Exercise :

Let $f:\mathbb C \to \mathbb C$ be an entire function.

(a) If $|f(z)| \geq 1$ $\forall z \in \mathbb C$, show that the function $f$ is constant in $\mathbb C.$

(b) If :

$$\lim_{|z|\to \infty} \frac{f(z)}{1 + |z|^{5/2}} = 0$$

show that $f$ is a polynomial of degree $\leq 2.$

Attempt :


Since $f: \mathbb C \to \mathbb C$ is an entire complex function and f :

It is :

$$|e^f| \leq e^{|f|} \geq e^1 \Leftrightarrow |e^f| \leq e $$

Thus we can say that $e^f$ is constant (by Liouville's Theorem). Now, I know I must show that this implies that $f$ is constant but I do not see how to do it, so I'd appreciate some help (I think I should try to see how $f$ could be constant and then prove that it contradicts the assumption that $f$ is entire).

For (b) I do not know how to grasp the work in such exercises and I would really like a thorough solution, because I just stepped on the Liouville's Theorem examples and exercises, so I'd like to figure out how such questions can be solved.

I would really appreciate any help !

Kind regars,

Charalampos Filippatos.

  • $\begingroup$ I do not understand your inequality. But any way, I'd just consider $1/f$ instead, which is entire as $f$ has not zero and bounded. // For b I believe you can use use Cauchy formula to show that the second derivative is constant, implying the claim. But it is really just a believe; I did not think it through. $\endgroup$
    – quid
    Commented Jun 16, 2017 at 9:29
  • 1
    $\begingroup$ Your argument for (a) is invalid . You have inadvertently reversed an inequality and concluded that $|e^{f(x)}|<e$ for all $x$. What if $f(x)=e^2$ for all $x$? $\endgroup$ Commented Jun 16, 2017 at 10:01

2 Answers 2


(a) If $(\forall z\in\mathbb{C}):\bigl|f(z)\bigr|\geqslant1$, then $(\forall z\in\mathbb{C}):\bigl|\frac1{f(z)}\bigr|\leqslant1$ and therefore $\frac1f$ is constant, by Liouville's theorem.

(b) If $f(z)=a_0+a_1z+a_2z^2+\cdots$, then, by Cauchy's inequalities,$$(\forall R>0):|a_n|\leqslant\frac1{R^n}\sup_{|z|=R}\bigl|f(z)\bigr|$$It follows from your inequality that this converges to $0$ when $R$ goes to $+\infty$ and $n>2$. Therefore, $f(z)=a_0+a_1z+a_2z^2$.


For the first one use $g=1/f$. Clearly g is entire because $|f(z)| \geq 1$. Now $|g(z)| \leq 1$. Hence by Lioville's theorem , g is constant, hence f is constant.

(b) Here observe that due to the limiting condition $$ \Big{|} \frac{f(z)}{1+|z|^{\frac{5}{2}}} \Big{|} \leq M$$.

Now use the following Cauchy's estimate: if $|f(z)| \leq M$ on $C(0,R)$, then $$ |f^{n}(0)| \leq \frac{n!M}{R^{n}}$$.

Use this to get from the above:

$$|f^{n}(0)| \leq \frac{n!M(1+R^{5/2})}{R^{n}}$$ Now, observe if $n\geq 3$, then as $R\to \infty$, the RHS of the above inequality approaches 0 , hence $f^{n}(0) =0$ for all $n \geq 3$. Now use the global power series representation of f around 0 to conclude the result.


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