# Prove that there exists complex numbers $a$ and $c$ such that $f(z)=ae^{cz}$.

Let $$f$$ be an entire function such that $$|f'(z)|\leq|f(z)|$$ for all $$z\in \mathbb{C}$$. Prove that there exist complex numbers $$a$$ and $$c$$ such that $$f(z)=ae^{cz}$$ for all $$z\in\mathbb{C}$$.

My try: take $$g(z)$$ such that $$f(z)=e^{g(z)}.$$ Then by chain rule $$f'(z)=e^{g(z)}g'(z).$$ Now we have $$|g'(z)||f(z)|=|g'(z)||e^{g(z)}|=|f'(z)|\leq |f(z)|$$ which implies $$|g'(z)|\leq 1.$$ Thus $$g(z)=cz$$, and then $$f(z)=e^{cz}$$. However, how to make $$g(z)$$ is entire function? and where is the constant $$a$$?

## 2 Answers

You have to justify writing $$f$$ as $$e^{g}$$. Here is a correct solution: If $$f \equiv 0$$ we can take $$a=0$$ and $$c=1$$. Otherwise the zeros of $$f$$ form a set $$\{z_n\}$$ with no limit points. On $$\mathbb C \setminus \{z_n\}$$ we have $$|\frac {f'(z)} {f(z)}| \leq 1$$. This implies that $$\frac {f'(z)} {f(z)}$$ has a removable singularity at each $$z_n$$. Hence it extends to a bounded entire function. By Louiville's Theorem we have $$\frac {f'(z)} {f(z)}=A$$ for some constant $$A$$. Now observe that $$(e^{-Az}f(z))'=0$$ so $$e^{-Az}f(z)=B$$ for some constant $$B$$. Hence $$f(z)=Be^{Az}$$.

Note that $$f$$ has not zeros. In fact, if $$a$$ is a zero of $$f$$ of order $$n\in\mathbb N$$, then $$a$$ is a zero of $$f'$$ os order $$n-1$$. But then, near $$a$$, the inequality $$\bigl\lvert f'(z)\bigr\rvert\leqslant\bigl\lvert f(z)\bigr\rvert$$ means$$\bigl\lvert na_n(z-a)^{n-1}+(n+1)a_{n+1}(z-a)^n+\cdots\bigr\rvert\leqslant\bigl\lvert a_n(z-a)^n+a_{n+1}(z-a)^{n+1}+\cdots\bigr\rvert.$$Now, dividing both sides by $$\bigl\lvert(z-a)^n\bigr\rvert$$, you get that$$\bigl\lvert na_n(z-a)^{-1}+(n+1)a_{n+1}+\cdots\bigr\rvert\leqslant\bigl\lvert a_n+a_{n+1}(z-a)+\cdots\bigr\rvert,$$which is impossible, because the LHS tends to $$+\infty$$ when $$z\to a$$, whereas the RHS tends to $$\lvert a_n\rvert$$.

So, since $$f$$ has no zeros and $$\mathbb C$$ is simply connected, there is indeed an entire function $$g$$ such that $$f=e^g$$. You proved that $$g'$$ is bounded. Therefore, by Liouville's therem, it is constant. So there are constants $$k$$ and $$c$$ such that $$(\forall z\in\mathbb{C}):g(z)=cz+k$$and therefore$$(\forall z\in\mathbb{C}):f(z)=e^{g(z)}=e^{cz+k}=e^ke^{cz}.$$