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?

  • $\begingroup$ Do you know Liouville's theorem? Try to show that $f(z)e^{-z}$ is constant. $\endgroup$ – Crostul Jan 25 '17 at 7:30
  • 1
    $\begingroup$ @Crostul I know it, but I cannot show$|f(z)e^{-z}|\leq |f(z)|e^{-|z|}$, so I cannot prove it is bound. $\endgroup$ – 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 $1$.

  • $\begingroup$ what theorem guarantee there is an entire function $g$ such that $f(z)= e^{g(z)}$ $\endgroup$ – noname1014 Jan 25 '17 at 10:50
  • $\begingroup$ 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. $\endgroup$ – ts375_zk26 Jan 25 '17 at 11:44
  • $\begingroup$ 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. $\endgroup$ – Alan Wang Jun 2 '18 at 2:40
  • 1
    $\begingroup$ @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 $1$. $\endgroup$ – ts375_zk26 Jun 2 '18 at 5:20

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.