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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?2012-04-15
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    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!2012-04-15
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    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.2012-04-16
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    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

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