Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\mathbb{D}$ be the open unit disk in $\mathbb{C}$, centre $0$. Write $\mathcal{S}$ for the holomorphic injective maps $\{ f : \mathbb{D} \to \mathbb{D} | f(z) = e^{-\lambda} z + O(z^2) \}$ i.e. those that are a (real) scaling near $0$. $\lambda$ is always non-negative and will be called $logcap(f)$ or the logarithmic capacity (from $0$). It is additive, i.e. $logcap(f \circ g) = logcap(f) + logcap(g)$.

I have a sequence $f_n \in \mathcal{S}$, where $f_n$ and $f_{n+1}$ are related as follows:

$f_n = g_1 \circ g_2 \circ \cdots \circ g_k$

$f_{n+1} = h_1 \circ g_1 \circ h_2 \circ g_2 \circ \cdots \circ h_k \circ g_k \circ h_{k+1}$

for some choices of $k$, $g_i$ and $h_i$ (depending on $n$) in $\mathcal{S}$ -- i.e. each $f_i$ is expressed as a sequence of compositions and to get to $f_{i+1}$ we interleave more maps amongst the compositions.

What topologies do we have on $\mathcal{S}$ that give us nice convergence criteria for sequences of the form $f_i$?

If we choose the compact-open topology, which in this case is just the uniform norm, then I can get a long way by assuming that $logcap(f_i)$ is convergent -- i.e. that it is bounded above.

In fact, what I'm specifically stuck on is the following problem

Is $\\|f \circ g - f\|$ bounded uniformly in $f$ in terms of $logcap(g)$?

But if there are any other approaches to the general question I'd be happy to hear them.

share|improve this question
add comment

1 Answer 1

Is there something missing in the statement of the question, or is some sign wrong? By the Cauchy integral formula, any holomorphic map of the disk to itself has first derivative at 0 with norm $\le 1$, and equality holds if and only if the map is a rigid rotation of the disk. More generally, any holomorphic map between domains in $\mathbb C$ is non-increasing on distances in the Poincaré metric, and it is an infinitesimal isometry at some point if and only if it is a covering map. Thus, your hypothesis that $\lambda$ is nonnegative implies $f$ is a rotation.

share|improve this answer
    
Yes I missed a minus sign, thanks! Now fixed. –  Tom Ellis Feb 7 '11 at 21:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.