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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!

Gerry Myerson
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Karthik C
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1 Answers1

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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 $\{w:|w|<r\}\subset f(\{z:|z|<1\})$. The function $$g(z)=\frac{f(z)-z}{zf(z)}$$ 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.

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    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. – Jonas Meyer Jul 09 '12 at 22:47
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    @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. –  Jul 09 '12 at 22:53
  • Great answer indeed, +1! I didn't know how to prove this without Casorati-Weierstrass... – Malik Younsi Jul 10 '12 at 02:50
  • Thank you very much Kovalev! – Karthik C Jul 10 '12 at 05:21
  • @Leonid could you tell me how to normalize and how we can assume $f'(0)=1$? – Myshkin Jul 26 '12 at 07:32
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    @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. –  Jul 27 '12 at 01:28
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    $f(z)=\frac{z}{1+cz}$ is linear? – Myshkin Mar 06 '13 at 23:16
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    or as $f$ is non-constant and has no poles so $c=0$ must? and $f(z)=z$? – Myshkin Mar 06 '13 at 23:22
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    also why $g$ has just removable singularity at $z=0$? I can see $f(0)=0$ so $g$ has a pole at $0$? – Myshkin Mar 07 '13 at 03:54
  • and the conclusion you made assuming $f$ is open map is true? – Myshkin Mar 07 '13 at 03:55
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    $g(z)$ 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 $|f(z)|<r$ 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!!

     – Karthik C Mar 07 '13 at 04:38
  • I am still not convinced about the statement concluded from OMT – Myshkin Mar 07 '13 at 09:40