0
$\begingroup$

If $f(z)$ is an entire function (analytic in the complex plane), with the following property:

There exist $r_0>0$ such that $|f(re^{it})|\leq g(r)$ for all $r>r_0$, and all $t\in [0,2\pi]$ ($g(r)$ is some continuous function of $r$, for all $r>0$).

How I can show that: Given 0 there exists $M_0>0$ such that $|f(re^{it})|\leq M_0g(r).$

Edit: $g(r)=e^{cr}, c>0$.

  • 0
    I have added that g(r)=e^{cr}, c>0, this was as a second part of the problem. Does it make any difference?2012-04-16

2 Answers 2

1

Let the theorem not be true. Then for any $M_0>0$, there is a $z\in B_{r_0}[0]$ such that $|f(z)|>M_0 g(|z|)$, and since $g(r)\neq 0$, this means that $f$ is unbounded (if $|f(z)|>M_0 g(|z|)$, then we can take $M_0'=M_0/g(|z|)$ and get that $|f(z)|>M_0'$ for any $M_0$), but any analytic function is bounded on a compact set.

  • 0
    Yes, your theorem is true. This is a proof by contradiction.2012-04-18
0

Here is a simpler way to express Auke's answer:

The function $f(re^{it})$ is bounded on the ball of radius $r_0$, so as long as $g(r)$ is bounded above $0$ when $0 \leq r \leq r_0$, we can find $M_0$ such that $M_0 g(r)$ (for any value of $r \leq r_0$) is greater than $f(re^{it})$ (for any $r \leq r_0$ and any $t$). In particular, this ensures that $f(re^{it}) \leq M_0 g(r)$ when $r \leq r_0$.

Nicole's particular function $g(r)$ is $e^{cr}$, which is $\geq 1$ when $r \geq 0,$ and so the preceding discussion applies in her case.