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Given an infinitely complex differentiable (at up to an open subset of points), periodic function

$$f(z):\mathbb{C}\to \mathbb{C}\\ f(z+a)=f(z),~~a\in\mathbb{R}$$

is there anything that can be said in general about its boundedness properties in the complex plane? In particular, is there a way to determine its asymptotic behavior as $z\to\pm i\infty$?

Kagaratsch
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  • What do you mean? Have you looked at examples....$e^z$, say. (if you insist on a real period, work with $e^{2\pi iz}$). – lulu Jan 30 '20 at 20:26
  • @lulu yeah, so, a function $e^{iz}$ is periodic under $z\to z+2\pi$, but if we consider the limit $z\to\pm i\infty$ it exponentially vanishes or exponential diverges in these directions. I am wondering if one can show this to be generic behavior for all periodic complex functions with real period? – Kagaratsch Jan 30 '20 at 20:30
  • What limit are you taking? Are you just looking at the imaginary axis? – lulu Jan 30 '20 at 20:33
  • @lulu , yes, imaginary axis for instance. However, I believe having a real part $0\leq$Re$(z)<a\ll\infty$ does not change the asymptotic limiting point into the complex direction, no? – Kagaratsch Jan 30 '20 at 20:36
  • For the general question, you can write any such periodic function as a Fourier Series, so are just looking at sums of functions like $e^{2\pi inz}$. – lulu Jan 30 '20 at 20:39
  • @lulu but does the Fourier series expansion hold for complex values of $z$? Since the expansion is defined for real functions. Furthermore, even a non-periodic function can be expanded in Fourier basis (giving an integral expansion instead of sum), then one would conclude that Fourier basis vectors blow up at complex infinity for all functions, which does not sound quite right. What do you think? – Kagaratsch Jan 30 '20 at 20:43
  • I would look into something like $e^{e^{2\pi iz}}$ and related stuff since there it does matter what $\Re z$ is as $|e^{e^{2\pi iz}}|=e^{e^{-2\pi \Im z}\cos{2\pi \Re z}}$, so in particular if $\Re z =\frac{\pi}{4}$, the function is bounded vertically and obvosuly you can construct examples that propagate the behavior – Conrad Jan 30 '20 at 20:46
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    here is a good description of Fourier expansions for holomorphic functions. – lulu Jan 30 '20 at 20:48
  • @Conrad Oh, wow, that thing looks scary, is it $L^p$ measurable? I should have probably specified more tame properties for the periodic functions I have in mind... – Kagaratsch Jan 30 '20 at 20:50
  • @lulu I see, so the Fourier series stays the same. But can we really conclude that any such function must blow up at complex infinity from this? After all, almost any function can be expanded in Fourier basis (integral instead of sum for non-periodic). But not all of these functions are unbounded at complex infinity... – Kagaratsch Jan 30 '20 at 21:05
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    I guess finite order would be a good condition as then by phragmen-lindelof the function must be unbounded on any vertical line, though not sure if it must go to infinity – Conrad Jan 30 '20 at 21:06
  • I don't know whether it must blow up on the imaginary axis or not. Certainly typical examples of holomorphic functions which are bounded on both the real and imaginary axis (see, e.g., this or this) are not periodic. – lulu Jan 30 '20 at 21:12
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    @lulu by periodicity the function must be unbounded on any vertical line if it is of finite order; otherwise see the counterxample above which of course can be translated so the function has constant modulus one on the imaginary axis (or any fixed vertical line ) – Conrad Jan 30 '20 at 21:17
  • @Conrad Ah, thank you. That example, in particular, is very useful. – lulu Jan 30 '20 at 21:20
  • @Conrad may I ask, what exactly do you mean by "finite order" in this context? It seems to me that what you are saying is very relevant, so if you could write an answer below with a bit more details, I will likely upvote and accept it. – Kagaratsch Jan 30 '20 at 21:24
  • @Conrad oh, you probably mean http://mathworld.wolfram.com/FiniteOrder.html – Kagaratsch Jan 30 '20 at 21:36
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    put a partial answer as asked – Conrad Jan 30 '20 at 21:39
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    Let $g$ be any holomorphic function on $\mathbb{C}\setminus {0}$. Then $f(z) = g\bigl(e^{\frac{2\pi i}{a}z}\bigr)$ is an entire function with period $a$, and every entire function with period $a$ is of that form. – Daniel Fischer Jan 30 '20 at 22:11
  • @DanielFischer Interesting! But can we have a periodic function that is holomorphic, but not entire? – Kagaratsch Jan 30 '20 at 22:24
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    Sure. If $g$ is holomorphic on an annulus $r < \lvert w\rvert < R$, then $f$ defined as above is holomorphic on the parallel strip $-\frac{a}{2\pi}\log R < \operatorname{Im} z < -\frac{a}{2\pi}\log r$ (assuming $a > 0$) and has period $a$. Again all such functions arise in that way. – Daniel Fischer Jan 30 '20 at 22:33
  • @DanielFisher wow, ok, but does that mean that taking a Laurent series of $g$ one has an expansion in terms of exponentials, out of which only the zero mode would not blow up at plus or minus complex infinity? (assuming $f$ has a holomorphic strip in its domain continuously connecting plus and minus complex infinity.) – Kagaratsch Jan 30 '20 at 22:48

1 Answers1

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As a partial answer per the comments above, we need to impose the restriction that $f$ has finite order - this means that there is some $a>0$ for which $\max_{|z|=R}|f(z)| \le C(a, \epsilon)e^{R^{a+\epsilon}}$, for all $\epsilon >0, R>0$ and the infimum of such is called the order

(polynomials have order zero but there are entire functions of order zero that are not polynomials, $e^{2\pi iz}, e^z$ have order $1$, $e^{z^2}$ has order $2$ and $\cos {\sqrt z}$ - defined formally by the Taylor series $\sum{(-1)^n\frac{z^n}{(2n)!}}$ has order $\frac{1}{2}$)

Then Phragmen-Lindelof theorem (which in one version states that if a finite order holomorphic function is bounded on the edges of a strip, it satisfies the usual maximum modulus theorem, i.e. it is bounded in the whole strip by the edge bounds) and periodicity implies that $f$ must be unbounded on any vertical line

(if $f$ with period $a>0$ would be bounded on a vertical line $\Re z= b$, it would be bounded on the vertical line $\Re z = b+a$ by periodicity, hence in the full strip bounded by those two lines by P-L, hence in the plane by periodicity, hence it would be constant!)

Outside finite order, things become dicey and the example $e^{e^{2\pi iz}}$ which has modulus $1$ on $\Re z =\frac{\pi}{4}$ as $|e^{e^{2\pi iz}}|=e^{e^{-2\pi \Im z}\cos{2\pi \Re z}}$ shows that you can concoct counterexamples that are bounded on any vertical line of one's choice by translating the example above which is obviously periodic with period $1$.This is the typical counterexample that shows that Phragmen Lindelof needs finite order to hold.

Conrad
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