Show this formula holds for an analytic, periodic function in each half plane Let $f(z)$ be an analytic function in the upper half plane that is periodic with real period $2\pi \lambda > 0$. Suppose that $\exists A,C > 0$ such that $|f(x + iy)| < Ce^{Ay}$ for $y > 0$. Show that
$$f(z) = \sum_{n \geq -A\lambda} a_n e^{\frac{inz}{\lambda}}$$
where the series converges uniformly in each half plane $\{y \geq \epsilon\}$ for $\epsilon >0$
So, I know that for an analytic function with a period, $T$, on the half plane, we can write it as
$$f(z) = \sum_{k = 0}^{\infty} a_k e^{\frac{2\pi z}{T}}$$
In our case, $T = 2 \pi \lambda$ meaning we have
$$f(z) = \sum_{k = 0}^{\infty} a_k e^{\frac{inz}{\lambda}}$$
From here is where I'm unsure of how to connect it. First off, we are told $2 \pi \lambda > 0$ meaning $\lambda > 0$. We are also told $y > 0$ and $A > 0$. However, under the summation we want to end up with, it has $n \geq -A\lambda$. Since $A > 0$ and $\lambda > 0$, this statement doesn't really make much sense to me. Why not just say $n \geq 0$? They're both positive, so the negative is kind of pointless isn't it? In general, though, I'm unsure of how to get to the conclusion series. Can anyone lend a hand here?
 A: Let's start with a $2\pi \lambda$-periodic, analytic function $f$ in the upper half-plane. Such a function can be written as $f(z) = g(e^{iz/\lambda})$ where $g$ is defined in $D= \{ z : 0 < |z| < 1 \}$ as
$$
 g(w) = f( -\lambda i \log w)  \, .
$$
$g$ is well-defined and holomorphic in $D$ because the expression on the right does not depend on the choice of the branch of the logarithm. $g$ can be developed into a Laurent series
$$
g(w) = \sum_{n = -\infty}^\infty a_n w^n
$$
which converges locally uniformly for $0 < |w| < 1$. It follows that
$$
 f(z) =  \sum_{n = -\infty}^\infty a_n e^{inz/\lambda} \, .
$$
Now assume that $f$ additionally satisfies a growth condition $|f(x + iy)| < Ce^{Ay}$ for $x \in \Bbb R$ and $y > 0$. Then
$$
 |g(w)| = |f(\lambda \arg(w)-\lambda i\log |w| )|
 \le C e^{-A\lambda \log|w|} = \frac{C}{|w|^{A \lambda}} \, ,
$$
so that $g$ has a removable singularity or a pole of order at most $\le A\lambda $ at $w=0$. It follows that the Laurent series of $g$ has only terms with exponent $\ge -A \lambda$, i.e.
$$
 g(w) = \sum_{n \ge -A \lambda} a_n w^n \, .
$$
and consequently
$$
 f(z) =  \sum_{n \ge -A \lambda} a_n e^{inz/\lambda} \, .
$$
Finally, if $f(x+iy)$ is bounded for $y \to \infty$, uniformly in $x$, then $g(w)$ is bounded for $w \to 0$, so that is has a removable singularity at $w=0$. In this case, the Laurent series of $g$ has only terms with non-negative exponents (i.e. it is a normal power series), and consequently
$$
 f(z) =  \sum_{n =0}^\infty a_n e^{inz/\lambda} \, .
$$
