- By rearranging the series
\[ \frac {1}{z ^ {2}} + \sum_ {\omega \in \Lambda^ {*}} \left[ \frac {1}{(z + \omega) ^ {2}} - \frac {1}{\omega^ {2}} \right], \]
show directly, without diferentiation, that \(\wp ( z + \omega ) = \wp ( z )\) whenever \(\omega \in \Lambda .\) [Hint: For R suficiently large, note that \(\wp ( z ) = \wp ^ { R } ( z ) + O ( 1 / R )\) where \(\begin{array} { r } { \dot { \wp } ^ { R } ( z ) = z ^ { - 2 } + \sum _ { 0 < \lvert \omega \rvert < R } ( ( z + \omega ) ^ { - 2 } - \omega ^ { - 2 } ) } \end{array}\) Next, observe that both \(\boldsymbol { \wp } ^ { R } ( z + 1 ) - \boldsymbol { \wp } ^ { R } ( z )\) and \(\wp ^ { R } ( z + \tau ) - \wp ^ { R } ( z )\) are \(\begin{array} { r } { O ( \sum _ { R - c < | \omega | < R + c } | \omega | ^ { - 2 } ) = O ( 1 / R ) . ] } \end{array}\)
Solution
It suffices to prove periodicity under an arbitrary fixed \(\lambda\in\Lambda\). For \(R>0\), define \[ \wp^R(z)=\frac1{z^2}+ \sum_{0<|\omega|<R} \left(\frac1{(z+\omega)^2}-\frac1{\omega^2}\right). \] For \(z\) in a fixed compact set disjoint from \(-\Lambda\), the tail satisfies \[ \wp(z)-\wp^R(z) =O\left(\sum_{|\omega|\ge R}|\omega|^{-3}\right) =O(R^{-1}), \tag{1} \] because \[ \frac1{(z+\omega)^2}-\frac1{\omega^2}=O(|\omega|^{-3}) \] uniformly on such compact sets.
Also \[ \wp^R(z)= \sum_{|\omega|<R}\frac1{(z+\omega)^2} -\sum_{0<|\omega|<R}\frac1{\omega^2}. \] Thus, after reindexing \(\eta=\omega+\lambda\), \[ \wp^R(z+\lambda)-\wp^R(z) = \sum_{|\eta-\lambda|<R}\frac1{(z+\eta)^2} -\sum_{|\eta|<R}\frac1{(z+\eta)^2}. \tag{2} \] Only lattice points in the symmetric difference of the two radius-\(R\) discs contribute. They lie in the annulus \[ R-|\lambda|<|\eta|<R+|\lambda|. \] This annulus contains \(O(R)\) lattice points, while each summand in (2) is \(O(R^{-2})\). Hence \[ \wp^R(z+\lambda)-\wp^R(z)=O(R^{-1}). \tag{3} \] Combining (1) and (3), \[ \wp(z+\lambda)-\wp(z)=O(R^{-1}). \] Letting \(R\to\infty\) gives \[ \wp(z+\lambda)=\wp(z) \] for all \(z\) away from the poles. By meromorphic continuation the identity holds everywhere. Thus every \(\lambda\in\Lambda\) is a period of \(\wp\).