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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?

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noname1014
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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 $1$.

ts375_zk26
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  • 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. – Wang Kah Lun Jun 02 '18 at 02:40
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    @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 02 '18 at 05:20