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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 $0<r\leq r_0$ there exists $M_0>0$ such that $$|f(re^{it})|\leq M_0g(r).$$

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

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Take $f(z)=e^z$ and $g$ is continuous such that $g(r)=e^r$ for $r\geq 2$ and $g(r)=0$ for $r\leq 1$. Take $r_0=2$, then $|f(re^{it})|=|e^{rit}|=e^r=g(r)$ for $r\geq r_0=2$. However, it is impossible find $M_0>0$ such that $|f(re^{it})|\leq M_0g(r)$ for $r\leq 1$ since $g(r)=0$ and $|f(z)|\neq 0$. Did I miss something? – Paul Apr 15 '12 at 22:18
I thought we can use the Mximum Modulus principle Theorem: So, for any $z$ with $|z|=r\leq r_{0}$ we have $|f(z)|\leq \max_{|z| \leq r_{o}} |f(z)|=|f(z_{o})|$, for some point $z; |z|=r_{o}$. Then, take any point $z_{1}$ with $|z_{1}|=r_{1}>r_{o}$, we get $|f(z_{o})|\leq |f(z_{1})|\leq g(r)$. But I feel like I miss something! – Nicole Apr 15 '12 at 23:38
Read again Paul's comment: there is no constraint on $g$ on $[0,r_0/2]$ except being less than $g(r_0)$ hence $g$ can be as small as desired on $[0,r_0/2]$, for example $g(r)=0$ for every $r\leqslant r_0/2$. This shows the inequality you seek cannot hold. – Did Apr 16 '12 at 9:43
I have added that $g(r)=e^{cr}, c>0$, this was as a second part of the problem. Does it make any difference? – Nicole Apr 16 '12 at 13:20

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.

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So you mean that my statement above is true!! I'm confused now!! – Nicole Apr 18 '12 at 3:15
Yes, your theorem is true. This is a proof by contradiction. – akkkk Apr 18 '12 at 12:46

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.

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