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 $$\limsup_{r\to\infty}\log|f(re^{it})|\leq \limsup_{|z|\to\infty}\log|f(z)|$$
$f(z)$ and $f(re^{it})$ ??
-
0The latter is the supremum over $t$ of the former. In particular, the value of the latter is at least the value of the former at any given $t$. – 2012-04-11
-
0Yes, that's right, but why this is true! $$\max_{t}\limsup_{r\to\infty}\log|f(re^{it})|= \limsup_{|z|\to\infty}\log|f(z)|$$ – 2012-04-11
-
0So, any comments!!? – 2012-04-11
1 Answers
$\limsup_{r\to\infty}\log|f(re^{it})|$ 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 $|z|$.