0
$\begingroup$

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 :

(a)

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.

$\endgroup$
  • $\begingroup$ 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 Jun 16 '17 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 $x$? $\endgroup$ – DanielWainfleet Jun 16 '17 at 10:01
3
$\begingroup$

(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, $f(z)=a_0+a_1z+a_2z^2$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.