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I am learning complex variables again with a greater depth, and I learnt the Great Picard Theorem, which is stated as follows:

If $z_{0}$ is an essential singularity of $f$, the image of any punctured neighborhood of $z_{0}$ via $f$ is the whole complex plane, except possibly for one point, infinitely often.

What does it mean by "infinitely often" here? What I know about i.o. is in the sense of probability: say we have $\{A_{n}\}_{n=1}^{\infty}$ a infinite sequence of events, then $$\{A_{i}\ \text{i.o}\}=\{\omega\in\Omega:\omega\in A_{i}\ \ \text{for infinitely many of}\ \ i\in \{1,2,3,\cdots\}\}.$$ I also know the related Borel-Cantelli.

However, what does i.o. imply here? Like, there are infinitely many points in any punctured neighborhood of $z_{0}$ that can be mapped by $f$ to a point in the complex plane? What does this even mean?

My second confusion is from this post: Entire function whose square or composition with itself is a polynomial

In this post, we have $f$ entire such that $f^{2}$ is a polynomial. The post says that as $f$ cannot achieves any values in $\mathbb{C}$ infinitely many times, it must be a polynomial by Great Picard Theorem.

I don't follow. It is indeed correct that as $f$ is entire, there is no singularity, so there is no essential singularity. But great picard theorem does not give a necessary consequence of the non-existence of singularity, like, for instance, no essential singularity implies no value can be achieved i.o.

I understand this theorem is really deep and really powerful, and it has many extension. But for now I am quite confused about it..

Thanks in advanced for any help!

Edit 1: Answers

Thanks to everybody that participated in this discussion, I know understand both the meaning of i.o. and the proof of the linked post.

Firstly,

i.o. means that for any $\omega\in\mathbb{C}\setminus\{\omega_{0}\}$ where $\omega_{0}$ possibly exists, $f^{-1}(\omega)$ is an infinite set. In other words, $f(z)=\omega$ has infinitely many solutions, for any $\omega\in\mathbb{C}\setminus\{\omega_{0}\}$ and for any $z$ that is not the essential singularity (because any punctured neighborhood can be mapped to the whole $\mathbb{C}\setminus\{\omega_{0}\}$).

Second, if $f^{2}(z)$ is a polynomial, we suppose $f(z)$ is not a polynomial, as $f(z)$ is entire, it means that $f(z)$ has an essential singularity at $\infty$. By Great Picard Theorem, any punctured neighborhood of $z_{0}:=\infty$ then will be mapped to the whole $\mathbb{C}\setminus\{\omega_{0}\}$ for some possibly existing $\omega_{0}$, and $f(z)=\omega$ has infinitely many solutions for any $\omega\in\mathbb{C}\setminus\{\omega_{0}\}$ and $z\neq\infty$. It then implies that $$f^{2}(z)=\omega^{2}$$ has infinitely many solutions for any $\omega\in\mathbb{C}\setminus\{\omega_{0}\}$ but this is not possible as $f^{2}(z)$ is a polynomial, a contradiction.

I am really grateful for everyone who helped clarify this theorem. Now i can see its power. :)

