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This question involves the following homework problem:

PROBLEM

Suppose $f$ is analytic in the upper half plane and periodic of period 1. Show that $f$ has an extension of the form

$$f(z)=\sum_{-\infty}^{\infty} c_n e^{2\pi inz}$$

with

$$c_n= \int_0^1 f(x+iy) e^{-2\pi i n (x+iy)}dx$$

for any value $y>0$.

HINT

Show that there is an analytic function $f^*$ on a disk from which the origin is deleted such that

$$f^*(e^{2\pi i z })=f(z)$$

what is the Laurent series for $f^*$? Abbreviate $q=e^{2\pi i z }$.

MY TRY

I thought that the extension to the real line should also be analytic:

$$g(x):= \lim_{y\text{ to }0}f(x+iy)$$

and then by periodicity and the fact that $f,g$ are analytic this function is equal to its Fourier series everywhere (with possibly complex coefficients). Then by again the fact that $f,g,\sin$ and cos are analytic this is then equal to the same series using the complex series definition for $\sin$ and $\cos$. I have the idea that I am still missing something, since the hint implies a totally different approach. Thanks!

user1043065
  • 938
  • We know that $f(z)$ has a power series expansion that converges for every y>0, with coefficients given by Cauchy's formula. Now we can compose with $h(q)=\log(q)/(2 \pi i)$ such that $(f \circ h)(e^{2 \pi i z})=f(z)$ giving us $f^\star$. Now we can expand this into a power series by using Cauchy's formula for the coefficients and the fact that the logarithm is a analytic function. – Pol van Hoften Jun 02 '14 at 14:02

0 Answers0