Prove or disprove that every Holomorphic function preserving unboundedness is a polynomial.

Question

Roughly speaking as the title says: Prove or disprove that every holomorphic function preserving unboundedness is a polynomial.

Let $f : \Bbb C \to \Bbb C$ be a holomorphic function such that for every unbounded set $ V \subset \Bbb C$ the image $f(V)$ is also unbounded.

Prove or disprove that $f$ must be a polynomial.

Answer 1

Recall :Picard's theorem Little Picard Theorem: If a function f : C → C is entire and non-constant, then the set of values that f(z) assumes is either the whole complex plane or the plane minus a single point.

Great Picard's Theorem: If an analytic function f has an essential singularity at a point w, then on any punctured neighborhood of w, f(z) takes on all possible complex values, with at most a single exception, infinitely often.

Without loss of generality, Up to a shifting we assume that $z=0$ is not an exceptional point of $f$ otherwise replace $f(z) $ by $g(z)= f(z)-a $. where a where a is chosen to be not an exceptional value of $f.$

Since $f$ is entire (then $f$ analytic with radius of convergence $R=\infty $) we can write $$f(z) =\sum_{n=0}^{\infty}a_n z^n$$ we let
$$ h(z) = f\left(\frac{1}{z}\right)=\sum_{n=0}^{\infty}\frac{a_n }{z^n}$$ the point $z=0$ is either a pole of finite oder to $h$ or an essential singularity.

Assuming, $h$ has an essential singularity then by great Picard theorem set $h^{-1}\{0\}$ is infinite in every neighborhood of $z=0$. Hence for $n>$ there exists $|z_n|\le \frac1n$ such that $$h(z_n)=f\left(\frac{1}{z_n}\right)= 0$$

Hence the set $$V=\left\{\frac{1}{z_n} : n\in \Bbb N\right\}$$ is unbounded but $f(V)= \{0\}$ which is contradictory.

Hence necessarily, $z=0$ is polar of finite oder to $h$ that is there is $N$ such that $a_n=0$ for $n\ge N+1$ since in this case $h$ can be only written as

$$h(z) = f\left(\frac{1}{z}\right)=a_0 +\frac{a_1 }{z}+\frac{a_2 }{z^2}+\cdots +\frac{a_N }{z^N}$$

which means that $f$ ia a polynomial and we have, $$f(z) = a_0 +a_1z +a_2 z^2+\cdots +a_N z^N $$

Thanks explanations from @N.S.

Answer 2

Hint At $z=\infty$ your function has either a removable singularity, essential singularity or a pole.

If it is removable, the function has a finite limit at $\infty$, but this contradicts the given condition.

If it is essential, by Picard's Theorem, the function achieves all values, excepting one, infinitely many times. Use this information to build an unbounded set which goes to one value.

It follows that $f$ must have a pole at infinity. Use this to prove that $f$ is polynomial.

Answer 3

Let $V=\{z; |f(z)|\leq 1\}$. As $f(V)$ is bounded, $V$ must be bounded, there exist $R>0$ such that $V\subset B(0,R)$. For $z$ such that $|z|\geq R$, we have $|f(z)|>1$. Hence $f$ has a finite number of zeroes. Write $f(z)=P(z)g(z)$, $P$ a polynomial, $g$ entire with no zeroes. Then $\frac{1}{g(z)}=\frac{P(z)}{f(z)}$ is entire, bounded by $|P(z)|$, if $|z|>R$, hence bounded by $c|z|^m$ for large $|z|$. Then $1/g$ is a polynomial, hence a constant, and we are done.

Answer 4

Casorati-Weierstrass gives a simple proof: We want to show that if $f$ is not a polynomial, then $f$ maps some unbounded set to a bounded set. But if $f$ is not a polynomial, then $f(1/z)$ has an essential singularity at $0.$ Thus by CW, $f(1/z)$ maps every $\{0<|z|<r\}$ onto a dense subset of the plane. Hence there is a sequence $z_k \to 0$ such that $|f(1/z_k)| < 1$ for all $k.$ Thus $f$ maps the unbounded set $\{1/z_k\}$ to a bounded set.