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This post is closely related to Entire function whose square or composition with itself is a polynomial.

In the linked post, the OP asked to prove the following:

  1. If $f:\mathbb{C}\longrightarrow\mathbb{C}$ is entire, and $f^{2}(z)$ is a polynomial, then $f$ is also a polynomial.

  2. If $f:\mathbb{C}\longrightarrow\mathbb{C}$ is entire, and $f(f(z))$ is a polynomial, then $f$ is also a polynomial.

Several proofs were given: see Proving $f(z)$ and $g(z)$ are polynomials for a proof using Casorati-Weierstrass Theorem, and Two confusions on Great Picard Theorem for a proof using Great Picard Theorem.

My question is what if $f$ is not entire in the first place, does $f^{2}(z)$ being a polynomial still imply $f$ being a polynomial? and does $f(f(z))$ being a polynomial imply $f$ be a polynomial?

As the proof depends heavily on the entire property of $f$, I believe the answer for both of my questions is NO.

But I cannot find a counterexample.

For the first one, I tried $f(z):=|z|$, but then $f(z)\neq z^{2}$, because $|z|$ is a complex norm, so that $$f(z)=|z|^{2}=x^{2}+y^{2}\neq z^{2}=x^{2}+2ixy-y^{2}.$$

Without absolute value, I am not sure what else I can try since $f^{2}(z)$ must be a polynomial (so all the fraction without removable singularities seems not work well..)

For the second one, perhaps we can use $f(z):=\dfrac{1}{z}$ which has a pole at $z=0$, and $f(f(z))=z$ is a polynomial.

Is my second example correct? Any idea about the first example? Thank you!

Eric Wofsey
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JacobsonRadical
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  • So, you are assuming nothing about $f$ other than that $f^2$ or $f\circ f$ is a polynomial functions? Is that it? – José Carlos Santos Dec 03 '20 at 23:17
  • $\sqrt z$ is analytic on $\mathbb C -\mathbb R_{-}$ – Conrad Dec 03 '20 at 23:17
  • @JoséCarlosSantos yes. – JacobsonRadical Dec 03 '20 at 23:19
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    If $f$ is not required to be continuous, then let $f(x)=x$ for every value except $x=1$ and let $f(1)=-1$. – lulu Dec 03 '20 at 23:19
  • @lulu oh.... right... then $f^{2}(x)=x^{2}$ for all $x$ – JacobsonRadical Dec 03 '20 at 23:19
  • @lulu nice example. – JacobsonRadical Dec 03 '20 at 23:19
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    Or$$f(z)=\begin{cases}1&\text{ if }z\in\Bbb Q\0&\text{ otherwise.}\end{cases}$$Then $f\circ f$ is the constant polynomial function $1$. – José Carlos Santos Dec 03 '20 at 23:20
  • if you require $f$ continuous on $\mathbb C$ and $f^2$ polynomial, it actually follows that $f$ is a polynomial – Conrad Dec 03 '20 at 23:22
  • @Conrad why is that? – JacobsonRadical Dec 03 '20 at 23:23
  • because $f$ is analytic except at the zeroes (compose $g=f^2$ with the appropriate branch of the analytic square root near any point where $f(w) \ne 0$) and they are isolated unless $f$ is identically zero, so by continuity they are removable, hence $f$ is entire – Conrad Dec 03 '20 at 23:26
  • @Conrad I am sorry for the silly question, but I don't follow why $f$ is analytic except at the zeros. So we say a function $f(z)$ is analytic at $z_{0}$ if $f(z)=\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}$ converges in $D_{z_{0}}(r)$ for some $r$. I don't even follow why we can express a continuous function in power series... – JacobsonRadical Dec 03 '20 at 23:34
  • also note that $f(z)=\bar z$ is not analytic but highly smooth (harmonic etc) and $f\circ f$ is a polynomial the second problem is trickier even in the continuous or even $C^{\infty}$ say case – Conrad Dec 03 '20 at 23:35
  • $f^2=P$ implies $f=\sqrt P$ (locally for an appropriate branch of the square root as long as we stay away from zeroes) and as noted the square root is analytic outside zeroes of $P$ – Conrad Dec 03 '20 at 23:36
  • @Conrad oh okay. I get it. You combined the continuity of $f$ and $f^{2}$ being a polynomial. Nice! – JacobsonRadical Dec 03 '20 at 23:38
  • if $f^2=g, g$ is analytic and $f$ continuous, then $f$ analytic; in other words a continuous square root of an analytic function is analytic - the crucial word being continuous of course – Conrad Dec 03 '20 at 23:40
  • @Conrad Thank you so much for these nice comments :)! – JacobsonRadical Dec 03 '20 at 23:42
  • @Conrad oh sorry for coming back. I was confused for a while by the complex square root, and searched it for a while. Okay, so the example you gave, $\sqrt{z}$, you mean the principal branch of the square root right? which is analytic on $\mathbb{C}\setminus\mathbb{R}_{\leq 0}$. And for the later discussion, $\sqrt{P}$ is analytic outside the zeros of the polynomial $P$. does this also hold for, say, some general analytic function $g$? – JacobsonRadical Dec 04 '20 at 00:02
  • yes it does hold - see my comment above; continuous roots of analytic functions are analytic (could be cube root too, etc); the point is that locally roots are "separated" outside zeroes, so continuity forces to choose the same branch locally, hence implies analyticity outside zeroes and then the zeroes are isolated etc; if you study the modern proof of the Riemann Mapping Theorem, you will see the square root trick which uses precisely this separation, to create functions with disjoint images – Conrad Dec 04 '20 at 00:08
  • @Conrad nice. Thank you for the patient clarification. Recently I found so many subtitles in complex analysis I didnt realize before.. – JacobsonRadical Dec 04 '20 at 00:10

1 Answers1

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What do you mean with 'not entire' ?

$f(z)=1/z^n$ gives $f(f(z))=z^{n^2}$

with $\phi$ biholomorphic (or just bijective ?) on the right open set then $g(z)= \phi^{-1}(1/\phi(z))$ gives $g(g(z)) = z$,

same for $h(z)= \phi^{-1}(1-\phi(z)),h(h(z))=z$.

reuns
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