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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?

Nolan P
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  • Related: https://math.stackexchange.com/q/14423/96384 with follow-up https://math.stackexchange.com/q/148027/96384 (stronger growth condition); https://math.stackexchange.com/q/818118/96384 (no growth condition). – Torsten Schoeneberg May 07 '24 at 03:19

1 Answers1

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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 $$ f(z) = \sum_{n =0}^\infty a_n e^{inz/\lambda} \, . $$

Martin R
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  • Hi there. This is detailed, and I do believe this makes sense. I do want to confirm something though. In my book, there is a theorem that says "Suppose $f(z)$ is analytic on the half plane ${Im(z) > \alpha}$ and $f(z)$ is periodic with period $1$. If $f(z)$. If $f(z)$ is bounded as $Im(z) \to \infty$, then $f(z)$ can be expanded in an absolutely convergent series of exponentials. $$f(z) = \sum_{k = 0}^{\infty} a_k e^{2\pi ikz}$$. Here, the sum does start at $k = 0$. Is this due to the period being $1$? That seems to be the only major difference in the theorem unless if I am missing something – Nolan P May 31 '21 at 21:12
  • Ah, and we don't assume boundedness here for $y \to \infty$. So is there a general formula for this then? My book only presented the bounded variant I listed, which ended up confusing me. Is there a general one for boundedness as $y \to a$ where $a \in \mathbb{C}$? – Nolan P Jun 01 '21 at 13:28
  • I'm looking for a general formula regardless of how the function is bounded. The formula being bounded as $y \to \infty$ got me confused, so I want a general one. – Nolan P Jun 01 '21 at 14:28
  • @bomb456: I have rewritten the answer, to make it (hopefully) more clear. – Martin R Jun 01 '21 at 17:47
  • That still isn't what I'm looking for. I guess I can't explain it well enough to ask, it seems like its an obvious quesiton to me but it's clearly not. My book gave me the formula starting at $0$. They did not give me a formula in general. So I am unsure of how I'd approach a different problem without it being bounded as $y \to \infty$ as I don't have a formula for it. As of right now, I feel I can only solve problems where $f$ is bounded as $y \to \infty$. I am looking for a formula outside of that specific case. – Nolan P Jun 01 '21 at 18:05
  • @bomb456: One only gets $f(z) = \sum_{n =0}^\infty a_n e^{inz/\lambda} $ if $f$ is bounded for $y \to \infty$. – You asked why $f(z) = \sum_{n \geq -A\lambda} a_n e^{\frac{inz}{\lambda}}$ if $f$ satisfies a growth condition. I have tried to answer that, and to explain the general case. I honestly don't know what else to say. – Martin R Jun 01 '21 at 18:07
  • I'll ask my professor. I'm either not understanding or I'm not asking my question clearly. I do appreciate the effort though! I will accept your answer as it does answer the asked question. – Nolan P Jun 01 '21 at 18:13