\rvert \geqslant c\cdot \sqrt{x^2+y^2}$ for all $x,y\in\mathbb{R}$, so we can estimate the last sum by

$$\frac{1}{c^3}\sum_{m^2+n^2 \geqslant c_1\cdot R^2} \frac{1}{(m^2+n^2)^{3/2}},$$

and estimating that sum by an integral, we obtain $\lvert \wp(z+a) - \wp^R(z+a)\rvert \leqslant \dfrac{C_2}{R}$ for all $R \geqslant 2(2+\lvert\tau\rvert)$, $\lvert z\rvert \leqslant 1$ and $a \in \{0,1,\tau\}$.

Further, for large enough $R$, we have

$$\begin{align} \wp^R(z+1) &- \wp^R(z)\\ &= \frac{1}{(z+1)^2} + \sum_{0 < \lvert \omega\rvert < R}\left(\frac{1}{(z+1+\omega)^2}- \frac{1}{\omega^2}\right) - \frac{1}{z^2} - \sum_{0 <\lvert\omega\rvert < R} \left(\frac{1}{(z+\omega)^2}- \frac{1}{\omega^2}\right)\\ &= \sum_{\lvert\omega\rvert < R} \frac{1}{(z+1+\omega)^2} - \sum_{\lvert \omega\rvert <R} \frac{1}{(z+\omega)^2}\\ &= \sum_{\lvert\omega-1\rvert < R} \frac{1}{(z+\omega)^2} - \sum_{\lvert\omega\rvert<R} \frac{1}{(z+\omega)^2}\\ &= \sum_{\lvert\omega-1\rvert < R \leqslant \lvert\omega\rvert} \frac{1}{(z+\omega)^2} - \sum_{\lvert\omega\rvert<R\leqslant \lvert\omega-1\rvert} \frac{1}{(z+\omega)^2}. \end{align}$$

We can thus estimate

$$\begin{align} \lvert \wp^R(z+1)-\wp^R(z)\rvert &\leqslant \sum_{\lvert\omega-1\rvert < R \leqslant \lvert\omega\rvert} \frac{1}{\lvert z+\omega\rvert^2} + \sum_{\lvert\omega\rvert<R\leqslant \lvert\omega-1\rvert} \frac{1}{\lvert z+\omega\rvert^2}\\ &\leqslant \sum_{R-1\leqslant \lvert\omega\rvert\leqslant R+1} \frac{1}{\lvert z+\omega\rvert^2}. \end{align}$$

Again, we can estimate $\lvert m+n\tau\rvert$ by $\sqrt{m^2+n^2}$ and obtain

$$\lvert \wp^R(z+1)-\wp^R(z)\rvert \leqslant \frac{C_3}{R}$$

for all $\lvert z\rvert \leqslant 1$ and $R\geqslant 2(2+\lvert\tau\rvert)$.

Basically the same computation shows

$$\lvert\wp^R(z+\tau)-\wp^R(z)\rvert \leqslant \frac{C_4}{R}$$

for all $\lvert z\rvert \leqslant 1$ and $R \geqslant 2(2+\lvert\tau\rvert)$.

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I tried a slightly different approach that I think exposes the main ideas a bit more clearly.

Define $$ \Omega(h,k)=\left\{\begin{array}{} 0&\text{if }h=k=0\\ 1&\text{otherwise} \end{array}\right.\tag{1} $$ and $w=h+k\tau$. Then, with $0/0=0$ whenever it occurs, $$ \wp(z)=\sum_{h=-\infty}^\infty\sum_{k=-\infty}^\infty\left[\frac1{(z+w)^2}-\frac{\Omega(h,k)}{w^2}\right]\tag{2} $$ Define $$ \wp_R(z)=\sum_{h=-R}^R\sum_{k=-R}^R\left[\frac1{(z+w)^2}-\frac{\Omega(h,k)}{w^2}\right]\tag{3} $$ Using $(2)$ and $(3)$, we get $$ \begin{align} \wp_R(z+\tau)-\wp_R(z) &=\sum_{h=-R}^R\left[\frac1{(z+h+(R+1)\tau)^2}-\frac1{(z+h-R\tau)^2}\right]\\ &=\sum_{h=-R}^RO\left(\frac1{R^2}\right)\\ &=O\left(\frac1R\right)\tag{4} \end{align} $$ and $$ \begin{align} \wp_R(z+1)-\wp_R(z) &=\sum_{k=-R}^R\left[\frac1{(z+R+1+k\tau)^2}-\frac1{(z-R+k\tau)^2}\right]\\ &=\sum_{h=-R}^RO\left(\frac1{R^2}\right)\\ &=O\left(\frac1R\right)\tag{5} \end{align} $$ Sending $R\to\infty$ in $(4)$ and $(5)$ yields $$ \wp(z+\tau)=\wp(z)=\wp(z+1)\tag{6} $$ Induction and $(6)$ shows that $\wp(z+w)=\wp(z)$ whenever $w\in\Lambda$.

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The hint says that the the function $\wp^R$ is uniformly close to $\wp,$ and also uniformly almost periodic. Which means that $\wp$ is uniformly almost periodic for any $R,$ hence periodic.

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