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What is the relation between $\limsup_{r\to\infty}\log|f(re^{it})|$ and $\limsup_{|z|\to\infty}\log|f(z)|$ where $z=re^{it}$, $r>0, 0.

I know that the first one is a function of $t$, but the second one is a constant (assuming both limits exist), I'm told that I have this relation but I don't know why?? any help

$\limsup_{r\to\infty}\log|f(re^{it})|\leq \limsup_{|z|\to\infty}\log|f(z)|$

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    So, any comments!!?2012-04-11

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  • $\limsup_{r\to\infty}\log|f(re^{it})|$ is the infimum of number $b$ such that $\log|f(re^{it})|\le b$ for all sufficiently large $r$.

  • $\limsup_{|z|\to\infty}\log|f(z)|$ is the infimum of number $b$ such that $\log|f(z)|\le b$ for all $z$ with sufficiently large modulus.

Any $b$ that works for the latter also works for the former. This is why $\limsup_{r\to\infty}\log|f(re^{it})|\leq \limsup_{|z|\to\infty}\log|f(z)|$


The formula stated in a comment, $\max_{t}\limsup_{r\to\infty}\log|f(re^{it})|= \limsup_{|z|\to\infty}\log|f(z)|$ is not true. For example, define $f$ so that $f(x+ix^2)=2$ for $x\in\mathbb R$, and $f=1$ elsewhere. Since every line meets the parabola $y=x^2$ at most twice, it follows that $ \limsup_{r\to\infty}\log|f(re^{it})|=\log 1=0$ for all $t$. On the other hand, $\limsup_{|z|\to\infty}\log|f(z)|=\log 2>0$ because $\log|f(z)|$ takes on the value $\log 2$ at some points $z$ with arbitrarily large $|z|$.