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Prove that $\lim\limits_{z \to \infty} |f(z)| = \infty$ where $f(z)$ is entire and has entire inverse $g(z)$.

I can show that the limit cannot be finite since if it were, then we can use Liouville's theorem to conclude that $f$ is constant. But how do I show that the limit is actually infinity?

Every answer in the other question uses some of Picard's theorem, the Open mapping theorem, Riemann's theorem. The only theorems I've covered which are relevant to the question are Casorati-Weierstrass and Liouville.

John
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  • @MartinR The answer uses the open mapping theorem as well which I cannot use. – John Jul 10 '19 at 19:43
  • @Saad There's more than one. Some don't use open mapping theorem – Jakobian Jul 10 '19 at 19:47
  • @Jakobian Which one are you referring to? All of them involve things I haven't learnt yet. The first two use open mapping, the third uses Picard's theorem, the fourth uses open mapping, the fifth uses Riemann's sphere, and the sixth uses Riemann's theorem. – John Jul 10 '19 at 19:50
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    @Saad I looked wrong, thought the second one doesn't. Actually, all of the solutions prove it using open mapping theorem. – Jakobian Jul 10 '19 at 19:53

3 Answers3

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Here is an elementary proof which does not even use Casorati-Weierstraß, only the (assumed) existence of a holomorphic inverse.

Assume that $\lim_{z \to \infty} |f(z)| = \infty$ does not hold. Then there is a sequence $(z_n)$ of complex numbers such that $z_n \to \infty$ and $w_n = f(z_n)$ is bounded. A bounded sequence has a convergent subsequence: $w_{n_k} \to w^* \in \Bbb C$.

But the inverse function $g$ is continuous, therefore $$ z_{n_k} = g(w_{n_k}) \to g(w^*) \in \Bbb C $$ in contradiction to the assumption that $z_n \to \infty$.

Martin R
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  • Why must there be a bounded sequence? Is it because if there is an unbounded sequence then it admits a subsequence converging to infinity? – John Jul 10 '19 at 21:19
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    @Saad: It follows directly from the definition. $\lim_{z \to \infty} |f(z)| = \infty$ means $\forall M> 0: \exists R>0 : (|z|> R \implies |f(z)| > M)$. The negation is $\exists M>0 : \forall R> 0: \exists z$ with $|z| > R$ and $|f(z)|\le M$. – Martin R Jul 10 '19 at 21:25
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Hint: If $f(z)$ is entire injective then $f(z)=a\cdot z+b$ (To show this, use the fact that the Taylor expansion must be infinite together with Casoratti-Weierstrass or Picard's theorem).

Jakobian
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miraunpajaro
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  • This almost certainly is too high tech for the OP. – A. Thomas Yerger Jul 10 '19 at 19:45
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    This is what I was originally trying to prove. I considered the Taylor expansion of $f(z)$ around $0$ and then replaced $z$ with $1/z$ to get a Laurent series of $f(1/z)$ around $0$ with only negative terms. Then if I can show there's a pole at $0$, I am done since then $f$ is a polynomial and its injective, so it must be linear. How do I show this? – John Jul 10 '19 at 19:46
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    @Saad Suppose you $f(\frac{1}{z})$ had an essential singularity, then $A=f(D(0,1))$ is dense and if $B=\mathbb{C}-D[0,1]$ then $f(B)$ is open then $A\cap B\neq0$ which means $f$ is not injective (becuase $\exists z\in A\cap B$ which means $z=f(a)=f(b)}$ for some $a\in A, b\in B$ – miraunpajaro Jul 10 '19 at 19:55
  • @Saad I add this here, there is one technical detail on the way I wrote, which is, what happens if $f$ has a zero in $B$? – miraunpajaro Jul 10 '19 at 19:58
  • @miraunpajaro I'm not allowed to use the open mapping theorem, so I can't conclude that $f(B)$ is open – John Jul 10 '19 at 20:00
  • @Saad can you use Picard? – miraunpajaro Jul 10 '19 at 20:00
  • No. But we are allowed to assume that $f$ has an entire inverse. – John Jul 10 '19 at 20:01
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If $f$ is not a polynomial then it has an essential singularity at infinity. According to Casorati-Weierstraß, $$ G = f(\{z: |z| > 1\}) $$ is dense in $\Bbb C$. Now consider the inverse function $g = f^{-1}$ (which is assumed to exist as an entire function). It follows that $$ g(G) = \{z: |z| > 1\} $$ and since $g$ is continuous and $G$ dense in $\Bbb C$ this implies $$ g(\Bbb C) \subset \{z: |z| \ge 1\} \, . $$ Then $1/g$ is a bounded, entire function and therefore constant, which is not possible (as the inverse of $f$).

So $f$ is necessarily a polynomial, and that implies $\lim_{z \to \infty} |f(z)| = \infty$.

Martin R
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