JacobsonRadical
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    I think its means that the pre-image $f^{-1}(z)$ is an infinite set for all $z$ with the exception of possibly one point. – Nicholas Roberts Dec 03 '20 at 20:08
  • @NicholasRoberts so basically it means that any point $z\in\mathbb{C}$ corresponds to (via $f$) infinitely many points in any neighborhood of the essential singularity? I am sorry for a general question, so why is this powerful? like, okay, we have infinitely many points, but then what?? – JacobsonRadical Dec 03 '20 at 20:09
  • Any point in $\mathbb{C} \setminus z_0$ corresponds to infinitely many points via the relation $f^{-1}$. As for your other question, I'll let someone else chime in. – Nicholas Roberts Dec 03 '20 at 20:13
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    For the second part, a (non-constant) polynomial is considered to have a pole "at infinity" and a non-polynomial entire function has an essential singularity "at infinity". What this means is that $f(1/z)$ has those types of singularities at $z = 0$. So you can apply Picard to $f(1/z)$ and check that everything works out. – Jair Taylor Dec 03 '20 at 20:24
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    entire functions which are not polynomials have an essential singularity at infinity, so Picard applies directly (the behavior of $f$ at infinity is taken as the behavior of $g(z)=f(1/z)$ at zero for example); in particular meromorphic functions in the plane which are not constant cannot omit more than two values is an equivalent form of Picard as analytic functions omit infinity by definition - $e^z/(e^z-1)$ obviously doesn't take zero and one – Conrad Dec 03 '20 at 20:25
  • @NicholasRoberts thank you so much for your help. – JacobsonRadical Dec 03 '20 at 22:15
  • @JairTaylor Thank you for the inspiring clarification! I just edited the post and gave a proof, would you mind taking a look at it? – JacobsonRadical Dec 03 '20 at 22:22
  • @Conrad Thank you so much. I didnt know that non-polynomial entire function has an essential singularity at infinity. And nice connection to the meromorphic function. Thank you! – JacobsonRadical Dec 03 '20 at 22:23
  • happy to help - there are beautiful results that strengthen the theorem; for example for an entire function there at most two values $a$ for which the equation $f(z)=a$ has only multiple roots (in other words $f(z)=a$ implies $f'(z)=0$), while for meromorphic functions $4$ such values can appear; and many more in this vein (something like if two meromorphic functions in the plane coincide on $5$ values - the sets $f(z)=a, g(z)=a$ coincide including multiplicities for $5$ such $a$ - then the function are equal, so in particular if they are entire and coinicde on $4$ finite values, they are equa – Conrad Dec 03 '20 at 22:39
  • @Conrad really nice. It is quite exciting that one theorem can have so many extensions and applications. – JacobsonRadical Dec 03 '20 at 22:40

1 Answers1

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When we say that the image of $f$ “is the whole complex plane, except possibly for one point, infinitely often”, what that means is that, for some $w_0\in\Bbb C$, whenever $w\in\Bbb C\setminus\{w_0\}$, then $f^{-1}(\{w\})$ is an infinite set.

On the other hand, a corollary of the Great Picard Theorem states that if $f$ is an entire non-polynomial function, then, for some $w_0\in\Bbb C$, the equation $f(z)=w$ has infinitely many solutions for any $w\in\Bbb C\setminus\{w_0\}$. Sometimes, people call “Great Picard Theorem” to this corollary.

José Carlos Santos
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  • Sorry for the late reply. So the proof of this corollary should be like: If $f$ is an entire function but is not polynomial, it means that $f$ has an essential singularity at $\infty$. It then follows from Great Picard Theorem that any puncture neighborhood of $z_{0}:=\infty$ will be mapped by $f$ to the whole $\mathbb{C}\setminus{\omega_{0}}$ for a possibly existing $\omega_{0}$. And $f^{-1}(\omega)$ is an infinite set for any $\omega\in\mathbb{C}\setminus{\omega_{0}}$. – JacobsonRadical Dec 03 '20 at 22:14
  • This means that $f(z)=\omega$ has infinitely many solutions for all $\omega\in\mathbb{C}\setminus{\omega_{0}}$, and for $z\neq \infty.$ Am I right? – JacobsonRadical Dec 03 '20 at 22:14
  • Yes, that is correct. – José Carlos Santos Dec 03 '20 at 22:18
  • Nice. thank you so much! – JacobsonRadical Dec 03 '20 at 22:22
  • @JoséCarlosSantos $e^z$ has essential singularity at $\infty$. For any $A\ne 0$, $e^z=A$ had infinitely many solutions of the sort $z=\log|z|+iArg(A)+i2\pi k$. How would that be different if the point was not essentail singularity? – Alexander Cska Feb 19 '24 at 21:48