7
$\begingroup$

Given $f$ entire function on $\mathbb C$ and $f$ one-one. Is it true that $f$ is linear?

At least among polynomials the only such functions are linear!

  • 1
    Why the hypothesis linear?2012-07-09
  • 0
    this might help you Aneesh http://math.stackexchange.com/questions/163780/transcendental-entire-function-aut-mathbbc2012-07-09
  • 0
    See also http://math.stackexchange.com/q/29758/2012-07-09

1 Answers 1

19

The link given by @Patience leads to a proof, but one can avoid the heavier things like Picard, Casorati-Weierstrass and the very notion of essential singularity. Liouville's theorem is enough.

Pick a point $a$ such that $f\,'(a)\ne 0$. (I don't even want to argue that $f\,'$ never vanishes). Normalize so that $a=0$, $f(0)=0$, and $f\,'(0)=1$. Since $f$ is an open map, there exists $r>0$ such that $\{w:|w|$$g(z)=\frac{f(z)-z}{zf(z)}$$ has a removable singularity at $0$. When $|z|\ge1$, we have $$|g(z)| = \left|\frac{1}{z}-\frac{1}{f(z)}\right|\le \frac{1}{|z|}+\frac{1}{|f(z)|}\le 1+\frac{1}{r}.$$ Thus, $g$ is a bounded entire function. By Liouville's theorem $g$ is constant, and it follows that $f$ is linear.

  • 2
    This is a great answer. Of the answers to this question on other threads which I have seen, this is most similar to [this answer of Zarrax](http://math.stackexchange.com/a/29822/).2012-07-09
  • 3
    @JonasMeyer Thanks. After writing this, I noticed that $f$ could be assumed to be merely meromorphic, since the poles of $f$ are removable for $g$. Then the conclusion is $f(z)=z/(1-cz)$, where $c$ is not necessarily zero anymore.2012-07-09
  • 0
    Great answer indeed, +1! I didn't know how to prove this without Casorati-Weierstrass...2012-07-10
  • 0
    Thank you very much Kovalev!2012-07-10
  • 0
    @Leonid could you tell me how to normalize and how we can assume $f'(0)=1$?2012-07-26
  • 1
    @Patience You have a function such that $f\,'(a)\ne 0$. Let $\tilde f(z)=(f(z+a)-f(a))/f\,'(a)$. This new function has $\tilde f(0)=0$, and $\tilde f\,'(0)=1$, as desired. Note that $f$ is linear if and only if $\tilde f$ is linear. So, we can simply work with $\tilde f$ from now on, and even rename it as $f$ to simplify writing.2012-07-27
  • 1
    $f(z)=\frac{z}{1+cz}$ is linear?2013-03-06
  • 1
    or as $f$ is non-constant and has no poles so $c=0$ must? and $f(z)=z$?2013-03-06
  • 1
    also why $g$ has just removable singularity at $z=0$? I can see $f(0)=0$ so $g$ has a pole at $0$?2013-03-07
  • 0
    and the conclusion you made assuming $f$ is open map is true?2013-03-07
  • 3
    $g(z)$ has an infinity plus infinity form so we don't know what it is unless we compute!! $zg(z)$ has the limit 0 as $z$ tends to 0, so the singularity at $0$ is a removable singularity and g extends to an entire function at 0! So don't bother about what $g(0)$ is! The open mapping theorem shows the existence of an $r$ such that $|f(z)| implies $|z|<1$ Equivalently $|z|\ge1$ implies $|f(z)|\ge r$ or equivalently $|z|\ge1$ implies $\frac1{|f(z)|}\le\frac1r$ At the end of it all, indeed $f(z)=z$ because the absence of poles implies $c=0$ So you are done!!2013-03-07
  • 0
    I am still not convinced about the statement concluded from OMT2013-03-07