# Prove an entire function is constant

Let $f$ be an entire function such that $f(0)=1, f'(0)=0$, and $$0<|f(z)|\leq e^{|z|}$$ for every $z\in \mathbb{C}$. Prove $f$ is constant $1$ on $\mathbb{C}$.

I am going to use Cauchy estimate similar to this . but I found it does not work, can you give me some hint?

• Do you know Liouville's theorem? Try to show that $f(z)e^{-z}$ is constant. – Crostul Jan 25 '17 at 7:30
• @Crostul I know it, but I cannot show$|f(z)e^{-z}|\leq |f(z)|e^{-|z|}$, so I cannot prove it is bound. – noname1014 Jan 25 '17 at 7:48

The condition $0<|f(z)|$ is very important. From this we see that $f(z)= e^{g(z)}$ for some entire function $g(z)$. Then the condition $$|f(z)|=e^{\operatorname{Re}g(z)}\le e^{|z|}$$ implies $$\operatorname{Re}g(z)\le |z|.$$ If you can conclude from this that $g(z)$ is a polynomial of degree at most $1$, that is, $g(z)=az+b$, then the condition $f(0)=1,f^\prime(0)=0$ yields that $f$ is constant $1$.
• what theorem guarantee there is an entire function $g$ such that $f(z)= e^{g(z)}$ – noname1014 Jan 25 '17 at 10:50
• See, for example, Conway's book, p.94, 6.17 Corollary which states: Let $G$ be simply connected and let $f:G\to \mathbb{C}$ be an analytic function such that $f(z)\ne 0$ for any $z$ in $G$. Then there is an analytic function $g:G\to \mathbb{C}$ such that $f(z)=\exp g(z)$. Outline of the proof: Consider a primitive $g$ of $f^\prime/f$ and $h=\exp g$. Then $f/h$ should be a constant. – ts375_zk26 Jan 25 '17 at 11:44
• Can you give further hint to conclude that $g(z)$ is polynomial of degree at most 1? I know that $\text{Re } g(z)\leq |z|$ but we don't have $|g(z)|\leq |z|$ so we can't use Cauchy's Estimate to prove that. – Alan Wang Jun 2 '18 at 2:40
• @Alan Wang Let $f$ be an entire function, and $f=u+iv=\sum a_n z_n$ is its Taylor series. If $A(r)=\max_{|z|=r}u$, then $$|a_n|r^n \leq \max\{4A(r),0\}-2u(0).$$ For the proof see here. The proof is valid for entire functions. Now if $g(z)=\sum a_n z_n$ satisfies $$\operatorname{Re}g(z)\le |z|,$$ then for $n\ge 2$ $$|a_n|\leq 4r^{1-n}-2u(0)r^{-n}\to 0\,(n\to \infty).$$ Therefore $g(z)$ is a polynomial of degree at most $1$. – ts375_zk26 Jun 2 '18 at 5:20