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Spring 2021 - Problem 8

construction, entire functions, Jacobi theta function

Show that there exists a non-zero entire function f ⁣:C→C\func{f}{\C}{\C} and constants b,c∈Cb, c \in \C satisfying

f(0)=0,f(z+1)=ebzf(z),andf(z+i)=eczf(z).f\p{0} = 0, \quad f\p{z + 1} = e^{bz}f\p{z}, \quad\text{and}\quad f\p{z + i} = e^{cz}f\p{z}.
Solution.

First, formally express

f(z)=∑n∈Zane2πinz,f\p{z} = \sum_{n \in \Z} a_n e^{2\pi inz},

Then f(z+1)=f(z)f\p{z + 1} = f\p{z} and

∑n∈Zane2πinze−2πn=f(z+i)=eczf(z)=∑n∈Zane2πi(n+c2πi)z\sum_{n \in \Z} a_n e^{2\pi inz} e^{-2\pi n} = f\p{z + i} = e^{cz} f\p{z} = \sum_{n \in \Z} a_n e^{2\pi i\p{n + \frac{c}{2\pi i}}z}

Let 2πik=c2\pi i k = c, and suppose k∈Zk \in \Z. Then

∑n∈Zane−2πne2πinz=∑n∈Zane2πi(n+k)z=∑n∈Zan−ke2πinz,\sum_{n \in \Z} a_n e^{-2\pi n} e^{2\pi inz} = \sum_{n \in \Z} a_n e^{2\pi i\p{n+k}z} = \sum_{n \in \Z} a_{n-k} e^{2\pi inz},

so we require ane−2πn=an−ka_n e^{-2\pi n} = a_{n-k}. If, say, k=−2k = -2, then if n=2mn = 2m is even,

a2m=e−4πma2(m−1)=e−2πm(m+1)a0a_{2m} = e^{-4\pi m} a_{2\p{m-1}} = e^{-2\pi m\p{m+1}} a_0

and if n=2m+1n = 2m + 1 is odd,

a2m+1=e−4πma2(m−1)+1=e−2πm(m+1)a1a_{2m+1} = e^{-4\pi m} a_{2\p{m-1}+1} = e^{-2\pi m\p{m+1}} a_1

Thus, if we split the sum into even and odd terms, we see

f(z)=a0∑n∈Ze−2πn2−2πn+2πi(2n)z+a1∑n∈Ze−2πn2−2πn+2πi(2n+1)z=a0∑n∈Ze−2πn2−2πn+4πinz+a1∑n∈Ze−2πn2−2πn+4πinz+2πiz\begin{aligned} f\p{z} &= a_0 \sum_{n \in \Z} e^{-2\pi n^2 - 2\pi n + 2\pi i\p{2n}z} + a_1 \sum_{n \in \Z} e^{-2\pi n^2 - 2\pi n + 2\pi i\p{2n+1}z} \\ &= a_0 \sum_{n \in \Z} e^{-2\pi n^2 - 2\pi n + 4\pi inz} + a_1 \sum_{n \in \Z} e^{-2\pi n^2 - 2\pi n + 4\pi inz + 2\pi iz} \\ \end{aligned}

satisfies f(z+1)=f(z)f\p{z + 1} = f\p{z} and f(z+i)=e−4πizf(z)f\p{z + i} = e^{-4\pi iz}f\p{z}. Hence, if we pick a0a_0 and a1a_1 accordingly, we can also ensure that f(0)=0f\p{0} = 0. Finally, the sum converges locally uniformly since e−2πn2e^{-2\pi n^2} is summable, so such an ff exists.