- #1

- 206

- 0

## Homework Statement

Show that f is 2-pi periodic and analytic on the strip [itex]\vert Im(z) \vert < \eta[/itex], iff it has a Fourier expansion [itex]f(z) = \sum_{n = -\infty}^{\infty} a_{n}z^{n}[/itex], and that [itex]a_n = \frac{1}{2 \pi i} \int_{0}^{2\pi} e^{-inx}f(x) dx[/itex]. Also, there's something about the lim sup of the modulus nth roots of the coefficients being less than [itex]e^{-\eta}[/itex]. But I think that follows pretty obviously from the Laurent series existing.

## Homework Equations

## The Attempt at a Solution

The direction assuming it has a Fourier expansion is pretty easy. Just set [itex]\omega = e^{iz}[/itex]. Then the series with [itex]\omega[/itex] is analytic on the strip, and that it's 2-pi periodic is clear.

For the other direction I think the idea here is to find a Laurent series for f(z) (or at least that agrees with f on a "big enough" subset of its domain) that's actually analytic on the annulus [itex]e^{-\eta} < \vert z \vert < e^{\eta}[/itex]. Then you just use that Laurent series with the variable replaced by [itex]e^{iz}[/itex].

However there doesn't seem to be any way to get such a series that I can see. I've tried a lot of ideas: I've tried looking for a Laurent series for f in the intersection of the annulus with the strip, I've tried finding an appropriate Laurent series for f for an annulus contained in the strip, and I can't get those to work. Also I've tried cheating and just appealing to the Fourier series existing for a real valued function and then using the fact that if analytic functions agree on a "big enough" part of their domain, they have to be the same, but that just seems like a cop out and not very convincing.

So, I'd appreciate any hints or help. Thanks.