Sufficient conditions for an entire function to be constant or polynomial

Question

Can you provide different sufficient conditions (with justification) for an entire function ($f:\mathbb{C}\longrightarrow \mathbb{C},$ $f$ is analytic functions on $\mathbb{C}$) to be a polynomial or constant function.

For example,

Boundedness is a condition for $f$ to be constant.

Or, having pole at infinity is a sufficient condition for $f$ to be a polynomial.

Any help is welcomed. Thank you.

Answer 1

If there exist a nonnegative integer $n$ and constant $M$ such that $$|f(z)| \le M|z|^n$$ for all $z\in \Bbb C$, then $f$ is a polynomial of degree at most $n$.

The proof is similar to showing that bounded entire functions are constant. Cauchy's estimate tells us that $$|f^{(n+1)}(0)| \le \sup_{|z| \le R}\dfrac{(n+1)!|f(z)|}{R^{n+1}} \le M(n+1)!\dfrac{1}{R}$$ for all $R > 0$. Letting $R\to\infty$ tells us that $f^{(n+1)}(0) = 0$.

A similar application shows that $f^{(k)}(0) = 0$ for all $k > n$ and thus, the power series at $0$ reduces to just the first $n$ terms. Since the function is entire, the function equals this power series everywhere.

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The above is a necessary condition if we modify it to be:
If $f$ is a polynomial, then there exists a nonnegative integer $n$ and constants $M, R > 0$ such that $$|f(z)| \le M|z|^n$$ for all $z$ satisfying $|z| > R$.

Answer 2

An entire function is a nonconstant polynomial if and only if it is proper (in other words if $K$ compact and $f^{-1}(K)=L$ then $L$ is compact)

Proof: exercise for the OP with the following hints:

A non constant polynomial satisfies $|f(z)| \to \infty, |z| \to \infty$ (why?) Show that this implies $f$ proper (use contradiction and the fact that compact is equivalent to closed and bounded in the plane)

Conversely, show that $f$ entire proper implies $|f(z)| \to \infty, |z| \to \infty$ (see hint above)

Conclude that $f$ is polynomial - one can do it using machinery like the density of image near essential singularities, but also one can simply consider $g(z)=1/f(1/z)$ which is holomorphic near zero and has a removable singularity at zero with value zero.

If $k \ge 1$ is the order of the zero of $g$ there, show that it then follows that $f$ is a polynomial of degree $k$ by showing that $f(z)/z^k \to c \ne 0, \infty, |z| \to \infty$

Answer 3

If for some finite $M$, an entire function$f$ satisfies $\mathrm{lim}_{z\rightarrow\infty}|f(z)|=M$ then $f$ is constant.

If $\mathrm{lim}_{z\rightarrow\infty}|f(z)|=M$ then $f$ is bounded in a neighborhood of $\infty$. That is, there is $R>0$ such that $f$ is bounded in $\{z\in\Bbb{C}:|z|>R\}$. And since $\{z\in\Bbb{C}:|z|\leq R\}$ is compact, so $f$ is bounded on it also. Hence $f$ is a bounded entire function.