1826 CHAPTER 58. ELLIPTIC FUNCTIONS

and so

c1 = ℘

(−w1

2+w1

)−℘

(−w1

2

)= ℘

(w1

2

)−℘

(−w1

2

)= 0

which shows the constant for ℘(z+w1)−℘(z) must equal zero. Similarly the constantfor ℘(z+w2)−℘(z) also equals zero. Thus ℘ is periodic having the two periods w1,w2.

Of course to justify this, you need to consider whether the series of 58.1.6 converges.Consider the terms of the series.∣∣∣∣∣ 1

(z−w)2 −1

w2

∣∣∣∣∣= |z|∣∣∣∣∣ 2w− z

(z−w)2 w2

∣∣∣∣∣If |w|> 2 |z| , this can be estimated more. For such w,∣∣∣∣∣ 1

(z−w)2 −1

w2

∣∣∣∣∣= |z|

∣∣∣∣∣ 2w− z

(z−w)2 w2

∣∣∣∣∣≤ |z| (5/2) |w||w|2 (|w|− |z|)2

≤ |z| (5/2) |w||w|2 ((1/2) |w|)2 = |z| 10

|w|3.

It follows the series in 58.1.6 converges uniformly and absolutely on every compact set, Kprovided ∑w̸=0

1|w|3

converges. This question is considered next.

Claim: There exists a positive number, k such that for all pairs of integers, m,n, notboth equal to zero,

|mw1 +nw2||m|+ |n|

≥ k > 0.

Proof of claim: If not, there exists mk and nk such that

limk→∞

mk

|mk|+ |nk|w1 +

nk

|mk|+ |nk|w2 = 0

However,(

mk|mk|+|nk|

, nk|mk|+|nk|

)is a bounded sequence in R2 and so, taking a subsequence,

still denoted by k, you can have(mk

|mk|+ |nk|,

nk

|mk|+ |nk|

)→ (x,y) ∈ R2

and so there are real numbers, x,y such that xw1 + yw2 = 0 contrary to the assumption thatw2/w1 is not equal to a real number. This proves the claim.