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Is there a $2\pi i$-periodic holomorphic function $f$ on the complex plane $\mathbb{C}$, $$f(z+2\pi i n)=f(z), \, \forall n\in \mathbb{Z} \, \forall z\in\mathbb{C}$$ that blows up in both directions of the real axis, $$|f(z)|^2 \to\infty$$ as $z\to+\infty$ and $z\to-\infty$ and whose derivative has no zeroes, i.e. $$f'(z)\neq 0$$ for all $z\in \mathbb{C}$?

I cannot find one, and my guess is that such a function does not exist. If I am correct, how can I proof it? If I am incorrect, what is an example for such a function (even better would be an iteration/classification of all such functions)?

Edit: I tried to make the divergence-requirement more precise.

user192049
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  • If $f$ is assumed to send $\mathbb R$ into $\mathbb R$, and the blowup at $\pm \infty$ is to $+\infty$, then the function has a minimum. – Dustan Levenstein Jul 06 '20 at 18:40
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    Are you considering $\infty=+\infty=-\infty$ a la Riemann sphere? – R. Burton Jul 06 '20 at 18:42
  • @R.Burton yes, $-\infty$ is okay as well. To be more precise, what I want is $|f(z)|^2 \to \infty$. I should probably edit my question. – user192049 Jul 06 '20 at 21:06
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    Note that the Riemann surface obtained by identifying $z$ with $z+2\pi i$ on $\mathbb C$ is biholomorphic with $\mathbb C^\times$, via the exponential function. So one way to interpret your question is you want a holomorphic function on $\mathbb C^\times$ with poles at $0$ and $\infty$, but whose derivative never vanishes on $\mathbb C^\times$.

    My complex analysis isn't strong enough to know whether there's a reason why such a function can't exist; certainly no rational function will do the trick, the derivative of a rational function is guaranteed to have a zero.

     – Dustan Levenstein Jul 06 '20 at 21:12
  • Another valid interpretation would be you merely want $\lim_{\substack{x \to \infty \ x \in \mathbb R}} |f(x)| = \lim_{\substack{x \to 0^+\ x \in \mathbb R}} |f(x)| = \infty$. Again in the $\mathbb C^\times$ picture. – Dustan Levenstein Jul 06 '20 at 21:16
  • @DustanLevenstein That's actually more where my problem is coming from. Ultimately I am interested in a Riemann sphere with $n$ punctures, but so far I don't even understand the $n=2$ case. (I am aware, that meromorphic functions won't do it, because they have as many zeroes as they have poles.) – user192049 Jul 06 '20 at 21:18
  • @DustanLevenstein regarding your third comment. I am apparently too ignorant to realize the difference. Does it matter whether I restrict to a line if I am talking about a holomorphic function? (parden my poor complex analysis understanding) – user192049 Jul 06 '20 at 21:29

2 Answers2

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Assuming that you want $$\lim_{\operatorname{Re}(z) \to \pm \infty} |f(z)| = \infty,$$ there is no solution; translating the problem via the exponential function into the corresponding problem of finding a function on $\mathbb C^\times$ which approaches infinity at both $0$ and $\infty$, that means the singularities at $0$ and $\infty$ are not essential, and therefore they must be poles. Which makes $f$ a meromorphic function, so $f'$ is also a meromorphic function, with poles of one degree higher at $0$ and $\infty$. Since meromorphic functions on the Riemann sphere have the same number of zeroes as poles, $f'$ therefore has a zero.

Dustan Levenstein
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  • Could you elaborate on why that implies that the singulartities are not essential? – user192049 Jul 06 '20 at 21:32
  • Are you saying that there exists an open neighborhood around $z=0$ on $\mathbb{C}^\times$ where the function doesn't take all complex values except one? I don't see that. $f(z)=e^{z} + e^{1/z}$ happily has essential singularities at $z=0$ and $z=\infty$. – user192049 Jul 06 '20 at 21:48
  • @user192049 Picard theorem is overkill (sorry, my complex analysis isn't strong enough to know what's elementary and what's not). Look at it another way: take the function $1/f(z)$ (where $f$ is defined on $\mathbb C^\times$ and notice that $0$ and $\infty$ are removable discontinuities. Therefore $1/f(z)$ has zeroes at $0$ and $\infty$, so $f(z)$ has poles at those places. – Dustan Levenstein Jul 06 '20 at 21:55
  • @user192049 Have a look at the behavior of of your $f(z) = e^z + e^{1/z}$ away from the real line. Take, e.g., a small circle around $0$, and you should be able to work out that $f$ oscillates wildly from small values to very large values along that circle. – Dustan Levenstein Jul 06 '20 at 22:04
  • @user192049 Note my assumption at the beginning; there is a difference between $\lim_{\operatorname{Re}(z) \to \pm \infty} |f(z)| = \infty$ and $\lim_{\substack{\operatorname{Re}(z) \to \pm \infty \ z \in \mathbb R}} |f(z)| = \infty$. As a separate note, I believe you might also be confusing the $2\pi i$ periodicity exposition of the problem with the $\mathbb C^\times$ exposition of the problem with your example of the function $f(z) = e^z+e^{1/z}$, which is not $2\pi i$-periodic. – Dustan Levenstein Jul 06 '20 at 22:14
  • I think I see now. The argument that the singularities of the function $f(z)$ cannot be essential directly depends on the requirement $\lim_{\operatorname{Re}(z) \to \pm \infty} |f(z)| = \infty$. This argument would not be valid if only $\lim_{\substack{\operatorname{Re}(z) \to \pm \infty \ z \in \mathbb R}} |f(z)| = \infty$. Am I correct? – user192049 Jul 07 '20 at 16:35
  • Yes, that is correct! – Dustan Levenstein Jul 07 '20 at 17:09
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The function $$ f(z) := e^{e^z} $$ has period $2\pi i$ and its derivative $$ f'(z) = e^z\;e^{e^z} = e^{z+e^z} $$ is never zero.

But this does not have the desired limit at $-\infty$.

GEdgar
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