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Hello all,

Let $\mathbb{H}$ be the upper half plane, and $i+\mathbb{H}$ be all points with imaginary part greater than 1. Suppose that $U$ is a domain with a nice boundary and with $i+\mathbb{H} \subseteq U \subseteq \mathbb{H}$. Let $f(z)$ be the conformal map from $\mathbb{H}$ to $U$ which fixes $2i$, which extends continuously to the real axis, and which maps $0$ to a point on the intersection of the imaginary axis and the boundary of $U$. I would like to show that there is a constant $K$ such that

$|f(z)| < K(1+|z|) $

for all $z \in \mathbb{H}$. I think that this is true because it is true if $U = i+\mathbb{H}$(so $f(z) = i+ z/2$) and if $U=\mathbb{H}$(so $f(z)=z$), so I am hoping that there is some kind of "pinch" theorem which works here, but I don't see how to get at this. If it helps, we can also assume that the boundary of $U$ gets arbitrarily close to the real axis near infinity, or even that it is something nice like the graph of $y=1/(1+x^2)$ if that helps. I hope that there is some kind of general theory that someone can point me to. Thanks.

Greg

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1 Answer 1

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In general it should be wrong. Your map is uniquely determined and there is no reason why infinity should go to infinity (in general it would not). This means that there is a point $x$ on the boundary which is mapped to infinity. At this point your inequality is wrong.

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  • $\begingroup$ Hmm, ok, you're right, and thanks for your answer. But let's look at the case determined by $y=1/(1+x^2)$. The map is, as you say, uniquely determined, and thus by symmetry we must have infinity going to infinity. Can we get a bound in this case? Actually even better would be $|f(z)/z|$ -> $1$ as $|z|$ -> $\infty$, or at least that the quotient is bounded away from 0 and $\infty$. $\endgroup$ Commented Nov 2, 2011 at 11:38

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