The set of entire function follow the condition ${|f'(z)| \leq |f(z)|}$

Question

let

$$F= \left \{f: \mathbb{C} \to \mathbb{C} \;\;|f \;\;\text{where f is entire function} \right \}\\ \text{such that }{|f'(z)| \leq |f(z)|} \;\;\; \forall z \in \mathbb{C} $$

Then

1). $F$ is an infinite set.

2). $F=\left \{\beta e^{\alpha z}:\beta \in \mathbb{C},|\alpha| \leq 1 \right \}$

The solution I Tried-

As we know that if $f$ s entire then its derivative is also entre so $f'(z)$ is also entire ,now according to question it is given that $$|f'(z) \leq |f(z)|$$ so it implies that $$\frac{|f'(z)|}{ |f(z)|} \leq 1\;\; \forall z \in \mathbb{C}$$but as we know that denominator should not be zero for fraction to be exist thus $f(z)$ should't be zero for any $z \in \mathbb{C}$ $\Rightarrow$ function $f$ should be of the form $e^{\alpha z}$ because exponential function doesn't have zero in complex Plane.Next I am not getting how to proceed .

Please help!

Answer 1

The first part is easy because all constant functions are in $F$.

For the second part, here is a proof.

The zero function is of the form $\beta e^{\alpha z}$ with $\beta=0$.

If $f$ is not the zero function, fix $z_0 \in \mathbb C$. Write $f(z)=(z-z_0)^m g(z)$, where $g(z_0)\ne0$ and $m\ge 0$.

If $m\ge 1$, then $|f'(z) \leq |f(z)|$ at $z=z_0$ gives $|mg(z_0)|=0$, a contradiction.

Thus, $f$ has no zeros. By Liouville's theorem $f'/f$ is a constant function, and so $f'(z)=\alpha f(z)$.

Let $E(z)=e^{-\alpha z}$. Then $(fE)'=0$ and so $fE=\beta$, a constant. Thus, $f(z)=\beta e^{\alpha z}$.

Finally, $|\alpha| = |f'(z)/f(z)| \le 1$.

Answer 2

Obvously, the constant zero function is $\in F$, so henceforth assume $f\not\equiv 0$. Let $z_0\in\Bbb C$. Then $f(z)=(z-z_0)^ng(z)$ for some $n\ge 0$ and some entire $g(z)$ with $g(z_0)\ne 0$. This makes $$f'(z)=n(z-z_0)^{n-1}g(z)+(z-z_0)^ng'(z)$$ and so for $z\ne z_0$, $|f'(z)|\le |f(z)|$ amounts to $$ |ng(z)+(z-z_0)g'(z)|\le |z-z_0||g(z)|$$ Take the limit as $z\to z_0$ to arrive at $$ |ng(z_0)|\le 0,$$ in other words $n=0$, which means that $f(z_0)\ne 0$.

Now we can write $f(z)=e^{h(z)}$ for some entire $h$, obtain $f'(z)=h'(z)e^{h(z)}$, and by the given inequality $|h'(z)|\le 1$. The bounded entire function $h'$ must be constant, hence $h$ is of the form $h(z)=a+bz$.

Answer 3

Assume $f$ is not identically zero. Then $g=f'/f$ is meromorphic in the plane (that is, entire except for (isolated) poles.

But $g$ is bounded (on the set where $f\ne0$), and if a function is bounded near an isolated singularity the singularity is removable. So $g$ has no poles, hence $f$ has no zero.

So $f=e^L$ for some entire function $L$. Now $|f'|\le|f|$ shows that $|L'|\le1$. So $L'$ is constant. If $L'=c$ then $L(z)=cz+a,$so $$f(z)=e^ae^{cz}$$(and it's clear that $|c|\le1$).